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http://13thmonkey.org/~boris/gammacorrection/ | 1,632,481,876,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057524.58/warc/CC-MAIN-20210924110455-20210924140455-00473.warc.gz | 359,309 | 1,932 | # Some notes on computer screens, gamma, and CIE Rec. 709
The fact that computer screens have a non-linear response to RGB values is not a bug, as some people seem to think, it's a feature! It is meant to give a broader dynamic range, in correspondence to the way the human eye works.
One function that is meant to describe the transformation between RGB values and the monitor's luminance response is the gamma function, described by a power function: `y = x**gamma`. This is just an approximation, and it should not be expected that it is very accurate.
In fact, there are norms for how a computer screen should respond, one well known norm is the CIE Rec.709. In various software, one specifies a Gamma of 2.2. For example, PNG specifies a default gamma of 2.2. This value is taken from Rec.709, which reads:
```PhotonCountToRGB(colour) {
if (colour <= 0.018) {
return 4.5 * colour; /* toe slope */
} else {
return 1.099*pow(colour,0.45) - 0.099; /* scaled and transposed gamma
* function */
}
}
RGBToPhotonCount(colour) {
if (colour <= 0.081) {
return colour / 4.5;
} else {
return pow((colour + 0.099) / 1.099, 1.0/0.45);
}
}
```
Basically it's a gamma function with a linear piece at the dark end (called the toe slope) to compensate for the sudden steepness at the dark end of the regular gamma function. So, 1/0.45 is 2.22. This is where the gamma of 2.2 comes from. However, the Rec.709 is also scaled and transposed. When one plots the Rec.709 function (forgetting about the toe slope for a moment, plotted in red), it is much closer to a gamma of 1.8 (blue) than 2.2 (green). See the plot below.
Concluding, we may say that a gamma of 2.2 is NOT the most accurate approximation of the Rec.709 function.
# A simple and general alternative for the `toe slope'
Rec.709 also indicates that a plain gamma function is not very good at the very darkest end of the brightness range. It suggests a very broad range of shades from near-black to black, while in reality the computer screen's actual response may even mean a flatter curve there, rather than a steeper one. I've been trying to make a generalised alternative to the Rec.709's strategy of piecing together two functions. A very simple alternative to get rid of the too-steep dark end, is to transpose the function to the left, and then scale it so it has domain [0:1] and range [0:1] (function drawn in blue, transposed by 0.15). It closely fits the Rec.709 function. See the plots below. The right plot shows a detail of the left one. The reason for the toe slope seems to be apparent in this plot as well: the gamma function is very steep here. The toe slope (green) ends when it intersects the 709's transposed/scaled gamma (red). | 697 | 2,705 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | longest | en | 0.91358 |
http://docplayer.net/25351374-Unit-5-chemical-quantities-the-mole.html | 1,547,995,713,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583722261.60/warc/CC-MAIN-20190120143527-20190120165527-00169.warc.gz | 58,851,059 | 24,010 | # Unit 5 Chemical Quantities & The Mole
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1 Unit 5 Chemical Quantities & The Mole
2 Molar mass is the mass of one mole of a substance. Molar Mass Other names for molar mass include *formula mass *gram formula mass *molecular weight
3 Molar Mass One of mole any element will have a mass in grams corresponding to the value of its. atomic mass o1 mol carbon = g/mol o1 mol calcium = g/mol o(its atomic mass from the periodic table = 1 mole)
4
5 Diatomic Elements never exist alone in nature!!! *Trick for remembering the 7 diatomics *
6 Molar Mass of Compounds mol One of any molecule/compound will have a mass in grams corresponding to the value of its molar mass (the sum of the masses of the elements that compose it).
7 Example: Water, H 2 O Element Molar Mass # of Atoms Total H X g O X g g/mol This is the mass of 1 mol of water!
8 Ex. Ca(NO 3 ) 2 Ca: x 1 = N: x 2 = O: x 6 = g/mole
9 Sample Problems K Potassium carbonate 2 CO 3 # of K atoms: x = # of C atoms: x = # of O atoms: x = g K 2 CO 3
10 Sample Problems (NH Ammonium sulfate 4 ) 2 SO 4 # of N atoms: x = # of H atoms: x = # of S atoms: x = # of O atoms: x = g (NH 4 ) 2 SO 4
11 Practice 1. What is the atomic mass of sodium? 2. Calculate the molar mass of Al 2 (SO 4 ) 3 3. Calculate the molar mass of nitrogen (hint: diatomic!?!?).
12 Mole Highway NOTES: To convert between units, follow the highway. Notice, there is no shortcut from grams to liters or between any of the three units surrounding the mole. This means you have to convert to moles before converting to another unit!
13 Draw the Mole Road Map!!
14 Molar mass: called gram atomic mass when single element is used. called gram formula unit when ionic compound is used. called gram molecular unit when molecular compound or diatomic molecules used. Diatomic molecules are atoms that bond with themselves. There are SEVEN of these that you need to remember: Br 2 I 2 N 2 Cl 2 H 2 O 2 F 2 (Remember this by the name BRINClHOF )
15 Determine the number of moles in. 25 g sodium 25 g 1 mol = 1.1 mol g Molar mass of Sodium = Na = 1 x = g
16 Determine the number of moles in. 85 g H 2 SO 4 85 g 1 mol = 0.87 mol g Mass in grams of H 2 SO 4 = H = 2 x = S = 1 x = O = 4 x = Add to get molar mass of g
17 Determine the number of grams in. 2.5 moles of sodium 2.5 mol g = 57 g I mol Molar mass of Sodium = Na = 1 x = g
18 Determine the number of grams in moles of H 2 SO mol g = 49 g I mol Mass in grams of H 2 SO 4 = H = 2 x = S = 1 x = O = 4 x = Add to get molar mass of g
19 How many moles are there in 27 g of ethanol (C 2 H 5 OH)? 27 g 1 mol = 0.59 mol g Mass of C 2 H 5 OH: C = 2 x = H = 6 x 1.008= O = 1 x = Add these all up for mass in grams = g
20 Homework page 7 Show all work and units to receive full credit.
21 Representative Particles Moles
22 It s a quantity, and it s a BIG one. Avogadro s Number = 6.02 x and is also called a. Mole Like a pair or dozen, a mole represents a set number of things; in chemistry, those things are particles. For example: 1 dozen = 12 items 3 dozen = 36 cookies.5 dozen = 6 doughnuts
23 It s a quantity, and it s a BIG one. So, a dozen is a counting unit equal to 12 of any object. Likewise, a Mole is a counting unit equal to 6.02 x of an object, even really small ones like, atoms, molecules or. formula units
24 What s so special about 6.02 x 10 23? Why do scientists use that number? oa conversion factor. oallows us to manipulate many small particles, as if they were one whole part. o For elements on the periodic table, there is a 1 to 1 relationship between the mass of a single atom (in amu) and the mass of 1 Mole of the same species of atom (in grams). o Remember, Mole is a quantity 6.02 x particles.
25 What s so special about 6.02 x 10 23? Atomic Number (number of protons) Atomic Mass (average number of protons and neutrons) Ar Using the periodic table, we ve learned that an Argon atom has a mass of amu. Unfortunately, manipulating a single atom of any element isn t reasonable, so taking its individual mass isn t possible. However, we do have the Mole So, a g sample of Argon contains 6.02 x atoms that s Avogadro s Number, the MOLE! It s a 1 to 1 relationship between a single atom in amu and a Mole of atoms 1 Atom of Argon in grams. 1 Mole of Argon
26 Representative Particles A representative particle is the smallest unit of a substance.
27 Types of Particles Monatomic elements = Diatomic elements = Ionic compounds = Covalent compounds = atoms Ions = molecules Acids = atoms molecules formula units molecules
28 Avogadro s number, which is, 6.02 x represents the number of chemical units in one mole of any substance. For the monatomic elements, the chemical unit is an. atom 1 mol of any chemical = 6.02 x particles.
29 Examples 1 mol CaCl 2 = 1 mol Ca 2+ = 6.02 x formula units 6.02 x atoms 1 mol HCl (aq) = 6.02 x molecules 1 mol P 2 O 5 = 1 mol Ca = 6.02 x molecules 6.02 x atoms 1 mol Cl 2 = 6.02 x molecules
30
31 Sample problems 1. How many moles are in 4.50 x 10 5 atoms of manganese? atoms 1 mol atoms = 74.8 moles
32
33 Sample problems 2. How many atoms are found in 3.27 mol of magnesium? 3.27 mol atoms 1 mol = atoms
34
35 Sample problems 3. Chalk is composed primarily of calcium carbonate. How many particles are in 3.4 moles of calcium carbonate? 3.4 mol formula units 1 mol = formula units
36 Moles of Chalk Lab
37 Homework page 10 Show all work and units to receive full credit.
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Name Date Class The Mole Section.1 Measuring Matter In your textbook, read about counting particles. In Column B, rank the quantities from Column A from smallest to largest. Column A Column B 0.5 mol 1. | 8,750 | 32,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.847852 |
http://www.textkit.com/greek-latin-forum/viewtopic.php?p=158273 | 1,527,295,510,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867254.84/warc/CC-MAIN-20180525235049-20180526015049-00081.warc.gz | 464,464,807 | 9,599 | sub-set propositions
Are you learning Koine Greek, the Greek of the New Testament and most other post-classical Greek texts? Whatever your level, use this forum to discuss all things Koine, Biblical or otherwise, including grammar, textbook talk, difficult passages, and more.
sub-set propositions
According to Wallace in GGBB, p. 41, there are two types of S-PN constructions.
(a) The subset proposition. >> This is where the S (subject) is a subset of the PN (predicate nominative).
(b) The convertible proposition. >> This construction indicates an identical exchange. This is the less frequent semantic relationship between S and PN. In this construction, both nouns have an identical referent. The mathematical formulas of A=B, B=A are applicable in such instances. There is complete interchange between the two nouns.
In this thread I'm mainly concerned about subset propositions , i.e. (a) above. As noted, this is a situatinon where the S is a subset of the PN. Thus "the word of the cross is foolishness" (1 Cor. 1:18) is not the same as "foolishness is the word of the cross," since there are other kinds of foolishness. "God is love" is not the same as "love is God," etc...
Question: Can we conclude that by definition, in a subset proposition the S and PN cannot both be definite nouns ? That is, the PN will either be qualitative, or indefinite but not definite ?
The answer seems fairly obvious to me, -- "yes" on both counts, -- but I would like the readers input.
Οὐαὶ οἱ λέγοντες τὸ πονηρὸν καλὸν καὶ τὸ καλὸν πονηρόν, οἱ τιθέντες τὸ σκότος φῶς καὶ τὸ φῶς σκότος, οἱ τιθέντες τὸ πικρὸν γλυκὺ καὶ τὸ γλυκὺ πικρόν
Isaac Newton
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Re: sub-set propositions
Isaac Newton wrote:According to Wallace in GGBB, p. 41, there are two types of S-PN constructions.
...
Question: Can we conclude that by definition, in a subset proposition the S and PN cannot both be definite nouns ? That is, the PN will either be qualitative, or indefinite but not definite ?
This is a bit of NT Greek "traditional grammar" found for example in Zerwick (171-175) and converted by Wallace into a RULE using some sort of math-like approach to the problem. Natural language is not math.
The syntactical analysis of the use of the article in NT Greek has a long tradition and much that tradition has been reevaluated in the last half century by linguists involved in bible translation. Quite recently Chapter 6 of Richard A. Hoyle, Scenarios, discourse and translation. SIL 2008.
http://www.sil.org/silepubs/Pubs/50670/ ... lation.pdf
The article with predicate nominative is used when the information is hearer old. It need not be discourse old. If the currently active scenario includes the idea represented by the substantive/noun then the article will be use even if it has not been referenced previously in the discourse. Failing a currently active scenario, if the idea is part of a shared cognitive framework then it can be assumed to be cognitively assessable and thus hearer old. That is the default pattern. On the other hand a hearer old substantive can be left anarthrous as a form of salience marking.
None of this addresses your question. But I think Wallace's approach is likely to get you headed down the wrong road on this. Smyth §1152:
Even in the predicate the article is used with a noun referring to a definite object (an individual or a class) that is well known, previously mentioned or hinted at, or identical with the subject ...
Notice the or. It doesn't have to be identical with the subject. Consider John 1:49:
John 1:49 ἀπεκρίθη αὐτῷ Ναθαναήλ· ῥαββί, σὺ εἶ ὁ υἱὸς τοῦ θεοῦ, σὺ βασιλεὺς εἶ τοῦ Ἰσραήλ.
Nathanael answered him, “Rabbi, you are the Son of God! You are the King of Israel!” ESV
RP: εἶ ὁ βασιλεὺς ♦ NA/UBS: βασιλεὺς εἶ
So why the article in English? The lack of the article βασιλεὺς marks it as salient, since βασιλεὺς εἶ τοῦ Ἰσραήλ is part of the shared cognitive framework thus always hearer old without being discourse old.
On the other hand, reading the Byz Textform:
Ἀπεκρίθη Ναθαναήλ καὶ λέγει αὐτῷ, Ῥαββί, σὺ εἶ ὁ υἱὸς τοῦ θεοῦ, σὺ εἶ ὁ βασιλεὺς τοῦ Ἰσραήλ. Robinson-Pierpont 2005
Is this really a convertible proposition? Would ὁ βασιλεὺς τοῦ Ἰσραήλ unambiguously point to the same referent as Ῥαββί in this context? I don't think so.
FOOTNOTE: Iver Larsen (Denmark/East Africa SIL) has had some favorable things to say about Hoyle's treatment of the article in Scenarios and Discourse.
C. Stirling Bartholomew
C. S. Bartholomew
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Re: sub-set propositions
C. S. Bartholomew wrote:
Isaac Newton wrote:According to Wallace in GGBB, p. 41, there are two types of S-PN constructions.
...
Question: Can we conclude that by definition, in a subset proposition the S and PN cannot both be definite nouns ? That is, the PN will either be qualitative, or indefinite but not definite ?
This is a bit of NT Greek "traditional grammar" found for example in Zerwick (171-175) and converted by Wallace into a RULE using some sort of math-like approach to the problem. Natural language is not math.
The syntactical analysis of the use of the article in NT Greek has a long tradition and much that tradition has been reevaluated in the last half century by linguists involved in bible translation. Quite recently Chapter 6 of Richard A. Hoyle, Scenarios, discourse and translation. SIL 2008.
http://www.sil.org/silepubs/Pubs/50670/ ... lation.pdf
The article with predicate nominative is used when the information is hearer old. It need not be discourse old. If the currently active scenario includes the idea represented by the substantive/noun then the article will be use even if it has not been referenced previously in the discourse. Failing a currently active scenario, if the idea is part of a shared cognitive framework then it can be assumed to be cognitively assessable and thus hearer old. That is the default pattern. On the other hand a hearer old substantive can be left anarthrous as a form of salience marking.
None of this addresses your question. But I think Wallace's approach is likely to get you headed down the wrong road on this. Smyth §1152:
Even in the predicate the article is used with a noun referring to a definite object (an individual or a class) that is well known, previously mentioned or hinted at, or identical with the subject ...
Notice the or. It doesn't have to be identical with the subject. Consider John 1:49:
John 1:49 ἀπεκρίθη αὐτῷ Ναθαναήλ· ῥαββί, σὺ εἶ ὁ υἱὸς τοῦ θεοῦ, σὺ βασιλεὺς εἶ τοῦ Ἰσραήλ.
Nathanael answered him, “Rabbi, you are the Son of God! You are the King of Israel!” ESV
RP: εἶ ὁ βασιλεὺς ♦ NA/UBS: βασιλεὺς εἶ
So why the article in English? The lack of the article βασιλεὺς marks it as salient, since βασιλεὺς εἶ τοῦ Ἰσραήλ is part of the shared cognitive framework thus always hearer old without being discourse old.
On the other hand, reading the Byz Textform:
Ἀπεκρίθη Ναθαναήλ καὶ λέγει αὐτῷ, Ῥαββί, σὺ εἶ ὁ υἱὸς τοῦ θεοῦ, σὺ εἶ ὁ βασιλεὺς τοῦ Ἰσραήλ. Robinson-Pierpont 2005
Is this really a convertible proposition? Would ὁ βασιλεὺς τοῦ Ἰσραήλ unambiguously point to the same referent as Ῥαββί in this context? I don't think so.
FOOTNOTE: Iver Larsen (Denmark/East Africa SIL) has had some favorable things to say about Hoyle's treatment of the article in Scenarios and Discourse.
Thanks for your thoughts Bartholomew.. I should be more specific as to exactly what I'm seeking in this thread. So my asking very specific questions , I think , is the best way to approach the problem.
(a) Would you agree that in a sub-set proposition the S and the PN cannot both be definite ? So for example, if we take both θεὸς (anarthrous) and ὁ λόγος (articular) at John 1:1c as definite nouns, then καὶ θεὸς ἦν ὁ λόγος is not a subset proposition but a convertible proposition . Agree / Disagree ?
(b) Also , my reading of Wallace's "pecking order" rule in p. 43 of his GGBB ("the subject will be articular") suggests that this rule is inapplicable where both S and PN are definite, more generally that is, that this rule is invalid where both nouns have the same semantic tag . Agree / disagree ?
(c) A construction with two definite nouns would be a convertible proposition ? Agree/disagree ?
(d) If however we take θεὸς qualitatively at John 1:1c but ὁ λόγος as a definite noun, then this is a subset proposition. Agree/disagree? .
(e) And only in subset constructions would Wallace's "pecking order" rule about the articular noun being the subject noun (see GGBB , p. 43) come into play ? Agree/disagree ?
Οὐαὶ οἱ λέγοντες τὸ πονηρὸν καλὸν καὶ τὸ καλὸν πονηρόν, οἱ τιθέντες τὸ σκότος φῶς καὶ τὸ φῶς σκότος, οἱ τιθέντες τὸ πικρὸν γλυκὺ καὶ τὸ γλυκὺ πικρόν
Isaac Newton
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Re: sub-set propositions
Isaac Newton wrote: (a) Would you agree that in a sub-set proposition the S and the PN cannot both be definite ? So for example, if we take both θεὸς (anarthrous) and ὁ λόγος (articular) at John 1:1c as definite nouns, then καὶ θεὸς ἦν ὁ λόγος is not a subset proposition but a convertible proposition . Agree / Disagree ?
Agree.
(d) If however we take θεὸς qualitatively at John 1:1c but ὁ λόγος as a definite noun, then this is a subset proposition. Agree/disagree?
Strongly agree. And I do indeed take the θεός there as indefinite/qualitative, so that "καὶ θεὸς ἦν ὁ λόγος" means essentially καὶ ὁ λόγος τις θεὸς ἦν or καὶ ὁ λόγος ἦν θεῖος. On the other hand,"καὶ ὁ Θεὸς ἦν ὁ λόγος" would mean "Jesus is Yahweh and Yahweh is Jesus."
χάρις σοι καὶ σοῖς ἐν ὀνόματι Ἰησοῦ, ὦ φίλε Ἰσαάκ!
Markos
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Re: sub-set propositions
Markos wrote:
Strongly agree. And I do indeed take the θεός there as indefinite/qualitative, so that "καὶ θεὸς ἦν ὁ λόγος" means essentially καὶ ὁ λόγος τις θεὸς ἦν or καὶ ὁ λόγος ἦν θεῖος. On the other hand,"καὶ ὁ Θεὸς ἦν ὁ λόγος" would mean "Jesus is Yahweh and Yahweh is Jesus."
Yes, we would end up with a statement of identity at clause c if that were the case, leading inextricably either to Modalism (at best) or to linguistic nonsense (at worse). What is required in this third clause is a statement of essential predication, and this can only be achieved by a qualitative [or an indefinite] Θεὸς . I think the ancient Greek grammarian Origen in his Commentary on John, Bk II, ch.3, explained the reasons for the anarthrous θεὸς very well.
DanielWallace in GGBB, p. 268 is instructive in this matter:
"Further, calling θεὸς in John 1:1c definite is the same as saying that if it had followed the verb it would have had the article. Thus it would be a convertible proposition with λόγος (i.e., "the Word" = "God" and "God" = "the Word"). The problem of this argument is that the θεὸς in 1:1b is the Father . Thus to say that the θεὸς in 1:1c is the same person is to say that "the Word was the Father." This, as the older grammarians and exegetes pointed out, is embryonic Sabellianism or modalism. The fourth Gospel is about the least likely place to find modalism in the NT."
χάρις σοι καὶ σοῖς ἐν ὀνόματι Ἰησοῦ, ὦ φίλε Ἰσαάκ!
May God's abundant blessings and favour rest upon you Markos.
Οὐαὶ οἱ λέγοντες τὸ πονηρὸν καλὸν καὶ τὸ καλὸν πονηρόν, οἱ τιθέντες τὸ σκότος φῶς καὶ τὸ φῶς σκότος, οἱ τιθέντες τὸ πικρὸν γλυκὺ καὶ τὸ γλυκὺ πικρόν
Isaac Newton
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Joined: Thu May 30, 2013 3:15 am | 3,350 | 11,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.91996 |
http://bettergradesfast.com/academy/multiplication-division-primary-math-ebook/ | 1,652,866,266,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521883.7/warc/CC-MAIN-20220518083841-20220518113841-00572.warc.gz | 5,873,169 | 22,881 | Multiplication and Division
Multiplication
To multiply means to add the multiplicand together as many times as the multiplier dictates.
Example:
4 x 6 means that 4 should be added together 6 times.
Therefore, 4 x 6 = 4 + 4 + 4 + 4 + 4 + 4 = 24
Example:
Multiply 78 and 29.
Therefore, your final answer to 78 x 29 should be equal to 2262.
Division
Division answers the question of how many times does a value “fit” into another value.
Example:
56 / 8 = 7 because 7 x 8 = 56.
The Long Division Method
Note Well:
written otherwise, we have 125 / 5 = 25 | 161 | 564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2022-21 | latest | en | 0.866425 |
http://mathhelpforum.com/math-software/52756-mathematica-replaceall.html | 1,524,526,945,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946256.50/warc/CC-MAIN-20180423223408-20180424003408-00569.warc.gz | 200,441,033 | 10,904 | 1. ## Mathematica - ReplaceAll
Can anyone help me out with this command that I am trying to get to work:
Sum[((Binomial[12, i]*Binomial[40, n - i])/Binomial[52, n]), {i, 1, n}] /. n -> {1, 2, 3, 4}
I would like it to carry out the sum four times and return the results for each one. If I go in and manually change the n on the top of the sum for each of the desired values, it works fine... however that defeats the purpose of using the /.
2. Use the Map operator. The shorthand symbol is /@. Note the use of the # symbols. These are placeholders for the Map function which maps the values from the set {1,2,3,4} into these places. The & symbol designates the Sum expression as a pure function with the list {1,2,3,4} mapped sequentially into the placeholder expressions.
Code:
In[15]:=
(Sum[(Binomial[12, i]*Binomial[40, #1 - i])/
Binomial[52, #1], {i, 1, #1}] & ) /@
{1, 2, 3, 4}
Out[15]=
{3/13, 7/17, 47/85, 2759/4165}
3. ## Table vs. Map
I would suggest using table for that instance:
Table[Sum[((Binomial[12, i]*Binomial[40, n - i])/Binomial[52, n]), {i, 1, n}] ,{n,1,4}]
So n becomes an iterator which varies from 1-4.
Actually, map and table are completely exchangable in most cases, however I prefer to use table where the iterator is a simple number. Problems that can arise with annonymous functions if you nest them, as you may often be tempted to do when shortening long pieces of code. It can become incredibly confusing to figure out which # means which thing coming from where, so it is best to use table whenever it makes sense. You can use table exactly like map if you set it up like
Table[f[n],{n,{List of values}}]
which is equivalent to
f/@list of values.
The problem with anonymous functions can, of course, easily be avoided by writing out the anonymous function:
Function[n,Sum[((Binomial[12, i]*Binomial[40, n - i])/Binomial[52, n]), {i, 1, n}]]/@{1,2,3,4}
which is the long hand of what shadowsend wrote, replacing # with n (the first argument of function)
when in doubt, never be reluctant to write out your anonymous functions.
A third option, which is my favorite, is to write
Array[Sum[((Binomial[12, i]*Binomial[40, # - i])/Binomial[52, #]), {i, 1,#}]&,4]
ReplaceAll is for changing the form of the answer, often to prepare it to be the argument of another function. You can make some really powerful algorithims out of it, but it can be hard to use. In this case, it does not work.
4. This /@ seems pretty powerful. I'll need to devote some time to learning how to use it. Thanks to both of you.
5. Map is quite powerful, as are all of the functional programming functions.
I suggest that you become familliar with Map, Apply, Fold, Select, and Pick. I have found these to be the staples of most programs. Fold is the hardest to grasp, and solves many problems almost instantly. Map and Apply I use in just about every program I make.
The way I learned to use mathematica was through lots of experimentation, and several ambitious programming assignments. Its fun, and totally worth it. Good luck =) | 836 | 3,044 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2018-17 | latest | en | 0.875668 |
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Page 1 of 2 5 7 graphing and solving quadratic inequalities 301 quadratic inequalities in one variable one way to solve a is to use a graph. 1 x2 9x 18 0 2 x2 5x 4 0 3 n2 64 0 4 b2 5b 0 5 35n2 22n 3 0 6 15b2 4b 4 0 7 7p2 38p 24 0 8 3×2 14x 49 0 9 3k2 18k 21 0 10 6k2 42k 72 0 11 x2 11x 28 12 k2 15k 56. Solution x2 x 2 2x x 2 0.
Solve the quadratic inequality. When does the object hit the ground. Answer all questions.
Solve each equation with the quadratic formula. To solve ax 2 bx c 0 or ax bx c 0 graph y ax bx cand identify the x values for which the graph lies below or on and below the x axis. Solve 2×2 6x 20 o.
Test each region using theinequality. And best of all they all well most come with answers. An object is launched at 4 9 meters per second from a 58 8 meter tall platform.
Solving quadratic inequalities algebraically worksheet name solve the following quadratic inequalities algebraically. An object is launched directly upward at 64 feet per second from a platform 80 feet.
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Inequalities Cheat Sheet It Includes Both Examples And General Guidance Good For Algebra Class Graphing Inequalities Homeschool Math Math Notebooks | 913 | 3,380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2022-21 | latest | en | 0.842746 |
https://hackage.haskell.org/package/monoid-extras-0.3.3.5/docs/Data-Monoid-Coproduct.html | 1,566,079,834,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027312025.20/warc/CC-MAIN-20190817203056-20190817225056-00223.warc.gz | 495,022,399 | 2,667 | monoid-extras-0.3.3.5: Various extra monoid-related definitions and utilities
Data.Monoid.Coproduct
Description
The coproduct of two monoids.
Synopsis
# Documentation
data m :+: n Source
`m :+: n` is the coproduct of monoids `m` and `n`. Values of type `m :+: n` consist of alternating lists of `m` and `n` values. The empty list is the identity, and composition is list concatenation, with appropriate combining of adjacent elements when possible.
Instances
(Show m, Show n) => Show ((:+:) m n) Monoid ((:+:) m n) The coproduct of two monoids is itself a monoid. Semigroup ((:+:) m n) (Action m r, Action n r) => Action ((:+:) m n) r Coproducts act on other things by having each of the components act individually.
inL :: m -> m :+: n Source
Injection from the left monoid into a coproduct.
inR :: n -> m :+: n Source
Injection from the right monoid into a coproduct.
mappendL :: m -> (m :+: n) -> m :+: n Source
Prepend a value from the left monoid.
mappendR :: n -> (m :+: n) -> m :+: n Source
Prepend a value from the right monoid.
killL :: Monoid n => (m :+: n) -> n Source
`killL` takes a value in a coproduct monoid and sends all the values from the left monoid to the identity.
killR :: Monoid m => (m :+: n) -> m Source
`killR` takes a value in a coproduct monoid and sends all the values from the right monoid to the identity.
untangle :: (Action m n, Monoid m, Monoid n) => (m :+: n) -> (m, n) Source
Take a value from a coproduct monoid where the left monoid has an action on the right, and "untangle" it into a pair of values. In particular,
`m1 <> n1 <> m2 <> n2 <> m3 <> n3 <> ...`
is sent to
`(m1 <> m2 <> m3 <> ..., (act m1 n1) <> (act (m1 <> m2) n2) <> (act (m1 <> m2 <> m3) n3) <> ...)`
That is, before combining `n` values, every `n` value is acted on by all the `m` values to its left. | 556 | 1,835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | longest | en | 0.78311 |
http://uncyclopedia.wikia.com/wiki/UnNews:Mathematicians:_42_really_the_same_number_as_23?oldid=5368650 | 1,462,161,493,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860122420.60/warc/CC-MAIN-20160428161522-00141-ip-10-239-7-51.ec2.internal.warc.gz | 304,493,163 | 14,310 | # UnNews:Mathematicians: 42 really the same number as 23
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Mathematicians: 42 really the same number as 23 Your A.D.D. news outl — Oooh, look at the pictures! Monday, May 2, 2016, 03:58:59 (UTC)
21 April 2007
PRINCETON, NJ -- Mathematicians have proved that 42 and 23, each long known to be the Ultimate and Second-to-Ultimate Answers to Life, the Universe and Everything due to their overuse, are actually both the same number.
Paul Johnston Dullford, a pure mathematician at Princeton University's Institute of Advanced Study of Advanced Study, uses the analogy of blind men examining a V-dub to explain the course of numerics until today. "One describes touching a tire, one describes touching a trunk, another describes the bonnet, yet another sadly is run over by the V-dub, because the secretly non-blind person drives off with the V-dub, but that's not important," he says. "They come up with different descriptions but they don't see the big picture. There is only one V-dub and they describe different parts of it."
Dullford's claim is this: 22 is the same number as 22 since 1) 2+2 = 4 = 2*2, and 2) 22 = 2*11 and 2+1+1 = 4 = 2+2. Likewise, 42 must be the same number as 23 because 1) 4+2 = 6 = 2*3 and 2) 42 = 2*3*7 and 2+3+7 = 5 = 2+3. Dullford explains that "7 is exempt from any addition" because 7 is what he calls a "lucky" number. He tried to explain the concept of "luck" to us, saying that "it's similar to Gel Man's concept of 'strange,' 'charm,' and 'beauty' quarks," but UnNews found it incomprehensible.
Dullford theorizes that 42 and 23 are part of a framework number (another incomprehensible concept) tentatively named M. When asked what M stood for by fellow mathematician Jack Wiston, he replied that M "stands for magic, mystery, or Monday, according to taste." He also added, "Some cynics have occasionally suggested that M also stands for 'murky,' because our level of understanding of the number is in fact so primitive."
Skeptics of M have joked that the "M" means "Moronic", "Microsoft" or "Grue", which really doesn't start with an M, which is the point since skeptics still think 42 and 23 are as different as M and G are. Some also suggest that the M is an upside down W, for "Wiston".
Skeptics also criticize that Dullford misspelled the physicist Murray Gell-Mann's name.
42 commented that "I told you so!" 23 commented that "That explains a lot."
Meanwhile Douglas Adams and Jim Carrey were unavailable for comment. Douglas Adams was busy pushing the daisies, while Jim Carrey was filming The Number 1337. | 711 | 2,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2016-18 | latest | en | 0.953189 |
https://intfiction.org/t/awkward-name-clash/11687 | 1,719,325,343,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866143.18/warc/CC-MAIN-20240625135622-20240625165622-00740.warc.gz | 263,796,659 | 6,070 | Awkward name clash?
I have two separate objects with unique names, in two separate rooms - but Inform sees this as a conflict. Here’s the error;
Problem. You wrote ‘The deep forest leaves are scenery in The Deep Forest’ , but also ‘The pile of sticks and leaves is scenery in The Bottom of the Hole’ : that seems to be saying that the same object (deep forest leaves) must be in two different places (Deep Forest and Bottom of the Hole). This looks like a contradiction.
I actually don’t understand the logic behind why it thinks they’re the same, if I knew I might try and avoid it. How do you commonly work around a problem like this?
I believe it is because your Thing is “deep forest leaves” and your Location is “The Deep Forest”. The actual conflict is between a thing and a room. Maybe change the Thing’s name to something else, maybe more descriptive?
If I’m wrong, someone will follow-up and let us know!
If you absolutely want the Thing’s name to be “deep forest leaves”, you can do a little trick like this:
```some deep_forest_leaves are scenery in The Deep Forest. understand "deep","forest","leaf","leaves" as the deep_forest_leaves. the printed name of the deep_forest_leaves is "deep forest leaves".```
There are a few ways to handle this. You could also write
A thing called the pile of sticks and leaves is scenery in The Bottom of the Hole.
You can see the problem by compiling this:
```The Bottom of the Hole is a room. The pile of sticks and leaves is scenery in The Bottom of the Hole.```
and looking at the Map section of the Index. Because of the “and,” Inform thinks that you’re trying to create one piece of scener called “the pile of sticks” and another called"leaves"! Then since you’ve already defined “The deep forest leaves,” it thinks “leaves” is meant to refer to the deep forest leaves, and you get the name clash.
zarf’s solution with “called” will ensure that you create one thing called “the pile of sticks and leaves,” instead of two. That should resolve the namespace clash. As MTW suggested, you can also give things funny internal names and use the printed name property and Understand statements to get them to behave the way you want.
Note that if you use the “unambiguous internal names overridden with Understand lines” solution, it’s best to make the thing privately-named (which means no default Understand lines will be created for it).
I disagree. Don’t bother with privately-named.
Hahah, you guys are right - matt w’s breakdown showed the problem, it indeed was the “and”. It’s like the old Panda joke - eats shoots and leaves
zarf’s solution worked perfectly And MTW I will keep the “the printed name of the…” solution on hand, I think that may come in handy some time, seeing as all object names are global… I’m going to run into clashes with similar objects that exist in different locations; I like to make a lot of scenery objects to give the player things to examine after reading the room description.
Worst case scenario, you can always fall back to "Frob is a room. The printed name of frob is “Elaborate Location West of the House and Containing Words You Shouldn’t Use”.
Yes exactly I’m glad this option exists. | 714 | 3,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.955353 |
https://thethoughtnow.com/difference-between-gravitation-and-gravity/ | 1,701,352,343,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100227.61/warc/CC-MAIN-20231130130218-20231130160218-00645.warc.gz | 646,567,220 | 16,973 | Difference Between Gravitation and Gravity
The terms gravity and gravitation refer to the phenomenon that causes weights to attract each other. These terms are used interchangeably, with permission.
In some fields of research, however, the distinction between gravity and gravitation is more pronounced. Gravity is used when the field of gravity is received, and the focus is on how the particle interacts with the field. In contrast, the term “gravity” is used when the center is the point of mutual attraction between the two masses.
What is Gravitation?
Gravity is the universal force of attraction acting between all masses and masses. It is the weakest of all the fundamental forces in nature. Gravity approximates well to Newton’s law of universal gravitation.
F ∝ (m1×m2) / r2
Sir Isaac Newton proposed the law of universal gravitation in 1687 and used it to explain the observed motion of the planets and moons.
F = G.(m1 × m2) / r2
Newton’s law of universal gravitation states that every particle attracts every other particle in space with a force directly proportional to mass times and a force inversely proportional to the square of the distance between them.
What is Gravity?
Basically, an adverb is similar to gravity, however, the adverb is often attached to objects, e.g. So in summary, we can say that there is a cause and effect relationship between gravity and gravity; The force of gravity or gravity is the cause, and gravity is the effect.
All objects on earth have a downward gravitational force that exerts the weight of the earth on them. The weight of the earth is measured by the speed of free-falling objects. The acceleration due to gravity at the surface of the earth is 9.8 ms-2.
F ∝ M
Thus, for every second an object is in free fall, its velocity increases by approximately 9.8 meters per second. The acceleration due to the earth’s gravity is indicated by the symbol ‘g’.
Difference Between Gravitation and Gravity
Gravitation Gravity It could be an attractive or repulsive force, depending on the nature of the force. It’s always a sort of enticing energy. The universal force is what it is referred to as. It isn’t thought of as a universal force. The gravitational force acts in a radial direction away from the masses. This force is directed along the line connecting the earth’s center and the body’s center. It is continually striving to reach the earth’s core. This is a very weak type of force. This is a powerful force. This force is a physical quantity with a vector. The gravitational force has its own vector field. When the distance between the bodies is infinite, the force will be zero. At the earth’s center, the gravitational force will be zero. It necessitates the use of objects with two masses. It simply needs one mass. | 596 | 2,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2023-50 | latest | en | 0.949689 |
https://cs.stackexchange.com/questions/21765/kleene-star-of-an-infinite-unary-language-always-yields-a-regular-language?noredirect=1 | 1,716,013,613,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057327.58/warc/CC-MAIN-20240518054725-20240518084725-00499.warc.gz | 166,534,930 | 36,419 | # Kleene star of an infinite unary language always yields a regular language [duplicate]
Let $L = \{a^n \mid n \ge 0\}$, where $a^0 = \epsilon$ and $a^n = a^{n-1}a$ for all $n \ge 1$.
Thus $L$ consists of sequences of $a$ of all lengths, including a sequence of length $0$. Let $L_2$ be any infinite subset of $L$. I need to show there always exists a DFA to recognize $L_2^*$.
If $L_2$ is a finite subset it is very obvious as $L_2$ would be a DFA and hence by Kleene closure $L_2^*$ would be recognized by a DFA. But I am unable to get it for infinite subset as $L_2$ may not be expressed as DFA when, e.g., string lengths are prime.
Suppose there are two words in the language whose lengths are relatively prime. Let these lengths be $x$ and $y$. We know (see this) that by adding these numbers to each other repeatedly, we can get any number greater than $(x - 1)(y - 1) - 1$. So if $x$ and $y$ are $13$ and $7$, we can write any number greater than $72$ as a linear combination of $7$ and $13$. What this means for us: $L_2^*$ consists of some arbitrary finite language (regular, as all finite languages), in union with the language $\{w \in a^* \mid |a| > (x-1)(y-1)-1\}$. This language is regular since it is the language of all words with a finite set of words removed. Since $L_2^*$ is a union of regular languages, it must also be regular.
If all words in $L_2^*$ have lengths which share a greatest common factor (call this common factor $m$), then repeat the above argument, but instead of using string lengths, use string lengths divided by $m$. In this case, $L_2^*$ will be the concatenation of an arbitrary finite language (regular) and the language $\{w \in (a^m)^* \mid |w| > m^2[(x/m - 1)(y/m - 1) - 1]\}$, also regular (since $(a^m)^* is regular and we are removing finitely many words from it). For example, suppose all words in$L$have a GCF of 2, and the language contains the words$a^4$and$a^{10}$. We have$m = 2$,$x/m = 4/2 = 2$, and$y/m = 10/2 = 5$, which are relatively prime. Therefore, we know that we can get any word whose length is multiple of$m$if the length is greater than$m^2[(x/m - 1)(y/m - 1) - 1] = 2^2[(2 - 1)(5 - 1) - 1] = (4)(3) = 12$by concatenating$a^4$and$a^{10}\$. | 692 | 2,213 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-22 | latest | en | 0.922794 |
https://mcqbooks.com/ncert-solutions-for-class-6-maths-chapter-12-ex-12-3/ | 1,701,445,251,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100290.24/warc/CC-MAIN-20231201151933-20231201181933-00197.warc.gz | 435,632,532 | 12,797 | # NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.3
These NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.3 Questions and Answers are prepared by our highly skilled subject experts.
## NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Exercise 12.3
Question 1.
If the cost of 7 m of cloth is ₹ 294, find the cost of 5 m of cloth.
Cost of 7 m of cloth = ₹ 294
∴ Cost of 1 m of cloth = $$\frac{294}{7}$$ = ₹ 42
∴ Cost of 5 m of cloth = 42 × 5 = ₹ 210
Thus, the cost of 5 m of cloth is ₹ 210.
Question 2.
Ekta earns ₹ 1500 in 10 days. How much will she earn in 30 days?
Earning of 10 days = ₹ 1500
∴ Earning of 1 day = $$\frac{1500}{10}$$ = ₹ 150
∴ Earning of 30 days = 150 × 30 = ₹ 4500
Thus, the earning of 30 days is ₹ 9000.
Question 3.
If it has rained 276 mm in the last 3 days, how many cm of rain will fall in one full week (7 days)? Assume that the rain continues to fall at the same rate.
Rain in 3 days = 276 mm
Rain in 1 day = $$\frac{276}{3}$$ = 92 mm
∴ Rain in 7 days = 92 × 7 = 644 mm
Thus, the rain in 7 days is 644 mm.
Question 4.
Cost of 5 kg of wheat is ₹ 30.50.
(a) What will be the cost of 8 kg of wheat?
(b) What quantity of wheat can be purchased in ₹ 183?
(a) Cost of 5 kg of wheat = ₹ 30.50
∴ Cost of 1 kg of wheat = $$\frac{30.50}{5}=\frac{3050}{500}$$ = ₹ 6.10
∴ Cost of 8 kg of wheat = 6.10 × 8 = ₹ 48.80
(b) From ₹ 30.50, quantity of wheat can be purchased = 5 kg
∴ From ₹ 1, quantity of wheat can be purchased = $$\frac{5}{30.50}$$
From ₹ 61, quantity of wheat can be purchased = $$\frac{5}{30.50}$$ × 61
= $$\frac{5}{3050}$$ × 6100 = 10 kg
Question 5.
The temperature dropped 15 degree Celsius in the last 30 days. If the rate of temperature drop remains the same, how many degrees will the temperature drop in the next ten days?
Degree of temperature dropped in last 30 days = 15 degrees
∴ Degree of temperature dropped in last 30 days = $$\frac{15}{30}=\frac{1}{2}$$ degree
∴ Degree of temperature dropped in last 10 days
= $$\frac{1}{2}$$ × 10 = 5 degree
Thus, 5 degree Celsius temperature dropped in 10 days.
Question 6.
Shaina pays ₹ 7500 as rent for 3 months. How much does she has to pay for a whole year, if the rent per month remains same?
Rent paid for 3 months = ₹ 7500
∴ Rent paid for 1 months = $$\frac{7500}{3}$$ = ₹ 2500
∴ Rent paid for 12 months = 2500 × 12 = ₹ 30,000
Thus, the total rent of one year is ₹ 30,000.
Question 7.
Cost of 4 dozens bananas is ₹ 60. How many bananas can be purchased for ₹ 12.50?
Cost of 4 dozen bananas = ₹ 60
Cost of 48 bananas = ₹ 60
[4 dozen = 4 × 12 = 48]
∴ From ₹ 60, number of bananas can be purchased = 48
∴ From ₹ 1, number of bananas can be purchased = $$\frac{48}{60}=\frac{4}{5}$$
= $$\frac{4}{5}$$ × 12.50 = $$\frac{4}{5} \times \frac{1250}{100}=\frac{250}{25}$$
= 10 bananas
∴ From ₹ 12.50, number of bananas can be purchased
Thus, 10 bananas can be purchased for ₹ 12.50.
Question 8.
The weight of 72 books is 9 kg what is the weight of 40 such books?
The weight of 72 books = 9 kg
The weight of 1 book = $$\frac{9}{12}=\frac{1}{8}$$
∴ The weight of 40 books
= $$\frac{1}{8}$$ × 40 = 5 kg
Thus, the weight of 40 books is 5 kg.
Question 9.
A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?
For covering 594 km, a truck will be required diesel = 108 litres
For covering 1 km, a truck will be required diesel = $$\frac{108}{594}=\frac{2}{11}$$
∴ For covering 1650 km, a truck will be required diesel
= $$\frac{2}{11}$$ × 1650 = 300 litres
Thus, 300 litres diesel required by the truck to cover a distance of 1650 km.
Question 10.
Raju purchases 10 pens for ₹ 150 and Manish buys 7 pens for ₹ 84. Can you say who got the pen cheaper?
Raju purchase 10 pens for = ₹ 150
∴ Raju purchases 1 pen for = $$\frac{150}{10}$$ = ₹ 15
Manish purchases 7 pens for = ₹ 84
∴ Manish purchases 1 pen for = $$\frac{84}{7}$$ = ₹ 12
Thus, Manish got the pens cheaper.
Question 11.
Anish made 42 runs in 6 overs and Anup made 63 runs in 7 overs. Who made more runs per over?
∴ Anish made in 1 overs = $$\frac{42}{6}$$ = 7 runs
∴ Anup made in 1 overs = $$\frac{63}{7}$$ = 9 runs. | 1,462 | 4,218 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2023-50 | latest | en | 0.884022 |
https://www.solipsys.co.uk/new/FindingPerrinPseudoPrimes_Part2.html?WhenTheTextAndHtmlDisagree | 1,621,270,488,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991258.68/warc/CC-MAIN-20210517150020-20210517180020-00224.warc.gz | 1,061,819,609 | 5,977 | # Finding Perrin Pseudo Primes_Part 2
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Additionally, some earlier writings:
# Finding Perrin Pseudo-primes: Part 2 - 2015/05/17
#!/usr/bin/python n = 2 t = 10 t_prime = 0.0 t_perrin = 0.0 while clock() < int(argv[1]): t_prime -= clock() is_prime = is_prime_q(n) t_prime += clock() t_perrin -= clock() is_perrin = is_perrin_q(n) t_perrin += clock() if is_prime != is_perrin: print n, is_prime, is_perrin if t < clock(): print 'Tested to %d in %d secs' % (n,t) t += 10 n += 1 print 'Time in prime :', t_prime print 'Time in Perrin:', t_perrin # ================ # Edited output: # # Tested to 42763 in 100 secs # # Time in prime : 0.11 # Time in Perrin: 99.72 # ===== # 99.83
So now we've got the scaffolding of a program to find these Perrin Pseudo-Primes. Here is the main loop of the code, with some simplistic timing added to it. (Note that this code is incomplete and won't run as is).
The output shows that when given 100 seconds to run it gets as far as n=42763, but more importantly, the timing shows that overwhelmingly it spends its time in the routine to test whether or not a number passes the "Perrin Test." So there are a few things we need to do:
• Make fewer calls to the Perrin Test
• Make the Perrin Test run faster.
In this post we'll look at a simple technique to accomplish the first of these.
There's a maxim in programming:
You can't make a program run faster, you can only make it do less.
One way of making the program do less is to pre-compute things and then use the result multiple times. Another way is to do a quick test to avoid a longer one. We'll do both of those.
Here's the critical observation:
• If m divides n,
• and n divides k(n),
• then m divides k(n).
The second part of the observation is that the Perrin sequence modulo m has to repeat itself after at most $m^3$ steps. There are only $m^3$ possible triples, and the sequence is reversable, so eventually we get the first triple appearing again. Here are the sequences for 2, 3, and 5, up to the point where they start to repeat
• mod 2 : 1 0 0 1 0 1 1 1 0 0 ...
• repeats after 7 terms
• mod 3 : 0 0 2 0 2 2 2 1 1 0 2 1 2 0 0 2 ...
• repeats after 13 terms
• mod 5 : 3 0 2 3 2 0 0 2 0 2 2 2 4 4 1 3 0 4 3 4 2 2 1 4 3 0 2 ...
• repeats after 24 terms
#!/usr/bin/python def compute_perrin_number(n): if n in [2, 3]: return 0 a0, a1, a2 = 3, 0, 2 i = n while i > 0: a0, a1, a2 = a1, a2, (a0+a1) % n i -= 1 return a0 def make_pre_filter(m): rtn = [ 3 % m, 0, 2 % m ] while True: rtn.append( (rtn[-3] + rtn[-2]) % m ) if rtn[:3] == rtn[-3:]: return rtn[:-3] pre_filters = {} pre_filter_elements = range(2,104) for m in pre_filter_elements: pre_filters[m] = make_pre_filter(m) def is_perrin_q(n): for m in pre_filter_elements: if n % m == 0: pre_filter_to_use = pre_filters[m] ndx = n % len(pre_filter_to_use) if 0 != pre_filter_to_use[ndx]: return False return compute_perrin_number(n) == 0
So what we can do is this. For each of several primes (and maybe even some composites) we precompute the Perrin sequence modulo that number. Then when we are looking at a number n we look at all the precomputed sequences that divide n and ask if the sequence has a 0 in the right place. If not, then n will not divide k(n). Of course, if it does have a 0 in the right place for every precomputed m that divides n then we will still have to do the computation, but with any luck that will be a lot less.
So here is the result.
• Without the sieving:
• Tested to 45761 in 100 secs
• Perrin computations: 45760
• With sieving from m=2 to 103:
• Tested to 109159 in 100 secs
• Perrin computations: 19004
As you can see, we get more than twice as far, so we're clearly doing a good thing. It's obvious why when we look at the number of Perrin computations we do. Without the sieving every number has the full calculation, but with sieving only 19 in every 109 gets the calculation, or roughly 17.4%, a saving of 82.6%.
But there's still a problem that we can see clearly when we look at the timing:
• Time in prime : 0.25 secs
• Time in Perrin : 99.47 secs
Despite saving over 80% we're still spending all our time in the Perrin test routine. In the next installment we'll see how we can speed that up further by using matrices and an adaptation of the Peasant Multiplication Algorithm.
<<<< Prev <<<< FindingPerrinPseudoPrimes Part1 : >>>> Next >>>> Russian Peasant Multiplication
I've decided no longer to include comments directly via the Disqus (or any other) system. Instead, I'd be more than delighted to get emails from people who wish to make comments or engage in discussion. Comments will then be integrated into the page as and when they are appropriate.
If the number of emails/comments gets too large to handle then I might return to a semi-automated system. That's looking increasingly unlikely. | 1,428 | 4,881 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2021-21 | latest | en | 0.874951 |
https://www.mapleprimes.com/posts/200903-The-Number-Of-Nonnegative-And-Positive | 1,716,689,974,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058861.60/warc/CC-MAIN-20240526013241-20240526043241-00121.warc.gz | 764,900,602 | 28,825 | :
## The number of non-negative and positive solutions of linear Diophantine equations
This post is related to the this thread
The recursive procedure PosIntSolve finds the number of non-negative or positive solutions of any linear Diophantine equation
a1*x1+a2*x2+ ... +aN*xN = n with positive coefficients a1, a2, ... aN .
Formal parameters: L is the list of coefficients of the left part, n is the right part, s (optional) is nonneg (by default) for nonnegint solutions and pos for positive solutions.
The basic ideas:
1) If you make shifts of the all unknowns by the formulas x1'=x1-1, x2'=x2-1, ... , xN'=xN-1 then the number of positive solutions of the first equation equals the number of non-negative solutions of the second equation.
2) The recurrence formula (penultimate line of the procedure) can easily be proved by induction.
The code of the procedure:
restart;
PosIntSolve:=proc(L::list(posint), n::nonnegint, s::symbol:=nonneg)
local N, n0;
option remember;
if s=pos then n0:=n-`+`(op(L)) else n0:=n fi;
N:=nops(L);
if N=1 then if irem(n0,L[1])=0 then return 1 else return 0 fi; fi;
end proc:
Examples of use.
Finding of the all positive solutions of equation 30*a+75*b+110*c+85*d+255*e+160*f+15*g+12*h+120*i=8000:
st:=time():
PosIntSolve([30,75,110,85,255,160,15,12,120], 8000, pos);
time()-st;
13971409380
2.125
To test the procedure, solve (separately for non-negative and positive solutions) the simple equation 2*x1+7*x2+3*x3=2000 in two ways (by the procedure and brute force method):
ts:=time():
PosIntSolve([2,7,3], 2000);
PosIntSolve([2,7,3], 2000, pos);
time()-ts;
47905
47334
0.281
ts:=time():
k:=0:
for x from 0 to 2000/2 do
for y from 0 to floor((2000-2*x)/7) do
for z from 0 to floor((2000-2*x-7*y)/3) do
if 2*x+7*y+3*z=2000 then k:=k+1 fi;
od: od: od:
k;
k:=0:
for x from 1 to 2000/2 do
for y from 1 to floor((2000-2*x)/7) do
for z from 1 to floor((2000-2*x-7*y)/3) do
if 2*x+7*y+3*z=2000 then k:=k+1 fi;
od: od: od:
k;
time()-ts;
47905
47334
50.063
Another example - the solution of the famous problem: how many ways can be exchanged \$ 1 using the coins of smaller denomination.
PosIntSolve([1,5,10,25,50],100);
292
Edit. The code has been slightly edited
| 769 | 2,261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-22 | latest | en | 0.599711 |
https://www.unitconverters.net/speed/centimeter-hour-to-kilometer-hour.htm | 1,660,593,550,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572198.93/warc/CC-MAIN-20220815175725-20220815205725-00032.warc.gz | 892,457,021 | 3,285 | Home / Speed Conversion / Convert Centimeter/hour to Kilometer/hour
# Convert Centimeter/hour to Kilometer/hour
Please provide values below to convert centimeter/hour [cm/h] to kilometer/hour [km/h], or vice versa.
From: centimeter/hour To: kilometer/hour
### Centimeter/hour to Kilometer/hour Conversion Table
Centimeter/hour [cm/h]Kilometer/hour [km/h]
0.01 cm/h1.0E-7 km/h
0.1 cm/h1.0E-6 km/h
1 cm/h1.0E-5 km/h
2 cm/h2.0E-5 km/h
3 cm/h3.0E-5 km/h
5 cm/h5.0E-5 km/h
10 cm/h0.0001 km/h
20 cm/h0.0002 km/h
50 cm/h0.0005 km/h
100 cm/h0.001 km/h
1000 cm/h0.01 km/h
### How to Convert Centimeter/hour to Kilometer/hour
1 cm/h = 1.0E-5 km/h
1 km/h = 100000 cm/h
Example: convert 15 cm/h to km/h:
15 cm/h = 15 × 1.0E-5 km/h = 0.00015 km/h | 297 | 742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.490455 |
https://numberworld.info/12985 | 1,624,077,120,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487643380.40/warc/CC-MAIN-20210619020602-20210619050602-00137.warc.gz | 376,231,920 | 3,834 | # Number 12985
### Properties of number 12985
Cross Sum:
Factorization:
5 * 7 * 7 * 53
Divisors:
1, 5, 7, 35, 49, 53, 245, 265, 371, 1855, 2597, 12985
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
32b9
Base 32:
clp
sin(12985)
-0.7156558879653
cos(12985)
-0.69845304066959
tan(12985)
1.0246299268441
ln(12985)
9.4715501240968
lg(12985)
4.1134419539653
sqrt(12985)
113.9517441727
Square(12985)
### Number Look Up
Look Up
12985 (twelve thousand nine hundred eighty-five) is a very unique figure. The cross sum of 12985 is 25. If you factorisate the number 12985 you will get these result 5 * 7 * 7 * 53. 12985 has 12 divisors ( 1, 5, 7, 35, 49, 53, 245, 265, 371, 1855, 2597, 12985 ) whith a sum of 18468. 12985 is not a prime number. The figure 12985 is not a fibonacci number. The figure 12985 is not a Bell Number. 12985 is not a Catalan Number. The convertion of 12985 to base 2 (Binary) is 11001010111001. The convertion of 12985 to base 3 (Ternary) is 122210221. The convertion of 12985 to base 4 (Quaternary) is 3022321. The convertion of 12985 to base 5 (Quintal) is 403420. The convertion of 12985 to base 8 (Octal) is 31271. The convertion of 12985 to base 16 (Hexadecimal) is 32b9. The convertion of 12985 to base 32 is clp. The sine of the number 12985 is -0.7156558879653. The cosine of the figure 12985 is -0.69845304066959. The tangent of 12985 is 1.0246299268441. The square root of 12985 is 113.9517441727.
If you square 12985 you will get the following result 168610225. The natural logarithm of 12985 is 9.4715501240968 and the decimal logarithm is 4.1134419539653. that 12985 is impressive number! | 667 | 1,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2021-25 | latest | en | 0.735334 |
https://www.physicsforums.com/threads/electric-field-of-a-thick-infinite-non-conducting-plate.919321/ | 1,606,563,385,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195417.37/warc/CC-MAIN-20201128095617-20201128125617-00177.warc.gz | 807,815,663 | 16,615 | # Electric field of a thick infinite non-conducting plate
gj2
## Homework Statement
An infinite non-conducting plate of thickness ##d## lies in the ##xy## plane. The bottom surface of the plate lies in the plane ##z=0##. The charge density of the plate is ##\rho=\rho_0 z /d~,~\rho_0>0##.
Find the electric field in the regions ##z<0~~,~~0<z<d~~,~~z>d## and the potential (assume it's zero at ##z=0##).
## Homework Equations
Poisson's equation ##{\displaystyle {\nabla }^{2}\varphi =-{\frac {\rho }{\varepsilon_0 }}}## and ##\vec{E}=-\operatorname{grad} \varphi##
## The Attempt at a Solution
The charge distribution does not depend on ##x## or ##y## and the plate is symmetric therefore both the potential and the electric field can't depend on ##x## or ##y##. We know that ##\vec{E}=-\operatorname{grad} \varphi##. The Poisson's equation for the region inside the plate is
$$\frac{\partial^2 \varphi}{\partial z^2}=-\frac{\rho_0 z}{\varepsilon_0 d}$$
Assuming ##\varphi(z=0)=0## we obtain
$$\varphi(z)=C_1z-\frac{\rho_0 z^3}{6\varepsilon_0 d}~~~,~~~0<z<d$$
Poisson's equation for the region outside the plate
$$\frac{\partial^2 \varphi}{\partial z^2}=0$$
Therefore
$$\varphi(z)=C_2z$$
The potential must be continuous everywhere and particularly at ##z=d## so
$$C_2=C_1-\frac{\rho_0 d}{6\varepsilon_0}$$
In other words
$$\varphi(z)=\left\{\begin{matrix} Cz-\frac{\rho_0 z^3}{6\varepsilon_0 d} & z\in(0,d)\\ Cz-\frac{\rho_0 dz}{6\varepsilon_0} & z\notin(0,d) \end{matrix}\right.$$
But I can't figure out what other condition should I impose in order to obtain the last missing constant.
Last edited by a moderator:
Delta2
Related Introductory Physics Homework Help News on Phys.org
ehild
Homework Helper
## Homework Statement
An infinite non-conducting plate of thickness ##d## lies in the ##xy## plane. The bottom surface of the plate lies in the plane ##z=0##. The charge density of the plate is ##\rho=\rho_0 z /d~,~\rho_0>0##.
Find the electric field in the regions ##z<0~~,~~0<z<d~~,~~z>d## and the potential (assume it's zero at ##z=0##).
## Homework Equations
Poisson's equation ##{\displaystyle {\nabla }^{2}\varphi =-{\frac {\rho }{\varepsilon_0 }}}## and ##\vec{E}=-\operatorname{grad} \varphi##
## The Attempt at a Solution
The charge distribution does not depend on ##x## or ##y## and the plate is symmetric therefore both the potential and the electric field can't depend on ##x## or ##y##. We know that ##\vec{E}=-\operatorname{grad} \varphi##. The Poisson's equation for the region inside the plate is
$$\frac{\partial^2 \varphi}{\partial z^2}=-\frac{\rho_0 z}{\varepsilon_0 d}$$
Assuming ##\varphi(z=0)=0## we obtain
$$\varphi(z)=C_1z-\frac{\rho_0 z^3}{6\varepsilon_0 d}~~~,~~~0<z<d$$
Poisson's equation for the region outside the plate
$$\frac{\partial^2 \varphi}{\partial z^2}=0$$
Therefore
$$\varphi(z)=C_2z$$
The potential must be continuous everywhere and particularly at ##z=d## so
$$C_2=C_1-\frac{\rho_0 d}{6\varepsilon_0}$$
In other words
$$\varphi(z)=\left\{\begin{matrix} Cz-\frac{\rho_0 z^3}{6\varepsilon_0 d} & z\in(0,d)\\ Cz-\frac{\rho_0 dz}{6\varepsilon_0} & z\notin(0,d) \end{matrix}\right.$$
But I can't figure out what other condition should I impose in order to obtain the last missing constant.
For a second-order equation, you need to give two boundary conditions. What can they be?
Do not forget that the Poisson equation is derived from the original Maxwell equations, so you can use Gauss' Law.
TSny
Homework Helper
Gold Member
If you are going to find ##\varphi## by integrating Poisson's equation, you have three separate regions to deal with. You cannot assume that the mathematical expression for ##\varphi(z)## when ##z>d## is the same as for when ##z<0##.
Adding a bit to ehild's hint, it might be easier to first find ##\mathbf{E}## and then use it to find ##\varphi(z)##.
gj2
For a second-order equation, you need to give two boundary conditions. What can they be?
Do not forget that the Poisson equation is derived from the original Maxwell equations, so you can use Gauss' Law.
If you are going to find ##\varphi## by integrating Poisson's equation, you have three separate regions to deal with. You cannot assume that the mathematical expression for ##\varphi(z)## when ##z>d## is the same as for when ##z<0##.
Adding a bit to ehild's hint, it might be easier to first find ##\mathbf{E}## and then use it to find ##\varphi(z)##.
Yes, thank you both. I realized beforehand that I did a mistake. In fact, it is much more complicated. But anyways I managed to solve it. In order to obtain the constants I used three things: 1) the fact that the electric field outside the plate is symmetrical w.r.t the plate (and not just constant) 2) Gauss law where the two bases of the Gaussian cylinder/box are outside the plate 3) Gauss law where one base is inside the plate and the other base outside it. | 1,480 | 4,862 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2020-50 | latest | en | 0.716001 |
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Senior Member
Join Date: May 2013
Posts: 1,365
April 9th, 2019, 07:33 PM
I've started yet another project that I'm not sure if I'll make it through, but for now, I'm wondering if I could get some scripting help.
If I'm reading the rules correctly, each power costs 1 point to add to the character. After that, every point spent raises the power level by 2. So in the portal, I set the incrementer to have an interval of 2, and that's working fine. But I'm having trouble calculating the point cost.
The script I'm using is below, and this is what's happening.
I add a power. Now I'm left with 9 power points. (This is correct so far.)
Now I increment the power level. It goes from 1 to 3. (This is also correct.)
The problem is, I still have 9 power points. Going from 1 to 3 doesn't subtract a point from my resources.
After that, it works fine. Going from 3 to 5 subtracts 1 point, then 5 to 7 and so on.
I've tried different ways of coding the script, but I can't get it to work how I want.
And yes, I sometimes miss obvious things.
Code:
``` <!-- Each power that is added by the user costs 1 point
Each point spent after that buys 2 points.
-->
<eval index="1" phase="Setup" priority="5000"><![CDATA[
~we must ignore bootstrapped abilities from races and abilities added via advances
if (isuser + origin.ishero >= 2) then
var uval as number
var cost as number
uval = field[trtUser].value
cost = (uval - 1) / 2
if (cost < 1) then
cost = 1
endif
hero.child[resPP].field[resSpent].value += cost
endif
]]></eval>```
#1
Senior Member
Join Date: Mar 2013
Location: Greater London, UK
Posts: 1,922
April 9th, 2019, 11:34 PM
So "(uval -1) / 2"
Does that give the correct subtraction, or should it be "(uval + 1) / 2"?
Farling
#2
Senior Member
Lone Wolf Staff
Join Date: May 2005
Posts: 12,485
April 10th, 2019, 07:30 AM
I would not trust an interval to protect what you're trying to do. Users can still edit the number and type in whatever they want. Instead, I would add a second field that is the power level, and is calculated at 2x field[trtUser].value. Then, you're having the user control the number of points they spent, and displaying the final power level separately. (Or adjust the finalize script on trtUser to display the power level, but that might look odd if the user edits it when it says 4, because they'd see a 2).
I'd read up on the Savage Worlds Wiki - they do a similar thing where the user needs to control dice sizes, and they also need to only show even numbers to the user, but still have a single click change the die size from d4 to d6.
#3
Senior Member
Lone Wolf Staff
Join Date: May 2005
Posts: 12,485
April 10th, 2019, 07:35 AM
cost = (uval - 1) / 2
So, at a uval of 0, the cost = -1/2 (but then you apply a minimum)
At a uval of 2, the cost = 1/2 (and the minimum applies)
at 4, the cost = 3/2
At 6, the cost = 5/2
That doesn't match the costs you were describing.
#4
EightBitz
Senior Member
Join Date: May 2013
Posts: 1,365
April 10th, 2019, 11:31 AM
Quote:
Originally Posted by Farling So "(uval -1) / 2" Does that give the correct subtraction, or should it be "(uval + 1) / 2"?
It gives the correct subtraction for the division by 2, but not, I now realize, for the final calculation. Your reply gave me this idea:
Code:
``` cost = (uval - 1) / 2
cost = round(cost,0,1) + 1```
And that works.
#5
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-- Default Style -- Majestic Contact Us - Lone Wolf Development Forums - Archive - Top | 1,607 | 5,993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | latest | en | 0.938891 |
http://math.stackexchange.com/questions/195064/chernoff-inequalities-for-the-sum-of-exponential-rvs | 1,466,969,040,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783395548.53/warc/CC-MAIN-20160624154955-00001-ip-10-164-35-72.ec2.internal.warc.gz | 200,495,917 | 18,049 | # Chernoff inequalities for the sum of Exponential RVs
These two well-known Chernoff bounds for the sum of RVs $X=\sum_{k=1}^{n}X_k$ in mulitplicative form,
$\mathbf{P}(X \leq (1- \delta)\mathbf{E}X) \leq e^{-\frac{\delta^2 \mathbf{E}X}{2}}\\ \mathbf{P}(X \geq (1+ \delta)\mathbf{E}X) \leq \big(\frac{e^{\delta}}{(1+\delta)^{(1+\delta)}} \big)^{\mu}$
apply in the case when $X_k$ take values 0 and 1. I'm wondering if and how these results can extend in a more general case, e.g. when $X_k$ are exponentially distributed iid.
-
Yes, start from the same exponential inequality and optimize over $t$. – Did Sep 13 '12 at 5:21
Thanks, could you give more details or explain this approach? – Alex Sep 13 '12 at 5:52
Show what you tried in this direction (the beginning merely copies the approach in the Bernoulli case explained on the WP page) and we will talk. – Did Sep 13 '12 at 7:52
If $\mathbf{E} X_k = \frac{1}{\mu}$, then $\mathbf{E}X= \frac{n}{\mu}$ and MGF of X is $\varphi_{X}(t) = \bigg(\frac{\mu}{\mu-t} \bigg)^{n}$, so te upper bound using Markov inequality is $\mathbf{P}(X>(1+\delta)\mathbf{E}X) < \bigg(\frac{\mu}{\mu-t} \bigg)^{n}e^{-t(1+\delta)\mathbf{E}X}$, but I do not understand the whole optimization part. – Alex Sep 14 '12 at 5:17
As explained by the OP in a comment, Markov inequality yields $$P(X\gt(1+\delta)n/\mu)\leqslant \exp(-nI(t/\mu,\delta)),$$ for every $t\gt0$, where $$I(s,\delta)=s(1+\delta)+\log(1-s).$$ The derivative is $\partial_sI(s,\delta)=1+\delta-1/(1-s)$ hence the optimal choice is $s_\delta=\delta/(1+\delta)$, which yields $I(s_\delta,\delta)=\delta-\log(1+\delta)$, and finally, $$P(X\gt(1+\delta)n/\mu)\leqslant\left(\frac{1+\delta}{\mathrm e^\delta}\right)^n.$$ To sum up, Markov (exponential) inequality yields a family of upper bounds, indexed by $t$ (or by $s$), and one chooses the value of $s$ (or of $t$) which minimizes this upper bound.
Thanks my only concern is that $\frac{\partial^2 I}{\partial s^2} = -\frac{1}{(1-s)^2}$, so $s_{\delta}$ is the maximum point when we want this point to be a minimum. What's wrong with this logic? – Alex Sep 15 '12 at 2:34
The upper bound is $\exp(-nI)$, with a minus sign, hence the most powerful upper bound is when $I$ is maximum. Let me suggest to draw the graph of $s\mapsto I(s,\delta)$. – Did Sep 15 '12 at 4:29
That's nice. – Did Sep 16 '12 at 12:09 | 828 | 2,357 | {"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} | 3.625 | 4 | CC-MAIN-2016-26 | latest | en | 0.683235 |
https://fr.mathworks.com/matlabcentral/cody/problems/262-swap-the-input-arguments/solutions/262432 | 1,580,058,283,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251690095.81/warc/CC-MAIN-20200126165718-20200126195718-00101.warc.gz | 449,859,097 | 15,648 | Cody
# Problem 262. Swap the input arguments
Solution 262432
Submitted on 17 Jun 2013 by Mandeep Singh
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% [q,r] = swapInputs(5,10); assert(isequal(q,10)); assert(isequal(r,5));
q = 10 r = 5
2 Pass
%% [q,r] = swapInputs(magic(3), 'hello, world'); assert(isequal(q,'hello, world')); assert(isequal(r,magic(3)));
q = hello, world r = 8 1 6 3 5 7 4 9 2
3 Pass
%% [q,r] = swapInputs({}, NaN); assert(isnan(q)); assert(iscell(r) && isempty(r));
q = NaN r = {} | 213 | 633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-05 | latest | en | 0.498872 |
http://mail-archives.apache.org/mod_mbox/lucene-dev/201305.mbox/%3CJIRA.12641676.1365530943467.294213.1368128717711@arcas%3E | 1,524,422,245,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945637.51/warc/CC-MAIN-20180422174026-20180422194026-00377.warc.gz | 196,785,974 | 3,143 | # lucene-dev mailing list archives
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From "John Berryman (JIRA)" <j...@apache.org>
Subject [jira] [Commented] (LUCENE-4922) A SpatialPrefixTree based on the Hilbert Curve and variable grid sizes
Date Thu, 09 May 2013 19:45:17 GMT
```
[ https://issues.apache.org/jira/browse/LUCENE-4922?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel&focusedCommentId=13653095#comment-13653095
]
John Berryman commented on LUCENE-4922:
---------------------------------------
Hmmm... integer representation huh. Well here's a thought then:
As a first got at this idea, let's define something like a geohash where x are interleaved,
but here's how we do it. At the top level, number squares from 0 to 3.
{noformat}
0 --- 1
| |
2 --- 3
{noformat}
At the next level, number thing similarly,
{noformat}
00 -- 01 -- 10 -- 11
| | | |
02 -- 03 -- 12 -- 13
| | | |
20 -- 21 -- 30 -- 31
| | | |
22 -- 23 -- 32 -- 33
{noformat}
Even though this *looks* like the hilbert thing I did above, notice that this is actually
the Z-ordering and it's much easier to compute.
In this case, the first two bits encodes which of the four big boxes the point is in, the
next two bits encodes which of the four sub boxes the point is in, etc. So for example [0.375,
0.625] would be encoded to a depth of 2 by "03" which can be stored in half a byte.
Got it? So... now since we have the original point encoded in z-ordering. We can create a
new hilbert_point algorithm that takes a byte array representing the z-ordering encoding of
a point rather than a 2-vector of doubles. And the code looks much the same except that instead
of the "val[0]/2" we're actually just iterating through the byte array 2 bits at a time (with
This would make for some exquisitely indecipherable code. But ultimately it might not help
that much - it largely depends upon how complex the z-ordering encoding is.
> A SpatialPrefixTree based on the Hilbert Curve and variable grid sizes
> ----------------------------------------------------------------------
>
> Key: LUCENE-4922
> URL: https://issues.apache.org/jira/browse/LUCENE-4922
> Project: Lucene - Core
> Issue Type: New Feature
> Components: modules/spatial
> Reporter: David Smiley
> Assignee: David Smiley
> Labels: gsoc2013, mentor, newdev
>
> My wish-list for an ideal SpatialPrefixTree has these properties:
> * Hilbert Curve ordering
> * Variable grid size per level (ex: 256 at the top, 64 at the bottom, 16 for all in-between)
> * Compact binary encoding (so-called "Morton number")
> * Works for geodetic (i.e. lat & lon) and non-geodetic
> Some bonus wishes for use in geospatial:
> * Use an equal-area projection such that each cell has an equal area to all others at
the same level.
> * When advancing a grid level, if a cell's width is less than half its height. then divide
it as 4 vertically stacked instead of 2 by 2. The point is to avoid super-skinny cells which
occurs towards the poles and degrades performance.
> All of this requires some basic performance benchmarks to measure the effects of these
characteristics.
--
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https://collegemathteaching.wordpress.com/category/topology/page/2/ | 1,569,059,643,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574377.11/warc/CC-MAIN-20190921084226-20190921110226-00073.warc.gz | 423,772,419 | 39,616 | # College Math Teaching
## March 17, 2015
### Compact Spaces and the Tychonoff Theorem IV: conclusion
Filed under: topology — Tags: , , — collegemathteaching @ 9:14 pm
We are finishing up a discussion of the Tychonoff Theorem: an arbitrary product of compact spaces is compact (in the product topology, of course). The genesis of this discussion comes from this David Wright article.
In the first post in this series, we gave an introduction to “compactness”.
In the second post, we gave a proof that the finite product of compact spaces is compact.
In the third post, we gave come equivalent definitions of compactness
In particular, we showed that:
1. A space is compact if and only if the space has the following property: if $A \subset X$ is an infinite union of open sets with no finite subcover, then $A$ is a proper subset of $X$; that is, $X -A \neq \emptyset$ and
2. A space is compact if and only if the space has the following property: every infinite subset $E$ has a perfect limit point. Note: a perfect limit point for a set $E$ is a point $x \in X$ such that, for every open $U, x \in U$, $|U \cap E| = |E|$ (the intersection of every open neighborhood of a perfect limit point with $E$ has the same cardinality as $E$.
Note the following about these two facts: each of these facts promises the existence of a specific point rather than the existence/non-existence of a cover of a particular type. Fact 1 promises the existence of an excluded point, and fact 2 promises the existence of a perfect limit point.
When it comes to a point in an infinite of topological spaces, constructing a point is really like constructing a sequence of points (in the case of countable products) or a net of points (in the case of uncountable products). That is, if one wants to construct a point in an infinite product of spaces, one can assume some well ordering of the index used in the product, then construct the first coordinate of the point from the first factor space, the second coordinate from the second factor space, and so on.
We’ll use fact one: the excluded point property to prove Tychonoff’s Theorem.
Proof. Assume that $X = \Pi_{\alpha \in I} X_{\alpha}$ and that $I$ is well ordered. We start out by showing that the product of two compact spaces is compact, and use recursion to get the general result.
Let $\mathscr{O}$ be an infinite union of open sets in $X_1 \times X_2$ with no finite subcover. First, we show that there is some $a \in X_1$ such that for each open set $U, a \in U$, no finite subcollection of $\mathscr{O}$ covers $U \times X_2$. Now if there is some open $U \subset X_1$ where $U$ is disjoint from every $\pi_1 (O), O \in \mathscr{O}$ we are done with this step. So assume not; assume that every $U$ is a subset of the first factor of some $O \in \mathscr{O}$. If it isn’t the case that there is $x \in X_1$ where $U_x \times X_2$ has no finite subcover of elements of $\mathscr{O}$, for each such $U_x \subset X_1, x \in X_1$ there is a finite number of elements of $\mathscr{O}$ that covers $U_x \times X_2$. Now since $X_1$ is compact, a finite number of $U_x$ covers $X_1$, hence a finite subcover of $\mathscr{O}$ covers ALL of $X_1 \times X_2$. Hence some point $a \in X_1$ exists such that no finite subcover of $\mathscr{O}$ covers $U \times X_2$ for any open $U \subset X_1, a \in U$.
Similarly, we can find $b \in X_2$ so that for all open $V \subset X_2, b \in V$, no finite subcollection of $\mathscr{O}$ covers $U \times V$ where $U$ is a basic open set in $X_1$ that contains $a$. This shows that $(a,b) \notin \cup_{O \in \mathscr{O}}$ because, if it were, this single point would lie in some basic open set $U \times V$ which, by definition, is a finite subcover.
Now given an arbitrary product with a well ordered index set $I$ we can now assume that there is some collection of open sets that lacks a finite subcover and inductively define $a_{\gamma} \in X_{\gamma}$ so that, if $U$ is any basic open set containing $\Pi_{\alpha \leq \gamma} \{a_{\alpha} \} \times \Pi_{\alpha > \gamma} X_{\alpha}$ then no finite subcollection of $\mathscr{O}$ covers $U$. The point $(a_{\gamma})$ thus constructed lies in no $\mathscr{O}$.
Note: if you are wondering why this “works”, note that we assumed NOTHING about the compactness of the remaining product space factors $\Pi_{\alpha > \gamma} X_{\alpha}$.
And remember that we are using the product topology: an open set in this topology has the entire space as factors for all but a finite number of indices. So we only exploit the compactness of the leading factors.
### Compact Spaces and the Tychonoff Theorem III
Filed under: topology — Tags: , , — collegemathteaching @ 2:49 am
We continue on our quest to prove the Tychonoff Theorem: an arbitrary product of compact spaces is compact. We just show that this is true for the FINITE product of compact spaces.
It is our goal to do this by using elementary tools and avoiding things like nets (for example, Willard uses ultranets)
We will basically adding background and commentary to David Wright’s excellent 1994 paper which appeared in the Proceedings of the American Mathematical Society. We will use a bit of cardinal arithmetic and facts about ordinals at times.
Yes, we do need some background, but the background we are providing is necessary for the understanding of any mathematics that uses topology anyway.
Conditions which are equivalent to compactness
1. If $X \subset R^n$ in the usual topology, then $X$ is compact if $X$ is closed and bounded.
Proof: Let $X$ be compact. Then $X$ is closed because the usual topology for $R^n$ is Hausdorff. $X$ is bounded as well, as if it weren’t, the if $x \in X, d(x,0) = M, cover$latex X \$ by $\cup_{x \in X} B_x(\frac{1}{M})$. This open cover has no finite subcover as $M$ is unbounded.
Now let $X$ be closed and bounded. Then $X \subset \Pi^n_{i=1} [a,b]$ for some real $a, b$, which is compact by our “junior” Tychonoff Theorem. So $X$ is a closed subset of a compact set and therefore compact.
2. I’ll call this the excluded point condition: Let $U = \cup_{\alpha \in I} U_{\alpha}$, where each $U_{\alpha}$ is open. We say that $U$ lacks a finite subcover if there is no finite subcollection of the $U_{\alpha}$ that covers U. A topological space $X$ is said to have the excluded point condition if any subset $U$ which has an open cover which lacks a finite subcover must exclude at least one point of $X$; that is, any set which has an open cover with no finite subcover must be a proper subset of $X$.
Example of an open cover which lacks a finite subcover: consider $[0,1]$ as a subset of $R^1$ in the usual topology, and let $U = \cup^{\infty}_{n=3}(\frac{1}{n}, \frac{n-1}{n})$; here $\{0, 1 \}$ are the excluded points from this open cover.
Theorem: a space is compact if and only if it has the excluded point condition for open covers.
Proof. If $X$ is compact then any open cover of $X$ has a finite subcover, hence any subset of $X$ which has an open cover which lacks a finite subcover cannot be all of $X$.
Now assume that $X$ has the excluded point condition. Let $U$ be any open cover of $X$. $U$ cannot lack a finite subcover as any subset which has an open cover lacking a finite subcover must exclude a point of $X$.
3. The finite intersection property condition: let $\mathscr{C}$ be any collection of closed sets. We say that $\mathscr{C}$ has the finite intersection property if the following holds: if the intersection of any finite subcollection of elements of $\mathscr{C}$ is non-empty.
Example: in $R^1$, $\mathscr{C} = \{ [1 - \frac{1}{n}, 1+ \frac{1}{n} ], n \in \{1, 2, ...\} \}$ has the finite intersection property. On the other hand, $\{ [n, n+1], n \in \{..-2, -1, 0, 1, 2, ..\} \}$ does not have this property as there are finite subcollections of this set that have an empty intersection.
Theorem: $X$ is compact if and only if the following holds: if $\mathscr{C}$ is a collection of closed sets with the finite intersection property, then an arbitrary intersection of elements of $\mathscr{C}$ is non-empty.
Proof: Let $X$ be compact. Let $\mathscr{C}$ be an infinte collection of closed sets with the finite intersection property. This means that no finite collection of the complements of these sets can cover $X$ Then $X - \cap_{C \in \mathscr{C}} C = \cup_{C \in \mathscr{C}}(X - C)$ is an open cover of a subset of $X$. Since no finite subcollection of these closed set complements (open sets) can cover all of $X$, then $X - \cup_{C \in \mathscr{C}}(X - C)$ is non-empty and therefore so is $\cap_{C \in \mathscr{C}} C$
Now let the finite intersection property hold for $X$. Let $\mathscr{U}$ be any open cover for $X$. This means that $X-\cup_{U \in \mathscr{U}} U = \cap_{U \in \mathscr{U}} (X -U)$ is empty. Hence the collection of closed sets $\{ (X-U), U \in \mathscr{U} \}$ cannot have the finite intersection property which means that there is some finite subcollection $F \subset \mathscr{U}$ where $\cap_{U \in F} (X-U)$ is empty which means $\cup_{U \in F} U$ covers $X$.
4. Limit point compactness: part I. Theorem: a space $X$ is compact if and only if every infinite subset $E \subset X$ has a limit point.
Note: we can actually prove a bit more; that will be in part II. This is a “junior theorem” which can lead the beginner to understanding the “varsity theorem”.
Proof. Let $X$ be compact and let $E$ be an infinite subset of $X$. Consider: $U_{x \in X} U_x$ where $U_x$ is some open set containing $x$. If $E$ has no limit point we can assume that the $U_x$ are chosen so that each $|E \cap U_x|$ is finite for each $x$. Now a finite subcollection of the $U_x$ covers..say $\cup_{i=1}^k U_{x_i}$ and $E = \cup_{i=1}^k (U_{x_i} \cap E)$. This is impossible as each $|(U_{x_i} \cap E|$ is finite but $E$ is infinite.
Now assume that $X$ is limit point compact in that every infinite subset has a limit point. Let $\mathscr{U}$ be an open cover which has no finite subcover. We assume that this open cover is efficient in that for each $U \in \mathscr{U}, U \not \subset \cup_{V \in \mathscr{U}, V \neq U} V$; that is, any $U$ in the open cover contains at least one point not contained in the union of the other open cover sets. Then the set $x_{U} \in U$ is an infinite set with no limit point.
5. Let $E$ be a set with cardinality $c$. We say that $x$ is a perfect limit point of $E$ if for all open sets $U_x$ containing $x$, $|U_x \cap E| = c$. Example: $[0,1]$ has every point as a perfect limit point (usual topology) as $[0,1]$ has the cardinality of the real numbers and if $U$ is open in $R^1$ and contains any point of $[0,1]$ then $U \cap [0,1]$ has the cardinality of the real numbers.
Now the stronger theorem is this: a topological space $X$ is compact if and only if every infinite subset $E$ has a perfect limit point.
Proof. First, assume that $X$ is compact. Let $E$ be an infinite subset with cardinality $c$. Cover $X$ by open sets $\cup_{x \in X} U_x$ where $x \in U_x$. Suppose that for all $U_x, |U_x \cap E| < c$. Now this open cover of $X$ has a finite subcover $U_1, U_2, ...U_k$ and so we have $E = \cup^k_{i=1} U_i \cap E$ and so $|E| \leq |U_1 \cap E| + |U_2 \cap E| + ...+|U_k \cap E|$ where each $|U_j \cap E| < c$. This is impossible because $c$ is an infinite cardinal (a limit cardinal) and it is impossible to reach a limit cardinal by a finite sum of strictly smaller cardinals.
If you are new to this and are a bit confused, start by assuming that $c$ is, say, the first countably infinite cardinal. ALL lesser cardinals are finite cardinals, and it is impossible for a finite sum of finite cardinals to add up to any infinite cardinal. Then, imagine $c$ being the first uncountable cardinal. One can not reach any uncountable cardinal by the finite sum of countable (or smaller) cardinals (the finite sum of countable cardinals is still countable). That is more or less what is going on here.
Now, suppose that every infinite set has a perfect limit point. Let $\cup_{\alpha \in I} U_{\alpha}$ be an open cover which has no finite subcover. We can assume that $I$ is the index of smallest cardinality for which this is true and that the cover is efficient: $U_{\beta} \not \subset \cup_{\alpha < \beta}U_{\alpha}$ that is, the open subcover is built by adding open sets which contain at least one point not contained by the previously added open sets. Also we put a well ordering on $I$ where the cardinality of $\{ \alpha \in I | \alpha < \beta \} < |I|$. If this confuses you a bit, think of a countable index set where the cardinality of the previous indices are finite, or of an uncountable index set where the smaller cardinals are all countable.
So, for each $\beta$ let $x_{\beta} \in U_{\beta} - \cup_{\alpha < \beta} U_{\alpha}$. Now let $E = \cup_{\alpha \in I} x_{\alpha}$ and note $|E| = |I|$ by design.
Now if $x \in X$, there is some $\alpha < I$ where $x \in U_{\alpha}$ but $|E \cap U_{\alpha}| \leq |I|$ as all $\alpha \in I$ have smaller cardinality than $I$ Therefore $E$ has no perfect limit point.
Again, the person new to topology can run through this with $I$ first being the countable ordinal (and every previous ordinal having finite cardinality) or $I$ being the first uncountable ordinal with every previous ordinal having at most countable cardinality.
We now have the background to give a simple proof of the full strength Tychonoff Theorem, which we will do in the next post.
## March 16, 2015
### Compact Spaces and Tychonoff’s Theorem II
Filed under: advanced mathematics, topology — Tags: , , — collegemathteaching @ 6:10 pm
Ok, now lets prove the following: If $X, Y$ are compact spaces, then $X \times Y$ is compact (in the usual product topology). Note: this effectively proves that the finite product of compact spaces is compact. One might call this a “junior” Tychonoff Theorem.
Proof. We will prove this theorem a couple of times; the first proof is the more elementary but less elegant proof. It can NOT be easily extended to show that the arbitrary product of compact spaces is compact (which is the full Tychonoff Theorem).
We will show that an open covering of $X \times Y$ by basis elements of the form $U \times V$, $U$ open in $X$ and $V$ open in $Y$ has a finite subcover.
So let $\mathscr{U}$ be an open cover of $X \times Y$. Now fix $x_{\beta} \in X$ and consider the subset $x_{\beta} \times Y$. This subset is homeomorphic to $Y$ and is therefore compact; therefore there is a finite subcollection of $\mathscr{U}$ which overs $x_{\beta} \times Y$, say $\cup^{\beta, k}_{i=1} U_{\beta, i} \times V_{\beta, i}$ Note that each $U_{\beta, i}$ is an open set in $X$ which contains $x_{\beta}$ and there are only a finite number of these. Hence $\cap^{\beta k}_{i=1} U_{\beta i} = U_{\beta}$ is also an open set which contains $x_{\beta}$. Also know that $U_{\beta} \times Y \subset \cup^{\beta, k}_{i=1} U_{\beta, i} \times V_{\beta, i}$
We can do this for each $x_{\beta} \in X$ and so obtain an open cover of $X$ by $\cup_{x_{\beta} \in X} U_{\beta}$ and because $X$ is compact, a finite subcollection of these covers $X$. Call these $U_1, U_2, U_3....U_m$. For each one of these, we have $U_j \times Y \subset \cup^{j, k}_{i=1} U_{j, i} \times V_{j, i}$.
So, our finite subcover of $X \times Y$ is $\cup^m_{j=1}\cup^{j, k}_{i=1} U_{j, i} \times V_{j, i}$.
Now while this proof is elementary, it doesn’t extend to the arbitrary infinite product case.
So, to set up such an extension, we’ll give some “equivalent” definitions of compactness. Note: at some point, we’ll use some elementary cardinal arithmetic.
## March 15, 2015
### Compact spaces and Tychonoff’s Theorem I
Note to the reader: when I first started this post, I thought it would be a few easy paragraphs. The size began to swell, so I am taking this post in several bite sized parts. This is part I.
Pretty much everyone knows what a compact space is. But not everyone is up on equivalent definitions and on how to prove that the arbitrary product of compact spaces is compact.
The genesis of this blog post is David Wright’s Proceedings of the American Mathematical Society paper (1994) on Tychonoff’s Theorem.
Since I am also writing for my undergraduate topology class, I’ll keep things elementary where possible and perhaps put in more detail than a professional mathematician would have patience for.
I should start by saying why this topic is dear to me: my research area is knot theory; in particular I studied embeddings of the circle into the 3-dimensional sphere $S^3$, which can be thought of as the “compactification” of $R^3$; basically one starts with $R^3$ and then adds a point $\infty$ and declares that the neighborhoods of this new point will be generated by sets of the following form: $\{ (x,y,z) | x^2 + y^2 + z^2 > M^2 \}$ The reason we do this: we often study the complement of the embedded circle, and it is desirable to have a compact set as the complement.
I’ve also studied (in detail) certain classes of embeddings of the real line into non-compact manifolds; to make this study a bit more focused, I insist that such embeddings be “proper” in that the inverse image of a compact set be compact. Hence compactness comes up again, even when the objects of study are not compact.
So, what do we mean by “compact”?
Instead of just blurting out the definition and equivalent formulations, I’ll try to give some intuition. If we are talking about a subset of a metric space, a compact subset is one that is both closed and bounded. Now that is NOT the definition of compactness, though it is true that:
Given a set $X \subset R^n$, $X$ is compact if and only if $X$ is both closed (as a topological subset) and bounded (in that it fits within a sufficiently large closed ball). In $R^1$ compact subsets can be thought of as selected finite unions and arbitrary intersections of closed intervals. In the higher dimensions, think of the finite union and arbitrary intersections of things like closed balls.
Now it is true that if $f:X \rightarrow Y$ is continuous, then if $X$ is a compact topological space, then $f(X)$ is compact (either as a space, or in the subspace topology, if $f$ is not onto.
This leads to a big theorem of calculus: the Extreme Value Theorem: if $f:R^n \rightarrow R$ is continuous over a compact subset $D \subset R^n$ then $f$ attains both a maximum and a minimum value over $D$.
Now in calculus, we rarely use the word “compact” but instead say something about $D$ be a closed, bounded subset. In the case where $n = 1$ we usually say that $D =[a,b]$, a closed interval.
So, in terms of intuition, if one is thinking about subsets of $R^n$, one can think of a compact space as a domain on which any continuous real valued function always attains both a minimum and a maximum value.
Now for the definition
We need some terminology: a collection of open sets $U_{\alpha}$ is said to be an open cover of a space $X$ if $\cup_{\beta \in I } U_{\beta} = X$ and if $A \subset X$ a collection of open sets is said to be an open cover of $A$ if $A \subset \cup_{\beta \in I } U_{\beta}$ A finite subcover is a finite subcollection of the open sets such that $\cup^k_{i=1} U_i = \cup_{\beta \in I} U_{\beta}$.
Here is an example: $(\frac{3}{4}, 1] \cup^{\infty}_{n=1} [0, \frac{n}{n+1})$ is an open cover of $[0,1]$ in the subspace topology. A finite subcover (from this collection) would be $[0, \frac{4}{5}) \cup (\frac{3}{4}, 1]$
Let $X$ be a topological space. We say that $X$ is a compact topological space if any open over of $X$ has a finite subcover. If $C \subset X$ we say that $C$ is a compact subset of $X$ if any open cover of $C$ has a finite subcover.
Prior to going through examples, I think that it is wise to mention something. One logically equivalent definition is this: A space (or a subset) is compact if every cover by open basis elements has a finite subcover. Here is why: if $X$ is compact, then ANY open cover has a finite subcover, and an open cover by basis elements is an open cover. On the other hand: if we assume the “every open cover by open basis elements has a finite subcover” condition: then if $\mathscr{U}$ is an open cover, then we can view this open cover as an open cover of the basis elements whose union is each open $U_{\beta} \in \mathscr{U}$. This open cover of basis elements has a finite subcover of basis elements..say $B_1, B_2, ....B_k$. Then for each basis element, choose a single $U_i \in \mathscr{U}$ for which $B_i \subset U_i$. That is the required open subcover.
Now, when convenient, we can assume that the open cover in question (during a proof) consists of basic open sets. That will simplify things at times.
So, what are some compact spaces and sets, and what are some basic facts?
Let’s see some compact sets, some non compact sets and see some basic facts.
1. Let $X$ be any topological space and $A \subset X$ a finite subset. Then $A$ is a compact subset. Proof: given any open cover of $A$ choose one open set per element of $A$ which contains said element.
2. Let $R$ have the usual topology. Then the integers $Z \subset R^1$ is not a compact subset; choose the open cover $\cup^{\infty}_{n = -\infty} (n - \frac{1}{4}, n+ \frac{1}{4})$ is an infinite cover with no finite subcover. In fact, ANY unbounded subset $A \subset R^n$ in the usual metric topology fails to be compact: for $a \in A$ with $d(a, 0) \geq n$ choose $B_a(\frac{1}{n})$; clearly this open cover can have no finite subcover.
3. The finite union of compact subsets is compact (easy exercise).
4. If $C \subset X$ is compact and $X$ is a Hausdorff topological space ($T_2$) then $C$ is closed. Here is why: let $x \notin C$ and for every $c \in C$ choose $U_c, V_c$ open where $x \in U_c, c \in V_c$. Now $\cup_{c \in C}V_c$ is an open set which contains $C$ and has a finite subcover $\cup_{i=1}^k V_i$ Note that each $U_i$ is an open set which contains $x$ and now we have only a finite number of these. Hence $x \in \cap^k_{i=1} U_i$ which is disjoint from $\cup_{i=1}^k V_i$ which contains $C$. Because $x$ was an arbitrary point in $X -C$, $X-C$ is open which means that $C$ is closed. Note: this proof, with one minor addition, shows that a compact Hausdorff space is regular ($T_3$) we need only show that a closed subset of a compact Hausdorff space is compact. That is easy enough to do: let $\mathscr{U}$ be an open cover for $C$; then the collection $\mathscr{U} \cup (X-C)$ is an open cover for $X$, which has a finite subcover. Let that be $\cup^k_{i=1} U_i \cup (X-C)$ where each $U_i \in \mathscr{U}$. Now since $X-C$ does not cover $C, \cup^k_{i=1} U_i$ does.
So we have proved that a closed subset of a compact set is compact.
5. Let $R$ (or any infinite set) be given the finite complement topology (that is, the open sets are the empty set together with sets whose complements consist of a finite number of points). Then ANY subset is compact! Here is why: let $C$ be any set and let $\mathscr{U}$ be any open cover. Choose $U_1 \in \mathscr{U}$. Since $X -U_1$ is a finite set of points, only a finite number of them can be in $C$, say $c_1, c_2, ...c_k$. Then for each of these, select one open set in the open cover that contains the point; that is the finite subcover.
Note: this shows that non-closed sets can be compact sets, if the underlying topology is not Hausdorff.
6. If $f: X \rightarrow Y$ is continuous and onto and $X$ is compact, then so is $Y$. Proof: let $\cup_{\beta \in I} U_{\beta}$ cover $Y$ and note that $\cup_{\beta}f^{-1}(U_{\beta})$ covers $X$, hence a finite number of these open sets cover: $X = \cup^{k}_{i=1}f^{-1}(U_i)$. Therefore $\cup^k_{i=1}U_i$ covers $Y$. Note: this shows that being compact is a topological invariant; that is, if two spaces are homeomorphic, then either both spaces are compact or neither one is.
7. Ok, let’s finally prove something. Let $R^1$ have the usual topology. Then $[0, 1]$ (and therefore any closed interval) is compact. This is (part) of the famous Heine-Borel Theorem. The proof uses the least upper bound axiom of the real numbers.
Let $\mathscr{U}$ be any open cover for $[0,1]$. If no finite subcover exists, let $x$ be the least upper bound of the subset $F$ of $[0,1]$ that CAN be covered by a finite subcollection of $\mathscr{U}$. Now $x > 0$ because at least one element of $\mathscr{U}$ contains $0$ and therefore contains $[0, \delta )$ for some $\delta > 0$. Assume that $x < 1$. Now suppose $x \in F$, that is $x$ is part of the subset that can be covered by a finite subcover. Then because $x \in U_{\beta}$ for some $U_{\beta} \in \mathscr{U}$ then $(x-\delta, x + \delta) \subset U_{\beta}$ which means that $x + \delta \in F$, which means that $x$ isn’t an upper bound for $F$.
Now suppose $x \notin F$; then because $x < 1$ there is still some $U_{\beta}$ where $(x-\delta, x+ \delta) \subset U_{\beta}$. But since $x = lub(F)$ then $x - \delta \in F$ and so $[0, x- \delta ) \subset F$. So if $F$ can be covered by $\cup^k_{i=1} U_i$ then $\cup^k_{i=1} U_i \cup U_{\beta}$ is a finite subcover of $[0, x + \delta )$ which means that $x$ was not an upper bound. It follows that $x = 1$ which means that the unit interval is compact.
Now what about the closed ball in $R^n$? The traditional way is to show that the closed ball is a closed subset of a closed hypercube in $R^n$ and so if we show that the product of compact spaces is compact, we would be done. That is for later.
8. Now endow $R^1$ with the lower limit topology. That is, the open sets are generated by basis elements $[a, b)$. Note that the lower limit topology is strictly finer than the usual topology. Now in this topology: $[0,1]$ is not compact. (note: none of $(0,1), [0,1), (0, 1]$ are compact in the coarser usual topology, so there is no need to consider these). To see this, cover $[0,1]$ by $\cup ^{\infty}_{n=1} [0, \frac{n}{n+1}) \cup [1, \frac{3}{2})$ and it is easy to see that this open cover has no finite subcover. In fact, with a bit of work, one can show that every compact subset is at most countable and nowhere dense; in fact, if $A$ is compact in the lower limit topology and $a \in A$ there exists some $y_a$ where $(y_a, a) \cap A = \emptyset$.
## March 12, 2015
### Radial plane: interesting topology for the plane
Filed under: advanced mathematics, topology — Tags: , — collegemathteaching @ 4:14 pm
Willard (in the book General Topology) defines something called the “radial plane”: the set of points is $R^2$ and a set $U$ is declared open if it meets the following property: for all $\vec{x} \in U$ and each unit vector $\vec{u}_{\theta} = \langle cos(\theta), sin(\theta) \rangle$ there is some $\epsilon_{\theta} > 0$ such that $\vec{x} + \epsilon_{\theta} \vec{u}_{\theta} \subset U$
In words: a set is open if, for every point in the set, there is an open line segment in every direction from the point that stays with in the set; note the line segments do NOT have to be of the same length in every direction.
Of course, a set that is open in the usual topology for $R^2$ is open in the radial topology.
It turns out that the radial topology is strictly finer than the usual topology.
I am not going to prove that here but I am going to show a very curious closed set.
Consider the following set $C = \{(x, x^4), x > 0 \}$. In the usual topology, this set is neither closed (it lacks the limit point $(0,0)$ ) nor open. But in the radial topology, $C$ is a closed set.
To see this we need only show that there is an open set $U$ that misses $C$ and contains the origin (it is easy to find an open set that shields other points in the complement from $C$. )
First note that the line $x = 0$ contains $(0,0)$ and is disjoint from $C$, as is the line $y = 0$. Now what about the line $y = mx$? $mx = x^4 \rightarrow x^4-mx = (x^3-m)x = 0$ and so the set $\{(x, mx) \}$ meets $C$ only at $x = m^{\frac{1}{3}}, y = m^{\frac{4}{3}}$ and at no other points; hence, by definition, $R^2 - C$ is an open set which contains $(0,0)$.
Of course, we can do that at ANY point on the usual graph of $f(x) = x^4$; the graphs of such “curvy” functions have no limit points.
Therefore such a graph, in the subspace topology…has the discrete topology.
On the other hand, the lattice of rational points in the plane form a countable, dense set (a line segment from a rational lattice point with a rational slope will intercept another rational lattice point).
So we have a separable topological space that lacks a countable basis: $R^2$ with the radial topology is not metric. Therefore it is strictly finer than the usual topology.
PS: I haven’t checked the above carefully, but I am reasonably sure it is right; a reader who spots an error is encouraged to point it out in the comments. I’ll have to think about this a bit.
## February 16, 2015
### Topologist’s Sine Curve: connected but not path connected.
Filed under: student learning, topology — Tags: , , — collegemathteaching @ 1:01 am
I wrote the following notes for elementary topology class here. Note: they know about metric spaces but not about general topological spaces; we just covered “connected sets”.
I’d like to make one concession to practicality (relatively speaking). When it comes to showing that a space is path connected, we need only show that, given any points $x,y \in X$ there exists $f: [a,b] \rightarrow X$ where $f$ is continuous and $f(a) = x, f(b) = y$. Here is why: $s: [0,1] \rightarrow [a,b]$ by $s(t) = a + (b-a)t$ maps $[0,1]$ to $[a,b]$ homeomorphically provided $b \neq a$ and so $f \circ s$ provides the required continuous function from $[0,1]$ into $X$.
Now let us discuss the topologist’s sine curve. As usual, we use the standard metric in $R^2$ and the subspace topology.
Let $S = \{(t, sin(\frac{1}{t}) | t \in (0, \frac{1}{\pi} \}$. See the above figure for an illustration. $S$ is path connected as, given any two points $(x_1, sin(\frac{1}{x_1}), (x_2, sin(\frac{1}{x_2})$ in $S$, then $f(x) = (x, sin(\frac{1}{x})$ is the required continuous function $[x_1, x_2] \rightarrow S$. Therefore $S$ is connected as well.
Note that $(0,0)$ is a limit point for $S$ though $(0,0) \notin S$.
Exercise: what other limit points does $S$ that are disjoint from $S$?
Now let $T = S \cup \{ (0,0) \}$, that is, we add in the point at the origin.
Fact: $T$ is connected. This follows from a result that we proved earlier but here is how a “from scratch” proof goes: if there were open sets $U, V$ in $R^2$ that separated $T$ in the subspace topology, every point of $S$ would have to lie in one of these, say $U$ because $S$ is connected. So the only point of $T$ that could lie in $V$ would be $(0,0)$ which is impossible, as every open set containing $(0,0)$ hits a point (actually, uncountably many) of $S$.
Now we show that $T$ is NOT path connected. To do this, we show that there can be no continuous function $f: [0, \frac{1}{\pi}] \rightarrow T$ where $f(0) = (0,0), f(\frac{1}{\pi}) = (\frac{1}{\pi}, 0 )$.
One should be patient with this proof. It will go in the following stages: first we show that any such function $f$ must include EVERY point of $S$ in its image and then we show that such a function cannot be extended to be continuous at $(0,0)$.
First step: for every $(z, sin(\frac{1}{z})),$ there exists $x \in (0,\frac{1}{\pi} ]$ where $f(x) = (z, sin(\frac{1}{z}) )$ Suppose one point was missed; let $z_0$ denote the least upper bound of all $x$ coordinates of points that are not in the image of $f$. By design $z_0 \neq \frac{1}{\pi}$ (why: continuity and the fact that $f(\frac{1}{\pi}) = (\frac{1}{\pi}, 0)$ ) So $(z_0, sin(\frac{1}{z_0})$ cuts the image of TS into two disjoint open sets $U_1, V_1$ (in the subspace topology): that part with x-coordinate less than and that part with x-coordinate greater than $x = z_0$. So $f^{-1}(U_1)$ and $f^{-1}(V_1)$ form separating open sets for $[0,\frac{1}{\pi}]$ which is impossible.
Note: if you don’t see the second open set in the picture, note that for all $(w, sin(\frac{1}{w})), w > z_0$ one can find and open disk that misses the part of the graph that occurs “before” the $x$ coordinate $z_0$. The union of these open disks (an uncountable union) plus an open disk around $(0,0)$ forms $V_1$; remember that an arbitrary union of open sets is open.
Second step: Now we know that every point of $S$ is hit by $f$. Now we can find the sequence $a_n \in f^{-1}(\frac{1}{n \pi}, 0))$ and note that $a_n \rightarrow 0$ in $[0, \frac{1}{\pi}]$. But we can also find $b_n \in f^{-1}(\frac{2}{1 + 4n \pi}, 1)$ where $b_n \rightarrow 0$ in $[0, \frac{1}{\pi}]$. So we have two sequences in the domain converging to the same number but going to different values after applying $f$. That is impossible if $f$ is continuous.
This gives us another classification result: $T$ and $[0,1]$ are not topologically equivalent as $T$ is not path connected.
## February 2, 2015
### The challenge of teaching undergraduate topology
Filed under: pedagogy, topology — Tags: — collegemathteaching @ 10:44 pm
I mentioned that I’d be teaching undergraduate topology for the first time ever. Yes, I’ve taught “proof required” courses before: on two separate occasions between 1991 and now. But in each case, I taught a senior level abstract algebra course in which every student came in having taken a course that required proof.
Technically speaking, this course doesn’t require a previous “proof” course.
Then there is how long I’ve been doing this; my first published paper appeared in 1991 (I did the work in 1989). That was well before any student in my class was born!!!
So, this process is going to be a bit like a native language speaker trying to teach someone else a new language.
This means: I’ll have to slow down; this book I’ve chosen won’t be too easy after all. 🙂 I am going to have to back way off of the pace. After all, I am teaching both the material AND proof writing.
## January 25, 2015
### An interesting topological space
Filed under: advanced mathematics, topology — Tags: , — collegemathteaching @ 2:46 pm
I am teaching an undergraduate topology course this semester. While we are still going through basic set theory, I’ve been racing ahead and looking at examples. Yes, Counterexamples in Topology (Steen and Seebach) is an excellent reference. In fact, I have two copies! 🙂
One space that caught my eye is Alexandroff Square. Take the usual closed $[0,1] \times [0,1]$ square in the plane. Now if $(x,y)$ is a non-diagonal point, let a local basis be open segments of the form $\{x \} \times (y - \epsilon, y + \epsilon )$, that is, small open vertical line segments that miss the diagonal. Open sets that include diagonal points $(x,x)$ are open horizontal strips $[0,1] \times (x + \epsilon, x - \epsilon)$ MINUS a finite number of vertical line segments $\{x_i \} \times (x + \epsilon, x - \epsilon)$.
This topological space is compact, Hausdorff, regular (that is, $T_3$) and normal (that is, $T_4$)
(quick reminder: Hausdorff (or $T_2$) means that any two points lie in disjoint open sets, Regular means that a point and a disjoint closed set can be separated by mutually disjoint open sets, and normal means that two disjoint closed sets can be separated by mutually disjoint open sets.)
One unusual aspect (to me) about this topology is how different the open sets are; there should be a way of characterizing this property. In a sense, the collection of open sets isn’t homogeneous.
I’ve decided to play with a simpler example that is based on the Alexandroff square:
consider the closed interval $[-1,1]$ For all $x \neq 0$ use the discrete topology. For $0$ use the entire interval minus any finite set as a local basis. Then one obtains many of the same features of the Alexandroff square, for similar reasons.
So, what do I mean by “different types” of open sets for different points?
This might work: let $x, y \in X$. Now I’d say that the open neighborhoods for $x, y$ are similar if, for every open neighborhood $U_x$ containing $x$ there is a continuous bijection $f|U_x \rightarrow X$ where $f(x) = y$ and $f(U_x)$ is an open neighborhood of $y$. That is, $f|U_x$ is a homeomorphism.
Let’s turn to the Alexandroff square for a minute. If $p$ is not a diagonal point, choose $U_p$ to be vertical open line segment. Now let $q$ be a diagonal element; every open $U_q$ has an open subset which is homeomorphic to the standard 2-disk in the plane (usual topology). Hence no local homemorphism from $U_p$ to $U_q$ can exist.
## January 17, 2015
### Convergence of functions and nets (from advanced calculus)
Sequences are a very useful tool but they are inadequate in some astonishingly elementary applications.
Take the case of pointwise convergence of functions (studied earlier in this blog here and here. )
Let’s look at the very simple example: $f(x) = 0$ for all $x \in R$. Yes, this is just the constant function.
Now consider a set $A$ which consists of all functions $g(x) = 0$ if $x \in F$ where $F$ is some finite set of points, and $g(x) =1$ otherwise. Remember $|F| = k 0$ there exists $m$ such that for all $k \geq m$ we have $|h_k(x) -h(x) | < \epsilon$. Note: the $m$ can vary with $x$.
Now recall that $A \subset R^R$. Now what topology are we using in $R^R$, since this is a product space? For pointwise convergence, we use the product topology in which the open sets are the usual open sets for a finite number of values in $R$ and the entire real line for the remaining values.
If this seems strange, consider the easier case $R^N$ where $N$ represents the natural numbers. Then we can view elements of $R^N$ as sequences, each of which takes a real value. And the open sets here will be the “sequences” $(U_1, U_2, U_3, ....U_k, ...)$ where the $U_i$ are copies of the real line, except for a finite number of indices in which case the $U_i$ can be some arbitrary open set in the real line.
For $R^R$, we index by the real numbers instead of by $N$.
So, in terms of pointwise convergence, this means for every $x \in R$ the $\epsilon_x$ is associated with the open set in the product topology where the real line factor associated with that value of $x$ has the usual real line open sets and the remaining factors of $R$ just have the whole real line; in short, they don’t really matter when figuring out of this function converges AT THIS VALUE OF $x$.
So with the product topology in place, look at our $f(x) = 0$ for all $x$. Given ANY open set $U \subset R^R, f \in U$, we see that $U \cap A \ne \emptyset$. For example, if $h(x) = 1$ for $x \ne 0$, $h(0) = 0$ and $U$ is the open set which corresponds to $(-\delta, \delta)$ in the $0$ factor and the real line elsewhere, then $h \in U \cap A$. So, we conclude that $f$ is in the topological closure of $A$.
But THERE IS NO SEQUENCE IN $A$ which converges to $f$. In fact, it is relatively easy to see that if $g_i$ is any sequence in $A$ and if $g_i \rightarrow g$, then $g$ is zero in at most a countable number of points. That is, if we use the Lebesgue integral, $\int^b_a g(x) dx = b-a.$ (of course, $g$ might not be Riemann integrable).
So sequences cannot reach a point in the closure. For the experts: this shows that $R^R$ in the product topology is not a first countable topological space; that is, its topology has no countable neighborhood basis. This also implies that it is not a metric space (or, more precisely, can’t be made into a metric space).
But I am digressing. The point is that, in situations like this, we want another tool to take the place of a sequence. That will be called a net.
Nets and Directed Sets
Roughly, a net is a “sequence like” thing that can be indexed by an uncountable set. But this indexing set needs to have a “direction like” quality to it. So, what works are “directed sets”.
A directed set is a collection of objects $a_{I}$ with a relation $\preccurlyeq$ that satisfy the following properties:
1) $a_I \preccurlyeq a_I$ (reflexive property)
2) If $a_I \preccurlyeq a_J$ and $a_J \preccurlyeq a_K$ then $a_I \preccurlyeq a_K$ (transitive property)
3) Given $a_I$ and $a_J$ there exists $a_K$ where $a_J \preccurlyeq a_K$ and $a_I \preccurlyeq a_K$ (direction)
Note: though a directed set could be an ordered set (say, $R$ with the usual order relation) or a partially ordered set (say, subsets ordered by inclusion), they don’t have to be. For example: one can form a directed set out of the complex numbers by declaring $w \preccurlyeq z$ if $|w| \leq |z|$. Then note $i \preccurlyeq 1$ and $1 \preccurlyeq i$ but $1 \ne i$.
Now a net is a map from a directed set into a space. It is often denoted by $x_I$ ($I$ is an element in the index set, which is a directed set). So, a real valued net indexed by the reals is, well, an object in $R^R$.
Now given a set $U$ in a topological space, we say that a net $x_I$ is eventually in $U$ if there is an index $J$ such that, for all $K \succcurlyeq J$, $x_K \in U$ and we say that $x_I$ is eventually in $U$. We say that a net $x_J \rightarrow x$ if for all open sets $U, x \in U$ we have $x_J$ is eventually in $U$.
Now getting back to our function example: we CAN come up with a net in $A$ that converges to our function $f$; we merely have to be clever at how we choose our index set though. One way: make a directed set $g_I \in A$ by declaring $g_I \preccurlyeq g_J$ if $g^{-1}_I (0) \subset g^{-1}_J(0)$. Now if we take any neighborhood of $f$ in the product topology, (remember that this consists of the product of a finite number of the usual open set in the real line with an infinite number of copies of the real line), we have elements of this net eventually in this open set, namely the functions which are zero for the values of $R$ that correspond to those open sets. (see here for a couple of ways of doing this)
This demonstrates the usefulness of nets. Note that trying to use a “sequence idea” by just starting with a function that is zero at exactly one point and then going to a function that is zero at two points, three points,…can only get you to a function that is zero at a countable number of points, which is NOT in $A$. That is, one “leaves $A$ prior to getting to where one wants to go, which is a function that is zero at all points of the real line.
On the other hand, a directed set can “start” at an uncountable number of elements of $A$ to begin with and get to being eventually in any basic open set containing $f$ in a finite number of steps. Of course, one must allow for an uncountable number of sequence like paths to get into any of the uncountable number of basic open sets, but each path consists of only a finite number of steps.
## January 16, 2015
### Power sets, Function spaces and puzzling notation
I’ll probably be posting point-set topology stuff due to my being excited about teaching the course…finally.
Power sets and exponent notation
If $A$ is a set, then the power set of $A$, often denoted by $2^A$, is a set that consists of all subsets of $A$.
For example, if $A = \{1, 2, 3 \}$, then $2^A = \{ \emptyset , \{1 \}, \{ 2 \}, \{3 \}, \{1, 2 \}, \{1,3 \}, \{2, 3 \}, \{1, 2, 3 \} \}$. Now is is no surprise that if the set $A$ is finite and has $n$ elements, then $2^A$ has $2^n$ elements.
However, there is another helpful way of listing $2^A$. A subset of $A$ can be defined by which elements of $A$ that it has. So, if we order the elements of $A$ as $1, 2, 3$ then the power set of $A$ can be identified as follows: $\emptyset = (0, 0, 0), \{1 \} = (1, 0, 0), \{ 2 \} = (0,1,0), \{ 3 \} = (0, 0, 1), \{1,2 \} = (1, 1, 0), \{1,3 \} = (1, 0, 1), \{2,3 \} = (0, 1, 1), \{1, 2, 3 \} = (1, 1, 1)$
So there is a natural correspondence between the elements of a power set and a sequence of binary digits. Of course, this makes the counting much easier.
The binary notation might seem like an unnecessary complication at first, but now consider the power set of the natural numbers: $2^N$. Of course, listing the power sets would be, at least, cumbersome if not impossible! But there the binary notation really shows its value. Remember that the binary notation is a sequence of 0’s and 1’s where a 0 in the i’th slot means that element isn’t an element in a subset and a 1 means that it is.
Since a subset of the natural numbers is defined by its list of elements, every subset has an infinite binary sequence associated with it. We can order the sequence in the usual order 1, 2, 3, 4, ….
and the sequence 1, 0, 0, 0…… corresponds to the set with just 1 in it, the sequence 1, 0, 1, 0, 1, 0, 1, 0,… corresponds to the set consisting of all odd integers, etc.
Then, of course, one can use Cantor’s Diagonal Argument to show that $2^N$ is uncountable; in fact, if one uses the fact that every non-negative real number has a binary expansion (possibly infinite), one then shows that $2^N$ has the same cardinality as the real numbers.
Power notation
We can expand on this power notation. Remember that $2^A$ can be thought of setting up a “slot” or an “index” for each element of $A$ and then assigning a $1$ or $0$ for every element of $A$. One can then think of this in an alternate way: $2^A$ can be thought of as the set of ALL functions from the elements of $A$ to the set $\{ 0, 1 \}$. This coincides with the “power set” concept as set membership is determined by being either “in” or “not in”. So, the set in the exponent can be thought of either as the indexing set and the base as the value each indexed value can take on (sequences, in the case that the exponent set is either finite or countably finite), OR this can be thought of as the set of all functions where the exponent set is the domain and the base set is the range.
Remember, we are talking about ALL possible functions and not all “continuous” functions, or all “morphisms”, etc.
So, $N^N$ can be thought of as either set set of all possible sequences of positive integers, or, equivalently, the set of all functions of $N$ to $N$.
Then $R^N$ is the set of all real number sequences (i. e. the types of sequences we study in calculus), or, equivalently, the set of all real valued functions of the positive integers.
Now it is awkward to try to assign an ordering to the reals, so when we consider $R^R$ it is best to think of this as the set of all functions $f: R \rightarrow R$, or equivalently, the set of all strings which are indexed by the real numbers and have real values.
Note that sequences don’t really seem to capture $R^R$ in the way that they capture, say, $R^N$. But there is another concept that does, and that concept is the concept of the net, which I will talk about in a subsequent post. | 13,253 | 46,668 | {"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": 627, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.908328 |
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Part 3: Chapter 25 | Deductive Logic | George William Joseph Stock, M.A. | Lit2Go ETC
# Lit2Go
## Deductive Logic
### by George William Joseph Stock, M.A.
#### Part 3: Chapter 25
• Year Published: 1888
• Language: English
• Country of Origin: England
• Source: Stock, G. W. J. (1888). Deductive Logic. Oxford, England; Pembroke College.
• Flesch–Kincaid Level: 11.0
• Word Count: 1,003
• Genre: Informational
• Keywords: math, math history
PART III.—OF INFERENCES
CHAPTER XXV.
The Disjunctive Syllogism regarded as an Immediate Inference.
770. If no stress be laid on the transition from disjunctive hypothesis to fact, the disjunctive syllogism will run with the same facility as its predecessor into the moulds of immediate inference.
771.
Denial of Antecedent. Subalternation.
Either A is B or C is D, Every case of A not being B
is a case of C being D.
.'. A not being B, C is D. .'. Some case of A not being B
is a case of C being D.
772.
Denial of Consequent. Conversion by Contraposition
+ Subalternation.
Either A is B or C is D. All cases of A not being B
are cases of C being D.
.'. C not being D, A is B .'. All cases of C not being D are
cases of A being B.
.'. Some case of C not being D is
a case of A being B.
773. Similarly the two invalid types of disjunctive syllogism will be found to coincide with fallacies of immediate inference.
774.
Affirmation of Antecedent. Contraposition without
Conversion.
Either A is B or C is D. All cases of A not being B are
cases of C being D.
.'. A being B, C is not D .'. All cases of A being B are
cases of C not being D.
775. The affirmation of the antecedent thus comes under the formula—
All not-A is B, .’. All A is not-B,
a form of inference which cannot hold except where A and B are known to be incompatible. Who, for instance, would assent to this?—
All non-boating men play cricket. .’. All boating men are non-cricketers.
776.
Affirmation of Consequent. Simple Conversion of A.
Either A is B or C is D. All cases of A not being B are
cases of C being D.
.'.C being D, A is not B. .'. All cases of C being D are
cases of A not being B.
777. We may however argue in this way—
Conversion of A per accidens.
Either A is B or C is D. All cases of A not being B
are cases of C being D.
.'. C being D, A is sometimes B. .'. Some cases of C being D are
cases of A not being B.
The men who pass this examination must have either talent or industry.
.’. Granting that they are industrious, they may be without talent. | 784 | 2,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-22 | longest | en | 0.851036 |
https://math.stackexchange.com/questions/840979/distribution-parameters-expectations | 1,643,320,273,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305288.57/warc/CC-MAIN-20220127193303-20220127223303-00112.warc.gz | 437,305,024 | 32,551 | # Distribution, parameters, expectations?
You regularly attend baseball games. At these games, you have two methods for receiving free souvenirs: either by catching a foul ball or by catching a t-shirt thrown by the team mascot. Assume that your success at catching a foul ball is independent of your success at catching a t-shirt. Further, each game you have a probability $p_1$ of catching a foul ball and a probability $p_2$ of catching a t-shirt, independent of all the other games. Let random variable $X$ be the number of games you attend before catching at least one souvenir (e.g. if you got no souvenir the first game, but caught a foul ball the second game, then $X = 2$).
1. What type of distribution does $X$ have, and what are its parameters?
2. If $p_1 = 0.01$ and $p_2 = 0.02$, what is the expected number of games you must attend before you catch at least one souvenir?
The probability of getting a souvenir at a game is $$p=P(\text{ball}\vee\text{t-shirt})=1-(1-p_1)(1-p_2)=p_1+p_2-p_1p_2$$ The probability of getting the first souvenir at game $k$ is $$P(X=k)=P(\text{nothing at k-1 games})P(\text{souvenir})=(1-p)^{k-1}p$$ This is called Negative binomial distribution $NB(1,1-p)$.
The mean ("expected number of games you must attend...") is $$E(X)=\sum_{n=1}^{\infty}n(1-p)^{n-1}p=\frac{1}{p}$$ | 380 | 1,317 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2022-05 | latest | en | 0.875447 |
https://www.americanexpress.com/en-us/credit-cards/credit-intel/how-to-calculate-interest-rates/?linknav=creditintel-article-article | 1,702,248,872,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102697.89/warc/CC-MAIN-20231210221943-20231211011943-00000.warc.gz | 699,752,681 | 60,539 | 6 Min Read | November 06, 2019
# How to Calculate Interest Rates
Calculating interest rate can be complicated and confusing. Here are a few simple steps to calculate interest rate and credit card interest.
This article contains general information and is not intended to provide information that is specific to American Express products and services. Similar products and services offered by different companies will have different features and you should always read about product details before acquiring any financial product.
## At-A-Glance
Interest rates go by different names and are calculated in different ways.
They come in two broad varieties: fixed and variable.
Calculators can help with one of the most complex forms: credit card interest.
Interest can be a funny thing. When you’re collecting it on a bank account or investment, it can make you smile. Not so much when you’re paying it to a lender. But either way, knowing how to calculate interest rates can help you know exactly how much a loan will truly cost, or how much income a particular investment will generate.
The challenge is, while simple interest rate calculations can be easy, in the real world how to calculate interest can get really complicated really fast.
## Interest and Interest Rate Basics
Interest is the amount of money a lender charges you to borrow a set amount of money (the principal). It is the lender’s incentive to lend, and it is customarily quoted as an annual percentage of the principal. Interest can be thought of as the lender’s rate of return—so the more risk the lender expects from a borrower, the higher its incentive typically needs to be. What you, as a borrower, usually must repay is the principal amount plus calculated interest charges.
Interest rates go by many names, including borrowing rate, lending rate, mortgage rate, and lease rate. But whatever the name, interest accumulates based on the stated interest rate of a loan or on the annual percentage rate (APR) of a credit card. By law, the interest rate must be disclosed to consumers when the lending relationship begins.1
## Where Calculating Interest Rates Starts Getting Complex
While interest rates usually are stated on an annual basis (called the nominal interest rate), they are often calculated at different frequencies depending on the terms of the loan. The different frequencies of Interest accrual tend to make real-world interest rate calculations more complicated. Some common frequencies:
• Annually
• Monthly (e.g., mortgages)
• Weekly
• Daily (e.g., credit cards)
For example, a 12 percent nominal interest rate translates to a 1 percent monthly periodic interest rate or a 0.033 percent daily periodic rate (DPR). That DPR is the 12 percent nominal rate divided by either 360 days (called “ordinary interest”) or 365 days (called “exact interest”), again, depending on the borrowing terms.
## So, How Do You Calculate Simple Interest?
There are two main methods of calculating interest charges—simple and compound—and they can result in significant differences in interest costs. The simple interest formula, which usually results in lower overall interest costs, looks like this:
Simple Interest (SI) = Principal (P) x Nominal Rate (R) x Loan Period in Years (T)
The simple interest method multiplies the principal times the nominal interest rate times the number of years the loan will be outstanding.2
What loans typically use the simple interest method?
• Personal loans
• Auto loans
• Some student loans
## Calculating Compound Interest
In the compound interest method, interest accrues on the outstanding principal plus any interest that was not paid during the previous compounding period—in other words, you’re charged interest on interest. The more frequent the compounding periods (represented by “N” in the formula below), the higher the resulting interest charges. Compounding is the most frequent method by which interest is calculated for mortgages, credit cards, and small business loans. The compound interest formula looks like this:5
Compound Interest (CI) = P (1 + R/N) (NT)
For a deeper understanding of compound interest calculations, see “What is Compound Interest?
## Fixed and Variable Interest Rates
The nominal interest rate in either the simple or compound calculations can be fixed or variable. Variable interest rates can further complicate interest rate calculations. Whether it’s fixed or variable is another up-front disclosure required by law. Here are some key differences that distinguish the two:
Fixed Interest Rates
• Stay the same for the entire borrowing term.
• Generally benefit you when interest rates are low.
• Can be a disadvantage when interest rates are high.
• Offer greater certainty that makes household budgeting easier.
Variable Interest Rates
• Fluctuate during the borrowing term at pre-determined intervals (e.g. monthly, quarterly, annually).
• Rise and fall with an associated benchmark, such as the prime rate.
• Can change significantly over time – your loan may end up costing much more or less than expected.
## How to Calculate Interest on Credit Cards
Credit card interest calculations are among the most complicated—they involve everything discussed thus far. Let’s break them down step by step.
Step 1: Understand APR and DPR
The credit card APR (interest rate) is stated on an annual basis, but interest is calculated daily using either the exact DPR (365 days) or the ordinary DPR (360 days), depending on the card issuer. The issuer charges the interest to you on a monthly basis, taking into account the number of days in each month. The APR is usually a variable interest rate that fluctuates based on the prime rate as well as other factors, most notably your creditworthiness and your payment history.
Each time you make a purchase, return, or payment, your outstanding principal changes. This moving outstanding principal goes into the average daily balance (ADB) calculation.6 ADB is determined by adding up the daily balances for each day the DPR is in effect and then dividing by that number of days.
Step 3: Apply the Formula
The formula for calculating monthly credit card interest looks like this:
Interest charged = ADB x DPR x Days the DPR is in effect.
Many online calculators can help you estimate the interest charges for credit cards. Check your card agreement to find the variables you’ll need to input into those calculators.
## Does Credit Card Interest Apply to Me?
For most credit card members, if you pay your full balance each month you can avoid credit card interest payments entirely. But if you pay anything less, such as the minimum payment, you will incur interest on those balances again and also on any previously charged interest. Of course, none of this applies during introductory period of a 0 percent APR credit card.
## The Takeaway
Most American household budgets rely on credit. Understanding interest rates and how they are calculated is one important step in evaluating your financing choices.
Kristina Russo is a CPA and MBA with over 20 years of business experience in firms of all sizes and across several industries, including media and publishing, entertainment, retail, and manufacturing.
All Credit Intel content is written by freelance authors and commissioned and paid for by American Express.
## Related Articles
What is Compound Interest & How is it Calculated?
Understand what compound interest is and how it works. Make interest work for you and grow your finances more quickly.
Tell me more
What is APR and How to Calculate It
Learn what annual percentage rate (APR) is, how to compare different types of APR and how to calculate it.
Tell me more
What Is an Interest Rate?
Interest is the amount of money a lender charges you to borrow, and interest rates are how they calculate how much to charge. Learn more about interest rates and what determines them.
Tell me more
The material made available for you on this website, Credit Intel, is for informational purposes only and intended for U.S. residents and is not intended to provide legal, tax or financial advice. If you have questions, please consult your own professional legal, tax and financial advisors. | 1,618 | 8,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.954289 |
https://math.stackexchange.com/questions/2716063/additional-assumption-needed-to-apply-pythagorean-theorem | 1,571,088,714,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655310.17/warc/CC-MAIN-20191014200522-20191014224022-00257.warc.gz | 581,513,781 | 33,372 | # Additional assumption needed to apply Pythagorean theorem.
Given triangle as given in the below image, assume that angles $BCA, CA'B$ and $AB'C$ have equal measure. Show that $(AC)^2 + (BC)^2 = AB(AB'+ A'B)$. What additional assumption gives the Pythagorean Theorem?
Need first show that if angles $BCA, CA'B$ and $AB'C$ have equal measure, show that $(AC)^2 + (BC)^2 = AB(AB'+ A'B)$.
Consider the $3$ triangles: $BCA, CA'B, AB'C$, with orientation shown in the same cck order, and the common angle in center, with base being respectively $AB, BC, CA$.
It is obvious that only similarity can be worked out here.
Edit : Kindly see my first comment below for doubts about whether only similarity can be worked out here.
*Edit 2: * Using the @GCab hint: triangles $BCA, CA'B$ common angles: $C=A', B=B$ (but, now the 2nd triangle would be listed as $BA'C$ to have the same cck orientation of angles and sides). Unable to find any (need one similar side between common angles, to apply similarity) common sides. Anyway the third angle of both triangles have to be equal to each other. Similarly, for triangles $BCA, CB'A$. Can use two ratios for these two pairs of slr. triangles (i.e., (i)$BCA, BA'C$. , (ii) $BCA, CB'A$) to get : $\frac{AB}{CB} = \frac{CB}{A'B}$, $\frac{AB}{AC} = \frac{AC}{AB'}$. This would lead to two equalities among products to be summed up, to solve the first part easily.
But, what is the intuition gained?
The intuition I expected was in algebraic terms, and not that for second question to be solved $A'=B'$, i.e. there can be only one position on $AB$ that can raise at $90^0$ to meet vertex $A$, even if I am correct.
• The additional assumption is, obviiously, that $\angle BCA=90^\circ$ (which makes $A'=B'$) – Hagen von Eitzen Mar 31 '18 at 13:56
• @HagenvonEitzen But, this would bring more understanding if the first question is completely answered. I have left it complete. Please help in that. Also, am I correct in stating that only similarity can be worked out, it should be case (triangle's dimensions', angles' based) dependent. But, not sure. – jiten Mar 31 '18 at 13:57
• sum of angles $=\pi$ for all the triangles – G Cab Mar 31 '18 at 14:01
• @GCab Using hint: triangles $BCA, CA'B$ common angles: $C=A', B=B$ (but, now the 2nd triangle would be listed as $BA'C$ to have the same cck orientation of angles and sides). Unable to find any (need one similar side betn. cmn. angles, to apply similarity) common sides. Anyway the third angle of both triangles have to be equal to each other. Similarly, for triangles $BCA, CB'A$. Can use two ratios for these two pairs of slr. triangles (i.e., (i)$BCA, BA'C$. , (ii) $BCA, CB'A$) to get : $\frac{AB}{CB} = \frac{CB}{A'B}$, $\frac{AB}{AC} = \frac{AC}{AB'}$. But, what is the intuition gained? – jiten Mar 31 '18 at 14:27
• @GCab Please help with the intuition gained with this question, as the issue is raised in detail in the second Edit of the OP. – jiten Mar 31 '18 at 14:41
The triangles $ACB$, $AB'C$ and $BA'C$ are similar, having two angles (and one edge) in common.
Thus establish the similarity proportionalities $$\left\{ \begin{gathered} \frac{{c_{\,a} }} {b} = \frac{d} {a} = \frac{b} {c} \hfill \\ \frac{{c_{\,b} }} {a} = \frac{d} {b} = \frac{a} {c} \hfill \\ \end{gathered} \right.$$ which then can be developed as follows \eqalign{ & \left\{ \matrix{ ac_{\,a} = bd \hfill \cr bc_{\,b} = ad \hfill \cr dc = ab \hfill \cr} \right. \cr & \left\{ \matrix{ cac_{\,a} = bab \hfill \cr cbc_{\,b} = aab \hfill \cr dc = ab \hfill \cr} \right. \cr & \left\{ \matrix{ cc_{\,a} = b^{\,2} \hfill \cr cc_{\,b} = a^{\,2} \hfill \cr dc = ab \hfill \cr} \right. \cr & c\left( {c_{\,a} + c_{\,b} } \right) = a^{\,2} + b^{\,2} \cr} to obtain the starting relation in your post.
Then to obtain that $c_a+c_b=c$ you shall clearly have that $\gamma = \pi/2$.
• Also, I hope the only mathematical intuition is that of the law of cosines, i.e. $c^2=a^2+b^2-2ab\cos C$, with angle $C$ changing from acute to obtuse, with being a right angle in between. Also, this applying for the special case of $3$ common angles, as shown above. However, the algebraic significance of the similarity of $3$ common angles is unknown still. May be application of ratios applies here, which couples with the law of cosines. – jiten Apr 1 '18 at 9:50
• @jiten: a) the drawing has been realized with Geogebra, which is open-source. b) Sure, you can rework the problem in many related ways, including cosine law, but also sine law etc. c) "..significance of the similarity of $3$ common angles is unknown still" : don't catch what you mean. – G Cab Apr 1 '18 at 11:54 | 1,443 | 4,631 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2019-43 | latest | en | 0.920443 |
http://gmatclub.com/forum/the-shipbuilding-industry-in-eighteenth-century-england-118569.html?fl=similar | 1,485,202,898,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282937.55/warc/CC-MAIN-20170116095122-00569-ip-10-171-10-70.ec2.internal.warc.gz | 113,336,547 | 57,762 | The shipbuilding industry in eighteenth century England : GMAT Sentence Correction (SC)
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The shipbuilding industry in eighteenth century England
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The shipbuilding industry in eighteenth century England [#permalink]
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08 Aug 2011, 13:07
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The shipbuilding industry in eighteenth – century England created a need that pine and flax from Russia be made into masts and sails.
that pine and flax from Russia be made into masts and sails
for pine and flax from Russia that are made into masts and sails
that there be a production of masts and sails out of pine and flax from Russia
that masts and sails are made out of pine and flax from Russia
for pine and flax from Russia to be made into masts and sails
If you have any questions
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Re: SC - 700 level - shipbuilding [#permalink]
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08 Aug 2011, 13:17
bschool83 wrote:
The shipbuilding industry in eighteenth – century England created a need that pine and flax from Russia be made into masts and sails.
that pine and flax from Russia be made into masts and sails
for pine and flax from Russia that are made into masts and sails
that there be a production of masts and sails out of pine and flax from Russia
that masts and sails are made out of pine and flax from Russia
for pine and flax from Russia to be made into masts and sails
My take E .. idiomatic usage case: NEED FOR
B changes the meaning , need is for the process of making, not for pine and flax from Russia THAT are made into masts and sails ...
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08 Aug 2011, 13:49
Definitely E, and doesn't feel like a 700-level question...
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Re: SC - 700 level - shipbuilding [#permalink]
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08 Aug 2011, 20:46
IMO A , Reason subjunctive.
Will explain in detail if answer is correct.
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Re: SC - 700 level - shipbuilding [#permalink]
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08 Aug 2011, 21:47
IMO E....based on only idioms...
As pointed by someone else does not look like a 700 level question unless I am missing something.
_________________
Re-taking GMAT. Hope the charm works this time..
Re: SC - 700 level - shipbuilding [#permalink] 08 Aug 2011, 21:47
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Display posts from previous: Sort by | 1,249 | 4,573 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-04 | latest | en | 0.916807 |
https://listvidz.com/what-does-regrouping-mean/ | 1,675,039,849,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499790.41/warc/CC-MAIN-20230130003215-20230130033215-00627.warc.gz | 380,529,823 | 28,300 | # what does regrouping mean
To regroup means to rearrange groups in place value to carry out an operation. We use regrouping in subtraction, when digits in the minuend are smaller than the digits in the same place in the subtrahend. Here’s how we regroup hundred and tens to subtract 182 from 427.
## Whats the definition of regrouping?
Definition of regroup
transitive verb. : to form into a new grouping regroup military forces. intransitive verb. 1 : to reorganize (as after a setback) for renewed activity. 2 : to alter the tactical formation of a military force.
## What is meant by regrouping in addition?
Answer: Regrouping means that we arrange the numbers according to their place values in order to perform an operation like addition. Regrouping, in addition, is known as carryover and in subtraction, it is known as borrowing.
## What does no regrouping mean?
In this lesson, we are adding numbers without regrouping. This means that when we add the digits in each place value column together, we will not get an answer in each place value column that is larger than 9.
## Is regrouping the same as carrying?
In math, regrouping is the process of making groups of tens when adding or subtracting two digit numbers (or more) and is another name for carrying and borrowing.
## What is another word for regrouping?
Regroup Synonyms – WordHippo Thesaurus.
What is another word for regroup?
rally reassemble
reunite reconvene
reorganiseUK reorganizeUS
gather together again re-form
get together again
## What does regroup mean in business?
to form into a new or restructured group or grouping. to become reorganized in order to make a fresh start: If the plan doesn’t work, we’ll have to regroup and try something else.
## What does it mean to regroup in a relationship?
To come together again in a group. verb.
## How do you explain regrouping in subtraction?
Regrouping in subtraction is a process of exchanging one tens into ten ones. We use regrouping in subtraction when the minuend is smaller than the subtrahend. We use subtraction with regrouping to work out different subtraction problems. For example, Ray buys chocolates worth \$47.
## What is the difference between grouping and regrouping?
As verbs the difference between group and regroup
is that group is to put together to form a group while regroup is to pause and get organized before trying again.
## How do you regroup ones and tens?
Learn to Regroup Ones into Tens
1. Start by writing the equation in column form. …
2. Now add the Ones digits.
3. Then move on to the Tens place value digits:
4. Start by writing the equation in column form:
5. That’s when you regroup.
6. This is also called “carrying the 1 over”:
7. After we carry the 1, we add all of the Tens place digits.
## What is renaming math?
Using renaming to subtract big numbers helps your child to understand what he or she is doing and why, as opposed to simply following a rule of thumb. … It may help your child to understand and use renaming if he or she could use objects such as cubes, counters, buttons, or money.
## What grade do you learn regrouping?
In first grade: Kids join single-digit and double-digit numbers for addition. They also subtract single-digit numbers and 10s. In second grade: Kids work on more complicated addition and subtraction. They also start learning regrouping, or “borrowing.”
## What is a regrouping strategy?
“Regrouping” is defined as the process of making groups of tens when adding or subtracting two digit numbers (or more) and is another name for carrying and borrowing. When first introducing regrouping it’s best to use concrete manipulatives* and relate it to place value.
## What is the opposite of regrouping?
Opposite of to come together again after being dispersed. disperse. disband. separate.
## What’s it called when you meet someone again?
Make (someone) acquainted again. reacquaint. reunite. catch up. get together.
## Is Reunitement a word?
verb (used with or without object), re·u·nit·ed, re·u·nit·ing. to unite again, as after separation.
## What part of speech is regroup?
REGROUP (verb) definition and synonyms | Macmillan Dictionary.
## Is regroup hyphenated?
regroup, re-group re-groupv. i. & t. To reorganize into new groups; to reform a group that has been dispersed. The spelling without the hyphen is more common.
## What does difference mean in math?
Difference is the result of subtracting one number from another. … So, difference is what is left of one number when subtracted from another. In a subtraction equation, there are three parts: The minuend (the number being subtracted from) The subtrahend (the number being subtracted)
## Is taking a step back breaking up?
While there is an implied finality with breaking up, taking a break is essentially making the decision to take a step back from the relationship rather than stepping away, with the goal of allowing each person some much needed perspective and clarity on if and how the relationship can continue.
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There are five main types of chemistry: organic, inorga... | 1,859 | 8,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2023-06 | latest | en | 0.91054 |
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vacuum - a concept widely used in physics and engineering.This word comes from the Latin vacuus, which means "empty".This sense of the word "vacuum" is stored in a general sense the vacuum space is free of substances.In physics and engineering sciences is considered vacuum environment in ... | 658 | 3,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-05 | latest | en | 0.944734 |
https://www.thoughtco.com/how-to-aim-cut-shots-in-pool-and-billiards-367994 | 1,553,140,941,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202484.31/warc/CC-MAIN-20190321030925-20190321052925-00056.warc.gz | 914,745,912 | 30,167 | # How To Aim Cut Shots In Pool And Billiards
01
of 03
### Aim Primer 6: Fraction Aim And Ghost Ball Aim
Continuing our multi-article aim primer, detailing every method top players use to play cut shots successfully in pool and billiards. Part of a multi-article series.
Need a refresher? Catch up on the primer so far in:
### A Quarter Ball Hit
Other convenient fractions for aim include hitting "quarter ball" as in the illustration above. Again the blue box and the red spot show the difference between center ball aim and actual contact point. The cue ball will eclipse one-quarter of the object ball's surface from the player's point of view.
The diagram can be flipped to show a three-quarters hit. Imagine the white ball covering three-quarters of the yellow ball and not one quarter, that is, a half hit plus an additional quarter of the object ball overlapped.
CAUTION: Pool players also use the terms thin and thick (or full) to describe varying degrees of hit. The ¾ hit is full or thick than the ¼ hit, while the ¼ hit is a far thinner hit than a ½ or ¾ ball hit. Thick hits absorb more of the cue ball's momentum into the object ball and are easier to aim than thin hits.
**Next page: Fraction Aim, "Perfect Aim", Edge-To-Edge Aim and Ghost Ball Aim**
02
of 03
### Limitations Of Fraction Aim
What are the limitations of the fraction aim system and how can they be overcome?
### A Fraction System Refinement
Gene Albrecht's "Perfect Aim Billiards" enhances typical fraction methods by placing the "inside" edge of the cue ball with its overlap on the object ball, and the player's eye nearest the object ball directly over this alignment. For example, for a half ball hit sending the object ball to the player's left, as shown in Diagram 10, focus from the cue ball's left edge to the base of the object ball for aim, while positioning the left eye above that line.
A quarter ball hit to the left would require sighting with the left eye from the cue ball's left edge to the point ½ radius from the object ball's left edge. Straight shots (full hits) would require ball edges to be sighted with either eye.
### Limitations of Fraction Aim
1. A half ball hit describes a cut shot angled at a bit less than 30 degrees, and a quarter-ball hit has a cut shot cue ball/object ball angle of 48.6 degrees. But pool is astounding in its complexity with millions of shots offered and at all angles between 0 and 90 degrees. How should a 53½ degree shot be measured, or a 75¾ degree shot? With what fractions of balls eclipsed?
2. Most players have trouble with sighting small fractions of hit. Even those players with sharper than normal eyesight have trouble aiming at one-eighth to one-sixteenth of a ball or less across the length of the table.
### Refining Fraction Aim - "Edge to Edge Aim"
Some billiards pros visualize the specific portions of the balls that will eclipse (or overlap) without naming them as real fractions, knowing that quite subtle degrees of shot are possible. They then "shoot this piece of the cue ball" directly into "that piece of the object ball" rather than worry over discerning ball edges or vanishing points. This method of "piece brushes piece" is called edge to edge aim for cut shots.
### Limitation of Edge to Edge Aim
Just one limitation here. Pros use this method because it works effectively. A high degree of creative visualization skill is needed, however, to be consistent. Other systems amateurs can employ more simply are presented in subsequent pages and articles.
**Next page: Ghost Ball Systems**
03
of 03
### Ghost Ball Systems
Again, if the cue ball and object ball impact along the aim line, as in Diagram 13, the object ball will be driven into the pocket. "Ghost ball" aim is a simple way to imagine the cue ball's arrival on the geometric aim line. Ghost ball is the method most often taught to beginners despite its flaws.
The ghost ball system has long been praised for its simplicity:
1. Plot the aim line, visualizing the ghosted cue ball on that line at impact.
2. Extend a line from the ghost ball's base to the cue ball's base.
3. Position the cue stick on this shot line behind the cue ball and stroke. The cue ball replaces the ghost ball as the player watches to assess their accuracy of aim.
CAUTION: The shot lines of ghost ball aim (Diagram 13) and half ball aim (Diagram 10) are indeed the same line. Marking the cloth or using measuring aids other than a cue stick are illegal in pool so aim systems aid the player in discerning invisible lines.
**Next time: Ghost Ball Aim Limitations And New, Better Systems** | 1,020 | 4,617 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | latest | en | 0.882105 |
https://www.expertsmind.com/library/for-what-wavelength-of-light-will-the-first-order-dark-fring-5245256.aspx | 1,720,780,866,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514387.30/warc/CC-MAIN-20240712094214-20240712124214-00046.warc.gz | 501,808,209 | 14,482 | ### For what wavelength of light will the first-order dark fring
Assignment Help Physics
##### Reference no: EM13245256
Coherent light with wavelength 601nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringe.
For what wavelength of light will the first-order dark fringe (the first dark fringe next to a central maximum) be observed at this same point on the screen?
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One runner covered the 100-m dash in 10.9 s. Another runner came in second at a time of 11.4 s, determine the separation between the two runners when the winner crossed the finish line | 670 | 2,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.909503 |
https://studylib.net/doc/9654430/energy-2 | 1,600,555,830,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400192887.19/warc/CC-MAIN-20200919204805-20200919234805-00434.warc.gz | 651,532,717 | 13,942 | # Energy 2
Chapter 6
Energy
Energy
Universe is made up of matter
and energy.
Energy is the mover of matter.
Energy has several forms:
– Kinetic, Potential, Electrical, Heat,
etc.
Work
Now instead of a force for how long
in time we consider a force for how
long in distance.
Work = Force Distance
W=Fd
The unit for work is the
Newton-meter which is also
called a Joule.
Questions:
How much work is done when a
weight lifter lifts a barbell weighing
1000 Newtons 1.5 meters above the
ground?
How much work is done when a
weight lifter pushes on a stationary
wall with a force of 1000 Newtons
for 15 seconds?
Power
Power is equal to the amount of
work done per unit time.
work done
Power
time interval
The unit for power is the
Joule/second which is also called a
Watt.
Light Bulbs and Appliances
electrical energy used
Power Rating
time interval
How much energy does a 100 Watt
light bulb use in one hour?
How about a 40 Watt light bulb?
Which bulbs shines brighter?
Mechanical Energy
When work is done on an object,
the object generally has acquired
the ability to do work.
This "ability to do work" is called
energy and it has the same units
as work….Joules.
Two Types of Mechanical Energy
– Potential Energy and Kinetic Energy
Potential Energy
The energy that is stored is called
potential energy.
Examples:
– Rubber bands
– Springs
– Bows
– Batteries
– Gravity?
Gravitational Potential Energy
PE = Weight height
PE = m g h
Question:
– How much potential energy does a
10kg mass have relative to the ground
if it is 5 meter above the ground?
Kinetic Energy
Kinetic Energy is the energy of
motion.
Kinetic Energy = ½ mass speed2
1
2
KE mv
2
Question: How much kinetic
energy does a 1kg mass have if it
is moving at 10 meters/second?
Work/Energy Relationship
If you want to move something, you
have to do work.
The work done is equal to the
change in kinetic energy.
Work = DKE
Example Question
When the brakes of a car going
90 km/h are locked, how much
farther will it skid than if the
brakes lock at 30 km/h?
Conservation of Energy
Energy cannot be created or
destroyed; it may be transformed
from one form into another, but the
total amount of energy never
changes.
Demos
– Galileo's incline
– Bowling ball pendulum
– Loop the loop
Example Problem
A 100 kg mass is dropped from rest
from a height of 1 meter.
How much potential energy does it
have when it is released?
How much kinetic energy does it
have just before it hits the ground?
What is its speed just before
impact?
How much work could it do if it were
to strike a nail before hitting the
ground?
100 kg
KE 12 mv 2 0
PE mgh (100kg)(9.8m / s 2 )(1m) 980J
1 meter
100 kg
nail
100 kg
KE 12 mv 2 980 Joules
PE mgh 0 Joules
Work Done Force Distace 980 Joules
Machines - An Application of
Energy Conservation
If there is no mechanical energy
losses then for a simple
machine...
work input = work output
(F
d)input = (F d)output
Examples - levers and tire jacks
Efficiency
work done
Efficiency
energy used
Useful energy becomes wasted
energy with inefficiency.
Heat is the graveyard of useful
energy.
Comparison of Kinetic Energy
and Momentum
Kinetic energy is a scalar quantity.
Momentum is a vector quantity.
Discuss rubber bullets as
Example Questions
A 10 lb weight is lifted 5 ft. A 20 lb weight
is lifted 2.5 ft. Which lifting required the
most work?
(a) 10 lb weight
(b) 20 lb weight
(c) same work for each lifting
(d) not enough information is given to work
the problem
Two cars, A and B, travel as fast as they can to
the top of a hill. If their masses are equal and
they start at the same time, which one does
the most work if A gets to the top first?
(a) A
(b) B
(c) they do the same amount of work
An object of mass 6 kg is traveling at a
velocity of 30 m/s. How much total work
was required to obtain this velocity starting
from a position of rest?
(a) 180 Joules
(b) 2700 Joules
(c) 36 Joules
(d) 5 Joules
(e) 180 N
A 20 Newton weight is lifted 4 meters.
The change in potential energy of the
weight in Newton.meters is
(a) 20
(b) 24
(c) 16
(d) 80
(e) 5 | 1,316 | 4,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2020-40 | latest | en | 0.931899 |
https://www.physicsforums.com/threads/euclidean-geometry-kiselevs-plainimetry-question-242.830179/ | 1,723,183,255,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640759711.49/warc/CC-MAIN-20240809044241-20240809074241-00472.warc.gz | 724,501,771 | 20,441 | # [Euclidean Geometry] Kiselev's Plainimetry Question 242
In summary, the conversation discusses a geometry problem involving tangents and a circle. It is stated that the perimeter of triangle DME and the angle DOE do not depend on the position of point C. Two solutions are presented, one using the concept of bisectors and the other using the properties of kites. It is mentioned that these solutions can be used as a method for similar problems in the future.
## Homework Statement
Two lines passing through a point Μ are tangent to a circle at the points A and B. Through a point С taken on the smaller of the arcs AB, a third tangent is drawn up to its intersection points D and Ε with MA and MB respectively. Prove that (1) the perimeter of DME, and (2) the DOE (where О is the center of the circle) do not depend on the position of the point C.
2. The attempt at a solution
In my first attempt, I imagined the point C sliding on the smaller arc AB like a pendulum and when C is at A or B, the ∠DOE will be half of the ∠AOB. Following the same imagination, because the tangent at C will swipe out equal areas in the s AOM and BOM, the perimeter of the ▲DME will remain constant.
In my second attempt, I followed a hint and proved that ∠DOE is half of ∠AOB and the perimeter of ▲DME is equal to MA+MB. However, I've proved this only when the tangent at C is perpendicular to OM. Even if this proof is all that is required, how shall I prove that the perimeter of ▲DME and the ∠DOE are independent of the position of point C?
Look at the arc AOD and compare it to the arc DOC, since OA and OC are both radii, OD is the bisector of that arc...no matter where C is.
Using this, you should be able to conclude, in the general sense that angle DOE is half of angle AOB, regardless of the location of C. And similarly, perimeter of DME = MA+MB.
RUber said:
Look at the arc AOD and compare it to the arc DOC, since OA and OC are both radii, OD is the bisector of that arc...no matter where C is.
Using this, you should be able to conclude, in the general sense that angle DOE is half of angle AOB, regardless of the location of C. And similarly, perimeter of DME = MA+MB.
Oh, I didn't think about that. If we consider D and E to be points from which tangents are drawn (A and C from D and B and C from E) the quadrilaterals OADC and OCEB will be kites. Therefore, OD will bisect AC and OE will bisect BC at 90 degrees. This will be true no matter where C is.
Similarly, the perimeter of DME has been proven to equal to MA+MB and (still) I can only reason that the ascent on one side will always be equal to the descent on the other side in every other case. I don't know if this is sufficient.
For the perimeter, you can see from the kites that DC = DA and CE=EB so when you add up the sides, you still get MA+MB.
RUber said:
For the perimeter, you can see from the kites that DC = DA and CE=EB so when you add up the sides, you still get MA+MB.
Ah. I see. I feel bad for not being able to solve it on my own.
These problems give you a lot of options and you have to find the best road to get the solution. I had to look at it for about an hour before I saw the right path. Don't feel bad, just add it to your bag of tricks for next time.
## 1. What is Kiselev's Plainimetry Question 242?
Kiselev's Plainimetry Question 242 is a famous problem in Euclidean Geometry that asks for the ratio of the areas of two triangles formed by the intersection of the diagonals of a quadrilateral.
## 2. How do you solve Kiselev's Plainimetry Question 242?
The solution to Kiselev's Plainimetry Question 242 involves using the properties of similar triangles, as well as the formula for the area of a triangle (A = 1/2 * base * height). It can also be solved using algebraic equations and the Pythagorean theorem.
## 3. What are the applications of Kiselev's Plainimetry Question 242?
Kiselev's Plainimetry Question 242 is a classic problem that helps students develop their understanding of geometric concepts and problem-solving skills. It also has real-world applications in fields such as architecture, engineering, and cartography.
## 4. Are there any variations of Kiselev's Plainimetry Question 242?
Yes, there are many variations of Kiselev's Plainimetry Question 242 that involve different types of quadrilaterals or ask for different relationships between the areas of the triangles. These variations can increase in difficulty and provide further practice in applying geometric principles.
## 5. Can Kiselev's Plainimetry Question 242 be solved using other methods?
Yes, there are multiple methods for solving Kiselev's Plainimetry Question 242, such as using trigonometric ratios, coordinate geometry, and even geometric constructions. However, the most common and efficient method involves using the properties of similar triangles.
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lewis140.brittadaudert.com | 1,623,641,634,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487611320.18/warc/CC-MAIN-20210614013350-20210614043350-00002.warc.gz | 349,805,474 | 23,981 | ## How Does A Roulette Table Work?
Posted on June 13, 2021
# How Does A Roulette Table Work?
A Roulette Table, or even more correctly, the Roulette Board, may be the most important part of a Roulette Machine. The Basics. When you place your bets, the wheel is situated on the left side of it, while the area where you place your bets is in the center. The latter is cloth-lined in green, as with all the tables for playing generally, and is sometimes covered with a felt material.
There are basically three colors applied to a roulette table, green, red, and black. Green indicates a win, red indicates a loss, while black implies that the player has lost or hasn’t won yet. The bets are put on the designated areas, and the amount bet depends upon the win or loss that has occurred. There are two forms of bets that a player could make on a roulette table: outside bets and inside bets. Outside bets are put on what is known as the house edge, while inside bets are put on the virtual machine, or the counter.
As mentioned above, there are different colored edges for the wheels, but the actual design of the wheel remains the same. All the variations of the roulette wheel have already been derived from the essential wheel layout that was created in the 18th century. Since that is a vintage design, the 에볼루션 카지노 wheel can happen to be smooth, despite the fact that it’s crafted from many curves and turns. This is because it must withstand the weight that’s placed upon it and the pressure from the person placing their bets on it, so the roulette wheel often appears smoother than it really is.
The layout of the roulette wheel. It begins at the middle with six numbers using one side and eight numbers on the other. These numbers spin around on the inner edges of the wheel. The numbers that spin around are called the numbers that are in a straight line, as the numbers which are curved are called the numbers that come in a non-straight line. The more numbers which are in the straight line, and the fewer curved numbers, the bigger the chances that one will win.
One will eventually lose if they bet once the number of numbers which are in a straight line will equal the total number of numbers that are in a curved line, plus the total number of chips which are in the pot. Therefore, in case a person gets the same amount of chips as the house has on their chips, they will win. However, they will lose if they bet a lot more than the house has on its chips. This can be a main reason why a roulette wheel includes a house edge.
The reason for the house edge is that a great deal of effort is put into the random number generator. An excellent one can predict with seventy percent accuracy whether a player will eventually lose or win. The house edge acts as the statistical offset that keeps the roulette system rougher, in order that a better player could have a better chance of winning. Most players find it difficult to call the bets of the house edge. However, it is important to note that no matter just how much a player bets, he cannot reduce the expected value of his winnings.
In most cases, the more outside bets a player makes, the lower the probability that he will win. For instance, if a player bets two outside bets and wins, that player increase the expected value of his winnings by two. Therefore, the easiest way to reduce the house edge would be to make as few outside bets as you possibly can. However, this reduces the opportunity of winning, as well. If a player bets around fifty cents on a hot hand, it is his ninety percent chance of winning, which makes the result of the game uneven.
Generally, it is best for players to bet the amount they can afford to reduce rather than betting large amounts of money that they can’t afford to lose. A good rule is to bet the amount you can afford to lose, and bet exactly the same amount on three of one’s first five bets. This gives you a concept of what you can afford to lose on any given day. After starting out slowly and gradually, an individual can make his winnings gradually and steadily increase. It is also important to keep an eye on the number wins and losses in order that a person can figure out the frequency with that they occur.
Posted on June 13, 2021
All online casinos offer some sort of online casino bonus to lure people to play there. The word bonus itself covers a wide array of bonuses. Many online casino sites use terms like subscribe bonus, deposit bonus, loyalty club or referral bonus when describing their online casino bonus. These are all different bonuses wanted to new players.
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Finally, it is very important remember that the best casino bonuses are not necessarily wanted to everyone. Before you take part in an online wager game, make sure that you have the money to back it up. Not merely is this vital for making certain you win, but you also want to make sure that you are playing at a satisfactory site. The last thing that you want is to play at a niche site that is not reliable and which might not permit you to play your preferred games.
## MAKING Money At A Casino With Slot Machines
Posted on June 12, 2021
# MAKING Money At A Casino With Slot Machines
Benefit from the best all in one NEVADA casino experience with casino slots that you always wished to try. Casino Saga offers a range of free casino slots with progressive jackpots, new Vegas traditional, old Vegas classic, Blackjack, Roulette, bonus games and much more. It is possible to play anytime from any the main world with free casino slots. Besides free casino slot machines you can also enjoy a range of casino promotions. The promotion is made to provide you with the maximum jackpot promotions, upping your chances of winning big jackpot prizes.
To win big at casino slots, you need to know choosing winning combinations that come up most often. This is one thing that lots of slot players fail to do because they often be determined by luck. There are three forms of slot machine strategies that you should use to increase your probability of winning. These are termed as slot machine game game strategy, casino slots game knowledge, and effective payout schedule or wagering plan.
Slots are played in single player versions and multi-player versions. To win at casino slots, you need to learn to understand that multi-player games have progressive jackpots, and those don’t. To determine, read casino slot machines reviews, study the bonus offers at different casinos, and make the most of any special promotions. Once you play at online casinos, you’ll be able to find information regarding bonus games and wagering plans.
It is possible to increase your chances of winning real cash at casinos by choosing slot machines that offer you generous bonus prizes. You will get great freebies once you play casino games free of charge. The slots with the largest prize tend to be in popular.
In a multi-player slot machine game game, each person in the game contributes different points to the jackpot. The ball player who ends with points after the timer runs out wins the jackpot. The benefit of playing slots with large prizes is that you will win more if you manage to hit the maximum amount of times. Choosing a slot machine that offers a progressive jackpot is preferred for those who desire to make the biggest profits.
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In many casinos today, they use slots to lure customers. Actually, lots of people say that playing slots is better still than betting. Casino goers nowadays enjoy slot machines because they offer a thrilling experience while winning money. In addition they provide good amusement and relaxation following a long day at work. It is no wonder that casinos around the globe are thriving nowadays.
## Reasons Why You Might Vape
Posted on June 12, 2021
# Reasons Why You Might Vape
Vaporizing an e-liquid, or electronic liquid, is simple to do when you know the steps. A vaporizer will produce the vapor you desire. The process is very much like smoking. Just like with a cigarette, but rather of inhaling a chemical (in cases like this nicotine), you burn off an alternative, or e-juice, during your vaporizer.
Most vaporizers are powered by batteries. They can be small electrical devices or larger, battery-operated ones. Batteries need to be recharged periodically. If you do not recharge your batteries regularly, they could not last so long as you would like them to. In general Puff Bar smaller devices have a higher power output than the larger ones.
Vaporizing your liquids through a vaporizer is simple. The vaporizer runs on the heating element to boil the liquid, then it inhales the steam that rises from the boiling liquid. A wide variety of vaporizers are available to meet your requirements. Vapes vary in proportions and cost.
A simple vaporizer should only work with a few herbs in a liquid to produce a good flavor. When you are new to using herbs, get one of these less potent one to begin with. Try several different types until you discover the one that suits your requirements best. You can find vaporizers at many online stores, even eBay.
Lots of people would rather use their vaporizers to create tea, oil, coffee, or other beverages. There are a wide variety of liquids to choose from and they are very affordable. A lot of them use propylene glycol or vegetable oil. They can be found in clear liquid, and in mist form. The liquids are safe to utilize online, however if you opt to buy one you ought to know there are serious dangers involved with the products.
A number of the products include additives that could be harmful to the health. There are also some forms of herbal liquids that may cause allergic reactions. The vaporizer ought to be properly maintained in order to avoid any problems. It should never be left on a hot surface.
You might experience a problem with your vaporizer if it does not take in enough liquid to operate properly. If this happens, you should clean out the pipes that these devices uses to take in the vapor. When cleaning up the pipes, you should never use soap water. Only use tepid to warm water. Do not use any detergent to clean out the device because it can be dangerous. The pipes may burst if that is done incorrectly.
Many devices have become easy to use and so are available in many different sizes. There are some vaporizers that require a grinder to slice the herbs and to produce the final product. These products can be quite expensive and they are easier to buy the normal one that you can use with the original grinder. Most vaporizers have user manuals which have information about how exactly to use and maintain the device.
Vaping is becoming popular and is considered to be a healthier alternative to smoking. When you smoke a cigarette, you are taking in thousands of chemicals into your system. By smoking an herbal vaporizer, you’re keeping these harmful chemicals out of your body. The vaporizer produces smaller amounts of herbal things that are combined into a concentrated liquid. This concentrated liquid is then inhaled by an individual.
Some vaporizers come with their very own containers which fit snugly onto the unit. Additionally, there are reusable jars that are being sold so that the user could have their own way to obtain liquids available to them. These jars have simple on lid and are created from glass. They are made to be strong and durable and are also dishwasher safe. The jars could be washed and reused multiple times.
Most vaporizers come in multiple colors and patterns. The user can choose their personal favorite pattern or logo to go along with their personalized message. The application of this type of personalization has been found to be very effective for many people. Some people prefer to use patterns and logos that assist to relieve stress and promote a wholesome lifestyle.
Another reason some users prefer to vapourize and vapes online is basically because it is an inexpensive alternative. Vaping can be purchased for very little money. This is much cheaper than investing in a normal bottle of the juice. When comparing the amount of money that is spent each month on e juice versus the price of a vaporizer, you will find that vaporizing is a better investment. Many companies sell vaporizers for under \$100. To be able to try this exciting approach to consuming and inhaling, you can purchase a vaporizer and try it for yourself!
## SELECTING E Cigarettes THAT WON’T Cause HEALTH THREATS
Posted on June 12, 2021
# SELECTING E Cigarettes THAT WON’T Cause HEALTH THREATS
The U.S. tobacco industry is fighting a hardcore battle contrary to the rising popularity of electronic cigarettes and their various derivatives like vaporizers, cigar removers and hookahs. Many public health experts are concerned about the possible health effects of long term use of the products. They argue that while smokers may decrease their threat of certain diseases such as lung cancer and heart disease, they also increase their risks of developing a great many other cancers as well. Electronic cigarettes are said to contain things that may cause a number of cancers including: leukemia, nasal cancer, mouth cancer, liver cancer, cervical cancer, kidney cancer and throat cancer. Additionally, there are concerns that users of electronic cigarettes may start to suffer from chronic diseases such as for example diabetes.
It’s been estimated that there are approximately four million ex-smokers who have gone on to create a serious addiction to cigarette smoking. The best way to combat this might be for manufacturers of electric cigarettes to create products that are completely non-toxic. E-juices and hookah pipes certainly are a few examples of these things that may help reduce the harm that nicotine can perform to your body. Actually, nicotine gums have already been developed and are obtainable in vapor form so you no longer need to smoke traditional cigarettes or chew gum to really get your nicotine fix.
Most public health officials believe that e-juices and other vaporizers might help fight smoking cessation efforts in the U.S., nonetheless it is still unclear whether or not they will succeed in the long run. One reason why it is hard for e-juice to take hold in the fight smoking is that it’s difficult to create something that is very like the actual taste of cigarettes. In fact, some cigarette companies have tried to patent their own variants but up to now these have been rejected. This might change with the introduction of huey said.
It is very important remember that smokers who wish to quit will have to search out an alternative solution to nicotine replacement therapy. They will also need to ensure that they don’t really introduce any new tobacco products to their lives. In many ways, that is a lot easier than attempting to stop smoking with nicotine replacement therapies. Associated with that e-juice and other vaporizers can only work if you never smoke another cigarette again which is something that can’t be said for nicotine replacement therapies.
The only method to be completely safe when it comes to vaporizing e-juices along with other nicotine products is to completely avoid them. There exists a lot of doubt that can be done since there are lots of flavored e-juices available and you may easily purchase them. The thing is that these flavored e-juices are created using artificial ingredients and as such they do contain a large amount of artificial substances. This consists of caffeine, propylene glycol along with other harmful chemicals which could prove harmful to the body. It is important to find a Smok Novo product that is manufactured from all natural ingredients. If you fail to find one then it might be best to leave the idea of favoring e-cigs until you will get a good product that you want.
One major concern that folks have about vapors may be the fear that they can replace nicotine. Nicotine is present in every e-cigarette products and the theory that are being pushed by the opponents of e-cigs is that by vaporizing it you are replacing nicotine. This may not be further from the reality. You will be able to continue to enjoy smoking if you want. In the event that you quit using e-cigs you won’t ever be able to smoke again. By quitting you will have replaced just about the most addictive substances known to mankind with a thing that is equally addictive and dangerous.
Actually there is no evidence at all to suggest that e-cigs have long-term health effects. The long-term health effects associated with nicotine are shown to be cancerous and this is the subject that has received many attention in recent years. Additionally it is well-known that long-term cigarette smoking is extremely hazardous for the health and can result in many different cancers along with other health problems over the long-term. It really is strongly advised that if you smoke to cut out as soon as possible. By choosing to use less powerful tobacco you can take advantage of the reduced smoking addiction and stop the dreaded pangs of withdrawal each time you light.
Plus the dangers of nicotine there are other things to consider when you start using these products. There are various health risks associated with used smoking and it is highly recommended you don’t expose yourself or others to secondhand smoke. Many people who have an issue with the smell of tobacco smoke discover that the vapor from vapes can ruin their homes. There are also many health risks to drinking the e-juice also to using the heating element of these devices. These things have to be carefully considered once you choose an electronic cigarette or when you are looking to buy the gear.
## Will Vaporwave E-Liquids Make Me Quit Smoking?
Posted on June 12, 2021
# Will Vaporwave E-Liquids Make Me Quit Smoking?
What’s vaporware juice fast becoming popular for? Many people are embracing Vaporwave Juice for helping them quit smoking cigarettes. While it might not seem as sexy as the new apple or some other gourmet fruit juices that can help you quit smoking, its benefits are undeniable. It tastes great and contains plenty of benefits you won’t find in other vapinger.com juices. So let’s get to know vaporwave juice and what it can do for you personally.
What’s Vaporwave Juice? Vaping juice is actually the liquid used in e cigarettes and makes real hot steam with actual nicotine in it. It still contains nicotine but is significantly less than the amount found in cigarette smoke. The biggest benefit to using Vaporwave Juice over other forms of quitting products is that it tastes great. Many smokers who quit cold turkey find it hard to cope with the taste of nicotine while they are quitting.
What are some of the options available with Vaporwave Juice? There are always a wide selection of different Vaporwave Juice flavors available. While it may appear to be your regular store brand juice, there are actually a large number of different flavors available that could appeal to you personally. If you enjoy fruit flavors, there are various kinds of fruit flavors which you can use, including grape, banana, and even pineapple.
Are there other ingredients used in combination with Vaporwave Juice? One of the most commonly used ingredients with Vaporwave Juice are vegetable glycerin, propylene glycol, and propylene glycol (PE). These three ingredients are combined to create the product we realize today as Vaporwave Juice. There are other ingredients such as for example cinnamon, buttermilk, vanilla, and more, which can all be utilized with Vaporwave E-Liquids.
Any kind of negative effects with Vaporwave Juice? No, you can find not. The thing that may be a concern for a few users is that some of the ingredients may transfer onto the fruit or vegetable in the juice, making them unpleasant to drink. So as to address this, many companies have added ingredients that do not affect the taste of the juice, such as stabilizing agents.
MUST I stop drinking this product? No, you should not stop drinking the juice. Vaporwave juice does not have tobacco in it, but it can still be addicting if you drink it on a regular basis. It is important to understand that Nicotine is really a highly addictive stimulant, and really should not be consumed in virtually any form, shape, or form. If you want a nicotine boost, the easiest method to get one is via an inhaler, lozenge, or sublingual tablet.
How come it taste like coffee when I put it in the tank? One of the reasons why it may take some time to obtain used to is because of the highly concentrated Nicotine content in the product. A lot of Nicotine is included so that you do not feel like you are taking a hit each time you puff. The concentration in the juice also may help with the coffee taste, mainly because that a massive amount caffeine is not needed to make it taste like coffee. That is perfect for a person who is new to smoking, since they usually do not yet have the nicotine content needed to get a high and is a great alternative to smoking.
How can I tell if Vaporwave E-Liquids is good for me? Your best bet to learn if the E-juice flavors are healthy is to try them. If you don’t want to wreck havoc on putting something in your mouth, you can always purchase the standard version and use that. The difference between your two products is simply the way they are packaged and made. You can always go back and buy the upgrade versions of Vaporwave E-Liquids, which gives similar benefits, but are a bit more expensive.
## Slots ARE EXCELLENT casino Game Toys
Posted on June 11, 2021
# Slots ARE EXCELLENT casino Game Toys
It’s amazing if you ask me how many casino slots there are on the planet. In fact, there is a minumum of one slot machine for each and every casino in Las Vegas and almost every major casino outside of Las Vegas. These days you just don’t get into a casino nowadays without bringing a collection of money with you. Whether you are considering something to do when you wait for that special someone or you’re playing on a budget, casinos have grown to be the place to be for entertainment at an inexpensive.
The slots are probably the most popular type of gambling at a casino. Some individuals enjoy the challenge of hitting the reels and developing more money than they devote, but slots offer another exciting game to play. What draws visitors to slots is the possibility to win a prize. Be it free rolls or jackpots which are hard to beat, slots have a way of developing a lot more than your initial entrance fee paid to play.
Slots are designed to fool the player into thinking that the reels are moving faster than they are actually. The faster the reels move the bigger the prize will be. When you look at 엠 카지노 쿠폰 an internal out slot machine, you’ll notice that the reels are lined up horizontally, not vertically like in the case of video slot machines. Because of this the probability of hitting a jackpot are much greater.
Once you look at a video slot machine, you’ll observe that it has both horizontal and vertical bars. As the machine spins the vertical bars make a noise as they pass on the button. The horizontal bars do not have this noise. Once you place your bet on a video slot machine game, your bet will be positioned on a coin that is inserted into the machine. The coin will be rolled and slot it in to the slot machine. The reels begin to move rapidly and soon you will have what is called an absolute ticket.
Video slots have no reels, and that means you cannot tell whenever your bet is a winner until you collect your winnings. It can take many hours before you collect your winnings from these machines. Some people claim that playing slots for money is a waste of money and time, because you will likely lose everything you devote. There are, however, many people who enjoy playing these slots. They may get a few bucks as a result every occasionally, or may end up raking in a huge selection of dollars a month.
Slots can be purchased in almost every casino that’s currently operating. There are many different types of slots. The slot machine that most of the people to play with may be the red lighted slot machine. The jackpot offered in these machines is normally very big and is enough to live on for a number of days.
Black jack slots are also obtainable in casinos. These machines pay out a smaller amount of money, but do not appear to give out a big amount of cash when the wheel is spinning. They are more suitable for players who can’t stand playing with much money. These machines may also be suitable for players who do not want to play an excessive amount of. However, it is more challenging to win with one of these machines, which means that there is a need to be patient with one of these slot machines.
There exists a wide range of different machines that you may play with, depending on which kind of casino you are at. way to make your casino experience better still, then you should consider playing at one of these slot machines. When you play with a slot machine game, you will have fun and could get some money. You may also win a prize from it if you play a certain number of times. Choosing a slot machine to play with can be a great idea to be able to make your casino experience worthwhile.
## BENEFITS OF Playing Free Slots Online
Posted on June 11, 2021
# BENEFITS OF Playing Free Slots Online
Free slots refer to online slot machines, which you can easily play and enjoy free without investing any income. The most prominent slot machine game providers offer these at no cost. The slot machines offering this kind of feature are the same ones that 샌즈 카지노 might be generally in most online casinos but will most likely only be accessible through a free or demo mode. These machines work on a random number generator (RNG) system and the goal is to spin the reels and hope that they will stop on the reels which come up. You do not have the money to place on bets, but since you are not actually betting you have all the likelihood of hitting it big.
This is usually a great way to test the real fun and excitement that the web offers. When you want to test a new casino game you can simply login to the casino and play free of charge. Most of these are based on the traditional slots games where you need to put a spin/stop combination on the reels. With the free slot games you are allowed to do that with almost any casino game including the classic slots games.
Although free online slots games provide a great opportunity to play free of charge, this is often associated with many risks. It is always better to play real money rather than playing free online slots games. It is because:
– It gives you the opportunity to check the real game. Online casinos permit players to play free slots games as a means of getting a feel of the way the online casinos operate. Players can get a feel for the way the games are designed and how the software works. This is useful if you are going to enter the world of online casinos and whether you need to have a risk in these casinos by risking your own money.
– It is possible to feel like you are in charge. If you are among those individuals who play free casino slots for the experience then there is nothing enjoy it. However, if you want to try your luck in real life and you desire to be sure that you won’t lose your money in it then it is best to play real money.
One of many benefits of playing free slots online is that it helps you learn without spending hardly any money. You learn by trying various things in the virtual environment. You may spend nothing and yet it is possible to test different strategies and techniques and ultimately decide what you like and what you do not. This is another reason people play free slots casinos. They would like to know whether they are wasting their amount of time in playing games which don’t have any value for them.
– Another advantage of playing free slots games is that they offer a lot of entertainment. You can choose from a multitude of slots games and you will always find something interesting to do. Playing this way can be good for people who desire to improve their skills. You can start with one reels and gradually move ahead to the next. You need to use the wild symbols to change the reel and thus winning or losing will not have any impact in the game.
– You may make use of the scatter reels in free slot machine game games. Scatter reels are used when you want to increase the quantity of spins in a slot machine game game. The essential idea behind utilizing the scatter reels is that the ball player really wants to gain more points by making as many spins as possible. However, the chances of winning are lower when compared to normal reels so the points gained aren’t that significant.
## WHAT’S Vaping and How Does It Work?
Posted on June 11, 2021
# WHAT’S Vaping and How Does It Work?
What is E-Cigarette? – this is the first question that pops up in our minds whenever we hear about e-cigs. Basically, electronic cigarette is a miniature version of cigarette smoking. It usually consists of a tube, an atomizer, and a protective container like a bottle or cartridge. Rather than tobacco, the user inhards nicotine.
Inhaling the vapor through an electronic cigarettes not merely helps smokers quit the habit of smoking, it also helps them in order to avoid the nasty health ramifications of smoking. Exactly like regular cigarettes, vapors released from electronic cigarettes can affect the respiratory system of the smoker. A lot of the harmful components of tobacco are present in vaporized forms of the drug. Therefore, the electric cigarettes help to protect the lungs from being adversely affected.
How Is the Long-Term Health Effects of Vaping? When the user first inhales the vapor, it can cause coughing, chest tightness and wheezing. The immediate aftereffect of these chemicals is to create a sensation to be full. The chemicals help to suppress the craving of an individual for tobacco cigarettes.
Regardless of the initial effects, there exists a flip side to the flip side of the coin. Once the smoker has stopped inhaling the vapor, he’ll experience the withdrawal symptoms. The longer the smoker remains without e-juices, the more pronounced the withdrawal symptoms can be.
But could it be safe for kids? Many doctors have raised doubts about the safety of the e-juices, especially for teenagers. They argue that vaporizing tobacco will expose young people to the same high levels of tar and nicotine as smokers.
Do you know the Alternatives to Why Use Vaping Products? Children can resort to using natural herbs such as for example SmokeRX or herbal throat sprays to create the Element Vape Coupon e-juices within their own homes. These methods may cause serious lung damage if they are used repeatedly. However, it can be just about the most effective substitutes to cigarette smoking.
It may not be as lethal as tobacco. Even so, long-term tobacco use can lead to many people afflicted with diseases such as lung cancer, bronchitis, emphysema and heart disease. If the vapor will not contain nicotine, it is completely safe. You can find no adverse reactions once the vapors containing no tobacco are inhaled.
What’s Vaping? Many say that we now have very few alternatives from what is smoking cigarettes but not only is it an excellent alternative for the individual who suffers from debilitating health risks when smoking cigarettes. It’s the choice of the average person smoker to obtain the option that’s best for her or him and what is said about the new technology by the experts in this field says Dr. taskiran. It is certainly safer than ingesting it.
This electronic cigarette contains liquid that is heated utilizing a microprocessor. The heating liquid is what gives the user the vapors that could be inhaled in an exceedingly safe fashion. There are numerous types of liquids from which to choose on the market today.
The sort of liquid it is advisable to select is not the only real factor to consider when coming up with the decision to quit. The volume of liquid to use can be an important consideration. There are several different levels of liquid that you can choose from. Many think that e-juice works much better than any liquid.
What is Vaping? Studies have shown that this new technology could be a great way to assist a person who is trying to quit the harmful effects of nicotine. The question still remains concerning whether this e-liquid is actually an alternative to regular cigarettes. It will most definitely help to reduce the cravings the smoker experiences. However, it is very unlikely it will take away the addictive drug that is present in regular cigarettes.
This electronic smoking device may help smokers to lessen their withdrawal symptoms, but it cannot completely remove all the harm that is present in smoke. E-juices containing only 25 percent nicotine can help reduce the symptoms. Lots of people who are struggling to quit tobacco use this product to help them visit any cost. It is very important to avoid inhaling harmful chemicals when quitting.
## Vaporizer Cigarettes
Posted on June 11, 2021
# Vaporizer Cigarettes
Vaporizer cigarettes or e Cigs are now one of the hottest nicotine delivery systems available. They have quickly become extremely popular as they essentially eliminate most of the major health hazards associated with regular tobacco and they also leave behind a number of the more common side effects connected with regular tobacco. This product is a wonderful way to stop smoking without sacrificing your budget or your lifestyle. There are a couple of different methods you can utilize when trying to kick the habit. These products are all available at your local drugstore and may be bought in multiple convenient ways.
One of the most common methods used by most vaporizer cigarettes and electronic smoking devices is through the use of patches. These patches are put on the skin and slowly release a specific quantity of electronic nicotine in to the body. This is often times an extremely slow release rate and requires that the individual place the patch on your skin for an extended time frame. Patches usually require that you use them approximately eight hours before you plan to retire for the night and again after every meal for 14 days straight.
Another option that a lot of vaporizers offer is through the use of a battery. These batteries are designed specifically to create the vapor that is necessary for successful smoking cessation. These kind of vaporizers tend to be more expensive than the previous two options but remain very reasonable. These vaporizers operate by using a battery that must definitely be changed regularly and generally last between four and five hours before having to be recharged.
Many vaporizer cigarettes and electronic smoking devices enable the user to add a variety of different flavors to the liquid that is produced. These flavors not only provide an exceptional degree of taste once you Smok Novo 2 smoke it, however they also tend to cause the smoker’s lungs to begin with to relax and obtain used to the feeling of smoking. This particular aspect of the procedure has been found to be extremely beneficial in the treatment of some types of cancer. Vaping has even been discovered to help alleviate symptoms in those who have been diagnosed with lung cancer.
The final type of vaporizer cigarettes that you will discover available to purchase are the ones that are powered by a heating element. This specific type of vaporizer is the mostly sold product on the market today. These units are generally powered by either a natural heating element or a battery. As the natural heating elements that are associated with these products are typically much less expensive than the batteries that power many of these units, the heating element is normally a lot less efficient. As a way to compensate for this, many manufacturers have designed their very own high-powered heating units that use a microwave or a propane gas ignition system.
The benefit of using vaporizers over traditional cigarettes is that it does not increase the nicotine level within your body like a traditional cigarette would. If you suffer from a nicotine addiction, you then probably know just how difficult it can be to avoid if your body hasn’t gotten familiar with the nicotine levels which you have. With vaporizers, the nicotine and any other harmful toxins are contained in a dilute state. This means that the body will not experience any kind of withdrawal symptoms when you make the switch from traditional cigarettes to vaporizers.
Another difference between your two types of vaporizer cigarettes is that while the e smokes do not need a heating element to operate, some models of vaporizers require a rechargeable battery in order to be used. Many people do not like the thought of having to constantly worry about changing batteries. That said, there are several models available that not require a battery to become used. These electronic cigarettes don’t have a heating element and don’t use a chemical such as nicotine. These electric cigarettes still require a power current to function.
One of the popular electronic cigarette brands may be the V2. The reason why they are so popular is because they are quite efficient at delivering the number of nicotine into your system that you’ll require. Most vaporizers take several puffs for a person to get the same level of nicotine that they receive in one electronic cigarette. The main advantage to these vaporizers is they are much easier on the throat than traditional cigarettes. In addition they do not produce any kind of smoke, so you need not be worried about anybody noticing your smoking. | 8,451 | 42,100 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-25 | latest | en | 0.967465 |
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# Questions tagged [notation]
Use this tag whenever you have a question about what some particular symbol/notation means.
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### Question on interpreting logical notation, relating to alphabets, theoretical comp sci
$\exists x \in \Sigma^* (t=sx)$ Have I interpreted the above into words correctly?: "There exists a symbol 'x', which is a member of the set which contains all possible strings of alphabet sigma, ...
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### On lambda calculus notation: FGa
If we've got this expression: FGa where F and G are functions (as well as a, of course; but let's treat a as a constant). It must be understood that: first apply F taking G as input; then apply the ...
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### Trouble understanding Grammar Rules
I am totaly new to theoretical computer science, so I am sorry if I get the terminology wrong: I was reading the definition of a grammar, what I didn't get was the formal definition of the production ...
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### There is any notation for a language that is empty infinite?
Assume that $L$ is a language, is there any established notation that means that $L$ is infinite or empty?
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### Principles of Programming Languages: Understanding Judgements
I am taking a principles of programming languages class right now and am trying to understand the following judgement form. ...
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### Function notation/mapping for this type of functions?
In short, what is the proper function notation/mapping for this type of functions? From Wikipedia regarding maps: In the communities surrounding programming languages that treat functions as first-...
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### Same notation/terminology for union of sets and concatenation (Kleene star)?
For the union of sets we use the union operator $\cup$ (or $\bigcup$). And for a concatenation (Kleene star) we also use the union operator. The operations are different, but why the same terminology ...
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### Clarification notation in CLRS runtime analysis
I am learning algorithms from the clrs book. I am having trouble understanding the notation the book has placed. The book wrote the following: $t_j$ denote the number of times the ...
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### Meaning of <==> in documentation? [closed]
Title says it all. I just need to confirm that... X.__add__(y) <==> x+y ...doesn't mean anything more than just 'same as' The problem is Google returns zero ...
### Why $\left\lceil lgn \right\rceil <lgn+1\le 2lgn\quad for\quad all\quad n\ge 2$
I have some confusion about 3.2-4 in CLRS. Here is the question : Is the function $\left\lceil \log { n } \right\rceil !$ polynomially bounded? Is the function \$\left\lceil \log { \log { n } } \... | 640 | 2,745 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-25 | latest | en | 0.904427 |
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how do i convert these decimals for trig into a radical
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Author Message
Roiman
Registered: 30.05.2006
From: Weert, Limburg, The Netherlands
Posted: Monday 29th of Oct 17:16 I think God would have been in a really bad mood that he came up with something called math to trouble us! I’ve spent days trying to figure out a solution to this algebra problem which relates to how do i convert these decimals for trig into a radical and I still can’t solve it. I’m particularly having problems with linear equations, radical inequalities and ratios. Can anyone show me the way on how to go about solving such problems? I’ve tried all ways that I could think of, but none helped. I need some urgent help now. Anybody?
IlbendF
Registered: 11.03.2004
From: Netherlands
Posted: Wednesday 31st of Oct 13:38 I have been in your place a few years ago when I was learning how do i convert these decimals for trig into a radical. What part of relations and adding numerators poses more difficulties? Because I am sure that what you really need is a good software to help you figure out the basic concepts and methods of solving the exercises. Did you ever use a software like that? I have tried several of those but I have to say that Algebrator is the best and the easiest to use. It's not like those other programs because it teaches you how to solve, it doesn't just give you the answers .
Sdefom Koopmansshab
Registered: 28.10.2001
From: Woudenberg, Netherlands
Posted: Thursday 01st of Nov 07:54 I must agree that Algebrator is a cool thing and the best software of this kind you can find. I was thrilled when after weeks of anger I simply typed in trigonometric functions and that was the end of my difficulties with algebra . It's also great that you can use the program for any level: I have been using it for several years now, I used it in Pre Algebra and in Intermediate algebra too ! Just try it and see it for yourself!
octeom_man
Registered: 22.05.2002
From: England Nr Europe
Posted: Friday 02nd of Nov 16:00 Is it really that helpful? I’m just concerned because it might not really help because it only solves the problem per ?e. I like to learn how a problem is answered and not only find out the answer. Nevertheless, could you give me a link for this product ?
TC
Registered: 25.09.2001
From: Kµlt °ƒ Ø, working on my time machine
Posted: Sunday 04th of Nov 07:32 Don’t worry my friend . As what I said, it shows the solution for the problem so you won’t really have to copy the answer only but it makes you know how did the software came up with the answer. Just go to this page https://softmath.com/ordering-algebra.html and prepare to learn and solve faster . | 931 | 3,846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2020-10 | latest | en | 0.910628 |
https://www.physicsforums.com/threads/1-to-the-power-of-infinity-why-is-it-indeterminate.966112/ | 1,597,457,433,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740423.36/warc/CC-MAIN-20200815005453-20200815035453-00086.warc.gz | 781,573,250 | 21,796 | # 1 to the power of infinity, why is it indeterminate?
• B
Gold Member
## Main Question or Discussion Point
I've been taught that $$1^\infty$$ is undetermined case. Why is it so? Isn't $$1*1*1...=1$$ whatever times you would multiply it? So if you take a limit, say $$\lim_{n\to\infty} 1^n$$, doesn't it converge to 1? So why would the limit not exist?
PeroK
Homework Helper
Gold Member
I've been taught that $$1^\infty$$ is undetermined case. Why is it so? Isn't $$1*1*1...=1$$ whatever times you would multiply it? So if you take a limit, say $$\lim_{n\to\infty} 1^n$$, doesn't it converge to 1? So why would the limit not exist?
The indeterminate case is ##\lim_{n \rightarrow \infty} a_n^{b_n}## where ##\lim_{n \rightarrow \infty} a_n =1## and ##\lim_{n \rightarrow \infty} b_n =\infty##
##\lim_{n\to\infty} 1^n = 1##
scottdave, Delta2 and fresh_42
fresh_42
Mentor
I can only guess, what the one who said "undetermined" has meant.
##\lim_{n \to \infty} 1^n = 1## is certainly true. But this doesn't allow us to write ##1^\infty##. What does it mean? It is just not defined and as such, undetermined. You read it as ##1^{\mathbb{N}}##, but why isn't it ##1^{\mathbb{R}}##? Fact is, we can only finitely often multiply ##1## by itself, so ##1^\infty## is not defined.
dextercioby
Gold Member
I can only guess, what the one who said "undetermined" has meant.
##\lim_{n \to \infty} 1^n = 1## is certainly true. But this doesn't allow us to write ##1^\infty##. What does it mean? It is just not defined and as such, undetermined. You read it as ##1^{\mathbb{N}}##, but why isn't it ##1^{\mathbb{R}}##? Fact is, we can only finitely often multiply ##1## by itself, so ##1^\infty## is not defined.
But doesn't 1^infty mean that we are multiplying 1 infinite times and hence the final product must be 1.
TachyonLord and Delta2
PeroK
Homework Helper
Gold Member
But doesn't 1^infty mean that we are multiplying 1 infinite times and hence the final product must be 1.
No. ##\infty## is not a number, so ##1^{\infty}## is simply not defined. You can't do anything an infinite numbers of times, which is why limits were developed.
TachyonLord, WWGD, QuantumQuest and 2 others
Delta2
Homework Helper
Gold Member
it would be useful to read again what @PeroK wrote at post #2, it is important to distinguish between the two cases
1) ##1^{\infty}## which is we could write it more formally as ##\lim_{n\to \infty}1^n## and which equals 1
2) ##\lim_{n\to \infty} a_n^{b_n}## with ##\lim a_n=1## and ##\lim b_n=\infty##. This case is different than the case 1) and it is indeterminate. Analogous is the difference where we have ##0\cdot\infty## . It is ##\lim 0\cdot n=0## but also ##\lim \frac{1}{n}##=0 ,##\lim n=\infty## but ##\lim \frac{1}{n}\cdot n=1##.
if we take the logarithm of ##a_n^{b_n}## we will have ##b_n\ln{a_n}## and it would be
##\lim b_n\ln{a_n}=\lim b_n\lim\ln{a_n}=\infty\cdot \ln1=\infty\cdot 0## that is it reduces to the case ##0\cdot\infty##
fresh_42
Mentor
##1^\infty## doesn't tell you what kind of infinity it is, and what to do with the number ##1##. You can write
$$\prod_{n=1}^\infty 1^n = 1$$
but this is an abbreviation of ##\lim_{n \to \infty}1^n = 1##
... which is why limits were developed.
I think one could still agree on ##1^\infty = 1## if everybody agrees on the notation ##1^\infty = \lim_{n \to \infty}1^n## but then in my opinion it will do more harm than good, because of the example @PeroK has given: You cannot substitute limits of sequences in arithmetic expressions if they are not numbers, or the expression isn't defined. ##1^\infty## would suggest that ##\lim_{n \to \infty}a_n^{b_n} = 1## if ##a_n \to 1## and ##b_n \to \infty## which is wrong.
I think this example as in post #2 is the reason why ##1^\infty## should not be used.
Mark44
Mentor
I've been taught that $$1^\infty$$ is undetermined case. Why is it so?
Perhaps you meant the indeterminate form ##[1^\infty]## (written in brackets to emphasize that this is an indeterminate form).
This limit -- ##\lim_{n \to \infty}(1 + \frac 1 n)^n## -- is an example of this indeterminate form. Naively, you might think that since the quantity in parentheses is approaching 1 as the exponent increases without bound, the value of the limit is just 1. This isn't true -- the actual limit is the natural number ##e##.
scottdave and suremarc
lurflurf
Homework Helper
I can only guess, what the one who said "undetermined" has meant.
If you were to guess it would be reasonable to guess the one who said "undetermined" has meant what calculus books mean when they use "undetermined" as it is standard terminology.
$$1^\infty$$
means
$$\lim a^b$$
where
$$\lim a=1$$
and
$$\lim b=\infty$$
You cannot substitute limits of sequences in arithmetic expressions if they are not numbers, or the expression isn't defined.
You have not defined numbers, nor is it a term with a widely agreed upon precise meaning.
Arithmetic expressions are not real numbers would be a more sensible statement.
We can define an arithmetic of limits of sequences.
We can compare this system to real number arithmetic or extended real number arithmetic.
For example by writing
a={a}
b={b}
where a and b are limits of {a} and {b} respectively.
Often
$$\{a\}^{\{b\}}=\{a^b\}=a^b$$
ie
if
{a}=3
{b}=5
$$\{a\}^{\{b\}}=\{a^b\}=3^5=243$$
but not always.
Indeterminate forms are the situations where this does not hold and more detailed analysis is needed.
$$1^\infty$$
is one example.
$$\{(1+1/n)^n\}=e\ne \{(1-1/n)^n\}=1/e$$
are unequal even though
$$\{1+1/n\}=1= \{1-1/n\}$$
To give one example.
What does $1\cdot 1\cdot \ldots$ mean exactly? By definition, the binary operation ##\cdot## is guaranteed to 'work' for finite expressions i.e you can apply it finitely many times and you have something that's well-defined. That's not the case for multiplication over arbitrary families so you have no guarantee that, for example ##\prod _{n\in\mathbb N} 1 ## or ##\prod _{x\in\mathbb R} 1 ## are well-defined.
For all positive x in ℝ, $\lim\limits_{n\to\infty} x^\frac{1}{n} = 1$.
WWGD
Gold Member
2019 Award
You can , in a sense, substitute:
##1^{a_n} = e^{a_nln(1)} ## and then try L'Hopital : ##e^{0/(1/a_n)}##. You need a few details worked out.
The indeterminate case is ##\lim_{n \rightarrow \infty} a_n^{b_n}## where ##\lim_{n \rightarrow \infty} a_n =1## and ##\lim_{n \rightarrow \infty} b_n =\infty##
##\lim_{n\to\infty} 1^n = 1##
That's fascinating!! I first encountered the statement "0x∞ is indeterminate" and never thought twice of that; but actually this is not indeterminate, right?
limn->∞ 0.n = 0
while the real indeterminate expression is this, right?
limn->∞ limm->0 m.n
Learning something new in my old years, yeah!
Hmmm... something not making sense in what I just posted. Specifically... shouldn't we write that n and m go to their limits simultaneously? Like, something like this, maybe (although I never saw this notation, I'm inventing):
limn->∞, m->0 m.n
Or is the "lim" thingie something that we can compound, just like a function? like, is this true?
limn->∞ limm->0 m.n = limn->∞ ( limm->0 m.n ) =
limm->0 ( limn->∞ m.n ) = limm->0 limn->∞ m.n
and all of that is the same as
limn->∞, m->0 m.n ?
suspect that expressions like the one below don't make sense, because it implies the variables are going to their limits separately (it assumes independent variables); there's no such thing as the "limit of two independent variables", right?
limn->∞ limm->0 f(n).g(m)
Therefore I feel I've butchered Mathematics with that notation; I guess the right notation (repeating DeltaK above) is this form:
limn->∞ f(n).g(n)
where limn->∞ f(n) = 0, limn->∞ g(n) = ∞
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WWGD
Gold Member
2019 Award
That's fascinating!! I first encountered the statement "0x∞ is indeterminate" and never thought twice of that; but actually this is not indeterminate, right?
limn->∞ 0.n = 0
while the real indeterminate expression is this, right?
limn->∞ limm->0 m.n
Learning something new in my old years, yeah!
limn->∞ 0.n = 0 this is true only of/when the limit exists, i.e., it is finite.
You can , in a sense, substitute:
##1^{a_n} = e^{a_nln(1)} ## and then try L'Hopital : ##e^{0/(1/a_n)}##. You need a few details worked out.
Isn't the following the actual reason? It's basically the exponential equivalent of dividing by 0 (taking something to the 1/infinitieth power):
For all positive x in ℝ, $\lim\limits_{n\to\infty} x^\frac{1}{n} = 1$.
jbriggs444
Homework Helper
2019 Award
Hmmm... something not making sense in what I just posted. Specifically... shouldn't we write that n and m go to their limits simultaneously? Like, something like this, maybe (although I never saw this notation, I'm inventing):
limn->∞, m->0 m.n
That is a reasonable notation. But now you have to worry about the path which is traversed as n and m each go toward their respective limiting values. If the result can depend on the path then the limit is indeterminate.
limn->∞ limm->0 m.n = limn->∞ ( limm->0 m.n ) =
limm->0 ( limn->∞ m.n ) = limm->0 limn->∞ m.n
Limits do not, in general, commute.
Matt Benesi
WWGD
Gold Member
2019 Award
Isn't the following the actual reason? It's basically the exponential equivalent of dividing by 0 (taking something to the 1/infinitieth power):
Yes, this is a good equivalence. But notice that we know the denominator of the exponent will not, however small, ever by zero, so that the exponent is of the form 0/b with b non-zero. And then we can use continuity of exp.
Yes, this is a good equivalence. But notice that we know the denominator of the exponent will not, however small, ever by zero, so that the exponent is of the form 0/b with b non-zero. And then we can use continuity of exp.
Ok, I'm not following what you're saying. Can you explain it more?
I thought in the case of ℝ, that $1^\infty$ was undefined because for all positive x in ℝ $\lim\limits_{n\to\infty} \, x^\frac{1}{n} \, = \, 1$.
So $\lim\limits_{n\to\infty} \, 2^\frac{1}{n} \, = \,\lim\limits_{n\to\infty} \, 3^\frac{1}{n} \, = \,\lim\limits_{n\to\infty} \, 4^\frac{1}{n} \, = \, b$ so $\lim\limits_{n\to\infty} \, b^n \, = \, 2 \, = \, 3 \, = \, 4$, etc. unless limits are supposed to be specific, so $\lim\limits_{n\to\infty} \, 2^\frac{1}{n} \, != \,\lim\limits_{n\to\infty} \, 3^\frac{1}{n} \,$.
What about ℂ? $1 \, = \, ( i^x )^\frac{4}{x} \, = \,e^{i \, \frac{\pi}{2} \cdot x \cdot \frac{4}{x}}$, for all x in ℝ.... but whats up with the 0 case?
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WWGD
Gold Member
2019 Award
Ok, I'm not following what you're saying. Can you explain it more?
I thought in the case of ℝ, that $\lim\limits_{n\to\infty}1^n$ was undefined because for all positive x in ℝ $\lim\limits_{n\to\infty} \, x^\frac{1}{n} \, = \, 1$.
So $\lim\limits_{n\to\infty} \, 2^\frac{1}{n} \, = \,\lim\limits_{n\to\infty} \, 3^\frac{1}{n} \, = \,\lim\limits_{n\to\infty} \, 4^\frac{1}{n} \, = \, b$ so $\lim\limits_{n\to\infty} \, b^n \, = \, 2 \, = \, 3 \, = \, 4$, etc. unless limits are supposed to be specific, so $\lim\limits_{n\to\infty} \, 2^\frac{1}{n} \, != \,\lim\limits_{n\to\infty} \, 3^\frac{1}{n} \,$.
What about ℂ? $1 \, = \, ( i^x )^\frac{4}{x} \, = \,e^{i \, \frac{\pi}{2} \cdot x \cdot \frac{4}{x}}$, for all x in ℝ.... but whats up with the 0 case?
Sorry if I was confusing, Matt, but we may have to spend some time unwrapping this: are you referring to EDIT ## Lim _{ a_n \rightarrow \infty} 1^{a_n} ## or just the expression ## Lim _{ n \rightarrow \infty} 1^n ##?
Sorry if I was confusing, Matt, but we may have to spend some time unwrapping this: are you referring to ## Lim _{ n \rightarrow \infty} 1^n ## or just the expression ## Lim _{ n \rightarrow \infty} 1^n ##?
Ohhh. I get what you said earlier now (the cases I mentioned weren't specific enough):
You can , in a sense, substitute:
##1^{a_n} = e^{a_nln(1)} ## and then try L'Hopital : ##e^{0/(1/a_n)}##. You need a few details worked out.
WWGD | 3,854 | 11,920 | {"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} | 3.78125 | 4 | CC-MAIN-2020-34 | latest | en | 0.907584 |
https://themeasureofthings.com/results.php?comp=data&unit=kb&amt=717&sort=cnt&p=1 | 1,627,579,642,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153892.74/warc/CC-MAIN-20210729172022-20210729202022-00268.warc.gz | 566,105,619 | 6,527 | Bluebulb Projects presents:
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It's about 0.00000000000000002 times as much as All Spoken Words in Human History (recorded) In other words, 717 kilobytes is 0.000000000000000015 times the amount of All Spoken Words in Human History (recorded), and the amount of All Spoken Words in Human History (recorded) is 67,000,000,000,000,000.000000000000000000 times that amount. (2003 figures) (assumes 16 Khz, 16-bit mono recording)Criticizing a 2002 estimate of 5,600,000,000,000,000.000000000000000000 kilobytes, linguist and University of Pennsylvania professor Mark Liberman asserted that it would actually require 48,000,000,000,000,000,000.0000000000000000000000 kilobytes to house a recording of all speech in human history, even at a relatively low level of quality. For the purposes of his calculations, Liberman estimated the total duration of such a project to include 416,390,367 years of continuous audio. It's about 0.00000000000010 times as much as The Internet In other words, 717 kilobytes is 0.000000000000100 times the amount of The Internet, and the amount of The Internet is 10,000,000,000,000 times that amount. (2005 figures) (estimated)Although the Internet is continuously changing, a 2005 estimate by Google CEO Eric Schmidt was that the total amount of data on the Internet would measure about 5,000,000,000,000,000.000000000000000000 kilobytes. An estimated 1 trillion web pages are published on the Internet, excluding photos, videos, and music content. It's about 0.000000000010 times as much as Mozy In other words, 717 kilobytes is 0.0000000000100 times the amount of Mozy, and the amount of Mozy is 100,000,000,000 times that amount. (2009 figures) (total file storage)Mozy, the online data backup service, stores about 50,000,000,000,000 kilobytes of data backed up its users. Founded in 2005, Mozy's customer base has grown to 1 million personal and 60,000 business subscribers in just 5 years. It's about 0.000000000010 times as much as The Books in the Library of Congress In other words, 717 kilobytes is 0.0000000000100 times the amount of The Books in the Library of Congress, and the amount of The Books in the Library of Congress is 100,000,000,000 times that amount. (2009 figures) (digitized entire collection)The total collection of books, photographs, and other media housed by the United States Library of Congress would occupy about 80,000,000,000,000 kilobytes if fully digitized. The collection contains a total of 142,544,498 items as of 2009. It's about 0.00000000010 times as much as The Letters Delivered by the US Postal Service in 2010 In other words, 717 kilobytes is 0.000000000100 times the amount of The Letters Delivered by the US Postal Service in 2010, and the amount of The Letters Delivered by the US Postal Service in 2010 is 10,000,000,000 times that amount. (2010 figures)All letters delivered by the United States Postal Service in 2010 will equate to 5,000,000,000,000 kilobytes of data if stored digitally. In delivering the mail, the Postal Service fleet travels a total of 1.25 billion miles annually. It's about 0.0000000002 times as much as Netflix's Watch Instantly catalog In other words, 717 kilobytes is 0.000000000208 times the amount of Netflix's Watch Instantly catalog, and the amount of Netflix's Watch Instantly catalog is 4,810,000,000 times that amount. (May, 2013 figures)The uncompressed source copies of all the movies available for instant viewing on Netflix total 3,450,000,000,000 kilobytes. According to estimates, videos streamed from Netflix account for about 29% of all Internet traffic It's about 0.0000000003 times as much as a Human Brain In other words, 717 kilobytes is 0.00000000026 times the amount of a Human Brain, and the amount of a Human Brain is 3,800,000,000 times that amount. According to Northwestern University psychology professor Paul Reber, the capacity of the human brain is a theoretical 2,700,000,000,000 kilobytes. Each lobe of the brain consists of folded neural tissue with a total area, if unfolded, of about 0.24 sq. m. | 1,113 | 4,190 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-31 | latest | en | 0.91238 |
https://www.physicsforums.com/threads/heat-transfer-in-a-nuclear-fuel-rod-with-steel-slabs.749249/ | 1,725,715,822,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00225.warc.gz | 903,442,206 | 22,633 | # Heat Transfer in a Nuclear Fuel Rod with Steel Slabs
• member 428835
Read MoreIn summary, the problem involves a nuclear fuel surrounded by two steel slabs, with heat being generated in the fuel and removed by a fluid at a certain temperature. The steel and fuel have different thermal conductivities and the question asks for an equation for the temperature distribution in the fuel. The heat equation is used for each material and boundary conditions are set at the edges of the slabs. The heat flux at the interface between the fuel and steel is determined by the temperature gradient and thermal conductivity of both materials. The boundary conditions for the fuel are found by analyzing the behavior of the steel plates. The solution involves solving a differential equation for the temperature in the fuel.
member 428835
## Homework Statement
A nuclear fuel of thickness ##2L## has a steel slab to the left and right, each slab of thickness ##b##. Heat generates within the rod at a rate ##\dot{q}## and is removed by a fluid at ##T_{\infty}## (the question doesn't say, but I believe ##T_{\infty}## is temperature of the fluid), which is to the right of the rightmost slab of steel and is convecting by a coefficient ##h##. The other surface (to the left of the leftmost steel slab) is well insulated, and the fuel and steel have thermal conductivities of ##k_f## and ##k_s##, respectively.
Obtain an equation for the temperature distribution ##T(x)## in the nuclear fuel. Express answers in the above variables.
## Homework Equations
the heat equation: $$\frac{\partial T}{\partial t} = \alpha \frac{\partial ^2 T}{\partial x^2} + \dot{q}$$
## The Attempt at a Solution
nothing is said about time, thus we may assume ##\frac{\partial T}{\partial t} = 0##. Additionally, we are only transferring heat in 1-dimension.
Before trying to solve anything, i believe i will have to use the heat equation for each of the 3 materials (the leftmost steel slab ##T_l##, the nuclear fuel in the middle ##T_m##, and the rightmost slab of steel ##T_r##).
First, i'll try prescribing boundary conditions, assuming the center of the fuel is ##x=0##, the end of the far right steel slab is ##x=L+b## and the end of the far left steel slab is ##x=-L-b##.
$$T_l '(-L-b) = 0$$
I'm just not sure how to deal with the other boundary conditions? My professor said something about, at ##x=L+b## we have that ##-k_s T_r'(x) = hA(T_r-T_{\infty})## where I think ##A=b## since we are in 1-dimension. Can someone explain this relation?
Also, what about ##x=L,-L##? Can't we prescribe some kind of flux similarity, such as ##T_l'(-L) = T_m'(-L)## and ##T_m'(L) = T_r'(L)##?
Similarly, can't we also say ##T_l(-L) = T_m(-L)## and ##T_m(L) = T_r(L)##?
Please help! I've been working very hard and think I just need a little push in the correct direction.
Thanks everyone!
joshmccraney said:
nothing is said about time, thus we may assume ##\frac{\partial T}{\partial t} = 0##. Additionally, we are only transferring heat in 1-dimension.
First, i'll try prescribing boundary conditions, assuming the center of the fuel is ##x=0##, the end of the far right steel slab is ##x=L+b## and the end of the far left steel slab is ##x=-L-b##.
$$T_l '(-L-b) = 0$$
OK so far!
I'm just not sure how to deal with the other boundary conditions? My professor said something about, at ##x=L+b## we have that ##-k_s T_r'(x) = hA(T_r-T_{\infty})## where I think ##A=b## since we are in 1-dimension. Can someone explain this relation?
If you have a steady temperature distribution in the system, the heat flux out of the slab must be the same as the internal heat generation. You correctly said there is no heat lost at the ##x=-L-b## boundary.
##hA(T_r-T_{\infty})## looks like Newton's law of cooling, and it is the heat flux out of the steel at ##x = L+b##, where ##A## is the area of the free surface. The question seems to imply this is a "large" slab of material and you can ignore the edge effects, so you might as well consider a piece of the slab with unit area.
Also, what about ##x=L,-L##? Can't we prescribe some kind of flux similarity, such as ##T_l'(-L) = T_m'(-L)## and ##T_m'(L) = T_r'(L)##?
You are right to think about the heat flux either side of the boundary between the fuel and the steel, but the flux depends on the temperature gradient and the thermal conductivity of the two materials.
There is no heat generation in the steel plates. What does that tell you about the heat flux through the steel, and the "shape" of the temperature distribution through the steel, as a function of ##x##?
When you have got a clear idea what happens in the steel plates (and you don't really need differential equations to solve that part of the question) you will have the boundary conditions for the fuel.
You will need to write a differential equation for the temperature in the fuel, and solve it.
Note: I'm not sure what ##\dot q## means in the question. It could be the heat generated per unit volume of the fuel, or the total heat generated in the slab. It doesn't make much difference to the thought process in the question either way.
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1 person
joshmccraney,
You did a nice job of analyzing the problem, and AlephZero correctly pointed out that you needed to include the thermal conductivities in matching the heat fluxes at the interfaces.
Regarding the equation ##-k_s T_r'(x) = hA(T_r-T_{\infty})## :
There shouldn't be an A on the right hand side of this equation. The units on both sides of the equation have to match, and they don't with the A present.
Chet
1 person
AlephZero said:
You are right to think about the heat flux either side of the boundary between the fuel and the steel, but the flux depends on the temperature gradient and the thermal conductivity of the two materials.
I'm not too sure what to do here. are you referring to fourier's law of heat transfer ##\vec{q} = -k \nabla T##, or in our 1-dimensional case, ##q = -k \frac{dT}{dx}##? If so, it seems that at each boundary of ##\pm L## we have that ##-k_f \frac{dT_m}{dx} = -k_s \frac{dT_i}{dx}## for the ##i^{th}## steel slab?
AlephZero said:
There is no heat generation in the steel plates. What does that tell you about the heat flux through the steel, and the "shape" of the temperature distribution through the steel, as a function of ##x##?
ahh, you're being clever, if i understand you correct. it would seem as though ##\frac{dT_l}{dx}=\frac{dT_m}{dx}=0## when ##x=-L## since we are in steady state. thus, no heat will transfer out, so ##T## is constant along ##[-L-b,-L]##. is this correct? As for the shape, it seems the temperature will decrease from left to right. do you agree?
AlephZero said:
When you have got a clear idea what happens in the steel plates (and you don't really need differential equations to solve that part of the question) you will have the boundary conditions for the fuel.
so, if the above reasoning is correct, our boundary conditions are: $$T'(-L)=0$$ $$k_f T_m'(L) = k_s T_r'(L)$$ $$-k_s T_r'(L+b) = h(T_r - T_{\infty})$$ but don't we need a B.C. for the temperature profile (not only derivatives)?
the equation we will use is first $$-T_r''(x) = \dot{q}$$ and next is $$-T_m''(x) = \dot{q}$$ and it seems we actually don't need the ##T_l## profile (if the above logic is correct)?
Chestermiller said:
Regarding the equation ##-k_s T_r'(x) = hA(T_r-T_{\infty})## :
There shouldn't be an A on the right hand side of this equation. The units on both sides of the equation have to match, and they don't with the A present.
Chet
good call! of course, what are the units associate with ##h##? i know ##A## has m^2 and ##T## has kelvins. I'm just confused why area is present at all? is it because the rate of cooling is proportional to surface area and the temperature gradient, which in this case is ambient temp minus surface temp? thus, ##hA(T_r-T_{\infty})## takes units Watts per square meter?
thanks! please let me know if this is correct.
thinking about this question more (after alephzero's and chet's comments) do you two think the heat is being generated throughout the entire fluid? if so, isn't temperature constant from ##[-L-b,L]## since the generation is taking place continuously, constantly, and everywhere through ##[-L,L]##?
joshmccraney said:
thinking about this question more (after alephzero's and chet's comments) do you two think the heat is being generated throughout the entire fluid? if so, isn't temperature constant from ##[-L-b,L]## since the generation is taking place continuously, constantly, and everywhere through ##[-L,L]##?
You are correct to say that the temperature from ##[-L-b,L]## is constant, because no heat is flowing through this region at steady state. This is going to be at the highest temperature in your system. So you might as well take the insulated boundary to be at -L; it will make no difference in the calculated temperature profiles to the right of -L.
Chet
thanks for the help guys! i think i have a solution, but can one of you comment on this brief analysis: ##\dot{q} + k_f T''(x) = 0## implies ##\dot{q}## takes units Watts per cubic meter. However, for my energy balance I have that ##\dot{q} = \frac{k_s}{b}A(T_m-T_r)##. how do i get rid of the area term? my professor said we have ##\dot{q}2L = \frac{k_s}{b}(T_m-T_r)## but i don't see how these units work out. how can we divide by area and multiply the left side by length?
nevermind, i see that the q term is my flux. thanks to you both!
## 1. What is heat transfer in a nuclear fuel rod with steel slabs?
Heat transfer is the process of thermal energy being transferred from one object or substance to another. In the case of a nuclear fuel rod with steel slabs, heat is generated by the nuclear reaction in the fuel rod and is transferred to the surrounding steel slabs.
## 2. How does heat transfer occur in a nuclear fuel rod with steel slabs?
Heat transfer in this system occurs through three main mechanisms: conduction, convection, and radiation. Conduction is the transfer of heat through direct contact between the fuel rod and steel slabs. Convection is the transfer of heat through the movement of fluids, such as coolant, around the fuel rod. Radiation is the transfer of heat through electromagnetic waves.
## 3. What factors affect heat transfer in a nuclear fuel rod with steel slabs?
The rate of heat transfer in this system can be affected by several factors, including the temperature difference between the fuel rod and steel slabs, the material properties of the fuel rod and steel slabs, the geometry of the system, and the presence of any insulating materials.
## 4. How is heat transfer in a nuclear fuel rod with steel slabs important?
Heat transfer is a crucial aspect of nuclear reactor design and operation. In a fuel rod with steel slabs, efficient heat transfer is necessary to prevent the fuel rod from overheating and potentially melting, which could lead to a nuclear meltdown. Therefore, understanding and accurately predicting heat transfer in this system is essential for ensuring the safe and efficient operation of a nuclear reactor.
## 5. How is heat transfer in a nuclear fuel rod with steel slabs studied and analyzed?
Scientists use various methods, such as mathematical modeling and experimental studies, to study and analyze heat transfer in a nuclear fuel rod with steel slabs. These methods allow for the prediction and optimization of heat transfer processes in this system, leading to advancements in nuclear reactor design and safety.
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1K | 2,970 | 11,748 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-38 | latest | en | 0.925845 |
https://podcast.mathisfigureoutable.com/1062400/10441274-ep-96-if-not-the-subtraction-algorithm-then-what?t=0 | 1,685,322,968,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644574.15/warc/CC-MAIN-20230529010218-20230529040218-00035.warc.gz | 502,645,211 | 16,985 | # Ep 96: If Not the Subtraction Algorithm, Then What?
April 19, 2022 Pam Harris Episode 96
Math is Figure-Out-Able with Pam Harris
Ep 96: If Not the Subtraction Algorithm, Then What?
It's time for subtraction! In this episode Pam and Kim name and give examples of the four main subtraction strategies students need to own to solve any subtraction problem that is reasonable to solve without a calculator.
Talking Points:
• Remove a Friendly Number
• Remove a Friendly Number - Over
• Remove to a Friendly Number
• Constant Difference
• How do students choose which one to use?
Pam Harris:
Hey fellow mathematicians, welcome to the podcast where math is Figure-Out-Able! I'm Pam.
Kim Montague:
And I'm Kim.
Pam Harris:
And we make the case that mathematizing is not about mimicking steps or rote memorizing facts. But it's about thinking and reasoning; about creating and using mental relationships. Y'all, we take the strong stance that not only are algorithms not particularly helpful in teaching, but that mimicking algorithms actually keeps students from being the mathematicians they can be. We answer the question, if not algorithms and step by step procedures, then what?
Kim Montague:
And we are answering that right now. For
Pam Harris:
Yeah, everybody's favorite, right?the last couple of weeks, we've been in the middle of a series
Kim Montague:
Yeah, everybody's favorite.
If not algorithms, then what? Last
Pam Harris:
Not so much. In fact, Kim, true story. The week we tackled addition and shared the four major addition strategies that we believe students need to own and have experience with. This week,we're going to tackle subtraction.second workshop I ever did with elementary teachers - I had done tons of professional learning workshops with secondary teachers, it was mostly centered around graphing calculators -but the very first one that I did for elementary teachers, or the second one I did, first one was kind of prescribed and I had to do certain things. But the second one I did was like me,you know, like, hey, come work with our teachers. And I said,"Great, happy to do that. What are your pain points? You know,what do you want me to do?" And they said, "Oh, so glad you
subtraction across rows."And I was like, "Absolutely, we will do that. What does that mean?" I had no idea. Like, what is that? As a secondary teacher that didn't even make any sense to me. So subtraction is a big deal. Like it's a big pain point for teachers. And let's talk about subtraction today. What are, and we're gonna say there are four, what are the four major subtraction strategies that students need to own, that we need to build in students so that we can give them then any subtraction problem that's reasonable to solve without a calculator, and they will be successful and fluent and efficient.
Kim Montague:
You ready for me to name them?
Pam Harris:
Yes.
Kim Montague:
Okay. All right,we've got 'Remove a Friendly Number'. We've got 'Remove a Friendly Number Over', 'Remove to a Friendly Number', and'Constant Difference'. And now everyone owns those right?
Pam Harris:
You got them!Alright, we're done. Thanks for joining us today. It doesn't quite work that way. Like, our names are not important. Other people have different names. If you're familiar with Cathy Fosnot's work, you can see the influence of her work on my work. I give her a lot of credit with her Landscapes of Learning and helping me sort of ferret these out. But I think in a huge way, we've taken sort of a step further from that and said, what are the major ones that kids need to own in order for them to be where we can give any problem that's reasonable without a calculator, and they can be successful and fluent and efficient. So you might have noticed that those four strategies might have sounded familiar. If you have not listened to last week's episode,you might want to go listen to the prior episode to this before you listen to this one, because there's going to be a lot of similarities. And we're going to kind of build on that. We're going to build from last week's addition episode into today.We're not going to maybe go as in-depth in some things, because we can build on those. So for these subtraction strategies,Kim, our problem of the day that we're going to use is 71 minus37.
Kim Montague:
Okay.
Pam Harris:
I'll bet you a Dairy Queen blizzard that you just wrote that problem down. Am I right?
Kim Montague:
Yes! Shoot.
Pam Harris:
We're just trying to make - no this is a good thing.We're just trying to make the point that mental math does not mean you did it all in your head. Mental math means you do it with your head, I just quoted Cathy Fosnot there. It is perfectly okay for you to not have to hold that in your head.You can free up that working memory so that you can actually think about what you want to think about, not just hold numbers. Totally cool, just made that little point right there.Alright, I'm gonna start. Would you please solve that problem?No, you have me start because then I want you to do the over one.
Kim Montague:
Okay. Alright. So you start by Removing a Friendly Number.
Pam Harris:
Okay, so I'm going to Remove a Friendly Number.Remove a Friendly Number. So I'm going to start at 71. And I'm going to remove. I'm going to subtract. I'm going to minus 37.Yuk! I'm going to minus 30. I'm gonna remove or subtract 30first. I can do that. 71 minus30. No problem, that's 41. 71minus 30 is 41. I've removed a friendly 30. But I was supposed to remove 37. So I've got to remove seven more. I'm gonna subtract off seven more. So I'm at 41. I got to subtract seven more. I'm thinking about that as one to 40 and six more to 34. So41 minus seven is 34. So yeah,so the entire problem 71 minus37, remove 30 first, then get rid of the seven, and I land on34.
Kim Montague:
And like we said last week, if you are familiar with some of the strategies, you could easily pick up a pencil,because that's what you write with a pencil -
Pam Harris:
Or a pen! I got a pen in my hand, do you literally have a pencil?
Kim Montague:
Yes.
Pam Harris:
I have a pen,always.
Kim Montague:
And you could model what Pam and I are saying to give yourself some more practice, just modeling thinking.
Pam Harris:
And you might look at the model that you just drew,and were your jumps the same height? I mentioned that because we have somebody right now helping us do some computer work where we are turning hand drawn models into computer generated models. And some of the bigger jumps like a longer jump are taller than the shorter jumps.No, don't do that. Make your jumps the same. Like all the jumps are the same height,they're just shorter and longer,not taller. That's not important. Okay. So that was'Remove a Friendly Number'. Kim.
Kim Montague:
Yep.
Pam Harris:
I would like you to solve the same problem. 71 minus37. But this time, I want you to Remove a Friendly Number Over.
Kim Montague:
This is good for me. Okay, so, I need to subtract37. But I don't want to do that.So I'm going to subtract 40.Because 37 is close to 40. And71 minus 40 is 31. But I subtracted too much. And I asked myself how much too much. And because I was only supposed to subtract 37 I subtracted three too much. So now I'm going to add three back on to 31 to get34.
Pam Harris:
Nice. Those of you that are practicing modeling,take a look at your models right now. Are your 71s lined up from the first strategy to the second one? Are your 34s lined up from the first strategy to the second one? Bam. That's a thing to look for. Because we want to represent the strategies in such a way that the relationships are a bit more visible, they're a bit more apparent. So we want the number lines to be in relation. Okay, so not what this episode is about. Sorry, moving on. Okay, so we've Removed a Friendly Number, we've Removed a Friendly Number Over now let's talk about 'Remove to a Friendly Number'. Can you keep going?Because I want to do the next one.
Kim Montague:
Sure. That's fine.Alright. So Remove to a Friendly Number. I'm starting at 71. And I need to remove 37. But I'm going to remove just one to get to 70. And then I still need to remove 36. And actually, it's not really maybe fair to the listeners, because I play 'I Have, You Need' a lot. And so I know what the next hop is. So I'm going to do that. Can I just make the next...?
Pam Harris:
Say it and then maybe we can pick back up.
Kim Montague:
So then because I need to remove 36 still, that I know that I'm going to land on34. Because I know the partner of 70 and 36.
Pam Harris:
The partner of 36 to70, you're saying, is 34. Like,
Kim Montague:
No but I've played with 100 a lot and with 10s and you play 'I Have, You Need' with70?20s a lot. So I feel like any10s number, any multiple of 10for me is not a bad combo.
Pam Harris:
Hmm, that's interesting. Alright, I'll work on that. Because then I can say I'm like you. Because I want to go up and be like Kim. Alright,totally cool. Alright, so'Remove to a Friendly Number'was all about starting with that first number, the 71. And then not removing a big friendly something, but removing to that friendly 70 and then getting rid of the rest of it. So that could have looked like, if it was more like me, once you remove the one to get to 70 and you have to remove 36, you could have removed 30 to get to 40 and then remove the six to get to the 34.That could have been how you do it.
Kim Montague:
And I think this is why we name it by the first move that you make, because what you just described was 'Remove to a Friendly Number'. And then when you removed 30 that was kind of 'Removing a Friendly Number'. Which is awesome,right? If you own both, then you can you them up.
Pam Harris:
Kim Montague:
Constant Difference. Okay, so if I put 37and 71, on a number line, I can find the distance between those.But I'm actually going to shift those down one. And so I'm going to shift the 71 down to 70. And in order to maintain the same distance between the numbers,I'm going to shift the 37 down to 36. And now I'm just going to find the distance between 36 and70. And I think I did that again, because I just said, I just know the distance between those is 34. Otherwise, I would have made a different shift a different way maybe.
Pam Harris:
Let's talk about the different shifts in just a second. So by shifting both of those numbers, and you kept the distance between them the same,it's like you created an equivalent problem that was easier to solve.
Kim Montague:
Yep. Because 71minus 37 is equivalent to 70minus 36.
Pam Harris:
Yeah, that sounds really interesting. So you could then make a choice. Do you want to solve the problem 71 minus37? Or do you want to solve an equivalent problem 70 minus 36?And you say, 70 minus 36, you got that. And I say, I don't want to solve either of those.So you mentioned that there was another way to shift, can I tell you which way I would shift?
Kim Montague:
Sure, yeah.
Pam Harris:
I would like to have a subtraction problem, where the second number in the subtraction problem is the nicer number. You made the first number nice, I want to make the second. You made the 71 nice, I'm gonna make the 37 nice.
Kim Montague:
Okay.
Pam Harris:
So on my number line, I've got 37 and I've got71. But I'm going to shift them both to the right three.
Kim Montague:
The right up a number line?
Pam Harris:
Up the number line.So I'm going to shift the 37 to40. And the 71 to 73. And create the equivalent problem -
Kim Montague:
Pause, pause,pause.
Pam Harris:
Okay, hang on... Oh,I was looking at the wrong number line. I'm going to shift them both up three. So I'm going to shift the 37 to 40 and the 71to 74. Alright, it's real, real live. Here we are recording as we go. So what I've turned now is the problem 71 minus 37, I've turned it into the equivalent 74minus 40. Bam. That is a problem I would prefer to solve. So I guess I could solve 70 minus 36.But not my first inclination, my first inclination is definitely to shift both up three and create the equivalent problem 74minus 40.
Kim Montague:
I like.
Pam Harris:
Yeah. And we would consider that the most sophisticated subtraction strategy. That's a little fun.Hey, Kim, I'm just acknowledging that if we were to give either of us a subtraction problem, I'm wondering if you have a go to.Like, if I were to give you a different subtraction problem, I don't know, like 82 minus 59, or something like that, what's your inclination? Are you like, all of these are the same? You might like flip a coin, and you're like, ah, I mean, they're all -no?
Kim Montague:
No, I really enjoy doing an over subtraction. So I would do 82 minus 60 to get 22.And then add back one to get to23.
Pam Harris:
And I would not, but you're the over-girl right now.I own over subtraction. I do.And I could do it with the best of them. But it's not my first inclination. And I'm okay with that. Because I own them all.Ya'll, that's the thing. Once we own them all, once we help students really develop all these, then we let them choose.Then we're like, "Hey, if you own those, I'm good, whichever one." And so for me, I would tend to look at the difference,the distance, between 59 and 82.And I would like to shift that to create the distance between60 and 83. I've shifted both up one, and now I have the problem83 minus 60. And that just, I can't not see 23 for that problem.
Kim Montague:
You know what,though? You just said, "Once I own them, it's what I want to choose." But I think I would also wonder in myself, if there's a reason why I don't choose Constant Difference if I never did. So I recognize that in myself, I do choose it sometimes. Because I do want to let the numbers influence. And then I think there have been other strategies for other operations where I've said to myself, "Oh, I almost never use that strategy." So I'm going to force myself for a while to use it so that it's as easily accessible as readily accessible.
Pam Harris:
And once we own them, then don't you and I both find that we tend to lean towards the most sophisticated strategy?
Kim Montague:
Yes.
Pam Harris:
Because they're more efficient. Yeah. And cool, and slick, and often easier to do.Though, there are times where if I don't have something to record my thinking, there are times I might do a less sophisticated strategy, because I can hang on to it in my working memory a little bit easier. It's a bit more sequential, which also means I can kind of hang on to the relationships when I can't write stuff down. So it's another reason why we want kids to own them all. So that given the circumstances, given the situation they could choose,they have the power to say,"Ooh, in this moment, which one am I inclined to do?" Bam. But if they own them all then we can also have this great conversation about which one do you wish you would have chosen?Now that you've looked at the numbers, you've stepped out of the relationships. It's a great place to be.
Kim Montague:
Pam Harris:
What are you waiting for?
Kim Montague:
What are you waiting for? It's exactly what I was gonna say. You need to get on over to the show notes or go to this site. You need to find this at mathisFigureOutAble.com/big,where we have a download available for you for free, that has all of the major strategies for each of the four operations,with examples, with problems,
with models, super easy to find:
mathisFigureOutAble.com/big.
Pam Harris:
Big. B-I-G. Because this is big. You're gonna want this download and then listen in for all these episodes, this series that we're doing, and we'll talk you through all the examples that you're looking at,you get to sort of hear them live, which makes it all the more - more what?Figure-Out-Able? Developable?Digestible? I'm going to stop trying to come up with words.Alright, y'all. If you want to learn more mathematics and refine your math teaching,download mathisFigureOutAble.com/big you're gonna like it so that you and students are mathematizing more and more, then join the Math is Figure-Out-Able Movement and help us spread the word that math is Figure-Out-Able! | 3,946 | 16,218 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-23 | latest | en | 0.951026 |
https://www.geeksforgeeks.org/multiply-two-numbers-represented-linked-lists-third-list/?ref=rp | 1,695,405,577,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506421.14/warc/CC-MAIN-20230922170343-20230922200343-00504.warc.gz | 874,459,384 | 53,435 | Open In App
# Multiply two numbers represented as linked lists into a third list
Given two numbers represented by linked lists, write a function that returns the head of the new linked list that represents the number that is the product of those numbers.
Examples:
```Input : 9->4->6
8->4
Output : 7->9->4->6->4
Input : 9->9->9->4->6->9
9->9->8->4->9
Output : 9->9->7->9->5->9->8->0->1->8->1
```
We have already discussed a solution in below post.
Multiply two numbers represented by Linked Lists
The solution discussed above stores result in an integer. Here we store result in a third list so that large numbers can be handled.
Remember old school multiplication? we imitate that process. On paper, we take the last digit of a number and multiply with the second number and write the product. Now leave the last column and same way each digit of one number is multiplied with every digit of other number and every time result is written by leaving one last column. then add these columns that forms the number. Now assume these columns as nodes of the resultant linked list. We make resultant linked list in reversed fashion.
Algorithm:
```Reverse both linked lists
Make a linked list of maximum result size (m + n + 1)
For each node of one list
For each node of second list
a) Multiply nodes
b) Add digit in result LL at corresponding
position
c) Now resultant node itself can be higher
than one digit
d) Make carry for next node
Leave one last column means next time start
From next node in result list
Implementation:
## C++
`// C++ program to Multiply two numbers``// represented as linked lists``#include ``using` `namespace` `std;` `// Linked list Node``struct` `Node {`` ``int` `data;`` ``struct` `Node* next;``};` `// Function to create a new Node``// with given data``struct` `Node* newNode(``int` `data)``{`` ``struct` `Node* new_node =`` ``(``struct` `Node*)``malloc``(``sizeof``(``struct` `Node));`` ``new_node->data = data;`` ``new_node->next = NULL;`` ``return` `new_node;``}` `// Function to insert a Node at the``// beginning of the Linked List``void` `push(``struct` `Node** head_ref, ``int` `new_data)``{`` ``// allocate Node`` ``struct` `Node* new_node = newNode(new_data);` ` ``// link the old list of the new Node`` ``new_node->next = (*head_ref);` ` ``// move the head to point to the new Node`` ``(*head_ref) = new_node;``}` `// Function to reverse the linked list and return``// its length``int` `reverse(``struct` `Node** head_ref)``{`` ``struct` `Node* prev = NULL;`` ``struct` `Node* current = *head_ref;`` ``struct` `Node* next;`` ``int` `len = 0;`` ``while` `(current != NULL) {`` ``len++;`` ``next = current->next;`` ``current->next = prev;`` ``prev = current;`` ``current = next;`` ``}`` ``*head_ref = prev;`` ``return` `len;``}` `// Function to make an empty linked list of``// given size``struct` `Node* make_empty_list(``int` `size)``{`` ``struct` `Node* head = NULL;`` ``while` `(size--)`` ``push(&head, 0);`` ``return` `head;``}` `// Multiply contents of two linked lists => store``// in another list and return its head``struct` `Node* multiplyTwoLists(``struct` `Node* first,`` ``struct` `Node* second)``{`` ``// reverse the lists to multiply from end`` ``// m and n lengths of linked lists to make`` ``// and empty list`` ``int` `m = reverse(&first), n = reverse(&second);` ` ``// make a list that will contain the result`` ``// of multiplication.`` ``// m+n+1 can be max size of the list`` ``struct` `Node* result = make_empty_list(m + n + 1);` ` ``// pointers for traverse linked lists and also`` ``// to reverse them after`` ``struct` `Node *second_ptr = second,`` ``*result_ptr1 = result, *result_ptr2, *first_ptr;` ` ``// multiply each Node of second list with first`` ``while` `(second_ptr) {` ` ``int` `carry = 0;` ` ``// each time we start from the next of Node`` ``// from which we started last time`` ``result_ptr2 = result_ptr1;` ` ``first_ptr = first;` ` ``while` `(first_ptr) {` ` ``// multiply a first list's digit with a`` ``// current second list's digit`` ``int` `mul = first_ptr->data * second_ptr->data`` ``+ carry;` ` ``// Assign the product to corresponding Node`` ``// of result`` ``result_ptr2->data += mul % 10;` ` ``// now resultant Node itself can have more`` ``// than 1 digit`` ``carry = mul / 10 + result_ptr2->data / 10;`` ``result_ptr2->data = result_ptr2->data % 10;` ` ``first_ptr = first_ptr->next;`` ``result_ptr2 = result_ptr2->next;`` ``}` ` ``// if carry is remaining from last multiplication`` ``if` `(carry > 0) {`` ``result_ptr2->data += carry;`` ``}` ` ``result_ptr1 = result_ptr1->next;`` ``second_ptr = second_ptr->next;`` ``}` ` ``// reverse the result_list as it was populated`` ``// from last Node`` ``reverse(&result);`` ``reverse(&first);`` ``reverse(&second);` ` ``// remove if there are zeros at starting`` ``while` `(result->data == 0) {`` ``struct` `Node* temp = result;`` ``result = result->next;`` ``free``(temp);`` ``}` ` ``// Return head of multiplication list`` ``return` `result;``}` `// A utility function to print a linked list``void` `printList(``struct` `Node* Node)``{`` ``while` `(Node != NULL) {`` ``cout << Node->data;`` ``if` `(Node->next)`` ``cout<<``"->"``;`` ``Node = Node->next;`` ``}`` ``cout << endl;``}` `// Driver program to test above function``int` `main(``void``)``{`` ``struct` `Node* first = NULL;`` ``struct` `Node* second = NULL;` ` ``// create first list 9->9->9->4->6->9`` ``push(&first, 9);`` ``push(&first, 6);`` ``push(&first, 4);`` ``push(&first, 9);`` ``push(&first, 9);`` ``push(&first, 9);`` ``cout<<``"First List is: "``;`` ``printList(first);` ` ``// create second list 9->9->8->4->9`` ``push(&second, 9);`` ``push(&second, 4);`` ``push(&second, 8);`` ``push(&second, 9);`` ``push(&second, 9);`` ``cout<<``"Second List is: "``;`` ``printList(second);` ` ``// Multiply the two lists and see result`` ``struct` `Node* result = multiplyTwoLists(first, second);`` ``cout << ``"Resultant list is: "``;`` ``printList(result);` ` ``return` `0;``}` `// This code is contributed by SHUBHAMSINGH10`
## C
`// C program to Multiply two numbers``// represented as linked lists``#include ``#include ` `// Linked list Node``struct` `Node {`` ``int` `data;`` ``struct` `Node* next;``};` `// Function to create a new Node``// with given data``struct` `Node* newNode(``int` `data)``{`` ``struct` `Node* new_node =`` ``(``struct` `Node*)``malloc``(``sizeof``(``struct` `Node));`` ``new_node->data = data;`` ``new_node->next = NULL;`` ``return` `new_node;``}` `// Function to insert a Node at the``// beginning of the Linked List``void` `push(``struct` `Node** head_ref, ``int` `new_data)``{`` ``// allocate Node`` ``struct` `Node* new_node = newNode(new_data);` ` ``// link the old list of the new Node`` ``new_node->next = (*head_ref);` ` ``// move the head to point to the new Node`` ``(*head_ref) = new_node;``}` `// Function to reverse the linked list and return``// its length``int` `reverse(``struct` `Node** head_ref)``{`` ``struct` `Node* prev = NULL;`` ``struct` `Node* current = *head_ref;`` ``struct` `Node* next;`` ``int` `len = 0;`` ``while` `(current != NULL) {`` ``len++;`` ``next = current->next;`` ``current->next = prev;`` ``prev = current;`` ``current = next;`` ``}`` ``*head_ref = prev;`` ``return` `len;``}` `// Function to make an empty linked list of``// given size``struct` `Node* make_empty_list(``int` `size)``{`` ``struct` `Node* head = NULL;`` ``while` `(size--)`` ``push(&head, 0);`` ``return` `head;``}` `// Multiply contents of two linked lists => store``// in another list and return its head``struct` `Node* multiplyTwoLists(``struct` `Node* first,`` ``struct` `Node* second)``{`` ``// reverse the lists to multiply from end`` ``// m and n lengths of linked lists to make`` ``// and empty list`` ``int` `m = reverse(&first), n = reverse(&second);` ` ``// make a list that will contain the result`` ``// of multiplication.`` ``// m+n+1 can be max size of the list`` ``struct` `Node* result = make_empty_list(m + n + 1);` ` ``// pointers for traverse linked lists and also`` ``// to reverse them after`` ``struct` `Node *second_ptr = second,`` ``*result_ptr1 = result, *result_ptr2, *first_ptr;` ` ``// multiply each Node of second list with first`` ``while` `(second_ptr) {` ` ``int` `carry = 0;` ` ``// each time we start from the next of Node`` ``// from which we started last time`` ``result_ptr2 = result_ptr1;` ` ``first_ptr = first;` ` ``while` `(first_ptr) {` ` ``// multiply a first list's digit with a`` ``// current second list's digit`` ``int` `mul = first_ptr->data * second_ptr->data`` ``+ carry;` ` ``// Assign the product to corresponding Node`` ``// of result`` ``result_ptr2->data += mul % 10;` ` ``// now resultant Node itself can have more`` ``// than 1 digit`` ``carry = mul / 10 + result_ptr2->data / 10;`` ``result_ptr2->data = result_ptr2->data % 10;` ` ``first_ptr = first_ptr->next;`` ``result_ptr2 = result_ptr2->next;`` ``}` ` ``// if carry is remaining from last multiplication`` ``if` `(carry > 0) {`` ``result_ptr2->data += carry;`` ``}` ` ``result_ptr1 = result_ptr1->next;`` ``second_ptr = second_ptr->next;`` ``}` ` ``// reverse the result_list as it was populated`` ``// from last Node`` ``reverse(&result);`` ``reverse(&first);`` ``reverse(&second);` ` ``// remove if there are zeros at starting`` ``while` `(result->data == 0) {`` ``struct` `Node* temp = result;`` ``result = result->next;`` ``free``(temp);`` ``}` ` ``// Return head of multiplication list`` ``return` `result;``}` `// A utility function to print a linked list``void` `printList(``struct` `Node* Node)``{`` ``while` `(Node != NULL) {`` ``printf``(``"%d"``, Node->data);`` ``if` `(Node->next)`` ``printf``(``"->"``);`` ``Node = Node->next;`` ``}`` ``printf``(``"\n"``);``}` `// Driver program to test above function``int` `main(``void``)``{`` ``struct` `Node* first = NULL;`` ``struct` `Node* second = NULL;` ` ``// create first list 9->9->9->4->6->9`` ``push(&first, 9);`` ``push(&first, 6);`` ``push(&first, 4);`` ``push(&first, 9);`` ``push(&first, 9);`` ``push(&first, 9);`` ``printf``(``"First List is: "``);`` ``printList(first);` ` ``// create second list 9->9->8->4->9`` ``push(&second, 9);`` ``push(&second, 4);`` ``push(&second, 8);`` ``push(&second, 9);`` ``push(&second, 9);`` ``printf``(``"Second List is: "``);`` ``printList(second);` ` ``// Multiply the two lists and see result`` ``struct` `Node* result = multiplyTwoLists(first, second);`` ``printf``(``"Resultant list is: "``);`` ``printList(result);` ` ``return` `0;``}`
## Java
`// Java program to multiply two numbers``// represented as linked lists``import` `java.io.*;` `// Node class``class` `Node {`` ``int` `data;`` ``Node next;` ` ``// Constructor to create a new node`` ``public` `Node(``int` `data)`` ``{`` ``this``.data = data;`` ``this``.next = ``null``;`` ``}``}` `// LinkedList class``class` `LinkedList {`` ``Node head;` ` ``// Constructor to initialize an empty linked list`` ``public` `LinkedList() { ``this``.head = ``null``; }` ` ``// Method to add a new node at the beginning of the list`` ``public` `void` `push(``int` `newData)`` ``{`` ``// Create a new node`` ``Node newNode = ``new` `Node(newData);` ` ``// Make next of new node as head`` ``newNode.next = head;` ` ``// Move the head to point to new node`` ``head = newNode;`` ``}` ` ``// Method to print the linked list`` ``public` `void` `printList()`` ``{`` ``// Object to iterate the list`` ``Node ptr = head;` ` ``if` `(head.data == ``0``) {`` ``ptr = head.next;`` ``}` ` ``// Loop to iterate the list`` ``while` `(ptr != ``null``) {`` ``System.out.print(ptr.data + ``" -> "``);` ` ``// Moving the iterating object to the next node`` ``ptr = ptr.next;`` ``}` ` ``System.out.println();`` ``}``}` `class` `GFG {` ` ``// Function to reverse the linked list and return its`` ``// length`` ``public` `static` `int` `reverse(LinkedList headRef)`` ``{`` ``// Initializing prev and current at null and`` ``// starting node respectively`` ``Node prev = ``null``;`` ``Node current = headRef.head;` ` ``int` `Len = ``0``;` ` ``// Loop to reverse the link of each node in the list`` ``while` `(current != ``null``) {`` ``Len++;`` ``Node next = current.next;`` ``current.next = prev;`` ``prev = current;`` ``current = next;`` ``}` ` ``// Assigning new starting object to main head object`` ``headRef.head = prev;` ` ``// Returning the length of linked list`` ``return` `Len;`` ``}` ` ``// Function to define an empty linked list of given size`` ``// and each element as zero`` ``public` `static` `LinkedList makeEmptyList(``int` `size)`` ``{`` ``LinkedList head = ``new` `LinkedList();` ` ``while` `(size-- > ``0``) {`` ``head.push(``0``);`` ``}` ` ``// Returns the head object`` ``return` `head;`` ``}` ` ``// Multiply contents of two linked lists and store it in`` ``// another list and return its head`` ``public` `static` `LinkedList`` ``multiplyTwoLists(LinkedList first, LinkedList second)`` ``{`` ``// Reverse the lists to multiply from the end and`` ``// calculate their lengths`` ``int` `m = reverse(first);`` ``int` `n = reverse(second);` ` ``// Make a list that will contain the result of the`` ``// multiplication m + n + 1 can be the max size of`` ``// the list`` ``LinkedList result = makeEmptyList(m + n + ``1``);` ` ``// Objects to iterate the lists and also to reverse`` ``// them after`` ``Node secondPtr = second.head;`` ``Node resultPtr1 = result.head;` ` ``// Multiply each node of second list with first`` ``while` `(secondPtr != ``null``) {`` ``int` `carry = ``0``;` ` ``// Each time we start from the next node from`` ``// which we started last time`` ``Node resultPtr2 = resultPtr1;`` ``Node firstPtr = first.head;` ` ``// Multiply a first list's digit with a current`` ``// second list's digit`` ``while` `(firstPtr != ``null``) {`` ``int` `mul = (firstPtr.data * secondPtr.data`` ``+ carry);` ` ``// Assign the product to corresponding node`` ``// of result`` ``resultPtr2.data += mul % ``10``;` ` ``// Now resultant node itself can have more`` ``// than one digit`` ``carry = (mul / ``10``) + (resultPtr2.data / ``10``);`` ``resultPtr2.data = resultPtr2.data % ``10``;` ` ``firstPtr = firstPtr.next;`` ``resultPtr2 = resultPtr2.next;`` ``}` ` ``// If carry is remaining from last`` ``// multiplication`` ``if` `(carry > ``0``) {`` ``resultPtr2.data += carry;`` ``}` ` ``resultPtr1 = resultPtr1.next;`` ``secondPtr = secondPtr.next;`` ``}` ` ``// Reverse the result list`` ``reverse(result);` ` ``// Return the head of result list`` ``return` `result;`` ``}` ` ``public` `static` `void` `main(String[] args)`` ``{` ` ``// Creating the first list`` ``LinkedList first = ``new` `LinkedList();`` ``first.push(``9``);`` ``first.push(``6``);`` ``first.push(``4``);`` ``first.push(``9``);`` ``first.push(``9``);`` ``first.push(``9``);`` ``System.out.print(``"First List is: "``);`` ``first.printList();` ` ``// Creating the second list`` ``LinkedList second = ``new` `LinkedList();`` ``second.push(``9``);`` ``second.push(``4``);`` ``second.push(``8``);`` ``second.push(``9``);`` ``second.push(``9``);`` ``System.out.print(``"Second List is: "``);`` ``second.printList();` ` ``// Multiplying the lists`` ``LinkedList result = multiplyTwoLists(first, second);`` ``System.out.print(``"Resultant List is: "``);`` ``result.printList();`` ``}``}` `// This code is contributed by lokesh.`
## Python3
`# Python3 program to multiply two numbers``# represented as linked lists` `# Node class``class` `Node:`` ` ` ``# Function to initialize the node object`` ``def` `__init__(``self``, data):` ` ``self``.data ``=` `data`` ``self``.``next` `=` `None`` ` `# Linked List Class``class` `LinkedList:` ` ``# Function to initialize the`` ``# LinkedList class.`` ``def` `__init__(``self``):` ` ``# Initialize head as None`` ``self``.head ``=` `None` ` ``# This function insert a new node at the`` ``# beginning of the linked list`` ``def` `push(``self``, new_data):`` ` ` ``# Create a new Node`` ``new_node ``=` `Node(new_data)` ` ``# Make next of new Node as head`` ``new_node.``next` `=` `self``.head` ` ``# Move the head to point to new Node`` ``self``.head ``=` `new_node`` ` ` ``# Method to print the linked list`` ``def` `printList(``self``):` ` ``# Object to iterate`` ``# the list`` ``ptr ``=` `self``.head` ` ``# Loop to iterate list`` ``while``(ptr !``=` `None``):`` ``print``(ptr.data, ``'->'``, end ``=` `'')` ` ``# Moving the iterating object`` ``# to next node`` ``ptr ``=` `ptr.``next`` ` ` ``print``()` `# Function to reverse the linked``# list and return its length``def` `reverse(head_ref):` ` ``# Initialising prev and current`` ``# at None and starting node`` ``# respectively.`` ``prev ``=` `None`` ``current ``=` `head_ref.head` ` ``Len` `=` `0` ` ``# Loop to reverse the link`` ``# of each node in the list`` ``while``(current !``=` `None``):`` ``Len` `+``=` `1`` ``Next` `=` `current.``next`` ``current.``next` `=` `prev`` ``prev ``=` `current`` ``current ``=` `Next` ` ``# Assigning new starting object`` ``# to main head object.`` ``head_ref.head ``=` `prev` ` ``# Returning the length of`` ``# linked list.`` ``return` `Len` `# Function to define an empty``# linked list of given size and``# each element as zero.``def` `make_empty_list(size):`` ` ` ``head ``=` `LinkedList()`` ` ` ``while``(size):`` ``head.push(``0``)`` ``size ``-``=` `1` ` ``# Returns the head object.`` ``return` `head` `# Multiply contents of two linked``# list store it in other list and``# return its head.``def` `multiplyTwoLists(first, second):` ` ``# Reverse the list to multiply from`` ``# end m and n lengths of linked list`` ``# to make and empty list`` ``m ``=` `reverse(first)`` ``n ``=` `reverse(second)` ` ``# Make a list that will contain the`` ``# result of multiplication.`` ``# m+n+1 can be max size of the list.`` ``result ``=` `make_empty_list(m ``+` `n ``+` `1``)` ` ``# Objects for traverse linked list`` ``# and also to reverse them after.`` ``second_ptr ``=` `second.head`` ``result_ptr1 ``=` `result.head` ` ``# Multiply each node of second`` ``# list with first.`` ``while``(second_ptr !``=` `None``):`` ``carry ``=` `0` ` ``# Each time we start from next`` ``# node from which we started last`` ``# time.`` ``result_ptr2 ``=` `result_ptr1`` ``first_ptr ``=` `first.head` ` ``while``(first_ptr !``=` `None``):`` ` ` ``# Multiply a first list's digit`` ``# with a current second list's digit.`` ``mul ``=` `((first_ptr.data) ``*`` ``(second_ptr.data) ``+` `carry)` ` ``# Assign the product to corresponding`` ``# node of result.`` ``result_ptr2.data ``+``=` `mul ``%` `10` ` ``# Now resultant node itself can have`` ``# more than one digit.`` ``carry ``=` `((mul ``/``/` `10``) ``+`` ``(result_ptr2.data ``/``/` `10``))`` ``result_ptr2.data ``=` `result_ptr2.data ``%` `10` ` ``first_ptr ``=` `first_ptr.``next`` ``result_ptr2 ``=` `result_ptr2.``next` ` ``# If carry is remaining from`` ``# last multiplication`` ``if``(carry > ``0``):`` ``result_ptr2.data ``+``=` `carry` ` ``result_ptr1 ``=` `result_ptr1.``next`` ``second_ptr ``=` `second_ptr.``next` ` ``# Reverse the result_list as it`` ``# was populated from last node`` ``reverse(result)`` ``reverse(first)`` ``reverse(second)` ` ``# Remove starting nodes`` ``# containing zeroes.`` ``start ``=` `result.head`` ``while``(start.data ``=``=` `0``):`` ``result.head ``=` `start.``next`` ``start ``=` `start.``next` ` ``# Return the resultant multiplicated`` ``# linked list.`` ``return` `result` `# Driver code``if` `__name__``=``=``'__main__'``:` ` ``first ``=` `LinkedList()`` ``second ``=` `LinkedList()` ` ``# Pushing elements at start of`` ``# first linked list.`` ``first.push(``9``)`` ``first.push(``6``)`` ``first.push(``4``)`` ``first.push(``9``)`` ``first.push(``9``)`` ``first.push(``9``)` ` ``# Printing first linked list`` ``print``(``"First list is: "``, end ``=` `'')`` ``first.printList()` ` ``# Pushing elements at start of`` ``# second linked list.`` ``second.push(``9``)`` ``second.push(``4``)`` ``second.push(``8``)`` ``second.push(``9``)`` ``second.push(``9``)` ` ``# Printing second linked list.`` ``print``(``"Second List is: "``, end ``=` `'')`` ``second.printList()` ` ``# Multiply two linked list and`` ``# print the result.`` ``result ``=` `multiplyTwoLists(first, second)`` ``print``(``"Resultant list is: "``, end ``=` `'')`` ``result.printList()`` ` `# This code is contributed by Amit Mangal`
## C#
`// C# program to multiply two numbers``// represented as linked lists``using` `System;` `// Node class``class` `Node``{`` ``public` `int` `data;`` ``public` `Node next;` ` ``// Constructor to create a new node`` ``public` `Node(``int` `data)`` ``{`` ``this``.data = data;`` ``this``.next = ``null``;`` ``}``}` `// LinkedList class``class` `LinkedList``{`` ``public` `Node head;` ` ``// Constructor to initialize an empty linked list`` ``public` `LinkedList()`` ``{`` ``this``.head = ``null``;`` ``}` ` ``// Method to add a new node at the beginning of the list`` ``public` `void` `Push(``int` `newData)`` ``{`` ``// Create a new node`` ``Node newNode = ``new` `Node(newData);` ` ``// Make next of new node as head`` ``newNode.next = head;` ` ``// Move the head to point to new node`` ``head = newNode;`` ``}` ` ``// Method to print the linked list`` ``public` `void` `PrintList()`` ``{`` ``// Object to iterate the list`` ``Node ptr = head;` ` ``if` `(head.data == 0)`` ``{`` ``ptr = head.next;`` ``}` ` ``// Loop to iterate the list`` ``while` `(ptr != ``null``)`` ``{`` ``Console.Write(ptr.data + ``" -> "``);` ` ``// Moving the iterating object to the next node`` ``ptr = ptr.next;`` ``}` ` ``Console.WriteLine();`` ``}``}` `class` `GFG``{` ` ``// Function to reverse the linked list and return its`` ``// length`` ``public` `static` `int` `Reverse(LinkedList headRef)`` ``{`` ``// Initializing prev and current at null and`` ``// starting node respectively`` ``Node prev = ``null``;`` ``Node current = headRef.head;` ` ``int` `Len = 0;` ` ``// Loop to reverse the link of each node in the list`` ``while` `(current != ``null``)`` ``{`` ``Len++;`` ``Node next = current.next;`` ``current.next = prev;`` ``prev = current;`` ``current = next;`` ``}` ` ``// Assigning new starting object to main head object`` ``headRef.head = prev;` ` ``// Returning the length of linked list`` ``return` `Len;`` ``}` ` ``// Function to define an empty linked list of given size`` ``// and each element as zero`` ``public` `static` `LinkedList MakeEmptyList(``int` `size)`` ``{`` ``LinkedList head = ``new` `LinkedList();` ` ``while` `(size-- > 0)`` ``{`` ``head.Push(0);`` ``}` ` ``// Returns the head object`` ``return` `head;`` ``}` ` ``// Multiply contents of two linked lists and store it in`` ``// another list and return its head`` ``public` `static` `LinkedList`` ``MultiplyTwoLists(LinkedList first, LinkedList second)`` ``{`` ``// Reverse the lists to multiply from the end and`` ``// calculate their lengths`` ``int` `m = Reverse(first);`` ``int` `n = Reverse(second);` ` ``// Make a list that will contain the result of the`` ``// multiplication m + n + 1 can be the max size of`` ``// the list`` ``LinkedList result = MakeEmptyList(m + n + 1);` ` ``// Objects to iterate the lists and also to reverse`` ``// them after`` ``Node secondPtr = second.head;`` ``Node resultPtr1 = result.head;` ` ``// Multiply each node of second list with first`` ``while` `(secondPtr != ``null``)`` ``{`` ``int` `carry = 0;` ` ``// Each time we start from the next node from`` ``// which we started last time`` ``Node resultPtr2 = resultPtr1;`` ``Node firstPtr = first.head;` ` ``// Multiply a first list's digit with a current`` ``// second list's digit`` ``while` `(firstPtr != ``null``)`` ``{`` ``int` `mul = (firstPtr.data * secondPtr.data`` ``+ carry);` ` ``// Assign the product to corresponding node`` ``// of result`` ``resultPtr2.data += mul % 10;` ` ``// Now resultant node itself can have more`` ``// than one digit`` ``carry = (mul / 10) + (resultPtr2.data / 10);`` ``resultPtr2.data = resultPtr2.data % 10;` ` ``firstPtr = firstPtr.next;`` ``resultPtr2 = resultPtr2.next;`` ``}` ` ``// If carry is remaining from last`` ``// multiplication`` ``if` `(carry > 0)`` ``{`` ``resultPtr2.data += carry;`` ``}` ` ``resultPtr1 = resultPtr1.next;`` ``secondPtr = secondPtr.next;`` ``}` ` ``// Reverse the result list`` ``Reverse(result);` ` ``// Return the head of result list`` ``return` `result;`` ``}` ` ``public` `static` `void` `Main(String[] args)`` ``{` ` ``// Creating the first list`` ``LinkedList first = ``new` `LinkedList();`` ``first.Push(9);`` ``first.Push(6);`` ``first.Push(4);`` ``first.Push(9);`` ``first.Push(9);`` ``first.Push(9);`` ``Console.Write(``"First List is: "``);`` ``first.PrintList();` ` ``// Creating the second list`` ``LinkedList second = ``new` `LinkedList();`` ``second.Push(9);`` ``second.Push(4);`` ``second.Push(8);`` ``second.Push(9);`` ``second.Push(9);`` ``Console.Write(``"Second List is: "``);`` ``second.PrintList();` ` ``// Multiplying the lists`` ``LinkedList result = MultiplyTwoLists(first, second);`` ``Console.Write(``"Resultant List is: "``);`` ``result.PrintList();`` ``}``}`
## Javascript
`// JavaScript program to multiply two numbers``// represented as linked lists``// linked list node``class Node{`` ``constructor(data){`` ``this``.data = data;`` ``this``.next = ``null``;`` ``}``}` `// function to create a new node``// with given data``function` `newNode(data){`` ``return` `new` `Node(data);``}` `// function to insert a node at the``// beginning of the linked list``function` `push(head_ref, new_data){`` ``// allocated node and put data`` ``let new_node = ``new` `Node(new_data);`` ` ` ``// link to old list of the new node`` ``new_node.next = head_ref;`` ` ` ``// move the head to point to the new node`` ``head_ref = new_node;`` ``return` `head_ref;``}` `// function to reverse the linked list``// and return its length``function` `reverse(head_ref){`` ``let prev = ``null``;`` ``let current = head_ref;`` ``let next;`` ``while``(current != ``null``){`` ``next = current.next;`` ``current.next = prev;`` ``prev = current;`` ``current = next;`` ``}`` ``head_ref = prev;`` ``return` `head_ref;``}` `// function to make an empty linked list of``// given size``function` `make_empty_list(size){`` ``let head = ``null``;`` ``while``(size--){`` ``head = push(head, 0);`` ``}`` ``return` `head;``}` `// function return the length of linked list``function` `len(head){`` ``let n = 0;`` ``while``(head != ``null``){`` ``n++;`` ``head = head.next;`` ``}`` ``return` `n;``}` `// multiply contents of two linked lists => store``// in another list and return its head``function` `multiplyTwoLists(first, second){`` ``// reverse the lists to multiply from end`` ``// m and n lengths of linked lists to make`` ``// and empty list`` ``first = reverse(first);`` ``second = reverse(second);`` ` ` ``let m = len(first);`` ``let n = len(second);`` ` ` ``// make a list that will contain the result`` ``// of multiplication.`` ``// m+n+1 can be max size of the list`` ``let result = make_empty_list(m+n+1);`` ` ` ``// pointers for traverse linked lists and also`` ``// to reverse them after`` ``let second_ptr = second;`` ``let result_ptr1 = result;`` ``let result_ptr2;`` ``let first_ptr;`` ` ` ``// multiply each node of second list with first`` ``while``(second_ptr){`` ``let carry = 0;`` ` ` ``// each time we start from the next of node`` ``// from which we started last time`` ``result_ptr2 = result_ptr1;`` ` ` ``first_ptr = first;`` ` ` ``while``(first_ptr){`` ``// multiply a first list's digit with a`` ``// current second list's digit`` ``let mul = first_ptr.data * second_ptr.data + carry;`` ` ` ``// assign the product to corresponding node`` ``// of result`` ``result_ptr2.data += mul % 10;`` ` ` ``// now resultant node itself can have more`` ``// than 1 digit`` ``carry = parseInt(mul / 10) + parseInt(result_ptr2.data/10);`` ``result_ptr2.data = result_ptr2.data % 10;`` ` ` ``first_ptr = first_ptr.next;`` ``result_ptr2 = result_ptr2.next;`` ``}`` ``// if carry is remaining from last multiplication`` ``if` `(carry > 0) {`` ``result_ptr2.data += carry;`` ``}`` ` ` ``result_ptr1 = result_ptr1.next;`` ``second_ptr = second_ptr.next;`` ``}`` ``// reverse the result_list as it was populated`` ``// from last Node`` ``result = reverse(result);`` ``first = reverse(first);`` ``second = reverse(second);`` ` ` ``// remove if there are zeros at starting`` ``while` `(result.data == 0) {`` ``let temp = result;`` ``result = result.next;`` ``}`` ` ` ``// Return head of multiplication list`` ``return` `result;``}` `// a utility function to print a linked list``function` `printList(Node){`` ``while``(Node != ``null``){`` ``console.log(Node.data);`` ``if``(Node.next)`` ``console.log(``"->"``);`` ``Node = Node.next;`` ``}`` ``console.log(``"\n"``);``}` `// driver code to test above functions``let first = ``null``;``let second = ``null``;` `// create first list 9->9->9->4->6->9``first = push(first, 9);``first = push(first, 6);``first = push(first, 4);``first = push(first, 9);``first = push(first, 9);``first = push(first, 9);``console.log(``"First List is : "``);``printList(first);` `// create second list 9->9->8->4->9``second = push(second, 9);``second = push(second, 4);``second = push(second, 8);``second = push(second, 9);``second = push(second, 9);``console.log(``"Second List is : "``);``printList(second);` `// multiply the two lists and see result``let result = multiplyTwoLists(first, second);``console.log(``"Resultant list is : "``);``printList(result);` `// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)`
Output:
```First List is: 9->9->9->4->6->9
Second List is: 9->9->8->4->9
Resultant list is: 9->9->7->9->5->9->8->0->1->8->1```
Time complexity: O(M+N) where M and N are size of given two linked lists respectively
Auxiliary Space: O(1)
Note: we can take care of resultant node that can have more than 1 digit outside the loop just traverse the result list and add carry to next digit before reversing. | 10,813 | 33,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.766746 |
https://www.geeksforgeeks.org/oyo-rooms-interview-experience-set-6-senior-software-developer/?ref=rp | 1,709,264,726,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00480.warc.gz | 774,833,265 | 54,291 | # OYO Rooms Interview Experience | Set 6 (For Senior Software Developer)
First Round: (Written)
1. Maximum sum of non-contiguous elements in array
```Input : 1 12 5 4 13
Output: 25```
2. Given an array of integers, find a combination of four elements in the array whose sum is equal to a given value X.
```Input Array : 1 5 1 0 6 0
Input Sum: 7
Output : 1 (1 if present, else 0)```
Second : (F2F)
1. Discussion of above two questions.
2. If a doubly linked list has pointers in the form of integers that represesnts memory location, and we want to manage only one reference for both prev and next node, how will the list be traversed.
3. Rotate right a binary tree
Third : (F2F)
1. Diameter of a binary tree
2. How DNS lookup works
3. SQL Query with 3 tables : Student, Class, Test
```Input : Student : SID, CID, Name
Class : CID, Cname
Test : TestId, WeekId, SID, Marks ```
Write a query to print average marks classwise for each week Output Example :
``` ClassName, WeekId, Avg_Marks
Tenth, 1, 33
Eleventh, 1, 34
Tenth, 2, 45
Eleventh, 2, 21 ```
Solution : select (select Cname from Class where CID = S.CID)ClassName, T.WeekId, AVG(T.Marks) from Test T LEFT_JOIN Student S on T.SID=S.SID group by ClassName, T.WeekId
Previous
Next | 371 | 1,236 | {"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} | 2.578125 | 3 | CC-MAIN-2024-10 | latest | en | 0.726315 |
bankhelpsite.com | 1,386,748,644,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164033438/warc/CC-MAIN-20131204133353-00041-ip-10-33-133-15.ec2.internal.warc.gz | 14,035,505 | 13,005 | ## Car Loan Payoff Calculator: 5 Simple Steps for Calculation Your Car Loan Payoff
Unlike in past, owning a car nowadays is no longer considered as a luxury but it has been already a necessity. Besides that, cars nowadays have a lot of variations and types, as well as their mode of function. Nonetheless, purchasing a car, especially new ones or even second-hand quality cars still to be expensive. Hence, most people would opt for car loans in order to finance the purchase.
Car loans are secured loan programs. Hence, in order for you to completely release the collateralized lien on a car, you would have to pay the loan in full. Most people tend to think that calculating a payoff can be challenging and complicated. However, this is entirely possible without even using a car loan payoff calculator. All you need to have are the auto loan paperwork and the loan statement, as well as a reliable scientific calculator.
1. To begin with, you can use this formula for computation like a car loan payoff calculator:
• M=P (J/1-[1+J]^-N)
• Where M is the monthly payment.
• P is the principal loan amount
• J is the interest in decimal form
• And N is the term of the loan in months (divide the interest rate by 1200 to get the decimal form)
2. Next, just simply substitute or put in the values denoted by the formula. For this scenario, let’s say you have a \$30,000 car loan for 50 months at 5% interest. Hence, your data should look like this:
• M=30,000 (.004/1 – [1+.004]^-50)
3. After that, multiply the monthly payment value by the total remaining months of the loan.
4. Just to make sure you don’t miss any other interest on your computation, it is best you check out with your lender and ask if there are any capitalized interests for the loan. The capitalized interest increases if there are any late payments you’ve made for the loan. If there is any, you should add this to the value we have in step 3.
5. comparing car loan payoff calculator values
6. Now that you computation is done, request for the official payoff statement from your lender. Ensure that your calculation is the same with the lender’s calculation. If there are discrepancies, be sure to talk with your loan officer to have the payoff statement reviewed.
### 104 Responses to “Car Loan Payoff Calculator: 5 Simple Steps for Calculation Your Car Loan Payoff”
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http://programming.cybercour.se/course/program-structure | 1,516,369,717,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887981.42/warc/CC-MAIN-20180119125144-20180119145144-00609.warc.gz | 281,477,069 | 16,357 | # Aussie rules
Australian rules is a type of football. You score points by getting the ball between posts at either end of the field:
If the ball goes between the center two posts, that’s a goal, and is worth six points. If the ball goes between one of the center and outside posts, that’s a behind, and is worth one point.
Here’s an Eva program for working that out.
Give it a try.
What do you notice?
Simon
It’s almost the same as the compute tip program.
Tara
But the variables are different, and the programs do different things.
Simon
Well, yes, but the programs are… organized like each other.
Tara
Good! You’re right.
# Program structure
Here’s the code for the Aussie rules program.
1. Dim goals As Integer
2. Dim behinds As Integer
3. Dim score As Integer
4. goals = Cells(1, 2)
5. behinds = Cells(2, 2)
6. score = goals * 6 + behinds
7. Cells(3, 2) = score
Tara
Hey, Jeremy! What one sentence would describe the first three lines?
Jeremy
Let me think… How about “Make some variables?”
Tara
Good! Geek speak for “make variables” is “declare variables.” Means the same thing.
Let’s add a comment to the code. A new first line.
1. 'Declare variables
2. Dim goals As Integer
3. Dim behinds As Integer
4. Dim score As Integer
5. goals = Cells(1, 2)
6. behinds = Cells(2, 2)
7. score = goals * 6 + behinds
8. Cells(3, 2) = score
Comments begin with that single quote. The CPU ignores the text after the ‘. You use comments to remind yourself how the code works.
Tara
Kurt, how would you describe the next two lines in one sentence? That’s lines 5 and 6.
Klaus
Umm…. how about “Get data from the worksheet”?
Tara
Sounds good!
1. 'Declare variables
2. Dim goals As Integer
3. Dim behinds As Integer
4. Dim score As Integer
5. 'Get data from the worksheet
6. goals = Cells(1, 2)
7. behinds = Cells(2, 2)
8. score = goals * 6 + behinds
9. Cells(3, 2) = score
Tara
Simon
1. 'Declare variables
2. Dim goals As Integer
3. Dim behinds As Integer
4. Dim score As Integer
5. 'Get data from the worksheet
6. goals = Cells(1, 2)
7. behinds = Cells(2, 2)
8. 'Compute results
9. score = goals * 6 + behinds
10. 'Put results into worksheet
11. Cells(3, 2) = score
Tara
Great! There’s a “Compute results” section, and a “Put results into worksheet” section.
Let’s look at the first program we did. Here’s the compute tip code:
1. Dim amount As Single
2. Dim tip As Single
3. Dim total As Single
4. amount = Cells(1, 2)
5. tip = amount * 0.15
6. total = amount + tip
7. Cells(2, 2) = tip
8. Cells(3, 2) = total
It has exactly the same sections:
• Declare variables
• Get data from the worksheet
• Compute results
• Put results into worksheet
1. 'Declare variables
2. Dim amount As Single
3. Dim tip As Single
4. Dim total As Single
5. 'Get data from the worksheet
6. amount = Cells(1, 2)
7. 'Compute results
8. tip = amount * 0.15
9. total = amount + tip
10. 'Put results into worksheet
11. Cells(2, 2) = tip
12. Cells(3, 2) = total
Here are the programs, side by side.
'Declare variables
Dim amount As Single
Dim tip As Single
Dim total As Single
'Get data from the worksheet
amount = Cells(1, 2)
'Compute results
tip = amount * 0.15
total = amount + tip
'Put results into worksheet
Cells(2, 2) = tip
Cells(3, 2) = total
'Declare variables
Dim goals As Integer
Dim behinds As Integer
Dim score As Integer
'Get data from the worksheet
goals = Cells(1, 2)
behinds = Cells(2, 2)
'Compute results
score = goals * 6 + behinds
'Put results into worksheet
Cells(3, 2) = score
If we just look at the comments, they’re exactly the same:
‘Declare variables
‘Get data from the worksheet
‘Compute results
‘Put results into worksheet
‘Declare variables
‘Get data from the worksheet
‘Compute results
‘Put results into worksheet
Two programs, doing different things, but they have the same structure.
# Levels
You can look at every program on multiple levels. The structure level only has comments. Each comment describes a piece of the program. It says what will happen, without the details of how.
1. 'Declare variables
2. 'Get data from the worksheet
3. 'Compute results
4. 'Put results into worksheet
The code level shows the structure comments and the code.
1. 'Declare variables
2. Dim goals As Integer
3. Dim behinds As Integer
4. Dim score As Integer
5. 'Get data from the worksheet
6. goals = Cells(1, 2)
7. behinds = Cells(2, 2)
8. 'Compute results
9. score = goals * 6 + behinds
10. 'Put results into worksheet
11. Cells(3, 2) = score
When programmers start writing code, they start with the structure. They fill in the details later.
We’ll come back to this idea of program structure throughout the course. Programs are much easier to write if you start with the structure first.
# Summary
We looked at two programs that have the same structure. They do different things (tips vs. Aussie rules scores), use different variables, and different calculations. They have the same structure, however. | 1,575 | 4,914 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-05 | longest | en | 0.58848 |
https://ssl.instructables.com/id/Egg-Drop-Physics/ | 1,558,904,270,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259757.86/warc/CC-MAIN-20190526205447-20190526231447-00505.warc.gz | 640,521,467 | 21,438 | # Egg Drop Physics
88,091
15
6
The egg drop experiment can have many variations. This is the one I like best. I have used it for a number of years, made some adjustments along the way, and think it's finally time to share with you.
I use this project as a way to make the mathematics part of physics relevant to my 7th graders. They are calculating mass, speed, velocity, momentum, force, and acceleration and having fun at the same time. They have the freedom to design their own project but are constrained by the materials provided and the time allowed.
The materials are cheap and easy to acquire which is a recurring theme on my teacher budget (materials are on the next page).
I usually give students about 45 minutes of pure build time. This does not include the time taken to hand out materials. I usually hand out the materials and give them some planning time... then start the clock. Because my school has 45 minute periods, we can't do it all in one class period. I force them to build and test in two.
I have an area to drop that is 5.3 meters. You will want to find an area that is at least 4. The higher the better!
*The new national science standards, if your state chooses to adopt them, will place greater emphasis on the process and application of topics than on recall. This is a simple, yet effective way to asses what they know and are able to do (and aligns perfectly with the motion and forces standards).
## Step 1: Title, Introduction, Materials, and Methods
Title:
How does the design of an egg contraption protect an egg from the combined forces acting on it when subjected to a drop of 5.3 meters?
Introduction:
a. Background - In class, the topics of speed, velocity, resultant velocity, acceleration, and momentum were explained and mathematical calculations were performed.
b. Purpose - This experiment is designed to review these forces by completing an egg drop lab.
c. Hypothesis - If an egg is dropped from a height of 5.3 meters and the egg shell must not crack, then the egg must be well protected from outside forces acting on it.
d. Prediction - Egg contraptions with the most speed and therefore the most velocity, acceleration, momentum, and force, will have the greatest chance of breaking.
Procedure
a. Materials: Teacher should supply a balance, a good place to test the contraption, a way to measure the height of the drop in meters, and a stopwatch.
1 egg (I don't always hand these out right away)
5 popsicle sticks
5 straws (I like bendy straws but it doesn't matter)
5 rubber bands
2 sheets of paper
100 cm of string
b. Methods -
1st: Create a detailed drawing of what you plan to build.
2nd:Gather materials
3rd: Build a contraption that can protect an egg from a fall of 5.3 meters
4th: Find the mass of the contraption (with the egg)
5th: Test the contraption (calculate time of fall with stopwatch)
6th: Calculate speed, velocity, acceleration, force, and momentum
7th: Analyze results
8th: Create a detailed drawing of what you actually tested.
## Step 2: A Fun Twist
After students start to build, I give them the option of trading materials with the "store".
Here is a short skit of what an exchange might sound like...
Student: What can I get for 3 popsicle sticks?
Teacher: What do you need?
Student: Another sheet of paper.
Teacher: That will cost you 4 popsicle sticks and a straw.
Student: Let me go ask my lab partner... You drive a hard bargain, but we'll take it!
Teacher: Make sure you record any transactions for when you write your formal lab reflection.
It's that simple. You can base what you trade on how you're feeling that day, the after market value, or how much you like the student. I'll admit that I even caved when a group had no materials left to trade but offered me a sweet pen. Bribery got the best of me that day (but I still have the pen :).
I also give students .5 point back for each whole, unused material they can turn in at the end.
## Step 3: Lab Reflection
Procedure Continued (written in paragraph form with NO personal pronouns) It should read much like a cookbook.
Describe how the contraption was assembled. Were any materials exchanged? If so, which ones, how many, what for? Were any materials omitted?
Results: Provide an account of what happened during the experiment.
Include the mass of your project with the egg. Include the speed of the project and show all the math needed to calculate the results. Calculate the velocity, momentum, acceleration, and force of your contraption. All mathematical calculations should be written neatly on a separate sheet of paper and attached to the lab.
Conclusion: This is the most important part of the lab! Restate the hypothesis first
Include the following: What features did the contraption have to help protect the egg? What did the egg look like after impact? What worked (either in your project or others)? What did not work (either in your project or others)? List three things you would do differently (in bullet form) if you could do this project again.
## Recommendations
• ### Lamps Class
9,295 Enrolled
## 6 Discussions
Awesome job, Biodynamic! How I wish I'd had a science teacher like you when I was a kid in 1950-60's! I would have loved your "hands-on, you build it" methods. Thank you for making these Instrutibles.
Martin | 1,233 | 5,344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2019-22 | latest | en | 0.955487 |
http://grahammitchell.com/assignments/space-boxing.html | 1,686,391,487,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657169.98/warc/CC-MAIN-20230610095459-20230610125459-00013.warc.gz | 21,014,392 | 1,793 | • Author: Graham Mitchell
• Filename: SpaceBoxing.java
### Space Boxing
Julio Cesar Chavez Mark VII is an interplanetary space boxer, who currently holds the championship belts for various weight categories on many different planets within our solar system. However, it is often difficult for him to recall what his "target weight" needs to be on earth in order to make the weight class on other planets. Write a program to help him keep track of this.
It should ask him what his earth weight is, and to enter a number for the planet he wants to fight on. It should then compute his weight on the destination planet based on the table below:
#PlanetRelative gravity
1Venus0.78
2Mars0.39
3Jupiter2.65
4Saturn1.17
5Uranus1.05
6Neptune1.23
So, for example, if Julio weighs 128 lbs. on earth, then he would weigh just under 50 lbs. on Mars, since Mars' gravity is 0.39 times earth's gravity. (128 * 0.39 is 49.92)
I have information for the following planets:
1. Venus 2. Mars 3. Jupiter
4. Saturn 5. Uranus 6. Neptune
Which planet are you visiting? 2
Your weight would be 49.92 pounds on that planet. | 298 | 1,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-23 | latest | en | 0.940465 |
https://www.maa.org/press/maa-reviews/teaching-money-applications-to-make-mathematics-meaningful-grades-7-12 | 1,632,789,762,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058589.72/warc/CC-MAIN-20210928002254-20210928032254-00607.warc.gz | 879,991,273 | 23,991 | # Teaching Money Applications to Make Mathematics Meaningful Grades 7- 12
###### Elizabeth Marquez and Paul Westbrook
Publisher:
Corwin Press
Publication Date:
2007
Number of Pages:
152
Format:
Paperback
Price:
29.95
ISBN:
9781412941396
Category:
Monograph
[Reviewed by
Mark Bollman
, on
10/12/2007
]
When I work with future teachers in my Mathematics for Elementary Teachers class, they often put together lesson plans dealing with money, on the sensible grounds that this concrete application will help draw their future students into learning mathematics. In Teaching Money Applications, we find that same mindset taken to the second half of the K-12 curriculum.
This is not a textbook; it falls more under the heading of a “resource guide” for teachers interested in infusing some lessons on money into their mathematics lessons. Toward that end, it is organized around financial topics rather than mathematical concepts, which gets a bit confusing at times. For example: The first chapter, which deals with buying a car, starts out with some simple arithmetic calculations and then leaps into quadratic regression of miles per gallon figures — with a graphing calculator used as a “black box” to perform the regression. Chapter 2, on savings plans, then is back doing only straightforward arithmetic.
To the authors’ credit, they have resisted the temptation to “black-box” every complicated statistical calculation. Correlation coefficients, for example, are calculated (for small data sets) with tables of values and the standard formula with all of its sums. The applications are good ones, with clear importance to students’ futures — savings plans, credit card debt, and taxes are all covered. The book then goes on to include a brief discussion of simple economic principles and what goes into a business plan, thus helping to drive home the message that mathematics is a subject that finds its way into all kinds of professions.
In the hands of a careful teacher, this book could be a fine addition to a middle or high school math class.
Mark Bollman (mbollman@albion.edu) is an associate professor of mathematics at Albion College in Michigan. His mathematical interests include number theory, probability, and geometry. His claim to be the only Project NExT fellow (Forest dot, 2002) who has taught both English composition and organic chemistry to college students has not, to his knowledge, been successfully contradicted. If it ever is, he is sure that his experience teaching introductory geology will break the deadlock.
Foreword by Charlotte Danielson
Introduction
Math Locator
1. Cars, Cars, and More Cars: Paying for Your Wheels
NCTM Standards Applied in This Chapter
Teaching Example 1.1. Car Costs
Teaching Example 1.2. MPG Plus
Teaching Example 1.3. Buy or Lease
2. Savings: Reality Check and the Difference Between Needs and Wants
NCTM Standards Applied in This Chapter
Background: Basics of Savings
Teaching Example 2.1. Budgets and Savings
Teaching Example 2.2. Future Value: A Geometric Sequence
Teaching Example 2.3. Annuities: A Geometric Series
3. Credit Cards and Debt Management: Using Credit Seems Unreal
NCTM Standards Applied in This Chapter
Background: Basics of Credit and Debit Cards
Teaching Example 3.1. Calculating Minimum Monthly Payments
Teaching Example 3.2. Credit Card Payoff: A Spreadsheet
4. Investments: The Basics and More
NCTM Standards Applied in This Chapter
Background: Basics of Investments
Teaching Example 4.1. Stock Prices and Percents
Teaching Example 4.2: Analyzing Graphs of Stock Prices
Teaching Example 4.3. Statistics of Stock I: Measures of Center, Standard Deviation, Normal Distribution
Teaching Example 4.4. Statistics of Stock II: Weighted Average, Correlation, Line of Best Fit
5. Buying a House Versus Renting: How Much Will It Cost?
NCTM Standards Applied in This Chapter
Background: Basics of Buying a House and Renting an Apartment
Teaching Example 5.1. Budgeting for a Down Payment
Teaching Example 5.2. Determining Mortgage Payments: Tables, Formulas, Spreadsheets
Teaching Example 5.3. Principal and Interest Functions
6. Taxes: How They Work
NCTM Standards Applied in This Chapter
Background: Basics of Income Taxes--Federal and State
Teaching Example 6.1. Income Tax and FICA
Teaching Example 6.2. Capital Gains and Losses
Teaching Example 6.3. How Uncle Sam Uses Tax Dollars
7. Economics: Supply and Demand, Inflation, and the GDP
NCTM Standards Applied in This Chapter
Background: Economics, Supply and Demand, Inflation, and the GDP
Teaching Example 7.1. Supply and Demand Functions
Teaching Example 7.2.: Inflation and the GDP | 1,046 | 4,661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | longest | en | 0.913511 |
https://www.physicsforums.com/threads/calculating-final-velocity-of-two-objects.688128/ | 1,532,058,446,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591481.75/warc/CC-MAIN-20180720022026-20180720042026-00604.warc.gz | 973,080,135 | 14,578 | # Homework Help: Calculating Final Velocity of Two Objects
1. Apr 26, 2013
### woolford85
A space vehicle travelling at a velocity of 100m/s separates by a controlled explosion into two sections of mass 850kg and 250kg. The two parts carry on in the same direction with the heavier rear section moving 120m/s slower than the lighter front section. Determine the final velocity of each section.
I have tried to look at a way of solving this using SUVAT but there are too many unknowns to use these equations. This question is part of an assignment I have to do, I need to know if it is even possible to do these calculations with the information that I have been given. If anyone could help me that would be great, thanks.
Michael
2. Apr 26, 2013
Use conservation of momentum
I guess we can use the law of perfect collision in 1-dimension,
Total mass initially be, M = 850 + 250 = 1100 kg
Total initial momentum will be, p = 1100*100 = 110000 kg.m/s
After the splitting, we have two masses. The velocities of heavier mass be v1and that of the lighter mass be v2.
Total momentum after splitting will be, 850v1 + 250v2
It is also given that heavier section moves at a speed 120 m/s slower than the lighter section. Therefore, v1 = v2 - 120
By the law of conservation of momentum, we have initial momentum = final momentum
110000 = 850v1 + 250v2
Substituting v1 = v2 - 120,
110000 = 850(v2 - 120) + 250v2
110000 = 1100v2 - 120*850
8000 = 1100v2
therefore, v2 = 7.273 m/s and
v1 = 127.273 m/s
Hope I am correct
3. Apr 26, 2013 | 431 | 1,539 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-30 | latest | en | 0.878717 |
https://getknowtrading.com/gbpaud-pip-value-calculator | 1,718,964,051,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862070.5/warc/CC-MAIN-20240621093735-20240621123735-00758.warc.gz | 245,053,250 | 44,988 | # GBPAUD Pip Value Calculator – How to Calculate
Aug 22, 2023Forex Calculator
The pip calculator in Forex represents a Forex calculator that calculates the value of a pip in the currency you want by defining following values:
• number of pips
• lot size used
• currency pair
• deposit currency
Why do you need GBPAUD pip value calculator?
You can calculate the value for a number of pips. And that means for example if you have 100 pips as a stop loss or take profit set, then you can calculate how much that will be in terms of a currency you select.
How to use GBPAUD pip value calculator?
Inside the calculator you have several fields you need to fill with the data. And those are the number of pips, currency pair, deposit currency and lot size. At the end you click the button Calculate and you get the value of a pip.
In this article I will show you all details you need to know about GBPAUD pip value calculator and what you can get with using it, why you should use it to speed up the process of calculating pip value and how to use it so you do not get confused when looking at pip calculator.
Read more: What is Pip Value Calculator
## GBPAUD Pip Value Calculator Example
The best way to explain how the GBPAUD calculator works is to show an example. I will make a few examples so you can see the difference in calculations.
First example is when you have GBPAUD currency pair with GBP as a deposit currency:
• Number of pips: 1
• Instrument: GBPAUD
• Lot size: 1.00 (100,000 units)
• Deposit currency: GBP
• GBPAUD pip size: 0.0001
Second example is when you have GBPAUD currency pair with AUD as a deposit currency:
• Number of pips: 1
• Instrument: GBPAUD
• Lot size: 1.00 (100,000 units)
• Deposit currency: AUD
• GBPAUD pip size: 0.0001
### How to Calculate Pips for GBPAUD
To calculate pips for GBPAUD you need to use following formula which defines the pip value:
For deposit currency which is equal to base currency, GBP:
Pip value = (Pip / Current market price) x Lot size
For deposit currency which is equal to quote currency, AUD:
Pip value = Pip x lot size
### GBPAUD Pip Value
Now with the formula you have you need to use the first formula where the deposit currency is GBP.
#### Pip Value for Base Currency
For deposit currency which is equal to base currency, GBP, pip value will be equal to:
Pip value = (Pip / Current market price) x Lot size
Here are other data you need:
• Number of pips: 1
• Instrument: GBPAUD = 0.8246
• Lot size: 1.00 (100,000 units)
• Deposit currency: GBP
• GBPAUD pip size: 0.0001
Now, when you put all the data in the formula you get:
Pip value = (Pip / Current market price) x Lot size
Pip value = (0.0001 / 0.8246) x 100,000
Pip value = (1.213e-4) x 100,000
Pip value = 12.13
#### Pip Value for Quote Currency
For deposit currency which is equal to quote currency, AUD, pip value will be equal to:
Pip value = Pip x Lot size
Here are other data you need:
• Number of pips: 1
• Instrument: GBPAUD = 0.8246
• Lot size: 1.00 (100,000 units)
• Deposit currency: AUD
• GBPAUD pip size: 0.0001
Now, when you put all the data in the formula you get:
Pip value = Pip x Lot size
Pip value = 0.0001 x 100,000
Pip value = AUD\$10,00
#### Pip Value for Third Currency
Third case is when you have a GBPAUD currency pair, but the deposit currency is USD. Which is not GBP or AUD.
This case requires that you make more calculations. If you use the GBPAUD calculator then the whole calculation is done by the pip calculator.
But, if you want to do it manually, then you need to use the following process.
First:
• decide in which currency you will calculate the pip value. Will that be GBP or AUD
Let’s use GBP. The formula for the pip value will be:
• Number of pips: 1
• Instrument: GBPAUD = 0.8246
• Lot size: 1.00 (100,000 units)
• Deposit currency: GBP
• GBPAUD pip size: 0.0001
Pip value = (Pip / Current market price) x Lot size
Pip value = (0.0001 / 0.8246) x 100,000
Pip value = (1.213e-4) x 100,000
Pip value = 12.13
Second:
Now, you need to use the GBP/USD currency pair so you can extract USD pip value from it.
Current market price for the GBP/USD = 0.6217. Which gives us GBP = 0.6217 USD.
So, the formula would be:
Pip value (USD) = Pip value (GBP) x (GBP/USD)
Pip value (USD) = 12.13 x 0.6217
Pip value (USD) = 7.54
## How Do You Calculate GBPAUD Pip Profits?
With the above calculated you can calculate GBPAUD pip profit.
Let’s say you have an open SELL order on the market with the GBPAUD = 0.8246.
And you want to close the trade at GBPAUD = 0.8226.
The price difference in pips is:
Pips = |Entry price – Exit price|
Pips = |0.8246 – 0.8226|
Pips = 0.20
You can see the difference is 20 pips between open and close price.
Now, the profit for 20 pips is:
Profit = Pip value x Pips
Profit = 7.54 x 20
Profit = \$150.8
If you want to use GBP as a deposit currency then the pip value is 12.13:
Profit = Pip value x Pips
Profit = 12.13 x 20
Profit = 242.6
## Conclusion
If you are beginner in Forex trading you will need this calculator because the calculation manually is not so easy.
Even the Forex basics is not so easy to understand, but in time you learn everything.
Use this calculator and other calculators to see how much you will make per pip so you can set stop loss and take profit properly.
### Disclaimer
Calculation`s made in the trading calculator are for informational purposes only. Whilst every effort is made to ensure the accuracy of this information, you should not rely upon it as being complete or up to date. Furthermore this information may be subject to change at any time.
Neither GETKNOWTRADING.com nor its employee or affiliates will be held responsible for the reliability or accuracy of this data. The service is provided in good faith.
However, there are no explicit or implicit warranties of accuracy. The user agrees not to hold GETKNOWTRADING.com or any of its affiliates, liable for trading decisions that are based on the calculators from this website.
“Disclosure: Some of the links in this post are “affiliate links.” This means if you click on the link and purchase the item, I will receive an affiliate commission. This does not cost you anything extra on the usual cost of the product, and may sometimes cost less as I have some affiliate discounts in place I can offer you”
#### Frano Grgić
A Forex trader since 2009. I like to share my knowledge and I like to analyze the markets. My goal is to have a website which will be the first choice for traders and beginners. Market analysis is featured by Forex Factory next to large publications like DailyFX, Bloomberg... GetKnowTrading is becoming recognized among traders as a website with simple and effective market analysis.
##### Forex Calculator
A list of all Forex Calculators | 1,790 | 6,820 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.85291 |
https://cstheory.stackexchange.com/questions/1590/machine-characterization-of-saci | 1,713,069,158,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816864.66/warc/CC-MAIN-20240414033458-20240414063458-00054.warc.gz | 178,791,078 | 40,039 | # Machine characterization of $SAC^i$
$SAC^i$ is the class of decision problems solvable by a family of $O({\log}^i{n})$ depth circuits with unbounded-fanin OR and bounded-fanin AND gates. Negations are only allowed at the input level. It is known that $SAC^i$ for $i \geq 1$ is closed under complement and $SAC^0$ is not. Also, $SAC^1 = LogCFL$ and hence has a machine characterization, since LogCFL is the set of languages accepted by an $O({\log}n)$ space bounded and polynomial time bounded auxiliary PDA. Are there similar machine characterizations of $SAC^i$ for $i \geq 2$ ?
• Are $k$ and $i$ meant to be the same thing? Sep 22, 2010 at 22:39
• Yes. Sorry for the typo. Fixed it now. Sep 23, 2010 at 0:29
## 1 Answer
Yes. Stack heights. $\mathsf{SAC^1} = \mathsf{NAuxPDA}(\log n, \log n)$, that is, with $O(\log n)$ space and $O(\log n)$ stack height; this implies $\log n$ configurations and therefore $\log^2(n)$ bits. We have
$$\mathsf{SAC^k} = \mathsf{NAuxPDA}(\log n, \log^k n);$$
these machines will run in time $2^{\log^{k}(n)}$. Without restriction on stack height, we will get exactly $\mathsf{P}$. The result should follow from: W. Ruzzo, Tree-size bounded alternation. JCSS 1980.
• Vinay, you can use regular latex in the answer: it might help make it a bit more readable Sep 23, 2010 at 7:06 | 429 | 1,317 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-18 | latest | en | 0.874395 |
https://classes.engineering.wustl.edu/2009/spring/mase5513/abaqus/docs/v6.6/books/usb/pt06ch23s03abo22.html | 1,675,029,259,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499768.15/warc/CC-MAIN-20230129211612-20230130001612-00670.warc.gz | 189,521,127 | 3,890 | ### 23.3.1 Beam modeling: overview
ABAQUS offers a wide range of beam modeling options.
### Overview
Beam modeling consists of:
### Determining whether beam modeling is appropriate
Beam theory is the one-dimensional approximation of a three-dimensional continuum. The reduction in dimensionality is a direct result of slenderness assumptions; that is, the dimensions of the cross-section are small compared to typical dimensions along the axis of the beam. The axial dimension must be interpreted as a global dimension (not the element length), such as
• distance between supports,
• distance between gross changes in cross-section, or
• wavelength of the highest vibration mode of interest.
In ABAQUS a beam element is a one-dimensional line element in three-dimensional space or in the XY plane that has stiffness associated with deformation of the line (the beam's “axis”). These deformations consist of axial stretch; curvature change (bending); and, in space, torsion. (Truss elements, Section 23.2.1, are one-dimensional line elements that have only axial stiffness.) Beam elements offer additional flexibility associated with transverse shear deformation between the beam's axis and its cross-section directions. Some beam elements in ABAQUS/Standard also include warping—nonuniform out-of-plane deformation of the beam's cross-section—as a nodal variable. The main advantage of beam elements is that they are geometrically simple and have few degrees of freedom. This simplicity is achieved by assuming that the member's deformation can be estimated entirely from variables that are functions of position along the beam axis only. Thus, a key issue in using beam elements is to judge whether such one-dimensional modeling is appropriate.
The fundamental assumption used is that the beam section (the intersection of the beam with a plane that is perpendicular to the beam axis; see the discussion in Choosing a beam cross-section, Section 23.3.2) cannot deform in its own plane (except for a constant change in cross-sectional area, which may be introduced in geometrically nonlinear analysis and causes a strain that is the same in all directions in the plane of the section). The implications of this assumption should be considered carefully in any use of beam elements, especially for cases involving large amounts of bending or axial tension/compression of non-solid cross-sections such as pipes, I-beams, and U-beams. Section collapse may occur and result in very weak behavior that is not predicted by beam theory. Similarly, thin-walled, curved pipes exhibit much softer bending behavior than would be predicted by beam theory because the pipe wall readily bends in its own section—another effect precluded by this basic assumption of beam theory. This effect, which must generally be considered when designing piping elbows, can be modeled by using shell elements to model the pipe as a three-dimensional shell (see Shell elements: overview, Section 23.6.1) or, in ABAQUS/Standard, by using elbow elements (see Pipes and pipebends with deforming cross-sections: elbow elements, Section 23.5.1).
In addition to beam elements, frame elements are provided in ABAQUS/Standard. These elements provide efficient modeling for design calculations of frame-like structures composed of initially straight, slender members. They operate directly in terms of axial force, bending moments, and torque at the element's end nodes. They are implemented for small or large displacements (large rotations with small strains) and permit the formation of plastic hinges at their ends through a “lumped” plasticity model that includes kinematic hardening. See Frame elements, Section 23.4.1, for details.
### Using beam elements in dynamic and eigenfrequency analysis
The rotary inertia of a beam cross-section is usually insignificant for slender beam structures, except for twist around the beam axis. Therefore, ABAQUS/Standard ignores rotary inertia of the cross-section for Euler-Bernoulli beam elements in bending. For thicker beams the rotary inertia plays a role in dynamic analysis, but to a lesser extent than shear deformation effects.
For Timoshenko beams the inertia properties are calculated from the cross-section geometry. The rotary inertia associated with torsional modes is different from that of flexural modes. For unsymmetric cross-sections the rotary inertia is different in each direction of bending. ABAQUS allows you to choose the rotary inertia formulation for Timoshenko beams. When an approximate isotropic formulation is requested, the rotary inertia associated with the torsional mode is used for all rotational degrees of freedom in ABAQUS/Standard, and a scaled flexural inertia with a scaling factor chosen to maximize the stable time increment is used for all rotational degrees of freedom in ABAQUS/Explicit. The center of mass of the cross-section is taken to be located at the beam node. When the exact (anisotropic) formulation is requested, the rotary inertia associated with bending and torsion differ and the coupling between the translational and rotational degrees of freedom is included for beam cross-section definitions where the beam node is not located at the center of mass of the cross-section. For Timoshenko beams with the exact (default) rotary inertia formulation, you can define an additional mass and rotary inertia contribution to the beam's inertia response that does not add to its structural stiffness; see Adding inertia to the beam section behavior for Timoshenko beams” in “Beam section behavior, Section 23.3.5. | 1,106 | 5,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-06 | latest | en | 0.929053 |
https://www.knowpia.com/knowpedia/Maximum_usable_frequency | 1,643,078,983,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304749.63/warc/CC-MAIN-20220125005757-20220125035757-00269.warc.gz | 892,364,060 | 15,566 | BREAKING NEWS
Maximum usable frequency
## Summary
In radio transmission maximum usable frequency (MUF) is the highest radio frequency that can be used for transmission between two points via reflection from the ionosphere (skywave or "skip" propagation) at a specified time, independent of transmitter power. This index is especially useful in regard to shortwave transmissions.
In shortwave radio communication, a major mode of long distance propagation is for the radio waves to reflect off the ionized layers of the atmosphere and return diagonally back to Earth. In this way radio waves can travel beyond the horizon, around the curve of the Earth. However the refractive index of the ionosphere decreases with increasing frequency, so there is an upper limit to the frequency which can be used. Above this frequency the radio waves are not reflected by the ionosphere but are transmitted through it into space.
The ionization of the atmosphere varies with time of day and season as well as with solar conditions, so the upper frequency limit for skywave communication varies on an hourly basis. MUF is a median frequency, defined as the highest frequency at which skywave communication is possible 50% of the days in a month, as opposed to the lowest usable high frequency (LUF) which is the frequency at which communication is possible 90% of the days, and the Frequency of optimum transmission (FOT).
Typically the MUF is a predicted number. Given the maximum observed frequency (MOF) for a mode on each day of the month at a given hour, the MUF is the highest frequency for which an ionospheric communications path is predicted on 50% of the days of the month.
On a given day, communications may or may not succeed at the MUF. Commonly, the optimal operating frequency for a given path is estimated at 80 to 90% of the MUF. As a rule of thumb the MUF is approximately 3 times the critical frequency.[1]
${\displaystyle {\text{MUF}}={\frac {\text{critical frequency}}{\cos \theta }}}$[2]
where the critical frequency is the highest frequency reflected for a signal propagating directly upward and θ is the angle of incidence.[3]
## Optimum Working Frequency
Another important parameter used in skywave propagation is the optimum working frequency (OWF), which estimates the maximum frequency that must be used for a given critical frequency and incident angle. It is the frequency chosen to avoid the irregularities of the atmosphere.
${\displaystyle \mathrm {OWF} =0.85\mathrm {MUF} =0.85f_{\mathrm {c} }/(cos\theta )}$ | 546 | 2,538 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-05 | latest | en | 0.920761 |
http://stackoverflow.com/questions/12041296/calculating-total-time-left-and-total-time-from-a-background-worker-onprogressch | 1,454,867,692,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701150206.8/warc/CC-MAIN-20160205193910-00181-ip-10-236-182-209.ec2.internal.warc.gz | 207,810,955 | 19,243 | # Calculating total time left and total time from a background worker OnProgressChanged event
I have a background worker reporting the progress of the computation. It returns the current line number it is processing in the ProgressChangedEventArgs. I can from there determine the percentage done by doing
``````(int) (100 * (double)e.ProgressPercentage / csTextLines)
``````
where csTextLines is the number of lines I am processing. The trouble I am running into is that I want to get a total estimated time and total time left also. There is nothing in background worker that I saw that would help me so I guess it has to be done with DateTime calculations. Here is what I have done so far to try and get TotalEstTime and TimeLeft
• TimeLeft = TotalEstTime - TimeSoFar;
• TotalEstTime = AvgTimeForEachLine * #OfLines;
Where
• TimeSoFar = DateTime.Now - m_startTime;
• AvgTimeForEachLine = (TimeFromLastLine + TotalTimeForAllLines) / #OfLines;
Where
• TimeFromLastLine = DateTime.Now - LastRecordedTime;
• TotalTimeForAllLines += TimeFromLastLine;
So if we turn this into code I have so far
``````// This event handler updates the progress.
private void backgroundWorker1_ProgressChanged(object sender, ProgressChangedEventArgs e)
{
TimeSpan difInCalls = DateTime.Now.Subtract(m_LastProgressCall);
m_LastProgressCall = DateTime.Now;
TimeSpan avgDif = new TimeSpan(0, 0, 0, 0, ((int)m_totalDif.TotalMilliseconds / m_csNumLines));
double totalTime = m_csNumLines * avgDif.TotalSeconds;
double timeLeft = totalTime - (DateTime.Now.Subtract(m_Form2Start).TotalSeconds);
Console.WriteLine("TotalEstTime: " + totalTime + " TimeLeft: " + timeLeft);
// Update the progress label
resultLabel.Text = "Line " + e.ProgressPercentage.ToString() + " of " + m_csNumLines + " at " + (int)(100 * (double)e.ProgressPercentage / m_csNumLines) + "% loaded";
}
``````
The problems that I am having are that my TotalTime grows significantly with how many lines are processed. In a 1437 line string it starts out at saying I have 1.4s and at the end it estimates 18.6s left. While actually taking 16.54s to complete.
My second problem is that the time left, does not really change. It fluctuates between about 1.1 and 2.5.
What could there be attributed to? What could be the problem?
Here is the full code: http://pastebin.com/x1CCceY7
-
The problem is that you're working off the last progress time, which will likely fluctuate as you go. If you work from the starting time and the current time, and use the percentage complete, then you'll find that the time "self adjusts" as you move along.
For example, if you store `DateTime.Now` into a variable (such as `m_operationStart`, using your naming), you could write:
``````private void backgroundWorker1_ProgressChanged(object sender, ProgressChangedEventArgs e)
{
if (e.ProgressPercentage != 0)
{
double percentageComplete = (double)e.ProgressPercentage / m_csNumLines;
TimeSpan timeSinceStart = DateTime.Now.Subtract(m_operationStart);
TimeSpan totalTime = TimeSpan.FromMilliseconds(timeSinceStart.TotalMilliseconds / percentageComplete);
TimeSpan timeLeft = totalTime - timeSinceStart;
Console.WriteLine("TotalEstTime: " + totalTime + " TimeLeft: " + timeLeft);
// Update the progress label
resultLabel.Text = "Line " + e.ProgressPercentage.ToString() + " of " + m_csNumLines + " at " + (int)(100.0 * percentageComplete) + "% loaded";
}
else
resultLabel.Text = "Line " + e.ProgressPercentage.ToString() + " of " + m_csNumLines;
}
``````
-
This code makes sense when I read it over, but whenever I try to execute the code I get a OverFlow exception. It is just when calculating the TotalTime. – AnotherUser Aug 20 '12 at 17:13
@user1596244 This should work - the ticks was overflowing since the numbers (in your case) were too large. I believe using Milliseconds should work fine, though. – Reed Copsey Aug 20 '12 at 17:14
Still overflowing, Ill try seconds.. -- edit I tried seconds and it too overflowed. – AnotherUser Aug 20 '12 at 17:22
@user1596244 Try my latest - I think that will work properly – Reed Copsey Aug 20 '12 at 17:27
Yeah it overflows with ticks, milliseconds, and seconds, and minutes. – AnotherUser Aug 20 '12 at 17:32 | 1,044 | 4,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-07 | latest | en | 0.7801 |
https://amara.org/ja/teams/udacity/videos/?page=1&project=cs046 | 1,619,032,885,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039550330.88/warc/CC-MAIN-20210421191857-20210421221857-00065.warc.gz | 202,096,184 | 25,264 | ## Udacity
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• one: ⏎ 5 言語⏎ other: ⏎ 5 言語⏎ | 600 | 1,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-17 | latest | en | 0.480425 |
https://proofwiki.org/wiki/Fourier_Series/Sawtooth_Wave/Special_Cases | 1,637,978,585,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358078.2/warc/CC-MAIN-20211127013935-20211127043935-00584.warc.gz | 560,266,975 | 10,368 | # Fourier Series/Sawtooth Wave/Special Cases
## Special Cases of Fourier Series for Sawtooth Wave
### Unit Half Interval
Let $\map S x$ be the sawtooth wave defined on the real numbers $\R$ as:
$\forall x \in \R: \map S x = \begin {cases} x & : x \in \openint {-1} 1 \\ \map S {x + 2} & : x < -1 \\ \map S {x - 2} & : x > +1 \end {cases}$
Then its Fourier series can be expressed as:
$\ds \map S x$ $\sim$ $\ds \frac 2 \pi \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin n \pi x$ $\ds$ $=$ $\ds \frac 2 \pi \paren {\sin \pi x - \frac {\sin 2 \pi x} 2 + \frac {\sin 3 \pi x} 3 + \dotsb}$
### Half Interval $\pi$
Let $\map S x$ be the sawtooth wave defined on the real numbers $\R$ as:
$\forall x \in \R: \map S x = \begin {cases} x & : x \in \openint {-\pi} \pi \\ \map S {x + 2 \pi} & : x < -\pi \\ \map S {x - 2 \pi} & : x > +\pi \end {cases}$
Then its Fourier series can be expressed as:
$\ds \map S x$ $\sim$ $\ds 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin n x$ $\ds$ $=$ $\ds 2 \paren {\sin x - \frac {\sin 2 x} 2 + \frac {\sin 3 x} 3 + \dotsb}$ | 440 | 1,101 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.327326 |
https://oeis.org/A000304 | 1,503,079,491,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105086.81/warc/CC-MAIN-20170818175604-20170818195604-00079.warc.gz | 812,327,483 | 4,446 | This site is supported by donations to The OEIS Foundation.
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A000304 a(n) = a(n-1)*a(n-2). 7
2, 3, 6, 18, 108, 1944, 209952, 408146688, 85691213438976, 34974584955819144511488, 2997014624388697307377363936018956288, 104819342594514896999066634490728502944926883876041385836544 (list; graph; refs; listen; history; text; internal format)
OFFSET 2,1 COMMENTS A038500(a(n)) = A010098(n-2); for n > 2: A006519(a(n)) = A000301(n-3); A001222(a(n)) = A000045(n-1). - Reinhard Zumkeller, Jul 06 2014 LINKS T. D. Noe, Table of n, a(n) for n = 2..18 Sergio Falcon, Fibonacci's multiplicative sequence, Int. J. Math. Edu. Sci. Technol. 34-2 (2003), 310-315. [Sergio Falcon, Nov 23 2009] FORMULA For n>=4, a(n) = 2^A000045(n-3)*3^A000045(n-2). - Benoit Cloitre, Sep 26 2003 For n > 2: a(n) = A000301(n-3) * A010098(n-2). - Reinhard Zumkeller, Jul 06 2014 MAPLE A000304 := proc(n) option remember; if n <=3 then n else A000304(n-1)*A000304(n-2); fi; end; a[0]:=2: a[1]:=3: for n from 2 to 13 do a[n]:=a[n-1]*a[n-2] od: seq(a[n], n=0..10); # Zerinvary Lajos, Mar 19 2009 MATHEMATICA a=2; b=3; lst={a, b}; Do[z=a*b; AppendTo[lst, z]; a=b; b=z, {n, 12}]; lst (* Vladimir Joseph Stephan Orlovsky, Feb 17 2010 *) t={2, 3}; Do[AppendTo[t, t[[-1]]*t[[-2]]], {n, 15}]; t (* Vladimir Joseph Stephan Orlovsky, Nov 23 2010 *) nxt[{a_, b_}]:={b, a*b}; Transpose[NestList[nxt, {2, 3}, 12]][[1]] (* Harvey P. Dale, Nov 16 2014 *) PROG (Haskell) a000304 n = a000304_list !! (n-2) a000304_list = 2 : 3 : zipWith (*) a000304_list (tail a000304_list) -- Reinhard Zumkeller, Jul 06 2014 CROSSREFS Cf. A000045, A000301, A001222, A006519, A010098, A038500, A249406. Sequence in context: A093468 A185625 A114302 * A000614 A233239 A018290 Adjacent sequences: A000301 A000302 A000303 * A000305 A000306 A000307 KEYWORD nonn AUTHOR EXTENSIONS More terms from Vladimir Joseph Stephan Orlovsky, Feb 17 2010 STATUS approved
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The OEIS Community | Maintained by The OEIS Foundation Inc. | 877 | 2,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2017-34 | longest | en | 0.410676 |
http://stackoverflow.com/questions/6809814/implementing-multicore-threading-on-this-algorithm?answertab=votes | 1,438,267,356,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042987228.91/warc/CC-MAIN-20150728002307-00016-ip-10-236-191-2.ec2.internal.warc.gz | 223,052,727 | 17,816 | # Implementing Multicore Threading on this algorithm
I'm trying to find a way to make the following algorithm being processed on multiple cores, but I don't get on a good point. Using a locked iterator shared between multiple processes wouldn't be the most efficient way I think.
`````` def sortCharset(set):
_set = ""
for c in set:
if c not in _set:
_set += c
set = _set
del _set
set = list(set)
set.sort()
return "".join(set)
def stringForInt(num, set, length):
setLen = len(set)
string = ""
string += set[num % setLen]
for n in xrange(1,length):
num //= setLen
string += set[num % setLen]
return string
def bruteforce(set, length, raw = False):
if raw is False:
set = sortCharset(set)
for n in xrange(len(set) ** length):
yield stringForInt(n, set, length)
``````
Short explanation: The code is used to create every possible combination from a set of chars, i.e. to hack a password. (Of course not my intention, just some Py-training. ;-)
What is a good way to run this algorithm on multiple cores ?
-
Thanks for the edit. I still don't get the StackExchange formatting. – Niklas R Jul 24 '11 at 21:25
Instead of `charset = sortCharset(charset)` just use `charset = sorted(set(charset))`. (There's no need for `del _set` because `_set` is local anyway.) Also, `not raw` is better than `raw is False`. – MRAB Jul 24 '11 at 21:37
Thanks, the sorted(.. one is nice ! I've always an ear for suggestions on how to make code better. – Niklas R Jul 24 '11 at 21:41
Also, as a general principle, don't call a variable `set`, because `set` is a built in type, and replacing it just leads to confusion. – Thomas K Jul 24 '11 at 22:40
I think you might be looking for itertools.permutations - have a look at the docs, see if it's doing what you want. docs.python.org/library/itertools.html#itertools.permutations – Thomas K Jul 24 '11 at 22:43
The question isn't really about naming style or how to get a sorted set of characters out of a string.
You might want to look into the multiprocessing module. I'm pretty much a n00b w/r/t multi-core parallelism but got something working:
``````import multiprocessing, itertools
def stringForInt(args):
num, charset, length = args ## hack hack hack
setlen = len(charset)
s = []
s.append(charset[num % setlen])
for n in xrange(1, length):
num //= setlen
s.append(charset[num % setlen])
return ''.join(s)
def bruteforce(charset, length, mapper, raw=False):
if not raw:
charset = sorted(set(charset))
return mapper(stringForInt, ((n,charset,length) for n in xrange(len(charset)**length)))
if __name__ == '__main__':
import time, sys
if len(sys.argv) == 1 or sys.argv[1] == 'map':
mapper = map
else:
p = multiprocessing.Pool()
pfunc = {'pmap':p.map,
'imap':p.imap,
'imapu':p.imap_unordered}[sys.argv[1]]
mapper = lambda f, i: pfunc(f, i, chunksize=5)
o = bruteforce('abcdefghijk',6,mapper)
if not isinstance(o, list):
list(o)
``````
The nature of the hack is that you need to use pickleable objects for the functions in `multiprocessing` and only functions that are defined at the top-level are can be pickled. (There would be other ways around this using `multiprocessing.Value` or `multiprocessing.Manager` but they aren't really worth going into for present purposes.)
Here's output for various runs:
``````\$ for x in map pmap imap imapu ; do time python mp.py \$x; done
real 0m9.351s
user 0m9.253s
sys 0m0.096s
real 0m10.523s
user 0m20.753s
sys 0m0.176s
real 0m4.081s
user 0m13.797s
sys 0m0.276s
real 0m4.215s
user 0m14.013s
sys 0m0.236s
``````
- | 1,018 | 3,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2015-32 | latest | en | 0.838525 |
http://html.rhhz.net/jckxjsgw/html/61918.htm | 1,721,392,337,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514902.63/warc/CC-MAIN-20240719105029-20240719135029-00646.warc.gz | 14,335,833 | 11,303 | 空投鱼雷水中弹道以及弹道散布模拟研究
舰船科学技术 2020, Vol. 42 Issue (6): 180-184 DOI: 10.3404/j.issn.1672-7649.2020.06.037 PDF
1. 中国船舶及海洋工程设计研究院 上海 200001;
2. 哈尔滨工程大学 船舶工程学院 黑龙江 哈尔滨 150001
The simulation research of underwater trajectory and trajectory distribution about airdropped torpedo
ZHU Xin1, SHI Zhang2, HUANG Li-min2
1. Marine Design and Research Institute of China, Shanghai 200001, China;
2. College of Shipbuilding Engineering, Harbin Engineering University, Harbin 150001, China
Abstract: The accurate prediction of underwater trajectory of airdropped torpedo is vital to attack specific targets. Lots of factors can influence the underwater trajectory of airdropped torpedo due to the complex marine environment. In order to analyze the influence of current velocity and torpedo attitude to the torpedo trajectory, based on six-degree motion equation, this thesis adds the influence of random current to the motion equation to stimulate the underwater trajectory of torpedo as well as the torpedo's trajectory distribution under different current velocity and attitude. And finally the corresponding simulation results are posed and the distribution regularity of torpedo trajectory is discussed as well.
Key words: airdropped torpedo underwater trajectory random current distribute regular
0 引 言
1 鱼雷水中运动模型
1.1 运动坐标系
图 1 运动坐标系 Fig. 1 Motion coordinate system
1.2 动力学方程
${{ A}_m}\left( \begin{array}{l} {{\dot v}_x} \\ {{\dot v}_y} \\ {{\dot v}_z} \\ {{\dot \omega }_x} \\ {{\dot \omega }_y} \\ {{\dot \omega }_z} \\ \end{array} \right)+{A_{vw}}\left( {{A_m}\left( \begin{array}{l} {v_x} \\ {v_y} \\ {v_z} \\ {\omega _x} \\ {\omega _y} \\ {\omega _z} \\ \end{array} \right)} \right) = {A_F}\text{。}$ (1)
${{ A}_{ m}} = \left[\!\!\! {\begin{array}{*{20}{c}} m&0&0&0&{m{z_c}}&{ - m{y_c}} \\ 0&m&0&{ - m{z_c}}&0&{m{x_c}} \\ 0&0&m&{m{y_c}}&{ - m{x_c}}&0 \\ 0&{ - m{z_c}}&{m{y_c}}&{{J_{xx}}}&{ - {J_{xy}}}&{ - {J_{xz}}} \\ {m{z_c}}&0&{ - m{x_c}}&{ - {J_{yx}}}&{{J_{yy}}}&{ - {J_{yz}}} \\ { - m{y_c}}&{m{x_c}}&0&{ - {J_{zx}}}&{ - {J_{zy}}}&{{J_{zz}}} \end{array}} \!\!\!\right]\text{，}$ (2)
${{ A}_{{vw}}}$ 是速度矩阵，且有
${{ A}_{{vw}}} = \left[ {\begin{array}{*{20}{c}} 0&{ - {\omega _z}}&{{\omega _y}}&0&0&0 \\ {{\omega _z}}&0&{ - {\omega _x}}&0&0&0 \\ { - {\omega _y}}&{{\omega _x}}&0&0&0&0 \\ 0&{ - {v_z}}&{{v_y}}&0&{ - {\omega _z}}&{{\omega _y}} \\ {{v_z}}&0&{ - {v_x}}&{{\omega _z}}&0&{ - {\omega _x}} \\ { - {v_y}}&{{v_x}}&0&{ - {\omega _y}}&{{\omega _x}}&0 \end{array}} \right]\text{，}$ (3)
${{ A}_{ F}}$ 是力矩阵，且有
${{ A}_{{F}}} = \left[ {\begin{array}{*{20}{c}} {{F_x}} \\ {{F_y}} \\ {{F_z}} \\ {{M_x}} \\ {{M_y}} \\ {{M_z}} \end{array}} \right] = \left[ \begin{array}{l} {F_{x1}} - {F_{x2}} - {F_{x3}} \\ {F_{y1}} - {F_{y2}} - {F_{y3}} \\ {F_g} - {F_f} - {F_{z2}} - {F_{z3}} \\ {M_{x1}} - {M_{x2}} \\ {M_{y1}} - {M_{y2}} \\ {M_{z1}} - {M_{z2}} \\ \end{array} \right]\text{。}$ (4)
\begin{aligned} {F_1} = & 0.5\rho {A_w}{C_d}{u^2},\;{F_2} = 0.5\rho {A_w}{C_d}{V^2},\; \hfill \\ {F_3} =& 0.5\rho {A_w}V{C_d}wL \text{，} \end{aligned} (5)
${M_1} = 0.5\rho {A_w}{V^2}{C_d}L, \;{M_2} = 0.5\rho {A_w}V{C_d}w{L^2}\text{。}$ (6)
2 数值模拟
2.1 均匀海流下鱼雷的水中弹道模拟分析
图 2 鱼雷水下弹道 Fig. 2 The underwater trajectory of torpedo
图 3 鱼雷速度变化曲线 Fig. 3 The curve of torpedo speed
2.2 随机海流下鱼雷的弹道散布模拟分析
图 4 海流速度幅值 Fig. 4 The velocity amplitude of current
图 5 随机海流作用下的鱼雷弹道落点散布图 Fig. 5 Torpedo trajectory sinking position distribution under random current
3 结 语
[1] CHU P C, BUSHNELL J M , FAN Chen-wu, et al. Modeling of underwater bomb trajectory for mine clearance[J]. The journal of defense modeling and simulation: Application, Methodology, Technology: JDMS, 2011, 8(1): 25-36. DOI:10.1177/1548512910387807 [2] DING Bao-Jun. Simulation analysis of underwater trajectory of air-dropped mine[J]. The journal of Machinery management and development, 2009, 24(5): 51-52. [3] GU Chuang. Research on the influence of sea current on underwater trajectory of submarine mine[J]. The journal of Ship electronic engineering, 2010, 30(2): 168-171. [4] FANG Yan-Fei. Modeling and simulation on mine’s trajectory of underwater based on ocean current velocity and direction[J]. The journal of ship science and technology, 2013, 35(8): 57-61. [5] TAO Li-run, ZHE Chao. Simulation and analysis of the underwater trajectory of small bag sand filling[J]. The journal of China harbour construction, 2017, 37(3): 18-21. [6] GANG Wen. Study of dispersion of a certain type hydrobomb under different directions and velocities of seawater[J]. The journal of Wuhan university of technology (Transportation Science& Engineering), 2007(01): 39-42. [7] ZHANG Rui, YUAN Zhi-yong, LIU Zhong-le, et al. Anti-torpedo-torpedo intercepting trajectory and analysis of hit probability under sliding mode guidance. In: Proceedings of 10th International Conference on Intelligent Human-Machine Systems and Cybernetics, IHMSC 2018, v 1, p 133-137, November 9, 2018. [8] DUAN Wen-yang, SHI Zhang, CHEN Yun-sai, et al. Research on the probability distribution of the underwater moving of the wrecked targets. In: 38th International Conference on Ocean, Offshore & Arctic Engineer, OMAE2019, June 12, 2019. | 1,948 | 5,174 | {"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} | 2.734375 | 3 | CC-MAIN-2024-30 | latest | en | 0.529704 |
http://www.nag.com/numeric/FL/nagdoc_fl24/html/E02/e02jdf.html | 1,386,309,520,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163049615/warc/CC-MAIN-20131204131729-00019-ip-10-33-133-15.ec2.internal.warc.gz | 446,904,692 | 24,565 | E02 Chapter Contents
E02 Chapter Introduction
NAG Library Manual
# NAG Library Routine DocumentE02JDF
Note: before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details.
Note: this routine uses optional parameters to define choices in the problem specification and in the details of the algorithm. If you wish to use default settings for all of the optional parameters, you need only read Sections 1 to 9 of this document. If, however, you wish to reset some or all of the settings please refer to Section 10 for a detailed description of the specification of the optional parameters produced by the routine.
## 1 Purpose
E02JDF computes a spline approximation to a set of scattered data using a two-stage approximation method.
The computational complexity of the method grows linearly with the number of data points; hence large datasets are easily accommodated.
## 2 Specification
SUBROUTINE E02JDF ( N, X, Y, F, LSMINP, LSMAXP, NXCELS, NYCELS, LCOEFS, COEFS, IOPTS, OPTS, IFAIL)
INTEGER N, LSMINP, LSMAXP, NXCELS, NYCELS, LCOEFS, IOPTS(*), IFAIL REAL (KIND=nag_wp) X(N), Y(N), F(N), COEFS(LCOEFS), OPTS(*)
Before calling E02JDF, E02ZKF must be called with OPTSTR set to ‘’. Settings for optional algorithmic parameters may be specified by calling E02ZKF before a call to E02JDF.
## 3 Description
E02JDF determines a smooth bivariate spline approximation to a set of data points $\left({x}_{\mathit{i}},{y}_{\mathit{i}},{f}_{\mathit{i}}\right)$, for $\mathit{i}=1,2,\dots ,n$. Here, ‘smooth’ means ${C}^{1}$.
The approximation domain is the bounding box $\left[{x}_{\mathrm{min}},{x}_{\mathrm{max}}\right]×\left[{y}_{\mathrm{min}},{y}_{\mathrm{max}}\right]$, where ${x}_{\mathrm{min}}$ (respectively ${y}_{\mathrm{min}}$) and ${x}_{\mathrm{max}}$ (respectively ${y}_{\mathrm{max}}$) denote the lowest and highest data values of the $\left({x}_{i}\right)$ (respectively $\left({y}_{i}\right)$).
The spline is computed by local approximations on a uniform triangulation of the bounding box. These approximations are extended to a smooth spline representation of the surface over the domain. The local approximation scheme is by least-squares polynomials (Davydov and Zeilfelder (2004)).
The two-stage approximation method employed by E02JDF is derived from the TSFIT package of O. Davydov and F. Zeilfelder.
Values of the computed spline can subsequently be computed by calling E02JEF or E02JFF.
## 4 References
Davydov O and Zeilfelder F (2004) Scattered data fitting by direct extension of local polynomials to bivariate splines Advances in Comp. Math. 21 223–271
## 5 Parameters
1: N – INTEGERInput
On entry: $n$, the number of data values to be fitted.
Constraint: ${\mathbf{N}}>1$.
2: X(N) – REAL (KIND=nag_wp) arrayInput
3: Y(N) – REAL (KIND=nag_wp) arrayInput
4: F(N) – REAL (KIND=nag_wp) arrayInput
On entry: the $\left({x}_{i},{y}_{i},{f}_{i}\right)$ data values to be fitted.
Constraint: ${\mathbf{X}}\left(j\right)\ne {\mathbf{X}}\left(1\right)$ for some $j=2,\dots ,n$ and ${\mathbf{Y}}\left(k\right)\ne {\mathbf{Y}}\left(1\right)$ for some $k=2,\dots ,n$; i.e., there are at least two distinct $x$ and $y$ values.
5: LSMINP – INTEGERInput
6: LSMAXP – INTEGERInput
On entry: are control parameters for the local approximations.
Each local approximation is computed on a local domain containing one of the triangles in the discretization of the bounding box. The size of each local domain will be adaptively chosen such that if it contains fewer than LSMINP sample points it is expanded, else if it contains greater than LSMAXP sample points a thinning method is applied. LSMAXP mainly controls computational cost (in that working with a thinned set of points is cheaper and may be appropriate if the input data is densely distributed), while LSMINP allows handling of different types of scattered data.
Setting ${\mathbf{LSMAXP}}<{\mathbf{LSMINP}}$, and therefore forcing either expansion or thinning, may be useful for computing initial coarse approximations. In general smaller values for these arguments reduces cost.
A calibration procedure (experimenting with a small subset of the data to be fitted and validating the results) may be needed to choose the most appropriate values for LSMINP and LSMAXP.
Constraints:
• $1\le {\mathbf{LSMINP}}\le {\mathbf{N}}$;
• ${\mathbf{LSMAXP}}\ge 1$.
7: NXCELS – INTEGERInput
8: NYCELS – INTEGERInput
On entry: NXCELS (respectively NYCELS) is the number of cells in the $x$ (respectively $y$) direction that will be used to create the triangulation of the bounding box of the domain of the function to be fitted.
Greater efficiency generally comes when NXCELS and NYCELS are chosen to be of the same order of magnitude and are such that N is $\mathit{O}\left({\mathbf{NXCELS}}×{\mathbf{NYCELS}}\right)$. Thus for a ‘square’ triangulation — when ${\mathbf{NXCELS}}={\mathbf{NYCELS}}$ — the quantities $\sqrt{{\mathbf{N}}}$ and NXCELS should be of the same order of magnitude. See also Section 8.
Constraints:
• ${\mathbf{NXCELS}}\ge 1$;
• ${\mathbf{NYCELS}}\ge 1$.
9: LCOEFS – INTEGERInput
10: COEFS(LCOEFS) – REAL (KIND=nag_wp) arrayOutput
On exit: if ${\mathbf{IFAIL}}={\mathbf{0}}$ on exit, COEFS contains the computed spline coefficients.
Constraint: ${\mathbf{LCOEFS}}\ge \left(\left(\left({\mathbf{NXCELS}}+2\right)×\left({\mathbf{NYCELS}}+2\right)+1\right)/2\right)×10+1$.
11: IOPTS($*$) – INTEGER arrayCommunication Array
On entry: the contents of IOPTS must not be modified in any way either directly or indirectly, by further calls to E02ZKF, before calling either or both of the evaluation routines E02JEF and E02JFF.
12: OPTS($*$) – REAL (KIND=nag_wp) arrayCommunication Array
On entry: the contents of OPTS must not be modified in any way either directly or indirectly, by further calls to E02ZKF, before calling either or both of the evaluation routines E02JEF and E02JFF.
13: IFAIL – INTEGERInput/Output
On entry: IFAIL must be set to $0$, $-1\text{ or }1$. If you are unfamiliar with this parameter you should refer to Section 3.3 in the Essential Introduction for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\text{ or }1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this parameter, the recommended value is $0$. When the value $-\mathbf{1}\text{ or }\mathbf{1}$ is used it is essential to test the value of IFAIL on exit.
On exit: ${\mathbf{IFAIL}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
## 6 Error Indicators and Warnings
If on entry ${\mathbf{IFAIL}}={\mathbf{0}}$ or $-{\mathbf{1}}$, explanatory error messages are output on the current error message unit (as defined by X04AAF).
Errors or warnings detected by the routine:
${\mathbf{IFAIL}}=2$
On entry, ${\mathbf{N}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{N}}>1$.
${\mathbf{IFAIL}}=4$
On entry, ${\mathbf{LSMINP}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{N}}=⟨\mathit{\text{value}}⟩$.
Constraint: $1\le {\mathbf{LSMINP}}\le {\mathbf{N}}$.
${\mathbf{IFAIL}}=5$
On entry, ${\mathbf{LSMAXP}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{LSMAXP}}\ge 1$.
${\mathbf{IFAIL}}=6$
On entry, ${\mathbf{NXCELS}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{NXCELS}}\ge 1$.
${\mathbf{IFAIL}}=7$
On entry, ${\mathbf{NYCELS}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{NYCELS}}\ge 1$.
${\mathbf{IFAIL}}=8$
On entry, ${\mathbf{LCOEFS}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{LCOEFS}}\ge \left(\left(\left({\mathbf{NXCELS}}+2\right)×\left({\mathbf{NYCELS}}+2\right)+1\right)/2\right)×10+1$.
${\mathbf{IFAIL}}=9$
Option arrays are not initialized or are corrupted.
${\mathbf{IFAIL}}=11$
${\mathbf{IFAIL}}=12$
On entry, all elements of X or of Y are equal.
${\mathbf{IFAIL}}=21$
The value of optional parameter Polynomial Starting Degree was invalid.
${\mathbf{IFAIL}}=-999$
Dynamic memory allocation failed.
## 7 Accuracy
Technical results on error bounds can be found in Davydov and Zeilfelder (2004).
Local approximation by polynomials of degree $d$ for $n$ data points has optimal approximation order ${n}^{-\left(d+1\right)/2}$.
The approximation error for ${C}^{1}$ global smoothing is $\mathit{O}\left({n}^{-2}\right)$.
Whether maximal accuracy is achieved depends on the distribution of the input data and the choices of the algorithmic parameters. The reference above contains extensive numerical tests and further technical discussions of how best to configure the method.
$n$-linear complexity and memory usage can be attained for sufficiently dense input data if the triangulation parameters NXCELS and NYCELS are chosen as recommended in their descriptions above. For sparse input data on such triangulations, if many expansion steps are required (see LSMINP) the complexity may rise to be loglinear.
## 9 Example
The Franke function
$fx,y = 0.75 exp - 9x-2 2 + 9y-2 2 / 4 + 0.75 exp - 9x+1 2 / 49 - 9y+1 / 10 + 0.5 exp - 9x-7 2 + 9y-3 2 / 4 - 0.2 exp - 9x-4 2 - 9y-7 2$
is widely used for testing surface-fitting methods. The example program randomly generates a number of points on this surface. From these a spline is computed and then evaluated at a vector of points and on a mesh.
### 9.1 Program Text
Program Text (e02jdfe.f90)
### 9.2 Program Data
Program Data (e02jdfe.d)
### 9.3 Program Results
Program Results (e02jdfe.r)
## 10 Optional Parameters
Several optional parameters in E02JDF control aspects of the algorithm, methodology used, logic or output. Their values are contained in the arrays IOPTS and OPTS; these must be initialized before calling E02JDF by first calling E02ZKF with OPTSTR set to ‘’.
Each optional parameter has an associated default value; to set any of them to a nondefault value, or to reset any of them to the default value, use E02ZKF. The current value of an optional parameter can be queried using E02ZLF.
The remainder of this section can be skipped if you wish to use the default values for all optional parameters.
The following is a list of the optional parameters available. A full description of each optional parameter is provided in Section 10.1.
### 10.1 Description of the Optional Parameters
For each option, we give a summary line, a description of the optional parameter and details of constraints.
The summary line contains:
• the keywords;
• a parameter value, where the letters $a$, $i\text{ and }r$ denote options that take character, integer and real values respectively;
• the default value.
Keywords and character values are case insensitive.
For E02JDF the maximum length of the parameter CVALUE used by E02ZLF is $6$.
Averaged Spline $a$ Default $\text{}=\text{'NO'}$
When the bounding box is triangulated there are 8 equivalent configurations of the mesh. Setting ${\mathbf{Averaged Spline}}=\text{'YES'}$ will use the averaged value of the $8$ possible local polynomial approximations over each triangle in the mesh. This usually gives better results but at (about 8 times) higher computational cost.
Constraint: ${\mathbf{Averaged Spline}}=\text{'YES'}$ or $\text{'NO'}$.
Minimum Singular Value LPA $r$ Default $\text{}=1.0$
A tolerance measure for accepting or rejecting a local polynomial approximation (LPA) as reliable.
The solution of a local least squares problem solved on each triangle subdomain is accepted as reliable if the minimum singular value $\sigma$ of the matrix (of Bernstein polynomial values) associated with the least squares problem satisfies ${\mathbf{Minimum Singular Value LPA}}\le \sigma$.
In general the approximation power will be reduced as Minimum Singular Value LPA is reduced. (A small $\sigma$ indicates that the local data has hidden redundancies which prevent it from carrying enough information for a good approximation to be made.) Setting Minimum Singular Value LPA very large may have the detrimental effect that only approximations of low degree are deemed reliable.
Minimum Singular Value LPA will have no effect if ${\mathbf{Polynomial Starting Degree}}=0$, and it will have little effect if the input data is ‘smooth’ (e.g., from a known function).
A calibration procedure (experimenting with a small subset of the data to be fitted and validating the results) may be needed to choose the most appropriate value for this parameter.
Constraint: ${\mathbf{Minimum Singular Value LPA}}\ge 0.0$.
Polynomial Starting Degree $i$ Default $\text{}=1$
The degree to be used in the initial step of each local polynomial approximation.
At the initial step the method will attempt to fit with local polynomials of degree Polynomial Starting Degree. If the approximation is deemed unreliable (according to Minimum Singular Value LPA), the degree will be decremented by one and a new local approximation computed, ending with a constant approximation if no other is reliable.
Polynomial Starting Degree is bounded from above by the maximum possible spline degree, $3$.
The default value gives a good compromise between efficiency and accuracy. In general the best approximation can be obtained by setting ${\mathbf{Polynomial Starting Degree}}=3$.
Constraint: $0\le {\mathbf{Polynomial Starting Degree}}\le 3$. | 3,792 | 13,415 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 94, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2013-48 | longest | en | 0.633702 |
https://rdrr.io/cran/Riemann/src/R/special_sphere.R | 1,675,246,885,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499919.70/warc/CC-MAIN-20230201081311-20230201111311-00034.warc.gz | 508,912,541 | 12,199 | # R/special_sphere.R In Riemann: Learning with Data on Riemannian Manifolds
#### Documented in sphere.geo2xyzsphere.runifsphere.utestsphere.xyz2geo
## SOME SPECIAL FUNCTIONS ON SPHERE
# (01) sphere.runif
# (02) sphere.utest
# (03) sphere.convert
# (01) sphere.runif ============================================================
#' Generate Uniform Samples on Sphere
#'
#' It generates \eqn{n} random samples from \eqn{\mathcal{S}^{p-1}}. For convenient
#' usage of users, we provide a number of options in terms of the return type.
#'
#' @param n number of samples to be generated.
#' @param p original dimension (of the ambient space).
#' @param type return type; \describe{
#' \item{\code{"list"}}{a length-\eqn{n} list of length-\eqn{p} vectors.}
#' \item{\code{"matrix"}}{a \eqn{(n\times p)} where rows are unit vectors.}
#' \item{\code{"riemdata"}}{a S3 object. See \code{\link{wrap.sphere}} for more details (\emph{Default}).}
#' }
#'
#' @return an object from one of the above by \code{type} option.
#'
#' @examples
#' #-------------------------------------------------------------------
#' # Draw Samples on Sphere
#' #
#' # Multiple return types on S^4 in R^5
#' #-------------------------------------------------------------------
#' dat.list = sphere.runif(n=10, p=5, type="list")
#' dat.matx = sphere.runif(n=10, p=5, type="matrix")
#' dat.riem = sphere.runif(n=10, p=5, type="riemdata")
#'
#' @references
#' \insertRef{chikuse_statistics_2003}{Riemann}
#'
#' @concept sphere
#' @export
sphere.runif <- function(n, p, type=c("list","matrix","riemdata")){
# PREPROCESSING
# parameters
N = round(n)
p = round(p) # S^{p-1} in R^p
# return object type
if (missing(type)){
retype = "riemdata"
} else {
retype = match.arg(tolower(type), c("list","matrix","riemdata"))
}
# GENERATION
outmat = runif_sphere(N,p)
if (all(retype=="matrix")){
return(outmat)
} else {
outobj = wrap.sphere(outmat)
if (all(retype=="list")){
return(outobj$data) } else { return(outobj) } } } # (02) sphere.utest ============================================================ #' Test of Uniformity on Sphere #' #' Given \eqn{N} observations \eqn{\lbrace X_1, X_2, \ldots, X_M \brace} on #' \eqn{\mathcal{S}^{p-1}}, it tests whether the data is distributed uniformly #' on the sphere. #' #' @param spobj a S3 \code{"riemdata"} class for \eqn{N} Sphere-valued data. #' @param method (case-insensitive) name of the test method containing \describe{ #' \item{\code{"Rayleigh"}}{original Rayleigh statistic.} #' \item{\code{"RayleighM"}}{modified Rayleigh statistic with better order of error.} #' } #' #' @return a (list) object of \code{S3} class \code{htest} containing: \describe{ #' \item{statistic}{a test statistic.} #' \item{p.value}{\eqn{p}-value under \eqn{H_0}.} #' \item{alternative}{alternative hypothesis.} #' \item{method}{name of the test.} #' \item{data.name}{name(s) of provided sample data.} #' } #' #' #' @examples #' #------------------------------------------------------------------- #' # Compare Rayleigh's original and modified versions of the test #' #------------------------------------------------------------------- #' # Data Generation #' myobj = sphere.runif(n=100, p=5, type="riemdata") #' #' # Compare 2 versions : Original vs Modified Rayleigh #' sphere.utest(myobj, method="rayleigh") #' sphere.utest(myobj, method="rayleighm") #' #' #' @references #' \insertRef{chikuse_statistics_2003}{Riemann} #' #' \insertRef{mardia_directional_1999}{Riemann} #' #' @seealso \code{\link{wrap.sphere}} #' @concept sphere #' @export sphere.utest <- function(spobj, method=c("Rayleigh","RayleighM")){ ## CHECK INPUT check_inputmfd(spobj, "sphere.utest") DNAME = deparse(substitute(spobj)) # borrowed from HDtest method.now = ifelse(missing(method),"rayleigh", match.arg(tolower(method), c("rayleigh","rayleighm"))) ## COMPUTE # prepare data in 3d array data3d <- aux_rvec2array3d(spobj) output <- switch(method.now, rayleigh = sp.utest.Rayleigh(data3d, DNAME, is.original = TRUE), rayleighm = sp.utest.Rayleigh(data3d, DNAME, is.original = FALSE)) return(output) } #' @keywords internal #' @noRd sp.utest.Rayleigh <- function(x, dname, is.original=TRUE){ # Take the version from RiemStiefel {will be deprecated} p = dim(x)[1] # r-frame in R^p with n observations r = dim(x)[2] n = dim(x)[3] xbar = array(0,c(p,r)) for (i in 1:p){ for (j in 1:r){ xbar[i,j] = base::mean(as.vector(x[i,j,])) } } S = p*n*sum(diag(t(xbar)%*%xbar)) if (is.original){ hname = "Rayleigh Test of Uniformity on Sphere" thestat = S pvalue = stats::pchisq(thestat, df=as.integer(p*r), lower.tail=FALSE) } else { hname = "Modified Rayleigh Test of Uniformity on Sphere" term1 = 1/(2*n) term2 = 1 - (S/((p*r) + 2)) thestat = S*(1 - term1*term2) pvalue = stats::pchisq(thestat, df=as.integer(p*r), lower.tail=FALSE) } # COMPUTATION : DETERMINATION Ha = paste("data is not uniformly distributed on ",p-1,"-sphere.",sep="") names(thestat) = "statistic" res = list(statistic=thestat, p.value=pvalue, alternative = Ha, method=hname, data.name = dname) class(res) = "htest" return(res) } # (03) sphere.convert ========================================================= # https://stackoverflow.com/questions/1185408/converting-from-longitude-latitude-to-cartesian-coordinates #' Convert between Cartesian Coordinates and Geographic Coordinates #' #' In geospatial data analysis, it is common to consider locations on the Earth as #' data. These locations, usually provided by latitude and longitude, are not directly #' applicable for spherical data analysis. We provide two functions - \code{sphere.geo2xyz} and \code{sphere.xyz2geo} - #' that convert geographic coordinates in longitude/latitude into a unit-norm vector on \eqn{\mathcal{S}^2}, and vice versa. #' As a convention, latitude and longitude are represented as \emph{decimal degrees}. #' #' @param lat latitude (in decimal degrees). #' @param lon longitude (in decimal degrees). #' @param xyz a unit-norm vector in \eqn{\mathcal{S}^{2}}. #' #' @return transformed data. #' #' @examples #' ## EXAMPLE DATA WITH POPULATED US CITIES #' data(cities) #' #' ## SELECT ALBUQUERQUE #' geo = cities$coord[1,]
#' xyz = cities\$cartesian[1,]
#'
#' ## CHECK TWO INPUT TYPES AND THEIR CONVERSIONS
#' sphere.geo2xyz(geo[1], geo[2])
#' sphere.xyz2geo(xyz)
#'
#' @name sphere.convert
#' @concept sphere
#' @rdname sphere.convert
NULL
#' @rdname sphere.convert
#' @export
sphere.geo2xyz <- function(lat, lon){
xlat = as.double(lat)*pi/180
xlon = as.double(lon)*pi/180
x = cos(xlat)*cos(xlon)
y = cos(xlat)*sin(xlon)
z = sin(xlat)
outvec = c(x,y,z)
return(outvec/sqrt(sum(outvec^2)))
}
#' @rdname sphere.convert
#' @export
sphere.xyz2geo <- function(xyz){
x = as.double(xyz[1])
y = as.double(xyz[2])
z = as.double(xyz[3])
latitude = 180*asin(z)/pi
longitude = 180*atan2(y,x)/pi
output = c(latitude, longitude)
return(output)
}
## Try the Riemann package in your browser
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Riemann documentation built on March 18, 2022, 7:55 p.m. | 2,013 | 7,030 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-06 | latest | en | 0.282262 |
https://www.philosophy-science-humanities-controversies.com/listview-details-economics-politics.php?id=1203696&a=a&first_name=Peter&author=Norvig&concept=Terminology | 1,623,840,173,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487623596.16/warc/CC-MAIN-20210616093937-20210616123937-00262.warc.gz | 719,135,714 | 6,773 | # Economics Dictionary of Arguments
Home
Author Item Summary Meta data
Peter Norvig on Terminology - Dictionary of Arguments
Norvig I 8
Terminology/Russell/Norvig: Although decidability and computability are important to an understanding of computation, the notion of tractability has had an even greater impact. Roughly speaking, a problem is called intractable if the time required to solve instances of the problem grows exponentially with the size of the instances. The distinction between polynomial and exponential growth in complexity was first emphasized in the mid-1960s (Cobham, 1964(1); Edmonds, 1965(2)). It is important because exponential growth means that even moderately large instances cannot be solved in any reasonable time.
Norvig I 106
Pattern databases: The idea behind them is to store these exact solution costs for every possible
subproblem instance (…)
Norvig I 108
Def Problem: A problem consists of five parts: the initial state, a set of actions, a transition model describing the results of those actions, a goal test function, and a path cost function. The environment of the problem is represented by a state space. A path through the state space from the initial state to a goal state is a solution.
Norvig I 135
And/or nodes/search trees: or: In a deterministic environment, the only branching is introduced by the agent’s own choices in each state. We call these nodes OR nodes
And: In a nondeterministic environment, branching is also introduced by the environment’s choice of outcome for each action. We call these nodes AND nodes. A solution for an AND–OR search problem is a subtree that (1) has a goal node at every leaf, (2) specifies one action at each of its OR nodes, and (3) includes every outcome branch at each of its AND nodes.
Norvig I 148
Competitive ratio: Typically, the agent’s objective is to reach a goal state while minimizing cost. (Another possible objective is simply to explore the entire environment.) The cost is the total path cost of the path that the agent actually travels. It is common to compare this cost with the path cost of the path the agent would follow if it knew the search space in advance—that is, the actual shortest path (or shortest complete exploration). In the language of online algorithms, this is called the competitive ratio; we would like it to be as small as possible. >Online search/Norvig.
Norvig I 162
Def Pruning: Pruning allows us to ignore portions of the search tree that make no difference to the final choice.
Def Heuristic evaluation functions: allow us to approximate the true utility of a state without doing a complete search.
Def utility function: (also called an objective function or payoff function), defines the final numeric value for a game that ends in terminal state s for a player p. In chess, the outcome is a win, loss, or draw, with values +1, 0, or 1 2 . Some games have a wider variety of possible outcomes; the payoffs in backgammon range from 0 to +192.
Def Zero-sum game: is (confusingly) defined as one where the total payoff to all players is the same for every instance of the game. Chess is zero-sum. “Constant-sum” would have been a better term, but zero-sum is traditional and makes sense if you imagine each player is charged an entry fee of 1/2 .
Norvig I 165
Minimax Algorithm/gaming: The minimax algorithm (…) computes the minimax decision from the current state. It uses a simple recursive computation of the minimax values of each successor state, directly implementing the defining equations. The recursion proceeds all the way down to the leaves of the tree, and then the minimax values are backed up through the tree as the recursion unwinds. The minimax algorithm performs a complete depth-first exploration of the game tree. For real games, of course, the time cost is totally impractical, but this algorithm serves as the basis for the mathematical analysis of games and for more practical algorithms.
Norvig I 208
Def node consistency: A single variable (corresponding to a node in the CSP network) is node-consistent if all the values in the variable’s domain satisfy the variable’s unary constraints.
Def arc consistency: A variable in a CSP is arc-consistent if every value in its domain satisfies the variable’s binary constraints. More formally, Xi is arc-consistent with respect to another variable Xj if for every value in the current domain Di there is some value in the domain Dj that satisfies the binary constraint on the arc (Xi,Xj). >Constraint Satisfaction Problems/CSP/Norvig.
Norvig I 210
Def Path consistency: Arc consistency tightens down the domains (unary constraints) using the arcs (binary constraints). To make progress on problems like map coloring, we need a stronger notion of consistency. Path consistency tightens the binary constraints by using implicit constraints that are inferred by looking at triples of variables.
Norvig I 211
Def K-consistency: Stronger forms of propagation can K-CONSISTENCY be defined with the notion of k-consistency. A CSP is k-consistent if, for any set of k − 1 variables and for any consistent assignment to those variables, a consistent value can always be assigned to any kth variable.
For forward chaining, backward chaining: see >Software agents/Norvig.
Norvig I 266
Propositions: The idea of associating propositions with time steps extends to any aspect of the world changes over time.
Fluent: We use the word fluent (from the Latin fluents, flowing) to refer an aspect of the world that changes. “Fluent” is a synonym for “state variable”.
Norvig I 346
Skolemize: Skolemization is the process of removing existential quantifiers by elimination. In the simple case, it is just like the Existential Instantiation rule (…):
translate ∃x P(x) into P(A), where A is a new constant.
Norvig I 410
Nondeterministic action: The programming languages community has coined the term demonic nondeterminism for the case where an adversary makes the DEMONIC choices, contrasting this with angelic nonde-
Norvig I 411
terminism, where the agent itself makes the choices. We borrow this term to define angelic semantics for HLA descriptions.
Norvig I 468
Closed-world assumption: as implemented in logic programs, provides a simple way to avoid having to specify lots of negative information. It is best interpreted as a default that can be overridden by additional information.
1. Cobham, A. (1964). The intrinsic computational difficulty of functions. In Proc. 1964 International
Congress for Logic, Methodology, and Philosophy of Science, pp. 24–30.
2. Edmonds, J. (1965). Paths, trees, and flowers. Canadian Journal of Mathematics, 17, 449–467.
_____________
Explanation of symbols: Roman numerals indicate the source, arabic numerals indicate the page number. The corresponding books are indicated on the right hand side. ((s)…): Comment by the sender of the contribution. Translations: Dictionary of Arguments
The note [Author1]Vs[Author2] or [Author]Vs[term] is an addition from the Dictionary of Arguments. If a German edition is specified, the page numbers refer to this edition.
Norvig I
Peter Norvig
Stuart J. Russell
Artificial Intelligence: A Modern Approach Upper Saddle River, NJ 2010
> Counter arguments against Norvig | 1,590 | 7,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.935399 |
http://physics.stackexchange.com/questions/tagged/general-relativity+gravitational-redshift | 1,406,312,604,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997894689.94/warc/CC-MAIN-20140722025814-00014-ip-10-33-131-23.ec2.internal.warc.gz | 228,805,542 | 16,368 | # Tagged Questions
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Has the gravitational red shift been proven for electromagnetic waves only or also for a single photon? | 1,006 | 4,261 | {"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} | 3.046875 | 3 | CC-MAIN-2014-23 | longest | en | 0.872104 |
https://www.vivran.in/post/dax-iterators-the-x-factors | 1,725,859,389,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651072.23/warc/CC-MAIN-20240909040201-20240909070201-00497.warc.gz | 982,516,958 | 199,884 | top of page
Search
# DAX Iterators: The X-Factors
To a beginner, one can find some similarities between the Excel functions and DAX. And the primary reason is that both Excel and DAX are functional languages.
The concepts of statements, loops, and jumps do not exist in Excel functions and DAX. So, aggregator functions like SUM, MIN, MAX, AVERAGE have similar structures or arguments in Excel and DAX.
The concept of iterators is new for Excel users. Consider the following example:
To calculate the sales amount for the year, we implement the following steps:
• Add a column – Sales Amount
• Apply the formula [Unit Sold]*[Unit Price]
• Apply aggregate function SUM on Sales Amount
With DAX iterators, we can perform the same operation in a single step. As the name suggests, a DAX iterator performs a calculation on each row of the table, aggregating the result to return the single value requested.
``` Sales Amount =
SUMX(
dtSales,
dtSales[Unit Price] * dtSales[Unit Sold]
)
```
In simple English, using DAX, we are saying:
· Go to the dtSales table
· Start evaluating each row of the dtSales table
· Apply the formula [Unit Price] * [Unit Sold] in each row
· Aggregate the value of the previous step
We can calculate the average instead of a sum by following similar steps.
``` Avg Sales Amount =
AVERAGEX(
dtSales,
dtSales[Unit Price] * dtSales[Unit Sold]
)
```
DAX first calculates the sales amount for each row by applying the expression [Unit Price] * [Unit Sold], and then calculate the average of the output.
The above steps are happening without making any physical change (read adding a column) in the table. Hence, the calculation is quicker and uses less memory. The flip side is a calculation in iterators that are not visual. We do not see the columns or tables it creates for the calculation; it exists only for the lifetime of the calculation.
A few common iterators in DAX are SUMX, AVERAGEX, MEDIANX, MINX, MAXX, COUNTX, COUNTAX.
All of them have the same arguments: (Table, Expression)
Iterators in DAX follow below-mentioned steps:
• Evaluates the first parameter in the existing context(read filter contexts) to determine the rows to scan
• Creates a new row context for each row of the table evaluated in the previous step
• Iterates the table and evaluates the second parameter in the existing evaluation context, including the newly created row context.
• And last aggregates the values computed during the previous step.
For details on Filter and Row contexts in DAX, follow the article
We can add filters and conditions in the iterators to further define the calculation:
``` Avg Sales (15 or more units) =
AVERAGEX(
//Filter sales table for Unit Sold >= 15
FILTER(
dtSales,
dtSales[Unit Sold] >= 15
),
dtSales[Unit Sold] * dtSales[Unit Price]
)
```
In simple English, we are saying DAX to:
• Apply filter on the dtSales table where Units Sold >= 15
• On that filtered table, insert a temporary column and apply the formula [Unit Price] * [Unit Sold]
• Calculate the average of the temporary column | 692 | 3,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-38 | latest | en | 0.83129 |
https://pronursingtutors.com/explain-the-central-theorem-limit/ | 1,652,959,401,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662527626.15/warc/CC-MAIN-20220519105247-20220519135247-00241.warc.gz | 534,510,011 | 30,070 | # Explain the central theorem limit.
## Explain the central theorem limit.
Explain the central theorem limit. 150 150 Nyagu
PUB-550 Topic 2: Probability and Confidence Intervals discussion questions
PUB-550 Topic 2: Probability and Confidence Intervals discussion questions
Topic 2: Probability and Confidence Intervals
Objectives:
Apply descriptive statistics to describe qualitative and quantitative characteristics of a sample.
Demonstrate how probability and normal distribution are used to determine significance.
Apply a sample standard deviation to describe a variable and its dispersion.
Explain the central theorem limit.
Evaluate the application of confidence intervals in public health research.
PUB-550 Topic 2: Probability and Confidence Intervals discussion questions
ORDER A PLAGIARISM-FREE PAPER HERE
Topic 2 DQ 1
P-values and confidence intervals are both used in hypothesis testing. Explain three reasons why it may be preferable to report a confidence interval over a P-value. Provide a specific example to justify your reasons.
Topic 2 DQ 1
The use of P-values and Confidence Intervals (CIs) are important for understanding evaluations of scientific studies. Specifically, they are used in evaluating hypothesis testing. A P value is the core of statistical inference method but is limited to dichotomous evaluation whether null is statistically significant or not (Lee, 2016). An alternative approach that is more informative that P value is confidence intervals. Whereas with P values an investigator could state P0.05, there is little information about the direction or size of the differences (Gardner & Altman, 1986). CIs gives a greater amount of information of the magnitude and precision of the measurement.
The first reason that confidence intervals may be preferable over P-values is that confidence intervals are interval estimates for population value rather than a single point value. Researchers are able to take a sample from a larger population, calculate the sample mean, and construct a range (Confidence interval), that has a greater probability of containing the population mean than a point estimate (Corty, 2016).
The second reason confidence intervals may be preferable over P-values, is that they are not limited to dichotomous significant or not significant, but rather are able to provide the magnitude of the effect. A 95% confidence interval means that the population mean is within the range of 95% of confidence interval of the calculated mean with a probability of 95% (Lee, 2016). With the increase of sample size, the confidence interval narrows while limits of significance remain unchanged. For example, statistical results that may have the same P-value, the estimated confidence interval narrows with a larger sample size indicating a more reliable result (Lee, 2016). PUB-550 Topic 2: Probability and Confidence Intervals discussion questions.
The third reason confidence intervals may be preferable over P-values, is that p values enable recognition of statistically noteworthy findings, whereas confidence intervals provide information about the range in which the true value of the population lies within a certain degree of probability, as well as direction and strength of the effect (du Prel, Hommel, Rohrig, &Blettner, 2009) PUB-550 Topic 2: Probability and Confidence Intervals discussion questions. The smaller the P-value the more statistically significant and stronger the evidence is. Confidence intervals reflect results at the level of data measurement, and include the desired parameter within a certain probability (du Prel et al., 2009).
References
Corty, E. Using and interpreting statistics: A practical text for behavioral, social, and health sciences (3rd ed.). New York, NY: Macmillan Learning.
du Prel, J-B., Hommel, G., Rohrig, B., & Blettner, M. (2009). Confidence interval or P-value. Deutschees Arzteblatt International, 106(19), 335-339. doi: 10.3238/arztebl.2009.0335 PUB-550 Topic 2: Probability and Confidence Intervals discussion questions
Gardner, M. J. & Altman, D. G. (1986). Confidence intervals rather than P values: estimation rather than hypothesis testing. British Medical Journal, 292, 746-750. Retrieved from https://www.bmj.com/content/bmj/292/6522/746.full.pdf
Lee, D. K. (2016). Alternatives to P value: confidence interval and effect size. Korean Journal of Anesthesiology, 69(6), 555-562. doi: 10.4097/kjae.2016.69.6.555
Topic 2 DQ 2
During the simulation the results of increasing the sample size really didn’t change the curve at all. The central limit theorem demonstrates how much sampling error exists in samples by taking repeated random samples from a population, calculating the mean for each sample and then making a frequency distribution of the mean (Corty, 2016). The theory predicts that the standard error of the mean can be calculated if one knows the population standard deviation and the size of the sample (Corty, 2016). The theory also predicts that the sampling distribution is normally distributed, and the mean of the sampling distribution is the mean of the population (Corty, 2016).
I found this example from an unofficial source, but I found it descriptive – Find an old stone step or lintel in front of a doorway (one that is old enough that it has been worn down by generations of people walking over it). If you look at how it is worn down, the wearing down won’t be uniform over the surface of the step: rather, it will be in a bell-curve. (People tend to walk through the middle of the doorway.) If the door doesn’t open all the way, the bell-curve won’t be perfect; there will a bias towards the side away from the hinges”. For some reason this helped cement the theory for me, the bell curve is “where most of the samples walk through the doorway”. PUB-550 Topic 2: Probability and Confidence Intervals discussion questions
References | 1,266 | 5,889 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.827613 |
https://engineering.purdue.edu/~ee538/MinVarforSOS.m | 1,547,916,144,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-04/segments/1547583671342.16/warc/CC-MAIN-20190119160425-20190119182425-00480.warc.gz | 505,343,193 | 1,667 | %Minimum Variance Spectral Estimation for Sum of %Sinusoids Example done during Session 30. %Forward-Backward Averaged Correlation Matrix Employed. clear all clf set(0,'defaultaxesfontsize',20); P=3; Nl=128; M=10; % %P=input('no. of complex sinewaves = '); %omegas=input('frequencies of the sinewaves = '); %amps=input('amplitudes of the sinewaves = '); %omegas=[.55*pi .6*pi .65*pi]; omegas=[-.2*pi .1*pi .3*pi]; poles=exp(j*omegas); amps=[.5 1 1]; x=zeros(Nl,1); % %create Nl values of SOS data % for i=1:P; x(1:Nl,1)=x(1:Nl,1)+amps(1,i)*exp(j*omegas(1,i)*(0:Nl-1)'); end; % %compute unconstrained LS estimate of autocorrelation matrix % Rb=zeros(M,M); for n=1:Nl-M+1; Rb=Rb+x(n:n+M-1,1)*x(n:n+M-1,1)'; end Rff=Rb(M:-1:1,:); Rf=conj(Rff(:,M:-1:1)); Rfb=0.5*(Rf+Rb)/(Nl-M+1)+.001*eye(M); Rinv=inv(Rfb); % %plot poles and zeroes % polar(0,1,'.') hold on plot(real(poles),imag(poles),'x','MarkerSize',16,'Linewidth',3); hold off pause % %compute estimate of spectrum, plot and compare %with true spectrum % freqomega=linspace(-pi,pi,1024); for n=1:1024; s=exp(j*freqomega(n)*[0:M-1]).'; specest(n)=1/(s'*Rinv*s); end fmag=10*log10(abs(specest)); fmag=fmag-max(fmag); plot(freqomega,fmag,'k','LineWidth',4); axis([-pi pi -60 0]); title('Min Var Spectral Estimate'),... xlabel('Omega'), ylabel('Magnitude'), grid pause hold specfft=20*log10(abs(fftshift(fft(x,1024)))); fmag1=specfft-max(specfft); freqomega=linspace(-pi,pi,1024); plot(freqomega,fmag1,'r','LineWidth',2); pause for p=1:P; plot([omegas(1,p) omegas(1,p)],[-60 0],'b','LineWidth',3); end legend('Min Var Estimate','Mag. Squared of FFT'); hold off for p=1:P+1; omega0=input('Input frequency of interest:') a0=exp(j*omega0*[0:M-1]).'; h0=(Rinv*a0)/(a0'*Rinv*a0); H0=20*log10(abs(fftshift(fft(h0,1024)))); H0mag=H0-max(H0); plot(freqomega,H0mag,'k','LineWidth',4); axis([-pi pi -50 10]); title('DTFT of h[n;omega0]'),... xlabel('Omega'), ylabel('Magnitude'), grid hold on for p=1:P; plot([omegas(1,p) omegas(1,p)],[-50 10],'b','LineWidth',3); end hold off pause y=real(conv(x,h0)); plot(y,'LineWidth',4); axis([0 (Nl+M-1) -1 1]); title('output when x[n] is input to h(n;omega0)'),... xlabel('Time'), ylabel('Amplitude'), grid end | 804 | 2,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-04 | latest | en | 0.215034 |
http://www.mathworks.com/help/symbolic/le.html?requestedDomain=www.mathworks.com&nocookie=true | 1,512,972,260,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512208.1/warc/CC-MAIN-20171211052406-20171211072406-00339.warc.gz | 393,570,826 | 13,651 | # Documentation
### This is machine translation
Translated by
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# le
Define less than or equal to relation
## Syntax
```A <= B le(A,B) ```
## Description
`A <= B` creates a less than or equal to relation.
`le(A,B)` is equivalent to `A <= B`.
## Input Arguments
`A` Number (integer, rational, floating-point, complex, or symbolic), symbolic variable or expression, or array of numbers, symbolic variables or expressions. `B` Number (integer, rational, floating-point, complex, or symbolic), symbolic variable or expression, or array of numbers, symbolic variables or expressions.
## Examples
Use `assume` and the relational operator `<=` to set the assumption that `x` is less than or equal to 3:
```syms x assume(x <= 3)```
Solve this equation. The solver takes into account the assumption on variable `x`, and therefore returns these three solutions.
`solve((x - 1)*(x - 2)*(x - 3)*(x - 4) == 0, x)`
```ans = 1 2 3```
Use the relational operator `<=` to set this condition on variable `x`:
```syms x cond = (abs(sin(x)) <= 1/2);```
```for i = 0:sym(pi/12):sym(pi) if subs(cond, x, i) disp(i) end end```
Use the `for` loop with step π/24 to find angles from 0 to π that satisfy that condition:
```0 pi/12 pi/6 (5*pi)/6 (11*pi)/12 pi```
## Tips
• Calling `<=` or `le` for non-symbolic `A` and `B` invokes the MATLAB® `le` function. This function returns a logical array with elements set to logical ```1 (true)``` where `A` is less than or equal to `B`; otherwise, it returns logical ```0 (false)```.
• If both `A` and `B` are arrays, then these arrays must have the same dimensions. ```A <= B``` returns an array of relations ```A(i,j,...) <= B(i,j,...)```
• If one input is scalar and the other an array, then the scalar input is expanded into an array of the same dimensions as the other array. In other words, if `A` is a variable (for example, `x`), and `B` is an m-by-n matrix, then `A` is expanded into m-by-n matrix of elements, each set to `x`.
• The field of complex numbers is not an ordered field. MATLAB projects complex numbers in relations to a real axis. For example, ```x <= i``` becomes `x <= 0`, and ```x <= 3 + 2*i``` becomes `x <= 3`.
## Alternatives
You can also define this relation by combining an equation and a less than relation. Thus, `A <= B` is equivalent to `(A < B) | (A == B)`. | 673 | 2,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-51 | latest | en | 0.772353 |
https://oeis.org/A304156 | 1,719,356,735,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866422.9/warc/CC-MAIN-20240625202802-20240625232802-00751.warc.gz | 365,739,026 | 4,143 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A304156 T(n,k)=Number of nXk 0..1 arrays with every element unequal to 2, 3, 4 or 6 king-move adjacent elements, with upper left element zero. 7
0, 0, 0, 0, 3, 0, 0, 5, 5, 0, 0, 18, 4, 18, 0, 0, 61, 42, 42, 61, 0, 0, 209, 130, 346, 130, 209, 0, 0, 702, 464, 1767, 1767, 464, 702, 0, 0, 2381, 1722, 10182, 13279, 10182, 1722, 2381, 0, 0, 8069, 6378, 60352, 100942, 100942, 60352, 6378, 8069, 0, 0, 27330, 22939, 350540 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,5 COMMENTS Table starts .0....0.....0.......0........0..........0...........0.............0 .0....3.....5......18.......61........209.........702..........2381 .0....5.....4......42......130........464........1722..........6378 .0...18....42.....346.....1767......10182.......60352........350540 .0...61...130....1767....13279.....100942......839935.......6803634 .0..209...464...10182...100942....1100030....12806826.....145611915 .0..702..1722...60352...839935...12806826...217640633....3520683600 .0.2381..6378..350540..6803634..145611915..3520683600...80851458613 .0.8069.22939.2034140.54261600.1634366796.56360283031.1830588080314 LINKS R. H. Hardin, Table of n, a(n) for n = 1..180 FORMULA Empirical for column k: k=1: a(n) = a(n-1) k=2: a(n) = 3*a(n-1) +a(n-2) +2*a(n-3) -2*a(n-4) -4*a(n-5) for n>6 k=3: [order 18] for n>19 k=4: [order 66] for n>67 EXAMPLE Some solutions for n=5 k=4 ..0..1..0..1. .0..1..0..0. .0..1..0..1. .0..1..1..0. .0..1..0..1 ..1..0..1..0. .1..0..1..1. .0..1..1..0. .1..0..0..1. .1..1..0..0 ..1..1..1..1. .1..0..0..0. .1..0..1..0. .1..0..0..1. .0..1..1..1 ..0..1..0..0. .1..0..0..1. .0..1..1..1. .1..0..0..1. .0..1..1..1 ..1..0..1..1. .0..1..0..1. .1..0..1..0. .1..0..0..1. .0..1..0..0 CROSSREFS Column 2 is A303684. Sequence in context: A072736 A135090 A303690 * A305509 A305175 A316763 Adjacent sequences: A304153 A304154 A304155 * A304157 A304158 A304159 KEYWORD nonn,tabl AUTHOR R. H. Hardin, May 07 2018 STATUS approved
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Last modified June 25 18:50 EDT 2024. Contains 373707 sequences. (Running on oeis4.) | 1,025 | 2,428 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.561819 |
https://www.physicsforums.com/threads/calculate-the-voltmeter-reading.887019/ | 1,508,329,571,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822930.13/warc/CC-MAIN-20171018104813-20171018124733-00004.warc.gz | 961,149,073 | 17,163 | 1. Sep 28, 2016
### moenste
1. The problem statement, all variables and given/known data
The light-dependent resistor (LDR) in the circuit below is found to have resistance 800 Ω in moonlight and resistance 160 Ω in daylight. Calculate the voltmeter reading, Vm, in moonlight with the switch S open.
If the reading of the voltmeter in daylight with the switch S closed is also equal to Vm what is the value of the resistance R?
2. The attempt at a solution
At first I would like to check whether I understand the scheme correctly. The switch S is a diode (open is a reverse-biased and closed is a forward-biased). The LDR is just a regular resistor, just need to take into account different values of resistance depending on the daytime. Should this be correct?
I this representation correct?
2. Sep 28, 2016
### Staff: Mentor
A switch is not a diode. A switch will pass current in either direction with equal facility and does not exhibit a potential drop when current passes through it (at least for an ideal switch). Diodes pass current in one direction only, and except for ideal diodes cause a potential drop on the order of a half volt when they do.
It's better to think of a switch as a zero resistance wire connection that can be inserted or removed manually.
3. Sep 28, 2016
### moenste
We need to find the voltmeter reading during moonlight when the resistance is 800 Ω and the switch is open. If the switch is open, the resistor R is shut off. Therefore we have only two resistors 200 + 800 = 1000 Ω. I = V / R = 10 / 1000 = 0.01 A. Now we find the 200 Ω has V200 = I R200 = 0.01 * 200 = 2 V, and the LDR resistor, as a result has 8 V. Since the voltmeter is connected to LDR, its reading is 8 V. Correct logic?
Update: for the second question we have: since the reading doesn't change, therefore V = I R and we can find I = V / R = 8 / 160 = 0.05 A of the whole circuit. Then, knowing V = 2 V and I = 0.05 A, we find the RParallel: R = V / I = 2 / 0.05 = 40 Ω. Then, by 1 / R = 1 / R + 1 / R → 1 / 40 = 1 / 200 + 1 / R → R = 50 Ω. Should be correct.
Last edited: Sep 28, 2016
4. Sep 28, 2016
### Staff: Mentor
That all looks good. Well done. | 615 | 2,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-43 | longest | en | 0.934292 |
http://zh.wikipedia.org/wiki/%E5%8D%81%E4%B8%83%E8%BE%B9%E5%BD%A2 | 1,432,577,114,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928562.33/warc/CC-MAIN-20150521113208-00087-ip-10-180-206-219.ec2.internal.warc.gz | 878,697,174 | 16,508 | # 十七边形
17
4.
a2cotπ.
17
22.735491898417a2
17..
o = 15814.
17
o
158.82352941176°
## 作圖方法
### 作圖
1796年高斯证明了可以用尺規作圖作出正十七邊形，同時發現了可作圖多邊形的條件。正十七邊形其中一个作圖方法如下：
$\operatorname{cos}{2\pi\over17}=\frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}}}{16}.$
### 證明
$16 \alpha = 360^\circ - \alpha$
$\sin 16\alpha = - \sin \alpha$，而
\begin{align} \sin 16\alpha & = 2\sin 8\alpha \cos 8\alpha \\ & = 2^2\sin 4\alpha \cos 4\alpha \cos 8\alpha \\ & = 2^4 \sin \alpha \cos \alpha \cos 2\alpha \cos 4\alpha \cos 8\alpha \\ \end{align}
$16 \cos \alpha \cos 2\alpha \cos 4\alpha \cos 8\alpha = -1$
$2(\cos \alpha + \cos 2\alpha + \cdots + \cos 8\alpha) = -1$
$x = \cos \alpha + \cos 2\alpha + \cos 4\alpha + \cos 8\alpha$
$y = \cos 3\alpha + \cos 5\alpha + \cos 6\alpha + \cos 7\alpha$
$x + y = - \frac {1}{2}$
\begin{align} xy & = (\cos \alpha + \cos 2\alpha + \cos 4\alpha + \cos 8\alpha)(\cos 3\alpha + \cos 5\alpha + \cos 6\alpha + \cos 7\alpha) \\ & = \frac {1}{2} (\cos 2\alpha + \cos 4\alpha + \cos 4\alpha + \cos 6\alpha + \cdots + \cos \alpha + \cos 15\alpha) \\ & = -1 \\ \end{align}
$x = \frac {-1 + \sqrt{17}}{4}$
$y = \frac {-1 - \sqrt{17}}{4}$
$x_1 = \cos \alpha + \cos 4\alpha$$x_2 = \cos 2\alpha + \cos 8\alpha$
$y_1 = \cos 3\alpha + \cos 5\alpha$$y_2 = \cos 6\alpha + \cos 7\alpha$
$x_1 + x_2 = \frac {-1 + \sqrt{17}}{4}$
$y_1 + y_2 = \frac {-1 - \sqrt{17}}{4}$
$\cos \alpha + \cos 4\alpha = x_1$
$\cos \alpha \cos 4\alpha = \frac{y_1}{2}$
$\cos \alpha=\frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}}}{16}$
Q.E.D | 815 | 1,645 | {"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": 28, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-22 | latest | en | 0.112655 |
https://www.sawaal.com/time-and-work-questions-and-answers/parvesh-is-three-times-as-efficient-as-dinesh-parvesh-takes-20-days-less-than-dinesh-to-complete-a-b_39293 | 1,600,550,819,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400192887.19/warc/CC-MAIN-20200919204805-20200919234805-00006.warc.gz | 1,087,498,754 | 14,241 | 0
Q:
# Parvesh is three times as efficient as Dinesh. Parvesh takes 20 days less than Dinesh to complete a book. If both of them work together, then in how many days the book will be completed?
A) 7.5 B) 8 C) 10 D) 6.5
Explanation:
Q:
Vikram and Vivek can finish a work in 50 days. They worked together for 20 days and then left. How much work has been done by them?
A) 3/5 B) 1/3 C) 1/2 D) 2/5
Explanation:
0 20
Q:
9 women can complete a work in 50 days. If more 6 women were added to them, how many days before the work can be completed?
A) 30 B) 20 C) 15 D) 25
Explanation:
0 53
Q:
In a camp 180 students had ration for 20 days. How many students should leave the camp if the ration should last for 25 days?
A) 36 B) 24 C) 28 D) 40
Explanation:
0 88
Q:
Smita can finish a work in 12 days and Sam can finish the same work in 9 days. After working together for 4 days they both leave the job. What is the fraction of unfinished work?
A) 2/9 B) 1/4 C) 1/2 D) 7/9
Explanation:
0 60
Q:
X is 25% more efficient than Y. If Y alone can make a chair in 10 days, then X alone can make the chair in how many days?
A) 8 B) 12 C) 12.5 D) 7.5
Explanation:
2 157
Q:
M can complete 3/4 part of a work in 12 days and N can complete 2/7 part of the same work in 8 days. In how many days will both complete 11/14 part of the total work?
A) 8 B) 9 C) 7 D) 6
Explanation:
0 256
Q:
P and Q together can make a door in 12 days. They started the work together but after 8 days P left the work. The remaining work is completed by Q alone in 7 more days. In how many days can P alone complete the entire work?
A) 21 B) 24 C) 28 D) 32
Explanation:
0 223
Q:
30 women do half of the total work in 30 days. How many more women will be required to complete the remaining work in 10 days?
A) 45 B) 30 C) 60 D) 80 | 598 | 1,825 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2020-40 | longest | en | 0.928722 |
https://cueflash.com/decks/91391/Med_Calc_Conversions | 1,529,671,556,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864482.90/warc/CC-MAIN-20180622123642-20180622143642-00502.warc.gz | 578,171,119 | 7,496 | This site is 100% ad supported. Please add an exception to adblock for this site.
# Med Calc Conversions
## Terms
undefined, object
copy deck
How many grams in a kilogram?
1000
How many milligrams in a gram?
1000
How many micrograms in a milligram?
1000
How many cc's in an ml?
1
How many ml in a cc?
1
How many drops (gtts) are there in a minum?
1
One drop (gtt) is how many minums?
1
How many gtts in an ml?
15 or 16
How many minums in an ml?
15 or 16
How many dr in a teaspoon?
1
How many minums in a gtt?
1
How many gtts in a minum
1
A gtt is a ______.
drop
1 minum = 1 ______.
1 minum = 1 drop
1 dram (dr) =
1 teaspoon
4 ml (adult)
5 ml (pediatric)
1 teaspoon (t) =
4 ml (adult) or
5 ml (pediatric) or
1 dram
How many mls in a dram?
4 (adult) or 5 (pediatric)
How many drams in a teaspoon?
1
How many mls in an ounce?
30
How many tablespoons (T) in an ounce?
2
One ounce (oz.) =
30 ml or 2 T
How many mls in a liter?
1000
How many ounces in a liter?
32
How many quarts in a liter?
1
How many mls in a quart?
1000
How many ounces in a quart?
32
How many ounces in a cup?
8
How many mls in a cup?
240
How many ounces in a pint?
16
How many mls in a pint?
480
How many mg in a gr?
60
how many grs in a gram (g)?
15
A gram is how many grains?
15
gr 1/2 = ______ mg.
30
gr 1/6 = ______ mg.
10
gr 1/150 = ______ mg.
0.4
gr 1/300 = ______ mg.
0.2
gtt is the abbreviation for...
drop
a cursive letter Z is the abbreviation for...
ounce or dram
oz is the abbreviation for...
ounce
dr is the abbreviation for...
dram
mcg is the abbreviation for...
microgram (1000th of a milligram)
SS is the abbreviation for...
1/2
g is the abbreviation for...
gram
gr is the abbreviation for...
grain
How many pounds is 1 kg?
2.2
In medication calculations, carry all answers to _______ (what decimal place?)
tenths
For drop or microdrop answers, round to _________.
a whole number (you can't have half a drop).
What two drugs are always rounded to hundredths?
Insulin and Heparin
For ml/hr, round to ________.
a whole number
Do not round pediatric calculations to tenths. Carry pediatric answers to hundredths and _____________.
drop (do not round) after two decimal places.
Example: 78.6799 = 78.67
Do not round until _________.
the final answer has been obtained.
What are the four exceptions to the First Rule of Rounding (carry all answers to tenths)?
1. Pediatric
2. Drop & Microdrop
3. ml/hr
4. Insulin & Heparin
53 | 749 | 2,406 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-26 | latest | en | 0.889413 |
https://gmatclub.com/forum/looking-for-suggestions-to-improve-quant-speed-score-108438.html?kudos=1 | 1,511,287,042,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806421.84/warc/CC-MAIN-20171121170156-20171121190156-00177.warc.gz | 620,917,742 | 48,121 | It is currently 21 Nov 2017, 10:57
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# Looking for suggestions to improve quant speed/score
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Intern
Joined: 01 Oct 2010
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Looking for suggestions to improve quant speed/score [#permalink]
### Show Tags
29 Jan 2011, 11:36
Hi guys,
I'm looking for advice on improving my quant score. Background: I always had stronger verbal skills than quant skills throughout my life. (English is my second language). I never liked math. However, my WE has been 100% in IT for a loooong time where I usually utilize analytical skills rather than quant skills. Being away from school for a long time, I knew I needed to work on math. My diagnostic test confirmed it - I couldn't even finish math section. The verbal section was OK. So I started with Kaplan Premier, went through a half of it and then switched to MGMAT. I went through Math Foundation first, then Number Properties, FDP's, Geometry, and half of Word Translations. I also completed CR Bible. Overall time for study - 4 months. Generally, I can solve most of the 600-700 questions in OG12 (without time limit). OG Quant Review feels a bit harder than OG12 to me. I also use MGMAT flash cards and Anki spaced repetition software.
CATs:
CAT1 MGMAT: 510 (Q27, V33) - My timing was bad on quant. I spent 8 min on one question and got it wrong. Also I had two 5 min questions and had to guess about 10-12 questions.
Week later
CAT2 MGMAT: 520 (Q24, V38) - This time around I selected an option for exam to move on after 2 min per question. The result is no answer on 2 questions, one string of 6 wrong answers and another one of 4 wrong answers. Plenty of guessed questions as well. At the same time, I felt good on verbal and didn't have any time pressure on it.
Overall impression: During the quant section I'm struggling with time (unless it is 400 level question) and I'm having a feeling as if all formulas were washed out of my brain.
However, when I review the questions I got wrong, in most cases I wonder how come I didn't solve them.
Proposed course of action: finish OG12 and Gmat Club tests. However, I don't think it can take me to mid-40s on quant. (700+ is overall goal)
Do you guys have any suggestions?
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Intern
Joined: 24 Dec 2010
Posts: 19
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Re: Looking for suggestions to improve quant speed/score [#permalink]
### Show Tags
29 Jan 2011, 13:54
1
KUDOS
Well, it seems that your main problem is lack of practice. At the current moment you need to solve as many problems as possible, bringing your quant skills to the level when you can recognize patterns immediately. Ideally, you should see how to solve the problem in first 10-15 secs after reading it. (that's my opinion) So the main thing is to bring all required for GMAT concepts in a system. Imagine that you have a garret in your head (like Sherlock Holmes ) and when you see a problem you simply search a tool that can solve it. So all needed to be done is to improve search speed. When you practice a lot, you get this improvement quite fast. I love the way MGMAT guides classify problems and tools for tackling 'em, just go through these books and solve OG/QR questions, I think everything will be alright after that. Also consider going through end-chapter questions, they help to memorize key concepts. I've noticed you've already completed some of the books. However, try re-doing questions which were hard for you. And, of course, check OA. So the general advice is: keep practicing and don't worry about your speed, it will improve quite automatically I believe.
When you obtain certain level of confidence, try going through GMATclub tests. This is a very useful tool, I've started using these tests recently, they are quite challenging but overall doable. Also my advice is to go untimed first 5-7 tests so that time issues won't distract you from searching patterns.
Finally, I don't think you should take many CAT's at this moment. Just focus on memorizing concepts and techniques.
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Posts: 168
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Re: Looking for suggestions to improve quant speed/score [#permalink]
### Show Tags
01 Feb 2011, 15:50
1
KUDOS
The beauty of DS is that you don't have to do any calculations at all. Sometimes, in the harder questions, I'll make some quick notes and throw around the numbers a little, but that's it.
Since GMAT throws only a limited number of problems at you it's mostly about learning the math behind the different kind of problems.
- Absolute values
- Basic equations
- Different kinds of rates
- Geometry
- Probability and some permutations and combinations
And so on.
What I've learned so far:
- Try to learn different approaches to the problems so that when you get stuck you'll have different methods.
- Always try to learn and practice on how to solve the problems faster. Many different problems can be solved a lot faster with different shortcuts. For instance, learn the different triangles (30,60,90 / x, xsqr(3), 2x) so that you don't have use the Pythagora theorem every time you see a problem with triangles.
- If you can solve the problem faster with another method - use that method instead.
- Work on any weakness you have until you master every topic.
- Practice, practice until you can solve most problems without any calculations at all.
The best of luck!
_________________
12/2010 GMATPrep 1 620 (Q34/V41)
01/2011 GMATPrep 2 640 (Q42/V36)
01/2011 GMATPrep 3 700 (Q47/V39)
02/2011 GMATPrep 4 710 (Q48/V39)
02/2011 MGMAT CAT 1 650 (Q46/V32)
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Re: Looking for suggestions to improve quant speed/score [#permalink]
### Show Tags
01 Feb 2011, 18:19
1
KUDOS
Hey guys,
Great discussion so far! If I could add a few suggestions...
1) Data Sufficiency
Another great way to get your mind in tune with DS questions is to train yourself to "think like the testmaker" by turning each DS question into a series of a few questions - ask yourself how they could change one word, symbol, etc. to elicit an entirely different answer. For example, there's an OG question that reads:
For nonzero integers m and n, is m/n > (mn)^4?
1) m/n > 0
2) m < 0
For this one, since we know that m and n are integers, there's no chance that m/n could ever be greater than (mn)^4. Thinking logically, on the left we take the same m and divide it, whereas on the right we multiply it by m^3 n^4. Even if either is negative, the right hand side will end up positive because of the even exponent. At worst we can construct a tie if all values are 1 or -1. The answer is D.
But to become a true master of DS, ask yourself how they can use the same setup to elicit a different answer. One way - if they just changed the direction of the inequality in the question to:
Is m/n < (mn)^4?
Usually the answer will be "yes" - take 4 and 2, 4/2 is much less than (4*2)^4. BUT, since we know that we can construct a "tie" with all 1s, then we can still get that answer "NO". Here, we learn that with inequalities you have to be careful because the yes/no question format can be tricky with the potential for a "tie".
2) Timing
One of my favorite timing drills is one I call "Quick First Step", in which you take 10 or 20 quant problems and start on each in 30 seconds, then move on to the next. When you're done with 30 seconds/question, go back and finish the question. Here, your goal is first to train yourself to begin questions quickly by drawing relationships and identifying your known quantities - so often people spend 30 seconds reading, 15 seconds worrying, and then it's almost a full minute before they start "doing". You want to actively read and begin work on a problem, both for pacing and confidence reasons. Second, you'll also learn which mistakes you tend to make when setting up a problem quickly. I'd argue that most quant mistakes come in the first 30 seconds and last 30 seconds you spend on a problem - you're either setting it up incorrectly or beginning with false assumptions, or you're missing steps or leaving a problem short when you finish. This drill allows you to focus on that first-30-seconds portion that can be so crucial.
_________________
Brian
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Re: Looking for suggestions to improve quant speed/score [#permalink]
### Show Tags
29 Jan 2011, 14:15
DaoEmo wrote:
Well, it seems that your main problem is lack of practice. At the current moment you need to solve as many problems as possible, bringing your quant skills to the level when you can recognize patterns immediately. Ideally, you should see how to solve the problem in first 10-15 secs after reading it. (that's my opinion) So the main thing is to bring all required for GMAT concepts in a system. Imagine that you have a garret in your head (like Sherlock Holmes ) and when you see a problem you simply search a tool that can solve it. So all needed to be done is to improve search speed. When you practice a lot, you get this improvement quite fast. I love the way MGMAT guides classify problems and tools for tackling 'em, just go through these books and solve OG/QR questions, I think everything will be alright after that. Also consider going through end-chapter questions, they help to memorize key concepts. I've noticed you've already completed some of the books. However, try re-doing questions which were hard for you. And, of course, check OA. So the general advice is: keep practicing and don't worry about your speed, it will improve quite automatically I believe.
When you obtain certain level of confidence, try going through GMATclub tests. This is a very useful tool, I've started using these tests recently, they are quite challenging but overall doable. Also my advice is to go untimed first 5-7 tests so that time issues won't distract you from searching patterns.
Finally, I don't think you should take many CAT's at this moment. Just focus on memorizing concepts and techniques.
Thanks for your response. In other words, the more practice questions I solve the faster is a response time. Make sense. I often wonder how on earth someone has time to rephrase a question and then evaluate it in DS
Kudos [?]: 6 [0], given: 12
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Posts: 19
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Re: Looking for suggestions to improve quant speed/score [#permalink]
### Show Tags
29 Jan 2011, 14:49
Yeah, DS questions can be harsh sometimes However, the more answer explanations you read, the more equipped you are. There are special chapters in MGMAT guides also, as I remember, you can check how to rephrase there. Keeping attention to small details is very important when solving DS questions. Always notice words like "positive", "integer" and so on, this may be crucial.
Forgot to say, I think your goal is quite achievable since you have decent verbal and quant is much easier to master. Good luck!
Kudos [?]: 10 [0], given: 5
Director
Affiliations: HBS Class of 2013
Joined: 16 Dec 2010
Posts: 729
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Re: Looking for suggestions to improve quant speed/score [#permalink]
### Show Tags
29 Jan 2011, 15:50
Don't just do OG problems--reread the MGMAT books and find out how to solve all the most common types of problems. You need to be able to recognize all the ways the GMAT uses to make simple questions look more obscure...for example, how complicated-looking questions can be simplified to (a^2 - b^2) which obviously equals (a+b)(a-b). And you MUST memorize all the equations and formulas, exponent rules, characteristics of geometric shapes, etc.
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Re: Looking for suggestions to improve quant speed/score [#permalink]
### Show Tags
31 Jan 2011, 09:33
DaoEmo wrote:
Yeah, DS questions can be harsh sometimes However, the more answer explanations you read, the more equipped you are. There are special chapters in MGMAT guides also, as I remember, you can check how to rephrase there. Keeping attention to small details is very important when solving DS questions. Always notice words like "positive", "integer" and so on, this may be crucial.
Forgot to say, I think your goal is quite achievable since you have decent verbal and quant is much easier to master. Good luck!
Yeah, I noticed that about DS questions. By the way, I find it time consuming to draw tables for work/rate and speed/distance problems. By the time I fill it in, like 60-80 seconds are gone.
Kudos [?]: 6 [0], given: 12
Intern
Joined: 01 Oct 2010
Posts: 33
Kudos [?]: 6 [0], given: 12
Location: US, NJ
Re: Looking for suggestions to improve quant speed/score [#permalink]
### Show Tags
02 Feb 2011, 07:41
VeritasPrepBrian wrote:
Hey guys,
Great discussion so far! If I could add a few suggestions...
1) Data Sufficiency
Another great way to get your mind in tune with DS questions is to train yourself to "think like the testmaker" by turning each DS question into a series of a few questions - ask yourself how they could change one word, symbol, etc. to elicit an entirely different answer. For example, there's an OG question that reads:
For nonzero integers m and n, is m/n > (mn)^4?
1) m/n > 0
2) m < 0
For this one, since we know that m and n are integers, there's no chance that m/n could ever be greater than (mn)^4. Thinking logically, on the left we take the same m and divide it, whereas on the right we multiply it by m^3 n^4. Even if either is negative, the right hand side will end up positive because of the even exponent. At worst we can construct a tie if all values are 1 or -1. The answer is D.
But to become a true master of DS, ask yourself how they can use the same setup to elicit a different answer. One way - if they just changed the direction of the inequality in the question to:
Is m/n < (mn)^4?
Usually the answer will be "yes" - take 4 and 2, 4/2 is much less than (4*2)^4. BUT, since we know that we can construct a "tie" with all 1s, then we can still get that answer "NO". Here, we learn that with inequalities you have to be careful because the yes/no question format can be tricky with the potential for a "tie".
2) Timing
One of my favorite timing drills is one I call "Quick First Step", in which you take 10 or 20 quant problems and start on each in 30 seconds, then move on to the next. When you're done with 30 seconds/question, go back and finish the question. Here, your goal is first to train yourself to begin questions quickly by drawing relationships and identifying your known quantities - so often people spend 30 seconds reading, 15 seconds worrying, and then it's almost a full minute before they start "doing". You want to actively read and begin work on a problem, both for pacing and confidence reasons. Second, you'll also learn which mistakes you tend to make when setting up a problem quickly. I'd argue that most quant mistakes come in the first 30 seconds and last 30 seconds you spend on a problem - you're either setting it up incorrectly or beginning with false assumptions, or you're missing steps or leaving a problem short when you finish. This drill allows you to focus on that first-30-seconds portion that can be so crucial.
Great post Brian! I think the 30 second "Quick First Step" drill is going to highlight one's weaknesses and gaps in the knowledge of basic math concepts. I'll definitively try that.
Kudos [?]: 6 [0], given: 12
Re: Looking for suggestions to improve quant speed/score [#permalink] 02 Feb 2011, 07:41
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# Looking for suggestions to improve quant speed/score
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 4,357 | 17,576 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | latest | en | 0.948571 |
https://www.crazy-numbers.com/en/20187 | 1,713,118,079,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.19/warc/CC-MAIN-20240414161724-20240414191724-00141.warc.gz | 680,362,689 | 3,500 | Warning: Undefined array key "numbers__url_substractions" in /home/clients/df8caba959271e8e753c9e287ae1296d/websites/crazy-numbers.com/includes/fcts.php on line 156
Number 20187: mathematical and symbolic properties | Crazy Numbers
Everything about number 20187
Discover a lot of information on the number 20187: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
Mathematical properties of 20187
Is 20187 a prime number? No
Is 20187 a perfect number? No
Number of divisors 6
List of dividers
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1, 3, 9, 2243, 6729, 20187
Sum of divisors 29172
Prime factorization 32 x 2243
Prime factors
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3, 2243
How to write / spell 20187 in letters?
In letters, the number 20187 is written as: Twenty thousand hundred and eighty-seven. And in other languages? how does it spell?
20187 in other languages
Write 20187 in english Twenty thousand hundred and eighty-seven
Write 20187 in french Vingt mille cent quatre-vingt-sept
Write 20187 in spanish Veinte mil ciento ochenta y siete
Write 20187 in portuguese Vinte mil cento oitenta e sete
Decomposition of the number 20187
The number 20187 is composed of:
1 iteration of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2
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1 iteration of the number 0 : ... Find out more about the number 0
1 iteration of the number 1 : The number 1 (one) represents the uniqueness, the unique, a starting point, a beginning.... Find out more about the number 1
1 iteration of the number 8 : The number 8 (eight) represents power, ambition. It symbolizes balance, realization.... Find out more about the number 8
1 iteration of the number 7 : The number 7 (seven) represents faith, teaching. It symbolizes reflection, the spiritual life.... Find out more about the number 7
Mathematical representations and links
Other ways to write 20187
In letter Twenty thousand hundred and eighty-seven
In roman numeral
In binary 100111011011011
In octal 47333 | 795 | 2,882 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-18 | latest | en | 0.700317 |
http://pari.math.u-bordeaux.fr/archives/pari-users-0406/msg00007.html | 1,406,857,370,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510273874.36/warc/CC-MAIN-20140728011753-00325-ip-10-146-231-18.ec2.internal.warc.gz | 220,669,813 | 2,434 | Karim Belabas on Wed, 16 Jun 2004 11:18:38 +0200
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Re: question about primitive roots
```* Justin Walker [2004-06-15 22:10]:
>
> On Jun 15, 2004, at 12:45, Mc Laughlin, James wrote:
>
> >Hello everyone,
> >
> >I have a question about primitive roots.
> >I have a prime p and I want a primitive root g.
> >
> >I know that znprimroot(79) will return a primitive root for the prime p
> >= 79 in the form Mod(3,79), meaning that if p=79, g=3 is a primitive
> >root.
> >
> >Is there anyway I can extract this "3" from the "Mod(3,79)" and set
> >"g=3;"
>
> Check 'lift()' (via "?lift").
Actually, ??lift should be even more helpful.
You also have centerlift() for symmetric residue system.
Karim.
P.S: Btw, if g = Mod(3,79), you can get the '79' via g.mod .
--
Karim Belabas Tel: (+33) (0)1 69 15 57 48
Dep. de Mathematiques, Bat. 425 Fax: (+33) (0)1 69 15 60 19
Universite Paris-Sud http://www.math.u-psud.fr/~belabas/
F-91405 Orsay (France) http://pari.math.u-bordeaux.fr/ [PARI/GP]
``` | 380 | 1,116 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-23 | longest | en | 0.689369 |
http://conversion.org/mass/grave/atomic-unit-of-mass-electron-rest-mass | 1,568,580,676,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572289.5/warc/CC-MAIN-20190915195146-20190915221146-00351.warc.gz | 48,078,830 | 7,025 | # grave to atomic unit of mass, electron rest mass conversion
Conversion number between grave [gv.] and atomic unit of mass, electron rest mass [me] is 1.0977691228099 × 10+30. This means, that grave is bigger unit than atomic unit of mass, electron rest mass.
### Contents [show][hide]
Switch to reverse conversion:
from atomic unit of mass, electron rest mass to grave conversion
### Enter the number in grave:
Decimal Fraction Exponential Expression
[gv.]
eg.: 10.12345 or 1.123e5
Result in atomic unit of mass, electron rest mass
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
### Calculation process of conversion value
• 1 grave = (1) / (9.10938356*10^-31) = 1.0977691228099 × 10+30 atomic unit of mass, electron rest mass
• 1 atomic unit of mass, electron rest mass = (9.10938356*10^-31) / (1) = 9.10938356 × 10-31 grave
• ? grave × (1 ("kg"/"grave")) / (9.10938356*10^-31 ("kg"/"atomic unit of mass, electron rest mass")) = ? atomic unit of mass, electron rest mass
### High precision conversion
If conversion between grave to kilogram and kilogram to atomic unit of mass, electron rest mass is exactly definied, high precision conversion from grave to atomic unit of mass, electron rest mass is enabled.
Since definition contain rounded number(s) too, there is no sense for high precision calculation, but if you want, you can enable it. Keep in mind, that converted number will be inaccurate due this rounding error!
### grave to atomic unit of mass, electron rest mass conversion chart
Start value: [grave] Step size [grave] How many lines? (max 100)
visual:
graveatomic unit of mass, electron rest mass
00
101.0977691228099 × 10+31
202.1955382456198 × 10+31
303.2933073684297 × 10+31
404.3910764912395 × 10+31
505.4888456140494 × 10+31
606.5866147368593 × 10+31
707.6843838596692 × 10+31
808.7821529824791 × 10+31
909.879922105289 × 10+31
1001.0977691228099 × 10+32
1101.2075460350909 × 10+32
Copy to Excel
## Multiple conversion
Enter numbers in grave and click convert button.
One number per line.
Converted numbers in atomic unit of mass, electron rest mass:
Click to select all
## Details about grave and atomic unit of mass, electron rest mass units:
Convert Grave to other unit:
### grave
Definition of grave unit: ≡ 1 kg. grave was the original name of the kilogram
Convert Atomic unit of mass, electron rest mass to other unit:
### atomic unit of mass, electron rest mass
Definition of atomic unit of mass, electron rest mass unit: ≈ 9.10938356 x 10-31 kg.
← Back to Mass units | 723 | 2,542 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2019-39 | latest | en | 0.596672 |
https://linearprogramminghelp.com/the-benefits-of-using-the-linear-programming-tableau-solver/ | 1,695,993,941,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510516.56/warc/CC-MAIN-20230929122500-20230929152500-00098.warc.gz | 401,675,702 | 20,430 | # The Benefits Of Using The Linear Programming Tableau Solver
Using a linear programming tableau solver can be very helpful in presenting information to others in a clear and organized manner. There are a number of benefits to using a tableau of this nature. Most importantly, it makes the information much more accessible and therefore much easier to understand. Also, it can help to develop a solid data analysis method. There are a number of ways that a person can use these tools to make a remarkable difference in their presentations and also in the bottom line of their projects.
One of the primary uses for a linear programming tableau solver is to show the impact of variable changes on a set of parameters over time. In doing so, you are able to create a much more dynamic presentation of the data that you have at your disposal. The primary benefit of using this type of tool is that you can design a linear programming system that functions in real-time without having to wait for any delays. The linear programming tableau type comes with a host of different advantages. For example, using one of these systems will allow you to display data from anywhere in the world and they provide a means for a user to input data and then to view it at any given moment.
When you choose to use one of these linear programming tableau solvers you will find that you are able to design a much more flexible linear model system. When it comes to linear models, there are generally four different options. These include quadratic, cubic, disc and so on. You can simply select the option that best meets your needs and then develop your linear programming model in whichever way suits your needs.
Another benefit to using a linear programming tableau solver is that you can make use of mathematical calculations without having to actually program the system. In other words, you can work with the linear programming tableau system without needing to write or modify anything. This makes the linear programming system very powerful and allows you to handle a wide range of different calculations. If you have a need to calculate the area, volume, area ratio, density, temperature or anything else, this type of programmable system can help you get the answers that you need. All you need to do is select the function and then input the required data.
A linear programming tableau control can also be programmed to give you an output in two ways. In one of these ways, the output from the linear algorithm can be automatically adjusted and another mode allows the user to manually adjust the output. If you need to manually adjust the output, you will find that the linear programming tableau control will be especially useful.
One of the benefits to using a linear programming tableau control is that you will find that it can save you a great deal of time. You will not have to wait for results and you can focus your attention on more important things. In other words, the linear tableau control can be an extremely valuable commodity. If you are interested in expanding your knowledge of various linear algorithms you may find that the linear programming tableau system is exactly what you need to make your job simple and your calculations more accurate.
If you are looking for a tool to help you with statistical analysis and graphing you may find that the linear programming tableau control is exactly what you need. If you are working with a statistical analysis, you may be able to get a great deal of information about relationships between factors. You may be able to examine whether or not a certain variable affects another and this can give you a huge benefit when you are trying to make statistical statements. If you would like to examine relationships between any variables you will find that the linear programming system is exactly what you need to help you examine these relationships.
If you are looking for a tool to help you prepare financial statements, you may find that the linear programming tableau is exactly what you need. This system can save you a great deal of time and money and if you are preparing financial statements this can allow you to print out very accurate results almost immediately. If you are interested in expanding your knowledge of linear programming and you would like to have a tool that is very easy to use, affordable, and accurate than the linear tableau software will meet all of your needs. You will be able to create some wonderful results very quickly using this excellent software program. | 849 | 4,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-40 | latest | en | 0.930107 |
https://zbmath.org/?q=an%3A0724.90048 | 1,660,580,548,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572192.79/warc/CC-MAIN-20220815145459-20220815175459-00172.warc.gz | 941,415,515 | 10,859 | ## Dual method for the solution of a one-stage stochastic programming problem with random RHS obeying a discrete probability distribution.(English)Zbl 0724.90048
A linear program with discretely distributed random right hand side is handled by incorporating a mean penalty term into the objective function to penalize the violations of the stochastic constraints; moreover, an additional chance constraint is used in case of inequality constraints in the original random program. Introducing some additional, linear constraints, the objective function can then be written in the form of a separable, piecewise linear, convex function. If the original random linear program contains only equality constraints, then, using the so- called $$\lambda$$-representation of univariate piecewise linear convex functions, the substituting problem can be represented by a deterministic LP having a special structure. This LP is then solved by a special dual type algorithm. In order to use this method also in case of additional chance constraints, first the notion of a “p level efficient point” of the probability distribution of the random elements is introduced, which implies then certain additional linear constraints. Thus, the same dual type linear programming method as in the first case can be used also here.
### MSC:
90C15 Stochastic programming 90-08 Computational methods for problems pertaining to operations research and mathematical programming
Full Text:
### References:
[1] E.M.L. Beale (1955) On Minimizing a Convex Function Subject to Linear Inequalities, Journal of the Royal Statistical Society Ser. B. 17, 173-184 · Zbl 0068.13701 [2] R.G. Bland (1977) New Finite Pivoting Rules for the Simplex Method, Mathematics of Operations Research 2, 103-107 · Zbl 0408.90050 [3] A. Charnes, W.W. Cooper and G.H. Symonds (1985) Cost Horizons and Certainty Equivalents: An Approach to Stochastic Programming of Heating Oil Production, Management Science 4, 236-263 [4] G.B. Dantzig (1955) Linear Programming under Uncertainty, Management Science 1, 197-206 · Zbl 0995.90589 [5] C.E. Lemke (1954) The Dual Method for Solving the Linear Programming Problem, Naval Research Logistic Quarterly 1, 36-47. · Zbl 0128.39605 [6] K. Murty (1968) Linear Programming under Uncertainty: A Basic Property of the Optimal Solution, Zeitschrift für Wahrscheinlichkeitstheorie ver. Geb. 10, 284-288 · Zbl 0169.51403 [7] A. Prékopa (1973) Contributions to the Theory of Stochastic Programming, Mathematical Programming 4, 202-221 · Zbl 0273.90045 [8] A. Prékopa (1990) The Discrete Moment Problem and Linear Programming. To appear in Discrete Applied Mathematics · Zbl 0712.90045 [9] R.J.B. Wets (1983) Solving Stochastic Programs with Simple Recourse, Stochastics 10, 219-242 · Zbl 0584.90067
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 743 | 3,156 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-33 | latest | en | 0.847333 |
zeyangwang.com | 1,534,385,023,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210408.15/warc/CC-MAIN-20180816015316-20180816035316-00140.warc.gz | 596,218,122 | 7,157 | This was a challenge posted by Aliyun, which is the cloud computing business of Alibaba Group, for the “shopping carnival” on 11/11 every year, which date also known as “Single Day” in China.
Here is the original chanllenge:
There were three hints:
1. Colorful anagram, green and red stand on opposite side
2. Ancient encryption technique, invented 300 years ago
3. In programmers’ world, 1+1 != 2
When looking at the second hint, we can almost certain that that was not any fancy encryption algorithm because the history of computer is only around 60 years. That should give us a lot information, back to that age, the most popular encryption method would be substitution cipher, like `simple substitution` or `polygraphic cipher`. What’s more, by taking a look at the question, space is also given, which is another import factor that tells us length of each word.
Noted that letter frequency analysis would not necessarily solve this problem because polygraphic cipher like `Vigenère cipher` could mask letter frequency with different key length. However, with modern computing power, and know word length, in theory we should be able to use brute force cracking this by dictionary lookups easily.
But why not use more intelligence instead of computer power? :)
With hint 1, all the green and red letters from questin string are:
Feels like an anagram. I googled some anagram solver, however, I didn’t get any reasonable result. It does look like a lot with `keyboard` with a b instead of d. This sounds a little wobble, but if we think b as a rotated d it’s good enough, heh.
Alright, so we have our first clue: `keyboard`.
We can try this in two way:
1. keyboard simple substitution cipher, or
2. key for Vigenère cipher
First one is not hard to understand, a keyboard substitution is just keyboard order to alphabetical order: qwert -> abcde etc.
Second one would require some effort, take a look at the following Vigenère cipher:
Say if you want to encrypt “apple” with key “dog”, you just need repeat “dog” to same length of input text and lookup each letter from the graph:
It turned out it was just a simple keyboard substitution. Here’s the decoded text:
In the room there are four identical basketballs and two identical footballs. Now if you want to put them in one line, how many solutions are there? Tips:please change the form of the number you get.The Programmers!
Now we are getting really close. We successfully decrypted the first level encryption, which is text encryption. Now it’s a basic probability question. Two ways to tackle this:
1. place 4 basketball in a row, then insert footballs between them, with cases two footballs are adjacent and seperated by basketball. C(5,2) + C(5,1) = 15
2. place 6 basketball in a row, pick either two as football. C(6,2) = 15
Second level encryption revealed. Anwser is 15.
It’s getting really obvious now. With the tip from decrypted message above, it’s asking you change form of the number iin programming world: decimal -> binary.
so: 15 -> 1111.
Awesome, the final anwser is 1111, which is the date for the shopping carnival. | 715 | 3,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | longest | en | 0.940924 |
https://essaysstudio.com/copy-machine-use-the-average-number-of-pages-a-person-copies/ | 1,670,264,375,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711042.33/warc/CC-MAIN-20221205164659-20221205194659-00859.warc.gz | 276,114,949 | 13,448 | # Copy machine use the average number of pages a person copies
Copy Machine Use A store manager hypothesizes that the average number of pages a person copies on
the store’s copy machine is less than 40. A sample of 50 customers’ orders is selected. At = 0.01, is there enough evidence to support the claim? Use the P-value hypothesis-testing method. Assume = 30.9.
2 2 2 5 32 5 29 8 2 49 21 1 24 72 70 21 85 61 8 42 3 15 27 113 36 37 5 3 58 82 9 2 1 6 9 80 9 51 2 122 21 49 36 43 61 3 17 17 4 1 | 191 | 499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-49 | latest | en | 0.696536 |
https://gamedev.stackexchange.com/questions/211237/why-is-my-perlin-noise-generated-texture-a-uniform-grey | 1,720,933,247,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514548.45/warc/CC-MAIN-20240714032952-20240714062952-00160.warc.gz | 235,583,107 | 36,817 | # Why is my perlin-noise generated texture a uniform grey?
I'm following Brackeys' tutorial for perlin noise and i'm at the point where he's just done the offset thing but this whole time all that i'm seeing in my project is a plain gray square, no matter what settings i change. Here's my code:
using UnityEngine;
public class PerlinNoise : MonoBehaviour
{
public int width = 256;
public int height = 256;
public float scale = 20f;
public float offsetX = 100f;
public float offsetY = 100f;
void Update ()
{
Renderer renderer = GetComponent<Renderer>();
renderer.material.mainTexture = GenerateTexture();
}
Texture2D GenerateTexture ()
{
Texture2D texture = new Texture2D(width, height);
for (int x = 0; x < width; x++)
{
for (int y = 0; y < height; y++)
{
Color color = CalculateColor(x, y);
texture.SetPixel(x, y, color);
}
}
texture.Apply();
return texture;
}
Color CalculateColor (int x, int y)
{
float xCoord = (float)x / width * scale + offsetX;
float yCoord = (float)y / height * scale + offsetY;
float sample = Mathf.PerlinNoise(x, y);
return new Color(sample, sample, sample);
}
}
Here's the result i get:
and here's the settings in the editor:
In the function CalculateColor, you calculate values for the variables xCoord and yCoord, but then you never use them. You are using the pixel coordinates x and y instead. The result is that you are generating perlin noise on integer values. Due to the way Perlin noise works, integer values will always give you the same results.
float sample = Mathf.PerlinNoise(xCoord, yCoord);
By the way, Perlin noise is actually obsolete. Over 20 years ago, Ken Perlin himself created Simplex Noise as a drop-in replacement which not only has less weird quirks and artifacts (including this one) but is also faster to compute. The reason why it didn't replace Perlin noise already is because Simplex noise used to be patented. But the patent ran out 2 years ago. So nowadays Simplex noise is almost always the better choice. You can find a simplex noise implementation in the Unity math library. If you want to give it a try, add using Unity.Mathematics; at the top of your script, and try this instead:
float sample = noise.snoise(new float2(xCoord, yCoord)) * 0.5f + 0.5f;
(the * 0.5f + 0.5f part is required because noise.snoise generates noise in the range of -1 to 1 instead of 0 to 1 like Mathf.PerlinNoise). | 584 | 2,375 | {"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} | 2.578125 | 3 | CC-MAIN-2024-30 | latest | en | 0.750651 |
https://rdrr.io/cran/RandomFields/man/RMmatrix.html | 1,624,020,848,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487636559.57/warc/CC-MAIN-20210618104405-20210618134405-00427.warc.gz | 447,146,139 | 10,787 | # RMmatrix: Matrix operator In RandomFields: Simulation and Analysis of Random Fields
## Description
`RMmatrix` is a multivariate covariance model depending on one multivariate covariance model, or one or several univariate covariance models C0,…. The corresponding covariance function is given by
C(h) = M phi(h) M^t
if a multivariate case is given. Otherwise it returns a matrix whose diagonal elements are filled with the univarate model(s) `C0`, `C1`, etc, and the offdiagonals are all zero.
## Usage
```1 2``` ```RMmatrix(C0, C1, C2, C3, C4, C5, C6, C7, C8, C9, M, vdim, var, scale, Aniso, proj) ```
## Arguments
`C0` a k-variate covariance `RMmodel` or a univariate model or a list of models joined by `c` `C1,C2,C3,C4,C5,C6,C7,C8,C9` optional univariate models `M` a k times k matrix, which is multiplied from left and right to the given model; M may depend on the location, hence it is then a matrix-valued function and C will be non-stationary with C(x, y) = M(x) phi(x, y) M(y)^t `vdim` positive integer. This argument should be given if and only if a multivariate model is created from a single univariate model and `M` is not given. (In fact, if `M` is given, `vdim` must equal the number of columns of `M`) `var,scale,Aniso,proj` optional arguments; same meaning for any `RMmodel`. If not passed, the above covariance function remains unmodified.
## Value
`RMmatrix` returns an object of class `RMmodel`.
## Note
• `RMmatrix` also allows variogram models are arguments.
`RMmodel`, `RFsimulate`, `RFfit`.
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50``` ```RFoptions(seed=0) ## *ANY* simulation will have the random seed 0; set ## RFoptions(seed=NA) to make them all random again ## Not run: ## first example: bivariate Linear Model of Coregionalisation x <- y <- seq(0, 10, 0.2) model1 <- RMmatrix(M = c(0.9, 0.43), RMwhittle(nu = 0.3)) + RMmatrix(M = c(0.6, 0.8), RMwhittle(nu = 2)) plot(model1) simu1 <- RFsimulate(RPdirect(model1), x, y) plot(simu1) ## second, equivalent way of defining the above model model2 <- RMmatrix(M = matrix(ncol=2, c(0.9, 0.43, 0.6, 0.8)), c(RMwhittle(nu = 0.3), RMwhittle(nu = 2))) simu2 <- RFsimulate(RPdirect(model2), x, y) stopifnot(all.equal(as.array(simu1), as.array(simu2))) ## third, equivalent way of defining the above model model3 <- RMmatrix(M = matrix(ncol=2, c(0.9, 0.43, 0.6, 0.8)), RMwhittle(nu = 0.3), RMwhittle(nu = 2)) simu3 <- RFsimulate(RPdirect(model3), x, y) stopifnot(all(as.array(simu3) == as.array(simu2))) ## End(Not run) ## second example: bivariate, independent fractional Brownian motion ## on the real axis x <- seq(0, 10, 0.1) modelB <- RMmatrix(c(RMfbm(alpha=0.5), RMfbm(alpha=1.5))) ## see the Note above print(modelB) simuB <- RFsimulate(modelB, x) plot(simuB) ## third example: bivariate non-stationary field with exponential correlation ## function. The variance of the two components is given by the ## variogram of fractional Brownian motions. ## Note that the two components have correlation 1. x <- seq(0, 10, 0.1) modelC <- RMmatrix(RMexp(), M=c(RMfbm(alpha=0.5), RMfbm(alpha=1.5))) print(modelC) simuC <- RFsimulate(modelC, x, x, print=1) #print(as.vector(simuC)) plot(simuC) ``` | 1,085 | 3,280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-25 | longest | en | 0.632866 |
https://www.chegg.com/homework-help/applied-differential-equations-3rd-edition-chapter-8.3.3-solutions-9780130400970 | 1,529,606,419,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864256.26/warc/CC-MAIN-20180621172638-20180621192638-00109.warc.gz | 786,131,393 | 13,588 | # Applied Differential Equations (3rd Edition) View more editions Solutions for Chapter 8.3.3
• 1531 step-by-step solutions
• Solved by professors & experts
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Chapter: Problem:
Sample Solution
Chapter: Problem:
• Step 1 of 3
a) We have to write geometric interpretation of the given equation
…… (1)
Suppose from eq. (1)
We have
…… (2)
The eq. (2) is the differential equation and its solution can be obtain by
…… (3)
And it’s the solution of .
The geometrical interpretation of is given by the following graph.
• Step 2 of 3
b) We have to write geometric interpretation of the given equation
…… (1)
Differentiate eq. (1) twice with respect to, we have
The geometrical interpretation of is given by the following graph.
• Step 3 of 3
c) We have to write geometric interpretation of the given equation
…… (1)
Where
Let , we have
and its solution is given by
, if we put it will give the same solution
Is the solution of .
Thus the geometrical interpretation of is given by the following graph.
Corresponding Textbook
Applied Differential Equations | 3rd Edition
9780130400970ISBN-13: 0130400971ISBN: Murray R SpiegelAuthors: | 299 | 1,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2018-26 | latest | en | 0.872418 |
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Calculate the absolute value of complex number -15-29i.
|-15-29i| = 32.6
### Step-by-step explanation:
$\mathrm{\mid }-15-29i\mathrm{\mid }=\sqrt{\left(-15{\right)}^{2}+\left(-29{\right)}^{2}}=32.6$
We will be pleased if You send us any improvements to this math problem. Thank you!
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The balls occupy between 30 and 50% of the volume of the cylinder. The diameter of the balls depends on the size of the feed and the diameter of the cylinder. The diameter of the balls varies from 2cm to 15cm. The balls can be made of metal, porcelain, or stainless steel. The ball acts. as a grinding medium. ... ratio as compared to Ball mill. ...
#### How much ball quantity should be controlled in the ball mill
Φ represents the ratio of the volume (V ball) of the ball mill steel ball (including pores) to the effective volume of the grinder (V machine). When the specification and rotating speed of the ball mill are fixed, the filling rate of the ball mill is in the range of less than 50%, and its production capacity will increase with the increase of ...
#### What's the Difference Between Ball Mill, Rod Mill and SAG ...
different grinding media of ball mills, rod mills and SAG mills. Crushing ratio Comparison. The biggest feature of SAG mills is the large crushing ratio, which can crush and grind the materials to 0.074mm at one time, accounting for …
#### Optimization of mill performance by using - SciELO
to the volumetric mill filling which influences grinding media wear rates, throughput, power draw, and product grind size from the circuit. Each of these performance parameters peaks at different filling values. In order to contin-uously optimize mill operation, it is vital to obtain regular measurements of the ball load and pulp position. | 2,689 | 12,423 | {"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} | 2.90625 | 3 | CC-MAIN-2023-23 | latest | en | 0.890151 |
https://math.stackexchange.com/questions/1513248/how-do-i-inverse-laplace-fracs13s4/1513270 | 1,582,817,255,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146714.29/warc/CC-MAIN-20200227125512-20200227155512-00389.warc.gz | 398,613,575 | 31,182 | # How do I Inverse Laplace $\frac{(s+1)^3}{s^4}$
I missed a class this week in maths and been a bit lost since with Inverse Laplace, how do I go about finding the Inverse laplace of: $$\frac{(s+1)^3}{s^4}$$
Do I simply expand the numerator? then inverse laplace? or is there a quicker way of doing it, expanding the brackets seems like a lengthy process...if someone could help me I'd be grateful as I have a test on this Friday!
• Doesn't look that onerous to me...and the inverse transforms of the $1/s^n$ are straightforward. – Simon S Nov 4 '15 at 18:48
• I thought I could handle it but surpisingly I'm stuck here, maybe my brain is fried from this weeks exams, I'm confused to as why there's an "s" in the numerator...if you can give me another hint at least I'd be grateful. – Modrisco Nov 4 '15 at 18:50
• So I expand it? I thought they're would be a quicker way which why I'm skeptical. – Modrisco Nov 4 '15 at 18:52
$$\frac{(s+1)^3}{s^4} = \frac 1s + \frac 3{s^2} + \frac 3{s^3} + \frac 1{s^4}$$ and the inverse Laplace transform of each of those terms should be standard to you. After you've found it, it may be possible to simplify the answer! | 351 | 1,159 | {"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} | 3.40625 | 3 | CC-MAIN-2020-10 | latest | en | 0.944671 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.03%3A_Exothermic_and_Endothermic_Processes | 1,631,809,733,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053657.29/warc/CC-MAIN-20210916145123-20210916175123-00561.warc.gz | 220,119,198 | 24,115 | # 17.3: Exothermic and Endothermic Processes
A campfire is an example of basic thermochemistry. The reaction is initiated by the application of heat from a match. The reaction converting wood to carbon dioxide and water (among other things) continues, releasing heat energy in the process. This heat energy can then be used to cook food, roast marshmallows, or simply to keep warm when it's cold outside.
## Exothermic and Endothermic Processes
When physical or chemical changes occur, they are generally accompanied by a transfer of energy. The law of conservation of energy states that in any physical or chemical process, energy is neither created nor destroyed. In other words, the entire energy in the universe is conserved. In order to better understand the energy changes taking place during a reaction, we need to define two parts of the universe: the system and the surroundings. The system is the specific portion of matter in a given space that is being studied during an experiment or an observation. The surroundings is everything in the universe that is not part of the system. In practical terms for a laboratory chemist, the system is the particular chemicals being reacted, while the surroundings is the immediate vicinity within the room. During most processes, energy is exchanged between the system and the surroundings. If the system loses a certain amount of energy, that same amount of energy is gained by the surroundings. If the system gains a certain amount of energy, that energy is supplied by the surroundings.
A chemical reaction or physical change is endothermic if heat is absorbed by the system from the surroundings. In the course of an endothermic process, the system gains heat from the surroundings, and so the temperature of the surroundings decreases. The quantity of heat for a process is represented by the letter $$q$$. The sign of $$q$$ for an endothermic process is positive because the system is gaining heat. A chemical reaction or physical change is exothermic if heat is released by the system into the surroundings. Because the surroundings are gaining heat from the system, the temperature of the surroundings increases. The sign of $$q$$ for an exothermic process is negative because the system is losing heat.
## Units of Heat
Heat flow is measured in one of two common units: the calorie and the joule. The joule $$\left( \text{J} \right)$$ is the SI unit of energy. The calorie is familiar because it is commonly used when referring to the amount of energy contained within food. A calorie $$\left( \text{cal} \right)$$ is the quantity of heat required to raise the temperature of 1 gram of water by $$1^\text{o} \text{C}$$. For example, raising the temperature of $$100 \: \text{g}$$ of water from $$20^\text{o} \text{C}$$ to $$22^\text{o} \text{C}$$ would require $$100 \times 2 = 200 \: \text{cal}$$.
Calories contained within food are actually kilocalories $$\left( \text{kcal} \right)$$. In other words, if a certain snack contains 85 food calories, it actually contains $$85 \: \text{kcal}$$ or $$85,000 \: \text{cal}$$. In order to make the distinction, the dietary calorie is written with a capital C.
$1 \: \text{kilocalorie} = 1 \: \text{Calorie} = 1000 \: \text{calories}$
To say that the snack "contains" 85 Calories means that $$85 \: \text{kcal}$$ of energy are released when that snack is processed by the human body.
Heat changes in chemical reactions are typically measured in joules rather than calories. The conversion between a joule and a calorie is shown below.
$1 \: \text{J} = 0.2390 \: \text{cal or} \: 1 \: \text{cal} = 4.184 \: \text{J}$
We can calculate the amount of heat released in kilojoules when a 400 Calorie hamburger is digested:
$400. \: \text{Cal} = 400. \: \text{kcal} \times \frac{4.184 \: \text{kJ}}{1 \: \text{kcal}} = 1.67 \times 10^3 \: \text{kJ}$
## Summary
• The law of conservation of energy states that in any physical or chemical process, energy is neither created nor destroyed.
• A specific portion of matter in a given space that is being studied during an experiment or an observation is the system.
• The surroundings is everything in the universe that is not part of the system.
• A chemical reaction or physical change is endothermic if heat is absorbed by the system from the surroundings.
• A reaction or change is exothermic if heat is released by the system into the surroundings. | 1,047 | 4,409 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-39 | longest | en | 0.951021 |
https://www.studymode.com/essays/Analytical-Tool-1338955.html | 1,566,468,809,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317037.24/warc/CC-MAIN-20190822084513-20190822110513-00233.warc.gz | 997,163,946 | 26,091 | # Analytical Tool
Topics: Arithmetic mean, Average, Mean Pages: 5 (1555 words) Published: January 3, 2013
Analytical tool
An analytical tool is something used to analyze or "take a closer look at" something. It is normally a way to review the effectiveness of something. For example, Google offers a free web analytics tool that is used by Web Masters to track visitors on a given site. It allows Web Masters to see where visitors are coming from, how long they stay, what links they are reviewing, etc. Analytic Tool help with Tracking and Reports!
Data flow diagram
-A data flow diagram (DFD) is a graphical representation of the "flow" of data through an information system, modeling its process aspects. Often they are a preliminary step used to create an overview of the system which can later be elaborated.[2]DFDs can also be used for the visualization of data processing (structured design). Diagram
HIPO
-The HIPO (Hierarchy plus Input-Process-Output) technique is a tool for planning and/or documenting a computer program. A HIPO model consists of a hierarchy chart that graphically represents the program’s control structure and a set of IPO (Input-Process-Output) charts that describe the inputs to, the outputs from, and the functions (or processes) performed by each module on the hierarchy chart. Using the HIPO technique, designers can evaluate and refine a program’s design, and correct flaws prior to implementation. Given the graphic nature of HIPO, users and managers can easily follow a program’s structure. The hierarchy chart serves as a useful planning and visualization document for managing the program development process. The IPO charts define for the programmer each module’s inputs, outputs, and algorithms. The HIPO technique is often used to plan or document a structured program (# 62). A variety of tools, including pseudocode (# 59) and structured English (# 60), can be used to describe processes on an IPO chart. System flowcharting symbols (# 37) are sometimes used to identify physical input, output, and storage devices on an IPO chart. A completed HIPO package has two parts. A hierarchy chart is used to represent the top-down structure of the program. For each module depicted on the hierarchy chart, an IPO (Input-Process-Output) chart is used to describe the inputs to, the outputs from, and the process performed by the module.
VTOC
-In the IBM mainframe storage architecture, Volume Table Of Contents, or VTOC, is a data structure that provides a way of locating the data sets that reside on a particular disk volume. It can reside within the first 64K tracks on the volume, and lists the names of each data set on the volume as well as size, location, and permissions. Additionally, it contains an entry for every area of contiguous free space on the volume. The third record on the first track of the first cylinder of any volume of DASD (i.e. disk pack) is known as the volume label and must contain a pointer to the location of the VTOC. A VTOC is added to a disk when it is initialized using the Device Support Facilities utility program, ICKDSF. VTOC was originally designed for removable disk packs. To locate a data set, a program will generally interrogate a z/OS catalog to find the volume where the data set resides. Having found the correct volume, the VTOC is searched to find out where on the disk the data set is stored.
Statist treatment of the data
Statistical treatment of data is essential in order to make use of the data in the right form. Raw data collection is only one aspect of any experiment; the organization of data is equally important so that appropriate conclusions can be drawn. This is what statistical treatment of data is all about.
Frequency distribution
In statistics, a frequency distribution is an arrangement of the values that one or more variables take in a sample. Each entry in the table contains the frequency or count of the occurrences of values within a particular group or interval, and in this way, the table summarizes the distribution of values in the sample A frequency distribution... | 888 | 4,092 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.84074 |
http://eclipseclp.org/docs/current/bips/kernel/typetest/nonground-3.html | 1,508,350,665,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823067.51/warc/CC-MAIN-20171018180631-20171018200631-00847.warc.gz | 109,244,436 | 1,457 | [ Type Testing | Reference Manual | Alphabetic Index ]
# nonground(+N, ?Term, -VarList)
Succeeds if Term contains at least N different variables, and returns N of them in the list VarList.
N
Integer.
Term
Prolog term.
VarList
List or variable.
## Description
Used to test whether Term contains at least N different variables. The argument VarList is unified with a list of exactly N of those variables. If Term contains more than N variables, it is not further specified which ones will be in the list and in which order. As usual, attributed variables are also considered variables.
Note that this predicate is a generalisation of nonground/1 and nonground/2 which could be written as:
``` nonground(Term) :- nonground(1, Term, _).
nonground(Term, Var) :- nonground(1, Term, [Var]).
```
This predicate can handle cyclic terms.
### Modes and Determinism
• nonground(+, ?, -) is semidet
### Fail Conditions
Fails if Term contains less than N variables
### Exceptions
(4) instantiation fault
N is not instantiated.
(5) type error
VarList instantiated but not to a list.
(6) out of range
N is not positive.
## Examples
```Success:
nonground(1, Term, L). % gives L = [Term]
nonground(1, f(a,B,c), L). % gives L = [B]
nonground(2, [X,Y,Z], L). % gives L = [Y,X]
nonground(2, [X,Y,Z], L). % gives L = [Y,X]
nonground(1, s(X{a}), L). % gives L = [X{a}]
Fail:
nonground(1, atom, L).
nonground(2, f(a,B,c), L).
nonground(2, [X,X,X], L).
```
## See Also
nonground / 1, nonground / 2, nonvar / 1, var / 1 | 463 | 1,532 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.77941 |
https://physics.stackexchange.com/questions/457704/flux-density-via-gauss-law-inside-sphere-cavity | 1,721,066,029,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00618.warc.gz | 402,739,281 | 41,217 | # Flux density via Gauss' Law inside sphere cavity
There are many similar questions on here but most seem to deal conceptually with electric field, not the mathematics behind it.
The setup is this: there is some sphere of radius b which has a volume charge density of ρv. However, there is a cavity inside the sphere, centered at the origin, of radius a, inside which ρv = 0. Also, ρv = 0 outside of the sphere of radius b. What we want to determine is the flux density D at r < a, a < r < b, and b < r.
Conceptually, it makes sense to me that the electric field inside the cavity is 0 since all charges on the inner surface are symmetrically distributed. However, there should be some D because flux would be transmitted along unit vector -ar. Same goes for outside the sphere: there should be flux radiating out along ar.
I know I have an error in thinking somewhere because when I use the divergence theorem to relate the forms of Gauss' Law (i.e. the surface integral of flux density is the volume integral of charge density), and I have charge density = 0 inside r < a, I get D = 0. The same happens for when I look at r > b; the integral equates to 0 so, regardless of limits of integration, D = 0.
Between a and b the integrals are obvious and straightforward, so I get a D = 4πρv( $$\frac{b^3 - a^3}{3}$$). What is not obvious about the cavity and outside the sphere?
• As a technical observation, the radius is a scalar I think you should be $b$ and not $\textbf{b}$. Same with $a$ rather than $\textbf{a}$. Commented Jan 30, 2019 at 9:20
• @ZeroTheHero Yup, you're correct. I was mixing vector notation and grammatical emphasis. Fixed. Commented Jan 30, 2019 at 23:01
Since $$\textbf{D}=\epsilon\textbf{E}$$ and there is no field, there will be no flux anywhere. Flux is not transmitted along vectors: it is the sum of $$\textbf{D}\cdot d\textbf{S}$$ evaluated on each of the surface elements $$d\textbf{S}$$,
Now it is true that it’s possible to have $$0$$ net flux through a closed surface but non-zero field in the region: if the closed surface contains no source charges, then every field line of $$\textbf{D}$$ that enters the closed surface will also exit the surface. This would happen for instance if you plunged a cube in the constant field between two infinite plates.
However, in your geometry, the flux must be in the $$\hat{\textbf{r}}$$ direction by symmetry, and the vector $$\textbf{D}$$ must also be along $$\hat{\textbf{r}}$$ by symmetry, so the contribution from all the charges on the inner surface of your hollow sphere will cancel out when summed over the whole surface.
This is qualitatively different from the situation where you want the flux through a surface containing the entire arrangement: the field outside the arrangement is identical to the field of a single charge of apppropriate magnitude located at the origin; the field lines outside the sphere will point along $$+\hat{\textbf{r}}$$ at every point of your surface, so that every contribution $$\textbf{D}\cdot d\textbf{S}$$ will be positive, resulting in a non-zero net flux.
Geometrically, the best way to understand this is by examining the figure below:
Of course with the situation on the left, where you look for the field at the center, the field is $$0$$ by symmetry.
For the situation on the right, where we want to field for a field off-center: the charges in angular opening on the right are closer but the area intercepted by the opening is small, whereas the charges in the larger opening opposite are farther but the area intercepted is larger. The effects exactly cancel out because the field goes like $$1/r^2$$ but the area intercepted grows like $$r^2$$. The result is that the net field at the off-centered point is still $$0$$.
• So to clarify, I think you're answer is telling me that E = D = 0 inside the cavity due to symmetrically opposing surface charge distribution (or $integral (rho_v * dV) = Q_enc$ inside cavity will = 0, which makes sense). However, outside the sphere-with-cavity, even though rho_v = 0, shouldn't there still be some D since the charge distribution on the outer sphere surface isn't symmetrically opposed? Like with a solid sphere with rho_v of radius a, where D outside radius a is $D = ((a^3)/(3r^2))*(rho-v)*(a_r)$. Commented Jan 30, 2019 at 22:56
• @PoGaMi added some material for a geometrical interpretation. Commented Jan 31, 2019 at 1:04
• Ohhhh, I think I see now. If I choose a Gaussian sphere to surround my whole sphere-with-cavity deal, the net flux on that surface is also zero due to a similar symmetry to the cavity within the sphere. The pictures really helped once I stared at them for a minute. Thank you! Commented Jan 31, 2019 at 3:57 | 1,173 | 4,717 | {"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": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-30 | latest | en | 0.936041 |
https://www.oreilly.com/library/view/tensorflow-for-deep/9781491980446/ch04.html | 1,709,006,144,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474670.19/warc/CC-MAIN-20240227021813-20240227051813-00497.warc.gz | 902,081,784 | 35,687 | Chapter 4. Fully Connected Deep Networks
This chapter will introduce you to fully connected deep networks. Fully connected networks are the workhorses of deep learning, used for thousands of applications. The major advantage of fully connected networks is that they are “structure agnostic.” That is, no special assumptions need to be made about the input (for example, that the input consists of images or videos). We will make use of this generality to use fully connected deep networks to address a problem in chemical modeling later in this chapter.
We delve briefly into the mathematical theory underpinning fully connected networks. In particular, we explore the concept that fully connected architectures are “universal approximators” capable of learning any function. This concept provides an explanation of the generality of fully connected architectures, but comes with many caveats that we discuss at some depth.
While being structure agnostic makes fully connected networks very broadly applicable, such networks do tend to have weaker performance than special-purpose networks tuned to the structure of a problem space. We will discuss some of the limitations of fully connected architectures later in this chapter.
What Is a Fully Connected Deep Network?
A fully connected neural network consists of a series of fully connected layers. A fully connected layer is a function from $double-struck upper R Superscript m$ to $double-struck upper R Superscript n$. Each output dimension depends on each input dimension. Pictorially, a fully connected layer is represented as follows in Figure 4-1.
Let’s dig a little deeper into what the mathematical form of a fully connected network is. Let $x\in {ℝ}^{m}$ represent the input to a fully connected layer. Let ${y}_{i}\in ℝ$ be the $i$-th output from the fully connected layer. Then ${y}_{i}\in ℝ$ is computed as follows:
${y}_{i}=\sigma \left({w}_{1}{x}_{1}+\cdots +{w}_{m}{x}_{m}\right)$
Here, $sigma$ is a nonlinear function (for now, think of $sigma$ as the sigmoid function introduced in the previous chapter), and the $w Subscript i$ are learnable parameters in the network. The full output y is then
$y=\left(\begin{array}{c}\sigma \left({w}_{1,1}{x}_{1}+\cdots +{w}_{1,m}{x}_{m}\right)\\ ⋮\\ \sigma \left({w}_{n,1}{x}_{1}+\cdots +{w}_{n,m}{x}_{m}\right)\end{array}\right)$
Note that it’s directly possible to stack fully connected networks. A network with multiple fully connected networks is often called a “deep” network as depicted in Figure 4-2.
As a quick implementation note, note that the equation for a single neuron looks very similar to a dot-product of two vectors (recall the discussion of tensor basics). For a layer of neurons, it is often convenient for efficiency purposes to compute y as a matrix multiply:
$y=\sigma \left(wx\right)$
where sigma is a matrix in $double-struck upper R Superscript n times m$ and the nonlinearity $sigma$ is applied componentwise.
“Neurons” in Fully Connected Networks
The nodes in fully connected networks are commonly referred to as “neurons.” Consequently, elsewhere in the literature, fully connected networks will commonly be referred to as “neural networks.” This nomenclature is largely a historical accident.
In the 1940s, Warren S. McCulloch and Walter Pitts published a first mathematical model of the brain that argued that neurons were capable of computing arbitrary functions on Boolean quantities. Successors to this work slightly refined this logical model by making mathematical “neurons” continuous functions that varied between zero and one. If the inputs of these functions grew large enough, the neuron “fired” (took on the value one), else was quiescent. With the addition of adjustable weights, this description matches the previous equations.
Is this how a real neuron behaves? Of course not! A real neuron (Figure 4-3) is an exceedingly complex engine, with over 100 trillion atoms, and tens of thousands of different signaling proteins capable of responding to varying signals. A microprocessor is a better analogy for a neuron than a one-line equation.
In many ways, this disconnect between biological neurons and artificial neurons is quite unfortunate. Uninitiated experts read breathless press releases claiming artificial neural networks with billions of “neurons” have been created (while the brain has only 100 billion biological neurons) and reasonably come away believing scientists are close to creating human-level intelligences. Needless to say, state of the art in deep learning is decades (or centuries) away from such an achievement.
As you read further about deep learning, you may come across overhyped claims about artificial intelligence. Don’t be afraid to call out these statements. Deep learning in its current form is a set of techniques for solving calculus problems on fast hardware. It is not a precursor to Terminator (Figure 4-4).
AI Winters
Artificial intelligence has gone through multiple rounds of boom-and-bust development. This cyclical development is characteristic of the field. Each new advance in learning spawns a wave of optimism in which prophets claim that human-level (or superhuman) intelligences are incipient. After a few years, no such intelligences manifest, and disappointed funders pull out. The resulting period is called an AI winter.
There have been multiple AI winters so far. As a thought exercise, we encourage you to consider when the next AI winter will happen. The current wave of deep learning progress has solved many more practical problems than any previous wave of advances. Is it possible AI has finally taken off and exited the boom-and-bust cycle or do you think we’re in for the “Great Depression” of AI soon?
Learning Fully Connected Networks with Backpropagation
The first version of a fully connected neural network was the Perceptron, (Figure 4-5), created by Frank Rosenblatt in the 1950s. These perceptrons are identical to the “neurons” we introduced in the previous equations.
Perceptrons were trained by a custom “perceptron” rule. While they were moderately useful solving simple problems, perceptrons were fundamentally limited. The book Perceptrons by Marvin Minsky and Seymour Papert from the end of the 1960s proved that simple perceptrons were incapable of learning the XOR function. Figure 4-6 illustrates the proof of this statement.
This problem was overcome with the invention of the multilayer perceptron (another name for a deep fully connected network). This invention was a formidable achievement, since earlier simple learning algorithms couldn’t learn deep networks effectively. The “credit assignment” problem stumped them; how does an algorithm decide which neuron learns what?
The full solution to this problem requires backpropagation. Backpropagation is a generalized rule for learning the weights of neural networks. Unfortunately, complicated explanations of backpropagation are epidemic in the literature. This situation is unfortunate since backpropagation is simply another word for automatic differentiation.
Let’s suppose that $f left-parenthesis theta comma x right-parenthesis$ is a function that represents a deep fully connected network. Here $x$ is the inputs to the fully connected network and $theta$ is the learnable weights. Then the backpropagation algorithm simply computes $StartFraction normal partial-differential f Over normal partial-differential theta EndFraction$. The practical complexities arise in implementing backpropagation for all possible functions f that arise in practice. Luckily for us, TensorFlow takes care of this already!
Universal Convergence Theorem
The preceding discussion has touched on the ideas that deep fully connected networks are powerful approximations. McCulloch and Pitts showed that logical networks can code (almost) any Boolean function. Rosenblatt’s Perceptron was the continuous analog of McCulloch and Pitt’s logical functions, but was shown to be fundamentally limited by Minsky and Papert. Multilayer perceptrons looked to solve the limitations of simple perceptrons and empirically seemed capable of learning complex functions. However, it wasn’t theoretically clear whether this empirical ability had undiscovered limitations. In 1989, George Cybenko demonstrated that multilayer perceptrons were capable of representing arbitrary functions. This demonstration provided a considerable boost to the claims of generality for fully connected networks as a learning architecture, partially explaining their continued popularity.
However, if both backpropagation and fully connected network theory were understood in the late 1980s, why didn’t “deep” learning become more popular earlier? A large part of this failure was due to computational limitations; learning fully connected networks took an exorbitant amount of computing power. In addition, deep networks were very difficult to train due to lack of understanding about good hyperparameters. As a result, alternative learning algorithms such as SVMs that had lower computational requirements became more popular. The recent surge in popularity in deep learning is partly due to the increased availability of better computing hardware that enables faster computing, and partly due to increased understanding of good training regimens that enable stable learning.
Is Universal Approximation That Surprising?
Universal approximation properties are more common in mathematics than one might expect. For example, the Stone-Weierstrass theorem proves that any continuous function on a closed interval can be a suitable polynomial function. Loosening our criteria further, Taylor series and Fourier series themselves offer some universal approximation capabilities (within their domains of convergence). The fact that universal convergence is fairly common in mathematics provides partial justification for the empirical observation that there are many slight variants of fully connected networks that seem to share a universal approximation property.
Universal Approximation Doesn’t Mean Universal Learning!
A critical subtlety exists in the universal approximation theorem. The fact that a fully connected network can represent any function doesn’t mean that backpropagation can learn any function! One of the major limitations of backpropagation is that there is no guarantee the fully connected network “converges”; that is, finds the best available solution of a learning problem. This critical theoretical gap has left generations of computer scientists queasy with neural networks. Even today, many academics will prefer to work with alternative algorithms that have stronger theoretical guarantees.
Empirical research has yielded many practical tricks that allow backpropagation to find good solutions for problems. We will go into many of these tricks in significant depth in the remainder of this chapter. For the practicing data scientist, the universal approximation theorem isn’t something to take too seriously. It’s reassuring, but the art of deep learning lies in mastering the practical hacks that make learning work.
Why Deep Networks?
A subtlety in the universal approximation theorem is that it in fact holds true for fully connected networks with only one fully connected layer. What then is the use of “deep” learning with multiple fully connected layers? It turns out that this question is still quite controversial in academic and practical circles.
In practice, it seems that deeper networks can sometimes learn richer models on large datasets. (This is only a rule of thumb, however; every practitioner has a bevy of examples where deep fully connected networks don’t do well.) This observation has led researchers to hypothesize that deeper networks can represent complex functions “more efficiently.” That is, a deeper network may be able to learn more complex functions than shallower networks with the same number of neurons. For example, the ResNet architecture mentioned briefly in the first chapter, with 130 layers, seems to outperform its shallower competitors such as AlexNet. In general, for a fixed neuron budget, stacking deeper leads to better results.
A number of erroneous “proofs” for this “fact” have been given in the literature, but all of them have holes. It seems the question of depth versus width touches on profound concepts in complexity theory (which studies the minimal amount of resources required to solve given computational problems). At present day, it looks like theoretically demonstrating (or disproving) the superiority of deep networks is far outside the ability of our mathematicians.
Training Fully Connected Neural Networks
As we mentioned previously, the theory of fully connected networks falls short of practice. In this section, we will introduce you to a number of empirical observations about fully connected networks that aid practitioners. We strongly encourage you to use our code (introduced later in the chapter) to check our claims for yourself.
Learnable Representations
One way of thinking about fully connected networks is that each fully connected layer effects a transformation of the feature space in which the problem resides. The idea of transforming the representation of a problem to render it more malleable is a very old one in engineering and physics. It follows that deep learning methods are sometimes called “representation learning.” (An interesting factoid is that one of the major conferences for deep learning is called the “International Conference on Learning Representations.”)
Generations of analysts have used Fourier transforms, Legendre transforms, Laplace transforms, and so on in order to simplify complicated equations and functions to forms more suitable for handwritten analysis. One way of thinking about deep learning networks is that they effect a data-driven transform suited to the problem at hand.
The ability to perform problem-specific transformations can be immensely powerful. Standard transformation techniques couldn’t solve problems of image or speech analysis, while deep networks are capable of solving these problems with relative ease due to the inherent flexibility of the learned representations. This flexibility comes with a price: the transformations learned by deep architectures tend to be much less general than mathematical transforms such as the Fourier transform. Nonetheless, having deep transforms in an analytic toolkit can be a powerful problem-solving tool.
There’s a reasonable argument that deep learning is simply the first representation learning method that works. In the future, there may well be alternative representation learning methods that supplant deep learning methods.
Activations
We previously introduced the nonlinear function $sigma$ as the sigmoidal function. While the sigmoidal is the classical nonlinearity in fully connected networks, in recent years researchers have found that other activations, notably the rectified linear activation (commonly abbreviated ReLU or relu) $sigma left-parenthesis x right-parenthesis equals max left-parenthesis x comma 0 right-parenthesis$ work better than the sigmoidal unit. This empirical observation may be due to the vanishing gradient problem in deep networks. For the sigmoidal function, the slope is zero for almost all values of its input. As a result, for deeper networks, the gradient would tend to zero. For the ReLU function, the slope is nonzero for a much greater part of input space, allowing nonzero gradients to propagate. Figure 4-7 illustrates sigmoidal and ReLU activations side by side.
Fully Connected Networks Memorize
One of the striking aspects about fully connected networks is that they tend to memorize training data entirely given enough time. As a result, training a fully connected network to “convergence” isn’t really a meaningful metric. The network will keep training and learning as long as the user is willing to wait.
For large enough networks, it is quite common for training loss to trend all the way to zero. This empirical observation is one the most practical demonstrations of the universal approximation capabilities of fully connected networks. Note however, that training loss trending to zero does not mean that the network has learned a more powerful model. It’s rather likely that the model has started to memorize peculiarities of the training set that aren’t applicable to any other datapoints.
It’s worth digging into what we mean by peculiarities here. One of the interesting properties of high-dimensional statistics is that given a large enough dataset, there will be plenty of spurious correlations and patterns available for the picking. In practice, fully connected networks are entirely capable of finding and utilizing these spurious correlations. Controlling networks and preventing them from misbehaving in this fashion is critical for modeling success.
Regularization
Regularization is the general statistical term for a mathematical operation that limits memorization while promoting generalizable learning. There are many different types of regularization available, which we will cover in the next few sections.
Regularization has a long history in the statistical literature, with entire sheaves of papers written on the topic. Unfortunately, only some of this classical analysis carries over to deep networks. The linear models used widely in statistics can behave very differently from deep networks, and many of the intuitions built in that setting can be downright wrong for deep networks.
The first rule for working with deep networks, especially for readers with prior statistical modeling experience, is to trust empirical results over past intuition. Don’t assume that past knowledge about techniques such as LASSO has much meaning for modeling deep architectures. Rather, set up an experiment to methodically test your proposed idea. We will return at greater depth to this methodical experimentation process in the next chapter.
Dropout
Dropout is a form of regularization that randomly drops some proportion of the nodes that feed into a fully connected layer (Figure 4-8). Here, dropping a node means that its contribution to the corresponding activation function is set to 0. Since there is no activation contribution, the gradients for dropped nodes drop to zero as well.
The nodes to be dropped are chosen at random during each step of gradient descent. The underlying design principle is that the network will be forced to avoid “co-adaptation.” Briefly, we will explain what co-adaptation is and how it arises in non-regularized deep architectures. Suppose that one neuron in a deep network has learned a useful representation. Then other neurons deeper in the network will rapidly learn to depend on that particular neuron for information. This process will render the network brittle since the network will depend excessively on the features learned by that neuron, which might represent a quirk of the dataset, instead of learning a general rule.
Dropout prevents this type of co-adaptation because it will no longer be possible to depend on the presence of single powerful neurons (since that neuron might drop randomly during training). As a result, other neurons will be forced to “pick up the slack” and learn useful representations as well. The theoretical argument follows that this process should result in stronger learned models.
In practice, dropout has a pair of empirical effects. First, it prevents the network from memorizing the training data; with dropout, training loss will no longer tend rapidly toward 0, even for very large deep networks. Next, dropout tends to slightly boost the predictive power of the model on new data. This effect often holds for a wide range of datasets, part of the reason that dropout is recognized as a powerful invention, and not just a simple statistical hack.
You should note that dropout should be turned off when making predictions. Forgetting to turn off dropout can cause predictions to be much noisier and less useful than they would be otherwise. We discuss how to handle dropout for training and predictions correctly later in the chapter.
How Can Big Networks Not Overfit?
One of the most jarring points for classically trained statisticians is that deep networks may routinely have more internal degrees of freedom than are present in the training data. In classical statistics, the presence of these extra degrees of freedom would render the model useless, since there will no longer exist a guarantee that the model learned is “real” in the classical sense.
How then can a deep network with millions of parameters learn meaningful results on datasets with only thousands of exemplars? Dropout can make a big difference here and prevent brute memorization. But, there’s also a deeper unexplained mystery in that deep networks will tend to learn useful facts even in the absence of dropout. This tendency might be due to some quirk of backpropagation or fully connected network structure that we don’t yet understand.
Early stopping
As mentioned, fully connected networks tend to memorize whatever is put before them. As a result, it’s often useful in practice to track the performance of the network on a held-out “validation” set and stop the network when performance on this validation set starts to go down. This simple technique is known as early stopping.
In practice, early stopping can be quite tricky to implement. As you will see, loss curves for deep networks can vary quite a bit in the course of normal training. Devising a rule that separates healthy variation from a marked downward trend can take significant effort. In practice, many practitioners just train models with differing (fixed) numbers of epochs, and choose the model that does best on the validation set. Figure 4-9 illustrates how training and test set accuracy typically change as training proceeds.
We will dig more into proper methods for working with validation sets in the following chapter.
Weight regularization
A classical regularization technique drawn from the statistical literature penalizes learned weights that grow large. Following notation from the previous chapter, let $script upper L left-parenthesis x comma y right-parenthesis$ denote the loss function for a particular model and let $theta$ denote the learnable parameters of this model. Then the regularized loss function is defined by
${ℒ}^{\text{'}}\left(x,y\right)=ℒ\left(x,y\right)+\alpha \parallel \theta \parallel$
where $parallel-to theta parallel-to$ is the weight penalty and $alpha$ is a tunable parameter. The two common choices for penalty are the L1 and L2 penalties
${\parallel \theta \parallel }_{2}=\sqrt{{\sum }_{i=1}^{N}{\theta }_{i}^{2}}$
${\parallel \theta \parallel }_{1}=\sum _{i=1}^{N}|{\theta }_{i}|$
where $parallel-to theta parallel-to$ and $parallel-to theta parallel-to$ denote the L1 and L2 penalties, respectively. From personal experience, these penalties tend to be less useful for deep models than dropout and early stopping. Some practitioners still make use of weight regularization, so it’s worth understanding how to apply these penalties when tuning deep networks.
Training Fully Connected Networks
Training fully connected networks requires a few tricks beyond those you have seen so far in this book. First, unlike in the previous chapters, we will train models on larger datasets. For these datasets, we will show you how to use minibatches to speed up gradient descent. Second, we will return to the topic of tuning learning rates.
Minibatching
For large datasets (which may not even fit in memory), it isn’t feasible to compute gradients on the full dataset at each step. Rather, practitioners often select a small chunk of data (typically 50–500 datapoints) and compute the gradient on these datapoints. This small chunk of data is traditionally called a minibatch.
In practice, minibatching seems to help convergence since more gradient descent steps can be taken with the same amount of compute. The correct size for a minibatch is an empirical question often set with hyperparameter tuning.
Learning rates
The learning rate dictates the amount of importance to give to each gradient descent step. Setting a correct learning rate can be tricky. Many beginning deep-learners set learning rates incorrectly and are surprised to find that their models don’t learn or start returning NaNs. This situation has improved significantly with the development of methods such as ADAM that simplify choice of learning rate significantly, but it’s worth tweaking the learning rate if models aren’t learning anything.
Implementation in TensorFlow
In this section, we will show you how to implement a fully connected network in TensorFlow. We won’t need to introduce many new TensorFlow primitives in this section since we have already covered most of the required basics.
Installing DeepChem
In this section, you will use the DeepChem machine learning toolchain for your experiments (full disclosure: one of the authors was the creator of DeepChem). Detailed installation directions for DeepChem can be found online, but briefly the Anaconda installation via the `conda` tool will likely be most convenient.
Tox21 Dataset
For our modeling case study, we will use a chemical dataset. Toxicologists are very interested in the task of using machine learning to predict whether a given compound will be toxic or not. This task is extremely complicated, since today’s science has only a limited understanding of the metabolic processes that happen in a human body. However, biologists and chemists have worked out a limited set of experiments that provide indications of toxicity. If a compound is a “hit” in one of these experiments, it will likely be toxic for a human to ingest. However, these experiments are often costly to run, so data scientists aim to build machine learning models that can predict the outcomes of these experiments on new molecules.
One of the most important toxicological dataset collections is called Tox21. It was released by the NIH and EPA as part of a data science initiative and was used as the dataset in a model building challenge. The winner of this challenge used multitask fully connected networks (a variant of fully connected networks where each network predicts multiple quantities for each datapoint). We will analyze one of the datasets from the Tox21 collection. This dataset consists of a set of 10,000 molecules tested for interaction with the androgen receptor. The data science challenge is to predict whether new molecules will interact with the androgen receptor.
Processing this dataset can be tricky, so we will make use of the MoleculeNet dataset collection curated as part of DeepChem. Each molecule in Tox21 is processed into a bit-vector of length 1024 by DeepChem. Loading the dataset is then a few simple calls into DeepChem (Example 4-1).
Example 4-1. Load the Tox21 dataset
````import` `deepchem` `as` `dc`
`_``,` `(``train``,` `valid``,` `test``),` `_` `=` `dc``.``molnet``.``load_tox21``()`
`train_X``,` `train_y``,` `train_w` `=` `train``.``X``,` `train``.``y``,` `train``.``w`
`valid_X``,` `valid_y``,` `valid_w` `=` `valid``.``X``,` `valid``.``y``,` `valid``.``w`
`test_X``,` `test_y``,` `test_w` `=` `test``.``X``,` `test``.``y``,` `test``.``w````
Here the `X` variables hold processed feature vectors, `y` holds labels, and `w` holds example weights. The labels are binary 1/0 for compounds that interact or don’t interact with the androgen receptor. Tox21 holds imbalanced datasets, where there are far fewer positive examples than negative examples. `w` holds recommended per-example weights that give more emphasis to positive examples (increasing the importance of rare examples is a common technique for handling imbalanced datasets). We won’t use these weights during training for simplicity. All of these variables are NumPy arrays.
Tox21 has more datasets than we will analyze here, so we need to remove the labels associated with these extra datasets (Example 4-2).
Example 4-2. Remove extra datasets from Tox21
````# Remove extra tasks`
`train_y` `=` `train_y``[:,` `0``]`
`valid_y` `=` `valid_y``[:,` `0``]`
`test_y` `=` `test_y``[:,` `0``]`
`train_w` `=` `train_w``[:,` `0``]`
`valid_w` `=` `valid_w``[:,` `0``]`
`test_w` `=` `test_w``[:,` `0``]````
Accepting Minibatches of Placeholders
In the previous chapters, we created placeholders that accepted arguments of fixed size. When dealing with minibatched data, it is often convenient to be able to feed batches of variable size. Suppose that a dataset has 947 elements. Then with a minibatch size of 50, the last batch will have 47 elements. This would cause the code in Chapter 3 to crash. Luckily, TensorFlow has a simple fix to the situation: using `None` as a dimensional argument to a placeholder allows the placeholder to accept tensors with arbitrary size in that dimension (Example 4-3).
Example 4-3. Defining placeholders that accept minibatches of different sizes
````d` `=` `1024`
`with` `tf``.``name_scope``(``"placeholders"``):`
`x` `=` `tf``.``placeholder``(``tf``.``float32``,` `(``None``,` `d``))`
`y` `=` `tf``.``placeholder``(``tf``.``float32``,` `(``None``,))````
Note `d` is 1024, the dimensionality of our feature vectors.
Implementing a Hidden Layer
The code to implement a hidden layer is very similar to code we’ve seen in the last chapter for implementing logistic regression, as shown in Example 4-4.
Example 4-4. Defining a hidden layer
````with` `tf``.``name_scope``(``"hidden-layer"``):`
`W` `=` `tf``.``Variable``(``tf``.``random_normal``((``d``,` `n_hidden``)))`
`b` `=` `tf``.``Variable``(``tf``.``random_normal``((``n_hidden``,)))`
`x_hidden` `=` `tf``.``nn``.``relu``(``tf``.``matmul``(``x``,` `W``)` `+` `b``)````
We use a `tf.name_scope` to group together introduced variables. Note that we use the matricial form of the fully connected layer. We use the form xW instead of Wx in order to deal more conveniently with a minibatch of input at a time. (As an exercise, try working out the dimensions involved to see why this is so.) Finally, we apply the ReLU nonlinearity with the built-in `tf.nn.relu` activation function.
The remainder of the code for the fully connected layer is quite similar to that used for the logistic regression in the previous chapter. For completeness, we display the full code used to specify the network in Example 4-5. As a quick reminder, the full code for all models covered is available in the GitHub repo associated with this book. We strongly encourage you to try running the code for yourself.
Example 4-5. Defining the fully connected architecture
````with` `tf``.``name_scope``(``"placeholders"``):`
`x` `=` `tf``.``placeholder``(``tf``.``float32``,` `(``None``,` `d``))`
`y` `=` `tf``.``placeholder``(``tf``.``float32``,` `(``None``,))`
`with` `tf``.``name_scope``(``"hidden-layer"``):`
`W` `=` `tf``.``Variable``(``tf``.``random_normal``((``d``,` `n_hidden``)))`
`b` `=` `tf``.``Variable``(``tf``.``random_normal``((``n_hidden``,)))`
`x_hidden` `=` `tf``.``nn``.``relu``(``tf``.``matmul``(``x``,` `W``)` `+` `b``)`
`with` `tf``.``name_scope``(``"output"``):`
`W` `=` `tf``.``Variable``(``tf``.``random_normal``((``n_hidden``,` `1``)))`
`b` `=` `tf``.``Variable``(``tf``.``random_normal``((``1``,)))`
`y_logit` `=` `tf``.``matmul``(``x_hidden``,` `W``)` `+` `b`
`# the sigmoid gives the class probability of 1`
`y_one_prob` `=` `tf``.``sigmoid``(``y_logit``)`
`# Rounding P(y=1) will give the correct prediction.`
`y_pred` `=` `tf``.``round``(``y_one_prob``)`
`with` `tf``.``name_scope``(``"loss"``):`
`# Compute the cross-entropy term for each datapoint`
`y_expand` `=` `tf``.``expand_dims``(``y``,` `1``)`
`entropy` `=` `tf``.``nn``.``sigmoid_cross_entropy_with_logits``(``logits``=``y_logit``,` `labels``=``y_expand``)`
`# Sum all contributions`
`l` `=` `tf``.``reduce_sum``(``entropy``)`
`with` `tf``.``name_scope``(``"optim"``):`
`train_op` `=` `tf``.``train``.``AdamOptimizer``(``learning_rate``)``.``minimize``(``l``)`
`with` `tf``.``name_scope``(``"summaries"``):`
`tf``.``summary``.``scalar``(``"loss"``,` `l``)`
`merged` `=` `tf``.``summary``.``merge_all``()````
Adding Dropout to a Hidden Layer
TensorFlow takes care of implementing dropout for us in the built-in primitive `tf.nn.dropout(x, keep_prob)`, where `keep_prob` is the probability that any given node is kept. Recall from our earlier discussion that we want to turn on dropout when training and turn off dropout when making predictions. To handle this correctly, we will introduce a new placeholder for `keep_prob`, as shown in Example 4-6.
Example 4-6. Add a placeholder for dropout probability
``keep_prob` `=` `tf``.``placeholder``(``tf``.``float32``)``
During training, we pass in the desired value, often 0.5, but at test time we set `keep_prob` to 1.0 since we want predictions made with all learned nodes. With this setup, adding dropout to the fully connected network specified in the previous section is simply a single extra line of code (Example 4-7).
Example 4-7. Defining a hidden layer with dropout
````with` `tf``.``name_scope``(``"hidden-layer"``):`
`W` `=` `tf``.``Variable``(``tf``.``random_normal``((``d``,` `n_hidden``)))`
`b` `=` `tf``.``Variable``(``tf``.``random_normal``((``n_hidden``,)))`
`x_hidden` `=` `tf``.``nn``.``relu``(``tf``.``matmul``(``x``,` `W``)` `+` `b``)`
`# Apply dropout`
`x_hidden` `=` `tf``.``nn``.``dropout``(``x_hidden``,` `keep_prob``)````
Implementing Minibatching
To implement minibatching, we need to pull out a minibatch’s worth of data each time we call `sess.run`. Luckily for us, our features and labels are already in NumPy arrays, and we can make use of NumPy’s convenient syntax for slicing portions of arrays (Example 4-8).
Example 4-8. Training on minibatches
````step` `=` `0`
`for` `epoch` `in` `range``(``n_epochs``):`
`pos` `=` `0`
`while` `pos` `<` `N``:`
`batch_X` `=` `train_X``[``pos``:``pos``+``batch_size``]`
`batch_y` `=` `train_y``[``pos``:``pos``+``batch_size``]`
`feed_dict` `=` `{``x``:` `batch_X``,` `y``:` `batch_y``,` `keep_prob``:` `dropout_prob``}`
`_``,` `summary``,` `loss` `=` `sess``.``run``([``train_op``,` `merged``,` `l``],` `feed_dict``=``feed_dict``)`
`print``(``"epoch ``%d``, step ``%d``, loss: ``%f``"` `%` `(``epoch``,` `step``,` `loss``))`
`train_writer``.``add_summary``(``summary``,` `step``)`
`step` `+=` `1`
`pos` `+=` `batch_size````
Evaluating Model Accuracy
To evaluate model accuracy, standard practice requires measuring the accuracy of the model on data not used for training (namely the validation set). However, the fact that the data is imbalanced makes this tricky. The classification accuracy metric we used in the previous chapter simply measures the fraction of datapoints that were labeled correctly. However, 95% of data in our dataset is labeled 0 and only 5% are labeled 1. As a result the all-0 model (which labels everything negative) would achieve 95% accuracy! This isn’t what we want.
A better choice would be to increase the weights of positive examples so that they count for more. For this purpose, we use the recommended per-example weights from MoleculeNet to compute a weighted classification accuracy where positive samples are weighted 19 times the weight of negative samples. Under this weighted accuracy, the all-0 model would have 50% accuracy, which seems much more reasonable.
For computing the weighted accuracy, we use the function `accuracy_score(true, pred, sample_weight=given_sample_weight)` from `sklearn.metrics`. This function has a keyword argument `sample_weight`, which lets us specify the desired weight for each datapoint. We use this function to compute the weighted metric on both the training and validation sets (Example 4-9).
Example 4-9. Computing a weighted accuracy
````train_weighted_score` `=` `accuracy_score``(``train_y``,` `train_y_pred``,` `sample_weight``=``train_w``)`
`print``(``"Train Weighted Classification Accuracy: ``%f``"` `%` `train_weighted_score``)`
`valid_weighted_score` `=` `accuracy_score``(``valid_y``,` `valid_y_pred``,` `sample_weight``=``valid_w``)`
`print``(``"Valid Weighted Classification Accuracy: ``%f``"` `%` `valid_weighted_score``)````
While we could reimplement this function ourselves, sometimes it’s easier (and less error prone) to use standard functions from the Python data science infrastructure. Learning about this infrastructure and available functions is part of being a practicing data scientist. Now, we can train the model (for 10 epochs in the default setting) and gauge its accuracy:
```Train Weighted Classification Accuracy: 0.742045
Valid Weighted Classification Accuracy: 0.648828```
In Chapter 5, we will show you methods to systematically improve this accuracy and tune our fully connected model more carefully.
Using TensorBoard to Track Model Convergence
Now that we have specified our model, let’s use TensorBoard to inspect the model. Let’s first check the graph structure in TensorBoard (Figure 4-10).
The graph looks similar to that for logistic regression, with the addition of a new hidden layer. Let’s expand the hidden layer to see what’s inside (Figure 4-11).
You can see how the new trainable variables and the dropout operation are represented here. Everything looks to be in the right place. Let’s end now by looking at the loss curve over time (Figure 4-12).
The loss curve trends down as we saw in the previous section. But, let’s zoom in to see what this loss looks like up close (Figure 4-13).
Note that loss looks much bumpier! This is one of the prices of using minibatch training. We no longer have the beautiful, smooth loss curves that we saw in the previous sections.
Review
In this chapter, we’ve introduced you to fully connected deep networks. We delved into the mathematical theory of these networks, and explored the concept of “universal approximation,” which partially explains the learning power of fully connected networks. We ended with a case study, where you trained a deep fully connected architecture on the Tox21 dataset.
In this chapter, we haven’t yet shown you how to tune the fully connected network to achieve good predictive performance. In Chapter 5, we will discuss “hyperparameter optimization,” the process of tuning network parameters, and have you tune the parameters of the Tox21 network introduced in this chapter.
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https://www.geeksforgeeks.org/proposition-logic/?ref=leftbar-rightbar | 1,702,275,736,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103558.93/warc/CC-MAIN-20231211045204-20231211075204-00270.warc.gz | 844,202,320 | 47,208 | # Propositional Logic
### What is Logic?
Logic is the basis of all mathematical reasoning, and of all automated reasoning. The rules of logic specify the meaning of mathematical statements. These rules help us understand and reason with statements such as –
such that where
Which in Simple English means “There exists an integer that is not the sum of two squares”. Importance of Mathematical Logic The rules of logic give precise meaning to mathematical statements. These rules are used to distinguish between valid and invalid mathematical arguments. Apart from its importance in understanding mathematical reasoning, logic has numerous applications in Computer Science, varying from design of digital circuits, to the construction of computer programs and verification of correctness of programs.
### Propositional Logic
What is a proposition? A proposition is the basic building block of logic. It is defined as a declarative sentence that is either True or False, but not both. The Truth Value of a proposition is True(denoted as T) if it is a true statement, and False(denoted as F) if it is a false statement. For Example,
1. The sun rises in the East and sets in the West.
2. 1 + 1 = 2
3. 'b' is a vowel.
All of the above sentences are propositions, where the first two are Valid(True) and the third one is Invalid(False). Some sentences that do not have a truth value or may have more than one truth value are not propositions. For Example,
1. What time is it?
2. Go out and play.
3. x + 1 = 2.
The above sentences are not propositions as the first two do not have a truth value, and the third one may be true or false. To represent propositions, propositional variables are used. By Convention, these variables are represented by small alphabets such as . The area of logic which deals with propositions is called propositional calculus or propositional logic. It also includes producing new propositions using existing ones. Propositions constructed using one or more propositions are called compound propositions. The propositions are combined together using Logical Connectives or Logical Operators.
### Truth Table
Since we need to know the truth value of a proposition in all possible scenarios, we consider all the possible combinations of the propositions which are joined together by Logical Connectives to form the given compound proposition. This compilation of all possible scenarios in a tabular format is called a truth table. Most Common Logical Connectives-
1. Negation – If is a proposition, then the negation of is denoted by , which when translated to simple English means- “It is not the case that ” or simply “not “. The truth value of is the opposite of the truth value of . The truth table of is-
Example, The negation of “It is raining today”, is “It is not the case that is raining today” or simply “It is not raining today”.
2. Conjunction – For any two propositions and , their conjunction is denoted by , which means “and “. The conjunction is True when both and are True, otherwise False. The truth table of is-
Example, The conjunction of the propositions – “Today is Friday” and – “It is raining today”, is “Today is Friday and it is raining today”. This proposition is true only on rainy Fridays and is false on any other rainy day or on Fridays when it does not rain.
3. Disjunction – For any two propositions and , their disjunction is denoted by , which means “or “. The disjunction is True when either or is True, otherwise False. The truth table of is-
Example, The disjunction of the propositions – “Today is Friday” and – “It is raining today”, is “Today is Friday or it is raining today”. This proposition is true on any day that is a Friday or a rainy day(including rainy Fridays) and is false on any day other than Friday when it also does not rain.
4. Exclusive Or – For any two propositions and , their exclusive or is denoted by , which means “either or but not both”. The exclusive or is True when either or is True, and False when both are true or both are false. The truth table of is-
Example, The exclusive or of the propositions – “Today is Friday” and – “It is raining today”, is “Either today is Friday or it is raining today, but not both”. This proposition is true on any day that is a Friday or a rainy day(not including rainy Fridays) and is false on any day other than Friday when it does not rain or rainy Fridays.
5. Implication – For any two propositions and , the statement “if then ” is called an implication and it is denoted by . In the implication is called the hypothesis or antecedent or premise and is called the conclusion or consequence. The implication is is also called a conditional statement. The implication is false when is true and is false otherwise it is true. The truth table of is-
You might wonder that why is true when is false. This is because the implication guarantees that when and are true then the implication is true. But the implication does not guarantee anything when the premise is false. There is no way of knowing whether or not the implication is false since did not happen. This situation is similar to the “Innocent until proven Guilty” stance, which means that the implication is considered true until proven false. Since we cannot call the implication false when is false, our only alternative is to call it true. This follows from the Explosion Principle which says- “A False statement implies anything” Conditional statements play a very important role in mathematical reasoning, thus a variety of terminology is used to express , some of which are listed below.
"if , then "" is sufficient for "" when ""a necessary condition for is "" only if "" unless "" follows from "
Example, “If it is Friday then it is raining today” is a proposition which is of the form . The above proposition is true if it is not Friday(premise is false) or if it is Friday and it is raining, and it is false when it is Friday but it is not raining.
6. Biconditional or Double Implication – For any two propositions and , the statement “if and only if(iff) ” is called a biconditional and it is denoted by . The statement is also called a bi-implicationhas the same truth value as The implication is true when and have same truth values, and is false otherwise. The truth table of is-
Some other common ways of expressing are-
" is necessary and sufficient for ""if then , and conversely"" if "
Example, “It is raining today if and only if it is Friday today.” is a proposition which is of the form . The above proposition is true if it is not Friday and it is not raining or if it is Friday and it is raining, and it is false when it is not Friday or it is not raining. Exercise:
1) Consider the following statements:
P: Good mobile phones are not cheap.
Q: Cheap mobile phones are not good.
L: P implies Q
M: Q implies P
N: P is equivalent to Q
Which one of the following about L, M, and N is CORRECT?(Gate 2014)
(A) Only L is TRUE.
(B) Only M is TRUE.
(C) Only N is TRUE.
(D) L, M and N are TRUE.
For solution, see GATE | GATE-CS-2014-(Set-3) | Question 11
2) Which one of the following is not equivalent to p?q (Gate 2015)
For solution, see GATE | GATE-CS-2015 (Set 1) | Question 65
Discrete Mathematics and its Applications, by Kenneth H Rosen
Read next part : Introduction to Propositional Logic – Set 2
If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. | 1,772 | 7,725 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2023-50 | latest | en | 0.93188 |
https://testfoodkitchen.com/is-a-cup-of-coffee-6-oz/ | 1,702,204,682,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101779.95/warc/CC-MAIN-20231210092457-20231210122457-00368.warc.gz | 615,795,453 | 16,953 | # is a cup of coffee 6 oz?
The jury is still out on whether a standard cup of coffee is 6 oz or not. Some people swear by the 6 oz measure, while others say that it can vary depending on the size and style of the cup. In general, most coffee shops will serve coffee in cups that range from 8-12 oz, so your best bet is to ask for a smaller or larger cup if you’re looking for a precise measurement.
How much coffee is in a cup? This is a question that many people may ask, and the answer may not be so straightforward. Depending on the size of the cup and the type of coffee, there could be anywhere from four to eight ounces in a cup. For example, a standard coffee mug holds twelve ounces, while an espresso shot glass only holds one or two. So, how do you measure a cup of coffee?
There are actually quite a few ways to measure out a cup of coffee. One way is to use volume measurements, such as teaspoons, tablespoons, or milliliters. Another way is to use weight measurements, such as ounces or grams. And yet another way is to use scales that measure both volume and weight. Ultimately, it depends on the person’s preference on which measurement they find more accurate.
## Is a cup of coffee 6 or 8 oz?
When you order a cup of coffee, how many ounces are you actually getting? Most people would say 8, but is that really the case? According to most coffee shops, a standard cup of coffee is 6 ounces. However, many people argue that a true cup of coffee is 8 ounces. So, which is it?
To answer this question, we need to look at the definition of an ounce. An ounce is equal to 1/16th of a pound. This means that there are 16 ounces in a pound. When we divide 8 by 16, we get 0.50, which means that a true cup of coffee is actually 8 ounces.
So next time you order a cup of coffee, make sure to specify that you want an 8-ounce cup!
## Why is a cup of coffee only 6 ounces?
A cup of coffee is typically only 6 ounces. The small size has become popular in recent years, but why is a cup of coffee only 6 ounces?
Coffee companies have been shrinking the size of their cups over time. In the early 1990s, the average cup of coffee was 10 ounces. But by 2008, that number had shrunk to 8 ounces. And today, the average cup of coffee is just 6 ounces.
There are a few reasons for this trend. First, Americans are drinking less coffee than they used to. In the 1990s, the average American drank 3 cups of coffee per day. Today, that number has dropped to 2 cups per day.
But shrinking cup sizes isn’t just about consumer behavior; it’s also about marketing strategy.
## Is a cup of coffee 4 or 8 oz?
In the United States, a cup of coffee is typically defined as 8 fluid ounces. However, in other parts of the world, a cup of coffee can be as small as 4 fluid ounces or as large as 12 fluid ounces. So, what is the real definition of a cup of coffee?
The size of a cup of coffee can vary depending on the country or region that you are in. In the United States, a cup is defined as 8 fluid ounces. However, in other countries such as Australia and the United Kingdom, a cup is defined as 4 fluid ounces. Similarly, in other countries such as Italy and Spain, a cup is defined as 12 fluid ounces.
So, what is the correct definition of a cup of coffee? The answer depends on where you are located.
## Why is a cup of coffee 6 oz and not 8?
In America, a cup of coffee is typically served in a 6-ounce cup. But why is that? Is there a scientific reason behind it or is it simply tradition?
The answer to that question is actually a bit of both. For many years, coffee was served in 8-ounce cups because that’s the size of the traditional coffee mug. However, over time, people began to prefer smaller cups of coffee and 6-ounce cups eventually became the norm.
There are several reasons why people might prefer smaller cups of coffee. For one, it’s easier to control your caffeine intake when you’re drinking smaller cups. Additionally, small cups tend to cool down faster than larger ones, which can be beneficial if you’re drinking coffee on a hot day. And finally, small cups just seem more manageable and less overwhelming than large ones.
## What’s 6 oz in cups?
When it comes to measuring out liquids, there are several different units of measurement that can be used. One of the most common is cups. Cups can be used to measure both volume and weight. When measuring volume, a cup is equal to 8 fluid ounces.
When measuring weight, a cup is equal to 6 ounces. This can be important to know if you are trying to bake or cook a recipe that calls for a specific amount of liquid. If you are not familiar with the conversion, it can be helpful to use a kitchen scale to measure out the desired amount of liquid instead of using cups.
## Is a cup of coffee 5 or 6 oz?
There is some discrepancy over how much a standard cup of coffee actually holds. Most sources say that a cup of coffee is around 5-6 ounces, but there are others who say it can be up to 8 or even 12 ounces.
The reason for the discrepancy may be because “cup” can refer to different sizes – a regular coffee cup, a mug, or even a pitcher. So the next time you’re at your favorite cafe and order a cup of coffee, be sure to ask for clarification on what they mean by “cup.
## Is a coffee cup 1 cup?
When you go to a coffee shop and order a coffee, you are typically given a cup with a lid. You may be wondering how many cups of coffee are in that cup. The answer is that it depends on the size of the cup. Most coffee cups hold between 8 and 12 ounces of liquid. This means that if you order an 8-ounce cup of coffee, you will be drinking 1 cup of coffee. If you order a 12-ounce cup of coffee, you will be drinking 1.5 cups of coffee.
## What is a standard cup of coffee?
A standard cup of coffee is generally defined as containing 6 fluid ounces or 177 milliliters. Coffee beans are roasted and ground to create a coffee powder. Hot water is then poured over the coffee powder and the coffee is brewed. The strength of the coffee can be varied by adding more or less water.
## Is a cup always 8 ounces?
The answer is no. A cup can hold anywhere from 4 to 12 ounces, depending on the size of the cup. When measuring, be sure to fill the cup up to the top in order to get an accurate measurement.
## How many Oz is a standard coffee mug?
A coffee mug is a cylindrical drinking vessel with a handle and a spout. They come in all sorts of sizes, from small espresso cups to large travel mugs. But how many ounces is a standard coffee mug?
Most coffee mugs hold between 8 and 12 ounces of liquid. This means that they can accommodate anywhere from 2 to 3 shots of espresso or 6 to 8 ounces of brewed coffee. If you like your coffee extra strong, go for the smaller mug size; if you prefer it weaker or want to add milk or cream, go for the larger size.
## How much coffee is a cup?
Some people believe that there is a standard size for a cup of coffee. However, the amount of coffee in a cup can vary depending on the type of coffee and the brewing method.
A traditional U.S. cup of coffee is 8 fluid ounces. However, a standard European coffee cup is 10 fluid ounces. And, an Australian or British cup of coffee is often 11 or 12 fluid ounces.
Brewed coffee generally contains between 100 and 200 milligrams of caffeine per 8-ounce serving, while espresso can contain up to 350 milligrams per 8-ounce serving.
## How many Oz is a small coffee?
Many people are curious about how much coffee they are drinking when they order a “small” coffee. Are you getting 12 ounces, or is the cup actually closer to 8 ounces?
It turns out that the definition of a small coffee can vary from place to place, but in general, you can expect to get around 8-10 ounces of coffee when you order a small. Of course, if you want something smaller or larger, most places will be happy to accommodate your request.
## What size is a small coffee cup?
Small coffee cups come in a variety of sizes, but typically hold between 8 and 12 ounces of liquid. They are often used for espresso drinks or to-go cups. Small coffee cups can be found in restaurants, cafes, and as part of to-go sets.
## How many ounces is 8 cups of coffee?
Brewing coffee is a science, and there are a lot of different ways to make the perfect pot. But even with all the possible variations, there are some constants that remain the same. For example, how much coffee should you use for a pot? And how many cups is that equivalent to?
The standard for brewing coffee is one tablespoon of ground coffee per cup of water. So if you’re using an 8-cup pot, you’ll need eight tablespoons of ground coffee. Keep in mind that this is just a general guideline; you might like your coffee stronger or weaker than this.
But what does 8 cups of coffee look like? It’s about 128 ounces, or just over 10 cups. If you’re not sure how much that is in pounds or kilograms, it’s just under 6 pounds or 2.7 kilograms.
## How many cups of coffee a day is healthy?
Coffee is one of the most popular drinks in the world, enjoyed by millions of people every day. But how many cups of coffee a day is healthy?
Studies have shown that drinking up to five cups of coffee a day is not harmful and may even have some health benefits. Caffeine has been shown to improve mental performance and alertness, and can also help to increase metabolism and burn fat.
However, drinking too much coffee can cause problems such as restlessness, anxiety and insomnia. It’s also important to remember that caffeine is a stimulant and can cause problems for people with heart conditions or other health issues.
## How much coffee do you use for 6 cups?
Brewing coffee is an art form that takes practice to perfect. The amount of coffee you use for 6 cups will vary depending on the roast, grind size, and brewing method you use. For a stronger cup of coffee, you can use more ground coffee. However, using too much coffee can make your coffee taste bitter.
## Is 6 oz the same as 1 cup?
This is a question that often comes up when cooking. The answer is, unfortunately, it depends. In general, 6 ounces is equivalent to 1/2 cup, but there are some recipes that call for a specific measurement like 1 cup and others that call for a different measurement, like 1/3 cup. If you’re not sure how much an ingredient should weigh or measure, it’s always best to consult a recipe.
## How much is 1oz in a cup?
One ounce is equivalent to 1/16th of a cup. This means that if you want to measure out one ounce of a substance, you would need to use a measuring device that measures in 16th increments. When measuring out an ingredient for a recipe, it is important to be precise so that the end result is as expected. If you are not comfortable with fractions, there are many online calculators that can help you convert between ounces and cups.
Categories FAQ | 2,523 | 10,875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2023-50 | latest | en | 0.971048 |
http://www.stata.com/statalist/archive/2005-03/msg00796.html | 1,495,648,443,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607849.21/warc/CC-MAIN-20170524173007-20170524193007-00429.warc.gz | 668,325,467 | 2,868 | # st: nested logit model: how to interpret insignificant nesting parameter?
From "Kelchtermans, Stijn" To Subject st: nested logit model: how to interpret insignificant nesting parameter? Date Thu, 24 Mar 2005 10:21:56 +0100
```All,
I have estimated a nested logit model explaining schooling choice, with all schooling options in one branch, and the outside option in the (degenerate) second branch. I have estimated the model sequentially because of the computational burden.
Question: the nesting parameter that measure the dissimilarity among the schooling options comes out slighly negative (-.0459305) and insignificant. Does the insignificance imply that the nesting is inappropriate?
The output (the estimates for the lower model were fixed to their first-step estimates):
Nested logit estimates
Levels = 2 Number of obs = 546468
Dependent variable = chosen LR chi2(14) = 11816.01
Log likelihood = -7198.8489 Prob > chi2 = 0.0000
----------------------------------------------------------------------------
| Coef. Std. Err. z P>|z| [95% Conf. Int.]
-------------+--------------------------------------------------------------
option |
LSIZE | .9400799 . . . . .
VOCL | -.6005096 . . . . .
VOCLmale | .2947062 . . . . .
VOCLfor | .4408798 . . . . .
[cut]
ctrav | -5.145888 . . . . .
-------------+--------------------------------------------------------------
STUDY |
NONE | 1.169447 .2569964 4.55 0.000 .6657433 1.673151
NONEmale | .1646824 .1107066 1.49 0.137 -.0522985 .3816633
[cut]
NONEcath | -.4957498 .1248174 -3.97 0.000 -.7403875 -.2511122
-------------+----------------------------------------------------------------
(incl. value |
parameters) |
STUDY |
/YES | -.0459305 .073946 -0.62 0.535 -.1908621 .099001
/NO | 1 . . . . .
------------------------------------------------------------------------------
LR test of homoskedasticity (iv = 1): chi2(-164)= -26.94 Prob > chi2 = .
----------------------------------------------------------------------------
Many thanks,
Stijn Kelchtermans
Katholieke Universiteit Leuven
Belgium
Stijn.kelchtermans@econ.kuleuven.ac.be
*
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
``` | 678 | 2,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | latest | en | 0.586065 |
http://necked.66.after.medienmaedchen.com/associative-property-worksheets/ | 1,603,391,543,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880014.26/warc/CC-MAIN-20201022170349-20201022200349-00674.warc.gz | 70,903,231 | 18,084 | # Associative Property Worksheets
## Associative Property Worksheets For Practice
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#### Commutative Associative And Distributive Properties
These Worksheets Review The Commutative Associative And Distributive Properties And Identify The Correct Property For Given Expressions While These Are Defined Students Should Have Some Prior Knowledge Commutative Associative And Distributive Properties Worksheets Click The Buttons To Print Each Worksheet And Answer Key
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Commutative Property And Associative Property Displaying Top 8 Worksheets Found For This Concept Some Of The Worksheets For This Concept Are Commutative Distributive And Associative Properties Multiplication Properties Multiplication Associative Property Of Addition 1 Notes Commutative Associative Properties Commutative Property Of Addition 1 Name In Numbers 1 9 Select The Property
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Associative Property Of Multiplication
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Associative Property Worksheets. The worksheet is an assortment of 4 intriguing pursuits that will enhance your kid's knowledge and abilities. The worksheets are offered in developmentally appropriate versions for kids of different ages. Adding and subtracting integers worksheets in many ranges including a number of choices for parentheses use.
You can begin with the uppercase cursives and after that move forward with the lowercase cursives. Handwriting for kids will also be rather simple to develop in such a fashion. If you're an adult and wish to increase your handwriting, it can be accomplished. As a result, in the event that you really wish to enhance handwriting of your kid, hurry to explore the advantages of an intelligent learning tool now!
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There isn't anything like a superb story, and nothing like being the person who started a renowned urban legend. Deciding upon the ideal approach route Cursive writing is basically joined-up handwriting. Practice reading by yourself as often as possible.
Research urban legends to obtain a concept of what's out there prior to making a new one. You are still not sure the radicals have the proper idea. Naturally, you won't use the majority of your ideas. If you've got an idea for a tool please inform us. That means you can begin right where you are no matter how little you might feel you've got to give. You are also quite suspicious of any revolutionary shift. In earlier times you've stated that the move of independence may be too early.
Each lesson in handwriting should start on a fresh new page, so the little one becomes enough room to practice. Every handwriting lesson should begin with the alphabets. Handwriting learning is just one of the most important learning needs of a kid. Learning how to read isn't just challenging, but fun too.
The use of grids The use of grids is vital in earning your child learn to Improve handwriting. Also, bear in mind that maybe your very first try at brainstorming may not bring anything relevant, but don't stop trying. Once you are able to work, you might be surprised how much you get done. Take into consideration how you feel about yourself. Getting able to modify the tracking helps fit more letters in a little space or spread out letters if they're too tight. Perhaps you must enlist the aid of another man to encourage or help you keep focused.
Associative Property Worksheets. Try to remember, you always have to care for your child with amazing care, compassion and affection to be able to help him learn. You may also ask your kid's teacher for extra worksheets. Your son or daughter is not going to just learn a different sort of font but in addition learn how to write elegantly because cursive writing is quite beautiful to check out. As a result, if a kid is already suffering from ADHD his handwriting will definitely be affected. Accordingly, to be able to accomplish this, if children are taught to form different shapes in a suitable fashion, it is going to enable them to compose the letters in a really smooth and easy method. Although it can be cute every time a youngster says he runned on the playground, students want to understand how to use past tense so as to speak and write correctly. Let say, you would like to boost your son's or daughter's handwriting, it is but obvious that you want to give your son or daughter plenty of practice, as they say, practice makes perfect.
Without phonics skills, it's almost impossible, especially for kids, to learn how to read new words. Techniques to Handle Attention Issues It is extremely essential that should you discover your kid is inattentive to his learning especially when it has to do with reading and writing issues you must begin working on various ways and to improve it. Use a student's name in every sentence so there's a single sentence for each kid. Because he or she learns at his own rate, there is some variability in the age when a child is ready to learn to read. Teaching your kid to form the alphabets is quite a complicated practice. | 1,435 | 7,492 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.593721 |
https://www.trophybookarchery.com/targets/how-archery-competition-is-scored-solved.html | 1,653,685,957,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663006341.98/warc/CC-MAIN-20220527205437-20220527235437-00551.warc.gz | 1,194,938,430 | 10,453 | # How Archery Competition Is Scored? (Solved)
Each ring of the Archery target that the archers shot at to get points is valued, and this is how archery is scored. The center ring is worth ten points, while the rest of the rings are worth nine points each, starting from the inside and working outward. The arrow does not get any points if it does not hit the target.
In archery, what is the greatest possible score?
• The largest number of points you may get is generally ten, and this is awarded for striking the inside gold circle on a given target. The lowest possible score is one, which is obtained by striking the outside white circle at the farthest end of the target. Right, with that in mind, let’s take a closer look at the scoring system for archery competitions.
## How is an archery score calculated?
In archery, scoring is fairly straightforward: you just total up the number of points earned based on where your arrows strike the target. The maximum possible score for a single arrow is 10 points for striking the inner gold ring, while the lowest possible score is one point for hitting the outside white ring.
## How do you win in a competition of archery?
The match is won by the first player to reach six set points. If the score is still tied after five sets, each player receives one arrow to shoot. Unless they both shoot a 10, in which case they both get a second arrow, the one who is closest to the bullseye wins.
## How does Olympic archery scoring work?
Archers aim at the five-color target, which consists of ten scoring zones in gold, red, blue, black, and white rings and is divided into five color groups. The innermost yellow rings receive 10 and 9 points, while the outermost white rings receive two and one point. The red rings receive eight and seven points, the blue rings receive six and five points, and the black rings receive four and three points.
## What is a perfect score in archery?
In this round, the maximum possible score is 10×30=300 points, which is derived from the word “round.” The 300 Round is a Target Round (as opposed to a Field Archery Round), which means that one fires at a single target for the duration of the round and from level ground.
## What are the rules in archery?
The range’s rules are as follows:
• Understand and abide by the range rules. Inspect your equipment before you begin shooting. On the range, stroll rather than run at all times. Shoot only with a bow and an arrow that has been properly nocked. Keep your arrows in your quiver until you’re ready to shoot them. Keep the arrows pointing down or in the direction of the objective. Shoot only the targets that are within your path of fire.
## What variables might affect the score of an archer?
When archers miss their target, they lose points based on characteristics that change either linearly or quadratically with distance from the target. With the use of a normal distribution, an archer’s arrow group on the target may be represented mathematically, with the standard deviation dependent on the archer’s ability level and distance from the target.
## How long does an archery competition last?
A normal target archery tournament may run between three and four hours, however some may go on for longer periods of time. A regular end can last anywhere from 2-4 minutes, depending on the quantity of arrows launched, and a typical tournament will include somewhere between 20 and 30 ends.
## How many arrows would an archer carry?
Each archer carried a sheaf of arrows, which included 24 arrows. When they were dispatched, further reinforcements were brought in from supply wagons. Traditionally, archers carried their arrows in a quiver or threaded them through a belt loop. Crossbows were used by certain soldiers to shoot small arrows known as bolts.
## Where is archery most popular?
Yes, you read it correctly: the United States of America has been the world’s leading archery nation since 2008, according to the World Archery Federation, which rates countries based on how athletes perform in international events.
## How much does an Olympic archery bow cost?
After a quick visit to the Lancaster website, a top-end Hoyt bow will cost approximately \$1200, 500 for a dozen X10 arrows and points, 300 for a set of stabilizers, 300 for a sight, and 150 for a rest, plus another 300 or so for a case and assorted tabs, armguards, quiver, tools and other miscellaneous items, and so on.
## What is follow through in archery?
THE TRANSACTION. Actually, the follow-up is only a lengthier expansion of the initial release. It’s just the portion of the release that continues after the arrow has exited the bow and continues until the arrow strikes the target (if applicable).
See also: What Is The Name Of Someone Who Loves Archery? (Solution found)
## What does 5 whistles mean in archery?
SING: FIVE OR MORE BLASTS OF VOICE (series of blasts) Stopping the fire using verbal commands such as “STOP, STOP, STOP, STOP” or “CEASE FIRE” Meaning: Put the arrows back in the quiver as soon as possible, and walk slightly behind the waiting line to avoid being caught. On the range, there is a medical emergency. Keep in mind to adhere to the unspoken norms of archery, which we refer to as Range Etiquette.
## Who was the greatest archer of all time?
Birmingham, Alabama, United States Lemuel Howard Hill, later known as Howard H. Hill, was an expert bowman who, for more than two decades, from the early 1930s to the 1950s, was frequently introduced or billed as “The World’s Greatest Archer.” Hill was born on November 13, 1899, and died on February 4, 1975, in Los Angeles, California.
## What skills are used in archery?
List of Archery Techniques
• Draw the arrow and load it. Load the arrow. Lock the arrow. Precision
• Balance/Coordination
• Composition
• Depth Perception
• DRAWN AND LOOMED THE ARCHER | 1,313 | 5,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2022-21 | latest | en | 0.954046 |
https://math.stackexchange.com/questions/2464511/single-summation-form | 1,582,425,187,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145742.20/warc/CC-MAIN-20200223001555-20200223031555-00480.warc.gz | 460,907,814 | 32,937 | # Single summation form
I want to write as a sigle summation of \begin{align*} \sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k}-\sum_{k=0}^{n+1}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k}. \end{align*} How to change upper limit of any one of the summations?
You can use Binomial theorem, then given expression will be:
$$(\pi+\cot \alpha)^n-(\pi+\cot \alpha)^{n+1}=(1-\pi-\cot \alpha)(\pi+\cot \alpha)^n$$
Now, use the Binomial theorem to get a single summation.
• +1 I totally didn't recognize the binomial theorem for some reason. – orlp Oct 9 '17 at 15:00
$$\sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k}-\sum_{k=0}^{n+1}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k} =$$
$$\sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k}-\sum_{k=0}^{n}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k} - \binom{n+1}{n+1}\pi^0(\cot \alpha)^{n+1} =$$
$$\sum_{k=0}^{n} \left[\binom{n}{k} \pi^{n-k} - \binom{n+1}{k} \pi^{n+1-k}\right ](\cot\alpha)^{k} - (\cot \alpha)^{n+1} =$$
$$- (\cot \alpha)^{n+1} + \sum_{k=0}^{n} \left[\binom{n}{k} - \pi\binom{n+1}{k}\right ]\pi^{n-k}(\cot\alpha)^{k}$$
Or, arguably better as Arun Badajena notices:
$$\sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k}-\sum_{k=0}^{n+1}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k} =$$
$$(\pi + \cot\alpha)^n - (\pi + \cot\alpha)^{n+1} =$$ $$(1 - \pi - \cot\alpha)(\pi + \cot\alpha)^n =$$ $$(1 - \pi - \cot\alpha)\sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k}$$
Second term, evaluated up to $n$ in the sum and then placing out the $k=n+1$ term.
\begin{align*} \sum_{k=0}^{n+1}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k} &= \sum_{k=0}^{n}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k} +\cot^{n+1} \alpha \end{align*}
You can just split off the last term of the second sum \begin{align*} \sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k}-\sum_{k=0}^{n+1}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k} \\ = \sum_{k=0}^{n}\binom{n}{k} \pi^{n-k}(\cot\alpha)^{k}-\sum_{k=0}^{n}\binom{n+1}{k} \pi^{n+1-k}(\cot\alpha)^{k}-(\cot \alpha)^{n+1}\\ =\sum_{k=0}^{n}\left[\binom{n}{k}+\pi\binom{n+1}k\right] \pi^{n-k}(\cot\alpha)^{k}-(\cot \alpha)^{n+1} \end{align*} | 1,019 | 2,123 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-10 | latest | en | 0.566314 |
http://www.jiskha.com/display.cgi?id=1378612028 | 1,498,222,042,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320057.96/warc/CC-MAIN-20170623114917-20170623134917-00026.warc.gz | 554,359,094 | 5,243 | # Quantum physics
posted by on .
If you Fourier sample the state 122√(|000>−|001>+|010>−|011>+|100>−|101>+|110>−|111>), which of the following states can be the outcome of your measurement? Select all that apply.
1. |000>
2. |001>
3. |010>
4. |011>
5. |100>
6. |101>
7. |110>
8. |111>
• Quantum physics - ,
what is the answer??
• Quantum physics - ,
did you got any other answer?
• Quantum physics - ,
maybe it was a problem to read it:
1/(2√2) * (|000>−|001>+|010>−|011>+|100>−|101>+|110>−|111>)
• Quantum physics - ,
how many answers ..its not all 8...
• Quantum physics - ,
how are you sure its not all 8???
• Quantum physics - ,
Help or any hint needed please.
• Quantum physics - ,
1 3 5 8 is the answer.....
• Quantum physics - ,
how many of you got 1 3 5 8 crt?
• Quantum physics - ,
@scholar its wrong
• Quantum physics - ,
then it must be 1 3 5 7.....
otherwise am screwed
• Quantum physics - ,
Don't know...I have no options left sorry cant help... can you help with Q1 please
• Quantum physics - ,
you tried 1 3 5 8 for 12? do you 4 or 5?
• Quantum physics - ,
i got 1 correct.....
• Quantum physics - ,
yes: 1, 3,5, 8 was marked wrong.
Q4:a)0 b) 1
• Quantum physics - ,
@scholar can you give some hint for Q1
• Quantum physics - ,
is it (sqrt5*sqrt2)/6 for q1
• Quantum physics - ,
q1 10/36
• Quantum physics - ,
any other 5 or 9 you got
• Quantum physics - ,
Thanks for your help scholar, I managed to pass.
Q5: 1/sqrt3 and sqrt2i/sqrt3 ...put brackets after sqrt.
Q9: CNOT12,CNOT23
good luck
• Quantum physics - ,
How do you get number 6????
• Quantum physics - ,
did anyone get problem 11 correct???
• Quantum physics - ,
11 is d
• Quantum physics - ,
thanks scholar, can you help with number 6 also??
• Quantum physics - ,
Hey can any1 help me with question 10, on my last try, i cant seem to figure it out
• Quantum physics - ,
problem 8 anyone? im on my last try!!
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Post a New Question | 621 | 2,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-26 | latest | en | 0.805204 |
http://robojax.com/course1/?vid=lecture10 | 1,721,470,385,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515079.90/warc/CC-MAIN-20240720083242-20240720113242-00205.warc.gz | 26,478,847 | 2,788 | # Lesson 10: Using Potentiometer reading voltage, Analog and Digital
We first learn the difference between Analog and digital. Then we learn what is Potentiometer and how to measure DC voltage with Arduino and the how to find the value of resistor.
``````
/*
* S01-10 Potentiometer
Please watch video instruction here https://youtu.be/wuNrzQ8rpYw
This code is available at http://robojax.com/course1/?vid=lecture10
with over 100 lectures Free On YouTube Watch it here http://robojax.com/L/?id=338
Get the code for the course: http://robojax.com/L/?id=339
* Feb 23, 2019
*
*/
const int potPin = A0;// pint name for reading voltage
void setup() {
Serial.begin(9600);// initialize the serial monitor
}
void loop() {
//Robojax Step By Step Arduino Course http://robojax.com/L/?id=338
Serial.print("potValue:");
Serial.print(potValue);
Serial.print(" Voltage:");
float voltage = potValue * (5.0 / 1023.0);
Serial.print(voltage);
Serial.println("V");
if(voltage >= 2.8 && voltage <= 4.1)
{
Serial.println("Voltage between 2.8 and 4.1");
}
delay(200);
}
``` ```
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https://www.numbersaplenty.com/180148 | 1,708,947,691,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474659.73/warc/CC-MAIN-20240226094435-20240226124435-00548.warc.gz | 916,341,513 | 3,111 | Search a number
180148 = 22291553
BaseRepresentation
bin101011111110110100
3100011010011
4223332310
521231043
63510004
71350133
oct537664
9304104
10180148
11113391
1288304
1363cc7
144991a
153859d
hex2bfb4
180148 has 12 divisors (see below), whose sum is σ = 326340. Its totient is φ = 86912.
The previous prime is 180137. The next prime is 180161. The reversal of 180148 is 841081.
It can be written as a sum of positive squares in 2 ways, for example, as 51984 + 128164 = 228^2 + 358^2 .
It is a congruent number.
It is an unprimeable number.
180148 is an untouchable number, because it is not equal to the sum of proper divisors of any number.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 661 + ... + 892.
It is an arithmetic number, because the mean of its divisors is an integer number (27195).
2180148 is an apocalyptic number.
It is an amenable number.
180148 is a deficient number, since it is larger than the sum of its proper divisors (146192).
180148 is a wasteful number, since it uses less digits than its factorization.
180148 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1586 (or 1584 counting only the distinct ones).
The product of its (nonzero) digits is 256, while the sum is 22.
The square root of 180148 is about 424.4384525464. The cubic root of 180148 is about 56.4776323082.
It can be divided in two parts, 1801 and 48, that added together give a square (1849 = 432).
The spelling of 180148 in words is "one hundred eighty thousand, one hundred forty-eight". | 472 | 1,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-10 | latest | en | 0.882526 |
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