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https://jp.maplesoft.com/support/help/maplesim/view.aspx?path=DifferentialGeometry/LieAlgebras/Query/RegularElement | 1,642,393,381,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300289.37/warc/CC-MAIN-20220117031001-20220117061001-00717.warc.gz | 401,191,167 | 37,844 | DifferentialGeometry/LieAlgebras/Query/RegularElement - Maple Help
Query[RegularElement] - check if an element of a Lie algebra is regular
Calling Sequences
Query()
Parameters
X - a vector in a Lie algebra
options - the keywords arguments rank = $m,$ algebratype = "semisimple"
Description
• Let g be a Lie algebra. For each , let This is the generalized null space of ). The rank of g is defined as rank{dimfor }. An element is called regular if dimrankIf is a regular element, then the centralizer is a Cartan subalgebra. Conversely, if $Z\left(x\right)$ is a Cartan subalgebra, then is a regular element.
• Alternatively, for each , set Then the rank of g is the smallest integer (as varies) such that the coefficient of in ${p}_{x}\left({\mathrm{\lambda }}_{}\right)$ is nonzero.
• With the calling sequence Query(), the centralizer $Z\left(X\right)$ of $X$ is computed. It is then determined if this centralizer is a Cartan subalgebra.
• With the calling sequence Query(), the calculations are simplified using the fact that the Cartan subalgebra must be Abelian (in general, it need only be nilpotent).
• With the calling sequence Query( rank = ${\mathrm{m}}{,}$"), the regularity of is determined by calculating the generalized null space of This is the fastest method for checking if an element is regular, assuming that the rank of is known.
Examples
> $\mathrm{with}\left(\mathrm{DifferentialGeometry}\right):$$\mathrm{with}\left(\mathrm{LieAlgebras}\right):$
Example 1.
We check for regular elements in the Lie algebra First use the command SimpleLieAlgebraData to obtain the structure equations for $\mathrm{so}\left(5\right)$.
> $\mathrm{LD}≔\mathrm{SimpleLieAlgebraData}\left("so\left(5\right)",\mathrm{so5}\right)$
${\mathrm{LD}}{:=}\left[\left[{\mathrm{e1}}{,}{\mathrm{e2}}\right]{=}{\mathrm{e5}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e3}}\right]{=}{\mathrm{e6}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e7}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e5}}\right]{=}{-}{\mathrm{e2}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e6}}\right]{=}{-}{\mathrm{e3}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e7}}\right]{=}{-}{\mathrm{e4}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e3}}\right]{=}{\mathrm{e8}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e9}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e5}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e8}}\right]{=}{-}{\mathrm{e3}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e9}}\right]{=}{-}{\mathrm{e4}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e10}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e6}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e8}}\right]{=}{\mathrm{e2}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e10}}\right]{=}{-}{\mathrm{e4}}{,}\left[{\mathrm{e4}}{,}{\mathrm{e7}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e4}}{,}{\mathrm{e9}}\right]{=}{\mathrm{e2}}{,}\left[{\mathrm{e4}}{,}{\mathrm{e10}}\right]{=}{\mathrm{e3}}{,}\left[{\mathrm{e5}}{,}{\mathrm{e6}}\right]{=}{\mathrm{e8}}{,}\left[{\mathrm{e5}}{,}{\mathrm{e7}}\right]{=}{\mathrm{e9}}{,}\left[{\mathrm{e5}}{,}{\mathrm{e8}}\right]{=}{-}{\mathrm{e6}}{,}\left[{\mathrm{e5}}{,}{\mathrm{e9}}\right]{=}{-}{\mathrm{e7}}{,}\left[{\mathrm{e6}}{,}{\mathrm{e7}}\right]{=}{\mathrm{e10}}{,}\left[{\mathrm{e6}}{,}{\mathrm{e8}}\right]{=}{\mathrm{e5}}{,}\left[{\mathrm{e6}}{,}{\mathrm{e10}}\right]{=}{-}{\mathrm{e7}}{,}\left[{\mathrm{e7}}{,}{\mathrm{e9}}\right]{=}{\mathrm{e5}}{,}\left[{\mathrm{e7}}{,}{\mathrm{e10}}\right]{=}{\mathrm{e6}}{,}\left[{\mathrm{e8}}{,}{\mathrm{e9}}\right]{=}{\mathrm{e10}}{,}\left[{\mathrm{e8}}{,}{\mathrm{e10}}\right]{=}{-}{\mathrm{e9}}{,}\left[{\mathrm{e9}}{,}{\mathrm{e10}}\right]{=}{\mathrm{e8}}\right]$ (2.1)
> $\mathrm{DGsetup}\left(\mathrm{LD}\right)$
${\mathrm{Lie algebra: so5}}$ (2.2)
The vector is not regular.
> $\mathrm{Query}\left(\mathrm{e1},"RegularElement"\right)$
${\mathrm{false}}$ (2.3)
> $\mathrm{Query}\left(\mathrm{e1},\mathrm{rank}=2,"RegularElement"\right)$
${\mathrm{false}}$ (2.4)
The element is regular.
> $\mathrm{Query}\left(\mathrm{evalDG}\left(\mathrm{e1}+\mathrm{e5}-2\mathrm{e8}\right),"RegularElement"\right)$
${\mathrm{true}}$ (2.5)
so5 > $\mathrm{Query}\left(\mathrm{evalDG}\left(\mathrm{e1}+\mathrm{e5}-2\mathrm{e8}\right),\mathrm{rank}=2,"RegularElement"\right)$
${\mathrm{true}}$ (2.6)
so5 > $\mathrm{Query}\left(\mathrm{evalDG}\left(\mathrm{e1}+\mathrm{e5}-2\mathrm{e8}\right),\mathrm{algebratype}="semisimple","RegularElement"\right)$
${\mathrm{true}}$ (2.7) | 1,736 | 4,470 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 23, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.556611 |
https://brilliant.org/discussions/thread/help-41/ | 1,537,771,399,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160145.76/warc/CC-MAIN-20180924050917-20180924071317-00197.warc.gz | 458,893,231 | 22,364 | # help
Can you help me with this mechanics problem?
Note by Kyle Finch
3 years, 5 months ago
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@Raghav Vaidyanathan the problem is in the middle.
- 3 years, 5 months ago
I have rotated the image. $$\huge\ddot\smile$$
- 3 years, 5 months ago
U can tell me the solution . ???
- 3 years, 5 months ago
A disc has an angular velocity?
- 3 years, 5 months ago
$$\text{angular velocity}=\dfrac{\text{Relative velocity perpendicular to line joining them}}{\text{Distance between them}}$$
By simple geometry, $$\angle AOP=30^\circ$$
Case I: $$A$$ is fixed to ground, so, the velocity of $$P$$ will be $$\omega R$$ at angle $$120^\circ$$ to the line joining $$AP$$, thus, $$\omega_1=\dfrac{\omega R\cos 30^\circ}{\sqrt{3}R}$$
Case II: $$A$$ is fixed to disc, in this case, velocity of $$A$$ will add up in relative velocity. In this case also, velocity of $$A$$ will be $$\omega R$$ at angle $$60^\circ$$ to the line joining $$AP$$. Thus, $$\omega_2=\dfrac{\omega R\cos 30^\circ+\omega R\cos 30^\circ}{\sqrt{3}R}=2\omega_1\Rightarrow \dfrac{\omega_1}{\omega_2}=\dfrac{1}{2}$$
- 3 years, 5 months ago
Well indeed very nice by the way how was ur mains exam
- 3 years, 5 months ago
Not too good. Getting just 275.
- 3 years, 5 months ago
Was the physics section tough??
- 3 years, 5 months ago
It was NCERTish. I must not blame questions for my marks. I don't like EM.
- 3 years, 5 months ago
So who among all of you is supposed to get the highest.
- 3 years, 5 months ago
Ronak Agrawal (on Brilliant)
- 3 years, 5 months ago
Ok just one last query what makes a problem popular on brilliant. Can u tell ronaks score
- 3 years, 5 months ago
He's getting like 315-320. The quality of problem makes it popular!
- 3 years, 5 months ago
Bhavya Is getting 322.I guess??
- 3 years, 5 months ago
Yes, he does
- 3 years, 5 months ago
No i m asking do the staff member decide or the moderators
- 3 years, 5 months ago
$$\omega$$
- 3 years, 5 months ago | 976 | 2,840 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2018-39 | longest | en | 0.758394 |
https://nrich.maths.org/public/topic.php?code=125&cl=4&cldcmpid=7808 | 1,580,295,616,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251796127.92/warc/CC-MAIN-20200129102701-20200129132701-00322.warc.gz | 571,948,209 | 9,654 | # Resources tagged with: Sine, cosine, tangent
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There are many different methods to solve this geometrical problem - how many can you find? | 2,259 | 9,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2020-05 | longest | en | 0.857982 |
https://www.quizzes.cc/metric/percentof.php?percent=81&of=281 | 1,638,613,185,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362969.51/warc/CC-MAIN-20211204094103-20211204124103-00495.warc.gz | 1,029,646,988 | 4,457 | #### What is 81 percent of 281?
How much is 81 percent of 281? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 81% of 281 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 81% of 281 = 227.61
Calculate another percentage below. Type into inputs
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Calculating eighty-one of two hundred and eighty-one How to calculate 81% of 281? Simply divide the percent by 100 and multiply by the number. For example, 81 /100 x 281 = 227.61 or 0.81 x 281 = 227.61
#### How much is 81 percent of the following numbers?
81% of 281.01 = 22761.81 81% of 281.02 = 22762.62 81% of 281.03 = 22763.43 81% of 281.04 = 22764.24 81% of 281.05 = 22765.05 81% of 281.06 = 22765.86 81% of 281.07 = 22766.67 81% of 281.08 = 22767.48 81% of 281.09 = 22768.29 81% of 281.1 = 22769.1 81% of 281.11 = 22769.91 81% of 281.12 = 22770.72 81% of 281.13 = 22771.53 81% of 281.14 = 22772.34 81% of 281.15 = 22773.15 81% of 281.16 = 22773.96 81% of 281.17 = 22774.77 81% of 281.18 = 22775.58 81% of 281.19 = 22776.39 81% of 281.2 = 22777.2 81% of 281.21 = 22778.01 81% of 281.22 = 22778.82 81% of 281.23 = 22779.63 81% of 281.24 = 22780.44 81% of 281.25 = 22781.25
81% of 281.26 = 22782.06 81% of 281.27 = 22782.87 81% of 281.28 = 22783.68 81% of 281.29 = 22784.49 81% of 281.3 = 22785.3 81% of 281.31 = 22786.11 81% of 281.32 = 22786.92 81% of 281.33 = 22787.73 81% of 281.34 = 22788.54 81% of 281.35 = 22789.35 81% of 281.36 = 22790.16 81% of 281.37 = 22790.97 81% of 281.38 = 22791.78 81% of 281.39 = 22792.59 81% of 281.4 = 22793.4 81% of 281.41 = 22794.21 81% of 281.42 = 22795.02 81% of 281.43 = 22795.83 81% of 281.44 = 22796.64 81% of 281.45 = 22797.45 81% of 281.46 = 22798.26 81% of 281.47 = 22799.07 81% of 281.48 = 22799.88 81% of 281.49 = 22800.69 81% of 281.5 = 22801.5
81% of 281.51 = 22802.31 81% of 281.52 = 22803.12 81% of 281.53 = 22803.93 81% of 281.54 = 22804.74 81% of 281.55 = 22805.55 81% of 281.56 = 22806.36 81% of 281.57 = 22807.17 81% of 281.58 = 22807.98 81% of 281.59 = 22808.79 81% of 281.6 = 22809.6 81% of 281.61 = 22810.41 81% of 281.62 = 22811.22 81% of 281.63 = 22812.03 81% of 281.64 = 22812.84 81% of 281.65 = 22813.65 81% of 281.66 = 22814.46 81% of 281.67 = 22815.27 81% of 281.68 = 22816.08 81% of 281.69 = 22816.89 81% of 281.7 = 22817.7 81% of 281.71 = 22818.51 81% of 281.72 = 22819.32 81% of 281.73 = 22820.13 81% of 281.74 = 22820.94 81% of 281.75 = 22821.75
81% of 281.76 = 22822.56 81% of 281.77 = 22823.37 81% of 281.78 = 22824.18 81% of 281.79 = 22824.99 81% of 281.8 = 22825.8 81% of 281.81 = 22826.61 81% of 281.82 = 22827.42 81% of 281.83 = 22828.23 81% of 281.84 = 22829.04 81% of 281.85 = 22829.85 81% of 281.86 = 22830.66 81% of 281.87 = 22831.47 81% of 281.88 = 22832.28 81% of 281.89 = 22833.09 81% of 281.9 = 22833.9 81% of 281.91 = 22834.71 81% of 281.92 = 22835.52 81% of 281.93 = 22836.33 81% of 281.94 = 22837.14 81% of 281.95 = 22837.95 81% of 281.96 = 22838.76 81% of 281.97 = 22839.57 81% of 281.98 = 22840.38 81% of 281.99 = 22841.19 81% of 282 = 22842
1% of 281 = 2.81 2% of 281 = 5.62 3% of 281 = 8.43 4% of 281 = 11.24 5% of 281 = 14.05 6% of 281 = 16.86 7% of 281 = 19.67 8% of 281 = 22.48 9% of 281 = 25.29 10% of 281 = 28.1 11% of 281 = 30.91 12% of 281 = 33.72 13% of 281 = 36.53 14% of 281 = 39.34 15% of 281 = 42.15 16% of 281 = 44.96 17% of 281 = 47.77 18% of 281 = 50.58 19% of 281 = 53.39 20% of 281 = 56.2 21% of 281 = 59.01 22% of 281 = 61.82 23% of 281 = 64.63 24% of 281 = 67.44 25% of 281 = 70.25
26% of 281 = 73.06 27% of 281 = 75.87 28% of 281 = 78.68 29% of 281 = 81.49 30% of 281 = 84.3 31% of 281 = 87.11 32% of 281 = 89.92 33% of 281 = 92.73 34% of 281 = 95.54 35% of 281 = 98.35 36% of 281 = 101.16 37% of 281 = 103.97 38% of 281 = 106.78 39% of 281 = 109.59 40% of 281 = 112.4 41% of 281 = 115.21 42% of 281 = 118.02 43% of 281 = 120.83 44% of 281 = 123.64 45% of 281 = 126.45 46% of 281 = 129.26 47% of 281 = 132.07 48% of 281 = 134.88 49% of 281 = 137.69 50% of 281 = 140.5
51% of 281 = 143.31 52% of 281 = 146.12 53% of 281 = 148.93 54% of 281 = 151.74 55% of 281 = 154.55 56% of 281 = 157.36 57% of 281 = 160.17 58% of 281 = 162.98 59% of 281 = 165.79 60% of 281 = 168.6 61% of 281 = 171.41 62% of 281 = 174.22 63% of 281 = 177.03 64% of 281 = 179.84 65% of 281 = 182.65 66% of 281 = 185.46 67% of 281 = 188.27 68% of 281 = 191.08 69% of 281 = 193.89 70% of 281 = 196.7 71% of 281 = 199.51 72% of 281 = 202.32 73% of 281 = 205.13 74% of 281 = 207.94 75% of 281 = 210.75
76% of 281 = 213.56 77% of 281 = 216.37 78% of 281 = 219.18 79% of 281 = 221.99 80% of 281 = 224.8 81% of 281 = 227.61 82% of 281 = 230.42 83% of 281 = 233.23 84% of 281 = 236.04 85% of 281 = 238.85 86% of 281 = 241.66 87% of 281 = 244.47 88% of 281 = 247.28 89% of 281 = 250.09 90% of 281 = 252.9 91% of 281 = 255.71 92% of 281 = 258.52 93% of 281 = 261.33 94% of 281 = 264.14 95% of 281 = 266.95 96% of 281 = 269.76 97% of 281 = 272.57 98% of 281 = 275.38 99% of 281 = 278.19 100% of 281 = 281 | 2,692 | 5,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2021-49 | latest | en | 0.867353 |
https://im.kendallhunt.com/HS/teachers/3/2/16/practice.html | 1,722,885,026,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640454712.22/warc/CC-MAIN-20240805180114-20240805210114-00057.warc.gz | 248,165,697 | 26,115 | # Lesson 16
Minimizing Surface Area
### Problem 1
There are many cylinders with a volume of $$144\pi$$ cubic inches. The height $$h(r)$$ in inches of one of these cylinders is a function of its radius $$r$$ in inches where $$h(r)=\frac{144}{r^2}$$.
1. What is the height of one of these cylinders if its radius is 2 inches?
2. What is the height of one of these cylinders if its radius is 3 inches?
3. What is the height of one of these cylinders if its radius is 6 inches?
### Problem 2
The surface area $$S(r)$$ in square units of a cylinder with a volume of 18 cubic units is a function of its radius $$r$$ in units where $$S(r)=2\pi r^2+\frac{36}{r}$$. What is the surface area of a cylinder with a volume of 18 cubic units and a radius of 3 units?
### Problem 3
Han finds an expression for $$S(r)$$ that gives the surface area in square inches of any cylindrical can with a specific fixed volume, in terms of its radius $$r$$ in inches. This is the graph Han gets if he allows $$r$$ to take on any value between -1 and 5.
1. What would be a more appropriate domain for Han to use instead?
2. What is the approximate minimum surface area for the can?
### Problem 4
The graph of a polynomial function $$f$$ is shown. Is the degree of the polynomial even or odd? Explain your reasoning.
### Solution
(From Unit 2, Lesson 8.)
### Problem 5
The polynomial function $$p(x)=x^4+4x^3-7x^2-22x+24$$ has known factors of $$(x+4)$$ and $$(x-1)$$.
1. Rewrite $$p(x)$$ as the product of linear factors.
2. Draw a rough sketch of the graph of the function.
### Solution
(From Unit 2, Lesson 12.)
### Problem 6
Which polynomial has $$(x+1)$$ as a factor?
A:
$$x^3+2x^2-19x-20$$
B:
$$x^3-21x+20$$
C:
$$x^3+8x+11x-20$$
D:
$$x^3-3x^2+3x-1$$ | 528 | 1,755 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-33 | latest | en | 0.867195 |
https://capitalsportsbet.com/ZuluCode5/how-does-sports-betting-work-football-in-football.html | 1,568,599,802,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572471.35/warc/CC-MAIN-20190916015552-20190916041552-00330.warc.gz | 401,307,025 | 7,276 | The 2-way moneyline is what most North American bettors would simply refer to as “the moneyline”. This is one of the most common wagering options where the user bets which side will win the game straight up. (A draw or tie results in a push with the 2-way moneyline.) The term is sometimes highlighted during soccer betting to differentiate from the 3-way moneyline - a more popular option with the draw added as a wagering option.
## Typically hockey is not a popular sport when it comes to point-spread betting because most games are decided by only a goal or two. Sports like football and basketball are better for point-spread betting mainly because they are higher scoring sports. However, Proline does offer NHL point-spread betting and it can get confusing so let’s take a look at how it works.
Let's use this formula to calculate the implied probability of the Celtics winning their game against the Grizzlies. We know the odds are -240, which means we'd have to risk \$240 for a total potential return of \$340 (the initial stake plus the \$100 winnings). So the calculation here is \$240 divided by \$340. This gives us an implied probability of 0.7059.
A spread is a range of outcomes and the bet is whether the outcome will be above or below the spread. Spread betting has been a major growth market in the UK in recent years, with the number of gamblers heading towards one million.[1] Financial spread betting (see below) can carry a high level of risk if there is no "stop".[2] In the UK, spread betting is regulated by the Financial Conduct Authority rather than the Gambling Commission.[3]
If the team is an underdog, then the moneyline number represents exactly how much you would get paid in profit for a correct pick. So if we were to bet \$100 on the Boston Celtics and won, we would get paid \$145 in profit. Seems easy enough, but you may already be asking what happens if you don’t want to bet in increments of \$100. This is totally fine and still straight forward to figure out.
The moneyline is different. First, with the moneyline whichever team wins the game pays out. There’s no giving or taking away of points. How do the bookies even the playing field with the moneyline? They do it by making bettors wager more on the favorite to win less and allowing them to bet less to win more on the dog. The favorite is posted with a minus sign and a number. That number represents the amount of cash that has to be wagered in order to win \$100. The underdog, on the other hand, is listed with a plus sign in front of a number. That number shows how much a bettor wins when they bet \$100.
For UK spread betting firms, any final outcome that finishes in the middle of the spread will result in profits from both sides of the book as both buyers and sellers will have ended up making unit point losses. So in the example above, if the cricket team ended up scoring 345 runs both buyers at 350 and sellers at 340 would have ended up with losses of five unit points multiplied by their stake.
The first number in the listing pertains to the order this game appears on a sportsbook’s board. The next NFL game would be listed as #103 for the road team and #104 for the home team. You can think of these numbers in the same way that each horse in a race has its own betting number. The next big takeaway from this listing is that the top team is always the road team (thus the odd number) and the bottom team is playing at home.
Copyright © 2019 Bleacher Report, Inc. Turner Broadcasting System, Inc. All Rights Reserved. BleacherReport.com is part of Bleacher Report – Turner Sports Network, part of the Turner Sports and Entertainment Network. Certain photos copyright © 2019 Getty Images. Any commercial use or distribution without the express written consent of Getty Images is strictly prohibited. AdChoices
The secret to winning sports bets is finding value and picking winners. There is absolutely zero correlation between the complexity of a bet and how likely you are to win. In fact, you could say that there is a negative correlation because a lot of bettors don’t fully understand the complex bets they are making, meaning they are more likely to make mistakes and incorrectly assess value.
The key here is to target the point spread five and seven, because these are virtually tied as the most common margins of victory. It's important to recognize that most betting sites are only willing to sell 2 or 3 half points for 10 cents each, after which point they start charging more. Some sites sell up to four half points at this price though.
The prop bets available are going to vary from game to game, and the number of options will increase during the playoffs and into the NBA Championships. The important thing you need to make sure you’re aware of is that not all prop bets are created equally. Some prop bets take a lot of skill to predict, some take some skill to predict, and some take absolutely no skill whatsoever and are just a complete guess.
Apply the spread. In point-spread betting, the actual final score of the game is only the starting point. Say Chicago beats Detroit 24-17. Because Chicago was the favorite, you subtract the point spread from its final score. That's the purpose of the minus sign in the spread. The spread was 6, so you take 6 points away from Chicago's point total, giving you an "adjusted" score of Chicago 18, Detroit 17. If you'd bet on Chicago, you'd have won the bet. Now, say Chicago won the game 20-17. Subtracting the 6 points from Chicago's total gives you a final score of Detroit 17, Chicago 14. If you'd bet on Chicago, you'd have lost.
We do want to make sure to point out that this is total money returned and not your profit. We went over the difference above when discussing the American odds. Let’s take a look at the same bet again, but this time with decimal odds. We should be expecting to see a profit of \$33.33 for an Eagles bet and \$240 for a Falcons bet since we know that the bets are the exact same and are just presented in a different format.
This may sound confusing, but spread betting is one of the easiest forms of sports betting offered. It also offers better odds when betting on the favorite team and the normal odds are 10/11 or -110. When one chooses to bet on the underdog that team does not have to win, they just have to cover the spread. Spread betting is pretty simple and it allows punters to enjoy an exciting form of betting that can lead to some significant wins.
When betting on the underdog, the first step is the same. Divide the positive moneyline by 100, which in the case of the Grizzlies in the above example would give you 2.10. Then, multiply your stake by that number to get your potential winnings. \$450 multiplied by 2.10 is \$945. Essentially, this means if you risked \$450 on the Grizzlies, you would stand to win \$945.
"Winning Margin" (aka Result Betting) is where it is possible to bet on the final result of a game or event and select the correct ‘band’ of points between the winning team and losing team. For example, if you think the Patriots will win, but the game will be close, pick the New England Patriots 1-6 Points Winning Margin (where the Patriots winning by 1, 2, 3, 4, 5, or 6 points results in a winning selection).
Disclaimer: This site is for informational and entertainment purposes only. Individual users are responsible for the laws regarding accessing gambling information from their jurisdictions. Many countries around the world prohibit gambling, please check the laws in your location. Any use of this information that may violate any federal, state, local or international law is strictly prohibited.
Betting on sporting events has long been the most popular form of spread betting. Whilst most bets the casino offers to players have a built in house edge, betting on the spread offers an opportunity for the astute gambler. When a casino accepts a spread bet, it gives the player the odds of 10 to 11, or -110. That means that for every 11 dollars the player wagers, the player will win 10, slightly lower than an even money bet. If team A is playing team B, the casino is not concerned with who wins the game; they are only concerned with taking an equal amount of money of both sides. For example, if one player takes team A and the other takes team B and each wager \$110 to win \$100, it doesn’t matter what team wins; the casino makes money. They take \$100 of the \$110 from the losing bet and pay the winner, keeping the extra \$10 for themselves. This is the house edge. The goal of the casino is to set a line that encourages an equal amount of action on both sides, thereby guaranteeing a profit. This also explains how money can be made by the astute gambler. If casinos set lines to encourage an equal amount of money on both sides, it sets them based on the public perception of the team, not necessarily the real strength of the teams. Many things can affect public perception, which moves the line away from what the real line should be. This gap between the Vegas line, the real line, and differences between other sports books betting lines and spreads is where value can be found.
As you can see, each team is listed, followed by the adjustment or line change for each team, and then the odds that you would be paid out. If you notice, Chelsea has the word scratch next to their name. This is because they are the league favorite to win and all other adjustments are made about them. If your team is in first place at the end of the regular season after the adjustments are made, you will win your bet and be paid the posted odds. As you can see, the odds pay out fairly well on these bets as they are season long and are more difficult to win.
In sports like football and basketball, the moneyline is considered as the secondary option next to point spreads. Points spreads are the way that most people get their action in on basketball betting and football betting because the payouts are near doubling your money and it’s a fun way to handicap the game. Betting the moneyline in those sports is less popular because you might have some big mismatches and then it becomes too challenging to have faith in the underdog winning outright or too costly to bet the favorite.
If you like favorites, you're going to be betting a lot to win a little. The money line will always be listed to the right of the point spread on the odds board in a sports book. In the above example, the money line would probably be Chicago -250 and Detroit +200. To bet Chicago simply to win, you must wager \$250 to win \$100, while a \$100 bet on Detroit would pay \$200 if the Lions come through.
Identify the type of line you are looking at. All online sports books offer you the chance to have your lines in an "American" or "Money line" version. If I were you, I would use this as my standard. An "American" line uses either a + or - before a number to indicate odds. So a -120 and a +120 are two very different odds on a team… I will explain the differences shortly. Two other less common variations exist: decimal odds and fractional odds.
The prop bets available are going to vary from game to game, and the number of options will increase during the playoffs and into the NBA Championships. The important thing you need to make sure you’re aware of is that not all prop bets are created equally. Some prop bets take a lot of skill to predict, some take some skill to predict, and some take absolutely no skill whatsoever and are just a complete guess.
While we're going to cover this extensively in the moneyline section of the Basketball Betting Guide, we'll touch briefly on what the -110 means here. -110 refers to the moneyline payout for a particular bet. If you were to bet \$100 on a basketball point spread paying -110 and win, you would be paid out a profit of \$90.90. The easy way to figure out the amount you will get paid with minus moneyline odds is to divide your bet size by the absolute value of the betting odds and multiply that number by 100.
```A point spread in sports is a figure set by oddsmakers to provide an advantage or disadvantage based on the margin of victory or defeat for a given team. The “favorite” team (labeled with a “-” sign) would be at the disadvantage as they would need to win the game by a set number of points while the “underdog” team (labeled with a “+” sign) would be given an advantage to not lose the game by a set number of points. The reason oddsmakers do this is to provide betting interest for both sides due to one team typically being better than the other.
``` | 2,741 | 12,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2019-39 | latest | en | 0.963455 |
https://www.nag.com/numeric/MB/manual64_24_1/html/F08/f08kbf.html | 1,516,425,022,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889325.32/warc/CC-MAIN-20180120043530-20180120063530-00348.warc.gz | 921,596,225 | 7,093 | Integer type: int32 int64 nag_int show int32 show int32 show int64 show int64 show nag_int show nag_int
Chapter Contents
Chapter Introduction
NAG Toolbox
# NAG Toolbox: nag_lapack_dgesvd (f08kb)
## Purpose
nag_lapack_dgesvd (f08kb) computes the singular value decomposition (SVD) of a real m$m$ by n$n$ matrix A$A$, optionally computing the left and/or right singular vectors.
## Syntax
[a, s, u, vt, work, info] = f08kb(jobu, jobvt, a, 'm', m, 'n', n)
[a, s, u, vt, work, info] = nag_lapack_dgesvd(jobu, jobvt, a, 'm', m, 'n', n)
Note: the interface to this routine has changed since earlier releases of the toolbox:
Mark 23: work is now an output parameter
.
## Description
The SVD is written as
A = UΣVT , $A = UΣVT ,$
where Σ$\Sigma$ is an m$m$ by n$n$ matrix which is zero except for its min (m,n)$\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left(m,n\right)$ diagonal elements, U$U$ is an m$m$ by m$m$ orthogonal matrix, and V$V$ is an n$n$ by n$n$ orthogonal matrix. The diagonal elements of Σ$\Sigma$ are the singular values of A$A$; they are real and non-negative, and are returned in descending order. The first min (m,n)$\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left(m,n\right)$ columns of U$U$ and V$V$ are the left and right singular vectors of A$A$.
Note that the function returns VT${V}^{\mathrm{T}}$, not V$V$.
## References
Anderson E, Bai Z, Bischof C, Blackford S, Demmel J, Dongarra J J, Du Croz J J, Greenbaum A, Hammarling S, McKenney A and Sorensen D (1999) LAPACK Users' Guide (3rd Edition) SIAM, Philadelphia http://www.netlib.org/lapack/lug
Golub G H and Van Loan C F (1996) Matrix Computations (3rd Edition) Johns Hopkins University Press, Baltimore
## Parameters
### Compulsory Input Parameters
1: jobu – string (length ≥ 1)
Specifies options for computing all or part of the matrix U$U$.
jobu = 'A'${\mathbf{jobu}}=\text{'A'}$
All m$m$ columns of U$U$ are returned in array u.
jobu = 'S'${\mathbf{jobu}}=\text{'S'}$
The first min (m,n)$\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left(m,n\right)$ columns of U$U$ (the left singular vectors) are returned in the array u.
jobu = 'O'${\mathbf{jobu}}=\text{'O'}$
The first min (m,n)$\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left(m,n\right)$ columns of U$U$ (the left singular vectors) are overwritten on the array a.
jobu = 'N'${\mathbf{jobu}}=\text{'N'}$
No columns of U$U$ (no left singular vectors) are computed.
Constraint: jobu = 'A'${\mathbf{jobu}}=\text{'A'}$, 'S'$\text{'S'}$, 'O'$\text{'O'}$ or 'N'$\text{'N'}$.
2: jobvt – string (length ≥ 1)
Specifies options for computing all or part of the matrix VT${V}^{\mathrm{T}}$.
jobvt = 'A'${\mathbf{jobvt}}=\text{'A'}$
All n$n$ rows of VT${V}^{\mathrm{T}}$ are returned in the array vt.
jobvt = 'S'${\mathbf{jobvt}}=\text{'S'}$
The first min (m,n)$\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left(m,n\right)$ rows of VT${V}^{\mathrm{T}}$ (the right singular vectors) are returned in the array vt.
jobvt = 'O'${\mathbf{jobvt}}=\text{'O'}$
The first min (m,n)$\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left(m,n\right)$ rows of VT${V}^{\mathrm{T}}$ (the right singular vectors) are overwritten on the array a.
jobvt = 'N'${\mathbf{jobvt}}=\text{'N'}$
No rows of VT${V}^{\mathrm{T}}$ (no right singular vectors) are computed.
Constraints:
• jobvt = 'A'${\mathbf{jobvt}}=\text{'A'}$, 'S'$\text{'S'}$, 'O'$\text{'O'}$ or 'N'$\text{'N'}$;
• jobvt and jobu cannot both be 'O'$\text{'O'}$.
3: a(lda, : $:$) – double array
The first dimension of the array a must be at least max (1,m)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$
The second dimension of the array must be at least max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The m$m$ by n$n$ matrix A$A$.
### Optional Input Parameters
1: m – int64int32nag_int scalar
Default: The first dimension of the array a.
m$m$, the number of rows of the matrix A$A$.
Constraint: m0${\mathbf{m}}\ge 0$.
2: n – int64int32nag_int scalar
Default: The second dimension of the array a.
n$n$, the number of columns of the matrix A$A$.
Constraint: n0${\mathbf{n}}\ge 0$.
### Input Parameters Omitted from the MATLAB Interface
lda ldu ldvt lwork
### Output Parameters
1: a(lda, : $:$) – double array
The first dimension of the array a will be max (1,m)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$
The second dimension of the array will be max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
ldamax (1,m)$\mathit{lda}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$.
If jobu = 'O'${\mathbf{jobu}}=\text{'O'}$, a is overwritten with the first min (m,n)$\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left(m,n\right)$ columns of U$U$ (the left singular vectors, stored column-wise).
If jobvt = 'O'${\mathbf{jobvt}}=\text{'O'}$, a is overwritten with the first min (m,n)$\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left(m,n\right)$ rows of VT${V}^{\mathrm{T}}$ (the right singular vectors, stored row-wise).
If jobu'O'${\mathbf{jobu}}\ne \text{'O'}$ and jobvt'O'${\mathbf{jobvt}}\ne \text{'O'}$, the contents of a are destroyed.
2: s( : $:$) – double array
Note: the dimension of the array s must be at least max (1,min (m,n)) $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left({\mathbf{m}},{\mathbf{n}}\right)\right)$.
The singular values of A$A$, sorted so that s(i)s(i + 1)${\mathbf{s}}\left(i\right)\ge {\mathbf{s}}\left(i+1\right)$.
3: u(ldu, : $:$) – double array
The first dimension, ldu, of the array u will be
• if jobu = 'A'${\mathbf{jobu}}=\text{'A'}$ or 'S'$\text{'S'}$, ldu max (1,m) $\mathit{ldu}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$;
• otherwise ldu1$\mathit{ldu}\ge 1$.
The second dimension of the array will be max (1,m)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$ if jobu = 'A'${\mathbf{jobu}}=\text{'A'}$, max (1,min (m,n))$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left({\mathbf{m}},{\mathbf{n}}\right)\right)$ if jobu = 'S'${\mathbf{jobu}}=\text{'S'}$, and at least 1$1$ otherwise
If jobu = 'A'${\mathbf{jobu}}=\text{'A'}$, u contains the m$m$ by m$m$ orthogonal matrix U$U$.
If jobu = 'S'${\mathbf{jobu}}=\text{'S'}$, u contains the first min (m,n)$\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left(m,n\right)$ columns of U$U$ (the left singular vectors, stored column-wise).
If jobu = 'N'${\mathbf{jobu}}=\text{'N'}$ or 'O'$\text{'O'}$, u is not referenced.
4: vt(ldvt, : $:$) – double array
The first dimension, ldvt, of the array vt will be
• if jobvt = 'A'${\mathbf{jobvt}}=\text{'A'}$, ldvt max (1,n) $\mathit{ldvt}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$;
• if jobvt = 'S'${\mathbf{jobvt}}=\text{'S'}$, ldvt max (1,min (m,n)) $\mathit{ldvt}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left({\mathbf{m}},{\mathbf{n}}\right)\right)$;
• otherwise ldvt1$\mathit{ldvt}\ge 1$.
The second dimension of the array will be max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$ if jobvt = 'A'${\mathbf{jobvt}}=\text{'A'}$ or 'S'$\text{'S'}$, and at least 1$1$ otherwise
If jobvt = 'A'${\mathbf{jobvt}}=\text{'A'}$, vt contains the n$n$ by n$n$ orthogonal matrix VT${V}^{\mathrm{T}}$.
If jobvt = 'S'${\mathbf{jobvt}}=\text{'S'}$, vt contains the first min (m,n)$\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left(m,n\right)$ rows of VT${V}^{\mathrm{T}}$ (the right singular vectors, stored row-wise).
If jobvt = 'N'${\mathbf{jobvt}}=\text{'N'}$ or 'O'$\text{'O'}$, vt is not referenced.
5: work(max (1,lwork)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,\mathit{lwork}\right)$) – double array
If ${\mathbf{INFO}}={\mathbf{0}}$, work(1)${\mathbf{work}}\left(1\right)$ returns the optimal lwork.
If ${\mathbf{INFO}}>{\mathbf{0}}$, work(2 : min (m,n))${\mathbf{work}}\left(2:\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left({\mathbf{m}},{\mathbf{n}}\right)\right)$ contains the unconverged superdiagonal elements of an upper bidiagonal matrix B$B$ whose diagonal is in s (not necessarily sorted). B$B$ satisfies A = UBVT$A=UB{V}^{\mathrm{T}}$, so it has the same singular values as A$A$, and singular vectors related by U$U$ and VT${V}^{\mathrm{T}}$.
6: info – int64int32nag_int scalar
info = 0${\mathbf{info}}=0$ unless the function detects an error (see Section [Error Indicators and Warnings]).
## Error Indicators and Warnings
info = i${\mathbf{info}}=-i$
If info = i${\mathbf{info}}=-i$, parameter i$i$ had an illegal value on entry. The parameters are numbered as follows:
1: jobu, 2: jobvt, 3: m, 4: n, 5: a, 6: lda, 7: s, 8: u, 9: ldu, 10: vt, 11: ldvt, 12: work, 13: lwork, 14: info.
It is possible that info refers to a parameter that is omitted from the MATLAB interface. This usually indicates that an error in one of the other input parameters has caused an incorrect value to be inferred.
INFO > 0${\mathbf{INFO}}>0$
If nag_lapack_dgesvd (f08kb) did not converge, info specifies how many superdiagonals of an intermediate bidiagonal form did not converge to zero. See the description of work above for details.
## Accuracy
The computed singular value decomposition is nearly the exact singular value decomposition for a nearby matrix (A + E) $\left(A+E\right)$, where
‖E‖2 = O(ε) ‖A‖2 , $‖E‖2 = O(ε) ‖A‖2 ,$
and ε $\epsilon$ is the machine precision. In addition, the computed singular vectors are nearly orthogonal to working precision. See Section 4.9 of Anderson et al. (1999) for further details.
The total number of floating point operations is approximately proportional to mn2 $m{n}^{2}$ when m > n$m>n$ and m2n ${m}^{2}n$ otherwise.
The singular values are returned in descending order.
The complex analogue of this function is nag_lapack_zgesvd (f08kp).
## Example
```function nag_lapack_dgesvd_example
jobu = 'Overwrite A by U';
jobvt = 'Singular vectors (V)';
a = [2.27, -1.54, 1.15, -1.94;
0.28, -1.67, 0.94, -0.78;
-0.48, -3.09, 0.99, -0.21;
1.07, 1.22, 0.79, 0.63;
-2.35, 2.93, -1.45, 2.3;
0.62, -7.39, 1.03, -2.57];
[aOut, s, u, vt, work, info] = nag_lapack_dgesvd(jobu, jobvt, a);
aOut, s, u, vt, info
```
```
aOut =
-0.2774 -0.6003 -0.1277 0.1323
-0.2020 -0.0301 0.2805 0.7034
-0.2918 0.3348 0.6453 0.1906
0.0938 -0.3699 0.6781 -0.5399
0.4213 0.5266 0.0413 -0.0575
-0.7816 0.3353 -0.1645 -0.3957
s =
9.9966
3.6831
1.3569
0.5000
u =
0
vt =
-0.1921 0.8794 -0.2140 0.3795
-0.8030 -0.3926 -0.2980 0.3351
0.0041 -0.0752 0.7827 0.6178
-0.5642 0.2587 0.5027 -0.6017
info =
0
```
```function f08kb_example
jobu = 'Overwrite A by U';
jobvt = 'Singular vectors (V)';
a = [2.27, -1.54, 1.15, -1.94;
0.28, -1.67, 0.94, -0.78;
-0.48, -3.09, 0.99, -0.21;
1.07, 1.22, 0.79, 0.63;
-2.35, 2.93, -1.45, 2.3;
0.62, -7.39, 1.03, -2.57];
[aOut, s, u, vt, work, info] = f08kb(jobu, jobvt, a);
aOut, s, u, vt, info
```
```
aOut =
-0.2774 -0.6003 -0.1277 0.1323
-0.2020 -0.0301 0.2805 0.7034
-0.2918 0.3348 0.6453 0.1906
0.0938 -0.3699 0.6781 -0.5399
0.4213 0.5266 0.0413 -0.0575
-0.7816 0.3353 -0.1645 -0.3957
s =
9.9966
3.6831
1.3569
0.5000
u =
0
vt =
-0.1921 0.8794 -0.2140 0.3795
-0.8030 -0.3926 -0.2980 0.3351
0.0041 -0.0752 0.7827 0.6178
-0.5642 0.2587 0.5027 -0.6017
info =
0
``` | 4,551 | 11,498 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 143, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-05 | latest | en | 0.501925 |
http://mathoverflow.net/feeds/user/5706 | 1,369,257,928,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702448584/warc/CC-MAIN-20130516110728-00045-ip-10-60-113-184.ec2.internal.warc.gz | 171,401,408 | 3,826 | User slobodan simi&#263; - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T21:25:33Z http://mathoverflow.net/feeds/user/5706 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75038/when-is-a-closed-differential-form-harmonic-relative-to-some-metric When is a closed differential form harmonic relative to some metric? Slobodan Simić 2011-09-09T21:34:09Z 2012-05-13T03:40:31Z <p>Let $\omega$ be a closed non-exact differential $k$-form ($k \geq 1$) on a closed orientable manifold $M$. </p> <p><strong>Question</strong>: Is there always a Riemannian metric $g$ on $M$ such that $\omega$ is $g$-harmonic, i.e., $\Delta_g \omega = 0$? </p> <p>Here $\Delta_g$ is the Laplace-deRham operator, defined as usual by $\Delta_g = d \delta + \delta d$, where $\delta$ is the $g$-codifferential. Note that non-exactness is important, since if $\omega$ were to be exact and harmonic, then by the Hodge decomposition theorem $\omega = 0$.</p> <p>For instance, if $\omega$ is a 1-form on the unit circle, then it is not hard to see that $\omega$ is harmonic with respect to some metric $g$ if and only if it is a volume form (i.e., it doesn't vanish). This observation generalizes to forms of top degree on any $M$.</p> <p>What can be said in general for forms which are not of top degree?</p> http://mathoverflow.net/questions/77425/failures-that-lead-eventually-to-new-mathematics/77510#77510 Answer by Slobodan Simić for Failures that lead eventually to new mathematics Slobodan Simić 2011-10-08T04:04:26Z 2011-10-08T04:04:26Z <p>In the early 1960's Smale published a paper containing a conjecture whose consequence was that (in modern language) "chaos didn't exist". He soon received a letter from Norman Levinson informing him of an earlier work of Cartwright and Littlewood which effectively contained a counterexample to Smale's conjecture. Smale "worked day and night to resolve the challenges that the letter posed to my beliefs" (in his own words), trying to translate analytic arguments of Levinson and Cartwright-Littlewood into his own geometric way of thinking. This led him to his seminal discovery of the <a href="http://www.scholarpedia.org/article/Smale_horseshoe" rel="nofollow">horseshoe map</a>, followed by the foundation of the field of hyperbolic dynamical systems. For more details, see Smale's popular article "Finding a horseshoe on the beaches of Rio", Mathematical Intelligencer 20 (1998), 39-44.</p> http://mathoverflow.net/questions/76580/famous-mathematicians-with-background-in-arts-humanities-law-etc/76689#76689 Answer by Slobodan Simić for Famous mathematicians with background in arts/humanities/law etc Slobodan Simić 2011-09-28T20:41:49Z 2011-09-28T20:41:49Z <p>Henri Poincaré was a mining engineer. His first job was at the Corps des Mines as an inspector of mines. He participated in the rescue of miners trapped after an explosion, himself descending the shaft into the mine to investigate the cause of the explosion! Check <a href="http://www.gap-system.org/~history/HistTopics/Poincare_mines.html" rel="nofollow">this link</a> for details.</p> http://mathoverflow.net/questions/75053/inner-products-on-differential-forms Inner products on differential forms Slobodan Simić 2011-09-09T23:45:34Z 2011-09-10T01:40:28Z <p>Given a Riemannian metric $g$ on a smooth manifold $M$, one defines an $L^2$-inner product on the space $\bigwedge^\ast(M)$ of differential forms by</p> <p>$$\langle \alpha, \beta \rangle_g = \int_M \alpha \wedge \ast_g \beta,$$</p> <p>where $\ast_g$ denotes the Hodge-star operator relative to $g$, and $\alpha, \beta$ are forms of the same degree.</p> <p><strong>Question</strong>: Does every inner product on $\bigwedge^\ast(M)$ as a graded vector space come from some metric $g$? How about inner products on $k$-forms $\bigwedge^k(M)$ for a single $k$, especially $0 < k < \dim M$?</p> http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/26071#26071 Answer by Slobodan Simić for Examples of common false beliefs in mathematics. Slobodan Simić 2010-05-26T21:39:37Z 2010-05-26T21:39:37Z <p>If $f$ is (Lebesgue) integrable on $\mathbb{R}$, then $f(x) \to 0$, as $x \to \infty$. False: there exists a <em>continuous</em> integrable function on $\mathbb{R}$ such that $\limsup_{\infty} f = \infty$ (an exercise in Stein and Shakarchi's <a href="http://www.amazon.com/Real-Analysis-Integration-Princeton-Lectures/dp/0691113866/ref=sr_1_2?ie=UTF8&s=books&qid=1274909846&sr=1-2" rel="nofollow">Real Analysis</a>).</p> http://mathoverflow.net/questions/10282/alternative-undergraduate-analysis-texts/25745#25745 Answer by Slobodan Simić for Alternative Undergraduate Analysis Texts Slobodan Simić 2010-05-24T04:04:46Z 2010-05-24T04:04:46Z <p>V. A. Zorich's <a href="http://www.springer.com/mathematics/analysis/book/978-3-540-40386-9?changeHeader" rel="nofollow">Mathematical Analysis I</a> and <a href="http://www.springer.com/mathematics/analysis/book/978-3-540-40633-4?changeHeader" rel="nofollow">II</a> (Springer). It covers undergraduate material from an advanced viewpoint, contains lots of good physically oriented examples, and is quite comprehensive. </p> http://mathoverflow.net/questions/4994/fundamental-examples/24058#24058 Answer by Slobodan Simić for Fundamental Examples Slobodan Simić 2010-05-10T03:48:59Z 2010-05-10T03:48:59Z <p><a href="http://www.scholarpedia.org/article/Smale_horseshoe" rel="nofollow">Smale's horseshoe map</a> in dynamical systems.</p> http://mathoverflow.net/questions/76580/famous-mathematicians-with-background-in-arts-humanities-law-etc/76689#76689 Comment by Slobodan Simić Slobodan Simić 2011-09-28T21:38:53Z 2011-09-28T21:38:53Z Sorry, didn't realize that. http://mathoverflow.net/questions/75053/inner-products-on-differential-forms/75056#75056 Comment by Slobodan Simić Slobodan Simić 2011-09-23T19:14:34Z 2011-09-23T19:14:34Z Thanks, Brian and Paul - that's exactly what I wanted to know. Thanks also to José for an entirely different (at least for me) point of view. http://mathoverflow.net/questions/75038/when-is-a-closed-differential-form-harmonic-relative-to-some-metric Comment by Slobodan Simić Slobodan Simić 2011-09-23T19:09:41Z 2011-09-23T19:09:41Z Thanks to all who helped elucidate this question! I was mostly interested in the degrees $k = 1$ and $k = n-1$, but it would certainly be very interesting to see what can be said about the intermediate $k$'s. Perhaps a topic for a future Ph.D. thesis. | 1,991 | 6,534 | {"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} | 3.171875 | 3 | CC-MAIN-2013-20 | latest | en | 0.872129 |
https://www.coursehero.com/tutors-problems/Statistics-and-Probability/20043056-In-a-marketing-survey-a-random-sample-of-1012-supermarket-shoppers-re/ | 1,579,627,585,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250604849.31/warc/CC-MAIN-20200121162615-20200121191615-00019.warc.gz | 833,349,940 | 25,726 | View the step-by-step solution to:
Question
# In a marketing survey, a random sample of 1012 supermarket shoppers revealed that 264 always stock up on an
item when they find that item at a real bargain price.
What is the margin of error based on a 95% confidence interval? (Enter a number. Round your answer to three decimal places.)
Here we are given population size n = 1012 sample size = 264 confidence interval = 95% ... View the full answer
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Browse Documents | 173 | 803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-05 | latest | en | 0.887571 |
https://www.eswick.com/how-is-ground-wire-used-as-a-neutral-conductor | 1,653,759,675,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016949.77/warc/CC-MAIN-20220528154416-20220528184416-00271.warc.gz | 858,341,706 | 17,906 | How is ground wire used as a neutral conductor?
Because they didn't have a large enough piece, the wire nutted a piece of blue MTW wire from a junction box to the hot wire only, then hose clamped the neutral conductor to the well pipe that came in immediately next to it. What about the ground wire, you may wonder. It was just severed. The answer is that because they had no way to connect them together, they left it out. If you look at the wiring diagram for this house, you can see that both the white and the red conductors are broken at one end. This means that they were never connected together.
The best way to do things properly is of course to use two separate wires for neutral and ground. But because most homes were built before this practice became common, they often use one conductor for both functions. In other words, ground is also used as a neutral conductor. This is not recommended because it leaves your home vulnerable to electrical problems. If there's a problem with a nearby circuit, it could cause current to flow through your neutral conductor to these other circuits instead. This could damage your appliances or even start a fire.
To prevent this from happening, all new houses should be designed to use separate neutrals and grounds. But since this house was built in 1988, its original wiring was done according to code at the time. They used one conductor for both functions because there was no way of connecting them together on the bus bar.
How can you tell which wire is positive and ground?
Recognize that the black wire is positive, the white wire is negative, and the green wire is ground. A copper wire may be used instead of a green wire to connect to the ground. For example, some power outlets have a red wire plus a white wire, instead of the usual red/black or white/green. This red wire is then connected to the metal frame of the outlet to make it harder to plug in multiple devices at once.
If you're working with a four-wire cable (red/white/black/ground), then the term "plus" or "+" applies to all four wires. So, for example, if you were to touch one end of the wire to your body, then it would be the "positive" end. If you touched the other end of the wire to your body, then it would be the "negative" end.
In general, the black wire is positive, the white wire is negative, and the two green wires are ground. However, on some power cables (especially older ones), the order of the colors may be reversed. You should check the manufacturer's documentation for any rules regarding wire coloration.
Also note that some installations may have five wires inside a single sheath: black, red, white, ground, and signal.
Can a ground rod be a neutral wire?
The ground rod does not function as a neutral wire! In most residential circuits, a ground rod is put in the ground and linked to the hot bus. Many electricians believe that when a ground fault occurs, electricity flows through the rod and into the earth. This is incorrect; a ground rod does not provide a path for current to flow when there's a problem with the wiring.
When you're working with old wiring, it's very possible that you will encounter broken or frayed wiring insulation. This can lead to problems with electrical connections, which may cause your appliances to work improperly or even break down completely. If this happens, call an experienced Baltimore electrician to repair the damage so you can get your home back up and running.
Old wiring also poses a risk of fire if there are any defects or openings in the casing where water can enter. Make sure all your wiring is protected from exposure to moisture, sunlight, and heat. The National Electric Code requires that all wiring be covered by at least one-and-a-half inches of material. Any material between the wiring and the outer covering of your house is considered adequate if it provides protection from weather conditions.
If you're lucky enough to have power going into your garage, there's a good chance that it's being fed into a circuit shared with someone else's electricity.
Can I connect a ground wire to another ground wire?
All ground connections must be completed by connecting the ground wires. So, in your case, you are correct. Using a twist-on connector, connect the ground from the supply to the ground from the fixture (or other approved connection).
If you do not connect the grounds, then you might get an electrical fire if there is a difference in voltage between the two grounds. The voltage will be discharged through the person touching the unconnected grounds. This could be serious if it is someone working on a power line right of way or at an industrial site where people can come into contact with ungrounded conductors regularly.
To test your ground wiring for continuity, use a ground-fault circuit interrupter (GFCI) or dedicated ground fault detector. These devices check all ground wires for continuity. If one is not connected, the device will not function properly.
The National Electric Code requires that all circuits supplying electricity directly from the street main contain a ground conductor. This ground conductor cannot be removed from the cable without breaking the code. However, if this ground conductor is not needed, it can be removed. It is recommended that you leave this ground conductor in place, even if you are not using any part of the circuit.
A ground fault occurs when current is leaking from another circuit into yours.
Are neutral wires dangerous?
The grounded power cable is referred regarded as the "neutral" wire since it poses no hazard to exposed metal parts or plumbing. The term "hot" wire refers to how hazardous it is. The grounding of the neutral wire has nothing to do with the operation of electrical equipment but is essential for safety reasons. If a hot wire were not connected to something conductive, it would be very dangerous because it could cause a spark when it gets wet or dirty.
In any house built before 1979, if you're in charge of repairing or replacing wiring, you must ensure that all wires are protected from damage by covering them with non-conductive material. This means installing metal conduit or armor cables for all live wires and neutral wires. Other types of wiring can be made into conductors by using aluminum or steel tape to wrap each line twice around a rod large enough to go through the wall cavity and into the next room. These rods can then be covered with more tape or filled with plaster. The third type of protection available before 1979 was called "continuity testing". This means using a voltmeter to check to see that each conductor is still attached to its terminal after it passes through a junction box. If one conductor comes off the pole or flooring, then it needs to be reattached. If it isn't, then someone might be using it as a ground.
After 1979, these requirements were put in place to prevent accidents caused by worn out or damaged wiring.
Tyrone Biddick
Tyrone Biddick is a mechanic and engineer. He has a degree in mechanical engineering with a minor in business administration. He likes to work with machines, and he is good at fixing them. Tyrone also enjoys working with people and solving problems.
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EsWick.com is a participant in the Amazon Services LLC Associates Program, an affiliate advertising program designed to provide a means for sites to earn advertising fees by advertising and linking to Amazon.com. | 1,516 | 7,471 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.975367 |
https://homework.cpm.org/category/MN/textbook/cc3mn/chapter/2/lesson/2.1.7/problem/2-71 | 1,607,208,935,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141750841.83/warc/CC-MAIN-20201205211729-20201206001729-00126.warc.gz | 324,319,820 | 15,851 | ### Home > CC3MN > Chapter 2 > Lesson 2.1.7 > Problem2-71
2-71.
CALCULATOR CHECK
Use your scientific calculator to compute the value of each expression in the left-hand column below. Match each result to an answer in the right-hand column.
1. $−3 + 16 − \left(−5\right)$
1. $−16$
1. $\left(3 − 5\right)\left(6 + 2\right)$
1. $327$
1. $17\left(−23\right) + 2$
1. $0.5$
1. $5 − \left(3 − 17\right)\left(−2 + 25\right)$
1. $18$
1. $\left(−4\right)\left(−2.25\right)\left(−10\right)$
1. $−90$
1. $-1.5-2.25-\left(-4.5\right)$
1. $0.75$
1. $\frac{4-5}{-2}$
1. $−389$
Remember to be cautious when entering negative signs.
A few matches have been made below, so that you can check some of your work.
$\text{a.}\ \Rightarrow\ 4.$
$\text{f.}\ \Rightarrow\ 6.$
$\text{g.}\ \Rightarrow\ 3.$ | 314 | 800 | {"found_math": true, "script_math_tex": 17, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | longest | en | 0.466679 |
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# Purpose of proof by minimum counter example? watch
1. So, my notes essentially have a proof;
defined are polynomials
let
Let P(n) be the statement
" and proven"
Assume p(n) is not true
Let minimal criminal be 2
Then it shows that; using the definitions given and .
This contradicts the assumption that p(n) is false so p(n) is true.
(some intermediate steps skipped above to make the post shorter).
But what is the point of assuming that p(n) is false just to prove that p(n) is true if it can be proven true without an assumption of the negation with "strong induction " anyway? It seems that the method of induction would make proof by minimum counter example redundant.
2. The two concepts are very closely related. To be honest, I prefer "smallest counterexample" to induction in many ways. We generally spend several hours on explaining how/why induction works, but we usually don't even bother explaining smallest counterexample at all, because it's kind of obvious that it works.
You could argue that it's induction that's the redundant concept...
3. (Original post by DFranklin)
The two concepts are very closely related. To be honest, I prefer "smallest counterexample" to induction in many ways. We generally spend several hours on explaining how/why induction works, but we usually don't even bother explaining smallest counterexample at all, because it's kind of obvious that it works.
You could argue that it's induction that's the redundant concept...
Thanks for the reply. I guess you could argue that, but I just thought about it the other way around because I learnt standard proof by induction first.
Could I ask why you prefer smallest counterexample?
4. (Original post by NotNotBatman)
Thanks for the reply. I guess you could argue that, but I just thought about it the other way around because I learnt standard proof by induction first.
Could I ask why you prefer smallest counterexample?
Say you have some rules, you know are true
By assuming these rules are false, you can show a lot of assertions but they end up implying something that contradicts your original rules.
Therefore the assertions are false, and if everything the rules being false implies is false, then the rules must be true.
5. (Original post by emperorCode)
Say you have some rules, you know are true
By assuming these rules are false, you can show a lot of assertions but they end up implying something that contradicts your original rules.
Therefore the assertions are false.
Ok, so are you saying that it's easier to disproof the negation of "if p(n) then p(n+1)" like in some other standard implications?
This is the only example I've ever looked at of the minimum counterexample, so I haven't come across those types yet.
6. (Original post by NotNotBatman)
Thanks for the reply. I guess you could argue that, but I just thought about it the other way around because I learnt standard proof by induction first.
Could I ask why you prefer smallest counterexample?
I thought I explained it in my previous post. People generally find it easier to understand, you don't have all the formal "boiler plate" to learn (All that "assume true for n=k" stuff) like you do for induction. And you don't have to invoke a "magic" principle ("by the principle of mathematical induction" - it makes me think of He-man saying "by the power of Grey skull"!) to justify the final conclusion.
7. (Original post by NotNotBatman)
Ok, so are you saying that it's easier to disproof the negation of "if p(n) then p(n+1)" like in some other standard implications?
This is the only example I've ever looked at of the minimum counterexample, so I haven't come across those types yet.
Exactly.
You're trying to prove those rules being false makes no sense
8. (Original post by DFranklin)
I thought I explained it in my previous post. People generally find it easier to understand, you don't have all the formal "boiler plate" to learn (All that "assume true for n=k" stuff) like you do for induction. And you don't have to invoke a "magic" principle ("by the principle of mathematical induction" - it makes me think of He-man saying "by the power of Grey skull"!) to justify the final conclusion.
Thank you - had a chuckle because of the he man reference too.
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Help me? AQA Core 2 maths question Watch
1. Could anyone help me with part C of this question? I've been trying to learn about series today and I'm a bit confused, how can you find the sum of a series that doesn't have an end? I've had a look at the mark scheme and am none the wiser
2. hmmmmmm... Is the answer 25.92?
3. (Original post by kellyalisonchow)
hmmmmmm... Is the answer 25.92?
Yeah it is
4. (Original post by lacee_)
Could anyone help me with part C of this question? I've been trying to learn about series today and I'm a bit confused, how can you find the sum of a series that doesn't have an end? I've had a look at the mark scheme and am none the wiser
1. The sum to infinity of a geometric series that converges can be found using the formula
s = a / (1-r)
Where a is the first term and r is the common ratio.
For part C, you simply minus the first 3 terms of the series from your answer to part B
5. it helps if you understand what the notation used in (c) actually means:
it says "add up all the terms of the sequence from the 4th term up to infinity"
Since you have already found the "sum to infinity" - which means from the first term up to infinity - in part (b), what you need to find now is the sum of all the terms less the first three. You are given the first term (48), and how to generate the next term (x0.6) and you have also found the third term in part (a), so you are ready to go!
6. (Original post by lacee_)
Yeah it is
SO... the question wants you to find the sum to infinty start from the 4th term
which would be this
Sum to infinity- sum of the first 3 terms
= ans(b) minus 94.0
=120-94.08
=25.92
7. Much clearer now, I guess I just didn't really understand the notation properly. Thanks everyone
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# Measuring Solubility - PowerPoint PPT Presentation
Measuring Solubility. Chapter 11. Solubility. The solubility of a substance refers to the maximum amount of that substance that can be dissolved in a given quantity of solvent at a certain temperature.
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### Measuring Solubility
Chapter 11
Solubility
• The solubility of a substance refers to the maximum amount of that substance that can be dissolved in a given quantity of solvent at a certain temperature.
• A solution in which no more solute can be dissolved at that temperature is described as a saturated solution.
Solubility
• One way of measuring solubility is to determine the maximum mass of solute that can be dissolved in 100 grams of solvent at a particular temperature.
• Solubility values allow us to compare the extent to which different substances dissolved.
• Look at table 11.1 on page 207 and the worked example on the same page
Solubility Curves
• The relationship between solubility and temperature can be represented by a solubility curve.
• Each point in the solubility curve represents a saturated solution.
• Any point below a curve represents an unsaturated solution for that solute.
Example
• An 80 g sample of NaNO3 is added to 200g of H2O at 20°C. Use the solubility curve in Figure 11.1 to calculate how much more NaNO3 needs to be added to make the solution saturated with NaNO3 at 20°C.
Crystallisation
• You might have noticed that honey often crystallises if you keep it in the refrigerator. The sugar becomes less soluble as the honey cools.
• The sugar that will no longer stay dissolved comes out of solution as crystals.
• This process is known as crystallisation.
Supersaturation
• With some substances it is possible to produce an unstable solution that contains more dissolved solute than in a saturated solution.
• Such a solution is said to be supersaturated.
• Any point above a solubility curve represents a supersaturated solution.
Solubility of gases
• Gases such as oxygen an carbon dioxide are much less soluble in water than solutes such as NaCl and sugars. Why?
• But oxygen and carbon dioxide are present in our oceans and waterways.
• The solubility of a particular gas in a liquid depends on the temperature of the liquid and the pressure of the gas.
Temperature and Gas Solubility
• Unlike most solids, gases become less soluble as the temperature increases.
• When you heat water, small bubbles of air form and escape the water.
Pressure
• Soft drinks contain carbon dioxide.
• It is forced into the cans under high pressure to increase the amount that can be dissolved. When the can or bottle is opened the carbon dioxide can escape.
• This is how drinks get flat after a certain time. As more and more carbon dioxide escapes.
• Look at worked example 11.1d on page 210.
• Page 211
• Question 1
• Question 3
• Question 5
• Question 7
Concentration of solutions
• Before we begin I am just warning you that this is the return of the mole.
• The mole will continue right through til the end of unit 4.
• It is vital you understand the mole, if you are unsure of anything stop me and ask.
• If you are unsure chances are someone else in the class is unsure too.
Concentration of Solutions
• When talking concentrate think of cordial.
• If I pour the same amount of cordial into two glasses but have different amounts of water their concentrations are the same even if their volumes are different.
• Volume and concentration are two different things.
Concentration of Solutions
• The concentration of a solution describes the relative amounts of solute and solvent present.
• A solution in which the ratio of solute to solvent is high is said to be concentrated.
• A solution in which the ratio of solute to solvent is low is said to be dilute.
Concentration of Solutions
• Chemists use different measures of concentration depending on the particular situation.
• Earlier, units of grams of solute per 100 grams of solvent were used to describe the concentration of a saturated solution.
• Other ways of expressing concentration describe the amount of solute in a given amount of solution.
• They vary only in units used to measure the amount of solute and the amount of solutions.
Concentration of Solutions
• For chemists, the most commonly used units for concentration are:
• Mass of solute per litre of solution
• Amount, in mol, of solute per litre of solution. (does this one look familiar)
Mass of solute per litre of solution
• This unit expressed concentration in terms of the mass of solute present in 1L of solution.
• It is important you know how to convert metric units of volume.
mass of sulfate ions (mg)
Concentration =
volume of mineral water (L)
A 250ml glass of orange-flavoured mineral water contains 4.0mg of sulfate ions. What is the concentration (in mg L-1) of sulfate ions in the mineral water?
Solution:
Remember
250 ml is 0.250 L
250/1000
4.0 mg
Concentration =
0.250 L
Concentration = 16 mg/L or 16 mg L-1
Other units
• Other units commonly used to measure volume are the cubic centimetre (cm3), the cubic decimetre (dm3) and the cubic metre (m3).
• Where 1 mL = 1cm3, 1 L = dm3 and
1 KL = 1 m3
• Page 215
• Question 9
Amount, in mol, of solute per litre of solution
• Expressing concentration in moles per litre of solution allows chemists to compare relative numbers of atoms, molecules or ions present in a given volume of solution.
• The measure of concentration, known as molarity or molar concentration, is an important one for chemists.
Molarity (M)
• Molarity is defined as the number of moles of solute particles per litre of solution.
• A one molar (1 M) solution contains one mole of solute dissolved in each litre of solution.
• A concentration of such a solution is said to be one mole per litre, 1 mol L-1 or 1M.
• We use the term molarity to mean ‘concentration measure in moles per litre.
Molarity (M)
• 1.0 L of a 1.0 M solution of ethanol contains 1 mol of C2H5OH
• 1.0 L of a 1.0 M solution of sodium chloride contains 1 mol of NaCl
• 2.0 L or a 0.5 mol solution of sodium chloride contains 1 mol of NaCl
• 0.25 L of a 4.0 M solutions of ammonia contains 1 mol of NH3.
• Each of these solutions contains 1 mole of the solutes dissolved in solution.
Amount, mol
Concentration, mol L-1 or M
The Equation
• The amount fo solute is linked to the concentration (molarity) and volume of the solution by the relationship:
n = c x V
Volume, L
Equation
n = cV
Or
n
n
c =
V =
V
c
Unit converstion
• The concentration units discussed here are g L-1 and mol L-1.
• We must be able to convert from one unit to the other at times.
• Since litres is common to both we are really just converting from grams to mole and vice versa.
• How do we convert from grams to mole again?
Don’t forget molar mass
• Both molarity and molar mass use M at times. Molarity uses it as units where as molar mass uses it as a symbol. Always look carefully to determine which one M means in each question.
÷ M
grams
moles
litre
litre
x M
Worked examples
• Lets do the worked examples together.
• Page 213
• Page 215
• Question 10, 11 and 12
Dilution
• If I don’t like my cordial strong I add more water.
• I am in effect adding more solvent (water) to a solution (cordial).
• This process is known as dilution.
• When a solute is diluted there is still the same amount of solute. The solute particles, however, are more widely spaced.
Dilution
• Because the amount (number of moles) of solute does not change during dilution, a useful mathematical relationship exists.
• Amount of solute before dilution is n1.
n1 = c1V1.
• Amount of solute after dilution is n2.
n2 = c2V2.
• But n1 = n2.
• So:
• c1V1 = c2V2.
Dilution
c1V1 = c2V2
• This is called the dilution formula.
• Since the amount of solute remains unchanged during dilution, so does the mass of the solute.
Page 214 Worked examples. | 2,125 | 8,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-39 | longest | en | 0.88972 |
http://www.velocityreviews.com/forums/t326546-integer-math-question.html | 1,386,625,358,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163999838/warc/CC-MAIN-20131204133319-00089-ip-10-33-133-15.ec2.internal.warc.gz | 587,368,371 | 10,243 | Velocity Reviews > Integer math question
# Integer math question
Frank
Guest
Posts: n/a
01-03-2004
Hi,
can anybody help with the following problem?
In C++
i = 5 / 10 and
i = -5 / 10 both have the same result 0.
In python
i = 5 / 10 gives me 0 as expected, but
i = -5 / 10 gives -1 as result.
Is this a feature or a bug? I remember Delphi gave me the same result as
C++.
TIA,
Frank
John Roth
Guest
Posts: n/a
01-03-2004
"Frank" <(E-Mail Removed)-bremen.de> wrote in message
news:(E-Mail Removed) om...
> Hi,
>
> can anybody help with the following problem?
>
> In C++
>
> i = 5 / 10 and
> i = -5 / 10 both have the same result 0.
>
> In python
>
> i = 5 / 10 gives me 0 as expected, but
> i = -5 / 10 gives -1 as result.
>
> Is this a feature or a bug? I remember Delphi gave me the same result as
> C++.
That's a feature. Integer division is explicitly defined in
Python to do exactly that.
The basic thing to remember is that the *correct*
mathematical result of dividing one integer by
another integer is a rational. Python does not
have a rational data type, so it has to pick one
of the multitude of possible ways of rounding a
non-integral result to an integer.
There is no universally right answer to this process:
the "right" answer to any rounding problem is
what the customer wants it to be.
John Roth
>
> TIA,
> Frank
Sean Ross
Guest
Posts: n/a
01-03-2004
"Frank" <(E-Mail Removed)-bremen.de> wrote in message
news:(E-Mail Removed) om...
> Hi,
>
> can anybody help with the following problem?
>
> In C++
>
> i = 5 / 10 and
> i = -5 / 10 both have the same result 0.
>
> In python
>
> i = 5 / 10 gives me 0 as expected, but
> i = -5 / 10 gives -1 as result.
>
> Is this a feature or a bug? I remember Delphi gave me the same result as
> C++.
>
> TIA,
> Frank
Using the division algorithm:
Let a,b be integers with b>0. Then there exist unique integers q and r such
that:
a = bq + r and 0<=r<b [1]
If we let a=5 and b=10, then a/b is represented by [1] as
a = bq + r
5 = (10) q + r
.... skipping some steps, we find that q=0 and r=5
5 = 10(0) + 5
5 = 0 + 5
5 = 5
LHS = RHS
so, 5/10 = 0 in integer division (with no remainder).
Now, let a = -5 and b = 10
-5 = (10)q + r
If we have q = -1, then
-5 = (10)(-1) + r
-5 = -10 + r
Recall [1] above, 0 <= r < b. r must be nonnegative.
In this case r=5,
-5 = -10 + 5
-5 = -5
LHS = RHS
So, q = -1, and r=5, in which case -5/10 = -1 in integer division (without
remainder).
Suppose we let q=0 for the second problem above. Then
-5 = (10)(0) + r
-5 = 0 + r
-5 = r
or
r = -5
But, by [1], r must be nonnegative. So, we have a contradiction. In which
case we can say that q cannot equal 0 in the algorithm above.
"This is a feature - Python does it properly, where the others do not."
HTH
Sean
Sean Ross
Guest
Posts: n/a
01-03-2004
"Sean Ross" <(E-Mail Removed)> wrote in message
news:11EJb.20461\$(E-Mail Removed) ...
>then a/b is represented by [1] as
'represented' is the wrong word here, but hopefully, you get the idea ...
Sean Ross
Guest
Posts: n/a
01-03-2004
Here's an interactive Python example to help clarify the previous response:
>>> a = 5
>>> b = 10
>>> q , r = divmod(a,b)
>>> q
0
>>> r
5
>>> a = -a
>>> a
-5
>>> q , r = divmod(a,b)
>>> q
-1
>>> r
5
>>> b*q + r # division algorithm
-5
>>> -5 == a
True
>>>
Rainer Deyke
Guest
Posts: n/a
01-03-2004
Sean Ross wrote:
> a = bq + r and 0<=r<b [1]
But 0 <= r < b is a contradiction when b < 0.
--
Rainer Deyke - http://www.velocityreviews.com/forums/(E-Mail Removed) - http://eldwood.com
Sean Ross
Guest
Posts: n/a
01-03-2004
"Rainer Deyke" <(E-Mail Removed)> wrote in message
news:kTEJb.724242\$HS4.5376202@attbi_s01...
> Sean Ross wrote:
> > a = bq + r and 0<=r<b [1]
>
> But 0 <= r < b is a contradiction when b < 0.
>
Right. But, the division algorithm states "Let a,b be integers with b>0"
(which I mentioned in that post).
Sean Ross
Guest
Posts: n/a
01-03-2004
"Rainer Deyke" <(E-Mail Removed)> wrote in message
news:kTEJb.724242\$HS4.5376202@attbi_s01...
> Sean Ross wrote:
> > a = bq + r and 0<=r<b [1]
>
> But 0 <= r < b is a contradiction when b < 0.
>
> --
> Rainer Deyke - (E-Mail Removed) - http://eldwood.com
>
>
Hmm....
>>> a = 5
>>> b = -10
>>> q,r = divmod(a,b)
>>> q
-1
>>> r
-5
>>>
Here, the division algorithm does not apply (b is a negative integer).
Perhaps there's some other theorem for this case?
b<r<=0, when b < 0? I don't know.
Sean Ross
Guest
Posts: n/a
01-03-2004
"Sean Ross" <(E-Mail Removed)> wrote in message
news:5sFJb.20923\$(E-Mail Removed) ...
> "Rainer Deyke" <(E-Mail Removed)> wrote in message
> news:kTEJb.724242\$HS4.5376202@attbi_s01...
> >>> a = 5
> >>> b = -10
> >>> q,r = divmod(a,b)
> >>> q
> -1
> >>> r
> -5
> >>>
>
> Here, the division algorithm does not apply (b is a negative integer).
> Perhaps there's some other theorem for this case?
> b<r<=0, when b < 0? I don't know.
>
I think you're supposed to do something like this
a = bq + r, 0<= r < |b|
5 = (-10)q + r
-5 = -(-10)q - r
-5 = 10q - r
But, then, q would be 0 and r would be 5. <shrug>
Elaine Jackson
Guest
Posts: n/a
01-04-2004
C rounds toward the nearest integer and Python rounds down. The behavior is
consistent in each case.
"Frank" <(E-Mail Removed)-bremen.de> wrote in message
news:(E-Mail Removed) om...
| Hi,
|
| can anybody help with the following problem?
|
| In C++
|
| i = 5 / 10 and
| i = -5 / 10 both have the same result 0.
|
| In python
|
| i = 5 / 10 gives me 0 as expected, but
| i = -5 / 10 gives -1 as result.
|
| Is this a feature or a bug? I remember Delphi gave me the same result as
| C++.
|
| TIA,
| Frank | 1,956 | 5,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2013-48 | longest | en | 0.89868 |
https://repository.mouau.edu.ng/work/view/a-mathematical-model-for-typhoid-fever-infection-efughi-mercy-e-7-2 | 1,723,781,208,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641333615.45/warc/CC-MAIN-20240816030812-20240816060812-00851.warc.gz | 377,181,461 | 8,370 | ## A Mathematical Model For Typhoid Fever Infection:- Efughi Mercy E.
EFUGHI MERCY EBELE | 43 pages (6106 words) | Projects
ABSTRACT
This project work focuses on how mathematical modeling can be used to reduce the rate ol typhoid fever in Michael Okpara University of Agriculture, Umudike. From the data results, it was observed that the rate of infection was fluctuating from 2009 -2012 but increased geometrically in 2013.We discovered that the prevalence of typhoid fever infection in MOUAU is high. Also, it was noticed that the prevalence is high between 15-44 years of age. In a bid to reduce the infectivity rate, an S.E.I.R model was developed. The entire population was divided into four compartments namely: susceptible (S), exposed (E), infective (I) and recoveries (R). The model was solved using ordinary differential equation and numerical approach. The equilibrium points of the model system are presented and their stability is investigated.
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APA
EFUGHI, E (2024). A Mathematical Model For Typhoid Fever Infection:- Efughi Mercy E.. Repository.mouau.edu.ng: Retrieved Aug 16, 2024, from https://repository.mouau.edu.ng/work/view/a-mathematical-model-for-typhoid-fever-infection-efughi-mercy-e-7-2
MLA 8th
EBELE, EFUGHI. "A Mathematical Model For Typhoid Fever Infection:- Efughi Mercy E." Repository.mouau.edu.ng. Repository.mouau.edu.ng, 20 Feb. 2024, https://repository.mouau.edu.ng/work/view/a-mathematical-model-for-typhoid-fever-infection-efughi-mercy-e-7-2. Accessed 16 Aug. 2024.
MLA7
EBELE, EFUGHI. "A Mathematical Model For Typhoid Fever Infection:- Efughi Mercy E.". Repository.mouau.edu.ng, Repository.mouau.edu.ng, 20 Feb. 2024. Web. 16 Aug. 2024. < https://repository.mouau.edu.ng/work/view/a-mathematical-model-for-typhoid-fever-infection-efughi-mercy-e-7-2 >.
Chicago
EBELE, EFUGHI. "A Mathematical Model For Typhoid Fever Infection:- Efughi Mercy E." Repository.mouau.edu.ng (2024). Accessed 16 Aug. 2024. https://repository.mouau.edu.ng/work/view/a-mathematical-model-for-typhoid-fever-infection-efughi-mercy-e-7-2 | 627 | 2,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.749309 |
http://support.sas.com/documentation/cdl/en/hostwin/63047/HTML/default/n0y23hlrqxekj3n1bcltk7ovu0rc.htm | 1,563,502,201,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525973.56/warc/CC-MAIN-20190719012046-20190719034046-00402.warc.gz | 144,433,681 | 5,268 | # IBw.d Informat: Windows
Category: numeric Windows specifics: native floating-point representation See: IBw.d Informat in SAS Formats and Informats: Reference
## Syntax
IBw.d
### Required Arguments
w
specifies the width of the input field.
Default:4
Range:1–8
d
specifies the power of 10 by which to divide the input value. SAS uses the d value even if the input data contain decimal points.
Range:0–10
## Details
For integer binary data, the high-order bit is the value's sign: 0 for positive values, 1 for negative. Negative values are represented in twos-complement notation. If the informat includes a d value, the data value is divided by 10d.
Using the IBw.d informat requires you to understand twos complements and byte-swapped data format.
## Comparisons
The IBw.d informat and the PIBw.d informat give you different results. The IBw.d informat processes both positive and negative numbers and it uses the high-order bit as the sign bit. In contrast, the PIBw.d informat is used only for positive numbers and it does not look for a sign bit. For example, suppose your data contain the following two-byte (byte-swapped) value:
`01 80`
When you read this value using the IB2. informat, the informat looks for the sign bit, sees that it is on, and reads the value as −32,767. However, if you read this value with the PIB2. informat, no sign bit is used, and the result is 32,769.
## Example
Suppose that your data contain the following 6-byte (byte-swapped) value:
`64 00 00 00 00 00`
If you read this value using the IB6. informat, it is read as the fixed-point value 100.0. Now suppose that your data contain the following (byte-swapped) value:
`01 80`
Because the sign bit is set, the value is read as −32,767. | 438 | 1,731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-30 | latest | en | 0.809289 |
http://clocktail.it/creating-and-solving-compound-inequalities.html | 1,618,133,174,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038061820.19/warc/CC-MAIN-20210411085610-20210411115610-00114.warc.gz | 21,211,587 | 16,121 | # Creating And Solving Compound Inequalities
In algebra, inequalities are equations in which the answer can be a range of values instead of only one correct solution. Compound Inequalities Date_____ Period____ Solve each compound inequality and graph its solution. Absolute inequalities can be solved by rewriting them using compound inequalities. 1) x +8 >3. Algebra: Solving Inequalities. Graph compound inequalities 2. 2 Create equations in two or more variables to represent relationships. (3)(-2) < (7)(-2). Write an inequality that has both a minimum and a maximum value from a contextual situation. Create your own worksheets like this one with Infinite Algebra 1. To solve an absolute value inequality, knowledge of absolute values and solving inequalities are necessary. It can either be written as x > -1 and x < 2 or as -1 < x < 2. Solving Compound Inequalities - PowerPoint PPT Presentation. 4 Extension - Solving Compound Inequalities Lesson 3. Creating and Solving Compound Linear Inequalities DRAFT. Get Started. Follow 150 views (last 30 days) Ben on 14 Mar 2012. Create equations and inequalities in one variable and use them to solve problems MACC. 00011142_08444. The word "and" or the logic symbol ∧ can be used to combine separate inequalities into a compound inequality. !Solve and graph the following compound inequalities. If there is no solution, enter. 6 Solving Compound Inequalities. When solving a compound inequality involving AND…. 6-4 Solving Compound Inequalities. If you need a good place to start, try listing some x values that make the compound inequalities true. 4 Solving Compound Inequalities Learning Targets: 1. Compound inequalities can be solved using the same methods that are used to solve single inequalities, the Addition and Multiplication Properties of Inequalities. Solve each compound inequality and graph the solution. Inequalities usually have a lot of solutions—in fact, infinitely many. Solving Compound Inequalities - Concept - Examples with step by step explanation. And best of all they all (well, most!) come with answers. Students will apply properties of real numbers to various situations, graph linear equations and interpret linear models. This lesson will demonstrate how to solve a compound inequality that contains the word "or. Tyrique Rosey. Coal is mainly carbon but it may also contain sulfur compounds, which produce sulfur dioxide when the coal is burned. How do you solve the compound inequality #-20≤-6m-2≤58# and graph its solution?. This 18- piece, 21 question puzzle provides students with practice solving and graphing multi-step & compound inequalities. To make a sign analysis chart, use the key/critical values found in Step 2 to divide the number line into sections. Inequalities on a Number Line; Worksheet. 5 - Compound inequalities; 2. The Fundamental Theorem of Algebra is examined. It is now time to switch gears a little and start thinking about solving inequalities. How To Solve Compound Inequalities Solve compound inequalities in the form of and and express the solution graphically. 1) x +8 >3. Since this problem is "or". Sometimes more than one answer is possible. The graph of a compound inequality involving “OR” is the union of the two simple graphs. The process to solve inequalities is the same as the process to solve equations, which uses inverse operations. 5: Creating and Solving Compound Inequalities 1. If an inequality would be true for all possible values, the answer is all real numbers. Create your own worksheets like this one with Infinite Algebra 2. The flexibility and text book quality of the math worksheets, makes Math-Aids. 3x + 1 ≥ −8 AND 2x − 3 < 5. In algebra, inequalities are equations in which the answer can be a range of values instead of only one correct solution. Graph solutions to two-step inequalities Create and interpret line plots with fractions 5. The rules for an essay contest say that entries can have 500 words with an absolute deviation of at most 30 words. 9 9 customer reviews. Algebraic properties. Write an inequality that has both a minimum and a maximum value from a contextual situation. It also factors polynomials, plots polynomial solution sets and inequalities and more. The best videos and questions to learn about Compound Inequalities. Compound inequalities are inequalities bounded by more than one condition. Create Study Groups. In this algebra worksheet, students solve compound inequalities and graph their answer on a number line. What compound inequality represents the phrase? Graph the solutions. In this Warm Up, I provide the students with two real world examples. Examples: Set up and solve a compound inequality. Write compound inequalities from graphs 3. " New to Sophia? Create an Account. You can represent compound inequalities using words, symbols, or graphs. Compound events: find the number of. This 18- piece, 21 question puzzle provides students with practice solving and graphing multi-step & compound inequalities. PurposeGames Create. Khan Academy: Solving Compound Inequalities Powered by Create your own unique website with customizable templates. 7) 6 £ 6a + 6 £ -308) -13 £ 5 - 3x < 2. First, graph the "equals" line, then shade in the correct area. a) 4r > – 12 and 2r < 10 b) z + 3 ≥ 1 and z – 2 < – 1 c) – 2 < x + 1 < 4 d) 3 < 5x – 2 < 13 EXAMPLE 3: SOLVING COMPOUND INEQUALITIES INVOLVING OR Solve each compound inequality. For this we use compound inequalities, inequalities with multiple inequality signs. Plot an inequality, write an inequality from a graph, or solve various types of linear inequalities with or without plotting the solution set. Teachers and researchers worked in partnership to create AlgebraByExample assignments that:. Unit 9 Quiz 1 Solving Inequalities Name: _____ Hour: _____ The price of a concert ticket is more than 30. Come to Solve-variable. Graphically, you can. Part A: 7≤5%+2<22. inequality: graph: For 12, write an inequality to describe the situation. 5 Creating and Solving Compound Inequalities Objective: We will be able to solve compound inequality and graph the solution set. Compound inequalities involving the word “or” require us to solve each inequality and form the union of each solution set. x-5 ≥ -2 OR x-5 ≤ -6 7. Let’s see a few examples below to understand this concept. In this algebra worksheet, students solve compound inequalities and graph their answer on a number line. , creating social. Download: 4. Creating And Solving Compound Inequalities. Try for free. A compound inequality is an equation with two or more inequalities joined together with either "and" or "or" (for example, and ; or ). If you multiply or divide both sides of an inequality by a negative number, reverse the direction of the inequality sign!. Graphing Linear Inequalities. To solve your inequality using the Inequality Calculator, type in your inequality like x+7>9. Download: 4. This gas is a cause of acid rain. Compound inequalities are two inequalities joined by and or or. Inequalities on the Number Line - 4 Goal: Understand graph of 2 inequalities on the number line. Download Mathleaks app to gain access to the solutions. If you will be needing support with math and in particular with compound inequalities or rational functions come pay a visit to us at Mymathtutors. Tyrique Rosey. Solving an Inequality Step-by-step Lesson- This is a great follow up to the visual inequalities that we saw in early standards. In other words, both statements must be true at the same time. And we know is less than or equal to 22 (which is the same as being less than 23). " New to Sophia? Create an Account. Solving Absolute Value Inequalities, MORE Examples - Example 2. The absolute value of a number is the positive value of the number. Create and use representations to organize, record, and communicate mathematical ideas. In terms of inequalities, we know two things. The graph of a compound inequality involving AND is the intersection, or the overlapping region, of the simple inequality graphs. manner by writing inequalities, accompanying statements describing the solution sets, accompanying graphs, and accompanying compound inequalities. Solving linear inequalities is very much like solving linear equations, with one important difference. Creating and Solving Compound Linear Inequalities DRAFT. This is a graph of a linear inequality How to Graph a Linear Inequality. A compound inequality contains at least two inequalities that are separated by either "and" or "or". Free trial available at KutaSoftware. Elouise is creating a rectangular garden in her. You solve them exactly the same way you solve the linear inequalities shown above , except you do the steps to three "sides" (or parts) instead of only two. More than 370 students went on a field trip. Inequalities joined by the word or are disjunctions. Natively compiled stored procedures are created using CREATE PROCEDURE (Transact-SQL). How do you think we can write this solution using interval n otation?. Solve compound inequalities H. The solution of a compound inequality that consists of two inequalities joined with the word and is the intersection of the solutions of each inequality. Solving and graphing inequalities worksheets pdf printable, inequalities word problems worksheets pdf with solutions and graphing inequalities worksheets for high school students also create great. and Writing Their Answers in. The solution of a compound inequality that consists of two inequalities joined with the word and is the intersection of the solutions of each inequality. Get the free "Solve Inequalities with Absolute Values" widget for your website, blog, Wordpress, Blogger, or iGoogle. Reset Selection. Download: 4. Sieling’s Signature: (A) Level 4 1. 1 Solving One-Step Linear Inequalities 6. The pseudoinverse is one way to solve linear least squares problems. Examples: Set up and solve a compound inequality. Low-rank matrix approximation. 2 Identify and apply the distributive, associative, and commutative properties of real numbers and the properties of equality. Solve linear equations and linear inequalities in one variable, including equations with coefficients represented by letters (literal that are linear in the variables being solved for). IXL brings learning to life with over 200 different algebra skills. When we solve a rational inequality, we will use many of the techniques we used solving linear. First of all, add both sides of the inequality by 2. Free kumon maths worksheets, fraction + worksheet + "least to greatest", online questions and answer of general apptitute, algebra 2 workbook answers. Solve each compound inequality and graph the solution. org Solve Compound Inequalities with “and” Now that we know how to solve linear inequalities, the next step is to look at compound inequalities. In this video the instructor shows how to shade a system of inequalities. When you solve compound inequalities , separate the inequalities and solve each of them separately. In terms of inequalities, we know two things. 5 (Part 1) Create Compound Inequalities from Graphs - Lesson 2. 9x ; m 17 lt (4m 13) 3 Cummulative Review. The graph of a compound inequality involving “OR” is the union of the two simple graphs. You also often need to flip the inequality sign when solving inequalities with absolute values. The compound inequality that I have been given to look at and solve first according to the last name chart for the letter R which is the first letter of my last name is -11≤ -5 + 6x < 13. You're going to see. Create free printable worksheets for linear inequalities in one variable (pre-algebra/algebra 1). Compound Inequalities Little League is a commercially sponsored baseball league for boys and girls. Graphically, you can. Create your own worksheets like this one with Infinite Algebra 2. Writing And Solving 2 Step Inequalities Video #124615 Pre-Algebra Worksheets | Inequalities Worksheets #124616 linear inequalities word problems - YouTube #124617. Compounding/Composition is the way of word building when a word is formed by joining two or more stems to form one word. This lesson will prepare students to access A-CED. What are the solutions of the compound inequality? Graph the solutions. y<2x+1 y≥− 1 3 x+4. Here are a few examples of compound inequalities: x > -2 and x < 5 -2 < x < 5 x < 3 or x > 6. 4 Creating and Solving Inequalities pg 77 Aug 2012:17 PM Page 5/27. com and learn about absolute, dividing polynomials and a large number of additional math subjects. This 18- piece, 21 question puzzle provides students with practice solving and graphing multi-step & compound inequalities. Includes example problems that demonstrate solving compound inequalities and graphing the solution on a number line. Download: 4. 1 Mathematical Practices: 5. When you are asked to solve inequality like x-3 > 0, you are asked to find all possible values of x that make the inequality true. Recognize that inequalities of the form x > c or x < c have infinitely many solutions; represent solutions of such inequalities on number line diagrams. Solve advanced problems in Physics, Mathematics and Engineering. 2 > t 2 > 1. 1) x +8 >3. Solve both parts at the same time by adding 2 to each part. math, inequalities, grade 8, jigsaw Students Collaborate to Solve Compound Inequalities - The website is not compatible for the version of the browser you are using. Write and solve an absolute value inequality that represents the acceptable number of words. Whenever you actually have to have guidance with algebra and in particular with solve compound inequalities with fractions or monomials come visit us at Polymathlove. 4 Solving Compound Inequalities Learning Targets: 1. 5-4 Solving Compound Inequalities using "and" & "or" When considered together, two inequalities form a compound inequality. SOLVING INEQUALITIES 01 Solving an “And” Compound Inequality: 3 −9≤12𝑎 3 −9≥−3 Also Written as 3 −9≤12∧3 −9≥−3 Or Written as-3≤3 −9≤12 6≤3 ≤21 2≤ ≤7 The Common statement is sandwiched between the two inequalities. Be careful solving inequalities, they are harder to solve than equations and require more attention. Referring to point (1,5) #5< or>2(1)+3# #5< or >5# Is false. Get Started. Solving Compound Inequalities. Solving Inequalities with Fractions and Parentheses. Start by solving each inequality: if 2x- 3> 15 then 2x> 18 so x> 9. 4 Use variables to represent quantities in a real-world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities. -7x < 21 -7 -7 x > -3 ! 2. Let’s see a few examples below to understand this concept. In algebra, inequalities are equations in which the answer can be a range of values instead of only one correct solution. Compound states [like (s) (aq) or (g)] are not required. This Solving Compound Inequalities Worksheet is suitable for 8th - 9th Grade. Absolute value inequalities can have one or two variables. Create equations that describe numbers or relationships Topic Lesson Page: Algebra I Solving Equations – Algebra I Solving Inequalities – Compound Inequalities:. Follow the conventions about how to draw the boundary curve (solid line or dotted line) and use a test point to determine the region that is the solution set. 5Creating and Solving Compound Inequalities Essential Question: How can you solve a compound inequality and © Houghton Mifflin Harcourt Publishing Company Compound Inequalities: OR Words Algebra All real numbers less than 2 OR greater than 6 x < 2 OR x > 6 0 1 2. 6 Solving Absolute-Value Equations 6. Which compound inequality represents the number of ounces of clay, c, that Elisa uses to make one vase of either size? Their poll has a margin of error of 4% both above and below the predicted percentage. , creating social. After studying this lesson, you will be able to: Solve inequalities with fractions and parentheses. The word “and” is also known as a conjunction. When solving inequalities using chained notation, it is possible and sometimes necessary to evaluate the terms independently. Purplemath—Solving Linear Inequalities: Introduction and Formatting Inequalities making you feel like a less-than-stellar student? Use this quiz to practice for that day. Matrix-TI89, online graphing calculator printable, free download aptitude book, free Practise Papers & Answers gcse maths linear non calculator. The absolute number of a number a is written as $$\left | a \right |$$ And represents the distance between a and 0 on a number line. Bear in mind that in the tripartite Complete page 35 of Wallace's workbook to practice solving inequalities. Solve |5x + 6| + 4 < 1. From a graph, write a compound inequality to represent the restrictions. He uses dotted lines for lesser than or greater than. Solving Compound Inequalities When solving compound inequalities, we are going to deal with two general cases or types. Sal solves several compound linear inequalities. First solve each inequality separately. We solve compound inequalities using the same techniques we used to solve linear inequalities. a less than) is very different from solving an inequality with a > (i. Solving equations and inequalities worksheet answers free math lessons 9th grade teaching algebra compound independent practice tessshlo completing the square worksheets land 1 business mathematics 5th review for primary classes multi step word problems 2 nidecmege awesome website a must have teacher that does not textbooks school goodin s 5 2018 ink notebook Solving Equations And Inequalities. Solve linear equations in one variable that include simplifying algebraic expressions. Inequalities joined by the word or are disjunctions. Compound Inequalities - AND Compound Inequality: AND: Solving Inequalities is just like solving _____, just be sure to _____ when we _____ by a _____ Example A 6 E5 O11 AND 7 E2 Q44 Example B 11 F10 P3 F2 AND 2 :9 E3 ; F2 R10 E52 Practice A Practice B. The solution is 2 ≤ ≤7,. A compound inequality is made up of two inequalities connected by the word “and” or the word “or. Math Expression Renderer, Plots, Unit Converter, Equation Solver, Complex Numbers, Calculation History. 1 - Create equations and inequalities in one variable and use them to solve problems. Use variables to represent quantities in a real-world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities. WORKSHEETS Regents-Solving Linear Equations 1a. Every point on the dark line is a member of the set {x : x < 7orx≥11}. A compound inequality is a sentence with two inequality statements joined either by the word "or" or by the word "and. To solve a compound inequality, first separate it into two inequalities. Inequalities usually have a lot of solutions—in fact, infinitely many. There are 2 types of compound inequalities, ANDandOR. To be neat, the smaller number should be on the left, and the larger on the. 2 - Creating and Solving Equations; 1. How do you think we can write this solution using interval n otation?. Compound inequalities are two inequalities joined by the word and or by the word or. Then divide each Write the compound inequality using the part by 3. The solution of a compound inequality that consists of two inequalities joined with the word and is the intersection of the solutions of each inequality. Solving Equations and Inequalities. The graph of a compound inequality with an "and" represents the intersection of the graph of the inequalities. 19: 2-3: Solving More Complex Equations: Exercises. When a compound inequality with “and” is written as a single inequality, you can solve the inequality by performing the same operation on each expression. 2 Create equations in two or more variables to represent relationships. 12D - Calculate and compare simple interest and compound interest earnings. July 7 feel free to create and share an alternate functions Compound interest Compound ratio Conjugates Constructions. Graph solutions to compound inequalities Create equations with no solutions or infinitely many solutions D. Solving Equations with variables on both sides # 2. Come to Solve-variable. Practice and Problem Solving: A/B. Quiz by dianaburnham08. Follow the conventions about how to draw the boundary curve (solid line or dotted line) and use a test point to determine the region that is the solution set. You also often need to flip the inequality sign when solving inequalities with absolute values. When a compound inequality with “and” is written as a single inequality, you can solve the inequality by performing the same operation on each expression. Some practical applications need to solve the problem of approximating a matrix M with another matrix. Then divide each Write the compound inequality using the part by 3. 6 Stem-and-Leaf Plots and Mean, Median, and Mode 6. The solutions to compound inequalities can be graphed on a number line, and can be expressed as intervals. Inequalities joined by the word and are called conjunctions. Solve as a single unit or solve each side separately. 6 - Another Lesson on Solving Compound Inequalities Powered by Create your own unique website with customizable templates. Solving Absolute Value Inequalities. 2) You can graph the solutions of a compound inequality involving AND by using the idea of an overlapping region. 5 Compound Inequalities #3 - 01/13/2014 7. Compound inequalities involving the word “and” require the intersection of the solution sets for each inequality. Here are two simple inequalities:x > 2 x is greater than 2. If you continue browsing the site, you agree to the use of cookies on this website. Solve Polynomial and Return Real Solutions Solve Multivariate Equations and Assign Outputs to Structure Solve Inequalities. com solving an equation for x, solvang and santa ynez, solving an inequality in interval notation, solving and writing linear equations, solving an ivp,. Creating and Solving Inequalities Algebraic Models solutions, Houghton Mifflin Harcourt Algebra 1, 2015. In other words, the solution of the compound inequality is a solution of either inequality, not necessarily both. ***Before you set up a positive and negative inequality, make sure every value around the absolute value bars is moved to the other side of the inequality/equation sign. 00011142_08444. Software for math teachers that creates exactly the worksheets you need in a matter of minutes. Math Expression Renderer, Plots, Unit Converter, Equation Solver, Complex Numbers, Calculation History. To solve an inequality, first solve each inequality separately just like you would solve an equation. Solving quadratic equations by quadratic formula. Solving Compound Inequalities. This online quiz is called Solving Compound Inequalities. Since the joining word is "and," this indicates that the overlap or intersection is the desired result. The purpose of this Warm Up is for students to be able to recognize the difference and importance between the words "and", and "or". To this point in this chapter we've concentrated on solving equations. It is important to note that the two methods of reformulation, Convex Hull and Big-M, are necessary for solving problems with disjunctive inequalities. This lesson ensures that students have a solid foundation in the uses and mechanics of single-variable, unbounded inequalities before they study compound inequalities in the next lesson. Nature of the roots of a quadratic equations. Recognize that inequalities of the form x > c or x < c have infinitely many solutions; represent solutions of such inequalities on number line diagrams. ” “And” indicates that both statements of the compound sentence are true at the same time. 7) 6 £ 6a + 6 £ -308) -13 £ 5 - 3x < 2. Solve word problems leading to equations of the form px + q = r and p ( x + q ) = r , where p , q , and r are specific rational numbers. Start by solving each inequality: if 2x- 3> 15 then 2x> 18 so x> 9. Businesses use inequalities to control inventory, plan production lines, produce pricing models, and for shipping. This lesson begins with Truth Tables involving "AND" & "OR" statements. Union of Inequalities Every point on the dark line is a member of the set {x: x < 7orx≥11}. Absolute Value Inequalities The forget the minus sign" definition of the absolute value is useless for our purposes. Start studying Creating and Solving Compound Inequalities. Think about the inequality x > 3. 5 Practice Set. You will need to be aware that every single undergraduate comes into the world different. The powerpoints have been made to go with exercises from 10 Ticks that are indicated on the title screen. One-step: S1. Follow the conventions about how to draw the boundary curve (solid line or dotted line) and use a test point to determine the region that is the solution set. When solving compound inequalities, we use some of the same methods used in solving multi-step inequalities. x ≤ -3/2 or x ≥ 5/2 Example 4. This is fine. y<2x+1 y≥− 1 3 x+4. B w ja clwl t krnisg hdtts 4 yrqeksoe1rqvhe kdzm 1 bmha6dye3 twgilt ah 9 ei ln ofkiqncivtfee ta cl6g 8efbsrka y 21s. 2-1: Writing and Solving Equations: Exercises: p. We solve compound inequalities using the same techniques we used to solve linear inequalities. Solving Compound Inequalities with “And”. In other words, sometimes doing the same thing to both sides of an inequality does not result in the same solution sets (it changes the answer). Solve each inequality Add or subtract rst Divide Dividing by negative ips sign Graph the inequalities separatly above number line. Create free printable worksheets for linear inequalities in one variable (pre-algebra/algebra 1). Includes example problems that demonstrate solving compound inequalities and graphing the solution on a number line. In this problem, we'll move the variables to the left side. If it is unclear whether the inequality is a union of sets or an intersection of sets, then test each region to see if it satisfies the compound inequality. A compound inequality is a special type of inequality that places both an upper and lower boundary on a variable. Tim and Moby risk life and limb to show you how to graph and solve inequalities. Disjunctive inequalities can be a powerful tool for decision making in a variety of fields. Our software turns any iPad or web browser into a recordable, interactive whiteboard, making it easy for teachers and experts to create engaging video lessons and share them on the web. Some practical applications need to solve the problem of approximating a matrix M with another matrix. Chapter Objectives By the end of this chapter, the student should be able to Solve linear equations (simple, dualside variables, infinitely many solutions or no -. Intersection- the graph of a compound inequality containing and the solution is the set of elements. Interval Notation. Download Mathleaks app to gain access to the solutions. Then, they solve and graph their solution on a number line. The following example shows a memory-optimized table and a natively compiled stored procedure used for inserting rows into the table. There are two forms of shapes that this discipline focuses on. But, sometimes we'll have inequalities with three parts. We solve compound inequalities using the same techniques we used to solve linear inequalities. 3 - Activities for teaching Reasoning with Equations & Inequalities, including Reasoning with Equations & Inequalities worksheets, Reasoning with Equations & Inequalities practice problems, questions, assessments, quizzes, tests, lesson plans - aligned to Common Core and state standards - Goalbook Pathways. 5 (Part 2) Unit 1 Practice Quiz Review. A video on compound inequalities and methods for solving them. Once you have solved each part, graph the inequalities on the same number line. 4 Solve and graph simple and compound inequalities in one variable and be able to justify each step in a solution. It takes two steps to solve an equation or inequality that has more than one operation: Simplify using the inverse of addition or subtraction. Solve the linear equations with variables in numerator and denominator, check the solution and determine the conditions of solvability. " Solve each inequality separately. All possible solution values will be located between two defined numbers, and if this is impossible, the compound inequality simply has no solutions. Solving Inequalities with Fractions and Parentheses. MAKE SURE TO WATCH PART 1 FIRST!!!. Represent, solve and interpret equations/inequalities and systems of equations/inequalities algebraically and graphically. Geometry is a branch of math that focuses on shapes and the various properties that are associated with them. ~list the 6 steps used when creating graphs of inequalities ~draw inequalities correctly ~read an inequality graph correctly ~solve one step inequalities ~follow the correct process for solving inequalities ~explain the special rule to follow when dividing by a negative ~use this rule correctly when solving inequalities ~solve multi-step equations. Plot an inequality, write an inequality from a graph, or solve various types of linear inequalities with or without plotting the solution set. PurposeGames Create. Author: Created by Maths4Everyone. You can combine two simple inequalities to create a compound inequality. Solving linear inequalities with division. Solve the linear equations with variables in numerator and denominator, check the solution and determine the conditions of solvability. In this time and age there are very few gender limited roles. 3 Practice Set. A compound statement is formed by combining two or more simple statements. Let’s look now at what happens when we have an absolute value inequality. 3, solving inequalities, and is an extension of the 7th grade standard, 7. com and learn adding and subtracting polynomials, basic mathematics and a great deal of other algebra topics. Then solve each inequality separately. April 13, feel free to create and share an alternate version that worked well for Inequalities Tagged Solving inequalities Post. 2 Solving Inequalities Using Multiplication or Division 6. Write a compound inequality that describes the Little League rule. 1 - Create equations and inequalities in one variable and use them to solve problems. In this video the instructor shows how to shade a system of inequalities. Sometimes the two inequalities will overlap. Compound states [like (s) (aq) or (g)] are not required. This lesson will demonstrate how to solve a compound inequality that contains the word "and. Name: Date: Period: Topic: Solving & Graphing Compound Inequalities Essential Question: How does a compound inequality differ from a regular inequality?. Compounding/Composition is the way of word building when a word is formed by joining two or more stems to form one word. AND inequalities require both statements to be true. This Inequality handout is a good resource for students in the 5th Grade through the 8th Grade. Recognize that inequalities of the form x > c or x < c have infinitely many solutions; represent solutions of such inequalities on number line diagrams. com - Your collection of math tasks. 7) 6 £ 6a + 6 £ -308) -13 £ 5 - 3x < 2. ! Essential Questions. Home About Algebra 1. Algebraic properties. Algebra 1 answers to Chapter 3 - Solving Inequalities - 3-3 Solving Inequalities Using Multiplication or Division - Practice and Problem-Solving Exercises - Page 198 29 including work step by step written by community members like you. Need more problem types? Try MathPapa Algebra Calculator. A detailed description is provided in each math worksheets section. Compound inequalities involving the word “and” require the intersection of the solution sets for each inequality. First of all, add both sides of the inequality by 2. 1 - Create equations and inequalities in one variable and use them to solve problems. Download: 4. Create your lesson! Email: Message: Send. Braingenie | Solving Problems Involving Compound Inequalities: Overview. We explain Solving Compound Inequalities that Contain the Word "Or" with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. The size of the PDF file is 54277 bytes. Home; Main; Kids' TV; Category. A compound inequality is two simple inequalities joined by the word "and" or "or". Equations and Inequalities Involving Signed Numbers. Compound Inequality Match Up Compound Inequalities School Algebra Teaching Algebra Write a compound inequality for the following graph. Write a compound inequality that describes the Little League rule. At the end of this lesson, student will be able to: Create inequalities in one variable and use to solve problems. Exercise 1: On the number lines below, shade in all the values of x that solve the compound inequality. Practice and Problem Solving: A/B. Coal is mainly carbon but it may also contain sulfur compounds, which produce sulfur dioxide when the coal is burned. Tim and Moby risk life and limb to show you how to graph and solve inequalities. Solving Compound Inequalities. When solving compound inequalities, we use some of the same methods used in solving multi-step inequalities. If you continue browsing the site, you agree to the use of cookies on this website. Notice that a compound inequality has three pieces to it, rather than the usual two sides of an equation or simple inequality. We solve each inequality separately and then consider the two solutions. Graph solutions to compound inequalities Create bar graphs, line graphs and. This results in a parabola when plotting the inequality on a coordinate plane. By using this website, you agree to our Cookie Policy. You will need to be aware that every single undergraduate comes into the world different. Graphing Linear Inequalities; Chapter Review; Chapter Test; Chapter 7: Systems of Linear Equations. In he wants to donate at least40. The set of all values which satisfy either inequality is the set of all points which satisfy one or the other or both--this includes the overlap. Solve an "And" Compound Inequality Solve 10 3y - 2 < 19. 7 Solving Absolute-Value Inequalities 6. Solve linear. Solving inequalities using a number line To aid the process of solving an inequality it is important to provide a clear writing frame to show how the inequality remains balanced as it is simplified. Create and Solve Inequalities - Lesson 2. Graph compound inequalities. 5 Your Turn Solve each compound inequality and graph the solutions. k H vMNaFdqeV UwYi0t0h1 RISnrfRiKnai4tFek. July 7 feel free to create and share an alternate functions Compound interest Compound ratio Conjugates Constructions. 2 - Create equations in two or more variables to represent relationships between quantities; graph equations on. Disjunctive inequalities can be a powerful tool for decision making in a variety of fields. Solving Compound Inequalities. We solve compound inequalities using the same techniques we used to solve linear inequalities. This activity allows students to graph solutions to compound inequalities and also shows the solutions. The "graph" is the empty set. How to solve linear equations, inequalities, and absolute value equations. When graphing a compound in equality graph each inequality separately and then follow the rules for and and or problems. Unions allow us to create a new set from two that may or may not have elements in common. Union of Inequalities. The graph of a compound inequality with an "and" represents the intersection of the graph of the inequalities. When creating and solving compound inequalities, remember to use the SELFIE checklist. Solving Compound Inequalities Involving AND. feel free to create and share an alternate version that worked well for your class following the guidance here. The class then turns its efforts to the Collaborative Work: Creating and Solving Equations and Inequalities. You can, however, solve for the real roots where x < 0 and t > 0 via. April 13, feel free to create and share an alternate version that worked well for Inequalities Tagged Solving inequalities Post. To solve a literal equation for one letter in terms of the others follow the same steps as in chapter 2. (a) and List some values: (b) List some values:. Steps for Graphing the Solutions to Inequalities with One Variable ** Make sure the variable is on the LEFT in all solutions. Inequalities are arguably used more often in "real life" than equalities. Solving Compound Inequalities online quiz. Inequalities joined by the word or are disjunctions. Our math worksheets are free to download, easy to use, and very flexible. Solving quadratic equations by completing square. Solving Compound Inequalities You can solve a compound inequality by solving two inequalities separately. Unit 9 Quiz 1 Solving Inequalities Name: _____ Hour: _____ The price of a concert ticket is more than $30. You can use parenthesis () or brackets []. Graph the solution set on a number line. ) If you multiply or divide by a negative number you must switch the sign. Graphing compound inequalities worksheet answers. This page provides educational supplement to practice solving and graphing compound inequalities. If the inequality states something untrue there is no solution. The methods used to solve linear inequalities are similar to those used to solve linear equations. k H vMNaFdqeV UwYi0t0h1 RISnrfRiKnai4tFek. 1 - Modeling with Expressions; Module 2 - "Are you ready?" 1. Unit 9 Quiz 1 Solving Inequalities Name: _____ Hour: _____ The price of a concert ticket is more than$30. 1) x +8 >3. A 3 UAilgl5 srhitgWhJtrsI Mr6ehsmexrcv0ewdU. If you continue browsing the site, you agree to the use of cookies on this website. Solve each inequality, and graph the solution on a number line. Include equations arising from linear and quadratic functions, and simple rational and exponential functions. Day 4 - Solving Compound Inequalities Warm – Up Directions: Solve each inequality and graph the solution. Chapter Objectives By the end of this chapter, the student should be able to Solve linear equations (simple, dualside variables, infinitely many solutions or no -. Solve the linear equations with variables in numerator and denominator, check the solution and determine the conditions of solvability. This is TRUE. 8/12: Solving and Graphing Compound Inequalities. 2 Create equations in two or more variables to represent relationships. Low-rank matrix approximation. Compound Inequalities - AND Compound Inequality: AND: Solving Inequalities is just like solving _____, just be sure to _____ when we _____ by a _____ Example A 6 E5 O11 AND 7 E2 Q44 Example B 11 F10 P3 F2 AND 2 :9 E3 ; F2 R10 E52 Practice A Practice B. 2 > t 2 > 1. Graph solutions to compound inequalities Create bar graphs, line graphs and. Under review Under Consideration Equations and Inequalities Reasoning with Equations and Inequalities Solve systems. Solve Polynomial and Return Real Solutions Solve Multivariate Equations and Assign Outputs to Structure Solve Inequalities. Note that compound or tripartite inequalities are very similar to linear inequalities, except for the way they are structured. No bickering here. org Solve Compound Inequalities with “and” Now that we know how to solve linear inequalities, the next step is to look at compound inequalities. To link to this page, copy the following code to your site. WORKSHEETS: AI: Regents-Direct Variation 1a IA/A MC: 2/17: TST PDF DOC TNS: Regents-Direct Variation 1b IA/A bimodal: TST PDF DOC: Regents-Modeling Linear. Here are two simple inequalities:x > 2 x is greater than 2. Create and use representations to organize, record, and communicate mathematical ideas. Detailed Description for All Inequalities Worksheets. Everything we’ve learned about solving inequalities still holds, but we must consider how the absolute value impacts our work. solving systems of inequalities by graphing worksheets, algebra 1 inequalities worksheets printable and graphing inequality worksheets are three of main things we want to show you based on the gallery title. Compound inequalities are two inequalities joined by the word and or by the word or. Solve linear or quadratic inequalities with our free step-by-step algebra calculator. 5 Your Turn Solve each compound inequality and graph the solutions. Please update your bookmarks accordingly. 8 Lesson 7 Write and Graph Inequalities 27 Main Idea Write and graph inequalities. Graphing and Solving Quadratic Inequalities. Math Online Lesson 7 6. A compound inequality consists of two inequalities that are joined together by the word "and" or the word "or". Through repeated reasoning, students develop fluency in writing, interpreting, and translating between various forms of linear equations and inequalities and make conjectures about the form that a linear equation might take in a solution to a problem. 4 Practice Set. In this final section of the Solving chapter we will solve inequalities that involve absolute value. Compound Inequalities joined by the word "and" represent intersections (where both. com and learn adding and subtracting polynomials, basic mathematics and a great deal of other algebra topics. It is important to note that the two methods of reformulation, Convex Hull and Big-M, are necessary for solving problems with disjunctive inequalities. Home Semester 1 > > > > > > Semester 2 > Pacing Guide About Flipped Mastery. Solving Compound Inequalities online quiz. 1 Create equations and inequalities in one variable and use them to solve problems. If there is no solution, enter. com solving an equation for x, solvang and santa ynez, solving an inequality in interval notation, solving and writing linear equations, solving an ivp,. 5 - Compound inequalities; 2. Practice: Solve each compound inequality and graph its solution. Every point on the dark line is a member of the set {x : x < 7orx≥11}. Displaying all worksheets related to - Creating And Solving Compound Inequalities. 5 Chapter Test . Solving Absolute Value Inequalities. Inverse operations. Students extend their understanding of solving linear equations, inequalities, and systems to include all the different function types mentioned in the standards. I also need to write solution set in interval notation and. Download: 4. Solving Compound Inequalities Involving And; Create an account to see this video. Isolate the absolute value. 1 - Modeling with Expressions; Module 2 - "Are you ready?" 1. You're going to see. Solving and graphing inequalities worksheets pdf printable, inequalities word problems worksheets pdf with solutions and graphing inequalities worksheets for high school students also create great. The following example shows a memory-optimized table and a natively compiled stored procedure used for inserting rows into the table. Absolute inequalities can be solved by rewriting them using compound inequalities. Now multiply both sides by −(1/5). What compound inequality represents the phrase? Graph the solutions. Solve compound inequalities containing the word and then graph the solution. In this problem, we'll move the variables to the left side. , in R2013b. 00011142_08444. Solving "OR" Compound Inequalities; Lesson 4. all real numbers that are greater than –8 and less than 8. 1 Quiz Review; 2. Compound inequalities involving the word “or” require us to solve each inequality and form the union of each solution set. Try to complete this exercise before watching the video in this subunit. Example 1: Compound Inequalities. 3,774,498,432 quizzes played. The graph of a compound inequality with an "and" represents the intersection of the graph of the inequalities. A compound statement is formed by combining two or more simple statements. Absolute Value Inequalities The forget the minus sign" definition of the absolute value is useless for our purposes. to keep the equation or inequality balanced. Compound inequalities are two inequalities joined by the word and or by the word or. 2: Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. Free inequality calculator - solve linear, quadratic and absolute value inequalities step-by-step This website uses cookies to ensure you get the best experience. Multiplying and Dividing Inequalities by Negative Numbers The main situation where you'll need to flip the inequality sign is when you multiply or divide both sides of an inequality by a negative number. The solution is 2 ≤ ≤7,. The second type of compound inequality is an AND inequality. –8 < x < 8. Here are two simple inequalities:x > 2 x is greater than 2. Properties of Inequality Handout This Inequality Worksheet will create a handout for the properties of inequalities. Plot an inequality, write an inequality from a graph, or solve various types of linear inequalities with or without plotting the solution set. Create equations and inequalities in one variable and use them to solve problems. Linear Equations and Inequalities Plotting points Slope Graphing absolute value equations Percents Percent of change Markup, discount, and tax Polynomials Adding and subtracting Dividing Multiplying Naming Quadratic Functions Completing the square by finding the constant Graphing Solving equations by completing the square Solving equations by. 8/12: Solving and Graphing Compound Inequalities. HW: pg 8 #10-16 evens. Practice problems (one per topic). Graph compound inequalities 2. The levels are classified based on the number of steps that required solving compound inequalities. The questions are carefully selected so that they help students to develop their understanding of inequalities that have both an upper and lower bound. Students work on the problems in small groups (3-5 students) and I am rotating between groups. Starts tomorrow ; Those who scored low on. 03 Compound Inequalities This task requires you to create a graph. ) If you multiply or divide by a negative number you must switch the sign. Solving Inequalities With Fractions Worksheet Pdf. So yep, they're equivalent. Download: 4. This activity allows students to graph solutions to compound inequalities and also shows the solutions. But, sometimes we'll have inequalities with three parts. Solve Absolute Value Inequalities with “Less Than”. Solve for y:. The "graph" is the empty set. Solve Compound Inequalities. A compound inequality contains at least two inequalities that are separated by either "and" or "or". You may choose which kind of inequality to utilize from the difficulties. Those that exist (child bearing etc) are. Part A: Using the graph above, create a system of inequalities that only contain points D and E in the overlapping shaded regions. Download Mathleaks app to gain access to the solutions. There are two different types we need to understand. First, let’s talk about the AND, also called an INTERSECTION, where both inequalities are true at the same time. x ≤ -3/2 or x ≥ 5/2 Example 4. 2) 2x-4 8and-3x+2<3 Write an inequality for each word problem and use it to solve. The graph of a compound inequality with an "or" represents the union of the graphs of the inequalities. You solve inequalities by making it easier for the individuals who have less to overcome their circumstances. Quiz by dianaburnham08. When we solve a rational inequality, we will use many of the techniques we used solving linear. Solving Compound Inequalities online quiz. Commented: filston Rukerandanga on 14 Jul 2020. Also, you have to be careful about what sort of inequality operator you are dealing with (<, >, <=, >=). Try to complete this exercise before watching the video in this subunit. Pre-Algebra Worksheets. Solving and Graphing Linear Inequalities Simple Inequalities (add/subtraction) Simple Inequalities (multi/division) Two-Step & Multi-Step Compound Inequalities Special Cases of Compound Ineq Graphing Linear Ineq in Slope-Intercept Form Solving Systems of Inequalities. Solving Compound and Absolute Value Inequalities 6. –8 < x < 8. 5 Solving Compound Inequalities Involving 'Or' 6. Conduct your Individuals Will Need Solving And Graphing Inequalities Worksheet Answers? Get Acquainted with the College Students Primary! Creating a teacher-student romantic relationship is in all likelihood one of the best factors a tutor might have. Compound inequalities is when there is two inequality signs. Compound inequalities 4 | Linear inequalities | Algebra I | Khan Academy. Can you name the answer to these compound inequalities? Test your knowledge on this science quiz and compare your score to others. 5: Creating and Solving Compound Inequalities. Then divide each Write the compound inequality using the part by 3. RESULTS Students use the absolute deviation and mean price to create an absolute value inequality. Create equations and inequalities in one variable and use them to solve problems. Create your own worksheets like this one with Infinite. Create Study Groups. Represent, solve and interpret equations/inequalities and systems of equations/inequalities algebraically and graphically. 5-4 Solving Compound Inequalities using "and" & "or" When considered together, two inequalities form a compound inequality. Don't fret, any question you may have, will be answered. 3 Solving Compound Inequalities 6. Include and or or in the solution. | 10,659 | 49,443 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2021-17 | latest | en | 0.917604 |
http://www.ask.com/question/how-many-edges-does-a-cylinder-have | 1,397,977,320,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609538022.19/warc/CC-MAIN-20140416005218-00257-ip-10-147-4-33.ec2.internal.warc.gz | 300,797,361 | 17,869 | # How Many Edges Does a Cylinder Have?
A cylinder is circular and circular objects do not have edges.
A cylinder has 2 bases and 1 curved surface, no vertices and technically does not have any edges.
Because a cylinder has a curved surface, it is technically not a polyhedron. The terms "faces" and "edges" only apply to polyhedra which have flat surfaces, straight edges and vertices.
The number of edges on a cylinder depends on the edge, which is usually defined as a straight line or edge. In this case, a cylinder has no edges, so the answer is zero. You can find more information here: http://mathforum.org/library/drmath/view/54701.html
Q&A Related to "How Many Edges Does a Cylinder Have?"
An edge in math is basically a corner more or less. Since a cylinder does not have a single corner, we can conclude that a cylinder does not have any edges. http://answers.ask.com/Science/Mathematics/how_man... | 208 | 910 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-15 | longest | en | 0.928337 |
https://se.mathworks.com/matlabcentral/answers/1699105-circular-cross-correlation-2d-using-fft | 1,657,203,530,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104692018.96/warc/CC-MAIN-20220707124050-20220707154050-00683.warc.gz | 537,200,564 | 23,384 | # Circular Cross Correlation 2D Using FFT
8 views (last 30 days)
Ziv Blum on 18 Apr 2022
Answered: David Goodmanson on 18 Apr 2022
Hello,
How to do Circular Cross Correlation In 2D Using fft ?
I checked this code for vectors and it seems that it works.
Z=ifft2(fft2(X1).*conj(fft2(X2)));
Is this code works also for matrix ?
Dims:
X1 : [m,n]
X2: [m,n]
Z: [m,n]
Thank You.
David Goodmanson on 18 Apr 2022
Hi Ziv,
yes, it works similarly.
m = 5;
n = 6;
q = 4;
A = randi(m,n,q);
B = randi(m,n,q);
C = zeros(size(A));
for j = 1:n
for k = 1:q
C(j,k) = sum(circshift(A,[1,1]).*circshift(B,[j,k]),'all'); % [1 1] adjusts the indexing
end
end
C
ifft2(fft2(A).*conj(fft2(B))) % agrees with C | 261 | 684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.743053 |
https://whatisconvert.com/424-gallons-in-imperial-pints | 1,586,297,955,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371806302.78/warc/CC-MAIN-20200407214925-20200408005425-00041.warc.gz | 755,202,236 | 7,006 | # What is 424 Gallons in Imperial Pints?
## Convert 424 Gallons to Imperial Pints
To calculate 424 Gallons to the corresponding value in Imperial Pints, multiply the quantity in Gallons by 6.661393505188 (conversion factor). In this case we should multiply 424 Gallons by 6.661393505188 to get the equivalent result in Imperial Pints:
424 Gallons x 6.661393505188 = 2824.4308461997 Imperial Pints
424 Gallons is equivalent to 2824.4308461997 Imperial Pints.
## How to convert from Gallons to Imperial Pints
The conversion factor from Gallons to Imperial Pints is 6.661393505188. To find out how many Gallons in Imperial Pints, multiply by the conversion factor or use the Volume converter above. Four hundred twenty-four Gallons is equivalent to two thousand eight hundred twenty-four point four three one Imperial Pints.
## Definition of Gallon
The gallon (abbreviation "gal"), is a unit of volume which refers to the United States liquid gallon. There are three definitions in current use: the imperial gallon (≈ 4.546 L) which is used in the United Kingdom and semi-officially within Canada, the United States (liquid) gallon (≈ 3.79 L) which is the commonly used, and the lesser used US dry gallon (≈ 4.40 L).
## Definition of Imperial Pint
The pint (symbol: pt) is a unit of volume or capacity in both the imperial and United States customary measurement systems. The imperial pint is equal to one-eighth of an imperial gallon. One imperial pint is equal to 568.26125 millilitres (≈ 568 ml).
### Using the Gallons to Imperial Pints converter you can get answers to questions like the following:
• How many Imperial Pints are in 424 Gallons?
• 424 Gallons is equal to how many Imperial Pints?
• How to convert 424 Gallons to Imperial Pints?
• How many is 424 Gallons in Imperial Pints?
• What is 424 Gallons in Imperial Pints?
• How much is 424 Gallons in Imperial Pints?
• How many uk pt are in 424 gal?
• 424 gal is equal to how many uk pt?
• How to convert 424 gal to uk pt?
• How many is 424 gal in uk pt?
• What is 424 gal in uk pt?
• How much is 424 gal in uk pt? | 536 | 2,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-16 | latest | en | 0.835123 |
https://blogs.mathworks.com/graphics/2015/12/31/interactive-graph-layout/?from=kr | 1,701,847,684,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.31/warc/CC-MAIN-20231206063543-20231206093543-00871.warc.gz | 166,890,447 | 26,487 | Mike on MATLAB GraphicsGraphics & Data Visualization
Note
Mike on MATLAB Graphics has been archived and will not be updated.
Interactive Graph Layout
I really like the new graph visualization functions that were introduced in R2015b. I particularly like the various options for laying out the graph so you can get a nice picture quickly and easily. But sometimes I like to be able to adjust the layout myself. The documentation for the graph functions don't provide much information about how to do this, but it's actually quite easy. Let's look at an example.
Here's a graph I really like. It comes from a problem called the square-pairs problem (although there's also another famous problem with that name involving primes). It creates nodes for several small integers, and then connects pairs of nodes with an edge if they sum to a square number.
function g = square_pairs(n)
s = [];
e = [];
for i=1:n
for j=(i+1):n
a = i + j;
q = round(sqrt(a));
if q^2 == a
s = [s, i];
e = [e, j];
end
end
end
g = graph(s,e);
g = square_pairs(24);
h = plot(g);
The automatic layout does a pretty good job with that, but it could be better. I know that this graph is planar, but there are some edge crossings in this layout. I'd like to get rid of them. There are other layout options, such as 'circle' ...
layout(h,'circle')
... or layered.
layout(h,'layered')
But none of them are really better than the one that plot chose automatically.
layout(h,'auto')
If I want a better result, I'll have to lay the graph out by hand. I could do that by setting the XData and YData until I find values that look good, but I'd really like an interactive way to place the nodes.
It's actually pretty easy to do this. We'll just need to write a function that uses the position of the mouse to set the XData and YData. Let's walk through how I would write a function like this.
I'm going to call it edit_graph, and I'm going to use it like this:
set(gcf,'WindowButtonDownFcn',@(f,~)edit_graph(f,h))
The WindowButtonDownFcn is a property on each of the figure. If you set that to a function handle, the function will get called everytime you click on that figure. It takes two arguments, but we're only going to use the first one. That's a handle to the figure we've clicked on. We'll also need a handle to the graphplot object, so I pass that into edit_graph as a second argument.
The first thing we need to do is figure out where we've clicked. That's stored in a property on the axes named CurrentPoint. We can use the ancestor function to get a handle to the axes.
a = ancestor(e.Source,'axes');
pt = a.CurrentPoint(1,1:2);
Now we need to find the node in the graph that is closest to where we've clicked. That's the one we're going to move.
dy = h.YData - pt(2);
len = sqrt(dx.^2 + dy.^2);
[lmin,idx] = min(len);
Now lmin is the distance from where I clicked to the nearest node, and idx is the index of that node.
Next we check lmin. If it's large, then we didn't click near a node at all. In that case, we don't want to do anything.
tol = max(diff(a.XLim),diff(a.YLim))/20;
if lmin > tol || isempty(idx)
return
end
Otherwise we can just use idx as the node. We do need to worry about the case where the mouse is equidistant from multiple nodes, so we'll use idx(1) to grab the first of these.
node = idx(1);
Now we're ready to start moving the node!
To do that, we'll need to add two more callback functions to the figure. One will get called each time the mouse moves to a new position. That's called WindowButtonMotionFcn, and it's the one which will actually change the graph. The other function will get called when I let go of the mouse button. It's called WindowButtonUpFcn, and we'll use that to turn the other one off.
f.WindowButtonMotionFcn = @motion_fcn;
f.WindowButtonUpFcn = @release_fcn;
I'm going to do these as nested functions. That means that I'm going to declare them inside the edit_graph function. That will make it easy for them to share information.
Let's write the harder one first. It looks like this:
function motion_fcn(~,~)
newx = a.CurrentPoint(1,1);
newy = a.CurrentPoint(1,2);
h.XData(node) = newx;
h.YData(node) = newy;
drawnow;
end
The first two lines get the current location of the mouse. The next two lines stick those values into the XData and YData arrays at the index we found earlier. Then we call drawnow to tell the graphics system that we'd like to see the result of this change.
And then we just need to write release_fcn. That's pretty easy. It just sets these two function handles to empty.
function release_fcn(~,~)
f.WindowButtonMotionFcn = [];
f.WindowButtonUpFcn = [];
end
And that's it. Now I can start dragging the nodes around to adjust the layout. Here's what I came up with.
You can get the completed edit_graph function here if you'd like to try to create a better layout of this graph. You could also use it to see if square_pairs(30) is also planar. And of course, you could use it to fiddle the layout of one of your own graphs.
Published with MATLAB® R2015b | 1,244 | 5,047 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-50 | latest | en | 0.939834 |
https://blog.csdn.net/qq_16334327/article/details/79965162 | 1,547,595,040,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583656530.8/warc/CC-MAIN-20190115225438-20190116011438-00590.warc.gz | 467,688,720 | 30,663 | # LeetCode小结
RE-RegularExpresstion规则化表达
Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'.
'.' Matches any single character.
'*' Matches zero or more of the preceding element.
The matching should cover the entire input string (not partial).
Note:
• s could be empty and contains only lowercase letters a-z.
• p could be empty and contains only lowercase letters a-z, and characters like . or *.
Example 1:
Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".
Example 2:
Input:
s = "aa"
p = "a*"
Output: true
Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".
Example 3:
Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".
Example 4:
Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".
Example 5:
Input:
s = "mississippi"
p = "mis*is*p*."
Output: false
Example1：P=a，意味着s中只能是a；
Example2：P=a*，意味着s可以是任意个a，包括0个a；
Example3：P='.*'，*代表是对.的重复，而不是对aaaa的重复；
Example4：P='c*a*b'，代表s可以是c任意个+a任意个+b，这里的任意个都可以包括0个，但必须有一个b；
Example5：P='mis*is*p*.'，mis*是对s的任意重复，而不是对mis的任意重复。
• 广告
• 抄袭
• 版权
• 政治
• 色情
• 无意义
• 其他
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# Advanced Mathematics
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The Jason Society: I found the dual basis for each of the following bases for R^3: (a) {(1,0,0),(0,1,0),(0,0,1)}, (b) {(1,-2,3),(1,-1,1),(2,-4,7)}. -Mark R. Rowe
I computed the quadratic form of the polynomial, 2(x1)^(2) + 10x1x2 + 2(x2)^2 And then in the next series of steps, (I took) I determined whether or not matrix A (represents the given polynomial) is "positive definitive" or not positive definitive. This is photo 1 of 4 photos. -Mark R. Rowe http://illuminati1.us/ #MarkRoweBlog #jj #roths #computing #ComputingQuadraticForms #advancedLinearAlgebra #advancedEngineering #advancedPhysics #computeralgebrasystems #mathModeling #fortran #matlab #p
I computed the quadratic form of the polynomial, 2(x1)^(2) + 10x1x2 + 2(x2)^2 And then in the next series of steps, (I took) I determined whether or not matrix A (represents the given polynomial) is "positive definitive" or not positive definitive. This is photo 2 of 4 photos. -Mark R. Rowe http://illuminati1.us/ #MarkRoweBlog #jj #roths #computing #ComputingQuadraticForms #advancedLinearAlgebra #advancedEngineering #advancedPhysics #computeralgebrasystems #mathModeling #fortran #matlab #p
I computed the quadratic form of the polynomial, 2(x1)^(2) + 10x1x2 + 2(x2)^2 And then in the next series of steps, (I took) I determined whether or not matrix A (represents the given polynomial) is "positive definitive" or not positive definitive. This is photo 3 of 4 photos. -Mark R. Rowe http://illuminati1.us/ #MarkRoweBlog #jj #roths #computing #ComputingQuadraticForms #advancedLinearAlgebra #advancedEngineering #advancedPhysics #computeralgebrasystems #mathModeling #fortran #matlab #p
I computed the quadratic form of the polynomial, 2(x1)^(2) + 10x1x2 + 2(x2)^2 And then in the next series of steps, (I took) I determined whether or not matrix A (represents the given polynomial) is "positive definitive" or not positive definitive. This is photo 4 of 4 photos. -Mark R. Rowe http://illuminati1.us/ #MarkRoweBlog #jj #roths #computing #ComputingQuadraticForms #advancedLinearAlgebra #advancedEngineering #advancedPhysics #computeralgebrasystems #mathModeling #fortran #matlab #p
-Mark R. Rowe | #MarkRoweBlog | http://illuminati1.us/ | 701 | 2,440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2023-23 | latest | en | 0.784391 |
http://www.zentralblatt-math.org/zmath/en/search/?q=an:1063.60106&format=complete | 1,369,460,272,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705559639/warc/CC-MAIN-20130516115919-00015-ip-10-60-113-184.ec2.internal.warc.gz | 824,319,095 | 4,840 | Language: Search: Contact
Zentralblatt MATH has released its new interface!
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Zbl 1063.60106
Cairns, Ben; Pollett, P.K.
Extinction times for a general birth, death and catastrophe process.
(English)
[J] J. Appl. Probab. 41, No. 4, 1211-1218 (2004). ISSN 0021-9002
The birth, death and catastrophe process $X(t), t\geq 0$, is a continuous-time Markov chain taking nonnegative integer values. The process $X(t)$ is interpreted as the number of individuals in the population at time $t$ and evolves in time according transition rates $q_{ij}=f_i\sum_{k\geq i}d_k,$ when $j=0, i\geq 1$; $=f_i d_{i-j},$ when $j=1,2, \ldots, i-1, i\geq 2$; $=-f_i,$ when $j=i, i\geq 1$; $=f_i a,$ when $j=i+1, i\geq 1,$ and $=0,$ otherwise, where $f_i>0$ is the rate at which the population size changes when there are $i$ individuals present; when a change occurs, it is a birth with probability $a>0$ or a catastrophe of size $k$ with probability $d_k, k\geq 1$. Theorem 1 proves that $X(t)$ is nonexplosive (i.e. it cannot reach infinity in a finite time) iff $\sum_{i=1}^{\infty}1/f_i=\infty$ or $\sum_{i=1}^{\infty}i d_i\geq a$. Theorem 2 considers the case when $X(t)$ is subcritical and provides (1) necessary and sufficient conditions which ensure that the expected extinction time is finite; (2) the explicit expression for the expected extinction time. Theorem 2 is illustrated by several examples. In particular, the authors point out explicit expressions for the expected extinction times when (a) $f_i=\rho i, \rho>0$ (the linear case); (b) the catastrophe size has a geometric distribution.
[Aleksander Iksanov (Kiev)]
MSC 2000:
*60J27 Markov chains with continuous parameter
92B05 General biology and biomathematics
60J80 Branching processes
Keywords: catastrophe process; persistence time; hitting time
Cited in: Zbl 1154.60064
Highlights
Master Server | 610 | 2,033 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2013-20 | latest | en | 0.735991 |
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## Autodesk Revit MEP
Active Member
Posts: 9
Registered: 09-01-2005
# Calculation with dgC - again
223 Views, 6 Replies
04-10-2012 10:38 PM
Revit MEP 2013 has the same problem as RME2012:
dgC calc. RME2012
Is it really so difficult to fix this bug?
Employee
Posts: 1,278
Registered: 11-16-2005
# Re: Calculation with dgC - again
04-11-2012 04:02 AM in reply to: mrimmer
This may appear to be a bug due to the 'oddity' of the results, but it is actually a request for new functionality.
Revit always 'converts' temperatures based on their values, not based on their differences. This has an impact in how we deal with temperatures in formulae in families.
For example, 80°F - 55°F = 25°F
When converting from °F to °C, the formula is:
°C = (5 / 9) * (°F - 32)
However, if you were to convert the value of 25°F, you would end up with the value -3.9°C.
For a temperature difference conversion, what we need is a conversion like:
Delta°C = (5 / 9) * Delta°F
However, Revit doesn't have such a conversion... however, there is another way.
When performing subtraction on two temperatures (Val1 - Val2), Revit is essentially doing the following:
Val1 - (Val2 + AbsoluteZero)
AbsoluteZero = -459.67°F = -273.15°C
In the case of our example:
80 °F - 55 °F is translated to 80 °F - (55 °F + 459.67°F) = - 434.67°F
25 °C - 10 °C is translated to 25 °C - (10 °C + 258.15°C) = - 258.15°C
To counter act this, we add 0 to the formula:
80 °F - 55 °F + 0°Fis translated to 80 °F - (55 °F + 459.67°F) + 459.67°F = 25°F
25 °C - 10 °C + 0°C is translated to 25 °C - (10 °C + 258.15°C) + 258.15°C = 15°C
Hopefully that helps you achieve your goal here.. In any event, I've logged a request logged to provide a 'Temperature Difference' unit type.
Martin Schmid, P.E.
Product Manager - Analysis and Countrification
Architecture, Engineering, and Construction
Autodesk, Inc.
Active Member
Posts: 9
Registered: 09-01-2005
# Re: Calculation with dgC - again
04-11-2012 11:28 PM in reply to: martin.schmid
Thank you for the answer. You explained it very well.
I originally wanted to use it in calculating water flow:
V = Q / (ro . c . dT)
Q [W]
C [J/kg.K]
ro [kg/m3]
dT [K] = dT [°C]
Unfortunatelly, the temperature in the formula made such a mess that I rarther used dimensionless temperature and multiplied the final formula by required units. It works.
But I still think that one apple plus one apple are two apples and the same should apply to temperatures.
Employee
Posts: 1,278
Registered: 11-16-2005
# Re: Calculation with dgC - again
04-12-2012 12:11 AM in reply to: mrimmer
Yes, I think you are right, in 2012, you need to cancel the temperature units to unitless, but this is more primarily because there is no units for Specific Heat. This article goes into great detail:
In 2013, there is a Specific Heat units type, which simplifies this quite a bit. You still need to deal with the 'temperatures', but you don't need to cancel all the units. (the article is not yet published for 2013 on the wiki, but i've attached a PDF 'print' of it... I noticed that the title was cut off in the PDF, hopefully the rest of the content is there).
As for the one apple plus one apple, this is how we tend to think about it for most things, but Revit isn't just summing, it needs to handle conversions...
In 5th grade math... if the teacher told you to convert 0degF to Celcius, would your answer be -17.77degC or 0degC? Without more context, you don't know the right answer.
Temperature is unique. For most things, 0 means 'the absense of..' so, if you have 0 apples, you could convert this to 0 applesauce. However, if you have 0degF, you don't have an absence of temperature... and it doesn't equal 0degC (it equals -17.77degC) UNLESS you are talking about temperature difference, which is why a different type of conversion is necessary... and right now, Revit only has one type of conversion for temperature.
Martin Schmid, P.E.
Product Manager - Analysis and Countrification
Architecture, Engineering, and Construction
Autodesk, Inc.
Active Member
Posts: 9
Registered: 09-01-2005
# Re: Calculation with dgC - again
04-12-2012 01:41 AM in reply to: martin.schmid
Thank you for the article and your patience.
I want to work with °C (I'm from europe) and that is why I set temperature units to °C and I don't want to convert it to other unit.
0°C is always 0°C
Absulute zero is -273,15°C and this is 0 K.
X . t°C is always X . t°C
I understand that dt=2°C does not equal to 2°F, but it is problem that must solve programmers.
I understand that there probably some reasons, why Revit calculate in another way, but if 1°C + 1°C = 275.15°C, it is simply not good even if you explained to me why this is so. Yes, it is good for programmers but not for users.
Employee
Posts: 1,278
Registered: 11-16-2005
# Re: Calculation with dgC - again
04-12-2012 02:12 AM in reply to: mrimmer
I understand.. you want it to 'just work', whether you are adding, subtracting, or perhaps even averaging temperatures, without a need to know why or how it works. I was only explaining because it doesn't currently work as you expect, and havng that understanding may help you better solve what it is that you are trying to do. This issue is logged for our team to consider for a future release.
Martin Schmid, P.E.
Product Manager - Analysis and Countrification
Architecture, Engineering, and Construction
Autodesk, Inc.
Active Member
Posts: 9
Registered: 09-01-2005
# Re: Calculation with dgC - again
04-12-2012 04:06 AM in reply to: martin.schmid
I'm glad you explained it to me because it is important and useful to know what is happening in the calculation.
Thank you for it. | 1,647 | 5,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.821157 |
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### Topic: Solenoid Artistic Function - Magnet and Glass Tube (Read 1 time)previous topic - next topic
#### liudr
#15
##### Feb 02, 2013, 12:55 am
I did not posted this therefore I never mentioned any dimensions of anything. The author was. If he wants to make the magnet "float" it needs to be about the same diameter to avoid spinning. By the way I wasn't answering to your post. I think I know now why Grumpy is in your name. Maybe a similar word beginning with J will describe you be better.
Mike,
This member is "brainstorming" in another post regarding specific gravity. He is insistent if not self-indulging at the same time. I am not certain he will understand the magnetic force on a magnetic dipole depends not on magnetic induction, but the gradient of it, which is lacking in the center of the solenoid. Varying current is not going to vary this gradient so much. If you are close to the solenoid opening, then you do have this gradient. You can shoot something with a solenoid but the op wants some control.
#### retrolefty
#16
##### Feb 02, 2013, 01:15 amLast Edit: Feb 02, 2013, 01:18 am by retrolefty Reason: 1
Quote
It would be awesome if this magnet had an led on it too.
Then I guess the magnet will also need a battery added to it to power the led?
I've seen some videos of trying to hold a magnet in mid air and be position controlled and it's pretty hard to just hold the magnet in one steady position, but wanting to be able to change it's position vertically it gets very unstable and difficult if not impossible to control. The magnetic forces involved are not linear so the 'tuning' parameters for the PID control loop are virtually impossible to optimize for smooth control.
Not saying it can't be done on some level, but so far many have failed to demonstrate a perfected example. But perhaps you will be successful anyway as sometimes the most likely to succeed are those that don't know what can't be done.
Lefty
#### arduinoadrian
#17
##### Feb 02, 2013, 01:30 am
Mike,
This member is "brainstorming" in another post regarding specific gravity. He is insistent if not self-indulging at the same time. I am not certain he will understand the magnetic force on a magnetic dipole depends not on magnetic induction, but the gradient of it, which is lacking in the center of the solenoid. Varying current is not going to vary this gradient so much. If you are close to the solenoid opening, then you do have this gradient. You can shoot something with a solenoid but the op wants some control.
Sorry to tell you; but you keep having it wrong...
The magnet is a real object with physical dimensions. As a cylinder it has height and volume. Its not a plane. That height makes it occupy a real space inside the solenoid and the top and bottom faces of it are not coincident with the solenoid center as will the ideal physical model you are describing. There is gradient from the top to the bottom face. There is also a force called gravity which will be pulling the magnet down and that will keep it off center if no energy is supplied to stabilize it there. On the other hand I never mentioned the center. That's you own creation; but that's Ok the magnet will stay at any position if enough energy is supplied.
Please stay proffessional, your insults will not make your opinions prevail.
Perseverance is 90% of the solution. The remaining 10% is more perseverance.
#### liudr
#18
##### Feb 02, 2013, 01:45 am
Lefty,
Even pid would be difficult to change position. The force is not proportional to distance. If you just want to stabilize constant position then it is doable.
#### PeterH
#19
##### Feb 02, 2013, 01:48 am
This does sound like a high energy project b/c I want the magnet to float.
I think you'll find that's quite hard to achieve, and if you're after that sort of thing I suggest you start off by leaving the glass out of the equation and go for a much smaller system.
I only provide help via the forum - please do not contact me for private consultancy.
#### liudr
#20
##### Feb 02, 2013, 02:09 am
OP, sorry for diverting so much from your topic. If you want this without a wire pulling, you can consider air flow. Using air flowing from below into a pipe that has openings on both end, small openings, you can have a better control of the the height. That is how flow meters work.
#### encryptor
#21
##### Feb 02, 2013, 07:38 pm
thanks for the ideas. I think suspension of a magnet spinning in one spot would be easier to perform. I've seen the magnetic barbell that would spin above a concave base. However back to the idea of making the magnet float. What if I just have sections of coiled wire and I power the sections on each side of the magnet with a repelling force. Then I change which sections repel to get the magnet moving.
peace*&^
#### Grumpy_Mike
#22
##### Feb 02, 2013, 07:51 pm
Suspending a magnet is a lot harder than suspending a magnetic piece of metal. This is because the two inverse laws combine to give a fourth law, which is even harder to control.
Side fields will just flip the magnet over.
Look at diamagnetic material, that is material that is repelled by a magnetic field. Bismuth is one such metal, then use the electro magnet to push against gravity, it is a much more stable set up.
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Arduino
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Torino, 10131
Italy | 1,339 | 5,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2015-14 | longest | en | 0.949319 |
http://cotpi.com/p/20/ | 1,534,259,455,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221209165.16/warc/CC-MAIN-20180814150733-20180814170733-00278.warc.gz | 93,528,233 | 7,372 | ## Tiling with triominoes
There is a square board made of at least 4 equal-sized squares. The number of squares in the board is a power of 2. An unlimited number of L-shaped triominoes are given. Each triomino is made of 3 equal-sized squares. Each equal-sized square that makes a triomino has the same dimensions as that of each equal-sized square that makes the square board.
The square board needs to be tiled with the L-shaped triominoes such that each square is covered at most once without parts of each triomino extending beyond the 3 squares it covers.
What is the minimum number of squares that can't be covered with the triominoes?
[SOLVED]
#### Indhu Bharathi solved this puzzle:
Here is the proof:
1. Given an n × n matrix where n is a power of 2 whose all cells except the top right corner one are filled with triominoes, it is possible to fill all cells of 2n × 2n matrix except the top right one with triominoes. Here is how to do it:
1. Let's call the n × n matrix as N.
2. Let's break the 2n × 2n matrix into four n × n matrices and call the bottom left one as A, the top left one as B, the top right as C and the bottom right as D.
3. Place N in A.
4. Rotate N clockwise 90° and place at B.
5. Place N at C.
6. Rotate N counterclockwise 90° and place at D.
7. Fill the three-celled void in the center with a triominoe.
2. 2 × 2 matrix has an obvious way to place a triominoes in such a way that the top right is empty.
From (1) and (2) it is possible to fill any n × n matrix where n is a power of 2 with triominoes in such a way that the top right cell alone is not filled.
#### Rajesh Balakrishnan solved this puzzle:
There will always be a minimum of one square left.
The board sizes grow in multiples of 4: 4, 16, 64, 256, etc.
22n − 1 is always divisible by 3 because 22n − 1 = (2n + 1)(2n − 1) and one of these 2 factors is a multiple of 3, as 2^n is even.
The following strategy will ensure that only one square is left out.
For convenience let's assume that the top-right square is uncovered. To get a board of the next size, 3 boards of same size are added (placed to its left, bottom, bottom-left). While placing the 3 additional boards ensure that they have their blank squares toward the bottom-left corner of the first board. Thus we will have 3 uncovered squares touching the bottom-left corner of the first board (centre of the new board) into which a triominoe can be fitted easily. So the top-right square is the only uncovered one.
#### David Robertson solved this puzzle:
The problem, as described on cotpi problem 20, is to fill a 2n × 2n square grid with L-shaped triominoes (which may not overlap). What is the minimum number of squares that can't be covered?
Figure 1: An 4 × 4 grid (n = 2) and an L-shaped triomino.
Claim The minimum number of un-coverable squares is at most 1.
Proof. (By Induction)
Basis Given a 2 × 2 grid, position a triomino so that:
• It's middle square is in the lower-left corner of the grid.
• It's two other squares are in the upper-left and lower-right corner of the grid.
Figure 2: 3 squares have been covered, leaving 1 uncovered square in the top-right corner.
So the minimum number of uncovered squares is either 1 or less on a 21 × 21 grid.
Hypothesis Suppose we can cover a 2n × 2n grid with triominoes, leaving only 1 corner square uncovered. Such a covering can be rotated or reflected, to position the uncovered corner square in any corner of our choice. Without loss of generality, we choose to leave the upper-right corner uncovered.
Figure 3: An 8 × 8 grid has been partitioned into four 4 × 4 quadrants.
Step Now we show that given such a 2n × 2n grid, we can construct a 2n + 1 × 2n + 1 grid with the same properties. To begin, we partition our 2n + 1 × 2n + 1 rid into four 2n × 2n quadrants (Figure 3).
Fill the lower-left and upper-right quadrants with the group of triominoes described in the hypothesis. Then place a triomino with its center square in the top-right corner of the lower-left quadrant (Figure 4).
Figure 4: We have filled the grid as described, covering approximately half of all squares.
Now the lower-left quadrant is completely filled. The upper-right quadrant is almost completely filled — except for the upper-right corner. We have 'overspilt' into the upper-left corner of the lower-right quadrant, and the lower-right corner of the upper-left quadrant.
However, this leaves us with two quadrants that are empty apart from one corner! We can fill the remaining space using the combination of triominoes described in the hypothesis (after reflecting them). Figure 5 illustrates this.
Figure 5: The remaining quadrants have been covered.
We're done! By induction, we can construct a covering of triominoes that leaves only one square uncovered for any 2n × 2n square grid. Therefore, for any grid size, the minimum number of uncovered squares is at most 1.
Corollary The minimum number of uncovered squares is exactly 1.
Proof. We can cover some 2n square grid with only 1 uncovered square. Covering this final square would require either:
• 2 further uncovered squares, or
• covering 2 squares already covered by some other triomino.
Neither of these options is permitted. Therefore we cannot cover every square! So the minimum number of uncovered squares has to be exactly 1.
#### Alex Yeilding solved this puzzle:
The answer is 1. By induction:
Base case: n = 1: Board with 22n = 22 = 4 squares is trivially tiled, leaving the upper-right square uncovered.
Assume board with 22n squares can be tiled leaving only the upper-right corner uncovered. Then for a board with 2[2(n+1)] squares, divide it into four boards of the previously solved size, with the upper-right board oriented with its uncovered square at the upper-right corner, and the other three oriented with their uncovered squares toward the center. These three uncovered squares can be covered with one additional triomino.
### Credit
This puzzle is based on: | 1,537 | 5,985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-34 | longest | en | 0.923229 |
https://thmdex.org/d/211 | 1,716,161,522,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058009.3/warc/CC-MAIN-20240519224339-20240520014339-00720.warc.gz | 505,749,508 | 3,100 | ThmDex – An index of mathematical definitions, results, and conjectures.
▼ Set of symbols ▼ Alphabet ▼ Deduction system ▼ Theory ▼ Zermelo-Fraenkel set theory ▼ Set ▼ Collection of sets ▼ Set union ▼ Successor set ▼ Inductive set ▼ Set of inductive sets ▼ Set of natural numbers ▼ Set of integers ▼ Set of rademacher integers ▼ Rademacher integer ▼ Rademacher random integer ▼ Standard rademacher random integer
Definition D211
Standard gaussian random real number
Let $X_1, X_2, X_3, \ldots \in \text{Random} \{ -1, 1 \}$ each be a D5075: Random integer such that
(i) $$\forall \, n \in \{ 1, 2, 3, \ldots \} : \mathbb{P}(X_n = -1) = \mathbb{P}(X_n = 1) = 1/2$$ (ii) $X_1, X_2, X_3, \ldots$ is an D2713: Independent random collection
A D3161: Random real number $Z \in \text{Random}(\mathbb{R})$ is a standard gaussian random real number if and only if $$Z \overset{d}{=} \lim_{N \to \infty} \sum_{n = 1}^N \frac{X_n}{\sqrt{N}}$$
Let $X_1, X_2, X_3, \ldots \in \text{Random} \{ -1, 1 \}$ each be a D5075: Random integer such that
(i) $$\forall \, n \in \{ 1, 2, 3, \ldots \} : \mathbb{P}(X_n = -1) = \mathbb{P}(X_n = 1) = 1/2$$ (ii) $X_1, X_2, X_3, \ldots$ is an D2713: Independent random collection
A D3161: Random real number $Z \in \text{Random}(\mathbb{R})$ is a standard gaussian random real number if and only if $$Z \overset{d}{=} \lim_{N \to \infty} \left( \frac{X_1}{\sqrt{N}} + \frac{X_2}{\sqrt{N}} + \cdots + \frac{X_N}{\sqrt{N}} \right)$$
Let $X_1, X_2, X_3, \ldots \in \text{Rademacher}(1 / 2)$ each be a D5287: Standard rademacher random integer such that
(i) $X_1, X_2, X_3, \ldots$ is an D2713: Independent random collection
A D3161: Random real number $Z \in \text{Random}(\mathbb{R})$ is a standard gaussian random real number if and only if $$Z \overset{d}{=} \lim_{N \to \infty} \sum_{n = 1}^N \frac{X_n}{\sqrt{N}}$$
Children
▶ Chi random unsigned real number ▶ Gaussian random real number | 728 | 1,918 | {"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": 5, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-22 | latest | en | 0.433608 |
https://communities.sas.com/t5/Base-SAS-Programming/inserting-zeros-for-remaining-values/td-p/471326?nobounce | 1,531,778,413,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589470.9/warc/CC-MAIN-20180716213101-20180716233101-00565.warc.gz | 634,772,932 | 29,438 | ## inserting zeros for remaining values
Frequent Contributor
Posts: 76
# inserting zeros for remaining values
Hi All,
I want to see the values in number column for the first 3 observations for each name and remaining has to be zero for duplicate names in number column.
```data test;
infile cards;
input name \$ re\$ number;
cards;
a store 2
a on 2
a total 2
a store 2
a on 2
a total 2
b store 4
b on 4
b total 4
c store 3
c on 3
c total 3
c store 3
c on 3
c total 3
c store 3
c on 3
c total 3
d store 1
d on 1
d total 1
d store 1
d on 1
d total 1
e store 2
e on 2
e total 2
;
run;
/* my output should be like this */
a store 2
a on 2
a total 2
a store 0
a on 0
a total 0
b store 4
b on 4
b total 4
c store 3
c on 3
c total 3
c store 0
c on 0
c total 0
c store 0
c on 0
c total 0
d store 1
d on 1
d total 1
d store 0
d on 0
d total 0
e store 2
e on 2
e total 2```
Thanks,
SS
Super Contributor
Posts: 339
## Re: inserting zeros for remaining values
[ Edited ]
Hi,
Does the following give you what you want:
``````data want(drop = counter);
set test;
by name;
if first.name then
counter = 1;
else
counter + 1;
if counter gt 3 then
number = 0;
run;``````
Edit: Assumes the data is sorted as presented.
Regards,
Amir.
Super User
Posts: 9,599
## Re: inserting zeros for remaining values
Just to note, the binary choice functions:
``````data want(drop = counter);
set test;
by name;
counter=ifn(first.name,1,counter+1);
if counter gt 3 then number=0;
run;``````
Simplify binary choices.
Frequent Contributor
Posts: 86
## Re: inserting zeros for remaining values
``````proc sort data=have out=want;
by name res;
run;
data want;
set want;
by name res;
if first.res then number=number;
else number=0;
run;``````
Super User
Posts: 10,766
## Re: inserting zeros for remaining values
``````data test;
infile cards;
input name \$ re\$ number;
cards;
a store 2
a on 2
a total 2
a store 2
a on 2
a total 2
b store 4
b on 4
b total 4
c store 3
c on 3
c total 3
c store 3
c on 3
c total 3
c store 3
c on 3
c total 3
d store 1
d on 1
d total 1
d store 1
d on 1
d total 1
e store 2
e on 2
e total 2
;
run;
data want;
if _n_=1 then do;
if 0 then set test;
declare hash h(dataset:'test');
h.definekey('name','re');
h.definedone();
end;
set test;
if h.check()=0 then h.remove();
else number=0;
run;
proc print;run;``````
PROC Star
Posts: 1,772
## Re: inserting zeros for remaining values
``````data test;
infile cards;
input name \$ re\$ number;
cards;
a store 2
a on 2
a total 2
a store 2
a on 2
a total 2
b store 4
b on 4
b total 4
c store 3
c on 3
c total 3
c store 3
c on 3
c total 3
c store 3
c on 3
c total 3
d store 1
d on 1
d total 1
d store 1
d on 1
d total 1
e store 2
e on 2
e total 2
;
run;
data want;
do _n_=1 by 1 until(last.name);
set test;
by name;
if _n_>3 then number=0;
output;
end;
run;``````
Discussion stats
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• 6 in conversation | 1,117 | 3,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-30 | latest | en | 0.697462 |
https://www.cnblogs.com/dialect/p/6517335.html | 1,708,607,845,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473819.62/warc/CC-MAIN-20240222125841-20240222155841-00750.warc.gz | 733,238,174 | 7,908 | # 基于控制台的四则运算
coding地址：https://git.coding.net/Dialect/c-sizeyunsuan.git
1. 除了整数以外，还要支持真分数的四则运算，真分数的运算，例如：1/6 + 1/8 = 7/24
2. 运算符为 +, −, ×, ÷
3. 并且要求能处理用户的输入，并判断对错，打分统计正确率。
4. 要求能处理用户输入的真分数， 如 1/2, 5/12 等
5. 使用 -n 参数控制生成题目的个数，例如执行下面命令将生成10个题目
void DealInt(int n, int a[])
{
srand(time(NULL));
for (int p = 0; p<n; p++)
{
int i = (int)rand() % 10;
int j = (int)rand() % 10;
int k = (int)rand() % 100 / 25;
switch (k)
{
case 0:
cout << i << "+" << j << "=";
a[p] = i + j;
check2(n, a);
break;
case 1:
cout << i << "-" << j << "=";
a[p] = i - j;
check2(n, a);
break;
case 2:
cout << i << "*" << j << "=";
a[p] = i*j;
check2(n, a);
break;
case 3:
try
{
a[p] = i / j;
cout << i << "/" << j << "=";
check2(n, a);
}
catch (...)
{
p--;
}
}
}
float t = (float)r / (float)n;
cout << "正确率:" << 100 * t << "%" << endl;
}
void DealFenshu(int n, int a[][2])
{
srand(time(NULL));
for (int p = 0; p<n; p++)
{
int i = (int)rand() % 10;
int j = (int)rand() % 10;
while (j == 0 || i >= j)
{
i = (int)rand() % 10;
j = (int)rand() % 10;
}
int x = (int)rand() % 10;
int y = (int)rand() % 10;
while (y == 0 || x >= y)
{
x = (int)rand() % 10;
y = (int)rand() % 10;
}
int k = (int)rand() % 100 / 25;
switch (k)
{
case 0:
cout << "(" << i << "/" << j << ")" << "+" << "(" << x << "/" << y << ")" << "=";
a[p][0] = i*y + x*j;
a[p][1] = j*y;
check1(n, a);
break;
case 1:
cout << "(" << i << "/" << j << ")" << "-" << "(" << x << "/" << y << ")" << "=";
a[p][0] = i*y - x*j;
a[p][1] = j*y;
check1(n, a);
break;
case 2:
cout << "(" << i << "/" << j << ")" << "*" << "(" << x << "/" << y << ")" << "=";
a[p][0] = i*x;
a[p][1] = j*y;
check1(n, a);
break;
case 3:
cout << "(" << i << "/" << j << ")" << "/" << "(" << x << "/" << y << ")" << "=";
a[p][0] = i*y;
a[p][1] = j*x;
check1(n, a);
break;
}
}
float t = (float)r / (float)n;
cout << "正确率:" << 100 * t << "%" << endl;
}
PSP2.1 Personal Software Process Stages Time (%) Senior Student Time (%) Planning 计划 8 6 · Estimate 估计这个任务需要多少时间 8 6 Development 开发 82 88 · Analysis 需求分析 (包括学习新技术) 6 10 · Design Spec 生成设计文档 5 18 · Design Review 设计复审 4 12 · Coding Standard 代码规范 3 30 · Design 具体设计 10 12 · Coding 具体编码 36 21 · Code Review 代码复审 7 15 · Test 测试（自我测试，修改代码，提交修改） 13 21 Reporting 报告 9 60 · 测试报告 3 20 · 计算工作量 2 10 · 并提出过程改进计划 3 30
posted @ 2017-03-07 22:58 Dialect 阅读(194) 评论(2编辑 收藏 举报 | 1,036 | 2,315 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-10 | latest | en | 0.268707 |
https://www.teacherspayteachers.com/Product/Concept-Sorts-A-Single-Sort-Resource-for-Prime-and-Composite-Numbers-1627810 | 1,488,081,045,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171933.81/warc/CC-MAIN-20170219104611-00211-ip-10-171-10-108.ec2.internal.warc.gz | 900,549,589 | 25,662 | Total:
\$0.00
# Concept Sorts: A Single Sort Resource for Prime and Composite Numbers
Subjects
Resource Types
Common Core Standards
Product Rating
4.0
File Type
PDF (Acrobat) Document File
2.87 MB | 10 pages
### PRODUCT DESCRIPTION
Are you familiar with concept sorts? Many people use "sorts" with their spelling or word work programs, but sorting and categorizing can be extremely effective learning strategies for MANY areas! I have found sorts to be particularly effective in my math instruction, and I am excited to begin offering some of these sorts to you! If you are unfamiliar with how sorts are used, I have included a full blog post with photos to help get you started! So...what's included in THIS resource?
The cards needed to do ONE concept sort on prime and composite numbers. This is a perfect extension lesson or replacement for a more computation based lesson in a textbook! The cards for one sort are included--and the headers to sort them two different ways.
A blog post with photos that explains EXACTLY how I completed a different sort with my own students. Feel free to get creative and try different approaches—but I have given one highly effective and efficient way to do this.
A “Show What You Know” sheet that follows the rule of the sort. Use as independent practice or as an assessment after you have done a sort to see what the students know and what they still need to learn. Many of these also ask students to explain their thinking—a key part of the CCSS!
A page of blank cards if you wish to extend the learning by having students create MORE examples that go in each category. This is a great way to differentiate for more capable learners! See each sort for other differentiation hints!
No answer key. Why? The important part about doing these sorts is the discussion rather than making sure every answer is instantly correct. Let the students discuss, prove their ideas, and develop understanding!
You may be familiar with my concept sort SETS which each include FIVE different sorts on a topic…from fractions to multiplication to algebra thinking to geometry. These “single sorts” are NOT the same sorts that are included in the sort sets…they are additional sorts that might meet the needs of your students.
I hope you find the resource thorough, relevant, and engaging--and that it will push your students to increase the depth of their understanding and their mathematical practices as well.
Interested in another set of concept sorts? My new set of angle sorts is now available! CLICK HERE!
All rights reserved by ©The Teacher Studio. Purchase of this problem set entitles the purchaser the right to reproduce the pages in limited quantities for single classroom use only. Duplication for an entire school, an entire school system, or commercial purposes is strictly forbidden without written permission from the author at fourthgradestudio@gmail.com. Additional licenses are available at a reduced price.
Total Pages
10
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N/A
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\$2.50 | 692 | 3,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-09 | longest | en | 0.944257 |
https://nabinsharma.wordpress.com/2012/12/26/linear-hough-transform-using-python/ | 1,679,582,464,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945168.36/warc/CC-MAIN-20230323132026-20230323162026-00093.warc.gz | 479,332,205 | 28,411 | # Linear Hough Transform Using Python
In this post I will explain the Hough transform for line detection. I will demonstrate the ideas in Python/SciPy.
Hough transform is widely used as a feature extraction tool in many image processing problems. The transform can be used to extract more complex geometric shapes like circles and ellipses but this post focuses on extracting lines in an image.
Quick Conceptual Review
In Cartesian coordinates, a line can be represented in slope-intercept form as
$y=mx+c.$
The left panel in Fig. 1 shows a line (red color) in Cartesian coordinate system. The y-axis is chosen to be positive downward so that it matches with the commonly used image coordinate convention (top-left corner of an image is its origin). The value of $\theta$ is assumed positive when measured in clockwise direction from x-axis (to make the results comparable to MATLAB).
Fig. 1: Representation of a line in Cartesian (left) and Hough (right) coordinates.
The equation of the line can be modified to get its alternate representation as
$\rho=x\cos\theta+y\sin\theta$,
where $\rho$ is the distance of the line from the origin and and $\theta$ is the angle of the vector, of least magnitude, to the line. The transformed coordinate space which has $\theta$ and $\rho$ as its axes, as shown in right panel of the above figure, is called Hough space. A point in image space is mapped to a sinusoidal curve in Hough space. A set of points belonging to a line in image space get mapped to a set of sinusoids intersecting at a point in Hough space. So the problem of detecting a line in an image becomes a problem of detecting a point in Hough space. Once we detect the points in Hough space, we can do inverse transform to get the corresponding line in image space.
That’s it! We are ready for a simple demo.
A Simple Demo in Python
This demo is implemented in Python for algorithm explanation without paying attention to computational speed/optimization. We will apply Hough transform on Gantry crane image and extract first few strongest lines in the image. First of all, lets load the required libraries:
import numpy as N
import scipy.ndimage as I
import matplotlib.image as IM
import matplotlib.pyplot as plt
Next, we read the image using imread. The input RGB image is not a matrix (2D array). We would like to convert it into an image that can be represented as 2D array. We do so by converting the RGB image into grayscale image:
def rgb2gray(img_array):
assert(img_array.shape[2] == 3)
img_gray_array = N.zeros((img_array.shape[0], img_array.shape[1]), dtype=N.float32)
for i in range(img_array.shape[0]):
for j in range(img_array.shape[1]):
img_gray_array[i][j] = 0.2989*img_array[i][j][0] + \
0.5870*img_array[i][j][1] + 0.1140*img_array[i][j][2]
return img_gray_array
The RGB image read using imread and corresponding grayscale image generated using the above given function rgb2gray are shown in Fig. 2 and Fig. 3, respectively.
Fig. 2: Input image.
Fig.3: Grayscale image obtained from RGB image in Fig. 2.
Next, we would like to do some pre-processing on the grayscale image and get an image with few line or line-like features in it. There are several image preprocessing techniques that are applied to an image so that feature extraction becomes much easier. I am not going to clutter this simple post with explainations on noise reduction, thresholding, edge detection, etc. As an one-step solution, I am just binarizing the grayscale image with a hard threshold, that is, discard all the pixels with intensities lower than a specified value. Note that, we will lose several line-like features after this oversimplified binarization. This gives us the binarized image as shown in Fig. 4.
Fig. 4: Image obtained by binarizing image in Fig. 3
We can see that we have some line-like features in the binarized image so that we can apply standard Hough transform to extract them. The advantage of using this binarized image is that we operate only on the white pixels (1’s) of the image. Now you can guess why people would like to apply pre-processing techniques before applying Hough transform on an image.
Computation of Hough transform is a simple voting procedure. We first define a $(\theta, \rho)$ grid in Hough space. This grid is used to define an array called accumulator. The accumulator is simply a 2-D array that has same size as $(\theta, \rho)$ grid. All the values in the accumulator are initialized to zero. Then, for each point $(x,y)$ in image space, we generate corresponding sinusoid in Hough space. For each $(\theta, \rho)$ pair we get for a point in Hough space (the sinusoids are in Hough sapce), we quantize (bin) the values to the nearest points in the accumulator grid and increment the corresponding accumulator value by one. The process is repeated for all the points in image space. Following snippet of code might be more understandable than the mess of words:
def hough_transform(img_bin, theta_res=1, rho_res=1):
nR,nC = img_bin.shape
theta = N.linspace(-90.0, 0.0, N.ceil(90.0/theta_res) + 1.0)
theta = N.concatenate((theta, -theta[len(theta)-2::-1]))
D = N.sqrt((nR - 1)**2 + (nC - 1)**2)
q = N.ceil(D/rho_res)
nrho = 2*q + 1
rho = N.linspace(-q*rho_res, q*rho_res, nrho)
H = N.zeros((len(rho), len(theta)))
for rowIdx in range(nR):
for colIdx in range(nC):
if img_bin[rowIdx, colIdx]:
for thIdx in range(len(theta)):
rhoVal = colIdx*N.cos(theta[thIdx]*N.pi/180.0) + \
rowIdx*N.sin(theta[thIdx]*N.pi/180)
rhoIdx = N.nonzero(N.abs(rho-rhoVal) == N.min(N.abs(rho-rhoVal)))[0]
H[rhoIdx[0], thIdx] += 1
return rho, theta, H
Fig.5: Hough transform of image in Fig. 4.
Figure 5 shows the final accumulator array (or Hough transform) obtained for our binary image. We can see several sinusoids interesecting at several places. Each such intersection corresponds to a line in image space. The image is in jet colormap, so more red the intersection point is, better is the line in image sapce. We can easily extract the intersection points and find corresponding values of $(\theta, \rho)$ pairs in Hough space from which we can compute the slope $m$ and y-intercept $c$ of the line in image space. That’s how we detect lines in an image using standard Hough transform.
I detected the two strongest intersection points in the Hough transform image and used the detected $(\theta, \rho)$ values to generate lines in image space. Fig. 6 shows the detected lines overlaid in the binarized image.
Fig. 6: Two detected lines (colored).
Conclusion
We learned to use standard Hough transform to detect lines in an image.
### 7 thoughts on “Linear Hough Transform Using Python”
1. Can you please explain what “q” and “nrho” represent, and how the theta and rho resolutions come into play?
2. also, what does the “if” statement do here?
• I am using “q” and “nrho” to generate “rho” and “theta” grid. “nrho” represents the number of points along “rho” dimension. The resolutions you mentioned define the “fineness” of the Hough matrix plot.
3. Could you post your complete code? I would like to try this but not bother repeating your work.
• Sorry. I tried to find it but could not. The only thing you need to do is load the image in Python and follow the steps. If I get some time, I will do that and update the post with link to the complete code.
4. H[rhoIdx[0], thIdx] += 1 ??? We have lost sight of the image! Should be adding img_bin[rowIdx][colIdx] at these points.
• Thats just a counter. You should be able to plot the matrix H whatever way you want as long as the data inside it makes sense. | 1,902 | 7,584 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 15, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2023-14 | latest | en | 0.897599 |
http://excel.instantgrades.com/include-include-using-namespace-std-int-main-float-a-b-c-d-float-la-lb-lc-ld-h-float-dl-w-float-sqft-cout-h-cout-la-cout-lb-cout-lc-cout/ | 1,603,488,461,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107865665.7/warc/CC-MAIN-20201023204939-20201023234939-00003.warc.gz | 43,297,037 | 6,025 | Interested in a PLAGIARISM-FREE paper based on these particular instructions?...with 100% confidentiality?
# #include #include using namespace std; int main() { float a, b, c, d; float la, lb, lc, ld, h; float dl, w; float sqft; cout << "enter room height h" << endl; cin >> h; cout << "enter wall length a" << endl; cin >> la; cout << "enter wall length b" << endl; cin >> lb; cout << "enter wall length c" << endl; cin >> lc; cout << "enter wall length d" << endl; cin >> ld; cout << "enter number of windows" << endl; cin >> w; cout << "enter number of doors" << endl; cin >> dl; a = (la); b = (lb); c = (lc); d = (ld); sqft = (c*d) + (c*h) + (d*h) – dl – w + (d – a)*h – dl – w + 2 * (b – c)*h – dl – w; cout << “total square footage” << sqft << endl; return 0; } there are some mistakes need to be fixed but i dont know how to fix it
#include<iostream>
#include<math.h>
using namespace std;
int main()
{
float a, b, c, d;
float la, lb, lc, ld, h;
float dl, w;
float sqft;
cout << “enter room height h” << endl;
cin >> h;
cout << “enter wall length a” << endl;
cin >> la;
cout << “enter wall length b” << endl;
cin >> lb;
cout << “enter wall length c” << endl;
cin >> lc;
cout << “enter wall length d” << endl;
cin >> ld;
cout << “enter number of windows” << endl;
cin >> w;
cout << “enter number of doors” << endl;
cin >> dl;
a = (la);
b = (lb);
c = (lc);
d = (ld);
sqft = (c*d) + (c*h) + (d*h) – dl – w + (d – a)*h – dl – w + 2 * (b – c)*h – dl – w;
cout << “total square footage” << sqft << endl;
return 0;
}
there are some mistakes need to be fixed but i dont know how to fix it
Interested in a PLAGIARISM-FREE paper based on these particular instructions?...with 100% confidentiality? | 550 | 1,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-45 | latest | en | 0.627564 |
https://prosa.mpi-sws.org/releases/v0.3/spec/with-proofs/rt.implementation.job.html | 1,696,453,434,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511406.34/warc/CC-MAIN-20231004184208-20231004214208-00708.warc.gz | 506,802,744 | 2,707 | # Library rt.implementation.job
Require Import rt.model.time rt.util.all.
From mathcomp Require Import ssreflect ssrbool ssrnat eqtype seq.
Module ConcreteJob.
Import Time.
Section Defs.
(* Definition of a concrete task. *)
Record concrete_job :=
{
job_id: nat;
job_arrival: time;
job_cost: time;
}.
(* To make it compatible with ssreflect, we define a decidable
equality for concrete jobs. *)
Definition job_eqdef (j1 j2: concrete_job) :=
(job_id j1 == job_id j2) &&
(job_arrival j1 == job_arrival j2) &&
(job_cost j1 == job_cost j2) &&
(* Next, we prove that job_eqdef is indeed an equality, ... *)
Lemma eqn_job : Equality.axiom job_eqdef.
Proof.
unfold Equality.axiom; intros x y.
destruct (job_eqdef x y) eqn:EQ.
{
apply ReflectT; unfold job_eqdef in ×.
move: EQ ⇒ /andP [/andP [/andP [/andP [/eqP ID /eqP ARR] /eqP COST] /eqP DL] /eqP TASK].
by destruct x, y; simpl in *; subst.
}
{
apply ReflectF.
unfold job_eqdef, not in *; intro BUG.
apply negbT in EQ; rewrite negb_and in EQ.
destruct x, y.
rewrite negb_and in EQ.
move: EQ ⇒ /orP [EQ | /eqP TASK].
move: EQ ⇒ /orP [EQ | /eqP DL].
rewrite negb_and in EQ.
move: EQ ⇒ /orP [EQ | /eqP COST].
rewrite negb_and in EQ.
move: EQ ⇒ /orP [/eqP ID | /eqP ARR].
by apply ID; inversion BUG.
by apply ARR; inversion BUG.
by apply COST; inversion BUG.
by apply DL; inversion BUG. | 445 | 1,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-40 | latest | en | 0.386495 |
https://goodriddlesnow.com/riddles/view/3447 | 1,669,481,899,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446708010.98/warc/CC-MAIN-20221126144448-20221126174448-00566.warc.gz | 320,471,816 | 10,002 | # 😏
Question: I am very strong and tough but can be broken in some ways. What am I?
Riddle Discussion
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##### The Famous Ship Puzzle (medium)
Question: A Japanese ship was sailing in the Pacific Ocean. The Japanese captain of the ship put his diamond chain and Rolex watch on a shelf, went to get a shower and returned ten minutes later. Now listen carefully, as I will only tell it once: When he returned, both the chain and the watch were missing!! He called the crew of his ship together. There were four of them. A British guy was the cook of the ship. The captain asked him: "Where were you the last ten minutes?" And the cook answered "I was in the cold storage room to select the meat for lunch". A Sri Lankan was the house keeping guy. The captain repeated his question to him, and learnt that the Sri Lankan was at the top of the ship correcting the flag which had been put upside down. An Indian guy was the engineer maintaining the ship. Same question, and the Indian told that the he was in the generator room checking the generator. A French guy also served on the house keeping crew. Same question, and the French told that he was sleeping after the night shift. Within ten seconds the smart captain caught the thief. Who was the thief? How did the captain find him? Source: Puzzlevilla | 428 | 1,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | longest | en | 0.982801 |
http://www.enotes.com/homework-help/ashleys-18th-birthday-she-receives-gift-10000-306862 | 1,477,637,685,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721558.87/warc/CC-MAIN-20161020183841-00542-ip-10-171-6-4.ec2.internal.warc.gz | 424,882,035 | 9,837 | # On Ashley's 18th birthday, she receives a gift of \$10000, the accumulated amount of an investment her grandfather made for her when she was born.Determine the amount of the investment if the...
On Ashley's 18th birthday, she receives a gift of \$10000, the accumulated amount of an investment her grandfather made for her when she was born.
Determine the amount of the investment if the interest rate was 6.75% compounded: annually, semi-annually and monthly.
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
Assume Ashley's grandfather invested an amount A when she was born. On her 18th birthday she receives \$10000.
The annual rate of interest on the amount invested is 6.75%.
This gives:
For compounding done annually
10000 = A*(1.0675)^18
=> A = `10000/1.0675^18`
=> A = \$ 3085.87
For compounding done semiannually
A = `10000/1.03375^36`
=> A= \$ 3027.2
For compounding done monthly
A = `10000/(1.005625)^216`
=> A = \$2977.21
The amount invested initially was \$ 3085.87 for annual compounding, \$3027.2 for semi-annual compounding and \$2977.21 for monthly compounding. | 308 | 1,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2016-44 | latest | en | 0.963514 |
https://oldschool.runescape.wiki/w/Run_energy | 1,660,633,334,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572221.38/warc/CC-MAIN-20220816060335-20220816090335-00366.warc.gz | 390,986,104 | 20,774 | Energy
(Redirected from Run energy)
Run energy (sometimes called Energy or Stamina) allows players to run rather than walk around the landscape. Players move one tile per game tick while walking but move two tiles per game tick while running, making running twice as fast as walking. Run energy depletes while running, and turning on Run is only possible while a player has 1% energy or more.
Each player has at fewest 0 units of energy and up to 10,000 units of energy, and the game only displays the percentage of this energy. For instance, the game shows 100% when a player has the full 10,000 units of energy and 0% when a player has fewer than 100 units of energy. The energy percentage is displayed in the game's Settings screen and by the minimap if the Data orb option is toggled. Players can toggle this option to run everywhere they go, or they can run to one particular destination by holding down the Control key when they click to move, if run energy is toggled to walk. (see Game controls).
Using energy
Energy starts at 100% by default, but decreases as the player runs around. When a player's energy reaches 0%, their "run" option is automatically switched off. Moving to an adjacent tile through walking will not drain energy, even if Run is turned on.
The rate at which energy depletes grows with the weight of the items the player is carrying (i.e. items in the inventory and any items they have equipped). Players with negative weight (which is possible with weight reducing equipment) will have their energy deplete at the same rate as players with zero weight. Similarly, players with weight greater than 64kg (whether 64.1kg or as much as 1024.2kg) will have their energy deplete at the same rate as players with 64kg. Contrary to popular belief, a player's Agility level does not impact how quickly the player's run energy depletes.
The number of ticks running to go from 10,000 energy (100%) to 0 energy (0%) with various boosts. Recall that 100 ticks is 1 minute.
The formula for run energy lost per game tick ran by default is
${\displaystyle L({\text{weight}})=67+\left\lfloor {\frac {67\,\cdot \,{\text{clamp}}_{[0,64]}({\text{weight}})}{64}}\right\rfloor ,}$
where the ${\textstyle {\text{clamp}}_{[0,64]}}$ function restricts the weight to be between 0kg and 64kg.
Run energy lost per tick can be modified by two effects beyond weight: having a stamina potion effect active or equipping a Ring of endurance with at least 500 charges. With the stamina potion effect, a player's run energy depletion rate is reduced to ${\textstyle \lfloor .3\,L({\text{weight}})\rfloor }$. With a charged ring of endurance equipped, a player's run energy depletion rate is reduced to ${\textstyle \lfloor .85\,L({\text{weight}})\rfloor }$.
Further, on the tick that a player's available energy hits 0 while running, their energy will only deplete by their current energy. For instance, if a player has 1 energy unit and attempts to run, they will run and only have their energy deplete by 1 to 0.
Recovering energy
Energy gradually recovers up to a maximum of 100% or 10,000 energy nearly any time that the player is not running (i.e. walking or standing still). Run energy tends to not recover when the player is doing something else. For example, run energy will not regenerate while the player is running, crossing an agility obstacle, or flying on a Magic carpet. Run energy also will not regenerate while the player is logged out.
The rate at which energy recovers generally increases with the player's Agility level. Notice that the energy recovery rate does not increase with each and every individual level increase. For instance, at level 5 Agility, a player's energy will not recover faster than if they had level 1 agility. Therefore if a player intends to pause Agility training on their account indefinitely, it is optimal to train their Agility level until they achieve the first, lowest level within any "group" of levels (e.g., pausing Agility training at level 30, 36, 48, 60, etc.).
The number of ticks not running to go from 0 energy (0%) to 10,000 energy (100%) with and without graceful. Recall that 100 ticks is 1 minute.
The amount of energy regenerated per game tick by default is
${\displaystyle R({\text{agility}})=\left\lfloor {\frac {\text{agility}}{6}}\right\rfloor +8,}$
where ${\displaystyle {\text{agility}}}$ is the player's agility level. Notice that the energy recovery rate ${\textstyle R(50)=16}$ at agility level 50 is double the rate ${\textstyle R(1)=8}$ at agility level 1. At agility level 99, the recovery rate ${\textstyle R(99)=24}$ is triple what it was at agility level 1.
Run energy regenerated per tick can be modified by only one effect beyond agility level. With the full graceful set equipped, the energy regenerated per tick is modified to ${\displaystyle \lfloor 1.3R(a)\rfloor }$. Without the full set equipped, the amount of run energy regenerated is modified according to the information in the graceful page.
Level Seconds per 1% Energy Minutes from 0% to 100% Energy Ticks per 1% Energy
Default Graceful Default Graceful Default Graceful
0-5 7.500 6.000 12:30 10:00 12.500 10.000
6-11 6.667 5.455 11:07 9:06 11.111 9.091
12-17 6.000 4.615 10:00 7:42 10.000 7.692
18-23 5.455 4.286 9:06 7:09 9.091 7.143
24-29 5.000 4.000 8:20 6:40 8.333 6.667
30-35 4.615 3.750 7:42 6:15 7.692 6.250
36-41 4.286 3.333 7:09 5:34 7.143 5.556
42-47 4.000 3.158 6:40 5:16 6.667 5.263
48-53 3.750 3.000 6:15 5:00 6.250 5.000
54-59 3.529 2.727 5:53 4:33 5.882 4.545
60-65 3.333 2.609 5:34 4:21 5.556 4.348
66-71 3.158 2.500 5:16 4:10 5.263 4.167
72-77 3.000 2.308 5:00 3:51 5.000 3.846
78-83 2.857 2.222 4:46 3:43 4.762 3.704
84-89 2.727 2.143 4:33 3:35 4.545 3.571
90-95 2.609 2.069 4:21 3:27 4.348 3.448
96-101 2.500 1.935 4:10 3:14 4.167 3.226
102+ 2.400 1.875 4:00 3:08 4.000 3.125
Full energy restore
Certain activities or actions will fully restore a player's energy and other attributes, these include:
Energy-restoring action
Energy can be recovered instantly in a number of ways:
Item Energy % Notes Members
White tree fruit 5–10% Picked in the Varrock palace gardens.
Winter sq'irkjuice 10% Made from Winter sq'irks, found within the Sorceress's Garden minigame.
Spring sq'irkjuice 20% Made from Spring sq'irks, found within the Sorceress's Garden minigame.
Autumn sq'irkjuice 30% Made from Autumn sq'irks, found within the Sorceress's Garden minigame.
Summer sq'irkjuice 40% Made from Summer sq'irks, found within the Sorceress's Garden minigame.
Bandages 30% Can only be used within the Castle Wars and Soul Wars minigame.
Explorer's ring 1 50% Can be used up to two times daily.
Explorer's ring 2 50% Can be used up to three times daily.
Explorer's ring 3 50% Can be used up to four times daily.
Explorer's ring 4 100% Can be used up to three times daily.
Agility cape 100% Can be used once a day. Additionally reduces energy depletion by 70% for 1 minute.
Item Energy % Cost Cost per 1% Energy
Guthix rest(4) 5% 2,517 125.85
Papaya fruit 5% 1,008 201.60
Energy potion(4)[fn 1] 10% 489 12.23
Purple sweets 10% 7,822 782.20
Summer pie[fn 2] 10% 652 32.60
Super energy(4) 20% 2,119 26.49
Stamina potion(4)[fn 3] 20% 5,917 73.96
Strange fruit 30% 437 14.57
Mint cake 50% 14,693 293.86
Gout tuber 50% 937,370 18,747.40
1. ^ Also available in free-to-play.
2. ^ Also boosts agility by 5.
3. ^ Additionally reduces energy depletion by 70% for 2 minutes.
Other
• Energy instantly restores by around 2% when completing a Rooftop Agility obstacle.
• The Vile Vigour spell can convert all of the player's remaining prayer points into run energy at a 1:1 ratio.
Energy-lowering items
Energy can be lost instantly through these items.
Changes
Date Changes
9 December 2021
(update)
Date unknown
• Using 'Ctrl+Click' now toggles between running and walking, instead of always turning on run.
• Clicking to walk somewhere after using 'Ctrl+Click' now resets run mode to the run setting, instead of remembering a Ctrl+Click action made earlier in the movement action.
28 April 2016
(update)
Run energy is now restored at the end of the tutorial.
21 March 2013
(update | poll)
Toggle run mode remembered between logins.
24 March 2014
(update)
Holding CTRL to run now functions correctly when clicking on an object or NPC.
22 February 2013
(update)
This content was included when the Old School RuneScape servers officially launched.
Trivia
• Agility affects the restoration of run energy even in F2P worlds. As such, even if your membership expires, you still benefit from the skill.[1]
• If the game begins to load while the player is walking to a destination, the client may display the player character starting to run some distance before returning to walk, even though run energy is not expended. This "animation stalling" effect is due to the client and server reconciling the player's position in the game while the client is loading new map chunks.
References
1. ^ Jagex. Mod Ash's Twitter account. 6 February 2015. (Archived from the original on 12 February 2021.) Mod Ash: "Run energy restores faster at high Agility, including on F2P worlds. Like it was in 2007. We weren't planning to change it." | 2,653 | 9,198 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-33 | latest | en | 0.961832 |
https://www.prepbharat.com/EntranceExams/QuantitativeReasoning/quantitative-reasoning-practice-questions.html | 1,709,349,912,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475727.3/warc/CC-MAIN-20240302020802-20240302050802-00864.warc.gz | 957,103,502 | 7,024 | ## Quantitative Reasoning Questions and Answers Part-6
Data for Questions 1 to 3 : Five women decided to go shopping to M.G. Road, Bangalore.They arrived at the designated meeting place in the following order: 1. Archana, 2. Chellama, 3. Dhenuka, 4. Helen and 5. Sahnaz.
Each woman spent at least ` 1000. Below are some additional facts about how much they spent during their shopping spree.
i. The woman who spent ` 2234 arrive before the lady who spent ` 1193.
ii. One woman spent ` 1340 and she was not Dhenuka.
iii. One woman spent ` 1378 more than Chellamma.
iv. One woman spent ` 2517 and she was not Archana.
v. Helen spent more than Dhenuka.
vi. Shahnaz spent the largest amount and Chellamma the smallest.
1. The woman who spent ` 1193 is:
a) Archana
b) Chellamma
c) Dhenuka
d) Helen
Explanation: From the given information it is clear that four numbers which must have been the values of money spent would be: 2517 (Clue 4), 2234 (Clue 1), 1340 (Clue 2) and 1193 (Clue 1 again).
We need to work out the fifth value.
Also, since Chellamma spent the least and Shahnaz the maximum and since one woman spent` 1378 more than Chellamma, a little bit of introspection will give you the following possibilities for the five numbers:
Possibility 1: If 1193 is the least value
The five numbers are:
1193, 1340, 2234, 2517 and 2571 (since 2571 = 1193 + 1378)
Possibility 2: If 2517 is the maximum value, the five numbers are:
1139 (since 1139 = 2517 – 1378), 1193, 1340, 2234 and 2517
Accordingly, we have the following possible arrangements for the five women and the amount they spent:
Note: The thought structure for placing the 5 values with 5 women in the case of possibility 1, goes as follows:
Step 1:
After placing the least and maximum
Step 2: 2234 should be before 1193 and Dhenuka cannot have spent ` 1340.
A close look at the above table shows that clue 5 (H > D) is not obeyed by this arrangement. Hence, this solution is wrong.
We thus move into possibility 2, i.e.:
1139, 1193, 1340, 2234, 2517, are the five values. The thought structure for placing the five numbers for the five women goes as:
Step 1: Place the maximum and least values for G and C respectively.
This leaves us with 1340, 1193 and 2234 to place.
Step 2: We need to keep 2 constraints in mind while doing this.
(a) 2234 has to come before 1193 (remember not immediately before). At the same time H > D. (Clue 5)
We can arrange 2234 before 1193 in 3 ways as shown below and then 1340 automatically falls into the vacant space.
The answer is Dhenuka. Option (c) is correct
2. What was the amount spent by Helen?
a) ` 1193
b) ` 1340
c) ` 2234
d) ` 2517
Explanation: 1340. Option (b) is correct (Refer explanation of Question 1)
3. Which of the following amounts was spent by one of them?
a) ` 1139
b) ` 1378
c) ` 2571
d) ` 2718
Explanation: 1139. Option (a) is correct (Refer explanation of Question 1)
4. Three travellers are sitting around a fire, and are about to eat a meal. One of them has five small loaves of bread, the second has three small loaves of bread. The third has no food, but has eight coins. He offers to pay for some bread. They agree to share the eight loaves equally among the three travellers, and the third traveller will pay eight coins for his share of the eight loaves. All loaves were of the same size. The second traveller (who had three loaves) suggests that he be paid three coins, and that the first traveller be paid five coins. The first traveller says that he should get more than five coins. How much the first traveler should get?
a) 5
b) 7
c) 1
d) None of these
Explanation: Each loaf of bread is divided into 3 parts. So we have 24 parts and each traveller gets 8 parts.
Ist traveller has 15 parts. He ate 8 parts and gave his 7 parts to the IIIrd traveller.
IInd traveller has 9 parts. He ate 8 parts and gave his1 part to the IIIrd traveller.
So 8 coins should be divided in the ratio 7: 1.
First traveller gets 7 coins.
Option (b) is correct
5. My bag can carry no more than ten books. I must carry at least one book each of management, mathematics, physics and fiction. Also, for every management book I carry I must carry two or more fiction books, and for every mathematics book I carry I must carry two or more physics books. I earn 4, 3, 2, and 1 points for each management, mathematics, physics and fiction book, respectively, I carry in my bag. I want to maximise the points I can earn by carrying the most appropriate combination of books in my bag.
The maximum points that I can earn are:
a) 20
b) 21
c) 22
d) 23
Explanation: Maximum points of 22 can be achieved by taking (1 Management + 2 Fiction + 2 Maths + 5 Physics) books.
4 + 2 + 6 + 10 = 22
Option (c) is correct
6. Eighty kilograms (kg) of store material is to be transported to a location 10 km away. Any number of couriers can be used to transport the material. The material can be packed in any number units of 10, 20 or 40 kg. Courier charges are ` 10 per hour. Couriers travel at the speed of 10 km/hr if they are not carrying any load, at 5 km/hr if carrying 10 kg, at 2 km/hr if carrying 20 kg and at 1 km/hr if carrying 40 kg. A courier cannot carry more than 40 kg of load.
The minimum cost at which 80 kg of store material can be transported to its destination will be:
a) ` 180
b) ` 160
c) ` 140
d) ` 120
Explanation: 160. Option (b) is correct
Data for Questions 7 & 8 : Elle is three times older than Yogesh; Zaheer is half the age of Wahida. Yogesh is older than Zaheer.
7. Which of the following can be inferred?
a) Yogesh is older than Wahida.
b) Elle is older than Wahida.
c) Elle may be younger than Wahida.
d) None of these
Explanation:
From the above table we can infer that Elle must be older than Wahida (as she is thrice a higher value (y) while Wahida’s age is twice a lower value (z)).
Option (b) is correct.
8. Which of the following information will be sufficient to estimate Elle’s age?
a) Zaheer is 10 years old.
b) Both Yogesh and Wahida are older than Zaheer by the same number of years.
c) Both 1 and 2 above.
d) None of these
Explanation: Using both pieces of information we get that if Zaheer = 10, then Wahida and Yogesh = 20 and hence Elle = 60 years
Option (c) is correct (Refer explanation of Question 7)
9. On the walk through the park, Hamsa collected 50 coloured leaves, all either maple or oak. She sorted them by category when she got home, and found the following:
(i) The number of red oak leaves with spots is even and positive.
(ii) The number of red oak leaves without any spot equals the number of red maple leaves without spots.
(iii) All non-red oak leaves have spots, and there are five times as many of them as there are red spotted oak leaves.
(iv) There are no spotted maple leaves that are not red.
(v) There are exactly 6 red spotted maple leaves.
(vi) There are exactly 22 maple leaves that are neither spotted nor red.
How many oak leaves did she collect?
a) 22
b) 17
c) 25
d) 18
Explanation: The answer is 17. Option (b) is correct
10. I have a total of ` 1000. Item A costs ` 110, item B costs ` 90, item C costs ` 70, item D costs ` 40 and item E costs ` 45. For every item D that I purchase, I must also buy only two items of B. For every item A, I must buy one item of C. For every item, E, I must also buy two of item D and one of item B. For every item purchased I earn 1000 points and for every rupee not spent I earn a penalty of 1500 points. My objective is to maximise the points earned. What is the number of items that I must purchase to maximise my points?
a) 13
b) 14
c) 15
d) 16 | 2,123 | 7,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-10 | latest | en | 0.923359 |
https://www.spoj.com/problems/UOFTCA/ | 1,571,272,643,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672431.45/warc/CC-MAIN-20191016235542-20191017023042-00171.warc.gz | 1,085,901,145 | 8,069 | ## UOFTCA - A Research Project
no tags
The school year has just begun, so it's time for Alice to find a suitable boyfriend! Naturally, this process will first require some careful research using a convenient online academic source known as Facebook.
Alice is considering $G$ ($1 \leq G \leq 100$) guys, and wants to estimate how well-matched she would be with each of them - in other words, how attractive each of them is. For each guy, Alice can find $N$ ($1 \leq N \leq 100$) pictures of him on Facebook, the $i$th of which has attractiveness $A_i$ ($1 \leq A_i \leq 100$). The guy might be as ugly as his least-attractive picture (the one with the smallest attractiveness value), or as hot as his most-attractive picture.
In making her important and complex decision, Alice would like to know the potential range of attractiveness of each of the $G$ potential guys!
### Input
Line 1: 1 integer, $G$
For each guy:
Line 1: 1 integer, $N$
Line 2: $N$ integers, $A_{1..N}$
### Output
For each guy:
2 integers, the guy's worst-case and best-case attractiveness, respectively
### Example
Input:
342 5 1 3198516 11 11 14 21
Output:
1 598 9811 21
Explanation of Sample:
The first guy's worst picture (his third) has attractiveness 1, while his best (his second) has attractiveness 5.
The second guy has only one picture, making his attractiveness definitely 98.
Finally, the third guy's worst-case attractiveness is 11 (with two of his pictures having this value), while his best is 21.
Added by: SourSpinach Date: 2014-02-18 Time limit: 1s Source limit: 50000B Memory limit: 1536MB Cluster: Cube (Intel G860) Languages: All except: ASM64 Resource: Own problem, used in the 2013 UofT ACM Tryouts | 460 | 1,709 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2019-43 | latest | en | 0.929825 |
https://us.metamath.org/mpeuni/qdassr.html | 1,721,057,109,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00240.warc.gz | 540,029,412 | 3,798 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > qdassr Structured version Visualization version GIF version
Theorem qdassr 4653
Description: Two ways to write an unordered quadruple. (Contributed by Mario Carneiro, 5-Jan-2016.)
Assertion
Ref Expression
qdassr ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴} ∪ {𝐵, 𝐶, 𝐷})
Proof of Theorem qdassr
StepHypRef Expression
1 unass 4096 . 2 (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
2 df-pr 4531 . . 3 {𝐴, 𝐵} = ({𝐴} ∪ {𝐵})
32uneq1i 4089 . 2 ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = (({𝐴} ∪ {𝐵}) ∪ {𝐶, 𝐷})
4 tpass 4651 . . 3 {𝐵, 𝐶, 𝐷} = ({𝐵} ∪ {𝐶, 𝐷})
54uneq2i 4090 . 2 ({𝐴} ∪ {𝐵, 𝐶, 𝐷}) = ({𝐴} ∪ ({𝐵} ∪ {𝐶, 𝐷}))
61, 3, 53eqtr4i 2834 1 ({𝐴, 𝐵} ∪ {𝐶, 𝐷}) = ({𝐴} ∪ {𝐵, 𝐶, 𝐷})
Colors of variables: wff setvar class Syntax hints: = wceq 1538 ∪ cun 3882 {csn 4528 {cpr 4530 {ctp 4532 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2114 ax-9 2122 ax-ext 2773 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1085 df-tru 1541 df-ex 1782 df-sb 2070 df-clab 2780 df-cleq 2794 df-clel 2873 df-v 3446 df-un 3889 df-sn 4529 df-pr 4531 df-tp 4533 This theorem is referenced by: en4 8744 ex-pw 28217
Copyright terms: Public domain W3C validator | 754 | 1,341 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-30 | latest | en | 0.203045 |
https://entertainment.howstuffworks.com/football.htm#pt7 | 1,696,111,956,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510730.6/warc/CC-MAIN-20230930213821-20231001003821-00000.warc.gz | 263,311,708 | 53,743 | # How American Football Works
By: Kevin Bonsor
Football Image Gallery
Paul Spinelli/Getty Images
Running back LaDainian Tomlinson #21 of the San Diego Chargers leaps high over a pile of players as he scores a one yard touchdown. See more football pictures.
American football is a unique sport -- it is a game about gaining territory as much as it is about scoring points. When two teams step onto a football field, each is battling for every inch it can take from the other. Each team wants to defend the field that is behind it and invade the field in front of it. Ultimately, they want to gain enough ground to score a touchdown or field goal.
In this article, you will learn about the field, the positions, and how points are scored, with a focus on professional football rules. We will also look at some basic techniques for the offense and defense. To learn more about football, read How NFL Equipment Works and How the Physics of Football Works.
Up Next Football Quiz How the Physics of Football Works Discovery.com: Helmet ImpactsÂ
Football is a game of inches played on a field measured in yards -- English measurements are used to track movements on the field. Teams succeed based on how many yards they accumulate or allow. An official NFL football field is a rectangle that is 120 yards (110 m) long and 53 yards, 1 foot (49 m) wide. Most fields are covered in grass and set in an outdoor (open-air) stadium. Some fields are made of artificial turf, which you'll find in many of the indoor stadiums.
Contents
## The Football Field and Football
Let's take a look at some of the standard features of a professional football field:
• Sideline - The sideline is the 6-foot-wide (1.8-meter-wide) boundary line that runs the length of each side of the football field.
• End line - The end line is a 6-foot-wide boundary line that connects the two parallel sidelines. The end line and sideline compose the rectangular shape of the field. Two pylons flank the end of the end line.
• End zone - The end zones are two 10-yard-wide (9-meter-wide) areas at each end of the field inside the end line. The end zone behind a team is that team's end zone, and the end zone ahead of a team is its opponent's end zone.
• Goal line - The goal line is an 8-inch-wide (20-cm-wide) line that runs across the front of the end zone. Two pylons flank the end of the goal line.
• Yard lines and hash marks - In the 100 yards (91 meters) between goal lines, hash marks on either side of the field mark each yard. At every fifth yard, a solid white line runs from sideline to sideline, and at every 10 yards those lines are numbered (i.e., 10, 20, 30, 40, etc.). This crosshatch of lines gives the field its "gridiron" nickname.
• Goalposts - Centered at the back of the end zone is a pole that extends 10 feet (3 meters) high and connects with a horizontal cross bar. On each end of the 18-foot, 6-inch (5.5-meter) cross bar is an upright post that rises to a height of 30 feet (9 meters) above the ground. A 4-inch by 42-inch (10-cm x 107-cm) ribbon is tied to the top of each upright.
ÂThe most essential piece of equipment in a football game is the ball. Official NFL footballs are handmade by Wilson Sporting Goods Co. The football is an oblong sphere and is 11 to 11.5 inches (27.9 - 29.2 cm) long. It has a lengthwise circumference of about 28.5 inches (72.4 cm) and a width-wise circumference of about 21.5 inches (54.6 cm) in the middle of the ball. It weighs between 14 and 15 ounces (397 - 425 grams). The ball consists of an inflated, polyurethane bladder placed in cowhide covering and laced with gridcord material. Gridcord is cotton thread covered with vinyl. A three-ply, synthetic lining is sewn inside the leather covering to protect the bladder and help the football keep its distinct, elongated shape. A valve connected to the bladder protrudes through the leather and allows air to be pumped into the ball.
With the equipment identified and the field set, we can now play a game. An NFL game is divided into four quarters with an extended halftime break between quarters two and three. Each quarter is 15 minutes long. If the teams are tied after four quarters of play, they play an additional overtime period of 15 minutes. In the overtime, the first team to score wins.
While the game time adds up to one hour, it usually takes three to four hours to play a game. Teams can stop the clock by running out of bounds, throwing an incomplete pass, or calling a time-out, of which they have three per half. Time also stops for each of the two-minute warnings, observed two minutes prior to the end of the second and fourth quarters.
In the next section, we'll look at all of the players on the field.
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## The Football Offense
An NFL roster allows for no more than 53 players on a team. At any one time, only 11 players per team are allowed on the field. To understand an NFL roster, you have to identify the three teams within a team: the offense, the defense and special teams. Each of these groups has specialized positions with a specific set of skills. Let's take a closer look at each unit.
A team's offense is responsible for taking the ball down the field toward its opponent's end zone. To do this, the offense throws the ball from one player to another or holds the ball and runs forward. Here are the basic offensive positions.
• Quarterback (QB) - This player throws the ball to receivers or hands it off to running backs. The quarterback is also known as the "field general," because he's the on-the-field leader.
• Offensive linemen - These players provide blocking for the quarterback and running backs. Individual lineman positions include Center (C), Guards (LG/RG) and Tackles (LT/RT). The Center is located in the middle of the line. This player hikes the ball to the quarterback by bringing the ball up between his legs. The Guards flank the center. The tackles are positioned on the outside of each guard. Teams have two guards and two tackles.
• Receivers - Receivers run down the field and catch balls thrown by the quarterback. Receivers are either wide receivers (WR) or tight ends (LTE/RTE), depending on where they are positioned on the field.
• Running backs - Running backs take the ball from the quarterback and run up the field. Depending on the formation (arrangement) of the offensive players, a running back might be called a tailback (TB), halfback (HB) or fullback (FB).
Photo courtesy Kansas City Chiefs team photographer Hank Young
## Football Defense and Special Teams
When a team does not have possession of the ball, it is on defense and uses various methods to prevent the other team's offense from scoring. These players must tackle the offensive player who has the ball to stop the offense from advancing. Defense will also try to take the ball away from the offense.
Here are the basic defensive positions:
• Defensive linemen - The linemen put pressure on the quarterback by trying to tackle him before he releases the ball. They also try to stop running backs. There are typically three or four defensive linemen. Individual positions include Ends (LE/RE), Nose tackle (NT) and Tackle (LDT/RDT). The ends line up on the outside of the line and try to rush around the offensive tackles. The nose tackle lines up over the football. The tackle lines up across from a guard and tries to knife through the offensive line.
• Linebackers - When there are four linemen, there is a middle linebacker (MLB) and two outside linebackers (OLB). When there are three linemen, there are two inside linebackers (ILB) and two outside linebackers. Their job is to back up the linemen, as well as contain runners and cover receivers on some plays.
Photo courtesy The Philadelphia Eagles #51 in this photo is a linebacker
• Cornerbacks (CB) - The cornerbacks prevent the wide receivers from catching the ball by breaking up passes from the quarterback.
• Safeties - The safeties play deep behind the rest of the defense to prevent a long pass or run. A strong safety (SS) lines up on the side of the field where there are more offensive players. The free safety (FS) plays a deep, middle position.
Special Teams
Photo courtesy The Philadelphia Eagles Placekicker
If a team has to kick the ball, it uses its special-teams unit. This unit includes the team's kickers, the offensive line, and players who run down the field to tackle a returner (see below).
• Placekicker - The placekicker kicks the ball through the goalposts to score points and kicks the ball to the other team to start the game and after each scoring possession.
• Punter - The punter free-kicks the ball if his team cannot advance the ball down the field.
• Returner - During a kickoff or punt, the returner tries to catch the ball and return it as far as he can. A player can score a touchdown on a return.
In the next section, you will learn more about how teams move the ball and how the down-and-distance system works.
## Moving the Football and Finding the End Zone
A football game begins with a coin toss to decide which team will receive the opening kickoff. From the opening kickoff, the two teams battle to take possession of the ball. Possession means that a team"s offensive unit has the ball.
A team can take possession of the ball in several ways:
• Receiving a kickoff - A team receives a kickoff at the beginning of each half and after the other team scores.
• Turnover - A team recovers a ball dropped by the other team (fumble) or picks off a ball thrown by the other team"s quarterback (interception).
• Safety - A player is tackled in his own end zone, meaning the end zone his team is defending, so the other team gets the ball though a free kick.
• Punt - The defensive team stops the offensive team from getting 10 yards in three downs, and the offensive team free-kicks, or punts, the ball to the other team on third down.
• Turnover on downs - The offensive team fails to advance the ball 10 yards in four downs and has to surrender the ball to the other team.
For those new to the sport, the last two scenarios on this list may not make sense. One of the most confusing concepts of American-style football is the down-and-distance system. Every time a team takes possession of the ball, it is given a set of four downs, or attempts, to move the ball 10 yards. If the team can move the ball 10 yards or more within four downs, the team gets another set of four downs to go another 10 yards, and so on. For instance, if a team advances 3 yards on first down, the next play is second down with 7 yards to go (second and 7); if the team then advances 5 yards on second down, the next play is third and 2; if the team then advances 2 or more yards on third down, the next play is back to first and 10, with a whole new set of four downs during which to advance the ball.
After each play, the officials determine how many yards a team has advanced or lost (a team can lose yards if the ball holder is tackled behind the line of scrimmage -- this line is discussed in a moment). The officials then place the ball at the point where the team has ended up. This point determines the line of scrimmage, which is an imaginary line that runs across the field and is the starting point for the offensive team on each play. On the sideline, a team of officials handles a 10-yard-long chain, which designates that 10-yard mark a team must reach to get a first down. On close plays, this chain is sometimes brought onto the field to measure the distance from the ball to the 10-yard mark. The nose of the ball must reach the bar connected to the end of the chain for a team to be awarded a first down.
If a team fails to gain 10 yards after three downs, it may choose to punt the ball to the other team. If it doesn"t punt and chooses to use its fourth down, or "go for it," it must reach the 10-yard mark or it surrenders the ball. A team often chooses to punt the ball in order to back the opposing team up so that it has to cover a greater distance to score. The team receiving the punt can return it, meaning it can catch and run it back down the field. The kicking team is hoping to kick the ball down the field and tackle the receiving team"s kick returner before he comes back down the field.
All of this pushing and shoving to move a cowhide-covered ball has one purpose: move the ball over the opponent"s goal line to score a touchdown.
Photo courtesy The Philadelphia Eagles Running the ball
The opponent"s goal line is the one a team is advancing toward. Once any part of the ball reaches the edge of the goal line, it is considered in the end zone, and a touchdown has been scored. You will often hear commentators say that a ball breaks the plane of the end zone, which means the ball has crossed over the goal line.
A touchdown is just one way of scoring points in football. After scoring a touchdown, a team can kick a field goal for an extra point or attempt to run or pass the ball into the end zone for a two-point conversion. The team has only one chance at the two-point conversion.
Here is a complete look at ways points are scored and how many points are awarded for each:
Method Description Points Touchdown (TD) A ball is carried into an opponent"s end zone or caught in the end zone. 6 points Extra point A ball is kicked through the uprights of the opponent"s goalpost after a touchdown. 1 points 2-point conversion A ball is carried into an opponent"s end zone or caught in the end zone. 2 points Field goal A ball is kicked through the uprights of the opponent"s goalpost. 3 points Safety A player tackles an opposing player in the opposing player"s own end zone. 2 points
After scoring a field goal or touchdown and completing the extra point or two-point conversion attempt, a team must kick the ball to the opposing team. The only exception is on a safety. A team that scores a safety gets the ball on a free kick.
## Football Officials
Photo courtesy Pittsburgh Steelers/Mike Fabus In this photo you can see an official wearing black and white, just behind the action.
A football game actually consists of three teams, with the third team being the officiating crew. They also have a uniform, which consists of a shirt with vertical black and white stripes, white pants and a white or black hat. These men are responsible for enforcing the rules of the game as outlined by the NFL rules committee. An NFL officiating crew consists of six men, and each has distinct responsibilities:
• Referee - This is the head official on the field. He is responsible for giving signals and serves as the final authority on rule interpretation. If you watch an NFL game, this will be the official making announcements.
• Umpire - The umpire rules on players' equipment and conduct. The umpire takes a position about five yards behind the line of scrimmage.
• Head Linesman - The head linesman is responsible for calling infractions of player movement when lined up on the line of scrimmage. He also keeps track of the downs and manages the chain crew.
• Line Judge - The line judge keeps time during the game to backup the official clock operator. Also, he backs up the head linesman on line-of-scrimmage calls. He straddles the line of scrimmage on the opposite side from the Head Linesman.
• Field Judge - The field judge makes calls regarding the wide receivers and backs on his side of the field. He also watches the defensive players that the back is blocking. He makes calls determining if a player is in or out of bounds. He stands 20 yards away from the line of scrimmage at the beginning of a play, on the same side of the field as the Line Judge.
• Side Judge - The side judge makes calls regarding the wide receivers and backs on his side of the field. He also watches the defensive players the back is blocking. He makes calls determining if a player is in or out of bounds. He stands 20 yards away from the line of scrimmage at the beginning of the play, on the same side of the field as the Head Linesman.
• Back Judge - The back judge makes calls regarding the tight end and the player the tight end might be blocking. He is also responsible for keeping the time for the 25-second play clock, time-outs, and intermissions. He stands 25 yards beyond the line of scrimmage.
## Penalties
Officials must memorize and be ready to call an infraction in a split second. An official signals an infraction by throwing a yellow flag. There are many rules in the NFL Rule Book; here are a few of the ones of which you might be unaware:
• Clipping - This is a block thrown in the back of the opposing player.
• Chop block - This is an illegal block thrown below the waist of an opposing player. These types of blocks have been known to cause severe leg injuries to the opposing player. The offensive team is penalized 15 yards for this infraction.
• Encroachment - A defending player moves into the neutral zone and makes contact with an offensive player before the ball is put in play. The neutral zone is a space the length of the ball that separates the offense and defense prior to a play. The only player who can legally enter the neutral zone is the center, who hands, or snaps, the ball to the quarterback to start a play. The offensive team is awarded 5 yards for this penalty.
• Excessive crowd noise - The referee determines that the crowd is too loud. The home team can be penalized 5 yards or can lose a time-out.
• Fair catch - A player receiving a kick or punt can signal that he does not intend to return the ball by putting his arm in the air. Once he signals for a fair catch, he cannot be tackled and cannot move beyond the spot where he catches the ball.
• Intentional grounding - A quarterback, who is in the pocket, intentionally throws the ball away to avoid being tackled behind the line of scrimmage for a loss of yards. The pocket is the rounded shape formed by the offensive linemen during a play when they are blocking for the quarterback.
• Leaping rule - While players can block kicks, they cannot run from more than 1 yard behind the line of scrimmage to do so. According to NFL rules, a defensive player can run forward and leap in attempt to block a kick if he was lined up within 1 yard of the line of scrimmage when the ball was snapped. But if the player is lined up more than 1 yard from the line of scrimmage, he cannot run up to the line, leap to block a kick and land on other players. A 15-yard penalty is assessed for this infraction.
• Tuck rule - A player, typically the quarterback, drops the ball when his arm is moving forward to tuck the ball away. The action is considered an incomplete pass rather than a fumble because his arm is moving forward.
• "Emmitt Smith" helmet rule - A player cannot remove his helmet on the field unless it is to adjust his equipment. This rule is dubbed the "Emmitt Smith rule" because Smith, who holds the record for most rushing touchdowns, was famous for ripping off his helmet to celebrate a touchdown. This rule was enacted to quell excessive celebrations. The team of the offending player is assessed a 15-yard penalty.
## Football Instant Replay
In a modern NFL game, there are as many as 20 cameras covering the fast-paced action of a game. In 1999, the NFL added an instant replay system to back up the officials. Each game camera catches a different view of each play, and those views can be used to review questionable calls. However, not every play is reviewable.
During certain plays, coaches can challenge an official"s call. The coach challenges the play by tossing a red flag on the field. Each team is allotted two challenges per game. If the team loses the challenge, it loses a time-out and the official"s call stands, according to NFL rules. If a team wins the challenge, it retains its time-out and the official"s call is overruled. A challenge must be made before the next play begins but cannot occur in the last two minutes of each half.
An official replay assistant can also initiate a review in the last two minutes of each half and in the overtime period. The replay assistant is not limited as to how many replays he can request.
When a play is challenged, the referee has 90 seconds to review the play. He reviews the play at a field-level monitor to the side of the field.
Here is a list of some reviewable plays:
• Scoring plays
• Pass complete, incomplete, or intercepted
• Out of bounds
• Recovery of a loose ball
Photo courtesy The Philadelphia Eagles There"s a loose ball in there somewhere.
• Illegal passes - illegal receiver or beyond line of scrimmage
• Quarterback incomplete forward pass or fumble
• Runner rule down by contact
• Touching of a kick
• Number of players on the field
Officials are not always 100 percent correct, but their well-trained eyes allow them to be correct the majority of the time. | 4,600 | 20,849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-40 | longest | en | 0.95566 |
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1. 1. Solusi Solusi OSN Matematika 2014 Part 3 f (x) = 2 2 + 4x maka: f 1006 2014 = 2 2 + 4 1006 2014 f 1008 2014 = 2 2 + 4 1008 2014 Edy Wihardjo Soal Solusi Olimpiade Matematika 6 Mei 2014 1 / 4
2. 2. Solusi f 1006 2014 + f 1008 2014 = 2 2 + 4 1006 2014 + 2 2 + 4 1008 2014 = 2 2 + 4 1008 2014 + 2 2 + 4 1006 2014 2 + 4 1006 2014 2 + 4 1008 2014 = 4 + 2 4 1008 2014 + 4 + 2 4 1006 2014 4 + 2 4 1008 2014 + 2 4 1006 2014 + 4 1006 2014 4 1008 2014 Edy Wihardjo Soal Solusi Olimpiade Matematika 6 Mei 2014 2 / 4
3. 3. Solusi f 1006 2014 + f 1008 2014 = 8 + 2 4 1008 2014 + 2 4 1006 2014 4 + 2 4 1008 2014 + 2 4 1006 2014 + 4( 1006 2014+1008 2014 ) = 8 + 2 4 1008 2014 + 2 4 1006 2014 4 + 2 4 1008 2014 + 2 4 1006 2014 + 41 = 8 + 2 4 1008 2014 + 2 4 1006 2014 8 + 2 4 1008 2014 + 2 4 1006 2014 = 1 Edy Wihardjo Soal Solusi Olimpiade Matematika 6 Mei 2014 3 / 4
4. 4. Solusi f 1006 2014 + f 1008 2014 = 1 Edy Wihardjo Soal Solusi Olimpiade Matematika 6 Mei 2014 4 / 4
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0 | 687 | 1,324 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-39 | latest | en | 0.259526 |
https://www.tutorela.com/math/power-of-a-quotient/examples-exercises | 1,718,469,013,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00703.warc.gz | 955,680,388 | 33,502 | # Fraction power - Examples, Exercises and Solutions
## Power of a Quotient
When we encounter an expression with a quotient (or division) inside parentheses and the entire expression is raised to a certain exponent, we can take the exponent and apply it to each of the terms in the expression.
Let's not forget to maintain the fraction bar between the terms.
Formula of the property:
$(\frac {a}{b})^n=\frac {a^n}{b^n}$
This property is also relevant to algebraic expressions.
## Practice Fraction power
### Exercise #1
$(\frac{4^2}{7^4})^2=$
### Step-by-Step Solution
$(\frac{4^2}{7^4})^2=\frac{4^{2\times2}}{7^{4\times2}}=\frac{4^4}{7^8}$
$\frac{4^4}{7^8}$
### Exercise #2
$(\frac{2}{6})^3=$
### Step-by-Step Solution
We use the formula:
$(\frac{a}{b})^n=\frac{a^n}{b^n}$
$(\frac{2}{6})^3=(\frac{2}{2\times3})^3$
We simplify:
$(\frac{1}{3})^3=\frac{1^3}{3^3}$
$\frac{1\times1\times1}{3\times3\times3}=\frac{1}{27}$
$\frac{1}{27}$
### Exercise #3
$5^4\cdot(\frac{1}{5})^4=\text{?}$
### Step-by-Step Solution
This problem can be solved using the properties of powers for a negative power, power over a power, and the property of powers for the product between terms with identical bases, which is the natural way of solving,
But here we prefer to solve it in another way that is a bit faster:
To this end, the power by power law is applied to the parentheses in which the terms are multiplied, but in the opposite direction:
$x^n\cdot y^n=(x\cdot y)^n$Since in the expression in the problem there is a multiplication between two terms with identical powers, this law can be used in its opposite sense, so we will apply this property to the problem:
$5^4\cdot(\frac{1}{5})^4=\big(5\cdot\frac{1}{5}\big)^4$Since the multiplication in the given problem is between terms with the same power, we could apply this law in the opposite direction and write the expression as the multiplication of the bases of the terms in parentheses to which the same power is applied.
We will continue and simplify the expression in parentheses, we will do it quickly if we notice that in parentheses there is a multiplication between two opposite numbers, then their product will give the result: 1, we will apply this understanding to the expression we arrived at in the last step:
$\big(5\cdot\frac{1}{5}\big)^4 = 1^4=1$When in the first step we apply the previous understanding, and then use the fact that raising the number 1 to any power will always give the result: 1, which means that:
$1^x=1$Summarizing the steps to solve the problem, we get that:
$5^4\cdot(\frac{1}{5})^4=\big(5\cdot\frac{1}{5}\big)^4 =1$Therefore, the correct answer is option b.
1
### Exercise #4
$7^4\cdot8^3\cdot(\frac{1}{7})^4=\text{?}$
### Step-by-Step Solution
We use the formula:
$(\frac{a}{b})^n=\frac{a^n}{b^n}$
We decompose the fraction in parentheses:
$(\frac{1}{7})^4=\frac{1^4}{7^4}$
We obtain:
$7^4\times8^3\times\frac{1^4}{7^4}$
We simplify the powers: $7^4$
We obtain:
$8^3\times1^4$
Remember that the number 1 in any power is equal to 1, so we obtain:
$8^3\times1=8^3$
$8^3$
### Exercise #5
$(\frac{2}{3})^{-4}=\text{?}$
### Step-by-Step Solution
We use the formula:
$(\frac{a}{b})^{-n}=(\frac{b}{a})^n$
Therefore, we obtain:
$(\frac{3}{2})^4$
We use the formula:
$(\frac{b}{a})^n=\frac{b^n}{a^n}$
Therefore, we obtain:
$\frac{3^4}{2^4}=\frac{3\times3\times3\times3}{2\times2\times2\times2}=\frac{81}{16}$
$\frac{81}{16}$
### Exercise #1
$9^?(\frac{1}{2})^{-4}=\frac{16}{3}$
### Step-by-Step Solution
$9^?(\frac{1}{2})^{-4}=\frac{16}{3}$as an equation for all (and of course it is indeed an equation),
Therefore, we replace the problem sign in the unknown x and solve it:
$9^x(\frac{1}{2})^{-4}=\frac{16}{3}$
Now we briefly discuss the solution technique:
Quite generally, the goal when solving exponential equations is to reach a situation where there is a term on each side of the equation so that both sides have the same base, in such a situation we can unequivocally state that the power exponents on both sides of the equation are equal, and solve a simple equation for the unknown,
Mathematically, we will perform a mathematical manipulation (according to the laws of course) on both sides of the equation (or development of one side with the help of power properties and algebra) and will arrive at the following situation:
$b^{m(x)}=b^{n(x)}$when$m(x),\hspace{4pt}n(x)$Algebraic expressions (actually functions of the unknown$x$) that can also exclude the unknowns ($x$) that we try to find in the problem, which is the solution to the equation,
It is then stated that:
$m(x)=n(x)$and we solve the simple equation we get,
We solve the equation in the given problem again:
$9^x(\frac{1}{2})^{-4}=\frac{16}{3}$In solving this equation, various power properties are used:
a. Power property with negative exponent:
$a^{-n}=\frac{1}{a^n}$b. Power property for a power of an exponent raised to another exponent:
$(a^m)^n=a^{m\cdot n}$
First we will arrive at a simple presentation of the terms of the equation, that is, "eliminate" fractions and roots (if there are any in the problem, there are none here)
To do this, we start by treating the fraction on the right side of the equation, this will be done using the power property of a negative exponent specified in A above and represent this fraction (in parentheses) as a term with a negative exponent:
$9^x(\frac{1}{2})^{-4}=\frac{16}{3} \\ 9^x(2^{-1})^{-4}=\frac{16}{3}\\ 9^x2^{(-1)\cdot(-4)}=\frac{16}{3}\\ 9^x2^{4}=\frac{16}{3}\\$When we perform the development on the left side of the equation as described above, and in the last step simplify the expression in the power exponent on the left side of the equation,
Later we would like to be able to obtain an identical base on both sides of the equation, the best way to achieve this is by decomposing all and every one of the numbers in the problem into prime factors (using powers as well), here in the problem we note that the numbers exist:
$16,\hspace{4pt}9,\hspace{4pt}3,\hspace{4pt}2$The numbers: 2, 3 are prime, so we will not touch them, we will note that the number 16 is a power of the number 2 and that the number 9 is a power of the number 3, that is:
$16=2^4\\ 9=3^2$This is the presentation (decomposition) of the numbers 16 and 9 with the help of their prime factors, so we will return to the equation we obtained in the previous step and replace these numbers in the decomposition of their prime factors:
$9^x2^{4}=\frac{16}{3}\\ (3^2)^x2^{4}=\frac{2^4}{3}\\$Now we will notice that we can get rid of the term.$2^4$By dividing both sides of the equation by it, we will also notice that this term does not depend on the unknown and is different from zero and therefore there is no limitation that says it is forbidden to divide it, we will do so:
$(3^2)^x2^{4}=\frac{2^4}{3} \hspace{8pt}\text{/:}2^{4}\\ \frac{(3^2)^x\cdot\not{2^4}}{\not{2^4}}=\frac{\not{2^4}}{3\cdot\not{2^4}} \\ (3^2)^x=\frac{1}{3}$When in the first step we divide both sides of the equation by the term we wanted to get rid of and then simplify the fractions obtained on both sides of the equation,
Now we go back to remember the power laws that we have already used and that were mentioned before:
a. Power property with negative exponent:
$a^{-n}=\frac{1}{a^n}$b. Power property for a power of an exponent raised to another exponent:
$(a^m)^n=a^{m\cdot n}$In the next step, we will apply the power raised to another power law specified in B above on the left side, to get rid of the parentheses, and in the next step we deal with the right side with the goal of undoing the fraction, for this purpose, we will use the power property with a negative exponent specified in A above, we will perform a step by development line:
$(3^2)^x=\frac{1}{3} \\ 3^{2x}=\frac{1}{3} \\ 3^{2x}=3^{-1}$We have reached our goal, we obtained an equation in which both sides have terms with the same base, therefore we can affirm that the power exponents of the terms on both sides are equal, and to solve the resulting equation for the unknown, we do the following:
$3^{2x}=3^{-1} \\ \downarrow\\ 2x=-1$We continue and solve the resulting equation, this we will do by isolating the unknown on the left side, we will achieve this by dividing both sides of the equation by its coefficient:
$2x=-1 \hspace{8pt}\text{/:}2 \\ \bm{x=-\frac{1}{2} }$We have thus solved the given equation, we briefly summarize the solution steps:$9^x(\frac{1}{2})^{-4}=\frac{16}{3} \\ 9^x2^{4}=\frac{16}{3}\\ (3^2)^x2^{4}=\frac{2^4}{3}\hspace{8pt}\text{/:}2^{4}\\ (3^2)^x=\frac{1}{3} \\ 3^{2x}=3^{-1} \\ \downarrow\\ 2x=-1\hspace{8pt}\text{/:}2 \\ \bm{x=-\frac{1}{2} }$Therefore, the correct answer is option c.
$-\frac{1}{2}$
### Exercise #2
$((\frac{1}{5})^2)^?:5=125$
### Step-by-Step Solution
$\big( \big(\frac{1}{5} \big)^2 \big)^?:5=125$as an equation for all (and of course it is indeed an equation),
Therefore, we replace the problem sign in the unknown x and solve it:
$\big( \big(\frac{1}{5} \big)^2 \big)^x:5=125$Later on we will remember that dividing by a certain number is multiplying by its inverse, so we will rewrite the given equation bearing this in mind:
$\big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125$
Now we briefly discuss the solution technique:
In quite a general way, the goal when solving exponential equations is to achieve a situation where there is a term on each of the two sides of the equation so that both sides have the same base, in such a situation we can unequivocally state that the power exponents on both sides of the equation are equal, and solve a simple equation for the unknown,
Mathematically, we will perform a mathematical manipulation (according to the laws of course) on both sides of the equation (or development of one of the sides with the help of power properties and algebra) and arrive at the following situation:
$b^{m(x)}=b^{n(x)}$when $m(x),\hspace{4pt}n(x)$Algebraic expressions (actually functions of the unknown $x$) that can also exclude the unknowns ($x$) that we are trying to find in the problem, which is the solution to the equation,
It is then stated that:
$m(x)=n(x)$and we solve the simple equation we obtain,
We go back to solving the equation in the given problem:
$\big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125$In solving this equation, various power properties are used:
a. Power property with negative exponent:
$a^{-n}=\frac{1}{a^n}$b. Power property for a power of an exponent raised to another exponent:
$(a^m)^n=a^{m\cdot n}$
First we will arrive at a simple presentation of the terms of the equation, that is, "eliminate" fractions and roots (if there are any in the problem, there are none here)
To do this, we will start by dealing with the fraction on the left side of the equation:
$\frac{1}{5}$That is, both the fraction inside the parenthesis and the fraction outside the parenthesis, this is done with the help of the power property with negative exponent specified in A above and we represent this fraction as a term with negative power and in the next step we will apply the power property for a power of an exponent raised to another exponent specified in B above and we will get rid of the parentheses, another step starting from the inner parenthesis to the outer, we will do this, step by step below:
$\big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125 \\ \big( (5^{-1})^2 \big)^x\cdot 5^{-1}=125 \\ (5^{(-1)\cdot 2} )^x\cdot 5^{-1}=125 \\ 5^{(-1)\cdot 2\cdot x} \cdot 5^{-1}=125 \\ 5^{-2x} \cdot 5^{-1}=125 \\$When we carry out the development of the left side of the equation as described above, we initially apply the power property with negative exponent mentioned above in A and in the following steps we apply the power property for a power of an exponent raised to another exponent mentioned above in B and we got rid of the parentheses: starting from the inner parenthesis to the outer, in the last step we simplify the expression in the power exponent on the left side of the equation,
c. Later we remember the power property for multiplying terms with identical bases:
$a^m\cdot a^n=a^{m+n}$And we will apply this law to the left side of the equation we obtained in the last step, this is to have a term on this side, we will do this:
$5^{-2x} \cdot 5^{-1}=125 \\ 5^{-2x+(-1)}=125 \\ 5^{-2x-1}=125 \\$When in the first step we apply the aforementioned power law to the product between members with identical bases mentioned above in C and in the following steps we simplify the expression in the power exponent on the left side,
Next, we would like to obtain the same base on both sides of the equation, the best way to achieve this is by decomposing each and every one of the numbers in the problem into prime factors (using powers as well), you will notice that the number 125 is a power of the number 5, that is:
$125=5^3$This is the presentation (factorization) of the number 125 using its prime factor, which is the number 5, so we go back to the equation we received in the previous step and replace this number with its decomposition into prime factors:
$5^{-2x-1}=125 \\ 5^{-2x-1}=5^3 \\$We have reached our goal, we have received an equation in which both sides have terms with the same base, therefore we can state that the power exponents of the terms on both sides are equal, and to solve the resulting equation for the unknown, we will do this:
$5^{-2x-1}=5^3 \\ \\ \downarrow\\ -2x-1=3$We will continue and solve the resulting equation, we will do this by isolating the unknown on the left side, we will achieve this in the usual way, moving the sections and dividing the final equation by the unknown's coefficient:
$-2x-1=3 \\ -2x=3+1\\ -2x=4 \hspace{8pt}\text{/:}(-2) \\ \frac{\not{-2}x}{\not{-2}}=\frac{4}{-2}\\ x=-\frac{4}{2}\\ \bm{x=-2 }$When in the first step we simplify the equation by moving the sides, remembering that when a term is moved its sign changes, then we complete the isolation by nullifying dividing both sides of the equation by its coefficient, in the last steps, we simplify the expression obtained by reducing the fractions,
We have thus solved the given equation, we briefly summarize the solution steps:
$\big( \big(\frac{1}{5} \big)^2 \big)^x\cdot \frac{1}{5}=125 \\ \big( (5^{-1})^2 \big)^x\cdot 5^{-1}=125 \\ 5^{-2x} \cdot 5^{-1}=125 \\ 5^{-2x-1}=5^3 \\ \downarrow\\ -2x-1=3 \\ -2x=4 \hspace{8pt}\text{/:}(-2) \\ \bm{x=-2 }$Therefore, the correct answer is option a.
$-2$
### Exercise #3
What is the result of the following power?
$(\frac{2}{3})^3$
### Video Solution
$\frac{8}{27}$
### Exercise #4
$4^5-4^6\cdot\frac{1}{4}=\text{?}$
0
### Exercise #5
$(\frac{13}{2})^0\cdot(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}=\text{?}$
### Video Solution
$(\frac{2}{13})^3$
### Exercise #1
$300^{-4}\cdot(\frac{1}{300})^{-4}=?$
1
### Exercise #2
$(\frac{7}{8})^{-2}=\text{?}$
### Video Solution
$1\frac{15}{49}$
### Exercise #3
$(\frac{ax}{b})^{-z}=\text{?}$
### Video Solution
$b^za^{-z}x^{-z}$
### Exercise #4
$(\frac{3}{7})^{-9}=\text{?}$
### Video Solution
$\frac{7^9}{3^9}$
### Exercise #5
Which value is greater?
### Video Solution
$(x^3)^5$ | 4,461 | 15,209 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 89, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2024-26 | latest | en | 0.822938 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1-common-core-15th-edition/chapter-1-foundations-for-algebra-chapter-review-page-71/73 | 1,537,445,721,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156471.4/warc/CC-MAIN-20180920120835-20180920141235-00498.warc.gz | 756,692,828 | 14,645 | Algebra 1: Common Core (15th Edition)
We follow the instructions to plug in 5 for p and -3 for q: $((-3)(5))^{2}$ Note, -3 times 5 is in parenthesis, so we multiply first: $((-3)(5))^{2}$=$(-15)^{2}$=225 (Note, 225 is positive because a -15$\times$-15=225). | 92 | 258 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-39 | longest | en | 0.739541 |
http://www.coppernblue.com/2011/3/8/1719715/why-is-even-strength-play-so-important | 1,409,623,615,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535921318.10/warc/CC-MAIN-20140909054252-00444-ip-10-180-136-8.ec2.internal.warc.gz | 800,961,989 | 23,204 | ## Why Is Even Strength Play So Important?
LOS ANGELES, CA - FEBRUARY 28: Pavel Datsyuk #13 of the Detroit Red Wings controls the puck againstg Jack Johnson #3 of the Los Angeles Kings at Staples Center on February 28, 2011 in Los Angeles, California. The Red Wings won 7-4. (Photo by Stephen Dunn/Getty Images)
Two of the most common requests I encounter when discussing possession metrics and advanced stats here, via e-mail or in person are "Can you use more charts? I love charts, they are so much easier to understand," and "When you talk about these stats, you're always talking about even strength numbers. Why is even strength play so important? I always hear announcers say that the power play and the penalty kill wins games. Can you explain that?"
I've decided to kill two birds with one stone and explain this in chart form. One of my favorite ways to look at how advanced stats impact winning teams is by stripping away team names and evaluating the finishing position of each team by the advanced stat we're interested in. After the jump, I'll give the same treatment to some traditional stats to show why even strength play rules the day.
Our hypothetical e-mailer is right - announcers do often talk about how important the power play is to a winning team, but is that really the case? In the line graph below, I've calculated the combined power play percentage of all of the teams in finishing ranks from 2006-07 through the end of the year (projected).
On top of the graph, I've used the linear regression trend line tool to show the trend. I've included the R-squared value (.525) for the people who often ask for that as well. Essentially, there is some correlation, but given the R-squared value, that correlation isn't particularly strong.
But special teams are very important, right? So if it's not the power play that drives a winning team, it must be the penalty kill. Below, in chart form, is the combined penalty kill percentage of all of the teams in finishing ranks from 2006-07 through the end of the year (projected).
The R-squared value here is .465, indicating even less correlation between finishing rank and penalty kill efficiency.
What happens when we look at special teams as a whole? I've looked at Special Teams Efficiency before and noted the Oilers' historically bad special teams. With yet another line graph, I've totaled the combined STE of all of the teams in finishing ranks from 2006-07 through the end of the year (projected).
There is a stronger correlation (.698) between STE and finishing rank, and it's significant enough to investigate (and yes, I promised Desjardins I would do this).
Now, on to even strength, the subject of this post. I've totaled the combined even strength goal differential of all of the teams in finishing ranks from 2006-07 through the end of the year (projected).
A R-squared of .89! Now we're on to something!
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We have our own Community Guidelines at The Copper & Blue. You should read them. | 926 | 4,223 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-35 | latest | en | 0.948167 |
https://jp.mathworks.com/matlabcentral/answers/440957-splitting-large-data-set-using-if-function | 1,603,960,197,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107903419.77/warc/CC-MAIN-20201029065424-20201029095424-00719.warc.gz | 377,152,207 | 23,190 | # splitting large data set using if function
1 ビュー (過去 30 日間)
pillowheaded 2019 年 1 月 22 日
コメント済み: jahanzaib ahmad 2019 年 1 月 22 日
Hi all!
i have some very large data sets which is 4 coloumns i need to seperate the data out into individual sets. The 3rd coloumn is a repeating set and i want to split the data based upon when it resets so i now have each group of data seperated for example
coloumn 1 coloumn 2 coloumn 3 coloumn 4
1 52 1 47
17 28 2 84
24 34 1 12
i want the data so it would now be
data set 1 = 1 52 1 47 data set 2 = 24 34 1 12
17 28 2 84
sorry i have written this so poorly, any help would be much appreciated!
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### 回答 (1 件)
jahanzaib ahmad 2019 年 1 月 22 日
if B is your data then C has all the sets in cell array
C={};
B=rand(10,4);
for k = 1 : length(B)
Ck =B(k,:);
C{k}=Ck;
end
#### 1 件のコメント
jahanzaib ahmad 2019 年 1 月 22 日
because u said its very large data i m not sure how much time it will take to make sets for each row
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Translated by | 385 | 1,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.602705 |
https://gamedev.stackexchange.com/questions/51397/how-should-i-calculate-the-new-angle-direction-of-my-ball-hitting-a-wall?noredirect=1 | 1,638,756,157,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363229.84/warc/CC-MAIN-20211206012231-20211206042231-00284.warc.gz | 341,996,123 | 31,666 | # How should I calculate the new angle/direction of my ball hitting a wall? [duplicate]
I'm building a Pong game and I am stuck at how the ball should bounce when hitting a horizontal wall. I tried a few methods but none seems to work.
So far I have an update method, which is called every 15ms and which does this:
float scale_x = Math.Cos(ball._angle).ToFloat();
float scale_y = Math.Sin(ball._angle).ToFloat();
//speed multiplied by scale_x is our x velocity
float velocity_x = (gameSpeed * scale_x);
//speed multiplied by scale_y is our y velocity
float velocity_y = (gameSpeed * scale_y);
ball._vector.x += velocity_x;
ball._vector.y += velocity_y;
and if the ball hits the horizontal wall
if (ball._vector.y + velocity_y < 0){
Vector n = new Vector(ball._vector.x, ball._vector.y);
n.nor();
float dot = (ball._vector.dot(n));
}
I've been using this formula R = 2 * (V dot N) * N - V to calculate the new direction. But I can't seem to get it to work. To be honest, I'm not sure what kind of formula I need. I just need to find the new angle when it hits the wall.
Here is the solution for my game. first i made velocity_x and y instance variables.
private float velocity_x;
private float velocity_y;
then i sat the start for my direction in my constructor
float scale_x = Math.Cos(ball._angle).ToFloat(); //The x scale is the cosine of the angle
float scale_y = Math.Sin(ball._angle).ToFloat(); //The y scale is the sin of the angle
velocity_x = (gameSpeed * scale_x); //speed multiplied by scale_x is our x velocity
velocity_y = (gameSpeed * scale_y); //speed multiplied by scale_y is our y velocity
Finally in my updateBall method i leave:
ball._vector.x += velocity_x;
ball._vector.y += velocity_y;
• N in the formula refers to the normal of your reflective surface, eg. your wall. A wall to the left would have a normal pointing to the right: x: 1, y: 0 etc. Mar 20 '13 at 14:09
• I know this might be a duplicate, have already checked that post. But I cant get it to work. So N at a horizontal wall, will have x: 0, y:1?
– Alex
Mar 20 '13 at 14:22
• @Alex Yes. A horizontal wall at the bottom has { x: 0, y: 1 }, a wall at the top has { x: 0, y: -1 } (assuming your coordinates start bottom-left and increase up and right). Mar 20 '13 at 14:33
• @Alex Unfortunately, not being able to implement an existing answer isn't really reason enough to open a new question. If you can get 20 rep answering questions you can come to chat and ask about it. Learn some more about normals in the mean time and see if that helps.
– House
Mar 20 '13 at 14:33
• Why not open Xna's built in method Vector2.Reflect() in your favorite assembly viewer like reflector or dotPeek and see how Xna does it? Mar 20 '13 at 14:48
If what you're looking for is a simple reflection to a horizontal surface, then all you have to do is flip the y-component of the velocity vector depending on the normal of the surface you're hitting.
if(collision_horizontal){
velocity_y = -velocity_y;
}
similarly:
if(collision_vertical){
velocity_x = -velocity_x;
}
• I tried to do this already, but I just end up getting a weird result, where the ball will follow the horizontal wall.
– Alex
Mar 21 '13 at 11:30
• this was actually the answer, but I have tried that formula a couple of time. The problem was my update ball, that would refresh my answer no matter what I've done. So i move my scale_y & y and velocity_x & y to constructor and made velocity x & y instance variables
– Alex
Mar 21 '13 at 12:16
The formula R = 2 * (V dot N) * N - V requires normalized vectors for N, in order to calculate the new direction.
V would be a 2-dimensional vector defining the velocity or direction the ball is travelling in (in your case scale_x/y or velocity_x/y)
N defines the normal vector (length=1) of the collision plane. E.g. (0,1) for a horizontal plane and (1,0) for a vertical plane.
This will yield R, the reflected velocity vector, or the new direction
Using this formula would be overkill however, since you are only colliding with horizontal (and I expect vertical) planes. So I would suggest flipping the respective component of your velocity vector. For a horizontal plane this would be the y-component, for a vertical plane the x-component.
Another thought: Is there a specific reason why you are working with an angle for the direction of the ball? Cos and Sin is usually not good for performance and should be avoided (although in the case of pong it is probably negligible). I would try to use a direction vector instead.
• There is no specific reason why im using sin and cos. I could try using a direction vector instead.
– Alex
Mar 21 '13 at 11:31
• What is this formula? Is there a name? A related wiki article or something? I would like to read up on what, why and how we're achieving here. Oct 18 '20 at 20:34
If you simply need a new angle when a barrier is hit then:
[new angle] = 2*[barrier angle] - [old angle]
In order to cap it to the normal range you should implement that as:
[new angle] = ( 2*PI + 2*[barrier angle] - [old angle] ) mod 2*PI
Assuming radians, if you use degrees then you should use 360 instead of 2*PI. | 1,321 | 5,155 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-49 | latest | en | 0.851334 |
https://fr.slideshare.net/VijeeshSoman1/central-limit-theorem | 1,566,671,672,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321351.87/warc/CC-MAIN-20190824172818-20190824194818-00146.warc.gz | 473,046,366 | 37,173 | Ce diaporama a bien été signalé.
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# Central limit theorem
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### Central limit theorem
1. 1. Central Limit Theorem Presented By Vijeesh S1-MBA (PT)
2. 2. Introduction The Central Limit Theorem describes the relationship between the sampling distribution of sample means and the population that the samples are taken from.
3. 3. Normal Populations Important Fact: If the population is normally distributed, then the sampling distribution of x is normally distributed for any sample size n.
4. 4. Sampling Distribution of x- normally distributed population n=10 /10 Population distribution: N( , ) Sampling distribution of x: N( , /10)
5. 5. Non-normal Populations What can we say about the shape of the sampling distribution of x when the population from which the sample is selected is not normal? 53 490 102 72 35 21 26 17 8 10 2 3 1 0 0 1 0 100 200 300 400 500 600 Frequency Salary (\$1,000's) Baseball Salaries
6. 6. The Central Limit Theorem If a random sample of n observations is selected from a population (any population), then when n is sufficiently large, the sampling distribution of x will be approximately normal. (The larger the sample size, the better will be the normal approximation to the sampling distribution of x.)
7. 7. • Suppose that a sample is obtained containing a large number of observations, each observation being randomly generated in a way that does not depend on the values of the other observations, and that the arithmetic average of the observed values is computed. • If this procedure is performed many times, the central limit theorem says that the computed values of the average will be distributed according to the normal distribution
8. 8. For eg: Suppose we have a population consisting of the numbers {1,2,3,4,5} and we randomly selected two numbers from the population and calculated their mean. For example, we might select the numbers 1 and 5 whose mean would be 3. Suppose we repeated this experiment (with replacement) many times. We would have a collection of sample means ( millions of them). We could then construct a frequency distribution frequency distribution of these sample means. The resulting distribution of sample means is called the sampling distribution of sample means. Now, having the distribution of sample means we could proceed to calculate the mean of all sample means (grand mean) and their
9. 9. The Central Limit Theorem predicts that regardless of the distribution of the parent population: [1] The mean of the population of means is always equal to the mean of the parent population from which the population samples were drawn. [2] The standard deviation (standard error ) of the population of means is always equal to the standard deviation of the parent population divided by the square root of the sample size (N). SD’= SD/√N
10. 10. If the population from which samples are drawn is normally distributed then the sampling distribution of sample means will be normally distributed regardless of the size of the sample, and the CLT is not needed. But, if the population is not normal, the CLT tells us that the sampling distribution of sample means will be normal provided the sample size is sufficiently large. How large must the sample size be so that the sampling distribution of the mean becomes a normal distribution? If the samples were drawn from a population with a high degree of skewness (not normal), the sample size must be 30 or more before the sampling distribution of the mean becomes a normal distribution. A sample of size 30 or more is called a large sample and as a sample of size less that thirty is called a small sample. | 919 | 4,035 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-35 | latest | en | 0.648954 |
https://astronomy.stackexchange.com/questions/44989/is-it-nonsense-to-even-talk-about-objects-outside-the-observable-universe-not | 1,660,383,244,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571911.5/warc/CC-MAIN-20220813081639-20220813111639-00683.warc.gz | 134,972,166 | 61,605 | # Is it "nonsense to even talk about" objects outside the observable universe not having gravitational influence on us? (finite speed of gravity)
In this supplemental answer to Is the zero gravity experienced in ISS the “artificial” kind? in Space Exploration SE I said:
1. Gravity moves at the speed of light so nothing outside out observable universe pulls on us.
After a series of comments below it this was said:
There are endless discussions all over the net about it. Usually they ask: if the earth was attracted to the retarded location of the sun it would it not slow down very quickly? And the answer is yet it would, but the scalar potential is instantaneous, see here https://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html About objects outside the observable universe, it is nonsense to even talk about it: thats not the way GR works.
Question: Focusing on the last sentence; is it really "nonsense to even talk about" objects currently outside the observable universe but may become visible in the future not having any gravitational influence on us because gravity travels at the speed of light, or is this a useful concept even if perhaps an inexact expression when GR is embraced?
First, some definitions.
The limits of observability in our Universe are actually set by cosmological horizons, of which the "observable Universe" is just one type. These depend on not only whether can we detect the thing from Earth, but also on when was the signal produced?
There are various types of cosmic horizons:
The particle horizon sets a limit on the distance that can be seen due to the finite age of the universe - this is likely what you meant by "observable Universe." That is, the particle horizon represents the largest comoving distance from which light could have reached the observer by a specific time. Due to the expansion of the space of the Universe, this is not the age of the universe times the speed of light, as in the Hubble horizon (see below), but rather the speed of light multiplied by the conformal time.
The Hubble horizon is a theoretical horizon defining the boundary between particles that are moving slower or faster than the speed of light relative to an observer at a given time. This does not mean the particle is unobservable, since the light from the past is able to reach the observer. In the current models of the expanding Universe, light emitted from the Hubble horizon would reach us in finite time. (these details depend on the sign of the Hubble parameter).
The cosmic event horizon is the largest comoving distance from which light emitted now can reach an observer in the future. The current distance to our cosmic event horizon is about 16 billion light-years, which is within the "observable" sphere given by the particle horizon.
There are also "practical horizons" such as the surface of last scattering for photons (ie. recombination, origin of the cosmic microwave background), and suspected surfaces of last scattering for early universe neutrinos and gravitational waves.
Lastly, depending on the expansion of the Universe and whether we continues expanding forever, there could be a "future horizon," which sets a limit on the farthest distance that can possibly be measured in units of today's proper distance.
Keep in mind that this terminology is not always used rigidly and consistently, for example:
"Sometimes astrophysicists distinguish between the visible universe, which includes only signals emitted since recombination (when hydrogen atoms were formed from protons and electrons and photons were emitted)—and the observable universe, which includes signals since the beginning of the cosmological expansion (the Big Bang in traditional physical cosmology, the end of the inflationary epoch in modern cosmology)."
is it really "nonsense to even talk about" objects currently outside the observable universe but may become visible in the future not having any gravitational influence on us because gravity travels at the speed of light, or is this a useful concept even if perhaps an inexact expression when GR is embraced?
I hope with the definitions above, it is clearly NOT nonsense and clearly depends on what we mean by "observable Universe." Even taking the simplest example of the particle horizon, the Copernican principle implies that every point in the Universe has its own particle horizon, which implies that the objects near the edge of our particle horizon could be effected by objects beyond that edge. We would indirectly observe those effects. One can imagine more complicated scenarios, but I think this makes the point. | 906 | 4,636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-33 | longest | en | 0.952188 |
http://www.ecnmag.com/blogs/2012/12/designing-just-fun?qt-recent_content=0 | 1,429,599,117,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246640550.27/warc/CC-MAIN-20150417045720-00276-ip-10-235-10-82.ec2.internal.warc.gz | 467,506,941 | 15,892 | Blogs
# Designing just for fun
Thu, 12/20/2012 - 3:58pm
M. Simon, Technical Contributor
I'm in the process of designing and building a 10 MHz time/frequency receiver to pick up the WWV signal and to see if I can accurately reproduce the signal frequency for general lab calibration purposes. Yes, there are better ways to get accurate frequency calibration. A GPS Disciplined Oscillator (GPSDO) is one way to go. But a 10 MHz receiver is something I always wanted to do, so I'm doing it. My expectation is to be able to get the frequency accuracy to within 1E7. My hope is to be able to get to within 1E8. Because of ionospheric shifts, that is probably the best that can be done. According to this calculator, I'm 1378.8 Km from Boulder, Colorado (close to the the site of WWV). Let us say 1500 Km for order of magnitude purposes. To hold frequency to 1E7, the path length to the ionosphere and back to ground level must hold to within 30 meters over 1 second. To get to 1E8, the path has to be constant within about 3 meters over 1 second. If I can get the receiver to work (no guarantee because the design is novel), I can use it to get ionospheric soundings between here and Boulder by comparing the receiver frequency output to a fixed oscillator. A topic for a later date if I can get the device working.
While doing research for the design, I came across a number of interesting www pages. I'm going to list a few of them. How to Use Analog Switches as Mixers is one. Analog switches are very good with low-level signals, but their frequency range is limited to about 100 MHz or less for off-the-shelf components. Note that not all the links on this page are working links.
Here is a whole list of circuits that captured my attention. I particularly liked the Ultra Low Noise High Input Impedance DC Amplifier and Using HCMOS Gates as Frequency Multipliers. The circuit diagram showing how to use a 74HC74 flip flop to multiply a frequency by 1.25 was particularly fascinating. It uses tuned components so it is not a general purpose multiplier.
This page of circuits which was linked here has a stunning picture of a Japanese temple on the top of the page. I particularly liked circuit #34. A one transistor AM modulated CW transmitter for code practice. The transistor generates both the RF signal and the signal that modulates it.
It is back to the bench for me now. I have some projects to complete.
M. Simon's e-mail can be found on the sidebar at Space-Time Productions
Engineering is the art of making what you want from what you can get at a profit. | 598 | 2,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-18 | latest | en | 0.954964 |
http://factoring-polynomials.com/factoring-polynomials-5.htm | 1,511,033,992,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805023.14/warc/CC-MAIN-20171118190229-20171118210229-00120.warc.gz | 103,517,945 | 8,757 | Home A Summary of Factoring Polynomials Factoring The Difference of 2 Squares Factoring Trinomials Quadratic Expressions Factoring Trinomials The 7 Forms of Factoring Factoring Trinomials Finding The Greatest Common Factor (GCF) Factoring Trinomials Quadratic Expressions Factoring simple expressions Polynomials Factoring Polynomials Fractoring Polynomials Other Math Resources Factoring Polynomials Polynomials Finding the Greatest Common Factor (GCF) Factoring Trinomials Finding the Least Common Multiples
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:
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Solve for:
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After studying this lesson, you will be able to:
• Factor various types of problems.
Steps of Factoring:
1. Factor out the GCF
2. Look at the number of terms:
• 2 Terms: Look for the Difference of 2 Squares
• 3 Terms: Factor the Trinomial
• 4 Terms: Factor by Grouping
3. Factor Completely
4. Check by Multiplying
This section is a review of the types of factoring we'vecovered so far. Follow the steps listed above to factor theproblems.
Example 1
Factor 3x 2 - 27
1 st : Look for a GCF....the GCF is 3 so we factorout 3: 3( x 2 - 9)
2 nd : Look at the number of terms in theparenthesis. There are 2 terms and it is the difference of 2squares. We factor the difference of 2 squares (keeping the 3).3(x + 3) ( x - 3)
3 rd : Now, make sure the problem is factoredcompletely. It is.
4 th : Check by multiplying.
Example 2
Factor 9y 2 - 42y + 49
1 st : Look for a GCF....the GCF is 1 so we don'thave to worry about that.
2 nd : Look at the number of terms. There are 3terms so we factor the trinomial.
-make 2 parentheses
-using the sign rules, we know the signs will be the samebecause the constant term is positive
- we also know they will be negative because theinside/outside combination must equal -58y
-find the factors of the 1 st term: 1y, 9y and 3y, 3y . Let'stry 3y, 3y
-find the factors of the constant term: 1, 49 and 7, 7. Let'stry 7, 7 (3y - 7) (3y - 7)
-check the inside/outside combination: inside we have -21y andoutside we have -21y which adds up to -42y
3 rd : Now, make sure the problem is factoredcompletely. It is.
4 th : Check by multiplying.
Example 3
Factor x 3 - 5x 2 - 9x + 45
1 st : Look for a GCF....the GCF is 1 so we don'thave to worry about that.
2 nd : Look at the number of terms. There are 4terms so we factor by grouping.
Group the terms (x 3 - 5x 2 ) + (- 9x +45 )
Take the GCF of the each group: x 2 (x -5 )(- 9(x - 5 ))
Take the GCF of the entire problem: (x - 5 )(x2 -9)
3 rd : Now, make sure the problem is factoredcompletely. It isn't. We can factor the second parenthesis.
(x - 5 )(x + 3)(x - 3)
4 th : Check by multiplying. | 957 | 3,153 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | latest | en | 0.817378 |
https://www.sanfoundry.com/structural-analysis-questions-answers-determinacy-stability/ | 1,717,066,340,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059632.19/warc/CC-MAIN-20240530083640-20240530113640-00719.warc.gz | 860,970,410 | 20,252 | # Structural Analysis Questions and Answers – Determinacy and Stability
This set of Structural Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Determinacy and Stability”.
1. For the validity of principle of superposition, materials should behave in which manner?
a) linear-elastic
b) non-linear-elastic
c) Non-linear- inelastic
d) Linear- inelastic
Explanation: They should behave in a linear elastic manner so that, Hooke’s law is valid.
2. If in planar system, X parts/members are there with Y no. of forces, then condition for statically determinacy is:-
a) Y < 3X
b) Y > 3X
c) Y = 3X
d) None of the mentioned
Explanation: There can be at max. 3 equilibrium equations for each part.
3. If Y > 3X (X and Y are from the above question) then, the system is:-
a) Statically indeterminate
b) Statically determinate
c) Can’t say
d) Depends on other conditions
Explanation: As there can be at max. 3X equations, at least one of the forces will be unsolvbale.
4. If in a planar system, only 2 reaction forces are acting, then the system is:-
a) Essentially unstable
b) Essentially stable
c) Can’t say
d) None of the mentioned
Explanation: If no. of reactions is less than 3 in any planar system, then the system is essentially unstable.
5. If all the reactions acting on a planar system are concurrent in nature, then the system is:-
a) Can’t say
b) Essentially stable
c) Essentially unstable
d) None of the mentioned
Explanation: The system can rotate about the point of concurrency. So, it is essentially unstable.
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6. If 4 reactions are acting on a beam, then the system is:-
a) Unstable & indeterminate
b) Stable & indeterminate
c) Stable & determinate
d) Can’t say
Explanation: 4 reactions mean that the system is definitely indeterminate. But stability would depend upon the nature of forces acting on the planar structure.
7. If a system has more equations of equilibrium than no. of forces, then the system is:-
a) Improperly constrained
b) Partially constrained
c) Stable
d) Solvable
Explanation: In these cases, system is unstable and unsolvable and is termed as partially constrained.
8. How many cases out of the following are improperly constrained?
Parallel forces, concurrent forces, perpendicular forces, only moment
a) 1
b) 2
c) 3
d) 4
Explanation: parallel and concurrent are improperly constrained.
9. If a structure has total 10 joints, then what should be the minimum no. of joints in which equilibrium equations should be concurrently satisfied for stability?
a) 7
b) 8
c) 9
d) 10
Explanation: For stability, equilibrium equations should be satisfied concurrently at each and every joint of the structure.
10. If a structure has 2j – r no. of members, then it will be:-
a) stable
b) unstable
c) depends upon structure
d) depends upon magnitude of load
Explanation: In these cases, structures can be stable as well unstable.
Here, j represent no. of joints and r represents no. of external forces.
11. If a truss consists of a non-triangular element, then it will essentially be unstable.
State whether the above statement is true or false.
a) true
b) false
Explanation: In these cases, instability is the most probable occurrence, but there are a lot of examples which are unstable.
Sanfoundry Global Education & Learning Series – Structural Analysis.
To practice all areas of Structural Analysis, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected] | 877 | 3,620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.868706 |
https://www.transtutors.com/questions/applied-statistical-modeling-must-know-sas-program-and-must-have-sas-program--28412.htm | 1,576,353,985,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541288287.53/warc/CC-MAIN-20191214174719-20191214202719-00046.warc.gz | 869,227,432 | 28,931 | # Applied statistical modeling (Must know SAS program and must have SAS program) 1 answer below »
This is an applied course on statistical modeling. The course will cover simple linear regression, multiple linear regression, analysis of variance(ANOVA), analysis of covariance(ANCOVA), generalized linear models, generalized estimating equations methods, and linear mixed effects models. Students will learn how to use SAS to perform statistical analyses. Textbook: Applied Linear Statistical Models, 5th Edition, by Kutner, Nachtsheim, Neter, and Li. SAS code you can use/find http://www.ats.ucla.edu/stat/sas/examples/alsm/
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Due: May, 2 2011 (Monday) (Must know SAS program/coding) This is an applied course on statistical modeling. The course will cover simple linear regression, multiple linear regression, analysis of variance(ANOVA), analysis of covariance(ANCOVA), generalized linear models, generalized estimating equations methods, and linear mixed effects models. Students will learn how to use SAS to perform statistical analyses. Textbook: Applied Linear Statistical Models, 5th Edition, by Kutner, Nachtsheim, Neter, and Li. SAS code you can use/find http://www.ats.ucla.edu/stat/sas/examples/alsm/ Assignment Problem 1: (simple/muti linear regression model problem) An investigator is studying the relationship between weight (in pounds) and height (in inches) using data from a sample of 126 high school students. A simple linear regression model is used to model the relationship between weight (Y) and height (X). Partial information from the SAS PROC REG output from fitting this model is provided bellow. a) Please complete the ANOVA table and related results by using the SAS output from PROC MEANS which gives the sample means and the sample standard deviations for the two variables involved. b) Obtain a 99% prediction interval for the weight of a 15 years-old student who has a height of 68 inches. Can we use our model to predict the weight of a 54 years-old teacher with a height of 68 inches? Why or why not? c) Calculate the coefficient of correlation between weight and height. Are they positively or negative correlated? d) How do you interpret that? Is the result statistically significant at the 0.05 level? ANOVA Table and related results from Proc reg Sum of Mean Source DF Squares Square F Value Pr > F Model ? 31126 ? ? ? Error ? ? ...
## 1 Approved Answer
Pradeepika S
3 Ratings, (9 Votes)
generalized estimating equation method mparing two treatments for a respiratory disorder. See "Gee Model for Binary Data" in the SAS/STAT Sample Program Library for the complete data set. These data are from Stokes, Davis, and Koch (1995), where a SAS macro is used to fit a GEE model. A GEE model is fit, using the REPEATED statement in the GENMOD procedure. Table: Respiratory Disorder Data Patients in each of two centers are randomly assigned to groups receiving the active treatment or a placebo. During treatment, respiratory status (coded here as 0=poor, 1=good) is determined for each of four visits. The variables center, treatment, sex, and baseline (baseline respiratory status) are classification variables with two levels. The variable age (age at time of entry into the study) is a continuous variable. Explanatory variables in the model are Intercept ( xij 1), treatment ( xij 2), center ( xij 3), sex ( xij 4), age ( xij 6), and baseline ( xij 6), so that x ij' = [ xij 1, xij 2, ... , xij 6] is the vector of explanatory variables. Indicator variables for the classification explanatory variables can be automatically generated by listing them in the CLASS statement in PROC GENMOD. However, in order to be consistent with the analysis in Stokes, Davis, and Koch (1995), the four classification explanatory variables are coded as follows: Suppose yij represents the respiratory status of patient i at the j th visit, j = 1, ... ,4, and represents the mean of the...
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Attach Files | 907 | 4,059 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2019-51 | latest | en | 0.833822 |
https://programming.vip/docs/abstract-base-class-abc.html | 1,638,273,548,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358973.70/warc/CC-MAIN-20211130110936-20211130140936-00506.warc.gz | 522,398,672 | 6,483 | # Abstract base class ABC
## Ellipse class
Develop a graphics program to display circles and ellipses. A Circle is a special case of an Ellipse. The major axis and minor axis are ellipses of equal length. Therefore, all circles are ellipses, and the Circle class can be derived from the Ellipse class. The data members include the coordinates of the Ellipse center, semi major axis, short semi axis and direction angle. It also includes some methods to move the Ellipse, return the area of the Ellipse, rotate the Ellipse, and scale the long and short semi axes.
```class Ellipse
{
private:
double x; //x coordinate of ellipse center
double y; //y coordinate of ellipse center
double a; //Semimajor axis
double b; //Short half axis
double angle; //Direction angle
...
public:
void Move(int nx, int ny) { x = nx; y = ny; }
virtual double Area() const { return 3.14159 * a * b; }
virtual void Rotable(double nang) { angle += nang; }
virtual void Scale(double sa, double sb) { a *= sa; b *= sb; }
...
};
```
Now derive a Circle class from the Ellipse class:
```class Circle : public Ellipse
{
...
};
```
Although a Circle is an ellipse, this derivation is clumsy. For example, a Circle needs only one value (radius) to describe its size and shape, and needs to have a long semi axis and a short semi axis. The Circle constructor can take care of this situation by assigning the same value to members a and b, but this will lead to information redundancy. The angle parameter and the Rotate() method have no practical meaning for circles; The Scale() method will scale the two axes differently and turn the Circle into an ellipse. Some techniques can be used to fix these problems, such as including the redefined Rotate() method in the private part of the Circle class, so that Rotate() cannot be used for circles publicly. In general, however, inheritance cannot be used. It is simpler to define the Circle class directly:
```class Circle : public Ellipse
{
private:
double x;
double y;
double r;
...
public:
...
void Move(int nx, int ny) { x = nx; y = ny; }
double Area() const { return 3.14159 * r * r; }
void Scale(double sr) { r *= sr; }
...
};
```
Class contains only the required members. But this solution is not efficient. The Circl and Ellipse classes have a lot in common. Defining them separately ignores this fact.
Another solution is to abstract their commonalities from the Ellipse and Circle classes and put these features into an ABC. The Circle and Ellipse classes are then derived from this ABC. You can use the base class pointer array to manage both Circle and Ellipse objects, that is, you can use polymorphic methods. What these two classes have in common are the central coordinates, the Move() method (which is the same for the two classes), and the Area() method (which is different for the two classes). You can't even implement the Area() method in ABC because it doesn't contain the necessary data members.
C + + provides unimplemented functions by using pure virtual functions. The declaration of a pure virtual function ends with = 0.
```class BaseEllipse //Abstract base class
{
private:
double x;
double y;
...
public:
BaseEllipse(double x0 = 0, double y0 = 0) : x(x0),y(y0) {}
virtual ~BaseEllipse() {}
void Move(int nx, int ny) { x = nx; y = ny; }
virtual double Area() const = 0; //Pure virtual function
...
};
```
When a class declaration contains a pure virtual function, you cannot create an object of the class. Classes containing pure virtual functions are used only as base classes. To be a true ABC, you must include at least one pure virtual function. In principle, = 0 makes the virtual function a pure virtual function. The method Area() here is not defined, but C + + even allows pure virtual functions to be defined.
For example, all base class methods, like Move(), can be defined in the base class, but this class still needs to be declared abstract. In this case, you can declare the prototype as virtual:
```void Move(int nx, int ny) = 0;
```
This will make the base class abstract, but you can still implement the definition of the method provided in the file:
```void BaseEllipse::Move(int nx, int ny) { x = nx; y = ny; }
```
Use = 0 in the prototype to indicate that the class is an abstract base class, and the function can not be defined in the class.
You can derive the Ellipse class and the Circle class from the BaseEllipse class, and add the required members to complete each class. Note that the Circle class always represents a Circle, while the Ellipse class always represents an Ellipse. However, the Ellipse class Circle can be rescaled to a non Circle, while the Ciecle class must always be a Circle.
Programs of these classes can create Ellipse objects and Circle objects, but cannot create BaseEllipse objects. Since the base classes of Circle and Ellipse objects are the same, you can use BaseEllipse pointer array to manage both objects at the same time. Classes like Circle and Ellipse are sometimes referred to as concrete classes, which means that these types of objects can be created.
ABC describes an interface that uses at least one pure virtual function. Classes derived from ABC will use conventional virtual functions to implement this interface according to the specific characteristics of the derived classes.
## ABC
First, define an ABC named AcctABC. This class contains all methods and data members common to Brass and Brass plus classes, and those methods that behave differently in Brass plus and Brass classes should be declared as virtual functions. At least one virtual function should be a pure virtual function, so that AcctABC can become an abstract class.
## acctabc.h
To help derived classes access base class data, AcctABC provides some protection methods; Derived class methods can call these methods, but they are not part of the public interface of the derived class object. AcctABC provides a protected member function to handle formatting. In addition, the AcctABC class has two pure virtual functions, so it is indeed an abstract class.
```#ifndef ACCTABC_H_
#define ACCTABC_H_
#include <iostream>
#include <string>
//Abstract base class
class AcctABC
{
private:
std::string fullName;
long acctNum;
double balance;
protected:
struct Formatting
{
std::ios_base::fmtflags flag;
std::streamsize pr;
};
const std::string & FullName() const { return fullName; }
long AcctNum() const { return acctNum; }
Formatting SetFormat() const;
void Restore(Formatting & f) const;
public:
AcctABC(const std::string & s = "Nullbody", long an = -1, double bal = 0.0);
void Deposit(double amt);
virtual void Withdraw(double amt) = 0;
double Balance() const { return balance; };
virtual void ViewAcct() const = 0;
virtual ~AcctABC() {}
};
//Brass Account class
class Brass :public AcctABC
{
public:
Brass(const std::string & s = "Nullbody", long an = -1, double bal = 0.0) : AcctABC(s, an, bal) {}
virtual void Withdraw(double amt);
virtual void ViewAcct() const;
virtual ~Brass() {}
};
//Brass Plus Account class
class BrassPlus :public AcctABC
{
private:
double maxLoan;
double rate;
double owesBank;
public:
BrassPlus(const std::string & s = "Nullbody", long an = -1, double bal = 0.0, double ml = 500, double r = 0.10);
BrassPlus(const Brass & ba, double ml = 500, double r = 0.1);
virtual void ViewAcct() const;
virtual void Withdraw(double amt);
void ResetMax(double m) { maxLoan = m; }
void ResetRate(double r) { rate = r; }
void ResetOwes() { owesBank = 0; }
};
#endif
```
## acctabc.cpp
```#include <iostream>
#include "acctabc.h"
using std::cout;
using std::ios_base;
using std::endl;
using std::string;
AcctABC::AcctABC(const std::string & s, long an, double bal)
{
fullName = s;
acctNum = an;
balance = bal;
}
void AcctABC::Deposit(double amt)
{
if (amt < 0)
cout << "Negative deposit not allowed; "
<< "deposit is cancelled.\n";
else
balance += amt;
}
void AcctABC::Withdraw(double amt)
{
balance -= amt;
}
AcctABC::Formatting AcctABC::SetFormat() const
{
Formatting f;
f.flag = cout.setf(ios_base::fixed, ios_base::floatfield);
f.pr = cout.precision(2);
return f;
}
void AcctABC::Restore(Formatting & f) const
{
cout.setf(f.flag, ios_base::floatfield);
cout.precision(f.pr);
}
void Brass::Withdraw(double amt)
{
if (amt < 0)
cout << "Withdrawal amount must be positive; "
<< "withdrawal canceled.\n";
else if (amt <= Balance())
AcctABC::Withdraw(amt);
else
cout << "Withdrawal amount of \$" << amt
<< "Withdrawal canceled.\n";
}
void Brass::ViewAcct() const
{
Formatting f = SetFormat();
cout << "Brass Client: " << FullName() << endl;
cout << "Account Number: " << AcctNum() << endl;
cout << "Balance: \$" << Balance() << endl;
Restore(f);
}
BrassPlus::BrassPlus(const string & s, long an, double bal, double ml, double r) : AcctABC(s, an, bal)
{
maxLoan = ml;
owesBank = 0.0;
rate = r;
}
BrassPlus::BrassPlus(const Brass & ba, double ml, double r) : AcctABC(ba)
{
maxLoan = ml;
owesBank = 0.0;
rate = r;
}
void BrassPlus::ViewAcct() const
{
Formatting f = SetFormat();
cout << "Brass Client: " << FullName() << endl;
cout << "Account Number: " << AcctNum() << endl;
cout << "Balance: \$" << Balance() << endl;
cout << "Maximum loan: \$" << maxLoan << endl;
cout << "Owed to bank: \$" << owesBank << endl;
cout.precision(3);
cout << "Loan Rate: " << 100 * rate << "%\n";
Restore(f);
Restore(f);
}
void BrassPlus::Withdraw(double amt)
{
Formatting f = SetFormat();
double bal = Balance();
if (amt <= bal)
AcctABC::Withdraw(amt);
else if (amt <= bal + maxLoan - owesBank)
{
double advance = amt - bal;
owesBank += advance * (1.0 + rate);
cout << "Finance charge: \$" << advance * rate << endl;
AcctABC::Withdraw(amt);
}
else
cout << "Credit limit exceeded. Transaction cancelled.\n";
Restore(f);
}
```
The protection methods FullName() and AcctNum() provide read-only access to the data members fullName and acctNum, making it possible to further customize the ViewAcct() of each derived class.
After improving the output format, a structure is defined to store two format settings; And use this structure to set and restore the format, so only two function calls are required:
```struct Formatting
{
std::ios_base::fmtflags flag;
std::streamsize pr;
};
....
Formatting f = SetFormat();
...
Restore(f);
```
These functions and structure Formatting are placed in a separate namespace, but they protect access rights, so these structures and functions are placed in the protection part of the class definition. This makes them available to base and derived classes while hiding them out.
## main.cpp
```#include <iostream>
#include <string>
#include "acctabc.h"
const int CLIENTS = 4;
int main()
{
using std::cin;
using std::cout;
using std::endl;
AcctABC * p_clients[CLIENTS];
std::string temp;
long tempnum;
double tempbal;
char kind;
for (int i = 0; i < CLIENTS; i++)
{
cout << "Enter client's name: ";
getline(cin, temp);
cout << "Enter client's account number: ";
cin >> tempnum;
cout << "Enter opening balance: \$";
cin >> tempbal;
cout << "Enter 1 for Brass Account or "
<< "2 for BrassPlus Account: ";
while (cin >> kind && (kind != '1' && kind != '2'))
cout << "Enter either 1 or 2: ";
if (kind == '1')
p_clients[i] = new Brass(temp, tempnum, tempbal);
else
{
double tmax, trate;
cout << "Enter the overdraft limit: \$";
cin >> tmax;
cout << "Enter the interest rate "
<< "as a decimal fraction: ";
cin >> trate;
p_clients[i] = new BrassPlus(temp, tempnum, tempbal, tmax, trate);
}
while (cin.get() != '\n')
continue;
}
cout << endl;
for (int i = 0; i < CLIENTS; i++)
{
p_clients[i]->ViewAcct();
cout << endl;
}
for (int i = 0; i < CLIENTS; i++)
{
delete p_clients[i];
}
cout << "Done.\n";
return 0;
}
```
Keywords: C++
Added by joelg on Mon, 08 Nov 2021 05:31:05 +0200 | 2,970 | 11,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.81918 |
http://www.zhishibo.com/articles/280203.html | 1,652,740,587,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662512249.16/warc/CC-MAIN-20220516204516-20220516234516-00586.warc.gz | 127,414,352 | 8,611 | # leetcode 3. Longest Substring Without Repeating Characters 无重复字符的最长子串(leetcode 3. Longest substring without repeating characters)-其他
## leetcode 3. Longest Substring Without Repeating Characters 无重复字符的最长子串(leetcode 3. Longest substring without repeating characters)
### 一、题目大意
https://leetcode.cn/problems/longest-substring-without-repeating-characters/
请注意，你的答案必须是 子串 的长度，”pwke” 是一个子序列，不是子串。
请注意，你的答案必须是 子串 的长度，”pwke” 是一个子序列，不是子串。
• 0 <= s.length <= 5 * 104
• s 由英文字母、数字、符号和空格组成
### 3.1 Java实现
``````public class Solution {
public int lengthOfLongestSubstring(String s) {
int[] chars = new int[128];
int left = 0;
int maxSize = 0;
// 窗口右边界向右滑
for (int right = 0; right < s.length(); right++) {
chars[s.charAt(right)]++;
if (chars[s.charAt(right)] == 1) {
maxSize = Math.max(right - left + 1, maxSize);
} else {
// 窗口左边界向右滑
while(chars[s.charAt(right)] != 1) {
chars[s.charAt(left)]--;
left++;
}
}
}
return maxSize;
}
}
``````
### 四、总结小记
• 2022/5/15 滑动窗口，重点是：窗口在滑动的过程中记录什么值最终要求什么值，以及窗口左边界向右滑的结束条件是什么
————————
### 1、 General idea of the topic
https://leetcode.cn/problems/longest-substring-without-repeating-characters/
Given a string s, please find out the length of * * longest substring * * which does not contain duplicate characters.
< strong > example 1: < / strong >
Enter: S = “abcabcbb”
Output: 3
Explanation: because the longest substring without repeated characters is “ABC”, its length is 3.
Enter: S = “abcabcbb”
Output: 3
Explanation: because the longest substring without repeated characters is “ABC”, its length is 3.
< strong > example 2: < / strong >
Input: S = “bbbbb”
Output: 1
Explanation: because the longest substring without repeated characters is “B”, its length is 1.
Input: S = “bbbbb”
Output: 1
Explanation: because the longest substring without repeated characters is “B”, its length is 1.
< strong > example 3: < / strong >
Input: S = “pwwkew”
Output: 3
Explanation: because the longest substring without repeated characters is “WKE”, its length is 3.
Please note that your answer must be the length of the substring, “pwke” is a substring, not a substring.
Input: S = “pwwkew”
Output: 3
Explanation: because the longest substring without repeated characters is “WKE”, its length is 3.
Please note that your answer must be the length of the substring, “pwke” is a substring, not a substring.
< strong > prompt: < / strong >
• 0 <= s.length <= 5 * 104
• S ， consists of English letters, numbers, symbols and spaces
### 2、 Problem solving ideas
Solution of sliding window
### 3.1 Java实现
``````public class Solution {
public int lengthOfLongestSubstring(String s) {
int[] chars = new int[128];
int left = 0;
int maxSize = 0;
// 窗口右边界向右滑
for (int right = 0; right < s.length(); right++) {
chars[s.charAt(right)]++;
if (chars[s.charAt(right)] == 1) {
maxSize = Math.max(right - left + 1, maxSize);
} else {
// 窗口左边界向右滑
while(chars[s.charAt(right)] != 1) {
chars[s.charAt(left)]--;
left++;
}
}
}
return maxSize;
}
}
``````
### 4、 Summary notes
• 2022 / 5 / 15 sliding window, focusing on: what value is recorded in the sliding process of the window, what value is finally required, and what is the end condition of sliding the left boundary of the window to the right | 983 | 3,221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-21 | latest | en | 0.588867 |
https://www.howcast.com/videos/396556-how-to-solve-a-maze/ | 1,524,498,123,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946077.4/warc/CC-MAIN-20180423144933-20180423164933-00037.warc.gz | 798,091,868 | 15,462 | # How to Solve a Maze
Deep inside a maze, twists and turns steer you off course until you are lost between the walls. Navigating a maze can be challenging, but finding your way through is rewarding.
### You will need
• Map or paper maze
• Pencil
• Paper
• Chalk
• String
Step 1 Keep your right hand on the right wall Keep your right hand on the right wall or your left hand on the left wall as you walk through the maze. This technique navigates you out of the maze, but only if the goal is to reach the exit, not the center.
Step 2 Visit and mark all possibilities Visit all possible points of the maze if you don’t have a paper representation. Mark where you have traveled and from what direction with chalk.
Step 3 Fill in dead ends Fill in dead ends all the way to the decision junction so you don’t waste time hitting a wall.
Step 4 Use string Take a tip from Greek mythology and trail a length of string behind you. The string will show you where you’ve gone before, and help you find your way out of the maze. With these tips, you’re certain not to get lost! | 247 | 1,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.88132 |
https://small-s.science/tag/decision-procedure/ | 1,716,581,249,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058736.10/warc/CC-MAIN-20240524183358-20240524213358-00692.warc.gz | 439,552,884 | 37,840 | ## What is NHST, anyway?
I am not a fan of NHST (Null Hypothesis Significance Testing). Or maybe I should say, I am no longer a fan. I used to believe that rejecting null-hypotheses of zero differences based on the p-value was the proper way of gathering evidence for my substantive hypotheses. And the evidential nature of the p-value seemed so obvious to me, that I frequently got angry when encountering what I believed were incorrect p-values, reasoning that if the p-value is incorrect, so must be the evidence in support of the substantive hypothesis.
For this reason, I refused to use the significance tests that were most frequently used in my field, i.e. performing a by-subjects analysis and a by-item analysis and concluding the existence of an effect if both are significant, because the by-subjects analyses in particular regularly leads to p-values that are too low, which leads to believing you have evidence while you really don’t. And so I spent a huge amount of time, coming from almost no statistical background – I followed no more than a few introductory statistics courses – , mastering mixed model ANOVA and hierarchical linear modelling (up to a reasonable degree; i.e. being able to get p-values for several experimental designs). Because these techniques, so I believed, gave me correct p-values. At the moment, this all seems rather silly to me.
I still have some NHST unlearning to do. For example, I frequently catch myself looking at a 95% confidence interval to see whether zero is inside or outside the interval, and actually feeling happy when zero lies outside it (this happens when the result is statistically significant). Apparently, traces of NHST are strongly embedded in my thinking. I still have to tell myself not to be silly, so to say.
One reason for writing this blog is to sharpen my thinking about NHST and trying to figure out new and comprehensible ways of explaining to students and researchers why they should be vary careful in considering NHST as the sine qua non of research. Of course, if you really want to make your reasoning clear, one of the first things you should do is define the concepts you’re reasoning about. The purpose of this post is therefore to make clear what my “definition” of NHST is.
My view of NHST is very much based on how Gigerenzer et al. (1989) describe it:
“Fisher’s theory of significance testing, which was historically first, was merged with concepts from the Neyman-Pearson theory and taught as “statistics” per se. We call this compromise the “hybrid theory” of statistical inference, and it goes without saying the neither Fisher nor Neyman and Pearson would have looked with favor on this offspring of their forced marriage.” (p. 123, italics in original).
Actually, Fisher’s significance testing and Neyman-Pearson’s hypothesis testing are fundamentally incompatible (I will come back to this later), but almost no texts explaining statistics to psychologists “presented Neyman and Pearson’s theory as an alternative to Fisher’s, still less as a competing theory. The great mass of texts tried to fuse the controversial ideas into some hybrid statistical theory, as described in section 3.4. Of course, this meant doing the impossible.” (p. 219, italics in original).
So, NHST is an impossible, as in logically incoherent, “statistical theory”, because it (con)fuses concepts from incompatible statistical theories. If this is true, which I think it is, doing science with a small s, which involves logical thinking, disqualifies NHST as a main means of statistical inference. But let me write a little bit more about Fisher’s ideas and those of Neyman and Pearson, to explain the illogic of NHST.
I will try to describe the main characteristics of the two approaches that got hybridized in NHST at a conceptual level. I will have to simplify a lot and I hope these simplifications do little harm. Let’s start with Fisher’s significance testing.
### Fisher’s significance testing
The main purpose of Fisher’s significance testing is gathering evidence about parameters in a statistical model on the basis of a sample of data. So, the nature of the approach is evidential. Crucially, the evidence the data provides can only be evidence against a statistical model, but it can not be evidence in favour of the model, much in line with Popper’s idea of progress in science by means of falsification. The statistical model to be nullified, i.e. the model one tries to obtain evidence against, is called the null-hypothesis.
Conceptually, the statistical model is a descriptive model of a population of possible values. An important part of Fisher’s approach is therefore to judge what kind of model provides an appropriate model of the population. For instance, this process of formulating the model (which, of course, involves a lot of thought and judgement) may lead one to assume that the random variable has a normal distribution, which is characterized by only two parameters, μ the expected value or mean of the distribution and σ, the square root of the variance of the distribution, which in the case of the normal distribution is it’s standard deviation (the standard deviation is the square root of the variance).
The values of μ and σ (or σ2) are generally unknown, but we may assume (again as a result of thinking and judging) that they have particular values. For reasons of exposition, I will now assume that the value of σ is known, say σ = 15, so that we only have to take the unknown value of μ into account. Let’s suppose that our thinking and judging has led us to assume that the unknown value of μ = 100. The null-hypothesis is therefore that the variable has a normal distribution with μ = 100, and σ = 15.
We can obtain evidence against this null-hypothesis, by determining a p-value. We first gather data, say we take a random sample of N = 225 participants, which enables us to obtain observed values of the variable. Next, we calculate a test statistic, for example by estimating the value of μ (on the basis of our data) subtracting the hypothesized value and dividing the estimate by it’s standard error. Our estimated value may for example be 103, and the standard error equals 15 / √225 = 1.0, so the value of the test statistic equals (103 – 100) / 1 = 3. And now we are ready to calculate the p-value.
The p-value is the probability of obtaining (when sampling repeatedly) a value of the test statistic as large as or larger than the one obtained in the study, provided that the null-hypothesis is true. This probability can be calculated because the exact distribution of the test statistic can be deduced from the specification of the null-hypothesis. In our example, the test statistic is approximately normally distributed with μ = 0, and σ = 1.0. (The distribution is approximately normal, assuming the null-hypothesis is true, so the p-value in our example not exact). The p-value equals 0.003. (This is the so-called two-sided p-value, it is the probability of obtaining a value equal or larger than 3 or equal of smaller than -3, but we will ignore the technicalities of two-sided tests).
The p-value tells us that if the null-hypothesis is true, and we repeatedly take random samples from the population (as described by the null-hypothesis) we will find a value of our test statistic or a larger value in 0.3% of these samples. Thus, the probability of obtaining a value equal to or larger than 3.0 is very small.
Following Fisher, this low p-value can be interpreted as that something “improbable” occurred (assuming the null-hypothesis is true) or as inductive evidence against the null-hypothesis, i.e. the null-hypothesis is not true.
In his early writings Fisher proposed a p-value smaller than .05 as inductive evidence against the null-hypothesis (keeping in mind the possibility that the null is true, but that something improbable happened), but later he thought using the fixed criterion of .05 to be non-scientific. If the p-value is smaller than the criterion (say .05), the result is statistically significant.
In sum, the approach by Fisher, significance testing, involves specifying a statistical model, and using the p-value to test the assumptions of the model, such as specific values for μ or σ. If the p-value is smaller than the criterion value, either something improbable occurred or the null-hypothesis is not true. Crucially, the p-value may provide inductive evidence against the assumptions of the null-hypothesis, but a large p-value (larger than the criterion value) is not inductive support for the null-hypothesis.
### Neyman-Pearson hypothesis testing
In contrast to Fisher’s evidential approach, Neyman and Pearson’s hypothesis testing is non-evidential. Its primary goal is to choose on the basis of repeated random sampling between two hypotheses (or more; but I will only consider two) in order to make behavioral decisions (so to speak) that will minimize decision errors and their associated costs (loss) in the long run. In stead of trying to figure out which of the two hypotheses is true, one decides to accept one (and reject the other) of the two hypothesis as if it were true, without actually having to believe it, and act accordingly.
As with Fisher, Neyman-Pearson hypothesis testing starts with formulating descriptive models of the population. We may for instance propose (after thinking and judging) that one model (hypothesis H1) assumes that the variable has a normal distribution with μ = 100 and one model (hypothesis H2) that assumes that the variable has a normal distribution with μ = 106. We will assume the value of σ is known, say it equals 15. We will have to choose one of the two hypothesis, by rejecting one (and accepting the other).
Let’s suppose that only one of the models is true and that they cannot both be false. This means that we can incorrectly decide to reject or accept each of the two hypotheses. That is, if we incorrectly reject H1, we incorrectly accept H2. So, there are two types of errors we can make. A type I error occurs when we incorrectly reject a true hypothesis and a type II error occurs when we incorrectly accept a false hypothesis.
In a previous post (here), I used the following conceptual descriptions of these errors: the type I error is the error of excessive skepticism, and the type II error is the error of extreme gullibility, but from the perspective of Neyman-Pearson hypothesis testing these conceptual descriptions may not make much sense, because these terms imply a relation between the decisions about a hypothesis and belief in the hypothesis, while in the Neyman-Pearson approach a rejection or non-rejection does not lead to commitment in believing or not believing the hypothesis, although the hypotheses themselves are based on beliefs (and judging and reasoning) that the descriptive model is suitable for the population at hand.
The crucial point is that the goal of Neyman-Pearson hypothesis testing is to base courses of action on the decision to reject or not-reject a statistical hypothesis. This entails minimizing the costs (loss) associated with type I and type II errors. In particular, the approach minimizes the probability (β) of a type II error bounded by the probability (α) of a type I error. We may also say that we want to maximize the probability (1 – β), the probability of rejecting a false hypothesis, the so called power of the test, while keeping α at a maximum (usually low) value.
Suppose, that our considerations of the loss associated with type I and type II errors, has led us to the insight that false rejection of H2 is the most costly error. And suppose that we have agreed/determined/reasoned/judged that the probability of falsely rejecting it should be at most .05. So, α = .05. Of course, we also “know” the loss associated with falsely accepting it, and we have determined that the probability β should not exceed .10. Now, suppose that we repeatedly sample N = 225 observations from the (unknown) population. We do not know whether H1 or H2 provides the correct description of the population, but we assume that one of them must be true if we select a particular sample, and they cannot both be false.
We will reject H2 (Normal distribution with μ = 106, and σ = 15) if the sample mean in our random sample equals 104.35 or less (this corresponds to a test statistic with value -1.65). Why, because the probability of obtaining a sample mean equal or smaller than 104.35 is approximately .05 when H2 is true. Thus, if we repeatedly sample from the population when H2 is true, we will incorrectly reject it in 5% of the cases. Which is the probability of a type I error that we want.
We have arranged things so, that when H2 is false, H1 is per definition true. If H1 is true (H2 is false), there is a probability of approximately .99 to obtain a sample mean of 104.35 or smaller. Thus, the probability to reject H2 when it it false is .99, this is the power of the test, and the probability is approximately .01 of incorrectly not rejecting H2 when it is false. The latter probability is the probability of a type II error, which we did not want to be larger than .10.
Now suppose the results is that the sample mean equals 103 (the value of the test statistic equals -3). According to the decision criterion we reject H2 (with α = .05) and accept H1 and act as if μ = 100 is true. Crucially, we do not have to believe it is actually true, nor do we consider the test statistic with value -3 as inductive evidence against H2. So, the test result provides neither support for H1 nor evidence against H2, but we know from the specification of the models and the assumptions about sampling that repeatedly using this procedure leads to 5% type I errors and 1% type II errors in the long run, depending on which of the two hypotheses is true (which is unknown to us). Given that we know the loss associated with each error, we are able to minimize the expected loss associated with acting upon the decisions we make about the hypotheses.
Note that Fisher’s significance testing would consider the p-value associated with the test statistic of -3, i.e. p < .01 either as inductive evidence against H2 or as an indication that something unusual (improbable) happened assuming H2 is true. Note also that in Fisher’s approach, it is not possible to reason from the inferred untruth of H2 to the truth of H1, because H1 does not exist in that approach.
It should be noted further that in the Neyman-Pearson approach, the importance of the value of the test statistic is restricted to whether or not the value exceeds a critical value (i.e. whether or not the value of the statistic is in the rejection region). That means that it is of no concern how much the test statistic exceeds the critical value, since all values larger than the critical value lead to the same decision: reject the hypothesis. In other words, because the approach is non-evidential, the magnitude of the test statistic is inconsequential as far as the truth of the hypothesis is concerned. Compare this to the Fisher approach, where the larger the test statistic is (the smaller the p-value), the stronger the inductive evidence is against the null-hypothesis.
### Null-hypothesis significance testing (NHST)
NHST combines Fisher’s significance testing with Neyman-Pearson hypothesis testing, without regard for the logical incompatibilities of the two approaches. Fisher’s p-value is used both as a measure of inductive evidence against the null-hypothesis, with smaller p-values considered to be stronger evidence against the null than larger p-values, and as a test statistic. In its latter use, the null-hypothesis is (usually) rejected if the p-value is smaller than .05.
Contrary to significance testing, NHST uses the p-value to decide between the null-hypothesis and an alternative hypothesis. But contrary to the Neyman-Pearson approach, α, the probability of a type I error is not based on judgement and careful consideration of loss-functions, but is mechanically set at .05 (or .01). And, contrary to the Neyman-Pearson approach, the probability of a type II error (β) is usually not considered.
One reason for the latter may be that specification of the null-hypothesis is also mechanized. In the case of differences between means or testing correlations or regression coefficients, etc, the standard null-hypothesis is that the difference, the correlation or the coefficient equals 0. This is also called the nil-hypothesis. As the alternative excludes the null, the standard alternative hypothesis is that the parameter in question is not equal to zero, which makes it hard to say something about the type II error, because determining the probability of a type II error requires thinking about a minimal consequential effect size (consequential in terms of decisions and associated loss) that can serve as the alternative hypothesis.
Specifying a non-nil alternative hypothesis, i.e. that the parameter value is not equal to zero, implies that results arbitrarily close to nil, but not equal to nil, are as consequential as effect sizes that are far away from the null-value, both in acting upon the value as in not-acting upon it. Crucially, not specifying a minimal consequential effect size, rules out determining β. So, even though NHST uses the concept of an alternative hypothesis (contrary to Fisher), the nil-hypothesis is such that the procedure of Neyman and Pearson can no longer work: it is impossible to strike a balance between loss associated with type I and type II errors, and so NHST is not a hypothesis testing procedure.
For these reasons I am very much inclined to characterize NHST as fixed-α significance testing. But using fixed-α in combination with an evidential interpretation of p-values leads to logical inconsistencies. (As always, I assume that being logically consistent is one of the characteristics of doing science, but maybe you disagree). Note, by the way, that I am talking about the p-value as measure of evidence against the nil-hypothesis, and not about the p-value as test statistic. (But remember that proper use of the p-value as test statistic requires being able to specify a non-nil alternative hypothesis).
One of the logical inconsistencies is that α and the p-value-as-evidence involve contradictory conceptualisations of probability. In terms of p-values, α is simply the probability that the p-value is smaller than .05 (the usual criterion) assuming the nil-hypothesis is true. That probability follows deductively from the specification of the null-hypothesis (including, of course, the statistical model underlying it). Note that α is completely independent of actually realized results: it an assertion about the p-value assuming repeated sampling from the null-population; α is about the test-procedure and not about actual data.
But the p-value-as-evidence against the null is not the result of deductive reasoning, but of inductive reasoning. The p-value is not a probability associated with the test-procedure. It is a random variable the value of which depends on the actual data, the null-value and the statistical model. Crucially, from a single realized result (a p-value) an inference is made about a probability distribution. But this is inconsistent with the frequency interpretation of probability that underlies the conceptualisation of α, because under this interpretation no probability statement can be made about realized single results (except that the probability is 100% that it happened) or about an unrealized single result (that probability is 0 if it does not happen or 1.0 if it happens). To make the point: using p-value-as-evidence and (fixed)-α requires both believing that probability statements can be made on the basis of a single result and believing that that is impossible. So, it boils down to believing that both A and not-A are true.
To me, logical inconsistencies like these disqualify NHST as a scientific means of statistical inference. I repeat that this is because I believe that doing science entails being logically consistent. Assuming or believing that A and not-A are both true, is not an example of logical consistency.
## Type I error probability does not destroy the evidence in your data
Have you heard about that experimental psychologist? He decided that his participants did not exist, because the probability of selecting them, assuming they exist, was very small indeed (p < .001). Fortunately, his colleagues were quick to reply that he was mistaken. He should decide that they do exist, because the probability of selecting them, assuming they do not exist, is very remote (p < .001). Yes, even unfunny jokes can be telling about the silliness of significance testing.
But sometimes the silliness is more subtle, for instance in a recent blog post by Daniel Lakens, the 20% Statistician with the title “Why Type I errors are more important than Type 2 errors (if you care about evidence).” The logic of his post is so confused, that I really do not know where to begin. So, I will aim at his main conclusion that type I error inflation quickly destroys the evidence in your data.
(Note: this post uses mathjax and I’ve found out that this does not really work well on a (well, my) mobile device. It’s pretty much unreadable).
Lakens seems to believe that the long term error probabilities associated with decision procedures, has something to do with the actual evidence in your data. What he basically does is define evidence as the ratio of power to size (i.e. the probability of a type I error), it’s basically a form of the positive likelihood ratio
which makes it plainly obvious that manipulating (for instance by multiplying it with some constant c) influences the PLR more than manipulating by the same amount. So, his definition of “evidence” makes part of his conclusion true, by definition: has more influence on the PLR than , But it is silly to reason on the basis of this that the type I error rate destroys the evidence in your data.
The point is that and (or the probabilities of type I errors and type II errors) have nothing to say about the actual evidence in your data. To be sure, if you commit one of these errors, it is the data (in NHST combined with arbitrary i,e, unjustified cut-offs) that lead you to these errors. Thus, even and , do not guarantee that actual data lead to a correct decision.
Part of the problem is that Lakens confuses evidence and decisions, which is a very common confusion in NHST practice. But, deciding to reject a null-hypothesis, is not the same as having evidence against it (there is this thing called type I error). It seems that NHST-ers and NHST apologists find this very very hard to understand. As my grandmother used to say: deciding that something is true, does not make it true
I will try to make plausible that decisions are not evidence (see also my previous post here). This should be enough to show you that error probabilities associated with the decision procedure tells you nothing about the actual evidence in your data. In other words, this should be enough to convince you that Type 1 error rate inflation does not destroy the evidence in your data, contrary to the 20% Statistician’s conclusion.
Let us consider whether the frequency of correct (or false) decisions is related to the evidence in the data. Suppose I tell you that I have a Baloney Detection Kit (based for example on the baloney detection kit at skeptic.com) and suppose I tell you that according to my Baloney Detection Kit the 20% Statistician’s post is, well, Baloney. Indeed, the quantitative measure (amount of Baloneyness) I use to make the decision is well above the critical value. I am pretty confident about my decision to categorize the post as Baloney as well, because my decision procedure rarely leads to incorrect decisions. The probability that I decide that something is Baloney when it is not is only and the probability that I decide that something is not-Baloney when it is in fact Baloney is only 1% as well ().
Now, the 20% Statistician’s conclusion states that manipulating , for instance by setting destroys the evidence in my data. Let’s see. The evidence in my data is of course the amount of Baloneyness of the post. (Suppose my evidence is that the post contains 8 dubious claims). How does setting have any influence on the amount of Baloneyness? The only thing setting does is influence the frequency of incorrect decisions to call something Baloney when it is not. No matter what value of (or , for that matter) we use, the amount of Baloneyness in this particular post (i.e. the evidence in the data) is 8 dubious claims.
To be sure, if you tell the 20% Statistician that his post is Baloney, he will almost certainly not ask you how many times you are right and wrong on the long run (characteristics of the decision procedure), he will want to see your evidence. Likewise, he will probably not argue that your decision procedure is inadequate for the task at hand (maybe it is applicable to science only and not to non-scientific blog posts), but he will argue about the evidence (maybe by simply deciding (!) that what you are saying is wrong; or by claiming that the post does not contain 8 dubious claims, but only 7).
The point is, of course, this: the long term error probabilities and associated with the decision procedure, have no influence on the actual evidence in your data. The conclusion of the 20% Statistician is simply wrong. Type I error inflation does not destroy the evidence in your data, nor does type II error inflation.
## Decisions are not evidence
The thinking that lead to this post began with trying to write something about what Kline (2013) calls the filter myth. The filter myth is the arguably – in the sense that it depends on who you ask – mistaken belief in NHST practice that the p-value discriminates between effects that are due to chance (null-hypothesis not rejected) and those that are real (null-hypothesis rejected). The question is whether decisions to reject or not reject can serve as evidence for the existence of an effect.
Reading about the filter myth made me wonder whether NHST can be viewed as a screening test (diagnostic test), much like those used in medical practice. The basic idea is that if the screening test for a particular condition gives a positive result, follow-up medical research will be undertaken to figure out whether that condition is actually present. (We can immediately see, by the way, that this metaphor does not really apply to NHST, because the presumed detection of the effect is almost never followed up by trying to figure out whether the effect actually exists, but the detection itself is, unlike the screening test, taken as evidence that the effect really exists; this is simply the filter myth in action).
Let’s focus on two properties of screening tests. The first property is the Positive Likelihood Ratio (PLR). The PLR is the ratio of the probability of a correct detection to the probability of a false alarm. In NHST-as-screening-test, the PLR equals the ratio of the power of the test to the probability of a type I error: PLR = (1 – β) / α. A high value of the PLR means, basically, that a rejection is more likely to be a rejection of a false null than a rejection of a true null, thus the PLR means that a rejection is more likely to be correct than incorrect.
As an example, if β = .20, and α = . 05, the PLR equals 16. This means that a rejection is 16 times more likely to be correct (the null is false) than incorrect (the null is true).
The second property I focused on is the Negative Likelihood Ratio (NLR). The NLR is the ratio of the frequency of incorrect non-detections to the frequency of correct non-detections. In NHST-as-screening-test, the NLR equals the ratio of the probability of a type II error to the probability of a correct non-rejection: NLR = β / (1 – α). A small value of the NLR means, in essence, that a non-rejection is less likely to occur when the null-hypothesis is false than when it is true.
As an example, if β = .20, and α = . 05, the NLR equals .21. This means that a non-rejection is .21 times more likely (or 4.76 (= 1/.21) times less likely) to occur when the null-hypothesis is false, than when it is true.
The PLR and the NLR can be used to calculate the likelihood ratio of the alternative hypothesis to the null-hypothesis given that you have rejected or given that you have not-rejected, the posterior odds of the alternative to the null. All you need is the likelihood ratio of the alternative to the null before you have made a decision and you multiply this by the PLR after you have rejected, and by the NLR after you have not rejected.
Suppose that we repeatedly (a huge number of times) take a random sample from a population of null-hypotheses in which 60% of them are false and 40% true. If we furthermore assume that a false null means that the alternative must be true, so that the null and the alternative cannot both be false, the prior likelihood of the alternative to the null equals p(H1)/p(H0) = .60/.40 = 1.5. Thus, of all the randomly selected null-hypotheses, the proportion of them that are false is 1.5 times larger than the proportion of null-hypotheses that are true. Let’s also repeatedly sample (a huge number of times) from the population of decisions. Setting β = .20, and α = . 05, the proportion of rejections equals p(H1)*(1 – β) + p(H0)*α = .60*.80 + .40*.05 = .50 and the proportion of non-rejections equals p(H1)*β + p(H0)*(1 – α) = .60*.20 + .40*.95 = .50. Thus, if we sample repeatedly from the population of decisions 50% of them are rejections and 50% of them are non-rejections.
First, we focus only on the rejections. So, the probability of a rejection is now taken to be 1.0. The posterior odds of the alternative to the null, given that the probability of a rejection is 1.0, is the prior likelihood ratio multiplied by the PLR: 1.5 * 16 = 24. Thus, we have a huge number of rejections (50% of our huge number of randomly sampled decisions) and within this huge number of rejections the proportion of rejections of false nulls is 24 times larger than the proportion of rejections of true nulls. The proportion of rejections of false nulls equals the posterior odds / (1 + posterior odds) = 24 / 25 = .96. (Interpretation: If we repeatedly sample a null-hypothesis from our huge number of rejected null-hypotheses, 96% of those samples are false null-hypotheses).
Second, we focus only on the non-rejections. Thus, the probability of a non-rejection is now taken to be 1.0. The posterior odds of the alternative to the null, given that the probability of a non-rejection is 1.0, is the prior odds multiplied by the NLR: 1.5 * 0.21 = 0.32. In other words, we have a huge number of non-rejections (50% of our huge sample of randomly selected decisions) and the proportion of non-rejections of false nulls is 0.32 times as large as the proportion of non-rejections of true nulls. The proportion of non-rejections of false nulls equals 0.32 / ( 1 + 0.32) = .24. (Interpretation: If we repeatedly sample a null-hypothesis from our huge number of non-rejected hypotheses, 24% of them are false nulls).
So, based on the assumptions we made, NHST seems like a pretty good screening test, although in this example NHST is much better at detecting false null-hypothesis than ruling out false alternative hypotheses. But how about the question of decisions as evidence for the reality of an effect? I will first write a little bit about the interpretation of probabilities, then I will show you that decisions are not evidence.
Sometimes, these results are formulated as follows. The probability that the alternative is true given a decision to reject is .96 and the probability that the alternative hypothesis is true given a decision to not-reject is .24. If you want to correctly interpret such a statement, you have to keep in mind what “probability” means in the context of this statement, otherwise it is very easy to misinterpret the statement’s meaning. That is why I included interpretations of these results that are consistent with the meaning of the term probability as it used in our example. (In conceptual terms, the limit of the relative frequency of an event (such as reject or not-reject) as the number of random samples (the number of decisions) goes to infinity).
A common (I believe) misinterpretation (given the sampling context described above) is that rejecting a null-hypothesis makes the alternative hypothesis likely to be true. This misinterpretation is easily translated to the incorrect conclusion that a significant test result (that leads to a rejection) makes the alternative hypothesis likely to be true. Or, in other words, that a significant result is some sort of evidence for the alternative hypothesis (or against the null-hypothesis).
The mistake can be described as confusing the probability of a single result with the long term (frequentist) probabilities associated with the decision or estimation procedure. For example, the incorrect interpretation of the p-value as the probability of a type I error or the incorrect belief that an obtained 95% confidence interval contains the true value of a parameter with probability .95.
A quasi-simple example may serve to make the mistake clear. Suppose I flip a fair coin, keep the result hidden from you, and let you guess whether the result is heads or tails (we assume that the coin will not land on it’s side). What is the probability that your guess is correct? | 7,337 | 33,676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-22 | latest | en | 0.946322 |
http://de.metamath.org/mpegif/3eqtr3ri.html | 1,600,771,476,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400205950.35/warc/CC-MAIN-20200922094539-20200922124539-00271.warc.gz | 35,937,688 | 3,533 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > 3eqtr3ri Structured version Unicode version
Theorem 3eqtr3ri 2467
Description: An inference from three chained equalities. (Contributed by NM, 15-Aug-2004.)
Hypotheses
Ref Expression
3eqtr3i.1
3eqtr3i.2
3eqtr3i.3
Assertion
Ref Expression
3eqtr3ri
Proof of Theorem 3eqtr3ri
StepHypRef Expression
1 3eqtr3i.3 . 2
2 3eqtr3i.1 . . 3
3 3eqtr3i.2 . . 3
42, 3eqtr3i 2460 . 2
51, 4eqtr3i 2460 1
Colors of variables: wff setvar class Syntax hints: wceq 1437 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1665 ax-4 1678 ax-ext 2407 This theorem depends on definitions: df-bi 188 df-cleq 2421 This theorem is referenced by: indif2 3722 dfif5 3931 resdm2 5345 co01 5370 funiunfv 6168 dfdom2 7602 1mhlfehlf 10832 crreczi 12394 rei 13198 bpoly3 14089 bpoly4 14090 cos1bnd 14219 rpnnen2lem3 14247 rpnnen2lem11 14255 m1bits 14388 6gcd4e2 14476 3lcm2e6 14652 karatsuba 15019 ring1 17765 sincos4thpi 23333 sincos6thpi 23335 1cubrlem 23632 cht3 23963 bclbnd 24071 bposlem8 24082 ex-ind-dvds 25744 ip1ilem 26312 mdexchi 27823 disjxpin 28037 xppreima 28088 df1stres 28124 df2ndres 28125 xrge0slmod 28446 cnrrext 28653 ballotth 29196 poimirlem3 31647 poimirlem30 31674 mbfposadd 31692 asindmre 31731 areaquad 35800 inductionexd 36230 stoweidlem26 37455 3exp4mod41 38306
Copyright terms: Public domain W3C validator | 749 | 1,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2020-40 | latest | en | 0.167163 |
https://proofwiki.org/wiki/Unbounded_Closed_Real_Interval/Examples/Example_1 | 1,621,129,372,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991659.54/warc/CC-MAIN-20210516013713-20210516043713-00397.warc.gz | 495,206,580 | 9,735 | # Unbounded Closed Real Interval/Examples/Example 1
## Example of Unbounded Closed Real Interval
Let $I$ be the unbounded closed real interval defined as:
$I := \hointl \gets 3$
Then $2 \in I$.
## Proof
By definition of open real interval:
$I = \set {x \in \R: x \le 3}$
As $2 \le 3$ it follows that $2 \in I$.
$\blacksquare$ | 107 | 334 | {"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} | 3.421875 | 3 | CC-MAIN-2021-21 | latest | en | 0.647958 |
http://blog.datumbox.com/the-dirichlet-process-mixture-model/ | 1,721,701,877,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517931.85/warc/CC-MAIN-20240723011453-20240723041453-00889.warc.gz | 4,589,210 | 18,148 | # The Dirichlet Process Mixture Model
This blog post is the fourth part of the series on Clustering with Dirichlet Process Mixture Models. In previous articles we discussed the Finite Dirichlet Mixture Models and we took the limit of their model for infinite k clusters which led us to the introduction of Dirichlet Processes. As we saw, our target is to build a mixture model which does not require us to specify the number of k clusters/components from the beginning. After presenting different representations of Dirichlet Processes, it is now time to actually use DPs to construct an infinite Mixture Model that enable us to perform clustering. The target of this article is to define the Dirichlet Process Mixture Models and discuss the use of Chinese Restaurant Process and Gibbs Sampling. If you have not read the previous posts, it is highly recommended to do so as the topic is a bit theoretical and requires good understanding on the construction of the model.
Update: The Datumbox Machine Learning Framework is now open-source and free to download. Check out the package com.datumbox.framework.machinelearning.clustering to see the implementation of Dirichlet Process Mixture Models in Java.
## 1. Definition of Dirichlet Process Mixture Model
Using Dirichlet Processes allows us to have a mixture model with infinite components which can be thought as taking the limit of the finite model for k to infinity. Let’s assume that we have the following model:
Equation 1: Dirichlet Process Mixture Model
Where G is defined as and used as a short notation for which is a delta function that takes 1 if and 0 elsewhere. The θi are the cluster parameters which are sampled from G. The generative distribution F is configured by cluster parameters θi and is used to generate xi observations. Finally we can define a Density distribution which is our mixture distribution (countable infinite mixture) with mixing proportions and mixing components .
Figure 1: Graphical Model of Dirichlet Process Mixture Model
Above we can see the equivalent Graphical Model of the DPMM. The G0 is the base distribution of DP and it is usually selected to be conjugate prior to our generative distribution F in order to make the computations easier and make use of the appealing mathematical properties. The α is the scalar hyperparameter of Dirichlet Process and affects the number of clusters that we will get. The larger the value of α, the more the clusters; the smaller the α the fewer clusters. We should note that the value of α expresses the strength of believe in G0. A large value indicates that most of the samples will be distinct and have values concentrated on G0. The G is a random distribution over Θ parameter space sampled from the DP which assigns probabilities to the parameters. The θi is a parameter vector which is drawn from the G distribution and contains the parameters of the cluster, F distribution is parameterized by θi and xi is the data point generated by the Generative Distribution F.
It is important to note that the θi are elements of the Θ parameter space and they “configure” our clusters. They can also be seen as latent variables on xi which tell us from which component/cluster the xi comes from and what are the parameters of this component. Thus for every xi that we observe, we draw a θi from the G distribution. With every draw the distribution changes depending on the previous selections. As we saw in the Blackwell-MacQueen urn scheme the G distribution can be integrated out and our future selections of θi depend only on G0: . Estimating the parameters θi from the previous formula is not always feasible because many implementations (such as Chinese Restaurant Process) involve the enumerating through the exponentially increasing k components. Thus approximate computational methods are used such as Gibbs Sampling. Finally we should note that even though the k clusters are infinite, the number of active clusters is . Thus the θi will repeat and exhibit a clustering effect.
## 2. Using the Chinese Restaurant Process to define an Infinite Mixture Model
The model defined in the previous segment is mathematically solid, nevertheless it has a major drawback: for every new xi that we observe, we must sample a new θi taking into account the previous values of θ. The problem is that in many cases, sampling these parameters can be a difficult and computationally expensive task.
An alternative approach is to use the Chinese Restaurant Process to model the latent variables zi of cluster assignments. This way instead of using θi to denote both the cluster parameters and the cluster assignments, we use the latent variable zi to indicate the cluster id and then use this value to assign the cluster parameters. As a result, we no longer need to sample a θ every time we get a new observation, but instead we get the cluster assignment by sampling zi from CRP. With this scheme a new θ is sampled only when we need to create a new cluster. Below we present the model of this approach:
Equation 2: Mixture Model with CRP
The above is a generative model that describes how the data xi and the clusters are generated. To perform the cluster analysis we must use the observations xi and estimate the cluster assignments zi.
## 3. Mixture Model Inference and Gibbs Sampling
Unfortunately since Dirichlet Processes are non-parametric, we can’t use EM algorithm to estimate the latent variables which store the cluster assignments. In order to estimate the assignments we will use the Collapsed Gibbs Sampling.
The Collapsed Gibbs Sampling is a simple Markov Chain Monte Carlo (MCMC) algorithm. It is fast and enables us to integrate out some variables while sampling another variable. Nevertheless this algorithms requires us to select a G0 which is a conjugate prior of F generative distribution in order to be able to solve analytically the equations and be able to sample directly from .
The steps of the Collapsed Gibbs Sampling that we will use to estimate the cluster assignments are the following:
• Initialize the zi cluster assignments randomly
• Repeat until convergence
• Select randomly a xi
• Keep the other zj fixed for every j≠i:
• Assign a new value on zi by calculating the “CRP probability” that depends on zj and xj of all j≠i:
In the next article we will focus on how to perform cluster analysis by using Dirichlet Process Mixture models. We will define two different Dirichlet Process Mixture Models which use the Chinese Restaurant Process and the Collapsed Gibbs Sampling in order to perform clustering on continuous datasets and documents. | 1,330 | 6,616 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.907889 |
https://braindump.jethro.dev/zettels/optimization/ | 1,586,370,840,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371821680.80/warc/CC-MAIN-20200408170717-20200408201217-00412.warc.gz | 385,733,522 | 7,995 | # Optimization
## What is Convex Optimization?
Convex optimization is a special class of mathematical optimization problems, which includes least-squares and linear programming problems.
There are many advantages to recognizing or formulating a problem as a convex optimization problem. First, the problem can be solved reliably and efficiently, using interior-point methods or other special methods for convex optimization. There are also theoretical or conceptual advantages of formulating a problem as a convex optimization problem.
## Mathematical Optimization
An optimization problem has the form:
\begin{align} \label{dfn:optimization} &\text{minimize} &f_0(x) \\\
&\text{subject to} &f_i(x) \le b_i, i = 1, \dots, m \end{align}
Here the vector $$x = (x_1, \dots, x_n)$$ is the optimization variable of the problem, the function $$f_0 : \mathbb{R^n} \rightarrow \mathbb{R}$$ is the objective function, $$f_i \mathbb{R^n} \rightarrow \mathbb{R}$$ are the (inequality) constraint functions, and the constants $$b_1, \dots, b_m$$ are the limits, or bounds, for the constraints.
We consider families or classes of optimization problems, characterized by particular forms of the objective and constraint functions. The optimization problem is a linear program if the objective and constraint functions $$f_0, \dots, f_m$$ are linear.
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https://jp.mathworks.com/matlabcentral/cody/problems/44666-draw-o/solutions/1923208 | 1,607,124,455,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141745780.85/warc/CC-MAIN-20201204223450-20201205013450-00312.warc.gz | 358,405,995 | 16,948 | Cody
# Problem 44666. Draw 'O' !
Solution 1923208
Submitted on 5 Sep 2019 by Doan Quang Manh
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; y_correct = 1; assert(isequal(your_fcn_name(x),y_correct))
a = 1 y = 1
2 Pass
x = 2; y_correct = [1 1;1 1]; assert(isequal(your_fcn_name(x),y_correct))
a = 1 1 1 1 y = 1 1 1 1
3 Pass
x = 3; y_correct = [1 1 1;1 0 1;1 1 1]; assert(isequal(your_fcn_name(x),y_correct))
a = 1 0 1 1 0 1 1 0 1 y = 1 1 1 1 0 1 1 1 1
4 Pass
x = 4; y_correct = [1 1 1 1 1 0 0 1 1 0 0 1 1 1 1 1]; assert(isequal(your_fcn_name(x),y_correct))
a = 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 y = 1 1 1 1 1 0 0 1 1 0 0 1 1 1 1 1
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 401 | 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} | 2.78125 | 3 | CC-MAIN-2020-50 | latest | en | 0.648824 |
https://brainmass.com/statistics/probability/probability-normal-distributions-494716 | 1,716,893,089,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059085.33/warc/CC-MAIN-20240528092424-20240528122424-00213.warc.gz | 120,158,015 | 6,602 | Purchase Solution
# Probability - Normal Distributions
Not what you're looking for?
A credit union claims that the credit card debts carried by its clients are normally distributed with a mean of \$3050 and a standard deviation of \$950.
a) What is the probability that a randomly selected client has a balance of less than \$2900 on his credit card?
b) The credit union wants to advertise some financial products to clients with low balances. For that purpose they want to identify the largest amount of debt held by the 10% of clients with lowest debts. Determine the value of this debt.
c) You randomly select 25 holders of the union's credit card. What is the probability that their mean credit card balance is between \$2500 and \$2900?
##### Solution Summary
Normal distributions of probability of randomly selected variables are examined.
##### Solution Preview
A credit union claims that the credit card debts carried by its clients are ...
##### Terms and Definitions for Statistics
This quiz covers basic terms and definitions of statistics.
##### Measures of Central Tendency
Tests knowledge of the three main measures of central tendency, including some simple calculation questions.
##### Measures of Central Tendency
This quiz evaluates the students understanding of the measures of central tendency seen in statistics. This quiz is specifically designed to incorporate the measures of central tendency as they relate to psychological research. | 274 | 1,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-22 | latest | en | 0.950853 |
https://ar5iv.labs.arxiv.org/html/1812.03097 | 1,726,358,892,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651601.85/warc/CC-MAIN-20240914225323-20240915015323-00077.warc.gz | 86,581,561 | 83,455 | # The Wigner flow on the sphere
Popo Yang1, Iván F Valtierra2, Andrei B Klimov2, Shin-Tza Wu3, Ray-Kuang Lee1, Luis L Sánchez-Soto4,5 and Gerd Leuchs5,6,7 1Department of Physics, National Tsing Hua University, Hsinchu 300, Taiwan 2Departamento de Física, Universidad de Guadalajara, 44420 Guadalajara, Jalisco, Mexico 3Department of Physics, National Chung Cheng University, Chiayi 621, Taiwan 4Departamento de Óptica, Facultad de Física, Universidad Complutense, 28040 Madrid, Spain 5 Max-Planck-Institut für die Physik des Lichts, Staudtstraße 2, 91058 Erlangen, Germany 6 Department of Physics, University of Erlangen-Nuremberg, Staudtstraße 7 B2, 91058 Erlangen, Germany 7 Physics Department, Centre for Research in Photonics, University of Ottawa, Advanced Research Complex, 25 Templeton, Ottawa ON Canada, K1N 6N5
###### Abstract
We derive a continuity equation for the evolution of the SU(2) Wigner function under nonlinear Kerr evolution. We give explicit expressions for the resulting quantum Wigner current, and discuss the appearance of the classical limit. We show that the global structure of the quantum current significantly differs from the classical one, which is clearly reflected in the form of the corresponding stagnation lines.
## 1 Introduction
In classical statistical mechanics, an ensemble of particles is described by a distribution function $f(x,p|t)$ that depends on the phase-space variables $x$ and $p$ and evolves in time. The corresponding dynamics is governed by the Liouville equation [1], which asserts that for conservative forces $f(x,p|t)$ is constant along the trajectories of the system. In other words, the local density of points traveling through phase-space is constant with time. This conservation can be succinctly summarized as a continuity equation
$\frac{\partial f(x,p|t)}{\partial t}=-\nabla\cdot\mathbf{J}(x,p|t)\,,$ (1.1)
where $\mathbf{J}(x,p|t)$ is a probability current. Indeed, this flow is regular [2] and largely determined by location and nature of its stagnation points; i.e, those points for which $\mathbf{J}=0$. For conservative systems, the form of $J(x,p|t)$ immediately follows from the corresponding Poisson brackets.
This scenario can be extended to more general systems admitting a dynamical symmetry group. This enables the construction of a phase space $\mathcal{M}$ as an appropriate homogeneous manifold [3, 4]. This classical formulation associates a probability with every point $\Omega\in\mathcal{M}$. However, in the quantum domain the uncertainty principle does not allow one to attribute a state to a single point in phase space [5, 6, 7, 8, 9]. Because of this fundamental difference, there is no unique way of defining a quantum probability distribution. The Wigner function $W_{\varrho}(\Omega)$ is perhaps the closest to the analogous counterpart. Note that, although the Wigner function has the correct marginal probability distributions, it can itself be negative.
In the phase-space approach, every observable $\hat{A}$ is mapped onto a function $W_{A}(\Omega)$ (called its Weyl symbol). In particular, the Weyl symbol of the density matrix is precisely the Wigner function and its time evolution reads
$\partial_{t}W_{\varrho}(\Omega|t)=\{W_{\varrho}(\Omega),W_{H}(\Omega)\}_{M}\,,$ (1.2)
where $W_{H}(\Omega)$ is the symbol of the Hamiltonian and the Moyal bracket $\{\cdot,\cdot\}_{M}$ is the image of the quantum commutator [times $(\rmi\hbar)^{-1}$] under the Weyl map [10]. The resulting partial differential equation contains, in general, higher-order derivatives, which significantly complicate the search for an exact solution. However, it admits a natural expansion in powers of a semiclassical parameter $\varepsilon\ll 1$ that characterizes the strength of quantum fluctuations in the system. This parameter depends on the dynamical symmetry and, roughly speaking, is the inverse of the number of excitations [11]. To the lowest order in $\varepsilon$, equation (1.2) is of the Liouvillian form
$\partial_{t}W_{\varrho}(\Omega)=\varepsilon\{W_{\varrho}(\Omega),W_{H}(\Omega)\}+O(\varepsilon^{3})\,,$ (1.3)
where now $\{\cdot,\cdot\}$ is the Poisson bracket in the manifold $\mathcal{M}$. The semiclassical or truncated Wigner approximation (TWA) [12, 13, 14, 15] consists in disregarding the higher order terms, so that $W_{\varrho}(\Omega|t)\simeq W_{\varrho}(\Omega(-t)|0)$, where $\Omega(t)$ are classical trajectories generated by $W_{H}(\Omega)$.
It has been pointed out [16, 17, 18], that one can construct a Wigner current 111We will mainly call the quantity $\mathbf{J}$ the Wigner current. However, it can also be interpreted as quasiprobability flow. For this reason, the designation Wigner flow has been used in the literature before. Note though that, as discussed in [19], no flow (in the sense of mapping of a distribution along trajectories) exists in the quantum domain. in such a way that the evolution can be mapped as a continuity equation very much analogous to (1.1). Surprisingly, this Wigner current, which is the equivalent of the classical Liouville flow, has, so far, not been studied in great detail [20, 21, 22, 23]. The form of the current, and especially the behavior in the vicinity of its stagnation points, can be used for the characterization of the quantumness of the evolution (see also references [24, 25, 26], where the stagnation points of the Husimi $Q$ function were studied).
In this paper, we extend these ideas to spinlike systems, where the classical phase space is the unit sphere. We stress that this is not a mere academic curiosity, since the underlying SU(2) symmetry plays a pivotal role in numerous models in physics [27].
In the spirit of equations (1.2) and (1.3), we introduce in a natural way the classical and quantum Wigner currents. We will show, using the simplest example of nonlinear Kerr dynamics, that the global structure of the quantum Wigner current significantly differs from the classical one. Such a difference is clearly observable even during the short-time evolution of semiclassical states (specified by localized distributions in phase space), when the Wigner distribution can still be well described in terms of the semiclassical approximation. In other words, the Wigner current allows us to distinguish between quantum and semiclassical dynamics, while the distributions evolved according to the Moyal and Poisson brackets are still quite similar. The stagnation points/lines of the classical Wigner current are basically determined by the zeros of the semiclassicaly evolved Wigner distribution. Therefore, the structure of the stagnation lines can be used for the analysis of the quantumness of the phase-space dynamics in the semiclassical limit. Furthermore, an extra benefit of bringing the Wigner current into play is that it can give a compelling visual representation of how nonclassical features arise during the evolution.
## 2 Wigner function on the sphere
We consider a system whose dynamical symmetry group is SU(2). As heralded in the Introduction, we follow the ideas in references [3, 4] to work out quasiprobability distributions on the sphere satisfying all the pertinent requirements. This construction was generalized by others [28, 29, 30, 31, 32] and has proved to be very useful in visualizing properties of spinlike systems [33, 34, 35, 36].
The corresponding Lie algebra $\mathfrak{su}(2)$ is spanned by the operators $\{\hat{S}_{x},\hat{S}_{y},\hat{S}_{z}\}$ satisfying the angular momentum commutation relations
$[\hat{S}_{x},\hat{S}_{y}]=\rmi\hat{S}_{z}\,,$ (2.1)
and cyclic permutations (in units $\hbar=1$, which will be used throughout). The Casimir operator is $\hat{\mathbf{S}}^{2}=\hat{S}_{x}^{2}+\hat{S}_{y}^{2}+\hat{S}_{z}^{2}=S(S+1)\leavevmode\hbox{\small 1\normalsize\kern-3.30002pt1}$, so the eigenvalue $S$ (which is a nonnegative integer or half integer) labels the irreducible representations (irreps). We take a fixed irrep of spin $S$, with a $2S+1$-dimensional carrier space $\mathcal{H}_{S}$ spanned by the standard angular momentum basis $\{|Sm\rangle,m=-S,\ldots,S\}$, whose elements are simultaneous eigenstates of $\hat{\mathbf{S}}^{2}$ and $\hat{S}_{z}$:
$\hat{\mathbf{S}}^{2}|S,m\rangle=S(S+1)|S,m\rangle\,,\qquad\hat{S}_{z}|S,m\rangle=m|S,m\rangle\,.$ (2.2)
The highest weight state is $|S,S\rangle$ and it is annihilated by the ladder operator $\hat{S}_{+}$ (with $\hat{S}_{\pm}=\hat{S}_{x}\pm\rmi\hat{S}_{y}$). The isotropy subgroup (i.e., the largest subgroup that leaves the highest weight state invariant) consists of all the elements of the form $\exp(i\chi\hat{S}_{z})$, so it is isomorphic to U(1). The coset space is then SU(2)$/$U(1), which is just the unit sphere $\mathcal{S}_{2}$ and it is the classical phase space, the natural arena to describe the dynamics.
The SU(2) coherent states $|\Omega\rangle$ (with $\Omega=(\theta,\phi)\in\mathcal{S}_{2}$) are defined, up to a global phase, by the action of the displacement operator [37]
$\hat{D}(\Omega)=\exp\left[{\textstyle\frac{1}{2}}\theta(\hat{S}_{+}\rme^{-\rmi\phi}-\hat{S}_{-}\rme^{\rmi\phi})\right]$ (2.3)
on the highest weight state, with explicit expression in terms of $\Omega$ given by
$\displaystyle|\Omega\rangle$ $\displaystyle=$ $\displaystyle\hat{D}(\Omega)|S,S\rangle$ $\displaystyle=$ $\displaystyle\sum_{m=-S}^{S}\sqrt{\frac{(2S)!}{(S-m)!(S+m)!}}[\cos(\theta/2)]^{S+m}[\sin(\theta/2)]^{S-m}\,\rme^{-\rmi m\phi}|S,m\rangle\,.$
Operators acting in a $\mathcal{H}_{S}$ can be mapped onto functions on $\mathcal{S}_{2}$ by means of the Stratonovich-Weyl kernel. It can be concisely defined as [38]
$\hat{w}(\Omega)=\sqrt{\frac{4\pi}{2S+1}}\sum_{K=0}^{2S}\sum_{q=-K}^{K}Y_{Kq}^{\ast}(\Omega)\,\hat{T}_{Kq}^{S}\,,$ (2.5)
where $Y_{Kq}(\Omega)$ are the spherical harmonics, $\ast$ indicates complex conjugation, and $\hat{T}_{Kq}^{S}$ are the irreducible tensor operators [39, 40]
$\hat{T}_{Kq}^{S}=\sqrt{\frac{2K+1}{2S+1}}\sum_{m,m^{\prime}=-S}^{S}C_{Sm,Kq}^{Sm^{\prime}}\,|S,m^{\prime}\rangle\langle S,m|\,,$ (2.6)
$C_{Sm,Kq}^{Sm^{\prime}}$ being the corresponding Clebsch-Gordan coefficient [41]. The symbol $W_{A}$ of an operator $\hat{A}$ is then defined as
$W_{A}(\Omega)=\Tr[\hat{A}\,\hat{w}(\Omega)]\,.$ (2.7)
Since the tensors $\hat{T}_{Kq}^{S}$ constitute an orthonormal basis for the operators acting on $\mathcal{H}_{S}$, any observable $\hat{A}$ can be expanded as
$\hat{A}=\sum_{K=0}^{2S}\sum_{q=-K}^{K}A_{Kq}\,\hat{T}_{Kq}^{S}\,,$ (2.8)
with $A_{Kq}=\Tr[\hat{A}\hat{T}_{Kq}^{S\dagger}]$, $\dagger$ standing for Hermitian conjugation. Therefore, the symbol of $\hat{A}$ can be expressed as the sum of symbols of the tensor components
$W_{A}(\Omega)=\sqrt{\frac{4\pi}{2S+1}}\sum_{K=0}^{2S}\sum_{q=-K}^{K}A_{Kq}\,Y_{Kq}^{\ast}(\Omega)\,.$ (2.9)
As some relevant examples we shall need in what follows we quote
$\begin{array}[]{rcl}\hat{S}_{i}&\mapsto&W_{S_{i}}(\Omega)=\sqrt{S(S+1)}\;n_{i},\\ \{\hat{S}_{i},\hat{S}_{j}\}&\mapsto&W_{\{S_{i},S_{j}\}}(\Omega)=C_{S}\,n_{i}n_{j}\,,\\ \hat{S}_{i}^{2}&\mapsto&W_{S_{i}^{2}}(\Omega)=\frac{1}{2}C_{S}\left(n_{i}^{2}-\frac{1}{3}\right)+\frac{1}{3}S(S+1),\end{array}$ (2.10)
where the Latin indexes run the values $\{i,j\}\in x,y,z$, $\mathbf{n}=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)^{t}$ is a unit vector in the direction of spherical angles $(\theta,\phi)\in\mathcal{S}_{2}$, and $C_{S}=[S(S+1)(2S-1)(2S+3)]^{1/2}$.
The Wigner function is the symbol of the density operator $\hat{\varrho}$. It is SU(2) covariant: under the action of a $2S+1$-dimensional irrep of SU(2) given by the matrix $\hat{R}(\Omega)$ [that is, $\hat{\varrho}^{\prime}=\hat{R}(\Omega^{\prime})\,\hat{\varrho}\,\hat{R}^{-1}(\Omega^{\prime})$], $W_{\varrho}(\Omega)$ experiences the transformation
$W_{\varrho^{\prime}}(\Omega)=W_{\varrho}(R^{-1}\Omega)\,,$ (2.11)
so that it follows rotations rigidly without changing its form. In addition, we have the overlap relation
$\Tr(\hat{\varrho}\hat{A})=\frac{2S+1}{4\pi}\int_{\mathcal{S}_{2}}\rmd\Omega\,W_{\varrho}(\Omega)\,W_{A}(\Omega)\,,$ (2.12)
where $\rmd\Omega=\sin\theta\rmd\theta\rmd\phi$ is the invariant measure in $\mathcal{S}_{2}$.
For a coherent state $|\Omega_{0}\rangle$, the Wigner function can be computed directly from the definition (2.7) by taking into account that
$\langle\Omega_{0}|\hat{T}_{Kq}^{S}|\Omega_{0}\rangle=(2S)!\sqrt{\frac{4\pi}{(2S-K)!(2S+K+1)!}}\,Y_{Kq}(\Omega_{0})\,.$ (2.13)
$W_{\Omega_{0}}(\Omega)=(2S)!\sum_{K=0}^{2S}\sqrt{\frac{2S+1}{(2S-K)!(2S+K+1)!}}\,P_{K}(\cos\zeta)\,,$ (2.14)
where $\cos\zeta=\cos\theta\cos\theta_{0}+\sin\theta\sin\theta_{0}\cos(\phi-\phi_{0})$ and $P_{K}(\omega)$ are the Legendre polynomials.
To conclude, we stress that this approach assume a fixed $S$. In some instances, as in polarization optics, a superposition of different $S$ arise [42, 43]. The formalism can be generalized to cover this more general situation [44].
## 3 Dynamics and Wigner current on the sphere
The exact evolution equation for the Wigner function $W_{\varrho}(\Omega)$ has been obtained in [13]. For arbitrary Hamiltonians [living in a $(2S+1)$-dimensional representation of the universal enveloping algebra of $\mathfrak{su}(2)$] the expressions are quite involved. For simplicity, in what follows, we restrict ourselves to two simple examples of great interest in applications.
### 3.1 Linear Hamiltonians
First of all, we consider the dynamics generated by linear Hamiltonians
$\hat{H}_{L}=\sum_{i}a_{i}\hat{S}_{i}$ (3.1)
whose symbol can be directly inferred from (2.10). The exact phase-space evolution is given by the first-order partial differential equation
$\partial_{t}W_{\varrho}(\Omega|t)=\sum_{i}a_{i}\{W_{\varrho}(\Omega|t),n_{i}\}\,,$ (3.2)
where
$\{f,g\}=\frac{1}{\sin\theta}\left(\partial_{\phi}f\,\partial_{\theta}g-\partial_{\theta}f\,\partial_{\phi}g\right)$ (3.3)
is the Poisson bracket on the sphere $\mathcal{S}_{2}$. The evolution for the Wigner function is
$W_{\varrho}(\Omega|t)=W_{\varrho}(\Omega(-t)|0)\,,$ (3.4)
where $\Omega(t)$ denotes classical trajectories, which are solutions of the classical Hamiltonian equations. It thus corresponds to a rotation of the initial distribution.
Next, we observe that if the evolution can be recast in the form $\partial_{t}W_{\varrho}(\Omega|t)=\{A,B\}$, it can be interpreted as a continuity equation with current given by
$J_{\phi}=-A\partial_{\theta}B\,,\qquad J_{\theta}=\frac{1}{\sin\theta}A\partial_{\phi}B\,.$ (3.5)
Accordingly, the linear dynamics is generated by
$\displaystyle J_{\theta}$ $\displaystyle=$ $\displaystyle\frac{1}{\sin\theta}W_{\varrho}(\Omega|t)\sum_{i}a_{i}\partial_{\phi}n_{i}\,,$ $\displaystyle J_{\phi}$ $\displaystyle=$ $\displaystyle-W_{\varrho}(\Omega|t)\sum_{i}a_{i}\,\partial_{\theta}n_{i}\,.$
Since for these linear Hamiltonians the exact evolution is the classical Liouville equation [45], the quantum and classical currents are just the same.
For the particular case of $\hat{H}_{L}=\omega\hat{S}_{z}$ the resulting components of Wigner current are:
$\displaystyle J_{\theta}$ $\displaystyle=$ $\displaystyle 0\,,$ $\displaystyle{J}_{\phi}$ $\displaystyle=$ $\displaystyle{\omega\sin\theta\;W_{\varrho}(\theta,\phi-\omega t|0)\,.}$
In the supplemental material, we present an animation of this current for an initial coherent state.
### 3.2 Kerr dynamics
For quadratic Hamiltonians, we content ourselves with the simplest case of the so-called Kerr medium [46, 47], which is described by
$\hat{H}=\chi\hat{S}_{z}^{2}\,.$ (3.8)
The ensuing dynamics has been examined in terms of the standard position-momentum phase space [48, 49] and the associated Wigner current has been recently discussed [50]. For the SU(2) Wigner function the evolution equation turns out to be [31, 13]
$\partial_{t}W_{\varrho}(\Omega|t)=-\frac{\chi}{\epsilon}\cos\theta\ \hat{\Gamma}(\theta,\mathcal{L}^{2})\,\partial_{\phi}W_{\varrho}(\Omega|t)\,,$ (3.9)
where $\varepsilon=1/(2S+1)$ and the operator $\hat{\Gamma}(\theta,\mathcal{L}^{2})$ is
$\hat{\Gamma}(\theta,\mathcal{L}^{2})=\frac{1}{2}\Phi(\mathcal{L}^{2})-\frac{\epsilon^{2}}{2}(1+2\tan\theta\,\partial_{\theta})\Phi^{-1}(\mathcal{L}^{2})\,.$ (3.10)
Here, $\Phi(\mathcal{L}^{2})$ is
$\Phi(\mathcal{L}^{2})=\left[2-\epsilon^{2}(2\mathcal{L}^{2}+1)+2\sqrt{1-\epsilon^{2}(2\mathcal{L}^{2}+1)+\epsilon^{4}\mathcal{L}^{4}}\right]^{1/2}\,,$ (3.11)
and $\Phi^{-1}(\mathcal{L}^{2})$ its inverse. Both are functions solely of $\mathcal{L}^{2}$, which is a differential realization of the Casimir operato on $\mathcal{S}_{2}$
$\mathcal{L}^{2}=-\left(\partial_{\theta\theta}+\cot\theta\,\partial_{\theta}+\frac{1}{\sin^{2}\theta}\partial_{\phi\phi}\right)\,,$ (3.12)
and, consequently, we have $\mathcal{L}^{2}Y_{Kq}(\Omega)=K(K+1)Y_{Kq}(\Omega)$. Note also that the term between parentheses in (3.12) is the Laplace-Beltrami operator in $\mathcal{S}_{2}$.
Equation (3.9) can be represented in terms of the Poisson brackets as follows
$\partial_{t}W_{\varrho}(\Omega|t)=2\epsilon\chi\left\{\hat{\Gamma}(\theta,\mathcal{L}^{2})\,W_{\varrho}(\Omega),\frac{1}{4\epsilon^{2}}\cos^{2}\theta\right\}\,.$ (3.13)
Actually, the operator $\hat{\Gamma}(\theta,\mathcal{L}^{2})$ is responsible for the quantum deformation of the distribution. The current can be immediately found from (3.5):
$\displaystyle J_{\phi}(t)$ $\displaystyle=$ $\displaystyle\chi\epsilon^{-1}\sin\theta\cos\theta\;\hat{\Gamma}(\theta,\mathcal{L}^{2})\;W_{\varrho}(\Omega|t)$ $\displaystyle J_{\theta}(t)$ $\displaystyle=$ $\displaystyle 0\,.$
The nonzero components of the current can be recast as
$J_{\phi}(t)=\sin\theta\hat{U}(t)\frac{1}{\sin\theta}J_{\phi}(t=0),$ (3.15)
where
$\hat{U}(t)=\exp\left[-\frac{\chi t}{\epsilon}\cos\theta\,\hat{\Gamma}(\theta,\mathcal{L}^{2})\ \partial_{\phi}\right]$ (3.16)
is the evolution operator in phase space; that is, $W_{\varrho}(\Omega|t)=\hat{U}(t)W_{\varrho}(\Omega|t=0)$.
In figure 1 we plot the quantum current for an initial coherent state on the equator ($\theta=\pi/2,\phi=0$). We have chosen three different dimensionless times $\tau=\chi t$ corresponding to $\tau=0$, $\tau=0.32$ (close to the best squeezing time) and $\tau=1.5$ (close to the appearance of two-component Schrödinger cats) for the case $S=10$. The white arrows represent the current $J_{\phi}$. At $\tau=0$ , the size and position of the arrows clearly indicate the direction of the deformation of the Wigner function: in the vicinity of the initial maximum, the laminar flow with increasing speed towards polar regions leads to the squeezing of the distribution along transverse directions for short times $\tau\sim S^{-1/2}$. Such a deformation is actually reflected in a real squeezing of $\hat{S}_{x}$ and $\hat{S}_{y}$ components. In addition, first signs of the quantum interference are observed. For cat times, $\tau\sim 1$ , the structure of the quantum current is quite complicated: multiple regions where the current changes direction can be easily noticed. In the supplemental material, the reader can find an animation of this current for an initial coherent state.
To better appreciate the stagnation lines (recall that $J_{\theta}=0$ identically), in figure 2 we plot the Wigner current of figure 1, but now in the plane. There is always a trivial zero line at $\theta=\pi/2$. At the initial moment, the stagnation lines separate regions of positive and negative values of the Wigner function, as well as the minima of the negative ripples. New zero lines around negative parts of the Wigner distribution appear at the best squeezing time. Finally, nontrivial stagnation lines take the form of closed curves rounding minima of the interference pattern. These stagnation lines thus provide a complementary picture of quantum interference in phase space.
### 3.3 Semiclassical limit
The semiclassical limit in spinlike systems is related to large value of spin, naturally characterized by the parameter $\epsilon\ll 1$. The semiclassical states are usually associated with smooth and localized (with extension of order $\sqrt{S}$) phase-space distributions [51]. Algebraically, the density matrix of semiclassical states is decomposed only on low rank tensors with $K\lesssim\sqrt{S}$ in equation (2.8). The typical semiclassical states are the spin coherent states (LABEL:CS).
The operator (3.10) in the semiclassical limit tends to $\hat{\Gamma}(\theta,\mathcal{L}^{2})=1+O(\epsilon^{2})$, and the evolution equation (3.13) takes the form of the classical equation of motion corresponding to the Hamiltonian (3.8); viz,
$\partial_{t}W_{\varrho}(\Omega|t)\simeq-\frac{\chi}{\epsilon}\cos\theta\partial_{\phi}W_{\varrho}(\Omega|t)=2\epsilon\{W_{\varrho}(\Omega),W_{H}(\Omega)\}\,,$ (3.17)
where the symbol of the Hamiltonian is
$W_{H}(\Omega)\simeq\frac{\chi}{4\epsilon^{2}}\cos^{2}\theta\,.$ (3.18)
The solution is defined by the classical trajectories according to equation (3.4). Nevertheless, in this case different points of the initial distribution evolve with different velocities, so that the classical motion leads to a semiclassical deformation of the initial distribution,
$W_{\varrho}\left(\Omega|t\right)\simeq W_{\varrho}\left(\theta,\phi-\frac{\chi t}{\epsilon}\cos\theta\Bigl{|}t=0\right)=W_{\varrho}(\Omega(-t)|0)\,.$ (3.19)
The evolution distorts the initial distribution but cannot convert positive regions of the Wigner function into negative regions (and vice versa) as follows from the conservation of local Poincareé invariants under the action of Poisson bracket.
Such a deformation represents, for instance, squeezing and is generated by the semiclassical current
$\displaystyle J_{\theta}^{\mathrm{scl}}(\Omega)$ $\displaystyle=$ $\displaystyle 0,$ $\displaystyle J_{\phi}^{\mathrm{scl}}(\Omega)$ $\displaystyle=$ $\displaystyle\frac{1}{2}\frac{\chi}{\epsilon}\sin(2\theta)\,W_{\varrho}(\Omega(-t)|0)\,.$
In the lower panel of figure 1 we plot this semiclassical current at the same times as for the quantum case. At the initial times, both semiclassical and quantum currents look quite similar. However, already for short times, $\tau=0.32$, the semicclassical distribution differs from the quantum one. The semiclassical current only produces a deformation of the initial distribution, as it follows from the (3.19). The semiclassically-evolved distribution is slightly narrower than the quantum one, but still describes very well the effect of spin squeezing [13]. For longer times, the semiclassical current keeps twisting the Wigner distribution, which obviously does not show any sort of interference pattern. The stagnation lines in the semiclassical case coincide with zeros of the evolved Wigner function, as it follows from (LABEL:cl_SL) and differ from the quantum case, even at the initial times. Such a difference is significant and can be in principle used for a detection of genuine quantum features.
The higher moments of the Wigner distribution
$\mathfrak{m}_{k}(t)=\left(\frac{2S+1}{4\pi}\right)^{k}\int_{\mathcal{S}_{2}}d\Omega\,W_{\varrho}^{k}(\Omega|-t)\,.$ (3.21)
Since in the semiclassical approximation the evolution is generated by canonical transformations, these higher moments are time-independent. The devia- tions from their initial values describe a spread of the initial distribution due to purely quantum effects. Actually, since $\partial_{t}\mathfrak{m}(t)|_{t=0}=0$, the widths of the $\mathfrak{m}_{k}(t)$ at $t=0$, given by $\partial_{t}^{2}\mathfrak{m}(t)|_{t=0}$ define the timescales over which the semiclassical approximation gives a bona fide description of the dynamics.
## 4 Concluding remarks
In summary, we have studied the dynamics of the Wigner function for spinlike systems and the associated phase-space flow. For linear Hamiltonians, the quantum and classical flows coincide. For nonlinear evolution, there are significant differences between quantum and classical flows, even for short times $\tau\sim S^{-1/2}$. From an experimental viewpoint, quantum effects can hardly be observed by measuring low-order moments of spin operators [51] when $S\gg 1$. Thus, by analyzing the Wigner current, in principle, it is possible to detect genuine quantum features of large spin systems arising in the course of nonlinear dynamics.
This work is partially supported by the Grant 254127 of CONACyT (Mexico). S. T. W. is supported by the Ministry of Science and Technology Taiwan (Grant MOST 107-2112-M-194-002. L. L. S. S. acknowledges the support of the Spanish MINECO (Grant FIS2015-67963-P).
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• [51] Valtierra I F, Romero J L and Klimov A B 2017 Ann. Phys. 383 620–634 | 8,894 | 28,546 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 182, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.721896 |
https://www.studypool.com/discuss/504725/solve-the-system-of-equations-using-substitution?free | 1,508,464,217,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823605.33/warc/CC-MAIN-20171020010834-20171020030834-00196.warc.gz | 994,794,133 | 14,320 | ##### Solve the system of equations using substitution.
label Mathematics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Solve the system of equations using substitution.
Solve the system of equations using substitution.3y = –1/2x + 2y = –x + 9
a. (10, –1)
b. (–1, 8)
c. (20, –4)
d. (3, 6)
Apr 29th, 2015
Something must be wrong here.
Only a. and d. obey the second equation y= -x + 9
but none obey the first equation : 3y=-1/(2x)+2
for a: -1/(2x)+2= -1/20+2 not equal 3y=-3
for d: -1/(2x)+2= -1/6+2 not equal 3y=18
So it can be solved but none of the solutions agree with it.
Apr 29th, 2015
the second equation is 3 y = - 1/2 x + 2
3 y - one half x + 2
Apr 29th, 2015
Ok it must be b then I'll send you the solution in a sec
Apr 29th, 2015
I meant a
one sec
Apr 29th, 2015
y = -x+9
substitute in the first: 3(-x+9)=(-1/2)x+2
-3x+27=-x/2+2
5x/2=25 => x=25*2/5=10 and y= -10+9=-1
so its a
Apr 29th, 2015
...
Apr 29th, 2015
...
Apr 29th, 2015
Oct 20th, 2017
check_circle | 416 | 1,026 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.745751 |
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### How to algorithmically calculate the adjacency matrix of platonic solids
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### How do i prove that there are at least 12 faces of degree 5 for a regular graph of degree 3 where there are no faces of degree less than five,
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### Induction: Every connected graph has a spanning tree
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### Number of all possible orientations in a graph
What is the number of all possible orientations for an undirected graph? I think it must be $2^{|E|}$, because we have $|E|$ edges, each of them can have 2 choices for it's direction. Is it true?
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### Prove that every strongly connected digraph has an odd directed cycle if its underlying graph has an odd cycle
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### Is the hypercube the only connected, regular, bipartite simple finite graph?
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### Proof that Laplacian spectrum is symmetric for bipartite graphs
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### Given a simple graph with $n = 4k + 2$ vertices. Can the vertices of this graph have distinct degrees?
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### Proof that graphs are not isomorphic
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### Using Euler's theorem to calculate the number of edges in a graph
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### The number of pendant vertices in a tree
Let $T$ be a tree with vertices $\{v_1, v_2, . . . , v_n \}$ for $n \geq 2$. Prove that the number of pendant vertices in $T$ is equal to \large{2 + \sum_{v_i,deg(v_i) \geq 3}\big( deg(v_i) - 2 ...
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### What does Ramsey theory tell us?
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Consider an unweighted and undirected graph $G=(V,E)$, where the vertices $V$ of $G$ lie on the unit n-sphere. If we choose a normal vector uniformly at random on this $n$-sphere, then the ...
### Prove that Graph $(V \cup W,E^{\prime})$ is connected
Suppose $(V,E)$ is a connected undirected graph, in which $V = \{v_1, v_2,..... , v_n\}$. Let $W = \{w_1,w_2,.....,w_n\}$. How can I prove that the undirected graph $(V \cup W,E^{\prime})$ is ... | 3,642 | 13,720 | {"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} | 3.5 | 4 | CC-MAIN-2016-18 | longest | en | 0.910324 |
https://blog.chungyc.org/2008/08/coloring-a-plane/ | 1,619,135,062,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039563095.86/warc/CC-MAIN-20210422221531-20210423011531-00173.warc.gz | 249,705,836 | 27,081 | # Coloring a plane
Skeptic's Play has a couple of interesting geometry puzzles I've repeated below:
1. Let's say I've painted a plane — the flat kind, not the flying kind. Every single point on the plane is assigned the color red or blue. Prove that there must exist two points of the same color that are exactly one unit apart.
2. Every single point on the plane is assigned the color red, blue or green. Prove that there must exist two points of the same color that are exactly one unit apart.
What makes this interesting is that you can't assume anything about how a plane could be painted. Mathematically, there can be all sorts of ways one could color a plane that are literally unimaginable. For example, if you color all points with rational number Cartesian coordinates with red, then the entire plane is colored red and yet it is at the same time almost completely colorless. Fractals are another example, where you can imagine large scale structures, but it's not really possible to imagine it with all the small scale structures repeating no matter how much you zoom in. To prove the above statements, one needs to use a method that would still apply no matter how bizarrely the plane is colored.
Anyways, the first statement is pretty easy to prove. Let's assume that there are no two points that are one unit apart that have the same color. Then there are both red and blue points in the plane. If there was only one color, then any two points at unit distance would obviously be the same color. Let's pick any blue point. Then the circle centered around the blue point with a unit radius must be completely red. Any two points on the circle that are a unit distance apart would both be red, which would be a contradiction.
That was easy, but would the second statement be as easy to solve? Let's try to repeat our approach and assume that no two points a unit distance apart has the same color. Without loss of generality, let's pick any point A and say that it's blue. Then the unit circle centered around A must be made of points that are either red or green. Again without loss of generality, let's pick any point B on the circle and say that it's green. Then the unit circle around B must be blue or red. Then point C at where the two circles intersect must be red. And the unit circle around C must be blue or green. I show this in the figure below, where a magenta circle is made up of points that can be either blue or red, a yellow circle is made up of points that can be either red or green, and a cyan circle is made up of points that can be either blue or green.
The obvious path to take is to repeatedly draw a circle around each intersection and hope there is a point which must result in a contradictory coloring. This would happen if magenta, yellow, and cyan circles all intersected at a single point. Unfortunately, this never happens no matter how far you go. In fact, we get a repeating pattern for the intersections that is consistent with our initial assumption. We need another approach.
Instead of trying to find two intersections from circles that are centered around existing intersections, we'll try making a unit circle that isn't centered around a pre-existing intersection. First we do what we did before until we get point D below. Then we move around a unit circle that passes through point D and the arc between A and A' below. This unit circle must be centered around a point O on the magenta circle centered on D. We want to change L by moving point O and its circle around until it's equal to the unit distance. This is possible since the smallest value for L is zero and the largest value is the square root of three or about 1.73, which is the distance between A and A'.
Once we move point O so that L becomes the unit distance as in below, we finally get our contradiction. Without loss of generality, assume that point O is colored red. Then points E and F, both being the intersection of the magenta and cyan circles, must both be blue. Since points E and F are a unit distance apart, this contradicts our initial assumption that no two points a unit distance apart are the same color.
This was a fun but frustrating puzzle to solve. Proving the first statement was really easy, but I got trapped in the dead-end approach for too long for the second statement. The ease of the first one kept me working on the second one since I thought it should also have a trivial solution. Unfortunately, I kept working with variations of the dead-end approach hoping to get an intersection with contradictory colorings, and it even reached the point that I wondered if there was a way to color a plane so that no two points a unit distance apart were the same color. But a slight detour in the approach helped me reach a simple solution, so I can now stop worrying about the puzzle.
It was a nice way to exercise my brain and prove that it hasn't completely ossified yet. However, when the Skeptic's Play gives its next puzzle, I should be wary of deceptively easy looking puzzles like this one, where the first half was so simple that it made me think the second half should also be as trivial. It should at least help prevent me from underestimating how much time I need to waste ...
## 4 Replies to “Coloring a plane”
1. Yoo says:
I forgot to reset the URL when commenting on your blog, so the anime blog URL that I use when commenting in other anime blogs managed to sneak in. ^_^;;I've got four blogs in different topics, but my personal blog and my general blog here are the ones I regularly update.
2. Stacy says:
Wow! I'm going to have to come back in the morning and look at this post again. My brain does not function properly this late at night!
Hey - (OT) How many blogs do you have?
I clicked on your name earlier and it took me to a site that looks as if it is about animation.
3. Stacy says:
"Anime" - that's the word I couldn't remember (slaps forehead)!
What a jerk I have over at my place huh? Thanks for setting him straight.
Stacy
4. Yoo says:
No problem. "Communists kill because of atheism" claims just irritate me to no end, especially because it's a real, though distant, possibility that I could actually be killed by a communist state and my atheism would not help one bit.
Having a totalitarian communist state to the north that is technically still at war in the state I currently live in tends to make one consider the possibility once in a while.
I just worry that my sarcasm would have been lost on the person. | 1,404 | 6,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | longest | en | 0.964151 |
https://www.sqlbi.com/topics/dax/page/41/?uf=23789568 | 1,597,217,352,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738878.11/warc/CC-MAIN-20200812053726-20200812083726-00112.warc.gz | 842,760,322 | 13,436 | Topic: DAX
Page 41 of 409 posts
blog post –
If you use PowerPivot and write some DAX formula, don’t miss this post on PowerPivotPro blog – if you want to get an external editor for your DAX formula, you can use Notepad++ for free – and adding the customization described in this post by Colin Banfield, Read more
blog post –
ALL, ALLEXCEPT and VALUES in DAX
When you use CALCULATE in DAX you are creating a new filter context for the calculation, based on the existing one. There are a few functions that are used to clear or preserve a column filter. These functions are: ALL – it can be used with one or more Read more
blog post –
How to relate tables in DAX without using relationships
UPDATE 2017-07-12: please note this article was written in 2010, there are now better ways to obtain the same result. Please read these articles: Physical and Virtual Relationships in DAX and Propagate filters using TREATAS in DAX PowerPivot supports Read more
blog post –
Avoiding calculated column in DAX
A calculated column is a DAX expression which is evaluated when the PowerPivot workbook is updated. It is very useful, but there are cases where you want to delay calculation at query time. For example: You want to make part of the calculation depending Read more
blog post –
Memory Considerations about PowerPivot for Excel
PowerPivot for Excel is an Add-In which uses a local version of Analysis Services (SSAS) to process data and make calculation to respond to user queries made using Excel 2010. The SSAS engine is loaded in-memory in the process of Excel. Especially with Read more
blog post –
ABC Analysis in PowerPivot
The calculation for ABC analysis can be made in PowerPivot using calculated columns. In this way each row can have an attribute with the appropriate ABC class. The ABC calculation has to be made considering a particular grouping and sort order. For example, Read more
blog post –
How CALCULATE works in DAX
The CALCULATE function in DAX is the magic key for many calculations we can do in DAX. However, it is not pretty intuitive how it works and I spent a lot of time trying to understand how it can be used. First of all, this is the syntax. CALCULATE( <expression>, Read more
blog post –
Distinct Count Measure in PowerPivot using DAX
UPDATE: PowerPivot for Excel 2010 in SQL Server 2012 and Excel 2013 support the DISTINCTCOUNT aggregation in DAX. This blog post is relevant only if you use PowerPivot for Excel 2010 in SQL Server 2008 R2. A PivotTable based on PowerPivot data doesn’t Read more
blog post –
Many-to-Many relationships in PowerPivot
UPDATE: This blog post is still good for learning DAX principles, but a better description of many-to-many patterns is available on The Many-to-Many Revolution whitepaper that is available here: http://www.sqlbi.com/articles/many2many/ – please download Read more | 673 | 2,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-34 | latest | en | 0.901216 |
https://yards-to-meters.appspot.com/3/pl/ | 1,716,851,995,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.34/warc/CC-MAIN-20240527205559-20240527235559-00872.warc.gz | 923,796,394 | 11,319 | Jard Na Metr
# yd na mJardy na Metry
## Jard na Metr przetwornik
Przelicznika jednostek możesz użyć, by przeliczać jednostki miary.
yd
=
m
## Jak konwertować jardy do metry?
1 yd * 0.9144 m = 0.9144 m 1 yd
## Konwersja 1 yd do wspólnych długości
Jednostka miaryDługości
Nanometr914400000.0 nm
Mikrometr914400.0 µm
Milimetr914.4 mm
Centymetr91.44 cm
Cal36.0 in
Stopa3.0 ft
Jard1.0 yd
Metr0.9144 m
Kilometr0.0009144 km
Mila0.0005681818 mi
Mila morska0.0004937365 nmi
## Konwersja yd na m
Jard (yd)Metr (m)
1 yd0.9144 m
2 yd1.8288 m
3 yd2.7432 m
4 yd3.6576 m
5 yd4.572 m
6 yd5.4864 m
7 yd6.4008 m
8 yd7.3152 m
9 yd8.2296 m
10 yd9.144 m
11 yd10.0584 m
12 yd10.9728 m
13 yd11.8872 m
14 yd12.8016 m
15 yd13.716 m
16 yd14.6304 m
17 yd15.5448 m
18 yd16.4592 m
19 yd17.3736 m
20 yd18.288 m
21 yd19.2024 m
22 yd20.1168 m
23 yd21.0312 m
24 yd21.9456 m
25 yd22.86 m
Jard (yd)Metr (m)
26 yd23.7744 m
27 yd24.6888 m
28 yd25.6032 m
29 yd26.5176 m
30 yd27.432 m
31 yd28.3464 m
32 yd29.2608 m
33 yd30.1752 m
34 yd31.0896 m
35 yd32.004 m
36 yd32.9184 m
37 yd33.8328 m
38 yd34.7472 m
39 yd35.6616 m
40 yd36.576 m
41 yd37.4904 m
42 yd38.4048 m
43 yd39.3192 m
44 yd40.2336 m
45 yd41.148 m
46 yd42.0624 m
47 yd42.9768 m
48 yd43.8912 m
49 yd44.8056 m
50 yd45.72 m
Jard (yd)Metr (m)
51 yd46.6344 m
52 yd47.5488 m
53 yd48.4632 m
54 yd49.3776 m
55 yd50.292 m
56 yd51.2064 m
57 yd52.1208 m
58 yd53.0352 m
59 yd53.9496 m
60 yd54.864 m
61 yd55.7784 m
62 yd56.6928 m
63 yd57.6072 m
64 yd58.5216 m
65 yd59.436 m
66 yd60.3504 m
67 yd61.2648 m
68 yd62.1792 m
69 yd63.0936 m
70 yd64.008 m
71 yd64.9224 m
72 yd65.8368 m
73 yd66.7512 m
74 yd67.6656 m
75 yd68.58 m
Jard (yd)Metr (m)
76 yd69.4944 m
77 yd70.4088 m
78 yd71.3232 m
79 yd72.2376 m
80 yd73.152 m
81 yd74.0664 m
82 yd74.9808 m
83 yd75.8952 m
84 yd76.8096 m
85 yd77.724 m
86 yd78.6384 m
87 yd79.5528 m
88 yd80.4672 m
89 yd81.3816 m
90 yd82.296 m
91 yd83.2104 m
92 yd84.1248 m
93 yd85.0392 m
94 yd85.9536 m
95 yd86.868 m
96 yd87.7824 m
97 yd88.6968 m
98 yd89.6112 m
99 yd90.5256 m
100 yd91.44 m
Jard (yd)Metr (m)
105 yd96.012 m
110 yd100.584 m
115 yd105.156 m
120 yd109.728 m
125 yd114.3 m
130 yd118.872 m
135 yd123.444 m
140 yd128.016 m
145 yd132.588 m
150 yd137.16 m
155 yd141.732 m
160 yd146.304 m
165 yd150.876 m
170 yd155.448 m
175 yd160.02 m
180 yd164.592 m
185 yd169.164 m
190 yd173.736 m
195 yd178.308 m
200 yd182.88 m
210 yd192.024 m
220 yd201.168 m
230 yd210.312 m
240 yd219.456 m
250 yd228.6 m
Jard (yd)Metr (m)
260 yd237.744 m
270 yd246.888 m
280 yd256.032 m
290 yd265.176 m
300 yd274.32 m
310 yd283.464 m
320 yd292.608 m
330 yd301.752 m
340 yd310.896 m
350 yd320.04 m
360 yd329.184 m
370 yd338.328 m
380 yd347.472 m
390 yd356.616 m
400 yd365.76 m
410 yd374.904 m
420 yd384.048 m
430 yd393.192 m
440 yd402.336 m
450 yd411.48 m
460 yd420.624 m
470 yd429.768 m
480 yd438.912 m
490 yd448.056 m
500 yd457.2 m
Jard (yd)Metr (m)
510 yd466.344 m
520 yd475.488 m
530 yd484.632 m
540 yd493.776 m
550 yd502.92 m
560 yd512.064 m
570 yd521.208 m
580 yd530.352 m
590 yd539.496 m
600 yd548.64 m
610 yd557.784 m
620 yd566.928 m
630 yd576.072 m
640 yd585.216 m
650 yd594.36 m
660 yd603.504 m
670 yd612.648 m
680 yd621.792 m
690 yd630.936 m
700 yd640.08 m
710 yd649.224 m
720 yd658.368 m
730 yd667.512 m
740 yd676.656 m
750 yd685.8 m
Jard (yd)Metr (m)
760 yd694.944 m
770 yd704.088 m
780 yd713.232 m
790 yd722.376 m
800 yd731.52 m
810 yd740.664 m
820 yd749.808 m
830 yd758.952 m
840 yd768.096 m
850 yd777.24 m
860 yd786.384 m
870 yd795.528 m
880 yd804.672 m
890 yd813.816 m
900 yd822.96 m
910 yd832.104 m
920 yd841.248 m
930 yd850.392 m
940 yd859.536 m
950 yd868.68 m
960 yd877.824 m
970 yd886.968 m
980 yd896.112 m
990 yd905.256 m
1000 yd914.4 m
Jard (yd)Metr (m)
1050 yd960.12 m
1100 yd1005.84 m
1150 yd1051.56 m
1200 yd1097.28 m
1250 yd1143.0 m
1300 yd1188.72 m
1350 yd1234.44 m
1400 yd1280.16 m
1450 yd1325.88 m
1500 yd1371.6 m
1550 yd1417.32 m
1600 yd1463.04 m
1650 yd1508.76 m
1700 yd1554.48 m
1750 yd1600.2 m
1800 yd1645.92 m
1850 yd1691.64 m
1900 yd1737.36 m
1950 yd1783.08 m
2000 yd1828.8 m
2100 yd1920.24 m
2200 yd2011.68 m
2300 yd2103.12 m
2400 yd2194.56 m
2500 yd2286.0 m
Jard (yd)Metr (m)
2600 yd2377.44 m
2700 yd2468.88 m
2800 yd2560.32 m
2900 yd2651.76 m
3000 yd2743.2 m
3100 yd2834.64 m
3200 yd2926.08 m
3300 yd3017.52 m
3400 yd3108.96 m
3500 yd3200.4 m
3600 yd3291.84 m
3700 yd3383.28 m
3800 yd3474.72 m
3900 yd3566.16 m
4000 yd3657.6 m
4100 yd3749.04 m
4200 yd3840.48 m
4300 yd3931.92 m
4400 yd4023.36 m
4500 yd4114.8 m
4600 yd4206.24 m
4700 yd4297.68 m
4800 yd4389.12 m
4900 yd4480.56 m
5000 yd4572.0 m
Jard (yd)Metr (m)
5100 yd4663.44 m
5200 yd4754.88 m
5300 yd4846.32 m
5400 yd4937.76 m
5500 yd5029.2 m
5600 yd5120.64 m
5700 yd5212.08 m
5800 yd5303.52 m
5900 yd5394.96 m
6000 yd5486.4 m
6100 yd5577.84 m
6200 yd5669.28 m
6300 yd5760.72 m
6400 yd5852.16 m
6500 yd5943.6 m
6600 yd6035.04 m
6700 yd6126.48 m
6800 yd6217.92 m
6900 yd6309.36 m
7000 yd6400.8 m
7100 yd6492.24 m
7200 yd6583.68 m
7300 yd6675.12 m
7400 yd6766.56 m
7500 yd6858.0 m
Jard (yd)Metr (m)
7600 yd6949.44 m
7700 yd7040.88 m
7800 yd7132.32 m
7900 yd7223.76 m
8000 yd7315.2 m
8100 yd7406.64 m
8200 yd7498.08 m
8300 yd7589.52 m
8400 yd7680.96 m
8500 yd7772.4 m
8600 yd7863.84 m
8700 yd7955.28 m
8800 yd8046.72 m
8900 yd8138.16 m
9000 yd8229.6 m
9100 yd8321.04 m
9200 yd8412.48 m
9300 yd8503.92 m
9400 yd8595.36 m
9500 yd8686.8 m
9600 yd8778.24 m
9700 yd8869.68 m
9800 yd8961.12 m
9900 yd9052.56 m
10000 yd9144.0 m
## Alternatywna pisownia
yd na Metry, yd do Metry, yd na m, yd do m, Jardy na Metr, Jardy do Metr, Jard na m, Jard do m, Jardy na Metry, Jardy do Metry, Jardy na m, Jardy do m, yd na Metr, yd do Metr | 3,115 | 5,626 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-22 | latest | en | 0.324396 |
http://funcall.blogspot.com/2011/04/exercises.html | 1,553,406,858,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203326.34/warc/CC-MAIN-20190324043400-20190324065400-00526.warc.gz | 82,603,950 | 17,671 | ## Friday, April 1, 2011
### Exercises
Apparently I was not as clear with my questions as I thought I was. I've edited the questions to make them more precise. The additional text is in this alternative color.
Exercise 1: Write a program that returns 1 if run on a any tail-recursive implementation of a language, but returns 0 if run on a any non tail-recursive implementation of that same language. The goal is not to find a program and language implementation pair that exhibit the behavior, but to design a program that can correctly infer whether on not any supplied language implementation supports tail recursion by examining its own behavior.
Exercise 1a (extra credit): Write a program that crashes or doesn't return if run on a any tail-recursive implementation of a language, but returns 0 if run on a any non tail-recursive implementation of that same language. The goal is not to find a program and language implementation pair that exhibit the behavior, but to design a program that can correctly infer that any supplied implementation does not support tail recursion (this is a weaker condition because we do not require that the program correctly infer support of tail recursion, but only lack of support).
Exercise 1b, essay (extra extra credit): Discuss the implications of your answers to exercises 1 and 1a to the semantics of the programs. In particular, briefly outline what changes to the various semantic models — denotational, operational, and/or axiomatic — take place upon introducing tail recursion.
Louis Reasoner wrote this program to sort numbers:
```(define (smallest list predicate)
;; Returns the smallest element of list.
(if (null? list)
#f
(if (predicate (car list) (cadr list))
(smallest (cons (car list) (cddr list)) predicate)
(smallest (cdr list) predicate))))
(define (louis-sort list predicate)
(do ((remainder list (cdr remainder))
(define test-list
(do ((i 0 (+ i 1))
```
He complains “I am unable to debug this because `smallest` tail calls itself and leaves no trace on the stack. Look!
```1 ]=> (louis-sort test-list <)
;The object (), passed as an argument to safe-car, is not a pair.
2 error> (debug)
There are 9 subproblems on the stack.
Subproblem level: 0 (this is the lowest subproblem level)
Expression (from stack):
(predicate (car list) ###)
subproblem being executed (marked by ###):
Environment created by the procedure: SMALLEST
applied to: ((0) #[arity-dispatched-procedure 39])
The execution history for this subproblem contains 1 reduction.
You are now in the debugger. Type q to quit, ? for commands.
3 debug> H
H
SL# Procedure-name Expression
0 smallest (predicate (car list) (cadr list))
1 smallest (if (predicate (car list) (cadr list)) (smalle ...
2 do-loop (cons (smallest remainder predicate) (quote ()))
3 do-loop (append answer (cons (smallest remainder predi ...
4 do-loop (do-loop (cdr remainder) (append answer (cons ...
5 %repl-eval (let ((value (hook/repl-eval s-expression envi ...
6 %repl-eval/write (hook/repl-write (%repl-eval s-expression envi ...
7 do-loop (begin (if (queue-empty? queue) (let ((environ ...
8 loop (loop (bind-abort-restart cmdl (lambda () (der ...
```
“I was expecting to see pending stack frames as the program recursively called `smaller`, but since they are all tail-recursive calls, I don't have a way of knowing how deep it went. I wish I could selectively enable or disable tail recursion for this one call...”
Exercise 2: What trivial change can Louis make to his code for `smaller` that will disable tail recursion?
Louis again has a problem. “I have a stack trace, but, but, well, just LOOK at it!
```1 ]=> (louis-sort test-list <)
;The object (), passed as an argument to safe-car, is not a pair.
2 error> (debug)
There are more than 50 subproblems on the stack.
Subproblem level: 0 (this is the lowest subproblem level)
Expression (from stack):
(predicate (car list) ###)
subproblem being executed (marked by ###):
Environment created by the procedure: SMALLEST
applied to: ((0) #[arity-dispatched-procedure 39])
The execution history for this subproblem contains 1 reduction.
You are now in the debugger. Type q to quit, ? for commands.
3 debug> H
H
SL# Procedure-name Expression
0 smallest (predicate (car list) (cadr list))
1 smallest (if (predicate (car list) (cadr list)) (smalle ...
102 do-loop (cons (smallest remainder predicate) (quote ()))
103 do-loop (append answer (cons (smallest remainder predi ...
104 do-loop (do-loop (cdr remainder) (append answer (cons ...
105 %repl-eval (let ((value (hook/repl-eval s-expression envi ...
106 %repl-eval/write (hook/repl-write (%repl-eval s-expression envi ...
107 do-loop (begin (if (queue-empty? queue) (let ((environ ...
108 loop (loop (bind-abort-restart cmdl (lambda () (der ...
```
“I wish I could start eliminating the tail call from the second call onwards...
Exercise 3: What trivial change can Louis make to his code for `smaller` that will enable tail recursion on all but the initial call to `smaller`? (Hint, the bold text in the following backtrace shows how this differs from the first backtrace.)
```1 ]=> (louis-sort test-list <)
;The object (), passed as an argument to safe-car, is not a pair.
2 error> (debug)
There are 10 subproblems on the stack.
Subproblem level: 0 (this is the lowest subproblem level)
Expression (from stack):
(predicate (car list) ###)
subproblem being executed (marked by ###):
Environment created by the procedure: SMALLEST
applied to: ((0) #[arity-dispatched-procedure 39])
The execution history for this subproblem contains 1 reduction.
You are now in the debugger. Type q to quit, ? for commands.
3 debug> H
H
SL# Procedure-name Expression
0 smallest (predicate (car list) (cadr list))
1 smallest (if (predicate (car list) (cadr list)) (smalle ...
3 do-loop (cons (smallest remainder predicate) (quote ()))
4 do-loop (append answer (cons (smallest remainder predi ...
5 do-loop (do-loop (cdr remainder) (append answer (cons ...
6 %repl-eval (let ((value (hook/repl-eval s-expression envi ...
7 %repl-eval/write (hook/repl-write (%repl-eval s-expression envi ...
8 do-loop (begin (if (queue-empty? queue) (let ((environ ...
9 loop (loop (bind-abort-restart cmdl (lambda () (der ...
```
Louis seems to have gotten his wish and is currently making progress despite the existence of tail recursion. Cy D. Fect, however, is unimpressed. He complains “Sure, it is trivial to disable tail recursion whenever you desire, but I don't like guessing whether the compiler is going to emit a tail call, and I'd simply rather not learn. I'd prefer some sort of declaration or decoration so I can explicitly tell the compiler to not emit a tail call. Something like this:
```(define (smallest list predicate)
;; Returns the smallest element of list.
(if (null? list)
#f
(if (predicate (car list) (cadr list))
;; This branch should leave a backtrace - CDF
(disable-tail-recursion
(smallest (cons (car list) (cddr list)) predicate))
(smallest (cdr list) predicate))))```
Exercise 4: Implement `disable-tail-recursion` as a macro.
Homework: Fix Louis's code.
1. I may be wrong, but it seems to me that anything you do to defeat tail recursion, short of calling write-char, can be removed by a compiler that does interprocedural optimization and effects analysis. For example, you can rewrite (f x) in tail position as (let ((y (f x))) (cons 'a 'b) y), but the SSC will see that the result of this cons is garbage no matter what silly procedures you nest it in, and remove the call to it. (I'll agree that even a SSC can't notice that the write-char is being done to a junk output file and remove it!)
2. John: You could easily set up a situation where, to determine whether the optimization is possible, the compiler would need to solve the halting problem. So any optimizer would have to give up in some cases.
3. Can your sufficiently smart compiler™ solve the halting problem?
If not, how can it ‘remove the call to [cons]’ without proving that every ‘silly procedure you nest it in’ does in fact halt?
Consider this code:
(define (solution-exists? solution-finding-program)
(let ((test-program
(sufficiently-smartly-compile
`(define (test)
(forward-reference
,solution-finding-program)
#t))))
(assembly-code/contains-call?
test-program
'forward-reference)))
If the SSC™ can optimize this, then we can solve the halting problem. On the other hand, if the SSC™ cannot optimize it, then we have a suitable mechanism for reliably and predictably defeating tail recursion.
4. GCC can optionally optimize tail calls using the -foptimize-sibling-calls switch.
I wrote a C function, is_tail_recursive,
that compares the addresses of local variables in different invocations. It works, at least for GCC 4.4.5.
So that solves exercise 1. Given the predicate is_tail_recursive, exercise 1a is trivial, so I'm not sure why that's an extra credit problem. (-:
5. Here it is.
https://gist.github.com/898702
6. kbob said...
I wrote a C function, is_tail_recursive,
that compares the addresses of local variables in different invocations. It works, at least for GCC 4.4.5.
Why do you think it works?
7. /*
* In a separate compilation unit, save is defined as:
*
* void save(int *p)
* {
* *p = (int)&p;
* }
*/
extern void save(int *);
void recurse(int n, int *p)
{
if (n == 0)
return;
save(p);
recurse(n - 1, p + 1);
}
int is_tail_recursive(void)
{
int a[2];
recurse(2, a);
return a[0] == a[1];
}
This code tests if the addresses of the arguments change during a recursive call. But you need two additional assumptions:
1. Without tail recursion a compiler is prohibited from moving stack frames.
2. With tail recursion, a compiler is required to use stack frames in a particular order.
It is unlikely someone would write a compiler that did not optimize tail recursion but nevertheless moved stack frames around, so the test for non tail recursion is unlikely to fail. (But it isn't inconceivable.) However, a compiler like Chicken, which most definitely supports tail recursion, would not pass this test.
Can you write a test that depends solely on the language definition and not on the quirks of a particular implementation of the compiler?
8. In the spirit of your 1a exercise and kbob's suggestion, here's a program that should work, but would crash if a compiler tried to naively apply TCO:
http://pastebin.com/c8QbsqYm
Happily, GCC is pretty smart! On -O2 and -O3, it will tail-call optimize acc_normal_rec, but it will refuse to tail-call optimize acc_evil_rec -- presumably it has a heuristic like "don't do TCO if the call takes any references to local variables."
9. mquander said...
here's a program that should work, but would crash if a compiler tried to naively apply TCO
You're adding the additional assumption that your compiler has a particular kind of bug. Can you write a test that depends solely on the language definition and not on the quirks of a particular implementation of the compiler?
Assume this: I have an ANSI C interpreter that is very pedantic and perverse. It rigidly follows the defined semantics of ANSI C, but it handles `unspecified behavior' (according to the language spec) by silently modifying memory in an undocumented way. (I believe that `unspecified behavior' covers that possibility.)
So why do you think your program would properly distinguish a tail-recursive interpreter from a non-tail-recursive one?
10. If tail-call optimization is defined only in terms of space used, then Exercise 1 requires a way to detect space usage without crashing unrecoverably. In a sufficiently underspecified language, this is impossible; e.g. in R5RS Scheme, memory might be infinite, and memory usage might vary arbitrarily for reasons unrelated to your program, and there's no way to ask about memory usage or catch out-of-memory errors — and any of these possibilities makes tail-recursion potentially unobservable. But anything is impossible in a sufficiently underspecified language, so this interpretation isn't very interesting. Given the realistic assumptions of upper and lower bounds on available memory, no arbitrary memory use (e.g. no more than a naive interpreter), and a way to catch out-of-memory errors, can't we always detect TCO by recursing to a depth exceeding the maximum possible memory, and catching out-of-memory errors? Or by measuring the maximum possible depth by nontail recursion, and seeing whether it varies when wrapped in different depths of tail recursion?
I think the argument that a SSC could rewrite the program to run in constant space is not a problem, because a compiler that does so is a tail-recursive compiler, as far as is observable. We can't determine the implementation's behavior on *all* programs, only what it does for specific programs.
It is odd that Scheme requires a space optimization without specifying anything else about space. Is the point you're trying to make here that tail-call optimization is meaningful only wrt the informal operational semantics? | 3,185 | 13,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | latest | en | 0.840992 |
https://philosophy-question.com/library/lecture/read/413613-what-does-reduction-in-biology-mean | 1,660,592,984,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572198.93/warc/CC-MAIN-20220815175725-20220815205725-00343.warc.gz | 415,500,139 | 7,531 | # What does reduction in biology mean?
## What does reduction in biology mean?
The act of decreasing or reducing something. ... Any process in which electrons are added to an atom or ion (as by removing oxygen or adding hydrogen); always occurs accompanied by oxidation of the reducing agent. The act of reducing complexity.
## What are the risk reduction strategies?
The four types of risk mitigating strategies include risk avoidance, acceptance, transference and limitation. Avoid: In general, risks should be avoided that involve a high probability impact for both financial loss and damage.
## What is risk reduction in statistics?
Risk Reduction statistics are a group of statistics that are increasingly used in clinical practice as more practitioners use evidence-based practice as their approach to clinical care. Their use involves the recognition that all treatments are prescribed to reduce the patient's risk of an adverse outcome.
## What is a good absolute risk reduction?
Absolute risk reduction (ARR) – also called risk difference (RD) – is the most useful way of presenting research results to help your decision-making. In this example, the ARR is 8 per cent (20 per cent - 12 per cent = 8 per cent).
## How do you assess relative risk reduction?
For example, if 60% of the control group died and 30% of the treated group died, the treatment would have a relative risk reduction of 0.
## What is the difference between absolute and relative risk reduction?
If something you do triples your risk, then your relative risk increases 300%. Absolute risk is the size of your own risk. Absolute risk reduction is the number of percentage points your own risk goes down if you do something protective, such as stop drinking alcohol.
## What does a negative relative risk reduction mean?
The interpretation of a negative value for NNT is as follows: if NNT patients are treated with the new treatment, one fewer patient will benefit than if they were all treated with the control. When NNT is negative, it is called NNH—the number needed to harm.
## How do you calculate NNT?
NNTs are always rounded up to the nearest whole number and accompanied as standard by the 95% confidence interval. Example: if a drug reduces the risk of a bad outcome from 50% to 40%, the ARR = 0.
## What is considered a good NNT?
Putting NNTs into perspective As a general rule of thumb, an NNT of 5 or under for treating a symptomatic condition is usually considered to be acceptable and in some cases even NNTs below 10.
## What is a good number needed to harm?
The lower the NNH, the more risk of harm; An NNH of 1 would mean that every patient treated is harmed. A different NNH is calculated for each specific adverse event.
## What is the NNT for statins?
Statins, which have become synonymous with “heart-attack-and-stroke-preventing,” have an NNT of 60 for heart attack and 268 for stroke: That's how many healthy people have to take statins for five years for those respective outcomes to be prevented.
## What is the absolute risk reduction of statins?
All-cause mortality after a mean follow-up of 4.
## What is the NNT for aspirin?
4 showed a small reduction in risk of ischemic stroke in patients allocated to the aspirin group, with an NNT value of 625 (Table 2). | 702 | 3,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-33 | latest | en | 0.954155 |
http://puzzle.queryhome.com/7961/if-1-x-1-y-1-3-1-x-1-z-1-5-1-y-1-z-1-7-then-x-y | 1,502,983,694,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103579.21/warc/CC-MAIN-20170817151157-20170817171157-00466.warc.gz | 340,134,016 | 30,657 | # If 1/x + 1/y = 1/3; 1/x + 1/z = 1/5; 1/y + 1/z = 1/7 then x/y = ??
519 views
If
1/x + 1/y = 1/3
1/x + 1/z = 1/5
1/y + 1/z = 1/7
then
x/y = ??
posted Jun 13, 2015
## 1 Solution
Let 1/x = a , 1/y = b , 1/z = c
The new equations are
a + b = 1/3 ----->1
a + c = 1/5 ------->2
b +c = 1/7 ------->3
By solving 2 and 3, a-b = 2/35 ---> 4
Solving for a , we get a= 41/210
Substitute a in equation 4 , we get value of b as 29/210
X/Y = 1/a * b = 29/210 * 210 /41 == 29/41
Ans : X/Y = 29/41
solution Jun 15, 2015
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http://lara.epfl.ch/w/sav08:summary_of_hoare_logic?do=diff&rev2%5B0%5D=1204710833&rev2%5B1%5D=&difftype=sidebyside | 1,563,916,131,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529664.96/warc/CC-MAIN-20190723193455-20190723215455-00452.warc.gz | 84,884,034 | 7,663 | • English only
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sav08:summary_of_hoare_logic [2008/03/05 10:53]vkuncak sav08:summary_of_hoare_logic [2015/04/21 17:30] (current) Both sides previous revision Previous revision 2008/03/05 10:55 vkuncak 2008/03/05 10:53 vkuncak 2008/03/04 22:31 vkuncak 2008/03/04 22:31 vkuncak 2008/03/04 22:10 vkuncak 2008/03/04 22:05 vkuncak 2008/03/04 21:40 vkuncak 2008/03/04 21:37 vkuncak 2008/03/04 21:15 vkuncak 2008/03/04 20:52 vkuncak 2008/03/04 20:49 vkuncak 2008/03/04 20:47 vkuncak 2008/03/04 20:43 vkuncak 2008/03/04 20:43 vkuncak created Next revision Previous revision 2008/03/05 10:55 vkuncak 2008/03/05 10:53 vkuncak 2008/03/04 22:31 vkuncak 2008/03/04 22:31 vkuncak 2008/03/04 22:10 vkuncak 2008/03/04 22:05 vkuncak 2008/03/04 21:40 vkuncak 2008/03/04 21:37 vkuncak 2008/03/04 21:15 vkuncak 2008/03/04 20:52 vkuncak 2008/03/04 20:49 vkuncak 2008/03/04 20:47 vkuncak 2008/03/04 20:43 vkuncak 2008/03/04 20:43 vkuncak created Line 8: Line 8: Strengthening precondition: Strengthening precondition: - $+ \begin{equation*} \frac{\models P_1 \rightarrow P_2 \ \ \ \{P_2\}c\{Q\}} \frac{\models P_1 \rightarrow P_2 \ \ \ \{P_2\}c\{Q\}} {\{P_1\}c\{Q\}} {\{P_1\}c\{Q\}} -$ + \end{equation*} Weakening postcondition: Weakening postcondition: - $+ \begin{equation*} \frac{\{P\}c\{Q_1\} \ \ \ \models Q_1 \rightarrow Q_2} \frac{\{P\}c\{Q_1\} \ \ \ \models Q_1 \rightarrow Q_2} {\{P\}c\{Q_2\}} {\{P\}c\{Q_2\}} -$ + \end{equation*} === Loop Free Blocks === === Loop Free Blocks === Line 23: Line 23: We can directly use rules we derived in [[lecture04]] for basic loop-free code. Either through weakest preconditions or strongest postconditions. We can directly use rules we derived in [[lecture04]] for basic loop-free code. Either through weakest preconditions or strongest postconditions. - $+ \begin{equation*} \{wp(c,Q)\} c \{Q\} \{wp(c,Q)\} c \{Q\} -$ + \end{equation*} or, or, - $+ \begin{equation*} \{P\} c \{sp(P,c)\} \{P\} c \{sp(P,c)\} -$ + \end{equation*} For example, we have: For example, we have: - $\begin{array}{l} + \begin{equation*}\begin{array}{l} \{Q[x:=e]\}\ (x=e)\ \{Q\} \\ \{Q[x:=e]\}\ (x=e)\ \{Q\} \\ \ \\ \ \\ Line 41: Line 41: \{P\}\ assume(F)\ \{P \land F\} \{P\}\ assume(F)\ \{P \land F\} \end{array} \end{array} -$ + \end{equation*} === Loops === === Loops === - $+ \begin{equation*} \frac{\{I\}c\{I\}} \frac{\{I\}c\{I\}} {\{I\}\ {\it l{}o{}o{}p}(c)\ \{I\}} {\{I\}\ {\it l{}o{}o{}p}(c)\ \{I\}} -$ + \end{equation*} === Sequential Composition === === Sequential Composition === - $+ \begin{equation*} \frac{\{P\}\, c_1 \{Q\} \ \ \ \{Q\}\, c_2\, \{R\}} \frac{\{P\}\, c_1 \{Q\} \ \ \ \{Q\}\, c_2\, \{R\}} {\{P\}\ c_1;c_2\ \{R\}} {\{P\}\ c_1;c_2\ \{R\}} -$ + \end{equation*} === Non-Deterministic Choice === === Non-Deterministic Choice === - $+ \begin{equation*} \frac{\{P\} c_1 \{Q\} \ \ \ \{P\} c_2 \{Q\}} \frac{\{P\} c_1 \{Q\} \ \ \ \{P\} c_2 \{Q\}} {\{P\} c_1 [] c_2 \{Q\}} {\{P\} c_1 [] c_2 \{Q\}} -$ + \end{equation*} ===== Applying Proof Rules given Invariants ===== ===== Applying Proof Rules given Invariants ===== Line 97: Line 97: we omit any assert statements in the middle, obtaining from c1,...,cK statements d1,...,dL. We call we omit any assert statements in the middle, obtaining from c1,...,cK statements d1,...,dL. We call - $+ \begin{equation*} \{P\} d1\ ;\ \ldots\ ;\ dL \{Q\} \{P\} d1\ ;\ \ldots\ ;\ dL \{Q\} -$ + \end{equation*} the Hoare triple of the assertion path. the Hoare triple of the assertion path. Line 118: Line 118: In a program of size $n$, what is the ++bound on the number of basic paths?| it can be $2^{O(n)}$++ In a program of size $n$, what is the ++bound on the number of basic paths?| it can be $2^{O(n)}$++ + + ++Choice of annotations?|If we put annotations after if statements, we avoid exponential explosion.++ + ===== Further reading ===== ===== Further reading ===== | 1,671 | 3,951 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-30 | latest | en | 0.451388 |
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# If y<>(not equal to) 3 and 3x/y is a prime integer
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If y<>(not equal to) 3 and 3x/y is a prime integer [#permalink]
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01 Oct 2003, 14:59
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If y<>(not equal to) 3 and 3x/y is a prime integer greater than 2
which of the following is true?
1. x=y
2.y=1
3.a and y are prime integers.
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03 Oct 2003, 04:59
consider 3x/y=7
(1) x=y can be true, but not necessarily
(2) y=1, can be true, but not necessarily
(3) never true
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03 Oct 2003, 08:39
stolyar wrote:
consider 3x/y=7
(1) x=y can be true, but not necessarily
(2) y=1, can be true, but not necessarily
(3) never true
How can 3x/y be a prime number if y = 1? Wouldn't it be equal to 3x, having two factors, thus a non prime number?
03 Oct 2003, 08:39
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http://curious.astro.cornell.edu/people-and-astronomy/careers-in-astronomy/138-physics/the-theory-of-relativity/general-questions/844-do-we-have-to-worry-about-relativity-when-studying-galactic-objects-intermediate | 1,670,142,548,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710968.29/warc/CC-MAIN-20221204072040-20221204102040-00095.warc.gz | 10,573,979 | 14,519 | ## Do we have to worry about relativity when studying galactic objects? (Intermediate)
I'm puzzled by the implication of the Einstein's theory about time being in a way a function of speed and/or acceleration. Doesn't it mean that time must be actually different on the bodies with different speed if this speed is big enough to notice it? Might be not noticeable on Mars compared to Earth, since the rotation of the Solar System probably compensates or balances somehow and their speed isn't big anyway. But what about time in the whole of the Solar System which rotates with a really huge speed around the centre of the Galaxy compared to time around Antares? What about time in our Local Group compared to the time in the M83, for instance, ot M31?
You ask a good question; it turns out that astronomers do have to worry about the time effects implied by Einstein's theories in certain circumstances.
As you mention, the time measured by two observers in two different reference frames depends on their relative speeds; in short, an observer that is observing an event in a reference frame moving at speed v will measure the time in that frame to move slower than in her frame by a factor equal to the square root of (1-(v/c)2), where c is the speed of light. So, we only need to worry about these effects when (v/c) squared is comparable to (but never greater than) 1. To be safe, let's assume that no correction is needed if v is 20% the speed of light.
Now, the speed of light is 300000 km/s, and so only objects with velocities of 60 000 km/s relative to the Earth need to take relativity into account. The Solar System does move quickly around the galactic centre by Earth standards, averaging a speed of 220 km/s, but this is nowhere near fast enough to require a correction for relativistic effects. So, the speed with which time elapses in the Solar System and, say, in Antares are very, very nearly the same. Nevertheless, astronomers do have to worry about these relativistic effects when studying so-called "high energy phenomena" in the Universe. A good example is the study of jets that are emitted in the vicinity of black holes, both at the centre of distant galaxies and in stellar systems in the solar neighbourhood. These jets are often observed to propagate through the interstellar medium at substantial fractions of the speed of light (some are even observed to move faster than c, but this is just an illusion caused by relativity). Estimates of the propagation speed and energetics of the jets may require substantial corrections for relativistic effects. Look here for a pretty picture of a jet in a nearby galaxy.
#### Kristine Spekkens
Kristine studies the dynamics of galaxies and what they can teach us about dark matter in the universe. She got her Ph.D from Cornell in August 2005, was a Jansky post-doctoral fellow at Rutgers University from 2005-2008, and is now a faculty member at the Royal Military College of Canada and at Queen's University.
Kristine's email:
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### Reinforcing 10 with a Ten Frame Game
Ten Frames are an important tool when it comes to building number sense, building a concept of what 10 is and helping students to explore strategies such as composing and decomposing of numbers. Having a solid understanding of ten and being able to visualize the number bonds that make up the number ten, is important for students. Ten frames are great for allowing students to visualize the numbers that make up ten and to explore those number bonds.
This 'Rolling 10 with Friends' activity, is a fun way to get your students 'hands on with the 10 frame.
How to Play:
1. Students roll a die. (The game has 2 dice that have the numbers from 1 - 9.) They record the number they roll on the recording sheet, and they make that number on the ten frame, using the 'Friends' number strips.
2. Students then use the student number strips to figure out how many more friends are needed to fill in the ten frame. Students could also count on, to figure out how many spaces are left on the ten frame first, before covering the remaining spaces with friends.
3. Students then write the number sentence on the recording sheet '3 + 7' = 10
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Anything modular synth related that is not format specific.
Moderators: lisa, luketeaford, Kent, Joe.
mcleinn
Learning to Wiggle
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### Linearise exponential CV?
Hello dear modular community,
my wife really likes to play the Doepfer Theremin patch I did today, but she says that the pitch scale is too
exponential (ie. the higher the pitch gets, the quicker it changes with the same amount of movement)
So what would be the easiest way to linearise the pitch CV before going into the VCO, so that changes in high
CV (as a result of the exponential Theremin controller) actually lead to less pitch change than changes in low CV,
resulting in a (more or less) linear function finger movement - pitch?
Maths 101 says I take that one can linearise an exponential function by applying a logarithmic function, but how to
do on my modular without a PC? Maybe using Maths or some custom Voltage Controlled Resistor circuit? Any
modules helping out in this (potentially allowing generic CV transformations)? I tried to google for VCAs and VCOs,
but it seems that a lot of times, when people say "logarithmic" they actually mean exponential, which makes
browsing for solutions somewhat difficult.
Thank you!
jorg
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https://www.tes.com/en-au/teaching-resources/hub/secondary/mathematics/advanced-statistics/hypothesis-testing | 1,532,259,966,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593208.44/warc/CC-MAIN-20180722100513-20180722120513-00448.warc.gz | 1,011,598,047 | 25,579 | #### Hypothesis Testing (Part 1)
Do you remember when you were a kid in science class, and you just wanted to blow something up already! But, your teacher wanted you write down all of your steps in accordance with the SCIENTIFIC METHOD? Yep, this is the statistics version of that…
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#### OCR MEI Statistics 2 Bivariate Data (Chapter 4)
This bundle brings together my resources for the bivariate data chapter/section of the S2 paper: Pearson's, Spearman's and Least Squares Regression Lines.
#### OCR MEI Statistics 2 : Bivariate Data
The introduction (Prove It) task asks students to re-cap hypothesis testing from Statistics 1 using the Binomial distribution, before introducing the Pearson Product Moment Correlation Coefficient and carrying out hypothesis tests to determine significance of correlation.
#### Hypothesis testing (new A level) - notes, examples, exercises. MCQs and a homework/test
This 17-page resource covers all the required knowledge and techniques for hypothesis testing in the AS part of the new A level. It contains detailed notes, examples to work through with your class, and exercises of questions for students to attempt themselves (answers included). The topics covered are: 1. Sampling - different methods of sampling, biased and representative samples 2. Unbiased estimators - estimating the population mean and variance from a sample 3. Setting up a hypothesis test - null and alternative hypotheses 4. Making a conclusion - p-values, significance levels, 1-tail and 2-tail tests 5. Critical regions - finding the critical region for a hypothesis test 6. Significance levels and errors - probability of incorrectly rejecting null hypothesis, nominal vs actual significance level This projectable and printable resource will save you having to write out or create any notes/examples when teaching this topic. It also increases how much you can get through in lessons as students don't have to copy notes/questions and can work directly onto spaces provided for solutions. You could also email/print some or all of this for students who have missed lessons or need additional notes/practice/revision. The second resource is a set of multiple-choice questions that can be used a quick assessment or as part of a revision/refresher lesson. Also included is a 2-page assessment that covers the whole topic. Fully worked solutions are included.
#### GCSE Statistics: Collecting Data
GCSE statistics presentations covering data types and data collection. to include - quantitative and qualitative data - continuous and discrete data - primary and secondary data - sampling (random and non random) - questionnaires and surveys - Investigations and experiments Very visual with problems embedded in presentation all images cc3.0 template slidescarnival
#### GCSE Statistics Collecting Data 6: Investigations and Experiments
Clear very visual presentation covering: 1. Good secondary data sources 2. Investigation/Experiment outlines 3. Hypothesis making 4. Identifying variables 5. Data collection methods to include a) before and after experiments b) matched pairs c) data logging d) control groups e) capture and recapture and problems for each of these. Images CC3.0 and presentation template from slidescarnival
#### GCSE Statistics Collecting Data 5: Primary and Secondary Data
Clear very visual presentation covering: 1. Primary and Secondary Data 2. Advantages and disadvantages of each 3. Examples of primary and secondary data 4. Methods of collecting data to include a) Questionnaires b) interviews .... and the advantages and disadvantages to each and problems for each of these. Images CC3.0 and presentation template from slidescarnival
#### Quantitative skills: Spearman's Rank and Chi Squared
This lesson is designed to meet requirements of the Geography AQA A-Level specification on quantitative and qualitative skills. As per my usual format this lesson consists of; - Starter activity to get them thinking about prior knowledge and using this to link to this lessons learning (the learning linked to this is found on slides 12-16 as I set this as for students to work on in study time/homework). - Learning objectives and success criteria that are clearly linked to the specification. - New information on Spearman's rank and chi squared with examples designed to work through as a class or in teams, with supporting worksheet at the very end of the PPT. - Plenary/application task - to consolidate learning from the lesson. - Further reading with hyperlinked websites. - Further study - hyperlink and consolidating exam style question on fieldwork techniques that would be used in coastal areas.
#### Statistics: Normal Distribution BUNDLE (4 Lessons) - Perfect for S1 and S2
This is four excellent lessons. All the lessons are approximately 25 slides and should be enough for a long double A-Level lesson (1hr 40 - 2hours) Lesson 1 - Standard Normal / Introduction Lesson 2 - Non Standard / Standardizing Lesson 3 - Backwards and Problems Lesson 4 - Approximation for Binomial Distribution NOTE: Feel free to browse my shop for more free and premium resources! As always please RATE and FEEDBACK (comments good or bad) thank you!
#### Statistics: (S1) Normal Distribution 3 - Backwards and Further Problems (+ worksheets)
This is the third of four lessons on the normal distribution, which builds on the first two lessons. This is your A/A* material, looking at different types of questions and problems. This is an excellent resource, with lots and lots of excellent teaching slides, 21 slides. Specifically, this lesson is for teaching how to work backwards and different questions. It does touch on linear combinations, so I recommend teaching that first (also available from my shop) and looks in detail at little issues/ mistakes that people can make. This is a vital lesson for your top end students and gives less able students more time to practise the normal distribution. I have therefore included two worksheets (exam questions) aimed at different abilities students and then come with accompanying mark schemes. NOTE: Feel free to browse my shop for more free and premium resources! As always please RATE and FEEDBACK, thank you!
#### S2 Hypothesis Testing Topic Mat
A mat to summarise key concepts from the S2 topic Hypothesis Testing (Edexcel). I also include a suggestion for how the mat could be filled out. I usually use it as a plenary – ask the pupils for the key ideas from the lesson and collectively fill in sections over a sequence of lessons. At the end of a topic they have a topic mat from which to revise from. I also photocopy them on different coloured paper so they act as dividers in students’ folders.
#### S3 Goodness of Fit and Contingency Tables Topic Mat
A mat to summarise key concepts from the S3 topic Goodness of Fit and Contingency Tables (Edexcel). This one is quite big so I put it onto A3 paper! I also include a suggestion for how the mat could be filled out. I usually use it as a plenary – ask the pupils for the key ideas from the lesson and collectively fill in sections over a sequence of lessons. At the end of a topic they have a topic mat from which to revise from. I also photocopy them on different coloured paper so they act as dividers in students’ folders.
#### S3 Estimation and Confidence Intervals Topic Mat
A mat to summarise key concepts from the S3 topic Estimation and Confidence Interval (Edexcel). This one is quite big so I put it onto A3 paper! I also include a suggestion for how the mat could be filled out. I usually use it as a plenary – ask the pupils for the key ideas from the lesson and collectively fill in sections over a sequence of lessons. At the end of a topic they have a topic mat from which to revise from. I also photocopy them on different coloured paper so they act as dividers in students’ folders. | 4,262 | 20,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-30 | latest | en | 0.923849 |
www.scalefour.org | 1,591,512,834,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348523564.99/warc/CC-MAIN-20200607044626-20200607074626-00558.warc.gz | 863,392,964 | 24,411 | ## Abstruse CSB Theory
Alan Turner
Posts: 559
Joined: Sun Jul 20, 2008 4:24 pm
### Re: Loco Suspension, fitting CSBs
Russ Elliott wrote:Just started reading your pdf, Alan:
The basic underlying assumption is that the wheel loads can be calculated before the calculation of
the deflections and that the wheel loads remain unaffected by the deflection.
No. The assumption is (like any spring) that wheel load will change according to the deflection.
But as the wheel load is an input into the spreadsheet how do you calculate it?
Alan
grovenor-2685
Posts: 3279
Joined: Sun Jun 29, 2008 8:02 pm
### Re: Loco Suspension, fitting CSBs
The naive assumption would be that each wheel carries 1/3 of
W. That cannot be correct. A more so realistic assumption would be that the outer wheels carry 1⁄4W
and the middle wheel carries 1/2W
Alan,
I have a problem with this, either of these results could be produced by appropriate positioning of the fulcrums, and we definitely want a result closer to the former than the latter, and preferably with the middle wheels carrying a bit less than 1/3. But the wheel load distribution is not something to be assumed but rather something that the calculations must account for.
The current programme although upside down, does allow the user to define the axleload distribution they want. What is missing is a means of relating these axleloads to the CofG position.
Changing from a definition of axleloads to one of loads carried by each fulcrum just makes this more difficult as there are more fulcrums to share the load and their positions are not available at the input to the problem but are actually the desired output.
The basic underlying assumption is that the wheel loads can be calculated before the calculation of
the deflections and that the wheel loads remain unaffected by the deflection.
Not necessarily calculated, but certainly specified, as the purpose of the calculation is to achieve the specified wheelload. That load will be the load at rest on level track, none of this addresses the dynamic situation.
The basic underlying assumption is that the wheel loads can be calculated before the calculation of
the deflections and that the wheel loads remain unaffected by the deflection.
How will you calculate the loads on the fulcrums in your version, seems to me to be the same problem?
Regards
Keith
Regards
Keith
Grovenor Sidings
Alan Turner
Posts: 559
Joined: Sun Jul 20, 2008 4:24 pm
### Re: Loco Suspension, fitting CSBs
Keith,
the underlying complexity with what we are trying to do is the fact that the load applied by each of the load points (fulcrum) is not independant of each other. This is because they are attached to a ridged beam (the chassis). In fact what is realy happening is that a fixed displacement is applied to the CSB by the fulcrum points. Now this is a very complex problem to solve (although any structural analysis programme will do it - if you have the odd £1k to spare) and it cannot be solved by naive assumptions about load distributions. My approach suffers from the same problem. the difference with mine is that PROVIDED you ensure equal displacements of the fulcrum points then I am suggesting that the relation to load distribution and displacement are a reasonable approximation of the actual displacement and load.
You can do it your way as well but you cannot dictate the wheel loads and again you have to ensure equal displacement (or lying on a straight line) of the fulcrum points.
Alan
grovenor-2685
Posts: 3279
Joined: Sun Jun 29, 2008 8:02 pm
### Re: Loco Suspension, fitting CSBs
You can do it your way as well but you cannot dictate the wheel loads and again you have to ensure equal displacement (or lying on a straight line) of the fulcrum points.
Alan,
I don't have a way I don't own the spreadsheet and had nothing to do with authoring it. But it does appear to give results that work when applied.
All I was trying to do with your paper was to try and understand it.
Surely the purpose of the spreadsheet is to assist with the location of the fulcrums, because of the complexity it does not try to calculate the positions for you but allows you by trial end error to see the effect of various locations. As I understand it the deflection is calculated from the wheelload you input, so you need to understand that the actual deflections will be equalised when the loco is on the track and consequently the wheelloads will not be as input. Super accuracy is not important, what is important is to be able to see when the wheelload on the centre axle is a little less than that on the ends so that the running will be stable. The spreadsheet illustrates this by the relative deflections but in the inverse sense so you aim for a larger deflection on the centre axle.
Perhaps it would be possible to turn this calculation around to use a fixed deflection and calculate the wheelload instead. However this alternative approach would need input of the position of the CofG would it not? Which is allowed for in the existing spreadsheet by the wheelload input.
Regards
Keith
Regards
Keith
Grovenor Sidings
Alan Turner
Posts: 559
Joined: Sun Jul 20, 2008 4:24 pm
### Re: Loco Suspension, fitting CSBs
Keith,
use of the collective you - bad habit of mine.
The basic problem comes down to this:
loads.png (24.9 KiB) Viewed 9859 times
They are two differnt problems.
Alan
grovenor-2685
Posts: 3279
Joined: Sun Jun 29, 2008 8:02 pm
### Re: Loco Suspension, fitting CSBs
They are two differnt problems.
Different, yes but related, are you saying that the answers given using the spreadsheet derived from the first problem will be significantly different from those that might be derived using the second? If so, by how much and does it matter?
Regards
Keith
Regards
Keith
Grovenor Sidings
Will L
Posts: 1770
Joined: Sun Jul 20, 2008 3:54 pm
### Re: Loco Suspension, fitting CSBs
Alan Turner wrote:I offered to set out some thoughts on CSBs. I do so here in the attached pdf file.
Alan
Sorry I haven't been ignoring you, my PC died and I've only just recovered it.
I'll have a look at you PDF tonight Alan but I have an area group meeting this afternoon.
A though for Keith/Rob. Given that this sort of technical discussion is enjoyed by some of us, but other find it very off putting, I'm wondering if we could split the thread down into the blow by blow posts, plus all the directly related posts, and a second "CSB Clinic" thread which could carry all the technical discussion and be clearly welcoming for people to ask their own questions
Alan Turner
Posts: 559
Joined: Sun Jul 20, 2008 4:24 pm
### Re: Loco Suspension, fitting CSBs
grovenor-2685 wrote:
If so, by how much and does it matter?
Does it matter? Within our context probably not because I think that any reasonably intuitive modeller could set up a CSB without recourse to the spreadsheet.
However if we are to mathematically model the CSB system then I think we should attempt to simulate it as best we can.
The problem I have with the present spreadsheet approach is its underlying assumption that the wheel loads are independent of each other and that they are an input. They are not independent of each other as they are connected by a ridged beam (the chassis) and they are an outcome not an input.
Having given this problem some thought over the week-end I believe a more appropriate model is this:
Beam dia.PNG (22.03 KiB) Viewed 9766 times
Here we represent the CSBs as springs acting on a ridged beam.
In order for the chassis to sit level then the centre of gravity of the weight must be coincident with the centre of action of the springs.
That is a relatively easy calculation.
The spring constants of the CSBs can be calculated by use of the standard equn's for beams as so:
Beam dia 2.PNG (4.32 KiB) Viewed 9766 times
and
Beam dia 3.PNG (4.53 KiB) Viewed 9766 times
The spring constants of both are easily calculated. I consider the encastra condition justified as the slope of the CSB will be horizontal at the fulcrum (due to the desire to have near equal defections at each wheel).
Hence once the springs are calculated it is a relatively easy matter to calculate the necessary spring rates to ensure coincident of the centre of action of the spring with the centre of action of the weight.
I will prepare a paper with the necessary equations and publish it later. I might even have a chance to write a spreadsheet that does the calculations.
Alan
Russ Elliott
Posts: 930
Joined: Thu Jun 02, 2011 6:38 pm
### Re: Abstruse CSB Theory
Alan Turner wrote:The problem I have with the present spreadsheet approach is its underlying assumption that the wheel loads are independent of each other and that they are an input. They are not independent of each other as they are connected by a ridged beam (the chassis) and they are an outcome not an input.
The wheelbase and wheel load inputs to the current spreadsheet are merely the means of establishing the longitudinal positions of the frame fulcrum points. They are not independent of each other nor are they assumed to be 'independent' of each other. The actual value(s) of the wheel loads are not particularly important, e.g. the spreadsheet will output the same frame fulcrum positions whether the wheel loads are input as 35g each or 40g each or 20g each. Wheel load inputs can be varied if desired, e.g. 35g (front wheel), 32g (middle wheel), 35g (rear wheel). Or alternatively, the frame fulcrum positions, after inputting an initial set of equal wheel load values, can be changed to effect a 32g middle axle, for example - the methodology is essentially the same.
What inputs would you envisage?
I consider the encastra condition justified as the slope of the CSB will be horizontal at the fulcrum (due to the desire to have near equal defections at each wheel).
No, no, no! Such a beam would be an entirely different type of CSB, where the end moments are fixed. We are not dealing with such a beast: it would be (beam diameter for diameter) about twice as long as what we have got currently! All one can assume, in the context of the type of CSB we are dealing with (i.e. M is zero at the outermost fulcrum points), is that the deflection of the beam will be zero at the frame fulcrum points. The slopes of the beam (the first differential of the curve) might not be exactly horizontal at the intermediate frame fulcrum positions, and they certainly won't be anywhere near horizontal at the outermost frame fulcrum positions.
It is true that the CSB can be replicated as a rigid thing ('the chassis') sitting on top of three springs (for the 3-axle case). The determination of the position of the CofG to implement chassis horizontality is easy, as you say. What I think is impossible to determine is the new set of spring forces if that CofG is intentionally moved, because the overall situation, albeit still in equilibrium, then becomes statically indeterminate, i.e. there isn't a unique solution to the resulting set of moment equations. An iterative (trig-based) reversal method could derive this, but I'm not sure of the mathematical methodology needed to address the general (non-trig) case. I'm currently consulting mathematicians on this matter, btw, but at the moment they seem to be a bit stumped!
It begs the question though as to what we are trying to achieve here. Are we simply saying that someone who does not attempt to balance a loco (within reasonable limits of course), or intentionally shifts the balance of a loco, will not achieve what was intended. Isn't that self-evident?
Does it matter? Within our context probably not because I think that any reasonably intuitive modeller could set up a CSB without recourse to the spreadsheet.
Been there 10 years ago, got the t-shirt, and had a bunch of rows about the disastrous 'guessing-game' approach. I felt strongly about properly reflecting the physics, Roger (likewise) did the spreadsheet gawd bless 'im, and the penny eventually dropped. But hey, if someone insists on having a CSB but also insists on not using a spreadsheet or the plots readily available, or insists on guessing everything because they don't know how it works or they aren't prepared to know how it works, that's their problem. You can hardly expect me to be sympathetic to that approach, can you?
Please note however I'm not saying there aren't different approaches (compared with the method used in the spreadsheet) to the solving of CSBs.
Alan Turner
Posts: 559
Joined: Sun Jul 20, 2008 4:24 pm
### Re: Abstruse CSB Theory
Russ Elliott wrote:No, no, no! Such a beam would be an entirely different type of CSB, where the end moments are fixed. We are not dealing with such a beast: it would be (beam diameter for diameter) about twice as long as what we have got currently! All one can assume, in the context of the type of CSB we are dealing with (i.e. M is zero at the outermost fulcrum points), is that the deflection of the beam will be zero at the frame fulcrum points. The slopes of the beam (the first differential of the curve) might not be exactly horizontal at the intermediate frame fulcrum positions, and they certainly won't be anywhere near horizontal at the outermost frame fulcrum positions.
The outer most fulcrum is a pin and therfore the moment is zero but if the slopes at intermediate fulcrums are horizontal then that is a condition of fixity. However, I agree it would be better to allow the moment to be free - it just brings an added complexity but I can see how it can be handled now after playing about with the equations. I am at a loss however to understand why it would result in a CSB twice as long. If anything I would have thought it should bring about a shorter or thiner wire.
However I recognise that my approach of having encastra ends was a simplification and it would be better to calculate the moments, as per the spreadsheet (by moment distribution) or directly by Clapeyron's equtions. However that cannot be done directly as the wheel loads are dependant of the spring stiffness supporting each wheel. There is therefore an iteration between spring stiffness (ie the fulcrum positions), fulcrum moments and wheel loads.
It may be easier to solve this problem by developing the stiffness matrix for the system and go that way. I shall however pursue my original approach a little longer before going down that road.
Nevertheless the problem I have with the present spreadsheet approach still remains, the wheel loads must be an outcome not an input.
The model applies its weight through the fulcrum points. If the centre of weight is coincident with the centre of stiffness of the springs bearing on the wheels then the fulcrum points will undergo an equal deflection and hence the CSB will bend up an equal amount above each wheel (assuming the the wheels are on straight track and the chassis does not itself bend - reasonable given the disparity in stiffness of the CSB and the chassis).
The load on each wheel will therefore be dependant on the stiffness of the spring above it. As that stiffness is a function of the geometry of the fulcrum points (and yes the EI of the wire) the wheel loads cannot be predetermined.
Alan
Will L
Posts: 1770
Joined: Sun Jul 20, 2008 3:54 pm
### Re: Abstruse CSB Theory
Personally I have enjoyed reading this immensely. I do have to say that quite a bit of it has gone over my head too, but I think, after a bit of a false start, we do seem to be getting somewhere.
Alan Turner wrote:...Nevertheless the problem I have with the present spreadsheet approach still remains, the wheel loads must be an outcome not an input.
While I have to agree with you, Alan, that one would instinctively like that to be true, I think it was the difficulty in achieving that result which gave rise to the upside down approach, which does allow weight as an input. What the upside down approach does do is to allow the interaction between adjacent sections of the CSB to be modelled and give a result as a deflection. As you yourself said:-
The load on each wheel will therefore be dependant on the stiffness of the spring above it. As that stiffness is a function of the geometry of the fulcrum points (and yes the EI of the wire) the wheel loads cannot be predetermined.
But the deflection can be, and if we calculates the deflection upside down with equal weights, the deflection calculated will be proportional to the weight carried when the right way up. Given the assumption of a CofG central over the wheelbase, and just for the moment that the centre axle is in the middle, it surely must be true that an axle deflected by the same amount on the spread sheet will carry the same weight. So if you can get the spreadsheet to show equal deflection on all wheels then the weight carried by each wheel will be equal. So not perhaps such a naive assumption.
Let us not forget, the function of the spread sheet is to allow us to find fulcrum solutions that produce the desirable equal deflection result. And this it does no matter what you may think of its structural analysis credentials, and I'm sorry but I don't think the fulcrum points it suggest are anything like as intuitive as you may think. The fact that there are many such solutions and that they all involve a minimum of 4 variables (the fulcrum points) suggests that numerical analysis may not provide any simple solutions. As it happens I do have a non spread sheet way of calculating results, but this involves yet more assumptions and a not very constant value somewhere between .56 and .6 which I have derived empirically. See this post.
So having got a perfectly good solution to a rather assumption bound case (both CofG and centre axle central), we need to consider real worlds systems where neither the CofG or the centre axle turn up in the middle.
As it turns out the off centre axle is never far enough off centre to cause too much trouble, particulay if you keep the centre axle slightly softer than the outer two. The slight variation in deflections at either end implies the chassis won't sit perfectly flat but the gradient front to back can be kept acceptably invisible in practice.
Which leaves us with the off centre CogG. I have practical experience which says that letting the CofG stray too far from the centre can produce unexpected results. Hence it would be nice to be able to find a way of dealing with this eventuality, but it will still need to be based on finding a fulcrum point solution which gives a level chassis, not in predicting the weight distribution once the fulcrum points are known.
Will
Alan Turner
Posts: 559
Joined: Sun Jul 20, 2008 4:24 pm
### Re: Abstruse CSB Theory
Will L wrote:
But the deflection can be, and if we calculates the deflection upside down with equal weights, the deflection calculated will be proportional to the weight carried when the right way up.
Ah but you can't do that because all wheel springs (i.e. the CSB above each wheel) is forced to deflect an equal amount (assuming coincidence of weight with centre of spring stiffness and rigidity of the chassis). The load carried by each wheel is therefore the product of this deflection and the stiffness of its spring.
In fact if you were to increase the deflection of say the centre wheel by putting a matchstick underneath it would attract more load and relive the outer wheels. The CSB is of course setup assuming the wheels are resting on level track.
Alan
Will L
Posts: 1770
Joined: Sun Jul 20, 2008 3:54 pm
### Re: Abstruse CSB Theory
Alan Turner wrote:Ah but you can't do that because all wheel springs (i.e. the CSB above each wheel) is forced to deflect an equal amount (assuming coincidence of weight with centre of spring stiffness and rigidity of the chassis). The load carried by each wheel is therefore the product of this deflection and the stiffness of its spring.
Ah but I have an ah but too. While I'm sure your right for any old set of fulcrum points with significantly differing deflections, what we are actually dealing with here would be a specific fulcrum set, devised so that the outer pair of wheels have a deflection under the same weight as close as identical as we can manage, and the inner wheel within a few percent the same. In these limited circumstances, which happen to be the ones we are operating in, I think you can say that the weight carried will be close enough to directly proportional to the calculated deflection as to make no practical odds.
Yes if you run over a match stick, the weight born by the various wheels will change significantly, meaning the one on the match stick will get more weight and depress more, while the ones on the track will receive less weight and depress less. But so long as the match stick is less than 1mm thick, they do all stay in contact with the track and isn't that just the sort of behaviour that we wanted?
The proportionality of deflection to weight carried is the nub of the argument I was trying my best to make last night. If you come to the conclusion I might have a point, you might want to re-consider if where I took it is valid as well.
Remember what we need here is a practical process which works well enough. Taking advantage of special cases is fine so long as we understand when the special case no longer applies. Which I think we have, so long as we can keep the CofG neatly in the middle of the chassis. Which I think is where I came in.
Will
Alan Turner
Posts: 559
Joined: Sun Jul 20, 2008 4:24 pm
### Re: Abstruse CSB Theory
Will L wrote: Which I think we have, so long as we can keep the CofG neatly in the middle of the chassis. Which I think is where I came in.
Will
but that is not necessary if the wheel loads are calculated as you can arrange for the resultant of the wheel loads to be coincident with the CG of the loco weight.
The calculation of the CSB is actually quite simple it is the determination of the loads being applied by the fulcrum points that is the difficulty. That is because they are not independent of each other but interdependent due to them all being connected by a ridged beam (the chassis).
Once the point loads are determined then the CSB calculations are:
You will note that for constant stiffness of CSB (EI) the wheel loads (and moments) are independant of CSB stiffness. Stiffness only affects resultant deflection.
If however you follow what I set out in my original "thoughts on CSBs" I propose a restriction which I believe overcomes the difficulty with determining the loads, namely you ensure that all fulcrum points have the same deflection. At that point the wheel load distribution is correct for level track.
what you then need to do is arrange things so that the centre of action of the wheel loads is coincident with the CG of the loco weight.
I am working on a spreadsheet, I will let you know how I get on.
Alan
Will L
Posts: 1770
Joined: Sun Jul 20, 2008 3:54 pm
### Re: Abstruse CSB Theory
Alan Turner wrote:but that is not necessary if the wheel loads are calculated as you can arrange for the resultant of the wheel loads to be coincident with the CG of the loco weight.
Yes I think that's what were after
... I propose a restriction which I believe overcomes the difficulty with determining the loads, namely you ensure that all fulcrum points have the same deflection. At that point the wheel load distribution is correct for level track.
Ok, but we have to be a little careful, to equalise the deflection exactly, fulcrum points need to be placed with considerable precision, while practical construction demands no greater accuracy than 0.5mm.
what you then need to do is arrange things so that the centre of action of the wheel loads is coincident with the CG of the loco weight.
I am working on a spreadsheet, I will let you know how I get on.
I look forward to it. I won't be able to comment further for a while, but I will be back.
Will
Russ Elliott
Posts: 930
Joined: Thu Jun 02, 2011 6:38 pm
### Re: Abstruse CSB Theory
Alan - on your diagram (and equations), I think you've got the direction of the MB moment the wrong way up.
Will L
Posts: 1770
Joined: Sun Jul 20, 2008 3:54 pm
### Re: Abstruse CSB Theory -In defence of the CSB Spread Sheet
Previously in this thread, the validity of designing CSB chassis using the spread sheet available from the CLAG web site, or else where on this forum from me, has been questioned. Better methods have been suggested but have not yet fully materialized.
In practice our use of the spread sheet has implied making assumptions, that is
1. assuming all wheels carry equal weight .
2. assuming that the loco Centre of Gravity (CofG) should be placed centrally over the CSBed wheelbase.
It has to be said that from a theoretical point of view these assumptions are not comfortable, particularly as the spread sheet comes at the problem from a counter intuitive direction. While practical experience suggests that these assumptions work, in as far as we get good working results, they do seem to artificially restrict ones options. By adopting the central CofG design approach are we failing to consider other, possibly better, solutions. After all it is perfectly practical to specify a CSB chassis which does have differently weighted wheels, yet sits level and runs happily. So what are you supposed to do with a loco where the CofG doesn’t naturally fall at the CSB midpoint? Then there is the whole thorny topic of load bearing carrying wheels.
In the CSBs and the Single Bogie thread, I have been thinking about how one would deal with chassis for which particularly the CofG location assumption does not appear to be helpful, e.g. chassis with load bearing carrying wheels. The result of all this thinking is that I feel I can now explain why the way we have been using the spreadsheet is fully justified.
So the purpose of this post is to show that an equal weight distribution on all axles is a perfectly reasonable aim for a CSB chassis, that to achieve it placing the CodG central on the chassis is the only way to go, and just why the spreadsheet is a practical tool for doing this. I can now also explain exactly what we are gaining from careful control of the CofG location and I have a method for dealing with chassis with more than just driving wheels, but as this post is already big enough, this additional stuff will come in a different post on another day.
Revision Needed
Also at this point I need to introduce you to, or remind you of, the “principle of moments”, which has a lot to say about they way a loco's weight is distributed across its wheels. I can do no better than point you at Scalefour Digest 41.0 “The principles of model locomotive suspension”, and particularly annex 2 on determining the location of a locos CofG. This is more of Russ’s work on the CLAG web site. It shows you how you can establish where a loco's CofG is, if you know the weights carried by each wheel.
From this you would have thought that, given a known position for the CofG, the same mathematics would allow you to establish how much weight is carried by each wheel/axle. Well, you can for a compensated chassis, see large tracts of digest 41.0, and you will note that it is not just CSB suspension that can appear mired down in complex mathematics! However springing muddies the maths rather, because the amount of weight carried by each wheel/axle now may also depend on the strength of the spring betwixt wheel and chassis (the spring rate). So for most CSB chassis you can’t calculate the weight carried by each wheel simply from knowing the CofG location.
Justifying the equal weight on each axle assumption – 2 axle CSB chassis
We will start with a 4 wheel/2axle chassis. This is actually the CSB trivial case. Assigning the fulcrum points does not require any fancy calculations. The centre fulcrum goes half way between the wheels and the outer two should be placed symmetrically (i.e. the same distance) either side. They shouldn’t be too close to the axle but otherwise you can put them more or less where you like. I do have a 2 axle version of that spread sheet, but the only useful thing you can do with it is to calculate what size of CSB wire you will need, given the weight carried and the position of the fulcrum points. You can probably manage very well without this, though it has its uses which I will be coming back to in due course, but not in this post.
This diagram was original used in the CSBs and the Single Bogie thread.
CSB C12 draw 3.jpg (79.63 KiB) Viewed 9366 times
In these simple circumstances the presence of springing makes no difference. if the CofG is central between the wheels, then the wheels will be equally loaded. You can use the principle of moments calculations to verify this. Conversely, if the CofG isn’t central then the wheels wont be equally loaded.
If the CofG is set centrally, and the fulcrums were not set symmetrically, i.e. making the spring rate different on each side, The chassis will not sit level but the wheels will be equally loaded (unless it sits so crookedly it materially affects the position of the CofG in which case…. Oh lets not go there, we want our chassis to sit level don't we?).
Thinking ahead, of course a 4 wheel chassis may well be a bogie, in which case the CofG is replaced by a load transmitted through the bogie pivot, but the principles don’t change.
Justifying the equal weight on each axle assumption – 3 axle CSB chassis
When we move on to a 6 wheel/ 3 axle chassis, things are definitely more complicated and the springs do matter. In these circumstances there are a range of possible weights that can be carried by each wheel that will satisfy the principle of moments calculations for any given CofG location. This gives an infinite number possible solutions to the question, how much weight is born by each wheel with the CofG just here. As in any circumstance where there is an infinity of possible answers, the next step has to be to limit some of the variables. So lets look at a symmetrical chassis and keep that CofG central.
CSB gravity 1.jpg (100.96 KiB) Viewed 9366 times
As the outer two CSB spans, and their relationship to the centre axle, are identical they will have the same spring rate. From that it should be clear that the outer two axles will carry equal weight. If this is not clear to you, consider what would happen if the spring rate over the central axle is reduced to zero. Now the centre axle would carry none of the weight and whole set up becomes equivalent to a 4 wheel/2 axle solution. In more general terms, how the weight is shared between the two outer axle and the inner one will depend on the relationship between their spring rates. A whole range of possible answers is available. Weights ranging from nothing to the total weight can be placed on w2. Any weight not placed on w2 will be shared between w1 and w3. All of these will give results that will satisfy the principle of moments calculation.
Now we are getting somewhere, because that the desired “equal weight on each wheel” result is in there among all the other possibilities, and the principle of moment calculation to demonstrate it is true is not hard to do. This calculation will also remind you that there is only one place the CofG can be for this to be true. That is dead central.
What the spreadsheet really does
At this stage we need to return to the spreadsheet. You will remember that this works by placing the chassis on its back, placing weights on each wheel and calculating the resulting deflection of the spring for each wheel. To help you remember, the following diagram shows the spreadsheet inputs and outputs, but if that doesn’t help then perhaps you might want to read the revision suggested earlier.
csb draw 07.jpg (39.55 KiB) Viewed 9296 times
The calculation the spread sheet performs is to determine the spring defection given the weights applied and a set of fixed fulcrum points. Most importantly it takes into account the interaction between adjacent wheels across the fixed fulcrums which are implicit in how CSBs work. Clearly when the calculated deflections vary considerably from wheel to wheel, they bear very little relationship to any real world situation when the chassis is right way up and standing on a (presumably) flat peace of rail. However these deflection results do have some validity in as far as they are directly proportional to the effective spring rate of the section of the CSB to which they apply, so equal defections imply that the effective spring rate is equal on each wheel.
If we have followed our assumption of applying equal weight to each wheel, and we have equal defections on each wheel, then the wheel rims will be as level as if they were the other way up and standing on a piece of track. Therefore, we have a chassis in the same state as if it was the right way up standing level on a flat peace of track, with the CofG dead central and a equal load on each wheel.
QED
It may be only one case out of many possibilities, but it is the one we want, its useful, and we don’t need any of the others.
You can get the same answers for 4 or more axles.
Deviations from perfection.
Yes I did assume that the 6 wheel chassis was symmetrical, and it is true that very few actually are. So what are the implications for asymmetrical chassis?
If we go back to the principle of moments, It is certainly possible to calculate a location for the CofG for three equally loaded axles which are not symmetrical, but perhaps not surprisingly this isn’t quite dead centre of the wheel base. Fortunately the maths works out relatively easily and gives a generalised result for both 3 and 4 axle cases (and probably 5, and 6 axle cases too but I haven’t chosen to go there). These results aren’t easy to explain in words but the following diagram should show how to calculate the answers clearly enough.
CSB gravity 2.jpg (137.98 KiB) Viewed 9366 times
If your inclined to doubt these results, it is easy enough to verify. Take a chassis and work out the CofG location assuming equal weight on each axle à la Digest 41.0. Then use the results above, you should get the same answer.
If you decide you want to use these results, I plan to reissue my version of the spread sheet shortly and that version will do the necessary sums for you. However I’m quite happy with the view that this may be overcomplicating things for the average CSB users, and going for the mid point of the chassis will be close enough in most cases.
Last edited by Will L on Thu Jun 30, 2011 4:44 pm, edited 1 time in total.
Alan Turner
Posts: 559
Joined: Sun Jul 20, 2008 4:24 pm
### Re: Abstruse CSB Theory
I really must get on and complete my spreadsheet as previously promised.
However I can't agree with your approach to the "equal weight" aspect of your explanation.
The fundamental problem is to think of weights as an input. They are an output because you are not in fact applying loads; you are applying deflections and the weight (or force more properly) that gets applied to the wheel axle is a result of that deflection i.e. F=dxK where K is the spring stiffness. In the case of flat track the deflections are equal.
csb%20draw%2007.jpg (55.88 KiB) Viewed 9333 times
Your weights need to be replaced by equal deflections (assuming flat track for the purposes of this argument). The weight then carried is calculated as above. The spring stiffness’s being derived from your spreadsheet or by taking a stiffness matrix approach (which is what I am trying to develop).
If you examine the diagram the spring stiffness of the outer two axles is by observation much less than the middle axle (if only because the outer two have pin ends). By applying equal deflection, which is what the flat track will do, the load carried by the middle axel will be much greater than the outer two. I would hazard a guess at a ratio of 1:2:1.
Regards
Alan
Will L
Posts: 1770
Joined: Sun Jul 20, 2008 3:54 pm
### Re: Abstruse CSB Theory
Alan Turner wrote:I really must get on and complete my spreadsheet as previously promised.
Alan Turner wrote:However I can't agree with your approach to the "equal weight" aspect of your explanation.
The fundamental problem is to think of weights as an input. They are an output because you are not in fact applying loads; you are applying deflections and the weight (or force more properly) that gets applied to the wheel axle is a result of that deflection i.e. F=dxK where K is the spring stiffness. In the case of flat track the deflections are equal.
I'm not pretending its a perfect model of any configuration of chassis Alan, a good enough proxy will do and this I think I have. Remember what we actually need from the spreadsheet is not the weights in or out, but the fixed fulcrum positions that will give us close to equal weight distribution. They aren't an output either. More than that, there isn't just one right answer, so the maths can't be expected to give us one. See this post a bit further back, and the diagram in the one that follows it. Using the weights as input and inspection to discover an acceptable set of fulcrum points that give us equal deflections as an output, we then have a recipe for a chassis design that performs as we require. What more would a more rigorous analysis give us ?
Alan Turner wrote:...If you examine the diagram the spring stiffness of the outer two axles is by observation much less than the middle axle (if only because the outer two have pin ends). By applying equal deflection, which is what the flat track will do, the load carried by the middle axel will be much greater than the outer two. I would hazard a guess at a ratio of 1:2:1.
Woops sorry, I used an old diagram which wasn't drawn with a equal weight outcome in mind or to scale for that matter. It just shows where the spreadsheet inputs and outputs fit against a generic chassis design . Your right, as drawn weight distribution would not accompany equal deflections. I'll correct the relative dimensions, but beyond being a bit misleading in the circumstance, I don't think this changes the underlying argument.
Anybody interested in spotting the difference should note that the original diagram has now changed, but the one in Alan's response has not.
Will
Alan Turner
Posts: 559
Joined: Sun Jul 20, 2008 4:24 pm
### Re: Abstruse CSB Theory
Well having threatened to write my version I have finally done it.
First of all let me acknowledge the assistance given to me by Will Litchfield who acted as both a guinea pig for the development versions and also in giving invaluable advice about the user interface.
What is attached is a beta development version and it is time limited (so if you are using it for real print off the results). With feed back I receive from now to Scaleforum I will endeavour to take it on board and then issue a version for use after that.
For this version macros have to be enabled.
This version (apart from details of the model) only needs to know the deflection that you want the CSB to have and what wire you are using.
You adjust the fulcrum points to achieve the axle load (note NOT wheel load) distribution you want and to ensure the Centre of Gravity of the model is coincident with the Centre of Action of the axle loads. This latter point is important; the results are not valid otherwise.
Regards
Alan
Attachments
CSB caculator - beta test v1.0.xls
Chris Mitton
Posts: 206
Joined: Sun Aug 31, 2008 1:18 pm
### Re: Abstruse CSB Theory
Hi Alan, thanks for this.....look forward to trying out the numbers for my engines.
But.....re the weight of the model, I assume that (a) you have to omit the weight of the wheels, axles, axleboxes and coupling rods since the springs don't carry them, and (b) you forgot to weigh them separately before you assembled the loco!
On the other hand, does this matter? What you're really interested in is the weight distribution, not the absolute weight.
Regards
Chris
Alan Turner
Posts: 559
Joined: Sun Jul 20, 2008 4:24 pm
### Re: Abstruse CSB Theory
Chris Mitton wrote:Hi Alan, thanks for this.....look forward to trying out the numbers for my engines.
But.....re the weight of the model, I assume that (a) you have to omit the weight of the wheels, axles, axleboxes and coupling rods since the springs don't carry them, and (b) you forgot to weigh them separately before you assembled the loco!
On the other hand, does this matter? What you're really interested in is the weight distribution, not the absolute weight.
Regards
Chris
You are strictly correct and the weight required is the sprung weight of the model.
However as you probably have some wheels, axles and hornblocks lying around why not weigh these and deduct their weight.
Regards
Alan
Russ Elliott
Posts: 930
Joined: Thu Jun 02, 2011 6:38 pm
### Re: Abstruse CSB Theory
Alan - I'm couldn't get your Excel file to fit into my modest Excel window screen. Is it possible for you to bring in the side margins a bit, particularly on the right-hand side? The big problem I have though is the way I can't get to the bottom of the file (in order to vertically resize the window).
Russ Elliott
Posts: 930
Joined: Thu Jun 02, 2011 6:38 pm
### Re: Abstruse CSB Theory -In defence of the CSB Spread Sheet
Btw, on Will's point about our spreadsheet assumption that "all wheels carry equal weight", this isn't necessarily true. The wheel/axle weight inputs can be made different on the spreadsheet if desired. Our 'slightly slackening off' the middle axle technique by means of allowing it some extra deflection could equally be achieved by giving it a lesser input weight - the resulting plot in either case will be identical. In explanation to Alan's initial theoretical objection, these 'input' values are what we are seeking as output values (forces) on the wheel. I liken it to hitting a certain key on a keyboard - it's both an 'input', and with a lot a technical wizardy inbetween, the desired character appears as an output.
P.S. Apologies for being incommunicado for so long. Effectively, I've been without a computer (at least, no Excel, no drawing software, no html editor, no reliable internet etc) for about 6 months.
Will L
Posts: 1770
Joined: Sun Jul 20, 2008 3:54 pm
### Re: Abstruse CSB Theory -In defence of the CSB Spread Sheet
Russ Elliott wrote:Btw, on Will's point about our spreadsheet assumption that "all wheels carry equal weight", this isn't necessarily true. The wheel/axle weight inputs can be made different on the spreadsheet if desired. Our 'slightly slackening off' the middle axle technique by means of allowing it some extra deflection could equally be achieved by giving it a lesser input weight - the resulting plot in either case will be identical. In explanation to Alan's initial theoretical objection, these 'input' values are what we are seeking as output values (forces) on the wheel. I liken it to hitting a certain key on a keyboard - it's both an 'input', and with a lot a technical wizardy inbetween, the desired character appears as an output.
P.S. Apologies for being incommunicado for so long. Effectively, I've been without a computer (at least, no Excel, no drawing software, no html editor, no reliable internet etc) for about 6 months.
Welcome back Russ I've been wondering where you were.
While it is indeed true that the original speed sheet does allow us to vary the weights, which could be used as you describe, trying to do anything more radical wasn't really on as it gave you no clue if the entered weight distribution was actuality reproducible for any possible CofG location, or what that CofG location would be.
The joy of Alan's spreadsheet is that it will allow you to specify a chassis with a CofG in any reasonable location, get it to sit level and and know what the weight distribution would be.
You'll be pleased to hear that both old and new spreadsheets give consistent answers for CofG's placed centrally to produce near equal axle loadings, and I did checked out a what I hope was a representative sample of the plots posted on the CLAG website to be sure this was true. Now, with Alan's spread sheet, if anybody badly wants a a loco with the CofG well away from the chassis centre, we can now do a plot for that too.
I also think that by hiding the works Alan has produced a tool which should be less intimidating for the non technical.
Will | 9,995 | 44,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-24 | longest | en | 0.949602 |
https://community.fabric.microsoft.com/t5/Desktop/Reiniciar-acumulado-por-determinado-numero-de-meses/td-p/3846580 | 1,718,590,492,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861674.39/warc/CC-MAIN-20240616233956-20240617023956-00054.warc.gz | 149,666,492 | 123,426 | cancel
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Helper III
Qué código le agrego a mi fórmula para que reinicie el contador, en este caso cada 12 meses.
Gracias por el apoyo. La fórmula de acumulado funciona correctamente pero quiero que reinicie en un determinado número de meses. para el ejemplo que presento en Marzo del 2021 debería volver a iniciar el acumulado.
Tengo tabla Calendario
VAR MaxFecha = MAXX(ALLSELECTED(Calendario),Calendario[Date])
VAR MinFecha = MINX(ALLSELECTED(Calendario),Calendario[Date])
CALCULATE([Ventas],
FILTER(ALLSELECTED(Calendario), Calendario[Date]<= MAX(Calendario[Date]) && Calendario[Date] >= MinFecha && Calendario[Date] <= MaxFecha))
RETURN
3 ACCEPTED SOLUTIONS
Community Support
Maybe you want something like this. We need to adjust the start date of each period to start a new running total. And if we change the start date, it will calculate the running total based on the new start date for the next 12 months.
Accumulative Total =
var firstStartDate = MINX(ALLSELECTED('calendar'),'calendar'[Date])
var curYear = MAX('calendar'[Year])
var startDate = DATE(IF(MAX('calendar'[Month])>=MONTH(firstStartDate),curYear,curYear-1), MONTH(firstStartDate), DAY(firstStartDate))
var endDate = EOMONTH(startDate,11)
RETURN
CALCULATE([Total],'calendar'[Date]>=startDate && 'calendar'[Date]<=endDate && 'calendar'[Date]<=MAX('calendar'[Date]))
Best Regards,
Jing
If this post helps, please Accept it as Solution to help other members find it. Appreciate your Kudos!
Helper III
Thank you so much.
Helper III
Muchas gracias, eres inteligente.
7 REPLIES 7
Helper III
Helper III
Muchas gracias, eres inteligente.
Helper III
Thank you so much.
Community Support
Do you have a fiscal year column in your calendar table? If not, you can first add a calendar year column to it like below.
DAX:
Fiscal Year = IF(MONTH([Date])<3, YEAR([Date])-1, YEAR([Date]))
Then add a filter on fiscal year to your current measure like below
VAR MaxFecha = MAXX(ALLSELECTED(Calendario),Calendario[Date])
VAR MinFecha = MINX(ALLSELECTED(Calendario),Calendario[Date])
CALCULATE([Ventas],
FILTER(ALLSELECTED(Calendario), Calendario[Date]<= MAX(Calendario[Date]) && Calendario[Date] >= MinFecha && Calendario[Date] <= MaxFecha && Calendario[Fiscal Year] = MAX(Calendario[Fiscal Year])))
RETURN
In addition, if the year-end month is not February, there is an easier method to calculate the running total with DATESYTD function as this function can define a different year-end date. However as in your case the year-end month is Febaruay, the year-end date is "2-29" in leap years and "2-28" in other years, it seems not work very well with DATESYTD. Computing running totals in DAX - SQLBI
Best Regards,
Jing
If this post helps, please Accept it as Solution to help other members find it. Appreciate your Kudos!
Helper III
No creo que necesite un anio fiscal, sólo necesito indicar cuántos meses después quiero reiniciar el acumulado a partir de una fecha inicial.
VAR MaxFecha = MAXX(ALLSELECTED(Calendario),Calendario[Date])
VAR MinFecha = MINX(ALLSELECTED(Calendario),Calendario[Date])
VAR ReiniciarAcumulado = EOMONTH(MinFecha,+11) // This is the number of months in which I want the monthly cumulative to reset, I don't know if the syntax to reset cumulative is correct
CALCULATE([Ventas],
FILTER(ALLSELECTED(Calendario), Calendario[Date]<= MAX(Calendario[Date]) && Calendario[Date] >= MinFecha && Calendario[Date] ) ) //How do I code that every 11 months I restart the accumulated in this period
RETURN
Helper III
Community Support
Maybe you want something like this. We need to adjust the start date of each period to start a new running total. And if we change the start date, it will calculate the running total based on the new start date for the next 12 months.
Accumulative Total =
var firstStartDate = MINX(ALLSELECTED('calendar'),'calendar'[Date])
var curYear = MAX('calendar'[Year])
var startDate = DATE(IF(MAX('calendar'[Month])>=MONTH(firstStartDate),curYear,curYear-1), MONTH(firstStartDate), DAY(firstStartDate))
var endDate = EOMONTH(startDate,11)
RETURN
CALCULATE([Total],'calendar'[Date]>=startDate && 'calendar'[Date]<=endDate && 'calendar'[Date]<=MAX('calendar'[Date]))
Best Regards,
Jing
If this post helps, please Accept it as Solution to help other members find it. Appreciate your Kudos!
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Top Kudoed Authors | 1,272 | 4,992 | {"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} | 2.890625 | 3 | CC-MAIN-2024-26 | latest | en | 0.307365 |
https://matplotlib.org/3.1.3/api/_as_gen/matplotlib.axes.Axes.pcolor.html | 1,623,669,503,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487612154.24/warc/CC-MAIN-20210614105241-20210614135241-00183.warc.gz | 338,756,662 | 8,281 | You are reading an old version of the documentation (v3.1.3). For the latest version see https://matplotlib.org/stable/api/_as_gen/matplotlib.axes.Axes.pcolor.html
# matplotlib.axes.Axes.pcolor¶
Axes.pcolor(self, *args, alpha=None, norm=None, cmap=None, vmin=None, vmax=None, data=None, **kwargs)[source]
Create a pseudocolor plot with a non-regular rectangular grid.
Call signature:
pcolor([X, Y,] C, **kwargs)
X and Y can be used to specify the corners of the quadrilaterals.
Hint
pcolor() can be very slow for large arrays. In most cases you should use the similar but much faster pcolormesh instead. See there for a discussion of the differences.
Parameters:
C : array_like
A scalar 2-D array. The values will be color-mapped.
X, Y : array_like, optional
The coordinates of the quadrilateral corners. The quadrilateral for C[i,j] has corners at:
(X[i+1, j], Y[i+1, j]) (X[i+1, j+1], Y[i+1, j+1])
+--------+
| C[i,j] |
+--------+
(X[i, j], Y[i, j]) (X[i, j+1], Y[i, j+1]),
Note that the column index corresponds to the x-coordinate, and the row index corresponds to y. For details, see the Notes section below.
The dimensions of X and Y should be one greater than those of C. Alternatively, X, Y and C may have equal dimensions, in which case the last row and column of C will be ignored.
If X and/or Y are 1-D arrays or column vectors they will be expanded as needed into the appropriate 2-D arrays, making a rectangular grid.
cmap : str or Colormap, optional
A Colormap instance or registered colormap name. The colormap maps the C values to colors. Defaults to rcParams["image.cmap"] (default: 'viridis').
norm : Normalize, optional
The Normalize instance scales the data values to the canonical colormap range [0, 1] for mapping to colors. By default, the data range is mapped to the colorbar range using linear scaling.
vmin, vmax : scalar, optional, default: None
The colorbar range. If None, suitable min/max values are automatically chosen by the Normalize instance (defaults to the respective min/max values of C in case of the default linear scaling).
edgecolors : {'none', None, 'face', color, color sequence}, optional
The color of the edges. Defaults to 'none'. Possible values:
• 'none' or '': No edge.
• None: rcParams["patch.edgecolor"] (default: 'black') will be used. Note that currently rcParams["patch.force_edgecolor"] (default: False) has to be True for this to work.
• 'face': Use the adjacent face color.
• An mpl color or sequence of colors will set the edge color.
The singular form edgecolor works as an alias.
alpha : scalar, optional, default: None
The alpha blending value of the face color, between 0 (transparent) and 1 (opaque). Note: The edgecolor is currently not affected by this.
snap : bool, optional, default: False
Whether to snap the mesh to pixel boundaries.
Returns:
collection : matplotlib.collections.Collection
Other Parameters:
antialiaseds : bool, optional, default: False
The default antialiaseds is False if the default edgecolors="none" is used. This eliminates artificial lines at patch boundaries, and works regardless of the value of alpha. If edgecolors is not "none", then the default antialiaseds is taken from rcParams["patch.antialiased"] (default: True), which defaults to True. Stroking the edges may be preferred if alpha is 1, but will cause artifacts otherwise.
**kwargs
Additionally, the following arguments are allowed. They are passed along to the PolyCollection constructor:
Property Description
agg_filter a filter function, which takes a (m, n, 3) float array and a dpi value, and returns a (m, n, 3) array
alpha float or None
animated bool
antialiased or aa or antialiaseds bool or sequence of bools
array ndarray
capstyle {'butt', 'round', 'projecting'}
clim a length 2 sequence of floats; may be overridden in methods that have vmin and vmax kwargs.
clip_box Bbox
clip_on bool
clip_path [(Path, Transform) | Patch | None]
cmap colormap or registered colormap name
color color or sequence of rgba tuples
contains callable
edgecolor or ec or edgecolors color or sequence of colors or 'face'
facecolor or facecolors or fc color or sequence of colors
figure Figure
gid str
hatch {'/', '\', '|', '-', '+', 'x', 'o', 'O', '.', '*'}
in_layout bool
joinstyle {'miter', 'round', 'bevel'}
label object
linestyle or dashes or linestyles or ls {'-', '--', '-.', ':', '', (offset, on-off-seq), ...}
linewidth or linewidths or lw float or sequence of floats
norm Normalize
offset_position {'screen', 'data'}
offsets float or sequence of floats
path_effects AbstractPathEffect
picker None or bool or float or callable
pickradius unknown
rasterized bool or None
sketch_params (scale: float, length: float, randomness: float)
snap bool or None
transform Transform
url str
urls List[str] or None
visible bool
zorder float
pcolormesh
for an explanation of the differences between pcolor and pcolormesh.
imshow
If X and Y are each equidistant, imshow can be a faster alternative.
Notes
X, Y and C may be masked arrays. If either C[i, j], or one of the vertices surrounding C[i,j] (X or Y at [i, j], [i+1, j], [i, j+1], [i+1, j+1]) is masked, nothing is plotted.
Grid orientation
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Objects passed as data must support item access (data[<arg>]) and membership test (<arg> in data). | 1,505 | 6,053 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-25 | longest | en | 0.666516 |
https://math.stackexchange.com/questions/1311228/what-is-the-most-unusual-proof-you-know-that-sqrt2-is-irrational/1311260 | 1,713,131,349,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.9/warc/CC-MAIN-20240414192536-20240414222536-00181.warc.gz | 363,354,600 | 60,998 | # What is the most unusual proof you know that $\sqrt{2}$ is irrational?
What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof: $3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of this result:
Theorem: If $n$ is a positive integer and there are positive integers $x$ and $y$ such that $x^2-ny^2 = 1$, then $\sqrt{n}$ is irrational.
The proof is in two parts, each of which has a one line proof.
### Part 1:
Lemma: If $x^2-ny^2 = 1$, then there are arbitrarily large integers $u$ and $v$ such that $u^2-nv^2 = 1$.
Proof of part 1:
Apply the identity $(x^2+ny^2)^2-n(2xy)^2 =(x^2-ny^2)^2$ as many times as needed.
### Part 2:
Lemma: If $x^2-ny^2 = 1$ and $\sqrt{n} = \frac{a}{b}$ then $x < b$.
Proof of part 2:
$1 = x^2-ny^2 = x^2-\frac{a^2}{b^2}y^2 = \frac{x^2b^2-y^2a^2}{b^2}$ or $b^2 = x^2b^2-y^2a^2 = (xb-ya)(xb+ya) \ge xb+ya > xb$ so $x < b$.
These two parts are contradictory, so $\sqrt{n}$ must be irrational.
First, this does not need Lagrange's theorem that for every non-square positive integer $n$ there are positive integers $x$ and $y$ such that $x^2-ny^2 = 1$.
Second, the key property of positive integers needed is that if $n > 0$ then $n \ge 1$.
• I wonder... If you could assume $\sqrt2$ rational and from there arrive at a solution to Fermat's last theorem. Jun 3, 2015 at 22:42
• Hmm...I'm afraid you'll have to elaborate on the $3^2-2\cdot2^2 = 1$ proof. I'm too obtuse to follow without some exposition. Jun 3, 2015 at 22:54
• @Arthur: Yes, because a false statement implies anything. Jun 3, 2015 at 22:58
• It does not make a lot of sense to ask «What is the most unusual proof you know?» and then pick one as an accepted answer... Jul 21, 2015 at 7:05
• It's.... Irrational Dec 21, 2017 at 5:44
Suppose that $\sqrt{2} = a/b$, with $a,b$ positive integers. Meaning $a = b\sqrt{2}$. Consider $$A = \{ m \in \Bbb Z \mid m > 0 \text{ and }m\sqrt{2} \in \Bbb Z \}.$$
Well, $A \neq \varnothing$, because $b \in A$. By the well-ordering principle, $A$ has a least element, $s$. And $s,s\sqrt{2} \in \Bbb Z_{>0}$. Then consider the integer: $$r= s\sqrt{2}-s.$$ We have $r =s(\sqrt{2}-1) < s$, and $r > 0$. But $r\sqrt{2} = 2s-s\sqrt{2}$ is again an integer. Hence $r \in A$ and $r < s$, contradiction.
• This is indeed a strange proof. Jun 3, 2015 at 22:49
• Isn't this just dressing up the usual proof with infinite descent? Jun 3, 2015 at 23:02
• @Asaf I'm not sure that it is the same as the usual one but I think it's pretty close philosophically. The usual uses the fundamental theorem of arithmetic but there's no mention of that here Jun 3, 2015 at 23:12
• @Matt The proof is very natural when viewed in proper light. It is simple Euclidean descent in the ideal of all possible denominators for the fraction $\,a/b = \sqrt{2}.\,$ See this answer for further discussion of such denominator and conductor ideals. Jun 3, 2015 at 23:42
• And this generalizes very nicely to all non-square integers. Jun 4, 2015 at 2:07
Here is something that I just came up with.
If $\sqrt2$ were rational, it would have been in every field of characteristics $0$.
It is also well-known that there are infinitely many prime numbers $p$, such that $x^2-2$ has no root over $\Bbb F_p$. Let $P$ be the set of these primes, and let $U$ be a free ultrafilter over $P$. Now consider $F=\prod_{p\in P}\Bbb F_p/U$.
Using Los theorem we have that:
1. $F$ is a field.
2. $F$ has characteristics $0$.
3. $\lnot\exists x(x^2-2=0)$.
This means exactly that we found a field which extends the rational numbers, but has no roots for $x^2-2$, which in other words means $\sqrt2\notin F$ and therefore $\sqrt2\notin\Bbb Q$.
• And there you have it friends, a proof using the axiom of choice! And in some models there are no free ultrafilters over a countable set, and in those models this method of proof fails (although this can be overcome by using all manners of absoluteness to repeat the proof in some inner model). Jun 3, 2015 at 22:57
• Very nice use of the ultrapower construction. There should be a contest for answering questions outside logic/model theory using Łoś's Theorem :) Jun 3, 2015 at 23:05
• Do you need this ultrafiler business? I thought you would say it would have to live in a p-adic field, but by Hensels lemma no solution over $F_p$ means no soultion in $Q_p$. Jun 3, 2015 at 23:06
• @Jonas: I think that Terry Tao has a blog post about connecting hard analysis with soft analysis using ultraproducts. Jun 3, 2015 at 23:09
• @Asvin: The ultraproduct proof is roughly equivalent to the alternative you suggest, just done in a simpler way that requires less fiddly attention to details (assuming, of course, you're used to ultraproducts). Another way to phrase it is in terms of nonstandard analysis; there is an infinite prime $P$ such that $\sqrt{2} \in \mathbb{F}_P$, but externally, $\mathbb{F}_P$ is a field of characteristic zero.
– user14972
Aug 16, 2015 at 8:18
Consider the linear application $A:\mathbb{R}^2\to \mathbb{R}^2$ given by $$A=\begin{pmatrix} -1&2 \\ 1&-1 \end{pmatrix} .$$ $A$ maps $\mathbb{Z}^2$ into itself and $V=\{y=\sqrt 2 x\}$ is an eigenspace relative to the eigenvalue $\sqrt 2-1$. But $A\mid_V$ is a contraction mapping, so $\mathbb{Z}^2\cap V=\emptyset$.
• Isn't this assuming the result in the last step? Why can't this contraction happen to map a $\mathbb{Z}^2$ to another $\mathbb{Z}^2$? Jun 9, 2015 at 2:58
• Suppose $p\in \mathbb{Z}^2\cap V$. There exists $k$ such that $A^k(p)\in B(0, 1/2)\setminus\{0\}$. Absurd. Jun 9, 2015 at 11:11
Ever so slightly off-topic, but I can't resist reminding folks of the proof that $\sqrt[n]{2}$ is irrational for $n \ge 3$ using Fermat's Last Theorem:
Suppose that $\sqrt[n]{2} = a/b$ for some positive integers $a$ and $b$. Then we have $2 = a^n / b^n$, or $b^n + b^n = a^n$. But Andrew Wiles has shown that there are no nonzero integers $a, b$ satisfying the last equation. Thus $\sqrt[n]{2}$ must be irrational. [This proof is due to W. H. Schultz and appeared in the May, 2003 issue of the American Mathematical Monthly.]
• The really funny part: FLT isn't powerful enough to prove the irrationality of $\sqrt 2$. Jun 15, 2015 at 8:00
• But are you sure that this is not a circular reasoning, namely the irrationality of $\sqrt[n]2$ is not used in the proof of Fermat's Last Theorem by Andrew Wiles? Without having thoroughly read the proof, I dare not say that ... Jul 1, 2021 at 21:11
• In fact the initial manipulations of the FLT proof, going from an FLT counterexample to a semistable modular elliptic curve which contradicts Ribet's theorem, freely uses the fact that if a^n+b^n=c^n is nontrivial then WLOG gcd(a,b,c)=1, so it's really just the same techniques as the proof that 2^{1/n} is irrational. Jul 3, 2021 at 8:31
• We can also prove $\sqrt 2$ is irrational similarly by FLT, e.g. see here Apr 28, 2022 at 14:00
• Thanks for the extension and references. Apr 28, 2022 at 19:44
$$\boxed{\text{If the boxed statement is true, then the square root of two is irrational.}}$$
Lemma. The boxed statement is true.
Proof. Assume for a contradiction that the boxed statement is false. Then it has the form "if $S$ then $T$" where $S$ is false, but a conditional with a false antecedent is true.
Theorem. The square root of two is irrational.
Proof.
1. The boxed statement is true. (By the Lemma.)
2. If the boxed statement is true, then the square root of two is irrational. (This is the boxed statement itself.)
3. The square root of two is irrational. (Modus ponens.)
• This has to be the funniest math joke I've ever read.
– user286485
Mar 26, 2018 at 12:11
Consider $\mathbb Z[\sqrt{2}]= \{a + b\sqrt 2 ; a,b \in \mathbb Z\}$. Take $\alpha = \sqrt 2 - 1 \in \mathbb Z [\sqrt 2]$, then $$0 < \alpha < 1 \implies \alpha ^k \to 0 \,\,\text{as} \,\, k \to \infty \tag {*}$$
Say $\sqrt 2 = \frac{p}{q}$, since $\mathbb Z[\sqrt 2]$ is closed under multiplication and addition we have
$$\alpha^k = e + f \sqrt 2 = \frac{eq + fp}{q} \geq \frac{1}{q}$$
which contradicts $(*)$.
• See here for a generalization, contrasting discretness of ring of integers vs. denseness of rings of fractions. Jun 3, 2015 at 23:55
• I'll definitely take a look at it. Jun 4, 2015 at 0:01
• @AaronMaroja could you please explain the last inequality? Can't $eq+fp$ be negative? If it can, and you meant absolute value, why can't $\lvert eq+fp \rvert$ be zero?
– vuur
Aug 10, 2015 at 11:32
• Notice that $\alpha > 0$ and $\sqrt 2 = \frac{p}{q}$. So $$e + f \sqrt 2 = e + f\frac{p}{q} = \frac{eq + fp}{q} \geq \frac{1}{q}> 0$$ Aug 10, 2015 at 13:11
I want to add the following figure. I might not call it unusual but perhaps not common enough to be widely available.
Update: I later checked the image from wikipedia article (linked in comments to the current question) and it presents the same proof in a bit complex fashion. There is no need for two arcs. Just note that if $m$ is hypotenuse of larger triangle and $n$ is one of the other sides then after this construction the length of side of smaller triangle is $(m - n)$ (this is the obvious part). The slightly hard part is to show that the hypotenuse of smaller triangle is $(2n - m)$ and for this wiki article draws two arcs.
However we can easily see that tangents drawn from external point to a circle are of equal length. Hence both the tangents are equal to side of smaller triangle i.e. $(m - n)$. Hence the hypotenuse of smaller triangle is $(n - (m - n)) = 2n - m$.
• Yep. That's the $\sqrt{2}-1$. Jun 14, 2015 at 5:36
• @martycohen: This is geometrical proof of the statement "If $\sqrt{2} = m/n$ then $\sqrt{2} = (2n - m)/(m - n)$" so that if $\sqrt{2}$ is a ratio then there is a ratio with smaller denominator which is also equal to $\sqrt{2}$. Jun 14, 2015 at 5:43
• Nice approach! G. H. Hardy uses this concept in his book "A Course of Pure Mathematics ". Jul 28, 2020 at 6:47
• @Learning: yes I have read in Hardy's Pure Mathematics also. Jul 28, 2020 at 11:31
I just thought of this one:
Consider the equation $x^2-n=0$ for natural $n$. Evidently, $\sqrt n$ is a solution to the equation. Now the rational root theorem implies that for a root to be rational for that equation, it must be a factor of $n$ (up to sign). If $\sqrt n$ is a factor of $n$ ($n$ is a perfect square), then $\sqrt n$ is rational. If $\sqrt n$ is not a factor of $n$ ($n$ is not a perfect square), then $\sqrt n$ must be irrational.
Now just plug in $n=2$.
Define the $2$-adic valuation $\nu_2(r)$ of a nonzero rational number $r = \frac{p}{q}$ to be the number of times $2$ divides $p$ minus the number of times $2$ divides $q$. The $2$-adic valuation of the square of a rational number is even. But the $2$-adic valuation of $2$ is odd. Hence $2$ is not the square of a rational number.
This argument generalizes with no difficulty to the following: if $n$ is a positive integer, then $\sqrt[k]{n}$ is rational iff the $p$-adic valuation $\nu_p(n)$ of $n$ is always divisible by $k$.
• +1, but I'm not sure I'd call this proof unusual; it's exactly the proof I'd give if someone asked me now to prove that $\sqrt{2}$ is irrational. Jun 14, 2015 at 6:13
Here is my favorite but perhaps a little too sophisticated. It make use of the theorem which says that if $K$ is an extension of some field $F$ then $[K:F]=1$ iff $K=F$.
From this fact consider the minimum polynomial of $\sqrt{2}$ over $\mathbb{Q}$ which is $m(x)=x^2-2$ by Eisenstein's Criterion. Thus, $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}] > 1$ and so by the contrapositive of the previous remarks $\mathbb{Q} (\sqrt{2}) \neq \mathbb{Q}$. That is, $\sqrt{2} \notin \mathbb{Q}$.
This uses the notion $[K:F]$ means the degree of the minimum polynomial of the extension(s) element(s).
• Related: Proof by Gauss lemma is one line. In fact, Gauss lemma is one way to prove the Eisenstein criterion. Jun 4, 2015 at 2:51
Below is a simple way to implement the (Euclidean) denominator descent implicit in Ivo Terek's answer (which was John Conway's favorite way to present this proof).
Theorem $$\quad \rm r = \sqrt{n}\:$$ is an $$\rm\color{#c00}{integer}$$ if rational,$$\:$$ for $$\:\rm n\in\mathbb{N}$$
Proof $$\ \ \$$ Put $$\ \ \displaystyle\rm r = \frac{A}B ,\;$$ least $$\rm\; B>0.\,$$ $$\ \displaystyle\rm\sqrt{n}\; = \frac{n}{\sqrt{n}} \ \Rightarrow\ \frac{A}B = \frac{nB}A.\$$ Taking fractional parts: $$\rm\displaystyle\ \frac{b}B = \frac{a}A\$$ for $$\rm\ 0 \le b < B.\$$ If $$\,\rm\displaystyle\ \color{#c00}{B\nmid A}\,$$ then $$\rm\ b\ne 0,\$$ so $$\rm\,\ \displaystyle \frac{A}B = \frac{a}b,\$$ contra leastness of $$\,\rm B$$.
Remark $$\$$ See here for a conceptual view of the proof (principality of denominator ideals). Though the proof may seem unusual to those who have not yet studied advanced number theory, it is quite natural once one learns about conductor and denominator ideals.
• What does it mean for a number to be integral? Jun 14, 2015 at 18:25
• @YoTengoUnLCD Here integral means "an integer", Jun 14, 2015 at 18:35
If $\sqrt{2}$ were rational it would have to live in every field of characteristic zero. There exist primes $p$ where $x^2-2$ has no roots. By Hensel's lemma, a polynomial with simple roots $f\in \mathbb{Q}_p[x]$ has a root in $\mathbb{Q}_p$ if and only if its reduction has a root in $\mathbb{F}_p$. For any $p$ such that $x^2-2$ has no roots, it then follows that $\sqrt{2}\notin \mathbb{Q}_p$, but the latter is a field of characteristic zero, contradiction.
• For concreteness, $p=5$ works. Also $p=11$. Jun 3, 2015 at 23:13
Tennenbaum gave a geometric proof of the irrationality of $\sqrt{2}$:
The large square has side length $a$, while the light purple/blue square has side length $b$, with $a, b$ positive integers such that $({a\over b})^2=2$. But then it's easy to see that the blue square has twice the area of the pink square - that is, $({2b-a\over a-b})^2=2$. Since the numerator and denominator are each positive integers integers, and are less than $a$ and $b$ respectively, we have an infinite descent.
• A nice variation. Oct 12, 2016 at 1:37
• For me, this proof is entirely useless. Not only I'm color blind, I have problems thinking about math in terms of geometry... :-P Dec 30, 2016 at 20:34
• @AsafKaragila That's fair. I definitely find it less easily understood than the usual one, but I do think it's an interesting different take. Dec 30, 2016 at 21:28
• This is the proof I posted a year earlier, only my skills didn't extend to including a picture. May 9, 2018 at 5:44
• @GerryMyerson Sorry, I just saw this comment. You're absolutely right, I missed yours when I was checking to see if this had already been posted. Would you like me to delete this answer (I can also add the picture to yours)? Jul 14, 2019 at 23:28
The irrationality of $$\sqrt{2}$$ can be deduced from the following
Theorem (Fermat, 1640): The number $$1$$ is not congruent.
Reasoning: If $$\sqrt{2}$$ were rational then $$\sqrt{2},\sqrt{2}$$,and $$2$$ would be the sides of a rational right triangle with area $$1$$. This is a contradiction of $$1$$ not being a congruent number.
A positive rational number $$n$$ is called a congruent number if there is a rational right triangle with area $$n$$: there are rational $$a,b,c>0$$ such that $$a^2+b^2=c^2\qquad\text{ and }\qquad\frac{1}{2}ab=n$$
A proof of this theorem based upon Fermat's method of descent is given in The congruent number problem Theorem 2.1 by Keith Conrad.
• First, I think you mean $\sqrt{2}$ and $1$, not $\sqrt{2}$ and $2$. Second, the proof about congruent numbers (which shows there are no solutions to $a^2+b^2 = c^4$) is harder than the proofs that $\sqrt{2}$ is irrational. Jun 20, 2015 at 19:33
• @martycohen: I'm referring to a right triangle with legs $\sqrt{2}$ and hypotenuse $2$. So, we get $(\sqrt{2})^2+(\sqrt{2})^2=2^2$ and $\frac{1}{2}\sqrt{2}\sqrt{2}=1$. Maybe you would also like to check the paper by Keith Conrad which I'm referring too. But I totally agree with your second argument! :-) Jun 20, 2015 at 19:45
• You are right. I read too fast. Jun 21, 2015 at 15:49
The final decimal digit of $$a^2$$ and the final decimal digit of $$2b^2$$ can't agree unless $$a$$ and $$b$$ are both multiples of five, leading to an infinite descent. (Check: the possible last digits of $$a^2$$ are 0,1,4,5,6,9 and the possible last digits of $$2b^2$$ are 0,2,8.)
• I've recently been told that this proof is actually quite common in French textbooks. Jul 6, 2021 at 13:06
Here is my favorite one. Suppose for the sake of contradiction that $\sqrt{2} = \frac{a}b$ for integers $a,b$. Then $2b^2 = a^2$. Let $p$ be an odd prime that is not congruent to $\pm 1 \pmod 8$. Then by quadratic reciprocity, $$\left( \frac{2b^2}p \right) = \left( \frac{2}p \right) = -1 \ne \left( \frac{a^2}p \right) = 1.$$
Suppose $\sqrt2$ is arational, then there exist $p,q$ two natural numbers such that $$\color{Red}{p^2=q^2+q^2.}$$ Then by the parametric solution of Pythagoras Equation there exist two natural numbers $a,b$ such that $a\gt b\ge 1$ and $p=a^2+b^2,$ $$\color{Green}{q=a^2-b^2=2ab.}$$
Now, if $r=a+b$ and $s=a,$ then $$\color{Red}{r^2=s^2+s^2}$$ with $r\lt p,\ s\lt q.$ Hence, by the Infinite Descent there are no such $p$ and $q$ natural numbers.
Using the rational root theorem on $x^2 - 2 = 0$ is a very simple and elegant way of proving the irrationality of $\sqrt{2}$. Peersonally, I like it beacuse it can be explained easily to high school students
This proof will look much better if someone can add a diagram to it.
Suppose $\sqrt2=a/b$ with $a$ and $b$ positive integers chosen as small as possible. Draw a square of side $a$. In the upper left corner, place a square of side $b$, and another one in the lower right corner. The two $b$-squares have total area $2b^2=a^2$, the same as the area of the $a$-square. Thus, the overlap of the two $b$-squares, which is a square we'll call a $c$-square, must have area equal to that of the two corners of the $a$-square not covered by the $b$-squares. Those corners are squares, call them $d$-squares. Then $c^2=2d^2$, so $\sqrt2=c/d$, where $c$ and $d$ are integers (indeed, $c=2b-a$, $d=a-b$) and are less than $a$ and $b$. Contradiction!
You can see this, with diagram, here. It also appears, with many other proofs, here. | 5,789 | 18,329 | {"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": 32, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-18 | latest | en | 0.86877 |
https://www.topperlearning.com/forums/home-work-help-19/find-the-height-above-the-surface-of-the-earth-where-acelera-physics-gravitation-25038/reply | 1,513,045,322,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948514250.21/warc/CC-MAIN-20171212021458-20171212041458-00320.warc.gz | 845,684,088 | 41,280 | Question
Sun August 24, 2008 By: Suvriti Bali
Find the height above the surface of the earth where aceleration due to gravity is half,one-third
Tue August 26, 2008
use the formula g' =g (1-d / R)
where g' is the gravity at depth d,
g,is the acceleration due to gravity on surface of earth,R is the radius of the earth,
d,is depth ,
put the value g' in terms of g and find the answer.
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a piece of iron hasa dimension 3cm x1.5cm x6cm.. if its mass is 205.2gms,its density is
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https://math.stackexchange.com/questions/131198/subset-of-mathbbq | 1,560,806,912,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998580.10/warc/CC-MAIN-20190617203228-20190617225228-00441.warc.gz | 539,467,720 | 35,680 | # Subset of $\mathbb{Q}$
Let $S= \{x_0,\dots,x_n\}$ be a finite subset of $[0,1]$ , $x_0=0$ and $x_1=1$ such that every distance between pair of elements of $S$ occurs at least twice, except for the distance $1$, then we are to show that $S$ is a subset of $\mathbb{Q}$.
Let $V$ be the $\mathbb{Q}$-vector space generated by $S$ and let $\leq$ be any total order on $V$ compatible with addition (forall $x,y,z \in V$, if $x \leq y$ then $x+z \leq y+z$, and more generally, any reasoning you are used to involving $+$ and $\leq$ is valid).
Then, for this particular total order, there is a unique pair $(x_i,x_j) \in S^2$ such that $x_i - x_j$ is the greatest distance between pair of elements of $S$ :
Since $S$ is finite and $\leq$ is total, the set $\{x-y, (x,y)\in S^2\}$ is finite so it has a maximal element $x_i - x_j$ for some pair $(x_i,x_j) \in S^2$.
Then we show it is unique : suppose there is $(x_k,x_l) \in V^2$ such that $(x_i - x_j) = (x_k - x_l)$.
Then $(x_i - x_j) + (x_i - x_j) = (x_k - x_l) + (x_i - x_j) = (x_k - x_j) + (x_i - x_l)$. Since $(x_i - x_j)$ is maximal, $(x_k - x_j) \leq (x_i - x_j)$ and $(x_i - x_l) \leq (x_i - x_j)$. If any of those two inequalities were strict, we would get the contradiction $(x_i - x_j) + (x_i - x_j) < (x_i - x_j) + (x_i - x_j)$, so they have to be equalities, which implies that $(x_k,x_l) = (x_i,x_j)$.
Now, the hypothesis on $S$ says that the only pairs $(x_i,x_j)$ such that $x_i-x_j$ is unique are the pairs $(0,1)$ and $(1,0)$. So it implies that for any total order on $V$ compatible with addition, this greatest distance is always $1$ or $-1$.
So we only need to show that if $V \neq \mathbb{Q}$, there must exist total orders $\leq$ on $V$ such that the greatest distance for $\leq$ is not $1$ or $-1$ :
Since $S$ generates $V$, and $(x_1 = 1)$ is free, we can add elements of $S$ to $(x_1)$ to form a basis $(e_1, \ldots, e_m) = (x_{i_1}, \ldots, x_{i_{m-1}},x_1)$ of $V$.
• Pick the lexicographical order induced by this basis. It is caracterised by the property that $0 < \sum a_i e_i$ if and only if the first nonzero coefficient is positive.
In particular, $(x_{i_1} - x_0) = x_{i_1} > \pm x_1 = \pm (x_1 - x_0) = \pm 1$, so neither $1$ nor $-1$ can be the greatest distance for $\leq$.
• Define a linear application $f : V \to \mathbb{R}$ with $f(1) = 1$ and $f(x_{i_k}) = \pi^k$. Since $\pi$ is transcendental, this map $f$ is injective. Pick the order defined by $x \leq y \Leftrightarrow f(x) \leq_\mathbb{R} f(y)$, where $\leq_\mathbb{R}$ is the usual order on $\mathbb{R}$. But then, since $f(x_{i_1}) >_\mathbb{R} f(x_1)$, we have again $(x_{i_1} - x_0) = x_{i_1} > \pm x_1 = \pm (x_1 - x_0) = \pm 1$.
• first of all In the problem it is given that the points are in [0,1] so the greatest no $(x_i-x_j)$ is the unique no 1, there is nothing to prove in it ,secondly I do not understand why in lexicographic order a number which is not a rational number then why that will not be in [0,1]? – Marso Apr 15 '12 at 19:13
• Over all I have not understood properly. – Marso Apr 15 '12 at 20:18 | 1,113 | 3,074 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-26 | latest | en | 0.824401 |
https://quantumai.google/reference/python/cirq/CZPowGate?hl=nl | 1,669,502,125,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446709929.63/warc/CC-MAIN-20221126212945-20221127002945-00158.warc.gz | 525,855,558 | 40,132 | # cirq.CZPowGate
A gate that applies a phase to the |11⟩ state of two qubits.
### Used in the notebooks
Used in the tutorials
The unitary matrix of CZPowGate(exponent=t) is:
$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & e^{i \pi t} \\ \end{bmatrix}$
cirq.CZ, the controlled Z gate, is an instance of this gate at exponent=1.
exponent The t in gate**t. Determines how much the eigenvalues of the gate are phased by. For example, eigenvectors phased by -1 when gate**1 is applied will gain a relative phase of e^{i pi exponent} when gate**exponent is applied (relative to eigenvectors unaffected by gate**1).
global_shift Offsets the eigenvalues of the gate at exponent=1. In effect, this controls a global phase factor on the gate's unitary matrix. The factor is:
exp(i * pi * global_shift * exponent)
For example, cirq.X**t uses a global_shift of 0 but cirq.rx(t) uses a global_shift of -0.5, which is why cirq.unitary(cirq.rx(pi)) equals -iX instead of X.
ValueError If the supplied exponent is a complex number with an imaginary component.
exponent
global_shift
## Methods
### controlled
View source
Returns a controlled CZPowGate, using a CCZPowGate where possible.
The controlled method of the Gate class, of which this class is a child, returns a ControlledGate. This method overrides this behavior to return a CCZPowGate or a ControlledGate of a CCZPowGate, when this is possible.
The conditions for the override to occur are:
• The global_shift of the CZPowGate is 0.
• The control_values and control_qid_shape are compatible with the CCZPowGate:
• The last value of control_qid_shape is a qubit.
• The last value of control_values corresponds to the control being satisfied if that last qubit is 1 and not satisfied if the last qubit is 0.
If these conditions are met, then the returned object is a CCZPowGate or, in the case that there is more than one controlled qudit, a ControlledGate with the Gate being a CCZPowGate. In the latter case the ControlledGate is controlled by one less qudit than specified in control_values and control_qid_shape (since one of these, the last qubit, is used as the control for the CCZPowGate).
If the above conditions are not met, a ControlledGate of this gate will be returned.
Args
num_controls Total number of control qubits.
control_values Which control computational basis state to apply the sub gate. A sequence of length num_controls where each entry is an integer (or set of integers) corresponding to the computational basis state (or set of possible values) where that control is enabled. When all controls are enabled, the sub gate is applied. If unspecified, control values default to 1.
control_qid_shape The qid shape of the controls. A tuple of the expected dimension of each control qid. Defaults to (2,) * num_controls. Specify this argument when using qudits.
Returns
A cirq.ControlledGate (or cirq.CCZPowGate if possible) representing self controlled by the given control values and qubits.
### num_qubits
View source
The number of qubits this gate acts on.
### on
View source
Returns an application of this gate to the given qubits.
Args
*qubits The collection of qubits to potentially apply the gate to.
Returns: a cirq.Operation which is this gate applied to the given qubits.
### on_each
View source
Returns a list of operations applying the gate to all targets.
Args
*targets The qubits to apply this gate to. For single-qubit gates this can be provided as varargs or a combination of nested iterables. For multi-qubit gates this must be provided as an Iterable[Sequence[Qid]], where each sequence has num_qubits qubits.
Returns
Operations applying this gate to the target qubits.
Raises
ValueError If targets are not instances of Qid or Iterable[Qid]. If the gate qubit number is incompatible.
TypeError If a single target is supplied and it is not iterable.
### qubit_index_to_equivalence_group_key
View source
Returns a key that differs between non-interchangeable qubits.
### validate_args
View source
Checks if this gate can be applied to the given qubits.
By default checks that:
• inputs are of type Qid
• len(qubits) == num_qubits()
• qubit_i.dimension == qid_shape[i] for all qubits
Child classes can override. The child implementation should call super().validate_args(qubits) then do custom checks.
Args
qubits The sequence of qubits to potentially apply the gate to.
Raises
ValueError The gate can't be applied to the qubits.
### with_probability
View source
Creates a probabalistic channel with this gate.
Args
probability floating point value between 0 and 1, giving the probability this gate is applied.
Returns
cirq.RandomGateChannel that applies self with probability probability and the identity with probability 1-p.
### wrap_in_linear_combination
View source
Returns a LinearCombinationOfGates with this gate.
Args
coefficient number coefficient to use in the resulting cirq.LinearCombinationOfGates object.
Returns
cirq.LinearCombinationOfGates containing self with a coefficient of coefficient.
View source
### __call__
View source
Call self as a function.
View source
View source
View source
View source
View source
View source
View source
### __truediv__
View source
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[{ "type": "thumb-up", "id": "easyToUnderstand", "label":"Easy to understand" },{ "type": "thumb-up", "id": "solvedMyProblem", "label":"Solved my problem" },{ "type": "thumb-up", "id": "otherUp", "label":"Other" }] | 1,452 | 5,900 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-49 | latest | en | 0.786859 |
https://measuringly.com/how-heavy-is-5-pounds/ | 1,712,967,862,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816465.91/warc/CC-MAIN-20240412225756-20240413015756-00696.warc.gz | 363,950,911 | 81,359 | # How Heavy is 5 Pounds? (With 14 Examples) +Pics
Are you curious about how heavy 5 pounds feels like?
Or do you want to know what item weighs approximately 5 pounds?
Or perhaps you want to understand the heaviness of 5 pounds in other units, especially if you come from countries that use the metric system of measurements?
Then you are in the right place.
Being familiar with weight measurements can make everyday activities more efficient and informed.
And five pounds is no exception.
Knowing how heavy it is can be valuable in many practical situations. It helps with tasks like cooking, packing, shopping, fitness, and more where a weight of 5 pounds is involved.
So, in this post, I’ll walk you through 14 ways to help you understand how heavy it is for whatever purpose you want to.
I’ve used generally everyday items to make it relatable to everyone.
Read: 8 Common Things That Weigh 6 Pounds (Examples Inside)
## 14 Familiar References For 5 Pounds
Five pounds is about 2,268 grams, 80 ounces, or 2.268 kilograms.
That’s relatively light in everyday life, as most people can easily lift and carry this weight without much effort.
You can compare it to the following items to understand it better.
### 1. A Standard Brick
Bricks are an excellent way to understand weights, from relatively light, such as two kilograms, to heavy, such as 80 kilograms. And so we can use it to understand five pounds since it belongs to the former category.
But before using them as references, always check the standard size in your country since they vary from country to country.
In this case, we refer to the standard brick in the United States, which is usually about 4.5 pounds.
It’s not precisely five pounds heavy, but it gives you a good perspective of that weight.
### 2. 2 Liters of Water
You can divide water into different quantities to suit whatever weight you want to determine. And the best part is it is universally available. That is why it’s one of the most commonly used references for lightweight measurements, such as one ounce, and heavy ones, such as 100 kilograms.
Two liters of water weigh 4.4 pounds, and when you factor in the weight of the container, it goes up to approximately five pounds.
### 3. 2 & 1/2 Pineapples
Did you know that the largest pineapple ever recorded weighed an astonishing 18.3 lb. (8.28kg)?
But that’s too much for a pineapple, as most in farmers’ markets, grocery stores, and supermarkets weigh about two pounds.
That means if you can get pineapples around, you can compare five pounds to two and a half.
### 4. 5 Loaves of Bread
Five loaves of bread may evoke the Biblical story where Jesus miraculously multiplied five loaves of bread and two fish to feed a large crowd of people who had gathered to hear him speak.
But let’s not go into that.
A standard loaf of bread weighs approximately one pound in many countries, including the United States, and is a reliable way to understand lightweight measurements.
For five pounds, you can compare it to five breads, and you can’t go wrong.
### 5. 2 & 1/2 Quarts of Milk
A quart of milk contains 32 fluid ounces or one-fourth of a gallon.
It’s a substantial amount and is commonly used for cooking or baking when you need a significant amount of milk.
Now, imagine something twice and half as heavy to get an impression of five pounds.
### 6. 7 Cups of Honey
Do you have some honey at home?
The sweetener is popular in many households, used as a natural alternative to sugar in various recipes, as a topping for foods like pancakes and biscuits, and as a remedy for sore throats and allergies.
If you use it, it can be a sweet way to determine how heavy five pounds is.
A cup of honey weighs 0.74 pounds, meaning all you need are seven cups.
### 7. 10 Cups of Butter
Butter is another household item you can rely on to understand and determine various lightweight weights, such as two ounces and two pounds.
It’s a staple in many households and is used in many recipes and for spreading on foods like bread and pancakes. And similarly to water, you can divide it into different quantities (tablespoons, cups, etc.).
Since five pounds is relatively heavy, going with cups is the most realistic option.
A cup of butter is precisely eight ounces (approximately 0.5 pounds), meaning you need as many as ten to realize a weight of five pounds.
### 8. 5 Size 5 Soccer Balls
Suitable for players aged 12 and older, Size 5 soccer balls are the standard size used in adult and professional soccer matches.
They typically have a circumference of 27-28 inches (68-70 cm) and a weight of 410-450 grams, approximately one pound.
So, if you can recall the weight of the last soccer ball you held, whether playing or passing it on to a friend, five pounds is five times heavier.
### 9. 5 American Footballs
American football is another sport-related way to understand how heavy five pounds is. The balls weigh approximately as much as the Size 5 soccer balls, so you need only five.
American football is, alongside soccer, among the most popular sports in the United States, played even in streets, so many people are familiar with the balls.
### 10. A Ream of Letter-sized Papers
Letter-sized paper is the standard size for many office and personal printing needs, and it is the default paper size in many regions, including North America. All printing documents, letters, and other materials usually use this paper.
These papers come in a ream containing 500 sheets and weighing precisely five pounds.
### 11. 6 Standard Bowling Pins
Bowling might not be as popular as soccer or American football.
However, if you are a fan or player of bowling, bowling pins are a reliable way to determine lightweight weights, starting with three pounds
They weigh just as much according to the governing body, United States Bowling Congress standards (USBC).
Thus, two regulation pins can give you an idea of how heavy five pounds is as they exceed the mark by only one kilogram.
### 12. Half a Gallon of Paint
Half a gallon of paint is typically equivalent to 64 fluid ounces or 1.89 liters. It’s a popular size for home improvement projects, like painting a small room or touch-ups around the house.
A gallon weighs approximately eleven pounds, depending on the type of paint and its specific formulation, meaning half is an excellent reference point for five pounds.
### 13. A Pair of Steel-toe Boots
Steel-toe boots are designed to protect the toes and feet in industrial and construction settings.
To make them up to their tasks, they typically have a reinforced steel cap in the toe area to protect against heavy objects, impacts, and compression.
And that makes them heavier than regular shoes, weighing approximately five pounds. | 1,485 | 6,767 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-18 | latest | en | 0.951322 |
https://www.ask-math.com/math-puzzle-7.html | 1,680,107,473,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949009.11/warc/CC-MAIN-20230329151629-20230329181629-00453.warc.gz | 704,212,540 | 12,833 | # Math Puzzle 7
This math puzzle 7 is on largest number,fraction,consecutive numbers etc.
1) Three numbers are in the ratio of 3:4:5 respectively. If the sum of the first and third numbers is more than the second number by 52, then which will be the largest number?
a)65 b)52 c)79 d)63 e)None of these
2) In New York, the fare for a taxi ride is paid according to a meter. However, since the meters are of old manufacture and are not up to date on current prices, the passenger is required to pay one and half times what the meter reads. In addition, after 10pm there is a night charge, which is calculated on the adjusted total (i.e. the meter + half). If Sam and Alley want to go to a late night costume party by taxi, what fraction of the meter reading will they have to pay in total?
3) Glass A contains 100ml of water and glass B contains 100ml of wine. A 10 ml spoonful of wine is taken from glass B and mixed thoroughly with water in glass A. A 10 ml spoonful of the mixture from A is returned to B. Is there now more wine in water or more water in wine?
4)Tariff of Hotel Shilton are:- 33/60 of the rooms have showers
12/60 of the rooms have single bed
24/60 of the rooms have cable tv
18/48 of the spaces are under cover
Convert the fractions into decimals.
5) There are three consecutive numbers a, b and c. The differences between the squares of these numbers are two consecutive prime numbers. If b2 is 5 more than a2 and a = 2 ,find b and c.
6) Alley invested a sum of money in shares. After three years she received $12100, which was the original sum. Sam heard about this and decided to invest half of what Alley had invested. How much did Sam get after 3 years? 7) Alley and Sam had a fight, so Sam buried Alley’s coin collection in his grandfather’s compound. He then drew a map so that he could find it when he wanted to. The coins were buried in an earthen pot to the south of the pond. It was at a distance of the third power of three-fourth the distance from the centre of the pond to the palm to the east. The map is given below. Find the distance from the centre of the pond where the treasure is buried. 8) Alley has given Sam a problem: The fourth power of a number is equal to the square of another number. If the square is 256 find the two numbers. Sam started off enthusiastically but then got stuck. Can you help him find the numbers? (Hint : If the two numbers are x and y. x4 = y2 = 256.) 9) It’s Alley’s lucky day: she has just found a huge diamond lying on the street! She knows that a 10g diamond is worth$9300, and that the value of diamonds varies directly with the square of their weight. If her newly found diamond weighs 18g, how much money can she sell it for?(math puzzle 7>
10) A farmer has a large orchard where he grows apple trees. He usually sells his apples for \$14.50 per kg. this year, however, a violent hailstorm ruins 3/8 of his crop. How much will he have to sell apples for if he wishes to make the same profit?
Math puzzle 7
From Math puzzle 7 to Math Teasers
Home | 761 | 3,046 | {"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} | 3.625 | 4 | CC-MAIN-2023-14 | latest | en | 0.974035 |
http://www.urbandictionary.com/define.php?term=pythagoras%20theorem | 1,501,002,233,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425339.22/warc/CC-MAIN-20170725162520-20170725182520-00100.warc.gz | 580,892,685 | 22,313 | Top definition
Using Pythagoras Theorem, the third side of a right-angled triangle can be calculated when two sides are given.
Suppose A = length of hypotenuse and
B & C = lengths of the sides containing the right angle
Then (A^2) = (B^2)+(C^2)
Proof:
If a = angle opposite side A ( =90 degrees)
b = angle opposite side B
c = angle opposite side C
then B = A sin a and C = A cos a
Squaring and adding,we get the result.
Pythagorean triplets:
3,4,5
5,12,13
8,15,17
by Jai Shri Ram May 20, 2005
### The Urban Dictionary Mug
One side has the word, one side has the definition. Microwave and dishwasher safe. Lotsa space for your liquids.
2
Simply put, in a triangle, the square of the hypoteneuse is equal to the sum of the squares of the other two sides. Simple as that.
the pythagorean theorem, in a simple mathematical formula, is: a² = b² + c²
where a is the hypoteneuse and b & c are the other two sides
pythagoras theorem
by DannoMack April 12, 2006
### The Urban Dictionary Mug
One side has the word, one side has the definition. Microwave and dishwasher safe. Lotsa space for your liquids.
3
A fucking useless maths thing that u will never need to use.
i can't believe we have a test on pythagoras theorem. its not like we will ever use it later on in life
by Christopher Mckay February 22, 2006
### The Urban Dictionary Mug
One side has the word, one side has the definition. Microwave and dishwasher safe. Lotsa space for your liquids. | 398 | 1,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | latest | en | 0.899776 |
http://mathhelpforum.com/math-topics/181714-another-projectile-questions-involves-distance-print.html | 1,529,316,416,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860168.62/warc/CC-MAIN-20180618090026-20180618110026-00549.warc.gz | 200,606,595 | 2,984 | # Another projectile questions involves distance
• May 26th 2011, 08:27 AM
silvercats
Another projectile questions involves distance
Object is projected from a speed of V and an angle of Θ
Horizontal displacement is 15m ,vertical displacement is 5m
Find V and Tan Θ
(right )
x=V cos Θ*t
t=x/V cos Θ
(up)
y=v Sin Θ*t-g/2*t²
from first equation y=V sinΘ(x/V cosΘ)-g/2*(x/v cosΘ)²
5=V sin Θ(15/V cos Θ)-g/2 *(15/V cosΘ)² <<<Stuck!!!!
help wanted
Books says 12.5 , 4/3
• May 26th 2011, 09:20 AM
abhishekkgp
Quote:
Originally Posted by silvercats
Object is projected from a speed of V and an angle of Θ
Horizontal displacement is 15m ,vertical displacement is 5m
Find V and Tan Θ
(right )
x=V cos Θ*t
t=x/V cos Θ
(up)
y=v Sin Θ*t-g/2*t²
instead of using 't' you should have used 't/2' in the above eqn. do you see why?
from first equation y=V sinΘ(x/V cosΘ)-g/2*(x/v cosΘ)²
5=V sin Θ(15/V cos Θ)-g/2 *(15/V cosΘ)² <<<Stuck!!!!
help wanted
Books says 12.5 , 4/3
did this help?
• May 26th 2011, 09:32 AM
silvercats
Quote:
Originally Posted by abhishekkgp
instead of using 't' you should have used 't/2' in the above eqn. do you see why?
why?????
• May 26th 2011, 11:11 AM
abhishekkgp
Quote:
Originally Posted by silvercats
why?????
't' is the total time of flight, that is the time taken by the projectile to hit the ground. it takes half the time of flight to achieve maximum height(here max height is 5m). now do you understand what i was trying to say?
• May 26th 2011, 06:27 PM
silvercats
ah.. yeah.can you guide me through the rest?
thanks | 518 | 1,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-26 | latest | en | 0.876704 |
http://vzombies.cf/forum9001-what-is-the-difference-between-dewpoint-and-relative-humidity.html | 1,527,321,422,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867374.98/warc/CC-MAIN-20180526073410-20180526093410-00137.warc.gz | 328,198,664 | 6,144 | what is the difference between dewpoint and relative humidity
# what is the difference between dewpoint and relative humidity
The difference between air temperature and dew point indicates whether the relative humidity is low or high.Record this value in Data Table 2. 7. Using the Dewpoint table on page 12 in your ESRT, determine the Dewpoint and record it in Data Table 2. Part III: Relative Humidity Calculation: Go to While the relative humidity has increased between 4pm and 10pm, the air outside your house is actually less humid because the dewpoint has dropped.An example of how dewpoint and relative humidity values changed in Cincinnati on August 12, 2013. Relative humidity is the amount of moisture in the air compared to what the air can hold at that temperature, Shaffer said.In fact, the difference is so great that clouds (which contain droplets of moisture as well as water vapor) readily stay in the air. Explaining Dewpoint and Relative Humidity to the Public is specifically found atThe explanation then goes on to tell us that dew point is described in degrees of temperature, while relative humidity is given as a percent. 1. Subtract Air Temperature and Wet Bulb for each time of day. 2. The difference will be the Wet Bulb Depression.Dew Point (C). Relative Humidity ().
CONCLUSION QUESTIONS. 1. What is the relationship between air temperature and moisture Many people confuse relative humidity with dew point. They often hear about dew point on the weather channel but arent really sure what it means. Lets take a closer look at these two concepts. What is the difference between Dew Point and Humidity.Dew point expresses the temperature at which a relative humidity of 100 would be reached and dew would begin to form. Humidity vs Relative Humidity The Earths atmosphere is composed of gases which are held together by gravity. It protects the Earth and all living things therein from solar radiation. It consists of What is the ideal relative humidity that is the most comfortable for Ratios - presenting one figure relative to another Relative Utility Modelling.A comfortable humidity level for most people is between 30 and 60 percent. | 444 | 2,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2018-22 | latest | en | 0.936833 |
https://www.mm.bme.hu/~gyebro/files/ans_help_v182/ans_cmd/Hlp_C_EORIENT.html | 1,652,951,480,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662526009.35/warc/CC-MAIN-20220519074217-20220519104217-00581.warc.gz | 1,053,514,794 | 3,080 | EORIENT
EORIENT, `Etype`, `Dir`, `TOLER`
Reorients solid element normals.
Compatible Products: – | Pro | Premium | Enterprise | Ent PP | Ent Solver | DYNA
`Etype`
Specifies which elements to orient.
LYSL — Specifies that certain solid elements (such as SOLID185 with KEYOPT(3) = 1, SOLID186 with KEYOPT(3) = 1, and SOLSH190) will be oriented. This value is the default.
`Dir`
The axis and direction for orientation, or an element number. If `Dir` is set to a positive number (`n`), then all eligible elements are oriented as similarly as possible to element `n`.
NEGX — The element face with the outward normal most nearly parallel to the element coordinate system’s negative x-axis is designated (reoriented) as face 1. POSX — The element face with the outward normal most nearly parallel to the element coordinate system’s positive x-axis is designated (reoriented) as face 1. NEGY — The element face with the outward normal most nearly parallel to the element coordinate system’s negative y-axis is designated (reoriented) as face 1. . POSY — The element face with the outward normal most nearly parallel to the element coordinate system’s positive y-axis is designated (reoriented) as face 1. NEGZ — (Default) The element face with the outward normal most nearly parallel to the element coordinate system’s negative z-axis is designated (reoriented) as face 1. POSZ — The element face with the outward normal most nearly parallel to the element coordinate system’s positive z-axis is designated (reoriented) as face 1.
`TOLER`
The maximum angle (in degrees) between the outward normal face and the target axis. Default is 90.0. Lower `TOLER` values will reduce the number of faces that are considered as the basis of element reorientation.
## Notes
EORIENT renumbers the element faces, designating the face most parallel to the XY plane of the element coordinate system (set with ESYS) as face 1 (nodes I-J-K-L, parallel to the layers in layered elements). It calculates the outward normal of each face and changes the node designation of the elements so the face with a normal most nearly parallel with and in the same general direction as the target axis becomes face 1.
The target axis, defined by `Dir`, is either the negative or positive indicated axis or the outward normal of face 1 of that element.
All SOLID185 Layered Structural Solid, SOLID186 Layered Structural Solid, and SOLSH190 solid shell elements in the selected set are considered for reorientation.
After reorienting elements, you should always display and graphically review results using the /ESHAPE command. When plotting models with many or symmetric layers, it may be useful to temporarily reduce the number of layers to two, with one layer being much thicker than the other.
You cannot use EORIENT to change the normal direction of any element that has a body or surface load. We recommend that you apply all of your loads only after ensuring that the element normal directions are acceptable.
Prisms and tetrahedrals are also supported, within the current limitations of the SOLID185, SOLID186, and SOLSH190 elements. (Layers parallel to the four-node face of the prism are not supported.) | 707 | 3,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-21 | latest | en | 0.862245 |
https://mapleprimes.com/questions/81049-Finding-The-Integration-Limit-Of-A-Very | 1,713,746,932,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818067.32/warc/CC-MAIN-20240421225303-20240422015303-00519.warc.gz | 341,870,392 | 24,091 | # Question:Finding the integration limit of a very hard integral
## Question:Finding the integration limit of a very hard integral
Maple
Hello,
I have defined an expression f. This expression f, when integrated from 18 to b, is equal to 20. My goal is to find this b value. Since the integral is very hard to solve symbolically. This is what I did:
with(Student[Calculus1]);
ApproximateInt(f, t = 18 .. b, method = simpson);
by trying some random numbers v, I have found the approximate value of b.
I want to write something that tries values automatically and stops when the absolute value of
ApproximateInt(f, t = 18 .. b, method = simpson) minus 20 is equal to, let's say 0.5
How would I do that? I know nothing about programmation
| 186 | 745 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-18 | latest | en | 0.891746 |
https://retrocomputing.stackexchange.com/questions/1709/atari-2600-high-voltage-output | 1,660,231,624,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571472.69/warc/CC-MAIN-20220811133823-20220811163823-00223.warc.gz | 448,488,156 | 65,209 | # Atari 2600 high voltage output
I've never worked on a Atari 2600 before. The first time I did, I wasn't getting anything on my old box TV. Not being able to do much at the time, I decided to check the voltage. The output voltage is 9v DC, but I was getting 15v DC. Yesterday I got my hands on another Atari 2600, so I decided to check the voltage again... it's also 15v DC. Is this the actual voltage, or do both power supplies have a short (or something)?
The Atari 2600 has an unregulated transformer. Transformers of this type only produce their design voltage under load; if you're just connecting a voltmeter between the output wires, it's going to read significantly higher than the output rating.
• Ah, so everything should be fine. I'm just worried about frying the bored. Nov 12, 2016 at 2:49
• The original power supply is also rather inefficient and prone to failure; there are modern switching replacements available, such as this one (in Europe). Nov 12, 2016 at 16:03
• I can confirm that 15 volts was what I would see on my original Atari 2600 power supply when unconnected to the system Dec 12, 2019 at 17:08
The design of the Atari 2600, and many other devices that use "9-volt DC" wall bricks, doesn't really care about receiving exactly 9 volts, subject to a few constraints:
1. The voltage must be far enough about 5 volts to allow a cheap regulator circuit to convert it into 5 volts.
2. There is an upper limit which will cause essentially immediate damage--typically somewhere between 15-35 volts.
3. Heat dissipation in the regulator will be proportional to product of load current times the amount by which the input exceeds 5 volts. A 13-volt supply will produce twice as much heat in the regulator as a 9-volt supply, but under conditions of light loading that won't be an issue.
The Atari 2600 draws a moderate amount of current itself, but cartridges draw additional current from the same supply, so total current draw may depend upon what cartridge is being used. On the other hand, a typical unregulated "9 volt" supply will only produce voltages significantly above 9 volts under conditions of light loading when heat dissipation isn't likely to pose a problem. | 515 | 2,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-33 | longest | en | 0.95634 |
http://www.numbersaplenty.com/227367888 | 1,524,159,461,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937015.7/warc/CC-MAIN-20180419165443-20180419185443-00104.warc.gz | 468,664,718 | 4,620 | Search a number
227367888 = 243112931479
BaseRepresentation
bin11011000110101…
…01101111010000
3120211211111002220
431203111233100
5431201233023
634321141040
75430411066
oct1543255720
9524744086
10227367888
11107385950
126418a780
133814a217
14222a8036
1514e633e3
hexd8d5bd0
227367888 has 160 divisors (see below), whose sum is σ = 685670400. Its totient is φ = 64243200.
The previous prime is 227367883. The next prime is 227367893. The reversal of 227367888 is 888763722.
It is an interprime number because it is at equal distance from previous prime (227367883) and next prime (227367893).
227367888 is an astonishing number since 227367888 = 7888 + ... + 22736.
It is a self number, because there is not a number n which added to its sum of digits gives 227367888.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (227367883) by changing a digit.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 474433 + ... + 474911.
It is an arithmetic number, because the mean of its divisors is an integer number (4285440).
Almost surely, 2227367888 is an apocalyptic number.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 227367888, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (342835200).
227367888 is an abundant number, since it is smaller than the sum of its proper divisors (458302512).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
227367888 is a wasteful number, since it uses less digits than its factorization.
227367888 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 561 (or 555 counting only the distinct ones).
The product of its digits is 1806336, while the sum is 51.
The square root of 227367888 is about 15078.7230228557. The cubic root of 227367888 is about 610.3463839395.
The spelling of 227367888 in words is "two hundred twenty-seven million, three hundred sixty-seven thousand, eight hundred eighty-eight". | 606 | 2,161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2018-17 | latest | en | 0.878784 |
http://www.perlmonks.org/index.pl?node_id=508728 | 1,516,479,113,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889681.68/warc/CC-MAIN-20180120182041-20180120202041-00528.warc.gz | 517,127,249 | 6,140 | No such thing as a small change PerlMonks
### Re: Algorithm to fit photos into spaces on pages
on Nov 15, 2005 at 18:52 UTC ( #508728=note: print w/replies, xml ) Need Help??
in reply to Algorithm to fit photos into spaces on pages
I'm still not sure about an efficient (polynomial-time) algorithm for your original question (minimizingmaximizing the number of distinct layouts used in representing a sequence of photos), but I can address your postscript:
P.S. This does beg the questions of how can one determine if a selection of layouts could deal with every possible selection of photos,
I'm in a formal-languages frame of mind right now, and this question is easily answerable using regular languages & automata. If your layouts are, say, {pll,lp,pp}, then the possible photo sequences you can represent are exactly those sequences which match /^(pll|lp|pp)*\$/.
If you want to know whether this allows you to get all possible photo sequences, then you are asking whether /^(pll|lp|pp)*\$/ is equivalent to /^[pl]*\$/. This is called the "regex universality" problem (does a given regex match all strings?), and it is PSPACE-complete in the general case. But for reasonable sized examples, you can check this by building a DFA from the regex, complementing it, and checking to see if it accepts any strings. I even happen to know of an upcoming Perl project that would make these regular expression & automata operations very easy (which happens to be why I'm in the formal-language frame of mind to begin with) {nudge nudge wink wink}.
and similarly, what is the smallest possible selection of layouts that are similarly "complete"? I don't need to know the answer, but it is intriguing!
Well, the smallest is {p,l} of course ;) To make the problem more interesting, if you are given a set of layouts (say, {pll,lp,pp} again), can you make this set smaller without losing expressivity? What you can do is calculate the minimal generating set of /^(pll|lp|pp)*\$/. This can also be done through some manipulation of the automata for the associated language.
Of course, this only address the question of whether or not you can represent a sequence of photos in a given basis of layouts. It does not address the optimality of the layouts according to your criteria (fewestmaximum number of distinct layouts used). This is where it gets a little harder....
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Notices? | 625 | 2,685 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-05 | latest | en | 0.905826 |
https://howtocreate.com/question-how-to-create-static-electricity-generator-71512/ | 1,716,547,616,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058709.9/warc/CC-MAIN-20240524091115-20240524121115-00142.warc.gz | 268,899,599 | 12,763 | ### Can a person generate static electricity?
Some objects such as wool, glass, human skin and hair are more likely to accumulate electric charges and have static electricity. Shuffling your feet across carpet, particularly in socks, is another way your body gains more electrons; they are released when you touch something such as a doorknob or another person.
### How does a static electricity generator work?
Electrostatic generators develop electrostatic charges of opposite signs rendered to two conductors, using only electric forces, and work by using moving plates, drums, or belts to carry electric charge to a high potential electrode.
### How much static electricity can a person generate?
One experimenter estimates the capacitance of the human body as high as 400 picofarads, and a charge of 50,000 volts, discharged e.g. during touching a charged car, creating a spark with energy of 500 millijoules. Another estimate is 100–300 pF and 20,000 volts, producing a maximum energy of 60 mJ.
### How do I make the most static electricity?
The best combinations of materials to create static electricity would be to have one material from the positive charge list and one from the negative charge list. Examples include combining human skin with polyester clothes, combing your hair with a plastic comb, and rubbing fur on a Plexiglas rod.
### How do you get rid of static electricity?
How to Get Rid of Static Electricity in Your Home
1. Install a Humidifier. The most effective way to minimize static electricity in the home is to install a humidifier.
2. Treat Your Rugs and Carpeting. A static charge in your rugs and carpeting can cause a shock when you walk across them.
3. Use Products on Clothing.
### What materials have the most static electricity?
Materials that gain a positive (+) electrical charge (or tend to give up electrons)
Dry human skin Greatest tendency to giving up electrons and becoming highly positive (+) in charge
Leather
Rabbit fur Fur is often used to create static electricity
Glass The glass on your TV screen gets charged and collects dust
### What materials Cannot conduct static electricity?
Some materials like plastic, cloth, and glass do not give up their electrons easily. These are called insulators. Materials such as metals lose their electrons more easily and are called conductors. Since plastics are insulators, they are poor conductors of electricity.
### Does foam create static electricity?
Yes, some types of styrofoam can carry a slight amount of static electrify. Just wrap the GPU is a plastic bag and you’ll be fine.
### What fabric does not cause static?
Fabrics That Don’t Cause Static
Anytime you need a guaranteed no-static zone, reach for your denim, chinos, tees, button-downs, cardigans and field jackets. 2. Leather. Somewhere in the tanning process, your moto jacket must have lost its conductivity.
### Does silk have static problems?
Silk is a poor conductor of electricity, so it is prone to static. Static can make clothes cling to the body in an unattractive way. When cleaning silk, static control is another important factor to consider.
### Why is polyester so static?
Polyester creates static electricity when it rubs up against another piece of synthetic material or another material with an opposite charge. There are several ways to reduce the static cling. Choose a method that best suits your needs and the type of polyester material.
### Is 100 cotton anti static?
In the past, the majority of protective clothing used in the field of ESD was 100% cotton. As with all natural fibres, this offers the advantage of being subject to only minimal antistatic build-up without the use of additional equipment. Cotton also poses a hazard in that its staple fibres can come loose.
### Does fleece cause static electricity?
Some materials, such as fleece, are more prone to static electricity than others. It can also build up between your skin and clothing because of friction. When static builds up to a point where it has nowhere to go, it will be released on the next oppositely charged item it comes into contact with.
### Why do I keep getting electric shocks off things?
Static shocks are more common when it’s cold and dry. This dry, cold air holds less water vapour than warm summer air. So, when you touch something like a metal doorknob or car door, those extra electrons will rapidly leave your body and give you the shock.
### What causes static electricity?
Static electricity is the result of an imbalance between negative and positive charges in an object. The rubbing of certain materials against one another can transfer negative charges, or electrons. For example, if you rub your shoe on the carpet, your body collects extra electrons.
### What are 3 examples of static?
What are three examples of static electricity? (Some examples might include: walking across a carpet and touching a metal door handle and pulling your hat off and having your hair stand on end.) When is there a positive charge? (A positive charge occurs when there is a shortage of electrons.)
### What are the 3 laws of static electricity?
Based on the same types of experiments like the one you performed, scientists were able to establish three laws of electrical charges: Opposite charges attract each other. Like charges repel each other. Charged objects attract neutral objects.
### What are some examples of static electricity at home?
Rub the surface of the rod with the cloth for 40 seconds. Flatten the plastic bag and rub the cloth against its surface for 40 seconds.
1. Flying Plastic Bag
• a plastic rod.
• a piece of cloth.
• a light plastic bag.
### Can static electricity eliminate you?
You might even see a spark if the discharge of electrons is large enough. The good news is that static electricity can‘t seriously harm you. Your body is composed largely of water and water is an inefficient conductor of electricity, especially in amounts this small. Not that electricity can‘t hurt or kill you.
### Can static electricity start a fire in bed?
A: Yes – but only if they are wet of something extremely flammable such as gasoline so the vapors will catch fire from a spark. Otherwise this kind of a static electricity has too little energy to cause thermal effects significant enough to set the fabric on fire.
### Can static electricity power a light bulb?
Static electricity can provide enough power to light up a light bulb. If you have ever experienced a little zap from static electricity, this amount of energy is capable of powering a fluorescent light bulb for a short time.
### Is it bad to sleep with static electricity?
Static electricity is a result of electrical equipment and the friction caused by synthetic furnishings. While they typically balance each other out without issue, the aforementioned friction could lead to sleep disruption as well as negative side effects such as stress or even anxiety.
### Can static electricity eliminate your PC?
Although it doesn’t happen often, a good zap of static electricity can eliminate a PC, either while it’s running or when you’re or working on it. The odds of a static discharge are so low, many of us will build tons of computers and never zap anything. | 1,467 | 7,250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-22 | latest | en | 0.906265 |
https://la.mathworks.com/matlabcentral/cody/problems/149-is-my-wife-right/solutions/1443856 | 1,596,773,678,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737152.0/warc/CC-MAIN-20200807025719-20200807055719-00383.warc.gz | 329,289,995 | 15,490 | Cody
# Problem 149. Is my wife right?
Solution 1443856
Submitted on 19 Feb 2018 by Polina Malinina
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 'But I''m actually right this time'; y_correct = 'yes'; assert(isequal(wiferight(x),y_correct))
2 Pass
x = 'But you just said that 2+2=3'; y_correct = 'yes'; assert(isequal(wiferight(x),y_correct)) | 140 | 474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-34 | latest | en | 0.758799 |
https://community.airtable.com/t5/formulas/creating-airtable-formulas-to-calculate-auction-house-commission/td-p/83209 | 1,675,702,324,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500356.92/warc/CC-MAIN-20230206145603-20230206175603-00268.warc.gz | 189,575,747 | 47,956 | Upcoming database upgrades. Airtable functionality will be reduced for ~15 minutes at 06:00 UTC on Feb. 4 / 10:00 pm PT on Feb. 3. Learn more here
# Creating Airtable formulas to calculate auction house commission rates
Topic Labels: Formulas
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5 - Automation Enthusiast
I am looking for some help please to bring a formula from Excel into Airtable in order to calculate Auction fees and commission based on certain variables.
First, I will explain how this example auction works:
The Hammer Price is the final amount someone bids on an item. The auction adds a “Buyers Premium” based on the final amount to be paid by the buyer. Here is how it is calculated.
X Auction charges a Buyer’s Premium calculated on the Hammer Price as follows: a rate of twenty-five percent (25%) of the Hammer Price of the Lot up to and including \$25,000; plus twenty percent (20%) on the part of the Hammer Price over \$25,000 and up to and including \$5,000,000; plus fifteen percent (15%) on the part of the Hammer Price over \$5,000,000.
In addition to the above charges, Auction X also charges the seller a fixed Selling Commission of 10% on the Hammer Price. This is amount is subtracted from the Hammer Price before the proceeds are remitted to the seller. Therefore, the total auction commission is the Selling Commission + Buyer’s Premium.
If I know the Hammer Price, I can easily calculate the Buyer’s Premium with the attached Array Formula in Excel. I am looking for some help please to bring this formula into Airtable.
=ARRAY_CONSTRAIN(ARRAYFORMULA(SUM(IF(S4>S7:S10,(IF(S4<S8:S11,S4,S8:S11)-S7:S10)*T8:T11,0))), 1, 1)
If I don’t know the Hammer Price, and perhaps only what the buyer paid, I would like to be able to reverse calculate the Hammer Price and what the seller received using the same variables described above.
I would really appreciate some help producing the various formulas in Airtable.
Thank you!
0 Replies 0 | 467 | 1,977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-06 | longest | en | 0.904127 |
https://www.educationindex.com/essay/Geometry-in-Golf-F3XQ874EZ | 1,580,249,260,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251783342.96/warc/CC-MAIN-20200128215526-20200129005526-00003.warc.gz | 857,594,467 | 10,767 | # Geometry in Golf
1194 words 5 pages
“Bringing it all Together: The Geometry of Golf”
Golf in Geometry?? No Way!
Geometry In The Game of Golf
For hundreds of years, golf has been an extremely popular and growing sport all around the world. Looking where golf is now, it is growing rapidly from the young to the elder population. The first round of gold was first played in the 15th century off the coast of Scotland, but it did not start to be played until around 1755. The standard rules of golf were written by a group of Edinburgh golfers. Today, people of the US, Scotland, and England, have been drawn to the game because it is fun, challenging, and hardly any athletic ability at all is required for amateurs. In breaking down the game, geometry plays a major
If one side of the putter were slanted or not parallel to the other, it would cause the balance to be off in weight and make it hard to swing the putter straight and make a proper stroke.
Triangles are also very important in golf, particularly in the swing. There should always be a triangle all throughout the swing. In the beginning, a triangle is formed between the shoulders and arms down to the hands. This triangle stays together until the club and left arm are perpendicular to each other. At this point, the elbows fold up and another triangle is formed with the elbows to the hands. At that point, the club is parallel to the ground. From there, the elbows release back down, and the original triangle is re-formed. At contact, the player should have triangle that he had when addressing the ball except with more weight on his or her back foot.
Another part of geometry in golf is the diameter of the golf ball. The ideal ball, ever since golf was invented, has been one that is small. The objective has always been finding the balance. The diameter really controls everything in the ball. If you increase the diameter, the ball is harder to put in the hole because the hole to ball ratio is increased. If the ball is heavier the ball will not go as far and new technology would need to come out so the ball is hit harder by the club. The final geometric topic is angle measurements. Each club that someone owns has a
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