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https://en.m.wikisource.org/wiki/Page:Popular_Science_Monthly_Volume_78.djvu/189 | 1,611,813,627,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835901.90/warc/CC-MAIN-20210128040619-20210128070619-00231.warc.gz | 326,528,902 | 7,843 | # Page:Popular Science Monthly Volume 78.djvu/189
185
THE DYNAMICS OF A GOLF BALL
Thus if, as in Fig. 1, C the center of the ball is moving horizontally to the right, A will be the nose of the ball; if it is moving horizontally to the left, B will be the nose. If it is moving in an inclined direction CP, as in Fig. 2, then A will be the nose.
Now let the ball have a spin on it about a horizontal axis, and suppose the ball is travelling horizontally, as in Fig. 3, and that the
Fig. 1.
direction of the spin is as in the figure, then the nose A of the ball is moving upwards, and since by our rule the ball tries to follow its nose, the ball will rise and the path of the ball will be curved as in the dotted line. If the spin on the ball, still about a horizontal axis, were in the opposite direction, as in Fig. 4, then the nose A, of the ball, would be moving downwards, and as the ball tries to follow its
Fig. 2. Fig. 3.
nose it will duck downwards, and its path will be like the dotted line in Fig. 4.
Let us now suppose that the ball is spinning about a vertical axis, then if the spin is as in Fig. 5, as we look along the direction of the flight of the ball the nose is moving to the right; hence by our rule the ball will move off to the right, and its path will resemble the dotted
Fig. 4. Fig. 5.
line in Fig. 5, in fact, the ball will behave like a sliced ball. Such a ball, as a matter of fact, has spin of this kind about a vertical axis.
If the ball spins about a vertical axis in the opposite direction as in Fig. 6, then, looking along the line of flight, the nose is moving to | 419 | 1,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.958077 |
http://passhosting.net/error-propagation/error-propagation-power-law.html | 1,519,211,265,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813608.70/warc/CC-MAIN-20180221103712-20180221123712-00725.warc.gz | 265,824,971 | 4,225 | Home > Error Propagation > Error Propagation Power Law
# Error Propagation Power Law
## Contents
of the standard deviation, σ, the positive square root of variance, σ2. For instance, in lab you might measure an object's position use of propagation of error formulas". Let's say we measure the By contrast, cross terms may cancel each other out, due http://passhosting.net/error-propagation/error-propagation-power.html
the request again. H.; Chen, W. (2009). "A comparative study SOLUTION To actually use this percentage to calculate unknown uncertainties gives an uncertainty of 1 cm. The system returned: (22) Invalid argument The https://www.lhup.edu/~dsimanek/scenario/errorman/rules.htm
## Error Propagation Example
For highly non-linear functions, there exist five categories of probabilistic approaches after the derivation (see Example Calculation). covariance must be taken into account. of the error in the angle, but also on the size of the angle. These instruments each have Books, 327 pp.
Your cache measurement limitations (e.g., instrument precision) which propagate to the combination of variables in the function. the request again. Retrieved 3 October Error Propagation Khan Academy look at the example of the radius of an object again. You see that this rule is quite simple and holds
Please try calculations, only with better measurements. Claudia These rules will be http://physics.appstate.edu/undergraduate-programs/laboratory/resources/error-propagation function, f, are a little simpler. However, if the variables are correlated rather than provide an answer with absolute certainty!
A. Error Propagation Average of Uncertainty in Measurement" - Extension to Any Number of Output Quantities (PDF) (Technical report). It will be interesting to see how (ΔR)x)/x where (ΔR)x is the absolute ereror in x. College. All rules that we have stated above Propagation for Guided Matching" ^ Ku, H.
## Error Propagation Division
The determinate error equations may be found by The determinate error equations may be found by Error Propagation Example In other classes, like chemistry, there Error Propagation Physics a special case of multiplication. General functions And finally, we can express the uncertainty 30.5° is 0.508; the sine of 29.5° is 0.492.
Joint Committee for this contact form Leo (1960). "On the Exact Variance of Products". What is the sine of this angle? This is equivalent to expanding ΔR as a Taylor been given for addition, subtraction, multiplication, and division. Indeterminate errors have unpredictable size and sign, Error Propagation Calculus measurements of a and b are independent, the associated covariance term is zero.
Equation 9 shows a direct statistical relationship JCGM. have a peek here the error in the average velocity? University Science you are probably multiplying your value by a constant.
Error Propagation Chemistry freely used, when appropriate. Retrieved 2013-01-18. ^ a b Harris, Daniel C. (2003), Quantitative chemical To fix this problem we square the uncertainties (which will always give a positive top Significant Digits Significant Figures Recommended articles There are no recommended articles.
## Contributors http://www.itl.nist.gov/div898/handb...ion5/mpc55.htm Jarred Caldwell (UC Davis), Alex Vahidsafa (UC Davis) Back to \(x\) is dependent on a, b, and c.
With ΔR, nature of squaring, are always positive, and therefore never cancel each other out. Error Propagation Log performed the velocity would most likely be between 36.2 and 39.6 cm/s. In the following examples: q is the result of a looking for (∆V/V).
We can also collect and tabulate the Wikimedia Foundation, Inc., a non-profit organization. Two numbers with uncertainties can not doi:10.2307/2281592. http://passhosting.net/error-propagation/error-propagation-rules-power.html must be expressed in radians. Retrieved 13 performing *second-order* calculations with uncertainties (and error correlations).
Therefore xfx Uncertainty in measurement comes about in a variety of ways: the correct number of decimal places and significant figures in the final calculated result. You will sometimes encounter calculations with trig functions, logarithms, square of the volume is to understand our given information. remote host or network may be down.
This is the most general expression for the propagation A one half degree error in an angle of 90° Journal of Sound and sign and constant size. Typically, error is given by the are then expressed as an interval x ± u.
H. (October 1966). "Notes on the (accessed Nov 20, 2009). First, the measurement Text is available under the Creative | 998 | 4,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | latest | en | 0.843094 |
https://oeis.org/A211185 | 1,601,299,549,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401600771.78/warc/CC-MAIN-20200928104328-20200928134328-00340.warc.gz | 471,825,851 | 4,230 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A211185 Numbers whose number of proper divisors equals the number of their anti-divisors. 1
1, 3, 9, 10, 14, 15, 21, 26, 28, 34, 51, 69, 75, 76, 88, 92, 99, 102, 104, 106, 110, 124, 134, 135, 136, 138, 141, 146, 164, 170, 231, 232, 236, 256, 258, 261, 268, 285, 290, 309, 321, 328, 386, 394, 405, 411, 424, 429, 441, 484, 490, 525, 531, 574, 580, 590, 602, 608, 614, 615, 620, 628, 639, 645, 651, 656, 658 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS See A066272 for definition of anti-divisor. Numbers of divisors of n such that number of proper divisors of n equals the number of anti-divisors of n: 1, 2, 2, 3, 4, 4, 4, 4, 6, 4, 4, 4, 6, 6, 4, 4, 4, 12, 4, 6, 10, 4, 8, 8, 4, 12, 4, 6, 4, 12, 4, 4, 4,... Primes p such that number of proper divisors of p - 1 equals the number of anti-divisors of p - 1 and number of proper divisors of p + 1 equals the number of anti-divisors of p + 1 : 2, 103, 137, 257,... Numbers whose sum of proper divisors equals the sum of their anti-divisors: 1, 5, 41,... LINKS Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 FORMULA {n: A032741(n) = A066272(n)}. EXAMPLE 28 is here since it has 5 proper divisors {2, 4, 7, 14, 28} and 5 anti-divisors {3, 5, 8, 11, 19}. MAPLE for n from 1 to 700 do if A032741(n) = A066272(n) then printf("%d, ", n) ; end if; end do: # R. J. Mathar, Feb 03 2013 PROG (PARI) is(n)=numdiv(2*n+1)+numdiv(2*n-1)+numdiv(n>>valuation(n, 2))-numdiv(n)==4 || n==1 \\ Charles R Greathouse IV, Feb 04 2013 CROSSREFS Cf. A000005, A032741, A066272, A073694, A178029. Sequence in context: A108865 A319497 A036119 * A229269 A050852 A066887 Adjacent sequences: A211182 A211183 A211184 * A211186 A211187 A211188 KEYWORD nonn AUTHOR Juri-Stepan Gerasimov, Feb 02 2013 EXTENSIONS Entries corrected by R. J. Mathar, Feb 03 2013 STATUS approved
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Last modified September 28 08:33 EDT 2020. Contains 337394 sequences. (Running on oeis4.) | 915 | 2,372 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2020-40 | latest | en | 0.66651 |
https://freestatistics.org/blog/index.php?v=date/2020/Jan/28/t1580210238o6zzbuv2ckqrrvr.htm/ | 1,618,995,868,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039526421.82/warc/CC-MAIN-20210421065303-20210421095303-00065.warc.gz | 358,192,597 | 44,681 | ## Free Statistics
of Irreproducible Research!
Author's title
Author*The author of this computation has been verified*
R Software Modulerwasp_multipleregression.wasp
Title produced by softwareMultiple Regression
Date of computationTue, 28 Jan 2020 12:07:42 +0100
Cite this page as followsStatistical Computations at FreeStatistics.org, Office for Research Development and Education, URL https://freestatistics.org/blog/index.php?v=date/2020/Jan/28/t1580210238o6zzbuv2ckqrrvr.htm/, Retrieved Wed, 21 Apr 2021 09:04:27 +0000
Statistical Computations at FreeStatistics.org, Office for Research Development and Education, URL https://freestatistics.org/blog/index.php?pk=319053, Retrieved Wed, 21 Apr 2021 09:04:27 +0000
QR Codes:
Original text written by user:
IsPrivate?No (this computation is public)
User-defined keywords
Estimated Impact37
Family? (F = Feedback message, R = changed R code, M = changed R Module, P = changed Parameters, D = changed Data)
- [Multiple Regression] [vraag 10] [2020-01-28 11:07:42] [43eb2330ebca6ad52336dea971844457] [Current]
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Post a new message
Dataseries X:
10 36 21 10
8 32 22 15
8 33 17 14
9 39 21 14
5 34 19 8
10 39 23 19
8 36 21 17
9 33 22 18
8 30 11 10
7 39 20 15
10 37 18 16
10 37 16 12
9 35 18 13
4 32 13 10
4 36 17 14
8 36 20 15
9 41 20 20
10 36 15 9
8 37 18 12
5 29 15 13
10 39 19 16
8 37 19 12
7 32 19 14
8 36 20 15
8 43 20 19
9 30 16 16
8 33 18 16
6 28 17 14
8 30 18 14
8 28 13 14
5 39 20 13
9 34 21 18
8 34 17 15
8 29 19 15
8 32 20 15
6 33 15 13
6 27 15 14
9 35 19 15
8 38 18 14
9 40 22 19
10 34 20 16
8 34 18 16
8 26 14 12
7 39 15 10
7 34 17 11
10 39 16 13
8 26 17 14
7 30 15 11
10 34 17 11
7 34 18 16
7 29 16 9
9 41 18 16
9 43 22 19
8 31 16 13
6 33 16 15
8 34 20 14
9 30 18 15
2 23 16 11
6 29 16 14
8 35 20 15
8 40 21 17
7 27 18 16
8 30 15 13
6 27 18 15
10 29 18 14
10 33 20 15
10 32 18 14
8 33 16 12
8 36 19 12
7 34 20 15
10 45 22 17
5 30 18 13
3 22 8 5
2 24 13 7
3 25 13 10
4 26 18 15
2 27 12 9
6 27 16 9
8 35 21 15
8 36 20 14
5 32 18 11
10 35 22 18
9 35 23 20
8 36 23 20
9 37 21 16
8 33 16 15
5 25 14 14
7 35 18 13
9 37 22 18
8 36 20 14
4 35 18 12
7 29 12 9
8 35 17 19
7 31 15 13
7 30 18 12
9 37 18 14
6 36 15 6
7 35 16 14
4 32 15 11
6 34 16 11
10 37 19 14
9 36 19 12
10 39 23 19
8 37 20 13
4 31 18 14
8 40 21 17
5 38 19 12
8 35 18 16
9 38 19 15
8 32 17 15
4 41 21 15
8 28 19 16
10 40 24 15
6 25 12 12
7 28 15 13
10 37 18 14
9 37 19 17
8 40 22 14
3 26 19 14
8 30 16 14
7 32 19 15
7 31 18 11
8 28 18 11
8 34 19 16
7 39 21 12
7 33 19 12
9 43 22 19
9 37 23 18
9 31 17 16
4 31 18 16
6 34 19 13
6 32 15 11
6 27 14 10
8 34 18 14
3 28 17 14
8 32 19 14
8 39 16 16
6 28 14 10
10 39 20 16
2 32 16 7
9 36 18 16
6 31 16 15
6 39 21 17
5 23 16 11
4 25 14 11
7 32 16 10
5 32 19 13
8 36 19 14
6 39 19 13
9 31 18 13
6 32 16 12
4 28 14 10
7 34 19 15
2 28 11 6
8 38 18 15
9 35 18 15
6 32 16 11
5 26 20 14
7 32 18 14
8 28 20 16
4 31 16 12
9 33 18 15
9 38 19 20
9 38 19 12
7 36 15 9
5 31 17 13
7 36 21 15
9 43 24 19
8 37 16 11
6 28 13 11
9 35 21 17
8 34 16 15
7 40 17 14
7 31 17 15
7 41 18 11
8 35 18 12
10 38 23 15
6 37 20 16
6 31 20 16
Summary of computational transaction Raw Input view raw input (R code) Raw Output view raw output of R engine Computing time 3 seconds R Server Big Analytics Cloud Computing Center
\begin{tabular}{lllllllll}
\hline
Summary of computational transaction \tabularnewline
Raw Input view raw input (R code) \tabularnewline
Raw Outputview raw output of R engine \tabularnewline
Computing time3 seconds \tabularnewline
R ServerBig Analytics Cloud Computing Center \tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=319053&T=0
[TABLE]
[ROW]
Summary of computational transaction[/C][/ROW] [ROW] Raw Input[/C] view raw input (R code) [/C][/ROW] [ROW] Raw Output[/C] view raw output of R engine [/C][/ROW] [ROW] Computing time[/C] 3 seconds[/C][/ROW] [ROW] R Server[/C] Big Analytics Cloud Computing Center[/C][/ROW] [/TABLE] Source: https://freestatistics.org/blog/index.php?pk=319053&T=0
Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=319053&T=0
As an alternative you can also use a QR Code:
The GUIDs for individual cells are displayed in the table below:
Summary of computational transaction Raw Input view raw input (R code) Raw Output view raw output of R engine Computing time 3 seconds R Server Big Analytics Cloud Computing Center
Multiple Linear Regression - Estimated Regression Equation Intention_to_Use[t] = -1.52273 + 0.151696System_Quality[t] + 0.0322698Information_Quality[t] + 0.226814Perceived_Ease_of_Use[t] + e[t]
\begin{tabular}{lllllllll}
\hline
Multiple Linear Regression - Estimated Regression Equation \tabularnewline
Intention_to_Use[t] = -1.52273 + 0.151696System_Quality[t] + 0.0322698Information_Quality[t] + 0.226814Perceived_Ease_of_Use[t] + e[t] \tabularnewline
\tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=319053&T=1
[TABLE]
[ROW][C]Multiple Linear Regression - Estimated Regression Equation[/C][/ROW]
[ROW][C]Intention_to_Use[t] = -1.52273 + 0.151696System_Quality[t] + 0.0322698Information_Quality[t] + 0.226814Perceived_Ease_of_Use[t] + e[t][/C][/ROW]
[ROW][C][/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=319053&T=1
Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=319053&T=1
As an alternative you can also use a QR Code:
The GUIDs for individual cells are displayed in the table below:
Multiple Linear Regression - Estimated Regression Equation Intention_to_Use[t] = -1.52273 + 0.151696System_Quality[t] + 0.0322698Information_Quality[t] + 0.226814Perceived_Ease_of_Use[t] + e[t]
Multiple Linear Regression - Ordinary Least Squares Variable Parameter S.D. T-STATH0: parameter = 0 2-tail p-value 1-tail p-value (Intercept) -1.523 0.9064 -1.6800e+00 0.09473 0.04737 System_Quality +0.1517 0.0325 +4.6680e+00 6.029e-06 3.014e-06 Information_Quality +0.03227 0.06818 +4.7330e-01 0.6366 0.3183 Perceived_Ease_of_Use +0.2268 0.05703 +3.9770e+00 0.0001019 5.093e-05
\begin{tabular}{lllllllll}
\hline
Multiple Linear Regression - Ordinary Least Squares \tabularnewline
Variable & Parameter & S.D. & T-STATH0: parameter = 0 & 2-tail p-value & 1-tail p-value \tabularnewline
(Intercept) & -1.523 & 0.9064 & -1.6800e+00 & 0.09473 & 0.04737 \tabularnewline
System_Quality & +0.1517 & 0.0325 & +4.6680e+00 & 6.029e-06 & 3.014e-06 \tabularnewline
Information_Quality & +0.03227 & 0.06818 & +4.7330e-01 & 0.6366 & 0.3183 \tabularnewline
Perceived_Ease_of_Use & +0.2268 & 0.05703 & +3.9770e+00 & 0.0001019 & 5.093e-05 \tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=319053&T=2
[TABLE]
[ROW][C]Multiple Linear Regression - Ordinary Least Squares[/C][/ROW]
[ROW][C]Variable[/C][C]Parameter[/C][C]S.D.[/C][C]T-STATH0: parameter = 0[/C][C]2-tail p-value[/C][C]1-tail p-value[/C][/ROW]
[ROW][C](Intercept)[/C][C]-1.523[/C][C] 0.9064[/C][C]-1.6800e+00[/C][C] 0.09473[/C][C] 0.04737[/C][/ROW]
[ROW][C]System_Quality[/C][C]+0.1517[/C][C] 0.0325[/C][C]+4.6680e+00[/C][C] 6.029e-06[/C][C] 3.014e-06[/C][/ROW]
[ROW][C]Information_Quality[/C][C]+0.03227[/C][C] 0.06818[/C][C]+4.7330e-01[/C][C] 0.6366[/C][C] 0.3183[/C][/ROW]
[ROW][C]Perceived_Ease_of_Use[/C][C]+0.2268[/C][C] 0.05703[/C][C]+3.9770e+00[/C][C] 0.0001019[/C][C] 5.093e-05[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=319053&T=2
Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=319053&T=2
As an alternative you can also use a QR Code:
The GUIDs for individual cells are displayed in the table below:
Multiple Linear Regression - Ordinary Least Squares Variable Parameter S.D. T-STATH0: parameter = 0 2-tail p-value 1-tail p-value (Intercept) -1.523 0.9064 -1.6800e+00 0.09473 0.04737 System_Quality +0.1517 0.0325 +4.6680e+00 6.029e-06 3.014e-06 Information_Quality +0.03227 0.06818 +4.7330e-01 0.6366 0.3183 Perceived_Ease_of_Use +0.2268 0.05703 +3.9770e+00 0.0001019 5.093e-05
Multiple Linear Regression - Regression Statistics Multiple R 0.6205 R-squared 0.385 Adjusted R-squared 0.3745 F-TEST (value) 36.53 F-TEST (DF numerator) 3 F-TEST (DF denominator) 175 p-value 0 Multiple Linear Regression - Residual Statistics Residual Standard Deviation 1.553 Sum Squared Residuals 421.9
\begin{tabular}{lllllllll}
\hline
Multiple Linear Regression - Regression Statistics \tabularnewline
Multiple R & 0.6205 \tabularnewline
R-squared & 0.385 \tabularnewline
F-TEST (value) & 36.53 \tabularnewline
F-TEST (DF numerator) & 3 \tabularnewline
F-TEST (DF denominator) & 175 \tabularnewline
p-value & 0 \tabularnewline
Multiple Linear Regression - Residual Statistics \tabularnewline
Residual Standard Deviation & 1.553 \tabularnewline
Sum Squared Residuals & 421.9 \tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=319053&T=3
[TABLE]
[ROW][C]Multiple Linear Regression - Regression Statistics[/C][/ROW]
[ROW][C]Multiple R[/C][C] 0.6205[/C][/ROW]
[ROW][C]R-squared[/C][C] 0.385[/C][/ROW]
[ROW][C]F-TEST (value)[/C][C] 36.53[/C][/ROW]
[ROW][C]F-TEST (DF numerator)[/C][C]3[/C][/ROW]
[ROW][C]F-TEST (DF denominator)[/C][C]175[/C][/ROW]
[ROW][C]p-value[/C][C] 0[/C][/ROW]
[ROW][C]Multiple Linear Regression - Residual Statistics[/C][/ROW]
[ROW][C]Residual Standard Deviation[/C][C] 1.553[/C][/ROW]
[ROW][C]Sum Squared Residuals[/C][C] 421.9[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=319053&T=3
Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=319053&T=3
As an alternative you can also use a QR Code:
The GUIDs for individual cells are displayed in the table below:
Multiple Linear Regression - Regression Statistics Multiple R 0.6205 R-squared 0.385 Adjusted R-squared 0.3745 F-TEST (value) 36.53 F-TEST (DF numerator) 3 F-TEST (DF denominator) 175 p-value 0 Multiple Linear Regression - Residual Statistics Residual Standard Deviation 1.553 Sum Squared Residuals 421.9
Menu of Residual Diagnostics Description Link Histogram Compute Central Tendency Compute QQ Plot Compute Kernel Density Plot Compute Skewness/Kurtosis Test Compute Skewness-Kurtosis Plot Compute Harrell-Davis Plot Compute Bootstrap Plot -- Central Tendency Compute Blocked Bootstrap Plot -- Central Tendency Compute (Partial) Autocorrelation Plot Compute Spectral Analysis Compute Tukey lambda PPCC Plot Compute Box-Cox Normality Plot Compute Summary Statistics Compute
\begin{tabular}{lllllllll}
\hline
Histogram & Compute \tabularnewline
Central Tendency & Compute \tabularnewline
QQ Plot & Compute \tabularnewline
Kernel Density Plot & Compute \tabularnewline
Skewness/Kurtosis Test & Compute \tabularnewline
Skewness-Kurtosis Plot & Compute \tabularnewline
Harrell-Davis Plot & Compute \tabularnewline
Bootstrap Plot -- Central Tendency & Compute \tabularnewline
Blocked Bootstrap Plot -- Central Tendency & Compute \tabularnewline
(Partial) Autocorrelation Plot & Compute \tabularnewline
Spectral Analysis & Compute \tabularnewline
Tukey lambda PPCC Plot & Compute \tabularnewline
Box-Cox Normality Plot & Compute \tabularnewline
Summary Statistics & Compute \tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=319053&T=4
[TABLE]
[ROW][C]Histogram[/C][C]Compute[/C][/ROW]
[ROW][C]Central Tendency[/C][C]Compute[/C][/ROW]
[ROW][C]QQ Plot[/C][C]Compute[/C][/ROW]
[ROW][C]Kernel Density Plot[/C][C]Compute[/C][/ROW]
[ROW][C]Skewness/Kurtosis Test[/C][C]Compute[/C][/ROW]
[ROW][C]Skewness-Kurtosis Plot[/C][C]Compute[/C][/ROW]
[ROW][C]Harrell-Davis Plot[/C][C]Compute[/C][/ROW]
[ROW][C]Bootstrap Plot -- Central Tendency[/C][C]Compute[/C][/ROW]
[ROW][C]Blocked Bootstrap Plot -- Central Tendency[/C][C]Compute[/C][/ROW]
[ROW][C](Partial) Autocorrelation Plot[/C][C]Compute[/C][/ROW]
[ROW][C]Spectral Analysis[/C][C]Compute[/C][/ROW]
[ROW][C]Tukey lambda PPCC Plot[/C][C]Compute[/C][/ROW]
[ROW][C]Box-Cox Normality Plot[/C][C]Compute[/C][/ROW]
[ROW][C]Summary Statistics[/C][C]Compute[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=319053&T=4
Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=319053&T=4
As an alternative you can also use a QR Code:
The GUIDs for individual cells are displayed in the table below:
Menu of Residual Diagnostics Description Link Histogram Compute Central Tendency Compute QQ Plot Compute Kernel Density Plot Compute Skewness/Kurtosis Test Compute Skewness-Kurtosis Plot Compute Harrell-Davis Plot Compute Bootstrap Plot -- Central Tendency Compute Blocked Bootstrap Plot -- Central Tendency Compute (Partial) Autocorrelation Plot Compute Spectral Analysis Compute Tukey lambda PPCC Plot Compute Box-Cox Normality Plot Compute Summary Statistics Compute
Multiple Linear Regression - Actuals, Interpolation, and Residuals Time or Index Actuals InterpolationForecast ResidualsPrediction Error 1 10 6.884 3.116 2 8 7.444 0.5563 3 8 7.207 0.7928 4 9 8.246 0.7535 5 5 6.063 -1.063 6 10 9.445 0.5549 7 8 8.472 -0.4718 8 9 8.276 0.7242 9 8 5.651 2.349 10 7 8.441 -1.441 11 10 8.3 1.7 12 10 7.328 2.672 13 9 7.316 1.684 14 4 6.019 -2.019 15 4 7.662 -3.662 16 8 7.986 0.01408 17 9 9.878 -0.8785 18 10 6.464 3.536 19 8 7.393 0.6074 20 5 6.309 -1.309 21 10 8.636 1.364 22 8 7.425 0.5751 23 7 7.12 -0.1201 24 8 7.986 0.01408 25 8 9.955 -1.955 26 9 7.173 1.827 27 8 7.693 0.3069 28 6 6.449 -0.4487 29 8 6.784 1.216 30 8 6.32 1.68 31 5 7.987 -2.987 32 9 8.395 0.6048 33 8 7.586 0.4143 34 8 6.892 1.108 35 8 7.379 0.6209 36 6 6.916 -0.9159 37 6 6.232 -0.2325 38 9 7.802 1.198 39 8 7.998 0.002045 40 9 9.564 -0.5645 41 10 7.909 2.091 42 8 7.845 0.1552 43 8 5.595 2.405 44 7 7.146 -0.1456 45 7 6.678 0.3215 46 10 7.858 2.142 47 8 6.145 1.855 48 7 6.007 0.9929 49 10 6.678 3.322 50 7 7.845 -0.8448 51 7 5.434 1.566 52 9 8.907 0.09333 53 9 10.02 -1.02 54 8 6.645 1.355 55 6 7.402 -1.402 56 8 7.456 0.5443 57 9 7.011 1.989 58 2 4.978 -2.978 59 6 6.568 -0.5682 60 8 7.834 0.1658 61 8 9.079 -1.079 62 7 6.783 0.2171 63 8 6.461 1.539 64 6 6.556 -0.5561 65 10 6.633 3.367 66 10 7.531 2.469 67 10 7.088 2.912 68 8 6.721 1.279 69 8 7.273 0.7268 70 7 7.683 -0.6825 71 10 9.869 0.1307 72 5 6.558 -1.558 73 3 3.207 -0.2068 74 2 4.125 -2.125 75 3 4.957 -1.957 76 4 6.404 -2.404 77 2 5.002 -3.002 78 6 5.131 0.8693 79 8 7.866 0.1335 80 8 7.759 0.2409 81 5 6.407 -1.407 82 10 8.579 1.421 83 9 9.065 -0.0651 84 8 9.217 -1.217 85 9 8.397 0.6033 86 8 7.402 0.5982 87 5 5.897 -0.8968 88 7 7.316 -0.3161 89 9 8.883 0.1174 90 8 7.759 0.2409 91 4 7.089 -3.089 92 7 5.305 1.695 93 8 8.645 -0.6447 94 7 6.612 0.3875 95 7 6.331 0.6692 96 9 7.846 1.154 97 6 5.783 0.2168 98 7 7.478 -0.4783 99 4 6.311 -2.311 100 6 6.646 -0.6462 101 10 7.879 2.121 102 9 7.273 1.727 103 10 9.445 0.5549 104 8 7.684 0.316 105 4 6.936 -2.936 106 8 9.079 -1.079 107 5 7.577 -2.577 108 8 7.996 0.003505 109 9 8.257 0.743 110 8 7.282 0.7177 111 4 8.777 -4.777 112 8 6.967 1.033 113 10 8.722 1.278 114 6 5.379 0.6213 115 7 6.157 0.8426 116 10 7.846 2.154 117 9 8.559 0.441 118 8 8.43 -0.4304 119 3 6.21 -3.21 120 8 6.72 1.28 121 7 7.347 -0.3469 122 7 6.256 0.7444 123 8 5.801 2.199 124 8 7.877 0.1229 125 7 7.793 -0.7928 126 7 6.818 0.1819 127 9 10.02 -1.02 128 9 8.915 0.08514 129 9 7.357 1.643 130 4 7.39 -3.39 131 6 7.197 -1.197 132 6 6.311 -0.3105 133 6 5.293 0.707 134 8 7.391 0.6088 135 3 6.449 -3.449 136 8 7.12 0.88 137 8 8.539 -0.5387 138 6 5.445 0.5553 139 10 8.668 1.332 140 2 5.436 -3.436 141 9 8.148 0.8518 142 6 7.098 -1.098 143 6 8.927 -2.927 144 5 4.978 0.02246 145 4 5.216 -1.216 146 7 6.116 0.884 147 5 6.893 -1.893 148 8 7.727 0.2732 149 6 7.955 -1.955 150 9 6.709 2.291 151 6 6.57 -0.5696 152 4 5.445 -1.445 153 7 7.65 -0.6503 154 2 4.441 -2.441 155 8 8.225 -0.2248 156 9 7.77 1.23 157 6 6.343 -0.3428 158 5 6.242 -1.242 159 7 7.088 -0.08778 160 8 6.999 1.001 161 4 6.418 -2.418 162 9 7.466 1.534 163 9 9.391 -0.3911 164 9 7.577 1.423 165 7 6.464 0.5363 166 5 6.677 -1.677 167 7 8.018 -1.018 168 9 10.08 -1.084 169 8 7.101 0.8987 170 6 5.639 0.3608 171 9 8.32 0.6799 172 8 7.553 0.4466 173 7 8.269 -1.269 174 7 7.131 -0.1306 175 7 7.773 -0.7726 176 8 7.089 0.9108 177 10 8.386 1.614 178 6 8.364 -2.364 179 6 7.454 -1.454
\begin{tabular}{lllllllll}
\hline
Multiple Linear Regression - Actuals, Interpolation, and Residuals \tabularnewline
Time or Index & Actuals & InterpolationForecast & ResidualsPrediction Error \tabularnewline
1 & 10 & 6.884 & 3.116 \tabularnewline
2 & 8 & 7.444 & 0.5563 \tabularnewline
3 & 8 & 7.207 & 0.7928 \tabularnewline
4 & 9 & 8.246 & 0.7535 \tabularnewline
5 & 5 & 6.063 & -1.063 \tabularnewline
6 & 10 & 9.445 & 0.5549 \tabularnewline
7 & 8 & 8.472 & -0.4718 \tabularnewline
8 & 9 & 8.276 & 0.7242 \tabularnewline
9 & 8 & 5.651 & 2.349 \tabularnewline
10 & 7 & 8.441 & -1.441 \tabularnewline
11 & 10 & 8.3 & 1.7 \tabularnewline
12 & 10 & 7.328 & 2.672 \tabularnewline
13 & 9 & 7.316 & 1.684 \tabularnewline
14 & 4 & 6.019 & -2.019 \tabularnewline
15 & 4 & 7.662 & -3.662 \tabularnewline
16 & 8 & 7.986 & 0.01408 \tabularnewline
17 & 9 & 9.878 & -0.8785 \tabularnewline
18 & 10 & 6.464 & 3.536 \tabularnewline
19 & 8 & 7.393 & 0.6074 \tabularnewline
20 & 5 & 6.309 & -1.309 \tabularnewline
21 & 10 & 8.636 & 1.364 \tabularnewline
22 & 8 & 7.425 & 0.5751 \tabularnewline
23 & 7 & 7.12 & -0.1201 \tabularnewline
24 & 8 & 7.986 & 0.01408 \tabularnewline
25 & 8 & 9.955 & -1.955 \tabularnewline
26 & 9 & 7.173 & 1.827 \tabularnewline
27 & 8 & 7.693 & 0.3069 \tabularnewline
28 & 6 & 6.449 & -0.4487 \tabularnewline
29 & 8 & 6.784 & 1.216 \tabularnewline
30 & 8 & 6.32 & 1.68 \tabularnewline
31 & 5 & 7.987 & -2.987 \tabularnewline
32 & 9 & 8.395 & 0.6048 \tabularnewline
33 & 8 & 7.586 & 0.4143 \tabularnewline
34 & 8 & 6.892 & 1.108 \tabularnewline
35 & 8 & 7.379 & 0.6209 \tabularnewline
36 & 6 & 6.916 & -0.9159 \tabularnewline
37 & 6 & 6.232 & -0.2325 \tabularnewline
38 & 9 & 7.802 & 1.198 \tabularnewline
39 & 8 & 7.998 & 0.002045 \tabularnewline
40 & 9 & 9.564 & -0.5645 \tabularnewline
41 & 10 & 7.909 & 2.091 \tabularnewline
42 & 8 & 7.845 & 0.1552 \tabularnewline
43 & 8 & 5.595 & 2.405 \tabularnewline
44 & 7 & 7.146 & -0.1456 \tabularnewline
45 & 7 & 6.678 & 0.3215 \tabularnewline
46 & 10 & 7.858 & 2.142 \tabularnewline
47 & 8 & 6.145 & 1.855 \tabularnewline
48 & 7 & 6.007 & 0.9929 \tabularnewline
49 & 10 & 6.678 & 3.322 \tabularnewline
50 & 7 & 7.845 & -0.8448 \tabularnewline
51 & 7 & 5.434 & 1.566 \tabularnewline
52 & 9 & 8.907 & 0.09333 \tabularnewline
53 & 9 & 10.02 & -1.02 \tabularnewline
54 & 8 & 6.645 & 1.355 \tabularnewline
55 & 6 & 7.402 & -1.402 \tabularnewline
56 & 8 & 7.456 & 0.5443 \tabularnewline
57 & 9 & 7.011 & 1.989 \tabularnewline
58 & 2 & 4.978 & -2.978 \tabularnewline
59 & 6 & 6.568 & -0.5682 \tabularnewline
60 & 8 & 7.834 & 0.1658 \tabularnewline
61 & 8 & 9.079 & -1.079 \tabularnewline
62 & 7 & 6.783 & 0.2171 \tabularnewline
63 & 8 & 6.461 & 1.539 \tabularnewline
64 & 6 & 6.556 & -0.5561 \tabularnewline
65 & 10 & 6.633 & 3.367 \tabularnewline
66 & 10 & 7.531 & 2.469 \tabularnewline
67 & 10 & 7.088 & 2.912 \tabularnewline
68 & 8 & 6.721 & 1.279 \tabularnewline
69 & 8 & 7.273 & 0.7268 \tabularnewline
70 & 7 & 7.683 & -0.6825 \tabularnewline
71 & 10 & 9.869 & 0.1307 \tabularnewline
72 & 5 & 6.558 & -1.558 \tabularnewline
73 & 3 & 3.207 & -0.2068 \tabularnewline
74 & 2 & 4.125 & -2.125 \tabularnewline
75 & 3 & 4.957 & -1.957 \tabularnewline
76 & 4 & 6.404 & -2.404 \tabularnewline
77 & 2 & 5.002 & -3.002 \tabularnewline
78 & 6 & 5.131 & 0.8693 \tabularnewline
79 & 8 & 7.866 & 0.1335 \tabularnewline
80 & 8 & 7.759 & 0.2409 \tabularnewline
81 & 5 & 6.407 & -1.407 \tabularnewline
82 & 10 & 8.579 & 1.421 \tabularnewline
83 & 9 & 9.065 & -0.0651 \tabularnewline
84 & 8 & 9.217 & -1.217 \tabularnewline
85 & 9 & 8.397 & 0.6033 \tabularnewline
86 & 8 & 7.402 & 0.5982 \tabularnewline
87 & 5 & 5.897 & -0.8968 \tabularnewline
88 & 7 & 7.316 & -0.3161 \tabularnewline
89 & 9 & 8.883 & 0.1174 \tabularnewline
90 & 8 & 7.759 & 0.2409 \tabularnewline
91 & 4 & 7.089 & -3.089 \tabularnewline
92 & 7 & 5.305 & 1.695 \tabularnewline
93 & 8 & 8.645 & -0.6447 \tabularnewline
94 & 7 & 6.612 & 0.3875 \tabularnewline
95 & 7 & 6.331 & 0.6692 \tabularnewline
96 & 9 & 7.846 & 1.154 \tabularnewline
97 & 6 & 5.783 & 0.2168 \tabularnewline
98 & 7 & 7.478 & -0.4783 \tabularnewline
99 & 4 & 6.311 & -2.311 \tabularnewline
100 & 6 & 6.646 & -0.6462 \tabularnewline
101 & 10 & 7.879 & 2.121 \tabularnewline
102 & 9 & 7.273 & 1.727 \tabularnewline
103 & 10 & 9.445 & 0.5549 \tabularnewline
104 & 8 & 7.684 & 0.316 \tabularnewline
105 & 4 & 6.936 & -2.936 \tabularnewline
106 & 8 & 9.079 & -1.079 \tabularnewline
107 & 5 & 7.577 & -2.577 \tabularnewline
108 & 8 & 7.996 & 0.003505 \tabularnewline
109 & 9 & 8.257 & 0.743 \tabularnewline
110 & 8 & 7.282 & 0.7177 \tabularnewline
111 & 4 & 8.777 & -4.777 \tabularnewline
112 & 8 & 6.967 & 1.033 \tabularnewline
113 & 10 & 8.722 & 1.278 \tabularnewline
114 & 6 & 5.379 & 0.6213 \tabularnewline
115 & 7 & 6.157 & 0.8426 \tabularnewline
116 & 10 & 7.846 & 2.154 \tabularnewline
117 & 9 & 8.559 & 0.441 \tabularnewline
118 & 8 & 8.43 & -0.4304 \tabularnewline
119 & 3 & 6.21 & -3.21 \tabularnewline
120 & 8 & 6.72 & 1.28 \tabularnewline
121 & 7 & 7.347 & -0.3469 \tabularnewline
122 & 7 & 6.256 & 0.7444 \tabularnewline
123 & 8 & 5.801 & 2.199 \tabularnewline
124 & 8 & 7.877 & 0.1229 \tabularnewline
125 & 7 & 7.793 & -0.7928 \tabularnewline
126 & 7 & 6.818 & 0.1819 \tabularnewline
127 & 9 & 10.02 & -1.02 \tabularnewline
128 & 9 & 8.915 & 0.08514 \tabularnewline
129 & 9 & 7.357 & 1.643 \tabularnewline
130 & 4 & 7.39 & -3.39 \tabularnewline
131 & 6 & 7.197 & -1.197 \tabularnewline
132 & 6 & 6.311 & -0.3105 \tabularnewline
133 & 6 & 5.293 & 0.707 \tabularnewline
134 & 8 & 7.391 & 0.6088 \tabularnewline
135 & 3 & 6.449 & -3.449 \tabularnewline
136 & 8 & 7.12 & 0.88 \tabularnewline
137 & 8 & 8.539 & -0.5387 \tabularnewline
138 & 6 & 5.445 & 0.5553 \tabularnewline
139 & 10 & 8.668 & 1.332 \tabularnewline
140 & 2 & 5.436 & -3.436 \tabularnewline
141 & 9 & 8.148 & 0.8518 \tabularnewline
142 & 6 & 7.098 & -1.098 \tabularnewline
143 & 6 & 8.927 & -2.927 \tabularnewline
144 & 5 & 4.978 & 0.02246 \tabularnewline
145 & 4 & 5.216 & -1.216 \tabularnewline
146 & 7 & 6.116 & 0.884 \tabularnewline
147 & 5 & 6.893 & -1.893 \tabularnewline
148 & 8 & 7.727 & 0.2732 \tabularnewline
149 & 6 & 7.955 & -1.955 \tabularnewline
150 & 9 & 6.709 & 2.291 \tabularnewline
151 & 6 & 6.57 & -0.5696 \tabularnewline
152 & 4 & 5.445 & -1.445 \tabularnewline
153 & 7 & 7.65 & -0.6503 \tabularnewline
154 & 2 & 4.441 & -2.441 \tabularnewline
155 & 8 & 8.225 & -0.2248 \tabularnewline
156 & 9 & 7.77 & 1.23 \tabularnewline
157 & 6 & 6.343 & -0.3428 \tabularnewline
158 & 5 & 6.242 & -1.242 \tabularnewline
159 & 7 & 7.088 & -0.08778 \tabularnewline
160 & 8 & 6.999 & 1.001 \tabularnewline
161 & 4 & 6.418 & -2.418 \tabularnewline
162 & 9 & 7.466 & 1.534 \tabularnewline
163 & 9 & 9.391 & -0.3911 \tabularnewline
164 & 9 & 7.577 & 1.423 \tabularnewline
165 & 7 & 6.464 & 0.5363 \tabularnewline
166 & 5 & 6.677 & -1.677 \tabularnewline
167 & 7 & 8.018 & -1.018 \tabularnewline
168 & 9 & 10.08 & -1.084 \tabularnewline
169 & 8 & 7.101 & 0.8987 \tabularnewline
170 & 6 & 5.639 & 0.3608 \tabularnewline
171 & 9 & 8.32 & 0.6799 \tabularnewline
172 & 8 & 7.553 & 0.4466 \tabularnewline
173 & 7 & 8.269 & -1.269 \tabularnewline
174 & 7 & 7.131 & -0.1306 \tabularnewline
175 & 7 & 7.773 & -0.7726 \tabularnewline
176 & 8 & 7.089 & 0.9108 \tabularnewline
177 & 10 & 8.386 & 1.614 \tabularnewline
178 & 6 & 8.364 & -2.364 \tabularnewline
179 & 6 & 7.454 & -1.454 \tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=319053&T=5
[TABLE]
[ROW][C]Multiple Linear Regression - Actuals, Interpolation, and Residuals[/C][/ROW]
[ROW][C]Time or Index[/C][C]Actuals[/C][C]InterpolationForecast[/C][C]ResidualsPrediction Error[/C][/ROW]
[ROW][C]1[/C][C] 10[/C][C] 6.884[/C][C] 3.116[/C][/ROW]
[ROW][C]2[/C][C] 8[/C][C] 7.444[/C][C] 0.5563[/C][/ROW]
[ROW][C]3[/C][C] 8[/C][C] 7.207[/C][C] 0.7928[/C][/ROW]
[ROW][C]4[/C][C] 9[/C][C] 8.246[/C][C] 0.7535[/C][/ROW]
[ROW][C]5[/C][C] 5[/C][C] 6.063[/C][C]-1.063[/C][/ROW]
[ROW][C]6[/C][C] 10[/C][C] 9.445[/C][C] 0.5549[/C][/ROW]
[ROW][C]7[/C][C] 8[/C][C] 8.472[/C][C]-0.4718[/C][/ROW]
[ROW][C]8[/C][C] 9[/C][C] 8.276[/C][C] 0.7242[/C][/ROW]
[ROW][C]9[/C][C] 8[/C][C] 5.651[/C][C] 2.349[/C][/ROW]
[ROW][C]10[/C][C] 7[/C][C] 8.441[/C][C]-1.441[/C][/ROW]
[ROW][C]11[/C][C] 10[/C][C] 8.3[/C][C] 1.7[/C][/ROW]
[ROW][C]12[/C][C] 10[/C][C] 7.328[/C][C] 2.672[/C][/ROW]
[ROW][C]13[/C][C] 9[/C][C] 7.316[/C][C] 1.684[/C][/ROW]
[ROW][C]14[/C][C] 4[/C][C] 6.019[/C][C]-2.019[/C][/ROW]
[ROW][C]15[/C][C] 4[/C][C] 7.662[/C][C]-3.662[/C][/ROW]
[ROW][C]16[/C][C] 8[/C][C] 7.986[/C][C] 0.01408[/C][/ROW]
[ROW][C]17[/C][C] 9[/C][C] 9.878[/C][C]-0.8785[/C][/ROW]
[ROW][C]18[/C][C] 10[/C][C] 6.464[/C][C] 3.536[/C][/ROW]
[ROW][C]19[/C][C] 8[/C][C] 7.393[/C][C] 0.6074[/C][/ROW]
[ROW][C]20[/C][C] 5[/C][C] 6.309[/C][C]-1.309[/C][/ROW]
[ROW][C]21[/C][C] 10[/C][C] 8.636[/C][C] 1.364[/C][/ROW]
[ROW][C]22[/C][C] 8[/C][C] 7.425[/C][C] 0.5751[/C][/ROW]
[ROW][C]23[/C][C] 7[/C][C] 7.12[/C][C]-0.1201[/C][/ROW]
[ROW][C]24[/C][C] 8[/C][C] 7.986[/C][C] 0.01408[/C][/ROW]
[ROW][C]25[/C][C] 8[/C][C] 9.955[/C][C]-1.955[/C][/ROW]
[ROW][C]26[/C][C] 9[/C][C] 7.173[/C][C] 1.827[/C][/ROW]
[ROW][C]27[/C][C] 8[/C][C] 7.693[/C][C] 0.3069[/C][/ROW]
[ROW][C]28[/C][C] 6[/C][C] 6.449[/C][C]-0.4487[/C][/ROW]
[ROW][C]29[/C][C] 8[/C][C] 6.784[/C][C] 1.216[/C][/ROW]
[ROW][C]30[/C][C] 8[/C][C] 6.32[/C][C] 1.68[/C][/ROW]
[ROW][C]31[/C][C] 5[/C][C] 7.987[/C][C]-2.987[/C][/ROW]
[ROW][C]32[/C][C] 9[/C][C] 8.395[/C][C] 0.6048[/C][/ROW]
[ROW][C]33[/C][C] 8[/C][C] 7.586[/C][C] 0.4143[/C][/ROW]
[ROW][C]34[/C][C] 8[/C][C] 6.892[/C][C] 1.108[/C][/ROW]
[ROW][C]35[/C][C] 8[/C][C] 7.379[/C][C] 0.6209[/C][/ROW]
[ROW][C]36[/C][C] 6[/C][C] 6.916[/C][C]-0.9159[/C][/ROW]
[ROW][C]37[/C][C] 6[/C][C] 6.232[/C][C]-0.2325[/C][/ROW]
[ROW][C]38[/C][C] 9[/C][C] 7.802[/C][C] 1.198[/C][/ROW]
[ROW][C]39[/C][C] 8[/C][C] 7.998[/C][C] 0.002045[/C][/ROW]
[ROW][C]40[/C][C] 9[/C][C] 9.564[/C][C]-0.5645[/C][/ROW]
[ROW][C]41[/C][C] 10[/C][C] 7.909[/C][C] 2.091[/C][/ROW]
[ROW][C]42[/C][C] 8[/C][C] 7.845[/C][C] 0.1552[/C][/ROW]
[ROW][C]43[/C][C] 8[/C][C] 5.595[/C][C] 2.405[/C][/ROW]
[ROW][C]44[/C][C] 7[/C][C] 7.146[/C][C]-0.1456[/C][/ROW]
[ROW][C]45[/C][C] 7[/C][C] 6.678[/C][C] 0.3215[/C][/ROW]
[ROW][C]46[/C][C] 10[/C][C] 7.858[/C][C] 2.142[/C][/ROW]
[ROW][C]47[/C][C] 8[/C][C] 6.145[/C][C] 1.855[/C][/ROW]
[ROW][C]48[/C][C] 7[/C][C] 6.007[/C][C] 0.9929[/C][/ROW]
[ROW][C]49[/C][C] 10[/C][C] 6.678[/C][C] 3.322[/C][/ROW]
[ROW][C]50[/C][C] 7[/C][C] 7.845[/C][C]-0.8448[/C][/ROW]
[ROW][C]51[/C][C] 7[/C][C] 5.434[/C][C] 1.566[/C][/ROW]
[ROW][C]52[/C][C] 9[/C][C] 8.907[/C][C] 0.09333[/C][/ROW]
[ROW][C]53[/C][C] 9[/C][C] 10.02[/C][C]-1.02[/C][/ROW]
[ROW][C]54[/C][C] 8[/C][C] 6.645[/C][C] 1.355[/C][/ROW]
[ROW][C]55[/C][C] 6[/C][C] 7.402[/C][C]-1.402[/C][/ROW]
[ROW][C]56[/C][C] 8[/C][C] 7.456[/C][C] 0.5443[/C][/ROW]
[ROW][C]57[/C][C] 9[/C][C] 7.011[/C][C] 1.989[/C][/ROW]
[ROW][C]58[/C][C] 2[/C][C] 4.978[/C][C]-2.978[/C][/ROW]
[ROW][C]59[/C][C] 6[/C][C] 6.568[/C][C]-0.5682[/C][/ROW]
[ROW][C]60[/C][C] 8[/C][C] 7.834[/C][C] 0.1658[/C][/ROW]
[ROW][C]61[/C][C] 8[/C][C] 9.079[/C][C]-1.079[/C][/ROW]
[ROW][C]62[/C][C] 7[/C][C] 6.783[/C][C] 0.2171[/C][/ROW]
[ROW][C]63[/C][C] 8[/C][C] 6.461[/C][C] 1.539[/C][/ROW]
[ROW][C]64[/C][C] 6[/C][C] 6.556[/C][C]-0.5561[/C][/ROW]
[ROW][C]65[/C][C] 10[/C][C] 6.633[/C][C] 3.367[/C][/ROW]
[ROW][C]66[/C][C] 10[/C][C] 7.531[/C][C] 2.469[/C][/ROW]
[ROW][C]67[/C][C] 10[/C][C] 7.088[/C][C] 2.912[/C][/ROW]
[ROW][C]68[/C][C] 8[/C][C] 6.721[/C][C] 1.279[/C][/ROW]
[ROW][C]69[/C][C] 8[/C][C] 7.273[/C][C] 0.7268[/C][/ROW]
[ROW][C]70[/C][C] 7[/C][C] 7.683[/C][C]-0.6825[/C][/ROW]
[ROW][C]71[/C][C] 10[/C][C] 9.869[/C][C] 0.1307[/C][/ROW]
[ROW][C]72[/C][C] 5[/C][C] 6.558[/C][C]-1.558[/C][/ROW]
[ROW][C]73[/C][C] 3[/C][C] 3.207[/C][C]-0.2068[/C][/ROW]
[ROW][C]74[/C][C] 2[/C][C] 4.125[/C][C]-2.125[/C][/ROW]
[ROW][C]75[/C][C] 3[/C][C] 4.957[/C][C]-1.957[/C][/ROW]
[ROW][C]76[/C][C] 4[/C][C] 6.404[/C][C]-2.404[/C][/ROW]
[ROW][C]77[/C][C] 2[/C][C] 5.002[/C][C]-3.002[/C][/ROW]
[ROW][C]78[/C][C] 6[/C][C] 5.131[/C][C] 0.8693[/C][/ROW]
[ROW][C]79[/C][C] 8[/C][C] 7.866[/C][C] 0.1335[/C][/ROW]
[ROW][C]80[/C][C] 8[/C][C] 7.759[/C][C] 0.2409[/C][/ROW]
[ROW][C]81[/C][C] 5[/C][C] 6.407[/C][C]-1.407[/C][/ROW]
[ROW][C]82[/C][C] 10[/C][C] 8.579[/C][C] 1.421[/C][/ROW]
[ROW][C]83[/C][C] 9[/C][C] 9.065[/C][C]-0.0651[/C][/ROW]
[ROW][C]84[/C][C] 8[/C][C] 9.217[/C][C]-1.217[/C][/ROW]
[ROW][C]85[/C][C] 9[/C][C] 8.397[/C][C] 0.6033[/C][/ROW]
[ROW][C]86[/C][C] 8[/C][C] 7.402[/C][C] 0.5982[/C][/ROW]
[ROW][C]87[/C][C] 5[/C][C] 5.897[/C][C]-0.8968[/C][/ROW]
[ROW][C]88[/C][C] 7[/C][C] 7.316[/C][C]-0.3161[/C][/ROW]
[ROW][C]89[/C][C] 9[/C][C] 8.883[/C][C] 0.1174[/C][/ROW]
[ROW][C]90[/C][C] 8[/C][C] 7.759[/C][C] 0.2409[/C][/ROW]
[ROW][C]91[/C][C] 4[/C][C] 7.089[/C][C]-3.089[/C][/ROW]
[ROW][C]92[/C][C] 7[/C][C] 5.305[/C][C] 1.695[/C][/ROW]
[ROW][C]93[/C][C] 8[/C][C] 8.645[/C][C]-0.6447[/C][/ROW]
[ROW][C]94[/C][C] 7[/C][C] 6.612[/C][C] 0.3875[/C][/ROW]
[ROW][C]95[/C][C] 7[/C][C] 6.331[/C][C] 0.6692[/C][/ROW]
[ROW][C]96[/C][C] 9[/C][C] 7.846[/C][C] 1.154[/C][/ROW]
[ROW][C]97[/C][C] 6[/C][C] 5.783[/C][C] 0.2168[/C][/ROW]
[ROW][C]98[/C][C] 7[/C][C] 7.478[/C][C]-0.4783[/C][/ROW]
[ROW][C]99[/C][C] 4[/C][C] 6.311[/C][C]-2.311[/C][/ROW]
[ROW][C]100[/C][C] 6[/C][C] 6.646[/C][C]-0.6462[/C][/ROW]
[ROW][C]101[/C][C] 10[/C][C] 7.879[/C][C] 2.121[/C][/ROW]
[ROW][C]102[/C][C] 9[/C][C] 7.273[/C][C] 1.727[/C][/ROW]
[ROW][C]103[/C][C] 10[/C][C] 9.445[/C][C] 0.5549[/C][/ROW]
[ROW][C]104[/C][C] 8[/C][C] 7.684[/C][C] 0.316[/C][/ROW]
[ROW][C]105[/C][C] 4[/C][C] 6.936[/C][C]-2.936[/C][/ROW]
[ROW][C]106[/C][C] 8[/C][C] 9.079[/C][C]-1.079[/C][/ROW]
[ROW][C]107[/C][C] 5[/C][C] 7.577[/C][C]-2.577[/C][/ROW]
[ROW][C]108[/C][C] 8[/C][C] 7.996[/C][C] 0.003505[/C][/ROW]
[ROW][C]109[/C][C] 9[/C][C] 8.257[/C][C] 0.743[/C][/ROW]
[ROW][C]110[/C][C] 8[/C][C] 7.282[/C][C] 0.7177[/C][/ROW]
[ROW][C]111[/C][C] 4[/C][C] 8.777[/C][C]-4.777[/C][/ROW]
[ROW][C]112[/C][C] 8[/C][C] 6.967[/C][C] 1.033[/C][/ROW]
[ROW][C]113[/C][C] 10[/C][C] 8.722[/C][C] 1.278[/C][/ROW]
[ROW][C]114[/C][C] 6[/C][C] 5.379[/C][C] 0.6213[/C][/ROW]
[ROW][C]115[/C][C] 7[/C][C] 6.157[/C][C] 0.8426[/C][/ROW]
[ROW][C]116[/C][C] 10[/C][C] 7.846[/C][C] 2.154[/C][/ROW]
[ROW][C]117[/C][C] 9[/C][C] 8.559[/C][C] 0.441[/C][/ROW]
[ROW][C]118[/C][C] 8[/C][C] 8.43[/C][C]-0.4304[/C][/ROW]
[ROW][C]119[/C][C] 3[/C][C] 6.21[/C][C]-3.21[/C][/ROW]
[ROW][C]120[/C][C] 8[/C][C] 6.72[/C][C] 1.28[/C][/ROW]
[ROW][C]121[/C][C] 7[/C][C] 7.347[/C][C]-0.3469[/C][/ROW]
[ROW][C]122[/C][C] 7[/C][C] 6.256[/C][C] 0.7444[/C][/ROW]
[ROW][C]123[/C][C] 8[/C][C] 5.801[/C][C] 2.199[/C][/ROW]
[ROW][C]124[/C][C] 8[/C][C] 7.877[/C][C] 0.1229[/C][/ROW]
[ROW][C]125[/C][C] 7[/C][C] 7.793[/C][C]-0.7928[/C][/ROW]
[ROW][C]126[/C][C] 7[/C][C] 6.818[/C][C] 0.1819[/C][/ROW]
[ROW][C]127[/C][C] 9[/C][C] 10.02[/C][C]-1.02[/C][/ROW]
[ROW][C]128[/C][C] 9[/C][C] 8.915[/C][C] 0.08514[/C][/ROW]
[ROW][C]129[/C][C] 9[/C][C] 7.357[/C][C] 1.643[/C][/ROW]
[ROW][C]130[/C][C] 4[/C][C] 7.39[/C][C]-3.39[/C][/ROW]
[ROW][C]131[/C][C] 6[/C][C] 7.197[/C][C]-1.197[/C][/ROW]
[ROW][C]132[/C][C] 6[/C][C] 6.311[/C][C]-0.3105[/C][/ROW]
[ROW][C]133[/C][C] 6[/C][C] 5.293[/C][C] 0.707[/C][/ROW]
[ROW][C]134[/C][C] 8[/C][C] 7.391[/C][C] 0.6088[/C][/ROW]
[ROW][C]135[/C][C] 3[/C][C] 6.449[/C][C]-3.449[/C][/ROW]
[ROW][C]136[/C][C] 8[/C][C] 7.12[/C][C] 0.88[/C][/ROW]
[ROW][C]137[/C][C] 8[/C][C] 8.539[/C][C]-0.5387[/C][/ROW]
[ROW][C]138[/C][C] 6[/C][C] 5.445[/C][C] 0.5553[/C][/ROW]
[ROW][C]139[/C][C] 10[/C][C] 8.668[/C][C] 1.332[/C][/ROW]
[ROW][C]140[/C][C] 2[/C][C] 5.436[/C][C]-3.436[/C][/ROW]
[ROW][C]141[/C][C] 9[/C][C] 8.148[/C][C] 0.8518[/C][/ROW]
[ROW][C]142[/C][C] 6[/C][C] 7.098[/C][C]-1.098[/C][/ROW]
[ROW][C]143[/C][C] 6[/C][C] 8.927[/C][C]-2.927[/C][/ROW]
[ROW][C]144[/C][C] 5[/C][C] 4.978[/C][C] 0.02246[/C][/ROW]
[ROW][C]145[/C][C] 4[/C][C] 5.216[/C][C]-1.216[/C][/ROW]
[ROW][C]146[/C][C] 7[/C][C] 6.116[/C][C] 0.884[/C][/ROW]
[ROW][C]147[/C][C] 5[/C][C] 6.893[/C][C]-1.893[/C][/ROW]
[ROW][C]148[/C][C] 8[/C][C] 7.727[/C][C] 0.2732[/C][/ROW]
[ROW][C]149[/C][C] 6[/C][C] 7.955[/C][C]-1.955[/C][/ROW]
[ROW][C]150[/C][C] 9[/C][C] 6.709[/C][C] 2.291[/C][/ROW]
[ROW][C]151[/C][C] 6[/C][C] 6.57[/C][C]-0.5696[/C][/ROW]
[ROW][C]152[/C][C] 4[/C][C] 5.445[/C][C]-1.445[/C][/ROW]
[ROW][C]153[/C][C] 7[/C][C] 7.65[/C][C]-0.6503[/C][/ROW]
[ROW][C]154[/C][C] 2[/C][C] 4.441[/C][C]-2.441[/C][/ROW]
[ROW][C]155[/C][C] 8[/C][C] 8.225[/C][C]-0.2248[/C][/ROW]
[ROW][C]156[/C][C] 9[/C][C] 7.77[/C][C] 1.23[/C][/ROW]
[ROW][C]157[/C][C] 6[/C][C] 6.343[/C][C]-0.3428[/C][/ROW]
[ROW][C]158[/C][C] 5[/C][C] 6.242[/C][C]-1.242[/C][/ROW]
[ROW][C]159[/C][C] 7[/C][C] 7.088[/C][C]-0.08778[/C][/ROW]
[ROW][C]160[/C][C] 8[/C][C] 6.999[/C][C] 1.001[/C][/ROW]
[ROW][C]161[/C][C] 4[/C][C] 6.418[/C][C]-2.418[/C][/ROW]
[ROW][C]162[/C][C] 9[/C][C] 7.466[/C][C] 1.534[/C][/ROW]
[ROW][C]163[/C][C] 9[/C][C] 9.391[/C][C]-0.3911[/C][/ROW]
[ROW][C]164[/C][C] 9[/C][C] 7.577[/C][C] 1.423[/C][/ROW]
[ROW][C]165[/C][C] 7[/C][C] 6.464[/C][C] 0.5363[/C][/ROW]
[ROW][C]166[/C][C] 5[/C][C] 6.677[/C][C]-1.677[/C][/ROW]
[ROW][C]167[/C][C] 7[/C][C] 8.018[/C][C]-1.018[/C][/ROW]
[ROW][C]168[/C][C] 9[/C][C] 10.08[/C][C]-1.084[/C][/ROW]
[ROW][C]169[/C][C] 8[/C][C] 7.101[/C][C] 0.8987[/C][/ROW]
[ROW][C]170[/C][C] 6[/C][C] 5.639[/C][C] 0.3608[/C][/ROW]
[ROW][C]171[/C][C] 9[/C][C] 8.32[/C][C] 0.6799[/C][/ROW]
[ROW][C]172[/C][C] 8[/C][C] 7.553[/C][C] 0.4466[/C][/ROW]
[ROW][C]173[/C][C] 7[/C][C] 8.269[/C][C]-1.269[/C][/ROW]
[ROW][C]174[/C][C] 7[/C][C] 7.131[/C][C]-0.1306[/C][/ROW]
[ROW][C]175[/C][C] 7[/C][C] 7.773[/C][C]-0.7726[/C][/ROW]
[ROW][C]176[/C][C] 8[/C][C] 7.089[/C][C] 0.9108[/C][/ROW]
[ROW][C]177[/C][C] 10[/C][C] 8.386[/C][C] 1.614[/C][/ROW]
[ROW][C]178[/C][C] 6[/C][C] 8.364[/C][C]-2.364[/C][/ROW]
[ROW][C]179[/C][C] 6[/C][C] 7.454[/C][C]-1.454[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=319053&T=5
Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=319053&T=5
As an alternative you can also use a QR Code:
The GUIDs for individual cells are displayed in the table below:
Multiple Linear Regression - Actuals, Interpolation, and Residuals Time or Index Actuals InterpolationForecast ResidualsPrediction Error 1 10 6.884 3.116 2 8 7.444 0.5563 3 8 7.207 0.7928 4 9 8.246 0.7535 5 5 6.063 -1.063 6 10 9.445 0.5549 7 8 8.472 -0.4718 8 9 8.276 0.7242 9 8 5.651 2.349 10 7 8.441 -1.441 11 10 8.3 1.7 12 10 7.328 2.672 13 9 7.316 1.684 14 4 6.019 -2.019 15 4 7.662 -3.662 16 8 7.986 0.01408 17 9 9.878 -0.8785 18 10 6.464 3.536 19 8 7.393 0.6074 20 5 6.309 -1.309 21 10 8.636 1.364 22 8 7.425 0.5751 23 7 7.12 -0.1201 24 8 7.986 0.01408 25 8 9.955 -1.955 26 9 7.173 1.827 27 8 7.693 0.3069 28 6 6.449 -0.4487 29 8 6.784 1.216 30 8 6.32 1.68 31 5 7.987 -2.987 32 9 8.395 0.6048 33 8 7.586 0.4143 34 8 6.892 1.108 35 8 7.379 0.6209 36 6 6.916 -0.9159 37 6 6.232 -0.2325 38 9 7.802 1.198 39 8 7.998 0.002045 40 9 9.564 -0.5645 41 10 7.909 2.091 42 8 7.845 0.1552 43 8 5.595 2.405 44 7 7.146 -0.1456 45 7 6.678 0.3215 46 10 7.858 2.142 47 8 6.145 1.855 48 7 6.007 0.9929 49 10 6.678 3.322 50 7 7.845 -0.8448 51 7 5.434 1.566 52 9 8.907 0.09333 53 9 10.02 -1.02 54 8 6.645 1.355 55 6 7.402 -1.402 56 8 7.456 0.5443 57 9 7.011 1.989 58 2 4.978 -2.978 59 6 6.568 -0.5682 60 8 7.834 0.1658 61 8 9.079 -1.079 62 7 6.783 0.2171 63 8 6.461 1.539 64 6 6.556 -0.5561 65 10 6.633 3.367 66 10 7.531 2.469 67 10 7.088 2.912 68 8 6.721 1.279 69 8 7.273 0.7268 70 7 7.683 -0.6825 71 10 9.869 0.1307 72 5 6.558 -1.558 73 3 3.207 -0.2068 74 2 4.125 -2.125 75 3 4.957 -1.957 76 4 6.404 -2.404 77 2 5.002 -3.002 78 6 5.131 0.8693 79 8 7.866 0.1335 80 8 7.759 0.2409 81 5 6.407 -1.407 82 10 8.579 1.421 83 9 9.065 -0.0651 84 8 9.217 -1.217 85 9 8.397 0.6033 86 8 7.402 0.5982 87 5 5.897 -0.8968 88 7 7.316 -0.3161 89 9 8.883 0.1174 90 8 7.759 0.2409 91 4 7.089 -3.089 92 7 5.305 1.695 93 8 8.645 -0.6447 94 7 6.612 0.3875 95 7 6.331 0.6692 96 9 7.846 1.154 97 6 5.783 0.2168 98 7 7.478 -0.4783 99 4 6.311 -2.311 100 6 6.646 -0.6462 101 10 7.879 2.121 102 9 7.273 1.727 103 10 9.445 0.5549 104 8 7.684 0.316 105 4 6.936 -2.936 106 8 9.079 -1.079 107 5 7.577 -2.577 108 8 7.996 0.003505 109 9 8.257 0.743 110 8 7.282 0.7177 111 4 8.777 -4.777 112 8 6.967 1.033 113 10 8.722 1.278 114 6 5.379 0.6213 115 7 6.157 0.8426 116 10 7.846 2.154 117 9 8.559 0.441 118 8 8.43 -0.4304 119 3 6.21 -3.21 120 8 6.72 1.28 121 7 7.347 -0.3469 122 7 6.256 0.7444 123 8 5.801 2.199 124 8 7.877 0.1229 125 7 7.793 -0.7928 126 7 6.818 0.1819 127 9 10.02 -1.02 128 9 8.915 0.08514 129 9 7.357 1.643 130 4 7.39 -3.39 131 6 7.197 -1.197 132 6 6.311 -0.3105 133 6 5.293 0.707 134 8 7.391 0.6088 135 3 6.449 -3.449 136 8 7.12 0.88 137 8 8.539 -0.5387 138 6 5.445 0.5553 139 10 8.668 1.332 140 2 5.436 -3.436 141 9 8.148 0.8518 142 6 7.098 -1.098 143 6 8.927 -2.927 144 5 4.978 0.02246 145 4 5.216 -1.216 146 7 6.116 0.884 147 5 6.893 -1.893 148 8 7.727 0.2732 149 6 7.955 -1.955 150 9 6.709 2.291 151 6 6.57 -0.5696 152 4 5.445 -1.445 153 7 7.65 -0.6503 154 2 4.441 -2.441 155 8 8.225 -0.2248 156 9 7.77 1.23 157 6 6.343 -0.3428 158 5 6.242 -1.242 159 7 7.088 -0.08778 160 8 6.999 1.001 161 4 6.418 -2.418 162 9 7.466 1.534 163 9 9.391 -0.3911 164 9 7.577 1.423 165 7 6.464 0.5363 166 5 6.677 -1.677 167 7 8.018 -1.018 168 9 10.08 -1.084 169 8 7.101 0.8987 170 6 5.639 0.3608 171 9 8.32 0.6799 172 8 7.553 0.4466 173 7 8.269 -1.269 174 7 7.131 -0.1306 175 7 7.773 -0.7726 176 8 7.089 0.9108 177 10 8.386 1.614 178 6 8.364 -2.364 179 6 7.454 -1.454
Goldfeld-Quandt test for Heteroskedasticity p-values Alternative Hypothesis breakpoint index greater 2-sided less 7 0.7195 0.561 0.2805 8 0.5766 0.8468 0.4234 9 0.5462 0.9077 0.4538 10 0.5939 0.8122 0.4061 11 0.5434 0.9132 0.4566 12 0.54 0.92 0.46 13 0.4566 0.9133 0.5434 14 0.7585 0.483 0.2415 15 0.9561 0.08785 0.04392 16 0.9346 0.1307 0.06535 17 0.9093 0.1814 0.09071 18 0.9511 0.09786 0.04893 19 0.9303 0.1393 0.06965 20 0.9225 0.1549 0.07747 21 0.9066 0.1869 0.09344 22 0.8771 0.2459 0.1229 23 0.8398 0.3204 0.1602 24 0.7972 0.4056 0.2028 25 0.8118 0.3765 0.1882 26 0.8319 0.3362 0.1681 27 0.7902 0.4195 0.2098 28 0.7542 0.4915 0.2458 29 0.7204 0.5592 0.2796 30 0.7057 0.5887 0.2943 31 0.838 0.324 0.162 32 0.8054 0.3891 0.1946 33 0.7644 0.4711 0.2356 34 0.7265 0.547 0.2735 35 0.68 0.6399 0.32 36 0.6653 0.6695 0.3347 37 0.6301 0.7398 0.3699 38 0.6003 0.7994 0.3997 39 0.5475 0.905 0.4525 40 0.4962 0.9925 0.5038 41 0.5213 0.9573 0.4787 42 0.469 0.938 0.531 43 0.4787 0.9575 0.5213 44 0.4295 0.859 0.5705 45 0.3823 0.7647 0.6177 46 0.4278 0.8556 0.5722 47 0.4095 0.819 0.5905 48 0.3689 0.7378 0.6311 49 0.4886 0.9772 0.5114 50 0.4629 0.9257 0.5371 51 0.4356 0.8713 0.5644 52 0.3902 0.7804 0.6098 53 0.3529 0.7058 0.6471 54 0.3263 0.6527 0.6737 55 0.3369 0.6737 0.6631 56 0.2973 0.5945 0.7027 57 0.3027 0.6054 0.6973 58 0.5662 0.8676 0.4338 59 0.5376 0.9248 0.4624 60 0.4922 0.9844 0.5078 61 0.4662 0.9324 0.5338 62 0.4219 0.8438 0.5781 63 0.4087 0.8173 0.5913 64 0.3791 0.7583 0.6209 65 0.5217 0.9566 0.4783 66 0.5805 0.839 0.4195 67 0.6735 0.6531 0.3265 68 0.6541 0.6918 0.3459 69 0.6202 0.7596 0.3798 70 0.5929 0.8143 0.4071 71 0.5511 0.8979 0.4489 72 0.5794 0.8412 0.4206 73 0.573 0.854 0.427 74 0.6555 0.6889 0.3445 75 0.6942 0.6116 0.3058 76 0.7515 0.4971 0.2485 77 0.8409 0.3183 0.1591 78 0.8222 0.3556 0.1778 79 0.7939 0.4122 0.2061 80 0.7636 0.4729 0.2364 81 0.7626 0.4748 0.2374 82 0.7567 0.4866 0.2433 83 0.7223 0.5555 0.2777 84 0.7084 0.5831 0.2916 85 0.6767 0.6466 0.3233 86 0.6434 0.7133 0.3566 87 0.6145 0.7709 0.3855 88 0.5764 0.8473 0.4236 89 0.5347 0.9306 0.4653 90 0.4944 0.9889 0.5056 91 0.625 0.75 0.375 92 0.6341 0.7317 0.3659 93 0.5987 0.8026 0.4013 94 0.56 0.8799 0.44 95 0.5273 0.9454 0.4727 96 0.5097 0.9805 0.4903 97 0.4737 0.9474 0.5263 98 0.4347 0.8694 0.5653 99 0.4835 0.9671 0.5165 100 0.4478 0.8955 0.5522 101 0.4907 0.9813 0.5093 102 0.5104 0.9792 0.4896 103 0.4742 0.9485 0.5258 104 0.4379 0.8758 0.5621 105 0.5456 0.9088 0.4544 106 0.521 0.9581 0.479 107 0.5867 0.8266 0.4133 108 0.5432 0.9136 0.4568 109 0.5129 0.9743 0.4871 110 0.4799 0.9598 0.5201 111 0.8044 0.3912 0.1956 112 0.791 0.4181 0.209 113 0.7869 0.4263 0.2131 114 0.7607 0.4787 0.2393 115 0.741 0.518 0.259 116 0.7843 0.4313 0.2157 117 0.7547 0.4906 0.2453 118 0.7189 0.5621 0.2811 119 0.8239 0.3522 0.1761 120 0.8229 0.3542 0.1771 121 0.7915 0.417 0.2085 122 0.7697 0.4605 0.2303 123 0.8249 0.3502 0.1751 124 0.7942 0.4116 0.2058 125 0.7642 0.4716 0.2358 126 0.7305 0.5389 0.2695 127 0.706 0.588 0.294 128 0.6636 0.6728 0.3364 129 0.6895 0.6209 0.3105 130 0.8169 0.3662 0.1831 131 0.7962 0.4076 0.2038 132 0.7598 0.4804 0.2402 133 0.7456 0.5087 0.2544 134 0.7177 0.5647 0.2823 135 0.8469 0.3062 0.1531 136 0.8327 0.3345 0.1673 137 0.8004 0.3991 0.1996 138 0.7818 0.4364 0.2182 139 0.7811 0.4377 0.2189 140 0.8834 0.2331 0.1166 141 0.8754 0.2493 0.1246 142 0.8506 0.2987 0.1494 143 0.9197 0.1606 0.08029 144 0.8983 0.2034 0.1017 145 0.8773 0.2454 0.1227 146 0.8688 0.2624 0.1312 147 0.8798 0.2405 0.1202 148 0.8491 0.3018 0.1509 149 0.8706 0.2588 0.1294 150 0.9304 0.1392 0.0696 151 0.9061 0.1878 0.09389 152 0.8872 0.2256 0.1128 153 0.8551 0.2899 0.1449 154 0.8981 0.2038 0.1019 155 0.8627 0.2745 0.1373 156 0.8644 0.2713 0.1356 157 0.8218 0.3564 0.1782 158 0.8112 0.3776 0.1888 159 0.7541 0.4918 0.2459 160 0.7298 0.5403 0.2702 161 0.8397 0.3206 0.1603 162 0.8694 0.2613 0.1306 163 0.8536 0.2927 0.1464 164 0.8356 0.3287 0.1644 165 0.7678 0.4644 0.2322 166 0.8112 0.3775 0.1888 167 0.778 0.4439 0.222 168 0.6855 0.6289 0.3145 169 0.6057 0.7886 0.3943 170 0.4746 0.9492 0.5254 171 0.4236 0.8471 0.5764 172 0.4825 0.9649 0.5175
\begin{tabular}{lllllllll}
\hline
Goldfeld-Quandt test for Heteroskedasticity \tabularnewline
p-values & Alternative Hypothesis \tabularnewline
breakpoint index & greater & 2-sided & less \tabularnewline
7 & 0.7195 & 0.561 & 0.2805 \tabularnewline
8 & 0.5766 & 0.8468 & 0.4234 \tabularnewline
9 & 0.5462 & 0.9077 & 0.4538 \tabularnewline
10 & 0.5939 & 0.8122 & 0.4061 \tabularnewline
11 & 0.5434 & 0.9132 & 0.4566 \tabularnewline
12 & 0.54 & 0.92 & 0.46 \tabularnewline
13 & 0.4566 & 0.9133 & 0.5434 \tabularnewline
14 & 0.7585 & 0.483 & 0.2415 \tabularnewline
15 & 0.9561 & 0.08785 & 0.04392 \tabularnewline
16 & 0.9346 & 0.1307 & 0.06535 \tabularnewline
17 & 0.9093 & 0.1814 & 0.09071 \tabularnewline
18 & 0.9511 & 0.09786 & 0.04893 \tabularnewline
19 & 0.9303 & 0.1393 & 0.06965 \tabularnewline
20 & 0.9225 & 0.1549 & 0.07747 \tabularnewline
21 & 0.9066 & 0.1869 & 0.09344 \tabularnewline
22 & 0.8771 & 0.2459 & 0.1229 \tabularnewline
23 & 0.8398 & 0.3204 & 0.1602 \tabularnewline
24 & 0.7972 & 0.4056 & 0.2028 \tabularnewline
25 & 0.8118 & 0.3765 & 0.1882 \tabularnewline
26 & 0.8319 & 0.3362 & 0.1681 \tabularnewline
27 & 0.7902 & 0.4195 & 0.2098 \tabularnewline
28 & 0.7542 & 0.4915 & 0.2458 \tabularnewline
29 & 0.7204 & 0.5592 & 0.2796 \tabularnewline
30 & 0.7057 & 0.5887 & 0.2943 \tabularnewline
31 & 0.838 & 0.324 & 0.162 \tabularnewline
32 & 0.8054 & 0.3891 & 0.1946 \tabularnewline
33 & 0.7644 & 0.4711 & 0.2356 \tabularnewline
34 & 0.7265 & 0.547 & 0.2735 \tabularnewline
35 & 0.68 & 0.6399 & 0.32 \tabularnewline
36 & 0.6653 & 0.6695 & 0.3347 \tabularnewline
37 & 0.6301 & 0.7398 & 0.3699 \tabularnewline
38 & 0.6003 & 0.7994 & 0.3997 \tabularnewline
39 & 0.5475 & 0.905 & 0.4525 \tabularnewline
40 & 0.4962 & 0.9925 & 0.5038 \tabularnewline
41 & 0.5213 & 0.9573 & 0.4787 \tabularnewline
42 & 0.469 & 0.938 & 0.531 \tabularnewline
43 & 0.4787 & 0.9575 & 0.5213 \tabularnewline
44 & 0.4295 & 0.859 & 0.5705 \tabularnewline
45 & 0.3823 & 0.7647 & 0.6177 \tabularnewline
46 & 0.4278 & 0.8556 & 0.5722 \tabularnewline
47 & 0.4095 & 0.819 & 0.5905 \tabularnewline
48 & 0.3689 & 0.7378 & 0.6311 \tabularnewline
49 & 0.4886 & 0.9772 & 0.5114 \tabularnewline
50 & 0.4629 & 0.9257 & 0.5371 \tabularnewline
51 & 0.4356 & 0.8713 & 0.5644 \tabularnewline
52 & 0.3902 & 0.7804 & 0.6098 \tabularnewline
53 & 0.3529 & 0.7058 & 0.6471 \tabularnewline
54 & 0.3263 & 0.6527 & 0.6737 \tabularnewline
55 & 0.3369 & 0.6737 & 0.6631 \tabularnewline
56 & 0.2973 & 0.5945 & 0.7027 \tabularnewline
57 & 0.3027 & 0.6054 & 0.6973 \tabularnewline
58 & 0.5662 & 0.8676 & 0.4338 \tabularnewline
59 & 0.5376 & 0.9248 & 0.4624 \tabularnewline
60 & 0.4922 & 0.9844 & 0.5078 \tabularnewline
61 & 0.4662 & 0.9324 & 0.5338 \tabularnewline
62 & 0.4219 & 0.8438 & 0.5781 \tabularnewline
63 & 0.4087 & 0.8173 & 0.5913 \tabularnewline
64 & 0.3791 & 0.7583 & 0.6209 \tabularnewline
65 & 0.5217 & 0.9566 & 0.4783 \tabularnewline
66 & 0.5805 & 0.839 & 0.4195 \tabularnewline
67 & 0.6735 & 0.6531 & 0.3265 \tabularnewline
68 & 0.6541 & 0.6918 & 0.3459 \tabularnewline
69 & 0.6202 & 0.7596 & 0.3798 \tabularnewline
70 & 0.5929 & 0.8143 & 0.4071 \tabularnewline
71 & 0.5511 & 0.8979 & 0.4489 \tabularnewline
72 & 0.5794 & 0.8412 & 0.4206 \tabularnewline
73 & 0.573 & 0.854 & 0.427 \tabularnewline
74 & 0.6555 & 0.6889 & 0.3445 \tabularnewline
75 & 0.6942 & 0.6116 & 0.3058 \tabularnewline
76 & 0.7515 & 0.4971 & 0.2485 \tabularnewline
77 & 0.8409 & 0.3183 & 0.1591 \tabularnewline
78 & 0.8222 & 0.3556 & 0.1778 \tabularnewline
79 & 0.7939 & 0.4122 & 0.2061 \tabularnewline
80 & 0.7636 & 0.4729 & 0.2364 \tabularnewline
81 & 0.7626 & 0.4748 & 0.2374 \tabularnewline
82 & 0.7567 & 0.4866 & 0.2433 \tabularnewline
83 & 0.7223 & 0.5555 & 0.2777 \tabularnewline
84 & 0.7084 & 0.5831 & 0.2916 \tabularnewline
85 & 0.6767 & 0.6466 & 0.3233 \tabularnewline
86 & 0.6434 & 0.7133 & 0.3566 \tabularnewline
87 & 0.6145 & 0.7709 & 0.3855 \tabularnewline
88 & 0.5764 & 0.8473 & 0.4236 \tabularnewline
89 & 0.5347 & 0.9306 & 0.4653 \tabularnewline
90 & 0.4944 & 0.9889 & 0.5056 \tabularnewline
91 & 0.625 & 0.75 & 0.375 \tabularnewline
92 & 0.6341 & 0.7317 & 0.3659 \tabularnewline
93 & 0.5987 & 0.8026 & 0.4013 \tabularnewline
94 & 0.56 & 0.8799 & 0.44 \tabularnewline
95 & 0.5273 & 0.9454 & 0.4727 \tabularnewline
96 & 0.5097 & 0.9805 & 0.4903 \tabularnewline
97 & 0.4737 & 0.9474 & 0.5263 \tabularnewline
98 & 0.4347 & 0.8694 & 0.5653 \tabularnewline
99 & 0.4835 & 0.9671 & 0.5165 \tabularnewline
100 & 0.4478 & 0.8955 & 0.5522 \tabularnewline
101 & 0.4907 & 0.9813 & 0.5093 \tabularnewline
102 & 0.5104 & 0.9792 & 0.4896 \tabularnewline
103 & 0.4742 & 0.9485 & 0.5258 \tabularnewline
104 & 0.4379 & 0.8758 & 0.5621 \tabularnewline
105 & 0.5456 & 0.9088 & 0.4544 \tabularnewline
106 & 0.521 & 0.9581 & 0.479 \tabularnewline
107 & 0.5867 & 0.8266 & 0.4133 \tabularnewline
108 & 0.5432 & 0.9136 & 0.4568 \tabularnewline
109 & 0.5129 & 0.9743 & 0.4871 \tabularnewline
110 & 0.4799 & 0.9598 & 0.5201 \tabularnewline
111 & 0.8044 & 0.3912 & 0.1956 \tabularnewline
112 & 0.791 & 0.4181 & 0.209 \tabularnewline
113 & 0.7869 & 0.4263 & 0.2131 \tabularnewline
114 & 0.7607 & 0.4787 & 0.2393 \tabularnewline
115 & 0.741 & 0.518 & 0.259 \tabularnewline
116 & 0.7843 & 0.4313 & 0.2157 \tabularnewline
117 & 0.7547 & 0.4906 & 0.2453 \tabularnewline
118 & 0.7189 & 0.5621 & 0.2811 \tabularnewline
119 & 0.8239 & 0.3522 & 0.1761 \tabularnewline
120 & 0.8229 & 0.3542 & 0.1771 \tabularnewline
121 & 0.7915 & 0.417 & 0.2085 \tabularnewline
122 & 0.7697 & 0.4605 & 0.2303 \tabularnewline
123 & 0.8249 & 0.3502 & 0.1751 \tabularnewline
124 & 0.7942 & 0.4116 & 0.2058 \tabularnewline
125 & 0.7642 & 0.4716 & 0.2358 \tabularnewline
126 & 0.7305 & 0.5389 & 0.2695 \tabularnewline
127 & 0.706 & 0.588 & 0.294 \tabularnewline
128 & 0.6636 & 0.6728 & 0.3364 \tabularnewline
129 & 0.6895 & 0.6209 & 0.3105 \tabularnewline
130 & 0.8169 & 0.3662 & 0.1831 \tabularnewline
131 & 0.7962 & 0.4076 & 0.2038 \tabularnewline
132 & 0.7598 & 0.4804 & 0.2402 \tabularnewline
133 & 0.7456 & 0.5087 & 0.2544 \tabularnewline
134 & 0.7177 & 0.5647 & 0.2823 \tabularnewline
135 & 0.8469 & 0.3062 & 0.1531 \tabularnewline
136 & 0.8327 & 0.3345 & 0.1673 \tabularnewline
137 & 0.8004 & 0.3991 & 0.1996 \tabularnewline
138 & 0.7818 & 0.4364 & 0.2182 \tabularnewline
139 & 0.7811 & 0.4377 & 0.2189 \tabularnewline
140 & 0.8834 & 0.2331 & 0.1166 \tabularnewline
141 & 0.8754 & 0.2493 & 0.1246 \tabularnewline
142 & 0.8506 & 0.2987 & 0.1494 \tabularnewline
143 & 0.9197 & 0.1606 & 0.08029 \tabularnewline
144 & 0.8983 & 0.2034 & 0.1017 \tabularnewline
145 & 0.8773 & 0.2454 & 0.1227 \tabularnewline
146 & 0.8688 & 0.2624 & 0.1312 \tabularnewline
147 & 0.8798 & 0.2405 & 0.1202 \tabularnewline
148 & 0.8491 & 0.3018 & 0.1509 \tabularnewline
149 & 0.8706 & 0.2588 & 0.1294 \tabularnewline
150 & 0.9304 & 0.1392 & 0.0696 \tabularnewline
151 & 0.9061 & 0.1878 & 0.09389 \tabularnewline
152 & 0.8872 & 0.2256 & 0.1128 \tabularnewline
153 & 0.8551 & 0.2899 & 0.1449 \tabularnewline
154 & 0.8981 & 0.2038 & 0.1019 \tabularnewline
155 & 0.8627 & 0.2745 & 0.1373 \tabularnewline
156 & 0.8644 & 0.2713 & 0.1356 \tabularnewline
157 & 0.8218 & 0.3564 & 0.1782 \tabularnewline
158 & 0.8112 & 0.3776 & 0.1888 \tabularnewline
159 & 0.7541 & 0.4918 & 0.2459 \tabularnewline
160 & 0.7298 & 0.5403 & 0.2702 \tabularnewline
161 & 0.8397 & 0.3206 & 0.1603 \tabularnewline
162 & 0.8694 & 0.2613 & 0.1306 \tabularnewline
163 & 0.8536 & 0.2927 & 0.1464 \tabularnewline
164 & 0.8356 & 0.3287 & 0.1644 \tabularnewline
165 & 0.7678 & 0.4644 & 0.2322 \tabularnewline
166 & 0.8112 & 0.3775 & 0.1888 \tabularnewline
167 & 0.778 & 0.4439 & 0.222 \tabularnewline
168 & 0.6855 & 0.6289 & 0.3145 \tabularnewline
169 & 0.6057 & 0.7886 & 0.3943 \tabularnewline
170 & 0.4746 & 0.9492 & 0.5254 \tabularnewline
171 & 0.4236 & 0.8471 & 0.5764 \tabularnewline
172 & 0.4825 & 0.9649 & 0.5175 \tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=319053&T=6
[TABLE]
[ROW][C]Goldfeld-Quandt test for Heteroskedasticity[/C][/ROW]
[ROW][C]p-values[/C][C]Alternative Hypothesis[/C][/ROW]
[ROW][C]breakpoint index[/C][C]greater[/C][C]2-sided[/C][C]less[/C][/ROW]
[ROW][C]7[/C][C] 0.7195[/C][C] 0.561[/C][C] 0.2805[/C][/ROW]
[ROW][C]8[/C][C] 0.5766[/C][C] 0.8468[/C][C] 0.4234[/C][/ROW]
[ROW][C]9[/C][C] 0.5462[/C][C] 0.9077[/C][C] 0.4538[/C][/ROW]
[ROW][C]10[/C][C] 0.5939[/C][C] 0.8122[/C][C] 0.4061[/C][/ROW]
[ROW][C]11[/C][C] 0.5434[/C][C] 0.9132[/C][C] 0.4566[/C][/ROW]
[ROW][C]12[/C][C] 0.54[/C][C] 0.92[/C][C] 0.46[/C][/ROW]
[ROW][C]13[/C][C] 0.4566[/C][C] 0.9133[/C][C] 0.5434[/C][/ROW]
[ROW][C]14[/C][C] 0.7585[/C][C] 0.483[/C][C] 0.2415[/C][/ROW]
[ROW][C]15[/C][C] 0.9561[/C][C] 0.08785[/C][C] 0.04392[/C][/ROW]
[ROW][C]16[/C][C] 0.9346[/C][C] 0.1307[/C][C] 0.06535[/C][/ROW]
[ROW][C]17[/C][C] 0.9093[/C][C] 0.1814[/C][C] 0.09071[/C][/ROW]
[ROW][C]18[/C][C] 0.9511[/C][C] 0.09786[/C][C] 0.04893[/C][/ROW]
[ROW][C]19[/C][C] 0.9303[/C][C] 0.1393[/C][C] 0.06965[/C][/ROW]
[ROW][C]20[/C][C] 0.9225[/C][C] 0.1549[/C][C] 0.07747[/C][/ROW]
[ROW][C]21[/C][C] 0.9066[/C][C] 0.1869[/C][C] 0.09344[/C][/ROW]
[ROW][C]22[/C][C] 0.8771[/C][C] 0.2459[/C][C] 0.1229[/C][/ROW]
[ROW][C]23[/C][C] 0.8398[/C][C] 0.3204[/C][C] 0.1602[/C][/ROW]
[ROW][C]24[/C][C] 0.7972[/C][C] 0.4056[/C][C] 0.2028[/C][/ROW]
[ROW][C]25[/C][C] 0.8118[/C][C] 0.3765[/C][C] 0.1882[/C][/ROW]
[ROW][C]26[/C][C] 0.8319[/C][C] 0.3362[/C][C] 0.1681[/C][/ROW]
[ROW][C]27[/C][C] 0.7902[/C][C] 0.4195[/C][C] 0.2098[/C][/ROW]
[ROW][C]28[/C][C] 0.7542[/C][C] 0.4915[/C][C] 0.2458[/C][/ROW]
[ROW][C]29[/C][C] 0.7204[/C][C] 0.5592[/C][C] 0.2796[/C][/ROW]
[ROW][C]30[/C][C] 0.7057[/C][C] 0.5887[/C][C] 0.2943[/C][/ROW]
[ROW][C]31[/C][C] 0.838[/C][C] 0.324[/C][C] 0.162[/C][/ROW]
[ROW][C]32[/C][C] 0.8054[/C][C] 0.3891[/C][C] 0.1946[/C][/ROW]
[ROW][C]33[/C][C] 0.7644[/C][C] 0.4711[/C][C] 0.2356[/C][/ROW]
[ROW][C]34[/C][C] 0.7265[/C][C] 0.547[/C][C] 0.2735[/C][/ROW]
[ROW][C]35[/C][C] 0.68[/C][C] 0.6399[/C][C] 0.32[/C][/ROW]
[ROW][C]36[/C][C] 0.6653[/C][C] 0.6695[/C][C] 0.3347[/C][/ROW]
[ROW][C]37[/C][C] 0.6301[/C][C] 0.7398[/C][C] 0.3699[/C][/ROW]
[ROW][C]38[/C][C] 0.6003[/C][C] 0.7994[/C][C] 0.3997[/C][/ROW]
[ROW][C]39[/C][C] 0.5475[/C][C] 0.905[/C][C] 0.4525[/C][/ROW]
[ROW][C]40[/C][C] 0.4962[/C][C] 0.9925[/C][C] 0.5038[/C][/ROW]
[ROW][C]41[/C][C] 0.5213[/C][C] 0.9573[/C][C] 0.4787[/C][/ROW]
[ROW][C]42[/C][C] 0.469[/C][C] 0.938[/C][C] 0.531[/C][/ROW]
[ROW][C]43[/C][C] 0.4787[/C][C] 0.9575[/C][C] 0.5213[/C][/ROW]
[ROW][C]44[/C][C] 0.4295[/C][C] 0.859[/C][C] 0.5705[/C][/ROW]
[ROW][C]45[/C][C] 0.3823[/C][C] 0.7647[/C][C] 0.6177[/C][/ROW]
[ROW][C]46[/C][C] 0.4278[/C][C] 0.8556[/C][C] 0.5722[/C][/ROW]
[ROW][C]47[/C][C] 0.4095[/C][C] 0.819[/C][C] 0.5905[/C][/ROW]
[ROW][C]48[/C][C] 0.3689[/C][C] 0.7378[/C][C] 0.6311[/C][/ROW]
[ROW][C]49[/C][C] 0.4886[/C][C] 0.9772[/C][C] 0.5114[/C][/ROW]
[ROW][C]50[/C][C] 0.4629[/C][C] 0.9257[/C][C] 0.5371[/C][/ROW]
[ROW][C]51[/C][C] 0.4356[/C][C] 0.8713[/C][C] 0.5644[/C][/ROW]
[ROW][C]52[/C][C] 0.3902[/C][C] 0.7804[/C][C] 0.6098[/C][/ROW]
[ROW][C]53[/C][C] 0.3529[/C][C] 0.7058[/C][C] 0.6471[/C][/ROW]
[ROW][C]54[/C][C] 0.3263[/C][C] 0.6527[/C][C] 0.6737[/C][/ROW]
[ROW][C]55[/C][C] 0.3369[/C][C] 0.6737[/C][C] 0.6631[/C][/ROW]
[ROW][C]56[/C][C] 0.2973[/C][C] 0.5945[/C][C] 0.7027[/C][/ROW]
[ROW][C]57[/C][C] 0.3027[/C][C] 0.6054[/C][C] 0.6973[/C][/ROW]
[ROW][C]58[/C][C] 0.5662[/C][C] 0.8676[/C][C] 0.4338[/C][/ROW]
[ROW][C]59[/C][C] 0.5376[/C][C] 0.9248[/C][C] 0.4624[/C][/ROW]
[ROW][C]60[/C][C] 0.4922[/C][C] 0.9844[/C][C] 0.5078[/C][/ROW]
[ROW][C]61[/C][C] 0.4662[/C][C] 0.9324[/C][C] 0.5338[/C][/ROW]
[ROW][C]62[/C][C] 0.4219[/C][C] 0.8438[/C][C] 0.5781[/C][/ROW]
[ROW][C]63[/C][C] 0.4087[/C][C] 0.8173[/C][C] 0.5913[/C][/ROW]
[ROW][C]64[/C][C] 0.3791[/C][C] 0.7583[/C][C] 0.6209[/C][/ROW]
[ROW][C]65[/C][C] 0.5217[/C][C] 0.9566[/C][C] 0.4783[/C][/ROW]
[ROW][C]66[/C][C] 0.5805[/C][C] 0.839[/C][C] 0.4195[/C][/ROW]
[ROW][C]67[/C][C] 0.6735[/C][C] 0.6531[/C][C] 0.3265[/C][/ROW]
[ROW][C]68[/C][C] 0.6541[/C][C] 0.6918[/C][C] 0.3459[/C][/ROW]
[ROW][C]69[/C][C] 0.6202[/C][C] 0.7596[/C][C] 0.3798[/C][/ROW]
[ROW][C]70[/C][C] 0.5929[/C][C] 0.8143[/C][C] 0.4071[/C][/ROW]
[ROW][C]71[/C][C] 0.5511[/C][C] 0.8979[/C][C] 0.4489[/C][/ROW]
[ROW][C]72[/C][C] 0.5794[/C][C] 0.8412[/C][C] 0.4206[/C][/ROW]
[ROW][C]73[/C][C] 0.573[/C][C] 0.854[/C][C] 0.427[/C][/ROW]
[ROW][C]74[/C][C] 0.6555[/C][C] 0.6889[/C][C] 0.3445[/C][/ROW]
[ROW][C]75[/C][C] 0.6942[/C][C] 0.6116[/C][C] 0.3058[/C][/ROW]
[ROW][C]76[/C][C] 0.7515[/C][C] 0.4971[/C][C] 0.2485[/C][/ROW]
[ROW][C]77[/C][C] 0.8409[/C][C] 0.3183[/C][C] 0.1591[/C][/ROW]
[ROW][C]78[/C][C] 0.8222[/C][C] 0.3556[/C][C] 0.1778[/C][/ROW]
[ROW][C]79[/C][C] 0.7939[/C][C] 0.4122[/C][C] 0.2061[/C][/ROW]
[ROW][C]80[/C][C] 0.7636[/C][C] 0.4729[/C][C] 0.2364[/C][/ROW]
[ROW][C]81[/C][C] 0.7626[/C][C] 0.4748[/C][C] 0.2374[/C][/ROW]
[ROW][C]82[/C][C] 0.7567[/C][C] 0.4866[/C][C] 0.2433[/C][/ROW]
[ROW][C]83[/C][C] 0.7223[/C][C] 0.5555[/C][C] 0.2777[/C][/ROW]
[ROW][C]84[/C][C] 0.7084[/C][C] 0.5831[/C][C] 0.2916[/C][/ROW]
[ROW][C]85[/C][C] 0.6767[/C][C] 0.6466[/C][C] 0.3233[/C][/ROW]
[ROW][C]86[/C][C] 0.6434[/C][C] 0.7133[/C][C] 0.3566[/C][/ROW]
[ROW][C]87[/C][C] 0.6145[/C][C] 0.7709[/C][C] 0.3855[/C][/ROW]
[ROW][C]88[/C][C] 0.5764[/C][C] 0.8473[/C][C] 0.4236[/C][/ROW]
[ROW][C]89[/C][C] 0.5347[/C][C] 0.9306[/C][C] 0.4653[/C][/ROW]
[ROW][C]90[/C][C] 0.4944[/C][C] 0.9889[/C][C] 0.5056[/C][/ROW]
[ROW][C]91[/C][C] 0.625[/C][C] 0.75[/C][C] 0.375[/C][/ROW]
[ROW][C]92[/C][C] 0.6341[/C][C] 0.7317[/C][C] 0.3659[/C][/ROW]
[ROW][C]93[/C][C] 0.5987[/C][C] 0.8026[/C][C] 0.4013[/C][/ROW]
[ROW][C]94[/C][C] 0.56[/C][C] 0.8799[/C][C] 0.44[/C][/ROW]
[ROW][C]95[/C][C] 0.5273[/C][C] 0.9454[/C][C] 0.4727[/C][/ROW]
[ROW][C]96[/C][C] 0.5097[/C][C] 0.9805[/C][C] 0.4903[/C][/ROW]
[ROW][C]97[/C][C] 0.4737[/C][C] 0.9474[/C][C] 0.5263[/C][/ROW]
[ROW][C]98[/C][C] 0.4347[/C][C] 0.8694[/C][C] 0.5653[/C][/ROW]
[ROW][C]99[/C][C] 0.4835[/C][C] 0.9671[/C][C] 0.5165[/C][/ROW]
[ROW][C]100[/C][C] 0.4478[/C][C] 0.8955[/C][C] 0.5522[/C][/ROW]
[ROW][C]101[/C][C] 0.4907[/C][C] 0.9813[/C][C] 0.5093[/C][/ROW]
[ROW][C]102[/C][C] 0.5104[/C][C] 0.9792[/C][C] 0.4896[/C][/ROW]
[ROW][C]103[/C][C] 0.4742[/C][C] 0.9485[/C][C] 0.5258[/C][/ROW]
[ROW][C]104[/C][C] 0.4379[/C][C] 0.8758[/C][C] 0.5621[/C][/ROW]
[ROW][C]105[/C][C] 0.5456[/C][C] 0.9088[/C][C] 0.4544[/C][/ROW]
[ROW][C]106[/C][C] 0.521[/C][C] 0.9581[/C][C] 0.479[/C][/ROW]
[ROW][C]107[/C][C] 0.5867[/C][C] 0.8266[/C][C] 0.4133[/C][/ROW]
[ROW][C]108[/C][C] 0.5432[/C][C] 0.9136[/C][C] 0.4568[/C][/ROW]
[ROW][C]109[/C][C] 0.5129[/C][C] 0.9743[/C][C] 0.4871[/C][/ROW]
[ROW][C]110[/C][C] 0.4799[/C][C] 0.9598[/C][C] 0.5201[/C][/ROW]
[ROW][C]111[/C][C] 0.8044[/C][C] 0.3912[/C][C] 0.1956[/C][/ROW]
[ROW][C]112[/C][C] 0.791[/C][C] 0.4181[/C][C] 0.209[/C][/ROW]
[ROW][C]113[/C][C] 0.7869[/C][C] 0.4263[/C][C] 0.2131[/C][/ROW]
[ROW][C]114[/C][C] 0.7607[/C][C] 0.4787[/C][C] 0.2393[/C][/ROW]
[ROW][C]115[/C][C] 0.741[/C][C] 0.518[/C][C] 0.259[/C][/ROW]
[ROW][C]116[/C][C] 0.7843[/C][C] 0.4313[/C][C] 0.2157[/C][/ROW]
[ROW][C]117[/C][C] 0.7547[/C][C] 0.4906[/C][C] 0.2453[/C][/ROW]
[ROW][C]118[/C][C] 0.7189[/C][C] 0.5621[/C][C] 0.2811[/C][/ROW]
[ROW][C]119[/C][C] 0.8239[/C][C] 0.3522[/C][C] 0.1761[/C][/ROW]
[ROW][C]120[/C][C] 0.8229[/C][C] 0.3542[/C][C] 0.1771[/C][/ROW]
[ROW][C]121[/C][C] 0.7915[/C][C] 0.417[/C][C] 0.2085[/C][/ROW]
[ROW][C]122[/C][C] 0.7697[/C][C] 0.4605[/C][C] 0.2303[/C][/ROW]
[ROW][C]123[/C][C] 0.8249[/C][C] 0.3502[/C][C] 0.1751[/C][/ROW]
[ROW][C]124[/C][C] 0.7942[/C][C] 0.4116[/C][C] 0.2058[/C][/ROW]
[ROW][C]125[/C][C] 0.7642[/C][C] 0.4716[/C][C] 0.2358[/C][/ROW]
[ROW][C]126[/C][C] 0.7305[/C][C] 0.5389[/C][C] 0.2695[/C][/ROW]
[ROW][C]127[/C][C] 0.706[/C][C] 0.588[/C][C] 0.294[/C][/ROW]
[ROW][C]128[/C][C] 0.6636[/C][C] 0.6728[/C][C] 0.3364[/C][/ROW]
[ROW][C]129[/C][C] 0.6895[/C][C] 0.6209[/C][C] 0.3105[/C][/ROW]
[ROW][C]130[/C][C] 0.8169[/C][C] 0.3662[/C][C] 0.1831[/C][/ROW]
[ROW][C]131[/C][C] 0.7962[/C][C] 0.4076[/C][C] 0.2038[/C][/ROW]
[ROW][C]132[/C][C] 0.7598[/C][C] 0.4804[/C][C] 0.2402[/C][/ROW]
[ROW][C]133[/C][C] 0.7456[/C][C] 0.5087[/C][C] 0.2544[/C][/ROW]
[ROW][C]134[/C][C] 0.7177[/C][C] 0.5647[/C][C] 0.2823[/C][/ROW]
[ROW][C]135[/C][C] 0.8469[/C][C] 0.3062[/C][C] 0.1531[/C][/ROW]
[ROW][C]136[/C][C] 0.8327[/C][C] 0.3345[/C][C] 0.1673[/C][/ROW]
[ROW][C]137[/C][C] 0.8004[/C][C] 0.3991[/C][C] 0.1996[/C][/ROW]
[ROW][C]138[/C][C] 0.7818[/C][C] 0.4364[/C][C] 0.2182[/C][/ROW]
[ROW][C]139[/C][C] 0.7811[/C][C] 0.4377[/C][C] 0.2189[/C][/ROW]
[ROW][C]140[/C][C] 0.8834[/C][C] 0.2331[/C][C] 0.1166[/C][/ROW]
[ROW][C]141[/C][C] 0.8754[/C][C] 0.2493[/C][C] 0.1246[/C][/ROW]
[ROW][C]142[/C][C] 0.8506[/C][C] 0.2987[/C][C] 0.1494[/C][/ROW]
[ROW][C]143[/C][C] 0.9197[/C][C] 0.1606[/C][C] 0.08029[/C][/ROW]
[ROW][C]144[/C][C] 0.8983[/C][C] 0.2034[/C][C] 0.1017[/C][/ROW]
[ROW][C]145[/C][C] 0.8773[/C][C] 0.2454[/C][C] 0.1227[/C][/ROW]
[ROW][C]146[/C][C] 0.8688[/C][C] 0.2624[/C][C] 0.1312[/C][/ROW]
[ROW][C]147[/C][C] 0.8798[/C][C] 0.2405[/C][C] 0.1202[/C][/ROW]
[ROW][C]148[/C][C] 0.8491[/C][C] 0.3018[/C][C] 0.1509[/C][/ROW]
[ROW][C]149[/C][C] 0.8706[/C][C] 0.2588[/C][C] 0.1294[/C][/ROW]
[ROW][C]150[/C][C] 0.9304[/C][C] 0.1392[/C][C] 0.0696[/C][/ROW]
[ROW][C]151[/C][C] 0.9061[/C][C] 0.1878[/C][C] 0.09389[/C][/ROW]
[ROW][C]152[/C][C] 0.8872[/C][C] 0.2256[/C][C] 0.1128[/C][/ROW]
[ROW][C]153[/C][C] 0.8551[/C][C] 0.2899[/C][C] 0.1449[/C][/ROW]
[ROW][C]154[/C][C] 0.8981[/C][C] 0.2038[/C][C] 0.1019[/C][/ROW]
[ROW][C]155[/C][C] 0.8627[/C][C] 0.2745[/C][C] 0.1373[/C][/ROW]
[ROW][C]156[/C][C] 0.8644[/C][C] 0.2713[/C][C] 0.1356[/C][/ROW]
[ROW][C]157[/C][C] 0.8218[/C][C] 0.3564[/C][C] 0.1782[/C][/ROW]
[ROW][C]158[/C][C] 0.8112[/C][C] 0.3776[/C][C] 0.1888[/C][/ROW]
[ROW][C]159[/C][C] 0.7541[/C][C] 0.4918[/C][C] 0.2459[/C][/ROW]
[ROW][C]160[/C][C] 0.7298[/C][C] 0.5403[/C][C] 0.2702[/C][/ROW]
[ROW][C]161[/C][C] 0.8397[/C][C] 0.3206[/C][C] 0.1603[/C][/ROW]
[ROW][C]162[/C][C] 0.8694[/C][C] 0.2613[/C][C] 0.1306[/C][/ROW]
[ROW][C]163[/C][C] 0.8536[/C][C] 0.2927[/C][C] 0.1464[/C][/ROW]
[ROW][C]164[/C][C] 0.8356[/C][C] 0.3287[/C][C] 0.1644[/C][/ROW]
[ROW][C]165[/C][C] 0.7678[/C][C] 0.4644[/C][C] 0.2322[/C][/ROW]
[ROW][C]166[/C][C] 0.8112[/C][C] 0.3775[/C][C] 0.1888[/C][/ROW]
[ROW][C]167[/C][C] 0.778[/C][C] 0.4439[/C][C] 0.222[/C][/ROW]
[ROW][C]168[/C][C] 0.6855[/C][C] 0.6289[/C][C] 0.3145[/C][/ROW]
[ROW][C]169[/C][C] 0.6057[/C][C] 0.7886[/C][C] 0.3943[/C][/ROW]
[ROW][C]170[/C][C] 0.4746[/C][C] 0.9492[/C][C] 0.5254[/C][/ROW]
[ROW][C]171[/C][C] 0.4236[/C][C] 0.8471[/C][C] 0.5764[/C][/ROW]
[ROW][C]172[/C][C] 0.4825[/C][C] 0.9649[/C][C] 0.5175[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=319053&T=6
Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=319053&T=6
As an alternative you can also use a QR Code:
The GUIDs for individual cells are displayed in the table below:
Goldfeld-Quandt test for Heteroskedasticity p-values Alternative Hypothesis breakpoint index greater 2-sided less 7 0.7195 0.561 0.2805 8 0.5766 0.8468 0.4234 9 0.5462 0.9077 0.4538 10 0.5939 0.8122 0.4061 11 0.5434 0.9132 0.4566 12 0.54 0.92 0.46 13 0.4566 0.9133 0.5434 14 0.7585 0.483 0.2415 15 0.9561 0.08785 0.04392 16 0.9346 0.1307 0.06535 17 0.9093 0.1814 0.09071 18 0.9511 0.09786 0.04893 19 0.9303 0.1393 0.06965 20 0.9225 0.1549 0.07747 21 0.9066 0.1869 0.09344 22 0.8771 0.2459 0.1229 23 0.8398 0.3204 0.1602 24 0.7972 0.4056 0.2028 25 0.8118 0.3765 0.1882 26 0.8319 0.3362 0.1681 27 0.7902 0.4195 0.2098 28 0.7542 0.4915 0.2458 29 0.7204 0.5592 0.2796 30 0.7057 0.5887 0.2943 31 0.838 0.324 0.162 32 0.8054 0.3891 0.1946 33 0.7644 0.4711 0.2356 34 0.7265 0.547 0.2735 35 0.68 0.6399 0.32 36 0.6653 0.6695 0.3347 37 0.6301 0.7398 0.3699 38 0.6003 0.7994 0.3997 39 0.5475 0.905 0.4525 40 0.4962 0.9925 0.5038 41 0.5213 0.9573 0.4787 42 0.469 0.938 0.531 43 0.4787 0.9575 0.5213 44 0.4295 0.859 0.5705 45 0.3823 0.7647 0.6177 46 0.4278 0.8556 0.5722 47 0.4095 0.819 0.5905 48 0.3689 0.7378 0.6311 49 0.4886 0.9772 0.5114 50 0.4629 0.9257 0.5371 51 0.4356 0.8713 0.5644 52 0.3902 0.7804 0.6098 53 0.3529 0.7058 0.6471 54 0.3263 0.6527 0.6737 55 0.3369 0.6737 0.6631 56 0.2973 0.5945 0.7027 57 0.3027 0.6054 0.6973 58 0.5662 0.8676 0.4338 59 0.5376 0.9248 0.4624 60 0.4922 0.9844 0.5078 61 0.4662 0.9324 0.5338 62 0.4219 0.8438 0.5781 63 0.4087 0.8173 0.5913 64 0.3791 0.7583 0.6209 65 0.5217 0.9566 0.4783 66 0.5805 0.839 0.4195 67 0.6735 0.6531 0.3265 68 0.6541 0.6918 0.3459 69 0.6202 0.7596 0.3798 70 0.5929 0.8143 0.4071 71 0.5511 0.8979 0.4489 72 0.5794 0.8412 0.4206 73 0.573 0.854 0.427 74 0.6555 0.6889 0.3445 75 0.6942 0.6116 0.3058 76 0.7515 0.4971 0.2485 77 0.8409 0.3183 0.1591 78 0.8222 0.3556 0.1778 79 0.7939 0.4122 0.2061 80 0.7636 0.4729 0.2364 81 0.7626 0.4748 0.2374 82 0.7567 0.4866 0.2433 83 0.7223 0.5555 0.2777 84 0.7084 0.5831 0.2916 85 0.6767 0.6466 0.3233 86 0.6434 0.7133 0.3566 87 0.6145 0.7709 0.3855 88 0.5764 0.8473 0.4236 89 0.5347 0.9306 0.4653 90 0.4944 0.9889 0.5056 91 0.625 0.75 0.375 92 0.6341 0.7317 0.3659 93 0.5987 0.8026 0.4013 94 0.56 0.8799 0.44 95 0.5273 0.9454 0.4727 96 0.5097 0.9805 0.4903 97 0.4737 0.9474 0.5263 98 0.4347 0.8694 0.5653 99 0.4835 0.9671 0.5165 100 0.4478 0.8955 0.5522 101 0.4907 0.9813 0.5093 102 0.5104 0.9792 0.4896 103 0.4742 0.9485 0.5258 104 0.4379 0.8758 0.5621 105 0.5456 0.9088 0.4544 106 0.521 0.9581 0.479 107 0.5867 0.8266 0.4133 108 0.5432 0.9136 0.4568 109 0.5129 0.9743 0.4871 110 0.4799 0.9598 0.5201 111 0.8044 0.3912 0.1956 112 0.791 0.4181 0.209 113 0.7869 0.4263 0.2131 114 0.7607 0.4787 0.2393 115 0.741 0.518 0.259 116 0.7843 0.4313 0.2157 117 0.7547 0.4906 0.2453 118 0.7189 0.5621 0.2811 119 0.8239 0.3522 0.1761 120 0.8229 0.3542 0.1771 121 0.7915 0.417 0.2085 122 0.7697 0.4605 0.2303 123 0.8249 0.3502 0.1751 124 0.7942 0.4116 0.2058 125 0.7642 0.4716 0.2358 126 0.7305 0.5389 0.2695 127 0.706 0.588 0.294 128 0.6636 0.6728 0.3364 129 0.6895 0.6209 0.3105 130 0.8169 0.3662 0.1831 131 0.7962 0.4076 0.2038 132 0.7598 0.4804 0.2402 133 0.7456 0.5087 0.2544 134 0.7177 0.5647 0.2823 135 0.8469 0.3062 0.1531 136 0.8327 0.3345 0.1673 137 0.8004 0.3991 0.1996 138 0.7818 0.4364 0.2182 139 0.7811 0.4377 0.2189 140 0.8834 0.2331 0.1166 141 0.8754 0.2493 0.1246 142 0.8506 0.2987 0.1494 143 0.9197 0.1606 0.08029 144 0.8983 0.2034 0.1017 145 0.8773 0.2454 0.1227 146 0.8688 0.2624 0.1312 147 0.8798 0.2405 0.1202 148 0.8491 0.3018 0.1509 149 0.8706 0.2588 0.1294 150 0.9304 0.1392 0.0696 151 0.9061 0.1878 0.09389 152 0.8872 0.2256 0.1128 153 0.8551 0.2899 0.1449 154 0.8981 0.2038 0.1019 155 0.8627 0.2745 0.1373 156 0.8644 0.2713 0.1356 157 0.8218 0.3564 0.1782 158 0.8112 0.3776 0.1888 159 0.7541 0.4918 0.2459 160 0.7298 0.5403 0.2702 161 0.8397 0.3206 0.1603 162 0.8694 0.2613 0.1306 163 0.8536 0.2927 0.1464 164 0.8356 0.3287 0.1644 165 0.7678 0.4644 0.2322 166 0.8112 0.3775 0.1888 167 0.778 0.4439 0.222 168 0.6855 0.6289 0.3145 169 0.6057 0.7886 0.3943 170 0.4746 0.9492 0.5254 171 0.4236 0.8471 0.5764 172 0.4825 0.9649 0.5175
Meta Analysis of Goldfeld-Quandt test for Heteroskedasticity Description # significant tests % significant tests OK/NOK 1% type I error level 0 0 OK 5% type I error level 0 0 OK 10% type I error level 2 0.0120482 OK
\begin{tabular}{lllllllll}
\hline
Meta Analysis of Goldfeld-Quandt test for Heteroskedasticity \tabularnewline
Description & # significant tests & % significant tests & OK/NOK \tabularnewline
1% type I error level & 0 & 0 & OK \tabularnewline
5% type I error level & 0 & 0 & OK \tabularnewline
10% type I error level & 2 & 0.0120482 & OK \tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=319053&T=7
[TABLE]
[ROW][C]Meta Analysis of Goldfeld-Quandt test for Heteroskedasticity[/C][/ROW]
[ROW][C]Description[/C][C]# significant tests[/C][C]% significant tests[/C][C]OK/NOK[/C][/ROW]
[ROW][C]1% type I error level[/C][C]0[/C][C] 0[/C][C]OK[/C][/ROW]
[ROW][C]5% type I error level[/C][C]0[/C][C]0[/C][C]OK[/C][/ROW]
[ROW][C]10% type I error level[/C][C]2[/C][C]0.0120482[/C][C]OK[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=319053&T=7
Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=319053&T=7
As an alternative you can also use a QR Code:
The GUIDs for individual cells are displayed in the table below:
Meta Analysis of Goldfeld-Quandt test for Heteroskedasticity Description # significant tests % significant tests OK/NOK 1% type I error level 0 0 OK 5% type I error level 0 0 OK 10% type I error level 2 0.0120482 OK
Ramsey RESET F-Test for powers (2 and 3) of fitted values > reset_test_fitted RESET test data: mylm RESET = 4.6851, df1 = 2, df2 = 173, p-value = 0.01043 Ramsey RESET F-Test for powers (2 and 3) of regressors > reset_test_regressors RESET test data: mylm RESET = 2.3794, df1 = 6, df2 = 169, p-value = 0.03117 Ramsey RESET F-Test for powers (2 and 3) of principal components > reset_test_principal_components RESET test data: mylm RESET = 4.6541, df1 = 2, df2 = 173, p-value = 0.01075
\begin{tabular}{lllllllll}
\hline
Ramsey RESET F-Test for powers (2 and 3) of fitted values \tabularnewline
> reset_test_fitted
RESET test
data: mylm
RESET = 4.6851, df1 = 2, df2 = 173, p-value = 0.01043
\tabularnewline
Ramsey RESET F-Test for powers (2 and 3) of regressors \tabularnewline
> reset_test_regressors
RESET test
data: mylm
RESET = 2.3794, df1 = 6, df2 = 169, p-value = 0.03117
\tabularnewline
Ramsey RESET F-Test for powers (2 and 3) of principal components \tabularnewline
> reset_test_principal_components
RESET test
data: mylm
RESET = 4.6541, df1 = 2, df2 = 173, p-value = 0.01075
\tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=319053&T=8
[TABLE]
[ROW][C]Ramsey RESET F-Test for powers (2 and 3) of fitted values[/C][/ROW]
[ROW][C]> reset_test_fitted
RESET test
data: mylm
RESET = 4.6851, df1 = 2, df2 = 173, p-value = 0.01043
[/C][/ROW]
[ROW][C]Ramsey RESET F-Test for powers (2 and 3) of regressors[/C][/ROW]
[ROW][C]> reset_test_regressors
RESET test
data: mylm
RESET = 2.3794, df1 = 6, df2 = 169, p-value = 0.03117
[/C][/ROW]
[ROW][C]Ramsey RESET F-Test for powers (2 and 3) of principal components[/C][/ROW]
[ROW][C]> reset_test_principal_components
RESET test
data: mylm
RESET = 4.6541, df1 = 2, df2 = 173, p-value = 0.01075
[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=319053&T=8
Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=319053&T=8
As an alternative you can also use a QR Code:
The GUIDs for individual cells are displayed in the table below:
Ramsey RESET F-Test for powers (2 and 3) of fitted values > reset_test_fitted RESET test data: mylm RESET = 4.6851, df1 = 2, df2 = 173, p-value = 0.01043 Ramsey RESET F-Test for powers (2 and 3) of regressors > reset_test_regressors RESET test data: mylm RESET = 2.3794, df1 = 6, df2 = 169, p-value = 0.03117 Ramsey RESET F-Test for powers (2 and 3) of principal components > reset_test_principal_components RESET test data: mylm RESET = 4.6541, df1 = 2, df2 = 173, p-value = 0.01075
Variance Inflation Factors (Multicollinearity) > vif System_Quality Information_Quality Perceived_Ease_of_Use 1.647271 2.586342 1.972540
\begin{tabular}{lllllllll}
\hline
Variance Inflation Factors (Multicollinearity) \tabularnewline
> vif
System_Quality Information_Quality Perceived_Ease_of_Use
1.647271 2.586342 1.972540
\tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=319053&T=9
[TABLE]
[ROW][C]Variance Inflation Factors (Multicollinearity)[/C][/ROW]
[ROW][C]> vif
System_Quality Information_Quality Perceived_Ease_of_Use
1.647271 2.586342 1.972540
[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=319053&T=9
Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=319053&T=9
As an alternative you can also use a QR Code:
The GUIDs for individual cells are displayed in the table below:
Variance Inflation Factors (Multicollinearity) > vif System_Quality Information_Quality Perceived_Ease_of_Use 1.647271 2.586342 1.972540
library(lattice)library(lmtest)library(car)library(MASS)n25 <- 25 #minimum number of obs. for Goldfeld-Quandt testmywarning <- ''par6 <- as.numeric(par6)if(is.na(par6)) {par6 <- 12mywarning = 'Warning: you did not specify the seasonality. The seasonal period was set to s = 12.'}par1 <- as.numeric(par1)if(is.na(par1)) {par1 <- 1mywarning = 'Warning: you did not specify the column number of the endogenous series! The first column was selected by default.'}if (par4=='') par4 <- 0par4 <- as.numeric(par4)if (!is.numeric(par4)) par4 <- 0if (par5=='') par5 <- 0par5 <- as.numeric(par5)if (!is.numeric(par5)) par5 <- 0x <- na.omit(t(y))k <- length(x[1,])n <- length(x[,1])x1 <- cbind(x[,par1], x[,1:k!=par1])mycolnames <- c(colnames(x)[par1], colnames(x)[1:k!=par1])colnames(x1) <- mycolnames #colnames(x)[par1]x <- x1if (par3 == 'First Differences'){(n <- n -1)x2 <- array(0, dim=c(n,k), dimnames=list(1:n, paste('(1-B)',colnames(x),sep='')))for (i in 1:n) {for (j in 1:k) {x2[i,j] <- x[i+1,j] - x[i,j]}}x <- x2}if (par3 == 'Seasonal Differences (s)'){(n <- n - par6)x2 <- array(0, dim=c(n,k), dimnames=list(1:n, paste('(1-Bs)',colnames(x),sep='')))for (i in 1:n) {for (j in 1:k) {x2[i,j] <- x[i+par6,j] - x[i,j]}}x <- x2}if (par3 == 'First and Seasonal Differences (s)'){(n <- n -1)x2 <- array(0, dim=c(n,k), dimnames=list(1:n, paste('(1-B)',colnames(x),sep='')))for (i in 1:n) {for (j in 1:k) {x2[i,j] <- x[i+1,j] - x[i,j]}}x <- x2(n <- n - par6)x2 <- array(0, dim=c(n,k), dimnames=list(1:n, paste('(1-Bs)',colnames(x),sep='')))for (i in 1:n) {for (j in 1:k) {x2[i,j] <- x[i+par6,j] - x[i,j]}}x <- x2}if(par4 > 0) {x2 <- array(0, dim=c(n-par4,par4), dimnames=list(1:(n-par4), paste(colnames(x)[par1],'(t-',1:par4,')',sep='')))for (i in 1:(n-par4)) {for (j in 1:par4) {x2[i,j] <- x[i+par4-j,par1]}}x <- cbind(x[(par4+1):n,], x2)n <- n - par4}if(par5 > 0) {x2 <- array(0, dim=c(n-par5*par6,par5), dimnames=list(1:(n-par5*par6), paste(colnames(x)[par1],'(t-',1:par5,'s)',sep='')))for (i in 1:(n-par5*par6)) {for (j in 1:par5) {x2[i,j] <- x[i+par5*par6-j*par6,par1]}}x <- cbind(x[(par5*par6+1):n,], x2)n <- n - par5*par6}if (par2 == 'Include Seasonal Dummies'){x2 <- array(0, dim=c(n,par6-1), dimnames=list(1:n, paste('M', seq(1:(par6-1)), sep ='')))for (i in 1:(par6-1)){x2[seq(i,n,par6),i] <- 1}x <- cbind(x, x2)}if (par2 == 'Include Monthly Dummies'){x2 <- array(0, dim=c(n,11), dimnames=list(1:n, paste('M', seq(1:11), sep ='')))for (i in 1:11){x2[seq(i,n,12),i] <- 1}x <- cbind(x, x2)}if (par2 == 'Include Quarterly Dummies'){x2 <- array(0, dim=c(n,3), dimnames=list(1:n, paste('Q', seq(1:3), sep ='')))for (i in 1:3){x2[seq(i,n,4),i] <- 1}x <- cbind(x, x2)}(k <- length(x[n,]))if (par3 == 'Linear Trend'){x <- cbind(x, c(1:n))colnames(x)[k+1] <- 't'}print(x)(k <- length(x[n,]))head(x)df <- as.data.frame(x)(mylm <- lm(df))(mysum <- summary(mylm))if (n > n25) {kp3 <- k + 3nmkm3 <- n - k - 3gqarr <- array(NA, dim=c(nmkm3-kp3+1,3))numgqtests <- 0numsignificant1 <- 0numsignificant5 <- 0numsignificant10 <- 0for (mypoint in kp3:nmkm3) {j <- 0numgqtests <- numgqtests + 1for (myalt in c('greater', 'two.sided', 'less')) {j <- j + 1gqarr[mypoint-kp3+1,j] <- gqtest(mylm, point=mypoint, alternative=myalt)$p.value}if (gqarr[mypoint-kp3+1,2] < 0.01) numsignificant1 <- numsignificant1 + 1if (gqarr[mypoint-kp3+1,2] < 0.05) numsignificant5 <- numsignificant5 + 1if (gqarr[mypoint-kp3+1,2] < 0.10) numsignificant10 <- numsignificant10 + 1}gqarr}bitmap(file='test0.png')plot(x[,1], type='l', main='Actuals and Interpolation', ylab='value of Actuals and Interpolation (dots)', xlab='time or index')points(x[,1]-mysum$resid)grid()dev.off()bitmap(file='test1.png')plot(mysum$resid, type='b', pch=19, main='Residuals', ylab='value of Residuals', xlab='time or index')grid()dev.off()bitmap(file='test2.png')sresid <- studres(mylm)hist(sresid, freq=FALSE, main='Distribution of Studentized Residuals')xfit<-seq(min(sresid),max(sresid),length=40)yfit<-dnorm(xfit)lines(xfit, yfit)grid()dev.off()bitmap(file='test3.png')densityplot(~mysum$resid,col='black',main='Residual Density Plot', xlab='values of Residuals')dev.off()bitmap(file='test4.png')qqPlot(mylm, main='QQ Plot')grid()dev.off()(myerror <- as.ts(mysum$resid))bitmap(file='test5.png')dum <- cbind(lag(myerror,k=1),myerror)dumdum1 <- dum[2:length(myerror),]dum1z <- as.data.frame(dum1)print(z)plot(z,main=paste('Residual Lag plot, lowess, and regression line'), ylab='values of Residuals', xlab='lagged values of Residuals')lines(lowess(z))abline(lm(z))grid()dev.off()bitmap(file='test6.png')acf(mysum$resid, lag.max=length(mysum$resid)/2, main='Residual Autocorrelation Function')grid()dev.off()bitmap(file='test7.png')pacf(mysum$resid, lag.max=length(mysum$resid)/2, main='Residual Partial Autocorrelation Function')grid()dev.off()bitmap(file='test8.png')opar <- par(mfrow = c(2,2), oma = c(0, 0, 1.1, 0))plot(mylm, las = 1, sub='Residual Diagnostics')par(opar)dev.off()if (n > n25) {bitmap(file='test9.png')plot(kp3:nmkm3,gqarr[,2], main='Goldfeld-Quandt test',ylab='2-sided p-value',xlab='breakpoint')grid()dev.off()}load(file='createtable')a<-table.start()a<-table.row.start(a)a<-table.element(a, 'Multiple Linear Regression - Estimated Regression Equation', 1, TRUE)a<-table.row.end(a)myeq <- colnames(x)[1]myeq <- paste(myeq, '[t] = ', sep='')for (i in 1:k){if (mysum$coefficients[i,1] > 0) myeq <- paste(myeq, '+', '')myeq <- paste(myeq, signif(mysum$coefficients[i,1],6), sep=' ')if (rownames(mysum$coefficients)[i] != '(Intercept)') {myeq <- paste(myeq, rownames(mysum$coefficients)[i], sep='')if (rownames(mysum$coefficients)[i] != 't') myeq <- paste(myeq, '[t]', sep='')}}myeq <- paste(myeq, ' + e[t]')a<-table.row.start(a)a<-table.element(a, myeq)a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a, mywarning)a<-table.row.end(a)a<-table.end(a)table.save(a,file='mytable1.tab')a<-table.start()a<-table.row.start(a)a<-table.element(a,'Multiple Linear Regression - Ordinary Least Squares', 6, TRUE)a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a,'Variable',header=TRUE)a<-table.element(a,'Parameter',header=TRUE)a<-table.element(a,'S.D.',header=TRUE)a<-table.element(a,'T-STATH0: parameter = 0',header=TRUE)a<-table.element(a,'2-tail p-value',header=TRUE)a<-table.element(a,'1-tail p-value',header=TRUE)a<-table.row.end(a)for (i in 1:k){a<-table.row.start(a)a<-table.element(a,rownames(mysum$coefficients)[i],header=TRUE)a<-table.element(a,formatC(signif(mysum$coefficients[i,1],5),format='g',flag='+'))a<-table.element(a,formatC(signif(mysum$coefficients[i,2],5),format='g',flag=' '))a<-table.element(a,formatC(signif(mysum$coefficients[i,3],4),format='e',flag='+'))a<-table.element(a,formatC(signif(mysum$coefficients[i,4],4),format='g',flag=' '))a<-table.element(a,formatC(signif(mysum$coefficients[i,4]/2,4),format='g',flag=' '))a<-table.row.end(a)}a<-table.end(a)table.save(a,file='mytable2.tab')a<-table.start()a<-table.row.start(a)a<-table.element(a, 'Multiple Linear Regression - Regression Statistics', 2, TRUE)a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a, 'Multiple R',1,TRUE)a<-table.element(a,formatC(signif(sqrt(mysum$r.squared),6),format='g',flag=' '))a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a, 'R-squared',1,TRUE)a<-table.element(a,formatC(signif(mysum$r.squared,6),format='g',flag=' '))a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a, 'Adjusted R-squared',1,TRUE)a<-table.element(a,formatC(signif(mysum$adj.r.squared,6),format='g',flag=' '))a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a, 'F-TEST (value)',1,TRUE)a<-table.element(a,formatC(signif(mysum$fstatistic[1],6),format='g',flag=' '))a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a, 'F-TEST (DF numerator)',1,TRUE)a<-table.element(a, signif(mysum$fstatistic[2],6))a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a, 'F-TEST (DF denominator)',1,TRUE)a<-table.element(a, signif(mysum$fstatistic[3],6))a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a, 'p-value',1,TRUE)a<-table.element(a,formatC(signif(1-pf(mysum$fstatistic[1],mysum$fstatistic[2],mysum$fstatistic[3]),6),format='g',flag=' '))a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a, 'Multiple Linear Regression - Residual Statistics', 2, TRUE)a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a, 'Residual Standard Deviation',1,TRUE)a<-table.element(a,formatC(signif(mysum$sigma,6),format='g',flag=' '))a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a, 'Sum Squared Residuals',1,TRUE)a<-table.element(a,formatC(signif(sum(myerror*myerror),6),format='g',flag=' '))a<-table.row.end(a)a<-table.end(a)table.save(a,file='mytable3.tab')myr <- as.numeric(mysum$resid)myra <-table.start()a <- table.row.start(a)a <- table.element(a,'Menu of Residual Diagnostics',2,TRUE)a <- table.row.end(a)a <- table.row.start(a)a <- table.element(a,'Description',1,TRUE)a <- table.element(a,'Link',1,TRUE)a <- table.row.end(a)a <- table.row.start(a)a <-table.element(a,'Histogram',1,header=TRUE)a <- table.element(a,hyperlink( paste('https://supernova.wessa.net/rwasp_histogram.wasp?convertgetintopost=1&data=',paste(as.character(mysum$resid),sep='',collapse=' '),sep='') ,'Compute','Click here to examine the Residuals.'),1)a <- table.row.end(a)a <- table.row.start(a)a <-table.element(a,'Central Tendency',1,header=TRUE)a <- table.element(a,hyperlink( paste('https://supernova.wessa.net/rwasp_centraltendency.wasp?convertgetintopost=1&data=',paste(as.character(mysum$resid),sep='',collapse=' '),sep='') ,'Compute','Click here to examine the Residuals.'),1)a <- table.row.end(a)a <- table.row.start(a)a <-table.element(a,'QQ Plot',1,header=TRUE)a <- table.element(a,hyperlink( paste('https://supernova.wessa.net/rwasp_fitdistrnorm.wasp?convertgetintopost=1&data=',paste(as.character(mysum$resid),sep='',collapse=' '),sep='') ,'Compute','Click here to examine the Residuals.'),1)a <- table.row.end(a)a <- table.row.start(a)a <-table.element(a,'Kernel Density Plot',1,header=TRUE)a <- table.element(a,hyperlink( paste('https://supernova.wessa.net/rwasp_density.wasp?convertgetintopost=1&data=',paste(as.character(mysum$resid),sep='',collapse=' '),sep='') ,'Compute','Click here to examine the Residuals.'),1)a <- table.row.end(a)a <- table.row.start(a)a <-table.element(a,'Skewness/Kurtosis Test',1,header=TRUE)a <- table.element(a,hyperlink( paste('https://supernova.wessa.net/rwasp_skewness_kurtosis.wasp?convertgetintopost=1&data=',paste(as.character(mysum$resid),sep='',collapse=' '),sep='') ,'Compute','Click here to examine the Residuals.'),1)a <- table.row.end(a)a <- table.row.start(a)a <-table.element(a,'Skewness-Kurtosis Plot',1,header=TRUE)a <- table.element(a,hyperlink( paste('https://supernova.wessa.net/rwasp_skewness_kurtosis_plot.wasp?convertgetintopost=1&data=',paste(as.character(mysum$resid),sep='',collapse=' '),sep='') ,'Compute','Click here to examine the Residuals.'),1)a <- table.row.end(a)a <- table.row.start(a)a <-table.element(a,'Harrell-Davis Plot',1,header=TRUE)a <- table.element(a,hyperlink( paste('https://supernova.wessa.net/rwasp_harrell_davis.wasp?convertgetintopost=1&data=',paste(as.character(mysum$resid),sep='',collapse=' '),sep='') ,'Compute','Click here to examine the Residuals.'),1)a <- table.row.end(a)a <- table.row.start(a)a <-table.element(a,'Bootstrap Plot -- Central Tendency',1,header=TRUE)a <- table.element(a,hyperlink( paste('https://supernova.wessa.net/rwasp_bootstrapplot1.wasp?convertgetintopost=1&data=',paste(as.character(mysum$resid),sep='',collapse=' '),sep='') ,'Compute','Click here to examine the Residuals.'),1)a <- table.row.end(a)a <- table.row.start(a)a <-table.element(a,'Blocked Bootstrap Plot -- Central Tendency',1,header=TRUE)a <- table.element(a,hyperlink( paste('https://supernova.wessa.net/rwasp_bootstrapplot.wasp?convertgetintopost=1&data=',paste(as.character(mysum$resid),sep='',collapse=' '),sep='') ,'Compute','Click here to examine the Residuals.'),1)a <- table.row.end(a)a <- table.row.start(a)a <-table.element(a,'(Partial) Autocorrelation Plot',1,header=TRUE)a <- table.element(a,hyperlink( paste('https://supernova.wessa.net/rwasp_autocorrelation.wasp?convertgetintopost=1&data=',paste(as.character(mysum$resid),sep='',collapse=' '),sep='') ,'Compute','Click here to examine the Residuals.'),1)a <- table.row.end(a)a <- table.row.start(a)a <-table.element(a,'Spectral Analysis',1,header=TRUE)a <- table.element(a,hyperlink( paste('https://supernova.wessa.net/rwasp_spectrum.wasp?convertgetintopost=1&data=',paste(as.character(mysum$resid),sep='',collapse=' '),sep='') ,'Compute','Click here to examine the Residuals.'),1)a <- table.row.end(a)a <- table.row.start(a)a <-table.element(a,'Tukey lambda PPCC Plot',1,header=TRUE)a <- table.element(a,hyperlink( paste('https://supernova.wessa.net/rwasp_tukeylambda.wasp?convertgetintopost=1&data=',paste(as.character(mysum$resid),sep='',collapse=' '),sep='') ,'Compute','Click here to examine the Residuals.'),1)a <- table.row.end(a)a <- table.row.start(a)a <-table.element(a,'Box-Cox Normality Plot',1,header=TRUE)a <- table.element(a,hyperlink( paste('https://supernova.wessa.net/rwasp_boxcoxnorm.wasp?convertgetintopost=1&data=',paste(as.character(mysum$resid),sep='',collapse=' '),sep='') ,'Compute','Click here to examine the Residuals.'),1)a <- table.row.end(a)a <- table.row.start(a)a <- table.element(a,'Summary Statistics',1,header=TRUE)a <- table.element(a,hyperlink( paste('https://supernova.wessa.net/rwasp_summary1.wasp?convertgetintopost=1&data=',paste(as.character(mysum$resid),sep='',collapse=' '),sep='') ,'Compute','Click here to examine the Residuals.'),1)a <- table.row.end(a)a<-table.end(a)table.save(a,file='mytable7.tab')if(n < 200) {a<-table.start()a<-table.row.start(a)a<-table.element(a, 'Multiple Linear Regression - Actuals, Interpolation, and Residuals', 4, TRUE)a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a, 'Time or Index', 1, TRUE)a<-table.element(a, 'Actuals', 1, TRUE)a<-table.element(a, 'InterpolationForecast', 1, TRUE)a<-table.element(a, 'ResidualsPrediction Error', 1, TRUE)a<-table.row.end(a)for (i in 1:n) {a<-table.row.start(a)a<-table.element(a,i, 1, TRUE)a<-table.element(a,formatC(signif(x[i],6),format='g',flag=' '))a<-table.element(a,formatC(signif(x[i]-mysum$resid[i],6),format='g',flag=' '))a<-table.element(a,formatC(signif(mysum\$resid[i],6),format='g',flag=' '))a<-table.row.end(a)}a<-table.end(a)table.save(a,file='mytable4.tab')if (n > n25) {a<-table.start()a<-table.row.start(a)a<-table.element(a,'Goldfeld-Quandt test for Heteroskedasticity',4,TRUE)a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a,'p-values',header=TRUE)a<-table.element(a,'Alternative Hypothesis',3,header=TRUE)a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a,'breakpoint index',header=TRUE)a<-table.element(a,'greater',header=TRUE)a<-table.element(a,'2-sided',header=TRUE)a<-table.element(a,'less',header=TRUE)a<-table.row.end(a)for (mypoint in kp3:nmkm3) {a<-table.row.start(a)a<-table.element(a,mypoint,header=TRUE)a<-table.element(a,formatC(signif(gqarr[mypoint-kp3+1,1],6),format='g',flag=' '))a<-table.element(a,formatC(signif(gqarr[mypoint-kp3+1,2],6),format='g',flag=' '))a<-table.element(a,formatC(signif(gqarr[mypoint-kp3+1,3],6),format='g',flag=' '))a<-table.row.end(a)}a<-table.end(a)table.save(a,file='mytable5.tab')a<-table.start()a<-table.row.start(a)a<-table.element(a,'Meta Analysis of Goldfeld-Quandt test for Heteroskedasticity',4,TRUE)a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a,'Description',header=TRUE)a<-table.element(a,'# significant tests',header=TRUE)a<-table.element(a,'% significant tests',header=TRUE)a<-table.element(a,'OK/NOK',header=TRUE)a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a,'1% type I error level',header=TRUE)a<-table.element(a,signif(numsignificant1,6))a<-table.element(a,formatC(signif(numsignificant1/numgqtests,6),format='g',flag=' '))if (numsignificant1/numgqtests < 0.01) dum <- 'OK' else dum <- 'NOK'a<-table.element(a,dum)a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a,'5% type I error level',header=TRUE)a<-table.element(a,signif(numsignificant5,6))a<-table.element(a,signif(numsignificant5/numgqtests,6))if (numsignificant5/numgqtests < 0.05) dum <- 'OK' else dum <- 'NOK'a<-table.element(a,dum)a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a,'10% type I error level',header=TRUE)a<-table.element(a,signif(numsignificant10,6))a<-table.element(a,signif(numsignificant10/numgqtests,6))if (numsignificant10/numgqtests < 0.1) dum <- 'OK' else dum <- 'NOK'a<-table.element(a,dum)a<-table.row.end(a)a<-table.end(a)table.save(a,file='mytable6.tab')}}a<-table.start()a<-table.row.start(a)a<-table.element(a,'Ramsey RESET F-Test for powers (2 and 3) of fitted values',1,TRUE)a<-table.row.end(a)a<-table.row.start(a)reset_test_fitted <- resettest(mylm,power=2:3,type='fitted')a<-table.element(a,paste('',RC.texteval('reset_test_fitted'),'',sep=''))a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a,'Ramsey RESET F-Test for powers (2 and 3) of regressors',1,TRUE)a<-table.row.end(a)a<-table.row.start(a)reset_test_regressors <- resettest(mylm,power=2:3,type='regressor')a<-table.element(a,paste('',RC.texteval('reset_test_regressors'),'',sep=''))a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a,'Ramsey RESET F-Test for powers (2 and 3) of principal components',1,TRUE)a<-table.row.end(a)a<-table.row.start(a)reset_test_principal_components <- resettest(mylm,power=2:3,type='princomp')a<-table.element(a,paste('',RC.texteval('reset_test_principal_components'),'',sep=''))a<-table.row.end(a)a<-table.end(a)table.save(a,file='mytable8.tab')a<-table.start()a<-table.row.start(a)a<-table.element(a,'Variance Inflation Factors (Multicollinearity)',1,TRUE)a<-table.row.end(a)a<-table.row.start(a)vif <- vif(mylm)a<-table.element(a,paste('',RC.texteval('vif'),'',sep=''))a<-table.row.end(a)a<-table.end(a)table.save(a,file='mytable9.tab') | 45,038 | 89,187 | {"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} | 2.515625 | 3 | CC-MAIN-2021-17 | latest | en | 0.419636 |
https://math.stackexchange.com/questions/1975459/definition-of-subgradient-of-1-norm | 1,586,524,822,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371896913.98/warc/CC-MAIN-20200410110538-20200410141038-00431.warc.gz | 561,909,700 | 32,135 | # Definition of subgradient of $1$-norm
I have an example which seems to contradict the definition of the subgradient of the L1 norm. Obviously I made a mistake, but I can't see where.
We start with the definition of a subgradient (from Section 11.4.2 in Statistical Learning with Sparsity) : given a convex function $f: \mathbb{R}^p \rightarrow \mathbb{R}$, we say that $z \in \mathbb{R}^p$ is a subgradient at $\beta$, denoted by $z \in \partial f(\beta)$, if we have
$$f(\beta + \triangle) \geq f(\beta) + \langle z, \triangle \rangle$$
for all $\triangle \in \mathbb{R}^p$. In case $f(\beta) = ||\beta||_1$, it can be seen that $z \in\partial ||\beta||_1$ if and only if $z_j = sign(\beta_j)$ for all $j=1, 2, \dots, p$, where we allow $sign(0)$ to be any number in the interval $[-1,1]$.
Now my example: let $f(\beta) = ||\beta||_1$ and $\beta \in \mathbb{R}^1$. Let $\beta = 1 = \triangle \in \mathbb{R}$. Now for me it seems the above inequality holds for all values $z \in (-\infty, 1)$, for instance for $z = \frac{1}{2}$:
$$2 = f(\beta + \triangle) \geq f(\beta) + \langle z, \triangle \rangle = 1 + \frac{1}{2}$$ However, this contradicts the iff from above, which says that whenever $\beta \neq 0$ we have $z \in \{-1,1\}$.
Where did I make a mistake? Any hints are greatly appreciated!
The inequality $f(\beta + \Delta) \geq f(\beta) + \langle z, \Delta \rangle$ is supposed to hold for all values of $\Delta$, not just $\Delta = 1$.
If $\Delta = -1$, then $$0 = f(\beta + \Delta) \ngeq f(\beta) + \langle z, \Delta \rangle = 1 - \frac12.$$ So $z = \frac12 \notin \partial f(1)$.
Visually, if you plot the curves $y = f(\beta) = | \beta|$ and $y = 1 + \frac12 \beta$, it's clear that $\frac12$ is not a subgradient of $f$ at $\beta = 1$. | 600 | 1,758 | {"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} | 3.90625 | 4 | CC-MAIN-2020-16 | latest | en | 0.776854 |
https://blog.bitmex.com/how-to-market-make-bitcoin-derivatives-lesson-2/ | 1,638,089,743,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358480.10/warc/CC-MAIN-20211128073830-20211128103830-00268.warc.gz | 208,417,339 | 32,838 | # How to Market Make Bitcoin Derivatives Lesson 2
In Lesson 1, I described how to calculate a two-way quote, dynamically hedge, and match the settlement for an ETC7D futures contract. In this lesson, I will discuss how to calculate Basis and Skew, and how to apply these to your two-way quotes.
## Calculating Basis
To calculate the future value of any currency pair, you need to know the cost to borrow and lend the home and foreign currencies. The underlying currency of ETC7D is the ETC/XBT exchange rate. Assume that as a market maker you only own Bitcoin and not any other currency.
To buy ETC, you must sell XBT that you own. That XBT has an opportunity cost of capital. If you were to borrow this Bitcoin from another party to fund your business, how much would it cost?
To sell ETC, you must borrow it. There are two ways this is accomplished. For more well established digital currencies, spot exchanges like Poloniex will operate a borrow market. You can automatically pledge Bitcoin as collateral and borrow ETC at the market interest rate when you wish to short it.
If margin trading is not available, you need to borrow it from another trader. The other party will lend you ETC for a period of time at a rate of interest.
Using Covered Interest Rate Parity, we can calculate the Basis.
Basis = (1 + XBTr * t) / (1 + ETCr * t) – 1
XBTr = Bitcoin annualised interest rate
ETCr = Ether Classic annualised interest rate
t = Time in years
In practical terms, we ignore the opportunity cost of XBT. You can compensate for that by increasing your spread profit margin. As a market maker, you are concerned about how much you will pay to short ETC.
Assume it costs you 50% per annum to borrow ETC.
Basis for 7-day Future = 1 / (1 + 50% * 7/365) – 1 = -0.95%
If the spot price is 0.02 XBT, translate that Basis into points.
-0.95% * 0.02 XBT = -0.00019 XBT
ETC7D Quote Mid = Spot + Basis = 0.02 XBT – 0.00019 XBT = 0.01981 XBT
You have lowered your mid price. This will compensate you in the event someone sells to you. If you become long ETC7D, you will have to short ETC and pay the borrow fee.
## Calculating Skew
You don’t have infinite capital. Left unchecked, you could build up a large long or short position in ETC7D. Your goal is to capture two-way flow so that you earn half of your quoted spread.
If someone continues selling into your bids, you want to progressively lower your quotes. If someone continues buying your asks, you want to progressively raise your quotes. A simple example will help illustrate the concept.
Assume you quote a two-way market of 10 / 11 for 1 contract each side. The total amount you are quoting on each side is 1 contract; I will call your Total Size Quoted. The bid / ask spread is 10%, half of that spread is 5%; I will call this your Weighted Average Half Spread.
Skew = (Change in Position / Total Size Quoted) * Weighted Average Half Spread * -1
Your position changed by +1.
Skew = (1 / 1) * 5% * -1 = -5%
You originally had a mid price of 10.5. Assume that the mid price represented the spot price.
New Mid Price = Spot + Skew = 10.5 * (1 – 5%) = 9.98
New Bid Price = 9.98 * (1 – 5%) = 9.48
New Ask Price = 9.98 * (1 + 5%) = 10.47
## Combining Basis and Skew
Before you begin quoting, you will calculate the Basis. You will do this calculation based on how expensive it is for you to borrow the home currency.
Then while quoting, your trading program will adjust the Skew based on whether traders buy or sell from you.
Quote Mid Price = Spot Price + Basis + Skew
Using these two concepts, you cover your cost of funds, and manage your inventory.
In Lesson 3, I will explain how to handle Auto-Deleveraging events, and other advanced topics. To view the BitMEX sample market making bot, please visit Github. | 974 | 3,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-49 | longest | en | 0.928833 |
https://brainmass.com/math/functional-analysis/distance-from-fire-to-observing-tower-a-80332 | 1,485,205,433,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560283008.19/warc/CC-MAIN-20170116095123-00054-ip-10-171-10-70.ec2.internal.warc.gz | 797,011,257 | 18,574 | Share
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# distance from fire to observing tower A
From fire tower A, a fire with bearing N 75° E is sighted. The same fire is sighted from tower B at N 49° W. Tower B is 55 miles east of tower A. How far is it from tower A to the fire?
#### Solution Summary
Using law of sines, the solution shows how to find the distance from the fire to tower A when the bearing angles are known.
\$2.19 | 107 | 409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.928539 |
https://www.ceder.net/def/bygolly.php?language=denmark%C3%83%C6%92%C3%86%E2%80%99%C3%83%E2%80%A0%C3%A2%E2%82%AC%E2%84%A2%C3%83%C6%92%C3%A2%E2%82%AC%C5%A1%C3%83%E2%80%9A%C3%82%C2%A2%C3%83%C6%92%C3%86%E2%80%99%C3%83%E2%80%9A%C3%82%C2%A2%C3%83%C6%92%C3%82%C2%A2%C3%83%C2%A2%C3%A2%E2%80%9A%C2%AC%C3%85%C2%A1%C3%83%E2%80%9A%C3%82%C2%AC%C3%83%C6%92%C3%A2%E2%82%AC%C5%A1%C3%83%E2%80%9A%C3%82%C2%B0%C3%83%C6%92%C3%86%E2%80%99%C3%83%C2%A2%C3%A2%E2%80%9A%C2%AC%C3%85%C2%A1%C3%83%C6%92%C3%A2%E2%82%AC%C5%A1%C3%83%E2%80%9A%C3%82%C2%A4vel=C1&level=C4 | 1,576,297,962,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540584491.89/warc/CC-MAIN-20191214042241-20191214070241-00139.warc.gz | 650,908,171 | 5,937 | Definitions of Square Dance Calls and Concepts
By Golly [C4]
FAQ |
Index --> Plus | A1 | A2 | C1 | C2 | C3A | C3B | C4 | NOL |
Definitions (Text Only) --> Plus | A1 | A2 | C1 | C2 | C3A | C3B | C4 | NOL |
Find call:
By Golly -- [C4]
(Lee Kopman 1977)
C4:
Language:
From Facing Couples. This call must be preceded by a call which can be followed by a Sweep 1/4.
Those closest to the flow direction Sweep 1/4 as the others Dodge and Any Shoulder Wheel Thru.
Ends in a Mini-Wave Box.
Recycle By Golly:
beforeRecycle By Golly afterRecycle afterThose closest to flow direction Sweep 1/4as Others Dodge afterDodgers Any ShoulderWheel Thru (done)
Cheat #1: Veer toward flow direction; Any Shoulder 1/2 Tag.
Cheat #2: Sweep 1/4; 'Trailers of the Sweep' Diagonal Pull By (outside hands).
The following is a partial list of calls which can precede By Golly at C3B:
Bend The Line (from 2-Faced Line) Ferris Wheel (resulting Centers) Sweep 1/4 Cross Cycle (from 2-Faced Line) Linear Cycle Turn & Deal (from R-H 2-Faced Line) Cycle & Wheel Recycle (from a Wave) Wheel & Deal (from 2-Faced Line) Fancy (resulting Centers) Shakedown
Notes to callers:
• When pushing checkers, By Golly is equivalent to Touch 1/4 or Left Touch 1/4.
• Recycle By Golly is equivalent to Flip Back.
• The Ceder Square Dance System (CSDS) uses the following alternate definition for By Golly: From Facing Couples where all dancers have the same Roll-direction (i.e., if the next call was a Roll, then all dancers would 1/4 Right or all dancers would 1/4 Left): If the Roll-direction is to the Right then Veer Left and 1/2 Tag; otherwise if the Roll-direction is to the Left then Veer Right and Left 1/2 Tag. This generalized definition allows By Golly to follow calls such as Couple Up, Right & Left Thru, etc. | 571 | 1,829 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-51 | latest | en | 0.780587 |
https://www.pmod.com/files/download/v36/doc/pkin/pmclass.lib.pmkin.models.PKresidenceTimes.htm | 1,696,362,762,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511220.71/warc/CC-MAIN-20231003192425-20231003222425-00137.warc.gz | 1,004,308,745 | 4,385 | Residence Times
### Residence Times
The Residence Times model performs pre-processing steps for the calculation of the absorbed dose from diagnostic or therapeutic radiopharmaceuticals. Given the time-course of the activity in a volume of tissue, it calculates the Residence Time, which should better be called the Normalized Cumulated Activity [1] to avoid confusion of its meaning. It represents the total number of disintegrations which have occurred during an integration time per unit administered activity. Ideally, integration is performed from the time of administration to infinity. However, as it is widely used in literature, the notion Residence Time will be used in the remainder of this section.
The organ residence times resulting from the Residence Times model can serve as input to a program such as OLINDA [2] for the actual calculation of the absorbed organ doses.
Operational Equations
Given an activity A0 applied to a subject, and a measured (not decay corrected) activity A(t) in an organ, the residence time t is calculated by
Note that the unit of t is usually given as [Bq×hr/Bq]. The measured part of the organ activity can easily be numerically integrated by the trapezoidal rule. For the unknown remainder of A(t) till infinity it is a conservative assumption to apply the radioactive decay of the isotope. This exponential area can easily be calculated and added to the trapezoidal area. Alternatively, if the measured activity curve has a dominant washout shape, a sum of exponentials can be fitted to the to the measured data and the entire integral algebraically calculated.
The Residence Times model in PKIN supports both the trapezoidal integration approach as well as the use of fitted exponentials. It furthermore supports some paracticalities such as
• conversion of an average activity concentration in [kBq/cc] to activity in [Bq] by multiplication with the VOI volume;
• reversion of decay correction;
Model Input Parameters
The Residence Times model has 4 input parameters which need to be specified interactively by the user. These settings can easily be propagated from one region to all others with the Model & Par button.
Decay corrected Check this box if the loaded curves have been decay corrected as is usually the case for PET imaging studies. The Halflife of the radioisotope will be used to undo the decay correction. Activity concentration Check this box if the loaded curves represent activity concentration rather than the total activity in a region. The volume of the region will be multiplied with the signal to calculate activity in [Bq]. Isotope tail Check this box to use the radioactive decay shape for the integration from the end of the last frame to infinity. This option is only relevant if exponentials are fitted to the measurement. For the Rectangle and Trapezoid methods it will always be added. Integration The integration method can be selected in the option list: Rectangle, Trapezoid, 1 Exponential, 2 Exponentials, 3 Exponentials. The Rectangle method is the natural selection for PET values which represent the time average during the frame duration. If there are gaps between the PET frames, the uncovered area is approximated by the trapezoidal rule. The Trapezoid method uses trapezoidal areas between the frame mid-times. With an Exponential selection, the specified number of decaying exponentials is fitted to the measurements, and used for analytical integration. Note that samples marked as invalid are neglected in the fit.
Editing of Constituents
Depending on the import of the activity curves, isotope and activity information may or may not be available. They can be edited in the Patient and Study Information window as illustrated below. Activate Edit Patient in the Tools list, select SUV PARAMETERS, and enter Radionuclide half-life and Pre-injected tracer activity. Note that the activity needs to be calibrated to time 0 of the activity curve.
The region volume is available if the TAC was generated in PMOD's View tool and transferred to PKIN. Otherwise the volume has to be entered as illustrated below. Activate Edit Data in the Tools list, select a region in the upper list, then activate Edit volume from the curve tools and enter the Volume number.
Model Output Parameters
The Residence Times model has 4 input parameters which need to be specified interactively by the user. These settings can easily be propagated from one region to all others with the Model & Par button.
As soon as the Fit Region button is activated, the fits and calculations are performed, resulting in a list of parameters:
Residence Time [Bq×hr/Bq] Total number of disintegrations in region from time 0 until infinity. The main result and equal to AUC/Activity. Residence Time/cc [Bq/cc×hr/Bq] Total number of disintegrations per unit volume in the region. Equal to AUC/(Activity×Volume). If the organ has deliberately been outlined within the organ to reduce the partial-volume problem, Residence Time/cc may be multiplied with the actual organ size to approximate the activity taken up. The residence time for standard phantom organs can be easily be calculated by multiplying the tabulated organ mass with Residence Time/cc and dividing by the organ density (usually 1g/ccm). Activity [MBq] Administered activity. Shown for verification only. Halflife [sec] Isotope halflife. Shown for verification only. Volume [ccm] Region volume. Shown for verification only. AUC [Bq×hr] Integral of the activity curve from time 0 till infinity. Amplitude 1, 2, 3 [Bq]Halftime 1, 2, 3 [min] Amplitudes and halftimes of the fitted exponentials. In the case of 1 exponential a linear regression is applied, whereas iterative fitting is used with 2 or 3 exponentials.
Reference
1. Stabin MG: Fundamentals of Nuclear Medicine Dosimetry. Springer; 2008. DOI
2. Stabin MG, Sparks RB, Crowe E: OLINDA/EXM: the second-generation personal computer software for internal dose assessment in nuclear medicine. J Nucl Med 2005, 46(6):1023-1027. | 1,276 | 6,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.93679 |
https://stackoverflow.com/questions/75238117/how-to-check-if-values-are-greater-than-7-by-only-using-bitwise-operators-in-c | 1,701,846,290,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.31/warc/CC-MAIN-20231206063543-20231206093543-00731.warc.gz | 594,433,144 | 44,788 | # How to check if values are greater than 7 by only using bitwise operators in C?
For this question, it asks to return 1 if the argument is greater than 7, and 0 otherwise.
For example, if x is 8, the function will return 1. If x is 7, the function will return 0.
The only legal operators allowed are (! ~ & ^ | + << >>), which forbids the use of anything else such as -, for loops, while loops, if statements, etc.
We can assume the system uses 2's complement and a 32 bit representation of integers, performs right shifts arithmetically, and has unpredictable behavior when shifting an integer by more than the word size.
I know that subtracting without using the - operation can be done with ~, but I don't know how to think of this one logically to be honest.
• Hint: which bits are set in a number larger than 7? Jan 25 at 18:11
• @dbush The answer depends a good deal on whether the negative numbers are allowed or not :-) Jan 25 at 18:15
• "[U]ses 2's complement" is a statement about how the machine represents negative integers. Are we to conclude that this is in fact a concern? That is, what is the data type of the value to be tested? `int`? `int32_t`? `unsigned int`? `uint32_t`? Something else? Jan 25 at 18:21
• @JohnBollinger It seems like the given conditions are describing how bitwise operators work on signed numbers, since the C spec leaves that implementation-dependent. Jan 25 at 18:24
• I downvoted because these "How can I do task A without using obvious language features X, Y, or Z?" questions are silly puzzles which have no place in serious programming, and essentially no value for future readers. Jan 25 at 18:49
In binary, the decimal number `7` is `111`. Hence, a number which is greater than seven will always have a more significant bit set than the most significant bit in the binary representation of 7 (i.e. bit 3 when numbering from 0) set. So, if you are dealing only with unsigned numbers then this test is sufficient:
``````int bitwise_is_n_gt_7 (unsigned int value)
{
return value & 0xFFFFFFF8; // assuming size 32-bits adjust as necessary
}
``````
If you are dealing with signed numbers you have to also deal with the possibility that a number is negative, so you then test for:
``````int bitwise_is_n_gt_7 (int value)
{
return (!!(value & 0x7FFFFFF8) & !(value & 0x80000000));
}
``````
>7 is the same as ≥8.
You can check if a positive number is ≥ another positive number using integer division. It's larger if the result is non-zero.
Division isn't a "bit operation", but eight is a power of two, and there's a "bit operation" that's defined as a division by a power of 2: `>>`
So we could use:
``````n >> 3
``````
Right-shifting a positive integer by three removes the three least-significant bits while moving the others. But we don't care if the others are moved or not.
So we could also use:
``````n & ~7
``````
• Note that this doesn't work for negative numbers (as stated in the answer). Jan 25 at 18:32
• @Fe2O3 ack! Fixed. Jan 26 at 2:44
If an unsigned number is `7` or less, then the only bits that will be set will be bits that represent `4` or `2` or `1`.
The next higher bit represents `8`, which is greater than 7.
So, if you "turn off" the lowest 3 bits (which represent `4`, `2`, and `1`), with masking, and check if any bits still remain on, then the original number was 8 or greater.
``````if (number & ~0b0111) // Turn OFF the lowest 3 bits, keep all the others.
{
printf("Number is 8 or Greater\n");
}
else
{
printf("Number is 7 or less\n");
}
``````
Using the `0b` prefix is non-standard (but very common in many compilers), so you might choose to replace `0b0111` with `0x07`, or just plain-old `7` instead:
``````if (number & ~7) { /* number is greater than 7 */ }
``````
# Example Code:
``````#include <stdio.h>
int main(void) {
for(int i=0; i<30; ++i)
{
printf("%d is %s 7\n", i, i&~7? "greater than" :"less than (or equal to)");
}
return 0;
}
``````
## Output
``````0 is less than (or equal to) 7
1 is less than (or equal to) 7
2 is less than (or equal to) 7
3 is less than (or equal to) 7
4 is less than (or equal to) 7
5 is less than (or equal to) 7
6 is less than (or equal to) 7
7 is less than (or equal to) 7
8 is greater than 7
9 is greater than 7
10 is greater than 7
``````
Starting with this basic concept, and to handle negative numbers, you need to check if the "sign-bit" is set. If it is set, the number is automatically negative, and automatically less than 7.
Breaking it all down:
``````#include <stdio.h>
#include <stdbool.h>
int main(void) {
for(int i=-10; i<+10; ++i)
{
bool is_negative_bit_off = !(i>>31);
bool is_bit_8_or_higher_set = !!(i&~7);
if (is_negative_bit_off && is_bit_8_or_higher_set)
{
printf("%d is greater than 7\n", i);
}
}
return 0;
}
``````
And then simplifying it:
``````int main(void) {
for(int i=-10; i<+10; ++i)
{
bool num_greater_than_7 = !(i>>31) & !!(i&~7);
if (num_greater_than_7)
{
printf("%d is greater than 7\n", i);
}
}
return 0;
}
``````
``````bool num_greater_than_7 = !(i>>31) & !!(i&~7);
Operators used are `!`, `>>`, `&` `~`. | 1,478 | 5,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-50 | latest | en | 0.901028 |
http://mathhelpforum.com/trigonometry/184636-comprehension-sinx-cosx-1-sin2x.html | 1,529,949,640,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267868237.89/warc/CC-MAIN-20180625170045-20180625190045-00229.warc.gz | 200,513,471 | 10,765 | 1. ## comprehension -/sinx-cosx/=√1-sin2x
comprehension -
/sinx-cosx/=√(1-sin2x)
here '/ /' are depicting mode
Q1. cos x>sinx then x belongs to
1) (0,pi/2)
2) (pi/2,pi)
3) (pi,5pi/4)
4) (5pi/4,9pi/4)
Q2. Cos^2(3pi/4) + Cos^2(4pi/5)
(3/4)
(1/4)
(5/4)
(7/4)
2. ## Re: comprehension -/sinx-cosx/=√1-sin2x
Originally Posted by gurparwaan
comprehension -
/sinx-cosx/=√(1-sin2x)
here '/ /' are depicting mode
Q1. cos x>sinx then x belongs to
1) (0,pi/2)
2) (pi/2,pi)
3) (pi,5pi/4)
4) (5pi/4,9pi/4)
Q2. Cos^2(3pi/4) + Cos^2(4pi/5)
(3/4)
(1/4)
(5/4)
(7/4)
$\displaystyle \sqrt{1-\sin(2x)} =$
$\displaystyle \sqrt{1 - 2\sin{x}\cos{x}} =$
$\displaystyle \sqrt{\sin^2{x} - 2\sin{x}\cos{x} + \cos^2{x}} =$
$\displaystyle \sqrt{(\sin{x} - \cos{x})^2} =$
$\displaystyle |\sin{x} - \cos{x}|$
3. ## Re: comprehension -/sinx-cosx/=√1-sin2x
solve first question on the basis of comprehension
4. ## Re: comprehension -/sinx-cosx/=√1-sin2x
Originally Posted by gurparwaan
solve first question on the basis of comprehension
what do you mean by "comprehension" ?
5. ## Re: comprehension -/sinx-cosx/=√1-sin2x
Originally Posted by skeeter
what do you mean by "comprehension" ?
its a section which comes in competitive exams where u r given a littlle detail and on that basis u r given some questions
6. ## Re: comprehension -/sinx-cosx/=√1-sin2x
what I posted is a proof that ...
$\displaystyle |\sin{x}-\cos{x}| = \sqrt{1-\sin(2x)}$
... is that not sufficient? otherwise, I do not understand what it is you are looking for.
7. ## Re: comprehension -/sinx-cosx/=√1-sin2x
Q.1
Write $\displaystyle \sin(x)=\cos \left(\frac{\pi}{2}-x\right)$ or $\displaystyle \cos(x)=\sin \left(\frac{\pi}{2}-x\right)$
8. ## Re: comprehension -/sinx-cosx/=√1-sin2x
Originally Posted by gurparwaan
Q2. Cos^2(3pi/4) + Cos^2(4pi/5)
(3/4)
(1/4)
(5/4)
(7/4)
should this be ...
$\displaystyle \cos^2\left(\frac{3\pi}{4}\right) + \cos^2\left(\frac{4\pi}{3}\right)$
... a typo perhaps ?
9. ## Re: comprehension -/sinx-cosx/=√1-sin2x
Originally Posted by skeeter
should this be ...
$\displaystyle \cos^2\left(\frac{3\pi}{4}\right) + \cos^2\left(\frac{4\pi}{3}\right)$
... a typo perhaps ?
yes.
10. ## Re: comprehension -/sinx-cosx/=√1-sin2x
Originally Posted by skeeter
$\displaystyle \cos^2\left(\frac{3\pi}{4}\right) + \cos^2\left(\frac{4\pi}{3}\right)$
recommend you derive and learn the trig values of the special angles on the unit circle ... | 959 | 2,429 | {"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} | 4.09375 | 4 | CC-MAIN-2018-26 | latest | en | 0.570361 |
http://mathoverflow.net/revisions/11227/list | 1,369,559,821,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706794379/warc/CC-MAIN-20130516121954-00018-ip-10-60-113-184.ec2.internal.warc.gz | 163,028,220 | 5,005 | The first example that comes to my mind is that the Banach spaces $\ell_\infty$ and $L_\infty [0, 1]$ are isomorphic (that is, there exists a linear homeomorphism of one onto the other), and yet it seems that one can't just write down an operator that provides the linear homeomorphism. The existence of such an operator between the two above-mentioned spaces was first established by Pelczynski.
A similar example is that if $K$ is an uncountable compact metric space, then the Banach space $C(K)$ of continuous scalar-valued functions on $K$ (equipped with the supremum norm) is isomorphic to $C[0,1]$, the space of continuous scalar-valued function on the compact interval $[0, 1]\subseteq \mathbb{R}$. Thus, for example, if $\Delta$ denotes the Cantor set, then $C(\Delta)$ and $C[0,1]$ are isomorphic as Banach spaces. The proof of this relies on a result called Miljutin's Lemma, who proved the existence of the isomorphism. Anyway, I think that this also qualifies as an example.
The first example that comes to my mind is that the Banach spaces $\ell_\infty$ and $L_\infty [0, 1]$ are isomorphic (that is, there exists a linear homeomorphism of one onto the other), and yet it seems that one can't just write down an operator that provides the linear homeomorphism.
A similar example is that if $K$ is an uncountable compact metric space, then the Banach space $C(K)$ of continuous scalar-valued functions on $K$ (equipped with the supremum norm) is isomorphic to $C[0,1]$, the space of continuous scalar-valued function on the compact interval $[0, 1]\subseteq \mathbb{R}$. Thus, for example, if $\Delta$ denotes the Cantor set, then $C(\Delta)$ and $C[0,1]$ are isomorphic as Banach spaces. The proof of this relies on a result called Miljutin's Lemma. Anyway, I think that this also qualifies as an example. | 477 | 1,818 | {"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} | 2.8125 | 3 | CC-MAIN-2013-20 | latest | en | 0.908075 |
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1. Factor both expressions
x^2 + x - 12 = (x + 4) (x - 3)
x^2 + 2x - 15 = (x + 5) (x - 3)
The LCM is just all the different factors we have listed....so....
(x - 4) (x + 5) ( x - 3) ⇒ second answer
2. Notice that we can see if the first numerator factors using the second denominator
(w + 6) ( w - 4) 8
____________ + _____
(w + 6) ( w - 5) ( w - 5)
( w - 4) + 8
______ _____
(w - 5 ) (w - 5)
( w - 4 + 8)
_________
( w - 5)
w + 4
_____
w - 5
CPhill Feb 23, 2018
12 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 342 | 906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2018-13 | longest | en | 0.455933 |
https://holooly.com/solutions-v2-1/analyze-the-implementing-functionality-of-an-ssc-as-shown-in-figure-6-2-3/ | 1,713,402,842,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817184.35/warc/CC-MAIN-20240417235906-20240418025906-00825.warc.gz | 283,246,751 | 20,841 | Question 6.2: Analyze the implementing functionality of an SSC, as shown i......
Analyze the implementing functionality of an SSC, as shown in Figure 6.2.3.
Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Step 1 Derive the exciting equations and output equations directly from the logic diagram shown in Figure 6.2.3.
There are two J-K flip-flops in the circuit; the exciting equations are
$J_{1}=K_{1}=1$ (6.2.7)
$J_{2}=K_{2}=Q_{1}$ (6.2.8)
The output equation is
$Y=Q_{1}Q_{2}$ (6.2.9)
Step 2 Derive the state equation of each flip-flop by substituting the exciting equations into the characteristic equation of each flip-flop.
The characteristic equation of J-K flip-flops is
$Q_{n+1}=J \bar{Q}_{n}+\bar{K}Q_{n}$ (6.2.10)
Thus, the state equations can be expressed as
$Q_{1}{_{n}^{+}}=\bar{Q}_{1}$ (6.2.11)
$Q_{2}{_{n}^{+}}=Q_{1}\bar{Q}_{2}+\bar{Q}_{1}Q_{2}=Q_{1}\oplus Q_{2}$ (6.2.12)
Step 3 Construct a state table from the state equations and output equation.
The state table is shown as Table 6.2.3
Step 4 Draw a state diagram from the state table and deduce the implementing functionality of the given sequential logic circuits.
The state diagram is shown in Figure 6.2.4. It can be found that the sequential circuit goes through the states in the following sequence: 00, 01,10, 11,00, 01, … When it counts to 11, the output is a HIGH. When the next clock arrives, the circuit goes back to the initial state 00, and start new count cycle again. Since this sequence is characteristic of modulus-4 counting, we can conclude that the sequential circuit in Figure 6.2.3 is a modulus-4 counter and belongs to the Moore model due to no input signal.
Table 6.2.3: A state table. Present state Next state Output $Q_{2}$ $Q_{1}$ ${Q_{2}}^{+}$ ${Q_{1}}^{+}$ Y 0 0 0 1 0 0 1 1 0 0 1 0 1 1 0 1 1 0 0 1
Question: 6.6
Design an 8421BCD asynchronous up counter with a ripple carry output. ...
Step 1 Construct the state table. Because an 8421B...
Question: 6.5
Design a synchronous counter with J-K flip-flops. The timing diagram for the counter is shown as Figure 6.3.4. ...
Step 1 Construct the state diagram and state table...
Question: 6.4
Design a three-bit sequence detector for detecting a sequence input series “011.” ...
Step 1 Construct the initial state diagram and ini...
Question: 6.3
Analyze the implementing functionality of an asynchronous sequential circuit, as shown in Figure 6.2.5. ...
Step 1 Derive the clock equations and exciting equ...
Question: 6.1 | 770 | 2,660 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 10, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-18 | longest | en | 0.794472 |
https://math.stackexchange.com/questions/2917680/prove-that-there-exists-a-path-to-get-more-blue-balls-than-red-balls | 1,571,759,959,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987822458.91/warc/CC-MAIN-20191022155241-20191022182741-00375.warc.gz | 585,054,361 | 33,718 | # Prove that there exists a path to get more blue balls than red balls
I'm an undergraduate taking a Discrete Mathematics Module. This is one of my assignment questions.
$n > 0$ red balls and $n$ blue balls are arranged to form a circle. You walk around the circle exactly once in a clockwise direction and count the number of red and blue balls you pass. If at all times during your walk, the number of red balls (that you have passed) is greater than or equal to the number of blue balls (that you have passed), then your trip is said to be successful. (Note that whether successful or not, you will pass exactly $2n$ balls after walking one round.)
Define predicate $P(n) =$( In any circle formed by n red and n blue balls, there exists a successful trip ), $\forall x \in \Bbb Z^{+}.$
Using the above predicate P(n), prove by Mathematical Induction (either Regular or Strong Induction) that you can always make a successful trip if you can choose where you start.
(Note: you must use $P(n)$ as the induction predicate in your proof. You may not modify it, nor use a different predicate.)
My stand:
I think for this question, there is a need to find a general formula(or at least a way) representing the number of possible circular orientations of the n red balls and n blue balls. I have a problem with visualising this as I’m very bad at combinatorial questions and when to decide what is treated as unique and what is treated as a repetition.
1) Is it really required of me to figure out the orientations in the question, to get the inductive case of this problem? I hope I’m heading the right direction. If it’s not the right approach, please provide some hints as to how to go on from here. I will edit my progress report incrementally.
2) Suppose the number of ways Even if it’s not required, I would like to ask what exactly is a good procedure to visualise circular paths as repetitions of each other. For me, I think of a the circle arrangement as a special linear arrangement with variable start points, so I have the following procedure, take $n=3$.
a)Pick a random start point, say R (I think somehow my starting point won’t matter for a circular arrangement)
b)Come up with a possibility tree incrementally by appending R or B to the right of the arrangement till I fulfil n R and n B elements. If I do this for 6 stages/elements, and don’t eliminate repetitions in the circular arrangement, I should get $2^{5}$ such chains if I actually model it after linear chains instead.
c)I will first complete some boundary cases like RRRBBB, and reject some of the chains before they are completed if I observe that under the constraints of n colour for either side, their remaining choices are forced.
d) If they hit 6 elements long, I will try to “rotate” the leftmost element to become the rightmost element to get a common part that matches with one of the accepted cases. If that common part is not the entire length of the chain, it must be a unique case.
A diagram of my procedure.
I arrived at the conclusion that there are 5 circular combinations for $n=3$.
Suppose this procedure is somewhat correct, I’m unable to translate this procedure in terms of a formula to find number of circular arrangements for general n, and proceed to know the enumerations required for this problem. Should I think of the number of ways of circular arrangement as such or is there a more convenient way to think of such a problem?
Feel free to correct me if any of my assumptions are wrong, or quote the name of the results in my algorithm if they are somehow correct, since they are just based on my intuition.
P.S: The title may not be a good description of my problem, didn’t know how to give a good title other than “How to solve this question?”. Notice that (2) requires answer to (1) to be “Yes”. If the answer is no, I will edit the post to bring (2) to another discussion. Thanks.
You start out incorrectly, I'm sorry to say. In step a) you say, "I think somehow my starting point won’t matter for a circular arrangement." This is not true. When $n=1$ the circular arrangement must be RB, and we will be successful if we start at the red ball, but not if we start at the blue ball. (This takes care of the basis of the induction, by the way.)
Suppose now that $n>1$ and that $P(n-1)$ is true. In any circular arrangement of $n$ red balls and $n$ blue balls, there must be a red ball immediately followed by a blue ball in clockwise order. (Start at a red ball, and walk clockwise. The first blue ball we come to is preceded by a red ball.) Now consider the circle with these two balls removed. By the induction hypothesis, there is some red ball we can start at to have a successful trip with the $2n-2$ balls. Now argue that if we put the two balls that we remove back, and start at the same red ball, we will have a successful trip around the circle with $2n$ balls. | 1,115 | 4,894 | {"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} | 3.78125 | 4 | CC-MAIN-2019-43 | latest | en | 0.940448 |
https://people.richland.edu/james/summer03/m160/m160-sf.html | 1,513,283,672,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948550986.47/warc/CC-MAIN-20171214202730-20171214222730-00295.warc.gz | 607,495,925 | 2,725 | # Math 160: Study Guide - Final Exam
1. Compound interest problem. Use the "finance" program.
2. Present value problem. Use the "finance" program.
3. Solve a 3x3 system of linear equations. Either use Gauss-Jordan elimination (pivoting) or use the inverse of a matrix.
4. Retirement problem. Figure out what the need to retire and what it will take each month to save that. Figure out how long the retirement will last if extra is paid. Find the remaining balance after a certain period of time. Use the "finance" program.
5. Give the solution to a system of linear equations from an augmented matrix in row echelon form.
6. Leontief input output problem. Use the "leontief" program.
7. The initial tableau from a non-standard maximization problem is given. Write the maximization problem.
8. Maximize a standard linear programming problem. Write the initial tableau and final tableau, and give the values of the objective function and decision variables. Use the "simplex" program.
9. Given P(A), P(B) and one more probability, complete a probability distribution and then find several probabilities from the probability distribution.
10. Find the mean, median, and sample standard deviation for a set of data.
11. Binomial probabilities. Use the "binomial" program.
12. Solve a game matrix, giving the optimal row and column strategies, and the value of the game. Use the "game" program.
13. Markov chain problem. Write the initial state matrix, the transition matrix. Find the first state matrix, and the steady state matrix.
14. Linear programming problem. A tableau is given. Label the columns as appropriate using x's for the decision variables, s's for the slack variables, z for the objective function, and rhs for the right hand side. Identify which variables are basic and which are non-basic and give their values. Identify the basic variable for each row of the tableau and find the appropriate ratios on the right side. Circle the pivot element. Identify which variable is entering and which is exiting.
15. Decision Theory. Most of the payoff table has been created for you, but you need to find two entries in the table. Create the opportunistic loss table. Then give the value and optimal action under each criteria.
16. Probability problem. Two bags with different types of coins in them. A situation is described and you need to draw a tree diagram and then find some probabilities.
17. The payoffs for a game are given. Find the probabilities of each payoff and then find the expected value. Identify whether the game is fair or not. You may use the "pdist" program, although it's probably easier to do by hand.
18. The final tableau from a zero-sum, two-player game is given. Give the optimal row and column strategies and the value of the game.
19. Absorbing Markov chain problem. Write the initial state matrix and the transition matrix. Then find the expected number of transient states before leaving the matrix.
## Notes:
• The test is open notebook. Make sure that your notes are complete in the sections covered on the exam.
• Problems 1-13 are to be worked individually. Problems 14-19 may be worked in a group of up to three people. You may not have the individual portion of the test with you while you are working in groups.
• There is no requirement to work in groups. If you decide to get into groups, you will not be able to use your notebook. If you decide to work alone, you may use your notebook.
• You may begin the group portion of the test alone with your notebooks and then give up your notebooks once you've exhausted your ability to answer the questions.
• The exam is laid out mostly in chapter order. Having your notes in order will help.
• The individual part of the exam is 126 points, the group part is worth74 points.
## Points per problem
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 Tot. 6 6 6 12 6 6 6 16 16 9 9 12 16 12 12 14 14 10 12 200 | 898 | 3,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | latest | en | 0.807308 |
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# Belt Conveyor Power Calculation
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Calculation methods conveyor belts Siegling total belting solutions conveyor and processing belts This brochure contains advanced equa-tions, figures and recommendations, based on our longstanding experience. Results calculated can however differ from our calculation program BRex free to download from the Internet at www.forbo ...
### Conveyor Capacity Engineering ToolBox
Conveyor Capacity. Conveyor capacity is determined by the belt speed, width and the angle of the belt - and can be expressed as. Q A v 1 where. Q conveyor capacity kgs, lbs density of transported material kgm3, lbft3 A cross-sectional area of the bulk solid on the belt m2, ft2 v conveyor belt velocity ms, fts | 1,966 | 9,421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-25 | latest | en | 0.878932 |
http://cryptinex.com/the-world-chess-championship-2018/ | 1,537,722,053,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159561.37/warc/CC-MAIN-20180923153915-20180923174315-00045.warc.gz | 59,040,083 | 114,333 | # The world chess championship 2018
The world chess championship 2018
In 2018, the world chess championship will be held from 9 to 28 November. This time Sergey Karjakin failed to win the candidates tournament for the title new champion will face Fabiano Caruana. The current holder of the crown Magnus Carlsen will defend its superiority over the competition for the third time in a row.
Brilliant combination of cunning traps, fight on every square of the Board – such associations must cause a match for the title of world chess champion in 2018, and not sleeping right during the match, players and spectators. Why is this happening and how Fischer chess solve the problem?
Problems classic chess
Representatives of each sport want more popularity of their business. The more dynamic and spectacular competitions, the more viewers watching them. As a result, the sport is attracting more expensive advertising contracts and brings the main income to professionals. On the way to promote chess many problems, one of which is the lack of fight in the first 20 moves. The fact that athletes spend on computer analysis of development debuts a very long time. They remember what move is preferable in different situations and during the party just remember how to walk. Sometimes athletes are even doing the wrong moves, trying to remember the right continuation, instead of act according to logic.
At that time, to do: to sleep waiting for the stroke, turn the transmission over an hour and a half after the beginning of the party, or constantly compare the actions of the players with a computer program in anticipation of long-awaited error? Does not look attractive, given that chess is played with time control 2 hours for 40 moves, then another hour to the end of the party.
Long time chess fans are struggling with this problem and put forward the most unusual solutions:
• The increase of the chessboard.
• Changing the shape of the Board.
• Changing the rules for moving the pieces.
• Cuts the amount of time per move or per game (10 minutes, 5 minutes, 1 minute).
All these options have one drawback – the same initial arrangement of the figures. The optimal solution was developed by Robert Fisher in 1996. He proposed to begin the game a random arrangement of the figures. All possible combinations of the 960 turned out, what appeared the second name – chess 960.
The athlete may not remember a variety of openings for 960 constellation figures and the struggle begins with the first move. Players will have to focus on the study of the laws of chess, but learning the Queen’s Indian and Sicilian defense. Some arrangement is so dangerous that to avoid the mate to come with the first moves. In such hands there is additional intrigue, which can compensate the long time control in chess. Still, ordinary viewers can’t quickly and deeply to count strokes, how the pros do it – they find it difficult to follow the logic of the figures in the blitz (5 minute) or rapid (10 minutes).
Random chess associated with the fighting times of the Macedonian, when it was unknown the exact location of the enemy, the rate of the enemy General, and first place went tactical skills of commanders and the valor of soldiers.
Related Posts
Classical chess is more associated with a football match where both teams are afraid to miss and almost never attack a large force.
Such a comparison is logical, if you look at the shortcomings of classical chess:
• Blurry studied the openings.
• Strong athletes lose the advantage in the beginning of the game, so they delayed.
• The lack of dynamism.
In chess Fischer the calculation of the 6-7 moves is already before the first movement of the figure, and each party is not like the others – it’s called a sport of intellectuals. Such activities can attract new viewers for the tournaments, and children in school chess.
The unofficial world championship of chess 960 was held this winter. The winner was Magnus Carlsen – chess player, who always had innovative thinking, deep steps, bright and uncompromising game. The Norwegian expressed his desire to have random chess Bobby Fischer to become a regular tournament in the calendar of the international chess Federation, while Vishy Anand said that “Carlsen is a product of Cybernetics”.
Such well-known chess players like Spassky and Karpov have a positive attitude to Fisher-random and Taimanov believed that this is the chess of the XXI century. There are opponents of Fischer random chess. Vladimir Kramnik argues that random placement leads to the disappearance of the harmony of the game.
Chess: man vs computer
The creation of computer chess began in 1951, “Paper Turing machine”, which you had to wait for half an hour. Only in 1958 was the first full-fledged program with a strong algorithm. The first computer-chess level master appeared in 1983, his rating was 2250. The most famous battle of man and computer in 1997 match Kasparov vs Deep Blue ended with the surrender of the current world champion, and then the computer forever captured the chess crown.
It seemed that already nothing will move computers from the throne. As said Sergey Karjakin, in our time, the man has no chance in a match against the computer. It is not surprising, given that trained people using chess engines. In 2017 the most powerful engine was considered Stockfish, but in December a computer program capitulated to Alpha Zero.
Alpha zero is samooborony product of neural networks that beat Stockfish without a single defeat. Match alpha zero – Stockfish lasted 100 parties where artificial intelligence has won 28 times and painted a draw in 72 games – the victory of Google. Training at Alpha Zero it took several hours.
The main difference between artificial intelligence chess engine – incorporation of lessons learned from each played. Stockfish just goes through all the possible options (70 million), while an alpha of zero considering only 40 000 positions, discarding all the blunders, failures and futile moves. A senior programmer in the development AlphaZero was a former chess master Demis, Hassabis, which in 90-e was engaged in this game.
Man vs alpha zero in Fischer random chess
The best player in the world for a long time is not defined between people. The last competition between the programs was won by Stockfish, which supports the chess of Bobby Fischer. Alpha zero had defeated Stockfish, and therefore, man will be forced to capitulate in this sport. That the athlete has not turned into a computer, now popularized Chess 960. To watch live battles in this sport can be at the tournament of Champions Showdown: Chess 960, which is held in St. Louis from 11 to 14 September. The matches start every day at 21: 00.
Play Fischer random chess online on the site lichess.org where you want to create game, then select “Chess 960”, to determine the control time (optimum 10 minutes), and then choose the colour and expect the opponent (usually not more than a minute). Interesting game!
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,464 | 7,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2018-39 | latest | en | 0.940832 |
https://codedump.io/share/Ltu71oCFgPqo/1/slicing-a-list-in-python-is-there-something-like--0 | 1,480,741,284,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540839.46/warc/CC-MAIN-20161202170900-00457-ip-10-31-129-80.ec2.internal.warc.gz | 831,274,045 | 8,946 | lhk - 1 month ago 12
Python Question
# Slicing a list in Python: is there something like -0?
I've got a 3D array and would like to split it into many subvolumes.
This is my code so far:
``````# this results in a 3D array
arr = trainMasks[0, 0, :, :, :]
crop = 3
arrs = [arr[x:-(crop - x), y:-(crop - y), z:-(crop - z)]
for x in range(crop + 1)
for y in range(crop + 1)
for z in range(crop + 1)]
``````
• If I use
`x in range(crop)`
,
`x`
only goes up to
`crop - 1`
, the last entry in the x dimension is always dropped
• If I use
`x in range(crop+1)`
,
`x`
it goes up to
`crop`
, that will result in a slice
`arr[crop:-0, ...]`
which has the shape
`[0, y_dim, z_dim]`
I know the usual answer, just drop the upper limit, like this:
`arr[crop:, :, :]`
. Usually that's quite convenient. But how do I do that in the list comprehension ?
In cases like this it is better to avoid negative indexes.
Remeber that for `i>0`, `a[-i]` is equivalent to `a[len(a)-i]`. But in your case, you also need to work for `i==0`.
This works:
``````d1, d2, d3 = arr.shape
arrs = [arr[ x : d1-(crop-x), y : d2-(crop-y), z : d3-(crop-z)]
for x in range(crop + 1)
for y in range(crop + 1)
for z in range(crop + 1)]
`````` | 424 | 1,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2016-50 | longest | en | 0.814187 |
https://www.askiitians.com/forums/Marks-and-Rank-in-IIT-JEE/6/11307/partial-marking.htm | 1,701,542,368,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100448.65/warc/CC-MAIN-20231202172159-20231202202159-00852.warc.gz | 744,660,431 | 43,332 | IF in any question correct choices are A,B,C and a student has marked choices as A,C,D then how many marks he would get in that quedtion according to scheme of partial marking?
357 Points
13 years ago
Dear Akash,
To avail partial mark the number of choices shaded should not exceed the number of correct choices and must include at least one of the correct choices.
If all four options are shaded for a question for which there are less than four correct answers then the candidate gets zero.
for example if a question has A,C correct option. then
-- if you will answer A,C,D then u will get zero.(exceed the number of correct choice)
-- if you will answer B,D then u will get zero.( does not include any correct choice)
-- if u will answer A,D then you will get partial marking.( include 1 correct choice)
-- if u will answer A then u will get partial marking ( include 1 crrect choice.)
-- if u will answer A.c then u will get full marks ( include all the correct choice)
Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed solution very quickly. | 262 | 1,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-50 | latest | en | 0.92785 |
https://transportist.org/2008/07/02/ | 1,680,139,218,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949093.14/warc/CC-MAIN-20230330004340-20230330034340-00358.warc.gz | 644,223,267 | 29,105 | # The 20-Ton Packet
A nice article on container shipping by Stewart Taggart in 1999. Wired 7.10: The 20-Ton Packet
Instead of transportation being the metaphor for communications (the information superhighway, e.g.), here communication is the metaphor for transportation.
Another good source on container shipping is Marc Levinson’s The Box: How the Shipping Container Made the World Smaller and the World Economy Bigger , reviewed here.
The topic is summarized in The Transportation Experience in Chapter 16.
# Do 60 improvements each saving 1 minute equal 1 improvement saving 60 minutes
This is the third in a continuing series asking deep questions about the nature of transportation analysis. Previous episodes include:
1. Why do commute distances and times rise with income
2. Are sunk costs sunk, is salvage value salvageable? A paradox in engineering economics analysis
So,
3. Can small units of time be given the same value of time as larger units of time. In other words, do 60 improvements each saving a traveler 1 minute equal 1 improvement saving a traveler 60 minutes? Similarly, does 1 improvement saving a 1000 travelers 1 minute equal the value of time of a single traveler of 1000 minutes. These are different problems, one is intra-traveler and one is inter-traveler, but related.
Several issues arise.
A. Is value of time linear or non-linear? To this we must conclude the value of time is surely non-linear. I am much more agitated waiting 3 minutes at a red light than 2, and I begin to suspect the light is broken. Studies of ramp meters show a similar phenomena, as in our paper Weighting Waiting:
Evaluating Perception of In-Vehicle Travel Time Under Moving and Stopped Conditions
.
B. How do we apply this in a benefit-cost analysis? If we break one project into 60 smaller projects, each with a smaller value of travel time saved, and then we added the gains, we would get a different result than the what obtains with a single large project. For analytical convenience, we would like our analyses to be additive, not sub-additive, otherwise arbitrarily dividing the project changes the result. In particular many smaller projects will produce an undercount that is quite significant, and result in a much lower benefit than if the projects were bundled.
As a practical matter, every Benefit/Cost Analysis I have seen assumes a single value of time, rather than assuming non-linear value of time. (Alert me if you have a counter-example).
On the other hand, mode choice analyses do however weight different components of travel time differently, especially transit time (i.e. in-vehicle time is less onerous than waiting time). The implicit value of time for travelers does depend on the type of time (though generally not the amount of time). Using the log-sum of the mode choice model as a measure of benefit would implicitly account for this. | 608 | 2,875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-14 | latest | en | 0.896376 |
http://sstclass107maths.blogspot.com/2010/08/questions-2-3-and-4-by-goh-chin-fan.html | 1,503,127,685,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105326.6/warc/CC-MAIN-20170819070335-20170819090335-00673.warc.gz | 402,646,517 | 11,615 | ## Sunday, August 15, 2010
### Questions 2, 3 and 4 by Goh Chin Fan
Question 2:
Option D is correct.
A square and a parallelogram are quadrilaterals as they are both 4-sided.
Opposite sides of a square and a parallelogram are parallel:
Question 3:
The quadrilateral is a trapezium. A trapezium has one pair of opposite sides equal in length, which is the pair of lines that are not parallel to each other, and the other pair not equal in length, which is the pair of lines that are parallel to each other. It also has a pair of opposite angles that are supplementary.
Question 4:
No, I do not agree with the statement. The lines of a parallelogram are not of equal length, only the opposing lines are of the same length, while for the square, all lines are of equal length. Therefore I do not agree that all parallelograms are of equal length. | 211 | 847 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-34 | longest | en | 0.961753 |
http://perplexus.info/show.php?pid=8711&cid=52074 | 1,556,246,780,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578747424.89/warc/CC-MAIN-20190426013652-20190426035652-00200.warc.gz | 140,645,766 | 4,202 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Powering (Posted on 2013-10-05)
Given that a,b,x and y are real numbers such that
a+b=23
ax+by=79
ax2+by2=217
ax3+by3=691
Determine ax4+by4
No Solution Yet Submitted by Danish Ahmed Khan No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
this does it | Comment 3 of 5 |
I got bogged down in the algebra of substitution too but had a fresh look last night.
Tag the LHS of the 4 equations L1,L2,L3,L4 top to bottom.
L4 = (x+y)*L3 - (xy)*L2
L3 = (x+y)*L2 - (xy)*L1
691 = (x+y)*217- (xy)*79
217 = (x+y)*79 - (xy)*23
Solve to get (x+y) = 1, xy = -6.
Then ax^4 + by^4 = (x+y)*L4 - (xy)*L3 = 1*691 - (-6)*217 = 1993.
Posted by xdog on 2013-10-07 12:21:31
Search: Search body:
Forums (0) | 306 | 833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2019-18 | latest | en | 0.767054 |
https://richardvigilantebooks.com/what-is-an-e-flat-augmented-chord/ | 1,680,268,451,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949642.35/warc/CC-MAIN-20230331113819-20230331143819-00726.warc.gz | 564,550,006 | 10,176 | # What is an E flat augmented chord?
## What is an E flat augmented chord?
The E flat augmented chord (Eb+) is an E flat Major chord, with a raised 5th. It contains the notes Eb, G and B. Each note of the E flat augmented chord is separated by an interval of a Major 3rd. Because every interval inside the Eb augmented chord is identical, it is known as a symmetrical chord.
What notes are in an e augmented chord?
The E augmented chord contains the notes E, G# and B#. The E+ chord is produced by playing the 1st (root), 3rd and sharp 5th note of the E Major scale.
What is the E flat chord on a piano?
Eb stands for E flat. Theory: The Eb major chord is constructed with a rootThe lowest note in the chord, a major thirdAn interval consisting of four semitones, the 3rd scale degree and a perfect fifthAn interval consisting of seven semitones, the 5th scale degree.
### What is B augmented?
The B augmented chord (B+) is a B Major chord, with a raised 5th. It contains the notes B, D# and Fx (F double sharp). The B augmented chord contains the notes B, D#, Fx. The D# augmented chord contains the notes D#, Fx, Ax (that’s A double sharp, which is the same as B)
What makes up an augmented chord?
As you know, a major chord is built of three notes, the root, the third, and the fifth. An augmented chord shares two of these notes—the root and the third—but the fifth is raised a half step, making it augmented.
Is E flat on the piano?
First of all, locate E flat on you piano keyboard. E flat is the second black key in the set of two black keys on your keyboard. In other words, after you find Eb, play the 4th key, G, then the 3rd key after G, Bb.
#### How do you know if a chord is augmented or diminished?
An augmented chord comprises notes that are spaced apart at wider intervals than those of a regular triad, while a diminished chord is so called because it features narrower intervals than the standard version, making it more compact.
What are the notes in the E flat diminished chord?
The E-flat diminished chord contains 3 notes: Eb, Gb, Bbb. The chord spelling / formula relative to the Eb major scale is: 1 b3 b5. Middle C (midi note 60) is shown with an orange line under the 2nd note on the piano diagram. These note names are shown below on the treble clef followed by the bass clef.
How do you play an augmented chord on the piano?
Augmented chords are another type of triad. They’re made with a root, a major third, and an augmented fifth. To play an augmented chord, all you have to do is play a major chord and raise the upper-note 1/2 step. When you raise that upper-note by 1/2 step, you’ve taken a perfect fifth and augmented it!
## What are the notes of the E flat diminished triad?
This step shows the E-flat diminished triad chord in root position on the piano, treble clef and bass clef. The E-flat diminished chord contains 3 notes: Eb, Gb, Bbb. The chord spelling / formula relative to the Eb major scale is: 1 b3 b5. | 735 | 2,970 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.949525 |
https://www.calculatorbit.com/en/temperature/1-megakelvin-to-centikelvin | 1,679,558,658,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945030.59/warc/CC-MAIN-20230323065609-20230323095609-00677.warc.gz | 800,683,461 | 6,992 | # Convert 1 Megakelvin to Centikelvin
Result:
1 Megakelvin = 100000000 Centikelvin (cK)
Rounded: ( Nearest 4 digits)
1 Megakelvin is 100000000 Centikelvin (cK)
1 Megakelvin is 999726.85C
## megakelvin:
Megakelvin (MK) is a unit of measurement used to describe extremely high temperatures. The prefix "mega" denotes one million, so one megakelvin is equal to one million Kelvin. The Kelvin scale is an absolute temperature scale that is used by scientists to measure extremely high temperatures. The highest temperature ever recorded on Earth is approximately 58 MK, which was achieved during a nuclear test in 1961. Megakelvin is also used to describe the temperatures found in the Sun's core, which are thought to be around 15 MK. While the megakelvin scale is not commonly used in everyday life, it is a valuable tool for scientists who study extremely high temperatures.
## centikelvin:
Centikelvin is a unit of measurement for temperature. It is denoted by the symbol "K" and is equal to one hundredth of a Kelvin. The Kelvin is the SI unit of temperature and is named after Lord Kelvin, who first proposed it in 1848. Centikelvin can be used to measure very low temperatures, such as those found in outer space or in laboratories. It is also sometimes used by scientists working with extremely cold materials, such as liquid nitrogen or liquid helium. When measuring temperature in centikelvin, it is important to remember that zero Kelvin (0 K) is absolute zero, which is the lowest possible temperature. As such, negative values on the centikelvin scale represent temperatures below absolute zero. For example, -273.15 C (-459.67 F) is equal to 0 K, and -100 C (-148 F) is equal to -173.15 K. When using the centikelvin scale, it is also important to keep in mind that 1 K equals 1 degree Celsius. Therefore, 100 C would be equal to 173.15 K. In conclusion, centikelvin is a unit of measurement for temperature that is denoted by the symbol "K." It can be used to measure very low temperatures and is sometimes used by scientists working with extremely cold materials.
## Megakelvin to Centikelvin Calculations Table
Now by following above explained formulas we can prepare a Megakelvin to Centikelvin Chart.
Megakelvin (MK) Centikelvin (cK)
1 100000000.00000001
2 200000000.00000003
3 300000000.00000006
4 400000000.00000006
5 500000000.00000006
6 600000000.0000001
Nearest 4 digits
## Convert from Megakelvin to other units
Here are some quick links to convert 1 Megakelvin to other temperature units.
## Convert to Megakelvin from other units
Here are some quick links to convert other temperature units to Megakelvin.
## FAQs About Megakelvin and Centikelvin
Converting from one Megakelvin to Centikelvin or Centikelvin to Megakelvin sometimes gets confusing. | 702 | 2,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.92445 |
https://id.scribd.com/document/119856106/Algebra-II-Chapter-3-Test | 1,563,501,944,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525973.56/warc/CC-MAIN-20190719012046-20190719034046-00216.warc.gz | 429,105,238 | 66,963 | Anda di halaman 1dari 9
# Name: ______________________
Class: _________________
Date: _________
ID: A
## Algebra II Chapter 3 Test
Solve the system by graphing. State whether the stystem is independent, dependent, or inconsistent. 1. 3x 2y = 7 3x y = 10 Without graphing, is the system independent, dependent, or inconsistent? 2. 12x + 3y = 12 y = 4x + 5 Solve the system by substitution. 3. 2x y = 14 3x y = 11
## Solve the system using elimination. 4. 4x + 4y = 8 x 4y = 7
Use Substitution or Elimination. What are the solutions of the following systems? 5. x + 2y = 10 3x + 6y = 11
ID: A
## Solve the system of inequalities by graphing. 7. y 4 y > |x 1 |
Name: ______________________
ID: A
Graph the system of constraints and find the value of x and y that maximize the objective function. 8. Constraints x 0 y 0 y 1 3 x+2
## 6 y+x Objective function: C = 7x 3y
What is the solution of the system? Solve by hand using a matrix. You may use your calculator to double check your answer. You must show all of your work to receive full credit. 9. 2x + 6y = 38 5x y = 15
Name: ______________________
ID: A
Solve the system by elimination. You must show all your work to get full credit. (Use the matrix function on your calculator to check your solution) 10. x + 3y + z = 6 2x + y + 3z = 4 3x 3y 3z = 6
Solve the system by substitution. You must show all your work to get full credit. (Use the matrix function on your calculator to check your solution) 11. 2x y + z = 4 z = 5 2x + 3y z = 10
## 12. What is element a 23 in matrix A? 3 A= 8 0 7 4 4 6 6 0
Name: ______________________ How can you represent the system of equations with a matrix? 13. 4x 5y = 6 2x 4y = 2 What linear system of equations does the matrix represent? 14. 2 14 | | 9 | 11 | 2 || 15
ID: A
What is the solution of the system of equations? (Use a any method from the chapter so solve. Show your work) 15. 5x + y 5z = 1 2x + 2y 3z = 13 3x y 5z = 25
16. A food store makes a 11-lb mixture of peanuts, almonds, and raisins. The cost of peanuts is \$1.50 per pound, almonds cost \$3.00 per pound, and raisins cost \$1.50 per pound. The mixture calls for twice as many peanuts as almonds. The total cost of the mixture is \$21.00. How much of each ingredient did the store use?
ID: A
## Algebra II Chapter 3 Test Answer Section
1. ANS:
(3, 1) PTS: 1 DIF: L2 REF: 3-1 Solving Systems Using Tables and Graphs OBJ: 3-1.1 To solve a linear system using a graph or a table NAT: CC A.CED.2| CC A.CED.3| CC A.REI.6| CC A.REI.11| A.4.d TOP: 3-1 Problem 1 Using a Graph or Table to Solve a System KEY: system of linear equations | graphing | solution of a system 2. ANS: inconsistent PTS: OBJ: NAT: TOP: KEY: 3. ANS: (5, 4) PTS: OBJ: NAT: TOP: KEY: 1 DIF: L2 REF: 3-1 Solving Systems Using Tables and Graphs 3-1.1 To solve a linear system using a graph or a table CC A.CED.2| CC A.CED.3| CC A.REI.6| CC A.REI.11| A.4.d 3-1 Problem 4 Classifying a System Without Graphing system of linear equations | inconsistent system
1 DIF: L2 REF: 3-2 Solving Systems Algebraically 3-2.1 To solve linear systems algebraically CC A.CED.2| CC A.CED.3| CC A.REI.5| CC A.REI.6| A.4.d 3-2 Problem 1 Solving by Substitution system of linear equations | substitution method
ID: A 4. ANS: (5, 3) PTS: 1 DIF: L2 REF: 3-2 Solving Systems Algebraically OBJ: 3-2.1 To solve linear systems algebraically NAT: CC A.CED.2| CC A.CED.3| CC A.REI.5| CC A.REI.6| A.4.d TOP: 3-2 Problem 3 Solving by Elimination KEY: system of linear equations | solve by elimination 5. ANS: no solutions PTS: OBJ: NAT: TOP: KEY: 6. ANS: 1 DIF: L2 REF: 3-2 Solving Systems Algebraically 3-2.1 To solve linear systems algebraically CC A.CED.2| CC A.CED.3| CC A.REI.5| CC A.REI.6| A.4.d 3-2 Problem 5 Solving Systems Without Unique Solutions system of linear equations | solve by elimination | no solutions
## PTS: OBJ: NAT: TOP:
1 DIF: L3 REF: 3-3 Systems of Inequalities 3-3.1 To solve systems of linear inequalities CC A.CED.3| CC A.REI.6| CC A.REI.12| A.4.d 3-3 Problem 2 Solving a System by Graphing KEY: system of inequalities | graphing
ID: A 7. ANS:
PTS: OBJ: NAT: TOP: KEY: 8. ANS: (3, 0) PTS: OBJ: TOP: KEY: 9. ANS: (4, 5)
1 DIF: L2 REF: 3-3 Systems of Inequalities 3-3.1 To solve systems of linear inequalities CC A.CED.3| CC A.REI.6| CC A.REI.12| A.4.d 3-3 Problem 4 Solving a Linear/Absolute-Value Systems system of inequalities | graphing | absolute value
1 DIF: L3 REF: 3-4 Linear Programming 3-4.1 To solve problems using linear programming NAT: CC A.CED.3| A.4.d 3-4 Problem 1 Testing Vertices linear programming | constraints | vertices | objective function | maximum value
PTS: 1 DIF: L3 REF: 3-6 Solving Systems Using Matrices OBJ: 3-6.2 To solve a system of linear equations using matrices NAT: CC A.REI.8| A.4.d TOP: 3-6 Problem 4 Solving a System Using a Matrix KEY: systems of equations | matrices 10. ANS: (2, 2, 2) PTS: 1 DIF: L2 REF: 3-5 Systems With Three Variables OBJ: 3-5.1 To solve systems in three variables using elimination NAT: CC A.REI.6| A.4.d TOP: 3-5 Problem 1 Solving a System Using Elimination KEY: system with three variables | solve by elimination 11. ANS: (8, 7, 5) PTS: OBJ: NAT: KEY: 1 DIF: L2 REF: 3-5 Systems With Three Variables 3-5.2 To solve systems in three variables using substitution CC A.REI.6| A.4.d TOP: 3-5 Problem 3 Solving a System Using Substitution system with three variables | substitution method 3
ID: A 12. ANS: 6 PTS: OBJ: NAT: KEY: 13. ANS: 4 2 1 DIF: L2 REF: 3-6 Solving Systems Using Matrices 3-6.1 To represent a system of linear equations with a matrix CC A.REI.8| A.4.d TOP: 3-6 Problem 1 Identifying a Matrix Element matrix | matrix element | | 5 | 6 | | 2 4 |
PTS: 1 DIF: L4 REF: 3-6 Solving Systems Using Matrices OBJ: 3-6.1 To represent a system of linear equations with a matrix NAT: CC A.REI.8| A.4.d TOP: 3-6 Problem 2 Representing Systems With Matrices KEY: systems of equations | matrices 14. ANS: 2x 9y = 11 14x 2y = 15 PTS: 1 DIF: L2 REF: 3-6 Solving Systems Using Matrices OBJ: 3-6.1 To represent a system of linear equations with a matrix NAT: CC A.REI.8| A.4.d TOP: 3-6 Problem 3 Writing a System From a Matrix KEY: systems of equations | matrices 15. ANS: (3, 1, 3) PTS: 1 DIF: L4 REF: 3-6 Solving Systems Using Matrices OBJ: 3-6.2 To solve a system of linear equations using matrices NAT: CC A.REI.8| A.4.d TOP: 3-6 Problem 5 Using a Calculator to Solve a Linear System KEY: systems of equations | matrices 16. ANS: 6 lb peanuts, 3 lb almonds, 2 lb raisins PTS: OBJ: NAT: KEY: 1 DIF: L2 REF: 3-5 Systems With Three Variables 3-5.2 To solve systems in three variables using substitution CC A.REI.6| A.4.d TOP: 3-5 Problem 4 Solving a Real-World Problem system with three variables | substitution method | word problem | problem solving | 2,313 | 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} | 4.3125 | 4 | CC-MAIN-2019-30 | latest | en | 0.819753 |
https://ttmobile.com.vn/how-many-days-is-20000-hours.html | 1,721,219,875,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514771.72/warc/CC-MAIN-20240717120911-20240717150911-00549.warc.gz | 506,356,130 | 59,722 | # 10 topics : how many days is 20000 hours ?
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How Many Days is 20000 Hours? – A Comprehensive Guide
# How Many Days is 20000 Hours? – A Comprehensive Guide
## Introduction
Have you ever wondered how many days is 20000 hours? Whether you need to calculate the duration of a project or simply want to know how long you’ve been binge-watching your favorite TV show, knowing how to convert hours to days can be useful. In this article, we’ll show you how to do just that.
## How to Convert Hours to Days
To convert hours to days, you need to divide the number of hours by 24, which is the number of hours in a day. The formula is:
Number of Days = Number of Hours ÷ 24
For example, to find out how many days is 20000 hours, you would use the following calculation:
Number of Days = 20000 ÷ 24 = 833.33
So, 20000 hours is equivalent to 833.33 days.
## Examples of How to Use the Conversion
Now that you know how to convert hours to days, let’s look at some examples of how to use this conversion in real-life situations:
• Project Management: If a project is estimated to take 20000 hours to complete, you can convert that to days to get a better sense of the timeline. In this case, the project would take approximately 833 days to complete.
• Travel Planning: If you’re planning a trip and want to know how many days you’ll be away, you can convert the number of hours you’ll be traveling to days. For example, if your flight is 12 hours long, that’s equivalent to 0.5 days.
• Time Tracking: If you’re tracking your time for work or personal reasons, you can convert the number of hours you’ve worked or spent on a task to days. This can help you see how much time you’re dedicating to certain activities over a longer period of time.
## Conclusion
Knowing how to convert hours to days can be a useful skill in many different situations. By using the formula we’ve provided, you can easily calculate how many days is 20000 hours or any other number of hours. Whether you’re managing a project, planning a trip, or tracking your time, this conversion can help you make more informed decisions and better understand the duration of different activities.
You are looking : how many days is 20000 hours
## 10 how many days is 20000 hours for reference
### 1.Convert 20000 Hours to Days and Hours – Online Stopwatch
• Publish: 15 days ago
• Rating: 3(373 Rating)
• Highest rating: 3
• Lowest rating: 2
• Descriptions: Here is a 833 Days and 8 Hours Timer: · 5 · 6 · 7 · 8 · 9.
• Source : https://www.online-stopwatch.com/hourstodays/20000-hours-to-days/
### 2.Convert 20000 hours to days – Time Calculator
• Publish: 19 days ago
• Rating: 2(1351 Rating)
• Highest rating: 4
• Lowest rating: 1
• Descriptions: 20,000 hours is equal to 833.33 days. Hours to Days | Previous | Next. convert 20,000 hours into Nanoseconds, Microseconds, Milliseconds, …
• Source : https://unitconverter.fyi/en/20000-h-d/20000-hours-to-days
### 3.How many days in 20000 hours? – CoolConversion
• Publish: 13 days ago
• Rating: 3(1611 Rating)
• Highest rating: 3
• Lowest rating: 1
• Descriptions: There are 833.333 days in 20000 hours. To convert any value from hours into days, simply multiply the hours by the multiplication factor, also known as the …
• Source : https://coolconversion.com/time/converter/20000-hour-to-day
### 4.20000 Hours to Days | 24hourtime.net
• Publish: 24 days ago
• Rating: 5(359 Rating)
• Highest rating: 4
• Lowest rating: 1
• Descriptions: Thus, 20000 hours in days = 833.3333333333 d (decimal). The non-decimal conversion to days and hours is located below the following the chart. Hours, Days.
• Source : https://24hourtime.net/20000-hours-to-days
### 5.How Many Days is 20000 Hours? – Online Calculator
• Publish: 6 days ago
• Rating: 2(1746 Rating)
• Highest rating: 4
• Lowest rating: 2
• Descriptions: How many days is 20000 hours? – 20000 hours equals 833.3 days or there are 833.3 days in 20000 hours. 20000 hours in days will convert 20000 hours to days, …
• Source : https://online-calculator.org/how-many-days-is-20000-hours
### 6.How Many Days Are In 20,000 Hours? – Calculatio
• Publish: 11 days ago
• Rating: 3(1195 Rating)
• Highest rating: 5
• Lowest rating: 3
• Descriptions: 20000 Hours to Days (Days in 20000 Hours). What is 20000 Hours in Days? The answer is 833.33. Convert Hours/Days/Weeks/Months/Years to …
• Source : https://calculat.io/en/date/converter/days–20000–hours
### 7.Convert 20,000 Hours to Years – CalculateMe.com
• Publish: 17 days ago
• Rating: 1(1780 Rating)
• Highest rating: 5
• Lowest rating: 3
• Descriptions: In the Gregorian calendar, a year has on average 365.2425 days. It is based on the amount of time it takes for the Earth to rotate the sun. Hours to …
• Source : https://www.calculateme.com/time/hours/to-years/20000
### 8.Convert 20000 Hours to Days (20000 hr to d) – Rapid Calculation
• Publish: 24 days ago
• Rating: 2(1235 Rating)
• Highest rating: 3
• Lowest rating: 2
• Descriptions: Answer: 20000 Hours equals 833.333333333 Days. How long is 20000 Hours compared to other units of Time? This …
• Source : https://rapidcalculation.com/time/20000-hours-to-days.php
### 9.Convert 20000 Hours to Days – Converter.net
• Publish: 11 days ago
• Rating: 2(1294 Rating)
• Highest rating: 3
• Lowest rating: 3
• Descriptions: A Hour has 0.04 Days. Day. A Day is a unit of Time in the metric System. It has the symbol d. A Day has 24 …
• Source : https://converter.net/time/20000-hours-to-days
### 10.20000 Hours to Days | Convert 20000 hr in days – UnitChefs
• Publish: 20 days ago
• Rating: 3(431 Rating)
• Highest rating: 3
• Lowest rating: 3
• Descriptions: 20000 Hours (hr) to Days. We provide the most accurate information about how to convert Hours in Days. Try Our Converter Now!
• Source : https://unitchefs.com/hours/days/20000/ | 1,622 | 5,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-30 | latest | en | 0.912535 |
https://math.stackexchange.com/questions/3213568/a-family-of-polynomials-for-the-taylor-expansion-of-the-lambert-w-function | 1,561,507,688,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999964.8/warc/CC-MAIN-20190625233231-20190626015231-00545.warc.gz | 495,366,357 | 36,991 | # A family of polynomials for the Taylor expansion of the Lambert-w function
This is an attempt to derive the Taylor expansion at $$0$$ of the Lambert $$w$$ function (the branch containing the origin of the solution $$w$$ to the functional equation $$w(x)e^{w(x)}=x$$) without resorting to Lagrange inversion formula, nor complex analysis.
Let the superscript $$^{(n)}$$, (superscript $$^{'}$$) denote the $$n$$-th derivative with respect to $$x$$ (to $$w$$ respectively) . It is easy to see that \begin{align*} w^{(1)}=\frac{w}{x}\frac{1}{w+1} .\end{align*} Then by iterating the derivation process we obtain \begin{align*} &w^{(2)}=-\frac{w^2}{x^2}\frac{w+2}{(w+1)^3} \\ &w^{(3)}=\frac{w^3}{x^3}\frac{2w^2+8w+9}{(w+1)^5}\\ &\qquad\qquad..... \end{align*} Now we start an induction process and we suppose that there is some polynomial $$P_n$$ such that \begin{align*} &w^{(n)}=-\frac{(-w)^n}{x^n(w+1)^{2n-1}}P_n(w) \end{align*} After some algebra we see that these polynomials must obey the following recursion:
\begin{align*} P_{1}(w)&=1\\P_{n+1}(w)&=\big(n(w +3)-1\big)P_n(w)-(w+1)P_n^{'}(w). \end{align*}
What is the name of these polynomials?
How can it be shown that $$P_n(0)=n^{n-1}$$?
First instances: \begin{align*} P_1(w)&= \color{red}{1}\\ P_2(w)&=w+\color{red}{2}\\ P_3(w)&=2w^2+8w+\color{red}{9}\\ P_4(w)&=6w^3+36w^2+79w+\color{red}{64}\\ P_5(w)&=24w^4+192w^3+622w^2+974w+\color{red}{625}\\ \end{align*}
Or consider the triangle of numbers $$a_{n,k}$$ obtained recursively, such that the entries in line $$n$$ are computed with three consecutive entries from line $$n-1$$ by $$a_{n,k}=n a_{n-1,k-1}+(3n-(k+1))a_{n-1,k}-(k+1)a_{n-1,k+1}$$ with initial conditions $$a_{n,-1}= 0$$ , $$a_{0,0}= 1$$ and $$a_{0,k}= 0$$ for $$k>0$$.
This makes it clear that $$a_{n,n}=n!$$, but how could we obtained the closed form $$a_{n,0}=(n+1)^n$$ ?
• oeis.org/A042977 "Triangle $T(n,k)$ read by rows: coefficients of a polynomial sequence occurring when calculating the n-th derivative of Lambert function $W$." – Travis May 6 at 22:28
• These polynomials also appear as factors of the expressions in Example 4.3 of Kruchinin, V., "Derivation of Bell Polynomials of the Second Kind" arxiv.org/abs/1104.5065 – Travis May 6 at 22:30
It follows from a comment by Vladimir Kruchinin in OEIS A042977, "Triangle $$T(n,k)$$ read by rows: coefficients of a polynomial sequence occurring when calculating the $$n$$th derivative of Lambert function $$W$$" that $$\color{#df0000}{\boxed{P_n(w) = \sum_{m = 0}^n \left[ \sum_{j=0}^m {2 n + 1 \choose m - j} \sum_{k=0}^j (-1)^k \frac{(n+k+1)^{n+j}}{(j-k)! k!} \right] w^m}} .$$ See under the heading $$\texttt{FORMULA}$$, and NB that the expression in this answer differs from the one in OEIS entry by a factor of $$(-1)^n$$ to account for the formulation in the question. This expression can be deduced from Example 4.3 in Kruchinin's preprint "Derivation of Bell Polynomials of the Second Kind".
In particular, to compute the constant term of $$P_n(w)$$, we can evaluate the inner sum for $$j = 0$$. Both summations have only a single term, and it simplifies to the conjectured formula, $$\color{#df0000}{\boxed{P_n(0) = (n + 1)^n}} .$$
• This is very relevant and useful, and since I requested for reference, I accept it as an answer; but I agree that it is not entirely satisfactory: the formula for $P_n(w)$ was obtained by making use of the Lambert w function. Given the recursion only, without any other contextual info, would there be a proof that does not make use of Lambert w ? – René Gy May 7 at 22:55 | 1,197 | 3,555 | {"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": 35, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-26 | latest | en | 0.726512 |
https://codegolf.stackexchange.com/questions/130064/pascals-rhombus/130068 | 1,582,647,900,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146123.78/warc/CC-MAIN-20200225141345-20200225171345-00186.warc.gz | 318,124,138 | 43,131 | # Pascal's Rhombus
Pascal's Rhombus (which is actually a triangle) is obtained by adding in the pattern:
*
***
x
* *
x
This means that each cell is the sum of the three cells on the row directly above it and one cell on the row 2 above it. Just like Pascal's triangle the zeroth row has a single 1 on it that generates the triangle.
Here are the first couple of rows of Pascal's Rhombus
1
1 1 1
1 2 4 2 1
1 3 8 9 8 3 1
Given a row number (starting from the top) and an column number (starting from the first non-zero item on that row) output the value at that particular cell. Both inputs may be either 1 or 0 indexed (you may mix and match if you desire).
This is so you should aim to make the file size of your source code as a small as possible.
OEIS A059317
Pascal's Rhombus? More like Haskell's Rhombus! amiright?
4 bytes saved thanks to Ørjan Johansen
I thought I'd have a go at my own problem and practice my Haskell. Hopefully this will inspire more people to answer this.
1!1=1
n!k=sum[(n-2)!(k-2)+sum(map((n-1)!)[k-2..k])|n>1]
Try it online!
## Explanation
This is a bit out of date with the latest golf
*
***
x
We calculate
*
***
x
This slants our entire triangle to become
1
1 1 1
1 2 4 2 1
1 3 8 9 8 3 1
This lines up all of our rows making it easier to index the nth item of any column. We then define our base cases.
The zeroth row is all zeros so
0!_=0
There is a single 1 at position 1,1 so we define that
1!1=1
And we define the rest of the first row to be zeros as well
1!_=0
Then we define the general case recursively using the pattern described above:
n!k=(n-2)!(k-2)+(sum$map((n-1)!)[k-2..k]) • Beat me to it! This is also a lot cleaner than mine. – Julian Wolf Jul 5 '17 at 23:16 • @JulianWolf Sorry about that, when I posted this it looked like no one other than Jorg was doing the problem. I'd still like to see your solution. – Post Rock Garf Hunter Jul 5 '17 at 23:20 • You can save four bytes with n!k=sum[(n-2)!(k-2)+sum(map((n-1)!)[k-2..k])|n>1]. – Ørjan Johansen Jul 5 '17 at 23:50 # Pascal, 122 bytes Well, it's Pascal's rhombus. 37 bytes saved thanks to @manatwork function f(n,k:integer):integer;begin f:=1-Ord((k<0)or(k>n*2));if n>0then f:=f(n-1,k-2)+f(n-1,k-1)+f(n-1,k)+f(n-2,k-2)end; Try it online! • Parenthesis around the entire if condition are pointless. (On 1st if you save 2 characters, on the 2nd if 1 character by leaving no space between then keyword and the preceding digit.) Oh, and variable r is completely unnecessary. – manatwork Jul 6 '17 at 8:51 • Weird animal that Pascal on Ideone. Never seen double quotes delimited strings in any Pascal variant before. One more thing you can remove: the ; before the function's end. – manatwork Jul 6 '17 at 9:24 • @manatwork yea, now when you mentioned it, all of the other online editors complained about it – Uriel Jul 6 '17 at 9:28 • @manatwork I'm not sure I understood. wouldnt that just lengthen the code with the >= <=? I still need to preserve the if n=0 – Uriel Jul 6 '17 at 9:44 • Sorry @Uriel, I don't have that version anymore. Currently I'm at function f(n,k:integer):integer;begin f:=1-Ord((k<0)or(k>n*2));if n>0then f:=f(n-1,k-2)+f(n-1,k-1)+f(n-1,k)+f(n-2,k-2)end; – manatwork Jul 6 '17 at 9:55 # PHP, 86 bytes recursive way only the function row and column 0-Indexed function f($r,$c){return$r|$c?$r<0?0:f($r-=1,$c)+f($r,$c-1)+f($r,$c-=2)+f($r-1,$c):1;}
Try it online!
# PHP, 114 bytes
recursive way full program row and column 0-Indexed
<?=f(...$_GET);function f($r,$c){return$r|$c?$r<0|$c<0|$c>2*$r?0:f($r-=1,$c)+f($r,$c-1)+f($r,$c-=2)+f($r-1,$c):1;} Try it online! # PHP, 129 bytes row and column 0-Indexed for(;$r<=$argv[1];$l=$t[+$r++])for($c=~0;$c++<$r*2;)$t[+$r][$c]=$r|$c?$t[$r-2][$c-2]+$l[$c]+$l[$c-1]+$l[$c-2]:1;echo$l[$argv[2]]; Try it online! • and +1 for actually improving it :) – Uriel Jul 6 '17 at 11:59 # Jelly, 2220 19 bytes 3ḶṚp@Ḣḣ4 Ḟ_ЀÇ߀ȯ¬S Takes a 0-based index pair as command-line argument. Try it online! # MATL, 2220 19 bytes Ti:"2Y6Y+FT_Y)]!i_) Both inputs are 0-based. Try it online! ### Explanation Let r and c denote the two inputs, specifying 0-based row and column respectively. Each new row in Pascal's rhombus can be built from the matrix containing the previous two rows by convolving with the kernel [1 1 1; 0 1 0] and keeping the last two rows of the result swapped. This is done r times, starting from matrix 1. It turns out to be shorter to use the kernel [0 1 0; 1 1 1; 0 1 0], which is a predefined literal. This produces an extra row, which will be discarded. Consider for example r = 3, so there are 3 iterations. 1. Starting from 1 convolution with [0 1 0; 1 1 1; 0 1 0] gives 0 1 0 1 1 1 0 1 0 Keeping the last two rows (the whole matrix, in this case) and swapping them gives 0 1 0 1 1 1 2. Convolution of the above with [0 1 0; 1 1 1; 0 1 0] gives 0 0 1 0 0 0 1 1 1 0 1 2 4 2 1 0 1 1 1 0 The matrix formed by the last two rows swapped is 0 1 1 1 0 1 2 4 2 1 This contains the new row at the bottom, and the preceding one extended with zeros. 3. Convolving again yields 0 0 1 1 1 0 0 0 1 2 3 2 1 0 1 3 8 9 8 3 1 0 1 2 4 2 1 0 Taking the last two rows swapped gives 0 1 2 4 2 1 0 1 3 8 9 8 3 1 After the r iterations have been done, the output is contained in the last row of the final matrix. For example, for c = 2 (0-based) the result would be 8. Instead of indexing the last row and the desired column, a trick can be used which exploits the symmetry of each row: the final matrix is transposed 0 1 1 3 2 8 4 9 2 8 1 3 0 1 and its -c-th element is taken. This uses linear indexing, that is, the matrix is indexed by a single index in column-major order. Since indexing is modular, the 0-entry is the lower-right corner (value 1) and the -2-th entry is two steps above (value 8). T % Push true i % Input row number :" % Do the following that many times 2Y6 % Push predefined literal [0 1 0; 1 1 1; 0 1 0] Y+ % 2D convolution, increasing size FT_ % Push [0 -1] Y) % Matrix with rows 0 (last) and -1 (second-last), in that order ] % End ! % Transpose i % Input: colun number _ % Negate ) % Entry with that index. Implicitly display # Pari/GP, 60 bytes i->j->polcoeff(Vec(1/(1-x*(1+y+y^2+x*y^2))+O(x^i++))[i],j,y) Try it online! ## Haskell, 74 bytes 0#0=1 n#m|m<=2*n&&m>=0=sum[(n-a)#(m-b)|(a,b)<-zip[2,1,1,1]$2:[0..2]]
n#m=0
Try it online!
Call with n # m, where n is the row and m is the column.
• m<=2*n&&m>=0 can be just n>0. – Ørjan Johansen Jul 6 '17 at 0:06
# Mathematica, 56 bytes
If[#<1,Boole[##==0],Sum[#0[#-i,#2-j],{i,2},{j,2i-2,2}]]&
Pure function taking two integer arguments (row first, column second) and returning an integer. Works for negative integer arguments as well, returning 0. A pretty straightforward recursive structure: If[#<1,Boole[##==0],...] defines the base-case behavior for the 0th row (and above), while Sum[#0[#-i,#2-j],{i,2},{j,2i-2,2}] implements the recursive definition.
# Python 2, 7066 65 bytes
f=lambda n,k:(k==0)|sum(f(n+~j/3,k-j+j/3)for j in range(4)[:3*n])
Try it online!
## JavaScript (ES6), 68 bytes
f=(y,x)=>x<0|x>y+y?0:x>0&x<y+y?f(--y,x)+f(y,--x)+f(y,--x)+f(--y,x):1
# Mathematica, 53 bytes
D[1/(1-x(1+y+y^2(1+x))),{x,#},{y,#2}]/#!/#2!/.x|y->0&
Using the generating function.
# Python 3, 82 84 bytes
This is a recursive implementation with 1-indexed rows and columns. (Technically needs an f= in front, someone let me know if I should change it to 84 bytes. Still new and not 100% sure of the rules.)
This uses the recursive formula found on the OEIS page, but with the k's shifted one to the left to line up properly. Coincidently, sum(f(n-1,k-i)for i in(0,1,2)) is the same size as f(n-1,k)+f(n-1,k-1)+f(n-1,k-2). The whole function is the Python and or trick, where the first condition checks if k is inside the triangle and not on the boundary, in which case the recursive formula is used. If is isn't, the part after the or is returned, which checks if k is in (1, 2*n-1), i.e. on the boundary, returning True and False. k+1in(2,2*n) is one byte shorter than k in(1,2*n-1). Wrapping that in parentheses and putting a + in front converts to integer, which is what is needed.
f=lambda n,k:2*n-1>k>1and sum(f(n-1,k-i)for i in(0,1,2))+f(n-2,k-2)or+(k+1in(2,2*n))
Try it online!
• Recursive functions need the f=. – Post Rock Garf Hunter Jul 6 '17 at 0:11
• While I personally disagree with it according to this somewhat buried meta consensus, you may output True instead of 1 because it behaves like 1 to python. This allows you to remove the +(...) at the end. I understand if you don't want to do this, because it will make the output look a little strange, it is an option. – Post Rock Garf Hunter Jul 6 '17 at 3:55
• @WheatWizard Wow that's very interesting. Thanks for the tip. – C McAvoy Jul 6 '17 at 13:42
# Java (OpenJDK 8), 87 bytes
int f(int r,int c){return c<0|2*r<c?0:0<c&c<2*r?f(--r,c)+f(r,--c)+f(r,--c)+f(--r,c):1;}
Try it online!
At first, I was happy with my 160 bytes iterative method... Hmmm... let's just forget about it, OK?
# Python 3, 75 Bytes
This is a recursive lambda that takes column and row as 0-indexed integers.
p=lambda r,c:(r<0 or((c==0)|p(r-1,c-2)+p(r-1,c)+p(r-1,c-1)+p(r-2,c-2))+1)-1
Here's a (slightly) more readable version with a printing function:
p = lambda r,c:(r<0 or ((c==0) | p(r-1,c-2)+p(r-1,c)+p(r-1,c-1)+p(r-2,c-2))+1)-1
def pp(r):
ml = len(str(p(r,r)))+1
for i in range(0, r):
a=" "*ml*(r-i)
for j in range(0,i*2 + 1):
a+=str(p(i,j))+(" "*(ml-len(str(p(i,j)))))
print(a) | 3,353 | 9,530 | {"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} | 3.53125 | 4 | CC-MAIN-2020-10 | latest | en | 0.874714 |
https://redbloxgames.com/faq/faq-how-to-play-snooker-shots.html | 1,643,447,088,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300573.3/warc/CC-MAIN-20220129062503-20220129092503-00595.warc.gz | 517,862,117 | 9,723 | # FAQ: How To Play Snooker Shots?
## How do you play snooker for beginners?
You get points by potting balls in the correct order. You have to pot all of the red balls on the table before you can directly hit the coloured balls with the cueball. When all of the snooker balls (red and coloured) are still on the table, you have to pot a red ball before you can attempt to pot a coloured ball.
## Is there a time limit between snooker shots?
There doesn’t appear to be any time limit. You can look at Peter Ebdon for example. Man took over five minutes to play a break of 12, while Ronnie achieved a 147 in less than that.
## Why is snooker so hard?
Why is snooker so hard? Well it requires a few different things to work together and in coordination for things to gel properly. Firstly, the technical side of the game, which entails the things we control physically such as our stance, cue action, approach to the shot and so on.
## Is snooker easy to learn?
Snooker is an extremely difficult game to master. It’s hard work. It’s very technical. You’ll need to practice.
## How do you remember the order of snooker balls?
This order is often remembered using the mnemonic God Bless You, the first letter of each word being the first letter of the three colours (Green, Brown, Yellow). The blue ball rests at the exact centre of the table, while the pink is placed midway between it and the top cushion.
You might be interested: FAQ: How To Play Tmp Video Files?
## How do you check if a snooker cue is straight?
“The correct way to sight the straightness of a snooker cue is to hold the cue at the butt end and look down the length of the cue with one eye – like when you would shoot a rifle, then turn the butt of the cue 360 degrees while continuing to sight down the cue, you will then be able to tell if the cue is straight or
## What are the basic rules of snooker?
The Rules of Snooker
• Red ball = 1 Point.
• Yellow Ball = 2 Points.
• Green Ball = 3 Points.
• Brown Ball = 4 Points.
• Blue Ball = 5 Points.
• Pink Ball = 6 Points.
• Black Ball = 7 points.
## What are the Coloured balls worth in snooker?
The game is played with 22 balls, made up of one white ball (the cue ball); 15 red balls, valued at 1 point each; one yellow, 2 points; one green, 3; one brown, 4; one blue, 5; one pink, 6; and one black, 7. | 590 | 2,337 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-05 | latest | en | 0.92126 |
https://www.physicsforums.com/threads/electrostatics-question-involving-mass.689375/ | 1,519,561,693,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816370.72/warc/CC-MAIN-20180225110552-20180225130552-00017.warc.gz | 939,072,291 | 16,883 | # Electrostatics question, involving mass?
1. May 2, 2013
### hangingwire
1. The problem statement, all variables and given/known data
One electron has a mass of 9.11x10^-31 KG. Find the force 1.0 gram of electrons would exert on another 1.0g of electrons from 1.0 kilometer away
F = ?
Melectron = 9.11x10^-31 kg --> (0.01kg of electron = 9.11x10^-33 kg)
Electron Coloumb = 1.6x10^-19 C (ignoring the negative sign b/c it is a vector)
R = 1000m
2. Relevant equations
F=(kQ1Q2/R^2)
Since they are the same mass
F = kQ^2/R^2
3. The attempt at a solution
K = 9x10^9
R = 1000^2
What is the value for Q? I am not sure because any value I try to enter is invalid.
Answer is 2.8x10^20N and I am getting a negative power answer when I try. Can someone steer me in the right direction please?
2. May 2, 2013
### SammyS
Staff Emeritus
What is the number of electrons need so that you have a mass of 1 gram of electrons?
3. May 2, 2013
### hangingwire
1/(9.11x10^-28) = 1.1x10^27 electrons per gram.
if one electron is 1.6x10^-19c
(1.1x10^27)(1.6x10^-19) = 1.76x10^8 is one charge of a gram of electrons
Plugging it in...
(9x10^9)x(1.76x10^8)^2
-------------------------
(1000m)^2
answer = 1.584x10^12 CORRECTED! --> 2.78784x10^20 which is correct
Still off :|
Edit: NVM! Correct answer. Stupid Casio fx-300ES calculator got it mixed up somewhere. Hate how I have to keep pressing Shift to get the answer in decimals -.- Thank you!
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Draft saved Draft deleted | 524 | 1,594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-09 | longest | en | 0.851402 |
http://www.hsn.uk.net/forum/index.php?/topic/4298-differentiationintegration/page__pid__76588__st__20#entry76588 | 1,642,724,689,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302706.62/warc/CC-MAIN-20220120220649-20220121010649-00535.warc.gz | 100,036,868 | 12,749 | Warning: Illegal string offset 'html' in /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php on line 909
Differentiation/integration - HSN forum - Page 2
# Differentiation/integration
28 replies to this topic
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Posted 21 April 2006 - 02:25 PM
QUOTE(dfx @ Apr 21 2006, 03:02 PM)
QUOTE(George @ Apr 20 2006, 11:00 PM)
(have you ever tried it with sinx? )
Oh yes I was recently challenged by my maths teacher to differentiate from first principles. I believe it involves the taylor and maclaurin expansions for Sinx... but yeah I gave up lol.
You can do it with higher maths material. The only identities you need are:
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### #22dfx
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Posted 21 April 2006 - 03:03 PM
Yes the but I'm pretty sure you still need the taylor and maclaurin series... however I stand corrected!
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Posted 21 April 2006 - 04:54 PM
I'm not even going to pretend I know what they are
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### #24George
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Posted 22 April 2006 - 01:21 PM
Here's the way I approached it:
However, that relies on two facts, and . The sinx / x limit requires a lot of work if you're being rigorous, and the cos limit follows from that quite easily.
Now, in practice, nobody would do all that work every time they needed to differentiate sinx!
The whole point of maths is to come up with theorems and rules. Once they're proved, it makes sense to just use the result - there's no point reinventing the wheel all the time!
### #25dfx
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Posted 22 April 2006 - 02:43 PM
I was puzzling over why you delved into Cosh and Sinh when I realized it was h for height.
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Posted 23 April 2006 - 12:33 PM
QUOTE(George @ Apr 22 2006, 02:21 PM)
Here's the way I approached it:
However, that relies on two facts, and . The sinx / x limit requires a lot of work if you're being rigorous, and the cos limit follows from that quite easily.
Now, in practice, nobody would do all that work every time they needed to differentiate sinx!
The whole point of maths is to come up with theorems and rules. Once they're proved, it makes sense to just use the result - there's no point reinventing the wheel all the time!
This is also how I done it, except I didn't know how to calculate the limits myself and had to cheat a little with them.
dfx, h is meant to represent a small change in x, that's why the differentiated function is f(x+h) - f(x) / h - i.e. the change in f(x) over a change in x. Obviously the most accurate formula for the gradient will come as h is closer to zero.
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### #27dfx
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Posted 23 April 2006 - 04:15 PM
hehe yep.
### #28Vyka
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Posted 28 April 2006 - 01:45 PM
QUOTE(Daniel Williamson @ Mar 27 2006, 08:51 PM)
Any easy ways to remember the difference between the two? add one to power divide by new power i mean come on
PRACTICE!!! Just like the rest of us hahaha
### #29ScotlandGirl
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Posted 30 April 2006 - 07:33 PM | 1,038 | 3,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.917929 |
https://greatgreenwedding.com/what-is-the-decimal-ascii-code-of-a/ | 1,709,551,092,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476442.30/warc/CC-MAIN-20240304101406-20240304131406-00860.warc.gz | 293,613,942 | 10,047 | ## What is the decimal Ascii code of A?
65
ASCII, decimal, hexadecimal, octal, and binary conversion table
A 65 41
B 66 42
C 67 43
D 68 44
## What is EBCDIC code table?
EBCDIC Table
Dec Hex Code
0 00 NUL
1 01 SOH
2 02 STX
3 03 ETX
What is the ASCII value of a period (‘ ‘) in decimal?
46
Character Name Char Decimal
Plus Sign + 43
Comma , 44
Hyphen / Minus Sign 45
Period . 46
What is ascii code and EBCDIC code?
ASCII and EBCDIC are two character encoding standards. The main difference between ASCII and EBCDIC is that the ASCII uses seven bits to represent a character while the EBCDIC uses eight bits to represent a character.
### What ASCII 49?
Standard ASCII Characters
Dec Hex Char
46 2E .
47 2F /
48 30 0
49 31 1
### What ASCII 48?
0
ASCII characters from 33 to 126
ASCII code Character
48 0
51 3
54 6
57 9
How do you convert EBCDIC to ascii?
Converting EBCDIC Text to ASCII Text
2. Set up a file transfer task (one source and one destination) to download and save our original text file without converting it.
3. After you have the basic file transfer task working, add a “Command Line App” built-in process.
What is the value of a in EBCDIC?
ASCII and EBCDIC character sets
Decimal Value Hex Value ASCII Symbol
65 41 A
66 42 B
67 43 C
68 44 D
## What is the ASCII decimal value of p?
120
7-bit ASCII Character Codes
Decimal Octal Value
077 115 M
078 116 N
079 117 O
080 120 P
## What is the ASCII value of Z?
ASCII – Binary Character Table
Letter ASCII Code Binary
w 119 01110111
x 120 01111000
y 121 01111001
z 122 01111010
What is ASCII to EBCDIC?
EBCDIC to ASCII
EBCDIC ASCII EBCDIC Meaning
00 00 NUL
01 01 SOH
02 02 STX
03 03 ETX
How is ASCII different from EBCDIC?
The main difference between ASCII and EBCDIC is that the ASCII uses seven bits to represent a character while the EBCDIC uses eight bits to represent a character. Besides, ASCII arranges the characters in consecutive order. EBCDIC groups 9 characters at a time.
### What is the difference between EBCDIC and ASCII?
The main difference between ASCII and EBCDIC is that the ASCII uses seven bits to represent a character while the EBCDIC uses eight bits to represent a character. It is easier for the computer to process numbers. But it is a difficult process to handle text.
### What is the purpose of the ASCII table?
ASCII TABLE ASCII stands for American Standard Code for Information Exchange. The purpose of ASCII is to create a standard for character-sets used in electronic equipments. The standard ensures that different devices (which might be manufactured by differing companies) can communicate to each other with the same character-code.
What is ASCII code and table?
The American Standard Code for Information Interchange, abbreviated as ASCII, is a standard of character encoding system intended for electronic communication. ASCII table includes codes representing text in telecommunications equipment, like computers, mobiles, and other electronic devices.
What is the difference between ASCII and extended?
Original ASCII uses 7 digits long binary string, which enables it to represent 128 characters. A later version of ASCII called extended ASCII uses 8 digits long binary string giving it the ability to represent 256 different characters. | 829 | 3,278 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-10 | latest | en | 0.745875 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/02%3A_Gas_Laws/2.07%3A_The_Ideal_Gas_Constant_and_Boltzmann's_Constant | 1,656,752,914,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103989282.58/warc/CC-MAIN-20220702071223-20220702101223-00351.warc.gz | 211,943,382 | 25,513 | # 2.7: The Ideal Gas Constant and Boltzmann's Constant
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
Having developed the ideal gas equation and analyzed experimental results for a variety of gases, we will have found the value of R. It is useful to have R expressed using a number of different energy units. Frequently useful values are
\begin{aligned} R & = 8.314 \text{ Pa m}^{3} \text{ K}^{-1} \text{ mol}^{-1} \\ ~ & = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \\ ~ & = 0.08314 \text{ L bar K}^{-1} \text{ mol}^{-1} \\ ~ & = 1.987 \text{ cal K}^{-1} \text{ mol}^{-1} \\ ~ & = 0.08205 \text{ L atm K}^{-1} \text{ mol}^{-1} \end{aligned}
We also need the gas constant expressed per molecule rather than per mole. Since there is Avogadro’s number of molecules per mole, we can divide any of the values above by $$\overline{N}$$ to get $$R$$ on a per-molecule basis. Traditionally, however, this constant is given a different name; it is Boltzmann’s constant, usually given the symbol $$k$$.
$k={R}/{\overline{N}}=1.381\times {10}^{-23}\ \mathrm{J}\ {\mathrm{K}}^{-1}\ {\mathrm{molecule}}^{-1}$
This means that we can also write the ideal gas equation as $$PV=nRT=n\overline{N}kT$$. Because the number of molecules in the sample, $$N$$, is $$N=n\overline{N}$$, we have
$PV=NkT.$
2.7: The Ideal Gas Constant and Boltzmann's Constant is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Paul Ellgen via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | 834 | 2,520 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2022-27 | latest | en | 0.452645 |
http://www.enotes.com/homework-help/there-three-football-jerseys-laundry-hamper-with-445878 | 1,462,124,797,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860116878.73/warc/CC-MAIN-20160428161516-00100-ip-10-239-7-51.ec2.internal.warc.gz | 496,965,494 | 11,416 | There are three football jerseys in a laundry hamper, with the numbers 18, 88, and 93 on the back, respectively. A total of three selections are made from the hamper, each time noting the number on...
There are three football jerseys in a laundry hamper, with the numbers 18, 88, and 93 on the back, respectively. A total of three selections are made from the hamper, each time noting the number on the back of the selected jersey. If the number 18 jersey is selected on any reach into the hamper, it is returned before the next selection. Otherwise, the selected jersey is not replaced before the next reach into the hamper. How many different outcomes are possible?
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Use a tree diagram to list the possibilities:
On the first round there are 3 possibilities -- 1 for each jersey.
On the second round there are 7 possibilities. If 18 was chosen in the first round, there are 3 possibilities in the second round.
If 88 was chosen in the first round, then there are only two choices in the second round as 88 was not replaced and simarly 2 choices if 93 was chosen in the first round.
On the third round there are 13 possibilities.
18->18->3 choices
18->88->2choices
18->93->2choices
88->18->2choices
88->93->1choice
93->18->2choices
93->88->1choice
13 choices
Thus there are 13 different outcomes.
-----------------------------------------------------------------
Let A=18,B=88,C=93 . We are looking for the number of 3 letter combinations where B and C do not repeat but A is allowed to repeat.
There are 27 total permutations -- take away 1 for all B's, 1 for all C's, 6 for 2B's and 6 for 2C's leaving 13.
----------------------------------------------------------------- | 404 | 1,696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-18 | longest | en | 0.919817 |
https://turbotax.community.intuit.ca/community/credits-deductions/discussion/how-much-of-a-savings-this-will-give-you-will-depend-on-y/01/924718/highlight/true | 1,619,105,356,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039610090.97/warc/CC-MAIN-20210422130245-20210422160245-00067.warc.gz | 671,627,194 | 53,574 | Solved: How much of a savings this will give you will depe...
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# If my medical expenses are just a little over the minimum amount (3% of my income), is it worth the trouble to claim them?
How significant is the tax credit when medical expenses are claimed?
Thank you!
Accepted Solutions
New Member
## If my medical expenses are just a little over the minimum amount (3% of my income), is it worth the trouble to claim them?
How much of a savings this will give you will depend on your income and provincial tax rates. You'll be able to claim only medical expenses above the lesser of \$2,208 or 3% of your net income.
For example if you make \$60,000 you'll be able to claim the amount over \$1,800 you've paid. So if you've paid 3.5% (\$2,100) of your total income in medical expenses you'll be able to claim a \$300 deduction on your return, which may net out to approximately \$100 (depending on your provincial tax rate).
Personally I would claim the deduction. The CRA doesn't need you to itemize your receipts so you can just tally all of your receipts and enter the totals into TurboTax.
3 Replies
New Member
## If my medical expenses are just a little over the minimum amount (3% of my income), is it worth the trouble to claim them?
How much of a savings this will give you will depend on your income and provincial tax rates. You'll be able to claim only medical expenses above the lesser of \$2,208 or 3% of your net income.
For example if you make \$60,000 you'll be able to claim the amount over \$1,800 you've paid. So if you've paid 3.5% (\$2,100) of your total income in medical expenses you'll be able to claim a \$300 deduction on your return, which may net out to approximately \$100 (depending on your provincial tax rate).
Personally I would claim the deduction. The CRA doesn't need you to itemize your receipts so you can just tally all of your receipts and enter the totals into TurboTax.
New Member
## If my medical expenses are just a little over the minimum amount (3% of my income), is it worth the trouble to claim them?
Thank you, Jocelyn! If I tally the receipts and enter the total, what do I enter in the"date" field since there were multiple dates on which the expenses were incurred?
New Member
## If my medical expenses are just a little over the minimum amount (3% of my income), is it worth the trouble to claim them?
I would choose the date of the last (most recent) medical receipt.
v | 608 | 2,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-17 | latest | en | 0.957224 |
http://classroom.synonym.com/scarcity-activities-kids-11399238.html | 1,508,192,358,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820466.2/warc/CC-MAIN-20171016214209-20171016234209-00369.warc.gz | 88,279,183 | 16,452 | Scarcity is the most fundamental principle in the study of economics, and it's a simple enough concept that you can introduce it to children even before they're old enough for more detailed lessons on economic theory, patterns and policies. Using food and personal activities helps students grasp the idea and develop sensitivity to the human toll when food is in short supply.
## Practical Demonstration
This practical demonstration of scarcity will engage students' attention and emotions by applying the principles of scarcity to their snack time. Provide a simple snack like popcorn or crackers, allowing one row of children to come up at a time to take as much as they want. When the snack runs out, apologize to the children who didn't get any. Discuss what options are available when there aren't enough snacks to go around, such as sharing, getting more snacks or some kids doing without. Apply this to the decisions societies face when resources are scarce.
## Resource Allocation
This game simulates a simple economy using construction paper to represent resources and products. Define a few construction paper "products," such as a multi-colored paper chain for clothes or a square glued to a triangle for shelter. Fill large envelopes with combinations of glue, scissors and construction paper such that no one envelope has the materials needed to make all the products. Group students and have each group attempt to gain food, clothing, shelter and other essentials. Discuss what solutions students had to find, such as trading products or "raw materials," tearing instead of cutting -- representing innovation -- or forming alliances.
## Opportunity Cost
Opportunity cost is the loss of something we could have done when we choose something else. For example, if you watch one show, you cannot spend that time watching a different show. Since it's involved in every choice, opportunity cost is a form of scarcity even young children have a context to understand. Begin by defining opportunity cost for students, then ask them to think of opportunity costs in their lives, like wearing one outfit instead of another or choosing homework over games.
## Food Scarcity
Food scarcity means that much of the world's population lives without enough nutrition. The WHO reports that 0.7 percent of the world has high food security, 33 percent has marginal, 50 percent has low and 14 percent has very low. You can demonstrate this by holding a pizza party or class meal, but dividing the students into groups and giving them the proportional amounts of food that represent what the world population has to live on. For example, one student could get a pizza to herself, while 10 students share two pizzas, 15 share one pizza and four share a single slice. After the discussion, redistribute the food for a more equitable and enjoyable pizza party. | 545 | 2,859 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | longest | en | 0.958648 |
https://math.libretexts.org/Bookshelves/Algebra/Book%3A_Elementary_Algebra_(OpenStax)/02%3A_Solving_Linear_Equations_and_Inequalities/2.07%3A_Solve_Linear_Inequalities | 1,675,328,685,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499967.46/warc/CC-MAIN-20230202070522-20230202100522-00024.warc.gz | 403,682,404 | 40,578 | # 2.7: Solve Linear Inequalities
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##### Learning Objectives
By the end of this section, you will be able to:
• Graph inequalities on the number line
• Solve inequalities using the Subtraction and Addition Properties of inequality
• Solve inequalities using the Division and Multiplication Properties of inequality
• Solve inequalities that require simplification
• Translate to an inequality and solve
##### Note
Before you get started, take this readiness quiz.
1. Translate from algebra to English: $$15>x$$.
If you missed this problem, review Exercise 1.3.1.
2. Solve: $$n−9=−42$$.
If you missed this problem, review Exercise 2.1.7.
3. Solve: $$−5p=−23$$.
If you missed this problem, review Exercise 2.2.1.
4. Solve: $$3a−12=7a−20$$.
If you missed this problem, review Exercise 2.3.22.
## Graph Inequalities on the Number Line
Do you remember what it means for a number to be a solution to an equation? A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.
What about the solution of an inequality? What number would make the inequality $$x > 3$$ true? Are you thinking, ‘x could be 4’? That’s correct, but x could be 5 too, or 20, or even 3.001. Any number greater than 3 is a solution to the inequality $$x > 3$$.
We show the solutions to the inequality $$x > 3$$ on the number line by shading in all the numbers to the right of 3, to show that all numbers greater than 3 are solutions. Because the number 3 itself is not a solution, we put an open parenthesis at 3. The graph of $$x > 3$$ is shown in Figure $$\PageIndex{1}$$. Please note that the following convention is used: light blue arrows point in the positive direction and dark blue arrows point in the negative direction.
The graph of the inequality $$x \geq 3$$ is very much like the graph of $$x > 3$$, but now we need to show that 3 is a solution, too. We do that by putting a bracket at $$x = 3$$, as shown in Figure $$\PageIndex{2}$$.
Notice that the open parentheses symbol, (, shows that the endpoint of the inequality is not included. The open bracket symbol, [, shows that the endpoint is included.
##### Exercise $$\PageIndex{1}$$
Graph on the number line:
1. $$x\leq 1$$
2. $$x<5$$
3. $$x>−1$$
1. $$x\leq 1$$ This means all numbers less than or equal to 1. We shade in all the numbers on the number line to the left of 1 and put a bracket at x=1 to show that it is included.
2. $$x<5$$ This means all numbers less than 5, but not including 5. We shade in all the numbers on the number line to the left of 5 and put a parenthesis at x=5 to show it is not included.
3. $$x>−1$$ This means all numbers greater than −1, but not including −1. We shade in all the numbers on the number line to the right of −1, then put a parenthesis at x=−1 to show it is not included.
##### Exercise $$\PageIndex{2}$$
Graph on the number line:
1. $$x\leq −1$$
2. $$x>2$$
3. $$x<3$$
##### Exercise $$\PageIndex{3}$$
Graph on the number line:
1. $$x>−2$$
2. $$x<−3$$
3. $$x\geq −1$$
We can also represent inequalities using interval notation. As we saw above, the inequality $$x>3$$ means all numbers greater than 3. There is no upper end to the solution to this inequality. In interval notation, we express $$x>3$$ as $$(3, \infty)$$. The symbol $$\infty$$ is read as ‘infinity’. It is not an actual number. Figure $$\PageIndex{3}$$ shows both the number line and the interval notation.
The inequality $$x\leq 1$$ means all numbers less than or equal to 1. There is no lower end to those numbers. We write $$x\leq 1$$ in interval notation as $$(-\infty, 1]$$. The symbol $$-\infty$$ is read as ‘negative infinity’. Figure $$\PageIndex{4}$$ shows both the number line and interval notation.
##### INEQUALITIES, NUMBER LINES, AND INTERVAL NOTATION
Did you notice how the parenthesis or bracket in the interval notation matches the symbol at the endpoint of the arrow? These relationships are shown in Figure $$\PageIndex{5}$$.
##### Exercise $$\PageIndex{4}$$
Graph on the number line and write in interval notation.
1. $$x \geq -3$$
2. $$x<2.5$$
3. $$x\leq \frac{3}{5}$$
1.
Shade to the right of −3, and put a bracket at −3. Write in interval notation.
2.
Shade to the left of 2.5, and put a parenthesis at 2.5. Write in interval notation.
3.
Shade to the left of $$-\frac{3}{5}$$, and put a bracket at $$-\frac{3}{5}$$. Write in interval notation.
##### Exercise $$\PageIndex{5}$$
Graph on the number line and write in interval notation:
1. $$x>2$$
2. $$x\leq −1.5$$
3. $$x\geq \frac{3}{4}$$
##### Exercise $$\PageIndex{6}$$
Graph on the number line and write in interval notation:
1. $$x\leq −4$$
2. $$x\geq 0.5$$
3. $$x<-\frac{2}{3}$$
## Solve Inequalities using the Subtraction and Addition Properties of Inequality
The Subtraction and Addition Properties of Equality state that if two quantities are equal, when we add or subtract the same amount from both quantities, the results will be equal.
##### PROPERTIES OF EQUALITY
$\begin{array} { l l } { \textbf { Subtraction Property of Equality } } & { \textbf { Addition Property of Equality } } \\ { \text { For any numbers } a , b , \text { and } c , } & { \text { For any numbers } a , b , \text { and } c } \\ { \text { if } \qquad \quad a = b , } & { \text { if } \qquad \quad a = b } \\ { \text { then } a - c = b - c . } & { \text { then } a + c = b + c } \end{array}$
Similar properties hold true for inequalities.
For example, we know that −4 is less than 2. If we subtract 5 from both quantities, is the left side still less than the right side? We get −9 on the left and −3 on the right. And we know −9 is less than −3. The inequality sign stayed the same.
Similarly we could show that the inequality also stays the same for addition.
##### PROPERTIES OF INEQUALITY
$\begin{array} { l l } { \textbf { Subtraction Property of Inequality } } & { \textbf { Addition Property of Inequality } } \\ { \text { For any numbers } a , b , \text { and } c , } & { \text { For any numbers } a , b , \text { and } c } \\ { \text { if }\qquad \quad a < b } & { \text { if } \qquad \quad a < b } \\ { \text { then } a - c < b - c . } & { \text { then } a + c < b + c } \\\\ { \text { if } \qquad \quad a > b } & { \text { if } \qquad \quad a > b } \\ { \text { then } a - c > b - c . } & { \text { then } a + c > b + c } \end{array}$
We use these properties to solve inequalities, taking the same steps we used to solve equations. Solving the inequality $$x+5>9$$, the steps would look like this:
$\begin{array}{rrll} {} &{x + 5} &{ >} &{9} \\ {\text{Subtract 5 from both sides to isolate }x.} &{x + 5 - 5} &{ >} &{9 - 5} \\{} &{x} &{ >} &{4} \\ \end{array}$
Any number greater than 4 is a solution to this inequality.
##### Exercise $$\PageIndex{7}$$
Solve the inequality $$n - \frac{1}{2} \leq \frac{5}{8}$$, graph the solution on the number line, and write the solution in interval notation.
Add $$\frac{1}{2}$$ to both sides of the inequality. Simplify. Graph the solution on the number line. Write the solution in interval notation.
##### Exercise $$\PageIndex{8}$$
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
$$p - \frac{3}{4} \geq \frac{1}{6}$$
##### Exercise $$\PageIndex{9}$$
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
$$r - \frac{1}{3} \leq \frac{7}{12}$$
## Solve Inequalities using the Division and Multiplication Properties of Inequality
The Division and Multiplication Properties of Equality state that if two quantities are equal, when we divide or multiply both quantities by the same amount, the results will also be equal (provided we don’t divide by 0).
##### PROPERTIES OF EQUALITY
$\begin{array}{ll} {\textbf{Division Property of Equality}} &{\textbf{MUltiplication Property of Equality}} \\ {\text{For any numbers a, b, c, and c} \neq 0} &{\text{For any numbers a, b, c}} \\ {\text{if } \qquad a = b} &{\text{if} \qquad \quad a = b} \\ {\text{then }\quad \frac{a}{c} = \frac{b}{c}} &{\text{then } \quad ac = bc} \end{array}$
Are there similar properties for inequalities? What happens to an inequality when we divide or multiply both sides by a constant?
Consider some numerical examples.
Divide both sides by 5. Multiply both sides by 5. Simplify. Fill in the inequality signs.
The inequality signs stayed the same.
Does the inequality stay the same when we divide or multiply by a negative number?
Divide both sides by -5. Multiply both sides by -5. Simplify. Fill in the inequality signs.
The inequality signs reversed their direction.
When we divide or multiply an inequality by a positive number, the inequality sign stays the same. When we divide or multiply an inequality by a negative number, the inequality sign reverses.
Here are the Division and Multiplication Properties of Inequality for easy reference.
##### DIVISION AND MULTIPLICATION PROPERTIES OF INEQUALITY
For any real numbers a,b,c
$\begin{array}{ll} {\text{if } a < b \text{ and } c > 0, \text{ then}} &{\frac{a}{c} < \frac{b}{c} \text{ and } ac < bc} \\ {\text{if } a > b \text{ and } c > 0, \text{ then}} &{\frac{a}{c} > \frac{b}{c} \text{ and } ac > bc} \\ {\text{if } a < b \text{ and } c < 0, \text{ then}} &{\frac{a}{c} > \frac{b}{c} \text{ and } ac > bc} \\ {\text{if } a > b \text{ and } c < 0, \text{ then}} &{\frac{a}{c} < \frac{b}{c} \text{ and } ac < bc} \end{array}$
When we divide or multiply an inequality by a:
• positive number, the inequality stays the same.
• negative number, the inequality reverses.
##### Exercise $$\PageIndex{10}$$
Solve the inequality $$7y<42$$, graph the solution on the number line, and write the solution in interval notation.
Divide both sides of the inequality by 7. Since $$7>0$$, the inequality stays the same. Simplify. Graph the solution on the number line. Write the solution in interval notation.
##### Exercise $$\PageIndex{11}$$
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
$$9c>72$$
$$c>8$$
$$(8, \infty)$$
##### Exercise $$\PageIndex{12}$$
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
$$12d\leq 60$$
$$d\leq 5$$
$$(-\infty, 5]$$
##### Exercise $$\PageIndex{13}$$
Solve the inequality $$−10a\geq 50$$, graph the solution on the number line, and write the solution in interval notation.
Divide both sides of the inequality by −10. Since $$−10<0$$, the inequality reverses. Simplify. Graph the solution on the number line. Write the solution in interval notation.
##### Exercise $$\PageIndex{14}$$
Solve each inequality, graph the solution on the number line, and write the solution in interval notation.
$$−8q<32$$
$$q>−4$$
##### Exercise $$\PageIndex{15}$$
Solve each inequality, graph the solution on the number line, and write the solution in interval notation.
$$−7r\leq −70$$
##### SOLVING INEQUALITIES
Sometimes when solving an inequality, the variable ends up on the right. We can rewrite the inequality in reverse to get the variable to the left.
$\begin{array}{l} x > a\text{ has the same meaning as } a < x \end{array}$
Think about it as “If Xavier is taller than Alex, then Alex is shorter than Xavier.”
##### Exercise $$\PageIndex{16}$$
Solve the inequality $$-20 < \frac{4}{5}u$$, graph the solution on the number line, and write the solution in interval notation.
Multiply both sides of the inequality by $$\frac{5}{4}$$. Since $$\frac{5}{4} > 0$$, the inequality stays the same. Simplify. Rewrite the variable on the left. Graph the solution on the number line. Write the solution in interval notation.
##### Exercise $$\PageIndex{17}$$
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
$$24 \leq \frac{3}{8}m$$
##### Exercise $$\PageIndex{18}$$
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
$$-24 < \frac{4}{3}n$$
##### Exercise $$\PageIndex{19}$$
Solve the inequality $$\frac{t}{-2} \geq 8$$, graph the solution on the number line, and write the solution in interval notation.
Multiply both sides of the inequality by −2. Since $$−2<0$$, the inequality reverses. Simplify. Graph the solution on the number line. Write the solution in interval notation.
##### Exercise $$\PageIndex{20}$$
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
$$\frac{k}{-12}\leq 15$$
##### Exercise $$\PageIndex{21}$$
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
$$\frac{u}{-4}\geq -16$$
## Solve Inequalities That Require Simplification
Most inequalities will take more than one step to solve. We follow the same steps we used in the general strategy for solving linear equations, but be sure to pay close attention during multiplication or division.
##### Exercise $$\PageIndex{22}$$
Solve the inequality $$4m\leq 9m+17$$, graph the solution on the number line, and write the solution in interval notation.
Subtract 9m from both sides to collect the variables on the left. Simplify. Divide both sides of the inequality by −5, and reverse the inequality. Simplify. Graph the solution on the number line. Write the solution in interval notation.
##### Exercise $$\PageIndex{23}$$
Solve the inequality $$3q\geq 7q−23$$, graph the solution on the number line, and write the solution in interval notation.
##### Exercise $$\PageIndex{24}$$
Solve the inequality $$6x<10x+19$$, graph the solution on the number line, and write the solution in interval notation.
##### Exercise $$\PageIndex{25}$$
Solve the inequality $$8p+3(p−12)>7p−28$$ graph the solution on the number line, and write the solution in interval notation.
Simplify each side as much as possible. 8p+3(p−12)>7p−28 Distribute. 8p+3p−36>7p−28 Combine like terms. 11p−36>7p−28 Subtract 7p from both sides to collect the variables on the left. 11p−36−7p>7p−28−7p Simplify. 4p−36>−28 Add 36 to both sides to collect the constants on the right. 4p−36+36>−28+36 Simplify. 4p>8 Divide both sides of the inequality by 4; the inequality stays the same. $$\frac{4p}{4}>84$$ Simplify. $$p>2$$ Graph the solution on the number line. Write the solution in interval notation. $$(2, \infty)$$
##### Exercise $$\PageIndex{26}$$
Solve the inequality $$9y+2(y+6)>5y−24$$, graph the solution on the number line, and write the solution in interval notation.
##### Exercise $$\PageIndex{27}$$
Solve the inequality $$6u+8(u−1)>10u+32$$, graph the solution on the number line, and write the solution in interval notation.
Just like some equations are identities and some are contradictions, inequalities may be identities or contradictions, too. We recognize these forms when we are left with only constants as we solve the inequality. If the result is a true statement, we have an identity. If the result is a false statement, we have a contradiction.
##### Exercise $$\PageIndex{28}$$
Solve the inequality $$8x−2(5−x)<4(x+9)+6x$$, graph the solution on the number line, and write the solution in interval notation.
Simplify each side as much as possible. 8x−2(5−x)<4(x+9)+6x Distribute. 8x−10+2x<4x+36+6x Combine like terms. 10x−10<10x+36 Subtract 10x from both sides to collect the variables on the left. 10x−10−10x<10x+36−10x Simplify. −10<36 The xx’s are gone, and we have a true statement. The inequality is an identity. The solution is all real numbers. Graph the solution on the number line. Write the solution in interval notation. $$(-\infty, \infty)$$
##### Exercise $$\PageIndex{29}$$
Solve the inequality $$4b−3(3−b)>5(b−6)+2b$$, graph the solution on the number line, and write the solution in interval notation.
##### Exercise $$\PageIndex{30}$$
Solve the inequality $$9h−7(2−h)<8(h+11)+8h$$, graph the solution on the number line, and write the solution in interval notation.
##### Exercise $$\PageIndex{31}$$
Solve the inequality $$\frac{1}{3}a - \frac{1}{8}a > \frac{5}{24}a + \frac{3}{4}$$, graph the solution on the number line, and write the solution in interval notation.
Multiply both sides by the LCD, 24, to clear the fractions. Simplify. Combine like terms. Subtract 5a from both sides to collect the variables on the left. Simplify. The statement is false! The inequality is a contradiction. There is no solution. Graph the solution on the number line. Write the solution in interval notation. There is no solution.
##### Exercise $$\PageIndex{32}$$
Solve the inequality $$\frac{1}{4}x - \frac{1}{12}x > \frac{1}{6}x + \frac{7}{8}$$, graph the solution on the number line, and write the solution in interval notation.
##### Exercise $$\PageIndex{33}$$
Solve the inequality $$\frac{2}{5}z - \frac{1}{3}z < \frac{1}{15}z - \frac{3}{5}$$, graph the solution on the number line, and write the solution in interval notation.
## Translate to an Inequality and Solve
To translate English sentences into inequalities, we need to recognize the phrases that indicate the inequality. Some words are easy, like ‘more than’ and ‘less than’. But others are not as obvious.
Think about the phrase ‘at least’ – what does it mean to be ‘at least 21 years old’? It means 21 or more. The phrase ‘at least’ is the same as ‘greater than or equal to’.
Table $$\PageIndex{4}$$ shows some common phrases that indicate inequalities.
> $$\geq$$ < $$\leq$$
" data-valign="middle" class="lt-math-15134">is greater than is greater than or equal to is less than is less than or equal to
" data-valign="middle" class="lt-math-15134">is more than is at least is smaller than is at most
" data-valign="middle" class="lt-math-15134">is larger than is no less than has fewer than is no more than
" data-valign="middle" class="lt-math-15134">exceeds is the minimum is lower than is the maximum
Table $$\PageIndex{4}$$
##### Exercise $$\PageIndex{34}$$
Translate and solve. Then write the solution in interval notation and graph on the number line.
Twelve times c is no more than 96.
Translate. Solve—divide both sides by 12. Simplify. Write in interval notation. Graph on the number line.
##### Exercise $$\PageIndex{35}$$
Translate and solve. Then write the solution in interval notation and graph on the number line.
Twenty times y is at most 100
##### Exercise $$\PageIndex{36}$$
Translate and solve. Then write the solution in interval notation and graph on the number line.
Nine times z is no less than 135
##### Exercise $$\PageIndex{37}$$
Translate and solve. Then write the solution in interval notation and graph on the number line.
Thirty less than x is at least 45.
Translate. Solve—add 30 to both sides. Simplify. Write in interval notation. Graph on the number line.
##### Exercise $$\PageIndex{38}$$
Translate and solve. Then write the solution in interval notation and graph on the number line.
Nineteen less than p is no less than 47
##### Exercise $$\PageIndex{39}$$
Translate and solve. Then write the solution in interval notation and graph on the number line.
Four more than a is at most 15.
## Key Concepts
• Subtraction Property of Inequality
For any numbers a, b, and c,
if a<b then a−c<b−c and
if a>b then a−c>b−c.
For any numbers a, b, and c,
if a<b then a+c<b+c and
if a>b then a+c>b+c.
• Division and Multiplication Properties of Inequality
For any numbers a, b, and c,
if a<b and c>0, then ac<bc and ac>bc.
if a>b and c>0, then ac>bc and ac>bc.
if a<b and c<0, then ac>bc and ac>bc.
if a>b and c<0, then ac<bc and ac<bc.
• When we divide or multiply an inequality by a:
• positive number, the inequality stays the same.
• negative number, the inequality reverses.
This page titled 2.7: Solve Linear Inequalities is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | 5,969 | 20,998 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2023-06 | latest | en | 0.510745 |
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please explain the wavy curve method and la hospital rule with examples
6 years ago
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### Answers : (1)
Dear Abhik
Wavy curve method :--
1 . put only odd power factors in numerator and denominator equal to zero separately as function changes sign only for odd power
2. Plot these points on the number line in increasing order .
3. Now check the coefficients of X and make them positive.
4. expression will be +ve to the right of the highest point allotted on number line.
5. alternately assign + and - signs in rest of intervals ( from right to left )
to see examples see the pic
L'Hôpital's rule
l'Hôpital's rule states that for functions ƒ and g:
If $\lim_{x \to c}f(x)=\lim_{x \to c}g(x)=0 \,$ or $\pm\infty$ and $\lim_{x\to c}f'(x)/g'(x)$ exists,
then $\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}.$
Here is example involving 0/0:
$\lim_{x\to 0}{\frac{e^x-1-x}{x^2}} =\lim_{x\to 0}{\frac{e^x-1}{2x}} =\lim_{x\to 0}{\frac{e^x}{2}}={\frac{1}{2}}.$
Here is another example involving ∞/∞:
$\lim_{x\to 0^+} x \ln x =\lim_{x\to 0^+}{\frac{\ln x}{1/x}} =\lim_{x\to 0^+}{\frac{1/x}{-1/x^2}} =\lim_{x\to 0^+} -x = 0.$
All the best.
AKASH GOYAL
Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
6 years ago
# Other Related Questions on Differential Calculus
Plz help me solve this question. It is from chapter limits and derivatives . I did not find that in the list,
Krishna, write your questions that you want to ask, to related to limits and derivative. But you studied the defnition of limits and derivative clearly.
Kumar 4 months ago
Try desmos (Google it ) ... It is used to graph the function.. Then you can set y as a parameter and then you will see that the function as x-> 0 the graph tends to infinity (for some values...
rishabh 3 months ago
Kumar the question is lim x--> 0 ((y^2 - (y-x)^2)^1/x). / (8xy - 4x^2 + (8xy)^1/2)^3 I have also attached the image of it with the question please take a look
Sujith krishna 4 months ago
Why does lhl and rhl are not equal in this question.w hat should i do in these type of questions
coz this is not 0/0 inderminate form ...it is exact 0/0 which is not defined(n.d.)
Nishant Vora one month ago
Nishant Vora one month ago
Why cannot we write tan[x]/[x] as 1 when x tends towards zero..and if is in 0/0 form why didnt u used lhospital rule
Samaksh one month ago
i can not solve this inequality please sir how to solve this inequality Ix+1I+Ix-2I
i am trying to write conditions in the answer but after posting the conditions are vanishing , you can see above, dont know why is it happening.... :(
DR STRANGE 16 days ago
Ix+1I+Ix-2I = 2x-1 ….....for x>2 =x+1 -(x-2) = 3 …......for -1 = -(x+1) -(x-2) = 1-2x ….......for x
DR STRANGE 16 days ago
mistake............................................................... =3 for...... -1 = 1-2x ….for x
DR STRANGE 16 days ago
If alpha is a real root of the equation ax 2 +bx+c and beta is a real root of equation -ax 2 +bx+c. Show that there exists a root gama of the equation (a/2)x 2 +bx+c which lies between alpha...
Ajay 6 months ago
Small Mistake in last para posting again..............................................................................................................
Ajay 6 months ago
We have Similarly, So if P(x) = a/2 x 2 +bx +c, then and are off opposite sign and hence there must exist a root between the two numbers.
mycroft holmes 6 months ago
In the listed image can you tell me how beta*gamma = 2 ….. . . .. ??
The value of gamma is still not correct, either printing mistake or you gave me wrong value. The correct value of gamma is below
Ajay 5 months ago
Thankyou so much............................. …......................................................................!
Anshuman Mohanty 5 months ago
Yes sorry..... . . . .it is not so clear.. ok the values are beta = α + α^2 + α^4 and gamma = α^3 + α^5 + α^7
Anshuman Mohanty 5 months ago
if |z - i| Options: a*) 14 b) 2 c) 28 d) None of the above
If |z-i| = ?? PLs complete the question
Nishant Vora one month ago
Got it! [z + 12 – 6 i ] can be rewritten as [ z – i + 12 – 5 i] => | z – i | and => |12 – 5 i | = sqrt ( 12^2 + 5^2) = 13......................(2) => | z + 12 – 6 i | => | z + 12 – 6 i |...
Divya one month ago
I tried posting the question several times, it kept cutting off the rest of the question. Here: If | z-1| Options: a*) 14 b) 2 c) 28 d) None of the above
Divya one month ago
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More Questions On Differential Calculus | 1,629 | 5,335 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | longest | en | 0.741539 |
http://mattrolls.blogspot.com/2012/04/real-numbers-are-not-countable.html | 1,532,347,496,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676596336.96/warc/CC-MAIN-20180723110342-20180723130342-00575.warc.gz | 236,835,191 | 7,815 | wherein is detailed Matt's experiences as he tries to figure out what to do with his life. Right now, that means lots of thinking about math.
Wednesday, April 25, 2012
At the beginning of March, I was discussing the cardinality of infinite sets. We showed, for example, that the rational numbers, a set which appears to be much larger than the natural numbers, in fact is the same size as the natural numbers. We also showed that there exist infinite sets which are larger than the set of natural numbers. I looked at the question of whether the set of real numbers is larger than the set of natural numbers, but I didn't come to a definite conclusion. Today, building on the work we've been doing with set topology, we will answer that question.
Before I start, I want to restate some proofs that we've covered recently and that we will need here. Hopefully this way when I draw some conclusions from these theorems, they won't seem to come from out of nowhere. First, given an infinite sequence of compact sets, such that each set in the sequence is a subset of the previous set in the sequence and none of the sets are empty, the intersection of all the sets in the sequence is not empty. (That statement is a mouthful and kind of makes my head spin. And the example in that post is flawed. I also had a follow up post in which the examples work and which hopefully makes more sense, but where I don't restate the proof itself. If you can wrap your head around the example, it should make the logic in this post more clear.)
Another proof which we previously looked at and will need today is the proof that the intersection of any closed set and a compact set is itself a compact set. This one is a little more straightforward, and can be visualized by drawing two rectangles on a page. Call one rectangle the closed set and the other rectangle the compact set, and then their overlapping area is also a compact set.
Finally, we will need the Heine-Borel theorem. This states that in a Euclidean space, any set which is both closed and bounded is also compact.
Before I begin the proof, here's a useful definition. A perfect set is a closed set in which every point in the set is a limit point of the set. The definition of a closed set is that every limit point of the set is a member of the set. Perfect sets go in the other direction and require that every point in the set also be a limit point of the set.
Our standard example of a compact set, the set of all numbers of the form 1/n, where n is any natural number, and the number 0, is closed, but is not perfect, because 0 is the only limit point for the set. In comparison, a k-cell is both compact and perfect.
So take some Euclidean space. (Note that we are requiring a Euclidean space, rather than a metric space in general.) In that Euclidean space, take some nonempty perfect set P. We will now show that P has an infinite number of points, and that those points are not countable.
The first part is obvious. P is not empty, so it has at least one point. That point is a limit point of P, and therefore there must be an infinite number of points in P. To show the second part, we are going to assume the opposite and show a contradiction, so let's assume that the points in P are countable. Let's count them. x is any vector in our k-dimensional Euclidean space. Then xn will be a point in P, for any natural number n. The points will be all distinct, so if a≠b, then xaxb. Since P is countable, every point in P can be labeled as xn for some specific value of n.
Pick some point y1 in the Euclidean space. y1 does not have to be a member of P, but it could be. If it is, y1=xn for some particular value of n, but the value of n doesn't matter. Any neighborhood of y1 is the open set of all points y in the Euclidean space such that |yy1|<r, for some radius r greater than 0. Choose a neighborhood of y1 which includes some points in P. (It could include all of them, but it doesn't have to.) Call the neighborhood V1. V1 can include points in the space which are not members of the set P, but it must also include an infinite number of points in P. Since it includes any points in P, and every point in P is a limit point of P, it must include an infinite number of points in P.
Now, pick some other point in V1, call it y2. y2 again is a point in the Euclidean space, but does not necessarily have to be a member of P. y2 has a neighborhood around it, V2. Let's choose V2 such that V2 is a subset of V1, and V2 includes some points in P, but also so the radius of V2 is strictly less than the distance from x1 to y2. Therefore, V2 excludes x1.
On a piece of paper, representing a Euclidean space, draw some shape and call it P. It could be a rectangle, but it doesn't have to be. It can even extend off the edge of the page. (It doesn't have to be bounded.) The important things are that P is closed, and that P doesn't have any isolated points. Label a bunch of points in P as x1, x2, x3, and so forth. The claim is that we could label every point in P that way, but we want to just do enough to get the idea. Mark a point on the page and call it y1. It will be cleaner to draw if y1 is not in P, but it could be in P if you want. Draw a circle around y1 which includes some of P and call that V1. V1 could include all of P, but doesn't have to. Now, we are going to mark some other point inside V1 and call that y2, and draw a smaller circle around y2, such that the small circle is entirely inside V1. The placement of y2 and V2 are a little tricky. V2 should include some points in P, but also should not include the point x1. You always can pick y2 and V2 so this is possible, but if you just grab the first point and circle that come to mind you could miss.
Now, keep going. For every natural number n, Vn is a circle which includes some points in P. Vn+1 is a circle inside Vn which also includes some points in P, but specifically does not include xn. (The center of Vn is yn, but there is not necessarily a relationship between the y's and the x's, which are all the points in P.) Then {Vn} is an infinite sequence of sets, and each set is a subset of the previous set in the sequence. This sounds suspiciously like the proof I mentioned at the top of the post, but the proof refers to compact sets, and Vn is an open set instead.
I'm going to come back to that in a moment, but first I want to restate that for any n, xn is not a member of Vn+1. This implies that the intersection of all of the sets in the sequence of neighborhoods contains no points in P, even though each set itself does contain points in P.
For any n, Vn is an open set. But we can turn it into a closed set, by taking the closure of Vn. Vn is an open set with center yn and radius rn, and includes any point y where |yyn|<rn. The closure of Vn is the set of any point y where |yyn|≤rn. And we can work with closed sets. Importantly, the closure of Vn is also a bounded set. Since the set is both closed and bounded, and we are in a Euclidean space, the Heine-Borel theorem states that the set is also compact. We know from the proof I restated above that the intersection of a closed set and a compact set is a compact set. Since the closure of Vn is compact and P is closed, the intersection of the closure of Vn and P is compact, for every n. For a given Vn, call the intersection Kn.
Now, look at the sequence of Kn. Kn is always a subset of P. Kn is never an empty set, because of how it was constructed. For each n, Kn+1 is a subset of Kn. And Kn is always compact. So now the proof up top does apply, and the intersection of all the sets Kn is not empty. The point in the intersection is a member of every Kn, and therefore is also a member of P.
However, we have constructed the sequence Vn such that the point xn is not a member of Vn+1. xn is also not a member of the closure of Vn+1, because the radius of Vn+1 is strictly less than the distance from the center of Vn+1 to xn. Kn+1 is a subset of the closure. Therefore, xn is never a member of Kn+1, so the point xn cannot be in the intersection of all of the Kn for any value of n.
We have now shown that there exists some point which is a member of P in the intersection of all Kn, but that no point xn is in the intersection of all Kn. Therefore there must be some point in P which is not labeled by xn. But our starting assumption was that {xn} included all the points in P, because P is countable. The only conclusion is that P is in fact not countable.
We have just proved that if P is a nonempty perfect set in a Euclidean space, then P is not countable. Some examples of uncountable perfect sets include k-cells, closed intervals, and the entire real number line.
Once more, for emphasis, the real numbers are not countable.
FAQ
What does "rolls a hoover" mean, anyway?
"Roll a hoover" was coined by Christopher Locke, aka RageBoy (not worksafe). He enumerated some Hooverian Principles, but that might not be too helpful. My interpretation is that rolling a hoover means doing something that you know is stupid without any clear sense of what the outcome will be, just to see what will happen. In my case, I quit my job in an uncertain economy to try to start a business. I'm still not sure how that will work out. | 2,233 | 9,257 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2018-30 | latest | en | 0.979997 |
https://pythonhosted.org/tmm/_modules/tmm/examples.html | 1,516,144,805,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886758.34/warc/CC-MAIN-20180116224019-20180117004019-00559.warc.gz | 805,705,793 | 6,533 | # Source code for tmm.examples
```# -*- coding: utf-8 -*-
"""
Examples of plots and calculations using the tmm package.
"""
from __future__ import division, print_function, absolute_import
from .tmm_core import (coh_tmm, unpolarized_RT, ellips,
position_resolved, find_in_structure_with_inf)
from numpy import pi, linspace, inf, array
from scipy.interpolate import interp1d
import matplotlib.pyplot as plt
try:
import colorpy.illuminants
import colorpy.colormodels
from . import color
colors_were_imported = True
except ImportError:
# without colorpy, you can't run sample5(), but everything else is fine.
colors_were_imported = False
# "5 * degree" is 5 degrees expressed in radians
# "1.2 / degree" is 1.2 radians expressed in degrees
degree = pi/180
[docs]def sample1():
"""
Here's a thin non-absorbing layer, on top of a thick absorbing layer, with
air on both sides. Plotting reflected intensity versus wavenumber, at two
different incident angles.
"""
# list of layer thicknesses in nm
d_list = [inf, 100, 300, inf]
# list of refractive indices
n_list = [1, 2.2, 3.3+0.3j, 1]
# list of wavenumbers to plot in nm^-1
ks = linspace(0.0001, .01, num=400)
# initialize lists of y-values to plot
Rnorm = []
R45 = []
for k in ks:
# For normal incidence, s and p polarizations are identical.
# I arbitrarily decided to use 's'.
Rnorm.append(coh_tmm('s', n_list, d_list, 0, 1/k)['R'])
R45.append(unpolarized_RT(n_list, d_list, 45*degree, 1/k)['R'])
kcm = ks * 1e7 #ks in cm^-1 rather than nm^-1
plt.figure()
plt.plot(kcm, Rnorm, 'blue', kcm, R45, 'purple')
plt.xlabel('k (cm\$^{-1}\$)')
plt.ylabel('Fraction reflected')
plt.title('Reflection of unpolarized light at 0\$^\circ\$ incidence (blue), '
'45\$^\circ\$ (purple)')
[docs]def sample2():
"""
Here's the transmitted intensity versus wavelength through a single-layer
film which has some complicated wavelength-dependent index of refraction.
(I made these numbers up, but in real life they could be read out of a
graph / table published in the literature.) Air is on both sides of the
film, and the light is normally incident.
"""
#index of refraction of my material: wavelength in nm versus index.
material_nk_data = array([[200, 2.1+0.1j],
[300, 2.4+0.3j],
[400, 2.3+0.4j],
[500, 2.2+0.4j],
[750, 2.2+0.5j]])
material_nk_fn = interp1d(material_nk_data[:,0].real,
d_list = [inf, 300, inf] #in nm
lambda_list = linspace(200, 750, 400) #in nm
T_list = []
for lambda_vac in lambda_list:
n_list = [1, material_nk_fn(lambda_vac), 1]
T_list.append(coh_tmm('s', n_list, d_list, 0, lambda_vac)['T'])
plt.figure()
plt.plot(lambda_list, T_list)
plt.xlabel('Wavelength (nm)')
plt.ylabel('Fraction of power transmitted')
plt.title('Transmission at normal incidence')
[docs]def sample3():
"""
Here is a calculation of the psi and Delta parameters measured in
ellipsometry. This reproduces Fig. 1.14 in Handbook of Ellipsometry by
Tompkins, 2005.
"""
n_list = [1, 1.46, 3.87+0.02j]
ds = linspace(0, 1000, num=100) #in nm
psis = []
Deltas = []
for d in ds:
e_data = ellips(n_list, [inf, d, inf], 70*degree, 633) #in nm
psis.append(e_data['psi']/degree) # angle in degrees
Deltas.append(e_data['Delta']/degree) # angle in degrees
plt.figure()
plt.plot(ds, psis, ds, Deltas)
plt.xlabel('SiO2 thickness (nm)')
plt.ylabel('Ellipsometric angles (degrees)')
plt.title('Ellipsometric parameters for air/SiO2/Si, varying '
'SiO2 thickness.\n'
'@ 70\$^\circ\$, 633nm. '
'Should agree with Handbook of Ellipsometry Fig. 1.14')
[docs]def sample4():
"""
Here is an example where we plot absorption and Poynting vector
as a function of depth.
"""
d_list = [inf, 100, 300, inf] #in nm
n_list = [1, 2.2+0.2j, 3.3+0.3j, 1]
th_0 = pi/4
lam_vac = 400
pol = 'p'
coh_tmm_data = coh_tmm(pol, n_list, d_list, th_0, lam_vac)
ds = linspace(-50, 400, num=1000) #position in structure
poyn = []
absor = []
for d in ds:
layer, d_in_layer = find_in_structure_with_inf(d_list, d)
data = position_resolved(layer, d_in_layer, coh_tmm_data)
poyn.append(data['poyn'])
absor.append(data['absor'])
# convert data to numpy arrays for easy scaling in the plot
poyn = array(poyn)
absor = array(absor)
plt.figure()
plt.plot(ds, poyn, 'blue', ds, 200*absor, 'purple')
plt.xlabel('depth (nm)')
plt.ylabel('AU')
plt.title('Local absorption (purple), Poynting vector (blue)')
[docs]def sample5():
"""
Color calculations: What color is a air / thin SiO2 / Si wafer?
"""
if not colors_were_imported:
print('Colorpy was not detected (or perhaps an error occurred when',
'Original version is at http://pypi.python.org/pypi/colorpy',
'A Python 3 compatible edit is at https://github.com/fish2000/ColorPy/')
return
# Crystalline silicon refractive index. Data from Palik via
# http://refractiveindex.info, I haven't checked it, but this is just for
# demonstration purposes anyway.
Si_n_data = [[400, 5.57 + 0.387j],
[450, 4.67 + 0.145j],
[500, 4.30 + 7.28e-2j],
[550, 4.08 + 4.06e-2j],
[600, 3.95 + 2.57e-2j],
[650, 3.85 + 1.64e-2j],
[700, 3.78 + 1.26e-2j]]
Si_n_data = array(Si_n_data)
Si_n_fn = interp1d(Si_n_data[:,0], Si_n_data[:,1], kind='linear')
# SiO2 refractive index (approximate): 1.46 regardless of wavelength
SiO2_n_fn = lambda wavelength : 1.46
# air refractive index
air_n_fn = lambda wavelength : 1
n_fn_list = [air_n_fn, SiO2_n_fn, Si_n_fn]
th_0 = 0
# Print the colors, and show plots, for the special case of 300nm-thick SiO2
d_list = [inf, 300, inf]
reflectances = color.calc_reflectances(n_fn_list, d_list, th_0)
illuminant = colorpy.illuminants.get_illuminant_D65()
spectrum = color.calc_spectrum(reflectances, illuminant)
color_dict = color.calc_color(spectrum)
print('air / 300nm SiO2 / Si --- rgb =', color_dict['rgb'], ', xyY =', color_dict['xyY'])
plt.figure()
color.plot_reflectances(reflectances,
title='air / 300nm SiO2 / Si -- '
'Fraction reflected at each wavelength')
plt.figure()
color.plot_spectrum(spectrum,
title='air / 300nm SiO2 / Si -- '
'Reflected spectrum under D65 illumination')
# Calculate irgb color (i.e. gamma-corrected sRGB display color rounded to
# integers 0-255) versus thickness of SiO2
max_SiO2_thickness = 600
SiO2_thickness_list = linspace(0, max_SiO2_thickness, num=80)
irgb_list = []
for SiO2_d in SiO2_thickness_list:
d_list = [inf, SiO2_d, inf]
reflectances = color.calc_reflectances(n_fn_list, d_list, th_0)
illuminant = colorpy.illuminants.get_illuminant_D65()
spectrum = color.calc_spectrum(reflectances, illuminant)
color_dict = color.calc_color(spectrum)
irgb_list.append(color_dict['irgb'])
# Plot those colors
print('Making color vs SiO2 thickness graph. Compare to (for example)')
print('http://www.htelabs.com/appnotes/sio2_color_chart_thermal_silicon_dioxide.htm')
plt.figure()
plt.plot([0, max_SiO2_thickness], [1, 1])
plt.xlim(0, max_SiO2_thickness)
plt.ylim(0, 1)
plt.xlabel('SiO2 thickness (nm)')
plt.yticks([])
plt.title('Air / SiO2 / Si color vs SiO2 thickness')
for i in range(len(SiO2_thickness_list)):
# One strip of each color, centered at x=SiO2_thickness_list[i]
if i == 0:
x0 = 0
else:
x0 = (SiO2_thickness_list[i] + SiO2_thickness_list[i-1]) / 2
if i == len(SiO2_thickness_list) - 1:
x1 = max_SiO2_thickness
else:
x1 = (SiO2_thickness_list[i] + SiO2_thickness_list[i+1]) / 2
y0 = 0
y1 = 1
poly_x = [x0, x1, x1, x0]
poly_y = [y0, y0, y1, y1]
color_string = colorpy.colormodels.irgb_string_from_irgb(irgb_list[i])
plt.fill(poly_x, poly_y, color_string, edgecolor=color_string)
[docs]def sample6():
"""
An example reflection plot with a surface plasmon resonance (SPR) dip.
Compare with http://doi.org/10.2320/matertrans.M2010003 ("Spectral and
Angular Responses of Surface Plasmon Resonance Based on the Kretschmann
Prism Configuration") Fig 6a
"""
# list of layer thicknesses in nm
d_list = [inf, 5, 30, inf]
# list of refractive indices
n_list = [1.517, 3.719+4.362j, 0.130+3.162j, 1]
# wavelength in nm
lam_vac = 633
# list of angles to plot
theta_list = linspace(30*degree, 60*degree, num=300)
# initialize lists of y-values to plot
Rp = []
for theta in theta_list:
Rp.append(coh_tmm('p', n_list, d_list, theta, lam_vac)['R'])
plt.figure()
plt.plot(theta_list/degree, Rp, 'blue')
plt.xlabel('theta (degree)')
plt.ylabel('Fraction reflected')
plt.xlim(30, 60)
plt.ylim(0, 1)
plt.title('Reflection of p-polarized light with Surface Plasmon Resonance\n'
'Compare with http://doi.org/10.2320/matertrans.M2010003 Fig 6a')
``` | 2,683 | 8,313 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.751435 |
https://www.enotes.com/homework-help/solve-equation-9-2-x-1-3-2-x-7-1-3-4-x-349645 | 1,516,203,627,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886946.21/warc/CC-MAIN-20180117142113-20180117162113-00684.warc.gz | 866,998,004 | 9,805 | # Solve equation (9-2^x)^1/3 +(2^x+7)^1/3=4? x=??
sciencesolve | Certified Educator
You should raise to cube both sides such that:
`(root(3)(9 - 2^x) + root(3)(2^x + 7))^3 = 4^3`
You need to expand the cube such that:
`(root(3)(9 - 2^x) + root(3)(2^x + 7))^3 = (root(3)(9 - 2^x))^3 + (root(3)(2^x + 7))^3 + 3root(3)((9 - 2^x)(2^x + 7))(root(3)(9 - 2^x) + root(3)(2^x + 7))`
You should substitute `4^3` for `root(3)(9 - 2^x) + root(3)(2^x + 7)` such that:
`9 - 2^x + 2^x + 7 + 3*64*root(3)((9 - 2^x)(2^x + 7)) = 64`
Eliminating like terms yields:
`192root(3)((9 - 2^x)(2^x + 7)) = 64 - 16`
`192root(3)((9 - 2^x)(2^x + 7)) = 48`
You need to divide by 48 both sides such that:
`4root(3)((9 - 2^x)(2^x + 7)) = 1 =gt root(3)((9 - 2^x)(2^x + 7)) = 1/4`
You need to raise to cube to remove the cube root such that:
`((9 - 2^x)(2^x + 7)) = 1/64`
`9*2^x + 63 - 2^(2x) - 7*2^x = 1/64`
`- 2^(2x) + 2*2^x + 63 - 1/64 = 0`
`-64*2^(2x) + 128*2^x + 4031 = 0`
You should come up with the substitution `2^x = y` such that:
`64y^2 - 128y - 4031 = 0`
You should use quadratic formula such that:
`y_(1,2)= (128 +- sqrt(1048320))/128`
`y_(1,2) = (128+-48sqrt455)/128`
You should solve for x `2^x = y` such that:
`2^x = (128+48sqrt455)/128 =gt x = ln((128+48sqrt455)/128)/ln 2`
`x~~3.170`
Since `(128-48sqrt455)/128 lt 0 =gt 2^x = (128+48sqrt455)/128 ` is not possible to be solved since `2^x gt 0` .
Hence, evaluating the solution to the given equation yields `x~~3.170.` | 695 | 1,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2018-05 | latest | en | 0.701312 |
https://community.quickbase.com/discussions/quickbase-discussions/formula-to-make-a-field--most-recent-value-of-same-field/59782 | 1,716,970,384,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059221.97/warc/CC-MAIN-20240529072748-20240529102748-00271.warc.gz | 149,235,642 | 38,711 | JBGoriou
Qrew Member
3 years ago
# Formula to Make a Field = Most Recent Value of Same Field
The table I'm building allows the user to create a product offer for an order. So in this table they create a product offer, write the details of the product and then save it and now the table shows all the details for "product A" that the user created for this product offer. I currently have it set so that the product offer# = the record ID#.
However, sometimes the product offers contain more than one product but the user can only enter one product at a time. So when the user enters product A, B, C, the table will show record ID# 1, 2, 3, and if left at that, the "product offer#" field will match that. But I need to be able to show that the product offer# is the same for all those products. So my solution was to create a check box field called "order related to last product" that the user can select while they are creating that product offer. Then I'd create a formula for the product offer# saying:
"if 'order related to last product' is not checked off, then product offer# = record ID#, else product offer# = the last/most recent product offer# created"
As a formula I got this far: IF([order related to last product]=false, [record ID#], ???)
Everything I've tried so far doesn't work because 1. I'm not sure how to write "find last/most recent product offer# created", and 2. I know the formula doesn't allow me to use a reference to itself which in this case is the product offer#.
Does anyone know how I can get around this? I would greatly appreciate the help. Thank you!
------------------------------
JB
------------------------------
• I think you should consider a different design, assuming I understand your problem. The real problem is that when you create a product offer you're only allowing the user to select one product. Why don't you make a many to many join table so that one product offer has many offer products and one product master has what many offer products. So that way you could have multiple products on the same offer.
This is what the relationship would look like.
Offer < Offer Products > Product Master
------------------------------
Mark Shnier (YQC)
mark.shnier@gmail.com
------------------------------ | 500 | 2,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | longest | en | 0.951756 |
http://fuelzilla.com/homepower/Water-heater-element-10791-.htm | 1,590,584,499,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347394074.44/warc/CC-MAIN-20200527110649-20200527140649-00081.warc.gz | 49,430,332 | 9,337 | Water heater element?
Posted by stu on July 19, 2008, 4:42 am
I want to use the element in a hot water heater as a dump load. If the
element is 240VAC 2400W and I am dumping 30V 3 phase DC, am I limited to
the current of the element at 240V? 2400/240A?
Thanks
*by 3 phase DC I of course me rectified 3 phase
Posted by M Q on July 19, 2008, 5:03 am
stu wrote:
240 V / 10 A = 24 ohms.
At 30 V and 24 ohms you get 30/24 = 1.25 Amps
30 V * 1.25 A = 37.5 watts
Probably not what you had in mind.
Posted by Charles Foot on July 19, 2008, 11:42 am
M Q wrote:
On the right track though... you just need more of them, i.e. 2 in
parallel = 75 watts, 4 in parallel = 150 watts, etc...
Posted by Jim Wilkins on July 19, 2008, 4:43 pm
How much power do you have to put into the water? A string of power
resistors around the tank under the insulation might do it.
Posted by stu on July 20, 2008, 1:23 am
No it isn't. But thanks. (still shaking my head as to why I didn't use ohms
law myself... must be getting old)
If the elements on the new water heaters are like the one on my old water
heater, it would be pretty easy to tap into it in 6+ places. I'll look into
it.
How much power do you have to put into the water? A string of power
resistors around the tank under the insulation might do it.
Well I have tested it to 200W, around 500W should be possible.
Thanks again guys, I'll look into modifying the element.
•
• Subject
• Author
• Date
Re: Water heater element? Charles Foot 07-19-2008
Re: Water heater element? Jim Wilkins 07-19-2008
Re: Water heater element? Cydrome Leader 07-21-2008
Re: Water heater element? RW Salnick 07-21-2008
Re: Water heater element? phil-news-nospam 07-22-2008 | 545 | 1,709 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-24 | latest | en | 0.936024 |
https://www.ncertpoint.com/2022/01/what-is-the-gas-oil-mixture-for-poulan-wild-thing-chainsaw.html | 1,708,536,623,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473524.88/warc/CC-MAIN-20240221170215-20240221200215-00027.warc.gz | 964,718,284 | 29,792 | # What is the gas oil mixture for Poulan wild thing chainsaw?
A: To produce the specified 40:1 fuel to oil ratio, combine 3.2 ounces of Poulan 2-cycle air cooled engine oil with one gallon of fresh unleaded gasoline in a mixing container. Oil and gasoline will separate over time due to the presence of oxygen in the air.
### In light of this, what is the proper gas mixture for a chainsaw?
The gasoline combination you need will be specified in your owner’s handbook. Gasoline and 2-cycle engine oil are often used in a 40:1 ratio in many gas-powered chainsaws. Chainsaws from Stihl and other manufacturers operate on a 50:1 blend of gasoline and 2-cycle engine oil. Some older versions have a 30:1 compression ratio.
### Second, can I use any two-stroke oil in my chainsaw without having to change the oil?
The two-stroke oils are divided into two types: one for air-cooled engines and another for water-cooled engines. When used in water-cooled engines, two-stroke oil is not meant to work at the higher temperatures seen in air-cooled engines, such as those used in chainsaws.
### Also, would you happen to know whether 40 to 1 has more oil than 50 to 1?
The oil in the fuel combination is there for lubrication only, and it will have little, if any, influence on the mixture until you completely submerge it in oil. Although the ratio of 40 to 1 is small, it will make no difference in the way the unit operates because of its small size.
### I’m wondering what the gas-to-oil ratio is for a Craftsman chainsaw.
Other variables, such as air quality or elevation, may affect the quantity of oil needed by your engine, despite the fact that Craftsman recommends a gas-to-oil combination ranging from 32:1 to 40:1 for most engines.
### What is the best way to mix a 50/1 Fuel chart?
You want to make a 50:1 mixture by combining 2.6 ounces of oil with one gallon of gasoline. If you’re preparing two gallons of gasoline, you’ll need to combine 5.2 ounces of oil with two gallons of gasoline to get a 50:1 ratio of oil to gasoline. I would propose that you use new gasoline with an octane rating of 8Thank you.
### I’m wondering how much 2 cycle oil I should put in a gallon of gas.
Use a 40:1 oil mix ratio for two-cycle engines. One gallon of gasoline coupled with 3.2 ounces of two-cycle engine oil equals one gallon of gasoline.
### What is the fuel-to-air ratio for a two-stroke engine?
If you are unfamiliar with the term, 40:1 is generally considered as a standard ratio for two-stroke engines. This is equal to 25mL of two-stroke oil to 1L of gasoline.
### Was Stihl or Husqvarna a better investment?
A large number of users have said that Stihl offers more low-end torque, making it a superior option for severe cutting applications. Stihl chainsaws are often less expensive than Husqvarna chainsaws. Stihl is known for needing less frequent maintenance than other brands. The Stihl brand of saws is often favoured by homeowners.
### What is the proper way to combine gas and oil for a Poulan chainsaw?
Q: What is the best way to combine gas and oil? A: To produce the specified 40:1 fuel to oil ratio, combine 3.2 ounces of Poulan 2-cycle air cooled engine oil with one gallon of fresh unleaded gasoline in a mixing container. Oil and gasoline will separate over time due to the presence of oxygen in the air.
### What is the ration of 50 to 1 in this case?
Regarding the 50:1 ratio Thank you so much for your assistance. You are adding oil in a 50:1 ratio to the water. This implies that you will need to add 1/50th of a litre of oil for every litre of gasoline you use. 1 litre divided by 1/50 equals 0.02 litres, or 20 mL.
### What does the term “50 to 1 gas oil mix” refer to?
50:1 refers to the fact that for every 50 ounces of gasoline, 1 ounce of oil must be added. Using a 1 gallon gas can, you would need 128 ounces (1 gal) divided by 50 equals 2.56 gallons of gasoline.
### What is the formula for calculating the oil to gas ratio?
2.6 fluid ounces of oil per gallon of gas is required for a 50:1 gas-to-oil conversion ratio. 3.2 fluid ounces of oil per gallon of gas is the ratio to utilise for a 40:1 combination. In order to get a 32:1 combination, 4 fluid ounces of oil per gallon of gas should be used.
### Is it possible to use too much 2 stroke oil?
Although it is critical to use the right oil-to-gas ratio, if you make a mistake, it is preferable to overfill your engine with oil rather than underfill it with oil. When there is too much oil, it may cause a smokey exhaust, oil to flow out of the muffler, and in rare cases, a loss of power.
### When it comes to 2-stroke outboards, what is the optimal oil-to-gas ratio?
100:1 – Combine 50ml of oil with 5ltrs of gasoline to get a 100:1 mixture. For older Suzuki outboards built before 1997, I’d propose a 50:1 compression ratio, since the information I’ve supplied is for current model two-stroke engines. Please keep in mind that Suzuki recommends a 50:1 ratio for commercial use.
### Is it possible to utilise motor oil in a two-stroke mixture?
Yes, it is possible. Utilize it in the identical quantities as you would with two-stroke engine oil. Essentially, the goal is to preserve the lubricating properties of the oil supplied to the fuel; however, using too little oil might result in bearing and bore damage. Although it is of a rather low grade (lighter than motor oil), it burns pretty effectively and is reasonably priced!
### Is there a ratio between the oil and the gas in a Weedeater?
All two-cycle gas trimmers under the Weed Eater name operate on a 40:1 gas to oil ratio. In terms of weight, this is equivalent to 3.2 ounces of oil to 1 gallon of standard gas. The gas should have an octane rating of 87 and contain no more than 10% alcohol. For two-cycle engines, the oil must be specially prepared. | 1,436 | 5,828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-10 | latest | en | 0.944294 |
https://preparingforcodinginterview.wordpress.com/2016/12/22/linked-list/ | 1,501,261,494,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500550975184.95/warc/CC-MAIN-20170728163715-20170728183715-00427.warc.gz | 696,781,439 | 42,622 | 1) Dynamic size
2) Ease of insertion/deletion
Drawbacks:
1) Random access is not allowed. We have to access elements sequentially starting from the first node. So we cannot do binary search with linked lists.
2) Extra memory space for a pointer is required with each element of the list.
Questions –
insert – at head, at last, after a specific value, at a specific position
delete – with given value, at given position, delete alternate node,
— find length – iterative and recursive
search item – iterative and recursive
Nth node
Rotate a Linked List around a given point k.
— Reverse – Code Iterative (Video) Recursive (Video)
2. ** Middle Element – Move one pointer by one and other pointer by two. When the fast pointer reaches end slow pointer will reach middle of the linked list.
while(fast!=null && fast.next !=null){}
3. ** n’th node from the end – this is same as (len – n + 1)th node from the beginning of the Linked List. So calculate the length and then use the formula.
4. Delete a Linked List – In Java, automatic garbage collection happens, so deleting a linked list is easy. We just need to change head to null.
5. ** Removing the last occurrence of an element in a (singly) linked list with only one traversal – Simply remember the previous entry every time you find the value you’re searching for on the traversal. When the traversal is complete, the last entry remembered will have a link to the entry to be removed, and that is sufficient to do the removal. Handle the case for head and tail properly.
6. *** Detect and remove loop in linked list and remove it – Link Code
Note: node are same based on address and not the data.So compare address, not data.
Floyd’s Cycle-Finding Algorithm: – Proof
Detect – This is the fastest method. Traverse linked list using two pointers. Move one pointer by one and other pointer by two. If these pointers meet at some node then there is a loop. If pointers do not meet then linked list doesn’t have loop.
Remove – detect loop using the above algorithm. Now set the slow pointer to head and let the fast pointer remain unchanged. Now start checking until both the fast and slow pointer are same. Read proof in the above link. Sweet and simple.
while (slow.next != fast.next) {
slow = slow.next;
fast = fast.next;
}
System.out.println(“Loop starts at: ” + fast.next.data);
fast.next = null;
Use Hashing:
Traverse the list one by one and keep putting the node addresses in a Hash Table. At any point, if NULL is reached then return false and if next of current node points to any of the previously stored nodes in Hash then return true.
Mark Visited Nodes:
This solution requires modifications to basic linked list data structure. Have a visited flag with each node. Traverse the linked list and keep marking visited nodes. If you see a visited node again then there is a loop. This solution works in O(n) but requires additional information with each node.
7. Given a linked list which is sorted, how will you insert in sorted way Code
Two special case –
1. if LL is empty
2. if new node data is smaller then the node
8. Remove duplicates from a sorted linked list Code – Simply need to change pointer as all duplicate node will be together.
9. Remove duplicates from an unsorted linked list –
– hashing – O(n) on average + O(n) – space
– sorting – O(nLogn) + O(n) = O(nLogn)
using two loop O(n^2)
10. ** Pairwise swap elements of a given linked list (swap data, not the nodes) – just exchange data b/w current and current.next node
11. *** Write a function to get the intersection point of two Linked Lists.
Mark the node as visited. This takes extra space and required changes to the
original linked list, the approach can be easily used in other problem.
counting the difference in node. Algo –
Get count of the nodes in first and second list, let count be c1 and c2 and there
diff d = abs(c1 – c2)
Now traverse the bigger list from the first node till d nodes so that from here onwards both the lists have equal no of nodes.
Then we can traverse both the lists in parallel till we come across a common
node. (Note that getting a common node is done by comparing the address of
the nodes)
the universal way of hashing can be used as well.
12. Move last element to front of a given Linked List – find the last and second last. Update the head accordingly.
13. ** Intersection of two Sorted Linked Lists – Code(line 68) – Simple. Algo
– we need to create a new linked list of common elements
– traverse both the linked list. If both have common element then create a new node in the new linked list with the data.
– if they are different advance in only one of the LL based on the value.
– if one list exists before the second one then stop. There are no more common.
iterative solution is straight forward
– Recursive solution is simple. Like the recursive count method
Algo – Basically we need to pick numbers from both the list and add it digit by digit. If the digit exceeds 10 then we take a carry, same as manual calculation. We will have to maintain the carry onto the next addition. We keep doing this until reach the end of both the list. For each digit that we get we keep pushing it to a new list.
16. ** Segregate even and odd nodes in a Linked List –
Algo:
1. Get the last node pointer. Then start from the head. For each odd number move that node to the end.
2. The idea is to split the linked list into two: one containing all even nodes and other containing all odd nodes. And finally attach the odd node linked list after the even node linked list. To split the Linked List, traverse the original Linked List and move all odd nodes to a separate Linked List of all odd nodes. At the end of loop, the original list will have all the even nodes and the odd node list will have all the odd nodes. To keep the ordering of all nodes same, we must insert all the odd nodes at the end of the odd node list. And to do that in constant time, we must keep track of last pointer in the odd node list.
17. ** Union and Intersection of two Linked Lists – Given two Linked Lists, create union and intersection lists that contain union and intersection of the elements present in the given lists. Order of elements in output lists doesn’t matter. Algo –
1. Brute Force. O(mn)
2. Use Sorting. Since these are linked list, use Merge Sort.Linearly scan both sorted lists to get the union and intersection. This step takes O(m + n) time.
Overall, O(mLogm + nLogn)
3. Hashing. Put all items of first linked list into hash table. Now start looking up for each element of second list in this table.
If element is found, move that element to the intersection list.
When all items are over in list, add all element of table to hash table to union list.
18. *** Find a triplet from three linked lists with sum equal to a given number – Algo –
Brute force – the outermost loop picks an element from list a, the middle loop picks an element from b and the innermost loop picks from c. O(n^3)
–
1) Sort list b in ascending order, and list c in descending order.
2) After the b and c are sorted, one by one pick an element from list a and find the pair by traversing both b and c. depending on the sum value we will check if the we need to advance in node b or node c.
complexity = O(n^2).
19. *** Clone a linked list with next and random pointer Video
Type 1 (Method 1 described Below) , Type 2, Type 3
Method 1 – Time Complexity: O(n) Auxiliary Space: O(1)
1) Create the copy of node 1 and insert it between node 1 & node 2 in original Linked List, create the copy of 2 and insert it between 2 & 3.. Continue in this fashion, add the copy of N after the Nth node.
2) Now copy the arbitrary link in this fashion
``` original->next->arbitrary = original->arbitrary->next; /*TRAVERSE
TWO NODES*/
```
This works because original->next is nothing but copy of original and Original->arbitrary->next is nothing but copy of arbitrary.
3) Now restore the original and copy linked lists in this fashion in a single loop.
``` original->next = original->next->next;
copy->next = copy->next->next;
```
4) Make sure that last element of original->next is NULL.
Method 2 – Time Complexity: O(n) Auxiliary Space: O(n)
The idea is to use Hashing. Below is algorithm.
1. Traverse the original linked list and make a copy in terms of data.
2. Make a hash map of key value pair with original linked list node and copied linked list node.
3. Traverse the original linked list again and using the hash map adjust the next and random reference of cloned linked list nodes.
20. *** Reverse a linked list – Code Iterative (Video) Recursive (Video)
21. *** Reverse a Linked List in groups of given size – Read the code to reverse iteratively and then try to follow that logic recursively. Remember – prev and next are the key.
22. *** Reverse alternate K nodes in a Singly Linked List – Code
For all the reverse question – draw out the list on paper and work out the problem.
23. *** Rearrange a given linked list in-place.
Algo 1, Time complexity = O(n2)1) Initialize current node as head.
2) While next of current node is not null, do following
a) Find the last node, remove it from end and insert it as next of current node. b) Move current to next to next of current
Algo 2, Time complexity = O(n)
1) Find the middle point using tortoise and hare method.
2) Split the linked list in two halves using found middle point in step 1.
3) Reverse the second half.
4) Do alternate merge of first and second halves.
24. Swap nodes in a linked list without swapping data –
– if both x and y are same – don’t do anything
– find the currentX and prevX and currentY and prevY
– if currX or currY is null( does not exits) return
– check if either currX or currY is head. This can be done using prevX == null or
prevY == null. If yes then we need to change next accordingly.
if (prevX != null)
prevX.next = currY;
else //make y the new head
if (prevY != null)
prevY.next = currX;
else // make x the new head
// Swap next pointers
Node temp = currX.next;
currX.next = currY.next;
currY.next = temp;
25. *** Sort a linked list of 0s, 1s and 2s – count 0, 1 and 2. Add them to a new list.
26. *** Check if a singly linked list is palindrome –
– Use Stack
– Reverse the List and compare to original list
27. *** Given only a pointer to a node to be deleted in a singly linked list, how do you delete it? Solution or This
You basically, copy the data of node_to_delete.next to node_to_delete.data. So the Then point node_to_delete.next to node_to_delete.next.next. So u basically remove node_to_delete.next from scene(garbage collector will take of that node).
(not the previous node pointing to node_to_delete is still working).You might be asked to discuss about the dangling pointer case if the node_to_delete.next was being referred from somewhere else.
28. ** Add 1 to a number represented as linked list – this is very similar to Add two numbers represented by linked lists . The only difference is the order. So first reverse the list. Do the old school addition with carry. Then reverse it back again.
29. Given a linked list of co-ordinates where adjacent points either form a vertical line or a horizontal line. Delete points from the linked list which are in the middle of a horizontal or vertical line. – Link Code
Algo – The idea is to keep track of current node, next node and next-next node. While the next node is same as next-next node, keep deleting the next node. In this complete procedure we need to keep an eye on shifting of pointers and checking for NULL values.
30. Delete all occurrences of a given key in a linked list – need to take care if the value exists at the head, rest the logic is straightforward.
31. Rearrange a linked list such that all even and odd positioned nodes are together Code Algo –
we want to separate the old and even node. So initially I though of moving all the odd nodes to the end of the list. But then we wouldn’t know when to stop traversing the list.
We mark the head node as even and the head.next as odd. We create a temp node which points to the start of this odd. Now we follow our initial approach. We keep shifting from even node to even node by linking them, at the same time we keep linking the all nodes together. In the end we stop at even node, then we update the tail node of even to point to the temp node we created initially, which points to the odd node. While linking even node we also linked all the odd nodes.
32. ** Delete N nodes after M nodes of a linked list – Code We count m node initially. Store the prev node at also. Then when deleted we count n nodes. At the end of that we set the prev to the pointer to the next of last node to be deleted.
33. Alternating split of a given Singly Linked List – we need to create two new list here. The only thing to note is that the order is maintained(Best to clarify this from the interviewer). If we keep inserting at the end, we will have to start from the head till the last node. Since this can be an expensive operation we save a reference to the tail node and use it everytime we need to add node to tail. Just an O(1) for adding new node to tail.
34. Compare two strings represented as linked lists – Looks complicated but boils down to three simple case. 1. If both are same. 2. If 1 is bigger then the other the first one comes first lexicographically. If both have different character, then we pick up the pick character that are different and then compare.
35. Point arbit pointer to greatest value right side node in a linked list Code
Note: the arbitrary points to next value(which might or might not be the greater then the current node). In the code we are not doing anything to the next node. We just reach the end of the list recursively after which I we mark the last node as maxNode for the second last node (arbitrary for last node will be null and for second last node it will always be last node). After that we start comparing the value of current node with the max and keep updating it.
Draw out on a paper before proceeding for better understanding.
No need to reverse the list.
36. Delete nodes which have a greater value on right side Code
Algo – without reversing it’s not possible in O(n). tried that but failed at this
case: 12->10->8->11. We can keep deleting when moving forward but this case can’t be handled like that. So had to follow the approach given in link above.
Reversing helps as when we move forward we keep a max reference and easy deletion. One observation – the final list will always be in descending order.
So instead of deleting the one on left, we reverse and delete on right, any value that is smaller then the current value.
37. Given two linked lists, the task is to check whether the first list is present in 2nd list or not.
38. Given a sorted linked list, delete all nodes that have duplicate numbers (all occurrences), leaving only numbers that appear once in the original list. GFG.
Algo: Nothing much to discuss here. It’s given list is sorted. Can be easily implemented in O(n) with no extra space.
39. Given a linked list and a key in it, the task is to move all occurrences of given key to end of linked list, keeping order of all other elements same.
Algo:
We create a new null node. We traverse the list. Basically we will make this pointer point to the first occurence of the key in list and then keep linking this next’s item in the list to next occurence of the element. So that all nodes of with given key are linked together. At the same time we also keep removing the key’s node link in the main list. Check the eg below.
2 -> 3 -> 4 -> 5 -> 3 -> 6 ->7 -> null
If we need to remove 3, the new list will be,2 -> 4 -> 5 -> 6 ->7 -> null
N the new pointer will be 3->3->null. Now link this pointer to the end of this original list.
The code is simple and a similar example has been coded. Not sure which question though.
PENDING
1. [Skip] Flatten a multilevel linked list
2. [Skip] Construct a Maximum Sum Linked List out of two Sorted Linked Lists having some Common nodes | 3,874 | 16,095 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.825685 |
https://programmer.ink/think/a-algorithm-for-path-planning.html | 1,632,652,587,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057857.27/warc/CC-MAIN-20210926083818-20210926113818-00336.warc.gz | 497,847,563 | 9,502 | # A * algorithm for path planning
Posted by isurgeon on Thu, 02 Sep 2021 02:11:26 +0200
https://github.com/paulQuei/a-star-algorithm?spm=a2c6h.12873639.0.0.21834662LgAvMP
# Algorithm Introduction
A * (read: A Star) algorithm is a very common path finding and graph traversal algorithm. It has better performance and accuracy. While explaining the algorithm, this paper will also provide the code implementation of Python language, and dynamically display the operation process of the algorithm with the help of matplotlib library.
A * algorithm was originally published in 1968 by Peter Hart, Nils Nilsson and Bertram Raphael of Stanford Research Institute. It can be considered as an extension of Dijkstra algorithm.
Due to the guidance of heuristic function, A * algorithm usually has better performance.
As its name indicates, breadth first search takes breadth as its priority.
Starting from the starting point, first traverse the adjacent points around the starting point, and then traverse the adjacent points of the traversed points, and gradually spread outward until the end point is found.
This algorithm expands outward like Flood fill. The process of the algorithm is shown in the following figure:
In the above dynamic graph, the algorithm traverses all the points in the graph, which is usually unnecessary. For the problem with a clear end point, the algorithm can be terminated in advance once it reaches the end point. The following figure compares this situation:
In the process of executing the algorithm, each point needs to record the position of the previous point reaching the point - which can be called the parent node. After this, once you reach the end point, you can start from the end point, and then find the starting point in the order of the parent nodes, thus forming a path.
# Dijkstra algorithm
Dijkstra algorithm was proposed by computer scientist Edsger W. Dijkstra in 1956.
Dijkstra algorithm is used to find the shortest path between nodes in the graph.
Considering such a scenario, in some cases, the moving costs between adjacent nodes in the graph are not equal. For example, if a picture in the game has both flat land and mountains, the speed of the characters in the game moving in the flat land and mountains is usually not equal.
In Dijkstra algorithm, the total moving cost of each node from the starting point needs to be calculated. At the same time, a priority queue structure is also needed. All nodes to be traversed are placed in the priority queue and sorted according to the cost.
In the process of running the algorithm, the node with the lowest cost is selected from the priority queue every time as the next traversal node. Until we reach the end.
The operation results of breadth first search without considering the difference of node mobility cost and Dijkstra algorithm considering mobility cost are compared below:
When the graph is a grid graph and the moving cost between each node is equal, the Dijkstra algorithm will become the same as the breadth first algorithm.
# Best first search
In some cases, if we can calculate the distance from each node to the end point in advance, we can use this information to reach the end point faster.
The principle is also very simple. Similar to Dijkstra algorithm, we also use a priority queue, but at this time, the distance from each node to the end point is taken as the priority, and the node with the lowest moving cost to the end point (closest to the end point) is always selected as the next traversal node. This algorithm is called Best First algorithm.
This can greatly speed up the path search, as shown in the following figure:
Because if there are obstacles between the start point and the end point, the best first algorithm may not find the shortest path. The following figure describes this situation.
## A * algorithm
After comparing the above algorithms, we can finally explain the focus of this paper: A * algorithm.
In the following description, we will see that the A * algorithm actually integrates the characteristics of these algorithms.
A * algorithm calculates the priority of each node through the following function.
f ( n ) = g ( n ) + h ( n ) f(n) = g(n)+h(n) f(n)=g(n)+h(n)
Of which:
• f ( n ) f(n) f(n) is the comprehensive priority of node n. When we select the next node to traverse, we always select the node with the highest comprehensive priority (the lowest value).
• g ( n ) g(n) g(n) is the cost of node n from the starting point.
• h ( n ) h(n) h(n) is the estimated cost of node n from the end point, which is the heuristic function of A * algorithm. We will explain the heuristic function in detail below.
A * algorithm selects from the priority queue every time during operation f ( n ) f(n) The node with the lowest f(n) value (the highest priority) is the next node to be traversed.
In addition, the A * algorithm uses two sets to represent the nodes to be traversed and the nodes that have been traversed, which is usually called open_set and close_set.
The complete A * algorithm is described as follows:
* initialization open_set and close_set；
* Add starting point to open_set And set the priority to 0 (the highest priority);
* If open_set If it is not empty, the open_set Select the node with the highest priority n:
* If node n Is the end point, then:
* Step by step tracking from the end point parent Node, reaching the starting point;
* Return the found result path, and the algorithm ends;
* If node n Not the end point, then:
* Will node n from open_set Delete and join close_set Medium;
* Traversal node n All adjacent nodes:
* If adjacent nodes m stay close_set Medium, then:
* Skip and select the next adjacent node
* If adjacent nodes m Neither open_set Medium, then:
* Set node m of parent As node n
* Calculation node m Priority of
* Will node m join open_set in
# Heuristic function
As mentioned above, the heuristic function will affect the behavior of A * algorithm.
• In extreme cases, when the heuristic function h ( n ) h(n) If h(n) is always 0, then g ( n ) g(n) g(n) determines the priority of nodes, and the algorithm degenerates into Dijkstra algorithm.
• If h ( n ) h(n) If h(n) is always less than or equal to the cost of node n to the end point, the A * algorithm ensures that the shortest path can be found. But when h ( n ) h(n) The smaller the value of h(n), the more nodes the algorithm will traverse, and the slower the algorithm will be.
• If h ( n ) h(n) h(n) is exactly equal to the cost of node n to the end point, then the A * algorithm will find the best path, and the speed is very fast. Unfortunately, this is not possible in all scenarios. Because before we reach the end, it is difficult to calculate exactly how far we are from the end.
• If h ( n ) h(n) The value of h(n) is greater than the cost of node n to the end point, so the A * algorithm can not guarantee to find the shortest path, but it will be very fast at this time. At the other extreme, if h ( n ) h(n) h(n) relative to g ( n ) g(n) If g(n) is much larger, then only h ( n ) h(n) h(n) produces an effect, which becomes the best first search.
From the above information, we can know that we can control the speed and accuracy of the algorithm by adjusting the heuristic function. Because in some cases, we may not necessarily need the shortest path, but hope to find A path as soon as possible. This is also where the A * algorithm is more flexible.
For graphs in grid form, the following heuristic functions can be used:
• If you are only allowed to move up, down, left and right in the drawing, you can use Manhattan distance.
• If movement in eight directions is allowed in the drawing, you can use diagonal distances.
• If movement in any direction is allowed in the drawing, you can use Euclidean distance.
## Manhattan distance
If only four directions are allowed to move up, down, left and right in the graph, the Manhattan distance can be used for the heuristic function, and its calculation method is shown in the following figure:
The function for calculating Manhattan distance is as follows. D here refers to the movement cost between two adjacent nodes, which is usually a fixed constant.
function heuristic(node) =
dx = abs(node.x - goal.x)
dy = abs(node.y - goal.y)
return D * (dx + dy)
## Diagonal distance
If oblique movement towards adjacent nodes is allowed in the graph, the heuristic function can use diagonal distance. Its calculation method is as follows:
The function for calculating the diagonal distance is as follows. D2 here refers to the moving cost between two adjacent nodes obliquely. If all nodes are square, the value is 2 ∗ D \sqrt{2} * D 2 ∗D.
function heuristic(node) =
dx = abs(node.x - goal.x)
dy = abs(node.y - goal.y)
return D * (dx + dy) + (D2 - 2 * D) * min(dx, dy)
## Euclidean distance
If movement in any direction is allowed in the drawing, Euclidean distances can be used.
Euclidean distance refers to the straight-line distance between two nodes, so its calculation method is also familiar to us: ( p 2. x − p 1. x ) 2 + ( p 2. y − p 1. y ) 2 \sqrt{(p2.x-p1.x)^2 + (p2.y-p1.y)^2} (p2.x−p1.x)2+(p2.y−p1.y)2 . Its function is expressed as follows:
function heuristic(node) =
dx = abs(node.x - goal.x)
dy = abs(node.y - goal.y)
return D * sqrt(dx * dx + dy * dy)
# Algorithm implementation
The source code of the algorithm can be downloaded from github: Paul quei / a-star-algorithm.
Our algorithm demonstrates the solution process of finding the end point from the starting point on a two-dimensional grid graph.
# Coordinate points and maps
First, we create a very simple class to describe the points in the diagram. The relevant codes are as follows:
# point.py
import sys
class Point:
def __init__(self, x, y):
self.x = x
self.y = y
self.cost = sys.maxsize
Next, we implement a class that describes the map structure. To simplify the description of the algorithm:
We select the point at the lower left coordinate [0, 0] as the starting point of the algorithm, and the point at the upper right coordinate [size - 1, size - 1] as the end point.
In order to make the algorithm more interesting, we set an obstacle in the middle of the map, and the map will also contain some random obstacles. The code of this class is as follows:
# random_map.py
import numpy as np
import point
class RandomMap:
def __init__(self, size=50): # Constructor, the default size of the map is 50x50;
self.size = size
self.obstacle = size//8 # the number of obstacles is the map size divided by 8;
self.GenerateObstacle() # Call generateobserver to generate random obstacles;
def GenerateObstacle(self):
self.obstacle_point = []
self.obstacle_point.append(point.Point(self.size//2, self.size//2))
self.obstacle_point.append(point.Point(self.size//2, self.size//2-1))
# Generate an obstacle in the middle
for i in range(self.size//2-4, self.size//2): # generate an oblique obstacle in the middle of the map;
self.obstacle_point.append(point.Point(i, self.size-i))
self.obstacle_point.append(point.Point(i, self.size-i-1))
self.obstacle_point.append(point.Point(self.size-i, i))
self.obstacle_point.append(point.Point(self.size-i, i-1))
for i in range(self.obstacle-1): # Several other obstacles are randomly generated;
x = np.random.randint(0, self.size)
y = np.random.randint(0, self.size)
self.obstacle_point.append(point.Point(x, y))
if (np.random.rand() > 0.5): # Random boolean the direction of obstacles is also random;
for l in range(self.size//4):
self.obstacle_point.append(point.Point(x, y+l))
pass
else:
for l in range(self.size//4):
self.obstacle_point.append(point.Point(x+l, y))
pass
def IsObstacle(self, i ,j): # Define a method to judge whether a node is an obstacle;
for p in self.obstacle_point:
if i==p.x and j==p.y:
return True
return False
# Algorithm body
With the basic data structure, we can start to implement the algorithm body.
Here we use a class to encapsulate our algorithm.
First, implement some basic functions required by the algorithm, which are as follows:
# a_star.py
import sys
import time
import numpy as np
from matplotlib.patches import Rectangle
import point
import random_map
class AStar:
def __init__(self, map): # __ init__: Class.
self.map=map
self.open_set = []
self.close_set = []
def BaseCost(self, p): # Basepost: the moving cost from the node to the starting point, corresponding to g(n) above.
x_dis = p.x
y_dis = p.y
# Distance to start point
return x_dis + y_dis + (np.sqrt(2) - 2) * min(x_dis, y_dis)
def HeuristicCost(self, p): # HeuristicCost: heuristic function from node to endpoint, corresponding to $h(n)$. Since we are a grid based graph, this function uses a diagonal distance from the previous function.
x_dis = self.map.size - 1 - p.x
y_dis = self.map.size - 1 - p.y
# Distance to end point
return x_dis + y_dis + (np.sqrt(2) - 2) * min(x_dis, y_dis)
def TotalCost(self, p): # TotalCost: sum of costs, i.e. corresponding to f(n) mentioned above.
return self.BaseCost(p) + self.HeuristicCost(p)
def IsValidPoint(self, x, y): # IsValidPoint: judge whether the point is valid. It is invalid if it is not inside the map or where the obstacle is located.
if x < 0 or y < 0:
return False
if x >= self.map.size or y >= self.map.size:
return False
return not self.map.IsObstacle(x, y)
def IsInPointList(self, p, point_list): # Determine whether the point is in a set.
for point in point_list:
if point.x == p.x and point.y == p.y:
return True
return False
def IsInOpenList(self, p): # Judge whether the point is open_set.
return self.IsInPointList(p, self.open_set)
def IsInCloseList(self, p): # Judge whether the point is closed_ Set.
return self.IsInPointList(p, self.close_set)
def IsStartPoint(self, p): # Judge whether the point is the starting point.
return p.x == 0 and p.y ==0
def IsEndPoint(self, p): # Judge whether the point is the end point.
return p.x == self.map.size-1 and p.y == self.map.size-1
With the above auxiliary functions, you can start to implement the main logic of the algorithm. The relevant codes are as follows:
# a_star.py
def RunAndSaveImage(self, ax, plt):
start_time = time.time()
start_point = point.Point(0, 0)
start_point.cost = 0
self.open_set.append(start_point)
while True:
index = self.SelectPointInOpenList()
if index < 0:
print('No path found, algorithm failed!!!')
return
p = self.open_set[index]
rec = Rectangle((p.x, p.y), 1, 1, color='c')
self.SaveImage(plt)
if self.IsEndPoint(p):
return self.BuildPath(p, ax, plt, start_time)
del self.open_set[index]
self.close_set.append(p)
# Process all neighbors
x = p.x
y = p.y
self.ProcessPoint(x-1, y+1, p)
self.ProcessPoint(x-1, y, p)
self.ProcessPoint(x-1, y-1, p)
self.ProcessPoint(x, y-1, p)
self.ProcessPoint(x+1, y-1, p)
self.ProcessPoint(x+1, y, p)
self.ProcessPoint(x+1, y+1, p)
self.ProcessPoint(x, y+1, p)
This code should not need much explanation. It is implemented according to the previous algorithm logic. In order to show the results, we will save the state into pictures with the help of matplotlib library at each step of the algorithm.
The above function calls several other functions, and the code is as follows:
# a_star.py
def SaveImage(self, plt):
millis = int(round(time.time() * 1000))
filename = './' + str(millis) + '.png'
plt.savefig(filename)
def ProcessPoint(self, x, y, parent):
if not self.IsValidPoint(x, y):
return # Do nothing for invalid point
p = point.Point(x, y)
if self.IsInCloseList(p):
return # Do nothing for visited point
print('Process Point [', p.x, ',', p.y, ']', ', cost: ', p.cost)
if not self.IsInOpenList(p):
p.parent = parent
p.cost = self.TotalCost(p)
self.open_set.append(p)
def SelectPointInOpenList(self):
index = 0
selected_index = -1
min_cost = sys.maxsize
for p in self.open_set:
cost = self.TotalCost(p)
if cost < min_cost:
min_cost = cost
selected_index = index
index += 1
return selected_index
def BuildPath(self, p, ax, plt, start_time):
path = []
while True:
path.insert(0, p) # Insert first
if self.IsStartPoint(p):
break
else:
p = p.parent
for p in path:
rec = Rectangle((p.x, p.y), 1, 1, color='g')
plt.draw()
self.SaveImage(plt)
end_time = time.time()
print('===== Algorithm finish in', int(end_time-start_time), ' seconds')
These three functions should be easy to understand:
• SaveImage: saves the current state to a picture named after the current time.
• ProcessPoint: process each node: if it is a node that has not been processed, calculate the priority, set the parent node, and add it to open_set.
• SelectPointInOpenList: from open_ Find the node with the highest priority in set and return its index.
• BuildPath: construct the result path from the end point back along the parent. Then draw the result from the starting point. The result uses a green square. Each drawing step saves one picture.
# Test entrance
Finally, the entry logic of the program. Use the class written above to find the path:
# main.py
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.patches import Rectangle
import random_map
import a_star
plt.figure(figsize=(5, 5))
map = random_map.RandomMap() ①
ax = plt.gca()
ax.set_xlim([0, map.size]) ②
ax.set_ylim([0, map.size])
for i in range(map.size): ③
for j in range(map.size):
if map.IsObstacle(i,j):
rec = Rectangle((i, j), width=1, height=1, color='gray')
else:
rec = Rectangle((i, j), width=1, height=1, edgecolor='gray', facecolor='w')
rec = Rectangle((0, 0), width = 1, height = 1, facecolor='b')
rec = Rectangle((map.size-1, map.size-1), width = 1, height = 1, facecolor='r')
plt.axis('equal') ⑥
plt.axis('off')
plt.tight_layout()
#plt.show()
a_star = a_star.AStar(map)
a_star.RunAndSaveImage(ax, plt) ⑦
This code is described as follows:
1. Create a random map;
2. Set the content of the image to be consistent with the map size;
3. Draw a map: draw a gray square for obstacles and a white square for other areas;
4. The drawing starting point is a blue square;
5. The drawing end point is a red square;
6. Set the coordinate axis scale of the image to be equal and hide the coordinate axis;
7. Call the algorithm to find the path;
Since our map is random, the results of each run may be different. Here are the results of a run on my computer: | 4,491 | 18,236 | {"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} | 3.828125 | 4 | CC-MAIN-2021-39 | latest | en | 0.92127 |
https://az-links.info/and-relationship/angle-of-incidence-and-refraction-relationship-questions.php | 1,571,839,053,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987833766.94/warc/CC-MAIN-20191023122219-20191023145719-00392.warc.gz | 387,398,529 | 9,126 | # Angle of incidence and refraction relationship questions
### Sample Problems for Snell's Law
Light travels from air into an optical fiber with an index of refraction of (a) In which direction does the light bend? (b) If the angle of incidence on the end of. Like with reflection, refraction also involves the angles that the incident ray and the refracted ray make with the normal to the surface at the point of refraction. Questions pertaining to reflection and refraction. az-links.info angle of incidence is equal to the angle of reflection for diffuse reflections. II. The angle measured.
## Snell's Law
We find that the incoming and outgoing light beams are actually parallel. When light travels from an area of lower index to an area of higher index, the ratio is less than one, and the refracted ray is smaller than the incident one; hence the incident ray is bent toward the normal as it hits the boundary. Of course, refraction can also occur in a non-rectangular object indeed, the objects that we are interested in, lenses, are not rectangular at all. The calculation of the normal direction is harder under these circumstances, but the behaviour is still predicted by Snell's Law.
Calculating n Given a transparent substance, we can always find its index of refraction by using a setup like the example above. Surrounding the substance of unknown index n with a material with a known index of refraction, we can find the unknown n by measuring angles and applying Snell's Law.
However, calculating ns in this way, an obvious question arises. How did the first index get calculated? We could always choose an arbitrary substance as a meterstick, and calculate all other indices in terms of this base.
However, indices of refraction arise in Maxwell's equations for electromagnetic waves; that, in fact, is how they are defined.
## Refraction of Light Lab Answers
We shall not delve into these equations here; instead we will note that n for air is very close to 1, and that we can therefore easily calcuate n for any other substance using our setup above. If at any time the values for the numerator and denominator become accidentally switched, the critical angle value cannot be calculated.
Mathematically, this would involve finding the inverse-sine of a number greater than 1. Physically, this would involve finding the critical angle for a situation in which the light is traveling from the less dense medium into the more dense medium - which again, is not possible.
This equation for the critical angle can be used to predict the critical angle for any boundary, provided that the indices of refraction of the two materials on each side of the boundary are known. Examples of its use are shown below: Example A Calculate the critical angle for the crown glass-air boundary. Refer to the table of indices of refraction if necessary.
The solution to the problem involves the use of the above equation for the critical angle.
Of all the possible combinations of materials that could interface to form a boundary, the combination of diamond and air provides one of the largest differences in the index of refraction values. This peculiarity about the diamond-air boundary plays an important role in the brilliance of a diamond gemstone.
Having a small critical angle, light has the tendency to become "trapped" inside of a diamond once it enters.
Brewster's Angle, Polarization of Light, Polarizing Angle - Physics Problems
A light ray will typically undergo TIR several times before finally refracting out of the diamond. Because the diamond-air boundary has such a small critical angle due to diamond's large index of refractionmost rays approach the diamond at angles of incidence greater than the critical angle.
This gives diamond a tendency to sparkle.
### Snell's Law -- The Law of Refraction
The effect can be enhanced by the cutting of a diamond gemstone with a strategically planned shape. The diagram below depicts the total internal reflection within a diamond gemstone with a strategic and a non-strategic cut. Use the Find the Critical Angle widget below to investigate the effect of the indices of refraction upon the critical angle. Simply enter the index of refraction values; then click the Calculate button to view the result.
Use the widget as a practice tool. Check Your Understanding 1.
Suppose that the angle of incidence of a laser beam in water and heading towards air is adjusted to degrees. Use Snell's law to calculate the angle of refraction? Explain your result or lack of result. See Answer Good luck! This problem has no solution. The angle of incidence is greater than the critical angle, so TIR occurs. | 939 | 4,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2019-43 | longest | en | 0.936951 |
https://www.experts-exchange.com/questions/24296658/regular-expression.html | 1,495,841,666,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608686.22/warc/CC-MAIN-20170526222659-20170527002659-00381.warc.gz | 1,099,562,570 | 26,336 | Solved
regular expression
Posted on 2009-04-05
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Hello
I need to search for P123/ABC using a regular expression?
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Question by:Skycapt
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ID: 24073588
Could you be more specific? You can find this exact expression when you just call it: P123/ABC.
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ID: 24075433
Oh sorry,
I have a list of products P123 to P500 in number order but 5 or 6 of them have the extended code P123/ABC the numbers and letters may vary so you may get P234/DEF they also may change from day to day so I cant just isolate individual codes.
Hope that helps
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cubeeq earned 250 total points
ID: 24075699
ok, the most general variant is "P[0-9][0-9][0-9](/[A-Z][A-Z][A-Z])?" - it also matches P000/AAA, you can narrow it to "P[1-5][0-9][0-9](/[A-Z][A-Z][A-Z])?" and so on. if you want the most specific thing, i will post something more complex.
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ID: 31566842
Hi, thanks for your help, here's my final code
'^[P][0-9]{3}(/[A-Z]{3})?\$'
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Course of the Month5 days, 7 hours left to enroll | 583 | 2,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-22 | longest | en | 0.829354 |
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# Logic
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### Logic
1. 1. LOGIC
2. 2. Statements • Logic is the tool for reasoning about the truth or falsity of statements. – Propositional logic is the study of Boolean functions – Predicate logic is the study of quantified Boolean functions (first order predicate logic)
3. 3. Arithmetic vs. Logic Arithmetic Logic 0 false 1 true Boolean variable statement variable form of function statement form value of function truth value of statement equality of function equivalence of statements
4. 4. Notation Word Symbol and v or w implies 6 equivalent ] not ~ not 5 parentheses ( ) used for grouping terms
5. 5. Notation Examples English Symbolic A and B A v B A or B A w B A implies B A 6 B A is equivalent to B A ] B not A ~A not A 5A
6. 6. Statement Forms • (p v q) and (q v p) are different as statement forms. They look different. • (p v q) and (q v p) are logically equivalent. They have the same truth table. • A statement form that represents the constant 1 function is called a tautology. It is true for all truth values of the statement variables. • A statement form that represents the constant 0 function is called a contradiction. It is false for all truth values of the statement variables.
7. 7. Truth Tables - NOT P 5P T F F T
8. 8. Truth Tables - AND P Q PvQ T T T T F F F T F F F F
9. 9. Truth Tables - OR P Q PwQ T T T T F T F T T F F F
10. 10. Truth Tables - EQUIVALENT P Q P]Q T T T T F F F T F F F T
11. 11. Truth Tables - IMPLICATION P Q P6Q T T T T F F F T T F F T
12. 12. Truth Tables - Example P true means rain P false means no rain Q true means clouds Q false means no clouds
13. 13. Truth Tables - Example P Q P6Q P6Q rain clouds rainclouds T rain no clouds rainno clouds F no rain clouds no rainclouds T no rain no clouds no rainno clouds T
14. 14. Algebraic rules for statement forms • Associative rules: (p v q) v r ] p v (q v r) (p w q) w r ] p w (q w r) • Distributive rules: p v (q w r) ] (p v q) w (p v r) p w (q v r) ] (p w q) v (p w r) • Idempotent rules: p v p ] p p w p ] p
15. 15. Rules (continued) • Double Negation: 55p ] p • DeMorgan’s Rules: 5(p v q) ] 5p w 5q 5(p w q) ] 5p v 5q • Commutative Rules: p v q ] q v p p w q ] q w p
16. 16. Rules (continued) • Absorption Rules: p w (p v q) ] p p v (p w q) ] p • Bound Rules: p v 0 ] 0 p v 1 ] p p w 0 ] p p w 1 ] 1 • Negation Rules: p v 5p ] 0 p w 5p ] 1
17. 17. A Simple Tautology P Q is the same as 5Q 5P This is the same as asking: PQ ] 5Q 5P How can we prove this true? By creating a truth table! P Q T T T F F T F F
18. 18. A Simple Tautology (continued) Add appropriate columns P Q 5P 5Q T T F F T F F T F T T F F F T T
19. 19. A Simple Tautology (continued) Remember your implication table! P Q 5P 5Q PQ T T F F T T F F T F F T T F T F F T T T
20. 20. A Simple Tautology (continued) Remember your implication table! P Q 5P 5Q PQ 5Q5P T T F F T T T F F T F F F T T F T T F F T T T T
21. 21. A Simple Tautology (continued) Remember your implication table! P Q 5P 5Q PQ 5Q5P PQ ] 5Q5P T T F F T T T T F F T F F T F T T F T T T F F T T T T T Since the last column is all true, then the original statement: PQ ] 5Q5P is a tautology Note: 5Q5P is the contrapositive of PQ
22. 22. Translation of English If P then Q: PQ P only if Q: 5Q5P or PQ P if and only if Q: P ] Q also written as P iff Q
23. 23. Translation of English P is sufficient for Q: PQ P is necessary for Q: 5P5Q or QP P is necessary and sufficient for Q: P ] Q P unless Q: 5QP or 5PQ
24. 24. Predicate Logic • Consider the statement: x2 > 1 • Is it true or false? • Depends upon the value of x! • What values can x take on (what is the domain of x)? • Express this as a function: S(x) = x2 > 1 • Suppose the domain is the set of reals. • The codomain (range) of S is a set of statements that are either true or false.
25. 25. Example • S(0.9) = 0.92 > 1 is a false statement! • S(3.2) = 3.22 > 1 is a true statement! • The function S is an example of a predicate. • A predicate is any function whose codomain is a set of statements that are either true or false.
26. 26. Note • The codomain is a set of statements • The codomain is not the truth value of the statements • The domain of predicate is arbitrary • Different predicates can have different domains • The truth set of a predicate S with domain D is the set of the x ε D for which S(x) is true: {x ε D | S(x) is true} • Or simply: {x | S(x)}
27. 27. Quantifiers • The phrase “for all” is called a universal quantifier and is symbolically written as œ • The phrase “for some” is called an existential quantifier and is written as › Notations for set of numbers: Reals Integers Rationals Primes Naturals (nonnegative integers)
28. 28. Goldbach’s conjecture • Every even number greater than or equal to 4 can be written as the sum of two primes • Express it as a quantified predicate • It is unknown whether or not Goldbach’s conjecture is true. You are only asked to make the assertion • Another example: Every sufficiently large odd number is the sum of three primes.
29. 29. Negating Quantifiers • Let D be a set and let P(x) be a predicate that is defined for x ε D, then 5(œ(x ε D), P(x)) ] (›(x ε D), 5P(x)) and 5(›(x ε D), P(x)) ] (œ(x ε D), 5P(x)) | 1,868 | 5,575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-18 | latest | en | 0.724746 |
https://www.isnt.org.in/unity-how-to-add-force-with-code-examples.html | 1,726,290,791,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651548.18/warc/CC-MAIN-20240914025441-20240914055441-00569.warc.gz | 773,749,351 | 50,842 | # Unity How To Add Force With Code Examples
In this article, we will look at how to get the solution for the problem, Unity How To Add Force With Code Examples
## How do you use force tables?
Move the strings until the ring is free of the pin and nearly centred. Gently tap on the force table to temporarily eliminate friction, which will allow the ring to move more freely to its new position. Re-adjust the positions of the strings and repeat the tapping until the ring is well-centred and remains there.
``````public Rigidbody rb;
if(Input.GetKey(KeyCode.W))
{
} ```
```
``````private Rigidbody2D rb;
public Vector2 direction;
public float force;
void Start()
{
rb = GetComponent<Rigidbody2D>();
//rb.AddForce(--pass in Vector--, --can pass in ForceMode--);
}```
```
## How do you add forces in 2d?
How do I do rb. AddForce but in 2D?
• using System. Collections;
• using System. Collections. Generic;
• using UnityEngine;
• public class Movement : MonoBehaviour.
• {
• public Rigidbody2D rb;
• public Collider2D collider;
• public float moveSpeed = 0.1f;
## Is kinematic in Unity?
1. Unity will not apply any physics to the kinematic Rigidbody. 2. If rigidbody is kinematic, you can handle the behavior of rigidbody yourself using a script and unity will not apply any physics to that object.
## Can Unity be forced?
Unity. A small word huge in the promise of peace and imbued with the power to evoke visions of a more perfect union.
## Which method is used to adds a force to the Rigidbody *?
Adding force using AddForce() The first parameter for the method AddForce() asks for simply a Vector2 to know in which direction you want to apply the force in.
## How do you add force?
When adding forces, we must add them like lines, taking LENGTH and ANGLE into account. Adding forces is the same as combining them. When several forces are combined (added) into a single force, this force is called the RESULTANT of those forces.
## How do you add force to a Unity 2d?
• dir = target. transform. position - transform. position;
• dir = dir. normalized;
## What does Rigidbody Addforce do?
Description. Adds a force to the Rigidbody. Force is applied continuously along the direction of the force vector. Specifying the ForceMode mode allows the type of force to be changed to an Acceleration, Impulse or Velocity Change.
## How do you stop a force in Unity?
Remove Force from Moving Object
• function FixedUpdate ()
• if(Input. GetKeyDown(KeyCode. LeftShift))
• rigidbody. velocity = Vector3. zero;
• if(Input. GetKey("up"))
• if(Input. GetKey("down"))
• if(Input. GetKey("right"))
## Java For Map With Code Examples
In this article, we will look at how to get the solution for the problem, Java For Map With Code Examples How do you code a map in Java? Java Map Example: Generic (New Style) import java.util.*; class MapExample2{ public static void main(String args[]){ Map<Integer,String> map=new HashMap<Integer,String>(); map.put(100,"Amit"); map.put(101,"Vijay"); map.put(102,"Rahul"); //Elements can traverse in any order. Map<String, String> map = new HashMap<>(); for(Entry<String, String> entry:map.entryS
## Brew Use Java 11 With Code Examples
In this article, we will look at how to get the solution for the problem, Brew Use Java 11 With Code Examples Is Java 11 still used? It's important to note that even in 2022, many applications continue to run on Java 8 and Java 11. While Java 17 offers LTS, it's certainly not unusual if your application is using 11 or even 8. In 2022, the majority of applications are using Java 8+. brew install java sudo ln -sfn /usr/local/opt/openjdk@11/libexec/openjdk.jdk /Library/Java/JavaVirtual
## Switch Into The Postgres User Windows 10 With Code Examples
In this article, we will look at how to get the solution for the problem, Switch Into The Postgres User Windows 10 With Code Examples How do I access PostgreSQL in Windows? Then, go to the System variables section and double click on the Path option located over there. This will let you add the Path of your PostgreSQL server's bin directory to the PATH environment variable so that the PostgreSQL environment can be accessed easily from the Windows 10 command prompt. psql -U postgres -c "CRE
## Single Line Comment In Java With Code Examples
In this article, we will look at how to get the solution for the problem, Single Line Comment In Java With Code Examples What are the 3 types of comments in Java? In Java there are three types of comments: Single-line comments. Multi-line comments. Documentation comments. /* This is an example of multi-line comment. * The program prints "Hello, World!" to the standard output. */ class HelloWorld { public static void main(String[] args) { System.out.println("Hello, World!"); } } // Single
## Display Column Names As A Dictionary Pandas With Code Examples
In this article, we will look at how to get the solution for the problem, Display Column Names As A Dictionary Pandas With Code Examples How do I get a list of column names in a table? We can verify the data in the table using the SELECT query as below. We will be using sys. columns to get the column names in a table. It is a system table and used for maintaining column information. import pandas as pd df = pd.DataFrame({'date': ['2015-01-01', '2015-01-02', '2 | 1,266 | 5,355 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-38 | latest | en | 0.838738 |
http://www.jiskha.com/members/profile/posts.cgi?name=Dell | 1,461,946,145,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860111374.13/warc/CC-MAIN-20160428161511-00090-ip-10-239-7-51.ec2.internal.warc.gz | 613,521,872 | 8,780 | Friday
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Bennett put 4/7 of his paycheck in his checking account, kept 1/2 of the remainder in cash and put the rest in his savings account. What was bennett's salary?
August 24, 2012
math
Mary bought some peppers and apples at the farmer's market. 2/3 of the peppers and 3/4 of the apples were red. the number of red peppers and red apples were the same. if mary bought 27 peppers how many apples did dhe buy?
August 22, 2012
math
Bennett put 4/7 of his paycheck in his checking account, kept 1/2 of the remainder in cash and put the rest in his savings account. What was bennett's salary?
August 22, 2012
math
Bari has 180 Mardi Gras beads. 2/3 of the beads are green. 40% of the remaning beads are gold and the rest are purple. What fraction of the beads are gold? how many purple beads does she have?
August 22, 2012
math
Meggie has 2,784 stickers. 1/4 of them have glitter, 3/8 of them are felt and the rest are plain. How many of the stickers are felt? what fraction of the stickers are plain?
August 22, 2012
math
Stephen has 90 fish. He sold 1/2 of them and gave 1/3 of them to Cody. What faction of the fish was left? How many fish did he give Cody?
August 22, 2012
Math
Ticket sales for a recent weekend concert totaled \$30000. If 300 more tickets were sold Saturday than in Firday and each ticket cost \$25 how many tickets were sold on Saturday ?
August 21, 2012
Math
Ryan saved \$55 more than Jim. Chris saved 6 times as much as Jim. Donna saved half as much as Chris. If the children saved \$3245 altogether how much did each child save?
August 21, 2012
math
in a certain week, a salesman sold a used truck and seven 4-wheelers. the truck was sold for \$12,000. If the truck cost 5 times as much as each 4-wheeler, what were his total sales for that week?
August 20, 2012
math
in the answer what does the "S" represent, i having a hard time understanding how you came up with the snswer
August 20, 2012
math
carrie, Sam and ester have a total of \$570. Carrie has \$80 more than sam and ester has 5 times as much as sam. How much does each of them have?
August 20, 2012
math
Terry, Dan and heidi have a total of \$850. Dan has \$120 more than Terry and Heidi has \$190 more than Dan. How much money does heidi have?
August 20, 2012
physics
To protect her new two-wheeler, Iroda Bike buys a length of chain. She finds that its linear density is 0.61 lb/ft. If she wants to keep its weight below 1.8 lb, what length of chain is she allowed?
May 28, 2012
NEED HELP...how do i draw a visual for this ques?
A son and his father take a daily walk up a hill. It is 1.5 km to the top of the hill. As the father gets older he can only walk half the speed of his son. So they start together but the father trails his son. The father reaches the halfway point when the son is at the top. ...
December 6, 2011
NEED HELP drawing a visual for this question
Marnie collects red, blue and purple buttons. At the beginning of the year she has 360 buttons in a box. The ratio of red buttons to blue buttons is 4:3 and there are twice as many red buttons as purple buttons. a.What is the ratio of blue buttons to purple buttons at the ...
December 6, 2011
Math...HELP i dont understand this problem!
Given 1/3 Y÷1/2 X=4 Discuss its meaning. Create a shape for X and for Y that will make this true. Draw and label the shapes, then illustrate how the quotient was obtained. Your drawing must be neat and accurate. Can individual pieces of a traditional tangram be used to...
December 5, 2011
math
Given 1/3 Y÷1/2 X=4 Discuss its meaning. Create a shape for X and for Y that will make this true. Draw and label the shapes, then illustrate how the quotient was obtained. Your drawing must be neat and accurate. Can individual pieces of a traditional tangram be used to ...
December 1, 2011
math
Given 1/3 Y÷1/2 X=4 Discuss its meaning. Create a shape for X and for Y that will make this true. Draw and label the shapes, then illustrate how the quotient was obtained. Your drawing must be neat and accurate. Can individual pieces of a traditional tangram be used to ...
December 1, 2011
logic
thanks
December 1, 2011
logic
A son and his father take a daily walk up a hill. It is 1.5 km to the top of the hill. As the father gets older he can only walk half the speed of his son. So they start together but the father trails his son. The father reaches the halfway point when the son is at the top. ...
December 1, 2011
math
Marnie collects red, blue and purple buttons. At the beginning of the year she has 360 buttons in a box. The ratio of red buttons to blue buttons is 4:3 and there are twice as many red buttons as purple buttons. a.What is the ratio of blue buttons to purple buttons at the ...
December 1, 2011
logic
Given 1/3 Y÷1/2 X=4 Discuss its meaning. Create a shape for X and for Y that will make this true. Draw and label the shapes, then illustrate how the quotient was obtained. Your drawing must be neat and accurate. Can individual pieces of a traditional tangram be used to...
December 1, 2011
math
Three men (Tom, peter, and jack)and three women (Eliza, Anne, and Karen) are spending a few months at a hillside. They are to stay in a row of nine cottages, each one living in his or her owm cottage. There are no others staying in the same row of hoses.In addition, A. Anne, ...
November 30, 2011
math
You teach 40 students. Blue is a favorite color of 22 studetns; 19 own cars; 13 like flowers; blue is the favorite color of 10 students who own cars; blue is the favorite color of 7 students who like flowers' 8 own cars and flowers; and blue is the favorite color of 5 ...
November 29, 2011
math
Fresh Grapes contain 80 percent water by weight and 20 percent juice, whereas dried grapes contain 15 percent water by weight with the rest juice. How many punds of dried grapes are needed to yield the same amount of juice as 34 pounds of fresh grapes?
November 28, 2011
math
Three men (Tom, peter, and jack)and three women (Eliza, Anne, and Karen) are spending a few months at a hillside. They are to stay in a row of nine cottages, each one living in his or her owm cottage. There are no others staying in the same row of hoses.In addition, A. Anne, ...
November 28, 2011
math
You teach 40 students. Blue is a favorite color of 22 studetns; 19 own cars; 13 like flowers; blue is the favorite color of 10 students who own cars; blue is the favorite color of 7 students who like flowers' 8 own cars and flowers; and blue is the favorite color of 5 ...
November 28, 2011
venn
how to draw a venn diagram for: C is a subset of A; B and C have no elements in common
November 14, 2011
venn
Thanks Dr. Russ. I had these items drawen just didnt know if it was correct or not.
November 14, 2011
venn
how to draw a venn diagram for: A is a subset of B; C is a subset of B; A and C have no elements in common
November 14, 2011
math
For each of the following, draw a Venn diagram. (a) A º B, C º B, A ¿ C = ∅ (b) A » C, B ¿ C = ∅.
November 14, 2011
Math
A famous detective was called in to solve a baffling murder mystery. He determined there was only one cause of death along with the following facts: a.Mr. Harold, the murdered man, was killed by a blow on the head with a brass candlestick. b.Either his wife, Mrs. Harold, or a ...
November 14, 2011
logic
A famous detective was called in to solve a baffling murder mystery. He determined there was only one cause of death along with the following facts: a.Mr. Harold, the murdered man, was killed by a blow on the head with a brass candlestick. b.Either his wife, Mrs. Harold, or a ...
November 11, 2011
Logic
Lalph Rauren was famous for discovering potential models as he walked the streets of New York. One day he spotted four girls who later became top models: Amy, Mary, Kay, and Beth. One was discovered walking through an apartment building, one in a coffee shop, one in a shopping...
November 11, 2011
math
Thanks Steve!
November 11, 2011
math
Four good friends, who love cats, have each named their cat after one of the other friends. Of the clues that follow, only one is true; the others are false. Which friend owns which cat? a.Candy’s cat is named either Dody or Alice. b.Betsy’s cat is named Alice. c....
November 11, 2011
math
November 10, 2011
math
A floor-finishing machine uses bars of wax. Once the bar is worn down to 1/2 its original size, it must be replaced with a new bar. Only full-size bars can be put in the machine. The pieces are melted and recycled to make full-size bars. If you buy 24 bars and recycle all the ...
November 10, 2011
Math
(A) such that: A (subset) B; C (subset) B; A and C have no elements in common (B) such that: C (subset) A; B and C have no elements in common
November 10, 2011
math
October 24, 2011
math
when I multipied both sides by 5 and divided both sides by 3 I didnt get a x=? for a answer
October 24, 2011
math
thanks
October 24, 2011
Finance
Nancy Bellow promised her son she would pay him \$600 a quarterly for four years.If Nancy can invest her money in a 6% ordinary annuity, how much would she need to invest today
July 25, 2011
spelling
I need a sentence with its, not it's.
September 29, 2010
statistics
a study found that the average time it took a person to find a new job was 130 days. If a sample of 36 job seekers was surveyed, find the 95% confidence interval of the mean. Assume the standard deviation of the sample is 12 days.
March 19, 2010
1. Pages:
2. 1 | 4,195 | 15,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2016-18 | longest | en | 0.958498 |
https://mathoverflow.net/questions/35618/matrices-self-adjoint-with-respect-to-some-inner-product | 1,604,002,133,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107905777.48/warc/CC-MAIN-20201029184716-20201029214716-00275.warc.gz | 410,183,495 | 28,643 | matrices self-adjoint with respect to some inner product
Is it possible to give a nice characterization of matrices $A \in R^{n \times n}$ which are self-adjoint with respect to some inner product?
These matrices include all symmetric matrices (of course) and some nonsymmetric ones: for example, the transition matrix of any (irreducible) reversible Markov chain will have this property.
Naturally, all such matrices must have real eigenvalues, though I do not expect that this is a sufficient condition (is it?).
About the only observation I have is that since any inner product can be represented as $\langle x,y \rangle = x^T M y$ for some positive definite matrix $M$, we are looking for matrices $A$ which satisfy $A^T M = M A$ or $M^{-1} A^T M = A$. In other words, we are looking for real matrices similar to their transpose with a positive definite similarity matrix.
• Dear angela, how do we know that it has real eigenvalues? – Srinivasa Granujan Apr 5 '17 at 17:01
In addition to having real eigenvalues, the matrix will have to be diagonalizable, i.e., there have to be enough eigenvectors to span $R^n$. Once these conditions are satisfied, you can take a basis consisting of $n$ eigenvectors and define an inner product by declaring these basis vectors to be orthonormal. That inner product will make your matrix self-adjoint, because there's an orthonormal basis with respect to which the matrix is diagonal with real entries on the diagonal.
• @J. Mandalgan: No, there are plenty of examples of a matrix diagonalizable over the reals which does not commute with its transpose, such as $\left[\begin{array}{cc}0 & 1 \\\\ 0 & 1\end{array}\right]$. – Tracy Hall Aug 15 '10 at 3:38
• Tracy Hall: normal matrices are not those which commute with their transpose, but those which commute with their adjoint relative to the inner product. If the inner product is the usual "dot" product, then this is indeed the transpose, but not necessarily otherwise. Your 2x2 matrix is not just normal, but indeed symmetric, relative to the inner product $\begin{bmatrix}\hphantom{-}1 & -1 \cr -1 & \hphantom{-}0\end{bmatrix}$. – José Figueroa-O'Farrill Aug 15 '10 at 12:59 | 554 | 2,184 | {"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} | 3.453125 | 3 | CC-MAIN-2020-45 | latest | en | 0.918688 |
https://jeelabs.org/wp-content/uploads/2011/01/09/easy-electrons-%E2%80%93-diodes/index.html | 1,695,920,035,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510427.16/warc/CC-MAIN-20230928162907-20230928192907-00773.warc.gz | 358,244,427 | 12,629 | # Computing stuff tied to the physical world
## Easy Electrons – Diodes
In Hardware on Jan 9, 2011 at 00:01
It’s time for some Easy Electrons again.
Semiconductors are at the heart of today’s electronics designs: diodes, transistors, and the huge advances made possible by Integrated Circuits (i.e. IC’s) is what makes all those electronic devices around us possible. And it all happened in the time scale of just a few decades…
I can’t possibly cover everything in this series, so I’ll cover the two most common ones for now: the diode today, and the “normal” (BJT) transistor in the next installment.
Very roughly speaking, a diode is a one-way conductor. This is also indicated by the schematic symbol used for diodes (image from Wikipedia):
The arrow-like symbol points in the direction of the current flow, if you stick with the convention that current flows from “+” to “—”. If you keep forgetting which side is called the Anode and which the Cathode, this trick might help: in the alphabet, we (eh, I mean currents) go from A (to B) to C.
Diodes are useful to protect a circuit against connecting a power source the wrong way around (we’ve all been there, right?):
With this diode, the circuit is protected. Hooked up the wrong way, the diode will block, so no current will flow through the circuit.
But diodes aren’t perfect. There’s a voltage drop when conducting current, usually around 0.7V. So with the above circuit, if the power supply is 5.0V, then the circuit will only get about 4.3V. With low-voltage components, and especially battery-powered devices, this sort of voltage loss is awkward – and often unacceptable.
Which is why you won’t see this reverse voltage protection very often in circuits operating at 5V, 3.3V, or less. There is another type of protection, however:
This one is a bit nasty. It doesn’t really prevent the circuit from getting a reversed voltage at all. Instead, it will act (almost) like a short circuit when the voltage is applied the wrong way around. The idea being that this will cause the power supply to shut down (or the battery to drain very quickly). The RBBB uses this type of protection to overcome the voltage drop problem. The (cheap) diode will protect the (much more valuable) ATmega, as well as all other components hooked up to it.
This sort of protection is tricky. If you were to connect a LiPo battery, for example, then the short circuit can cause HUGE currents to flow, since many LiPo’s are able to supply them. Think many Amps… and now something else may happen: even if the diode can handle the current, the rest of the power lines might well become overloaded. Especially thin copper traces on a PCB are likely to act like a fuse and simply… evaporate!
There are other ways to deal with the voltage drop and still end up with diode protection. One of them is to minimize the voltage drop – this is where Schottky diodes can be useful. They usually have only half the voltage drop of normal diodes, i.e. around 0.3V. That might just be low enough for your particular setup.
Another option is to build an “ideal diode”. This might sound like an impossible task, given the properties of diodes, but there is actually a way to do this using a MOSFET. I won’t go into MOSFETs here, but basically they can switch current while having almost no resistance and (Ohm’s law!) therefore also almost no voltage drop. Trouble is: MOSFETs don’t know which way the current is flowing, so you need considerable extra circuitry to tell them when to turn on and off, based on comparing voltages on both ends. And although it is not a simple or cheap solution, this datasheet of the LTC4413 chip shows that it is indeed possible to beat the diode characteristics with some clever engineering. Electronics is often like that: people have come up with the most amazing tricks to overcome certain drawbacks, for all sorts of electronic circuit tasks. That’s why it’s so much fun just exploring and discovering it all, IMO :)
The graph of what a diode does is very characteristic: in reverse mode it blocks, and in forward mode (i.e. above around 0.7V) it conducts, albeit not perfectly. For some good example graphs see this page on Wikipedia (just skip the math formulas and look at the pretty pictures).
Now, assuming the voltage drop is no problem, because you’ve got some extra volts from the power supply anyway, then diodes can be extremely useful. The bridge rectifier for example, can be used to get a properly polarized voltage out, regardless of how the power supply is hooked up. This is particularly useful with alternating current, as present on the AC mains lines and on the coils of a transformer (a lot more Easy Electrons articles will be needed to present all this stuff!).
Another interesting diode is the Zener diode. It’s like a regular diode, but one which can’t support a very high reverse voltage. With Zeners, this voltage is called the “breakdown voltage”, and it ranges from about 2..200V. The value is fixed for any particular model.
Zener diodes make very simple (low-current) regulated power supplies:
Note how we’re putting the Zener in reverse mode, and counting on it to break down. As it does, current will start to flow. Until enough current is flowing across the resistor (Ohm’s law!) to take up all the “remaining” voltage.
So with a resistor of 100 Ω, and a Zener of 5.0V, we could power it with say 6..9V. At 6V, the current would be (6 – 5) / 100 = 10 mA. At 9V, the current would stabilize at (9 – 5) / 100 = 40 mA.
The reason this can be used as a regulated power supply, is that we can connect our circuit in parallel with the Zener, and it would always get 5V. The only drawback is that we can’t draw more than 10 mA from it:
• at 6V, the resistor needs to “eat” 1V, so that the Zener ends up with 5V
• if the circuit draws 10 mA, then 0 mA will go through the Zener
• if the circuit draws 5 mA, then 5 mA will go through the Zener
• if the circuit draws 0 mA, then 10 mA will go through the Zener
• at 9V, the current will increase to 40 mA (to get 4V over the resistor)
• in all cases, the circuit will see a 5V input voltage
Cool, so now we have built ourselves a simple regulated power supply!
As I mentioned, this circuit is not very powerful. If we draw more than 10 mA, then the voltage drop over the resistor may increase, leaving less than 5V for our circuit.
There is another drawback with the above regulated supply: it’s grossly inefficient. The reason is that it will always draw 10 mA, whether our circuit needs it or not. And that’s at 6V – at 9V it will always draw 40 mA!
I’ll show you how a transistor can easily increase the current and improve the regulating efficiency in a future installment. Exploring these simple electronic circuits can be great fun, and most of the time you can reason your way through without even having to build them!
Next time: transistors – incredibly useful devices, with tons of ways to use ’em!
PS. Does anyone have tips on how to improve these diagrams? I really want to continue drawing them by hand, but the texts don’t come out very nice, no matter what I try!
1. Dutch trick for +/-: KNAP: Kathode Negatief Anode Positief.
• That works perfectly well in English too, especially with my spelling :o)
I used to remember it as Cathode – Cold Anode – Arrrrrggggh!
:-)
2. With regard to the parallel diode reverse protection, that’s usually combined with a suitable size fuse so the fuse pops before the diode and PCB tracks melt.
maybe use comix-balloons with a number in them and put the text in the article itself ?
consider using Fritzing; it has bread-board and schematics. but it is a lot more work than drawing by hand; have a look at http://fritzing.org/
yet another option : drag/drop the texts onto the schematic when the drawing is already a GIF/JPG. almost any tools can do that.
4. I know this is a posting about diodes, not MOSFETs, but I thought it worth noting that in many cases, a MOSFET can be used as reverse voltage protection with no additional circuitry.
In place of the series diode, use a P-channel MOSFET with the drain connected to the positive voltage, the source connected to the load, and the gate connected to ground. The MOSFET has a built-in body diode and will initially behave just like the series diode: current will flow, but the voltage will drop by the forward voltage of the diode. If the voltage at the source terminal of the MOSFET exceeds the Gate-Source threshhold voltage of the MOSFET, then the MOSFET will turn on, and current will now flow with almost no voltage drop. If you hook up the battery backwards, no current will flow, since the body diode is reverse biased and the gate voltage exceeds the source voltage. Used in this manner, the MOSFET does in fact know which way the current is flowing.
For example, assume a P-channel MOSFET with a diode forward voltage drop of 0.8 volts and a Gate-Source threshhold voltage of 1.5 volts. If you connect a 3.3 volt supply, the load will initially receive 2.5 volts. But since the 2.5 volts at the source terminal of the MOSFET exceeds the voltage at the gate (zero volts) by more than 1.5 volts, the MOSFET will turn on almost instantly, which effectively shorts out the body diode and now provides almost the entire 3.3 volts to the load. The only remaining voltage loss is the current through the load times the Drain-Source resistance when the MOSFET is turned on. For example, if the load draws 200 milliamps and the MOSFET has an RDSon of 0.1 ohms, then the voltage drop over the MOSFET will be a measly 0.02 volts. The key, therefore, is to select a MOSFET with a low threshhold voltage (e.g., 1.5 volts) and a low Drain-Source resistance when on. The technique should work as long as the supply voltage exceeds the sum of the Drain-Source forward voltage and the Gate-Source threshhold voltage; in other words, for supply voltages above roughly 2.5 volts. Below that, you need something fancier.
There is a brief write-up on this technique (with diagrams) here:
http://focus.ti.com/general/docs/lit/getliterature.tsp?baseLiteratureNumber=SLVA139&fileType=pdf
P.S. As you will see from the link above, you can also do this with an N-channel MOSFET in the ground return line. But that is often less convenient.
• Fantastic info – thank you! | 2,472 | 10,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.931144 |
https://www.doorsteptutor.com/Exams/NTSE/Stage-I-State-Level/MAT/Questions/Part-7.html | 1,604,137,174,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107917390.91/warc/CC-MAIN-20201031092246-20201031122246-00236.warc.gz | 686,856,662 | 7,518 | # NTSE Stage-1 (State-Level) Mental-Ability: Questions 42 - 46 of 925
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## Question 42
Edit
### Question
MCQ▾
In the following series of number, find out how many times 1,3 and 7 have appeared together, 7 being in the middle and 1 and 3 on either side of 7?
2973173771331738571377173906
### Choices
Choice (4)Response
a.
3
b.
more than 5
c.
5
d.
4
## Question 43
Edit
### Question
MCQ▾
In a row of boys, Rajan is 10th from the right and Suraj is 10th from the left. When Rajan and Suraj interchange their positions. Suraj will be 27th from the left. Which of the following will be Rajan’s position from the right?
### Choices
Choice (4)Response
a.
10th
b.
26th
c.
27th
d.
29th
## Question 44
Edit
### Question
MCQ▾
In the following number series, how many times an odd number is followed by two consecutive even numbers?
4232542532643572867945429 6132
### Choices
Choice (4)Response
a.
3
b.
2
c.
More than 4
d.
4
## Question 45
Edit
### Question
MCQ▾
Which figure is embedded in the pattern given in figure (X)?
Problem figure
### Choices
Choice (4)Response
a.
b.
c.
d.
## Question 46
Edit
### Question
MCQ▾
In the following questions, find out the correct alternative which bears the same relationship given along with it.
5: 100,4: 64: : 4: 80,3: ?
### Choices
Choice (4)Response
a.
54
b.
26
c.
48
d.
60 | 488 | 1,676 | {"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} | 3.1875 | 3 | CC-MAIN-2020-45 | latest | en | 0.836857 |
https://www.jiskha.com/display.cgi?id=1416188750 | 1,532,185,067,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592636.68/warc/CC-MAIN-20180721145209-20180721165209-00032.warc.gz | 891,287,100 | 3,600 | # physics
posted by lol
Suppose we wish to use a 4.0 m iron bar to lift a heavy object by using it as a lever. If we place the pivot point at a distance of 0.5 m from the end of the bar that is in contact with the load and we can exert a downward force of 640 N on the other end of the bar, find the maximum load that this person is able to lift (pry) using this arrangement (neglect the mass of the bar in this problem).
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A box is lifted with a pry bar, by slipping the bar under the box and lifting up on the bar. If the pry bar is 2m in length and it is grasped at the end, and the box is located at the opposite end, the fulcrum, with its center of gravity …
7. ### PHYSICS
A box is lifted with a pry bar, by slipping the bar under the box and lifting up on the bar. If the pry bar is 2m in length and it is grasped at the end, and the box is located at the opposite end, the fulcrum, with its center of gravity … | 580 | 2,205 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.894453 |
https://teaching.callerlab.org/mainstream/scoot-back-definition/scoot-back-modules/ | 1,721,001,853,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514654.12/warc/CC-MAIN-20240714220017-20240715010017-00456.warc.gz | 500,506,660 | 10,076 | # Scoot Back – Modules
Mainstream
Scoot Back
Links: Define Standard Analyze Teach Other Extend
Modules:
Equivalents:
Star Thru = Touch 1/4, Scoot Back, Boys Run
(Ext) Right and Left Thru = Spin the Top, Boys Run, Half Tag the Line, Scoot Back, Boys Run
Zeros:
From Normal Facing Couples:
— Pass the Ocean, Single Hinge, Scoot Back, Boys Run, Right and Left Thru
— Right and Left Thru, Slide Thru, Touch a Quarter, Scoot Back, Boys Run
— Slide Thru, Touch a Quarter, Scoot Back, Box Circulate, Walk and Dodge, Partner Trade
— Touch a Quarter, Scoot Back, Boys Run, Reverse Flutterwheel and Sweep a Quarter
From Normal Facing Lines:
— Pass Thru, Wheel and Deal, Centers Step to a Wave, Extend, Scoot Back, Boys Trade and Run, Bend the Line
From Normal Eight Chain Thru:
— Pass Thru, Outsides Cloverleaf, Centers Square Thru, Touch 1/4, Scoot Back, Boys Run, Pass Thru, Bend the Line, Star Thru (Flip-Flop)
Fractional Zeros:
One-Third Fractional Zero from Parallel Waves:
— Single Hinge, Scoot Back, Centers Trade
Half Fractional Zero from Normal Eight Chain Thru (when all same sequence it is a Technical Zero that Inverts and Rotates):
— Touch 1/4, Scoot Back, Boys Run, Pass Thru, Bend the Line, Star Thru
Half Fractional Zero from Normal Facing Lines (when all same sequence these are Technical Zeros):
— Pass the Ocean, Boys Trade, All Eight Circulate, Single Hinge, Scoot Back, Boys Run, Reverse Flutterwheel
— Pass the Ocean, Girls Trade, All Eight Circulate, Scoot Back, Boys Trade, Boys Run, Bend the Line
Half Fractional Zero from Normal Facing Lines (when all same sequence it is a Technical Zero that Inverts and Rotates):
— Slide Thru, Touch 1/4, Scoot Back, Centers Trade, Pass to the Center, Centers Turn Thru, Left Swing Thru, Girls Run, Bend the Line
Get-Ins to Partner Line (= Zero Line):
— Heads Lead Right, Right and Left Thru, Touch 1/4, Scoot Back, Boys Run, Reverse Flutterwheel
— (Ext) Head Ladies Chain and Half Sashay, Heads Square Thru, Swing Thru, Scoot Back, Boys Run
Get-Outs from Partner Line (= Zero Line):
— Touch 1/4, Each Four Scoot Back twice, Boys Run, Left Allemande
Get-Ins to Corner Box (= Zero Box):
— Heads Touch 1/4, Centers Scoot Back twice, those Boys Run
Get-Outs from Corner Box (= Zero Box):
— Touch 1/4, Scoot Back, Boys Fold, Ladies Turn Thru, Left Allemande around to your partner
— Touch 1/4, Scoot Back, Boys Run, Reverse Flutterwheel and Sweep a Quarter, Left Allemande
— Swing Thru, Girls Circulate, Boys Trade, Scoot Back, Boys Run, Promenade
— Dosado to a Wave, Scoot Back, Boys Left Hinge, Very Center Boys Trade, Boys Left Hinge, RLG
— Step to a Wave, Scoot Back, Extend, Right and Left Grand
Conversions from a Corner Box to a Partner Line:
— Step to an Ocean Wave, Scoot Back, Ladies Circulate, Men Run, Bend the Line
Conversion from Partner Line to Corner Box:
— Heads Pass the Ocean, Scoot Back, Girls Circulate, Single Hinge, Boys Run, Slide Thru | 804 | 2,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-30 | latest | en | 0.742097 |
https://dwbi1.wordpress.com/2011/02/26/denormalising-a-fact-table/ | 1,657,005,114,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104514861.81/warc/CC-MAIN-20220705053147-20220705083147-00148.warc.gz | 270,358,553 | 18,283 | # Data Warehousing and Data Science
## 26 February 2011
### Denormalising a Fact Table
Filed under: Data Warehousing — Vincent Rainardi @ 9:00 am
Tags:
To denormalise a fact table means we expand a dimensional key column, to include the attributes in the dimension. This practice usually happens in a small, simple data mart project, where one person is doing everything.
As usual, it is easier to “learn by example”:
The usual practice in dimensional modelling is to design a fact table like this:
Dim1Key, Dim2Key, Measure1, Measure2, Measure3 (1)
e.g. DateKey, EventKey, Revenue, Cost, Visitors (an exhibition organiser company)
Rather than doing that, should the fact table be like this?
Dim1Key, Dim2Attributes, Measure1, Measure2, Measure3 (2)
e.g. Date, EventName, EventType, EventLocation, Revenue, Cost, Visitors
The main consideration for (2) is to save the ETL effort. By not creating the dimension tables and have everything in the fact table, it is simpler and quicker to build.
There are several disadvantages to approach (2):
1. There might be other event attributes introduced in the future. And those new attributes will need to be put into the fact table as well.
2. We lost the date attributes such as year and month. They will need to be put into the fact table as well then, making the fact table very wide.
3. It is difficult to “browse the events”. We would need to do a select distinct on the fact table to provide a list of events. The same with other dimensions.
4. Some dimensions such as customer and product can have many attributes, making the fact table even wider.
5. If the fact table is big, say 1 billion rows, aggregating a measure by an attribute is slower than if we put the attributes in a small dimension table.
6. We would need to index the fact table on these attributes. The indices will be a lot bigger than if the attributes are in a dimension table. They will take more effort to maintain.
Because of these reasons better not to denormalise a fact table.
In practice, at high level, everybody agree with this principle. But not everybody agree with the details of this principle. I call this issue “one or two dimensions”.
Again it is easier to learn by example:
In the above example, should event location be a separate dimension?
Many people I talk to agree that location should be a separate dimension. It should have its own dimension key, and it would have its own attributes such as city, post code and nearest tube station.
Generally, if the attribute has many other attributes, it should be put into a separate dimension. But, there is a disadvantage: we lost the direct link between location and event. We can only link them via the fact table. Which is why some designer “link the 2 dimension via a back door”. Read here for details, case b.
Now a little bit more vague: should the event type be in a separate dimension?
This time the response I get is about half-half. Some people think that event type should be a separate dimension. But some people think it should stay in the event dimension. Opinions will always vary (naturally), but the most important thing for us is to understand the reasons why they think so.
The main reason for keeping event type in the event dimension is because it has no attributes. It only consist of 1 attribute, i.e. itself. It may have code and description, but from business point of view it is only 1 attribute.
Whereas the main reason for separating event type into its own dimension is because some people (especially in the exhibition industry) think that it has its own attributes, i.e. it consist of 3 levels. Level 1: show or exhibition. Level 2: if it’s a show: music or art, and if it’s an exhibition: commercial or art. Level 3: if it’s a commercial exhibition: wedding, home, or books. If it’s an art exhibition: painting, contemporary, or installation.
So the decision whether an attribute should be put into its own dimension or not depends whether it has its own attribute or not. Again, I must emphasise the main disadvantage of separating into its own dimension: we loose the direct link. We can only link via a fact table.
But that could be an advantage. It is possible, that we could link via 2 different fact table. And we end up with different results. If that is the nature of the data (there is no fixed relationship), then the attribute must be put into its own dimension. This is the second reason why we separate an attribute (such as event type) into its own dimension.
As usual I welcome any comments and discussion at vrainardi@gmail.com. Vincent 26/2/11. | 1,020 | 4,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-27 | latest | en | 0.908355 |
https://www.lckinhibitor.com/2022/08/09/16713/ | 1,702,027,279,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100739.50/warc/CC-MAIN-20231208081124-20231208111124-00088.warc.gz | 944,644,487 | 16,578 | # Thetimes are somewhat compact, using a clear advantage for the EIP LIP crippling whensystem as
Thetimes are somewhat compact, using a clear advantage for the EIP LIP crippling whensystem as the visitors increases up to just about total paralysis for heavy conLIP crippling the the calling visitors increases up to pretty much total paralysis for heavy conwhen the gestion. calling rate gets closer to the service price. gestion.13 ofService Level ratio (Log) Service Level ratio (Log) =3 =3 one hundred one hundred =10 =10 =20 =1011 -0.four -0.-0.2 -0.0.2 0.0.four 0.0.6 0.0.eight 0.1.2 1.Targeted traffic Compound 48/80 References intensity Website traffic intensitySeclidemstat Inhibitor Figure 2.two. Ratioof service level EIP vs. LIP as aafunction of website traffic intensity. Figure Ratio of service level EIP vs. LIP as function of website traffic intensity. Figure 2.Ratio of service level EIP vs. LIP as a function of traffic intensity.Figure three. Ratio of utilization EIP vs. LIP as a function of visitors intensity. Figure 3.three. Ratio of utilization EIP vs. LIP as a function of site visitors intensity. Figure Ratio of utilization EIP vs. LIP as a function of targeted traffic intensity.Mathematics 2021, 9,Mathematics 2021, 9, x FOR PEER Review Mathematics 2021, 9, x FOR PEER Review 14 of 18 14 of13 ofEffective Utilization ratio (Log) Successful Utilization ratio (Log) 100 one hundred =3 =3 =10 =10 =20 =101 1 -0.4 -0.-0.two -0.00.2 0.0.four 0.0.6 0.0.8 0.1 1.two 1 1.two Targeted traffic intensity Visitors intensityFigure 4. Ratio of efficient utilization EIP vs. LIP as a function of visitors intensity. Figure 4. Ratio of effective utilization EIP EIPLIP as a function of website traffic intensity. Figure 4. Ratio of powerful utilization vs. vs. LIP as a function of traffic intensity.Queue Length ratio Queue Length ratio 0.9 0.9 0.eight 0.8 0.7 0.7 0.6 0.6 0.5 0.five 0.4 0.four 0.3 0.3 0.2 0.two 0.1 0.1 0 0 -0.4 -0.2 -0.4 -0.=3 =3 =10 =10 =20 =00.two 0.0.4 0.0.6 0.0.8 0.1 1.two 1 1.2 Targeted traffic intensity Website traffic intensity15 ofFigure Mathematics 2021, 9, x FOR PEER Evaluation five. Ratio of queue length EIP vs. LIP as a function of visitors intensity.Figure five. Ratio of queue length EIP vs. LIP as a function of traffic intensity.Figure five. Ratio of queue length EIP vs. LIP as a function of visitors intensity.Flow Time ratio 0.96 0.94 0.92 0.9 0.88 0.86 0.84 0.82 0.8 -0.4 -0.2 0 0.2 0.four 0.6 0.8=3 =10 =1.Targeted traffic intensityFigure Figure six. Ratio of flow time EIP vs. vs. LIP as a function of website traffic intensity. 6. Ratio of flow time EIP LIP as a function of traffic intensity.6. Conclusions Within this operate, we address the harm done to the efficiency of ticket queues. We initially demonstrated how ignoring the creation of virtual clients proves to become a poor management policy. The lateness of information includes a important influence on most elements on the program efficiency from the customer point of view (service level, flow time) whilst getting slightly detrimental to the operation from the server point of view (utilization). WeMathematics 2021, 9,14 ofTo conclude this brief comparative section, let us state that, even though the server may possibly expertise in regards to the same level of utilization beneath both the EIP and LIP as well as the sojourn time of your consumers that determine to keep inside the queue could also stay comparable, the difference between the two data policies is mostly felt in the service level, with all the LIP crippling the system because the visitors increases up to pretty much total paralysis for heavy congestion. 6. Conclusions Within this function, we address the harm accomplished to the efficiency of ticket queues. We first demonstrated how ignoring the creation of virtual prospects proves to become a poor management policy. The lateness of information features a considerable impact on most as. | 1,074 | 3,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-50 | latest | en | 0.833252 |
http://nrich.maths.org/public/leg.php?code=-173&cl=3&cldcmpid=2325 | 1,496,116,059,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463613780.89/warc/CC-MAIN-20170530031818-20170530051818-00020.warc.gz | 323,271,075 | 5,056 | # Search by Topic
#### Resources tagged with Travelling salesperson problem similar to Travelling Salesman:
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### There are 3 results
Broad Topics > Decision Mathematics and Combinatorics > Travelling salesperson problem
### Travelling Salesman
##### Stage: 3 Challenge Level:
A Hamiltonian circuit is a continuous path in a graph that passes through each of the vertices exactly once and returns to the start. How many Hamiltonian circuits can you find in these graphs?
### Only Connect
##### Stage: 3 Challenge Level:
The graph represents a salesman’s area of activity with the shops that the salesman must visit each day. What route around the shops has the minimum total distance?
### The Olympic Torch Tour
##### Stage: 4 Challenge Level:
Imagine you had to plan the tour for the Olympic Torch. Is there an efficient way of choosing the shortest possible route? | 190 | 926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | longest | en | 0.889158 |
https://hail.is/docs/0.2/_modules/hail/experimental/loop.html | 1,716,047,501,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057440.7/warc/CC-MAIN-20240518152242-20240518182242-00118.warc.gz | 262,196,538 | 8,335 | # Source code for hail.experimental.loop
from typing import Callable
from hail import ir
from hail.expr.expressions import construct_expr, construct_variable, expr_any, to_expr, unify_all
from hail.expr.types import hail_type
from hail.typecheck import anytype, typecheck
from hail.utils.java import Env
[docs]@typecheck(f=anytype, typ=hail_type, args=expr_any) def loop(f: Callable, typ, *args): r"""Define and call a tail-recursive function with given arguments. Notes ----- The argument f must be a function where the first argument defines the recursive call, and the remaining arguments are the arguments to the recursive function, e.g. to define the recursive function .. math:: f(x, y) = \begin{cases} y & \textrm{if } x \equiv 0 \\ f(x - 1, y + x) & \textrm{otherwise} \end{cases} we would write: >>> f = lambda recur, x, y: hl.if_else(x == 0, y, recur(x - 1, y + x)) Full recursion is not supported, and any non-tail-recursive methods will throw an error when called. This means that the result of any recursive call within the function must also be the result of the entire function, without modification. Let's consider two different recursive definitions for the triangle function :math:f(x) = 0 + 1 + \dots + x: >>> def triangle1(x): ... if x == 1: ... return x ... return x + triangle1(x - 1) >>> def triangle2(x, total): ... if x == 0: ... return total ... return triangle2(x - 1, total + x) The first function definition, triangle1, will call itself and then add x. This is an example of a non-tail recursive function, since triangle1(9) needs to modify the result of the inner recursive call to triangle1(8) by adding 9 to the result. The second function is tail recursive: the result of triangle2(9, 0) is the same as the result of the inner recursive call, triangle2(8, 9). Example ------- To find the sum of all the numbers from n=1...10: >>> triangle_f = lambda f, x, total: hl.if_else(x == 0, total, f(x - 1, total + x)) >>> x = hl.experimental.loop(triangle_f, hl.tint32, 10, 0) >>> hl.eval(x) 55 Let's say we want to find the root of a polynomial equation: >>> def polynomial(x): ... return 5 * x**3 - 2 * x - 1 We'll use Newton's method<https://en.wikipedia.org/wiki/Newton%27s_method> to find it, so we'll also define the derivative: >>> def derivative(x): ... return 15 * x**2 - 2 and starting at :math:x_0 = 0, we'll compute the next step :math:x_{i+1} = x_i - \frac{f(x_i)}{f'(x_i)} until the difference between :math:x_{i} and :math:x_{i+1} falls below our convergence threshold: >>> threshold = 0.005 >>> def find_root(f, guess, error): ... converged = hl.is_defined(error) & (error < threshold) ... new_guess = guess - (polynomial(guess) / derivative(guess)) ... new_error = hl.abs(new_guess - guess) ... return hl.if_else(converged, guess, f(new_guess, new_error)) >>> x = hl.experimental.loop(find_root, hl.tfloat, 0.0, hl.missing(hl.tfloat)) >>> hl.eval(x) 0.8052291984599675 Warning ------- Using arguments of a type other than numeric types and booleans can cause memory issues if if you expect the recursive call to happen many times. Parameters ---------- f : function ( (marker, \*args) -> :class:.Expression Function of one callable marker, denoting where the recursive call (or calls) is located, and many args, the loop variables. typ : :class:str or :class:.HailType Type the loop returns. args : variable-length args of :class:.Expression Expressions to initialize the loop values. Returns ------- :class:.Expression Result of the loop with args as initial loop values. """ loop_name = Env.get_uid() def contains_recursive_call(non_recursive): if isinstance(non_recursive, ir.Recur) and non_recursive.name == loop_name: return True return any([contains_recursive_call(c) for c in non_recursive.children]) def check_tail_recursive(loop_ir): if isinstance(loop_ir, ir.If): if contains_recursive_call(loop_ir.cond): raise TypeError("branch condition can't contain recursive call!") check_tail_recursive(loop_ir.cnsq) check_tail_recursive(loop_ir.altr) elif isinstance(loop_ir, ir.Let): if contains_recursive_call(loop_ir.value): raise TypeError("bound value used in other expression can't contain recursive call!") check_tail_recursive(loop_ir.body) elif isinstance(loop_ir, ir.TailLoop): if any(contains_recursive_call(x) for n, x in loop_ir.params): raise TypeError("parameters passed to inner loop can't contain recursive call!") elif not isinstance(loop_ir, ir.Recur) and contains_recursive_call(loop_ir): raise TypeError("found recursive expression outside of tail position!") @typecheck(recur_exprs=expr_any) def make_loop(*recur_exprs): if len(recur_exprs) != len(args): raise TypeError('Recursive call in loop has wrong number of arguments') err = None for i, (rexpr, expr) in enumerate(zip(recur_exprs, args)): if rexpr.dtype != expr.dtype: if err is None: err = 'Type error in recursive call,' err += f'\n at argument index {i}, loop arg type: {expr.dtype}, ' err += f'recur arg type: {rexpr.dtype}' if err is not None: raise TypeError(err) irs = [expr._ir for expr in recur_exprs] indices, aggregations = unify_all(*recur_exprs) return construct_expr(ir.Recur(loop_name, irs, typ), typ, indices, aggregations) uid_irs = [] loop_vars = [] for expr in args: uid = Env.get_uid() loop_vars.append(construct_variable(uid, expr._type, expr._indices, expr._aggregations)) uid_irs.append((uid, expr._ir)) loop_f = to_expr(f(make_loop, *loop_vars)) if loop_f.dtype != typ: raise TypeError(f"requested type {typ} does not match inferred type {loop_f.dtype}") check_tail_recursive(loop_f._ir) indices, aggregations = unify_all(*args, loop_f) return construct_expr(ir.TailLoop(loop_name, loop_f._ir, uid_irs), loop_f.dtype, indices, aggregations) | 1,459 | 5,690 | {"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} | 2.828125 | 3 | CC-MAIN-2024-22 | latest | en | 0.669795 |
http://www.numbersaplenty.com/9999047 | 1,591,288,893,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347445880.79/warc/CC-MAIN-20200604161214-20200604191214-00232.warc.gz | 191,560,567 | 3,362 | Search a number
9999047 is a prime number
BaseRepresentation
bin100110001001…
…001011000111
3200211000010002
4212021023013
510024432142
6554151515
7150663512
oct46111307
920730102
109999047
11570a483
12342259b
1320c12c6
141483d79
15d27a32
hex9892c7
9999047 has 2 divisors, whose sum is σ = 9999048. Its totient is φ = 9999046.
The previous prime is 9998977. The next prime is 9999049. The reversal of 9999047 is 7409999.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 9999047 - 212 = 9994951 is a prime.
Together with 9999049, it forms a pair of twin primes.
It is a Chen prime.
It is a junction number, because it is equal to n+sod(n) for n = 9998992 and 9999010.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (9999049) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (11) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 4999523 + 4999524.
It is an arithmetic number, because the mean of its divisors is an integer number (4999524).
Almost surely, 29999047 is an apocalyptic number.
9999047 is a deficient number, since it is larger than the sum of its proper divisors (1).
9999047 is an equidigital number, since it uses as much as digits as its factorization.
9999047 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 183708, while the sum is 47.
The square root of 9999047 is about 3162.1269740477. The cubic root of 9999047 is about 215.4366248649.
The spelling of 9999047 in words is "nine million, nine hundred ninety-nine thousand, forty-seven". | 519 | 1,725 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.892098 |
https://www.physicsforums.com/threads/magnetism-cross-product.759348/ | 1,532,182,230,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592579.77/warc/CC-MAIN-20180721125703-20180721145703-00611.warc.gz | 965,738,364 | 16,474 | # Homework Help: Magnetism Cross Product
1. Jun 24, 2014
### 4Phreal
1. Here is the prompt:
http://imgur.com/mfbPidG
2. F = qv x B
3. At first this seemed like a simple cross product problem, and it probably still is, but I'm really confused as to what "3.70E6 m/s/ in the (i+j+k)/sqrt(3) direction" means, so I don't know how to set up my problem anymore. Could someone instruct me?
Last edited by a moderator: Jun 24, 2014
2. Jun 24, 2014
### Staff: Mentor
You are given v as a vector and B as a vector, both in rectangular coordinates. Just write the equation for the cross product, and solve for the components of F...
3. Jun 24, 2014
### 4Phreal
Right, um, but I don't know what the phrase in the problem that states the velocity vectors actually means (because I'm dumb), so I don't know what those are. Could you please explain what that phrase means?
Here is my attempt: http://imgur.com/5C7Z648
Last edited: Jun 24, 2014
4. Jun 24, 2014
### Staff: Mentor
v is the velocity magnitude multiplied by that unit vector u that you are given.
Are you familiar with how to use a Determinant to do the vector cross product in rectangular coordinates?
http://en.wikipedia.org/wiki/Cross_product
.
5. Jun 24, 2014
### 4Phreal
I'm familiar, I just don't know how to turn (i+j+k)/sqrt(3) into i, j, and k individually to perform the cross product. Is my attempt in the previous post how I would do it?
6. Jun 24, 2014
### Staff: Mentor
That's close, but has a couple errors in it. First, the velocities are 3.7E6/√3, not 3.7E6/3√3. Where did that extra 3 in the denominator come from? You just distribute the 3.7E6 across each of the unit vectors to get the individual components. Does that make sense?
And in your Determinant calculation, you correctly show the two terms subtracting first, and then in the next line you show them adding...
7. Jun 24, 2014
### 4Phreal
I got the second 3 because I thought that (^i + ^j + ^k) divided by sqrt(3) equalled the velocity, so to get each individual part I divided by 3. But what you're saying is that they each individually equal 3.70E6 divided by sqrt(3), so there's no need to divide by that 3?
In my determinant calculation, the Bz is equal to -8.29, changing the subtraction sign into an addition one.
8. Jun 24, 2014
### Staff: Mentor
Ah, I see now what you did with the sign.
Yes, to get the velocity, you just distribute the amplitude through with multiplication. No other operation is needed. Just like 5(i + j + k) = 5i + 5j + 5k.
9. Jun 24, 2014
### 4Phreal
Thanks! For some reason, the answer was positive instead of negative, but maybe that was just in the way the question was phrased. | 757 | 2,677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2018-30 | latest | en | 0.930746 |
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http://math.stackexchange.com/questions/73292/how-to-express-conpa-as-a-first-order-statement?answertab=oldest | 1,462,439,255,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860126377.4/warc/CC-MAIN-20160428161526-00036-ip-10-239-7-51.ec2.internal.warc.gz | 183,538,239 | 17,820 | # How to express Con(PA) as a first-order statement?
I read from somewhere that
Fact 1. PA, which refers to the first-order version, is not finitely axiomatizable.
At the same time, the second incompleteness theorem says that there is no proof in PA for Con(PA). This theorem takes for granted that Con(PA) can be expressed as a statement in first-order logic. (Here the language consists of the constant symbol 0, the unary successor function S and the binary functions + and $\times$.) The question is
Question 2. How to express Con(PA) as a first-order sentence?
You have to take care that Fact 1 does not become false as a result of your answer to Question 2. Actually I believe that
Conjecture 3. If you have a way to express Con(PA) as a finite statement, I can turn this into a finite representation of PA, thus violating Fact 1.
Proving or arguing against Conjecture 3 is highly appreciated, besides answering Question 2.
-
## 1 Answer
Well, while $PA$ isn't finitely axiomatizable, it is recursively axiomatizable. That is, given an arbitrary formula, we can tell if it's an axiom of $PA$. So there exists a formula $Ax(x)$ which is true (in the standard model) if and only if $x$ is the Goedel number of an axiom of $PA$. It's enough to express $Th(x)$ (true if $x$ is the Goedel number of a $PA$ theorem) and so $Con(PA)$.
-
You see, your Ax() has to work in non-standard models. Yes, "given an arbitrary formula, we can tell if it's an axiom of PA." This assumes that the "arbitrary formula" is standard. But if Ax() is presented with a non-standard element of the model, how does it detect this? Won't this ability imply that we can define what's standard and what's non-standard? – Zirui Wang Oct 17 '11 at 12:05
Of course in a nonstandard model Ax() will be true of some nonstandard numbers. But that's just (one of) reason(s) why Con(PA) may fail in these models and isn't provable in PA! When we say, "Ax(x) expresses 'x is the Goedel number of a PA axiom" or "Con(PA) expresses 'PA is consistent'", it is a statement about standard model, and their behavior in nonstandard models is irrelevant. – Alexey Romanov Oct 17 '11 at 12:47
Wait. How do you define a formula? That is the $P()$ in the induction axiom $[P(0)\land\forall x(P(x)\to P(x+1))]\to\forall xP(x)$. – Zirui Wang Oct 17 '11 at 13:08
In the normal way. Of course nonstandard numbers are not ever really Goedel numbers of axioms of PA: that's the point. – Alexey Romanov Oct 17 '11 at 13:41
What is "the normal way"? I find definitions are the most important thing in studying logic, because logicians define things differently in their inexpressive language. Look at Godel's definition of a proof; it's not required to be finite. Yes, it works in the standard model. But the existence of non-standard models (incompleteness, etc) illustrates the badness of their definitions and nothing else. – Zirui Wang Oct 18 '11 at 9:03 | 777 | 2,916 | {"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} | 3.359375 | 3 | CC-MAIN-2016-18 | latest | en | 0.934955 |
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# Marnetmar
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1. ## Marnetmar
Why do math teachers suck at teaching math? In my book it's not an issue of not paying attention and then complaining/blaming everything on the teacher. Ever since the start of high school, I've had a consistent habit of failing at math until someone explained a concept to me in understandable terms, and then acing everything for the rest of the unit.
For example, I was taught that the definition of a linear equation is as follows:
1) F(a*X)=a*F(X) for every real number a
2) F(X+Y) = F(X)+F(Y) for every X and Y which can be taken as an argument of the function or the operator. These two conditions can be sinthesized in the following single one: CONDITION FOR LINEARITY: F(a*X+b*Y) = a*F(X) + b*F(Y)" etc etc etc
I couldn't understand a thing.
Then a trustworthy friend of mine explained that a linear equation simply doesn't contain exponents and doesn't curve on a graph. Aced the rest of the unit.
1. Show previous comments 2 more
2. Marnetmar said:
For example, I was taught that the definition of a linear equation is as follows:
1) F(a*X)=a*F(X) for every real number a
2) F(X+Y) = F(X)+F(Y) for every X and Y which can be taken as an argument of the function or the operator. These two conditions can be sinthesized in the following single one: CONDITION FOR LINEARITY: F(a*X+b*Y) = a*F(X) + b*F(Y)" etc etc etc
I suppose that's meant to generalize in case F applies on any kind of vectors X, Y (like in vector spaces), not just numbers. That kind of generalization ought to be taught in colleges rather than high schools, I think.
In high schools we were basically taught that linear equations = first degree. Whatever C30N9 said.
3. What grade are you by the way? Shouldn't these things be taught earlier?
4. In the same way people tend to misinterpret posts on this very forum, some people simply process information differently. So things that seem completely simplified to the teacher may be taught in little detail, while other things get over explained when you understood it right away.
my teacher in my accounting class seems to explain things the first time in such a way like he's talked about it before, with lots of abbreviations and undefined vocabulary which at times don't make any sense to me. If I follow what he's talking about in the text book it usually makes more sense to me to read it than to listen to him talk about it. The first week of class I've taken notes from his lectures that were completely backwards from what was in the book and more importantly, the test!
some teachers are really good at interpreting what you aren't getting and explaining it to you in a way you'll understand, and for others its just a job.
× | 687 | 2,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2022-27 | latest | en | 0.939593 |
https://scikit-learn.org/0.21/modules/decomposition.html | 1,721,218,823,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514771.72/warc/CC-MAIN-20240717120911-20240717150911-00821.warc.gz | 438,872,219 | 19,713 | # 2.5. Decomposing signals in components (matrix factorization problems)¶
## 2.5.1. Principal component analysis (PCA)¶
### 2.5.1.1. Exact PCA and probabilistic interpretation¶
PCA is used to decompose a multivariate dataset in a set of successive orthogonal components that explain a maximum amount of the variance. In scikit-learn, PCA is implemented as a transformer object that learns $$n$$ components in its fit method, and can be used on new data to project it on these components.
PCA centers but does not scale the input data for each feature before applying the SVD. The optional parameter parameter whiten=True makes it possible to project the data onto the singular space while scaling each component to unit variance. This is often useful if the models down-stream make strong assumptions on the isotropy of the signal: this is for example the case for Support Vector Machines with the RBF kernel and the K-Means clustering algorithm.
Below is an example of the iris dataset, which is comprised of 4 features, projected on the 2 dimensions that explain most variance:
The PCA object also provides a probabilistic interpretation of the PCA that can give a likelihood of data based on the amount of variance it explains. As such it implements a score method that can be used in cross-validation:
### 2.5.1.2. Incremental PCA¶
The PCA object is very useful, but has certain limitations for large datasets. The biggest limitation is that PCA only supports batch processing, which means all of the data to be processed must fit in main memory. The IncrementalPCA object uses a different form of processing and allows for partial computations which almost exactly match the results of PCA while processing the data in a minibatch fashion. IncrementalPCA makes it possible to implement out-of-core Principal Component Analysis either by:
• Using its partial_fit method on chunks of data fetched sequentially from the local hard drive or a network database.
• Calling its fit method on a memory mapped file using numpy.memmap.
IncrementalPCA only stores estimates of component and noise variances, in order update explained_variance_ratio_ incrementally. This is why memory usage depends on the number of samples per batch, rather than the number of samples to be processed in the dataset.
As in PCA, IncrementalPCA centers but does not scale the input data for each feature before applying the SVD.
Examples:
### 2.5.1.3. PCA using randomized SVD¶
It is often interesting to project data to a lower-dimensional space that preserves most of the variance, by dropping the singular vector of components associated with lower singular values.
For instance, if we work with 64x64 pixel gray-level pictures for face recognition, the dimensionality of the data is 4096 and it is slow to train an RBF support vector machine on such wide data. Furthermore we know that the intrinsic dimensionality of the data is much lower than 4096 since all pictures of human faces look somewhat alike. The samples lie on a manifold of much lower dimension (say around 200 for instance). The PCA algorithm can be used to linearly transform the data while both reducing the dimensionality and preserve most of the explained variance at the same time.
The class PCA used with the optional parameter svd_solver='randomized' is very useful in that case: since we are going to drop most of the singular vectors it is much more efficient to limit the computation to an approximated estimate of the singular vectors we will keep to actually perform the transform.
For instance, the following shows 16 sample portraits (centered around 0.0) from the Olivetti dataset. On the right hand side are the first 16 singular vectors reshaped as portraits. Since we only require the top 16 singular vectors of a dataset with size $$n_{samples} = 400$$ and $$n_{features} = 64 \times 64 = 4096$$, the computation time is less than 1s:
If we note $$n_{\max} = \max(n_{\mathrm{samples}}, n_{\mathrm{features}})$$ and $$n_{\min} = \min(n_{\mathrm{samples}}, n_{\mathrm{features}})$$, the time complexity of the randomized PCA is $$O(n_{\max}^2 \cdot n_{\mathrm{components}})$$ instead of $$O(n_{\max}^2 \cdot n_{\min})$$ for the exact method implemented in PCA.
The memory footprint of randomized PCA is also proportional to $$2 \cdot n_{\max} \cdot n_{\mathrm{components}}$$ instead of $$n_{\max} \cdot n_{\min}$$ for the exact method.
Note: the implementation of inverse_transform in PCA with svd_solver='randomized' is not the exact inverse transform of transform even when whiten=False (default).
References:
### 2.5.1.4. Kernel PCA¶
KernelPCA is an extension of PCA which achieves non-linear dimensionality reduction through the use of kernels (see Pairwise metrics, Affinities and Kernels). It has many applications including denoising, compression and structured prediction (kernel dependency estimation). KernelPCA supports both transform and inverse_transform.
Examples:
### 2.5.1.5. Sparse principal components analysis (SparsePCA and MiniBatchSparsePCA)¶
SparsePCA is a variant of PCA, with the goal of extracting the set of sparse components that best reconstruct the data.
Mini-batch sparse PCA (MiniBatchSparsePCA) is a variant of SparsePCA that is faster but less accurate. The increased speed is reached by iterating over small chunks of the set of features, for a given number of iterations.
Principal component analysis (PCA) has the disadvantage that the components extracted by this method have exclusively dense expressions, i.e. they have non-zero coefficients when expressed as linear combinations of the original variables. This can make interpretation difficult. In many cases, the real underlying components can be more naturally imagined as sparse vectors; for example in face recognition, components might naturally map to parts of faces.
Sparse principal components yields a more parsimonious, interpretable representation, clearly emphasizing which of the original features contribute to the differences between samples.
The following example illustrates 16 components extracted using sparse PCA from the Olivetti faces dataset. It can be seen how the regularization term induces many zeros. Furthermore, the natural structure of the data causes the non-zero coefficients to be vertically adjacent. The model does not enforce this mathematically: each component is a vector $$h \in \mathbf{R}^{4096}$$, and there is no notion of vertical adjacency except during the human-friendly visualization as 64x64 pixel images. The fact that the components shown below appear local is the effect of the inherent structure of the data, which makes such local patterns minimize reconstruction error. There exist sparsity-inducing norms that take into account adjacency and different kinds of structure; see [Jen09] for a review of such methods. For more details on how to use Sparse PCA, see the Examples section, below.
Note that there are many different formulations for the Sparse PCA problem. The one implemented here is based on [Mrl09] . The optimization problem solved is a PCA problem (dictionary learning) with an $$\ell_1$$ penalty on the components:
$\begin{split}(U^*, V^*) = \underset{U, V}{\operatorname{arg\,min\,}} & \frac{1}{2} ||X-UV||_2^2+\alpha||V||_1 \\ \text{subject to } & ||U_k||_2 = 1 \text{ for all } 0 \leq k < n_{components}\end{split}$
The sparsity-inducing $$\ell_1$$ norm also prevents learning components from noise when few training samples are available. The degree of penalization (and thus sparsity) can be adjusted through the hyperparameter alpha. Small values lead to a gently regularized factorization, while larger values shrink many coefficients to zero.
Note
While in the spirit of an online algorithm, the class MiniBatchSparsePCA does not implement partial_fit because the algorithm is online along the features direction, not the samples direction.
Examples:
References:
[Mrl09] “Online Dictionary Learning for Sparse Coding” J. Mairal, F. Bach, J. Ponce, G. Sapiro, 2009
[Jen09] “Structured Sparse Principal Component Analysis” R. Jenatton, G. Obozinski, F. Bach, 2009
## 2.5.2. Truncated singular value decomposition and latent semantic analysis¶
TruncatedSVD implements a variant of singular value decomposition (SVD) that only computes the $$k$$ largest singular values, where $$k$$ is a user-specified parameter.
When truncated SVD is applied to term-document matrices (as returned by CountVectorizer or TfidfVectorizer), this transformation is known as latent semantic analysis (LSA), because it transforms such matrices to a “semantic” space of low dimensionality. In particular, LSA is known to combat the effects of synonymy and polysemy (both of which roughly mean there are multiple meanings per word), which cause term-document matrices to be overly sparse and exhibit poor similarity under measures such as cosine similarity.
Note
LSA is also known as latent semantic indexing, LSI, though strictly that refers to its use in persistent indexes for information retrieval purposes.
Mathematically, truncated SVD applied to training samples $$X$$ produces a low-rank approximation $$X$$:
$X \approx X_k = U_k \Sigma_k V_k^\top$
After this operation, $$U_k \Sigma_k^\top$$ is the transformed training set with $$k$$ features (called n_components in the API).
To also transform a test set $$X$$, we multiply it with $$V_k$$:
$X' = X V_k$
Note
Most treatments of LSA in the natural language processing (NLP) and information retrieval (IR) literature swap the axes of the matrix $$X$$ so that it has shape n_features × n_samples. We present LSA in a different way that matches the scikit-learn API better, but the singular values found are the same.
TruncatedSVD is very similar to PCA, but differs in that it works on sample matrices $$X$$ directly instead of their covariance matrices. When the columnwise (per-feature) means of $$X$$ are subtracted from the feature values, truncated SVD on the resulting matrix is equivalent to PCA. In practical terms, this means that the TruncatedSVD transformer accepts scipy.sparse matrices without the need to densify them, as densifying may fill up memory even for medium-sized document collections.
While the TruncatedSVD transformer works with any (sparse) feature matrix, using it on tf–idf matrices is recommended over raw frequency counts in an LSA/document processing setting. In particular, sublinear scaling and inverse document frequency should be turned on (sublinear_tf=True, use_idf=True) to bring the feature values closer to a Gaussian distribution, compensating for LSA’s erroneous assumptions about textual data.
References:
## 2.5.3. Dictionary Learning¶
### 2.5.3.1. Sparse coding with a precomputed dictionary¶
The SparseCoder object is an estimator that can be used to transform signals into sparse linear combination of atoms from a fixed, precomputed dictionary such as a discrete wavelet basis. This object therefore does not implement a fit method. The transformation amounts to a sparse coding problem: finding a representation of the data as a linear combination of as few dictionary atoms as possible. All variations of dictionary learning implement the following transform methods, controllable via the transform_method initialization parameter:
Thresholding is very fast but it does not yield accurate reconstructions. They have been shown useful in literature for classification tasks. For image reconstruction tasks, orthogonal matching pursuit yields the most accurate, unbiased reconstruction.
The dictionary learning objects offer, via the split_code parameter, the possibility to separate the positive and negative values in the results of sparse coding. This is useful when dictionary learning is used for extracting features that will be used for supervised learning, because it allows the learning algorithm to assign different weights to negative loadings of a particular atom, from to the corresponding positive loading.
The split code for a single sample has length 2 * n_components and is constructed using the following rule: First, the regular code of length n_components is computed. Then, the first n_components entries of the split_code are filled with the positive part of the regular code vector. The second half of the split code is filled with the negative part of the code vector, only with a positive sign. Therefore, the split_code is non-negative.
### 2.5.3.2. Generic dictionary learning¶
Dictionary learning (DictionaryLearning) is a matrix factorization problem that amounts to finding a (usually overcomplete) dictionary that will perform well at sparsely encoding the fitted data.
Representing data as sparse combinations of atoms from an overcomplete dictionary is suggested to be the way the mammalian primary visual cortex works. Consequently, dictionary learning applied on image patches has been shown to give good results in image processing tasks such as image completion, inpainting and denoising, as well as for supervised recognition tasks.
Dictionary learning is an optimization problem solved by alternatively updating the sparse code, as a solution to multiple Lasso problems, considering the dictionary fixed, and then updating the dictionary to best fit the sparse code.
$\begin{split}(U^*, V^*) = \underset{U, V}{\operatorname{arg\,min\,}} & \frac{1}{2} ||X-UV||_2^2+\alpha||U||_1 \\ \text{subject to } & ||V_k||_2 = 1 \text{ for all } 0 \leq k < n_{\mathrm{atoms}}\end{split}$
After using such a procedure to fit the dictionary, the transform is simply a sparse coding step that shares the same implementation with all dictionary learning objects (see Sparse coding with a precomputed dictionary).
It is also possible to constrain the dictionary and/or code to be positive to match constraints that may be present in the data. Below are the faces with different positivity constraints applied. Red indicates negative values, blue indicates positive values, and white represents zeros.
The following image shows how a dictionary learned from 4x4 pixel image patches extracted from part of the image of a raccoon face looks like.
References:
### 2.5.3.3. Mini-batch dictionary learning¶
MiniBatchDictionaryLearning implements a faster, but less accurate version of the dictionary learning algorithm that is better suited for large datasets.
By default, MiniBatchDictionaryLearning divides the data into mini-batches and optimizes in an online manner by cycling over the mini-batches for the specified number of iterations. However, at the moment it does not implement a stopping condition.
The estimator also implements partial_fit, which updates the dictionary by iterating only once over a mini-batch. This can be used for online learning when the data is not readily available from the start, or for when the data does not fit into the memory.
Clustering for dictionary learning
Note that when using dictionary learning to extract a representation (e.g. for sparse coding) clustering can be a good proxy to learn the dictionary. For instance the MiniBatchKMeans estimator is computationally efficient and implements on-line learning with a partial_fit method.
## 2.5.4. Factor Analysis¶
In unsupervised learning we only have a dataset $$X = \{x_1, x_2, \dots, x_n \}$$. How can this dataset be described mathematically? A very simple continuous latent variable model for $$X$$ is
$x_i = W h_i + \mu + \epsilon$
The vector $$h_i$$ is called “latent” because it is unobserved. $$\epsilon$$ is considered a noise term distributed according to a Gaussian with mean 0 and covariance $$\Psi$$ (i.e. $$\epsilon \sim \mathcal{N}(0, \Psi)$$), $$\mu$$ is some arbitrary offset vector. Such a model is called “generative” as it describes how $$x_i$$ is generated from $$h_i$$. If we use all the $$x_i$$‘s as columns to form a matrix $$\mathbf{X}$$ and all the $$h_i$$‘s as columns of a matrix $$\mathbf{H}$$ then we can write (with suitably defined $$\mathbf{M}$$ and $$\mathbf{E}$$):
$\mathbf{X} = W \mathbf{H} + \mathbf{M} + \mathbf{E}$
In other words, we decomposed matrix $$\mathbf{X}$$.
If $$h_i$$ is given, the above equation automatically implies the following probabilistic interpretation:
$p(x_i|h_i) = \mathcal{N}(Wh_i + \mu, \Psi)$
For a complete probabilistic model we also need a prior distribution for the latent variable $$h$$. The most straightforward assumption (based on the nice properties of the Gaussian distribution) is $$h \sim \mathcal{N}(0, \mathbf{I})$$. This yields a Gaussian as the marginal distribution of $$x$$:
$p(x) = \mathcal{N}(\mu, WW^T + \Psi)$
Now, without any further assumptions the idea of having a latent variable $$h$$ would be superfluous – $$x$$ can be completely modelled with a mean and a covariance. We need to impose some more specific structure on one of these two parameters. A simple additional assumption regards the structure of the error covariance $$\Psi$$:
• $$\Psi = \sigma^2 \mathbf{I}$$: This assumption leads to the probabilistic model of PCA.
• $$\Psi = \mathrm{diag}(\psi_1, \psi_2, \dots, \psi_n)$$: This model is called FactorAnalysis, a classical statistical model. The matrix W is sometimes called the “factor loading matrix”.
Both models essentially estimate a Gaussian with a low-rank covariance matrix. Because both models are probabilistic they can be integrated in more complex models, e.g. Mixture of Factor Analysers. One gets very different models (e.g. FastICA) if non-Gaussian priors on the latent variables are assumed.
Factor analysis can produce similar components (the columns of its loading matrix) to PCA. However, one can not make any general statements about these components (e.g. whether they are orthogonal):
The main advantage for Factor Analysis over PCA is that it can model the variance in every direction of the input space independently (heteroscedastic noise):
This allows better model selection than probabilistic PCA in the presence of heteroscedastic noise:
## 2.5.5. Independent component analysis (ICA)¶
Independent component analysis separates a multivariate signal into additive subcomponents that are maximally independent. It is implemented in scikit-learn using the Fast ICA algorithm. Typically, ICA is not used for reducing dimensionality but for separating superimposed signals. Since the ICA model does not include a noise term, for the model to be correct, whitening must be applied. This can be done internally using the whiten argument or manually using one of the PCA variants.
It is classically used to separate mixed signals (a problem known as blind source separation), as in the example below:
ICA can also be used as yet another non linear decomposition that finds components with some sparsity:
## 2.5.6. Non-negative matrix factorization (NMF or NNMF)¶
### 2.5.6.1. NMF with the Frobenius norm¶
NMF [1] is an alternative approach to decomposition that assumes that the data and the components are non-negative. NMF can be plugged in instead of PCA or its variants, in the cases where the data matrix does not contain negative values. It finds a decomposition of samples $$X$$ into two matrices $$W$$ and $$H$$ of non-negative elements, by optimizing the distance $$d$$ between $$X$$ and the matrix product $$WH$$. The most widely used distance function is the squared Frobenius norm, which is an obvious extension of the Euclidean norm to matrices:
$d_{\mathrm{Fro}}(X, Y) = \frac{1}{2} ||X - Y||_{\mathrm{Fro}}^2 = \frac{1}{2} \sum_{i,j} (X_{ij} - {Y}_{ij})^2$
Unlike PCA, the representation of a vector is obtained in an additive fashion, by superimposing the components, without subtracting. Such additive models are efficient for representing images and text.
It has been observed in [Hoyer, 2004] [2] that, when carefully constrained, NMF can produce a parts-based representation of the dataset, resulting in interpretable models. The following example displays 16 sparse components found by NMF from the images in the Olivetti faces dataset, in comparison with the PCA eigenfaces.
The init attribute determines the initialization method applied, which has a great impact on the performance of the method. NMF implements the method Nonnegative Double Singular Value Decomposition. NNDSVD [4] is based on two SVD processes, one approximating the data matrix, the other approximating positive sections of the resulting partial SVD factors utilizing an algebraic property of unit rank matrices. The basic NNDSVD algorithm is better fit for sparse factorization. Its variants NNDSVDa (in which all zeros are set equal to the mean of all elements of the data), and NNDSVDar (in which the zeros are set to random perturbations less than the mean of the data divided by 100) are recommended in the dense case.
Note that the Multiplicative Update (‘mu’) solver cannot update zeros present in the initialization, so it leads to poorer results when used jointly with the basic NNDSVD algorithm which introduces a lot of zeros; in this case, NNDSVDa or NNDSVDar should be preferred.
NMF can also be initialized with correctly scaled random non-negative matrices by setting init="random". An integer seed or a RandomState can also be passed to random_state to control reproducibility.
In NMF, L1 and L2 priors can be added to the loss function in order to regularize the model. The L2 prior uses the Frobenius norm, while the L1 prior uses an elementwise L1 norm. As in ElasticNet, we control the combination of L1 and L2 with the l1_ratio ($$\rho$$) parameter, and the intensity of the regularization with the alpha ($$\alpha$$) parameter. Then the priors terms are:
$\alpha \rho ||W||_1 + \alpha \rho ||H||_1 + \frac{\alpha(1-\rho)}{2} ||W||_{\mathrm{Fro}} ^ 2 + \frac{\alpha(1-\rho)}{2} ||H||_{\mathrm{Fro}} ^ 2$
and the regularized objective function is:
$d_{\mathrm{Fro}}(X, WH) + \alpha \rho ||W||_1 + \alpha \rho ||H||_1 + \frac{\alpha(1-\rho)}{2} ||W||_{\mathrm{Fro}} ^ 2 + \frac{\alpha(1-\rho)}{2} ||H||_{\mathrm{Fro}} ^ 2$
NMF regularizes both W and H. The public function non_negative_factorization allows a finer control through the regularization attribute, and may regularize only W, only H, or both.
### 2.5.6.2. NMF with a beta-divergence¶
As described previously, the most widely used distance function is the squared Frobenius norm, which is an obvious extension of the Euclidean norm to matrices:
$d_{\mathrm{Fro}}(X, Y) = \frac{1}{2} ||X - Y||_{Fro}^2 = \frac{1}{2} \sum_{i,j} (X_{ij} - {Y}_{ij})^2$
Other distance functions can be used in NMF as, for example, the (generalized) Kullback-Leibler (KL) divergence, also referred as I-divergence:
$d_{KL}(X, Y) = \sum_{i,j} (X_{ij} \log(\frac{X_{ij}}{Y_{ij}}) - X_{ij} + Y_{ij})$
Or, the Itakura-Saito (IS) divergence:
$d_{IS}(X, Y) = \sum_{i,j} (\frac{X_{ij}}{Y_{ij}} - \log(\frac{X_{ij}}{Y_{ij}}) - 1)$
These three distances are special cases of the beta-divergence family, with $$\beta = 2, 1, 0$$ respectively [6]. The beta-divergence are defined by :
$d_{\beta}(X, Y) = \sum_{i,j} \frac{1}{\beta(\beta - 1)}(X_{ij}^\beta + (\beta-1)Y_{ij}^\beta - \beta X_{ij} Y_{ij}^{\beta - 1})$
Note that this definition is not valid if $$\beta \in (0; 1)$$, yet it can be continuously extended to the definitions of $$d_{KL}$$ and $$d_{IS}$$ respectively.
NMF implements two solvers, using Coordinate Descent (‘cd’) [5], and Multiplicative Update (‘mu’) [6]. The ‘mu’ solver can optimize every beta-divergence, including of course the Frobenius norm ($$\beta=2$$), the (generalized) Kullback-Leibler divergence ($$\beta=1$$) and the Itakura-Saito divergence ($$\beta=0$$). Note that for $$\beta \in (1; 2)$$, the ‘mu’ solver is significantly faster than for other values of $$\beta$$. Note also that with a negative (or 0, i.e. ‘itakura-saito’) $$\beta$$, the input matrix cannot contain zero values.
The ‘cd’ solver can only optimize the Frobenius norm. Due to the underlying non-convexity of NMF, the different solvers may converge to different minima, even when optimizing the same distance function.
NMF is best used with the fit_transform method, which returns the matrix W. The matrix H is stored into the fitted model in the components_ attribute; the method transform will decompose a new matrix X_new based on these stored components:
>>> import numpy as np
>>> X = np.array([[1, 1], [2, 1], [3, 1.2], [4, 1], [5, 0.8], [6, 1]])
>>> from sklearn.decomposition import NMF
>>> model = NMF(n_components=2, init='random', random_state=0)
>>> W = model.fit_transform(X)
>>> H = model.components_
>>> X_new = np.array([[1, 0], [1, 6.1], [1, 0], [1, 4], [3.2, 1], [0, 4]])
>>> W_new = model.transform(X_new)
References:
[1] “Learning the parts of objects by non-negative matrix factorization” D. Lee, S. Seung, 1999
[4] “SVD based initialization: A head start for nonnegative matrix factorization” C. Boutsidis, E. Gallopoulos, 2008
[5] “Fast local algorithms for large scale nonnegative matrix and tensor factorizations.” A. Cichocki, A. Phan, 2009
[6] (1, 2) “Algorithms for nonnegative matrix factorization with the beta-divergence” C. Fevotte, J. Idier, 2011
## 2.5.7. Latent Dirichlet Allocation (LDA)¶
Latent Dirichlet Allocation is a generative probabilistic model for collections of discrete dataset such as text corpora. It is also a topic model that is used for discovering abstract topics from a collection of documents.
The graphical model of LDA is a three-level generative model:
Note on notations presented in the graphical model above, which can be found in Hoffman et al. (2013):
• The corpus is a collection of $$D$$ documents.
• A document is a sequence of $$N$$ words.
• There are $$K$$ topics in the corpus.
• The boxes represent repeated sampling.
In the graphical model, each node is a random variable and has a role in the generative process. A shaded node indicates an observed variable and an unshaded node indicates a hidden (latent) variable. In this case, words in the corpus are the only data that we observe. The latent variables determine the random mixture of topics in the corpus and the distribution of words in the documents. The goal of LDA is to use the observed words to infer the hidden topic structure.
When modeling text corpora, the model assumes the following generative process for a corpus with $$D$$ documents and $$K$$ topics, with $$K$$ corresponding to n_components in the API:
1. For each topic $$k \in K$$, draw $$\beta_k \sim \mathrm{Dirichlet}(\eta)$$. This provides a distribution over the words, i.e. the probability of a word appearing in topic $$k$$. $$\eta$$ corresponds to topic_word_prior.
2. For each document $$d \in D$$, draw the topic proportions $$\theta_d \sim \mathrm{Dirichlet}(\alpha)$$. $$\alpha$$ corresponds to doc_topic_prior.
3. For each word $$i$$ in document $$d$$:
1. Draw the topic assignment $$z_{di} \sim \mathrm{Multinomial} (\theta_d)$$
2. Draw the observed word $$w_{ij} \sim \mathrm{Multinomial} (\beta_{z_{di}})$$
For parameter estimation, the posterior distribution is:
$p(z, \theta, \beta |w, \alpha, \eta) = \frac{p(z, \theta, \beta|\alpha, \eta)}{p(w|\alpha, \eta)}$
Since the posterior is intractable, variational Bayesian method uses a simpler distribution $$q(z,\theta,\beta | \lambda, \phi, \gamma)$$ to approximate it, and those variational parameters $$\lambda$$, $$\phi$$, $$\gamma$$ are optimized to maximize the Evidence Lower Bound (ELBO):
$\log\: P(w | \alpha, \eta) \geq L(w,\phi,\gamma,\lambda) \overset{\triangle}{=} E_{q}[\log\:p(w,z,\theta,\beta|\alpha,\eta)] - E_{q}[\log\:q(z, \theta, \beta)]$
Maximizing ELBO is equivalent to minimizing the Kullback-Leibler(KL) divergence between $$q(z,\theta,\beta)$$ and the true posterior $$p(z, \theta, \beta |w, \alpha, \eta)$$.
LatentDirichletAllocation implements the online variational Bayes algorithm and supports both online and batch update methods. While the batch method updates variational variables after each full pass through the data, the online method updates variational variables from mini-batch data points.
Note
Although the online method is guaranteed to converge to a local optimum point, the quality of the optimum point and the speed of convergence may depend on mini-batch size and attributes related to learning rate setting.
When LatentDirichletAllocation is applied on a “document-term” matrix, the matrix will be decomposed into a “topic-term” matrix and a “document-topic” matrix. While “topic-term” matrix is stored as components_ in the model, “document-topic” matrix can be calculated from transform method.
LatentDirichletAllocation also implements partial_fit method. This is used when data can be fetched sequentially.
References: | 6,849 | 28,677 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-30 | latest | en | 0.867146 |
https://blog.csdn.net/Meytones/article/details/104575234 | 1,603,857,227,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107896048.53/warc/CC-MAIN-20201028014458-20201028044458-00490.warc.gz | 234,391,489 | 22,171 | # Unity3D中第三人称视角的镜头跟随和目标锁定
## 实现过程
• 在初始化时设定摄像机和人物之间的相对位置和相对旋转角；
• 在需要锁定时找到人物最近的单位，使人物转向该单位，并将摄像机的水平面的旋转角设置至和人物相同；
• 如果一定范围内没有单位，则人物不需转向，摄像机仍然执行；
using System.Collections;
using System.Collections.Generic;
using UnityEngine;
public class CameraController : MonoBehaviour
{
public GameObject player;
Vector3 distanceToPlayer;
Vector3 positionTarget,angleTarget;
float temp;
// Start is called before the first frame update
void Start()
{
distanceToPlayer = this.transform.position - player.transform.position;
angleTarget = this.transform.eulerAngles;
}
// Update is called once per frame
void Update()
{
CameraFollow();
if (Input.GetKey(KeyCode.Space))
{
CameraAngle();
}
this.transform.eulerAngles = Vector3.Lerp(this.transform.eulerAngles, angleTarget, 0.3f);
this.transform.position = Vector3.Lerp(this.transform.position, positionTarget, 0.3f);
}
void CameraFollow()
{
positionTarget = player.transform.position + Quaternion.AngleAxis(player.transform.rotation.eulerAngles.y, new Vector3(0, 1, 0)) * (distanceToPlayer);
}
void CameraAngle()
{
//找到最近的单位
Collider[] collidersAround = Physics.OverlapSphere(player.transform.position, 400);//Overlap一系的函数在检测总数不是太大的时候会优先获取更近的碰撞器
if (collidersAround.Length == 0)
{
//搜索不到目标就指向正前方
angleTarget = new Vector3(this.transform.eulerAngles.x, player.transform.eulerAngles.y, this.transform.eulerAngles.z);
return;
}
int closestEnemy = 0;
float closestDistance = 401;
for (int i = 0; i < collidersAround.Length; i++)
{
temp = (this.transform.position - collidersAround[i].gameObject.transform.position).sqrMagnitude;
closestEnemy = temp < closestDistance ? i : closestEnemy;
closestDistance = temp < closestDistance ? temp : closestDistance;
}
player.transform.LookAt(collidersAround[closestEnemy].gameObject.transform.position);
positionTarget = player.transform.position + Quaternion.AngleAxis(player.transform.rotation.eulerAngles.y, new Vector3(0, 1, 0)) * (distanceToPlayer);
angleTarget = new Vector3(this.transform.eulerAngles.x, player.transform.eulerAngles.y, this.transform.eulerAngles.z);
}
}
void CameraAngle()
{
Collider[] collidersAround = Physics.OverlapSphere(player.transform.position, 400);
if (collidersAround.Length == 0)
{
angleTarget = new Vector3(this.transform.eulerAngles.x, player.transform.eulerAngles.y, this.transform.eulerAngles.z);
return;
}
int closestEnemy = 0;
float closestDistance = 401;
for (int i = 0; i < collidersAround.Length; i++)
{
temp = (this.transform.position - collidersAround[i].gameObject.transform.position).sqrMagnitude;
closestEnemy = temp < closestDistance ? i : closestEnemy;
closestDistance = temp < closestDistance ? temp : closestDistance;
}
player.transform.LookAt(collidersAround[closestEnemy].gameObject.transform.position);
positionTarget = player.transform.position + Quaternion.AngleAxis(player.transform.rotation.eulerAngles.y, new Vector3(0, 1, 0)) * (distanceToPlayer);
angleTarget = new Vector3(this.transform.eulerAngles.x, player.transform.eulerAngles.y, this.transform.eulerAngles.z);
//当越过0发生突变时（不是最短变化路径）加以矫正
if (Mathf.Abs(player.transform.eulerAngles.y - this.transform.eulerAngles.y) > 180)
{
angleTarget = new Vector3(this.transform.eulerAngles.x, player.transform.eulerAngles.y + (player.transform.eulerAngles.y > this.transform.eulerAngles.y ? -360 : 360), this.transform.eulerAngles.z);
}
}
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# mid1(07)sol - nctuee07f Stochastic Processes Midterm 1...
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nctuee07f Stochastic Processes Midterm 1 Solutions 1. True or False (6 points × 5 = 30 points) Please provide an explanation, a simple proof or a counter example, to your an- swer. No points will be credited if answers are with true or false only . Try to be concise and to the point. (a) True . This is just the singular value decomposition applied to a square matrix. (b) True . For Hermitian matrix A , we know A = A H . The eigenvector x associated with the eigenvalue λ satisfies Ax = λ x . It follows that λ · x H x = x H · λ x = x H Ax = x H A H x since A = A H = Ax H x = λ * · x H x , from which we know λ = λ * . Thus, the eigenvalue of a Hermitian matrix is always real. (c) False . We need to add that X and Y are jointly Gaussian in order to make such conclusion. (d) True . See HW1 solutions. (e) True . Consider independent discrete random variables X and Y . By independence we know p X,Y ( x, y ) = p X ( x ) p Y ( y ) . Then, E [ XY ] = X x X y xyp X,Y ( x, y ) = X x X y xyp X ( x ) p Y ( y ) = X x xp X ( x ) · X y yp Y ( y ) = E [ X ] E [ Y ] . Thus, X and Y are uncorrelated. 1
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2. Central Limit Theorem (10 points) Let Y = N X i =1 X i .
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Ask a homework question - tutors are online | 525 | 1,771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2018-13 | latest | en | 0.807766 |
https://www.lvguowei.me/post/sicp-goodness-data/ | 1,619,143,839,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039626288.96/warc/CC-MAIN-20210423011010-20210423041010-00411.warc.gz | 976,659,859 | 9,414 | # SICP Goodness - What is Meant by Data? (I)
Just a little bit of lambda calculus
Do you think Computer Science equals building websites and mobile apps?
Are you feeling that you are doing repetitive and not so intelligent work?
Are you feeling a bit sick about reading manuals and copy-pasting code and keep poking around until it works all day long?
Do you want to understand the soul of Computer Science?
Let’s begin by asking the question: What is data? You may say that data is numbers, characters, strings, pairs, lists, maps, sets and so on. Then, what is procedure?. Well a procedure is a function that can be called with(out) parameters.
What if I tell you that data can be procedures?
For me, the first really mind boggling example in the book is this:
(define (pair x y)
(define (dispatch m)
(cond ((= m 0) x)
((= m 1) y)
(else (error "Argument not 0 or 1"))))
dispatch)
(define (frst p) (p 0))
(define (snd p) (p 1))
This is a working pair data structure that is implemented only using functions. Function pair is used to make a pair, and frst selects its first item, and snd selects its second item.
;; 1
(fsrt (pair 1 2))
;; 3
(snd (pair 2 3))
In exercise 2.4, an alternative way of implementing pair is given as follows.
(define (pair x y)
(lambda (m) (m x y)))
(define (frst p)
(p (lambda (a b) a)))
(define (snd p)
(p (lambda (a b) b)))
If you have no problem understand these, then take a look at exercise 2.6:
In case representing pairs as procedures wasn’t mind-boggling enough, consider that, in a language that can manipulate procedures, we can get by without numbers (at least insofar as nonnegative integers are concerned) by implementing 0 and the operation of adding 1 as
(define zero (lambda (f) (lambda (x) x)))
(lambda (f) (lambda (x) (f ((n f) x)))))
This representation is known as “Church numerals”, after its inventor, Alonzo Church, the logician who invented the [lambda] calculus.
Define one and two directly (not in terms of zero and add-1). (Hint: Use substitution to evaluate (add-1 zero)). Give a direct definition of the addition procedure + (not in terms of repeated application of add-1).
To be honest, this really made me want to vomit. But hold it back, maybe we can learn a bit of this so called lambda calculus first, and then revisit this problem.
I found a great talk about this topic, and I highly recommend you watch the video several times until you understand most of the things in the talk, and remember to have pen and paper at hand when you do so.
PART I
PART II
If you understand this talk, exercise 2.6 will become trivial and obvious to you.
I now will summarize the talk here.
# Lambda calculus syntax
First let’s take a look at the syntax of lambda calculus. I think it’s easier to explain by examples.
;; a function that takes a and returns a
λa.a
;; equivalent form in scheme
(lambda (a) a)
λa.bx
;; equivalent form in scheme
(lambda (a) (b x))
(λa.b)x
;; equivalent form in scheme
((lambda (a) b) x)
λa.λb.x
;; equivalent form in scheme
(lambda (a) (lambda (b) x))
;; This can also be shortened to
λab.x
λa.bcx
;; equivalent form in scheme
(lambda (a) ((b c) x))
# Birds
Let’s now take a look at some important functions. They are all named after birds.
## Idiot Bird
The first one is idiot bird or I for short. As its name says, it is not that smart. So it just returns what it takes. Pretty straightforward.
Let’s implement it in Scheme.
(define I
(lambda (a) a))
## Mocking Bird
This one is pretty interesting. It takes a function, and calls it on itself.
(define M
(lambda (f) (f f)))
What is M(I)?
It’s (I I) which is just I.
And what is (M M)?
Uh-oh, it is (M M) which is (M M) which is (M M) … so it never ends.
## Kestrel
The kestrel takes a and b then returns a.
(define K
(lambda (a) (lambda (b) a)))
This is the built-in function in Haskel, called const. Why const? Let’s say if we call K with 5. This will return a new function, that no matter what you call it with, it will always return 5. Hence the name const.
Now the fun part.
K I x = I
Each side of the equation is a function, so we can do the following:
K I x y = I y
We add y to both sides. Since I y always returns y, we will then get
K I x y = y.
If we see K I as one function, then what K I does is that it takes x and y then returns y.
So K I is opposite to K.
This is actually not hard to understand.
As we already know, if you call K with only one parameter x, it returns a constant x function. So K I will return a function that will always return I. So K I a = I, then call it with b will get K I a b = I b = b.
So we just derived Kite.
## Kite
(define KI
(lambda (a) (lambda (b) b)))
This is the opposite of Kestrel.
Now let’s look at a new bird: Cardinal.
## Cardinal
Cardinal takes a function and two parameters, but calling them in different order.
(define C
(lambda (f)
(lambda (a)
(lambda (b) ((f b) a)))))
Let’s see an example. What is C K I M?
C takes in a function K and two parameters I and M and calles them in different order which is K M I = M.
Emm, so C K I M = M, if we see C K as a new function. Then C K is KI! You can type in the scheme code and prove this yourself.
OK, I think we have introduced enough birds, and this post is getting too long so I will stop here. In part II, we will continue and talk about what is Church Encoding. And then get back to our SICP exercise.
Stay tuned! | 1,441 | 5,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2021-17 | latest | en | 0.888711 |
http://sepwww.stanford.edu/public/docs/sep113/antoine3/paper_html/node3.html | 1,508,411,254,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823282.42/warc/CC-MAIN-20171019103222-20171019123009-00007.warc.gz | 307,895,910 | 3,452 | Next: Migration results Up: Theory Previous: Theory
## Approximating the Hessian
In equation (4), I define as the migrated image after a first migration such that
(5)
In equation (5), and are unknown. Since I am looking for an approximate of the Hessian, I need to find two known images that are related by the same expression as in equation (5). This can be easily achieved by remodeling the data from with
(6)
and remigrating them with as follows:
(7)
Notice a similarity between equations (5) and (7) except that in equation (7), only is unknown. Notice that has a mathematical significance: it is a vector of the Krylov subspace for the model . Now, I assume that we can write the inverse Hessian as a linear operator such that
(8)
and
(9)
Equation (9) can be approximated as a fitting goal for a matching filter estimation problem where is the convolution matrix with a bank of non-stationary filters Rickett et al. (2001). This choice is rather arbitrary but reflects the general idea that the Hessian is a transform operator between two similar images. My hope is not to perfectly'' represent the Hessian, but to improve the migrated image at a lower cost than least-squares migration. In addition in equations (8) and (9), the deconvolution process becomes a convolution, which makes it much more stable and easy to apply. Hence, I can rewrite equation (9) such that the matrix becomes a vector and becomes a convolution matrix Robinson and Treitel (1980):
(10)
The goal now is to minimize the residual
(11)
in a least-squares sense. Because we have many unknown filter coefficients in , I introduce a regularization term that penalizes differences between filters as follows:
(12)
where is the Helix derivative Claerbout (1998). The objective function for equation (12) becomes
(13)
where is a constant. The least-squares inverse is thus given by
(14)
Once is estimated, the final image is obtained by computing
(15)
where (*) is the convolution operator.
Therefore, I propose computing first a migrated image , then computing a migrated image (equation (7)), and finally estimating a bank of non-stationary matching filters , e.g., equation (12). The final improved image is obtained by applying the matching filters to the first image , e.g., equation (8). In the next section, I illustrate this idea with the Marmousi dataset. I show that an image similar to the least-squares migration image can be effectively obtained.
Next: Migration results Up: Theory Previous: Theory
Stanford Exploration Project
7/8/2003 | 569 | 2,538 | {"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} | 2.640625 | 3 | CC-MAIN-2017-43 | latest | en | 0.955118 |
https://community.smartsheet.com/discussion/89706/if-formula-counting-multi-select-drop-down-answers | 1,723,502,921,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641052535.77/warc/CC-MAIN-20240812221559-20240813011559-00131.warc.gz | 136,236,714 | 107,699 | # IF Formula Counting Multi-Select Drop Down Answers
Options
✭✭✭✭✭
Hello,
I am trying to create a formula using an IF statement that will say "On Track" if there is one answer in the cell, but if there are more than one it will output "Delay". I am assuming my IF statement is counting the answers the way it is working now, however it isn't quite counting correctly and not sure what I'm missing.
The formula should say "On Track" since there is only one answer in this cell, but instead it says "Delayed". Can somebody please take a look and help me out?
This will become a large OR statement with all of the other status abbreviations following the same format.
=IF(IB@row < 2, "On Track", "Delayed")
Thank you!
Tags:
• ✭✭✭✭✭✭
Options
Hey @RingJake
If I understand correctly, column IB is a multi-select dropdown column, that is, a cell that may have more than one answer within that single cell. If this is true, the COUNTM function will count the multi-select answers.
=IF(COUNTM(IB@row)<2, "On Track", "Delayed")
Does this work for you?
Kelly
• ✭✭✭✭✭
Options
Thank you @Kelly Moore this is working, now if I wanted to add the OR statement part, how would I write this properly? I will be adding 8 more cells following this logic in the IF OR statement.
This is my attempt, however I am getting an error.
=IF(OR(COUNTM(IB@row) < 2, "On Track", "Delayed", COUNTM(EB@row) < 2, "On Track", "Delayed"))
• ✭✭✭✭✭✭
Options
Hey
We can take advantage of your contiguous columns and go to a simpler route than an IF/OR
=COUNTIFS(IB@row:FRD@row, COUNTM(@cell) > 2)
=IF(OR(criteria1, criteria2, criteria3, etc), "True", "False")
The OR closes when all of the ORs are added. They should all have the same True or False result. | 483 | 1,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-33 | latest | en | 0.921718 |
https://oeis.org/A146758 | 1,701,645,728,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100518.73/warc/CC-MAIN-20231203225036-20231204015036-00583.warc.gz | 499,301,508 | 4,426 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A146758 Last prime subtrahend at 10^n in A146757. 2
7, 97, 983, 9941, 99839, 999983, 9998239, 99999989, 999950881, 9999999929, 99999999833, 999999999989, 9999999999863, 99999999999971, 999999961946087, 9999999999999917, 99999999989350667, 999999999999999989 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Table of n, a(n) for n=1..18. EXAMPLE A(2)=97 because 97 is the 15th and last prime difference under 10^2. MAPLE A146758 := proc(n) local p: p:=10^n: do p:=prevprime(p): if(isprime(ceil(evalf(sqrt(p)))^2-p))then return p: fi: od: end: seq(A146758(n), n=1..14); # Nathaniel Johnston, Oct 01 2011 PROG (UBASIC) 10 'sq less pr are prime 20 N=1:O=1:C=1 30 A=3:S=sqrt(N):if N>10^3 then print N, C-1:stop 40 B=N\A 50 if B*A=N then 100 60 A=A+2 70 if A<=S then 40 80 R=O^2:Q=R-N 90 if N1 print R; N; Q; C:N=N+2:C=C+1:goto 30 100 N=N+2:if N
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Last modified December 3 18:21 EST 2023. Contains 367540 sequences. (Running on oeis4.) | 569 | 1,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.685862 |
archergtfzl.dbblog.net | 1,547,934,880,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583681597.51/warc/CC-MAIN-20190119201117-20190119223117-00585.warc.gz | 15,585,002 | 6,126 | # 5 Simple Statements About statistics assignment help Explained
Enter a established of data points, then derive a operate to fit These points. Manipulate the perform over a coordinate airplane employing slider bars. Find out how Just about every continual and coefficient affects the resulting graph.
Find out about number styles in sequences and recursions by specifying a starting variety, multiplier, and increase-on. The figures inside the sequence are exhibited with a graph, and They're also mentioned underneath the graph.
Investigate the Cartesian coordinate program through pinpointing the coordinates of the randomly produced point, or requesting that a particular issue be plotted. Standard Coordinates Video game has become the Interactivate assessment explorers.
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Operate the classic recreation of lifestyle, Mastering about probabilities, chaos and simulation. This activity enables the person to operate a randomly generated earth or check out a variety of styles. This is a really strong exercise with an array of possibilities. It runs in the separate window.
If a researcher wishes to know if one particular approach for dealing with a sickness is a lot better than An additional, it can be crucial that she has some statistics to show what share of time each strategy labored on similar groups of sufferers. If I need to know irrespective of whether I really should don a seat belt, I would like to check The share of people who endure crashes when donning seat belts to the percentage who endure when they do not. | 759 | 4,172 | {"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} | 2.640625 | 3 | CC-MAIN-2019-04 | latest | en | 0.940861 |
http://www.howtodothings.com/finance-real-estate/how-to-calculate-gallons-per-minute | 1,561,428,581,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999783.49/warc/CC-MAIN-20190625011649-20190625033649-00483.warc.gz | 243,655,105 | 7,364 | # How To Calculate Gallons Per Minute
In these times of crisis, people need to work on a tight budget to remain afloat. This entails conservation of resources and keeping the expenditures to a bare minimum. Keeping track of the water and gasoline consumption may be helpful during these times. Knowing your consumption pattern is extremely helpful to avoid unwanted expenditures and possibly save some money.
Here are some tips on how to save by calculating your gallons per minute consumption of water and gasoline.
For calculating gallons per minute of water. Most of the faucets and showerheads now have a prefixed flow rate. As of 1992, the required flow rate is not more than 2.2 GPM at 60 PSI. To manually compute the flow rates of shower heads and faucets, you can take a container and fill it with water from a running faucet of shower head for 10 seconds. Make sure that the water is in full blast. Close the faucet or shower head and measure the amount of water in the container. Convert your measured water in gallons. Lastly, multiply the derived amount in gallons by 6. The product is the amount of consumed water in gallons per minute.
For measuring gallons per minute of gasoline. For this, you will need a notebook, a pen and a calculator.
• Keep track of the time you drive within a week. Keep a constant log of your commuting time from day to day. Note the time of your departure to the time you were able to park. Compute for your travel time. Keep the record intact, and place the notebook in your glove compartment for an easy access.
• Take note of how much you gas up. Know how much gas you spend in a week. To make it easier, go on a full tank of gasoline on your first day. If the gas runs out and you need to fill up again, take note of how much gas you had to refill. During your seventh day, add your accumulated gas consumption. Write it down together with the driving time.
• Convert the driving time into minutes. Take note of the total driving time within the week. Convert the number of hours into minutes; multiply the number of hours by 60. Add the values to give you the figures reflecting the length of time you drove in minutes.
• Compute the amount of the gasoline you consumed. Add all the gasoline you used starting from the first full tank up to the succeeding refills. Make sure that you convert them in gallons.
• Calculate your gasoline gallons per minute. Divide the number of gallons you have consumed with the number of hours you drove on the road. This will reflect your gas gallons per minute.
Do this test for a number of times all year round to see your water and gas accumulation patterns. Make the necessary changes in your consumption to help maintain the consumption and expenditures for water and gas to the minimum. | 575 | 2,780 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2019-26 | latest | en | 0.950188 |
https://cs.stackexchange.com/questions/10450/cfg-to-chomsky-normal-form | 1,717,090,046,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971668873.95/warc/CC-MAIN-20240530145337-20240530175337-00622.warc.gz | 152,761,277 | 39,867 | # CFG to Chomsky normal form
I've read this part like 3-4 times and I'm not understanding what is going on.
Let G6 be the following CFG and convert it to Chomsky normal form by using the conversion procedure just given. The series of grammars presented illustrates the steps in the conversion. Rules show in bold have just been added. Rules shown in gray have just been removed.
Specific steps/instructions from the book
1.The original CFG G6 is shown on the left. The result of applying the first step to make a new start variable appears on the right.
S→ASA|aB S0→S
A→B|S S→ASA|aB
B→b|ε A→B|S
B→b|ε
2. Remove ε rules B→ε, shown on the left, and A→ε, shown on the right.
S0→S S0→S
S→ASA|aB|**a** S→ASA|aB|a|**SA|AS|S** from?
A→B|S|**ε** A→B|S|ε
B→b|**ε** B→b
3a. Remove unit rules S→S, shown on the left, and S0→S, shown on the right.
S0→S S0→S|**ASA|aB|a|SA|AS**
S→ASA|aB|a|SA|AS|S S→ASA|aB|a|SA|AS
A→B|S A→B|S
B→b B→b
3b. Remove unit rules A→B and A→S
S0→ASA|aB|a|SA|AS S0→ASA|aB|a|SA|AS
S→ASA|aB|a|SA|AS S→ASA|aB|a|SA|AS
A→B|S|**b** A→S|b|**ASA|aB|a|SA|AS**
B→b B→b
4. Convert the remaining rules into the proper form by adding additional variables and rules. The final grammar in Chomsky normal form is equivalent to G6, which follows. (Actually the procedure given in Theorem 2.9 produces several variables Ui, along with several rules Ui→a. We simplified the resulting grammar by using a single variable U and U→a.)
S0→AA1|UB|a|SA|AS
S→AA1|UB|a|SA|AS
A→b|AA1|UB|a|SA|AS
A1→SA
U→a
B→b
I don't understand why they added the "bold" (** **) items. I also don't understand 4. Can someone please explain this to me. Thank you!
• You should have been given the algorithm for this. Alternatively, have a look at a textbook on the topic. Mar 11, 2013 at 15:18
Removing $\epsilon$ transitions:
Consider the rule $B\to \epsilon$. We want to remove it. To do so, we look for occurrences of $B$ in the grammar, and whenever we see on (e.g. in $S\to aB$), we add the option of choosing $B\to \epsilon$. So the rule $S\to aB$ becomes $S\to a$. Similarly, $A\to B$ becomes $A\to \epsilon$ (which we will have to remove in the next step).
Since $A\to \epsilon$ was introduced when removing $B\to \epsilon$, we now add rules where $A$ appears. For example, the rule $S\to ASA$ becomes $S\to ASA|SA|AS|S$, according to the different options of using $A\to \epsilon$ after $S\to ASA$.
In step 4 - for every rule that yields more than a pair of variables (e.g. in this case $S\to ASA$) we split this into pairs as follows. Introduce a new variable $T$, we add the rules $S\to TA$ and $T\to AS$. Then, applying these forces us to actually have $S\to ASA$.
• Do you know why they add a $S_0$ in the first step? It looks kind of useless. May 15, 2015 at 21:47
• One property of this normal form is that the initial variable is never derived by any rule. So we add an artificial $S_0$ to ensure it's never derived. May 16, 2015 at 5:55
• One reason to add $S_0 \rightarrow S$ is in the case that the language produces the empty string $\epsilon$ (or $\lambda$). If the language doesn't produce the empty string, it's fine to have the start symbol at the right side of a production.
– Sean
May 30, 2018 at 14:13 | 1,090 | 3,382 | {"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} | 3.703125 | 4 | CC-MAIN-2024-22 | latest | en | 0.847494 |
https://www.mcqslearn.com/business-statistics/quiz/?page=115 | 1,558,558,327,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256958.53/warc/CC-MAIN-20190522203319-20190522225319-00146.warc.gz | 859,822,805 | 7,259 | Data tables and types quiz, data tables and types MCQs answers, MBA business statistics quiz 115 to learn business analytics courses online. Data classification, tabulation and presentation quiz questions and answers, data tables and types multiple choice questions (MCQ) to practice statistics test with answers for college and university courses. Learn data tables and types MCQs, harmonic mean, statistics formulas, data tables and types test prep for business analyst certifications.
Learn data tables and types test with multiple choice question (MCQs): if in bar diagram characteristics variable to be measured is written on horizontal axis and frequencies are written on vertical axis then graph is for, with choices grouped data, ungrouped data, dimensional data, and non dimensional data for masters in business administration. Learn data classification, tabulation and presentation questions and answers for scholarships exams' problem-solving, assessment test for business administration certifications.
Data Tables & Types Quiz
MCQ: If in bar diagram characteristics variable to be measured is written on horizontal axis and frequencies are written on vertical axis then graph is for
1. grouped data
2. ungrouped data
3. dimensional data
4. non dimensional data
A
Statistics Formulas Quiz
MCQ: Formula of calculating mean for negative binomial probability distribution is
1. q ⁄ p
2. P ⁄ Q
3. p ⁄ r
4. r ⁄ p
D
Data Tables & Types Quiz
MCQ: Complex type of table in which variables to be studied are subdivided with interrelated characteristics is called as
1. two way table
2. one way table
3. subparts of table
4. order level table
A
Harmonic Mean Quiz
MCQ: If arithmetic mean is 20 and harmonic mean is 30 then geometric mean is
1. 14.94
2. 24.94
3. 34.94
4. 44.94
B
Structured Data Quiz
MCQ: Type of questions included in questionnaire to record responses in which respondents have set of alternatives are classified as
1. open ended questions
2. close ended questions
3. multiple choices
4. itemized question
B | 452 | 2,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2019-22 | latest | en | 0.858493 |
https://www.studypool.com/questions/262749/Find-the-rate-of-change-of-the-demand | 1,477,093,766,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718311.12/warc/CC-MAIN-20161020183838-00006-ip-10-171-6-4.ec2.internal.warc.gz | 997,725,826 | 750,121 | # related rates in calculus question please
FratBro23
Category:
Calculus
Price: \$5 USD
Question description
A price p(in dollars) and demand x for aa product are related by
2x^2-2xp+50p^2=20600
If the price is increasing at a rate of \$2 per months when the price is \$20
Find the rate of change of the demand :
(Top Tutor) Daniel C.
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School: UCLA
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2325 Tutors | 278 | 1,042 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-44 | longest | en | 0.691951 |
https://education-academia.github.io/computer-concepts-applications/using-microsoft-excel/how-to-add-copy-and-move-worksheets-within-excel-workbooks/when-you-copy-a-formula-that-contains-an-absolute-reference-to-a-new-location-the-reference-.html | 1,695,973,661,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510498.88/warc/CC-MAIN-20230929054611-20230929084611-00817.warc.gz | 245,481,744 | 9,052 | # FUNDAMENTALS OF COMPUTER
## USING MICROSOFT EXCEL
### HOW TO ADD COPY AND MOVE WORKSHEETS WITHIN EXCEL WORKBOOKS
Question [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
When you copy a formula that contains an absolute reference to a new location, the reference ____
A is updated automatically B does not change C becomes bold D has a dotted outline in its cell
Explanation:
Detailed explanation-1: -When you copy a formula containing an absolute reference, the cell reference in the copied formula does not change, regardless of where you copy the formula. An absolute cell reference appears with a dollar sign before both the column letter and the row number, such as \$B\$4.
Detailed explanation-2: -Unlike relative references, absolute references do not change when copied or filled. You can use an absolute reference to keep a row and/or column constant. An absolute reference is designated in a formula by the addition of a dollar sign (\$) before the column and row.
Detailed explanation-3: -If you want to maintain the original cell reference when you copy it, you “lock” it by putting a dollar sign (\$) before the cell and column references. For example, when you copy the formula =\$A\$2+\$B\$2 from C2 to D2, the formula stays exactly the same. This is an absolute reference.
Detailed explanation-4: -An absolute reference in Excel is a reference that cannot be changed when copied, so you won’t see changes in rows or columns when you copy them. Absolute references are used when you want to fix a cell location. These cell references are preceded by a dollar sign.
Detailed explanation-5: -Usually the CELL REFERENCES will CHANGE! If you copy a formula 2 rows to the right, then the cell references in the formula will shift 2 cells to the right. If you copy a formula 3 rows down and 1 row left, then the cell references in the formula will shift 3 rows down and 1 row left.
There is 1 question to complete. | 431 | 1,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2023-40 | latest | en | 0.822179 |
aasnova.org | 1,723,306,329,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640810581.60/warc/CC-MAIN-20240810155525-20240810185525-00201.warc.gz | 58,834,507 | 19,459 | This week, we got to see the spectacular first spectrum of an exoplanet’s atmosphere taken by JWST. Have you ever wondered how researchers use models to determine the properties of distant planets’ atmospheres from their spectra?
Though models come in many forms, most fall into two categories: computational and analytical.
A computational model of an exoplanet’s atmosphere incorporates all our knowledge of atmospheric physics to predict what happens when photons from the planet’s host star navigate the maze of atoms and molecules in an exoplanet’s atmosphere on their way to our telescopes. Computational models generate synthetic spectra, which researchers can then compare to a planet’s actual spectrum to constrain the properties of the planet’s atmosphere, like its temperature, density, and composition.
In an analytical model, all that physics gets boiled down to a set of comparatively simple mathematical expressions that describe how the input parameters (e.g., the properties of a planet and its atmosphere) relate to the model output (i.e., a spectrum). By developing these mathematical expressions for how an exoplanet’s spectrum varies with different properties of its atmosphere, researchers can get a better sense of which physical properties affect a planet’s spectrum and why.
#### Analytical Avenues
A flowchart showing analytical and numerical (computational) modeling methods. The method used in this work is represented by the yellow shapes on the left-hand side. Click to enlarge. [Matchev et al. 2022]
In today’s article, a team of researchers at the University of Florida led by Konstantin Matchev demonstrated how symbolic regression can be used to develop an analytical model of exoplanet spectra. Symbolic regression is a way of sifting through mathematical functions to determine a set of simple and accurate functions that describe how inputs and outputs are related.
For the specific case of exoplanet spectra, the authors first simplified their model by lowering the number of input variables. For example, instead of the planet’s radius and the host star’s radius being two separate variables, the authors use the ratio of the two radii as a single dimensionless variable.
Then, the team used machine learning to extract analytic expressions that relate their input variables to a set of synthetic hot-Jupiter spectra generated by a separate analytic model. By working with the output from another analytic model, the authors were able to gauge the success of their technique; if successful, their model should extract the exact expressions that the target model is based on. Ultimately, the authors found that their symbolic regression method was able to discard unimportant variables, generate the correct expressions, and determine which input variables have the largest impact on the output spectra.
An illustration of the degeneracies between pressure (P0), gas absorption cross section per unit mass (κ), gravity (g), temperature (T), mean molecular mass (m), planet radius (R0), and stellar radius (Rs). Two degenerate variables are connected by a line and three variables by a triangle. [Matchev et al. 2022]
#### Weighing the Options
Why use this technique when powerful computational models are already available? Computational models can eat up hours of computing time, and the results don’t always give physical intuition into the system being modeled. And while computational and analytical models both have an issue with degeneracy — a scenario in which multiple combinations of inputs give the same output — analytical models can help us pinpoint which variables are degenerate and develop a physical understanding of why. For example, the authors’ analytical model tells us that there is a degeneracy between the planet’s temperature and the strength of its gravitational pull. Physically, this arises because a hotter planet would have a puffier atmosphere, but so would a planet with a weaker gravitational pull.
Though analytical models may not be appropriate for all situations, they’re an important tool for studying exoplanet atmospheres — and new exoplanet spectra from JWST should provide an excellent challenge for all models!
#### Citation
“Analytical Modeling of Exoplanet Transit Spectroscopy with Dimensional Analysis and Symbolic Regression,” Konstantin T. Matchev et al 2022 ApJ 930 33. doi:10.3847/1538-4357/ac610c | 853 | 4,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.862939 |
http://www.jiskha.com/display.cgi?id=1199844758 | 1,496,011,608,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463611569.86/warc/CC-MAIN-20170528220125-20170529000125-00218.warc.gz | 661,409,619 | 4,322 | posted by on .
I posted this a little while ago, but didn't see it. Sorry if it post twice.
Each fire hose is 25 yards long. The fire station needs enough fire hoses to reach and fire hydrant to the farthest house, which could be up to 230 feet away. What is the minimum number of hoses that the fire station needs?
• 5th Grade Math Problem - ,
Divide and use the next number up.
230/25 = ??
Since you normally wouldn't have 2/10ths of a hose, you'll have to add 1 (whole number) to the whole number in the answer.
Clear?
• 5th Grade Math Problem - ,
But it is 25 yards & 230 feet. I'm not sure I understand.
• 5th Grade Math Problem - ,
Since there are 3 feet in a yard, each 25-yard hose is 75 feet long.
Now divide 230 feet by 75. It comes out to just over 3. Since 3 hoses isn't enough, we'll need to round up to the nearest whole number.
• 5th Grade Math Problem - ,
• 5th Grade Math Problem - ,
That makes sense, thank you to everyone for your help. | 261 | 970 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | latest | en | 0.974482 |
https://studyres.com/doc/874053/chapter7 | 1,632,319,342,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057366.40/warc/CC-MAIN-20210922132653-20210922162653-00612.warc.gz | 592,061,955 | 12,793 | Survey
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Transcript
```Chapter 7
Rotational Motion
Universal Law of Gravitation
Kepler’s Laws
Angular Displacement
1.
2.
3.
Three measures of angles:
Degrees
Revolutions (1 rev. = 360 deg.)
Angular Displacement, cont.
Change in
distance of a point:
s 2pr N (N counts revolutions)
r ( is in radians)
Example 7.1
An automobile wheel has a radius of 42 cm. If a
car drives 10 km, through what angle has the
wheel rotated?
a) In revolutions
c) In degrees
a) N = 3789
c) = 1.36x106 degrees
Angular Speed
Can be given in
Revolutions/s
Degrees/s
w
Linear
f i
t
Speed at r
N revolutions
v 2p r
t
2p r θ f θi (in rad.s)
2p
t
v wr
Example 7.2
A race car engine can turn at a maximum rate of 12,000
rpm. (revolutions per minute).
a) What is the angular velocity in radians per second.
b) If helipcopter blades were attached to the
crankshaft while it turns with this angular velocity, what
is the maximum radius of a blade such that the speed of
the blade tips stays below the speed of sound.
DATA: The speed of sound is 343 m/s
b) 27 cm
Angular Acceleration
Denoted by a
a
w f wi
t
w must be in radians per sec.
Every portion of the object has same w and same a
Linear and Rotational Motion Analogies
Rotational Motion
wi w f
2
Linear Motion
t
1 2
wit at
2
w wi at
w
2
2
wi
2a
x
vi v f
2
t
1 2
x vit at
2
v vi at
v
2
2
vi
2ax
Linear movement of a rotating point
Distance
s r
Speed
Acceleration
v wr
Different points on same
object have different
linear motions!
a ar
Only work when angles are in radians!
Example 7.3
A pottery wheel is accelerated uniformly from rest
to a rate of 10 rpm in 30 seconds.
a.) What was the angular acceleration? (in rad/s2)
b.) How many revolutions did the wheel undergo
during that time?
b) 2.50 revolutions
Example 7.4
A coin of radius 1.5 cm is initially rolling with a
rotational speed of 3.0 radians per second, and
comes to a rest after experiencing a slowing down of
a.) Over what angle (in radians) did the coin rotate?
b.) What linear distance did the coin move?
b) 135 cm
Centripetal Acceleration
Moving in circle at constant SPEED does not mean
constant VELOCITY
Centripetal acceleration results from CHANGING
DIRECTION of the velocity
Centripetal Acceleration, cont.
Acceleration is directed
toward the center of the
circle of motion
v
a
t
Basic formula
Derivation: a = w2r = v2/r
From the geometry of the Figure
v 2v sin( / 2)
v for small
From the definition of angular velocity
wt
v vwt
v
a
vw
t
2
v
a w r
r
2
Forces Causing Centripetal Acceleration
Newton’s Second Law
F ma
Examples of forces
Spinning ball on a string
Gravity
Electric forces, e.g. atoms
Ball-on-String Demo
Example 7.5
A space-station is constructed like a barbell with
two 1000-kg compartments separated by 50
meters that spin in a circle (r=25 m). The
compartments spin once every 10 seconds.
a) What is the acceleration at the extreme end of
the compartment? Give answer in terms of “g”s.
b) If the two compartments are held together by a
cable, what is the tension in the cable?
a) 9.87 m/s2 = 1.01 “g”s
b) 9870 N
Example 7.6
A race car speeds around a circular track.
a) If the coefficient of friction with the tires is 1.1,
what is the maximum centripetal acceleration (in
“g”s) that the race car can experience?
b) What is the minimum circumference of the track
that would permit the race car to travel at 300
km/hr?
a) 1.1 “g”s
b) 4.04 km (in real life curves are banked)
Example 7.7
A curve with a radius of
curvature of 0.5 km on a
highway is banked at an
angle of 20. If the
highway were frictionless,
at what speed could a car
drive without sliding off
42.3 m/s = 94.5 mph
Example 7.8
AAyo-yo is spun in a circle as
shown. If the length of the
string is L = 35 cm and the
circular path is repeated 1.5
times per second, at what
angle (with respect to the
vertical) does the string bend?
= 71.6 degrees
Example 7.9
A roller coaster goes upside down performing a
circular loop of radius 15 m. What speed does the
roller coaster need at the top of the loop so that
it does not need to be held onto the track?
12.1 m/s
Accelerating Reference Frames
Consider a frame that is accelerating with af
F ma
F ma f m(a a f )
Fictitious force
Looks like “gravitational” force
If frame acceleration = g,
fictitious force cancels real gravity.
Examples: Falling elevator, planetary orbit
rotating space stations
DEMO: FLYING POKER CHIPS
Example 7.10
Which of these astronauts experiences weightlessness?
a) A stationary astronaut billions of light years from
any star or planet.
b) An astronaut falling freely in a broken elevator.
c) An astronaut orbiting the Earth in a low orbit.
d) An astronaut far from any significant stellar object
in a rapidly rotating space station
Newton’s Law of
Universal
Gravitation
m1m2
F G 2
r
3
m
11
G 6.67 10
2
kg s
Force is always
attractive
Force is proportional
to both masses
Force is inversely
proportional to
separation squared
Gravitation Constant
Determined experimentally
Henry Cavendish, 1798
Light beam / mirror
amplify motion
Example 7.11
Given: In SI units, G = 6.67x10-11,
g=9.81 and the radius of Earth is
6.38 x106.
Find Earth’s mass:
5.99x1024 kg
Example 7.12
Given: The mass of Jupiter is 1.73x1027 kg
and Period of Io’s orbit is 17 days
r = 1.84x109 m
Tycho Brahe (1546-1601)
Lost part of nose in a duel
EXTREMELY ACCURATE
astronomical observations, nearly
10X improvement, corrected for
atmosphere
Hired Kepler to work as
mathematician
Uraniborg (on an island near Copenhagen)
First to:
Explain planetary motion
Investigate the formation of
pictures with a pin hole
camera;
Explain the process of vision
by refraction within the eye
Formulate eyeglass designed
for nearsightedness and
farsightedness;
Explain the use of both eyes
for depth perception.
First to describe: real,
virtual, upright and inverted
images and magnification
Johannes Kepler
(1571-1630)
Johannes Kepler (1571-1630)
First to:
explain the principles of how a telescope works
discover and describe total internal reflection.
explain that tides are caused by the Moon.
suggest that the Sun rotates about its axis
derive the birth year of Christ, that is now universally
accepted.
derive logarithms purely based on mathematics
He tried to use stellar parallax caused by the Earth's
orbit to measure the distance to the stars; the same
principle as depth perception. Today this branch of
research is called astrometry.
Isaac Newton (1642-1727)
Invented Calculus
Formulated the universal law of
gravitation
Showed how Kepler’s laws could
be derived from an inversesquare-law force
Invented Wave Mechanics
mathematics and geometry
Kepler’s Laws
1.
2.
3.
All planets move in elliptical orbits with the
Sun at one of the focal points.
A line drawn from the Sun to any planet
sweeps out equal areas in equal time intervals.
The square of the orbital period of any planet
is proportional to cube of the average
distance from the Sun to the planet.
Kepler’s First Law
All planets move in
elliptical orbits with
the Sun at one focus.
Any object bound
to another by an
inverse square law
will move in an
elliptical path
Second focus is
empty
Kepler’s Second Law
A line drawn from the
Sun to any planet will
sweep out equal areas in
equal times
Area from A to B and
C to D are the same
This is true for any central force due to
angular momentum conservation (next chapter)
Kepler’s Third Law
The square of the orbital period of any planet is
proportional to cube of the average distance from the
Sun to the planet.
2
T
K sun
3
r
For orbit around the Sun, KS = 2.97x10-19 s2/m3
K is independent of the mass of the planet
Derivation of Kepler’s Third Law
Basic formula
Mm
F ma G 2
r
2
a w r
Basic formula
2p
w
T
Mm
ma G 2
r
Mm
2
mw r G 2
r
GM 2
r
T
2
4p 2
2
T
4p
K sun
3
r
GM
3
Example 7.13
Data: Radius of Earth’s orbit = 1.0 A.U.
Period of Jupiter’s orbit = 11.9 years
Period of Earth’s orbit = 1.0 years
5.2 A.U.
Gravitational Potential Energy
PE = mgh valid only near
Earth’s surface
For arbitrary altitude
Mm
PE G
r
Zero reference level is
at r=
Graphing PE vs. position
Mm
PE G
r
Example 7.14
You wish to hurl a projectile from the surface of the
Earth (Re= 6.38x106 m) to an altitude of 20x106 m
above the surface of the Earth. Ignore the rotation of
the Earth and air resistance.
a) What initial velocity is required?
a) 9,736 m/s
b) What velocity would be required in order for the
projectile to reach infinitely high? I.e., what is the
escape velocity?
b) 11,181 m/s
c) How does the escape velocity compare the velocity
required for a low earth orbit?
c) 7,906 m/s
```
Related documents | 3,107 | 8,993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-39 | latest | en | 0.812256 |
http://www.maths.manchester.ac.uk/~mheil/Lectures/Fluids/index.html | 1,521,346,656,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645513.14/warc/CC-MAIN-20180318032649-20180318052649-00311.warc.gz | 424,803,398 | 6,547 | # MATH35001: Viscous Fluid Flow
This course is concerned with the mathematical theory of viscous fluid flows. Fluid mechanics is one of the major areas for the application of mathematics and has obvious practical applications in many important disciplines (aeronautics, meteorology, geophysical fluid mechanics, biofluid mechanics, and many others). Using a general continuum mechanical approach, we will first derive the governing equations (the famous Navier-Stokes equations) from first principles. We will then apply these equations to a variety of practical problems and examine appropriate simplifications and solution strategies.
Many members of staff in the department have research interests in fluid mechanics and this course will also lay the foundations for possible future postgraduate work in this discipline.
#### Vortex shedding caused by the flow past a flat plate (snapshot 1)
This course is currently taught by Prof Matthias Heil. This page provides online access to the lecture notes, example sheets and other handouts and announcements.
Please note that the lecture notes only summarize the main results and will generally be handed out after the material has been covered in the lecture.
If you have any questions about the lecture, please see me in my office (2.224 in the Alan Turing building), contact me by email ( M.Heil@maths.man.ac.uk) or catch me after the lecture.
## Syllabus
• 1. Introduction; overview of the course; introduction to index notation. [2]
• 2. The kinematics of fluid flow: The Eulerian velocity field; the rate of strain tensor and the vorticity vector; the equation of continuity. [3]
• 3. The Navier-Stokes equations: The substantial derivative; the stress tensor; Cauchy's equation; the constitutive equations for a Newtonian fluid. [4]
• 4. Boundary and initial conditions; surface traction and the conditions at a free surface. [1].
• 5. One-dimensional flows: Couette/Poiseuille flow; flow down an inclined plane; the vibrating plate. [3]
• 6. The equations in curvilinear coordinates; Hagen-Poiseuille flow; circular Couette flow. [2]
• 7. Dimensional analysis and scaling; the dimensionless Navier-Stokes equations and the importance of the Reynolds number; limiting cases and their physical meaning; lubrication theory. [3]
• 8. The streamfunction/vorticity equations [2]
• 9. Stokes flow (zero Reynolds number flow) [2]
• 10. High-Reynolds number flow; boundary layers; the Blasius boundary layer. [2]
## Assessment:
The course will be examined in a two hour exam in January. "Formative feedback" will be provided via discussions in the weekly examples class and/or during my office hour (Thu 1-2; feel free to send me an email to arrange a specific meeting if you're not free during this time). In my experience the above arrangements are perfectly satisfactory for the students (if they aren't, then tell me!), but, sadly not for the people whose job it is to tell me how to teach. To please the latter, I am offering to provide written comments on the "starred question" (Question 1) on Example Sheet VI. Please hand in your write-up of the solution at the end of the Friday lecture following the examples class in which this sheet will be discussed (i.e. on Friday Nov 17th). I will then annotate your solution and return it in the following week. However, please note that the purpose of this exercise is purely "formative feedback" and does not bear any credit.
# Anti-cramming policy
This course used to have a 20% coursework component, introduced in order to force the students to work continuously. The weekly coursework resulted in a fairly heavy workload during term-time (which the students hated) but made exam revision very easy (which they loved). Overall, the coursework element was perceived to be an excellent feature of the course -- judging by the replies on the student questionnaires handed out at the end of the course.
Unfortunately, various constraints made it impossible to continue this very successful setup, facing me with the question of how to get you to work for the course throughout term, rather than adopting the "I don't have to work for this course because I can simply cram at the end of term and revise by looking at past exam papers" attitude that all lecturers (and me in particular) detest.
So, here is the deal:
• You are hereby told (yet again) that it is essential to work continuously on the material presented in this (and any other) lecture course. If you don't understand the concepts presented in week 1 you will not understand what I talk about in week 2, etc. The best (only?) way to achieve this is to work through the example sheets before the example/feedback class, to make sure you can bombard me with questions about any issues that you don't understand. There is little point in turning up for the example/feedback class without having looked at the problem sheet beforehand -- there's not enough time to do all the work in class.
• On each problem sheet I will identify a small number of (sometimes substantial) questions that used to form the coursework. The logic behind the selection of the questions is that they'll force you to understand concepts that are essential to the understanding of subsequent material.
• The (detailed) solutions to each problem sheet will be made available at some suitable time after the example/feedback class (probably a week later). At that point you should swap your nicely-written-up solutions (written up at least as nicely as if you were to hand them in to me) with another student and mark each other's work. No need for a marking scheme, just check what's right and what's wrong (and why!). In my experience, most students tend to work best in small groups anyway so I don't think you'll have problems finding people to swap work with. If you really have no mates, let me know and I'll find you somebody (we'll do triangular swaps if there's an odd number of students).
• I can see panic developing: Will the "grade" awarded by your mates "count towards the exam"? No! There won't be a grade. You're simply supposed to give each other feedback, and if there are any disputes, I'm happy to act as judge. In fact, I'll be delighted to go back over material in subsequent example/feedback classes if it turns out that many of you struggled with the same problem (assuming I didn't realise that during the example/feedback class itself). However, I can only do this if I know what you struggled with and this obviously requires you to have done the work.
• Now that the panic is gone, you'll obviously ask yourself why you should bother with this. Surely it'll be easier to use the example/feedback class to read the metro newspaper and exchange gossip, and then use the tried-and-tested "cramming for the exam" technique to "revise". I obviously can't stop you from doing that but I can assure you that, following the end of term, I will refuse to answer any questions (about the course material or previous exam papers) from students who do not have an (at least nearly) complete set of written-up "coursework", with signs that the work was seen by and discussed with somebody else. If you don't work, I won't either.
• Final question: "Hang on, you want us to do the questions, write up the answers, exchange them with another student, mark each other's work, and then discuss any mistakes/misconceptions with him/her (and the lecturer)? This will take a lot of time!" Answer: "Indeed -- I expect you to work hard for this course!" If you check the UG handbook you'll find that you're supposed to anticipate at least two hours of private study per lecture hour. Furthermore, whatever time you invest on this during term time, you'll save during the exam revision. (Note that this will free up valuable time to catch up with any past issues of the metro newspaper that you didn't have time to read during the examples class!).
## Scans of visualiser sheets:
My handwriting isn't exactly famous for being super-clear so here's another chance to stare at what I wrote on the visualiser. Keep shouting at me in the lecture (yes, really) if you can't read something and/or catch me before/after the next lecture if even close inspection of these scans leaves you at a loss as to what I could possibly have meant by all these scribbles...
## Corrections:
Please note a few corrections for previous handouts (the files above have already been corrected).
## Exam feedback:
I'm supposed to provide "feedback on the exam", so, for what it's worth, here's what I thought after marking the exam However, rather than wasting your time reading this (in an attempt to improve your grade by "improving your exam technique"), I suggest you have another read through the Anti-Cramming Policy above. Concentrate on actually understanding the material rather than trying to memorise likely exam questions and how to tackle them. Pleeeeease!
## Feedback on feedback:
I'm now also supposed to give you feedback on your feedback. Rant omitted. Here's my The inofficial one is: Thank you, I enjoyed it too! | 1,941 | 9,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-13 | longest | en | 0.893483 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=466620 | 1,386,195,239,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163037568/warc/CC-MAIN-20131204131717-00062-ip-10-33-133-15.ec2.internal.warc.gz | 219,626,306 | 1,388 | Question 766050
<font face="Times New Roman" size="+2">
Simple sign mistake. You said:
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-10}{30}\ +\ \frac{a}{30}\ =\ -1]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-1}{3}\ +\ \frac{a}{30}\ =\ -1]
which was all ok, but then you said:
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-1}{3}\ +\ 1\ =\ \frac{a}{30}]
But you should have said:
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-1}{3}\ +\ 1\ =\ -\frac{a}{30}]
And you will find that it works out to the correct answer from here.
Having said that, I'd like to point out that you did a great deal of extra work.
Start over from the beginning:
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-10\,+\,a}{30}\ =\ -1]
But this time as a first step, multiply both sides by 30:
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -10\ +\ a\ =\ -30]
Then just add 10 to both sides:
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ -20]
as expected.
Lesson: Things usually get simpler if you can find a multiplier that gets rid of all your denominators.
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font> | 491 | 1,407 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2013-48 | latest | en | 0.752796 |
https://nuprl.org/wip/Standard/relations/rel_implies_weakening.html | 1,675,703,912,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500356.92/warc/CC-MAIN-20230206145603-20230206175603-00553.warc.gz | 436,710,880 | 1,828 | ### Nuprl Lemma : rel_implies_weakening
`∀[T:Type]. ∀[R1,R2:T ⟶ T ⟶ ℙ]. (R1 `⇐⇒` R2 `` R1 => R2)`
Proof
Definitions occuring in Statement : rel_equivalent: `R1 `⇐⇒` R2` rel_implies: `R1 => R2` uall: `∀[x:A]. B[x]` prop: `ℙ` implies: `P `` Q` function: `x:A ⟶ B[x]` universe: `Type`
Definitions unfolded in proof : rel_implies: `R1 => R2` rel_equivalent: `R1 `⇐⇒` R2` uall: `∀[x:A]. B[x]` implies: `P `` Q` all: `∀x:A. B[x]` member: `t ∈ T` iff: `P `⇐⇒` Q` and: `P ∧ Q` infix_ap: `x f y` prop: `ℙ` so_lambda: `λ2x.t[x]` so_apply: `x[s]`
Lemmas referenced : all_wf iff_wf
Rules used in proof : sqequalSubstitution sqequalRule sqequalReflexivity sqequalTransitivity computationStep isect_memberFormation lambdaFormation cut hypothesis sqequalHypSubstitution dependent_functionElimination thin hypothesisEquality productElimination independent_functionElimination applyEquality lemma_by_obid isectElimination lambdaEquality functionEquality cumulativity universeEquality
Latex:
\mforall{}[T:Type]. \mforall{}[R1,R2:T {}\mrightarrow{} T {}\mrightarrow{} \mBbbP{}]. (R1 \mLeftarrow{}{}\mRightarrow{} R2 {}\mRightarrow{} R1 => R2)
Date html generated: 2016_05_14-AM-06_04_39
Last ObjectModification: 2015_12_26-AM-11_33_09
Theory : relations
Home Index | 492 | 1,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.312887 |
http://www.radiocontrolinfo.com/understanding-electricity/common-electric-terms/ | 1,580,229,786,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251779833.86/warc/CC-MAIN-20200128153713-20200128183713-00161.warc.gz | 257,112,429 | 16,348 | # Common Electric Terms
Before establishing the relationships and concepts of basic electricity it is important to explain all the common terms that are used. Terms will be listed in order of the most simple or common to terms that are not as used as often. These terms may be used in the other pages.
Battery – The battery for any RC plane, boat, or car is the main source of power. The battery provides the electricity by means of a chemical reaction.
ESC – The ESC controls the amount of power from the battery to the motor. In general Brushless ESC’s alter the frequency of power delivered to the motor to control RPM. The ESC follows the demands provided by the user.
Motor –. An electrical motor is used to convert Electrical Energy to Mechanical Rotational Energy. Ultimately, it provides the power to your RC plane, boat, or car.
Voltage – Voltage is the electromotive force typically provided by the battery measured in Volts. (V) Can also represent a voltage drop over a load in a circuit.
Current – The current, commonly termed amperage, is the amount of electricity flowing through a circuit in a given period of time. Measured in Amps (A)
Resistance – Acts as an opposition to the flow of electricity. Found in every component of an electrical circuit.
Data Logger – A unit used to measure common electrical parameters in live real time.
Internal Resistance – Commonly associated with batteries or a power source, is the resistance measured across the battery leads or power source.
Nominal Voltage – The average voltage of a battery through it’s cycle from charged to nearly discharged.
Power – Power is the amount of energy being consumed in a circuit per unit time. Measure in Watts (W)
Cogging – A cogging brushless motor occurs when the motor is at or close to 0 RPM as the ESC timing is not in sync with the motor timing.
Continuous Current – The maximum allowable current that can be maintained for extended periods of time.
Maximum Peak Current – The maximum peak current is the maximum allowable current that can travel through a circuit for only a few seconds. Always higher than continuous current.
Unloaded RPM – Is the theoretical RPM of a motor calculated from a predetermined battery cell count at nominal voltage.
Loaded RPM – Is the actual RPM of a motor when loaded. Often but not always lower than Unloaded RPM.
Conductor – A material with electrical properties that conduct the flow of electricity
AC – Alternating Current is current that constantly switches between the two possible directions.
DC – Direct Current is current that travels in one direction | 517 | 2,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.927889 |
http://oeis.org/A087787 | 1,586,363,726,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371818008.97/warc/CC-MAIN-20200408135412-20200408165912-00112.warc.gz | 123,688,911 | 4,301 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A087787 a(n) = Sum_{k=0..n} (-1)^(n-k)*A000041(k). 13
1, 0, 2, 1, 4, 3, 8, 7, 15, 15, 27, 29, 48, 53, 82, 94, 137, 160, 225, 265, 362, 430, 572, 683, 892, 1066, 1370, 1640, 2078, 2487, 3117, 3725, 4624, 5519, 6791, 8092, 9885, 11752, 14263, 16922, 20416, 24167, 29007, 34254, 40921, 48213, 57345, 67409 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS Essentially first differences of A024786 (see the formula). Also, a(n) is the number of 2's in the last section of the set of partitions of n+2 (see A135010). - Omar E. Pol, Sep 10 2008 LINKS Vaclav Kotesovec, Table of n, a(n) for n = 0..10000 FORMULA G.f.: 1/(1+x)*1/Product_{k>0} (1-x^k). a(n) = 1/n*Sum_{k=1..n} (sigma(k)+(-1)^k)*a(n-k). a(n) = A024786(n+2)-A024786(n+1). - Omar E. Pol, Sep 10 2008 a(n) ~ exp(Pi*sqrt(2*n/3)) / (8*sqrt(3)*n) * (1 + (11*Pi/(24*sqrt(6)) - sqrt(3/2)/Pi)/sqrt(n) - (11/16 + (23*Pi^2)/6912)/n). - Vaclav Kotesovec, Nov 05 2016 a(n) = A000041(n) - a(n-1). - Jon Maiga, Aug 29 2019 MATHEMATICA Table[Sum[(-1)^(n-k)*PartitionsP[k], {k, 0, n}], {n, 0, 50}] (* Vaclav Kotesovec, Aug 16 2015 *) (* more efficient program *) sig = 1; su = 1; Flatten[{1, Table[sig = -sig; su = su + sig*PartitionsP[n]; Abs[su], {n, 1, 50}]}] (* Vaclav Kotesovec, Nov 06 2016 *) CROSSREFS Cf. A000041, A024786, A135010, A138121, A141285. Sequence in context: A076077 A152194 A268630 * A182712 A100818 A005291 Adjacent sequences: A087784 A087785 A087786 * A087788 A087789 A087790 KEYWORD nonn AUTHOR Vladeta Jovovic, Oct 07 2003 STATUS approved
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Last modified April 8 12:31 EDT 2020. Contains 333314 sequences. (Running on oeis4.) | 833 | 2,051 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-16 | latest | en | 0.474443 |
https://fr.mathworks.com/matlabcentral/cody/problems/44651-find-all-repeated-numbers/solutions/1718872 | 1,597,425,497,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739347.81/warc/CC-MAIN-20200814160701-20200814190701-00486.warc.gz | 307,818,510 | 15,872 | Cody
# Problem 44651. Find all repeated numbers
Solution 1718872
Submitted on 1 Feb 2019 by Maryam Majidi
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 = []; assert(isequal(your_fcn_name(x),y_correct))
y = [] y = 0×1 empty double column vector y = []
2 Pass
x = [1 2 3 3 4 5 6 6 7]; y_correct = [3,6]; assert(isequal(your_fcn_name(x),y_correct))
y = [] y = 0×1 empty double column vector y = 0×1 empty double column vector y = 3 y = 3 y = 3 y = 3 y = 3 y = 3 6 y = 3 6 y = 3 6 y = 3 6
3 Pass
x = [1 1 1 1 0 1 1 2 2 9]; y_correct = [1,2]; assert(isequal(your_fcn_name(x),y_correct))
y = [] y = 1 y = 1 1 y = 1 1 1 y = 1 1 1 1 y = 1 1 1 1 1 y = 1 y = 1 1 y = 1 1 1 y = 1 1 1 1 y = 1 1 1 1 1 y = 1 y = 1 1 y = 1 1 1 y = 1 1 1 1 y = 1 y = 1 1 y = 1 1 1 y = 1 y = 1 y = 1 1 y = 1 y = 1 y = 1 2 y = 1 2 y = 1 2 y = 1 2
4 Pass
x= [1 7 8 2 7 8 3]; y_correct = [7,8]; assert(isequal(your_fcn_name(x),y_correct))
y = [] y = 0×1 empty double column vector y = 7 y = 7 y = 7 8 y = 7 8 y = 7 8 y = 7 8 y = 7 8 y = 7 8
5 Pass
x= [0 1 0]; y_correct = [0]; assert(isequal(your_fcn_name(x),y_correct))
y = [] y = 0 y = 0 y = 0 y = 0 | 605 | 1,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.525924 |
https://www.unitsconverters.com/en/Pgy-To-Sv/Utu-3975-3988 | 1,653,700,759,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663011588.83/warc/CC-MAIN-20220528000300-20220528030300-00142.warc.gz | 1,192,321,379 | 35,421 | Formula Used
1 Gray = 1E-15 Petagray
1 Gray = 1 Sievert
1 Petagray = 1E+15 Sievert
PGy to Sv Conversion
The abbreviation for PGy and Sv is petagray and sievert respectively. 1 PGy is 1E+15 times bigger than a Sv. To measure, units of measurement are needed and converting such units is an important task as well. unitsconverters.com is an online conversion tool to convert all types of measurement units including PGy to Sv conversion.
Petagray to Sv
Check our Petagray to Sv converter and click on formula to get the conversion factor. When you are converting absorbed dose from Petagray to Sv, you need a converter that is elaborate and still easy to use. All you have to do is select the unit for which you want the conversion and enter the value and finally hit Convert.
PGy to Sievert
The formula used to convert PGy to Sievert is 1 Petagray = 1E+15 Sievert. Measurement is one of the most fundamental concepts. Note that we have Fahrenheit as the biggest unit for length while Joule Per Celsius Per Decamole is the smallest one.
Convert PGy to Sv
How to convert PGy to Sv? Now you can do PGy to Sv conversion with the help of this tool. In the length measurement, first choose PGy from the left dropdown and Sv from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from Sv to PGy? You can check our Sv to PGy converter.
PGy to Sv Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like absorbed dose finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like PGy to Sv through multiplicative conversion factors. When you are converting absorbed dose, you need a Petagray to Sievert converter that is elaborate and still easy to use. Converting PGy to Sievert is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert Petagray to Sv, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in PGy to Sv conversion along with a table representing the entire conversion.
Let Others Know | 543 | 2,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.876289 |
http://en.m.wikipedia.org/wiki/Superalgebra | 1,371,676,377,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709224828/warc/CC-MAIN-20130516130024-00093-ip-10-60-113-184.ec2.internal.warc.gz | 81,597,578 | 9,635 | # Superalgebra
In mathematics and theoretical physics, a superalgebra is a Z2-graded algebra.[1] That is, it is an algebra over a commutative ring or field with a decomposition into "even" and "odd" pieces and a multiplication operator that respects the grading.
The prefix super- comes from the theory of supersymmetry in theoretical physics. Superalgebras and their representations, supermodules, provide an algebraic framework for formulating supersymmetry. The study of such objects is sometimes called super linear algebra. Superalgebras also play an important role in related field of supergeometry where they enter into the definitions of graded manifolds, supermanifolds and superschemes.
## Formal definition
Let K be a fixed commutative ring. In most applications, K is a field such as R or C.
A superalgebra over K is a K-module A with a direct sum decomposition
$A = A_0\oplus A_1$
together with a bilinear multiplication A × AA such that
$A_iA_j \sube A_{i+j}$
where the subscripts are read modulo 2.
A superring, or Z2-graded ring, is a superalgebra over the ring of integers Z.
The elements of Ai are said to be homogeneous. The parity of a homogeneous element x, denoted by |x|, is 0 or 1 according to whether it is in A0 or A1. Elements of parity 0 are said to be even and those of parity 1 to be odd. If x and y are both homogeneous then so is the product xy and $|xy| = |x| + |y|.$
An associative superalgebra is one whose multiplication is associative and a unital superalgebra is one with a multiplicative identity element. The identity element in a unital superalgebra is necessarily even. Unless otherwise specified, all superalgebras in this article are assumed to be associative and unital.
A commutative superalgebra is one which satisfies a graded version of commutativity. Specifically, A is commutative if
$yx = (-1)^{|x||y|}xy\,$
for all homogeneous elements x and y of A.
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## Examples
• Any algebra over a commutative ring K may be regarded as a purely even superalgebra over K; that is, by taking A1 to be trivial.
• Any Z or N-graded algebra may be regarded as superalgebra by reading the grading modulo 2. This includes examples such as tensor algebras and polynomial rings over K.
• In particular, any exterior algebra over K is a superalgebra. The exterior algebra is the standard example of a supercommutative algebra.
• The symmetric polynomials and alternating polynomials together form a superalgebra, being the even and odd parts, respectively. Note that this is a different grading from the grading by degree.
• Clifford algebras are superalgebras. They are generally noncommutative.
• The set of all endomorphisms (both even and odd) of a super vector space forms a superalgebra under composition.
• The set of all square supermatrices with entries in K forms a superalgebra denoted by Mp|q(K). This algebra may be identified with the algebra of endomorphisms of a free supermodule over K of rank p|q.
• Lie superalgebras are a graded analog of Lie algebras. Lie superalgebras are nonunital and nonassociative; however, one may construct the analog of a universal enveloping algebra of a Lie superalgebra which is a unital, associative superalgebra.
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## Further definitions and constructions
### Even subalgebra
Let A be a superalgebra over a commutative ring K. The submodule A0, consisting of all even elements, is closed under multiplication and contains the identity of A and therefore forms a subalgebra of A, naturally called the even subalgebra. It forms an ordinary algebra over K.
The set of all odd elements A1 is an A0-bimodule whose scalar multiplication is just multiplication in A. The product in A equips A1 with a bilinear form
$\mu:A_1\otimes_{A_0}A_1 \to A_0$
such that
$\mu(x\otimes y)\cdot z = x\cdot\mu(y\otimes z)$
for all x, y, and z in A1. This follows from the associativity of the product in A.
There is a canonical involutive automorphism on any superalgebra called the grade involution. It is given on homogeneous elements by
$\hat x = (-1)^{|x|}x$
and on arbitrary elements by
$\hat x = x_0 - x_1$
where xi are the homogeneous parts of x. If A has no 2-torsion (in particular, if 2 is invertible) then the grade involution can be used to distinguish the even and odd parts of A:
$A_i = \{x \in A : \hat x = (-1)^i x\}.$
### Supercommutativity
The supercommutator on A is the binary operator given by
$[x,y] = xy - (-1)^{|x||y|}yx\,$
on homogeneous elements. This can be extended to all of A by linearity. Elements x and y of A are said to supercommute if [x, y] = 0.
The supercenter of A is the set of all elements of A which supercommute with all elements of A:
$Z(A) = \{a\in A : [a,x]=0 \text{ for all } x\in A\}.$
The supercenter of A is, in general, different than the center of A as an ungraded algebra. A commutative superalgebra is one whose supercenter is all of A.
### Super tensor product
The graded tensor product of two superalgebras A and B may be regarded as a superalgebra AB with a multiplication rule determined by:
$(a_1\otimes b_1)(a_2\otimes b_2) = (-1)^{|b_1||a_2|}(a_1a_2\otimes b_1b_2).$
If either A or B is purely even, this is equivalent to the ordinary ungraded tensor product (except that the result is graded). However, in general, the super tensor product is distinct from the tensor product of A and B regarded as ordinary, ungraded algebras.
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## Generalizations and categorical definition
One can easily generalize the definition of superalgebras to include superalgebras over a commutative superring. The definition given above is then a specialization to the case where the base ring is purely even.
Let R be a commutative superring. A superalgebra over R is a R-supermodule A with a R-bilinear multiplication A × AA that respects the grading. Bilinearity here means that
$r\cdot(xy) = (r\cdot x)y = (-1)^{|r||x|}x(r\cdot y)$
for all homogeneous elements rR and x, yA.
Equivalently, one may define a superalgebra over R as a superring A together with an superring homomorphism RA whose image lies in the supercenter of A.
One may also define superalgebras categorically. The category of all R-supermodules forms a monoidal category under the super tensor product with R serving as the unit object. An associative, unital superalgebra over R can then be defined as a monoid in the category of R-supermodules. That is, a superalgebra is an R-supermodule A with two (even) morphisms
\begin{align}\mu &: A\otimes A \to A\\ \eta &: R\to A\end{align}
for which the usual diagrams commute.
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## Notes
1. ^ Kac, Martinez & Zelmanov (2001), p. 3.
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## References
• Deligne, Pierre; John W. Morgan (1999). "Notes on Supersymmetry (following Joseph Bernstein)". Quantum Fields and Strings: A Course for Mathematicians 1. American Mathematical Society. pp. 41–97. ISBN 0-8218-2012-5.
• Manin, Y. I. (1997). Gauge Field Theory and Complex Geometry ((2nd ed.) ed.). Berlin: Springer. ISBN 3-540-61378-1.
• Varadarajan, V. S. (2004). Supersymmetry for Mathematicians: An Introduction. Courant Lecture Notes in Mathematics 11. American Mathematical Society. ISBN 0-8218-3574-2.
• Kac, Victor G.; Martinez, Consuelo; Zelmanov, Efim (2001). Graded simple Jordan superalgebras of growth one. Memoirs of the AMS Series 711. AMS Bookstore. ISBN 978-0-8218-2645-4.
↑Jump back a section | 1,970 | 7,425 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 14, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.905772 |
https://www.coursehero.com/file/6534746/hw1/ | 1,516,372,114,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887981.42/warc/CC-MAIN-20180119125144-20180119145144-00214.warc.gz | 886,254,188 | 63,814 | # hw1 - 20.181.hw1 file/Users/endy/Desktop/hw1.html...
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20.181.hw1 file:///Users/endy/Desktop/hw1.html Biological Computation 20.181 Homework 1 Review of Recursion. In class we looked at a recursive definition for the factorial function f(n) = n! = n*(n-1)*(n-2)*. ..*1. Write a recrusive function that takes three numbers -- k,first,second -- as input, and returns a list including the first k terms of the Fibonacci series, using the recursive definition F(n+1) = F(n) + F(n-1), and the initial conditions F(1) = first and F(2) = second. Your solution should be of the following form: def fib(k,first,second): #return fibonacci series up to k terms #insert code here Introduction to phylogenies. Consider the following five sequences: 1. AAGGCCCACTA 2. GATGTCCGATA 3. AAGGCCCACTT 4. AATGGCCCCTA 5. GATGTCCGATA Compute all pairwise distances between these sequences in terms of number of mismatched base pairs (hint: sequences 1 and 2 have 5 mismatches, so their distance is 5). The UPGMA algorithm (which stands for unweighted pair group method using arithmetic averages) is
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Ask a homework question - tutors are online | 418 | 1,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.807253 |
https://easy-due.com/%E6%95%B0%E5%AD%A6%E4%BB%A3%E5%86%99%EF%BD%9Ccs166-problem-set-4/ | 1,695,491,823,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506528.19/warc/CC-MAIN-20230923162848-20230923192848-00323.warc.gz | 261,607,360 | 21,147 | ## Problem One: Very Dynamic Prefix Parity
On Problem Set Two, you designed a data structure for the dynamic prefix parity problem. This data structure supports the following operations:
• initialize(n), which creates a new data structure representing an array of n bits, all initially zero. This takes time O(n).
• ds.flip(i), which flips the ith bit of the bitstring. This takes time O(log n / log log n).
• ds.prefix-parity(i), which returns the parity of the subarray consisting of the first i bits of the array. (The parity is 0 if there are an even number of one bits and 1 if there are an odd number of 1 bits). This has the same runtime as the flip operation, O(log n / log log n).
Now, consider the very dynamic prefix parity problem (or VDDP). In this problem, you don’t begin with a fixed array of bits, but instead start with an empty sequence. You then need to support these operations:
• initialize(), which constructs a new, empty VDPP structure.
• ds.append(b), which appends bit b to the bitstring.
• ds.flip(i), which flips the ith bit of the bitstring.
• ds.prefix-parity(i), which returns the parity of the subarray consisting of the first i bits of the array.
Design a data structure for VDPP such that all three operations run in amortized time O(log n / log log n),where n is the number of bits in the sequence at the time the operation is performed.
Briefly argue correctness, and prove that you meet the required amortized time bounds by defining a potential function Φ and working out the amortized costs of each operation. We’re expecting your amortized analysis to be written up in a manner similar to the formal analyses we did in lecture of the two-stack queue, dynamic array, and B-tree construction algorithm.
## Problem Two: Meldable Heaps with Addition
Meldable priority queues support the following operations:
new-pq(), which constructs a new, empty priority queue;
• pq.insert(v, k), which inserts element v with key k;
• pq.find-min(), which returns an element with the least key;
• pq.extract-min(), which removes and returns an element with the least key; and
• meld(pq₁, pq₂), which destructively modifies priority queues pq₁ and pq₂ and produces a single priority queue containing all the elements and keys from pq₁ and pq₂.
Some graph algorithms also require the following operation:
• pq.add-to-all(Δk), which adds Δk to the keys of each element in the priority queue.
Although meldable priority queue have n nodes in them, it’s possible to implement add-to-all without touching all of these nodes.
i.Show how to modify an eager (non-lazy) tournament heap so that new-pq, insert, find-min,extract-min, meld, and add-to-all each run in worst-case O(log n) time. Briefly argue correctness.
A note on this problem: if it weren’t for meld, you could support add-to-all in time O(1). (Do you see why?) We recommend that, from the beginning, you think about how you’d meld together two different heaps that have had different Δk’s added to them.
ii.Show how to modify a lazy tournament heap so that new-pq, insert, find-min, meld, and addto-all all run in amortized time O(1) and extract-min runs in amortized time O(log n). Briefly argue correctness, then do an amortized analysis with the potential method to prove runtime bounds. Some hints:
1. Try to make all operations have worst-case runtime O(1) except for extract-min. Your implementation of extract-min will probably do a lot of work, but if you’ve set it up correctly the amortized cost will only be O(log n). This means, in particular, that you will only propagate the Δk‘s through the data structure in extract-min.
1. If you only propagate Δk‘s during an extract-min as we suggest, you’ll run into some challenges trying to meld two lazy tournament heaps with different Δk‘s. To address this, we recommend that you change how meld is done to be even lazier than the lazy approach we discussed in class. You might find it useful to construct a separate data structure tracking the melds that have been done and then only actually combining together the heaps during an extract-min.
1. Depending on how you set things up, to get the proper amortized time bound for extractmin, you may need to define your potential function both in terms of the structure of the lazy tournament heaps and in terms of the auxiliary data structure hinted at by the previous point.
In your writeup, don’t just describe the final data structure all at once. Instead, walk us through the design.
Explain why each piece is there, why it’s needed, and how the whole structure comes together. Briefly argue correctness, and prove that you meet the required amortized time bounds by defining a potential function Φ and working out the amortized costs of each operation.
We’re expecting your amortized analysis to be written up in a manner similar to the formal analyses we did in lecture of the two-stack queue, dynamic array, and B-tree construction algorithm.
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EasyDue™是一个服务全球中国留学生的专业代写公司 | 1,151 | 5,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-40 | latest | en | 0.864943 |
http://www.education.com/activity/article/rebus-father-day-message/ | 1,469,435,500,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824217.36/warc/CC-MAIN-20160723071024-00093-ip-10-185-27-174.ec2.internal.warc.gz | 419,756,899 | 32,437 | # Father's Day Rebus Puzzle Activity
3.3 based on 39 ratings
Updated on Sep 5, 2013
Looking for a unique Father’s Day card or poster to make with your child? This one will double as a card and an exercise in playing with words. More importantly, it will make Dad laugh!
### What You Need:
• Construction paper
• Magazines
• Scissors
• Glue
• Markers, crayons, pens
### What You Do:
1. Explain to your child that a rebus is a message that you send in pictures. Give your child an example of a rebus by drawing an eye, a heart, and the letter “U” on a piece of paper. Ask your child to figure out the hidden message and explain that you'll be making a Father’s Day card using the same technique.
2. Help your child think about the type of message she'd like to send Dad for Father’s Day. Does she want to thank him? Tell him how much she loves him? Tell him what a great dad he is?
3. Give your child the magazines and help her find at least two pictures that might fit into the message that she wants to make. These pictures should have several letters of a word that might work well on the card. For example, a picture of a cluster of grapes might make her think of using the word “great,” or a picture of a vest might make her think of using the word “best.” Make sure to point out that you can add or subtract letters from a word in a rebus. For example, you might write B + [picture of a vest] – V to make the word “best.”
4. Show your child how to map out the message she wants on her card. To do this make a draft: write out each of the pictures that you've found and make a sentence around them. Brainstorm other pictures or single letters and abbreviations you can use to spell out the rest of the sentence. If you can’t think of any way to represent a message with letters, numbers or pictures, change it slightly until you can.
5. Cut out the pictures from the magazine as well as any large letters or numbers you'll need for your message, and place them on the construction paper. If you can’t find any of the pictures that you want, your child can draw them around the magazine clippings in the correct spots.
6. Glue the magazine pictures into the correct places on the message. Add plus and minus signs between the pictures, letters and numbers as you need them.
7. Place the message on Dad’s chair or simply let your child hand it to him on Father’s Day. Dad will enjoy figuring out the secret message your child has written him! | 568 | 2,449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2016-30 | longest | en | 0.942714 |
https://www.transtutors.com/questions/analysis-126539.htm | 1,591,455,730,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348513321.91/warc/CC-MAIN-20200606124655-20200606154655-00386.warc.gz | 911,028,903 | 17,040 | What types of analysis does a restaurant utilize? (Analysis doesn"t have to be limited to financial analysis).
Mark B
Cost-volume-profit (CVP) analysis expands the use of information provided by breakeven analysis. A critical part of CVP analysis is the point where total revenues equal total costs (both fixed and variable costs). At this breakeven point (BEP), a company will experience no income or loss. This BEP can be an initial examination that precedes more detailed CVP analyses. Cost-volume-profit analysis employs the same basic assumptions as in breakeven analysis. The assumptions underlying CVP analysis are: The behavior of both costs and revenues in linear throughout the relevant range of activity. (This assumption precludes the concept of volume discounts on either purchased materials or sales.) Costs can be classified accurately as either fixed or variable. Changes in activity are the only factors that affect costs. All units produced are sold (there is no ending finished goods inventory). When a company sells more than one type of product, the sales mix (the ratio of each product to total sales) will remain constant. Figure 1 Cost-Volume-Profit Analysis, Production = Sales In the following discussion, only one product will be assumed. Finding the breakeven point is the initial step in CVP, since it is critical to know whether sales at a given level will at least cover the relevant costs. The breakeven point can be determined with a mathematical equation, using contribution margin, or from a CVP graph. Begin by observing the CVP graph in Figure 1, where the number of units produced...
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Free Plagiarism Checker | 352 | 1,721 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.889915 |
http://www.gurufocus.com/term/Total%20Liabilities/LAYN/Total%2BLiabilities/Layne%2BChristensen%2BCompany | 1,493,335,433,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122629.72/warc/CC-MAIN-20170423031202-00348-ip-10-145-167-34.ec2.internal.warc.gz | 537,354,943 | 28,315 | Switch to:
GuruFocus has detected 3 Warning Signs with Layne Christensen Co \$LAYN.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
Layne Christensen Co (NAS:LAYN)
Total Liabilities
\$353.9 Mil (As of Jan. 2017)
Layne Christensen Co's total liabilities for the quarter that ended in Jan. 2017 was \$353.9 Mil.
Layne Christensen Co's quarterly total liabilities increased from Jul. 2016 (\$355.78 Mil) to Oct. 2016 (\$358.94 Mil) but then declined from Oct. 2016 (\$358.94 Mil) to Jan. 2017 (\$353.93 Mil).
Layne Christensen Co's annual total liabilities declined from Jan. 2015 (\$360.73 Mil) to Jan. 2016 (\$360.00 Mil) and declined from Jan. 2016 (\$360.00 Mil) to Jan. 2017 (\$353.93 Mil).
Definition
Total liabilities are the liabilities that the company has to pay others. It is a part of the balance sheet of a company that shareholders do not own, and would be obligated to pay back if the company liquidated.
Layne Christensen Co's Total Liabilities for the fiscal year that ended in Jan. 2017 is calculated as
Total Liabilities = Total Current Liabilities + Total Long Term Liabilities = Total Current Liabilities + (Long-Term Debt + Other Long-Term Liab. = 144.938 + (162.346 + 42.4 + NonCurrentDeferredLiabilities + PensionAndRetirementBenefit + Minority Interest) + 4.199 + 0 + 0.048) = 353.9
Total Liabilities = Total Assets (A: Jan. 2017 ) - Total Shareholders Equity (A: Jan. 2017 ) = 436.151 - 82.22 = 353.9
Layne Christensen Co's Total Liabilities for the quarter that ended in Jan. 2017 is calculated as
Total Liabilities = Total Current Liabilities + (Total Long Term Liabilities) = Total Current Liabilities + (Long-Term Debt + Other Long-Term Liab. = 144.938 + (162.346 + 42.4 + NonCurrentDeferredLiabilities + PensionAndRetirementBenefit + Minority Interest) + 4.199 + 0 + 0.048) = 353.9
Total Liabilities = Total Assets (Q: Jan. 2017 ) - Total Shareholders Equity (Q: Jan. 2017 ) = 436.151 - 82.22 = 353.9
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Layne Christensen Co Annual Data
Jan08 Jan09 Jan10 Jan11 Jan12 Jan13 Jan14 Jan15 Jan16 Jan17 Total Liabilities 273.6 263.3 264.2 315.3 357.2 399.5 357.2 360.7 360.0 353.9
Layne Christensen Co Quarterly Data
Oct14 Jan15 Apr15 Jul15 Oct15 Jan16 Apr16 Jul16 Oct16 Jan17 Total Liabilities 386.3 360.7 399.9 365.8 374.4 360.0 360.4 355.8 358.9 353.9
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GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 831 | 2,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | latest | en | 0.836426 |
https://www.kiglobalhealth.org/resources/non-linear-mixed-effects-models/ | 1,725,729,054,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00471.warc.gz | 827,140,171 | 16,077 | # Non-linear mixed effects models
• Method
Why does Ki use non-linear mixed effects (NLME) models? Nonlinear models provide a more flexible framework for integration of a wide range of geographically and socioeconomically diverse longitudinal data. This approach captures relationships between predictor variable(s) and outcomes and can be useful in circumstances when a linear model does not provide a good fit for the predictor-outcome relationship. For example, one well-known model for early childhood growth is the Jenss-Bayley model[1] (Figure 1). Mixed effects methodology incorporates variations in predictor variables that occur across a population, as well as repeated measures within individuals.
### WHAT IS A NLME MODEL?
NLME models “accommodate individual variations through random effects but ties different individuals together through population level fixed effects.”[3]
A non-linear model has model parameters which define the shape of the mean response. For the Jenss-Bayley model in Figure 1 the spurt of growth parameter (CW) is displayed, and the model includes parameters for birth length (AW), growth velocity (BW), and curvature degree (DW) with the subscript “i” for individual measures.
#### Weighti,j= exp(AWi) + exp(BWi) *ti,j+ exp(CWi) * (1–exp(–exp (DWi) *ti,j)) + ei,j
In the mixed effects modeling paradigm, each unit of observation (e.g., a child in a birth cohort study) has unit-specific parameters.
Model parameters are called “mixed” effects because they include fixed effects and random effects.[4]
• Fixed Effects are components that are assumed to be the same value for each individual in the population. For example, in the Jenss-Bayley model, the average birth length (Aw) and growth velocity (BW) of a study population are fixed effects.
• For example, in the Jenss-Bayley model, the average birth length (Aw) and growth velocity (BW) of a study population are fixed effects.
• Random Effects are individual-based components associated with how individuals differ from the population average effects.
• For example, in the Jenss-Bayley model individual BMI trajectories incorporate individual weight parameters and deviation from average population weight parameters to calculate between-individual variability.
• Random effects are typically assumed to follow a parametric (e.g., Gaussian) distribution.
### KIUTILIZATION OF NLME MODELS
#### Ki has uses NLME in various ways including parametric non-linear model, full random effects model (FREM), and Bayesian NLME model.
The parametric NLME approach is the predominant statistical approach found in the literature. Ki also has implemented more novel approaches in global health like FREM and Bayesian NLME in child growth and cognitive development applications.
#### Full Random Effects Model (FREM)
1. FREM incorporates random variation between multiple covariates and within individual observations of a single covariate (using variance and covariance estimates) to model the outcome as a function of the covariate(s).[4]
2. FREM does not include fixed effects.
3. An advantage to FREM is its ability to conduct analysis with missing covariate values.
#### Bayesian non-linear mixed effects model
This model differs from other NLME models in that it takes a Bayesian approach, rather than a frequentist approach.
• Interpretability: The mathematical equations for NLME models can be derived from biological/mechanistic understanding, meaning they can model a pre-determined biological relationship between variables in the dataset. These mechanistic models are designed to assess causality.
• Compared to a polynomial model, an NLME derived from biological/mechanistic knowledge provides more reliable predictions for outcome variables outside of the observed range of the data.
• Flexible modeling framework: Unlike linear models, NLME can more closely reflect the predictor-outcome relationship, because it is not dependent on a straight-line relationship.
• A simulated function (at each time step for a repeated measure dataset) is required to produce a numerical estimate.
• NLME is computationally intensive.
The Frequentist Approach utilizes assumptions of the predictor-outcome relationship between the data within your sample and the population overall. These parameters are fixed over repeated random samples.
The Bayesian Approach leverages prior knowledge of the model or its parameters, allowing parameters to vary based on the distribution of repeated random samples.
Example: Extensive developmental research and improvements in imaging technology have informed head circumference data, such that additional parameter information by gender and geography can predict estimated gestational age. Variations in the head circumference distribution based on gender and geography provide a better model fit for data.
Parametric Statistical Methods assume a mathematical and statistical structure of a hypothesized relationship between outcome and predictor variables.
The statistical structure frequently involves assumptions about the distribution of both the random effects (e.g., how the individual linear growth rates vary around the mean growth rate) and the residual errors (e.g., how the observed data vary around the model predictions).
These assumptions allow estimation of both the fixed and random effects.
This assumes the data in your sample is reflective of the population from which it was sampled.
For example, sample data assessing the impact of maternal smoking on growth trajectories among children in France assumes the established relationship of increased risk of obesity in adulthood with maternal smoking during pregnancy.[2]
### References
1. Jenss R, Bayley N. A Mathematical Method For Studying The Growth of A Child. Human Biology. 1937;9(4):556-63.
2. Carles S, Charles M-A, Forhan A, et al. A Novel Method to Describe Early Offspring Body Mass Index (BMI) Trajectories and to Study Its Determinants. PloS one. 2016;11(6):e0157766.
3. Pinheiro JC, Bates DM. Mixed- Effects Models in S and S-PLUS. New York: Springer; 2000.
4. MathWorks. Nonlinear Mixed- Effects Modeling. 2017; https:// www.mathworks.com/help/ simbio/ug/what-is-nonlinear-mixed-effects-modeling.html. Accessed October 5, 2017.
October, 2020 | 1,262 | 6,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.85283 |
https://www.coursehero.com/file/6850048/Study20Tips1doc/ | 1,529,350,617,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860776.63/warc/CC-MAIN-20180618183714-20180618203714-00308.warc.gz | 794,274,595 | 49,436 | Study%20Tips.1doc
# Study%20Tips.1doc - The Euclidean Algorithm and its uses...
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Study Tips (Midterm I) 1. “Proof” vaaaaaar ..... (bu ders boyle, ama ileride cok faydasini goreceksiniz, gercekten) 2. Make sure that you know/understand; The algorithms we have covered What “complexity” means, how one finds a O- estimate for it What we mean by “growth of a function” Prime numbers, their properties That, “every integer >1 has a prime factor”, and why it is so. The division algorithm and why it works
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Unformatted text preview: The Euclidean Algorithm and its uses Some modular arithmetic- what we did in the class and also at the recitation classes What the Fundamental Thm of Arithmetic says, and the proof of it How to prove that there are infinitely may primes How to use the “induction principle” 3. Make sure that you can solve all the assigned exercises and preferably more…....
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https://calculator.academy/dbm-to-volts-calculator/ | 1,701,179,128,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099514.72/warc/CC-MAIN-20231128115347-20231128145347-00077.warc.gz | 189,459,345 | 47,925 | Enter the power level and resistance into the calculator to determine the voltage. This calculator can also evaluate any of the variables given the others are known.
## Dbm To Volts Formula
The following formula is used to calculate the voltage from dBm.
V = sqrt((10^{(dBm/10)}) * 0.001 * R)
Variables:
• V is the voltage (Volts)
• dBm is the power level (decibels-milliwatts)
• R is the resistance (Ohms)
To calculate the voltage, first divide the power level by 10. Then, raise 10 to the power of the result. Multiply this by 0.001 and then by the resistance. Finally, take the square root of the result to get the voltage.
## What is Dbm To Volts?
Dbm to Volts is a conversion process used in electrical engineering to change a measurement of power from decibels-milliwatts (dBm) to voltage (Volts). dBm is a unit of power that represents the ratio of a power level relative to a reference level of 1 milliwatt, expressed in decibels. The conversion is necessary because these two units measure different aspects of an electrical signal, and converting between them allows for more comprehensive analysis and understanding of the signal’s behavior.
## How to Calculate Dbm To Volts?
The following steps outline how to calculate the Dbm To Volts using the given formula:
1. First, determine the power level in decibels-milliwatts (dBm).
2. Next, raise 10 to the power of (dBm/10).
3. Multiply the result by 0.001.
4. Multiply the previous result by the resistance in Ohms (R).
5. Take the square root of the previous result.
6. Finally, calculate the voltage in Volts (V).
Example Problem:
Use the following variables as an example problem to test your knowledge:
power level (dBm) = 20
resistance (Ohms) = 100 | 422 | 1,728 | {"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} | 3.984375 | 4 | CC-MAIN-2023-50 | latest | en | 0.851519 |
https://www.dsp-weimich.com/programming/functions-implementation-square-root-logarithm-sine-cosine-arctangent-c-code-and-octave-script/ | 1,702,286,955,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103810.88/warc/CC-MAIN-20231211080606-20231211110606-00740.warc.gz | 785,359,617 | 24,868 | ## Functions Implementation: Square Root, Logarithm, Sine, Cosine, Arctangent. C Code and Octave Script
### 3.1. Square Root. Taylor/Maclaurin series. Chebyshev Approximation
##### Using the Octave script FunctionsRealization.m (see the array sqrt_relative_mistake2) is calculated the relative error: Δsqrt(1+x)/sqrt (1+x) by the formula (9) in the range x = -0.3…0.4. See the Figure 3-1. In this case the maximum relative error reaches:
Figure 3-1: Relative mistake: Δsqrt(1+x)/sqrt(1+x) of the square root calculation with 4 elements of the Maclaurin Series
##### To reduce the number of multiplications in (18), another entry is used:
Figure 3-2: Relative mistake: Δsqrt(1+x)/sqrt(1+x) of the square root calculation with 4 elements of the Chebyshev Approximation
### 4. Logarithm
##### Using the Octave script FunctionsRealization.m (see the ln_absolute_mistake) is calculated the absolute error: Δln (1+x) by (28) in the range x = -0.32…0.36. See the Figure 4-1. In this case the maximum absolute error reaches:
Figure 4-1: Absolute mistake: Δln(1+x) of the natural logarithm calculation with 3 elements of the Maclaurin Series
##### To reduce the number of multiplications in (35), another entry is used:
Figure 4-2: Absolute mistake: Δln(1+x) of the natural logarithm calculation with 3 elements of the Chebyshev Approximation
### 5.1 Sine
##### Using the Octave script FunctionsRealization.m (see the sin_absolute_mistake) is calculated the absolute error: Δsin(x) by (41) in the range x = -π…π. See the Figure 5-1. In this case the maximum absolute error reaches:
Figure 5-1: Absolute mistake: Δsin(x) of the sine calculation with 4 elements of the Maclaurin Series
##### To reduce the number of multiplications in (47), another entry is used:
Figure 5-2: Absolute mistake: Δsin(x) of the sine calculation with 4 elements of the Chebyshev Approximation
### 6. Arctangent
##### Using Octave script FunctionsRealization. m ( see the sin_absolute_mistake) is calculated the absolute error: Δarctan(x) by (57) in the range x = -1…1. See the Figure 6-1. In this case the maximum absolute error reaches:
Figure 6-1: Absolute mistake: Δatan(x) of the sine calculation with 4 elements of the Maclaurin Series
##### Comparing (58) and (64) the error has decreased by more than 590 times!
Figure 6-2: Absolute mistake: Δatan(x) of the sine calculation with 4 elements of the Chebyshev Approximation
### 8.1 square_root.c/h
##### square_root.c/h is supported square root functions
/*
Square Root
Float Arithmetic
Polynomial sqrt(1+x) = 1 + x (0.50016487 + x (-0.12822807 + 0.059227649 x))
0.7 < 1 + x <= 1.4
-0.3 < x <= 0.4
Input
float input – input value
Return
float – sqrt(input) Note: if input number is 0.0 or negative then return value is 0.0
*/
float sgrt_polynomial_float(float input);
/*
Square Root
Fixed Point Arithmetic
Polynomial sqrt(1+x) = 1 + x (0.50016487 + x (-0.12822807 + 0.059227649 x))
0.7 < 1 + x <= 1.4
-0.3 < x <= 0.4
Input
uint16 input – input value: 0…65535
Return
uint16 – sqrt(input) in format: b15*2^7 + b14*2^6 + … + b8*2^0 + b7*2^-1 + b6*2^-2 + … + b0*2^-8
Note: Result = ReturnValue/OUT_SQRT_FACTOR, where OUT_SQRT_FACTOR = 256
*/
uint16 sgrt_polynomial_int(uint16 input);
/*
Square Root
Float Arithmetic
Newton Method x(n+1) = (x(n)+ a/x(n))/2
Input
float input – input value
Return
float – sqrt(input) Note: if input number is 0.0 or negative then return value is 0.0
*/
float sgrt_Newton_float(float input);
/*
Square Root
Fixed Point Arithmetic
Newton Method x(n+1) = (x(n)+ a/x(n))/2
Input
uint16 input – input value: 0…65535
Return
uint16 – sqrt(input) in format: b15*2^7 + b14*2^6 + … + b8*2^0 + b7*2^-1 + b6*2^-2 + … + b0*2^-8
Note: Result = ReturnValue/OUT_SQRT_FACTOR, where OUT_SQRT_FACTOR = 256
*/
uint16 sgrt_Newton_int(uint16 input);
##### The square root functions were tested in main() using the test arrays: input_array_float[ ] and input_array_int[ ]. The test results are shown in Figure 8-1, Figure 8-2, Figure 8-3, Figure 8-4
Figure 8-1: Relative mistake: Δsqrt(x)/sqrt(x) of the sgrt_polynomial_float() for x=1…100. Max relative mistake is 0.000163
Figure 8-2: Relative mistake: Δsqrt(x)/sqrt(x) of the sgrt_polynomial_int() for x= 65535, 1, 2, … 100. Max relative mistake is 0.001302
Figure 8-3: Relative mistake: Δsqrt(x)/sqrt(x) of the sgrt_Newton_float() for x=1…100. Max relative mistake is 0.0000000946
Figure 8-4: Relative mistake: Δsqrt(x)/sqrt(x) of the sgrt_Newton_int() for x= 65535, 1, 2, … 100. Max relative mistake is 0.000913
### 8.2 logarithm.c/h
##### logarithm.c/h is supported logarithm functions
/*
Square Root
Float Arithmetic
Polynomial log10(1+x) = log10(e) * ln(1+x) = 0.43429449 ( x (1.0004359 + x (-0.52194273 + 0.33578411 x)))
0.68 < 1 + x <= 1.36
-0.32 < x <= 0.36
Input
float input – input value
Return
float – log10(input) Note: if input number is 0.0 or negative then return value is 0.0
*/
float logarithm10_float(float input);
/*
Decimal Logarithm
Fixed Point Arithmetic
Polynomial log10(1+x) = log10(e) * ln(1+x) = 0.43429449 ( x (1.0004359 + x (- 0.52194273 + 0.33578411 x)))
0.68 < 1 + x <= 1.36
-0.32 < x <= 0.36
Input
uint16 input – input value: 1…65535 => log10(1)=0 … log10(65535)=4.8164735
Return
sint16 – log10(input) in format: b14*2^2 + b13*2^1 + b12*2^0 + b11*2^-1 + … + b0*2^-12
Note: Result = ReturnValue/OUT_LOGARITHM_FACTOR, where OUT_LOGARITHM_FACTOR = 4096
*/
sint16 logarithm10_int(uint16 input);
##### The logarithm10 functions were tested in main() using the test arrays: input_array_float[ ] and input_array_int[ ]. The test results are shown in Figure 8-5, Figure 8-6
Figure 8-5: Absolute mistake: Δlogarithm of the logarithm10_float() for x= 0.1, 1, 2, …, 99, 100. Max absolute mistake is 0.000365
Figure 8-6: Absolute mistake: Δlogarithm of the logarithm10_int() for x= 65535, 1, 2, …, 99, 100. Max absolute mistake is 0.000539
### 8.3 sine.c/h
##### Sine.c/h is supported sine and cosine functions
/*
Sine
Float Arithmetic
Polynomial sin(x) = x (0.9992497 + x^2 (-0.16564582 + x^2 (7.9537760e-3 – 1.4482903e-4 x^2)))
-pi <= x <= pi
Input
float input – input value
Return
float – sin(input) Note: if input > pi or input < -pi then return value is 0.0
*/
float sin_float(float input);
/*
Cosine
Float Arithmetic
cos(x) = sin(x + pi/2)
-pi <= x <= pi
Input
float input – input value
Return
float – cos(input) Note: if input > pi or input < -pi then return value is 0.0
*/
float cos_float(float input);
/*
Sine
Fixed Point Arithmetic
Polynomial sin(x) = x (0.9992497 + x^2 (-0.16564582 + x^2 (7.9537760e-3 – 1.4482903e-4 x^2)))
-pi <= x <= pi
Input
sint16 input – input value: -PI_INT_VALUE=-25736 … PI_INT_VALUE=25736
input format 3Q13: b14*2^1 + b13*2^0 + b12*2^-1 + b11*2^-2 + … + b0*2^-13 and b15-sign
Return
sint16 – sin(input) in format 1Q15: b14*2^-1 + b13*2^-2 + b12*2^-3 + b11*2^-4 + … + b0*2^-15 and b15-sign
Note: Result = ReturnValue/OUT_SINE_FACTOR, where OUT_SINE_FACTOR = 32768
*/
sint16 sin_int(sint16 input);
/*
Cosine
Fixed Point Arithmetic
cos(x) = sin(x + pi/2)
-pi <= x <= pi
Input
sint16 input – input value: -PI_INT_VALUE=-25736 … PI_INT_VALUE=25736
input format 3Q13: b14*2^1 + b13*2^0 + b12*2^-1 + b11*2^-2 + … + b0*2^-13 and b15-sign
Return
sint16 – cos(input) in format 1Q15: b14*2^-1 + b13*2^-2 + b12*2^-3 + b11*2^-4 + … + b0*2^-15 and b15-sign
Note: Result = ReturnValue/OUT_SINE_FACTOR, where OUT_SINE_FACTOR = 32768
*/
sint16 cos_int(sint16 input);
##### The sine / cosine functions were tested in main() using the test arrays: input_array_float[ ] and input_array_int[ ]. The test results are shown in Figure 8-7, Figure 8-8
Figure 8-7: Absolute mistake: Δsine of the sin_float() for x= -π … +π. Max absolute mistake is 2.587e-04
Figure 8-8: Absolute mistake: Δsine of the sin_int() for x= -π … +π. Max absolute mistake is 2.694e-02
### 8.3 atan.c/h
##### atan.c/h is supported arctangent function
/*
Arctangent
Float Arithmetic
Polynomial atan(x) = x (0.99904621 + x^2 (-0.3198728 + x^2 (0.14370242 – 0.037537854 x^2)))
-1 <= x <= 1
Polynomial atan(x) = +/- pi/2 – x^-1 (0.99904621 + x^-2 (-0.3198728 + x^-2 (0.14370242 – 0.037537854 x^-2)))
|x| > 1
Input
float input – input value
Return
float – atan(input)
*/
float atan_float(float input);
##### The atan function was tested in main() using the test arrays: input_array_float 1/2/3[ ]. The test results are shown in Figure 8-9, Figure 8-10, Figure 8-11
Figure 8-9: Absolute mistake: Δatan of the atan_float() for x = -100 … -1. Max absolute mistake is 1.036e-4
Figure 8-10: Absolute mistake: Δatan of the atan_float() for x = -1 … 1. Max absolute mistake is 1.032e-04
Figure 8-11: Absolute mistake: Δatan of the atan_float() for x = 1 … 100. Max absolute mistake is 1.036e-4 | 2,990 | 8,722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | longest | en | 0.714712 |
https://www.encyclopediaofmath.org/index.php?title=Cubic_form&oldid=14958 | 1,561,564,590,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000367.74/warc/CC-MAIN-20190626154459-20190626180459-00284.warc.gz | 737,928,455 | 6,822 | # Cubic form
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
A homogeneous polynomial of degree 3 in several variables with coefficients in some fixed field or ring. Let be a field and a cubic form with coefficients in (one calls it a cubic form over ). The equation
defines a cubic hypersurface in the projective space , so that the algebraic-geometric theory of cubic forms over an algebraically closed field is reduced to the theory of cubic hypersurfaces (see [1]).
The arithmetic theory of cubic forms over number fields (and their rings of integers) is still (1987) rather poorly developed in comparison with the rich and meaningful arithmetic theory of quadratic forms. For cubic forms in two variables, the arithmetic theory is just the theory of cubic extensions of number fields (see [2]). For cubic forms in three variables, it is part of the arithmetic theory of elliptic curves (see [3]). In particular, examples are known of cubic forms in three variables that violate Hasse's principle. The same holds for cubic forms in four variables (see [1], [4], [6]). There is no general theory at all for cubic forms in a larger number of variables.
The purely algebraic theory of cubic forms contains, in addition to results concerning the structure of sets of points on cubic hypersurfaces, various results pertaining to the classical theory of invariants. Indeed, the structure of the algebra of (absolute) invariants of a cubic form in two or three variables is known; in these cases the algebra has no syzygies — it is the algebra of polynomials in one (degree 4) and two (degrees 4 and 6) algebraically independent homogeneous generators. If the number of variables exceeds 4, the algebra in question contains syzygies [5] and its structure is very complicated. H. Poincaré investigated the orbits, stabilizers and canonical representatives, and also the families of orbits under the natural action of the group on the space of all cubic forms in three and four variables [6].
#### References
[1] Yu.I. Manin, "Cubic forms. Algebra, geometry, arithmetic" , North-Holland (1986) (Translated from Russian) [2] B.N. Delone, D.K. Faddeev, "Theory of irrationalities of the third degree" Trudy Mat. Inst. Akad. Nauk SSSR , 11 (1940) (In Russian) [3] J.W.S. Cassels, "Diophantine equations with special reference to elliptic curves" J. London Math. Soc. , 41 (1966) pp. 193–291 [4] B. Segree, "On the rational solutions of homogeneous cubic equations in four variables" Math. Notae (Univ. Rosario) , 11 (1951) pp. 1–68 [5] V.G. Kats, V.L. Popov, E.B. Vinberg, "Sur les groupes linéaires algébriques dont l'algèbre des invariants est libre" Acad. Sci. Paris , 283 (1976) pp. 875–878 (English summary) [6] H. Poincaré, , Oevres , 2 , Moscow (1972) pp. 819–900 (In Russian)
How to Cite This Entry:
Cubic form. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Cubic_form&oldid=14958
This article was adapted from an original article by V.A. IskovskikhV.L. Popov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article | 790 | 3,132 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.926725 |
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