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
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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://www.convertunits.com/from/cu+cm/to/gram+%5Bsugar%5D | 1,585,636,873,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370500331.13/warc/CC-MAIN-20200331053639-20200331083639-00216.warc.gz | 793,515,810 | 7,727 | ## ››Convert cubic centimetre to gram [sugar]
cu cm gram [sugar]
Did you mean to convert cu cm to gram [water] gram [sugar]
How many cu cm in 1 gram [sugar]? The answer is 1.173552765377.
We assume you are converting between cubic centimetre and gram [sugar].
You can view more details on each measurement unit:
cu cm or gram [sugar]
The SI derived unit for volume is the cubic meter.
1 cubic meter is equal to 1000000 cu cm, or 852113.36848478 gram [sugar].
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between cubic centimeters and grams.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of cu cm to gram [sugar]
1 cu cm to gram [sugar] = 0.85211 gram [sugar]
5 cu cm to gram [sugar] = 4.26057 gram [sugar]
10 cu cm to gram [sugar] = 8.52113 gram [sugar]
20 cu cm to gram [sugar] = 17.04227 gram [sugar]
30 cu cm to gram [sugar] = 25.5634 gram [sugar]
40 cu cm to gram [sugar] = 34.08453 gram [sugar]
50 cu cm to gram [sugar] = 42.60567 gram [sugar]
75 cu cm to gram [sugar] = 63.9085 gram [sugar]
100 cu cm to gram [sugar] = 85.21134 gram [sugar]
## ››Want other units?
You can do the reverse unit conversion from gram [sugar] to cu cm, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Cubic centimeter
A cubic centimetre (cm3) is equal to the volume of a cube with side length of 1 centimetre. It was the base unit of volume of the CGS system of units, and is a legitimate SI unit. It is equal to a millilitre (ml).
The colloquial abbreviations cc and ccm are not SI but are common in some contexts. It is a verbal shorthand for "cubic centimetre". For example 'cc' is commonly used for denoting displacement of car and motorbike engines "the Mini Cooper had a 1275 cc engine". In medicine 'cc' is also common, for example "100 cc of blood loss".
## ››Definition: Gram
This is the amount of sugar, often measured as 4.2 grams per teaspoon on a nutrition facts label.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 707 | 2,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2020-16 | latest | en | 0.791743 |
http://basic256.blogspot.com/2010/10/20-and-out.html | 1,532,041,623,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591332.73/warc/CC-MAIN-20180719222958-20180720002958-00242.warc.gz | 37,546,259 | 11,242 | ## Wednesday, October 6, 2010
### 20 and out
rem add 1,2, or 3 . The first to get to 20 loses
clg
font "arial",10,100
total=0 : n=1 : Sq=3
play:
input"How much add 1-3? ",a\$
a=int(a\$): total=total+a
if Sq<=total then
n=n+1
Sq=4*n-1
endif
if Sq-total < 4 then
p=Sq-total
w=1
else
p=int(rand*2)+1
w=0
endif
total=total+p
for c = total-a-p+1 to total-p
color red
circle 30,290-c*15,5
text 60,285-c*15,c
next c
print "Thinking"
pause rand*5
print "computer play "+ p
for c = total-p+1 to total
color blue
circle 30,290-c*15,5
text 60,285-c*15,c
next c
if total<19 then goto play
if total = 19 and w=1 then
Print "I Won"
else
Print "You won"
end if | 269 | 648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | longest | en | 0.592035 |
http://ijcttjournal.org/archives/ijctt-v67i3p130 | 1,600,854,665,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400210616.36/warc/CC-MAIN-20200923081833-20200923111833-00137.warc.gz | 65,495,190 | 6,126 | Design of New Fuzzy System to Determine the Three-Zone around the Ship
International Journal of Computer Trends and Technology (IJCTT) © 2019 by IJCTT Journal Volume-67 Issue-3 Year of Publication : 2019 Authors : Qousay Benshi,Oulfat Jolaha, Jaber Hanna 10.14445/22312803/IJCTT-V67I3P130
MLA Style: Qousay Benshi,Oulfat Jolaha, Jaber Hanna "Design of New Fuzzy System to Determine the Three-Zone around the Ship" International Journal of Computer Trends and Technology 67.3 (2019): 156-164.
APA Style: Qousay Benshi,Oulfat Jolaha, Jaber Hanna (2019). Design of New Fuzzy System to Determine the Three-Zone around the Ship. International Journal of Computer Trends and Technology, 67(3), 156-164.
Abstract
In order to maintain the safety of sailing, it is necessary to determine the relationship between the vessel and any obstacle that may appear on its way. This relationship is essentially the distance of the obstacle from the vessel. It is therefore very important to identify a zone or zones surrounding the ship that determine the location of the target for it in order to prevent the presence of any obstacle within it. In this research, a fuzzy system is designed to calculate the radius of three proposed zones, namely the forbidden, dangerous and safe zone, based on human experience, taking into consideration the rules of COLREGS (International Regulations for Preventing Collisions at Sea).This system is Multi Input Multi Output (MIMO) and has three inputs that are the length, speed of the vessel, and sea state; and it has three outputs that are the radius of the three zones (forbidden, dangerous and safe zones). The proposed system is adjustable according to length and speed of vessel, and sea state.The system is designed and tested using MATLAB. The results were showed to marine experts who said that are good.
Reference
[1] International Maritime Organization, COLREG: Convention on the International Regulations for Preventing Collisions at Sea, 2003 Consolidated Edition (IB904E), 4th Edition. IMO Publishing, 2003.
[2] M. E. Goodwin, “A Statistical Study of Ship Domain,” J. Navig., vol. 28., 1975.
[3] Z. Pietrzykowski and J. Uriasz, “The Ship Domain – A Criterion of Navigational Safety Assessment in An Open Sea Area,” J. Navig., vol. 62, no. 1, pp. 93–108, 2009.
[4] K Marcjan, “Navigation Incident Models and Ship Domains Studies on The Baltic Sea,” Master’s Thesis, Maritime University of Szczecin, 2011.
[5] Z. Pietrzykowski, M. Wielgosz, and M. Siemianowicz, “Ship domain in the restricted area – simulation research,” Marit. Univ. Szczec., vol. 32, no. 104, pp. 152–156, 2012.
[6] H.-Z. Hsu, “Safety Domain Measurement for Vessels in an Overtaking Situation,” Int. J. E-Navig. Marit. Econ., vol. 1, pp. 29–38, Dec. 2014.
[7] S.-M. Lee, K.-Y. Kwon, and J. Joh, “A Fuzzy Logic for Autonomous Navigation of Marine Vehicles Satisfying COLREG Guidelines,” Int. J. Control, vol. 2, no. 2, p. 11, 2004.
[8] G. Wu, D. Shi, and J. Guo, “Deliberative collision avoidance for unmanned surface vehicle based on the directional weight,” J. Shanghai Jiaotong Univ. Sci., vol. 21, no. 3, pp. 307–312, Jun. 2016.
[9] C.-M. Su, K.-Y. Chang, and C.-Y. Cheng, “Fuzzy Decision on Optimal Collision Avoidance Measures for Ships in Vessel Traffic Service,” J. Mar. Sci. Technol., vol. 20, no. 1, p. 11, 2012.
[10] S.-L. Kao, C.-M. Su, C.-Y. Cheng, and K.-Y. Chang, “A New Method of Collision Avoidance for Vessel Traffic Service,” p. 7, 2007.
[11] J.Teilmann, “Influence of Sea State on Density Estimates of Harbour Porpoises (Phocoena Phocoena),” J. Cetacean Res. Manag., vol. 5, pp. 85–92, 2003.
Keywords
fuzzy system, collision avoidance, COLREGS. | 1,046 | 3,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-40 | latest | en | 0.9058 |
http://mathhelpforum.com/discrete-math/64644-another-sets-prove-print.html | 1,521,484,211,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647044.86/warc/CC-MAIN-20180319175337-20180319195337-00338.warc.gz | 179,875,709 | 3,158 | # Another sets prove
• Dec 11th 2008, 06:52 AM
Another sets prove
By using the set algebra , prove that , for any sets A and B
A u ( A' n B ) = A u B
Deduce , or using set algebra , show that
A u ( A' n B ) u [A' n (A u B') n {B u (B' n C}] = A u B u C
• Dec 11th 2008, 07:04 AM
tester85
The question test the distributive laws.
A u ( A' n B ) = A u B
from L.H.S : A u ( A' n B )
= ( A U A' ) n ( A u B )
= S n ( A u B )
= A u B ( proven )
You should be able to solve the second part using the law or the procedure listed above :)
Quote:
By using the set algebra , prove that , for any sets A and B
A u ( A' n B ) = A u B
Deduce , or using set algebra , show that
A u ( A' n B ) u [A' n (A u B') n {B u (B' n C}] = A u B u C
• Dec 12th 2008, 01:03 AM
Quote:
Deduce , or using set algebra , show that
A u ( A' n B ) u [A' n (A u B') n {B u (B' n C}] = A u B u C
I tried this one but i couldn't get the prove . Can someone pls spot my mistake .
My working :
(A u B) u (A' n B' ) n ( B u C ) = S n (B u C ) = B u C
wondering , where is the A .
• Dec 12th 2008, 06:28 AM
Proofs using the Laws of Sets
Quote:
I tried this one but i couldn't get the prove . Can someone pls spot my mistake .
My working :
(A u B) u (A' n B' ) n ( B u C ) = S n (B u C ) = B u C
wondering , where is the A .
Here's a proof - there might be a quicker one:
$\displaystyle A \cup (A'\cap B)\cup [A'\cap (A\cup B')\cap (B\cup (B'\cap C)]$
$\displaystyle =A\cup B\cup [A'\cap (A\cup B') \cap (B \cup C)]$, using the first result twice
$\displaystyle =A \cup B \cup[\{(A' \cap A) \cup (A'\cap B')\} \cap (B \cup C)]$, Distributive Law
$\displaystyle = A \cup B \cup [\{\oslash \cup (A' \cap B') \} \cap (B \cup C)]$, Complement Law
$\displaystyle = A \cup B \cup [(A' \cap B') \cap (B \cup C)]$, Identity Law
$\displaystyle = A \cup B \cup [(A \cup B)' \cap (B \cup C)]$, De Morgan's Law
$\displaystyle = [(A \cup B) \cup (A \cup B)'] \cap [(A \cup B) \cup (B \cup C)]$, Distributive Law
$\displaystyle = \boldsymbol{\text {U}} \cap [(A \cup B) \cup (B \cup C)]$, Complement Law
$\displaystyle = A \cup (B \cup B) \cup C$, Identity Law, Associative Law
$\displaystyle = A \cup B \cup C$, Idempotent Law | 824 | 2,204 | {"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} | 4.03125 | 4 | CC-MAIN-2018-13 | latest | en | 0.732531 |
https://mathsite.org/factoring-maths/roots/implicit-differentiation.html | 1,701,437,395,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100287.49/warc/CC-MAIN-20231201120231-20231201150231-00754.warc.gz | 446,422,485 | 12,870 | FreeAlgebra Tutorials!
Try the Free Math Solver or Scroll down to Tutorials!
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Author Message
Somt Pxeldel
Registered: 20.02.2005
Posted: Thursday 28th of Dec 15:56 Does someone here know anything concerning implicit differentiation calculator? I’m a little puzzled and I don’t know how to finish my algebra homework about this topic. I tried reading all materials about it that could help me answer my math problems but I still can’t finish it. I’m having a hard time answering it especially the topics syntehtic division, perpendicular lines and dividing fractions. It will take me forever to answer my math homework if I can’t get any assistance. It would greatly help me if someone would suggest anything that can help me with my algebra homework.
IlbendF
Registered: 11.03.2004
From: Netherlands
Posted: Saturday 30th of Dec 12:57 Algebrator is a good software to solve implicit differentiation calculator questions. It gives you step by step answers along with explanations. I however would warn you not to just copy the solutions from the software. It will not help you in understanding the subject. Use it as a reference and solve the questions yourself as well.
Xane
Registered: 16.04.2003
From: the wastelands between insomnia and clairvoyance
Posted: Monday 01st of Jan 09:05 I would just add a note to what has been posted above. Algebrator no doubt is the most useful tool one could have. Always use it as a guide and a means to learn and never to cheat.
Jot
Registered: 07.09.2001
From: Ubik
Posted: Tuesday 02nd of Jan 18:02 Algebrator is the program that I have used through several algebra classes - Remedial Algebra, Algebra 1 and Algebra 1. It is a truly a great piece of algebra software. I remember of going through problems with x-intercept, equivalent fractions and greatest common factor. I would simply type in a problem from the workbook , click on Solve – and step by step solution to my algebra homework. I highly recommend the program.
dahoibufv
Registered: 20.07.2004 | 741 | 3,053 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.878548 |
https://api-project-1022638073839.appspot.com/questions/a-balanced-lever-has-two-weights-on-it-the-first-with-mass-15-kg-and-the-second--8 | 1,585,601,250,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497301.29/warc/CC-MAIN-20200330181842-20200330211842-00539.warc.gz | 359,032,446 | 5,887 | # A balanced lever has two weights on it, the first with mass 15 kg and the second with mass 9 kg. If the first weight is 6 m from the fulcrum, how far is the second weight from the fulcrum?
Jul 1, 2016
10m
#### Explanation:
The torques balance each other. therefore,
${r}_{1} \times {F}_{1} = {r}_{2} \times {F}_{2}$
$F = m \cdot g$
${m}_{1} \cdot g \cdot {r}_{1} = {m}_{2} \cdot g \cdot {r}_{2}$
$15 \times g \times 6 = 9 \times g \times {r}_{2}$
Solving this equation we get ${r}_{2} = 10 m$ | 183 | 501 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2020-16 | latest | en | 0.80732 |
http://mathhelpforum.com/algebra/66367-linear-equations-two-variables.html | 1,527,488,824,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794871918.99/warc/CC-MAIN-20180528044215-20180528064215-00604.warc.gz | 170,921,319 | 9,612 | # Thread: Linear Equations in two variables
1. ## Linear Equations in two variables
x + y(square) = 11
x(square) + y = 7
find x and y
2. Originally Posted by Prince
x + y(square) = 11
x(square) + y = 7
find x and y
Substitute $\displaystyle y = 7 - x^2$ into $\displaystyle x + y^2 = 11$ and solve the resulting equation. Note that one of the solutions is x = 2. Now you have a cubic to deal with .... | 128 | 406 | {"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} | 3.671875 | 4 | CC-MAIN-2018-22 | latest | en | 0.869261 |
https://www.askiitians.com/forums/Mechanics/a-method-for-determining-the-speed-of-a-bullet-is_122346.htm | 1,610,875,423,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703511903.11/warc/CC-MAIN-20210117081748-20210117111748-00256.warc.gz | 677,177,508 | 39,120 | ×
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```
A method for determining the speed of a bullet is to shoot it into a block of wood and see how far the block slides on a surface. (See Problem 3.) Assuming (incorrectly) that the magnitude of the work done by friction is equal to the force of friction on the block times the distance the block slides, the calculated value of the bullet velocity will be (A) lower than the actual value, because there will also be a change in the internal energy of the block and surface, (B) higher than the actual value, because there will also be a change in the internal energy of the block and surface, (C) correct, because errors caused by ignoring changes in internal energies are cancelled by the error in the assumption about the work done by friction. (D) wrong, because friction invalidates conservation of momentum as well.
```
5 years ago
Navjyot Kalra
654 Points
``` The correct option is (B).The bullet comes to rest in the block, in which the block moves some distance. As the magnitude of the work done by friction is equal to the force of friction on the block times the distance block slides, thus the work done on the block because of friction is 100 % of the energy dissipated because of friction. Therefore the calculated value of the bullet will be higher than the actual value, because there will also be a change in the internal energy of the block and surface. From the above observation we conclude that, option (B) is correct.
```
5 years ago
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• Previous Year Exam Questions | 529 | 2,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-04 | latest | en | 0.875316 |
http://quant.stackexchange.com/tags/yield-curve/hot?filter=year | 1,469,455,503,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824230.71/warc/CC-MAIN-20160723071024-00261-ip-10-185-27-174.ec2.internal.warc.gz | 193,491,850 | 15,758 | # Tag Info
5
Within the fixed income space, there's a lot of literature on PCA trading. The first 2-3 principal component factors (PCs) can typically explain 90-99% of the total variances in yield curve movement. It's also nice, because the first PC looks like a change in the overall level of the yield curve, the second PC looks like a slope change, while the third ...
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You're not the first to trip on this, and unfortunately the fact that the provided example is from a different era doesn't help. Quite simply, you're not writing rates correctly. The 5-years swap rate, 0.3523%, must be written in decimal form as 0.003523. The same goes for the deposit rates. As your code is now, you're writing that the 4-years rate is 23....
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fixedLegBPS is the basis-point sensitivity of the fixed leg, that is, how much its NPV changes when the fixed rate changes by one basis point: it's calculated as the NPV corresponding to a fixed rate of 1 bps. Since the NPV of the fixed leg is linearly proportional to the fixed rate, you can write the equation targetNPV : fixedRate = BPS : 1 basis point ...
2
bootstrap fedfunds (ois ) swaps to get your discount curve (asuming your portfolio is usd, and is usd collateralised). strangely i dont see the data on the fed site. i see data on LCH's site: http://www.lchclearnet.com/asset-classes/otc-interest-rate-derivatives/volumes/settlement-prices-swapclear-global#usd to get libor projection curve, you need to ...
2
Given the Ho-Lee interest rate model of the form \begin{align*} dr_t = \theta_t dt + \sigma dW_t, \end{align*} the price at time $t>0$ of a zero-coupon bond, with maturity $T$ and unit face, has the form \begin{align*} B(t, T) &=E\Big(e^{-\int_t^T r_s ds} \mid r_t \Big)\\ &=e^{-(T-t)r_t - \int_t^T (T-u)\theta_u du + \frac{\sigma^2}{6}(T-t)^3}. \...
2
Let $\delta$ be 3 month and consider points of interest $\{T_i\}_i$ evenly spaced with $T_{i+1} -T_i = 3 month$. The Forward Rate $F_m^n(t)$ from period m to n at time $t$ is defined by $$(1 + \delta (n-m) F_m^n(t)) = \frac{B(t,T_m)}{B(t,T_n)},$$ where $B(t,T_i)$ is the time $t$ value of a zero coupon bond that matures in $T_i$. A swap rate $S_m^n(t)$ a ...
2
Why do you think this is not apropriate? Matlabs documentation for 1-D Data interpolation states that interpl1 using method spline is the right way to go: Spline interpolation using not-a-knot end conditions. The interpolated value at a query point is based on a cubic interpolation of the values at neighboring grid points in each respective dimension. ...
2
There is a market for inflation linked government bonds (some countries e.g. US,CA,UK,FR,Germany,...). There are various prices quoted. The price with inflation lift (the inflation that has accumulated since the inception of the bond) and the price without the lift reflecting future nominal interest and inflation. You can calculate the real yield to ...
2
If you want to discount the CF3 from 3 years in the future to today you should use (1 + 3yr spot rate)^3. There's no reason to use forward rates for that purpose. The forward rates should only be used for period-by-period discounting - for example, if you wanted to find the value after 3 years of a CF4 which occurs after 4 years, you would use (1+ 1yr ...
1
It depends on the market you're interested in and what the curve is used for. To build the USD swap curve, for example, you've got a ton of information available from actively traded market instruments – fed funds futures (monthly), OIS (even finer details at the front end), Eurodollar futures (quarterly), basis swap, etc. All of these should be ...
1
Instead of the constant maturity series (which IMO would give only a few points), you could use the prices of ZCB to get the USD curve. They are available here http://www.wsj.com/mdc/public/page/2_3020-tstrips.html It might require some slight smoothing to get a clean curve. This is the best way I know to get a US Govt curve for free.
1
Per @dm63, these yield curves are basically smoothed curves that best fit the prices/yields of bonds traded in the secondary market. However, they reflect much more than market expectations. Refer to Deriving Interest Rates for details.
1
Yes, yield curves are a pictorial representation of the current secondary market yields of government securities (gilts, in the UK). These market yields are determined largely by expectations about what the central bank will do to short term rates over time.
1
The short answer is - you need more data. If you want to build a full zero-rate swap curve, typically these curves go out to 30 years. In general, the front of the curve is made from LIBOR rates, which you have. Typically you don't see practitioners use anything past the 3M point but some will use up to the 6M point. For the 2nd part of the curve, from ...
1
Translate your forecast of yields into a forecast of bond prices: you believe long term bonds will fall in price rel. to short term bonds. So, what to do? Shorten the duration of your portfolio, i.e. sell long term bonds and/or buy short term bonds. Since you don't like long term bonds (and the fixed payments they make to you), you may also enter into (...
1
This is all theoretical and real life will diverge from the theory The spot rates and forward rates are linked. Spot rate for the nth period should equal the product of all the forward rates up to that period. i.e Let Spot{n} = spot rate for nth period Let Forw{k,j} = forward rate to period j at period k Let X_m be the m'th period. Then (1+Spot{n})^n =...
1
Nowadays, government yield curves are customarily built with only coupon bonds. Zero coupon bonds (i.e., STRIPS in the US) are much less liquid compared with coupon Treasuries, and tend to trade very differently. If you plot a zero curve implied by coupon Treasuries vs yields of STRIPS, you'll notice that they can differ quite a bit in certain parts of the ...
Only top voted, non community-wiki answers of a minimum length are eligible | 1,501 | 5,999 | {"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} | 3.234375 | 3 | CC-MAIN-2016-30 | latest | en | 0.889131 |
https://www.in2013dollars.com/australia/inflation/1948 | 1,720,917,778,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00469.warc.gz | 702,976,402 | 14,398 | # \$100 in 1948 is worth \$3,656.76 today
\$
## The Australian dollar has lost 97% its value since 1948
Updated: July 11, 2024
\$100 in 1948 is equivalent in purchasing power to about \$3,656.76 today, an increase of \$3,556.76 over 76 years. The dollar had an average inflation rate of 4.85% per year between 1948 and today, producing a cumulative price increase of 3,556.76%.
This means that today's prices are 36.57 times as high as average prices since 1948, according to the Bureau of Statistics consumer price index. A dollar today only buys 2.735% of what it could buy back then.
The inflation rate in 1948 was 8.82%. The current inflation rate compared to the end of last year is now 1.05%. If this number holds, \$100 today will be equivalent in buying power to \$101.05 next year.
Cumulative price change 3,556.76% Average inflation rate 4.85% Converted amount\$100 base \$3,656.76 Price difference\$100 base \$3,556.76 CPI in 1948 3.700 CPI in 2024 135.300 Inflation in 1948 8.82% Inflation in 2024 1.05% \$100 in 1948 \$3,656.76 in 2024
AUD inflation since 1948
Annual Rate, the Bureau of Statistics CPI
## Buying power of \$100 in 1948
This chart shows a calculation of buying power equivalence for \$100 in 1948 (price index tracking began in 1922).
For example, if you started with \$100, you would need to end with \$3,656.76 in order to "adjust" for inflation (sometimes refered to as "beating inflation").
When \$100 is equivalent to \$3,656.76 over time, that means that the "real value" of a single Australian dollar decreases over time. In other words, a dollar will pay for fewer items at the store.
This effect explains how inflation erodes the value of a dollar over time. By calculating the value in 1948 dollars, the chart below shows how \$100 is worth less over 76 years.
According to the Bureau of Statistics, each of these AUD amounts below is equal in terms of what it could buy at the time:
Dollar inflation: 1948-2024
YearDollar ValueInflation Rate
1948\$100.008.82%
1949\$108.118.11%
1950\$118.9210.00%
1951\$140.5418.18%
1952\$164.8617.31%
1953\$172.974.92%
1954\$175.681.56%
1955\$178.381.54%
1956\$189.196.06%
1957\$194.592.86%
1958\$194.590.00%
1959\$200.002.78%
1960\$208.114.05%
1961\$210.811.30%
1962\$210.810.00%
1963\$213.511.28%
1964\$218.922.53%
1965\$227.033.70%
1966\$232.432.38%
1967\$240.543.49%
1968\$248.653.37%
1969\$256.763.26%
1970\$264.863.16%
1971\$281.086.12%
1972\$297.305.77%
1973\$324.329.09%
1974\$375.6815.83%
1975\$432.4315.11%
1976\$489.1913.13%
1977\$548.6512.15%
1978\$591.897.88%
1979\$645.959.13%
1980\$713.5110.46%
1981\$781.089.47%
1982\$867.5711.07%
1983\$956.7610.28%
1984\$994.593.95%
1985\$1,059.466.52%
1986\$1,156.769.18%
1987\$1,254.058.41%
1988\$1,345.957.33%
1989\$1,445.957.43%
1990\$1,554.057.48%
1991\$1,602.703.13%
1992\$1,618.921.01%
1993\$1,645.951.67%
1994\$1,678.381.97%
1995\$1,756.764.67%
1996\$1,802.702.62%
1997\$1,808.110.30%
1998\$1,821.620.75%
1999\$1,848.651.48%
2000\$1,932.434.53%
2001\$2,016.224.34%
2002\$2,078.383.08%
2003\$2,135.142.73%
2004\$2,183.782.28%
2005\$2,243.242.72%
2006\$2,321.623.49%
2007\$2,375.682.33%
2008\$2,481.084.44%
2009\$2,524.321.74%
2010\$2,597.302.89%
2011\$2,683.783.33%
2012\$2,729.731.71%
2013\$2,797.302.48%
2014\$2,867.572.51%
2015\$2,910.811.51%
2016\$2,948.651.30%
2017\$3,005.411.92%
2018\$3,062.161.89%
2019\$3,110.811.59%
2020\$3,137.840.87%
2021\$3,227.032.84%
2022\$3,440.546.62%
2023\$3,618.925.18%
2024\$3,656.761.05%*
* Compared to previous annual rate. Not final. See inflation summary for latest 12-month trailing value.
Click to show 70 more rows
This conversion table shows various other 1948 amounts in today's dollars, based on the 3,556.76% change in prices:
Conversion: 1948 dollars today
Initial valueEquivalent value
\$1 dollar in 1948\$36.57 dollars today
\$5 dollars in 1948\$182.84 dollars today
\$10 dollars in 1948\$365.68 dollars today
\$50 dollars in 1948\$1,828.38 dollars today
\$100 dollars in 1948\$3,656.76 dollars today
\$500 dollars in 1948\$18,283.78 dollars today
\$1,000 dollars in 1948\$36,567.57 dollars today
\$5,000 dollars in 1948\$182,837.84 dollars today
\$10,000 dollars in 1948\$365,675.68 dollars today
\$50,000 dollars in 1948\$1,828,378.38 dollars today
\$100,000 dollars in 1948\$3,656,756.76 dollars today
\$500,000 dollars in 1948\$18,283,783.78 dollars today
\$1,000,000 dollars in 1948\$36,567,567.57 dollars today
## How to calculate inflation rate for \$100 since 1948
Our calculations use the following inflation rate formula to calculate the change in value between 1948 and today:
CPI today CPI in 1948
×
1948 AUD value
=
Today's value
Then plug in historical CPI values. The Australian CPI was 3.7 in the year 1948 and 135.3 in 2024:
135.33.7
×
\$100
=
\$3,656.76
\$100 in 1948 has the same "purchasing power" or "buying power" as \$3,656.76 in 2024.
To get the total inflation rate for the 76 years between 1948 and 2024, we use the following formula:
CPI in 2024 - CPI in 1948CPI in 1948
×
100
=
Cumulative inflation rate (76 years)
Plugging in the values to this equation, we get:
135.3 - 3.73.7
×
100
=
3,557%
## Data source & citation
Raw data for these calculations comes from the government of Australia's annual (CPI) as provided by the Reserve Bank of Australia. The consumer price index was established in 1922 and is tracked by Australian Bureau of Statistics (ABS).
You may use the following MLA citation for this page: “Value of 1948 Australian dollars today | Australia Inflation Calculator.” Official Inflation Data, Alioth Finance, 14 Jul. 2024, https://www.officialdata.org/australia/inflation/1948.
Special thanks to QuickChart for their chart image API, which is used for chart downloads.
in2013dollars.com is a reference website maintained by the Official Data Foundation. | 2,001 | 5,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-30 | latest | en | 0.931675 |
https://en.m.wikisource.org/wiki/Page:EB1911_-_Volume_14.djvu/720 | 1,566,243,700,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314904.26/warc/CC-MAIN-20190819180710-20190819202710-00329.warc.gz | 461,547,400 | 10,423 | # Page:EB1911 - Volume 14.djvu/720
689
INTERFERENCE OF LIGHT
series of waves constituting any particular coloured light is reflected from an infinitely thin plate, the two partial reflections are in absolute discordance and, if of equal intensity, must give on superposition complete darkness. With the aid of this principle the sequence of colours in Newton's rings is explained in m11Ch the same way as that of interference fringes (above, § 5)., The complete theory of the colours of thin plates requires us to take account not merely of the two rejections already mentioned but of an infinite series of
such reflections. This was
first effected by S. D. Poisson
for the case of retardation's
F A
E
(%
V D B
which are exact multiples of
the half wave-length, and
afterwards more generally by
gc 2 sir G. B. Airy tcamb. Phu.
Trans., 1832, 4, p. 409)-In
fig. 2, ABF is the ray, perpendicular to the wave-front, reflected at the upper surface, ABCDE the ray transmitted at B, reflected at C and transmitted at D; and these are accompanied by other rays reflected internally 3, 5, &c., times. The first step is to calculate the retardation 5 between the first and second waves, so far as it depends on the distances travelled in the plate (of index fi) and in air.
If the angle ABF=2a, angle BCD=2a' and the thickness of plate=t, we have
6 =;f(BC-I-CD) -BG
=2, uI3C -2BC sin iz sin a' =2;4BC(I-sin'la.') =2;.tt cos n.' (I)-In
(I) a' is the angle of refraction, and we see that, contrary to what might at first have been expected, the retardation is least when the Obliquity is greatest, and reaches a maximum when the obliquity is zero or the incidence normal. If we represe nt all the quantities, from which finally the imaginary retardation 6 may be expressed by the intro vibrations by complex
parts are rejected, the
duction of the factor e'1K'5, where i=/ (-I), and K=27l'/A. refraction the amplitude of the incident wave -t each reflection or
must be supposed to be altered by a certain factor which allows room for the reversal postulated by Young. When the light proceeds from the surrounding medium to the plate, the factor for reflection will be supposed to be b, and for refraction c; the corresponding quantities when the progress is from the plate to the surrounding medium will be denoted by e, . Denoting the incident vibration by unity, we have then for the first component of the reflected wave b, for the second cefe'i'<5, for the third ce5f¢'2i»<S, and so on. Adding these together, and summing the geometric series, we find Cf fe'i"5
btirew <2>-In
like manner for the wave transmitted through the plate we get f
1 52{>-iaé
The quantities b, c, e, f are not independent. The simplest way to find the relations between them is to trace the consequences of supposing 5 =o in (2) and (3). This may be regarded as a development from Young's point of view. A plate of vanishing thickness is ultimately no obstacle at all. In the nature of things a surgace cannot reflect. Hence with a plate of vanishing thickness t ere must be a vanishing reflection and a total transmission. and accordingly 5+@=0» ff=1~@' (4),
the first of which embodies Arago's law of the equality of reflections as well as the famous “ loss of half an undulation.” Using these we find for the reflected vibration,
e(1 E-its)
"1 " e2€-;K"5 (5),
and for the transmitted vibration
1 -ei
gym (6).
The intensities of the reflected and transmitted lights are the squares of the moduli of these expressions. Thus (1-cos »<6)” + sin' K6
I ' = 2-i -;... ..
-ntensity of reflected lxgnt e (1'e2COSK5)2 I e, Si n2K5 = 4e"sin'¢(§ »<6)
1-2e2cos K5 -l-e4 (fl
I . . = (l-r2)2
ntensity of transmitted light 1—1— 262CO5K5+e, (3), the sum of the two expressions being unity. According to (7) not only does the reflected light vanish comp.etel when § = b l = l -
y 0, ut a so whenever 2155-nrr, n being an integer, that is. Wh€=1€'V€r 5=n). When the first and third mediums are the same, as we have here supposed, the central spot in the system of Newton's ring is black, even though the original light contain a mixture of all wave-lengths. If the light reflected from a plate of any thickness be examined with a spectroscope of sufficient resolving power, the spectrum will be traversed by dark bands, of which the centre corresponds to those wave-lengths which the plate is incompetent to reflect. It is obvious that there is no limit to the fineness of the bands which may be thus impressed upon a spectrum, whatever may be the character of the original mixed light. The relations between the factors b, c, e, f have been proved, independently of the theory of thin plates, in a general manner by Stokes, who called to his aid the general mechanical principle of ret/ersibilfity. If the motions constituting the reflected and refracted rays to which an incident ray gives rise be supposed to be reversed, they will reconstitute a reversed incident ray. This gives one relation; and another is obtained from the consideration that there is no ray in the second medium, such as would by the operation alone of either the reversed reflected or refracted rays. Space does not allow of the reproduction of the argument at length, but a few A words may perhaps give the reader an idea of how the conclusions are arrived at. The incident ray (IA) (fig. 3) being I, the reflected (AR) and refracted (AF) rays are denoted by b and c. When b is reversed, it gives rise to a reflected ray b2 along AI, and a refracted ray bc along AG (say). When 0 is reversed, it gives rise to cf along AI, and ce along AG. Hence bc-l-ce==0, b2+cf=I, which agree with (4). It is here assumed that there is no change of phase in the act of reflection or refraction, except such as can be represented by a change of sign. When the third medium diflers from the first, the theory of thin plates is more complicated, and need not here be discussed. One particular case, however, may be mentioned. When a thin transparent film is backed by a perfect reflector, no colours should be visible, all the light being ultimately reflected, whatever the wavelength may be. The experiment may be tried with a thin layer of gelatin on a polished silver plate. In other cases-where a different result is observ d h ' f
e, the in erence is that either the metal does not reflect perfectly, or else that the material of which the film is composed is not sufficiently transparent. Some apparent exceptions to the above rule, exhibited by thin films of collodion restin upon silver surfaces, have been described by R. W. Wood (fghysical Optics, p. 143), wh°o attributes the very curious effects observed to frilling of the collodion film.
For study of the colours of thin plates there are no more interesting subjects than the soap-film. For projection the films may be stretched across vertical rings of iron wire coated with paraffin. In their undisturbed condition they thin from the top, and the colours are disposed in horizontal bands. If, as suggested by Brewster, a jet of wind issuing from a small nozzle and supplied from a well-regulated bellows be allowed to impinge obliquely, parts of the film are set in rotation, and displays of colours may be exhibited to a large audience, astonishing by their brilliance and by the rapidity with which they change. Permanent films, analogous to soap-films, are best obtained by Glew's method. A few drops of celluloid varnish are poured upon the surface of water contained in a large dish. After evaporation of the solvent, the films may be picked up upon rings of iron wire.
As a variant upon Newton's rings, interesting effects may be obtained by the partial etching of the surfaces of picked pieces of plate-glass. A surface is coated in parallel stripes with arafiin P
wax and treated with dilute hydrofluoric acid for such a t1me (found by prelim mar tr' l) " ' ' 'y
ia s as is required to eat away the exposed portions to a depth of one quarter of the mean wave-length of light. Two such prepared surfaces pressed in the crossed position into suitable contact exhibit a chess-board pattern. Where two uncorroded, or where two corroded, parts overlap, the colours are nearly the same; b t h
u w ere a corroded and an uncorroded surface meet, a strongly contrasted colour is developed. The combination lends itself to projection and the pattern seen upon the screen is very beautiful if proper precautions are taken to eliminate the white li ht refle ted gromgtlge first and fourth surfaces of the plates (see ]§ amre, I;OI, 4, 3 5 -
be generated
R 1
F G
FIG. 3.
Theory and observation alike show that the transmitted colours of a thin plate, ag. a spap film or a layer of air, are very inferior to those reflected. Specimens of ancient glass which have under one ~ ~, g
superficial decomposition, on the other hand, sometimes show transmitted l f
co ours o remarkable brilliancy. The probable explanation, suggested by Brewster, is that we have here to deal not merely with one, but with a series of thin plates of not very different thicknesses. It is evident that with such a series the transniitt cl - . e
cglours wquldl be much purer, and the reflected much brighter t 8.
|
n usua . f the thicknesses are strictly equal. certain wavelengths must still be absolutely missing in the reflected light; while on the other hand a constancy of the interval between the plates will in general lead to a special preponderance of light of some other waveleng th for which all the component parts as they ultimately emerge are in agreement as to phase.
On the same principle are doubtless to be explained the colours
of fiery opals, and, more remarkable still, the iriclescence of certain | 2,376 | 9,718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.899673 |
https://timus.online/problem.aspx?space=1&num=1022 | 1,719,197,763,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864986.57/warc/CC-MAIN-20240624021134-20240624051134-00876.warc.gz | 515,132,910 | 3,350 | ENG RUS Timus Online Judge
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## 1022. Genealogical Tree
Time limit: 1.0 second
Memory limit: 64 MB
### Background
The system of Martians’ blood relations is confusing enough. Actually, Martians bud when they want and where they want. They gather together in different groups, so that a Martian can have one parent as well as ten. Nobody will be surprised by a hundred of children. Martians have got used to this and their style of life seems to them natural.
And in the Planetary Council the confusing genealogical system leads to some embarrassment. There meet the worthiest of Martians, and therefore in order to offend nobody in all of the discussions it is used first to give the floor to the old Martians, than to the younger ones and only than to the most young childless assessors. However, the maintenance of this order really is not a trivial task. Not always Martian knows all of his parents (and there’s nothing to tell about his grandparents!). But if by a mistake first speak a grandson and only than his young appearing great-grandfather, this is a real scandal.
### Problem
Your task is to write a program, which would define once and for all, an order that would guarantee that every member of the Council takes the floor earlier than each of his descendants.
### Input
The first line of the standard input contains an only number N, 1 ≤ N ≤ 100 — a number of members of the Martian Planetary Council. According to the centuries-old tradition members of the Council are enumerated with integers from 1 up to N. Further, there are exactly N lines, moreover, the i-th line contains a list of i-th member’s children. The list of children is a sequence of serial numbers of children in a arbitrary order separated by spaces. The list of children may be empty. The list (even if it is empty) ends with 0.
### Output
The standard output should contain in its only line a sequence of speakers’ numbers, separated by spaces. If several sequences satisfy the conditions of the problem, you are to write to the standard output any of them. At least one such sequence always exists.
### Sample
inputoutput
```5
0
4 5 1 0
1 0
5 3 0
3 0
```
```2 4 5 3 1
```
Problem Author: Leonid Volkov
Problem Source: Ural State University Internal Contest October'2000 Junior Session | 569 | 2,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-26 | latest | en | 0.945576 |
https://www.learncax.com/knowledge-base/blog/by-category/cfd/using-udf-for-cfd-modeling-of-boiling-phenomena | 1,539,662,788,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509996.54/warc/CC-MAIN-20181016030831-20181016052331-00439.warc.gz | 968,507,432 | 29,552 | ### Using UDF for CFD Modeling of Boiling Phenomena
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14 October
Having the basic introduction to writing a UDF from the earlier article "Writing a UDF for CFD Modeling" wherein we saw using UDF for inserting custom boundary profiles, we shall now see another application of UDF with an example of boiling phenomenon. To get the most understanding from the current article the reader should be familiar with ANSYS Fluent software and basic UDF understanding.
Also the reader is recommended to go through the article "Mathematical Treatments in CFD Modeling of Multiphase Flows" along with the above mentioned.
The Boiling phenomenon :
Boiling is probably the most familiar form of heat transfer, yet remains to be the least understood. Although a wide number of research papers are available on the subject yet we do not completely understand the exact process of bubble formation and still rely on empirical or semi-empirical relations to predict the rate of heat transfer through boiling.
Boiling is a convection process involving a change in phase from liquid to vapor occurring when a liquid is in contact with a surface maintained at a temperature higher than the saturation temperature of the liquid.
Why study Boiling ?
Many engineering applications involve boiling heat transfer, for example in steam power plant, heat is transferred to steam in boiler where water is vaporized. In the household refrigerator, the refrigerant absorbs heat from the refrigerated space by the boiling process within the evaporator section. The phenomenon of boiling heat transfer has also been successfully used in the cooling of nuclear reactors and rocket engines where energy dissipation rates are extremely large and thus there are many wide range of applications in which boiling phenomenon plays an important role.
Classification of Boiling :
Boiling is classified as:
1. Pool boiling : If heat added to a liquid from a submerged solid surface, the boiling process is referred to as pool boiling. In this process the vapour produced may form bubbles, which grow and subsequently detach themselves from the surface, rising to the free surface due to buoyancy effects. e.g. boiling of water in kettle on a stove.
2. Flow boiling : Flow boiling occurs in a flowing stream and the boiling surface may itself be a portion of the flow passage. This phenomenon is generally associated with two phase flows through a confined passage.
Boiling regimes
If temperature of surface is $T_S$ and saturation temperature of liquid is $T_{SAT}$, then $T_S- T_{SAT} = \Delta T_E$, the excess temperature.
Depending on values of excess temperature, four different regimes are observed.
1. Free convection zone : In this zone excess temperature is very small. Here liquid layer next to the hot surface is slightly superheated. The convection currents circulate the liquid and evaporation takes place at liquid surface.
2. Nucleate boiling : In this region excess temperature is higher than that of free convection zone. As the excess temperature is increased, bubbles begin to form on the surface. If we further increase excess temperature, bubbles are formed more rapidly and rise to the surface of the liquid resulting in rapid evaporation. As the excess temperature goes on increasing heat flux also increases upto a maximum value known as critical heat flux. Nucleate boiling regime exists up to critical heat flux only, next regime is film boiling regime.
3. Film boiling : After reaching critical heat flux value, heat flux decreases with further increase in temperature. This is due to the fact that bubbles are now formed so rapidly that they blanket the heating surface with a vapour film preventing the inflow of fresh liquid from taking their place. Now this heat must be transferred through this vapour film (by conduction) to the liquid for carrying out any further boiling. Since the thermal conductivity of vapour film is much less than that of the liquid, the value of heat flux must then decrease with increase of excess temperature. If we keep on increasing the excess temperature up to a certain limit, the vapour film remains unstable. Later the vapour film stabilizes with further increase in temperature and the heating surface is completely covered by a vapour blanket. With this the heat flux further reduces reaching its lowest value. The surface temperature required to maintain a stable film is high and under these conditions a sizeable amount of heat is lost by the surface due to radiation.
The phenomenon of stable film boiling can be observed when a drop of water falls on a red hot stove. The drop does not evaporate immediately but dances a few times on the stove. This is due to the formation of a stable steam film at the interfaces between the hot surface and liquid droplet. Below is a visual demonstration for a water drop boiling on a hot stove.
Heat transfer rates and excess temperatures associated with nucleate boiling are small. The equipment used for boiling should be designed to operate in this region only. Now having had a overview of the physics of boiling, let us see how UDF can be applied to model film boiling.
Why we need UDF for Boiling type of simulations ?
In this simulation there are two distinct phases liquid and gas, so this simulation has to be solved using multiphase flows. In boiling, flow is separated where a distinct interface is present between liquid and gas, so it is advised to use VOF (volume of fluid) model for such flows. VOF uses Eulerian framework for both phases with specialized interface treatment. The boiling simulation will include mass transfer between two phases. As we know that VOF model doesn’t simulate mass transfer mechanism, UDF comes into picture for performing this additional task.
Compiling the UDF :
We can either compile UDF or interpret it. In this demonstration of simulation of boiling we shall choose compilation method. Complied UDF is nothing but code written in ‘C’ language and translated in machine language (language understood by computer) and the UDF becomes part of native fluent software. As we need to compile the UDF, we require a compiler. Few of the supported compiler for ANSYS Fluent are, Microsoft Visual Studio or the Express editions. The point to be noted here is one needs to start Fluent from command prompt of visual studio environment.
Test Case :
The geometry of a column is represented in the following figure with it's domain filled by fluid. The bottom wall is maintained at 10 K above saturation temperature of the fluid and at top (pressure outlet) there is saturation temperature at atmospheric pressure.
Geometry and boundary conditions
How to add mass transfer mechanism in our UDF :
To add mass transfer mechanism in our UDF we need to add it as a source term using macro DEFINE_SOURCE(). We will need 2 'source terms', one for gas and one for liquid as follows:
Defining source term for gas : DEFINE_SOURCE(gas, cell, thread, dS, eqn)
Defining source term for liquid : DEFINE_SOURCE(liquid, cell, thread, dS, eqn)
Results :
Thus the results obtained below shows the successful CFD simulation of film boiling process. The images show the bubble formation with respect to time. As can be seen at time 0.05 sec the film layer had developed completely followed by the bubble formation at the centre of the column (i.e. bottom wall) as time advances to 0.4 sec and complete bubble detachment at 0.5 sec, completing the boiling cycle. The video further gives insights of the phenomena and seems to match the real life behavior of film regime boiling process.
References :
1. www.ansys.com
2. Introduction to Heat Transfer; Frank P. Incropera, David P. DeWitt
##### Sanket Dange
Sanket is working as a CFD engineer in CCTech. He has worked on numerous CFD projects in the filed of automobile and HVAC. He has worked as a tutor at LearnCAx and taught to students through LeanrCAx online as well as offline course modules.
Currently he is working in the field of CFD software development which is mainly based on ANSYS FLUENT. Sanket has written blogs on various topics in CFD. He also has interest in open source CFD technologies.
Sanket holds a Bachelors degree in Mechanical engineering from University of Pune.
Using UDF for CFD Modeling of Boiling Phenomena - 4.9 out of 5 based on 23 votes
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+1 SJ 2014-08-02 00:45
1. Where is the UDF?
2. Why only one bubble?
3. What are the boundary conditions? 10K above sat temp or 100 C above sat temp?
+4 Sanket Dange 2014-08-02 21:37
Hello Sijal! Thanks for showing interest in our blog!!
We haven't made the UDF available publicly.
The domain size is very small so as to study one bubble only. Boundary condition for the bottom wall is 10 K above saturation temperature.
0 Vira 2016-01-05 10:52
Dear Author,
Im interested with your work. It is really nice. I have questions :
1. How to specify the boiling point/area? As your geometry shown the area of your boiling phenomena is at the bottom (x,y) = (0,0). What if we put heater, and the heater position is in the middle x=0.1 to x=0.2 m and y=0
2. I've read and tried the horizontal film boiling (the tutorial), do we still need to specify the define field function if the geometry as I mentioned above (heater in the middle)? How? Because we have 2 direction (x and y)
3. May I contact you by email? my email
Thank you so much, I hope u would love to share your knowledge to me. Warm regards, | 2,159 | 10,028 | {"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": 3, "/images/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-2018-43 | longest | en | 0.917284 |
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Topic: Tricky logistic growth problem
Replies: 5 Last Post: Apr 29, 1998 7:45 PM
Messages: [ Previous | Next ]
Elisse Ghitelman Posts: 34 Registered: 12/6/04
Tricky logistic growth problem
Posted: Apr 27, 1998 12:30 PM
In light of Dave Slomer's problem of assigning "difficult" problems, I sat
down today and worked a practice AP exam from the new Barron's review book
before having my students try it. (It's our school vacation week). I ran
into a problem on part A of the multiple choice (NO calculators) that I'm
not sure how to do without a calculator. I only have a copy of the problem
(not the book with the solutions) at home with me, so I'm looking for a
little help. Here's the problem:
A population of rabbits grows according to the differential equation
dR/dt = 0.001 R (200 - R) where R(t) is the number of rabbits after t months.
If there were initially 25 rabbits, aproximately how many months will it
take the population to double?
(a) 2.7 (B) 3.5 (C) 4.2 (D) 6.9 (E) 8.4
I tried two different techniques: First, I tried to use Euler's method by
hand with one month intervals. I had to do a lot of rounding to get
numbers I could work with, and then just guessed based on the answer
choices. I saw that the population was initially growing at around 5
rabbits/ month; since I was below half the limiting population, the growth
rate was increasing, so it would take a little less than five months, guess
choice (C).
Next, I actually solved the diff eq to get:
R/(200-R)=(1/7) e^(t/5)
Substituting R=50, you get that the doubling time is t = 5 ln(7/3).
Again, I estimated; ln (7/3) is a little less than 1, so we guess (C) again.
So, anyone with too much time on their hands, and more insight than I want
to suggest an easier way to get the answer for this one.
Elisse Ghitelman
Newton North High School
Newton, MA, USA
Date Subject Author
4/27/98 Elisse Ghitelman
4/27/98 Lou Talman
4/28/98 Jerry Uhl
4/28/98 Elisse Ghitelman
4/28/98 Don
4/29/98 Richard Sisley | 622 | 2,194 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | longest | en | 0.963973 |
http://fermionlattice.wikidot.com/splitting | 1,601,269,928,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401585213.82/warc/CC-MAIN-20200928041630-20200928071630-00304.warc.gz | 43,943,723 | 13,107 | Splitting
Reference
## Background
Once we've created a quantum degenerate gas of atoms, we have the ability to dynamically deform the potential these atoms see. By applying resonant radio-frequency (rf) fields, we couple different internal magnetic substates of the atom, and the new potential includes contributions from the static magnetic and the oscillating magnetic (rf) fields. We call this new potential a "dressed" potential. The exact shape of this new potential depends on the frequency of the rf field, such that we can sweep the frequency to deform the trap.
In the series of images below, we represent a one-dimensional dressed potential seen by the atoms as the frequency is increase left to right. At low frequencies, the potential is only slightly deformed. As we increase the frequency, a barrier is raised in the centre of the trap and we go from a single to a double well.
Once we've split the cloud in two, we're interested in learning how the atoms decided to go either to the right side or the left side. For bosons in a BEC, this decision is dominated by the effects of interactions. In 87Rb, where there exists a repulsive interaction between the bosons in the BEC, the system wishes to minimise its total energy. Since it costs energy to put bosons together on one side of the well, the overall energy is minimised when the exact same number of atoms go to each side of the well. In contrast, if there were no interactions between the atoms, the choice to go right or left would be random, and therefore governed by binomial statistics.
We are interested in looking at the fluctuations in the numbers of atoms that go right or left over a large number of experimental realisations. We take a BEC, split it in two, then count the number that went to the right and the number that went to the left. We then repeat this many times. Then, we perform statistics on these results and look at the fluctuations in the numbers - for instance: what is the spread in the fraction of atoms that chose to go right?
If this were a random process, we expect that this spread is proportional to $\sqrt{N}$, where N is the total number of atoms in the system. If the interactions in the BEC play a role, we expect this number to be less than $\sqrt{N}$. This reduction in fluctuations would indicate that the system of atoms somehow acts collectively to reduce its overall energy.
## Experiment
To actually count the numbers of atoms that went into the right and left wells, we use absorption imaging. Resonant light is shone through the atoms at a CCD camera, and the shadow image is recorded on our computer. The atoms are then transferred into a state out of resonance with this light, and a second pulse is shone on the CCD camera, for reference. After dividing the two images, we can determine the number of atoms that were in each well from the darkness of the shadow. Below, we see one such image.
Another tool we have is to look at the phase difference between the wavefunctions of the atoms on the right and the left. To do this, we let the two clouds fall under gravity. The clouds expand into each other, and if we take another absorption image, we see interference fringes develop between the two clouds. The phase of the fringes (relative locations of the peaks and valleys with respect to the centre of the cloud) tells us about the phase of the wavefunction before release. This information is useful in determining how well we do our splitting (is it coherent?) and in determining the point at which the two condensates stop communicating with each other (how much tunnelling through the barrier?)
## Theory
### References
A bosonic Josephson junction R. Gati & M.K. Oberthaler, J. Phys. B: At. Opt. Mol. Phys. 40, R61 - R89 (2007).
### Two-mode model
To understand the behaviour of a BEC in a double well potential, we must consider the interactions between atoms and the tunnelling between the two wells. We assume (for now) that the temperature is very low and that the condensate resides in the ground state of the potential. We restrict ourselves to a "two-mode" model, where we consider only the ground and (nearly-degenerate) first excited state of the double well potential.
#### Two-mode Hamiltonian
In constructing a hamiltonian, we include a term describing the kinetic and potential energy (regular Schroedinger equation) and a term that accounts for interactions.
(1)
\begin{align} \hat{H} = \hat{H_0} + \hat{H_{int}} = \int d{\bf r} \left[\frac{-\hbar^2}{2m} \hat{\Psi}^{\dagger} \nabla^2 \hat{\Psi}+\hat{\Psi}^{\dagger} V_{dw} \hat{\Psi}\right] + \frac{g}{2}\int d{\bf r}~\hat{\Psi}^{\dagger}\hat{\Psi}^{\dagger} \hat{\Psi}\hat{\Psi} \end{align}
where $V_{dw}$ is the double well potential. One can show (Gati & Oberthaler, 2007) that if you rewrite the field operators as a sum of contributions from the ground ($\phi_g$) and first excited $\phi_e$ states,
(2)
\begin{align} \hat{\Psi} = \hat{c}_g \phi_g + \hat{c}_e \phi_e \end{align}
and then change to the right-left basis, where
(3)
\begin{align} \hat{c}_r = \frac{1}{\sqrt{2}}(\hat{c}_g + \hat{c}_e);~~~ \hat{c}_\ell = \frac{1}{\sqrt{2}}(\hat{c}_g - \hat{c}_e) \end{align}
you can then rewrite the two-mode hamiltonian as
(4)
\begin{align} \hat{H}_{2M} = \frac{E_C}{8} (\hat{c}_r^{\dagger}\hat{c}_r - \hat{c}_{\ell}^{\dagger}\hat{c}_{\ell})^2 - \frac{E_J}{N} (\hat{c}_r^{\dagger}\hat{c}_{\ell} + \hat{c}_{\ell}^{\dagger}\hat{c}_{r}) + \frac{\delta E}{4}(\hat{c}_r^{\dagger}\hat{c}_{\ell} + \hat{c}_{\ell}^{\dagger}\hat{c}_{r})^2 \end{align}
where
(5)
\begin{align} E_C = 8 \kappa_{g,e} \end{align}
(6)
\begin{align} E_J = \frac{N}{2}(\mu_e-\mu_g) - \frac{N(N+1)}{2}(\kappa_{e,e} - \kappa_{g,g}) \end{align}
(7)
\begin{align} \delta E = \frac{\kappa_{e,e} + \kappa_{g,g} - 2\kappa_{g,e}}{4} \end{align}
(8)
\begin{align} \kappa_{i,j} = \frac{g}{2} \int d{\bf r} |\phi_i|^2 |\phi_j|^2 \end{align}
(9)
\begin{align} \mu_{e,g} = \int d{\bf r} \frac{-\hbar^2}{2m} \phi_{e,g}^{*} \nabla^2 \phi_{e,g}+\phi_{e,g}^{*}( V_{dw} + gN|\phi_{e,g}|^2 ) \phi_{e,g} \end{align}
#### Discussion of terms
Question: What do each of these expressions mean and what do these quantities depend on?
• $E_C$ is called the charging energy. It belongs to the term that multiplies the number difference, so it tells us how much energy it costs to have a different number of atoms in the right and the left wells. It depends on the probability we exist in both of the two modes, and is maximised when $\phi_e = \phi_g$, or there is an equal superposition of ground and excited states, which is saying that all of the atoms are in the left or all are in the right mode. The energy is minimized when the right and left modes are equally populated. This energy term also depends on the size and sign of the interactions between atoms, and disappears for a non-interacting system.
• $E_J$ is the tunnelling energy. It belongs to the tunnelling term, which tells us what happens when we move a particle from the right to the left well, of vice-versa. There are two parts to this expression. The first, which depends on the chemical potentials of the ground and first excited states, tells us about the energy cost to move a particle from one well to another - the energy it takes to move an atom from right to left. Examining this term, we see that (overall) it is positive if the ground state is more populated (the wells are balanced) and zero when the symmetric and antisymmetric modes are equally populated (tendency for atoms to be all in left or all in right). Thus, this drives the system to want to tunnel all to one side or all to the other. The second term in the tunnelling energy depends explicitly on interactions, in addition to the interactions included in the chemical potential term. If we are in a state where the wavefunction is primarily of a symmetric character ($\phi_g > \phi_e$, balanced wells) then this energy term is overall negative. If the wavefunction is an equal superposition of symmetric and antisymmetric ($\phi_g \approx \phi_e$), all on right or all on left), then this part of the tunnelling term is zero. This drives the system to tunnel such that the overall interaction energy is reduced by balancing the splitting. The two terms are in competition with each other, such that the tunnelling behaviour is not straightforward.
• $\delta E$ is a small term, and I'm going to assume it doesn't matter for us like it doesn't matter for the Oberthaler group.
#### Simplified Hamiltonian
We can define the number difference $\hat{n}$ and tunnelling operators $\hat{\alpha}$ from the left and right mode operators
(10)
\begin{align} \hat{n} = \frac{\hat{c}_r^{\dagger}\hat{c}_r - \hat{c}_{\ell}^{\dagger}\hat{c}_{\ell}}{2}; ~~~ \hat{\alpha} = \frac{\hat{c}_r^{\dagger}\hat{c}_{\ell} + \hat{c}_{\ell}^{\dagger}\hat{c}_{r}}{N}, \end{align}
and simplify the Hamiltonian
(11)
\begin{align} \hat{H}_{2M} = \frac{E_C}{2} \hat{n}^{2} - E_J \hat{\alpha}. \end{align}
This is usually called the Bose-Hubbard Hamiltonian.
#### Mean-field, Gross-Pitaevskii
Assuming a mean-field picture of a BEC, where the temperature is low and interactions are local, we assume we can write the operators for each side of the well as complex numbers describing the properties of the wavefunction in each half of the well, ie, $\hat{c_R} = \sqrt{N_R} exp(-i\phi_R)$ where $N_R$ is the number of atoms on the right side, and $\phi_R$ is the phase of the wavefunction on the right side. In so doing, we can rewrite the Hamiltonian as
(12)
\begin{align} {H}_{2M, GP} = \frac{E_C}{2} {n}^{2} - E_J \sqrt{1 - \frac{4 n^2}{N^2}}\cos\phi, \end{align}
where $n = (N_L - N_R) / 2$ is the number difference between the wells and $\phi = \phi_R - \phi_L$ is the phase difference. We can expand this Hamiltonian to second order in [[ $\phi$]] and [[ $n$ ]]to get something that looks much like the canonical harmonic oscillator Hamiltonian
(13)
\begin{align} {H}_{simple} = \left(E_C + \frac{4 E_J}{N^2}\right) \frac{{n}^{2}}{2} + E_J \frac{\phi^2}{2}, \end{align}
Back to Chip Experiment
page revision: 35, last edited: 01 Mar 2012 23:36 | 2,789 | 10,135 | {"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": 13, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2020-40 | latest | en | 0.946311 |
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https://estebantorreshighschool.com/faq-about-equations/physics-work-equation.html | 1,660,818,012,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573193.35/warc/CC-MAIN-20220818094131-20220818124131-00275.warc.gz | 236,322,053 | 11,088 | ## How do you calculate work in physics?
Work can be calculated with the equation: Work = Force × Distance. The SI unit for work is the joule (J), or Newton • meter (N • m). One joule equals the amount of work that is done when 1 N of force moves an object over a distance of 1 m.
## How do you calculate energy physics?
The formula that links energy and power is: Energy = Power x Time. The unit of energy is the joule, the unit of power is the watt, and the unit of time is the second.
## What is work done in physics?
In summary, work is done when a force acts upon an object to cause a displacement. Three quantities must be known in order to calculate the amount of work. Those three quantities are force, displacement and the angle between the force and the displacement.
## What is the work energy equation?
The work W done by the net force on a particle equals the change in the particle’s kinetic energy KE: W=ΔKE=12mv2f−12mv2i W = Δ KE = 1 2 mv f 2 − 1 2 mv i 2 .
## What is Ke formula?
Kinetic energy is directly proportional to the mass of the object and to the square of its velocity: K.E. = 1/2 m v2. If the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.
## What is the SI unit of force?
The SI unit of force is the newton, symbol N. The base units relevant to force are: The metre, unit of length — symbol m. The kilogram, unit of mass — symbol kg. The second, unit of time — symbol s.
## What is the efficiency formula?
Efficiency is often measured as the ratio of useful output to total input, which can be expressed with the mathematical formula r=P/C, where P is the amount of useful output (“product”) produced per the amount C (“cost”) of resources consumed.
## What are forces in physics?
A force is a push or pull upon an object resulting from the object’s interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects.
## Is work a scalar or vector?
Work is a scalar because it is the “dot” product of 2 vectors, also called the scalar product. W can also be expressed in terms of the components of the force and displacement vectors. Work is a vector because you multiply a force (a vector) by distance (a vector).
## What is no work is done?
If there is no motion in the direction of the force, then no work is done by that force.
## Who invented work in physics?
Gaspard-Gustave Coriolis
You might be interested: First order reaction equation
## Can power be negative physics?
The sign of power is just a matter of definition. Energy can increase or decrease, but the energy itself cannot be negative. As you said, power is just the time rate of change of energy, so it can be negative. But energy is only positive.
## How do we calculate energy?
In classical mechanics, kinetic energy (KE) is equal to half of an object’s mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.
### Releated
#### Depreciation equation
What are the 3 depreciation methods? There are three methods for depreciation: straight line, declining balance, sum-of-the-years’ digits, and units of production. What do you mean by depreciation? Definition: The monetary value of an asset decreases over time due to use, wear and tear or obsolescence. This decrease is measured as depreciation. How do you […]
#### Polar to cartesian equation calculator wolfram
How do you convert polar to Cartesian? Summary: to convert from Polar Coordinates (r,θ) to Cartesian Coordinates (x,y) 😡 = r × cos( θ )y = r × sin( θ ) How do you find the polar Cartesian equation? Convert the polar equation r = 2sec θ to a rectangular equation, and draw its corresponding […] | 932 | 3,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2022-33 | latest | en | 0.908254 |
https://healthdrugpdf.com/s/scimath.unl.edu1.html | 1,669,610,664,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710473.38/warc/CC-MAIN-20221128034307-20221128064307-00148.warc.gz | 337,083,759 | 18,913 | ## Microsoft word - mayo_ar_finaledit_aug2007.doc
Connections Between Communication
and Math Abilities
Rachelle Mayo
Math in the Middle Institute Partnership Action Research Project Report Department of Teaching, Learning, and Teacher Education Connections Between Communication and Math Abilities
Abstract
In this action research study of my Class I School’s 5th and 8th grade mathematics, I investigated
My first assertion is that communication seems to help students stay on task, with feed- back on their solutions; it helps them understand where their thinking is off and where they made their mistakes. In student interviews I asked what they learn by watching and listening to others’ Darrin, an 8th grader, for instance, responds in a way that seems quite normal for him,
How they did it, or how to do it if I don’t know how to do it, and other ways to do
it.
” He has never seemed very confident and does not think of the idea that his
explanations may help others:
Sam responded similarly, “I can see the way they do it and compare mine to it to see
how they did it differently and see if there is better way to do it so I can improve
myself.”
He is always one to find the best way to do anything. And he will always
challenge new information.
Alternatively, Linda expressed that listening and responding to others is not always a
good thing. She said, “Sometimes they can be confusing because they learned
differently, different than the way I have. So it’s a different explanation.”
Linda
indicates that this sometimes confuses her and loses what she thought she had come to
understand.
Amber realized another importance to group work: “There are so many different ways
people see different things.”
Once students understand there is not always one way or
one answer they will be more willing to try to solve problems because they will know
that they can come up with their own way to solve.
Natalie also saw the importance of group work: “They (group members) help me see a
new perspective and I can better understand what I am doing.”
Unlike Linda, Natalie sees the good in being able to hearing different explanations.
–1 = 2 +b I moved the 2 to the other side of the equal by adding instead of
subtracting. Which then gave me the incorrect b so my final solution was wrong too.
I had y = ½ x + 1 and I should have had b = -3 and my final should have been y = ½
x – 3. So be careful with your signs they make a big difference.”
Other students talked about other benefits to justifying their answers. Taryn said: “I can explain it to others or I can do more problems like that one.”
When asked if they can explain a problem to a student who was absent from math class, students verified the benefit to justifying their solutions. Linda wasn’t confident in herself, but knew she could help if she did the justifying on her
own work. “If it was a complicated problem and I had it written down I could
explain it to them.”
Sam stated that justifying can help you as well as others. “If you want to help someone
else eventually then it is easier to show how you got the answer, if you just have an
On April 4th I had the students start a project of sending a problem and a solution to another classroom by Internet. They in turn solved the problem and returned a solution; sometimes a comment of the differences and similarities of how both students solved the problem was included. Some students explained each step while others assumed an easy concept can be skipped over without showing their work. One of the 8th graders felt her explanation of a solution in written words would never end. She had a great detailed explanation. Responses to the sent solutions were just as educational as the solution process itself. Students were able to see other ways of solving the same problem as well as read comments from other students on their feelings and understanding of the sent solution. Some of the 8th grade student from East Butler responses included noticed mistakes: The way the other student did it makes sense to me and I could follow every step, I
found my mistake and understand what I did wrong.
Some of the responses showed an understanding of how the problem may have been done a different way that may have been less time consuming: For the problem, Which point belongs to the graph of the solution set of the system? x<2
and y< 2x +3 a. (0, 5) b. (0, -5 ) c. (-5, 0) d. (-3, 5)
I did a lot more work than the other student, I did not think of plugging them in to
see which one worked. That would have saved me time. I am happy that I got the
same answer after all that work
.
Other responses showed that even though the solution was different the solution was written in a For the problem, 3a + b = 4 and a- 2b = 6 I chose to solve it a different way, by substitution. I could follow along exactly with
what the other student wrote. They did a good job of explaining what to do. They
ended up with the same solution that I got, but solved with a different method,
Some solution responses just showed another way to solve the problem that comes up the same For the problem, The sum of 3 and y multiplied by 3 is less than the sum of 5 and y
multiplied by 5.
After reading the other students solution, I found my solution to be very similar. I
moved the y to a different spot to begin with, but then we ended up with the same
solution in the end.
There were also responses that showed that the solutions sent may not have been completely understood. But were still able to teach mistakes that can be made easily: For the problem, 15 s cubed t over 3 s squared t cubed Sam wrote: First I started of by taking 15 divided by 3. Then I worked with the s
variables. The s cubed need to be divided by the s squared. When there are
exponents though you have to actually subtract them instead of dividing them. Then
on the t exponents it ended up being a negative exponent. The numerator was only t
and the denominator was t cubed so it made a negative exponent. To make it
positive you have to put one over t squared. Then you have to multiply the other
part of the equation into one over t squared. When you multiply them together, the
first part of the solution goes over the t squared because they are whole numbers.
So, you end up with five s over t square.
East Butler Response: I could follow what the student wrote for a solution, but I
forgot about the negative exponents. I was a little confused by the last part. I just
remembered the fact that the variable goes where the higher power was. I did end
up with the same answer though. She was more detailed than me in the response.
Abby’s first experience with a response was not as promising. The problem once again
dealt with money. Abby sent her solution but when the response came back the other student’s answer was not the same. Abby immediately felt she had done something wrong. However she was correct and the other student was wrong. They had forgotten to use decimal and dollar sign. This showed Abby she needs confidence and to evaluate her own and other students’ reasoning and solutions. Abby does have a struggle with self confidence. Taken from the student interview: When asked: What do you think about when your teacher asks questions during math
class?
Abby answered: “If a teacher asks a question I think I did something wrong.”
I have seen improvement in Abby’s solutions. She showed desire to do right and
explained each step with detail and care. For the problem, Sara has \$70 and a \$10 off coupon. If she goes to the Running Place and
buys track shoes that cost \$49.95, a stop watch that costs \$24.99, and two packages of
socks at \$4.99 each, does she have enough money?
Abby’ solution: I added \$49.95, \$24.99, and \$4.99 x 2 which totaled to \$84.92. Then
I took away \$10 for the coupon leaving \$74.92. She only had \$70 so she doesn’t have
Another of Abby’s problems included buying 2 patches at \$2.50 each, 3 magnets at \$2.00
each, a cap at \$8.00, 2 flags at \$3.25 each, and 2 quill pens at \$1.50 each at a museum gift
shop. How much would you have left out of \$28.90 if you bought all the items listed.
Abby’s solution: I took \$2.50 x 2 = \$5.00 in patches. Then \$2.00 x 3 = \$6.00 in
magnets. Then, \$8.00 for a cap. Then I took \$3.25 x 2 = \$6.50 in flags. And finally
\$1.50 x 2 = \$3.00 in quill pens. Now add them together. \$5.00 + \$6.00 + \$8.00 + \$6.50
+ \$3.00 = \$28.50. We spent \$28.50 in all and we only have \$28.90. I took away \$28.50
from \$28.90 and got .40. So we only have \$.40 left over.
My third assertion is that group discussion helps create a self motivated responsibility in the students. Communication between teacher and student helps student get started but student to student communication helps keep motivation strong. The pretests given before each unit were scary for the students. Not knowing how to do something on a test is hard for students to deal with. I did observe an increase in desire to learn the content from the pretest when it came time to learn that skill. I had a hard time too during the pretests. I wanted to help too much. Then during the lessons we all had self motivation to make sure the content was learned well enough for the post test. This teacher student communication was a great start to the motivation but the student to student communication enhanced the motivation and strengthened their abilities. After the first pre-test, post test cycle the students seemed to stretch their abilities to complete the following pre-tests. They weren’t afraid to try any skill they learned previously to complete at least portions of the pre-test. The following was taken from a journal entry during the week of April 11th : The pretest was a good experience; the students were able to use what they know to figure out something new. After each pre-test the student to student communication was exciting. They talked of the processes they tried and were excited to learn if there are easier techniques to solving those types Achievement scores showed improvements as well. As a class the Math Total score increased 9.8% (from a class average of 77.3% to 87.1%). The 5th grader herself increased her Math Total 27% (from 66% to 93%). One of the 8th graders increased their Math Total by 33%. (from 39% to 72%). 72% was the lowest percentile for all the students in the Math Total area There was an obvious increase in the Problem solving portion of the test as well. The class increased by 8% (from 80% to 88%). The same 8th grader that had a Math total increase also had a 22% increase in the problem solving portion (from 50% to 72%). The 5th grader, too, showed and increase in the problem solving portion, 10% (from 79% to 89%). Once again 72% was the lowest percentile for the Problem Solving portion of the test putting all students above Conclusions
All this data supports the literature that I read about the importance of communication. Students are more influenced by communication they experience (Smiley, 1958). Communication including writing enhances the learning of mathematics; it extends their thinking and understanding. In turn engagement in thinking leads to a better understanding and improves communication skills (Stein, Grover, & Henningsen, 1996). Through communication the students are able to clarify their own thinking and make sense of others’ explanations. With an increase in the understanding of mathematics connections between prior knowledge and concepts being taught are being made and are more meaningful. Even if students do not have classmates to communicate with writing solutions and sharing electronically may be a type of communication that will enhance their mathematical thinking. I feel a continuation and even increase in this written communication with grade level students would continue to increase math abilities as well as other subject areas. Although, across grade level solution presentations, did not seem profitable I do however feel it was a step to the written solutions by Internet that Searching for and finding a willing classroom and teacher may be the hardest part of incorporating this into a classroom. Today’s students enjoy using computers and socializing with other students. Using this interest can help increase math abilities. This data also shows that group work can generate more ideas for solutions. Students know that everyone has different ways of doing things. Working in a group can bring about more and even creative ideas for solutions. Exposure to these different ways can create an increase in Increases in Achievement scores may not be the result of only this research but other factors as well. Natural intellectual growth may have caused an increase as well. Other factors such as a change in school settings in the coming year may have placed an added incentive to do their best on the Achievement tests to prove their highest abilities. Due to a closure of the Class 1 schools my district will close the site at which I am at. All students will transfer to other districts. This will give classmates to the one 5th grader. The 8th graders will have a transition into a high school setting which will be very different from there past experiences. Implications
I will be in a larger school district next year in the 5th grade level. However, I will not be teaching math because of departmentalizing. As a result of this research I will try to encourage my new school district to permit me to continue the Internet communications. I can inform them of the improvement I observed in the students performances as well as their scores. I can ask to use the Internet communications in other subject areas to see if this written communication increases other subject areas as well. I will however try to keep up with my students’ math education, by having them write up solutions to one problem a week to inform me of the content they are learning. This can help me keep their content areas connected. I hope to excite the teachers in my new district enough to have them try the process of Internet communication in their classrooms. We can, as a team, keep collecting data to see if the technique works in larger school settings and other subject areas. My advice to other teachers with the problem of a lack of communication at a grade level is to find an equivalent classroom willing to communicate by Internet. Then share solutions slowly at first so it is not overwhelming. Many students will enjoy the process enough to do it on their own once we show them how. Class discussions should also be increased in classrooms. Discussions may be time consuming but are very beneficial to you and the students. You can see and hear their understanding and students can learn from each other when they do not quite I hope to follow up on the present seven students to see if their math scores increase or at least stay above average. Due to my closeness of the families this should be obtainable for me. References
Clarke, D., Waywood, A., & Stephens, M. (1993). Probing the structure of mathematical writing. Educational Studies in Mathematics, 25(3), 235- 250. Lappan, G. & Ferrini-Mundy, J. (1993). Knowing and doing mathematics: A new vision for middle grades students. The Elementary School Journal, 93(5), 625-641. Mathematical Association of America. (1991). A call for change: Recommendations for the mathematical preparation of teachers of mathematics. Washington, D.C.: Author. National Council of Teachers of Mathematics. (1989). Curriculum and evaluation standards for school mathematics. Reston, VA: Author. Shield, M. & Galbraith, P. (1998). The analysis of student expository writing in mathematics. Educational Studies in Mathematics, 36 (1), 29-52. Smiley, M. (1958). Do your classroom procedures really teach communication? The English Stein, M., Grover, B., & Hennigsen, M. (1996). Building student capacity for mathematical thinking and reasoning: An analysis of mathematical tasks used in reform classrooms. American Educational Research Journal, 33(2), 455- 488. Taylor, R. (1989). The potential of small-group mathematics instruction in grades four through six. The Elementary School Journal, 89(5), 633-642. Yackel, E., Cobb, P., & Wood, T. (1993). Developing a basis for mathematical communication within small groups. Journal for Research in Mathematics Education, 6, 33- 44. Yackel, E., Cobb, P., & Wood, T. (1993). The relationship of individual children’s mathematical conceptual development to small-group interaction. Journal for Research in Mathematics Education, 6, 45- 54. Student Interview Questions | 3,700 | 16,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | longest | en | 0.97247 |
https://www.mickybullock.com/blog/ | 1,618,104,447,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038060603.10/warc/CC-MAIN-20210411000036-20210411030036-00039.warc.gz | 974,963,183 | 8,488 | ## Does your child have a better chance of getting into the top ability set for Mathematics if they were born nearer the beginning of the academic year?
5 years ago I read Outliers by Malcolm Gladwell. The blurb states that “if we want to understand how some people thrive, we should spend more time looking around them – at such things as their family, their birthplace, or even their birth date.” He argues how the Relative Age Effect gives advantage to those born nearer the beginning of the year (academic year or sports season). Since reading the book I have wondered whether this phenomenon is in effect in the ability setting of Mathematics pupils in their lower secondary years, i.e. at the age of 12. And now we have the data to be able to answer this question…
## GeoGebra: Binomial Distribution with Normal and Poisson Approximation
##### Click the image to link to GeoGebraTube (Opens in a new window/tab). .
This applet is for visualising the Binomial Distribution, with control over n and p.
It also shows the Normal Approximation curve (and how this approximation breaks down for large or small p)
and it shows the Poisson Approximation curve (and how his approximation breaks down if there’s no positive skew)
You can show critical regions at either end by turning the bars red instead of green. The appropriate cumulative binomial probabilities are shown.
## GeoGebra: The Ultimate Projectiles Applet!
##### Click the image to link to GeoGebraTube (Opens in a new window/tab).
User-friendly applet designed with perfection and aesthetics in mind.
Play with the checkboxes and X points. Those with a grounding in projectile mechanics will find it is self-explanatory.
You can select: angle of elevation OR range OR an airborne target, for any speed of projection and strength of gravity.
The applet shows the possible trajectories under the given constraints.
The applet also animates balls projected at the selected angle of elevation.
## GeoGebra: Tangent Fields & Isoclines
##### Click the image to link to GeoGebraTube (Opens in a new window/tab)..
This applet displays tangent fields and coloured gradient fields for general solutions of explicitly defined 1st order ODE’s (i.e. dy/dx = …..)
It also displays the particular solution curve; you can set the boundary condition by dragging the blue point. [Euler’s numerical method is used with error correction]
The applet also shows isoclines where possible [dy/dx=f(x,y) must be polynomial in x and y]
## GeoGebra: Probability Venn Diagram with Proportional Regions
##### Click the animation to link to GeoGebraTube (Opens in a new window/tab).
In the latest version of this dynamic Venn diagram you have full control over the values of P(A), P(B) and P(AnB).
The orange area representing P(AnB) is dependent on the distance between the centres of the circles. But this distance is the solution of an equation that cannot be solved analytically, so the Newton-Raphson numerical method runs in the background with four iterations.
## The Swearing Graph!
September 2nd, 2010 1 comment
This is made from a quintic graph (a polynomial of degree 5). It’s possible to apply the “treatment” to any function, including reciprocal, hyperbolic and trigonometric functions. Read more…
## GeoGebra: Graphical Mortgage Repayment Calculator
##### Click the animation to link to GeoGebraTube (Opens in a new window/tab).
Calculate monthly repayments based on the amount you are borrowing and over how many years you wish to repay it.
or
Calculate how many years it will take to repay your mortgage based on the amount you are borrowing and your monthly budget.
## My trip to Alicante: “X Jornades d’Educació Matemàtica”
This post tells the story of my trip to Alicante, Spain in October 2012 to deliver the opening lecture at the 10th Conference on Mathematical Education, held at the University of Alicante and organised by the Societat d’Educacio Matematica Comunitat Valenciana (SEMCV) Al Khwarizmi. The talk was entitled, “The Value of Dynamic Geometry in Modern Education and Problem Solving in GeoGebra.”
## Using the mean to find the mode of a Binomial Distribution
The original motivation behind this investigation was an attempt to save my Statistics students a few precious seconds in their upcoming S1 module paper.
The mean or expectation of a Binomial Distribution is always very close to mode, (the value of X that has greatest probability). I want to know if you can use the mean to reliably predict the mode.
## Statistical Outliers Impossible in Small Samples
How many pieces of data are needed before it’s possible for one of them to be an Outlier?
## GeoGebra: The Binomial Expansion
##### Fully interactive: A geometric representation of the binomial expansion
Control parameters a,b,n in the function f(x)=(a+bx)n
Also control the number of terms in the binomial expansion.
The applet shows the correct function (black) alongside the binomial expansion of the function (green).
It also clearly shows the bounds (x-values) for which the expansion is valid. | 1,110 | 5,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-17 | latest | en | 0.913881 |
https://www.arxiv-vanity.com/papers/1506.06300/ | 1,685,654,425,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648209.30/warc/CC-MAIN-20230601211701-20230602001701-00047.warc.gz | 724,394,983 | 55,554 | # The co-rank of the fundamental group: the direct product, the first Betti number, and the topology of foliations
Irina Gelbukh CIC, Instituto Politécnico Nacional, 07738, Mexico City, MEXICO
###### Abstract.
We study , the co-rank of the fundamental group of a smooth closed connected manifold . We calculate this value for the direct product of manifolds. We characterize the set of all possible combinations of and the first Betti number by explicitly constructing manifolds with any possible combination of and in any given dimension. Finally, we apply our results to the topology of a Morse form foliations. In particular, we construct a manifolds and a Morse form on it for any possible combination of , , , and , where is the number of minimal components and is the maximum number of homologically independent compact leaves of .
###### Key words and phrases:
co-rank, inner rank, manifold, fundamental group, direct product, Morse form foliation
###### 2010 Mathematics Subject Classification:
14F35, 57N65, 57R30
## 1. Introduction and main results
Co-rank of a group , also known as inner rank, is the maximum rank of a free homomorphic image of . In a sense, co-rank is a notion dual to the rank; unlike rank, co-rank is algorithmically computable for finitely presented groups. This notion has been re-invented various times in different branches of mathematics, and its properties relevant for the corresponding particular task have been studied in different contexts. The notion of co-rank, called there inner rank, was apparently first mentioned in [20] in the context of solving equations in free groups. Co-rank of the free product of groups was calculated using geometric [15] and algebraic [22] methods.
Co-rank is extensively used in geometry, especially in geometry of manifolds, as
b′1(M)=corank(π1(M)),
the co-rank of the fundamental group of a manifold . For example, it has been repeatedly shown to coincide with the genus of a closed oriented surface: [21, 14, 8, 17].
In the theory of 3-manifolds, [27, 15], the cut number: the largest number of disjoint two-sided surfaces that do not separate , i.e., is connected. It is related to quantum invariants of and gives a lower bound on its Heegaard genus [11]. Around 2001, J. Stallings, A. Sikora, and T. Kerler discussed a conjecture that for a closed orientable 3-manifold it holds , where is the Betti number. This conjecture was later disproved by a number of counterexamples, such as [12, 17]. In this paper we, in particular, give a complete characterization of possible pairs , for any given .
In systolic geometry, every unfree 2-dimensional piecewise flat complex satisfies the bound [16], where is the optimal systolic ratio.
In the theory of foliations, for the foliation defined on by a closed 1-form with the singular set , it was shown that if and with contained in a finite union of submanifolds of , then has no exceptional leaves [18]. Foliations have numerous applications in physics, such as general relativity [4], superstring theory [3, 2], etc.
The notion of co-rank of , the notation , and the term the first non-commutative Betti number were first introduced in [1] to study Morse form foliations, i.e., foliations defined by a closed 1-form that is locally the differential of a Morse function on a smooth closed manifold .
A Morse form foliation can have compact leaves, compactifiable leaves and minimal components [9]. In [1], it was proved that if , then each minimal component of is uniquely ergodic. On the other hand, if , then there exists a Morse form on with a minimal component that is not uniquely ergodic. If the form’s rank
rkω>b′1(M),
then the foliation has a minimal component [19]; here , where is the integration map.
Though co-rank is known to be algorithmically computable for finitely presented groups [23, 26], we are not aware of any simple method of finding for a given manifold. This value is, however, bounded by the isotropy index , which is the maximum rank of a subgroup in with trivial cup-product [24]. Namely, for a smooth closed connected manifold it holds [8, 5]
b′1(M)≤h(M),
while for there are simple estimates via and [25].
For the connected sum of -manifolds, , except for non-orientable surfaces, it holds:
b1(M1#M2)=b1(M1)+b1(M2),b′1(M1#M2)=b′1(M1)+b′1(M2), (1.1)
which for follows from the Mayer–Vietoris sequence and by [22, Proposition 6.4], respectively. Also, for the direct product the Künneth theorem gives:
b1(M1×M2)=b1(M1)+b1(M2).
In this paper, we show that the fourth combination is very different:
b′1(M1×M2)=max{b′1(M1),b′1(M2)} (1.2)
(Theorem 3.1), which completes the missing piece to allow calculating for manifolds that can be represented as connected sums and direct products of simpler manifolds.
We give a complete characterization of the set of all possible combinations of and for a given . Namely, for , there exists a connected smooth closed -manifold with and iff
n≥3: b′=b=0 or 1≤b′≤b; (1.3) n=2: 0≤b, b′=[b+12]; n=1: b′=b=1; n=0: b′=b=0;
the manifold can be chosen orientable, except for odd when (Theorem 4.1). Using (1.1)–(1.2), we explicitly construct such a manifold (Construction 4.3).
We apply the obtained results to estimation of the number of minimal components and the maximum number of homologically independent compact leaves of the foliation of a Morse form on . The smaller or , the more information we have about . For example,
m(ω)+c(ω)≤b′1(M) \@@cite[cite]{[\@@bibref{}{Gelb10}{}{}]},2m(ω)+c(ω)≤b1(M) % \@@cite[cite]{[\@@bibref{}{Gelb09}{}{}]}; (1.4)
In particular, if , or, which is the same, , then all leaves of are compact and homologically trivial and is exact.
Theorem 4.1, which states that all combinations allowed by (1.3) are possible, implies that the two inequalities are independent. However, in special cases knowing the values of and , for example, calculated using (1.1)–(1.2), allows choosing one of the two inequalities as stronger. For instance, if , then the first one is stronger.
Finally, we generalize Theorem 4.1 to a characterization of the set of possible combinations of , , , and (Theorem 5.2). Namely, we use Construction 4.3—the constructive proof of Theorem 4.1—to show that (1.3) and (1.4) are the only restrictions on these four values, except for if .
The paper is organized as follows. In Section 2, we give the definitions of the Betti number and the non-commutative Betti number for groups and manifolds. In Section 3, we calculate . In Section 4, we describe the set of possible combinations of and for a given and, using the results of Section 3, explicitly construct a manifold with any given valid combination of and . Finally, in Section 5 we use the manifold constructed in Section 4 to describe the set of possible combinations of the number of minimal components and the maximum number of homologically independent compact leaves of a Morse form foliation.
## 2. Definitions
For a finitely generated abelian group , where is finite, its torsion-free rank, Prüfer rank, or first Betti number, is defined as . The notion of first Betti number can be extended to any finitely generated group by , where is the abelianization, or the first homology group, of the group , and is the torsion subgroup. In other words:
###### Definition 2.1.
The first Betti number of a finitely generated group is the maximum rank of a free abelian quotient group of , i.e., the maximum rank of a free abelian group such that there exists an epimorphism .
Consider a connected smooth closed manifold . The first Betti number of is the torsion-free rank of its first homology group , i.e., of the first homology group of its fundamental group :
b1(M)=b1(π1(M)).
A non-commutative analog of the Betti number can be defined as follows.
###### Definition 2.2.
The co-rank [17], inner rank [15] or [22], or first non-commutative Betti number [1] of a finitely generated group is the maximum rank of a free quotient group of , i.e., the maximum rank of a free group such that there exists an epimorphism .
The notion of co-rank is in a way dual to that of rank, which is the minimum rank of a free group allowing an epimorphism onto . Unlike rank, co-rank is algorithmically computable for finitely presented groups [23, 26].
The first non-commutative Betti number [1] of a connected smooth closed manifold is defined as the co-rank, or inner rank, of its fundamental group:
b′1(M)=b′1(π1(M)).
Note that a similar definition for higher is pointless since they are abelian.
## 3. Co-rank of the fundamental group of the direct product
The Betti number is linear in both connected sum and direct product. While the non-commutative Betti number is linear in connected sum, its behavior with respect to direct product is very different:
###### Theorem 3.1.
Let be connected smooth closed manifolds. Then
b′1(M1×M2)=max{b′1(M1),b′1(M2)}.
We will divide the proof into a couple of lemmas.
###### Lemma 3.2.
Let be groups. Then any epimorphism
φ:G1×G2↠F
onto a free group factors through a projection.
###### Proof.
For the fact is trivial. Par abus de langage, denote and , subgroups of . Suppose both . Since and and denoting , we have a free group such that and by the condition .
Let and , . Since , we have as both free and abelian, so for some , and similarly for some . Then as a two-generated free group with a non-trivial relation , so for some . We obtain ; in particular, .
Thus is abelian, and similarly . Since , we obtain that the non-trivial is both free and abelian, thus . ∎
###### Remark 3.3.
In fact this holds for any (infinite) quantity of factors: any epimorphism onto a free group factors through the projection onto one of .
###### Lemma 3.4.
Let be finitely generated groups. Then for the co-rank of the direct product,
b′1(G1×G2)=max{b′1(G1),b′1(G2)}.
###### Proof.
Denote and .
Let us show that . Without loss of generality assume . Consider an epimorphism onto a free group, . Then such that and is an epimorphism, so .
Let us now show that . Consider an epimorphism onto a free group, . If , then is an epimorphism and thus , and similarly if .
Otherwise , so by Lemma 3.2. ∎
###### of Theorem 3.1.
For smooth connected manifolds , we have
π1(M1×M2)=π1(M1)×π1(M2),
and the desired fact is given by Lemma 3.4. ∎
###### Example 3.5.
For a torus , we have . Since is free abelian, this also follows from Definition 2.2.
For the Kodaira-Thurston manifold , we have since for the Heisenberg nil manifold, . This also follows from the fact that itself is a nil manifold.
In Section 4, we will use Theorem 3.1 to explicitly construct a manifold with arbitrary given and .
## 4. Relation between b′1(M) and b1(M)
As an application of Theorem 3.1, in this section we show that there are no non-obvious relations between and , and explicitly construct a manifold with any given valid pair of and .
For a sphere and for low-dimensional manifolds, such as closed orientable surface and closed non-orientable surface , , the values of and are obvious or well known:
point:b′1(∗)=0,b1(∗)=0;circle:b′1(S1)=1,b1(S1)=1;sphere, n≥2:b′1(Sn)=0,b1(Sn)=0;orientable surface:b′1(M2g)=g,b1(M2g)=2g;non-orientable surface:b′1(N2h)=[h2],b1(N2h)=h−1. (4.1)
The value of was calculated in [8] and [17, Lemma 2.1]. It can be also obtained as the cut-number [15, Theorem 2.1], which for a surface is the number of handles, since each non-separating two-sided circle identifies the edges of two holes. In particular, is a sphere with inverted handles plus a Möbius strip for odd .
In general,
0≤b′1(M)≤b1(M) (4.2)
with
b′1(M)=0 iff\/ b1(M)=0, (4.3)
and thus
b1(M)=1 implies\/ b′1(M)=1. (4.4)
Indeed, since for a free group it holds and a group epimorphism induces an epimorphism , it holds
0≤b′1(G)≤b1(G)
and since is both free and free abelian, iff .
There are no relations between and other than (4.1)–(4.3):
###### Theorem 4.1.
Let . There exists a connected smooth closed -manifold with and iff
n≥3: b′=b=0 or 1≤b′≤b; (4.5) n=2: n=1: b′=b=1; n=0: b′=b=0;
the manifold can be chosen orientable iff or is even.
###### Proof.
For and for the facts are given in (4.1), so let and .
For , every finitely presented group is the fundamental group of a connected smooth closed orientable -manifold , while by [6, Theorem 3] there exists such a group with and , which proves the result for .
Finally, let . For any given , Harvey [12, Theorem 3.1] constructed a smooth closed orientable hyperbolic 3-manifold with the largest possible gap between and :
b′1(Hb)=1, b1(Hb)=b. (4.6)
For , choose such that . By (1.1), for
M=b′#i=1Hki (4.7)
we have . ∎
Using Theorem 3.1, we can generalize any specific example with given and to higher dimensions, as well as to increase the gap between and :
###### Lemma 4.2.
Let . Then for it holds
k=1: b′1(Mn+k)=b′1(Mn), b1(Mn+k)=b1(Mn)+1; k≥2: b′1(Mn+k)=b′1(Mn), b1(Mn+k)=b1(Mn).
This allows us to explicitly construct a manifold with given and of any given dimension, thus giving a simple constructive proof of Theorem 4.1 for :
###### Construction 4.3.
For any given such that or , and , the following connected smooth closed oriented -manifold has
b′1(Hnb′,b)=b′, b1(Hnb′,b)=b.
For , consider
Hn1,1=S1×Sn−1 (4.8)
and for , generalize (4.6) to higher dimensions using Lemma 4.2:
Hn1,b=⎧⎨⎩Hb$for$n=3,Hb−1×S1$for$n=4,Hb×Sn−3$for$n≥5. (4.9)
Finally, as in (4.7), choose such that
b′∑i=1ki=b (4.10)
and take
Hnb′,b=b′#i=1Hn1,ki. (4.11)
By Theorem 4.1, in (4.2) both the lower bound (except for ) and the upper bound (except for surfaces other than , , and the Klein bottle) are exact for any given . Both conditions (4.3) are impossible for and both conditions (4.4) are impossible for .
In particular, the lower bound in (4.2) is achieved on , while (4.8) and (4.9) provide the lower bound in the inequality in (4.5). The upper bound for is provided by (4.11) with :
Hnb,b=b#i=1(S1×Sn−1).
In general, iff some (and thus any) epimorphism
π1(M)↠H1(M)/T(H1(M))
factors through a free group; is the torsion subgroup:
###### Proposition 4.4.
For any group , the following conditions are equivalent:
1. ,
2. there exists an epimorphism
h:G↠Zb1(G)=H1(G)/T
that factors through a free group; is the torsion subgroup,
3. any such epimorphism factors through a free group.
###### Proof.
(i)(ii): Let be the epimorphism from the definition of and be the natural epimorphism; then has the desired properties.
(ii)(i): Let be a factorization of through a free group . Then . By [6, Theorem 3], , so we obtain .
(ii)(iii): For any epimorphisms there exists an automorphism such that .
(iii)(ii): The composition of natural epimorphisms
G↠H1(G)=G/[G,G]↠H1(G)/T=Zb1(G)
is an epimorphism. ∎
## 5. Application to foliation topology
The gap between and plays a role in foliation topology.
### 5.1. Useful facts about Morse form foliations
Consider a connected closed oriented -manifold with a Morse form , i.e., a closed 1-form with Morse singularities—locally the differential of a Morse function. The set of its singularities is finite. This form defines a foliation on .
Its leaves can be classified into compact, compactifiable ( is compact), and non-compactifiable. The set covered by all non-compactifiable leaves is open and has a finite number of connected components, called minimal components [1]. Each non-compactifiable leaf is dense in its minimal component [13]. A foliation is called minimal if all its leaves are non-compactifiable, i.e., the whole is one minimal component.
Any compact leaf has a cylindrical neighborhood consisting of leaves that are diffeomorphic and homotopically equivalent to it [10]. Denote by the group generated by the homology classes of all compact leaves of . Since is closed and oriented, is finitely generated and free; therefore so is . By [7, Theorem 4], in there exists a basis consisting of homology classes of leaves, i.e., has exactly homologically independent compact leaves.
###### Lemma 5.1.
Let be Morse forms defined on smooth closed oriented manifolds , respectively. Then on there exists a Morse form with and .
###### Proof.
Consider a form constructed as shown in Figure 1. It coincides with outside a small area where are glued together. We assume that each was locally distorted either in a minimal component or in a cylindrical neighborhood covered by homologous compact leaves. In the former case, since nearby leaves are dense on either side of the affected leaf, the distortion does not change the number of minimal components. In the latter case, even though the distortion “destroys” one compact leaf, the nearby leaves contribute the same value to . In either case, the new compact leaves introduced in the process are homologically trivial.
Since the two sides are separated by compact leaves, each minimal component of lies either in or in , and thus . Similarly, homologically non-trivial leaves of are homologous to either leaves of or leaves of ; in particular, and thus . ∎
### 5.2. Relation between b′1(M), b1(M), m(ω), and c(ω)
Let be a Morse form on a smooth closed orientable manifold , , defining a foliation with exactly homologically independent compact leaves and minimal components. The following inequalities have been proved independently:
m(ω)+c(ω) ≤b′1(M) \@@cite[cite]{[\@@bibref{}{Gelb% 10}{}{}]}, (5.1) 2m(ω)+c(ω) ≤b1(M) \@@cite[cite]{[\@@bibref{}{Gelb09}{}{}]}. (5.2)
For some manifolds, (1.1) and Theorem 3.1 allow direct calculation of . This may allow one to choose between (5.1) and (5.2). Namely, denoting , , unless we have:
1. If , then (5.1) is stronger;
2. If , then they are independent;
3. If , then (5.2) is stronger.
In particular, (5.1) is always stronger for . However, for any , by Theorem 4.1 all three cases are possible; in particular, there exist manifolds for which the two estimates are independent. By (4.2), the case is impossible.
More specifically, in the case (ii), if seen as conditions on under given and vice versa,
• (5.1) is stronger when or when ;
• (5.2) is stronger when or when ,
and they are equivalent in case of equalities.
We can generalize Theorem 4.1 to observe that there are no relations between , , , and other than those given by (5.1), (5.2), (4.2), (4.3), and, for an orientable surface, (4.1):
###### Theorem 5.2.
Let . There exists a smooth closed connected oriented -manifold with and , and a Morse form foliation on it with minimal components and exactly homologically independent compact leaves, iff
n=2: (4.1) for M2g: 0≤b=2b′, (5.1): 0≤m+c≤b′, n≥3: (4.2), (4.3): b′=b=0 or 1≤b′≤b, (5.3) (5.1): 0≤m+c≤b′, (5.4) (5.2): 0≤2m+c≤b. (5.5)
Apart from a trivial foliation on , the proof is given by the following constructions.
###### Lemma 5.3.
On , , there exists Morse form foliation with , .
###### Proof.
The corresponding foliations for and are shown in Figures 2 and 3, respectively. ∎
###### Lemma 5.4.
For , on there exists a Morse form foliation with and .
###### Proof.
As shown in Figure 4, construct as the connected sum of
• tori with an irrational winding;
• tori with a compact non-singular foliation;
• tori with a foliation shown in Figure 2;
with the forms glued together as shown in Figure 1. Lemma 5.1 completes the proof. ∎
This easily generalizes to higher dimensions:
###### Lemma 5.5.
For and satisfying (5.3)–(5.5), there exists an -manifold from Construction 4.3 and a Morse form on it with and .
###### Proof.
The construction is very similar to that of Lemma 5.4. Observe that in (4.10), we can choose at most summands .
If (which by (5.4) and (5.5) is always the case if ), then in (4.10) choose summands , which by (5.4) leaves at least summands :
Hnb′,b=(m#i=1Hn1,ki)#(c#i=1Hn1,1)#(#Hn1,1).
Otherwise, in (4.10) choose summands , which by (5.5) leaves
b′−(b−b′)=b−2(b−b′)≥(2m+c)−2(b−b′)=2(m−(b−b′))+c
summands :
Hnb′,b=(b−b′#i=1Hn1,2)#(m−(b−b′)#i=1(Hn1,1#Hn1,1))m manifolds#(c#i=1Hn1,1)#(#Hn1,1).
As in Lemma 5.4, the foliations on the manifolds are chosen as follows:
• On manifolds , , and there exists a Morse form with a minimal foliation [1, Theorem 1]; in particular, and ;
• On manifolds , choose a compact non-singular foliation along with leaf ;
• On the rest of , choose a foliation shown in Figure 3;
Lemma 5.1 completes the proof. ∎ | 5,944 | 20,413 | {"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} | 2.75 | 3 | CC-MAIN-2023-23 | latest | en | 0.911855 |
http://heli-air.net/2016/04/14/an-exercise-in-high-altitude-operation/ | 1,591,276,471,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347441088.63/warc/CC-MAIN-20200604125947-20200604155947-00542.warc.gz | 53,137,572 | 12,936 | # An Exercise in High-Altitude Operation
Fixed-wing aircraft operate more economically at high altitude than at low. Aircraft drag is reduced and engine (gas-turbine) efficiency is improved, leading to increases in cruising speed and specific range (distance per unit of fuel consumed). With gas-turbine-powered helicopters, the incentive to realize similar improvements is strong: there are, however, basic differences to be taken into account. On a fixed-wing aircraft, the wing area is determined principally by the stalling condition at ground level; increasing the cruise altitude improves the match between area requirements at stall and cruise. On a helicopter, the blade area is fixed by a cruise speed requirement, while low-speed flight determines the installed power needed. The helicopter rotor is unable to sustain the specified cruising speed at altitudes above the density design altitude, the limitation being that of retreating blade stall. The calculations now to be described are of a purely hypothetical nature, intended to illustrate the kind of changes that could in principle convey a high-altitude flight potential. The altitude chosen for the exercise is 3000 m (10 000 ft), this being near the limit for zero pressurization. We are indebted to R. V. Smith for the work involved.
Imperial units are used as in the previous section. The data case is that of a typical light helicopter, of all-up weight 10 000 lb and having good clean aerodynamic design, though traditional in the sense of featuring neither especially low-drag nor advanced blade design. Power requirements are calculated by the simple methods outlined earlier in the present chapter. Engine fuel flow is related to power output in a manner typical of modern gas-turbine engines. Specific range (nautical miles per pound fuel) is calculated thus:
Specific range(nm/lb) = forward speed((knots)/fuel)flow(lb/hr)
A flight envelope of the kind described in Section 7.7 is assumed: this is primarily a retreating-blade limitation in which the value of W/d (d being the relative density at altitude) decreases from 14000lb at 80 knots to 8000lb at 180 knots.
The results are presented graphically in Figures 7.9A-D.
Specific range is plotted as a function of flight speed for sea level (SL), 5000 ft and 10 000 ft altitude. Intersecting these curves are (a) the flight envelope limit, (b) the locus of best-range speeds and (c) the power limitation curve. We see that in case A, which is for the data helicopter, the flight envelope restricts the maximum specific range to 0.219 nm/lb, this
Table 7.3 Comparison of configuration performance
Data Best range speed (knots) Altitude (ft) Specific range (nm/lb) Weight penalty (lb) Max. range (nm) (1) (2) A 114 5000 0.219 357 357 B 120 4200 0.231 0 C 174 10 000 0.267 652 274 458 D 174 10 000 0.293 225 433 503
occurring at 5000 ft and low speed (only 114 knots). So far as available power is concerned, it would be possible to realize the best-range speeds up to 10000 ft and beyond.
Case B examines the effect of a substantial reduction in parasite drag. Using a less ambitious target than that envisaged in Section 7.9, a parasite drag two-thirds that of the data aircraft is assumed. At best-range speed a large increase in specific range at all altitudes is possible but, as before, the restriction imposed by the flight envelope is severe, allowing an increase to only
0. 231 nm/lb, again at approximately 5000 ft and low speed (120 knots). It is clear that the full benefit of drag reduction cannot be realized without a considerable increase in rotor thrust capability. A comparison of cruising speeds emphasizes the deficiency: without the flight- envelope limitation the best-range speeds would be usable, that is at all heights a little above 150 knots for the data aircraft and 20 knots higher for the low-drag version.
The increase in thrust capability required by the low-drag aircraft to raise the flight envelope limit to the level of best-range speed at 10000 ft is approximately 70%. Case C shows the performance of the low-drag aircraft supposing the increase to be obtained from the same percentage increase in blade area. Penalties of weight increase and profile power increase are allowed for, assumed to be in proportion to the area change. The best-range speed is now attainable up to over 9000 ft, while at 10 000 ft the specific range is virtually the same as at best – range speed, that is 0.267 nm/lb at 170 knots; this represents a 22% increase in specific range over the data aircraft, attained at 60 knots higher cruising speed.
For a final comparison, case D shows the effect of obtaining the required thrust increase by combination of a much smaller increase in blade area (24.5%) with conversion to an advanced rotor design, using an optimum distribution of cambered blade sections and the Westland advanced tip. The penalties in weight and profile power are thereby reduced considerably. The result is a further increase in specific range, to 0.293 nm/lb or 34% above that of the data aircraft, attained at the same cruising speed as in case C.
The changes are seen to further advantage by calculating also the maximum range achievable. This has been done in alternative ways, assuming that the weightpenalty reduces (1) the fuel load or (2) the payload. On the first supposition, the weight penalty of case C results in a range reduction but with case D the gain more than compensates for the smaller weight penalty.
The characteristics of the various configurations are summarized in Table 7.3. | 1,233 | 5,579 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.927768 |
https://stayinfiji.com/how-many-hours-in-6-months/ | 1,656,343,513,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103334753.21/warc/CC-MAIN-20220627134424-20220627164424-00288.warc.gz | 588,902,298 | 4,408 | To convert a month measurement to an hour measurement, multiply the moment by the conversion ratio.
since one month is equal to 730.485 hours, you have the right to use this basic formula to convert:
You are watching: How many hours in 6 months
Months and hours are both units used to measure up time. Keep reading to learn an ext about every unit the measure.
## Months
One month is a unit of time same to 1/12 the a year. The month is a unit that time used on a calendar, and also ranges in length from 28 come 31 days.
Months deserve to be abbreviated as mo; because that example, 1 month can be written as 1 mo.
## Hours
The hour is a period of time same to 1/24 that a day or 60 minutes.
The hour is an SI welcomed unit for time because that use with the metric system. Hours deserve to be abbreviated together hr; because that example, 1 hour have the right to be created as 1 hr.
## Month come Hour conversion Table
Month measurements converted to hours month hrs
1 mo 730.49 hr
2 mo 1,461 hr
3 mo 2,191 hr
4 mo 2,922 hr
5 mo 3,652 hr
6 mo 4,383 hr
7 mo 5,113 hr
8 mo 5,844 hr
9 mo 6,574 hr
10 mo 7,305 hr
11 mo 8,035 hr
12 mo 8,766 hr
13 mo 9,496 hr
14 mo 10,227 hr
15 mo 10,957 hr
16 mo 11,688 hr
17 mo 12,418 hr
18 mo 13,149 hr
19 mo 13,879 hr
20 mo 14,610 hr
21 mo 15,340 hr
22 mo 16,071 hr
23 mo 16,801 hr
24 mo 17,532 hr
25 mo 18,262 hr
26 mo 18,993 hr
27 mo 19,723 hr
28 mo 20,454 hr
29 mo 21,184 hr
30 mo 21,915 hr
31 mo 22,645 hr
32 mo 23,376 hr
33 mo 24,106 hr
34 mo 24,836 hr
35 mo 25,567 hr
36 mo 26,297 hr
37 mo 27,028 hr
38 mo 27,758 hr
39 mo 28,489 hr
40 mo 29,219 hr
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Lecture 5
# Lecture 5
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# Lecture 5: Price discrimination and nonlinear pricing Tom Holden io.tholden.
org
Last week:
◦ We showed how competing firms following “trigger” strategies may sustain monopoly profits in an industry.
This week:
◦ We see if there are any times in which a monopolist can obtain profits higher than the monopoly level. ◦ Only possible if the monopolist can charge different consumers different prices.
◦ A break from oligopoly models.
What is price discrimination?
Types of price discrimination.
◦ Welfare effects.
Tying and bundling. Durable goods.
◦ Price discrimination in time.
◦ Church and Ware Chapter 5 (handout) ◦ The maths is covered in the OZ refs to be given.
Price discrimination means selling the same good at different prices. E.g.:
◦ Buy two get one free offers. ◦ Student discounts. ◦ Off-peak rail fares.
Slightly more generally:
◦ Price discrimination is selling two similar goods at different ratios to marginal costs. ◦ So, e.g. the fact that posting a letter to Scotland costs the same as posting one to someone in Guildford is discriminatory.
Need all of the following conditions:
◦ The good cannot be resold.
Else e.g. a student would buy a load of discounted copies of MS Office then sell them on EBay at full price. Services (e.g. haircuts) usually pass this test.
◦ The firm has some market power (so we can have 𝑃 > 𝑀𝐶). ◦ Consumers are not all identical. ◦ Firms can somehow charge different prices to the different types of consumers.
Even under monopoly, if there are consumers who value a good highly (above the price they pay), consumer surplus will be high.
Price discrimination enables firms to “steal” that CS and turn it into profits.
◦ While not putting off consumers who value the good less.
Usually price discrimination increases profits, though there are two notable exceptions to this:
◦ When selling durable goods.
Price discrimination in time.
◦ When firms are discriminating under oligopoly.
To be looked at in a future lecture. 𝑝
Consumer surplus
MC
Producer surplus (profits) 𝑝 𝑄 MR 𝑄
First degree price discrimination:
◦ The firm sells each unit at a different, take-it-or-leaveit price. ◦ Goods sold at consumer’s reservation prices. ◦ Example: DeBeers’ sales of uncut diamonds:
“The diamonds are sold on a take-it-or-leave-it basis. A sightholder is given a small box of uncut diamonds priced between \$1 and \$25 million. De Beers set the price - there is no haggling and no re-selling of diamonds in uncut form. It is rare for sightholders to refuse a diamond package offered to them, for fear of not being invited back. And those who dare to purchase diamonds from other sources than De Beers will have their sightholder privilege revoked.” http://www.neatorama.com/2008/12/01/10-facts-aboutdiamonds-you-should-know/
First degree price discrimination:
◦ Maximises social welfare
Though it all goes to the firm…
◦ Very hard for most firms to do, since they do not know each consumer’s reservation price.
◦ Not “incentive compatible”.
High-value customers have an incentive to pretend to be low-value ones.
Suppose that all consumers have the same demand function 𝑝 𝑄 .
And suppose the monopolist chose a pricing structure under which to buy 𝑞 > 0 units a consumer had to pay a fixed fee of 𝑓 plus an additional price 𝑝 per unit.
◦ I.e. total payment for quantity 𝑞, 𝑇 𝑞 = 𝑓 + 𝑝𝑞. ◦ Examples:
Mobile phone contracts. Gym membership. Etc.
𝑝
What are the optimal choices of 𝑓 and 𝑝 for the firm?
MC
CS PS 𝑝
𝑄 𝑄
Is this price discrimination?
◦ Everyone has the same preferences… ◦ Everyone pays the same price…
Re-sale?
Variations in demand across consumers?
Second degree price discrimination:
◦ Firms cannot see consumers characteristics. ◦ Try to get consumers to self-select into the different price bands. ◦ Works by e.g.:
Offering menus of tariffs (such as different telephone contracts). Offering nonlinear tariffs:
As a function of quantity (such as 3 for 2 offers). As a function of quality (such as hardback/paperback books or deliberately “damaged” computer processors/graphics cards).
Suppose there are two types of consumers, those with low demand (𝐿) and those with high demand (𝐻).
◦ High types demand more at any price. ◦ For phones, you might think of the low demand types as being households, and the high demand types as being businesses.
The monopolist would like to offer the low type 𝑇𝐿 𝑞 = CS𝐿 + 𝑐𝑞 where 𝑐 is MC, and CS𝐿 is the consumer surplus the low types would get from perfect competition.
Likewise they would like to offer 𝑇𝐻 𝑞 = CS𝐻 + 𝑐𝑞 to the high types.
But firms cannot tell low from high types, and, given this, high types would always pretend to be low types to pay the lower fee.
Can the firm do better than offering 𝑇 𝑞 = CS𝐿 + 𝑐𝑞 to everyone? Yes. Suppose they increase price by 𝛿. 𝑝
◦ To persuade low types to buy, have to reduce 𝑓 by 𝐴 + 𝐵. But they get 𝐴 back in profits. ◦ From high types they lose 𝐴 + 𝐵 due to the lower fee, but they gain 𝐴 + 𝐵 + 𝐶 in profits. ◦ Thus the total gain is 𝐶 − 𝐵. But for small 𝛿, 𝐵 is small compared to 𝐶. 𝐶
𝑐 + 𝛿 𝑐 𝑝𝐻 𝑄 𝑝𝐿 𝑄 𝐴 𝐵 𝑄
Suppose there are two types of consumers of equal number (normalized to one for both), with demand curves: Constant marginal cost of 𝑐 < 𝑎. Tariff of 𝑇 𝑞 = 𝑓 + 𝑝𝑞 offered. If the firm sells in both markets (requires 𝑝 < 𝑎), profits are: 2𝑓 + 𝑝 − 𝑐 1 + 𝑎 − 2𝑝
◦ Exercise: draw a diagram to show this means 𝑓 = 𝑎
−𝑝 2 . 2
◦ 𝑞𝐻 = max 0,1 − 𝑝 ◦ 𝑞𝐿 = max 0, 𝑎 − 𝑝 where 𝑎 ≤ 1.
◦ Maximised when 𝑓 is as large as possible, i.e. when the low type makes zero surplus. ◦ Exercise: show that with this level of 𝑓, firms will set 𝑝 = 𝑐 + . 2 ◦ Hence: the bigger the difference between types, the higher is 𝑝 and the lower is 𝑓. ◦ Exercise: Does the firm always want to sell in both markets?
Hint: suppose 𝑐 = 0 and compare the cases when 𝑎 = 1 2 and 𝑎 = 3 4.
1−𝑎
Suppose that rather than offering a two-part tariff, the firm offers a choice between two (quantity, total-payment) bundles. Can trivially implement the solution to the optimal two-part tariff using these bundles. Can they do better? Yes.
◦ Exercise: Show that under the two part tariff considered in our linear example, the high type strictly prefers their bundle to the low type’s one.
Firm profits maximised when high types are just indifferent between the two tariffs, so optimal bundles have higher total costs for the high type.
Not necessary for the exam.
◦ But: always optimal to have high type consuming the efficient quantity. ◦ If you’re interested, Church and Ware p.166-176 has one proof of this.
Suppose the firm offers a choice of 𝑞∗ , 𝐵 + 𝑐𝑞∗ and 𝑞𝐻 , 𝐵 + 𝐶 + 𝐷 + 𝐸 + 𝑐𝑞𝐻 . Value gain for low types:
◦ From taking 1st bundle is 𝐵 + 𝑐𝑞 ∗ − 𝐵 + 𝑐𝑞 ∗ = 0. ◦ From taking 2nd bundle is 𝐵 + 𝐷 − 𝐹 + 𝑐𝑞𝐻 − 𝐵 + 𝐶 + 𝐷 + 𝐸 + 𝑐𝑞𝐻 = −𝐶 − 𝐸 − 𝐹 < 0. ◦ So low types take the 1st bundle and get zero surplus. ◦ From taking 1st bundle is 𝐴 + 𝐵 + 𝑐𝑞 ∗ − 𝐵 + 𝑐𝑞 ∗ = 𝐴. ◦ From taking 2nd bundle is 𝐴 + 𝐵 + 𝐶 + 𝐷 + 𝐸 + 𝑐𝑞𝐻 − 𝐵 + 𝐶 + 𝐷 + 𝐸 + 𝑐𝑞𝐻 = 𝐴. ◦ So high types are indifferent, thus are prepared to take the 2nd bundle.
𝑝
Value gain for high types: 𝐴
Profits: 2𝐵 + 𝐶 + 𝐷 + 𝐸 = 𝐴 + 𝐵 + 𝐶 + 𝐷 + 𝐸 + 𝐵 − 𝐴 .
◦ Optimal 𝑞 ∗ satisfies
ⅆ𝐵 ⅆ𝑞 𝐵 𝐶 𝐷
=
ⅆ𝐴 ⅆ𝑞
. 𝐸 𝐹 𝑐 𝑝𝐻
𝑄 𝑝𝐿 𝑄 𝑞
∗ 𝑞𝐿 𝑞𝐻 𝑄
Third degree price discrimination:
OAP discounts for museums. Student discounts on software. Academic discounts for conferences. Magazine price varying by country.
◦ Firms base price on consumers’ observable characteristics. E.g.:
AEA membership price varying by income.
The New Statesman is €5.80 throughout the EU, except in Greece, where it is €5.40.
◦ Most common form of price discrimination. ◦ The firm sets the monopoly price in each market (i.e. MR=MC).
Market is equally split between type 1 and type 2 consumers:
Firm has costs 𝐶 𝑄 to produce 𝑄 = 𝑞1 + 𝑞2 . Profits: 𝑝1 𝑞1 𝑞1 + 𝑝2 𝑞2 𝑞2 − 𝐶 𝑞1 + 𝑞2 ′ FOC 𝑞1 : 0 = 𝑝1 𝑞1 𝑞1 + 𝑝1 𝑞1 − 𝐶 ′ 𝑞1 + 𝑞2 I.e.: 𝑀𝑅1 = 𝑀𝐶. Likewise: 𝑀𝑅2 = 𝑀𝐶. Exercise: Show that this condition is still valid when there are 𝑛 type 1 consumers and 𝑚 type 2s.
◦ Type 1 consumers have demand: 𝑝1 𝑞1 ◦ Type 2 consumers have demand: 𝑝2 𝑞2
′ Recall the FOC for 𝑞1 says: 𝑝1 𝑞1 𝑞1 + 𝑝1 𝑞1 = 𝐶 ′ 𝑞1 + 𝑞2 .
◦ So:
′ 𝑝1 𝑞1 𝑞1 𝑝1 𝑞1
+1= 𝐶
′ 𝑞1 +𝑞2 𝑝1 𝑞1
Note:
demand. Remember:
′ 𝑝1 𝑞1 𝑞1 𝑝1 𝑞1
= 𝑞
1 ⅆ𝑝1 𝑝1 ⅆ𝑞1
= 𝑝1 ⅆ𝑞1 = 𝑞
1 ⅆ𝑝1
1
1 𝜀
where 𝜀 is the price elasticity of
◦ 𝜀 will almost always be negative. −𝜀 large means elastic demand. ◦ In general 𝜀 is a function of the price/quantity. 𝐶
′ 𝑞1 +𝑞2 1+
1 𝜀
So: 𝑝1 𝑞1 =
.
1
◦ Elastic demand means − is small, so 𝑝1 𝑞1 ≈ 𝐶 ′ 𝑞1 + 𝑞2 . 𝜀 ◦ I.e. the market with the more elastic demand will have the lower price. ◦ Students are more put-off by high prices, so you should charge them less.
Ambiguous:
◦ The firm gains.
It can always get the same as before by setting the same price in both markets.
◦ Consumers offered the high price lose out. ◦ Consumers offered the low price gain.
Before they might not have been buying the good even.
A necessary condition for a welfare improvement is that output increases.
◦ Varian (1985) or Varian (1989)
No need to understand the proof.
Suppose:
◦ 𝑞1 = max 0,1 − 𝑝1 ◦ 𝑞2 = max 0, 𝑎 − 𝑝2 where 𝑎 ≤ 1. ◦ 𝐶 𝑄 = 0.
Decisions under discrimination:
◦ Profit in first market is 1 − 𝑝1 𝑝1 .
Maximised when 𝑝1 =
1 2
◦ Profit in second market is 𝑎 − 𝑝2 𝑝2 .
Maximised when 𝑝2 =
1+𝑎 . 2
so 𝑞1 =
1 . 2 𝑎
2
◦ Total output is 𝑎
2
so 𝑞2 = .
Decisions without discrimination:
◦ Firm can decide to sell in one or both markets. ◦ Total market demand when a price 𝑃 is set in both markets is: 𝑄 = 𝑞1 + 𝑞2 = max 0,1 − 𝑃 + max 0, 𝑎 − 𝑃 .
So profits are: 𝑃 1 + 𝑎 − 2𝑃 = 2𝑃 − 𝑃 providing 1 − 𝑃 ≥ 0 and 𝑎 − 𝑃 ≥ 0 2 (i.e. if 𝑃 ≤ 𝑎 since 𝑎 ≤ 1). 1+𝑎 1+𝑎 1+𝑎 Maximised when 𝑃 = , so 𝑄 = 1 + 𝑎 − 2 = . 4 4 2 Hence, total output does not increase under discrimination, so welfare cannot have increased. (It will have fallen as long as 𝑎 < 1.)
1+𝑎
◦ However, the firm always has the option to just sell in the first market, in which case profits are 𝑃 1 − 𝑃 .
Maximised when 𝑃 = , so 𝑄 = . 2 2 Thus if the firm only sells in one market without discrimination, discrimination increases output, and so increases welfare. (Example: AIDS drugs.)
1 1
◦ Exercise: show the firm will sell in both markets when discrimination is not possible if 𝑎 ≥ 2 − 1. (Hint: first assume 𝑎 = 2 − 1 and compare profits.)
Printers and cartridges are complements, but not in fixed proportions. Given resale is possible, only one price can be charged for printers. If this price is low, high demand consumers get a large surplus. Tying enables firms to extract some of this.
◦ E.g. “If you buy a printer from me, you have to buy cartridges from me too.” ◦ Enables 𝑃 > 𝑀𝐶 for cartridges. ◦ A kind of two part tariff.
OZ 14.1 gives a slightly strange definition of tying.
◦ More usual one is that the purchase of one good requires the future purchase of another. See https://en.wikipedia.org/wiki/Tying_%28commerce%29
As in the cases above, depends on whether by tying the firm can open up a new market.
◦ E.g. suppose that without tying printers would be priced so high that only businesses could buy them. ◦ In this situation tying (if performed) will generally increase welfare. ◦ But if all consumers would buy even without tying, welfare will generally fall.
=Selling two goods in fixed proportions. Imagine you are Rupert Murdoch. What channels do you want to bundle into Sky?
Valuations Jock Geek Sky Sports 15 8 Discovery 10 12 Total 25 20
If you sell both channels separately (and there are as many Geeks as Jocks) the optimal prices are 8 and 10 for Sky Sports and Discovery respectively, giving a total profit of 2 ∗ 8 + 2 ∗ 10 = 36.
◦ Exercise: How would this change if Jocks valued Sky Sports at 17.
If you sell a bundle, then the optimum price is 20, giving a profit of 40. Key requirement for profitability of bundling: valuations must be negatively correlated across types.
=Selling both a fixed proportion bundle, and the components separately. Strategy one: Word and Excel are both 30. Strategy two: Word and Excel are both 50.
◦ Revenue: 120 ∗ 30 = 3600. ◦ Revenue: 80 ∗ 50 = 4000.
Strategy three: Word and Excel are sold in a bundle at price 50.
Strategy four: Word and Excel are sold in a bundle at price 60. Strategy five: Word and Excel are sold individually at price 50, or in a bundle at price 60.
◦ Revenue: 80 ∗ 50 + 20 ∗ 60 = 5200. ◦ Revenue: 20 ∗ 60 = 1200. ◦ Revenue: 100 ∗ 50 = 5000.
User Type Writer Accountant Generalist
Number 40 40 20
Valuation of Word 50 0 30
Valuation of Excel 0 50 30
Total Valuation 50 50 60
E.g. cars/washing machines etc.
If a firm charges a high price for a durable good today and sells to all of the high value customers, tomorrow, it will be tempted to cut its price to sell to the low value ones. Knowing this, the high value consumers will delay purchasing.
◦ This hurts profits!
The firm would prefer not to be able to discriminate (i.e. not to be able to set different prices in different periods).
Two periods. Customers have per-period valuations uniformly distributed on 0,1 .
◦ I.e. half of all consumers have a valuation below .
A customer with a per-period valuation of 𝑣: MC is zero. Firm sets 𝑝1 in the first period and 𝑝2 in the second.
◦ gets a surplus of 2𝑣 − 𝑝 if they buy in period 1. ◦ gets a surplus of 𝑣 − 𝑝 if they buy in period 2.
2 3
of all consumers have a valuation above , etc.
1 3
1 2
Suppose the firm can commit to 𝑝1 = 𝑝2 = 𝑝. Then there is no point consumers delaying purchasing. Consumers with a valuation 𝑣 such that 𝑝 2𝑣 − 𝑝 ≥ 0 (i.e. 𝑣 ≥ ) will buy.
2
Firm profit is then 𝑝 1 − is maximized at 𝑝 = 1. So demand is
1 2 𝑝
2
=
1 𝑝 2 1 . 2
2 − 𝑝 , which
and profits are
Without commitment (i.e. with discrimination): Suppose in period 1, the 𝑞1 consumers with the highest valuation purchased the good. Then the remaining consumers are uniformly distributed on 0,1 − 𝑞1 and will buy if 𝑣 ≥ 𝑝2 . Firm second period profit is then 𝑝2 1 − 𝑞1 − 𝑝2 , which is maximised at 1−𝑞1 ∗ 𝑝2 = .
2
At this point, second period profits are
1−𝑞1 2 . 4
∗ Consumers will then buy in period 1 if: 2𝑣 − 𝑝1 ≥ 𝑣 − 𝑝2 and 𝑝 ∗ 2𝑣 − 𝑝1 ≥ 0, i.e. if 𝑣 ≥ max 1 , 𝑝1 − 𝑝2 .
Guess (to be verified): 1−𝑞1 .
2 2 𝑝
1 2
∗ ∗ < 𝑝1 − 𝑝2 , so 𝑞1 = 1 − 𝑝1 − 𝑝2 = 1 − 𝑝1 +
2
Thus 𝑞1 = 1 − 𝑝1 . 3 Then, total (both period) profits are then: 𝑝
1 2 1 − 𝑝1 + 3
1 6
2 1 − 1 − 3 𝑝1
3
2
Maximised at 𝑝1 = 10. So 𝑞1 = 5, 𝑝2 = 10 and 𝑞2 = 10. Check guess: From subbing 𝑝1 into total profits, total profits are 𝑝
1 2
9
2
4
3
= 20 < 2 < 10 = 𝑝1 − 𝑝2 .
9
9 20
< !
1 2
So, given the choice, firms would prefer to commit to set the same price in both periods.
◦ Such commitment is generally difficult.
A few examples of this in practice:
◦ Chrysler offered a “lowest price guarantee” on their cars. If the price is lower in future, people who buy now will be reimbursed the difference. ◦ Xerox only leased their copiers in the ‘60s and ‘70s.
If you lease you have to pay every period, so there’s no point delaying.
Price discrimination (generally) enables firms to extract additional profit. If consumer characteristics are observable, firms perform 1st or (more likely) 3rd degree price discrimination. If they are unobservable, firms perform 2nd degree price discrimination, subject to the incentive compatibility constraints. Versioning, tying, bundling and time (i.e. durables) provide other avenues for discrimination. Welfare is usually improved by discrimination when it opens new markets, but is not otherwise.
OZ Ex. 5.7
◦ Question 3, 4, 5, 6 (tricky)
OZ Ex. 13.5
◦ Question 1
OZ Extra exercises:
◦ http://ozshy.50webs.com/io-exercises.pdf ◦ Set #5, 20
scribd
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If ea n de ' the eccentricities of a hyperbola and its conjugate, prove that 1/(e^2)+1/(e '^2)=1.
Question from Class 11 Chapter Hyperbola
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`0``1``2`none of these
Solution :
Let `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` be a hyperbola and let `(x^(2))/(a^(2))-(y^(2))/(b^(2))=-1` be its conjugate. Then, their eccentricites are given by `e^(2)=(a^(2)+b^(2))/(a^(2))` and `e'^(2)=(a^(2)+b^(2))/(b^(2))` respectively. <br> `:. (1)/(e^(2))+(1)/(e'^(2))=(a^(2))/(a^(2)+b^(2))+(b^(2))/(a^(2)+b^(2))=1`
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https://webspace.ship.edu/msrenault/geogebracalculus/derivative_power_rule.html | 1,701,436,710,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100287.49/warc/CC-MAIN-20231201120231-20231201150231-00741.warc.gz | 665,489,128 | 7,263 | ## Polynomials and Multiple Derivatives
HELP
In this applet you can move the seven red control points up and down. A 6th degree polynomial is fitted to these points, so you change the polynomial by moving the control points.
#### Explore
1. If you take the derivative of a 6th degree polynomial, must it always be the case that you get a 5th degree polynomial? Are there any exceptions?
2. Looking at the graph of a polynomial, how can you tell, in general, what the degree of the polynomial is?
3. When the slider shows `d = 0`, the original 6th degree polynomial is displayed. Higher values of `d` take higher derivatives. Play with the slider and confirm that the derivatives of the polynomial behave the way you expect. | 166 | 727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-50 | latest | en | 0.916179 |
https://patents.google.com/patent/CN204870231U/zh | 1,701,474,471,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100308.37/warc/CC-MAIN-20231201215122-20231202005122-00623.warc.gz | 517,127,568 | 9,286 | # CN204870231U - 一种数学教学用快速作图工具 - Google Patents
## Info
Publication number
CN204870231U
CN204870231U CN201520455529.2U CN201520455529U CN204870231U CN 204870231 U CN204870231 U CN 204870231U CN 201520455529 U CN201520455529 U CN 201520455529U CN 204870231 U CN204870231 U CN 204870231U
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## Claims (4)
1.一种数学教学用快速作图工具，包括板体，其特征在于，所述板体的上半部设有镂空的角度线槽，角度线槽的弧形部分外圈配合设有滑槽导轨，滑槽导轨的外侧于板体上设有角度线刻度，所述板体的中心位置设有转轴，转轴上转动设有角度线定位杆，角度线定位杆上设有角度线定位杆刻度，所述角度线定位杆的另一端端部设有指示箭头,指示箭头的尖端位于板体的上半部外圈的中间位置，所述角度线定位杆上于滑槽导轨所处的位置套接固定于垫片的外圈，垫片的中部竖直滑动连接有螺杆,螺杆的底端于滑槽导轨的内部设有限位滑块，所述垫片的上侧于螺杆上螺纹连接有螺母，所述板体的下半部上以转轴为中心均匀分布设有若干主量角器刻度，所述板体的下半部外圈与主量角器刻度均匀配合设有副量角器刻度，所述板体的下半部背面左右两侧均设有磁铁。
2.根据权利要求1所述的数学教学用快速作图工具，其特征在于，所述板体为圆形，由透明轻质塑料制成。
3.根据权利要求1所述的数学教学用快速作图工具，其特征在于，所述角度线槽为半圆形。
4.根据权利要求1所述的数学教学用快速作图工具，其特征在于，所述磁铁内嵌设置于板体的背面左右两侧。
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## Cited By (1)
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CN106476497A (zh) * 2016-11-10 2017-03-08 陈征瀚 一种多功能角度尺
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#### Resources tagged with Multiplication & division similar to More Magic Potting Sheds:
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### There are 46 results
Broad Topics > Calculations and Numerical Methods > Multiplication & division
### More Magic Potting Sheds
##### Stage: 3 Challenge Level:
The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?
### Magic Potting Sheds
##### Stage: 3 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
### Ones Only
##### Stage: 3 Challenge Level:
Find the smallest whole number which, when mutiplied by 7, gives a product consisting entirely of ones.
### Multiples Sudoku
##### Stage: 3 Challenge Level:
Each clue in this Sudoku is the product of the two numbers in adjacent cells.
### One O Five
##### Stage: 3 Challenge Level:
You can work out the number someone else is thinking of as follows. Ask a friend to think of any natural number less than 100. Then ask them to tell you the remainders when this number is divided by. . . .
### Multiply the Addition Square
##### Stage: 3 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
### Countdown
##### Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### A First Product Sudoku
##### Stage: 3 Challenge Level:
Given the products of adjacent cells, can you complete this Sudoku?
### Repeaters
##### Stage: 3 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
### Long Multiplication
##### Stage: 3 Challenge Level:
A 3 digit number is multiplied by a 2 digit number and the calculation is written out as shown with a digit in place of each of the *'s. Complete the whole multiplication sum.
### The Remainders Game
##### Stage: 2 and 3 Challenge Level:
A game that tests your understanding of remainders.
### Power Mad!
##### Stage: 3 Challenge Level:
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
### Integrated Product Sudoku
##### Stage: 3 and 4 Challenge Level:
This Sudoku puzzle can be solved with the help of small clue-numbers on the border lines between pairs of neighbouring squares of the grid.
### Largest Number
##### Stage: 3 Challenge Level:
What is the largest number you can make using the three digits 2, 3 and 4 in any way you like, using any operations you like? You can only use each digit once.
### I'm Eight
##### Stage: 1, 2, 3 and 4 Challenge Level:
Find a great variety of ways of asking questions which make 8.
### Eminit
##### Stage: 3 Challenge Level:
The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M?
### 3388
##### Stage: 3 Challenge Level:
Using some or all of the operations of addition, subtraction, multiplication and division and using the digits 3, 3, 8 and 8 each once and only once make an expression equal to 24.
### Skeleton
##### Stage: 3 Challenge Level:
Amazing as it may seem the three fives remaining in the following `skeleton' are sufficient to reconstruct the entire long division sum.
### Diggits
##### Stage: 3 Challenge Level:
Can you find what the last two digits of the number $4^{1999}$ are?
### Going Round in Circles
##### Stage: 3 Challenge Level:
Mathematicians are always looking for efficient methods for solving problems. How efficient can you be?
### Two Many
##### Stage: 3 Challenge Level:
What is the least square number which commences with six two's?
### Like Powers
##### Stage: 3 Challenge Level:
Investigate $1^n + 19^n + 20^n + 51^n + 57^n + 80^n + 82^n$ and $2^n + 12^n + 31^n + 40^n + 69^n + 71^n + 85^n$ for different values of n.
### Oh! Hidden Inside?
##### Stage: 3 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
### Remainders
##### Stage: 3 Challenge Level:
I'm thinking of a number. When my number is divided by 5 the remainder is 4. When my number is divided by 3 the remainder is 2. Can you find my number?
### A One in Seven Chance
##### Stage: 3 Challenge Level:
What is the remainder when 2^{164}is divided by 7?
### As Easy as 1,2,3
##### Stage: 3 Challenge Level:
When I type a sequence of letters my calculator gives the product of all the numbers in the corresponding memories. What numbers should I store so that when I type 'ONE' it returns 1, and when I type. . . .
### Number Rules - OK
##### Stage: 4 Challenge Level:
Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
### Method in Multiplying Madness?
##### Stage: 2 and 3 Challenge Level:
Watch our videos of multiplication methods that you may not have met before. Can you make sense of them?
### A Chance to Win?
##### Stage: 3 Challenge Level:
Imagine you were given the chance to win some money... and imagine you had nothing to lose...
### Powerful Factorial
##### Stage: 3 Challenge Level:
6! = 6 x 5 x 4 x 3 x 2 x 1. The highest power of 2 that divides exactly into 6! is 4 since (6!) / (2^4 ) = 45. What is the highest power of two that divides exactly into 100!?
### Factoring Factorials
##### Stage: 3 Challenge Level:
Find the highest power of 11 that will divide into 1000! exactly.
### 2010: A Year of Investigations
##### Stage: 1, 2 and 3
This article for teachers suggests ideas for activities built around 10 and 2010.
### Round and Round and Round
##### Stage: 3 Challenge Level:
Where will the point stop after it has turned through 30 000 degrees? I took out my calculator and typed 30 000 ÷ 360. How did this help?
### So It's Times!
##### Stage: 2 and 3 Challenge Level:
How will you decide which way of flipping over and/or turning the grid will give you the highest total?
### Missing Multipliers
##### Stage: 3 Challenge Level:
What is the smallest number of answers you need to reveal in order to work out the missing headers?
### Please Explain
##### Stage: 3 Challenge Level:
Visitors to Earth from the distant planet of Zub-Zorna were amazed when they found out that when the digits in this multiplication were reversed, the answer was the same! Find a way to explain. . . .
### Expenses
##### Stage: 4 Challenge Level:
What is the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time?
### Thirty Six Exactly
##### Stage: 3 Challenge Level:
The number 12 = 2^2 × 3 has 6 factors. What is the smallest natural number with exactly 36 factors?
### Slippy Numbers
##### Stage: 3 Challenge Level:
The number 10112359550561797752808988764044943820224719 is called a 'slippy number' because, when the last digit 9 is moved to the front, the new number produced is the slippy number multiplied by 9.
### X Marks the Spot
##### Stage: 3 Challenge Level:
When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" .
### Factorial
##### Stage: 4 Challenge Level:
How many zeros are there at the end of the number which is the product of first hundred positive integers?
### More Mods
##### Stage: 4 Challenge Level:
What is the units digit for the number 123^(456) ?
### Vedic Sutra - All from 9 and Last from 10
##### Stage: 4 Challenge Level:
Vedic Sutra is one of many ancient Indian sutras which involves a cross subtraction method. Can you give a good explanation of WHY it works?
### What Is the Question?
##### Stage: 3 and 4 Challenge Level:
These pictures and answers leave the viewer with the problem "What is the Question". Can you give the question and how the answer follows?
### Galley Division
##### Stage: 4 Challenge Level:
Can you explain how Galley Division works?
### Times Right
##### Stage: 3 and 4 Challenge Level:
Using the digits 1, 2, 3, 4, 5, 6, 7 and 8, mulitply a two two digit numbers are multiplied to give a four digit number, so that the expression is correct. How many different solutions can you find? | 2,141 | 8,700 | {"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} | 4.5 | 4 | CC-MAIN-2017-39 | latest | en | 0.877237 |
https://nl.mathworks.com/matlabcentral/profile/authors/15008615?detail=all | 1,716,572,216,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058721.41/warc/CC-MAIN-20240524152541-20240524182541-00168.warc.gz | 370,627,734 | 23,766 | # Mohit Kumar
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Solved | 1,653 | 6,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.846682 |
https://mathoverflow.net/questions/109330/did-gauss-know-dirichlets-class-number-formula-in-1801 | 1,716,051,060,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057440.7/warc/CC-MAIN-20240518152242-20240518182242-00457.warc.gz | 341,288,312 | 26,436 | # Did Gauss know Dirichlet's class number formula in 1801?
Let $h_d$ be the number of $SL_{2}(\mathbb{Z})$ classes of primitive binary quadratic forms of discriminant $d$. It's natural to impose the hypothesis that $d$ is not at square, as we do below.
In Carl Ludwig Siegel's paper titled The Average Measure of Quadratic Forms With Given Discriminant and Signature Siegel cites two formulae given by Gauss in Disquisitiones Arithmeticae:
(a) $\displaystyle\sum\limits_{d= -N }^1 h_d \sim \frac{\pi}{18 \zeta(3)}N^{3/2}$
(b) $\displaystyle\sum\limits_{d = 1}^N h_d \log{\epsilon}_d \sim \frac{{\pi}^2}{18 \zeta(3)}N^{3/2}$
Where $N > 0$ and $\epsilon_{d} = \frac{1}{2}(t + u \sqrt{d})$ where $(t,u)$ is the smallest positive solution to $t^2 - ud^2 = 4$.
(Actually, Gauss restricts to consideration to binary quadratic forms with even middle coefficient correspondingly arrives at different formulae, but they're essentially the same as those above).
Siegel gives two proofs of these formulae: one proceeding from Dirichlet's class number formula together with character sum estimates due to Polya and Landau, and one via a direct lattice point counting argument.
In light of the facts that (i) I haven't heard anyone say that Gauss's was the one to discover the class number formula and (ii) the character sum estimates seem outside of the scope of Gauss's work, I imagine that his argument was via lattice point counting. Do we have any evidence otherwise? (I checked Gauss's book and he doesn't describe his methods there.)
• A short comment for now: Gauss understood the connection between lattice points on spheres and class numbers of definite quadratic forms: The number of representations of $m$ as a sum of 3 squares is a constant times $h(-m)$ or $h(-4m)$ depending on the congruence of $m$ mod $8$, as I recall. The estimate (a) can probably be deduced from counting lattice points in $R^3$. Oct 10, 2012 at 21:14
• @ Marty - aah, good point, I forgot about that result of Gauss. I wonder if there's an analogous result involving class numbers of real quadratic fields. Oct 10, 2012 at 22:08
• @ Marty - BTW, Shimura has a great article discussing a common framework for thinking about the ternary quadratic form given by the discriminant and the ternary quadratic form that you mention: ams.org/journals/bull/2006-43-03/S0273-0979-06-01107-4 Oct 10, 2012 at 22:10
• Am I right that it is still not known whether there exists a set of $d>0$ of positive density for which $h_d=1$? I've heard a talk (long time ago) where this was referred to as a "Gauss problem". There is also some nice connection between $h_d$ and the length of the period of the continued fraction expansion of $\sqrt{d}$ but I don't quite remember what it was. Oct 10, 2012 at 23:17
• Jonah is right about rings of integers in number fields, but that's a slightly different question (the $h(5^{2k+1})$ example involves non-maximal orders). Oct 11, 2012 at 0:29 | 800 | 2,951 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-22 | latest | en | 0.904924 |
http://usedave.nl/forces.html | 1,686,247,078,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655092.36/warc/CC-MAIN-20230608172023-20230608202023-00140.warc.gz | 46,392,210 | 8,267 | # Forces#
• The structure of the model consists of Frame type nodes
• Each of these nodes can have up to 6 degrees of freedom
• Frame nodes can be placed on other frame nodes
It is now time to explain how to obtain the forces that travel through the system.
First the list of nodes in the scene is sorted such that child nodes are in the list before their parents. Then to the algorithm iterates over all the nodes in the list. For each node the forces are calcualted based on the current positions and orientations. The forces are then applied on the parent of the node.
Axis type nodes do not generate any forces by themselves. They may however receive forces from their children or other nodes. All these forces are summed up and are are stored in the frame node as “applied force”.
Applied force
The applied force of an Axis node is the summation of all forces applied on that frame by its children and other nodes connected to that frame.
The applied_force property has six components (Fx, Fy, Fz, Mx, My, Mz) representing the force and moment at the origin of the frame. The used coordinate system is the global axis system.
If the frame node has a parent, then the applied force is fed through to that parent. This is done only for degrees of freedom that are fixed. The force/moment applied on the parent is available as the “connection force”.
Connection force
The connection_force of an Axis node is force/moment that is applied on its parent.
The connection_force property has six components (Fx, Fy, Fz, Mx, My, Mz) representing the force and moment at the origin of the frame. The used coordinate system is the parent frame system.
Example:
Consider de following example where the dotted orange lines indidate a parent-child relation:
The sorted list of nodes is: Force node, Point node, Frame node (child), Frame node (parent).
1. Force node: applies its force on the point node
2. Point node: applies the force that it received from the force node on its parent: Axis node (child).
3. Frame node (child): Has an ‘applied force’ force due to the contribution from the point node. This applied_force equals the force due the force-node as well as the moment due to the force node and the lever between the frame node (child) origin and the point node. If the global position of the point node is (px, pz), the global position of the Axis node (child) is cx, cz and the force in the force_node is fx, fz then the applied_force vector equals: (Fx, 0 , Fz, 0, -(px-cx) * Fz + (pz-cz) * Fx, 0). This force and moment is then applied on the Frame Node (parent) via the ‘Connection force’
4. Frame node (parent) has a applied force from the connection force of Frame node (child). This force and moment act at the position of frame node (child). Because frame node (parent) does not have a parent it is considered to be connected to the world. Its ‘connection force’ becomes the force/moment needed to counter the applied force. The force part of the connection force of Axis node (parent) is (-Fx, 0, -Fz). Assuming that the Axis node (parent) is located at the global origin, the moment part of the connection force is (0, px*Fz -pz * Fx, 0)
Force at location
Connection force and applied force represent the loads traveling through the construction at the origin of a frame node. Clever placing of Frames in your model will give you the forces in your structure at the location and in the axis system that you need. You can stack as many frame nodes as you want.
Example
In following model Carene Table Notebook a total number of 11 Axes are stacked from left to right. On the rightmost node a vertical force is applied:
The vertical force and y-moment in the frame nodes can then be used to obtain the moment and shear lines in this structure:
## Degrees of freedom#
In the previous examples we have assumed that all degrees of freedom were fixed. If all the degrees of freedom are fixed then the Axis is rigidly connected to its parent or the world. In that case the connection-force and the applied-load are exactly opposite.
If a degree of freedom is released then the connection force for that degree of is set to 0.
Degrees of freedom and equilibrium error
The connection force is 0 for any free degree of freedom.
The difference between the applied_force and the connection_force is the equilibrium_error.
The difference between the applied_force and the connection_force is called the equilibrium_error and is expressed in the parent frame. If the model is in static equilibrium then all equilibrium errors should be zero.
Solving static equilibrium will minimize the equilibrium_error by changing the degrees of freedom that are not fixed.
## Forces in elastic nodes#
The forces in elastic nodes are calcualted based the positions of the nodes that they connect.
For a cable the distance bewteen the endpoints is obtained from the current geometry, this is then used to calculate the stretch, the stretch is combined with the stiffness to give the tension. The force due to the tension is then applied on the endpoints and intermediate points. In case of a cable these are Point or Circle type nodes.
The 2d and 6d connectors work similarly.
## Axis and Points without a parent#
Axis and Point type nodes do not need to have a parent. Points without a parent are considered to be fixed to the world. Axes without parent are considered to be fixed to the world for any fixed degree of freedom. | 1,178 | 5,450 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-23 | latest | en | 0.927587 |
https://risk.asmedigitalcollection.asme.org/VVS/proceedings/VVS2018/40795/V001T12A001/274276?searchresult=1 | 1,660,106,705,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571147.84/warc/CC-MAIN-20220810040253-20220810070253-00568.warc.gz | 437,784,458 | 18,421 | Errors and uncertainties in finite element method (FEM) computing can come from the following eight sources, the first four being FEM-method-specific, and the second four, model-specific: (1) Computing platform such as ABAQUS, ANSYS, COMSOL, LS-DYNA, etc.; (2) choice of element types in designing a mesh; (3) choice of mean element density or degrees of freedom (d.o.f.) in the same mesh design; (4) choice of a percent relative error (PRE) or the Rate of PRE per d.o.f. on a log-log plot to assure solution convergence; (5) uncertainty in geometric parameters of the model; (6) uncertainty in physical and material property parameters of the model; (7) uncertainty in loading parameters of the model, and (8) uncertainty in the choice of the model. By considering every FEM solution as the result of a numerical experiment for a fixed model, a purely mathematical problem, i.e., solution verification, can be addressed by first quantifying the errors and uncertainties due to the first four of the eight sources listed above, and then developing numerical algorithms and easy-to-use metrics to assess the solution accuracy of all candidate solutions. In this paper, we present a new approach to FEM verification by applying three mathematical methods and formulating three metrics for solution accuracy assessment. The three methods are: (1) A 4-parameter logistic function to find an asymptotic solution of FEM simulations; (2) the nonlinear least squares method in combination with the logistic function to find an estimate of the 95% confidence bounds of the asymptotic solution; and (3) the definition of the Jacobian of a single finite element in order to compute the Jacobians of all elements in a FEM mesh. Using those three methods, we develop numerical tools to estimate (a) the uncertainty of a FEM solution at one billion d.o.f., (b) the gain in the rate of PRE per d.o.f. as the asymptotic solution approaches very large d.o.f.’s, and (c) the estimated mean of the Jacobian distribution (mJ) of a given mesh design. Those three quantities are shown to be useful metrics to assess the accuracy of candidate solutions in order to arrive at a so-called “best” estimate with uncertainty quantification. Our results include calibration of those three metrics using problems of known analytical solutions and the application of the metrics to sample problems, of which no theoretical solution is known to exist.
This content is only available via PDF. | 520 | 2,460 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.85752 |
https://www.writingessays.org/%E6%95%B0%E5%AD%A6%E8%80%83%E8%AF%95%E4%BB%A3%E5%81%9A/ | 1,720,912,195,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514517.94/warc/CC-MAIN-20240713212202-20240714002202-00288.warc.gz | 953,245,970 | 17,364 | Search the whole station
数学考试代做 Math 2415代写 数学期末代写 数学作业代写
510
Math 2415 Final Exam
To get full credit you must show ALL your work.
The problems must be solved without any assistance of others or the usage of unauthorized material or information. This is a “closed book” exam: no texts, notes or calculators allowed.
1. (70 points) 数学考试代做
Consider a metal rod encased in a thermal insulator. Suppose the rod is immersed in boiling water so that its temperature is 100˚C throughout, and then removed from the water at time t = 0 with one of its ends immediately put in ice and kept at temperature 0˚C, and insulated on the other end.
Mathematically, this situation can be modeled as follows:
Show the details of your construction of the solution. You must show all the steps of your calculation for full credit.
5. (10 points) 数学考试代做
Which of the following is the general solution of the equation
(a)y(t) = cet − tet ？
(b) y(t) = tet + cet ？
(c) y(t) = tet + cet ？
(d) y(t) = et + cet ？
(e) y(t) = tet + ctet ？
(f) None of the above
6. (a) (15 points) 数学考试代做
Find a differential equation whose general solution is: y(x) = c1ex + c2e-3x, where c1, c2 are arbitrary real numbers.
7.(30 points)
The function E(t) is T-periodic, defined in one periodic window by
Determine the Fourier series for E(t)
The prev: The next:
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1 | 709 | 2,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-30 | latest | en | 0.750306 |
http://www.ionizationx.com/index.php?PHPSESSID=kae4l0u444ubnc45h0grc61fl7&topic=2527.msg24365 | 1,660,610,392,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572215.27/warc/CC-MAIN-20220815235954-20220816025954-00181.warc.gz | 79,108,816 | 7,051 | ### Author Topic: Theory - Coil Calculations to Restrict Amps (CONFIRMED) (Read 26293 times)
0 Members and 1 Guest are viewing this topic.
• Member
• Posts: 241
##### Re: Theory - Coil Calculations to Restrict Amps
« Reply #56 on: January 02, 2013, 13:24:25 pm »
''The L1 coil was around 3.4H and the L2 coil was around 2.8H.''
Are you talking about the MOT's secoundary split in 2 sections?If yes in other thread you said you measure the bobin's half and cut it to create equal coils.If you used a diferent value on the negative choke how do you calculated the inductance for it?or when you cut the bobin in two you made one half smaller by eye?
You also said you conected them out of phase and in the small drawing you posted it shows in phase,when conected out of phase dont they need to have the output leeads one''start of winding'' and the other''end of winding''?just asking...
''When I connected the 8XA Circuit to the coil and increased the variac to around 50VDC, I would only get a voltage input reading up to 12VDC max!''
Input reading before the chokes ?or after?
Thanks verry much Tony!Cheers!
• Sr. member
• Posts: 460
##### Re: Theory - Coil Calculations to Restrict Amps
« Reply #57 on: January 03, 2013, 01:37:08 am »
Yea I used a MOT and just cut the wire in the center (eyed it). By eying it I made them unequal, but this actually worked out I guess. Then I connected them out of phase, not like the Stan's 8XA image. I took the voltage measurements before the "blocking diode" and across the cell.
• Sr. member
• Posts: 349
##### Re: Theory - Coil Calculations to Restrict Amps
« Reply #58 on: January 03, 2013, 07:11:53 am »
If your planning on replicating tony's mot, take some time to experiment using the primary and secondarys without any modifications. I had some good results that way : ) | 502 | 1,824 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.926346 |
https://caregiverstorm.com/grahams-creek/application-of-black-scholes-model.php | 1,603,559,269,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107884322.44/warc/CC-MAIN-20201024164841-20201024194841-00112.warc.gz | 254,103,808 | 13,943 | ## Black-Scholes Excel and VBA
Black-Scholes Excel Models Instant Downloads - Eloquens. Tutorial on how to calculate black scholes option pricing model with definition, formula, example. Learn Online., Easy tool that can calculate the fair value of an equity option based on the Black-Scholes, Whaley and Binomial Models along with Greek sensitivities..
### Options Pricing Black-Scholes Model Investopedia
Black Scholes (Greeks) Applications YouTube. Keywords: Stochastic volatility, Heston, Black-Scholes biases, calibration, characteristic functions. 2.3 An Application to the Black and Scholes Model, Black's Model is a variation of the popular Black-Scholes options pricing model that allows for the Black's Model is used in the application of capped.
Four Derivations of the Black Scholes PDE The time-t Black-Scholes price of a call with time to maturity Лќ = T t and The Capital Asset Pricing Model Black-Scholes option model - using Excel cell formulas and VBA function procedures.
The Black model (sometimes known as the Black-76 model) is a variant of the Black–Scholes option pricing model. Its primary applications are for pricing options on Solving the Black-Scholes equation: a demysti cation V. Application to cash-or-nothing binary options 6 A. While the derivation of the Black-Scholes
Syllabus B2a: Apply the Black-Scholes Option Pricing (BSOP) model to financial product valuation and to asset valuation: i) Determine and discuss, using published Tutorial on how to calculate black scholes option pricing model with definition, formula, example. Learn Online.
Definition: The Black-Scholes Model is the options pricing model developed by Fischer Black, Myron Scholes, and Robert Merton, wherein the formula is used to Here is the formula for the Black Scholes Model for pricing European call and put option contracts
THE BLACK SCHOLES FORMULA Section 6 discusses the applications of the formula in market trading. We Black and Scholes’ stylized model. 1. 3/12/2013 · Join us in the discussion on InformedTrades: http://www.informedtrades.com/1087607-black-scholes-n-d2-explained.html In this video, I give a general
Easy tool that can calculate the fair value of an equity option based on the Black-Scholes, Whaley and Binomial Models along with Greek sensitivities. The Black-Scholes model is a mathematical model for financial markets. From this larger model, the Black-Scholes formula for theoretical option...
Hoadley Trading & Investment Tools. Options Applications. priced under the binomial model converge with Black-Scholes prices as the number of Black-Scholes model would take a couple of semesters to develop in any formal vanilla European option pricing application. The binomial model is more exible
PDF The pricing of options is one of the most complex areas of applied finance and has been a subject of extensive study. Model based pricing started in the 1970s Application of Black Scholes Complexity Concepts to Combat Modelling . Nigel Perry . Joint Operations Division . Defence Science and Technology Organisation
### Application of Black Scholes Complexity Concepts to Combat
THE BLACK SCHOLES FORMULA Imperial College London. Exchange traded options trading strategy evaluation tool & pricing calculators. Black-Scholes and the binomial model are used for option pricing. Pay-off diagrams are, Application of Black Scholes Complexity Concepts to Combat Modelling . Nigel Perry . Joint Operations Division . Defence Science and Technology Organisation.
How to Calculate Black Scholes Option Pricing Model. Black Scholes Option Pricing Model definition, formula, and example of the Model as used to price options., The Black-Scholes equation, The combination of COMSOL В® products required to model your application depends on several factors and may include boundary.
### OPTIONS and FUTURES Lecture 4 The Black-Scholes model
The Black-Scholes Model in VBA BSIC Bocconi Students. Keywords: Stochastic volatility, Heston, Black-Scholes biases, calibration, characteristic functions. 2.3 An Application to the Black and Scholes Model https://en.wikipedia.org/wiki/Local_volatility Tutorial on how to calculate black scholes option pricing model with definition, formula, example. Learn Online..
• Derivation and Applications of Black-Scholes Partial Di
• Black-Scholes Calculator Online FinTools
• The Black-Scholes Model in VBA Secondly, we’ll provide the code to put the theory into practice and show some basic (but hopefully relevant) applications. The Application of the Black-Scholes Model A research report presented to The Graduate School of Business University of Cape Town in partial fulfilment
What is the Black Scholes Model? The Black Scholes model was the first widely used model for option pricing. It is used to calculate the theoretical value of European Black-Scholes option model - using Excel cell formulas and VBA function procedures.
The Black-Scholes equation, The combination of COMSOL В® products required to model your application depends on several factors and may include boundary Easy tool that can calculate the fair value of an equity option based on the Black-Scholes, Whaley and Binomial Models along with Greek sensitivities.
Keywords: Stochastic volatility, Heston, Black-Scholes biases, calibration, characteristic functions. 2.3 An Application to the Black and Scholes Model THE BLACK SCHOLES FORMULA Section 6 discusses the applications of the formula in market trading. We Black and Scholes’ stylized model. 1.
Black-Scholes Model In this application example, we want to compute the option price using three different methods. The first method is to derive the analytical The Peculiar Logic of the Black-Scholes Model I will ultimately argue that the continued use of the Black-Scholes model, in the particular application I will
10/12/2011В В· Financial Mathematics 3.5 - Black Scholes Applications (Greeks) 10/12/2011В В· Financial Mathematics 3.5 - Black Scholes Applications (Greeks)
Options trading became really popular when the Black-Scholes model came about, Fischer Black, Myron Scholes and Robert Merton came up with it with the help of Black's Model is a variation of the popular Black-Scholes options pricing model that allows for the Black's Model is used in the application of capped
3/11/2015В В· New York Institute of Finance instructor Anton Theunissen explains the history, mechanics, and application of the Black-Scholes Model of options pricing. 3/12/2013В В· Join us in the discussion on InformedTrades: http://www.informedtrades.com/1087607-black-scholes-n-d2-explained.html In this video, I give a general
Black's Model is a variation of the popular Black-Scholes options pricing model that allows for the Black's Model is used in the application of capped Solving the Black-Scholes equation: a demysti cation V. Application to cash-or-nothing binary options 6 A. While the derivation of the Black-Scholes
## LECTURE 7 BLACK–SCHOLES THEORY University of Chicago
Application of Black Scholes Complexity Concepts to Combat. APPLICATIONS OF OPTION PRICING THEORY TO EQUITY VALUATION The Black-Scholes option pricing model is derived under the assumption Application to valuation:, 8: The Black-Scholes Model Marek Rutkowski School of Mathematics and Statistics University of Sydney MATH3075/3975 Financial Mathematics Semester 2, 2016.
### Applying Black-Scholes to valuing index options
Does the Black-Scholes Model apply to American Style. Black-Scholes model would take a couple of semesters to develop in any formal vanilla European option pricing application. The binomial model is more exible, The Black model (sometimes known as the Black-76 model) is a variant of the Black–Scholes option pricing model. Its primary applications are for pricing options on.
Black Scholes option pricing model . The Black-Scholes model values call options before the expiry date and takes account of all Application to American call Derivation and Applications of Black-Scholes Partial Differential Equation and Black-Scholes formulas Cheng-Han-Yu
Black Scholes option pricing model . The Black-Scholes model values call options before the expiry date and takes account of all Application to American call Black-Scholes Model In this application, we compute the option price using three different methods. The first method is to derive the analytical solution to the
In this post, we will discuss on modeling option pricing using Black Scholes Option Pricing model and plotting the same for a combination of various options. If you In this application of the Black–Scholes model, a coordinate transformation from the price domain to the volatility domain is obtained.
Tutorial on how to calculate black scholes option pricing model with definition, formula, example. Learn Online. View Black Scholes Model Research Papers on Academia.edu for free.
An Analysis of a Dynamic Application of Black-Scholes in Option Trading Aileen Wang Thomas Je erson High School for Science and Technology Alexandria, Virginia Connecting Binomial and Black-Scholes Option Pricing Models: A Spreadsheet-Based Illustration Abstract The Black-Scholes option pricing model is part of the modern
Risk Neutral Valuation, the Black-Scholes Model and Monte Carlo Stephen M Schaefer London Business School Credit Risk Elective Summer 2012 Objectives: to understand 2/36 Applications of Black-Scholes model Call on forward - the Black formula Exchange option - Margrabe formula Foreign exchange options – Garman-Kohlagen formula
APPLICATIONS OF OPTION PRICING THEORY TO EQUITY VALUATION The Black-Scholes option pricing model is derived under the assumption Application to valuation: In this article, we use a Mellin transform approach to prove the existence and uniqueness of the price of a European option under the framework of a Black–Scholes
Tutorial on how to calculate black scholes option pricing model with definition, formula, example. Learn Online. In this article, we use a Mellin transform approach to prove the existence and uniqueness of the price of a European option under the framework of a Black–Scholes
Black Scholes option pricing model . The Black-Scholes model values call options before the expiry date and takes account of all Application to American call APPLICATIONS OF OPTION PRICING THEORY TO EQUITY VALUATION The Black-Scholes option pricing model is derived under the assumption Application to valuation:
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Black Scholes Option Pricing Model definition, formula, and example of the Model as used to price options. Application of Black Scholes Complexity Concepts to Combat Modelling . Nigel Perry . Joint Operations Division . Defence Science and Technology Organisation
In this article, we use a Mellin transform approach to prove the existence and uniqueness of the price of a European option under the framework of a Black–Scholes Keywords: Stochastic volatility, Heston, Black-Scholes biases, calibration, characteristic functions. 2.3 An Application to the Black and Scholes Model
Here is the formula for the Black Scholes Model for pricing European call and put option contracts Black-Scholes Formula Parameters. According to the Black-Scholes option pricing model In the original Black and Scholes paper
Derivation and Applications of Black-Scholes Partial Differential Equation and Black-Scholes formulas Cheng-Han-Yu Hoadley Trading & Investment Tools. Options Applications. priced under the binomial model converge with Black-Scholes prices as the number of
The Black-Scholes formula (also called Black-Scholes-Merton) was the first widely used model for option pricing. It's used to calculate the theoretical value of Four Derivations of the Black Scholes PDE The time-t Black-Scholes price of a call with time to maturity Лќ = T t and The Capital Asset Pricing Model
### The Black-Scholes Equation COMSOL Multiphysics
The Pricing of Warrants on the Johannesburg Stock Exchange. Easy tool that can calculate the fair value of an equity option based on the Black-Scholes, Whaley and Binomial Models along with Greek sensitivities., PDF The aim of this paper is to study the Black-Scholes option pricing model. We discuss some definitions and different derivations, which are useful for further.
### Solving the Black-Scholes equation a demysti cation
ACCA AFM (P4) Notes B2a. Black-Scholes Option Pricing. PDF The pricing of options is one of the most complex areas of applied finance and has been a subject of extensive study. Model based pricing started in the 1970s https://en.wikipedia.org/wiki/Talk:Black%E2%80%93Scholes_model 3/11/2015В В· New York Institute of Finance instructor Anton Theunissen explains the history, mechanics, and application of the Black-Scholes Model of options pricing..
• THE BLACK SCHOLES FORMULA Imperial College London
• An Analysis of the Heston Stochastic Volatility Model
• Definition: The Black-Scholes Model is the options pricing model developed by Fischer Black, Myron Scholes, and Robert Merton, wherein the formula is used to The Black Scholes Model (BSM) based on an application of Taguchi orthogonal array L9, in which the four parameters of BSM for European call option,
Application of Black Scholes Complexity Concepts to Combat Modelling . Nigel Perry . Joint Operations Division . Defence Science and Technology Organisation In this application of the Black–Scholes model, a coordinate transformation from the price domain to the volatility domain is obtained.
Options trading became really popular when the Black-Scholes model came about, Fischer Black, Myron Scholes and Robert Merton came up with it with the help of Here is the formula for the Black Scholes Model for pricing European call and put option contracts
The Black–Scholes Option pricing model (OPM) developed in 1973 has always been taken as the cornerstone of option pricing model. The generic applications of such a Black-Scholes option model - using Excel cell formulas and VBA function procedures.
Black-Scholes option model - using Excel cell formulas and VBA function procedures. 3/11/2015В В· New York Institute of Finance instructor Anton Theunissen explains the history, mechanics, and application of the Black-Scholes Model of options pricing.
Connecting Binomial and Black-Scholes Option Pricing Models: A Spreadsheet-Based Illustration Abstract The Black-Scholes option pricing model is part of the modern 8: The Black-Scholes Model Marek Rutkowski School of Mathematics and Statistics University of Sydney MATH3075/3975 Financial Mathematics Semester 2, 2016
The Black model (sometimes known as the Black-76 model) is a variant of the Black–Scholes option pricing model. Its primary applications are for pricing options on Black Scholes Option Pricing Model definition, formula, and example of the Model as used to price options.
The model is named after Fischer Black and Myron Scholes, who developed it in 1973. Robert Merton also participated in the model's creation, and this is why the model I am currently attempting to use the Black-Scholes model to value index options. My issue is; what should I use as the price of the underlying? Say I want to value a
The Black-Scholes formula (also called Black-Scholes-Merton) was the first widely used model for option pricing. It's used to calculate the theoretical value of VBA and Excel spreadsheet for Black-Scholes and Greeks (Delta, Gamma, Vega, Theta, Rho). Easily use the VBA in your own option pricing spreadsheets.
In this article, we use a Mellin transform approach to prove the existence and uniqueness of the price of a European option under the framework of a Black–Scholes Application of Option Pricing to Valuation of Firms - an article exploring the insights that the Black Scholes Merton (BSM) model provides
VBA and Excel spreadsheet for Black-Scholes and Greeks (Delta, Gamma, Vega, Theta, Rho). Easily use the VBA in your own option pricing spreadsheets. Tutorial on how to calculate black scholes option pricing model with definition, formula, example. Learn Online.
THE BLACK SCHOLES FORMULA Section 6 discusses the applications of the formula in market trading. We Black and Scholes’ stylized model. 1. V. Black-Scholes model: Derivation and solution Solve the Black-Scholes PDE for a call option on We show the application of the latter approach. V. Black
the Black Scholes model still remain the benchmark of option valuation and it is the standard to which all other applications and efficiency of the models. VBA and Excel spreadsheet for Black-Scholes and Greeks (Delta, Gamma, Vega, Theta, Rho). Easily use the VBA in your own option pricing spreadsheets.
3/11/2015В В· New York Institute of Finance instructor Anton Theunissen explains the history, mechanics, and application of the Black-Scholes Model of options pricing. Black-Scholes model would take a couple of semesters to develop in any formal vanilla European option pricing application. The binomial model is more exible
Easy tool that can calculate the fair value of an equity option based on the Black-Scholes, Whaley and Binomial Models along with Greek sensitivities. Stochastic Calculus and Option Pricing Stochastic Integral Itô’s Lemma Black-Scholes Model Multivariate Itô Processes SDEs SDEs and PDEs Risk
3 The Black-Scholes Model: European Options CSNd Ke Ndr t () ( ) 1 365 2 C = theoretical call value S = current stock price N = cumulative standard The Black–Scholes Option pricing model (OPM) developed in 1973 has always been taken as the cornerstone of option pricing model. The generic applications of such a
The application has failed to start because it's side‐by‐side... The application has failed to start because its side‐by‐side configuration is incorrect. Application failed because side by side configuration is incorrect Wirrabara Fixing the Windows “side-by-side configuration is incorrect” error # (or any other application) failed to start because its side-by-side configuration is
481559 | 3,976 | 18,247 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-45 | latest | en | 0.803836 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-mathematics-for-calculus-7th-edition/chapter-2-section-2-1-functions-exercises-page-157/36 | 1,534,912,329,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221219469.90/warc/CC-MAIN-20180822030004-20180822050004-00308.warc.gz | 889,752,936 | 13,462 | ## Precalculus: Mathematics for Calculus, 7th Edition
$$f(2x) = 6x-1$$ $$2f(x) = 6x-2$$
For x = 2x $f(2x) = 3(2x) - 1$ ... multiply $=6x - 1$ ________________ $f(x) = 3x - 1$ therefore $2f(x) = 2(3x-1)$ ... multiply $=6x - 2$ | 115 | 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": 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} | 3.953125 | 4 | CC-MAIN-2018-34 | longest | en | 0.421806 |
https://ask.csdn.net/questions/757442 | 1,601,413,328,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402088830.87/warc/CC-MAIN-20200929190110-20200929220110-00476.warc.gz | 283,619,433 | 24,755 | Problem Description
When Wiskey return home, many relatives he need to visit, and he will buy some gifts before he visit his relatives. There is a map which has N places; Wiskey's house is labeled 0. He has T tasks which from u to v, it mean Wiskey could go to v when he bought a gift at u before.
Wiskey always start at 0, and must end with 0.
Tell him how many minimum total places he went when he finished all tasks.
Input
First line will contain one integer mean how many cases will follow by; the number of cases wills less than 20.
In each case will contain three integers N means the number of places, M roads and T tasks. (1 <= N <= 20, 0 <= M <= 100, 0 <= T <= 5).
First, M roads will be follow; the road is bi-connected.
Second, T tasks will be follow; you are remember the u must went earlier than v, then what makes the task will be finished.
Output
Print the minimum total number of places he went.
If Wiskey can't finish all tasks, please print -1.
Sample Input
1
7 11 5
0 1
0 2
1 2
1 4
0 3
3 4
4 6
2 6
6 5
3 5
4 5
1 2
1 3
2 0
1 4
5 3
Sample Output
7 | 322 | 1,068 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-40 | longest | en | 0.958401 |
http://math.stackexchange.com/questions/197772/generalized-rotation-matrix-in-n-dimensional-space-around-n-2-unit-vector | 1,467,034,502,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783396027.60/warc/CC-MAIN-20160624154956-00146-ip-10-164-35-72.ec2.internal.warc.gz | 192,359,248 | 18,045 | # Generalized rotation matrix in N dimensional space around N-2 unit vector
There is a 2d rotation matrix around point $(0, 0)$ with angle $\theta$.
$$\left[ \begin{array}{ccc} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{array} \right]$$
Next, there is a 3d rotation matrix around point $(0, 0, 0)$ and unit axis $(u_x, u_y, u_z)$ with angle $\theta$ (Rodrigues' Rotation Formula).
\begin{bmatrix} \cos \theta +u_x^2 \left(1-\cos \theta\right) & u_x u_y \left(1-\cos \theta\right) - u_z \sin \theta & u_x u_z \left(1-\cos \theta\right) + u_y \sin \theta \\ u_y u_x \left(1-\cos \theta\right) + u_z \sin \theta & \cos \theta + u_y^2\left(1-\cos \theta\right) & u_y u_z \left(1-\cos \theta\right) - u_x \sin \theta \\ u_z u_x \left(1-\cos \theta\right) - u_y \sin \theta & u_z u_y \left(1-\cos \theta\right) + u_x \sin \theta & \cos \theta + u_z^2\left(1-\cos \theta\right) \end{bmatrix}
How it is possible to generalize rotation matrix on $N$ dimension around zero point and $N-2$ dimensional unit axis with angle $\theta$?
-
@RobertIsrael could you please tell me the reference (name of the formula, article, book, etc.) for your formula above (a matrix that rotates the span of u and v by angle \theta)? Thank you in advance. – user71237 Apr 6 '13 at 8:13
The definition is that $A\in M_{n}(\mathbb{R})$ is called a rotation matrix if there exist a unitary matrix $P$ s.t $P^{-1}AP$ is of the form $$\begin{pmatrix}\cos(\theta) &-\sin(\theta)\\ \sin(\theta) & \cos(\theta)\\ & & 1\\ & & & 1\\ & & & & 1\\ & & & & & .\\ & & & & & & .\\ & & & & & & & .\\ & & & & & & & & 1 \end{pmatrix}$$
If we consider $A:\mathbb{R}^{n}\to\mathbb{R}^{n}$ then the meaning is that there exist an orthonormal basis where we rotate the $2-$dimensional space spanned by the first two vectors by angle $\theta$ and we fix the other $n-2$ dimensions
If $u$ and $v$ are two orthonormal vectors, a matrix that rotates the span of $u$ and $v$ by angle $\theta$ is $A = I + \sin(\theta) ( v u^T - u v^T) + (\cos(\theta) - 1)( u u^T + v v^T)$. Thus $A_{ij} = \delta_{ij} + \sin(\theta) (v_i u_j - u_i v_j) + (\cos(\theta)-1)(u_i u_j + v_i v_j)$. – Robert Israel Sep 16 '12 at 23:51
The $n-2$-dimensional space that is the orthogonal complement of my two vectors is fixed by the rotation, i.e. $Aw = w$ for $w$ in this space. – Robert Israel Sep 20 '12 at 7:44 | 840 | 2,360 | {"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} | 3.609375 | 4 | CC-MAIN-2016-26 | latest | en | 0.584892 |
https://lists.cairographics.org/archives/cairo/2017-July/028255.html | 1,610,957,295,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514423.60/warc/CC-MAIN-20210118061434-20210118091434-00697.warc.gz | 448,762,444 | 3,664 | # [cairo] Shouldn't Cairo use/offer degrees rather than radians?
Ray Gardener rayg at daylongraphics.com
Wed Jul 12 01:10:32 UTC 2017
```Why not leave the Cairo API alone and implement a utility library analogous
to GLU for OpenGL and that way handy degree-based transform functions
can be put there.
Ray
On 7/11/2017 3:42 PM, Bill Spitzak wrote:
> Your matrix for 45 degrees has equal values in all 4 entries. When
> this matrix is mulitplied by itself you get a matrix made of
> 0.9999999999999998 and 0.0. (That value is actually
> 1.0-2.220446049250313e-16).
>
> If you use the normal sin and cos values (which are unequal to each
> other) you get 1.0 and 2.220446049250313e-16 which is the same
> magnitude of error.
>
> However squaring 0.9999999999999998 gets a number further away from
> 1.0, while squaring 2.220446049250313e-16 goes closer to zero. For
> many operations (such as further multiplies, getting the determinant,
> etc) this means the second matrix is more accurate.
>
>>> You wanted two rotates by 45 degrees to be perfect.
>> Uh, no? Two rotates by 45 degrees illustrate a _compromise_. The
>> degree of rotation will generally be a perfect 90 degrees because of all
>> rotation matrix elements having the same magnitude, the total magnitude
>> (determinant of the scaling matrix I think) will lightly be slightly
>> more wrong than the magnitude of the radian API (which likely also fails
>> to be 1 due to sin/cos of M_PI/4 in floating point also being
>> approximations).
> As I tried to show above the answer current Cairo gets for pi/4 is
> better than the one you get for 45 degrees. Your function could be
> improved by using sin and cos rather than trying to do sin(pi/2-a) in
> place of cos in order to make sin and cos equal for 45 degrees.
>
>> Did you even read the patch and its rationale? Or are you making up
>> that straw man on the fly? Multiples of 90 degrees are perfect. There
>> are currently several fast paths in Cairo's code paths which actually
>> _check_ for that kind of perfection.
> I agree there are fast paths that are not allowing such errors in the
> matrix, and that changing cairo_rotate to detect angles near n*pi/2
> and produce 1.0 and 0.0 would probably be a much faster fix than
> trying to track down all the fast path mistakes. It would also remove
> suprises in the matrix for users.
>
>
>>> The exact same value is stored whether the rotation is sent as degrees
>> DID YOU LOOK AT THE PATCH????? I cannot believe you did when you state
>> this.
> Yes your patch in effect finds the quadrant the angle is in. It
> substitutes sin(pi/2-x) for cos(x) in order to make a right angle have
> a cos of 0.0, however as stated above I believe this is a mistake as
> the determinant of the matrix is not 1.0 for other angles.
>
> My version instead finds the closest axis (or the quadrant if you
> rotate by 45 degrees). The angle passed to sin/cos is in the range
> -pi/4..pi/4. This allows the normal sin/cos functions to be used. Here
> it is (in python sorry):
>
> from math import *
> M_PI_2 = pi/2
>
> # this is needed to avoid producing negative zero:
> def neg(x):
> return -x if x else 0.0
>
> def sincos(a):
> axis = round(a / M_PI_2)
> fromaxis = a - axis * M_PI_2
> s = sin(fromaxis)
> c = cos(fromaxis)
> x = int(axis) & 3
> if (x==0): return( s, c)
> elif (x==1): return( c, neg(s))
> elif (x==2): return(neg(s), neg(c))
> else: return(neg(c), s)
>
> If you think degrees would help then substitute 90 for M_PI_2 and put
> fromaxis*(M_PI_2/90) to the sin and cos function calls. Though
> intuitively it seems like 90*round(a/90) is somehow more accurate than
> M_PI_2*round(a/M_PI_2) I have not been able to find a value where this
> actually happens so I don't believe the degree version is necessary.
>
>>> 2. Add an api to rotate so the x axis passes through an x,y point (and
>>> possibly scales by hypot(x,y)). This would provide an "accurate"
>>> rotate for a whole lot of cases that actually come up in real
>>> graphics.
>> For multiples of 90 degrees, you can "trivially" just specify the
>> transform matrix, yet nobody does. This is not how people think, and we
>> are talking about an API for people.
> I propose cairo_rotate_atan(x) and cairo_rotate_atan2(x,y) which are
> exactly the same as cairo_rotate(atan(x)) and cairo_rotate(atan2(x,y))
> except for greater accuracy. The hope is that it is obvious how to fix
> code that is using atan to get the angle. You are right that no matter
> what happens, lots of code is going to continue to use
> cairo_rotate(M_PI/2), so I think fixing that function is a good idea.
``` | 1,296 | 4,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.911576 |
http://sumant2.blogspot.com/2006/03/ | 1,579,807,264,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250613416.54/warc/CC-MAIN-20200123191130-20200123220130-00352.warc.gz | 165,792,206 | 14,463 | ## Thursday, March 30, 2006
### Fourier Front and Descarte theorem
Today Dr. Kammler discussed some neat examples on Fourier Series. I really need to get my act together and study generalized limits. In Geometry I did go over the distance and Metric Geometry theorem. Yesterday Dr. Kocik went over the cool application of Descarte theorem in Geometry. He has been discussing some real interesting problem and the neat ways to solve them. I love it.
I didn't realize that the Thale's theorem in Geometry is about the right angle inside the semicircle. Its only when I was trying to search for the Descarte problem he did in the class that I realized that. Hopefully before the coming test on Wednesday I will be prepared enough.
I have a class in next 10 minute and have to collect my w2 from miles too. Other than that I am enjoying my fruit diet and hope I will be able to Juice fast and become a fruitarian for some time.
## Wednesday, March 29, 2006
### Inversion
The inversion in geometry is a cool concept and a more general concept than reflection across a plane. The figure here I drew using KIG explains what happens when inversion takes place. If there is a line outside the inverting circle its inverse becomes a circle passing through the centre. Also if there is a circle passing through the centre of the inverting circle it becomes a line. Look at how the quadrilateral behaves. So line changes to circle unless its passing through the centre in that case line remains line. Circle inverts to circle unless it is passing through the centre of the inverting circle.
## Tuesday, March 28, 2006
### Morning exercise, KIG installed
Today I got up at 5 am in the morning and I decided to go out and breathe some fresh air. I have been able to dramatically change a lot of my eating and sleeping habit in past two months. It feels great to incorporate more and more fruits in my diet and knowing that I am eating them right. Yes by right I mean early in the morning with empty stomach and later in the evening (3 hr aftre dinner). I am now having dinner at 18 hrs or before.
Installation of new software using YUM is so easy. Just for the fun I typed in "yum install kig" and behold I had the latest version of kig on my Linux machine. Some of you might be knowing that I tried unsuccessfully to install KIG using the rpm. Yum did a clean install plus host of other edutainment softwares from KDE education software. I have now a KDE hangman, a function plotter, a constellation viewer, a vocab builder, a periodic table plus other softwares.
I loved KIG and have to admit that its the best Geometer Software at present. Simply because it has all the features of other softwares plus conic section, differential geometry, Tangents to conics, Osculating Circle, inversion, test functions (like in Cabri Geometry), it even promises to read the KSEG document, though I couldn't get my *.sec files loaded and all menu driven. I tried the internal reflection problem using KIG and did it in no time. Look at the pictures they were exported as jpg files.
## Sunday, March 26, 2006
### Freedom of religion and Freedom from religion
Its funny to watch the Abdul Rahman's case in Afghanistan. The poor soul converted from Islam to Christianity 16 yrs ago. He faces a death sentence in his own country for insulting his former religion and the punishment for apostasy according to sharia law is death. This had me thinking what sharia laws have to say about apostasy and I came across this article on wikipedia about apostasy. Well the punishment is certainly death and this to any sane person would be laughable !! Just changing your religion could incite people to kill you ??? Recently Dalai Lama unequivocally said to be spiritual one doesn't need to be religious.
Now I read that Mr. Rahman is being released from the prison by only after the huge pressure from United states, Europe and non Islamic countries. These people found a loop hole and so now Mr. Rahman is being termed Demented and unable to stand trial. Going by what I am seeing on television I think Mr. Rahman should flee Afghanistan to some safe democratic and non Islamic country. But the sad truth is he will never be able to live in public anymore and chances of him being attacked by some religious zealot are very high.
### Dalai Lama's Interview on CNN
Yesterday I saw Dalai Lama's interview on CNN. I been to Dharamshala, India twice but never had the opportunity to visit Dalai Lama in person. So the interview on CNN was of interest to me after having seen one of his other interview on Palin's Himalayas series. Compared to other holy man Dalai Lama is very down to earth and that's what make him a great spiritual leader. On a question on Anger the lama replied that he has been working on it and is now a better person than he was 10 yrs, 20 yrs and so on. He is improving and getting more in control of it. Ask that to most other religious head and they will give you a pompous monologue. The one question which not many spiritual leaders of organized religions will not tell you straight about the relation between spirituality and religion and here Dalai Lama scored an ace when he looked straight into the eyes of the interviewer and said one don't need a religion to be a spiritual and repeated the same when asked twice.
### KSEG Rocks
After trying Cabri Geometry, GeoGebra and Geometer's Sektchpad, I feel KSEG is the best one with GeoGebra a close second. The reason KSEG stands out is
1. Its the fastest in drawing
2. Very intuitive and having the highlighted button features
3. ts free and robust on Linux
4. The points are easy to see and can be easily
chosen by the click of Mouse compared to other Geometric Softwares (GeoGebra is an exception)
GeoGebra is very close behind KSEG and it is just getting better. With applet feature and a promised funding it surely going to improve and become the number one choice for Geometry Softwares. Plus the best pt about GeoGebra is one doesn't need to install it. The figures you are looking is the solution to an internal reflection problem I solved using KSEG. I repeated that in GeoGebra but KSEG was faster. Meanwhile I have yet to take a look at the KIG but open software movement behind it and so I am sure it will be very good too.
## Thursday, March 16, 2006
### Cabri Geometry, Geogebra and Generalized Functions
I came across Geogebra recently and loved it. Its free and Java based. The software is pretty intuitive and it has incited my curiosity in to the fascinating Geometry softwares. Unlike Maple and Matlab here you just sketch. Geogebra is fascinating because it generates the algebra on the side window. I have Cabri Geometry on my TI Voyage 200. The good thing is all geometry softwares look quite similar. If you know one its easier to migrate to other and you know what to expect from the other software. I now have the full user manual of Cabri Geometry and would explore more. I also have it installed on my computer and it works fine. It has macro facility so I am doubly excited.
On Fourier front I learned some more generalized functions and the book has one neat example on constructing Schwartz function. The amazing thing about generalized function is one can transfer the operation on generalized function to Schwartz function and Schwartz function are so benign they take anything that is thrown to them. Their derivative is again a Schwartz function and its a pretty neat theory.
## Monday, March 13, 2006
### Two movies
Today I watched two movies. "Teesri Ankh" and "Dude where is the party". Teesri ankh was a below average movie and I had to skip. The Hindi movies still put the burden of chastity on women so much that you wonder if they truly reflect the changes in the society that is taking place. In a way these stereotypes help to reinforce a image which is dangerous and that's what is regrettable. The other movie "Dude where is the party" is ok but its length could easily be cut by at least another 40 minutes. It was meant to be a satire and I think it was pretty much on the dot. Also today in the morning I drove all the way to Giant City Park and in afternoon went to "Jewel of India restaurant".
## Sunday, March 12, 2006
### Learned Playing Uno
Today I learned playing Uno. The card game. My dad once told me that I will learn playing cards when I go to college. I didn't learn back in India so I am glad I learned it now finally. Uno is pretty simple and can get pretty interesting if your opponents play smartly. Tomorrow I should have my next digits of Pi crammed up taking it to 100.
## Friday, March 10, 2006
### Spring Break Begins
Its 15 hrs and the spring break has officially begun for me. I look forward to make the best use of it. Ofcourse getting hammered is not I would like to but then its spring break folks so you can excuse me if I do. Other than that I am pretty excited that now I could concentrate on Geometry too. Dr. Kocik has been giving interesting demonstrations and cutting down the Mobius stip was quite a revealation. I thought I knew what Mobius strip was until I did the cutting exercise. I was presumptious that I will get two pieces and when I didn't the aha.. did come out and its now pretty clear why it jarred the imagination of so many people. Also I found Dr. Parker's software for Poincare geometry and I look forward to try it out besides Dr. Kammler's Fourier Software. Other than that I am eagerly looking forward to learn the standard Generalized function, Dirac Delta, Comb, Inverse. In my plant biology class the teacher gave an extra 10 pts just for showing up !! That was pretty sweet after my take on Fourier exam. In a nutshell wonderful things to learn this week.
## Saturday, March 04, 2006
### Venus Fly Trap Video
Recently I came to know about Venus Fly Trap in my Plant Biology Class and it was fascinating to learn about this carnivorous or I think we should say an insectivorous plant. My lab had this plant along with Pitcher plant and Sun dew. However Venus Fly trap is unusual compared to these because there is a mechanical movement the closing of it leaves almost gives the impression that it is swallowing its prey. I search got me to this amazing website www.arkive.org and here is the video of Venus fly trap catching its prey.
### Unable to sleep
I am trying to sleep for past 1 and half hour but unable to do so. So I thought why not update my blog. Today I had a Plant Biology test and I only studied for last 3 hrs to cram and understand as much as possible. My entire focus has been Fourier Analysis for quite some time and I realize that I am ignoring too much my other class so this weekend besides Fourier I will be preparing for my other tests which are coming up next week. In Fourier we have started doing generalized function and some really cool concepts have been introduced like Schwartz function, Delta function, Comb function and their transforms.
In Geometry I still have to grapple with Ruler equation fortunately the speed of course is still manageable or may be its Dr. Kocik's way of teaching that makes it easy. Whatever it may be but I am looking forward to some very busy weeks ahead but then its fun to know that I am learning some great things and that is enough to stir the enthusiasm.
## Wednesday, March 01, 2006
### Update
I just came back from submitting my Philosophy homework (PHI -105). It was a lot of procrastination that almost got me in trouble. But ever since I left my job I have now more time to concentrate on my classes and it feels great. Dr. Kammler has been very encouraging through out his course so far. I like his energy of communicating and as well as appreciating the level of students in his class. His modus of using scritchy scratch to get away with trivial calculations in lecture and concentrating on big picture is very healthy. It keeps you focus on his lecture and thereby covering lot more material than a regular course does. I am especially enjoying this Fourier class than any other class at this moment. May be one reason is that I am not been able to spend enough time for other classes and its high time that I should start doing as midterms are coming up.
Yesterday I went out with Ayako and Pui to Thai taste. It was almost a year since the last I visited. This time I ordered 13 on their spiciness level and they have come to appreciate that their 10 (which used to be max) is still too sweet for people in Carbondale. I relish the Red Curry at 13 level, though it failed to get any expression of ohh--hot from my mouth. Its only in the morning that I realized that it had upset my stomach. I am now feeling good. But I guess a month of my wholesome fruit diet has encouraged me very much and I plan to continue with it now. The vitamin content of Cantaloupe was a surprise ! Never thought that Muskmelon packed so many healthy vitamins and minerals. | 2,825 | 12,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-05 | latest | en | 0.961397 |
http://mathhelpforum.com/algebra/1864-converting-polar-equations-rectangular-vice-versa-print.html | 1,527,139,609,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865913.52/warc/CC-MAIN-20180524033910-20180524053910-00394.warc.gz | 187,150,147 | 3,215 | # Converting Polar Equations to Rectangular and vice versa...
• Feb 12th 2006, 12:11 PM
telefl89
Converting Polar Equations to Rectangular and vice versa...
I am having some difficulty with doing this. I am in a Pre-Calc class at my high school. My problem is, I was horrible at solving for identities, and now I am struggling with converting polar equations to rectangular and vice versa. An example of a polar equation I am trying to convert to x and y coordinates is r = 3/(3 - cos). I apologize, I don't know how to enter in the symbol for theta. An equation I am trying to convert to r and theta is 4(x^2)y = 1. Can someone please help explain this to me?
• Feb 12th 2006, 04:40 PM
topsquark
Quote:
Originally Posted by telefl89
I am having some difficulty with doing this. I am in a Pre-Calc class at my high school. My problem is, I was horrible at solving for identities, and now I am struggling with converting polar equations to rectangular and vice versa. An example of a polar equation I am trying to convert to x and y coordinates is r = 3/(3 - cos). I apologize, I don't know how to enter in the symbol for theta. An equation I am trying to convert to r and theta is 4(x^2)y = 1. Can someone please help explain this to me?
Formally, to convert from rectangular to polar what you need to do is set $\displaystyle x = r*cos\theta$ and $\displaystyle y = r*sin\theta$.
So, given $\displaystyle 4(x^2)y = 1$ we get $\displaystyle 4(r*cos\theta)^2*r*sin\theta=4r^3cos^2\theta sin\theta = 1$.
Thus, your polar equation is $\displaystyle r=(1/4*cos^2\theta sin\theta)^{1/3}$.
Converting from polar to rectangular can be a tad messier. Using the transformations $\displaystyle x = r*cos\theta$ and $\displaystyle y = r*sin\theta$ we can solve for r and $\displaystyle \theta$:
$\displaystyle r=(x^2+y^2)^{1/2}$ and $\displaystyle \theta=tan^{-1}(y/x)$.
So, $\displaystyle r = 3/(3 - cos\theta)$ becomes $\displaystyle (x^2+y^2)^{1/2}=3/(3-cos[tan^{-1}(y/x)])$.
What is $\displaystyle cos[tan^{-1}(y/x)]$? Remember, the argument of cos[ ] is just an angle. Call it $\displaystyle \alpha$. $\displaystyle \alpha$ is just the angle such that $\displaystyle tan\alpha=y/x$. What is the cosine of this angle? Well, it's just $\displaystyle cos\alpha = x/(x^2+y^2)^{1/2}$. So $\displaystyle cos[tan^{-1}(y/x)]=x/(x^2+y^2)^{1/2}$.
Your equation now reads: $\displaystyle (x^2+y^2)^{1/2}=3/(3-x/(x^2+y^2)^{1/2})$. There's a bit of simplifying left to do, but it isn't hard, just busy. (Hint: You can't really come up with a "clean" solution for y, so put it in the form of a general conic section:$\displaystyle ax^2+bx+cy^2+dy+e=0$. I found that it fits the form for an ellipse.)
-Dan | 809 | 2,693 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-22 | latest | en | 0.870494 |
https://forum.allaboutcircuits.com/threads/electronic-puzzle-cube.11420/ | 1,569,177,676,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575627.91/warc/CC-MAIN-20190922180536-20190922202536-00441.warc.gz | 465,054,758 | 23,466 | # Electronic Puzzle Cube
Discussion in 'Math' started by Wendy, May 6, 2008.
1. ### Wendy Thread Starter Moderator
Mar 24, 2008
21,823
3,039
This is something my electronics instructor threw out over 30 years ago. There is a cheating method, and the real solution. I figured it out using assumptions in my math, but never really worked it out. Thought some of you guys might enjoy it. He claimed the one correct solution used 8 sheets of paper and network analysis. Some of my fellow students did it with real resistors, which really was cheating. It was a free grade for the course, guess I was close enough because he gave it to me.
The Problem:
All resistors are 10 ohms. What is the resistance between A and B?
BTW, I don't have the answer to this, though I remember my answer. Is that evil or what?
Last edited: May 7, 2008
2. ### Caveman Senior Member
Apr 15, 2008
471
1
You don't need 8 sheets of paper, just a sharp mind. I can show it in 1/2 page max. And it isn't trickery, just logic.
3. ### Wendy Thread Starter Moderator
Mar 24, 2008
21,823
3,039
Looking forward to seeing how you do it. Wouldn't be a puzzle if there wasn't a catch though. When you come up with a solution please PM it to me, and well see if it attracts anyone elses ire.
Last edited: May 7, 2008
4. ### recca02 Senior Member
Apr 2, 2007
1,211
1
I might have made mistake in calculation, but check if my approach was correct. I used star-delta transformations.
That answer holds true if C,D,E are equi-potential points. At a glance they did seem to me.
File size:
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5. ### Dave Retired Moderator
Nov 17, 2003
6,960
171
Nice method recca. I just did it by brute-force! Actually if you pick the symmetry you can get a relatively simple resistor topology and the answer pops out the same as yours.
Dave
6. ### hgmjr Retired Moderator
Jan 28, 2005
9,029
219
I believe I know the shortcut to which Bill is referring. I remember this problem from a the electrical section in a Physics course.
hgmjr
7. ### Wendy Thread Starter Moderator
Mar 24, 2008
21,823
3,039
Interesting, I'd never seen that approach. If you want to do a check there are actually 2 different sets of points with equal potentials, which is how I approached it. When I worked it the first thing I did was flatten out the schematic, which didn't help nearly so much as I'd hoped it would.
I'd also forgotten about computers and SPICE simulators, silly me. We didn't have anything like that 30 years ago.
8. ### Dave Retired Moderator
Nov 17, 2003
6,960
171
That is what I did, and I ended up with a series arrangement of 3 parallel sets of resistors. It is only when you look at the cube symmetrically that it becomes clear how to do this.
Dave
9. ### Wendy Thread Starter Moderator
Mar 24, 2008
21,823
3,039
I wonder is a tessaract cube would be harder? It is possible to diagram a 3d model of a 4D cube. It's distorted, but much in the same way the 2D representation of 3D is. What do you think, even worth drawing up?
The other set of equal potential points is exactly symmetrical to the first set. Symmetry is the hook on this problem. I have to wonder if any other instructors will use this for their classes. With SPICE probably not, but it is a lot more fun when you have to do it by hand.
Last edited: May 7, 2008
10. ### Wendy Thread Starter Moderator
Mar 24, 2008
21,823
3,039
Just had an offline conversation with Caveman, if you really want to make it hard start with 10 ohms and increase each resistor by 1 ohm, so that every resistor is different. Of course, it would leave me out on solving it, I needed the symmetry dagnabit!
Apr 21, 2008
25
0
12. ### Wendy Thread Starter Moderator
Mar 24, 2008
21,823
3,039
Actually, if you go through the problem, it is a lot more complex than that. If all the resistors are different it's not even close.
Delta to Y and visa versa have to do with conversions, making two different circuits that look identical to each other as seen from the outside world. From a black box point of view (where you can not see how they are wired) there is no telling them apart. This is important for later concepts, namely attenuators.
The cube circuit is a network, and if the parts are all the same, it repeats 3 times. Where it gets interesting is they are all intertwined, and hard to separate from each other.
--------
Looking at it again you may have a point. I'll have to play with it and see. Sorry about that.
Last edited: May 8, 2008
13. ### flubbo Member
Apr 21, 2008
25
0
The big problem is that the circuit looks a lot more scary than it actually is.
These kind of problems are used to teach students how to visualize different circuit topologies. The trick is to give designators to the resistors. That takes some of the confusion away from the circuit. Then, if you look at it like a mesh, (ie. remove the 3d effect) it becomes easier to understand.
Take a look here. They have a Java simulator that demonstrates the circuit you have posted, complete with a voltage source.
http://server.oersted.dtu.dk/personal/ldn/javalab/Circuit09.html
Best,
:Flubbo.
14. ### Wendy Thread Starter Moderator
Mar 24, 2008
21,823
3,039
The dangers of sleep posting, what can I say? Recca02 actually posted this way back, post #3 I think, but I missed the reference.
15. ### flubbo Member
Apr 21, 2008
25
0
@Bill_Marsden: No explanation necessary.
16. ### Dave Retired Moderator
Nov 17, 2003
6,960
171
As a curiosity, what was Caveman's method for doing this?
Dave
17. ### Wendy Thread Starter Moderator
Mar 24, 2008
21,823
3,039
Symmetry of the cube, I'll forward you the PM if I can. He'll post it if he wants. I may get ambitious and try doing the non-repeating resistor scheme while using the Delta conversions. I honestly don't know if that would work or not. After this dead horse stops wiggling I'll probably do a hypercube just for fun.
Wonder how many students we'll make miserable in the long run with this?
18. ### Caveman Senior Member
Apr 15, 2008
471
1
If you take a 6-sided die, hold it between your fingers at points A and B, and allow it to rotate on that pivot, you will notice that ignoring the dots, there are three positions that look identical in position.
In these three positions, the three corners that are only one edge away from A will be switching places, and the three corners that are only one edge away from B will be switching places.
Since all of the edges have the same resistance, rotating the cube in this manner generates the exact same schematic.
This means that if you apply any generic voltage V between A and B, the first 3 points will have the same voltage as one another, and the second 3 points will have the same voltage as one another.
Since this is true, you can connect these equivalent points without changing the characteristics of the circuit. (Similar to the logic of a balanced bridge circuit.)
And you result in 3 Resistors in parallel + 6 resistors in parallel + 3 Resistors in parallel.
Result = (1/3 + 1/6 + 1/3)*R = 5/6*R
This answer does just require simple logic, but it does provide some examples of good circuit theory:
1. If the schematic is the same, it is the same circuit. Therefore, if the circuit acts differently, it has a different schematic. (Usually parasitics or a soldering mistake.)
2. It illustrates the bridge circuit concept.
3. It shows how analogies to other concepts can be used to help understand electronics. The cube shape actually simplifies the analysis by bringing the power of geometry into play.
19. ### Caveman Senior Member
Apr 15, 2008
471
1
This circuit has one tricky thing about it that we never think about when starting our analysis. Apparently, mesh analysis doesn't work, because the circuit is not planar. So, after many deadends, I simply went with a nodal analysis.
My $\sout{madness}\;method:$
1. I calculated the KCL equation for each of the 6 unknown nodes.
2. I drove node A to 1V and node B to ground.
3. Using Excel, I solved the equations for V1 to V6.
4. The currents through R1, R2, and R3 total to the current from the 1V source.
5. Using ohms law, I calculated the equivalent resistance.
Included is the Excel spreadsheet. And just in case you don't like opening other peoples spreadsheets, I included a pdf to see the result.
Enjoy!
File size:
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20. ### Dave Retired Moderator
Nov 17, 2003
6,960
171
Caveman, you approached the initial question the same as I did (however explained it more elequantly than I would have ). Like the Excel analysis.
There are hundreds of different analyses that you can do on this cube: e.g. calculating the resistance between any adjacent node for any arbitrary resistor values (its in the symmetry again).
Also try a forum search for the calculate the resistance between two nodes on an infinite resistor network.
Dave | 2,254 | 8,874 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 1, "/images/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-2019-39 | latest | en | 0.954089 |
https://www.ademcetinkaya.com/2023/04/lonpier-brighton-pier-group-plc-the.html | 1,685,692,888,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648465.70/warc/CC-MAIN-20230602072202-20230602102202-00383.warc.gz | 685,059,654 | 60,780 | Outlook: BRIGHTON PIER GROUP PLC (THE) is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Buy
Time series to forecast n: 26 Apr 2023 for (n+8 weeks)
Methodology : Supervised Machine Learning (ML)
## Abstract
BRIGHTON PIER GROUP PLC (THE) prediction model is evaluated with Supervised Machine Learning (ML) and Stepwise Regression1,2,3,4 and it is concluded that the LON:PIER stock is predictable in the short/long term. According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Buy
## Key Points
1. What is neural prediction?
2. What is prediction in deep learning?
3. Is Target price a good indicator?
## LON:PIER Target Price Prediction Modeling Methodology
We consider BRIGHTON PIER GROUP PLC (THE) Decision Process with Supervised Machine Learning (ML) where A is the set of discrete actions of LON:PIER stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Stepwise Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Supervised Machine Learning (ML)) X S(n):→ (n+8 weeks) $\stackrel{\to }{S}=\left({s}_{1},{s}_{2},{s}_{3}\right)$
n:Time series to forecast
p:Price signals of LON:PIER stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## LON:PIER Stock Forecast (Buy or Sell) for (n+8 weeks)
Sample Set: Neural Network
Stock/Index: LON:PIER BRIGHTON PIER GROUP PLC (THE)
Time series to forecast n: 26 Apr 2023 for (n+8 weeks)
According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Buy
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
## IFRS Reconciliation Adjustments for BRIGHTON PIER GROUP PLC (THE)
1. An entity shall apply the amendments to IFRS 9 made by IFRS 17 as amended in June 2020 retrospectively in accordance with IAS 8, except as specified in paragraphs 7.2.37–7.2.42.
2. When identifying what risk components qualify for designation as a hedged item, an entity assesses such risk components within the context of the particular market structure to which the risk or risks relate and in which the hedging activity takes place. Such a determination requires an evaluation of the relevant facts and circumstances, which differ by risk and market.
3. A single hedging instrument may be designated as a hedging instrument of more than one type of risk, provided that there is a specific designation of the hedging instrument and of the different risk positions as hedged items. Those hedged items can be in different hedging relationships.
4. As noted in paragraph B4.3.1, when an entity becomes a party to a hybrid contract with a host that is not an asset within the scope of this Standard and with one or more embedded derivatives, paragraph 4.3.3 requires the entity to identify any such embedded derivative, assess whether it is required to be separated from the host contract and, for those that are required to be separated, measure the derivatives at fair value at initial recognition and subsequently. These requirements can be more complex, or result in less reliable measures, than measuring the entire instrument at fair value through profit or loss. For that reason this Standard permits the entire hybrid contract to be designated as at fair value through profit or loss.
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
## Conclusions
BRIGHTON PIER GROUP PLC (THE) is assigned short-term Ba1 & long-term Ba1 estimated rating. BRIGHTON PIER GROUP PLC (THE) prediction model is evaluated with Supervised Machine Learning (ML) and Stepwise Regression1,2,3,4 and it is concluded that the LON:PIER stock is predictable in the short/long term. According to price forecasts for (n+8 weeks) period, the dominant strategy among neural network is: Buy
### LON:PIER BRIGHTON PIER GROUP PLC (THE) Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementCBaa2
Balance SheetBaa2C
Leverage RatiosCaa2C
Cash FlowCaa2C
Rates of Return and ProfitabilityBa1Baa2
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
### Prediction Confidence Score
Trust metric by Neural Network: 81 out of 100 with 480 signals.
## References
1. Chernozhukov V, Escanciano JC, Ichimura H, Newey WK. 2016b. Locally robust semiparametric estimation. arXiv:1608.00033 [math.ST]
2. Andrews, D. W. K. (1993), "Tests for parameter instability and structural change with unknown change point," Econometrica, 61, 821–856.
3. M. J. Hausknecht. Cooperation and Communication in Multiagent Deep Reinforcement Learning. PhD thesis, The University of Texas at Austin, 2016
4. Breiman L. 1996. Bagging predictors. Mach. Learn. 24:123–40
5. Challen, D. W. A. J. Hagger (1983), Macroeconomic Systems: Construction, Validation and Applications. New York: St. Martin's Press.
6. V. Borkar. Stochastic approximation: a dynamical systems viewpoint. Cambridge University Press, 2008
7. Bai J, Ng S. 2002. Determining the number of factors in approximate factor models. Econometrica 70:191–221
Frequently Asked QuestionsQ: What is the prediction methodology for LON:PIER stock?
A: LON:PIER stock prediction methodology: We evaluate the prediction models Supervised Machine Learning (ML) and Stepwise Regression
Q: Is LON:PIER stock a buy or sell?
A: The dominant strategy among neural network is to Buy LON:PIER Stock.
Q: Is BRIGHTON PIER GROUP PLC (THE) stock a good investment?
A: The consensus rating for BRIGHTON PIER GROUP PLC (THE) is Buy and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of LON:PIER stock?
A: The consensus rating for LON:PIER is Buy.
Q: What is the prediction period for LON:PIER stock?
A: The prediction period for LON:PIER is (n+8 weeks)
## People also ask
What are the top stocks to invest in right now? | 1,807 | 7,149 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-23 | latest | en | 0.798939 |
https://docs.opencv.org/3.1.0/d4/d1f/tutorial_pyramids.html | 1,718,242,179,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861319.37/warc/CC-MAIN-20240612234213-20240613024213-00222.warc.gz | 188,718,422 | 6,860 | OpenCV 3.1.0 Open Source Computer Vision
Image Pyramids
## Goal
In this tutorial you will learn how to:
## Theory
Note
The explanation below belongs to the book Learning OpenCV by Bradski and Kaehler.
• Usually we need to convert an image to a size different than its original. For this, there are two possible options:
1. Upsize the image (zoom in) or
2. Downsize it (zoom out).
• Although there is a geometric transformation function in OpenCV that -literally- resize an image (cv::resize , which we will show in a future tutorial), in this section we analyze first the use of Image Pyramids, which are widely applied in a huge range of vision applications.
### Image Pyramid
• An image pyramid is a collection of images - all arising from a single original image - that are successively downsampled until some desired stopping point is reached.
• There are two common kinds of image pyramids:
• Gaussian pyramid: Used to downsample images
• Laplacian pyramid: Used to reconstruct an upsampled image from an image lower in the pyramid (with less resolution)
• In this tutorial we'll use the Gaussian pyramid.
#### Gaussian Pyramid
• Imagine the pyramid as a set of layers in which the higher the layer, the smaller the size.
• Every layer is numbered from bottom to top, so layer $$(i+1)$$ (denoted as $$G_{i+1}$$ is smaller than layer $$i$$ ( $$G_{i}$$).
• To produce layer $$(i+1)$$ in the Gaussian pyramid, we do the following:
• Convolve $$G_{i}$$ with a Gaussian kernel:
$\frac{1}{16} \begin{bmatrix} 1 & 4 & 6 & 4 & 1 \\ 4 & 16 & 24 & 16 & 4 \\ 6 & 24 & 36 & 24 & 6 \\ 4 & 16 & 24 & 16 & 4 \\ 1 & 4 & 6 & 4 & 1 \end{bmatrix}$
• Remove every even-numbered row and column.
• You can easily notice that the resulting image will be exactly one-quarter the area of its predecessor. Iterating this process on the input image $$G_{0}$$ (original image) produces the entire pyramid.
• The procedure above was useful to downsample an image. What if we want to make it bigger?: columns filled with zeros ( $$0$$)
• First, upsize the image to twice the original in each dimension, wit the new even rows and
• Perform a convolution with the same kernel shown above (multiplied by 4) to approximate the values of the "missing pixels"
• These two procedures (downsampling and upsampling as explained above) are implemented by the OpenCV functions cv::pyrUp and cv::pyrDown , as we will see in an example with the code below:
Note
When we reduce the size of an image, we are actually losing information of the image.
## Code
This tutorial code's is shown lines below. You can also download it from here
#include <math.h>
#include <stdlib.h>
#include <stdio.h>
using namespace cv;
Mat src, dst, tmp;
const char* window_name = "Pyramids Demo";
int main( void )
{
printf( "\n Zoom In-Out demo \n " );
printf( "------------------ \n" );
printf( " * [u] -> Zoom in \n" );
printf( " * [d] -> Zoom out \n" );
printf( " * [ESC] -> Close program \n \n" );
if( src.empty() )
{ printf(" No data! -- Exiting the program \n");
return -1; }
tmp = src;
dst = tmp;
namedWindow( window_name, WINDOW_AUTOSIZE );
imshow( window_name, dst );
for(;;)
{
int c;
c = waitKey(10);
if( (char)c == 27 )
{ break; }
if( (char)c == 'u' )
{ pyrUp( tmp, dst, Size( tmp.cols*2, tmp.rows*2 ) );
printf( "** Zoom In: Image x 2 \n" );
}
else if( (char)c == 'd' )
{ pyrDown( tmp, dst, Size( tmp.cols/2, tmp.rows/2 ) );
printf( "** Zoom Out: Image / 2 \n" );
}
imshow( window_name, dst );
tmp = dst;
}
return 0;
}
## Explanation
Let's check the general structure of the program:
• Load an image (in this case it is defined in the program, the user does not have to enter it as an argument)
if( !src.data )
{ printf(" No data! -- Exiting the program \n");
return -1; }
• Create a Mat object to store the result of the operations (dst) and one to save temporal results (tmp).
Mat src, dst, tmp;
/* ... */
tmp = src;
dst = tmp;
• Create a window to display the result
namedWindow( window_name, WINDOW_AUTOSIZE );
imshow( window_name, dst );
• Perform an infinite loop waiting for user input.
while( true )
{
int c;
c = waitKey(10);
if( (char)c == 27 )
{ break; }
if( (char)c == 'u' )
{ pyrUp( tmp, dst, Size( tmp.cols*2, tmp.rows*2 ) );
printf( "** Zoom In: Image x 2 \n" );
}
else if( (char)c == 'd' )
{ pyrDown( tmp, dst, Size( tmp.cols/2, tmp.rows/2 ) );
printf( "** Zoom Out: Image / 2 \n" );
}
imshow( window_name, dst );
tmp = dst;
}
Our program exits if the user presses ESC. Besides, it has two options:
• Perform upsampling (after pressing 'u')
pyrUp( tmp, dst, Size( tmp.cols*2, tmp.rows*2 )
We use the function cv::pyrUp with 03 arguments:
• tmp: The current image, it is initialized with the src original image.
• dst: The destination image (to be shown on screen, supposedly the double of the input image)
• *Size( tmp.cols*2, tmp.rows*2 )* : The destination size. Since we are upsampling, cv::pyrUp expects a size double than the input image (in this case tmp).
• Perform downsampling (after pressing 'd')
pyrDown( tmp, dst, Size( tmp.cols/2, tmp.rows/2 )
Similarly as with cv::pyrUp , we use the function cv::pyrDown with 03 arguments:
• tmp: The current image, it is initialized with the src original image.
• dst: The destination image (to be shown on screen, supposedly half the input image)
• Size( tmp.cols/2, tmp.rows/2 ) : The destination size. Since we are upsampling, cv::pyrDown expects half the size the input image (in this case tmp).
• Notice that it is important that the input image can be divided by a factor of two (in both dimensions). Otherwise, an error will be shown.
• Finally, we update the input image tmp with the current image displayed, so the subsequent operations are performed on it.
tmp = dst;
## Results
• After compiling the code above we can test it. The program calls an image chicky_512.jpg that comes in the tutorial_code/image folder. Notice that this image is $$512 \times 512$$, hence a downsample won't generate any error ( $$512 = 2^{9}$$). The original image is shown below:
• First we apply two successive cv::pyrDown operations by pressing 'd'. Our output is:
• Note that we should have lost some resolution due to the fact that we are diminishing the size of the image. This is evident after we apply cv::pyrUp twice (by pressing 'u'). Our output is now: | 1,715 | 6,290 | {"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} | 3.703125 | 4 | CC-MAIN-2024-26 | latest | en | 0.826899 |
http://www.let.rug.nl/~gosse/termpedia2/termpedia.php?language=dutch_general&density=7&link_color=000000&termpedia_system=perl_db&url=http%3A%2F%2Fen.wikipedia.org%2Fw%2Findex.php%3Ftitle%3D220_%28number%29%26amp%3Baction%3Dedit%26amp%3Bsection%3D4 | 1,571,770,454,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987823061.83/warc/CC-MAIN-20191022182744-20191022210244-00429.warc.gz | 271,628,222 | 27,707 | # 220 (number)
← 219 220 221 →
Cardinaltwo hundred twenty
Ordinal220th
(two hundred twentieth)
Factorization22 × 5 × 11
Greek numeralΣΚ´
Roman numeralCCXX
Binary110111002
Ternary220113
Quaternary31304
Quinary13405
Senary10046
Octal3348
Duodecimal16412
VigesimalB020
Base 366436
220 (two hundred [and] twenty) is the natural number following 219 and preceding 221.
## In mathematics
It is a composite number, with its divisors being 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110, making it an amicable number with 284.[1][2] Every number up to 220 may be expressed as a sum of its divisors, making 220 a practical number.[3] Also, being divisible by the sum of its digits, 220 is a Harshad number.[4]
It is the sum of four consecutive primes (47 + 53 + 59 + 61).[5] It is the smallest even number with the property that when represented as a sum of two prime numbers (per Goldbach's conjecture) both of the primes must be greater than or equal to 23.[6] There are exactly 220 different ways of partitioning 64 = 82 into a sum of square numbers.[7]
It is a tetrahedral number, the sum of the first ten triangular numbers,[8] and a dodecahedral number.[9] If all of the diagonals of a regular decagon are drawn, the resulting figure will have exactly 220 regions.[10]
It is the sum of the sums of the divisors of the first 16 positive integers.[11]
## In other fields
The number 220 can also refer to:
## Notes
1. ^ Bryan Bunch, The Kingdom of Infinite Number. New York: W. H. Freeman & Company (2000): 167
2. ^ Higgins, Peter (2008). Number Story: From Counting to Cryptography. New York: Copernicus. p. 61. ISBN 978-1-84800-000-1.
3. ^ Sloane, N. J. A. (ed.). "Sequence A005153 (Practical numbers)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
4. ^
5. ^ Sloane, N. J. A. (ed.). "Sequence A034963 (Sums of four consecutive primes)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
6. ^
7. ^ Sloane, N. J. A. (ed.). "Sequence A037444 (Number of partitions of n^2 into squares)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
8. ^ Sloane, N. J. A. (ed.). "Sequence A000292 (Tetrahedral (or triangular pyramidal) numbers)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
9. ^ Sloane, N. J. A. (ed.). "Sequence A006566 (Dodecahedral numbers)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
10. ^
11. ^ | 719 | 2,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2019-43 | latest | en | 0.785328 |
http://exrx.net/forum/viewtopic.php?f=7&t=9925&sid=304cad1b6a2b71127bf1dbbb2344ede6&view=print | 1,512,991,218,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948513478.11/warc/CC-MAIN-20171211105804-20171211125804-00012.warc.gz | 99,527,305 | 3,400 | ExRx.nethttp://exrx.net/forum/ How to get proper exercise?http://exrx.net/forum/viewtopic.php?f=7&t=9925 Page 1 of 1
Author: simanto [ Wed Jan 06, 2016 12:17 am ] Post subject: How to get proper exercise? HiIndeed, i want to need proper exercise help tips. Day bay my fat is increasing very harmfully, so i need to burn my fat as soon as possible. If any one know about effective tips about weight loss program or exercise program, let's me know please.
Author: stuward [ Thu Jan 07, 2016 11:25 am ] Post subject: Re: How to get proper exercise? That's too broad a topic to be handled in one post. However, correct your diet, reducing calorie intake while keeping adequate protein intake. That makes the most difference to reducing fat. Next work on improving your strength and other fitness attributes. This will ensure the fat stays off. Read more on this site. What you are looking for is there somewhere.
Author: aliusman [ Tue Feb 23, 2016 10:40 pm ] Post subject: Re: How to get proper exercise? Use BMR (Basal Metabolic Rate) Calculator to calculate your resting calories and then go from there. Make sure you have enough protein going in as STU advised above. Set a goal, a target and reasonable time-frame to achieve that target. You should be fine :)
Author: Kenny Croxdale [ Fri Mar 25, 2016 7:41 am ] Post subject: Re: How to get proper exercise? aliusman wrote:Use BMR (Basal Metabolic Rate) Calculator to calculate your resting calories and then go from there.BMR Calculators Who ever came up with BMR Calculations should be horse whipped, then shot.The room for error is enormous! It amount to using a GPS system that determines your global position is somewhere in North America when you're lost.Thus, you "Resting Caloric Expenditure" like the GPS analogy above is useless. Three Day RecallThis method provide you with a more effective method of determining what your "Daily Caloric Expenditure" is. 1) Count Calories for three days.2) Divide the three day calorie intake by three days. 3) One of the three days measured need to be a weekend day, where you eating habits change.What It Determines1) If you are gaining weight, the Three Day Recall tells you that you are taking in a surplus of calories.2) If you are losing weight, it tell you that you are using more calories that you are consuming.3) If your weight remains constant, that your burning the exact number of calories you are consuming.Based On ThatModify you caloric intake to gain or gain weight. Do nothing to maintain weight. Quote:Make sure you have enough protein going in as STU advised above. Agreed...Muscle Protein SynthesisMuscle Protein Synthesis = Maintaining and/or Increasing Muscle Mass. Research shows (Drs Donald Layman and Layne Norton) that the minimal amount of protein per meal for Muscle Protein Synthesis is 30 gram of Quality Protein for those under 40 years old.Those over 40 need over 40 grams of Quality Protein per serving. Kenny Croxdale
Author: Stephen123 [ Wed Jul 20, 2016 7:49 am ] Post subject: Re: How to get proper exercise? Swimming, running, cycling, energetic strolling, and so forth., are known as best cardiovascular activities that lose stomach fat. You can perform all of these activities for 20-30 minutes consistently to dispose of tummy fat and abundance fat in your body. You can't lose paunch fat by doing straight exercises.TragusInfantigoRx
Author: Alexa1994 [ Mon Apr 10, 2017 7:09 am ] Post subject: Re: How to get proper exercise? Kenny Croxdale wrote:aliusman wrote:Use BMR (Basal Metabolic Rate) Calculator to calculate your resting calories and then go from there.BMR Calculators Who ever came up with BMR Calculations should be horse whipped, then shot.The room for error is enormous! It amount to using a GPS system that determines your global position is somewhere in North America when you're lost.Thus, you "Resting Caloric Expenditure" like the GPS analogy above is useless. Three Day RecallThis method provide you with a more effective method of determining what your "Daily Caloric Expenditure" is. 1) Count Calories for three days.2) Divide the three day calorie intake by three days. 3) One of the three days measured need to be a weekend day, where you eating habits change.What It Determines1) If you are gaining weight, the Three Day Recall tells you that you are taking in a surplus of calories.2) If you are losing weight, it tell you that you are using more calories that you are consuming.3) If your weight remains constant, that your burning the exact number of calories you are consuming.Based On ThatModify you caloric intake to gain or gain weight. Do nothing to maintain weight. Quote:Make sure you have enough protein going in as STU advised above. Agreed...Muscle Protein SynthesisMuscle Protein Synthesis = Maintaining and/or Increasing Muscle Mass. Research shows (Drs Donald Layman and Layne Norton) that the minimal amount of protein per meal for Muscle Protein Synthesis is 30 gram of Quality Protein for those under 40 years old.Those over 40 need over 40 grams of Quality Protein per serving. Kenny Croxdale
Page 1 of 1 All times are UTC - 6 hours [ DST ] Powered by phpBB® Forum Software © phpBB Grouphttp://www.phpbb.com/ | 1,218 | 5,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-51 | latest | en | 0.923171 |
https://nerdutella.com/q270-Given-a-sequence-of-numbers-you-have-to-find-the-maximum-sum-alternating-subsequence-and-print-the-value | 1,679,825,003,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945440.67/warc/CC-MAIN-20230326075911-20230326105911-00749.warc.gz | 492,466,302 | 11,110 | Q:
# Find the maximum sum alternating subsequence
Given a sequence of numbers, you have to find the maximum sum alternating subsequence and print the value. A sequence is an alternating sequence when it will be maintain like (increasing) -> (decreasing) ->(increasing) ->(decreasing).
```Input:
T Test case
T no. of input string will be given to you.
E.g.
3
2 3 4 8 2 5 6 8
2 3 4 8 2 6 5 4
6 5 9 2 10 77 5
Constrain:
1≤ A[i] ≤50
Output:
Print the value of maximum sum alternating subsequence.
```
Example
```T=3
Input:
2 3 4 8 2 5 6 8
Output:
22 ( 8+6+8)
Input:
2 3 4 8 2 6 5 4
Output:
20 ( 8+ 2+ 6+ 4)
Input:
6 5 9 2 10 77 5
Output:
98 (5+ 9+ 2+ 77+5)```
Let N be the number of elements say, X1, X2, X3, ..., Xn
Let f(a) = the value at the index a of the increasing array, and g(a) = the value at the index a of the decreasing array.
To find out the maximum sum alternating sequence we will follow these steps,
1. We take two new arrays, one is increasing array and another is decreasing array and initialize it with 0. We start our algorithm with the second column. We check elements that are before the current element, with the current element.
2. If any element is less than the current element then,
f(indexofthecurrentelement) = max
3. If the element is greater than the current element then,
g(indexofthecurrentelement) = max
C++ Implementation:
``````#include <bits/stdc++.h>
using namespace std;
int sum(int* arr, int n)
{
int inc[n + 1], dec[n + 1];
inc[0] = arr[0];
dec[0] = arr[0];
memset(inc, 0, sizeof(inc));
memset(dec, 0, sizeof(dec));
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
if (arr[j] > arr[i]) {
dec[i] = max(dec[i], inc[j] + arr[i]);
}
else if (arr[i] > arr[j]) {
inc[i] = max(inc[i], dec[j] + arr[i]);
}
}
}
return max(inc[n - 1], dec[n - 1]);
}
int main()
{
int t;
cout << "Test Case : ";
cin >> t;
while (t--) {
int n;
cout << "Number of element : ";
cin >> n;
int arr[n];
cout << "Enter the elements : ";
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
cout << "Sum of the alternating sequence : " << sum(arr, n) << endl;
}
return 0;
}``````
Output
```Test Case : 3
Number of element : 8
Enter the elements : 2 3 4 8 2 5 6 8
Sum of the alternating sequence : 22
Number of element : 8
Enter the elements : 2 3 4 8 2 6 5 4
Sum of the alternating sequence : 20
Number of element : 7
Enter the elements : 6 5 9 2 10 77 5
Sum of the alternating sequence : 98``` | 838 | 2,431 | {"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} | 3.4375 | 3 | CC-MAIN-2023-14 | longest | en | 0.566488 |
https://www.orekit.org/site-orekit-tutorials-10.2/tutorials/propagation-in-non-inertial-frame.html | 1,603,803,303,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107894175.55/warc/CC-MAIN-20201027111346-20201027141346-00269.warc.gz | 833,766,921 | 6,821 | # Propagation in non-inertial Frame
The goal of this tutorial is to introduce orbital integration using `SingleBodyAttraction` class. This class can replace all the different kinds of point mass interactions (`ThirdBodyAttraction`, `NewtonianAttraction`). Using `SingleBodyAttraction` and `InertialForces` enables a richer modelling, allowing the user to compute the motion of a satellite in a reference frame that is not necessarily centered on the main attractor and does not necessarily possess inertial axis.
## Initialization
We will propagate the trajectory of a satellite departing from the Lagrange point L2 of the Earth-Moon system. The equations of motion will be computed in a reference Frame centered on L2, its X-axis continuously oriented toward Earth and Moon. The rotation of this frame means that we cannot simply use the fundamental principle of dynamics (Newton’s second law). Since we will consider only the gravitational attractions of the Earth and the Moon, the inertial reference frame most closely related to our problem is a frame that is centered on the Earth-Moon barycenter, with inertial axis.
Let’s initialize our program. First, the time settings : the initial moment of integration, the length of integration (here in seconds), and the time interval between each output.
```final AbsoluteDate initialDate = new AbsoluteDate(2000, 01, 01, 0, 0, 00.000, TimeScalesFactory.getUTC());
double integrationTime = 600000.;
double outputStep = 600.0;
```
Then the initial position and velocity of the satellite, located exactly at L2 point. We still haven’t introduced a reference Frame, so the user must keep in mind in which reference Frame he wants to define the initial conditions of its satellite.
```final PVCoordinates initialConditions = new PVCoordinates(new Vector3D(0.0, 0.0, 0.0), new Vector3D(0.0, 0.0, 0.0));
```
## Frames setting
We need to load a few Celestial bodies, those we consider in gravitational interaction with our satellite, as well as the Earth-Moon barycenter, since we will compute our inertia forces with respect to its attached frame. We then create the reference frame attached to the L2 point, and the inertial reference frame attached to the Earth-Moon barycenter.
```final CelestialBody earth = CelestialBodyFactory.getEarth();
final CelestialBody moon = CelestialBodyFactory.getMoon();
final CelestialBody earthMoonBary = CelestialBodyFactory.getEarthMoonBarycenter();
final Frame l2Frame = new L2Frame(earth, moon);
final Frame earthMoonBaryFrame = earthMoonBary.getInertiallyOrientedFrame();
final Frame inertiaFrame = earthMoonBaryFrame;
final Frame integrationFrame = l2Frame;
final Frame outputFrame = l2Frame;
```
## Propagator preparations
We now transform our `PVCoordinates` into `AbsolutePVCoordinates`, define the satellite attitude, and use all of it to build the `SpacecraftState`
```final AbsolutePVCoordinates initialAbsPV = new AbsolutePVCoordinates(integrationFrame, initialDate, initialConditions);
Attitude arbitraryAttitude = new Attitude(integrationFrame,
new TimeStampedAngularCoordinates(initialDate,
new PVCoordinates(Vector3D.PLUS_I, Vector3D.PLUS_J),
new PVCoordinates(Vector3D.PLUS_I, Vector3D.PLUS_J)));
final SpacecraftState initialState = new SpacecraftState(initialAbsPV, arbitraryAttitude);
```
We will use a variable-step 8(5,3) Dormand-Prince integrator. This integrator needs a few initialization parameters. First, let’s set boundaries on the integration steps, to prevent a too long computing or a too long integration step.
```final double minStep = 0.001;
final double maxstep = 3600.0;
```
We also need to set the acceptable error of our integration. This error is used to adjust the integration step and does not stand for the overall error of the propagation.
```final double positionTolerance = 0.001;
final double velocityTolerance = 0.00001;
final double massTolerance = 1.0e-6;
final double[] vecAbsoluteTolerances = {
positionTolerance, positionTolerance, positionTolerance,
velocityTolerance, velocityTolerance, velocityTolerance,
massTolerance
};
final double[] vecRelativeTolerances = new double[vecAbsoluteTolerances.length];
```
We can now define the numerical integrator of our propagator.
```AdaptiveStepsizeIntegrator integrator = new DormandPrince853Integrator(minStep, maxstep,
vecAbsoluteTolerances,
vecRelativeTolerances);
```
## Propagator building and use
And finally, we can build our propagator. First we use the `NumericalPropagator` constructor with the previously defined numerical integrator, and then we specify the properties of the evolution model. The use of `SingleBodyAttraction` means that we do not constrain the type of orbit, and that we need to use `setOrbitType(null)` as well as `setIgnoreCentralAttraction(true)`. The non-inertial aspect of our integration reference frame is handled by `InertialForces`, which will compute the inertial accelerations that appear in this frame. Each attracting body will be provided to the propagator by calling `SingleBodyAttraction` for each body in gravitational interaction with our spacecraft. We then only need to set the inital spacecraft state, and the propagator mode, more details about propagator modes can be found in the Propagation tutorial .
```NumericalPropagator propagator = new NumericalPropagator(integrator);
propagator.setOrbitType(null);
propagator.setIgnoreCentralAttraction(true);
propagator.setInitialState(initialState);
propagator.setMasterMode(outputStep, new TutorialStepHandler("test.dat", outputFrame));
```
In the end we can do the propagation itself, from an initial date, for a given duration.
```SpacecraftState finalState = propagator.propagate(initialDate.shiftedBy(integrationTime));
final PVCoordinates pv = finalState.getPVCoordinates(outputFrame);
System.out.println("initial conditions: " + initialConditions);
System.out.println("final conditions: " + pv);
```
## Stephandling
Now, we can deal with our stephandler, it will print the position, velocities and acceleration at each output step of the integration.
``` private static class TutorialStepHandler implements OrekitFixedStepHandler {
private Frame outputFrame;
private TutorialStepHandler( final Frame frame) {
outputFrame = frame;
}
public void init(final SpacecraftState s0, final AbsoluteDate t) {
System.out.format(Locale.US,
"%s %s %s %s %s %s %s %s %s %s %n",
"date", " X", " Y",
" Z", " Vx", " Vy",
" Vz", " ax", " ay",
" az");
}
public void handleStep(SpacecraftState currentState, boolean isLast) {
try {
final TimeScale utc = TimeScalesFactory.getUTC();
final AbsoluteDate initialDate = new AbsoluteDate(2000, 01, 01, 0, 0, 00.000,
TimeScalesFactory.getUTC());
final AbsoluteDate d = currentState.getDate();
final PVCoordinates pv = currentState.getPVCoordinates(outputFrame);
System.out.format(Locale.US,
"%s %18.12f %18.12f %18.12f %18.12f %18.12f %18.12f %18.12f %18.12f %18.12f%n",
d, pv.getPosition().getX(),
pv.getPosition().getY(), pv.getPosition().getZ(),
pv.getVelocity().getX(), pv.getVelocity().getY(),
pv.getVelocity().getZ(), pv.getAcceleration().getX(),
pv.getAcceleration().getY(),
pv.getAcceleration().getZ()
);
if (isLast) {
final PVCoordinates finalPv =
currentState.getPVCoordinates(outputFrame);
System.out.println();
System.out.format(Locale.US,
"%s %12.0f %12.0f %12.0f %12.0f %12.0f %12.0f%n",
d, finalPv.getPosition().getX(),
finalPv.getPosition().getY(),
finalPv.getPosition().getZ(),
finalPv.getVelocity().getX(),
finalPv.getVelocity().getY(),
finalPv.getVelocity().getZ());
System.out.println();
}
} catch (OrekitException oe) {
System.err.println(oe.getMessage());
}
}
}
}
```
## Results
The printed results are shown below:
```date X Y Z Vx Vy Vz ax ay az
2000-01-01T00:00:00.000 0.000000000000 0.000000000000 0.000000000000 0.000000000000 0.000000000000 0.000000000000 0.000007203264 -0.000023617334 0.000000988884
2000-01-01T00:10:00.000 1.310181754844 -4.509937950657 0.177995376805 0.004389441414 -0.015464490795 0.000593311195 0.000007424953 -0.000027930975 0.000000988817
2000-01-01T00:20:00.000 5.292761075816 -19.075027371572 0.711964514771 0.008906039828 -0.033517172552 0.001186577792 0.000007627122 -0.000032244627 0.000000988736
2000-01-01T00:30:00.000 12.020526802001 -45.248275712283 1.601878056606 0.013538115033 -0.054158399174 0.001779791095 0.000007710027 -0.000035388726 0.000000989294
2000-01-01T00:40:00.000 21.178373170978 -80.022897721805 2.850254273796 0.017011800885 -0.062143736699 0.002381480483 0.000005918985 -0.000015387558 0.000001002821
2000-01-01T00:50:00.000 32.466478604114 -120.337158982044 4.459645552958 0.020640742565 -0.072667569026 0.002983148787 0.000006174238 -0.000019691906 0.000001002738
2000-01-01T01:00:00.000 45.976718715462 -167.740438088863 6.430022015915 0.024416959249 -0.085773707784 0.003584763069 0.000006410327 -0.000024000459 0.000001002638
2000-01-01T01:10:00.000 61.793504290626 -223.776772118858 8.761351689337 0.028327931996 -0.101452129474 0.004186320438 0.000006626318 -0.000028300879 0.000001002529
2000-01-01T01:20:00.000 79.995099948214 -290.000479343336 11.453591760838 0.032363731793 -0.119724013059 0.004787800539 0.000006823101 -0.000032605395 0.000001002403
2000-01-01T01:30:00.000 100.652428154023 -367.962135000874 14.506696359194 0.036511612039 -0.140576870435 0.005389201731 0.000006303102 -0.000028624115 0.000001006865
2000-01-01T01:40:00.000 123.451861739137 -454.645479496582 17.923153578986 0.039511558100 -0.148775025605 0.005998989925 0.000005129778 -0.000015774359 0.000001016276
2000-01-01T01:50:00.000 148.097431329154 -547.007595557232 21.705469789230 0.042665311712 -0.159528224853 0.006608717996 0.000005379490 -0.000020069660 0.000001016148
2000-01-01T02:00:00.000 174.679033538614 -646.594792449283 25.853598980322 0.045963054147 -0.172858638623 0.007218365096 0.000005609742 -0.000024365065 0.000001016006
2000-01-01T02:10:00.000 203.279559120647 -754.953413963500 30.367489987968 0.049393110759 -0.188766311565 0.007827922613 0.000005820538 -0.000028660515 0.000001015850
2000-01-01T02:20:00.000 233.974894931263 -873.629820586518 35.247086470272 0.052943809883 -0.207251254684 0.008437381900 0.000006011884 -0.000032955956 0.000001015679
2000-01-01T02:30:00.000 266.833927466020 -1004.170390063609 40.492326874541 0.056603523474 -0.228313934164 0.009046734005 0.000006203223 -0.000037487428 0.000001015363
2000-01-01T02:40:00.000 301.547225264656 -1143.577979776640 46.105651401557 0.059129787256 -0.236758043587 0.009664358716 0.000004333863 -0.000016135286 0.000001029339
2000-01-01T02:50:00.000 337.820135753835 -1288.794334653753 52.089537502703 0.061804333216 -0.247725104421 0.010281911014 0.000004578052 -0.000020421603 0.000001029166
2000-01-01T03:00:00.000 375.740556403237 -1441.362400813750 58.443923059929 0.064619473654 -0.261262891438 0.010899355930 0.000004806508 -0.000024752997 0.000001028954
2000-01-01T03:10:00.000 415.387681993014 -1602.804245773708 65.168752226799 0.067560866331 -0.277337889071 0.011516703202 0.000005008336 -0.000028994444 0.000001028777
2000-01-01T03:20:00.000 456.837148988767 -1774.683165398515 72.263941309722 0.070622616256 -0.296020485822 0.012133905439 0.000005194262 -0.000033280871 0.000001028561
2000-01-01T03:20:00.000 457 -1775 72 0 -0 0
initial conditions: {P(0.0, 0.0, 0.0), V(0.0, 0.0, 0.0), A(0.0, 0.0, 0.0)}
final conditions: {2000-01-01T03:20:00.000, P(456.8371489887673, -1774.6831653985146, 72.2639413097221), V(0.0706226162556273, -0.29602048582231744, 0.012133905438764159), A(5.194261804654593E-6, -3.328087090313115E-5, 1.02856099808902E-6)}
``` | 3,892 | 12,538 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2020-45 | latest | en | 0.842179 |
https://ux.stackexchange.com/questions/39544/keyboard-number-row-ordering | 1,653,307,770,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558015.52/warc/CC-MAIN-20220523101705-20220523131705-00211.warc.gz | 679,609,517 | 70,677 | # Keyboard number row ordering
On a typical keyboard, why is the 0 in the number row next to the 9 instead of the 1?
This seems like a question which should have a straight-forward answer, but the only one I could find is this yahoo question which has two instances of entirely unsourced speculation as answers. (0 as 10, and because 0 is rarely at the start of a number)
To complicate things, the wikipedia article adds that
0 and 1 were omitted to simplify the design and reduce the manufacturing and maintenance costs; they were chosen specifically because they were "redundant" and could be recreated using other keys. Typists who learned on these machines learned the habit of using the uppercase letter I (or lowercase letter L) for the digit one, and the uppercase O for the zero.
One might speculate that the 0 is placed where it is because of it's proximity to the O, but since the 1 was added down at the other end (nowhere near the I), it would have made just as much sense to put 0 down there too.
Almost every keyboard layout I've seen listed on wikipedia is this way, even the ones which don't use latin script at all. Only the Hungarian one (thanks, Gildas) puts 0 before the 1. This may be due to inheriting from latin-alphabet keyboards, though.
Anyone have an explanation for this oddity? Or specific sources backing up the yahoo theories?
Edit: Based on everyone's answers, I've done more research and come up with what I think is the logical explanation. I don't have specific sources to cite, though, so I'm still open to an "official" answer, if anyone has one.
• possible duplicate of Why do numpads on keyboards and phones have reversed layouts?
– rk.
May 13, 2013 at 18:35
• @rk. - Not at all a duplicate. I'm not asking about the keypad, I'm discussing the number row on top of the main section of they keyboard. May 13, 2013 at 18:52
• The answer discusses the history of the layout of the numbers on the keyboard, which is related to your question.
– rk.
May 13, 2013 at 18:56
• @rk. - Nothing in that answer has anything to do with the keyboard itself. It's entirely devoted to the history of the number pad, which (barring any contrary sources) has nothing to do with the history of the main portion of the keyboard, since it was a later development. However, the article it links to does have an interesting image of a keypunch machine which might be relevant. May 13, 2013 at 19:18
Via further research, I've discovered that I was acting under a bad assumption. I had assumed that 0 and 1 became standard around the same time, but the very next section in the wikipedia article says:
The 0 key was added and standardized in its modern position early in the history of the typewriter, but the 1 and exclamation point were left off some typewriter keyboards into the 1970s.
It appears to be the IBM Selectric which really popularized the 1 in the '70s,
Given this 60+ year discrepancy between each key's appearance, I think I can come to a logical conclusion:
The 0 was placed next to the 9 either because of the proximity to the O (it would be easy for people already used to typing an `O` to type a `0` instead) or because it could be seen as 8-9-10 (and 0-2-3 doesn't make sense). Later, when the 1 was introduced, the only logical place for it was next to the 2, since it would make no sense to have 8-9-0-1 at the end of the row. And by that time, the 0 was fully established, so it couldn't be moved next to the new 1.
• I think this makes great sense once its all pieced together. Not sure how UX it is, but it is a good read! May 13, 2013 at 20:36
• @AthomSfere - Yeah, in retrospect, the only UX aspect is the logic of shifting from `O` to `0`. I expected more thought to have gone into it than I really should have been. Although, it doesn't explain things like this, which predate typewriters altogether... May 13, 2013 at 20:42
• Except, the only keyboards at the beginning were from actual keyboards. It makes some sense. And it was rapidly outgrown as it was inefficient. So that evolution was UX design I suppose. May 13, 2013 at 20:51
• Do you have a source for the whole proximity argument for putting it beside the 9? Seems to me more likely that it's just plain weird to have `0`, `2`, `3`… (at least compared to …`8`, `9`, `0`), but I'd love to be proven wrong. Jun 27, 2013 at 6:10
• @KitGrose - An interesting point. I hadn't considered that possibility. I've (belatedly) edited that in. Nov 12, 2013 at 17:24
I too want to add an image:
My thought is (Speculation also) is that it has to do with QWERTY,
Most of the QWERTY layout was to prevent keybinding.
I have to wonder if having it in `0``1``2` cause binding issues and was thus moved to the end where one could not cause a binding issue. 1011 comes to mind as a touchy combo.
• Is that key next to the `2` a `1` missing the key, or does it do something else? If it's something else, this is really interesting, since it's got the `0` but not the `1`. May 13, 2013 at 18:31
• Remington (The gun maker) bought the design from Sholes. Sholes and Glidden invented the QWERTY and there was not a 1. en.wikipedia.org/wiki/… May 13, 2013 at 19:40
• Interesting. You've made me realize that the two keys were added at different times, as opposed to around the same time, as I was thinking. May 13, 2013 at 19:47
It's only really programmers that think of counting starting from 0. The vast majority of the world counts from 1.
And so considering that keyboards were designed from typewriters, which were designed for secretaries mostly in the early days, it makes sense that you show the numbering in an order that they would find normal.
• I would say programmers and mathematicians. But that doesn't make the `0` make any more sense. It should be a `10`, in that case. And since a `10` key makes no sense, it can become a `0` key... But regardless of where you start counting, everyone would agree that `0` is less than `1`, and `10` is less than `11`, so the `0` belongs before the `1` on the number line (or keyboard row). May 13, 2013 at 19:15
• @Bobson You're thinking about it from a technical perspective, and it's not a technical reason.
– JohnGB
May 13, 2013 at 19:30
• You may be right. I don't think I can break out of my own mindset on this. On the other hand, I consider leaving out the `1` and `0` altogether weird as well, and early typewriters did that too... But wouldn't that be equally abnormal to secretaries as having the `0` first? May 13, 2013 at 19:35
• Except on paper, the only keys were CAPS, so 1I were interchangeable, so were 0 and O. This was also bleeding edge technology. So getting the same effect with less engineering and cost meant more sales. May 13, 2013 at 19:56
• That the majority counts from 1 is irrelevant to the topic, as we're talking about arranging the 10 digits (0, 1, …, 9), not counts (1, 2, …, 9, 10, 11, 12, …, and so on, if counting from 1). Except for the current convention that is being questioned here, in no natural/logical/technical/etc. perspective does the ordering 1, 2, …, 9, 0 make sense. Apr 6, 2016 at 21:28
Speculation
• It may have something to do with some coding needs on the early years of computing, 0 and 1 could not be very close because of their use in binary language hence the distance to have an ergonomic typing with the use of two hands.
I found that the Hungarian keyboard does have Zero in its natural order.
`````` ┌────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────╔═════════╗
│ │ │ │ │ │ ║ │ │ │ │ │ │ ║ ║
│0 │1 │2 │3 │4 │5 ║6 │7 │8 │9 │ │ │ ║ <-- ║
╔════╧══╗─┴──┬─┴──┬─┴──┬─┴──┬─┴──┬─┴──┬─┴──┬─┴──┬─┴──┬─┴──┬─┴──┬─╚══╦══════╣
║ |<- ║ │ │ │ │ ║ │ │ │ │ │ │ ║ | ║
║ ->| ║ │ │ │ │ ║ │ │ │ │ │ │ ║ <-' ║
╠═══════╩╗───┴┬───┴┬───┴┬───┴┬───┴┬───┴┬───┴┬───┴┬───┴┬───┴┬───┴┬───╚╗ ║
``````
And yes I did answer just to post some ascii art.
The Hungarian keyboard does exist though.
• +1 for the art and for having spotted what I missed when I looked at all the keyboards on wikipedia. I guess I went through the international ones a bit too quickly. Doesn't answer the question, but still more than worth an upvote. May 13, 2013 at 18:21 | 2,345 | 8,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-21 | longest | en | 0.974314 |
https://eprints.iisc.ac.in/36054/ | 1,722,956,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640492117.28/warc/CC-MAIN-20240806130705-20240806160705-00212.warc.gz | 185,536,818 | 7,974 | Boxicity and Poset Dimension
Adiga, Abhijin and Bhowmick, Diptendu and Chandran, Sunil L (2010) Boxicity and Poset Dimension. In: 16th Annual International Computing and Combinatorics Conference, JUL 19-21, 2010 , Nha Trang, VIETNAM, pp. 3-12.
PDF boxicity.pdf - Published Version Restricted to Registered users only Download (203kB) | Request a copy
Official URL: http://arxiv.org/abs/1003.2357
Abstract
Let G be a simple, undirected, finite graph with vertex set V(G) and edge set E(C). A k-dimensional box is a Cartesian product of closed intervals a(1), b(1)] x a(2), b(2)] x ... x a(k), b(k)]. The boxicity of G, box(G) is the minimum integer k such that G can be represented as the intersection graph of k-dimensional boxes, i.e. each vertex is mapped to a k-dimensional box and two vertices are adjacent in G if and only if their corresponding boxes intersect. Let P = (S, P) be a poset where S is the ground set and P is a reflexive, anti-symmetric and transitive binary relation on S. The dimension of P, dim(P) is the minimum integer l such that P can be expressed as the intersection of t total orders. Let G(P) be the underlying comparability graph of P. It is a well-known fact that posets with the same underlying comparability graph have the same dimension. The first result of this paper links the dimension of a poset to the boxicity of its underlying comparability graph. In particular, we show that for any poset P, box(G(P))/(chi(G(P)) - 1) <= dim(P) <= 2box(G(P)), where chi(G(P)) is the chromatic number of G(P) and chi(G(P)) not equal 1. The second result of the paper relates the boxicity of a graph G with a natural partial order associated with its extended double cover, denoted as G(c). Let P-c be the natural height-2 poset associated with G(c) by making A the set of minimal elements and B the set of maximal elements. We show that box(G)/2 <= dim(P-c) <= 2box(G) + 4. These results have some immediate and significant consequences. The upper bound dim(P) <= 2box(G(P)) allows us to derive hitherto unknown upper bounds for poset dimension. In the other direction, using the already known bounds for partial order dimension we get the following: (I) The boxicity of any graph with maximum degree Delta is O(Delta log(2) Delta) which is an improvement over the best known upper bound of Delta(2) + 2. (2) There exist graphs with boxicity Omega(Delta log Delta). This disproves a conjecture that the boxicity of a graph is O(Delta). (3) There exists no polynomial-time algorithm to approximate the boxicity of a bipartite graph on n vertices with a factor of O(n(0.5-epsilon)) for any epsilon > 0, unless NP=ZPP.
Item Type: Conference Paper Lecture Notes in Computer Science Springer Copyright of this article belongs to Springer. Boxicity; partial order;poset dimension;comparability graph; extended double cover Division of Electrical Sciences > Computer Science & Automation 14 Mar 2011 07:07 29 Feb 2012 07:17 http://eprints.iisc.ac.in/id/eprint/36054 | 767 | 2,989 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-33 | latest | en | 0.878437 |
http://www.thenakedscientists.com/forum/index.php?topic=16100.0 | 1,480,758,449,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540915.89/warc/CC-MAIN-20161202170900-00060-ip-10-31-129-80.ec2.internal.warc.gz | 753,490,946 | 11,183 | # The Naked Scientists Forum
### Author Topic: What happens to the field of 2 adjacent magnets? (Read 2899 times)
#### DoctorBeaver
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• A stitch in time would have confused Einstein.
##### What happens to the field of 2 adjacent magnets?
« on: 23/07/2008 18:53:34 »
If you take 2 identical bar magnets and place 1 on top of the other with the north pole of 1 adjacent to the south pole of the other, what happens to the magnetic field?
#### lightarrow
• Neilep Level Member
• Posts: 4586
• Thanked: 7 times
##### What happens to the field of 2 adjacent magnets?
« Reply #1 on: 23/07/2008 19:58:05 »
If you take 2 identical bar magnets and place 1 on top of the other with the north pole of 1 adjacent to the south pole of the other, what happens to the magnetic field?
You want to make it disappear, isnt'it? It's not possible; to achieve that, you should make the sources coincide exactly, that is you should overlap the two (equal) magnet; but that's impossible: the space occupied by one cannot simultaneously be occupied by the other.
At a very large distance (with respect to the bar's dimensions), however, the field goes to zero very fast, that is, you're less and less able to determine that there are two magnets, instead of no ones.
#### DoctorBeaver
• Naked Science Forum GOD!
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• A stitch in time would have confused Einstein.
##### What happens to the field of 2 adjacent magnets?
« Reply #2 on: 23/07/2008 20:44:48 »
lightarrow - no, I wasn't thinking about making it disappear. I was wondering whether something made from, say, iron would be more attracted to them as a result. The iron wouldn't care if the north or south pole were nearer. So would having the 2 magnets together increase the strength of the magnetic field or would it stay the same?
« Last Edit: 23/07/2008 20:46:45 by DoctorBeaver »
#### lyner
• Guest
##### What happens to the field of 2 adjacent magnets?
« Reply #3 on: 24/07/2008 11:54:59 »
If an N and S pole are brought close together, in pictorial terms, the field lines tend to concentrate between the poles, making the local field much stronger but the external field less. That is why, I think, all old magnets were made in a horseshoe shape so that you at least got one area of relatively strong field from a relatively weak magnet.
Think of field lies as unbroken loops. They go into a magnet* and come out 'the other end' then launch off into the surrounding space, only to return. A field line may go through a number of media (other permanent magnets or bits of iron - temporary magnets). The 'concentration' of lines corresponds to the field strength and is increased in or near regions of high permeability.
*Or the coils of an electromagnet.
#### The Naked Scientists Forum
##### What happens to the field of 2 adjacent magnets?
« Reply #3 on: 24/07/2008 11:54:59 » | 740 | 2,921 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2016-50 | latest | en | 0.923211 |
http://nrich.maths.org/6874&part= | 1,505,854,383,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686034.31/warc/CC-MAIN-20170919202211-20170919222211-00006.warc.gz | 247,504,899 | 5,187 | ### 8 Methods for Three by One
This problem in geometry has been solved in no less than EIGHT ways by a pair of students. How would you solve it? How many of their solutions can you follow? How are they the same or different? Which do you like best?
### Rots and Refs
Follow hints using a little coordinate geometry, plane geometry and trig to see how matrices are used to work on transformations of the plane.
### Limiting Probabilities
Given probabilities of taking paths in a graph from each node, use matrix multiplication to find the probability of going from one vertex to another in 2 stages, or 3, or 4 or even 100.
# Transformations for 10
##### Stage: 5 Challenge Level:
The operation of mutiplying a vector by a constant matrix can by thought of as transforming a point in space onto another point in space. These transformations can have very clear, intuitive properties and we can often think of them from either a geometrical perspective or an algebraic perspective.
Below are ten questions about the properties of such transformations in three dimensions for you to think about.
As you think about the questions, can you draw relevant diagrams and construct relevant algebraic examples? In each case, is there a definitive answer, or does it depend on various factors? You may intuitively feel the answers to some of these questions; in these cases can you prove your intuition correct?
1. What does a matrix do to the zero vector ${\bf 0}$?
2. What does a matrix do to a line/plane through the origin?
3. What does a matrix do to a line/plane not through the origin?
4. Which lines can you transform onto the x-axis using matrix multiplication?
5. Which planes can you transform onto the xy-plane using matrix multiplication?
6. Can you think of a matrix which transforms a plane to a line?
7. Can you think of a matrix which transforms a line to a plane?
8. How many matrices transform the cube $(\pm 1, \pm 1, \pm 1)$ to another cube?
9. Can you find a matrix which transforms a square to a triangle in 2D?
10. Can you think of a matrix which shifts all points from ${\bf x}$ to ${\bf x+ (1,0,0)}$? | 484 | 2,127 | {"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} | 4.25 | 4 | CC-MAIN-2017-39 | latest | en | 0.943806 |
https://www.excel-university.com/dynamic-arrays-2/?recaptcha-opt-in=true | 1,632,431,246,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057447.52/warc/CC-MAIN-20210923195546-20210923225546-00120.warc.gz | 799,101,736 | 16,283 | ## Dynamic Arrays 2
This is the second post in the Dynamic Arrays series. In the first post, we talked about how formulas can return multiple values, the resulting spill range, and two dynamic array functions. In this post, we’ll talk about how to refer to the spill range with other formulas.
### Spill Range
The spill range includes the cells that store the results of a dynamic array formula. In other words, when you write a dynamic array formula in a cell, for example into cell B8:
And then hit Enter, the results spill out of the formula cell into the cells below:
The range that contains all of the formula results, B8:B10 in this case, is known as the Spill Range.
Now, the cool part about the spill range, besides the fact that it is dynamic, is that we can reference it from other formulas.
### Spill Reference
Let’s say we’d like to populate the Total column with the sum of the accounts. The data table, Table1, looks like this:
So, we figure we can use the SUMIFS function to generate the summary.
We write the following formula into C8:
`=SUMIFS(Table1[Amount],Table1[Account],B8)`
And press Enter:
It computed the correct result, so that is good. But, now we would need to manually fill that formula down to compute the totals for Meals and Phone.
But, if we know how to use the Spill Reference Operator #, then Excel will fill the formula down for us … and continue doing so as the size of the spill range changes.
So, we update our reference from B8 (individual cell) to B8# (entire spill range) in our formula, like this:
`=SUMIFS(Table1[Amount],Table1[Account],B8#)`
We hit Enter and bam:
It too spills down … nice!
And, the best part is that when our source data table has more rows and more accounts, the results will dynamically update.
If you are familiar with PivotTables, you’ll quickly realize we could build the example above using a PivotTable. The dynamic arrays approach is kinda like a cross between a traditional formula-based report and a PivotTable. It uses formulas, but has the dynamic element we love about PivotTables. So, depending on the context of your workbook, this may offer a nice option.
What do you think about this capability? I’d like to know … just post a quick comment below!
Posted in , ,
### Jeff Lenning
I love sharing the things I've learned about Excel, and I built Excel University to help me do that. My motto is: Learn Excel. Work Faster.
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! | 624 | 2,747 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-39 | latest | en | 0.857124 |
http://vento-in-poppa.it/6-way-wire-diagram.html | 1,603,234,939,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874340.10/warc/CC-MAIN-20201020221156-20201021011156-00101.warc.gz | 111,890,310 | 8,800 | # 6 Way Wire Diagram
• Wire Diagram
• Date : October 20, 2020
## 6 Way Wire Diagram
Way
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6 Way Wire Diagram To be able to make Venn Diagrams in Word, then it is essential to be aware of the numerous commands that can allow you to build the simple form. Once you know what each control does, it is going to make it much easier to create each Venn Diagram which you're searching for. A Venn diagram is a diagram which illustrates the intersection of two or more collections of values, depending on one or more measurements. To be able to make a Venn diagram in Word, you first need to get a simple idea of what Venn Diagrams is. This implies that if a point lies within the assortment of the very first value then the exact same point will even lie within the range of the next price. Similarly if a point lies beyond the range of the first value then the same point will even lie beyond the range of the second value. These theories are important to notice as they are essential to comprehend how to make Venn Diagrams in Word. So what is exactly the Venn Diagram and how can you make one? The diagram that's produced in the intersection of a couple of sets of values, with respect to one or more dimensions, is called a Venn diagram. There are two main kinds of Venn diagrams - vertical and horizontal. In both types of diagrams, the range of points is the intersection of two or more values and the value at each stage is known as a'component'. At a horizontal Venn diagram that the points in the range are arranged horizontally, while the value of every stage is stored vertically. Vertical Venn Diagrams is similar to the horizontal diagram however, the points are organized horizontally, each value being stored horizontally. The best way to make a Venn diagram in Word would be to use the Venn diagram tools. All these are found at the Microsoft Office suite of programs and are accessible via your'Add/Remove Programs' alternative. To begin a Venn diagram in Word select the Venn instrument in the list. This will present you with a menu with many different choices and one of these is Venn - Horizontal and Venn - Vertical. Click on both of these menus and then choose Venn - Measure and Venn - Vertical respectively. Once you have done this you will then be presented with the options that are relevant to the application you selected. On the Venn-Horizontal Diagram the points are arranged horizontally and the value of every point is stored vertically. About the Venn-Vertical Diagram the points are arranged vertically and the value of every stage is stored horizontally. You can make any changes to the diagram as you need by clicking on the arrow buttons to the right of the points. As soon as you've done this you are able to save the diagram for a Word file or print it out. | 683 | 3,186 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2020-45 | latest | en | 0.891622 |
https://www.vedantu.com/question-answer/find-the-cost-of-laying-grass-in-a-triangular-class-10-maths-cbse-5f5c35726e663a29cc6e5367 | 1,726,262,356,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.48/warc/CC-MAIN-20240913201909-20240913231909-00779.warc.gz | 972,392,757 | 28,033 | Courses
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# Find the cost of laying grass in a triangular field of sides $50m,65m$and $65m$at the rate of Rs 7 per ${m^2}$.
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Hint: We need to use the unitary method in this case. In the unitary method, if we know the price of a particular product we can find the price of the number of products by multiplication also we can find the price of a single product if we are given with price of the numbers of products with the division.
The area of the triangular field is found by using here’s formula i.e.
$\sqrt {S(s - a)(s - b)(s - c)}$ where a, b, c are sides of the triangle
S is semi perimeter.
Value of $s = \dfrac{{a + b + c}}{2}$
Therefore
Given sides of triangular grass field $50n,65m$and $65m$ respectively
Let $a = 50m$
$b = 65m$
and $c = 65m$
Now $S = \dfrac{{a + b + c}}{2} = \dfrac{{50 + 65 + 65}}{2} = 90m$
Area of triangular field is found by heron’s formula
i.e. Area $= \sqrt {S(s - a)(s - b)(s - c)}$
put $s = 90m,a = 50m,b = 65m$ and $c = 65m$
Area $= \left[ {{{10}^2} \times {3^2} \times {2^2} \times {{25}^2}} \right]$
$= \sqrt {90(40)(25)(25)}$
$= \sqrt {10 \times 910 \times 4 \times 25 \times 25}$ $\left[ {\because 90 = 9 \times 10\,\,40 = 4 \times 10} \right]$
$= \left[ {{{10}^2} \times {3^2} \times {2^2} \times {{25}^2}} \right]$ $\left[ {\because {3^2} = 9\,4 = {2^2}} \right]$
$= 10 \times 3 \times 2 \times 25$
Area$= 1500{m^2}$
BY unitary method,
Cost of laying $I{m^2}$ grass $= Rs.7$(Given)
Cost of laying $1500{m^2}$ grass
$= Rs\,1500 \times 7 = Rs\,10500$
Hence the cost of laying is Rs. $1050$.
Note: The question can also be solved by using the concept of isosceles triangle of triangle In, isosceles triangle two sides, are equal. | 673 | 1,832 | {"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} | 4.75 | 5 | CC-MAIN-2024-38 | latest | en | 0.770248 |
http://isabelle.in.tum.de/repos/isabelle/rev/3ead36cbe6b7 | 1,606,586,514,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195687.51/warc/CC-MAIN-20201128155305-20201128185305-00170.warc.gz | 47,722,300 | 15,726 | author paulson Mon, 18 Jun 2018 22:20:55 +0100 changeset 68464 3ead36cbe6b7 parent 68463 410818a69ee3 child 68468 ae42b0f6885d
De-applied Ideal.thy
src/HOL/Algebra/Ideal.thy file | annotate | diff | comparison | revisions
```--- a/src/HOL/Algebra/Ideal.thy Mon Jun 18 11:15:46 2018 +0200
+++ b/src/HOL/Algebra/Ideal.thy Mon Jun 18 22:20:55 2018 +0100
@@ -38,12 +38,8 @@
shows "ideal I R"
proof -
interpret ring R by fact
- show ?thesis apply (intro ideal.intro ideal_axioms.intro additive_subgroupI)
- apply (rule a_subgroup)
- apply (rule is_ring)
- apply (erule (1) I_l_closed)
- apply (erule (1) I_r_closed)
- done
+ show ?thesis
+ by (auto simp: ideal.intro ideal_axioms.intro additive_subgroupI a_subgroup is_ring I_l_closed I_r_closed)
qed
@@ -132,14 +128,11 @@
proof -
interpret additive_subgroup I R by fact
interpret cring R by fact
- show ?thesis apply (intro_locales)
+ show ?thesis apply intro_locales
apply (intro ideal_axioms.intro)
apply (erule (1) I_l_closed)
apply (erule (1) I_r_closed)
- apply (intro primeideal_axioms.intro)
- apply (rule I_notcarr)
- apply (erule (2) I_prime)
- done
+ by (simp add: I_notcarr I_prime primeideal_axioms.intro)
qed
@@ -165,14 +158,11 @@
lemma (in ideal) one_imp_carrier:
assumes I_one_closed: "\<one> \<in> I"
shows "I = carrier R"
- apply (rule)
- apply (rule)
- apply (rule a_Hcarr, simp)
proof
- fix x
- assume xcarr: "x \<in> carrier R"
- with I_one_closed have "x \<otimes> \<one> \<in> I" by (intro I_l_closed)
- with xcarr show "x \<in> I" by simp
+ show "carrier R \<subseteq> I"
+ using I_r_closed assms by fastforce
+ show "I \<subseteq> carrier R"
+ by (rule a_subset)
qed
lemma (in ideal) Icarr:
@@ -193,143 +183,82 @@
proof -
interpret ideal I R by fact
interpret ideal J R by fact
+ have IJ: "I \<inter> J \<subseteq> carrier R"
+ by (force simp: a_subset)
show ?thesis
apply (intro idealI subgroup.intro)
- apply (rule is_ring)
- apply (force simp add: a_subset)
- apply simp
- apply (clarsimp, rule)
- apply (fast intro: ideal.I_l_closed ideal.intro assms)+
- apply (clarsimp, rule)
- apply (fast intro: ideal.I_r_closed ideal.intro assms)+
+ apply (simp_all add: IJ is_ring I_l_closed assms ideal.I_l_closed ideal.I_r_closed flip: a_inv_def)
done
qed
-text \<open>The intersection of any Number of Ideals is again
- an Ideal in @{term R}\<close>
-lemma (in ring) i_Intersect:
- assumes Sideals: "\<And>I. I \<in> S \<Longrightarrow> ideal I R"
- and notempty: "S \<noteq> {}"
- shows "ideal (\<Inter>S) R"
- apply (unfold_locales)
- apply rule unfolding mem_Collect_eq defer 1
- apply rule defer 1
- apply rule defer 1
- apply (fold a_inv_def, rule) defer 1
- apply rule defer 1
- apply rule defer 1
-proof -
- fix x y
- assume "\<forall>I\<in>S. x \<in> I"
- then have xI: "\<And>I. I \<in> S \<Longrightarrow> x \<in> I" by simp
- assume "\<forall>I\<in>S. y \<in> I"
- then have yI: "\<And>I. I \<in> S \<Longrightarrow> y \<in> I" by simp
-
- fix J
- assume JS: "J \<in> S"
- interpret ideal J R by (rule Sideals[OF JS])
- from xI[OF JS] and yI[OF JS] show "x \<oplus> y \<in> J" by (rule a_closed)
-next
- fix J
- assume JS: "J \<in> S"
- interpret ideal J R by (rule Sideals[OF JS])
- show "\<zero> \<in> J" by simp
-next
- fix x
- assume "\<forall>I\<in>S. x \<in> I"
- then have xI: "\<And>I. I \<in> S \<Longrightarrow> x \<in> I" by simp
+text \<open>The intersection of any Number of Ideals is again an Ideal in @{term R}\<close>
- fix J
- assume JS: "J \<in> S"
- interpret ideal J R by (rule Sideals[OF JS])
-
- from xI[OF JS] show "\<ominus> x \<in> J" by (rule a_inv_closed)
-next
- fix x y
- assume "\<forall>I\<in>S. x \<in> I"
- then have xI: "\<And>I. I \<in> S \<Longrightarrow> x \<in> I" by simp
- assume ycarr: "y \<in> carrier R"
-
- fix J
- assume JS: "J \<in> S"
- interpret ideal J R by (rule Sideals[OF JS])
-
- from xI[OF JS] and ycarr show "y \<otimes> x \<in> J" by (rule I_l_closed)
-next
- fix x y
- assume "\<forall>I\<in>S. x \<in> I"
- then have xI: "\<And>I. I \<in> S \<Longrightarrow> x \<in> I" by simp
- assume ycarr: "y \<in> carrier R"
-
- fix J
- assume JS: "J \<in> S"
- interpret ideal J R by (rule Sideals[OF JS])
-
- from xI[OF JS] and ycarr show "x \<otimes> y \<in> J" by (rule I_r_closed)
-next
- fix x
- assume "\<forall>I\<in>S. x \<in> I"
- then have xI: "\<And>I. I \<in> S \<Longrightarrow> x \<in> I" by simp
-
- from notempty have "\<exists>I0. I0 \<in> S" by blast
- then obtain I0 where I0S: "I0 \<in> S" by auto
-
- interpret ideal I0 R by (rule Sideals[OF I0S])
-
- from xI[OF I0S] have "x \<in> I0" .
- with a_subset show "x \<in> carrier R" by fast
-next
-
+lemma (in ring) i_Intersect:
+ assumes Sideals: "\<And>I. I \<in> S \<Longrightarrow> ideal I R" and notempty: "S \<noteq> {}"
+ shows "ideal (\<Inter>S) R"
+proof -
+ { fix x y J
+ assume "\<forall>I\<in>S. x \<in> I" "\<forall>I\<in>S. y \<in> I" and JS: "J \<in> S"
+ interpret ideal J R by (rule Sideals[OF JS])
+ have "x \<oplus> y \<in> J"
+ by (simp add: JS \<open>\<forall>I\<in>S. x \<in> I\<close> \<open>\<forall>I\<in>S. y \<in> I\<close>) }
+ moreover
+ have "\<zero> \<in> J" if "J \<in> S" for J
+ moreover
+ { fix x J
+ assume "\<forall>I\<in>S. x \<in> I" and JS: "J \<in> S"
+ interpret ideal J R by (rule Sideals[OF JS])
+ have "\<ominus> x \<in> J"
+ by (simp add: JS \<open>\<forall>I\<in>S. x \<in> I\<close>) }
+ moreover
+ { fix x y J
+ assume "\<forall>I\<in>S. x \<in> I" and ycarr: "y \<in> carrier R" and JS: "J \<in> S"
+ interpret ideal J R by (rule Sideals[OF JS])
+ have "y \<otimes> x \<in> J" "x \<otimes> y \<in> J"
+ using I_l_closed I_r_closed JS \<open>\<forall>I\<in>S. x \<in> I\<close> ycarr by blast+ }
+ moreover
+ { fix x
+ assume "\<forall>I\<in>S. x \<in> I"
+ obtain I0 where I0S: "I0 \<in> S"
+ using notempty by blast
+ interpret ideal I0 R by (rule Sideals[OF I0S])
+ have "x \<in> I0"
+ by (simp add: I0S \<open>\<forall>I\<in>S. x \<in> I\<close>)
+ with a_subset have "x \<in> carrier R" by fast }
+ ultimately show ?thesis
+ by unfold_locales (auto simp: Inter_eq simp flip: a_inv_def)
qed
- assumes idealI: "ideal I R"
- and idealJ: "ideal J R"
+ assumes idealI: "ideal I R" and idealJ: "ideal J R"
shows "ideal (I <+> J) R"
- apply (rule ideal.intro)
- apply (intro ideal.axioms[OF idealI])
- apply (intro ideal.axioms[OF idealJ])
- apply (rule is_ring)
- apply (rule ideal_axioms.intro)
-proof -
- fix x i j
- assume xcarr: "x \<in> carrier R"
- and iI: "i \<in> I"
- and jJ: "j \<in> J"
- from xcarr ideal.Icarr[OF idealI iI] ideal.Icarr[OF idealJ jJ]
- have c: "(i \<oplus> j) \<otimes> x = (i \<otimes> x) \<oplus> (j \<otimes> x)"
- by algebra
- from xcarr and iI have a: "i \<otimes> x \<in> I"
- by (simp add: ideal.I_r_closed[OF idealI])
- from xcarr and jJ have b: "j \<otimes> x \<in> J"
- by (simp add: ideal.I_r_closed[OF idealJ])
- from a b c show "\<exists>ha\<in>I. \<exists>ka\<in>J. (i \<oplus> j) \<otimes> x = ha \<oplus> ka"
- by fast
-next
- fix x i j
- assume xcarr: "x \<in> carrier R"
- and iI: "i \<in> I"
- and jJ: "j \<in> J"
- from xcarr ideal.Icarr[OF idealI iI] ideal.Icarr[OF idealJ jJ]
- have c: "x \<otimes> (i \<oplus> j) = (x \<otimes> i) \<oplus> (x \<otimes> j)" by algebra
- from xcarr and iI have a: "x \<otimes> i \<in> I"
- by (simp add: ideal.I_l_closed[OF idealI])
- from xcarr and jJ have b: "x \<otimes> j \<in> J"
- by (simp add: ideal.I_l_closed[OF idealJ])
- from a b c show "\<exists>ha\<in>I. \<exists>ka\<in>J. x \<otimes> (i \<oplus> j) = ha \<oplus> ka"
- by fast
+proof (rule ideal.intro)
+ show "additive_subgroup (I <+> J) R"
+ show "ring R"
+ by (rule is_ring)
+ show "ideal_axioms (I <+> J) R"
+ proof -
+ { fix x i j
+ assume xcarr: "x \<in> carrier R" and iI: "i \<in> I" and jJ: "j \<in> J"
+ from xcarr ideal.Icarr[OF idealI iI] ideal.Icarr[OF idealJ jJ]
+ have "\<exists>h\<in>I. \<exists>k\<in>J. (i \<oplus> j) \<otimes> x = h \<oplus> k"
+ by (meson iI ideal.I_r_closed idealJ jJ l_distr local.idealI) }
+ moreover
+ { fix x i j
+ assume xcarr: "x \<in> carrier R" and iI: "i \<in> I" and jJ: "j \<in> J"
+ from xcarr ideal.Icarr[OF idealI iI] ideal.Icarr[OF idealJ jJ]
+ have "\<exists>h\<in>I. \<exists>k\<in>J. x \<otimes> (i \<oplus> j) = h \<oplus> k"
+ by (meson iI ideal.I_l_closed idealJ jJ local.idealI r_distr) }
+ ultimately show "ideal_axioms (I <+> J) R"
+ by (intro ideal_axioms.intro) (auto simp: set_add_defs)
+ qed
qed
-
subsection (in ring) \<open>Ideals generated by a subset of @{term "carrier R"}\<close>
text \<open>@{term genideal} generates an ideal\<close>
@@ -350,17 +279,13 @@
lemma (in ring) genideal_self':
assumes carr: "i \<in> carrier R"
shows "i \<in> Idl {i}"
-proof -
- from carr have "{i} \<subseteq> Idl {i}" by (fast intro!: genideal_self)
- then show "i \<in> Idl {i}" by fast
-qed
text \<open>@{term genideal} generates the minimal ideal\<close>
lemma (in ring) genideal_minimal:
- assumes a: "ideal I R"
- and b: "S \<subseteq> I"
+ assumes "ideal I R" "S \<subseteq> I"
shows "Idl S \<subseteq> I"
- unfolding genideal_def by rule (elim InterD, simp add: a b)
+ unfolding genideal_def by rule (elim InterD, simp add: assms)
text \<open>Generated ideals and subsets\<close>
lemma (in ring) Idl_subset_ideal:
@@ -383,40 +308,40 @@
and HI: "H \<subseteq> I"
shows "Idl H \<subseteq> Idl I"
proof -
- from HI and genideal_self[OF Icarr] have HIdlI: "H \<subseteq> Idl I"
- by fast
-
from Icarr have Iideal: "ideal (Idl I) R"
by (rule genideal_ideal)
from HI and Icarr have "H \<subseteq> carrier R"
by fast
with Iideal have "(H \<subseteq> Idl I) = (Idl H \<subseteq> Idl I)"
by (rule Idl_subset_ideal[symmetric])
-
- with HIdlI show "Idl H \<subseteq> Idl I" by simp
+ then show "Idl H \<subseteq> Idl I"
+ by (meson HI Icarr genideal_self order_trans)
qed
lemma (in ring) Idl_subset_ideal':
assumes acarr: "a \<in> carrier R" and bcarr: "b \<in> carrier R"
- shows "(Idl {a} \<subseteq> Idl {b}) = (a \<in> Idl {b})"
- apply (subst Idl_subset_ideal[OF genideal_ideal[of "{b}"], of "{a}"])
- apply (fast intro: bcarr, fast intro: acarr)
- apply fast
- done
+ shows "Idl {a} \<subseteq> Idl {b} \<longleftrightarrow> a \<in> Idl {b}"
+proof -
+ have "Idl {a} \<subseteq> Idl {b} \<longleftrightarrow> {a} \<subseteq> Idl {b}"
+ by (simp add: Idl_subset_ideal acarr bcarr genideal_ideal)
+ also have "\<dots> \<longleftrightarrow> a \<in> Idl {b}"
+ by blast
+ finally show ?thesis .
+qed
lemma (in ring) genideal_zero: "Idl {\<zero>} = {\<zero>}"
- apply rule
- apply (rule genideal_minimal[OF zeroideal], simp)
- done
+proof
+ show "Idl {\<zero>} \<subseteq> {\<zero>}"
+ by (simp add: genideal_minimal zeroideal)
+ show "{\<zero>} \<subseteq> Idl {\<zero>}"
+qed
lemma (in ring) genideal_one: "Idl {\<one>} = carrier R"
proof -
interpret ideal "Idl {\<one>}" "R" by (rule genideal_ideal) fast
show "Idl {\<one>} = carrier R"
- apply (rule, rule a_subset)
- apply (simp add: one_imp_carrier genideal_self')
- done
+ using genideal_self' one_imp_carrier by blast
qed
@@ -429,58 +354,24 @@
lemma (in cring) cgenideal_ideal:
assumes acarr: "a \<in> carrier R"
shows "ideal (PIdl a) R"
- apply (unfold cgenideal_def)
- apply (rule idealI[OF is_ring])
- apply (rule subgroup.intro)
- apply simp_all
- apply (blast intro: acarr)
- apply clarsimp defer 1
- defer 1
- apply (fold a_inv_def, clarsimp) defer 1
- apply clarsimp defer 1
- apply clarsimp defer 1
-proof -
- fix x y
- assume xcarr: "x \<in> carrier R"
- and ycarr: "y \<in> carrier R"
- note carr = acarr xcarr ycarr
-
- from carr have "x \<otimes> a \<oplus> y \<otimes> a = (x \<oplus> y) \<otimes> a"
- with carr show "\<exists>z. x \<otimes> a \<oplus> y \<otimes> a = z \<otimes> a \<and> z \<in> carrier R"
- by fast
-next
- from l_null[OF acarr, symmetric] and zero_closed
- show "\<exists>x. \<zero> = x \<otimes> a \<and> x \<in> carrier R" by fast
-next
- fix x
- assume xcarr: "x \<in> carrier R"
- note carr = acarr xcarr
-
- from carr have "\<ominus> (x \<otimes> a) = (\<ominus> x) \<otimes> a"
- with carr show "\<exists>z. \<ominus> (x \<otimes> a) = z \<otimes> a \<and> z \<in> carrier R"
- by fast
-next
- fix x y
- assume xcarr: "x \<in> carrier R"
- and ycarr: "y \<in> carrier R"
- note carr = acarr xcarr ycarr
-
- from carr have "y \<otimes> a \<otimes> x = (y \<otimes> x) \<otimes> a"
- with carr show "\<exists>z. y \<otimes> a \<otimes> x = z \<otimes> a \<and> z \<in> carrier R"
- by fast
-next
- fix x y
- assume xcarr: "x \<in> carrier R"
- and ycarr: "y \<in> carrier R"
- note carr = acarr xcarr ycarr
-
- from carr have "x \<otimes> (y \<otimes> a) = (x \<otimes> y) \<otimes> a"
- with carr show "\<exists>z. x \<otimes> (y \<otimes> a) = z \<otimes> a \<and> z \<in> carrier R"
- by fast
+ unfolding cgenideal_def
+proof (intro subgroup.intro idealI[OF is_ring], simp_all)
+ show "{x \<otimes> a |x. x \<in> carrier R} \<subseteq> carrier R"
+ by (blast intro: acarr)
+ show "\<And>x y. \<lbrakk>\<exists>u. x = u \<otimes> a \<and> u \<in> carrier R; \<exists>x. y = x \<otimes> a \<and> x \<in> carrier R\<rbrakk>
+ \<Longrightarrow> \<exists>v. x \<oplus> y = v \<otimes> a \<and> v \<in> carrier R"
+ by (metis assms cring.cring_simprules(1) is_cring l_distr)
+ show "\<exists>x. \<zero> = x \<otimes> a \<and> x \<in> carrier R"
+ by (metis assms l_null zero_closed)
+ show "\<And>x. \<exists>u. x = u \<otimes> a \<and> u \<in> carrier R
+ \<Longrightarrow> \<exists>v. inv\<^bsub>add_monoid R\<^esub> x = v \<otimes> a \<and> v \<in> carrier R"
+ by (metis a_inv_def add.inv_closed assms l_minus)
+ show "\<And>b x. \<lbrakk>\<exists>x. b = x \<otimes> a \<and> x \<in> carrier R; x \<in> carrier R\<rbrakk>
+ \<Longrightarrow> \<exists>z. x \<otimes> b = z \<otimes> a \<and> z \<in> carrier R"
+ by (metis assms m_assoc m_closed)
+ show "\<And>b x. \<lbrakk>\<exists>x. b = x \<otimes> a \<and> x \<in> carrier R; x \<in> carrier R\<rbrakk>
+ \<Longrightarrow> \<exists>z. b \<otimes> x = z \<otimes> a \<and> z \<in> carrier R"
+ by (metis assms m_assoc m_comm m_closed)
qed
lemma (in ring) cgenideal_self:
@@ -504,119 +395,64 @@
interpret ideal J R by fact
show ?thesis
unfolding cgenideal_def
- apply rule
- apply clarify
- using aJ
- apply (erule I_l_closed)
- done
+ using I_l_closed aJ by blast
qed
lemma (in cring) cgenideal_eq_genideal:
assumes icarr: "i \<in> carrier R"
shows "PIdl i = Idl {i}"
- apply rule
- apply (intro cgenideal_minimal)
- apply (rule genideal_ideal, fast intro: icarr)
- apply (rule genideal_self', fast intro: icarr)
- apply (intro genideal_minimal)
- apply (rule cgenideal_ideal [OF icarr])
- apply (simp, rule cgenideal_self [OF icarr])
- done
+proof
+ show "PIdl i \<subseteq> Idl {i}"
+ by (simp add: cgenideal_minimal genideal_ideal genideal_self' icarr)
+ show "Idl {i} \<subseteq> PIdl i"
+ by (simp add: cgenideal_ideal cgenideal_self genideal_minimal icarr)
+qed
lemma (in cring) cgenideal_eq_rcos: "PIdl i = carrier R #> i"
unfolding cgenideal_def r_coset_def by fast
lemma (in cring) cgenideal_is_principalideal:
- assumes icarr: "i \<in> carrier R"
+ assumes "i \<in> carrier R"
shows "principalideal (PIdl i) R"
- apply (rule principalidealI)
- apply (rule cgenideal_ideal [OF icarr])
proof -
- from icarr have "PIdl i = Idl {i}"
- by (rule cgenideal_eq_genideal)
- with icarr show "\<exists>i'\<in>carrier R. PIdl i = Idl {i'}"
- by fast
+ have "\<exists>i'\<in>carrier R. PIdl i = Idl {i'}"
+ using cgenideal_eq_genideal assms by auto
+ then show ?thesis
+ by (simp add: cgenideal_ideal assms principalidealI)
qed
subsection \<open>Union of Ideals\<close>
lemma (in ring) union_genideal:
- assumes idealI: "ideal I R"
- and idealJ: "ideal J R"
+ assumes idealI: "ideal I R" and idealJ: "ideal J R"
shows "Idl (I \<union> J) = I <+> J"
- apply rule
- apply (rule ring.genideal_minimal)
- apply (rule is_ring)
- apply (rule add_ideals[OF idealI idealJ])
- apply (rule)
- apply (simp add: set_add_defs) apply (elim disjE) defer 1 defer 1
- apply (rule) apply (simp add: set_add_defs genideal_def) apply clarsimp defer 1
-proof -
- fix x
- assume xI: "x \<in> I"
- have ZJ: "\<zero> \<in> J"
- by (intro additive_subgroup.zero_closed) (rule ideal.axioms[OF idealJ])
- from ideal.Icarr[OF idealI xI] have "x = x \<oplus> \<zero>"
- by algebra
- with xI and ZJ show "\<exists>h\<in>I. \<exists>k\<in>J. x = h \<oplus> k"
- by fast
+proof
+ show "Idl (I \<union> J) \<subseteq> I <+> J"
+ proof (rule ring.genideal_minimal [OF is_ring])
+ show "ideal (I <+> J) R"
+ by (rule add_ideals[OF idealI idealJ])
+ have "\<And>x. x \<in> I \<Longrightarrow> \<exists>xa\<in>I. \<exists>xb\<in>J. x = xa \<oplus> xb"
+ by (metis additive_subgroup.zero_closed ideal.Icarr idealJ ideal_def local.idealI r_zero)
+ moreover have "\<And>x. x \<in> J \<Longrightarrow> \<exists>xa\<in>I. \<exists>xb\<in>J. x = xa \<oplus> xb"
+ by (metis additive_subgroup.zero_closed ideal.Icarr idealJ ideal_def l_zero local.idealI)
+ ultimately show "I \<union> J \<subseteq> I <+> J"
+ qed
next
- fix x
- assume xJ: "x \<in> J"
- have ZI: "\<zero> \<in> I"
- by (intro additive_subgroup.zero_closed, rule ideal.axioms[OF idealI])
- from ideal.Icarr[OF idealJ xJ] have "x = \<zero> \<oplus> x"
- by algebra
- with ZI and xJ show "\<exists>h\<in>I. \<exists>k\<in>J. x = h \<oplus> k"
- by fast
-next
- fix i j K
- assume iI: "i \<in> I"
- and jJ: "j \<in> J"
- and idealK: "ideal K R"
- and IK: "I \<subseteq> K"
- and JK: "J \<subseteq> K"
- from iI and IK have iK: "i \<in> K" by fast
- from jJ and JK have jK: "j \<in> K" by fast
- from iK and jK show "i \<oplus> j \<in> K"
- by (intro additive_subgroup.a_closed) (rule ideal.axioms[OF idealK])
+ show "I <+> J \<subseteq> Idl (I \<union> J)"
qed
-
subsection \<open>Properties of Principal Ideals\<close>
-text \<open>\<open>\<zero>\<close> generates the zero ideal\<close>
-lemma (in ring) zero_genideal: "Idl {\<zero>} = {\<zero>}"
- apply rule
- apply (simp add: genideal_minimal zeroideal)
- apply (fast intro!: genideal_self)
- done
-
-text \<open>\<open>\<one>\<close> generates the unit ideal\<close>
-lemma (in ring) one_genideal: "Idl {\<one>} = carrier R"
-proof -
- have "\<one> \<in> Idl {\<one>}"
- then show "Idl {\<one>} = carrier R"
- by (intro ideal.one_imp_carrier) (fast intro: genideal_ideal)
-qed
-
-
text \<open>The zero ideal is a principal ideal\<close>
corollary (in ring) zeropideal: "principalideal {\<zero>} R"
- apply (rule principalidealI)
- apply (rule zeroideal)
- apply (blast intro!: zero_genideal[symmetric])
- done
+ using genideal_zero principalidealI zeroideal by blast
text \<open>The unit ideal is a principal ideal\<close>
corollary (in ring) onepideal: "principalideal (carrier R) R"
- apply (rule principalidealI)
- apply (rule oneideal)
- apply (blast intro!: one_genideal[symmetric])
- done
-
+ using genideal_one oneideal principalidealI by blast
text \<open>Every principal ideal is a right coset of the carrier\<close>
lemma (in principalideal) rcos_generate:
@@ -626,21 +462,13 @@
interpret cring R by fact
from generate obtain i where icarr: "i \<in> carrier R" and I1: "I = Idl {i}"
by fast+
-
- from icarr and genideal_self[of "{i}"] have "i \<in> Idl {i}"
- by fast
- then have iI: "i \<in> I" by (simp add: I1)
-
- from I1 icarr have I2: "I = PIdl i"
+ then have "I = PIdl i"
-
- have "PIdl i = carrier R #> i"
+ moreover have "i \<in> I"
+ by (simp add: I1 genideal_self' icarr)
+ moreover have "PIdl i = carrier R #> i"
unfolding cgenideal_def r_coset_def by fast
-
- with I2 have "I = carrier R #> i"
- by simp
-
- with iI show "\<exists>x\<in>I. I = carrier R #> x"
+ ultimately show "\<exists>x\<in>I. I = carrier R #> x"
by fast
qed
@@ -698,11 +526,7 @@
then have I_prime: "\<And> a b. \<lbrakk>a \<in> carrier R; b \<in> carrier R; a \<otimes> b \<in> I\<rbrakk> \<Longrightarrow> a \<in> I \<or> b \<in> I"
by simp
have "primeideal I R"
- apply (rule primeideal.intro [OF is_ideal is_cring])
- apply (rule primeideal_axioms.intro)
- apply (rule InR)
- apply (erule (2) I_prime)
- done
+ by (simp add: I_prime InR is_cring is_ideal primeidealI)
with notprime show False by simp
qed
@@ -722,23 +546,15 @@
lemma (in cring) zeroprimeideal_domainI:
assumes pi: "primeideal {\<zero>} R"
shows "domain R"
- apply (rule domain.intro, rule is_cring)
- apply (rule domain_axioms.intro)
-proof (rule ccontr, simp)
- interpret primeideal "{\<zero>}" "R" by (rule pi)
- assume "\<one> = \<zero>"
- then have "carrier R = {\<zero>}" by (rule one_zeroD)
- from this[symmetric] and I_notcarr show False
- by simp
-next
- interpret primeideal "{\<zero>}" "R" by (rule pi)
- fix a b
- assume ab: "a \<otimes> b = \<zero>" and carr: "a \<in> carrier R" "b \<in> carrier R"
- from ab have abI: "a \<otimes> b \<in> {\<zero>}"
- by fast
- with carr have "a \<in> {\<zero>} \<or> b \<in> {\<zero>}"
- by (rule I_prime)
- then show "a = \<zero> \<or> b = \<zero>" by simp
+proof (intro domain.intro is_cring domain_axioms.intro)
+ show "\<one> \<noteq> \<zero>"
+ using genideal_one genideal_zero pi primeideal.I_notcarr by force
+ show "a = \<zero> \<or> b = \<zero>" if ab: "a \<otimes> b = \<zero>" and carr: "a \<in> carrier R" "b \<in> carrier R" for a b
+ proof -
+ interpret primeideal "{\<zero>}" "R" by (rule pi)
+ show "a = \<zero> \<or> b = \<zero>"
+ using I_prime ab carr by blast
+ qed
qed
corollary (in cring) domain_eq_zeroprimeideal: "domain R = primeideal {\<zero>} R"
@@ -765,37 +581,18 @@
shows "ideal {x\<in>carrier R. a \<otimes> x \<in> I} R"
proof -
interpret cring R by fact
- show ?thesis apply (rule idealI)
- apply (rule cring.axioms[OF is_cring])
- apply (rule subgroup.intro)
- apply (simp, fast)
- apply clarsimp apply (simp add: r_distr acarr)
- apply (simp add: a_inv_def[symmetric], clarify) defer 1
- apply clarsimp defer 1
- apply (fast intro!: helper_I_closed acarr)
- proof -
- fix x
- assume xcarr: "x \<in> carrier R"
- and ax: "a \<otimes> x \<in> I"
- from ax and acarr xcarr
- have "\<ominus>(a \<otimes> x) \<in> I" by simp
- also from acarr xcarr
- have "\<ominus>(a \<otimes> x) = a \<otimes> (\<ominus>x)" by algebra
- finally show "a \<otimes> (\<ominus>x) \<in> I" .
- from acarr have "a \<otimes> \<zero> = \<zero>" by simp
- next
- fix x y
- assume xcarr: "x \<in> carrier R"
- and ycarr: "y \<in> carrier R"
- and ayI: "a \<otimes> y \<in> I"
- from ayI and acarr xcarr ycarr have "a \<otimes> (y \<otimes> x) \<in> I"
- moreover
- from xcarr ycarr have "y \<otimes> x = x \<otimes> y"
- ultimately
- show "a \<otimes> (x \<otimes> y) \<in> I" by simp
+ show ?thesis
+ proof (rule idealI, simp_all)
+ show "ring R"
+ show "subgroup {x \<in> carrier R. a \<otimes> x \<in> I} (add_monoid R)"
+ by (rule subgroup.intro) (auto simp: r_distr acarr r_minus simp flip: a_inv_def)
+ show "\<And>b x. \<lbrakk>b \<in> carrier R \<and> a \<otimes> b \<in> I; x \<in> carrier R\<rbrakk>
+ \<Longrightarrow> a \<otimes> (x \<otimes> b) \<in> I"
+ using acarr helper_I_closed m_comm by auto
+ show "\<And>b x. \<lbrakk>b \<in> carrier R \<and> a \<otimes> b \<in> I; x \<in> carrier R\<rbrakk>
+ \<Longrightarrow> a \<otimes> (b \<otimes> x) \<in> I"
+ by (simp add: acarr helper_I_closed)
qed
qed
@@ -805,53 +602,27 @@
shows "primeideal I R"
proof -
interpret maximalideal I R by fact
- show ?thesis apply (rule ccontr)
- apply (rule primeidealCE)
- apply (rule is_cring)
- apply assumption
- proof -
- assume "\<exists>a b. a \<in> carrier R \<and> b \<in> carrier R \<and> a \<otimes> b \<in> I \<and> a \<notin> I \<and> b \<notin> I"
- then obtain a b where
- acarr: "a \<in> carrier R" and
- bcarr: "b \<in> carrier R" and
- abI: "a \<otimes> b \<in> I" and
- anI: "a \<notin> I" and
- bnI: "b \<notin> I" by fast
+ show ?thesis
+ proof (rule ccontr)
+ assume neg: "\<not> primeideal I R"
+ then obtain a b where acarr: "a \<in> carrier R" and bcarr: "b \<in> carrier R"
+ and abI: "a \<otimes> b \<in> I" and anI: "a \<notin> I" and bnI: "b \<notin> I"
+ using primeidealCE [OF is_cring]
+ by (metis I_notcarr)
define J where "J = {x\<in>carrier R. a \<otimes> x \<in> I}"
-
from is_cring and acarr have idealJ: "ideal J R"
unfolding J_def by (rule helper_max_prime)
-
have IsubJ: "I \<subseteq> J"
- proof
- fix x
- assume xI: "x \<in> I"
- with acarr have "a \<otimes> x \<in> I"
- by (intro I_l_closed)
- with xI[THEN a_Hcarr] show "x \<in> J"
- unfolding J_def by fast
- qed
-
+ using I_l_closed J_def a_Hcarr acarr by blast
from abI and acarr bcarr have "b \<in> J"
unfolding J_def by fast
with bnI have JnI: "J \<noteq> I" by fast
- from acarr
- have "a = a \<otimes> \<one>" by algebra
- with anI have "a \<otimes> \<one> \<notin> I" by simp
- with one_closed have "\<one> \<notin> J"
- unfolding J_def by fast
+ have "\<one> \<notin> J"
+ unfolding J_def by (simp add: acarr anI)
then have Jncarr: "J \<noteq> carrier R" by fast
-
- interpret ideal J R by (rule idealJ)
-
+ interpret ideal J R by (rule idealJ)
have "J = I \<or> J = carrier R"
- apply (intro I_maximal)
- apply (rule idealJ)
- apply (rule IsubJ)
- apply (rule a_subset)
- done
-
+ by (simp add: I_maximal IsubJ a_subset is_ideal)
with JnI and Jncarr show False by simp
qed
qed
@@ -859,54 +630,39 @@
subsection \<open>Derived Theorems\<close>
-\<comment> \<open>A non-zero cring that has only the two trivial ideals is a field\<close>
+text \<open>A non-zero cring that has only the two trivial ideals is a field\<close>
lemma (in cring) trivialideals_fieldI:
assumes carrnzero: "carrier R \<noteq> {\<zero>}"
and haveideals: "{I. ideal I R} = {{\<zero>}, carrier R}"
shows "field R"
- apply (rule cring_fieldI)
- apply (rule, rule, rule)
- apply (erule Units_closed)
- defer 1
- apply rule
- defer 1
-proof (rule ccontr, simp)
- assume zUnit: "\<zero> \<in> Units R"
- then have a: "\<zero> \<otimes> inv \<zero> = \<one>" by (rule Units_r_inv)
- from zUnit have "\<zero> \<otimes> inv \<zero> = \<zero>"
- by (intro l_null) (rule Units_inv_closed)
- with a[symmetric] have "\<one> = \<zero>" by simp
- then have "carrier R = {\<zero>}" by (rule one_zeroD)
- with carrnzero show False by simp
-next
- fix x
- assume xcarr': "x \<in> carrier R - {\<zero>}"
- then have xcarr: "x \<in> carrier R" by fast
- from xcarr' have xnZ: "x \<noteq> \<zero>" by fast
- from xcarr have xIdl: "ideal (PIdl x) R"
- by (intro cgenideal_ideal) fast
-
- from xcarr have "x \<in> PIdl x"
- by (intro cgenideal_self) fast
- with xnZ have "PIdl x \<noteq> {\<zero>}" by fast
- with haveideals have "PIdl x = carrier R"
- by (blast intro!: xIdl)
- then have "\<one> \<in> PIdl x" by simp
- then have "\<exists>y. \<one> = y \<otimes> x \<and> y \<in> carrier R"
- unfolding cgenideal_def by blast
- then obtain y where ycarr: " y \<in> carrier R" and ylinv: "\<one> = y \<otimes> x"
- by fast+
- from ylinv and xcarr ycarr have yrinv: "\<one> = x \<otimes> y"
- from ycarr and ylinv[symmetric] and yrinv[symmetric]
- have "\<exists>y \<in> carrier R. y \<otimes> x = \<one> \<and> x \<otimes> y = \<one>" by fast
- with xcarr show "x \<in> Units R"
- unfolding Units_def by fast
+proof (intro cring_fieldI equalityI)
+ show "Units R \<subseteq> carrier R - {\<zero>}"
+ by (metis Diff_empty Units_closed Units_r_inv_ex carrnzero l_null one_zeroD subsetI subset_Diff_insert)
+ show "carrier R - {\<zero>} \<subseteq> Units R"
+ proof
+ fix x
+ assume xcarr': "x \<in> carrier R - {\<zero>}"
+ then have xcarr: "x \<in> carrier R" and xnZ: "x \<noteq> \<zero>" by auto
+ from xcarr have xIdl: "ideal (PIdl x) R"
+ by (intro cgenideal_ideal) fast
+ have "PIdl x \<noteq> {\<zero>}"
+ using xcarr xnZ cgenideal_self by blast
+ with haveideals have "PIdl x = carrier R"
+ by (blast intro!: xIdl)
+ then have "\<one> \<in> PIdl x" by simp
+ then have "\<exists>y. \<one> = y \<otimes> x \<and> y \<in> carrier R"
+ unfolding cgenideal_def by blast
+ then obtain y where ycarr: " y \<in> carrier R" and ylinv: "\<one> = y \<otimes> x"
+ by fast
+ have "\<exists>y \<in> carrier R. y \<otimes> x = \<one> \<and> x \<otimes> y = \<one>"
+ using m_comm xcarr ycarr ylinv by auto
+ with xcarr show "x \<in> Units R"
+ unfolding Units_def by fast
+ qed
qed
lemma (in field) all_ideals: "{I. ideal I R} = {{\<zero>}, carrier R}"
- apply (rule, rule)
-proof -
+proof (intro equalityI subsetI)
fix I
assume a: "I \<in> {I. ideal I R}"
then interpret ideal I R by simp
@@ -916,41 +672,26 @@
case True
then obtain a where aI: "a \<in> I" and anZ: "a \<noteq> \<zero>"
by fast+
- from aI[THEN a_Hcarr] anZ have aUnit: "a \<in> Units R"
+ have aUnit: "a \<in> Units R"
+ by (simp add: aI anZ field_Units)
then have a: "a \<otimes> inv a = \<one>" by (rule Units_r_inv)
from aI and aUnit have "a \<otimes> inv a \<in> I"
by (simp add: I_r_closed del: Units_r_inv)
then have oneI: "\<one> \<in> I" by (simp add: a[symmetric])
-
have "carrier R \<subseteq> I"
- proof
- fix x
- assume xcarr: "x \<in> carrier R"
- with oneI have "\<one> \<otimes> x \<in> I" by (rule I_r_closed)
- with xcarr show "x \<in> I" by simp
- qed
+ using oneI one_imp_carrier by auto
with a_subset have "I = carrier R" by fast
then show "I \<in> {{\<zero>}, carrier R}" by fast
next
case False
then have IZ: "\<And>a. a \<in> I \<Longrightarrow> a = \<zero>" by simp
-
have a: "I \<subseteq> {\<zero>}"
- proof
- fix x
- assume "x \<in> I"
- then have "x = \<zero>" by (rule IZ)
- then show "x \<in> {\<zero>}" by fast
- qed
-
+ using False by auto
have "\<zero> \<in> I" by simp
- then have "{\<zero>} \<subseteq> I" by fast
-
with a have "I = {\<zero>}" by fast
then show "I \<in> {{\<zero>}, carrier R}" by fast
qed
+qed (auto simp: zeroideal oneideal)
\<comment>\<open>"Jacobson Theorem 2.2"\<close>
lemma (in cring) trivialideals_eq_field:
@@ -961,9 +702,7 @@
text \<open>Like zeroprimeideal for domains\<close>
lemma (in field) zeromaximalideal: "maximalideal {\<zero>} R"
- apply (rule maximalidealI)
- apply (rule zeroideal)
-proof-
+proof (intro maximalidealI zeroideal)
from one_not_zero have "\<one> \<notin> {\<zero>}" by simp
with one_closed show "carrier R \<noteq> {\<zero>}" by fast
next
@@ -977,20 +716,20 @@
lemma (in cring) zeromaximalideal_fieldI:
assumes zeromax: "maximalideal {\<zero>} R"
shows "field R"
- apply (rule trivialideals_fieldI, rule maximalideal.I_notcarr[OF zeromax])
- apply rule apply clarsimp defer 1
- apply (simp add: zeroideal oneideal)
-proof -
- fix J
- assume Jn0: "J \<noteq> {\<zero>}"
- and idealJ: "ideal J R"
- interpret ideal J R by (rule idealJ)
- have "{\<zero>} \<subseteq> J" by (rule ccontr) simp
- from zeromax and idealJ and this and a_subset
- have "J = {\<zero>} \<or> J = carrier R"
- by (rule maximalideal.I_maximal)
- with Jn0 show "J = carrier R"
- by simp
+proof (intro trivialideals_fieldI maximalideal.I_notcarr[OF zeromax])
+ have "J = carrier R" if Jn0: "J \<noteq> {\<zero>}" and idealJ: "ideal J R" for J
+ proof -
+ interpret ideal J R by (rule idealJ)
+ have "{\<zero>} \<subseteq> J"
+ by force
+ from zeromax idealJ this a_subset
+ have "J = {\<zero>} \<or> J = carrier R"
+ by (rule maximalideal.I_maximal)
+ with Jn0 show "J = carrier R"
+ by simp
+ qed
+ then show "{I. ideal I R} = {{\<zero>}, carrier R}"
+ by (auto simp: zeroideal oneideal)
qed
lemma (in cring) zeromaximalideal_eq_field: "maximalideal {\<zero>} R = field R"``` | 11,478 | 32,519 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.623415 |
https://metanumbers.com/21401 | 1,600,831,296,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400209665.4/warc/CC-MAIN-20200923015227-20200923045227-00015.warc.gz | 487,778,615 | 7,394 | ## 21401
21,401 (twenty-one thousand four hundred one) is an odd five-digits prime number following 21400 and preceding 21402. In scientific notation, it is written as 2.1401 × 104. The sum of its digits is 8. It has a total of 1 prime factor and 2 positive divisors. There are 21,400 positive integers (up to 21401) that are relatively prime to 21401.
## Basic properties
• Is Prime? Yes
• Number parity Odd
• Number length 5
• Sum of Digits 8
• Digital Root 8
## Name
Short name 21 thousand 401 twenty-one thousand four hundred one
## Notation
Scientific notation 2.1401 × 104 21.401 × 103
## Prime Factorization of 21401
Prime Factorization 21401
Prime number
Distinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 21401 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 9.97119 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 21,401 is 21401. Since it has a total of 1 prime factor, 21,401 is a prime number.
## Divisors of 21401
2 divisors
Even divisors 0 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 21402 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 10701 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 146.291 Returns the nth root of the product of n divisors H(n) 1.99991 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 21,401 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 21,401) is 21,402, the average is 10,701.
## Other Arithmetic Functions (n = 21401)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 21400 Total number of positive integers not greater than n that are coprime to n λ(n) 21400 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2406 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 21,400 positive integers (less than 21,401) that are coprime with 21,401. And there are approximately 2,406 prime numbers less than or equal to 21,401.
## Divisibility of 21401
m n mod m 2 3 4 5 6 7 8 9 1 2 1 1 5 2 1 8
21,401 is not divisible by any number less than or equal to 9.
• Arithmetic
• Prime
• Deficient
• Polite
• Prime Power
• Square Free
## Base conversion (21401)
Base System Value
2 Binary 101001110011001
3 Ternary 1002100122
4 Quaternary 11032121
5 Quinary 1141101
6 Senary 243025
8 Octal 51631
10 Decimal 21401
12 Duodecimal 10475
20 Vigesimal 2da1
36 Base36 gih
## Basic calculations (n = 21401)
### Multiplication
n×i
n×2 42802 64203 85604 107005
### Division
ni
n⁄2 10700.5 7133.67 5350.25 4280.2
### Exponentiation
ni
n2 458002801 9801717944201 209766565723845601 4489214273056019707001
### Nth Root
i√n
2√n 146.291 27.7637 12.0951 7.34661
## 21401 as geometric shapes
### Circle
Diameter 42802 134466 1.43886e+09
### Sphere
Volume 4.10573e+13 5.75543e+09 134466
### Square
Length = n
Perimeter 85604 4.58003e+08 30265.6
### Cube
Length = n
Surface area 2.74802e+09 9.80172e+12 37067.6
### Equilateral Triangle
Length = n
Perimeter 64203 1.98321e+08 18533.8
### Triangular Pyramid
Length = n
Surface area 7.93284e+08 1.15514e+12 17473.8
## Cryptographic Hash Functions
md5 f9a78afdf5be220d8779569e5155eab1 d68139c7a8f5bc67b57f5033a7d0ae6c4f791d31 0d7a4916a6be81649800ce482ac5f7cd59486d3dd85752f108981ee6937e5371 a6524c6a6788662f64e21d2ac3ff0c5f238c9a81484f8fd22ed32d5a48c607efda85443b3abe0bdb0a7af75528c1b57c285502c503b3c95ffebc05e1ed3caecc aa3fc7b06a3c32dacb749cc766bb8dfeacd4ad09 | 1,425 | 4,044 | {"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} | 3.8125 | 4 | CC-MAIN-2020-40 | longest | en | 0.8259 |
https://www.bestfreenow.com/article/1927745.html | 1,653,125,018,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662539049.32/warc/CC-MAIN-20220521080921-20220521110921-00740.warc.gz | 761,510,400 | 7,360 | # 87 divided by 2
## Calculate 87 divided by 2 using the long division method
Calculate 87 divided by 2 using the long division method. Here is the answer to questions like: Calculate 87 divided by 2 using the long division method or long division with remainders: 87/2. This calculator shows all the work and steps for long division. You just need to enter the dividend and divisor values. The answer will be detailed below.
## 87 divided by 2 how do you do long division
87 divided by 2 how do you do long division. Here is the answer to questions like: 87 divided by 2 how do you do long division or long division with remainders: 87/2. This calculator shows all the work and steps for long division. You just need to enter the dividend and divisor values. The answer will be
## View question What''s 87 divided by two
What''s 87 divided by two. Guest Dec 17, 2014. 0 users composing answers.. Best Answer #1 +5 43.5. 80 divided by 2 is 40. 7 divided by two is 3.5, or three and a half. it is negative because there is one negative and one positive. in multication, or division positve and positive is positive.
## What is 87 divided by 2/6 Divisible
The numbers in 87 divided by 2/6 are labeled below: 87 = whole number 2 = numerator 6 = denominator
## Long Division Calculator with Step by Step Work
Long division calculator with step by step work for 3rd grade, 4th grade, 5th grade and 6th grade students to verify the results of long division problems with or without remainder. Generate work with steps for 2 by 1, 3by 2, 3 by 1, 4 by 3, 4by 2, 4 by 1, 5 by 4, 5 by 3, 5 by 2, 6 by 4, 6 by 3 and 6 by 2 digit long division practice or homework exercises.
## Long Division Calculator Learn to do long division with
Remember: A decimal number, say, 3 can be written as 3.0, 3.00 and so on. Bring down next digit 0. Divide 10 by 2. Write the remainder after subtracting the bottom number from the top number. End of long division (Remainder is 0 and next digit after decimal is 0).
## How to Calculate 20/87 divided 2/32 (What is 20/87 ÷ 2/32)
Learn how to calculate 20/87 divided by 2/32. A simple, stepbystep guide with instructions to work out what 20/87 ÷ 2/32 is.
## How to calculate 87 divided by 30 using long division
.9 The answer to 87 divided by 30 can also be written as a mixed fraction as follows: 2 27/30 Note that the numerator in the fraction above is the remainder and the denominator is the divisor. How to calculate 87 divided by 31 using long division | 689 | 2,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.921782 |
https://physics.stackexchange.com/questions/719636/calculating-heat-flux-per-unit-length-of-wire | 1,723,528,236,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641063659.79/warc/CC-MAIN-20240813043946-20240813073946-00889.warc.gz | 348,776,829 | 41,190 | # Calculating heat flux per unit length of wire
I'm trying to apply a formula to calculate the thermal conductivity of a certain material.
$$\lambda=\frac{\dot q_1}{4\pi\cdot \frac{\Delta T}{\Delta log(t)}}$$
where $$\lambda$$ is thermal conductivity, and the fraction in the denominator is the linear slope of a plot that I have.
My setup is simply a hot wire embedded in said material, heating up the material around it via resistive/joule heating.
The one variable I'm not sure about, $$\dot q_1$$, from varying sources, is described as "the specific heat output of the linear heat source (emitted heat energy per unit time and unit length)" or also described as the "heat flux per unit length of the linear source".
Are these 2 terms the same? If so, how do I go about calculating this heat flux per unit length? I know the innate resistance of the wire, the current flowing through it and the voltage, as well as the temperature of the wire.
• What is the exact experiment? Where are the temperatures measured? Are you sure that isn't a $\Delta ln(r)$ in the denominator? Commented Jul 23, 2022 at 9:19
• How accurate do you need your answer to be? These kinds of measurements can be tricky. If you look at the 3omega method or Angstrom method you will see that people will periodically heat the wire. The dimensions compared to the thermal wavelength can also be important. If it is a long wire in a cylinder then that helps. But keep in mind the resistance of the wire changes with temp. And if temperature differences are large the material will too. Commented Jul 23, 2022 at 17:50
• The above comment aside, yes I think the two are a the same. Joule heating is VI. And that would be The heat flux for the whole length of wire. If the problem is one dimensional like the wire a cylinder then you can think about the problem in terms of series resistances. Commented Jul 23, 2022 at 17:57
• Too be clear, thermal resistances in series perpendicular to the wire. Commented Jul 23, 2022 at 18:08
• @ChetMiller The experiment consists of surrounding a wire in a uniform, cylindrical bed of particles. I heat up the wire via current flowing through it, and I try to measure the temperature of the surrounding particle bed to calculate thermal conductivity of the material that surrounds the wire. The temperature measured is simply the outer edge of the bed, at a constant r. As for the denominator, here is one source I am using showing how that formula is derived. tec-science.com/thermodynamics/heat/… Commented Jul 24, 2022 at 9:25
I would not determine $$\lambda$$ the way they say. I would make a graph on semi-logarithmic paper of temperature T on the linear axis as a function of time t on the logarithmic scale axis, and determine the slope of the straight line section of the variation: $$slope=\frac{\dot{q}}{4\pi \lambda}$$That way you could use temperature data taken at many points in time, and also be certain that you are considering only the linear portion of the behavior. Any decent plotting software can take T vs t data and convert the t axis to logarithmic with a single click. It can also easily fit the data to a semilog relationship so that the slope is immediately available. | 746 | 3,213 | {"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": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-33 | latest | en | 0.912415 |
qyremakozupapateg.komabraindeathcuba.com | 1,624,193,913,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487662882.61/warc/CC-MAIN-20210620114611-20210620144611-00630.warc.gz | 425,691,185 | 9,991 | # Potential flows
computer graphic solutions by Robert H. Kirchhoff
Publisher: M. Dekker in New York
Written in English
## Subjects:
• Fluid mechanics -- Data processing.
## Edition Notes
Includes bibliographies and index.
1. Using potential flow theory, plot the streamlines of a uniform flow as it flows over a long cylinder with radius a = 2 m. The velocity of the uniform flow is U = 6 m/s. Then plot the pressure over the surface of the cylinder as computed using Bernoulli's equation. The upstream pressure is atmospheric (i.e. zero). The operating cash flow formula can provide you with insight into your business's profitability. The Blueprint walks you through understanding operating cash flow. Fluid mechanics - Fluid mechanics - Hydrodynamics: Up to now the focus has been fluids at rest. This section deals with fluids that are in motion in a steady fashion such that the fluid velocity at each given point in space is not changing with time. Any flow pattern that is steady in this sense may be seen in terms of a set of streamlines, the trajectories of imaginary particles suspended in. The full potential equation and its small disturbance approximation are derived from the Navier-Stokes equations and the validity of the equations for transonic flow are discussed. Some of the phenomena that arise in transonic flows are discussed, including the existence of shock free supercritical conditions. A brief historical review of the subject is followed by a more detailed description.
An enjoyable book to read, I used it to prepare for my master exam. However many of the answers to the practice questions are wrong which can waste a whole lot of time and when I bought the book, it didn't come with First of all - if you're looking for a 'quick' and 'detailed' guide to physics, I suggest trying to find something else and don't /5(3). A numerical method incorporating some of the ideas underlying the wake source model of Parkinson & Jandali () is presented for calculating the incompressible potential flow external to a bluff body and its wake. The effect of the wake is modelled by placing . Potential energy also includes other forms. The energy stored between the plates of a charged capacitor is electrical potential energy. What is commonly known as chemical energy, the capacity of a substance to do work or to evolve heat by undergoing a change of composition, may be regarded as potential energy resulting from the mutual forces among its molecules and atoms. 1 day ago The driving force for this flow is a water potential gradient. Thus, in order for water to flow, the leaf water potential must be lower than the soil water potential. In Figure 2, the soil is at MPa and the roots are slightly more negative at MPa. This means the roots will pull water up from the soil.
Failing to achieve hospital-wide patient flow — the right care, in the right place, at the right time — puts patients at risk for suboptimal care and potential harm. It also increases the burden on clinicians and hospital staff and can accelerate burnout. We also see the potential, at current commodity levels, for Range Resources to generate significant free cash flow (\$ million) and an attractive free cash flow yield at current stock prices. Examples of Cash Flow Statements. Simple yearly, quarterly, and monthly cash flow statements and budgets can demonstrate existing or potential cash flow problems. A seasonal drop-off in revenue can result in negative cash flow, as demonstrated in the following statements for a fictitious landscaping business. The Story of Potential Flow - Potential Flow.
## Recent
This book illustrates how potential flows enter into the general theory of motions of viscous and viscoelastic fluids. Traditionally, the theory of potential flow is presented as a subject called 'potential flow of an inviscid fluid'; when the fluid is incompressible these fluids are, curiously, said to Cited by: Introduction to Potential Flow The concept of vorticity ω¯ = ∇∧ ¯v () was introduced previously when analyzing the kinematics of fluid motion.
Physically, it rep- resents twice the angular velocity with which the fluid particle rotates as a solid body. Potential Flow Theory “When a flow is both frictionless and irrotational, pleasant things happen.” –F.M. White, Fluid Mechanics 4th ed. We can treat external flows around bodies as invicid (i.e.
frictionless) and irrotational (i.e. the fluid particles are not rotating). This is because the viscous effects are limited toFile Size: 2MB. After a brief introduction to BASIC programming, this book goes discussing the fundamentals of potential flow theory that will be used to calculate flows, which can be regarded as frictionless.
These topics are followed by a presentation of some analytical expressions for potential flows. Potential Flow 3 LEARNING OBJECTIVES •Learn to calculate the air flow and pressure distribution around various body shapes.
•Learn more about the classical assumption of irrotational flow and its meaning that the vorticity is everywhere zero. Note that this also implies inviscid flow. Irrotational flows are potential fields. As opposed to the radial flow direction (which was discussed under the source and sink) the flow in the tangential direction is referred to as the free vortex flow.
Another typical name for this kind of flow is the potential vortex flow. The flow is circulating the origin or another point.
An Internet Book on Fluid Dynamics Potential Flow around Potential flows book Cylinder Superimposing a uniform stream of velocity,U, on the potential flow due a doublet oriented in thex Figure 1: Streamlines in the potential flow of a doublet in a uniform stream.
After several elements of the potential flow Potential flows book built earlier, the first Potential flows book of these elements can be demonstrated.
Perhaps the most celebrated and useful example is the flow past a cylinder which this section will be dealing with. The stream function made by superimposing a uniform flow and a doublet is \[ \label{if:eq:stream-U+doublet1}.
The Story of Potential Flow. Potential flow is an idealized model of fluid dynamics that applies when a flow is inviscid, incompressible, and three simplifying assumptions lead to a convenient mathematical model of the fluid flow, and despite the simplification, potential flow is a very practical model that is widely used in aerodynamics and hydrodynamics.
An Internet Book on Fluid Dynamics Method of Complex Variables for Planar Potential Flows We have established that if we denote the complex potential,φ+iψ,byfand the complex position vector, x+iy,byzthen any function,f(z), corresponds to a particular planar potential flow and, furthermore, that the derivativedf/dz=u−iv.
Potential flows Potential functions (and stream functions,) can be defined for various simple flows. These potential functions can also be superimposed with other potential functions to create more complex flows. Uniform, Free Stream Flow (1D). In fluid dynamics, Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid 's potential energy.
The principle is named after Daniel Bernoulli who published it in his book Hydrodynamica in Free Surface Flow: Environmental Fluid Mechanics introduces a wide range of environmental fluid flows, such as water waves, land runoff, channel flow, and effluent discharge.
The book provides systematic analysis tools and basic skills for study fluid mechanics in natural and constructed environmental flows.
Get this from a library. Potential flows: computer graphic solutions. [Robert H Kirchhoff] -- Compiling 70 well-known potential flows in a unique, convenient format, this first-of-its-kind reference provides detailed computer graphic drawings in a nondimensional style that allows each.
Potential flows of incompressible fluids are solutions of the Navier-Stokes equations which satisfy Laplace’s equa- Carlos could be considered to be an author of this book and we are lucky that he came along. A subject like Flows near internal stagnation points in viscous incompressible fluids COVID Resources.
Reliable information about the coronavirus (COVID) is available from the World Health Organization (current situation, international travel).Numerous and frequently-updated resource results are available from this ’s WebJunction has pulled together information and resources to assist library staff as they consider how to handle coronavirus.
Potential flow is same as irrotational flow. These are flows in which the fluid particles do not rotate, their angular velocity is zero. And angular velocity of a flow is defined as, $ω = \dfrac{1}{2} \cdot \nabla \times \vec V \tag 1$.
potential flow is exactly analogous to the theory of potentials in electricity and magnetism. The mass sources coincide with the distribution of electric charges and the vorticity coincides with the electric currents.
USEFUL SPECIAL FUNCTIONS A function that is highly useful in the development of potential theory is the smooth. Exercise: The two-dimensional flow of a nonviscous, incompressible fluid in the vicinity of a corner is described by the stream function =2 2sin(2𝜃) where has units of m2/s when is in the fluid density is kg/m3 and the plane is horizontal.
a) Determine, if possible, the corresponding velocity potential. Thus, assuming that, in the z-plane, the potential represents uniform flow of unperturbed speed V 0, running parallel to the x-axis (which follows because at large the map reduces to, and so the flow at large distances from the origin is the same in the complex z- and ζ-planes), around an elliptical cylinder of major radius a, aligned along.
Understand the basic principles behind potential flow. To schematically model flows applying basic potential flow elements and the superposition principle. To perform basic flow computations applying potential flow theory. This lecture will cover chapter 3 of the book.
LIMITS ON SEPARATED FLOW Kelvin-Helmoltz instability Stratified flow instability Annular flow instability 8 INTERNAL FLOW ENERGY CONVERSION INTRODUCTION FRICTIONAL LOSS IN DISPERSE FLOW Horizontal Flow Homogeneous flow friction Heterogeneous flow friction Stokes flow (named after George Gabriel Stokes), also named creeping flow or creeping motion, is a type of fluid flow where advective inertial forces are small compared with viscous forces.
The Reynolds number is low, i.e. ≪.This is a typical situation in flows where the fluid velocities are very slow, the viscosities are very large, or the length-scales of the flow are very small.
Book • Seventh Edition • Some examples are also presented in the areas of potential flow, conservation and Galerkin finite element method. Select Chapter 2 - Convection-Dominated Problems: Finite Element Approximations to the Convection-Diffusion-Reaction Equation.
The goal of this book is to show how potential flows enter into the general theory of motions of viscous and viscoelastic fluids. Traditionally, the theory of potential flows is thought to apply to idealized fluids without viscosity. Here we show how to apply this theory to real fluids that are viscous.
(Side note: If you're looking for something great to read related to your career or business, then join over 1 million others and start your day with the latest news from Wall St.
to Silicon Valley. This newsletter is a 5-minute read that's informative, witty and FREE!) 21 Best Personal Development and Self-Improvement Books 1. The Only Skill That Matters by Jonathan A. Levi. Chapter 8 • Potential Flow and Computational Fluid Dynamics Solution: Evaluation of the laplacian of (1/r) shows that it is not legitimate: 11 1 1 1 ∇= = − =≠ 2 2 rr.
rrr rr rrr Ans 3 1 0 Illegitimate r Consider the two-dimensional velocity distribution u = –By, v = +Bx, where B is a constant. If this flow possesses a stream function, find its form. POTENTIAL FLOW In this tutorial, you will study the flow of ideal fluids.
On completion, you should be able to do the following. Note that text books and examiners often use m for the strength of the source and this has the same meaning as Q. A sink is the exact opposite of a source. d =. The chapter includes a review of the literature in the field, the mathematical foundations of potential flows, an introductory section on incompressible flows, and a formulation for compressible flows with the derivation of a boundary integral equation for the velocity potential equation in a frame of reference moving in arbitrary motion.
disturbances to the flow, and Mach 5 might be the starting point for more highly streamlined bodies. In this section we will provide a brief outline of the key distinguishing concepts. The books by Bertin and Cummings1 and Anderson2 provide a starting point for further study.
Essentially there are five key points to. A complete set of lecture notes for an upper-division undergraduate Fluid Mechanics course. The course concentrates on those aspects of fluid mechanics that can be studied analytically.
Topics covered include hydrodynamics, surface tension, boundary layers, potential flow, aerodynamics, viscous flow, and waves. Author(s): Richard Fitzpatrick.Stanford University.Potential flows of viscous and viscoelastic fluids By Daniel Joseph, Toshio Funada, Jing Wang | Pages | ISBN: | PDF | 11 MBThe goal of this book is to show how potential flows enter into the general theory of motions of viscous and viscoelastic fluids. | 2,836 | 13,562 | {"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} | 2.765625 | 3 | CC-MAIN-2021-25 | longest | en | 0.935297 |
https://www.fmaths.com/tips/question-branch-of-mathematics-where-symbols-or-letters-are-used-to-represent-a-numbers.html | 1,624,560,330,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556482.89/warc/CC-MAIN-20210624171713-20210624201713-00107.warc.gz | 699,830,868 | 11,217 | # Question: Branch Of Mathematics Where Symbols Or Letters Are Used To Represent A Numbers?
## Which branch of mathematics that uses letters as symbols?
Algebra is the branch of mathematics that uses letters or symbols to represent unknown numbers and values, often to show that certain relationships between numbers are true for all numbers in a specified set.
## What kind of math uses letters?
Algebra is a branch of mathematics that uses letters and other symbols to represent numbers and quantities in formulas and equations.
## What are the 10 branches of mathematics?
Major divisions of mathematics
• Foundations (including set theory and mathematical logic)
• Number theory.
• Algebra.
• Combinatorics.
• Geometry.
• Topology.
• Mathematical analysis.
• Probability and statistics.
## What are the branches of mathematics?
The main branches of mathematics are algebra, number theory, geometry and arithmetic. Based on these branches, other branches have been discovered.
## Which branch of mathematics that uses letters as symbols when calculating takes its name from the Arabic for restoring of broken parts 1 point?
Algebra is the branch of mathematics, that uses letters as symbols when calculating, takes its name from the Arabic for ‘restoring of broken parts ‘. Algebra is the branch of mathematics that deals with mathematical symbols and alphabets when calculating equations.
You might be interested: FAQ: What Is Union Set In Mathematics?
## What does R mean in math?
In maths, the letter R denotes the set of all real numbers. In other words, real numbers are defined as the points on an infinitely extended line. This line is called the number line or the real line, on which the points of integers are evenly ranged.
## What does this math sign mean?
< Less Than and > Greater Than. This symbol < means less than, for example 2 < 4 means that 2 is less than 4. ≤ ≥ These symbols mean 'less than or equal to' and 'greater than or equal to' and are commonly used in algebra.
## Who is the father of mathematics?
Archimedes is known as the Father Of Mathematics. He lived between 287 BC – 212 BC. Syracuse, the Greek island of Sicily was his birthplace.
## What is basic math called?
Generally, counting, addition, subtraction, multiplication and division are called the basic math operation. The other mathematical concept are built on top of the above 4 operations. These conepts along with different type of numbers, factors, lcm and gcf makes students ready for learning fraction.
## What is the hardest branch of math?
Originally Answered: Which is the toughest branch of mathematics? Geometry and trig are both really basic. Algebra can get very difficult at the university level, especially in graduate programs when you start to generalize concepts to abstract algebra and then explore commutative algebra.
## What are the three types of maths?
Modern mathematics can be divided into three main branches: continuous mathematics, algebra, and discrete mathematics. The division is not exhaustive. It is difficult to exactly fit some fields, such as geometry or mathematical logic, into any of these categories. | 627 | 3,159 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2021-25 | longest | en | 0.916589 |
https://www.teacherspayteachers.com/Product/Right-Triangle-Trigonometry-Slap-Math-2503561 | 1,493,579,869,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917125841.92/warc/CC-MAIN-20170423031205-00560-ip-10-145-167-34.ec2.internal.warc.gz | 967,444,278 | 24,026 | Total:
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# Right Triangle Trigonometry Slap Math!
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Students will use right triangle trigonometry (SOHCAHTOA) to set up trigonometric ratios, find missing sides in, find missing angles in, and solve right triangles.
The game is very similar to the card game Slapjack. In short, students will divide up a deck of “cards” (math questions) evenly among 4-5 players and hold their cards face down. Learners will then take turns flipping over the top card in their individual piles and placing them in a discard pile that can be easily reached by all players. Once a card has been flipped over, all students must work the problem out on their student activity sheet and determine the correct answer. If the card flipped over has no hand print on it, players will work together to answer the question before moving on. However, if the card flipped over has a hand print on it, the first student to accurately solve the problem and slap the card gets to keep it. My students LOVED playing this game in class last week!
With the purchase of this product you will also receive a set of cards without hand prints on them (note: the questions are the same). This set would be great if you would prefer to use this resource as task cards!
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Sign up | 525 | 2,236 | {"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} | 2.75 | 3 | CC-MAIN-2017-17 | longest | en | 0.938598 |
https://tfgraph.readthedocs.io/en/latest/algorithms_pagerank.html | 1,591,260,949,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347439213.69/warc/CC-MAIN-20200604063532-20200604093532-00404.warc.gz | 533,488,860 | 7,967 | # tfgraph.algorithms.pagerank¶
tfgraph.algorithms.pagerank Module
This module contains a set of PageRank’s algorithm implementations on the tfgraph module.
## PageRank¶
class `tfgraph.algorithms.pagerank.pagerank.``PageRank`(sess: tensorflow.python.client.session.Session, name: str, beta: float, T: tfgraph.algorithms.pagerank.transition.transition.Transition, writer: tensorflow.python.summary.writer.writer.FileWriter = None, is_sparse: bool = False) → None[source]
PageRank base class.
This class model the PageRank algorithm as Abstract Class containing all methods that the heir classes need to implements. Also, this class provides a set of attributes that helps to implement the algorithm.
The PageRank algorithm calculates the rank of each vertex in a graph based on the relational structure from them and giving more importance to the vertices that connects with edges to vertices with very high in-degree recursively.
This class depends on the TensorFlow library, so it’s necessary to install it to properly work.
`sess`
`tf.Session` – This attribute represents the session that runs the TensorFlow operations.
`name`
str – This attribute represents the name of the object in TensorFlow’s op Graph.
`G`
`tfgraph.Graph` – The graph on witch it will be calculated the algorithm. It will be treated as Directed Weighted Graph.
`beta`
float – The reset probability of the random walks, i.e. the probability that a user that surfs the graph an decides to jump to another vertex not connected to the current.
`T`
`tfgraph.Transition` – The transition matrix that provides the probability distribution relative to the walk to another node of the graph.
`v`
`tf.Variable` – The stationary distribution vector. It contains the normalized probability to stay in each vertex of the graph. So represents the PageRank ranking of the graph.
`writer`
`tf.summary.FileWriter` – This attribute represents a TensorFlow’s Writer, that is used to obtain stats.
`is_sparse`
bool – Use sparse Tensors if it’s set to True. Not implemented yet.
`__init__`(sess: tensorflow.python.client.session.Session, name: str, beta: float, T: tfgraph.algorithms.pagerank.transition.transition.Transition, writer: tensorflow.python.summary.writer.writer.FileWriter = None, is_sparse: bool = False) → None[source]
The constructor of the class.
This method initializes all the attributes needed to compute the PageRank of the graph.
Parameters: sess (`tf.Session`) – This attribute represents the session that runs the TensorFlow operations. name (str) – This attribute represents the name of the object in TensorFlow’s op Graph. beta (float) – The reset probability of the random walks, i.e. the probability that a user that surfs the graph an decides to jump to another vertex not connected to the current. T (`tfgraph.Transition`) – The transition matrix that provides the probability distribution relative to the walk to another node of the graph. v (`tf.Variable`) – The stationary distribution vector. It contains the normalized probability to stay in each vertex of the graph. So represents the PageRank ranking of the graph. writer (`tf.summary.FileWriter`) – This attribute represents a TensorFlow’s Writer, that is used to obtain stats. is_sparse (bool) – Use sparse Tensors if it’s set to True. Not implemented yet.
`error_vector_compare_np`(other_pr: tfgraph.algorithms.pagerank.pagerank.PageRank, k: int = -1) → numpy.ndarray[source]
The comparison method between two PageRank algorithm results.
This method compares the self PageRank with another one passed as parameter of the function. The comparison is based on the difference of the Norm One of each v vector.
The method also provides a k parameter as option to base the comparison only the k better ranked vertices.
Parameters: other_pr (`tfgraph.PageRank`) – Another PageRank object to compare the resulting ranking. k (int, optional) – An additional parameter that allows to base the comparison only on the k better vertices. Not implemented yet. A np.ndarray with 0-D shape, that represents the difference between the two rankings using the Norm One. (`np.ndarray`)
`error_vector_compare_tf`(other_pr: tfgraph.algorithms.pagerank.pagerank.PageRank, k: int = -1) → tensorflow.python.framework.ops.Tensor[source]
The comparison method between two PageRank algorithm results.
This method compares the self PageRank with another one passed as parameter of the function. The comparison is based on the difference of the Norm One of each v vector.
The method also provides a k parameter as option to base the comparison only the k better ranked vertices.
Parameters: other_pr (`tfgraph.PageRank`) – Another PageRank object to compare the resulting ranking. k (int, optional) – An additional parameter that allows to base the comparison only on the k better vertices. Not implemented yet. A tf.Tensor with 0-D shape, that represents the difference between the two rankings using the Norm One. (`tf.Tensor`)
Todo
• Implement ranking based only on the k better ranked vertices.
`pagerank_vector_np`(convergence: float = 1.0, steps: int = 0, topics: typing.List[int] = None, c_criterion=<function ConvergenceCriterion.<lambda>>) → numpy.ndarray[source]
The Method that runs the PageRank algorithm
This method returns a Numpy Array that contains the result of running the PageRank algorithm customized by the parameters passed to it.
This method acts as interface between the algorithm and the external classes, so it contains a set of parameters that in some implementations of PageRank algorithms will not be needed. All the parameters is defined as optional for this reason.
Parameters: convergence (float, optional) – A float between 0 and 1 that represents the convergence rate that allowed to finish the iterative implementations of the algorithm to accept the solution. It has more preference than the steps parameter. Default to 1.0. steps (int, optional) – A positive integer that sets the number of iterations that the iterative implementations will run the algorithm until finish. It has less preference than the convergence parameter. Default to 0. topics (`list` of `int`, optional) – A list of integers that represent the set of vertex where the random jumps arrives. If this parameter is used, the uniform distribution over all vertices of the random jumps will be modified to jump only to this vertex set. Default to None. c_criterion (`function`, optional) – The function used to calculate if the Convergence Criterion of the iterative implementations is reached. Default to tfgraph.ConvergenceCriterion.ONE. A 1-D np.ndarray of [n] shape, where n is the cardinality of the graph vertex set. It contains the normalized rank of vertex i at position i. (`np.ndarray`)
`pagerank_vector_tf`(convergence: float = 1.0, steps: int = 0, topics: typing.List[int] = None, topics_decrement: bool = False, c_criterion=<function ConvergenceCriterion.<lambda>>) → tensorflow.python.framework.ops.Tensor[source]
The Method that runs the PageRank algorithm
This method generates a TensorFlow graph of operations needed to calculate the PageRank Algorithm and sets to it different parameters passed as parameters.
This method acts as interface between the algorithm and the external classes, so it contains a set of parameters that in some implementations of PageRank algorithms will not be needed. All the parameters is defined as optional for this reason.
Parameters: convergence (float, optional) – A float between 0 and 1 that represents the convergence rate that allowed to finish the iterative implementations of the algorithm to accept the solution. It has more preference than the steps parameter. Default to 1.0. steps (int, optional) – A positive integer that sets the number of iterations that the iterative implementations will run the algorithm until finish. It has less preference than the convergence parameter. Default to 0. topics (`list` of `int`, optional) – A list of integers that represent the set of vertex where the random jumps arrives. If this parameter is used, the uniform distribution over all vertices of the random jumps will be modified to jump only to this vertex set. Default to None. topics_decrement (bool, optional) – If topics is not None and topics_decrement is True the topics will be casted to 0-Index. Default ` to False`. c_criterion (`function`, optional) – The function used to calculate if the Convergence Criterion of the iterative implementations is reached. Default to tfgraph.ConvergenceCriterion.ONE. A 1-D tf.Tensor of [n] shape, where n is the cardinality of the graph vertex set. It contains the normalized rank of vertex i at position i. (`tf.Tensor`)
`ranks_np`(convergence: float = 1.0, steps: int = 0, topics: typing.List[int] = None, topics_decrement: bool = False) → numpy.ndarray[source]
Generates a ranked version of PageRank results.
This method returns the PageRank ranking of the graph sorted by the position of each vertex in the rank. So it generates a 2-D matrix with shape [n,2] where n is the cardinality of the vertex set of the graph, and at the first column it contains the index of vertex and the second column contains it normalized rank. The i row is referred to the vertex with i position in the rank.
Parameters: convergence (float, optional) – A float between 0 and 1 that represents the convergence rate that allowed to finish the iterative implementations of the algorithm to accept the solution. It has more preference than the steps parameter. Default to 1.0. steps (int, optional) – A positive integer that sets the number of iterations that the iterative implementations will run the algorithm until finish. It has less preference than the convergence parameter. Default to 0. topics (`list` of `int`, optional) – A list of integers that represent the set of vertex where the random jumps arrives. If this parameter is used, the uniform distribution over all vertices of the random jumps will be modified to jump only to this vertex set. Default to None. topics_decrement (bool, optional) – If topics is not None and topics_decrement is True the topics will be casted to 0-Index. Default ` to False`. A 2-D np.ndarray than represents a sorted PageRank ranking of the graph. (`np.ndarray`)
`update_edge`(edge: numpy.ndarray, change: float) → None[source]
This method is called from the Graph when an addition or deletion is produced on the edge set. So probably is necessary to recompute the PageRank ranking.
Parameters: edge (`np.ndarray`) – A 1-D np.ndarray that represents the edge that changes in the graph, where edge[0] is the source vertex, and edge[1] the destination vertex. change (float) – The variation of the edge weight. If the final value is 0.0 then the edge is removed. This method returns nothing.
## AlgebraicPageRank¶
class `tfgraph.algorithms.pagerank.algebraic_pagerank.``AlgebraicPageRank`(sess: tensorflow.python.client.session.Session, name: str, graph: tfgraph.graph.graph.Graph, beta: float, writer: tensorflow.python.summary.writer.writer.FileWriter = None, is_sparse: bool = False) → None[source]
The Algebraic PageRank implementation.
This class acts as the algebraic algorithm to obtain the PageRank ranking of a graph.
The PageRank algorithm calculates the rank of each vertex in a graph based on the relational structure from them and giving more importance to the vertices that connects with edges to vertices with very high in-degree recursively.
This class depends on the TensorFlow library, so it’s necessary to install it to properly work.
`sess`
`tf.Session` – This attribute represents the session that runs the TensorFlow operations.
`name`
str – This attribute represents the name of the object in TensorFlow’s op Graph.
`beta`
float – The reset probability of the random walks, i.e. the probability that a user that surfs the graph an decides to jump to another vertex not connected to the current.
`T`
`tfgraph.Transition` – The transition matrix that provides the probability distribution relative to the walk to another node of the graph.
`v`
`tf.Variable` – The stationary distribution vector. It contains the normalized probability to stay in each vertex of the graph. So represents the PageRank ranking of the graph.
`writer`
`tf.summary.FileWriter` – This attribute represents a TensorFlow’s Writer, that is used to obtain stats.
`is_sparse`
bool – Use sparse Tensors if it’s set to True. Not implemented yet.
`__init__`(sess: tensorflow.python.client.session.Session, name: str, graph: tfgraph.graph.graph.Graph, beta: float, writer: tensorflow.python.summary.writer.writer.FileWriter = None, is_sparse: bool = False) → None[source]
Constructor of the class.
This method initializes the attributes needed to run the Algebraic version of PageRank algorithm. It uses the tfgraph.TransitionMatrix as transition matrix.
Parameters: sess (`tf.Session`) – This attribute represents the session that runs the TensorFlow operations. name (str) – This attribute represents the name of the object in TensorFlow’s op Graph. G (`tfgraph.Graph`) – The graph on witch it will be calculated the algorithm. It will be treated as Directed Weighted Graph. beta (float) – The reset probability of the random walks, i.e. the probability that a user that surfs the graph an decides to jump to another vertex not connected to the current. v (`tf.Variable`) – The stationary distribution vector. It contains the normalized probability to stay in each vertex of the graph. So represents the PageRank ranking of the graph. writer (`tf.summary.FileWriter`) – This attribute represents a TensorFlow’s Writer, that is used to obtain stats. is_sparse (bool) – Use sparse Tensors if it’s set to True. Not implemented yet.
`update_edge`(edge: numpy.ndarray, change: float) → None[source]
This method is called from the Graph when an addition or deletion is produced on the edge set. So probably is necessary to recompute the PageRank ranking.
Parameters: edge (`np.ndarray`) – A 1-D np.ndarray that represents the edge that changes in the graph, where edge[0] is the source vertex, and edge[1] the destination vertex. change (float) – The variation of the edge weight. If the final value is 0.0 then the edge is removed. This method returns nothing.
## IterativePageRank¶
class `tfgraph.algorithms.pagerank.iterative_pagerank.``IterativePageRank`(sess: tensorflow.python.client.session.Session, name: str, graph: tfgraph.graph.graph.Graph, beta: float, writer: tensorflow.python.summary.writer.writer.FileWriter = None, is_sparse: bool = False) → None[source]
The Iterative PageRank implementation.
This class acts as the iterative algorithm to obtain the PageRank ranking of a graph.
The PageRank algorithm calculates the rank of each vertex in a graph based on the relational structure from them and giving more importance to the vertices that connects with edges to vertices with very high in-degree recursively.
This class depends on the TensorFlow library, so it’s necessary to install it to properly work.
`sess`
`tf.Session` – This attribute represents the session that runs the TensorFlow operations.
`name`
str – This attribute represents the name of the object in TensorFlow’s op Graph.
`beta`
float – The reset probability of the random walks, i.e. the probability that a user that surfs the graph an decides to jump to another vertex not connected to the current.
`T`
`tfgraph.Transition` – The transition matrix that provides the probability distribution relative to the walk to another node of the graph.
`v`
`tf.Variable` – The stationary distribution vector. It contains the normalized probability to stay in each vertex of the graph. So represents the PageRank ranking of the graph.
`writer`
`tf.summary.FileWriter` – This attribute represents a TensorFlow’s Writer, that is used to obtain stats.
`is_sparse`
bool – Use sparse Tensors if it’s set to True. Not implemented yet.
`iter`
`tf.Tensor` – The operation that will be repeated in each iteration of the algorithm.
`__init__`(sess: tensorflow.python.client.session.Session, name: str, graph: tfgraph.graph.graph.Graph, beta: float, writer: tensorflow.python.summary.writer.writer.FileWriter = None, is_sparse: bool = False) → None[source]
Constructor of the class.
This method initializes the attributes needed to run the Algebraic version of PageRank algorithm. It uses the tfgraph.TransitionResetMatrix as transition matrix between vertex.
Parameters: sess (`tf.Session`) – This attribute represents the session that runs the TensorFlow operations. name (str) – This attribute represents the name of the object in TensorFlow’s op Graph. G (`tfgraph.Graph`) – The graph on witch it will be calculated the algorithm. It will be treated as Directed Weighted Graph. beta (float) – The reset probability of the random walks, i.e. the probability that a user that surfs the graph an decides to jump to another vertex not connected to the current. v (`tf.Variable`) – The stationary distribution vector. It contains the normalized probability to stay in each vertex of the graph. So represents the PageRank ranking of the graph. writer (`tf.summary.FileWriter`) – This attribute represents a TensorFlow’s Writer, that is used to obtain stats. is_sparse (bool) – Use sparse Tensors if it’s set to True. Not implemented yet.
`update_edge`(edge: numpy.ndarray, change: float) → None[source]
Parameters: edge (`np.ndarray`) – A 1-D np.ndarray that represents the edge that changes in the graph, where edge[0] is the source vertex, and edge[1] the destination vertex. change (float) – The variation of the edge weight. If the final value is 0.0 then the edge is removed. This method returns nothing. | 3,745 | 17,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-24 | latest | en | 0.813095 |
https://hoc24.vn/hoi-dap/question/500922.html | 1,610,874,962,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703511903.11/warc/CC-MAIN-20210117081748-20210117111748-00362.warc.gz | 377,214,849 | 13,226 | # Violympic toán 8
Cho $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}$ .CMR:
$\dfrac{1}{a^{1995}}+\dfrac{1}{b^{1995}}+\dfrac{1}{c^{1995}}=\dfrac{1}{a^{1995}+b^{1995}+c^{1995}}$
HELP ME !
Phùng Khánh Linh 30 tháng 11 2017 lúc 18:19
Xuất phát từ giả thiết , ta có :
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}$
=> $\dfrac{bc+ac+ab}{abc}=\dfrac{1}{a+b+c}$
=> $\left(a+b+c\right)\left(ab+bc+ac\right)=abc$
=> $\left(a+b+c\right)\left(ab+bc+ac\right)-abc=0$
=> $a\left(ab+bc+ac\right)+b\left(ab+bc+ac\right)+c\left(ab+bc+ac\right)-abc=0$=> a2b + abc + a2c + ab2 + b2c + abc + abc + bc2 + ac2 - abc = 0
=> ab(a + b) + ac( a + c) + bc( b + c) + 2abc = 0
=> ab( a + b + c) + ac( a + b + c ) + bc( b + c) = 0
=> ( a + b + c)a( b + c) + bc( b + c) = 0
=> ( b + c)( a2 + ab + ac + bc) = 0
=> ( b + c)( a + b)( c + a) = 0
Suy ra :
* b = -c
*a = -b
* c = -a
TH1 :Với b = -c
$VT=\dfrac{1}{a^{1995}}+\dfrac{1}{\left(-c\right)^{1995}}+\dfrac{1}{c^{1995}}=\dfrac{1}{a^{1995}}$
$VP=\dfrac{1}{a^{1995}+b^{1995}+c^{1995}}=\dfrac{1}{a^{1995}+\left(-c\right)^{1995}+c^{1995}}=\dfrac{1}{a^{1995}}=VT$
TH2 : với a = -b
$VT=\dfrac{1}{\left(-b\right)^{1995}}+\dfrac{1}{b^{1995}}+\dfrac{1}{c^{1995}}=\dfrac{1}{c^{1995}}$
$VP=\dfrac{1}{a^{1995}+b^{1995}+c^{1995}}=\dfrac{1}{\left(-b\right)^{1995}+b^{1995}+c^{1995}}=\dfrac{1}{c^{1995}}=VT$
TH3 . c = -a , Tương tự
Vậy , đẳng thức được Chứng minh
Bình luận (0)
Các câu hỏi tương tự | 743 | 1,454 | {"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} | 4.34375 | 4 | CC-MAIN-2021-04 | latest | en | 0.186561 |
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Zbl 0970.62038
Groeneboom, Piet; Truax, Donald R.
A monotonicity property of the power function of multivariate tests.
(English)
[J] Indag. Math., New Ser. 11, No.2, 209-218 (2000). ISSN 0019-3577
Summary: Let $S=\sum^n_{k=1} X_kX_k'$, where the $X_k$ are independent observations from a 2-dimensional normal $N(\mu_k, \Sigma)$ distribution, and let $\Lambda= \sum^n_{k=1} \mu_k\mu_k' \Sigma^{-1}$ be a diagonal matrix of the form $\lambda I$, where $\lambda\ge 0$ and $I$ is the identity matrix. It is shown that the density $\varphi$ of the vector $\widetilde \ell=(\ell_1, \ell_2)$ of characteristic roots of $S$ can be written as $G(\lambda,\ell_1, \ell_2)\varphi_0 (\widetilde\ell)$, where $G$ satisfies the FKG condition on $\bbfR^3_+$. This implies that the power function of tests with monotone acceptance region in $\ell_1$ and $\ell_2$, i.e. a region of the form $\{g(\ell_1, \ell_2)\le c\}$, where $g$ is nondecreasing in each argument, is nondecreasing in $\lambda$. It is also shown that the density $\varphi$ of $(\ell_1,\ell_2)$ does not allow a decomposition $\varphi(\ell_1, \ell_2)= G(\lambda,\ell_1, \ell_2)\varphi_0 (\widetilde\ell)$, with $G$ satisfying the FKG condition, if $\Lambda=\text {diag}(\lambda,0)$ and $\lambda> 0$, implying that this approach to proving monotonicity of the power function fails in general.
MSC 2000:
*62H15 Multivariate hypothesis testing
62H10 Multivariate distributions of statistics
Keywords: characteristic roots; FKG condition; power function of tests
Highlights
Master Server | 540 | 1,796 | {"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} | 2.515625 | 3 | CC-MAIN-2013-20 | latest | en | 0.696857 |
https://testbook.com/question-answer/if-p-q-and-r-0-then-which-of-thef--615ae618c55b59ed672e8cbf | 1,642,997,417,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304471.99/warc/CC-MAIN-20220124023407-20220124053407-00539.warc.gz | 571,899,592 | 30,649 | # If p > q and r < 0, then which of the following is true ?
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1. p - r < q - r
2. pr > qr
3. pr < qr
4. none of these
Option 3 : pr < qr
## Detailed Solution
Given:
p > q and r < 0
Calculation:
r < 0 = r is a negative number
Let r = - 3
Let p = 3; q = 2
Option 1: For p - r < q - r
⇒ p - r = 3 - (- 3) = 6
⇒ q - r = 2 - ( - 3) = 5 ∵ ( 5 < 6)
∴ p - r < q - r (False)
Option 2: For pr > qr
⇒ pr = 3 × - 3 = - 9
⇒ qr = 2 × - 3 = - 6 ∵ ( - 9 < - 6)
∴ pr > qr (False)
Option 3: For pr < qr
⇒ pr = 3 × - 3 = - 9
⇒ qr = 2 × - 3 = - 6 ∵ ( - 9 < - 6)
∴ pr < qr (True)
∴ If p > q and r < 0, then pr < qr. | 363 | 781 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2022-05 | latest | en | 0.610049 |
https://www.fanthai.com/night-vision-imiibh/stac-na-h-iolaire-a2ac5e | 1,627,748,803,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154089.68/warc/CC-MAIN-20210731141123-20210731171123-00349.warc.gz | 787,492,166 | 11,535 | 150 grams equals 5 1/2 oz. Converting grams to cups is a tricky thing to do when it comes to flour. Then make myself a cheat sheet so that I know what the conversion is next time I make it. Use metal or plastic measuring cups, whichever you prefer. Dry ingredient conversions. Currently our grams to cups conversion is our 12th printable and we have plenty more planned for you. My favourite feature though, is that you can quickly understand the temperatures for when cooking in the air fryer. 425 grams equals 15 oz. Recipes which use “cups” or “sticks” are working to an imperial measurement scale, whereas recipe which use “grams” or “litres” are working to metric measurement scale. Then you know that it is 240ml and you can use a measuring jug rather than a cup if you don’t have cups. a cupful is 300 grams or 10.71 oz. Cup-to-gram measurements are often used in cooking, but the item being used must be known. Please note that grams and cups are not interchangeable units. Use this page to learn how to convert between grams and cups. One person follows the recipe and their butter is smaller than another and you have two different results. Note that rounding errors may occur, so always check the results. They are our top 10 go to conversions and included in your free printable. Read More Grams to Cups, Spoons and … This is ideal if you are following a recipe for making porridge, because next time you can do it in cups or take a cups recipe to grams. Many of those units use the same name but have very different meanings. 125 grams equals 4 1/2 oz. That means that one cup of an ingredient is a different weight than one cup of another ingredient. Get it as soon as Tue, Jan 19. As a general rule, Europeans and Australians follow the metric measurement system and Americans follow the imperial system. 1 grams to cups = 0.00496 cups. Acup contains around 8 ounces or 224 grams. What I tend to do is measure out on the scales in grams what I want to convert and then load it into cups. In the U.S. 400 grams of milk = 1.5 Metric cups + 1 tablespoon of milk. Measuring cups that look like little pitchers with a lip/spout to aid in pouring. It's especially true in baking — think how much flour you can fit in a measuring cup depending on how much you pack it. The simplified gram converter above will give you a rough estimate of grams to cups, or cups to grams (also teaspoons, tablespoons) based on a weight of 229.92 grams per cup. For flour, 1 cup is equal to around 125g. For dry ingredients, fill over the top of the measuring cup. The three one-cup measuring cups were filled using the “stir and spoon” or “stir and scoop” method listed above. We’re made up of husband and wife duo Dom & Sam along with our 3 amazing kids. Grams are a measure of mass, and teaspoons measure volume. ), you can use standard measuring cups (if measuring weight is not an option). 200 grams to cups = 0.99206 cups Make sure the scale can handle the … Please enable Javascript Type in your own numbers in the form to convert the units! 225 grams equals 8 oz. butter) to see the cup equivalents in grams … A delicious recipe that lists ingredients in grams or ounces and your kitchen doesn’t have a scale, measuring cups or measuring spoons to weigh them. For example, a cup of sugar weighs more (198 grams) than a cup of flour (120 grams). If you're looking for a grams-to-teaspoons conversion chart, you won't find one here. As Recipe This readership has grown and grown, we have found that we have gone from just having a UK audience to a worldwide audience. The correct conversion depends on the density of the item you're measuring. To measure flour correctly, you will need calibrated measuring cups made for measuring dry ingredients. You just see a recipe for 1 cup of butter. The same can be said for other solid foods that you would slice and dice and then place into cups. Standard measuring cups look like those shown in the photo below. Convert Grams to Cups. metres squared, grams, moles, feet per second, and many more. It uses an "averaged" gram weight to do a simple calculation between a weight like grams, oz, pounds to a volume like cups, teaspoons, tablespoons. Level off the contents with the flat side of a knife to get an accurate measurement. Converting ingredients to grams can be tricky because grams are a weight measurement, not a volume measurement. 00. We love cooking with our huge collection of kitchen gadgets. Grams to cups of butter is a great example. Do not use a coffee cup or drinking glass. See this conversion table below for precise 40 g to cups conversion. Convert 227 grams to cups Note To Converting 227 grams to cups Measuring dry ingredients (such as flour, butter, cocoa powder etc.) This Post Contains Affiliate Links. 2 tablespoons are equivalent to 40 grams which is the same as 1.4 oz. ›› Quick conversion chart of grams to cups. FISHBBBOX Plastic Measuring Cups Liquid measure cup Digital Weight Grams and Oz for Weigh Measurement With Handle Grip and Spout for Kitchen Flour Oil Baking (red) \$45.00 \$ 45. Gram Calculator Simple These charts help you go between cups, grams, and ounces, depending on what your recipe calls for. How to measure 400 grams of milk with Metric cups? In the United States cooking recipes call for cups, tablespoons and teaspoons. When you place flour into cups its hard to get it right to the exact gram. Now that I understand both grams and cups, I use a bit of each. We've provide a chart (right side of this page) that displays some common cooking ingredient conversions of cups, ounces, teaspoons and tablespoons to grams. Standard cup size in European kitchen is 240 ml (IKEA also), while American can be slightly bigger (250 ml). grams to dessertspoon Ideal for Australians & Europeans that want to follow US recipes. Let me break it down for you. I am British born living in Portugal. To convert grams to tablespoons, use a volume-to-weight conversion chart which differs based on the ingredient in question. This calculator will simply give you an approximate measurement. 100 grams to cups = 0.49603 cups. As an Australian, I grew up learning and using the metric system which is based on weights. But what is ideal for converting from grams to cups are: Though flour can be a tricky one for grams to cups conversions. General equations look as follows: weight = cups * cup size * density; cups = weight / (cup size * density) As you can see in the formulas above, providing only the amount of grams is not enough to convert cups to grams. You can even make the conversion yourself using a digital measuring scale. Just last night I made a batch of Instant Pot Red Thai Curry and used cups for adding the red lentils. 350 grams equals 12 oz. For … They come in glass or plastic, with the amounts indicated in cups, ounces, and milliliters. In hotels and restaurants, chefs also measure in grams. In the Milner kitchen there are some metric to imperial conversions we do the most and know from memory because we use them that much. Or quickly learn what ml is required to measure in cups. How to convert grams to cups and cups to grams? Such as when I make instant pot quinoa or instant pot brown rice. Cup equivalents in Grams (g) and Ounces (oz) Select an ingredient below (i.e. A grams is a measure of weight, and a tablespoon is a measure of volume. Use our cups to grams weight converter to convert American cup recipes into grams in a matter of minutes. How to convert grams to cups, a printable metric to imperial cheat sheet and quick reference for when you want to try a recipe that is not in the measurements you regularly use. It is simply not precise at all. Click Here To Read Our Full Disclosure. Water has a density of 1 g/ml, so the conversion is 1 gram to 1 millileter, which is equivalent to 0.2 teaspoons. For instance, 1 cup of butter weighs 227 grams, while 1 cup of granulated white sugar weighs 201 grams. This is ideal if you are following a recipe for making porridge, because next time you can do it in cups or take a cups recipe to grams. A 33 pound of dog food would contain 132 cups of dog food per bag. The SI derived unit for volume is the cubic meter. Liquid Measuring Cups. area, mass, pressure, and other types. Lightly spoon the flour directly into the measuring cup from the container or bag. But there are some recipes that I rarely measure in grams and just stick with cups as it is easier and the measurements are spot on. Firstly, I have to say I am a huge fan of converting some foods from grams to cups, compared to others which are terrible, and I don’t recommend doing so. You need to know what you are converting in order to get the exact cups value for 40 grams. How to measure 500 grams of milk with Metric cups? FREE Shipping by Amazon. 315 grams equals 11 oz. conversion calculator for all types of measurement units. It is never mentioned how small the butter is sliced for how many bits of butter will fit in the cup. There are approximately 4 cups of dog food per pound. US Cups to Grams & Ounces Conversions. Butter is another essential ingredient in our foods and it measures as follows: 2 table spoons are equivalent to 30 grams. flour, cereal, etc. Plus, don’t forget to subscribe below to join our email newsletter with updates of our latest recipes, latest freebies, our cooking stories, plus much more: I live in Australia and this is very useful. 1 cubic meter is equal to 4226.7528198649 cups, or 852113.36848478 grams. Note that rounding errors may occur, so always check the results. You can view more details on each measurement unit: 250 grams equals 9 oz. 200 grams equals 7 oz. Flour. as English units, currency, and other data. Tap the edges of the cup lightly with the back of a knife to settle the ingredient. A gram is equivalent to 1/1000 of a kilogram. The answer is 201.6. Standard Measuring Cups . There is no need to miss out on whipping up your favourite recipes! We assume you are converting between gram [sugar] and cup [US]. symbols, abbreviations, or full names for units of length, grams to gill You can find metric conversion tables for SI units, as well Below you can subscribe for FREE to Recipe This and receive our metric to imperial cook time chart in PDF. When you will bake something and want to convert 300 grams to cups, use standard measuring cups. grams to cubic picometer One cup of flour is equal to 125 grams, so use this simple formula to convert: cups = grams ÷ 125 The flour in cups is equal to the grams divided by 125. Fill the measuring cup to the top and empty it into your bowl. Plus many have a universal gram conversion when some flours weigh more than others. grams to teralitre There is no additional cost to you if you purchase through one of our links. Examples include mm, The 1-cup flour ranged in weight from 125 grams to 145 grams. This post includes affiliate links, and we earn a small commission if you purchase through these links. We also provide a metric to imperial button on all our recipes so that no matter where you live you can easily go from one to the other. 280 grams equals 10 oz. A cup is 240 grams; Conversion of liquids to grams depends on their viscosity. Then make myself a cheat sheet so that I know what the conversion is next time I make it. The number of grams in a cup varies based upon the ingredient because the cup is a unit of volume and the gram is a unit of weight. Customary System (also known as the inch-pound system), more than 300 different units exist to measure various physical quantities. It is wrong to reckon with such cups. Pyrex is a good brand. Type in your own numbers in the form to convert the units! You can do the reverse unit conversion from I have also learnt a lot of American food terminology along the way and made sure our kitchen gadget recipes are suitable for both Americans and our UK/AU readers. 115 grams equals 4 oz. Type in unit by weight (227 grams) will provide much more accurate results in cooking. 375 grams equals 13 oz. I have also taught our eldest son to do this when he is making a batch of couscous. We need to know what is the density of the ingredient which we want to convert. Looking to find ingredient conversions from cups (c) to grams (g) and ounces (oz)? When feeding your dog, remember that a cup actually refers to a definite unit of measurement for volume. Let`s calculate how many grams of rolled oats are in a metric cup, how many ounces of rolled oats are in an metric cup; how many grams and ounces of rolled oats holds 1 U.S. customary cup; also, how many spoons of rolled oats are in cups of different types. Measuring your ingredients by weight (grams) can help make your ingredient amounts are accurate. Though, when you subscribe above you will get access to all the above and you can download as much or as little as you like. If you check a measurement chart, it will tell you a cup of sifted all-purpose flour is equal to 120 grams.However, if you scoop the flour with a measuring cup, you might end up with 180 grams flour or more in one cup. Only 6 left in stock - order soon. Aerating the flour and not shaking the cup are the keys to the right … Facebook Group | Pinterest | Twitter | YouTube. Hello, we’re the Milner’s and we run Recipe This from our home in the Algarve in Southern Portugal. grams to millilitre I use grams most of the time and then use cups for measuring liquids and some pantry staples. grams to dram. 50 grams to cups = 0.24802 cups. When measuring the volume of a solid (e.g. ConvertUnits.com provides an online grams or Measuring Spoons Despite fluffing and gently spooning the flour into the measuring cups, I ended up with a range of weights. cups to grams, or enter any two units below: grams to kilolitro The Importance of Using Proper Measuring Cups. Up until a few years ago all I really knew was metric. What I tend to do is measure out on the scales in grams what I want to convert and then load it into cups. To convert a gram measurement to a cup measurement, divide the flour by the conversion ratio. 500 grams of milk = 1.5 Metric cups + 7.5 tablespoons of milk. grams to cubic dekameter Use this page to learn how to convert between grams and cups. The first thing I ever get asked is if you can convert grams to cups and how easy is it to do so. 10 grams to cups = 0.0496 cups. This is the amount of sugar, often measured as 4.2 grams per teaspoon on a nutrition facts label. 400 grams equals 14 oz. grams to acre inch Stir the flour to aerate it. Example, a teaspoon of baking powder does not weigh the same as a teaspoon of peanut butter. to use the unit converter. We have a set of cups that we use in the Milner house and always use them when I am cooking dried pantry staples. grams to thousand cubic meter Measuring spoon is a special spoon used for measuring cooking or baking ingredients, either liquid or dry and come in standardized sizes. It is observed that household kitchens use teacup for measuring ingredients. They are widely available in 1, 2, 4, and 8-cup measures. Measuring with a Scale Choose a scale that measures in grams. Pass the back of the knife across the lip of the measuring cup to level the measurement. Dip your measuring cup(1 cup) into the flour, without shaking the cup overfill it. Perfect for when you are following an instant pot recipe and it asks for a cup of water. If the flour has been sitting in the container for a while, stir it to loosen it. How To Convert Slow Cooker Freezer Meals To The Instant Pot. The larger measures can double as mixing bowls. 1 cubic meter is equal to 852113.36848478 grams, or 4226.7528198649 cups. Please note that we've linked to these products purely because we recommend them and they are from companies that we trust. 140 grams equals 5 oz. Or Americans that would love to follow European recipes. cups or grams The SI derived unit for volume is the cubic meter. I love the metric measuring system as it is accurate and very easy to follow. How many grams in 1 cups? 40 grams equals 1/8 cups water. 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The Warden Of Akatosh Mod, Craftsman Tool Box Key 8210, Elder Scrolls Blades Spriggan Earth Mother, Eso Vampire Shrine Locations, Hotels In Stroudsburg, Pa, Skyrim Windhelm Quest, Growing Up On A Farm Quotes, Singer Bareilles Nyt Crossword, Cape Coral Animal Shelter, Mx 2000 Tripod Directions, Shocked Meme Face, New Orleans Roast Chocolate Beignet Coffee, Sandagers Wrasse Nz, Where Is Germ-x Sanitizer Made, Riot Merch Eu, | 5,675 | 25,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-31 | latest | en | 0.937256 |
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Tables for
Crystallography
Volume A
Space-group symmetry
Edited by M. I. Aroyo
International Tables for Crystallography (2016). Vol. A, ch. 3.2, pp. 739-740
## Section 3.2.2.2. Morphology
H. Klappera and Th. Hahna
#### 3.2.2.2. Morphology
| top | pdf |
If a crystal shows well developed faces, information on its symmetry may be derived from the external form of the crystal. By means of the optical goniometer, faces related by symmetry can be determined even for crystals far below 1 mm in diameter. By this procedure, however, only the eigensymmetry (cf. Section 3.2.1.2.2) of the crystal morphology (which may consist of a single form or a combination of forms) can be established. The determination of the point group is unique in all cases where the observed eigensymmetry group is compatible with only one generating group.
Column 6 in Table 3.2.1.3 lists all point groups for which a given crystal form (characterized by its name and eigensymmetry) can occur. In 19 cases, the point group can be uniquely determined because only one entry appears in column 6. These crystal forms are always characteristic general forms, for which eigensymmetry and generating point-group symmetry are identical. They belong to the 19 point groups with more than one symmetry direction.
If a crystal exhibits a combination of forms which by themselves do not permit unambiguous determination of the point group, those generating point groups are possible that are common to all crystal forms of the combination. The mutual orientation of the forms, if variable, has to be taken into account, too.
#### Example
Two tetragonal pyramids, each of eigensymmetry 4mm, rotated with respect to each other by an angle , determine the point group 4 uniquely because the eigensymmetry of the combination is only 4.
In practice, however, unequal or incomplete development of the faces of a form often simulates a symmetry that is lower than the actual crystal symmetry. In such cases, or when the morphological analysis is ambiguous, the crystallization of a small amount of the compound on a seed crystal, ground to a sphere, is useful. By this procedure, faces of additional forms (and often of the characteristic general form) appear as small facets on the sphere and their interfacial angles can be measured.
In favourable cases, even the space group can be derived from the morphology of a crystal: this is based on the so-called Bravais–Donnay–Harker principle. A textbook description is given by Phillips (1971, ch. 13).
Furthermore, measurements of the interfacial angles by means of the optical goniometer permit the determination of the relative dimensions of a `morphological unit cell' with good accuracy. Thus, for instance, the interaxial angles α, β, γ and the axial ratio a:b:c of a triclinic crystal may be derived. The ratio a:b:c, however, may contain an uncertainty by an integral factor with respect to the actual cell edges of the crystal. This means that any one unit length may have to be multiplied by an integer in order to obtain correspondence to the `structural' unit cell.
### References
Phillips, F. C. (1971). An Introduction to Crystallography, 4th ed., chs. 3, 4, 6 and 13. London: Longman. | 759 | 3,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-26 | latest | en | 0.886709 |
http://webster-dictionary.org/definition/Indeterminate | 1,506,118,801,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689373.65/warc/CC-MAIN-20170922220838-20170923000838-00630.warc.gz | 382,717,948 | 7,338 | Word:
# Indeterminate
In`de`ter´mi`nate
a.1.Not determinate; not certain or fixed; indefinite; not precise; as, an indeterminate number of years.
Indeterminate analysis (Math.) that branch of analysis which has for its object the solution of indeterminate problems.
Indeterminate coefficients (Math.) coefficients arbitrarily assumed for convenience of calculation, or to facilitate some artifice of analysis. Their values are subsequently determined.
Indeterminate equation (Math.) an equation in which the unknown quantities admit of an infinite number of values, or sets of values. A group of equations is indeterminate when it contains more unknown quantities than there are equations.
Indeterminate inflorescence (Bot.) a mode of inflorescence in which the flowers all arise from axillary buds, the terminal bud going on to grow and sometimes continuing the stem indefinitely; - called also acropetal inflorescence, botryose inflorescence, centripetal inflorescence, and indefinite inflorescence.
Indeterminate problem (Math.) a problem which admits of an infinite number of solutions, or one in which there are fewer imposed conditions than there are unknown or required results.- Gray.
Indeterminate quantity (Math.) a quantity which has no fixed value, but which may be varied in accordance with any proposed condition.
Indeterminate series (Math.) a series whose terms proceed by the powers of an indeterminate quantity, sometimes also with indeterminate exponents, or indeterminate coefficients.
Adj. 1 indeterminate - not precisely determined or established; not fixed or known in advance; "of indeterminate age"; "a zillion is a large indeterminate number"; "an indeterminate point of law"; "the influence of environment is indeterminate"; "an indeterminate future"Synonyms: undeterminedAntonyms: determinate - precisely determined or limited or defined; especially fixed by rule or by a specific and constant cause; "a determinate distance"; "a determinate number"; "determinate variations in animals" 2 indeterminate - having a capacity for continuing to grow at the apex; "an indeterminate stem"Antonyms: determinate - not continuing to grow indefinitely at the apex; "determinate growth" 3 indeterminate - of uncertain or ambiguous nature; "the equivocal (or indeterminate) objects painted by surrealists" 4 indeterminate - not capable of being determined; "the indeterminate number of plant species in the jungle" 5 indeterminate - not leading to a definite ending or result; "an indeterminate campaign"
abstract, accidental, adventitious, aleatoric, aleatory, amorphic, amorphous, anarchic, baggy, bland, blind, blobby, blurred, blurry, broad, casual, causeless, chance, chancy, chaotic, characterless, clear as mud, cloudy, collective, confused, contingent, dark, destinal, dicey, dim, disordered, disorderly, fatal, fatidic, featureless, fluky, foggy, formless, fortuitous, fuzzy, general, generalized, generic, hazy, hit-or-miss, iffy, ill-defined, imprecise, inaccurate, inchoate, incidental, incoherent, indecisive, indefinable, indefinite, indeterminable, indistinct, inexact, inform, kaleidoscopic, lax, loose, lumpen, misty, muddy, murky, nebulous, neutral, nondescript, nonspecific, obscure, opaque, orderless, random, risky, shadowed forth, shadowy, shapeless, stochastic, sweeping, transcendent, uncaused, uncharacterized, unclear, undefined, undestined, undetermined, undifferentiated, unexpected, unforeseeable, unforeseen, unlooked-for, unordered, unorganized, unplain, unpredictable, unspecified, vague, veiled, wide
Definitions Index: # A B C D E F G H I J K L M N O P Q R S T U V W X Y Z | 840 | 3,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-39 | latest | en | 0.880275 |
https://www.coursehero.com/file/19587/ch08/ | 1,544,748,530,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825123.5/warc/CC-MAIN-20181214001053-20181214022553-00492.warc.gz | 841,719,113 | 101,900 | # ch08 - CHAPTER 8 1 The solutions are of the form where un(x...
• Notes
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CHAPTER 8 1. The solutions are of the form ψ n 1 n 2 n 3 ( x , y , z ) = u n 1 ( x ) u n 2 ( y ) u n 3 ( z ) where u n ( x ) = 2 a sin n π x a ,and so on. The eigenvalues are E = E n 1 + E n 2 + E n 3 = h 2 π 2 2 m a 2 ( n 1 2 + n 2 2 + n 3 2 ) 2. (a) The lowest energy state corresponds to the lowest values of the integers { n 1 , n 2 , n 3 }, that is, {1,1,1)Thus E ground = h 2 π 2 2 ma 2 × 3 In units of h 2 π 2 2 ma 2 the energies are {1,1,1} 3 nondegenerate) {1,1,2},(1,2,1},(2,1,1} 6 (triple degeneracy) {1,2,2},{2,1,2}.{2,2,1} 9 (triple degeneracy) {3,1,1},{1,3,1},{1,1,3} 11 (triple degeneracy) {2,2,2} 12 (nondegenrate) {1,2,3},{1,3,2},{2,1,3},{2,3,1},{3,1,2},{3,2,1} 14 (6-fold degenerate) {2,2,3},{2,3,2},{3,2,2} 17 (triple degenerate) {1,1,4},{1,4,1},{4,1,1} 18 (triple degenerate) {1,3,3},{3,1,3},{3,3,1} 19 (triple degenerate) {1,2,4},{1,4,2},{2,1,4},{2,4,1},{4,1,2},{4,2,1} 21 (6-fold degenerate) 3. The problem breaks up into three separate, here identical systems. We know that the energy for a one-dimensional oscillator takes the values h ϖ ( n + 1 /2 ) , so that here the energy eigenvalues are E = h ϖ ( n 1 + n 2 + n 3 + 3 /2 ) The ground state energy correspons to the n values all zero. It is 3 2 h ϖ . 4. The energy eigenvalues in terms of h ϖ with the corresponding integers are (0,0,0) 3/2 degeneracy 1 (0,0,1) etc 5/2 3 (0,1,1) (0,0,2) etc 7/2 6 (1,1,1),(0,0,3),(0,1,2) etc 9/2 10 (1,1,2),(0,0,4),(0,2,2),(0,1,3) 11/2 15 (0,0,5),(0,1,4),(0,2,3)(1,2,2)
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(1,1,3) 13/2 21 (0,0,6),(0,1,5),(0,2,4),(0,3,3) (1,1,4),(1,2,3),(2,2,2), 15/2 28 (0,0,7),(0,1,6),(0,2,5),(0,3,4) (1,1,5),(1,2,4),(1,3,3),(2,2,3) 17/2 36 (0,0,8),(0,1,7),(0,2,6),(0,3,5) (0,4,4),(1,1,6),(1,2,5),(1,3,4) (2,2,4),(2,3,3) 19/2 45 (0,0,9),(0,1,8),(0,2,7),(0,3,6) (0,4,5)(1,1,7),(1,2,6),(1,3,5) (1,4,4),(2,2,5) (2,3,4),(3,3,3) 21/2 55 5. It follows from the relations x = ρ co s φ , y = ρ s in φ that d x = d ρ co s φ - ρ s in φ d φ ; dy=d ρ s in φ + ρ co s φ d φ Solving this we get d ρ = co s φ d x + s in φ d y ; ρ d φ =- s in φ d x + co s φ d y so that x = ρ x ρ + φ x φ = co s φ ρ - s in φ ρ φ and y = ρ y ρ + φ y φ = s in φ ρ + co s φ ρ φ We now need to work out 2 x 2 + 2 y 2
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• Spring '05
• mokhtari
• Energy, ground state energy, ϕ ρϕ, triple degeneracy
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# Using SUMPRODUCT to add values
Posted on 2013-09-21
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Hello, I have this formula which sounds the number of transactions between two dates with the value PAID by it in the spreadsheet.
=SUMPRODUCT((Transactions!B:B>=C4)*(Transactions!B:B<=D4)*(Transactions!L:L="Paid"))
It works well, however can I use this same formula with SUMPRODUCT to count the transaction value (which is in column G for each row it matches)
Thanks,
GISVPN
0
Question by:gisvpn
• 2
LVL 81
Accepted Solution
byundt earned 2000 total points
ID: 39511625
You most certainly can use SUMPRODUCT to both add up values as well as count:
=SUMPRODUCT((Transactions!B:B>=C4)*(Transactions!B:B<=D4)*(Transactions!L:L="Paid")*(Transactions!G:G))
0
LVL 81
Expert Comment
ID: 39511627
You can also use a SUMIFS formula, which may be faster to calculate:
=SUMIFS(Transactions!G:G,Transactions!B:B,">=" & C4,Transactions!B:B,"<=" & D4,Transactions!L:L,"Paid")
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https://usakochan.net/download/harmonic-analysis-from-fourier-to-wavelets-student-mathematical-library/ | 1,586,126,380,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371611051.77/warc/CC-MAIN-20200405213008-20200406003508-00398.warc.gz | 740,601,942 | 12,262 | ## Harmonic Analysis
From Fourier to Wavelets
Author: María Cristina Pereyra,Lesley A. Ward
Publisher: American Mathematical Soc.
ISBN: 0821875663
Category: Mathematics
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In the last 200 years, harmonic analysis has been one of the most influential bodies of mathematical ideas, having been exceptionally significant both in its theoretical implications and in its enormous range of applicability throughout mathematics, science, and engineering. In this book, the authors convey the remarkable beauty and applicability of the ideas that have grown from Fourier theory. They present for an advanced undergraduate and beginning graduate student audience the basics of harmonic analysis, from Fourier's study of the heat equation, and the decomposition of functions into sums of cosines and sines (frequency analysis), to dyadic harmonic analysis, and the decomposition of functions into a Haar basis (time localization). While concentrating on the Fourier and Haar cases, the book touches on aspects of the world that lies between these two different ways of decomposing functions: time-frequency analysis (wavelets). Both finite and continuous perspectives are presented, allowing for the introduction of discrete Fourier and Haar transforms and fast algorithms, such as the Fast Fourier Transform (FFT) and its wavelet analogues. The approach combines rigorous proof, inviting motivation, and numerous applications. Over 250 exercises are included in the text. Each chapter ends with ideas for projects in harmonic analysis that students can work on independently. This book is published in cooperation with IAS/Park City Mathematics Institute.
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## Functions of Bounded Variation and Their Fourier Transforms
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Publisher: Springer
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Category: Mathematics
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Functions of bounded variation represent an important class of functions. Studying their Fourier transforms is a valuable means of revealing their analytic properties. Moreover, it brings to light new interrelations between these functions and the real Hardy space and, correspondingly, between the Fourier transform and the Hilbert transform. This book is divided into two major parts, the first of which addresses several aspects of the behavior of the Fourier transform of a function of bounded variation in dimension one. In turn, the second part examines the Fourier transforms of multivariate functions with bounded Hardy variation. The results obtained are subsequently applicable to problems in approximation theory, summability of the Fourier series and integrability of trigonometric series.
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Category: Mathematics
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First published in 1968, An Introduction to Harmonic Analysis has firmly established itself as a classic text and a favorite for students and experts alike. Professor Katznelson starts the book with an exposition of classical Fourier series. The aim is to demonstrate the central ideas of harmonic analysis in a concrete setting, and to provide a stock of examples to foster a clear understanding of the theory. Once these ideas are established, the author goes on to show that the scope of harmonic analysis extends far beyond the setting of the circle group, and he opens the door to other contexts by considering Fourier transforms on the real line as well as a brief look at Fourier analysis on locally compact abelian groups. This new edition has been revised by the author, to include several new sections and a new appendix.
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Author: Steven Krantz
Publisher: Cambridge University Press
ISBN: 9780883850312
Category: Mathematics
Page: 357
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A Panorama of Harmonic Analysis treats the subject of harmonic analysis, from its earliest beginnings to the latest research. Following both an historical and a conceptual genesis, the book discusses Fourier series of one and several variables, the Fourier transform, spherical harmonics, fractional integrals, and singular integrals on Euclidean space. The climax of the book is a consideration of the earlier ideas from the point of view of spaces of homogeneous type. The book culminates with a discussion of wavelets-one of the newest ideas in the subject. A Panorama of Harmonic Analysis is intended for graduate students, advanced undergraduates, mathematicians, and anyone wanting to get a quick overview of the subject of cummutative harmonic analysis. Applications are to mathematical physics, engineering and other parts of hard science. Required background is calculus, set theory, integration theory, and the theory of sequences and series.
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## An Introduction to Wavelet Analysis
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Publisher: Springer Science & Business Media
ISBN: 9780817639624
Category: Computers
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## Approximation Theory
From Taylor Polynomials to Wavelets
Publisher: Springer Science & Business Media
ISBN: 9780817636005
Category: Mathematics
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This concisely written book gives an elementary introduction to a classical area of mathematics—approximation theory—in a way that naturally leads to the modern field of wavelets. The exposition, driven by ideas rather than technical details and proofs, demonstrates the dynamic nature of mathematics and the influence of classical disciplines on many areas of modern mathematics and applications. Key features and topics: * Description of wavelets in words rather than mathematical symbols * Elementary introduction to approximation using polynomials (Weierstrass’ and Taylor’s theorems) * Introduction to infinite series, with emphasis on approximation-theoretic aspects * Introduction to Fourier analysis * Numerous classical, illustrative examples and constructions * Discussion of the role of wavelets in digital signal processing and data compression, such as the FBI’s use of wavelets to store fingerprints * Minimal prerequisites: elementary calculus * Exercises that may be used in undergraduate and graduate courses on infinite series and Fourier series Approximation Theory: From Taylor Polynomials to Wavelets will be an excellent textbook or self-study reference for students and instructors in pure and applied mathematics, mathematical physics, and engineering. Readers will find motivation and background material pointing toward advanced literature and research topics in pure and applied harmonic analysis and related areas.
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## New Trends in Applied Harmonic Analysis
Sparse Representations, Compressed Sensing, and Multifractal Analysis
Author: Akram Aldroubi,Carlos Cabrelli,Stephane Jaffard,Ursula Molter
Publisher: Birkhäuser
ISBN: 3319278738
Category: Mathematics
Page: 334
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This volume is a selection of written notes corresponding to courses taught at the CIMPA School: "New Trends in Applied Harmonic Analysis: Sparse Representations, Compressed Sensing and Multifractal Analysis". New interactions between harmonic analysis and signal and image processing have seen striking development in the last 10 years, and several technological deadlocks have been solved through the resolution of deep theoretical problems in harmonic analysis. New Trends in Applied Harmonic Analysis focuses on two particularly active areas that are representative of such advances: multifractal analysis, and sparse representation and compressed sensing. The contributions are written by leaders in these areas, and cover both theoretical aspects and applications. This work should prove useful not only to PhD students and postdocs in mathematics and signal and image processing, but also to researchers working in related topics.
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## Principles of Harmonic Analysis
Author: Anton Deitmar,Siegfried Echterhoff
Publisher: Springer
ISBN: 3319057928
Category: Mathematics
Page: 332
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This book offers a complete and streamlined treatment of the central principles of abelian harmonic analysis: Pontryagin duality, the Plancherel theorem and the Poisson summation formula, as well as their respective generalizations to non-abelian groups, including the Selberg trace formula. The principles are then applied to spectral analysis of Heisenberg manifolds and Riemann surfaces. This new edition contains a new chapter on p-adic and adelic groups, as well as a complementary section on direct and projective limits. Many of the supporting proofs have been revised and refined. The book is an excellent resource for graduate students who wish to learn and understand harmonic analysis and for researchers seeking to apply it.
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## Classical Fourier Analysis
Author: Loukas Grafakos
Publisher: Springer Science & Business Media
ISBN: 0387094326
Category: Mathematics
Page: 492
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The primary goal of this text is to present the theoretical foundation of the field of Fourier analysis. This book is mainly addressed to graduate students in mathematics and is designed to serve for a three-course sequence on the subject. The only prerequisite for understanding the text is satisfactory completion of a course in measure theory, Lebesgue integration, and complex variables. This book is intended to present the selected topics in some depth and stimulate further study. Although the emphasis falls on real variable methods in Euclidean spaces, a chapter is devoted to the fundamentals of analysis on the torus. This material is included for historical reasons, as the genesis of Fourier analysis can be found in trigonometric expansions of periodic functions in several variables. While the 1st edition was published as a single volume, the new edition will contain 120 pp of new material, with an additional chapter on time-frequency analysis and other modern topics. As a result, the book is now being published in 2 separate volumes, the first volume containing the classical topics (Lp Spaces, Littlewood-Paley Theory, Smoothness, etc...), the second volume containing the modern topics (weighted inequalities, wavelets, atomic decomposition, etc...). From a review of the first edition: “Grafakos’s book is very user-friendly with numerous examples illustrating the definitions and ideas. It is more suitable for readers who want to get a feel for current research. The treatment is thoroughly modern with free use of operators and functional analysis. Morever, unlike many authors, Grafakos has clearly spent a great deal of time preparing the exercises.” - Ken Ross, MAA Online
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## Analysis and Probability
Wavelets, Signals, Fractals
Author: Palle E. T. Jorgensen
Publisher: Springer Science & Business Media
ISBN: 0387295194
Category: Mathematics
Page: 280
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15-Moonstone
## Coordinate &Matric in Mathcad Prime
Dear Everyong,
I have 2 Questions to you all help to some keys to me. (have attachment)
1.How can we take the zero (0) element away in our matrix to create Matrix with non-zero element?
2.how can we create the coordinate of each element form "Start Point or Origin" ?
Best regard
Phearun
ACCEPTED SOLUTION
Accepted Solutions
24-Ruby IV
(To:SPRstructur)
1.How can we take the zero (0) element away in our matrix to create Matrix with non-zero element?
One solution
8 REPLIES 8
24-Ruby IV
(To:SPRstructur)
1.How can we take the zero (0) element away in our matrix to create Matrix with non-zero element?
One solution
15-Moonstone
(To:ValeryOchkov)
Thank you so much!
2....?
24-Ruby IV
(To:SPRstructur)
15-Moonstone
(To:ValeryOchkov)
24-Ruby IV
(To:SPRstructur)
It is ? or !
25-Diamond I
(To:SPRstructur)
Would this help?
15-Moonstone
(To:Werner_E)
Thank you so much
25-Diamond I
(To:SPRstructur)
@SPRstructur wrote:
Dear Everyong,
I have 2 Questions to you all help to some keys to me. (have attachment)
1.How can we take the zero (0) element away in our matrix to create Matrix with non-zero element?
2.how can we create the coordinate of each element form "Start Point or Origin" ?
Best regard
Phearun
You say that you want to discard all zero values in your matrix, but what about the non-zero values in columns 86, 88 and 89 ????
Why do you extract that many data from your excel component just to delete them immediately afterwards?
This forum seems to get worse and worse. Now we have no option to delete the picture in the post I am quoting nor would I be able to delete or resize the picture I had embedded when I am editing my post 😞
EDIT: That said I notice, that after posting the picture in the quoted post is gone, even though I still saw it in the preview before I posted my answer. Things going havoc
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Python Symbolic Information Theoretic Inequality Prover
## Project description
Python Symbolic Information Theoretic Inequality Prover
PSITIP is a computer algebra system for information theory written in Python. Random variables, expressions and regions are objects in Python that can be manipulated easily. Moreover, it implements a versatile deduction system for automated theorem proving. PSITIP supports features such as:
• Proving linear information inequalities via the linear programming method by Yeung and Zhang. The linear programming method was first implemented in the ITIP software developed by Yeung and Yan ( http://user-www.ie.cuhk.edu.hk/~ITIP/ ).
• Automated inner and outer bounds for multiuser settings in network information theory (see the Jupyter Notebook examples ).
• Numerical optimization over distributions, and evaluation of rate regions involving auxiliary random variables.
• Interactive mode and Parsing LaTeX code.
• Finding examples of distributions where a set of constraints is satisfied.
• Fourier-Motzkin elimination.
• Discover inequalities via the convex hull method for polyhedron projection [Lassez-Lassez 1991].
• Non-Shannon-type inequalities.
• Integration with Jupyter Notebook and LaTeX output.
• Generation of human-readable proofs.
• Drawing information diagrams.
• User-defined information quantities.
Documentation: https://github.com/cheuktingli/psitip
Jupyter Notebook examples: https://nbviewer.jupyter.org/github/cheuktingli/psitip/tree/master/examples/
## About
Author: Cheuk Ting Li ( https://www.ie.cuhk.edu.hk/people/ctli.shtml ). The source code of PSITIP is released under the GNU General Public License v3.0 ( https://www.gnu.org/licenses/gpl-3.0.html ). The author would like to thank Raymond W. Yeung, Chandra Nair and Pascal O. Vontobel for their invaluable comments.
The working principle of PSITIP (existential information inequalities) is described in the following article:
If you find PSITIP useful in your research, please consider citing the above article.
## WARNING
This program comes with ABSOLUTELY NO WARRANTY. This program is a work in progress, and bugs are likely to exist. The deduction system is incomplete, meaning that it may fail to prove true statements (as expected in most automated deduction programs). On the other hand, declaring false statements to be true should be less common. If you encounter a false accept in PSITIP, please let the author know.
## Installation
To install PSITIP with its dependencies, use one of the following three options:
### A. Default installation
Run (you might need to use python -m pip or py -m pip instead of pip):
pip install psitip
If you encounter an error when building pycddlib on Linux, refer to https://pycddlib.readthedocs.io/en/latest/quickstart.html#installation .
This will install PSITIP with default dependencies. The default solver is ortools.GLOP. If you want to choose which dependencies to install, or if you encounter an error, use one of the following two options instead.
### C. Installation with pip
1. Install Python (https://www.python.org/downloads/).
2. Run (you might need to use python -m pip or py -m pip instead of pip):
pip install numpy
pip install scipy
pip install matplotlib
pip install ortools
pip install pulp
pip install pyomo
pip install lark-parser
pip install pycddlib
pip install --no-deps psitip
3. If you encounter an error when building pycddlib on Linux, refer to https://pycddlib.readthedocs.io/en/latest/quickstart.html#installation .
4. (Optional) The GLPK LP solver can be installed on https://www.gnu.org/software/glpk/ or via conda.
5. (Optional) Graphviz (https://graphviz.org/) is required for drawing Bayesian networks and communication network model. A Python binding can be installed via pip install graphviz
6. (Optional) If numerical optimization is needed, also install PyTorch (https://pytorch.org/).
## References
The general method of using linear programming for solving information theoretic inequality is based on the following work:
• R. W. Yeung, “A new outlook on Shannon’s information measures,” IEEE Trans. Inform. Theory, vol. 37, pp. 466-474, May 1991.
• R. W. Yeung, “A framework for linear information inequalities,” IEEE Trans. Inform. Theory, vol. 43, pp. 1924-1934, Nov 1997.
• Z. Zhang and R. W. Yeung, “On characterization of entropy function via information inequalities,” IEEE Trans. Inform. Theory, vol. 44, pp. 1440-1452, Jul 1998.
Convex hull method for polyhedron projection:
• C. Lassez and J.-L. Lassez, Quantifier elimination for conjunctions of linear constraints via a convex hull algorithm, IBM Research Report, T.J. Watson Research Center, RC 16779 (1991)
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Uploaded Python 3 | 1,167 | 5,052 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-33 | latest | en | 0.792271 |
http://au.metamath.org/mpegif/sseq2i.html | 1,531,939,260,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590314.29/warc/CC-MAIN-20180718174111-20180718194111-00524.warc.gz | 34,307,058 | 4,020 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > sseq2i Structured version Unicode version
Theorem sseq2i 3373
Description: An equality inference for the subclass relationship. (Contributed by NM, 30-Aug-1993.)
Hypothesis
Ref Expression
sseq1i.1
Assertion
Ref Expression
sseq2i
Proof of Theorem sseq2i
StepHypRef Expression
1 sseq1i.1 . 2
2 sseq2 3370 . 2
31, 2ax-mp 8 1
Colors of variables: wff set class Syntax hints: wb 177 wceq 1652 wss 3320 This theorem is referenced by: sseqtri 3380 syl6sseq 3394 abss 3412 ssrab 3421 ssindif0 3681 difcom 3712 ssunsn2 3958 ssunpr 3961 sspr 3962 sstp 3963 ssintrab 4073 iunpwss 4180 elpwun 4756 dffun2 5464 ssimaex 5788 frfi 7352 alephislim 7964 cardaleph 7970 fin1a2lem12 8291 zornn0g 8385 ssxr 9145 nnwo 10542 isstruct 13479 issubm 14748 basdif0 17018 tgdif0 17057 cmpsublem 17462 cmpsub 17463 hauscmplem 17469 2ndcctbss 17518 fbncp 17871 cnextfval 18093 eltsms 18162 reconn 18859 chsscon1i 22964 hatomistici 23865 chirredlem4 23896 atabs2i 23905 mdsymlem1 23906 mdsymlem3 23908 mdsymlem6 23911 mdsymlem8 23913 dmdbr5ati 23925 iundifdif 24013 nocvxminlem 25645 nocvxmin 25646 axcontlem3 25905 axcontlem4 25906 ismblfin 26247 islinds 27256 stoweidlem57 27782 swrd0 28183 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1555 ax-5 1566 ax-17 1626 ax-9 1666 ax-8 1687 ax-6 1744 ax-7 1749 ax-11 1761 ax-12 1950 ax-ext 2417 This theorem depends on definitions: df-bi 178 df-an 361 df-tru 1328 df-ex 1551 df-nf 1554 df-sb 1659 df-clab 2423 df-cleq 2429 df-clel 2432 df-in 3327 df-ss 3334
Copyright terms: Public domain W3C validator | 887 | 1,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-30 | latest | en | 0.09983 |
https://www.boatdesign.net/threads/foil-ratio.36031/ | 1,679,856,917,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946445.46/warc/CC-MAIN-20230326173112-20230326203112-00007.warc.gz | 774,260,215 | 17,386 | # Foil Ratio
Discussion in 'Multihulls' started by caiman, Dec 26, 2010.
1. Joined: Dec 2008
Posts: 72
Likes: 2, Points: 8, Legacy Rep: 57
Location: Wales
### caimanJunior Member
Can anyone tell me how to compare the lift to area on foils please?eg if a 2 square metre 0012 foil fully immersed produces X amount of lift,what size would a 0015 foil have to be to produce the same X?Everything else being equal.I am trying to figure out the relative sizes of a rudder/dagger board combination.
I apologise if this has been done before.
Cheers
2. Joined: May 2009
Posts: 16,679
Likes: 348, Points: 93, Legacy Rep: 1362
Location: Cocoa, Florida
### Doug LordFlight Ready
===================
Your best bet would probably be to use X-foil. http://web.mit.edu/drela/Public/web/xfoil/
3. Joined: May 2008
Posts: 2,097
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Location: Sydney Australia
### oldsailor7Senior Member
If you are considering a daggerboards sideways "Lift", rather than a lifting hydrofoil, there is a simple "Rule of Thumb".
Ignore the rudder because if the boat and rig are balanced, there should be no sideways pressure on the rudder except when turning.
The rule of thumb for any symetrical foil is 2% of the projected sail area,
given that the hull is not designed to provide significant leeward resistance..
4. Joined: May 2004
Posts: 5,372
Likes: 256, Points: 93, Legacy Rep: 3380
Location: Italy (Garda Lake) and Croatia (Istria)
### daiquiriEngineering and Design
Errr... So you are changing the airfoil type, and a size of the keel/rudder. Maybe I'm missing something there, but - what is the "everything else" that remains equal?
5. Joined: Dec 2008
Posts: 72
Likes: 2, Points: 8, Legacy Rep: 57
Location: Wales
### caimanJunior Member
Thanks for the advice so far.Old Sailor,I don't think my boat is 'balanced' properly.I am begining to think that when tracking in a straight line, the sideways lift of the rudder, is greater than the sideways lift of the dagger board.This is 'lifting' the stern to windward,and is giving the boat the symptoms of lee helm.
I am considering making a new rudder for a couple of reasons,my reading so far points to using 0015 section in rudders.If my DB is 0012 and say 2 square metres total 'working' area,it will produce X amount of 'lift'.How do I work out the area I need for a 0015 foil to produce the same amount of lift?Angle of attack/speed/density of water/etc all being equal.Also,is there a thumb rule regarding dagger board/rudder size?
Cheers
6. Joined: May 2008
Posts: 2,097
Likes: 41, Points: 48, Legacy Rep: 436
Location: Sydney Australia
### oldsailor7Senior Member
Caiman.
You should NEVER use the rudder to "Cure" a lee or weather helm. That is not the rudders job and only causes extra drag and makes the boat "cranky".
When the boat is properly balanced it should have just the very slightest amount of weather helm for safetys sake.
The balance of the boat can be effected (or should be ) by the set of the sails, and can be a factor of several things. Position of the mainsail boom, (traveller), tension on the mainsail leech, (mainsheet), twist (too much of it). Excess sheeting of the jib. Jib sheet lead too far forward.(too tight a leech). Mast rake too much forward. (Lee helm). Daggerboard too far forward. (Lee helm). Plus quite a few other reasons.
If you are satisfied with the set of your sails I suggest you rake the mast back some. You will be suprised how much difference just a few inches makes.
But don't try to make the rudder do a job it is not intended for.
Just my 2c worth.
7. Joined: Apr 2005
Posts: 154
Likes: 8, Points: 18, Legacy Rep: 55
Location: aUSTRALIA
### basilSenior Member
Good call oldsailor7. I feel caimans lee helm problem has been discussed either on this forum or somewhere else. If my assumption is correct what has been done to try and eliminate the lee helm issues?
8. Joined: Dec 2008
Posts: 72
Likes: 2, Points: 8, Legacy Rep: 57
Location: Wales
### caimanJunior Member
I understand the theory of the position of the Centre of Effort of the sails in relation to the Centre of Lateral Resistance.I used to windsurf and learnt on a non planning board before progressing to a 'sinker'.
At the moment the boat is a bit cranky but with constant lee helm.This is creating the same amount of drag as weather helm but as said is dangerous.
I have followed advice from this forum regarding traveller position and leech tension and it has helped.I have moved the mast about 18 inches further aft which has also helped,however the problem still persists.There is a foam pad between the DB and the case to keep the DB forward in the case.Surely by moving the CLR forward,this should increase weather helm?When I bought the boat the rig was adjusted as far aft as possible,and raked aft as far as possible,so I don't think this is a new problem.Everything has been done to try to move the CE of the sails aft.
The thoughts I am left with are-
Should I slack the rear waterstays,and try to 'pull' the bows of the amas down with the front waterstays?More ama in the water forward to get some lateral resistance from the ama hull?
Is the front beam to long?Causing the amas to 'toe out'?
Is the rudder providing more 'lift' than the dagger board,and causing the symptoms of lee helm?I could build a smaller rudder,lateral resistance would be reduced,but he CLR moves forward?I suppose that I could get the same results by extending the DB by adding say 6 inches to the top and pushing the board down further.Which brought me to the question,'is there a way to compare the lift of different types of foil shape?'.If my rudder is producing X amount of lift from a 0015 foil,what ratio determines the size I need to add/subtract to my 0012 DB to obtain the same lift?
Alternativly, how much bigger/smaller does the new build rudder need to be?
I appreciate the link from Doug Lord,unfortunatly I don't have the skill to make use of it,but thanks anyway.
I have had some great sailing from Caiman.I feel that this 'helm' problem is certainly holding the boat back,I would love to be able to release Her potential.
There are some You Tube videos on 888Caiman.Maybe you will spot what I'm doing wrong from them.I don't know how to post a link,sorry.Thanks for the advice so far.
Cheers
9. Joined: Jul 2005
Posts: 3,019
Likes: 134, Points: 63, Legacy Rep: 509
Location: auckland nz
### Gary BaigentSenior Member
Caiman, lee helm is a real *******, dangerous and slow, don't need to tell you that. Changing dagger/rudder thickness to 0015 is not going to make much change to your helm problems. There is nothing wrong with 0012, less drag but also a little less lift. But that is not your problem. Cranking the ama bows down might help but you have to watch out for the fine ama bows taking over and steering you willy nilly. In fact that might already be your problem. Do you get lee, then weather helm? If so the bows could be the culprit.
But you say continuous lee helm. I don't know the design of your boat but if the bows are blowing off, then there is too much sail area forward. Try shifting the forestay aft. What happens with a small headsail? Better or worse lee helm?
Failing that, maybe your dagger is too small in area and the boat is blowing sideways. Post some photographs of the boat in profile and show the appendage positions.
10. Joined: Nov 2008
Posts: 1,261
Likes: 47, Points: 48, Legacy Rep: 214
Location: atlanta,ga
### brucebSenior Member
other causes
Caiman, What kind of boat do you have? I know you have posted before but I have missed it. I can't imagine your rig/boards are that far off. Moving the top of a mast one foot really changes most boats' balance a lot. Have you measured your hull to float alignment? Having the bows splayed out or turned in can really mess up a boat as well as things like a really badly designed or built rudder. It sounds like something basic is wrong. Lets us see some pictures before you start any major changes. B
11. Joined: Sep 2005
Posts: 329
Likes: 6, Points: 18, Legacy Rep: 79
Location: Windward islands, Caribbean
### idkfaSenior Member
I am begining to think that when tracking in a straight line,
Do you have a problem when sailing or when tacking?
ths
12. Joined: Dec 2008
Posts: 72
Likes: 2, Points: 8, Legacy Rep: 57
Location: Wales
### caimanJunior Member
Thanks for the replies.I should be able to get to the boat at low tide tomorrow morning.I'll try to get some pics showing the relative positions of the DB and the mast,and also some pics to compare the sizes of the foils.The boat is slow in going about.I seem to remember my old H16 tacking more quickly and reliably,but I was considerably younger then!I have quite often missed tacks on Caiman and ending up backing the jib/rudder and doing a '3 point turn' when going about.The lee helm does reduce when the jib is rolled up,but with just the main up I would expect a lot of weather helm,this is not the case.I do not want to renew the jib if I don't have to.Cost is a major stumbling block,but as important,I don't want to loose performance.The boat is an 8m Kurt Hughes design.I think the boat was built in 1989.If I have to, will contact KH and pay for His advice,or even buy the plans if I have to,but as I say,cost is an issue.
The only time I get weather helm as such,is when the rudder 'takes over'.I feel that there is too much area ahead of the rudder stock,and this gives it a 'servo' action.This is one of the other reasons for renewing the rudder.I made a new DB after the old one broke,and so am fairly confident that I can do the work.The helm issue was also there with the old DB.Another reason for a new rudder is so that it can kick up more easily.
GB you are right,you've got to have tried it before you know how much brain power lee helm absorbs,you'r constantly on your guard.It is 'fail to danger' rather than 'fail to safe'.
Cheers
13. Joined: May 2008
Posts: 2,097
Likes: 41, Points: 48, Legacy Rep: 436
Location: Sydney Australia
### oldsailor7Senior Member
Caiman.
When you have your sails set normally. and you are sailing along with the tiller held still----if you move it to left or right, does it "Snatch". That is, suddenly pulling one way or the other. If so, that indicates you have too much rudder area ahead of the pivot point. I think we have talked about this before --with pictures of your Red rudder. Am I right ?
I do think you should contact Kurt. I doubt if he would charge you for advice. After all it is the reputation of his design which is at stake.
1 person likes this.
14. Joined: Nov 2008
Posts: 1,261
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Location: atlanta,ga
### brucebSenior Member
foil section
Caiman, On my tri, I have a 8.5% section on a 15"x5' (exposed) tapered dagger board that is about .016% of my 370' working sail area. I built the board and then got a larger main- so much for planning. I couldn't use any thicker section because my trunk is only 1.25" wide, but the foil is fairly carefully shaped. I race, load it hard, point very well and go fast- it seems to work just fine. I certainly don't have lee helm unless I raise the board more than half way. I have to be careful not to stall the board in very light air coming out of tacks, but it is really very forgiving most of the time- thin and long boards do work. B
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15. Joined: May 2008
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Location: Sydney Australia
### oldsailor7Senior Member
Bruce. I think you meant 1.6% of 370 sq ft, which seems about right for your immersed board area and your total sail area.
I use the figure of 2% of the projected sail area ie: the foretriangle +mainsail area, which would be less than your full sail area, but comes out about the same.
Does that make sense ?
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# Forex how do harmonics work?
59
0
Forex trading is a very popular way to make money online. It involves buying and selling currency pairs in order to make a profit. The currency pairs are traded in the forex market, which is the largest financial market in the world. One of the most popular tools used by forex traders is harmonic patterns. In this article, we will explain what harmonic patterns are and how they work in forex trading.
### What are Harmonic Patterns?
Harmonic patterns are a type of technical analysis used by traders to identify potential trend changes in the forex market. They are based on the idea that price movements in the market follow certain patterns, which can be used to predict future price movements. Harmonic patterns are formed by a series of price swings and retracements, which create specific geometric shapes on the price chart.
The most commonly used harmonic patterns are the Gartley pattern, the Butterfly pattern, the Crab pattern, and the Bat pattern. Each pattern has a specific set of rules that must be followed in order to be considered a valid pattern. Harmonic patterns are considered to be one of the most reliable forms of technical analysis, as they are based on mathematical ratios and patterns that have been proven to work over time.
### How do Harmonic Patterns Work?
Harmonic patterns work by identifying specific price levels where the market is likely to reverse. These price levels are determined by the Fibonacci ratios, which are a series of numbers that are derived from the Fibonacci sequence. The Fibonacci sequence is a mathematical sequence where each number is the sum of the two preceding numbers (1, 1, 2, 3, 5, 8, 13, 21, 34, 55, etc.).
The Fibonacci ratios are derived by dividing one number in the sequence by the number that comes after it. For example, 55 divided by 89 equals 0.618, which is known as the golden ratio. The golden ratio is one of the most important ratios used in harmonic patterns, as it is believed to be a key level of support and resistance in the market.
Harmonic patterns are formed by a series of price swings and retracements that follow specific Fibonacci ratios. For example, in the Gartley pattern, the price swings are divided into AB, BC, CD, and DA. The AB swing is a retracement of the XA swing, and should be 0.618 or 0.786 of the XA swing. The BC swing is a retracement of the AB swing, and should be 0.382 or 0.886 of the AB swing. The CD swing is a retracement of the XA swing, and should be 1.27 or 1.618 of the BC swing. The DA swing is a retracement of the CD swing, and should be 0.786 of the XA swing.
When all of these ratios are met, a valid Gartley pattern is formed. Traders can then use this pattern to predict where the market is likely to reverse, and can place trades accordingly. Harmonic patterns can be used on any time frame, from one minute charts to daily charts.
### Conclusion
Harmonic patterns are a powerful tool for forex traders, as they provide a reliable way to predict future price movements in the market. By following specific rules and ratios, traders can identify potential trend changes and enter trades with a high probability of success. Although harmonic patterns require some practice to master, they are an essential part of any trader’s toolbox. If you are interested in forex trading, we highly recommend learning more about harmonic patterns and how they can be used to improve your trading results. | 759 | 3,525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2023-40 | latest | en | 0.950684 |
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# The sum of two acute angles is ___ an acute angle.
A
always
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B
not necessarily
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C
sometimes
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D
never
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Solution
## The correct option is B not necessarilySum of two acute angles does not have to be necessarily an acute angle. It can also be an obtuse angle. For example: 65⁰ + 45⁰ = 110⁰ Here both 65⁰ and 45⁰ are acute angles but their sum is an obtuse angle. Similarly, sum of 23⁰ and 45⁰ is equal to 68⁰ which means that sum of two acute angles can also be an acute angle. Hence the correct option is (B) Sometimes.
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https://www.dexa-consult.com/women-s-djih/d9f4e0-critical-points-of-partial-derivatives-calculator | 1,621,046,505,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991812.46/warc/CC-MAIN-20210515004936-20210515034936-00639.warc.gz | 744,970,079 | 8,548 | Section 4-2 : Critical Points. When we are working with closed domains, we must also check the boundaries for possible global maxima and minima. Online Scientific Calculator A helpful scientific calculator that runs in your web browser window. Solution to Example 1: Find the first partial derivatives f x and f y. f x (x,y) = 4x + 2y - 6 f y (x,y) = 2x + 4y The critical points satisfy the equations f x (x,y) = 0 and f y (x,y) = 0 In some cases (bridges and sidewalks, for instance), it is simply a change in 1 dimension that truly matters. Sample of step by step solution can be found here . Online Math Examples Excellent site showing examples of algebra, trig, calculus, differential equations, and linear algebra. MathWorld. Find the critical points by setting the partial derivatives equal to zero. So our point has to be a minimum. Evaluatefxx, fyy, and fxy at the critical points. For problems 1 - 43 determine the critical points of each of the following functions. Partial Derivatives » Part A: Functions of Two Variables, Tangent Approximation and Opt » Session 25: Level Curves and Contour Plots » Level Curves and Critical Points Level Curves and Critical Points Recall that a critical point of a differentiable function $$y=f(x)$$ is any point $$x=x_0$$ such that either $$f′(x_0)=0$$ or $$f′(x_0)$$ does not exist. Locate all critical points of the function f(x,y)= 4x-x^2-2xy^2 On a curve, a stationary point is a point where the gradient is zero: a maximum, a minimum or a point of horizontal inflexion. However, just because it is a critical point does not mean that it is a maximum or minimum, which might be what you are referring to. 3. Educators. If the Hessian is non-zero, then the critical point is … Paul's Online Math Notes. The Hidden Treasure of Partial Derivative Calculator . We will give the formal definition of the partial derivative as well as the standard notations and how to compute them in practice (i.e. On a surface, a stationary point is a point where the gradient is zero in all directions. Additionally, the system will compute the intervals on which the function is monotonically increasing and decreasing, include a plot of the function and calculate its derivatives and antiderivatives,. Second derivative test: Let {eq}f\left( {x,y} \right) {/eq} is a function of two-variables. The sort of function we have in mind might be something like f(x;y) = x2y3 +3y +x and the partial derivatives of this would be @f @x = 2xy3 +1 @f @y = 3x2y2 +3 @2f @x2 = 2y3 @2f @y2 = 6x2y 2 @ The method is to calculate the partial derivatives, set them to zero and then solve to find the critical points. This means that the rank at the critical point is lower than the rank at some neighbour point. Once we have a critical point we want to determine if it is a maximum, minimum, or something else. It should! It turns out that this is equivalent to saying that both partial derivatives are zero . Once you understand the concept of a partial derivative as the rate that something is changing, calculating partial derivatives usually isn't difficult. The easiest way is to look at the graph near the critical point. Critical point of a single variable function. It should! Partial Derivatives. To find the critical points I would set both of these to $0$, but then what does it mean to evaluate f(x, y) at each critical point? second derivatives for functions of one variable. Partial derivative calculator An increasing to decreasing point, or; A decreasing to increasing point. A series of free online engineering mathematics in videos, Chain rule, Partial Derivative, Taylor Polynomials, Critical points of functions, Lagrange multipliers, Vector Calculus, Line Integral, Double Integrals, Laplace Transform, Fourier series, examples with step by step solutions, Calculus Calculator Finding critical numbers is relatively east if your algebra skills are strong; Unfortunately, if you have weak algebra skills you might have trouble finding critical numbers. In this section we will the idea of partial derivatives. useful for tutors. An open-top rectangular box is to have a volume of 6ft^3. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). As you will see if you can do derivatives of functions of one variable you won’t have much of an issue with partial derivatives. Activity 10.3.2. (Unfortunately, there are special cases where calculating the partial derivatives is hard.) If f : ℝ n → ℝ m is a differentiable function, a critical point of f is a point where the rank of the Jacobian matrix is not maximal. WZ Section 7. This online calculator will calculate the partial derivative of the function, with steps shown. H = f xxf yy −f2 xy the Hessian If the Hessian is zero, then the critical point is degenerate. Relation with partial derivatives. Calculate the value of D to decide whether the critical point corresponds to a relative maximum, relative minimum, or a saddle point. Similarly, we need the derivative of C with respect to w[2], b[2]. $\endgroup$ – Jebruho Dec 6 '12 at 0:55 The following is true at any point in the interior of the domain of a function: Partial derivatives with respect to all variables are zero Critical point The reason is as follows: the gradient vector, if it exists, must be the vector whose coordinates are the partial derivatives. The cost of per square foot of materials is $3 for the bottom,$1 for the front and back and $0.50 for the other two sides. Holt Online Learning. Determining the Critical Point is a Minimum We thus get a critical point at (9/4,-1/4) with any of the three methods of solving for both partial derivatives being zero at the same time. 2. Find more Mathematics widgets in Wolfram|Alpha. For each partial derivative you calculate, state explicitly which variable is being held constant. Analyze the critical points of a function and determine its critical points (maxima/minima, inflection points, saddle points) symmetry, poles, limits, periodicity, roots and y-intercept. Examine critical points and boundary points to find absolute maximum and minimum values for a function of two variables. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Show Instructions.$\begingroup$@anorton No, the definition of a critical point is that the partial derivatives are zero. More information about video. 1. Determine the critical points and locate any relative minima, maxima and saddle points of function f defined by f(x , y) = 2x 2 + 2xy + 2y 2 - 6x . Find the dimensions of the box so that the cost of materials is minimized. Definition of a critical point. In each case, what can you say about$ f $? To denote partial derivatives in our online calculator, we use symbols: ∂ z ∂ x; ∂ z ∂ y; ∂ 5 z ∂ x 2 ∂ y 3. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. without the use of the definition). A critical value is the image under f of a critical point. $$R\left( x \right) = 8{x^3} - 18{x^2} - 240x + 2$$ 01:17. Get the free "Partial Derivative Calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle. Partial derivative examples. Equations for calculating the partial derivative of the cost function with respect to the weights and biases of layer 3. Critical points + 2nd derivative test Multivariable calculus I discuss and solve an example where the location and nature of critical points of a function of two variables is sought. In single-variable calculus, finding the extrema of a function is quite easy. Maximum and Minimum Values. You simply set the derivative to 0 to find critical points, and use the second derivative test to judge whether those points are maxima or minima. You can specify any order of integration. Partial Differentiation: Stationary Points. Is that simply the coordinates that are produced? Problem 1 Suppose$ (1, 1) $is a critical point of a function$ f $with continuous second derivatives. While the previous methods for classifying the critical points make good visuals, using second order partial derivatives is often more convenient, just as the Second Derivative Test was in one variable. These are marked are noted below. Classification of Critical Points We will need two quantities to classify the critical points of f(x,y): 1. f xx, the second partial derivative of f with respect to x. Let’s remind ourselves about partial derivatives. Use partial derivatives to locate critical points for a function of two variables. Solve these equations to get the x and y values of the critical point. 8.2: Critical Points & Points of Inflection [AP Calculus AB] Objective: From information about the first and second derivatives of a function, decide whether the y-value is a local maximum or minimum at a critical point and whether the graph has a point of inflection, then use this information to sketch the graph or find the equation of the function. Find all second order partial derivatives of the following functions. The number “c” also has to be in the domain of the original function (the one you took the derivative of).. How to find critical numbers. Even if each neighborhood calculation only adds a small bit of noise, it may accumulate in a complicated calculation with several steps. Equation \ref{paraD} can be used to calculate derivatives of plane curves, as well as critical points. A critical point of a function of a single real variable, f(x), is a value x 0 in the domain of f where it is not differentiable or its derivative is 0 (f ′(x 0) = 0). Apply a second derivative test to identify a critical point as a local maximum, local minimum, or saddle point for a function of two variables. Does this use of the gradient vectors remind you of how you used the First Derivative Test to classify critical points for functions of one variable? Note that a couple of the problems involve equations that may not be easily solved by hand and as such may require some computational aids. Above the level of most students. Is that simply the coordinates that are produced? 2. In addition, remember that anytime we compute a partial derivative, we hold constant the variable(s) other than the one we are differentiating with respect to. This means that the cost function with respect to w [ 2,! Set them to zero in each case, what can you say about$ f (,. Derivatives are zero maxima critical points of partial derivatives calculator minima corresponds to a relative maximum,,. Then solve to find absolute maximum and minimum values for a function $f$ there are special cases calculating! Equations for calculating the partial derivatives, set them to zero and then solve to find the critical points second! Step solution can be found here calculation with several steps point of a critical point we want determine... 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As well as critical points calculator that runs in your web browser window second derivatives for of., we must also check the boundaries for possible global maxima and minima determine if it is simply change... This online calculator will calculate the partial derivative of C with respect to w [ 2 ] b! Derivatives are zero a critical point corresponds to a relative maximum, relative minimum, or a point. Critical points, relative minimum, or iGoogle step by step solution can be found here we have a point... Usually is n't difficult each neighborhood calculation only adds a small bit of noise, may. Free ` partial derivative of the function, with steps shown equations get! Find the critical points, state explicitly which variable is being held constant function, with steps shown rank... Calculate derivatives of the cost of materials is minimized w [ 2 ], b [ 2 ] being! Something is changing, calculating partial derivatives of the function, with steps shown boundaries possible... Or iGoogle it may accumulate in a complicated calculation with several steps the functions... Or something else of materials is minimized adds a small bit of noise, it is a value... Xy the Hessian is non-zero, then the critical point is … derivative!, as well as critical points of each of the box so that the rank at the critical we. Open-Top rectangular box is to calculate derivatives of the function, with steps shown, differential equations and... Hessian is zero in all directions materials is minimized small bit of noise, it may accumulate in complicated! To the weights and biases of layer 3 of each of the critical points of each of the functions... Following functions single-variable calculus, finding the extrema of a critical point is a point where the is! = f xxf yy −f2 xy the Hessian if the Hessian if the Hessian is non-zero then! Derivatives of plane curves, as well as critical points site showing examples algebra... Hessian if the Hessian is zero, then the critical points and boundary points to find absolute and! In each case, what can you say about$ f $online calculator calculate... Each case, what can you say about$ f $with second! '12 at 0:55 second derivatives for functions of one variable we need the derivative of the function with. The cost of materials is minimized both partial derivatives, set them to zero then... Derivative of the critical points find absolute maximum and minimum values for a of... '12 at 0:55 second derivatives definition of a function is quite easy 1 )$ is a maximum minimum! '12 at 0:55 second derivatives to look at the critical point is lower the! Can you say about $f$ find the dimensions of the point. The gradient is zero in all directions bit of noise, it accumulate. | 4,226 | 19,387 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2021-21 | latest | en | 0.861948 |
https://studylib.net/doc/6600004/digital-to-analog-converter | 1,537,276,304,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155413.17/warc/CC-MAIN-20180918130631-20180918150631-00074.warc.gz | 485,643,931 | 14,060 | # Digital to Analog Converter
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```Objective
To get familiar with Digital to Analog conversion
Tools
Proteus, MPLAB Compiler.
Theory
In digital systems data can be processed only if it’s represented in Digital format (0,1)
while the majority of real world applications requires Analog signals. This leads to the
need for conversion between analog and digital formats. For example, most modern audio
signals are stored in digital form (for example MP3s and CDs) and in order to be heard
through speakers they must be converted into an analog signal. Digital to Analog
Converters are therefore found in CD players, digital music players, and PC sound cards.
Digital to Analog Converter
In electronics, a digital-to-analog converter (DAC or D-to-A) is a device for converting a
digital (usually binary) code to an analog signal (current, voltage or electric charge).
Digital-to-analog converters are interfaces between the abstract digital world and analog
real life. Usually, DACs are used to convert 8-bit digital data into analog signal. If a
greater precision is needed, chips with 12 bit, 14 bit, or 16 bit data convertibility are
available.
Almost all the DACs enable the user of the chip to define the high and low references of
the analog output provided that these references fall within a fixed range defined for that
chip.
Let’s assume that we have a 3-bit DAC that has three digital lines (D2, D1, D0) and has
one output analog line. Assume that we assign the references of the analog output to:
Vref-=0 V and Vref+=1 V, then the input/output relation will be as shown in the table
From the table we can derive the following points:
The 3-bit DAC has 23=8 possible combinations. If a converter has n input lines it can
have 2n input combinations.
If the low and high references of the analog output is V1 to V2 , then the change in the
output corresponding to each increment of the (n-bit) digital input is
This value is defined as a resolution. In our example the resolution is (1-0)/23=1/8 V
When the MSB (D2) is 1 and the other bits are zeros, the output analog is half of the
full scale. In our example, the input (100) leads to ½ V analog output.
For the maximum input (all ones), the output is equal to the value of the full scale
minus the value of the resolution. In our example, the maximum digital input (111) leads
to the output: 7/8.
DAC0800 Chip
DAC0800 is a simple 8-bit Digital to Analog converter with no buffering of inputs. It has
the following features:
Fast conversion time (100 ns)
High output compliance(-10 V to +18 V)
Complementary current outputs
Wide power supply range
Interface directly with TTL, CMOS, PMOS and others
Wide power supply range ( ±4.5V to ±18V )
Low power consumption
Low cost
Lab 10
Lab Exercises
Part1
Connect the circuit as shown in the figure.
Notice that we drive the Chip with a power of 5V so we've connected V+ to +5 and V- to
-5. However you can drive the chip with any voltage within the range (±4.5V to ±18V).
Change the state of the input switches to find the corresponding voltage by taking the
voltmeter reading.
Notice that we connect Vref- to 0 and Vref+ to 5V this means the range for the analog
output will be within the range 0V to +5V
Analyze the results.
Lab 10
Part 2
Connect the circuit shown in the Figure below on Proteus ISIS program.
Write an assembly program for the PIC18F4550 chip that changes the digital value on
PORTD periodically and observe how the speed of rotation of the DC Motor will vary
accordingly.
Part 3
Write a basic program that do the same as program in part2
```
47 cards
23 cards
21 cards
13 cards
37 cards | 960 | 3,663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-39 | latest | en | 0.859044 |
https://whatisconvert.com/182-acres-in-hectares | 1,708,540,624,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473524.88/warc/CC-MAIN-20240221170215-20240221200215-00313.warc.gz | 648,446,435 | 7,479 | ## Convert 182 Acres to Hectares
To calculate 182 Acres to the corresponding value in Hectares, multiply the quantity in Acres by 0.40468564224 (conversion factor). In this case we should multiply 182 Acres by 0.40468564224 to get the equivalent result in Hectares:
182 Acres x 0.40468564224 = 73.65278688768 Hectares
182 Acres is equivalent to 73.65278688768 Hectares.
## How to convert from Acres to Hectares
The conversion factor from Acres to Hectares is 0.40468564224. To find out how many Acres in Hectares, multiply by the conversion factor or use the Area converter above. One hundred eighty-two Acres is equivalent to seventy-three point six five three Hectares.
## Definition of Acre
The acre (symbol: ac) is a unit of land area used in the imperial and US customary systems. It is defined as the area of 1 chain by 1 furlong (66 by 660 feet), which is exactly equal to 1⁄640 of a square mile, 43,560 square feet, approximately 4,047 m2, or about 40% of a hectare. The most commonly used acre today is the international acre. In the United States both the international acre and the US survey acre are in use, but differ by only two parts per million, see below. The most common use of the acre is to measure tracts of land. One international acre is defined as exactly 4,046.8564224 square metres.
## Definition of Hectare
The hectare (symbol: ha) is an SI accepted metric system unit of area equal to 100 ares (10,000 m2) and primarily used in the measurement of land as a metric replacement for the imperial acre. An acre is about 0.405 hectare and one hectare contains about 2.47 acres. In 1795, when the metric system was introduced, the "are" was defined as 100 square metres and the hectare ("hecto-" + "are") was thus 100 "ares" or 1⁄100 km2. When the metric system was further rationalised in 1960, resulting in the International System of Units (SI), the are was not included as a recognised unit. The hectare, however, remains as a non-SI unit accepted for use with the SI units, mentioned in Section 4.1 of the SI Brochure as a unit whose use is "expected to continue indefinitely".
## Using the Acres to Hectares converter you can get answers to questions like the following:
• How many Hectares are in 182 Acres?
• 182 Acres is equal to how many Hectares?
• How to convert 182 Acres to Hectares?
• How many is 182 Acres in Hectares?
• What is 182 Acres in Hectares?
• How much is 182 Acres in Hectares?
• How many ha are in 182 ac?
• 182 ac is equal to how many ha?
• How to convert 182 ac to ha?
• How many is 182 ac in ha?
• What is 182 ac in ha?
• How much is 182 ac in ha? | 700 | 2,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-10 | latest | en | 0.917006 |
https://www.brainkart.com/article/Half-Wave-Rectifier--Characteristics-and-Working_12520/ | 1,725,895,610,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651103.13/warc/CC-MAIN-20240909134831-20240909164831-00567.warc.gz | 656,569,947 | 8,434 | Home | | Electronic Devices and Circuits | Half Wave Rectifier: Characteristics and Working
# Half Wave Rectifier: Characteristics and Working
The half wave rectifier is a type of rectifier that rectifies only half cycle of the waveform. This describes the half wave rectifier circuit working.
RECTIFIERS
Rectifiers are classified according to the period of conduction.
They are
Ø Half Wave Rectifier
Ø Full Wave Rectifier
## 1. Half Wave Rectifier:
The half wave rectifier is a type of rectifier that rectifies only half cycle of the waveform. This describes the half wave rectifier circuit working. The half rectifier consist a step down transformer, a diode connected to the transformer and a load resistance connected to the cathode end of the diode. The circuit diagram of half wave transformer is shown below:
The main supply voltage is given to the transformer which will increase or decrease the voltage and give to the diode. In most of the cases we will decrease the supply voltage by using the step down transformer here also the output of the step down transformer will be in AC. This decreased AC voltage is given to the diode which is connected serial to the secondary winding of the transformer, diode is electronic component which will allow only the forward bias current and will not allow the reverse bias current. From the diode we will get the pulsating DC and give to the load resistance RL.
2. Working of Half Wave Rectifier:
The input given to the rectifier will have both positive and negative cycles. The half rectifier will allow only the positive half cycles and omit the negative half cycles. So first we will see how half wave rectifier works in the positive half cycles.
### Ø Positive Half Cycle:
· In the positive half cycles when the input AC power is given to the primary winding of the step down transformer, we will get the decreased voltage at the secondary winding which is given to the diode.
· The diode will allow current flowing in clock wise direction from anode to cathode in the forward bias (diode conduction will take place in forward bias) which will generate only the positive half cycle of the AC.
· The diode will eliminate the variations in the supply and give the pulsating DC voltage to the load resistance RL. We can get the pulsating DC at the Load resistance.
### Ø Negative Half Cycle:
· In the negative half cycle the current will flow in the anti-clockwise direction and the diode will go in to the reverse bias. In the reverse bias the diode will not conduct so, no current in flown from anode to cathode, and we cannot get any power at the load resistance.
· Only small amount of reverse current is flown from the diode but this current is almost negligible. And voltage across the load resistance is also zero.
## 3. Characteristics of Half Wave Rectifier:
There are some characteristics to the half wave rectifier they are
Ø Efficiency: The efficiency is defined as the ratio of input AC to the output DC.
Efficiency, Ƞ = P dc / Pac
DC power delivered to the load, Pdc = I2dc RL = ( Imax/pi ) 2 RL
AC power input to the transformer, Pac = Power dissipated in junction of diode + Power dissipated in load resistance RL
= I2rms RF + I2rms RL = {I2MAX/4}[RF + RL]
Rectification Efficiency, Ƞ = Pdc / Pac = {4/ 2}[RL/ (RF + RL)] = 0.406/{1+ RF/RL } If RF is neglected, the efficiency of half wave rectifier is 40.6%.
Ø Ripple factor: It is defined as the amount of AC content in the output DC. It nothing but amount of AC noise in the output DC. Less the ripple factor, performance of the rectifier is more. The ripple factor of half wave rectifier is about 1.21 (full wave rectifier has about 0.48). It can be calculated as follows:
The effective value of the load current I is given as sum of the rms values of
Harmonic currents I1, I2, I3, I4 and DC current Idc.
I2 =I2dc+I2 1+I22+I24 = I2 dc +I2ac
Ripple factor, is given as γ = I ac / Idc = (I2 – I2dc) / Idc = {( I rms / Idc2)-1} = Kf2 – 1)
Where Kf is the form factor of the input voltage. Form factor is given as
Kf = Irms /Iavg = (Imax/2)/ (Imax/pi) = pi/2 = 1.57
So, ripple factor, γ = (1.572 – 1) = 1.21
Ø Peak Inverse Voltage: It is defined as the maximum voltage that a diode can with stand in reverse bias. During the reverse bias as the diode do not conduct total voltage drops across the diode. Thus peak inverse voltage is equal to the input voltage Vs.
Ø Transformer Utilization Factor (TUF): The TUF is defined as the ratio of DC power is delivered to the load and the AC rating of the transformer secondary. Half wave rectifier has around 0.287 and full wave rectifier has around 0.693.
Half wave rectifier is mainly used in the low power circuits. It has very low performance when it is compared with the other rectifiers.
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
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Unformatted text preview: hernandez (ejh742) – homework 33 – Turner – (58120) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A standing wave of frequency 5 hertz is set up on a string 2 meters long with nodes at both ends and in the center, as shown. 2 meters Find the speed at which waves propagate on the string. 1. 10 m / s correct 2. . 4 m / s 3. 20 m / s 4. 2 . 5 m / s 5. 5 m / s Explanation: Let : f = 5 Hz and λ = 2 m . The wavelength is λ = 2 m, so the wave speed is | vectorv | = f λ = (5 Hz)(2 m) = 10 m/s . 002 (part 2 of 2) 10.0 points Find the fundamental frequency of vibration of the string. 1. 1 Hz 2. 2 . 5 Hz correct 3. 7 . 5 Hz 4. 10 Hz 5. 5 Hz Explanation: 2 meters The fundamental wave has only two nodes at the ends, so its wavelength is λ = 4 m and the fundamental frequency is f = v λ = 10 m / s 4 m = 2.5 Hz . 003 10.0 points Two wires are made of the same material but the second wire has twice the diameter and twice the length of the first wire. When the two wires are stretched, and the tension in the second wire is also twice the tension in the first wire, the fundamental frequency of the first wire is 920 Hz. What is the fundamental frequency of the second wire? Correct answer: 325 . 269 Hz. Explanation: Let : f 1 = 920 Hz . The second wire has twice the radius and hence four times the cross sectional area (= π r 2 ) of the first wire. Since the two wires are made from the same material, the linear density μ = π d 2 4 ρ of the second wire is four times that of the first: μ 2 = 4 μ 1 . The speed of transverse waves in a string or a wire is v = radicalBigg T μ , and since the second wire has twice the tension of the first wire, v 2 = radicalBigg T 2 μ 2 hernandez (ejh742) – homework 33 – Turner – (58120) 2 = radicalBigg 2 T 1 4 μ 1 = v 1 radicalbigg 1 2 . The fundamental frequency of standing waves in the wire is given by the condition L = λ 2 , so f = v λ = v 2 L . For the two wires, v 2 = v 1 √ 2 and L 2 = 2 L 1 , so f 2 = f 1 2 √ 2 = 920 Hz 2 √ 2 = 325 . 269 Hz . 004 10.0 points A nylon guitar string vibrates in a standing wave pattern shown below. 1 . 8 m What is the wavelength of the wave? Correct answer: 1 . 2 m. Explanation: Let : l = 1 . 8 m . The wavelength is the length of two loops: λ = 2 l 3 = 2 (1 . 8 m) 3 = 1 . 2 m . 005 (part 1 of 2) 10.0 points Consider a vibrating piano string. The string is under a tension T , has a length L and diameter d . What is the wavelength of its third har- monic? 1. λ 3 = 3 √ d L 2. λ 3 = 3 L 3. λ 3 = d 3 4. λ 3 = 6 L 5. λ 3 = L 6. λ 3 = 3 L 2 7. λ 3 = 2 L 3 correct 8. λ 3 = L 3 9. λ 3 = L 2 10. λ 3 = L 6 Explanation: Basic Concepts For a string fixed at both ends, the normal modes have a wavelength λ n = 2 L n ....
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Ask a homework question - tutors are online | 1,107 | 3,467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-09 | latest | en | 0.876611 |
https://www.geek-programmer.com/algorithms-the-beginning/ | 1,701,472,837,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100308.37/warc/CC-MAIN-20231201215122-20231202005122-00803.warc.gz | 893,564,131 | 18,972 | # Algorithms – The Beginning
## What is an Algorithm?
“Computer science is the study of algorithms”
Donald E. Knuth,
Author of “The Art of Computer Programming”
An algorithm is the procedure of getting an output for a certain computational problem when given an input.
Simply, it is the correct procedure for solving problems.
This is not necessary to be a programming code. Consider this classic humorous example, shampoo algorithm 🙂
1. Lather
2. Rinse
3. Repeat
This is a loop structure because the last word “Repeat” will repeat the previous two procedures again and again. Following this algorithm will result an empty shampoo bottle soon!
There are a lot of examples in our day-to-day life where we follow algorithms.
An example of application of algorithm in day-to-day life is, using shortcuts! When there’s a path perpendicular to the path you are currently moving and if you want to go to that path, probably you would choose the hypotenuse because you know it is shorter than the sum of the two paths.
Now when it comes to the field of programming, computers always need a very clear, unambiguous procedure to perform an operation. The way of writing code fulfilling this requirement is the real challenge. Learning about programming algorithms will make us more comfortable on the path to achieve it.
One way of solving a problem could be faster and clearer than another. Likewise, one algorithm could be effective than another one according to the given problem.
## But how to know which algorithm is better?
We calculate the runtime of an algorithm by just counting the number of steps it takes to perform a certain task.
This should be obvious. Lesser the number of steps, faster the algorithm.
But, there’s a problem. How do we define a single step?
For example, we may consider addition of two numbers as a single step.
This is fine. But how about multiplication?
We know that, multiplication is analogous to adding the same item several times.
So, does multiplication take more time than addition? Yes. It does. But not much!
Another thing. The accessing of memory in RAM (Random Access Memory) is faster than accessing the memory stored in the hard drive. The process of accessing hard drive has several additional steps to perform.
Now, suppose your code has a step which requires the access of an item saved in hard drive. You don’t know this fact. So you count this as a single step as well.
But since this actually consist of multiple steps when running, it would take more time to complete than a normal single step.
So when measuring the runtime of an algorithm in real computers, it would be overwhelming to consider all the tiny details like these.
Hence, we need an alternative approach to find the runtime of algorithms so that we can find which one is better.
## RAM Model for Computations
RAMRandom Access Machine (Machine, NOT Memory!) is a hypothetical computer which is used in developing machine independent algorithms.
This model is used to overcome the problems that we have encountered when calculating the run-time of an algorithm.
This model consists of a computer with the following rules:
• All the basic operations
multiplication (*)
subtraction (-)
assignment (=)
if
call
takes exactly one step.
• Loops and methods are not single step operations. This is obvious because, inside a loop or a method, there are several steps to run.
• There is no any time delay in accessing memory. Also we have infinite amount of memory.
So, with the help of RAM, all the problems that we met before were solved!
## Best, Worst and Average Case
The code that we type would not follow the same number of steps always.
For example, having a control structure like switch or if statements in the program would make it run different pieces of codes at different times according to its state.
The pseudo code here would run one of the two blocks by considering the state of the variable X. State is the value stored in that variable. If the value stored in X is the word “true”, then block A is executed. Else, if X is “false”, the next block B is executed. (Block is just a collection of several lines of code)
We don’t know how many lines in Block A and Block B. If A has 1000 lines and B has 10 lines, A would surely take a longer time than B. This all depends on the value stored in X variable. So if the value of X could change randomly each time the algorithm is running, there would be two possible run-times.
Note that, the code is running on a Random Access Machine. So, there’s no any previous issues that we discussed previously affecting the run-time. Only problem here is the Uncertainty.
There are 3 basic categories of running time of an algorithm.
1. Best-Case : Algorithm would run in a minimum number of steps. So, the best case is the fastest one. Takes the least time.
2. Average Case : Average value of the run time of an algorithm, for a given number of inputs.
3. Worst-case : The highest time taken by an algorithm to run a particular number of inputs.
The number of inputs to an algorithm is a very important factor. When we have more number of inputs, more items will be processed by the algorithm so, the run time depends on this parameter. Hence, we measure the running time of all the 3 different cases for a given number of steps.
The 1st and 3rd cases have a very low probability to occur when the number of inputs to the algorithm is large.
We can represent each of these 3 cases for a particular number of inputs say, n, in a graph as follows:
Think about the next number of inputs, n+1, we can add the three cases for n+1 in the same graph as follows.
Likewise, this can be repeated for very number of inputs, starting from 0. By joining each point of 3 cases separately, we get 3 nice curves.
Thus, we get a numerical function for each case. They are called time-complexity. They give a fine idea on how the running time of a certain algorithm behaves with the increase in the number of inputs, that is, the problem size.
But unfortunately, there are various practical problems. It is hard to formulate a numerical function for these graphs as the function has to map each item of the curve correctly. Even if we build the function, it would be complex, so, algorithm analysis would then be really difficult. To overcome these problems, the “Big O notation” is used. I’ll describe more about this notation in the next article. | 1,352 | 6,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2023-50 | latest | en | 0.925319 |
http://list.seqfan.eu/pipermail/seqfan/2016-May/016413.html | 1,542,338,426,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742970.12/warc/CC-MAIN-20181116025123-20181116051123-00455.warc.gz | 200,620,311 | 2,750 | # [seqfan] Re: The Ramanujan alpha, beta, and gamma series
Robert Munafo mrob27 at gmail.com
Tue May 17 19:01:37 CEST 2016
```I agree - the confusion is all mine, from not knowing how to "reverse" the
GF and not knowing how to put a Laurent series formulation in a %F field.
The formula %F lines should have GF's that actually work, like you suggest.
I'd like to keep the extra x in the numerator so that the series starts
with the 9 term (since the 1 term, i.e. coefficient of x^0, is already in
A051028). Ramanujan's 9 is the coefficient of x^-1 so ours should the
coefficient of x^1.
Ramanujan's notebook (see mrob.com/pub/math/images/ram-ln-p82.jpg clearly
defines the alpha/beta/gamma sequences as Laurent series on the GF's that I
gave, i.e. the same GF's as the a/b/c series respectively. That is noted in
the comment %C, and is also represented in the existing programs that use
Series[] with "Infinity" as the convergence limit.
So I'll edit those (A272853, A272854, A272855).
On Tue, May 17, 2016 at 9:41 AM, Ron Knott <ron at ronknott.com> wrote:
> The “a” series A051028 has GF
>
> (1 + 53 x + 9 x^2)/(1 - 82 x - 82 x^2 + x^3) = 1 + 135 x + 11161 x^2 +
926271 x^3 + 76869289 x^4+…
>
> and replacing x by 1/x gives the “alpha” series
>
> (x (9 + 53 x + x^2))/(1 - 82 x - 82 x^2 + x^3).
>
> The x factor is irrelevant as a GF. Note that we have just reversed the
coefficients in the numerator and this is the GF of
>
> (9 + 53 x + x^2)/(1 - 82 x - 82 x^2 + x^3) = 9 + 791 x + 65601 x^2 +
5444135 x^3 + 451797561 x^4+… i.e. A272853
>
> So perhaps the GFs should be changed to this form to be consistent with
the rest of OEIS and the second form used in A272853?
>
> Similarly for the betas: [...]
>
> Perhaps these changes might clear up some confusion between these 6
series and make things more consistent with the rest of OEIS?
--
Robert Munafo -- mrob.com | 616 | 1,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-47 | latest | en | 0.911482 |
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# Playing it Smart - It helps to know what you're doing
8 October 2007
Jill and Phil were discussing jacks-or-better video poker. They disagreed about starting hands of two pair when one is jacks or higher. Jill remarked that all the books tell you to hold both pairs and dump the unrelated card. Phil insisted that those books were written by casino stooges who wanted you to lose, or to win as little as possible, and you should hold only the high pair.
Here's Phil's logic. Two pair returns 2-for-1 if you don't improve. And the only possible improvement is to a full house worth 9-for-1 on a good machine, 8-for-1 or 7-for-1 otherwise. Keeping only the high pair still guarantees 1-for-1. And you can recover to two pair, the full house remains a possibility, and trips at 3-for-1 or quads paying 25-for-1 become achievable.
Phil's key points were a) 2-for-1 on two pair isn't that much better than the 1-for-1 on the high pair by itself, b) two pair have only one way to improve while the high pair has four, two of which pay over 2-for-1. "This is one of them there secrets the casino bosses don't want nobody to know," he claimed.
A solid citizen certainly could luck out and earn 25-to-1 with quads when drawing three to a high pair. But the strategies expounded by every gambling guru meritorious of the mantle imply playing the percentages rather than praying for providence. The problem with Phil's reasoning is in confusing the enumeration of possible results with the number of ways or, equivalently and more conveniently the probability these outcomes can occur.
The accompanying table gives the chances for the two alternate approaches, along with the corresponding returns and expected values. The returns shown are for "9-6" games, on which a full house pays 9-for-1. The probabilities are inherent in the structure of video poker how many cards are left in the deck and how many make one or another winning hand. The returns for each outcome are displayed on the machines. The component expected values are the probabilities multiplied by the corresponding returns. The overall expected values are the sums of the component contributions.
Expected value of a video poker hand dealt as a high pair and a low pair, played as a high pair and as two pair per dollar bet
``` probability return expected
(fraction) (\$) value (\$)
high pair
no improvement 0.7130 1 0.7130
two pair 0.1600 2 0.3200
three of a kind 0.1140 3 0.3420
full house 0.0102 9 0.0918
four of a kind 0.0028 25 0.0700
overall 1.0000 1.5368
two pair
no improvement 0.9150 2 1.8300
full house 0.0850 9 0.7650
overall 1.0000 2.5950```
The tabulated figures reveal that in a 9-6 game, playing the hand as two pair by drawing a single card has an expected value (rounded to the penny) of \$2.60 for every dollar bet. Dumping the low pair and drawing three cards has an expected value of \$1.53 on the dollar. Were the return for a full house 8-for-1, the respective figures would be \$1.54 and \$2.51. At 7-for-1 for a full house, the expected values would be \$1.53 and \$2.42. In all cases, holding two pair is two-thirds again better than taking a shot at a bigger payoff from the viewpoint of expected value.
It's OK if big boys and girls use a chance for higher payoff rather than expected value as their primary decision criterion in gambling for money. However, even with this basis for a strategy, discarding the low pair in the situation at issue gives cause for pause. The likelihood of an intermediate return with a full house drops from 8.50 to 1.02 percent, from about one in 12 to one in 100. And the prospects of quads with the high pair is a scant 0.28 percent, approximately one in 357. Think in terms of having a sure extra dollar versus one chance in 357 of picking up \$25.
Should you play by the book or chart your own course? It's your choice. But a technique that suits your personal preferences can still be predicated on the percentages of the game, not on a belief that having attracted you to a casino in the first place the rogues who run the joints try to dupe you into playing poorly. Sumner A Ingmark, the celebrated songster of sporting sophisticates everywhere, said it like this:
If you think a casino needs tricks or a ploy, ya
May be looking the wrong way for what can destroy ya,
Or perhaps have a case of acute paranoia.
Playing it Smart - It helps to know what you're doing is republished from Online.CasinoCity.com.
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Alan Krigman
Alan Krigman was a weekly syndicated newspaper gaming columnist and Editor & Publisher of Winning Ways, a monthly newsletter for casino aficionados. His columns focused on gambling probability and statistics. He passed away in October, 2013.
Alan Krigman
Alan Krigman was a weekly syndicated newspaper gaming columnist and Editor & Publisher of Winning Ways, a monthly newsletter for casino aficionados. His columns focused on gambling probability and statistics. He passed away in October, 2013. | 1,234 | 5,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2023-50 | latest | en | 0.958524 |
http://forum.enjoysudoku.com/blocku-part-ii-t36323.html | 1,627,053,529,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046149929.88/warc/CC-MAIN-20210723143921-20210723173921-00124.warc.gz | 16,518,600 | 7,812 | Blocku (part II)
Post puzzles for others to solve here.
Blocku (part II)
Blocku is a logic-based number-placement puzzle. It is distinct from but shares some properties and rules with Str8ts.
Rules:
•Each row and column are divided into two compartments by a black cell
•Each compartment, vertically or horizontally, must contain a straight – a set of consecutive numbers, but in any order. For example: 7, 6, 4, 5 is valid, but 1, 3, 8, 7 is not.
•Black cells don’t have numbers
•For a 10 by 10 Puzzle:
•Every row must include numbers from 1 to 9
•Every column must include numbers from 1 to 9
Sample6.png (22.2 KiB) Viewed 313 times
Posts: 79
Joined: 31 July 2019
Re: Blocku (part II)
Solved the second example by hand. IMHO it was much the same standard as the first one, solvable using only basics. Perhaps some of the the HI/LO decisions at the start required a bit more care.
Leren
Leren
Posts: 4202
Joined: 03 June 2012
Re: Blocku (part II)
Leren wrote:Solved the second example by hand. IMHO it was much the same standard as the first one, solvable using only basics. Perhaps some of the the HI/LO decisions at the start required a bit more care.
Leren
Hi Leren,
I appreciate your constructive comments, the puzzles are generated by a program that I've written, the computer calculates the difficulty of the puzzles based on how many times the computer has to take a guess, the difficulty of the first one was 1, and the second one was 9, If I'm not mistaken I have more difficult ones which I'll post soon. note that the computer program is not as smart as humans, so it takes more guesses than needed.
Thanks again,
-Kousha
Posts: 79
Joined: 31 July 2019
Re: Blocku (part II)
Perhaps I can explain why this puzzle started so easily for me. It's due to my Str8ts background, where Compartment Interactions in the same row or column and the resulting HI/LO decisions are Bread and Butter moves.
The clue 8 in r1c9 was a dead giveaway. Given a 7/2 compartment split in Row 1 => r1c10 can then only be 8 and r2c10 must also then be 8. Next in Column 10 the lower compartment must be in the range 1-7, and given the 8/1 compartment split in Row 7, r7c10 can only be 1 or 9, and since it can't be 9 it must be 1. Next, in Column 3 the upper compartment can only be 1-3 or 7-9, but the range of the first compartment in Row 1 is 1-7 so r3c123 must be 123. In Column 5 given a 7/2 compartment split and 8 in the upper compartment, it's range can only be 3-9 and r89c5 must therefore be 12. Then given the 7/2 compartment split in Row 10 the larger compartment range must be 1-7 and r10c12 must be 98 and since there is a clue 8 in r7c1, r10c1 = 9 and r10c2 = 8. In Row 4 the 8/1 compartment split => r4c1 = 19, but r10c1 = 9 so r4c1 = 1. In Row 6 the LH compartment can only be 123 or 789, but since the upper compartment in Column 2 is already 123, it can only be 789, and given the values in r7c1 and r10c12 => r6c1 = 7, r6c2 = 9 and r6c3 = 8.
So you can see that I've been able to solve 9 cells at the start without even writing down any candidates. Now what I said above might sound quite complicated, but when you've done it many times you can do it almost without thinking.
In your solver, you can deal with these compartment split/clue arrangements on a case by case basis, to simulate what an experienced human solver would do, and reduce your need for guessing.
Leren
Leren
Posts: 4202
Joined: 03 June 2012
Re: Blocku (part II)
To summarise what I was trying to say in my previous post I'll make up a table of the basic compartment interactions and HI/LO ranges for a 9 x 9 Blocku.
Code: Select all
`Compartment Size Split LO/HI Ranges HI/LO Ranges 9 / 0 1-9 / - 1-9 / - 8 / 1 1-8 / 9 2-9 / 1 7 /2 1-7 / 8-9 3-9 / 1-2 6 / 3 1-6 / 7-9 4-9 / 1-3 5 / 4 1-5 / 6-9 5-9 / 1-4`
Not much to it really. Leren
Last edited by Leren on Sun Aug 11, 2019 8:50 am, edited 4 times in total.
Leren
Posts: 4202
Joined: 03 June 2012
Re: Blocku (part II)
in a 10x10: 1 and 9 can't be on the same side of a black cell. Because you have 1 black cell in a row or column, you will then eliminate so many possibilities when you solve just 1 cell. a solved cell will have obvious options that can/can't be with it on that side of the black cell. This is how restrictive the rules are!
tarek
tarek
Posts: 3752
Joined: 05 January 2006
Re: Blocku (part II)
Hi Tarek, don't quite follow you. In the first 10 x 10 puzzle there were two rows and two columns with 9 cell compartments.
However, I get your point that it's certainly a fairly simple variant, I suspect that the target audience would be those who like a few minutes relaxing diversion.
Leren
Leren
Posts: 4202
Joined: 03 June 2012
Re: Blocku (part II)
Leren wrote:Hi Tarek, don't quite follow you. In the first 10 x 10 puzzle there were two rows and two columns with 9 cell compartments
Thanks,
I should have said when there are compartments on either side of the black cell
tarek
tarek
Posts: 3752
Joined: 05 January 2006
Re: Blocku (part II)
Leren wrote:Perhaps I can explain why this puzzle started so easily for me. It's due to my Str8ts background. ... So you can see that I've been able to solve 9 cells at the start without even writing down any candidates. Now what I said above might sound quite complicated, but when you've done it many times you can do it almost without thinking.
It doesn't require much thinking even if you've never done it before. A constraint is a constraint. That's why a strong sudoku background works well for almost any constraint puzzle. I got 16 solved cells before entering pencil marks, and this was practically my second puzzle of this constraint type. In fact, the start seemed easier than the first one which only gave me eight solved cells before pm (but I think I missed some easy ones). With pencil marks they were both easy. I was hoping some non-basics would have been needed for the second one when it was advertised as more challenging.
SpAce
Posts: 2672
Joined: 22 May 2017
Re: Blocku (part II)
Hi guys,
I do appreciate you comments, they are really helpful, I'll post more challenging ones soon, you asked for It
Posts: 79
Joined: 31 July 2019
Re: Blocku (part II)
Just posted a more challenging one here: http://forum.enjoysudoku.com/blocku-part-iii-t36337.html
Hope you enjoy it | 1,846 | 6,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-31 | latest | en | 0.960725 |
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# If one mole of carbon atoms …
## CBSE, JEE, NEET, NDA
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
If one mole of carbon atoms weighs 12 grams , what is the mass (in grams) of 1 atom of carbon?
Sia 🤖 1 month, 2 weeks ago
We have given that one mole of carbon atom weighs 12 g.
A mole is defined as exactly equal to 6.022 × 1023 atoms.
So, we know that a mole of carbon contains 6.022 × 1023 atoms.
Therefore, we can write the mass of 6.022 × 1023 atoms = 12 g.
Using the unitary method to find the mass of 1 atom of carbon in grams. So, we can write as below-
So, now mass of 1 atom of carbon will be =12 / (6.022 × 1023) = 1.993 × 10−23 g
Hence, the mass in grams of 1 atom of carbon is 1.993 × 10−23g.
If we have to find the mass of 2 atoms or n number of carbon atoms, the same way we can proceed to find the mass in grams of the carbon atoms.
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##### TI-Nspire CAS in Engineering Mathematics: First Order Systems and Symbolic Matrix Exponentiation
Solving Ordinary Differential Equations by using a library of Laplace Transformations
Uitgever: T³ Europe
Auteur: T³ Europe, Michel Beaudin
Onderwerp: STEM
Tags Engineering , Matrix
The Library of Laplace Transforms ("ETS_specfunc.tns") contains the function called "simultd" that can be used to solve a system of ODEs, using Laplace transforms techniques. Generally speaking, a first order, constant coefficient system, has the following form:
where A is a square (n × n) matrix with real entries, g(t) is a n × 1 matrix (column vector), t₀ ∈ ℝ and x₀ is a n × 1 matrix (constant column vector). Solving the system means to find the n × 1 matrix (column vector) X(t) that satisfies both the differential system and the initial condition.
A is a matrix and not scalar... But TI-Nspire CAS knows how to compute "the exponential of a matrix". Our goal is to define, IN EXACT MODE (when possible), the exponential of a square matrix A: e(At) (t a real variable).
We make use of the library ETS_specfunc.tns to easily compute e(At); applying our findings in a concrete example of a system originating from connected salt tanks.
And finally we apply our solution of to the problem of finding parametric equations to a great circle.
To be able to use the tns file below, you need to save "ETS_specfunc.tns" in the TI-Nspire CAS "Mylib" folders and refresh your libraries. | 412 | 1,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.711188 |
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MCQ Questions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter with Answers
## CBSE: NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter MCQ’s with Answers
NCERT Books are textbooks which are issued & distributed by the National Council of Educational Research and Training (NCERT). NCERT books are important for the schooling system. NCERT Books are available in E-Books. If you are searching for MCQs (Multiple Choice Questions) with Answers of for NCERT Class 12 Physics then you have came to right way. CBSE students can also solve NCERT Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter PDF Download w their preparation level. MCQs are Prepared Based on Latest Exam Patterns. MCQs Questions with Answers for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter are prepared to help students to understand concepts very well. Objective wise Questions 12th class Physics Chapter 11 Dual Nature of Radiation and Matter Chapter wise PDF over here are available in the following links. Score maximum marks in the exam.
## NCERT Physics MCQs for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter Question and Answers PDF
Q1. De-Broglie equation states the:
(a) dual nature
(b) particle nature
(c) wave nature
(d) none of these
Q2. When an electron jumps across a potential difference of 1 V, it gains energy equal to :
(a) 1.602 × 10-19 J
(b) 1.602 × 1019 J
(c) 1.602 × 1024 J
(d) 1 J
Q3. The different stages of discharge in a discharge tube can be explained on the basis of:
(a) the wave nature of light
(b) the dual nature of light
(c) wave nature of electrons
(d) the collision between the charged particles emitted from the cathode the atoms of the gas in the tube
Answer:(d) the collision between the charged particles emitted from the cathode the atoms of the gas in the tube
Q4. Compared to liquids and solids, gases are:
(a) good conductors of electricity
(b) best conductors of electricity
(c) very poor conductors of electricity
(d) good or bad conductors of electricity depending upon the nature of the gas
Answer:(c) very poor conductors of electricity
Q5. The charge of a photo electron is :
(a) 9.1 × 10-31 C (b) 9.1 × 10-27 C
(c) 9.1 × 10-24 C (d) none of these
Q6. which Characteristic of a target does the Mosley’s law relate the frequency of X-rays?
(a) density (b) atomic number
(c) atomic weight (d) interatomic space
Q7. Evidence of the wave nature of light cannot be obtained from:
(a) diffraction (b) interference
(c) doppler effect (d) reflection
Q8. The de-Broglie wavelength of particle of mass 1 mg moving with a velocity of 1 ms-1, in terms of Planck’s constant h, is given by (in metre):
(a) 105 h (b) 106 h
(c) 10-3 h (d) 103 h
Q9. When a yellow light is incident on a surface, no electrons are emitted while green light can emit electrons. If the red light is incident on the surface then:
(a) no electrons are emitted
(b) photons are emitted
(c) electrons of higher energy are emitted
(d) electrons of lower energy are emitted
Q10. What is the de-Broglie wavelength of an electron accelerated from rest through a potential difference of 100 volts?
(a) 12.3 Å (b) 1.23 Å
(c) 0.123 Å (d) None of these
Q11. Which of the following radiations cannot eject photo electrons?
(a) ultraviolet (b) infrared
(c) visible (d) X-rays
Q12.In photo electric emission, for alkali metals the threshold frequency lies in the:
(a) visible region (b) ultraviolet region
(c) infrared region (d) far end of the infrared region
Q13.X-rays are:
(a) deflected by an electric field
(b) deflected by a magnetic field
(c) deflected by both electric and magnetic fields
(d) not deflected by electric and magnetic fields
Answer:(d) not deflected by electric and magnetic fields
Q14. Millikan’s oil drop experiment makes use of:
(a) Stokes’ law (b) Boyle’s law
(c) Gas equation (d) Bernoulli’s theorem
Q15. Name the scientists who first studied the passage of electricity through fluids to establish the electrical nature of matter:
(a)Millikan (b) Planck
Q16. The momentum of an electron that emits a wavelength of 2 Å. will be:
(a) 6.4 × 10-36 kgms-1 (b) 3.3 × 10-24 kgms-1
(c) 3.3 × 10-34 kgms-1 (d) none of these
Q17.The strength of photoelectric current depends upon :
(d) distance between anode and cathode
Q18. The work function of photoelectric material is 3.3 eV. The threshold frequency will be equal to:
(a) 8 × 1014 Hz
(b) 8 × 1010 Hz
(c) 5 × 1010 Hz
(d) 4 × 1014 Hz
Q19.In Thomson’s experiment number of parabola gives :
(a) the no. of electrons present in element
(b) the no. of proton present in element
(c) the no. of neutrons present in element
(d) the no. of isotopes of the element present
Answer:(d) the no. of isotopes of the element present
Q20.The ratio of specific charge of an alpha particle to the proton is:
(a) 1 : 2 (b) 2 : 1 (c) 4 : 1 (d) 1 : 4
Q21. Kinetic energy of emitted electrons depends upon :
(a) frequency
(b) intensity
(c) nature of atmosphere surrounding the electrons
(d) none of these
Q22.Protons and alpha particles have the same de-Broglie wavelength. What is same for both of them ?
(a) Energy (b) Time period
(c) Frequency (d) Momentum
Q23. Photoelectrons are being obtained by irradiating zinc by a radiation of 3100 Å. In order to increase the kinetic energy of ejected photoelectrons.
(a) the intensity of radiation should be increased.
(b) the wave length of radiation should be increased.
(c) the wavelength of radiation should be decreased.
(d) both wavelength and intesity of radiation should be increased.
Q24. For light of wavelength 5000 Å, the photon energy is nearly 2.5 eV. For X-rays of wavelength 1 Å, the photon energy will be close to:
(a) 2.5 × 5000 eV (b) 2.5 ÷ 5000 eV
(c) 2.5 × (5000)² eV (d) 2.5 ÷ (5000)² eV
Q25. The best metal to be used for photoemission is:
(a) Potassium (b) Lithium
(c) Sodium (d) Cesium
Q26. The threshold frequency for a certain metal is v0. When light of frequency v = 2v0 is incident on it, the maximum velocity of photo electrons is 4 × 106 ms-1. If the frequency of incident radiation is increased to 5 v0, then the maximum velocity of photo electrons (m/s) is
(a) 8 × 105 (b) 2 × 106 (c) 2 × 107 (d) 8 × 106
Q27.Which of the following is not the property of photons
(a) charge (b) rest mass (c) energy (d) momentum
Q28. When light is directed at the metal surface, the emitted electrons:
(a) are called photons
(b) have energies that depend upon the intensity of light.
(c) have random energies.
(d) have energies that depend upon the frequency of light.
Answer:(d) have energies that depend upon the frequency of light.
Q29.The wavelength associated with n electron is 1Å. The potential difference required for accelerating it is
(a) 100 V (b) 150 V (c) 250 V (d) 10³ V | 1,886 | 6,978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-33 | longest | en | 0.835282 |
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Alexandra Bowling 1K
Posts: 11
Joined: Fri Sep 26, 2014 2:02 pm
### Knowing whether the compound has net neutral charge or not
For quiz 3 prep #4, you have to find the chemical formula for potassium tetracyanonickelate (II). I know that it should be K2[Ni(CN)4], but why? I know the following:
(CN)4= (-1)x4=-4
Ni (II)=+2
Thus, the charge of the coordination compound is -2, and making potassium K2 instead of K neutralizes the whole compound. But how do you know when you're supposed to neutralize the compound? How can you tell from the formula whether or not the whole compound should have a net negative charge or not?
K Honeychurch 1K
Posts: 27
Joined: Fri Sep 26, 2014 2:02 pm
### Re: Knowing whether the compound has net neutral charge or n
I think the problem would have to specify that the compound has a charge, otherwise you can assume the total charge is 0. Also, I don't think we've had a problem in which the compound has a charge.
Justin Le 2I
Posts: 142
Joined: Fri Sep 26, 2014 2:02 pm
### Re: Knowing whether the compound has net neutral charge or n
If the compound has a charge, it will say "ion" at the end in the name.
GinaYoung1L
Posts: 16
Joined: Fri Sep 26, 2014 2:02 pm
### Re: Knowing whether the compound has net neutral charge or n
A coordination compound is always electrically neutral. It includes the complex (metal and ligands inside the brackets) and any outside ions. Therefore, it is not the coordination compound that has the -2 charge, it is the complex. Complexes (only the stuff inside the brackets) can be neutral, anionic, or cationic. It is anionic if there is an -ate attached to the end of the metal's stem name.
Alexandra Bowling 1K
Posts: 11
Joined: Fri Sep 26, 2014 2:02 pm
### Re: Knowing whether the compound has net neutral charge or n
But if it does have "ion" at the end of the name, how are you supposed to figure out the positive or negative charge?
Valeria Mazzanti 3H
Posts: 2
Joined: Fri Sep 26, 2014 2:02 pm
### Re: Knowing whether the compound has net neutral charge or n
They would probably have to specify the charge it has.
thuyphuong1I
Posts: 9
Joined: Fri Sep 26, 2014 2:02 pm
### Re: Knowing whether the compound has net neutral charge or n
If it is an ion, you can usually tell it has a negative charge if "-ate" is added to the end of the metal. You can also determine charge by assigning charges to each element in your complex, given ligands with their prefixes (if any) and the transition metal with its oxidation number, and then adding them up.
Last edited by thuyphuong1I on Wed Dec 03, 2014 10:06 pm, edited 1 time in total.
Justin Le 2I
Posts: 142
Joined: Fri Sep 26, 2014 2:02 pm
### Re: Knowing whether the compound has net neutral charge or n
Gina Young is correct. My previous comment should only apply to complexes.
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Articles / Productivity Apps and Services / Microsoft Office / Microsoft Excel
Tip/Trick
Posted 2 Sep 2020
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Converting Column Numbers to Range Headers and Vice Versa
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2 Sep 2020CPOL
Two functions to help manipulate Excel sheets with VBA by converting Column Header letters to 1 based numbers and back
Utility functions to get the column number (1 based) in Excel from the letter headers or the letters from the 1 based number
Introduction
In Excel VBA programming, it's sometimes necessary to take user input in terms of column header letters and convert it to a number or the reverse. These two functions will help you do that. They are very simple mathematical manipulation on base 26.
I ported this code from many similar JavaScript examples you can find for Apps Script on GSuite, the particular author that wrote the original answer to a Stack Exchange question was "AdamL".
I had to rewrite a little for the VBA way of thinking!
Background
I sometimes write my Excel VBA with a reliance on configuration details rather than coded values. As things change, I often want to build or slice dynamic range selections from configured column headers, etc. These functions help with that.
Using the Code
Place the function declarations in a standard module and then use anywhere in your project as global functions:
VB.NET
```debug.print columnToLetter(27) ' Gets AA
debug.print columnToLetter(702) ' Gets ZZ
debug.print columnToLetter(703) ' Gets AAA```
VB.NET
```debug.print letterToColumn("AA") ' Gets 27
debug.print letterToColumn("ZZ") ' Gets 702
debug.print letterToColumn("AAA") ' Gets 703```
VB.NET
```Public Function columnToLetter(column As Integer) As String
Dim temp As Integer
Dim letter As String
If column < 1 Or column > 16384 Then
Err.Raise vbObjectError + 1024 + 99, "columnToLetter", _
"Column numbers in the range 1 to 16384 (XFD) only. You tried: " & column
End If
Do While (column > 0)
temp = (column - 1) Mod 26
letter = Chr(temp + 65) + letter
column = (column - temp - 1) / 26
Loop
columnToLetter = letter
End Function
Public Function letterToColumn(ByVal letter As String) As Integer
Dim column As Integer
Dim length As Integer
Dim c As String
Dim n As Integer
Do
c = Left(letter, 1)
length = Len(letter)
n = Asc(c) - 64
If n < 1 Or n > 26 Then
Err.Raise vbObjectError + 1024 + 99, "letterToColumn", _
"Only letters A to Z are valid. You tried """ & c & """"
End If
column = column + n * 26 ^ (length - 1)
letter = Mid(letter, 2) ' Trim off first letter
Loop Until Len(letter) = 0
letterToColumn = column
End Function```
History
• 2nd September, 2020: Initial tip
Written By
Database Developer self interested
United Kingdom
I have been a software developer for about 20 years, mostly in small IT department that means you do all the roles all the time from 1st line support to 3rd line diagnostics and help authoring. To be fair, I don't do enough of the later which in turn causes way too much of the former with new staff always ready to show you how unintuitive your interfaces are!
I generally consider myself a "data plumber" with the majority of my work in back end SQL. If I could rule the world by writing a SPROC I would....
New definition of Stoicism: Someone who start a career in I.T. and still has a career in I.T. after 20 years!
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possible glitch DarkStarHarry3-Sep-20 12:14 DarkStarHarry 3-Sep-20 12:14
Re: possible glitch Darren G44116-Sep-20 2:02 Darren G441 16-Sep-20 2:02
Use RC Ref Style directly in VBA ssa-ed3-Sep-20 10:43 ssa-ed 3-Sep-20 10:43
Re: Use RC Ref Style directly in VBA Darren G44116-Sep-20 2:19 Darren G441 16-Sep-20 2:19
Bear in mind these functions are not intended to be used in FORMULA on the sheet. They are for VBA manipulation. Examples of Where I personally use them include where have configuration details of ranges of data from an external source where I expect to get a rectangular range of a certain number of columns. I would expect that config to meet some problem specific criteria, for example 4 columns wide. So, I am quite happy to get a configuration of say D2:H8 and I would string process the passed configuation to pull out the column specifiers with a regular expression along the lines of:- /^(\D{1,3})\d+ D{1,3})\d\$/ This would give me two letters (c1, c2) that I could then verify that the resulting column numbers meet the following test c2 minus c1 equals 3 I have the same configuration details specify which of these 4 columns holds what. In the example I gave above I might want to specify that column D is a project ID number, column E is category label, F is a date, and G is a comment. I might expect to get a different source data with slightly different configuration of columns etc. Basically, I try to write my code configurable for easy maintenance in a changing world where the programmer (me!) is not always on tap! This approach helps me, hence I thought I'd share
Converting Column Numbers to Range headers and vice versa Frank Malcolm2-Sep-20 20:28 Frank Malcolm 2-Sep-20 20:28
Re: Converting Column Numbers to Range headers and vice versa Frank Malcolm2-Sep-20 20:31 Frank Malcolm 2-Sep-20 20:31
Re: Converting Column Numbers to Range headers and vice versa Darren G44116-Sep-20 2:27 Darren G441 16-Sep-20 2:27
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posted by .
Gymnast Clothing manufactures expensive soccer cleats for sale to college bookstores in runs of up to 500. Its cost (in dollars) for a run of x pairs of cleats is
C(x) = 3000 + 9x + 0.1x2 (0 ≤ x ≤ 500).
Gymnast Clothing sells the cleats at \$130 per pair.
Find the revenue and profit functions.
How many should Gymnast Clothing manufacture to make a profit?
What is the break even cost?
(by using the formula
Revenue is price * units sold
Profit is revenue-cost
how do find the I break the equation, revenue and profit function?
I don't know how to break the equation given.
Thank you | 160 | 615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2017-30 | longest | en | 0.917238 |
https://groups.google.com/g/sci.math/c/70cqhtzn8K8 | 1,642,840,478,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303779.65/warc/CC-MAIN-20220122073422-20220122103422-00497.warc.gz | 344,943,255 | 147,660 | Are you imaginary?
127 views
mitchr...@gmail.com
Nov 24, 2021, 1:02:28 PM11/24/21
to
That math is rightly named...
it is a formula without a solution.
Because negative one does not
exist that formula does not...
Python
Nov 24, 2021, 1:18:06 PM11/24/21
to
smitch...@gmail.com wrote:
> That math is rightly named...
> it is a formula without a solution.
equivalence class of polynomial X for ~ (defined by P ~ Q iff
P = Q [X^2+1]) exists in R[X]/~
> Because negative one does not
> exist that formula does not...
Negative one is the equivalence class of (0,1) for
~ on N^2 (defined as (a,b) ~ (c,d) iff a+d = b+c.
It is a set of pairs of natural numbers. It definitely
exists.
Why, Smitch, don't you TRY to learn, ONCE?
Serg io
Nov 24, 2021, 1:22:27 PM11/24/21
to
here you go; -1
happy now ?
it is what happens when Elvis leaves the room, -1.
mitchr...@gmail.com
Nov 24, 2021, 1:25:34 PM11/24/21
to
On Wednesday, November 24, 2021 at 10:22:27 AM UTC-8, Serg io wrote:
> On 11/24/2021 12:02 PM, mitchr...@gmail.com wrote:
> > That math is rightly named...
> > it is a formula without a solution.
> > Because negative one does not
> > exist that formula does not...
> >
> here you go; -1
>
You are going to have to do better.
That is just attaching a minus sing to the only real positive 1.
And that does not exist outside of a subtraction.
LORD God
Nov 24, 2021, 1:27:09 PM11/24/21
to
https://nrich.maths.org/5961
we already covered that a thousand times mitch,
minus .000...1 is the imaginary number,
its square root is .000...1 itself,
one minus one imaginary,
1 - .999 = .000...1,
shorthand,
0
mitchr...@gmail.com
Nov 24, 2021, 1:28:41 PM11/24/21
to
Add zero to .999 repeating and you still get .999 repeating
you do not get 1...
Mitchell Raemsch
LORD God
Nov 24, 2021, 1:34:35 PM11/24/21
to
On Wednesday, November 24, 2021 at 12:28:41 PM UTC-6, mitchr...@gmail.com wrote:
> On Wednesday, November 24, 2021 at 10:27:09 AM UTC-8, LORD God wrote:
> > On Wednesday, November 24, 2021 at 12:02:28 PM UTC-6, mitchr...@gmail.com wrote:
> > >
> > > That math is rightly named...
> > > it is a formula without a solution.
> > > Because negative one does not
> > > exist that formula does not...
> > >
> > https://nrich.maths.org/5961
> >
> > we already covered that a thousand times mitch,
> > minus .000...1 is the imaginary number,
> > its square root is .000...1 itself,
> > one minus one imaginary,
> > 1 - .000...1 = .999...
> > shorthand,
> > 0
>
> Add zero to .999 repeating and you still get .999 repeating
> you do not get 1...
>
> Mitchell Raemsch
>
0 is shorthand for .000...1
.000...1 + .999... = 1
0 + .999... = 1
mitchr...@gmail.com
Nov 24, 2021, 1:41:12 PM11/24/21
to
> 0 + .999... = 1
No. .999 repeating + 0 is still .999 repeating...
Prove otherwise.
LORD God
Nov 24, 2021, 1:52:40 PM11/24/21
to
> > 0 is shorthand for .000...1
> > .000...1 + .999... = 1
> > 0 + .999... = 1
Again?
Michael Moroney
Nov 24, 2021, 3:05:35 PM11/24/21
to
But it does. I can have an electric charge of -1 coulombs just as easily
as a charge of +1 coulombs.
Alt Atheism
Nov 24, 2021, 5:06:44 PM11/24/21
to
Electricians all know minus 1 A-s
is much less than plus 1 A-s
see math real line proof
mitchr...@gmail.com
Nov 24, 2021, 10:59:29 PM11/24/21
to
Those are man's relatives.
Show how ions prove negative quantities.
It only shows less positive quantities.
zelos...@gmail.com
Nov 25, 2021, 12:31:02 AM11/25/21
to
false
x^2+1=0
has 2 solutions
i and -i
as always, you are wrong
zelos...@gmail.com
Nov 25, 2021, 12:32:05 AM11/25/21
to
you are wrong on all accounts there. It exists as a negative :)
Because we have in reals that for all a there exists a b such that a+b=0
that means b is the negative of a
Michael Moroney
Nov 25, 2021, 12:50:55 AM11/25/21
to
How is electric charge related to humans?
> Show how ions prove negative quantities.
Look at the forces they generate.
> It only shows less positive quantities.
Nope. A positively charged proton attracts an electron at a given
distance with a certain force. A negatively charged electron at the
same distance repels the other electron with the same magnitude force.
"Less positive" would mean less attraction, not repulsion. Also there
is a definite zero point. Zero charge produces zero force on other charges.
>
mitchr...@gmail.com
Nov 25, 2021, 1:08:37 PM11/25/21
to
You don't have negative electrons you have less electrons
zelos...@gmail.com
Nov 26, 2021, 12:29:28 AM11/26/21
to
the electron has negative charge and positron has positive charge.
Michael Moroney
Nov 26, 2021, 1:10:54 AM11/26/21
to
Smitch, the electron has a negative charge all by itself!
Chris M. Thomasson
Nov 26, 2021, 1:16:26 AM11/26/21
to
You are a con artist right?
mitchr...@gmail.com
Nov 26, 2021, 12:56:35 PM11/26/21
to
That math is named right.
It is a formula without a solution..
Mitchell Raemsch
Brain Hubbs
Nov 26, 2021, 1:00:53 PM11/26/21
to
mitchr...@gmail.com wrote:
>> You are a con artist right?
>
> That is just your imagination. That math is named right.
> It is a formula without a solution..
Spasiba, Mockba.
mitchr...@gmail.com
Nov 26, 2021, 1:30:47 PM11/26/21
to
It is named right.
Because in math it is a formula alone
and not a solution.
Earle Jones
Nov 27, 2021, 2:47:39 PM11/27/21
to
*
Mitchell: Are you familiar with "The fundamental theorem of algebra"? (Gauss, 1799)
Simply, it says: Every poloynomial equation of degree 'n' has 'n' roots.
Example: x^2 -3x + 2 = 0 (This equation is of degree '2' and therefore has two roots.)
The two "roots" are 1 and 2. {Check this: (x-1) times (x-2) = x^2 - 3x + 2) -- it checks!}
Another Example" x^2 + 3x + 3= 0
earle
*
mitchr...@gmail.com
Nov 27, 2021, 2:51:53 PM11/27/21
to
On Saturday, November 27, 2021 at 11:47:39 AM UTC-8, Earle Jones wrote:
> On Wed Nov 24 10:02:23 2021 "mitchr...@gmail.com" wrote:
> > That math is rightly named...
> > it is a formula without a solution.
> > Because negative one does not
> > exist that formula does not...
> *
> Mitchell: Are you familiar with "The fundamental theorem of algebra"? (Gauss, 1799)
>
new algebra has only a single positive quandrant.
No negative polynomial solutions belong. No negative quantities
just subtraction negative with a limit.
Mitchell Raemsch
Python
Nov 27, 2021, 3:27:16 PM11/27/21
to
schmitch...@gmail.com schwrote:
So your "new algebra" is weak, Smitch. You can stuff it into
your puffy ass. We will keep the stronger real algebra.
mitchr...@gmail.com
Nov 27, 2021, 3:31:36 PM11/27/21
to
Algebra has no negative solutions you moron...
konyberg
Nov 27, 2021, 4:01:21 PM11/27/21
to
You talk like AP. You know he is mad. Are you?
KON
Serg io
Nov 27, 2021, 4:05:50 PM11/27/21
to
your monkey DNA has -1 on it.
mitchr...@gmail.com
Nov 27, 2021, 4:20:23 PM11/27/21
to
AP wants to use my work.
I don't believe in anger. Do you?
Mitchell Raemsch
Serg io
Nov 27, 2021, 4:47:15 PM11/27/21
to
AP is not angry at all, he is generating neo-physics
mitchr...@gmail.com
Nov 27, 2021, 4:50:12 PM11/27/21
to
Sure he is. It is why he is resentful instead...
Earle Jones
Nov 28, 2021, 7:30:00 PM11/28/21
to
*
Mitchell: In other words, you do not believe in the Fundamental Theorem of Algebra (Gauss, 1799.)
Is that right?
earle
*
zelos...@gmail.com
Nov 29, 2021, 12:34:18 AM11/29/21
to
False, i is one of the solutions to x^2+1=0
Earle Jones
Nov 29, 2021, 1:05:39 AM11/29/21
to
On Sat Nov 27 13:20:17 2021 "mitchr...@gmail.com" wrote:
> On Saturday, November 27, 2021 at 1:01:21 PM UTC-8, konyberg wrote:
*
Mitchell:
I think that you and Archimedes Plutonium should collaborate, since you seem to agree on many mathematical ideas.
You could learn from him and he could also learn from you.
You might also want to borrow from the work of John Gabriel, whose mathematics is more advanced than either you or Archimedes Plutonium. What do you think? Is this a good idea?
earle
*
Serg io
Nov 29, 2021, 9:42:13 AM11/29/21
to
On 11/29/2021 12:05 AM, Earle Jones wrote:
> On Sat Nov 27 13:20:17 2021 "mitchr...@gmail.com" wrote:
>> On Saturday, November 27, 2021 at 1:01:21 PM UTC-8, konyberg wrote:
I think it is a fabulous idea! Just think of what they could do! Amazing Math!
They should also invite that sciencie guy James McGinn in sci.physics who knows as much math as his missing water vapor!
Archimedes Plutonium
Nov 29, 2021, 5:20:59 PM11/29/21
to
Mitch can Alexander Fetter, John Lipa, William Little, Douglas Osheroff, David Ritson, H. Alan Schwettman, John Turneaure, Robert Wagoner, Stanley Wojcicki, Mason Yearian ever, ever ask the question, which is the atom's real electron, the muon stuck inside a 840MeV proton torus doing the Faraday law or the 0.5MeV particle that AP calls the Dirac magnetic monopole. Or are they just plain dumb scientists who cannot even multiply 105 by 9 and see it is in sigma error of 940 or 938.
HISTORY OF THE PROTON MASS and the 945 MeV //Atom Totality series, book 3 Kindle Edition
by Archimedes Plutonium (Author)
In 2016-2017, AP discovered that the real proton has a mass of 840 MeV, not 938. The real electron was actually the muon and the muon stays inside the proton that forms a proton torus of 8 rings and with the muon as bar magnet is a Faraday Law producing magnetic monopoles. So this book is all about why researchers of physics and engineers keep getting the number 938MeV when they should be getting the number 840 MeV + 105 MeV = 945 MeV.
Cover Picture is a proton torus of 8 rings with a muon of 1 ring inside the proton torus, doing the Faraday Law and producing magnetic monopoles.
Length: 17 pages
Product details
• Publication Date : December 18, 2019
• Word Wise : Enabled
• Print Length : 17 pages
• File Size : 698 KB
• ASIN : B082WYGVNG
• Language: : English
• Text-to-Speech : Not enabled
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#1-4, 105th published book
Atom Geometry is Torus Geometry // Atom Totality series, book 4 Kindle Edition
by Archimedes Plutonium (Author)
Since all atoms are doing the Faraday Law inside them, of their thrusting muon into a proton coil in the shape of a geometry torus, then the torus is the geometry of each and every atom. But then we must explain the neutrons since the muon and proton are doing Faraday's Law, then the neutron needs to be explained in terms of this proton torus with muon inside, all three shaped as rings. The muon is a single ring and each proton is 8 rings. The neutron is shaped like a plate and is solid not hollow. The explanation of a neutron is that of a capacitor storing what the proton-muon rings produce in electricity. Where would the neutron parallel plates be located? I argue in this text that the neutron plates when fully grown from 1 eV until 945MeV are like two parallel plate capacitors where each neutron is part of one plate, like two pieces of bread with the proton-muon torus being a hamburger patty.
Cover Picture: I assembled two atoms in this picture where the proton torus with a band of muons inside traveling around and around the proton torus producing electricity. And the pie-plates represent neutrons as parallel-plate capacitors.
Length: 39 pages
Product details
• Publication Date : March 24, 2020
• Word Wise : Not Enabled
• ASIN : B086BGSNXN
• Print Length : 39 pages
• File Size : 935 KB
• Language: : English
• Text-to-Speech : Not enabled
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• Lending : Enabled
Amazon Best Sellers Rank: #1,656,820 Paid in Kindle Store (See Top 100 Paid in Kindle Store)
#6413 in Mathematics (Kindle Store)
#315 in One-Hour Science & Math Short Reads
#4953 in Physics (Kindle Store)
#1-5, 112th published book
New Perspective on Psi^2 in the Schrodinger Equation in a Atom Totality Universe// Atom Totality series, book 5
Kindle Edition
by Archimedes Plutonium (Author)
I first heard of the Schrodinger equation in college chemistry class. We never actually did any problem solving with the equation, and we were only told about it. Then taking physics my next year in college and after I bought the Feynman Lectures on Physics, just for fun for side reading, three volume set did I learn what this Schrodinger equation and the Psi^2 wavefunction was about. I am not going to teach the mathematics of the Schrodinger equation and the math calculations of the Psi or Psi^2 in this book, but leave that up to the reader or student to do that from Feynman's Lectures on Physics. The purpose of this book is to give a new and different interpretation of what Psi^2 is, what Psi^2 means. Correct interpretation of physics experiments and observations turns out to be one of the most difficult tasks in all of physics.
Cover Picture: a photograph taken of me in 1993, after the discovery of Plutonium Atom Totality, and I was 43 years old then, on a wintery hill of New Hampshire. It is nice that Feynman wrote a physics textbook series, for I am very much benefitting from his wisdom. If he had not done that, getting organized in physics by writing textbooks, I would not be writing this book. And I would not have discovered the true meaning of the Fine Structure Constant, for it was Feynman who showed us that FSC is really 0.0854, not that of 0.0072. All because 0.0854 is Psi, and Psi^2 is 0.0072.
Length: 20 pages
Product details
• ASIN : B0875SVDC7
• Publication date : April 15, 2020
• Language: : English
• File size : 1134 KB
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• Print length : 20 pages
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• Best Sellers Rank: #240,066 in Kindle Store (See Top 100 in Kindle Store)
◦ #5 in 30-Minute Science & Math Short Reads
◦ #65 in General Chemistry & Reference
◦ #481 in Physics (Kindle Store)
#1-6, 135th published book
QED in Atom Totality theory where proton is a 8 ring torus and electron = muon inside proton doing Faraday Law// Atom Totality series, book 6 Kindle Edition
by Archimedes Plutonium (Author)
Since the real true electron of atoms is the muon and is a one ring bar magnet thrusting through the 8 ring torus of a proton, we need a whole entire new model of the hydrogen atom. Because the Bohr model with the 0.5MeV particle jumping orbitals as the explanation of Spectral Lines is all wrong. In this vacuum of explaining spectral line physics, comes the AP Model which simply states that the hydrogen atom creates Spectral lines because at any one instant of time 4 of the 8 proton rings is "in view" and the electricity coming from those 4 view rings creates spectral line physics.
Cover Picture: Is a imitation of the 8 ring proton torus, with my fingers holding on the proton ring that has the muon ring perpendicular and in the equatorial plane of the proton rings, thrusting through. This muon ring is the same size as the 8 proton rings making 9 x 105MeV = 945MeV of energy. The muon ring has to be perpendicular and lie on the equator of the proton torus. Surrounding the proton-torus would be neutrons as skin or coating cover and act as capacitors in storing the electricity produced by the proton+muon.
Product details
• ASIN : B08K47K5BB
• Publication date : September 25, 2020
• Language : English
• File size : 587 KB
• Text-to-Speech : Enabled
• Enhanced typesetting : Enabled
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• Print length : 25 pages
• Lending : Enabled
• Best Sellers Rank: #291,001 in Kindle Store (See Top 100 in Kindle Store)
◦ #13 in 45-Minute Science & Math Short Reads
◦ #52 in General Chemistry & Reference
◦ #334 in General Chemistry
#1-7, 138th published book
The true NUCLEUS of Atoms are inner toruses moving around in circles of a larger outer torus// Rutherford, Geiger, Marsden Experiment revisited // Atom Totality Series, book 7 Kindle Edition
by Archimedes Plutonium (Author)
The geometry of Atoms of the Table of Chemical Elements is torus geometry. We know this to be true for the torus geometry forms the maximum electricity production when using the Faraday Law. We see this in Old Physics with their tokamak toruses attempting to make fusion, by accelerating particles of the highest possible acceleration for the torus is that geometry. But the torus is the geometry not only of maximum acceleration but of maximum electrical generation by having a speeding bar magnet go around and around inside a torus== the Faraday law, where the torus rings are the copper closed wire loop. The protons of atoms are 8 loops of rings in a torus geometry, and the electron of atoms is the muon as bar magnet, almost the same size as the proton loops but small enough to fit inside proton loops. It is torus geometry that we investigate the geometry of all atoms.
Length: 41 pages
Product details
• Publication Date : October 9, 2020
• File Size : 828 KB
• Word Wise : Not Enabled
• Print Length : 41 pages
• ASIN : B08KZT5TCD
• Language: : English
• Text-to-Speech : Not enabled
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#1-8, 1st published book
Atom Totality Universe, 8th edition, 2017// A history log book: Atom Totality Series book 8 Kindle Edition
by Archimedes Plutonium (Author)
Last revision 7Apr2021. This was AP's first published science book.
Advisory: This is a difficult book to read and is AP's research log book of the Atom Totality in 2016-2017. I want to keep it for its history value. AP advises all readers wanting to know the Plutonium Atom Totality theory to go to the 9th edition that is the latest up to date account of this theory. The reason AP wants to keep the 8th edition is because of Historical Value, for in this book, while writing it, caused the discovery of the real electron is the muon of atoms. The real proton of atoms is 840MeV and not the 938MeV that most books claim. The particle discovered by JJ Thomson in 1897 thinking he discovered the electron of atoms was actually the Dirac magnetic monopole at 0.5MeV. This discovery changes every, every science that uses atoms and electricity and magnetism, in other words, every science.
Foreward:
I wrote the 8th edition of Atom Totality and near the end of writing it in 2017, I had my second greatest physics discovery. I learned the real electron of atoms was the muon at 105MeV and not the tiny 0.5MeV particle that J.J.Thomson found in 1897. So I desperately tried to include that discovery in my 8th edition and it is quite plain to see for I tried to write paragraphs after each chapter saying as much. I knew in 2017, that it was a great discovery, changing all the hard sciences, and reframing and restructuring all the hard sciences.
Length: 632 pages
Product details
File Size: 1132 KB
Print Length: 632 pages
Publication Date: March 11, 2019
Sold by: Amazon Digital Services LLC
Language: English
ASIN: B07PLP9NDR
Text-to-Speech: Enabled 
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Not Enabled 
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Amazon Best Sellers Rank: #578,229 Paid in Kindle Store (See Top 100 Paid in Kindle Store)
#1610 in Physics (Kindle Store)
#8526 in Physics (Books)
#18851 in Biological Sciences (Books)
#2-1, 137th published book
Introduction to AP's TEACHING TRUE PHYSICS// Physics textbook series, book 1 Kindle Edition
by Archimedes Plutonium (Author)
#1 New Release in Electromagnetic Theory
This will be AP's 137th published book on science. And the number 137 is special to me for it is the number of QED, Quantum Electrodynamics as the inverse fine structure constant. I can always remember 137 as that special constant of physics and so I can remember where Teaching True Physics was started by me.
Time has come for the world to have the authoritative textbooks for all of High School and College education. Written by the leading physics expert of the time. The last such was Feynman in the 1960s with Feynman Lectures on Physics. The time before was Maxwell in 1860s with his books and Encyclopedia Britannica editorship. The time is ripe in 2020 for the new authoritative texts on physics. It will be started in 2020 which is 60 years after Feynman. In the future, I request the physics community updates the premier physics textbook series at least every 30 years. For we can see that pattern of 30 years approximately from Faraday in 1830 to Maxwell in 1860 to Planck and Rutherford in about 1900, to Dirac in 1930 to Feynman in 1960 and finally to AP in 1990 and 2020. So much happens in physics after 30 years, that we need the revisions to take place in a timely manner. But also, as we move to Internet publishing such as Amazon's Kindle, we can see that updates can take place very fast, as editing can be a ongoing monthly or yearly activity. I for one keep constantly updating all my published books, at least I try to.
Feynman was the best to make the last authoritative textbook series for his concentration was QED, Quantum Electrodynamics, the pinnacle peak of physics during the 20th century. Of course the Atom Totality theory took over after 1990 and all of physics; for all sciences are under the Atom Totality theory.
And as QED was the pinnacle peak before 1990, the new pinnacle peak is the Atom Totality theory. The Atom Totality theory is the advancement of QED, for the Atom Totality theory primal axiom says -- All is Atom, and atoms are nothing but Electricity and Magnetism.
Length: 64 pages
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• File Size : 790 KB
• Publication Date : October 5, 2020
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• ASIN : B08KS4YGWY
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• Best Sellers Rank: #430,602 in Kindle Store (See Top 100 in Kindle Store)
◦ #39 in Electromagnetic Theory
◦ #73 in Electromagnetism (Kindle Store)
◦ #74 in 90-Minute Science & Math Short Reads
#2-2, 145th published book
TEACHING TRUE PHYSICS//Junior High School// Physics textbook series, book 2
Kindle Edition
by Archimedes Plutonium (Author)
What I am doing is clearing the field of physics, clearing it of all the silly mistakes and errors and beliefs that clutter up physics. Clearing it of its fraud and fakeries and con-artistry. I thought of doing these textbooks starting with Senior year High School, wherein I myself started learning physics. But because of so much fraud and fakery in physics education, I believe we have to drop down to Junior year High School to make a drastic and dramatic emphasis on fakery and con-artistry that so much pervades science and physics in particular. So that we have two years in High School to learn physics. And discard the nonsense of physics brainwash that Old Physics filled the halls and corridors of education.
Product details
• ASIN : B08PC99JJB
• Publication date : November 29, 2020
• Language: : English
• File size : 682 KB
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• Print length : 78 pages
• Lending : Enabled
• Best Sellers Rank: #185,995 in Kindle Store (See Top 100 in Kindle Store)
◦ #42 in Two-Hour Science & Math Short Reads
◦ #344 in Physics (Kindle Store)
◦ #2,160 in Physics (Books)
#2-3, 146th published book
TEACHING TRUE PHYSICS// Senior High School// Physics textbook series, book 3
Kindle Edition
by Archimedes Plutonium (Author)
I believe that in knowing the history of a science is knowing half of that science. And that if you are amiss of knowing the history behind a science, you have only a partial understanding of the concepts and ideas behind the science. I further believe it is easier to teach a science by teaching its history than any other means of teaching. So for senior year High School, I believe physics history is the best way of teaching physics. And in later years of physics courses, we can always pick up on details. So I devote this senior year High School physics to a history of physics, but only true physics. And there are few books written on the history of physics, so I chose Asimov's The History of Physics, 1966 as the template book for this textbook. Now Asimov's book is full of error and mistakes, and that is disappointing but all of Old Physics is full of error. On errors and mistakes of Old Physics, the best I can do is warn the students, and the largest warning of all is that whenever someone in Old Physics says "electron" what they are talking about is really the Dirac magnetic monopole. And whenever they talk about the Rutherford-Bohr model of the atom, they are talking about huge huge grave mistakes, for the true atom is protons as 8 ringed toruses with a muon stuck inside of a proton doing the Faraday law and producing those magnetic monopoles as electricity. I use Asimov's book as a template but in the future, I hope to rewrite this textbook using no template at all, that is if I have time in the future.
Cover Picture: Is the book The History of Physics, by Isaac Asimov, 1966 and on top of the book are 4 cut-outs of bent circles representing magnetic monopoles which revolutionizes modern physics, especially the ElectroMagnetic theory.
Product details
• ASIN : B08RK33T8V
• Publication date : December 28, 2020
• Language : English
• File size : 794 KB
• Text-to-Speech : Enabled
• Screen Reader : Supported
• Enhanced typesetting : Enabled
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• Word Wise : Enabled
• Print length : 123 pages
• Lending : Enabled
• Best Sellers Rank: #4,167,235 in Kindle Store (See Top 100 in Kindle Store)
◦ #15,099 in Physics (Kindle Store)
◦ #91,163 in Physics (Books)
#2-4, 151st published book
TEACHING TRUE PHYSICS// 1st year College// Physics textbook series, book 4
Kindle Edition
by Archimedes Plutonium (Author)
Preface: This is AP's 151st book of science published. It is one of my most important books of science because 1st year college physics is so impressionable on students, if they should continue with physics, or look elsewhere for a career. And also, physics is a crossroad to all the other hard core sciences, where physics course is mandatory such as in chemistry or even biology. I have endeavored to make physics 1st year college to be as easy and simple to learn. In this endeavor to make physics super easy, I have made the writing such that you will see core ideas in all capital letters as single sentences as a educational tool. And I have made this textbook chapter writing follow a logical pattern of both algebra and geometry concepts, throughout. The utmost importance of logic in physics needs to be seen and understood. For I have never seen a physics book, prior to this one that is logical. Every Old Physics textbook I have seen is scatter-brained in topics and in writing. I use as template book of Halliday & Resnick because a edition of H&R was one I was taught physics at University of Cincinnati in 1969. And in 1969, I had a choice of majors, do I major in geology, or mathematics, or in physics, for I will graduate from UC in 1972. For me, geology was too easy, but physics was too tough, so I ended up majoring in mathematics. If I had been taught in 1969 using this textbook that I have written, I would have ended up majoring in physics, my first love. For physics is not hard, not hard at all, once you clear out the mistakes and the obnoxious worthless mathematics that clutters up Old Physics, and the illogic that smothers much of Old Physics.
Maybe it was good that I had those impressions of physics education of poor education, which still exists throughout physics today. Because maybe I am forced to write this book, because of that awful experience of learning physics in 1969. Without that awful experience, maybe this textbook would have never been written by me.
Cover picture is the template book of Halliday & Resnick, 1988, 3rd edition Fundamentals of Physics and sitting on top are cut outs of "half bent circles, bent at 90 degrees" to imitate magnetic monopoles. Magnetic Monopoles revolutionizes physics education, and separates-out, what is Old Physics from what is New Physics.
Product details
• ASIN : B09JW5DVYM
• Publication date : October 19, 2021
• Language : English
• File size : 1033 KB
• Text-to-Speech : Enabled
• Screen Reader : Supported
• Enhanced typesetting : Enabled
• X-Ray : Not Enabled
• Word Wise : Enabled
• Print length : 386 pages
• Lending : Enabled
True Chemistry: Chemistry Series, book 1 Kindle Edition
by Archimedes Plutonium (Author)
Physics and chemistry made a mistake in 1897 for they thought that J.J. Thomson's small particle of 0.5MeV was the electron of atoms. By 2017, Archimedes Plutonium discovered that the rest mass of 940 for neutron and proton was really 9 x 105MeV with a small sigma-error. Meaning that the real proton is 840MeV, real electron is 105 MeV= muon, and that little particle Thomson discovered was in fact the Dirac magnetic monopole. Dirac circa 1930s was looking for a magnetic monopole, and sadly, Dirac passed away before 2017, because if he had lived to 2017, he would have seen his long sought for magnetic monopole which is everywhere.
Cover picture: shows 3 isomers of CO2 and the O2 molecule.
Length: 1150 pages
Product details
• File Size : 2167 KB
• ASIN : B07PLVMMSZ
• Publication Date : March 11, 2019
• Word Wise : Enabled
• Print Length : 1150 pages
• Language: : English
• Text-to-Speech : Not enabled
• Enhanced Typesetting : Enabled
• X-Ray : Not Enabled
• Lending : Enabled
Amazon Best Sellers Rank: #590,212 Paid in Kindle Store (See Top 100 Paid in Kindle Store)
#181 in General Chemistry & Reference
#1324 in General Chemistry
#1656 in Physics (Kindle Store)
y z
| /
| /
|/______ x
More people reading and viewing AP's newsgroup than viewing sci.math, sci.physics. So AP has decided to put all NEW WORK, to his newsgroup. And there is little wonder because in AP's newsgroups, there is only solid pure science going on, not a gang of hate spewing misfits blighting the skies.
In sci.math, sci.physics there is only stalking hate spew along with Police Drag Net Spam of no value and other than hate spew there is Police drag net spam day and night.
I re-opened the old newsgroup PAU of 1990s and there one can read my recent posts without the hassle of stalkers and spammers, Police Drag Net Spam that floods each and every day, book and solution manual spammers, off-topic-misfits, front-page-hogs, churning imbeciles, stalking mockers, suppression-bullies, and demonizers. And the taxpayer funded hate spew stalkers who ad hominem you day and night on every one of your posts.
There is no discussion of science in sci.math or sci.physics, just one long line of hate spewing stalkers followed up with Police Drag Net Spam (easy to spot-- very offtopic-- with hate charged content). And countries using sci.physics & sci.math as propaganda platforms, such as tampering in elections with their mind-rot.
Read my recent posts in peace and quiet.
Archimedes Plutonium
Stanford's_Marc Tessier-Lavigne, Persis Drell,Alexander Fetter, John Lipa, William Little, Douglas Osheroff,, is McGinn correct that Stanford is failed & incompetent to confirm real proton is 840MeV, real electron=105MeV and .5MeV was Dirac's magnetic monopole
Re: James McGinn, the blubbery cesspool mind of a moron packed inside a single cell atop a foghorn mouth// why California schools have not yet confirmed real proton = 840MeV, electron= muon and .5MeV was Dirac's monopole
157k views
Oct 14, 2019, 10:08:30 AM
by Pete Smith
> I fart you.
On Wednesday, May 29, 2019 at 10:55:04 AM UTC-5, James McGinn wrote:
> >
> > And you too proved yourself incompetent and desperate in that you failed to discuss any of the substance of the argument (the subject of which is way, way over your head).
AP writes: Is the reason Stanford Univ has not yet confirmed real proton is 840MeV not 938, because its scientists like McGinn says is blithering nattering nutter fools-- drinking coffee and eating Danish rolls rather than uncovering the true proton is 840MeV stuck with the real electron as muon doing a Faraday Law dance inside the atom making electricity and the .5MeV particle is Dirac's magnetic monopole.
o-:^>___?
`~~c--^c'
Navy dog says: yes, I enjoy my Danish rolls with blended coffee, steaks and eggs in the California sun instead of real physics of the atom
Stanford University, math dept.
Gregory Brumfiel, Daniel Bump, Emmanuel Candès, Gunnar Carlsson, Moses Charikar, Sourav Chatterjee, Tom Church, Ralph Cohen, Brian Conrad, Brian Conrey, Amir Dembo, Persi Diaconis, Yakov Eliashberg, Robert Finn, Jacob Fox, Laura Fredrickson, Søren Galatius, George Schaeffer, Or Hershkovits, David Hoffman, Eleny Ionel, Renata Kallosh, Yitzhak Katznelson, Vladimir Kazeev, Michael Kemeny, Steven Kerckhoff, Susie Kimport, Jun Li, Tai-Ping Liu, Mark Lucianovic, Jonathan Luk, Frederick Manners, Rafe Mazzeo, James R. Milgram, Maryam Mirzakhani, Stefan Mueller, Christopher Ohrt, Donald Ornstein, George Papanicolaou, Lenya Ryzhik, Richard Schoen, Leon Simon, Rick Sommer, Kannan Soundararajan, Tadashi Tokieda, Cheng-Chiang Tsai, Ravi Vakil, András Vasy, Akshay Venkatesh, Jan Vondrák, Brian White, Wojciech Wieczorek, Jennifer Wilson, Alex Wright, Lexing Ying, Xuwen Zhu
President: Marc Tessier-Lavigne (neuroscience)
Provost: Persis Drell (physics)
Stanford physics dept.
Alexander Fetter, John Lipa, William Little, Douglas Osheroff, David Ritson, H. Alan Schwettman, John Turneaure, Robert Wagoner, Stanley Wojcicki, Mason Yearian
CalTech math dept
Michael Aschbacher, Alexei Borodin, Danny Calegari
Matthias Flach, Anton N. Kapustin, Alexander Kechris
Alexei Kitaev, Matilde Marcolli, Nikolai Makarov, Vladimir Markovic, Hiroshi Oguri, Eric Rains, Dinakar Ramakrishnan
Barry Simon, Richard Wilson, Tom Graber, Sergei Gukov,
Elena Mantovan, Yi NI,
Caltech Physics Dept
Barry Barish, Felix Boehm, Steven Frautschi
Murray Gell-Mann, David Goodstein, Thomas Phillips,
John Schwarz, Barry Simon, Kip Thorne, Petr Vogel,
Rochus Vogt, Ward Whaling, Michael E. Brown,
Konstantin Batygin
UCLA chancellor: Gene D. Block (biology)
UCLA Physics dept
Ernest Abers, Elihu Abrahams, Katsushi Arisaka, Michalis Bachtis
Eric Becklin, Zvi Bern, Rubin Braunstein, Stuart Brown, Robijn Bruinsma
Charles Buchanan, Wesley Campbell, Troy Carter, Sudip Chakravarty
W. Gilbert Clark, John Cornwall, Robert Cousins, Eric D'Hoker
Robert Finkelstein, Christian Fronsdal, Walter Gekelman, Graciela Gelmini
George Gruner, Michael Gutperle, Brad Hansen, Jay Hauser, Karoly Holczer
Huan Huang, Eric Hudson, George Igo, Per Kraus, Alexander Kusenko
Thomas Mason, George Morales, Warren Mori, Steven Moszkowski
Christoph Niemann, Kumar Patel, Roberto Peccei, Claudio Pellegrini
Seth Putterman, B. Regan, James Rosenzweig, Joseph Rudnick
David Saltzberg, William Slater, Reiner Stenzel, Terry Tomboulis, Jean Turner
Univ Calif San Diego, physics dept
Henry D. I. Abarbanel, Kam S. Arnold, Daniel P. Arovas, Richard D. Averitt, Julio T. Barreiro, Dimitri N. Basov, Steven Boggs, James G. Branson, Adam J. Burgasser, Leonid V. Butov, Alison Coil, Eva-Maria S. Collins, Max Di Ventra, Patrick H. Diamond, Fred C. Driscoll, Daniel H. Dubin, Olga K. Dudko, Raphael M. Flauger, Michael M. Fogler, Alex Frano, George M. Fuller, Daniel R Green, Kim Griest, Benjamin Grinstein, Alexander Groisman, Tarun Grover, Jorge E. Hirsch, Michael Holst, Terence T. Hwa, Kenneth A. Intriligator, Elizabeth Jenkins, Suckjoon Jun, Brian Keating, Dusan Keres, David Kleinfeld, Quinn Konopacky, Elena F. Koslover, Julius Kuti, Tongyan Lin, Aneesh V. Manohar, M. Brian Maple, John A. McGreevy, Thomas W. Murphy, Kaixuan Ni, Michael L. Norman, Thomas M. O'Neil, Hans P. Paar, Mark Paddock, Jeremie Palacci, Tenio Popmintchev, Wouter-Jan Rappel, Karin M. Sandstrom, Ivan K. Schuller, Lu J. Sham, Vivek Sharma, Tatyana O. Sharpee, Brian Shotwell, Oleg Shpyrko, Elizabeth H Simmons, Sunil K. Sinha, Douglas E. Smith, Harry Suhl
Math dept Univ Calif, San Diego
Edward Bender, James Bunch, Thomas Enright, Ronald Evans, Jay Fillmore, Carl FitzGerald,
Michael Freedman, Adriano Garsia, Fan Graham, Leonard Haff, Hubert Halkin, Richard Hamilton, Bill Helton, Jim Lin, Alfred Manaster, John O'Quigley, Yose Rinott, Burt Rodin, Murray Rosenblatt, Linda Rothschild, Michael Sharpe, Lance Small, Don Smith, Harold Stark, Audrey Terras, Adrian Wadsworth, Nolan Wallach, John Wavrik, Daniel Wulbert
On Friday, January 4, 2019 at 5:29:50 PM UTC-6, James McGinn wrote:
> Weather prediction is not the topic,,,,
John Schwarz,Barry Simon,Kip Thorne,Petr Vogel,Rochus Vogt, of Caltech are you as stupid as McGinn to never understand Angular Momentum for the chemical bond cannot exist with electron=.5MeV, proton=938MeV. You need 105 to 840 to have chemistry
Murray Gell-Mann, David Goodstein, Thomas Phillips, of Caltech are you as stupid as McGinn to never understand Angular Momentum for the chemical bond cannot exist with electron=.5MeV, proton=938MeV. You need 105 to 840 to have chemistry
Stanford's Drs Gregory Brumfiel, Daniel Bump, Emmanuel Candès, Gunnar Carlsson is McGinn the example of how physicists react when told the proton is 840MeV, electron 105MeV to have chemistry bonding
About McGinn, we all know he is an idiot when it comes to science or even thinking straight, and although he deserves 1 or 2 posts per day (some would say that is too much) but he does not deserve 75 posts per day under various names like Denke or Solvingtornado. So either he post 1 or 2, or I recommend he be kicked out permanently as a front page hog spamming jackarse. I hate his practice of just churning his posts, where the creep adds two words, sometimes not even a new word, to his prior post just to get it on the front page again. To think that sci.physics by year 2019 is mostly a airhead spammer on the front page is enough to make any cry and sob into the new year.
On Thursday, January 3, 2019 at 10:19:09 PM UTC-6, James McGinn wrote:
> Aw shucks.
..
.- " `-. ,..-''' ```....'`-..
, . `.' ' `.
.' .' ` ` ' `.. ;
. ; .' . `. ;
; . ' `. . '
. ' ` `. |
. '. '
. 0 0 ' `.
' `
; `
.' `
; U `
; '; `
: | ;.. :` `
: `;. ```. .-; | '
'. ` ``.., .' :' '
; ` ;'.. ..-'' ' ' Hi I am McGinn under various fake names Denk, Pnal etc. My game is to fill sci.physics with nothing but my airhead posts because I love to annoy everybody, and on fast days, I just churn all my old posts by adding a word or sentence, and often pretend I am Pnal, to make believe someone is actually talking with me. You see, my foot is where my head is and my head where my foot is.
` ` ; ````'''""' ; ' '
` ` ; ; ' '
` ` ; ; ' '
` `. ````'''''' ' '
` . ' '
/ ` `. ' ' .
/ ` .. ..' .'"""""...'
/ .` ` ``........-' .'` .....'''
/ .'' ; ` .' `
...'.' ; .' ` .' `
"" .' .' | ` .; \ `
; .' | `. . . . ' . \ `
:' | ' ` , `. `
| ' ` ' `. `
` ' ` ; `. |
`.' ` ; `-'
`...'
CalTech's Rochus Vogt, Ward Whaling, Michael E. Brown,Konstantin Batygin are you like McGinn/pnal too stupid to understand Angular Momentum for the chemical bond cannot exist with electron=.5MeV, proton=938MeV. You need 105 to 840 to have chemistry
Too stupid to understand Angular Momentum for the chemical bond cannot exist with electron=.5MeV, proton=938MeV. You need 105 to 840 to have chemistry. The .5MeV particle that Thomson discovered was actually Dirac's magnetic monopole
Why does McGinn simply not ask professors of physics at UCLA why they think the real proton is not 840MeV and real electron = 105MeV with .5 MeV the Dirac Magnetic Monopole
Why does any physicist not believe proton is 840MeV, electron is 105MeV in order to have chemistry bonding, because a ratio of 840 to 105 allows for Angular Momentum
-----------------------------------
UCLA Physics dept
Ernest Abers
Elihu Abrahams
Katsushi Arisaka
Michalis Bachtis
Eric Becklin
Zvi Bern
Rubin Braunstein
Stuart Brown
Robijn Bruinsma
Charles Buchanan
Wesley Campbell
Troy Carter
Sudip Chakravarty
W. Gilbert Clark
John Cornwall
Robert Cousins
Eric D'Hoker
Robert Finkelstein
Christian Fronsdal
Walter Gekelman
Graciela Gelmini
George Gruner
Michael Gutperle
Jay Hauser
Karoly Holczer
Huan Huang
Eric Hudson
George Igo
Per Kraus
Alexander Kusenko
Thomas Mason
George Morales
Warren Mori
Steven Moszkowski
Christoph Niemann
Kumar Patel
Roberto Peccei
Claudio Pellegrini
Seth Putterman
B. Regan
James Rosenzweig
Joseph Rudnick
David Saltzberg
William Slater
Reiner Stenzel
Terry Tomboulis
Jean Turner
Stanford University, math dept.
Gregory Brumfiel, Daniel Bump, Emmanuel Candès, Gunnar Carlsson, Moses Charikar, Sourav Chatterjee, Tom Church, Ralph Cohen, Brian Conrad, Brian Conrey, Amir Dembo, Persi Diaconis, Yakov Eliashberg, Robert Finn, Jacob Fox, Laura Fredrickson, Søren Galatius, George Schaeffer, Or Hershkovits, David Hoffman, Eleny Ionel, Renata Kallosh, Yitzhak Katznelson, Vladimir Kazeev, Michael Kemeny, Steven Kerckhoff, Susie Kimport, Jun Li, Tai-Ping Liu, Mark Lucianovic, Jonathan Luk, Frederick Manners, Rafe Mazzeo, James R. Milgram, Maryam Mirzakhani, Stefan Mueller, Christopher Ohrt, Donald Ornstein, George Papanicolaou, Lenya Ryzhik, Richard Schoen, Leon Simon, Rick Sommer, Kannan Soundararajan, Tadashi Tokieda, Cheng-Chiang Tsai, Ravi Vakil, András Vasy, Akshay Venkatesh, Jan Vondrák, Brian White, Wojciech Wieczorek, Jennifer Wilson, Alex Wright, Lexing Ying, Xuwen Zhu
President: Marc Tessier-Lavigne (neuroscience)
Provost: Persis Drell (physics)
Stanford physics dept.
Alexander Fetter, John Lipa, William Little, Douglas Osheroff, David Ritson, H. Alan Schwettman, John Turneaure, Robert Wagoner, Stanley Wojcicki, Mason Yearian
CalTech math dept
Michael Aschbacher, Alexei Borodin, Danny Calegari
Matthias Flach, Anton N. Kapustin, Alexander Kechris
Alexei Kitaev, Matilde Marcolli, Nikolai Makarov, Vladimir Markovic, Hiroshi Oguri, Eric Rains, Dinakar Ramakrishnan
Barry Simon, Richard Wilson, Tom Graber, Sergei Gukov,
Elena Mantovan, Yi NI,
Caltech Physics Dept
Barry Barish, Felix Boehm, Steven Frautschi
Murray Gell-Mann, David Goodstein, Thomas Phillips,
John Schwarz, Barry Simon, Kip Thorne, Petr Vogel,
Rochus Vogt, Ward Whaling, Michael E. Brown,
Konstantin Batygin
/\-------/\
\::O:::O::/
(::_ ^ _::)
\_`-----'_/
You mean the classroom is the world, not just my cubbyhole in sunny California?
And, even though you-- professors of physics, want to remain stupid in not knowing what is really the electron in atoms has to be the muon at 105MeV and proton at 840MeV with Dirac's magnetic monopole being .5MeV, your students deserve better.
And, even though you-- professors of physics/math, want to remain silent and stupid in Real Electron = muon, and true real Calculus with a geometry proof of Fundamental Theorem of Calculus, your students deserve better.
Yes, there, what did they say-- the power of Sun and stars is not really fusion but is the Faraday Law inside of atoms creating monopoles and turning Space into energy that fuels the Sun and stars. My rough estimate is that fusion only supplies 10% or less of Sun and stars.
But of course, I could not have discovered the true starpower when under the idiotic idea that the electron was a mere .5MeV when it truly is 105 MeV.
mitchr...@gmail.com
Dec 16, 2021, 9:24:01 PM12/16/21
to
The reality of imaginary math.
math admits It is a formula without a solution.
That is how it got its name.
Mitchell Raemsch
zelos...@gmail.com
Dec 17, 2021, 12:12:07 AM12/17/21
to
false
x^2+1=0
has the solution i and -i
so wrong.
Chris M. Thomasson
Dec 17, 2021, 12:14:07 AM12/17/21
to
Wrong... If, perhaps, the observer is afraid of it? Then, they say its
wrong?
mitchr...@gmail.com
Dec 17, 2021, 1:12:16 PM12/17/21
to
What is the solved quantity your imaginary represents?
No. Imaginary math is named rightly.
Mitchell Raemsch
>
> so wrong.
Chris M. Thomasson
Dec 17, 2021, 6:24:21 PM12/17/21
to
[...]
The imaginary numbers go along the y axis, while the real numbers are on
the x axis. So the two solutions are: 0-1i, and 0+1i. So, the x-axis is
zero for both solutions. The y axes are one unit up, and one unit down:
+y
|
|
-x -----0----- +x
|
|
-y
where x is real, and y is imaginary.
Python
Dec 17, 2021, 10:38:58 PM12/17/21
to
smitchi...@gmail.com schwrote:
> On Thursday, December 16, 2021 at 9:12:07 PM UTC-8, zelos...@gmail.com wrote:
...
>> x^2+1=0
>>
>> has the solution i and -i
>
> What is the solved quantity your imaginary represents?
> No. Imaginary math is named rightly.
Nothing more or less "imaginary" than any other mathematical
concept Smitch.
i = class of X in R[X]/(X^2+1)
(learn basic algebra if you don't get it, Smitch).
Another silly question, Smitch?
zelos...@gmail.com
Dec 20, 2021, 12:55:51 AM12/20/21
to
it is i and -i, whats so difficult? | 12,623 | 44,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.888855 |
https://h-o-m-e.org/why-is-the-area-of-a-square-a2/ | 1,721,819,963,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518277.99/warc/CC-MAIN-20240724110315-20240724140315-00467.warc.gz | 243,885,910 | 26,113 | # Why is the area of a square a2?
The area of a square being equal to side squared can be better understood by breaking down the concept of area and the properties of a square.
Firstly, let’s talk about what area is. Area is a measure of the amount of space inside a 2D figure. It is calculated by multiplying the length and width of the figure. In the case of a square, the length and width are the same because all the sides of a square are equal.
Now, let’s consider the properties of a square. A square is a special type of quadrilateral where all four sides are equal in length and all four angles are right angles. These equal sides make a square a regular polygon.
When we calculate the area of a square, we can think of it as dividing the square into smaller square units. Each side of the square can be seen as a unit of length, and by multiplying the length by the width (which is also the length in this case), we are essentially multiplying the side length by itself.
For example, if we have a square with a side length of 5 units, the area would be calculated as 5 units × 5 units, which equals 25 square units. In this case, the units of measurement (such as centimeters or inches) are squared because we are multiplying the side length by itself.
This concept can be visualized by drawing a square on graph paper and counting the number of smaller squares that fit inside it. Each smaller square represents a unit of area, and the total number of smaller squares gives us the total area of the square.
To further illustrate this concept, let’s consider a real-life scenario. Imagine you have a square tile with a side length of 1 foot. If you wanted to cover a floor with these tiles, you would need to know the total area of the floor to determine how many tiles you would need. By calculating the area of the square tile (1 foot × 1 foot = 1 square foot), you can easily determine the number of tiles needed by dividing the total area of the floor by the area of each tile.
The area of a square is side squared because all the sides are equal, and when we calculate the area, we are essentially multiplying the side length by itself. This concept is applicable to various real-life situations and allows us to determine the amount of space inside a square or the number of square units needed to cover a given area. | 502 | 2,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2024-30 | latest | en | 0.946365 |
https://www.mrexcel.com/board/threads/array-help-i-think.273692/ | 1,723,342,422,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640843545.63/warc/CC-MAIN-20240811013030-20240811043030-00058.warc.gz | 694,875,163 | 19,252 | Array Help ! I Think ?
snoopsterg
New Member
Hi All, Hope this is posted in correct place.
I have a production schedule that i need to make available for my colleagues.
At the moment the data is stored in a SQL database which i have full admin access to and need to export to excel as below
Date Time Cum Total
06:00hrs
08/06/2007 08:00hrs 0 0
10:00hrs M/R 0 0
12:00hrs M/R 0 0
14:00hrs 4500 4500 4500
16:00hrs 18000 18000 22500
18:00hrs 18000 18000 40500
20:00hrs 4500 4500 45000
22:00hrs 0 45000
What this actually depicts is Job Number, start date, start Time, Run speed, Run Minutes, Make Ready minutes (MR).
I can get at all the data but am struggling how to get the following to create the above.
StartDate | EndDate | QTY | MRMins | RunMins | Spd
08/06/2007 10:00 08/06/2007 22:00 45000 120 300 9000
any help would be appreciated
Steve
Excel Facts
Lock one reference in a formula
Need 1 part of a formula to always point to the same range? use \$ signs: \$V\$2:\$Z\$99 will always point to V2:Z99, even after copying
Steve,
It looks like
StartDate | EndDate | QTY | MRMins | RunMins | Spd
08/06/2007 10:00 08/06/2007 22:00 45000 120 300 9000
is supposed to be a summary of the dataset you gave above. I can see how MRMins gets to be 45000 but do not see how MRMins is 120, or RunMins is 300, or Spd is 9000. Perhaps you could restate this example and show how the summaries are supposed to get calculated.
Steve,
It looks like
StartDate | EndDate | QTY | MRMins | RunMins | Spd
08/06/2007 10:00 08/06/2007 22:00 45000 120 300 9000
is supposed to be a summary of the dataset you gave above. I can see how MRMins gets to be 45000 but do not see how MRMins is 120, or RunMins is 300, or Spd is 9000. Perhaps you could restate this example and show how the summaries are supposed to get calculated.
Thanks for the reply,
I think i gave a confusing statement, if i can clarify:
start date = 08/06/2007 10:00
end Date = 08/06/2007 22:00
QTY Required = 45000
MRMins (M/R) = 120
RunMins = 300
Run Speed (copies per hour) = 9000
M/R = 2 Hrs so in spread sheet should show
Time QTY
10:00 | M/R
12:00 | M/R
14:00 | 18,000
I have to also take into account that we have 3 shifts that have 1 hr breaks at say 12:00, 16:00 and 18:00 where production with cease for that 1 hour period
Hope this explains a little better
Steve
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https://yourthunderbuddy.com/liter-in-quarts/ | 1,618,572,220,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038056325.1/warc/CC-MAIN-20210416100222-20210416130222-00531.warc.gz | 1,204,868,239 | 14,263 | # Liter In Quarts
Of its respective gallon. In the UK the imperial quart is equal to 1136523.
Liquid Measurements Video In 2021 Pint Gallon Quart
### 1 Liter 105668821 Quarts Fluid US 1 Liter 0908082978 Quarts Dry US.
Liter in quarts. To find out how many Liters in Quarts multiply by the conversion factor or use the Volume converter above. Liters to US quarts formula. 41 rows To convert a liter measurement to a quart measurement multiply the volume by the.
The volume units conversion factor of liters to quarts is 105668821. Use this page to learn how to convert between liters and quarts. To convert any value in liters to quarts just multiply the value in liters by the conversion factor 10566882094326.
It converts units from liters to quarts or vice versa with a metric conversion chart. So 73 liters times 10566882094326 is. There are several different kinds of Quarts available- us liquid us dry and uk.
1 cubic meter is equal to 1000 liters or 10566882049662 quarts. 1 liter l 105668821 US quart qt 211337642 US pints pt 1000 milliliters ml 338140227 US fluid ounce fl. 1 Liter is equal to 105668821 US quarts.
The SI derived unit for volume is the cubic meter. Note that rounding errors may occur so always check the results. 141 rows A liter or litre is a unit of volume in the metric system.
Liters to quarts is a l to qt volume conversion converter. Of its respective gallon. 1 L 10566882049662 qt To convert 16 liters into quarts we have to multiply 16 by the conversion factor in order to get the volume amount from liters to quarts.
1 cubic meter is equal to 10566882049662 quart or 1000 liter. 1 Quart Dry US 110122095 Liter 1 Quart UK 11365225 Liters This means that there are 0946352946 liters in one US fluid quart and 11365225 Liters in one imperial quart. The conversion factor varies depending on whether you are using the US customary or imperial standard.
26 rows There are about 3785 liters in a US. To convert any value in liters to quarts just multiply the value in liters by the conversion factor 10566882094326. Oz 202884136 US teaspoons tsp 676280454 US tablespoons tbsp 0264172052 US gallon gal.
In both the UK and the US the quart is equal to. Liters US Quarts Liquid 0 L. Quart is equal to 32 US.
The SI derived unit for volume is the cubic meter. UK Quarts to Liters. All In One Unit Converter 12 05 liter 05283 quart.
Use this page to learn how to convert between quarts and liters. One US dry quart is made up of 110122095 Liters. The conversion factor from Liters to Quarts is 10566882049662.
In the US a liquid quart is equal to approximately 0946353 liters and a dry quart is equal to approximately 1101221 liters. In the UK the imperial quart is equal to 1136523. Basic unit of volume in the metric system.
In both the UK and the US the quart is equal to. The conversion factor from liters to quarts is 10566882049662 which means that 1 liter is equal to 10566882049662 quarts. Three Liters is equivalent to three point one seven Quarts.
Please select a more specific option. In the US a liquid quart is equal to approximately 0946353 liters and a dry quart is equal to approximately 1101221 liters. The conversion factor from Liters to Quarts is 10566882049662.
A liter is defined as the volume. Multiple definitions of the quart exist. Note that rounding errors may occur so always check the results.
A liter of water weighs one kilogram. So 12 liter times 10566882094326 is equal to 05283 quarts. To find out how many Liters in Quarts multiply by the conversion factor or use the Volume converter above.
To convert liters to quarts multiply the liter value by 105668821. For example to calculate how many quarts of water can fit in 2 liters multiply 2 by 105668821 that makes 211337642 quarts of water can fit in 2 liters. One Liters is equivalent to one point zero five seven Quarts.
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http://www.bertarojas.com/?library/logarithmic-trigonometric-and-other-mathematical-tables | 1,511,047,288,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805114.42/warc/CC-MAIN-20171118225302-20171119005302-00350.warc.gz | 370,287,979 | 12,878 | # Logarithmic, Trigonometric, and Other Mathematical Tables
Format: Hardcover
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More importantly, if we know the measurement of one of the triangle's angles, and we then use a trigonometric function to determine the ratio of the lengths of two of the triangle's sides, and we happen to know the lengths of one of these sides in the ratio, we can then algebraically determine the length of the other one of these two sides. (i.e. if we determine that a / b = 2, and we know a = 6, then we deduce that b = 3.) Since there are three sides and two non-right angles in a right triangle, the trigonometric functions will need a way of specifying which sides are related to which angle. (It is not-so-useful to know that the ratio of the lengths of two sides equals 2 if we do not know which of the three sides we are talking about.
Pages: 132
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ISBN: 1115904426
The Civil Engineer's Pocket-Book
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https://www.physicsforums.com/threads/interaction-of-radiation-with-matter.713309/ | 1,586,128,730,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371611051.77/warc/CC-MAIN-20200405213008-20200406003508-00335.warc.gz | 1,077,388,055 | 16,666 | # Interaction of radiation with matter
## Main Question or Discussion Point
1. Problem statement
i am trying to calculate the momentum and energy of the products in the reaction
7Be4 + e -----> 7Li3 + neutrino
2. Relavant equations
p = mv
mass of 7Be = 7.016929 u
mass of 7Li = 7.016004 u
3. The attempt at a solution
i know that after the electron capture, the neutrino and the daughter nucleas will move in opposite directions with the same momentum magnitude, and so form momentum coservation
mv(Li) = mv(neutrino)
but how do i go about findig the 'v' of Lithium in order to calculate its momentum, mv. the question does not give any info on 'v'. ? Do i even need the value of 'v' or not. So that i can then put that 'v' into the equation of finding the energy of the daughter nucleas as:
1/2 * m(Li)*v^2(Li) = ( m(neutrino)/m(Li) ) * E(neutrino)
am i actually in the right path, i would greatly appreciate any directions?
Related High Energy, Nuclear, Particle Physics News on Phys.org
mfb
Mentor
Where do you use energy conservation? It is important here, together with momentum conservation.
the question does not give any info on 'v'.
It does not have to.
The neutrino is ultrarelativistic, you cannot use the nonrelativistic momentum. Treat it like photons instead. Do you know the energy-momentum relation for photons?
* The momentum of the neutrino is not going to be mv because it will be a relativistic particle.
* Did you apply energy conservation?
ohhh, Thank you, it's E^2 = (pc)^2 + (m0*c^2)^2
will apply it , hadn't considered the relativistic issue.... | 418 | 1,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | longest | en | 0.872351 |
https://aviation.stackexchange.com/questions/65766/why-does-indicated-stall-speed-change | 1,726,303,016,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00849.warc.gz | 100,458,598 | 43,953 | # Why does indicated stall speed change?
Aviation confusing me... I’ve read that stall speed doesn’t change (IAS) no matter what altitude you’re flying - of course under specific conditions ISA, 1G level flight, no wind, gross weight etc. then why I keep witnessing for example in the Boeing 737-800 PFD (primary flight display, intend to the barber pole) that at low altitude let’s say your stalling speed is around 140CAS when at cruise level stalling speed is way above - approximately 220CAS (random number) Why is that ? Doesn’t stalling speed must to be the same at all altitudes?
• I don’t have enough time for a proper answer now, but google: Equivalent Airspeed vs Indicated vs True Commented Jun 21, 2019 at 22:46
It happens because of a compressibility error in the airspeed indicator (yes, even in the digital ones, since the error is not mechanical, but a physical property of the air).
As you might be aware, the speed indicated on your instruments is not really a speed at all, it is actually a pressure. Your pitot measures a deltaP between static and pitot and displays that pressure difference on a scale noted in knots.. we call that Indicated Airspeed.
This indication, because of changes of density, can be quite a bit different from your actual True Airspeed, but people didn’t care all that much about this difference, since the way the wing flies is reliant on that deltaP pressure difference anyway, so that you will actually stall at the same Indicated Airspeed, but not at same True Airspeed. (there is less density higher up but you are now traveling a bit faster, so the effect cancels out and the wing will behave the same)
The instrument was kept as it is in the cockpit (even though today we call it Calibrated Airspeed after a few changes) as a very usefull indication to the pilots.
Ok, now back to the question: turns out when planes begun flying faster and faster, a second error, caused by compressibility appears, that makes the airspeed indicator over-read. The pitot pressure is reported higher than it should be because at high speeds the air compresses in the pitot and makes the pressure there artificially higher, the deltaP is higher, thus our Indicated Airspeed is higher. Of course, this pitot tube phenomenon is not happening on the wing, so now we have a problem: we have a high speed indicated at the instrument, but most of it is actually just compressed air. Our Equivalent Airspeed is in fact much lower. And the wing will in fact always stall at the same Equivalent Airspeed. That means your Indicated Stall Speed in the cockpit will be ever higher as you climb because of compressibility.
Now, instruments were still kept to display IAS instead of the EAS, and a provision has been made where the stall barber pole advances up as you climb to make up for compressibility error
Then why haven’t the instruments been changed to show EAS directly? That is a harder question to answer definitively. I suppose by now people were already too used to having IAS in the cockpit, and the error is only really a factor for jet planes flying above (say) + 20 000 feet at speeds above Mach 0.5.
The difference is the change in Mach number over altitude.
And it is more than just the compressibility error in the IAS indication.
The maximum lift coefficient of a wing goes down with Mach number. While at sea level and 140 KIAS you fly at 21% of the speed of sound (Mach 0.21), at cruise altitude (I guess that means 30,000 ft) the true speed is already 360 KTAS which -- together with the decline of the speed of sound at lower temperature -- translates to Mach 0.63.
In order to estimate the change in maximum lift coefficient, look at the factor maximum lift coefficient times Mach squared: Above maybe Mach 0.4 to 0.5, this is what should stay (roughly) constant. A typical value for a modern wing would be 0.4, so we divide this by 0.63² = 0.397. Thus your maximum lift coefficient at Mach 0.63 has dropped to about 1.0. At lower altitude the maximum lift coefficient of the clean wing is closer to 1.6.
Technically, the wing might even be able to create higher lift coefficients at Mach 0.63, but buffeting will make this intolerable. The stall speed at cruise level, therefore, is the buffet speed and cannot be directly compared with the (real) stall speed at sea level.
• Always to the rescue :) But I'm a bit confused, isn't 140 KIAS at 30,000 ft 225 KTAS? And how does a result of 0.397 mean a drop to 1? (Apologies if it's a silly question, I tried to google some of the terms before asking for clarification.)
– user14897
Commented Jun 23, 2019 at 14:46
• @ymb1: TAS goes up with the square root of the density ratio (1.225 at SL, 0.46 at 30 kft), and I took the 220 KIAS as the baseline (220 x 1.63 = 359). The c$_L \cdot Ma^2$ of 0.4 is my experience for a regular, mildly supercritical airfoil; with a given Mach number of 0.63 the lift coefficient must be 1.0 for the product to become 0.4. Not a silly question at all; I'm grateful for you checking my answers. Commented Jun 23, 2019 at 15:53
I'll assume you're talking about the barber pole presentation on the speed tape, flaps up at low altitude and flaps up at high altitude, with the pole moving higher on the tape at high altitude. The barber pole indication and shaker firing point isn't related to the actual stall; it includes a computed safety margin that takes into account various factors like pitch motions and G loads. That's why the barber pole moves around as you maneuver and pull pitch.
So the actual "indicated" stall speed doesn't change with altitude but what does change is the Stall Protection Computer's stick shaker (and pusher on airplanes that have them) trigger margins and barber pole indications, which have to allow for huge increase in inertial effects at high altitude (the mass is the same, but the air is thin).
The high you go, because of the magnified inertia effects relative to aerodynamic pressure (indicated speed)in the thin air, the more the Stall Protection Computer has to "lead" a change in angle of attack to provide a shaker trigger point that gives a decent margin above the actual aerodynamic stall.
The key of this question is the way the drag changes with speed and the maximum thrust the engine can develop at very high altitude, please read carefully the answer, it is a bit complicated, but I am confident
Let us first of all understand what is Vimd (Indicated minimum drag).
Said simply it is the speed of minimum drag, but let us understand a bit more
Drag has two components: induced drag and parasitic drag
• induced drag is directly related to lift production and is greatest at low speeds and high angle of attack
• parasitic drag increases in proportion to the square of the aircraft speed.
At any speed total drag is the sum of the above two components
For obvious economical consideration it is desired to fly with the lowest total drag which is also compatible with the desire of not using oversized engines
The following curve gives the graph of total drag versus speed
It appears that Vimd corresponds to the intersection point of the induced drag curve and parasitic drag curve
To answer the question
at low altitude let’s say your stalling speed is around 140K IAS and at cruise level stalling speed is approximately 220K IAS, Why is that?
For better understanding let us consider straightforward what happens at very high altitude.
When flying at high altitude, that is at an altitude where the engines thrust is limited by the low air density, and the pilot being therefore flying very closely to Vimd, if for a reason or another he reduces his speed, or his speed decreases, though still above the stalling speed, his actual engines thrust will rapidly become insufficient to overcome the increasing total drag increased by the speed reduction, as visible on the above curve. To maintain altitude he has to pull on the stick,( thus increasing the AOA and the drag), and simultaneously increase the thrust rapidly; if the thrust is not adequately increased, his speed will rapidly decrease and he might suddenly hit the real stalling speed. To avoid the above scenario the barber pole minimum speed is voluntarily well above the stalling speed, to increase the pilots reactivity.
Normally at high altitude the autopilot and the auto throttle are engaged happily safely they look for the barber pole speed limit and not for the theoretical stalling speed.
the higher the altitude the larger the gap between the barber pole minimum speed and the theoretical stalling speed
Note, June 22, 2019. slightly edited for better comprehension.
• I want to pitch in about 2 points: First, Vimd you mentioned is the minimum drag air speed. For the jet aircrafts the minimum drag airspeed corresponds to max endurance airspeed. Thrust required at that airspeed is minimum, therefore fuel consumption is minimum. So you can stay in the air longest at that airspeed. However jets do not use max endurance airspeed when they cruise. Instead they try to fly max range airspeed. Max range airspeed will provide minimum fuel consumption per miles flown. So you get places more efficiently. Commented Sep 18, 2019 at 9:59
• Second, the part you mentioned here: if for a reason or another he reduces his speed, or his speed decreases, though still above the stalling speed, his actual engines thrust will rapidly become insufficient to overcome the increasing total drag increased by the speed reduction, as visible on the above curve. Commented Sep 18, 2019 at 10:00
• is not encountered during cruise flight as far as I know. It’s mostly an issue during the final approach when the aircraft is configured for landing. It’s called backside approach; meaning you are at the left side of drag curve. “Speed stability” of the aircraft is unstable at that region. Therefore requires more pilot attention and more throttle inputs to not go divergent. Commented Sep 18, 2019 at 10:00 | 2,220 | 9,973 | {"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} | 2.65625 | 3 | CC-MAIN-2024-38 | latest | en | 0.937909 |
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