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Matter & Interactions

Book edition 4th edition
Author(s) Ruth W. Chabay, Bruce A. Sherwood
Pages 1135 pages
ISBN 9781118875865

You hang a 10 kg mass from a copper wire, and the wire stretches by . (a) If you suspend the same mass from two copper wires, identical to the original wire, what happens? (b) If you suspend the same mass from a copper wire with half the cross-sectional area but the same length as the original wire, what happens? (c) If you suspend the same mass from a copper wire with the same cross-sectional area but twice the length of the original wire, what happens?

(a) The stretch in the wire when two wires are used instead of one is 4 mm.

(b) The stretch in the wire when the cross-sectional area is halved is 16 mm.

(b) The stretch in the wire when the initial length is doubled is 16 mm.

See the step by step solution

Step 1: Given data

A 10 kg mass is hung from a copper wire, and the wire stretches by 8 mm.

Step 2: Change in length of the wire

The change in length of a wire of initial length L and cross-sectional area A when force F is applied to it can be expressed as,

${\mathbf{∆}}{\mathbf{L}}{\mathbf{=}}\frac{\mathbf{FL}}{\mathbf{AY}}$ (I)

Here L is Young's modulus of the material of the wire.

Step 3: Determining the change in length when two wires are used instead of one

a)

When two wires are used, the cross-sectional area doubles. From equation (I), the new change in length can be calculated as,

$\begin{array}{rcl}∆\mathrm{L}& =& \frac{\mathrm{FL}}{2\mathrm{AY}}\\ & =& \frac{1}{2}\left(\frac{\mathrm{FL}}{\mathrm{AY}}\right)\\ & =& \frac{1}{2}×8\mathrm{mm}\\ & =& 4\mathrm{mm}\end{array}$

Thus, the new change in length is 4 mm.

Step 4: Determining the change in length when the cross-sectional area is halved

b)

From equation (I), the new change in length when the cross-sectional area is halved can be calculated as,

Thus, the new change in length is 16 mm.

Step 5: Determining the change in length when the initial length is doubled

c)

From equation (I), the new change in length when the initial length is doubled can be calculated as,

$\begin{array}{rcl}∆\mathrm{L}& =& \frac{\mathrm{F}2\mathrm{L}}{\mathrm{AY}}\\ & =& 2\left(\frac{\mathrm{FL}}{\mathrm{AY}}\right)\\ & =& 2×8\mathrm{mm}\\ & =& 16\mathrm{mm}\end{array}$

Thus, the new change in length is 16 mm.


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location:  Publications → journals
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Search: MSC category 32A25 ( Integral representations; canonical kernels (Szegoo, Bergman, etc.) )

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1. CJM 2010 (vol 62 pp. 1276)

El Wassouli, Fouzia
 A Generalized Poisson Transform of an $L^{p}$-Function over the Shilov Boundary of the $n$-Dimensional Lie Ball Let $\mathcal{D}$ be the $n$-dimensional Lie ball and let $\mathbf{B}(S)$ be the space of hyperfunctions on the Shilov boundary $S$ of $\mathcal{D}$. The aim of this paper is to give a necessary and sufficient condition on the generalized Poisson transform $P_{l,\lambda}f$ of an element $f$ in the space $\mathbf{B}(S)$ for $f$ to be in $L^{p}(S)$, $1 > p > \infty.$ Namely, if $F$ is the Poisson transform of some $f\in \mathbf{B}(S)$ (i.e., $F=P_{l,\lambda}f$), then for any $l\in \mathbb{Z}$ and $\lambda\in \mathbb{C}$ such that $\mathcal{R}e[i\lambda] > \frac{n}{2}-1$, we show that $f\in L^{p}(S)$ if and only if $f$ satisfies the growth condition $$\|F\|_{\lambda,p}=\sup_{0\leq r < 1}(1-r^{2})^{\mathcal{R}e[i\lambda]-\frac{n}{2}+l}\Big[\int_{S}|F(ru)|^{p}du \Big]^{\frac{1}{p}} < +\infty.$$ Keywords:Lie ball, Shilov boundary, Fatou's theorem, hyperfuctions, parabolic subgroup, homogeneous line bundleCategories:32A45, 30E20, 33C67, 33C60, 33C55, 32A25, 33C75, 11K70

2. CJM 2009 (vol 62 pp. 439)

Sundhäll, Marcus; Tchoundja, Edgar
 On Hankel Forms of Higher Weights: The Case of Hardy Spaces In this paper we study bilinear Hankel forms of higher weights on Hardy spaces in several dimensions. (The Schatten class Hankel forms of higher weights on weighted Bergman spaces have already been studied by Janson and Peetre for one dimension and by SundhÃ¤ll for several dimensions). We get a full characterization of Schatten class Hankel forms in terms of conditions for the symbols to be in certain Besov spaces. Also, the Hankel forms are bounded and compact if and only if the symbols satisfy certain Carleson measure criteria and vanishing Carleson measure criteria, respectively. Keywords:Hankel forms, Schattenâvon Neumann classes, Bergman spaces, Hardy spaces, Besov spaces, transvectant, unitary representations, MÃ¶bius groupCategories:32A25, 32A35, 32A37, 47B35

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Shown below are selected data from a recent annual report of Kimberly-Clark Corporation , a large...

Shown below are selected data from a recent annual report of Kimberly-Clark Corporation, a large consumer products provider. (Dollar amounts are in millions.)

 Beginning of the Year End of the Year Total assets  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $17,660$18,840 Total stockholders’ equity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4,280 5,690 Operating income . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3,210 Net income . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1,880

a.       Compute for the year Kimberly-Clark’s return on average total assets. (Round computations to the nearest two-tenths of 1 percent.)

b.       Compute for the year Kimberly-Clark’s return on average total stockholders’ equity. (Round computations to the nearest two-tenths of 1 percent.)

c.       What is the most likely explanation why Kimberly-Clark’s total stockholders’ equity for the year increased?

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# cpython / Doc / lib / emailcharsets.tex

The branch 'legacy-trunk' does not exist.
  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 \declaremodule{standard}{email.charset} \modulesynopsis{Character Sets} This module provides a class \class{Charset} for representing character sets and character set conversions in email messages, as well as a character set registry and several convenience methods for manipulating this registry. Instances of \class{Charset} are used in several other modules within the \module{email} package. Import this class from the \module{email.charset} module. \versionadded{2.2.2} \begin{classdesc}{Charset}{\optional{input_charset}} Map character sets to their email properties. This class provides information about the requirements imposed on email for a specific character set. It also provides convenience routines for converting between character sets, given the availability of the applicable codecs. Given a character set, it will do its best to provide information on how to use that character set in an email message in an RFC-compliant way. Certain character sets must be encoded with quoted-printable or base64 when used in email headers or bodies. Certain character sets must be converted outright, and are not allowed in email. Optional \var{input_charset} is as described below; it is always coerced to lower case. After being alias normalized it is also used as a lookup into the registry of character sets to find out the header encoding, body encoding, and output conversion codec to be used for the character set. For example, if \var{input_charset} is \code{iso-8859-1}, then headers and bodies will be encoded using quoted-printable and no output conversion codec is necessary. If \var{input_charset} is \code{euc-jp}, then headers will be encoded with base64, bodies will not be encoded, but output text will be converted from the \code{euc-jp} character set to the \code{iso-2022-jp} character set. \end{classdesc} \class{Charset} instances have the following data attributes: \begin{datadesc}{input_charset} The initial character set specified. Common aliases are converted to their \emph{official} email names (e.g. \code{latin_1} is converted to \code{iso-8859-1}). Defaults to 7-bit \code{us-ascii}. \end{datadesc} \begin{datadesc}{header_encoding} If the character set must be encoded before it can be used in an email header, this attribute will be set to \code{Charset.QP} (for quoted-printable), \code{Charset.BASE64} (for base64 encoding), or \code{Charset.SHORTEST} for the shortest of QP or BASE64 encoding. Otherwise, it will be \code{None}. \end{datadesc} \begin{datadesc}{body_encoding} Same as \var{header_encoding}, but describes the encoding for the mail message's body, which indeed may be different than the header encoding. \code{Charset.SHORTEST} is not allowed for \var{body_encoding}. \end{datadesc} \begin{datadesc}{output_charset} Some character sets must be converted before they can be used in email headers or bodies. If the \var{input_charset} is one of them, this attribute will contain the name of the character set output will be converted to. Otherwise, it will be \code{None}. \end{datadesc} \begin{datadesc}{input_codec} The name of the Python codec used to convert the \var{input_charset} to Unicode. If no conversion codec is necessary, this attribute will be \code{None}. \end{datadesc} \begin{datadesc}{output_codec} The name of the Python codec used to convert Unicode to the \var{output_charset}. If no conversion codec is necessary, this attribute will have the same value as the \var{input_codec}. \end{datadesc} \class{Charset} instances also have the following methods: \begin{methoddesc}[Charset]{get_body_encoding}{} Return the content transfer encoding used for body encoding. This is either the string \samp{quoted-printable} or \samp{base64} depending on the encoding used, or it is a function, in which case you should call the function with a single argument, the Message object being encoded. The function should then set the \mailheader{Content-Transfer-Encoding} header itself to whatever is appropriate. Returns the string \samp{quoted-printable} if \var{body_encoding} is \code{QP}, returns the string \samp{base64} if \var{body_encoding} is \code{BASE64}, and returns the string \samp{7bit} otherwise. \end{methoddesc} \begin{methoddesc}{convert}{s} Convert the string \var{s} from the \var{input_codec} to the \var{output_codec}. \end{methoddesc} \begin{methoddesc}{to_splittable}{s} Convert a possibly multibyte string to a safely splittable format. \var{s} is the string to split. Uses the \var{input_codec} to try and convert the string to Unicode, so it can be safely split on character boundaries (even for multibyte characters). Returns the string as-is if it isn't known how to convert \var{s} to Unicode with the \var{input_charset}. Characters that could not be converted to Unicode will be replaced with the Unicode replacement character \character{U+FFFD}. \end{methoddesc} \begin{methoddesc}{from_splittable}{ustr\optional{, to_output}} Convert a splittable string back into an encoded string. \var{ustr} is a Unicode string to unsplit''. This method uses the proper codec to try and convert the string from Unicode back into an encoded format. Return the string as-is if it is not Unicode, or if it could not be converted from Unicode. Characters that could not be converted from Unicode will be replaced with an appropriate character (usually \character{?}). If \var{to_output} is \code{True} (the default), uses \var{output_codec} to convert to an encoded format. If \var{to_output} is \code{False}, it uses \var{input_codec}. \end{methoddesc} \begin{methoddesc}{get_output_charset}{} Return the output character set. This is the \var{output_charset} attribute if that is not \code{None}, otherwise it is \var{input_charset}. \end{methoddesc} \begin{methoddesc}{encoded_header_len}{} Return the length of the encoded header string, properly calculating for quoted-printable or base64 encoding. \end{methoddesc} \begin{methoddesc}{header_encode}{s\optional{, convert}} Header-encode the string \var{s}. If \var{convert} is \code{True}, the string will be converted from the input charset to the output charset automatically. This is not useful for multibyte character sets, which have line length issues (multibyte characters must be split on a character, not a byte boundary); use the higher-level \class{Header} class to deal with these issues (see \refmodule{email.header}). \var{convert} defaults to \code{False}. The type of encoding (base64 or quoted-printable) will be based on the \var{header_encoding} attribute. \end{methoddesc} \begin{methoddesc}{body_encode}{s\optional{, convert}} Body-encode the string \var{s}. If \var{convert} is \code{True} (the default), the string will be converted from the input charset to output charset automatically. Unlike \method{header_encode()}, there are no issues with byte boundaries and multibyte charsets in email bodies, so this is usually pretty safe. The type of encoding (base64 or quoted-printable) will be based on the \var{body_encoding} attribute. \end{methoddesc} The \class{Charset} class also provides a number of methods to support standard operations and built-in functions. \begin{methoddesc}[Charset]{__str__}{} Returns \var{input_charset} as a string coerced to lower case. \method{__repr__()} is an alias for \method{__str__()}. \end{methoddesc} \begin{methoddesc}[Charset]{__eq__}{other} This method allows you to compare two \class{Charset} instances for equality. \end{methoddesc} \begin{methoddesc}[Header]{__ne__}{other} This method allows you to compare two \class{Charset} instances for inequality. \end{methoddesc} The \module{email.charset} module also provides the following functions for adding new entries to the global character set, alias, and codec registries: \begin{funcdesc}{add_charset}{charset\optional{, header_enc\optional{, body_enc\optional{, output_charset}}}} Add character properties to the global registry. \var{charset} is the input character set, and must be the canonical name of a character set. Optional \var{header_enc} and \var{body_enc} is either \code{Charset.QP} for quoted-printable, \code{Charset.BASE64} for base64 encoding, \code{Charset.SHORTEST} for the shortest of quoted-printable or base64 encoding, or \code{None} for no encoding. \code{SHORTEST} is only valid for \var{header_enc}. The default is \code{None} for no encoding. Optional \var{output_charset} is the character set that the output should be in. Conversions will proceed from input charset, to Unicode, to the output charset when the method \method{Charset.convert()} is called. The default is to output in the same character set as the input. Both \var{input_charset} and \var{output_charset} must have Unicode codec entries in the module's character set-to-codec mapping; use \function{add_codec()} to add codecs the module does not know about. See the \refmodule{codecs} module's documentation for more information. The global character set registry is kept in the module global dictionary \code{CHARSETS}. \end{funcdesc} \begin{funcdesc}{add_alias}{alias, canonical} Add a character set alias. \var{alias} is the alias name, e.g. \code{latin-1}. \var{canonical} is the character set's canonical name, e.g. \code{iso-8859-1}. The global charset alias registry is kept in the module global dictionary \code{ALIASES}. \end{funcdesc} \begin{funcdesc}{add_codec}{charset, codecname} Add a codec that map characters in the given character set to and from Unicode. \var{charset} is the canonical name of a character set. \var{codecname} is the name of a Python codec, as appropriate for the second argument to the \function{unicode()} built-in, or to the \method{encode()} method of a Unicode string. \end{funcdesc}


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# Relationship between the t-statistic and F-statistic in simple linear regression

Consider the simple linear regression model $$y_i = \beta_0 + \beta_1{x_i} + \epsilon_i$$

Suppose in an OLS, the $$t$$-statistic for the null hypothesis $$\beta_1 = 0$$ is $$1.92$$, what is the $$F$$-statistic for the overall significance of the model?

I am aware that $$F$$ statistic for testing exclusion of a single variable is equal to the square of the corresponding $$t$$ statistic: however, I can't seem to find how to link that information to this question. Am I missing information?

• Consider what the null hypothesis would be for the “significance of the entire model”.
– Dave
Nov 28 '21 at 22:45
• Oh it would just be $\beta_1$ = 0, wouldn't it? Nov 28 '21 at 23:11
• That’s what I would say, and that’s what R gives you in a “summary” of a linear model. And then the alternative hypothesis would be $\ne$. // This is a subtle point, but good for you for not writing $\hat\beta_1=0$.
– Dave
Nov 28 '21 at 23:16
• Thank you! I spent way too much time looking for an answer when it was sitting right in front me Nov 28 '21 at 23:19
• Since you pretty much solved the problem yourself, you might consider posting a self-answer. (Cross Validated does not consider it arrogant or poor etiquette to do so. In fact, I have posted several self-answers.)
– Dave
Nov 28 '21 at 23:37

In the simple linear regression model, testing the significance of the model requires testing the hypothesis: $$H_0 \colon \beta_1 = 0 \\ H_0 \colon \beta_1 \neq 0$$
Given that we are whether a single parameter is equal to 0, the $$F$$ statistic will simply equal the square of the corresponding $$t$$ statistic. This means that if the $$t$$-statistic for the null is $$1.92$$, the $$F$$-statistic will simply be $$1.92^2$$.


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# Optimising a football team rating system for match prediction accuracy

While I’ve been quite happy with the performance of my Predictaball football rating system, one thing that that’s bothered me since its inception last summer is the reliance on hard-coded parameters.

Similar to many other football rating methods, it’s an adaptation of the Elo system that was designed for Chess matches by Arpad Elo in the 1950s. His aim was to devise an easily implementable system to rate competitors in a 2-person zero-sum game.

Predictaball, on the other hand, is more concerned with accurately predicting future matches; model interpretability and portability is far less of a requirement. Furthermore, we have the advantage of an abundance of historical performance data and cheap computation to fit our models.

This post summarises the implementation of a rating system that simultaneously optimises parameters used in both the match prediction model and the rating update function in order to maximise match prediction accuracy.

## Current Elo implementation

I’ll firstly run a quick refresher on how the standard Elo method is used by Predictaball.

### Elo expected outcome

The formula for calculating the expected outcome $E$ of a match in Elo’s original formulation is shown below, where $\delta$ is the difference in rating between the home and away team, i.e. $\delta = R_{H} - R_{A}$.

$$E = \frac{1}{1+10^{- \delta /400}}$$

Note that this is the inverse logit function in base 10 rather than the standard natural base, and is therefore equivalent to logistic regression with 1 predictor ($\delta$) and the coefficient $\beta = \frac{1}{400}$. The version used by Predictaball (and others including 538) also adds in a factor to account for home advantage, i.e. $\delta = R_{H} - R_{A} + \alpha$, where $\alpha$ is estimated from the historical data as the rating difference associated with the proportion of matches that are home wins.

This expected outcome framed as logistic regression is therefore

$$\text{logit}(P(\text{home win})) = \alpha + \beta(R_{H} - R_{A})$$

However, this model isn’t overly useful on its own as it doesn’t tell us whether an away win or draw is more likely, and we might be more interested in other outcomes such as the number of goals scored (what Predictaball currently outputs). Furthermore, we might want to include additional predictor variables, such as if a team is missing a key player, or has had a new manager recently. We definitely have sufficient training data to fit more complex predictive models and not rely upon hardcoded simplifications.

### Rating update equation

The Elo rating update equation is a function of the difference between the actual outcome $O$ ($O \in \left\{0, 0.5, 1\right\}$ for away win, draw, and home win respectively), the expected outcome $E$ from the above equation, and a gain factor $K$ that controls how much rating is won/lost per game. $K$ is traditionally fixed at 20 for a long-running competition or set higher for a short tournament.

$$R^{\prime}_{H} = R_{H} + K(O - E)$$

Predictaball also scales the output to account for the margin of victory (MoV), similar to the method used by 538

$$R^{\prime}_{H} = R_{H} + KG(\text{MoV})(O - E)$$

The functional form of G is shown below, where the multiplier decays as the margin of victory increases. Also note the adjustment by $\delta$ (labelled $\text{dr}$ in the figure) to stop higher rating teams getting further and further ahead. The parameters of $G$ were derived by hand by tweaking them until the resulting shape looked appropriate, rather than using a data-driven approach.

### Secondary prediction network

The final limitation with my implementation is that currently the ratings are retrofitted to my dataset using the above (hardcoded) parameters, and then a secondary predictive model is fitted using the $\delta$ as input predictors to a Bayesian model that predicts how many goals each team will score. This model is far more flexible and useful than the logistic regression used in the Elo system, but it still seems a waste to fit a second predictive-model separately from the rating system.

## Rating System Optimised for Predictive Accuracy (RSOPA)

The properties of my ideal rating system are:

• Be entirely data driven, i.e. parameters fitted using historical data
• Flexible to choice of predictive model
• Primary objective is predictive accuracy

I’ll call this method Rating System Optimised for Predictive Accuracy (RSOPA), just because there aren’t enough acronyms in the world already, and will now detail its implementation.

### Predictive model

The crux of the Elo system is the $O-E$ term that is in $[-1, 1]$. Any predictive model for $E$ and any measure of outcome $O$ can be used provided a statistic in this range can be calculated.

For an initial proof-of-concept I’m going to use an ordinal multinomial regression for the match prediction model. This estimates probabilities for each of the 3 possible outcomes while taking the natural ordering into account (away win < draw < home win). The model is specified as follows

$$\mu = \alpha_{league} + \beta \delta$$ $$\text{logit}(q_1) = \eta_1 - \mu$$ $$\text{logit}(q_2) = \eta_2 - \mu$$ $$P(\text{away win}) = q_1$$ $$P(\text{draw}) = q_2 - q_1$$ $$P(\text{home win}) = 1 - q_2$$

Where $\alpha_{league}$ is a league specific intercept, while the coefficient $\beta$ and the two thresholds $\eta_1$ and $\eta_2$ are constant across leagues.

A value for $E$ in $[0,1]$ can be obtained by either scaling the linear predictor $\mu$, or by using a more hacky solution as the normalised weighted average of the 3 outcomes.

$$E = \frac{(1P(\text{away win}) + 2P(\text{draw}) + 3P(\text{home win}) - 1)}{3-1}$$

Using the same output formulation as before ($O \in \left\{0, 0.5, 1\right\}$), we have a more useful predictive model that can still be incorporated into the Elo rating system.

With 4 leagues in this dataset, this is a total of 7 parameters to be estimated for the predictive model.

### Rating system

The rating system is much as before

$$R^{\prime}_{H} = R_{H} + KG(\text{MoV})(O - E)$$

Where $G(x)$ now takes the general form $G(x) = A(1-exp(-Bx))+1$, where $A$ controls the maximum output and $B$ the decay rate, as shown in the plot below.

Alongside $A$ and $B$, the gain parameter $K$ can also be treated as free, and I’ll let it vary by league. Therefore, there are 6 free parameters to optimise from the rating update model.

### Parameter optimisation using Genetic Algorithm

In total there are 13 parameters to be solved: 7 from the match prediction model and 6 from the rating update equation. For an initial proof-of-concept I’m going to solve for these using a Genetic Algorithm - in particular the R interface to CMA-ES. As I’m primarily interested in obtaining the most accurate match predictions, my fitness function is the Ranked Probability Score (RPS), the specific case of the Brier score when the output is ordered.

I’m training the network on the 13,014 games in the top 4 European leagues (Premiership, Bundesliga, La Liga, Serie A) between the 2005-2006 and 2013-2014 seasons, although I’m allowing 3 seasons for the ratings to settle before the RPS is counted. To obtain an idea of how well the model works on unseen data I’ll form a test set comprising the seasons from 2014-2015 up to and including 2018-2019.

The hardcoded aspects of this model are the functional forms of the 2 models and the starting rating, which I’m setting at 2,000 completely arbitrarily.

## Final model

The table below compares the RPS from RSOPA against two Bayesian models that were fitted to the original hand-implemented rating system. The ‘Goals’ method predicts the goals scored by each team and was used in the Predictaball forecasts for the 2018-2019 season while ‘Ordinal’ is an ordinal multinomial regression and was used prior to 2018.

Unsurprisingly, RSOPA does far better on the training set - an improvement of 0.001 rps would be enough to move me up to 5th from 11th in 2017-2018 season - but there is less of an difference on the test set. This could be one area for future work, adding in regularisation methods or early-stopping to minimise the risk of over-fitting.

Model Training data Test data
RSOPA 0.1998 0.1974
Bayesian Ordinal 0.2007 0.1977
Bayesian Goals 0.2009 0.1978

The plot below shows the modelled outcome probabilities for a range of $\delta$ across the 4 leagues. It immediately highlights the benefits of using a more complex match prediction model than the standard Elo logistic regression as the away win and draw probabilities aren’t linearly related. Incidentally, the peak in the draw probability at around $\delta = -200$ suggests that the home advantage is worth 200 rating points. Finally, the known fact that the Bundesliga has a lower proportion of home wins can be seen by the red curve lying lower than the others, resulting from the league specific $\alpha$.

Moving onto the rating network, and the fitted $G(\text{MoV})$ function is shown below. It allows for very high multipliers which concerns me slightly as it could allow for stronger teams’ ratings to skyrocket. If this becomes a problem I could reshape $G$ to include a penalty term for the rating difference to reduce this auto-correlation, like what 538 use and I’ve borowed for the current version of Predictaball.

The team ratings at the end of the 2018-2019 season are shown below for both the current implementation (red) and RSOPA (blue). RSOPA definitely produces more dispersed teams due to the large multiplier effect of $G$, which is around 3x for a 5-goal win in the current method but 4.4x in the new system.

This has some advantages, in that it’s clear to see who the best and worst-rated teams are, for example the top 6 in the Premier League are far easier to identify than under the old system. But it could be easy to overinterpret these ratings by reading too much into the large differences between the teams, when really these need to be taken into consideration alongside the $\beta$ coefficient.

For example, if we were to predict the outcome of Liverpool vs Everton under the old system we would get H/D/A of 72% / 17% / 11%, which isn’t that much less than the 78% / 14% / 8% under RSOPA, so even though Liverpool’s rating is far closer to Everton’s under the old method the $\beta$ was larger to compensate.

## Further work

I’ve run out of time to tinker with this concept further this Summer, but the next thing I’d be looking at improving would be to remove $G$ entirely and encapsulate the predicted scale of victory in $E$, so that the rating update is simply $K(O-E)$ as before.

In particular, I’d like to use my current goal prediction model rather than the ordinal multinomial regression I’m using here and I’d like to go back to using a Bayesian model rather than just using point estimates of parameters. However, this is computationally more challenging to fit so I’d need to find time to sit down and properly think about it.

As for now, I’m going to use these ratings on The Predictaball’s website for the coming 2019-2020 season and see how they get on.


text:
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# Derivation of the Triangle Area Formula

## Derive and apply area equals half the product of two sides and the sine of the included angle.

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Derivation of the Triangle Area Formula

While in the lunch room with your friends one day, you're discussing different ways you can use the things you've learned in math class. You tell your friends that you've been learning a lot about triangles, such as how to find their area. One of your friends looks down at your plate and starts to smile.

"Alright," he says. "If you're so good at things involving triangles, I dare you to find something simple. Tell me the area of your slice of pizza." He points down at the pizza on your plate.

The pizza is shaped like a triangle. But unfortunately its not a right triangle. The outer edge is 5 inches long, and the long sides are 7 inches long. The angle between the edge and the long side of the slice is \begin{align*}69^\circ\end{align*}. Is there any way to tell the area of your pizza slice?

### Deriving the Triangle Area Formula

We can use the area formula from Geometry, \begin{align*}A = \frac{1}{2} bh\end{align*}, as well as the sine function, to derive a new formula that can be used when the height, or altitude, of a triangle is unknown.

In \begin{align*}\triangle{ABC}\end{align*} below, \begin{align*}BD\end{align*} is altitude from \begin{align*}B\end{align*} to \begin{align*}AC\end{align*}. We will refer to the length of \begin{align*}BD\end{align*} as \begin{align*}h\end{align*} since it also represents the height of the triangle. Also, we will refer to the area of the triangle as \begin{align*}K\end{align*} to avoid confusing the area with \begin{align*}\angle{A}\end{align*}.

\begin{align*}k & = \frac{1}{2}bh && \text{Area of a triangle} \\ k & = \frac{1}{2}b (c \sin A) && \sin A = \frac{h}{c}\ \text{therefore}\ c \sin A = h \\ k & = \frac{1}{2}bc \sin A && \text{Simplify}\end{align*}

We can use a similar method to derive all three forms of the area formula, regardless of the angle:

\begin{align*}K & = \frac{1}{2}bc \sin A \\ K & = \frac{1}{2}ac \sin B \\ K & = \frac{1}{2}ab \sin C \end{align*}

The formula \begin{align*}K = \frac{1}{2} \ bc \sin A \end{align*} requires us to know two sides and the included angle (SAS) in a triangle. Once we know these three things, we can easily calculate the area of an oblique triangle.

#### Finding the Area of a Triangle

1. In \begin{align*}\triangle{ABC}, \angle{C} = 62^\circ, b = 23.9\end{align*}, and \begin{align*}a = 31.6\end{align*}. Find the area of the triangle.

Using our new formula, \begin{align*}K = \frac{1}{2} \ ab \sin C \end{align*}, plug in what is known and solve for the area.

\begin{align*}K & = \frac{1}{2}(31.6)(23.9)\sin 62 \\ K & \approx 333.4\end{align*}

2. The Pyramid Hotel recently installed a triangular pool. One side of the pool is 24 feet, another side is 26 feet, and the angle in between the two sides is \begin{align*}87^\circ\end{align*}. If the hotel manager needs to order a cover for the pool, and the cost is \begin{align*}\ 35\end{align*} per square foot, how much can he expect to spend?

In order to find the cost of the cover, we first need to know the area of the cover. Once we know how many square feet the cover is, we can calculate the cost. In the illustration above, you can see that we know two of the sides and the included angle. This means we can use the formula \begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*}.

\begin{align*}& K = \frac{1}{2} \ (24)(26)\sin 87 \\ & K \approx 311.6 \\ & 311.6 \ sq. ft. \times \ 35/sq. ft. = \10,905.03 \end{align*}

The cost of the cover will be \begin{align*}\10,905.03\end{align*}.

3. In \begin{align*}\triangle{GHI}, \angle{I} = 15^\circ, g = 14.2\end{align*}, and \begin{align*}h = 7.9\end{align*}. Find the area of the triangle.

Using our new formula, \begin{align*}K = \frac{1}{2} \ ab \sin C \end{align*}, which is the same as \begin{align*}K = \frac{1}{2} \ gh \sin I \end{align*}, plug in what is known and solve for the area.

\begin{align*}K & = \frac{1}{2}(14.2)(7.9)\sin 15 \\ K & \approx 14.52\end{align*}

### Examples

#### Example 1

Earlier, you were asked if there was any way to tell the area of your pizza slice.

Use the formula \begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*}

where in this case, one of the sides is equal to 5, the other is equal to 7, and the angle is \begin{align*}69^\circ\end{align*}.

\begin{align*}K = \frac{1}{2} (5)(7) \sin 69^\circ = 16.34 in^2\end{align*}

#### Example 2

A farmer needs to replant a triangular section of crops that died unexpectedly. One side of the triangle measures 186 yards, another measures 205 yards, and the angle formed by these two sides is \begin{align*}148^\circ\end{align*}.

What is the area of the section of crops that needs to be replanted?

Use \begin{align*}K = \frac{1}{2} \ bc \sin A, K= \frac{1}{2}(186)(205)\sin 148^\circ\end{align*}. So, the area that needs to be replaced is 10102.9 square yards.

#### Example 3

The farmer goes out a few days later to discover that more crops have died. The side that used to measure 205 yards now measures 288 yards. How much has the area that needs to be replanted increased by?

\begin{align*}K = \frac{1}{2}(186)(288)\sin 148^\circ = 14193.4\end{align*}, the area has increased by 4090.5 yards.

#### Example 4

Find the perimeter of the quadrilateral at the left If the area of \begin{align*}\triangle{DEG} = 56.5\end{align*} and the area of \begin{align*}\triangle{EGF} = 84.7\end{align*}.

You need to use the \begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*} formula to find \begin{align*}DE\end{align*} and \begin{align*}GF\end{align*}.

\begin{align*}56.5 = \frac{1}{2}(13.6)DE \sin 39^\circ \rightarrow DE = 13.2 && 84.7 = \frac{1}{2}(13.6)EF \sin 60^\circ \rightarrow EF = 14.4\end{align*}

Second, you need to find sides \begin{align*}DG\end{align*} and \begin{align*}GF\end{align*} using the Law of Cosines.

\begin{align*}DG^2 & = 13.2^2 + 13.6^2 - 2 \cdot 13.2 \cdot 13.6 \cdot \cos 39^\circ \rightarrow DG = 8.95 \\ GF^2 & = 14.4^2 + 13.6^2 - 2 \cdot 14.4 \cdot 13.6 \cdot \cos 60^\circ \rightarrow GF = 14.0\end{align*}

The perimeter of the quadrilateral is 50.55.

### Review

Find the area of each triangle.

1. \begin{align*}\triangle ABC\end{align*} if a=13, b=15, and \begin{align*}m\angle C=71^\circ\end{align*}.
2. \begin{align*}\triangle ABC\end{align*} if b=8, c=4, and \begin{align*}m\angle A=67^\circ\end{align*}.
3. \begin{align*}\triangle ABC\end{align*} if b=34, c=29, and \begin{align*}m\angle A=138^\circ\end{align*}.
4. \begin{align*}\triangle ABC\end{align*} if a=3, b=7, and \begin{align*}m\angle C=80^\circ\end{align*}.
5. \begin{align*}\triangle ABC\end{align*} if a=4.8, c=3.7, and \begin{align*}m\angle B=43^\circ\end{align*}.
6. \begin{align*}\triangle ABC\end{align*} if a=12, b=5, and \begin{align*}m\angle C=20^\circ\end{align*}.
7. \begin{align*}\triangle ABC\end{align*} if a=3, b=10, and \begin{align*}m\angle C=50^\circ\end{align*}.
8. \begin{align*}\triangle ABC\end{align*} if a=5, b=9, and \begin{align*}m\angle C=14^\circ\end{align*}.
9. \begin{align*}\triangle ABC\end{align*} if a=5, b=7, and c=11.
10. \begin{align*}\triangle ABC\end{align*} if a=7, b=8, and c=9.
11. \begin{align*}\triangle ABC\end{align*} if a=12, b=14, and c=4.
12. A farmer measures the three sides of a triangular field and gets 114, 165, and 257 feet. What is the measure of the largest angle of the triangle?
13. Using the information from the previous problem, what is the area of the field?

Another field is a quadrilateral where three sides measure 30, 50, and 60 yards, and two angles measure \begin{align*}130^\circ\end{align*} and \begin{align*}140^\circ\end{align*}, as shown below.

1. Find the area of the quadrilateral. Hint: divide the quadrilateral into two triangles and find the area of each.
2. Find the length of the fourth side.
3. Find the measures of the other two angles.

To see the Review answers, open this PDF file and look for section 5.4.

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

### Vocabulary Language: English

Oblique Triangle

An oblique triangle is a triangle without a right angle as one of its internal angles.

SAS Triangle

An SAS triangle is a triangle where two sides and the angle in between them are known quantities.


text:
B. Equalize Prices

Description:
There are n products in the shop. The price of the i-th product is ai. The owner of the shop wants to equalize the prices of all products. However, he wants to change prices smoothly.

In fact, the owner of the shop can change the price of some product i in such a way that the difference between the old price of this product ai and the new price bi is at most k. In other words, the condition |ai−bi|≤k should be satisfied (|x| is the absolute value of x).

He can change the price for each product not more than once. Note that he can leave the old prices for some products. The new price bi of each product i should be positive (i.e. bi>0 should be satisfied for all i from 1 to n).

Your task is to find out the maximum possible equal price B of all productts with the restriction that for all products the condiion |ai−B|≤k should be satisfied (where ai is the old price of the product and B is the same new price of all products) or report that it is impossible to find such price B.

Note that the chosen price B should be integer.

You should answer q independent queries.

Input
The first line of the input contains one integer q (1≤q≤100) — the number of queries. Each query is presented by two lines.

The first line of the query contains two integers n and k (1≤n≤100,1≤k≤108) — the number of products and the value k. The second line of the query contains n integers a1,a2,…,an (1≤ai≤108), where ai is the price of the i-th product.

Output
Print q integers, where the i-th integer is the answer B on the i-th query.

If it is impossible to equalize prices of all given products with restriction that for all products the condition |ai−B|≤k should be satisfied (where ai is the old price of the product and B is the new equal price of all products), print -1. Otherwise print the maximum possible equal price of all products.

Example
input

4
5 1
1 1 2 3 1
4 2
6 4 8 5
2 2
1 6
3 5
5 2 5

output

2
6
-1
7

Note

In the first example query you can choose the price B=2. It is easy to see that the difference between each old price and each new price B=2 is no more than 1.
In the second example query you can choose the price B=6 and then all the differences between old and new price B=6 will be no more than 2.
In the third example query you cannot choose any suitable price B. For any value B at least one condition out of two will be violated: |1−B|≤2, |6−B|≤2.
In the fourth example query all values B between 1 and 7 are valid. But the maximum is 7, so it’s the answer.

Intentional analysis:
An easy problem.First we should judge the difference between the maximum and minimum values in the series and the size relationship of 2*k.If the difference is greater than 2*k,we should output “-1”,else output the sum of the minimum in the series and k.

Click to see Chinese Intentional analysis 我们只需要判断数列中最大最小值的差跟2\*k的大小关系，如果差大于2\*k，就输出“-1”，否则就输出数列中的最小值与k的和。

Code:

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e4 + 10;
int       a[maxn];
int main()
{
int q, n, k;
cin >> q;
while (q--)
{
cin >> n >> k;
for (int i = 0; i < n; i++)
cin >> a[i];
sort(a, a + n);
if ((a[n - 1] - a[0]) > 2 * k)
{
puts("-1");
continue;
}
else
{
cout << a[0] + k << endl;
}
}
return 0;
}

C. Computer Game

Description:
Vova is playing a computer game. There are in total n turns in the game and Vova really wants to play all of them. The initial charge of his laptop battery (i.e. the charge before the start of the game) is k.

During each turn Vova can choose what to do:

If the current charge of his laptop battery is strictly greater than a, Vova can just play, and then the charge of his laptop battery will decrease by a;
if the current charge of his laptop battery is strictly greater than b (b<a), Vova can play and charge his laptop, and then the charge of his laptop battery will decrease by b;
if the current charge of his laptop battery is less than or equal to a and b at the same time then Vova cannot do anything and loses the game.
Vova wants to complete the game (Vova can complete the game if after each of n turns the charge of the laptop battery is strictly greater than 0). Among all possible ways to complete the game, Vova wants to choose the one where the number of turns when he just plays is the maximum possible. It is possible that Vova cannot complete the game at all.

Your task is to find out the maximum possible number of turns Vova can just play or report that Vova cannot complete the game.

You have to answer q independent queries.

Input
The first line of the input contains one integer q (1≤q≤105) — the number of queries. Each query is presented by a single line.

The only line of the query contains four integers k,n,a and b (1≤k,n≤1e9,1≤b<a≤1e9) — the initial charge of Vova’s laptop battery, the number of turns in the game and values a and b, correspondingly.

Output
For each query print one integer: -1 if Vova cannot complete the game or the maximum number of turns Vova can just play otherwise.

Example
input

6
15 5 3 2
15 5 4 3
15 5 2 1
15 5 5 1
16 7 5 2
20 5 7 3

output

4
-1
5
2
0
1

Note

In the first example query Vova can just play 4 turns and spend 12 units of charge and then one turn play and charge and spend 2 more units. So the remaining charge of the battery will be 1.
In the second example query Vova cannot complete the game because even if he will play and charge the battery during each turn then the charge of the notebook battery will be 0 after the last turn.

Intentional analysis:
When the laptop battery is greater than a(b)(equal is not allowed to reduce),it is going to reduce a(b).The number of reductions is greater than n to win.
We should find the maximum number of times “a” if Vova can finish the game.This requires a two-point answer.

Click to see Chinese Intentional analysis 如果能获胜，就输出a能得到的最大值。说人话就是设减a的次数为x，求出可以满足ax+b(n-x)>k的a的值。然后二分就行了。

Code:

#include <bits/stdc++.h>
using namespace std;
#define ll    long long
ll k, n, a, b;
bool cal(ll m)
{
return (a * m + b * (n - m)) < k;
}
int main()
{
ll q;
cin >> q;
while (q--)
{
cin >> k >> n >> a >> b;
ll l = 0, r = n;
while (l <= r)
{
ll m = (l + r) / 2;
if (cal(m))
l = m + 1;
else
r = m - 1;
}
if (l > 0)
cout << min(l - 1, n) << endl;
else
cout << "-1" << endl;
}
return 0;
}

Thanks for my master @boctorio

D. Candy Box (easy version)

Description:
This problem is actually a subproblem of problem G from the same contest.

There are n candies in a candy box. The type of the i-th candy is ai (1≤ai≤n).

You have to prepare a gift using some of these candies with the following restriction: the numbers of candies of each type presented in a gift should be all distinct (i. e. for example, a gift having two candies of type 1 and two candies of type 2 is bad).

It is possible that multiple types of candies are completely absent from the gift. It is also possible that not all candies of some types will be taken to a gift.

Your task is to find out the maximum possible size of the single gift you can prepare using the candies you have.

You have to answer q independent queries.

If you are Python programmer, consider using PyPy instead of Python when you submit your code.

Input
The first line of the input contains one integer q (1≤q≤2⋅105) — the number of queries. Each query is represented by two lines.

The first line of each query contains one integer n (1≤n≤2⋅105) — the number of candies.

The second line of each query contains n integers a1,a2,…,an (1≤ai≤n), where ai is the type of the i-th candy in the box.

It is guaranteed that the sum of n over all queries does not exceed 2⋅105.

Output
For each query print one integer — the maximum possible size of the single gift you can compose using candies you got in this query with the restriction described in the problem statement.

Example
input

3
8
1 4 8 4 5 6 3 8
16
2 1 3 3 4 3 4 4 1 3 2 2 2 4 1 1
9
2 2 4 4 4 7 7 7 7

output

3
10
9

Note

In the first query, you can prepare a gift with two candies of type 8 and one candy of type 5, totalling to 3 candies.
Note that this is not the only possible solution — taking two candies of type 4 and one candy of type 6 is also valid.

Intentional analysis:
Count the number of occurrences of each number and sort it out. Add from the biggest beginning, pay attention to if you can’t continue to add the same, you should add one smaller than him.

Click to see Chinese Intentional analysis 统计出每个数字出现的次数，排一下序。从最大的开始往下加，注意如果出现相同的不能继续加，应该加上一个比它小一的（比如说4，4加出来应该是4+3，如果是4，2，加起来应该是4+2）。难点就是如何实现。

Code:

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
int       num[maxn];
int       n, u, ans;
int main()
{
int q;
cin >> q;
while (q--)
{
cin >> n;
for (int i = 1; i <= n; i++)
num[i] = 0;
ans = 0;
for (int i = 1; i <= n; i++)
{
cin >> u;
num[u]++;
}
sort(num + 1, num + 1 + n);
int nw = n;
for (int i = n; i >= 1; i--)
{
if (nw > num[i])
{
nw = num[i];
}
ans += nw;
nw--;
if (nw <= 0)
break;
}
cout << ans << endl;
}
}


text:
## Totem Tribe Cheats, Codes, Hints and Walkthroughs for PC Games.

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Hints and Tips for: Totem Tribe

# Totem Tribe Cheats

 Totem Tribe Cheat Codes: ------------ Submitted by: RM If you cant find an item, go to "Totem Tribedata exturegameitem folder" and find your item. Open it with paint, and paint white/black areas with red (its your choice). Save it, and enter game. It become more visible. Secret Pirate Island: --------------------- On island worship, there are four areas which are pink sparklings, you need to put any unit every circle, for example put scouts on SW, put warriors on SE, sent the workers to build a tower on NW, and then picked one of the workers and put on the circle at NE. Hints: ------ For the area in the comet I found building what is needed to get everything built (put fighters in between those buildings or they will be easily destroyed) then filling all the other space with fighters, healers, enough builders to be able to repair quickly all the time, and a few towers in areas you don't see many troops is the best way to go. Do NOT keep your army clumped at a flag and the whole area is filled with fighters who provide enough protection to get 0 to minimal damage when you're being attacked every minute. Once you build up your army get the combat spell ready and get them together in the area right before the (boss) then move to flag to attack then cast your spell when they begin fighting (so you don't lose spell time) and make another one quickly and use it. CANCEL your fighting flag when you start to lose most of your men (usually while making 2nd spell) so regenerated troops stay to protect buildings, and let the rest die out fighting. After you have repaired all your buildings and army, do it again and again til the boss is dead. If you're having problems finding the boss it's because it's mainly secret passages. (look top middle and use your minimap). Others open up afte you defeat the boss and so on. Open Chest: ----------- To open up the chest, navigate the hunter to the chest. Cradle of the North Totem Locations: ------------------------------------ Dolphin - In the water on the left. Whale - In the water on the right. Seagull - Upper left near where are the seagulls are flying. Monkey - Top center (tower needed to see it clearly). Polar Bear - Near your base in the icy area. Lizard - NE from base, surrounded by pebbles. How to find every single item: ------------------------------ Go to your folder: Totem Tribe/date/texture/game/item. You'll find ".png" images of the items. Open them in GIMP or Paint and change the background to a red or pink and save. Start the game. You'll be able to find the items quite quickly then. In later levels you'll search for crystals. They can be found in sub-folder "structure" instead of "item". Turn off the Fog Of War: ------------------------ Go to "C:/Program Files/Totem Tribe/datacfg" and make a backup copy of "gamevisual.cfg" then open it with wordpad and scroll to the fow colors and change the dense so it matches the clear color as seen below. 00FFFFFF 00FFFFFF Note: The fog of war is still used on minimap and you still have to expand by building, but this will show you the entire screen so you know where to go. (I also changed the mist and cloud colors I found in another config file so all of them are light as well. Build anywhere: --------------- To build pretty much anywhere (in trees and mountains and middle of water) backup and open "gamelogic.cfg" with wordpad and change the 0.1 to 0.0 as seen below. 0.0 note builders do get stuck sometimes but can be fixed by destroying the bld. or build anything close enough they get unstuck then you can cancel it. Remember not to block pathways so you don't get a traffic jam. Last note. When building somewhere out of reach it still takes that extra time to auto complete itself. There are many more tweaks you can do by just looking around the config as well as "C:/Program Files/Totem Tribe/dataleveloriginal" where you can see exactly how many items and monsters as well as solutions to puzzles. Don't forget to back up files. Hints: ------ Code I found in level folder. At the oracle lake press "Blue", "Yellow", "Violet", "Green", "Red" (colors) to get 3 meteors to use as well as. Press "Red", "Blue", "Yellow", "Violet", "Green" to get a chest of jewels. Hints: ------ Watch for the guide stones and make sure you read them all. They are filled with useful information about your location, hints about where to find things and clues about how to solve the puzzles! Hints: ------ Submitted by: David K. * While workers only fulfill one task at a time, training more workers quickly will get the tasks completed much sooner. * If Aruku or Goru are killed you will have to restart the game. To keep them safe you can click on them and drag them to safety. * Scouts are the best unit to send out to explore the land. They are quicker, have special abilities and this allows you to keep your army near to defend the village. * Guiding stones provide worthwhile information, hints and tips to complete puzzles. * Always click on the Guiding Stones to see what valuable information they contain. Gem painting: ------------- Many people have experienced difficulty finding all the gems that are hidden among the many islands of Totem Tribe. Since many times the gems are positioned to "blend in" with the background, here is a procedure whereby you can "paint" the gem images the game uses so they are easier to see. * First, exit out of the game completely. * Next, locate the directory in which the gem images are stored: C:\Program Files\Totem Tribe\data\texture\game\item * Make a backup of this directory for safe keeping. * The gems (there are 7 of them) are all stored in files named like this one: "gem_green.png" * Right-click on each and select "Open With" and choose Paint. (the program included with Windows). * Paint will come up with the picture you chose. * The easiest thing to do is to choose the spray paint option and the larger "speckle" and choose a color like burgundy or purple or something that more than likely isn't in the level. * When done, choose "save". * Launch the game, and your gems should be much easier to see now! Purple pyramids puzzle: ----------------------- (WinXP, your mileage may vary) Go to "..\Totem Tribe\data\level\original" then COPY and SAVE "level19.xml". Open "level19.xml2 with notepad or any other editor that can handle xml files. Find the text that says: structure hint="cr_#" There will be 5 of these, and they take a little searching because they're mixed into a very long list of structure commands. Here's the one for cr_2: structure hint="cr_2" type="Crystal" x="457" y="1416" What we're going to do is change the x and y values for cr_3, cr_4, and cr_5 (1 and 2 are the crystals near the comet and will be left alone.) Orignally these values read: cr_3: x="2272" y="1253" cr_4: x="1780" y="1675" cr_5: x="2369" y="2236" Change them as follows: cr_3: x="672" y="1553" cr_4: x="680" y="1375" cr_5: x="669" y="1136" Find the spot that reads: area name="cr_#". Here's a sample: area name="cr_3" radius="32" x="672" y="1553" These are a little easier because they're right next to each other. Change the x and y values to match the ones you used in the last set. Save the file. Open Totem Tribe. You probably have to restart the level. The 5 crystals should now form a rough sort of V around the comet/totem crystal, and all fit on the screen at the same time. It should now be possible to click them all in the allotted time. Addendum to Purple Pyramids Puzzle: ----------------------------------- Submitted by: liobhain If you continue searching the "level19.xml" file for: [area name="cr_3" radius="32" x="2272" y="1253" /] you can change these to match the new x,y values changed in "Purple pyramids puzzle" in order to move effects, etc. (In my file, the area names are defined in this order 3, 1, 2, 4, 5.) How to make your hut able to recharge magic spell: -------------------------------------------------- Submitted by: trangttuong open C:\Program Files\Totem Tribe\data\cfg, find the file called building(backup it just in case you mess up). Open it with notepad. Now use press ctrl+f to find the phrase "temple" a few lines below there is a bunch of sentence begin with "ability" Ex:[ability type="charge"]Purification [/ability]. This line means "allow the building to charge Purification spell". Now copy all of those ability you want ( there are 6 of em there) then press ctrl+f again. now find with the direction is up the phrase "[ability type="spawn"] Worker[/ability]". when you found it. press enter then paste all the ability you want there (remember, the symbol "<" is need to be equal with the line above, or it will make a bug and you cant play the game). Then check and make sure there is no empty line, and all "<" symbol is equal. Then save. Now your hut have the abikity to charge every spell you want (Note: each hut can only charge 1 spell at a time, then it will not charge any more until you already used that spell). To find king shade: ------------------- Submitted by: Maddie On dreamshore, you will notice every time you battle and defeat a prince shade, you earn a dark glow. Add the four glows to the boxes near the village. A purple glow will show. When enyone enters this, they will be teleported to the king shades abyss. When you defeat the king shade, the island in the abyss falls and kills most of your fighters. Then the lodges will reproduce.  Visit Cheatinfo for more Cheat Codes, FAQs or Tips!


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# Behavior Research Methods

## List of Papers (Total 5,402)

#### Gaze tracking accuracy in humans: One eye is sometimes better than two

Most modern video eye trackers deliver binocular data. Many researchers take the average of the left and right eye signals (the version signal) to decrease the variable error (precision) up to a factor of $$\sqrt {2}$$. What happens to the systematic error (accuracy) if the left and right eye signals are averaged? To determine the systematic error, we conducted a calibration...

#### Moderation analysis in two-instance repeated measures designs: Probing methods and multiple moderator models

Moderation hypotheses appear in every area of psychological science, but the methods for testing and probing moderation in two-instance repeated measures designs are incomplete. This article begins with a short overview of testing and probing interactions in between-participant designs. Next I review the methods outlined in Judd, McClelland, and Smith (Psychological Methods 1...

#### Fast and slow errors: Logistic regression to identify patterns in accuracy–response time relationships

Understanding error and response time patterns is essential for making inferences in several domains of cognitive psychology. Crucial insights on cognitive performance and typical behavioral patterns are disclosed by using distributional analyses such as conditional accuracy functions (CAFs) instead of mean statistics. Several common behavioral error patterns revealed by CAFs are...

#### Toward the markerless and automatic analysis of kinematic features: A toolkit for gesture and movement research

Action, gesture, and sign represent unique aspects of human communication that use form and movement to convey meaning. Researchers typically use manual coding of video data to characterize naturalistic, meaningful movements at various levels of description, but the availability of markerless motion-tracking technology allows for quantification of the kinematic features of...

#### Replication Bayes factors from evidence updating

We describe a general method that allows experimenters to quantify the evidence from the data of a direct replication attempt given data already acquired from an original study. These so-called replication Bayes factors are a reconceptualization of the ones introduced by Verhagen and Wagenmakers (Journal of Experimental Psychology: General, 143(4), 1457–1475 2014) for the common...

#### Sit still and pay attention: Using the Wii Balance-Board to detect lapses in concentration in children during psychophysical testing

During psychophysical testing, a loss of concentration can cause observers to answer incorrectly, even when the stimulus is clearly perceptible. Such lapses limit the accuracy and speed of many psychophysical measurements. This study evaluates an automated technique for detecting lapses based on body movement (postural instability). Thirty-five children (8–11 years of age) and 34...

#### The EU-Emotion Voice Database

In this study, we report the validation results of the EU-Emotion Voice Database, an emotional voice database available for scientific use, containing a total of 2,159 validated emotional voice stimuli. The EU-Emotion voice stimuli consist of audio-recordings of 54 actors, each uttering sentences with the intention of conveying 20 different emotional states (plus neutral). The...

#### The Visual Analogue Scale for Rating, Ranking and Paired-Comparison (VAS-RRP): A new technique for psychological measurement

Traditionally, the visual analogue scale (VAS) has been proposed to overcome the limitations of ordinal measures from Likert-type scales. However, the function of VASs to overcome the limitations of response styles to Likert-type scales has not yet been addressed. Previous research using ranking and paired comparisons to compensate for the response styles of Likert-type scales...

#### Picture perfect: A stimulus set of 225 pairs of matched clipart and photographic images normed by Mechanical Turk and laboratory participants

The present study provides normative measures for a new stimulus set of images consisting of 225 everyday objects, each depicted both as a photograph and a matched clipart image generated directly from the photograph (450 images total). The clipart images preserve the same scale, shape, orientation, and general color features as the corresponding photographs. Various norms (modal...

#### Development and validation of a high-speed stereoscopic eyetracker

Traditional video-based eyetrackers require participants to perform an individual calibration procedure, which involves the fixation of multiple points on a screen. However, certain participants (e.g., people with oculomotor and/or visual problems or infants) are unable to perform this task reliably. Previous work has shown that with two cameras one can estimate the orientation...

#### Modeling competence development in the presence of selection bias

A major challenge for representative longitudinal studies is panel attrition, because some respondents refuse to continue participating across all measurement waves. Depending on the nature of this selection process, statistical inferences based on the observed sample can be biased. Therefore, statistical analyses need to consider a missing-data mechanism. Because each missing...

#### Safe and sensible preprocessing and baseline correction of pupil-size data

Measurement of pupil size (pupillometry) has recently gained renewed interest from psychologists, but there is little agreement on how pupil-size data is best analyzed. Here we focus on one aspect of pupillometric analyses: baseline correction, i.e., analyzing changes in pupil size relative to a baseline period. Baseline correction is useful in experiments that investigate the...

#### General mixture item response models with different item response structures: Exposition with an application to Likert scales

This article proposes a general mixture item response theory (IRT) framework that allows for classes of persons to differ with respect to the type of processes underlying the item responses. Through the use of mixture models, nonnested IRT models with different structures can be estimated for different classes, and class membership can be estimated for each person in the sample...

#### Correction to: Estimating effect size when there is clustering in one treatment group

Equation (26) is formatted incorrectly in the pdf version. It should appear as follows.

#### All for one or some for all? Evaluating informative hypotheses using multiple N = 1 studies

Analyses are mostly executed at the population level, whereas in many applications the interest is on the individual level instead of the population level. In this paper, multiple N = 1 experiments are considered, where participants perform multiple trials with a dichotomous outcome in various conditions. Expectations with respect to the performance of participants can be...

#### PredPsych: A toolbox for predictive machine learning-based approach in experimental psychology research

Recent years have seen an increased interest in machine learning-based predictive methods for analyzing quantitative behavioral data in experimental psychology. While these methods can achieve relatively greater sensitivity compared to conventional univariate techniques, they still lack an established and accessible implementation. The aim of current work was to build an open...

#### Semantic ambiguity effects on traditional Chinese character naming: A corpus-based approach

Words are considered semantically ambiguous if they have more than one meaning and can be used in multiple contexts. A number of recent studies have provided objective ambiguity measures by using a corpus-based approach and have demonstrated ambiguity advantages in both naming and lexical decision tasks. Although the predictive power of objective ambiguity measures has been...

#### The Bangor Voice Matching Test: A standardized test for the assessment of voice perception ability

Recognising the identity of conspecifics is an important yet highly variable skill. Approximately 2 % of the population suffers from a socially debilitating deficit in face recognition. More recently the existence of a similar deficit in voice perception has emerged (phonagnosia). Face perception tests have been readily available for years, advancing our understanding of...

#### Detecting treatment-subgroup interactions in clustered data with generalized linear mixed-effects model trees

Identification of subgroups of patients for whom treatment A is more effective than treatment B, and vice versa, is of key importance to the development of personalized medicine. Tree-based algorithms are helpful tools for the detection of such interactions, but none of the available algorithms allow for taking into account clustered or nested dataset structures, which are...

#### Is human classification by experienced untrained observers a gold standard in fixation detection?

Manual classification is still a common method to evaluate event detection algorithms. The procedure is often as follows: Two or three human coders and the algorithm classify a significant quantity of data. In the gold standard approach, deviations from the human classifications are considered to be due to mistakes of the algorithm. However, little is known about human...

#### Examining reproducibility in psychology: A hybrid method for combining a statistically significant original study and a replication

The unrealistically high rate of positive results within psychology has increased the attention to replication research. However, researchers who conduct a replication and want to statistically combine the results of their replication with a statistically significant original study encounter problems when using traditional meta-analysis techniques. The original study’s effect...

#### The Box Task: A tool to design experiments for assessing visuospatial working memory

The present paper describes the Box Task, a paradigm for the computerized assessment of visuospatial working memory. In this task, hidden objects have to be searched by opening closed boxes that are shown at different locations on the computer screen. The set size (i.e., number of boxes that must be searched) can be varied and different error scores can be computed that measure...

#### GestuRe and ACtion Exemplar (GRACE) video database: stimuli for research on manners of human locomotion and iconic gestures

Human locomotion is a fundamental class of events, and manners of locomotion (e.g., how the limbs are used to achieve a change of location) are commonly encoded in language and gesture. To our knowledge, there is no openly accessible database containing normed human locomotion stimuli. Therefore, we introduce the GestuRe and ACtion Exemplar (GRACE) video database, which contains...

#### The cognitive reflection test is robust to multiple exposures

The cognitive reflection test (CRT) is a widely used measure of the propensity to engage in analytic or deliberative reasoning in lieu of gut feelings or intuitions. CRT problems are unique because they reliably cue intuitive but incorrect responses and, therefore, appear simple among those who do poorly. By virtue of being composed of so-called “trick problems” that, in theory...


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Place the mouse over a yellow county. It will turn to a hand click and you will be taken to the pictures within that county.

Counties with pictures on this map are: , , , , , , , , , , , , , , Jasper , , , , , , , , , Putnam , , , , Walton, and Whitfield .

Full County Alphabetical Index

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# Proving that the function $f(x)=\sum_n\frac{1}{10^n}\{10^nx\}$ is everywhere continuous but nowhere differentiable.

Theorem: The function $$f(x)=\sum_{n=1}^{\infty}\frac{1}{10^n}\{10^nx\}$$ is everywhere continuous but nowhere differentiable, where $$\{.\}$$ represents distance from nearest integer.
This theorem has been taken from chapter $$23$$ of Spivak's Calculus book.

By Weirstrauss M test, $$f$$ is uniformly continuous.
In the book, the theorem is proven by showing that the limit $$L=\lim_{m\to \infty}\frac{f(a+h_m)-f(a)}{h_m}=\lim_{m\to \infty}\sum_{n=1}^{\infty}\frac{\{10^n(a+h_m)\}-\{10^na\}}{10^nh_m}$$ does not exist when $$h_m\to 0$$, where $$a\in (0,1]$$. Let $$a=0.a_1a_2\cdots$$

Let $$h_m=10^{-m}$$ if $$a_m\ne 4,9$$ and $$h_m=-10^{-m}$$ if $$a_m=4,9$$.$$\tag{1}$$
No. of terms in summation in the above limit $$L$$ is finite as if $$n\ge m, 10^nh_m$$ is integer and hence numerator of the summations becomes zero. So for $$n\lt m$$, $$\{10^na\}=\text{integer}+0.a_{n+1}a_{n+2}\cdots a_m\cdots$$ and $$10^n \{a+h_m\}=\text{integer}+0.a_{n+1}a_{n+2}\cdots (a_m\pm1)\cdots$$ and this representation into decimals is correct as $$h_m=-10^{-m}$$ if $$m=9$$.
**

Then, Spivak makes a statement that if $$0.a_{n+1}a_{n+2}\cdots a_m\cdots \le 0.5$$, then we also have $$0.a_{n+1}a_{n+2}\cdots (a_m\pm 1)\cdots \le 0.5$$ as $$h_m=-10^{-m}$$ if $$a_m=4$$.

**

And I think that this is not true at all because in the special case $$m=n+1$$, if $$a_m=5$$ then clearly $$0.a_m\le 0.5$$ but $$0.(a_m\pm 1)\le 0.5$$ is not true! and hence condition on $$h_m$$ should be $$h_m=-10^{-m}$$ when $$a_m=5,9$$ and $$h_m=10^{-m}$$ when $$a_m\ne 5,9$$. Is my understanding correct?

Another doubt that I have is:

Is the following alternative way correct?
Let's choose $$h_m=10^{-m}$$ if $$a_m\ne 9$$ and $$h_m=-10^{-m}$$ if $$a_m=9$$. Then it's clear by writing decimal representation that, $$\{10^n(a+h_m)\}-\{10^na\}=\pm 10^{n-m}$$ and then $$L=\lim_{m\to \infty}\frac{f(a+h_m)-f(a)}{h_m}=\lim_{m\to \infty}\sum_{n=1}^{\infty}\frac{\{10^n(a+h_m)\}-\{10^na\}}{10^nh_m}=\lim_{m\to \infty}\sum_{n=1}^{m-1}\frac{\{10^n(a+h_m)\}-\{10^na\}}{10^nh_m}\lim_{m\to \infty}\sum_{n=1}^{m-1}\pm 1=\lim_{m\to \infty}\pm (m-1)$$,
which doesn't exist. Hence proved.

• @OliverDiaz: Even in the book in your link, it's not made clear why $h_m=-10^{-m}$ when $a_m=4$, which I think should have been $a_m=5$. And why do we have to distinguish cases $\lt 1/2$ , $\gt 1/2$? – Koro Sep 16 at 21:46
• Do you agree that with $h_m=-10^{m}$ for $a_m=4,9$ and $h_m=10^{-m}$ things work? All you need is a particular sequence $h_m\rightarrow0$ for which the sum does not converge. notice that there is a choice to make for the decimal expansion of $x$ when $x$ has a a possible finite decimal expansion. – Oliver Diaz Sep 16 at 21:56
• @OliverDiaz: I disagree with $h_m=-10^{-m}$ when $a_m=4$ as it will create problems for $a_m=5$. – Koro Sep 16 at 22:04

This is a comment that is to long for the comment section.

The idea of the construction is to keep preserved fractional parts on the same side of $$\frac{1}{2}$$, that is if $$10^na-[10^na]<\frac{1}{2}$$, then $$10^n(a+h_m)-[10^na]<\frac12$$ and if $$10^na-[10^na]\geq\frac{1}{2}$$, then $$10^n(a+h_m)-[10^na]\geq\frac12$$

Try to convince yourself that

• If $$a_m\notin\{4,9\}$$, then for all $$n=1,\ldots,m-1$$, $$\{10^n(a+10^{-m})\}-\{10^na\}=10^{n-m}\,(-1)^{\mathbb{1}_{a_n<5}(a)}$$ For $$n\geq m$$ $$\{10^n(a+10^{-m})\}=\{0.a_{m+1}a_{m+2}\ldots\}=\{10^na\}$$ so $$\Delta_m:=\frac{1}{10^{-m}}\sum^\infty_{n=1}\frac{\{10^n(a+10^{-m})\}-\{10^na\}}{10^n}\equiv m\mod 2$$

• If $$a_m\in\{4,9\}$$, then for all $$n=1,\ldots,m-1$$, $$\{10^n(a-10^{-m})\}-\{10^na\}=10^{n-m}\,(-1)^{1+\mathbb{1}_{a_n<5}(a)}$$ For $$n\geq m$$ $$\{10^n(a-10^{-m})\}=\{0.a_{m+1}a_{m+2}\ldots\}=\{10^na\}$$ so $$\Delta_m:=-\frac{1}{10^{-m}}\sum^\infty_{n=1}\frac{\{10^n(a-10^{-m})\}-\{10^na\}}{10^n}\equiv m\mod 2$$

If you were to modify the construction and set $$h_m=-10^{-m}$$ if $$a_m\in\{5,9\}$$ and $$h_m=10^{-m}$$ otherwise then, for $$a$$ with $$a_m=4$$ you get

$$\{10^{m-1}(a+10^{m})\}-\{10^{m-1}a\}=1-10^{-m} -2(0.4a_{m+1}\ldots)$$ and so $$\frac{1}{10^m}\sum^\infty_{n=1}\frac{\{10^n(a+10^{-m})\}-\{10^na\}}{10^n}\notin\mathbb{Z}$$

The genius of the construction is to keep all differential increments $$\Delta_m$$ as integers that vary on the parity of $$m$$.

• In "that is if $10^na-[10^na]<\frac{1}{2}$, then $10^n(a+h_m)-[10^na]<\frac12$", did you mean $10^n(a+h_m)-[10^n(a+h_m)]<\frac12$? I suppose $[.]$ is greatest integer function. – Koro Sep 17 at 5:45
• No, I wrote what I meant. The construction maintains which side of of 1/2 the fractional part is. – Oliver Diaz Sep 17 at 14:09
• @Koro: In the construction, the only crucial setp is when $a_m=4$ and $n=m-1$. then $\{10^{m-1}(0.a_1\ldots a_{m-1}(4-1)a_{m+1}\ldots\}-\{10^{m-1}(0.a_1\ldots a_{m-1}4a_{m+1}\ldots\}=-0.1$ – Oliver Diaz Sep 17 at 14:59


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## M13.1.1 Practice Exercise

_ Do these practice questions on your own, then click on the questions to see the step-by-step solution. Question 1: What is the value of…

## M13.6.4 Practice Exercise

_ Do these practice questions on your own, then click on the questions to see the step-by-step solution. Question 31: Express $$\frac{2x+3}{x^{2}+x-6}$$ as…

## M13.6.3 Practice Exercise

_ Do these practice questions on your own, then click on the questions to see the step-by-step solution. Question 21: Expand \( ~ \left( 3x-\frac{1}{2x^{2}}…

## M13.6.2 Practice Exercise

_ Do these practice questions on your own, then click on the questions to see the step-by-step solution. Question 11: By mathematical induction prove that…

## M13.6.1 Practice Exercise

_ Do these practice questions on your own, then click on the questions to see the step-by-step solution. Question 1 : If we define a…


text:
## anonymous 4 years ago what is the maxium value that the graph of y=sin x assumes?

1. anonymous

it was 1

2. Zarkon

for all x $-1\le \sin(x)\le 1$

3. anonymous

the term that describes a function that is symmetric about the y axis is ____________?


text:
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Milena Venkova3 years ago

Created this as a copy of Vicky's copy of Differentiation 11 - Product Rule (Basic).
Differentiation 11 - Product Rule (Basic) Ready to use Katie Lester 01/06/2016 09:47
Differentiation 12 - Product Rule (with Chain Rule) Ready to use Katie Lester 01/06/2016 09:47
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CINCO UNO Derivada producto draft Marlon Arcila 17/11/2016 09:56
CINCO DOS Derivada producto draft Marlon Arcila 19/11/2016 14:56
Differentiation 11 - Product Rule (Basic) draft steve kilgallon 21/12/2016 13:08
Differentiation 12 - Product Rule (with Chain Rule) draft steve kilgallon 21/12/2016 13:09
Product Rule draft cormac breen 31/01/2017 13:25
QM103_test2_2017_Q3 draft steve kilgallon 22/02/2017 12:34
cormac's copy of Product Rule draft cormac breen 17/04/2020 19:43
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Milena's copy of cormac's copy of cormac's copy of Product Rule draft Milena Venkova 08/03/2018 12:26
Milena's copy of Vicky's copy of Differentiation 11 - Product Rule (Basic) draft Milena Venkova 27/02/2018 14:56
Product Rule draft Milena Venkova 06/03/2018 14:43
Product Rule 2 draft Milena Venkova 08/03/2018 15:36
Milena's copy of Product Rule (Basic) draft Milena Venkova 08/03/2018 15:45
Product Rule (Basic)3 draft Milena Venkova 13/03/2018 14:34
Differentiation 12 - Product Rule (with Chain Rule) draft Xiaodan Leng 11/07/2019 05:33
Differentiation 11 - Product Rule (Basic) draft Xiaodan Leng 11/07/2019 06:56
Differentiation 12 - Product Rule (with Chain Rule) draft Xiaodan Leng 11/07/2019 06:56
Differentiation 12 - Product Rule (with Chain Rule) draft Timur Zaripov 11/10/2019 15:57
Differentiation - Product Rule (with Chain Rule) draft Andrew Kerridge 30/10/2019 09:38
Differentiation - Product Rule (Basic) draft Andrew Kerridge 31/03/2020 11:50
W5b - Differentiation 11 - Product Rule (Basic) Ready to use Timur Zaripov 31/10/2019 11:08
W5b - Differentiation 12 - Product Rule (with Chain Rule) Ready to use Timur Zaripov 31/10/2019 10:58
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Musa's copy of Differentiation - Product Rule (Basic) draft Musa Mammadov 25/04/2020 22:49
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Musa's copy of 3 Differentiation - Product Rule (Basic) draft Musa Mammadov 13/05/2020 09:14
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text:
# Synopsis: Element $Z=117$ Confirmed

Researchers have independently confirmed the existence of element $117$ and have discovered a new isotope along the way.

Nuclear physicists have invested huge effort in creating superheavy elements, which consist of enough neutrons to provide enhanced stability from nuclear decay. For the past $30$ years, experiments have been marching towards this “island of stability” with a new elemental discovery every $2$ to $3$ years. Part of the discovery process includes the confirmation by an independent experimental collaboration—it is only at this point that an element obtains its official status.

An international team using an intense ${}^{48}\text{Ca}$ beam provided by GSI research facility in Darmstadt, Germany, and a target material of radioactive ${}^{249}\text{Bk}$ supplied by Oak Ridge National Lab in Tennessee has produced two atoms of the superheavy element with atomic number $Z=117$, confirming the initial observation published in 2010 (see 9 April 2010 Viewpoint). In the process, a new isotope ${}^{266}\text{Lr}$ was discovered from the previously unknown alpha-decay branch of ${}^{270}\text{Db}$. With a half-life of $1$ hour, ${}^{270}\text{Db}$ is the longest-lived alpha emitter having an atomic number, $Z$, greater than $102$.

The experiment is a tour de force in superheavy element research and required a detailed reconstruction of a seven-step alpha-decay chain followed by the spontaneous fission of the newly discovered ${}^{266}\text{Lr}$. The difficulty stems from the large variation in decay lifetimes along the alpha chain. The discovery was made feasible by the use of TASCA, a gas-filled recoil separator specifically designed for a high selectivity of superheavy or transactinide elements.

The confirmation by the TASCA team serves as a much-needed step on the long road towards the island of stability. An easier feat will be deciding on a name for $Z=117$. – Kevin Dusling

More Features »

### Announcements

More Announcements »

Nuclear Physics

Astrophysics

Fluid Dynamics

## Related Articles

Gravitation

### Viewpoint: Moiré Effect Could Enhance Neutron Interferometry

A new and more flexible neutron interferometer design relies on the moiré effect, in which two periodic patterns are combined to give a longer-period pattern. Read More »

Nuclear Physics

### Synopsis: Intel on Stellar Element Production from Accelerator Data

Measurements of a nuclear reaction relevant to the synthesis of calcium, potassium, and argon in stars boost the accuracy of models for predicting the elements’ abundances. Read More »

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### Viewpoint: Heaviest Element Has Unusual Shell Structure

Calculations of the structure in oganesson—the element with the highest atomic number—reveal a uniform, gas-like distribution of its electrons and nucleons. Read More »


text:
## Prove conjugation law for Exp(z) using only Exp(x + y) = Exp(x)Exp(y)

Starting with just the property $$E(x + y) = E(x)E(y)$$, one can prove quite a lot of the main properties of the exponential function on real numbers. For example, $$E(0) = 1$$, and $$E'(X) = E(X)$$, and $$E(nx) = E(x)^n$$ all follow from a straight-forward application of definitions.

To move this in to complex analysis and Euler’s formula, it seems to me that the key property that needs to be proved is $$E(\overline{z})$$ = $$\overline{E(z)}$$.

Is there a nice way to prove this that is in the spirit of this particular line of reasoning?

Edit: For example, Chapter 8 of Baby Rudin follows this reasoning, but resorts to the definition $$E(x) = \Sigma_{n\ge0}(\frac{z^n}{n!})$$ to prove the conjugation property.


text:
## Collatz Tool Box --- any speed ups possible?

4

3

I have created a tool box for experimenting with the Collatz conjecture. Info here: Wolfram MathWorld and here: Wikipedia The patterns I have found allow us to make the conjecture jump through some hoops.

Final Edit countOrbit has been re-cast as a Do Until loop because the logic behaves similar to a one-directional linked list and I don't think it can be improved.

Note: the number of sub orbits $\leq$ (total steps) / 6

Edit for the down-voters: the current maximum that collatz projects have reached is around $10^{18}$. The example below has two numbers $10^{301030}$ and $10^{477119}$ that my functions can use quickly and accurately. That is many magnitudes greater that what is currently being done. So, why the down-votes?

This is one example: m is the count of numbers in an ascending sequence. uniqueRank[1,m] returns the first number of the first unique sub orbit of that length. omegaSubOrbit[x,m] returns the last number of that sub orbit. We display the digit counts of these numbers. Then we count the complete orbit down to 1. This takes a bit over 12 minutes on my AMD 1100T. The left-hand number is the count of multiplies (we don't count divisions by two) and the right-hand count is the number of sub orbits used to get the counts.

m = 1000000; x = uniqueRank[1, m];
y = omegaSubOrbit[x, m];
{IntegerLength[x], IntegerLength[y]}
countOrbit[1, x]

{301030, 477119}
{4805005, 1903828}
{13420758, 1903828} 7.8 minutes to count all multiplies and divides

My question: Since I'm using For and While in a few places, is there a way to speed up those functions? Especially, the countOrbit function?

ascendingQ[x_] := 3 == Mod[x, 4]
uniqueQ[x_] := 0 != Mod[2 x - 1, 3]
subOrbit[x_] :=
Block[{y = x},
Flatten[Join[
Table[y \[DirectedEdge] (y = 1/2 (3 y + 1)), {n, 1,
IntegerExponent[x + 1, 2] - 1}], {y \[DirectedEdge] (3 y + 1)/
2^IntegerExponent[3 y + 1, 2]}]]]
createSubOrbitGraph[n_] :=
Block[{j, v = {}},
For[j = 1, j <= n, j++,
If[uniqueQ[2 j + 1], AppendTo[v, subOrbit[2 j + 1]]];
];
Flatten[v]
]
uniqueRank[n_, m_] := Block[{a = If[1 != n && OddQ[n], n - 1, n]},
(n + IntegerExponent[a, 2]) 2^m - 1]
nonUniqueRank[n_, m_] := (2 n - 1)   2^m - 1
createOrbit[w_, x_] := Block[{u = subOrbit[x]},
While[w < u[[-1, 2]],
u = Flatten[AppendTo[u, subOrbit[u[[-1, 2]]]]]];
u]
createOrbitGraph[n_] :=
Block[{j, u = {}},
For[j = 1, j <= n, j++,
If[uniqueQ[2 j + 1], u = Union[u, createOrbit[2 j + 1, 2 j + 1]]];
];
Flatten[u]
]
omegaSubOrbit[x_, m_] :=
Block[{z = (-1 + (3/2)^m (1 + x))}, 2^(1 - IntegerExponent[2 z, 2]) z]
omegaSubOrbit[x_] :=
Block[{m = IntegerExponent[x + 1, 2], z = (-1 + (3/2)^m (1 + x))},
2^(1 - IntegerExponent[2 z, 2]) z]
countOrbit[w_, x_] :=
Block[{h = x, t, c, d = 0, s = 0, l = 0, m, n, a},
While[True,
m = IntegerExponent[h + 1, 2];
t = -1 + (3/2)^m (1 + h);
n = IntegerExponent[t, 2];
l += m;
d += n;
s += 1;
h = t/2^n;
If[h <= w, Break[]];
];
c = 2 l + d;
a = If[1 == w, "Full ", "Extended Sub "];
Print["---Counts ", a, "Orbit---",
"\nNumber of Sub Orbits: ", s,
"\nMultiplications by 3: ", l,
"\nDivisions by 2:       ", l,
"\nExtra Divisions by 2: ", d,
"\nTotal for Orbit:      ", c]
]

Some brief documentation:
ascendingQ[x] Returns True if odd x is ascending, False if descending
uniqueQ[x] Returns True if odd x is not embedded in any longer sequence
subOrbit[x] Returns graph data in the form {Head[DirectedEdge]Tail}
createSubOrbitGraph[n] Returns graph data of sub orbits up through n
uniqueRank[n, m] Returns the first number of the n-th unique occurance of a sub orbit of size m
nonUniqueRank[n, m] Returns the first number of the n-th occurance of a sub orbit of size m, which may not be unique.
createOrbit[w, x] Returns graph data for w=1 complete orbit or w=x extended sub orbit
Note: for a graph, both are equivalent. w=x produces fewer duplicates.
createOrbitGraph[n] Returns graph data of extended sub orbits up through n
omegaSubOrbit[x] Returns the last odd number of a sub orbit. (The one that required multiple divides.)
omegaSubOrbit[x, m] Same as above, but uses m to speed up the first step
countOrbit[w, x] Returns the count of multiply and division steps and the count of sub orbits processed, for w=1 complete orbit or w=x extended sub orbit

Edit Replaced countOrbit[w,x] with refined version that counts divisions and multiplication. Reduced the time from 15 minutes to 7.8 minutes for the above example.

1

many more basic collatz ideas here & hope to hear from others in [chat]

– vzn – 2015-06-12T17:22:43.997

The Collatz toolbox is useful but it could be improved if it included optimized code for merging Collatz sequences to generate tree graphs of the sequences (ending in 4 -> 2 -> 1), with a few bells and options for coloring even and odd $n$ nodes, or placements based on iteration time, or $n$ value... – David G. Stork – 2017-10-10T16:22:45.200

I didn't downvote this, but it may be that some "speed up my code" questions aren't always well-received by everyone. relevant Meta question...

– cormullion – 2013-07-31T11:41:04.270

@cormullion, Because I'm an old procedural programmer, I feel most comfortable using For and While and I don't usually see how to eliminate them. This post is two-fold---One, to let others grab this stuff and play with it and Two, to find out how to speed up the countOrbit function to maybe get it below 10 minutes duration on my pc. – Fred Kline – 2013-07-31T12:07:37.783

@cormullion, and I had to show all the code because if you tweak one function, most likely you would have to tweak another to keep everything working together. – Fred Kline – 2013-07-31T12:10:47.827

1

I posted this link recently. Might be of interest...

– cormullion – 2013-07-31T12:22:37.540

I admit I am not familiar with this problem and do not really understand your code, and this understanding will certainly be necessary for anyone wishing to contribute any serious improvements. Do you really systematically check every number up to $10^{477119}$ in 12 minutes? – Oleksandr R. – 2013-07-31T12:46:22.050

@OleksandrR.,no, I use my short-cut, which is what this is all about. Notice these numbers from the example: $4805005, 1903828$. The first is the count of multiplies required to get down to $1$ and the second is the number of sub orbits processed to get to that number. Each sub orbit contains its own length, so the first sub orbit adds $1000000$ to the count and moves on to the next sub orbit. So, the whole thing is done with a series of additions. The key is: IntegerExponent[x+1,2] applied to the first number of the sub orbit provides that count. – Fred Kline – 2013-07-31T12:59:07.870

@OleksandrR., also the $10^{477119}$ number is calculated in omegaSubOrbit[x] with x being the uniqueRank number. Both calculations are very quick--under 2 seconds. – Fred Kline – 2013-07-31T13:05:51.527

This Q and A is better for learning how to eliminate inefficient procedural habits. – cormullion – 2013-07-31T14:39:15.660

4The reason for my downvoting this is: on the one hand the "speed up my code" reason that cormuillion mentioned, and that you give too many functions without explaining what they do to someone who doesn't know about the Collatz problem.

I am happy to retract the vote if you make this into a more minimal question with an explanation for what the conjecture is, why subOrbit[n] skips the even steps and some indication this is done correctly and so on.. – gpap – 2013-07-31T14:58:35.710

1

Take a look at Leonid's great answer in this question, it will give you an idea how to improve your code.

– sebhofer – 2013-08-03T08:26:17.307

As stated in the OP, this is a "linked-list" type of algorithm that I don't feel can be speeded up using the usual methods. So, I figured that compiling the functions to use Integers Of Unusual Size would be the only way to speed up things. As noted in an answer on that thread, Mathematica compiles are limited to machine-precision so, I'm debating with myself about crafting a C program using GMP to test some timings. I will accept this answer to clear this thread from the unanswered queue.