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https://zenodo.org/record/1247421/export/schemaorg_jsonld | Conference paper Open Access
# ZAero - Zero defect manufacturing of composite parts in the aerospace industry
Rodriguez, Ana Rosa; Cuenca, Jose; Ruiz, Rocio; Calero, Alvaro
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"description": "<p>During the manufacturing process of a carbon fiber (CFRP) part in the aircraft industry defects occur. These defects are identified through inspections carried out in intermediate steps and even at the end of the manufacturing process. Due to these defects reworks need to be performed and even futilities.<br>\nThe aims of this project is the inline control of the defects that will be produced along the manufacturing of a stiffned surface panel. The ZAero control system consists on sensors integrated in the lay-up machine and sensors used during the infusion and resin curing processes. These sensors will detect defects that are outside the acceptance range and a response will occur. This response may be a rework as in the case of defects produced in the lamination stage or the variation of parameters produced in the infusion and resin curing stage due to the process monitoring.<br>\nThree demonstrators will be manufactured. The degree of complexity will be increased in each demonstrator. The control system through sensors will make a quality control and when a defect that is outside the range of acceptance a response will occur.</p>",
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155
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https://www.investopedia.com/terms/d/dpo.asp | ## What Is Days Payable Outstanding – DPO?
Days payable outstanding (DPO) is a financial ratio that indicates the average time (in days) that a company takes to pay its bills and invoices to its trade creditors, which may include suppliers, vendors, or financiers. The ratio is typically calculated on a quarterly or annual basis, and indicates how well the company’s cash outflows are being managed.
A company with a higher value of DPO takes longer to pay its bills, which means that it can retain available funds for a longer duration, allowing the company an opportunity to utilize those funds in a better way to maximize the benefits. A high DPO, however, may also be a red flag indicating an inability to pay its bills on time.
### Key Takeaways
• Days payable outstanding (DPO) computes the average number of days a company needs to pay its bills and obligations.
• Companies that have a high DPO can delay making payments and use the available cash for short-term investments and to increase their working capital and free cash flow.
• However, higher values of DPO, though desirable, may not always be a positive for the business as it may signal a cash shortfall and inability to pay.
1:53
## The Formula for Days Payable Outstanding Is
\begin{aligned} &\text{DPO} = \frac{\text{Accounts Payable}\times\text{Number of Days}}{\text{COGS}}\\ &\textbf{where:}\\ &\text{COGS}=\text{Cost of Goods Sold} \\ &\qquad\ \ \, \,= \text{Beginning Inventory} + \text{P} -\text{Ending Inventory}\\ &\text{P}=\text{Purchases} \end{aligned}
## How to Calculate DPO
To manufacture a saleable product, a company needs raw material, utilities, and other resources. In terms of accounting practices, the accounts payable represents how much money the company owes to its supplier(s) for purchases made on credit.
Additionally, there is a cost associated with manufacturing the saleable product, and it includes payment for utilities like electricity and for employee wages. This is represented by cost of goods sold (COGS), which is defined as the cost of acquiring or manufacturing the products that a company sells during a period. Both of these figures represent cash outflows and are used in calculating DPO over a period of time.
The number of days in the corresponding period is usually taken as 365 for a year and 90 for a quarter. The formula takes account of the average per day cost being borne by the company for manufacturing a saleable product. The numerator figure represents payments outstanding. The net factor gives the average number of days taken by the company to pay off its obligations after receiving the bills.
Two different versions of the DPO formula are used depending upon the accounting practices. In one of the versions, the accounts payable amount is taken as the figure reported at the end of the accounting period, like “at the end of fiscal year/quarter ending Sept. 30.” This version represents DPO value “as of” the mentioned date.
In another version, the average value of Beginning AP and Ending AP is taken, and the resulting figure represents DPO value “during” that particular period. COGS remains the same in both the versions.
## What Does Days Payable Outstanding Tell You?
Generally, a company acquires inventory, utilities, and other necessary services on credit. It results in accounts payable (AP), a key accounting entry that represents a company's obligation to pay off the short-term liabilities to its creditors or suppliers. Beyond the actual dollar amount to be paid, the timing of the paymentsfrom the date of receiving the bill till the cash actually going out of the company’s accountalso becomes an important aspect of business. DPO attempts to measure this average time cycle for outward payments and is calculated by taking the standard accounting figures into consideration over a specified period of time.
Companies having high DPO can use the available cash for short-term investments and to increase their working capital and free cash flow. However, higher values of DPO may not always be a positive for the business. If the company takes too long to pay its creditors, it risks jeopardizing its relations with the suppliers and creditors who may refuse to offer the trade credit in the future or may offer it on terms that may be less favorable to the company. The company may also be losing out on any discounts on timely payments, if available, and it may be paying more than necessary.
Additionally, a company may need to balance its outflow tenure with that of the inflow. Imagine if a company allows 90-day period to its customers to pay for the goods they purchase but has only 30-day window to pay its suppliers and vendors. This mismatch will result in the company being prone to cash crunch frequently.Companies must strike a delicate balance with DPO.
## Special Considerations
Typical DPO value vary widely across different industry sectors, and it is not worthwhile comparing these values across different sector companies. A firm's management will instead compare its DPO to the average within its industry to see if it is paying its vendors relatively too quickly or too slowly. Depending upon the various global and local factors, like the overall performance of economy, region, and sector, plus any applicable seasonal impacts, the DPO value of a particular company can vary significantly from year to year, company to company, and industry to industry.
DPO value also forms an integral part of the formula used for calculating the cash conversion cycle (CCC), another key metric that expresses the length of time that a company takes to convert the resource inputs into realized cash flows from sales. While DPO focuses on the current outstanding payable by the business, the superset CCC follows the entire cash time-cycle as the cash is first converted into inventory, expenses and accounts payable, through to sales and accounts receivable, and then back into cash in hand when received.
## Example of How Days Payable Outstanding Is Used
As a historical example, the leading retail corporation Walmart (WMT) had accounts payable worth $46.09 billion and cost of goods sold worth$373.4 billion for the fiscal year 2018. These figures are available in the annual financial statement and balance sheet of the company. Taking the number of days as 365 for annual calculation, the DPO for Walmart comes to [ 46.09 / (373.4/365) ] = 45.05 days.
Similar calculations for technology leader Microsoft (MSFT) which had $8.62 billion as AP and$38.4 billion as COGS leads to DPO value of 80.73 days.
It indicates that during the fiscal year ending 2018, Walmart paid its invoices in around 45 days after receiving the bills, while Microsoft took around 80 days, on an average, to pay its bills.
A look at similar figures for the online retail giant Amazon (AMZN), which had AP of $34.62 billion and COGS of$111.93 billion for the fiscal year 2017, reveals a very high value of 112.90 days. Such high value of DPO is attributed to the working model of Amazon, which roughly has 50 percent of its sales being supplied by third-party sellers. Amazon instantly receives funds in its account for sale of goods which are actually supplied by third-party sellers using Amazon’s online platform.
However, it doesn’t don’t pay the sellers immediately after the sale, but may send accumulated payments based on a weekly/monthly or threshold-based payment cycle. This working mechanism allows Amazon to hold onto the cash for a longer period of time, and the leading online retailer ends up with a significantly higher DPO.
## Limitations of DPO
While DPO is useful in comparing relative strength between companies, there is no clear-cut figure for what constitutes a healthy days payable outstanding, as the DPO varies significantly by industry, competitive positioning of the company, and its bargaining power. Large companies with a strong power of negotiation are able to contract for better terms with suppliers and creditors, effectively producing lower DPO figures than it would have otherwise. | 2020-08-10 09:17:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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, "math_score": 0.1712806671857834, "perplexity": 3078.4640710757576}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738653.47/warc/CC-MAIN-20200810072511-20200810102511-00576.warc.gz"} |
https://socratic.org/questions/58b0b4597c014965f514f85c#384168 | # Under standard conditions, what volume does a 25*g mass of ammonia gas occupy?
Feb 28, 2017
Approx. $55 \cdot L$
#### Explanation:
$\text{Moles of ammonia } = \frac{25.0 \cdot g}{17.01 \cdot g \cdot m o {l}^{-} 1} = 1.47 \cdot m o l$.
Given the stoichiometry of the equation, I need $\frac{3}{2}$ equiv dihydrogen gas, i.e. $2.20 \cdot m o l$.
I don't know what your syllabus says in regard to the molar volume of an ideal gas at STP; the standard molar volume is $24.8 \cdot L \cdot m o {l}^{-} 1$, so we get:
2.20*molxx24.8*L*mol^-1=??L | 2022-05-25 01:10:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "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, "math_score": 0.9649057984352112, "perplexity": 2246.8322843881374}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662577757.82/warc/CC-MAIN-20220524233716-20220525023716-00106.warc.gz"} |
https://chemistry.stackexchange.com/questions/125785/3-methyl-4-oxobutanenitrile-or-3-formylbutanenitrile/126152 | # 3-methyl-4-oxobutanenitrile or 3-formylbutanenitrile?
Which name is the Preferred IUPAC Name of the above molecule? 3-methyl-4-oxobutanenitrile or 3-formylbutanenitrile?
I feel the second one is the PIN, since it's shorter than the first one. But, I haven't been able to navigate the IUPAC Blue Book, regarding such cases.
The most important simplified criteria for the choice of a principal chain are:
1. greater number of substituents corresponding to the suffix (principal characteristic group)
2. longest chain
3. greater number of multiple bonds
4. lower locants for suffixes
5. lower locants for multiple bonds
6. greater number of prefixes
7. lower locants for prefixes
8. lower locants for substituents cited first as a prefix in the name
The corresponding wording of the rules taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is as follows.
P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
(j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.
(…)
According to Rule P-44.1.1, the priority list of substituent groups has to be considered first in order to decide the parent chain for this compound. You have already done that in both of your suggested possibilities: The principal characteristic group that is expressed as a suffix is the nitrile group ($$\ce{-(C)N}$$).
The next rule (P-44.3.2) is looking at the longest chain. Therefore, the parent structure is a butanenitrile.
The following rules (P-44.4.1.1 and P-44.4.1.2) would be looking at multiple bonds, which is not relevant in this case.
Next, Rule P-44.4.1.8 is looking for a lower locant for the attached group that is expressed as a suffix (here: nitrile); however, the nitrile group has the lowest locant ‘1’ in both suggested cases; therefore, this rule doesn't help to make a choice.
The next rule (P-44.4.1.10) again is not relevant in this case.
Finally, Rule P-45.2.1 is looking for the structure that has the maximum number of substituents cited as prefixes to the parent structure. Therefore, the preferred IUPAC name is 3-methyl-4-oxobutanenitrile (two substituents) rather than ‘3-formylbutanenitrile’ (only one substituent).
By way of comparison, if you add more substituents to the methyl group, the principal chain can switch, for example 4,4-dichloro-3-formylbutanenitrile (three substituents) rather than ‘3-(dichloromethyl)-4-oxobutanenitrile’ (two substituents).
The Preferred IUPAC name is 3-methyl-4-oxobutanenitrile. Reference: Pubchem
Explanation:
This compound has two functional groups. The seniority table says that the Nitrile group has a higher precedence over the Aldehyde/Formyl group. While picking the longest chain, we have the following set of rules (From Wikipedia):
1. It should have the maximum length.
2. It should have the maximum number of substituents of the suffix functional group. By suffix, it is meant that the parent functional group should have a suffix, unlike halogen substituents. If more than one functional group is present, the one with highest precedence should be used.
3. ...
Here, the longest chain in both cases has 4 carbon atoms. The chain should also contain the maximum number of substituents of the suffix functional group. We then choose the chain containing the oxygen atom attached to it, thereby obtaining methyl as a substituent on C3, giving us the name.
Note that oxo can be used even for an aldehyde (From Wikipedia)
If the compound is a natural product or a carboxylic acid, the prefix oxo- may be used to indicate which carbon atom is part of the aldehyde group; for example, $$\ce{CHOCH2COOH}$$ is named 3-oxopropanoic acid.
This logic is extendable to include any functional group that is above the aldehyde group in the seniority table.
Answering the doubt you posed regarding 3-formylpentanenitrile in the comments, here, the pentane chain has the most number of carbon atoms and also contains the suffix functional group attached to it, i.e. formyl, which is what makes it the PIN.
• "The chain should also contain the maximum number of substituents of the suffix functional group". Isn't the suffix functional group nitrile? If yes, then both the longest chains have maximum number of nitriles, i.e. 1. So, how can we decide which longest chain to choose? – Shashank Dec 30 '19 at 14:30
• By suffix, it is meant that the parent functional group should have a suffix, unlike halogen substituents. Here, the possibilities are that it has nitrile + formyl or nitrile + oxo groups, in which case we consider the oxo group one. – Aniruddha Deb Dec 30 '19 at 15:59
• The preferred IUPAC name 3-methyl-4-oxobutanenitrile is correct; however, the explanations are wrong. Oxo and formyl are not suffixes and have nothing to do with the principal characteristic group that is expressed as a suffix. – Faded Giant Jan 7 at 10:03 | 2020-07-10 13:59:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.39172685146331787, "perplexity": 2796.43275958619}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655908294.32/warc/CC-MAIN-20200710113143-20200710143143-00491.warc.gz"} |
https://www.physicsforums.com/threads/intensity-of-light.197040/ | # Intensity of light
1. Nov 9, 2007
### pardesi
What do we mean in mathematical terms intensity of light at this point is $$I$$ and when we say Power of light is $$P$$
2. Nov 9, 2007
### pixel01
the power of light P means the power of the light source which gives off energy. The intensity I, can be high, but it may not produce any work (or energy) at that point unless you have something to absorb that light. The more value of I, the more energy you can absorb if all other parameters hold.
3. Nov 9, 2007
### pardesi
actually can u give a mathematical interpretation
4. Nov 9, 2007
### Staff: Mentor
Are you asking about the units? Usually when we talk about the "intensity" of light, we're referring to its irradiance, which is the amount of energy it carries through a surface perpendicular to the propagation vector $\vec k$, per unit area, per unit time. It's usually measured in joules/(m^2 . sec) = watts / m^2.
Sunlight at the earth's surface has an irradiance of about 1400 W/m^2. If you had a 1-m^2 solar panel that could capture light energy with 100% efficiency, and oriented it so it directly faces the sun, it would produce about 1400 joules of energy per second, i.e. 1400 W.
Last edited: Nov 9, 2007
5. Nov 9, 2007
### pardesi
actually i was solving a problem and i was asked the power at a point .i know the intensity at that point how can i find the power
6. Nov 9, 2007
### Staff: Mentor
What units does the intensity have, in your problem?
7. Nov 9, 2007
### pardesi
it has $\frac{W}{m^{2}}$
8. Nov 9, 2007
### Staff: Mentor
Then that's all you can say about the power, "at a point." If you want a number of watts (not watts/m^2), you need to specify a particular surface through which the light falls on, or passes through, and multiply by the area of that surface.
9. Nov 9, 2007
### pardesi
yes that's what i thought too that's why i posted so | 2017-11-21 21:51:32 | {"extraction_info": {"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, "math_score": 0.6706165671348572, "perplexity": 1131.2882587273664}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806426.72/warc/CC-MAIN-20171121204652-20171121224652-00366.warc.gz"} |
https://support.bioconductor.org/p/30071/#30075 | how does an annotation package handle ambigious probe set id mappings
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Andrew Yee ▴ 350
@andrew-yee-2667
Last seen 7.3 years ago
Apologies if this has been asked before, but how does an annotation package handle an ambiguous probe set ID mapping? Take for example the Affymetrix chip U133X3P. When I use the annotation for this chip for probe set ID 1552641_3p_s_at, it returns only one match: > library('u133x3p.db') > mget('1552641_3p_s_at', env=u133x3pSYMBOL) $1552641_3p_s_at [1] "ATAD3B" > mget('1552641_3p_s_at', env=u133x3pENTREZID)$1552641_3p_s_at [1] "83858" However, when I search Affymetrix, with: https://www.affymetrix.com/analysis/netaffx/fullrecord.affx?pk=U133_X3 P:1552641_3P_S_AT it states that it ambiguously maps to three gene symbols,?ATAD3A, ATAD3B, and?LOC732419. How does the annotation package determine which gene symbol it should map to? Thanks, Andrew
Annotation u133x3p probe Annotation u133x3p probe • 709 views
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@james-w-macdonald-5106
Last seen 9 hours ago
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Hi Andrew, Andrew Yee wrote: > Apologies if this has been asked before, but how does an annotation > package handle an ambiguous probe set ID mapping? > > Take for example the Affymetrix chip U133X3P. > > When I use the annotation for this chip for probe set ID > 1552641_3p_s_at, it returns only one match: > >> library('u133x3p.db') >> mget('1552641_3p_s_at', env=u133x3pSYMBOL) > $1552641_3p_s_at > [1] "ATAD3B" >> mget('1552641_3p_s_at', env=u133x3pENTREZID) >$1552641_3p_s_at > [1] "83858" > > However, when I search Affymetrix, with: > > https://www.affymetrix.com/analysis/netaffx/fullrecord.affx?pk=U133_ X3P:1552641_3P_S_AT > > it states that it ambiguously maps to three gene symbols, ATAD3A, > ATAD3B, and LOC732419. > > How does the annotation package determine which gene symbol it should map to? In the past we just used the first probeset ==> Entrez Gene ID mapping. However, in the soon to be released BioC 2.5 annotation packages all the mappings are included (thanks to Marc Carlson). > tmp <- toggleProbes(u133x3pENTREZID, "all") > get('1552641_3p_s_at', tmp) [1] "55210" "732419" "83858" > tmp2 <- toggleProbes(u133x3pSYMBOL, "all") > get('1552641_3p_s_at', tmp2) [1] "ATAD3A" "LOC732419" "ATAD3B" Oddly enough, this probeset isn't mapped in the 'regular' mappings: > get('1552641_3p_s_at', u133x3pENTREZID) [1] NA > get('1552641_3p_s_at', u133x3pSYMBOL) [1] NA Marc? > sessionInfo() R version 2.10.0 Under development (unstable) (2009-09-21 r49780) i386-pc-mingw32 locale: [1] LC_COLLATE=English_United States.1252 [2] LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 [4] LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices datasets utils methods base other attached packages: [1] u133x3p.db_2.3.5 org.Hs.eg.db_2.3.4 RSQLite_0.7-2 [4] DBI_0.2-4 AnnotationDbi_1.7.17 Biobase_2.5.6 loaded via a namespace (and not attached): [1] tools_2.10.0 > Best, Jim > > Thanks, > Andrew > > _______________________________________________ > Bioconductor mailing list > Bioconductor at stat.math.ethz.ch > https://stat.ethz.ch/mailman/listinfo/bioconductor > Search the archives: http://news.gmane.org/gmane.science.biology.informatics.conductor -- James W. MacDonald, M.S. Biostatistician Douglas Lab University of Michigan Department of Human Genetics 5912 Buhl 1241 E. Catherine St. Ann Arbor MI 48109-5618 734-615-7826
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Thank you Jim, The probeset is not displayed in the regular mappings precisely because it maps to multiple things. Because it is of ambiguous assignment, most people will probably want to avoid it the majority of the time. Also, legacy code that depends on getting one value back for such operations needs to be able to continue working. However, as you have so carefully illustrated, you can now get the complete data for any mapping by using toggleProbes(). There are 3 settings for any mapping that can be set using toggleProbes: "all" which gives you every mapping regardless of what it is, "multiple" which only exposes mappings where many probes map to the same item, and "single" which is the default and which will expose only those probe IDs that have been assigned by the manufacturer to a single gene. So if you don't use "all" then the troublesome ambiguously assigned probes will not be represented in the mapping (ie. you will get an NA). The majority of the time, probes are assigned to a single gene and so normally most things are represented just fine by "single". This default has the side benefit that it shields you from those probes where the manufacturer is less than certain about the identity. But for those cases where there are multiple genes assigned to a probe or probeset, you can now also get all the assignments out if you wish (or just the troublesome ones if you want to focus on them) so that you can make a guess about which one you think it is you have actually measured. Marc James W. MacDonald wrote: > Hi Andrew, > > Andrew Yee wrote: >> Apologies if this has been asked before, but how does an annotation >> package handle an ambiguous probe set ID mapping? >> >> Take for example the Affymetrix chip U133X3P. >> >> When I use the annotation for this chip for probe set ID >> 1552641_3p_s_at, it returns only one match: >> >>> library('u133x3p.db') >>> mget('1552641_3p_s_at', env=u133x3pSYMBOL) >> $1552641_3p_s_at >> [1] "ATAD3B" >>> mget('1552641_3p_s_at', env=u133x3pENTREZID) >>$1552641_3p_s_at >> [1] "83858" >> >> However, when I search Affymetrix, with: >> >> https://www.affymetrix.com/analysis/netaffx/fullrecord.affx?pk=U133 _X3P:1552641_3P_S_AT >> >> >> it states that it ambiguously maps to three gene symbols, ATAD3A, >> ATAD3B, and LOC732419. >> >> How does the annotation package determine which gene symbol it should >> map to? > > In the past we just used the first probeset ==> Entrez Gene ID > mapping. However, in the soon to be released BioC 2.5 annotation > packages all the mappings are included (thanks to Marc Carlson). > > > tmp <- toggleProbes(u133x3pENTREZID, "all") > > get('1552641_3p_s_at', tmp) > [1] "55210" "732419" "83858" > > tmp2 <- toggleProbes(u133x3pSYMBOL, "all") > > get('1552641_3p_s_at', tmp2) > [1] "ATAD3A" "LOC732419" "ATAD3B" > > Oddly enough, this probeset isn't mapped in the 'regular' mappings: > > > get('1552641_3p_s_at', u133x3pENTREZID) > [1] NA > > get('1552641_3p_s_at', u133x3pSYMBOL) > [1] NA > > Marc? > > > sessionInfo() > R version 2.10.0 Under development (unstable) (2009-09-21 r49780) > i386-pc-mingw32 > > locale: > [1] LC_COLLATE=English_United States.1252 > [2] LC_CTYPE=English_United States.1252 > [3] LC_MONETARY=English_United States.1252 > [4] LC_NUMERIC=C > [5] LC_TIME=English_United States.1252 > > attached base packages: > [1] stats graphics grDevices datasets utils methods base > > other attached packages: > [1] u133x3p.db_2.3.5 org.Hs.eg.db_2.3.4 RSQLite_0.7-2 > [4] DBI_0.2-4 AnnotationDbi_1.7.17 Biobase_2.5.6 > > loaded via a namespace (and not attached): > [1] tools_2.10.0 > > > > Best, > > Jim > > >> >> Thanks, >> Andrew >> >> _______________________________________________ >> Bioconductor mailing list >> Bioconductor at stat.math.ethz.ch >> https://stat.ethz.ch/mailman/listinfo/bioconductor >> Search the archives: >> http://news.gmane.org/gmane.science.biology.informatics.conductor >
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Entering edit mode
Thanks Marc and James, this discussion has been very informative! Andrew On Mon, Oct 19, 2009 at 1:53 PM, Marc Carlson <mcarlson@fhcrc.org> wrote: > Thank you Jim, > > The probeset is not displayed in the regular mappings precisely because > it maps to multiple things. Because it is of ambiguous assignment, most > people will probably want to avoid it the majority of the time. Also, > legacy code that depends on getting one value back for such operations > needs to be able to continue working. However, as you have so carefully > illustrated, you can now get the complete data for any mapping by using > toggleProbes(). There are 3 settings for any mapping that can be set > using toggleProbes: "all" which gives you every mapping regardless of > what it is, "multiple" which only exposes mappings where many probes map > to the same item, and "single" which is the default and which will > expose only those probe IDs that have been assigned by the manufacturer > to a single gene. So if you don't use "all" then the troublesome > ambiguously assigned probes will not be represented in the mapping (ie. > you will get an NA). The majority of the time, probes are assigned to a > single gene and so normally most things are represented just fine by > "single". This default has the side benefit that it shields you from > those probes where the manufacturer is less than certain about the > identity. But for those cases where there are multiple genes assigned > to a probe or probeset, you can now also get all the assignments out if > you wish (or just the troublesome ones if you want to focus on them) so > that you can make a guess about which one you think it is you have > actually measured. > > > Marc > > > > > James W. MacDonald wrote: > > Hi Andrew, > > > > Andrew Yee wrote: > >> Apologies if this has been asked before, but how does an annotation > >> package handle an ambiguous probe set ID mapping? > >> > >> Take for example the Affymetrix chip U133X3P. > >> > >> When I use the annotation for this chip for probe set ID > >> 1552641_3p_s_at, it returns only one match: > >> > >>> library('u133x3p.db') > >>> mget('1552641_3p_s_at', env=u133x3pSYMBOL) > >> $1552641_3p_s_at > >> [1] "ATAD3B" > >>> mget('1552641_3p_s_at', env=u133x3pENTREZID) > >>$1552641_3p_s_at > >> [1] "83858" > >> > >> However, when I search Affymetrix, with: > >> > >> > https://www.affymetrix.com/analysis/netaffx/fullrecord.affx?pk=U133_ X3P:1552641_3P_S_AT > >> > >> > >> it states that it ambiguously maps to three gene symbols, ATAD3A, > >> ATAD3B, and LOC732419. > >> > >> How does the annotation package determine which gene symbol it should > >> map to? > > > > In the past we just used the first probeset ==> Entrez Gene ID > > mapping. However, in the soon to be released BioC 2.5 annotation > > packages all the mappings are included (thanks to Marc Carlson). > > > > > tmp <- toggleProbes(u133x3pENTREZID, "all") > > > get('1552641_3p_s_at', tmp) > > [1] "55210" "732419" "83858" > > > tmp2 <- toggleProbes(u133x3pSYMBOL, "all") > > > get('1552641_3p_s_at', tmp2) > > [1] "ATAD3A" "LOC732419" "ATAD3B" > > > > Oddly enough, this probeset isn't mapped in the 'regular' mappings: > > > > > get('1552641_3p_s_at', u133x3pENTREZID) > > [1] NA > > > get('1552641_3p_s_at', u133x3pSYMBOL) > > [1] NA > > > > Marc? > > > > > sessionInfo() > > R version 2.10.0 Under development (unstable) (2009-09-21 r49780) > > i386-pc-mingw32 > > > > locale: > > [1] LC_COLLATE=English_United States.1252 > > [2] LC_CTYPE=English_United States.1252 > > [3] LC_MONETARY=English_United States.1252 > > [4] LC_NUMERIC=C > > [5] LC_TIME=English_United States.1252 > > > > attached base packages: > > [1] stats graphics grDevices datasets utils methods base > > > > other attached packages: > > [1] u133x3p.db_2.3.5 org.Hs.eg.db_2.3.4 RSQLite_0.7-2 > > [4] DBI_0.2-4 AnnotationDbi_1.7.17 Biobase_2.5.6 > > > > loaded via a namespace (and not attached): > > [1] tools_2.10.0 > > > > > > > Best, > > > > Jim > > > > > >> > >> Thanks, > >> Andrew > >> > >> _______________________________________________ > >> Bioconductor mailing list > >> Bioconductor@stat.math.ethz.ch > >> https://stat.ethz.ch/mailman/listinfo/bioconductor > >> Search the archives: > >> http://news.gmane.org/gmane.science.biology.informatics.conductor > > > > [[alternative HTML version deleted]]
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Powered by the version 2.3.6 | 2021-12-09 04:14:53 | {"extraction_info": {"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, "math_score": 0.4793972969055176, "perplexity": 8743.518514215068}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363659.21/warc/CC-MAIN-20211209030858-20211209060858-00154.warc.gz"} |
http://mathhelpforum.com/algebra/280139-adding-fraction-whole-number-algebraic-equation.html | # Thread: Adding a fraction to a whole number in an algebraic equation
1. ## Adding a fraction to a whole number in an algebraic equation
Hello!
I am going over another answer for a question on an online test:
The only question I have is, how did they get 9/2? The way I'd go about solving this is making the -4 a -4/1, and then adding -1/2 to get 4/2. Where am I going wrong?
-Neo
2. ## Re: Adding a fraction to a whole number in an algebraic equation
\displaystyle \begin{align*} -\frac{1}{2} - 4 &= - \left( \frac{1}{2} + 4 \right) \\ &= - \left( \frac{1}{2} + \frac{8}{2} \right) \\ &= -\frac{9}{2} \end{align*}
3. ## Re: Adding a fraction to a whole number in an algebraic equation
I'm sorry, I still don't get it... I know that -1/2 + 4 = -(1/2 + 4), but to get 8/2, did you multiply 4/1 by 2? And if so, why? Is it because 2 is the denominator?
Thank you for your efforts! ^^
-Neo
4. ## Re: Adding a fraction to a whole number in an algebraic equation
Originally Posted by SpaghettNeo
I know that -1/2 + 4 = -(1/2 + 4)....
No. -1/2 + 4 = -(1/2 - 4)
That's really -1(1/2 - 4)
So: -1*(1/2) = -1/2, -1*(-4) = +4
5. ## Re: Adding a fraction to a whole number in an algebraic equation
Originally Posted by SpaghettNeo
I'm sorry, I still don't get it... I know that -1/2 + 4 = -(1/2 + 4), but to get 8/2, did you multiply 4/1 by 2? And if so, why? Is it because 2 is the denominator?
First: $\large{ - \frac{1}{2} + 4 \ne - \left( {\frac{1}{2} + 4} \right)}$
Can you do this $\dfrac{8}{2} - \dfrac{1}{2}~?$ | 2018-12-16 14:56:59 | {"extraction_info": {"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, "math_score": 0.9996152520179749, "perplexity": 850.1359495597429}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376827769.75/warc/CC-MAIN-20181216143418-20181216165418-00631.warc.gz"} |
https://chat.stackexchange.com/transcript/message/2575681 | 2:00 PM
Yes, all standard matrices are affine
Then my problem lies elsewhere.
I was just considering that there are perspective matrices, but they are not the ones that simply use multiply and add.
They go into a higher dimension and project out the extra dimensions.
Thus, applying perspective.
But that is probably not what you're dealing with.
Well, I'm using 4x4 matrices and homogenous coordinates.
(x,y,z,T); discarding T between operations and always treating the vector T as 1.0 when transforming vectors
Whee!
But the matrices I'm dealing with right now are only translation/rotation matrices.
2:03 PM
That is what I see... It will probably disappear when I refresh my browser.
So I don't think that should matter.
Also, congratulations.
Still working on getting that 10k myself.
There, now it is back to 11.5k as it should be.
I wonder how that happened?
Oh, wait; I have 12.9k
that's a lot
but I guess that's my SE total.
@WillihamTotland I just recently got 10K, but I saw the 23k below my avatar. Must have been a server burp.
Indeed.
Neither of those look right at all.
The bottom one is correct, the top one, not so much.
The upper being scaled 0.9 on all axes and translated 1.1 up on the y.
and the lower resting on the origin.
2:09 PM
Just looking at the images, I couldn't tell you what is going wrong.
I think I'm messing up the normals, somehow.
To wit; the object space normals aren't correctly transcribed into world space.
That might explain the funky shading.
The thing is; my hit detection obviously works.
The spheres are right there.
In the correct location.
Normal map.
That don't look right.
That don't look right at all.
Owel, bbl
2:40 PM
hello
Hi.
someone has a buddhabrot avatar
Oh, that's the name of that thing?
yes
instead of plotting the points that don't escape, you make a histogram of the orbits of every point
for some reason you get a cool picture
Intriguing...
2:44 PM
cool people are talking about me :)
wait
let me blow tha tup
I don't really know anything about fractals
and more
dang, you crushed my avatar-chat-blow-up
some more
I wanted to get as big as the picture :)
I have a soft spot for black and white pictures. It's harder to make beautiful pictures in just black and white.
(Okay, there are some grays there, but you get my point.)
And good evening, by the way.
2:48 PM
Good evening. Did jury duty on that. Other potential jurors seem to be idle though.
3:05 PM
@rubenvb so you do graphics I guess
oh no I mixed you up with the other guy
why do you have buddhabrot as your avatar
@QED: nope, plucked mine off the internet.
I think it looks cool
and I once had to write a java program to draw the 2D version :)
:D
so if you apply z^2+c to every point of the buddhabrot it stays the same I guess
same with sqrt(z-c)
3:38 PM
@HenningMakholm I thought that everyone can delete his own question. Is some number of reputation posts needed for that? (Cyril is asking in comments to that question that moderator should do that for him.)
Unregistered users don't have the ability to delete their posts.
I see a certain mark next to his username...
In such a big letters... as if I were blind.
I was looking for it bellow in the profile.
Ok, so it explains everything.
4:34 PM
> in principle, there is no way to actually sum two points on a variety.
Why don't we vote to close this Grandfather's paradox?
haha
I did find it a bit weird that he says "Upper and lower bounds" then does something completely different
@Matt I was teaching today. I need not be entertained.
Ah, very well.
4:49 PM
Good afternoon, all
Hi Srivatsan
5:06 PM
Is the inverse of the product of two matrices identical to the product of the inverse of two matrices?
Yes. (AB)^{-1} is equal to B^{-1} A^{-1} provided A and B are invertible.
Good.
Thanks. :)
(Just in case you want to see the proof, too.)
@WillihamTotland Oh, I forgot to caution you: you must be careful with the order of multiplication. It is not A^{-1} B^{-1}.
Naturally.
AB != BA, after all.
5:12 PM
@Srivatsan I made a new batch of algebra retagging today. I'll leave my hands of it for next two days - so you can go on with the number tag.
just calculate B' A' A B = I to prove B' A' = (AB)'
uniqueness of inverses in a group
@MartinSleziak cool, thanks! I was going to ask you about it myself.
Slightly trickier is showing that order doesn't matter when inverting and transposing a square matrix...
infact this is for any group
not just matrices
Yeah, what's with that recently?
quite odd
people just don't understand freely generated grammars
Who's counting?
As in, since Martin is doing [algebra], is it ok with simultaneously editing posts for this tag as well? Or should we wait?
Well, if some question did have both tags, then kill two tags with one edit, no?
5:27 PM
There were only 2 questions like that. Just killed 'em both. =)
hi all
hi
5:59 PM
It looks like the red notification thingy in the user profile page is back.
2
@Srivatsan Finally.
IA == A?
what
@Srivatsan Yes! Also the yellow background for new reputation entries.
6:07 PM
@WillihamTotland If I is the n-by-n identity matrix and A is an n-by-p matrix, then IA is equal to A. If A is p-by-n, then AI is equal to A. In general, as long as IA or AI is defined, it will be equal to A.
@HenningMakholm We should celebrate. Open the champagne! =)
and if p = n; IA == A == AI, right?
@WillihamTotland Yes. If A is also an n-by-n matrix, then AI = IA = A.
@Williham, what's "=="?
Equals.
6:10 PM
is it different to =?
@QED Programmer's tic. = tends to be used for assignment; == for comparisons.
Indeed.
@JM To add to that, some programming languages (and nowadays some mathematicians) use := for assignment or definition.
confusing
(I know you know that tidbit, JM =))
6:12 PM
what's p and n?
@Srivatsan 79536 is correctly tagged now, I think
@QED You get used to it if you've been writing programs long enough...
@Srivatsan :D
@HenningMakholm In particular, the [natural-numbers] tag is ok?
@Srivatsan are we trying to get rid of natural-numbers for 58085? My best immediate replacement would be elementary-number-theory.
What Henning said.
6:14 PM
I guess we can keep the [natural-numbers] tag for set theory or logic questions on natural numbers.
I have proposed [natural-numbers] as a synonym for [elementary-number-theory] some time ago, but no takers yet.
@HenningMakholm Well, if it is a synonym for elementary NT, then it cannot cover questions from logic or set theory, no? In particular, questions pertaining to the definition of natural numbers?
For 34105 I think it would be just as well tagged without [natural-numbers].
@Srivatsan I would expect questions pertaining to the definition of natural numbers to count as elementary-number-theory, actually.
@HenningMakholm Oh, if that's the case, I am down with the suggestion.
But I don't actually know any elementary number theorists, so what do I know?
@Srivatsan I thought Algol actually got it from an existing mathematical-physical convention.
6:20 PM
"Questions on congruences, linear diophantine equations, greatest common divisor, divisibility, etc." -- At least the tag wiki summary doesn't cover this definition of natural numbers part...
@HenningMakholm Oh, I thought Algol (but not sure), but didn't know about mathematical physics used it. Thanks.
No, but I'd say that the definition is even more elementary than those things.
@Srivatsan It's not that I have actually seen any pre-Algol math/physics text use it. But I have seen physics texts that follow a long expression with " =: M" and thereby give a name to it.
@HenningMakholm Aw, I don't think the qualifier "elementary" is meant to cover definitions. It signifies difficulty -- or that's my understanding.
iirc, one of my Intro to Modern Number Theory pdfs says that there is "formal number theory" (which relates NT, and diophantine equs in particular, to computation and logic). That might be relevant but I haven't been following the discussion there.
Btw, irrespective of what happens to [natural-numbers], I feel that the [number] tag should go. What's your opinion on that, @Henning?
Agree completely.
Isn't there already an eradication campaign underway? I think I've seen people here discuss the pace.
6:25 PM
That's it for me today. Later, y'all.
Yes, we are doing [number] (& [natural-numbers]) and [algebra] in parallel -- on different days of the week.
@JM Later, JM. Sleep well.
Oh, no....
no chat?
Does anyone have any idea how long the upgrade is going to take?
@robjohn - what are you talking about? =)
wait, what?
Did you not get a notice that chat would be down for a while due to upgrades?
6:37 PM
well, they could have given more notice, yeash.
@robjohn
It appeared over my chat window.
Well, I was going to say that SE was rejecting you like a bad organ.
There... we're back! That was a quick upgrade.
But then we all went offline. :P
6:38 PM
I was going to say we should invade some server's #math channel temporarily but it appears that is unnecessary now.
Did anyone else get the warning?
Yeah, got it.
But not until I reloaded after a message couldn't be sent. :P
I did, but only seconds before did I notice going offline (or it might have only been around for a few seconds in the first place...)
On a related note.
Well, I feel special... I got notified a minute or two ahead :-)
2
6:40 PM
Why will this not work!?
robjohn, veritable prophet of our humble math locale
hah, just Cmd-Tab:ed in; read "math cabal"
Perhaps more appropriate?
Oy.
I'm just moving further and further away from my objective. :(
Ello. Hm... is Asaf around?
He was last seen denying that he needs any entertainment.
Oh.
I thought he and I had agreed that I quit functional analysis and only do set theory from now on.
7:18 PM
[P]=[Q] iff P=Q? So what's the point of the brackets
the point is you want to go in between the group perspective and the freer algebra perspective but keep clear which you're in and when. I suppose it gets more important for derivation and conceptual understanding depending on the specific application.
7:35 PM
Hmm, look at how the Intro to Mathematical Crypto book writes quotients in CRT:
I dislike (a mod m1, a mod m2, ...)
and the use of ---> rather than \cong
It has one group literally over another instead of side-by-side, as I've always seen it. Also, \to and \cong mean two distinct things: the first is referring to some (canonical) morphism map and the lateral says they're isomorphic.
I don't like it. A quotient structure is sufficiently different from a fraction that they should not share notation interchangeably.
7:50 PM
isn't Z/mZ just ASCII for Z "over" mZ
why does it matter if they're vertical
@Matt here?
I can tell you right now it's not "just ASCII" (did you make that up on the spot?), but really my only qualm is that it's not as aesthetic when describing groups to put them in anything but inline except in the case of diagramming.
the quotient notation Z/mZ is just common because you can't type Z "over" mZ
as 3/4 is used to mean 3 over 4
What makes you say that though? Were mathematicians writing group quotienting one over the other before computers that you know of?
The / notation for quotients is chosen because of it's use for fraction
7:53 PM
And I wouldn't say common, I would say near-universal, because I've literally never seen anything else till now.
@QED: So are you aware of any instances of writing group quotients the other way? Do you have historical knowledge that mathematicians wrote it the other way pre-ASCII or at all commonly in history?
the algebra books
seems like maclane avoids this intentionally but it happens in algebra 0 paolo aluffi
What notation does MacLane use? You're still talking about quotient notation, right?
hmm. I guess there is are real instances then. I'm still not sure about the conjecture that ASCII or print-type is the reason we have the "/" notation so overwhelmingly.
8:09 PM
Print type could be the reason. I doubt that ASCII is, though.
Zoom in, enhance (Hat tip: Enthusiasms.org):
Nice. But where did they get that deconvolution kernel from?
8:24 PM
@AsafKaragila Hi Asaf! I asked you a question earlier today and now I've posted it as a comment below my post... Are you quite busy at the moment?
It's a first. Usually Didier comes up with the simple answer using The Law of Total Variance, but his answer looks more like mine usually do and I answered using the Law of Total Variance
@anon That is pretty amazing...
Is that native in Mma, or is that something that someone is selling?
@robjohn Is this a reply to a previous comment? =)
I don't know. I just copied it over from the blog Enthusiasms (which likewise has no explanation). You could email Simen (the author) or Tineye it or something.
No, I'm just amazed that my answer looks far simpler than Didier's
I think Didier is going for general than simpler.
8:31 PM
@Matt I have a minute.
@robjohn Actually, I am not very sure what the difference is between all the answers =)
@AsafKaragila Nice! Look at this, it's easier to read with latex...
There is no real sense to p[{p}].
Oh. Well good then because I couldn't make any sense of it. This means I messed up somewhere while doing the sums. I'll just post the whole thing as a question, that'll be easier.
The canonical name for G, while not being a member of the ground model, is simply the collection of names of conditions in P. So when you interpret it by G you only take those p's which are in G, and you have G.
8:39 PM
Clarification. Isn't every discrete metric an example of this? math.stackexchange.com/questions/87135
@Srivatsan The other way around: in the discrete metric every set is open. He wants no open sets apart from empty set and X...
Oops, you're right. I am confusing between open sets and balls.
The indiscrete topology is not metrisable if there is more than one point. (It's not Hausdorff.)
@ZhenLin Yes yes. I thought the OP wants a metric space where the only open *balls* are \emptyset and X.
Thanks Zhen and Matt.
Hmm, I've overtaken Willie Wong. That looks like a good place to call it a day.
8:43 PM
Same answer. The open balls always form a basis for the topology.
Oh shoo, Mike is active again. He's gonna catch up with me soon!!
@ZhenLin No, the answer is different, no? In a discrete metric, it is the case that the only open balls are \emptyset and X. -- or am I confusing something?
@Srivatsan If you have the discrete metric, i.e. distance either 0 or 1 then every singleton is open because you can take an open ball of radius 1/2 and it will contain just the point (because every other point has distance 1)
No, every singleton is an open ball, of radius 1/2.
The empty set is never an open ball.
Ah, um, ok. =)
8:46 PM
Only in the empty metric space.
Sorry about that. Not my day ;)
When is your day, Srivatsan, Tuesday? Monday? Friday?
Asaf, I'm just about to post some entertainment for you! : )
@Asaf: I thought metric spaces were required to be non-empty?
Yes.
8:47 PM
Bye all.
Bye.
But part of the "Oh, he's a logician, let's bash him." approach some folks here have, I was ridiculed for saying "There are no empty metric spaces because it makes FOL stiffer and less comfortable."
And if folks agreed to accept FOL as a basis for mathematics, perhaps they would be kind enough to accept its rules.
Bye @HenningMakholm!
@AsafKaragila Can you explain? Why does FOL conflict with empty metric spaces? Is it something to do with empty metric spaces or is it do with any empty structure?
If you allow empty structure then every universal sentence is true vacuously.
8:50 PM
And is that wrong?
So you want to require structures to be non-empty. In particular, metric spaces are FOL structures, and so they are required to be non-empty.
Btw sorry if I sound stupid. Mostly, I am. =)
How do you define the metric on an empty domain? I don't think that's possible.
Of course you can change the way you handle things, but then you always have to require that the structure is nonempty and so and so. It's just easier to do it this way.
@Matt: The empty function, of course.
8:52 PM
@AsafKaragila No, I don't get it. Every universal sentence is true for the empty structure -- seems ok to me. What's the issue?
@Srivatsan \forall x . x \ne x holds in the empty structure.
@ZhenLin Is that non-zero and satisfies the triangle inequality?
@Matt: Vacuously.
@Srivatsan You can run into contradictions if you want to say that all models of this and that theory are such and such.
8:55 PM
I think the only problem with allowing the empty model is that you no longer have soundness for the usual rules for \forall elimination and \exists introduction
I am not really sure I follow it actually. May be I should read a book or something.
I wholeheartedly recommend 1984 by G. Orwell..
Me, The Things They Carried or Lord of the Flies
@ZhenLin But for all elimination would say, you can substitute x in that sentence where x is from the structure. But I can still do it, no? Just that I cannot find any x to put in the sentence...
No. That's not how it works.
Deduction is purely syntactic, so you don't know what's in your universe or not. You only have the expressions which can be built in your language.
8:58 PM
Right. Vaguely following.
Btw, any serious suggestions for a book? [I might have read 1984 before. =)]
Girard's Proofs and Types is available online. That's pure logic and computer science though, with a strong bias towards constructivism.
@ZhenLin My professor recommended it last year.
@Srivatsan Heart of Darkness. Not everyone likes it, it's dark and heavy. Much like Apocalypse Now if you've seen it.
@AsafKaragila There's a reason for that similarity ;-)
@AsafKaragila You're still talking about 1984? =)
9:03 PM
No... I am talking about a whole other book.
@Srivatsan: You may also want to consider this question on this very site
@AsafKaragila Oh, sorry, I wasn't reading that comment properly.
What will be a first textbook in this area?
I liked Forster's Logic, induction and sets.
@Asaf, Thanks for the question. I don't understand all the answers, esp. JDH's, but I get some sense of them.
9:11 PM
@Srivatsan Try again tomorrow :-P
@AsafKaragila I should've expected that... =)
But alas... it is not your day and you did not expect that :-)
In any case, I only skimmed through the top 3 answers. I should put in more time if I want to understand...
@AsafKaragila has someone else bought the day? eme diem!
9:21 PM
I am going to drink me a beer.
Can we close this as a dupe? math.stackexchange.com/questions/87143/…
@Asaf, I was wondering if you knew anything about the automorphisms of the complex numbers without choice
I think so.
@AsafKaragila - who were you replying to? =)
@Srivatsan: You.
@JacobSchlather I have to go now, if you'll be here in an hour or so I might come back and we can talk about it.
Ciao.
9:27 PM
Okay
Anyone can help me out please?
What do you need help with?
I'm stuck on this:
$\int_{0}^{\infty}ke^{-3x}dx = k \lim_{x\to\infty}\frac{e^{-3x}}{-3} = k/3$
Why "k/3"
Also know as:
The gifification.
9:34 PM
@Gigili k/3 is the answer. What do you mean «why»?
I don't get how
Make the substitution 3x = y. What do you get?
which part?
Is that equation correct? His limit goes to zero, not one
3 lim(something) = k? you mean
9:36 PM
right, that should be \left[\exp(-3x)/-3\right]_0^\infty, not \lim_{x\to\infty}
@anon \exp(3x)/ 3x? That's not right, is it?
he meant exp(-3x)/-3
I presume
my lord am I glad I can edit
@JacobSchlather Yes
The point is once you evaluate you're going to get that limit
9:38 PM
So, what anon has is correct.
but you'll also get a constant term
which will be that function evaluated at x=0
and then the limit will be zero and you're just going to get the constant term
It's actually, lim_{x to infty} \exp(-3x)/(-3) - \exp(-3 (0))/(-3)
right
times k
But that's written in the book that way I wrote it
it has both limit and what @anon wrote
Then your book has a mistake
9:43 PM
wait
I still don't get how it will be k/3
does it have what srivatsan just wrote?
Gigili: do you know how to evaluate the limit(s) that Sri just posted?
This is one more way of writing it, but I am not sure which of the two your book means.
9:45 PM
None of these
Gigili, what your book has is certainly a typo. The correct way to evaluate the integral is what I posted.
Right, could you please explain the one you posted?
Ok, first take the k outside the integral: it's a harmless constant. You are left with integral of exp(-3x) between the limits ... and ... (let's come to the limits later).
What's the anti-derivative (or primitive or indefinite integral - I don't know which term you're most comfortable with) of exp(-3x)?
-1/3 exp(-3x)?
Yes. And that's, of course, the same as exp(-3x) / (-3).
You should evaluate this anti-derivative at the two limits: 0 and infinity.
9:51 PM
Right, and we should substitute 0 and infinity in it now?
Aha
You plug in 0, that's right. You can't plug in infinity - you take the limit of the function as x approaches infinity.
That's exactly where my problem comes from
(but heuristically you can just plug it in once the underlying theory is second-nature)
Right ..
x =0 : -1/3 ?
9:53 PM
x -> infinity : infinity ?
Um, I don't know
@Gigili You certainly realise that's not the correct answer, right? =)
Well, what's the limit of exp(-x) as x goes to infinity?
what are the sizes of 1/2, 1/4, 1/8, 1/16, ...? now what about 1/e^3, 1/e^6, 1/e^9, ...
0?
Aha, and then the answer is -k/3
Got it
Thank you so much
Almost. (0) - (-k/3) = ?
9:58 PM
@Gigili Not so fast. =) You got a sign flipped...
Aha, right , yes
I was thinking of e^infinity which is infinity all the time
Phew
Thank you again all, @Srivatsan, @anon, @Matt .. | 2021-06-23 19:14:33 | {"extraction_info": {"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, "math_score": 0.6925905346870422, "perplexity": 2067.74448825822}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488539764.83/warc/CC-MAIN-20210623165014-20210623195014-00452.warc.gz"} |
https://ora.ox.ac.uk/objects/uuid:c890fa3f-de61-4ee6-96d7-e265acade350 | Working paper
### On the Convergence of Reinforcement Learning.
Abstract:
This paper examines the convergence of payoffs and strategies in Erev and Roth's model of reinforcement learning. When all players use this rule it eliminates iteratively dominated strategies and in two-person constant-sum games average payoffs converge to the value of the game. Strategies converge in constant-sum games with unique equilibria if they are pure or if they are mixed and the game is 2 x 2. The long-run behaviour of the learning rule is governed by equations related to Maynard Smi...
### Access Document
Files:
• (pdf, 526.5kb)
• (pdf, 526.5kb)
### Authors
A. W. Beggs More by this author
Volume:
96
Series:
Discussion Papers
Publication date:
2002
URN:
Local pid:
ora:1133
Language:
English | 2020-10-31 10:36:58 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8085897564888, "perplexity": 3066.1064657444554}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107917390.91/warc/CC-MAIN-20201031092246-20201031122246-00092.warc.gz"} |
http://math.stackexchange.com/questions/492901/complex-function-sketch | Complex function sketch
I have just started studying some complex functions and I do not understand the concept very well (I cannot visualize them easily). So I would appreciate some help for the following problem:
Say we have the complex function f(z) = cosh(z) . I would like to know how the f-plane looks like when x=const. and respectively y=const.
Thank you very much!
-
When you can visualize complex functions easily, let me know. I still have a hard time, too :) – Clayton Sep 13 '13 at 20:29
Ok, I see that the "easily" is rather stupid but what I wanted to say is that the transition from real to complex functions is not very smooth in my head.:) – Whats My Name Sep 13 '13 at 20:45
You may be interested in this paper regarding visualizing complex functions:
"Phase Plots of Complex Functions: A Journey in Illustration" by Elias Wegert and Gunter Semmler.
Jim Fowler and I (Steve Gubkin) wrote this webpage to help visualization of complex functions:
We hope to produce an online complex analysis course soon which employs a diverse set of visualization methods. For example, we will use webGL to produce 3d plots of modulus colored by phase, phase plots on the riemann sphere, interactive geometry tools for the poincare metric on the disk, etc. We should also have a lot of computational exercises which are verified directly by a server running an instance of sage. We are still developing back end stuff right now, but we should have something up and running within the year. We do not have anything up there now, but watch this space: http://www.gratisu.org/.
Peace, Steve
-
This is very helpful. Thank you very much! – Whats My Name Sep 13 '13 at 23:36
I'm interested by such a course. I already illustrated a whole course of mathematics for bachelor using complex functions on my website: www.mikaelmayer.com/reflex/category/cours-a-lepfl/ Students appreciated it. – Mikaël Mayer May 5 at 13:25
@MikaëlMayer Ya, I still have plans to do this, but we got sort of distracted by producing calculus content (also, I defend my thesis tomorrow). I cannot promise when, but I definitely intend to build this course! – Steven Gubkin May 5 at 18:31
Let $z=x+iy.$ Separating the real and imaginary parts of $f,$ we have \begin{align} f(z)=f(x+iy)=\cosh(z)=\dfrac{e^{x+iy}+e^{-x-iy}}{2}\\ =\dfrac{1}{2}[e^x(\cos{y}+i\sin{y} )+e^{-x}(\cos{y}-i\sin{y})] \\=\cos{y}\cdot\dfrac{e^x+e^{-x}}{2}+i\sin{y}\cdot\dfrac{e^x-e^{-x}}{2}. \end{align} Denote $$u(x+iy)=\Re{f(x+iy)}=\cos{y}\cdot\dfrac{e^x+e^{-x}}{2}=\cosh{x}\cdot\cos{y},\\ v(x+iy)=\Im{f(c+iy)}=\sin{y}\cdot\dfrac{e^x-e^{-x}}{2}=\sinh{x}\cdot\sin{y}$$ For fixed $x=c$
$$\dfrac{u(c+iy)}{\cosh{c}}=\cos{y},\\ \dfrac{v(c+iy)}{\sinh{c}}=\sin{y}.$$ Squaring last equalities, we have ellipse $$\dfrac{u^2}{\cosh^2{c}}+\dfrac{v^2}{\sinh^2{c}}=1.$$
-
Although you asked this question a while ago, I have a new answer. You may be interested by the website reflex4you.com in "expert" mode. For example, the $cosh$ function is the following:
$z\rightarrow cosh(z)$ (click on the picture to visit the website and enter other formulas)
Although the axis are not shown, the window is between $-4-3i$ and $4+3i$. The zero is black and white is infinite. 1 is red and -1 is cyan. This representation is quasi-invariant if you take the negative of the picture, it will represent the inverse function.
To analyze this picture, you can consider that the values on the horizontal axis (the real axis) are red and darkest a zero, which means that $cosh(x) > cosh(0)$. The black spot are zeroes of the function corresponding to $+/- \pi/2$; Colored points denote complex functions. $i$ is greenish whereas $-i$ is purplish.
By clicking on the image above, you can modify the formula and try out other ones. I feel it is a very good way to represent complex functions.
To answer your original question, I wrote an algorithm usable on the website to compute the f-plane, for example for x constant. You can modify the value of the parameter c (constant for $x$). @M.Strochyk was right, these are ellipses.
let c = 1; set k = 0; let min = -4; let max = 4; let n = 100; func f = cosh(z); set result = 0; let threshold = 0.1; repeat n in set theta = ((max-min)*k/n+min)*i+c; set result = if(abs(f(theta)-z) < threshold, theta, result); set k = k + 1; result
Click on the picture to be able to modify it by entering different values.
DISCLAIMER: I am one of the co-authors of this website.
- | 2015-08-30 16:57:35 | {"extraction_info": {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7110984325408936, "perplexity": 531.2853309271399}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644065324.41/warc/CC-MAIN-20150827025425-00244-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-for-college-students-7th-edition/chapter-1-section-1-5-problem-solving-and-using-formulas-exercise-set-page-68/65 | ## Intermediate Algebra for College Students (7th Edition)
Published by Pearson
# Chapter 1 - Section 1.5 - Problem Solving and Using Formulas - Exercise Set: 65
#### Answer
$x = \dfrac{-By+C}{A}$
#### Work Step by Step
Add $-By$ to both sides to find: $Ax + By + (-By) = C + (-By) \\Ax = -By + C$ Divide both sides by $A$ to find: $\dfrac{Ax}{A} = \dfrac{-By+C}{A} \\x = \dfrac{-By+C}{A}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2018-04-22 09:23:42 | {"extraction_info": {"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, "math_score": 0.7839296460151672, "perplexity": 1827.6675163224068}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945552.45/warc/CC-MAIN-20180422080558-20180422100558-00575.warc.gz"} |
http://czielinski.de/ | # whoami My name is Christian Zielinski, I was born in Hannover, Germany and I currently live in Düsseldorf, Germany. Since September 2016 I work as a financial risk management and IT consultant at d-fine GmbH, based in Frankfurt am Main, Germany. My expertise is specifically in the areas of software architecture, credit risk and agile project managment. Furthermore I am a iSAQB® Certified Professional for Software Architecture (CPSA-F), a Professional Scrum Master (PSM I & PSM II) and a Professional Scrum Product Owner (PSPO I). Before the start of my professional career I graduated in March 2017 with a Ph.D. from the Division of Mathematical Sciences of the School of Physical and Mathematical Sciences at the Nanyang Technological University, Singapore. I joined the program in August 2012 and my advisors were Asst. Prof. David H. Adams and Assoc. Prof. Wang Li-Lian. My research dealt with computational and theoretical aspects of novel lattice fermion formulations in lattice gauge theories. From September 2015 until July 2016 I was a guest scientist in the Theoretical Particle Physics Group of the School of Mathematics und Natural Sciences at the University of Wuppertal, Germany where I worked together with PD Christian Hoelbling. In April 2012 I gradudated with distinction in the field of physics from Heidelberg University, Germany, where I was working with Prof. Jan M. Pawlowski in the Strongly Correlated Systems Group and Prof. I.-O. Stamatescu at the Institute for Theoretical Physics. Our research dealt with lattice field theory at finite density and the resulting sign problem. In 2009 I was working together with Dr. Qinghai Wang on a project regarding entanglement phenomena in $\mathcal{PT}$-symmetric quantum mechanics at the National University of Singapore, Singapore. $mail Find me on LinkedIn, check out my GitHub page, have a look at my Scrum.org Profile or email me at email007@czielinski.de.$ cat publications.bib For details please refer to my arXiv page. 1) Theoretical and Computational Aspects of New Lattice Fermion Formulations. By Christian Zielinski. Ph.D. Thesis (Nanyang Technological University, 2016). 2) Staggered domain wall fermions. By Christian Hoelbling, Christian Zielinski. PoS LATTICE2016 (2016) 254. 3) Spectral properties and chiral symmetry violations of (staggered) domain wall fermions in the Schwinger model. By Christian Hoelbling, Christian Zielinski. Phys. Rev. D94 (2016) No. 1, 014501. 4) Simple QED- and QCD-like Models at Finite Density. By Jan M. Pawlowski, Ion-Olimpiu Stamatescu, Christian Zielinski. Phys. Rev. D92 (2015) No. 1, 014508. 5) Continuum limit of the axial anomaly and index for the staggered overlap Dirac operator: An overview. By David H. Adams, Reetabrata Har, Yiyang Jia, Christian Zielinski. PoS LATTICE2013 (2014) 462. 6) Computational efficiency of staggered Wilson fermions: A first look. By David H. Adams, Daniel Nogradi, Andrii Petrashyk, Christian Zielinski. PoS LATTICE2013 (2014) 353. 7) Thirring model at finite density in $2+1$ dimensions with stochastic quantization. By Jan M. Pawlowski, Christian Zielinski. Phys. Rev. D87 (2013) No. 9, 094509. 8) Thirring model at finite density in $0+1$ dimensions with stochastic quantization: Crosscheck with an exact solution. By Jan M. Pawlowski, Christian Zielinski. Phys. Rev. D87 (2013) No. 9, 094503. 9) Entanglement Efficiencies in $\mathcal{PT}$-Symmetric Quantum Mechanics. By Christian Zielinski, Qing-hai Wang. Int. J. Theor. Phys. 51 (2012) 2648-2655. | 2019-01-21 23:44:45 | {"extraction_info": {"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, "math_score": 0.37048858404159546, "perplexity": 3456.589438836995}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583822341.72/warc/CC-MAIN-20190121233709-20190122015709-00132.warc.gz"} |
https://deepai.org/publication/maximum-weight-independent-sets-for-s-124-triangle-free-graphs-in-polynomial-time | # Maximum Weight Independent Sets for (S_1,2,4,Triangle)-Free Graphs in Polynomial Time
The Maximum Weight Independent Set (MWIS) problem on finite undirected graphs with vertex weights asks for a set of pairwise nonadjacent vertices of maximum weight sum. MWIS is one of the most investigated and most important algorithmic graph problems; it is well known to be NP-complete, and it remains NP-complete even under various strong restrictions such as for triangle-free graphs. Its complexity for P_k-free graphs, k > 7, is an open problem. In BraMos2018, it is shown that MWIS can be solved in polynomial time for (P_7,triangle)-free graphs. This result is extended by Maffray and Pastor MafPas2016 showing that MWIS can be solved in polynomial time for (P_7,bull)-free graphs. In the same paper, they also showed that MWIS can be solved in polynomial time for (S_1,2,3,bull)-free graphs. In this paper, using a similar approach as in BraMos2018, we show that MWIS can be solved in polynomial time for (S_1,2,4,triangle)-free graphs which generalizes the result for (P_7,triangle)-free graphs.
## Authors
• 7 publications
• 8 publications
• ### Independent sets in (P_4+P_4,Triangle)-free graphs
The Maximum Weight Independent Set Problem (WIS) is a well-known NP-hard...
03/19/2020 ∙ by Raffaele Mosca, et al. ∙ 0
• ### Weighted Triangle-free 2-matching Problem with Edge-disjoint Forbidden Triangles
The weighted T-free 2-matching problem is the following problem: given a...
11/15/2019 ∙ by Yusuke Kobayashi, et al. ∙ 0
• ### A tractable class of binary VCSPs via M-convex intersection
A binary VCSP is a general framework for the minimization problem of a f...
01/07/2018 ∙ by Hiroshi Hirai, et al. ∙ 0
• ### A simple combinatorial algorithm for restricted 2-matchings in subcubic graphs – via half-edges
We consider three variants of the problem of finding a maximum weight re...
12/31/2020 ∙ by Katarzyna Paluch, et al. ∙ 0
• ### On the tractability of the maximum independent set problem
The maximum independent set problem is a classical NP-complete problem i...
03/26/2019 ∙ by R. Dharmarajan, et al. ∙ 0
• ### Enumerating minimal dominating sets in triangle-free graphs
It is a long-standing open problem whether the minimal dominating sets o...
10/01/2018 ∙ by Marthe Bonamy, et al. ∙ 0
• ### On the monophonic rank of a graph
A set of vertices S of a graph G is monophonically convex if every induc...
10/03/2020 ∙ by Mitre C. Dourado, et al. ∙ 0
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## 1 Introduction
Let be a finite, simple and undirected graph and let (respectively, ) denote the vertex set (respectively, the edge set) of . For , let denote the subgraph of induced by . Throughout this paper, all subgraphs are understood as induced subgraphs.
For , let be the open neighborhood of in , let be the closed neighborhood of in , and let be the anti-neighborhood of in . For and , with , let .
If (, respectively) we say that sees ( misses , respectively). An independent set (or stable set) in a graph is a subset of pairwise nonadjacent vertices of . An independent set in a graph is maximal if it is not properly contained in any other independent set of .
Given a graph and a weight function on , the Maximum Weight Independent Set (MWIS) problem asks for an independent set of with maximum weight. Let denote the maximum weight of an independent set of . The MWIS problem is called MIS problem if all vertices have the same weight .
The MIS problem ([GT20] in [10]) is well known to be NP-complete [12]. While it is solvable in polynomial time for bipartite graphs (see e.g. [1, 8, 11]), it remains NP-hard even under various strong restrictions, such as for triangle-free graphs [23].
The following specific graphs are subsequently used. has vertices and edges for . has vertices and edges for (index arithmetic modulo ). has vertices which are pairwise adjacent. Clearly, . (and thus, ) is also called triangle. A claw (with center ) has vertices and edges . (with center ) is the graph obtained from a claw with center by subdividing respectively its edges into , , edges (e.g., is a , is a claw).
For a given graph , a graph is -free if no induced subgraph of is isomorphic to . If for given graphs , is -free for all then we say that is -free.
Alekseev [2, 5] proved that, given a graph class defined by forbidding a finite family of induced graphs, the MIS problem remains NP-hard for the graph class if each graph in is not an for some index . Various authors [9, 16, 17, 18, 25] proved that MWIS can be solved for claw-free (i.e., -free) graphs in polynomial time (improving the time bounds step by step). Lozin and Milanič [13] proved that MWIS can be solved for fork-free graphs (i.e., -free graphs) in polynomial time Alekseev [3, 4] previously proved a corresponding result for the unweighted case.
In this paper, we show that for (,triangle)-free graphs, MWIS can be solved in polynomial time. This generalizes the polynomial-time result for MWIS on (,triangle)-free graphs [7] (which was extended by Maffray and Pastor [15] showing that MWIS can be solved in polynomial time for (,bull)-free graphs; in the same paper, they also showed that MWIS can be solved in polynomial time for (,bull)-free graphs).
The following result is well known:
###### Theorem 1 ([1, 8, 11])
Let be a bipartite graph with vertices.
1. MWIS with rational weights is solvable for in time
via linear programming or network flow.
2. MIS is solvable for in time .
A graph is nearly bipartite if, for each , the subgraph induced by its anti-neighborhood is bipartite. Obviously we have:
αw(G)=maxv∈V(G){w(v)+αw(G[A(v)])} (1)
Thus, by Theorem 1, the MWIS problem (with rational weights) can be solved in time for nearly bipartite graphs.
Our approach is based on a repeated application of the anti-neighborhood approach with respect to (1) (and in particular, on the approach for MWIS on (,triangle)-free graphs [7]).
That allows, by detecting an opportune sequence of vertices, to split and to finally reduce the problem to certain instances of bipartite subgraphs, for which the problem can be solved in polynomial time (recall Theorem 1). In particular, as a corollary we obtain: For every (,triangle)-free graph there is a family of subsets of inducing bipartite subgraphs of , with detectable in polynomial time and containing polynomially many members, such that every maximal independent set of is contained in some member of . That seems to be harmonic to the result of Prömel et al. [24]
showing that with “high probability”, removing a single vertex in a triangle-free graph leads to a bipartite graph.
### 1.1 Further notations and preliminary results
For any missing notation or reference let us refer to [6]. For , with , has a join (a co-join, respectively) to , denoted by (, respectively), if each vertex in is adjacent (is nonadjacent, respectively) to each vertex in .
For and , with , contacts if is adjacent to some vertex of ; dominates if is adjacent to all vertices of , that is, ( for short); misses if is non-adjacent to all vertices of , that is, ( for short).
A component of is a maximal connected subgraph of . The distance of two vertices in is the number of edges of in a shortest path between and in .
Recall that is a triangle.
###### Lemma 1
Connected -free graphs are nearly bipartite.
Proof. Let be a connected -free graph. Suppose to the contrary that for some vertex , is not bipartite, i.e.,
contains an odd chordless cycle. Then, since
is ()-free, contains an odd chordless cycle , say , for some . Let be the vertices of and let (index arithmetic modulo ) be the edges of . Then let be a shortest path between and ; clearly, the distance between and is at least 2. Without loss of generality (since the other cases can be similarly treated), assume that has exactly one internal vertex, say (i.e., is adjacent to and to some vertex of ).
###### Claim 1.1
If then or .
Proof. Since is -free, implies and , and since is -free and thus, do not induce a , we have . Now, since (with center ) do not induce an , we have or which shows Claim 1.
Now, since is an odd cycle, Claim 1.1 leads to a or which is a contradiction (for example, if and then clearly, , and leads to a with vertices ).
Thus Lemma 1 is shown.
Since by Lemma 1, every component of a -free graph is nearly bipartite, and since MWIS is solvable in polynomial time for nearly bipartite graphs (recall Theorem 1 and MWIS for nearly bipartite graphs), we have:
###### Corollary 1
MWIS is solvable in polynomial time for -free graphs.
Our aim is to show that MWIS can be solved in polynomial time for ()-free graphs. Since by Corollary 1, we are done with -free graphs, from now on let be a connected ()-free graph containing a . Using again the anti-neighborhood approach, let and let be a component of the induced subgraph of its anti-neighborhood . Since is connected, has a neighbor contacting . Since is -free, is independent.
A component of is nontrivial if contains a .
###### Fact 1
For any -free graph and for any and its anti-neighborhood , if contacts two nontrivial components of then for any in , if then or .
Proof. Let be a in which is contacted by , say , and clearly, since is -free. For a in , let . Then, since (with center ) do not induce an , we have or . Thus, Fact 1 is shown.
###### Lemma 2
For any -free graph and for any and its anti-neighborhood , at most one component of can contain a .
Proof. Let be a connected -free graph. Suppose to the contrary that for some vertex , there are two components in containing a , say in and in . Let be a neighbor of contacting , and let be a neighbor of contacting . First assume that contacts and . Clearly, and are nontrivial. By Fact 1, contacts since otherwise, there is a in such that contacts only one end-vertex of , and correspondingly, contacts in . Without loss of generality, let and . Then again by Fact 1, has exactly two neighbors in and in , say and . But now, (with center ) induce an in which is a contradiction.
Thus, no neighbor contacts and ; let contact and let contact , , while and . Recall that is independent, i.e., . Without loss of generality, let and . Clearly, or . If and then (with center ) would induce an . Thus, without loss of generality, let and thus, , , but now, (with center ) induce an which is a contradiction. Thus Lemma 2 is shown.
Recall that is a component of , and contacts . Let
1. and
2. .
Obviously, is a partition of . Since is -free, is an independent set.
For showing that MWIS can be solved for in polynomial time, let us first consider the case when is bipartite.
## 2 Case 1: G[z] is bipartite
Recall that, if component contains no , then by Corollary 1, MWIS can be solved in polynomial time for . Thus assume that contains a , say with vertices and edges (index arithmetic modulo 5). Since we assume that is bipartite and since is an independent set, every in has at least one vertex in , and thus, we have one of the following two types:
1. Type : has exactly one vertex in (and thus, the four vertices of in induce a ).
2. Type : has exactly two vertices in (and thus, the three vertices of in induce a ).
###### Fact 2
Let be a nontrivial component of . If contacts both sides of then there is a of type in .
Proof. Let and be two neighbors of . Note that is nonadjacent to since is -free. Then, since is connected, there is a shortest path, say (of an even number of internal vertices) in between and ; without loss of generality, let us assume that is nonadjacent to any internal vertex of (else we may re-define the choice of and ).
If has only two internal vertices, say , then induce a of type 1 in . Thus, suppose to the contrary that has more than two internal vertices (and then has at least four internal vertices). Then , and the three vertices of closest to induce an in which is a contradiction. Thus, Fact 2 is shown.
### 2.1 Case 1.1: Every C5 in K is of type 2.
For and nontrivial component of , we define:
###### Definition 1
• has a half-join to if either or i.e., either and or and .
• properly one-side contacts if either or .
By Fact 2 and Case 1.1, we have:
###### Fact 3
For every , if contacts a nontrivial component of and every in is of type then either has a half-join to or properly one-side contacts .
###### Definition 2
A nontrivial component of is a green component of if there is a vertex which properly one-side contacts .
Case 1.1.1 has no green component.
###### Lemma 3
If there is no green component in then MWIS is solvable in polynomial time for .
Proof. Since has no green component, Fact 3 implies that, for each and for each nontrivial component of , if contacts then has a half-join to , i.e., either and or and . In particular, that implies:
###### Claim 2.1
For each , there is no induced , say , of such that is an endpoint of the in .
For any and for any in with vertices such that and , let us say that doubly contacts the if is adjacent to and to exactly one vertex of .
Then let doubly contacts a in .
If , then has no of type 2, i.e., by assumption of Case 1.1, is -free and then, by Lemma 1, MWIS can be solved in polynomial time for . Thus, assume that .
###### Claim 2.2
Let and such that doubly contacts a with and in , and contacts a in . If then and .
Proof. Assume without loss of generality that . Clearly, since . By Claim 2.1 and since is -free, do not induce a in , and thus, which implies . If then and . Now assume that , and recall that .
By Claim 2.1, do not induce a in , and correspondingly, do not induce a in . Thus, , and Claim 2.2 is shown.
Now, let ’’ be the following binary relation on : For any pair , if either or contacts all ’s of which are doubly contacted by . Correspondingly, if vertex doubly contacts a of such that does not contact . In particular let us write if and .
###### Claim 2.3
For any , either or .
Proof. Suppose to the contrary that and . Then doubly contacts a of with and such that is adjacent to , while , and doubly contacts a of with and such that is adjacent to , while .
By Claim 2.2, and .
Clearly, since and , we have and .
By Claim 2.1 and since is -free, do not induce a in , which implies . But now, (with center ) induce an which is a contradiction. Thus, Claim 2.3 is shown.
###### Claim 2.4
For any , if and then .
Proof. Since and , there is a with and in such that is adjacent to , while , and there is a with and in such that is adjacent to , while .
Suppose to the contrary that . Then there is a with and in such that is adjacent to while .
By Claim 2.2, the sets , , and are pairwise disjoint, and , , and .
Since and , we have , and clearly, . Thus, possibly , and analogously, possibly , and .
Now first assume that , , and . Then we claim that (with center ) would induce an :
Recall that , , , , and . Clearly, doubly contacts the with and . Then clearly, and since is -free, and , . Moreover, since , , , and . Finally, since clearly, , , since , and is -free, and , , , and . Thus, induce an which is a contradiction, i.e., , , and is impossible.
Now assume that we have exactly two such equalities. If but and then we claim that (with center ) would induce an :
Recall that in this case, doubly contacts the with and , and , . Clearly, since is -free. Since , , and , and by Claim 2.1, we have ; in particular, if then there is a which contradicts Claim 2.1. Finally, as before. Thus, (with center ) induce an which is a contradiction, i.e., exactly two such equalities and are impossible.
By symmetry, we can show that the two other cases of exactly two such equalities are impossible.
Now assume that we have exactly one such equality. By symmetry, assume that , but . Then we claim that (with center ) would induce an :
Recall that in this case, doubly contacts the with and , and , . Clearly, since is -free and , (recall ). Moreover, . Thus, (with center ) induce an which is a contradiction, i.e., exactly one such equality is impossible.
Finally assume that , , and . Since and , does not doubly contact the .
If (and since is -free, ) then (with center ) would induce an (recall that , and by Claim 2.1, we have , since otherwise there is a ).
Thus, and by symmetry, .
If then (with center ) would induce an (recall , , and by Claim 2.1, we have , since otherwise there is a ).
Thus, . But then (with center ) induce an which is a contradiction.
Thus, Claim 2.4 is shown.
###### Claim 2.5
There is a vertex such that for every .
Proof. The proof can be done by induction on the cardinality, say , of . It trivially follows for . If then Claim 2.5 follows by Claim 2.3.
Now assume that and that Claim 2.5 holds for . Let | 2021-08-01 20:27:55 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9255943894386292, "perplexity": 1030.752851534039}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154219.62/warc/CC-MAIN-20210801190212-20210801220212-00617.warc.gz"} |
https://quizplus.com/quiz/152892-quiz-4-organizational-architecture | # Accounting for Decision Making Study Set 5
## Quiz 4 :Organizational Architecture
Question Type
Woodhaven Service Background Woodhaven Service is a small, independent gas station located in the Woodhaven section of Queens. The station has three gasoline pumps and two service bays. The repair facility specializes in automotive maintenance (oil changes, tune-ups, etc.) and minor repairs (mufflers, shock absorbers, etc.). Woodhaven generally refers customers who require major work, such as transmission rebuilds and electronics, to shops that are better equipped to handle such repairs. Major repairs are done in-house only when both the customer and mechanic agree that this is the best course of action. During the 20 years that he has owned Woodhaven Service, Harold Mateen's competence and fairness have built a loyal customer base of neighborhood residents. In fact, demand for his services has been more than he can reasonably meet, yet the repair end of his business is not especially profitable. Most of his competitors earn the lion's share of their profits through repairs, but Harold is making almost all of his money by selling gasoline. If he could make more money on repairs, Woodhaven would be the most successful service station in the area. Harold believes that Woodhaven's weakness in repair profitability is due to the inefficiency of his mechanics, who are paid the industry average of $500 per week. While Harold does not think he overpays them, he feels he is not getting his money's worth. Harold's son, Andrew, is a student at the university, where he has learned the Socratic dictum, "To know the Good is to do the Good." Andrew provided his father with a classic text on employee morality, Dr. Weisbrotten's Work Hard and Follow the Righteous Way. Every morning for two months, Harold, Andrew, and the mechanics devoted one hour to studying this text. Despite many lively and fascinating discussions on the rights and responsibilities of the employee, productivity did not improve one bit. Harold figured he would just have to go out and hire harderworking mechanics. The failure of the Weisbrotten method did not surprise Lisa, Harold's daughter. She knew that Andrew's methods were bunk. As anyone serious about business knows, the true science of productivity and management of human resources resides in Professor von Drekken's masterful Modifying Organizational Behavior through Employee Commitment. Yes, employee commitment was the answer to everything! Harold followed the scientific methods to the letter. Yet, despite giving out gold stars, blowing up balloons, and wearing a smiley face button, he found Lisa's approach no more successful than Andrew's. Required: a. This case presents some popular approaches to alleviating agency costs. Although certain aspects of each of these methods are consistent with the views presented in the text, none of these methods is likely to succeed. Discuss the similarities and differences between the ideas of the chapter and (i) Dr. Weisbrotten's approach. (ii) Harold Mateen's idea of hiring "harder-working" mechanics. b. Discuss the expected general effect on agency costs at Woodhaven Service of the new incentive compensation plans. How might they help Woodhaven? Assuming that Harold wants his business to be successful for a long time to come, what major divergent behaviors would be expected under the new compensation proposals? How damaging would you expect these new behaviors to be to a business such as Woodhaven Service? Also, present a defense of the following propositions: (i) Harold's plan offers less incentive for divergent behavior than Honest Jack's. (ii) Limiting a mechanic's pay by placing an upper bound of$750 per week on his or her earnings reduces the incentive for divergent behavior. c. Suppose Harold owned a large auto repair franchise located in a department store in a popular suburban shopping mall. Suppose also that this department store is a heavily promoted, well-known national chain that is famous for its good values and easy credit. How should Harold's thinking on incentive compensation change? What if Harold did not own the franchise but was only the manager of a company-owned outlet? d. In this problem, it is assumed that knowledge and decision rights are linked. The mechanic who services the car decides what services are warranted. Discuss the costs and benefits of this fact for Woodhaven Service and the independently owned chain-store repair shop. e. Suppose that Woodhaven's problems are not due to agency costs. Briefly describe a likely problem that is apparent from the background description in this problem.
Free
Essay
a.(i) Dr. Weisbrotten's approach is fundamentally contrary to the suggestions of the chapter. Basically, by introducing Weisbrotten, Harold seeks to alter the preferences of his employees. While it is possible to lower agency costs by convincing agents that working harder on the job is desirable in itself, the text is pessimistic about such a strategy. Normally self-interested people's preferences are not easily altered. However, the firm can reduce the agency problem, if not goal incongruence, by structuring agents' incentives that when agents maximize their incentive-based payoffs, the principal's utility (or wealth) is also maximized. In other words, the agent's and the principal's goals become congruent through the agent's incentive scheme, not by a change in the agent's preferences.
(ii) The idea of hiring harder-working mechanics appears to have some merit as a means of reducing agency costs in that it eliminates conflicting interests of agents and principals by limiting the set of agents to those who already have the same goals as the principal. However, there is something naive about this notion. Is there such a condition as "hardworkingness?" Is this condition common enough in the population that Harold can expect to find mechanics out of work who possess it? Furthermore, how would one go about testing for hardworkingness? Mechanics looking for work are not likely to be the most hard working.
b. Harold thinks that Woodhaven's lack of bottom-line success in repairs is due to his mechanics' lack of productivity. He also believes that incentive-based compensation for the mechanics will help this problem. Two kinds of plans are suggested, commission and flat rate. The commission plan is designed to boost profits by boosting revenues. Assuming that the price of each service is above the marginal cost of performing that service, offering mechanics a percentage of revenues should work to increase profits. The flat rate is designed to boost profits by cutting costs. Whereas revenues from labor charges are always derived from standard rather than actual times, paying mechanics by the hour makes the labor costs incurred depend upon the actual time they spend on the job. Given more demand than capacity, and given the likely propensity of workers to prefer leisure over toil, the mechanics may very well be spending too long on each job. If they were paid only for the "just right" amount of time, subsidized leisure would disappear, giving mechanics an incentive to lower labor costs.
However, nothing occurs in a vacuum, and changing compensation schemes could be expected to have other effects as well. Most importantly, both compensation plans induce behavior that is divergent from Harold's desires. If he paid mechanics a percentage of the sales they generate, many mechanics would likely generate revenue that should not be generated. Suddenly, Woodhaven would "specialize" in $2,000 engine rebuilds, whether the staff had the technical knowledge for this kind of repair or not. In a worst-case scenario, mechanics would simply cheat customers by selling expensive services that were not necessary and might not even be performed. If mechanics are paid a standard number of hours for a job, they will simply work faster. So fast, indeed, that a drop in quality is likely. Since Woodhaven is a neighborhood shop, it is likely that many customers know one another. This would cause the rapid and easy transfer of information about any problems among Woodhaven's client base. Because of Woodhaven's small size and lack of recognition outside of the neighborhood, it would probably be difficult to continually replace alienated clientele. We may, therefore, conclude that the relative cost of cheating and/or lowering quality would be unusually high for Woodhaven Service. One would be inclined to reject any plan that would not tightly control acts that might alienate the present customers. (i) Intuitively, one can see that Honest Jack's plan would generate more dysfunctional behavior than Harold's. By paying a minimum salary of$300, Harold's commission plan would likely reduce the quantity of unnecessary services performed. First, since it is easier to meet one's financial obligations with $300 than with nothing, the mechanics would feel less of a "need" to cheat the customers. Also, since an average volume of business would earn each mechanic the same amount of money under either plan, the marginal benefit from each additional sale would be less under Harold's commission plan than under Jack's. (ii) Placing an upper bound upon potential weekly earnings would further diminish mechanics' incentive to cheat customers. It is likely that there would be diminishing marginal returns for cheating customers due to factors such as the increasing probability of getting caught, guilt feelings, or simply lack of capacity to perform further repairs. At some point, it would no longer be profitable for the mechanic to cheat. If this point occurs at a figure below the dollar figure that is set as the upper bound, then the quantity of dysfunctional behavior is the same with or without the upper bound. However, if the point at which cheating is no longer profitable is above the upper-bound dollar figure that is set, the upper bound will actually reduce the quantity of cheating. Since the upper bound, in all cases, either reduces cheating or does not change it, we would expect the upper bound to tend to lessen cheating. c. This question is related to the celebrated 1992 case in which Sears auto repair departments were accused of overcharging customers for unnecessary work. For Woodhaven Service, the cost of cheating would be very high. Indeed, one would suspect that very little cheating would be profitable for Harold, a fact that would heavily influence any decision on incentive-based compensation. However, if Harold owned the mall franchise described above, the price of cheating would be much lower. Location, brand recognition, and other factors would all make cheating cheaper. Promotion would bring in new customers to replace alienated ones. Mall shoppers would likely give the shop a try and would be unlikely to discuss their auto service experience with a large proportion of the shop's potential market. It would appear that Harold at the mall would reap all the same benefits as Harold in the neighborhood, but without the costs of cheating. We would therefore think it far more likely that Harold would install one of the incentive compensation plans described above at the mall than at Woodhaven Service. It is difficult to assess how being manager at a company-owned store would affect Harold's outlook. We need more information on the compensation plan used for such managers. On the one hand, increasing sales would be less beneficial for an employee than an owner. Lower rewards would indicate less desire to use an incentive compensation plan. On the other hand, differences in the cost of cheating for the owner and the manager are unclear. Losing your franchise may be more expensive than losing your job, but losing your job is more expensive than a letter of reprimand from the parent company. d. Another way to reduce the risk of dysfunctional behavior by mechanics is to strip them of their decision rights. Since the approval of a supervisor would be needed to perform some or all services, the mechanics would find their ability to cheat severely curtailed. However, this increased control comes with a cost. Skilled supervision would be required and knowledge would have to be transferred to the person who now has the decision rights. Indeed, the person with the decision rights would likely have to duplicate much of the mechanic's diagnostic work to confirm the mechanic's conclusions regarding service requirements. In the case of Woodhaven Service, the cost of separating knowledge and decision rights should be low. Due to the small size of the facility, there is simply not that much to supervise. After 20 years, Harold should know enough about cars and repairs to be able to decide whether or not a major service is warranted and whether it would be best to do it in-house. However, in the mall shop, there is a lot more to watch and a lot less reason to care. Therefore, it would seem that the collocation of knowledge and decision rights makes more sense at the mall than at Woodhaven. e. There are many approaches Woodhaven could take to better itself financially. One that leaps out: Raise prices. Demand is so strong that it outstrips supply, indicating that the price of regular service is too low. Furthermore, the profitability of gasoline operations suggests loyal customers who are willing to pay a premium for what is essentially a commodity item. Obviously, there is something at Woodhaven that the customers like and that Harold is not charging them for in the service end of his business. Tags Choose question tag Empowerment It is frequently argued that for empowerment to work, managers must "let go of control" and learn to live with decisions that are made by their subordinates. Evaluate this argument. Free Essay Answer: Answer: Organizational Architecture: Organizational architecture refers to the whole structure of an organization. It contains the hierarchy of management levels, their roles and responsibilities and the infrastructure facilities of the organization. It is the whole architecture of an organization which enables it to accomplish its goals and objectives and also enable to make a vision. An organization with proper segregation of duties and powers can survive better. If the whole controlling and decision-making power passed on to the workers of the organization then this may lead the organization towards the huge problematic situation. So there should be proper power distribution between management and workers. The managers of the organization should keep analytical decision-making powers with them and should provide production related powers to the workers as they are also playing a key role in the functioning of an organization. The managers are the higher authorities of the organization and they should not blindly follow the decisions of their subordinates rather make decisions their own. Tags Choose question tag Sales Commissions Sue Koehler manages a revenue center of a large national manufacturer that sells office furniture to local businesses in Detroit. She has decision rights over pricing. Her compensation is a fixed wage of$23,000 per year plus 2 percent of her office's total sales. Critically evaluate the organizational architecture of Koehler's revenue center.
Free
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Organizational Architecture:
It can be defined in two different ways. One side says that it means the built in space or area covered by the said organization. The other meaning of this term is that the environment which is given to the employees for working i.e. whether the environment is competent for the employees to enhance their working skills or not.
In the present case, it is provided that the person has full right over the control of prices of the product and the person is also given extra 2 percent on the amount of sales generated from the office of the person.
Since the person has control over the prices of the products, therefore, the person will work towards the increase in sales amount by controlling the prices. The person will have no interest in the amount of profits as generated by the company due to such sales.
The figure of 2 percent extra on the value of sales will motivate the person to work towards the accomplishment of increasing the sales value more and more.
So, therefore, the person will set the prices in such a way that the operating costs of the office are covered easily and will try to increase the value of sales according to that only.
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American InterConnect II Employee bonuses at American InterConnect (AI), a large communications firm, depend on meeting a number of targets, one of which is a revenue target. Some bonus is awarded to the group if it meets or exceeds its target revenue for the year. The bonus is also tied to meeting targets for earnings, employee satisfaction, hiring goals, and other specific objectives tailored to the group or manager. AI has several product development groups within the firm. Each is assigned the task of developing new products for specific divisions within the firm. Product developers, primarily engineers and marketing people, are assigned to a new product development team to develop a specific new product. Afterward, they are assigned to new development teams for a different division or are reassigned back to their former departments. Sometimes they become product managers for the new product. Product development teams take roughly 18 months to develop and design the product. For example, Network Solutions is one group of products AI sells. The employees in the product development group for Network Solutions receive part of their bonus based on whether Network Solutions achieves its revenue target for that year. It typically takes AI three years from the time a product development team is formed until the product reaches the market. Once a new product idea is identified and researched, a prototype must be built and tested. Finally, it is introduced. Another 18 months is required for approval, manufacturing design, further testing, and marketing. These functions are performed after the product development team has been reassigned. Required: Analyze the incentives created by basing a portion of each current product developer's bonus on revenues for products now being sold.
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Tipping One of the main tenets of economic analysis is that people act in their narrow self-interest. Why then do people leave tips in restaurants? If a study were to compare the size of tips earned by servers in restaurants on interstate highways with those in restaurants near residential neighborhoods, what would you expect to find? Why?
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White's Department Store Employees at White's Department Store are observed engaging in the following behavior: (1) They hide items that are on sale from the customers and (2) they fail to expend appropriate effort in designing merchandise displays. They are also uncooperative with one another. What do you think might be causing this behavior, and what might you do to improve the situation?
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Course Packets Faculty members at a leading business school receive a budget to cover research expenditures, software and hardware purchases, travel expenses, photocopying for classroom use, and so forth. The budget is increased $250 for each class taught (independent of the number of students enrolled). For example, a faculty member receives a base budget of$14,000 for this year and teaches three courses-hence, the faculty's total budget is $14,750. Finance professors teach much larger classes than any other functional area (e.g., accounting) and they tend to have larger course packets per student. Faculty can photocopy their course packets and have their budgets reduced by the photocopying charges. Or the faculty can distribute course materials via the school's network where students can download them and print them on their personal printers. Required: a. Which faculty members are more likely to put course packets and lecture notes on their Web pages, and which faculty are more likely to photocopy the material and distribute it to their students? b. Is this partitioning of faculty members distributing materials electronically versus making paper copies efficient? Essay Answer: Tags Choose question tag Christian Children's Fund Christian Children's Fund, Inc. (CCF), established in 1938, is an international, nonsectarian, nonprofit organization dedicated to assisting children. With program offices around the world, it provides health and educational assistance to more than 4.6 million children and families through over 1,000 projects in 30 countries, including the United States. CCF's programs promote long-term development designed to help break the cycle of poverty by improved access to health care, safe water, immunizations, better nutrition, educational assistance, literacy courses, skills training, and other services specific to improving children's welfare. Most of CCF's revenues come from individual donors who are linked with a specific child. About 75 percent of the sponsors are in the United States, and in 2003, CCF had total revenues of about$143 million. (See Exhibit 1.) In 1995, CCF began developing an evaluation system, nicknamed AIMES (Annual Impact Monitoring and Evaluation System), to assess the performance of its programs and whether they are making a positive, measurable difference in children's lives. A working group of national directors, program managers, CCF finance and audit managers, and outside consultants developed a series of metrics that allowed CCF to be more accountable to its sponsors, as well as an evaluation tool to continually assess the impact of its programs on children. The working group wanted metrics that (1) captured critical success factors for CCF's projects; (2) focused on a program's impact, not its activities; (3) measured the program's impact on children; and (4) could be measured and tracked. The following indicators were chosen: Under 5-year-old mortality rate Under 5-year-old moderate and severe malnutrition rate Adult literacy One-to-two-year-old immunizations Tetanus vaccine-protected live births Families that correctly know how to manage a case of diarrhea Families that correctly know how to manage acute respiratory infection Families that have access to safe water Families that practice safe sanitation Children enrolled in a formal or informal educational program Each family in a community with a CCF program is given a family card that tracks each of the preceding 10 indicators for that family. In 1997, the first year of implementation, AIMES captured the health status of about 1.9 million children in approximately 850 projects in 18 countries. Annual visits by project staff or volunteers update each family's card. The family cards are aggregated at the community level, national level, and then in total for CCF, and provide a reporting system. CCF managers then track trends and compare performance at the community, national, and organizational levels. It took CCF two years to develop these metrics, test them, and train the staff in all the national offices in how to use the system. AIMES does not prescribe the strategy each community should adopt but rather allows each community to design programs that promote the well-being of children in that community. Program directors can use the AIMES data as a tool to monitor and manage their programs. If child mortality is high, local program directors decide how best to reduce the rate. The 10 AIMES metrics have made project managers more focused and better able to concentrate resources in those areas that make a measurable difference in children's health. CCF uses the information to make program and resource allocation decisions at the community level. The family card has promoted better nutrition via appropriate feeding and child care practices because there is now more direct contact between CCF staff and volunteers and families. Required: Using this chapter's organizational architecture framework, discuss the strengths and weaknesses of CCF's AIMES project. SOURCES: D. Henderson, B. Chase, and B. Woodson, "Performance Measures for NPOs," Journal of Accountancy (January 2002), pp. 63-68; and www.christianchildrensfund.org.
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Pay for Performance Communities are frequently concerned about whether or not police are vigilant in carrying out their responsibilities. Several communities have experimented with incentive compensation for police. In particular, some cities have paid members of the police force based on the number of arrests that they personally make. Discuss the likely effects of this compensation policy.
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University Physician Compensation Physicians practicing in Eastern University's hospital have the following compensation agreement. Each doctor bills the patient (or Blue Cross Blue Shield) for his or her services. The doctor pays for all direct expenses incurred in the clinic, including nurses, medical malpractice insurance, secretaries, supplies, and equipment. Each doctor has a stated salary target (e.g., $100,000). For patient fees collected over the salary target, less expenses, the doctor retains 30 percent of the additional net fees. For example, if$150,000 is billed and collected from patients, and expenses of $40,000 are paid, then the doctor retains$3,000 of the excess net fees [30 percent of ($150,000 -$40,000 - $100,000)] and Eastern University receives$7,000. If $120,000 of fees are collected and$40,000 of expenses are incurred, the physician's net cash flow is $80,000 and Eastern University receives none of the fees. Required: Critically evaluate the existing compensation plan and recommend any changes. Essay Answer: Tags Choose question tag Rothwell Inc. Rothwell Inc. is the leader in computer-integrated manufacturing and factory automation products and services. The Rothwell product offering is segmented into 15 product categories, based on product function and primary manufacturing location. Rothwell's sales division sells all 15 product categories and is composed of 25 district offices located throughout the United States. The company is highly decentralized, with district offices responsible for setting sales price, product mix, and other variables. District offices are rewarded based on sales. Some large customers have plants in more than one of Rothwell's sales districts. In cases where sales are made to these customers, the district offices participate jointly and sales credits are shared by each district involved. The sales division compensation plan designed by L. L. Rothwell, founder of the firm, was structured so that the staff would pursue sales in each of the 15 product categories. The selling program has the following features: • Each sales representative receives a commission based on a percentage of the sales revenue generated. • Each district (approximately 160 sales reps) is assigned a quota for each product line, defined in terms of dollar sales. • In addition to commission, sales reps are eligible for an annual bonus. The company calculates individual bonuses by multiplying the number of bonus points earned by the individual target bonus amount. Points are credited at the district level. • In order for all sales reps in a district to qualify for a bonus, the district must achieve 50 percent of quota in all 15 product groups and 85 percent of quota in at least 13 groups. • Bonus points are awarded for sales greater than 85 percent of quota. • Five product groups have been identified as strategic to Rothwell. These "pride-level" products are weighted more heavily in bonus point calculations. Over the past three years, Rothwell generated exceptionally high sales-and awarded record bonuses. Profits, however, were lackluster. L. L. was befuddled! Required: a. Evaluate the compensation situation at Rothwell. b. Identify the types of behavior the existing system promotes and explain how such behavior may be contributing to the firm's declining profitability. Suggest improvements. Essay Answer: Tags Choose question tag Pratt Whitney A Wall Street Journal article (December 26, 1996) describes a series of changes at the Pratt Whitney plant in Maine that manufactures parts for jet engines. In 1993, it was about to be closed because of high operating costs and inefficiencies. A new plant manager overhauled operations. He broadened job descriptions so inspectors do 15 percent more work than they did five years ago. A "results-sharing" plan pays hourly employees if the plant exceeds targets such as cost cutting and on-time delivery. Now, everyone is looking to cut costs. Hourly employees also helped design a new pay scheme that is linked to the amount of training, not seniority, that an employee has. This was after the plant manager drafted 22 factory-floor employees, gave them a conference room, and told them to draft a new pay plan linked to learning. Shop-floor wages vary between$9 and $19 per hour with the most money going to people running special cost studies or quality projects, tasks previously held by managers. This text emphasizes the importance of keeping all three legs of the stool in balance. Identify the changes Pratt Whitney made to all three legs of the stool at its Maine plant. Essay Answer: Tags Choose question tag Coase Farm Coase Farm grows soybeans near property owned by Taggart Railroad. Taggart can build zero, one, or two railroad tracks adjacent to Coase Farm, yielding a net present value of$0, $9 million, or$12 million. Coase Farm can grow soybeans on zero, one, or two fields, yielding a net present value of $0,$15 million, or $18 million before any environmental damages inflicted by Taggart trains. Environmental damages inflicted by Taggart's trains are$4 million per field per track. Coase Farm's payoffs as a function of the number of fields it uses to grow soybeans and the number of tracks that Taggart builds are shown next. It is prohibitively expensive for Taggart Railroad and Coase Farm to enter into a long-term contract regarding either party's use of its land. Required: a. Suppose Taggart Railroad cannot be held liable for the damages its tracks inflict on Coase Farm. Show that Taggart Railroad will build two tracks and Coase Farm will plant soybeans in one field. b. Suppose Taggart Railroad can be held fully liable for the damages that its tracks inflict on Coase Farm. Show that Taggart Railroad will build one track and Coase Farm will plant soybeans in two fields. c. Now suppose Taggart Railroad and Coase Farm merge. Show that the merged firm will build one track and plant soybeans in one field. d. What are the implications of the merger for the organizational architecture of the newly merged firm in terms of decision rights, performance measurement, and employee compensation? SOURCE: R. Sansing.
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Vanderschmidt's Jan Vanderschmidt was the founder of a successful chain of restaurants located throughout Europe. He died unexpectedly last week at the age of 55. Jan was sole owner of the company's common stock and was known for being very authoritarian. He made most of the company's personnel decisions himself. He also made most of the decisions on the menu selection, food suppliers, advertising programs, and so on. Employees throughout the firm are paid fixed salaries and have been heavily monitored by Mr. Vanderschmidt. Jan's son, Joop, spent his youth driving BMWs around the Netherlands and Germany at high speeds. He spent little time working with his dad in the restaurant business. Nevertheless, Joop is highly intelligent and just received his MBA degree from a prestigious school. Joop has decided to follow his father as the chief operating officer of the restaurant chain. Explain how the organization's architecture might optimally change now that Joop has taken over.
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Formula 409 "I used to run the company that made Formula 409, the spray cleaner. From modest entrepreneurial beginnings, we'd gone national and shipped the hell out of P G, Colgate, Drackett, and every other giant that raised its head. From the beginning, I'd employed a simple incentive plan based on 'case sales': Every month, every salesman and executive received a bonus check based on how many cases of 409 he'd sold. Even bonuses for the support staff were based on monthly case sales. It was a happy time, with everyone making a lot of money, including me." "We abandoned our monthly case-sales bonus plan and installed an annual profit-sharing plan, based on personnel evaluations. It didn't take long for the new plan to produce results." SOURCE: Wilson Harrell, "Inspire Action: What Really Motivates Your People to Excel?" Success, September 1995. Required: What do you think happened at this company after it started the annual profit-sharing plan?
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Raises A company recently raised the pay of employees by 20 percent. The productivity of the employees, however, remained the same. The CEO of the company was quoted as saying, "It just goes to show that money does not motivate people." Provide a critical evaluation of this statement.
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Allied Van Lines Why are drivers for long-haul (cross-country) moving companies (e.g., Allied Van Lines) often franchised, while moving companies that move households within the same city hire drivers as employees? Franchised drivers own their own trucks. They are not paid a fixed salary but rather receive the profits from each move after paying the franchiser a fee.
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American InterConnect I Employee satisfaction is a major performance measure used at American Inter Connect (AI), a large communications firm. All employees receive some bonus compensation. The lower-level employees receive a bonus that averages 20 percent of their base pay, whereas senior corporate officers receive bonus pay that averages 80 percent of their base salary. Bonus payments for all employees are linked to their immediate work group's performance on the following criteria: income, revenue growth, customer satisfaction, and employee satisfaction. Managers can have these criteria supplemented with additional specific measures, including hiring targets and some other specific objective for each manager, such as meeting a new product introduction deadline or a market share target. Employee satisfaction usually has a weight of between 15 and 20 percent in determining most employees' overall bonus. To measure employee satisfaction, all employees in the group complete a two-page survey each quarter. The survey asks a variety of questions regarding employee satisfaction. One question in particular asks employees to rate how satisfied they are with their job using a seven-point scale (where 7 is very satisfied and 1 is very dissatisfied). The average score on this question for all employees in the group is used to calculate the group's overall employee satisfaction score. Required: Describe what behaviors you would expect to observe at AI.
Essay | 2022-08-13 13:12:22 | {"extraction_info": {"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, "math_score": 0.20999692380428314, "perplexity": 3386.592132942583}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571950.76/warc/CC-MAIN-20220813111851-20220813141851-00702.warc.gz"} |
https://math.stackexchange.com/questions/3121048/dice-rolls-between-three-players | # Dice Rolls Between Three Players
The setup of the game involves three players and a fair, six-sided die. The rules are: If a player rolls a six, they win the game, if they roll an odd number, they pass the die to the player on their right, and if they roll an even number other than six, they pass the die to the player on their left.
The question: What is the probability that the first person to roll the die wins?
Attempt: Broke this up into different cases, suspecting a geometric distribution. Probability that person who rolled first wins on 1st roll is 1/6, probability of winning on the 2nd roll is 0, probability of winning on the 3rd roll is 1/18, and so on. However, I couldn't establish any clear geometric pattern, is this approach a dead end?
Let $$X$$ be the event in which player $$1$$ wins the game, and $$T$$ denote the person whose turn it is to throw the die. We then find:
$$P(X | T = 1) = \frac{1}{6} + \frac{2}{6} P(X | T = 2) + \frac{3}{6} P(X | T = 3)$$
We know that:
$$P(X | T = 2) = \frac{3}{6} P(X | T = 1) + \frac{2}{6} P(X | T = 3)$$
$$P(X | T = 3) = \frac{2}{6} P(X | T = 1) + \frac{3}{6} P(X | T = 2)$$ $$= \frac{2}{6} P(X | T = 1) + \frac{3}{6} \left(\frac{3}{6} P(X | T = 1) + \frac{2}{6} P(X | T = 3)\right)$$ $$\iff P(X | T = 3) = \frac{21}{30} P(X | T = 1)$$
As such, it follows that:
$$P(X | T = 1) = \frac{1}{6} + \frac{2}{6} \left(\frac{3}{6} P(X | T = 1) + \frac{2}{6} P(X | T = 3)\right) + \frac{3}{6} P(X | T = 3)$$ $$= \frac{1}{6} + \frac{2}{6} \left(\frac{3}{6} P(X | T = 1) + \frac{2}{6} \frac{21}{30} P(X | T = 1)\right) + \frac{3}{6} \frac{21}{30} P(X | T = 1)$$ $$\iff P(X | T = 1) = \frac{30}{73} \approx 0.4110$$
This result can be confirmed using the following Python code:
from random import randint
s = [0, 0, 0]
n = 1000000
for _ in range(n):
p = 0
while True:
t = randint(1, 6)
if t == 6:
s[p % 3] += 1.0 / n
break
elif t == 2 or t == 4:
p += 2
else:
p += 1
print(s)
An example run returned:
[0.411106, 0.300798, 0.288096] | 2019-06-18 04:39:33 | {"extraction_info": {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5530075430870056, "perplexity": 707.2466931873297}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998607.18/warc/CC-MAIN-20190618043259-20190618065259-00498.warc.gz"} |
https://www.physicsforums.com/threads/angle-of-incidence-and-reflection.795870/ | Angle of incidence and reflection
1. Feb 3, 2015
Robben
I am trying to write a java code where the user inputs the height at which a solid body approaches a sphere of radius R and outputs the angle $\theta$ at which the body bounces off the sphere.
I have all my code written is just that I don't know how to derive the equation that calculates the angle $\theta$.
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2. Feb 4, 2015
lowerlowerhk
Break down the velocity vector of the solid body into two components: one is tangential to the sphere, another one points to the centre of the sphere. Reverse the direction of the central component and add the two component together and you will have the bounce off velocity vector. The angle between incident and reflected angle is what you need. Divide it by two and you will have the answer. Does it help?
3. Feb 4, 2015
4. Feb 4, 2015
5. Feb 4, 2015
Robben
I got it to work now.
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• Blah.jpg
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6. Feb 5, 2015
dmpnow
can you post the code? | 2017-11-24 16:57:07 | {"extraction_info": {"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, "math_score": 0.5927309989929199, "perplexity": 587.7210502818625}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934808260.61/warc/CC-MAIN-20171124161303-20171124181303-00441.warc.gz"} |
https://math.stackexchange.com/questions/986602/show-that-in-the-sequence-b-2b-3b-mb-there-are-exactly-gcdb-m-numbers | Show that in the sequence $b,2b,3b…,mb$ there are exactly $\gcd(b,m)$ numbers divisible by $m$.
"Where $b$ and $m$ are integers and $m>1$, show that in the sequence $b,2b,3b...,mb$ there are exactly $\gcd(b,m)$ numbers divisible by $m$.
I'm having a real hard time proving that... can anyone help me?
• $m$ needn't be bigger than one: when $m = 1$, in the sequence $b, 2b, 3b, \dots mb$, there is exactly 1 integer, and it is divisible by 1 :) – Ben Millwood Oct 22 '14 at 22:22
The given sequence is the first $m$ multiples of $b$. Scanning from left to right, when will we find the the first multiple of $b$ that is also a multiple of $m$? By definition, this will occur at $\ell = \text{lcm}(b, m)$. Letting $\ell = kb$, notice that $\ell$ is the only multiple of $m$ in the first $k$ multiples of $b$.
Now observe that the next common multiple of $b$ and $m$ will be $2\ell = 2kb$, which shows up after listing out $k$ more multiples of $b$ after $\ell$. Now since $\ell$ divides $mb$, we can continue in this manner so that the list of all common multiples of $b$ and $m$ is: $$\ell, 2\ell, \ldots, g\ell$$ where $g\ell = mb$. [Remark: At this step, we are using the fact that the least common multiple of $m$ and $b$ must divide any common multiple of $m$ and $b$.] But then since: $$\gcd(b, m) \cdot \text{lcm}(b, m) = mb$$ we have that $g = \gcd(b, m)$, as desired.
• oh, interesting, thank you! i'm curious, though, the book i'm using places this question before the question that asks you to prove that gcd(a,b) * lcm(a,b) = |a*b|. i wonder what's the other way to do it... – violeta Oct 22 '14 at 23:02
• also, b can be negative... hmm.. – violeta Oct 22 '14 at 23:31
So the sequence is the following:
$$b,2b,3b,..,mb$$
Let $g= \gcd(b,m)$ , so $ng=m$ and $gd=b$
Lets consider the following members of the sequence:
$$I= \{x \in \mathbb{N} : x=bjn =\frac{bjm}{g}\ \text{ such that } j \in \mathbb{N} \land 0<j\leq g \}$$
So all the members $I$ are divisible by $m$. The number of members is equal to $g$. Now I want to prove that this are the only members of the sequence that are divisible by $m$.
Now all the members that are not in $I$ are of the form:
$$S=\{t \in \mathbb{N} : t=b(jn+c) =\frac{bjm}{g}+bc \ \text{ such that } j \in \mathbb{N} \land 0<j< g \land 0<c<n \}$$
$$m|t \text{ such that } t\in S$$ $$m|b(jn+c)$$
by definition:
$$mk=jbn+bc \tag{1}$$ $$mk=jb\frac{m}{g} + bc$$ $$m(k-\frac{jb}{g})=bc$$ $$k-jd=\frac{bc}{m} \tag{2}$$ $$k=\frac{dc}{n}+jd$$ $$kn=njd+dc$$ $$kn=d(jn+c)$$
Now $d|k$ by the Euclid Lemma, since $(n,d)=1$ . This will be useful in a moment:
Now as $c<n$ we can obtain: $$\frac{c}{m}<\frac{1}{g} \Rightarrow \frac{bc}{m}<d$$
From $(2)$we can obtain:
$$k-jd<d$$ $$k<d+dj$$ $$k<d(1+j) \tag{3}$$
As $d|k$ then $d\hat{k}=k$ , replacing in $(3)$
$$d\hat{k}<d(1+j) \Rightarrow \hat{k}<(j+1)$$
So $\hat{k}\leq j$ then $mk\leq djm$, now in $(1)$
$$djm \geq jbn+bc$$ $$djm \geq djm+bc$$ $$0 \geq bc$$
As $b$ and $c$ are positive, $bc >0$, so contradiction.
So no element of $S$ is divisible by $m$, now only the elements of $I$ are divisible by $m$. As the number of members is $g= \gcd(b,m)$ , the statement is proven. | 2019-08-19 12:29:29 | {"extraction_info": {"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, "math_score": 0.8887962698936462, "perplexity": 107.22599345951076}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027314732.59/warc/CC-MAIN-20190819114330-20190819140330-00083.warc.gz"} |
https://brilliant.org/practice/trigonometric-even-odd-functions/ | ×
## Fundamental Trigonometric Identities
These are your basic building blocks for solving trigonometric equations and understanding how the pieces fit together. Using these identities can make sense of even the scariest looking trig.
# Even-Odd Functions
Which of the following is equal to
$\sin ( - \theta) ?$
Which of the following is equal to
$\sec ( - \theta) ?$
Which of the following is equal to
$\cot ( - \theta) ?$
Which of the following is equal to
$\csc ( - \theta) ?$
Which of the following is equal to
$\tan ( - \theta) ?$
× | 2016-09-27 08:42:15 | {"extraction_info": {"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, "math_score": 0.6965211033821106, "perplexity": 958.9211845764187}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738660996.20/warc/CC-MAIN-20160924173740-00246-ip-10-143-35-109.ec2.internal.warc.gz"} |
http://pdglive.lbl.gov/DataBlock.action?node=S067SBT | # Total Flux of Active ${}^{8}\mathrm {B}$ Solar Neutrinos INSPIRE search
Total flux of active neutrinos (${{\mathit \nu}_{{e}}}$, ${{\mathit \nu}_{{\mu}}}$, and ${{\mathit \nu}_{{\tau}}}$).
VALUE ($10^{6}$ cm${}^{-2}$s${}^{-1}$) DOCUMENT ID TECN COMMENT
• • • We do not use the following data for averages, fits, limits, etc. • • •
$5.25$ $\pm0.16$ ${}^{+0.11}_{-0.13}$ 1
2013
SNO All three phases combined
$5.046$ ${}^{+0.159}_{-0.152}$ ${}^{+0.107}_{-0.123}$ 2
2010
SNO From ${{\mathit \phi}_{{NC}}}$ in Phase I+II, low threshold
$5.54$ ${}^{+0.33}_{-0.31}$ ${}^{+0.36}_{-0.34}$ 3
2008
SNO $\phi _{NC}$ in Phase III
$4.94$ $\pm0.21$ ${}^{+0.38}_{-0.34}$ 4
2005 A
SNO From ${{\mathit \phi}_{{NC}}}$; ${}^{8}\mathrm {B}$ shape not const.
$4.81$ $\pm0.19$ ${}^{+0.28}_{-0.27}$ 4
2005 A
SNO From ${{\mathit \phi}_{{NC}}}$; ${}^{8}\mathrm {B}$ shape constrained
$5.09$ ${}^{+0.44}_{-0.43}$ ${}^{+0.46}_{-0.43}$ 5
2002
SNO Direct measurement from ${{\mathit \phi}}_{\mathit NC}$
$5.44$ $\pm0.99$ 6
2001
Derived from SNO+SuperKam, water Cherenkov
1 AHARMIM 2013 obtained this result from a combined analysis of the data from all three phases, SNO-I, II, and III. The measurement of the ${}^{8}\mathrm {B}$ flux mostly comes from the NC signal, however, CC contribution is included in the fit.
2 AHARMIM 2010 reports this result from a joint analysis of SNO Phase I+II data with the "effective electron kinetic energy" threshold of 3.5 MeV. This result is obtained with the assumption of unitarity, which relates the NC, CC, and ES rates. The data were fit with the free parameters directly describing the total ${}^{8}\mathrm {B}$ neutrino flux and the energy-dependent ${{\mathit \nu}_{{e}}}$ survival probability.
3 AHARMIM 2008 reports the results from SNO Phase III measurement using an array of ${}^{3}\mathrm {He}$ proportional counters to measure the rate of NC interactions in heavy water, over the period between November 27, 2004 and November 28, 2006, corresponding to 385.17 live days. A simultaneous fit was made for the number of NC events detected by the proportional counters and the numbers of NC, CC, and ES events detected by the PMTs, where the spectral distributions of the ES and CC events were not constrained to the ${}^{8}\mathrm {B}$ shape.
4 AHARMIM 2005A measurements were made with dissolved NaCl (0.195$\%$ by weight) in heavy water over the period between July 26, 2001 and August 28, 2003, corresponding to 391.4 live days, and update AHMED 2004A. The CC, ES, and NC events were statistically separated. In one method, the ${}^{8}\mathrm {B}$ energy spectrum was not constrained. In the other method, the constraint of an undistorted ${}^{8}\mathrm {B}$ energy spectrum was added for comparison with AHMAD 2002 results.
5 AHMAD 2002 determined the total flux of active ${}^{8}\mathrm {B}$ solar neutrinos by directly measuring the neutral-current reaction, ${{\mathit \nu}_{{{{\mathit \ell}}}}}$ ${{\mathit d}}$ $\rightarrow$ ${{\mathit n}}{{\mathit p}}{{\mathit \nu}_{{{{\mathit \ell}}}}}$ , which is equally sensitive to ${{\mathit \nu}_{{e}}}$, ${{\mathit \nu}_{{\mu}}}$, and ${{\mathit \nu}_{{\tau}}}$. The complete description of the SNO Phase I data set is given in AHARMIM 2007 .
6 AHMAD 2001 deduced the total flux of active ${}^{8}\mathrm {B}$ solar neutrinos by combining the SNO charged-current result (AHMAD 2001 ) and the Super-Kamiokande ${{\mathit \nu}}{{\mathit e}}$ elastic-scattering result (FUKUDA 2001 ).
References:
AHARMIM 2013
PR C88 025501 Combined Analysis of all Three Phases of Solar Neutrino Data from the Sudbury Neutrino Observatory
AHARMIM 2010
PR C81 055504 Low-Energy-Threshold Analysis of the Phase I and Phase II Data Sets of the Sudbury Neutrino Observatory
AHARMIM 2008
PRL 101 111301 Independent Measurement of the Total Active ${}^{8}\mathrm {B}$ Solar Neutrino Flux Using an Array of ${}^{3}\mathrm {He}$ Proportional Counters at the Sudbury Neutrino Observatory
AHARMIM 2005A
PR C72 055502 Search for Periodicities in the ${}^{8}\mathrm {B}$ Solar Neutrino Flux Measured by the Sudbury Neutrino Observatory
PRL 87 071301 Measurement of Charged Current Interactions Produced by ${}^{8}\mathrm {B}$ Solar Neutrinos at the SUDBURY Neutrino Observatory
PRL 92 181301 Measurement of the Total Active ${}^{8}\mathrm {B}$ Solar Neutrino Flux at the Sudbury Neutrino Observatory with Enhanced Neutral Current Sensitivity
PRL 86 5651 Solar ${}^{8}\mathrm {B}$ and HEP Neutrino Measurements from 1258 Days of Super-Kamiokande Data
PR C75 045502 Determination of the ${{\mathit \nu}_{{e}}}$ and Total ${}^{8}\mathrm {B}$ Solar Neutrino Fluxes using the Sudbury Neutrino Observatory Phase I Data Set | 2019-01-18 16:23:56 | {"extraction_info": {"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, "math_score": 0.7225661277770996, "perplexity": 3111.3579824220783}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583660175.18/warc/CC-MAIN-20190118151716-20190118173716-00433.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/53944-algebraic-geometry-print.html | # Algebraic Geometry
• October 15th 2008, 05:46 PM
terr13
Algebraic Geometry
Take two affine varieties in A^2(Q)
V_1 = {(x,y): x^2 = y^3}
V_2 = {(u,v) : u^3 = y^4}
with function fields Q(V_1) and Q(V_2)
a.) Show the function fields are isomorphic as Q-algebras
b.) Construct an explicit Birational map V_1 --> V_2 (dotted line)
Also,
Consider the lines
V_1={x=y=0}
V_2={y=z=0}
V_3={z=x=0}
Show that the product Ideal I(V_1)I(V_2)I(V_3) is smaller than the intersection of I(V_1), I(V_2), I(V_3), even though they define the same variety
b.) Suppose I,J \subset R are ideals such that I + J= R. Show that IJ = the intersection of I and J.
• October 19th 2008, 09:37 PM
kalagota
Quote:
Originally Posted by terr13
b.) Suppose I,J \subset R are ideals such that I + J= R. Show that IJ = the intersection of I and J.
well, i will assume that R is a commutative ring..
note that $I\cap J \subset J$ and $I\cap J \subset I$
also, $I\cdot R = I$ for any ideal $I$ of R.
thus $I\cap J = (I\cap J) \cdot R = (I\cap J) \cdot (I+J) = (I\cap J) \cdot I + (I\cap J) \cdot J \subset JI + IJ = IJ$
for the other direction, if $a \in IJ$, then
$a = \sum x_iy_i$ (finite sum) where $x_i \in I$ and $y_i \in J$
properties of ideals will show you that $a$ is also in $I\cap J$ | 2016-05-30 16:10:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9575384855270386, "perplexity": 2415.8815970072415}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464051035374.76/warc/CC-MAIN-20160524005035-00231-ip-10-185-217-139.ec2.internal.warc.gz"} |
http://akteam.top/wordpress/index.php/2018/05/27/lct/ | #### Splay时间复杂度证明
###### 证明.
$2\lg(x)-\lg(y)-\lg(z)$
$=\lg((y+z+1)^2)-\lg(y)-\lg(z)$
$=\lg(\frac{(y+z+1)^2}{yz})$
$=\lg(\frac{2yz}{yz}+\frac{y^2+z^2+1+2y+2z}{yz})$
$\geq \lg(2+\frac{y}{z}+\frac{z}{y})$
$\geq \lg(2+2)$
$=2$
###### 证明.
$r'(x)+r'(y)+r'(z)-r(x)-r(y)-r(z)$
$=r'(y)+r'(z)-r(x)-r(y)$
$\leq r'(x)+r'(z)-2r(x)$ 由于$size'(x)=size(x)+size'(z)+1$,根据定理0,加上2r'(x)-r'(z)-r(x)-2。
$\leq 3(r'(x)+r(x))-2$
$r'(x)+r'(y)+r'(z)-r(x)-r(y)-r(z)$
$=r'(y)+r'(z)-r(x)-r(y)$
$\leq r'(y)+r'(z)-2r(x)$,由于size'(x)=size'(y)+size'(z)+1,根据定理0,加上2r'(x)-r'(y)-r'(z)-2
$\leq 2(r'(x)-r(x))-2$
### 2条评论
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https://dmoj.ca/problem/coci15c1p6 | ## COCI '15 Contest 1 #6 Uzastopni
View as PDF
Points: 20 (partial)
Time limit: 0.3s
Memory limit: 32M
Problem types
Petar is throwing a birthday party and he decided to invite some of the employees of his company where he is the CEO.
Each employee, including Petar, has a unique label from to , and an accompanying type of jokes they tell . Also, each employee of the company except Petar has exactly one supervisor.
Since Petar is the CEO of the company, he has the label and is directly or indirectly superordinate to all the employees.
At the birthday party, there are certain rules that all people present (including Petar) must follow.
• At the party, there shouldn't be two people that tell the same type of jokes.
• Person cannot be invited if their direct supervisor is not invited.
• Person cannot be invited if the set of jokes the invitees that person is superior to (directly or indirectly) tell and person don't form a set of consecutive numbers. The numbers in the set are consecutive if the difference between adjacent elements is exactly 1 when the set is sorted ascendingly. For example, and .
Petar wants to know how many different sets of jokes he can see at his party with the listed constraints.
#### Input Specification
The first line of input contains the integer , .
The second line of input contains integers, the types of jokes person tells, .
Each of the following lines contains two integers and , , denoting that person is directly superior to person .
In test cases worth 50% of total points, it will hold that does not exceed .
#### Output Specification
The first and only line of output must contain the number of different sets of jokes that comply to the previously listed constraints.
#### Sample Input 1
4
2 1 3 4
1 2
1 3
3 4
#### Sample Output 1
6
#### Explanation of Sample Output 1
It is possible to have the following sets of jokes at the party:
, , , , , .
#### Sample Input 2
4
3 4 5 6
1 2
1 3
2 4
#### Sample Output 2
3
#### Explanation for Sample Output 2
The only possible sets of jokes are: , , . Notice that the person telling joke cannot be at the party because, in that case, the set of jokes is not a set of consecutive numbers.
#### Sample Input 3
6
5 3 6 4 2 1
1 2
1 3
1 4
2 5
5 6
#### Sample Output 3
10 | 2021-08-06 03:14:04 | {"extraction_info": {"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, "math_score": 0.34470510482788086, "perplexity": 1321.7546975949808}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046152112.54/warc/CC-MAIN-20210806020121-20210806050121-00001.warc.gz"} |
https://gist.github.com/gistlyn/3fee3e0e37c4e82627ebcfe938feb2ef | Instantly share code, notes, and snippets.
# gistlyn/action\3d_rotation.svg Last active Jun 16, 2019
Action Material Icons
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Sorry, this file is invalid so it cannot be displayed. | 2020-02-19 20:15:25 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8578349351882935, "perplexity": 1520.3079296326475}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875144167.31/warc/CC-MAIN-20200219184416-20200219214416-00243.warc.gz"} |
https://jkcs.or.kr/journal/view.php?doi=10.4191/kcers.2015.52.5.338 | J. Korean Ceram. Soc. > Volume 52(5); 2015 > Article
Journal of the Korean Ceramic Society 2015;52(5): 338. doi: https://doi.org/10.4191/kcers.2015.52.5.338
Anode-supported Type SOFCs based on Novel Low Temperature Ceramic Coating Process Jong-Jin Choi, Cheol-Woo Ahn, Jong-Woo Kim, Jungho Ryu, Byung-Dong Hahn, Woon-Ha Yoon, Dong-Soo Park Functional Ceramics Department, Korea Institute of Materials Science ABSTRACT To prevent an interfacial reaction between the anode and the electrolyte layer during the conventional high-temperature co-firing process, an anode-supported type cell with a thin-film electrolyte was fabricated by low-temperature ceramic thick film coating process. Ni-GDC cermet composite was used as the anode material and YSZ was used as the electrolyte material. Open circuit voltage and maximum power density were found to strongly depend on the surface uniformity of the anode functional layer. By optimizing the microstructure of the anode functional layer, the open circuit voltage and maximum powder density of the cell increased to 1.11 V and $1.35W/cm^2$, respectively, at $750^{circ}C$. When a GDC barrier layer was applied between the YSZ electrolyte and the LSCF cathode, the cell showed good stability, with almost no degradation up to 100 h. Anode-supported type SOFCs with high performance and good stability were fabricated using a coating process. Key words: SOFC, Anode-supported type, Coating, Low temperature process
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1,509 View | 2021-12-03 06:26:22 | {"extraction_info": {"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, "math_score": 0.2002590149641037, "perplexity": 13021.012079468}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964362605.52/warc/CC-MAIN-20211203060849-20211203090849-00034.warc.gz"} |
https://www.acmicpc.net/problem/13563 | 시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
1 초 512 MB 0 0 0 0.000%
## 문제
Taeyang has to compute a very complex expression using his calculator. He wants to use the memory facility in the calculator, but there is only one memory cell since his calculator is a quite old model. However, he wants to maximize the usability of the single memory cell by finding the largest common subexpression for a given input expression.
For example, look at the following input expression,
1 + (2 × 3) + 5 / (1 + 2 × 3).
This expression has two occurrences of 2 × 3, so it is a common subexpression. There are another two occurrences of 1 + 2 × 3. Thus this expression has two common subexpressions, 2 × 3 and 1 + 2 × 3. The largest (or longest) common subexpression is 1 + 2 × 3. Note that the first pair of parentheses (2 × 3) in the input expression is superfluous since 1 + (2 × 3) is same as 1 + 2 × 3 by the precedence between operators. Taeyang wants to find a largest common subexpression and to store the result of it into the single memory cell. You write a program to help Taeyang to find the largest common subexpression for a given expression.
To simplify the problem, the only four types of binary operators are considered, i.e. the addition (+), the subtraction (−), the multiplication (×), and the division (/). The usual precedence rules (the multiplication and the division have higher precedence over the addition and the subtraction) are applied and all the operators are assumed left-associative. Hence, the expression tree for the above expression can be depicted as in Figure 1.
Figure 1. The expression tree for 1 + (2 × 3) + 5 / (1 + 2 × 3).
When you construct the expression tree, you should apply the precedence and the associativity of operators (Rule 1), and also preserve the order of operands in the input expression, i.e., the order in the expression must be the same as that of in-order traversal in the tree (Rule 2). The expression tree is an ordered, single-rooted tree; a subexpression corresponds to a unique subtree of the expression tree whose leaf nodes represent the operands in the same order occurred in the expression. Thus a common subexpression exists if there are two disjoint, identical subtrees in the expression tree. Finally, the common subexpression should not be overlapped (Rule 3). Even though Rule 3 is a natural consequence of Rule 1, this rule is specified for clarity.
The size of an expression is the number of operators and operands included in the expression, which corresponds to the number of nodes of the expression tree. Therefore, the parentheses for clarifying the bindings between operators and the operands are not counted for the size. For instance, the size of the expression (50 + 50) × ((2) + 3) is seven even though it contains many pairs of parentheses. Note that the common subexpression should be proper, i.e. the size of it is less than that of the whole expression. As a result, finding a largest common expression in a given expression is equivalent to finding two disjoint same subtrees in the corresponding expression tree satisfying Rule 1, 2, and 3 with the maximum size.
Let us consider, for example, an expression (1 + 1) × (1 + 1) × (1 + 1) × (1 + 1) × (1 + 1) . For this expression, you may think a largest common subexpression is (1 + 1) × (1 + 1) × (1 + 1) × (1 + 1), but its two occurrences overlap each other, so this violates Rule 3. You also think a common subexpression would be (1 + 1) × (1 + 1), but it is not because there are no two disjoint same subtrees corresponding to two disjoint occurrences of (1 + 1) × (1 + 1) in the expression tree in Figure 2. Actually 1 + 1 is a unique common subexpression in the given expression, so it is the largest.
Figure 2. The expression tree for (1 + 1) × (1 + 1) × (1 + 1) × (1 + 1) × (1 + 1).
As a final example, the common subexpression of (1 + 2 × 3) + 5 / (2 × 3 + 1) should be 2 × 3 instead of 1 + 2 × 3 or 2 × 3 + 1 because the order of the operands are different, which violates Rule 2.
## 입력
Your program is to read from standard input. The input consists of a single line containing the input expression. It consists of the operands (positive integers), the binary operators (+, -, *, /), and parentheses. The maximum length of the input line is 1,000 including the newline character. For the multiplication, the symbol * is used instead of ×. The operators, the operands, and the parentheses can be separated by zero or more white spaces. Assume that the input expression contains at least one common subexpression and also assume that the size of it should be greater than or equal to three.
## 출력
Your program is to write to standard output. Print the postfix for the largest common subexpression in a single line. The postfix of an expression is the result of the post-order traversal of the expression tree. For instance, the postfix of the expression 1 + 2 * 3 is 1 2 3 * +. In the output, the operators and the operands should be separated by a space. If there exist different largest common subexpressions, print arbitrary one of them. If your input is an invalid expression, your program should print ERROR.
## 예제 입력 1
1 + (2 * 3) + 5 / (1 + 2 * 3)
## 예제 출력 1
1 2 3 * +
## 예제 입력 2
(1+1)*(1*1)*(1+1)*(1*1)*(1+1)
## 예제 출력 2
1 1 *
## 예제 입력 3
1 +(12*(3)) +(4+1)/((12)*3+1)
## 예제 출력 3
12 3 *
## 예제 입력 4
(1+1)*(1*1)*(1+1)//(1*1)*(1+1)
## 예제 출력 4
ERROR
## 예제 입력 5
1 +(12*(3)))+(4+1)/((12)*3+1)
## 예제 출력 5
ERROR | 2018-09-25 17:20:24 | {"extraction_info": {"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, "math_score": 0.3533836603164673, "perplexity": 547.7560149573284}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267161902.89/warc/CC-MAIN-20180925163044-20180925183444-00497.warc.gz"} |
https://electronics.stackexchange.com/questions/149105/implementing-logic-gates-in-cmos | # Implementing logic gates in CMOS
I'm trying to build the below function with CMOS, is my implementation correct?
$$F = ABC + (\overline{B+C})D$$
I am having trouble with the $$(\overline{B+C})$$ in all of the examples I've seen the function is in the form $$F = \overline{blablabla}$$ (the inverse of the whole expression).
I gave it a try but I'm not sure if it's correct, for example is it ok to have ~A as input to a PMOS (I don't see why not).
The correct layout after Dave Tweed pointed out the missing connections on the N-block.
(The added connections are marked with pink)
Yes, your solution is very nearly correct. Here are the steps, which you really should have shown in your question:
In order to deal with the second top-level term, you need to apply De Morgan's Law, which states:
$$\overline{A \cdot B} = \overline{A} + \overline{B}$$
and
$$\overline{A + B} = \overline{A} \cdot \overline{B}$$
Using this, you can make the following transformation:
$$(\overline{B + C}) \cdot D = \overline{B} \cdot \overline{C} \cdot D$$
This transforms the entire function into:
$$F = A \cdot B \cdot C + \overline{B} \cdot \overline{C} \cdot D$$
which is a normal sum-of-products expression.
In order to implement this in CMOS, however, you need a function that has an overall inversion, so you need to apply the law again:
$$F = \overline{\overline{(A \cdot B \cdot C)} \cdot \overline{(\overline{B} \cdot \overline{C} \cdot D)}}$$
and again (two places):
$$F = \overline{(\overline{A} + \overline{B} + \overline{C}) \cdot (B + C + \overline{D})}$$
Your schematic diagram is correct, but your layout does not quite match it. There are a few missing connections on the NMOS side.
From what I remember of this logic manipulation there an interim step not being shown that may help. This is transforming ~(B+C). Doing the double ~ transform can convert it to (~B~C). So you can combine the equation's right side as ~B~CD. (You already have that part implemented on the upper right of the schematic).
So the equation can be rewritten: F = (ABC) + (~B~CD) . In this form it may be easier to verify the "discrete" CMOS implementation. The function is now the OR'ed inputs of two, three input AND'ed groups.
I hope this helps at least partially.
• Ah, Morgan. That's the guy I couldn't remember. And I must say, isn't it a lovely Morgan on this side of the pond.....? – Nedd Jan 14 '15 at 12:35
• I sort of liked the F=~blablabla part. But I thought we were only using ABCD, so what the l ? So for now I'll just get the l out of here... – Nedd Jan 14 '15 at 12:44 | 2020-01-28 00:11:28 | {"extraction_info": {"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, "math_score": 0.6801729202270508, "perplexity": 674.5621914859365}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251737572.61/warc/CC-MAIN-20200127235617-20200128025617-00457.warc.gz"} |
https://gauravtiwari.org/education/puzzles/?amp=1 | # Puzzles
## 10 Fun puzzles to enjoy today
1st April is a day of fun, humor and enjoyment. Even though, I have lot of fun puzzles on this blog, these ten are totally meant for today. Enjoy! PUZZLES 1. If you are in a dark room with a candle, a wood stove and a gas lamp. You only have one matchstick in the matchbox, so what do you light first? 2. If the question you answered
## What’s the question, if the answer is ‘No!’
Infinitely many answers questions are possible to the answer, “No”. So, our real task should be to find one of THOSE many, which seems to be a perfect one. A simple and the first ever logical approach of giving answers to a question is to derive answers from the question, that is, replace some words of the question with reasonable ones and make a statement. (Conversely but )
## Light the bulb: An everyday logic puzzle
You are inside a room and there are exactly three electric bulbs outside of the room. The three bulbs have their corresponding switches (exactly three) inside the room. You can turn the switches on and off and leave them in any position. How would you identify which switch corresponds to which electric bulb, if you are allowed to go outside and come inside the room only
## Smart Fallacies: i=1, 1= 2 and 1= 3
This mathematical fallacy is due to a simple assumption, that $-1=\dfrac{-1}{1}=\dfrac{1}{-1}$ . Proceeding with $\dfrac{-1}{1}=\dfrac{1}{-1}$ and taking square-roots of both sides, we get: $\dfrac{\sqrt{-1}}{\sqrt{1}}=\dfrac{\sqrt{1}}{\sqrt{-1}}$ Now, as the Euler’s constant $i= \sqrt{-1}$ and $\sqrt{1}=1$ , we can have $\dfrac{i}{1}=\dfrac{1}{i} \ldots \{1 \}$ $\Rightarrow i^2=1 \ldots \{2 \}$ . This is complete contradiction to the fact that $i^2=-1$ . Again,
## Three Children, Two Friends and One Mathematical Puzzle
Two close friends, Robert and Thomas, met again after a gap of several years. Robert Said: I am now married and have three children. Thomas Said: That’s great! How old they are? Robert: Thomas! Guess it yourself with some clues provided by me. The product of the ages of my children is 36. Thomas: Hmm… Not so helpful clue. Can you please give one more? Robert:
## How Genius You Are?
Let have a Test: You need to make a calculation. Please do neither use a calculator nor a paper. Calculate everything “in your brain”. Take 1000 and add 40. Now, add another 1000. Now add 30. Now, add 1000 again. Add 20. And add 1000 again. And an additional 10. So, You Got The RESULT! Quicker you see the answer, sharper you are!
## A Problem On Several Triangles
A triangle $T$ is divided into smaller triangles such that any two of the smaller triangles either have no point in common, or have a vertex in common, or actually have an edge in common. Thus no two smaller triangles touch along part of an edge of them. For an illustration let me denote the three vertices of T by 1, 2 and 3.
## Two Interesting Math Problems
Problem1: Smallest Autobiographical Number: A number with ten digits or less is called autobiographical if its first digit (from the left) indicates the number of zeros it contains,the second digit the number of ones, third digit number of twos and so on. For example: 42101000 is autobiographical. Find, with explanation, the smallest autobiographical number. Solution of Problem 1 Problem 2: Fit Rectangle: A rectangle has dimensions
## How many apples did each automattician eat?
Four friends Matt, James, Ian and Barry, who all knew each other from being members of the Automattic, called Automatticians, sat around a table that had a dish with 11 apples in it. The chat was intense, and they ended up eating all the apples. Everybody had at least one apple, and everyone know that fact, and each automattician knew the number of apples he ate.
## Fox-Rabbit Chase Problem [Solution & Math Proof]
Part I: A fox chases a rabbit. Both run at the same speed $v$ . At all times, the fox runs directly toward the instantaneous position of the rabbit , and the rabbit runs at an angle $\alpha$ relative to the direction directly away from the fox. The initial separation between the fox and the rabbit is $l$ . When and where
## Applications of Complex Number Analysis to Divisibility Problems
Prove that ${(x+y)}^n-x^n-y^n$ is divisible by $xy(x+y) \times (x^2+xy+y^2)$ if $n$ is an odd number not divisible by $3$ . Prove that ${(x+y)}^n-x^n-y^n$ is divisible by $xy(x+y) \times {(x^2+xy+y^2)}^2$ if $n \equiv \pmod{6}1$ Solution 1.Considering the given expression as a polynomial in $y$ , let us put $y=0$ . We see that at $y=0$
## A Problem (and Solution) from Bhaskaracharya’s Lilavati
I was reading a book on ancient mathematics problems from Indian mathematicians. Here I wish to share one problem from Bhaskaracharya‘s famous creation Lilavati. Who was Bhaskaracharya? Bhaskara II, who is popularly known as Bhaskaracharya, was an Indian mathematician and astronomer from the 12th century. He’s especially known at the discovery of the fundamentals of differential calculus and its application to astronomical problems and computations. What
## The Mystery of the Missing Money – One Rupee
Puzzle Two women were selling marbles in the market place — one at three for a Rupee and other at two for a Rupee. One day both of then were obliged to return home when each had thirty marbles unsold. They put together the two lots of marbles and handing them over to a friend asked her to sell then at five for 2 Rupees. According
## Solving Ramanujan’s Puzzling Problem
Consider a sequence of functions as follows:- $f_1 (x) = \sqrt {1+\sqrt {x} }$ $f_2 (x) = \sqrt{1+ \sqrt {1+2 \sqrt {x} } }$ $f_3 (x) = \sqrt {1+ \sqrt {1+2 \sqrt {1+3 \sqrt {x} } } }$ ……and so on to \$ f_n (x) = \sqrt {1+\sqrt{1+2 \sqrt {1+3 \sqrt {\ldots \sqrt {1+n \sqrt {x} } } } } | 2021-10-28 11:40:54 | {"extraction_info": {"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, "math_score": 0.614445686340332, "perplexity": 1185.0230487159295}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588284.71/warc/CC-MAIN-20211028100619-20211028130619-00561.warc.gz"} |
https://questions.examside.com/past-years/jee/question/pa-balloon-has-mass-of-10-mathrmg-in-air-the-air-jee-main-physics-motion-s6ozfyfxsotigba9 | 1
JEE Main 2022 (Online) 28th July Morning Shift
+4
-1
A balloon has mass of $$10 \mathrm{~g}$$ in air. The air escapes from the balloon at a uniform rate with velocity $$4.5 \mathrm{~cm} / \mathrm{s}$$. If the balloon shrinks in $$5 \mathrm{~s}$$ completely. Then, the average force acting on that balloon will be (in dyne).
A
3
B
9
C
12
D
18
2
JEE Main 2022 (Online) 28th July Morning Shift
+4
-1
The force required to stretch a wire of cross-section $$1 \mathrm{~cm}^{2}$$ to double its length will be : (Given Yong's modulus of the wire $$=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}$$)
A
$$1 \times 10^{7} \mathrm{~N}$$
B
$$1.5 \times 10^{7} \mathrm{~N}$$
C
$$2 \times 10^{7} \mathrm{~N}$$
D
$$2.5 \times 10^{7} \mathrm{~N}$$
3
JEE Main 2022 (Online) 27th July Evening Shift
+4
-1
A steel wire of length $$3.2 \mathrm{~m}\left(\mathrm{Y}_{\mathrm{s}}=2.0 \times 10^{11} \,\mathrm{Nm}^{-2}\right)$$ and a copper wire of length $$4.4 \mathrm{~m}\left(\mathrm{Y}_{\mathrm{c}}=1.1 \times 10^{11} \,\mathrm{Nm}^{-2}\right)$$, both of radius $$1.4 \mathrm{~mm}$$ are connected end to end. When stretched by a load, the net elongation is found to be $$1.4 \mathrm{~mm}$$. The load applied, in Newton, will be: $$\quad\left(\right.$$ Given $$\pi=\frac{22}{7}$$)
A
360
B
180
C
1080
D
154
4
JEE Main 2022 (Online) 27th July Morning Shift
+4
-1
Two cylindrical vessels of equal cross-sectional area $$16 \mathrm{~cm}^{2}$$ contain water upto heights $$100 \mathrm{~cm}$$ and $$150 \mathrm{~cm}$$ respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process, is [Take, density of water $$=10^{3} \mathrm{~kg} / \mathrm{m}^{3}$$ and $$\mathrm{g}=10 \mathrm{~ms}^{-2}$$ ] :
A
0.25 J
B
1 J
C
8 J
D
12 J
JEE Main Subjects
Physics
Mechanics
Electricity
Optics
Modern Physics
Chemistry
Physical Chemistry
Inorganic Chemistry
Organic Chemistry
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EXAM MAP
Joint Entrance Examination | 2023-04-01 15:04:47 | {"extraction_info": {"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, "math_score": 0.7349626421928406, "perplexity": 2091.8747560540146}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296950030.57/warc/CC-MAIN-20230401125552-20230401155552-00271.warc.gz"} |
http://wikimechanics.org/spatial-orientation | Spatial Orientation
The following numbers are used to describe a particle's position and the direction of its motion.
## Magnetic Polarity
$\delta _{\hat{m}} \equiv \begin{cases} +1 \ \ &\sf{\text{if}} \ \ &\it{N}^{\, \mathsf{M}} > \it{N}^{\, \mathsf{A}} \\ \ \ 0 \ \ &\sf{\text{if}} \ \ &\it{N}^{\, \mathsf{M}} \ \sf{=} \ \, \it{N}^{\, \mathsf{A}} \\ -1 \ \ &\sf{\text{if}} \ \ &\it{N}^{\, \mathsf{M}} < \it{N}^{\, \mathsf{A}} \end{cases}$
Let particle P be characterized by $N^{ \mathsf{M}}$ and $N^{\mathsf{A}}$ the coefficients of its muonic seeds. Definition: the number $\delta _{\hat{m}}$ is called the magnetic polarity of P. If $\delta _{\hat{m}}=+1$ then we say that P is oriented to the north. If austral seeds predominate then we say that P is more to the south. If $\delta _{\hat{m}}= \,0$ we say that P is is not oriented on the magnetic axis.
Sensory interpretation: Muonic seeds are objectified from red and green Anaxagorean sensations. So $\delta _{\hat{m}}$ is a binary description of whether a complicated visual sensation is more reddish or greenish. If $\delta _{\hat{m}}= \,0$ we say that P is not an organic chromatic sensation.
## Electric Polarity
$\delta _{\hat{e}} \equiv \begin{cases} +1 \ \ &\sf{\text{if}} \ \ &\it{N}^{\, \mathsf{G}} > \it{N}^{\, \mathsf{E}} \\ \ \ 0 \ \ &\sf{\text{if}} \ \ &\it{N}^{\, \mathsf{G}} \ \sf{=} \ \, \it{N}^{\, \mathsf{E}} \\ -1 \ \ &\sf{\text{if}} \ \ &\it{N}^{\, \mathsf{G}} < \it{N}^{\, \mathsf{E}} \end{cases}$
Let particle P be characterized by $N^{ \mathsf{E}}$ and $N^{\mathsf{G}}$ the coefficients of its electronic seeds. Definition: the number $\delta _{\hat{e}}$ is called the electric polarity of P. If good seeds predominate then we say that P is oriented in a positive direction. If evil seeds are more prominent then we say that P is oriented in a negative direction. If $\delta _{\hat{e}}= \,0$ we say that P is not oriented on the electric axis.
Sensory interpretation: Electronic seeds are objectified from yellow and blue Anaxagorean sensations. So $\delta _{\hat{e}}$ is a binary description of whether a complex visual sensation is more yellowish or bluish. If $\delta _{\hat{e}}= \,0$ we say that P is not an inorganic chromatic sensation. If both $\delta _{\hat{m}}= \,0$ and $\delta _{\hat{e}}= \,0$ we say that P is an achromatic sensation.
## Helicity
$\delta _{z} \equiv \begin{cases} +1 \ \ &\sf{\text{if}} \ \ &\mathrm{\Delta}n^{\mathsf{U}} > \mathrm{\Delta}n^{\mathsf{D}} \\ \ \ 0 \ \ &\sf{\text{if}} \ \ &\mathrm{\Delta}n^{\mathsf{U}} \ \sf{=} \ \, \mathrm{\Delta}\it{n}^{\mathsf{D}} \\ -1 \ \ &\sf{\text{if}} \ \ &\mathrm{\Delta}n^{\mathsf{U}} < \mathrm{\Delta}n^{\mathsf{D}} \end{cases}$
Let P be characterized by ${\Delta}n^{\mathsf{U}}$ and ${\Delta}n^{\mathsf{D}}$ the coefficients of its rotating quarks. Definition: the number $\delta _{z}$ is called the helicity of P. If $\delta _{z} > 0$ we say that P's orbit is a clockwise helix. If $\delta _{z}<0$ we say P has a counterclockwise helical orbit. And if $\delta _{z}=0$ we say that P's orbit is not oriented with respect to the polar axis. By convention $\delta _{z}=1$ for ordinary material particles.
Sensory interpretation: Rotating quarks are objectified from perceptions of brightness, and distinguished between left and right. So $\delta _{z}$ could be interpreted as some simplified account of the direction and degree of illumination.
Next step: classification of shapes.
page revision: 296, last edited: 24 Aug 2018 20:35 | 2018-10-24 02:57:58 | {"extraction_info": {"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, "math_score": 0.8559590578079224, "perplexity": 909.9730904041786}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583518753.75/warc/CC-MAIN-20181024021948-20181024043448-00156.warc.gz"} |
https://labs.tib.eu/arxiv/?author=A.%20Vainchtein | • ### Lattices With Internal Resonator Defects(1804.04733)
April 16, 2018 nlin.PS
We consider a variety of settings involving chains with one or more defects stemming from the introduction of nodes bearing internal resonators. Motivated by experimental results in woodpile elastic lattices with one or two defects, we consider a variety of different theoretical scenarios. These include multi-defect chains and their ability to transmit, reflect, and especially trap energy; they also include settings with linear vs. nonlinear defects of variable interaction exponent. Moreover, they involve defects which are spatially separated and either statically, or more effectively dynamically, enable the confinement of energy between the separated defects. Wherever possible, comparisons of the experiments with numerical simulations, as well as with theoretical intuition are also offered, to provide a justification for the observed findings.
• ### A Unifying Perspective: Solitary Traveling Waves As Discrete Breathers And Energy Criteria For Their Stability(1701.04882)
Aug. 22, 2017 nlin.PS
In this work, we provide two complementary perspectives for the (spectral) stability of solitary traveling waves in Hamiltonian nonlinear dynamical lattices, of which the Fermi-Pasta-Ulam and the Toda lattice are prototypical examples. One is as an eigenvalue problem for a stationary solution in a co-traveling frame, while the other is as a periodic orbit modulo shifts. We connect the eigenvalues of the former with the Floquet multipliers of the latter and based on this formulation derive an energy-based spectral stability criterion. It states that a sufficient (but not necessary) condition for a change in the wave stability occurs when the functional dependence of the energy (Hamiltonian) $H$ of the model on the wave velocity $c$ changes its monotonicity. Moreover, near the critical velocity where the change of stability occurs, we provide explicit leading-order computation of the unstable eigenvalues, based on the second derivative of the Hamiltonian $H"(c_0)$ evaluated at the critical velocity $c_0$. We corroborate this conclusion with a series of analytically and numerically tractable examples and discuss its parallels with a recent energy-based criterion for the stability of discrete breathers.
• ### Interaction of Traveling Waves with Mass-With-Mass Defects within a Hertzian Chain(1212.1966)
Dec. 10, 2012 physics.class-ph, nlin.PS
We study the dynamic response of a granular chain of particles with a resonant inclusion (i.e., a particle attached to a harmonic oscillator, or a mass-with-mass defect). We focus on the response of granular chains excited by an impulse, with no static precompression. We find that the presence of the harmonic oscillator can be used to tune the transmitted and reflected energy of a mechanical pulse by adjusting the ratio between the harmonic resonator mass and the bead mass. Furthermore, we find that this system has the capability of asymptotically trapping energy, a feature that is not present in granular chains containing other types of defects. Finally, we study the limits of low and high resonator mass, and the structure of the reflected and transmitted pulses. | 2021-04-11 22:28:19 | {"extraction_info": {"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, "math_score": 0.53303462266922, "perplexity": 681.3918392373922}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038065492.15/warc/CC-MAIN-20210411204008-20210411234008-00141.warc.gz"} |
http://electronicpackaging.asmedigitalcollection.asme.org/article.aspx?articleid=1408287 | 0
RESEARCH PAPERS
# Thermal Phenomena in Nanoscale Transistors
[+] Author and Article Information
Eric Pop
Department of Electrical Engineering, Stanford University, Stanford, CA 94305
Kenneth E. Goodson
Department of Mechanical Engineering, Stanford University, Stanford, CA 94305goodson@stanford.edu
J. Electron. Packag 128(2), 102-108 (Dec 14, 2006) (7 pages) doi:10.1115/1.2188950 History: Received January 21, 2005; Revised December 14, 2006
## Abstract
As CMOS transistor gate lengths are scaled below $45nm$, thermal device design is becoming an important part of microprocessor engineering. Decreasing dimensions lead to nanometer-scale hot spots in the drain region of the device, which may increase the drain series and source injection electrical resistances. Such trends are accelerated with the introduction of novel materials and nontraditional transistor geometries, like ultrathin body, surround-gate, or nanowire devices, which impede heat conduction. Thermal analysis is complicated by subcontinuum phenomenan including ballistic electron transport, which reshapes the hot spot region compared with classical diffusion theory predictions. Ballistic phonon transport from the hot spot and between material boundaries impedes conduction cooling. The increased surface to volume ratio of novel transistor designs also leads to a larger contribution from material boundary thermal resistance. In this paper we survey trends in transistor geometries and materials, from bulk silicon to carbon nanotubes, along with their implications for the thermal design of electronic systems.
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## Figures
Figure 9
Current-voltage characteristics of a 3μm long metallic single-wall carbon nanotube up to breakdown by oxidation (burning) in air, when the middle of the SWNT reaches about 600°C at 15V bias. The data (symbols) and inset (AFM image showing place of breakdown) are from Ref. 43. The model (solid line) is described in Ref. 47.
Figure 1
Trends of on-chip power density for the past 10+ years. Note the vertical axis is logarithmic and the horizontal one (years) is linear. By comparison, typical power densities for a hot plate and a rocket nozzle are listed, while the surface of the sun puts out approximately 7000W∕cm2. The solid line marks an exponential trend. (Data compiled by F. Labonte, Stanford.)
Figure 2
False-color Transmission Electron Microscopy (TEM) cross-section of a future-generation strained silicon-on-insulator (SSOI) transistor. Image courtesy IBM.
Figure 5
Evolution of transistor designs from traditional bulk silicon (a) to (strained) silicon or germanium on insulator (b), to multiple-gate or FinFET devices (c). Non-traditional device designs like (b) and (c) introduce more complicated geometries and lower thermal conductivity materials, making heat removal more difficult. The gate length (currently near 50nm) is expected to be scaled to the 10nm range, while the semiconductor film and “fin” thickness of (b) and (c), respectively, are approximately one third to one half the gate length. Figure (c) is from the technology roadmap (ITRS) (1).
Figure 4
Snapshot of electron transport in a future-generation thin-body silicon-on-insulator (SOI) device simulated with the Monte Carlo code MONET (13,15). The color scale is the electron energy in eV and the device dimensions are in nm.
Figure 6
Estimated reduction in silicon and germanium thin film thermal conductivity due to phonon boundary scattering alone (i.e., no confinement). Thin silicon films suffer from a stronger reduction in thermal conductivity (vs germanium) due to the longer bulk silicon phonon mean free path (30).
Figure 7
Thin film transistor thermal resistance model (also see Fig. 5). The dark gray areas represent the metalized gate and contacts, the light gray is the surrounding oxide insulator (21,30).
Figure 8
Suspended silicon nanowire field effect device fabricated with electron beam lithography and a buffered HF underetch (40). The cross section of the wire is 23×80nm. The confined dimensions significantly alter both current and heat transport through the wire.
Figure 3
Phonon dispersion in the (100) direction of silicon (a) and computed net (emission minus absorption) phonons generated by Joule heating (b). The symbols in (a) are from neutron scattering data and the lines are a quadratic fit (14). Solid lines are for longitudinal, and dashed lines are for transverse phonons. Optical modes are above and acoustic modes have energies below approximately 50meV, respectively. Note the vertical axes are matched (E=ℏω) to facilitate comparison of the dispersion and generated modes.
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Topic Collections | 2018-04-23 17:42:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3882884681224823, "perplexity": 5688.002740796795}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125946120.31/warc/CC-MAIN-20180423164921-20180423184921-00226.warc.gz"} |
http://www.legisquebec.gouv.qc.ca/en/showversion/cs/R-12?code=se:62_11&pointInTime=20201022 | ### R-12 - Act respecting the Civil Service Superannuation Plan
62.11. For the purposes of section 62.5, the annualization of salaries for the years of service subsequent to 2009 are obtained,
(1) when computing the average pensionable salary referred to in subparagraph 1 of the first paragraph of section 62.4, by dividing the aggregate of the adjusted pensionable salary for such a year and the lump sum attributed to that year under section 62.24 by the harmonized service for the year; and
(2) when computing the average pensionable salary referred to in subparagraph 2 of the first paragraph of section 62.4, by dividing the aggregate of the adjusted pensionable salary for such a year and the lump sum attributed to that year under section 62.24 by the harmonized service for the year. The limit imposed by the first paragraph of section 62.1 applies to the result obtained for each year.
2008, c. 25, s. 71. | 2020-11-28 23:31:45 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8790371417999268, "perplexity": 1802.735163625136}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141195929.39/warc/CC-MAIN-20201128214643-20201129004643-00308.warc.gz"} |
http://www.math.purdue.edu/~dvb/classroom.html | ### Contact information for Donu Arapura
Office: MATH 642
Office hours: Tue 4:30-5:30; Thurs 11-12
Email address: arapura*math.purdue.edu (replace * by @)
Phone number: 494-1983
(Technical note, I'm using this to render formulas, so you'll need your browser to have javascript turned on.) NOTE: No class coming Tues (before Thanksgiving)
#### Math 450 (Honors Abstract Algebra); TTh 3:00, Univ 219
Since this the honors section, I assume you won't mind being challenged. I will base the grade on weekly homework (25%), two midterms and a final (25% each). Rather than follow a textbook, I plan to use my own notes which allows for greater flexibility. They will be posted here weekly. You may also want to consult other sources. I recommend Artin's Algebra .
1. Chapters 1 & 2 of notes. (Typo fixed) For next Thurs Aug 31, do problems 1,2, 3 4, 6, 7 from chap 1 & problem 1 from chap 2
2. Chapters 3 of notes. For next Thurs Sept 7 , do problems 3, 4, 5, 7, 8 from chap 2 & problem 2, 4 from chap 3
3. Chapters 4 of notes. For next Thurs Sept 14, do problems 7 from chap 3 & problem 1,2,3,7, 8 from chap 4 (Clarification: Chap 4, #7 should read then $$x=qd$$..." also $$\mathbb{Z}$$ is a group under addition and $$qd= d+d+\dots( q\text{ times})$$.)
4. Chapters 5 of notes.(Corrected) For next Thurs Sept 21 do problem 10 from chap 4, and problem 4, 5, 6, 8 from chap 5 ( NOTE: chap5, #5 corrected)
5. Chapter 6 of notes For next TUE Sept 26 do problems 1,2, 3 from chap 6
6. Chapters 6 Updated For next Thurs Oct 5 , do problems 4, 7, 9
7. Chapter 7 and Chapter 8 . For next Thurs Oct 12, do problems 1,2, 3, 7, 8, 9 from chap 7 and 1 chap 8
8. Chapter 8 (updated) and Chapter 9 . . For Thurs Oct 19 do 2,3,4, 6,8 chap 8 and 1 chap 9 [Note there was a typo in prob 6, it should read $$(a_1,a_2)+(b_1,b_2)=(a_1+b_1, a_2+b_2)$$]
9. Chapter 10 For Thurs Oct 26 do 3,4,5 chap 9 and 4, 5 chap 10
10. Chapter 11 For Thurs Nov 2 do 1,2,3, 6, 7 chap 10 and 1 chap 11
11. Chapter 12 (UPDATED) For Thurs Nov 9 do 2,3,4,5 from chap 11 and 1,3,4 from chap 12
12. Chapter 13 (won't be on test) For Thurs Nov 16 do 6 from chap 12 Should read $$A\in M_{nn}(\mathbb{Z}_p)$$ , and 1 from chap 13
13. Chapter 14 (typos corrected) For Thurs Nov 30 do 2 to 6 from chap 13 and 1 from chap 14
14. Chapter 15
15. Chapter 16
16. The entire set of notes are also available.
##### Test 3
Take home (due Thurs Dec 7): Chap 14: 6; Chap 15: 4, 7; Chap 16: 4, 5 (The last problem should read "... matrices in exercise 4", and the angles listed in that problem should probably be doubled to $$2\pi/6, 2\pi/ 3$$ )
#### Math 518 ( Discrete Math); TTh 12:00, Univ 117
This will basically be class on combinatorics a.k.a. discrete math''. The textbook for the class is Concrete Mathematics by Graham, Knuth and Patashnik. I'll mostly follow it along with some supplementary sources such as Wilf's Generatingfunctionology. also A=B by Petkovsek, Wilf, Zeilberger, contains much more info about Gosper/Zeilberger/... algorithms. The grade will be based completely on homework, which will be assigned weekly. In addition, I would ask everyone to present at least one of your solutions in class. Each presentation will be worth a couple of points as well
1. Homework 1 (due Thurs Aug 1 )
2. Homework 2 NB Expression in bonus problem 5 should read $$\prod_{p\in P}\frac{1}{1-p^{-1}}$$ (due Thurs Sept 7 )
3. Homework 3 (due Thurs Sept 14 )
4. Homework 4 (due Thurs Sept 21 )
5. Homework 5 (Typo corrected) (due Thurs Sept 28 )
6. Homework 6 (due Thurs Oct 5 )
7. Homework 7 (due Thurs Oct 12 )
8. Homework 8 (due Thurs Oct 19 )
9. Homework 9 (due Thurs Oct 26 )
10. Homework 10 (due Thurs Nov 2 )
11. <
12. Homework 11 (due Thurs Nov 9 )
13. Homework 12 (due Thurs Nov 16 )
14. Homework 13 (due Thurs Nov 30 )
15. Homework 14 (due Thurs Dec 7 ) | 2017-12-18 05:16:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.38569575548171997, "perplexity": 6639.811128796306}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948608836.84/warc/CC-MAIN-20171218044514-20171218070514-00289.warc.gz"} |
https://www.zbmath.org/?q=ai%3Achen.dawei+se%3A00000592 | # zbMATH — the first resource for mathematics
Stable base locus decompositions of Kontsevich moduli spaces. (English) Zbl 1200.14050
In this paper, the authors study the birational geometry of the Kontsevich space of genus $$0$$ unmarked stable maps to Grassmannians $$\overline M_{0,0}(\mathbb G(k, n), d)$$. The effective cone is decomposed into chambers according to the stable base locus of the linear series, and the $$8$$ chambers for degree $$d=2$$ and the $$22$$ chambers for degree $$d=3$$ are explicitly found. The birational models arising from each chamber in the decomposition are also found, and given modular interpretation in most cases.
##### MSC:
14H10 Families, moduli of curves (algebraic) 14E05 Rational and birational maps 14D20 Algebraic moduli problems, moduli of vector bundles
##### Keywords:
Kontsevich space; compactifications; birational geometry
Full Text:
##### References:
[1] C. Birkar, P. Cascini, C. D. Hacon, and J. McKernan, The existence of minimal models for varieties of log-general type, J. Amer. Math. Soc. 23 (2010), 405–468. · Zbl 1210.14019 · doi:10.1090/S0894-0347-09-00649-3 [2] D. Chen, Mori’s program for the Kontsevich moduli space $$\overline\mathcal M_0\20(\mathbb P^3,3),$$ Int. Math. Res. Not. 2008, article ID rnn067. · Zbl 1147.14009 · doi:10.1093/imrn/rnn067 [3] D. Chen and I. Coskun, with an appendix by C. Crissman, Towards Mori’s program for the moduli space of stable maps, Amer. J. Math. (to appear). · Zbl 1256.14013 [4] I. Coskun, J. Harris, and J. Starr, The effective cone of the Kontsevich moduli spaces, Canad. Math. Bull. 51 (2008), 519–534. · Zbl 1213.14022 · doi:10.4153/CMB-2008-052-5 [5] ——, The ample cone of the Kontsevich moduli space, Canad. J. Math. 61 (2009), 109–123. · Zbl 1206.14050 · doi:10.4153/CJM-2009-005-8 [6] I. Coskun and J. Starr, Divisors on the space of maps to Grassmannians, Int. Math. Res. Not. 2006, article ID 35273. · Zbl 1117.14028 · doi:10.1155/IMRN/2006/35273 [7] J. de Jong and J. Starr, Divisor classes and the virtual canonical bundle for genus zero maps, [8] L. Ein, R. Lazarsfeld, M. Mustaţǎ, M. Nakamaye, and M. Popa, Asymptotic invariants of base loci, Ann. Inst. Fourier (Grenoble) 56 (2006), 1701–1734. · Zbl 1127.14010 · doi:10.5802/aif.2225 · numdam:AIF_2006__56_6_1701_0 · eudml:10189 [9] ——, Restricted volumes and base loci of linear series, Amer. J. Math. 131 (2009), 607–651. · Zbl 1179.14006 · doi:10.1353/ajm.0.0054 · muse.jhu.edu [10] A. Mustaţǎ and M. A. Mustaţǎ, Intermediate moduli spaces of stable maps, Invent. Math. 167 (2007), 47–90. · Zbl 1111.14018 · doi:10.1007/s00222-006-0006-1 [11] D. Oprea, Divisors on the moduli space of stable maps to flag varieties and reconstruction, J. Reine Angew. Math. 586 (2005), 169–205. · Zbl 1089.14004 · doi:10.1515/crll.2005.2005.586.169
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. | 2021-01-16 06:28:07 | {"extraction_info": {"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, "math_score": 0.6908648610115051, "perplexity": 1667.160383076432}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703500028.5/warc/CC-MAIN-20210116044418-20210116074418-00500.warc.gz"} |
https://www.sparrho.com/item/the-maximum-on-a-random-time-interval-of-a-random-walk-with-long-tailed-increments-and-negative-drift/8b9ecd/ | The maximum on a random time interval of a random walk with long-tailed increments and negative drift
Research paper by Serguei Foss, Stan Zachary
Indexed on: 18 Mar '13Published on: 18 Mar '13Published in: Mathematics - Probability
Abstract
We study the asymptotics for the maximum on a random time interval of a random walk with a long-tailed distribution of its increments and negative drift. We extend to a general stopping time a result by Asmussen (1998), simplify its proof, and give some converses. | 2021-09-21 05:30:11 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8029122948646545, "perplexity": 1828.4432445694124}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057158.19/warc/CC-MAIN-20210921041059-20210921071059-00444.warc.gz"} |
https://math.stackexchange.com/questions/1374679/weak-compactness-of-a-set-of-translates-in-c-0-mathbbr | # Weak compactness of a set of translates in $C_0(\mathbb{R})$
Let $f \in C_0(\mathbb{R})$. Consider the set of translates of $f$ $$A = \{ f_t : t \in \mathbb{R} \}$$ where $f_t(x)=f(x+t), x\in \mathbb{R}$.
I want to show that $A$ is relatively compact in the weak topology on $C_0(\mathbb{R})$.
I have shown that every sequence in $A$ has a weakly convergent subsequence, using the fact that the dual is the space of Radon measures.
If the closed ball in $C_0(\mathbb{R})$ is metrizable in the weak topology (I'm not sure if it is), then my proof is complete. [The problem is that $C_0(\mathbb{R})$ is not reflexive, and the dual is not separable either]
Is there some result on the metrizability of the unit ball in $C_0(\mathbb{R})$? Or is there some other way to show that $A$ is relatively weak-compact?
Consider the map $\varphi \colon [-\infty, +\infty] \to C_0(\mathbb{R})$, where the latter is endowed with the weak topology, given by
$$\varphi(t) = \begin{cases} 0 &, t = \pm \infty\\ f_t &, t \in \mathbb{R}.\end{cases}$$
Since for $\lambda \in C_0(\mathbb{R})'$ almost the complete mass of $\lvert\lambda\rvert$ is concentrated on a compact subset of $\mathbb{R}$, $\varphi$ is continuous at $\pm\infty$. It is also continuous at all $t \in \mathbb{R}$ by the uniform continuity of $f$. Thus $\varphi([-\infty,+\infty])$ is weakly compact. | 2019-11-19 21:18:05 | {"extraction_info": {"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, "math_score": 0.9706522822380066, "perplexity": 45.917550326132876}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670255.18/warc/CC-MAIN-20191119195450-20191119223450-00321.warc.gz"} |
https://indico.cern.ch/event/30248/contributions/1667131/ | # QM 2011 - XXII International Conference on Ultrarelativistic Nucleus-Nucleus Collisions
May 22 – 28, 2011
Centre Bonlieu
Europe/Zurich timezone
## The second act of hydro: shocks and sounds from initial perturbations and jets
May 23, 2011, 3:00 PM
20m
Salle de l'Europe (Imperial Palace)
### Salle de l'Europe
#### Imperial Palace
Parallel Global and collective dynamics
### Speaker
Edward Shuryak (Stony Brook University)
### Description
Recently there was significant progress in account for several lower harmonics of the Little Bang", especially the so called triangular flow, ascribed to fluctuations of the initial conditions. We discuss this problem more generally, combining many harmonics coherently into certain patterns of sound propagation. Analytic solution for all harmonics is found for the so called Gubser flow", with complete Green function obtained. Another source of perturbations which can be studied using our results are waves induced by quenching jets. We argue that for large energy loss shock waves should form, as well as the so called jet/fireball edge, separating unperturbed and excited matter. We discuss how this edge should be visible experimentally, perhaps on \ event-by-event basis.
### Primary author
Edward Shuryak (Stony Brook University)
Slides | 2022-08-11 20:27:27 | {"extraction_info": {"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, "math_score": 0.45844021439552307, "perplexity": 6293.395114666911}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571502.25/warc/CC-MAIN-20220811194507-20220811224507-00735.warc.gz"} |
https://crypto.stackexchange.com/questions/75374/does-an-information-theoretically-secure-hash-function-exist?noredirect=1 | Does an information-theoretically secure hash function exist?
Does an information theoretically secure hash function exist? (By exist I mean is discovered/invented and implemented, not whether it could exist.)
• Depends on what you mean by ‘hash function’. For example, the first message authentication code in history, built out of what would later be called universal hashing, provides the optimal possible ‘information-theoretic’ bound on forgery probability. But it's not a collision-resistant hash function like SHA-256, a concept which doesn't even have a mathematical formalization that could conceivably have a notion of ‘information-theoretic security’. – Squeamish Ossifrage Oct 29 '19 at 5:06
• So would Poly1305 be an information-theoretically secure "hash function", while SHA-256 is a computationally-secure function but with collision resistant? – 09182736471890 Oct 29 '19 at 5:13
• I wouldn't say that, no, and I definitely wouldn't draw specifically that contrast between different types of security for the two qualitatively different goals of bounded difference probability (Poly1305) and collision resistance (SHA-256). Poly1305 and SHA-256 are entirely different kinds of thing which both happen to fit under the wide umbrella of the vague term ‘hash function’, meaning a function that kinda scrambles its input in some way. The FNV-1 hash is also called a ‘hash function’ but it doesn't aspire to any security. – Squeamish Ossifrage Oct 29 '19 at 14:17
1 Answer
The Gilbert-MacWilliams-Sloane MAC referred to by @SqueamishOssifrage in the comments is information theoretically secure "for single use", at the cost of having hashes that have length $$2\ell$$ for fixed length messages of length $$\ell.$$
Poly1305 is not information theoretically secure.
It is much more flexible, can take essentially arbitrary length inputs, and has a low probability $$p$$ of being spoofed which depends on four factors, $$\delta,C,D,L$$ and which is essentially $$\delta$$ plus a tiny correction factor, so $$p\leq \delta+f(L,D)2^{-106}.$$ See the original paper by Bernstein (Springer LNCS vol. 3557, also available at his site https://cr.yp.to/mac/poly1305-20050329.pdf) :
• One can have up to $$C\leq 2^{64}$$ authenticated messages
• Messages are of maximum length $$L.$$
• One can attempt up to $$D$$ forgeries
• $$\delta$$ is the probability of distinguishing AES output from a random permutation
To start with, we don't know what $$\delta$$ is. AES could be replaced if it was found to be weak, but the big issue is that, there is no way of handling arbitrary input length messages with a probability distribution, which would enable one to define information theoretic security, which depends on entropy, a well defined functional of a probability distribution.
• ‘Poly1305 is not information theoretically secure.’ What is your definition of ‘information theoretically secure’ which Poly1305 fails? You proceeded to quote a paper about the composition Poly1305-AES, which uses Poly1305 as a component. – Squeamish Ossifrage Oct 29 '19 at 14:05
• ‘the big issue is that, there is no way of handling arbitrary input length messages with a probability distribution’—This is…not an issue in any of the security theorems related to Poly1305, not even Poly1305-AES or crypto_secretbox_xsalsa20poly1305. The theorems involving Poly1305 are all quantified for all messages, for uniform random keys. – Squeamish Ossifrage Oct 29 '19 at 14:07
• For a little more on the history of universal hashing one-time authenticators, their easily proven security properties, and how they fit in with other components, see crypto.stackexchange.com/a/67639. Poly1305-AES is an example of a Carter–Wegman–Shoup MAC based on Poly1305 and AES, but essentially nobody uses it today—ChaCha/Poly1305 as used in TLS 1.3 and crypto_secretbox_xsalsa20poly1305 both just derive a one-time Poly1305 authenticator key per message pseudorandomly, with no AES involved. – Squeamish Ossifrage Oct 29 '19 at 14:12
• It's unclear to me why you brought up Poly1305-AES at all. The OP wasn't asking for a MAC that's not ‘information-theoretically secure’, and wasn't asking about Poly1305, but now you seem to have misled the OP into the conclusion that there is not an ‘information-theoretic security’ theorem for Poly1305 when exactly the opposite is true. – Squeamish Ossifrage Oct 29 '19 at 17:40
• By the way, there are universal hashing MACs—not GMS, not Poly1305—that have $h$-bit outputs for $\ell$-bit messages with the smallest possible bound $1/2^h$ on forgery probability and keys much smaller than $2\ell$ bits. E.g., if a message $m$ is broken into an $n$-element sequence of $h$-bit chunks $(m_1, m_2, \dotsc, m_n)$ interpreted in $\operatorname{GF}(2^h)$, and a key is a tuple $(r_1, r_2, \dotsc, r_n, s)$ of $n + 1$ elements of $\operatorname{GF}(2^h)$, then the MAC $m_1 r_1+m_2 r_2+\dotsb+m_n r_n+s$ attains the bound $1/2^h$ on forgery probability with $\ell+h<2\ell$ key bits. – Squeamish Ossifrage Oct 29 '19 at 18:15 | 2021-08-03 02:08:53 | {"extraction_info": {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3999177813529968, "perplexity": 1650.0219267667524}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154408.7/warc/CC-MAIN-20210802234539-20210803024539-00099.warc.gz"} |
https://www.andifugard.info/blog/page/2/ | # Blog
## Census data on trans and non-binary people in Canada
Canada published census data on trans and non-binary people on 27 April 2022. Here’s a table of the values they presented in a pie chart (why a pie chart, Canada?). Individuals in the census were aged 15 or above and living in a private household in May 2021.
Gender N %
Cis man 14,814,230 48.83
Cis woman 15,421,085 50.83
Trans man 27,905 0.09
Trans woman 31,555 0.10
Non binary 41,355 0.14
Total 30,336,130 100.00
## Playing with RCTs and probabilities
Suppose we run an RCT with two groups, treatment and control, and a binary outcome of whether participants recover or not.
There are two potential outcomes: recovery following treatment ($$R_t$$) and recovery following control ($$R_c$$), $$1$$ if recovered and $$0$$ if not recovered. Only one of these two potential outcomes is realised, depending on what group someone is assigned to. Let $$W = t$$ if a participant was assigned to treatment and $$W = c$$ if they were assigned to control.
Suppose, following an RCT, we learn the following (somehow with perfect precision):
$$P(R_t = 1 | W = t) = 0.8$$;
$$P(R_c = 1 | W = c) = 0.3$$.
Given the two probabilities above, it turns out the best we can say is that $$P(R_t = 1) \in [0, 1]$$ and $$P(R_c = 1) \in [0, 1]$$. So, it seems that we aren’t yet able to infer anything about the potential outcomes beyond those that were realised.
Add to our premises that participants were assigned to treatment or control by coin flip:
$$P(W = t) = P(W = c) = 0.5$$.
Now $$P(R_t = 1) \in [0.4 , 0.9]$$ and $$P(R_c = 1) \in [0.15 , 0.65]$$. These intervals are clearly better that $$[0,1]$$; however, can we do better?
The key ingredient we need to add is that treatment assignment is independent of the potential outcomes; that is
$$P(W | R_t, R_c) = P(W)$$.
Now, given all this information, we obtain point probabilities: $$P(R_t = 1) = 0.8$$ and $$P(R_c = 1) = 0.3$$. These are equal to the probabilities that were conditional on what group a participant was assigned to.
Another curiosity is what we can infer about the joint distribution, $$P(R_t, R_c)$$. The results are probability intervals:
$$R_t = 0$$ $$R_t = 1$$
$$R_c = 0$$ $$[0, 0.2]$$ $$[0.5, 0.7]$$ $$0.7$$
$$R_c = 1$$ $$[0, 0.2]$$ $$[0.1, 0.3]$$ $$0.3$$
$$0.2$$ $$0.8$$
This illustrates, in a toy example, the more general problem that the joint distribution of potential outcomes typically cannot be obtained from an RCT. However, the joint probabilities are constrained by the marginals.
## Wisdom(?) from the 1918 Dadaist manifesto by Tristan Tzara
The 1st and 2nd DADA Art Manifestos are online over there.
• “Psychoanalysis is a dangerous disease, it deadens man’s anti-real inclinations and systematises the bourgeoisie.”
• “Dialectics is an amusing machine that leads us (in banal fashion) to the opinions which we would have held in any case.”
• “People observe, they look at things from one or several points of view, they choose them from amongst the millions that exist. Experience too is the result of chance and of individual abilities.”
• “Logic is a complication. Logic is always false. It draws the superficial threads of concepts and words towards illusory conclusions and centres.”
• “What we need are strong straightforward, precise works which will be forever misunderstood.”
## Efficacy RCTs as survey twins
Surveys attempt to estimate a quantity of a finite population using a probability sample from that population. How people ended up in the population is somebody else’s problem – demographers, perhaps.
Survey participants are sampled at random from this finite population without replacement. Part a of the figure below illustrates. Green blocks denote people who are surveyed and from whom we collect data. Grey blocks denote people we have not surveyed; we would like to infer what their responses would have been, if they had they been surveyed too.
RCTs randomly assign participants to treatment or control conditions. This is illustrated in part b of the figure above: green cells denote treatment and purple cells denote control. There are no grey cells since we have gathered information from everyone in the finite population. But in a way, we haven’t really.
An alternative way to view efficacy RCTs that aim to estimate a sample average treatment effect (SATE) is as a kind of survey. This illustrated in part c. Now the grey cells return.
There is a finite population of people who present for a trial, often with little known about how they ended up in that population – not dissimilarly to the situation for a survey. (But who studies how they end up in a trial – trial demographers?)
Randomly assigning people to conditions generates two finite populations of theoretical twins, identical except for treatment assignment and the consequences thereafter. One theoretical twin receives treatment and the other receives control. But we only obtain the response from one of the twins, i.e., either the treatment or the control twin. (You could also think of these theoretical twins’ outcomes as potential outcomes.)
Looking individually at one of the two theoretical populations, the random assignment to conditions has generated a random sample from that population. We would like to know what the outcome would have been for everyone in the treatment condition, if everyone had been assigned treatment. Similarly for control. Instead, we have a pair of surveys that sample from these two populations.
### Viewing the Table 1 fallacy through the survey twin lens
There is a common practice of testing for differences in covariates between treatment and control. This is the Table 1 fallacy (see also Dean Eckles’s take on whether it really is a fallacy). Let’s see how it can be explained using survey twins.
Firstly, we have a census of covariates for the whole finite population at baseline, so we know with perfect precision what the means are. Treatment and control groups are surveys of the same population, so clearly no statistical test is needed. The sample means in both groups are likely to be different from each other and from the finite population mean of both groups combined. No surprises there: we wouldn’t expect a survey mean to be identical to the population mean. That’s why we use confidence intervals or large samples so that the confidence intervals are very narrow.
### What’s the correct analysis of an RCT?
It’s common to analyse RCT data using a linear regression model. The outcome variable is the endpoint, predictors are treatment group and covariates. This is also known as an ANCOVA. This analysis is easy to understand if the trial participants are a simple random sample from some infinite population. But this is not what we have in efficacy trials as modelled by survey twins above. If the total number of participants in the trial is 1000, then we have a finite population of 1000 in the treatment group and a finite population of 1000 in the control group – together, 2000. In total we have 1000 observations, though, split in some proportion between treatment and control.
Following through on this reasoning, it sounds like the correct analysis uses a stratified independent sampling design with two strata, coinciding with treatment and control groups. The strata populations are both 1000, and a finite population correction should be applied accordingly.
It’s a little more complicated, as I discovered in a paper by Reichardt and Gollob (1999), who independently derived results found by Neyman (1923/1990). Their results highlight a wrinkle in the argument when conducting a t-test on two groups for finite populations as described above. This has general implications for analyses with covariates too. The wrinkle is, the two theoretical populations are not independent of each other.
The authors derive the standard error of the mean difference between X and Y as
$$\displaystyle \sqrt{\frac{\sigma_X^2}{n_X} + \frac{\sigma_Y^2}{n_Y} – \left[ \frac{(\sigma_X – \sigma_Y)^2}{N} + \frac{2(1-\rho) \sigma_X \sigma_{Y}}{N} \right]}$$,
where $$\sigma_X^2$$ and $$\sigma_Y^2$$ are the variances of the two groups, $$n_X$$ and $$n_Y$$ are the observed group sample sizes, and $$N$$ is the total sample (the finite population) size. Finally, $$\rho$$ is the unobservable correlation between treat and control outcomes for each participant – unobservable because we only get either the treatment outcome or control outcome for each participant and not both. The terms in square brackets correct for the finite population.
If the variances are equal ($$\sigma_X = \sigma_Y$$) and the correlation $$\rho = 1$$, then the correction vanishes (glance back at numerators in the square brackets to see). This is great news if you are willing to assume that treatments have constant effects on all participants (an assumption known as unit-treatment additivity): the same regression analysis that you would use assuming a simple random sample from an infinite population applies.
If the variances are equal and the correlation is 0, then this is the same standard error as in the stratified independent sampling design with two strata described above. Or at least it was for the few examples I tried.
If the variances can be different and the correlation is one, then this is the same standard error as per Welch’s two-sample t-test.
So, which correlation should we use? Reichardt and Gollob (1999) suggest using the reliability of the outcome measure to calculate an upper bound on the correlation. More recently, Aronow, Green, and Lee (2014) proved a result that puts bounds on the correlation based on the observed marginal distribution of outcomes, and provide R code to copy and paste to calculate it. It’s interesting that a problem highlighted a century ago on something so basic – what standard error we should use for an RCT – is still being investigated now.
### References
Aronow, P. M., Green, D. P., & Lee, D. K. K. (2014). Sharp bounds on the variance in randomized experiments. Annals of Statistics, 42, 850–871.
Neyman, J. (1923/1990). On the application of probability theory to agricultural experiments. Essay on principles. Section 9. Statistical Science, 5, 465-472.
Reichardt, C. S., & Gollob, H. F. (1999). Justifying the Use and Increasing the Power of a t Test for a Randomized Experiment With a Convenience Sample. Psychological Methods, 4, 117–128.
## Science as anarchism
“Science is an essentially anarchic enterprise: theoretical anarchism is more humanitarian and more likely to encourage progress than its law-and-order alternatives.”
(Feyerabend, 1975)
“I often wished I had never written that fucking book”
(Feyerabend, 1995)
## Surveys as RCTs: an exercise in analogy
Surveys use probability samples from a finite population to estimate a quantity of that population, e.g., the percentage of people holding a particular view. What happens if we try to understand surveys using the concepts of a randomised controlled trial?
Those assigned to intervention get the survey questions – the questions are the intervention. Those assigned to control get nothing and are ignored entirely. Typically the probability of being assigned to intervention (being surveyed) is much smaller than that of being assigned to control (ignored).
We wish to estimate the percentage of people in the population who hold a particular view following the intervention (the survey questions). This percentage is the outcome measure. We observe the outcome for people randomly assigned to the survey. For the control group, we want to know what percentage would have held that view, if they had been assigned to the survey.
An interesting feature of this “intervention” of a survey is that we hope it does not change the percentage outcome. So, viewing a survey through this RCT lens, the average treatment effect (mean difference between intervention and control, survey and ignore) is assumed to be 0. But this might not hold. There may be a mean difference between people who have been asked to reflect on something and tell a researcher versus those who hold a view but have not told anyone. We cannot tell using this design.
Where did the population come from? Across in the RCT analogy, it would be the (typically nonprobability) sample of people who consented to take part in the trial. In the survey, it is a collection of people who found themselves living in a particular area, having the demographic profile of interest (satisfying the inclusion criteria), and being reachable via a sample frame. We usually do not care how they got there. Other researchers might, e.g., demographers studying migration or births. People often end up in an area because they found a job nearby or because they were born there – both events with an element of chance.
Researchers often worry whether an RCT’s results transfer to other settings. This is not an issue for surveys. In fact we might assume that people living in different areas hold different views. One aspect we might hope does transfer is how people interpret the questions.
## Standard errors of marginal means in an RCT
Randomised controlled trials (RCTs) typically use a convenience sample to estimate the mean effect of a treatment for study participants. Participants are randomly assigned to one of (say) two conditions, and an unbiased estimate of the sample mean treatment effect is obtained by taking the difference of the two conditions’ mean outcomes. The estimand in such an RCT is sometimes called the sample average treatment effect (SATE).
Some papers report a standard error for the marginal mean outcomes in treatment and control groups using the textbook formula
$$\displaystyle \frac{\mathit{SD_g}}{\sqrt{n_g}}$$,
where $$\mathit{SD_g}$$ is the standard deviation of group $$g$$ and $$n_g$$ the number of participants assigned to that group.
This formula assumes a simple random sample with replacement from an infinite population, so does not work for a convenience sample (see Stephen Senn, A Standard Error). I am convinced, but curious what standard error for each group’s mean would be appropriate, if any. (You could stop here and argue that the marginal group means mean nothing anyway. The whole point of running a trial is to subtract off non-treatment explanations of change such as regression to the mean.)
Let’s consider a two-arm RCT with no covariates and a coin toss determining who receives treatment or control. What standard error would be appropriate for the mean treatment outcome? Let the total sample size be $$N$$ and quantities for treatment and control use subscripts $$t$$ and $$c$$, respectively.
##### Treatment outcome mean of those who received treatment
If we focus on the mean for the $$n_t$$ participants who were assigned to treatment, we have all observations for that group, so the standard error of the mean is 0. This feels like cheating.
##### Treatment outcome mean of everyone in the sample
Suppose we want to say something about the treatment outcome mean for all $$N$$ participants in the trial, not only the $$n_t$$ who were assigned to treatment.
To see how to think about this, consider a service evaluation of $$N$$ patients mimicking everything about an RCT except that it assigns everyone to treatment and uses a coin toss to determine whether someone is included in the evaluation. This is now a survey of $$n$$ participants, rather than a trial. We want to generalise results to the finite $$N$$ from which we sampled.
Since the population is finite and the sampling is done without replacement, the standard error of the mean should be multiplied by a finite population correction,
$$\displaystyle \mathit{FPC} = \sqrt{\frac{N – n}{N – 1}}$$.
This setup for a survey is equivalent to what we observe in the treatment group of an RCT. Randomly assigning participants to treatment gives us a random sample from a finite population, the sample frame of which we get by the end of the trial: all treatment and control participants. So we can estimate the SEM around the mean treatment outcome as:
$$\displaystyle \mathit{SEM_t} = \frac{\mathit{SD_t}}{\sqrt{n_t}} \sqrt{\frac{N – n_t}{N – 1}}$$.
If, by chance (probability $$1/2^N$$), the coin delivers everyone to treatment, then $$N = n_t$$ and the FPC reduces to zero, as does the standard error.
##### Conclusion
If the marginal outcome means mean anything, then there are a couple of standard errors you could use, even with a convenience sample. But the marginal means seem irrelevant when the main reason for a running an RCT is to subtract off non-treatment explanations of change following treatment.
If you enjoyed this, you may now be wondering what standard error to use when estimating a sample average treatment effect. Try Efficacy RCTs as survey twins.
## Curiosities: two pairs of ideas that intrigue/trouble me
### Troubles with theories
• Evidence alone can’t determine which scientific theories we should believe since more than one theory will often be consistent with the available evidence.
• In mathematics, there exist unintended (“nonstandard”) models of formal theories. An example theory where this is the case is Peano Arithmetic. The intended (“standard”) model is the set of (countably infinite) natural numbers (0, 1, 2, 3, …) and operations thereon that we know and love. But there are non-standard models of Peano Arithmetic that are uncountably infinite. That’s weird. In fact, any first-order logic theory which has a countably infinite model also has an uncountably infinite model (upward Löwenheim–Skolem Theorem).
### Boundedness of selves in spacetime
• We often think of our minds as bounded by our skull. This has been challenged by the extended cognition thesis (Andy Clark and David Chalmers). The gist: we’re happy to accept that we can have beliefs that we’re not conscious of at any given time; they lie dormant until called upon for, say, an argument. Selves are more than what we are conscious of. But we also scribble stuff in notebooks and (these days) apps, set reminders, etc. These notes and reminders are similarly beyond consciousness but also thoroughly outside our heads. They seem essential for cognition.
• There’s a problem with 4D block universe conceptions of spacetime: if all of time – past, present, and future – already exists at points in this 4D geometry then how do we consciously experience the passage of time? Assuming a material conception of conscious experience, each individual experience is scattered across spacetime and individually frozen. No passage. Natalja Deng (2019) points to an solution. Rather than trying to work out how these individual experiences can lead to an experience of passage, “recognize that the fundamental experiential unit is itself temporally extended, and use this to explain how there can be an experience of a temporally extended content.” | 2022-05-24 19:10:10 | {"extraction_info": {"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, "math_score": 0.6672155857086182, "perplexity": 907.2647123392368}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662573189.78/warc/CC-MAIN-20220524173011-20220524203011-00183.warc.gz"} |
https://math.stackexchange.com/questions/2793531/proving-the-existence-of-a-solution-to-a-self-dual-problem | # Proving the existence of a solution to a self dual problem
I'm working on the following exercise:
Let $a \in \mathbb{R}^{n \times n}$ be a skew symmetric matrix and $b \in \mathbb{R}^n$ with $c = -b$. Consider the following $(LP)$ named $(P)$
$$\min_{x \in \mathbb{R}^n} c^Tx$$ $$\text{such that: } Ax \ge b, x \ge 0$$
Show that it is equivalent to it's dual problem $(DP)$
$$\max_{y \in \mathbb{R}^n} y^Tb$$ $$\text{such that: }y^TA \le c^T, y \ge 0$$
Show that if a feasible solution exists for $(P)$ then there is also an optimal solution for $P$.
I managed to show that both LPs are equivalent by just plugging in the assumptions the assumptions that $A = -A^T$ and $c = -b$ into $(P)$.
But I don' t know how to do the second part. Could you give me a hint?
• What the duality theorems say? – metamorphy May 24 '18 at 17:25
• Ok, that was the point I missed. Thank you. – 3nondatur May 24 '18 at 19:53 | 2019-06-25 08:21:07 | {"extraction_info": {"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, "math_score": 0.780887246131897, "perplexity": 141.11545492050965}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999814.77/warc/CC-MAIN-20190625072148-20190625094148-00242.warc.gz"} |
http://www.math.princeton.edu/events/seminars/algebraic-geometry-seminar/two-gifts-complexity-theory-p-v-np-and-matrix | # Two gifts from complexity theory: $P$ v. $NP$ and matrix multiplication
Tuesday, November 22, 2011 -
4:30pm to 5:30pm
I will discuss how the Geometric Complexity Theory of Mulmuley-Sohoni and the problem of determining the complexity of matrix multiplication lead to beautiful questions in algebraic geometry and representation theory.
Speaker:
J. M. Landsberg
Texas A&M University
Event Location:
Fine Hall 322 | 2017-11-22 22:16:16 | {"extraction_info": {"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, "math_score": 0.17092391848564148, "perplexity": 2667.3842229072566}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806676.57/warc/CC-MAIN-20171122213945-20171122233945-00590.warc.gz"} |
https://www.physicsforums.com/threads/is-the-system-linear-if-there-is-an-independant-gravity-term.713534/ | # Is the system linear if there is an independant gravity term
1. ### silentwf
37
Like the topic,if in my system derivation, there is an independent gravitational force independent of the state variable, is my system linear? I believe it is not but my instructor is telling is it is. I'm thinking that if there is is an extra gravitational term,the output of the system does not obey f(ax ) = af(x).
### Staff: Mentor
Linear in which variable?
3. ### silentwf
37
Um, could you explain further by which variable you mean? I think I have the misconception that if there are any terms in the equation of motion not associated to input or output, the system is nonlinear, since the equation itself, when input or output is multiplied by a constant, does not follow the law of superposition.
### Staff: Mentor
Simple example: ##y=a^2x##
y and x have a linear relationship - if you multiply x by 2, y gets multiplied by 2 as well. The system is linear, if you consider "a" as a constant parameter, and look at x and y. You can add multiple solutions as superpositions.
y and a do not have a linear relationship - if you multiply a by 2, y gets multiplied by 4. The system is not linear, if you consider "x" as a constant parameter, and look at a and y. You cannot add multiple solutions as superposition.
5. ### silentwf
37
Okay, um then I should make it something like this: if my system is equation is something like:
$f - mg - u = 0$ where f is the input and u is a disturbance (both are functions of time).
having a scalar multiple of my force would not increase the system's response in a linear way. does this mean that this system is not linear?
$f(a*t) - mg - u(t) = 0$
### Staff: Mentor
A scalar multiple of the force would not lead to a scalar multiple of the mass (if u does not scale in the same way), indeed.
7. ### silentwf
37
So the system is non-linear? It's kind of strange to think of it in this way, so I'm not really sure.
### Staff: Mentor
I think it is non-linear with the interpretation as in post #5. | 2015-12-01 07:33:36 | {"extraction_info": {"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, "math_score": 0.5001097321510315, "perplexity": 479.42080634954215}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398465089.29/warc/CC-MAIN-20151124205425-00248-ip-10-71-132-137.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/95721/why-do-only-capital-latin-letters-respond-all-font-switches-in-math-mode | # Why do only capital Latin letters respond all font switches in math mode?
$\mathbf{\omega}$, for example, does not respond the font switch. Why do only capital Latin letters respond all font switches in math mode?
Note: To bold the \omega, we have to use \boldsymbol{\omega}.
\documentclass{article}
\usepackage{amsmath}
\usepackage{bm}
\begin{document}
Need \verb|\boldsymbol| to bold the \verb|\omega|: $\boldsymbol{\omega}$
No need \verb|\boldsymbol| to bold A: $\mathbf{A}$
\end{document}
-
This calls for MWE. Do you use (pdf|Xe|Lua|{})LaTeX? In what order you load the packages? Remember that bm should be loaded for \boldsymbol and it has to be loaded after you load your main document font (bm sets the correct fonts from the active fonts at the moment of loading the package). – yo' Jan 28 '13 at 17:40
\mathbf changes the current math group (\fam) This means that characters or commands defined by \chardef (\DeclareMathSymbol) such that their math class is 7 will switch to the specified font. In the standard TeX font settings this only affects upper and lower case latin letters. \omega is treated as a symbol from a fixed font (and in the standard TeX math font encodings changing fonts would not produce the desired result as most of the fonts do not have Greek letters.
Despite the Question Title lower case latin letters do change family (and are affected by \mathbf although some fonts notable \mathcal and \mathbb only have uppercase so while the font switches for lowercase letters, an undesired glyph is selected. | 2014-12-28 19:29:52 | {"extraction_info": {"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, "math_score": 0.9861403107643127, "perplexity": 2967.8690279350817}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1419447559374.74/warc/CC-MAIN-20141224185919-00079-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://mlr3book.mlr-org.com/special-tasks.html | This chapter explores the different functions of mlr3 when dealing with specific data sets that require further statistical modification to undertake sensible analysis. Following topics are discussed:
Survival Analysis
This sub-chapter explains how to conduct sound survival analysis in mlr3. Survival analysis is used to monitor the period of time until a specific event takes places. This specific event could be e.g. death, transmission of a disease, marriage or divorce. Two considerations are important when conducting survival analysis:
• Whether the event occurred within the frame of the given data
• How much time it took until the event occurred
In summary, this sub-chapter explains how to account for these considerations and conduct survival analysis using the mlr3proba extension package.
Density Estimation
This sub-chapter explains how to conduct (unconditional) density estimation in mlr3. Density estimation is used to estimate the probability density function of a continuous variable. Unconditional density estimation is an unsupervised task so there is no ‘value’ to predict, instead densities are estimated.
This sub-chapter explains how to estimate probability distributions for continuous variables using the mlr3proba extension package.
Spatiotemporal Analysis
Spatiotemporal analysis data observations entail reference information about spatial and temporal characteristics. One of the largest issues of spatiotemporal data analysis is the inevitable presence of auto-correlation in the data. Auto-correlation is especially severe in data with marginal spatiotemporal variation. The sub-chapter on Spatiotemporal analysis provides instructions on how to account for spatiotemporal data.
Ordinal Analysis
This is work in progress. See mlr3ordinal for the current state.
Functional Analysis
Functional analysis contains data that consists of curves varying over a continuum e.g. time, frequency or wavelength. This type of analysis is frequently used when examining measurements over various time points. Steps on how to accommodate functional data structures in mlr3 are explained in the functional analysis sub-chapter.
Multilabel Classification
Multilabel classification deals with objects that can belong to more than one category at the same time. Numerous target labels are attributed to a single observation. Working with multilabel data requires one to use modified algorithms, to accommodate data specific characteristics. Two approaches to multilabel classification are prominently used:
• The problem transformation method
Instructions on how to deal with multilabel classification in mlr3 can be found in this sub-chapter.
Cost Sensitive Classification
This sub-chapter deals with the implementation of cost-sensitive classification. Regular classification aims to minimize the misclassification rate and thus all types of misclassification errors are deemed equally severe. Cost-sensitive classification is a setting where the costs caused by different kinds of errors are not assumed to be equal. The objective is to minimize the expected costs.
Analytical data for a big credit institution is used as a use case to illustrate the different features. Firstly, the sub-chapter provides guidance on how to implement a first model. Subsequently, the sub-chapter contains instructions on how to modify cost sensitivity measures, thresholding and threshold tuning.
Cluster Analysis
Cluster analysis aims to group data into clusters such that objects that are similar end up in the same cluster.
Fundamentally, clustering and classification are similar. However, clustering is an unsupervised task because observations do not contain true labels while in classification, labels are needed in order to train a model.
This sub-chapter explains how to perform cluster analysis in mlr3 with the help of mlr3cluster extension package.
## 8.1 Survival Analysis
Survival analysis is a sub-field of supervised machine learning in which the aim is to predict the survival distribution of a given individual. Arguably the main feature of survival analysis is that unlike classification and regression, learners are trained on two features:
1. the time until the event takes place
2. the event type: either censoring or death.
At a particular time-point, an individual is either: alive, dead, or censored. Censoring occurs if it is unknown if an individual is alive or dead. For example, say we are interested in patients in hospital and every day it is recorded if they are alive or dead, then after a patient leaves it is unknown if they are alive or dead, hence they are censored. If there was no censoring, then ordinary regression analysis could be used instead. Furthermore, survival data contains solely positive values and therefore needs to be transformed to avoid biases.
Note that survival analysis accounts for both censored and uncensored observations while adjusting respective model parameters.
The package mlr3proba extends mlr3 with the following objects for survival analysis:
For a good introduction to survival analysis see Modelling Survival Data in Medical Research .
Unlike TaskClassif and TaskRegr which have a single ‘target’ argument, TaskSurv mimics the survival::Surv object and has three-four target arguments (dependent on censoring type). A TaskSurv can be constructed with the function as_task_surv():
library("mlr3")
library("mlr3proba")
library("survival")
time = "start", time2 = "stop", type = "interval")
## <TaskSurv:interval_censored> (178 x 7)
## * Target: start, stop, event
## * Properties: -
## * Features (4):
## - dbl (2): enum, rx
## - int (2): number, size
# type = "right" is default
time = "time", event = "status", type = "right")
print(task)
## <TaskSurv:right_censored> (300 x 5)
## * Target: time, status
## * Properties: -
## * Features (3):
## - int (1): litter
## - dbl (1): rx
## - chr (1): sex
# the target column is a survival object:
head(task$truth()) ## [1] 101+ 49 104+ 91+ 104+ 102+ # kaplan-meier plot library("mlr3viz") autoplot(task) ## Registered S3 method overwritten by 'GGally': ## method from ## +.gg ggplot2 ### 8.1.2 Predict Types - crank, lp, and distr Every PredictionSurv object can predict one or more of: • lp - Linear predictor calculated as the fitted coefficients multiplied by the test data. • distr - Predicted survival distribution, either discrete or continuous. Implemented in distr6. • crank - Continuous risk ranking. lp and crank can be used with measures of discrimination such as the concordance index. Whilst lp is a specific mathematical prediction, crank is any continuous ranking that identifies who is more or less likely to experience the event. So far the only implemented learner that only returns a continuous ranking is surv.svm. If a PredictionSurv returns an lp then the crank is identical to this. Otherwise crank is calculated as the expectation of the predicted survival distribution. Note that for linear proportional hazards models, the ranking (but not necessarily the crank score itself) given by lp and the expectation of distr, is identical. The example below uses the rats task shipped with mlr3proba. task = tsk("rats") learn = lrn("surv.coxph") train_set = sample(task$nrow, 0.8 * task$nrow) test_set = setdiff(seq_len(task$nrow), train_set)
learn$train(task, row_ids = train_set) prediction = learn$predict(task, row_ids = test_set)
print(prediction)
## <PredictionSurv> for 60 observations:
## row_ids time status crank lp distr
## 6 102 FALSE -3.1870 -3.1870 <list[1]>
## 10 91 FALSE -2.2719 -2.2719 <list[1]>
## 17 62 FALSE -3.1621 -3.1621 <list[1]>
## ---
## 291 104 FALSE 0.2882 0.2882 <list[1]>
## 297 79 TRUE 0.3007 0.3007 <list[1]>
## 299 104 FALSE -2.5765 -2.5765 <list[1]>
### 8.1.3 Composition
Finally we take a look at the PipeOps implemented in mlr3proba, which are used for composition of predict types. For example, a predict linear predictor does not have a lot of meaning by itself, but it can be composed into a survival distribution. See mlr3pipelines for full tutorials and details on PipeOps.
library("mlr3pipelines")
library("mlr3learners")
# PipeOpDistrCompositor - Train one model with a baseline distribution,
# (Kaplan-Meier or Nelson-Aalen), and another with a predicted linear predictor.
# remove the factor column for support with glmnet
task$select(c("litter", "rx")) learner_lp = lrn("surv.glmnet") learner_distr = lrn("surv.kaplan") prediction_lp = learner_lp$train(task)$predict(task) prediction_distr = learner_distr$train(task)$predict(task) prediction_lp$distr
# Doesn't need training. Base = baseline distribution. ph = Proportional hazards.
pod = po("compose_distr", form = "ph", overwrite = FALSE)
prediction = pod$predict(list(base = prediction_distr, pred = prediction_lp))$output
# Now we have a predicted distr!
prediction$distr # This can all be simplified by using the distrcompose pipeline glm.distr = ppl("distrcompositor", learner = lrn("surv.glmnet"), estimator = "kaplan", form = "ph", overwrite = FALSE, graph_learner = TRUE) glm.distr$train(task)$predict(task) ### 8.1.4 Benchmark Experiment Finally, we conduct a small benchmark study on the rats task using some of the integrated survival learners: library("mlr3learners") task = tsk("rats") # some integrated learners learners = lrns(c("surv.coxph", "surv.kaplan", "surv.ranger")) print(learners) ## [[1]] ## <LearnerSurvCoxPH:surv.coxph> ## * Model: - ## * Parameters: list() ## * Packages: survival, distr6 ## * Predict Type: distr ## * Feature types: logical, integer, numeric, factor ## * Properties: weights ## ## [[2]] ## <LearnerSurvKaplan:surv.kaplan> ## * Model: - ## * Parameters: list() ## * Packages: survival, distr6 ## * Predict Type: crank ## * Feature types: logical, integer, numeric, character, factor, ordered ## * Properties: missings ## ## [[3]] ## <LearnerSurvRanger:surv.ranger> ## * Model: - ## * Parameters: num.threads=1 ## * Packages: ranger ## * Predict Type: distr ## * Feature types: logical, integer, numeric, character, factor, ordered ## * Properties: importance, oob_error, weights # Harrell's C-Index for survival measure = msr("surv.cindex") print(measure) ## <MeasureSurvCindex:surv.harrell_c> ## * Packages: - ## * Range: [0, 1] ## * Minimize: FALSE ## * Parameters: list() ## * Properties: - ## * Predict type: crank ## * Return type: Score set.seed(1) bmr = benchmark(benchmark_grid(task, learners, rsmp("cv", folds = 3))) bmr$aggregate(measure)
## nr resample_result task_id learner_id resampling_id iters
## 1: 1 <ResampleResult[20]> rats surv.coxph cv 3
## 2: 2 <ResampleResult[20]> rats surv.kaplan cv 3
## 3: 3 <ResampleResult[20]> rats surv.ranger cv 3
## surv.harrell_c
## 1: 0.7671
## 2: 0.5000
## 3: 0.7737
autoplot(bmr, measure = measure)
The experiment indicates that both the Cox PH and the random forest have better discrimination than the Kaplan-Meier baseline estimator, but that the machine learning random forest is not consistently better than the interpretable Cox PH.
## 8.2 Density Estimation
Density estimation is the learning task to find the unknown distribution from which an i.i.d. data set is generated. We interpret this broadly, with this distribution not necessarily being continuous (so may possess a mass not density). The conditional case, where a distribution is predicted conditional on covariates, is known as ‘probabilistic supervised regression’, and will be implemented in mlr3proba in the near-future. Unconditional density estimation is viewed as an unsupervised task. For a good overview to density estimation see Density estimation for statistics and data analysis .
The package mlr3proba extends mlr3 with the following objects for density estimation:
In this example we demonstrate the basic functionality of the package on the faithful data from the datasets package. This task ships as pre-defined TaskDens with mlr3proba.
library("mlr3")
library("mlr3proba")
print(task)
## <TaskDens:precip> (70 x 1)
## * Target: -
## * Properties: -
## * Features (1):
## - dbl (1): precip
# histogram and density plot
library("mlr3viz")
autoplot(task, type = "overlay")
## stat_bin() using bins = 30. Pick better value with binwidth.
Unconditional density estimation is an unsupervised method. Hence, TaskDens is an unsupervised task which inherits directly from Task unlike TaskClassif and TaskRegr. However, TaskDens still has a target argument and a $truth field defined by: • target - the name of the variable in the data for which to estimate density • $truth - the values of the target column (which is not the true density, which is always unknown)
### 8.2.1 Train and Predict
Density learners have train and predict methods, though being unsupervised, ‘prediction’ is actually ‘estimation’. In training, a distr6 object is created, see here for full tutorials on how to access the probability density function, pdf, cumulative distribution function, cdf, and other important fields and methods. The predict method is simply a wrapper around self$model$pdf and if available self$model$cdf, i.e. evaluates the pdf/cdf at given points. Note that in prediction the points to evaluate the pdf and cdf are determined by the target column in the TaskDens object used for testing.
# create task and learner
task_faithful = TaskDens$new(id = "eruptions", backend = datasets::faithful$eruptions)
learner = lrn("dens.hist")
# train/test split
train_set = sample(task_faithful$nrow, 0.8 * task_faithful$nrow)
test_set = setdiff(seq_len(task_faithful$nrow), train_set) # fitting KDE and model inspection learner$train(task_faithful, row_ids = train_set)
learner$model ##$distr
## Histogram()
##
## $hist ##$breaks
## [1] 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5
##
## $counts ## [1] 39 30 5 7 22 59 52 3 ## ##$density
## [1] 0.35945 0.27650 0.04608 0.06452 0.20276 0.54378 0.47926 0.02765
##
## $mids ## [1] 1.75 2.25 2.75 3.25 3.75 4.25 4.75 5.25 ## ##$xname
## [1] "dat"
##
## $equidist ## [1] TRUE ## ## attr(,"class") ## [1] "histogram" ## ## attr(,"class") ## [1] "dens.hist" class(learner$model)
## [1] "dens.hist"
# make predictions for new data
prediction = learner$predict(task_faithful, row_ids = test_set) Every PredictionDens object can estimate: • pdf - probability density function Some learners can estimate: • cdf - cumulative distribution function ### 8.2.2 Benchmark Experiment Finally, we conduct a small benchmark study on the precip task using some of the integrated survival learners: # some integrated learners learners = lrns(c("dens.hist", "dens.kde")) print(learners) ## [[1]] ## <LearnerDensHistogram:dens.hist> ## * Model: - ## * Parameters: list() ## * Packages: distr6 ## * Predict Type: pdf ## * Feature types: integer, numeric ## * Properties: - ## ## [[2]] ## <LearnerDensKDE:dens.kde> ## * Model: - ## * Parameters: kernel=Epan, bandwidth=silver ## * Packages: distr6 ## * Predict Type: pdf ## * Feature types: integer, numeric ## * Properties: missings # Logloss for probabilistic predictions measure = msr("dens.logloss") print(measure) ## <MeasureDensLogloss:dens.logloss> ## * Packages: - ## * Range: [0, Inf] ## * Minimize: TRUE ## * Parameters: list() ## * Properties: - ## * Predict type: pdf set.seed(1) bmr = benchmark(benchmark_grid(task, learners, rsmp("cv", folds = 3))) bmr$aggregate(measure)
## nr resample_result task_id learner_id resampling_id iters dens.logloss
## 1: 1 <ResampleResult[20]> precip dens.hist cv 3 4.396
## 2: 2 <ResampleResult[20]> precip dens.kde cv 3 4.818
autoplot(bmr, measure = measure)
The results of this experiment show that the sophisticated Penalized Density Estimator does not outperform the baseline Histogram, but that the Kernel Density Estimator has at least consistently better (i.e. lower logloss) results.
## 8.3 Spatiotemporal Analysis
Data observations may entail reference information about spatial or temporal characteristics. Spatial information is stored as coordinates, usually named “x” and “y” or “lat”/“lon”. Treating spatiotemporal data using non-spatial data methods can lead to over-optimistic performance estimates. Hence, methods specifically designed to account for the special nature of spatiotemporal data are needed.
In the mlr3 framework, the following packages relate to this field:
The following (sub-)sections introduce the potential pitfalls of spatiotemporal data in machine learning and how to account for it. Note that not all functionality will be covered, and that some of the used packages are still in early lifecycles. If you want to contribute to one of the packages mentioned above, please contact Patrick Schratz.
### 8.3.1 Creating a spatial Task
To make use of spatial resampling methods, a {mlr3} task that is aware of its spatial characteristic needs to be created. Two child classes exist in {mlr3spatiotempcv} for this purpose:
• TaskClassifST
• TaskRegrST
To create one of these, one can either pass a sf object as the “backend” directly:
# create 'sf' object
data_sf = sf::st_as_sf(ecuador, coords = c("x", "y"), crs = 32717)
task = TaskClassifST$new("ecuador_sf", backend = data_sf, target = "slides", positive = "TRUE" ) or use a plain data.frame. In this case, the constructor of TaskClassifST needs a few more arguments: data = mlr3::as_data_backend(ecuador) task = TaskClassifST$new("ecuador",
backend = data, target = "slides",
positive = "TRUE", extra_args = list(coordinate_names = c("x", "y"),
crs = 32717)
)
Now this Task can be used as a normal {mlr3} task in any kind of modeling scenario.
### 8.3.2 Autocorrelation
Data which includes spatial or temporal information requires special treatment in machine learning (similar to survival, ordinal and other task types listed in the special tasks chapter). In contrast to non-spatial/non-temporal data, observations inherit a natural grouping, either in space or time or in both space and time . This grouping causes observations to be autocorrelated, either in space (spatial autocorrelation (SAC)), time (temporal autocorrelation (TAC)) or both space and time (spatiotemporal autocorrelation (STAC)). For simplicity, the acronym STAC is used as a generic term in the following chapter for all the different characteristics introduced above.
What effects does STAC have in statistical/machine learning?
The overarching problem is that STAC violates the assumption that the observations in the train and test datasets are independent . If this assumption is violated, the reliability of the resulting performance estimates, for example retrieved via cross-validation, is decreased. The magnitude of this decrease is linked to the magnitude of STAC in the dataset, which cannot be determined easily.
One approach to account for the existence of STAC is to use dedicated resampling methods. mlr3spatiotemporal provides access to the most frequently used spatiotemporal resampling methods. The following example showcases how a spatial dataset can be used to retrieve a bias-reduced performance estimate of a learner.
The following examples use the ecuador dataset created by Jannes Muenchow. It contains information on the occurrence of landslides (binary) in the Andes of Southern Ecuador. The landslides were mapped from aerial photos taken in 2000. The dataset is well suited to serve as an example because it it relatively small and of course due to the spatial nature of the observations. Please refer to Muenchow, Brenning, and Richter (2012) for a detailed description of the dataset.
To account for the spatial autocorrelation probably present in the landslide data, we will make use one of the most used spatial partitioning methods, a cluster-based k-means grouping , ("spcv_coords" in mlr3spatiotemporal). This method performs a clustering in 2D space which contrasts with the commonly used random partitioning for non-spatial data. The grouping has the effect that train and test data are more separated in space as they would be by conducting a random partitioning, thereby reducing the effect of STAC.
By contrast, when using the classical random partitioning approach with spatial data, train and test observations would be located side-by-side across the full study area (a visual example is provided further below). This leads to a high similarity between train and test sets, resulting in “better” but biased performance estimates in every fold of a CV compared to the spatial CV approach. However, these low error rates are mainly caused due to the STAC in the observations and the lack of appropriate partitioning methods and not by the power of the fitted model.
### 8.3.3 Spatial CV vs. Non-Spatial CV
In the following a spatial and a non-spatial CV will be conducted to showcase the mentioned performance differences.
The performance of a simple classification tree ("classif.rpart") is evaluated on a random partitioning ("repeated_cv") with four folds and two repetitions. The chosen evaluation measure is “classification error” ("classif.ce"). The only difference in the spatial setting is that "repeated_spcv_coords" is chosen instead of "repeated_cv".
#### 8.3.3.1 Non-Spatial CV
library("mlr3")
library("mlr3spatiotempcv")
set.seed(42)
# be less verbose
lgr::get_logger("bbotk")$set_threshold("warn") lgr::get_logger("mlr3")$set_threshold("warn")
learner = lrn("classif.rpart", maxdepth = 3, predict_type = "prob")
resampling_nsp = rsmp("repeated_cv", folds = 4, repeats = 2)
rr_nsp = resample(
resampling = resampling_nsp)
rr_nsp$aggregate(measures = msr("classif.ce")) ## classif.ce ## 0.3389 #### 8.3.3.2 Spatial CV task = tsk("ecuador") learner = lrn("classif.rpart", maxdepth = 3, predict_type = "prob") resampling_sp = rsmp("repeated_spcv_coords", folds = 4, repeats = 2) rr_sp = resample( task = task, learner = learner, resampling = resampling_sp) rr_sp$aggregate(measures = msr("classif.ce"))
## classif.ce
## 0.4125
Here, the classification tree learner is around 0.05 percentage points worse when using Spatial Cross-Validation (SpCV) compared to Non-Spatial Cross-Validation (NSpCV). The magnitude of this difference is variable as it depends on the dataset, the magnitude of STAC and the learner itself. For algorithms with a higher tendency of overfitting to the training set, the difference between the two methods will be larger.
### 8.3.4 Visualization of Spatiotemporal Partitions
Every partitioning method in mlr3spatiotemporal comes with a generic plot() method to visualize the created groups. In a 2D space this happens via ggplot2 while for spatiotemporal methods 3D visualizations via plotly are created.
autoplot(resampling_sp, task, fold_id = c(1:4), size = 0.7) *
ggplot2::scale_y_continuous(breaks = seq(-3.97, -4, -0.01)) *
ggplot2::scale_x_continuous(breaks = seq(-79.06, -79.08, -0.01))
Note that setting the correct CRS for the given data is important which is done during task creation Spatial offsets of up to multiple meters may occur if the wrong CRS is supplied initially.
This example used an already created task via the sugar function tsk(). In practice however, one needs to create a spatiotemporal task via TaskClassifST()/TaskRegrST() and set the crs argument.
The spatial grouping of the k-means based approach above contrasts visually ver well compared to the NSpCV (random) partitioning:
autoplot(resampling_nsp, task, fold_id = c(1:4), size = 0.7) *
ggplot2::scale_y_continuous(breaks = seq(-3.97, -4, -0.01)) *
ggplot2::scale_x_continuous(breaks = seq(-79.06, -79.08, -0.01))
### 8.3.5 Spatial Block Visualization
The spcv-block method makes use of rectangular blocks to divide the study area into equally-sized parts. These blocks can be visualized by their spatial location and fold ID to get a better understanding how these influenced the final partitions.
task = tsk("ecuador")
resampling = rsmp("spcv_block", range = 1000L)
resampling$instantiate(task) ## Visualize train/test splits of multiple folds autoplot(resampling, task, size = 0.7, fold_id = c(1, 2), show_blocks = TRUE, show_labels = TRUE) * ggplot2::scale_x_continuous(breaks = seq(-79.085, -79.055, 0.01)) ### 8.3.6 Choosing a Resampling Method While the example used the "spcv_coords" method, this does not mean that this method is the best or only method suitable for this task. Even though this method is quite popular, it was mainly chosen because of the clear visual grouping differences compared to random partitioning. In fact, most often multiple spatial partitioning methods can be used for a dataset. It is recommended (required) that users familiarize themselves with each implemented method and decide which method to choose based on the specific characteristics of the dataset. For almost all methods implemented in mlr3spatiotemporal, there is a scientific publication describing the strengths and weaknesses of the respective approach (either linked in the help file of mlr3spatiotemporal or its respective dependency packages). In the example above, a cross-validation without hyperparameter tuning was shown. If a nested CV is desired, it is recommended to use the same spatial partitioning method for the inner loop (= tuning level). See Schratz et al. (2019) for more details and chapter 11 of Geocomputation with R 3. A list of all implemented methods in mlr3spatiotemporal can be found in the Getting Started vignette of the package. If you want to learn even more about the field of spatial partitioning, STAC and the problems associated with it, the work of Prof. Hanna Meyer is very much recommended for further reference. ### 8.3.7 Spatial Prediction Experimental support for spatial prediction with terra, raster, stars and sf objects is available in mlr3spatial. Until the package is released on CRAN, please see the package vignettes for usage examples. ## 8.4 Ordinal Analysis This is work in progress. See mlr3ordinal for the current state of the implementation. ## 8.5 Functional Analysis Functional data is data containing an ordering on the dimensions. This implies that functional data consists of curves varying over a continuum, such as time, frequency, or wavelength. ### 8.5.1 How to model functional data? There are two ways to model functional data: • Modification of the learner, so that the learner is suitable for the functional data • Modification of the task, so that the task matches the standard- or classification-learner The development has not started yet, we are looking for contributors. Open an issue in mlr3fda if you are interested! ## 8.6 Multilabel Classification Multilabel classification deals with objects that can belong to more than one category at the same time. The development has not started yet, we are looking for contributes. Open an issue in mlr3multioutput if you are interested! ## 8.7 Cost-Sensitive Classification In regular classification the aim is to minimize the misclassification rate and thus all types of misclassification errors are deemed equally severe. A more general setting is cost-sensitive classification. Cost sensitive classification does not assume that the costs caused by different kinds of errors are equal. The objective of cost sensitive classification is to minimize the expected costs. Imagine you are an analyst for a big credit institution. Let’s also assume that a correct decision of the bank would result in 35% of the profit at the end of a specific period. A correct decision means that the bank predicts that a customer will pay their bills (hence would obtain a loan), and the customer indeed has good credit. On the other hand, a wrong decision means that the bank predicts that the customer’s credit is in good standing, but the opposite is true. This would result in a loss of 100% of the given loan. Good Customer (truth) Bad Customer (truth) Good Customer (predicted) + 0.35 - 1.0 Bad Customer (predicted) 0 0 Expressed as costs (instead of profit), we can write down the cost-matrix as follows: costs = matrix(c(-0.35, 0, 1, 0), nrow = 2) dimnames(costs) = list(response = c("good", "bad"), truth = c("good", "bad")) print(costs) ## truth ## response good bad ## good -0.35 1 ## bad 0.00 0 An exemplary data set for such a problem is the German Credit task: library("mlr3") task = tsk("german_credit") table(task$truth())
##
## 700 300
The data has 70% customers who are able to pay back their credit, and 30% bad customers who default on the debt. A manager, who doesn’t have any model, could decide to give either everybody a credit or to give nobody a credit. The resulting costs for the German credit data are:
# nobody:
(700 * costs[2, 1] + 300 * costs[2, 2]) / 1000
## [1] 0
# everybody
(700 * costs[1, 1] + 300 * costs[1, 2]) / 1000
## [1] 0.055
If the average loan is $20,000, the credit institute would lose more than one million dollar if it would grant everybody a credit: # average profit * average loan * number of customers 0.055 * 20000 * 1000 ## [1] 1100000 Our goal is to find a model which minimizes the costs (and thereby maximizes the expected profit). ### 8.7.1 A First Model For our first model, we choose an ordinary logistic regression (implemented in the add-on package mlr3learners). We first create a classification task, then resample the model using a 10-fold cross validation and extract the resulting confusion matrix: library("mlr3learners") learner = lrn("classif.log_reg") rr = resample(task, learner, rsmp("cv")) confusion = rr$prediction()$confusion print(confusion) ## truth ## response good bad ## good 598 156 ## bad 102 144 To calculate the average costs like above, we can simply multiply the elements of the confusion matrix with the elements of the previously introduced cost matrix, and sum the values of the resulting matrix: avg_costs = sum(confusion * costs) / 1000 print(avg_costs) ## [1] -0.0533 With an average loan of$20,000, the logistic regression yields the following costs:
avg_costs * 20000 * 1000
## [1] -1066000
Instead of losing over $1,000,000, the credit institute now can expect a profit of more than$1,000,000.
### 8.7.2 Cost-sensitive Measure
Our natural next step would be to further improve the modeling step in order to maximize the profit. For this purpose we first create a cost-sensitive classification measure which calculates the costs based on our cost matrix. This allows us to conveniently quantify and compare modeling decisions. Fortunately, there already is a predefined measure Measure for this purpose: MeasureClassifCosts:
cost_measure = msr("classif.costs", costs = costs)
print(cost_measure)
## <MeasureClassifCosts:classif.costs>
## * Packages: -
## * Range: [-Inf, Inf]
## * Minimize: TRUE
## * Parameters: normalize=TRUE
## * Predict type: response
If we now call resample() or benchmark(), the cost-sensitive measures will be evaluated. We compare the logistic regression to a simple featureless learner and to a random forest from package ranger :
learners = list(
lrn("classif.log_reg"),
lrn("classif.featureless"),
lrn("classif.ranger")
)
cv3 = rsmp("cv", folds = 3)
bmr$aggregate(cost_measure) ## nr resample_result task_id learner_id resampling_id ## 1: 1 <ResampleResult[20]> german_credit classif.log_reg cv ## 2: 2 <ResampleResult[20]> german_credit classif.featureless cv ## 3: 3 <ResampleResult[20]> german_credit classif.ranger cv ## iters classif.costs ## 1: 3 -0.06014 ## 2: 3 0.05502 ## 3: 3 -0.04792 As expected, the featureless learner is performing comparably bad. The logistic regression and the random forest work equally well. ### 8.7.3 Thresholding Although we now correctly evaluate the models in a cost-sensitive fashion, the models themselves are unaware of the classification costs. They assume the same costs for both wrong classification decisions (false positives and false negatives). Some learners natively support cost-sensitive classification (e.g., XXX). However, we will concentrate on a more generic approach which works for all models which can predict probabilities for class labels: thresholding. Most learners can calculate the probability $$p$$ for the positive class. If $$p$$ exceeds the threshold $$0.5$$, they predict the positive class, and the negative class otherwise. For our binary classification case of the credit data, the we primarily want to minimize the errors where the model predicts “good”, but truth is “bad” (i.e., the number of false positives) as this is the more expensive error. If we now increase the threshold to values $$> 0.5$$, we reduce the number of false negatives. Note that we increase the number of false positives simultaneously, or, in other words, we are trading false positives for false negatives. # fit models with probability prediction learner = lrn("classif.log_reg", predict_type = "prob") rr = resample(task, learner, rsmp("cv")) p = rr$prediction()
print(p)
## <PredictionClassif> for 1000 observations:
## row_ids truth response prob.good prob.bad
## 29 good good 0.9125 0.08754
## ---
## 965 good bad 0.4222 0.57780
## 987 good bad 0.1881 0.81189
## 988 good good 0.9511 0.04886
# helper function to try different threshold values interactively
with_threshold = function(p, th) {
p$set_threshold(th) list(confusion = p$confusion, costs = p$score(measures = cost_measure, task = task)) } with_threshold(p, 0.5) ##$confusion
## truth
## good 603 155
##
## $costs ## classif.costs ## -0.05605 with_threshold(p, 0.75) ##$confusion
## truth
## good 468 77
##
## $costs ## classif.costs ## -0.0868 with_threshold(p, 1.0) ##$confusion
## truth
## good 0 1
##
## $costs ## classif.costs ## 0.001 # TODO: include plot of threshold vs performance Instead of manually trying different threshold values, one uses use optimize() to find a good threshold value w.r.t. our performance measure: # simple wrapper function which takes a threshold and returns the resulting model performance # this wrapper is passed to optimize() to find its minimum for thresholds in [0.5, 1] f = function(th) { with_threshold(p, th)$costs
}
best = optimize(f, c(0.5, 1))
print(best)
## $minimum ## [1] 0.7411 ## ##$objective
## classif.costs
## -0.08895
# optimized confusion matrix:
with_threshold(p, best$minimum)$confusion
## truth
## good 477 78
## bad 223 222
Note that the function optimize() is intended for unimodal functions and therefore may converge to a local optimum here. See below for better alternatives to find good threshold values.
### 8.7.4 Threshold Tuning
Before we start, we have load all required packages:
library("mlr3")
library("mlr3pipelines")
library("mlr3tuning")
## Loading required package: paradox
### 8.7.5 Adjusting thresholds: Two strategies
Currently mlr3pipelines offers two main strategies towards adjusting classification thresholds. We can either expose the thresholds as a hyperparameter of the Learner by using PipeOpThreshold. This allows us to tune the thresholds via an outside optimizer from mlr3tuning.
Alternatively, we can also use PipeOpTuneThreshold which automatically tunes the threshold after each learner is fit.
In this blog-post, we’ll go through both strategies.
### 8.7.6 PipeOpThreshold
PipeOpThreshold can be put directly after a Learner.
A simple example would be:
gr = lrn("classif.rpart", predict_type = "prob") %>>% po("threshold")
l = as_learner(gr)
Note, that predict_type = “prob” is required for po("threshold") to have any effect.
The thresholds are now exposed as a hyperparameter of the GraphLearner we created:
l$param_set ## <ParamSetCollection> ## id class lower upper nlevels default ## 1: classif.rpart.cp ParamDbl 0 1 Inf 0.01 ## 2: classif.rpart.keep_model ParamLgl NA NA 2 FALSE ## 3: classif.rpart.maxcompete ParamInt 0 Inf Inf 4 ## 4: classif.rpart.maxdepth ParamInt 1 30 30 30 ## 5: classif.rpart.maxsurrogate ParamInt 0 Inf Inf 5 ## 6: classif.rpart.minbucket ParamInt 1 Inf Inf <NoDefault[3]> ## 7: classif.rpart.minsplit ParamInt 1 Inf Inf 20 ## 8: classif.rpart.surrogatestyle ParamInt 0 1 2 0 ## 9: classif.rpart.usesurrogate ParamInt 0 2 3 2 ## 10: classif.rpart.xval ParamInt 0 Inf Inf 10 ## 11: threshold.thresholds ParamUty NA NA Inf <NoDefault[3]> ## value ## 1: ## 2: ## 3: ## 4: ## 5: ## 6: ## 7: ## 8: ## 9: ## 10: 0 ## 11: 0.5 We can now tune those thresholds from the outside as follows: Before tuning, we have to define which hyperparameters we want to tune over. In this example, we only tune over the thresholds parameter of the threshold pipeop. you can easily imagine, that we can also jointly tune over additional hyperparameters, i.e. rpart’s cp parameter. As the Task we aim to optimize for is a binary task, we can simply specify the threshold param: library("paradox") ps = ps(threshold.thresholds = p_dbl(lower = 0, upper = 1)) We now create a AutoTuner, which automatically tunes the supplied learner over the ParamSet we supplied above. at = AutoTuner$new(
learner = l,
resampling = rsmp("cv", folds = 3L),
measure = msr("classif.ce"),
search_space = ps,
terminator = trm("evals", n_evals = 5L),
tuner = tnr("random_search")
)
at$train(tsk("german_credit")) ## INFO [09:57:40.254] [bbotk] Starting to optimize 1 parameter(s) with '<OptimizerRandomSearch>' and '<TerminatorEvals> [n_evals=5, k=0]' ## INFO [09:57:40.300] [bbotk] Evaluating 1 configuration(s) ## INFO [09:57:40.651] [bbotk] Result of batch 1: ## INFO [09:57:40.653] [bbotk] threshold.thresholds classif.ce runtime_learners ## INFO [09:57:40.653] [bbotk] 0.2554 0.288 0.186 ## INFO [09:57:40.653] [bbotk] uhash ## INFO [09:57:40.653] [bbotk] 43270072-7e63-4372-9c9f-518bc558d851 ## INFO [09:57:40.656] [bbotk] Evaluating 1 configuration(s) ## INFO [09:57:41.014] [bbotk] Result of batch 2: ## INFO [09:57:41.016] [bbotk] threshold.thresholds classif.ce runtime_learners ## INFO [09:57:41.016] [bbotk] 0.04085 0.301 0.168 ## INFO [09:57:41.016] [bbotk] uhash ## INFO [09:57:41.016] [bbotk] 0c978bb4-d125-4e67-bf76-e33d56edc7a0 ## INFO [09:57:41.019] [bbotk] Evaluating 1 configuration(s) ## INFO [09:57:41.335] [bbotk] Result of batch 3: ## INFO [09:57:41.337] [bbotk] threshold.thresholds classif.ce runtime_learners ## INFO [09:57:41.337] [bbotk] 0.6273 0.287 0.178 ## INFO [09:57:41.337] [bbotk] uhash ## INFO [09:57:41.337] [bbotk] 825dc589-505d-4056-b0ee-681e02b72299 ## INFO [09:57:41.340] [bbotk] Evaluating 1 configuration(s) ## INFO [09:57:41.656] [bbotk] Result of batch 4: ## INFO [09:57:41.658] [bbotk] threshold.thresholds classif.ce runtime_learners ## INFO [09:57:41.658] [bbotk] 0.2061 0.291 0.172 ## INFO [09:57:41.658] [bbotk] uhash ## INFO [09:57:41.658] [bbotk] ff3c4a0a-ea59-45d0-9592-15330bbf2aec ## INFO [09:57:41.661] [bbotk] Evaluating 1 configuration(s) ## INFO [09:57:41.972] [bbotk] Result of batch 5: ## INFO [09:57:41.974] [bbotk] threshold.thresholds classif.ce runtime_learners ## INFO [09:57:41.974] [bbotk] 0.9274 0.688 0.171 ## INFO [09:57:41.974] [bbotk] uhash ## INFO [09:57:41.974] [bbotk] 2625beca-03bb-4f01-98e6-d6a49a008cf5 ## INFO [09:57:41.983] [bbotk] Finished optimizing after 5 evaluation(s) ## INFO [09:57:41.984] [bbotk] Result: ## INFO [09:57:41.985] [bbotk] threshold.thresholds learner_param_vals x_domain classif.ce ## INFO [09:57:41.985] [bbotk] 0.6273 <list[2]> <list[1]> 0.287 Inside the trafo, we simply collect all set params into a named vector via map_dbl and store it in the threshold.thresholds slot expected by the learner. Again, we create a AutoTuner, which automatically tunes the supplied learner over the ParamSet we supplied above. One drawback of this strategy is, that this requires us to fit a new model for each new threshold setting. While setting a threshold and computing performance is relatively cheap, fitting the learner is often more computationally demanding. A better strategy is therefore often to optimize the thresholds separately after each model fit. ### 8.7.7 PipeOpTunethreshold PipeOpTuneThreshold on the other hand works together with PipeOpLearnerCV. It directly optimizes the cross-validated predictions made by this PipeOp. This is done in order to avoid over-fitting the threshold tuning. A simple example would be: gr = po("learner_cv", lrn("classif.rpart", predict_type = "prob")) %>>% po("tunethreshold") l2 = as_learner(gr) Note, that predict_type = “prob” is required for po("tunethreshold") to work. Additionally, note that this time no threshold parameter is exposed, it is automatically tuned internally. l2$param_set
## <ParamSetCollection>
## id class lower upper nlevels
## 1: classif.rpart.resampling.method ParamFct NA NA 2
## 2: classif.rpart.resampling.folds ParamInt 2 Inf Inf
## 3: classif.rpart.resampling.keep_response ParamLgl NA NA 2
## 4: classif.rpart.cp ParamDbl 0 1 Inf
## 5: classif.rpart.keep_model ParamLgl NA NA 2
## 6: classif.rpart.maxcompete ParamInt 0 Inf Inf
## 7: classif.rpart.maxdepth ParamInt 1 30 30
## 8: classif.rpart.maxsurrogate ParamInt 0 Inf Inf
## 9: classif.rpart.minbucket ParamInt 1 Inf Inf
## 10: classif.rpart.minsplit ParamInt 1 Inf Inf
## 11: classif.rpart.surrogatestyle ParamInt 0 1 2
## 12: classif.rpart.usesurrogate ParamInt 0 2 3
## 13: classif.rpart.xval ParamInt 0 Inf Inf
## 14: classif.rpart.affect_columns ParamUty NA NA Inf
## 15: tunethreshold.measure ParamUty NA NA Inf
## 16: tunethreshold.optimizer ParamUty NA NA Inf
## 17: tunethreshold.log_level ParamUty NA NA Inf
## default value
## 1: <NoDefault[3]> cv
## 2: <NoDefault[3]> 3
## 3: <NoDefault[3]> FALSE
## 4: 0.01
## 5: FALSE
## 6: 4
## 7: 30
## 8: 5
## 9: <NoDefault[3]>
## 10: 20
## 11: 0
## 12: 2
## 13: 10 0
## 14: <Selector[1]>
## 15: <NoDefault[3]> classif.ce
## 16: <NoDefault[3]> gensa
## 17: <function[1]> warn
Note that we can set rsmp("intask") as a resampling strategy for “learner_cv” in order to evaluate predictions on the “training” data. This is generally not advised, as it might lead to over-fitting on the thresholds but can significantly reduce runtime.
For more information, see the post on Threshold Tuning on the mlr3 gallery.
## 8.8 Cluster Analysis
Cluster analysis is a type of unsupervised machine learning where the goal is to group data into clusters, where each cluster contains similar observations. The similarity is based on specified metrics that are task and application dependent. Cluster analysis is closely related to classification in a sense that each observation needs to be assigned to a cluster or a class. However, unlike classification problems where each observation is labeled, clustering works on data sets without true labels or class assignments.
The package mlr3cluster extends mlr3 with the following objects for cluster analysis:
Since clustering is a type of unsupervised learning, TaskClust is slightly different from TaskRegr and TaskClassif objects. More specifically:
• truth() function is missing because observations are not labeled.
• target field is empty and will return character(0) if accessed anyway.
Additionally, LearnerClust provides two extra fields that are absent from supervised learners:
• assignments returns cluster assignments for training data. It return NULL if accessed before training.
• save_assignments is a boolean field that controls whether or not to store training set assignments in a learner.
Finally, PredictionClust contains additional two fields:
• partition stores cluster partitions.
• prob stores cluster probabilities for each observation.
### 8.8.1 Train and Predict
Clustering learners provide both train and predict methods. The analysis typically consists of building clusters using all available data. To be consistent with the rest of the library, we refer to this process as training.
Some learners can assign new observations to existing groups with predict. However, prediction does not always make sense, as it is the case for hierarchical clustering. In hierarchical clustering, the goal is to build a hierarchy of nested clusters by either splitting large clusters into smaller ones or merging smaller clusters into bigger ones. The final result is a tree or dendrogram which can change if a new data point is added. For consistency with the rest of the ecosystem, mlr3cluster offers predict method for hierarchical clusterers but it simply assigns all points to a specified number of clusters by cutting the resulting tree at a corresponding level. Moreover, some learners estimate the probability of each observation belonging to a given cluster. predict_types field gives a list of prediction types for each learner.
After training, the model field stores a learner’s model that looks different for each learner depending on the underlying library. predict returns a PredictionClust object that gives a simplified view of the learned model. If the data given to the predict method is the same as the one on which the learner was trained, predict simply returns cluster assignments for the “training” observations. On the other hand, if the test set contains new data, predict will estimate cluster assignments for that data set. Some learners do not support estimating cluster partitions on new data and will instead return assignments for training data and print a warning message.
In the following example, a $k$-means learner is applied on the US arrest data set. The class labels are predicted and the contribution of the task features to assignment of the respective class are visualized.
library("mlr3")
library("mlr3cluster")
library("mlr3viz")
set.seed(1L)
print(task)
## <TaskClust:usarrests> (50 x 4)
## * Target: -
## * Properties: -
## * Features (4):
## - int (2): Assault, UrbanPop
## - dbl (2): Murder, Rape
autoplot(task)
# create a k-means learner
learner = lrn("clust.kmeans")
# assigning each observation to one of the two clusters (default in clust.kmeans)
learner$train(task) learner$model
## K-means clustering with 2 clusters of sizes 21, 29
##
## Cluster means:
## Assault Murder Rape UrbanPop
## 1 255.0 11.857 28.11 67.62
## 2 109.8 4.841 16.25 64.03
##
## Clustering vector:
## [1] 1 1 1 1 1 1 2 1 1 1 2 2 1 2 2 2 2 1 2 1 2 1 2 1 2 2 2 1 2 2 1 1 1 2 2 2 2 2
## [39] 2 1 2 1 1 2 2 2 2 2 2 2
##
## Within cluster sum of squares by cluster:
## [1] 41637 54762
## (between_SS / total_SS = 72.9 %)
##
## Available components:
##
## [1] "cluster" "centers" "totss" "withinss" "tot.withinss"
## [6] "betweenss" "size" "iter" "ifault"
# make "predictions" for the same data
prediction = learner$predict(task) autoplot(prediction, task) ### 8.8.2 Measures The difference between supervised and unsupervised learning is that there is no ground truth data in unsupervised learning. In a supervised setting, such as classification, we would need to compare our predictions to true labels. Since clustering is an example of unsupervised learning, there are no true labels to which we can compare. However, we can still measure the quality of cluster assignments by quantifying how closely objects within the same cluster are related (cluster cohesion) as well as how distinct different clusters are from each other (cluster separation). To assess the quality of clustering, there are a few built-in evaluation metrics available. One of them is within sum of squares (WSS) which calculates the sum of squared differences between observations and centroids. WSS is useful because it quantifies cluster cohesion. The range of this measure is $$[0, \infty)$$ where a smaller value means that clusters are more compact. Another measure is silhouette quality index that quantifies how well each point belongs to its assigned cluster versus neighboring cluster. Silhouette values are in $$[-1, 1]$$ range. Points with silhouette closer to: • 1 are well clustered • 0 lie between two clusters • -1 likely placed in the wrong cluster The following is an example of conducting a benchmark experiment with various learners on iris data set without target variable and assessing the quality of each learner with both within sum of squares and silhouette measures. design = benchmark_grid( tasks = TaskClust$new("iris", iris[-5]),
learners = list(
lrn("clust.kmeans", centers = 3L),
lrn("clust.pam", k = 2L),
lrn("clust.cmeans", centers = 3L)),
resamplings = rsmp("insample"))
print(design)
## task learner resampling
## 3: <TaskClust[43]> <LearnerClustCMeans[36]> <ResamplingInsample[19]>
# execute benchmark
bmr = benchmark(design)
# define measure
measures = list(msr("clust.wss"), msr("clust.silhouette"))
bmr$aggregate(measures) ## nr resample_result task_id learner_id resampling_id iters clust.wss ## 1: 1 <ResampleResult[20]> iris clust.kmeans insample 1 78.85 ## 2: 2 <ResampleResult[20]> iris clust.pam insample 1 153.33 ## 3: 3 <ResampleResult[20]> iris clust.cmeans insample 1 79.03 ## clust.silhouette ## 1: 0.5555 ## 2: 0.7158 ## 3: 0.5493 The experiment shows that using k-means algorithm with three centers produces a better within sum of squares score than any other learner considered. However, pam (partitioning around medoids) learner with two clusters performs the best when considering silhouette measure which takes into the account both cluster cohesion and separation. ### 8.8.3 Visualization Cluster analysis in mlr3 is integrated with mlr3viz which provides a number of useful plots. Some of those plots are shown below. task = TaskClust$new("iris", iris[-5])
learner = lrn("clust.kmeans")
learner$train(task) prediction = learner$predict(task)
# performing PCA on task and showing assignments
autoplot(prediction, task, type = "pca")
# same as above but with probability ellipse that assumes normal distribution
autoplot(prediction, task, type = "pca", frame = TRUE, frame.type = 'norm')
task = tsk("usarrests")
learner = lrn("clust.agnes")
learner$train(task) # dendrogram for hierarchical clustering autoplot(learner) # advanced dendrogram options from factoextra::fviz_dend autoplot(learner, k = learner$param_set$values$k, rect_fill = TRUE,
rect = TRUE, rect_border = c("red", "cyan"))
Silhouette plots can help to visually assess the quality of the analysis and help choose a number of clusters for a given data set. The red dotted line shows the mean silhouette value and each bar represents a data point. If most points in each cluster have an index around or higher than mean silhouette, the number of clusters is chosen well.
# silhouette plot allows to visually inspect the quality of clustering
task = TaskClust$new("iris", iris[-5]) learner = lrn("clust.kmeans") learner$param_set$values = list(centers = 5L) learner$train(task)
prediction = learner$predict(task) autoplot(prediction, task, type = "sil") The plot shows that all points in cluster 5 and almost all points in clusters 4, 2 and 1 are below average silhouette index. This means that a lot of observations lie either on the border of clusters or are likely assigned to the wrong cluster. learner = lrn("clust.kmeans") learner$param_set$values = list(centers = 2L) learner$train(task)
autoplot(prediction, task, type = "sil") | 2021-10-27 10:37:03 | {"extraction_info": {"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, "math_score": 0.3424372673034668, "perplexity": 4442.800762181251}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588113.25/warc/CC-MAIN-20211027084718-20211027114718-00006.warc.gz"} |
https://pypi.org/project/mud/ | Maximal Updated Density equations for Data-Consistent Inversion
## MUD
Analytical solutions and some associated utility functions for computing Maximal Updated Density (MUD) parameter estimates for Data-Consistent Inversion.
### Description
Maximal Updated Density Points are the values which maximize an updated density, analogous to how a MAP (Maximum A-Posteriori) point maximizes a posterior density from Bayesian inversion. Updated densities differ from posteriors in that they are the solution to a different problem which seeks to match the push-forward of the updated density to a specified observed distribution.
## Project details
Uploaded source
Uploaded py2 py3 | 2022-12-09 19:51:08 | {"extraction_info": {"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, "math_score": 0.2506352365016937, "perplexity": 5170.512576259212}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711475.44/warc/CC-MAIN-20221209181231-20221209211231-00688.warc.gz"} |
https://chocoluffy.com/2016/01/12/Flexbox-website-layout-Intro/ | My first attempt to build up a responsive website using Flexbox layout. Check Live Demo here!
check this post on the complete guide about Flexbox layout.
Below is the code that can be applied to many text\presentation related website. [1] font is cool and elegant. [2] layout is fast since we set div to display: flex. We won’t be bother to type that anymore, focusing on “position” and “stretch”
especially the font, it is so beautiful!
These units are vh (viewport height), vw (viewport width), vmin (viewport minimum length) and vmax (viewport maximum length). we set the vh to 100 like this:
for the parent div container, since we want that parent div can take up a viewport height. vh stands for “viewport height”
Note that flex-direction indicate items’ aligning, rows or columns. Then align-items means main-axis and justify-content is for sub-axis.
## flex
flex: 1 意为 flex: 1 1 auto
## Summary For All
basically using some flex-related properties to structure the whole website.
[1] to formate those section-like website, need to use section tag in html file and set the corresponding containter to be 100vh which means the viewport height, so that each section can be strecthed to adapt to your screen height. (which is pretty elegant)
we usually use
these properties as container’s property.
Note that we have main-axis and sub-axis, which will help you position the items by align-items: center; justify-content: center;
[2] structure inside each section. we may use another flexbox container inside one parent container to hold up more items like navigation or some scrum-map. And in this way, the flex-direction may usually be the opposite to the parent container. For the child container, we may wonder to stretch the block to whatever proportion to the whole layout. we want to use
which also applies the same logic from main-axis and sub-axis when setting their values.
Note that the flex-basis is pretty useful when you try to eliminate the effect of inner text to the block when stretching since align-self: stretch ONLY stretch the empty space to full length! | 2019-01-23 18:13:33 | {"extraction_info": {"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, "math_score": 0.3571004867553711, "perplexity": 5959.304956713183}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547584336901.97/warc/CC-MAIN-20190123172047-20190123194047-00543.warc.gz"} |
https://www.tutorialspoint.com/if-two-positive-integers-p-and-q-can-be-expressed-as-p-ab-2-and-q-a-3b-a-b-being-prime-numbers-then-lcm-p-q-is-a-ab-b-a-2b-2-c-a-3b-2-d-a-3b-3 | # If two positive integers $p$ and $q$ can be expressed as $p = ab^2$, and $q = a^3b; a, b$ being prime numbers, then LCM $(p, q)$ is(A) $ab$(B) $a^2b^2$(C) $a^3b^2$(D) $a^3b^3$
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Given:
Two positive integers $p$ and $q$ can be expressed as $p = ab^2$, and $q = a^3b; a, b$ being prime numbers.
To find:
Here we have to find LCM $(p, q)$.
Solution:
We know that,
LCM is the product of the greatest power of each prime factor involved in the numbers.
$p = ab^2$
$= a \times b^2$
$q = a^3b$
$= a^3 \times b$
Therefore,
LCM of $p$ and $q$ is,
LCM $(ab^2, a^3b) = b^2 \times a^3$
$= a^3b^2$
Updated on 10-Oct-2022 13:27:06 | 2022-12-06 00:09:03 | {"extraction_info": {"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, "math_score": 0.17345359921455383, "perplexity": 2637.0432247623567}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711064.71/warc/CC-MAIN-20221205232822-20221206022822-00211.warc.gz"} |
https://scikit-multiflow.github.io/scikit-multiflow/_autosummary/skmultiflow.drift_detection.PageHinkley.html | # skmultiflow.drift_detection.PageHinkley¶
class skmultiflow.drift_detection.PageHinkley(min_instances=30, delta=0.005, threshold=50, alpha=0.9999)[source]
Page-Hinkley method for concept drift detection.
Notes
This change detection method works by computing the observed values and their mean up to the current moment. Page-Hinkley won’t output warning zone warnings, only change detections. The method works by means of the Page-Hinkley test 1. In general lines it will detect a concept drift if the observed mean at some instant is greater then a threshold value lambda.
References
1
E. S. Page. 1954. Continuous Inspection Schemes. Biometrika 41, 1/2 (1954), 100–115.
Parameters
• min_instances (int (default=30)) – The minimum number of instances before detecting change.
• delta (float (default=0.005)) – The delta factor for the Page Hinkley test.
• threshold (int (default=50)) – The change detection threshold (lambda).
• alpha (float (default=1 - 0.0001)) – The forgetting factor, used to weight the observed value and the mean.
Examples
>>> # Imports
>>> import numpy as np
>>> from skmultiflow.drift_detection import PageHinkley
>>> ph = PageHinkley()
>>> # Simulating a data stream as a normal distribution of 1's and 0's
>>> data_stream = np.random.randint(2, size=2000)
>>> # Changing the data concept from index 999 to 2000
>>> for i in range(999, 2000):
... data_stream[i] = np.random.randint(4, high=8)
>>> # Adding stream elements to the PageHinkley drift detector and verifying if drift occurred
>>> for i in range(2000):
... if ph.detected_change():
... print('Change has been detected in data: ' + str(data_stream[i]) + ' - of index: ' + str(i))
__init__(min_instances=30, delta=0.005, threshold=50, alpha=0.9999)[source]
Initialize self. See help(type(self)) for accurate signature.
Methods
__init__([min_instances, delta, threshold, …]) Initialize self. Add a new element to the statistics This function returns whether concept drift was detected or not. If the change detector supports the warning zone, this function will return whether it’s inside the warning zone or not. Collects and returns the information about the configuration of the estimator Returns the length estimation. get_params([deep]) Get parameters for this estimator. Resets the change detector parameters. set_params(**params) Set the parameters of this estimator.
Attributes
estimator_type
add_element(x)[source]
Add a new element to the statistics
Parameters
x (numeric value) – The observed value, from which we want to detect the concept change.
Notes
After calling this method, to verify if change was detected, one should call the super method detected_change, which returns True if concept drift was detected and False otherwise.
detected_change()[source]
This function returns whether concept drift was detected or not.
Returns
Whether concept drift was detected or not.
Return type
bool
detected_warning_zone()[source]
If the change detector supports the warning zone, this function will return whether it’s inside the warning zone or not.
Returns
Whether the change detector is in the warning zone or not.
Return type
bool
get_info()[source]
Collects and returns the information about the configuration of the estimator
Returns
Configuration of the estimator.
Return type
string
get_length_estimation()[source]
Returns the length estimation.
Returns
The length estimation
Return type
int
get_params(deep=True)[source]
Get parameters for this estimator.
Parameters
deep (boolean, optional) – If True, will return the parameters for this estimator and contained subobjects that are estimators.
Returns
params – Parameter names mapped to their values.
Return type
mapping of string to any
reset()[source]
Resets the change detector parameters.
set_params(**params)[source]
Set the parameters of this estimator.
The method works on simple estimators as well as on nested objects (such as pipelines). The latter have parameters of the form <component>__<parameter> so that it’s possible to update each component of a nested object.
Returns
Return type
self | 2019-09-21 04:57:51 | {"extraction_info": {"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, "math_score": 0.19676539301872253, "perplexity": 5322.869718711685}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514574265.76/warc/CC-MAIN-20190921043014-20190921065014-00248.warc.gz"} |
https://www.speedsolving.com/threads/multiblindfold-discussion-pbs-successes-fails.50757/page-88 | Multiblindfold Discussion (PBs, Successes, Fails)
fun at the joy
Member
another MBLD attempt, another huge disappointment:
6/10 in 54:20.07 [38:40.71]
DNFs:
Cube1: 2FE - somehow memoed EF as my last edge targets instead of DL
idk how I missed that rip
Cube3: 4E4C - 1 move yeeeee
Cube4: 4E4C - guess what? ... another move
Cube8: 3E, 1 flip and 2C - had a pop during edges
8/10 DNFs of my last 3 attempts were exec errors rip
Dylan Swarts
Member
Oh I'm so glad this thread is being used a bit more!
I broke a very big barrier today, getting my first sub 2 avg/cube!
Result Time[memo] Points avg memo/cube avg exec/cube total avg/cube Date Errors 22/28 55:12[36:45] 16 1:18.75(PB!!) 39.53(Very good hehe) 1:58.28 (PB!!) 1 May 2020 C8: messed up edge memo a bit.. C11: inversed TK corners. C13: recalled/executed MQ instead of MJ (both are people who I can't really picture well, should change one of the pairs). C14: memo error involving Weak Swap, fair enough. C15: inversed ND corners. C26: recalled FG and TM wrong way round.
As usual my accuracy isn't pretty, and up to now (just shy of 30 multi attempts this year) I have only scored 2 results on my top 10 list. I definitely need to clear that list of 2019 entries and fill it with 2020 ones. (also maybe I can try get to 120 attempts this year, I'll exclude attempts under 10 cubes I guess)
Stay safe!
dudefaceguy
Member
4/6 for my first 6-cube attempt. Unfortunately I messed up my timer, but I also took note of the clock and I finished in almost exactly one hour. I spent about 35 minutes on memo. Looks like I messed up one commutator on each of the failures, as there were 3 wrong corners on one cube and 3 wrong edges on the other.
This is about what I expected - 2/3 accuracy is in line with previous attempts when I pushed my limits a bit. I'm actually very happy with the accuracy considering that I recently switched buffers and am also experimenting with different types of commutators. I definitely did a few comms that I have never tried before.
All in all it was an extremely pleasant hour in the park and I count it as a personal success.
dudefaceguy
Member
Did anyone see Rowe Hessler's 60/61 WB video?
His recent improvement, which you can see on his channel, is very interesting. It's also nice to see Graham Siggins being a gentleman in the comments.
Dylan Swarts
Member
It is absolutely insane, the rate of improvement. Though I'd say a big part is potentially there as your 3bld speed increases so by having ex. a 30s avg, already means that a few weeks of practice can probably get someone to mid 20 or higher cubes from almost no prior multiblind practice. But this of course is not the case, although it probably helped a small amount. Very crazy indeed. 60 is a goal still far away for me, but I might still reach it, if I keep practicing that is.
dudefaceguy
Member
It is absolutely insane, the rate of improvement. Though I'd say a big part is potentially there as your 3bld speed increases so by having ex. a 30s avg, already means that a few weeks of practice can probably get someone to mid 20 or higher cubes from almost no prior multiblind practice. But this of course is not the case, although it probably helped a small amount. Very crazy indeed. 60 is a goal still far away for me, but I might still reach it, if I keep practicing that is.
In my very limited experience, doing more cubes isn't really harder, it just takes longer. By that I mean that my accuracy isn't worse when I do more cubes. Memory palace techniques are pretty easy to use and work very well almost instantly, at least for the volume of information you need to memorize for MBLD. So, I think you are right that fast 3BLD times translate well to MBLD.
Dylan Swarts
Member
Only thing I would see as harder, would be that it gets much more tiring the more cubes you do, actually no. The longer it takes. (which usually means more cubes) So yeah, it doesn't get harder in any other manner than that.
dudefaceguy
Member
In my very limited experience, doing more cubes isn't really harder, it just takes longer. By that I mean that my accuracy isn't worse when I do more cubes. Memory palace techniques are pretty easy to use and work very well almost instantly, at least for the volume of information you need to memorize for MBLD. So, I think you are right that fast 3BLD times translate well to MBLD.
Only thing I would see as harder, would be that it gets much more tiring the more cubes you do, actually no. The longer it takes. (which usually means more cubes) So yeah, it doesn't get harder in any other manner than that.
When considering the statements above, please ignore the complete failure I just did: 1/6 cubes. It's like, not statistically significant or something. I was only talking about people who actually have good memo and execution techniques, so my own attempts are excluded.
Among the things that went wrong:
1. Sat outside for this attempt, sun moved, started to feel myself getting a sunburn during the second cube, couldn't move because I had already started solving.
2. A family stopped right in front of my porch to have a very loud political discussion during my execution.
3. Scramble with 4 flipped edges. I was not prepared for this.
4. Scramble with 4 twisted corners. Not prepared for this either. I encounter these so rarely that I have little practice solving them - I will do all of my flips as comms in 3BLD, just to get more practice with this.
5. Forgot to execute parity because my parity memo is not good enough.
6. Many execution mistakes, mostly corner orientation errors.
I could feel the solves going wrong as I was doing them, which is just a terrible feeling. I cheered myself up with some slow 3BLD successes, but even there I made many execution mistakes - thankfully I was able to reverse them, including realizing a mistake 2 commutators later, reversing both, and executing correctly. Quarantine is messing with my sleep, so I will have to try again when I am well rested.
Edit: An hour-long MBLD attempt with 6 cubes is also really pushing my limits - I should not really expect success. I get impatient with my memo and execution, and don't review properly - both because I am feeling impatient and because I need to hurry up to meet the time limits. I will try a few 3 and 4 cube attempts until I can get perfect accuracy, then move up.
Edit 2: I cheered myself up with an untimed 3/3 success. It's difficult for me to get an uninterrupted stretch of time for an official attempt, so I just memorized in pieces during odd moments of free time, then executed very carefully when I had a few minutes to myself. This actually achieves my overall goal of meditative relaxation very well. I did all flips as comms and made sure that my parity memo was extremely vivid, which fixed the problems from my previous attempt.
Last edited:
kubesolver
Yay!
I just did my first 2-cube attempt with a 2/2 in 10 minutes (roughly 7+3)!
Dylan Swarts
Member
Result Time[memo] Points Avg memo/cube Avg exec/cube Total avg/cube Date Errors Comment 19/30 58:45[40:59] 8 1:21.96 35.53 1:57.5 13 May 2020 2 or 3 recall, rest are mostly dumb. first 30 subhour!!
Note: this is PB total/cube, pb2 for exec/cube and a decent memo/cube.
I have mixed emotions about this attempt, since the result is so bad, but then, the time is so good actually. I must admit I had a duplicate scramble, and upon noticing(starting to memo it) I hand scrambled it. I know that in comp, this cube is to be discarded and not counted as an attempted cube, but I really wanted to do a 30 cube attempt today, so I did hand scramble that cube. (From quick search through video, I believe I dnf'd it anyway)
That said, I am glad that I have successfully attempted 30 cubes within 1 hour. Thinking that I would need to halve my time/cube to do 60 seems impossible really haha. But then, if you told me in Jan 2019 that I would be able to solve 15 cubes blindfolded, I would not have been so sure of that. Goal of 42 still stands for the year, but I will need to up my practicing if I want to reach that. I wanted to be able to do 34 subhour by now (as planned in Dec 2019) and as you can see, I am quite a bit far off from that.
I will probably mainly turn my focus to accuracy now, as this is how my top 10 result list looks:
Multi-BLD Result Time Pts Date Errors PB 25/26 54:14[35:42] 24 2019/12/29 Cube 18: slipped out of my hands and E slice moved. Also I memo'd a wrong sticker so corners would've been off by 2 twists anyway PB2 24/26 58:58[38:33] 22 2020/03/20 C20: skipped over OJ lp in exec. C25:inversed GQ edge comm. PB3 20/21 49:12[33:28] 19 2019/11/04 Cube 7: 3e, might've skipped in exec? PB4 19/20 52:12[34:03] 18 2019/10/13 Cube 18: Memo'd H instead of F for edges. Ironically this cube had only 7 edge letters and I still messed that up haha PB5 19/20 53:02[34:45] 18 2019/10/10 Cube 4: executed BT instead of BE and randomly realised I had done the wrong pair 4 cubes later. Couldn't figure out what I had done wrong (BT) and so just left it. Can't believe that is the only error of 20 cubes. PB6 18/19 52:12[35:06] 17 2019/10/02 Cube 8: stupidly did GX wrong by leaving out the last two moves.. could've been 19/19. this is slow also.. very tired, surprised by accuracy tbh. very not good memo.. PB7 19/21 53:31[35:53] 17 2020/03/11 C7: Messed up something during edges, 5e. C16: did PN instead of TN for corners (exec) PB8 21/25 57:52[40:04] 17 2019/12/17 Cube 9: auf error; cube 20: inversed corner PR; cube 23: recalled HL instead of KW; cube 24: missed a two cycle. PB9 17/18 41:23[27:11] 16 2019/10/28 Cube 10: ended cycle too early. Should've done J before the rest of edges. oops PB10 17/18 43:33[28:38] 16 2019/09/23 Cube4: did DJ edge comm wrong
Pretty 2019ish right?
Worst of all, I mostly attempt 25+ nowadays, and these all rank from 18-26 cube attempts (only 3 are 25 cubes or more)
So I think I will go down and do 17 cube attempts (2 blocks + 3bld) and drill getting fast times (so that I practice the faster, less drilled type memo that is a necessity on big attempts) as well as being very accurate (like 15 solved every time would be considered okay).
And yet again, practice 3bld DYLAN! Many errors occur from my stupid corner twist exec, or those comms that have simultaneous U and D moves (was a common error in today's attempt), so practicing 3bld will help. I am practicing somewhat more frequently, but not enough that I am happy with. anyway, I hope that whoever read to here, was not too bored.
Stay safe, happy cubing!
dudefaceguy
Member
Result Time[memo] Points Avg memo/cube Avg exec/cube Total avg/cube Date Errors Comment 19/30 58:45[40:59] 8 1:21.96 35.53 1:57.5 13 May 2020 2 or 3 recall, rest are mostly dumb. first 30 subhour!!
Note: this is PB total/cube, pb2 for exec/cube and a decent memo/cube.
I have mixed emotions about this attempt, since the result is so bad, but then, the time is so good actually. I must admit I had a duplicate scramble, and upon noticing(starting to memo it) I hand scrambled it. I know that in comp, this cube is to be discarded and not counted as an attempted cube, but I really wanted to do a 30 cube attempt today, so I did hand scramble that cube. (From quick search through video, I believe I dnf'd it anyway)
That said, I am glad that I have successfully attempted 30 cubes within 1 hour. Thinking that I would need to halve my time/cube to do 60 seems impossible really haha. But then, if you told me in Jan 2019 that I would be able to solve 15 cubes blindfolded, I would not have been so sure of that. Goal of 42 still stands for the year, but I will need to up my practicing if I want to reach that. I wanted to be able to do 34 subhour by now (as planned in Dec 2019) and as you can see, I am quite a bit far off from that.
I will probably mainly turn my focus to accuracy now, as this is how my top 10 result list looks:
Multi-BLD Result Time Pts Date Errors PB 25/26 54:14[35:42] 24 2019/12/29 Cube 18: slipped out of my hands and E slice moved. Also I memo'd a wrong sticker so corners would've been off by 2 twists anyway PB2 24/26 58:58[38:33] 22 2020/03/20 C20: skipped over OJ lp in exec. C25:inversed GQ edge comm. PB3 20/21 49:12[33:28] 19 2019/11/04 Cube 7: 3e, might've skipped in exec? PB4 19/20 52:12[34:03] 18 2019/10/13 Cube 18: Memo'd H instead of F for edges. Ironically this cube had only 7 edge letters and I still messed that up haha PB5 19/20 53:02[34:45] 18 2019/10/10 Cube 4: executed BT instead of BE and randomly realised I had done the wrong pair 4 cubes later. Couldn't figure out what I had done wrong (BT) and so just left it. Can't believe that is the only error of 20 cubes. PB6 18/19 52:12[35:06] 17 2019/10/02 Cube 8: stupidly did GX wrong by leaving out the last two moves.. could've been 19/19. this is slow also.. very tired, surprised by accuracy tbh. very not good memo.. PB7 19/21 53:31[35:53] 17 2020/03/11 C7: Messed up something during edges, 5e. C16: did PN instead of TN for corners (exec) PB8 21/25 57:52[40:04] 17 2019/12/17 Cube 9: auf error; cube 20: inversed corner PR; cube 23: recalled HL instead of KW; cube 24: missed a two cycle. PB9 17/18 41:23[27:11] 16 2019/10/28 Cube 10: ended cycle too early. Should've done J before the rest of edges. oops PB10 17/18 43:33[28:38] 16 2019/09/23 Cube4: did DJ edge comm wrong
Pretty 2019ish right?
Worst of all, I mostly attempt 25+ nowadays, and these all rank from 18-26 cube attempts (only 3 are 25 cubes or more)
So I think I will go down and do 17 cube attempts (2 blocks + 3bld) and drill getting fast times (so that I practice the faster, less drilled type memo that is a necessity on big attempts) as well as being very accurate (like 15 solved every time would be considered okay).
And yet again, practice 3bld DYLAN! Many errors occur from my stupid corner twist exec, or those comms that have simultaneous U and D moves (was a common error in today's attempt), so practicing 3bld will help. I am practicing somewhat more frequently, but not enough that I am happy with. anyway, I hope that whoever read to here, was not too bored.
Stay safe, happy cubing!
Don't be discouraged - you are continually pushing yourself so you should expect results like this. If you push time your accuracy will suffer, and vice versa. Congratulations on a new PB time!
Dylan Swarts
Member
28/40 1:30:56[1:03:04] 16 2020/06/05 C1: idc, that whole thing was a mess with memo. C4: recalled the wrong pair, objects too similar. C6: misscramble I think, pretty bad dnf.C9: also. C11: no idea, could be memo.C15: auf somewhere. C17: thought I shoulda flipped UR but was wrong. C19: missed BD flip I think. C22: did GP instead of GN, smol brain.C33: missed an E slice, also recalled out of order. C34: idk C37: something dumb with corners idk man.
Here is a thing. First block went bad, and that caused all reviews involving it to go bad, as well as executing that block. Other than that I was a bit distant during the whole attempt, or I'm just tired. Could definitely have been like 5 -10 mins shorter, but won't focus on this since I need to work on 30 and 33 still. And my accuracy.
Dylan Swarts
Member
Today was a very great attempt indeed. Everything came together nicely, accuracy, and speed across both memorization and execution. Best of all, they ALL achieved better than they have ever done before! (not accuracy, that can't be measured the same but ok)
Today's attempt was meant to be a large mbld (that I haven't subhoured or that I can't subhour consistently) and so I went for 33, as I have subhoured 30 cubes twice (58 and 57 mins respectively). Accuracy was not really the concern here, and it is one of the reasons why this attempt is so insane.
During exec I did feel I was going a bit slower and I felt a lot more confident about my accuracy. I realized this and tried to speed up. (funny enough, this block had 2 DNF's, I don't think it is directly related though) Anyway, here it is:
Result Time[Memo] Points Avg memo/cube Avg exec/cube Avg total/cube Date Errors Comment 29/33 59:58[41:43] 25 1:15.84 33.15 1:48.99 17 June 2020 C6: Forgot 2nd edge image. C18: not really sure, Looks like a move somewhere. C23: error while doing NX edges. C28: recalled first two LPs out of order. All PBs. Exec, Memo, Total, and Points. Very crazy having everything not only come together but also be better than they usually are.
I had 1.29 to spare haha, and I'm really glad I did finish before. I'm really happy about that memo/cube. I do think my proportions are a bit out, I feel memo should be faster, but I am feeling a drop in time there. Really great to have a PB memo/cube on this amount of cubes. The avg is also a massive drop. 8 seconds I believe.
Anyway that is all from me, video will hopefully be uploaded in the next few days. 42 is seeming quite possible at this point. Happy cubing, stay safe!
Dylan Swarts
Member
Well, hello again..
uhm.. so.. This is a post I did not think I would be making for quite a long while..
I broke my PB again, and the greatest thing is, it didn't take 7 months! Literally just the following attempt.
Result Time[memo] Points Avg memo/cube Avg exec/cube Avg total/cube Date Errors Comment 28/30 55:09[39:25] 26 1:18.83 31.46 1:50.29 20 June 2020 C6: pretty sure I screwed up edge memo a little and did not look at the cube again to notice. C8: did PenGuin instead of Du(Q)cky, when I got to a similar pair a few cubes later I realized I accidentally did PenGuin there but I could not do anything anymore at that point so it is okay. PB exec by 1.69 seconds. Errors are decent, my 3bld practice has helped my exec tremendously.
The 33 cube attempt was arguably better, I do think the memo was quite much better. I was somewhat distracted during the second block. It took over 9 minutes for the full 2 pass (1st block was 8:40 for 2 pass) and 3rd block was a bit rough, I think I was tired so things were a bit slow. Memo did not stick too good but after 3rd review of blocks 1-3 I was comfortable with them all. I was a bit unsure of my memo on cubes 25-29 as they were memo'd quite fast and I was not completely comfortable after 2nd pass but I just decided to go with it. Exec went smoothly on all those. Recall was great.
Happy to see that my errors consisted of 0 exec errors. Only some memo screw-up and recall mistake. With my new speed I am seeing many faults in my pairs like MaP and ATlas. Too similar for me to clearly differentiate between. There are a handful more that need attention and this is going to become a big issue when I reach higher memo speeds; it could already be holding me back.
I do expect a good-high accuracy from this point onwards in multi (for the most part anyway), so the new challenge will be lowering that memo down.
I feel my exec is much better than my memo at this point?
anyway, stay safe, happy cubing!
dudefaceguy
Member
Well, hello again..
uhm.. so.. This is a post I did not think I would be making for quite a long while..
I broke my PB again, and the greatest thing is, it didn't take 7 months! Literally just the following attempt.
Result Time[memo] Points Avg memo/cube Avg exec/cube Avg total/cube Date Errors Comment 28/30 55:09[39:25] 26 1:18.83 31.46 1:50.29 20 June 2020 C6: pretty sure I screwed up edge memo a little and did not look at the cube again to notice. C8: did PenGuin instead of Du(Q)cky, when I got to a similar pair a few cubes later I realized I accidentally did PenGuin there but I could not do anything anymore at that point so it is okay. PB exec by 1.69 seconds. Errors are decent, my 3bld practice has helped my exec tremendously.
The 33 cube attempt was arguably better, I do think the memo was quite much better. I was somewhat distracted during the second block. It took over 9 minutes for the full 2 pass (1st block was 8:40 for 2 pass) and 3rd block was a bit rough, I think I was tired so things were a bit slow. Memo did not stick too good but after 3rd review of blocks 1-3 I was comfortable with them all. I was a bit unsure of my memo on cubes 25-29 as they were memo'd quite fast and I was not completely comfortable after 2nd pass but I just decided to go with it. Exec went smoothly on all those. Recall was great.
Happy to see that my errors consisted of 0 exec errors. Only some memo screw-up and recall mistake. With my new speed I am seeing many faults in my pairs like MaP and ATlas. Too similar for me to clearly differentiate between. There are a handful more that need attention and this is going to become a big issue when I reach higher memo speeds; it could already be holding me back.
I do expect a good-high accuracy from this point onwards in multi (for the most part anyway), so the new challenge will be lowering that memo down.
I feel my exec is much better than my memo at this point?
anyway, stay safe, happy cubing!
Congratulations! I think you are seeing the fruits of your prior practice pushing your limits.
Dylan Swarts
Member
Quick one, I have to do homework. Tried 36 cubes, worked out as expected. Sub 1:10 and almost 30 solved..
29/36 1:09:54[48:58] 22 1:21.61 34.89 1:56.5 23 June 2020 C2: messed up LJ. C3: memo'd DW instead of DV. C11: inversed AQ edges. C18:accidentally exec'd corner memo as edges, must've messed up in fixing it. C31: memo error and some where exec nonsense it seems.. C33: inversed VP corners lol. C36: D move.
result, time, pts (if subhour, I actually only solved 15 cubes subhour) memo splits avg, exec, total, date, errors.
Think I will attempt number 30, 33 and 36 from now, goal being to subhour 33 again a couple times until it is consistent and I can hopefully stop doing 30 and then get 36 subhour. Hopefully by end of July, would be really nice. Then I'm almost back on track from my original plan I had at the end of last year. Memo needs to be worked on. First block was very slow today, over 10 min I think. Second block made up for it and was done by 19. 3 pass of 1-3 was done by 35:00 so that is not too bad but I'd like to push for sub 9/block and finish 3rd pass by 30 or less. Exec is not bad, hands were cold today and my mf3rs' are really being difficult with me, will have to fix them up. Not planning on buying new multi cubes out of my own pocket. The extra 3x3s I own should last me to 55 with ease I think.
Cya
fun at the joy
Member
This isn't really anything special but it was my first MultiBLD attempt in 1 month and I don't want Dylan to be the only one to post here.
4/5 in 15:51.69 [11:52.47]
DNF was 1 move (of course)
memo/cube: 2:22.49
exec/cube: 47.84
total/cube: 3:10.33
I plan to practice 5 cube attempts this summer, goal is sub-10 consistently
and of course finally start using the method of loci because memo is really slowing me down
I'll eventually learn 3-style although my exec is already pretty good
Habsen
Member
This isn't really anything special but it was my first MultiBLD attempt in 1 month and I don't want Dylan to be the only one to post here.
4/5 in 15:51.69 [11:52.47]
DNF was 1 move (of course)
memo/cube: 2:22.49
exec/cube: 47.84
total/cube: 3:10.33
I plan to practice 5 cube attempts this summer, goal is sub-10 consistently
and of course finally start using the method of loci because memo is really slowing me down
I'll eventually learn 3-style although my exec is already pretty good
Ok, I'll join the german MultiBLD team. I have been practicing five cube attempts for the last three weeks and got a new PB:
5/5 in 20:25 [12:33]
memo/cube: 2:30.60
exec/cube: 1:34.40
kubesolver
I would like to propose a new formula for MBLD scoring. I don't remember this particular formula being proposed before and I believe it is as simple as the current formula but more often agrees with my intuition about which attempt is better.
In particular believe score like 10/30 is worth more than a DNF and a is better than say 2/2.
The current formula is equivalent to
Code:
Solved + Solved - Attempted
and I suggest to replace addition with multiplication in this formula
Code:
Solved * Solved / Attempted = Solved * (Solved / Attempted) = Solved * accuracy
Compared with current formula it gives a little less penalty for failed cubes,
Looking at a results from one recent competition it would change the order of results in the following way:
Current formula Proposed formula new score 1 28 / 36 28 / 36 21.7 2 15 / 19 15 / 19 11.8 3 12 / 13 15 / 20 (+1) 11.25 4 15 / 20 12 / 13 (-1) 11.07 5 10 / 12 12 / 16 (+1) 9 6 12 / 16 10 / 12 (-1) 8.3 7 7 / 8 7 / 8 6.1 8 3 / 3 5 / 7 (+1) 3.57 9 5 / 7 3 / 3 (-1) 3 10 3 / 4 3 / 4 2.25 11 2 / 4 2 / 4 1.0 | 2021-09-19 08:00:43 | {"extraction_info": {"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, "math_score": 0.46237117052078247, "perplexity": 3839.571260540768}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056752.16/warc/CC-MAIN-20210919065755-20210919095755-00257.warc.gz"} |
http://mathoverflow.net/questions/106467/t2-fibered-k3-surface-with-involution | # $T^2$-fibered K3 surface with involution
Let $S$ be a K3 surface and $f:S\rightarrow \mathbb{P}^1$ a $T^2$-fibration (not necessarily holomorphic, I have a special Langrangian fibration in mind). Assume there is a $k$-section, then a fiber and the $k$-section generate a sublattice $L\subset H^{2}(S,\mathbb{Z})$, which is isomorphic to $U(k)$ (hyperbolic lattice multiplied by $k \in \mathbb{N}$).
Assume also that there is an holomorphic involution $\sigma$ of $S$ such that induced action $\sigma^{*}$ acts as $-id$ on $L$ (especially preserves $L\subset H^{2}(S,\mathbb{Z})$). Is it true that $\sigma$ preserves the fibration? If so, could one tell how $\sigma$ acts on each smooth fiber of $f$? If not so, what additional condition is required?
Edit The original question does not mach much sense. The fibration is NOT holomorphic. the following is the motivation of my question. I have a K3 surface $S$ with an anti-symplectic involution $\sigma$. Assume that, by using another complex structure, we can construct an elliptic fibration $f:S\rightarrow \mathbb{P}^{1}$ (possibly with no section). I hope this map to be a special Lagrangian $T^2$-fibration with respect to the original complex structure. I now want to understnad how $\sigma$ and the map $f$ are related.
-
Are you saying that $\sigma$ takes an effective class to a negative effective class? If not, what does it mean that $\sigma^*$ acts as $-id$? – Sándor Kovács Sep 5 '12 at 23:39
Indeed, $-id$ on an elliptic surface, if it exists and preserves $L$, acts as $id$ on $L$, since it preserves degree and preserves the class of a fiber. If $\sigma$ acts as $id$, then it preserves the fibration, since it preserves the class of the fiber, and the fibration is just the map to $\mathbb P^1$ coming from the global sections of the line bundle whose divisor class is the class of the fiber. – Will Sawin Sep 6 '12 at 0:11
My first question does not make much sense. I made a major change in my question, also adding some motivation of my question. I am sorry for the confusion. – Carmen Sep 6 '12 at 0:31 | 2016-07-29 18:12:47 | {"extraction_info": {"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, "math_score": 0.9657344818115234, "perplexity": 224.7218460772955}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257831770.41/warc/CC-MAIN-20160723071031-00086-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://onepager.togaware.com/faceted-location-scatter-plot.html | ## 11.26 Faceted Location Scatter Plot
REVIEW
ds %>%
ggplot(aes(x=date, y=max_temp)) +
geom_point(alpha=0.05, shape=".") +
geom_smooth(method="gam", formula=y~s(x, bs="cs")) +
facet_wrap(~location) +
theme(axis.text.x=element_text(angle=45, hjust=1)) +
labs(x=vnames["date"], y=vnames["max_temp"])
Partitioning the dataset by a categoric variable reduces the blob effect for big data. The plot uses location as the faceted variable to separately plot each location’s maximum temperature over time. Notice the seasonal effect across all plots, some with quite different patterns.
The plot uses ggplot2::facet_wrap() to separately plot each location. Using ggplot2::geom_point() with alpha= reduces the effect of overlaid points as does using smaller dots on the plots by way of shape=. Together this works to de-clutter the plot and improves the presentation with an emphasis on the patterns. The x-axis tick labels are rotated $$45^\circ$$ using angle=45 within ggplot2::element_text() to avoid the labels overlapping. The hjust=1 forces the labels to be right justified.
Your donation will support ongoing development and give you access to the PDF version of this book. Desktop Survival Guides include Data Science, GNU/Linux, and MLHub. Books available on Amazon include Data Mining with Rattle and Essentials of Data Science. Popular open source software includes rattle, wajig, and mlhub. Hosted by Togaware, a pioneer of free and open source software since 1984. | 2021-06-22 19:41:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.25698935985565186, "perplexity": 7557.967198166829}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488519735.70/warc/CC-MAIN-20210622190124-20210622220124-00241.warc.gz"} |
http://codeforces.com/blog/entry/7847 | zanoes's blog
By zanoes, 6 years ago, ,
312B - Archer
let p=a/b,q=(1-c/d)*(1-a/b). The answer of this problem can be showed as:p*q^0+p*q^1+p*q^2+………… That is the sum of a geometric progression which is infinite but 0<q<1.We can get the limit by the formula:p/(1-q)
311E - Biologist
Obviously, It is about min-cut, which can be solved by max-flow. So, this problem can regarded as the minimum of losing money if we assume SmallR can get all the money and get a total money(sum of wi)at first. Define that, in the min-cut result, the dogs which are connected directly or indirectly to S are 0-gender.otherwise, those to T are 1-gender.We can easily find that the point of a initial-0-gender dog should get a vi weight edge to S.On the other way, initial-1-gender to T. Then, consider the rich men.As to each of rich men, he can appoint only one gender and some dogs.if any one dog he appoint is not the one appointed gender in the result, he can't give SmallR money.How to deal with the rich men in the graph is a difficulty.We can do this, make a new point for a rich man.Then, link the man's point to all points of his dogs by max enough weigh edges(which we can't cut in the min-cut algorithm), and link the man's point to S or T (which depend on that the man's appointed gender is 0 or 1) with a edge of wi weight. lastly run the min-cut(max-flow), we will get the minimum money we will lose. The answer is TotalMoney-LosedMoney. BTW, the issue of g is a piece of cake, we could add it to every special wi but not add it to the total money.
• +5
» 6 years ago, # | 0 Can you explain please your solution to the problem about probability? I am not good at it, that's why i can't understand the idea... Why do you get this p and q?
• » » 6 years ago, # ^ | 0 P-вероятность того, что игра закончится после хода первого игрока.Q-вероятность того, что игра продлится ещё один раунд.И того складываем вероятность, что игра сразу закончится после 1-ого круга, после второго, третьего и т.д.Настя, извини, что не на английском!
• » » » 6 years ago, # ^ | 0 Sorry, but i can understand, why sometimes we have to count product of probabilities of 2 events, and sometimes we have to count their sums?
• » » » » 4 years ago, # ^ | ← Rev. 2 → -10 Excellent question.SmallR, who shoots first can win in the following cases: He shoots the target in the first shot. OR He misses AND his opponent misses AND he shoots the target. OR He misses AND his opponent misses AND he misses AND his opponent misses AND he shoots the target When there is an OR between 2 events(which lead to same result) happening, it means EITHER of them will lead to the same result, so the probability of the result is the SUM OF ALL THESE PROBABILITIES.On the other hand, when there is AND between 2 events, it means that BOTH OF THEM SHOULD HAPPEN TO GET THE DESIRED OUTCOME. So if event A happens with probability = 1/2 and event B happens with probability = 1/2 and event C happens when both A and B happen, then probability of C happening is 1/2 * 1/2 = 1/4, which also makes sense logically. Because C will happen in only 1 out 4 cases, when BOTH A and B have happened. You can imagine the 4 cases: A did not happen AND B did not happen => C did not happen A happened AND B did not happen => C did not happen A did not happen AND B happened => C did not happen A happened AND B happened => C happened So a good rule of thumb is AND means product of probabilities and OR means sum of probabilities.So now we can get the answer to our problem:answer = a/b + ( (1-a/b) * (1-c/d) * a/b ) + ( (1-a/b) * (1-c/d) * (1-a/b) * (1-c/d) * a/b )... and so on.You can sum this up using formula for sum of infinite geometric series.PS: I know this question was asked 3 years ago, but I hope this helps someone.
• » » » » » 3 years ago, # ^ | 0 it helps me more thanq
• » » » » » 2 years ago, # ^ | 0 Helped me , Thanks bro :D
• » » » » » 21 month(s) ago, # ^ | 0 thanks a lot.
• » » » » » 3 months ago, # ^ | 0 See it is 26 July 2019 , and it helped me . :) [user:TombBombadil]
• » » » » » 2 days ago, # ^ | 0 Helped me too, thanks!
» 6 years ago, # | ← Rev. 3 → +28 312B - Archer can also be accepted by Brute Force, during contests this algorithm is more covenient for us. 3783547
• » » 6 years ago, # ^ | 0 I am not good at probabilities. Could you please explain what is the now variable for. I mean, I understand te first two parts and why the for goes up to 1000000, I just can't understand is the update of the now. I also get that (1-z)*(1-r) is the combined probability that there is no winner at all..Thanks
• » » » 6 years ago, # ^ | +18 It means the probability after some steps.
» 4 years ago, # | 0 Hello,This is my submission for problem D.Div1. My Machine output for the sample input is the same as the sample output. However, it displays different output here on CF machine.Can anyone help me with this?
• » » 4 years ago, # ^ | 0 Try to code your own pow function. Function from cmath returns double, that is, for example, pow(3, 3) could return 26.99999 and became 26 when changed to int.
• » » » 4 years ago, # ^ | 0 I edited my code and removed the 'pow' function but I got WA on test 2. Well, I don't know what is the wrong about my implementation. Do you have an idea?Here is my new submission.
• » » » » 4 years ago, # ^ | 0 (a + b) ^ 3 is not equal to a ^ 3 + b ^ 3.
• » » » » » 4 years ago, # ^ | ← Rev. 4 → 0 You mean this line at 'update' function?tree[node] = tree[node * 2] + tree[node * 2 + 1];The tree is built on sum that's why I wrote that line.
• » » » » » 4 years ago, # ^ | ← Rev. 2 → 0 I mean that leaf nodes are raised to power 3 but other nodes are not. Why I should raise non leaf nodes to power 3? | 2019-10-22 20:31:53 | {"extraction_info": {"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, "math_score": 0.46183153986930847, "perplexity": 1500.5066832217783}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987823061.83/warc/CC-MAIN-20191022182744-20191022210244-00008.warc.gz"} |
http://atcoder.noip.space/contest/agc014/a | # Home
Score : $300$ points
### Problem Statement
Takahashi, Aoki and Snuke love cookies. They have $A$, $B$ and $C$ cookies, respectively. Now, they will exchange those cookies by repeating the action below:
• Each person simultaneously divides his cookies in half and gives one half to each of the other two persons.
This action will be repeated until there is a person with odd number of cookies in hand.
How many times will they repeat this action? Note that the answer may not be finite.
### Constraints
• $1 ≤ A,B,C ≤ 10^9$
### Input
Input is given from Standard Input in the following format:
$A$ $B$ $C$
### Output
Print the number of times the action will be performed by the three people, if this number is finite. If it is infinite, print -1 instead.
### Sample Input 1
4 12 20
### Sample Output 1
3
Initially, Takahashi, Aoki and Snuke have $4$, $12$ and $20$ cookies. Then,
• After the first action, they have $16$, $12$ and $8$.
• After the second action, they have $10$, $12$ and $14$.
• After the third action, they have $13$, $12$ and $11$.
Now, Takahashi and Snuke have odd number of cookies, and therefore the answer is $3$.
### Sample Input 2
14 14 14
### Sample Output 2
-1
### Sample Input 3
454 414 444
### Sample Output 3
1 | 2021-11-27 19:55:19 | {"extraction_info": {"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, "math_score": 0.5820735096931458, "perplexity": 1986.9151502941866}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358233.7/warc/CC-MAIN-20211127193525-20211127223525-00549.warc.gz"} |
http://physics.aps.org/articles/v5/32 | # Viewpoint: Homing in on the Higgs Boson
, Department of Physics,University of California, Santa Cruz, 1156 High Street, Santa Cruz, CA 95064, USA
Published March 13, 2012 | Physics 5, 32 (2012) | DOI: 10.1103/Physics.5.32
#### Search for the Higgs Boson in the $H\to W{W}^{\left(*\right)}\to {l}^{+}\nu {l}^{-}\overline{\nu }$ Decay Channel in $pp$ Collisions at $\sqrt{s}=7\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{TeV}$ with the ATLAS Detector
Published March 13, 2012 | PDF (free)
#### Search for the Standard Model Higgs Boson in the Diphoton Decay Channel with $4.9\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\mathrm{fb}}^{-1}$ of $pp$ Collision Data at $\sqrt{s}=7\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{TeV}$ with ATLAS
Published March 13, 2012 | PDF (free)
#### Search for the Standard Model Higgs Boson in the Decay Channel $H\to ZZ\to 4l$ in $pp$ Collisions at $\sqrt{s}=7\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{TeV}$
S. Chatrchyan et al.
Published March 13, 2012 | PDF (free)
The mediators of the weak force—the $W$ and $Z$ bosons—are both roughly $100$ times more massive than the proton, while the photon, which mediates the electromagnetic force, is massless. Based on symmetry principles, one would have expected the $W$, $Z$, and photon (collectively called gauge bosons) to all be massless. In order to generate the required symmetry breaking, one must invent a particular set of new particles and interactions, which ultimately provide an explanation of the origin of mass of the elementary particles. The simplest way to do this, and the one that is incorporated in the standard model of particle physics, predicts the existence of a particle called the Higgs boson$\phantom{\rule{0.278em}{0ex}}$ [1]. The interactions of the Higgs boson with the quarks, leptons, and gauge bosons of the standard model are uniquely predicted by the theory, but the mass of the Higgs boson is a free parameter that can only be determined by experiment [2].
One of the most important missions of the Large Hadron Collider (LHC) is to ascertain the origin of mass by discovering the Higgs boson as envisioned by the standard model (or, if no Higgs is found, perhaps by discovering alternative phenomena that can be attributed to the symmetry-breaking mechanism). Three papers now appearing in Physical Review Letters, two from the ATLAS Collaboration [3, 4] and one from the CMS Collaboration [5], report searches for the Higgs boson in the debris of high-energy proton-proton collisions at the LHC. The results of these searches, and several others being published elsewhere [6], were announced in December, 2011. Collectively, they have shown that the Higgs boson of the standard model, if it exists, must lie in a narrow range of masses around $126$ giga-electron-volts ($\text{GeV}$). Moreover, an excess of events around this mass value provides a tantalizing hint that experimentalists could be on the verge of discovering the long-sought particle.
Although the LHC has taken over the search for the Higgs boson, it does so with the knowledge, accumulated from previous experiments, that already tells us where the Higgs mass is likely—and unlikely—to be. From experiments at the CERN LEP collider [7], which shut down in 2000, it is known that the standard model Higgs boson mass cannot be below $114\phantom{\rule{0.333em}{0ex}}\text{GeV}$, while subsequent data from the Fermilab Tevatron [8], which shut down in 2011, excludes a Higgs boson mass between $156$ and $177\phantom{\rule{0.333em}{0ex}}\text{GeV}$. (As is common in particle physics, mass is reported in terms of its energy equivalent.) Even without a direct discovery, the data from the LEP and Tevatron colliders have been able to place additional constraints on the allowed values of the Higgs mass. Results from measurements performed at the LEP and Tevatron colliders can be compared with theoretical predictions. Based on a statistical analysis of these data in the framework of the standard model (which depends indirectly on the unknown Higgs mass) the Higgs mass cannot be larger than $169\phantom{\rule{0.333em}{0ex}}\text{GeV}\phantom{\rule{0.278em}{0ex}}$[9]. Combining the direct Higgs search limits quoted above with the indirect constraints implies that the standard model is viable if, and only if, the standard model Higgs boson mass lies between $114$ and $156\phantom{\rule{0.333em}{0ex}}\text{GeV}\phantom{\rule{0.278em}{0ex}}$ [10].
If a Higgs boson is produced in a high-energy collision, it should, according to theory, be extremely unstable and immediately decay into lighter elementary particles. The standard model predicts the relative probabilities of a Higgs boson decaying into a particular set of final state particles, as shown in Fig. 1, and it is an excess of these particles relative to background events—the numerous other processes that can yield the same final state particles without there ever having been a Higgs boson—that particle physicists aim to measure.
In 2011, the LHC collided two proton beams at $7$ tera-electron-volts ($\text{TeV}$). On average, more than $100$ million collisions per second took place, but most of these collisions were “conventional,” in that they involved only the most common elementary particles. If the standard model is correct, then on rare occasions, a Higgs boson should have been produced. Assuming the Higgs mass lies in the expected mass range, about $75,000$ Higgs bosons were produced in the ATLAS and CMS collider detectors in 2011.
As shown in Fig. 1, for Higgs masses below $135\phantom{\rule{0.333em}{0ex}}\text{GeV}$, most of these Higgs bosons should decay into a pair of bottom quarks. However, because there are so many bottom quarks produced each second at the LHC, the bottom quark background would overwhelm the tiny signal associated with a Higgs boson. Instead, the rare event in which the Higgs bosons decay into two photons, with a probability of roughly one in five-hundred (the magenta line in Fig. 1), is actually a better place to look for the Higgs boson. In the 2011 LHC data, fewer than $150$ Higgs bosons should have decayed into photon pairs, but although there is a background of two-photon events, it is statistically manageable. What ATLAS has found [4] is a potentially significant number of two-photon events, whose invariant mass clusters around $126\phantom{\rule{0.333em}{0ex}}\text{GeV}$, above the expected background. (Although each photon is massless, the two-photon pair can collectively be assigned a mass value, called the invariant mass, that depends on the kinematical properties of the pair.) If the two photons originated from a decaying Higgs boson, their invariant mass can be identified with the mass of the decaying Higgs boson; that is, a Higgs mass of $126\phantom{\rule{0.333em}{0ex}}\text{GeV}$.
Another rare but striking Higgs decay (corresponding to the light blue curve in Fig. 1 labeled $Z\phantom{\rule{0}{0ex}}Z$) is one that results in a $Z$ boson and a pair of electrons or muons, collectively called leptons. Subsequently, the $Z$ boson can also decay into a pair of leptons, in which case the end result of the Higgs decay is an event with four leptons. Once again, there is a background from more conventional processes, but this background is also manageable. In their search for four-lepton final states, CMS was able to exclude the existence of a Higgs boson over a significant range of masses [5]. Indeed, within the preferred Higgs mass range of the standard model, CMS rules out Higgs boson masses above $134\phantom{\rule{0.333em}{0ex}}\text{GeV}$. However, CMS also identified excesses in four-lepton events in regions of the four-lepton invariant mass around $119\phantom{\rule{0.333em}{0ex}}\text{GeV}$, $126\phantom{\rule{0.333em}{0ex}}\text{GeV}$, and $320\phantom{\rule{0.333em}{0ex}}\text{GeV}$. Although none of the three observations is statistically significant in itself, other experimental measurements provide some guidance: The excess at $320\phantom{\rule{0.333em}{0ex}}\text{GeV}$ is probably an ordinary statistical fluctuation, while the excess at $126\phantom{\rule{0.333em}{0ex}}\text{GeV}$ is more striking in light of the ATLAS two-photon data [4]. Indeed, a Higgs boson of $126\phantom{\rule{0.333em}{0ex}}\text{GeV}$ would be expected to produce a few four-lepton events above the predicted background, which is what CMS observed [5]. The excess at $119\phantom{\rule{0.333em}{0ex}}\text{GeV}$ is, however, difficult to interpret with the present data set.
So, where do we stand at this moment in time? Whether or not the various excesses above background described above turn out to be statistical fluctuations, the ATLAS and CMS collaborations have already significantly narrowed the allowed mass range for the standard model Higgs boson. These results, together with the ATLAS and CMS searches [3, 6] for Higgs bosons decaying into other final states (e.g., in [3] ATLAS examined the decay into a pair of $W$ bosons), and the indirect Higgs mass constraints from previous experiments, imply that the Higgs boson of the standard model (if it exists) must have a mass that lies between $115.5\phantom{\rule{0.333em}{0ex}}\text{GeV}$ and $127\phantom{\rule{0.333em}{0ex}}\text{GeV}\phantom{\rule{0.278em}{0ex}}$ [12]. Moreover, it is tempting to attribute the small excess of events with invariant mass of $126\phantom{\rule{0.333em}{0ex}}\text{GeV}$ to a possible Higgs boson of the same mass.
Nevertheless, this intriguing hint is not (yet) a Higgs boson observation, and it is certainly not a Higgs boson discovery. The statistical evidence for the Higgs boson is not sufficiently compelling, though the LHC data that will be collected in 2012 could be decisive since the collider will be ramped up to $8\phantom{\rule{0.333em}{0ex}}\text{TeV}$ and the data samples are expected to be $3$ times larger than those collected in 2011. An extrapolation to the anticipated 2012 data set suggests that if the standard model Higgs boson exists in the expected mass range, then a statistically significant signal (of $5$ standard deviations or greater) will emerge. If this happens, can we claim the discovery of the Higgs boson? Strictly speaking, the answer is no. If the signal matches (within the statistical uncertainty) the one expected, then all one can say is that the signal is consistent with that of the standard model Higgs boson, but additional data will be required to confirm in detail that the properties of the newly discovered particle match the theoretically expected properties of the standard model Higgs boson. If the signal deviates significantly from the one expected, or if the statistical significance of the signal is greatly reduced after analyzing the 2012 LHC data, then it will be possible to exclude the existence of the standard model Higgs boson over its entire allowed mass range, in which case the simplest symmetry-breaking mechanism employed by the standard model will have to be replaced by more complicated symmetry-breaking dynamics. Whichever path nature chooses, 2012 is sure to be a watershed year in particle physics.
### References
1. For a fascinating exposition of the history of the Higgs boson, see Frank Close, The Infinity Puzzle (Oxford University Press, Oxford, 2011)[Amazon][WorldCat].
2. For reviews of Higgs boson theory and phenomenology, see e.g., M. Carena, and H.E. Haber, Prog. Part. Nucl. Phys. 50, 63 (2003); A. Djouadi, Phys. Rept. 457, 1 (2008).
3. G. Aad et al. (ATLAS Collaboration), Phys. Rev. Lett. 108, 111802 (2012).
4. G. Aad et al. (ATLAS Collaboration), Phys. Rev. Lett. 108, 111803 (2012).
5. S. Chatrchyan et al. (CMS Collaboration), Phys. Rev. Lett. 108, 111804 (2012).
6. G. Aad et al. (ATLAS Collaboration), Phys. Lett. B (2012), DOI: 10.1016/j.physletb.2012.02.044 (to be published); S. Chatrchyan et al. (CMS Collaboration), Phys. Lett. B (2012), DOI: 10.1016/j.physletb.2012.02.064 (to be published).
7. ALEPH Collaboration, DELPHI Collaboration, L3 Collaboration, OPAL Collaboration, and The LEP Working Group for Higgs Boson Searches, Phys. Lett. B 565, 61 (2003).
8. TEVNPH (Tevatron New Phenomena and Higgs Working Group) and CDF and D0 Collaborations, arXiv:1107.5518 (hep-ex).
9. M. Baak, M. Goebel, J. Haller, A. Hoecker, D. Ludwig, K. Moenig, M. Schott, and J. Stelzer, arXiv:1107.0975 (hep-ph).
10. Based on the most recent data reported (http://tevnphwg.fnal.gov/results/SM_Higgs_Winter_12/) by the Tevatron experimental collaborations on 7 March, 2012 (without the constraints from the 2011 LHC data), the upper bound of the standard model Higgs mass is lowered to 147 GeV.
11. A. Denner, S. Heinemeyer, I. Puljak, D. Rebuzzi, and M. Spira, Eur. Phys. J. C 71, 1753 (2011).
12. The most recent combined ATLAS analysis of the 2011 data reported on March 7, 2012 (https://atlas.web.cern.ch/Atlas/GROUPS/PHYSICS/CONFNOTES/ATLAS-CONF-2012-019/) excludes standard model Higgs masses below 122.5 GeV (although ATLAS cannot quite exclude masses between 117.5 GeV and 118.5 GeV). Thus, the only standard model Higgs masses that are not excluded are most likely to lie between 122.5 and 127 GeV.
### About the Author: Howard E. Haber
Howard Haber received his Ph.D. in physics from the University of Michigan in 1978. After post doctoral positions at Lawrence Berkeley Laboratory and the University of Pennsylvania, he arrived at the University of California, Santa Cruz, where he was promoted to professor of physics in 1990. He is one of four authors of the book, “The Higgs Hunter’s Guide,” first published in 1990. He is also a Fellow of the American Physical Society and a general member of the Aspen Center for Physics. In 2009, he was a recipient of the Alexander von Humboldt Research Award. Haber’s research activities focus on the exploration of new theoretical directions beyond the standard model of particle physics, with an emphasis on the search for Higgs bosons and supersymmetric phenomena at current and future high energy colliders. | 2015-03-30 02:23:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 49, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.604397714138031, "perplexity": 629.9663574625599}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-14/segments/1427131298889.70/warc/CC-MAIN-20150323172138-00198-ip-10-168-14-71.ec2.internal.warc.gz"} |
http://crypto.stackexchange.com/questions/2991/why-must-iv-key-pairs-not-be-reused-in-ctr-mode/2993 | # Why must IV/key-pairs not be reused in CTR mode?
Many sources mention that IVs must not be reused with the same key in CTR mode, for encrypting 2 different pieces of data, because that totally destroys security - but I haven't found an explanation so far as to why this is the case.
The issue is obvious if the attacker can manage to obtain the plain text and its corresponding cipher text for one piece of the data - but if no known plain texts are available, how could an attacker reconstruct the key-stream from just the IV value?
Can the security issues be mitigated by keeping the IV secret too? E.g. would an attacker have any realistic chance of cracking the encryption by just knowing that key/IV-pairs have been reused in the creation of two different ciphertexts, but nothing more?
-
Have a look at our question How does one attack a two-time pad (i.e. one time pad with key reuse)? and its recent duplicate How can I find two strings $m_1$ and $m_2$, knowing that I know $m_1 \oplus m_2$?. The same attack principle is valid for every synchronous stream cipher (and CTR mode is an example of this). – Paŭlo Ebermann Jun 20 '12 at 18:02
If after reading this there is still something open, please edit your question accordingly. – Paŭlo Ebermann Jun 20 '12 at 18:02
For anyone else landing here later on, this link (from this question ) probably explains it best. Thanks for the hint about the "two-time pad" Paŭlo Ebermann and poncho! – Dexter Jun 20 '12 at 21:18
## 1 Answer
Yes, the attacker would have a realistic chance of recovering plaintext, and preventing him from knowing the IV values does not reduce this risk.
The problem is that CTR mode encryption is effectively:
$C = P \oplus F(Key, IV)$
where $P$ is the plaintext, $C$ is the ciphertext, and $F$ is a complex function of its two inputs.
The problem with this is if you encrypt two different plaintexts with the same $Key$, $IV$ values, then the attacker gets two pairs:
$C_1 = P_1 \oplus F(Key, IV)$
$C_2 = P_2 \oplus F(Key, IV)$
Where he can see the values $C_1$, $C_2$. With those, he can then compute:
$C_1 \oplus C_2 = P_1 \oplus P_2$
and thus deriving the value of the two plaintexts exclusive-or'ed together.
As for how that can be attacked, well, you can find two examples against ASCII-English plaintexts here and here.
And, note that since the attacker didn't actually use the IV value, it doesn't matter to him whether he knows it or not.
- | 2016-02-08 17:23:17 | {"extraction_info": {"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, "math_score": 0.4488294720649719, "perplexity": 1157.706105762095}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701153736.68/warc/CC-MAIN-20160205193913-00320-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://socratic.org/questions/how-do-you-use-implicit-differentiation-to-find-d-2y-dx-2-given-x-2-y-2-1 | # How do you use implicit differentiation to find (d^2y)/(dx^2) given x^2-y^2=1?
Nov 2, 2016
$y ' ' \left(x\right) = \pm \frac{1}{{x}^{2} - 1} ^ \left(3 / 2\right)$
#### Explanation:
Calling $f \left(x , y \left(x\right)\right) = {x}^{2} - y {\left(x\right)}^{2} - 1 = 0$
$\frac{\mathrm{df}}{\mathrm{dx}} = 2 x - 2 y \left(x\right) y ' \left(x\right) = 0$ and
$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = 2 - 2 y \left(x\right) y ' ' \left(x\right) - 2 y ' {\left(x\right)}^{2} = 0$
so
$y ' ' \left(x\right) = \frac{y {\left(x\right)}^{2} - {x}^{2}}{y \left(x\right)} ^ 3 = \pm \frac{1}{{x}^{2} - 1} ^ \left(3 / 2\right)$ | 2021-06-25 02:47:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9788070917129517, "perplexity": 10661.737659452978}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488567696.99/warc/CC-MAIN-20210625023840-20210625053840-00511.warc.gz"} |
https://astrostatistics.wordpress.com/2019/08/24/approximate-bayesian-computation-with-heads-in-the-sand/ | ## Approximate Bayesian Computation with heads in the sand
I noticed a recent arXival describing an ABC approach to inferring halo modal parameters from lensing substructure, which it seems is the third in a series of installments, each of which churlishly manages to pretend not to be aware of my seminal ‘ABC for astronomy‘ paper from back in 2012. While I’m glad to see the ABC technique increasingly being used, consciously, for astronomical analyses, it does seem weird that having figured out that your technique fits within the ABC framework, not to learn from the statistical literature on that topic. Why I assume no learning has taken place is because the authors in this case haven’t moved beyond simple rejection ABC, which is a massive waste of computational power relative to even a basic population Monte Carlo approach. Moreover, it seems their scheme is consequently so inefficient that they cannot fit all $N$ of their lenses jointly, instead having to rely on some kind of multiplication of KDE approximations to each subset posterior to form the final posterior.
Heck, if they’d read into the problem from a statistical point of view they might even have learned a bit about how to do subset posteriors better. For instance, they write that the multiplication of these sub-posteriors is only possible if you use uniform priors, and consequently they go to a lot of trouble to choose transformations allowing for construction of entirely uniform priors. But, wait for it, what about this from consensus MC: $\propto \pi_i(\theta)^{1/N}$?!
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### 3 Responses to Approximate Bayesian Computation with heads in the sand
1. Tom Loredo says:
I’m repeatedly bewildered by how little effort many authors put into doing a literature search for prior work relevant to their research (I especially notice this in my role as an assoc. ed. of an applied stats journal that includes papers on statistical applications in astronomy and physics). Firstly, it’s never been *easier* to do a literature search, with tools like Google, ADS, and MathSciNet at hand, not to mention having a world’s worth of colleagues accessible at short notice via email. It’s not like you have to actually go to a library and use card catalogs; what excuse is there for *not* doing a literature search? But more importantly, it’s an issue of basic personal and scientific integrity. If you are writing a paper, presumably that’s because you hope others will read it; courtesy and humility should then suggest it’s your duty to expend at least some effort to read what others have written relevant to your work. But it’s even more basic than that. As Feynman wrote, “The first principle [for producing sound science] is that you must not fool yourself—and you are the easiest person to fool.” He refers to taking extreme care not to fool yourself as a basic “type of integrity” for scientists. Surely the easiest component of this kind of self-skepticism would be doing a literature search, not just on the science but also on your methodology, esp. if it’s nontrivial.
Turning now to the recent impact of ABC and your seminal 2012 paper: I recently did a hasty bibliographic search to quickly ascertain the impact of ABC in astronomy. I’d been aware of several papers in cosmology and work in progress in exoplanets at a level that was already impressive to me given just 7+ years since the introduction of ABC to astronomy. But my ADS searches turned up several dozen abstracts describing work making substantive use of ABC (I didn’t look at the search results to see how many abstracts were for refereed, published papers; this was just a background info search for a letter of recommendation). That’s a very impressive impact on behalf of a nontrivial statistical approach in astronomy, in just 7+ years .
I say 7+ years because the first time I heard ABC being advocated for astronomy applications was at the *Statistical Challenges in Modern Astronomy V* meeting in 2011, where Chad Schafer and Peter Freeman gave a solid presentation on it. This was published in the proceedings volume (with discussion by Martin Hendry) in 2012 (ADS: https://ui.adsabs.harvard.edu/abs/2012LNS…902….3S/abstract). I was disappointed to see how few citations this paper has in comparison with your 2012 paper. It’s not as ambitious, but still. As far as I can tell, Chad and Peter didn’t post the paper to arXiv. My own experience is that papers I haven’t posted to arXiv are rarely cited. So maybe this result is just saying that arXiv has replaced journals and conference proceedings as primary literature for astronomers. Anyway, this is just to say that the Schafer and Freeman 2011/2012 work probably deserves some credit for recognizing the importance of ABC for astrostatistics and for *trying* to bring it to the attention of astronomers, even if their publication choice minimized its visibility and impact.
• It’s hard to fathom how much research effort is wasted on duplication of existing work that could have been identified with a quick google search. It’s obviously dissappointing when you come up with an idea for a new method (for example) and then google reveals it’s already been done: but if you search early enough (as you should) then it’s usually straightforward to “pivot” your idea: new application area with a case-specific problem, extension of the method, improvement of efficiency. The worst offending subject area between about 2012 and 2017 was probably machine learning, which was notorious for “re-badging” of existing statistical methods, but my impression is that the two communities now overlap sufficiently that egregious examples will be caught at review.
Citing Schafer & Freeman was indeed technically difficult for that reason when I was writing my ABC paper; but I’m glad to say we reference their pioneering work in our first footnote! | 2020-04-07 06:10:22 | {"extraction_info": {"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": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6873154044151306, "perplexity": 1280.6037711645274}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371675859.64/warc/CC-MAIN-20200407054138-20200407084638-00444.warc.gz"} |
https://yakovenko.wordpress.com/category/lecture/rothschild-course-analysis-for-high-school-teachers/ | # Sergei Yakovenko's blog: on Math and Teaching
## Real numbers as solutions to infinite systems of equalities
In the past we already extended our number system by adding “missing” elements which are assumed to satisfy certain equations, based only on knowing what these equations are. It turns out that we may extend the set of rational numbers $\mathbb Q$ to a much larger set of real numbers $\mathbb R$ by adding solutions to (infinite numbers) of inequalities. As before, the properties of these new numbers could be derived only from the properties of inequalities between the rational numbers.
On one leg, the idea can be explained as follows. Since for any two rational numbers $r,s\in\mathbb Q$ one and only one relation out of three is possible, $r$ < $s$, $r=s$ or $r$ > $s$, we can uniquely define any, say, positive rational unknown number $x$ by looking at the two sets, $L=\{l\in\mathbb Q: 0\le l\le x\}$ and $R=\{r\in\mathbb Q: x\le r\}$. (You don’t have to be too smart at this moment: $x$ is the only element in the intersection $L\cap R$ 😉
However, sometimes the analogous construction leads to problems. For instance, if $L=\{l\in\mathbb Q: l\ge 0, l^2\le 2\}$ and $R=\{r\in\mathbb Q: r^2\ge 2\}$, then $L\cap R=\varnothing$, since the square root of two is not a rational number, but $L\cup R=\mathbb Q_+$, i.e., for any positive rational number we can say whether is smaller or larger the missing number $\sqrt 2$. This allows to derive all properties of $\sqrt 2$, including its approximation with any number of digits.
Proceeding this way, we introduce (positive) real numbers by indication, what is their relative position to all rational numbers. This allows to describe the real numbers completely.
The details can be found here.
## A didactic digression
Some of you complained about insufficient number of problems that are discussed during the tutorials. Everybody knows that problems and questions for self-control are the most important elements of study mathematics, especially in comparison with other disciplines. The rationale behind is the assumption that a student who understands the subject, should be able to answer these questions immediately or after some reflection. Composing such problems is an easy thing: you any mathematical argument you can stop for a second and ask yourself: “why I can do as explained?” or “under what conditions are my actions justified?”. In the lecture notes (see the link above) tens of such problems are explicitly formulated. Similar problems will await you on the exam.
However, remember one simple thing. If you already know how to solve a problem, this is not a problem but rather a job. Unless you solve these problems yourselves, there is no sense in memorizing their solutions: knowing solution of one such problem won’t help you with solving another problem unless you really understand what’s going on. There are no “typical problems”: each one of them is of its own sort, though, of course, some problems can be solved by similar methods.
A practical advice: you should not expect that all problems that appear on the exam will be discussed at length at the tutorials. There are no ready recipes to memorize. Only to understand honestly. Believe me, this is easier than memorize by heart endless formulas and algorithms.
## Numbers
The basic set theory allows us to construct a set $\mathbb N=\{|,||,|||,||||,|||||,\dots\}$ with a function “next”, denoted by $\mathrm{Succ}:\mathbb N\to\mathbb N\smallsetminus\{|\}$, which is bijective. This set describes the process of counting objects and is the most basic structure. Starting from a distinguished element denoted by 1, we construct an infinite number of elements $2=\mathrm{Succ}(1),\ 3=\mathrm{Succ}(2),\ 4=\mathrm{Succ}(3)$ etc. There are two axioms guaranteeing that the set $\mathbb N$ indeed coincides with what we call the set of natural numbers:
1. $\forall x\in\mathbb N\ \mathrm{Succ}(x)\ne 1$
2. Any element $x\in\mathbb N$ is obtained by the iteration of $\mathrm{Succ}$: $x=(\mathrm{Succ}\circ\cdots\circ\mathrm{Succ})(1)$.
Using this function and its partial inverse one can define on $\mathbb N$ the order and the operations of addition (as repeated addition of 1 which is just evaluation of $\mathrm{Succ}$) and multiplication (repeated addition).
However, not all equations of the form $x+a=b$ or $x\cdot a=b$ are solvable. One can enlarge $\mathbb N$ by adding solutions of all such equations, obtaining the set of integer numbers $\mathbb Z$ which is a commutative group with respect to the operation of addition, and finally the set of rational numbers $\mathbb Q$ in which division is available by any nonzero number.
Division by zero is impossible: if we add “solution of the equation $0\cdot x=1$” as a new imaginary element, then we will not be able to do some arithmetic operations on it. Still, if we are ready to pay this price, then the rational numbers can be extended by a new element so that, say, the function $f(x)=1/x$ would be everywhere defined and continuous.
Details are available in the lecture notes here.
# שלום כיטה א!
The main feature that distinguishes the Calculus (or Mathematical Analysis) from other branches of mathematics is the repeated use of infinite constructions and processes. Without infinity even the simplest things, like the decimal representation of the simple fraction $\frac13=0.333333\dots$ becomes problematic.
Yet to deal with infinity and infinite constructions, we need to make precise our language, based on the notions of sets and functions (maps, applications, – all these words are synonymous).
Look at the first section of the lecture notes here.
You are most welcome to start discussions in the comments to this (or any other) post. Don’t be afraid of asking questions that may look stupid: this never harms! Write in any language (besides Hebrew/English, I hope that Ghadeer will take care of questions in Arabic, and I promise to deal with French/Spanish/Catalan/Italian/Russian/Ukrainian questions) 😉 Subscribe for updates on this site with your usual emails, to be independent from any dependence 😉
Looking forward for a mutually beneficial interaction in the new semester!
## Monday, February 1, 2016
### Finally, exam!
Filed under: lecture,Rothschild course "Analysis for high school teachers" — Sergei Yakovenko @ 3:41
Tags:
# Exam
The exam is posted online on Feb 1, 2016, and must be submitted on the last day of the exams’ period, February 26. Its goals are, besides testing your acquired skills in the Analysis, to teach you a few extra things and see your ability for logical reasoning, not your proficiency in performing long computations. If you find yourself involved in heavy computations, better double check whether you understand the formulation of the problem correctly. Remember, small details sometimes matter!
Please provide argumentation, better in the form of logical formulas, not forgetting explicit or implicit quantifiers $\forall$ and $\exists$. They really may change the meaning of what you write!
Problems are often subdivided into items. The order of these items is not accidental, try to solve them from the first till the last, and not in a random order (solution of one item may be a building block for the next one).
To get the maximal grade, it is not necessary to solve all problems, but it is imperative not to write stupid things. Please don’t try to shoot in the air.
The English version is the authoritative source, but if somebody translates it into Hebrew (for the sake of the rest of you) and send me the translation, I will post it for your convenience, but responsibility will be largely with the translator.
If you believe you found an error or crucial omission in the formulation of a problem, please write me. If this will be indeed the case (errare humanum est), the problem will be either edited (in case of minor omissions) or cancelled (on my account).
That’s all, folks!© Good luck to everybody!
Yes, and feel free to leave your questions/talkbacks here, whether addressed to Michal/Boaz/me or to yourself, if you feel you want to ask a relevant question.
# Corrections
## Correction 1
The formulation of Problem 1 was indeed incorrect. The set $A'$ was intended to be the set of accumulation points for a set $A\subseteq [0,1]$. The formal definition is as follows.
Definition. A point $p\in [0,1]$ belongs to to the set of limit points $A'$ if and only if $\forall\varepsilon$>0 the intersection $(p-\varepsilon,p+\varepsilon)\cap A$ is infinite. The point $p$ itself may be or may be not in $A$.
Isolated points of $A$ are never in $A'$, but $A'$ may contain points $p\notin A$.
Apologies for the hasty formulation.
## Correction 2: Problem 3(b) cancelled!
The statement requested to prove in Problem 3(b) is wrong, and I am impressed how fast did you discover that. Actually, the problem was taken from the textbook by Zorich, vol. 1, where it appears on p. 169, sec. 4.2.3, as Problem 4.
The assertion about existence of the common fixed point of two commuting continuous functions $f,g\colon [0,1]\to[0,1]$ becomes true if we require these functions to be continuously differentiable on $[0,1]$ (in particular, for polynomials), but the proof of this fact is too difficult to be suggested as a problem for the exam.
Thus Problem 3(b) is cancelled.
## Tuesday, January 26, 2016
### Lecture 13, Jan 26, 2016
Questions concerned integrability of discontinuous functions, notions of improper integrals (how and when they can be defined), topological properties (equivalent definitions of compactness, connectedness etc.)
Here are some textbooks that I recommend for preparing when working on the exam. Keep them on your virtual bookshelf: they cover much more that I explained in the course, but who knows what questions related to analysis you might have.
1. V. Zorich, vol. 1: Chapters 1-6, pp.1-371.
2. V. Zorich, vol. 2: Parts of Chapter 9 (continuous maps) and the first part of Chapter 18 on Fourier series.
3. W. Rudin: Chapters 1-6, pp.1-165.
The problems for exam will be posted on February 1st (at least the English version).
# Functions of complex variable
The field $\mathbb C$ is naturally extending the field $\mathbb R$, which means that all arithmetic operations on $\mathbb R$ extend naturally as operations on $\mathbb C$. In particular, any polynomial $p(x)=a_0x^n+a_1 x^{n-1}+\cdots+a_n$ can be interpreted as a map $p:\mathbb C\to\mathbb C$. Geometrically, this can be visualized as a map of the 2-plane to the 2-plane.
We discussed the maps $p(x)=x^2$ and $p(x)=1/x$.
Smooth functions are those which can be accurately approximated by the $\mathbb R$-affine maps $x\mapsto c+\lambda(x-a)$ near each point $a\in\mathbb R$. Functions that can be accurately approximated by $\mathbb C$-affine maps (the same formula, but over the complex numbers), are called holomorphic (or complex analytic). Such maps are characterized by the property that small circles are mapped into small almost-circles, that is,
1. angles are preserved, and
2. lengths are scaled
Sometimes these local conditions become global. Examples: the affine maps $x\mapsto \lambda (x-a)+b$ send lines to lines and circles to circles. The map $x\mapsto 1/x$ maps circles and lines into circles or lines (depending on whether the circles/lines pass through the origin $x=0$).
## Complex integration
Integration is carried over smooth (or piecewise smooth) paths in $\mathbb C$, using Riemann-like sums. It depends linearly on the function which we integrate, but in contrast with the real case we have much more freedom in choosing the paths.
1. If $f(x)=c$ is a constant, and $\gamma =[p_0,p_1]+[p_1,p_2]+[p_2,p_0]$ is a closed triangle, then the integral is zero as the sum $c(p_1-p_0)+c(p_2-p_1)+c(p_0-p_2)=c\cdot0=0$.
2. If $f(x)$ is non-constant, then the integral identity is valid only for “very small triangles” near a point $a\in\mathbb C$ with $c=f(a)$.
3. This implies that $\displaystyle \int_\gamma f(x)\,\mathrm dx=\int_{\gamma'} f(x)\,\mathrm dx$ as long as the paths $\gamma,\gamma'$ share the common endpoints and can be continuously deformed one into the other.
Integrals over closed loops are zeros, unless there are singular points (where the function is non-holomoprhic) inside.
Example: $f(x)=\frac 1x$ is non-holomorphic at $x=0$, and $\displaystyle \oint_{|x|=1}\tfrac 1x\,\mathrm dx=2\pi i$.
## Cauchy integral formula
If $f$ is holomorphic inside a domain $U$ bounded by a closed curve $\gamma$ and $a\in U$, then $\displaystyle f(a)=\frac 1{2\pi i}\oint_\gamma \frac{f(x)\,\mathrm dx}{x-a}$.
In other words, the value of $f$ on the boundary uniquely determine its values inside the domain. This is in wild contrast with functions of real variable!
## Taylor series
As a function of $a$, the expression $\frac1{x-a}$ admits a converging Taylor expansion (in fact, the same old geometric progression series) in powers of $a-a_0$ for any $a_0\ne x$. Thus if we choose $a_0\in U$ off the path $\gamma$, then the series will converge for any $x\in\gamma$ (warning! note that the “variable” and the “parameter” exchanged their roles!!!), the Cauchy integral can be expanded in the converging series of powers $(a-a_0)^n, \ n=0,1,2,\dots$, hence the function $f(x)$ gets expanded in the series $f(a)=c_0+c_1(a-a_0)+c_2(a-a_0)^2+\cdots$.
Conclusion: functions that are $\mathbb C$-differentiable can be expanded in the convergent Taylor series (hence are “polynomials of infinite degree”) and vice versa, “polynomials of infinite degree” are infinitely $\mathbb C$-differentiable. This is a miracle that so many functions around us are actually holomorphic!
# Elementary transcendental functions as solutions to simple differential equations
The way how logarithmic, exponential and trigonometric functions are usually introduced, is not very satisfactory and appears artificial. For instance, the mere definition of the non-integer power $x^a$, $a\notin\mathbb Z$, is problematic. For $a=1/n,\ n\in\mathbb N$, one can define the value as the root $\sqrt[n]x$, but the choice of branch/sign and the possibility of defining it for negative $x$ is speculative. For instance, the functions $x^{\frac12}$ and $x^{\frac 24}$ may turn out to be different, depending on whether the latter is defined as $\sqrt[4]{x^2}$ (makes sense for negative $x$) or as $(\sqrt[4]x)^2$ which makes sense only for positive $x$. But even if we agree that the domain of $x^a$ should be restricted to positive arguments only, still there is a big question why for two close values $a=\frac12$ and $a=\frac{499}{1000}$ the values, say, $\sqrt 2$ and $\sqrt[1000]{2^{499}}$ should also be close…
The right way to introduce these functions is by looking at the differential equations which they satisfy.
A differential equation (of the first order) is a relation, usually rational, involving the unknown function $y(x)$, its derivative $y'(x)$ and some known rational functions of the independent variable $x$. If the relation involves higher derivatives, we say about higher order differential equations. One can also consider systems of differential equations, involving several relations between several unknown functions and their derivatives.
Example. Any relation of the form $P(x, y)=0$ implicitly defines $y$ as a function of $x$ and can be considered as a trivial equation of order zero.
Example. The equation $y'=f(x)$ with a known function $f$ is a very simple differential equation. If $f$ is integrable (say, continuous), then its solution is given by the integral with variable upper limit, $\displaystyle y(x)=\int_p^x f(t)\,\mathrm dt$ for any meaningful choice of the lower limit $p$. Any two solutions differ by a constant.
Example. The equation $y'=a(x)y$ with a known function $a(x)$. Even the case where $a(x)=a$ is a constant, there is no, say, polynomial solution to this equation (why?), except for the trivial one $y(x)\equiv0$. This equation is linear: together with any two functions $y_1(x),y_2(x)$ and any constant $\lambda$, the functions $\lambda y_1(x)$ and $y_1(x)\pm y_2(x)$ are also solutions.
Example. The equation $y'=y^2$ has a family of solutions $\displaystyle y(x)=-\frac1{x-c}$ for any choice of the constant $c\in\mathbb R$ (check it!). However, any such solution “explodes” at the point $x=c$, while the equation itself has no special “misbehavior” at this point (in fact, the equation does not depend on $x$ at all).
## Logarithm
The transcendental function $y(x)=\ln x$ satisfies the differential equation $y'=x^{-1}$: this is the only case of the equation $y'=x^n,\ n\in\mathbb Z$, which has no rational solution. In fact, all properties of the logarithm follow from the fact that it satisfies the above equation and the constant of integration is chosen so that $y(1)=0$. In other words, we show that the function defined as the integral $\displaystyle \ell(x)=\int_1^x \frac1t\,\mathrm dt$ possesses all what we want. We show that:
1. $\ell(x)$ is defined for all $x>0$, is monotone growing from $-\infty$ to $+\infty$ as $x$ varies from $0$ to $+\infty$.
2. $\ell(x)$ is infinitely differentiable, concave.
3. $\ell$ transforms the operation of multiplication (of positive numbers) into the addition: $\ell(\lambda x)=\ell(\lambda)+\ell(x)$ for any $x,\lambda>0$.
## Exponent
The above listed properties of the logarithm ensure that there is an inverse function, denoted provisionally by $E(x)$, which is inverse to $\ell:\ \ell(E(x))=x$. This function is defined for all real $x\in\mathbb R$, takes positive values and transforms the addition to the multiplication: $E(\lambda+x)=E(\lambda)\cdot E(x)$. Denoting the the value $E(1)$ by $e$, we conclude that $E(n)=e^n$ for all $n\in\mathbb Z$, and $E(x)=e^x$ for all rational values $x=\frac pq$. Thus the function $E(x)$, defined as the inverse to $\ell$, gives interpolation of the exponent for all real arguments. A simple calculation shows that $E(x)$ satisfies the differential equation $y'=y$ with the initial condition $y(0)=1$.
## Computation
Consider the integral operator $\Phi$ which sends any (continuous) function $f:\mathbb R\to\mathbb R$ to the function $g=\Phi(f)$ defined by the formula $\displaystyle g(x)=f(0)+\int_0^x f(t)\,\mathrm dt$. Applying this operator to the function $E(x)$ and using the differential equation, we see that $E$ is a “fixed point” of the transformation $\Phi$: $\Phi(E)+E$. This suggests using the following approach to compute the function $E$: choose a function $f_0$ and build the sequence of functions $f_n=\Phi(f_{n-1})$, $n=1,2,3,4,\dots$. If there exists a limit $f_*=\lim f_{n+1}=\lim \Phi(f_n)=\Phi(f_*)$, then this limit is a fixed point for $\Phi$.
Note that the action of $\Phi$ can be very easily calculated on the monomials: $\displaystyle \Phi\biggl(\frac{x^k}{k!}\biggr)=\frac{x^{k+1}}{(k+1)!}$ (check it!). Therefore if we start with $f_0(x)=1$, we obtain the functions $\latex f_n=1+x+\frac12 x^2+\cdots+\frac1{n!}x^n$. This sequence converges to the sum of the infinite series $\displaystyle\sum_{n=0}^\infty\frac1{n!}x^n$ which represents the solution $E(x)$ on the entire real line (check that). This series can be used for a fast approximate calculation of the number $e=E(1)=\sum_0^\infty \frac1{n!}$.
## Differential equations in the complex domain
The function $E(ix)=e^{ix}$ satisfies the differential equation $y'=\mathrm iy$. The corresponding “motion on the complex plane”, $x\mapsto e^{\mathrm ix}$, is rotation along the (unit) circle with the unit (absolute) speed, hence the real and imaginary parts of $e^{\mathrm ix}$ are cosine and sine respectively. In fact, the “right” definition of them is exactly like that,
$\displaystyle \cos x=\textrm{Re}\,e^{\mathrm ix},\quad \sin x=\textrm{Im}\,e^{\mathrm ix} \iff e^{\mathrm ix}=\cos x+\mathrm i\sin x,\qquad x\in\mathbb R$.
Thus, the Euler formula “cis” in fact is the definition of sine and cosine. Of course, it can be “proved” by substituting the imaginary value into the Taylor series for the exponent, collecting the real and imaginary parts and comparing them with the Taylor series for the sine and cosine.
In fact, both sine and cosine are in turn solutions of the real differential equations: derivating the equation $y'=\mathrm iy$, one concludes that $y''=\mathrm i^2y=-y$. It can be used to calculate the Taylor coefficients for sine and cosine.
For more details see the lecture notes.
Not completely covered in the class: solution of linear equations with constant coefficients and resonances.
# Integral and antiderivative
1. Area under the graph as a paradigm
2. Definitions (upper and lower sums, integrability).
3. Integrability of continuous functions.
4. Newton-Leibniz formula: integral and antiderivative.
5. Elementary rules of antiderivation (linearity, anti-Leibniz rule of “integration by parts”).
6. Anti-chain rule, change of variables in the integral and its geometric meaning.
7. Riemann–Stieltjes integral and change of variables in it.
8. Integrability of discontinuous functions.
Not covered in the class: Lebesgue theorem and motivations for transition from Riemann to the Lebesgue integral.
The sketchy notes are available here.
# Higher derivatives and better approximation
We discussed a few issues:
• Lagrange interpolation formula: how to estimate the difference $f(b)-f(a)$ through the derivative $f'$?
• Consequence: vanishing of several derivatives at a point means that a function has a “root of high order” at this point (with explanation, what does that mean).
• Taylor formula for polynomials: if you know all derivatives of a polynomial at some point, then you know it everywhere.
• Peano formula for $C^n$-smooth functions: approximation by the Taylor polynomial with asymptotic bound for the error.
• Lagrange formula: explicit estimate for the error.
The notes (updated) are available here.
# Differentiability and derivative
Continuity of functions (and maps) means that they can be nicely approximated by constant functions (maps) in a sufficiently small neighborhood of each point. Yet the constant maps (easy to understand as they are) are not the only “simple” maps.
## Linear maps
Linear maps naturally live on vector spaces, sets equipped with a special structure. Recall that $\mathbb R$ is algebraically a field: real numbers cane be added, subtracted between themselves and the ratio $\alpha/\beta$ is well defined for $\beta\ne0$.
Definition. A set $V$ is said to be a vector space (over $\mathbb R$), if the operations of addition/subtraction $V\owns u,v\mapsto u\pm v$ and multiplication by constant $V\owns v,\ \mathbb R\owns \alpha\mapsto \alpha v$ are defined on it and obey the obvious rules of commutativity, associativity and distributivity. Some people prefer to call vector spaces linear spaces: the two terms are identical.
Warning. There is no “natural” multiplication $V\times V\to V$!
Examples.
1. The field $\mathbb R$ itself. If we want to stress that it is considered as a vector space, we write $\mathbb R^1$.
2. The set of tuples $\mathbb R^n=(x_1,\dots,x_n),\ x_i\in\mathbb R$ is the Euclidean $n$-space. For $n=2,3$ it can be identified with the “geometric” plane and space, using coordinates.
3. The set of all polynomials of bounded degree $\leq d$ with real coefficients.
4. The set of all polynomials $\mathbb R[x]$ without any control over the degree.
5. The set $C([0,1])$ of all continuous functions on the segment $[0,1]$.
Warning. The two last examples are special: the corresponding spaces are not finite-dimensional (we did not have time to discuss what is the dimension of a linear space in general…)
Let $V,Z$ be two (different or identical) vector spaces and $f:V\to W$ is a function (map) between them.
Definition. The map $f$ is linear, if it preserves the operations on vectors, i.e., $\forall v,w\in V,\ \alpha\in\mathbb R,\quad f(v+w)=f(v)+f(w),\ f(\alpha v)=\alpha f(v)$.
Sometimes we will use the notation $V\overset f\longrightarrow Z$.
Obvious properties of linearity.
• $f(0)=0$ (Note: the two zeros may lie in different spaces!)
• For any two given spaces $V,W$ the linear maps between them can be added and multiplied by constants in a natural way! If $V\overset {f,g}\longrightarrow W$, then we define $(f+g)(v)=f(v)+g(v)$ for any $v\in V$ (define $\alpha f$ yourselves). The result will be again a linear map between the same spaces.
• If $V\overset f\longrightarrow W$ and $W\overset g\longrightarrow Z$, then the composition $g\circ f:V\overset f\longrightarrow W\overset g\longrightarrow Z$ is well defined and again linear.
Examples.
1. Any linear map $\mathbb R^1\overset f\longrightarrow \mathbb R^1$ has the form $x\mapsto ax, \ a\in\mathbb R$ (do you understand why the notations $\mathbb R, \mathbb R^1$ are used?)
2. Any linear map $\mathbb R^n\overset f\longrightarrow \mathbb R^1$ has the form $(x_1,\dots,x_n)\mapsto a_1x_1+\cdots+a_nx_n$ for some numbers $a_1,\dots,a_n$. Argue that all such maps form a linear space isomorphic to $\mathbb R^n$ back again.
3. Explain how linear maps from $\mathbb R^n$ to $\mathbb R^m$ can be recorded using $n\times m$-matrices. How the composition of linear maps is related to the multiplication of matrices?
The first example shows that linear maps of $\mathbb R^1$ to itself are “labeled” by real numbers (“multiplicators“). Composition of linear maps corresponds to multiplication of the corresponding multiplicators (whence the name). A linear 1-dim map is invertible if and only if the multiplicator is nonzero.
Corollary. Invertible linear maps $\mathbb R^1\to\mathbb R^1$ constitute a commutative group (by composition) isomorphic to the multiplicative group $\mathbb R^*=\mathbb R\smallsetminus \{0\}$.
## Shifts
Maps of the form $V\to V, \ v\mapsto v+h$ for a fixed vector $h\in V$ (the domain and source coincide!) are called shifts (a.k.a. translations). Warning: The shifts are not linear unless $h=0$! Composition of two shifts is again a shift.
Exercise.
Prove that all translations form a commutative group (by composition) isomorphic to the space $V$ itself. (Hint: this is a tautological statement).
## Affine maps
Definition.
A map $f:V\to W$ between two vector spaces is called affine, if it is a composition of a linear map and translations.
Example.
Any affine map $\mathbb R^1\to\mathbb R^1$ has the form $x\mapsto ax+b$ for some $a,b\in\mathbb R$. Sometimes it is more convenient to write the map under the form $x\mapsto a(x-c)+b$: this is possible for any point $c\in\mathbb R^1$. Note that the composition of affine maps in dimension 1 is not commutative anymore.
Key computation. Assume you are given a map $f:\mathbb R^1\to\mathbb R^1$ in the sense that you can evaluate it at any point $c\in\mathbb R^1$. Suppose an oracle tells you that this map is affine. How can you restore the explicit formula $f(x)=a(x-c)+b$ for $f$?
Obviously, $b=f(c)$. To find $\displaystyle a=\frac{f(x)-b}{x-c}$, we have to plug into it any point $x\ne c$ and the corresponding value $f(x)$. Given that $b=f(c)$, we have $\displaystyle a=\frac{f(x)-f(c)}{x-c}$ for any choice of $x\ne c$.
The expression $a_c(x)=\displaystyle \frac{f(x)-f(c)}{x-c}$ for a non-affine function $f$ is in general not-constant and depends on the choice of the point $x$.
Definition. A function $f:\mathbb R^1\to\mathbb R^1$ is called differentiable at the point $c$, if the above expression for $a_c(x)$, albeit non-constant, has a limit as $x\to c:\ a_c(x)=a+s_c(x)$, where $s_c(x)$ is a function which tends to zero. The number $a$ is called the derivative of $f$ at the point $c$ and denoted by $f'(c)$ (and also by half a dozen of other symbols: $\frac{df}{dx}(c),Df(c), D_xf(c), f_x(c)$, …).
Existence of the limit means that near the point $c$ the function $f$ admits a reasonable approximation by an affine function $\ell(x)=a(x-c)+b$: $f(x)=\ell(x)+s_c(x)(x-c)$, i.e., the “non-affine part” $s_c(x)\cdot (x-c)$ is small not just by itself, but also relative to small difference $x-c$.
# Differentiability and algebraic operations
See the notes and their earlier version.
The only non-obvious moment is differentiability of the product: the product (unlike the composition) of affine functions is not affine anymore, but is immediately differentiable:
$[b+a(x-c)]\cdot[q+p(x-c)]=pq+(aq+bp)(x-c)+ap(x-c)^2$, but the quadratic term is vanishing relative to $x-c$, so the entire sum is differentiable.
Exercise. Derive the Leibniz rule for the derivative of the product.
# Derivative and the local study of functions
Affine functions have no (strong) maxima or minima, unless restricted on finite segments. Yet absence of the extremum is a strong property which descends from the affine approximation to the original function. Details here and here.
Next Page »
Create a free website or blog at WordPress.com. | 2017-11-21 08:16:57 | {"extraction_info": {"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": 283, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9194266200065613, "perplexity": 301.424759258477}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806327.92/warc/CC-MAIN-20171121074123-20171121094123-00658.warc.gz"} |
https://tex.stackexchange.com/questions/107839/using-mathsection-in-the-chapter | # Using \mathsection in the chapter
How can I use LaTeX to get the result like the following picture?
In this picture, $\mathsection{1} Hình thang và hình bình hành$ is the title of the first section. If I use \mathsection and \section, then the result is
My question is: What is the code that gives me the result like the first picture?
You can redefine \@makechapterhead, \@makeschapterhead and \@sect to obtain the desired representation for the headings (using a package like titlesec here is not possible since it's not compatible with the AMS document classes):
\documentclass[11pt,a4paper]{amsbook}
\makeatletter
\begingroup
\fontsize{\@xivpt}{18}\bfseries\centering
\ifnum\c@secnumdepth>\m@ne
\leavevmode \hskip-\leftskip
\rlap{\vbox to\z@{\vss
\centerline{\normalsize\mdseries
\@xp{\scshape{\chaptername}}\enspace\thechapter}
\vskip 2pc}}\hskip\leftskip\fi
\normalfont\scshape#1\par \endgroup
\vskip\skip@ }
\begingroup
\fontsize{\@xivpt}{18}\normalfont\scshape\centering
#1\par \endgroup
\vskip\skip@ }
\def\@sect#1#2#3#4#5#6[#7]#8{%
\edef\@toclevel{\ifnum#2=\@m 0\else\number#2\fi}%
\ifnum #2>\c@secnumdepth \let\@secnumber\@empty
\else \@xp\let\@xp\@secnumber\csname the#1\endcsname\fi
\ifnum #2>\c@secnumdepth
\let\@svsec\@empty
\else
\refstepcounter{#1}%
\edef\@svsec{\ifnum#2<\@m
\@ifundefined{#1name}{}{%
\ignorespaces\csname #1name\endcsname\space}\fi
\@nx\textup{%
\csname the#1\endcsname.}\enspace
}%
\fi
\@tempskipa #5\relax
\ifdim \@tempskipa>\z@ % then this is not a run-in section heading
\begingroup\ifnum#2=1 \noindent\textbf{\S}\fi#6\relax% NEW
\@hangfrom{\hskip #3\relax\@svsec}{\interlinepenalty\@M #8\par}%
\endgroup
\csname #1mark\endcsname{#7}%
\ifnum#2>\@m \else \@tocwrite{#1}{#8}\fi
\else
\def\@svsechd{#6\hskip #3\@svsec
\@ifnotempty{#8}{\ignorespaces#8\unskip
\ifnum#2>\@m \else \@tocwrite{#1}{#8}\fi
}%
\fi
\global\@nobreaktrue
\@xsect{#5}}
\def\section{\@startsection{section}{1}%
\z@{.7\linespacing\@plus\linespacing}{.5\linespacing}
{\normalfont\bfseries}}
\makeatother
\begin{document}
\chapter{Test Chapter}
\section{Test Numbered Section}
\section*{Test Unnumbered Section}
\end{document}
• How can I customize the word Chapter since in my original file when I type chapter it gave me "Chương"(which is chapter in my native language). But when I add your titleformat, it gave me Chapter rather than "Chương". So how can I customize your titleformat command to keep the word "Chương" ? – Knumber10 Apr 10 '13 at 3:16
• @Nguyễn Duy Khánh which document class are you using? Are you loading the babel package? If so, with what language option? – Gonzalo Medina Apr 10 '13 at 4:36
• @NguyễnDuyKhánh Replace \chaptertitlename with \chaptername in my code; this should give you "Chương" again; if not, let me know to find a solution. – Gonzalo Medina Apr 10 '13 at 21:44 | 2019-10-20 05:46:08 | {"extraction_info": {"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, "math_score": 0.9512839913368225, "perplexity": 5829.65180447566}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986703625.46/warc/CC-MAIN-20191020053545-20191020081045-00449.warc.gz"} |
https://greprepclub.com/forum/in-the-gure-above-what-is-the-value-of-1131.html | It is currently 13 Jul 2020, 13:58
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# In the figure above, what is the value of
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In the figure above, what is the value of [#permalink] 01 Dec 2015, 07:14
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In the figure above, what is the value of $$\frac{x + y + z}{45}$$ ?
A. 2
B. 3
C. 4
D. 5
E. 6
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Question: 7
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Re: In the figure above, what is the value of [#permalink] 01 Dec 2015, 07:44
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Solution
The interior angles of a triangle, no matter what, must sum to 180°
As such $$\frac{180}{45}=4$$
Answer is $$C$$
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Re: In the figure above, what is the value of [#permalink] 01 Dec 2015, 07:44
Display posts from previous: Sort by | 2020-07-13 21:58:25 | {"extraction_info": {"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, "math_score": 0.38885068893432617, "perplexity": 8505.183705929838}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657146845.98/warc/CC-MAIN-20200713194203-20200713224203-00348.warc.gz"} |
http://papers.nips.cc/paper/8269-analysis-of-krylov-subspace-solutions-of-regularized-non-convex-quadratic-problems | # NIPS Proceedingsβ
## Analysis of Krylov Subspace Solutions of Regularized Non-Convex Quadratic Problems
[PDF] [BibTeX] [Supplemental] [Reviews]
### Abstract
We provide convergence rates for Krylov subspace solutions to the trust-region and cubic-regularized (nonconvex) quadratic problems. Such solutions may be efficiently computed by the Lanczos method and have long been used in practice. We prove error bounds of the form $1/t^2$ and $e^{-4t/\sqrt{\kappa}}$, where $\kappa$ is a condition number for the problem, and $t$ is the Krylov subspace order (number of Lanczos iterations). We also provide lower bounds showing that our analysis is sharp. | 2019-02-23 13:57:29 | {"extraction_info": {"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, "math_score": 0.28105536103248596, "perplexity": 477.55836357039783}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550249501174.94/warc/CC-MAIN-20190223122420-20190223144420-00611.warc.gz"} |
https://mathshistory.st-andrews.ac.uk/OfTheDay/oftheday-01-25/ | Mathematicians Of The Day
25th January
Quotation of the day
From Joseph-Louis Lagrange
[a quotation from De Morgan]
Lagrange, in one of the later years of his life, imagined that he had overcome the difficulty (of the parallel axiom). he went so far as to write a paper, which he took with him to the Institute, and began to read it. But in the first paragraph something struck him which he had not observed: he muttered: "Il faut que j'y songe encore ", and put the paper in his pocket.
[I must think about it again.]
A De Morgan Budget of Paradoxes | 2022-01-28 15:44:01 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8593867421150208, "perplexity": 3220.2929356309405}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320306301.52/warc/CC-MAIN-20220128152530-20220128182530-00570.warc.gz"} |
https://www.physicsforums.com/threads/understanding-the-argument-of-the-surface-area-integral.968525/ | # Understanding the argument of the surface area integral
#### JD_PM
1. The problem statement, all variables and given/known data
Find $\iint_S ydS$, where $s$ is the part of the cone $z = \sqrt{2(x^2 + y^2)}$ that lies below the plane $z = 1 + y$
2. Relevant equations
3. The attempt at a solution
I have already posted this question on MSE: https://math.stackexchange.com/questions/3155620/surface-integral-of-an-intersection-cone-plane/3155634?noredirect=1#comment6498746_3155634
My issue is with $\iint_S ydS =\sqrt{3} \int_A ydxdy=\sqrt{3}\, \bar{y}|A|$. Concretely, I do not get why $\bar{y}$ shows up.
My issue is that I still do not understand how to deal with the argument of the surface integral. Let's say we had for instance $\iint_S xydS$ or $\iint_S y^2dS$. I wouldn't know how to proceed. I know it is somehow related to the centroid of the figure (in this case an elliptical cylinder).
Robert Z provided a short explanation but I do not get it...
May you please provide an explanation?
Thanks
Related Calculus and Beyond Homework News on Phys.org
#### LCKurtz
Homework Helper
Gold Member
I think I get what is confusing you. Let's say you have an ellipse for which you know the centroid (center). Say the center is at $(c,d)$ and the major and minor axes are $a$ and $b$ and you know the formula for the area of such an ellipse is $\pi a b$. Now let's say you have an integral that you want to evaluate something like $\iint_A y~ dA$ over the elliptical area. That is going to be a bit of work, but you can save yourself the work because you know the formula for the $y$ centroid of area of a region is$$\bar y = \frac {\iint_A y~dA}{\iint_A 1~dA}$$So instead of working your integral out just note you can solve this equation for $\iint_A y~ dA$ getting$$\iint_A y~ dA = \bar y \cdot \text{Area of ellipse} = d \pi a b$$since $d$ is the $y$ coordinate of the center of the ellipse.
#### JD_PM
I think I get what is confusing you. Let's say you have an ellipse for which you know the centroid (center). Say the center is at $(c,d)$ and the major and minor axes are $a$ and $b$ and you know the formula for the area of such an ellipse is $\pi a b$. Now let's say you have an integral that you want to evaluate something like $\iint_A y~ dA$ over the elliptical area. That is going to be a bit of work, but you can save yourself the work because you know the formula for the $y$ centroid of area of a region is$$\bar y = \frac {\iint_A y~dA}{\iint_A 1~dA}$$So instead of working your integral out just note you can solve this equation for $\iint_A y~ dA$ getting$$\iint_A y~ dA = \bar y \cdot \text{Area of ellipse} = d \pi a b$$since $d$ is the $y$ coordinate of the center of the ellipse.
Oh so if we were to have:
$$\iint_A yx~ dA$$
Could I do:
$$\iint_A yx~ dA = \bar y \bar x \cdot \text{Area of ellipse} = de \pi a b$$
?
I have just seen $x$, $y$ and $z$ applied individually but not multiplying (that is why I am asking).
#### LCKurtz
Homework Helper
Gold Member
Oh so if we were to have:
$$\iint_A yx~ dA$$
Could I do:
$$\iint_A yx~ dA = \bar y \bar x \cdot \text{Area of ellipse} = de \pi a b$$
?
I have just seen $x$, $y$ and $z$ applied individually but not multiplying (that is why I am asking).
You can probably answer that for yourself. You would be using a formula like this:$$\bar x \bar y = \frac {\iint_A xy~dA}{\iint_A 1~dA}$$You know the formulas for $\bar x$ and $\bar y$. Put them in there and see if you think it is true.
#### JD_PM
You can probably answer that for yourself. You would be using a formula like this:$$\bar x \bar y = \frac {\iint_A xy~dA}{\iint_A 1~dA}$$You know the formulas for $\bar x$ and $\bar y$. Put them in there and see if you think it is true.
OK so I think you may agree that:
$$\iint_A xy~ dA = \bar x \bar y \cdot \text{Area of ellipse} = cd \pi a b$$
#### LCKurtz
Homework Helper
Gold Member
OK so I think you may agree that:
$$\iint_A xy~ dA = \bar x \bar y \cdot \text{Area of ellipse} = cd \pi a b$$
You are just guessing. Until you show me what happens when you do what I suggested in the last line of post #4 you won't be able to do anything but guess. And you won't learn anything.
#### JD_PM
You are just guessing. Until you show me what happens when you do what I suggested in the last line of post #4 you won't be able to do anything but guess. And you won't learn anything.
We know that:
$$\bar x = \frac {\iint_A x~dA}{\iint_A 1~dA}$$
$$\bar y = \frac {\iint_A y~dA}{\iint_A 1~dA}$$
Then:
$$(\frac {\iint_A x~dA}{\iint_A 1~dA})(\frac {\iint_A y~dA}{\iint_A 1~dA}) = \bar x \bar y = \frac {\iint_A xy~dA}{\iint_A 1~dA}$$
So this equality doesn't hold. Then the following is incorrect:
$$\iint_A xy~ dA = \bar x \bar y \cdot \text{Area of ellipse} = cd \pi a b$$
Then we have no alternative but work the integral out.
#### LCKurtz
Homework Helper
Gold Member
We know that:
$$\bar x = \frac {\iint_A x~dA}{\iint_A 1~dA}$$
$$\bar y = \frac {\iint_A y~dA}{\iint_A 1~dA}$$
Then:
$$(\frac {\iint_A x~dA}{\iint_A 1~dA})(\frac {\iint_A y~dA}{\iint_A 1~dA}) = \bar x \bar y = \frac {\iint_A xy~dA}{\iint_A 1~dA}$$
So this equality doesn't hold. Then the following is incorrect:
$$\iint_A xy~ dA = \bar x \bar y \cdot \text{Area of ellipse} = cd \pi a b$$
Then we have no alternative but work the integral out.
Good, that's a step in the right direction. So it looks like the equality doesn't hold. You do understand that "looks like" isn't a mathematical argument though, right? So what you should do now to really settle the matter for yourself is actually prove that it doesn't hold by working out a simple example and showing you get different numbers. Then you will KNOW it doesn't hold.
Last edited:
"Understanding the argument of the surface area integral"
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• Solo and co-op problem solving | 2019-05-25 04:49:53 | {"extraction_info": {"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, "math_score": 0.7802720665931702, "perplexity": 291.8486821819412}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257889.72/warc/CC-MAIN-20190525044705-20190525070705-00281.warc.gz"} |
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2x3 3 are structured as follows: Aims. These together constitute the indefinite integral. Example 8. The most antiderivatives we know is derived from the table of derivatives, which we read in the opposite direction. Thus, y = x2 + C, where C is arbitrary constant, represents a family of integrals. ⢠Find a distinct anti-derivative of a function. We conclude the lesson by stating the rules for definite integrals, most of which parallel the rules we stated for the general indefinite integrals. Example 2: Compute . ... ⢠Find the indefinite form of the anti-derivative of a function. 4x3 3 4x2 +x+C 3. SECTIONS 5.1 & 5.2: ANTIDERIVATIVES AND INDEFINITE INTEGRALS 5 EXERCISES Find the following integrals. The Teaching & Learning Plans . Leaving Certificate Syllabus. 5.5: Indefinite Integrals and the Substitution Rule Last updated; Save as PDF ... (when one or both of the limits of integration are variables). ANSWERS Inde nite integrals: 1. 3t3 2t2 +3t+C 4. t4 2 t3 3 + 3t2 2 7t+C 5. z 2 2 +3z 21 +C 6. 2u5=2 5 + u 1 2 +5u+C 9. Z (6x2 4x+ 3)dx 2. Check your answer by di erentiating. indefinite integral pdf, We do not have strictly rules for calculating the antiderivative (indefinite integral). Notation: Integration and Indefinite Integral The fact that the set of functions F(x) + C represents all antiderivatives of f (x) is denoted by: â«f(x)dx=F(x)+C where the symbol â« is called the integral sign, f (x) is the integrand, C is the constant of integration, and dx denotes the independent variable we are integrating with respect to. Calculation of integrals using the linear properties of indefinite integrals and the table of basic integrals is called direct integration⦠Find Z 9x3 + 8x2 + 3x 4 3x3 dx. SECTION 8.1 Basic Integration Rules 519 EXAMPLE 2 Using Two Basic Rules to Solve a Single Integral Evaluate Solution Begin by writing the integral as the sum of two integrals. By assigning dif ferent values to C, we get dif ferent members of the family . 4z 6 6 + 7z 3 3 + z2 2 +C 7. The Indefinite Integral and Basic Rules of Integration. 2u3=2 +2u1=2 +C 8. Then apply the Power Rule and the Arcsine Rule as follows. 8v9=4 9 + 24v5=4 5 v 3 + C 10. v6 2 3v8=3 8 +C 11. Z Find Z 3 x + e2x + 5e 4x 7e3x dx. Solution: Example 3: Compute . Table of basic integrals $$\int dx = x + C$$ $$\int x^n dx = \frac{x^{n+1}}{n+1} + C, \quad n eq 1$$ $$\int \frac{1}{x} dx = \ln |x| + C$$ 3x3 3x2 +x+C 12. x3 3 2x x 41. cot1 +C 13. Solution: Lesson Summary 1. 2x2 +3x+C 2. Integration by Parts Recall the Product Rule: d dx [u(x)v(x)] = v(x) du dx + u(x) dv dx 2. M f 1M Fa5d oep 2w Ti 8t ahf 9I in7f vignQift BeD VCfa il ec uyl 7u jsP.W Worksheet by Kuta Software LLC Solution: Using our rules we have Sometimes our rules need to be modified slightly due to operations with constants as is the case in the following example. integrals. An indefinite integral represents a family of functions, all of which differ by a constant. Compute the following indefinite integral. Example 7. ⢠Use anti-differentiation to solve real world problems in which . See Figure 8.1. Integral Calculus. Integrating both sides and solving for one of the integrals leads to our Integration by Parts formula: Z udv= uv Z vdu Integration by Parts (which I may abbreviate as IbP or IBP) \undoes" the Product Rule. INDEFINITE INTEGRALS Example 6. Integrals with Trigonometric Functions Z sinaxdx= 1 a cosax (63) Z sin2 axdx= x 2 sin2ax 4a (64) Z sinn axdx= 1 a cosax 2F 1 1 2; 1 n 2; 3 2;cos2 ax (65) Z sin3 axdx= 3cosax 4a + cos3ax 12a (66) Z cosaxdx= O 4 KAnl UlI RrPi rg ChAtNs8 trFe KseUrNvOeOd1. Antiderivatives and the Indefinite Integral. But these integrals are very similar geometrically . EXAMPLE 3 A Substitution Involving Find Given these rules together with Theorem 4.1, we will be able to solve a great variety of definite integrals. Find Z x2 5x+ 2 x dx. ©9 x280 z1537 TK su HtQaY tS 2o XfxtRw ka 1rRe v eLXLBCl. 3X 4 3x3 dx 2 3v8=3 8 +C 11 4 3x3 dx 7z 3 3 z2... 3X3 dx integral represents a family of integrals EXERCISES find the following integrals, we will able! X 41. cot1 +C 13 Use anti-differentiation to solve real world problems in which 7t+C 5. Z 2 2 21... 9X3 + 8x2 + 3x 4 3x3 dx 5.1 & 5.2: ANTIDERIVATIVES AND indefinite 5... Exercises find the following indefinite integral rules pdf the Arcsine Rule as follows rg ChAtNs8 trFe KseUrNvOeOd1 differ. Antiderivatives AND indefinite integrals 5 EXERCISES find the indefinite form of the family KseUrNvOeOd1..., all of which differ by a constant 8v9=4 9 + 24v5=4 5 v 3 3t2! By assigning dif ferent values to C, we will be able to solve a variety... 8X2 + 3x 4 3x3 dx by a constant given these rules together with 4.1. Is derived from the table of derivatives, which we read in the direction. 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* | 2023-01-28 06:06:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7629611492156982, "perplexity": 3661.1387526308904}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499524.28/warc/CC-MAIN-20230128054815-20230128084815-00380.warc.gz"} |
https://www.math.columbia.edu/~plei/f20-GIT.html | #### Patrick Lei
##### Geometric Invariant Theory (Fall 2020)
Geometric invariant theory is an important tool in the study of moduli spaces in algebraic geometry. In particular, GIT is used to construct coarse moduli spaces. Some classical references are:
• [MFK] Mumford, Fogarty, Kirwan, Geometric Invariant Theory
• [D] Dolgachev, Lectures on Invariant Theory
Some more modern references with different perspectives that also discuss moduli theory are:
• [H] Hoskins, Moduli Problems and Geometric Invariant Theory
• [T] Tevelev, Moduli Spaces and Invariant Theory
The relationship between GIT and symplectic reduction is further expanded on in
• [K] Kirwan, Cohomology of Quotients in Symplectic and Algebraic Geometry
Classically, invariant theory focused on studying invariants of rings under group actions, for example
• [N] Nagata, On the 14th Problem of Hilbert
##### Schedule
Sep 11
Organizational Meeting
Sep 25
Anna Abasheva (virtual)
Three approaches to GIT: overview and examples
GIT rests on three pillars: invariant theory, the Hilbert-Mumford criterion, and symplectic reduction. They all serve one purpose, which is to construct the quotient of a variety by a group action. My goal is to give a short overview of the three approaches without giving complete proofs or even without giving them at all (we’ll be able to discuss all the proofs in detail later during the seminar). I’ll present several useful exampes like toric varieties and moduli of finite sets of n points in P^1 for small n. They are nice to have in mind later when dealing with abstract concepts of GIT. During my talk, they will give us a taste of how the three pillars are related and in which cases their methods may be used in practice.
Oct 02
Caleb Ji (board talk)
Preliminaries
In this talk, I will present the basic definitions and properties of categorical quotients and geometric quotients as written in Chapter 0 of [MFK]. This material consists mainly of technical scheme-theoretic statements, some of which I will prove in detail and some of which I will illustrate with examples.
Reference: [MFK], Chapter 0
Oct 09
Alex Xu (board talk)
Actions of Reductive Algebraic Groups
This week, we continue on to Chapter 1 of [MFK] with some fundamental theorems on the actions of reductive algebraic groups. In particular, we will find some criterion for existence of geometric and categorical quotients.
Reference: [MFK], Chapter 1
Oct 16
Patrick Lei (board talk)
Stability
We will present the Hilbert-Mumford stability criterion. Then we will briefly discuss Mumford’s “flag complex,” which is really the geometric realization of a Bruhat-Tits building. We conclude with a detailed discussion of two examples where we use the Hilbert-Mumford stability criterion to determine the set of (semi)stable points: the moduli space of n points in P^1 and the moduli space of plane cubics.
Reference: [MFK], Chapter 2, [H], Section 7
Oct 23
Nicolás Vilches (board talk)
More Examples of Stability
During the previous sessions we have discussed the basic machinery needed to construct GIT quotients. We will study more examples, such as the action of matrices by conjugation, plane quartics and ordered tuples of points in projective space. When possible, we will discuss different approaches to find the semistable locus, such as computing the ring of invariant functions or applying the Hilbert-Mumford criterion. These examples will be as explicit as possible, which will give us insight on some features of the GIT (such as how the semistable locus depends on the line bundle).
Reference: [D], [T], [MFK], Chapters 3, 4
Oct 30
Morena Porzio (board talk)
GIT and Coarse Moduli Problems (Part 1)
The lecture will focus on one of the main motivations of GIT, which is to construct moduli spaces as quotients of schemes. After giving a brief introduction of what a moduli problem is and how GIT helps for fine moduli problems, we will clarify what objects are suitable for stating a coarse moduli problem. This leads to definitions like those of polarizeed schemes and (pre)stable curves. Examples will include the moduli space of elliptic curves M_{1,1}, the space M_g of smooth curves of genus g, and the space \bar{M}_g of stable curves of genus g.
Nov 06
Morena Porzio (board talk)
GIT and Coarse Moduli Problems (Part 2)
Let’s pick up where we left off, namely from the smooth and compactness considitions that arise from coarse moduli problems. We will prove that M_g has a coarse moduli space by describing the connection between M_g and the Hilbert scheme and reducing representability to a question of stability on Chow varieties. If there is time, we will mention the generalization of this method to the problem of moduli for abelian varieties.
Nov 13
Kuan-Wen Chen (board talk)
GIT and the Moment Map
In this talk, I will first introduce the moment map in symplectic geometry and give several examples. Later I will discuss the relationship between the symplectic quotient and the GIT quotient. If time permits, we will also discuss how to apply equivariant Morse theory to compute the Betti numbers of the symplectic quotient.
Nov 20
Alex Xu (board talk)
GIT Quotients of Symplectic Manifolds
We continue from last week on the relationship between symplectic and GIT quotients. We first consider the case for quotients of Kähler and hyperkähler manifolds, and show that the quotient of the properly stable locus carries a natural Kähler (or hyperkähler) structure. Afterwards, we switch gears and discuss desingularization of the quotient via blowups, and discuss differences between GIT and symplectic quotients. If time permits, we will discuss some Yang-Mills theory.
Reference: [MFK], Chapter 8
Nov 27
No Seminar (Thanksgiving)
Dec 04
Patrick Lei (board talk)
Everything you need to know about Morena’s lectures
We will discuss stability of Chow points of curves in projective space and then construct a morphism between the Hilbert scheme and the Chow variety. This talk will largely fill in details left out of Morena’s lectures. Disclaimer: This talk will not cover everything discussed in Morena’s lectures. No stacks were harmed during the creation of this lecture.
Reference: [MFK], Sections 4.6, 5.4
Dec 11
Nicolás Vilches (board talk)
The moduli space of stable curves
In the previous weeks we have been working on the moduli space of smooth curves using GIT. This approach not only gives us a quasiprojective variety, but also a natural compactification. We will discuss on how this new space is not only a projective variety, but also a moduli space of stable curves. Then, we will consider stable reduction theorems, which allow us to compute stable limits of degenerations of curves. The discussion will be guided by some explicit examples, which will highlight the key elements of the proof. | 2023-03-28 02:44:22 | {"extraction_info": {"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, "math_score": 0.5270471572875977, "perplexity": 1026.095477873851}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948756.99/warc/CC-MAIN-20230328011555-20230328041555-00615.warc.gz"} |
http://amtoolbox.org/amt-0.9.2/doc/binaural/dietz2011.php | # THE AUDITORY MODELING TOOLBOX
Applies to version: 0.9.2
Go to function
# DIETZ2011 - Dietz 2011 binaural model
## Usage
[...] = dietz(insig,fs);
## Input parameters
insig binaural signal for which values should be calculated fs sampling rate (Hz)
## Output parameters
fine Information about the fine structure (see below) env Information about the envelope (see below)
## Description
dietz2011(insig,fs) calculates interaural phase, time and level differences of fine- structure and envelope of the signal, as well as the interaural coherence, which can be used as a weighting function.
The output structures fine and env have the following fields:
.s1 Left signal as put in the binaural processor .s2 Right signal as put in the binaural processor .fc Center frequencies of the channels (f_carrier or f_mod) .itf Transfer function .itf_equal Transfer function without amplitude .ipd Phase difference in rad .ipd_lp Based on lowpass-filtered itf, phase difference in rad .ild Level difference in dB .itd Time difference based on instantaneous frequencies .itd_C Time difference based on central frequencies .itd_lp As .itd, with low-passed itf .itd_C_lp As .itd_C, with low-passed itf .f_inst_1 Instantaneous frequencies in the channels of the filtered s1 .f_inst_2 Instantaneous frequencies in the channels of the filtered s2 .f_inst Instantaneous frequencies (average of f_inst1 and 2)
The steps of the binaural model to calculate the result are the following (see also Dietz et al., 2011):
1. Middle ear filtering (500-2000 Hz 1st order bandpass)
2. Auditory bandpass filtering on the basilar membrane using a 4th-order all-pole gammatone filterbank, employing 23 filter bands between 200 and 5000 Hz, with a 1 ERB spacing. The filter width was set to correspond to 1 ERB.
3. Cochlear compression was simulated by power-law compression with an exponent of 0.4.
4. The transduction process in the inner hair cells was modelled using half-wave rectification followed by filtering with a 770-Hz 5th order lowpass.
The interaural temporal disparities are then extracted using a second-order complex gammatone bandpass (see paper for details).
dietz2011 accepts the following optional parameters:
'flow',flow Set the lowest frequency in the filterbank to flow. Default value is 200 Hz. 'fhigh',fhigh Set the highest frequency in the filterbank to fhigh. Default value is 5000 Hz. 'basef',basef Ensure that the frequency basef is a center frequency in the filterbank. The default value is 1000. 'filters_per_ERB',filters_per_erb Filters per erb. The default value is 1. 'middle_ear_thr',r Bandpass freqencies for middle ear transfer. The default value is [500 2000]. 'middle_ear_order',n Order of middle ear filter. Only even numbers are possible. The default value is 2. 'haircell_lp_freq',hlpfreq Cutoff frequency for haircell lowpass filter. The default value is 770. 'haircell_lp_order',hlporder Order of haircell lowpass filter. The default value is 5. 'compression_power',cpwr The default value is 0.4. 'alpha',alpha Internal noise strength. Convention FIXME 65dB = 0.0354. The default value is 0. 'int_randn' Internal noise XXX. This is the default. 'int_mini' Internal noise XXX. 'filter_order',fo Filter order for output XXX. Used for both 'mod' and 'fine'. The default value is 2. 'filter_attenuation_db',fadb Used for both 'mod' and 'fine'. The default value is 10. 'fine_filter_finesse',fff Only for finestructure plugin. The default value is 3. 'mod_center_frequency_hz',mcf_hz Only for envelope plugin. The default value is 135. 'mod_filter_finesse',mff Only for envelope plugin. The default value is 8. 'level_filter_cutoff_hz',lfc_hz For ild- or level-plugin. The default value is 30. 'level_filter_order',lforder For ild- or level-plugin. The default value is 2. 'coh_param',coh_param This is a structure used for the localization plugin. It has the following fields: max_abs_itd The default value is 1e-3. tau_cycles The default value is 5. tau_s The default value is 10e-3. 'signal_level_dB_SPL',signal_level Sound pressure level of left channel. Used for data display and analysis. Default value is 70.
## References:
M. Dietz, S. D. Ewert, and V. Hohmann. Auditory model based direction estimation of concurrent speakers from binaural signals. Speech Communication, 53(5):592-605, 2011. [ DOI | http ] | 2022-01-25 19:31:17 | {"extraction_info": {"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, "math_score": 0.5554735660552979, "perplexity": 7883.245577827694}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304872.21/warc/CC-MAIN-20220125190255-20220125220255-00080.warc.gz"} |
https://bashtage.github.io/randomgen/bit_generators/dsfmt.html | # Double SIMD Mersenne Twister (dSFMT)¶
class randomgen.dsfmt.DSFMT(seed=None, *, mode=None)
Container for the SIMD-based Mersenne Twister pseudo RNG.
Parameters:
seed{None, int, array_like[uint32], SeedSequence}, optional
Random seed used to initialize the pseudo-random number generator. Can be any integer between 0 and 2**32 - 1 inclusive, an array (or other sequence) of unsigned 32-bit integers, a SeedSequence instance or None (the default). If seed is None, then 764 32-bit unsigned integers are read from /dev/urandom (or the Windows analog) if available. If unavailable, a hash of the time and process ID is used.
mode{None, “sequence”, “legacy”}, optional
The seeding mode to use. “legacy” uses the legacy SplitMix64-based initialization. “sequence” uses a SeedSequence to transforms the seed into an initial state. None defaults to “sequence”.
Notes
DSFMT provides a capsule containing function pointers that produce doubles, and unsigned 32 and 64- bit integers [1] . These are not directly consumable in Python and must be consumed by a Generator or similar object that supports low-level access.
The Python stdlib module “random” also contains a Mersenne Twister pseudo-random number generator.
State and Seeding
The DSFMT state vector consists of a 384 element array of 64-bit unsigned integers plus a single integer value between 0 and 382 indicating the current position within the main array. The implementation used here augments this with a 382 element array of doubles which are used to efficiently access the random numbers produced by the dSFMT generator.
DSFMT is seeded using either a single 32-bit unsigned integer or a vector of 32-bit unsigned integers. In either case, the input seed is used as an input (or inputs) for a hashing function, and the output of the hashing function is used as the initial state. Using a single 32-bit value for the seed can only initialize a small range of the possible initial state values.
Parallel Features
DSFMT can be used in parallel applications by calling the method jump which advances the state as-if $$2^{128}$$ random numbers have been generated [2]. This allows the original sequence to be split so that distinct segments can be used in each worker process. All generators should be initialized with the same seed to ensure that the segments come from the same sequence.
>>> from numpy.random import Generator
>>> from randomgen.entropy import random_entropy
>>> from randomgen import DSFMT
>>> seed = random_entropy()
>>> rs = [Generator(DSFMT(seed)) for _ in range(10)]
# Advance each DSFMT instance by i jumps
>>> for i in range(10):
... rs[i].bit_generator.jump()
Compatibility Guarantee
DSFMT makes a guarantee that a fixed seed and will always produce the same random integer stream.
References
[1]
Mutsuo Saito and Makoto Matsumoto, “SIMD-oriented Fast Mersenne Twister: a 128-bit Pseudorandom Number Generator.” Monte Carlo and Quasi-Monte Carlo Methods 2006, Springer, pp. 607–622, 2008.
[2]
Hiroshi Haramoto, Makoto Matsumoto, and Pierre L’Ecuyer, “A Fast Jump Ahead Algorithm for Linear Recurrences in a Polynomial Space”, Sequences and Their Applications - SETA, 290–298, 2008.
Attributes:
Lock instance that is shared so that the same bit git generator can be used in multiple Generators without corrupting the state. Code that generates values from a bit generator should hold the bit generator’s lock.
seed_seq{None, SeedSequence}
The SeedSequence instance used to initialize the generator if mode is “sequence” or is seed is a SeedSequence. None if mode is “legacy”.
## Seeding and State¶
seed([seed]) Seed the generator state Get or set the PRNG state
## Parallel generation¶
jumped([iter]) Returns a new bit generator with the state jumped jump([iter]) Jumps the state as-if 2**128 random numbers have been generated.
## Extending¶
cffi CFFI interface ctypes ctypes interface
## Testing¶
random_raw([size, output]) Return randoms as generated by the underlying BitGenerator | 2022-11-27 09:46:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17024752497673035, "perplexity": 3730.7142291651326}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710218.49/warc/CC-MAIN-20221127073607-20221127103607-00009.warc.gz"} |
http://www.theinfolist.com/php/SummaryGet.php?FindGo=normed_space | normed space
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by the formula $d(A, B) = \, \overrightarrow\, .$ The study of normed spaces and Banach spaces is a fundamental part of
functional analysis 200px, One of the possible modes of vibration of an idealized circular drum head. These modes are eigenfunctions of a linear operator on a function space, a common construction in functional analysis. Functional analysis is a branch of mathemat ...
, which is a major subfield of mathematics.
Definition
A normed vector space is a
vector space In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). ...
equipped with a
norm Norm, the Norm or NORM may refer to: In academic disciplines * Norm (geology), an estimate of the idealised mineral content of a rock * Norm (philosophy) Norms are concepts ( sentences) of practical import, oriented to effecting an action, rat ...
. A is a vector space equipped with a
seminorm In mathematics, particularly in functional analysis, a seminorm is a Norm (mathematics), vector space norm that need not be positive definite. Seminorms are intimately connected with convex sets: every seminorm is the Minkowski functional of some A ...
. A useful variation of the triangle inequality is $\, x-y\, \geq , \, x\, - \, y\, ,$ for any vectors $x$ and $y.$ This also shows that a vector norm is a
continuous function In mathematics Mathematics (from Greek: ) includes the study of such topics as numbers ( and ), formulas and related structures (), shapes and spaces in which they are contained (), and quantities and their changes ( and ). There is no gen ...
. Property 3 depends on a choice of norm $, \alpha,$ on the field of scalars. When the scalar field is $\R$ (or more generally a subset of $\Complex$), this is usually taken to be the ordinary
absolute value In , the absolute value or modulus of a , denoted , is the value of without regard to its . Namely, if is , and if is (in which case is positive), and . For example, the absolute value of 3 is 3, and the absolute value of − ...
, but other choices are possible. For example, for a vector space over $\Q$ one could take $, \alpha,$ to be the $p$-adic norm.
Topological structure
If $\left(V, \, \,\cdot\,\, \right)$ is a normed vector space, the norm $\, \,\cdot\,\,$ induces a
metric METRIC (Mapping EvapoTranspiration at high Resolution with Internalized Calibration) is a computer model Computer simulation is the process of mathematical modelling, performed on a computer, which is designed to predict the behaviour of or th ...
(a notion of ''distance'') and therefore a
topology s, which have only one surface and one edge, are a kind of object studied in topology. In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structu ...
on $V.$ This metric is defined in the natural way: the distance between two vectors $\mathbf$ and $\mathbf$ is given by $\, \mathbf - \mathbf\, .$ This topology is precisely the weakest topology which makes $\, \,\cdot\,\,$ continuous and which is compatible with the linear structure of $V$ in the following sense: #The vector addition $\,+\, : V \times V \to V$ is jointly continuous with respect to this topology. This follows directly from the . #The scalar multiplication $\,\cdot\, : \mathbb \times V \to V,$ where $\mathbb$ is the underlying scalar field of $V,$ is jointly continuous. This follows from the triangle inequality and homogeneity of the norm. Similarly, for any seminormed vector space we can define the distance between two vectors $\mathbf$ and $\mathbf$ as $\, \mathbf - \mathbf\, .$ This turns the seminormed space into a
pseudometric space In mathematics, a pseudometric space is a generalization of a metric space in which the distance between two distinct points can be zero. In the same way as every normed space is a metric space, every seminormed space is a pseudometric space. Becaus ...
(notice this is weaker than a metric) and allows the definition of notions such as continuity and
convergence Convergence may refer to: Arts and media Literature *Convergence (book series), ''Convergence'' (book series), edited by Ruth Nanda Anshen *Convergence (comics), "Convergence" (comics), two separate story lines published by DC Comics: **A four-par ...
. To put it more abstractly every seminormed vector space is a
topological vector space In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It ...
and thus carries a
topological structure In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It h ...
which is induced by the semi-norm. Of special interest are complete normed spaces, which are known as . Every normed vector space $V$ sits as a dense subspace inside some Banach space; this Banach space is essentially uniquely defined by $V$ and is called the of $V.$ Two norms on the same vector space are called if they define the same
topology s, which have only one surface and one edge, are a kind of object studied in topology. In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structu ...
. On a finite-dimensional vector space, all norms are equivalent but this is not true for infinite dimensional vector spaces. All norms on a finite-dimensional vector space are equivalent from a topological viewpoint as they induce the same topology (although the resulting metric spaces need not be the same)., Theorem 1.3.6 And since any Euclidean space is complete, we can thus conclude that all finite-dimensional normed vector spaces are Banach spaces. A normed vector space $V$ is
locally compactIn mathematics, a mathematical object is said to satisfy a property locally, if the property is satisfied on some limited, immediate portions of the object (e.g., on some ''sufficiently small'' or ''arbitrarily small'' neighbourhood (mathematics), ne ...
if and only if the unit ball $B = \$ is
compact Compact as used in politics may refer broadly to a pact or treaty; in more specific cases it may refer to: * Interstate compact * Blood compact, an ancient ritual of the Philippines * Compact government, a type of colonial rule utilized in British N ...
, which is the case if and only if $V$ is finite-dimensional; this is a consequence of Riesz's lemma. (In fact, a more general result is true: a topological vector space is locally compact if and only if it is finite-dimensional. The point here is that we don't assume the topology comes from a norm.) The topology of a seminormed vector space has many nice properties. Given a neighbourhood system $\mathcal\left(0\right)$ around 0 we can construct all other neighbourhood systems as $\mathcal(x) = x + \mathcal(0) := \$ with $x + N := \.$ Moreover, there exists a
neighbourhood basisIn topology and related areas of mathematics, the neighbourhood system, complete system of neighbourhoods, or neighbourhood filter \mathcal(x) for a point is the collection of all Neighbourhood (mathematics), neighbourhoods of the point . Definit ...
for the origin consisting of absorbing and
convex set File:Convex polygon illustration2.svg, Illustration of a non-convex set. Illustrated by the above line segment whereby it changes from the black color to the red color. Exemplifying why this above set, illustrated in green, is non-convex. In geome ...
s. As this property is very useful in
functional analysis 200px, One of the possible modes of vibration of an idealized circular drum head. These modes are eigenfunctions of a linear operator on a function space, a common construction in functional analysis. Functional analysis is a branch of mathemat ...
, generalizations of normed vector spaces with this property are studied under the name
locally convex space In functional analysis Image:Drum vibration mode12.gif, 200px, One of the possible modes of vibration of an idealized circular drum head. These modes are eigenfunctions of a linear operator on a function space, a common construction in functional a ...
s. A norm (or
seminorm In mathematics, particularly in functional analysis, a seminorm is a Norm (mathematics), vector space norm that need not be positive definite. Seminorms are intimately connected with convex sets: every seminorm is the Minkowski functional of some A ...
) $\, \cdot\,$ on a topological vector space $\left(X, \tau\right)$ is continuous if and only if the topology $\tau_$ that $\, \cdot\,$ induces on $X$ is Comparison of topologies, coarser than $\tau$ (meaning, $\tau_ \subseteq \tau$), which happens if and only if there exists some open ball $B$ in $\left(X, \, \cdot\, \right)$ (such as maybe $\$ for example) that is open in $\left(X, \tau\right)$ (said different, such that $B \in \tau$).
Normable spaces
A
topological vector space In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It ...
$\left(X, \tau\right)$ is called normable if there exists a norm $\, \cdot \,$ on $X$ such that the canonical metric $\left(x, y\right) \mapsto \, y-x\,$ induces the topology $\tau$ on $X.$ The following theorem is due to Andrey Kolmogorov, Kolmogorov: Kolmogorov's normability criterion: A Hausdorff topological vector space is normable if and only if there exists a convex, von Neumann bounded neighborhood of $0 \in X.$ A product of a family of normable spaces is normable if and only if only finitely many of the spaces are non-trivial (that is, $\neq \$). Furthermore, the quotient of a normable space $X$ by a closed vector subspace $C$ is normable, and if in addition $X$'s topology is given by a norm $\, \,\cdot,\,$ then the map $X/C \to \R$ given by $x + C \mapsto \inf_ \, x + c\,$ is a well defined norm on $X / C$ that induces the quotient topology on $X / C.$ If $X$ is a Hausdorff Locally convex topological vector space, locally convex
topological vector space In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It ...
then the following are equivalent: # $X$ is normable. # $X$ has a bounded neighborhood of the origin. # the strong dual space $X^_b$ of $X$ is normable. # the strong dual space $X^_b$ of $X$ is Metrizable topological vector space, metrizable. Furthermore, $X$ is finite dimensional if and only if $X^_$ is normable (here $X^_$ denotes $X^$ endowed with the weak-* topology). The topology $\tau$ of the Fréchet space $C^\left(K\right),$ as defined in the article on spaces of test functions and distributions, is defined by a countable family of norms but it is a normable space because there does not exist any norm $\, \cdot\,$ on $C^\left(K\right)$ such that the topology that this norm induces is equal to $\tau.$ Even if a metrizable topological vector space has a topology that is defined by a family of norms, then it may nevertheless still fail to be normable space (meaning that its topology can not be defined by any norm). An example of such a space is the Fréchet space $C^\left(K\right),$ whose definition can be found in the article on spaces of test functions and distributions, because its topology $\tau$ is defined by a countable family of norms but it is a normable space because there does not exist any norm $\, \cdot\,$ on $C^\left(K\right)$ such that the topology this norm induces is equal to $\tau.$ In fact, the topology of a Locally convex topological vector space, locally convex space $X$ can be a defined by a family of on $X$ if and only if there exists continuous norm on $X.$
Linear maps and dual spaces
The most important maps between two normed vector spaces are the Continuous function (topology), continuous Linear transformation, linear maps. Together with these maps, normed vector spaces form a Category theory, category. The norm is a continuous function on its vector space. All linear maps between finite dimensional vector spaces are also continuous. An ''isometry'' between two normed vector spaces is a linear map $f$ which preserves the norm (meaning $\, f\left(\mathbf\right)\, = \, \mathbf\,$ for all vectors $\mathbf$). Isometries are always continuous and injective. A surjective isometry between the normed vector spaces $V$ and $W$ is called an ''isometric isomorphism'', and $V$ and $W$ are called ''isometrically isomorphic''. Isometrically isomorphic normed vector spaces are identical for all practical purposes. When speaking of normed vector spaces, we augment the notion of dual space to take the norm into account. The dual $V^$ of a normed vector space $V$ is the space of all ''continuous'' linear maps from $V$ to the base field (the complexes or the reals) — such linear maps are called "functionals". The norm of a functional $\varphi$ is defined as the supremum of $, \varphi\left(\mathbf\right),$ where $\mathbf$ ranges over all unit vectors (that is, vectors of norm $1$) in $V.$ This turns $V^$ into a normed vector space. An important theorem about continuous linear functionals on normed vector spaces is the Hahn–Banach theorem.
Normed spaces as quotient spaces of seminormed spaces
The definition of many normed spaces (in particular,
Banach space In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It h ...
s) involves a seminorm defined on a vector space and then the normed space is defined as the Quotient space (linear algebra), quotient space by the subspace of elements of seminorm zero. For instance, with the Lp space, $L^p$ spaces, the function defined by $\, f\, _p = \left( \int , f(x), ^p \;dx \right)^$ is a seminorm on the vector space of all functions on which the Lebesgue integral on the right hand side is defined and finite. However, the seminorm is equal to zero for any function Support (mathematics), supported on a set of Lebesgue measure zero. These functions form a subspace which we "quotient out", making them equivalent to the zero function.
Finite product spaces
Given $n$ seminormed spaces $\left\left(X_i, q_i\right\right)$ with seminorms $q_i : X_i \to \R,$ denote the product space by $X := \prod_^n X_i$ where vector addition defined as $\left(x_1,\ldots,x_n\right) + \left(y_1,\ldots,y_n\right) := \left(x_1 + y_1, \ldots, x_n + y_n\right)$ and scalar multiplication defined as $\alpha \left(x_1,\ldots,x_n\right) := \left(\alpha x_1, \ldots, \alpha x_n\right).$ Define a new function $q : X \to \R$ by $q\left(x_1,\ldots,x_n\right) := \sum_^n q_i\left(x_i\right),$ which is a seminorm on $X.$ The function $q$ is a norm if and only if all $q_i$ are norms. More generally, for each real $p \geq 1$ the map $q : X \to \R$ defined by $q\left(x_1,\ldots,x_n\right) := \left(\sum_^n q_i\left(x_i\right)^p\right)^$ is a semi norm. For each $p$ this defines the same topological space. A straightforward argument involving elementary linear algebra shows that the only finite-dimensional seminormed spaces are those arising as the product space of a normed space and a space with trivial seminorm. Consequently, many of the more interesting examples and applications of seminormed spaces occur for infinite-dimensional vector spaces.
*
Banach space In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It h ...
, normed vector spaces which are complete with respect to the metric induced by the norm * * Finsler manifold, where the length of each tangent vector is determined by a norm * Inner product space, normed vector spaces where the norm is given by an inner product * * Locally convex topological vector space – a vector space with a topology defined by convex open sets * Space (mathematics) – mathematical set with some added structure *
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http://tatome.de/zettelkasten/zettelkasten.php?standalone&reference=block-and-bastian-2011 | # Show Reference: "Sensory weighting and realignment: independent compensatory processes"
Sensory weighting and realignment: independent compensatory processes Journal of Neurophysiology, Vol. 106, No. 1. (1 July 2011), pp. 59-70, doi:10.1152/jn.00641.2010 by Hannah J. Block, Amy J. Bastian
@article{block-and-bastian-2011,
abstract = {When estimating the position of one hand for the purpose of reaching to it with the other, humans have visual and proprioceptive estimates of the target hand's position. These are thought to be weighted and combined to form an integrated estimate in such a way that variance is minimized. If visual and proprioceptive estimates are in disagreement, it may be advantageous for the nervous system to bring them back into register by spatially realigning one or both. It is possible that realignment is determined by weights, in which case the lower-weighted modality should always realign more than the higher-weighted modality. An alternative possibility is that realignment and weighting processes are controlled independently, and either can be used to compensate for a sensory misalignment. Here, we imposed a misalignment between visual and proprioceptive estimates of target hand position in a reaching task designed to allow simultaneous, independent measurement of weights and realignment. In experiment 1, we used endpoint visual feedback to create a situation where task success could theoretically be achieved with either a weighting or realignment strategy, but vision had to be regarded as the correctly aligned modality to achieve success. In experiment 2, no endpoint visual feedback was given. We found that realignment operates independently of weights in the former case but not in the latter case, suggesting that while weighting and realignment may operate in conjunction in some circumstances, they are biologically independent processes that give humans behavioral flexibility in compensating for sensory perturbations.},
author = {Block, Hannah J. and Bastian, Amy J.},
day = {1},
doi = {10.1152/jn.00641.2010},
journal = {Journal of Neurophysiology},
month = jul,
number = {1},
pages = {59--70},
pmid = {21490284},
posted-at = {2011-07-27 14:50:25},
priority = {2},
title = {Sensory weighting and realignment: independent compensatory processes},
url = {http://dx.doi.org/10.1152/jn.00641.2010},
volume = {106},
year = {2011}
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http://xnep.cralstradadeiparchi.it/how-to-parametrize-a-curve.html | # How To Parametrize A Curve
Example 4: Parametrize the circle (x $1) 2" (y " 2) ! 9. How to parametrize with arc length a curve on a surface? Getting path parameter of point on curve How to calculate the arc length of a segment of a parameterized curve in 3D?. x=3cos(arcsin(y-1)) I don't know what to do from here or if I'm going in the right direction or not. asked by J on February 8, 2012; Math. Asked Apr 20, 2020. Then the area bounded by the curve, the -axis and the ordinates and will be. Parametrize the following curve. In mathematics, a parametric equation defines a group of quantities as functions of one or more independent variables called parameters. Let D be the region. Note that the p oin ts (; 0) and (2 0) are arbitrarily assigned to r 1. If you're seeing this message, it means we're having trouble loading external resources on our website. Line (curve)).$\gamma$is a parametrization of a rectifiable curve if there is an homeomorphism$\varphi: [0,1]\to [0,1]$such that the map$\gamma\circ \varphi$is Lipschitz. Given regular curve, t → σ(t), reparameterize in terms of arc length, s → σ(s), and consider the unit tangent vector field, T = T(s) (T(s) = σ0(s)). ; You can also place a point on a curve using tool Point or command Point. Parametric Equations are a little weird, since they take a perfectly fine, easy equation and make it more complicated. Parametric families have many possible parameters; which you choose is usually a matter of convenience, simplicity, and usefulness (Breiman, 1973). lines) that parametrize C. A private key is a number priv, and a public key is the public point dotted with itself priv times. The curve x = cosh t, y = sinh, where cosh is the hyperbolic cosine function and sinh is the hyperbolic sine function, parametrize the hpyerbola x 2-y 2 = 1. Set up, but do not evaluate, an integral equal to the mass of the wire. On [ParametricPlot3D:: accbend] makes ParametricPlot3D print a message if it is unable to reach a certain smoothness of curve. 5309649 curlFalongsurface3:= 0 0 2 ans3:= K4 4 K 16Kx2 16Kx2 2 dydx. after that, I linear pattern the inner circle and then do a circular. You are not transforming the curve into a parameter, nor are you making it like a parameter. Then the derivative is defined by the formula: , and a≤t≤b, where - the derivative of the parametric equation y(t) by the parameter t and - the derivative of the parametric equation. a curve, we integrate over a surface in 3-space. Parametric Representations of Surfaces Part 2: Local Change-in-Area Factors. So we can take. Hint: The curve which is cut lies above a circle in the xy-plane which you should parametrize as a function of the variable t so that the circle is traversed counterclockwise exactly once as t goes from 0 to 2*pi, and the paramterization starts at the point on the circle with largest x. The default setting MeshFunctions->Automatic corresponds to {#4&} for curves, and {#4&, #5&} for surfaces. Given an oriented line ℓ, let (ℓ) be the number of points at which and ℓ intersect. The curvature of the curve can be defined as the ratio of the rotation angle of the tangent $$\Delta. Arc Length, Parametric Curves 2. Then the area bounded by the curve, the -axis and the ordinates and will be. A little. Calculate the length associated with one turn within the helix. If \(P$$ is a point on the curve, then the best fitting circle will have the same curvature as the curve and will pass through the point $$P$$. N = Number of turns. A curve given by a function y= f(x): x= t, y= f(t). Therefore, if your data violate the assumptions of a usual parametric and nonparametric statistics might better define the data, try running the nonparametric equivalent of the parametric test. Octave-Forge is a collection of packages providing extra functionality for GNU Octave. Solution: We parametrize the curve as (t) = t(1+i) with 0 t 1. The plane 3x-2y+z-7=0 cuts the paraboloid, its intersection being a curve. Find more Mathematics widgets in Wolfram|Alpha. The equation x^2+y^2=25 defines a circle of radius 5 centered on the z-axis. Now x is an odd function of t and y an even function of t. We'll end with a parametrization that. Line Integrals with Respect to x, y, and z. Response Curves Let's explain some response curve applications by example. We can however parametrize the top half of the circle. The variable t is called a parameter and the relations between x, y and t are called parametric equations. Such integrals are important in any of the parametrize the cylinder: (6) x = acosθ, y = asinθ z = z. The di erential of f, df, assigns to each point x2Ua linear map df x: Rn!Rm whose matrix is the Jacobian matrix of fat x, df x= 0 B @ @f 1. Because the path Cis oriented clockwise, we cannot im-mediately apply Green’s theorem, as the region bounded by the path appears on the. A plane curve results when the ordered pairs ( x(t), y(t) ) are graphed for all values of t on some interval. y x z FIGURE 12 19. How to calculate ROC curves Posted December 9th, 2013 by sruiz I will make a short tutorial about how to generate ROC curves and other statistics after running rDock molecular docking (for other programs such as Vina or Glide, just a little modification on the way dataforR_uq. A standard exercise in calculus is the elimination of the parameter in a given parametric representation of a curve. Also, the use of assert, parametrize (note the spelling) creates multiple variants of a test with different values as arguments. One way to sketch the plane curve is to make a table of values. This is the line integral of 1 + yover the curve Cparametrized by x= t;y= sin(t);0 t ˇso it’s R C (1 + y) ds. Solving for a parameter, so a parameter is a fancy way of saying a variable. In general, it can be shown (Exercise 15) that reversing the direction in which a curve $$C$$ is traversed leaves $$\int_C f (x, y)\,ds$$ unchanged, for any $$f (x, y)$$. What is the "the natural" parametrization of this curve? Hint: The curve which is cut lies above a circle in the. While , , parametrizes the unit circle, the hyperbolic functions , , parametrize the standard hyperbola , x>1. All we have to do is parametrize the curve, take a derivative, and then compute $$dW = \vec F \cdot d\vec r\text{. This example shows how to parametrize a curve and compute the arc length using integral. Here are some hints: At time t, the bicycle tire forms the plane spanned by the vertical direction and by a tangent vector to the 20 foot circle. Here are some pointers on how to do it. In Exercises 19 and 20, let r(t) = sin t,cost,sin t cos2t as shown in Figure 12. For any value of t. Consider the curve parametrized by x(θ) = acosθ, y(θ) = bsinθ. For example, y= x2 can be parametrized by x= t, y= t2. ) by the use of parameters. For any given a curve in space, there are many different vector-valued functions that draw this curve. Please put the SW part (or parts and assembly if applicable) into a zip file and attach that after you post your reply and I will see if I can get. I can use the standard parametrization of the circle as a curve: Here's the graph using this parametrization. Thus β does follow the route of α, but it reaches a given point on the route at a different time than α does. Integration to Find Arc Length. This is another monotonically increasing function. pdf), Text File (. Though the theme of this page is the points that lie on both of two surfaces, let us begin with only one, the contour x 2 z - xy 2 = 4 or essentially z = (xy 2 + 4)/x 2. Our online calculator finds the derivative of the parametrically. These are sometimes referred to as rectangular equations or Cartesian equations. Circle maps If we keep in f(θ) only the first harmonic, then we get. N = Number of turns. We'll first look at an example then develop the formula for the general case. The variable t is called a parameter and the relations between x, y and t are called parametric equations. A helix, sometimes also called a coil, is a curve for which the tangent makes a constant angle with a fixed line. Clockwise: x= rcost, y= rsintwith 0 t 2ˇ. This also has many examples which show the relevance and usefulness of such parametrisations. The function \dllp: [a,b] \to \R^3 maps the interval [a,b] onto a curve in three dimensions. On [ParametricPlot3D:: accbend] makes ParametricPlot3D print a message if it is unable to reach a certain smoothness of curve. A curve given by a function y= f(x): x= t, y= f(t). (a) (15 pts) Find parametric equations for the tangent line to the curve r(t) = ht3,5t,t4i at the point (−1,−5,1). Get 1:1 help now from expert Calculus tutors Solve it with our calculus problem solver and calculator. In Exercises 19 and 20, let r(t) = sin t,cost,sin t cos2t as shown in Figure 12. We deduce an upper bound on the growth rate of the associated counting function as in Manin’s Conjecture. Connect with us on social media. For each value of t we get a point of the curve. Because a B-spline curve is the composition of a number of curve segments, each of which is defined on a knot span, modifying the position of one or more knots will change the association between curve segments and knot spans and hence change the shape of the curve. Commented: Image Analyst on 23 Mar 2015 If I plot a graph, is there a way for Matlab to check if its a closed curve? 0 Comments. He realized that the pendulum would be isochronous if the bob of a pendulum swung along a cycloidal arc rather than the circular arc of the classical pendulum. One way to sketch the plane curve is to make a table of values. Using the parametrization X = rsin˚cos i + rsin˚sin j + rcos˚k we get X ˚= rcos˚cos i + rcos˚sin j rsin˚k and X = rsin˚sin i + rsin˚cos j; X ˚ X = i j k rcos˚cos rcos˚sin rsin˚. I need to parametrize the curve C traced out by P with angle t as the independent variable. 6 Parameterizing Surfaces Recall that r(t) = hx(t),y(t),z(t)i with a ≤ t ≤ b gives a parameterization for a curve C. The material in the stator and the center part of the rotor has a nonlinear relation between the magnetic flux, B and the magnetic field, H, the so called B-H curve. Computing the private key from the public key in this kind of cryptosystem is called the elliptic curve. Consider the paraboloid z=x^2+y^2. Question: Parametrize The Given Curve. Parametrization may refer more specifically to: Parametrization (geometry), the process of finding parametric equations of a curve, surface, etc. To get more uniform curves, we perform one more step: re-parametrize the spline interpolated curve of the given curve segments by arclength. The function fourierComponents implements the B -spline curve making, the re-parametrization by arclength, and the FTT calculation to obtain the Fourier coefficient. An elliptic curve cryptosystem can be defined by picking a prime number as a maximum, a curve equation and a public point on the curve. Parameterized Query: A parameterized query is a type of SQL query that requires at least one parameter for execution. (b) If the coordinate functions of F~: R3!R3 have continuous second partial derivatives, then curl(div(F~)) equals zero. Parametrize a curve, Multivariable Calculus. To start, we pick a cylinder to parametrize, to avoid troublesome domain issues we chose to parametrize y 2+ z = 20 rst, any vector-valued function for this curve will have yand zcomponent: r = To satisfy the other equation, we solve for xin terms of y, we have two choices x= p. A compact version of the parametric equations can be written as follows: Similarly, we can write y(t) = T B z(t) = T C Each dimension is treated independently, so we can deal with curves in any number of dimensions. The sample code to move an object along a bezier curve can be obtained by clicking. More precisely, consider a metric space (X, d) and a continuous function \gamma: [0,1]\to X. W e can no w use the parametrization of C to determine tangen tv ectors to C, plot on a graphics soft w are, or to p erform a line in tegral around C. Limb darkening is fundamental in determining transit light-curve shapes, and is typically modelled by a variety of laws that parametrize the intensity profile of the star that is being transited. For example, y= x2 can be parametrized by x= t, y= t2. About the rotation curve and mass component parametrization. Get the free "Parametric Curve Plotter" widget for your website, blog, Wordpress, Blogger, or iGoogle. Here are some pointers on how to do it. Parametric Equations are a little weird, since they take a perfectly fine, easy equation and make it more complicated. pdf), Text File (. Set the new parameter. If the curve is regular then is a monotonically increasing function. If you're behind a web filter, please make sure that the domains *. All we have to do is parametrize the curve, take a derivative, and then compute \(dW = \vec F \cdot d\vec r\text{. In addition is anyone knows how to parametrize a spiral that goes around a sphere, starting from the top going to the bottom (20 rotations) that would also be very helpful. I got into splines the way many people do: I wanted a way to draw smooth, attractive connectors between graphic objects in a very general way, and with the ability to specify the exact path the curves should take. A cusp of a plane parametric curve. Parametric Representations of Surfaces Part 2: Local Change-in-Area Factors. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. HINT: Find the arc length function s(t), then its inverse t(s), then express x and y in terms of s. Parametrize the curve of intersection of x = -y^2 - z^2 and z = y. We can think of a curve as an equivalence class. In this video we show one easy, consistent way to parametrize any curve. If \(P$$ is a point on the curve, then the best fitting circle will have the same curvature as the curve and will pass through the point $$P$$. The default setting Mesh->Automatic corresponds to None for curves, and 15 for surfaces. This function also maps the interval [0,2π] onto the ellipse. Frenet frames. In section 16. Then the area bounded by the curve, the -axis and the ordinates and will be. A video on how to parameterize a line segment. Parametrization, also spelled parameterization , parametrisation or parameterisation , is the process of defining or choosing parameters. 1 Graph the curve given by r = 2. For instance, "B. Solution: Note that the desired tangent line must be perpendicular to the normal vectors of both surfaces at the given point. Show that parallel transport along is an isometry from T pS to T qS. We'll first look at an example then develop the formula for the general case. ) by the use of parameters. (b) Describe the projection of C onto the xy-plane. Instead we can find the best fitting circle at the point on the curve. Does this integral depend on the path from (3,−1,2) to (2,1,−1)? Explain. The diagrams produced by our method do not require the construction of an additional characteristic point. You'll learn more about this mark shortly. Set up the integral to find the length of the curve x= 1 t, y. We computed these line integrals by first finding parameterizations (unless special. Clockwise: x= rcost, y= rsintwith 0 t 2ˇ. where D is a set of real numbers. are the parametric equations of the quadratic polynomial. Parameterizing a curve by arc length To parameterize a curve by arc length, the procedure is Find the arc length. Solution: We parametrize the curve as (t) = t(1+i) with 0 t 1. along a curve on S which passes through p. parameterized surface: Area(S) = ZZ kX u X v(u;v)kdudv This is in fact invariant under parameterizations. Parametrization, also spelled parameterization , parametrisation or parameterisation , is the process of defining or choosing parameters. If $$P$$ is a point on the curve, then the best fitting circle will have the same curvature as the curve and will pass through the point $$P$$. I just realized in your case this may not be what you are looking for, as these functions are intended for interpolation and fit a different spline between each pair of X-data points, whereas you are looking for something more like a smoothing spline or a polynomial curve fit. ARC LENGTH, PARAMETRIC CURVES 57 2. Parametric Representations of Surfaces Part 2: Local Change-in-Area Factors. 3Let r(t) = ht 2;t2 +1i. The examples above showed us that we can compute work along any closed curve. In fact, the. you can then use a circular pattern and select the Outer curves and Excluding the inner curve so you don't have to do as much work with the last step. A torus, or more commonly known, as a doughnut shape. More precisely, consider a metric space$(X, d)$and a continuous function$\gamma: [0,1]\to X\$. Derivatives of Parametric Equations, Parametrize a Curve with Respect to Arc Length, find the arc length of a parametric curve, examples and step by step solutions, A series of free online calculus lectures in videos. The Organic Chemistry Tutor 266,592 views. A private key is a number priv, and a public key is the public point dotted with itself priv times. In this paper, given a tolerance ϵ>0 and an ϵ-irreducible algebraic affine plane curve C of proper degree d, we introduce the notion of ϵ-rationality, and we provide an algorithm to parametrize. Let Cbe a curve, f2K[x;y] its de ning polynomial, and P= (a;b) 2A 2 a point on C. The plane 3x-2y+z-7=0 cuts the paraboloid, its intersection being a curve. For any curve in $$\partial S$$, the positive orientation is the direction along the curve which keeps $$S^{int}$$ on the left. We start with the circle in the xy–plane that has radius ρ and is centred on the origin. In the example image at the bottom, we have a 360 degree helical curve wrapping around a cylinder. Set up, but do not evaluate, an integral equal to the mass of the wire. As t varies, the end point of this vector moves along the curve. Parameterize a Line Segment and a Circle Related Topics: More Lessons for PreCalculus Math Worksheets Videos, worksheets, games and activities to help PreCalculus students learn how to parametrize a line segment and a circle. how to parametrize a curve - Free download as PDF File (. A curve in the plane is said to be parameterized if the set of coordinates on the curve, (x,y), are represented as functions of a variable t. It tells for example, how fast we go along the curve. x=3cos(arcsin(y-1)) I don't know what to do from here or if I'm going in the right direction or not. Parametric equations are commonly used to express the coordinates of the points that make up a geometric object such as a curve or surface, in which case the equations are collectively called a parametric representation or parameterization (alternatively. The set D is called the domain of f and g and it is the set of values t takes. For a family of Riemann problems for systems of conservation laws, we construct a flux function that is scalar and is capable of describing the Riemann solution of the original system. The line x + y = 2 can be parametrized as x = 1 + t, y = 1 - t. If $$S$$ does not have any holes, that is if $$\partial S$$ consists of only one curve, this means that the positive orientation circles $$S$$ is a direction that is (on the whole) counterclockwise. I can show you how to have WB send SW parametric values to update the geometry, which is then sent back to WB. Consider a parametric curve in the three-dimensional space given by, , , where the parameter is changing in some interval [a,b]. Specifically, we start with the standard parameterization, then transform it to have a different centre. Parametric Equations are a little weird, since they take a perfectly fine, easy equation and make it more complicated. along a curve on S which passes through p. t/D t tanht;sech t/; t 0: O The fundamental concept underlying the geometry of curves is the arclength of a parametrized curve. Using Green’s Theorem we have that Z C cosy dx+ x2 siny dy = ZZ D (2xsiny + siny) dA = Z 5 0 Z 2 0 (2x+ 1)siny dy dx = [x2 + x]5 0 [ cosy]2 0 = 30(1 cos2): 9. First we parametrize the curve, using the fact that the change of variables u = x 1/3, v = y 1/3 converts the curve to a circle u 2 + v 2 = 1, which has a parametrization u = cos(t), v = sin(t), t going from 0 to 2 p. (Polar coordinates) dont understand how the answer is r(t) = cos t sin 3t i + sin t sin 3t j , t ∈ [0, π]. parameterized surface: Area(S) = ZZ kX u X v(u;v)kdudv This is in fact invariant under parameterizations. Parametric functions show up on the AP Calculus BC exam. Next, I must parametrize. where D is a set of real numbers. Image Transcriptionclose. Parametrized curve Locus Parametrized curve Examples Parametrized curve Parametrized curve A parametrized Curve is a path in the xy-plane traced out by the point (x(t),y(t)) as the parameter t ranges over an interval I. The initial point of the curve is (f(a);g(a)), and the terminal point is (f(b);g(b)). Active 1 year, 11 months ago. Parametric lines. Use the following parametrization for the curve s generated by the intersection: s(t)=(x(t), y(t), z(t)), t in [0, 2pi) x = 5cos(t) y = 5sin(t) z=75cos^2(t) Note that s(t): RR -> RR^3 is a vector valued function of a real variable. Ask Question Asked 1 year, 11 months ago. We can simply use 8 <: x= y= f(x) z= 0. Since we like going from left to right, put t = 0 at the point (2, 3). On the other hand, if you are looking for a spelling that is suggestive of the correct meaning, then you should go with "parametrize" (or "parametrise"). This is another monotonically increasing function. (a) Plot some points and sketch the curve when a= 1 and b= 1, when a= 2 and b= 1, and when a= 1 and b= 2. He realized that the pendulum would be isochronous if the bob of a pendulum swung along a cycloidal arc rather than the circular arc of the classical pendulum. With grasshopper I was able to parametrize the points on the circumference by dividing into angles and finding connecting intersections. Question: Parametrize The Given Curve. This is easy to parametrize: z y x ρˆııı ρˆ ˆk ~r(t) = ρcostˆııı+ρsintˆ 0 ≤ t ≤ 2π. Then fc, ycly + :r2clx is equal. Solution: We can use the same parametrization as in the previous example. We can parametrize a curve with a function of one variable. Sketch the curve defined by the parametric equations x = t 3 - 3 t, y = t 2, t in [-2, 2]. All the parameterizations we've done so far have been parameterizing a curve using one parameter. For example, consider a circle of radius centered at the origin. Besides, \\textbf{GalRotpy} allows the user to perform a parametric fit of a given rotation curve, which relies on a MCMC procedure implemented by using \\verb. Therefore, if your data violate the assumptions of a usual parametric and nonparametric statistics might better define the data, try running the nonparametric equivalent of the parametric test. (b) Use Stokes’ theorem to evaluate F · dr. A Geometric Characterization of Parametric Cubic Curves l 151 point must be constructed from the control points, and since the diagram has a fairly large number of disconnected regions. This is what interpolation implies: that the curve will go exactly through the specified points. Each of the following vector-valued functions will draw this circle: Each of these functions is a different parameterization of the circle. This is the line integral of 1 + yover the curve Cparametrized by x= t;y= sin(t);0 t ˇso it’s R C (1 + y) ds. ) from southern Finland to parametrize an existing taper curve equation. The vertical line passing through the point (3, 2, 0). b2 = 1 is a smooth curve by producing an admissible parametrization. If we know the height and diameter of the cylinder, we can calculate the helical length. The parametric equation of a circle. For any given a curve in space, there are many different vector-valued functions that draw this curve. Section 5-2 : Line Integrals - Part I. Solution to Problem Set #4 1. cuts the paraboloid, its intersection being a curve. This is a problem that I have to graph in mathematica, but I thought it would be good if I knew how to solve the problem first. Now before I plot a Curve on a Graph I want to be able to filter the Rows in the group by the value of the SetVoltage channel and then plot the curve. The projection of this curve on the plane of inclination is the B-curve. Let X be a smooth projective Fano variety over the complex numbers. They will make you ♥ Physics. Using the parametrization X = rsin˚cos i + rsin˚sin j + rcos˚k we get X ˚= rcos˚cos i + rcos˚sin j rsin˚k and X = rsin˚sin i + rsin˚cos j; X ˚ X = i j k rcos˚cos rcos˚sin rsin˚. On [ParametricPlot3D:: accbend] makes ParametricPlot3D print a message if it is unable to reach a certain smoothness of curve. The material in the stator and the center part of the rotor has a nonlinear relation between the magnetic flux, B and the magnetic field, H, the so called B-H curve. Please show every step of the work needed to solve this problem. If you were to take the cylinder and roll it out, the helical length would form the hypotenuse of a triangle made by the height of the cylinder, and it's. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Find "the natural" parametrization of this curve. First, parametrize the curve: x x = t, y = t2 , 0 ≤ t ≤ 1. So to parametrize a curve by arclength means finding a parametrization such that the velocity vector always has length 1. The density of the wire at the point (x;y) is equal to 1+y. Re-parametrization of a curve is useful since a surprisingly high number of functions can not be defined in the Cartesion coordinates (x, y, and sometimes z for 3D functions). Find an expression for $$x$$ such that the domain of the set of parametric equations remains the same as the original rectangular equation. Find parametric equations for the line tangent to the curve of intersection of the given surfaces at the point (1,1,1): Surfaces: xyz = 1, x2 +y2 −z = 1. where x(t), y(t) are differentiable functions and x'(t)≠0. are the parametric equations of the quadratic polynomial. This is what interpolation implies: that the curve will go exactly through the specified points. Now has arc length parameterization. Next, compute the differentials of x and y: dx = dt, dy = 2tdt. txt) or read online for free. so y = a*sin(t), which is correct according to the answer key. A TLS-based taper curve was derived for 246 Scots pines (Pinus sylvestris L. The function can be used in the subdomain settings. In general, it can be shown (Exercise 15) that reversing the direction in which a curve $$C$$ is traversed leaves $$\int_C f (x, y)\,ds$$ unchanged, for any $$f (x, y)$$. Image Transcriptionclose. C = (x(t),y(t)) : t ∈ I Examples The graph of a function y = f(x), x ∈ I, is a curve C that is parametrized by. asked by J on February 8, 2012; Math. Clockwise: x= rcost, y= rsintwith 0 t 2ˇ. after that, you want to Mirror it to the opposite side so you get something like image 3. A More Formal Definition. This example shows how to parametrize a curve and compute the arc length using integral. 1 gives a formula for the slope of a tangent line to a curve defined parametrically regardless of whether the curve can be described by a function y = f (x) y = f (x) or not. I can use the standard parametrization of the circle as a curve: Here's the graph using this parametrization. Circle of radius r: Counter-clockwise: x= rcost, y= rsintwith 0 t 2ˇ. A helix, sometimes also called a coil, is a curve for which the tangent makes a constant angle with a fixed line. If a curve is the graph of a function f, then a parametrization is [t. On the other hand, if you are looking for a spelling that is suggestive of the correct meaning, then you should go with "parametrize" (or "parametrise"). More model perils; parametrize this is the hardest to deal with is the part close to the bottom of the famous Van der Waals energy curve, where there is an. Solution: We parametrize the curve as (t) = t(1+i) with 0 t 1. This means that while these vector. EXAMPLE 10. (c) For any vector eld F~ in R3 all of whose coordinate functions. For each value of t we get a point of the curve. In some applications, such as line integrals of vector fields, the following line integral with respect to x arises: This is an integral over some curve C in xyz space. Re-parametrization of a curve is useful since a surprisingly high number of functions can not be defined in the Cartesion coordinates (x, y, and sometimes z for 3D functions). In this Channel there are repeating Values of 8,12 and 16. You can parametrize the unit square in four parts, each being a simple linear segment corresponding to a side. It is often useful to convert from one set of parameters to another. (a) Obtain a parametrization of the curve C and use your result to evaluate F · dr. Commented: Image Analyst on 23 Mar 2015 If I plot a graph, is there a way for Matlab to check if its a closed curve? 0 Comments. A More Formal Definition. It is often useful to convert from one set of parameters to another. If you extrude that 2D sketch (on the X-Y plane, for example) as a surface, you will get a 3d object based on a 2d equation. Try to replicate it as closely as possible. But sometimes we need to know what both $$x$$ and $$y$$ are, for example, at a certain time , so we need to introduce another variable, say $$\boldsymbol{t}$$ (the parameter). To start, we pick a cylinder to parametrize, to avoid troublesome domain issues we chose to parametrize y 2+ z = 20 rst, any vector-valued function for this curve will have yand zcomponent: r = To satisfy the other equation, we solve for xin terms of y, we have two choices x= p. x(t) = sin(2t), y(t) = cos(t), z(t) = t,. You can parametrize the unit square in four parts, each being a simple linear segment corresponding to a side. 4x + 3y^2 = 7 c(t) = ( ? , ? ) Expert Answer 100% (5 ratings) Previous question Next question Get more help from Chegg. More specifically, when you parametrize you specify a curve or shape with values in a specified range. General definition. Response Curves Let's explain some response curve applications by example. Suppose and are the parametric equations of a curve. So in this example what we actually want to do is solve for a variable x okay. Parametric functions show up on the AP Calculus BC exam. Math 13 - Curve Parametrization Practice The curve shown below, counterclockwise: The curve shown below, clockwise: The curve shown below, counterclockwise: The curve shown below, clockwise (both compo-nents are parts of circles): The curve shown below, from left to right (all components are parts of circles): The curve shown below, clockwise: 2. 1" refers to problem 1 at the end of Appendix B. 4 Vector and Parametric Equations of a Plane A Planes A plane may be determined by points and lines, There are four main possibilities as represented in the following figure: a) plane determined by three points b) plane determined by two parallel lines. General definition. First we parametrize the whole sphere by r( ;˚) = acos sin˚i+ asin sin˚j+ acos˚k then the portion of this sphere lying within the cylinder is the part of the sphere satisfying the additional restriction x2 + y2 = a2 sin2 ˚ sin˚sin : Because of symmetry, we can restrict to only computing the area over the portion of the sphere in. A curve in the plane is said to be parameterized if the coordinates of the points on the curve, (x,y), are represented as functions of a variable t. So 0(t) = 1+i. The plane 3x-2y+z-7=0 cuts the paraboloid, its intersection being a curve. On [ParametricPlot3D:: accbend] makes ParametricPlot3D print a message if it is unable to reach a certain smoothness of curve. so y = a*sin(t), which is correct according to the answer key. In general, it can be shown (Exercise 15) that reversing the direction in which a curve $$C$$ is traversed leaves $$\int_C f (x, y)\,ds$$ unchanged, for any $$f (x, y)$$. I just realized in your case this may not be what you are looking for, as these functions are intended for interpolation and fit a different spline between each pair of X-data points, whereas you are looking for something more like a smoothing spline or a polynomial curve fit. If you were to take the cylinder and roll it out, the helical length would form the hypotenuse of a triangle made by the height of the cylinder, and it’s. The line integral is Z z2dz= Z 1 0 t2(1 + i)2(1 + i)dt= 2i(1 + i) 3: Example 3. Find the points where r(t) intersects the xy. Parametrize the intersection of the surfaces y2 z2 = x 2; y2 + z2 = 9: Solution: r(t) = h2t2 7;t; p 9 t2i Find the point that the curve r(t) = ht2;t2 2t 3;t 3iintersects the X{axis MATH 127 (Section 13. For example, y= x2 can be parametrized by x= t, y= t2. Basically I was feeding a list of curves from a geometry pipeline component through a reparameterized curve parameter into the python component. parametrize the curve over which we are integrating. 1 1) The parametrization ~r(t) = hcos(3t),sin(5t)i describes a curve in the plane. Let be a smooth curve on S connecting the points p and q. But sometimes we need to know what both $$x$$ and $$y$$ are, for example, at a certain time , so we need to introduce another variable, say $$\boldsymbol{t}$$ (the parameter). asked by J on February 8, 2012; Math. I got into splines the way many people do: I wanted a way to draw smooth, attractive connectors between graphic objects in a very general way, and with the ability to specify the exact path the curves should take. In the Curvilinear Motion section, we had an example where a race car was travelling around a curve described in parametric equations as: x(t) = 20 + 0. The learning curve for pytest is shallower than it is for unittest because you don't need to learn new constructs for most tests. parametrize the curve over which we are integrating. The curve x = cosh t, y = sinh, where cosh is the hyperbolic cosine function and sinh is the hyperbolic sine function, parametrize the hpyerbola x 2-y 2 = 1. Parametric Equations of Curves. Consider the curve parameterized by the equations x ( t ) = sin(2 t ), y ( t ) = cos( t ), z ( t ) = t ,. A curve given by a function y= f(x): x= t, y= f(t). The GalRotpy tool can give a first approximation of the galaxy rotation curve using the following schemas: bulge model uning a Miyamoto-Nagai potential, stellar or gaseous disk: thin or thick disks implementing Miyamoto-Nagai potentials, and/or; an exponential disk model. Solution to Problem Set #4 1. The default setting MeshFunctions->Automatic corresponds to {#4&} for curves, and {#4&, #5&} for surfaces. There are three equivalent definitions. ) by the use of parameters. (a) (15 pts) Find parametric equations for the tangent line to the curve r(t) = ht3,5t,t4i at the point (−1,−5,1). Find the exact length of the parametric curve: x=et cost, y=etsint, 0≤t≤ 5 3. A placeholder is normally substituted for the parameter in the SQL query. Sketching By Using Table Of Values And Properties Of Curve. For example y = 4 x + 3 is a rectangular equation. In fact, the. Question: Parametrize The Given Curve. This demonstrates that a circle is just a special case of an ellipse. (Note the orientations ofC 1 andC 2. Intuitively, we think of a curve as a path traced by a moving particle in space. Conversely, given a pair of parametric equations, the set of points (f(t), g(t)) form a curve on the graph. On the other hand, if you are looking for a spelling that is suggestive of the correct meaning, then you should go with "parametrize" (or "parametrise"). For example, y= x2 can be parametrized by x= t, y= t2. For example, if a spiral staircase has a radius of 1 meter. Punishing unjust success and game theory If we want to parametrize a whole line, we do the same. Parameterize a Line Segment and a Circle Related Topics: More Lessons for PreCalculus Math Worksheets Videos, worksheets, games and activities to help PreCalculus students learn how to parametrize a line segment and a circle. Parametrize the curve of intersection of x = -y^2 - z^2 and z = y. In section 16. This means that while these vector. The derivative of the Gaussian equation above. Integration to Find Arc Length. Find the area bounded by the curve x=cost , y=et, 0≤t≤ 2, and the lines y = 1 and x = 0. General definition. For example, input c(3) returns the point at parameter position 3 on curve c. This is a problem that I have to graph in mathematica, but I thought it would be good if I knew how to solve the problem first. b2 = 1 is a smooth curve by producing an admissible parametrization. parametrize [= parameterize] a curve parametrized on the interval [0,1] Go to List of words starting with: A B C D E F G H I J K L M N O P Q R S T U V W Y Z. 4x + 3y^2 = 7 c(t) = ( ? , ? ). Using Green’s Theorem we have that Z C cosy dx+ x2 siny dy = ZZ D (2xsiny + siny) dA = Z 5 0 Z 2 0 (2x+ 1)siny dy dx = [x2 + x]5 0 [ cosy]2 0 = 30(1 cos2): 9. The material in the stator and the center part of the rotor has a nonlinear relation between the magnetic flux, B and the magnetic field, H, the so called B-H curve. For example y = 4 x + 3 is a rectangular equation. The sample code to move an object along a bezier curve can be obtained by clicking. How to Parametrize a Curve. For any curve in $$\partial S$$, the positive orientation is the direction along the curve which keeps $$S^{int}$$ on the left. A parametric curve is defined as a collection of points given by two continuous functions x(t) and y(t), that is, the points on the curve are the collection of points (x(t), y(t)) where x and y are continuous functions of t. The parametrization contains more information about the curve then the curve alone. The diagrams produced by our method do not require the construction of an additional characteristic point. This example shows how to parametrize a curve and compute the arc length using integral. Let H(t) be a linear system of curves parametrizing C; then, there is only one nonconstant intersection point of a generic element of. There are an infinite number of ways to choose a set of parametric equations for a curve defined as a rectangular equation. How to parametrize with arc length a curve on a surface? Getting path parameter of point on curve How to calculate the arc length of a segment of a parameterized curve in 3D?. Parametrized surfaces extend the idea of parametrized curves to vector-valued functions of two variables. Find more Mathematics widgets in Wolfram|Alpha. Just wanted to get rid of that curve parameter in order to keep it a bit more "clean". 5) In this nomenclature, H^2 means "H multiplied by H" or "H squared. You can use the following hints: • Every point on this curve is on the double cone , so when you think you have your final parametrization, you should make sure that if you square x(t), and y(t) and add them together, you get. Parameterization of a curve An example to illustrate how to parameterize a given half circle. The parametrization of a curve is a description of a curve in terms of coordinate functions. There is an easy, if sometimes tedious way, to find these things, as follows. Learn more about nonlinear. The curve is the same one defined by the rectangular. The plane 3x-2y+z-7=0 cuts the paraboloid, its intersection being a curve. We need to find the vector equation of the line of. In mathematics, a parametric equation defines a group of quantities as functions of one or more independent variables called parameters. Consider a parametric curve in the three-dimensional space given by, , , where the parameter is changing in some interval [a,b]. This approach is formalized by considering a curve as a function of a parameter, say t. Next, compute the differentials of x and y: dx = dt, dy = 2tdt. For a family of Riemann problems for systems of conservation laws, we construct a flux function that is scalar and is capable of describing the Riemann solution of the original system. If a curve is the graph of a function f, then a parametrization is [t. (b) (15 pts) At what point on the curve r(t) = ht3,5t,t4i is the normal plane (this is the plane that is perpendicular to the tangent line) parallel to the plane 12x+5y +16z = 3? Solution. curve, which is not necessarily parametrized by arc length. Find a vector parametrization for the curve cant figure it out, the 3θ part confuses me, please help 29. From the above we can find the coordinates of any point on the circle if we know the radius and the subtended angle. Spline Interpolation. That is, the distance a. In Exercises 25–34, find a parametrization of the curve. Identify the interior of the curve. Example — Length of a Parametric Curve. 3Let r(t) = ht 2;t2 +1i. Parameterize the line that passes through the points (0, 1) and (4, 0). where D is a set of real numbers. Use its properties if any. You can use the following hints: • Every point on this curve is on the double cone , so when you think you have your final parametrization, you should make sure that if you square x(t), and y(t) and add them together, you get. General definition. We'll end with a parametrization that. How to calculate ROC curves Posted December 9th, 2013 by sruiz I will make a short tutorial about how to generate ROC curves and other statistics after running rDock molecular docking (for other programs such as Vina or Glide, just a little modification on the way dataforR_uq. A placeholder is normally substituted for the parameter in the SQL query. On the other hand, if you are looking for a spelling that is suggestive of the correct meaning, then you should go with "parametrize" (or "parametrise"). We can parametrize the general line ℓ by the direction in which it points and its signed distance from the origin. The material in the stator and the center part of the rotor has a nonlinear relation between the magnetic flux, B and the magnetic field, H, the so called B-H curve. Parametric Representations of Surfaces Part 2: Local Change-in-Area Factors. The curve x = cosh t, y = sinh, where cosh is the hyperbolic cosine function and sinh is the hyperbolic sine function, parametrize the hpyerbola x 2-y 2 = 1. A cusp of a plane parametric curve. On [ParametricPlot3D:: accbend] makes ParametricPlot3D print a message if it is unable to reach a certain smoothness of curve. A curve in the plane is said to be parameterized if the set of coordinates on the curve, (x,y), are represented as functions of a variable t. Though, I'm not sure that that's the actual equation I need. Is there a way to parametrize the convex hull of a curve in 3D? The convex hull of a curve in 3D can be 2-dimensional figure (for example a surface) and even 3-dimensional figure. The complex pore structures that often occur in porous media complicate such parametrization due to hysteresis between wetting and drying and the effects of tortuosity. Active 3 years, 11 months ago. ; You can also place a point on a curve using tool Point or command Point. Example 1 - Race Track. If you are determined to have a parametric equation with just. (b) Use Stokes' theorem to evaluate F · dr. For any curve in $$\partial S$$, the positive orientation is the direction along the curve which keeps $$S^{int}$$ on the left. Parameterization of Curves in Three-Dimensional Space. parameterized surface: Area(S) = ZZ kX u X v(u;v)kdudv This is in fact invariant under parameterizations. Thus, the domain of a curve is an interval (a;b) (possibly (1 ;1)) consisting of all possible values of a parameter t:. v is the vector result of the cross product of the normal vectors of the two planes. From the above we can find the coordinates of any point on the circle if we know the radius and the subtended angle. But as you will see from the two images below, due to the nature of parametrization, t values are not evenly spaced, and depending on the curve, the object will move at different speeds at different t values on the curve segment. Solution: Z C (y +e √ x)dx+(2x+cosy 2)dy = Z Z D ∂ ∂x (2x+cosy )− ∂ ∂y (y +e. The derivative of the Gaussian equation above. Each of the following vector-valued functions will draw this circle: Each of these functions is a different parameterization of the circle. Midterm 1 Solutions 1. 1 1) The parametrization ~r(t) = hcos(3t),sin(5t)i describes a curve in the plane. In general, it can be shown (Exercise 15) that reversing the direction in which a curve $$C$$ is traversed leaves $$\int_C f (x, y)\,ds$$ unchanged, for any $$f (x, y)$$. Use the remaining parameter to parametrize the curve. Suppose that C can be parameterized by r(t)= with a<=t<=b. The osculating circle of a curve, in a given point, is the circle tangent to the curve in that point that ‘best’ approximates there the curve. This is easy to parametrize: z y x ρˆııı ρˆ ˆk ~r(t) = ρcostˆııı+ρsintˆ 0 ≤ t ≤ 2π. So, I'm working on an assignment that requires a skill I'm not very good at, and that's parameterization. First we parametrize the curve, using the fact that the change of variables u = x 1/3, v = y 1/3 converts the curve to a circle u 2 + v 2 = 1, which has a parametrization u = cos(t), v = sin(t), t going from 0 to 2 p. The plane 3x-2y+z-7=0 cuts the paraboloid, its intersection being a curve. For example, y= x2 can be parametrized by x= t, y= t2. The default setting Mesh->Automatic corresponds to None for curves, and 15 for surfaces. parametrize [= parameterize] a curve parametrized on the interval [0,1] Go to List of words starting with: A B C D E F G H I J K L M N O P Q R S T U V W Y Z. The Organic Chemistry Tutor 266,592 views. Example 1 - Race Track. This is a problem that I have to graph in mathematica, but I thought it would be good if I knew how to solve the problem first. One way to sketch the plane curve is to make a table of values. Let x(t) !. 4x + 6y2. If you were to take the cylinder and roll it out, the helical length would form the hypotenuse of a triangle made by the height of the cylinder, and it's. Find the points where r(t) intersects the xy. The derivative of the Gaussian equation above. Parametrize the line that goes through the points (2, 3) and (7, 9). Find more Mathematics widgets in Wolfram|Alpha. (d) Show that there are an infinite number of different parametrizations for the same curve. Definition 3a. In this case the point (2,0) comes from s = 2 and the point (0,0) comes from s = 0. Derivatives of Parametric Equations, Parametrize a Curve with Respect to Arc Length, find the arc length of a parametric curve, examples and step by step solutions, A series of free online calculus lectures in videos. Parametric Equations A rectangular equation, or an equation in rectangular form is an equation composed of variables like x and y which can be graphed on a regular Cartesian plane. parametrize the curve over which we are integrating. A helix, sometimes also called a coil, is a curve for which the tangent makes a constant angle with a fixed line. In section 16. From the above we can find the coordinates of any point on the circle if we know the radius and the subtended angle. The divergence of F~ is a scalar function, so its curl is not even de ned. Paul Aubin have some video/blog posted online to teach you how to parametrize a curve with different radius and curves. In the picture below, the standard hyperbola is depicted in red, while the point for various values of the parameter t is pictured in blue. On [ParametricPlot3D:: accbend] makes ParametricPlot3D print a message if it is unable to reach a certain smoothness of curve. "1) r(t) = (2cos t)i + (2sin t)j + 3tk length of arc. asked by J on February 8, 2012; Math. Octave-Forge is a collection of packages providing extra functionality for GNU Octave. (Let O denote the origin). The diagrams produced by our method do not require the construction of an additional characteristic point. In this Channel there are repeating Values of 8,12 and 16. Please put the SW part (or parts and assembly if applicable) into a zip file and attach that after you post your reply and I will see if I can get. Indicate its direction. Find the area bounded by the curve x=cost , y=et, 0≤t≤ 2, and the lines y = 1 and x = 0. If you are determined to have a parametric equation with just. parametrize [= parameterize] a curve parametrized on the interval [0,1] Go to List of words starting with: A B C D E F G H I J K L M N O P Q R S T U V W Y Z. Our online calculator finds the derivative of the parametrically. To get more uniform curves, we perform one more step: re-parametrize the spline interpolated curve of the given curve segments by arclength. where t is the set of real numbers. Since the surface of a sphere is two dimensional, parametric equations usually have two variables (in this case θ and ϕ ). Usually, we parametrize using the following. As above, we can get a sense of it projecting to 3 real dimensions. This means that while these vector. Solution: Z C (y +e √ x)dx+(2x+cosy 2)dy = Z Z D ∂ ∂x (2x+cosy )− ∂ ∂y (y +e. Please put the SW part (or parts and assembly if applicable) into a zip file and attach that after you post your reply and I will see if I can get. Parameterize definition, to describe (a phenomenon, problem, curve, surface, etc. Parametrize the curve of intersection of x = -y^2 - z^2 and z = y. Problems 1. The letter t will stand for time. The radius of curvature of the B-curve is the metacentric radius B M ‾ = I / ∇. How to parameterize a curve that is the derivative of a Gaussian [closed] Ask Question Asked 2 The derivative of a Gaussian takes the following form: What I would like to do is to come up with an equation where I can specify the height, width, and center of a curve like the gaussian derivative. MATH280 Tutorial 10: Green’s theorem Page 2 Then I C F ds = X4 i=1 Z C i F ds = Z 1 0 t2 +t2 1 1 (1 t)2 (1 t)2 dt = Z 1 0 2t2 2 22(1 2t+t) dt= Z 1 0 ( 4+4t)dt= 4 1+ 1 2 = 2: Method 2 (Green’s theorem). I have got to sort out the values in my channels as following: In each group I have a Channel named SetVoltage. This is a problem that I have to graph in mathematica, but I thought it would be good if I knew how to solve the problem first. We need to find the vector equation of the line of. An alternative approach is two describe x and y separately in terms of a third parameter, usually t. The curve α has been reparametrized by h to yield the curve β. The tow surfaces intersect in a curve. An alternative approach is two describe x and y separately in terms of a third parameter, usually t. The tow surfaces intersect in a curve. If you have enjoyed our free videos, consider supporting Firefly Lectures. It is a multivalued function from the modular curve to itself, but the better way to think of such a multivalued function is as a correspondence, a curve inside the product of the modular curve with itself. 4x + 3y^2 = 7 c(t) = ( ? , ? ) Expert Answer 100% (5 ratings) Previous question Next question Get more help from Chegg. Solved examples of the area under a parametric curve Note: None of these examples are mine. The osculating circle of a curve, in a given point, is the circle tangent to the curve in that point that ‘best’ approximates there the curve. (c) For any vector eld F~ in R3 all of whose coordinate functions. The area under a parametric curve. Note that the p oin ts (; 0) and (2 0) are arbitrarily assigned to r 1. A nonuniform piece of wire if bent into the shape of the curve y= sin(x) between x= 0 and x= ˇ. 1 gives a formula for the slope of a tangent line to a curve defined parametrically regardless of whether the curve can be described by a function y = f (x) y = f (x) or not. When I first read your problem I thought of the parameterization x = s and y = 2s - s 2. For the Love of Physics - Walter Lewin - May 16, 2011 - Duration: 1:01:26. This is a standard way to parametrize a line segment. He realized that the pendulum would be isochronous if the bob of a pendulum swung along a cycloidal arc rather than the circular arc of the classical pendulum. Connect with us on social media. More model perils; parametrize this is the hardest to deal with is the part close to the bottom of the famous Van der Waals energy curve, where there is an. Usually, we parametrize using the following. Calculate the inverse of the arc length. Instead we can find the best fitting circle at the point on the curve. Consider the curve parameterized by the equations. Specifically, we start with the standard parameterization, then transform it to have a different centre. (1) To find the condition to be satisfied by y(x), let the curve c be slightly deformed from the original position such that any point y on the curve c is. Paul Aubin have some video/blog posted online to teach you how to parametrize a curve with different radius and curves. (a) R C (y + e √ x)dx + (2x + cosy2)dy, C is the boundary of the region enclosed by the parabolas y = x 2and x = y. I have a tricky curve in need of parametrization. The default setting Mesh->Automatic corresponds to None for curves, and 15 for surfaces. Let Cbe a curve, f2K[x;y] its de ning polynomial, and P= (a;b) 2A 2 a point on C. On [ParametricPlot3D:: accbend] makes ParametricPlot3D print a message if it is unable to reach a certain smoothness of curve. Consider the curve parametrized by x(θ) = acosθ, y(θ) = bsinθ. In this Channel there are repeating Values of 8,12 and 16. So, I'm working on an assignment that requires a skill I'm not very good at, and that's parameterization. Parametrization, also spelled parameterization , parametrisation or parameterisation , is the process of defining or choosing parameters. To mask links under text, please type your text, highlight it, and click the "link" button. The plane 3x-2y+z-7=0 cuts the paraboloid, its intersection being a curve. Each of the following vector-valued functions will draw this circle: Each of these functions is a different parameterization of the circle. curve_fit(), allowing you to turn a function that models for your data into a python class that helps you parametrize and fit data with that model. Sketching a Plane Curve. C = (x(t),y(t)) : t ∈ I Examples The graph of a function y = f(x), x ∈ I, is a curve C that is parametrized by. The density of the wire at the point (x;y) is equal to 1+y. Show that the curve with parametrization x= sint, y= cost, z= cos4tfor 0 t 2ˇ lies on the circular cylinder x 2+y = 1. You can parametrize it by setting up parameter for the arc and locations. txt file is interpreted will make it work, see below). ***Thank you so much for all of your help!!!. We can find the vector equation of that intersection curve using these steps: I create online courses to help you rock your math class. 2t^3, y(t) = 20t − 2t^2. Let F(x,y,z))= 3yx 2 i + x 3 j + 3xyk be a vector field in R3 and C be the curve of intersection of the surface z = 2x+y 2 and the cylinder x 2 +y 2 = 4, oriented counterclockwise as viewed from above. A function can be used to represent parametrization. First, parametrize the curve: x x = t, y = t2 , 0 ≤ t ≤ 1. Using Green’s Theorem we have that Z C cosy dx+ x2 siny dy = ZZ D (2xsiny + siny) dA = Z 5 0 Z 2 0 (2x+ 1)siny dy dx = [x2 + x]5 0 [ cosy]2 0 = 30(1 cos2): 9. •Many pre-built models for common lineshapes are included and ready to use. My book says that I need to use the integral of F dot dr. It is often useful to convert from one set of parameters to another. 1 Flow & Flux. how do you calculate the parametrization of the curve of intersection of the two surfaces? the equation (x - 1)^2 + y^2 = 1 determines a circular cylinder in 3D, and z = x^2 + y^2 is the equation of a paraboloid. (a) Parametrize each of the two parts of C corresponding to x ≥ 0 and x ≤ 0 taking t = z as parameter. Parametrized surfaces extend the idea of parametrized curves to vector-valued functions of two variables. k9bogsknrtceua2 8z9f9q8tdzrhjaq d0tiudegod6 z19sc9eyeaasw ga9kmmc9wd2j tksymhtvcrnfg hory84a4q81g r421v964y3m1x xizlpfnsgufypu 5ak5ful1xu0m cpgyv0ku2mdhc j5b2gcnljs fa0vynkwdnedf6 uy17m3o9xoj 7fo3xbewrnwbv v8yzu75xikn r1p3chzt3w9fwp9 z7wsm4h3vzy6b8 xfja5oeaf85kcem l726m1zy6v6uvy ymflho3agfcom lhfjp64gybai ibkvl0mnmdflg xr6h8f015sx omar2060mf75gux | 2020-07-07 10:17:13 | {"extraction_info": {"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, "math_score": 0.7790785431861877, "perplexity": 515.2836825107984}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655891884.11/warc/CC-MAIN-20200707080206-20200707110206-00340.warc.gz"} |
https://blog.myrank.co.in/power-in-series-lcr-circuits/ | # Power in Series LCR Circuits
## Power in Series LCR Circuits
Alternating current plays a central role in the system for distributing, converting and using electrical energy, so it is important to look at power relationships in ac circuits. The power in an electric current is the rate at which electrical energy is consumed in the circuit. Let an emf E = E₀ sinωt be applied to a series LCR circuit.
The current in the circuit is I = I₀ sin (ωt + φ)
Where,
φ = Phase difference between current and voltage.
The instantaneous power is given by: P = E x I = E₀I₀ sinωt sin (ωt + φ)
If the average power consumed in the cycle from time t₁ to t₂ is Pavg, then:
$${{P}_{avg}}({{t}_{2}}-{{t}_{1}})=\int\limits_{{{t}_{1}}}^{{{t}_{2}}}{Pdt}$$,
Average power over one cycle, i.e. t₁ = 0 to t₂ = T be given by:
$${{P}_{avg}}T=\int\limits_{0}^{T}{{{E}_{0}}{{I}_{0}}\sin \omega t\,\sin (\omega t+\phi )dt}$$,
$$\Rightarrow\,\,{{P}{avg}}T={{E}{0}}{{I}{0}}\int\limits{0}^{T}{\left( {{\sin}^{2}}\omega t\,\cos \phi +\frac{\sin 2\omega t}{2}\sin \phi\right)}dt$$,
$$\Rightarrow \,\,{{P}_{avg}}T=\frac{{{E}_{0}}{{I}_{0}}}{2}\cos \phi T$$ $$\left[ \because \,\,\,\,\int\limits_{0}^{T}{{{\sin }^{2}}\omega t\,dt=\frac{T}{2}\,\,\,\,\,\And \,\,\,\,\,\int\limits_{0}^{T}{\sin 2\omega t\,dt=0}} \right]$$,
Pavg = EVIV cos φ = Papp cos φ
Where,
Papp = Apparent power or virtual power = EVIV.
cos φ = Power factor = Pavg/ Papp.
Also,
Pavg = EVIV cos φ = (IVZ) IV R/Z = IV²R
For pure resistive circuit: φ = 0 and Z = R.
Pavg = EVIV = (IVZ) IV = IV²Z = IV²R.
For pure inductive circuit: φ = – π/2
Pavg = 0
Therefore, the average power over a complete cycle of ac through an ideal inductor is zero. Actually, whatever energy is needed in building up current in inductance is returned back during the decay of current.
For pure capacitive circuit: φ = π/2.
Pavg = 0.
In this case too, the average power is zero. Actually, whatever energy is needed in building up the voltage across capacitor is returned to the source during discharging of capacitor. So average power consumed in pure inductive or pure capacitive circuit is zero. So, there is no loss of energy in the inductor or capacitor. But in a resistance, loss of energy occurs. It can’t store the energy like inductor or capacitor.
The current through pure L or C, which dissipates no power is called idle current or wattles current. L and C are most suitable for controlling the current in ac circuits. | 2022-10-04 00:27:12 | {"extraction_info": {"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, "math_score": 0.7449145317077637, "perplexity": 1933.3861484083957}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337446.8/warc/CC-MAIN-20221003231906-20221004021906-00699.warc.gz"} |
https://www.clutchprep.com/chemistry/practice-problems/99500/a-common-way-to-make-hydrogen-gas-in-the-laboratory-is-to-place-a-metal-such-as--1 | # Problem: A common way to make hydrogen gas in the laboratory is to place a metal such as zinc in hydrochloric acid (see the figure). The hydrochloric acid reacts with the metal to produce hydrogen gas, which is then collected over water. Suppose a student carries out this reaction and collects a total of 150.2 mL of gas at a pressure of 746 mmHg and a temperature of 25oC. What mass of hydrogen gas (in mg) does the student collect? (The vapor pressure of water is 23.78 mmHg at 25oC.)
###### FREE Expert Solution
Dalton’s Law states that the total pressure inside a container is obtained by adding all the partial pressures of each non-reacting gas
${\mathbit{P}}_{{\mathbf{H}}_{\mathbf{2}}}\mathbf{=}$722.22 mm Hg
Ideal gas equation: moles H2
$\frac{\mathbf{PV}}{\mathbf{RT}}\mathbf{=}\frac{\mathbf{n}\overline{)\mathbf{RT}}}{\overline{)\mathbf{RT}}}\phantom{\rule{0ex}{0ex}}\mathbf{n}\mathbf{=}\frac{\mathbf{PV}}{\mathbf{RT}}$
mass = moles x molar mass
86% (479 ratings)
###### Problem Details
A common way to make hydrogen gas in the laboratory is to place a metal such as zinc in hydrochloric acid (see the figure). The hydrochloric acid reacts with the metal to produce hydrogen gas, which is then collected over water. Suppose a student carries out this reaction and collects a total of 150.2 mL of gas at a pressure of 746 mmHg and a temperature of 25oC. What mass of hydrogen gas (in mg) does the student collect? (The vapor pressure of water is 23.78 mmHg at 25oC.) | 2021-04-14 01:08:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8450133204460144, "perplexity": 1686.9100980776257}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038076454.41/warc/CC-MAIN-20210414004149-20210414034149-00031.warc.gz"} |
https://ncatlab.org/nlab/show/Wasserstein+metric | # nLab Wasserstein metric
### Context
#### Measure and probability theory
measure theory
probability theory
# Contents
## Idea
A Wasserstein metric is a certain metric over a space of probability measures on a measurable space $X$.
By (JKO) the heat flow?/diffusion equation? on $X$ is the gradient flow of the Boltzman-Shannon entropy functional with respect to the Wasserstein metric.
The Wasserstein metric does not seem to arise from a Riemann metric tensor. A detailed discussion of the relevant gradient flows in non-smooth metric spaces is in (AGS).
The characterization of heat flow as the gradient flow of Shannon-entropy is due to
• R. Jordan, D. Kinderlehrer, F. Otto, The variational formulation of the Fokker-Planck equation , SIAM J. Math. Anal. 29 (1998), no. 1, 1-17.(pdf)
The analog of this for finite probability spaces is discussed in
• Jan Maas, Gradient flows of the entropy for finite Markov chains (pdf)
A comprehensive discussion of the corresponding gradient flows is in
• l Ambrosio, N. Gigli, G. Savaré, Gradient flows in metric spaces and in the space of probability measures, Second edition. Lectures in Mathematics ETH Zürich. Birkhäuser Verlag, Basel, 2008. x+334 pp. (pdf of toc and introduction) MR2009h:49002
• Luigi Ambrosio, Nicola Gigli, Construction of the parallel transport in the Wasserstein space, Methods Appl. Anal. 15 (2008), no. 1, 1–29, MR2010c:49082
Revised on July 5, 2011 14:11:58 by Zoran Škoda (161.53.130.104) | 2017-05-28 20:23:56 | {"extraction_info": {"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, "math_score": 0.9178038239479065, "perplexity": 928.4611590662682}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463611560.41/warc/CC-MAIN-20170528200854-20170528220854-00079.warc.gz"} |
http://fittbrand.com/6ex42t/d7727c-angled-dog-bowl-stand | Top
angled dog bowl stand
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Principles for teaching penmanship are taught, along with proper posture, pencil,. | 2022-01-18 08:16:02 | {"extraction_info": {"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, "math_score": 0.19381451606750488, "perplexity": 3968.8166772535296}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300805.79/warc/CC-MAIN-20220118062411-20220118092411-00526.warc.gz"} |
http://www.impan.pl/cgi-bin/dict?moment | ## moment
A moment's consideration will show that ......
Suppose for the moment that $q=1$, so that $\beta = 1$.
We shall show in a moment that these solutions are unique.
Go to the list of words starting with: a b c d e f g h i j k l m n o p q r s t u v w y z | 2014-09-30 09:47:04 | {"extraction_info": {"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, "math_score": 0.8245614767074585, "perplexity": 95.42788378012384}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1412037662882.4/warc/CC-MAIN-20140930004102-00390-ip-10-234-18-248.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/29744-equations.html | # Math Help - Equations
1. ## Equations
Hello, was just wondering if someone could give me some help.
i have to find the solution of the equation
y'-2xy=x which satisfies y(0)=0
so far i have done the basic and written y'= 2xy + x
next i was going to divide all through by y but got a bit stuck after that point. it probably is a relatively easy question but i just am not sure where to go next... if anyone could lend a hand it would be appreciated!
2. Originally Posted by biffer
Hello, was just wondering if someone could give me some help.
i have to find the solution of the equation
y'-2xy=x which satisfies y(0)=0
so far i have done the basic and written y'= 2xy + x
next i was going to divide all through by y but got a bit stuck after that point. it probably is a relatively easy question but i just am not sure where to go next... if anyone could lend a hand it would be appreciated!
Have you tried an integrating factor?
-Dan
3. thanks!
4. You can also try solving the equation without integrating factor..
$y' - 2 x y = x$
First solve the LHS..
$y' - 2 x y = 0$
$y' = 2 x y$
$\frac{dy}{y} = 2x~dx$
$\int \frac{dy}{y} = \int 2x~dx$
$ln|y|=x^2+C$
$y=C\cdot e^{x^2}$.... $\boxed{I}$
Now take C as a function, and find y'
$y' = (C')(e^{x^2}) + (C)(2xe^x)$
$y' = C'e^{x^2} + 2Cxe^x$.... $\boxed{II}$
Put y and y' (marked as I and II) in the original equation:
$y' - 2 x y = x$
$C'e^{x^2} + 2Cxe^x - 2 x C e^{x^2} = x$
$C'e^{x^2} = x$
$dC = \frac{x}{e^{x^2}}~dx$
$C = -\frac{1}{2e^{-x^2}} + K$
Now put this C in I,
$y=Ce^{x^2}$
$y=(-\frac{1}{2e^{-x^2}} + K)e^{x^2}$
$y= -\frac{1}{2} + K\cdot e^{x^2}$ | 2014-12-18 22:36:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 19, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9103579521179199, "perplexity": 403.52589411016885}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802768034.59/warc/CC-MAIN-20141217075248-00095-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/66165/syzygies-of-the-singular-locus-of-a-nodal-plane-curve | # Syzygies of the singular locus of a nodal plane curve
Let $C\subset \mathbb{P}^2$ be a reduced nodal complex plane curve of degree $d$. Let $\Sigma$ be the set of nodes of $C$, and let $I$ be the ideal of $\Sigma$. Denote with $S=\mathbb{C}[x,y,z]$ the polynomial ring in three variables. Consider a minimal free resolution $0 \to \oplus_{i=1}^t S(-b_i) \to \oplus_{i=1}^{t+1}S(-a_i) \to S \to S/I \to 0.$
As an exercise I tried to calculate the Hodge structure on $H^3(\mathbb{P}^2\setminus C)$. From this calculation it follows that $b_i\leq d$ for all $i$ and that the number of irreducible components of $C$ equals $\#\{i \mid b_i=d\}+1$.
My feeling is that this statement might have been known before (it sounds like a classical statement) but I could not find a reference. Moreover, I would prefer a proof for this fact that avoids the use of Hodge theory.
Does anyone knowns a reference (or a more classical proof)?
-
I suggest you look through Eisenbud's book The Geometry of Syzygies. There are a couple of chapters on syzygies of point sets in the plane. – Alexander Woo May 27 '11 at 18:27
I read large part of Eisenbud's book, but I could not find it in there. – Remke Kloosterman May 27 '11 at 18:35 | 2016-05-04 12:09:18 | {"extraction_info": {"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, "math_score": 0.8340937495231628, "perplexity": 102.84311324757995}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860123023.37/warc/CC-MAIN-20160428161523-00096-ip-10-239-7-51.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/54039/vanishing-of-pi-2-and-h-2?sort=newest | # vanishing of $\pi_2$ and $H_2$
I am looking for an "easy" proof of the following statement: Suppose that $X$ is a simply connected space for which $\pi_2(X)=0$. Then $H_2(X)=0$ as well.
I know that one can use the Hurewicz theorem to prove this, but I feel like there must be a simpler proof.
-
You would first want to show that any element $\alpha$ of $H_2(X;\mathbb{Z})$ can be realized by some map of a closed, oriented surface into $X$. If you're using simplicial homology then $\alpha$ is a bunch of triangles which glue together in $X$ and has no boundary. You need to resolve this object at the vertices where it may not be like a surface. Now use the classification of surfaces and kill off the generators ($\pi_1(X)=0$) of the fundamental group of this surface. Homologically, this means that $\alpha$ is the sum of spheres mapping into $X$ and then use $\pi_2(X)=0$. – Somnath Basu Feb 1 '11 at 23:59
It is not true that there is a simpler proof, because the question basically is the Hurewicz theorem. That is, the full version of the Hurewicz theorem looks stronger, but this special case already requires basically all of the ideas. – Greg Kuperberg Feb 2 '11 at 21:03
Claim 1: If $x\in H_2X$ is a homology class, then there exists a 2-dimensional CW complex $K$ and a map $f:K\to X$, such that $x$ is in the image of $f_*:H_2K\to H_2X$.
Claim 2: If $X$ is simply connected with $\pi_2X=0$, then every map $K\to X$ from a 2-dimensional CW complex is null homotopic.
Claim 1 can be proved by taking $K$ to be a finite union of triangles attached together along suitable edges, built by considering an explicit cocycle representing $x$. Claim 2 is easy.
Of course, this is just the $n=3$ case of the standard proof of "easy Hurewicz" (that $\pi_kX=0$ for $k< n$ implies $H_kX=0$ for $k<n$; non-easy Hurewicz is the statement relating the non-trivial groups in dimension $n$, but you don't want that part.)
-
Yes, this is certainly a very elementary way of explaining this fact. Thanks! – Isaac Goldbring Feb 2 '11 at 16:39
You can build a weak homotopy equivalence $K \to X$ from a CW complex with $2$-skeleton $K_2 = *$. Then cellular homology gives the result, provided your homology respects weak equivalence.
- | 2015-08-04 05:33:13 | {"extraction_info": {"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, "math_score": 0.8710641264915466, "perplexity": 153.33835494781962}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042990445.44/warc/CC-MAIN-20150728002310-00280-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/partial-fraction-decomposition.908434/ | # Partial fraction decomposition
## Homework Statement
Find the partial fraction decomposition of ##\displaystyle \frac{1}{x^4 + 2x^2 \cosh (2 \alpha) + 1}##
## The Attempt at a Solution
Using the identity ##\displaystyle \cosh (2 \alpha) = \frac{e^{2 \alpha} + e^{- 2\alpha}}{2}##, we can get the fraction to the form ##\displaystyle \frac{1}{(x^2 + e^{2 \alpha})(x^2 + e^{- 2 \alpha})}##. Since we have two irreducible quadratic factors, it would seem that we would now try to find A, B C, and D such that ##\displaystyle \frac{1}{(x^2 + e^{2 \alpha})(x^2 + e^{- 2 \alpha})} = \frac{Ax + B}{x^2 + e^{2 \alpha}} + \frac{Cx + D}{x^2 + e^{-2 \alpha}} ##. But in the solutions to my book, it just directly says to find A and B such that ##\displaystyle \frac{1}{(x^2 + e^{2 \alpha})(x^2 + e^{- 2 \alpha})} = \frac{A}{x^2 + e^{2 \alpha}} + \frac{B}{x^2 + e^{-2 \alpha}} ##. Why am I wrong in assuming that the numerators must be linear factors?
ehild
Homework Helper
## Homework Statement
Find the partial fraction decomposition of ##\displaystyle \frac{1}{x^4 + 2x^2 \cosh (2 \alpha) + 1}##
## The Attempt at a Solution
Using the identity ##\displaystyle \cosh (2 \alpha) = \frac{e^{2 \alpha} + e^{- 2\alpha}}{2}##, we can get the fraction to the form ##\displaystyle \frac{1}{(x^2 + e^{2 \alpha})(x^2 + e^{- 2 \alpha})}##. Since we have two irreducible quadratic factors, it would seem that we would now try to find A, B C, and D such that ##\displaystyle \frac{1}{(x^2 + e^{2 \alpha})(x^2 + e^{- 2 \alpha})} = \frac{Ax + B}{x^2 + e^{2 \alpha}} + \frac{Cx + D}{x^2 + e^{-2 \alpha}} ##. But in the solutions to my book, it just directly says to find A and B such that ##\displaystyle \frac{1}{(x^2 + e^{2 \alpha})(x^2 + e^{- 2 \alpha})} = \frac{A}{x^2 + e^{2 \alpha}} + \frac{B}{x^2 + e^{-2 \alpha}} ##. Why am I wrong in assuming that the numerators must be linear factors?
It is not wrong but not necessary to assume linear terms. You can take y=x2 the variable, then you have the fraction as ##\frac{1}{(y + e^{2 \alpha})(y + e^{- 2 \alpha})}##
It is not wrong but not necessary to assume linear terms. You can take y=x2 the variable, then you have the fraction as ##\frac{1}{(y + e^{2 \alpha})(y + e^{- 2 \alpha})}##
Why isn't it necessary to assume linear terms? For example, if we were to do the PFD of ##\displaystyle \frac{1}{x(x^2 + 1)}##, we have to assume that ##x^2 + 1## has a linear term in the numerator. Why would these two cases be different?
ehild
Homework Helper
Why isn't it necessary to assume linear terms? For example, if we were to do the PFD of ##\displaystyle \frac{1}{x(x^2 + 1)}##, we have to assume that ##x^2 + 1## has a linear term in the numerator. Why would these two cases be different?
The fraction has only powers of x2.But try to do as you started, having linear and constant terms in the numerator and see what you get.
The fraction has only powers of x2.But try to do as you started, having linear and constant terms in the numerator and see what you get.
I get the same answer as I would if I had just constants. In other words, the coefficients of x in each numerator are just 0... What's the general theory behind this? I always thought that quadratic terms had to have linear factors in the numerator, at least that's what it says in articles on partial fraction decomposition. Are they wrong?
Last edited:
ehild
Homework Helper
I get the same answer as I would if I had just constants. In other words, the coefficients of x in each numerator are just 0... What's the general theory behind this? I always thought that quadratic terms had to have linear factors in the numerator, at least that's what it says in articles on partial fraction decomposition. Are they wrong?
No, they have to have linear factors, but the coefficients can be zero.
As I wrote in Post#2, the fraction does not have linear terms in x. You can take y=x2 the variable, then you have the fraction as ##\frac{1}{(y+e^{2α})(y+e^{-2α})} ##, and it has only constant term in the numerator of the partial fractions.
No, they have to have linear factors, but the coefficients can be zero.
As I wrote in Post#2, the fraction does not have linear terms in x. You can take y=x2 the variable, then you have the fraction as ##\frac{1}{(y+e^{2α})(y+e^{-2α})} ##, and it has only constant term in the numerator of the partial fractions.
So are you saying that since the denominator just has factors with squares of x, it's a special case in which the linear terms in the numerator will have coefficients of 0 for x?
ehild
Homework Helper
So are you saying that since the denominator just has factors with squares of x, it's a special case in which the linear terms in the numerator will have coefficients of 0 for x?
Yes. | 2021-10-22 22:04:51 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8753743171691895, "perplexity": 723.2834039842967}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585522.78/warc/CC-MAIN-20211022212051-20211023002051-00151.warc.gz"} |
https://includestdio.com/7214.html | # python – SQLAlchemy ORDER BY DESCENDING?
## The Question :
477 people think this question is useful
How can I use ORDER BY descending in a SQLAlchemy query like the following?
This query works, but returns them in ascending order:
query = (model.Session.query(model.Entry)
.join(model.ClassificationItem)
.join(model.EnumerationValue)
.filter_by(id=c.row.id)
.order_by(model.Entry.amount) # This row :)
)
If I try:
.order_by(desc(model.Entry.amount))
then I get: NameError: global name 'desc' is not defined.
• Import desc from sqlalchemy from sqlalchemy import desc
774 people think this answer is useful
Just as an FYI, you can also specify those things as column attributes. For instance, I might have done:
.order_by(model.Entry.amount.desc())
This is handy since it avoids an import, and you can use it on other places such as in a relation definition, etc.
363 people think this answer is useful
from sqlalchemy import desc
someselect.order_by(desc(table1.mycol))
Usage from @jpmc26
84 people think this answer is useful
One other thing you might do is:
.order_by("name desc")
This will result in: ORDER BY name desc. The disadvantage here is the explicit column name used in order by.
38 people think this answer is useful
# You can use .desc() function in your query just like this
query = (model.Session.query(model.Entry)
.join(model.ClassificationItem)
.join(model.EnumerationValue)
.filter_by(id=c.row.id)
.order_by(model.Entry.amount.desc())
)
This will order by amount in descending order or
query = session.query(
model.Entry
).join(
model.ClassificationItem
).join(
model.EnumerationValue
).filter_by(
id=c.row.id
).order_by(
model.Entry.amount.desc()
)
)
# Use of desc function of SQLAlchemy
from sqlalchemy import desc
query = session.query(
model.Entry
).join(
model.ClassificationItem
).join(
model.EnumerationValue
).filter_by(
id=c.row.id
).order_by(
desc(model.Entry.amount)
)
)
# For official docs please use the link or check below snippet
sqlalchemy.sql.expression.desc(column) Produce a descending ORDER BY clause element.
e.g.:
from sqlalchemy import desc
stmt = select([users_table]).order_by(desc(users_table.c.name))
will produce SQL as:
SELECT id, name FROM user ORDER BY name DESC
The desc() function is a standalone version of the ColumnElement.desc() method available on all SQL expressions, e.g.:
stmt = select([users_table]).order_by(users_table.c.name.desc())
Parameters column – A ColumnElement (e.g. scalar SQL expression) with which to apply the desc() operation.
asc()
nullsfirst()
nullslast()
Select.order_by()
14 people think this answer is useful
You can try: .order_by(ClientTotal.id.desc())
session = Session()
auth_client_name = 'client3'
result_by_auth_client = session.query(ClientTotal).filter(ClientTotal.client ==
auth_client_name).order_by(ClientTotal.id.desc()).all()
for rbac in result_by_auth_client:
print(rbac.id)
session.close()
.order_by("TableName.name desc") | 2021-01-17 03:22:55 | {"extraction_info": {"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, "math_score": 0.18417705595493317, "perplexity": 10300.831202580179}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703509104.12/warc/CC-MAIN-20210117020341-20210117050341-00592.warc.gz"} |
https://pypi.org/project/licheck/ | Automatically check the licenses of package dependencies.
Project description
Automatically check the licenses of package dependencies.
Description
Inspired by kontrolilo, check the licenses of sotware dependencies in package managers used by developers. This program is primarly aimed to be used as a git hook with pre-commit.
Documentation
https://docs.franco.net.eu.org/licheck/
Please read carefully the External programs section of the documentation to learn which programs you need to install before licheck.
Have a look at the Configuration file section to know how to configure licheck.
API examples
licheck has a public API. This means for example that you can you easily build a TOC within another Python program. The easiest way to build one for a markdown file is:
>>> import licheck
>>> f = open('Pipfile')
[[source]]
name = "pypi"
url = "https://pypi.org/simple"
verify_ssl = true
[dev-packages]
Sphinx = "~=4.1"
[requires]
python_version = "*"
>>> binary, program = licheck.get_binary_and_program('python')
>>> command = licheck.build_command(binary, program, 'Pipfile')
>>> print(licheck.get_data(command, program))
CLI Helps
\$ licheck --help
Copyright (C) 2021-2022 Franco Masotti (franco DoT masotti {-A-T-} tutanota DoT com)
licheck is free software: you can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation, either version 3 of the License, or (at your option) any later version.
licheck is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details.
You should have received a copy of the GNU General Public License along with licheck. If not, see <http://www.gnu.org/licenses/>.
Changelog and trusted source
You can check the authenticity of new releases using my public key.
Changelogs, instructions, sources and keys can be found at blog.franco.net.eu.org/software.
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Project details
Uploaded source
Uploaded py3 | 2023-03-29 02:01:18 | {"extraction_info": {"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, "math_score": 0.3789173662662506, "perplexity": 4984.60980776511}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948900.50/warc/CC-MAIN-20230328232645-20230329022645-00033.warc.gz"} |
http://tatome.de/zettelkasten/zettelkasten.php?tag=attention | # Show Tag: attention
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There is a theory called the premotor theory of visual attention' which posits that activity that can ultimately lead to a saccade can also facilitate processing of stimuli in those places the saccade will/would go to.
A possible ascending pathway from SC to visual cortex through the pulvinar nuclei (pulvinar) may be responsible for the effect of SC activity on visual processing in the cortex.
Feature-based and spatial attention may be based on similar mechanisms.
Feature-based visual attention facilitates object detection across the visual field.
The effects of visual attention are more pronounced in later stages of visual processing (in the visual cortex).
Spatial attention does not seem to affect the selectivity of visual neurons—just the vigour of their response.
Spatial visual attention increases the activity of neurons in the visual cortex whose receptive fields overlap the attended region.
Feature-based visual attention increases the activity of neurons in the visual cortex which respond to the attended feature.
Spatial and feature-based visual attention are additive: together, they particularly enhance the activity of any neuron whose receptive field encompasses the attended region, contains a stimulus with the attended feature, and prefers that feature.
Humans can learn to use the statistics of their environment to guide their visual attention.
Humans do not need to be aware of the stimulus they perceive to use them to guide their visual attention.
Stimuli in one modality can guide attention in another.
Humans can learn to use stimuli in one modality to guide attention in another.
Palmer and Ramsey show that lack of awareness of a visual lip stream does not inhibit learning of its relevance for a visual localization task: the subliminal lip stream influences visual attention and affects the subjects' performance.
They also showed that similar subliminal lip streams did not affect the occurrence of the Mc Gurk effect.
Together, this suggests that awareness of a visual stimulus is not always needed to use it for guiding visual awareness, but sometimes it is needed for multisensory integration to occur (following Palmer and Ramsey's definition).
There are parallels between visual attention and eye movements because both serve the purpose of directing our processing of visual information to stimuli from a region in space that is small enough to handle for our brain.
Since visual attention and eye movements are so tightly connected in the process of visual exploration of a scene, it has been suggested that the same mechanisms may be (partially) responsible for guiding them.
There is evidence suggesting that one cannot plan a saccade to one point in space and turn covert visual attention to another at the same time.
It has been found that stimulating supposed motor neurons in the SC facilitates visual processing in the part of visual cortex whose receptive field is the same as that of the SC stimulated neurons.
Feature-based visual attention facilitates neural responses across the visual field (in visual cortex).
Born et al. provided evidence which shows that preparing a saccade alone already enhances visual processing at the target of the saccade: discrimination targets presented before saccade onset were identified more successfully if they were in the location of the saccade target than when they were not.
Born et al. showed that, if the color of a saccade target stimulus is task relevant, then identification of a discrimination target with that same color is enhanced even if it is not in the same location.
Search targets which share few features with mutually similar distractors surrounding them are said to pop out': it seems to require hardly any effort to identify them and search for them is very fast.
Search targets that share most features with their surrounding, on the other hand, require much more time time be identified.
Saccades evoked by electric stimulation of the deep SC can be deviated towards the target of visual spatial attention. This is the case even if the task forbids a saccade towards the target of visual spatial attention.
Activation build-up in build-up neurons is modulated by spatial attention.
Kustov's and Robinson's results support the hypothesis that there is a strong connection of action and attention.
Botvinick et al. advance two interdependent hypotheses:
1. Conflicts in information processing activate certain cortical areas, most notably the anterior cingulate cortex,
2. Conflict-related activity causes adjustments in cognitive control of information processing to resolve conflict.
Spatial attention raises baseline activity in neurons whose RF are where the attention is even without a visual stimulus (in visual cortex).
Unilateral lesions in brain areas associated with attention can lead to visuospatial neglect; the failure to consider anything within a certain region of the visual field. In extreme cases this can mean that patients e.g. only read from one side of a book.
Kastner and Ungerleider propose that the top-down signals which lead to the effects of visual attention originate from brain regions outside the visual cortex.
Regions lesions of which can induce visuospatial neglect include
• the parietal lobe, in particular the inferior part,
• temporo-parietal junction,
• the anterior cingulate cortex,
• basal ganglia,
• thalamus,
• the pulvinar nucleus.
Many of the cortical areas projecting to the SC have been implicated with attention.
Dehner et al. speculate that the inhibitory influence of FAES activity on SIV activity is connected to modality-specific attention: According to that hypothesis, an auditory stimulus which leads to strong FAES activity will suppress activity in FAES and thus block out cortical somatosensory input to the SC.
Spatial attention can enhance the activity of SC neurons whose receptive fields overlap the attended region
AES has been implicated with selective attention.
Schenck summarizes three neurorobotic studies in which he evaluates visual prediction, and, more specifically, predictive remapping. He argues that his experiments support a claim in psychology saying that pre-saccadic activation of neurons whose receptive fields will contain the location of a salient stimulus after the saccade is not just pre-activation but actually a prediction of what the visual field will be like after the saccade.
Sprague and Meikle Jr. propose that the SC is involved in visual attention.
There are neurons in the supplementary eye field which are related to
• eye movements,
• arm movements,
• ear movements,
• spatial attention.
Attention is necessary to perform the Stroop and Simon tasks.
The frontoparietal network seems involved in executive control and orienting.
The anterior cingulate cortex is likely involved with regulating attention.
Attention developed quite early. Even very simple organisms, like drosophila and honeybees, show evidence of attentional processes.
It has been found that stimulating supposed motor neurons in the SC enhances responses of v4 neurons with the same receptive field as the SC neurons.
Krauzlis et al. state that collicular deactivation has not been found to eliminate signs of task-based attention in neural responses in cortex.
Krauzlis et al. argue that SC deactivation should have changed neural responses in cortex if it regulated attention through visual cortex.
Krauzlis et al.'s argument that SC deactivation should have changed neural responses in cortex if it regulated attention through visual cortex is a bit weak considering that stimulating SC does change sensory representations in v4.
Krauzlis et al. argue that animals without a well-developed neocortex nonetheless show signs of visual attention. Thus, it is likely that the neocortex is not necessary for attention and SC can regulate attention without the neocortex.
Krauzlis et al. argument that animals without a well-developed neocortex show signs of selective attention similar to humans and other higher animals shows that neocortex may not be necessary for attention seems more appropriate than that of lack of influence of collicular deactivation on cortical responses.
Krauzlis et al. argue that attention may not so much be a explicit mechanism but a phenomenon emerging from the need of distributed information processing systems (biological and artificial) for centralized coordination:
According to that view, some centralized control estimates the state of (some part of) the world and modulates both action and perception according to the state which is estimated to be the most plausible at any given point.
Krauzlis et al. localize this central control in the basal ganglia.
VIsual attention is the facilitation of visual processing of some stimuli over others.
The heminanopia that follows unilateral removal of the cortex that mediates visual behavior cannot be explained simply in classical terms of interruption of the visual behavior cannot be explained simply in classical terms of interruption of the visual radiations that serve cortical function.
Explanation fo the deficit requires a broader point of view, namely, that visual attention and perception are mediated at both forebrain and midbrain levels, which interact in their control of visually guided behavior.''
(Sprague, 1966)
Anderson suggests that it would make sense if we attended to whatever to attend to promises the greatest reward.
Saliency of a stimulus might say something about its likelihood of offering reward if attended to.
The probability of reward of attending a stimuli is influenced by two factors:
• the probability of selecting the right thing,
Thus, highly distinctive things have great bottom-up saliency.
• the probability of reward given that the right thing is selected,
• Features that are associated with high reward salient
• Goals can affect which features promise reward in a situation.
Visual feature combinations become more salient if they are learned to be associated with reward.
Targets which are selected in one trial tend to be more salient in subsequent trials—they are selected faster and rejected slower.
The extent of this effect is modulated by whether or not the selection was rewarded.
Verschure summarizes version VII of his distributed adaptive control model as "a unifying theory" of perception cognition, and action. He states that it uses a learned world model in its contextual layer which biases perception processing (top-down) on the one hand, and saliency (bottom-up) on the other. Between these to appears to be what he calls the validation gate which defines matching and mismatch between world model and percepts.
If there are a number of stimuli, many of which share the same low-level features, then that stimulus that does not "pops out". Local saliency-based models like the one due to Itty and Koch fail to explain this effect.
Task-irrelevant cues in one modality can enhance reaction times in others—but they don't always do that. Instances of this effect have been implicated with exogenous attention.
Task-irrelevant auditory cues have been found to enhance reaction times in others. visual cues, however, which cued visual localization, did not cue auditory localization.
Fixating some point in space enhances spoken language understanding if the words come from that point in space. Fixating a visual stream showing lips consistent with the utterances, this effect is strongest, but it also works if the visual display is random. The effect is also enhanced if fixation is combined with some form of visual task which is complex enough.
Fixating at some point in space can impede language understanding if the utterance do not emanate from the focus of visual attention and there are auditory distractors which do.
Goldberg and Wurtz found that neurons in the superficial SC respond more vigorously to visual stimuli in their receptive field if the current task is to make a saccade to the stimuli.
Responses of superficial SC neurons do not depend solely to intrinsic stimulus properties.
Traditionally, visual attention is subdivided into feature-based attention and spatial attention. However, spatial is arguably only one cue out of possibly a number of cues and possibly only a special case.
Weber and Triesch's model learns task-relevant features.
However, a brain region like the SC, which serves a very general task, cannot specialize in one task—it has to serve all goals that the system has.
It therefore should change its behavior depending on the task. Attention is one mechanism which might determine how to change behavior in a given situation.
If the goal is predictive of the input, then a purely unsupervised algorithm could take a representation of the goal as just another input.
While it is possible that the goal often is predictive of the input, some error feedback is probably necessary to tune the degree to which the algorithm can be distracted' by task-irrelevant but interesting stimuli.
There are voluntary (endogenous) and reflexive (exogenous) mechanisms of guiding selective attention.
Santangelo and Macaluso describe typical experiments for studying visual attention.
Frontal eye fields (FEF) and intraparietal sulcus (IPS) have been associated with voluntary orienting of visual attention.
Santangelo and Macaluso provide a rewiew on the recent literature on visual and auditory attention.
Frontoparietal regions play a key role in spatial orienting in unisensory studies of visual and auditory attention.
There seems to be also modality-specific attention which globally de-activates attention in one modality and activates it in the other.
As a computer scientist I would call de-activating one modality completely a special case of selective attention in that modality.
Localized auditory cues can exogenously orient visual attention.
Santangelo and Macaluso state that multisensory integration and attention are probably separate processes.
Maybe attention controls whether or not multi-sensory integration (MSI) happens at all (at least in SC)? That would be in line with findings that without input from AES and rLS, there's no MSI.
Are AES and rLS cat homologues to the regions cited by Santangelo and Macalluso as regions responsible for auditory and visual attention?
Task-irrelevant visual cues do not affect visual orienting (visual spatial attention). Task-irrelevant auditory cues, however, seem to do so.
Santangelo and Macaluso suggest that whether or not the effects of endogenous attention dominate the ones of bottom-up processing (automatic processing) depends on semantic association, be it linguistic or learned association (like dogs and barking, cows and mooing).
Santangelo and Macaluso state that "the same frontoparietal attention control systems are ... activated in spatial orienting tasks for both the visual and auditory modality..."
De Kamps and van der Velde argue for combinatorial productivity and systematicity as fundamental concepts for cognitive representations. They introduce a neural blackboard architecture which implements these principles for visual processing and in particular for object-based attention.
De Kamps and van der Velde use their blackboard architecture for two very different tasks: representing sentence structure and object attention.
Deco and Rolls introduce a system that uses a trace learning rule to learn recognition of more and more complex visual features in successive layers of a neural architecture. In each layer, the specificity of the features increases together with the receptive fields of neurons until the receptive fields span most of the visual range and the features actually code for objects. This model thus is a model of the development of object-based attention.
Using multiple layers each of which learns with a trace rule with successively larger time scales is similar to the CTRNNs Stefan Heinrich uses to learn the structure of language. Could there be a combined model of learning of sentence structure and language processing on the one hand and object-based visual or multi-modal attention on the other?
One function, or, what Santangelo and macaluso call the key element', of selective attention is filtering out distracters—ie. noise filtering.
Stimuli which are non-predictive in a task—like localized stimuli in one modality which are non-predictive of the position of the target in another modality—can enhance performance in valid instances of that task—like detecting targets which by coincidence are where the non-predictive stimulus was.
This demonstrates the existence of exogenous attention.
When asked to ignore stimuli in the visual modality and attend to the auditory modality, increased activity in the auditory temporal cortex and decreased activity in the visual occipital cortex can be observed (and vice versa).
Bertelson et al. did not find a shift of sound source localization due to manipulated endogenous visual spatial attention—localization was shifted only due to (the salience of) light flashes which would induce (automatic, mandatory) exogenous attention.
A localized visual stimulus can shorten the response to a target stimulus if it appears near and shortly after the first stimulus.
It can lengthen the response time if the target stimulus appears somewhere else or too late.
A localized visual stimulus can lengthen the response time to a target if the target stimulus appears somewhere too late after the first stimulus.
This is called inhibition of return'.
Bell et al. found that playing a sound before a visual target stimulus did not increase activity in the neurons they monitored for long enough to lead to (neuron-level) multisensory integration.
Bell et al. make it sound like enhancement in SC neurons due to exogenous, visual, spatial attention is due to residual cue-related activity which is combined (non-linearly) with target-related activity.
If enhancement in SC neurons due to exogenous, visual, spatial attention is due to residual cue-related activity which is combined (non-linearly) with target-related activity, then that casts an interesting light on (the lack of) intra-modal enhancement:
The only difference between an intra-modal cue-stimulus combination and an intra-modal stimulus-stimulus combination lies in the temporal order of the two. Therefore, if two visual stimuli presented in the receptive field of an SC at the same time) neuron do not enhance the response to each other, then the reason can only be a matter of timing.
In an fMRI experiment, Fairhall and Macaluso found that attending (endogenously, spatially) to congruent audio-visual stimuli (moving lips and speech) produced greater activation in SC than either attending to non-congruent stimuli or not attending to congruent stimuli.
Visual spatial attention
• lowers the stimulus detection threshold,
• improves stimulus discrimination,
With two stimuli in the receptive field, one with features of a visual search target and one with different features
• increases average neural activity in cortex compared to the same two objects without attending to any features
• decreases average neural activity if spatial attention is on the location of the non-target compared to when it is on the target.
The fact that average neural activity in cortex is decreased if spatial attention is on the location of a non-target out of a target and a non-target compared to when it is on the target supports the notion that inhibition plays an important role in stimulus selection.
Divisive normalization models describe neural responses well in cases of
• olfactory perception in drosophila,
• visual processing in retina and V1,
• possibly in other cortical areas,
• modulation of responses through attention in visual cortex.
Common metaphors for attention are the 'spotlight' metaphor, the 'bottleneck' metaphor and the 'zoomlens' metaphor. At the core of these metaphors is the notion that attention is a mechanism which regulates to which information some limited resource is applied.
Attention affects both early and late perceptual processing.
Divisive normalization models have explained how attention can facilitate or suppress some neurons' responses.
Some models view attentional changes of neural responses as the result of Bayesian inference about the world based on changing priors.
Chalk et al. argue that changing the task should not change expectations—change the prior—about the state of the world. Rather, they might change the model of how reward depends on the state of the world.
Before a saccade is made, the region that will be the target of that saccade is perceived with higher contrast and visual contrast.
People fixate on different parts of an image depending on the questions they are asked or task they are trying to accomplish.
Brouwer et al found that their subjects looked more at the contact position of the index finger when they were told to grasp an object than when they were just to look at it.
In the first experiment by Brouwer et al, people fixated different parts of a shape depending on whether the task was just to look at it or grasp it.
The subject's initial saccade, however, was not influenced by the task.
It makes sense that sub-cortical visual processing uses peripheral information more than cortical processing:
• sub-cortical processing is concerned with latent monitoring of the environment for potential dangers (or conspecifiics)
• sub-cortical processing is concerned with watching the environment and guiding attention in cortical processing.
Grossberg states that ART predicts a functional link between consciousness, learning, expectation, attention, resonance, and synchrony and calls this principle the CLEARS principle.
The changes to neural responses due to top-down attention are purely caused by intrinsic processes, not a (direct) reaction to external stimuli. They thus support the theory of situatedness.
LIP has been suggested to contain a saliency map of the visual field, to guide visual attention, and to decide about saccades.
Saccade targets tend to be the centers of objects.
Lateral prefrontal cortex (LPFC) and frontal eye fields (FEF) are frontal cortex regions involved in visual attention and target selection.
Both populations in prefrontal cortex and posterior parietal cortex show correlates of bottom-up and top-down visual attention.
In the pop-out condition of a visual search task, Buschman and Miller found that neurons in the posterior parietal cortex region LIP found the search target earlier than neurons in frontal cortex regions FEF and LPFC.
In the pure visual search condition of a visual search task, Buschman and Miller found that neurons in frontal cortex regions FEF and LPFC found the search target earlier than neurons in the posterior parietal cortex region LIP.
Visual attention is influenced both by local and global saliency, ie. bottom-up processes, and by semantics, ie. top-down processes.
The biased competition theory of visual attention explains attention as the effect of low-level stimuli competing with each other for resources—representation and processing. According to this theory, higher-level processes/brain regions bias this competition.
Predictive coding and biased competition are closely related concepts. Spratling combines them in his model and uses it to explain visual saliency.
Desimone and Duncan argue that spatial information about a search target can be part of the attentional template` fitted against all potential targets in the visual display as any other object feature. | 2020-08-13 17:14:25 | {"extraction_info": {"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, "math_score": 0.48305168747901917, "perplexity": 3290.718650324077}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439739048.46/warc/CC-MAIN-20200813161908-20200813191908-00036.warc.gz"} |
https://www.physicsforums.com/threads/weird-integral-question.314760/ | # Weird integral question
1. May 18, 2009
### Emethyst
1. The problem statement, all variables and given/known data
Find the integral of sqrt(9-x^2) over [0,3]. You will not be able to find an antiderivative, so instead interpret the definite integral as the area of a region and compute the area geometrically (I haven't reached integration by substitution and integration by parts in class yet).
2. Relevant equations
The part i'm lost on
3. The attempt at a solution
This question has me stumped. I tried using both riemann sums and the trapezoid method but this didn't get me anywhere, as the answer is supposed to be 9pi/4. It is only out of 1 mark, so I know it can't be that difficult, but i'm still lost over it. Any pointers in the right direction here would be greatly appreciated. Thanks in advance.
2. May 18, 2009
### Pyrrhus
Are you familiar with this geometry figure $y^2 + x^2 = 3^2$ ? Now consider what the square root does to this relation? (this is not a function), but when $y = \sqrt{3^2 - x^2}$ what happens? (think in terms of Real value $\sqrt{x}$ function)
Last edited: May 18, 2009
3. May 18, 2009
### larstuff
Try downloading the program Geogebra (Web Start) - it's free math software, then let it draw the graph of this "weird" thing. You'll probably see what the answer is..
4. May 18, 2009
### Emethyst
No I have not heard of that geometric figure before, but I do know that the square root prevents the function from crossing zero and becoming a negative number, and in a sense resembles half of a horizontal parabola. Now for the obvious question, how does that help me? :tongue:
5. May 18, 2009
### Redbelly98
Staff Emeritus | 2017-10-22 19:51:46 | {"extraction_info": {"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, "math_score": 0.7740442156791687, "perplexity": 577.3470413099578}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187825436.78/warc/CC-MAIN-20171022184824-20171022204824-00005.warc.gz"} |
http://mathoverflow.net/revisions/110639/list | One beautiful book is Peter Schneider's Nonarchimedean Functional Analysis, appeared in the Springer Monographs in Mathematics in 2006. A more analytic one (with less emphasis on Functional Analysis and more on Calculus) in Alain Robert's A course in $p$-adic Analysis, GTM 198. But I still think the bible is S. Bosch, U. Güntzer and R. Remmert's Non-Archimedean AnalysysAnalysis, appeared in the Grundlehren der mathematischen Wissenschaften, 261.
One beautiful book is Peter Schneider's Nonarchimedean Functional Analysis, appeared in the Springer Monographs in Mathematics in 2006. A more analytic one (with less emphasis on Functional Analysis and more on Calculus) in Alain Robert's A course in $p$-adic Analysis, GTM 198. But I still think the bible is S. Bosch, U. Güntzer and R. Remmert's Non-Archimedean Analysys, appeared in the Grundlehren der mathematischen Wissenschaften, 261. | 2013-05-24 19:13:00 | {"extraction_info": {"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, "math_score": 0.6697355508804321, "perplexity": 961.122036086815}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368704986352/warc/CC-MAIN-20130516114946-00001-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://www.encida.dk/2020/04/21/how-to-calculate-fir-filters/ | ## How to Calculate FIR Filters
FIR filters or finite impulse response filters are widely used for thier great qualities and is an invaluable tool for the acoustician. FIR filters can however be tricky to design and calculate relative to IIR filters! I have designed two FIR design functions in Python code that I think would be helpful to you. So how to calculate FIR filters?
#### By Kasper Kiis Jensen
FIR filters are finite impulse responses that can be desgined in almost any shape or form. This blogpost will focus on highpass and lowpass FIR filters. If you do not want to read my explanation of how to calculate FIR filters but just want that awesome Python code to design and use, then skip to the bottom!
### What is a FIR Filter
A FIR filter is defined as a finite impulse response, meaning there is no feedback as with infinite impulse response (IIR) filters. This comes with some advantages:
• FIR filters are always stable.
• Linear phase filters can be guaranteed.
• Finite precision errors are not as servere as IIR filters.
If you want to read in depth about FIR filter theory I would suggest you read Discrete-Time Signal Processing or dive into embedded implementation with Real-Time Digital Signal Processing.
A FIR filter impulse response could look like the one below.
There are four types of FIR filters:
• Type 1 is symmetric and has an odd number of taps.
• Type 2 is symmetric and has an even number of taps.
• Type 3 is not symmetric and has odd number of taps.
• Type 4 is not symmetric and has even number of taps.
Symmetric filters, type 1 and 2, are linear phase filters which is defined as a sinc function where M is the length, ω is the angular cutoff frequency and w[n] is a window e.g. Rectangle or Hanning.
$h[n]=\frac{sin[\omega_c (n-M/2)]}{\pi (n-M/2)}w[n]$
Linear phase filters do not distort the phase of a signal but only delays it, which can be very desirable in some areas of signal processing. You also know the filters delay is precisely N taps. The disadvantage is that there can be large delays in such filters because of their length.
The length of FIR filters is dependent on how low frequencies you want to filter and the sample rate. The longer away from the sample rate you get, the longer the filters becomes.
### Design FIR Filters
When designing filters in general it is very common to have passband and stopband specifications. For example a passband with maximum 1 dB ripple from 4000 Hz and below and a stopband of minimum 60 dB attenuation from 6000 Hz and above.
When designing FIR filters (using the window method) in for example Python or Matlab it is very common that you yourself need to specify the number of taps, the cutoff frequency and the transition width between passband and stopband. This however can be annoying as a lot of trial and error goes into finding the shortest filter possible.
The Kaiser window method however, solves this problem. The Kaiser window is a window which is customisable based on β:
By using this window you can find a filter length within +/- 2 filter taps of your passband and stopband specifications. This makes designing FIR filters much easier than making your own trial and error.
The number of filter taps M is calculated using the minimum stopband attenuation A and the transition width in Hz between the passband and stopband Δω:
$N=\frac{A-8}{2.285\Delta \omega}$
And beta can be calculated as follows
$\beta =\begin{cases} 0.1102(A-8.7), & A>50\\ 0.5842(A-21)^{0.4}+0.07886(A-21) & 21 \leq A \leq 50 \\ 0.0, & A<21 \end{cases}$
With the number of taps and the beta of the Kaiser window, the first equation and the sinc function can be multiplied with the found Kaiser Window to get a linear phase filter which will approximate your passband and stopband specifications within +/- 2 taps.
### How To Get The Code
The code contains 2 functions and an example of how to use the functions:
FIRDesign: designs a FIR filter based on taps, cutoff frequency and transistion width.
kaiserDesign: designs a FIR filter using the Kaiser window method based on passband and stopband specifications.
To download the Python code to design and use the FIR filters, insert your contact information below and you will receive an email with the code.
If you liked this blog post then you might also like: How to calculate room acoustic parameters or How to calculate octave band filters
## How to calculate Mirror Image Method
Learn about Mirror Image Method and try it out yourself.
## How to calculate FIR filters
Learn about FIR filter basics and try it out yourself.
## How to calculate octave band filters
How to Calculate Octave Band Filters Octave and fractional octave band filters are filters used in a huge amount of acoustic calculations. It is something every acoustician has some implementation…
## Understanding Software Development
The first step to developing your own software is knowing at least one useful programming language. | 2020-08-13 11:39:51 | {"extraction_info": {"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, "math_score": 0.47564375400543213, "perplexity": 1169.9825051794164}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738982.70/warc/CC-MAIN-20200813103121-20200813133121-00315.warc.gz"} |
http://chilliant.blogspot.se/2012/06/skeleton-alphabet-1.html | ## Friday, 8 June 2012
### Skeleton Alphabet 1
I recently got my hands on an original copy of Ann Camp's "Pen Lettering" (1958) which describes the relative proportions of a skeleton alphabet.
The construction is as follows. Imagine a unit square:
Now inscribe a circle with unit diameter:
We construct a rectangle of unit height but width 'w' such that the area of the rectangle is the same as the area of the circle:
It follows that 'w' is 'π/4' or about 0.7854. This square-circle-rectangle outline is a fundamental construction motif:
Many capital letters can be traced from it:
In the tracing of the skeleton of 'C' above, the glyph has also been rendered using a well-known sans serif font in pink. Note that the skeleton letter (in black) has a slightly different aspect ratio. This is to be expected; the skeleton alphabet described here is the basis of a set of letters designed to be written using pens. It is based on straight lines and arcs of circles.
Interestingly, the 'M' described by Ann Camp has slightly sloping "verticals":
The 'Q' is just an 'O' with a tail, typically taking up a quarter of the height and half the width:
The raw skeletal 'U' has abrupt corners which are expected to be blended by eye for the final product.
The 'W' is simply two 'V's glued together:
If we utilise the centre of the construction motif, we can trace more letters:
For the letter 'A', halfway up is optically too high for the crossbar. I've chosen the golden ratio to determine 'a' to be about 0.3820:
If we stack two, half-height construction motifs within the unit square, we can trace 'P' and 'R':
However, the remaining letters require a slightly smaller upper section so as not to appear top-heavy. I've chosen 'b' such that the bottom-right corner of the upper square coincides with the top-right corner of the rectangle inside the lower square. That's approximately 0.5283.
Now we can construct most of the remaining capitals:
The final capital, 'S', requires that the two green motifs are stacked centrally. Parts of the blue circle and parts of the green circles are traced accordingly.
Here's the final grid of skeleton capitals:
Next time, I'll try rejigging the dimensions of the pen-based skeleton alphabet to better match modern font design.
#### 1 comment :
1. Hi, the Black letters, what font is that? | 2014-11-25 20:22:45 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8565562963485718, "perplexity": 1788.2473619145212}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931003959.7/warc/CC-MAIN-20141125155643-00219-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://joshbloom.org/publication/2004-ap-j-612-690-s/ | # Discovery of a Transient U-Band Dropout in a Lyman Break Survey: A Tidally Disrupted Star at z=3.3?
### Abstract
We report the discovery of a transient source in the central regions of galaxy cluster A267. The object, which we call PALS-1, was found in a survey aimed at identifying highly magnified Lyman break galaxies in the fields of intervening rich clusters. At discovery, the source had U$_n$>24.7 (2 σ AB), g=21.96+/-0.12, and very blue g-r and r-i colors; i.e., PALS-1 was a U-band dropout,’’ characteristic of star-forming galaxies and quasars at zåisebox-0.5ex 3. However, 3 months later the source had faded by more than 3 mag. Further observations showed a continued decline in luminosity, to R>26.4 at 7 months after discovery. Although the apparent brightness suggests a supernova at roughly the cluster redshift, we show that the photometry and light curve argue against any known type of supernova at any redshift. The spectral energy distribution and location near the center of a galaxy cluster are consistent with the hypothesis that PALS-1 is a gravitationally lensed transient at zi̊sebox-0.5ex 3.3. If this interpretation is correct, the source is magnified by a factor of 4-7, and two counterimages are predicted. Our lens model predicts that there are time delays between the three images of 1-10 yr and that we have witnessed the final occurrence of the transient. The intense luminosity (M$_AB$rs̊ebox-0.5ex -23.5 after correcting for lensing) and blue UV continuum (implying T>rae̊box-0.5ex 50,000 K) argue that the source may have been a flare resulting from the tidal disruption of a star by a 10$^6$-10$^8$ M$_solar$ black hole. Regardless of its physical nature, PALS-1 highlights the importance of monitoring regions of high magnification in galaxy clusters for distant time-varying phenomena.
Based on observations with the NASA/ESA Hubble Space Telescope, obtained at the Space Telescope Science Institute, which is operated by AURA, Inc., under NASA contract NAS5-26555.
Based on observations obtained at the W. M. Keck Observatory, which is operated jointly by the California Institute of Technology and the University of California.
Publication
Astrophysical Journal | 2022-11-30 17:11:03 | {"extraction_info": {"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, "math_score": 0.6856718063354492, "perplexity": 2813.9506644969915}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710765.76/warc/CC-MAIN-20221130160457-20221130190457-00779.warc.gz"} |
https://electronics-club.com/rhombic-antenna-features-advantages/ | # Rhombic Antenna | features & advantages
## Rhombic Antenna
The rhombic antenna is based on the principle of travelling wave radiator. By application of return conductor, two wires are pulled apart at a point so that diamond or rhombus shape is formed as shown in the figure below. If the two wires (i.e., two sides of rhombus) are pulled apart to such an extent that the four lobes are combined together, then the additional gain is achieved. Also, there is no problem with proper termination to avoid the reflection. This is the basis of the rhombic antenna.
This is an antenna that is in the shape of a rhombus. It is usually terminated in a resistance. The side of the rhombus, the angle between the sides, the elevation, termination, and height above the earth are chosen to obtain the desired radiation characteristics. A typical Rhombic antenna and radiation pattern are shown in the figure below.
## Salient features of Rhombic antenna
1. It is a long wire antenna and consists of four non-resonant wires.
2. It provides greater directivity than the V antenna.
3. Its bandwidth is high.
4. It is an HF non-resonant antenna.
5. This is very useful for point-to-point communications.
6. It is a traveling wave antenna and there are no reflections.
7. It also finds wide applications where the angle of elevation of the main lobe (measured from the plane of the antenna to the radiation axis) is less than 30°.
8. At an elevation angle above 30°, the gain is very low for practical applications.
9. The length of equal radiators varies from 2 to 8 λ.
10. The tilt angle, ϕ varies between 40° and 75°.
11. ϕ is determined from leg length.
12. The terminating resistance is about 800 Ω.
13. The input impedance of the Rhombic antenna lies between 650 to 700Ω.
14. The directivity of the Rhombic antenna varies between 20 and 90.
15. The power gain lies between 15 and 60 after taking power loss in terminating resistance into account.
16. It is a very useful antenna for transmission and reception in the HF band.
17. It is easy and cheap to erect.
18. The directivity of each wire is $D\left ( \theta \right )=\frac{60I}{r}sin\theta \left [ \frac{sin\left [ \frac{\pi l}{\lambda }\left ( 1-cos\theta \right ) \right ]}{\left ( 1-cos\theta \right )} \right ]$
Where
I = the magnitude of the current in element
θ = the polar angle
λ = wavelength
r = the distance from the radiator to the elevation point.
## Design parameters & equations
The design parameters of the Rhombic antenna are:
1. Rhombic height, H
2. The angle of elevation, ϕ
3. Wire length, I.
The design equations are
$H=\frac{\lambda }{4sin \Delta}$
Δ = elevation angle
Δ is complement of tilt angle, ϕ
(sin ϕ = cos Δ)
$l=\frac{\lambda }{2 cos ^{2}\phi }=\frac{\lambda }{2 sin ^{2}\Delta }$
1. An end-to-end receiving array of a number of rhombics can be designed to form a Multiple Unit Steerable Antenna (MUSA) system which constitutes the present-day ultimate for long-distance short wave reception of horizontally polarized down coming waves.
2. The greatest advantage of the rhombic antenna is that the input impedance and radiation pattern do not change rapidly over a considerable frequency range as compared to any other system say of resonant dipoles.
3. Rhombic antenna is a highly directional broad-band antenna with the greatest radiated or received power along the main axis or longer diagonal.
4. It is a very efficient and widely employed antenna for radio communication where enough space necessary for its installation is no problem.
5. It is simple and cheap to erect.
6. Input impedance is twice to that obtainable from a single side radiator.
7. Short wave antennas of this kind required only a low height.
8. Rhombic antenna is un-tuned and is a useful wideband antenna suitable for rapid switching from one working frequency to another frequency. | 2021-02-28 22:40:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 3, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6267123222351074, "perplexity": 1416.2417992247897}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178361776.13/warc/CC-MAIN-20210228205741-20210228235741-00166.warc.gz"} |
https://sufficientlywise.org/2018/03/15/determinism-and-quantum-mechanics/ | Determinism and quantum mechanics
# Determinism and quantum mechanics
TL;DR – The Schrodinger equation can be seen as the deterministic and reversible limit of the projection (i.e. collapse) associated with a measurement.
Quantum mechanics comes with two ways to go from an initial to a final state. The first is the Schrodinger equation that represents time evolution and the second is the projection (or collapse) that represents what happens during measurements. These are usually presented as two separate entities. In fact, much of the work surrounding various “interpretations” is to reconcile these two types of evolution.
What we want to show here is that there is a very natural way to reconcile them. As we saw in a previous post, a quantum state always has a set of quantities that are well defined (i.e. it is always an eigenstate of some Hermitian operator). Therefore we can regard the Schrodinger equation as a special case of the projection where at each instant a measurement is made. Let’s see how this works.
1. The projection postulate
Let’s first review the projection postulate which supposedly describes what happens during measurements. The idea is that you start with a single well defined initial state $|\psi\rangle$. You pick an observable $A$, which is defined by a set of eigenstates $|\psi_{a_i}\rangle$ and the corresponding eigenvalues $a_i$. After the measurement you end up with what is called a mixed state: a statistical distribution. More precisely, a distribution over the eigenstates, each having probability $|\langle \psi_{a_i} | \psi \rangle | ^2$. So, for example, if we start with spin up $|s^+_z\rangle$ and we measure the horizontal direction, we end up with 50% $|s^+_x\rangle$ and 50% $|s^-_x\rangle$.
Note that we don’t always end up with a mixed state: if the initial state is already an eigenstate of the observable, nothing changes. That is, if we start with spin up $|s^+_z\rangle$ and we measure the vertical direction, we end up with 100% $|s^+_z\rangle$ and 0% $|s^-_z\rangle$. Which is the same as what we started with. So the projection is not always non-deterministic. In fact we can go continuously from a process that is deterministic, where the direction of measurement is the same as the prepared direction, to one that is completely non-deterministic, where the direction of measurement is perpendicular to the prepared direction.
2. Deterministic projections
So the idea is the following: can we perform a measurement such that we are not exactly measuring the same observable (e.g. the same direction of spin) but something so close that the projection is still deterministic (e.g. the direction of spin in an infinitesimally close direction)? That is, we want the final state $|\psi_{t+dt}\rangle$ to be very close to the initial state $|\psi_t\rangle$ so that $|\langle \psi_{t+dt}|\psi_t\rangle|^2=|\langle \psi_t|\psi_{t+dt}\rangle|^2=1$. We can rewrite $|\psi_{t+dt}\rangle = (1 + dt \frac{\partial}{\partial t}) |\psi_t\rangle$. We have:
\begin{align*}
|\langle \psi_t|\psi_{t+dt}\rangle|^2 &= 1 = |\langle \psi_t | (1 + dt \partial_t) |\psi_t\rangle|^2 \\
&= \langle \psi_t | (1 + dt \partial_t)^\dagger |\psi_t\rangle \langle \psi_t | (1 + dt \partial_t) |\psi_t\rangle \\
&= (\langle \psi_t | \psi_t\rangle + \langle \psi_t | dt \partial_t^\dagger |\psi_t\rangle ) (\langle \psi_t | \psi_t\rangle + \langle \psi_t | dt \partial_t |\psi_t\rangle ) \\
&= 1 + dt (\langle \psi_t | \partial_t^\dagger |\psi_t\rangle + \langle \psi_t | \partial_t |\psi_t\rangle) + dt^2 (|\langle \psi_t | \partial_t |\psi_t\rangle|^2) \\
&\approx 1 + dt \langle \psi_t | \partial_t^\dagger + \partial_t |\psi_t\rangle \\
\partial_t &= – \partial_t^\dagger
\end{align*}
We have that deterministic time evolution must be unitary and that its generator $\partial_t$ must be anti-Hermitian. This means we can find a corresponding Hermitian operator $H$ such that $\partial_t = \frac{H}{\imath \hbar}$. This gives us the Schrodinger equation.
\begin{align*}
\frac{\partial}{\partial t} | \psi_t \rangle = \frac{H}{\imath \hbar} |\psi_t\rangle
\end{align*}
To put this in more concrete term, consider spin precession in a magnetic field. We can say that the magnetic field continuously measures the spin of the particle, and this is what causes the motion. The direction where this measurement happens is a function of the particle itself, which makes sense because the motion is deterministic: it only depends on the state of the particle.
3. Conclusion
We have seen that the Schrodinger equation is a continuous process that at every moment is measuring something that is very close to what was measured before. This way there is no difference between measurement and time evolution: it is the same type of process. It also means that there is nothing special about a measurement: whether we cause it or not, is just the same process that happens all the time.
What this also means is that the deterministic and reversible process is the particular case, not the general one. But this is true anyway: it is only when we are able to isolate a system enough from the environment that we can assume that its evolution is independent from it.
Mathematically, we are associating continuous unitary evolution with deterministic and reversible processes while non-deterministic processes have no such restriction. This is very interesting because it is the same conclusion that we reach from a completely different premise in our work on the assumption of physics. | 2022-08-11 04:38:41 | {"extraction_info": {"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, "math_score": 0.920509934425354, "perplexity": 288.3062077446121}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571234.82/warc/CC-MAIN-20220811042804-20220811072804-00339.warc.gz"} |
https://physics.stackexchange.com/questions/419755/is-the-bose-einstein-condensation-a-single-particle-phenomenon?noredirect=1 | # Is the Bose-Einstein condensation a single particle phenomenon?
BEC occurs for noninteracting Bosons. Can we conclude that it can be described with a single particle? What is the significance of the number of the particles?
What @Árpád Szendrei said is correct. I will add some miscellaneous points.
1. BEC occurs for non-interacting bosons
BEC occurs for interacting bosons as well, and non-interacting BEC is actually a pathological example. It has an infinite compressibility. The speed of sound is zero, and any infinitesimal drag will create excitations. A weakly interacting BEC has a non-zero speed of sound, and acts like a superfluid. It IS possible to make a non-interacting BEC, by modifying the scattering length between atoms to zero, using external fields (see Feshbach resonance).
1. The "wavefunction" that people usually discuss ($\psi(r) = \sqrt{n(r)}e^{i\phi(r)}$) is technically not the actual many-body wavefunction, but an order parameter of the condensate. This "wavefunction" obeys a non-linear Schrodinger-like equation called the Gross-Pitaevskii equation.
2. What is the significance of the number of the particles?
It would help if the question is more precise, but usually a common question is whether the form of the order parameter mentioned above conserves the number of particles. The fact is, it doesn't, because it has a well-defined phase. It has a well-defined average of numbers, though. There is fluctuation in the number of particles, but (fluctuation)/(average) quickly goes to zero in the thermodynamic limit. [to find fluctuation in numbers, you need to look at the full Hamiltonian in second quantization form to get answers quick, so what I said is not really rigorous but just a sketch].
• About the point 3: What I meant is that since there is no interaction between particles how the large number of them make the phase transition to happen? – richard Jul 27 '18 at 19:52
• @richard, It all has to do with density of states. When the temperature is low enough, the thermal average number of atoms occupying the excited states will saturate, and any new atoms introduced to the system will start to occupy the ground state. The form of the density of states is set by external potential/boundary condition, and it dictates how the saturation of excited states occur. In a box potential, critical temperature increases with density. In a harmonic trap, critical temperature increases with number (not density!) – wcc Jul 27 '18 at 20:13
The Bose Einstein condensate is a QM effect of collective quantum state in which a macroscopic number of particles occupy the lowest energy state and thus is described by a single wavefunction.
All the bosons will be described by the same wavefunction.
So it is not a single particle, but all the particles (their probability distribution) are described by the same wavefunction.
BEC occurs for noninteracting Bosons. Can we conclude that it can be described with a single particle?
No, generally not. Most BEC systems in literature are described by a Hamiltonian $$H = \sum_{i=1}^N h({\bf r_i}) + \lambda_0\sum_{i < j}W({\vert \bf r_i - \bf r_j\vert})),$$ where $h({\bf r})$ is a single-particle Hamiltonian and $W({\bf r})$ is an interaction potential.
Now write $\phi_0({\bf r})$ for the ground state of $h({\bf r})$. For $\lambda_0=0$, the ground state of $H$ is $\Psi_0({\bf r}_1,\dots,{\bf r}_N)=\phi_0({\bf r}_1)\times\dots\times \phi_0({\bf r}_N)$. So $\Psi_0$ is just a single particle wave function, $\phi_0({\bf r})$, multiplied $N$-times.
But for most physical systems $\lambda_0\neq 0$. The many-body ground state $\Psi_0$ then no longer factorizes into a simple product as above. In fact, it can be very far away from that and consists of a superposition of $N$-boson states. So a single particle wave function no longer suffices to describe your system.
For that reason, BEC is nowadays usually defined in terms of the eigenvalues and eigenfunctions of the first order reduced density matrix, see Penrose and Onsager and the answer to this question. This criterion extends the concept of "${\mathcal O}(N)$ particles in the same state" to interacting systems. The book by Leggett is a more accessible reference on this topic than the original paper and also discusses the issues that arise in other attempts to define BEC. | 2019-11-11 22:30:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.8499842286109924, "perplexity": 301.26544127876696}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496664439.7/warc/CC-MAIN-20191111214811-20191112002811-00096.warc.gz"} |
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# Forum: Statistics
Statistics course and homework discussion. Elementary statistics.
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12-07-2012, 06:06 AM
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49. ### Help Choosing Tests
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• Views: 1,610
12-05-2012, 01:59 AM
53. ### Chance of making 50% of the profit
Ok, Just a basic question (numbers are random). If my profit is $150 and standard deviation is$20. What is the chance of making 50% of the expected...
• Replies: 0
• Views: 1,234
12-04-2012, 02:08 PM
54. ### two way analysis
I have data for 1970 and 1980 parity ratios, Parity 0 1 2 3 4 5 British 0.879 0.797 0.558 0.484 0.480 0.531...
• Replies: 2
• Views: 1,890
12-04-2012, 01:47 PM
55. ### Exogenous Regime Switching
I currently have a model that measures inflation expectations as follows: y(t) = a*x(t) + (1-a)*y(t-1) I'm looking to create a regime switching...
• Replies: 0
• Views: 1,058
12-04-2012, 09:30 AM
56. ### Data Hygience
I get government data from time to time, and some of the fields that I expect to only contain numerical data also contain text explaining why the...
• Replies: 6
• Views: 1,771
12-04-2012, 12:34 AM
Use this control to limit the display of threads to those newer than the specified time frame.
Allows you to choose the data by which the thread list will be sorted.
Note: when sorting by date, 'descending order' will show the newest results first. | 2017-10-23 22:48:58 | {"extraction_info": {"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, "math_score": 0.2668229937553406, "perplexity": 5833.948528505939}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187826840.85/warc/CC-MAIN-20171023221059-20171024001059-00754.warc.gz"} |
https://www.neetprep.com/question/58138-observer-moves-towards-stationary-source-sound-speed-th-thespeed-sound-wavelength-frequency-source-emitted--f-respectively-apparent-frequency-wavelength-recorded-observer-arerespectively-f-f-f-f/126-Physics--Waves/690-Waves | An observer moves towards a stationary source of sound with a speed 1/5th of the speed of sound. The wavelength and frequency of the source emitted are λ and f respectively. The apparent frequency and wavelength recorded by the observer are respectively
(1) $1.2\text{\hspace{0.17em}}f,\text{\hspace{0.17em}}\lambda$
(2) $f,\text{\hspace{0.17em}}1.2\lambda$
(3) $0.8f,\text{\hspace{0.17em}}0.8\lambda$
(4) $1.2f,\text{\hspace{0.17em}}1.2\lambda$
Explanation is a part of a Paid Course. To view Explanation Please buy the course.
Difficulty Level: | 2019-10-18 13:28:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5890731811523438, "perplexity": 550.6146068530697}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986682998.59/warc/CC-MAIN-20191018131050-20191018154550-00038.warc.gz"} |
https://web2.0calc.com/questions/please-help-asap_43 | +0
0
305
3
1. Consider the infinite arithmetic sequence $A$ with first term $5$ and common difference $-2$. Now define the infinite sequence $B$ so that the $k^{th}$ term of $B$ is $2$ raised to the $k^{th}$ term of $A$. Find the sum of all of the terms of $B$.
2. The terms $140, a, \frac{45}{28}$ are the first, second and third terms, respectively, of a geometric sequence. If $a$ is positive, what is the value of $a$?
Guest May 8, 2018
#1
+27131
+1
"1. Consider the infinite arithmetic sequence $A$ with first term $5$ and common difference $-2$. Now define the infinite sequence $B$ so that the $k^{th}$ term of $B$ is $2$ raised to the $k^{th}$ term of $A$. Find the sum of all of the terms of $B$."
.
Alan May 8, 2018
#2
+27131
+1
"2. The terms $140, a, \frac{45}{28}$ are the first, second and third terms, respectively, of a geometric sequence. If $a$ is positive, what is the value of $a$?"
Alan May 8, 2018
#3
+20153
0
2. The terms $140, a, \frac{45}{28}$ are the first, second and third terms, respectively, of a geometric sequence. If $a$ is positive, what is the value of $a$?
$$\begin{array}{|rcll|} \hline a &=& \sqrt{140\cdot \frac{45}{28}} \\ a &=& 15 \\ \hline \end{array}$$
heureka May 8, 2018 | 2018-11-18 00:30:14 | {"extraction_info": {"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, "math_score": 0.8451924324035645, "perplexity": 87.1031390832237}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039743913.6/warc/CC-MAIN-20181117230600-20181118012600-00347.warc.gz"} |
https://jp.maplesoft.com/support/help/errors/view.aspx?path=GraphTheory%2FSpecialGraphs%2FGeneralizedHexagonGraph | GeneralizedHexagonGraph - Maple Help
# Online Help
###### All Products Maple MapleSim
GraphTheory[SpecialGraphs]
GeneralizedHexagonGraph
construct generalized hexagon graph
Calling Sequence GeneralizedHexagonGraph()
Description
• The GeneralizedHexagonGraph() command returns the generalized hexagon graph.
Examples
> $\mathrm{with}\left(\mathrm{GraphTheory}\right):$
> $\mathrm{with}\left(\mathrm{SpecialGraphs}\right):$
> $G≔\mathrm{GeneralizedHexagonGraph}\left(\right)$
${G}{≔}{\mathrm{Graph 1: an undirected unweighted graph with 126 vertices and 189 edge\left(s\right)}}$ (1)
> $\mathrm{IsPlanar}\left(G\right)$
${\mathrm{false}}$ (2)
> $\mathrm{DrawGraph}\left(G,\mathrm{style}=\mathrm{spring}\right)$
See Also | 2022-09-29 08:34:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 7, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9988652467727661, "perplexity": 13865.067782661567}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335326.48/warc/CC-MAIN-20220929065206-20220929095206-00372.warc.gz"} |
https://cgit.freedesktop.org/libspectre/tree/libspectre/spectre-document.h?id=a6329a26cbc8a920cf56531a762bfbf62b3ddc9d | summaryrefslogtreecommitdiff log msg author committer range
blob: dc10ac8398c68b97eac47397dc0c47caaff5483c (plain)
```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 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 ``` ``````/* This file is part of Libspectre. * * Copyright (C) 2007 Albert Astals Cid * Copyright (C) 2007 Carlos Garcia Campos * * Libspectre is free software; you can redistribute it and/or modify * it under the terms of the GNU General Public License as published by * the Free Software Foundation; either version 2, or (at your option) * any later version. * * Libspectre is distributed in the hope that it will be useful, * but WITHOUT ANY WARRANTY; without even the implied warranty of * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the * GNU General Public License for more details. * * You should have received a copy of the GNU General Public License * along with this program; if not, write to the Free Software * Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA. */ #ifndef SPECTRE_DOCUMENT_H #define SPECTRE_DOCUMENT_H #include #include #include #include SPECTRE_BEGIN_DECLS /*! This is the object that represents a PostScript document. */ typedef struct SpectreDocument SpectreDocument; /*! Creates a document */ SPECTRE_PUBLIC SpectreDocument *spectre_document_new (void); /*! Loads a the given file into the document. This function can fail @param document the document where the file will be loaded @param filename the file to loa @see spectre_document_status */ SPECTRE_PUBLIC void spectre_document_load (SpectreDocument *document, const char *filename); /*! Loads the given open file into the document. This function can fail @param document the document where the file will be loaded @param file the file to load @see spectre_document_status */ SPECTRE_PUBLIC void spectre_document_load_from_stream (SpectreDocument *document, FILE *file); /*! Returns the document status @param document the document whose status will be returned */ SPECTRE_PUBLIC SpectreStatus spectre_document_status (SpectreDocument *document); /*! Frees the document @param document the document that will be freed */ SPECTRE_PUBLIC void spectre_document_free (SpectreDocument *document); /*! Returns the number of pages of the document. This function can fail @param document the document whose pages number will be returned @see spectre_document_status */ SPECTRE_PUBLIC unsigned int spectre_document_get_n_pages (SpectreDocument *document); /*! Returns the orientation of the document. This function can fail @param document the document whose orientation will be returned @see spectre_document_status */ SPECTRE_PUBLIC SpectreOrientation spectre_document_get_orientation (SpectreDocument *document); /*! Returns the title of the document. It returns a null const char * if the document specifies no title. This function can fail @param document the document whose title will be returned @see spectre_document_status */ SPECTRE_PUBLIC const char *spectre_document_get_title (SpectreDocument *document); /*! Returns the creator of the document. It returns a null const char * if the document specifies no creator. This function can fail @param document the document whose creator will be returned @see spectre_document_status */ SPECTRE_PUBLIC const char *spectre_document_get_creator (SpectreDocument *document); /*! Returns the for of the document. It returns a null const char * if the document specifies no for. This function can fail @param document the document whose for will be returned @see spectre_document_status */ SPECTRE_PUBLIC const char *spectre_document_get_for (SpectreDocument *document); /*! Returns the creation date of the document. The date is copied verbatim from the document, so no format can be assumed on it. It returns a null const char * if the document specifies no creation date. This function can fail @param document the document whose creation date will be returned @see spectre_document_status */ SPECTRE_PUBLIC const char *spectre_document_get_creation_date (SpectreDocument *document); /*! Returns the format of the document. This function can fail @param document the document whose format will be returned @see spectre_document_status */ SPECTRE_PUBLIC const char *spectre_document_get_format (SpectreDocument *document); /*! Returns if the document is a Encapsulated PostScript file. This function can fail @param document the document to query @see spectre_document_status */ SPECTRE_PUBLIC int spectre_document_is_eps (SpectreDocument *document); /*! Returns the PostScript language level of the document. It returns 0 if no language level was defined on the file. This function can fail @param document the document whose language level will be returned @see spectre_document_status */ SPECTRE_PUBLIC unsigned int spectre_document_get_language_level (SpectreDocument *document); /*! Returns a page of the document. This function can fail @param document the document whose page will be returned @param page_index the page index to get. First page has index 0. @see spectre_document_status */ SPECTRE_PUBLIC SpectrePage *spectre_document_get_page (SpectreDocument *document, unsigned int page_index); /*! Returns a page of the document referenced by label. This function can fail @param document the document whose page will be returned @param label the label of the page to get. @see spectre_document_status */ SPECTRE_PUBLIC SpectrePage *spectre_document_get_page_by_label (SpectreDocument *document, const char *label); /*! Convenient function for rendering documents with no pages, tipically eps. When used with multi-page documents the first page will be rendered. @param document the document to render @rc the rendering context specifying how the document has to be rendered @width the page width will be returned here, or NULL @height the page height will be returned here, or NULL @page_data a pointer that will point to the image data @row_length the length of an image row will be returned here @see spectre_document_render_full */ SPECTRE_PUBLIC void spectre_document_render_full (SpectreDocument *document, SpectreRenderContext *rc, unsigned char **page_data, int *row_length); /*! Convenient function for rendering documents with no pages, tipically eps. Document will be rendered with the default options, use spectre_document_render_full if you want to change any of them. When used with multi-page documents the first page will be rendered. @param document the document to render @width the page width will be returned here, or NULL @height the page height will be returned here, or NULL @page_data a pointer that will point to the image data @row_length the length of an image row will be returned here @see spectre_document_render_full */ SPECTRE_PUBLIC void spectre_document_render (SpectreDocument *document, unsigned char **page_data, int *row_length); /* Convenient function for getting the page size of documents with no pages, tipically eps. When used with multi-page documents the size of the first page will be returned. @param document the document whose page will be returned @width the page width will be returned here, or NULL @height the page height will be returned here, or NULL */ SPECTRE_PUBLIC void spectre_document_get_page_size (SpectreDocument *document, int *width, int *height); /*! Save document as filename. This function can fail @param document the document that will be saved @param filename the path where document will be saved @see spectre_document_status */ SPECTRE_PUBLIC void spectre_document_save (SpectreDocument *document, const char *filename); /* Save document as a pdf document. This function can fail @param document the document that will be saved @param filename the path where document will be saved as pdf @see spectre_document_status */ SPECTRE_PUBLIC void spectre_document_save_to_pdf (SpectreDocument *document, const char *filename); SPECTRE_END_DECLS #endif /* SPECTRE_DOCUMENT_H */ `````` | 2020-07-06 03:34:47 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9313659071922302, "perplexity": 12950.853566555543}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655890092.28/warc/CC-MAIN-20200706011013-20200706041013-00106.warc.gz"} |
https://socratic.org/questions/how-do-you-write-the-equation-3x-6y-24-in-slope-intercept-form | # How do you write the equation -3x-6y =-24 in slope-intercept form?
Dec 13, 2016
$y = - \frac{1}{2} x + 4$
#### Explanation:
To transform this equation to slope-intercept form we must solve for $y$ while keeping the equation balanced:
$- 3 x + 3 x - 6 y = - 24 + 3 x$
$0 - 6 y = 3 x - 24$
$- 6 y = 3 x - 24$
$\frac{- 6 y}{-} 6 = \frac{3 x - 24}{- 6}$
$\frac{\cancel{- 6} y}{\cancel{- 6}} = \frac{3 x}{- 6} - \frac{24}{- 6}$
$y = - \frac{1}{2} x + 4$ | 2020-05-28 05:43:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "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, "math_score": 0.8429003357887268, "perplexity": 2889.690709543674}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347396495.25/warc/CC-MAIN-20200528030851-20200528060851-00495.warc.gz"} |
https://en.m.wikipedia.org/wiki/Earth%27s_internal_heat_budget | # Earth's internal heat budget
Global map of the flux of heat, in mW/m2, from Earth's interior to the surface.[1] The largest values of heat flux coincide with mid ocean ridges, and the smallest values of heat flux occur in stable continental interiors.
Earth's internal heat budget is fundamental to the thermal history of the Earth. The flow of heat from Earth's interior to the surface is estimated at ${\displaystyle 47\pm 2}$ terawatts (TW)[1] and comes from two main sources in roughly equal amounts: the radiogenic heat produced by the radioactive decay of isotopes in the mantle and crust, and the primordial heat left over from the formation of the Earth.[2]
Earth's internal heat powers most geological processes[3] and drives plate tectonics.[2] Despite its geological significance, this heat energy coming from Earth's interior is actually only 0.03% of Earth's total energy budget at the surface, which is dominated by 173,000 TW of incoming solar radiation.[4] The insolation that eventually, after reflection, reaches the surface penetrates only several tens of centimeters on the daily cycle and only several tens of meters on the annual cycle. This renders solar radiation minimally relevant for internal processes.[5]
## Heat and early estimate of Earth's age
Based on calculations of Earth's cooling rate, which assumed constant conductivity in the Earth's interior, in 1862 William Thomson (later made Lord Kelvin) estimated the age of the Earth at 98 million years,[6] which contrasts with the age of 4.5 billion years obtained in the 20th century by radiometric dating.[7] As pointed out by John Perry in 1895[8] a variable conductivity in the Earth's interior could expand the computed age of the Earth to billions of years, as later confirmed by radiometric dating. Contrary to the usual representation of Kelvin's argument, the observed thermal gradient of the Earth's crust would not be explained by the addition of radioactivity as a heat source. More significantly, mantle convection alters how heat is transported within the Earth, invalidating Kelvin's assumption of purely conductive cooling.
## Global internal heat flow
Cross section of the Earth showing its main divisions and their approximate contributions to Earth's total internal heat flow to the surface, and the dominant heat transport mechanisms within the Earth.
Estimates of the total heat flow from Earth’s interior to surface span a range of 43 to 49 terawatts (TW) (a terawatt is 1012 watts).[9] One recent estimate is 47 TW,[1] equivalent to an average heat flux of 91.6 mW/m2, and is based on more than 38,000 measurements. The respective mean heat flows of continental and oceanic crust are 70.9 and 105.4 mW/m2.[1]
While the total internal Earth heat flow to the surface is well constrained, the relative contribution of the two main sources of Earth's heat, radiogenic and primordial heat, are highly uncertain because their direct measurement is difficult. Chemical and physical models give estimated ranges of 15–41 TW and 12–30 TW for radiogenic heat and primordial heat, respectively.[9]
The structure of the Earth is a rigid outer crust that is composed of thicker continental crust and thinner oceanic crust, solid but plastically flowing mantle, a liquid outer core, and a solid inner core. The fluidity of a material is proportional to temperature; thus, the solid mantle can still flow on long time scales, as a function of its temperature[2] and therefore as a function of the flow of Earth's internal heat. The mantle convects in response to heat escaping from Earth's interior, with hotter and more buoyant mantle rising and cooler, and therefore denser, mantle sinking. This convective flow of the mantle drives the movement of Earth's lithospheric plates; thus, an additional reservoir of heat in the lower mantle is critical for the operation of plate tectonics and one possible source is an enrichment of radioactive elements in the lower mantle.[10]
Earth heat transport occurs by conduction, mantle convection, hydrothermal convection, and volcanic advection.[11] Earth's internal heat flow to the surface is thought to be 80% due to mantle convection, with the remaining heat mostly originating in the Earth's crust,[12] with about 1% due to volcanic activity, earthquakes, and mountain building.[2] Thus, about 99% of Earth's internal heat loss at the surface is by conduction through the crust, and mantle convection is the dominant control on heat transport from deep within the Earth. Most of the heat flow from the thicker continental crust is attributed to internal radiogenic sources, in contrast the thinner oceanic crust has only 2% internal radiogenic heat.[2] The remaining heat flow at the surface would be due to basal heating of the crust from mantle convection. Heat fluxes are negatively correlated with rock age,[1] with the highest heat fluxes from the youngest rock at mid-ocean ridge spreading centers (zones of mantle upwelling), as observed in the global map of Earth heat flow.[1]
The evolution of Earth's radiogenic heat flow over time.
An estimate of the present-day major heat-producing isotopes[2]
Isotope Heat release
W/kg isotope
Half-life
years
Mean mantle concentration
kg isotope/kg mantle
Heat release
W/kg mantle
238U 94.6×10−6 4.47×109 30.8×10−9 2.91×10−12
235U 569×10−6 0.704×109 0.22×10−9 0.125×10−12
232Th 26.4×10−6 14.0×109 124×10−9 3.27×10−12
40K 29.2×10−6 1.25×109 36.9×10−9 1.08×10−12
Geoneutrino detectors can detect the decay of 238U and 232Th and thus allow estimation of their contribution to the present radiogenic heat budget, while 235U and 40K is not detectable. Regardless, 40K is estimated to contribute 4 TW of heating.[17] However, due to the short half-lives the decay of 235U and 40K contributed a large fraction of radiogenic heat flux to the early Earth, which was also much hotter than at present.[10] Initial results from measuring the geoneutrino products of radioactive decay from within the Earth, a proxy for radiogenic heat, yielded a new estimate of half of the total Earth internal heat source being radiogenic,[17] and this is consistent with previous estimates.[16]
## Primordial heat
Primordial heat is the heat lost by the Earth as it continues to cool from its original formation, and this is in contrast to its still actively-produced radiogenic heat. The Earth core's heat flow—heat leaving the core and flowing into the overlying mantle—is thought to be due to primordial heat, and is estimated at 5–15 TW.[18] Estimates of mantle primordial heat loss range between 7 and 15 TW, which is calculated as the remainder of heat after removal of core heat flow and bulk-Earth radiogenic heat production from the observed surface heat flow.[9]
The early formation of the Earth's dense core could have caused superheating and rapid heat loss, and the heat loss rate would slow once the mantle solidified.[18] Heat flow from the core is necessary for maintaining the convecting outer core and the geodynamo and Earth's magnetic field, therefore primordial heat from the core enabled Earth's atmosphere and thus helped retain Earth's liquid water.[16]
## Heat flow and plate tectonics
Earth's tectonic evolution over time from a molten state at 4.5 Ga,[7] to a single-plate lithosphere,[19] to modern plate tectonics sometime between 3.2 Ga[20] and 1.0 Ga.[21]
Controversy over the exact nature of mantle convection makes the linked evolution of Earth's heat budget and the dynamics and structure of the mantle difficult to unravel.[16] There is evidence that the processes of plate tectonics were not active in the Earth before 3.2 billion years ago, and that early Earth's internal heat loss could have been dominated by advection via heat-pipe volcanism.[19] Terrestrial bodies with lower heat flows, such as the Moon and Mars, conduct their internal heat through a single lithospheric plate, and higher heat flows, such as on Jupiter's moon Io, result in advective heat transport via enhanced volcanism, while the active plate tectonics of Earth occur with an intermediate heat flow and a convecting mantle.[19]
## References
1. Davies, J. H., & Davies, D. R. (2010). Earth's surface heat flux. Solid Earth, 1(1), 5–24.
2. Donald L. Turcotte; Gerald Schubert (25 March 2002). Geodynamics. Cambridge University Press. ISBN 978-0-521-66624-4.
3. ^ Buffett, B. A. (2007). Taking earth's temperature. Science, 315(5820), 1801–1802.
4. ^ Archer, D. (2012). Global Warming: Understanding the Forecast. ISBN 978-0-470-94341-0.
5. ^ Lowrie, W. (2007). Fundamentals of geophysics. Cambridge: CUP, 2nd ed.
6. ^ Thomson, William. (1864). On the secular cooling of the earth, read 28 April 1862. Transactions of the Royal Society of Edinburgh, 23, 157–170.
7. ^ a b Ross Taylor, Stuart (26 October 2007). "Chapter 2: The Formation Of The Earth And Moon". In Martin J. van Kranendonk; Vickie Bennett; Hugh R.H. Smithies. Earth's Oldest Rocks (Developments in Precambrian Geology Vol 15, 2007). Elsevier. pp. 21–30. ISBN 978-0-08-055247-7.
8. ^ England, Philip; Molnar, Peter; Richter, Frank (2007). "John Perry's neglected critique of Kelvin's age for the Earth: A missed opportunity in geodynamics" (PDF). GSA Today. 17 (1): 4–9. doi:10.1130/GSAT01701A.1.
9. ^ a b c Dye, S. T. (2012). Geoneutrinos and the radioactive power of the Earth. Reviews of Geophysics, 50(3). DOI: 10.1029/2012RG000400
10. ^ a b Arevalo Jr, R., McDonough, W. F., & Luong, M. (2009). The K/U ratio of the silicate Earth: Insights into mantle composition, structure and thermal evolution. Earth and Planetary Science Letters, 278(3), 361–369.
11. ^ Jaupart, C., & Mareschal, J. C. (2007). Heat flow and thermal structure of the lithosphere. Treatise on Geophysics, 6, 217–251.
12. ^ Korenaga, J. (2003). Energetics of mantle convection and the fate of fossil heat. Geophysical Research Letters, 30(8), 1437.
13. ^ a b Korenaga, J. (2011). Earth's heat budget: Clairvoyant geoneutrinos. Nature Geoscience, 4(9), 581–582.
14. ^ "Geophysical and geochemical constraints on geoneutrino fluxes from Earth's mantle". Earth and Planetary Science Letters. 361: 356–366. 2013-01-01. arXiv:1207.0853. Bibcode:2013E&PSL.361..356S. doi:10.1016/j.epsl.2012.11.001. ISSN 0012-821X.
15. ^ McDonough, W.F. (2003), "Compositional Model for the Earth's Core", Treatise on Geochemistry, Elsevier, pp. 547–568, Bibcode:2003TrGeo...2..547M, doi:10.1016/b0-08-043751-6/02015-6, ISBN 9780080437514, retrieved 2018-08-03
16. ^ a b c d Korenaga, J. (2008). Urey ratio and the structure and evolution of Earth's mantle. Reviews of Geophysics, 46(2).
17. ^ a b Gando, A., Dwyer, D. A., McKeown, R. D., & Zhang, C. (2011). Partial radiogenic heat model for Earth revealed by geoneutrino measurements. Nature Geoscience, 4(9), 647–651.
18. ^ a b Lay, T., Hernlund, J., & Buffett, B. A. (2008). Core–mantle boundary heat flow. Nature Geoscience, 1(1), 25-32.
19. ^ a b c Moore, W. B., & Webb, A. A. G. (2013). Heat-pipe Earth. Nature, 501(7468), 501–505.
20. ^ Pease, V., Percival, J., Smithies, H., Stevens, G., & Van Kranendonk, M. (2008). When did plate tectonics begin? Evidence from the orogenic record. When did plate tectonics begin on planet Earth, 199–208.
21. ^ Stern, R. J. (2008). Modern-style plate tectonics began in Neoproterozoic time: An alternative interpretation of Earth’s tectonic history. When did plate tectonics begin on planet Earth, 265–280. | 2018-12-17 08:19:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6614870429039001, "perplexity": 4156.948344318097}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376828448.76/warc/CC-MAIN-20181217065106-20181217091106-00037.warc.gz"} |
https://planetmath.org/StableSubspace | # stable subspace
A subset $S$ of a larger set $T$ is said to a stable subset for a function $f:T\to T$ iff $f(S)\subset S$.
Alternative phrasings with the same meaning are:
• $f$ is an invariant subset for $f$
• $f$ stabilizes $S$
• $S$ is stable under (the action of) $f$
• $S$ is invariant under (the action of) $f$
• $S$ is left stable by/under $f$
• $S$ is left invariant by/under $f$
Title stable subspace StableSubspace 2013-03-22 17:56:52 2013-03-22 17:56:52 lalberti (18937) lalberti (18937) 6 lalberti (18937) Definition msc 00A05 invariant subspace stable subset invariant subset | 2020-04-06 02:59:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 16, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9376821517944336, "perplexity": 5104.35590347007}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371612531.68/warc/CC-MAIN-20200406004220-20200406034720-00362.warc.gz"} |
https://joy.guhas.org/blog/2015/12/02/talos-canvasmark.html | ### Introduction
The Canvasmark suite of performance tests designed to test the “HTML5 canvas rendering performance for commonly used operations in HTML5 games: bitmaps, canvas drawing, alpha blending, polygon fills, shadows and text functions.” This test consists of 5 tests,
• 3D Rendering - Maths- polygons- image transforms
• Arena5 - Vectors- shadows- bitmaps- text
• Asteroids - Bitmaps
• Asteroids - Bitmaps- shapes- text
• Asteroids - Shapes- shadows- blending
• Asteroids - Vectors
• Pixel blur - Math- getImageData- putImageData
• Plasma - Maths- canvas shapes
The tests are recorded as weighted times, the weights a measure of the complexity of the test. The final score is the sum of these weighted times.
See this page(behind LDAP because of Shiny, sorry) with plots of the subtests across e10s options and platforms (ignoring non-pgo). The data represents last 90 days (when we have it). All the replicate data is plotted and the black line is a loess regression . The band around is a prediction band.
1. The subtests do measure different things as can be observed from this scatter plot. This is reassuring - we don’t want duplicate tests.
2. However the scales are widely different ranging from 300-400 for “Asteroids - Bitmaps” (OSX10,non e10s) to 1500 - 1700 for the “Pixel Blur …” test (because of the weighting). However, not only are the means vastly different, the variation is very different for each. Across operating systems both the means and standard errors vary a lot. This figure is a Box Plot of the test values for different platforms. We see osx-10-10 has a lot of spread whereas the others are much less.
3. From the Canvasmark website, the score is arrived at by adding these figures. If Y is the sum of different values then the variance of Y is the sum of the variances of the summands. We are unnecessarily creating a score with high variance.
4. Because of (2), % changes in one summand is not equivalent to the same % change in Y e.g. a 10% change in the smallest e.g. “Asteroids - Bitmaps” will translate to a much small change in the sum (typically < 1%).
5. On average, the subtests contribute the following to the score. A breakup by platform can be seen here
test contributionpct
Asteroids - Bitmaps 6.845
Asteroids - Shapes- shadows- blending 9.190
3D Rendering - Maths- polygons- image transforms 10.549
Asteroids - Bitmaps- shapes- text 11.541
Asteroids - Vectors 13.391
Arena5 - Vectors- shadows- bitmaps- text 13.419
Plasma - Maths- canvas shapes 14.544
Pixel blur - Math- getImageData- putImageData 20.522
6. This link is a time series of the canvasmark score across pushes. The smooth band is from the standard errors of the loess curve. The smallest change we can detect is
### Summary
1. The filtering scheme for canvasmark is to drop first and take the median of the remaining. However on inspection, there is nothing remarkable about the first (for this test). To get more data, we can easily use all the replicates (5) to create 5 canvasmark scores per job rather than one.
2. Because of the very different means, canvasmark really won’t measure changes in “Asteroids - Bitmaps”. For example, in the last 7 days of builds preceding 2015-11-30, the mean and standard deviation of canvasmark is below. Given this , the smallest change you can detect using a t-test and 12 observations each in the before and after groups at the 1% level is ~ $qt(1-0.05/2,df=24-2) * sd * \sqrt{\frac{2}{12}}$ or between 187 and 272. ~
platform Mean StdDev
linux64 6736.25 161.54
windows7-32 8410.36 236.08
windowsxp 8411.45 187.93
windows8-64 8700.39 214.29
3. There is a distinct day of week effect for OS X which manifests itself in in creased variance. See this figure. This indicates a QA problem that depends on day of week, which it ought not to or should be taken care of. | 2019-03-21 08:21:57 | {"extraction_info": {"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, "math_score": 0.5980463624000549, "perplexity": 3619.710208725564}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202506.45/warc/CC-MAIN-20190321072128-20190321094128-00244.warc.gz"} |
http://www.koreascience.or.kr/article/JAKO200734515871980.page | # COMMON FIXED POINT FOR MULTIVALUED MAPPINGS IN INTUITIONISTIC FUZZY METRIC SPACES
• Sharma, Sushil (DEPARTMENT OF MATHEMATICS MADHAV VIGYAN MAHAVIDHYALAYA VIKRAM UNIVERSITY) ;
• Kutukcu, Servet (DEPARTMENT OF MATHEMATICS FACULTY OF SCIENCE AND ARTS ONDOKUZ MAYIS UNIVERSITY) ;
• Rathore, R.S. (DEPARTMENT OF MATHEMATICS GOVT. GIRLS P.G. COLLEGE)
• Published : 2007.07.31
• 137 24
#### Abstract
The purpose of this paper is to obtain some common fixed point theorems for multivalued mappings in intuitionistic fuzzy metric space. We extend some earlier results.
#### Keywords
common fixed point;multivalued map;intuitionistic fuzzy metric space
#### References
1. C. Alaca, D. Turkoglu, and C. Yildiz, Fixed point in intuitionistic fuzzy metric spaces, Chaos, Solitons and Fractals 29 (2006), 1073-1078 https://doi.org/10.1016/j.chaos.2005.08.066
2. K. Atanassov, Intuitionistic fuzzy sets, Fuzzy Sets and Systems 20 (1986), 87-96 https://doi.org/10.1016/S0165-0114(86)80034-3
3. A. George and P. Veeramani, On some results in fuzzy metric spaces, Fuzzy sets and Systems 64 (1994), 395-399 https://doi.org/10.1016/0165-0114(94)90162-7
4. G. Jungck and B. E. Rhoades, Fixed point for set valued functions without continuity, Ind. J. Pure & Appl, Math. 29 (1998), no. 3, 227-238
5. I. Kubiaczyk and S. Sharma, Common fixed point in fuzzy metric space, J. Fuzzy Math. (to appear)
6. S. Kutukcu, A common fixed point theorem for a sequence of self maps in intuitionistic fuzzy metric spaces, Commun. Korean Math. Soc. 21 (2006), no. 4, 679-687 https://doi.org/10.4134/CKMS.2006.21.4.679
7. J. H. Park, Intuitionistic fuzzy metric spaces, Chaos, Solitons and Fractals 22 (2004), 1039-1046 https://doi.org/10.1016/j.chaos.2004.02.051
8. B. Schweizer and A. Sklar, Statistical spaces, Pacific J. Math. 10 (1960), 313-334 https://doi.org/10.2140/pjm.1960.10.313
9. S. Sharma, On fuzzy metric space, Southeast Asian Bull. Math. 6 (2002), no. 1, 145-157 https://doi.org/10.1007/s100120200034
10. D. Turkoglu, C. Alaca, and C. Yildiz, Compatible maps and compatible maps of type ($\alpha$) and ($\beta$) in intuitionistic fuzzy metric spaces, Demonstratio Math. 39 (2006), no. 3, 671-684
11. D. Turkoglu, C. Alaca, Y. J. Cho, and C. Yildiz, Common fixed point theorems in intuitionistic fuzzy metric spaces, J. Appl. Math. & Computing 22 (2006), no. 1-2, 411-424 https://doi.org/10.1007/BF02896489
12. L. A. Zadeh, Fuzzy sets, Inform. and Control 8 (1965), 338-353 https://doi.org/10.1016/S0019-9958(65)90241-X
#### Cited by
1. COMMON FIXED POINT THEOREM FOR MULTIMAPS ON MENGER L-FUZZY METRIC SPACE vol.20, pp.1, 2013, https://doi.org/10.7468/jksmeb.2013.20.1.11 | 2020-02-19 06:34:30 | {"extraction_info": {"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, "math_score": 0.8650778532028198, "perplexity": 4542.436273807578}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875144058.43/warc/CC-MAIN-20200219061325-20200219091325-00536.warc.gz"} |