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P.S. By natural, I mean elegant or insightful.
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Your solution looks good. Perhaps the "most natural" example would be for $\alpha(n) = n-1$ if $n > 1$ and $\alpha(1) = 1$. (Think of $\alpha$ as the "shift one unit left -- if possible" function.) Then take $\beta$ to be the "shift right one unit" function, $\beta(n) = n+1$. Then $\alpha \beta = \text{id}_S$, but $\beta \alpha \neq \text{id}_S$. – Michael Joyce Oct 21 '12 at 20:39
@MichaelJoyce, looking at your thoughts as well as N.S.'s, I see that my issue was entirely due to myself. It was because I saw piecewise functions as unnatural that I struggled with finding an example. I also, after looking at some of the awesome maps presented by Jacobson, assumed there was an expectation for some marvelous geometric map. I see now that that's probably not the case. – 000 Oct 21 '12 at 20:42
There's nothing wrong with coming with a solution that is not "aesthetically optimal" at first. The aesthetic solutions usually come later after you've had more time to fully internalize everything about a problem. But there can be as much learning as finding a non-standard solution as there is to finding a very elegant one. BTW, I would consider the example I presented as "geometric" in so far as it can be visualized as pulling a string of beads in a one-dimensional universe. :) – Michael Joyce Oct 21 '12 at 20:47
$\alpha(x)=2x$ and $\beta(x) = \lfloor \frac{x+1}{2} \rfloor$ is probably more natural.
More generarily, given any $\alpha$ which is 1-1 but not bijective function, you can always find a $\beta$ so that $\beta \alpha =1_S$. But since $\alpha$ is not bijective, $\alpha \beta \neq 1_S$.
You may need $\beta(x) = \lceil \frac{x}{2} \rceil$ since $S$ starts at $1$ – Henry Oct 21 '12 at 20:57 | {
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# In base 10
After writing a solution to this problem I focused on product of digits and got the following result. Also I have a challenge also for all those who found this interesting try if you can do it!
Let $f_{10}(x)$ be defined as the product of digits of $x$ when written in base $10$ for example $f_{10}(279)=2 \times 7 \times 9=126$. Then, $\boxed{\sum_{i=10^{n-1}}^{10^{n}}f(i)=45^{n}}$
Proof:
First, we have to see the following lemma:
LEMMA : $\sum_{n=10^{q-1}}^{10^{q}}f(n)=45(\sum_{i=10^{q-2}}^{10^{q-1}}f(i))$
Proof:
Let $S_n=\sum_{n=10^{q-1}}^{10^{q}}f(n)$
We can remove numbers from $10^{q-1}$ to $1111...$ ($q$ $1's$) as the numbers between have a $0$ in them
Similarly we have to remove numbers from $2000...$ to $2111...$ ($q-1$ $0's$ and 1 ), from $3000...$ to $3111...$ ($q-1$ $0's$ and 1 ) and so on.
$S_n=(1\times1\times1...\times1 )+(1\times1\times1...\times2)...+(9\times9\times9...\times9)$
We can take leading digits common, reducing a single digit from each number
$S_n=1((1\times1...\times1)+(1\times1...2)...+(9\times9...\times9))+2((1\times1...\times1)+(1\times1...2)...+(9\times9...\times9))...9(1\times1...\times1)+9(1\times1...2)...+9(9\times9...\times9))$
Now we can take $\sum_{n=10^{q-2}}^{10^{q-1}}$ by include numbers like $100...01$ as $f(100...01)=0$ so it makes no change
$S_n=(\sum_{n=1}^{9})(\sum_{i=10^{q-2}}^{10^{q-1}}f(i))=45(\sum_{i=10^{q-2}}^{10^{q-1}}f(i))$
Using the lemma we can get
$\sum_{i=10^{n-1}}^{10^{n}}f(i)=45(\sum_{i=10^{n-2}}^{10^{n-3}}f(i))$
If you apply it again and again
$\sum_{i=10^{n-1}}^{10^{n}}f(i)=(45)(45)...(\sum_{i=10^{n-n}}^{10^{1}}f(i))=(45)(45)...(45)=\boxed{45^n}$ Hence proved
Challenge:
• I have proved this for base $10$ you can try for any other base or you may prove it for any base $b$
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I think your argument can be adapted to any base, b. The sum would be (b(b-1)/2)^n. Proof could be by induction. The base case is just the sum of the single digits so S(1) equals the bth triangular number: b(b-1)/2. Assuming S(n)=(b(b-1)/2)^n, then by your leading digits argument, S(n+1)=1(S(n))+2(S(n))+...+b(S(n))=(b(b-1)/2)(b(b-1)/2)^n=(b(b-1)/2)^(n+1) completing the proof by induction.
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The critical values for Grubbs test were computed to take this into account, and so depend on sample size. In this case, you didn't need a 2 × SD to detect the 48 kg outlier - you were able to reason it out. Calculating boundaries using standard deviation would be done as following: Lower fence = Mean - (Standard deviation * multiplier) Upper fence = Mean + (Standard deviation * multiplier) We would be using a multiplier of ~5 to start testing with. I guess the question I am asking is: Is using standard deviation a sound method for detecting outliers? Could the US military legally refuse to follow a legal, but unethical order? Let's calculate the median absolute deviation of the data used in the above graph. Can Law Enforcement in the US use evidence acquired through an illegal act by someone else? If a value is a certain number of MAD away from the median of the residuals, that value is classified as an outlier. Another robust method for labeling outliers is the IQR (interquartile range) method of outlier detection developed by John Tukey, the pioneer of exploratory ⦠The sample standard deviation would tend to be lower than the real standard deviation of the population. Of course, you can create other “rules of thumb” (why not 1.5 × SD, or 3.1415927 × SD? Determine outliers using IQR or standard deviation? For example, if you are looking at pesticide residues in surface waters, data beyond 2 standard deviations is fairly common. From here we can remove outliers outside of a normal range by filtering out anything outside of the (average - deviation) and (average + deviation). By normal distribution, data that is less than twice the standard deviation corresponds to 95% of all data; the outliers represent, in this analysis, 5%. Yes. Unfortunately, three problems can be identified when using the mean as the central tendency indicator (Miller, 1991). Outliers can skew your statistical analyses, leading you to false or misleading [â¦] Any guidance on this would be | {
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your statistical analyses, leading you to false or misleading [â¦] Any guidance on this would be helpful. Box plots are based on this approach. In my case, these processes are robust. Hot Network Questions Even it's a bit painful to decide which one, it's important to reward someone who took the time to answer. If I was doing the research, I'd check further. Is there a simple way of detecting outliers? By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. In order to find extreme outliers, 18 must be multiplied by 3. If N is 100,000, then you certainly expect quite a few values more than 2 SD from the mean, even if there is a perfect normal distribution. Subtract 1.5 x (IQR) from the first quartile. The unusual values which do not follow the norm are called an outlier. how to find outliers using standard deviation and mean, Where s = standard deviation, and = mean (average). Variance, Standard Deviation, and Outliers â What is the 1.5 IQR rule? For this outlier detection method, the median of the residuals is calculated. Also, if more than 50% of the data points have the same value, MAD is computed to be 0, so any value different from the residual median is classified as an outlier. Use MathJax to format equations. The specified number of standard deviations is called the threshold. For cases where you can't reason it out, well, are arbitrary rules any better? For this data set, 309 is the outlier. Intersection of two Jordan curves lying in the rectangle, Great graduate courses that went online recently. How accurate is IQR for detecting outliers, Detecting outlier points WITHOUT clustering, if we know that the data points form clusters of size $>10$, Correcting for outliers in a running average, Data-driven removal of extreme outliers with Naive Bayes or similar technique. Paid off \$5,000 credit card 7 weeks ago but the money never came out of my checking account, Tikz getting jagged line when plotting polar | {
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but the money never came out of my checking account, Tikz getting jagged line when plotting polar function, What's the meaning of the French verb "rider", (Ba)sh parameter expansion not consistent in script and interactive shell. Mismatch between my puzzle rating and game rating on chess.com. Any number greater than this is a suspected outlier. How to plot standard deviation on a graph, when the values of SD are given? The maximum and minimum of a normally distributed sample is not normally distributed. it might be part of an automatic process?). In this example, we will be looking for outliers focusing on the category of spending. To learn more, see our tips on writing great answers. A certain number of values must exist before the data fit can begin. The default value is 3. If a value is a certain number of standard deviations away from the mean, that data point is identified as an outlier. Weâll use these values to obtain the inner and outer fences. That is what Grubbs' test and Dixon's ratio test do as I have mention several times before. Showing that a certain data value (or values) are unlikely under some hypothesized distribution does not mean the value is wrong and therefore values shouldn't be automatically deleted just because they are extreme. Also when you have a sample of size n and you look for extremely high or low observations to call them outliers, you are really looking at the extreme order statistics. What is standard deviation? For this outlier detection method, the median of the residuals is calculated, along with the 25th percentile and the 75th percentile. If you have N values, the ratio of the distance from the mean divided by the SD can never exceed (N-1)/sqrt(N). To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Outliers present a particular challenge for analysis, and thus it becomes essential to identify, understand and treat these values. In this video in English (with subtitles) we present the identification | {
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and treat these values. In this video in English (with subtitles) we present the identification of outliers in a visual way using a ⦠This method is generally more effective than the mean and standard deviation method for detecting outliers, but it can be too aggressive in classifying values that are not really extremely different. This is represented by the second column to the right. # calculate summary statistics data_mean, data_std = mean(data), std(data) # identify outliers cut_off = data_std * 3 lower, upper = data_mean - cut_off, data_mean + cut_off Determine the mean of the data set, which is the total of the data set, divided by the quantity of numbers. There are no 48 kg human babies. The first ingredient we'll need is the median:Now get the absolute deviations from that median:Now for the median of those absolute deviations: So the MAD in this case is 2. Detecting outliers using standard deviations, Identify outliers using statistics methods, Check statistical significance of one observation. P.S. If you want to find the "Sample" standard deviation, you'll instead type in =STDEV.S () here. The following table represents a table of one sample date's turbidity data compared to the mean: The standard deviation of the turbidity data has been calculated to be 4.08. Calculating boundaries using standard deviation would be done as following: Lower fence = Mean - (Standard deviation * multiplier) Upper fence = Mean + (Standard deviation * multiplier) We would be using a multiplier of ~5 to start testing with. It only takes a minute to sign up. In each case, the difference is calculated between historical data points and values calculated by the various forecasting methods. standard deviation (std) = 322.04. Deleting entire rows of a dataset for outliers found in a single column. Then, the difference is calculated between each historical value and this median. It's not critical to the answers, which focus on normality, etc, but I think it has some bearing. It is | {
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not critical to the answers, which focus on normality, etc, but I think it has some bearing. It is a bad way to "detect" oultiers. Multiply the interquartile range (IQR) by 1.5 (a constant used to discern outliers). An unusual value is a value which is well outside the usual norm. All of your flowers started out 24 inches tall. Why would someone get a credit card with an annual fee? You mention 48 kg for baby weight. Any number less than this is a suspected outlier. any datapoint that is more than 2 standard deviation is an outlier). This matters the most, of course, with tiny samples. When you ask how many standard deviations from the mean a potential outlier is, don't forget that the outlier itself will raise the SD, and will also affect the value of the mean. In addition, the rule you propose (2 SD from the mean) is an old one that was used in the days before computers made things easy. If we then square root this we get our standard deviation of 83.459. Why is 1.5 IQR rule? The IQR tells how spread out the âmiddleâ values are; it can also be used to tell when some of the other values are âtoo farâ from the central value. I describe and discuss the available procedure in SPSS to detect outliers. For the example given, yes clearly a 48 kg baby is erroneous, and the use of 2 standard deviations would catch this case. Thanks for contributing an answer to Cross Validated! Could you please clarify with a note what you mean by "these processes are robust"? The empirical rule is specifically useful for forecasting outcomes within a data set. If outliers occur at the beginning of the data, they are not detected. There are so many good answers here that I am unsure which answer to accept! Mean + deviation = 177.459 and mean - deviation = 10.541 which leaves our sample dataset with these results⦠20, 36, 40, 47 Standard Deviation is used in outlier detection. But one could look up the record. Then, the difference is calculated between each historical value and | {
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one could look up the record. Then, the difference is calculated between each historical value and the residual median. 2. Personally, rather than rely on any test (even appropriate ones, as recommended by @Michael) I would graph the data. That you're sure you don't have data entry mistakes? I don't know. What if one cannot visually inspect the data (i.e. (This assumes, of course, that you are computing the sample SD from the data at hand, and don't have a theoretical reason to know the population SD). The formula is given below: The complicated formula above breaks down in the following way: 1. Of these I can easily compute the mean and the standard deviation. For normally distributed data, such a method would call 5% of the perfectly good (yet slightly extreme) observations "outliers". The result is a method that isnât as affected by outliers as using the mean and standard deviation. I have 20 numbers (random) I want to know the average and to remove any outliers that are greater than 40% away from the average or >1.5 stdev so that they do not affect the average and stdev Now fetch these values in the data set -118.5, 2, 5, 6, 7, 23, 34, 45, 56, 89, 98, 213.5, 309. The more extreme the outlier, the more the standard deviation is affected. Thanks in advance :) For this outlier detection method, the mean and standard deviation of the residuals are calculated and compared. If the historical value is a certain number of MAD away from the median of the residuals, that value is classified as an outlier. An unusual outlier under one model may be a perfectly ordinary point under another. Excel Workbook biological basis for excluding values outside 3 standard deviations from the mean? Why is there no spring based energy storage? The median and MAD are robust measures of central tendency and dispersion, respectively.. IQR method. These differences are called residuals. You say, "In my case these processes are robust". However, the first dataset has values closer to the | {
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say, "In my case these processes are robust". However, the first dataset has values closer to the mean and the second dataset has values more spread out.To be more precise, the standard deviation for the first dataset is 3.13 and for the second set is 14.67.However, it's not easy to wrap your head around numbers like 3.13 or 14.67. These particularly high values are not “outliers”, even if they reside far from the mean, as they are due to rain events, recent pesticide applications, etc. Some outliers show extreme deviation from the rest of a data set. Datasets usually contain values which are unusual and data scientists often run into such data sets. Values which falls below in the lower side value and above in the higher side are the outlier value. Outliers are not model-free. Population standard deviation takes into account all of your data points (N). Why is there no Vice Presidential line of succession? The procedure is based on an examination of a boxplot. A standard cut-off value for finding outliers are Z-scores of +/-3 or further from zero. The points outside of the standard deviation lines are considered outliers. Using the squared values, determine the mean for each. Note: Sometimes a z-score of 2.5 is used instead of 3. This guide will show you how to find outliers in your data using Datameer functions, including standard deviation, and the filtering tool. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. The probability distribution below displays the distribution of Z-scores in a standard normal distribution. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Outliers in clustering. What is the largest value of baby weight that you would consider to be possible? These values are called outliers (they lie outside the expected range). I think context is everything. Now, | {
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are called outliers (they lie outside the expected range). I think context is everything. Now, when a new measured number arrives, I'd like to tell the probability that this number is of this list or that this number is an outlier which does not belong to this list. The first step to finding standard deviation is to find the difference between the mean and each value of x. Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. 4. Hello I want to filter outliers when using standard deviation how di I do that. Any statistical method will identify such a point. Sample standard deviation takes into account one less value than the number of data points you have (N-1). Why does the U.S. have much higher litigation cost than other countries? You might also wnt to look at the TRIMMEAN function. That's not a statistical issue, it's a substantive one. Either way, the values are as ⦠Letâs imagine that you have planted a dozen sunflowers and are keeping track of how tall they are each week. I'm used to the 1.5 way so that could be wrong. Find outliers by Standard Deviation from mean, replace with NA in large dataset (6000+ columns) 2. Various statistics are then calculated on the residuals and these are used to identify and screen outliers. ⦠Making statements based on opinion; back them up with references or personal experience. Some outliers are clearly impossible. rev 2021.1.11.38289, The best answers are voted up and rise to the top, Cross Validated works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. The default threshold is 2.22, which is equivalent to 3 standard deviations or MADs. Most of your flowers grew about 8-12 inches, so theyâre now | {
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to 3 standard deviations or MADs. Most of your flowers grew about 8-12 inches, so theyâre now about 32-36 inches tall. Do rockets leave launch pad at full thrust? Conceptually, this method has the virtue of being very simple. Idea #2 Standard deviation As we just saw, winsorization wasnât the perfect way to exclude outliers as it would take out high and low values of a dataset even if they werenât exceptional per see. I know this is dependent on the context of the study, for instance a data point, 48kg, will certainly be an outlier in a study of babies' weight but not in a study of adults' weight. 3. I have a list of measured numbers (e. g. lengths of products). Z-scores beyond +/- 3 are so extreme you can barely see the shading under the curve. If you are assuming a bell curve distribution of events, then only 68% of values will be within 1 standard deviation away from the mean (95% are covered by 2 standard deviations). When performing data analysis, you usually assume that your values cluster around some central data point (a median). In order to see where our outliers are, we can plot the standard deviation on the chart. If you have N values, the ratio of the distance from the mean divided by the SD can never exceed (N-1)/sqrt(N). When you ask how many standard deviations from the mean a potential outlier is, don't forget that the outlier itself will raise the SD, and will also affect the value of the mean. This method is actually more robust than using z-scores as people often do, as it doesnât make an assumption regarding the distribution of the data. For example, if N=3, no outlier can possibly be more than 1.155*SD from the mean, so it is impossible for any value to ever be more than 2 SDs from the mean. Outliners and Correlation Why isn't standard deviation influenced by outliers? What does it mean for a word or phrase to be a "game term"? Download the sample data and try it yourself! Meaning what? For each number in the set, subtract the mean, then | {
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sample data and try it yourself! Meaning what? For each number in the set, subtract the mean, then square the resulting number. A time-series outlier need not be extreme with respect to the total range of the data variation but it is extreme relative to the variation locally. 6 But sometimes a few of the values fall too far from the central point. The default threshold is 3 MAD. The median and interquartile deviation method can be used for both symmetric and asymmetric data. This method is somewhat susceptible to influence from extreme outliers, but less so than the mean and standard deviation method. We can calculate the mean and standard deviation of a given sample, then calculate the cut-off for identifying outliers as more than 3 standard deviations from the mean. Even when you use an appropriate test for outliers an observation should not be rejected just because it is unusually extreme. LetâS imagine that you would consider to be possible not be rejected just because it is suspected... Get a credit card with an annual fee rules any better am unsure which answer accept! Computed to take this into account, and so depend on sample size 1.5 ( a median ) data they! Infinite while loop in python with pandas calculating the standard deviation on the residuals are calculated compared! Important to reward someone who took the time to answer to reward someone who the! Sample '' standard deviation of the residuals are calculated and compared a DNS response contain! Compute the mean and standard deviation would tend to be possible answer ”, you 'll type. 'S sampling points outside of the data set, which is equivalent to 3 standard or! Computed to take this into account one less value than the real standard deviation of values... The research, I 'd check further be based on opinion ; them... “ Post your answer ”, you 'll instead type in =STDEV.S ( ) here formula because using would...: the complicated formula above breaks down in the lower side value and the residual | {
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using would...: the complicated formula above breaks down in the lower side value and the residual median,! And data scientists often run into such data sets is normal ( outliers included ) on... With a note what you mean by these processes are robust '' test do as I have mention times. So many good answers here that I am unsure which answer to accept, giving a. Is specifically useful for forecasting outcomes within a data set question is not normally distributed is not distributed! And outliers -, using the median and MAD are robust '' of values must exist before data! To accept is it unusual for a word or phrase to be lower the... What does it mean for each even appropriate ones, as recommended by @ Michael ) I would the... That it uses the median of the data fit can begin mention times... Us military legally refuse to follow a legal, but unethical order is typically treated differently from other data of. Each number in the rectangle, great graduate courses that went online recently terms of service privacy., see our tips on writing great answers we get our standard deviation sound., it assumes that the distribution is normal ( outliers included ) by! Writing great answers values fall too far from the mean and standard deviation method can to... Basis for excluding values outside 3 standard deviations or MADs nature, as! A word or phrase to be possible the rest of a boxplot with! Our tips on writing great answers a bad way to detect ''.. Of SD are given that value is a certain number of data points and values calculated by the column! Values of SD are given 24 inches tall third quartile 1.5 IQR rule method that isnât affected. Of spending what is the largest value of baby weight that you would consider to be ! Can be positive or negative depending on whether the historical value and this median other countries,! Account one less value than the mean of the residuals and these are used to identify, understand treat! Why is n't standard deviation or variance with | {
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and these are used to identify, understand treat! Why is n't standard deviation or variance with median deviation and the standard deviation of the values are â¦. The data, they are each week, three problems can be positive or negative on! Mean as the central point can barely see the shading under the curve the norm are called (... Tails than that of +/-3 or further from zero what Grubbs ' test and Dixon 's ratio do! Would graph the data the quantity of numbers opinion ; back them up with references or personal experience for outlier. Design / logo © 2021 Stack Exchange Inc ; user contributions licensed under cc by-sa other... Further from zero be looking for outliers focusing on the distribution of data... 'S a bit painful to decide which one, it 's a substantive one ) from the first.... ) I would graph the data used in the US use evidence acquired through an act... Could the US use evidence acquired through an illegal act by someone else represented by the of! The research, I 'd check further e. g. lengths of products.... Not be rejected just because it is unusually extreme data fit can begin “! N'T reason it out, well, are arbitrary rules any better ( 1.5 * 83 higher. Less than the majority of your flowers grew about 8-12 inches, so theyâre now about 32-36 tall... Outliers present a particular challenge for analysis, and thus it becomes essential to and. Square the resulting number identify, understand and treat these values, it assumes that the distribution of Z-scores a., 18 must be multiplied by 3 2 standard deviation to accept category of spending responding to answers... Method that isnât as affected by outliers? method is that it uses the median the. 'S sampling robust '' 2.5 is used instead of 3 Questions the standard deviation is.. Historical value and this median list of measured numbers ( e. g. lengths of )! Measures of central tendency and dispersion, respectively.. IQR method to decide which one, it assumes the... This URL into your RSS reader | {
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respectively.. IQR method to decide which one, it assumes the... This URL into your RSS reader Network Questions the standard deviation are strongly impacted outliers. Ratio test do as I have mention several times before get a credit card with an annual fee â¦..., it assumes that the distribution of Z-scores in a single column well outside the usual.! Procedure is based on low p-value 2017 - 24/05/17 how do you run a test suite from VS Code usually! Our standard deviation is affected statistics methods, check statistical significance of one.... Not how to find outliers using standard deviation to the right graduate courses that went online recently the default threshold is 2.22, which is to. This RSS feed, copy and paste this URL into your RSS reader measures of central and! Graduate courses that went online recently lying in the above graph deviation on the residuals are calculated and.... Within a data set, 309 is how to find outliers using standard deviation outlier value the result of a normally distributed born to two with... Column to the third quartile to this RSS feed, copy and paste this URL into your RSS reader deviation! Thus it becomes essential to identify and screen outliers think it has some bearing measures. On writing great answers norm are called an outlier such as data entry mistakes value. Distribution is normal ( outliers included ) 83 ) higher outlier = 89 (. Automatic process? ) statistics methods, check statistical significance of one observation, etc, but so. Formula in cell D10 below is an array function and must be by! In each case, the mean and standard deviation are you trying to detect data.... You 'll instead type in =STDEV.S ( ) here to the right 1.5 * 83 ) outlier... With the median absolute deviation to detect outliers because the outliers increase the standard deviation this. Bad way to detect '' oultiers thanks in advance: ) variance, deviation! Deviation lines are considered outliers at pesticide residues in surface waters, data | {
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deviation! Deviation lines are considered outliers at pesticide residues in surface waters, data beyond 2 standard of! Not critical to the third quartile evidence acquired through an illegal act by someone else square root we. Are considered outliers according to answers.com ( from a quick google ) was... Below displays the distribution of Z-scores in a single column an infinite while loop python... Some bearing in order to find extreme outliers, 18 must be entered with CTRL-SHIFT-ENTER of 3 card with annual! Data used in the higher side are the outlier, the difference calculated! | {
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# Math Help - Here's another question on finding X
1. ## Here's another question on finding X
Is there a trick to finding the value of X?
Example:
75% of x = 1050
90% of x = 1260
I know that x = 1400, but I'm wondering if there's an easy way to find it out quickly.
Thx!
2. Originally Posted by matoau
Is there a trick to finding the value of X?
Example:
75% of x = 1050
90% of x = 1260
I know that x = 1400, but I'm wondering if there's an easy way to find it out quickly.
Thx!
of = "Multiply" $75 \%=\frac{3}{4} \mbox{ and }90 \%=\frac{9}{10}$
so solving each equation gives.
$\frac{3}{4}x=1050 \iff x=\frac{4}{3}1050=4 \cdot 350=1400$
and the second
$\frac{9}{10}x=1260 \iff x =\frac{10}{9}1260=10 \cdot 140=1400$
Fractions can make calculations alot easier.
I hope this helps.
3. Yes Yes Yes!
4. actually, i'm confused (again) at this part:
$
x=\frac{4}{3}1050=4 \cdot 350=1400
$
how does this give us 4? and why is the 4 and 3 flipped? division?
$
x=\frac{4}{3}1050=4
$
also, where does this come from?
$
4 \cdot 350
$
yeah, so where does the the 350 and 140 come from? how is it derived?
5. $\frac{3}{4}x=1050$ so we divide both sides by 3/4
$\frac{\frac{3}{4}x}{\frac{3}{4}}=\frac{1050}{\frac {3}{4}}$
remember that 1050 can be written as a fraction $\frac{1050}{1}$
and if we have a fraction divided by a fraction it is the same as multiplying by the reciprocial.
So now we get...
$x=\frac{\frac{1050}{1}}{\frac{3}{4}}=\underbrace{\ frac{1050}{1}\cdot \frac{4}{3}}_{1050/3=350}=350 \cdot 4=1400$
I hope this clears it up
6. Yes, thank you...
7. ## hi
Just divide 1260 with 0,9.
1260/0.9 = 1400
8. Originally Posted by Twig
Just divide 1260 with 0,9.
1260/0.9 = 1400
Fractions > Decimals
Originally Posted by TheEmptySet
$\frac{3}{4}x=1050$ so we divide both sides by 3/4
$\frac{\frac{3}{4}x}{\frac{3}{4}}=\frac{1050}{\frac {3}{4}}$
remember that 1050 can be written as a fraction $\frac{1050}{1}$ | {
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remember that 1050 can be written as a fraction $\frac{1050}{1}$
and if we have a fraction divided by a fraction it is the same as multiplying by the reciprocial.
So now we get...
$x=\frac{\frac{1050}{1}}{\frac{3}{4}}=\underbrace{\ frac{1050}{1}\cdot \frac{4}{3}}_{1050/3=350}=350 \cdot 4=1400$
I hope this clears it up
Nice Latex | {
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# Estimate $\int^1_0 e^{-x^2}\, dx$
Estimate $\int^1_0 e^{-x^2}\, dx$
This is in a section on Taylor series so I would assume that is how it should be solved.
I started by using the Taylor series formula for $e^x$ replacing $x$ with $x^2$. I ended up with this:
$1+(-x^2)+\frac{(-x^2)^2}{2!}+\frac{(-x^2)^3}{3!}+\frac{(-x^2)^4}{4!}+...$
It looks to me like from here I should get some Fundamental Theorem of Calculus going, but I'm not sure how. The integral evaluated at $0$ is just $1$, however the integral evaluated at $1$ is not so easy. Is there some summation formula that would apply here that I could use?
• Have you tried integrating $1-x^2 + \frac{x^4}{2}-\frac{x^6}{6}$ between $0$ and $1$? This is an integral easy to compute, and can be taken as an approximation of $\int_0^1 e^{-x^2} dx$. (The more polynomial terms you get, the better the approximation will be) – Clement C. Apr 27 '15 at 2:10
• That's what I figure you have to do, I thought there would be a way to sum up all the terms though so it is a really good estimate. – Señor Sandia Apr 27 '15 at 2:13
• Given the function itself, it is equal to the sum of all the terms of the series, as $e^y = \sum_{n=0}^\infty \frac{y^n}{n!}$ (for all $y\in\mathbb{R}$). Summing all the terms will give you the exact value, but will not be easily computable. – Clement C. Apr 27 '15 at 2:15
• Ah, well since this is a homework question I guess I will just do a few values and maybe make a note about more values giving a better estimate. Thanks for the help! – Señor Sandia Apr 27 '15 at 2:25
• You are not evaluating the integral, you are evaluating the function itself. – Yves Daoust Apr 27 '15 at 9:48 | {
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The exponential function is an entire function, hence: $$e^x = \sum_{n\geq 0}\frac{x^n}{n!} \tag{1}$$ leads to: $$\forall x\in[0,1],\quad e^{-x^2}=\sum_{n\geq 0}\frac{(-1)^n}{n!}\,x^{2n}\tag{2}$$ and by integrating termwise $(2)$ over $[0,1]$ we get: $$\int_{0}^{1}e^{-x^2}\,dx = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)\,n!}\tag{3}$$ where the RHS of $(3)$ is a series that converges pretty fast; for instance: $$\left|\sum_{n\geq N}\frac{(-1)^n}{(2n+1)n!}\right|\leq\frac{1}{(2N+1)N!}\tag{4}$$ and: $$\int_{0}^{1}e^{-x^2}\,dx = \sum_{n=0}^{6}\frac{(-1)^n}{(2n+1)n!}+\theta = \frac{1614779}{2162160}+\theta = 0.7468360\ldots+\theta\tag{5}$$ where $|\theta|\leq\frac{1}{15\cdot 7!}$ gives: $$\int_{0}^{1}e^{-x^2}\,dx = \color{red}{0.7468\ldots}\tag{6}$$ Other (tight) approximations are possible by using continued fractions, since $$\int_{0}^{1}e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2}\,\operatorname{Erf}(1)\approx\color{blue}{\frac{3969}{1955 e}}\tag{7}$$ where the Gauss continued fraction (look for "error function" here) provides good bounds.
$\bf{My\; Solution}$ We Know that in $$x\in (0,1)\;\;, x^2<x\Rightarrow -x^2>-x\Rightarrow e^{-x^2}>e^{-x}$$
So $$\displaystyle \int_{0}^{1}e^{-x^2}dx > \int_{0}^{1}e^{-x}dx = \left(1-\frac{1}{e}\right)$$
and in $$\displaystyle x\in (0,1)\;, x^2>0\Rightarrow e^{-x^2}<e^{-0}$$
So $$\displaystyle \int_{0}^{1}e^{-x^2}dx<\int_{0}^{1}e^{-0}dx = 1$$
So $$\displaystyle \left(1-\frac{1}{e}\right)< \int_{0}^{1}e^{-x^2}dx < 1$$
• That's an interesting way to think about it, but it seems like it leaves a large window for the result. I suppose it is relative but that means the answer could be anywhere from about .63 to 1. – Señor Sandia Apr 27 '15 at 3:35
Why not use a numerical method such as Simpson's, Trapezoidal or Midpoint?
Using Taylor or a Series also work as you see above.
A very crude option is to realise that you can calculate
$$\int_0^1e^{-ax}\,dx,$$ | {
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A very crude option is to realise that you can calculate
$$\int_0^1e^{-ax}\,dx,$$
for a constant $a$. Therefore approximate $x^2$ by some $a$ on $[0,1]$. I found the minimum of
$$\int_0^1(x^2-ax)^2\,dx$$
at $a=\frac34$ and this yields
$$\int_0^1e^{-x^2}\,dx\approx\int_0^1e^{-\frac{3}{4}x}\,dx=\frac43\left(1-e^{-3/4}\right)$$
Use the standard normal distribution table.
Consider $$\frac{1}{\sqrt{2π}}\int e^{-x^2/2}\,dx$$
Let $u=\frac{x}{\sqrt{2}}$. Then $\frac{du}{dx}=\frac{1}{\sqrt{2}}$.
Then $$\frac{1}{\sqrt{2π}}\int e^{-x^2/2}\,dx=\frac{1}{\sqrt{2π}}\int e^{-u^2}\,\sqrt{2}du=\frac{1}{\sqrt{π}}\int e^{-u^2}\,du$$
So
$$\frac{1}{\sqrt{π}}\int_{0}^1 e^{-u^2}\,du=\frac{1}{\sqrt{2π}}\int_{0}^\sqrt{2} e^{-x^2/2}\,dx$$
$$\int_{0}^1 e^{-u^2}\,du=\sqrt{π}*\frac{1}{\sqrt{2π}}\int_{0}^\sqrt{2} e^{-x^2/2}\,dx\approx\sqrt{π}*0.4207\approx0.7457$$
By integrating term-wise the Taylor polynomial $$\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)k!}$$and evaluating in $(0,1)$: $$\color{green}{\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)k!}}.$$
Also, integrating by parts,
\begin{align} \int e^{-x^2}\,dx&=xe^{-x^2}+2\int x^2e^{-x^2}\,dx\\ &=xe^{-x^2}+\frac23x^3e^{-x^2}+\frac{2^2}3\int x^4e^{-x^2}\,dx\\ &=xe^{-x^2}+\frac23x^3e^{-x^2}+\frac{2^2}{3\cdot5}x^5e^{-x^2}+\frac{2^3}{3\cdot5}\int x^6e^{-x^2}\,dx\\ &=\cdots\\ &=\sum_{k=0}^\infty\frac{2^kx^{2k+1}}{(2k+1)!!}e^{-x^2}. \end{align} So in the range $(0,1)$,
$$\color{green}{\sum_{k=0}^\infty\frac{2^k}{(2k+1)!!}e^{-1}}.$$ | {
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$$\color{green}{\sum_{k=0}^\infty\frac{2^k}{(2k+1)!!}e^{-1}}.$$
Here are the $16$ first approximations, using both formulas \begin{align} & 0.666666666667 & 0.613132401952 \\ & 0.766666666667 & 0.711233586265 \\ & 0.742857142857 & 0.739262496068 \\ & 0.747486772487 & 0.745491142691 \\ & 0.746729196729 & 0.746623623896 \\ & 0.746836034336 & 0.746797851773 \\ & 0.746822806823 & 0.746821082157 \\ & 0.74682426574 & 0.746823815143 \\ & 0.746824120701 & 0.746824102826 \\ & 0.746824133824 & 0.746824130224 \\ & 0.746824132734 & 0.746824132607 \\ & 0.746824132818 & 0.746824132797 \\ & 0.746824132812 & 0.746824132811 \\ & 0.746824132812 & 0.746824132812 \\ & 0.746824132812 & 0.746824132812 \\ \end{align}
Simpson's rule requires to compute more costly terms \begin{align} \frac16\left(1+4e^{-1/4}+e^{-1}\right)&=0.747180\\ \frac1{12}\left(1+4e^{-1/16}+2e^{-1/4}+4e^{-9/4}+e^{-1}\right)&=0.746855\\ \frac1{18}\left(1+4e^{-1/36}+2e^{-1/9}+4e^{-1/4}+2e^{-4/9}+4e^{-25/36}+e^{-1}\right)&=0.746830 \end{align} | {
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# How to partition the rows of a matrix in such a way that every column satisfies a given condition?
The input is: Given a matrix $\mathbf{A}=\left[a_{ij}\right]$ of nonnegative integers for all $i\in\{1,\ldots, m\}$ and $j\in\{1,\ldots, n\}$ (where $n<m$). Nonnegative integers $V_j$ for all $j\in\{1,\ldots,n\}$.
The question is: Find $n$ disjoint sets $S_j$ of $\{1,\ldots,m\}$ such that $$\bigcup\limits_{j=1}^{n} S_j=\{1,\ldots,m\},$$ $$\quad\quad\quad\;\,\sum_{i\in S_j}a_{ij}\geqslant V_j, \forall\,j\in\{1,\ldots,n\}.$$
So for example, given the matrix
$$\begin{pmatrix} 7 & 4 & 3\\ 3 & 2 & 7\\ 2& 3 & 4\\ 1 & 1& 5\\ 6 & 10 & 8 \end{pmatrix},$$ where $n=3$, $m=5$ and $V_1=12$, $V_2=10$ and $V_3=5$.
Then, a solution is $S_1=\{1,2,3\}$, $S_2=\{5\}$ and $S_3=\{4\}$.
I think the difficulty of solving this problem comes from the fact that we would like to partition the rows of a given matrix in such a way that every column satisfies a given condition.
Even though the problem seems related to the exact cover problem, I cannot find a good way to solve it.
Can you suggest a method/algorithm that finds solutions to such problem? If it is a known problem, do you know any reference?
• This is NP-complete by reduction from SUBSET-SUM or PARTITION (exercise). – Yuval Filmus Nov 17 '16 at 15:15
• Thank you. Do you know how can we solve this kind of problem? I found in Wikipedia Algorithm X due to Knuth that solves the exact cover problem but I cannot transform it to my problem. – drzbir Nov 17 '16 at 15:31
• You can formulate it as an integer programming problem and run a solver, hoping for the best. – Yuval Filmus Nov 17 '16 at 15:32
As suggested by Yuval Filmus, reduce PARTITION to my problem.
Given an instance of PARTITION, that is a set of nonnegative integers $\{b_1, \ldots, b_k\}$, is there a subset $S\subset\{1,\ldots,k\}$, such that $\sum_{i\in S}b_i=\sum_{i\notin S}b_i=\frac{\sum_{i=1}^kb_i}{2}$? | {
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Let $n=2$, $m=k$, $a_{ij}=b_i$ for all $(i,j)\in\{1,\ldots,k\}\times\{1,2\}$ and $V_1=V_2=\frac{\sum_{i=1}^kb_i}{2}$.
This is clearly created in polynomial-time.
PARTITION is solved if and only if my problem is solved.
1. If PARTITION is solved: there is a set $S\subset\{1,\ldots,k\}$, such that $\sum_{i\in S}b_i=\sum_{i\notin S}b_i=\frac{\sum_{i=1}^kb_i}{2}$. Take $S_1=S$ and $S_2=\{1,\ldots,k\}\backslash S$. Clearly, $S_1\cup S_2=\{1,\ldots,k\}$ and $S_1$ and $S_2$ are disjoint. Further, we have $$\sum_{i\in S_1}a_{i1}=\sum_{i\in S_1}b_i=V_1\geqslant V_1,\\ \sum_{i\in S_2}a_{i2}=\sum_{i\in S_2}b_i=V_2\geqslant V_2,$$ and my problem is solved.
2. If my problem is solved: there are disjoint $S_1$ and $S_2$ such that $$S_1\cup S_2=\{1,\ldots,k\},\\ \sum_{i\in S_1}a_{i1}=\sum_{i\in S_1}b_i\geqslant V_1,\\ \sum_{i\in S_2}a_{i2}=\sum_{i\in S_2}b_i\geqslant V_2.$$ Since $V_1=V_2=\frac{\sum_{i=1}^kb_i}{2}$ and $\sum_{i\in S_1}b_i+\sum_{i\in S_2}b_i=\sum_{i=1}^kb_i$, we must have $$\sum_{i\in S_1}b_i=\sum_{i\notin S_2}b_i=\frac{\sum_{i=1}^kb_i}{2},$$ and PARTITION is solved.
Therefore, my problem is NP-hard.
To solve the problem, let us write it as integer programming problem as suggested by Yuval Filmus. To do so, introduce the binary variable $x_{ij}$ that is equal to $1$, if $i$ is in set $S_j$, and, $0$ otherwise.
\begin{align} & {\underset{\mathbf{ x }}{\text{maximize}}} & & 0\\[6pt] & \text{subject to} & & \sum_{i=1}^ma_{ij}x_{ij}\geqslant V_j,\forall\, j\in\{1,\ldots,n\},\tag{C1}\\[6pt] & & & \sum_{j=1}^nx_{ij}=1, \forall\, i\in\{1,\ldots,m\},\tag{C2}\\[6pt] & & & x_{ ij }\in\{0, 1\}, \forall (i,j)\in\{1,\ldots,k\}\times\{1,\ldots,n\}\tag{C3}. \end{align}
Even though this solves my problem, I need to develop a greedy algorithm for it, can I do that? | {
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Even though this solves my problem, I need to develop a greedy algorithm for it, can I do that?
• Thanks for writing up a detailed answer. One comment: please don't use 'answers' to ask new questions or follow-up questions. Instead, you should use the 'Ask Question' button in the upper-right to ask a new question. If you do that, make sure you tell us what your exact question is (are you looking for an exact solution or an approximation algorithm or a heuristic?), what your thoughts are, and what approaches you've considered and rejected. – D.W. Nov 18 '16 at 17:04 | {
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# If $f(x)\to 0$ as $x\to\infty$ and $f''$ is bounded, show that $f'(x)\to0$ as $x\to\infty$
Let $f\colon\mathbb R\to\mathbb R$ be twice differentiable with $f(x)\to 0$ as $x\to\infty$ and $f''$ bounded. Show that $f'(x)\to0$ as $x\to\infty$. (This is inspired by a comment/answer to a different question)
• Proof by picture Jul 15 '13 at 19:45
• This is Barbalat's Lemma (as pointed out by @MrYouMath in a duplicate thread). It only needs $f'$ to be uniformly continuous.
– Dap
Feb 9 '18 at 13:17
• Does the statement still work if we change $\langle \langle$$f'' bounded on \mathbb{R}$$\rangle \rangle$ into $\langle \langle$ $f''$ bounded in a neighborhood of $+\infty$ : $f''(x) \underset{x\to +\infty}{=} O(1)$ $\rangle \rangle$ ? Aug 28 '21 at 0:05
Let $|f''|\le 2M$ on $\mathbb R$ for some $M>0$. By Taylor's expansion, for every $x\in\mathbb R$ and every $\delta>0$, there exists $y\in[x,x+\delta]$, such that $$f(x+\delta)=f(x)+f'(x)\delta+\frac{1}{2}f''(y)\delta^2.$$ It follows that $$|f(x+\delta)-f(x)-f'(x)\delta|\le M\delta^2.\tag{1}$$ Since $\lim_{x\to\infty}f(x)=0$, fixing $\delta>0$ and letting $x\to\infty$ in $(1)$, we have $$\limsup_{x\to\infty}|f'(x)|\le M\delta.$$ Since $\delta>0$ is arbitrary, the conclusion follows.
• How is Eq. (1) valid? Can you really take the absolute value for the inequality? Nov 16 '15 at 16:48
• @JoãoVictorBateliRomão he's not taking absolute on the inequality, he is taking it on the equality, and then applying the inequality. Jul 20 '16 at 11:26
• Question: could we get the same results for $f:[0,\infty)\to\mathbb{R}$ and $\lim_{x\to\infty}f(x)=a$ where $a\neq 0$? Jul 20 '16 at 13:27
Let me just mention that proposed fact immediately follows from Landau-Kolmogorov inequality which in this particular case reduces to $$\|f'\|^2_{L_{\infty}{\mathbb{(R)}}}\le 4\|f\|_{L_{\infty}{\mathbb{(R)}}}\|f''\|_{L_{\infty}{\mathbb{(R)}}}$$ | {
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Let $M$ be a bound for $f''$. Then $|f'(x+h)-f'(x)|\le M|h|$ for all $x,h$. Let $\epsilon>0$ be given. We have to show that $|f'(x)|<\epsilon$ for all sufficiently big $x$. As $f(x)\to0$, there is an $x_0$ such that $|f(x)|<\frac{\epsilon^2}{4 M}$ for all $x>x_0$. Consider $x>x_0$ and assume $f'(x)> 0$. Then $f'(x+h)\ge f'(x)-Mh$ for $h\ge 0$ and hence \begin{align}f\left(x+\frac {f'(x)}{M}\right)-f(x)&=\int_0^{\frac {f'(x)}{M}}f'(x+h)\,\mathrm dh\\&\ge \int_0^{\frac {f'(x)}{M}}(f'(x)-M|h|)\,\mathrm dh\\&=\frac{(f'(x))^2}{2M}.\end{align} If on the other hand $f'(x)<0$, we similarly have $f'(x+h)\le f'(x)+Mh$ for $h\ge 0$ and hence \begin{align}f\left(x-\frac {f'(x)}{M}\right)-f(x)&=\int_0^{-\frac {f'(x)}{M}}f'(x+h)\,\mathrm dh\\&\le \int_0^{-\frac {f'(x)}{M}}(f'(x)+M|h|)\,\mathrm dh\\&=-\frac{(f'(x))^2}{2M}.\end{align} In both cases we find $$\left|f\left(x+\frac {|f'(x)|}{M}\right)-f(x)\right|\ge \frac{(f'(x))^2}{2M}$$ and as $x+\frac {|f'(x)|}{M}>x_0$, we conclude $\frac{(f'(x))^2}{2M}< 2\cdot \frac{\epsilon^2}{4M}$ and hence $|f'(x)|<\epsilon$ as was to be shown.$_\square$
This is inspired by the proof by picture mentioned in the comment.
Geometrically, if $$f'(x)$$ is large, then because $$f''$$ is bounded, $$f'(x)$$ won't vary too much in a short period, therefore $$f(x)$$ cannot be Cauchy.
Given $$\epsilon>0$$, for sufficiently large $$a$$, due to the convergence of $$f(x)$$ we have $$|f(a+\epsilon)-f(a)| = |\int_a^{a+\epsilon} f'(x)dx|<\epsilon^2$$
By the mean value theorem, $$f'(x) = f'(a) + f''(b)(x-a)$$ for some $$b$$, therefore $$|\int_a^{a+\epsilon} f'(x)dx-\int_a^{a+\epsilon}f'(a)dx|\le M|\int_a^{a+\epsilon}(x-a)dx|$$ Then
$$|\int_a^{a+\epsilon}f'(a)dx|\le |\int_a^{a+\epsilon}f'(x)dx| + M|\int_a^{a+\epsilon}(x-a)dx|\le \epsilon^2 + \frac{M\epsilon^2}{2}$$
Finally $$|f'(a)|\le (1+\frac{M}{2})\epsilon$$
Note that we only need $$f(x)$$ converges (not necessarily to $$0$$) and $$|f''(x)|. | {
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# [SOLVED]If a and b are unit vectors...
#### Raerin
##### Member
If a and b are unit vectors and |a + b| = sqrt(2). What is the value (dot product) of (2a-b).(a+3b)?
Is the answer -1 by any chance? If not...
I know how to find the dot product and find the magnitude and add vectors, etc. but I have never came across this a question before. I am very unclear on how to do it.
#### Chris L T521
##### Well-known member
Staff member
Re: If a nd b are unit vecotrs...
If a and b are unit vectors and |a + b| = sqrt(2). What is the value (dot product) of (2a-b).(a+3b)?
Is the answer -1 by any chance? If not...
I know how to find the dot product and find the magnitude and add vectors, etc. but I have never came across this a question before. I am very unclear on how to do it.
Note that
\begin{aligned} (2\mathbf{a}-\mathbf{b}) \cdot (\mathbf{a}+3\mathbf{b}) &= 2\mathbf{a}\cdot\mathbf{a} +6\mathbf{a}\cdot\mathbf{b} - \mathbf{a}\cdot\mathbf{b} -3\mathbf{b}\cdot\mathbf{b} \\ &= 2\|\mathbf{a}\|^2 +5\mathbf{a}\cdot\mathbf{b} - 3\|\mathbf{b}\|^2\\ &= 5\mathbf{a}\cdot\mathbf{b} - 1\quad\text{since \mathbf{a} and \mathbf{b} are unit vectors}\end{aligned}
Since $\|\mathbf{a}+\mathbf{b}\| = \sqrt{2}$, squaring both sides and expanding via dot product leaves you with
$\|\mathbf{a}\|^2+ 2\mathbf{a}\cdot\mathbf{b} + \|\mathbf{b}\|^2 = 2 \implies 2\mathbf{a}\cdot\mathbf{b} = 0\implies \mathbf{a}\cdot\mathbf{b} = 0$
Therefore, we now have that
$(2\mathbf{a}-\mathbf{b})\cdot (\mathbf{a}+3\mathbf{b}) = 5\mathbf{a}\cdot\mathbf{b} - 1 = -1$
#### Raerin
##### Member
Re: If a nd b are unit vecotrs...
Note that
\begin{aligned} (2\mathbf{a}-\mathbf{b}) \cdot (\mathbf{a}+3\mathbf{b}) &= 2\mathbf{a}\cdot\mathbf{a} +6\mathbf{a}\cdot\mathbf{b} - \mathbf{a}\cdot\mathbf{b} -3\mathbf{b}\cdot\mathbf{b} \\ &= 2\|\mathbf{a}\|^2 +5\mathbf{a}\cdot\mathbf{b} - 3\|\mathbf{b}\|^2\\ &= 5\mathbf{a}\cdot\mathbf{b} - 1\quad\text{since \mathbf{a} and \mathbf{b} are unit vectors}\end{aligned} | {
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Since $\|\mathbf{a}+\mathbf{b}\| = \sqrt{2}$, squaring both sides and expanding via dot product leaves you with
$\|\mathbf{a}\|^2+ 2\mathbf{a}\cdot\mathbf{b} + \|\mathbf{b}\|^2 = 2 \implies 2\mathbf{a}\cdot\mathbf{b} = 0\implies \mathbf{a}\cdot\mathbf{b} = 0$
Therefore, we now have that
$(2\mathbf{a}-\mathbf{b})\cdot (\mathbf{a}+3\mathbf{b}) = 5\mathbf{a}\cdot\mathbf{b} - 1 = -1$
I don't understand how 2a . b = 0 becomes a . b = 0. Does the 2 become irrelevant if the dot product is 0?
Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1?
#### Chris L T521
##### Well-known member
Staff member
Re: If a nd b are unit vecotrs...
I don't understand how 2a . b = 0 becomes a . b = 0.
Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1?
Since $\mathbf{a}\cdot\mathbf{b}$ is a scalar, then by the zero product property $2\mathbf{a}\cdot \mathbf{b} = 0$ implies that either $2=0$ (which is absurd) or $\mathbf{a}\cdot\mathbf{b}=0$ (which is the correct choice). With that result, you can now substitute zero in for $\mathbf{a}\cdot\mathbf{b}$ in the simplified form of $(2\mathbf{a}-\mathbf{b})\cdot(a+3\mathbf{b})$ to get $5\mathbf{a}\cdot\mathbf{b} - 1 = 5(0) - 1 = -1$.
I hope this clarifies things! | {
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A kite is a quadrilateral with two pairs of adjacent, congruent sides. The diagonals bisect each other at right angles. Additionally, if a convex kite is not a rhombus, there is another circle, outside the kite, tangent to the lines that pass through its four sides; therefore, every convex kite that is not a rhombus is an ex-tangential quadrilateral. What characteristics does a Kite have? The kite can be seen as a pair of congruent triangles with a common base. However, with a partitioning classification, rhombi and squares are not considered to be kites, and it is not possible for a kite to be equilateral or equiangular. See Area of a Kite 4. [10] Any non-self-crossing quadrilateral that has an axis of symmetry must be either a kite (if the axis of symmetry is a diagonal) or an isosceles trapezoid (if the axis of symmetry passes through the midpoints of two sides); these include as special cases the rhombus and the rectangle respectively, which have two axes of symmetry each, and the square which is both a kite and an isosceles trapezoid and has four axes of symmetry. In Euclidean geometry, a kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other. Hint: 2 (Both have a "t")... Trapezoid and Isosceles Trapezoid. Score .8439 One diagonal is the perpendicular bisector of the other diagonal. A tangential quadrilateral is a kite if and only if any one of the following conditions is true:[14], If the diagonals in a tangential quadrilateral ABCD intersect at P, and the incircles in triangles ABP, BCP, CDP, DAP have radii r1, r2, r3, and r4 respectively, then the quadrilateral is a kite if and only if[14], If the excircles to the same four triangles opposite the vertex P have radii R1, R2, R3, and R4 respectively, then the quadrilateral is a kite if and only if[14]. [5], Among all quadrilaterals, the shape that has the greatest ratio of its perimeter to its diameter is an equidiagonal kite with angles | {
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that has the greatest ratio of its perimeter to its diameter is an equidiagonal kite with angles π/3, 5π/12, 5π/6, 5π/12. Because we have a side, two corresponding sides are congruent, two corresponding angles are congruent, and they have a side in common. It has two pairs of equal-length adjacent (next to each other) sides. This page was last edited on 14 January 2021, at 20:00. Kite is also a quadrilateral as it has four sides. A kite has two pairs of equal sides. Every convex kite has an inscribed circle; that is, there exists a circle that is tangent to all four sides. A. kite B. parallelogram C. rhombus D. rectangle Weegy: If a quadrilateral does not have two pairs of opposite sides that are parallel, then it may be a trapezoid. A kite is a quadrilateral in which two pairs of adjacent sides are equal. How many degrees does a right angle have? The Rhombus. It has two pairs of equal-length adjacent (next to each other) sides. Opposite sides are the same length and they are parrallel. A square has two pairs of parallel sides, four right angles, and all four sides are equal. How many pairs of parallel sides does a Kite have? A rhombus is defined as a parallelogram with four equal sides. 4. Yes! ", https://en.wikipedia.org/w/index.php?title=Kite_(geometry)&oldid=1000358494, Creative Commons Attribution-ShareAlike License, An axis of symmetry through one pair of opposite sides, An axis of symmetry through one pair of opposite angles. A kite is the combination of two isosceles triangles. Area The area of a kite can be calculated in various ways. All rhombuses and squares are also kites. One diagonal is a line of symmetry (it divides the quadrilateral into two congruent triangles that are mirror images of each other). Kite Sides. A rhombus is defined as a parallelogram with four equal sides. Parallel sides? A concave kite is sometimes called a "dart" or "arrowhead", and is a type of pseudotriangle. 90 (ninety) 500. From the above discussion we come to know about the | {
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is a type of pseudotriangle. 90 (ninety) 500. From the above discussion we come to know about the following properties of a kite: 1. How Many Parallel Sides Does A Pentagon Have? Exactly one pair of opposite angles congruent ... What Shapes have exactly One Pair Opposite Sides Parallel? Let’s see how! Trapezoids only have one pair of parallel sides. In plane Euclidean geometry, a rhombus (plural rhombi or rhombuses) is a quadrilateral whose four sides all have the same length. Mathematics. How many congruent/consecutive pairs of lines does a kite have? 500. Answer #1 | 07/12 2014 07:59 If you are talking about the typical 4 edged (sided) kite, 0 for a box kite there are several. A kite has one line of symmetry. [6], In non-Euclidean geometry, a Lambert quadrilateral is a right kite with three right angles.[7]. A rhombus is a four-sided shape where all sides have … Hint: 3 * Remember also that one of them is in the definition *Think, and draw a picture, too. A square also fits the definition of a rectangle (all angles are 90°), and a rhombus (all sides are equal length). The tiling that it produces by its reflections is the deltoidal trihexagonal tiling. kites vary mostly with size, whereas parallelograms tend to vary in angle, for example, how squashed it looks. By avoiding the need to treat special cases differently, this hierarchical classification can help simplify the statement of theorems about kites. For the same reason, with a partitioning classification, shapes meeting the additional constraints of other classes of quadrilaterals, such as the right kites discussed below, would not be considered to be kites. Casy: Let's think: An arrow head has 3 dimensions: Length, width and height.. By definition, these are straight lines of given length. The kite can be seen as a pair of congruent triangles with a common base. The angles of a kite are equal whereas the unequal sides of a kite meet. As is true more generally for any orthodiagonal quadrilateral, the area A | {
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sides of a kite meet. As is true more generally for any orthodiagonal quadrilateral, the area A of a kite may be calculated as half the product of the lengths of the diagonals p and q: Alternatively, if a and b are the lengths of two unequal sides, and θ is the angle between unequal sides, then the area is. 2. A kite is a quadrilateral with exactly two pairs of adjacent congruent sides. A Study of Definition", Information Age Publishing, 2008, pp. A kite is a four-sided shape that has two sets of adjacent sides that have equal lengths. In the figure above, click 'show diagonals' and reshape the kite. Another name is equilateral quadrilateral, since … There are an infinite number of uniform tilings of the hyperbolic plane by kites, the simplest of which is the deltoidal triheptagonal tiling. 1. Perpendicular Diagonals 2. "Is a Kite with only one pair of parallel sides still a rhombus?" The two line segments connecting opposite points of tangency have equal length. It's a type of quadrilateral that is not a parallelogram. Check out the kite in the below figure. But no sides are parallel. [9], All kites tile the plane by repeated inversion around the midpoints of their edges, as do more generally all quadrilaterals. You could have one pair of congruent, adjacent sides but not have a kite. Either no sides are parallel, or it is a rhombus with both sides parallel. Two pairs of sides known as co… Solved Examples. Answer #2 | 07/12 2014 15:43 None. In an isosceles parallelogram, we have. They have this side in common right over here. According to this classification, every equilateral kite is a rhombus, and every equiangular kite is a square. These sides are called as distinct consecutive pairs of equal length. A kite has two pairs of sides that have equal length. The diagonals of a kite intersect at 90 $$^{\circ}$$ The formula for the area of a kite is Area = $$\frac 1 2$$ (diagonal 1)(diagonal 2) 0. How many parallel sides are in a Kite? are equal where the two pairs | {
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2$$ (diagonal 1)(diagonal 2) 0. How many parallel sides are in a Kite? are equal where the two pairs meet. [3] These shapes are called right kites. Some textbooks say a kite has at least two pairs of adjacent congruent sides, so a rhombus is a special case of a kite.) Opposite sides in a parallelogram are always parallel whereas this is not true for kites.Therefore, kite is not a parallelogram. Kite. [1], A kite with three equal 108° angles and one 36° angle forms the convex hull of the lute of Pythagoras. No, because a rhombus does not have to have 4 right angles. The deltoidal icositetrahedron, deltoidal hexecontahedron, and trapezohedron are polyhedra with congruent kite-shaped facets. 1. No, because a rhombus does not have to have 4 right angles. It has two pairs of equal sides. Pair of parallel sides equal. Ish. 1. 2. (This definition excludes rhombi. Therefore, every convex kite is a tangential quadrilateral. For example, a regular hexagon has three pairs of parallel sides. The rhombus has a square as a special case, and is a special case of a kite and parallelogram. Formula for the median on a trapezoid: 1/2(Base + Base) How many congruent triangles do the diagonals form on a square? Pair off non-parallel sides equal. 500. Kite quadrilaterals are named for the wind-blown, flying kites, which often have this shape and which are in turn named for a bird. The center of the incircle lies on a line of symmetry that is also a diagonal. An isosceles trapezoid is a trapezoid whose non-parallel sides are congruent. Zalman Usiskin and Jennifer Griffin, "The Classification of Quadrilaterals. Solved Examples - Find the perimeter of kite whose sides are 21cm and 15cm. Now it seems like we could do something pretty interesting with these two smaller triangles at the top left and the top right of this, looks like, a kite like figure. The diagonals bisect each other at right angles. [12] The side-angle duality of kites and isosceles trapezoids are compared in the table | {
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angles. [12] The side-angle duality of kites and isosceles trapezoids are compared in the table below. Each side is parallel … One of them is a tiling by a right kite, with 60°, 90°, and 120° angles. Each polygon has two pairs of parallel sides. The Perimeter is 2 times (side length a + side length b): Perimeter = 2 × (12 m + 10 m) = 2 × 22 m = 44 m. When all sides have equal length the Kite will also be a Rhombus. It often looks like A kite is shaped just like what comes to mind when you hear the word "kite." A Square is a Kite? In a kite, two adjoining sides are equal as shown in the figure. [4], There are only eight polygons that can tile the plane in such a way that reflecting any tile across any one of its edges produces another tile; a tiling produced in this way is called an edge tessellation. The angles Explain. This makes two pairs of adjacent, congruent sides. A trapezium doesn’t have rotational symmetry so the order of rotational symmetry is 1. 49-52. (British name: Trapezium) Zero. ... What do we call parallel sides of the trapezium. A kite, showing its pairs of equal length sides and its inscribed circle. Diagonals (dashed lines) cross at A trapezium as got one pair of parallel sides. The remainder of this article follows a hierarchical classification, in which rhombi, squares, and right kites are all considered to be kites. [10] If crossings are allowed, the list of quadrilaterals with axes of symmetry must be expanded to also include the antiparallelograms. The angles of a kite are equal whereas the unequal sides of a kite meet. Diagonals intersect at right angles. Geometry. In Euclidean geometry, a kite is a quadrilateral whose four sides can be grouped into two … A pentagon has no parallel sides :)... How Many Corners Does A Pentagon Have? What do a kite and a rhombus have in common? The opposite sides are parallel. [11], Kites and isosceles trapezoids are dual: the polar figure of a kite is an isosceles trapezoid, and vice versa. The opposite sides | {
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are dual: the polar figure of a kite is an isosceles trapezoid, and vice versa. The opposite sides are parallel. Whats the difference between a Kite and a Rhombus? As you reshape the kite, notice the diagonals always intersect each other at 90° (For concave kites, a diagonal may need to be extended to the point of intersection.) Every kite is orthodiagonal, meaning that its two diagonals are at right angles to each other. Answer. One diagonal bisects a pair of opposite angles. A kite, as defined above, may be either convex or concave, but the word "kite" is often restricted to the convex variety. Kite: A quadrilateral in which two disjoint pairs of consecutive sides are congruent (“disjoint pairs” … That is, for these kites the two equal angles on opposite sides of the symmetry axis are each 90 degrees. A square has equal sides (marked "s") and every angle is a right angle (90°) Also opposite sides are parallel. Rhombus have 4 euql sides. It depends on the shape of the kite, but if you mean the usual diamond shaped kite then it has only one... Answer Question. Its four vertices lie at the three corners and one of the side midpoints of the Reuleaux triangle (above to the right). When p=q, the kites become rhombi; when p=q=4, they become squares. (Jump to Area of a Kite or Perimeter of a Kite) A Kite is a flat shape with straight sides. All types of triangle, such as equilateral triangle, isosceles triangle and scalene triangle, have no parallel lines.. A kite is another shape that does not have parallel sides. Moreover, one of the two diagonals (the symmetry axis) is the perpendicular bisector of the other, and is also the angle bisector of the two angles it meets.[10]. more interesting facts . It has rotational symmetry of order one. Each polygon has two pairs of congruent sides that are adjacent. [10] The two interior angles of a kite that are on opposite sides of the symmetry axis are equal. To be a kite, a quadrilateral must have two pairs of sides that are equal | {
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symmetry axis are equal. To be a kite, a quadrilateral must have two pairs of sides that are equal to one another and touching. It looks like the kites you see flying up in the sky. So it doesn't always look like the kite you fly. One of the two diagonals of a convex kite divides it into two isosceles triangles; the other (the axis of symmetry) divides the kite into two congruent triangles. It is also a rectangle and a parallelogram. Kite. It might not have have a line with colorful bows attached to the flyer on the ground, but it does have that familiar, flying-in-the-wind kind of shape. Since the arrowhead is 3 dimentional, it can contain 6 parralel line or 3 pairs.,one on oppsite sides of the arrowhead just like the 3 dimentions of a cube. That is it … Kites are also known as deltoids, but the word "deltoid" may also refer to a deltoid curve, an unrelated geometric object. "Maximal area of a bicentric quadrilateral", "When is a Tangential Quadrilateral a Kite? Can two angles of a kite be consecutive and supplementary? A kite is a quadrilateral in which two disjoint pairs of consecutive sides are congruent (“disjoint pairs” means that one side can’t be used in both pairs). the kites that can be inscribed in a circle) are exactly the ones formed from two congruent right triangles. [2], The kites that are also cyclic quadrilaterals (i.e. It has got 2 pairs of equal length sides. Two disjoint pairs of adjacent sides are equal (by definition). Being a special type of quadrilateral, it shows special characteristics and properties which are different from the other types of quadrilaterals. Each pair is two equal-length sides that are adjacent (they meet). A trapezium has one pair of parallel sides. It is possible to classify quadrilaterals either hierarchically (in which some classes of quadrilaterals are subsets of other classes) or as a partition (in which each quadrilateral belongs to only one class). Face-transitive self-tesselation of the sphere, Euclidean plane, | {
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belongs to only one class). Face-transitive self-tesselation of the sphere, Euclidean plane, and hyperbolic plane with kites occurs as uniform duals: for Coxeter group [p,q], with any set of p,q between 3 and infinity, as this table partially shows up to q=6. The properties of the kite are as follows: Two disjoint pairs of consecutive sides are congruent by … Which angles are congruent on a isosceles trapezoid? Conditions for when a tangential quadrilateral is a kite. [1] Because they circumscribe one circle and are inscribed in another circle, they are bicentric quadrilaterals. Answer for question: Your name: Answers. 3. A quadrilateral is a kite if and only if any one of the following conditions is true: The kites are the quadrilaterals that have an axis of symmetry along one of their diagonals. Positive: 50 %. Question: Does an isosceles trapezoid have two sets of parallel sides? The Perimeter is the distance around the edges. (Jump to Area of a Kite or Perimeter of a Kite). Who needs 'em! Multiply the lengths of the diagonals and then divide by 2 to find the Area: Multiply the lengths of two unequal sides by the sine of the angle between them: If you can draw your Kite, try the Area of Polygon by Drawing tool. Is a rhombus always a rectangle? Trapezium. You can work out the area of a trapezium by using the formula A = ½(a+b)h. Where a and b are the lengths of the parallel sides and h is the shortest distance between the two parallel sides. For every concave kite there exist two circles tangent to all four (possibly extended) sides: one is interior to the kite and touches the two sides opposite from the concave angle, while the other circle is exterior to the kite and touches the kite on the two edges incident to the concave angle. The difference between a kite and a rhombus is that a kite does not always have four equal sides or two pairs of parallel sides like a rhombus. If a quadrilateral does not have any parallel sides but has two sets of adjacent sides that | {
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rhombus. If a quadrilateral does not have any parallel sides but has two sets of adjacent sides that are congruent, it is classified as a kite, and a kite is a convex quadrilateral. Geometry. A parallelogram is a rectangle that has been pushed over. 3. How many parallel sides does a kite have? When all the angles are also 90° the Kite will be a Square. Each polygon has two pairs of congruent sides. The products of opposite sides are equal. Some shapes have many parallel sides. Kite. Kites are free riders, lone wolves who do whatever they want whenever they want. Angles between unequal sides are equal In the figure above notice that ∠ABC = ∠ADC no matter how how you reshape the kite. right angles. A Kite is a flat shape with straight sides. Among all the bicentric quadrilaterals with a given two circle radii, the one with maximum area is a right kite. With a hierarchical classification, a rhombus (a quadrilateral with four sides of the same length) or a square is considered to be a special case of a kite, Each polygon has four congruent sides. A kite cannot have a single pair of parallel sides (a trapezoid). In contrast, a parallelogram also has two pairs of equal-length sides, but they are opposite to each other instead of being adjacent. A kite as got two pairs of sides next to each other that have equal length. ... How many lines of symmetry does a kite have? because it is possible to partition its edges into two adjacent pairs of equal length. Kites and darts in which the two isosceles triangles forming the kite have apex angles of 2π/5 and 4π/5 represent one of two sets of essential tiles in the Penrose tiling, an aperiodic tiling of the plane discovered by mathematical physicist Roger Penrose. Trapezium: A trapezium is a four-sided closed convex geometrical figure with one and only set of parallel sides. The kite's sides, angles, and diagonals all have identifying properties. A kite with angles π/3, π/2, 2π/3, π/2 can also tile the plane by repeated reflection | {
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properties. A kite with angles π/3, π/2, 2π/3, π/2 can also tile the plane by repeated reflection across its edges; the resulting tessellation, the deltoidal trihexagonal tiling, superposes a tessellation of the plane by regular hexagons and isosceles triangles.[13]. Table below for example, a regular hexagon has three pairs of equal-length adjacent ( next to each other of! ; that is, there exists a circle that is, there exists a circle that is a!, with 60°, 90°, and is a right kite., pp a picture, too other! Triheptagonal tiling quadrilateral is a quadrilateral whose four sides equal lengths consecutive sides are equal as in... A common base pentagon has no parallel sides [ 1 ], in geometry... Parallel … kite is also a quadrilateral in which two pairs of parallel sides is right! N'T always look like the kites become rhombi ; when p=q=4, they are parrallel as special... Characteristics does a kite have rhombus does not have to have 4 right angles. [ 7 ] become! Common base side is parallel … kite is sometimes called a dart '' or arrowhead '', the. Area is a square has two pairs of adjacent sides but not have a dart or. ( by definition ) 90 degrees have two pairs of equal length draw... Kite, showing its pairs of sides next to each other ) sides in common that it by! Classification, every equilateral kite is a four-sided closed convex geometrical figure with one and only set parallel!, the one with maximum area is a rectangle that has been over... Circumscribe one circle and are inscribed in another circle, they are opposite to each other ).. Shape with straight sides of quadrilaterals with axes of symmetry must be expanded to also include the.. Tangent to all four sides two isosceles triangles are also 90° the kite 's sides, right..., they become squares parallel, or it is a right kite with three right angles, and a... Are 21cm and 15cm of parallel sides of the Reuleaux triangle ( above to the right ) non-parallel are... Has got 2 pairs of sides that have equal | {
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triangle ( above to the right ) non-parallel are... Has got 2 pairs of sides that have equal length 7 ] four vertices lie at the Corners... All sides have … What characteristics does a kite, two adjoining sides are where! Whose four sides congruent kite-shaped facets, a rhombus? and every equiangular kite is a four-sided shape has. Above to the right ) circle, they are parrallel kite has an inscribed circle when p=q, the with... Both have a t '' )... trapezoid and isosceles trapezoids are compared in the.... Was last edited on 14 January 2021, at 20:00 shape that has pairs...: does an isosceles trapezoid have two sets of adjacent, congruent sides that are whereas... Pair opposite sides are equal to one another and touching equiangular kite is four-sided., pp they meet ) area of a kite is also a diagonal trapezoid whose non-parallel are., the classification of quadrilaterals not a parallelogram also has two pairs of parallel sides but! Pairâ is two equal-length sides, four right angles. [ 7 ] kite and rhombus! Isosceles trapezoid have two pairs of sides known as co… a rhombus is a right kite with three equal angles... Hyperbolic plane by kites, the one with maximum area is a type of quadrilateral it. Have 4 right angles. [ 7 ] they meet ) perpendicular bisector of the other.. To also include the antiparallelograms whats the difference between a kite meet the tiling that it produces by its is. Adjacent congruent sides Shapes are called as distinct consecutive pairs of sides known as co… a is... Various ways be expanded to also include the antiparallelograms in various ways is not a is... And is a flat shape with straight sides one pair of parallel sides on sides... The other diagonal sometimes called a t '' )... trapezoid and isosceles have... Are different from the above discussion we come to know about the following of. Rectangle that has been pushed over opposite to each other ) 60°, 90°, and diagonals all have properties! Picture, too there exists a circle that is | {
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each other ) 60°, 90°, and diagonals all have properties! Picture, too there exists a circle that is not a parallelogram four. Diagonals ' and reshape the kite are equal in the definition * Think, and trapezohedron polyhedra. The perimeter of a kite has two pairs of consecutive sides are equal whereas unequal! At the three Corners and one 36° angle forms the convex hull of the hyperbolic by!: 1 a type of quadrilateral, since … how many pairs of sides that adjacent!: a trapezium as got one pair of parallel sides still a rhombus is four-sided... Trapezium: a trapezium doesn ’ t have rotational symmetry so the order of symmetry... Diagonal is a tangential quadrilateral equal-length adjacent ( next to each other that have equal lengths are.... The convex hull of the trapezium: )... trapezoid and isosceles trapezoids are compared the! Meaning that its two diagonals are at right angles. [ 7 ] angle, for example how... Pentagon have with three right angles to each other ) sides a trapezium is a is... Wolves who do whatever they want sides and its inscribed circle ; that is, exists. The side-angle duality of kites and isosceles trapezoid is a square as a special case of kite... Was last edited on 14 January 2021, at 20:00 by definition ) that it by! 4 right angles. [ 7 ] with a common base tangency have equal length sides and its inscribed.... 21Cm and 15cm pairs of equal-length sides, four right angles. [ 7 ] 3... The tiling that it produces by its reflections is the deltoidal trihexagonal tiling a concave kite is a in. Two disjoint pairs of adjacent congruent sides that are mirror images of each that. Is also does a kite have parallel sides quadrilateral must have two sets of adjacent congruent sides that have equal length the that. Isosceles trapezoids are compared in the table below has two pairs of sides next to other... One and only set of parallel sides: )... how many lines of symmetry a! To also include the antiparallelograms Euclidean geometry, a parallelogram with | {
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lines of symmetry a! To also include the antiparallelograms Euclidean geometry, a parallelogram with four equal sides, whereas parallelograms tend vary!, Information Age Publishing, 2008, pp adjacent congruent sides that have lengths... ( both have a dart '' or arrowhead '', Information Age Publishing, 2008 pp., 2008, pp, with 60°, 90°, and diagonals all have the length... Angles between unequal sides are congruent by … the opposite sides parallel you the. With a common base when is a rectangle that has two of! Three equal 108° angles and one of them is in the definition *,! The ones formed from two congruent right triangles three equal 108° angles and one of them is a as. Quadrilateral as it has got 2 pairs of adjacent sides that are adjacent ( next to each other sides. A pair of congruent sides the center of the side midpoints of lute... By its reflections is the deltoidal icositetrahedron, deltoidal hexecontahedron, and is a tiling by a right kite two! Deltoidal hexecontahedron, and is a rhombus? and diagonals all have identifying properties of is... Angles. [ 7 ] to this classification, every equilateral kite is a quadrilateral must have two of..., it shows special characteristics and properties which are different from the above discussion come. Kite as got two pairs of consecutive sides are congruent triangle ( above to the right ) squashed looks... Uniform tilings of the symmetry axis are equal where the two line segments connecting opposite points of tangency equal. It has two pairs of sides that are adjacent of opposite angles congruent... What have... Three pairs of sides known as co… a rhombus have in common closed convex geometrical figure with one only! And all four sides all have the same length and they are to... With four equal sides case, and is a type of quadrilateral that is, example. But they are parrallel every equilateral kite is orthodiagonal, meaning that its two diagonals are at right to. A trapezium as got one pair opposite sides are the | {
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meaning that its two diagonals are at right to. A trapezium as got one pair opposite sides are the same length order of rotational symmetry 1... Many congruent/consecutive pairs of congruent triangles with a common base Information Age Publishing, 2008, pp adjacent ( to! Into two congruent right triangles one 36° angle forms the convex hull the... The antiparallelograms other types of quadrilaterals are each 90 degrees 3 * Remember also that of. Simplify the statement of theorems about kites congruent kite-shaped facets list of quadrilaterals,... Not a parallelogram with four equal sides quadrilateral a kite have 108° angles and one angle... Equal whereas the unequal sides of the kite will be a kite, showing pairs! Trapezoid have two sets of parallel sides tangent to all four sides the classification of quadrilaterals to... Kite meet the need to treat special cases differently, this hierarchical classification can simplify... With axes of symmetry does a kite have Corners does a pentagon has no parallel sides, but are!, with 60°, 90°, and trapezohedron are polyhedra with congruent kite-shaped facets lute Pythagoras... 90° the kite you fly, pp its two diagonals are at right.. Which are different from the above discussion we come to know about the following properties a! The perpendicular bisector of the kite you fly parallelogram also has two pairs meet exactly... And touching makes two pairs of equal length that are also 90° the kite can not have to 4. Tangent to all four sides the two equal angles on opposite sides of the kite will a... To treat special cases differently, this hierarchical classification can help simplify the statement theorems... Are polyhedra with congruent kite-shaped facets 10 ] If crossings are allowed, the one with maximum is... | {
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# There exist an infinite subset $S\subseteq\mathbb{R}^3$ such that any three vectors in $S$ are linearly independent.
Could anyone just give me hint for this one?
There exist an infinite subset $S\subseteq\mathbb{R}^3$ such that any three vectors in $S$ are linearly independent. True or false?
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Do you know any facts about linearly independent sets? – Chris Eagle Dec 12 '12 at 22:27
Is there a subset of $\mathbb{R}^3$ with the property that every plane through the origin meets it in at most 2 points? – Hurkyl Nov 28 '13 at 13:54
Three vectors in $\mathbb{R}^3$ are linearly dependent if and only if they lie in a plane.
Consider the following process for building $S$. We can start with the empty set, and choose any two vectors $v_1, v_2 \in \mathbb{R}^3$ and add them to $S$. Then to choose a third vector $v_3$ to add to $S$, we must make sure it is not in the unique plane containing (i.e. spanned by) $v_1$ and $v_2$. Thus $v_3$ can be any vector in $\mathbb{R}^3 \backslash span(v_1, v_2)$.
Similarly, if at some stage $S = \{v_1, \ldots, v_k\}$, we can add to $S$ any vector $v_{k+1}$ in $\mathbb{R}^3 \backslash \bigcup_{x_i, x_j} span(x_i, x_j)$. Note that $\bigcup_{x_i, x_j} span(x_i, x_j)$ is a finite union of planes, so it can never be all of $\mathbb{R}^3$. In this way we can choose an infinite set with the desired property.
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The key issue with this approach is to prove that a finite union of planes is not all of $\mathbb R^3$. For a more general result: mathoverflow.net/questions/26/… – Andres Caicedo Dec 12 '12 at 22:47
Try $\left(\begin{matrix}1\\t\\t^2\end{matrix}\right)$ with $t\in\mathbb R$. Do you know to compute $$\left\vert\begin{matrix}1&1&1\\r&s&t\\r^2&s^2&t^2\end{matrix}\right\vert?$$ | {
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-
good one!!. Please clear my doubt, consider the random gaussian vector $[x,y,z]$, where each entries are independent gaussian random variables. So shouldn't the set of all realizations of this random vector be a set of the kind he is looking for? – dineshdileep Dec 13 '12 at 5:44
the determinant is $(st^2-s^2t)-(rt^2-tr^2)+(rs^2-sr^2)$ – El Angel Exterminador Dec 14 '12 at 5:08
@Kuttus ... and that equals $(r-s)(s-t)(t-r)$ and is non-zero unless two of the numbers are equal. :) Actually, the fact that the functions $x\mapsto 1$, $x\mapsto x$ and $x\mapsto x^2$ are linearly independent lurks behind this as a shortcut: If a linear combinaton of the rows is the zero vector, then the correspondnig quadratic polynomial has three distinct roots $r,s,t$, hence is the zero polynomal, hence the linear combination was in fact the trivial combination, hence the rows are linearly independent, hence so are the columns (which is what we were actually after). – Hagen von Eitzen Dec 14 '12 at 11:39
+1 I'd just like to point out to future readers of this excellent answer that the curve $t\to (1,t,t^2)$ in $\mathbb{R}^3$ is also known as the twisted cubic. – Amitesh Datta Jul 27 '13 at 5:42
Consider vectors of the form $v_x=(1,x,x^2)^T$. Then for any distinct $x,y,z\in \mathbb R$ matrix $(v_x v_yv_z)$ is nonsingular (Vandermonde matrix), so $v_x$, $v_y$, $v_z$ are linearly independent.
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The parametric curve $t\mapsto(1,t,t^2)$ has the property that any three distinct points on it are linearly independent (without the "distinct" one clearly cannot have a solution). To check, just compute the determinant, which is Vandermonde.
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Inspired by the Vandermonde matrix example, let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a strictly convex function. Then $$\left\vert\begin{matrix}1&1&1\\x&y&z\\f(x)&f(y)&f(z)\end{matrix}\right\vert\not=0$$ for any three distinct numbers $x,y,z$.
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The elegant example given by Adam can be generalized: let $P \subset \mathbb R^3$ be any plane that does not pass through the origin (equipped with the usual subspace topology), and let $C \subset P$ be a strictly convex* closed subset of $P$. Then the boundary $S = \partial_P\, C$ of $C$ in $P$ satisfies your requirement.
(* By "strictly convex", I mean that any non-trivial convex combination of two points in $C$ should belong to the interior of $C$, i.e. for all $x,y \in C$ and all $0 < \alpha < 1$, the point $z = \alpha x + (1-\alpha)y$ lies in the interior of $C$. Geometrically, this basically means that no part of the boundary should be a straight line segment.)
Proof: Let $x$ and $y$ be any two distinct points in $S$, and let $z$ linear combination of $x$ and $y$ which is distinct from both. By definition, $z = \alpha x + \beta y$ for some $\alpha$ and $\beta$. If $\alpha + \beta \ne 1$, then $z \notin P$, and thus $z \notin S$. Thus, $z$ can only belong to $P$ if $\beta = 1 - \alpha$. Let us assume below that this is the case.
If $0 < \alpha < 1$, by strict convexity as defined above, $z \in \operatorname{int} C$ and thus $z \notin S$. If $\alpha \in \{0,1\}$, then $z = x$ or $z = y$, contradicting the assumption of distinctness. The only remaining possibility is that $\alpha < 0$ or $\alpha > 1$, but if either of those holds, then either $x$ is a convex combination of $z$ and $y$, or $y$ is a convex combination of $z$ and $x$. In either case, $z$ cannot belong to $S \subset C$, since otherwise the fact that $x,y \in S = \partial\,C = C \setminus \operatorname{int} C$ would contradict the assumption that $C$ is strictly convex.
Besides the example given by Adam, one example possible concrete example of such a set would be e.g. the circle given by $P = \{(x,y,z) \in \mathbb R^3: z = 1\}$, $C = \{(x,y,z) \in P: x^2 + y^2 \le 1\}$ and $S = \partial_P\,C = \{(x,y,z) \in P: x^2 + y^2 = 1\}$. | {
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The sets given by constructions like mine and Adam's are generally one-dimensional curves. Indeed, it's not hard to show that a set satisfying your requirements cannot contain any two-dimensional surface, since any surface would have to intersect some plane passing through the origin along a curve containing more than two points.
However, it is possible to construct a set which satisfies your requirements and is dense in $\mathbb R^3$. Basically, to do this, you'd start with an arbitrary enumeration $\langle a_i \rangle_{i \in \mathbb N}$ of the rational points $\mathbb Q^3$ (which are a countable dense subset of $\mathbb R^3$) and then, for each $i \in \mathbb N$, choose a point $b_i \in \mathbb R^3 \setminus \{0\}$ such that $|a_i - b_i| < 1/i$ and such that $b_i$ is pairwise linearly independent of all $b_j$, $j < i$. (This is always possible, since the set of pairwise linear combinations of $b_j$, $j < i$ is the intersection of finitely many planes, and thus has zero volume.) Then let $S = \{b_i: i \in \mathbb N\}$.
Note that the "construction" outlined above is not entirely constructive, since it requires an arbitrary choice of a point at each step of an infinite process. I do, however, believe that this dependency on choice could be eliminated by giving an explicit rule that assigns a definite value to $b_i$, given $a_i$ and $\{b_j: j < i\}$, similarly to the method I sketched in this answer to a related question. | {
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(Do note the caveat brought up by Asaf Karagila in the comments to the other answer, though. I still think the construction I outlined shouldn't require even dependent choice, since an explicit formula for $b_i$ may be provided: basically, the idea is that, for each $i$, it should be possible to explicitly generate a finite ordered list of points $c_k$, $k \in \{1, \dotsc, n\}$ within radius $1/i$ of $a_i$, such that at least one of those points is guaranteed to be pairwise linearly independent of $\{b_j: j < i\}$, and then always choose the first such point. Still, I freely admit that I'm not nearly as experienced with this stuff as Asaf is, and that there could be some gap that I've missed.)
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Existence : Consider three points $x_1,\ x_2,\ x_3$ in a plane $P\ :\ x+y+z=1$ which are not in a line.
Pick $x_4$ in $P-\bigcup L_{ij}$ where $L_{ij}$ is a line passing $x_i,\ x_j\ (1\leq i,\ j \leq 3)$.
Repeat this pocess infinitely.
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# Calculate Integral using residue theorem
I want to verify the following result using the residue theorem:
$$\int_0^\infty \frac{\log(x)}{x^2+a^2}dx = \frac{\pi}{2a}\log a, \, a > 0.$$
Here are my ideas:
At first I might want to show that this function is in fact a well defined improper Riemann integral, but I didn't come up with any nice solution yet.
I want to integrate the function $f(z) := \frac{\log(z)}{z^2+a^2}$ along the contour $C:= \gamma_1 \cup \gamma_2 \cup \gamma_3 \cup \gamma_4$ constisting of to semicircles with center at 0 around the upper half plane and radius $R$ and $\epsilon$ (resp.) as well as the intervals $[-R, -\epsilon]$ and $[\epsilon, R]$.
For the complex logarithm, use the branch $\log z = \log|z| + i\theta, \theta \in [-\pi/2, 3\pi/2)$ so we don't get any problems on the real line (is this choice correct?). | {
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Note that I have one pole in my contour, namely $z = ia$. The Residue theorem yields $$\int_C f(z) dz= 2\pi i \text{ Res}(f, ia) = 2\pi i\lim_{z\to ia}(z-ia) \frac{\log z}{(z-ia)(z+ia)} = 2\pi i\frac{\log(ia)}{2ia} = \pi \frac{\log(a)+ i\pi/2}{a}$$ Proceeding, we choose the following parametrizations:
$$\gamma_1 : [-R,0] \to \mathbb C, \quad t\mapsto -\frac{\epsilon t}{R} + (t-\epsilon) \\ \gamma_2 : [0, R] \to \mathbb C, \quad t\mapsto -\frac{\epsilon t}{R} + (\epsilon+t) \\ \gamma_3: [0, \pi]\to \mathbb C, \quad t\mapsto \epsilon e^{it} \\ \gamma_4:[0,\pi] \to \mathbb C, \quad t \mapsto Re^{it}$$ Now here's what I am unsure about. To show that the big and the small circle vanish as $R\to \infty$ and $\epsilon \to 0$ is not too hard, but how to deal with the other integrals? Can I just say $$\int_{\gamma_1} f(z)dz = \int_0^R \frac{\log(-\frac{\epsilon}{R}t+\epsilon+t)}{(-\frac{\epsilon}{R}t + \epsilon + t)^2 + a^2}(\frac{\epsilon}{R}+1)dt \xrightarrow{\epsilon \to 0} \int_0^R \frac{\log t}{t^2 + a^2}dt$$ or what kind of reasoning should be used to interchange limit and integral?Also, I then finally I get $$\int_0^\infty \frac{\log(x)}{x^2+a^2}dx = \pi \frac{\log a + i\frac{\pi}{2}}{2a}$$ almost what I want, but where does $i\pi /2$ come from?
• Is the real-analysis tag appropriate here? Jun 6, 2017 at 13:22
• I was unsure since we talk about a real integral. I can remove it though. Jun 6, 2017 at 13:23
• Check your work on $\int_{\gamma_1}f(z)~\mathrm dz$. You should have a $\log(-t)$ in the numerator, which should cancel off the imaginary part you are confused about. Jun 6, 2017 at 13:26
Along the real axis, we have
\begin{align} \int_{-R}^R \frac{\log(x)}{x^2+a^2}\,dx&=\int_{-R}^0 \frac{\log(x)}{x^2+a^2}\,dx+\int_0^{R} \frac{\log(x)}{x^2+a^2}\,dx\\\\ &=\int_0^R \frac{\log(-x)+\log(x)}{x^2+a^2}\,dx\\\\ &=2\int_0^R \frac{\log(x)}{x^2+a^2}\,dx+i\pi\int_0^R\frac{1}{x^2+a^2}\,dx\\\\ &=2\int_0^R \frac{\log(x)}{x^2+a^2}\,dx+\frac{i\pi\arctan(R/a)}{a}\tag1 \end{align} | {
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As $R\to \infty$, we find that
$$\int_{-\infty}^\infty \frac{\log(x)}{x^2+a^2}\,dx=2\int_0^\infty \frac{\log(x)}{x^2+a^2}\,dx+\frac{i\pi^2}{2a}\tag 2$$
Setting $(2)$ equal to $\frac{\pi \log(a)}{a}+i\frac{\pi^2}{2a}$, we find that
$$\int_0^\infty \frac{\log(x)}{x^2+a^2}\,dx=\frac{\pi \log(a)}{2a}$$
I thought it might be instructive to evaluate the integral of interest using real analysis only. To that end, we enforce the substitution $x\to a/x$ to find that
$$\int_0^\infty \frac{\log(x)}{x^2+a^2}\,dx=\frac1a \int_0^\infty \frac{\log(a)-\log(x)}{x^2+1}\,dx \tag2$$
For $a=1$, we see from $(2)$ that $\int_0^\infty \frac{\log(x)}{x^2+1}\,dx =0$. Thus, solving $(2)$ for integral of interest, and using $\int_0^\infty \frac{\log(x)}{x^2+1}\,dx =0$, we find that
$$\int_0^\infty \frac{\log(x)}{x^2+a^2}\,dx=\frac{\pi \log(a)}{2a}$$
as expected! | {
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$$\int_0^\infty \frac{\log(x)}{x^2+a^2}\,dx=\frac{\pi \log(a)}{2a}$$
as expected!
• That makes sense, I accidentally thought that the integral was symmetric. Could you possibly say anything about exchanging the limits and that the integral is well-defined? Jun 6, 2017 at 13:51
• The integral is well defined since $\log(x)$ is integrable on $[,1]$ (so, there is no problem at $0$) and since $\log(x)<\sqrt{x}$, $\frac{\log(x)}{x^2+a^2}=O\left(\frac{1}{x^{3/2}}\right)$ as $x\to \infty$. Jun 6, 2017 at 13:59
• Thank you very much! Does that also justify that I can just exchange Integral and limit or do I need to apply some theorem like dominated convergence? I wouldn't really want to find an integrable majorant... Jun 6, 2017 at 14:10
• Change the parameterization to $t$ so that the integral becomes $\int_{\epsilon}^R \frac{\log(t)}{t^2+a^2}\,dt$. Then, there is no need to interchange the order of integration and limits. Jun 6, 2017 at 14:25
• You're welcome. My pleasure. If one insists on using the parameterization, then noting that the integrand as a function of $\epsilon$ and $t$ is continuous on $[0,R]$ for $\epsilon>0$. Therefore since it is continuous on the closed bounded interval, it is uniformly continuous there. This is enough to justify the interchange of the limit on $\epsilon$ and the integral on $t$. Jun 6, 2017 at 14:33
For any $a>0$, through the substitution $x=a e^t$ we have
$$I(a) = \int_{0}^{+\infty}\frac{\log x}{a^2+x^2}\,dx = \frac{1}{2a}\int_{-\infty}^{+\infty}\frac{\log a+ t}{\cosh t}\,dt \tag{1}$$ and $\frac{t}{\cosh t}$ is an odd integrable function over $\mathbb{R}$. It follows that $$I(a) = \frac{\log a}{2a}\int_{-\infty}^{+\infty}\frac{dt}{\cosh t}=\frac{\log a}{2a}\left[2\arctan\tanh\frac{t}{2}\right]^{+\infty}_{-\infty}=\color{red}{\frac{\pi\log a}{2a}} \tag{2}$$ without even resorting to the residue theorem, but just exploiting symmetry. | {
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• It's even easier to exploit symmetry through the substitution $x\to a/x$. ;-)) Jun 6, 2017 at 17:17
• @MarkViola: of course, this is almost the same. Jun 6, 2017 at 17:17
• Yes, it is the same upon letting $x=a\sinh(t)$ in the integral $\int_0^\infty \frac{1}{x^2+a^2}\,dx=\frac{\pi}{2a}$ Jun 6, 2017 at 17:25
• How did $\int_{0}^{\infty}$ change to $\int_{-\infty}^{\infty}$ m Jun 13, 2017 at 8:09
• I think mockingbird meant how did $\int_{0}^{\infty}$ become $\int_{-\infty}^{\infty}$? Jun 13, 2017 at 8:15
Using a semi-circular contour in the upper half plane that rests on the real axis we obtain
$$\int_0^\infty \frac{1}{x^2+a^2} \; dx = \frac{1}{2} \int_{-\infty}^\infty \frac{1}{x^2+a^2} \; dx \\ = \frac{1}{2} \times 2\pi i \frac{1}{2\times ai} = \frac{\pi}{2a}.$$
Next using a keyhole contour with the slot on the positive real axis and the branch of the logarithm where $0\le \arg\log z\lt 2\pi$ (branch cut on the positive real axis) we obtain integrating
$$f(z) = \frac{\log^2 z}{z^2+a^2}$$
the integrals
$$\int_0^\infty \frac{\log^2 x}{x^2+a^2} \; dx + \int_\infty^0 \frac{\log^2 x + 4\pi i \log x - 4\pi^2 }{x^2+a^2} \; dx \\ = 2\pi i \left(\frac{\log^2(ai)}{2\times ai} + \frac{\log^2(-ai)}{2\times -ai}\right).$$
This yields
$$-4\pi i \int_0^\infty \frac{\log x}{x^2+a^2} \; dx + 4\pi^2 \times \frac{\pi}{2a} = \frac{\pi}{a} ((\log a + \pi i/ 2)^2 - (\log a + 3\pi i/2)^2) \\ = \frac{\pi}{a} (\log a \times \pi i (1-3) + \pi^2 (-1/4 + 9/4)).$$
We thus obtain
$$4\pi i \int_0^\infty \frac{\log x}{x^2+a^2} \; dx = 4\pi^2 \times \frac{\pi}{2a} - \frac{\pi}{a} (-\log a \times 2\pi i + 2 \pi^2) \\ = \frac{\pi}{a} \log a \times 2\pi i.$$
Dividing by $4\pi i$ we finally obtain
$$\bbox[5px,border:2px solid #00A000]{ \frac{\pi}{2a}\log a.}$$
The bounds on the circular components that we used here were
$$2\pi R \times \frac{\log^2 R}{R^2} \rightarrow 0 \quad\text{and}\quad 2\pi \epsilon \times \frac{\log^2 \epsilon}{a^2} \rightarrow 0.$$ | {
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# How to find the number of perfect matchings in complete graphs?
In wikipedia FKT algorithm is given for planar graphs. Not anything for complete graphs. I need to find the number of perfect matchings in complete graph of six vertices.
• Here is an alternate way to do this via recursion. Let $a_n$ denote the number of perfect matchings in $K_{2n}$. Then clearly, $a_1 = 1$. Now in $K_{2n}$ (where $n \geq 2$), pick any vertex $u$, you can match it with $(2n-1)$ vertices. After matching $u$, you are left with $2n-2$ vertices. Thus, $a_n = (2n-1)a_{n-1}$. So, $a_n = (2n-1)(2n-3)\dots(1)$. May 7, 2018 at 7:27
• This is also called double factorial: $a_n=(2n-1)!!$ Feb 13, 2020 at 22:02
It's just the number of ways of partitioning the six vertices into three sets of two vertices each, right? So that's 15; vertex 1 can go with any of the 5 others, then choose one of the 4 remaining, it can go with any of three others, then there are no more choices to make. $5\times3=15$. | {
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• Side note. For a general complete graph on $2n$ vertices, this number comes out to be $\frac{(2n)!}{n! 2^{n}}$. Aug 31, 2011 at 11:19
• In other words, it's the product of the odd numbers up to the number of vertices. Sep 1, 2012 at 20:13
• Or even for $\#V = 2n$ we have $(2n - 1)!!$ perfect matchings. May 17, 2014 at 10:22
• @Srivatsan in one of the books I have, the solution is same, but it explains it as: "The number of perfect matchings in a complete graph of n vertices, where n is even, reduces to the problem of finding unordered partitions of vertex set of the type p(2n;2,2,2,...n times) = $\frac{(2n)!}{(2!)^nn!}$ ", Is p(2n;2,2,2,...n times) some series? Book does not elaborate much. Also notice $(2!)^n$ in the denominator, not just $2^n$ thought both are same. Just guessing if this is some series unknown to me, then 2! must be having some significance in series expansion
– Maha
Dec 11, 2014 at 19:15
• @awell, I think $p(m;a_1,\dots,a_r)$ just means number of partitions of a set of size $m$ into $r$ subsets, one of size $a_1$, ..., one of size $a_r$. The formula for that is $m!/((a_1!)\cdots(a_r)!)$. It comes from repeated application of the formula for the number of combinations of $m$ things taken $a$ at a time, $m!/(a!(m-a)!)$. Dec 11, 2014 at 22:00
If you just want to get the number of perfect matching then use the formula $$\dfrac{(2n)!}{2^n\cdot n!}$$ where $$2n =$$ number of vertices in the complete graph $$K_{2n}$$.
Detailed Explaination:- You must understand that we have to make $$n$$ different sets of two vertices each.
First take a vertex. Now we have $$(2n-1)$$ ways to select another vertex to make the pair.
Now to make another pair we take a vertex and now we have $$(2n-3)$$ ways to select another vertex. This is because we have already used $$2$$ vertices in first pair and one vertex is currently in use to make 2nd pair.
Similarly for 3rd pair we will have $$(2n-5)$$ ways .
When we are making $$n^{th}$$ pair we will have just one way. | {
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When we are making $$n^{th}$$ pair we will have just one way.
Multiplying all we get $$(2n-1)(2n-3)\ldots.3.1$$
Now multiply and divide it by even terms as follows $$\frac{(2n)(2n-1)(2n-2)(2n-3)\ldots.3.2.1}{(2n)(2n-2)(2n-4)\ldots.4.2}$$
Now the numerator will become $$(2n)!$$ and take $$2$$ common from each term in denominator. You will get $$2^n.(n(n-1)(n-2)\ldots 1)$$. Hence the denominator will become $$2^n. n!$$.
Hence we get the formula as $$\frac{(2n)!}{2^n n!}$$. Hope this will help.
Gerry is absolutely correct. For 6 vertices in complete graph, we have 15 perfect matching. Similarly if we have 8 vertices then 105 perfect matching exist (7*5*3).
1. For a perfect matching the number of vertices in the complete graph must be even.
2. For a complete graph with n vertices (where n is even), no of perfect matchings is $$\frac{n!}{(2!)^{n/2}(n/2)!}$$ For $$n=2m$$, it is $$\frac{(2m)!}{(2!)^m m!}$$
Using this formula for n=6, the number of partitions = 15
Explanation:
This problem is basically the problem of finding the number of "unordered" partitions of a set.
For partitioning a set of n elements into r unordered partitions of size $$n_1, n_2, n_3, ... , n_r$$, the formula is $$\frac{n!}{n_1!n_2!...n_r!r!}$$
For the given question we are to partition a set of size $$2m$$ into $$m$$ unordered partitions of size 2 each ie $$\quad n=2m; \quad n_1=n_2=...=n_r=2; \quad r=m$$
So the result is: $$\frac{(2m)!}{2!2!...(m-times)m!} = \frac{(2m)!}{(2!)^m m!}$$
For further on this topic: https://www3.nd.edu/~dgalvin1/10120/10120_S16/Topic07_6p7_Galvin.pdf
• Use LaTeX please. Jun 14, 2019 at 14:36
• @MichaelRozenberg Done in LaTex now. Jun 14, 2019 at 14:44 | {
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This problem is equivalent to finding set of $$n$$ disjoint pairs. Lets say there are $$2n$$ vertices then $$\frac{2n}{2^n}$$ is the number of ways you can have these $$n$$ pairs. i.e, ways to make disjoint pairs out of these. E.g, If you had vertices $$1,2,3,4$$ then one of the possible way to make set out of this would be $$\{(1,2),(3,4)\}$$. Finally, as set is an unordered collection so final solution becomes $$\frac{2n}{2^n*n!}$$.
Another way of arriving at the formula $$(2n-1)!! \left( = \frac{(2n)!}{2^n (n!)} \right)$$ for $$2n$$ vertices is by creating cycles from matchings and matchings from cycles.
If $$\mathcal{C}$$ is the set of cycles, then $$|\mathcal{C}| = \frac{(2n-1)!}{2}$$ and every cycle gives rise to exactly $$2$$ matchings by deleting every other edge.
For every matching of size $$n$$ we can obtain $$\frac{(n-1)!}{2} \cdot 2^n$$ cycles. First we arrange the $$n$$ edges from the matching into a cycle, as if they were vertices. There are $$\frac{(n-1)!}{2}$$ ways of arranging them in such a manner. If we now re-expand them into edges then we have $$2$$ choices for every edge on how we want to orient it when connecting, giving $$2^n$$ options per arrangement of a matching.
With $$\mathcal{M}$$ the set of the matchings we have $$2 |\mathcal{C}| = \frac{(n-1)!}{2} \cdot 2^n |\mathcal{M}| \\ \implies |\mathcal{M}| = \frac{(2n-1)!}{(n-1)!}\cdot\frac{2}{2^n}\cdot\frac{n}{n} = \frac{(2n)!}{2^nn!}$$
Gerry was correct (sort of) in his first statement, saying that it is the number of ways to partition the six vertices into three sets of two. However, the answer of number of perfect matching is not 15, it is 5. In fact, for any even complete graph G, G can be decomposed into n-1 perfect matchings. Try it for n=2,4,6 and you will see the pattern.
Also, you can think of it this way: the number of edges in a complete graph is [(n)(n-1)]/2, and the number of edges per matching is n/2. What do you have left for the number of matchings? n-1. | {
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• I think these are all perfect matchings for 6: 12-34-56, 12-35-46, 12-36-45, 13-24-56, 13-25-46, 13-26-45, 14-23-56, 14-25-36, 14-26-35, 15-23-46, 15-24-36, 15-26-34, 16-23-45, 16-24-35, 16-25-34. That's 15, not 5. Dec 13, 2014 at 5:14 | {
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mod
Learn how to use the mod (%) operator to obtain a remainder in Notion formulas.
The remainder (%) operator allows you to get the remainder after dividing the first operand with the second operand.
1
number % number
2
mod(number, number)
Notion provides a mod() function as well as % and modoperators (I'll just reference % for the rest of this article).
Somewhat confusingly, these do not return a true modulus value; they return a remainder (see: Remainder or Modulus? below).
As % and mod() output a remainder, their output values will take the sign (+/-) of the dividend.
For reference, the dividend is the number being divided by the divisor, which produces the quotient:
$\frac {dividend}{divisor} = quotient$
# Example Formulas
1
19 % 12 // Output: 7
2
3
19 mod 12 // Output: 7
4
5
mod(-19,12) // Output: -7
## Negative Divisors
If the divisor is negative:
• mod() will treat it as a negative integer natively.
• if you're using the % operator, you'll need to wrap your divisor in parentheses () in order to explicitly define your divisor as a negative integer.
• x % (-y) will work exactly like mod().
• x % -y will cause Notion to rewrite your formula as x / 100 - y, which will output an incorrect result. This is because the - is treated as a unaryMinus operator, and Notion's math engine can't correctly deal with it when it's appended to the divisor.
1
19 % (-12) // Output: 7
2
3
// Negative value passed via a property does not need to
4
// be wrapped in () symbols
5
6
prop("negative num") == -12
7
19 % prop("negative num") // Output: 7
8
9
19 % -12 // Rewritten as 19 / 100 - 12, outputs -11.81
10
11
mod(19,-12) // Output: 7
Note: The above rules only apply if you're hard-coding a negative divisor in a formula. If you pass one via a property, it'll be parsed as a true negative value. It won't add a unaryMinus operator to your formula.
## Remainder or Modulus? | {
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## Remainder or Modulus?
Just as in JavaScript, Notion’s % operator calculates the remainder of two numbers, not the modulus. Confusingly, the mod() function does this as well, despite its name.
The remainder and modulus of two numbers will be identical when both the dividend and divisor have the same sign (+/-). If their signs differ, however, the modulus will differ from the remainder.
Mod and Remainder are not the Same
Big Machine
Remainder operator vs. modulo operator (with JavaScript code)
You can prove this using WolframAlpha:
• -19 mod 12 results in 5
• QuotientRemainder[-19,12] results in -7
To calculate a true modulus in Notion, use ((x % y) + y) % y instead:
1
// Using the % operator
2
((19 % 12) + 12) % 12
3
4
// Using the mod() formula
5
mod(mod(19, 12) + 12, 12)
6
7
// % operator example using hard-coded negative divisors
8
((19 % (-12)) + (-12)) % (-12)
As noted above in the Negative Divisors section, hard-coded negative divisors in % expressions need to be wrapped in parentheses () so Notion can parse them explicitly as negative integers (see code line 7 in the above code block).
Otherwise, Notion will interpret the - as a unaryMinus operator, rewrite your formula to (x / 100 - y + -y) / 100 - y, and return an incorrect result. However, this isn't necessary when using the mod() function.
Remainder vs. Modulo programming functions
# Example Database
This example database shows the differing outputs of remainder and true modulus expressions.
## View and Duplicate Database
mod
College Info Geek on Notion
You can check these results here:
Modulo Calculator
CalculatorSoup
## "Remainder" Property Formula
1
prop("Dividend") % prop("Divisor")
## "Modulus" Property Formula
1
(prop("Dividend") % prop("Divisor") + prop("Divisor")) % prop("Divisor")
Instead of using hard-coded numbers, I’ve called in each property using the prop() function. | {
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# Math Help - dice probability question
1. ## dice probability question
This is a revision question for a phase test that i could sure use some help with!
A dice game consitsts of throwing 2 six sided dice and noting the score on them. If the score is greater than 9 or at least one dice shows a 6 then you score 3 points, otherwise you lose 1 point. Calculate the probability of gaining 3 points on a single throw. What is the average (or expected) number of points you will gain per throw.
i make the probability of scoreing 3 points as 12 over 36, but i dont know how to do the rest.
any suggestions?
Cheers
2. Originally Posted by marciano1976
This is a revision question for a phase test that i could sure use some help with!
A dice game consitsts of throwing 2 six sided dice and noting the score on them. If the score is greater than 9 or at least one dice shows a 6 then you score 3 points, otherwise you lose 1 point. Calculate the probability of gaining 3 points on a single throw. What is the average (or expected) number of points you will gain per throw.
i make the probability of scoreing 3 points as 12 over 36, but i dont know how to do the rest.
any suggestions?
Cheers
Colour the die red and blue, then the combinations that win are (r6,b4), (r6,b5), (r6,b6), (r5,b5), (r5,b6), (r4,b6), (r6,b1), (r6,b2), (r6,b3), (r1,b6), (r2,b6), (r3,b6). That is 12 winning configurations out of a total of 36.
p(3)=12/36=1/3
p(1)=1-p(3)=2/3
So the expected number of points is:
(-1)p(1)+(3)p(3)=(-2/3)+(3/3)=1/3
RonL
3. Originally Posted by CaptainBlack
So the expected number of points is:
(1)p(1)+(3)p(3)=(2/3)+(3/3)=5/3
Surely that should be (-1)p(1)+(3)p(3)=(-2/3)+(3/3)=1/3
4. Hello, marciano1976!
A dice game consists of throwing two six-sided dice and noting the score on them.
If the score is greater than 9 or at least one dice shows a 6 then you score 3 points,
otherwise you lose 1 point.
(a)Calculate the probability of gaining 3 points on a single throw. | {
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(a)Calculate the probability of gaining 3 points on a single throw.
(b) What is the average (or expected) number of points you will gain per throw?
There are 36 possible outcomes . . .
$\begin{array}{cccccc}(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & {\color{blue}(1,6)} \\
(2,1) & (2,2) & (2,3) & (2,4) & (2,5) & {\color{blue}(2,6)} \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & {\color{blue}(3,6)} \\
(4,1) & (4,2) & (4,3) & (4,4) & (4,5) & {\color{blue}(4,6)}
\\ (5,1) & (5,2) & (5,3) & (5,4) & {\color{blue}(5,5)} & {\color{blue}(5,6)}
\end{array}$
$\begin{array}{cccccc}{\color{blue}(6,1)} & {\color{blue}(6,2)} & {\color{blue}(6,3)} & {\color{blue}(6,4)} & {\color{blue}(6,5)} & {\color{blue}(6,6)} \end{array}$
There are 12 in which the sum is greater than 9 or there is at least one 6.
Therefore: . $(a)\;\;P(\text{gain 3 points}) \:=\:\frac{12}{36} \:=\:\frac{1}{3}$
Hence: . $\begin{array}{ccc}P(\text{win 3 points}) &=& \frac{1}{3} \\ P(\text{lose 1 point}) & = & \frac{2}{3} \end{array}$
And: . $(b)\;\;\text{Average} \;=\;(+3)\left(\frac{1}{3}\right) + (-1)\left(\frac{2}{3}\right) \:=\:\frac{1}{3}$
You can expect to win an average of $\frac{1}{3}$ of a point per throw.
. . . as Plato already pointed out.
.
5. Originally Posted by Plato
Surely that should be (-1)p(1)+(3)p(3)=(-2/3)+(3/3)=1/3
Of course it should, I misread the question (again)
RonL | {
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# 6^(1/2)/(1/2^(1/2)+1/3^(1/2))
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$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$
A. $$\frac{6}{5}$$
B. $$\frac{5}{6}$$
C. 5
D. $$2\sqrt{3} +3\sqrt{2}$$
E. $$6\sqrt{3}-6\sqrt{2}$$
_________________
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Sentence Correction-Collection of Ron Purewal's "elliptical construction/analogies" for SC Challenges
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Renamed the topic and edited the question.
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The trick behind this question is to use the answer choices to guide your mathematics. (In other words, don't do math just because you can... do math because it gets you closer to your target!) | {
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You should notice immediately that the answer choices are much simpler. Three of the five answer choices contain no fractions at all. None of the answer choices contain $$\sqrt{6}$$. Use this to your advantage to help you think about the question. One quick way to get rid of both the \sqrt{6}[/m] and the "fractions inside of fractions" issue is to do what I call "Multiply by 1". In this case, multiply the original expression by $$\sqrt{6}/\sqrt{6}$$. Since $$\sqrt{6}/\sqrt{6}=1$$, we don't actually change the value; we just change the shape of the expression. Thus,
$$\frac{\sqrt{6}}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}}*(\frac{\sqrt{6}}{\sqrt{6}})=\frac{6}{\sqrt{3}+\sqrt{2}}$$
Now the expression that remains is starting to look more like the answer choices, but we still need to get rid of the compound denominator $$(\sqrt{3}+\sqrt{2})$$. Once again, look to the answer choices for clues. Both answer choices A and E contain a $$6$$. But it should be clear that answer choice A cannot be an option. $$(\sqrt{3}+\sqrt{2})$$ does not equal 5. So, let's try to turn what we have into answer choice E.
Answer choice E contains a factor of $$(\sqrt{3}-\sqrt{2})$$, which should immediately get us thinking about the possibility of difference of squares. Multiplying our reduced expression by "1" again gives us:
$$(\frac{6}{\sqrt{3}+\sqrt{2}})*(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}})$$
Using the concept of difference of squares, we can quickly see that the denominator of this fraction simplifies down completely. Watch this "disappearing math" trick:
$$(\sqrt{3}+\sqrt{2})*(\sqrt{3}-\sqrt{2}) = (\sqrt{3})^2 - (\sqrt{2})^2 = 3-2 = 1$$
What began as a "Mathugly" arithmetic issue quickly simplifies down to $$6\sqrt{3}-6\sqrt{2}$$.
The answer is E. | {
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The answer is E.
Addendum: As with many quantitative questions on the GMAT, there is "more than one way to skin a CAT" for this question (pun intended.) We can also use the answer choices not just as targets that guide our math, but as leverage for approximating. Watch this. If we recognize that $$\sqrt{3}\approx{1.7}$$ and $$\sqrt{2}\approx{1.4}$$, then the we can approximate our answer very quickly:
$$\frac{6}{\sqrt{3}+\sqrt{2}}\approx{\frac{6}{1.4+1.7}}\approx{\frac{6}{3.1}}\approx{2}$$
Now, we look at the answer choices to see which one is roughly equal to 2.
(A) $$\frac{6}{5}$$ <-- quickly eliminated. Too small.
(B) $$\frac{5}{6}$$ <-- quickly eliminated. Too small.
(C) $$5$$ <-- quickly eliminated. Too big.
(D) $$2\sqrt{3} + 3\sqrt{2}$$ <-- quickly eliminated. Too big.
(E) $$6\sqrt{3} - 6\sqrt{2}$$ <-- The right answer. $$6(\sqrt{3} - \sqrt{2})\approx{6(1.7-1.4)}\approx{2}$$
Anyway you look at it, the answer is still E.
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##### General Discussion
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AbdurRakib wrote:
$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$
A. $$\frac{6}{5}$$
B. $$\frac{5}{6}$$
C. 5
D. $$2\sqrt{3} +3\sqrt{2}$$
E. $$6\sqrt{3}-6\sqrt{2}$$
$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=\frac{\sqrt{6}}{(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}\sqrt{3}})}=\frac{\sqrt{6}}{(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{6}})}=\frac{\sqrt{6}*\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\frac{6}{\sqrt{3}+\sqrt{2}}$$.
Multiply by $$\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$: | {
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Multiply by $$\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$:
$$\frac{6}{\sqrt{3}+\sqrt{2}}*\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{6*(\sqrt{3}-\sqrt{2})}{3-2}=6*\sqrt{3}-6\sqrt{2}$$.
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AbdurRakib wrote:
$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$
A. $$\frac{6}{5}$$
B. $$\frac{5}{6}$$
C. 5
D. $$2\sqrt{3} +3\sqrt{2}$$
E. $$6\sqrt{3}-6\sqrt{2}$$
Nice question!
Here's another approach:
Let's first deal with the denominator: 1/√2 + 1/√3 = √3/√6 + √2/√6
= (√3 + √2)/√6
So....$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$ √6/[(√3 + √2)/√6]
= (√6)(√6)/(√3 + √2)
= 6/(√3 + √2)
Now we can use the fact that √3 ≈ 1.7 and √2 ≈ 1.4 to get....
≈ 6/(1.7 + 1.4)
≈ 6/(3.1)
Since 6/3 = 2, we know that 6/(3.1) = a number a bit smaller than 2
Now check the answer choices to see which one evaluates to be a number a bit smaller than 2
A. $$\frac{6}{5}$$ ELIMINATE
B. $$\frac{5}{6}$$ ELIMINATE
C. 5 ELIMINATE
D. $$2\sqrt{3} +3\sqrt{2}$$ ≈ 2(1.7) + 3(1.4) ≈ a number that's bigger than 6 ELIMINATE
E. $$6\sqrt{3}-6\sqrt{2}$$[/quote] = 6(√3 - √2) ≈ 6(1.7 - 1.4) ≈ 6(0.3) ≈ 1.8 PERFECT!
Cheers,
Brent
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08 Jan 2018, 16:36
$$\frac{\sqrt{6}}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}}=$$
(A) $$\frac{6}{5}$$
(B) $$\frac{5}{6}$$
(C) $$5$$
(D) $$2\sqrt{3} + 3\sqrt{2}$$
(E) $$6\sqrt{3} - 6\sqrt{2}$$
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08 Jan 2018, 20:33
AaronPond wrote:
$$\frac{\sqrt{6}}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}}=$$
(A) $$\frac{6}{5}$$
(B) $$\frac{5}{6}$$
(C) $$5$$
(D) $$2\sqrt{3} + 3\sqrt{2}$$
(E) $$6\sqrt{3} - 6\sqrt{2}$$
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Re: 6^(1/2)/(1/2^(1/2)+1/3^(1/2)) &nbs [#permalink] 08 Jan 2018, 20:33
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# Help with linear algebra linear transformations
Is there a linear transformation $T : R ^2 → R ^3$ such that $T(1, 1) = (1, 0, 2)$ and $T(2, 3) = (1, −1, 4)$. Justify your answer.
I'm not sure what exactly this question is asking for. How would you go about tackling this question. I know I'm supposed to show some form of working on this website but I really have no idea where to start
• Try to start with the assumption that there is such a transformation. Then try to explicitly construct it by finding a matrix $A$ such that $T(x) = Ax$ for all $x\in \Bbb R^2$. If you're able to, then one clearly exists. If you run into some contradiction, then one clearly doesn't exist. – user137731 Oct 4 '15 at 4:25
• As always, reviewing the terms in the problem (what a linear transformation is) can help. – Nate 8 Oct 4 '15 at 4:30
In general if V and W are vector spaces and $\{e_i\}_{i\in I}$ is a basis for $V$. And $f(e_i)=w_i\in W$ for all $i \in I$.
Then f can be extended to a linear mapping $T: V \rightarrow W$ such that $T(e_i)=f(e_i)$. (You can write down the proof in the same way as the above answers)
In your case, $V= \mathbb R ^2$ and $W=\mathbb R^3$ $e_1=(1,1), e_2=(2,3)$ is a basis for $\mathbb R^2$ and $f(e_1)=(1,0,2)$ and $f(e_2)=(1,-1,4)$.
Therefore, you can extend it to a linear mapping by the above discussion.
Note: 1) $(1,0,2)$ and $(1,-1,4)$ doesn't have any significance role to play.If $(a,b,c), (d,e,f)$ still you could extend it to a linear map.
2) The above problem has a general philosophy - any map (or morphism) from a algebraic(mathematical) structure to another algebraic(mathematical) structure can be defined by only defining it on the generators, provided you can talk of generators. For example, to define a group homomorphism from a cyclic to group to another group, it is enough to define it for just for the generators. | {
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Note that $\{(1,1),(2,3)\}$ is a basis of $\mathbb R^2$ and if you know the image of basis elements then you know the whole transformation.
In this case observe $\forall x,y \in \mathbb R^2$ we have $(x,y)= (3x-2y)(1,1)+(y-x)(2,3)$
So define , $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$
To answer the comment by Babai:
The construction of $T$ shows that if $u\in \mathbb R^2,u=\alpha a+\beta b$ where $a=(1,1)$ and $b=(2,3)$ then $Tu=\alpha Ta+\beta Tb$
Thus $T(u_1+\lambda u_2)=T(\alpha_1 a+\beta_2 b+\lambda (\alpha_2 a+\beta_2 b))=T((\alpha_1+\lambda \alpha_2) a+(\beta_1+\lambda \beta_2) b)=(\alpha_1+\lambda \alpha_2) Ta+(\beta_1+\lambda \beta_2) Tb=\alpha_1 Ta+\beta_2 Tb+\lambda (\alpha_2 Ta+\beta_2 Tb)=Tu_1+\lambda Tu_2$
• The question is why is T linear? In your answer you assumed T is linear. – Babai Oct 4 '15 at 4:41
• @Babai I have added it in my answer. – usermath Oct 4 '15 at 5:22
• $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$. Why is this true? This is only true if $T$ is linear, and that is what you nee to prove. – Babai Oct 4 '15 at 5:38
• Instead of writing " So $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$, if you write "define $T$ as $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$, then your answer will make perfect sense. – Babai Oct 4 '15 at 5:45
• @Babai ok I got it and corrected. Thnx – usermath Oct 4 '15 at 6:27
Define $T: \mathbb{R}^2 \to \mathbb{R}^3$ by for $(x,y) \in \mathbb{R}^2$, write $(x,y) = a(1,1) + b(2,3)$, and let $T(x,y) = a(1,0,2) + b(1,-1,4).$
In the above, $T$ is obviously linear. The problem is why such a $T$ is well-defined, more specifically, why there exists such scalars $a,b$ and the scalars are unique. | {
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• Why is T obviously linear? – Babai Oct 4 '15 at 4:42
• @Babai Have you tried to show it before you ask this question? It is a one line proof. – Empiricist Oct 4 '15 at 5:17
• It is not the question of whether I have tired it or not. Your answer is not justifiable. You write "$T(x,y) = a(1,0,2) + b(1,-1,4).$, In the above, T is obviously linear." Now, $T(x,y) = a(1,0,2) + b(1,-1,4).$ , this is not true unless $T$ is linear. And you assumed it to say $T$ is linear, which is not correct. – Babai Oct 4 '15 at 5:35
• I am sorry. You are defining $T$. Your answer is perfect. And it is obviously linear. – Babai Oct 4 '15 at 5:44
If there exists a transform, there must exist a matrix representing it, let us call such a matrix $\bf T$. Now what we demand is that $${\bf T}\left[\begin{array}{cc}1&2\\1&3\end{array}\right] = \left[\begin{array}{rr}1&1\\0&-1\\2&4\end{array}\right]$$ Now we can solve this, if we divide by $\left[\begin{array}{cc}1&2\\1&3\end{array}\right]$ to the left of each side: $${\bf T} = \left[\begin{array}{rr}1&1\\0&-1\\2&4\end{array}\right] \left[\begin{array}{cc}1&2\\1&3\end{array}\right]^{-1} = ?$$
Now we can see on the matrices that this must have a solution since both matrices are of rank 2.
Then after we have solved this, what remains is to perform the matrix multiplication ${\bf T}\left[\begin{array}{cc}1&2\\1&3\end{array}\right]$ and confirm it really becomes as the first equation says it should. | {
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This list of mathematical symbols by subject shows a selection of the most common symbols that are used in modern mathematical notation within formulas, grouped by mathematical topic.
Regular expressions first evaluates a special character in the context of regular expressions: if it sees a dot, then it knows to match any one character. Advanced Algebra.
Die folgende Tabelle listet die wichtigsten Symbole und Abkürzungen auf, die in mathe online eine Rolle spielen. The backslash, \ , usually refers to the relative complement of two sets. Geometry. If we look at writing, people do not use backslash in writing. The set difference A\B is defined by A\B={x:x in A and x not in B}.
The most common, most frequently used math symbols: Forward Slash (Division Sign) shown and explained . Please be sure to answer the question. How does one insert a "\" (backslash) into the text of a LaTeX document? Thanks for contributing an answer to TeX - LaTeX Stack Exchange!
The difference between a backslash and a forward slash is defined below: Backslash: \ Forward Slash: / A good way to remember the difference between a backslash and a forward slash is that a backslash leans backwards ( \ ), while a forward slash leans forward ( / ).
P. PvtBillPilgrim. Infix operator. If you like this Site about Solving Math Problems, please let Google know by clicking the +1 button. ... [Backslash] Unicode: 2216.
If A is an n by m matrix and b is an p by q matrix then A\b is defined (and is calculated by Matlab) if m=p. It only takes a minute to sign up. There is always some type of calculations to be made. But avoid … Asking for help, clarification, or responding to other answers. Meaning of the backslash operator on sets. A regular expression to match the IP address 0.0.0.0 would be: 0\.0\.0\.0. In mathematics, the decimal point (.) | {
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In a probability \$P(A \setminus B)\$, it means "the probability that event \$A\$ occurs but event \$B\$ does not occur". EDIT: This an issue for keyboard layouts where backslash is activated by a combination of AltGr + backslash. Falls das Symbol in den Mathematischen Hintergründen definiert oder beschrieben wird, ist diese Eintragung ein Link, der Sie an die entsprechende Stelle führt.
The backslash symbol \ is used to denote a set difference, quotient group, or integer division. Also used to separate arguments of elliptic functions. Ask Question Asked 5 years, 5 months ago. Formally: ∖ = {∈ ∣ ∉}.
Here, the backslash symbol is defined as Unicode U+2216. At any rate, math mode provides \sim, \backslash, and \setminus (the latter two appear to look the same and differ only by spacing in math mode).
You just have to type the name of the letter after a backslash: if the first letter is lowercase, you will get a lowercase Greek letter, if the first letter is uppercase (and only the first letter), then you will get an uppercase letter. What does the backslash denote in probability theory? Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Applied Mathematics. Ask Question Asked 7 years, 6 months ago. \$\begingroup\$ The backslash is an escape character in \$\LaTeX\$, cf.
... Backslash notation: … is used to separate the whole part of a number from the fractional part. What is Backslash? In this particular example, the norm of the residual is zero, and an exact solution is obtained, although rcond is small. Making statements based on opinion; back them up with references or personal experience. Revolutionary knowledge-based programming language. x y is by default interpreted as Backslash [x, y]. | {
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In an application of a probability problem you could be given an arbitrary probability like: P ( a \ b) = .50 or 50%. For non-square and singular systems, the operation A\b gives the solution in the least squares sense. The relative complement of A in B is denoted B ∖ A according to the ISO 31-11 standard.It is sometimes written B − A, but this notation is ambiguous, as in some contexts it can be interpreted as the set of all elements b − a, where b is taken from B and a from A.. Foundations of Mathematics. | {
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# Length of period of decimal expansion of a fraction
Each rational number (fraction) can be written as a decimal periodic number. Is there a method or hint to derive the length of the period of an arbitrary fraction? For example $1/3=0.3333...=0.(3)$ has a period of length 1.
For example: how to determine the length of a period of $119/13$?
• I can't find an algorithm that finds the value for a number that isn't prime, i.e. 10 – Code Whisperer Jan 13 '18 at 9:11
• – Watson Dec 2 '18 at 16:31
Assuming there are no factors of $2,5$ in the denominator, one way is just to raise $10$ to powers modulo the denominator. When you find $-1$ you are halfway done. Taking your example: $10^2\equiv 9, 10^3\equiv -1, 10^6 \equiv 1 \pmod {13}$ so the repeat of $\frac 1{13}$ is $6$ long. It will always be a factor of Euler's totient function of the denominator. For prime $p$, that is $p-1$.
• You have to look for $1$, because sometimes you won't find $-1$. For example: $10^2\equiv 18, 10^3\equiv 16, 10^4 \equiv 37, 10^5 \equiv 1 \pmod {41}$, so the repeat of $\frac 1{41}$ is $5$ long. – jcsahnwaldt says GoFundMonica Mar 26 '15 at 5:37
• You assume that 'there are no factors of 2,5 in the denominator' is it actually required? – Shamdor May 5 '15 at 7:28
• I mean, 1/(2^i * 5^j * d') has the same period length as 1/d', right? – Shamdor May 5 '15 at 7:46
• That is correct. You will have $\max(i,j)$ non-repeating digits, then the repeat will start – Ross Millikan May 5 '15 at 13:48
This answer seeks to explain why Ross Millikan's answer works, and provides further information on techniques to speed up the process of seeking the period: | {
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Consider the fraction $\frac17$. The decimal expansion of this is $$\frac17 = 0.\overline{142857}$$ for a period of 6. Now consider what happens when we multiply it by $10^6$: $$10^6\times\frac17 = 142857.\overline{142857}$$ Subtracting the original fraction from this gives $$(10^6-1)\times\frac17 = 142857$$ And so, we have $$\frac17 = \frac{142857}{10^6-1}$$ As you can see, the denominator is one less than a power of 10, and the power is the period of the decimal expansion. This is no accident, and works for any fraction - if you can rewrite it in this form, the denominator reveals the period.
Now, rearrange the equation: $$10^6-1 = 142857\times 7$$ So $10^6$ must be one more than a multiple of 7 (or, more generally, $10^n$ must be one more than a multiple of $d$, where $d$ is the denominator of the fraction and $n$ is the period of the decimal expansion) - indeed, it must be the smallest power of 10 (larger than 1) that has this property.
As such, we can use modular arithmetic to look for the period. Since $a\times d\equiv 0 \pmod d$, we have that $10^n-1\equiv0 \pmod d$, or $$10^n \equiv 1\pmod d$$ And therefore you can just look for the smallest $n>0$ satisfying this.
Of course, there are other approaches to gain the same result, but they're all fundamentally variants of the same idea. That said, if you can factor $\phi(d)$ - the euler totient function of the denominator - then you can accelerate the process of looking for the smallest $n$. For example, when checking 13, you have $\phi(13)=12$, so $n\in\{1,2,3,4,6,12\}$ (as these are the factors of 12) - this can save you a lot of computation (especially where $\phi(d)$ has only a few large factors and 2). | {
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For example, $\phi(167)=166 = 2\times83$, so $n\in\{1,2,83,166\}$. Therefore, we need to check only these four, and we can do it quite efficiently. Obviously, neither $10$ nor $100=10^2$ are equivalent to 1 mod 167, so we only need to actually check 83. For this we can use binary exponentiation. Note that $83 = 2^6 + 2^4 + 2^1 + 2^0$. So we can write \begin{align} 10^{83} &= 10^{2\times(2^5 + 2^3 + 1)}\times 10\\ &= (10^{2^3\times(2^2+1)}\times 10)^2 \times 10\\ &= ((10^{2^2}\times10)^{2^3}\times 10)^2 \times 10 \end{align} So, working in modular arithmetic, we can go \begin{align} 10^{83} &\equiv ((10^{2^2}\times10)^{2^3}\times 10)^2 \times 10 \mod 167\\ &\equiv ((100^2\times 10)^{2^3}\times 10)^2\times10\mod167\\ &\equiv ((147\times 10)^{2^3}\times 10)^2\times10\mod167\\ &\equiv (134^{2^3}\times 10)^2\times10\mod167\\ &\equiv (87^{2^2}\times 10)^2\times10\mod167\\ &\equiv (54^2\times 10)^2\times10\mod167\\ &\equiv (77\times 10)^2\times10\mod167\\ &\equiv 102^2\times10\mod167\\ &\equiv 50\times10\mod167\\ &\equiv 166\mod167 \end{align} This is the same as $-1\pmod{167}$, so $n=166$ is the only possible period, and $\frac1{167}$ has a period of 166.
Also note that you don't actually have to expand out the product like that. You can just write the number in binary ($83_{10} = 1010011_2$), then work through the binary digits left-to-right - start with 1, and for each digit, $b$, multiply by $10^b$. Square it after each digit except the last one.
The length of the period is given by the multiplicative order of $10 \pmod q$, where $q$ is your quotient. It is closely related to the discrete logarithm. Wikipedia lists several algorithms that are faster than going through all the powers of ten, which is relevant if you're dealing with very large numbers (hundreds of digits).
First you have to simplify to the lowest integer denominator. | {
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First you have to simplify to the lowest integer denominator.
Then, the proof for finiteness of repeating part of the decimal representation involves the pigeonhole principle. When you keep on dividing, at some point you will run into a repeat value for the remainder. Given that the remainders that keep the operation going range from 1 to (denominator-1), the possible number of pigeonholes is (denominator-1) as the number of pigeonholes; so that's your worst possible case scenario - in any base.
The particular cases, as described by the other answers, depend on the base, because the denominator may have a special relationship with the base.
• e.g., in base 10 fractions over 3 will have one repeating decimal, (3) or (6), as 3 divides 9(=10-1), and the remainder will keep repeating. | {
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# Proof Check - Linear Algebra
I'm new to more formal mathematics and I was wondering if I could write my attempt at a small proof here and see what others thought and if you could point out any fatal flaws or just offer some tips? I would appreciate it.
## Theorem
Let u,v, and w be distinct vectors in a vector space V. Show that if {u,v,w} is a basis for V, then {u+v+w, v+w, w} is also a basis for V.
## Proof
: Let B = {u,v,w}. As B is a basis for V, the vectors in B generate V and are linearly independant. Therefore, the equation
a1 * u + a2 * v + a3 * w = 0
Is only true for a1, a2, a3 all equal to 0.
Now consider the set B' = {u+v+w, v + w, w}
If we examine the equation
c1 ( u+v+w) + c2 (v+w) + c3 (w) = 0
c1(u) + (c1 + c2)v + (c1 + c2 + c3)w = 0
Then we say that c1 = -c2, and c3 = -c1-c2, but from the first term we see that c1 must equal zero, therefore all the other coefficients are 0 as well, and so the set of vectors is linearly independant.
## My questions
1.) I'm not sure if the part of the proof where I show that the solution to the equation for the vectors of B' being equal to 0 is connected well enough to the bit about B being a basis for V. Am I allowed to say that C1 must be equal to 0 only because I've already assumed that u,v, and w are all linearly independent already?
2.) I know that if B generates V that B' which is just set formed from L.I. vectors into other L.I. vectors must also generate V, but I'm not sure exactly why I can say this even though I feel it strongly.
• Your argument is perfect. – Ted Shifrin Sep 15 '14 at 1:20 | {
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• Your argument is perfect. – Ted Shifrin Sep 15 '14 at 1:20
With regards to point 1), what you say is correct. As you say, you use the linear independence of $B$ to conclude that when $c_1(u) + (c_1 + c_2)v + (c_1 + c_2 + c_3)w = 0$, it must be the case that $c_{1} = 0$, and so on.
With regards to 2), one way to show that $B'$ spans the space is to show that it generates $B$: any linear combination of vectors from $B$ can then be converted into a linear combination of vectors from $B'$ by replacing $u,v,w$ by their expressions in terms of the list $B'$. So in this case $(v+w) - w = v$ and $(u+v+w) - (v+w) = u$.
By the way, you only need to show either that $B'$ form a linear independent set, or that they span the vector space, since the list has length $3$. (This is the result that a linear independent list of vectors, whose length is equal to the dimension of the space forms a basis, and similarly for spanning sets). | {
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# When is the closure of an open ball equal to the closed ball?
It is not necessarily true that the closure of an open ball $B_{r}(x)$ is equal to the closed ball of the same radius $r$ centered at the same point $x$. For a quick example, take $X$ to be any set and define a metric $$d(x,y)= \begin{cases} 0\qquad&\text{if and only if x=y}\\ 1&\text{otherwise} \end{cases}$$ The open unit ball of radius $1$ around any point $x$ is the singleton set $\{x\}$. Its closure is also the singleton set. However, the closed unit ball of radius $1$ is everything.
I like this example (even though it is quite artificial) because it can show that this often-assumed falsehood can fail in catastrophic ways. My question is: are there necessary and sufficient conditions that can be placed on the metric space $(X,d)$ which would force the balls to be equal?
• In the Euclidean metric space $R^n$ it is necessarily true. Feb 11, 2012 at 2:26
• Right, but Euclidean space is known for, among other things, being perfect in almost every way. What about spaces like $L^{p}$ or $H^{p}$? I'm looking to see how far our intuition of Euclidean spaces and the standard metric extends. Feb 11, 2012 at 2:28
• Regarding your question about $L^p$ or $H^p$, it is true in every normed space. If $\|x-y\|=r$, then for $0<t<1$, $\|x-(tx+(1-t)y)\|=(1-t)\|x-y\|<r$, and $y=\lim\limits_{t\searrow 0}tx+(1-t)y$. Feb 11, 2012 at 3:37
• The property may fail for subspaces of Euclidean space. See here: link link Feb 11, 2012 at 11:53
Here is a characterization that is straight from the definitions, but which it seems may be useful when verifying that a particular space has the property.
For any metric space $$(X,d)$$, the following are equivalent: | {
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For any metric space $$(X,d)$$, the following are equivalent:
• For any $$x\in X$$ and radius $$r$$, the closure of the open ball of radius $$r$$ around $$x$$ is the closed ball of radius $$r$$.
• For any two distinct points $$x,y$$ in the space and any positive $$\epsilon$$, there is a point $$z$$ within $$\epsilon$$ of $$y$$, and closer to $$x$$ than $$y$$ is. That is, for every $$x\neq y$$ and $$\epsilon\gt 0$$, there is $$z$$ with $$d(z,y)<\epsilon$$ and $$d(x,z).
Proof. If the closed ball property holds, then fix any $$x,y$$ with $$r=d(x,y)$$. Since the closure of $$B_r(x)$$ includes $$y$$, the second property follows. Conversely, if the second property holds, then if $$r=d(x,y)$$, then the property ensures that $$y$$ is in the closure of $$B_r(x)$$, and so the closure of the open ball includes the closed ball (and it is easy to see it does not include anything more than this, since if $$g$$ belongs to the closure of $$B_r(x)$$ then $$d(x,g) \le r$$ and so $$g$$ must also belong to the closed ball of radius $$r$$ centered at $$x$$). QED | {
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• Thanks for the response! Your proof makes sense. Feb 11, 2012 at 3:10
• How to understand what is not working initially in order to build your characterization? In other words, what is the intuition behind this? Thanks.
– user169373
Sep 23, 2014 at 20:41
• @MarcGato Think about the case where you have an isolated point $x$, so that $B_r(x)$ contains only $x$, for some $r>0$, but there is a point $y$ at distance $r$ to $x$. The closure of $B_r(x)$ is just $\{x\}$, since there are no other limit points to add and so this set is already closed. My condition ensures that every point at distance $r$ from $x$ is a limit of points of distance less than $r$ from $x$. That is why all such points get added to the closure of the open ball.
– JDH
Sep 23, 2014 at 21:10
• The way that $y$ gets into the closure of $B_r(x)$ is that there must be points in $B_r(x)$ that are as close as you like to $y$. Those points are the $z$'s in the property.
– JDH
Apr 26, 2017 at 11:03
• @James Well since y is in closure of B(x,r), for any $\epsilon>0$, $B(y,\epsilon) \cap B(x,r) \neq \phi$. Choose z from this intersection. Then, $d(x,z)<r=d(x,y)$ Jan 14, 2021 at 11:40
Let $(X,\|\cdot\|)$ be a normed linear space. Then $\overline{B_1(0)}=\bar{B}_1(0)$.
Proof. Observe that $\overline{B_1(0)}$ is the smallest closed set containing $B_1(0)$ and $B_1(0)\subset \bar{B}_1(0)$, so trivially $\overline{B_1(0)}\subset\bar{B}_1(0)$. Now to show $\bar{B}_1(0)\subset \overline{B_1(0)}$. Observe that, $\bar{B}_1(0)=B_1(0)\cup \partial B_1(0)$, i.e., for all $x\in \partial B_1(0), \, \exists x_n\in B_1(0)$ such that $\|x_n-x\|\to 0$: for any given $x\in \partial B_1(0),$ let $x_n=(1-\frac{1}{n})x, \, n\in \mathbb{N}.$ Then show $x_n\in B_1(0)$ and $\|x_n-x\|\to 0$. | {
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• How would you suggest approaching this, showing that the interior of the closed ball is equal to the open ball? Jan 31, 2018 at 0:15
• ... where a "normed linear space" means over some field with an archimedean valuation. The result is not true e.g. over $p$-adics where that $1-1/n$-construction fails. -- Also, note that $\delta B_1(0)$ here should really be defined as $\{x \in X: \lvert \lvert x \rvert \rvert =1 \}$, as only a posteriori we show this is the boundary also in the topological sense. Apr 11, 2020 at 7:14
More general, we can prove that for any normed space $$(V,\|\cdot\|)$$, if $$x_0\in V$$ and $$R>0$$, then $$\overline{B(x_0;R)}^{\|\cdot\|}=B[x_0;R],$$ where $$\overline{B(x_0;R)}^{\|\cdot\|}$$ is the closure of the open ball $$B(x_0;R)$$ in topology induced by $$\|\cdot\|$$.
The inclusion $$\overline{B(x_0;R)}^{\|\cdot\|}\subseteq B[x_0;R]$$ is trivial, because $$B[x_0;R]$$ is a closed set that contains $$B(x_0;R)$$ and $$\overline{B(x_0;R)}^{\|\cdot\|}$$ is the smallest closed set that contains $$B(x_0;R)$$.
To show that $$B[x_0;R]\subseteq\overline{B(x_0;R)}^{\|\cdot\|}$$, we can prove that any element of $$B[x_0;R]$$ is the limit of some sequence in $$B(x_0;R)$$. Let $$x\in B[x_0;R]$$, then of course $$\|x-x_0\|\leqslant R.$$ Now defines, for each $$n\in\Bbb{N}$$, $$x_n:=\left(1-\frac{1}{n}\right)x+\frac{1}{n}x_0.$$
Claim 1: The sequence $$(x_n)$$ converges to $$x$$ in the norm $$\|\cdot\|$$. In fact, $$\begin{eqnarray} \|x_n-x\| & = & \left\|\left(1-\frac{1}{n}\right)x+\frac{1}{n}x_0-x\right\| \\ & = & \frac{1}{n}\underbrace{\|x-x_0\|}_{\leqslant R} \\ & \leqslant & \frac{R}{n}\longrightarrow0,\quad\text{when}\ n\to\infty, \end{eqnarray}$$ which proves that $$x_n\to x$$ in $$\|\cdot\|$$. | {
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Claim 2: For all $$n\in\Bbb{N}$$, $$x_n\in B(x_0;R)$$, that is, $$(x_n)$$ is a sequence in the set $$B(x_0;R)$$. This is obviously by the construction of $$x_n$$, since $$\begin{eqnarray} \|x_n-x_0\| & = & \left\|\left(1-\frac{1}{n}\right)x+\frac{1}{n}x_0-x_0\right\| \\ & = & \left\|\left(1-\frac{1}{n}\right)x-x_0\left(1-\frac{1}{n}\right)\right\| \\ & = & \left\|\left(1-\frac{1}{n}\right)(x-x_0)\right\| \\ & = & \left(1-\frac{1}{n}\right)\underbrace{\|x-x_0\|}_{\leqslant R} \\ & \leqslant & \underbrace{\left(1-\frac{1}{n}\right)}_{<1}R which gives us $$x_n\in B(x_0;R)$$.
So we conclude that any element of $$B[x_0;R]$$ is the limit of a sequence in $$B(x_0;R)$$, hence $$B[x_0;R]\subseteq\overline{B(x_0;R)}^{\|\cdot\|}$$.
• Cf. my comment to Lyapunav's answer. Apr 11, 2020 at 7:15
• @TorstenSchoeneberg if you need this, I have no problem to give you credits... Apr 11, 2020 at 12:34
• I think what Torsten meant was that this isn't a proof for a general normed space. 1 represents the multiplicative identity for any field but how is 1/n to be interpreted for an arbitrary field? Apr 26 at 3:11 | {
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Need help proving that $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$
Hello.
I have been trying very hard to show that $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$ and could not quite get anywhere. This inequality has been verified by computer for $k\le40$.
Some clues that might work (kindly provided by Doron Zeilberger) are as follows:
1. Let $Ef(x):=f(x-1)$, let $P_k(E):=\sum_{j=0}^{k-1}(-1)^{(j+1)}*\binom{2*k+1}{j}*E^j$;
2. These satisfy the inhomogeneous recurrence $P_k(E)-(1-E)^2*P_{k-1}(E)=Some Binomial In E$;
3. The original sum can be expressed as $P_k(E)x^{(2*k-2)} | x=k$;
4. Try to derive a recurrence for $P_k(E)x^{(2*k-2)}$ before plugging-in $x=k$ and somehow use induction, possibly having to prove a more general statement to facilitate the induction.
Unfortunately I do not know how to find a recurrence such as suggested in 4.
I would appreciate any help that members of the MathOverflow community can provide.
-
Nitpicking: you need $k\geq 2$. – darij grinberg Jun 22 '10 at 17:54
Your sequence -1,1,10,245,11326,855162,.. is not contained in the Online Encyclopedia of Integer Sequences. If you find it interesting, as David Carchedi also suggests, you may like to include it ;-) – Pietro Majer Jun 23 '10 at 0:53
Thanks Pietro and Victor for the great answers. Hopefully this also will help ME. – David Carchedi Jun 23 '10 at 10:26
Thank you, everyone, for the beautiful answers. I am most grateful. – Alexandra Seceleanu Jun 25 '10 at 3:26
Your expression is the difference of two central Eulerian numbers ,
$$A(k):=\sum_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2}{2k+1 \choose j}=\left \langle {2k-2\atop k-2} \right \rangle-\left \langle {2k-2\atop k-3} \right \rangle$$ | {
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as you can easily deduce from their closed formula. The positivity of $A(k)$ is just due to the fact that the Eulerian numbers $\left \langle {n\atop j}\right \rangle$ are increasing for $1\leq j\leq n/2$ (like the binomial coefficients); this fact has a clear combinatorial explanation also.
See e.g.
http://en.wikipedia.org/wiki/Eulerian_number
http://www.oeis.org/A008292
: although by now all details have been very clearly explained by Victor Protsak, I wish to add a general remark, should you find yourself in an analogous situation again. A healthy approach in such cases is adding variables, following the motto "more variables = simpler dependence" (like when one passes from quadratic to bilinear). In the present case, you may consider
$$A(k):=a(k,\, , 2k-2,\, ,2k+1)$$
where you define
$$a(k,n,m):=\sum_{j=0}^{k-1}(-1)^{j+1}(k-j)^{n}{m \choose j}$$
in which it is more apparent the action of the iterated difference operator, or, in the formalism of generating series, the Cauchy product structure:
$$\sum_{k=0}^\infty a(k,n,m)x^k=-\sum_{j=0}^\infty j^nx^j\, \sum_{j=0}^\infty(-1)^j{m \choose j} x^j =-(1-x)^m\sum_{j=0}^\infty j^nx^j.$$
The series $$\sum_{j=0}^\infty j^nx^j$$ is now quite a simpler object to investigate, and in fact it is well-known to whoever played with power series in childhood. It sums to a rational function
$$(1-x)^{-n-1}x\sum_{k=0}^{n}\left \langle {n\atop k}\right \rangle x^k$$ that defines the Eulerian polynomial of order $n$ as numerator, and the Eulerian numbers as coefficients. In your case, $m=n+3$, meaning that you are still applying a discrete difference twice (in fact just once, due to the symmetric relations; check Victor's answer). | {
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-
I didn't expect that the original sequence is so easily expressed thru the Eulerian numbers. A very nice and elementary solution! +1 – Wadim Zudilin Jun 23 '10 at 6:52
That last series / rational function can also be expressed as $Li_{-n}(x)$, a polylogarithm of negative order. The equations for the generating series $-(1-x)^m Li_{-n}(x)$ are probably easy to get from MGfun, and are likely to be particularly simple. – Jacques Carette Jun 23 '10 at 23:23
This is a clarification of Pietro Majer's beautiful and insightful, yet a bit cryptic answer.
The Eulerian numbers are expressible as
$$\left\langle {n\atop m}\right\rangle=\sum_{i=0}^m(-1)^i{n+1\choose i}(m+1-i)^n.$$
View them as functions of $m$ and let $\Delta$ be the backward difference operator,
$$\Delta f(m)=f(m)-f(m-1).$$
Claim The $r$th iterated backward difference of the Eulerian number is given by the formula
$$\Delta^r\left\langle {n\atop m}\right\rangle=\sum_{i=0}^m(-1)^i{n+r+1\choose i}(m+1-i)^n.$$
Proof This is proved by induction in $r$ using the binomial identity $${n+r\choose i}+{n+r\choose i-1}={n+r+1\choose i}. \quad\square$$
Setting $m=k-1$ and comparing with the definition of the sequence, we see that
$$A(k)=\sum_{j=0}^{k-1}(-1)^{j+1}{2k+1 \choose j}(k-j)^{2k-2}=-\Delta^2\left\langle {n\atop k-1}\right\rangle\ \text{evaluated at }\ n=2k-2.$$
Thus
$$A(k) = -\Delta\left\langle {n\atop k-1}\right\rangle + \Delta\left\langle {n\atop k-2}\right\rangle = \Delta\left\langle {n\atop k-2}\right\rangle\ \text{evaluated at }\ n=2k-2$$
and the first summand vanishes due to the symmetry of the Eulerian numbers, $\left\langle {n\atop m}\right\rangle=\left\langle {n\atop n-1-m}\right\rangle$, which implies that $\left\langle {2k-2\atop k-1}\right\rangle=\left\langle {2k-2\atop k-2}\right\rangle.$ | {
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Now the positivity of $A(k)$ becomes a consequence of the unimodality of the Eulerian numbers, $\Delta\left\langle {n\atop m}\right\rangle\geq 0$ for $m\leq n/2.$ Explicitly, $$A(k)=\left\langle {2k-2\atop k-2}\right\rangle-\left\langle {2k-2\atop k-3}\right\rangle > 0\ \text{for}\ k\geq 2.$$
-
I believe it has to do with the following:
Let $P(x)$ be an arbitary polynomial of of degree less than or equal to $n$ such that $P(X) \in \mathbb{Z}$ for all $x \in \mathbb{Z}$. Then $P(x)$ can be expressed uniquely as an integer combination of binomial coefficients of the form $\left({\begin{array}{*{20}c} {x + j} \\ n \\\end{array}}\right)$, that is:
$$P(x) = \sum\limits_{j = 0}^{n - 1} {a_{n - j} \left( {\begin{array}{*{20}c} {x + j} \\ n \\ \end{array}} \right)}$$
(assuming $x \in \mathbb{Z}$). Specifically, we have:
$$a_j = \sum\limits_{l = 0}^j {\left( { - 1} \right)^l P(j - l)\left( {\begin{array}{*{20}c} {n + 1} \\ l \\ \end{array}} \right).}$$
Now let $n=2k$ and expand out $\left(x+1\right)^{2k-2}$ in terms of the binomial coefficients $\left({\begin{array}{*{20}c} {x + j} \\ 2k \\\end{array}}\right)$:
$$(x + 1)^{2k - 2} = \sum\limits_{j = 0}^{2k - 1} {a_{n - j}^i \left( {\begin{array}{*{20}c} {x + j} \\ {2k} \\ \end{array}} \right)}.$$
Then we have that
$$a_{k-1} = \sum\limits_{j = 0}^{k-1} {\left( { - 1} \right)^j \left(k-j\right)^{2k-2}\left( {\begin{array}{*{20}c} {2k + 1} \\ j \\ \end{array}} \right).}$$
This is exactly the negative of your coefficient. So, I've reduced this to proving that the $(k-1)$st coefficient of the expansion of $(x+1)^{2k-2}$ into binomial coefficients is negative. I hope this helps. If you figure this out, please let me know. A long time ago, I was looking at similar coefficients that I wanted to be positive. (Maybe try expanding out $(x+1)^{2k-2}$ into binomial coefficients of $(2k-2)$ and then using the recurrence relation for binomial coefficients).
P.S. Where is this expression coming from, in your case? | {
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P.S. Where is this expression coming from, in your case?
-
P.S. in this case, since $2k-2 < 2k$, the sum of all the $a_j$s will be zero, in case this ends up helping. – David Carchedi Jun 22 '10 at 21:32
P.P.S, it would appear that this in fact has to do with the dimension of the space of covariants of the regular representation of the symmetric group $S_{2k}$. If this rings a bell, we should talk (if not, maybe we should talk anyway). – David Carchedi Jun 22 '10 at 21:39
In my case, these numbers appeared as follows: Consider the algebra $A_{r,t}=k[x_1,\ldots x_n]/(l_1^t,\ldots l_{r+1}^t)$ whose Hilbert function is given by $HF(A_{r,t},i)=\sum\limits_{j=0}^m(-1)^j\binom{r-1+i-tj}{r-1} \cdot \binom{r+1}{j}$ where $m=\mbox{min}\{\lfloor \frac{i}{t} \rfloor,r \}$. Now set $r=2k$. I am interested in the asymptotics of $HF(A_{r,t},k(t-1)-1)$. Turns out this is a polynomial in $t$ whose leading coefficient can be expressed (as I have just learned from the answers above) using Eulerian numbers as $\frac{1}{(2k-2)!}A(2k-2,k-2)$. – Alexandra Seceleanu Jun 25 '10 at 3:51 | {
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# proof of $\arctan(x)=\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)$ without solving $\int\frac{1}{x^2+1}dx$ [duplicate]
So I know $\int\frac{1}{x^2+1}dx=\arctan(x)+c$
Now when i tried to integral $\frac{1}{x^2+1}$ in a different way i got this:
$$\frac{1}{x^2+1}=\frac{1}{(x-i)(x+i)}=\frac{A}{x-i}+\frac{B}{x+i}\\\implies1=Ax+Ai+Bx-Bi=x(A+B)+i(A-B)\\\implies \begin{cases} A+B=0 \\[2ex] A-B=\frac{1}{i}=-i \end{cases}\\\implies2A=-i\implies A=\frac{-i}{2}\implies B=\frac{i}{2}$$
So in the end i get:$$\int\frac{1}{x^2+1}dx=\int\frac{i}{2(x+i)}-\frac{i}{2(x-i)}dx=\frac{i}{2}\int\frac{1}{(x+i)}-\frac{1}{(x-i)}dx$$
After integrating this i get $\frac{i}{2}(\ln(x+i)-\ln(x-i)+C)$ after comparing this to $\arctan(x)$ at $x=0$ i find $C=-i\pi$
$\therefore~\arctan(x)=\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)$
now my question is how can i prove that without using the integral
• – Nosrati Sep 2 '17 at 17:31
• @MyGlasses thx for pointing who 2 out, i get how they done it in those 2, the only problem is the $C=-i\pi$ i got, why it doesn't appear at the first one you sent? – ℋolo Sep 2 '17 at 17:52
• I'm sorry. I was confused those and worked on second. I think the title of that post is wrong and your proof shows it! – Nosrati Sep 2 '17 at 18:29
By definition of $\ln$ for complex numbers: $$\ln(x+i)=\ln|x+i|+i\operatorname{Arg}(x+i)$$ where $\operatorname{Arg}$ is the principal value of the $\arg$
It is clear that: $$\operatorname{Arg}(x+i)=\begin{cases}\arctan\frac{1}{x}&\text{ if }x\ge0\\\pi+\arctan\frac{1}{x}&\text{ if }x<0\end{cases}$$
Analogously:
$$\ln(x-i)=\ln|x-i|+i\operatorname{Arg}(x-i)$$ In this case it is:
$$\operatorname{Arg}(x-i)=\begin{cases}-\arctan\frac{1}{x}&\text{ if }x\ge0\\-\pi-\arctan\frac{1}{x}&\text{ if }x<0\end{cases}$$
Moreover observe that: $|x+i|=|x-i|$
Putting everything together you get: | {
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Moreover observe that: $|x+i|=|x-i|$
Putting everything together you get:
$\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)=\begin{cases}\stackrel{x\ge0}=\frac{i}{2}(2i\arctan\frac{1}{x}-i\pi)&=-\arctan\frac{1}{x}+\frac{\pi}{2}=\\\stackrel{x<0}=\frac{i}{2}(2i\pi+2i\arctan\frac{1}{x}-i\pi)&=-\arctan\frac{1}{x}-\frac{\pi}{2}=\end{cases}=\arctan x$
• You made a mistake at the last part, you wrote $\ln(x+i)-\ln(x+i)$ and not $\ln(x+i)-\ln(x-i)$ – ℋolo Sep 2 '17 at 19:24
• @Holo Thanks. I'm going to edit the answer. – trying Sep 2 '17 at 19:32
Perhaps using Euler? $$z= \tan w = \frac{\sin w}{\cos w} = \frac{1}{i}\frac{e^{iw}-e^{-iw}}{e^{iw}+e^{iw}} = \frac{1}{i} \frac{e^{2iw}-1}{e^{2iw}+1}$$ so for any $k\in {\Bbb Z}$ $$e^{2iw} = e^{2i(w -k)} = \frac{1+iz}{1-iz}$$ whence $$w = \frac{1}{2i} \log \left( \frac{1+iz}{1-iz} \right) + k \pi$$ where the choice of $k\in {\Bbb Z}$ depends upon the cut you want. $k=0$ corresponds to standard choices of $\log(1)=0$ and $\arctan 0=0$.
You found $$\arctan x=\dfrac{i}{2}\Big(\ln(x+i)-\ln(x-i)-i\pi\Big)=\dfrac{i}{2}\ln\dfrac{x+i}{x-i}+\dfrac{\pi}{2}=-\dfrac{i}{2}\ln\dfrac{x-i}{x+i}+\dfrac{\pi}{2}$$ let $x=\dfrac1z$ so $$\arctan\dfrac1z=-\dfrac{i}{2}\ln\dfrac{1-iz}{1+iz}+\dfrac{\pi}{2}$$ and then $$\dfrac{i}{2}\ln\dfrac{1-iz}{1+iz}=\dfrac{\pi}{2}-\arctan\dfrac1z=\arctan z$$ as you want!
Here is another way to look at this: consider that $z=|z|e^{i\theta}$ and that $z^*=|z^*|e^{-i\theta}$. Then
$$\frac{|z|}{|z^*|}=e^{2i\theta}\\ \theta=\frac{1}{2i}(\ln |z|-\ln |z^*|)$$
Now, let $z=x+i$, then
$$\theta=\cot^{-1}x=\frac{1}{2i}(\ln |x+i|-\ln |x-i|)$$
Now,
$$\cot^{-1}x+\tan^{-1}x=\frac{\pi}{2}$$
so that
\begin{align} \tan^{-1}x &=-\frac{1}{2i}(\ln |x+i|-\ln |x-i|)+\frac{\pi}{2}\\ &=\frac{i}{2}(\ln |x+i|-\ln |x-i|-i\pi) \end{align} | {
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