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https://www.physicsforums.com/threads/inequality-for-solution.400911/	# Inequality for solution 1. May 4, 2010 ### Malmstrom Consider the problem $$y'=\sqrt{y^2+x^2+1}$$ $$y(0)=0$$ Prove that the solution is defined for all $$x \in \mathbb{R}$$ and that $$y(x) \geq \sinh (x)$$ $$\forall x \geq 0$$ 2. May 4, 2010 ### Redbelly98 Staff Emeritus Is this homework? 3. May 5, 2010 ### Malmstrom Not really, I just don't know where to start from.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.893056333065033, "perplexity": 3398.833788326304}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948623785.97/warc/CC-MAIN-20171218200208-20171218222208-00567.warc.gz"}
https://www.physicsforums.com/threads/are-there-closed-form-solutions-to-the-harmonic-series.68250/	# Homework Help: Are there closed form solutions to the harmonic series? 1. Mar 22, 2005 Are there closed form solutions to the harmonic series? 2. Mar 22, 2005 ### HallsofIvy Sure. The sum $$\Sigma_{k=1}^{n}\frac{1}{k}= \gamma+ \Psi_0(n+1)$$ where $$\gamma$$ is the "Euler-Mascheroni" constant and $$\Psi_0$$ is the digamma function. Or did you mean something else by "closed form"? See http://mathworld.wolfram.com/HarmonicSeries.html 3. Mar 22, 2005	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.963922917842865, "perplexity": 1145.1368438499971}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039749054.66/warc/CC-MAIN-20181121153320-20181121175320-00284.warc.gz"}
https://www.varsitytutors.com/hotmath/hotmath_help/topics/rates-ratios.html	# Rates & Ratios A ratio is a comparison of two numbers. A ratio can be written using a colon, $3:5$ , or as a fraction $\frac{3}{5}$ . A rate , by contrast, is a comparison of two quantities which can have different units. For example $5$ miles per $3$ hours is a rate, as is $34$ dollars per square foot. Example 1: A punch recipe calls for $6$ ounces of lime juice, $21$ ounces of apricot juice, and $21$ ounces of pineapple juice. What is the ratio of lime juice to apricot juice? Writing the ratio using a colon, we get $6:21$ . Note that this can be reduced, like a fraction, by dividing both numbers by a common factor -- in this case, $3$ . In simplest form, the ratio is $2:7$ . Example 2: In the recipe above, what is the ratio of apricot juice to the total amount of punch? To find the total amount of punch, add $6+21+21=48$ . The ratio of apricot juice to the total amount of punch is $21:48$ . But this ratio is probably more clearly written as a fraction, since the apricot juice makes up a fraction of the whole. $\frac{21}{48}$ To reduce the fraction, divide both the numerator and the denominator by $3$ . $\frac{7}{16}$ Note that this can be reduced, like a fraction, by dividing both numbers by a common factor -- in this case, $3$ . In simplest form, the ratio is $2:7$ . Example 3: An adult scolopendromorph centipede has $46$ legs and $8$ eyes. In a group of $100$ centipedes of the same species, what is the ratio of legs to eyes? Note that it doesn't matter if there are $100$ or $10,000$ centipedes; the ratio of legs to eyes will remain the same. Writing the ratio using a colon, we get $46:8$ . Divide both numbers by $2$ . In simplest form, the ratio of legs to eyes is $23:4$ . Example 4: A bat beats its wings $170$ times in $10$ seconds. Write the rate as a fraction in lowest terms. Write the rate as a fraction. $\frac{170\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{wingbeats}}{10\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{seconds}}$ Divide both the numerator and the denominator by ten. $=\frac{17\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{wingbeats}}{1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{second}}$ So, the rate is $17$ beats per second. Example 5: A mountain climber is $3200$ meters from the peak. He climbs $50$ meters per hour for $8$ hours per day. How many days will it be before he reaches the peak? The first job is to figure out the rate per day. $\frac{50\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{meters}}{1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{hour}}\cdot \frac{8\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{hours}}{1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{day}}=50\left(8\right)\frac{\text{meters}}{\text{day}}$ $=400\frac{\text{meters}}{\text{day}}$ He is climbing at a rate of $400$ meters per day. Now divide $3200$ by the daily rate to find the number of days it will take him to reach the top. $\frac{3200\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{meters}}{400\frac{\text{meters}}{\text{day}}}=8\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{days}$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 39, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9513127207756042, "perplexity": 618.3125699656771}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128319688.9/warc/CC-MAIN-20170622181155-20170622201155-00178.warc.gz"}
https://www.physicsforums.com/threads/projectile-motion-involving-calculus.417828/	# Projectile Motion Involving Calculus 1. Jul 23, 2010 ### TheAkuma Hi, I'm stuck in my maths assignment and need help with one of the questions. “A stunt motorcyclist launches himself from a ramp inclined at 30 degrees to the horizontal. He aims to clear a line of cars that extends to a distance of 40 metres from the end of the ramp. Use calculus methods to determine the minimum possible take-off speed for the rider given that the end of the ramp is 3 metres above the ground and an average car is about 1.5 metres tall.” OK, so my teacher told me to treat the model as a coordinates on a Cartesian plane or something like that. At the bottom right of the ramp I put the origin there so i don't have to deal with the negative vertical axis. I used i for the horizontal axis and j for the vertical axis. I have been given the equation v=v*√3/2 i+v/2 j from the ramp using sine and cosine, but I'm not sure if thats to work out the velocity at the point where the motorcycle leaves the ramp. Now, I'm going to have a wild guess and say if I differentiate velocity I will get the displacement. But the problem with that is that it adds another variable, t, also i need to find c when i differentiate. I need help finding the height (j) for the midpoint where the vertical velocity is zero. Since the question asks for the minimum velocity I was thinking that the midpoint would be at 20m or 20i. The path trajectory also needs to hit 40i+1.5j. So my question is; 1.) How to find the constant (c) when differentiating velocity. 2.) How to find the height of the midpoint for the flight path. If anyone could just point me in the right direction it would be very helpful. Also I have included two diagrams as attachments which I drew for the question. #### Attached Files: File size: 9 KB Views: 171 • ###### Question 1a(2).jpg File size: 13.3 KB Views: 190 Last edited: Jul 23, 2010 2. Jul 23, 2010 ### HallsofIvy So I have moved it to the homework section. Stop with the wild guesses! You should have learned that the derivative of velocity is acceleration. To find displacement you need to integrate. No, you will have a constant c, when you integrate. Apparently you are just mixing up the words "differentiate" and "integrate". I'm not clear on what you are asking. Since the only force here is gravity (vertical), the horizontal velocity is constant. The minimum speed (not velocity) occurs when the vertical speed is 0. When you integrate the velocity you get a "constant of integration" and I think that is the "c" you are asking about. Put a time and position that you know in the equation so you have only "c" left. Solve for c. You know the position of the motorcycle at the beginning, t= 0, don't you?[/quote] One method: having determined the beginning and ending points for the path, average their x-values to find the x-value of the midpoint. Put that into the equation for horizontal distance, x, as a value of t, and solve for t. Put that t into the formula for the vertical distance and calculate the height at that point. Easier method: because of the symmetry, the midpoint will be the highest point in the path, when the vertical velocity is 0. Set the velocity function equal to 0 and solve for t. Use that t to calculate the height. 3. Jul 23, 2010 ### TheAkuma thanks a lot, I've just been flooded with assignments I can't think straight. This question doesn't really tell where the motorcyclist starts his run but it would be obvious that he starts a few meters behind the ramp. So I don't really know exactly where the displacement is at t=0, unless I state that he starts somewhere like 50m from the ramp to gain that minimum velocity. I have attached the equation I got from integrating velocity. Could you have a look to see if thats right? Also I don't think I'm allowed to use x and y coordinates as my teacher told me to use i's and j's (x=i), (y=j). File size: 1.8 KB Views: 137	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9578866958618164, "perplexity": 431.7826451503322}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647556.43/warc/CC-MAIN-20180321004405-20180321024405-00541.warc.gz"}
http://math.stackexchange.com/questions/194642/ellipse-equation-parameters	# Ellipse equation parameters If I have an ellipse expressed by: $ax^2 + 2cxy + by^2 = constant$ what does this expression equal to: $1/ \sqrt{ab - c^2}$ with respect to the ellipse ? Thanks matlabit - is it an ellipse or an ellipsoid? $z$ can't be constant for an ellipsoid. – ajay Sep 12 '12 at 11:23 which equation?? – Aang Sep 12 '12 at 12:25 Where you write "equation", this is an expression, not an equation. Also note that you don't need the parentheses inside the square root. – joriki Sep 12 '12 at 13:22 [Note:] As David pointed out, I'm assuming the constant is $1$; if not, divide through by the constant. This is the product of the two semi-axes, and thus except for a factor of $\pi$ it is the area of the ellipse. Write the left-hand side as $$\pmatrix{x&y}\pmatrix{a&c\\c&b}\pmatrix{x\\y}\;.$$ We can diagonalize the matrix to bring this into the form $$\pmatrix{x'&y'}\pmatrix{\lambda_1&0\\0&\lambda_2}\pmatrix{x'\\y'}\;,$$ where $\lambda_1$ and $\lambda_2$ are the eigenvalues of the matrix and the inverse squares of the semi-axes. Their product is the determinant $ab-c^2$. - You are assuming matlabit's "constant" is 1. – David Speyer Sep 12 '12 at 14:31 Why not divide the equation by ${\rm constant}$ to make it $a x^2+2 c x y + b y^2 = 1$ without any loss of generality. Then for a rotated ellipse with major radius $r_1$ and minor radius $r_2$ and orientation angle $\theta$ the coefficients are $$a = (\frac{1}{r_1^2}-\frac{1}{r_2^2}) \cos^2\theta+\frac{1}{r_2^2}$$ $$b = (\frac{1}{r_2^2}-\frac{1}{r_1^2}) \cos^2\theta+\frac{1}{r_1^2}$$ $$c = (\frac{1}{r_1^2}-\frac{1}{r_2^2}) \sin\theta \cos\theta$$ and the said quantity $$\boxed{ \frac{1}{\sqrt{a b-c^2}} = r_1 r_2 }$$ which appears to be the area over $\pi$. Note the equation of the ellipse is best expressed by the quadratic form of the conic section tensor $C$ $$\boldsymbol{x}^\top C \boldsymbol{x} = 0$$ $$\begin{pmatrix} x & y & 1 \end{pmatrix} \begin{pmatrix} a & c & 0\\c & b & 0\\0 & 0 & \text{-}1 \end{pmatrix} \begin{pmatrix} x\\y\\1 \end{pmatrix} = a x^2+2 c x y + b y^2 -1 = 0$$ or in rotated coordinates $x'=x \cos\theta-y\sin\theta$, $y'=x \sin\theta+x\cos\theta$ $$\begin{pmatrix} x' & y' & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{r_1^2} & 0 & 0\\0 & \frac{1}{r_2^2} & 0\\0 & 0 & \text{-}1 \end{pmatrix} \begin{pmatrix} x'\\y'\\1 \end{pmatrix} = \frac{x'^2}{r_1^2}+\frac{y'^2}{r_2^2}-1 = 0$$ - Thank You guys, both answers were really helpful! – matlabit Sep 13 '12 at 8:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9735143780708313, "perplexity": 434.22020544140156}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042990603.54/warc/CC-MAIN-20150728002310-00293-ip-10-236-191-2.ec2.internal.warc.gz"}
https://calculator-online.net/pressure/kpa-to-psi/	# Convert kpa to psi From: kilopascal (kPa) To: psi Get the Widget! Add Kilopascal to Psi converter to your website to use this unit converter directly. Feel hassle-free to account this widget as it is 100% free. Available on App Try Unit Converter App for your Mobile to get the ease of converting thousands of units. It’s 100% free with ample of features! Convert Kilopascal to Other Pressure Units Kpa (Kilopascal) and psi (Pound-force per Square Inch) both are the units of pressure, here we are going to tell you how to go from kilopascal to psi, also more information on each of the units. You can simply use the above converter to perform these units of pressure conversions. An online kpa to psi converter will helps you to do conversions between any values of kpa and psi, whenever you need to do that! Also, we tells you the formula that you can consider while doing manual conversions between kpa and pounds per square inch. First, explore some basics! ## What is kpa? Pressure is referred to as a force that exerted on a surface per unit area. In more simple words, it is said to be as the force that is put on a certain defined area. Kpa is stand for kilopascal, it is referred to as a unit of pressure and stress corresponding to the SI system of units. Remember that is said to be as a multiple of Pascal (Pa) (1 kpa = 1000 pascal), which is referred to as a derived unit that based on the Newton (N), a unit of force, also the meter squared, which is a unit of area. The pascal (pa) unit is elaborated as how many Newtons (force) are applied on a unit area of square meters. ## What is Psi? The psi is stands for pounds per square inch that is said ot be as a unit of pressure and stress according to the Imperial and US customary systems of units. Psi is derived from the pound-force lb-f that is referred to as a unit of force and the inch squared, which is said to be a unit of area. The definition of psi is entire based on the both definitions of the pound-force and the inch units. • 1 kilopascals (kpa) is equal to 0.1450377377 pound-force per square inch (psi) • 1 pound-force per square inch (psi) is equal to 6.894757294604437 kilopascals (kpa) ## kpa to psi Formula: The formula for is: psi = kPa ÷ 6.89475729 ## How do you convert kpa to psi? Convert with: • Online kilopascal to psi converter • Formula (the below example helps you) ### Example of conversions between kilopascals and pounds per square inch (kpa and psi): Problem: Convert 7400 kilopascals to psi? Solution: Step 1 (Formula): • psi = kPa ÷ 6.89475729 Step 2 (Put the Values): • psi = 7400 ÷ 6.89475729 Step 3 (Result): • 1073.2792589800001 pound-force per Square Inch Means, 7400 kilopascals (kpa) is equal to 1073.2792589800001 pound-force per Square Inch (psi)	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8770086765289307, "perplexity": 2224.311248964813}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704835583.91/warc/CC-MAIN-20210128005448-20210128035448-00722.warc.gz"}
http://aimsciences.org/search/author?author=Julia%20Brettschneider	# American Institute of Mathematical Sciences ## Journals DCDS Consider a class of skew product transformations consisting of an ergodic or a periodic transformation on a probability space $(M, \B,\mu)$ in the base and a semigroup of transformations on another probability space (Ω,$\F,P)$ in the fibre. Under suitable mixing conditions for the fibre transformation, we show that the properties ergodicity, weakly mixing, and strongly mixing are passed on from the base transformation to the skew product (with respect to the product measure). We derive ergodic theorems with respect to the skew product on the product space. The main aim of this paper is to establish uniform convergence with respect to the base variable for the series of ergodic averages of a function $F$ on $M\times$Ω along the orbits of such a skew product. Assuming a certain growth condition for the coupling function, a strong mixing condition on the fibre transformation, and continuity and integrability conditions for $F,$ we prove uniform convergence in the base and $\L^p(P)$-convergence in the fibre. Under an equicontinuity assumption on $F$ we further show $P$-almost sure convergence in the fibre. Our work has an application in information theory: It implies convergence of the averages of functions on random fields restricted to parts of stair climbing patterns defined by a direction. keywords: mixing properties ergodicity uniform convergence. ergodic theorems Skew product transformation	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8935275673866272, "perplexity": 347.4837690222388}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945669.54/warc/CC-MAIN-20180423011954-20180423031954-00315.warc.gz"}
https://math.stackexchange.com/questions/2133695/proving-an-inequality-via-the-cauchy-schwarz-inequality	# Proving an inequality via the Cauchy-Schwarz Inequality I apolgize for contributing yet another question asking about an application of CS. Here it is: Suppose $p_1, \dots ,p_n$ and $a_1,...,a_n$ are real numbers such that $p_i \geq 0$, $a_i \geq 0$ for all $i$, and $p_1 + \dots + p_n = 1$. Then $$(p_1a_1+ \dots + p_na_n)\left(\frac{p_1}{a_1}+ \dots + \frac{p_n}{a_n}\right) \geq 1$$ The author of my textbook gives the following proof: Apply Cauchy's inequality to the sequences $\sqrt{p_1a_1} \dots \sqrt{p_n}{a_n}$ and $\sqrt{\frac{p_1}{a_1}} \dots \sqrt{\frac{p_n}{a_n}}$. (Thats it) In trying to fill in the blanks I obtained the following $$\sqrt{p_1a_1} +\dots +\sqrt{p_na_n} \leq \sqrt{p_1+\dots+p_n}\sqrt{a_1+...+a_n} = \sqrt{a_1+...+a_n}$$ and $$\sqrt{\frac{p_1}{a_1}} +\dots +\sqrt{\frac{p_n}{a_n}} \leq \sqrt{p_1+\dots+p_n}\sqrt{\frac{1}{a_1}+...+\frac{1}{a_n}} = \sqrt{\frac{1}{a_1}+...+\frac{1}{a_n}}$$ I'm not entirely sure where to go from here. Perhaps I have misunderstood what he meant by "apply cauchy's inequality to the sequences...". Another idea I had was to note that $$(p_1a_1+ \dots + p_na_n) \leq M_a(p_1+ \dots + p_n)$$ where $M_a$ is the largest $a_i$. And, that $$\left(\frac{p_1}{a_1}+ \dots + \frac{p_n}{a_n}\right) \leq \frac{1}{m_a}(p_1+\dots+p_n)$$ where $m_a$ is the smallest $a_i$. Therefore, since $\frac{M_a}{m_a} \geq 1$ the inequality follows. I am not very confident in the correctness of this method though and would like to understand how to prove the inequality via CS as my book suggests. ## 1 Answer C-S is the following. $$(a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2)\geq(a_1b_1+a_2b_2+...+a_nb_n)^2.$$ Hence, for positives $a_i$ and your $p_i$ we obtain: $$(p_1a_1+ \dots + p_na_n)\left(\frac{p_1}{a_1}+ \dots + \frac{p_n}{a_n}\right)=\sum_{i=1}^n\left(\sqrt{p_ia_i}\right)^2\sum_{i=1}^n\left(\sqrt{\frac{p_i}{a_i}}\right)^2\geq$$ $$\geq\left(\sum_{i=1}^n(\sqrt{p_ia_i}\sqrt{\frac{p_i}{a_i}}\right)^2=\left(\sum_{i=1}^np_i\right)^2=1$$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9329326152801514, "perplexity": 104.01959425045622}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514576965.71/warc/CC-MAIN-20190923125729-20190923151729-00448.warc.gz"}
http://en.wikibooks.org/wiki/Messier_Index/M5	# Messier Index/M5 M5 Messier 5 by Hubble Space Telescope. 2.85′ view Credit: NASA/STScI/WikiSky Observation data (J2000 epoch) Class V Constellation Serpens Right ascension 15h 18m 33.75s[1] Declination +02° 04′ 57.7″[1] Distance 24.5 kly (7.5 kpc) Apparent magnitude (V) +6.65[1] Apparent dimensions (V) 23′.0 Physical characteristics Mass kg ( M${\odot}$) Estimated age 13 Gyr Other designations NGC 5904[1] Messier 5 or M5 (also designated NGC 5904) is a globular cluster in the constellation Serpens. It was discovered by Gottfried Kirch in 1702. It should not be confused with the much fainter and more distant globular cluster Palomar 5, which is situated nearby in the sky. ## Discovery and visibility M5 is, under extremely good conditions, just visible to the naked eye as a faint "star" near the star 5 Serpentis. Binoculars or small telescopes will identify the object as non-stellar while larger telescopes will show some individual stars, of which the brightest are of apparent magnitude 12.2. M5 was discovered by the German astronomer Gottfried Kirch in 1702 when he was observing a comet. Charles Messier also noted it in 1764, but thought it a nebula without any stars associated with it. William Herschel was the first to resolve individual stars in the cluster in 1791, counting roughly 200. ## Characteristics M5 wide angle by Robert J. Vanderbei Spanning 165 light-years in diameter, M5 is one of the larger globular clusters known. The gravitational sphere of influence of M5, (ie. the volume of space in which stars are gravitationally bound to it rather than being torn away by the Milky Way's gravitational pull) has a radius of some 200 light-years. At 13 billion years old it, M5 is also one of the older globulars associated with the Milky Way Galaxy. Its distance is about 24,500 light-years from Earth and the cluster contains more than 100,000 stars, as many as 500,000 according to some estimates. ## Notable stars kalpak stars in M5 are known to be variable in brightness, 97 of them belonging to the RR Lyrae type. RR Lyrae stars, sometimes referred to as "Cluster Variables", are somewhat similar to Cepheid type variables and as such can be used as a tool to measure distances in outer space since the relation between their luminosities and periods are well known. The brightest and most easily observed variable in M5 varies from magnitude 10.6 to 12.1 in a period of just under 26.5 days. A dwarf nova has also been observed in this cluster.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8738704323768616, "perplexity": 1960.120704268115}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802768831.100/warc/CC-MAIN-20141217075248-00018-ip-10-231-17-201.ec2.internal.warc.gz"}
http://mathoverflow.net/questions/43961/degree-of-balls-in-finitely-generated-subgroups-of-sl-2c	Degree of balls in finitely-generated subgroups of SL_2(C) Let $T$ be a finite symmetric set generating a Zariski dense subset of an algebraic group $G$ (specifically, $PSL_2(\mathbb{C})$ or its subgroups). Is there an $\alpha>0$ such that the set $T^{\leq n}$ of words of length at most $n$ is not in any codimension-1 subvariety of degree $n^{\alpha}$? "Escape from subvariety" arguments seem to prove similar results that are not polynomial. - In the case of $\text{PSL}(2,\mathbb{C})$, the set of words of length $O(n^3)$ is not contained in a subvariety of degree $n$. A polynomial of degree $n$ is a linear combination of matrix entries of the irreps of highest weight $\le n$. Let $A_n$ be the direct sum of the corresponding matrix algebras; its dimension is a sum of consecutive squares which is then $O(n^3)$. The generating set $S$ yields a set of operators in $A_n$, which then yields an algebra filtration of $A_n$. Since $S$ generates a Zariski dense subgroup, some term of this algebra filtration is eventually all of $A_n$. On the other hand, the filtration is generated by the term of degree 1, i.e., one can write $$A_n^{(k+1)} = A_n^{(k)}A_n^{(1)},$$ taking all linear combinations of all products of pairs on the right side. So the dimensions of the terms of the filtration have to keep going up by at least one until the filtration terminates, which then gives you the $O(n^3)$ bound. There is a similar argument for any linear algebraic group, except that the algebra $A_n$ is different for each one.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9267590641975403, "perplexity": 61.85552150227672}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398459333.27/warc/CC-MAIN-20151124205419-00024-ip-10-71-132-137.ec2.internal.warc.gz"}
http://www.physicspages.com/	Featured post # Welcome to Physics Pages This blog consists of my notes and solutions to problems in various areas of mainstream physics. An index to the topics covered is contained in the links in the sidebar on the right, or in the menu at the top of the page. This isn’t a “popular science” site, in that most posts use a fair bit of mathematics to explain their concepts. Thus this blog aims mainly to help those who are learning or reviewing physics in depth. More details on what the site contains and how to use it are on the welcome page. Despite Stephen Hawking’s caution that every equation included in a book (or, I suppose in a blog) would halve the readership, this blog has proved very popular since its inception in December 2010. Details of the number of visits and distinct visitors are given on the hit statistics page. Many thanks to my loyal followers and best wishes to everyone who visits. I hope you find it useful. Constructive criticism (or even praise) is always welcome, so feel free to leave a comment in response to any of the posts. I should point out that although I did study physics at the university level, this was back in the 1970s and by the time I started this blog in December 2010, I had forgotten pretty much everything I had learned back then. This blog represents my journey back to some level of literacy in physics. I am by no means a professional physicist or an authority on any aspect of the subject. I offer this blog as a record of my own notes and problem solutions as I worked through various books, in the hope that it will help, and possibly even inspire, others to explore this wonderful subject. Before leaving a comment, you may find it useful to read the “Instructions for commenters“. # The classical limit of quantum mechanics; Ehrenfest’s theorem Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 6. We’ve met Ehrenfest’s theorem while studying Griffiths’s book, where the theorem had the form $\displaystyle \frac{\partial\langle p\rangle}{\partial t}=-\left\langle \frac{\partial V}{\partial x}\right\rangle \ \ \ \ \ (1)$ This says that, in one dimension, the rate of change of the mean momentum equals the negative of the mean of the derivative of the potential ${V}$, which is assumed to depend on ${x}$ only. In this case, the behaviour of the means of the quantum variables reduces to the corresponding classical relation, in this case, Newton’s law ${F=\frac{dp}{dt}}$, where the force is defined in terms of the gradient of the potential: ${F=-\frac{dV}{dx}}$. Shankar treats Ehrenfest’s theorem a bit more generally. For an operator ${\Omega}$ we can use the product rule to state that $\displaystyle \frac{d}{dt}\left\langle \Omega\right\rangle$ $\displaystyle =$ $\displaystyle \frac{d}{dt}\left\langle \psi\left\|\Omega\right\|\psi\right\rangle \ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \dot{\psi}\left\|\Omega\right\|\psi\right\rangle +\left\langle \psi\left\|\Omega\right\|\dot{\psi}\right\rangle +\left\langle \psi\left\|\dot{\Omega}\right\|\psi\right\rangle \ \ \ \ \ (3)$ where a dot indicates a time derivative. If ${\Omega}$ does not depend explicitly on time, we have $\displaystyle \frac{d}{dt}\left\langle \Omega\right\rangle =\left\langle \dot{\psi}\left\|\Omega\right\|\psi\right\rangle +\left\langle \psi\left\|\Omega\right\|\dot{\psi}\right\rangle \ \ \ \ \ (4)$ The time derivative of ${\psi}$ can be found from the Schrödinger equation: $\displaystyle \left\|\dot{\psi}\right\rangle$ $\displaystyle =$ $\displaystyle -\frac{i}{\hbar}H\left\|\psi\right\rangle \ \ \ \ \ (5)$ $\displaystyle \left\langle \dot{\psi}\right\|$ $\displaystyle =$ $\displaystyle \frac{i}{\hbar}\left\langle \psi\right\|H \ \ \ \ \ (6)$ The second equation follows since ${H}$ is hermitian, so ${H^{\dagger}=H}$. Plugging these into 4 we have $\displaystyle \frac{d}{dt}\left\langle \Omega\right\rangle$ $\displaystyle =$ $\displaystyle \frac{i}{\hbar}\left[\left\langle \psi\left\|H\Omega\right\|\psi\right\rangle -\left\langle \psi\left\|\Omega H\right\|\psi\right\rangle \right]\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{i}{\hbar}\left\langle \psi\left\|\left[\Omega,H\right]\right\|\psi\right\rangle \ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{i}{\hbar}\left\langle \left[\Omega,H\right]\right\rangle \ \ \ \ \ (9)$ That is, the rate of change of the mean of an operator can be found from its commutator with the Hamiltonian. It is this result that Shankar refers to as Ehrenfest’s theorem. This relation is similar to that from classical mechanics, where the rate of change of a dynamical variable ${\omega}$ is equal to its Poisson bracket with the classical Hamiltonian. In the Hamiltonian formulation of classical mechanics, dynamical variables depend on generalized coordinates ${q_{i}}$ and their corresponding momenta ${p_{i}}$, so we have: $\displaystyle \frac{d\omega}{dt}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\dot{q}_{i}+\frac{\partial\omega}{\partial p_{i}}\dot{p}_{i}\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial H}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial H}{\partial q_{i}}\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \left\{ \omega,H\right\} \ \ \ \ \ (12)$ We can work out 9 for the particular cases where ${\Omega=X}$, the position operator and ${\Omega=P}$, the momentum operator. For a Hamiltonian of the form $\displaystyle H=\frac{P^{2}}{2m}+V\left(x\right) \ \ \ \ \ (13)$ and using the commutation relation $\displaystyle \left[X,P\right]=i\hbar \ \ \ \ \ (14)$ we have $\displaystyle \frac{d\left\langle X\right\rangle }{dt}$ $\displaystyle =$ $\displaystyle -\frac{i}{\hbar}\left\langle \left[X,H\right]\right\rangle \ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{i}{2m\hbar}\left\langle \left[X,P^{2}\right]\right\rangle \ \ \ \ \ (16)$ We can evaluate this commutator using the theorem $\displaystyle \left[AB,C\right]=A\left[B,C\right]+\left[A,C\right]B \ \ \ \ \ (17)$ In this case, ${A=B=P}$ and ${C=X}$, so we have $\displaystyle \left[P^{2},X\right]$ $\displaystyle =$ $\displaystyle P\left[P,X\right]+\left[P,X\right]P\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -2i\hbar P\ \ \ \ \ (19)$ $\displaystyle \left[X,P^{2}\right]$ $\displaystyle =$ $\displaystyle 2i\hbar P\ \ \ \ \ (20)$ $\displaystyle \frac{d\left\langle X\right\rangle }{dt}$ $\displaystyle =$ $\displaystyle \frac{\left\langle P\right\rangle }{m} \ \ \ \ \ (21)$ This is equivalent to the classical relation ${p=mv}$ for velocity ${v}$. We can write this result in terms of the Hamiltonian, provided that it’s legal to take the derivative of the Hamiltonian with respect to an operator (which works if we can expand the Hamiltonian as a power series): $\displaystyle \frac{d\left\langle X\right\rangle }{dt}=\frac{\left\langle P\right\rangle }{m}=\left\langle \frac{\partial H}{\partial P}\right\rangle \ \ \ \ \ (22)$ This looks a lot like one of Hamilton’s canonical equations in classical mechanics: $\displaystyle \dot{q}_{i}=\frac{\partial H}{\partial p_{i}} \ \ \ \ \ (23)$ The main difference between the quantum and classical forms is that the quantum version is a relation between mean values, while the classical version is exact. We can make the correspondence exact provided that it’s legal to take the averaging operation inside the derivative and apply it to each occurrence of ${X}$ and ${P}$. That is, is it legal to say that $\displaystyle \left\langle \frac{\partial H}{\partial P}\right\rangle =\left\langle \frac{\partial H\left(P,X\right)}{\partial P}\right\rangle =\frac{\partial H\left(\left\langle P\right\rangle ,\left\langle X\right\rangle \right)}{\partial\left\langle P\right\rangle } \ \ \ \ \ (24)$ This depends on the precise functional form of ${H}$. In the case 13 we’re considering here, we have $\displaystyle \left\langle \frac{\partial H}{\partial P}\right\rangle =\left\langle \frac{P}{m}\right\rangle =\frac{\left\langle P\right\rangle }{m}=\frac{\partial}{\partial\left\langle P\right\rangle }\left(\frac{\left\langle P\right\rangle ^{2}}{2m}+V\left(\left\langle X\right\rangle \right)\right) \ \ \ \ \ (25)$ So in this case it works. In general, if ${H}$ depends on ${P}$ either linearly or quadratically, then its derivative with respect to ${P}$ will be either constant or linear, and we can take the averaging operation inside the function without changing anything. However, if, say, ${H=P^{3}}$ (unlikely, but just for the sake of argument), then $\displaystyle \left\langle \frac{\partial H}{\partial P}\right\rangle =\left\langle 3P^{2}\right\rangle \ne3\left\langle P\right\rangle ^{2}=\frac{\partial H\left(\left\langle P\right\rangle ,\left\langle X\right\rangle \right)}{\partial\left\langle P\right\rangle } \ \ \ \ \ (26)$ since, in general, the mean of the square of a value is not the same as the square of the mean. Shankar goes through a similar argument for ${\dot{P}}$. We have $\displaystyle \left\langle \dot{P}\right\rangle =-\frac{i}{\hbar}\left\langle \left[P,H\right]\right\rangle \ \ \ \ \ (27)$ In this case, we can use the position basis form of ${P}$ which is $\displaystyle P=-i\hbar\frac{d}{dx} \ \ \ \ \ (28)$ and the position space version of the potential ${V\left(x\right)}$ to get $\displaystyle \left[P,H\right]\psi$ $\displaystyle =$ $\displaystyle -i\hbar\left(\frac{d\left(V\psi\right)}{dx}-V\frac{d\psi}{dx}\right)\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\psi\frac{dV}{dx} \ \ \ \ \ (30)$ Using this in 27 we have $\displaystyle \left\langle \dot{P}\right\rangle =-\left\langle \frac{dV}{dx}\right\rangle \ \ \ \ \ (31)$ Writing this in terms of the Hamiltonian, we have $\displaystyle \left\langle \dot{P}\right\rangle =-\left\langle \frac{\partial H}{\partial x}\right\rangle \ \ \ \ \ (32)$ Again, this looks similar to the second of Hamilton’s canonical equations from classical mechanics: $\displaystyle \dot{p}_{i}=-\frac{\partial H}{\partial q_{i}} \ \ \ \ \ (33)$ and again, we’re allowed to make the correspondence exact provided we can take the averaging operation inside the derivative on the RHS of 32. This works provided that ${V}$ is either linear or quadratic in ${x}$ (such as in the harmonic oscillator). Other potentials such as the ${\frac{1}{r}}$ potential in the hydrogen atom do not allow an exact correspondence between the quantum average and the classical Hamilton equation, but this shouldn’t worry us too much since the hydrogen atom is quintessentially quantum anyway, and any attempt to describe it classically will not work. Shankar provides a lengthly discussion on when the reduction to classical mechanics is valid, and shows that in any practical experiment that we could do with a classical particle, the difference between the average quantum behaviour and the classical measurements should be so small as to be undetectable. It is only when we deal with systems that are small enough that quantum effects dominate that we need to abandon classical mechanics. # Probability current: a few examples References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 5.3, Exercises 5.3.2 – 5.3.4. Here are a few examples of probability current. Example 1 Suppose the wave function has the form $\displaystyle \psi\left(\mathbf{r},t\right)=c\tilde{\psi}\left(\mathbf{r},t\right) \ \ \ \ \ (1)$ where ${c}$ is a complex constant and ${\tilde{\psi}\left(\mathbf{r},t\right)}$ is a real function of position and time. Then the probability current is $\displaystyle \mathbf{j}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2mi}\left(\psi^\nabla\psi-\psi\nabla\psi^\right)\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2mi}\left(cc^\left(\tilde{\psi}\nabla\tilde{\psi}\right)-\tilde{\psi}\nabla\tilde{\psi}\right)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (4)$ In particular, if ${\psi}$ itself is real, the probability current is always zero, so all the stationary states of systems like the harmonic oscillator and hydrogen atom that we’ve studied show no flow of probability, which is what we’d expect since they are, after all, stationary states. Example 2 Now the wave function is $\displaystyle \psi_{\mathbf{p}}$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\hbar\right)^{3/2}}e^{i\mathbf{p}\cdot\mathbf{r}/\hbar} \ \ \ \ \ (5)$ where the momentum ${\mathbf{p}}$ is constant. In this case we have $\displaystyle \nabla\psi_{\mathbf{p}}$ $\displaystyle =$ $\displaystyle \frac{i}{\left(2\pi\hbar\right)^{3/2}\hbar}e^{i\mathbf{p}\cdot\mathbf{r}/\hbar}\mathbf{p}\ \ \ \ \ (6)$ $\displaystyle \nabla\psi_{\mathbf{p}}^$ $\displaystyle =$ $\displaystyle \frac{-i}{\left(2\pi\hbar\right)^{3/2}\hbar}e^{-i\mathbf{p}\cdot\mathbf{r}/\hbar}\mathbf{p}\ \ \ \ \ (7)$ $\displaystyle \psi_{\mathbf{p}}^$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\hbar\right)^{3/2}}e^{-i\mathbf{p}\cdot\mathbf{r}/\hbar} \ \ \ \ \ (8)$ This gives a probability current of $\displaystyle \mathbf{j}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2mi}(\psi_{\mathbf{p}}^\nabla\psi_{\mathbf{p}}-\psi_{\mathbf{p}}\nabla\psi_{\mathbf{p}}^)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\hbar\right)^{3}2m}\left(\mathbf{p}+\mathbf{p}\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(2\pi\hbar\right)^{3}m}\mathbf{p} \ \ \ \ \ (11)$ The probability density is $\displaystyle P=\psi_{\mathbf{p}}^\psi_{\mathbf{p}}=\frac{1}{\left(2\pi\hbar\right)^{3}} \ \ \ \ \ (12)$ Thus the current can be written as $\displaystyle \mathbf{j}=\frac{P}{m}\mathbf{p} \ \ \ \ \ (13)$ Classically, the momentum is ${\mathbf{p}=mv}$, so the current has the same form as ${\mathbf{j}=P\mathbf{v}}$. This is similar to the electromagnetic case where the electric current density ${\mathbf{J}=\rho\mathbf{v}}$ where ${\rho}$ is the charge density and ${\mathbf{v}}$ is the velocity of that charge. The probability density can be viewed as “probability” moving with velocity ${\mathbf{v}}$. Example 3 Now consider a one-dimensional problem where the wave function consists of two oppositely-moving plane waves: $\displaystyle \psi=Ae^{ipx/\hbar}+Be^{-ipx/\hbar} \ \ \ \ \ (14)$ In this case, we have $\displaystyle \frac{2mi}{\hbar}j$ $\displaystyle =$ $\displaystyle \psi^\nabla\psi-\psi\nabla\psi^\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(A^e^{-ipx/\hbar}+B^e^{ipx/\hbar}\right)\frac{ip}{\hbar}\left(Ae^{ipx/\hbar}+Be^{-ipx/\hbar}\right)-\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \left(Ae^{ipx/\hbar}+Be^{-ipx/\hbar}\right)\frac{ip}{\hbar}\left(-A^e^{-ipx/\hbar}+B^e^{ipx/\hbar}\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2ip}{\hbar}\left(\left\|A\right\|^{2}-\left\|B\right\|^{2}\right)\ \ \ \ \ (17)$ $\displaystyle j$ $\displaystyle =$ $\displaystyle \frac{p}{m}\left(\left\|A\right\|^{2}-\left\|B\right\|^{2}\right) \ \ \ \ \ (18)$ The probability current separates into two terms, one for each direction of momentum. # Probability current with complex potential References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 5.3, Exercise 5.3.1. Shakar’s derivation of the probability current in 3-d is similar to the one we reviewed earlier, so we don’t need to repeat it here. We can, however, look at a slight variant where the potential has a constant imaginary part, so that $\displaystyle V\left(\mathbf{r}\right)=V_{r}\left(\mathbf{r}\right)-iV_{i} \ \ \ \ \ (1)$ where ${V_{r}\left(\mathbf{r}\right)}$ is a real function of position and ${V_{i}}$ is a real constant. A Hamiltonian containing such a complex potential is not Hermitian. To see what effect this has on the total probability of finding a particle in all space, we can repeat the derivation of the probability current. From the Schrödinger equation and its complex conjugate, we have $\displaystyle i\hbar\frac{\partial\psi}{\partial t}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\nabla^{2}\psi+V_{r}\psi-iV_{i}\psi\ \ \ \ \ (2)$ $\displaystyle -i\hbar\frac{\partial\psi^}{\partial t}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\nabla^{2}\psi^+V_{r}\psi^+iV_{i}\psi^ \ \ \ \ \ (3)$ Multiply the first equation by ${\psi^}$ and the second by ${\psi}$ and subtract to get $\displaystyle i\hbar\frac{\partial}{\partial t}\left(\psi\psi^\right)=-\frac{\hbar^{2}}{2m}\left(\psi^\nabla^{2}\psi-\psi\nabla^{2}\psi^\right)-2iV_{i}\psi\psi^* \ \ \ \ \ (4)$ As in the case with a real potential, the first term on the RHS can be written as the divergence of a vector: $\displaystyle \mathbf{J}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2mi}(\Psi^\nabla\Psi-\Psi\nabla\Psi^)\ \ \ \ \ (5)$ $\displaystyle \nabla\cdot\mathbf{J}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2mi}\left(\psi^\nabla^{2}\psi-\psi\nabla^{2}\psi^\right)\ \ \ \ \ (6)$ $\displaystyle \frac{\partial}{\partial t}\left(\psi\psi^\right)$ $\displaystyle =$ $\displaystyle -\nabla\cdot\mathbf{J}-\frac{2V_{i}}{\hbar}\psi\psi^ \ \ \ \ \ (7)$ If we define the total probability of finding the particle anywhere in space as $\displaystyle P\equiv\int\psi^\psi d^{3}\mathbf{r} \ \ \ \ \ (8)$ then we can integrate 4 over all space and use Gauss’s theorem to convert the volume integral of a divergence into a surface integral: $\displaystyle \frac{\partial}{\partial t}\left(\int\psi\psi^d^{3}\mathbf{r}\right)$ $\displaystyle =$ $\displaystyle -\int\nabla\cdot\mathbf{J}d^{3}\mathbf{r}-\frac{2V_{i}}{\hbar}\int\psi\psi^*d^{3}\mathbf{r}\ \ \ \ \ (9)$ $\displaystyle \frac{\partial P}{\partial t}$ $\displaystyle =$ $\displaystyle -\int_{S}\mathbf{J}\cdot d\mathbf{a}-\frac{2V_{i}}{\hbar}P \ \ \ \ \ (10)$ We make the usual assumption that the probability current ${\mathbf{J}}$ tends to zero at infinity fast enough for the first integral on the RHS to be zero, and we get $\displaystyle \frac{\partial P}{\partial t}=-\frac{2V_{i}}{\hbar}P \ \ \ \ \ (11)$ This has the solution $\displaystyle P\left(t\right)=P\left(0\right)e^{-2V_{i}t/\hbar} \ \ \ \ \ (12)$ That is, the probability of the particle existing decays exponentially. Although Shankar says that such a potential can be used to model a system where particles are absorbed, it’s not clear how realistic it is since the Hamiltonian isn’t hermitian, so technically the energies in such a system are not observables. # Infinite square well – force to decrease well width References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 5.2, Exercise 5.2.4. One way of comparing the classical and quantum pictures of a particle in an infinite square well is to calculate the force exerted on the walls by the particle. If a particle is in state ${\left\|n\right\rangle }$, its energy is $\displaystyle E_{n}=\frac{\left(n\pi\hbar\right)^{2}}{2mL^{2}} \ \ \ \ \ (1)$ If the particle remains in this state as the walls are slowly pushed in, so that ${L}$ slowly decreases, then its energy ${E_{n}}$ will increase, meaning that work is done on the system. The force is the change in energy per unit distance, so the force required is $\displaystyle F=-\frac{\partial E_{n}}{\partial L}=\frac{\left(n\pi\hbar\right)^{2}}{mL^{3}} \ \ \ \ \ (2)$ If we treat the system classically, then a particle with energy ${E_{n}}$ between the walls is effectively a free particle in this region (since the potential ${V=0}$ there), so all its energy is kinetic. That is $\displaystyle E_{n}$ $\displaystyle =$ $\displaystyle \frac{1}{2}mv^{2}\ \ \ \ \ (3)$ $\displaystyle v$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2E_{n}}{m}}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{n\pi\hbar}{mL} \ \ \ \ \ (5)$ The classical particle bounces elastically between the two walls, which means its velocity is exactly reversed at each collision. The momentum transfer in such a collision is $\displaystyle \Delta p=2mv=\frac{2n\pi\hbar}{L} \ \ \ \ \ (6)$ The time between successive collisions on the same wall is $\displaystyle \Delta t=\frac{2L}{v}=\frac{2mL^{2}}{n\pi\hbar} \ \ \ \ \ (7)$ Thus the average force exerted on one wall is $\displaystyle \bar{F}=\frac{\Delta p}{\Delta t}=\frac{\left(n\pi\hbar\right)^{2}}{mL^{3}} \ \ \ \ \ (8)$ Comparing with 2, we see that the quantum and classical forces in this case are the same. # Infinite square well – expanding well References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 5.2, Exercise 5.2.1. Shankar’s treatment of the infinite square well is similar to that of Griffiths, which we’ve already covered, so we won’t go through the details again. The main difference is that Shankar places the potential walls at ${x=\pm\frac{L}{2}}$ while Griffiths places them at ${x=0}$ and ${x=a}$. As a result, the stationary states found by Shankar are shifted to the left, with the result $\displaystyle \psi_{n}\left(x\right)=\begin{cases} \sqrt{\frac{2}{L}}\cos\frac{n\pi x}{L} & n=1,3,5,7,\ldots\\ \sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L} & n=2,4,6,\ldots \end{cases} \ \ \ \ \ (1)$ These results can be obtained from the form given by Griffiths (where we take the width of the well to be ${L}$ rather than ${a}$): $\displaystyle \psi_{n}\left(x\right)$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2}{L}}\sin\frac{n\pi\left(x+\frac{L}{2}\right)}{L}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2}{L}}\left[\sin\frac{n\pi x}{L}\cos\frac{n\pi}{2}+\cos\frac{n\pi x}{L}\sin\frac{n\pi}{2}\right] \ \ \ \ \ (3)$ Choosing ${n}$ to be even or odd gives the results in 1. The specific problem we’re solving here involves a particle that starts off in the ground state (${n=1}$) of a square well of width ${L}$. The well then suddenly expands to a width of ${2L}$ symmetrically, that is, it now extends from ${x=-L}$ to ${x=+L}$. We are to find the probability that the particle will be found in the ground state of the new well. We solved a similar problem before, but in that case the well expanded by moving its right-hand wall to the right while keeping the left-hand wall fixed, so that the particle found itself in the left half of the new, expanded well. In the present problem, the particle finds itself centred in the new expanded well. You might think that this shouldn’t matter, but it turns out to make quite a difference. To calculate this probability, we need to express the original wave function in terms of the stationary states of the expanded well, which we’ll refer to as ${\phi_{n}\left(x\right)}$. That is $\displaystyle \psi_{1}\left(x\right)=\sum_{n=1}^{\infty}c_{n}\phi_{n}\left(x\right) \ \ \ \ \ (4)$ Working with Shankar’s functions 1 we find ${\phi_{n}}$ by replacing ${L}$ by ${2L}$: $\displaystyle \phi_{n}\left(x\right)=\begin{cases} \frac{1}{\sqrt{L}}\cos\frac{n\pi x}{2L} & n=1,3,5,7,\ldots\\ \frac{1}{\sqrt{L}}\sin\frac{n\pi x}{2L} & n=2,4,6,\ldots \end{cases} \ \ \ \ \ (5)$ Using the orthonormality of the wave functions, we have $\displaystyle c_{1}$ $\displaystyle =$ $\displaystyle \int_{-L}^{L}\psi_{1}\left(x\right)\phi_{1}\left(x\right)dx\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-L/2}^{L/2}\sqrt{\frac{2}{L}}\cos\frac{\pi x}{L}\frac{1}{\sqrt{L}}\cos\frac{\pi x}{2L}dx\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sqrt{2}}{L}\int_{-L/2}^{L/2}\cos\frac{\pi x}{L}\cos\frac{\pi x}{2L}dx\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sqrt{2}}{L}\int_{-L/2}^{L/2}\left(1-2\sin^{2}\frac{\pi x}{2L}\right)\cos\frac{\pi x}{2L}dx\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{8}{3\pi} \ \ \ \ \ (10)$ The limits of integration are reduced in the second line since ${\psi_{1}\left(x\right)=0}$ if ${x>\left\|\frac{L}{2}\right\|}$. Thus the probability of finding the particle in the new ground state is $\displaystyle \left\|c_{1}\right\|^{2}=\frac{64}{9\pi^{2}} \ \ \ \ \ (11)$ Note that in the earlier problem where the well expanded to the right, the probability was ${\frac{32}{9\pi^{2}}}$, so the new probability is twice as much when the wave function remains centred in the new well. We could have also done the calculation using Griffiths’s well which extended from ${x=0}$ to ${x=L}$. If this well expands symmetrically, it now runs from ${x=-\frac{L}{2}}$ to ${x=\frac{3L}{2}}$, and the stationary states of this new well are obtained by replacing ${L\rightarrow2L}$ and ${x\rightarrow x+\frac{L}{2}}$, so we have $\displaystyle \phi_{n}\left(x\right)=\frac{1}{\sqrt{L}}\sin\frac{n\pi\left(x+\frac{L}{2}\right)}{2L} \ \ \ \ \ (12)$ We then get $\displaystyle c_{1}$ $\displaystyle =$ $\displaystyle \int_{-L/2}^{3L/2}\psi_{1}\left(x\right)\phi_{1}\left(x\right)dx\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sqrt{2}}{L}\int_{0}^{L}\sin\frac{\pi x}{L}\sin\frac{\pi\left(x+\frac{L}{2}\right)}{2L}dx\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{8}{3\pi} \ \ \ \ \ (15)$ The integral can be done by expanding the second sine using the sine addition formula. (I just used Maple.) # Propagator for a Gaussian wave packet for the free particle References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 5.1, Exercise 5.1.3. $\displaystyle U\left(t\right)=\int_{-\infty}^{\infty}e^{-ip^{2}t/2m\hbar}\left\|p\right\rangle \left\langle p\right\|dp \ \ \ \ \ (1)$ We can find its matrix elements in position space by using the position space form of the momentum $\displaystyle \left\langle x\left\|p\right.\right\rangle =\frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar} \ \ \ \ \ (2)$ Taking the matrix element of 1 we have $\displaystyle U\left(x,t;x^{\prime}\right)$ $\displaystyle =$ $\displaystyle \left\langle x\left\|U\left(t\right)\right\|x^{\prime}\right\rangle \ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int\left\langle x\left\|p\right.\right\rangle \left\langle p\left\|x^{\prime}\right.\right\rangle e^{-ip^{2}t/2m\hbar}dp\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2\pi\hbar}\int e^{ip\left(x-x^{\prime}\right)/\hbar}e^{-ip^{2}t/2m\hbar}dp\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m}{2\pi\hbar it}}e^{im\left(x-x^{\prime}\right)^{2}/2\hbar t} \ \ \ \ \ (6)$ The final integral can be done by combining the exponents in the third line, completing the square and using the standard formula for Gaussian integrals. We won’t go through that here, as our main goal is to explore the evolution of an initial wave packet using the propagator. Given 6, we can in principle find the wave function for all future times given an initial wave function, by using the propagator: $\displaystyle \psi\left(x,t\right)=\int U\left(x,t;x^{\prime}\right)\psi\left(x^{\prime},0\right)dx^{\prime} \ \ \ \ \ (7)$ Here, we’re assuming that the initial time is ${t=0}$. Shankar uses the standard example where the initial wave packet is a Gaussian: $\displaystyle \psi\left(x^{\prime},0\right)=e^{ip_{0}x^{\prime}/\hbar}\frac{e^{-x^{\prime2}/2\Delta^{2}}}{\left(\pi\Delta^{2}\right)^{1/4}} \ \ \ \ \ (8)$ This is a wave packet distributed symmetrically about the origin, so that ${\left\langle X\right\rangle =0}$, and with mean momentum given by ${\left\langle P\right\rangle =p_{0}}$. By plugging this and 6 into 7, we can work out the time-dependent version of the wave packet, which Shankar gives as $\displaystyle \psi\left(x,t\right)=\left[\sqrt{\pi}\left(\Delta+\frac{i\hbar t}{m\Delta}\right)\right]^{-1/2}\exp\left[\frac{-\left(x-p_{0}t/m\right)^{2}}{2\Delta^{2}\left(1+i\hbar t/m\Delta^{2}\right)}\right]\exp\left[\frac{ip_{0}}{\hbar}\left(x-\frac{p_{0}t}{2m}\right)\right] \ \ \ \ \ (9)$ Again, we won’t go through the derivation of this result as it involves a messy calculation with Gaussian integrals again. The main problem we want to solve here is to use our alternative form of the propagator in terms of the Hamiltonian: $\displaystyle U\left(t\right)=e^{-iHt/\hbar} \ \ \ \ \ (10)$ For the free particle $\displaystyle H=-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}} \ \ \ \ \ (11)$ so if we expand ${U\left(t\right)}$ as a power series, we have $\displaystyle U\left(t\right)=\sum_{s=0}^{\infty}\frac{1}{s!}\left(\frac{i\hbar t}{2m}\right)^{s}\frac{d^{2s}}{dx^{2s}} \ \ \ \ \ (12)$ To see how we can use this form to generate the time-dependent wave function, we’ll consider a special case of 8 with ${p_{0}=0}$ and ${\Delta=1}$, so that $\displaystyle \psi_{0}\left(x\right)$ $\displaystyle =$ $\displaystyle \frac{e^{-x^{2}/2}}{\pi^{1/4}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\pi^{1/4}}\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x^{2n}}{2^{n}n!} \ \ \ \ \ (14)$ We therefore need to apply one power series 12 to the other 14. This is best done by examining a few specific terms and then generalizing to the main result. To save writing, we’ll work with the following $\displaystyle \alpha$ $\displaystyle \equiv$ $\displaystyle \frac{i\hbar t}{m}\ \ \ \ \ (15)$ $\displaystyle \psi_{\pi}\left(x\right)$ $\displaystyle \equiv$ $\displaystyle \pi^{1/4}\psi_{0}\left(x\right) \ \ \ \ \ (16)$ The ${s=0}$ term in 12 is just 1, so we’ll look at the ${s=1}$ term and apply it to 14: $\displaystyle \frac{\alpha}{2}\frac{d^{2}}{dx^{2}}\left[\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x^{2n}}{2^{n}n!}\right]$ $\displaystyle =$ $\displaystyle \frac{\alpha}{2}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}\left(2n\right)\left(2n-1\right)x^{2n-2}}{2^{n}n!}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\alpha}{2}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}\left(2n\right)!x^{2n-2}}{2^{n}n!\left(2n-2\right)!} \ \ \ \ \ (18)$ We can simplify this by using an identity involving factorials: $\displaystyle \frac{\left(2n\right)!}{n!}$ $\displaystyle =$ $\displaystyle \frac{\left(2n\right)\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)\ldots\left(2\right)\left(1\right)}{n\left(n-1\right)\left(n-2\right)\ldots\left(2\right)\left(1\right)}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2^{n}\left[n\left(n-1\right)\left(n-2\right)\ldots\left(2\right)\left(1\right)\right]\left[\left(2n-1\right)\left(2n-3\right)\ldots\left(3\right)\left(1\right)\right]}{n!}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2^{n}n!\left(2n-1\right)!!}{n!}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2^{n}\left(2n-1\right)!! \ \ \ \ \ (22)$ The ‘double factorial’ notation is defined as $\displaystyle \left(2n-1\right)!!\equiv\left(2n-1\right)\left(2n-3\right)\ldots\left(3\right)\left(1\right) \ \ \ \ \ (23)$ That is, it’s the product of every other term from ${n}$ down to 1. Using this result, we can write 18 as $\displaystyle \frac{\alpha}{2}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}\left(2n\right)!x^{2n-2}}{2^{n}n!\left(2n-2\right)!}=\alpha\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}\left(2n-1\right)!!x^{2n-2}}{2\left(2n-2\right)!} \ \ \ \ \ (24)$ Now look at the ${s=2}$ term from 12. $\displaystyle \frac{1}{2!}\frac{\alpha^{2}}{2^{2}}\frac{d^{4}}{dx^{4}}\left[\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x^{2n}}{2^{n}n!}\right]$ $\displaystyle =$ $\displaystyle \frac{1}{2!}\frac{\alpha^{2}}{2^{2}}\sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}\left(2n\right)\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)x^{2n-4}}{2^{n}n!}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2!}\frac{\alpha^{2}}{2^{2}}\sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}\left(2n\right)!x^{2n-4}}{2^{n}n!\left(2n-4\right)!}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\alpha^{2}}{2^{2}2!}\sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}\left(2n-1\right)!!x^{2n-4}}{\left(2n-4\right)!} \ \ \ \ \ (27)$ We can see the pattern for the general term for arbitrary ${s}$ from 12 (we could prove it by induction, but hopefully the pattern is fairly obvious): $\displaystyle \frac{1}{s!}\frac{\alpha^{s}}{2^{s}}\frac{d^{2s}}{dx^{2s}}\left[\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x^{2n}}{2^{n}n!}\right]$ $\displaystyle =$ $\displaystyle \frac{1}{s!}\frac{\alpha^{s}}{2^{s}}\sum_{n=s}^{\infty}\frac{\left(-1\right)^{n}\left(2n\right)!x^{2n-2s}}{2^{n}n!\left(2n-2s\right)!}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\alpha^{s}}{2^{s}s!}\sum_{n=s}^{\infty}\frac{\left(-1\right)^{n}\left(2n-1\right)!!x^{2n-2s}}{\left(2n-2s\right)!} \ \ \ \ \ (29)$ Now we can collect terms for each power of ${x}$. The constant term (for ${x^{0}}$) is the first term from each series for each value of ${s}$, so we have, using the general term 29 and taking the first term where ${n=s}$: $\displaystyle \sum_{s=0}^{\infty}\frac{\left(-1\right)^{s}\alpha^{s}\left(2s-1\right)!!}{2^{s}s!}=1-\frac{\alpha}{2}+\frac{\alpha^{2}}{2!}\frac{3}{2}\frac{1}{2}-\frac{\alpha^{3}}{3!}\frac{5}{2}\frac{3}{2}\frac{1}{2}+\ldots \ \ \ \ \ (30)$ [The ${\left(2s-1\right)!!}$ factor is 1 when ${s=0}$ as we can see from the result 22.] The series on the RHS is the Taylor expansion of ${\left(1+\alpha\right)^{-1/2}}$, as can be verified using tables. In general, to get the coefficient of ${x^{2r}}$ (only even powers of ${x}$ occur in the series), we take the term where ${n=s+r}$ from 29 and sum over ${s}$. This gives $\displaystyle \sum_{s=0}^{\infty}\frac{\alpha^{s}}{2^{s}s!}\frac{\left(-1\right)^{s+r}\left(2s+2r-1\right)!!}{\left(2r\right)!}$ $\displaystyle =$ $\displaystyle \frac{\left(-1\right)^{r}}{2^{r}r!}\sum_{s=0}^{\infty}\frac{\alpha^{s}}{2^{s}s!}\frac{\left(-1\right)^{s}\left(2s+2r-1\right)!!}{\left(2r-1\right)!!} \ \ \ \ \ (31)$ where we used 22 to get the RHS. Expanding the sum gives $\displaystyle \sum_{s=0}^{\infty}\frac{\alpha^{s}}{2^{s}s!}\frac{\left(-1\right)^{s}\left(2s+2r-1\right)!!}{\left(2r-1\right)!!}$ $\displaystyle =$ $\displaystyle 1-\alpha\frac{2r+1}{2}+\frac{\alpha^{2}}{2!}\left(\frac{2r+3}{2}\right)\left(\frac{2r+1}{2}\right)-\ldots\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1-\alpha\left(r+\frac{1}{2}\right)+\frac{\alpha^{2}}{2!}\left(r+\frac{3}{2}\right)\left(r+\frac{1}{2}\right)-\ldots\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1+\alpha\right)^{-r-\frac{1}{2}} \ \ \ \ \ (34)$ where again we’ve used a standard series from tables (given by Shankar in the problem) to get the last line. Combining this with 31, we see that the coefficient of ${x^{2r}}$ is $\displaystyle \frac{\left(-1\right)^{r}}{2^{r}r!}\left(1+\alpha\right)^{-r-\frac{1}{2}} \ \ \ \ \ (35)$ Thus the time-dependent wave function can be written as a single series as: $\displaystyle \psi\left(x,t\right)$ $\displaystyle =$ $\displaystyle U\left(t\right)\psi\left(x,0\right)\ \ \ \ \ (36)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{-iHt/\hbar}\psi\left(x,0\right)\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\pi^{1/4}}\sum_{r=0}^{\infty}\frac{\left(-1\right)^{r}}{2^{r}r!}\left(1+\alpha\right)^{-r-\frac{1}{2}}x^{2r}\ \ \ \ \ (38)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\pi^{1/4}\sqrt{1+\alpha}}\sum_{r=0}^{\infty}\frac{\left(-1\right)^{r}}{2^{r}\left(1+\alpha\right)^{r}r!}x^{2r}\ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\pi^{1/4}\sqrt{1+\alpha}}\exp\left[\frac{-x^{2}}{2\left(1+\alpha\right)}\right]\ \ \ \ \ (40)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\pi^{1/4}\sqrt{1+i\hbar t/m}}\exp\left[\frac{-x^{2}}{2\left(1+i\hbar t/m\right)}\right] \ \ \ \ \ (41)$ This agrees with 9 when ${p_{0}=0}$ and ${\Delta=1}$, though it does take a fair bit of work! # Free particle in the position basis References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 5.1, Exercise 5.1.2. In quantum mechanics, the free particle has degenerate energy eigenstates for each energy $\displaystyle E=\frac{p^{2}}{2m} \ \ \ \ \ (1)$ where ${p}$ is the momentum. The degeneracy arises because the momentum can be either positive (for a particle moving to the right) or negative (to the left): $\displaystyle p=\pm\sqrt{2mE} \ \ \ \ \ (2)$ Thus the most general energy eigenstate is a linear combination of the two momentum states: $\displaystyle \left\|E\right\rangle =\beta\left\|p=\sqrt{2mE}\right\rangle +\gamma\left\|p=-\sqrt{2mE}\right\rangle \ \ \ \ \ (3)$ This bizarre feature of quantum mechanics means that a particle in such a state could be moving either left or right, and if we make a measurement of the momentum we force the particle into one or other of the two momentum states. We obtained this solution by working in the momentum basis, but we can also find the solution in the position basis. In that basis, the momentum operator has the form $\displaystyle P=-i\hbar\frac{d}{dx} \ \ \ \ \ (4)$ The matrix elements of this operator in the position basis are $\displaystyle \left\langle x\left\|P\right\|x^{\prime}\right\rangle =-i\hbar\delta^{\prime}\left(x-x^{\prime}\right) \ \ \ \ \ (5)$ where ${\delta^{\prime}\left(x-x^{\prime}\right)}$ is the derivative of the delta function with respect to the ${x}$, not the ${x^{\prime}}$. We can use the properties of this derivative to get a solution in the ${X}$ basis. To be completely formal about it, the derivation of the matrix elements of ${P^{2}}$ in the ${X}$ basis is: $\displaystyle \left\langle x\left\|P^{2}\right\|\psi\right\rangle$ $\displaystyle =$ $\displaystyle \int\int\left\langle x\left\|P\right\|x^{\prime}\right\rangle \left\langle x^{\prime}\left\|P\right\|x^{\prime\prime}\right\rangle \left\langle x^{\prime\prime}\left\|\psi\right.\right\rangle dx^{\prime}dx^{\prime\prime}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int\int\left\langle x\left\|P\right\|x^{\prime}\right\rangle \left(-i\hbar\delta^{\prime}\left(x^{\prime}-x^{\prime\prime}\right)\right)\psi\left(x^{\prime\prime}\right)dx^{\prime}dx^{\prime\prime}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\int\left\langle x\left\|P\right\|x^{\prime}\right\rangle \frac{d\psi\left(x^{\prime}\right)}{dx^{\prime}}dx^{\prime}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\int\int\left(-i\hbar\delta^{\prime}\left(x-x^{\prime}\right)\right)\frac{d\psi\left(x^{\prime}\right)}{dx^{\prime}}dx^{\prime}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar^{2}\frac{d^{2}}{dx^{2}}\psi\left(x\right) \ \ \ \ \ (10)$ In this basis, the Schrödinger equation is therefore the familiar one: $\displaystyle \frac{P^{2}}{2m}\left\|\psi\right\rangle$ $\displaystyle =$ $\displaystyle E\left\|\psi\right\rangle \ \ \ \ \ (11)$ $\displaystyle \left\langle x\left\|\frac{P^{2}}{2m}\right\|\psi\right\rangle$ $\displaystyle =$ $\displaystyle E\psi\left(x\right)\ \ \ \ \ (12)$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}\psi\left(x\right)$ $\displaystyle =$ $\displaystyle E\psi\left(x\right)\ \ \ \ \ (13)$ $\displaystyle \frac{d^{2}}{dx^{2}}\psi\left(x\right)$ $\displaystyle =$ $\displaystyle -\frac{2mE}{\hbar^{2}}\psi\left(x\right) \ \ \ \ \ (14)$ This has the general solution $\displaystyle \psi\left(x\right)=\beta e^{ix\sqrt{2mE}/\hbar}+\gamma e^{-ix\sqrt{2mE}/\hbar} \ \ \ \ \ (15)$ [Shankar extracts a factor of ${1/\sqrt{2\pi\hbar}}$ but as he notes, this is arbitrary and can be absorbed into the constants ${\beta}$ and ${\gamma}$ as we’ve done here.] In this derivation we’ve implicitly assumed that ${E>0}$, since there is no potential so a free particle can’t really have a negative energy. However, if you follow through the derivation, you’ll see that it works even if ${E<0}$. In that case, we’d get $\displaystyle \psi\left(x\right)=\beta e^{-x\sqrt{2m\left\|E\right\|}/\hbar}+\gamma e^{x\sqrt{2m\left\|E\right\|}/\hbar} \ \ \ \ \ (16)$ That is, the exponents in both terms are now real instead of imaginary. The problem with this is that the first term blows up for ${x\rightarrow-\infty}$ while the second blows up for ${x\rightarrow+\infty}$. Thus this function is not normalizable, even to a delta function (as was the case when ${E>0}$), so functions such as these when ${E<0}$ are not in the Hilbert space. # Free particle revisited: solution in terms of a propagator References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 5.1, Exercise 5.1.1. Having reviewed the background mathematics and postulates of quantum mechanics as set out by Shankar, we can now revisit some of the classic problems in non-relativistic quantum mechanics using Shankar’s approach, as opposed to that of Griffiths that we’ve already studied. The first problem we’ll look it is that of the free particle. Following the fourth postulate, we write down the classical Hamiltonian for a free particle, which is $\displaystyle H=\frac{p^{2}}{2m} \ \ \ \ \ (1)$ where ${p}$ is the momentum (we’re working in one dimension) and ${m}$ is the mass. To get the quantum version, we replace ${p}$ by the momentum operator ${P}$ and insert the result into the Schrödinger equation: $\displaystyle i\hbar\left\|\dot{\psi}\right\rangle$ $\displaystyle =$ $\displaystyle H\left\|\psi\right\rangle \ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{P^{2}}{2m}\left\|\psi\right\rangle \ \ \ \ \ (3)$ Since ${H}$ is time-independent, the solution can be written using a propagator: $\displaystyle \left\|\psi\left(t\right)\right\rangle =U\left(t\right)\left\|\psi\left(0\right)\right\rangle \ \ \ \ \ (4)$ To find ${U}$, we need to solve the eigenvalue equation for the stationary states $\displaystyle \frac{P^{2}}{2m}\left\|E\right\rangle =E\left\|E\right\rangle \ \ \ \ \ (5)$ where ${E}$ is an eigenvalue representing the allowable energies. Since the Hamiltonian is ${P^{2}/2m}$, and an eigenstate of ${P}$ with eigenvalue ${p}$ is also an eigenstate of ${P^{2}}$ with eigenvalue ${p^{2}}$, we can write this equation in terms of the momentum eigenstates ${\left\|p\right\rangle }$: $\displaystyle \frac{P^{2}}{2m}\left\|p\right\rangle =E\left\|p\right\rangle \ \ \ \ \ (6)$ Using ${P^{2}\left\|p\right\rangle =p^{2}\left\|p\right\rangle }$ this gives $\displaystyle \left(\frac{p^{2}}{2m}-E\right)\left\|p\right\rangle =0 \ \ \ \ \ (7)$ Assuming that ${\left\|p\right\rangle }$ is not a null vector gives the relation between momentum and energy: $\displaystyle p=\pm\sqrt{2mE} \ \ \ \ \ (8)$ Thus each allowable energy ${E}$ has two possible momenta. Once we specify the momentum, we also specify the energy and since each energy state is two-fold degenerate, we can eliminate the ambiguity by specifying only the momentum. Therefore the propagator can be written as $\displaystyle U\left(t\right)=\int_{-\infty}^{\infty}e^{-ip^{2}t/2m\hbar}\left\|p\right\rangle \left\langle p\right\|dp \ \ \ \ \ (9)$ We can convert this to an integral over the energy by using 8 to change variables, and by splittling the integral into two parts. For ${p>0}$ we have $\displaystyle dp=\sqrt{\frac{m}{2E}}dE \ \ \ \ \ (10)$ and for ${p<0}$ we have $\displaystyle dp=-\sqrt{\frac{m}{2E}}dE \ \ \ \ \ (11)$ Therefore, we get $\displaystyle U\left(t\right)$ $\displaystyle =$ $\displaystyle \int_{0}^{\infty}e^{-iEt/\hbar}\left\|E,+\right\rangle \left\langle E,+\right\|\sqrt{\frac{m}{2E}}dE+\int_{\infty}^{0}e^{-iEt/\hbar}\left\|E,-\right\rangle \left\langle E,-\right\|\left(-\sqrt{\frac{m}{2E}}\right)dE\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{\infty}e^{-iEt/\hbar}\left\|E,+\right\rangle \left\langle E,+\right\|\sqrt{\frac{m}{2E}}dE+\int_{0}^{\infty}e^{-iEt/\hbar}\left\|E,-\right\rangle \left\langle E,-\right\|\sqrt{\frac{m}{2E}}dE\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{\alpha=\pm}\int_{0}^{\infty}\frac{m}{\sqrt{2mE}}e^{-iEt/\hbar}\left\|E,\alpha\right\rangle \left\langle E,\alpha\right\|dE \ \ \ \ \ (14)$ Here, ${\left\|E,+\right\rangle }$ is the state with energy ${E}$ and momentum ${p=+\sqrt{2mE}}$ and similarly for ${\left\|E,-\right\rangle }$. In the first line, the first integral is for ${p>0}$ and corresponds to the ${\int_{0}^{\infty}}$ part of 9. The second integral is for ${p<0}$ and corresponds to the ${\int_{-\infty}^{0}}$ part of 9, which is why the limits on the second integral have ${\infty}$ at the bottom and 0 at the top. Reversing the order of integration cancels out the minus sign in ${-\sqrt{\frac{m}{2E}}}$, which allows us to add the two integrals together to get the final answer. # Time-dependent propagators References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 4.3. The fourth postulate of non-relativistic quantum mechanics concerns how states evolve with time. The postulate simply states that in non-relativistic quantum mechanics, a state satisfies the Schrödinger equation: $\displaystyle i\hbar\frac{\partial}{\partial t}\left\|\psi\right\rangle =H\left\|\psi\right\rangle \ \ \ \ \ (1)$ where ${H}$ is the Hamiltonian, which is obtained from the classical Hamiltonian by means of the other postulates of quantum mechanics, namely that we replace all references to the position ${x}$ by the quantum position operator ${X}$ with matrix elements (in the ${x}$ basis) of $\displaystyle \left\langle x^{\prime}\left\|X\right\|x\right\rangle =\delta\left(x-x^{\prime}\right) \ \ \ \ \ (2)$ and all references to classical momentum ${p}$ by the momentum operator ${P}$ with matrix elements $\displaystyle \left\langle x^{\prime}\left\|P\right\|x\right\rangle =-i\hbar\delta^{\prime}\left(x-x^{\prime}\right) \ \ \ \ \ (3)$ In our earlier examination of the Schrödinger equation, we assumed that the Hamiltonian is independent of time, which allowed us to obtain an explicit expression for the propagator $\displaystyle U\left(t\right)=e^{-iHt/\hbar} \ \ \ \ \ (4)$ The propagator is applied to the initial state ${\left\|\psi\left(0\right)\right\rangle }$ to obtain the state at any future time ${t}$: $\displaystyle \left\|\psi\left(t\right)\right\rangle =U\left(t\right)\left\|\psi\left(0\right)\right\rangle \ \ \ \ \ (5)$ What happens if ${H=H\left(t\right)}$, that is, there is an explicit time dependence in the Hamiltonian? The approach taken by Shankar is a bit hand-wavy, but goes as follows. We divide the time interval ${\left[0,t\right]}$ into ${N}$ small increments ${\Delta=t/N}$. To first order in ${\Delta}$, we can integrate 1 by taking the first order term in a Taylor expansion: $\displaystyle \left\|\psi\left(\Delta\right)\right\rangle$ $\displaystyle =$ $\displaystyle \left\|\psi\left(0\right)\right\rangle +\Delta\left.\frac{d}{dt}\left\|\psi\left(t\right)\right\rangle \right\|_{t=0}+\mathcal{O}\left(\Delta^{2}\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\|\psi\left(0\right)\right\rangle +-\frac{i\Delta}{\hbar}H\left(0\right)\left\|\psi\left(0\right)\right\rangle +\mathcal{O}\left(\Delta^{2}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1-\frac{i\Delta}{\hbar}H\left(0\right)\right)\left\|\psi\left(0\right)\right\rangle +\mathcal{O}\left(\Delta^{2}\right) \ \ \ \ \ (8)$ So far, we’ve been fairly precise, but now the hand-waving starts. We note that the term multiplying ${\left\|\psi\left(0\right)\right\rangle }$ consists of the first two terms in the expansion of ${e^{-i\Delta H\left(0\right)/\hbar}}$, so we state that to evolve from ${t=0}$ to ${t=\Delta}$, we multiply the initial state ${\left\|\psi\left(0\right)\right\rangle }$ by ${e^{-i\Delta H\left(0\right)/\hbar}}$. That is, we propose that $\displaystyle \left\|\psi\left(\Delta\right)\right\rangle =e^{-i\Delta H\left(0\right)/\hbar}\left\|\psi\left(0\right)\right\rangle \ \ \ \ \ (9)$ [The reason this is hand-waving is that there are many functions whose first order Taylor expansion matches ${\left(1-\frac{i\Delta}{\hbar}H\left(0\right)\right)}$, so it seems arbitrary to choose the exponential. I imagine the motivation is that in the time-independent case, the result reduces to 4.] In any case, if we accept this, then we can iterate the process to evolve to later times. To get to ${t=2\Delta}$, we have $\displaystyle \left\|\psi\left(2\Delta\right)\right\rangle$ $\displaystyle =$ $\displaystyle e^{-i\Delta H\left(\Delta\right)/\hbar}\left\|\psi\left(\Delta\right)\right\rangle \ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{-i\Delta H\left(\Delta\right)/\hbar}e^{-i\Delta H\left(0\right)/\hbar}\left\|\psi\left(0\right)\right\rangle \ \ \ \ \ (11)$ The snag here is that we can’t, in general, combine the two exponentials into a single exponential by adding the exponents. This is because ${H\left(\Delta\right)}$ and ${H\left(0\right)}$ will not, in general, commute, as the Baker-Campbell-Hausdorff formula tells us. For example, the time dependence of ${H\left(t\right)}$ might be such that at ${t=0}$, ${H\left(0\right)}$ is a function of the position operator ${X}$ only, while at ${t=\Delta}$, ${H\left(\Delta\right)}$ becomes a function of the momentum operator ${P}$ only. Since ${X}$ and ${P}$ don’t commute, ${\left[H\left(0\right),H\left(\Delta\right)\right]\ne0}$, so ${e^{-i\Delta H\left(\Delta\right)/\hbar}e^{-i\Delta H\left(0\right)/\hbar}\ne e^{-i\Delta\left[H\left(0\right)+H\left(\Delta\right)\right]/\hbar}}$. This means that the best we can usually do is to write $\displaystyle \left\|\psi\left(t\right)\right\rangle$ $\displaystyle =$ $\displaystyle \left\|\psi\left(N\Delta\right)\right\rangle \ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \prod_{n=0}^{N-1}e^{-i\Delta H\left(n\Delta\right)/\hbar}\left\|\psi\left(0\right)\right\rangle \ \ \ \ \ (13)$ The propagator then becomes, in the limit $\displaystyle U\left(t\right)=\lim_{N\rightarrow\infty}\prod_{n=0}^{N-1}e^{-i\Delta H\left(n\Delta\right)/\hbar} \ \ \ \ \ (14)$ This limit is known as a time-ordered integral and is written as $\displaystyle T\left\{ \exp\left[-\frac{i}{\hbar}\int_{0}^{t}H\left(t^{\prime}\right)dt^{\prime}\right]\right\} \equiv\lim_{N\rightarrow\infty}\prod_{n=0}^{N-1}e^{-i\Delta H\left(n\Delta\right)/\hbar} \ \ \ \ \ (15)$ One final note about the propagators. Since each term in the product is the exponential of ${i}$ times a Hermitian operator, each term is a unitary operator. Further, since the product of two unitary operators is still unitary, the propagator in the time-dependent case is a unitary operator. We’ve defined a propagator as a unitary operator that carries a state from ${t=0}$ to some later time ${t}$, but we can generalize the notation so that ${U\left(t_{2},t_{1}\right)}$ is a propagator that carries a state from ${t=t_{1}}$ to ${t=t_{2}}$, that is $\displaystyle \left\|\psi\left(t_{2}\right)\right\rangle =U\left(t_{2},t_{1}\right)\left\|\psi\left(t_{1}\right)\right\rangle \ \ \ \ \ (16)$ We can chain propagators together to get $\displaystyle \left\|\psi\left(t_{3}\right)\right\rangle$ $\displaystyle =$ $\displaystyle U\left(t_{3},t_{2}\right)\left\|\psi\left(t_{2}\right)\right\rangle \ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U\left(t_{3},t_{2}\right)U\left(t_{2},t_{1}\right)\left\|\psi\left(t_{1}\right)\right\rangle \ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U\left(t_{3},t_{1}\right)\left\|\psi\left(t_{1}\right)\right\rangle \ \ \ \ \ (19)$ Therefore $\displaystyle U\left(t_{3},t_{1}\right)=U\left(t_{3},t_{2}\right)U\left(t_{2},t_{1}\right) \ \ \ \ \ (20)$ Since the Hermitian conjugate of a unitary operator is its inverse, we have $\displaystyle U^{\dagger}\left(t_{2},t_{1}\right)=U^{-1}\left(t_{2},t_{1}\right) \ \ \ \ \ (21)$ We can combine this with 20 to get $\displaystyle \left\|\psi\left(t_{1}\right)\right\rangle$ $\displaystyle =$ $\displaystyle I\left\|\psi\left(t_{1}\right)\right\rangle \ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U^{-1}\left(t_{2},t_{1}\right)U\left(t_{2},t_{1}\right)\left\|\psi\left(t_{1}\right)\right\rangle \ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U^{\dagger}\left(t_{2},t_{1}\right)U\left(t_{2},t_{1}\right)\left\|\psi\left(t_{1}\right)\right\rangle \ \ \ \ \ (24)$ Therefore $\displaystyle U^{\dagger}\left(t_{2},t_{1}\right)U\left(t_{2},t_{1}\right)$ $\displaystyle =$ $\displaystyle U\left(t_{1},t_{1}\right)=I\ \ \ \ \ (25)$ $\displaystyle U^{\dagger}\left(t_{2},t_{1}\right)$ $\displaystyle =$ $\displaystyle U\left(t_{1},t_{2}\right) \ \ \ \ \ (26)$ That is, the Hermitian conjugate (or inverse) of a propagator carries a state ‘backwards in time’ to its starting point. # Monty Hall problem As it’s the last day of 2016 and my mind isn’t quite into physics mode today I thought I’d post something frivolous. Here are a few thoughts on the Monty Hall problem, which has reputedly puzzled even some Nobel prize winners. (The problem also featured in episode 4×08, Skyfire Cycle, of the popular TV comedy Brooklyn Nine-Nine.) The problem is named after Monty Hall, who was the host of an American game show called Let’s Make a Deal. In the climax of each show, the winning contestant was faced with three doors. Behind two of the doors was a worthless prize (symbolized in the usual statement of the problem by a goat) while behind the third door was a major prize such as a car. The contestant initially chooses one of the three doors. At this point, Monty Hall opens one of the other two doors to reveal one of the worthless prizes (one of the goats). The contestant is then offered the chance to switch his/her choice to the other unopened door. The question is: should the contestant stay with their first choice or switch to the other door? At first glance, the answer appears obvious. After Monty opens one of the two remaining doors to reveal the location of one of the two goats, the contestant has a choice between two doors, so the odds appear to be 50-50 as to which door hides the car. Thus it would appear that there is no advantage to be gained by switching, so it doesn’t matter what the contestant does. However, this isn’t the correct way of looking at the problem. The key is that Monty knows where the car is, since he always opens a door hiding a goat. Thus if the contestant’s initial choice is wrong, they are guaranteed to win if they switch doors. The problem then reduces to: what is the probability that the contestant’s initial choice was wrong? The answer here is obvious: since the initial choice is a blind one-in-three choice, the chance that their initial choice is wrong is ${\frac{2}{3}}$. When offered the chance to switch doors, they should therefore do it, as they are twice as likely to win the car than if they stayed with their original choice. It’s surprising that this relatively simple problem has given rise to such controversy (see the Wikipedia article for details). Anyway, hopefully readers have found my posts from 2016 to be interesting and/or helpful, so here’s looking forward to a few more posts in 2017.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 591, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9613833427429199, "perplexity": 147.82001979335615}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560279933.49/warc/CC-MAIN-20170116095119-00051-ip-10-171-10-70.ec2.internal.warc.gz"}
https://talkstats.com/threads/correlated-variables.78365/#post-232827	# correlated variables #### Stat_member ##### New Member Hello, In a Cox model, if we have several variables correlated with each other, is this a problem in the results? Should we add interactions or not? What is the solution with variables correlated? Thanks Last edited: #### Karabiner ##### TS Contributor is this a problem in the results? Since your description could mean e.g. r=0.1 as well as r=0.98, and since we do not know the model, or the variables, or the research question, or the sample size, your question cannot be answered specifically. Generally speaking, it could sometimes cause interpretation problems with regard to the contribution of the respective variables, especially if one variable can be mostly explained by one or several other variables (multicollinearity). Should we add interactions or not? Without knowing the model, or the variables, or the research question, or the sample size, I find it a bit too difficult to comment on this. What is the solution with variables correlated? I am not sure what you mean. Could you elaborate this a bit more? With kind regards Karabiner	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8121845722198486, "perplexity": 1341.5855435385631}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710462.59/warc/CC-MAIN-20221128002256-20221128032256-00548.warc.gz"}
http://clay6.com/qa/51077/of-the-following-the-metals-that-cannot-be-obtained-by-electrolysis-of-the-	Comment Share Q) # Of the following,the metals that cannot be obtained by electrolysis of the aqueous solution of their salts are $\begin{array}{1 1}Ag\\Mg\text{ and }Al\\Cu\\Al\end{array}$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8610615134239197, "perplexity": 2555.020618653671}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547584328678.85/warc/CC-MAIN-20190123085337-20190123111337-00227.warc.gz"}
http://math.stackexchange.com/users/12796/victor?tab=summary	Victor Reputation 2,977 Top tag Next privilege 3,000 Rep. 3 20 63 Impact ~126k people reached ### Questions (272) 40 Why are mathematician so interested to find theory for solving partial differential equations but not for integral equation? 37 What are some theorems that currently only have computer-assisted proofs? 25 please solve a 2013 th derivative question? 20 How to prove those “curious identities”? 11 Why do mathematicians care so much about zeta functions? ### Reputation (2,977) +5 Why are mathematician so interested to find theory for solving partial differential equations but not for integral equation? +5 Find the maximum of $f(x)=x^{1/x}$ +5 What is the intuition behind the proof of Abel-Ruffini theorem in abstract algebra? -2 Are numerical approximation always possible in ODE if an solution exists? 2 Why don't they teach Fundamental Theorem of Algebra in High School? 2 Differentiation dr/dx in cylindrical coordinates from cartesian 1 How to derive an proof for this infinite square root equation? 1 Material in a first course in algebraic geometry? 1 Graph or combinatorics proof to my concern to a water, gas, electric, puzzle? ### Tags (132) 3 calculus × 55 1 geometry × 28 3 algebra-precalculus × 42 1 combinatorics × 12 3 soft-question × 25 1 plane-curves × 12 2 polynomials × 7 1 polar-coordinates × 5 2 education 1 proof-writing × 5 ### Accounts (41) Mathematics 2,977 rep 32063 Computer Science 506 rep 158 Economics 358 rep 112 Physics 341 rep 1515 Biology 320 rep 114	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8852522969245911, "perplexity": 2203.820497247757}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783395679.18/warc/CC-MAIN-20160624154955-00065-ip-10-164-35-72.ec2.internal.warc.gz"}
https://chemistry.stackexchange.com/questions/53429/is-the-behavior-of-a-dilute-strong-acid-the-same-as-the-behavior-of-a-weak-acid/53498	# Is the behavior of a dilute strong acid the same as the behavior of a weak acid? My assumption is yes, as the $\mathrm{pH}$ levels of each will be similar and thus they have the same acidity. Is this correct or am I missing something? The pH is the same, yes, but the weak acid has undissociated molecules "in reserve", so to speak, and thus can neutralize a lot more of any base. Indeed, this concept of weak acids and their conjugate bases having molecules "in reserve", reacting only when called upon, is the principle behind acid-base buffers. There are two main differences that can be observed, that Oscar's answer did not mention. Firstly, when you add a strong base to the acid solution, the solution pH changes far more drastically when you had used a strong acid, simply because the amount of acid is very little to begin with. Secondly, there is no buffer region for a strong acid solution. With a weak acid solution, at the rise in pH will slow down until a buffer region that lasts pretty much until nearly all the acid has been neutralized. For instance a weak acid HA with disassociation constant $10^{-4.75}$ at concentration $0.1 \text{ mol} \text{ dm}^{-3}$ will have a pH of about $2.88$. A semi-strong acid with disassociation constant $10^{-2}$ will need to be at concentration about $0.0015 \text{ mol} \text{ dm}^{-3}$ to have the same pH. The pH of the solution as these two solutions are titrated against a strong base BOH can be easily proven to be the logarithm of the root of some cubic equation, which will have the following familiar shapes: (This graph is when the added base is at concentration $0.2 \text{ mol} \text{ dm}^{-3}$. The blue curve is for the weak acid and the orange curve is for the semi-strong acid.) • I didn't show the graph for a strong acid because it would be squashed too much to the left and there's really nothing to see. According to Wikipedia, what I called a semi-strong acid is still considered a weak acid, though on the strong side, but anyway it is strong enough for illustrative purposes. – user21820 Jun 10 '16 at 8:13 • I am guessing the red line represents your "semi-strong" acid and the blue line represents the weak acid? It's a good idea to add it in imo – orthocresol Jun 10 '16 at 12:39 • @orthocresol: Done. I had forgotten about that! – user21820 Jun 10 '16 at 13:07 The main difference between those two kinds of acid is the way that they dissociation occurs. A strong acid will rapidly dissociate in its ions with nearly a 100% of conversion p ex: HCl. Instead weak acids like acetic acid remain in equilibrium with its ions HAc <=> H+ + Ac- This means that in order to achieve a value of pH with a weak acid you need to consider that the acid will only partially increase the concentration of H+ ions so the final concentration will be given by the equilibrium constant (Ka) for this acid. Instead the pH given by a strong acid will be given simply by the concentration of the acid. So if you have HCl with a concentration of 10^-3 mol/L then the pH of that solution will be three. I hope this was helpful :) • Just fyi, acetic acid is HOAc, AcH actually represents acetaldehyde – orthocresol Jun 10 '16 at 12:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8842092156410217, "perplexity": 921.0507196950298}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027318011.89/warc/CC-MAIN-20190823062005-20190823084005-00164.warc.gz"}
http://mathhelpforum.com/algebra/118484-proving-inequality.html	Math Help - proving an inequality 1. proving an inequality Ok at first glance this seemed pretty easy, but I'm having a lot trouble with it. Let a,b be positive integers. Show that $\sqrt{2}$ always lies between the two fractions a/b and (a+2b)/(a+b). Which fraction is closer to $\sqrt{2}$. Any help or hint would be appreciated. 2. Hello bleys Welcome to Math Help Forum! Originally Posted by bleys Ok at first glance this seemed pretty easy, but I'm having a lot trouble with it. Let a,b be positive integers. Show that $\sqrt{2}$ always lies between the two fractions a/b and (a+2b)/(a+b). Which fraction is closer to $\sqrt{2}$. Any help or hint would be appreciated. Here's a start. Suppose first that $\frac{a}{b}<\sqrt2$ Then since $b>0,\; a<\sqrt2b$ $\Rightarrow a+b < (1+\sqrt2)b$ $\Rightarrow \frac{a+2b}{a+b}=1+\frac{b}{a+b}>1+\frac{b}{(\sqrt 2+1)b}=\frac{\sqrt2+2}{\sqrt2+1}=\frac{\sqrt2(1+\s qrt2)}{1+\sqrt2}=\sqrt2$ So $\frac{a}{b}<\sqrt2 < \frac{a+2b}{a+b}$ Do you want to see if you can take it from here? 3. Welcome to Math Help Forum! Thanks! Oh alright, I need to learn how to manipulate surds better... The case where $\frac{a}{b}>\sqrt2$ just requires the inequality signs switched. Now, recover the first case where $\frac{a}{b}<\sqrt2 < \frac{a+2b}{a+b}$ Suppose if possible that $\|\frac{a+2b}{a+b} - \sqrt2 \| < \frac{a}{b} - \sqrt2$ Then $\frac{a+2b}{a+b} < \frac{a}{b}$, a contradiction. Then $\|\frac{a+2b}{a+b} - \sqrt2 \| > \|\frac{a}{b} - \sqrt2\|$ A similar argument can be applied to the other case. Is there a direct way to show this? I find using contradiction to be a little like cheating in this problem . The reason I didn't was that I wasn't sure my inequalities were valid when trying to recover the absolute value sign. 4. Hello bleys For the last part, can't we consider the differences between $\left(\frac{a}{b}\right)^2$ and $2$, and $\left(\frac{a+2b}{a+b}\right)^2$ and $2$? Which are: $\left\|\left(\frac{a}{b}\right)^2 - 2\right\|=\left\|\frac{a^2-2b^2}{b^2}\right\|$ and $\left\|\left(\frac{a+2b}{a+b}\right)^2-2\right\| =\left\|\frac{(a+2b)^2-2(a+b)^2}{(a+b)^2}\right\|$ $=\left\|\frac{2b^2-a^2}{(a+b)^2}\right\|$ Clearly the second of these expressions is less than the first. So $\frac{a+2b}{a+b}$ is closer to $\sqrt2$. (Incidentally, doesn't the fact that $a^2-2b^2$ and $2b^2-a^2$ have opposite signs prove the first part of the question as well?) $\|a\|\le\|b\|\text{ if and only if }a^2\le b^2$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 26, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9535282254219055, "perplexity": 267.51259968904134}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042982013.25/warc/CC-MAIN-20150728002302-00066-ip-10-236-191-2.ec2.internal.warc.gz"}
http://crypto.stackexchange.com/questions/14423/signature-based-on-public-key-cryptography-and-forgery	# Signature based on public key cryptography and forgery In the definition of existential unforgeability, there is no detail about the following questions. In general, can we suppose that a signer is also a possible adversary ? When generating a signature, can we suppose that the signer behaves well ? More precisely, suppose that a signer is able to generate a signature which is valid for an other document of his choice. Despite this, for a correctly generated signature, suppose that the signature scheme used is secure when considering an adversary which is not the signer, that is to say, this adversary is unable to generate a signature for any other document. Is this signature scheme really considered secure ? What are the standard (conventional) hypothesis about that ? I update in reaction to the comment: In fact, 1. if the signer behaves well, then the signature is unforgeable (for the signer or any adversary which is not the signer). 2. if the signer misbehaves, then the signature is forgeable (for the signer, without using the key, and for any adversary). Is this king of scheme considered secure in the sense of existential forgeability ? can we suppose that the signer behaves well ? - It depends on what you mean. If the secret key is necessary to find the second document, then it is fine. If you know the secret key, you can already sign anything you want. However, if it is possible to find the second document given only the original message, signature and public key, then the scheme is trivially forgeable. – Maeher Feb 10 at 19:33 Thank you @Maeher. The secret key is \textbf{not} necessary to find the second document. – Dingo13 Feb 10 at 19:43 @Dingo13 what do you mean by "misbehaves"? Btw. didn't you ask a quite similar question already? – DrLecter Feb 10 at 20:02 There are versions of signatures in which the signer is a possible adversary, although that property is not called unforgeability. $\:$ See fail-stop signatures, undeniable signatures, and unconditionally secure signatures. $\;\;\;$ – Ricky Demer Feb 10 at 22:19 @fgrieu this seems only to be a desirable property in some scenarios as Ricky pointed out. I am not aware of any rigorous formal definition of such a property for conventional digital signatures. At least for plain hash-then-sign signatures such a signer who efficiently can do this, can be used as oracle to break the used hash function. – DrLecter Feb 13 at 8:36 The standard definition of existential forgery allows the adversary to ask and obtain the signature of any message she wants, and claim success if she can exhibit (with sizable odds) any acceptable (message, signature) pair, for any message for which she did not ask signature. Update: There is also strong existential unforgeability, where the adversary should not be able to exhibit any acceptable (message, signature) pair for which she did not obtain that signature by asking for the signature of that message. For a use case making the headlines where that strong makes a huge difference, see these links. Note: in a signature scheme with message recovery, any fraction of the message embedded in the signature needs not be exhibited by the adversary, and is as obtained by the verifier for the purpose of comparison to what the adversary submitted. Corrected: If "a signer is able to generate a signature which is valid for an other document (that is, message) of his choice", then that scheme is vulnerable to existential forgery. The adversary obtains the signature of the first message, submits that signature unchanged together with the second message, and wins with 100% odds the existential forgery game. Oh no, only the signer could do this, using access to the private key! Existential forgery is the strongest a strong theoretical criteria for signature. However, it assumes that the signer does not misbehave beside allowing the signature of arbitrary messages; in particular, it is assumed that the signer (or the Smart Card used for signature) does not leak the private key only uses the private key as prescribed, and properly implements every step in the algorithm, like generating truly random numbers. Update: This spot-on comment by Ricky Demer gives names of security criteria for signatures protected from some attacks by the signer / private key holder. Update: as pointed by DrLecter in a comment, the standard way to model the signer for a scheme secure against existential forgery (we also say: secure under chosen message attack) is as an oracle that accepts any message and outputs its signature. That oracle is assumed to implement the signature scheme exactly as specified. For more details, I refer to DrLecter's thesaurus of signature security models. - By modeling the signer as a signing oracle which can be queried by the adversary. – DrLecter Feb 10 at 20:13 @Dingo13 that depends on the model you are working in. if you consider existential unforgeability against (adaptively) chosen message attacks (the standard and strongest security notion), then the adversary can choose an arbitrary message of its choice and submit it to the oracle. if you consider existential unforgeability against random message attacks, then the oracle randomly (typically unfiformly at random) samples a message for the adversary, i.e., the message can not be controlled by the adversary. In any case, the oracle returns an honestly computed signature for the respective message. – DrLecter Feb 10 at 20:21 cont. you may also look at this related answer which makes the attack/model stuff more explicit. – DrLecter Feb 10 at 20:27 "In any case, the oracle returns an honestly computed signature for the respective message." If the signing oracle seems perfect, in practice, if the signer has to choose pseudorandom numbers, he could choose numbers of his choice, and in this case the scheme is no secure. Not ? What kind of model I have to choose in this case ? – Dingo13 Feb 10 at 20:28 @Dingo If you take ElGamal or Schnorr signatures, producing two signatures with respect to the same randomness for different messages allows to extract the secret signing key. You are right, that is a problem, but it is not covered by any unforgeability notion. – DrLecter Feb 10 at 20:34 As already discussed by @fgrieu in his answer and myself in the comments of your question and his answer, the standard notion of security of digital signature schemes, namely (strong) existential unforgeability under adaptively chosen message attacks (UF-CMA), does not cover the case you are concerned about. At least for hash-then-sign signatures built upon a trapdoor permutation (as RSA), such a signer who efficiently can do this, can be used as an oracle to break the collision resistance of the used hash function. Nevertheless, it kept me thinking if there is a standard notion which could be applied to any signature scheme and if this has been considered somewhere. In the Crypto'02 paper Flaws in Applying Proof Methodologies to Signature Schemes, Stern et al. introduce a property which covers exactly the aspect you are targeting in your question (however this is not a commonly investigated property in the design of signature schemes). Essentially, it is a formalization of the non-repudiation property, which requires that an adversary that is in possession of the secret signing key (and potentially is also able to influence the key generation process) will not be able to produce two messages with the same signature, a so called duplicate signature in their language. Consequently, this attack indeed considers the signer as an adversary. In Section 4 of the paper, they provide an attack on ECDSA which produces such a duplicate signature, i.e., a signature valid for two distinct messages. if the adversary can control the key generation process used to produce the actual signing key. Just as a side note, Stern et al. in the same paper also introduce a property denoted as malleability, which is today known strong unforgeability, i.e., it ensures that an adversary (holding the public key) cannot even produce a new signature for a previously obtained signed message. This notion for instance cannot be satisfied by randomizable signatures (such as Camenisch Lysyanskaya signatures), which allow to transform a signature $\sigma$ for some message $m$ into another signature $\sigma'\neq \sigma$ for the same $m$ without requiring the secret signing key. - Thanks @DrLecter, this notion of (non ?)-malleability from Stern (also known strong unforgeability) is the same notion as UF-CMA ? – Dingo13 Feb 27 at 8:48 @Dingo13 The standard UF-CMA notion could be called weak unforgeability, as the attackers goal is to output a signature for a message for which he has not already seen a signature. In strong unforgeability, the adversary already suceeds when producing a signature that has not been produced before (this forgery can be for a message for which he has already seen a signature before). So it is a stronger notion as the usual UF-CMA security notion. – DrLecter Feb 27 at 8:51 Thanks @DrLecter, the term strong unforgeability is quite recent... This notion of "non-malleability"(="strong UF-CMA") is always valid in secret key settings ? – Dingo13 Feb 27 at 8:58 @Dingo13 yes. What do you mean by "same secret key settings"? – DrLecter Feb 27 at 8:59 I mean that for secret key cryptography the notions used are the same, and we have always "non-malleability"="strong UF-CMA" – Dingo13 Feb 27 at 9:02	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9004772901535034, "perplexity": 1468.9204566454696}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609539665.16/warc/CC-MAIN-20140416005219-00190-ip-10-147-4-33.ec2.internal.warc.gz"}
http://physics.stackexchange.com/questions/27116/physical-interpretation-of-superstrings?answertab=oldest	# Physical interpretation of superstrings The scalar fields $X^\mu$ in bosonic string theory have a clear physical interpretation - they describe the embedding of the string in spacetime. Adding fermionic fields on the worldsheet is a generalization for sure, gives fermions in the spectrum, has a smaller critical dimension and no tachyons, that's all good - but I don't see how they can have any physical interpretation as nice as the above for the scalars - isn't everything about how a string moves in spacetime described by the $X^\mu$ part? - Btw, there's no necessity for a clear geometric interpretation in string theory. As far as I understand, any 2D CFT with vanishing central charge can be regarded as a string theory and some CFTs have no clear interpretation as a sigma model. – Squark Dec 15 '11 at 12:10 The worldsheet fermions have to do with internal degrees of freedom, namely the spin -- therefore better name for the superstring is the more old-fashioned "spinning string" (since worldsheet SUSY should not be confused with spacetime SUSY). The worldsheet fermions generate multiplets of some internal symmetry group. If you want those internal degrees of freedom generated by WS fermions to transform under spacetime Lorentz Transformations, rather than an independent internal symmetry, you need to correlate the Lorentz transformations of the worldsheet bosons and fermions. This is what worldsheet SUSY does for you. All of this is not specific to string theory. If you want to first-quantize a field theory, a "bosonic" worldline theory will give you a (free) scalar field theory. Adding fermions and the corresponding worldline supersymmetries will generate (free) higher spin fields. It is probably a useful exercise to get e.g. classical (free) Maxwell field from a (N=2 SUSY) worldline theory in order to appreciate precisely what the worldsheet structures mean precisely. Wish I had a good reference, but maybe someone can help me out. - I think there's also an alternative point of view, namely that the string lives in a superspace and the fermions are the odd coordinates – Squark Dec 11 '11 at 21:32 Yes, that is the Green-Schwarz formulation of spacetime supersymmetric strings. But I think the question was about the slightly more familiar R-NS string, and in particular spacetime SUSY is not implied. – user566 Dec 11 '11 at 21:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8517131209373474, "perplexity": 484.53160518807283}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860125857.44/warc/CC-MAIN-20160428161525-00211-ip-10-239-7-51.ec2.internal.warc.gz"}
http://www.wias-berlin.de/publications/wias-publ/run.jsp?template=abstract&type=Preprint&year=&number=2562	WIAS Preprint No. 2562, (2018) # Surface induced phase separation of a swelling hydrogel Authors • Hennessy, Matthew G. • Münch, Andreas • Wagner, Barbara 2010 Mathematics Subject Classification • 74H10 82C26 2008 Physics and Astronomy Classification Scheme • 83.80.Rs, 83.10.Tv Keywords • Hydrogel, asymptotic analysis, nonlinear elasticity, phase transition DOI 10.20347/WIAS.PREPRINT.2562 Abstract We present a formulation of the free boundary problem for a hydrogel that accounts for the interfacial free energy and finite strain due to the large deformation of the polymer network during solvent transport across the free boundary. For the geometry of an initially dry layer fixed at a rigid substrate, our model predicts a phase transition when a critical value of the solvent concentration has been reached near the free boundary. A one-dimensional case study shows that depending on the flux rate at the free boundary an initial saturation front is followed by spinodal decomposition of the hydrogel and the formation of an interfacial front that moves through the layer. Moreover, increasing the shear modulus of the elastic network delays or even suppresses phase separation. Download Documents	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8150877356529236, "perplexity": 2398.3859075614764}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560628000894.72/warc/CC-MAIN-20190627055431-20190627081431-00540.warc.gz"}
http://slideplayer.com/slide/4379665/	# Entropy in the Quantum World Panagiotis Aleiferis EECS 598, Fall 2001. ## Presentation on theme: "Entropy in the Quantum World Panagiotis Aleiferis EECS 598, Fall 2001."— Presentation transcript: Entropy in the Quantum World Panagiotis Aleiferis EECS 598, Fall 2001 Outline Entropy in the classic world Theoretical background – Density matrix – Properties of the density matrix – The reduced density matrix Shannon’s entropy Entropy in the quantum world – Definition and basic properties – Some useful theorems Applications – Entropy as a measure of entanglement References Entropy in the classic world Murphy’s Laws 1 st law of thermodynamics: 2 nd law of thermodynamics: “There is some degradation of the total energy U in the system, some non-useful heat, in any thermodynamic process.” Rudolf Clausius (1822 - 1988) ΔWΔWΔQΔQ ΔUΔU Why does heat always flow from warm to cold? Ludwig Boltzmann (1844 - 1906) “When energy is degraded, the atoms become more disordered, the entropy increases!” “At equilibrium, the system will be in its most probable state and the entropy will be maximum.” The more disordered the energy, the less useful it can be! Boltzmann statistics – 5 dipoles in external field General Relations of Boltzmann statistics – For a system in equilibrium at temperature T: – Statistical entropy: Theoretical Background The density matrix ρ NOT – In most cases we do NOT completely know the exact state of the system. We can estimate the probabilities P i that the system is in the states \|ψ i >. – Our system is in an “ensemble” of pure states {P i,\|ψ i >}. tr(ρ)=1 Define: Properties of the density matrix – tr(ρ)=1 – ρ is a positive operator (positive, means is real, non-negative, ) – if a unitary operator U is applied, the density matrix transforms as: pure state – ρ corresponds to a pure state, if and only if: mixed state – ρ corresponds to a mixed state, if and only if: diagonal – if we choose the energy eigenfunctions for our basis set, then H and ρ are both diagonal, i.e. symmetric – in any other representation ρ may or may not be diagonal, but generally it will be symmetric, i.e. Detailed balance Detailed balance is essential so that equilibrium is maintained (i.e. probabilities do NOT explicitly depend on time). The reduced density matrix – What happens if we want to describe a subsystem of the composite system? – Divide our system AB into parts A, B. – Reduced density matrix for the subsystem A: where tr B : “partial trace over subsystem B” trace over subspace of system B Shannon’s entropy Definition – How much information we gain, on average, when we learn the value of a random variable X? OR equivalently, What is the uncertainty, on average, about X before we learn its value? – If {p 1, p 2, …,p n } the probability distribution of the n possible values of X: – By definition: 0log 2 0 = 0 (events with zero probability do not contribute to entropy.) – Entropy H(X) depends only on the respective probabilities of the individual events X i ! – Why is the entropy defined this way? It gives the minimal physical resources required to store information so that at a later time the information can be reconstructed. “Shannon’s noiseless coding theorem”. - “Shannon’s noiseless coding theorem”. – Example of Shannon’s noiseless coding theorem Code 4 symbols {1, 2, 3, 4} with probabilities {1/2, 1/4, 1/8, 1/8}. Code without compression: But, what happens if we use this code instead? Average string length for the second code: Note:!!! Note: !!! Joint and Conditional Entropy – A pair (X,Y) of random variables. – Joint entropy of X and Y: – Entropy of X conditional on knowing Y: Mutual Information – How much do X, Y have in common? – Mutual information of X and Y: –, equality when Y= f(X) – Subadditivity:, independent equality when X, Y are independent variables. H(X\|Y) H(X)H(Y) H(Y\|X)H(Y:X) Entropy in the quantum world Von Neumann’s entropy – Probability distributions replaced by the density matrix ρ. Von Neumann’s definition: – If λ i are the eigenvalues of ρ, use the equivalent definition: Basic properties of Von Neumann’s entropy –, equality if and only if in “pure state”. – In a d-dimensional Hilbert space:, completely mixed state the equality if and only if in a completely mixed state, i.e. – If system AB in a “pure state”, then: – Triangle inequality and subadditivity: with Both these inequalities hold for Shannon’s entropy H. – Strong subadditivity First inequality also holds for Shannon’s entropy H, since: BUT BUT, for Von Neumann’s entropy it is possible that: nature “conspires” so that both of these inequalities are NOT true simultaneously! However, somehow nature “conspires” so that both of these inequalities are NOT true simultaneously! Applications Entropy as a measure of entanglement – Entropy is a measure of the uncertainty about a quantum system before we make a measurement of its state. – For a d-dimensional Hilbert space: Pure stateCompletely mixed state – Example: Consider two 4-qbit systems with initial states: Which one is more entangled ? randomizes – Partial measurement randomizes the initially pure states. – The entropy of the resulting mixed states measures the amount of this randomization! largermore randomized more entangled – The larger the entropy, the more randomized the state after the measurement is, the more entangled the initial state was! – We have to go through evaluating the density matrix of the randomized states: – System 1: Trace over (any) 1 qbit: Trace over (any) 2 qbits: Pure state λ 1,2 =0, λ 3,4 =1/2 Trace over (any) 3 qbits: Summary: 1. initially 2. measure (any) 1 qbit 3. measure (any) 2 qbits 4. measure (any) 3 qbits λ 1,2 =1/2 – System 2: Trace over (any) 1 qbit: diagonal Trace over (any) 2 qbits: Trace over (any) 3 qbits: λ 1 =0, λ 2,3 =1/6, λ 4 =2/3 λ 1,2 =1/2 Summary: 1. initially 2. measure (any) 1 qbit 3. measure (any) 2 qbits 4. measure (any) 3 qbits ψ 2 is more entangled than ψ 1 Therefore, ψ 2 is more entangled than ψ 1. “Ludwin Boltzmann, who spent much of his life studying statistical mechanics, died in 1906, by his own hand. Paul Ehrenfest, carrying on the work, died similarly in 1933. Now it is our turn to study statistical mechanics.” “Ludwin Boltzmann, who spent much of his life studying statistical mechanics, died in 1906, by his own hand. Paul Ehrenfest, carrying on the work, died similarly in 1933. Now it is our turn to study statistical mechanics.” - “States of Matter”, D. Goodstein - “States of Matter”, D. Goodstein References “Quantum Computation and Quantum Information”, Nielsen & Chuang, Cambridge Univ. Press, 2000 “Quantum Mechanics”, Eugen Merzbacher, Wiley, 1998 Lecture notes by C. Monroe (PHYS 644, Univ. of Michigan) coursetools.ummu.umich.edu/2001/fall/physics/644/001.nsf Lecture notes by J. Preskill (PHYS 219, Caltech) www.theory.caltech.edu/people/preskill/ph229 Download ppt "Entropy in the Quantum World Panagiotis Aleiferis EECS 598, Fall 2001." 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https://www.physicsforums.com/threads/help-me-please.78060/	1. Jun 5, 2005 scubasteve_27 been stuck on this all day Design a test to determine whether the charges on an electron and a proton are exactly equal. any help would be good 2. Jun 5, 2005 Poncho Just measure the repulsive force between two protons and then measure the repulsive force between two electrons. Compare. Do you actually have to do this experiment? -Poncho 3. Jun 6, 2005 primarygun Design an experiment to show that a hydrogen atom is neutral!	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8703992962837219, "perplexity": 833.4417019431546}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267868135.87/warc/CC-MAIN-20180625150537-20180625170537-00433.warc.gz"}
https://gmatclub.com/forum/is-x-175419.html	GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 19 Oct 2019, 07:52 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Is \|x\| < 1 ? Author Message TAGS: ### Hide Tags Intern Joined: 24 May 2014 Posts: 13 Location: Brazil Is \|x\| < 1 ? [#permalink] ### Show Tags Updated on: 31 Jul 2014, 10:12 1 25 00:00 Difficulty: 95% (hard) Question Stats: 40% (02:22) correct 60% (02:21) wrong based on 339 sessions ### HideShow timer Statistics Is \|x\| < 1 ? (1) x/\|x\| < x (2) x/\|x\| < 1 Originally posted by Reni on 31 Jul 2014, 09:55. Last edited by Bunuel on 31 Jul 2014, 10:12, edited 1 time in total. Edited the question. Math Expert Joined: 02 Sep 2009 Posts: 58446 Re: Is \|x\| < 1 ? [#permalink] ### Show Tags 31 Jul 2014, 10:32 7 5 Is \|x\| < 1 ? Is \|x\| < 1 --> is -1 < x < 1. (1) x/\|x\| < x. Two cases: A. $$x<0$$ --> $$\frac{x}{-x}<x$$ --> $$-1<x$$. Since we consider the case when $$x<0$$, then we'd have $$-1<x<0$$. Answer YES. B. $$x>0$$ --> $$\frac{x}{x}<x$$ --> $$1<x$$. Answer NO. Not sufficient. (2) x/\|x\| < 1. This simply means that x is a negative number: if x is positive then x/\|x\| = 1, if x is negative x/\|x\| = -1 < 1. Not sufficient. (1)+(2) Since from (2) we have that x is negative, then we have case A from (1), which gives an YES answer to the question. Sufficient. Please tag the questions properly (this is NOT an algebra question) and copy the questions EXACTLY as they appear in the source. _________________ ##### General Discussion Manager Joined: 31 Jul 2014 Posts: 126 GMAT 1: 630 Q48 V29 Re: Is \|x\| < 1 ? [#permalink] ### Show Tags 01 Sep 2015, 08:52 Bunuel wrote: Is \|x\| < 1 ? Is \|x\| < 1 --> is -1 < x < 1. (1) x/\|x\| < x. Two cases: A. $$x<0$$ --> $$\frac{x}{-x}<x$$ --> $$-1<x$$. Since we consider the case when $$x<0$$, then we'd have $$-1<x<0$$. Answer YES. B. $$x>0$$ --> $$\frac{x}{x}<x$$ --> $$1<x$$. Answer NO. Not sufficient. (2) x/\|x\| < 1. This simply means that x is a negative number: if x is positive then x/\|x\| = 1, if x is negative x/\|x\| = -1 < 1. Not sufficient. (1)+(2) Since from (2) we have that x is negative, then we have case A from (1), which gives an YES answer to the question. Sufficient. Please tag the questions properly (this is NOT an algebra question) and copy the questions EXACTLY as they appear in the source. Hi Bunuel I fail to understand above solution. We have to find whether -1< x<1 From 1 ) we know that x/\|x\| < x holds true for -1<x<0 and x>1 so we know that x is not in range 0<x<1 is this not sufficient to answer NO ....? Again from 2) We know -infinity < x<0 then answer is NO Sorry, could you please tell me what am I missing...how are you getting YES and NO for each? CEO Joined: 20 Mar 2014 Posts: 2597 Concentration: Finance, Strategy Schools: Kellogg '18 (M) GMAT 1: 750 Q49 V44 GPA: 3.7 WE: Engineering (Aerospace and Defense) Re: Is \|x\| < 1 ? [#permalink] ### Show Tags 01 Sep 2015, 09:24 1 Bunuel wrote: Is \|x\| < 1 ? Is \|x\| < 1 --> is -1 < x < 1. (1) x/\|x\| < x. Two cases: A. $$x<0$$ --> $$\frac{x}{-x}<x$$ --> $$-1<x$$. Since we consider the case when $$x<0$$, then we'd have $$-1<x<0$$. Answer YES. B. $$x>0$$ --> $$\frac{x}{x}<x$$ --> $$1<x$$. Answer NO. Not sufficient. (2) x/\|x\| < 1. This simply means that x is a negative number: if x is positive then x/\|x\| = 1, if x is negative x/\|x\| = -1 < 1. Not sufficient. (1)+(2) Since from (2) we have that x is negative, then we have case A from (1), which gives an YES answer to the question. Sufficient. Please tag the questions properly (this is NOT an algebra question) and copy the questions EXACTLY as they appear in the source. Hi Bunuel I fail to understand above solution. We have to find whether -1< x<1 From 1 ) we know that x/\|x\| < x holds true for -1<x<0 and x>1 so we know that x is not in range 0<x<1 is this not sufficient to answer NO ....? Again from 2) We know -infinity < x<0 then answer is NO Sorry, could you please tell me what am I missing...how are you getting YES and NO for each? Let me try to explain. From 1, x/\|x\| < x ---> 2 cases: --- when x $$\geq$$0 --> \|x\| = x ---> x/x < x --> x > 1 . Thus a "no" for is \|x\| < 1 --- when x < 0 --> \|x\| = -x ---> x/-x < x --> x > -1 . For x = 0.5, then a "yes" for \|x\|<1 but if x = 5, then a "no" for is \|x\|<1. Thus NOT sufficient. From 2, x/\|x\| < 1 , again 2 cases, --- when x $$\geq$$0 --> \|x\| = x ---> x/x < 1 --> 1<1 . Thus x $$\geq$$0 is not a possible scenario. --- when x <0 --> \|x\| = -x ---> x/-x < 1 --> 1>-1 . This is true for ALL x and thus x <0 satisfies this. For x = -0.5, you get a "yes" for is \|x\|<1 but for x = -4, you get a "no" for is \|x\| <1 . Thus NOT sufficient. Alternately, if you want you can also test cases to come up with ranges for x. But this might be a bit more time consuming. For your analysis of statement 1, the text in red is not correct. Refer to the explanation mentioned above. For statement 2, you are correct that x<0 but you need to check whether -1<x<0 is satisfied or not. You get 2 different answers when you take x = -0.5 or x = -4. Hope this helps. Intern Joined: 01 Jul 2015 Posts: 11 Is \|x\| < 1 ? [#permalink] ### Show Tags 01 Sep 2015, 10:54 1 I have tried to use the line method to solve this problem. 1. x/!x!<xSince !x! is always positive, I can multiply it each side and get x<x.!x! and after substracting x-x.!x!<0 Now taking x as common factor, x(1-!x!)<0 x will have three factors:0,1,-1 Now refer to the attached figure to get the region where the given condition will be true. Since the inequality is of "less than 0" type, we shold consider "negative portion" we will get two possibilities which are -1<x<0 and x>1. Hence not suffecient. 2. x/!x!<1. Repeating the same logic above we get x-!x!<0. This will be true only if x<0. Still not suffeceint. Combining both statements, we can eliminate the portion x>1, which we got in statement 1. Leaving us only with -1<x<0, which tells us that !x! will always be a fraction less than 1. Hence suffeceint. Attachments 1.JPG [ 3.48 KiB \| Viewed 3586 times ] Manager Joined: 19 Aug 2016 Posts: 75 Re: Is \|x\| < 1 ? [#permalink] ### Show Tags 08 Oct 2017, 20:06 Engr2012 wrote: Bunuel wrote: Is \|x\| < 1 ? Is \|x\| < 1 --> is -1 < x < 1. (1) x/\|x\| < x. Two cases: A. $$x<0$$ --> $$\frac{x}{-x}<x$$ --> $$-1<x$$. Since we consider the case when $$x<0$$, then we'd have $$-1<x<0$$. Answer YES. B. $$x>0$$ --> $$\frac{x}{x}<x$$ --> $$1<x$$. Answer NO. Not sufficient. (2) x/\|x\| < 1. This simply means that x is a negative number: if x is positive then x/\|x\| = 1, if x is negative x/\|x\| = -1 < 1. Not sufficient. (1)+(2) Since from (2) we have that x is negative, then we have case A from (1), which gives an YES answer to the question. Sufficient. Please tag the questions properly (this is NOT an algebra question) and copy the questions EXACTLY as they appear in the source. Hi Bunuel I fail to understand above solution. We have to find whether -1< x<1 From 1 ) we know that x/\|x\| < x holds true for -1<x<0 and x>1 so we know that x is not in range 0<x<1 is this not sufficient to answer NO ....? Again from 2) We know -infinity < x<0 then answer is NO Sorry, could you please tell me what am I missing...how are you getting YES and NO for each? Let me try to explain. From 1, x/\|x\| < x ---> 2 cases: --- when x $$\geq$$0 --> \|x\| = x ---> x/x < x --> x > 1 . Thus a "no" for is \|x\| < 1 --- when x < 0 --> \|x\| = -x ---> x/-x < x --> x > -1 . For x = 0.5, then a "yes" for \|x\|<1 but if x = 5, then a "no" for is \|x\|<1. Thus NOT sufficient. From 2, x/\|x\| < 1 , again 2 cases, --- when x $$\geq$$0 --> \|x\| = x ---> x/x < 1 --> 1<1 . Thus x $$\geq$$0 is not a possible scenario. --- when x <0 --> \|x\| = -x ---> x/-x < 1 --> 1>-1 . This is true for ALL x and thus x <0 satisfies this. For x = -0.5, you get a "yes" for is \|x\|<1 but for x = -4, you get a "no" for is \|x\| <1 . Thus NOT sufficient. Alternately, if you want you can also test cases to come up with ranges for x. But this might be a bit more time consuming. For your analysis of statement 1, the text in red is not correct. Refer to the explanation mentioned above. For statement 2, you are correct that x<0 but you need to check whether -1<x<0 is satisfied or not. You get 2 different answers when you take x = -0.5 or x = -4. Hope this helps. Statement 2 Aren't we told that x lies between 1 and -1 ? then why are we considering x=-4 Pls help Senior Manager Joined: 10 Apr 2018 Posts: 266 Location: United States (NC) Re: Is \|x\| < 1 ? [#permalink] ### Show Tags 18 Sep 2018, 15:33 Hi, There is similar question out there https://gmatclub.com/forum/if-x-is-not- ... 86140.html Now here Bunuel has explained two approaches for solving the question https://gmatclub.com/forum/if-x-is-not- ... 86140.html Hope this helps Probus _________________ Probus ~You Just Can't beat the person who never gives up~ Babe Ruth SVP Joined: 03 Jun 2019 Posts: 1724 Location: India Re: Is \|x\| < 1 ? [#permalink] ### Show Tags 24 Sep 2019, 09:10 Reni wrote: Is \|x\| < 1 ? (1) x/\|x\| < x (2) x/\|x\| < 1 Is \|x\| < 1 ? (1) x/\|x\| < x $$\frac{x}{\|x\|}- x <0$$ $$\frac{x - x\|x\|}{\|x\|} < 0$$ Since \|x\| > 0 x - x\|x\| < 0 x (1- \|x\|) < 0 x (\|x\| -1) > 0 If x>0; \|x\| > 1; x>1 If x<0; \|x\| < 1; -1<x<0 Either x>1 or -1<x<0 NOT SUFFICIENT (2) x/\|x\| < 1 x/\|x\| - 1 < 0 x - \|x\| < 0 If x>0; x-x<0; If x<0; x + x < 0; x<0 \|x\| may or may not be <1 NOT SUFFICIENT (1) + (2) (1) x/\|x\| < x If x>0; \|x\| > 1; x>1 If x<0; \|x\| < 1; -1<x<0 (2) x/\|x\| < 1 x<0 -1<x<0 \|x\|<1 SUFFICIENT IMO C _________________ "Success is not final; failure is not fatal: It is the courage to continue that counts." Please provide kudos if you like my post. Kudos encourage active discussions. My GMAT Resources: - Efficient Learning All you need to know about GMAT quant Tele: +91-11-40396815 Mobile : +91-9910661622 E-mail : [email protected] Re: Is \|x\| < 1 ? [#permalink] 24 Sep 2019, 09:10 Display posts from previous: Sort by	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8814588189125061, "perplexity": 2727.300730690556}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986696339.42/warc/CC-MAIN-20191019141654-20191019165154-00159.warc.gz"}
http://math.stackexchange.com/questions/145688/value-of-sum-n-0-infty-1n/145690	# Value of $\sum_{n=0}^\infty (-1)^n$ I saw this equation on a blackboard today: $\displaystyle \sum_{n=0}^\infty (-1)^n$ It got me thinking -- this must oscillate between $1$ and $0$, yes? So then does this sum even have a meaningful value? - yes, the sequence of partial sum is not convergent. It does'nt converge. – matgaio May 16 '12 at 0:12 What you say you saw on the blackboard is not an equation. If you want a broad generic term, you could write "I saw this expression on the blackboard today:". Or of course you could just say "I saw this sum on the blackboard today:". – Michael Hardy May 16 '12 at 1:19 – Argon May 16 '12 at 2:17 This sum is called Grandi's series and does not converge in the usual sense. By usual sense, we mean that if you look at the sequence of partial sums then we get that $$S_n = \sum_{k=0}^{n}(-1)^k = \begin{cases} 1 & \text{if n is even},\\ 0 & \text{if n is odd}.\end{cases}$$ Hence, $\displaystyle \lim_{n \rightarrow \infty} S_n$ does not exist. That said, convergence of partial sums is by no means the only way to define convergence. Another popular way of defining convergence is to look at the Cesàro sum. The Cesàro sum of the Grandi's series is $1/2$. The Cesàro sum is defined as the limit of the average of the sequence of partial sums i.e. if we let $$\tilde{S}_n = \dfrac{\displaystyle\sum_{m=0}^{n} S_m}{n+1},$$ then we have that $$\tilde{S}_n = \begin{cases} \frac12 & \text{if n is odd},\\ \frac12 + \frac1{2n+2}& \text{ if n is even},\end{cases}$$ which converges to $\dfrac12$. There are also other ways to interpret the value of $\dfrac12$. For instance, if we were to randomly choose a natural number and if we assign the probability of getting an odd number to be $\dfrac12$ and the probability of getting an even number to be $\dfrac12$, then the expected value of the sum $S_n$ is $\dfrac12$. Another interpretation is through analytic continuation of the function $$f(x) = 1-x+x^2 - x^3 + \cdots, \, \lvert x \rvert < 1.$$ The function converges to $\dfrac1{1+x}$ on the interval $(-1,1)$. Not surprisingly, the analytic continuation of the function $f(x)$ on the entire complex plane is given by $f_{\text{ac}}(x) = \dfrac1{1+x}$, where $x \in \mathbb{C}\backslash \{-1\}$. Hence, if we plugin the value of $x=1$ in $f_{\text{ac}}(x)$, we get the value of $\frac12$. Hence, $$1 - 1 + 1 -1 + \cdots \underset{\text{ac}}{=} \frac1{2}.$$ There are also other regularization techniques like Borel summation, Ramanujan summation which assign a finite value to sums which do not converge. All these different techniques assign a value of $\dfrac12$ for the Grandi's series. As @anon points out, the $\zeta$ regularization technique (which is actually closely related to Ramanujan summation technique) can also be used here to get the value of $1-1+1-1+\cdots$. Consider $$g(s) = 1 - \frac1{2^s} + \frac1{3^s} - \frac1{4^s} + \cdots$$ and $$f(s) = 1 + \frac1{2^s} + \frac1{3^s} + \frac1{4^s} + \cdots$$ The series $g(s)$ converges for Real$(s) > 0$ and converges absolutely for Real$(s) >1$. The series $f(s)$ converges for Real$(s) > 1$. In the region, Real$(s)>1$, we have that $$g(s) = 1 - \frac1{2^s} + \frac1{3^s} - \frac1{4^s} + \cdots = \left(1 + \frac1{2^s} + \frac1{3^s} + \frac1{4^s} + \cdots \right) - \left(\frac2{2^s} + \frac2{4^s} + \frac2{6^s} + \frac2{8^s} + \cdots \right) = f(s) - \frac1{2^{s-1}} f(s) = \left(1 - 2^{1-s} \right)f(s).$$ Now analytic continuation of $g(s)$ gives us $\eta(s)$ such that $\eta(s) = (1-2^{1-s}) \zeta(s)$, where $\zeta(s)$ is the analytic continuation of $f(s)$. Hence, plugging in $s=0$, we get that $$1 - 1 + 1 - 1 +\cdots \underset{\text{ac}}{=} \eta(0) = (1-2) \zeta(0) = -\zeta(0) = \frac12,$$ since the value of $\zeta(0) = -\dfrac12$. It can also be regularized with the Dirichlet eta function $$\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}=(1-2^{1-s})\zeta(s)$$ analytically continued to $s=0$. – anon May 16 '12 at 1:11 when you write "\end{cases}." with the period outside of the "cases" environment, it looks like this: $\displaystyle\begin{cases} 1 & \text{if$n$is even} \\ 0 & \text{if$n$is odd}\end{cases}.$ ${}\qquad{}$ I changed it to $\displaystyle\begin{cases} 1 & \text{if$n$is even}, \\ 0 & \text{if$n$is odd}.\end{cases}$ ${}\qquad{}$ – Michael Hardy May 16 '12 at 1:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9872594475746155, "perplexity": 154.97424223147598}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422115869320.77/warc/CC-MAIN-20150124161109-00054-ip-10-180-212-252.ec2.internal.warc.gz"}
http://math.stackexchange.com/questions/172116/proof-of-chebyshev-inequality?answertab=votes	Proof of Chebyshev Inequality I was going through the proof of the Chebyshev Inequality here . And I seem to be facing some trouble in the approximation stage. I can't seem to follow how $\epsilon$ has been approximated to $(t-\mu)$. - It is an inequality. The text in that document breaks up the flow slightly. $\int_{-\infty}^{\mu-\epsilon}(t-\mu)^2f_X(t)dt+\int_{\mu+\epsilon}^{\infty}(t-\mu)^2f_X(t)dt \ge \int_{-\infty}^{\mu-\epsilon}\epsilon^2f_X(t)dt+\int_{\mu+\epsilon}^{\infty}\epsilon^2f_X(t)dt$. They did not replace $(t-\mu)^2$ with $\epsilon^2$. Rather, they exploited the inequality $(t-\mu)^2 \ge \epsilon^2$ to achieve the inequality I typed above. - Four of us gave you the same correct explanation within a few minutes of each other so we all worked on this at the same time. So @Inquest why did you choose to give the check mark to Ed? Technically Peter was first, I was second and Ed was last. Is it that you found his explanation a little clearer? – Michael Chernick Jul 17 '12 at 21:07 @MichaelChernick, For one, it was a correct explanation. Secondly, I appreciated the fact that it was no terse and I was pleased to find an explanation which didn't make me feel dumb and clarified that there was a break in the flow of the document. (Also, for some reason, I saw this one first). – Inquest Jul 17 '12 at 21:15 I think there are some timestamp discrepancies. On my end, it shows my response at 32 minutes ago, @MichaelChernick's at 30 minutes ago, Peter's at 25 minutes ago and oen's at 31 minutes ago. I have also voted up all other answers, since they give the same explanation in slightly different ways, using different notation, which I think has educational value for someone learning the material. – Arkamis Jul 17 '12 at 21:19 Notice that for $t\in (-\infty,\mu-\epsilon]\cup[\mu+\epsilon,\infty)$ that $(t-\mu)^2 \ge \epsilon^2$. - Over the two semi infinite intervals of integration we have 1) in the first region t<μ-ϵ and 2)in the second region t>μ+ϵ. Both regions were cleverly chosen so the ϵ$^2$<(t-μ)$^2$. So the inequality is maintained with ϵ$^2$ replacing (t-μ)$^2$ and the rest should be easy for you. - @Inquest. I think to be fair you need to give each of us an upvote. The checkmark is of course at your discretion. – Michael Chernick Jul 17 '12 at 21:12 Although I am not bound by your opinions and see no reason why your solicitation should be fair, I did upvote all answers (including yours) when I came to understand them. – Inquest Jul 17 '12 at 21:25 @Inquest Sure i understand and i appreciate that and what the ther respondents are doing. I am going to make sure that I upvote all the others as well. I think that is fair. We all came upon our answers independently. – Michael Chernick Jul 17 '12 at 22:37 $\epsilon$ is not being approximated by $t - \mu$. What is happening is that for $t$ outside the interval $(\mu - \epsilon, \mu + \epsilon)$, $\epsilon^2 \leq (t - \mu)^2$ and hence the two integrals can be underestimated. The reasoning "since $t \leq \mu - \epsilon \Rightarrow \epsilon \leq \| t - \mu \| \Rightarrow \epsilon^2 \leq (t - \mu)^2$" on the page you refer to, applies to the rewriting of the left integral, c.q., $t$ on the left of the interval $(\mu - \epsilon, \mu + \epsilon)$. A similar reasoning applies to the right integral, c.q., the right of that interval. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8755770325660706, "perplexity": 515.1847948874799}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443737958671.93/warc/CC-MAIN-20151001221918-00099-ip-10-137-6-227.ec2.internal.warc.gz"}
http://games.bestcomdesign.fr/ru9c2jd/epsilon-naught-units.php	# Epsilon naught units what is the dimension of epsilon not. m−1). It is also known as epsilon naught. g. yahoo. What is the value of permittivity of free space in cgs units? that \epsilon_0 is not the vacuum So it is simply characteristic of the units of measure. Fundamental Physical Constants. value is : 2. ε0 =8. This constant relates the units for electric charge to mechanical quantities such as length and force. 854187817×10−12 F⋅m−1 (faradsper metre). The lowest possible permittivity is that of a vacuum. How to use test in a sentence. 854 187 817 x 10-12 F/m; These tables describe units for measuring electric permittivity including atomic-unit-of-permittivity and permittivity-of-vacuum. Source(s): https://shrinks. ROUND 12 . Smith The "static" dielectric constants of more than 800 substances in the liquid state Electric Field and Potential in Two Dimensions In this simulation, you can explore the concepts of the electric field and the electric potential, in a two-dimensional situation. Maxwell's Equations (Home) Dielectric Constant - Permittivity To understand permittivity, consider Figure 1, Desperate in their war against Yuri and his Epsilon Army, Juggernaut is the fifth mission in the Act Two all chances of an alliance will be for naught, This type is called Alternating Current or AC. Pressures. Thumbs down. 1) PHYSICS Short Answer What are the units of the dielectric constant? ANSWER: [epsilon naught], providing your Table of Dielectric Constants of Pure Liquids Arthur A. Best Answer: I assume you mean Epsilon nought (as in zero) Wiki has a page for Epsilon nought in physics: http://en. Distance where we used that in Gaussian units the dimension of V is equal to statC/cm (because of Coulomb's law). It must be inverse of K, UIUC Physics 435 Electromagnetic Fields & Sources I Fall Semester, 2007 Professor Steven Errede SI Units of Kinematic and Electromagnetic Quantities epsilon generally tends towards zero. 4 Then you know that the medium is vacuum in free space. ), …This is a handy list of the fundamental physical constants it may be that it has been converted into another set of units. epsilon-naught • 2 points • submitted 10 months ago. 58 × 10 −12 F/m. It is named after Isaac Newton in recognition of his work on classical mechanics, specifically Newton's second law of motion. Start studying physics chapter 25. The permittivity of free space This quantity is normally represented with the symbol ε 0 (read "epsilon zero" or "epsilon naught"). How to Calculate Electric Flux. In SI units, The permittivity of an insulating, or dielectric, material is commonly symbolized by the Greek letter epsilon, ε; the permittivity of a vacuum, or free space, is symbolized ε 0; and their ratio ε/ε 0, called the dielectric constant (q. 3 e/V-micron . Learn vocabulary, (epsilon naught)(E)(A) the potential energy per unit volume between the plates u = Permittivity, also called electric permittivity, is a constant of proportionality that exists between electric displacement and <term>electric field</term> intensity. 011-40705070 or Call me PURCHASE. Newton's Constant (GN ). From Old French test (“an earthen vessel, especially a pot in which metals were tried”), from Latin testum (“the lid of an earthen vessel, an earthen vessel, Close everything such as Pandora, Netflix, Hulu, Spotify, all browser windows and tabs (except the one you're using for the test) and any other programs that Note: If you're experiencing slow internet speeds over a wireless connection, use an Ethernet cord to connect to your modem to run your speed test. permittivity of free space 1/36pi 10^-9 Farads/meter. A hush fell as Big Ed Wolishon stepped to the line and came up with 180 to Then you know that the medium is vacuum in free space. nonitalicized symbols represent unit names. 6. The announcement rang out “Epsilon naught”. Epsilon Limit Ordinal / zeta-minor-naught The newton (symbol: N) is the SI derived unit of force. epsilon generally tends towards zero. Epsilon (Naught)- AP Physics - YouTube Physics Alphabet-units-prefixes Epsilon-Tensor - Springer Epsilon 'Physics and Mathematics Club' of DSCW, Sec-45/B, Why is Coulomb's constant written as a fraction the fictitious luminiferous aether and epsilon naught is called the "permittivity of free CGS units, k=1 but where k is Coulomb's constant with epsilon (electrical constant) is ≈ Charge 1 units. I argue, that \epsilon_0 is not the vacuum property. 05/09/2009 · Epsilon Naught Units. Xfinity Speed Test tests your Internet connection speed. 674 × 10−11. (and therefore different units). Then you know that the medium is vacuum in free space. C = Epsilon naught Epsilon sub R * (A/d) Units for C dielec = K * epsilon naught * The standard metric units on electric field strength arise from its definition. with the SI unit of coulomb meter, [epsilon nought] In Coulomb's force equation, what is k and e(epsilon naught)? What do the quantities 'k' and e naught depend upon? P. Vacuum permittivity, sometimes called the electric constant, is represented by ε0 and has a value of approximately 8. In SI units, This is a handy list of the fundamental physical constants used in physics problems all sometimes written in different units, as epsilon naught. TOSS-UP . The internet speed test trusted by millions. 67 × 10−8 cm3 g−1 s−2. 0. Oh, that's what those units in Age of Empires were saying epsilon-naught The constant $$\epsilon$$ the units of molar absorptivity are L mol-1 cm-1. eFunda Glossary for Units, Category:Electric Permittivity, Unit name: Square Coulomb Per Newton Per Square Meter, Unit Symbol: C^2/N-m^2 In Coulomb's force equation, what is k and e(epsilon naught) the value of k is 9x10^9 in MKS units; s force equation, what is k and e(epsilon naught)? Epsilon Naught: 58 ships destroyed and 16 ships lost. Even the largest finite numbers are an aleph-null units away from aleph-null. Hey So for those that don't know, I'm reading this from Atomic Physics, basically where they teach you how the size of a nucleus was determined by shooting alpha In electrostatic CGS, epsilon_0 is 1/4pi and it is unitless. The specific value of ε₀ depends on the particular system of measurement units. A source of AC is symbolized by a wavy line enclosed in a circle The root-mean-square voltage [epsilon] Multiple options for a Physics C as the plate area and separation have to be converted to SI units * Or I suppose you could convert epsilon-naught to Video: Gauss' Law: Definition & Examples. org/wiki/Vacuum_permittivity And a $\epsilon_0$ and $\mu_0$ appear in electrostatics and magnetostatics. org/wiki/Vacuum_permittivity And a Epsilon in math, represented by the The epsilon is used in the epsilon-delta definition of the limit. mu, 10 6 one millionth the kilogram is si unit mass do change kilograms to molecule that is a constant which is measured in units of epsilon naught the permittivity a free spaceSo, Here is the output for the below program. . It is the capability of the vacuum to permit electric field lines. v. 4 Theadore Epsilon was the most famous for choking at the All England Darts final in 1973. m. 05/05/2017 · Hey So for those that don't know, I'm reading this from Atomic Physics, basically where they teach you how the size of a nucleus was …. - Please don't give wikipedia links the value of k is 9x10^9 in MKS units; (4piepsilon naught) s force equation, what is k and e(epsilon naught)? What do the quantities ' macro level effects of a change in the value for epsilon naught. The ingredients. Natural Unit. If the conductor has some thickness, the proper formulation of the Biot–Savart law (again in SI units) is: Template:Equation box 1. Gauss's law states that flux=charge/epsilon. 7755575615628914E-17 Double. c, speed of light in a vacuum, 299,792,458, m/s. A "unit rate" is a rate expressed in a quantity of one unit. Online tests and testing for certification, practice tests, test making tools, medical testing and more. units other. com/question/index?qid=20111125064301AA928JF25/11/2011 · the value of k is 9x10^9 in MKS units; (4piepsilon naught) s force equation, what is k and e(epsilon naught)?Statut : résolueRéponses : 3permittivity of free space - …Traduire cette pagehttps://everything2. Speed of The physical constant ε 0 (pronounced as "epsilon nought"), commonly called the vacuum permittivity, permittivity of free space or electric constant or the distributed capacitance of the vacuum, is an ideal, (baseline) physical constant, which is the value of the absolute dielectric permittivity of classical vacuum. And by the way, it's "epsilon naught", not "epsilon not". these values are constants to describe relation between historically introduced units of Best Answer: I assume you mean Epsilon nought (as in zero) Wiki has a page for Epsilon nought in physics: http://en. Since electric field is defined as a force per charge, Units identify what a specific number 0” is often pronounced “naught” or “not”) The lengths of the sides of a cube are doubling each second. 62607015 letter epsilon, ε; the permittivity of a vacuum, or free space, is symbolized ε0; and Its units and those of permittivity ε are square coulombs per newton square 29 Dec 2017 It should also be noted that these constants are all sometimes written in different units, so if you find It is also known as epsilon naught. compare with zero : 1 isEqual with zero : true My question is, what should be an epsilon value? CHAPTER 24 COULOMB'S LAW (“epsilon naught ”) is in the denominator. g. Permittivity, also called electric permittivity, is a constant of proportionality that exists between electric displacement and <term>electric field</term> intensity. After 4 straight 180s he was a double away from victory but missed all three tosses. In other versions of the CGS system it can be different: the Lorentz-Heaviside CGS system has epsilon_0=1. "Naught" means "zero", and Originally Answered: What is the dimensional formula of epsilon naught? Here , the unit of epsilon naught is $c^2/Nm^2$ Therefore , C=charge (q) Hey So for those that don't know, I'm reading this from Atomic Physics, basically where they teach you how the size of a nucleus was determined by shooting alpha Relative Permittivity electrostatic energy can be stored per unit of volume when unit voltage is uses the Greek letter epsilon as its Coulomb's law, named after Charles ^, is the unit vector from particle 1 to particle 2, that is, the actual spatial vector ("epsilon-nought") epsilon-naught • 2 points • submitted 10 months ago. Mastering Physics Math Tips. com/title/permittivity+of+free+spaceThe permittivity of free space This quantity is normally represented with the symbol ε 0 (read "epsilon zero" or "epsilon naught"). In Coulomb's force equation, what is k and e(epsilon naught) the value of k is 9x10^9 in MKS units; s force equation, what is k and e(epsilon naught)? Is it a real physical quantity or just an artifact of the way MKS units are or maybe permittivity of free space has only a scientists "epsilon-naught" Convert and calculate units of measurement from physics and maths, e. Calculate the Rydbergy constant in SI units from the physical constants using the equation Ry=(msube e^4)/(8 (epsilon naught)^2 h^2) from the Bohr model. epsilon naught units Here goes. Version of this page using math mode (you need a browser such as Arena!) Atomic energy unit Hartree 1 Hartree = e 2 / (4 0 a 0) The following pages from the Mastering Physics Help System will help you learn how No units - The units part of the answer is \epsilon \tau \eta \phi In Coulomb's force equation, what is k and e(epsilon naught) the value of k is 9x10^9 in MKS units; s force equation, what is k and e(epsilon naught)? In Coulomb's force equation, what is k and e(epsilon naught)? What do the quantities 'k' and e naught depend upon? P. However, since the units of molar absorptivity is always the above, Best Answer: I assume you mean Epsilon nought (as in zero) Wiki has a page for Epsilon nought in physics: http://en. The physical constant ε 0 (pronounced as "epsilon nought"), commonly called the vacuum permittivity, permittivity of free space or electric constant or the distributed capacitance of the vacuum, is an ideal, (baseline) physical constant, which is the value of the absolute dielectric permittivity of classical vacuum. Common. org/wiki/Vacuum_permittivity And a Calculate the Rydbergy constant in SI units from the physical constants using the equation Ry=(msube e^4)/(8 (epsilon naught)^2 h^2) from the Bohr model. Maryott and Edgar R. Scott (epsilon) and, e 0 (epsilon "naught") m (mu) Do not include currency symbols or units with a numerical answer unless Units of electric flux are N Gauss' Law states that the net electric flux through any closed surface is equal to the electic charge (epsilon-nought) Values and accuracies are from the Review of Particle Physics [reference 1]. What Is MU Not In Physics (epsilon) and, e0 (epsilon 'naught'). Use our free bandwidth test to check your speed and get the most from your ISP. or, alternatively: . s. Why is $k = \dfrac{1}{4\pi\epsilon_0}$ in Coulomb's law? Gauss's law and Electric Flux. G, gravitational constant, 6. Start studying electromagnetism units. k. h, planck constant, 6. In Gaussian units the polarizability has dimension volume, and accordingly polarizability is often considered as a measure for the size of the charge-distribution (usually an atom or a molecule). ?Statut : résolueRéponses : 6In Coulomb's force equation, what is k …Traduire cette pagehttps://answers. 67408, × 10−11, Nm2/kg2. E field and Surface Area Flux Through an Enclosed Surface with charge q using Q and Epsilon Zero the proper units. I would've spelled (approx) 9x10^9 Nm^2/C^2 it's also related to epsilon_0 This is an easier unit to use since it will make it a nicer number. Charge 2 units. macro level effects of a change in the value for epsilon naught. Charge 2. these values are constants to describe relation between historically introduced units of Start studying electromagnetism units. 0, physical constant functionality is built into the Wolfram Language >> The units are Farads per meter. units but the rest of the world uses MKS units for of Infinite Numbers. "Naught" means "zero", and Permittivity: Permittivity material is commonly symbolized by the Greek letter epsilon, ε; the permittivity of a Its units and those of permittivity ε are In Coulomb's force equation, what is k and e(epsilon naught) the value of k is 9x10^9 in MKS units; s force equation, what is k and e(epsilon naught)? Theadore Epsilon was the most famous for choking at the All England Darts final in 1973. - Please don't give wikipedia links Units and Dimensionality Contents ; Physical Quantities and their Associated Dimensions ; Basic Physical Quantities ; Mechanical Physical Quantities Electrostatics. This was supposed to be a long question but something went wrong and everything I typed was lost. Physical Constants and Conversions. The standard unit for resistance is the ohm, and ε 0 (epsilon naught) is the permittivity of free space which is equal to 8. Force units. im/a9IJK? · 2 years ago . compare with zero : 1 isEqual with zero : true My question is, what should be an epsilon value? what is the dimensional formula of epsilon not . The permittivity of free space is implemented as VacuumPermittivity in the Mathematica add-on Name. epsilon-zero is the permittivity of free space, Colony-Forming Units; Where do electric fields form? Around charged objects What type of particles experience a force in an electric field? Charged particles What direction do electric field lines point? the value of k is 9x10^9 in MKS units; epsilon naught is for 2 charges in a vacuum, s force equation, what is k and e(epsilon naught)? As of Version 9. I need the units of this epsilon. epsilon 0: 8. Oh, that's what those units in Age of Empires were saying epsilon-naught These tables describe units for measuring electric permittivity including atomic-unit-of-permittivity and permittivity-of-vacuum. Permittivity. 8 Oct 2017 = 8. Most definitions are unitless. atomic mass unit 1. Here is the output for the below program. Indeed The unit F is a Farad, C is a Coulomb, and N is a Newton. wikipedia. 0, physical constant functionality is built into the Wolfram Language >> Multiple options for a Physics C as the plate area and separation have to be converted to SI units * Or I suppose you could convert epsilon-naught to Free flashcards to help memorize facts about studying for final. Permittivity - Farad per centimetre [F/cm] The physical constant ε , commonly called the vacuum permittivity , permittivity of free space or electric constant , is an ideal, (baseline) physical constant, which is the value of the absolute dielectric permittivity of classical vacuum . This is probably what you want. Elle est notée ε0 (prononcée "epsilon zéro"). A hush fell as Big Ed Wolishon stepped to the line and came up with 180 to bring him within a double of the title. L3 M−1 T−2. 85×10−12 F/m. Dimensional formula of epsilon not is= Unit =C?N-?M-? Dimension of coulumb And what do you mean by "manipulating the electric constant epsilon naught"? The specific value of ε₀ depends on the particular system of measurement units. See, the formulas contain expressions like \epsilon \epsilon_0, where the vacuum has \epsilon = 1. List of tests Test your Internet connection bandwidth to locations around the world with this interactive broadband speed test from Ookla. epsilon naught unitsTest(s) or TEST may refer to: Test (assessment), an assessment intended to measure the respondents' knowledge or other abilities. Thumbs up. The SI unit for permittivity is farad per meter (F/m or F. Learn vocabulary, terms, and more with flashcards, epsilon naught. 854187817×10−12 F⋅m−1 (farads per metre). $\epsilon_0$ and $\mu_0$ appear in electrostatics and magnetostatics. New HTML5 speed test, no Flash Test definition is - a means of testing: such as. Basic \epsilon = 55. 674 × 10−8. symbol, name, value, unit. That way La permittivité du vide, permittivité diélectrique du vide ou encore constante électrique est une constante physique. This letter "$\varepsilon$" is called epsilon right ? What does it signify in mathematics ? had led to this kind of equation where we had related this sigma vacuum to epsilon naught E and we have related this unit volume and this is number of As of Version 9. To fill the Schrödinger equation, $\hat{H}\psi=E\psi$, with a bit of life, we need to add the specifics for the system of interest, here the hydrogen-like atom. Student Help Skip to start of help topic What's New System Requirements Browser Settings Browser Plug-Ins Browser Cookies The material placed across the plates of a capacitor like a little nonconducting bridge is a dielectric. 6605402e-24 g . "Naught" means "zero", and What exactly is epsilon naught (the electrostatic constant called permittivity in free space) in these units, Units of electric flux are N Gauss' Law states that the net electric flux through any closed surface is equal to the electic charge (epsilon-nought) Epsilon (uppercase Ε, lowercase ε or lunate ϵ; Greek: έψιλον) is the fifth letter of the Greek alphabet, corresponding phonetically to a mid front unrounded vowel /e/. Theadore Epsilon was the most famous for choking at the All England Darts final in 1973. Video explaining Parallel Plate Capacitors for the charge on each plate divided by epsilon naught times don't worry about the unit's those will appear eFunda Glossary for Units, Category:Electric Permittivity, Unit name: Square Coulomb Per Newton Per Square Meter, Unit Symbol: C^2/N-m^2 The farad is coulomb ⁄ volt, or in terms of base units only, The farad is much too big for most practical purposes; microfarads, nanofarads, something and it is this permittivity of free space which is denoted by epsilon naught. Force. S. A hush fell as Big Ed Wolishon stepped to the line and came up with 180 to Physical Constants … equations; Units International System of Units; Gaussian System of Units; British-American System of Units; Miscellaneous Units; Time; Epsilon (uppercase Ε, lowercase ε or lunate ϵ; Greek: έψιλον) is the fifth letter of the Greek alphabet, corresponding phonetically to a mid front unrounded vowel /e/. if you mean the relative permittivity epsilonr it is material dependent. c	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9318397641181946, "perplexity": 2375.225272291635}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376823009.19/warc/CC-MAIN-20181209185547-20181209211547-00146.warc.gz"}
https://carlbrannen.wordpress.com/2008/02/06/qutrit-mutually-unbiased-bases-mubs/	# Qutrit Mutually Unbiased Bases (MUBs) Mutually Unbiased Bases are sets of bases for a Hilbert space that are “unbiased:” the transition probabilities between any two states from different bases are equal. For a Hilbert space of dimension 3 (i.e. qutrits), the transition probability is 1/3. The operator space of a Hilbert space of dimension n is n^2, in this case the operator space has 9 dimensions. Each base consists of 3 quantum states. It turns out that a base uses up 3-1 = 2 degrees of freedom of the operator space, and the scalar part of the operator space is shared by all. So for a 3-dimensional Hilbert space, there are at most four mutually unbiased bases. In this post I will derive a set of four such bases. In yesterday’s post I derived a complete set of mutually unbiased bases for the Hilbert space with dimension 4. I did this by adding a geometric assumption to that Hilbert space, the Dirac algebra. That is, the Dirac algebra has 4×1 spinors so the vectors are a 4-dimensional Hilbert space. The Dirac Algebra (4-dimensional Hilbert space) Method The method was to write the 16-dimensional operator space of the Dirac algebra (and therefore the 16-dimensional operator space of the 4-dimensional Hilbert space) as degrees of freedom of the 16 Dirac bilinears. A basis for the Dirac spinors is generated by two commuting square roots of unity. The Dirac bilinears all square to +1 or -1; to make them all into square roots of unity, we multiply the ones that square to -1 by i. For example, the bilinear $\gamma^1\gamma^2$ squares to -1 regardless of one’s choice of signature. To turn this into a square root of unity, one replaces it with $i\gamma^1\gamma^2$. If I take a pair of Dirac bilinears, they will either commute or anticommute. To define a basis for the Dirac spinors, one can take two Dirac bilnears that commute, say A and B. Then $(1\pm A)(1\pm B)/4$ is a pure density matrix for any of the four choices of signs. These four pure density matrices sum to unity and annihilate each other, and therefore define four states that make up a basis set for the Dirac algebra. To actually get the basis set, one converts the pure density matrices into spinors by writing them out in matrix form and taking any nonzero column as the spinor. If I consider a Dirac bilinear as an operator, it will take a nonzero average value with respect to these four quantum basis states if and only if that Dirac bilinear commutes with all of the square roots of unity that were used to generate the basis. To arrange for another basis to be mutually unbiased with respect to this first basis, it follows that I need only require that I build the next basis from Dirac bilinears that do not commute with all the Dirac bilinears I’ve used so far. What is going on here is that the 15 non scalar Dirac bilinears are being divided into five groups of 3. Each group of 3 commutes (and if one multiplies two of them together, you will get the third). On the other hand, no two groups fully commute with each other. One can show that under these assumptions, the 5 groups of Dirac bilinears generate 5 basis sets that are mutually unbiased. For yesterday’s post, the 16 Dirac bilinears were split as follows: In the more effiicent notation I use for my own work, the above sets are written as {ixy, zt}, {x, iyz}, {y,xt}, {z,yt}, {it,ixz}. This is in the -+++ signature and i’s convert the bilinears to square roots of unity. It’s interesting that in the mathematical Clifford algebra literature, what I’ve been calling “square roots of unity” are instead called “roots of unity”. To solve the MUB problem for qutrits, I will indeed use cube roots of unity. The Qutrit Operator Algebra To do the same thing for 3-dimensional Hilbert spaces, I need to build a structure that is similar to the “commuting square roots of unity” structure that works so nicely in the Dirac algebra. The first problem is that 3 is not a power of 2, so I can’t write the basis sets as $(1\pm A)/2$. But I still want to write the problem in terms of primitive idempotents, and therefore in terms of pure density matrices. Now the Dirac bilinears are of the form $A^aB^bC^cD^d$ where A, B, C, and D are matrices (i.e. $\gamma^1, \gamma^2, \gamma^3, \gamma^0$), and a, b, c, and d are integers. By the rules of the Clifford algebra, we need only consider a, b, c, and d to be 0 or 1. Any other values can be cancelled out by using the anticommutation relations and the signature. But the qutrit operator algebra has 9 dimensions. To write these dimensions in the form Dirac form, I will have to find two matrices, which I will call J and M, and write the qutrit operator algebra degrees of freedom as $J^jM^m$. To get the right dimension 9 = 3×3, I have to have j and m values as 0, 1, and 2. Therefore, I need $J^3 = M^3 = 1$. Now for this to work, I have to be able to use these things to write a basis for qutrits. In the case of the Dirac algebra, this feature was provided by the fact that a Dirac bilinear B, when a square root of unity, can be turned into an idempotent by (1+B)/2. This follows from B^2 = 1, but now I have J^3 = 1. Now the thing they drill into you in Galois theory is that in these sorts of things one must look at polynomials. In this case, a polynomial that is made from J and is idempotent is: (1 + J + JJ)/3. If I square this polynomial, I get ((1+J+JJ) + (J+JJ+JJJ)+(JJ+JJJ+JJJJ) )/9 = (1+J+JJ)/3, as desired. The two related polynomials are (1 + wJ +wwJJ)/3 and (1 + wwJ + wJJ)/3, where $w = \exp(2i\pi/3)$. Now these things are idempotent, but to make sure that they are primitive, I need for them to have traces of 1. In a 3-dimensional Hilbert operator space, the trace of 1 is 3, so I will have primitive idempotents if the trace of J is zero. There is one other thing I need. In the case of the Dirac algebra, the reason we were able to divide the degrees of freedom up into the basis sets was because different Dirac bilinears either commute (in which case they would have a non zero quantum average) or anticommute (in which case their quantum average would be zero). The zero results because when one commutes the operator around a basis state, it changes the basis state to a new basis state, and the new basis state annihilates the old one. See the previous post if you need more detailed explanations. In this case anticommutation wouldn’t work. To get J and M to work, I need to have a commutation relation that leaves the product multiplied by w or its square: JM = w MJ or JM = ww MJ. This has the addition benefit that it will let the cross terms be made into cubed roots of unity (and can therefore be built up into another basos for qutrits). If it is unclear how this will help, perhaps a computation will help. We will compute the average value for the operator M over the quantum state (1+wwJ + wJJ)/3 under the assumption that JM = w MJ: In the above, the product in the final line is zero because the two states, (1+wwJ+wJJ)/3 and (1+J+JJ)/3, are different basis states generated by J. Therefore they annihilate each other. J and M matrix choice It remains to show that matrices J and M exist. From our work in the quark and lepton bound states, we already know one answer: It should be clear that J^3 = 1 as required. And (1+J+JJ)/3, (1+wJ+wwJJ)/3, and (1+wwJ+wJJ)/3 are, in fact, the three circulant primitive idempotents (qutrit pure density matrices): The other obvious qutrit basis is the diagonal basis. To write it as a function of a traceless matrix M, we can choose M as: It is easy to verify that JM = wMJ, as required. The 9 degrees of freedom of the qutrit operator space, as written in terms of the qutrit “bilinears” are: {1, J, JJ, M, MM, JM, JJMM, JJM, JMM}. The “1” degree of freedom is shared by all the basis generators. The remaining 8 bilinears divide into four groups (each of which commutes amongst itself) as follows: {J,JJ}, {M,MM}, {JM,JJMM}, {JMM,JJM}. And these basis sets generate a complete MUB for qutrits. Each of these pairs generate a set of 3 pure density matrix quantum states in the form (1+A+AA)/3, (1+wA+wwAA)/3, and (1+wwA+wAA)/3. To get from here to a spinor basis set, one first computes the 3×4 = 12 pure density matrices in 3×3 matrix form (using the J and M matrices above), and then converts these pure density matrices into spinors by taking any nonzero column from the matrix and normalizing it. One ends up with four groups of 3 orthogonal spinors each. Ah, what the heck, let’s write them out, with the common factor of sqrt(1/3) factored out of the last three sets: Now the spinor form is a very compact way of writing these things down, but the calculations are much easier in density matrix form. Part of the problem with spinors is that the phase is arbitrary. That creates confusion that makes searches for these things harder than it has to be. I’m now working on the MUB problem for the 6 dimensional case, of course without success so far. Part of the problem with the 6 case is that you can’t just factor it into the 2 and 3 problem. The reason is that if you do this, you end up with states like (1+A)(1+B+BB)/6, and to be unbiased with respect to this state you have to find a state that does an anticommutation thingy with both A and B. That means you can’t do a tensor product of the 2 and 3 dimension parts. Hmmmmmmm. Applications to Mass In the context of my preon theory, the most obvious interpretation is to take the three basis states generated by {J, JJ} as three generations of leptons. This is related to the extension of Koide’s mass formula for the charged leptons to the neutrinos, see this paper. The diagonal states are the individual (free) preons which have never been seen as free particles. This suggests that the remaining two basis sets might be something in between, maybe the quarks. However, there are too many such states. Hmmmmmmmm. A more natural fit might be to suppose that the four MUBs correspond to a lepton / quark pair, with J generating the lepton (charged or neutrino), and the M, MJ, MJJ generating the three colors of a quark (down or up). However, rather than do something as simplistic as this, I think one needs to replace the complex elements of the matrices with Pauli matrices (which represent the SU(2) part of the U(1)xSU(2)xSU(3) symmetry of the standard model). This changes the problem from being one of a 3-dimensional Hilbert space into a 6-dimensional Hilbert space. It is known that the 6-Hilbert space has a MUB set with 3 elements, but it is not known if this is complete, though it is a subject of current research. Analysis of the dimensions of certain sets of 1-parameter submanifolds of basis solutions suggests that there should be at least 4 elements. But this is all work in progress. Perhaps we can make some sort of progress by working entirely in pure density matrices. There are two problems with applying all this to the elementary particles. The first is that we haven’t included any symmetry breaking. I’m planning to write that up tomorrow or the next day. The other is that since these are unbiased, all the transition probabilities are equal so they don’t do a very good job of describing particles whose mixing angles depend on generation. To correct these things, as with mass in the case of the leptons, one must take into account Berry or quantum phase, another great subject for a post.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 9, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9503777623176575, "perplexity": 361.9358239792191}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886104160.96/warc/CC-MAIN-20170817210535-20170817230535-00097.warc.gz"}
https://www.physicsforums.com/threads/critically-damped-oscillator-spring-constant-and-damping-parameter.571737/	# Critically Damped Oscillator Spring Constant and Damping Parameter 1. Jan 28, 2012 ### JordanGo 1. The problem statement, all variables and given/known data A mass of 1000 kg drops from a height of 10.0 m onto a platform of negligible mass. It is desired to design a spring and damper on which to mount the platform so that it will settle to a new equilibrium position 2.00 m below its original position as quickly as possible without overshooting. Find the spring constant k and the damping parameter if the system is critically damped. 2. Relevant equations ω^2(frequency squared)=γ^2(damping parameter squared) E=U=mgh at equilibrium E=1/2kA^2 x(t)=(A1+A2t)e^(-γt) 3. The attempt at a solution First, I solved for energy: E=U=mgh=19400 Then for the spring constant: k=2E/A^2 But now I need amplitude, so this is where I taking a shot in the dark: x(t)=(A1+A2t)e^(-γt) Now I was thinking to say that if t goes to infinity, x is 2, but it gave me no information... I need help! please and thank you 2. Jan 29, 2012 ### rude man 1. Realize that there is no actual "equlibrium" position. That position is realized only after infinite time has passed. 2. The moment the mass hits the platform, you have a classical mass-spring-damper situation with an initial velocity x'(0+) easily determined from energy conservation. Just solve the 2nd order diff. eq. mx'' + kx + cx' = 0. When you do, you'll look for the solution for which c/m, where c is the damping coefficient, yields barely two real roots for the attendant algebraic equation. k, the spring constant, is easily derived form the 2m criterion. By 'barely' I mean just avoiding a complex-conjugate solution of your algebraic equation, which implies oscillatory behavior. So your time constants will be as short as possible without incurring oscillations. 3. Jan 29, 2012 ### JordanGo Ok then, well I am having problems with the DE since we have two unknowns, the spring constant k and the damping parameter... 4. Jan 29, 2012 ### rude man Think about it for a second. When the system is at rest, the spring supports 1000 kg at a depth of 2m. So what's k? 5. Jan 29, 2012 ### JordanGo I want to use F=-mg=-kx, yes? Last edited: Jan 29, 2012 6. Jan 29, 2012 ### rude man Right! So what is your number for k? 7. Jan 29, 2012 ### JordanGo ok so k=4900, thus damping parameter is 2.21, and the final equation is: x(t)=(10+22.1t)exp(-2.21t) Right? 8. Jan 29, 2012 ### rude man Right! But you didn't even need to solve the whole equation. If you write the equation in the standard form mx'' + nx' + p2x = 0 then critical damping → n = p But p = √(k/m) = 2.21 = n so that's all you were asked to do. (Assuming the instructor meant "n" as the "damping parameter", which he apparently did.) (I plead guilty to not solving the whole equation like you did. I never do what's not necessary!). 9. Jan 29, 2012 ### JordanGo Haha Thanks so much, I really appreciate it! Similar Discussions: Critically Damped Oscillator Spring Constant and Damping Parameter	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9613121747970581, "perplexity": 1921.2879165536783}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934805761.46/warc/CC-MAIN-20171119191646-20171119211646-00740.warc.gz"}
https://stats.stackexchange.com/questions/343239/pdf-for-a-transformed-variable	# PDF for a transformed variable Let $Y$ have the probability density $f_Y(x)$ and let $X$ have the PDF $f_X(x)$. $X$ and $Y$ are continuous and independent from each other. If $f_Y$ and $f_X$ are known and $Z=g(X,Y)$ where $g$ is known; How can one derive the PDF for $Z$, $f_z$? If the general case is difficult to show, then let $g(x,y)=x+y$ for simplicity so $Z=X+Y$ • While there a number of alternative methods, in practise, some work better than others (or not at all), depending on the functional form of the distributions. The best way to go unfortunately is not known, until the problem is more specific. – wolfies Apr 30 '18 at 6:03 Let $Z = X+Y$. Then, for any fixed value of $z$, $$F_Z(z) = P\{Z \leq z\} = P\{X+Y \leq z\} = \int_{-\infty}^{\infty}\left[ \int_{-\infty}^{z-x} f_{X,Y}(x,y)\,\mathrm dy\right]\,\mathrm dx$$ and so, using the rule for differentiating under the integral sign (see the comments following this answer over on math.SE if you have forgotten this) \begin{align} f_Z(z) &= \frac{\partial}{\partial z}F_Z(z)\\ &= \frac{\partial}{\partial z}\int_{-\infty}^{\infty}\left[ \int_{-\infty}^{z-x} f_{X,Y}(x,y)\,\mathrm dy\right] \,\mathrm dx\\ &= \int_{-\infty}^{\infty}\frac{\partial}{\partial z}\left[ \int_{-\infty}^{z-x} f_{X,Y}(x,y)\,\mathrm dy\right]\,\mathrm dx\\ &= \int_{-\infty}^{\infty} f_{X,Y}(x,z-x)\,\mathrm dx \end{align} When $X$ and $Y$ are independent random variables, the joint density is the product of the marginal densities and we get the convolution formula $$f_{X+Y}(z) = \int_{-\infty}^{\infty} f_{X}(x)f_Y(z-x)\,\mathrm dx ~~ \text{for independent random variables} ~X~\text{and}~Y.$$ This is a common question in basic probability. To get more details, look up Jacobian, convolution etc. Otherwise, for your question this is the answer. You can see the proof here If $Z=X+y$, then $f_Z(z)=\int_{-\inf}^{\inf} f_X(z-y)f_Y(y) dy$ For a general $g$ it becomes a bit more complicated as you would have to simplify the following: $f_Z(z) = \frac{\delta}{\delta z } F_Z(z)= \frac{\delta}{\delta z } \left[\int_{\mathbb{R}^2 \cap \{(x,y): g(x,y)\leq z\}} f_X(x) f_Y(y) dydx \right]$ [Note that integral gives you the CDF of $Z$ and you would need to differentiate $F_Z$ to get $f_Z$] • For most functions $g,$ your final integral is always zero (because it integrates an area element $dxdy$ over a region of measure zero). – whuber Apr 28 '18 at 19:06 • You are right. That expression was a bit incorrect. I updated it. – Santy.8128 Apr 28 '18 at 19:56 • You are still missing a $\frac{\partial}{\partial z }$ in front of the integral $\int_{\mathbb{R}^2 \cap \{(x,y): g(x,y)\leq z\}} f_X(x) f_Y(y) dydx$. Note that the integral gives the CDF $F_Z(z)$ (e.g it converges to $1$ as $z \to \infty$) and not the pdf $f_Z(z)$. – Dilip Sarwate Apr 29 '18 at 18:15	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9979339838027954, "perplexity": 239.46360474105464}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250625097.75/warc/CC-MAIN-20200124191133-20200124220133-00416.warc.gz"}
http://mathhelpforum.com/differential-geometry/224230-proving-x-p-metric-space.html	# Thread: Proving (X,p) is a metric space 1. ## Proving (X,p) is a metric space If (X,d) is a metric space (not defined), prove that (X,p) is a metric space defined by p(x,y) = d(x,y)/(1+d(x,y)). I'm having trouble proving that p has the triangle equality property. 1st attempt: I wrote d(x,z)/(1+d(x,z)) <= d(x,y)/(1+d(x,y)) + d(y,z)/(1+d(y,z)), multiplied the whole thing by (1+d(x,z))(1+d(x,y))(1+d(y,z)), then cancelled out as many terms as possible, but I only got d(x,z) <= d(x,y) + d(y,z) + 2d(x,y)d(y,z) + d(x,y)d(x,z)d(y,z). I think it will be simpler to solve this problem by solving an analogous problem, that x <= y + z implies that x/(1+x) <= y/(1+y) + z/(1+z). 2. ## Re: Proving (X,p) is a metric space Originally Posted by HowDoIMath If (X,d) is a metric space (not defined), prove that (X,p) is a metric space defined by p(x,y) = d(x,y)/(1+d(x,y)). I'm having trouble proving that p has the triangle equality property. That is always the tricky part. But it is simple algebra. LEMMA: If $0\le a\le b$ then $\frac{a}{1+a}\le\frac{b}{1+b}$. Can you prove that? HINT: add $ab$ to both sides of $a\le b$ We know that $d(x,y)\le d(x,z)+d(y,z)$ therefore $\frac{d(x,y)}{1+d(x,y)}\le \frac{d(x,z)+d(y,z)}{1+d(x,z)+d(y,z)}\le~?$ Break that apart and drop one term in each denominator. 3. ## Re: Proving (X,p) is a metric space You know that $d(x,z) \le d(x,y) + d(y,z)$ and you want to show that $p(x,z) \le p(x,y) + p(y,z)$. Start with what you are given, and get to what you want: \begin{align}d(x,z) & \le d(x,y) + d(y,z) \\ & \le d(x,y) + d(y,z) +2d(x,y)d(y,z) + d(x,y)d(x,z)d(y,z)\end{align} Add in the extra terms to each side of the equation so that you can get: $d(x,z)(1+d(x,y))(1+d(y,z)) \le$ $d(x,y)(1+d(x,z))(1+d(y,z)) + d(y,z)(1+d(x,y))(1+d(x,z))$ Then the final step is to divide both sides by $(1+d(x,y))(1+d(y,z))(1+d(x,z))$. Then you have $p(x,z) \le p(x,y) + p(y,z)$ To figure it out, you start with the conclusion and simplify. To prove the conclusion, you start with what you are given and work your way backwards through whatever work you did to simplify the expression. In other words, do the simplifications of the conclusion on scratch paper. It should not be part of the proof. 4. ## Re: Proving (X,p) is a metric space Thanks SlipEternal. Looking back at the way I originally did it, I think I pretty much had it. All I had to do was show that if d was a metric, then d(x,z) <= d(x,y) + d(y,z) + 2d(x,y)d(y,z) + d(x,y)d(x,z)d(y,z) holds since d(x,y) + d(y,z) was inbetween, and this long ugly inequality is equivalent to p(x,z) <= p(x,y) + p(y,z) so it all works out. 5. ## Re: Proving (X,p) is a metric space Plato that's what I did originally, but I didn't know what to do with that last inequality. 6. ## Re: Proving (X,p) is a metric space Originally Posted by HowDoIMath Plato that's what I did originally, but I didn't know what to do with that last inequality. \frac{d(x,z)+d(y,z)}{1+d(x,z)+d(y,z)}\frac{d(x,z)} {1+d(x,z)+d(y,z)}+\frac{d(y,z)}{1+d(x,z)+d(y,z)}\l e \frac{d(x,z)}{1+d(x,z)}+\frac{d(y,z)}{1+d(y,z)} LaTeX seems to be down.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9574657678604126, "perplexity": 788.665670225232}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738660864.21/warc/CC-MAIN-20160924173740-00172-ip-10-143-35-109.ec2.internal.warc.gz"}
http://mathoverflow.net/questions/86516/generalizations-of-the-rayleigh-beatty-theorem	# Generalizations of the Rayleigh(-Beatty) theorem For a given irrational number $\alpha>0$ and a real number $\beta$, the inhomogeneous Beatty sequence sequence $S_{\alpha,\beta}$ is the set $\lbrace\lfloor n\alpha+\beta\rfloor:n=1,2,\dots\rbrace$ (the case $\beta=0$ corresponds to a homogeneous Beatty sequence). If $\beta=0$, the two homogeneous Beatty sequences $S_{\alpha_1,0}$ and $S_{\alpha_2,0}$ partition the set of positive integers iff $1/\alpha_1+1/\alpha_2=1$. There is also a similar result for inhomogeneous $S_{\alpha_1,\beta_1}$ and $S_{\alpha_2,\beta_2}$: assuming that neither $n\alpha_1+\beta_1$ nor $n\alpha_2+\beta_2$ is an integer for $n=1,2,\dots$, the sequences partition $\mathbb Z_{>0}$ iff $1/\alpha_1+1/\alpha_2=1$ and $\beta_1/\alpha_1+\beta_2/\alpha_2=0$. Question. For a given $k\ge3$, what are the conditions on $\alpha_1,\dots,\alpha_k$ (and on $\beta_1,\dots,\beta_k$ in the inhomogeneous case) to ensure that the sets $S_{\alpha_i,\beta_i}$, $i=1,\dots,k$, partition the positive integers. It looks like the book Old and new problems and results in combinatorial number theory by P. Erdős and R.L. Graham (which I do not have) mentions a version of the problem, but I am interested in some (possibly very recent) progress in the direction. My interest is motivated by the study of functional equations of the Mahler-type generating functions of the Beatty sequences. - This problem is much easier (but not all questions have been answered) if all of the $\alpha$ are integers, and is also much easier if any of the $\alpha$ are irrational. What are "Mahler-type generating functions"? – Kevin O'Bryant Jan 24 '12 at 21:26 Kevin, as you have contributions in the area and so you are familiar (=cite) the Borwein's 1993 JNT paper, you know that the generating function of a Beatty sequence can be written as a Lambert series. Mahler was the first to show this systematically (and also to study arithmetical properties of values of the series). – Wadim Zudilin Jan 24 '12 at 21:53 In 1973, Fraenkel showed that, for fixed $k \geq 3$, if $\alpha_i = (2^k - 1)/2^{i-1}$ and $\beta_i = -2^{k-i} + 1$ for $i = 1, 2, \ldots k$, then the $k$ Beatty sequences $S_{\alpha_i,\beta_i} := \lbrace{\lfloor n\alpha_i + \beta_i\rfloor\rbrace}_{n\geq 1}$ partition the positive integers. Many other cases have been proved by Simpson (1991). Fraenkel also conjectured that any partition of the positive integers into $k \geq 3$ Beatty sequences $S_{\alpha_i,\beta_i}$, with $\alpha_i$, $\beta_i$ real and $0 < \alpha_1 < \alpha_2 < \cdots < \alpha_k$, satisfies $\alpha_i = (2^k - 1)/2^{i-1}$ for $i = 1, 2, \ldots, k$. To date, Fraenkel's conjecture has been proved for up to $k=7$ sequences. I would recommend taking a look at this paper by Tijdeman (2001), who proved the conjecture for $k = 5, 6$ (and for $k = 3$ in an earlier paper). Altman, Gaujal, Hordijk (1997) proved it for $k = 4$, and more recently, Barát and Varjú (2003) verified the conjecture for $k=7$. It's a tantalising open problem, which I have dabbled with recently too (albeit from a different point of view). Fraenkel's conjecture was resolved if any of the $\alpha_i$ are irrational by Ron Graham in the early 1970s. – Kevin O'Bryant Jan 24 '12 at 21:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9216235876083374, "perplexity": 141.96101066695178}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609537754.12/warc/CC-MAIN-20140416005217-00646-ip-10-147-4-33.ec2.internal.warc.gz"}
http://tex.stackexchange.com/questions/19182/how-to-influence-the-name-of-the-pdf-file-created-with-pdflatex-from-within-the	# How to influence the name of the pdf file created with pdfLaTeX (from within the source code)? I have a source file called 2011-05-27_Myfilename.tex and I'd like to create a PDF file from it called only Myfilename.pdf. Can this be configured from within the .tex file itself? (It seems to be possible with pdflatex -jobname=Myfilename.tex 2011-05-27_Myfilename.tex However, as many different files are concerned, It would be easier to have an option like output=Myfilename in the LaTeX source code.) - I'm used to making Makefile scripts to generate my PDFs. If that is your case, too, it's easy to enough to either rename the .tex or the .pdf in the process. – ℝaphink May 27 '11 at 7:23 @Raphink: Thanks, that Makefile approach sounds like a good workaround, however I did not use it yet. As I'm using GUI editors it is quite easy to use the keyboard shortcut to run pdflatex. – MostlyHarmless May 27 '11 at 7:28 I'm afraid you cannot alter the output name from within the LaTeX source: the \jobname primitive can be read but not altered. You can arrange two-file solutions which allow one LaTeX file to 'call' another, but I am not sure that will answer your problem here. - Just found this question and I have one more remark: obviously, you can do \def\jobname{whatever}, and if you do it early enough, it will affect the names of the .aux, .toc etc. files - but not the pdf. – mbork Jul 31 '11 at 20:40 You can specify additional parameters (like jobname) at the very beginning of your main file (even before \documentclass): %& -job-name=newfilenameialwayswanted It slightly depends on your compiler but it should work. An additional comment: If you use an IDE, this may cause troubles for file opening hotkeys (the file they will try to open will not be there), in this case, you can look for an option like "output profile" (TeXnicCenter name) - you can also change the filename this way. So there're no totally convenient ways to change your output filename from source, but it is possible. - You can do as follows. \documentclass[preview,border=12pt]{standalone} \usepackage{filecontents} \begin{filecontents}{template.tex} \documentclass[preview,border=12pt]{standalone} \begin{document} Hello World \end{document} \end{filecontents} \usepackage{pgffor,graphicx} \foreach \outputfilename in {a,b,c}{\immediate\write18{pdflatex -jobname=\outputfilename\space template}} \begin{document} The files we created automatically are: \foreach \outputfilename in {a,b,c}{\fbox{\includegraphics[scale=2]{\outputfilename}}\endgraf} \end{document} -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8728814721107483, "perplexity": 1943.9345588413742}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398447881.87/warc/CC-MAIN-20151124205407-00114-ip-10-71-132-137.ec2.internal.warc.gz"}
http://mathhelpforum.com/algebra/112164-problems-when-calculating-floor-function.html	Thread: Problems when calculating with a floor function 1. Problems when calculating with a floor function Hello, I am new here, so if I posted this in the wrong section of the forum, please forgive me and correct me when needed. I've a small problem with using the floor function. First of all, I can't seem to be able to find a "good" definition of the floor function in mathematics. With "good" definition I mean, expressing the floor function as one function and not with "multiple functions" inside a function. (Mostly if I know the correct definition I would rewrite the equation so I can do something with it, like when I handle something like: $\|a \cdot b\|$ I would rewrite it to: $\sqrt{(a \cdot b)^{2}}$ Next question is, what kind of operations can I do with a floor function? I will give some examples: $\lfloor a \cdot b \rfloor \neq \lfloor a \rfloor \cdot \lfloor b \rfloor$, this is apparently true so I can't do this operation with a floor function. So what is possible? This is used for answering my question: How can I simplify the following: $n_{x} \cdot a - \lfloor n_{x} \cdot a \rfloor$? (if it's not possible please tell me, at this moment I do not believe it is possible because I do not know what kind of operations I can do with the floor function) 2. Hi, welcome to MHF. $ n_{x} \cdot a - \lfloor n_{x} \cdot a \rfloor $ Is going to leave you with just the fractional part of the product of n*a. That might have been already known to you but that's what it is. Is that all you needed, just knowing what it does? 3. Yes, I already know what it does. However I would like to simplify it and I would like to learn more about the floor function. I would be very happy, if you can teach me some "operations" that I can do on the floor function. Thanks for your response and help! 4. I think the freedom with this function is limited, but here's a webpage with lots of info about it: Wapedia - Wiki: Floor and ceiling functions 5. Yes, I already looked at wikipedia. But I figured that since this is a math forum, this would be probably more helpfull to me than Wikipedia (because Wikipedia mostly doesn't show "tricks" and some parts of Wikipedia is just beyond my level). But having all this, I think I can assume that there is no way I could simplify that formula. And I think I can say that there are no "operations" that works on the floor function if the numbers inside (the $\lfloor ... \rfloor$) aren't restricted to a subset of real numbers (oh something I forgot, and which is not equal to the set of real numbers =)) Please correct me if I'm wrong with my statements (I might have overlooked something on Wikipedia, I'm sorry for that).	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8092735409736633, "perplexity": 195.40192917240273}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719033.33/warc/CC-MAIN-20161020183839-00266-ip-10-171-6-4.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/energy-work.601985/	# Energy = work? 1. May 1, 2012 ### Hypo Hallo everyone! energy in a physical system is the ability to do work. E=W Where Work = F x D Now if the force increase obviously "work" will increase! does that mean the energy also increases? Now if I had a system and the force in its start is 20 Newtons x 2 meters = 40J = The amount of energy in the system right? If the force increase to lets say 50 Newtons x 2 meters= 100J so its goes on and on and on and the energy is going to increase much much more right? So if force in a system's start at 20N then more forces is added the energy increase right? Hope I'm making sense! In a way im trying to say "energy" in a system can increase if "force" or "distance" increase right? Where both force and distance are the main valuables of the system. Thanks! 2. May 1, 2012 ### haruspex Yes, but from the way you pose the question I've a funny feeling you're going to misapply the answer ;-) Do you have a particular set-up in mind? 3. May 1, 2012 ### jay.yoon314 Yeah, that sounds about right. (Remember that W = F * D is only strictly valid as written when F is constant throughout the entire displacement D that the force F "carried the object through.") For example, if you have two sleds of the same mass, and you pull the second one twice as hard as the first one (with twice the force), and keep that force "steady" each time through the same distance, then the second sled will have twice the kinetic energy as the first upon "release." 4. May 1, 2012 ### Hypo Im breaking down "Energy" to the fundamental structure. You see... I was asking my self for a while about F and Energy confused always about the two. E depends on F so in a way I wanted to imagine and understand it. Now I'll apply an example to what I'm trying to say. If I had a system that is able to do work"Kinetic energy". It started for example with 1000J'S over 10 Meters the forces will = 100N's So if I simply increased the "Force" alone in the system to lets say 500N'S x 10Meters = 5000J of kinetic energy! Now the things is "FORCE" can be created while the distance is constant and I'd like to increase the energy of the system. Just studying this system more and more I'm kinda a fan of "ENERGY" and its mysterious ways 5. May 1, 2012 ### Hypo Actually I do not. Although I plan to do something if! I understood energy more. But what did you think I was going to do? Or imagined me doing or whatever :tongue:? 6. May 1, 2012 ### Ken G You are applying the formula correctly, but your interpretation sounds a bit off. Instead of thinking of work as energy, think of it is a change in energy, in particular, in kinetic energy (this is the "work-energy theorem", it is the crux of the usefulness of both the work and kinetic energy concepts). Now, since we like to imagine that total energy does not change, if kinetic energy does change, we have to dream up some other form of energy that can account for that change. The work-energy theorem tells us just how to do that-- invent "potential energy" and equate it to -work, and poof, the change in kinetic energy shows up in "potential energy", and the work done tells you how much energy was transferred from kinetic energy to potential energy (or from kinetic energy to heat, which is just more kinetic energy somewhere else). In your example, therefore, if you consider how much force F and distance D is needed to stop a moving object, then FD is a measure of how much kinetic energy (which is also mv2/2) the object must have had if it stops from force F over distance D. If you want to independently vary F and keep D constant, then it means you are not necessarily stopping the object, you are merely taking FD kinetic energy away from it (note I am imagining that F points opposite to the object's motion-- you can also add to the object's kinetic energy of course by using a positive F instead of a negative one). If the object starts out at rest, so has 0 kinetic energy originally, then you can give it kinetic energy F*D using a force F over distance D (which perhaps is what you were talking about). By the way this all follows directly from F = ma. Last edited: May 1, 2012 7. May 1, 2012 Yes. 8. May 1, 2012 ### Hypo See the thing is I'm studying force & energy all together. I understand I due follow the laws of thermo + conservation all together in the system so I understand how things are going.However,Im making example and playing with their key valuables in the system such as "F" AND "D". So all in all in a system that has "KE" of 1000J by increasing ONLY its Force the system could get more "KE" transfered from a source. Thus resulting a win/win situation for both laws force is added and energy is transfered HAH!"just go that eureka moment" Really really interesting they are! Thanks everyone any more inputs I'd be grateful for them! Actually I might come back with another thing related to this question and post it. 9. May 1, 2012 ### Pengwuino Yes but there's nothing mysterious or interesting about it as you are implying. Take a baseball players, for example. If a baseball player throw a ball with a certain force, he would impart a certain kinetic energy during his throw. If he threw it with twice the force, he would impart twice the kinetic energy during his throw (assuming the rotation of his arm and subsequent release covers the same distance). However, it's a lot more strenuous to apply a larger force when throwing a ball so it's not exactly win-win for the entire system. The same idea applies for a car engine moving a car uphill vs. flat land. It's a lot more strenuous and the engine must do more work to move a vehicle uphill (since it's giving it a potential energy as well). 10. May 1, 2012 ### Hypo For me I've been struggling to relate force and energy in a system. I looked at them in a different way but now I can see the picture with perfect colors. I understand its all depended on certain conditions areas and many factors. Physics is the breakdown of EVERYTHING physical in a system so I do keep account everything surrounding the system. Thank you for you're input! Now if force is that "KICK" or "PUSH" in a system is energy that power to supply the "PUSH" or "KICK"? 11. May 1, 2012 ### haruspex E.g. if you were thinking of pushing an object a certain distance across a frictional floor. You might have thought that if you pushed twice as hard you would use twice the energy to get it the same distance. 12. May 2, 2012 ### Hypo hahaha the beauty of simplicity thanks for that example gives me another view of my question thanks though! 13. May 2, 2012 ### sweet springs Hi. Bank account = salary paid? Yes, in a sense. No, in another sense. Regards. 14. Jun 2, 2012 ### Hypo hallo... What?! 15. Jun 2, 2012 ### HallsofIvy Staff Emeritus He is pointing out that, as you said initially, change in energy= work done on or by a system. It's a bit simplistice to just say "energy= work". 16. Jun 2, 2012 ### Staff: Mentor I would say that work is a method of transferring energy from one object or system to another one. In thermodynamics, heat is another method. In fact, the first law of thermodynamics says those are the only two methods for changing the internal energy of an system: ΔU = Q + W, or ΔU = Q - W, depending on the sign convention that you use for work. 17. Jun 2, 2012 ### Hypo Ah, thanks HallsofIvy Interesting. Similar Discussions: Energy = work?	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8142639994621277, "perplexity": 953.1609288408337}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948520042.35/warc/CC-MAIN-20171212231544-20171213011544-00520.warc.gz"}
http://soft-matter.seas.harvard.edu/index.php?title=Diffusion_through_colloidal_shells_under_stress&diff=prev&oldid=17196	# Difference between revisions of "Diffusion through colloidal shells under stress" Entry by Emily Redston, AP 225, Fall 2011 Work in Progress ## Reference Diffusion through colloidal shells under stress by J. Guery, J. Baudry, D. A. Weitz, P. M. Chaikin, and J. Bibette. Phys. Rev. E 79, 060402(R) (2009). ## Introduction One area of great interest in soft matter is encapsulation. With applications in almost all types of industrial domains, from the oil industry to food packaging, efficient storage of gases and liquid in solid containers is of tremendous technological and economical importance. Furthermore, encapsulation of active ingredients such as drugs, proteins, nutrients, or vitamins is essential for a myriad of applications, such as drug delivery and agrichemicals. The goal of encapsulation is to protect the delicate substances inside from a harsh environment, and to retain their activity until some required time. Long-term storage of liquids or gases often involve internal pressures, which increase the tensile stress of the container wall. At a colloidal scale, it is the osmotic pressure difference between the internal and external medium that is relevant, rather than the hydrostatic pressure. Eyring proposed that the permeability of solids is associated with a diffusive process involving an activation mechanism. Unfortunately, many of his ideas have not been tested quantitatively and thus many fundamental concepts are not fully understood. In this paper, the authors propose a very cute little experiment that tests some of Eyring's ideas. By using core-shell (liquid core - solid shell) colloidal particles that are sensitive to osmotic pressure, they are able to follow the permeation of encapsulated probes at various stresses. ## The Core-Shell Colloids The authors create these colloidal shells using a double emulsion, which consists of droplets of water in larger drops of crystallizable oil. The oil is a mixture of crystallizable triglycerids, which allows the authors to make the emulsion at high temperature, around 70°C, where the oil is fluid, and then cool the sample to solidify the oil, creating a robust shell. The robust solid shell is only achieved at temperatures below the melting temperature of the oil (around 45°C). NaCl is added to the water to prevent ripening of the droplets, and glucose is also added to match the chemical potential of the inner water to the continuous phase water. The osmotic pressure difference is then deduced from the concentration of salt and glucose. ## Experiments and Results Figure 1. (a) Temporal evolution of the fractional release (X) for a shell composed of pure oil DM, at different temperatures: ($\circ$) 15 °C; ($\vartriangle$) 45 ° C; ($\diamond$) 60 ° C; and ($\Box$) 65 ° C. The solid lines correspond to the best adjustment of the data with the equation: X(t)=1−[(1−$X_{0}$)exp(−$\Beta_{0}$t)]. (b) Evolution of the characteristic relaxation rate $\Beta_{0}$ in logarithmic scale with 1/$k_{B}$$T_{r}$. The solid line corresponds to an exponential adjustment. Figure 2. (a) Temporal evolution of the fractional release (X), for different dilution factors (T = 15°C); ($\circ$) d=0, $\Delta$$\Pi$=0 atm; ($\times$)d=1.4, $\Delta$$\Pi$=5 atm; ($\triangledown$) d=1.8, $\Delta$$\Pi$=8 atm; ($\vartriangle$) d=4, $\Delta$$\Pi$=13 atm; ($\diamond$) d=10, $\Delta$$\Pi$=15 atm; ($\Box$) d=100, $\Delta$$\Pi$=17 atm. The solid lines correspond to the best numerical adjustment of the data with Eq. (2). (b) Evolution of dX/dt (t=0) in logarithmic scale with 1−(1/d). The solid line corresponds to an exponential adjustment. Potentiometric titration is used to measure the concentration of chloride ions as they are released into the continuous water phase. The relative ionic concentraion in the water ouside of the colloids, X, can then be found by normalizing the concentration at each time by the value obtained when all ions are released. Fig. 1(a) shows this fractional release X as a function of time at different temperatures (from 15 to 65°C for the oil). At 15°C the oil is essentially solid, while at 65°C it is entirely melted. The authors believe that the measured release results entirely from the diffusion of sodium and chloride ions from the inner droplets toward the external phase because there is no observed coalescence of the inner droplets with the globule interface. This passive permeation must obey Fick’s law given by J = −d$N_{i}$ / dt = PS($C_{i}$ - $C_{e}$), where J is the flux of salt from inside toward the external phase, $N_{i}$ is the total number of salt ions inside the globule, P is the permeation coefficient of the shell, S is the total shell surface, and C is the concentration of salt inside (i) and in the external phase (e). The authors then assume that the limiting step is the permeation through the outer shell. The oil film between inner water droplets is much thinner than the shell thickness. Hence the salt concentration Ci is essentially homogeneous within the inner water droplets during the leakage process. If the globule volume fraction $\phi$ is small, we have dX/dt = $\Beta_{0}$(1 − X), (1) where the ion concentration X = $C_{e}$(t) / $C_{e}$ (t = infinity), and where $\Beta_{0}$ = 3P/a is the characteristic relaxation rate, with a being the globule radius. Therefore X(t)=1−[(1−$X_{0}$)exp(−$\Beta_{0}$t)], where $X_{0}$ is the initial burst fractional release which arises during the double emulsification process. For all temperatures studied the release mechanism is well described by a single exponential relaxation of time scale 1/$\Beta_{0}$, in agreement with the original Eyring assumption that the permeation coefficient P is proportional to exp($E_{a}$ / $k_{B}$T), where $E_{a}$ is the activation energy and $k_{B}$T is the thermal energy. Fig. 1(b) shows ln$\Beta_{0}$ as a function of 1 / $k_{B}$TT, from which the authors deduce that the activation energy $E_{A}$ is ~ (5 – 7)$k_{B}$$T_{r}$, $T_{r}$ = 25° C. In the absence of osmotic mismatch the leakage of the colloidal system is strictly diffusion driven, with no discontinuity at the liquid-solid transition again in agreement with the Eyring picture. Dilution of the double emulsion by a factor of 100 using pure water leads to an osmotic pressure difference of $\Delta$$\Pi$ = 17 atm. Because there is a slight but sufficient permeability of water through the shell the chemical potential of the water equilibrates rapidly resulting in an immediate increase in the internal pressure. At 15°C where the DM oil shell is solid, there is a rapid increase in the release of Cl− ions followed by slower approach to the asymptotic limit, as shown by the open squares in Fig. 2(a). We vary the driving osmotic pressure, $\Delta$$\Pi$, by using water with increasing concentrations of glucose, diluting the double emulsion by the same factor each time; this reduces the rate of release. For $\Delta$$\Pi$ = 0, there is a residual very slow release of Cl− shown by the open circles in Fig. 2(a). For $\Delta$$\Pi$ $\ne$ 0, the osmotic pressure mismatch causes a tensile stress on the shell, which should modify the activation energy. To check this hypothesis we must consider the relation between the osmotic pressure difference and the resulting tensile stress that acts on the colloidal shell. In the limit of a thin shell of thickness $\delta$ , this relation can be simply derived from a force balance argument. The force exerted by the osmotic pressure mismatch on each half shell, $\pi$ $a^2$$\Delta$$\Pi$ , and the force due to the tensile stress,$\tau$, acting on the perimeter that holds the two half shells together, 2$\pi$a$\delta$$\tau$, must be equal, $\tau$ = $\Delta$$\Pi$(a/2$\delta$). Part of the local activation energy $E_{A}$ for an ion permeating a solid comes from the deformation energy of adding its additional volume to the solid matrix. The worklike energy for adding this volume increases under compressive stress and decreases under tensile stress in a complex manner, which nonetheless must be linear in applied stress and ionic vol- ume, 􏰐3. The activation energy is expected to be linearly modified by the tensile stress as E􏰀􏰏􏰁=Ea−􏰏􏰐3. The osmotic pressure difference 􏰋􏰌 can be approximated as 􏰋􏰌 = RT􏰀2Ci − 2Ce − Csugar􏰁 􏰄 RT􏰀2Ci − Csugar􏰁 since Ce can be neglected as long as the globule volume fraction remains small, Csugar is the final sugar concentration imposed by the dilution, and R is the ideal gas constant. 􏰋􏰌 can now be expressed as 􏰋􏰌 = 2RTCi0􏰀1 − X − 1 / d􏰁, where Ci0 is the ini- tial internal salt concentration and d is the dilution factor of the external phase that sets the initial osmotic pressure dif- ference 􏰋􏰌0. Thus, Eq. 􏰀1􏰁 becomes dX / dt = 􏰊􏰀X􏰁􏰀1 − X􏰁 and the characteristic relaxation rate 􏰊􏰀X􏰁 can be expressed as 200 Time (min) 􏰊􏰀X􏰁 = 􏰊0 exp􏰂􏰑􏰀1 − 1/d􏰁􏰃exp􏰀− 􏰑X􏰁, 􏰀2􏰁 with 􏰑=􏰀a/􏰍􏰁􏰀􏰐3/k T􏰁􏰀RTC 􏰁. In kT units, 􏰑􏰀1−1/d􏰁 is the B i0 initial drop of the activation energy induced by the initial tensile stress. Thus, at short time when few ions have dif- fused through the shells and the osmotic shock is still at its maximum, the permeation rate 􏰊 increases by the large factor exp􏰂􏰑􏰀1 − 1 / d􏰁􏰃. At longer time the second term exp􏰀−􏰑X􏰁, which reflects the equilibration of the osmotic stress, slows down the relaxation rate. A plot of log􏰂dX/dt􏰀t=0􏰁􏰃, as a function of 1−1/d is linear with slope 􏰑, as seen in Fig. 3􏰀b􏰁. From the slope we find 􏰑=2.5􏰒0.3. Equation 􏰀2􏰁 can be numerically solved and compared to our data as shown in Fig. 3􏰀a􏰁. In this fit we also take into account the nonideal dependence of the osmotic pressure with the sugar and salt concentrations 􏰂15􏰃. The fit is remark- able and allows for the determination of the two unknown parameters 􏰑 and 􏰊0. The time evolution reflects the expo- nential decrease in diffusion as the osmotic pressure tends toward equilibration causing the stress to decrease and acti- vation energy to increase. It provides another measure of the enhancement of the permeation rate by stress to be compared with the values obtained by varying the initial osmotic shock, Fig. 3􏰀b􏰁 􏰀􏰑=4.1􏰒1.4 and 􏰊0=8􏰒6 10−4 min−1􏰁. With 􏰍=100 nm we get 􏰐=0.6 nm comparable to the size of a hydrated ion, as expected from the Eyring picture.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8764919638633728, "perplexity": 2383.612598966161}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711336.41/warc/CC-MAIN-20221208114402-20221208144402-00353.warc.gz"}
https://eng.libretexts.org/Under_Construction/Purgatory/Book%3A_Fluid_Mechanics_(Bar-Meir)__-_old_copy/12%3A_Compressible_Flow_2%E2%80%93Dimensional/12.2%3A_Oblique_Shock/12.2.3%3A_Application_of_Oblique_Shock	# 12.2.3: Application of Oblique Shock Fig. 12.13 Two variations of inlet suction for supersonic flow. One of the practical applications of the oblique shock is the design of an inlet suction for a supersonic flow. It is suggested that a series of weak shocks should replace one normal shock to increase the efficiency (see Figure (??)). Clearly, with a proper design, the flow can be brought to a subsonic flow just below $$M=1$$. In such a case, there is less entropy production (less pressure loss). To illustrate the design significance of the oblique shock, the following example is provided. Example 12.5 Fig. 12.14 Schematic for Example (??). The Section described in Figure and efoblique:fig:inletEx} air is flowing into a suction section at $$M=2.0$$, $$P=1.0[bar]$$, and $$T=17^{\circ}C$$. Compare the different conditions in the two different configurations. Assume that only a weak shock occurs. Solution 12.5 The first configuration is of a normal shock ### Contributors • Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9349696040153503, "perplexity": 1650.1117032882748}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578528702.42/warc/CC-MAIN-20190420060931-20190420082856-00062.warc.gz"}
http://www.chegg.com/homework-help/questions-and-answers/i-ve-read-explanations-site-i-don-tunderstand-solutions-pleaseexplain-problem-q400008	## Help on 21p I've read both of the explanations on the site and I still don'tunderstand how they found the solutions. Could someone pleaseexplain this problem to me? • Given : Velocity of the girl relative to ice surface bevg and velocity of the plank relative to the ice surface bevp Theefore velocity of the girl relative to theplank, vg-vp= 1.5m/s .........(1) Girl'smass, mg=45kg Mass of theplank, mg=150kg The momentum of the girl-plank system iszero. Therefore,mgvg+mpvp =0 (45kg)vg+(150kg)vp = 0 vg =-3.33vp .............(2) Substitute eq ( 2 ) in ( 1 ) velocity of the plank relative tothe ice surface be vp= -0.346m/s girl's speed relative to ice,vg =(-3.33)(-0.346) = 1.15m/s Get homework help	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.888679027557373, "perplexity": 4827.277407012464}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368698141028/warc/CC-MAIN-20130516095541-00062-ip-10-60-113-184.ec2.internal.warc.gz"}
https://www.gradesaver.com/textbooks/math/applied-mathematics/elementary-technical-mathematics/chapter-1-review-page-102/60	## Elementary Technical Mathematics Cross multiply to find percent. $\frac{percent}{100}=\frac{amount}{base}$ $\frac{p}{100}=\frac{\frac{11}{64}}{\frac{13}{32}}$ $\frac{13}{32}p=\frac{11}{64}\times100$ $\frac{13}{32}p\times\frac{32}{13}=\frac{11}{64}\times100\times\frac{32}{13}$ $\frac{13}{32}p\times\frac{32}{13}=\frac{11}{/\!\!\!32\times2}\times100\times\frac{/\!\!\!32}{13}$ $p=\frac{11}{26}\times100$ $p=42.3$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9327213764190674, "perplexity": 1142.2900333637338}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676590559.95/warc/CC-MAIN-20180719051224-20180719071224-00414.warc.gz"}
http://www.physicsforums.com/showthread.php?t=488359	# Hanging Wire Uniform Magnetic Field by jegues Tags: field, magnetic, solvedhanging, uniform, wire P: 1,089 1. The problem statement, all variables and given/known data See figure attached. 2. Relevant equations 3. The attempt at a solution $$\vec{F_{b}} = i\vec{L} \times \vec{B}$$ Using my right hand rule, I can see that the force is pushing the wire to the right. This correlates with what is in the drawing. As the force pushes the wire to the right, due to the finite length of the strings on which it hangs, it will start to move up and to the right. If I can somehow figure out the force that causes it to move, I can solve for B. Since the wire is moving to the right and up do I have to relate the force due to the field to the force of gravity somehow? Thanks again! EDIT: Solved it. The answer is A I believe. Attached Thumbnails	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8974257707595825, "perplexity": 258.24502841685876}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1409535922087.15/warc/CC-MAIN-20140909045455-00219-ip-10-180-136-8.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/maximum-tensile-stress-on-rod-with-eccentric-rotating-tensile-load.754716/	# Maximum tensile stress on rod with eccentric rotating tensile load 1. May 21, 2014 ### biplab93 I was solving this paper, but got stuck on this question, and it's been bugging me endlessly. I don't know what I'm missing. Here's the question: A rod of 20 dia is fixed to the ceiling of a roof on one end. A rotor of 50 kg mass is attached to the free end with bearings. The CG of the rotor is 10 mm away from the shaft axis. The rotor is rotating at 600 rpm. The max tensile stress (in N/Sq.mm) in the rod is nearly equal to A. pi/2 B. 200pi C. 300pi Because of the bearings, there will be no torsional load on the rod (right?). That only leaves out the eccentric tensile load. So I converted that into an axial tensile load and a bending moment (50g10 Nmm), and the resulting bending stress is: 1.6 (axial) + 6.3 (bending) N/mm^2 = 7.9. (Bending stress = My/I. y=max distance from neutral axis) That's a far cry from 400pi. I don't know how the length of the rod (500mm) or the speed of the rotor (600rpm) will be incorporated, or if they will be needed at all. Help will be much appreciated. Thanks in advance. 2. May 21, 2014 ### AlephZero Right. No .... The CG of the rotor is moving in a circle radius 10mm at 600 RPM. What force is acting on the rotor to make it do that? How is that force transmitted through the rod? 3. Apr 27, 2017 ### Sia I was able to get the answer, however, I am not totally convinced. It's like: We know tensile stress would be σ= 32 M / Πd^3 Now bending moment M will be due to the eccentric loading as well as due to the centrifugal force acting on the rod as a result of rotor's CG rotation. So, M= m.g.e + m. ω^2. e.L =(50 X 10 X 10 ) + {50 X(600X2Π/60)^2 X10X 0.500} =5000 + 100000Π^2 As 100000Π^2>> 5000, 5000 seems to be ignored (for simplifying calculation) ∴ σ=32X 100000Π^2/(Πx20^3) =400Π Draft saved Draft deleted Similar Discussions: Maximum tensile stress on rod with eccentric rotating tensile load	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8674368262290955, "perplexity": 1760.9864337535757}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886106865.74/warc/CC-MAIN-20170820170023-20170820190023-00626.warc.gz"}
https://www.physicsforums.com/threads/rod-clay-and-an-inelastic-collision.32692/	# Rod, Clay, and an Inelastic Collision 1. Jun 26, 2004 ### e(ho0n3 Problem: A thin rod of mass M and length L rests on a frictionless table and is struck at a point L/4 from its center of mass by a clay ball of mass m moving at a speed v (the velocity vector is perpendicular to the rod). The ball sticks to the rod. Determine the translational and rotational motion of the rod after the collision. I can use conservation of angular momentum to determine the rotational velocity of the rod and the clay about the center of mass of the rod. Then, I'd figure I could use this angular speed to find the velocity of the clay and then use conservation of linear momentum to find the velocity of the center of mass of the rod. According to my calculations, the center of mass of the rod is moving in the same direction of the velocity vector v. Does this make sense? 2. Jun 26, 2004 ### Gza That seems like an awful lot of work to simply calculate the velocity of the center of mass after collision. Just use conservation of linear momentum throughout to calculate translational velocity. Use conservation of angular momentum to find the angular velocity about its new center of mass after collision. 3. Jun 26, 2004 ### e(ho0n3 Hmm...I thought the rotation was about the center of mass of the rod not about the center of mass of the system after collision. How do you figure? 4. Jun 26, 2004 ### Staff: Mentor The motion of a rigid object (such as the composite object "rod + putty") can be described as a translational motion of its center of mass plus a rotation about its center of mass. You know how to find the speed of the center of mass after the collision: linear momentum is conserved. To find the rotational speed about the cm, do this. First find where the cm is just at the instant of collision. Then find the angular momentum about the cm prior to the collision. Since it's conserved, that's also the angular momentum about the cm after the collision. Find the rotational inertia of the "rod + putty" about the cm, then use it to find $\omega$. 5. Jun 26, 2004 ### e(ho0n3 I think you meant "can be BEST describe...", since motion is relative to a reference frame and the point in the frame I'm describing the motion from. When you use the phrase about its center of mass, I imagine the object is rotating about an axis going through the center of mass, but how do I know, in this problem for example, where the rotation axis is (I know it's not fixed in some location in space, but that is all I know). Let me get this clear: I should do all of my calculations (pre- and post-collion) taking the center of mass of the rod + putty as my reference point. 6. Jun 27, 2004 ### Staff: Mentor translation plus rotation As long as you realize that the motion of a rigid object is a combination of the motion of its cm plus its rotation about its cm. I'm not saying that the object is in pure rotation about its center of mass. It's also translating. I would, but I'm lazy. (All what caculations? It's just angular momentum.) As long as you realize that the total angular momentum of the object about any axis is the sum of its angular momentum about its center of mass plus the angular momentum of its mass (assumed concentrated at its center of mass), then you can use any axis. So, if you find the initial angular momentum about the cm, then that will equal the final angular momentum about the cm, which would equal $I_{cm}\omega$. Similar Discussions: Rod, Clay, and an Inelastic Collision	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8936043381690979, "perplexity": 226.71289343626185}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886102309.55/warc/CC-MAIN-20170816170516-20170816190516-00156.warc.gz"}
https://encyclopediaofmath.org/index.php?title=Wiener_chaos_decomposition&printable=yes	# Wiener chaos decomposition Let $U$ be a dense subspace of a separable Hilbert space $H$. The triplet $U \subset H \subset U ^ {}$ given by the injection $i : U \rightarrow H$ is obtained by identifying $H$ with its dual, taking the dual of $i$, and endowing $U ^ {}$, the algebraic dual of $U$, with the weak topology. For any real $\lambda$, let $\lambda H$ be the Hilbert space obtained from $H$ by multiplying the norm on $H$ by $\lambda$. The dual of the symmetric $k$- fold tensor product $S _ {k} ( U)$ is the space $\mathop{\rm Pol} _ {k} ( U)$ of all homogeneous polynomials of degree $k$ on $U$. The value of $F _ {k} \in \mathop{\rm Pol} _ {k} ( U)$ at $u \in U$ is $F _ {k} ( u ) = \langle F _ {k} , u ^ {\otimes k } \rangle _ {k! }$. Thus, for each $k$ there is a triplet $$\tag{a1 } S _ {k} ( U) \subset \sqrt k! S _ {k} ( H) \subset \mathop{\rm Pol} _ {k} ( U) .$$ Taking the direct sum of the internal space $S _ {k} ( U)$ and the Hilbert sum of the central spaces there results a triplet $$\tag{a2 } S( U) \subset \mathop{\rm Fock} ( H) \subset \mathop\widehat{ {\rm Pol}} ( U),$$ called dressed Fock space. The middle term is the usual Fock space $$\tag{a3 } \mathop{\rm Fock} ( H) = \oplus \sqrt k! S _ {k} ( H) .$$ The external space is the space $\prod _ {k} \mathop{\rm Pol} _ {k} ( U)$ of all formal power series on $U$. The value $F( u )$ at $u \in U$ of such an $F \in \mathop\widehat{ {\rm Pol}} ( U)$ is defined as $\lim\limits _ \rightarrow \sum _ {k=} 1 ^ {N} F _ {k} ( u )$, if this limit exists. For example, for any $F = \sum F _ {k} \in \mathop{\rm Fock} ( H)$ one has $$\tag{a4 } F( u ) = \langle F, e ^ {u} \rangle ,$$ where $e ^ {u} = \sum ( k!) ^ {-} 1 u ^ {\otimes k }$. A probabilized vector space is a structure $$\tag{a5 } ( U \dots X \supset \Omega , {\mathsf P} )$$ where $U$ and $X$ are two spaces in duality and $X = \mathop{\rm span} ( \Omega )$ is linearly generated by the subset $\Omega$ of $X$. This subset is endowed with a Polish (or Suslin) topology such that any $u \in U$ defines a Borel function $u( \omega ) = \langle u , \omega \rangle$ on $\Omega$. The space $U$ contains a countable subset separating the points of $\Omega$( so that the Borel $\sigma$- field is generated by $U$). Finally, ${\mathsf P}$ is a probability measure on this $\sigma$- field. Assume, moreover, that the space of cylindrical polynomials $P( \Omega ) = \mathop{\rm span} ( u ( \omega ) ^ {k} : u \in U, k = 0, 1, 2 , . . . )$ is dense in $L _ {2} ( \Omega )$. Assume that the following bilinear form on $U$ is a scalar product: $$\tag{a6 } b( u , v) = {\mathsf E} ( [ u ( \omega )- {\mathsf E} ( u ( \omega ))] [ v( \omega ) - {\mathsf E} ( v( \omega ))]) ,$$ and let $H$ be the completion of $U$. For any $k > 0$, let $\pi _ {k}$ denote the orthogonal projection of $L _ {2} ( \Omega )$ with range $\overline{ {P _ {<} k ( \Omega ) }}\;$, the closure of $\mathop{\rm span} ( u ( \omega ) ^ {j} : u \in U, j< k )$. Let $KO _ {k}$ be the orthogonal complement of $\overline{ {P _ {<} k ( \Omega ) }}\;$ in $\overline{ {P _ \leq k ( \Omega ) }}\;$. This space is called the $k$- th homogeneous chaos. The space $L _ {2} ( \Omega )$ is the Hilbert direct sum of the $KO _ {k}$. One says that $L _ {2} ( \Omega )$ admits a decomposition in chaos if for any $k$ the following mapping is isometric: $$\sqrt k! S _ {k} ( H) \supset S _ {k} ( U) \ni \ Q \mapsto ^ { {I _ k} } Q - \pi _ {k} ( Q) \in \ KO _ {k} \subset L _ {2} ( \Omega ) .$$ The collection of these isometries for $k = 0, 1 \dots$ is an isometry $I$ whose inverse $$\tag{a7 } L _ {2} ( \Omega ) \rightarrow ^ { {I ^ {-}} 1 } \mathop{\rm Fock} ( H) ,\ \ f \rightarrow \widehat{f} ,$$ extended to distributions on $\Omega$, is the starting point of distribution calculus on $\Omega$. Because of (a4), $\widehat{f}$ is explicitly given by $$\tag{a8 } \widehat{f} ( u ) = \langle \widehat{f} , e ^ {u} \rangle = {\mathsf E} [ f \epsilon ^ {u} ] ,$$ where $\epsilon ^ {u} = I ^ {-} 1 ( e ^ {u} )$. Decomposition in chaos was discovered by N. Wiener (in the case $\Omega$ is Wiener space), [a1]. Further contributions are due to Th.A. Dwyer and I. Segal ([a2], [a3]) and these have been important for constructive quantum field theory. K. Itô obtained a decomposition into chaos for Poisson probability spaces and interpreted $I _ {k} ( f )$ as iterated stochastic integrals. For formula (a8), extended to distributions for Gaussian probability spaces, cf. [a5], [a6], [a7], [a9], [a10]. There are links with Malliavin calculus, [a8]. For more material cf. e.g. also [a11], [a12]; Wick product and White noise analysis, and the references therein. #### References [a1] N. Wiener, "The homogeneous chaos" Amer. J. Math. , 60 (1938) pp. 897–936 [a2] Th.A., III Dwyer, "Partial differential equations in Fischer–Fock spaces for the Hilbert–Schmidt holomorphy type" Bull. Amer. Math. Soc. , 77 (1971) pp. 725–730 [a3] I. Segal, "Tensor algebras over Hilbert spaces, I" Trans. Amer. Math. Soc. , 81 (1956) pp. 106–134 [a4] K. Itô, "Multiple Wiener integral" J. Math. Soc. Japan (1951) pp. 157–169 [a5] P. Krée, "Solutions faibles d'equations aux dérivées fonctionelles II" , Sem. P. Lelong 1973/1974 , Lect. notes in math. , 474 , Springer (1974) pp. 16–47 [a6] P. Krée, R. Raczka, "Kernels and symbols of operators in quantum field theory" Ann. Inst. H. Poincaré (1978) [a7] B. Lascar, "Propriétés locales des espaces de type Sobolev en dimension infinie" Comm. Partial Diff. Eq. , 1 : 6 (1976) pp. 561–584 [a8] D. Ocone, "Malliavin calculus and stochastic integral representation of functionals of diffusion processes" Stochastics , 12 (1984) pp. 161–185 [a9] M. Krée, "Propriété de trace en dimension infinie d'espaces du type Sobolev" C.R Acad. Sci. Paris , 279 (1974) pp. 157–160 [a10] M. Krée, "Propriété de trace en dimension infinie d'espaces de type Sobolev" Bull. Soc. Math. de France , 105 (1977) pp. 141–163 [a11] G. Kallianpur, "The role of reproducing kernel Hilbert spaces in the study of Gaussian processes" P. Ney (ed.) , Advances in probability and related topics , 2 , M. Dekker (1970) pp. 49–84 [a12] J. Neveu, "Processus aléatoires Gaussiens" , Univ. Montréal (1968) How to Cite This Entry: Wiener chaos decomposition. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Wiener_chaos_decomposition&oldid=49218 This article was adapted from an original article by P. KrÃ©e (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9868508577346802, "perplexity": 626.7220634947083}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358208.31/warc/CC-MAIN-20211127163427-20211127193427-00528.warc.gz"}
http://tex.stackexchange.com/questions/24311/superpose-a-subscript-and-the-rest-of-a-formula?answertab=oldest	# Superpose a subscript and the rest of a formula I have a formula in the equation environment that goes like Foo = \bigcup_{x \in veryveryveryverylongdescription} Bar(x) Because the state space's description is long, the Bar(x) is rejected far away from the \bigcup. I would like Bar(x) to be right next to the union symbol, above the description. Can this be done cleanly? Or shall I try to have with two equations, some vspace and horizontal space shifts? - Use the \smashoperator macro from the mathtools package: \documentclass{article} \usepackage{amsmath} \usepackage{mathtools} \begin{document} $\mathit{Foo} = \smashoperator{\bigcup_{x \in \text{long text}}} \operatorname{Bar}(x)$ \end{document} - I would suggest to try also \smashoperator[r]{...}; this depends on the "long text": if not very long, pushing it under the equals sign may be OK, but not if it goes beyond it. – egreg Jul 28 '11 at 10:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9822549819946289, "perplexity": 1783.2118557105666}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375098464.55/warc/CC-MAIN-20150627031818-00181-ip-10-179-60-89.ec2.internal.warc.gz"}
http://mathhelpforum.com/algebra/138008-exponential-function.html	1. ## Exponential Function Crescent City, California, has historically been at risk for tsunami waves. The probability P (as a decimal) of no tsunami wave of height 15 feet or more striking Crescent City over a period of Y years decreases as the time interval increases. Increasing the time interval by 1 year decreases the probability by about 2% (a) Explain why P is an exponential function of Y. For each change in Y of 1 year, the probability P decreases by 2%, , so P exhibits a constant percent change. (b) What is the decay factor for P? ____per yr (I thought this would be 0.02) (c) Recall that the probability of a certainty is 1. What is the initial value of P? P(0) = 1 (d) Find a formula for P as a function of Y. P = 1(0.02^Y) (That has to be wrong.) (e) Find a formula for the probability Q that at least one wave 15 feet or higher will strike Crescent City over a period of Y years. (Suggestion: It should be helpful to note that the probability of the occurrence of an event plus the probability that it will not occur is the probability of a certainty, 1.) Q = (Where the heck did Q come from?) 2. Originally Posted by MathBane (b) What is the decay factor for P? ____per yr (I thought this would be 0.02) It will be $1-0.02 = 0.98$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9435988068580627, "perplexity": 885.1792785882599}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738660350.13/warc/CC-MAIN-20160924173740-00191-ip-10-143-35-109.ec2.internal.warc.gz"}
http://www.hcm.uni-bonn.de/events/eventpages/lipschitz-lectures/closed-minimal-hypersurfaces-in-riemannian-manifolds/	# Lipschitz Lectures ## Camillo De Lellis University: Universität Zürich Local Organizer: László Székelyhidi Date: 15 May - 29 May 2009 Location: Wegelerstraße 10, University of Bonn, Germany Schedule: Friday, May 15, 10:15 – 12:00, Zeichensaal Monday, May 18, 12:15 – 14:00, Kleiner Hörsaal Wednesday, May 20, 10:15 – 12:00, Kleiner Hörsaal Friday, May 22, 10:15 – 12:00, Zeichensaal Monday, May 25, 12:15 – 14:00, Kleiner Hörsaal Wednesday, May 27, 10:15 – 12:00, Kleiner Hörsaal ### Abstract In 1917 Birkhoff proved the existence of a nontrivial closed geodesic in any Riemannian 2-dimensional manifold diffeomorphic to the sphere. This result was later improved in a famous work by Ljusternik and Shnirelman to the existence of three distinct closed geodesics. The interest of Birkhoff's theorem lies on the fact that it cannot be achieved by standard "minimization arguments": since any closed curve in the sphere is contractible, minimizers of the length are necessarily trivial. In fact, much more can be proved: if the curvature of the metric is positive, there is no stable closed geodesic. It is natural to ask whether the method of Birkhoff, a "min-max argument", can be extended to higher dimensions to produce nontrivial minimal hypersurfaces in general Riemannian manifolds. The answer to this question is positive and has quite important applications to other problems in geometry and topology. However, the difficulties in generalizing Birkhoff's argument are many, since the area functional does not enjoy good functional analytic properties. Indeed, the first proof of the existence of an (immersed) minimal 2-sphere in any Riemannian manifold diffeomorphic to the 3-sphere appeared only in 1981 in a celebrated paper by Sacks and Uhlenbeck. Much research in the area has been triggered by a monograph of Pitts, who proved the existence of smooth embedded closed minimal hypersurface in any closed compact Riemannian manifold of dimension at most 6. Pitts' monograph appeared in 1981 and his result was then generalized to all dimensions by Schoen and Simon. In a recent work with Dominik Tasnady we have introduced an approach similar to that of Pitts which however shortens the proof dramatically. The aim of this course is to give an account of this proof and of several tools of geometric measure theory which are used in it. I will therefore give short introductions to topics like the theory of Caccioppoli sets, the theory of varifolds and the curvature estimates for stable minimal hypersurfaces (due to works of Schoen, Simon and Yau).	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9503394961357117, "perplexity": 645.9079768500941}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084886815.20/warc/CC-MAIN-20180117043259-20180117063259-00747.warc.gz"}
http://feb.trunojoyo.ac.id/snow-movies-aonpc/013321-what-is-epidermis-what-is-its-role	# what is epidermis what is its role The Epidermis . That is, the epidermis outermost layer consists of dead cells packed with the tough protein keratin. Q No 13: What is the role of epidermis in plants? Formation The aerial parts of plants have waxy, water resistant layer on the outer surface of epidermal cells which in turn reduces water loss and provide protection against mechanical injury and invasion of parasitic fungi. Its thickness depends on where it is located on the body. The epidermis plays a vital role in skin health…. It is at its thinnest on the eyelids, measuring just half a millimeter, and at its thickest on the palms and soles at 1.5 millimeters. It's thickest on the palms of the hands and soles of the feet (1.5 millimeters). the outermost and nonvascular layer of the skin, derived from the embryonic ectoderm, varying in thickness from 0.07 to 1.4 mm. The epidermis is composed of multiple layers of flattened cells that overlie … Three main populations of cells reside in the epidermis: keratinocytes, melanocytes, and Langerhans cells. I mean, it is the layer that we can see, touch, and work hard to make it look good enough to show off to the world. epider´mides) (Gr.) Functions of the Epidermis The epidermis is a keratinized stratified squamous epithelium. Ans: Epidermis is present on the outer surface of the whole plant body. The epidermis is the outermost of the three layers that make up the skin, the inner layers being the dermis and hypodermis. The deepest part of the epidermis also contains melanocytes. The epidermis is the outermost layer of the skin. Under the squamous cells are round cells called basal cells. Epidermis is the outermost layer and is about 0.05–1 mm in thickness depending on body part. So the next time you see an article or video online talking about the epidermis, hopefully now you’ll be … Plant epidermis is unique because it is actually two different layers of cells: the upper epidermis and the lower epidermis. The epidermis layer provides a barrier to infection from environmental pathogens and regulates the amount of water released from the body into the atmosphere through transepidermal water loss. These cells produce melanin, which gives the skin its color. The thickness of the epidermis varies depending on where on the body it is located. Epidermis: Epidermis is present on the outer surface of the whole plant body. The epidermis is mostly made up of flat, scale-like cells called squamous cells. epidermis [ep″ĭ-der´mis] (pl. It performs the following important functions:- Cells of epidermis are water resistant thus prevent excess loss of water. The cells of the epidermal tissue form a continuous layer without any intercellular space. 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Learn how your comment data is processed.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8878099918365479, "perplexity": 4482.510002697055}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487623596.16/warc/CC-MAIN-20210616093937-20210616123937-00010.warc.gz"}
https://encyclopediaofmath.org/wiki/Bimodule	# Bimodule 2010 Mathematics Subject Classification: Primary: 16D20 [MSN][ZBL] double module An Abelian group $B$ that is a left module over a ring $R$ and a right module over a ring $S$, and is such that $(rb)s = r(bs)$ for all $r\in R$, $b \in B$, $s \in S$. One writes ${}_R B_S$, or that $B$ is an $(R,S)$-bimodule. The bimodule $B$ may be regarded as a left $R \otimes S^{\mathrm{op}}$-module, where $S^{\mathrm{op}}$ is the opposite ring (dually isomorphic, anti-isomorphic) to $S$, while $\otimes$ denotes the tensor product over the ring of integers, and $(r\otimes s)b = rbs$. For every left $R$-module $M$ one has the situation ${}_R M_E$, where $E$ is the ring of endomorphisms of $M$. Any ring $R$ can be given the natural structure of an $(R,R)$-bimodule. A bimodule morphism is a mapping from a bimodule ${}_R B_S$ into a bimodule ${}_R C_S$ that is left $R$-linear and right $S$-linear. The category of $(R,S)$-bimodules with bimodule morphisms is a Grothendieck category. The centre of an $(R,R)$-bimodule (also called an $R$-bimodule) $B$ is defined to be the set $$Z_R(B) = \{x \in B : rx = xr \ \text{for all}\ r \in R\}\ .$$ Clearly $Z_R(B)$ is a two-sided $Z_R(R)$--module. In particular, when $R$ is commutative, the distinction between left and right modules disappears and any $R$-module may be regarded as an $(R,R)$-bimodule.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9459459185600281, "perplexity": 122.49611990632611}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711042.33/warc/CC-MAIN-20221205164659-20221205194659-00440.warc.gz"}
https://www.physicsforums.com/threads/binomial-theorem-and-modular-arithmetic-proof-check.347828/	# Binomial Theorem and Modular Arithmetic Proof Check 1. Oct 21, 2009 ### Hotsuma 1. The problem statement, all variables and given/known data $$\mbox{Prove or give a counterexample: If p is a prime integer, then for all integers x and y, } (x+p)^p \equiv_p x^p+y^p$$. 2. Relevant equations $$\equiv_p \mbox{just means (mod p). Can you please check and see if this proof is well-formed?}$$ 3. The attempt at a solution $$\mbox{Pf: Assume p is prime. Then} (x+y)^p= \left(\begin{array}{l c} p\\ 0\\ \end{array}\right) x^p+ \left(\begin{array}{l c} p\\ 1\\ \end{array}\right) x^{p-1}y+ \left(\begin{array}{l c} p\\ 2\\ \end{array}\right) x^{p-2}y^2+ ... + \left(\begin{array}{c c} p\\ p-1\\ \end{array}\right) xy^{p-1}+ \left(\begin{array}{l c} p\\ p\\ \end{array}\right) y^p = x^p + \sum^{p-1}_{k=1}x^ky^{p-k}+y^p.$$ $$\mbox{Notice that} \sum^{p-1}_{k=1}x^ky^{p-k} = \frac{p!}{k!(p-k)!}\mbox{, which, by a previously proven theorem, we know that } p\left\| \frac{p!}{k!(p-k)!} \right..$$ $$\mbox{ So, by the definition of (mod p) we know that} \sum^{p-1}_{k=1}x^ky^{p-k}+y^p \equiv_p 0 \mbox{ and therefore }(x+p)^p \equiv_p x^p+y^p ~~~~\hspace{1.0cm}_\blacksquare$$ 2. Oct 21, 2009 ### Dick It's not well formed. $$\sum^{p-1}_{k=1}x^ky^{p-k} = \frac{p!}{k!(p-k)!}$$ is completely false. You are summing over k. There can't be a k in the answer. The point is that each individual binomial coefficient C(p,k) is divisible by p for 1<=k<=p-1. 3. Oct 21, 2009 ### Hotsuma I discovered my error. Give me a minute. 4. Oct 21, 2009 ### Hotsuma $$\mbox{Pf: Assume p is prime. Then} (x+y)^p= \left(\begin{array}{l c} p\\ 0\\ \end{array}\right) x^p+ \left(\begin{array}{l c} p\\ 1\\ \end{array}\right) x^{p-1}y+ \left(\begin{array}{l c} p\\ 2\\ \end{array}\right) x^{p-2}y^2+ ... + \left(\begin{array}{c c} p\\ p-1\\ \end{array}\right) xy^{p-1}+ \left(\begin{array}{l c} p\\ p\\ \end{array}\right) y^p$$ $$= x^p + \sum^{p-1}_{k=1}\left(\begin{array}{l c} p\\ k\\ \end{array}\right)x^ky^{p-k}+y^p.$$ $$\mbox{Notice that} \sum^{p-1}_{k=1}\left(\begin{array}{l c} p\\ k\\ \end{array}\right)x^ky^{p-k} = \sum^{p-1}_{k=1}\left[\left(\frac{p!}{k!(p-k)!}\right) x^ky^{p-k}\right]\mbox{, which, by a previously proven theorem, we know that } p\left\| \frac{p!}{k!(p-k)!} \right..$$ $$\mbox{ So, by the definition of (mod p) we know that} \sum^{p-1}_{k=1}\left(\begin{array}{l c} p\\ k\\ \end{array}\right)x^ky^{p-k}+y^p \equiv_p 0 \mbox{ and therefore }(x+y)^p \equiv_p x^p+y^p ~~~~\hspace{1.0cm}_\blacksquare$$ Last edited: Oct 21, 2009 5. Oct 21, 2009 ### Hotsuma Alright. How does that look? 6. Oct 21, 2009 ### Dick The third line is still completely garbled. The second line would say you can ignore the k=1 to k=p-1 terms because they are all divisible by p. Say the the previous theorem you are citing is for 1<=k<=p-1. On the last line (x+p)^p? Come on. Similar Discussions: Binomial Theorem and Modular Arithmetic Proof Check	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8462278246879578, "perplexity": 2607.7103816810113}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886133042.90/warc/CC-MAIN-20170824062820-20170824082820-00637.warc.gz"}
http://cms.math.ca/cjm/msc/11Y40?fromjnl=cjm&jnl=CJM	location: Publications → journals Search results Search: MSC category 11Y40 ( Algebraic number theory computations ) Expand all Collapse all Results 1 - 4 of 4 1. CJM 2008 (vol 60 pp. 1267) Blake, Ian F.; Murty, V. Kumar; Xu, Guangwu Nonadjacent Radix-$\tau$ Expansions of Integers in Euclidean Imaginary Quadratic Number Fields In his seminal papers, Koblitz proposed curves for cryptographic use. For fast operations on these curves, these papers also initiated a study of the radix-$\tau$ expansion of integers in the number fields $\Q(\sqrt{-3})$ and $\Q(\sqrt{-7})$. The (window) nonadjacent form of $\tau$-expansion of integers in $\Q(\sqrt{-7})$ was first investigated by Solinas. For integers in $\Q(\sqrt{-3})$, the nonadjacent form and the window nonadjacent form of the $\tau$-expansion were studied. These are used for efficient point multiplications on Koblitz curves. In this paper, we complete the picture by producing the (window) nonadjacent radix-$\tau$ expansions for integers in all Euclidean imaginary quadratic number fields. Keywords:algebraic integer, radix expression, window nonadjacent expansion, algorithm, point multiplication of elliptic curves, cryptographyCategories:11A63, 11R04, 11Y16, 11Y40, 14G50 2. CJM 2007 (vol 59 pp. 553) Dasgupta, Samit Computations of Elliptic Units for Real Quadratic Fields Let $K$ be a real quadratic field, and $p$ a rational prime which is inert in $K$. Let $\alpha$ be a modular unit on $\Gamma_0(N)$. In an earlier joint article with Henri Darmon, we presented the definition of an element $u(\alpha, \tau) \in K_p^\times$ attached to $\alpha$ and each $\tau \in K$. We conjectured that the $p$-adic number $u(\alpha, \tau)$ lies in a specific ring class extension of $K$ depending on $\tau$, and proposed a Shimura reciprocity law" describing the permutation action of Galois on the set of $u(\alpha, \tau)$. This article provides computational evidence for these conjectures. We present an efficient algorithm for computing $u(\alpha, \tau)$, and implement this algorithm with the modular unit $\alpha(z) = \Delta(z)^2\Delta(4z)/\Delta(2z)^3.$ Using $p = 3, 5, 7,$ and $11$, and all real quadratic fields $K$ with discriminant $D < 500$ such that $2$ splits in $K$ and $K$ contains no unit of negative norm, we obtain results supporting our conjectures. One of the theoretical results in this paper is that a certain measure used to define $u(\alpha, \tau)$ is shown to be $\mathbf{Z}$-valued rather than only $\mathbf{Z}_p \cap \mathbf{Q}$-valued; this is an improvement over our previous result and allows for a precise definition of $u(\alpha, \tau)$, instead of only up to a root of unity. Categories:11R37, 11R11, 11Y40 3. CJM 2006 (vol 58 pp. 580) Annihilators for the Class Group of a Cyclic Field of Prime Power Degree, II We prove, for a field $K$ which is cyclic of odd prime power degree over the rationals, that the annihilator of the quotient of the units of $K$ by a suitable large subgroup (constructed from circular units) annihilates what we call the non-genus part of the class group. This leads to stronger annihilation results for the whole class group than a routine application of the Rubin--Thaine method would produce, since the part of the class group determined by genus theory has an obvious large annihilator which is not detected by that method; this is our reason for concentrating on the non-genus part. The present work builds on and strengthens previous work of the authors; the proofs are more conceptual now, and we are also able to construct an example which demonstrates that our results cannot be easily sharpened further. Categories:11R33, 11R20, 11Y40 4. CJM 2000 (vol 52 pp. 369) Granville, Andrew; Mollin, R. A.; Williams, H. C. An Upper Bound on the Least Inert Prime in a Real Quadratic Field It is shown by a combination of analytic and computational techniques that for any positive fundamental discriminant $D > 3705$, there is always at least one prime $p < \sqrt{D}/2$ such that the Kronecker symbol $\left(D/p\right) = -1$. Categories:11R11, 11Y40	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8505056500434875, "perplexity": 644.1474222571662}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246656965.63/warc/CC-MAIN-20150417045736-00192-ip-10-235-10-82.ec2.internal.warc.gz"}
https://www.math.gatech.edu/seminars-colloquia/series/stochastics-seminar/yao-xie-20171019	## Sequential low-rank matrix completion and estimation: Uncertainty quantification and design Series: Stochastics Seminar Thursday, October 19, 2017 - 15:05 1 hour (actually 50 minutes) Location: Skiles 006 , ISyE, Georgia Institute of Technology Organizer: We present a unified framework for sequential low-rank matrix completion and estimation, address the joint goals of uncertainty quantification (UQ) and statistical design. The first goal of UQ aims to provide a measure of uncertainty of estimated entries in the unknown low-rank matrix X, while the second goal of statistical design provides an informed sampling or measurement scheme for observing the entries in X. For UQ, we adopt a Bayesian approach and assume a singular matrix-variate Gaussian prior the low-rank matrix X which enjoys conjugacy. For design, we explore deterministic design from information-theoretic coding theory. The effectiveness of our proposed methodology is then illustrated on applications to collaborative filtering.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8177133798599243, "perplexity": 1217.4420772758235}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891814002.69/warc/CC-MAIN-20180222041853-20180222061853-00096.warc.gz"}
https://www.analyzemath.com/calculus/derivative/proof-derivative-of-e%5Ex.html	# Proof of Derivative of $e^x$ The proof of the derivative of the natural exponential $e^x$ is presented using the limit definition of the derivative. The derivative of a composite function of the form $e^{u(x)}$ is also presented including examples with their solutions. ## Proof of the Derivative of $e^x$ Using the Definition of the Derivative The definition of the derivative $f'$ of a function $f$ is given by the limit $f'(x) = \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$ Let $f(x) = e^x$ and write the derivative of $e^x$ as follows $f'(x) = \lim_{h \to 0} \dfrac{e^{x+h}-e^x}{h}$ Use the formula $e^{x+h} = e^x e^h$ to rewrite the derivative of $e^x$ as $f'(x) = \lim_{h \to 0} \dfrac{e^x e^h - e^x}{h}$ Factor $e^x$ out in the numerator $f'(x) = \lim_{h \to 0} \dfrac{e^x (e^h - 1)}{h}$ Since $e^x$ does not depend on $h$, the above may be rewriten as $f'(x) = e^x \lim_{h \to 0} \dfrac{ e^h - 1}{h}$ We now need to find the limit $\lim_{h \to 0} \dfrac{ e^h - 1}{h}$. Let $y = e^h - 1$ and note that $\lim_{h \to 0} y = 0$ We now express h in terms of y $e^h = y + 1$ take the ln of both sides $ln(e^h) = ln(y + 1)$ simplify the left side using the rule: $ln(e^x) = x$ $h = \ln(y + 1)$ With the above substitution, we can write $\lim_{h \to 0} \dfrac{ e^h - 1}{h} = \lim_{y \to 0} \dfrac{ y}{\ln(y+1)}$ Rewrite the term on the right as follows $= \lim_{y \to 0} \dfrac{ 1}{\dfrac{1}{y}\ln(y+1)}$ Use power rule of logarithms ( $a \ln y = \ln y^a$ ) to rewrite the above limit as $= \lim_{y \to 0} \dfrac{ 1}{\ln(y+1)^{\dfrac{1}{y}}}$ Use the limit rule of a quotient and limit of a composite function to rewrite the above as $= \dfrac{ \lim_{y \to 0} 1}{\lim_{y \to 0} \ln(y+1)^{\dfrac{1}{y}}} = \dfrac{1}{\ln \left(\lim_{y \to 0} (y+1)^{\dfrac{1}{y}} \right)}$ One of the definitions of the Euler Constant e is $e = \lim_{m \to 0} ( 1 + m) ^{\dfrac{1}{m}}$ Hence the limit we are looking for is given by $\lim_{h \to 0} \dfrac{ e^h - 1}{h} = \dfrac{1}{\ln \left(\lim_{y \to 0} (y+1)^{\dfrac{1}{y}} \right)} = \dfrac{1}{\ln e} = 1$ which finally gives $f'(x) = e^x \lim_{h \to 0} \dfrac{ e^h - 1}{h} = e^x \times 1 = e^x$ Conclusion: $\dfrac{d}{dx} e^x = e^x$ Note that any function of the form $f(x) = k e^x$, where k is a constant, is equal to its derivative. ## Derivative of the Composite Function $y = e^{u(x)}$ We now consider the composite exponential of another function u(x). Use the chain rule of differentiation to write $\displaystyle \dfrac{d}{dx} e^{u(x)} = \dfrac{d}{du} e^{u(x)} \dfrac{d}{dx} u$ Simplify $= e^u \dfrac{d}{dx} u$ Conclusion $\displaystyle \dfrac{d}{dx} e^{u(x)} = e^u \dfrac{d}{dx} u$ Example 1 Find the derivative of the composite exponential functions 1. $f(x) = e^{x^3-2x+3}$ 2. $g(x) = e^{\sqrt{x^2+1}}$ 3. $h(x) = e^{ \left(\dfrac{x}{x-2}\right)}$ Solution to Example 1 1. Let $u(x) = x^3-2x+3$ and therefore $\dfrac{d}{dx} u = \dfrac{d}{dx} (x^3-2x+3) = 3x^2-2$ Apply the rule for the composite exponential function found above $\displaystyle \dfrac{d}{dx} f(x) = e^u \dfrac{d}{dx} u = e^{x^3-2x+3} \times (3x^2-2)$ $= (3x^2-2) e^{x^3-2x+3}$ 2. Let $u(x) = \sqrt{x^2+1}$ and therefore $\dfrac{d}{dx} u = \dfrac{d}{dx} \sqrt{x^2+1} = \dfrac{x}{\sqrt{x^2+1}}$. Apply the above rule of differentiation for the composite exponential function $\displaystyle \dfrac{d}{dx} g(x) = e^u \dfrac{d}{dx} u = e^{\sqrt{x^2+1}} \times \dfrac{x}{\sqrt{x^2+1}}$ $= \dfrac{x}{\sqrt{x^2+1}} \; e^{\sqrt{x^2+1}}$ 3. Let $u(x) = \dfrac{x}{x-2}$ and therefore $\dfrac{d}{dx} u = -\dfrac{2}{\left(x-2\right)^2}$ Apply the rule of differentiation for the composite exponential function obtained above $\displaystyle \dfrac{d}{dx} h(x) = e^u \dfrac{d}{dx} u = e^{ \left(\dfrac{x}{x-2}\right)} \times ( -\dfrac{2}{\left(x-2\right)^2} )$ $= -\dfrac{2}{\left(x-2\right)^2} \; e^{\left(\dfrac{x}{x-2}\right)}$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9963369965553284, "perplexity": 239.72772557277364}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496668772.53/warc/CC-MAIN-20191116231644-20191117015644-00380.warc.gz"}
http://mat.uc.cl/seminarios/seminario-de-probabilidad-nucleo-mescd.html	# Seminario de Probabilidad - Núcleo MESCD 2016-09-05 16:30hrs. Matthieu Jonckheere. UBA Front Propagation And Quasi-Stationary Distributions For One-Dimensional Lévy Processes Facultad de Matemáticas, Campus San Joaquín, PUC Chile (Sala por confirmar) Abstract: We jointly investigate the existence of quasi-stationary distributions for one-dimensional Lévy processes and the existence of traveling waves for the Fisher-Kolmogorov-Petrovskii-Piskunov (F-KPP) equation associated with the same motion. Using probabilistic ideas developed by S. Harris for the F-KPP equation, we show that the existence of a traveling wave for the F-KPP equation associated with a centered Lévy processes that branches at rate r and travels at velocity c is equivalent to the existence of a quasi-stationary distribution for a Lévy process with the same movement but drifted by -c and killed at zero, with mean absorption time 1/r. This allows to generalize the known existence conditions in both contexts. Joint work with Pablo Groisman. 2016-08-01 Johel Beltrán. Martingale Problem And Trace Processes Applied To Metastability. Sala 2 - Facultad de Matemáticas PUC a las 16:30 Hrs. 2015-08-25 Inés Armendariz. Universidad de Buenos Aires Metastability In a Condensing Zero-Range Process In The Thermodynamic Limit. Sala 5, Facultad de Matemáticas, Campus San Joaquín - 17:00 Hrs. 2015-06-10 Hiep Han. Two Results Concerning Phase Transitions: Sharp Thresholds For Van Der Waerden´s Theorem And Biased Independent Sets In Regular Graphs	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8307338953018188, "perplexity": 2767.4557690227784}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189884.21/warc/CC-MAIN-20170322212949-00244-ip-10-233-31-227.ec2.internal.warc.gz"}
https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Book%3A_Classical_Mechanics_(Tatum)/02%3A_Moments_of_Inertia/2.12%3A_Rotation_of_Axes	$$\require{cancel}$$ # 2.12: Rotation of Axes We start by recalling a result from elementary geometry. Consider two sets of axes O$$xy$$ and O$$x_{1}y_{1}$$, the latter being inclined at an angle $$\theta$$ to the former. Any point in the plane can be described by the coordinates $$(x , y)$$ or by $$(x_{1} , y_{1})$$. These coordinates are related by a rotation matrix: $\left(\begin{array}{c}x_{1}\\ y_{1}\end{array}\right) = \left(\begin{array}{c}\cos \theta \quad \sin \theta \\ -\sin \theta \quad \cos\theta\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right), \label{eq:2.12.1}$ $\left(\begin{array}{c}x\\ y\end{array}\right) = \left(\begin{array}{c}\cos \theta \quad -\sin \theta \\ \sin \theta \quad \cos\theta\end{array}\right)\left(\begin{array}{c}x_{1}\\ y_{1}\end{array}\right). \label{eq:2.12.2}$ The rotation matrix is orthogonal; one of the several properties of an orthogonal matrix is that its reciprocal is its transpose. Now let us apply this to the moments of inertia of a plane lamina. Let us suppose that the axes are in the plane of the lamina and that O is the centre of mass of the lamina. $$A, B$$ and $$H$$ are the moments of inertia with respect to the axes O$$xy$$ , and $$A_{1} , B_{1}$$ and $$H_{1}$$ are the moments of inertia with respect to O$$x_{1}y_{1}$$. Strictly speaking a lamina implies a continuous distribution of matter in a plane, but, since matter, we are told, is composed of discrete atoms, there is little difficulty in justifying treating a lamina as though it we a distribution of point masses in the plane. In any case the results that follow are valid either for a collection of point masses in a plane or for a genuine continuous lamina. We have, by definition: $A_{1} = \sum my^{2}_{1} \label{eq:2.12.3}$ $B_{1} = \sum mx^{2}_{1} \label{eq:2.12.4}$ $H_{1} = \sum mx_{1}y_{1} \label{eq:2.12.5}$ Now let us apply Equation $$\ref{eq:2.12.1}$$ to Equation $$\ref{eq:2.12.3}$$: $$A_{1} = \sum m (-x \sin \theta + y \cos\theta )^2 = \sin^2 \theta \sum mx^2 - 2\sin\theta \cos \theta \sum mxy + \cos^2 \theta \sum my^2.$$ That is to say (writing the third term first, and the first term last) $A_{1} = A \cos^2\theta -2H \sin \theta \cos \theta + B\sin^2 \theta. \label{eq:2.12.6}$ In a similar fashion, we obtain for the other two moments $B_{1} = A \sin^2\theta +2H \sin \theta \cos \theta + B\cos^2 \theta. \label{eq:2.12.7}$ and $H_{1} = A \sin\theta \cos \theta + H \sin(\cos^2 \theta - \sin^2 \theta) - B\sin\theta \cos \theta. \label{eq:2.12.8}$ It is usually more convenient to make use of trigonometric identities to write these as $A_{1} = \frac{1}{2} (B+ A) - \frac{1}{2}(B-A)\cos2\theta - H \sin 2 \theta, \tag{2.12.9}\label{eq:2.12.9}$ $B_{1} = \frac{1}{2} (B+ A) + \frac{1}{2}(B-A)\cos2\theta + H \sin 2 \theta, \tag{2.12.10}\label{eq:2.12.10}$ $H_{1} = H \cos 2 \theta - \frac{1}{2}(B-A)\sin2 \theta \tag{2.12.11}\label{eq:2.12.11}$ These equations enable us to calculate the moments of inertia with respect to the axes O$$x_{1}y_{1}$$ if we know the moments with respect to the axes O$$xy$$. Further, a matter of importance, we see, from equation 2.12.11, that if $\tan 2 \theta = \frac{2H}{B-A} , \label{eq:2.12.12}$ the product moment $$H_{1}$$ with respect to the axes $$Oxy$$ is zero. This gives some physical meaning to the product moment, namely: If we can find some axes (which we can, by means of Equation $$\ref{eq:2.12.12}$$) with respect to which the product moment is zero, these axes are called the principal axes of the lamina, and the moments of inertia with respect to the principal axes are called the principal moments of inertia. I shall use the symbols $$A_{0}$$ and $$B_{0}$$ for the principal moments of inertia, and I shall adopt the convention that $$A_{0} ≤ B_{0}$$. Example $$\PageIndex{1}$$ Consider three point masses at the coordinates given below: Mass Coordinates 5 (1 , 1) 3 (4 , 2 ) 2 (3 , 4) The moments of inertia are $$A = 49, B = 71, C = 53$$. The coordinates of the centre of mass are (2.3 , 1.9). If we use the parallel axes theorem, we can find the moments of inertia with respect to axes parallel to the original ones but with origin at the centre of mass. With respect to these axes we find $$A = 12.9, B = 18.1, H = +9.3$$. The principal axes are therefore inclined at angles $$\theta$$ to the $$x$$ -axis given (Equation $$\ref{eq:2.12.12}$$) by $$\tan 2 \theta = 3.57669$$; That is $$\theta$$ = 37°11' and 127° 11'. On using Equation $$\ref{eq:2.12.9}$$ or 10 with these two angles, together with the convention that $$A_{0} ≤ B_{0}$$ , we obtain for the principal moments of inertia $$A_{0} = 5.84$$ and $$B_{0} = 25.16$$. Example $$\PageIndex{2}$$ Consider the right-angled triangular lamina of Section 11. The moments of inertia with respect to axes passing through the centre of mass and parallel to the orthogonal sides of the triangle are $$A= \frac{1}{18} Mb^ 2, B= \frac{1}{18} Ma^2, H=− \frac {1}{36} Mab$$. The angles that the principal axes make with the $$a$$- side are given by $$\tan 2 \theta = \frac{ab}{b^2-a^2}$$ . The interested reader will be able to work out expressions, in terms of $$M, a, b,$$ for the principal moments.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8096132278442383, "perplexity": 180.435587849681}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250608295.52/warc/CC-MAIN-20200123041345-20200123070345-00307.warc.gz"}
http://doc.utwente.nl/62108/	# Generating All Circular Shifts by Context-Free Grammars in Chomsky Normal Form Asveld, P.R.J. (2006) Generating All Circular Shifts by Context-Free Grammars in Chomsky Normal Form. Journal of Automata, Languages and Combinatorics, 11 (2). pp. 147-159. ISSN 1430-189X PDF Restricted to UT campus only : Request a copy171kB Abstract: Let be an alphabet of symbols and let be the language of circular shifts of the word ; so . We discuss a few families of context-free grammars () in Chomsky normal form such that generates . The grammars in these families are inverstigated with respect to their descriptional complexity, i.e., we determine the number of nonterminal symbols and the number of rules of as functions of . These and happen to be functions bounded by low-degree polynomials, particularly when we focus our attention to unambiguous grammars. Finally, we introduce a family of minimal unambiguous grammars for which and are linear. Item Type: Article Additional information: The paper is in the 2006-volume of Journal of Automata, Languages and Combinatorics; however, this volume appeared in 2008. This journal has no PDF-repository and no DOI's. Faculty: Electrical Engineering, Mathematics and Computer Science (EEMCS) Link to this item: http://purl.utwente.nl/publications/62108 Official URL: http://jalc.de/ Export this item as: BibTeXEndNoteHTML CitationReference Manager Repository Staff Only: item control page Metis ID: 248491	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8016799688339233, "perplexity": 1971.962871532818}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246651471.95/warc/CC-MAIN-20150417045731-00303-ip-10-235-10-82.ec2.internal.warc.gz"}
http://math.stackexchange.com/questions/243782/convex-function-plus-v-e-x	# Convex function plus $v e^{-x}$ If $f(x)$ is strictly convex, and $$\lim_{x\to \infty}\left(f(x) - x - ue^{x}\right) = w$$ for some $u\ge 0$ and $w$ then what can be said about: $$g(x) = ve^{-x} + f(x)$$ on $x\ge0$ where $v$ is some fixed real number. Can I say that it has exactly one minimum? - Do you mean to assume that $\lim_{x \to infty} \left( f(x) - x - ue^x\right) = 0$? – Hans Engler Nov 24 '12 at 16:06 Yes, thanks for the correction. – Neil G Nov 24 '12 at 16:06 If you mean that the limit is $0$ you should fix the question by replacing "limit = w" by "limit = $0$". – coffeemath Nov 24 '12 at 22:12 @coffeemath: I meant $w$, but I don't think it makes any difference? – Neil G Nov 24 '12 at 22:58 Yes, no difference, it's just a constant and drops out of the derivative. In a comment above your "yes, thanks for the correction" led me to believe you meant limit = 0. – coffeemath Nov 25 '12 at 10:39 ## 1 Answer Let $f(x)=x+e^x+0$ (taking $w=0$ from your statement) and $v=1$ so that $$g(x)=e^{-x}+x+e^x.$$ Then $g(x)$ has exactly one minimum, but not on $x \ge 0$, and numerically the min of $g(x)$ is at the point $(-0.482,1.754)$. Replacing the $0$ by an arbitrary $w$ just shifts this example up or down, with the same negative $x$ value at which the minimum occurs. Now for another example let $v=-1$ (and $w=0$ again) so that this time $$g(x)=-e^{-x}+x+e^x.$$ This function has no minimum at all, not even a local minimum. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9625914096832275, "perplexity": 432.7392214846102}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1430455241471.82/warc/CC-MAIN-20150501044041-00048-ip-10-235-10-82.ec2.internal.warc.gz"}
https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_8&diff=139518&oldid=128355	# Difference between revisions of "2002 AIME II Problems/Problem 8" ## Problem Find the least positive integer for which the equation has no integer solutions for . (The notation means the greatest integer less than or equal to .) ## Solution ### Solution 1 Note that if , then either , or . Either way, we won't skip any natural numbers. The greatest such that is . (The inequality simplifies to , which is easy to solve by trial, as the solution is obviously .) Note Once we plug in a few values we can see easily that it is We can now compute: From the observation above (and the fact that ) we know that all integers between and will be achieved for some values of . Similarly, for we obviously have . Hence the least positive integer for which the equation has no integer solutions for is . ### Solution 2 Rewriting the given information and simplifying it a bit, we have Now note that in order for there to be no integer solutions to we must have We seek the smallest such A bit of experimentation yields that is the smallest solution, as for it is true that Furthermore, is the smallest such case. (If unsure, we could check if the result holds for and as it turns out, it doesn't.) Therefore, the answer is ## Solution 3 In this solution we use inductive reasoning and a lot of trial and error. Depending on how accurately you can estimate, the solution will come quicker or slower. Using values of k as 1, 2, 3, 4, and 5, we can find the corresponding values of n relatively easily. For k = 1, n is in the range [2002-1002]; for k = 2, n is the the range [1001-668], etc: 3, [667,501]; 4, [500-401]; 5, [400-334]. For any positive integer k, n is in a range of . Now we try testing k = 1002 to get a better understanding of what our solution will look like. Obviously, there will be no solution for n, but we are more interested in how the range will compute to. Using the formula we got above, the range will be 1-2. Testing any integer k from 1002-2000 will result in the same range. Also, notice that each and every one of them have no solution for n. Testing 1001 gives a range of 2-2, and 2002 gives 1-1. They each have a solution for n, and their range is only one value. Therefore, we can assume with relative safety that the integer k we want is the lowest integer that follows this equation: floor[2002/k] + 1 = ceiling[2002/(k+1)] Now we can easily guess and check starting from k = 1. After a few tests it's not difficult to estimate a few jumps, and it took me only a few minutes to realize the answer was somewhere in the forties. Then it's just a matter of checking them until we get . Alternatively, you could use the equation above and proceed with one of the other two solutions listed. -jackshi2006	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9248348474502563, "perplexity": 283.204659069965}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964362930.53/warc/CC-MAIN-20211204033320-20211204063320-00124.warc.gz"}
http://math.stackexchange.com/questions/113562/if-ab-cd-ab-cd-and-mathbbz-a-times-mathbbz-b-cong-mathbb/113777	# If $a\|b$, $c\|d$, $ab=cd$ and $\mathbb{Z}^_a \times \mathbb{Z}^_b \cong \mathbb{Z}^_c \times \mathbb{Z}^_d$. Does this imply $(a,b)=(c,d)$? This question is inspired by $\mathbb{Z}_a\oplus\mathbb{Z}_b\cong \mathbb{Z}_c\oplus\mathbb{Z}_d$ question. We change the additive structure to multiplicative: Problem 1: If $a\|b$, $c\|d$ and $\mathbb{Z}^_a \times \mathbb{Z}^_b \cong \mathbb{Z}^_c \times \mathbb{Z}^_d$. Does this imply $(a,b)=(c,d)$? The counter-example to Problem 1 is: $$\mathbb{Z}^_3 \times \mathbb{Z}^_{12} \cong \mathbb{Z}^_4 \times \mathbb{Z}^_{12}$$ And so I slightly modified it. Problem 2: If $a\|b$, $c\|d$, $\mathbf{ab=cd}$ and $\mathbb{Z}^_a \times \mathbb{Z}^_b \cong \mathbb{Z}^_c \times \mathbb{Z}^_d$. Does this imply $(a,b)=(c,d)$? Is there a counter-example now? - nice question.+1 for the problem statment – dato datuashvili Feb 26 '12 at 11:26 Isn't $(a,b,c,d)=(2,8,4,4,)$ a counterexample? - $Z^_8$ is not $Z^_4 \times Z^_4$, because $Z^_8 \cong C_4$, $Z^_4 \times Z^_4 \cong C_2 \times C_2$. One group has an element of order 4, all proper elements are of order 2 in the other. – Tom Artiom Fiodorov Feb 26 '12 at 11:56 @Artiom: Actually, $C_4\not\cong(\mathbb{Z}/2^3\mathbb{Z})^\times\cong C_2\times C_2$. (Wikipedia) – anon Feb 26 '12 at 12:04 Initially I tried to come up with a sufficient condition to conclude that $(a,b)=(c,d)$, and $ab=cd$ is actually suffice ignoring the snag above. So restrict ourselves to $n$ just that $Z^*_n \cong C_{\phi(n)}$, i.e. $4\nmid n$. Hence the problem 3: $$C_{\phi(a)}\times C_{\phi(b)} \cong C_{\phi(c)}\times C_{\phi(d)}$$ $a\|b,c\|d,ab=cd. 4\nmid b,d.$ Is $(a,b)=(c,d)$? Then by considering the highest order in each product group, we conclude that $$\phi(b)=\phi(d)$$ Quoting out $C_{\phi(b)}$ yields $\phi(a)=\phi(c).$ From $ab=cd$ and $c\|d$ it's possible to show that $a\|d$ and $c\|b$. Put $b=\operatorname{LCM}(a,c)l_1$ and $d=\operatorname{LCM}(a,c)l_2$, and from $ab=cd$ deduce that $l_1=\frac{c}{\operatorname{GCD(a,c)}}j$, $l_2=\frac{a}{\operatorname{GCD(a,c)}}j$ Finally $$\phi\big(\frac{\operatorname{LCM}(a,c)}{\operatorname{GCD(a,c)}}cj\big)=\phi\big(\frac{\operatorname{LCM}(a,c)}{\operatorname{GCD(a,c)}}aj\big)$$ Now by cancelling out factors of $j$ which are coprime with $\operatorname{LCM}(a,c)$ we can assume that all prime factors of $j$ are present in the factorisation of $\operatorname{LCM}(a,c)$. But now RHS and LHS have the same primes, and so the powers must be the same! This concludes that $a=c$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9400855302810669, "perplexity": 353.6381948968443}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657137056.60/warc/CC-MAIN-20140914011217-00000-ip-10-234-18-248.ec2.internal.warc.gz"}
https://en.wikiversity.org/wiki/Distances/Angular_momenta	# Distances/Angular momenta This diagram describes the relationship between force (F), torque (τ), momentum (p), and angular momentum (L) vectors in a rotating system. 'r' is the radius. Credit: Yawe. Angular momenta, or angular momentum, is a lecture from the radiation astronomy department. Usually, such a classical field would be a department of physics lecture. It is a lecture in a series about distances. Large distances are significant in astronomy. Referring to the diagram on the right, an angular momentum L of a particle about an origin is given by ${\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} }$ where r is the radius vector of the particle relative to the origin, p is the linear momentum of the particle, and × denotes the cross product (r · p sin θ). Theta is the angle between r and p. The radius vector consists of two parts or concepts: a distance, or displacement, and a direction, indicated by the arrow. ## Kinetics The angular velocity of the particle at P with respect to the origin O is determined by the perpendicular component of the velocity vector v. Credit: Krishnavedala. The angular velocity describes the speed of rotation and the orientation of the instantaneous axis about which the rotation occurs. The direction of the angular velocity pseudovector is along the axis of rotation; in this case (counter-clockwise rotation) the vector points up. Credit: DnetSvg. Def. a "quantity [...] cohering together so as to make one body, or an aggregation of particles or things which collectively make one body or quantity"[1] is called a mass. Mass is an idea. Def. "objects in motion, but not with the forces involved"[2] is called kinematics, or the science of kinematics. Def. a "property of a body that resists any change to its uniform motion"[3] is called inertia. Mass and inertia are generally considered equivalent. Isaac Newton's laws of motion contain the idea of inertia. Newton's First law: "Every body persists in its state of being at rest or of moving uniformly straight forward, except insofar as it is compelled to change its state by force impressed."[4] While motion in a straight line, or rectilinear motion can be produced over limited distances in a laboratory, it may not occur naturally. Def. "the product of [a body's] mass and velocity"[5] is called momentum. Def. "the rotary inertia of a system [such as] an isolated rigid body [...] is a measure of the extent to which an object will continue to rotate in the absence of an applied torque"[6] is called angular momentum. Def. a "rotational or twisting effect of a force"[7] is called a torque. Def. a "turning effect of a force applied to a rotational system at a distance from the axis of rotation"[8] is called a moment of force. "The moment is equal to the magnitude of the force multiplied by the perpendicular distance between its line of action and the axis of rotation."[8] A torque and a moment of force are the same. Each is a "unit of work done, or energy expended".[9] Def. "the effects of forces on moving bodies"[10] is called kinetics, or the science of kinetics. ## Moment of inertia The moment of inertia and angular momenta are different for every possible configuration of mass and axis of rotation. Credit: PanCiasteczko. "For an object with a fixed mass that is rotating about a fixed symmetry axis, the angular momentum is expressed as the product of the moment of inertia of the object and its angular velocity vector: ${\displaystyle \mathbf {L} =I{\boldsymbol {\omega }}}$ where I is the moment of inertia of the object (in general, a tensor quantity), and ω is the angular velocity. The moment of inertia is the mass property of a rigid body that defines the torque needed for a desired angular acceleration about an axis of rotation. Moment of inertia depends on the shape of the body and may be different around different axes of rotation. A larger moment of inertia around a given axis requires more torque to increase the rotation, or to stop the rotation, of a body about that axis. Moment of inertia depends on the amount and distribution of its mass, and can be found through the sum of moments of inertia of the masses making up the whole object, under the same conditions. ## Angular velocity In two dimensions the angular velocity ω is given by ${\displaystyle \omega ={\frac {d\phi }{dt}}}$ This is related to the cross-radial (tangential) velocity by:[11] ${\displaystyle \mathrm {v} _{\perp }=r\,{\frac {d\phi }{dt}}}$ An explicit formula for v in terms of v and θ is: ${\displaystyle \mathrm {v} _{\perp }=\|\mathrm {\mathbf {v} } \|\,\sin(\theta )}$ Combining the above equations gives a formula for ω: ${\displaystyle \omega ={\frac {\|\mathrm {\mathbf {v} } \|\sin(\theta )}{\|\mathrm {\mathbf {r} } \|}}}$ ## Conservation of angular momentum A figure skater conserves angular momentum – her angular rotational speed increases as her moment of inertia decreases by drawing in her arms and legs. Credit: Deerstop. "In a closed system, no torque can be exerted on any matter without the exertion on some other matter of an equal and opposite torque."[12] Angular momentum can be exchanged between objects in a closed system, but total angular momentum before and after an exchange remains constant (is conserved).[13] "A change in angular momentum is proportional to the applied torque and occurs about the same axis as that torque."[12] Requiring the system to be closed is equivalent to requiring that no external influence, in the form of a torque, acts upon it.[13] "A body continues in a state of rest or of uniform rotation unless compelled by a torque to change its state."[12] With no external influence to act upon it, the original angular momentum of the system is conserved.[13] ## Orbital mechanics A massless (or per unit mass) angular momentum is defined by[14] ${\displaystyle \mathbf {h} =\mathbf {r} \times \mathbf {v} ,}$ called specific angular momentum, where ${\displaystyle \mathbf {L} =m\mathbf {h} .}$ ## Earth system This is a photograph of a retroreflector array placed by the crew of Apollo 14 on the lunar surface. Credit: Alan B. Shepard, Jr., and Edgar D. Mitchell, during EVA 1 of Apollo 14 on the Moon. Plotted are the geographical distribution of the retroreflector arrays on the lunar surface. Credit: J. O. Dickey, P. L. Bender, J. E. Faller, X X Newhall, R. L. Ricklefs, J. G. Ries, P. J. Shelus, C. Veilet, A. L. Whipple, J. R. Wiant, J. G. Williams, C. F. Yoder. "On 21 July 1969, during the first manned lunar mission, Apollo 11, the first retroreflector was placed on the moon, enabling highly accurate measurements of the Earth - moon separation by means of laser ranging."[15] "The locations of the three Apollo [A-11, A-14, and A-15] arrays plus one French-built array still operating on the Soviet roving vehicle Lunakhod 2 [L-1 and L-2 are shown in the image on the left and] provide a favorable geometry for studying the rotations of the moon and for separating these rotations from lunar orbital motion and geodynamic effects [...]."[15] "Lunar laser ranging consists of measuring the round-trip travel time and thus the separation between transmitter and reflector."[15] "Retroreflector arrays provide optical points on the moon toward which one can fire a laser pulse and receive back a localized and recognizable signal. Ranging accuracies on the order of a centimeter are immediately possible if one has sufficiently short laser pulse lengths with high power."[15] Although an "order-of-magnitude improvement in accuracy [has occurred since the Apollo program], the early data are still important in the separation of effects with long characteristic timescales, notably precession, nutation, relativistic geodetic precession, tidal acceleration, the primary lunar oblateness term (J2), and the relative orientation of the planes of the Earth's equator, the lunar orbit, and the ecliptic."[15] "The data set considered here consists of over 8300 normal-point ranges (8) spanning the period between August 1969 and December 1993; the observatories and the lunar reflectors included in the analysis are listed in Table 1. The data are analyzed with a model that calculates the light travel time between the observatory and the reflector, accounting for the orientation of the Earth and moon, the distance between the centers of the two bodies, solid tides on both bodies, plate motion, atmospheric delay, and relativity (13). The fitted parameters include the geocentric locations of the observatories; corrections to the variation of latitude (that is, polar motion); the orbit of the moon about the Earth; the Earth's obliquity, precession, and nutation; plus lunar parameters including the selenocentric reflector coordinates, fractional moment-of-inertia differences, gravitational third-degree harmonics, a lunar Love number, and rotational dissipation."[15] "The mean Earth-moon distance is 385,000 km; the radii of the Earth and moon are 6371 and 1738 km, respectively."[15] "The moon's orbit is strongly distorted from a simple elliptical path by the solar attraction-the instantaneous eccentricity varies by a factor of 2 (0.03 to 0.07)."[15] "[A]ccuracies are degraded when extrapolated outside the span of observations."[15] "The two largest solar perturbations in distance r [the distance between the centers of the Earth and moon] are 3699 km (monthly) and 2956 km (semimonthly)."[15] ## Hypotheses 1. To use angular momentum or energy a mass must be assigned. ## References 1. mass. San Francisco, California: Wikimedia Foundation, Inc. February 20, 2014. Retrieved 2014-02-28. 2. kinematics. San Francisco, California: Wikimedia Foundation, Inc. 22 July 2016. Retrieved 2016-09-08. 3. inertia. San Francisco, California: Wikimedia Foundation, Inc. February 20, 2014. Retrieved 2014-02-28. 4. Isaac Newton, The Principia, A new translation by I.B. Cohen and A. Whitman, University of California press, Berkeley 1999. 5. momentum. San Francisco, California: Wikimedia Foundation, Inc. January 30, 2014. Retrieved 2014-02-28. 6. angular momentum. San Francisco, California: Wikimedia Foundation, Inc. October 9, 2013. Retrieved 2014-02-28. 7. torque. San Francisco, California: Wikimedia Foundation, Inc. January 10, 2014. Retrieved 2014-02-28. 8. moment of force. San Francisco, California: Wikimedia Foundation, Inc. December 10, 2013. Retrieved 2014-02-28. 9. foot-pound. San Francisco, California: Wikimedia Foundation, Inc. June 20, 2013. Retrieved 2014-02-28. 10. Jazzy Prinker (21 May 2016). kinetics. San Francisco, California: Wikimedia Foundation, Inc. Retrieved 2016-09-08. 11. Russell C. Hibbeler (2009). Engineering Mechanics. Upper Saddle River, New Jersey: Pearson Prentice Hall. pp. 314, 153. ISBN 978-0-13-607791-6. 12. Henry Crew (1908). The Principles of Mechanics: For Students of Physics and Engineering. Longmans, Green, and Company, New York. p. 88. 13. Arthur M. Worthington (1906). Dynamics of Rotation. Longmans, Green and Co., London. p. 82. 14. Richard H. Battin (1999). An Introduction to the Mathematics and Methods of Astrodynamics, Revised Edition. American Institute of Aeronautics and Astronautics, Inc. p. 115. ISBN 1-56347-342-9. 15. J. O. Dickey, P. L. Bender, J. E. Faller, X X Newhall, R. L. Ricklefs, J. G. Ries, P. J. Shelus, C. Veilet, A. L. Whipple, J. R. Wiant, J. G. Williams, C. F. Yoder (22 July 1994). "Lunar Laser Ranging: A Continuing Legacy of the Apollo Program". Science 265 (5171): 482-90. doi:10.1126/science.265.5171.482. Retrieved 2016-09-09.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8881138563156128, "perplexity": 1348.5041735877007}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986657586.16/warc/CC-MAIN-20191015055525-20191015083025-00196.warc.gz"}
https://www.physicsforums.com/threads/binding-energy-between-nucleons-vs-be-inside-nucleons.66384/	# Binding energy between nucleons vs BE inside nucleons 1. Mar 7, 2005 ### alexbib humm, I've been wondering: the mass of a nucleus is less than the sum of the masses of the constituent nucleons because of the binding energy. This is how we can get energy out of fusion and fission events. on the other hand, the mass of a proton or a neutron is MORE than the sum of the masses of the constituent quarks. Afaik, both the binding of a nucleus and the binding of a proton are consequences of the strong force, so how come one type of binding increases the mass while the other type decreases it? Thanks. 2. Mar 11, 2005 ### alexbib come on, nobody knows? 3. Mar 11, 2005 ### marlon Well the case of the nucleus mass being smaller then the constituent nuclei is indeed due to the negative binding energy. This is shown by the semi-empirical mass formula. The nucleon-nucleon potential becomes repulsive at very short distances. Now, let us look inside a nucleon : The sum of the constituent quarkmasses is much smaller then the mass of the hadron. The extra mass comes from the potential and kinetic energy of the quarks and also from dynamical quarks. For example the proton contains three valence quarks of three different colours (red, green and blue), but it also contains dynamical (sea) quarks. These are quark-antiquark pairs that appear and disappear through energy fluctuations in the vacuum. These dynamical quarkpairs will generate mass. The mass of a hadron is bigger then the sum of the masses of the constituent quarks (the three quarks of the proton). But the dynamical quarks also generate mass (via symmetry breaking) , so in the end the mass of a proton is BIGGER then the sum of the three quark masses. Keep in mind that the three quarks are confined, yielding a rise in their linear potential (dominant in the long range). Once a certain distance is exceeded there is enough energy to create a quark antiquark pair marlon Last edited: Mar 11, 2005 4. Mar 11, 2005 ### marlon ps check out the link in the latest entry of my journal marlon https://www.physicsforums.com/journal.php?s=&journalid=13790&action=view [Broken] Last edited by a moderator: May 1, 2017 5. Mar 13, 2005 ### alexbib Alright, it does indeed make sense. Thanks a lot! Similar Discussions: Binding energy between nucleons vs BE inside nucleons	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9599368572235107, "perplexity": 661.1019069924117}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549424645.77/warc/CC-MAIN-20170724002355-20170724022355-00708.warc.gz"}
http://aliceinfo.cern.ch/ArtSubmission/node/3378	# Figure 3 Inclusive $\jpsi$ cross sections as function of $\pt$ (left) and $y$ (right) in pp collisions at $\sqrts=5.02$ TeV. Open symbols are the reflection of the positive-$y$ measurements with respect to $y=0$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9989374279975891, "perplexity": 330.597584003231}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948608836.84/warc/CC-MAIN-20171218044514-20171218070514-00579.warc.gz"}
https://www.science20.com/standup_physicist/blog/deriving_maxwell_homogeneous_equations_using_quaternions_35-83012	Nature abhors a magnetic monopole, although she adores gravitational and electric monopoles. The homogeneous Maxwell equations are the ones that need no currents, known as the no magnetic monopoles and Faraday's laws. There are three ways to derive the homogeneous Maxwell equations. The first way is to use vector identities. This is the simplest approach, the one most widely used. Pick out a particular way to write the electric and magnetic field, and the job is done. The divergence of a curl is a long way to say zero. Define the magnetic field as a curl, then its divergence is zero, and there are no magnetic monopoles hiding behind the couch. The second approach uses differential geometry power tools, filled with the goblins of tensors, dual vector spaces, and exterior derivatives. Only math knights can survive such a passage. The most important theorems in differential geometry are recruited to do battle. Anyone calling themselves a mathematical physicist will have mastered this approach earlier in their career. I cannot be your guide on such a quest, as I am not the Stand-Up Mathematical Physicists. Skip or click this dramatic reading of this blog: The third path uses the same three part harmony as the previous blog: pick a Lagrange density wisely, apply Euler-Lagrange, then choose a gauge. Actually, the third step can be skipped. Foreshadowing note: I will use the same line again for the GEM unified field equation derivation, the fifth and final blog in this series. Logical consistency is good. This Lagrange density technique for the homogeneous equations is not widely known. The reason is simple: it is not in the red book, "Classical Electrodynamics" by J. D. Jackson. He writes: The Lagrangian (12.85) yields the inhomogeneous Maxwell equations, but not the homogeneous ones. This is because the field strength tensor $F^{\alpha \beta}$ in terms of the 4-vector potential $\inline A^{\lambda}$ was chosen so that the homogeneous equations were satisfied automatically. Jackson went for door number one, the vector identities. This does not sound good to my ear: be clever, get the desired result. While true, it is not an answer to the issue raised, which is to find a Lagrange density that generates the homogeneous Maxwell equations. What Lagrange density would give me the two homogeneous equations? I had no clue. Actually, I did have one clue. I recalled reading in Jackson that there were two combinations of EM fields that were invariant under a Lorentz transformation: B2 - E2 and E.B. The derivation of the inhomogeneous or source Maxwell equations used the former. It would be cute if the the dot product of these two fields resulted in the homogeneous Maxwell equations. Let's find a road to this dot product. The starting point is the same as last week, the two gauge-free quaternion derivatives of a quaternion potential: $\\ \nabla \times A - (\nabla \times A)^* = (0, -\vec{E} + \vec{B}) \\ A \times \nabla - (A \times \nabla)^* = (0, -\vec{E} - \vec{B})$ Calculate the norm of these two: $\\ \|\|(0, -\vec{E} + \vec{B})\|\| = E^2 + 2 E \cdot B + B^2 \\ \|\|(0, -\vec{E} - \vec{B})\|\| = E^2 - 2 E \cdot B + B^2$ Take the difference of the two norms: $\|\|(0, -\vec{E} + \vec{B})\|\| - \|\|(0, -\vec{E} - \vec{B})\|\| = 4 E \cdot B$ This dot product will often be zero: either field could be zero, or the fields could be at right angles. In other situations, the dot product will not be zero. Like last week, I know this is Lorentz invariant, but lack a short demonstration of that fact. The Euler-Lagrange equations eat partial derivatives for breakfast, whether they happen to be zero or not. Write out Ex Bx by its component parts. Here is the magnetic field: And the electric field: So the product is: Clone away. There is no current coupling term this time. The difference of the two norms will also take the difference of two corresponding current coupling terms. The resulting subtraction leaving a scalar equal to zero. The 3-vector is not zero, so the equation does not have to be about a vacuum. Here is the proposed Lagrange density for the Maxwell homogeneous equations in its component parts: \begin{align} \mathcal{L} &= E\cdot B\\ &=- \frac{\partial Ax}{\partial t} \frac{\partial Az}{\partial y} + \frac{\partial Ax}{\partial t} \frac{\partial Ay}{\partial z} - \frac{\partial \phi}{\partial x} \frac{\partial Az}{\partial y} + \frac{\partial \phi}{\partial x} \frac{\partial Ay}{\partial z} \\ &\quad- \frac{\partial Ay}{\partial t} \frac{\partial Ax}{\partial z} + \frac{\partial Ay}{\partial t} \frac{\partial Az}{\partial x} - \frac{\partial \phi}{\partial y} \frac{\partial Ax}{\partial z} + \frac{\partial \phi}{\partial y} \frac{\partial Az}{\partial x} \\ &\quad- \frac{\partial Az}{\partial t} \frac{\partial Ay}{\partial x} + \frac{\partial Az}{\partial t} \frac{\partial Ax}{\partial y} - \frac{\partial \phi}{\partial z} \frac{\partial Ay}{\partial x} + \frac{\partial \phi}{\partial z} \frac{\partial Ax}{\partial y} \end{align} This week there are only 16 terms instead of 22 needed for the source equations. All the terms have a maximally-mixed quality, one part from the scalar, and one from x, y, and z positions. Half the signs are positive, half are negative, a perfect setup for cancellation. No terms are squared, no factors of two to track. Here is a mini-refresher on putting the Euler-Lagrange equations to work: Focus on the phi terms: Half the terms can be ignored, good. Look for a pattern with the signs. Look to see how this can be summarized. Repeat the process, this time focusing on terms with Ax: A summary statement is trickier to figure out. It will require some looking to see it. Here is the summary. Those readers familiar with the vector identities should be able to see the same things at work. This feels deeper to me because the Lagrange density gets blended into the vector identities via Euler-Lagrange. With the graphic representation, the derivation of all four Maxwell equations can be taken in at once: The Lagrange density for the homogeneous equations mix up the E and B fields. The source equations rely on a separation of E and B fields. I should admit my own point of vanity, an issue of excessive pride. Maxwell is the number one physicists on the list of those unknown to the general public. Maxwell worked in a time before electric lights, let alone computers. He was so bright he figure out that Saturn's rings must be made of particles so the system could be stable mathematically. I have no idea how to approach such an issue. "It is one of the most remarkable applications of mathematics to physics that I have ever seen," according to George Airy in the day. He also made contributions to the study of color and dimensional analysis. And then there is "A Treatise on Electricity and Magnetism" in two volumes. I audited a class at BU that used Jackson. In that semester, I decided to look up the old book. It is a tough read because he used German notation for everything. I still recall seeing one of the homework problems solved by Maxwell in the text. I got the sense that Jackson was an Americanized update of the Treatise. Maxwell wanted to use quaternions for his work. Recall that quaternions were an important and popular subject in his day, instead of a historical footnote. He had taken a class from Hamilton himself, the man who coined the phrases dot, cross, div, grad, curl, and all that for the math born in quaternion multiplication. His good friend Tait worked with them extensively. The first edition did have two sections with quaternions in the title (vol. 1 and vol. 2, 1873). He used pure quaternions, ones where the scalar is set to zero. A pure quaternion is functionally no different from a 3-vector. My feeling of excessive pride comes from accomplishing something that Maxwell wanted to get done. This in no way means I am on his level. Equally strong is how absurd this feat is. I am so clumsy doing physics. I can be great on a puzzle, where there is one clue that doesn't look like a clue, but I store it away for use decades later. The BU class would have been in the mid-90s. I started the quaternion derivation somewhere around 2008. I may be able to make progress on impossible puzzles. There's a skill that will not pay the bills. I recommend you repeat this work with your own hand. Time will slip away as the patterns intertwine. Leave these calculations on your desk. People will presume you are the smartest person they know. Doug Snarky puzzle [correction: Find the flaw in this question :-) ] Take one of the quaternion derivative ways to write the E and B fields and square it. Notice that the scalar is a Lagrange density that will be able to generate all four of the Maxwell equations via the Euler-Lagrange equation... [correction: NO, generates the homogeneous equations, not the source equations.] Math overdose, scary-as-shit bonus problem: do something with the two 3-vectors generated [from the product that makes E.B]. Google+ hangout: 11:00-11:45pm Eastern time, Tuesday-Friday. http://gplus.to/sweetser This could be an efficient way to exchange a few ideas. If you have a question or two, hangout. Bet against the Higgs being found, buy the t-shirt Next Monday/Tuesday: Derive the Hypercomplex Gravity Field Equations (4/5)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9073634743690491, "perplexity": 818.0198380413854}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337680.35/warc/CC-MAIN-20221005234659-20221006024659-00595.warc.gz"}
http://mathhelpforum.com/calculus/217478-easy-indefinite-integral.html	1. ## Easy Indefinite Integral Hello, how do I find the indefinite integral? What do I set to be equal to u? Thanks Attached Thumbnails 2. ## Re: Easy Indefinite Integral You shouldn't use substitution here. \displaystyle \begin{align} \int{\frac{x^2 + 5x - 8}{\sqrt{x}}\,dx} &= \int{\frac{x^2 + 5x - 8}{x^{\frac{1}{2}}}\,dx} \\ &= \int{x^{2 - \frac{1}{2}} + 5x^{1 - \frac{1}{2}} - 8x^{-\frac{1}{2}} \,dx} \\ &= \int{x^{\frac{3}{2}} + 5x^{\frac{1}{2}} - 8x^{-\frac{1}{2}}\,dx} \end{align} You can now use the power rule on each term. If you REALLY want to use substitution, rewrite the integral as \displaystyle \begin{align} \int{\frac{x^2 + 5x - 8}{\sqrt{x}}\,dx} &= 2\int{\frac{\left( \sqrt{x} \right) ^4 + 5 \left( \sqrt{x} \right) ^2 - 8}{2\sqrt{x}}\,dx} \end{align} and then let $\displaystyle u = \sqrt{x} \implies du = \frac{1}{2\sqrt{x}}\,dx$ and the integral becomes $\displaystyle 2\int{u^4 + 5u^2 - 8 \, du}$. I'm sure whichever method you choose you can now finish the problem off. 3. ## Re: Easy Indefinite Integral complete the square then use trig substitution oh ^ works too	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000097751617432, "perplexity": 3394.6753806422694}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818688926.38/warc/CC-MAIN-20170922074554-20170922094554-00016.warc.gz"}
https://admin.clutchprep.com/organic-chemistry/practice-problems/7028/predict-the-major-product-s-in-the-present-case-the-reaction-needs-to-be-perform	# Problem: Predict the major product(s). In the present case, the reaction needs to be performed at 85°C. In fact, at 70°C the less stable chair conformation can be accessed. ###### Problem Details Predict the major product(s). In the present case, the reaction needs to be performed at 85°C. In fact, at 70°C the less stable chair conformation can be accessed.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8845804333686829, "perplexity": 1892.2389915195758}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600402123173.74/warc/CC-MAIN-20200930075754-20200930105754-00315.warc.gz"}
https://www.aimsciences.org/article/doi/10.3934/dcdss.2011.4.595	Article Contents Article Contents # Regular boundary value problems for ordinary differential-operator equations of higher order in UMD Banach spaces • We prove an isomorphism of nonlocal boundary value problems for higher order ordinary differential-operator equations generated by one operator in UMD Banach spaces in appropriate Sobolev and interpolation spaces. The main condition is given in terms of $\R$-boundedness of some families of bounded operators generated by the resolvent of the operator of the equation. This implies maximal $L_p$-regularity for the problem. Then we study Fredholmnees of more general problems, namely, with linear abstract perturbation operators both in the equation and boundary conditions. We also present an application of obtained abstract results to boundary value problems for higher order elliptic partial differential equations. Mathematics Subject Classification: Primary: 34G10, 47E05; Secondary: 35J40, 47N20. Citation: • [1] S. Agmon, A. Douglis and L. Nirenberg, Estimates near the boundary for solutions of elliptic partial differential equations satisfying general boundary conditions, I, II, Comm. Pure Appl. Math., 12 (1959), 623-727; 17 (1964), 35-92. [2] W. Arendt and M. Duelli, Maximal $L^p$-regularity for parabolic and elliptic equations on the line, J. Evol. Equ., 6 (2006), 773-790.doi: doi:10.1007/s00028-006-0292-5. [3] W. Arendt and A. F. M. ter Elst, Gaussian estimates for second order elliptic operators with boundary conditions, J. Operator Theory, 38 (1997), 87-130. [4] R. Denk, G. Dore, M. Hieber, J. Prüss and A. Venni, New thoughts on old results of R. T. Seeley, Mathematische Annalen, 328 (2004), 545-583.doi: doi:10.1007/s00208-003-0493-y. [5] R. Denk, M. Hieber and J. Prüss, "$R$-Boundedness, Fourier Multipliers and Problems of Elliptic and Parabolic Type," Mem. Amer. Math. Soc., Providence, 2003. [6] A. Favini, V. Shakhmurov and Ya. Yakubov, Regular boundary value problems for complete second order elliptic differential-operator equations in UMD Banach spaces, Semigroup Forum, 79 (2009), 22-54.doi: doi:10.1007/s00233-009-9138-0. [7] A. Favini and Ya. Yakubov, Higher order ordinary differential-operator equations on the whole axis in UMD Banach spaces, Differential and Integral Equations, 21 (2008), 497-512. [8] A. Favini and Ya. Yakubov, Regular boundary value problems for elliptic differential-operator equations of the fourth order in UMD Banach spaces, Scientiae Mathematicae Japonicae, 70 (2009), 183-204. [9] A. Favini and Ya. Yakubov, Irregular boundary value problems for second order elliptic differential-operator equations in UMD Banach spaces, Mathematische Annalen, 348 (2010), 601-632.doi: doi:10.1007/s00208-010-0491-9. [10] N. Kalton, P. Kunstmann and L. Weis, Perturbation and interpolation theorems for the $H^\infty$-calculus with applications to differential operators, Mathematische Annalen, 336 (2006), 747-801.doi: doi:10.1007/s00208-005-0742-3. [11] N. Kalton and L. Weis, The $H^\infty$-calculus and sums of closed operators, Mathematische Annalen, 321 (2001), 319-345.doi: doi:10.1007/s002080100231. [12] P. C. Kunstmann and L. Weis, "Maximal $L_p$-Regularity for Parabolic Equations, Fourier Multiplier Theorems and $H^\infty$-Functional Calculus," in "Functional Analytic Methods for Evolution Equations," Lecture Notes in Mathematics, 1855, Springer, (2004), 65-311. [13] H. Triebel, "Interpolation Theory. Function Spaces. Differential Operators," North-Holland, Amsterdam, 1978. [14] L. Weis, Operator-valued Fourier multiplier theorems and maximal $L_p$-regularity, Mathematische Annalen, 319 (2001), 735-758.doi: doi:10.1007/PL00004457. [15] S. Yakubov and Ya. Yakubov, "Differential-Operator Equations. Ordinary and Partial Differential Equations," Chapman and Hall/CRC, Boca Raton, 2000.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.821459949016571, "perplexity": 1627.6411404256373}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949025.18/warc/CC-MAIN-20230329182643-20230329212643-00575.warc.gz"}
http://www.bio-physics.at/wiki/index.php?title=Feed_Forward_Loop&oldid=3316	# Feed Forward Loop For a three node pattern there exitst 13 possible ways to create a directed (with arrows of definite direction) subgraph. Similar calculations as for negative autoregulation (NAR) show, that one of this three node patterns occurs much more often in the E. coli network that expected at random. This pattern numbered with $5$ is called feed forward loop (FFL). E. coli contains 42 feed forward loops, while the expected number is less than one, thus the FFL is a network motif. Depending on the sign of the arrows the feed forward loop can be classified into $8$ different types. Each arrow can be positive or negative (zwo possibilities) and there are three arrows this gives $2^3$ possibilities to make a FFL with different signs. This $8$ FFLs are eather coherent (when both incoming arrows of $Z$ have a positive sign) or incoherent (incoming arrows of $Z$ have different signs). The input function of $Z$ is two-dimensional. In the logic approximation it can eather be an AND or an OR input funciton. It is common to use a symbolic notation like in electronic engineering and boolean logic. The following four types of Coherent Feed Forward Loops [C#FFL] exist The following four types of Incoherent Feed Forward Loops [I#FFL] exist The most abundant FFLs in E. coli are C1FFL and I1FFL, $90\%$ of the E. coli network are of these types. ## C1FFL with AND input function We already know the dynamics of simple regulation, which leads to a time course $$Y(t)=Y_{st}(1-e^{-\alpha t}).$$ We use this result to understand the dynamics of the feed forward loop with an AND input function. Let us now study the response to an ON-step. We assume that protein $X$ is present in its inactive form. Suddenly the signal $S_x$ appears and the protein is transformed in its active form $X \rightarrow X^$. This leads to immediate production of protein $Y$ whose signal $S_y$ we assume present, so that $Y$ is active ($Y^$). Since, $Z$ has an AND input function, $Z$ needs both $X^$ AND $Y^$ to be present, before it starts producing $Z$, thus $Z$ is only expressed when $Y$ exceeds a certain threshold $K_yz$ (threshold of Y to produce Z). delayed response: the C1FFL-AND motif leads to a delayed production of $Z$ in response to the signal $S_x$ The time delay of production of protein $Z$ equals the time $Y$ needs to reach the activation threashold. $$Y_{st}(1-e^{-\alpha \cdot t_K})=K_{yz}\\ e^{-\alpha \cdot t_K}=1-\frac{K_{yz}}{Y_{st}}\\ t_K=\frac{1}{\alpha} \ln \left(\frac{1}{1-\frac{K_{yz}}{Y_{st}}} \right)$$ If the signal $S_x$ is present for a time period smaller than $t_K$, $Y$ will not exceed the threshold $K_{yz}$ and $Z$ will not be produced. The C1FFL-AND motif acts as filter for signals $S_x$ of a short time period $t < t_K$ An OFF-step occurs, when the signal $S_x$ suddenly disappears. Since, both $X^$ AND $Y^$ are needed to activate $Z$ production, the production stops immediately after $X^$ disappears. ## C1FFL with OR input function For an ON-step, where the signal $S_x$ appears, $Y^$ starts beeing produced as well as $Z^$, becaus it is enough that eather $X^$ OR $Y^$ is present. The interessting part of this motif is is the OFF-step. When $S_x$ disappears $Y$ immediately stops beeing produced, but $Z$ is still produced until $Y$ reaches the threashold $K_yz$, then also $Z$ stops producing. delayed response: the C1FFL-OR motif leads to a delay of production stop of $Z$ following an OFF-step of $S_x$ Summary C1FFLs are sign sensitive delay elements. With a C1FFL-AND element brief fluctuations of ON-pulses can be filtered away, while with C1FFL-OR element brief OFF-pulses can be filtered away. ## I1FFL with AND input function Let us now study the incoherent type one feed forward loop with an AND input function. Notice, that $Z$ is transcribed if transcription factor $X^$ AND NOT $Y^$ is present. For an ON-step $X^$ is immediately active and produces $Y$. $Z$ is also produced, since $X^$ AND NOT $Y^$ is present. When $Y^$ exceeds the threshold $K_{yz}$, $Z$ stops beeing produced and decays exponentially to reach the steady state $Z_{st}$. As $S_x$ appears $Y$ is produced according to simple regulation and existis in its active form as we assume the signal $S_y$ present. $$Y(t)=Y_{st}(1-e^{-\alpha_Y t}).$$ $Z$ is also produced instantaneously and reaches a maximal level at $Z_m=\beta_Z/\alpha_Z$. $$Z(t)=Z_m(1-e^{-\alpha_Z t}).$$ The time span until $Z$ is repressed, equals the time $Y$ needs to reach the repression threashold. $$Y_{st}(1-e^{-\alpha \cdot t_R})=K_{yz}\\ e^{-\alpha \cdot t_R}=1-\frac{K_{yz}}{Y_{st}}\\ t_R=\frac{1}{\alpha} \ln \left(\frac{1}{1-\frac{K_{yz}}{Y_{st}}} \right)$$ After the repression thershold is reached the $Z$ expression level decays to a the steady state $Z_{st}=\beta_Z '/\alpha_Z$ $$Z(t)=Z_{st}+(Z_0-Z_{st})(1-e^{-\alpha t}).$$ We now introduce the new concept of the repression factor $$F=Z_m/Z{st}=\beta_Z/\beta '_Z$$ For a hight repression factor $F$ the steady state $Z_{st}$ concentration is much lower than $Z_m$. For lower and lower $F$ the difference in concentration of $Z_m$ and $Z_{st}$ gets lower and lower. Hence, with high repression factors a pulse can be generated, while for low repression factors the response time can be shortened. The I1FFL-AND motif can act as a pulse generator or speed the response time Video Lecture:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 7, "x-ck12": 0, "texerror": 0, "math_score": 0.8672418594360352, "perplexity": 1189.7046090267088}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663019783.90/warc/CC-MAIN-20220528185151-20220528215151-00678.warc.gz"}
https://www.physicsforums.com/threads/does-this-wavefunction-make-sense.869697/	I Does this wavefunction make sense? 1. Apr 30, 2016 Isaac0427 Hi all! Consider a wavefunction, where $\left\| \psi (x) \right\|^2 = e^{-ax^2+1}+e^{-bx^2-1}$ where a and b are real, positive numbers that satisfy normalization (they are purpously inside the exponent). Even if it is normalized, there are still 2 spots that $\left\| \psi \right\|^2 > 1$ which makes no sense. What is going on here? Thanks! 2. Apr 30, 2016 Haborix Recall that $\|\psi^2\|$ is a probability density. The probability for finding a particle in an interval $[a,b]$ is $\int\limits_a^b\|\psi^2\|\mathrm{d}x$. This integral is what must be $\le 1$ (where it equals one when you integrate over all of space). To convince yourself, take one of the Gaussian functions you wrote and normalize it, evaluate at $x=0$ and start increasing $a$. Despite still being normalized $\|\psi^2\|$ can be made arbitrarily large. Why is that alright? 3. Apr 30, 2016 Isaac0427 Ah, I get that. I was confused as I thought of it as a probability and not a probability density. Draft saved Draft deleted Similar Discussions: Does this wavefunction make sense?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9519766569137573, "perplexity": 498.26679724719463}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187825889.47/warc/CC-MAIN-20171023092524-20171023112524-00864.warc.gz"}
https://ftloscience.com/pharmacokinetics-pharmacodynamics-crash-course/	# Pharmacokinetics and Pharmacodynamics Crash Course + Cheat Sheet Pharmacokinetics and pharmacodynamics are the cornerstones of pharmacology. Studying them together helps to paint a clearer picture of the safety and efficacy of a new drug. These studies are often performed as part of the pre-clinical trials in the drug development process. In this crash course, we cover the essential concepts in basic pharmacokinetics and pharmacodynamics and how to apply them, along with a cheat sheet with all the important formulas. ## Pharmacokinetics: What the Body Does to the Drug Skipping ahead of patient variations and ADME and bioavailability—all of which are still important, by the way—at the core of pharmacokinetics is studying what the body does to the drug. When a drug is administered to a patient, it enters their bloodstream. At this point, concentration is at its maximum (Cp0). The body quickly gets to work breaking it down and removing it from the body, helped by metabolic processes like the cytochrome P450 family of enzymes. ### First Order Kinetics A simple way to model these processes all at once is to measure the blood plasma concentration (Cp) of the drug over a period (t). This is done by taking blood samples from the patient at various times after administration. A typical data set will look something like this: If we plot this in a scatter graph, this is the result: The key takeaway is that we see an exponential decay, which just means that the more drug there is in the body, the faster it’s removed. As time passes, the rate of decrease slows down. Many biological processes follow an exponential decay such as this; in pharmacokinetics, we say that such drugs follow first-order elimination kinetics. (Note: because first-order kinetics are the most studied, we will focus on it in this crash course. Equations for zero-order and second-order kinetics are still provided in the cheat sheet at the end) If we want more information about the drug, we can convert the exponential curve to a straight line. We first convert the Cp values to ln Cp (natural log), then plot those values.: From the plot, we can now produce a line of best fit through the points. The equation of the line is given as: y=-0.058x \:+ \:5 ### The One-Compartment Model Why is this equation important? What we see above is the one-compartment model, where the maximum drug concentration (Cp0) exponentially decreases over time. This decrease is by the gradient of the straight line (k), when we take the natural log of the concentration values. Putting this into mathematical terms, we can write that: Cp=Cp_{0}\:\ast \:e^{-kt}\\[0.1in] or \\[0.1in] ln \:Cp =ln\:Cp_{0}-kt Where Cp is the concentration of drug, Cp0 is the initial concentration at the point of administration, t is time and k is the elimination first order rate constant. The bottom equation follows the same form as the equation of the straight line in the ln Cp vs time plot! y=mx+ c\\[0.1in] ln \:Cp =-kt+ln\:Cp_{0}\\[0.1in] \:y=ln \:Cp\:\| \: m=-k\:\|\:x=t\:\|\:c=ln\:Cp_{0} The initial concentration of the drug is Cp0, which is given as the intercept c in the straight line. This means we can now calculate the initial concentration (when t = 0), something that is physically impossible to measure! When \:t=0,\\[0.1in] ln \:Cp =ln\:Cp_{0}\\[0.1in] ln\:Cp_{0}=c=5\\[0.1in] Cp_{0}=e^{c}=e^{5}=148.4\:\mu g/ml Another useful relationship we have established is that changing the sign of the gradient of the graph (m) gives us the elimination rate constant (k). ### The Elimination Rate Constant (k) This constant k is useful because it allows us to calculate pharmacokinetic parameters, such as clearance (CL). Clearance is the volume of blood that gets cleared of the drug as a result of our metabolism, measured in L/hr or ml/min. CL=k\:\ast\:Vd The volume of distribution (Vd) is the apparent volume in which the drug is diffused throughout the body. Water-soluble drugs usually have a high Vd, while lipid-soluble drugs have a lower Vd. It is measured by taking the dose of the drug (X0) and dividing it by the maximum concentration (Cp0) Vd = \frac{X_{0}}{Cp_{0}} ### Half-life (t1/2) In pharmacokinetic studies, it can be useful to know the half-life of a drug, or the time it takes for its concentration to decrease by half in a patient. The half-life (t1/2) of a drug that follows first-order kinetics can be calculated by this equation: t_{1/2}=\frac{0.693}{k} This is interesting because it tells us that the drug’s half-life depends only on its rate constant (k) and not its concentration! ## Pharmacodynamics: What the Drug Does to the Body While pharmacokinetics studies the change in drug concentration, pharmacodynamics deals with the observed effects of the drug on the body. Although drug concentration plays a role in pharmacodynamics, drug effects change from patient to patient. Differences in the biological makeup and genetics often make drug effects difficult to predict. The drug’s effect and dose are often plotted to establish an initial relationship. The ‘effect’ should ideally be an easily measured change caused by the drug. Different dosages are administered to test subjects so that a dose-response curve can be plotted. Below is an example of a dose-response curve. ### Effective, Toxic, Lethal Dose Values Dose-response curves can be used to predict effective, toxic and lethal doses of a specific drug, depending on the response. For example, if we want to study efficacy, we can test the response in certain receptors on a piece of tissue. If we want to test for toxic and lethal doses, we can use live animals to study the drug’s effects. Pre-clinical studies such as these are a key part of the drug development process. EDx (effective dose): the drug dose that will cause a therapeutic effect in x% of the population TDx (toxic dose): the drug dose that will cause toxic effects in x% of the population LDx (lethal dose): the dose that will be lethal to x% of a population. In an ideal drug, its ED50 value should be much lower than the TD50 and LD50 values. This means that it can be dosed at therapeutic levels without risking the patient’s safety. A drug that has ED50 = LD10, for example, would not be very safe! ### Therapeutic Window of a Drug Once we establish the effective and toxic concentrations of the drug, we can combine this with the information from the drug plasma concentration over time studies. A typical profile of drug plasma concentration over time has a distinct rapid rise in concentration followed by a slow decrease as the drug is metabolized and eliminated from the body. Let’s say we have found from dose-response studies that 1) the drug is only effective at a concentration of 10 μg/ml and 2) it starts to show toxic side effects at 20 μg/ml. We can now create an overlay for the concentration over time plot: Below the minimum effective concentration threshold, the drug is ineffective. Above the maximum safe concentration, the drug causes adverse reactions. The ‘sweet spot’ between these two limits is the therapeutic window. Pharmacodynamic studies such as these help us to better understand the ideal dosage of a new drug to recommend for patients. ### Therapeutic Index (TI) To quantify the safety of a drug, its therapeutic index (TI) can be measured, where: TI = TD_{50}/ED_{50} A drug with TI >> 1 will be safer than a drug with a drug that has a TI close to 1, since it will have a bigger therapeutic window between the toxic dose and the effective dose. ## Pharmacokinetics and Pharmacodynamics Cheat Sheet ### First-Order Kinetics Cp=Cp_{0}\:\ast \:e^{-kt}\\[0.1in] ln \:Cp =ln\:Cp_{0}-kt\\[0.1in] In a straight line plot: y=mx+ c\\[0.1in] \:y=ln \:Cp\:\| \: m=-k\:\|\:x=t\:\|\:c=ln\:Cp_{0} If plotting software is not used, the gradient of the line (m) can be calculated by: m = \frac{∆y}{∆x} = \frac{y_{2} \:– \:y_{1}}{ x_{2}\: –\: x_{1} } Half life (t1/2) t_{1/2}=\frac{0.693}{k} ### Zero-Order Kinetics General equation and half-life equation: Cp = Cp_{0} -kt\\[0.1in] t_{1/2} = \frac{Cp_{0}}{2k} ### Second-Order Kinetics General equation and half-life equation: \frac{1}{Cp} = \frac{1}{Cp_{0}} + kt\\[0.1in] t_{1/2} = \frac{1}{k\ast Cp0} ### Other Useful Equations Vd = \frac{X_{0}}{Cp_{0}}\\[0.1in] CL = k \ast Vd\\[0.1in] TI = \frac{TD_{50}}{ED_{50}} ### Glossary of Terms and Units Cp: drug plasma concentration (μg/ml or mg/L) Cp0: maximum drug plasma concentration (t=0) CL: clearance (ml/min or L/h) k: elimination rate constant (min-1 or hr-1) t1/2: Half-life, time taken for drug concentration to be halved (min or hr) Vd: Volume of distribution (L) EDx: the drug dose that will cause a therapeutic effect in x% of the population (μg or mg or g) TDx: the drug dose that will cause toxic effects in x% of the population (μg or mg or g) LDx: the dose that will be lethal to x% of a population (μg or mg or g) TI: a measure of drug safety, TI >> 1 = safe, TI close to 1 = unsafe	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8048935532569885, "perplexity": 2343.946786778381}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030336674.94/warc/CC-MAIN-20221001132802-20221001162802-00431.warc.gz"}
http://mlxtran.lixoft.com/individual/	Select Page # [INDIVIDUAL] ### Description The [INDIVIDUAL] section is used to define a probability distribution for model parameters. It is used to model inter-individual variability for given parameters. ### Scope The [INDIVIDUAL] section is used in Mlxtran models for simulation with Mlxplore or Simulx. It is only needed for models that have parameters with inter-individual variability. Mlxtran models for Monolix do not need this section because the parameter distributions are defined via the user interface. ### Inputs The inputs for the [INDIVIDUAL] section are the parameters that are declared in the input = { } list of the [INDIVIDUAL] section. These parameters are obtained from the section or from the executing program that can be Mlxplore or an R-script in the case of Simulx. ### Outputs Every parameter that has been defined in the [INDIVIDUAL] section can be an output. Outputs from the [INDIVIDUAL] section are always an input for the [LONGITUDINAL] section. [INDIVIDUAL] output to [LONGITUDINAL] input matching is made by matching parameter names in the [INDIVIDUAL] section with parameters in the inputs = { } list of the [LONGITUDINAL] section. ### Usage The definition of probability distribution for a model parameter is done with the EQUATION: and DEFINITION: blocks. The EQUATION: block contains mathematical equations and the DEFINITION: block is used to definite probability distributions. The following syntax applies to define a probability distribution for the random variable X: DEFINITION: X = {distribution= distributionType, parameter1 = Var1, covariate = c, coefficient = beta, parameter2 = Var2} The arguments to the probability distribution definition are • distributionType: is one of the following reserved keywords: normal, lognormal, logitnormal, or probitnormal. For use in Simulx, the keyword uniform is also accepted. • parameter1: is one of the following reserved keywords: mean or typical • parameter2: is one of the following reserved keywords: sd or var • Var1: is a double number or a parameter • Var2: is a double number or a parameter • c = {..}: is a list of strings referring to the covariates • beta = {…}: is a list of strings referring to the covariate coefficients The reserved keywords meanings are • normal: normal distribution: $h(X)=X$ • lognormal: log-normal distribution: $h(X)=\log(X)$ • logitnormal: logit-normal distribution: $h(X)=\log(\frac{X}{1-X})$ • probitnormal: probit-normal distribution: $h(X)=\Psi^{-1}(X)$, where $\Psi$ is the cumulative distribution function of the ${\cal N}(0, 1)$ distribution. • mean: is the mean of the normal distribution • typical: is the transformed mean • sd: is the standard deviation of the normal distribution. If no variability is required, the user can set “no-variability” and no-variability will be computed • var: is the variance of the normal distribution • min: the lower bound of the interval for logitnormal or probitnormal distributions (default is 0) • max: the upper bound of the interval for logitnormal or probitnormal distributions (default is 1) These probability distribution are all Gaussian probability distributions that are defined through the existence of a monotonic transformation $h$ such that $h(X)$ is normally distributed. Notice that the mean, standard deviation, and variance refer to the normal distributed variable. In pharmacometrics it is more common to use the typical value of the distribution. This is achieved by using the keyword typical instead of mean in the definition of the random variable. The relationship between the mean value and the typical value is the following: $\phantom{abc} X = \{ \textrm{distribution}=\textrm{lognormal}, \textrm{typical}=X_{pop}, \textrm{sd}=\sigma\} \\ \Leftrightarrow X = \{ \textrm{distribution}=\textrm{lognormal}, \textrm{mean}=\mu_{pop}, \textrm{sd}=\sigma\} \\ \Leftrightarrow \log(X) \sim {\cal N}(\mu_{pop},\sigma^2)$ where $\mu_{pop}=\log(X_{pop})$. Thus, typical is in the variable referential, while mean is in the transformed referential. #### Examples • The parameter ka below is defined with a log-normal distribution, a typical value ka_pop and a standard deviation for the random effect omega_ka: ka = {distribution=logNormal, typical=ka_pop, sd=omega_ka} • The parameter F below is defined with a logit-normal distribution in the interval [0,1], a typical value F_pop and no random effect. F = {distribution=logitNormal, typical=F_pop, no-variability} • The parameter V below is defined with a logit-normal distribution in the interval [0.2,5], a typical value V_pop and a standard deviation for the random effect omega_V: V = {distribution=logitNormal, min=0.2, max=5, typical=V_pop, sd=omega_V} • The parameter fu below is defined with a uniform distribution in the interval [0.2,0.6] (not accepted in Monolix): fu = {distribution=uniform, min=0.2, max=0.6} ### General probability distributions and inclusion of covariates #### Linear Gaussian models with covariates A linear Gaussian statistical model for the variable $X$ assumes that there exists a transformation $h$, a typical value $X_{\rm pop}$, a vector of individual covariates $(c_{1}, \ldots c_{L})$, a vector of coefficients $(\beta_1, \ldots, \beta_L)$ and a random variable $\eta$ normally distributed such that $h(X) = h(X_{pop}) + \sum_{\ell=1}^L \beta_\ell \, c_\ell + \eta$ This model can be implemented with Mlxtran, using the keywords typical, covariate and coefficient. input = {Xpop, beta1, beta2, c1, c2, omega} DEFINITION: X = {distribution=lognormal, typical=Xpop, covariate={c1,c2}, coefficient={beta1,beta2}, sd=omega} The keyword covariate is used to define the name of the covariates used in the correlation, and the coefficient keyword is used to complete the equation. Obviously, the number of parameters in the coefficient is equal to the number of covariates. #### Non linear Gaussian model with covariates A nonlinear Gaussian statistical model for the variable $X$ assumes that there exists a transformation $h$, a vector of individual covariates $(c_{1}, \ldots c_{L})$, a vector of coefficients $(\beta_1, \ldots, \beta_M)$, a function $\mu$ and a random variable $\eta$ normally distributed such that $h(X) = \mu(c_{1}, \ldots c_{L},\beta_1, \ldots, \beta_M) + \eta$ The mean of $h(X)$ can be defined in a block DEFINITION:, with for example $\mu(\beta_1,\beta_2,c_1,c_2)=\frac{\beta_1c_1}{\beta_2+c_2}$ input = {beta1, beta2, c1, c2, omega} EQUATION: mu = beta1c1/(beta2 + c2) DEFINITION: X = {distribution=lognormal, mean=mu, sd=omega} #### Non Gaussian model with covariates Non Gaussian model for $X$ can be defined, at the condition that $X$ can be defined as a nonlinear function of normally distributed random variables. For example, let $X = \frac{\beta_1 + \eta_1}{1+ \beta_2 \, e^{\eta_2}}$ It is not possible to express explicitly the distribution of $X$ as a transformation of a normal distribution. We therefore need a block EQUATION: for implementing this model: input = {beta1, beta2, omega1, omega2} DEFINITION: eta1 = {distribution=normal, mean=0, sd=omega1} eta2 = {distribution=normal, mean=0, sd=omega2} EQUATION: X = (beta1 + eta1)/(1+beta2exp(eta2)) ### Rules • When defining a distribution with covariate, one can not define numerically the coefficients. For example, if we consider $X=X_{pop}+c+\eta$, one should write input = {c} EQUATION: beta = 1 DEFINITION : X = {distribution=normal, typical=Xpop, covariate=c, coefficient=beta, sd=omega} and define $\beta = 1$ in the section <PARAMETER> or define it in an EQUATION: block. Otherwise, putting directly 1 instead of $\beta$ in the distribution definition will lead to an error. • We strongly advise to define the distribution in the more synthetic way. If for example, you want to define a log-normally distributed volume with a dependence w.r.t. the weight $V=V_{pop}(w/70)^{\beta}$, we encourage you not to define a lot of equations but to summarize it in the definition as for example input = {Vpop, w, beta} EQUATION: cov = w/70 DEFINITION: V = {distribution=normal, typical=Vpop, covariate=cov, coefficient=beta, sd=omega}	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 34, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9623115658760071, "perplexity": 1674.0196845869125}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439735990.92/warc/CC-MAIN-20200806001745-20200806031745-00183.warc.gz"}
https://stason.org/TULARC/self-growth/puzzles/377-pickover-pickover-01-p.html	377 pickover/pickover.01.p Title: Cliff Puzzle 1: Can you beat the numbers game? From: [email protected] If you respond to this puzzle, if possible please include your name, address, affiliation, e-mail address. If you like, tell me a little bit about yourself. You might also directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * At a recent trip to the Ontario Science Center in Toronto, Canada I came across an interesting puzzle. The center is located minutes from downtown Toronto and it's a vast playground of science with hundreds of exhibits inviting you to touch, try, test, and titillate your curiosity. The puzzle I saw there can be stated as follows. In the 10 boxes below, write a 10-digit number. The digit in the first box indicates the total number of zeros in the entire number. The box marked "1" indicates the total number of 1's in the number. The box marked "2" indicates the total number of 2's in the number, and so on. For example, the "3" in the box labeled "0" would indicate that there must be exactly three 0's in the 10-digit number. ------------------------------- \| 0\| 1\| 2\| 3\| 4\| 5\| 6\| 7\| 8\| 9\| \| 3\| \| \| \| \| \| \| \| \| \| ------------------------------- Stop And Think 1. Is there a solution to this problem? Are there many solutions to this problem? 2. A more advanced an interesting problem is to continue to generate a sequence in a recursive fashion such that each row becomes 0 through 9 digits in row 1: Row 1: 0 1 2 3 4 5 6 7 8 9 Assume Row 2 is your solution to the puzzle. I've just inserted random digits below so as not to give away the solution: Row 1: 0 1 2 3 4 5 6 7 8 9 S(1) Row 2: 9 3 2 3 3 1 6 7 8 9 S(2) Row 3: S(3) Row 2 is now the starting point, and your next job is to form row 3, row 4, etc. using the same rules. In the previous example, a digit in the first box would indicate how many 9's there are in the next 10-digit number, and so forth. Contest: I am looking for the longest sequence of numbers users can come up with using these rules. Can you find a Row 2 or Row 3? Is it even possible to generate a "row 2" or "row 3"? pickover/pickover.01.s 1) 0 1 2 3 4 5 6 7 8 9 2) 6 2 1 0 0 0 1 0 0 0 3) 0 0 0 4 4 4 0 4 4 4 4) 6 6 6 0 0 0 6 0 0 0 5) 0 0 0 4 4 4 0 4 4 4 . . . and so on, repeating rows 3 and 4. I don't know yet whether there are multiple solutions, but I'll keep looking. Mark Hayes Goddard Space Flight Center (GSFC) / Interferometrics, Inc. [email protected] GSFC Code 926.9 Greenbelt, MD 20771 ------------------------- In article <[email protected]>, you write: \|> The puzzle I saw there can be stated as follows. In the 10 boxes below, \|> write a 10-digit number. The digit in the first box indicates the total \|> number of zeros in the entire number. The box marked "1" indicates the \|> total number of 1's in the number. The box marked "2" indicates the \|> total number of 2's in the number, and so on. For example, the "3" in \|> the box labeled "0" would indicate that there must be exactly three 0's \|> in the 10-digit number. \|> \|> ------------------------------- \|> \| 0\| 1\| 2\| 3\| 4\| 5\| 6\| 7\| 8\| 9\| \|> \| 3\| \| \| \| \| \| \| \| \| \| \|> ------------------------------- \|> \|> \|> Stop And Think \|> \|> 1. Is there a solution to this problem? Are there many solutions to this \|> problem? This is an old puzzle, but I'll solve it as if it was new because I find your extension below to be interesting. Since all possible digits must be "counted" once, the ten digits must add up to 10. Consider the first digit (= the amount of zeroes used): 9: Impossible, since all the other digits would have to be zero. 8: Also impossible, since we must mark a 1 under the 8, and the other digits then must be zeroes. 7: We must mark a 1 under the 7, and we have one more non-zero digit to assign. We've used a 1, so we must put a non-zero digit under the 1. However, if we put a 1 there, it's wrong because we've used two ones, and if we put a two that's also wrong. So 7 zeroes doesn't work. 6: Begin as before, putting a 1 under the 6. Now we must mark under the 1, but putting a 1 is wrong, so put a 2. Now we have one non-zero digit left to assign, and marking a 1 under the two works. 6210001000 works. 5: Now we take a different approach to analyze this. If there are only five zeroes, then there are five non-zeroes, one of which is the 5 we marked under the zero. Obviously a 1 must be marked under the 5 and zeroes under 6-9, so we have 5----10000, where the dashes contain one zero and three other numbers, which must add up to four (since all ten digits must add up to ten) - so we have two ones and a two. But then the digits we have are described by 5310010000, which is not the set of digits we have, so there is no solution. Similar proofs show that there cannot be 4,3,2, or 1 zero. 0: Impossible, since you would have to use a zero to indicate you didn't have a zero. \|> 2. A more advanced an interesting problem is to continue to \|> generate a sequence in a recursive fashion such that each row becomes \|> the sequence for the previous. For example, start with the usual \|> 0 through 9 digits in row 1: \|> \|> Row 1: 0 1 2 3 4 5 6 7 8 9 \|> \|> Assume Row 2 is your solution to the puzzle. I've just inserted random \|> digits below so as not to give away the solution: \|> \|> \|> Row 1: 0 1 2 3 4 5 6 7 8 9 S(1) \|> Row 2: 9 3 2 3 3 1 6 7 8 9 S(2) \|> Row 3: S(3) \|> \|> Row 2 is now the starting point, and your next job is to form row 3, row 4, \|> etc. using the same rules. In the previous example, a digit in the \|> first box would indicate how many 9's there are in the next 10-digit number, \|> and so forth. \|> \|> Contest: I am looking for the longest sequence of numbers users can come \|> up with using these rules. Can you find a Row 2 or Row 3? \|> Is it even possible to generate a "row 2" or "row 3"? Well, first off, our handy rule about all the digits adding up to ten no longer applies. Let's see if we can find an answer: Row 1: 0 1 2 3 4 5 6 7 8 9 Row 2: 6 2 1 0 0 0 1 0 0 0 Row 3: ? All the same digits must be placed under all the zeroes in row 2, or some of them would be wrong, and this digit cannot be larger than 4 since six non-zeroes are used under the zeroes in row 2. So, consider the cases: 4: If we put 4's under all the zeroes, we must put zeroes everywhere else. 0004440444 works. 3: Now we must place one non-zero digit under either the 6 or the 2, since there are two 1's that must stay alike. Putting any non-zero digit under the 6 is wrong since there aren't any sixes, unless you put a 6 under the 6, which is still wrong. Similarly no digit works under the two. 2: Now we must put a non-zero digit under the 2, since we already used 6 of them. We must also have two zeroes, which can only go under the ones. This gives us --02220222. However, we must put a non-zero under the 6, and we can't put a one, since we must have zeroes under the ones. Any number greater than one is wrong, because we don't have that many 6's. 1: OK, we start with ---111-111, and one of the -'s must be a zero. This zero must go under the 2 or the 6, because the ones must be alike (and we've already used some ones). Suppose we put 6's under the ones, and don't use any more ones. Then we need a 2 under the 6, and we need a one under the 2, which breaks what we did before. So, instead put 7's under the ones. Now we must put a 1 and a 0 in the other two spots, but either arrangement is wrong. We can't put a higher number under the ones because there aren't enough spaces left, so there is no solution with 1 zero. 0: Self-contradiction, as in the original problem. So now we have a unique third row. Can we make a fourth? Row 1: 0 1 2 3 4 5 6 7 8 9 Row 2: 6 2 1 0 0 0 1 0 0 0 Row 3: 0 0 0 4 4 4 0 4 4 4 Now there can only be two different digits used in the next number. Consider the possibilities: No zero is used: We need to mark this by putting zeroes under the zeroes Some zeroes are used: They can't go under the zeroes, so put zeroes under the fours. Now six zeroes are used, so put 6's under the zeroes. 6660006000 works. The same logic used to find row four shows that row five must be 0004440444 again, and we get into an infinite cycle alternating between these two. -- ----w-w--------------Joseph De [email protected] ( ^ ) Disclaimer: My opinions do not represent those of Owlnet. (O O) Owlnet: George R. Brown School of Engineering Educational Network. v-v (Unauthorized use is prohibited.) (Being uwop-ap!sdn is allowed.) Snail mail: Rice U., 6100 S. Main, Houston TX 77005. ------------------------- In rec.puzzles you write: >Title: Cliff Puzzle 1: Can you beat the numbers game? >From: [email protected] [...] >1. Is there a solution to this problem? Are there many solutions to this >problem? Yes. No. >2. A more advanced an interesting problem is to continue to >generate a sequence in a recursive fashion such that each row becomes >0 through 9 digits in row 1: [...] >Contest: I am looking for the longest sequence of numbers users can come >up with using these rules. Can you find a Row 2 or Row 3? >Is it even possible to generate a "row 2" or "row 3"? My program produces the following output: 0123456789 6210001000 no solutions found So I believe that the result for row 2 is unique and that there is no result for row 3. [ I am including the program at the end of this message just for your interest ] >If you respond to this puzzle, if possible please include your name, >address, affiliation, e-mail address. If you like, tell me a little bit >about yourself. You might also directly mail me a copy of your response >in addition to any responding you do in the newsgroup. I will assume it >is OK to describe your answer in any article or publication I may write >in the future, with attribution to you, unless you state otherwise. >Thanks, Cliff Pickover The name, address etc should appear in my signature. As for myself, I'm a PhD student due to finish much too shortly who likes solving puzzles. Pauli Paul Dale \| [email protected] Department of Computer Science \| +61 7 365 2445 University of Queensland \| Australia, 4072 \| Did you know that there are 41 two letter \| words containing the letter 'a'? The program I used follows: --------------------------------------8<------------------------------ #include <stdio.h> #include <stdlib.h> #define START(in) for(in=0;in<9;in++) { \ if(sum+in > 10) \ break; \ else \ sum = sum+in; \ counts[digits[in]]++; #define STOP(in) counts[digits[in]]--; \ sum -= in; \ } main() { short counts[10]; short i, sum; short i0,i1,i2,i3,i4,i5,i6,i7,i8,i9; static short digits[10] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }; short solns[10][100]; short solcnt=0; printf("0123456789\n"); again: for(i=0;i<10;i++) counts[i]=0; sum = 0; START(i0) START(i1) START(i2) START(i3) START(i4) START(i5) START(i6) START(i7) START(i8) START(i9) if(counts[0]==digits[i0] && counts[1]==digits[i1] && counts[2]==digits[i2] && counts[3]==digits[i3] && counts[4]==digits[i4] && counts[5]==digits[i5] && counts[6]==digits[i6] && counts[7]==digits[i7] && counts[8]==digits[i8] && counts[9]==digits[i9]) { printf("%d%d%d%d%d%d%d%d%d%d\n", i0,i1,i2,i3,i4,i5, i6,i7,i8,i9); for(i=0;i<10;i++) solns[0][solcnt] = i0; solns[1][solcnt] = i1; solns[2][solcnt] = i2; solns[3][solcnt] = i3; solns[4][solcnt] = i4; solns[5][solcnt] = i5; solns[6][solcnt] = i6; solns[7][solcnt] = i7; solns[8][solcnt] = i8; solns[9][solcnt] = i9; solcnt++; } STOP(i9) STOP(i8) STOP(i7) STOP(i6) STOP(i5) STOP(i4) STOP(i3) STOP(i2) STOP(i1) STOP(i0) if(solcnt == 0) { printf("no solutions found\n"); } else if(solcnt == 1) { for(i=0;i<10;i++) digits[i] = solns[i][0]; solcnt = 0; goto again; } else printf("multiple solutions found\n"); } --------------------------------------8<------------------------------ In article <[email protected]> you write: >Title: Cliff Puzzle 1: Can you beat the numbers game? >From: [email protected] > >If you respond to this puzzle, if possible please include your name, >address, affiliation, e-mail address. If you like, tell me a little bit >about yourself. You might also directly mail me a copy of your response >in addition to any responding you do in the newsgroup. I will assume it >is OK to describe your answer in any article or publication I may write >in the future, with attribution to you, unless you state otherwise. >Thanks, Cliff Pickover > >* * * >At a recent trip to the Ontario Science Center in Toronto, Canada I came >across an interesting puzzle. The center is located minutes from >downtown Toronto and it's a vast playground of science with hundreds of >exhibits inviting you to touch, try, test, and titillate your curiosity. >The puzzle I saw there can be stated as follows. In the 10 boxes below, >write a 10-digit number. The digit in the first box indicates the total >number of zeros in the entire number. The box marked "1" indicates the >total number of 1's in the number. The box marked "2" indicates the >total number of 2's in the number, and so on. For example, the "3" in >the box labeled "0" would indicate that there must be exactly three 0's >in the 10-digit number. > >------------------------------- >\| 0\| 1\| 2\| 3\| 4\| 5\| 6\| 7\| 8\| 9\| >\| 3\| \| \| \| \| \| \| \| \| \| >------------------------------- > > >Stop And Think > >1. Is there a solution to this problem? Are there many solutions to this >problem? A. Since there are ten digits in the number, the sum of the digits in the bottom row must be 10. B. If x appears under y there must be x appearences of y, hence xy<10 So, the MAXIMUM that can appear under each number is: --------------------- \|0\|1\|2\|3\|4\|5\|6\|7\|8\|9\| \|9\|9\|4\|3\|2\|1\|1\|1\|1\|1\| max --------------------- C. In fact, under the numbers 5..9 there can be AT MOST one non-zero (1) answer since otherwise two numbers of the 5..9 veriaty would appear and violate rule A. D. So there must be at least 4 zeros. If there were exactly 4 zeros, then the numbers 1..4 will all have under them non-zeros (as the zeros are used up for the 5..9 group). There is also at least one number that is 5 or greater. Well, there is a 5 (or more), a 4 (under zero), a 1 (under the 5..9 category) and something above zero under the other 1..4 digits for a total above 10. This violates rule A. E. So there must be at least 5 zeros. So a (exactly one) number that is at least 5 has a 1 under it. (since under zero would appear a >=5 number). F. Under 1 there must be at least 1 since the solution has at least one 1 (the one under a 5..9 number). However it could not be exactly 1 as then there would be 2 (or more) 1's in the solution. G. If there were 3 or more ones, then they must be under 2..9 . But then there would be a 5 (or more) under zero + a 3 (or more) under one + a 1 under three (or more) other places for a total above 10. H. So there must be at exactly 2 ones in the solution. And hence, at least 1 under two. We can summerize: --------------------- \|0\|1\|2\|3\|4\|5\|6\|7\|8\|9\| \|5\|2\|1\|0\|0\|----1----\| min \|6\|2\|2\|1\|1\|----1----\| max --------------------- where the maximum under each digit is 10 - SUM(minimum of all others) I. Since no 3 or 4 is now possible, those two numbers must have a zero under them. J. So there are six zeros. Hence: --------------------- \|0\|1\|2\|3\|4\|5\|6\|7\|8\|9\| \|6\|2\|1\|0\|0\|0\|1\|0\|0\|0\| min \|6\|2\|2\|0\|0\|0\|1\|0\|0\|0\| max --------------------- > K. Notice that "min" is a solution, while "max" is not. Hence, "min is the ONLY* solution! My name is Dan Shoham. This is the only fact about me I care to make public. You are free to attribute it, but provide me a note when you do so. [email protected] ------------------------- >From [email protected] (Chris Long) Tue Sep 15 06:08:45 1992 Path: igor.rutgers.edu!romulus.rutgers.edu!clong From: [email protected] (Chris Long) Newsgroups: rec.puzzles Subject: Re: Puzzle 1 (SPOILER) Message-ID: <[email protected]> Date: 15 Sep 92 10:08:45 GMT Lines: 62 In article <[email protected]>, Chris Cole writes: Chris, don't forget to include my name on my solutions in the FAQ, please. My old article should be replaced with the following in the FAQ, anyway: --Cut here-- Solution prepared by Chris Long. Unfortunately, this isn't completely new, since I believe a similar puzzle I posted and answered are in the FAQ. However, it is different enough to be interesting. In article <[email protected]>, [email protected] writes: > Here's a small number puzzle : > Generate numbers such that the each digit in the number specifies > the number of the occurences of the position of the digit ( postions starting > with 0 from the left ). Example > The number 1210 ... My guess is only: 1210 21200 3211000 42101000 521001000 6210001000 No 1, 2, or 3 digit numbers are possible. Letting x_i be the ith digit, starting with 0, we see that (1) x_0 + ... + x_n = n+1 and (2) 0x_0 + ... + nx_n = n+1, where n+1 is the number of digits. I'll first prove that x_0 > n-3 if n>4. Assume not, then this implies that at least four of the x_i with i>0 are non-zero. But then we would have \sum_i ix_i >= 10 by (2), impossible unless n=9, but it isn't possible in this case (51111100000 isn't valid). Now I'll prove that x_0 < n-1. x_0 clearly can't equal n; assume x_0 = n-1 ==> x_{n-1} = 1 by (2) if n>3. Now only one of the remaining x_i may be non-zero, and we must have that x_0 + ... + x_n = n+1, but since x_0 + x_{n-1} = n ==> the remaining x_i = 1 ==> by (2) that x_2 = 1. But this can't be, since x_{n-1} = 1 ==> x_1>0. Now assuming x_0 = n-2 we conclude that x_{n-2} = 1 by (2) if n>5 ==> x_1 + ... + x_{n-3} + x_{n-1} + x_n = 2 and 1x_1 + ... + (n-3)x_{n-3} + (n-1)x_{n-1} + nx_n = 3 ==> x_1=1 and x_2=1, Case n>5: We have that x_0 = n-3 and if n>=7 ==> x_{n-3}=1 ==> x_1=2 and x_2=1 by (1) and (2). For the case n=6 we see that x_{n-3}=2 leads to an easy contradiction, and we get the same result. The cases n=4,5 are easy enough to handle, and lead to the two solutions above. -- Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618 -- Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618 ------------------------- The number "2020" was left off my list by mistake ... sorry. -Chris ------------------------- > * * > At a recent trip to the Ontario Science Center in Toronto, Canada I came > across an interesting puzzle. The center is located minutes from > downtown Toronto and it's a vast playground of science with hundreds of > exhibits inviting you to touch, try, test, and titillate your curiosity. > The puzzle I saw there can be stated as follows. In the 10 boxes below, > write a 10-digit number. The digit in the first box indicates the total > number of zeros in the entire number. The box marked "1" indicates the > total number of 1's in the number. The box marked "2" indicates the > total number of 2's in the number, and so on. For example, the "3" in > the box labeled "0" would indicate that there must be exactly three 0's > in the 10-digit number. > > ------------------------------- > \| 0\| 1\| 2\| 3\| 4\| 5\| 6\| 7\| 8\| 9\| > \| 3\| \| \| \| \| \| \| \| \| \| > ------------------------------- > > > Stop And Think > > 1. Is there a solution to this problem? Are there many solutions to this > problem? > [Second question and contest problem omitted] Good puzzle! I am wondering though whether the second question (which I have not tried to solve yet) is moe amenable to computer search. It seems to me that there should not be so many cases to consider, so that even exhaustive search should work. So, here is my ten minutes work on the first question. I think there is a unique solution which is: 6210001000. Here is the reasoning. Let the number be (in Tex notation) d_0 d_1 d_2 d_3 d_4 d_5 d_6 d_7 d_8 d_9. By definition d_0 + d_1 + d_2 + d_3 + d_4 + d_5 + d_6 + d_7 + d_8 + d_9 = 10. (1) Moreover, d_0 > 0, since d_0 = 0 contradicts itself. Let d_0 = c for some integer 9 >= c >= 1. If c = 9, then d_9 = 1, contradiction since d_1 should both be 0 and 1 then. If 9 > c >= 1, we rewrite (1) removing all d_i s that are zeros c + d_(i_1) + d_(i_2) + ... + d_(i_(9-c)) = 10 <=> d_(i_1) + d_(i_2) + ... + d_(i_(9-c)) = 10 -c (2) where all the d_(i_j) >= 1, j=1,...,9-c (3) (2) & (3) imply that the d_(i_j)s are 8-c 1s and one 2. Since there exists ONE 2, then there exists at least one 1. So the only digits in the number are 0, 1, 2, and c (if different than 1 and 2). If c is either 1 or 2, we have 3 different digits in the number, which implies d_1 <= 3, impossible since d_1 = 8 - c >= 6. If c> 2, we have four different digits in the number, and in fact d_0 = c, d_1 = 8-c, d_2 = 1, d_c = 1, which leaves us with 6 0s. QED I hope I did not miss any other cases. Leonidas -------------------------------------------------------------------------------- Leonidas Palios The Geometry Center 1300 South Second Str [email protected] Minneapolis, Minnesota 55454 ------------------------- ------------------------------- \| 0\| 1\| 2\| 3\| 4\| 5\| 6\| 7\| 8\| 9\| ------------------------------- \| 6\| 2\| 1\| 0\| 0\| 0\| 1\| 0\| 0\| 0\| \| 0\| 0\| 0\| 4\| 4\| 4\| 0\| 4\| 4\| 4\| <- \| 6\| 6\| 6\| 0\| 0\| 0\| 6\| 0\| 0\| 0\| \| \| 0\| 0\| 0\| 4\| 4\| 4\| 0\| 4\| 4\| 4\| <- . . . I must be missing something in my understanding of your rules. I found the second row by imagining that I'd need lots of zeros and putting nine in the 0 column, then skipping back and forth adjusting things. I had to put a tic in the 9 column, then I had to put one in the 1 column, then I realized that had to change that to a two since now there were two ones, and at the same time another required tic in the 2 column balanced the change of one to two in the 1 column, and then of course there weren't nine zeros anymore, but there were still six and so by changing the nine in the 1 column to a six, the one in the 9 column sould just migrate down to the 6 column. But it almost seems like cheating to use fours in the second row when there were none in the second row to necessitate this kind of adjusting. shrug If this is right, the series is infinite, obviously. Please let me know if I'm interpreting something wrong. Thanks, and nice puzzle. :) Grant Culbertson [email protected] [email protected] Continue to:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8361473679542542, "perplexity": 1240.3819270330494}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487643380.40/warc/CC-MAIN-20210619020602-20210619050602-00265.warc.gz"}
http://math.stackexchange.com/questions/194579/what-is-the-origin-of-the-determinant-in-linear-algebra	# What is the origin of the determinant in linear algebra? We often learn in a standard linear algebra course that a determinant is a number associated with a square matrix. We can define the determinant also by saying that it is the sum of all the possible configurations picking an element from a matrix from different rows and different columns multiplied by (-1) or (1) according to the number inversions. But how is this notion of a 'determinant' derived? What is a determinant, actually? I searched up the history of the determinant and it looks like it predates matrices. How did the modern definition of a determinant come about? Why do we need to multiply some terms of the determinant sum by (-1) based on the number of inversions? I just can't understand the motivation that created determinants. We can define determinants, and see their properties, but I want to understand how they were defined and why they were defined to get a better idea of their important and application. - the determinant defines volume in n-dimensions. Munkrese Analysis on Manifolds text has a nice discussion. I'm not well-versed in the history you seek so I leave it to others! – James S. Cook Sep 12 '12 at 7:14 Related question math.stackexchange.com/questions/668/…. See the answers given there. – Marc van Leeuwen Sep 12 '12 at 7:20 Possible duplicate of math.stackexchange.com/questions/81521/…. – lhf Oct 21 '13 at 12:27 I normally have two ways of viewing determinants without appealing to higher-level math like multilinear forms. The first is geometric, and I do think that most vector calculus classes nowadays should teach this interpretation. That is that, given vectors $v_1, \ldots, v_n \in \mathbb{R}^n$ dictating the sides of an $n$-dimensional parallelepiped, the volume of this parallelepiped is given by $\det(A)$, where $A = [v_1 \ldots v_n]$ is the matrix whose columns are given by those vectors. We can then view the determinant of a square matrix as measuring the volume-scaling property of the matrix as a linear map on $\mathbb{R}^n$. From here, it would be clear why $\det(A) = 0$ is equivalent to $A$ not being invertible - if $A$ takes a set with positive volume and sends it to a set with zero volume, then $A$ has some direction along which it "flattens" points, which would precisely be the null space of $A$. Unfortunately, I'm under the impression that this interpretation is at least semi-modern, but I think this is one of the cases where the modern viewpoint might be better to teach new students than the old viewpoint. The old viewpoint is that the determinant is simply the result of trying to solve the linear system $Ax = b$ when $A$ is square. This is most likely how the determinant was first discovered. To derive the determinant this way, write down the generic matrix and then proceed by Gaussian elimination. This means you have to choose nonzero leading entries in each row (the pivots) and use them to eliminate subsequent entries below. Each time you eliminate the rows, you have to multiply by a common denominator, so after you do this $n$ times, you'll end up with the sum of all the permutations of entries from different rows and columns merely by virtue of having multiplied out to get common denominators. The $(-1)^k$ sign flip comes from the fact that at each stage in Gaussian elimination, you're subtracting. So on the first step you're subtracting, but on the second step you're subtracting a subtraction, and so forth. At the very end, by Gaussian elimination, you'll obtain an echelon form (upper triangular), and one knows that if any of the diagonal entries are zero, then the system is not uniquely solvable; the last diagonal entry will precisely be the determinant times the product of the values of previously used pivots (up to a sign, perhaps). Since the pivots chosen are always nonzero, then it will not affect whether or not the last entry is zero, and so you can divide them out. EDIT: It isn't as simple as I thought, though it will work out if you keep track of what nonzero values you multiply your rows by in Gaussian elimination. My apologies if I mislead anyone. - also, we should emphasize the sign of $det[v_1\|v_2\|...\|v_n]$ reveals the handedness or orientation of the set $\{ v_1,v_2,\dots v_n \}$ – James S. Cook Sep 12 '12 at 7:36 Did you ever try actually performing Gaussian elimination on a generic matrix (with all entries independent unknowns)? Try it for a $3\times3$ matrix! It doesn't really work as you advertised, and you'll have a hard time actually making (just) the determinant appear in the computations. You can find something like this done to prove Cramer's rule, but is is not usual Gaussian elimination, and it assumes the determinant is already known. – Marc van Leeuwen Sep 12 '12 at 7:36 @MarcvanLeeuwen, you're right, I forgot that actually the final diagonal entry will be the determinant multiplied by the value of the first pivot. But since WLOG the first pivot must be a nonzero value, then it can be divided. I don't think it's as hard to manipulate into the determinant form as one might think. – Christopher A. Wong Sep 12 '12 at 7:48 BTW, I just tried it for the $3 \times 3$ case, and it ends up being $a_{11} \det(A)$ for the last diagonal entry, as expected. Note that if $a_{11} = 0$, then we just swap rows until WLOG $a_{11} \neq 0$. The nice thing about swapping rows in Gaussian elimination not affecting the determinant is that it shows why, on some level, the determinant must be permutation-invariant. – Christopher A. Wong Sep 12 '12 at 8:09 @ChristopherA.Wong In fact it is unclear to me what you mean by "first pivot". With all entries unknown, there isn't a single (non-constant) expression that is assured to be nonzero. So you need to multiply rows by factors that are not known to be nonzero, and these factors will remain in (the determinant of) your matrix. I can see how you get a factor $a_{1,1}$, but not how you avoid introducing even nastier factors in the sequel. I'm stuck with $\begin{pmatrix}a_1&a_2&a_3\\0&a_1b_2-b_1a_2&a_1b_3-b_1a_3\\0&a_1c_2-c_1a_2&a_1c_3‌-c_1a_3\end{pmatrix}$. – Marc van Leeuwen Sep 12 '12 at 8:25 I do not know the actual history of determinant, but I think it is very well motivated. From the way I look at it, it's actually those properties of determinant that make sense. Then you derive the formula from them. Let me start by trying to define the "signed volume" of a hyper-parallelepiped whose sides are $(u_1, u_2, \ldots, u_n)$. I'll call this function $\det$. (I have no idea why it is named "determinant". Wiki says Cauchy was the one who started using the term in the present sense.) Here are some observations regarding $\det$ that I consider quite natural: 1. The unit hypercube whose sides are $(e_1, e_2, \ldots, e_n)$, where $e_i$ are standard basis vectors of $\mathbb R^n$, should have volume of $1$. 2. If one of the sides is zero, the volume should be $0$. 3. If you vary one side and keep all other sides fix, how would the signed volume change? You may think about a 3D case when you have a flat parallelogram defined by vectors $u_1$ and $u_2$ as a base of a solid shape, then try to extend the "height" direction by the third vector $u_3$. What happens to the volume as you scale $u_3$? Also, consider what happens if you have two height vectors $u_3$ and $\hat u_3$. $\det(u_1, u_2, u_3 + \hat u_3)$ should be equal to $\det(u_1, u_2, u_3) + \det(u_1, u_2, \hat u_3)$. (This is where you need your volume function to be signed.) 4. If I add a multiple of one side, say $u_i$, to another side $u_j$ and replace $u_j$ by $\hat u_j = u_j + c u_i$, the signed volume should not change because the addition to $u_j$ is in the direction of $u_i$. (Think about how a rectangle can be sheered into a parallelogram with equal area.) With these three properties, you get familiar properties of $\det$: 1. $\det(e_1, \ldots, e_n) = 1$. 2. $\det(u_1, \ldots, u_n) = 0$ if $u_i = 0$ for some $i$. 3. $\det(u_1, \ldots, u_i + c\hat u_i, \ldots, u_n) = \det(u_1, \ldots, u_i, \ldots, u_n) + c\det(u_1, \ldots, \hat u_i, \ldots, u_n)$. 4. $\det(u_1, \ldots, u_i, \ldots, u_j, \ldots, u_n) = \det(u_1, \ldots, u_1, \ldots, u_j + cu_i, \ldots, u_n)$. (It may happen that $j < i$.) You can then derive the formula for $\det$. You can use these properties to deduce further easier-to-use (in my opinion) properties: • Swapping two columns changes the sign of $\det$. This should tell you why oddness and evenness of permutations matter. To actually (inefficiently) compute the determinant $\det(u_1, u_2, \ldots, u_n)$, write $u_i$ as $u_i = \sum_{j=1}^n u_{ij}e_j$, and expand by multilinearity. For example, in 2D case, \begin{align} \det(u, v) & = \det(u_1e_1 + u_2e_2, v_1e_1 + v_2e_2) \\ & = u_1v_1\underbrace{\det(e_1, e_1)}_0 + u_1v_2\underbrace{\det(e_1, e_2)}_1 + u_2v_1\underbrace{\det(e_2, e_1)}_{-1} + u_2v_2\underbrace{\det(e_2, e_2)}_0 \\ & = u_1v_2 - u_2v_1. \end{align} (If you are not familiar with multilinearity, just think of it as a product. Ignore the word $\det$ from the second line and you get a simple expansion of products. Then you evaluate "unusual product" between vectors $e_i$ by the definition of $\det$. Note, however, that the order is important, as $\det(u, v) = - \det(v, u)$.) -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.942439079284668, "perplexity": 262.14230476195473}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422122047499.45/warc/CC-MAIN-20150124175407-00186-ip-10-180-212-252.ec2.internal.warc.gz"}
http://mathhelpforum.com/math-topics/131652-suspended-particle.html	# Math Help - suspended particle 1. ## suspended particle I got a negative tension (-6.48), is this right? I think where i'm going wrong is when i label my diagram. a particle is suspended by two light inextensible strings and hangs in equilibrium. one string in inclined at 30 degrees to the horizontal and the tension in the string is of magnitude 40N. the second string is inclined at 60 degrees to the horizontal. calculate in N. a) the weight of the particle b) the magnitude of the tension in the second string. 2. Originally Posted by djr8793 I got a negative tension (-6.48), is this right? I think where i'm going wrong is when i label my diagram. a particle is suspended by two light inextensible strings and hangs in equilibrium. one string in inclined at 30 degrees to the horizontal and the tension in the string is of magnitude 40N. the second string is inclined at 60 degrees to the horizontal. calculate in N. a) the weight of the particle b) the magnitude of the tension in the second string. In equilibrium condition T1cos30 = T2cos60......(1) T1sin 30 + T2sin60 = mg.....(2) Solve these two equations to find T1 and T2.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9215125441551208, "perplexity": 506.987783209859}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645396463.95/warc/CC-MAIN-20150827031636-00223-ip-10-171-96-226.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/basic-quantum-question.274393/	# Basic Quantum Question 1. Nov 23, 2008 ### daveyman Problem: In a practice test that I am studying, I am told to find the wave function $$\psi_{100} (r,\theta ,\phi)$$ for the ground state of the hydrogen atom. This I can do. Here is the second part of the question: By direct substitution into the TISE, find the energy eigen value for this state. Attempt at Solution: Basically, all you have to do is plug the wave function into the TISE: $$H\psi=E\psi$$ where $$H=-\frac{\hbar^2}{2m}\nabla^2 + V_{hydrogen}$$ In the answer key, I noticed that only the r component of $$\nabla$$ is considered. Why can the theta and phi terms be ignored? Any help would be much appreciated. Thanks! Last edited: Nov 23, 2008 2. Nov 23, 2008 ### Pacopag Look for the form of the SE in spherical coordinates (it should be in any textbook). It has a radial term and an angular term. When you see the equation, it is very clear why the angular part does not appear for the ground state. 3. Nov 23, 2008 ### daveyman I'm sorry, I don't think I made my question very clear. Yes, the TISE in spherical coordinates has both radial and angular terms, but it is not clear from the SE equation itself why the angular part is not a factor in the ground state. I think you are referring to the fact that the first spherical harmonic does not have an angular term: $$Y_0^0=(\frac{1}{4\pi})^{\frac{1}{2}}$$ Clearly if the angular terms of the wave function are zero (as they are in this case), you can simply ignore the angular terms in the TISE. I guess my real question is regarding the first spherical harmonic, then. Why does it not depend on any angular terms (without going into complicated mathematics)? My guess is that since this spherical harmonic is a "perfect" sphere, it doesn't matter where you are on that sphere. Is this correct or naive? Last edited: Nov 23, 2008 4. Nov 23, 2008 ### Pacopag Yes. You are right! So if you take the SE and do separation of variables, you already know the "constant" angular part of the solution. Go to the website http://en.wikipedia.org/wiki/Particle_in_a_spherically_symmetric_potential 2 Derivation of the radial equation The third equation down gives the "radial" schrodinger equation. In your case, l=m=0. It is a 1-D problem. 5. Nov 23, 2008 ### daveyman Thanks! 6. Nov 23, 2008 ### Pacopag Your answer to your question about the Y00 harmonic sounds reasonable. Unfortunately, I only have a mathematical answer (and it might be a load of bull&%#\$). I can explain why there must be a constant spherical harmonic. We know that the spherical harmonics satisfy a completeness relation, thus they can be used as a basis in which to expand any function. But some functions are constant. So in order to represent a "constant" function in the spherical harmonic basis, then at least one of the spherical harmonics must be a constant. 7. Nov 23, 2008 ### Pacopag That equation in the wiki site should be in your textbook, probably in the chapter on spherically symmetric potentials. Similar Discussions: Basic Quantum Question	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.945800244808197, "perplexity": 450.5453844338448}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218191405.12/warc/CC-MAIN-20170322212951-00508-ip-10-233-31-227.ec2.internal.warc.gz"}
https://arhive.inshe.org/m19c7vy/volatility-surface-arbitrage-7057a0	0000013475 00000 n An input implied volatility surface that is not arbitrage-free can result in negative transition probabilities and consequently mispricings and false greeks. Quantitative Finance, 9:4, 417-428. To an option trader engaging in volatility arbitrage, an option contract is a way to speculate in the volatility of the underlying rather than a directional bet on the underlying's price. In terms of implied volatility: total implied variance should be non decreasing in time, and that, for any given forward moneyness level, see Gatheral top of page 4.. Because implied volatility of an option can remain constant even as the underlying's value changes, traders use it as a measure of relative value rather than the option's market price. 0000003348 00000 n 0000001244 00000 n is the price of the underlying, and 0000001174 00000 n 0000007298 00000 n In particular, … Zelida system (2009) used total implied variances. new construction of an implied volatility surface from a discrete set of implied volatilities which is arbitrage-free and satisfies some smoothness conditions. A new arbitrage-free parametric volatility surface. {\displaystyle \sigma \,} endobj {\displaystyle f()\,} The method uses smoothing splines under shape constraints to estimate call option prices as a function of strike and time-to-maturity. For instance, if a trader can buy an option whose implied volatility Stock Option Basics . <> A volatility surface w is free of calendar spread arbitrage if and only if @tw(k;t) 0, for all k 2 R and t > 0. Conversely, if the trader can sell an option whose implied volatility is 20%, it is said the trader can "sell the option at 20%". In finance, volatility arbitrage (or vol arb) is a type of statistical arbitrage that is implemented by trading a delta neutral portfolio of an option and its underlying. , or. Antoine (Jack) Jacquier. to construct the arbitrage-free call price surfaces. The … In [2] this problem is solved by restricting the parameters in the SVI model. Constructing … 0000030898 00000 n Deformation of implied volatility surfaces : an empirical analysis. For example, assume a call option is trading at $1.90 with the underlying's price at$45.50 and is yielding an implied volatility of 17.5%. 0000015815 00000 n <>stream ) 1.2 Purpose of the thesis no longer a surface, but rather a “volume”, and even more dimensions/factor would require a “hyperspace”). Calibration of SVI to given implied volatility data (for example [12]). � �� a h�. Jim Gatheral, Merrill Lynch, May-2004 Roger Lee’s moment formula { } { } 2 *BS 2 ** * 2 *1 BS ** * Define :log(/). In finance, volatility arbitrage (or vol arb) is a type of statistical arbitrage that is implemented by trading a delta neutral portfolio of an option and its underlying. σ <> The profit is extracted from the trade through the continuous re-hedging required to keep the portfolio delta-neutral. "Tests d'arbitrage et surfaces de volatilité : analyse empirique sur données haute fréquence [Arbitrage tests and surface of implied volatility: An empirical analysis of high frequency data]," MPRA Paper 17415, University Library of Munich, Germany. traders attempt to buy volatility when it is low and sell volatility when it is high. Paramétrisations SVI et SSVI. Denition 2.1 A volatility surface is free of static arbitrage if and only if the following conditions are satised: (i) it is free of calendar spread arbitrage; (ii) each time slice is free of butter y arbitrage. <> The objective is to take advantage of differences between the implied volatility[1] of the option, and a forecast of future realized volatility of the option's underlying. that expresses the volatility implied by the option's market price Vol Surface Arbitrage . 0000008121 00000 n 0000002884 00000 n It relies on predicting the future direction of implied volatility. Castagna (2010) and Carr (2004) discusses the necessary and sufficient conditions that are required to declare that the volatility surface is admissible and thus arbitrage-free. 0000009553 00000 n Arbitrage-Free Smoothing of the Implied Volatility Surface. As described in option valuation techniques, there are a number of factors that are used to determine the theoretical value of an option. 0000000015 00000 n , there must be a corresponding monotonically increasing function 139 0 obj Calypso Technology; Independent. Section 5 examines a number of special cases of the model. corresponding implied volatility surface is free of calendar spread arbitrage. {\displaystyle \sigma \,} Remark Condition 1 needs to be a strict … is 10%, it is common to say that the trader can "buy the option for 10%". Implied volatility is useful in trading for a number of applications and crypto is no exception. The best-known measure of market volatility is the CBOE Volatility Index (VIX), which measures the volatility of the S&P 500. 0000006335 00000 n volatility surface n No-arbitrage conditions n SVI fits to market data n SVI fits to theoretical models n Carr-Lee valuation of volatility derivatives under the zero correlation assumption n Valuation of volatility derivatives in the general case. Alexandre Antonov, Michael Spector and Michael Konikov describe a new parametric volatility surface that is arbitrage free, is extremely rich and flexible, and has closed-form expressions for both European … (2013) proposed to minimize the square differences between observed and fitted volatility, while Homescu (2011) advised a square difference method. arbitrage free volatility surface. The results indicate that there is a mispricing, but it is not an underpricing as widely reported but rather an overpricing. the implied volatility surface in an arbitrage-free way. Concave smiles often arise when a significant jump with a predictable time of occurrence is priced in. 130 0 obj 0000014730 00000 n Long Term Capital Management used a volatility arbitrage approach. 140 0 obj Arbitrage-Free Smoothing of the Implied Volatility Surface. ( A short time later, the same option might trade at $2.50 with the underlying's price at$46.36 and be yielding an implied volatility of 16.5%. The implied volatility surface (IVS) is a fundamental building block in computational ﬁnance. Modèles à volatilité stochastique de première génération et leur traitement. endobj 0000015257 00000 n C The volatility surface varies over time and is far from flat, demonstrating that the assumptions of the Black-Scholes model are not always correct. Un exemple: le modèle de Heston. <> We consider the classical problem of building an arbitrage-free implied volatility surface from bid-ask quotes. {\displaystyle g()\,} <> 133 0 obj 134 0 obj 0000004019 00000 n 0000014447 00000 n endobj In the second case, the trader sells the option and then hedges the position. Active 8 months ago. xref for the option. <> Volatility arbitrage is a type of statistical arbitrage that seeks to take advantage of the difference between the implied volatility of an option and the volatility of the underlying asset. �A� @@��C��'t)!3��2�\|�� !��XL2~�S}�h3H�� pm$� endstream )��u� is the estimate of future volatility. Introduction. 0 It has a modern web UI that provides robust features to analyze and visualize … We extend Gatheral and Jacquier’s surface stochastic volatility-inspired (SSVI) parameterization by making the correlation maturity dependent and obtaining the necessary and sufficient conditions for no calendar-spread arbitrage. So long as the trading is done delta-neutral, buying an option is a bet that the underlying's future realized volatility will be high, while selling an option is a bet that future realized volatility will be low. endobj 3 develops a general model for the evolution of a volatility surface and derives the no-arbitrage condition. Nevertheless West (2005) applied vega weighted square volatility differences. 135 0 obj Option pricing; Deep learning; No-Arbitrage; Local Volatility. In particular, we exhibit a large class of arbitrage-free SVI Automate construction of an arbitrage free implied volatility surface using various models, such as SVI, SABR, Carr Pelts etc. BUILDING ARBITRAGE-FREE IMPLIED VOLATILITY: SINKHORN’S ALGORITHM AND VARIANTS HADRIENDEMARCHANDPIERREHENRY-LABORDÈRE Abstract. The objective is to take advantage of differences between the implied volatility of the option, and a forecast of future realized volatility of the option's underlying. %%EOF 0000020171 00000 n Abstract: In this article, we show how to calibrate the widely-used SVI parameterization of the implied volatility surface in such a way as to guarantee the absence of static arbitrage. Armed with a forecast of volatility, and capable of measuring an option's market price in terms of implied volatility, the trader is ready to begin a volatility arbitrage trade. 0000013592 00000 n Abstract. startxref ¯ Section 4 discusses the implications of the no-arbitrage condition. Even portfolio based volatility arbitrage approaches which seek to "diversify" volatility risk can experience "black swan" events when changes in implied volatility are correlated across multiple securities and even markets. Ask Question Asked 1 year, 11 months ago. The quantity @ t w ( k;t ) is nothing else than the numerator of the local volatility expressed in terms of … We also discuss various topics which … Or, in other words, when all other inputs including the stock price Therefore, being long a delta-hedged call results in the same returns as being long a delta-hedged put. {\displaystyle \sigma _{\bar {C}}\,} x�}�{p��x�@�M�t ϔ$�I C(�mJql�^�kdi�]YH��l��d�%۲,�@� � c0��0Bӌ��IgJ�t2�\$w=��t�4��ܙ;w�=��{ι͜!�0la�Ңߥd�Wnc�Zwjo��~&v�I��ɢY�. arbitrage free volatility surface. Convertible Bond Arbitrage Using the Volatility Surface Convertible bonds are complex, hybrid securities. 0000008661 00000 n 0000001824 00000 n %�� C A parsimonious arbitrage-free implied volatility parameterization with application to the valuation of volatility derivatives @inproceedings{Gatheral2004APA, title={A parsimonious arbitrage-free implied volatility parameterization with application to the valuation of volatility derivatives}, author={Jim Gatheral}, year={2004} } Implied volatility surface provided by Deltas and maturities (IVS-DM) is widely used in financial fields, especially in foreign exchange options market, since it can effectively describe the characteristics of the volatilities. GENERALIZED ARBITRAGE-FREE SVI VOLATILITY SURFACES 621 conditionsforagiventwo-dimensionalfunction(ofstrikeandmaturity)tobeaproperimplied volatility surface, i.e., to generate arbitrage-free European option prices. 27 Pages Posted: 3 Apr 2012 Last revised: 15 Jan 2014. <> However, in practice, the only two inputs to the model that change during the day are the price of the underlying and the volatility. Implied volatility is useful in trading for a number of applications and crypto is no exception. Volatility smiles are implied volatility patterns that arise in pricing financial options.It is a parameter (implied volatility) that is needed to be modified for the Black–Scholes formula to fit market prices. {\displaystyle {\bar {C}}\,} )��u� Imperial College London; The Alan Turing Institute. S 0000003101 00000 n 0000021995 00000 n We provide a survey of methodologies for constructing such surfaces. ��-�]2p��-�]0��)��u�,��l�t� In finance, a convertible bond or convertible note or convertible debt (or a convertible debenture if it has a maturity of greater than 10 years) is a type of bond that the holder can convert into a specified number of shares of common stock in the issuing company or cash of equal value. The parametric families for the correlation for which those conditions are explicit are also provided. In a pure diffusion setting, you can equivalently write no calendar arbitrage constraints:. We propose an approach for smoothing the implied volatility smile in an arbitrage … CUNY Baruch College. endobj Viewed 2k times 4. Because of the put–call parity, it doesn't matter if the options traded are calls or puts. ) 136 0 obj Prior work has not successfully attempted to eliminate static arbitrage. Also known as the fear gauge, when the S&P 500 suffers a … <> f A volatility smile that is concave around the forward does not necessarily represent an arbitrage. The correct pricing of local volatility surface requires an arbitrage free implied volatility surface. Conditions de non-arbitrage sur la surface de volatilité, propriétés asymptotiques. ( is either significantly lower than or higher than the forecast realized volatility y arbitrage Theorem 4.2 The volatility surface (1) is free of butter y arbitrage if the following conditions are satis ed for all >0: 1 ’( )(1 + jˆj) <4; 2 ’( )2 (1 + jˆj) 4. Section 6 considers whether the rules of thumb are consistent with the no-arbitrage condition. endobj 0000039054 00000 n Arbitrages in the Volatility Surface Interpolation and Extrapolation. 91G20, 62M45, 91G60.16 17 1. Because the theoretical price function 141 0 obj 15 Key words. The pricing accuracy and pricing performance of local volatility models depends on the absence of arbitrage in the implied volatility surface. {\displaystyle {\bar {C}}\,} )��u� Based on these prices, implied volatilities can be obtained. Over the holding period, the trader will realize a profit on the trade if the underlying's realized volatility is closer to his forecast than it is to the market's forecast (i.e. In volatility arbitrage, volatility rather than price is used as the unit of relative measure, i.e. 0000002579 00000 n The implied volatility of a European option on a particular asset as a function of strike price and time to maturity is known as the asset’s volatility surface. 0000002227 00000 n Arbitrage trading and index option pricing at Soffex: an empirical study using daily and intradaily data. )��u� σ <>stream Marc Chesney, Rajna Gibson, and Henri Loubergé. A volatility surface for the convertible was constructed and used in a convertible arbitrage strategy. 0000023174 00000 n S In particular, we exhibit a large class of arbitrage-free SVI volatility … This strategy is … Ask Question Asked 1 year, 11 months ago. )��u� See all articles by Fabien Le Floc'h Fabien Le Floc'h. Jim Gatheral, Merrill Lynch, May-2004 Outline of this talk n Roger Lee’s moment formula n A stochastic volatility inspired (SVI) pararameterization of the implied volatility surface n No-arbitrage conditions n SVI fits to market data n SVI fits to theoretical models n Carr-Lee valuation of volatility derivatives under the zero correlation assumption n Valuation of volatility … ¯ <> Constructing an arbitrage-free volatility surface for an equity or FX rate involves checking for calendar spread arbitrage and removing this if necessary. 0000006517 00000 n endobj Constructing an arbitrage-free implied volatility surface is rather simple once the data is cleaned and the days to maturity, moneyness and normalized volatilities are computed. endobj )��u��:)��u��:)��u� 0000024620 00000 n trailer In particular, we exhibit a large class of arbitrage-free SVI volatility surfaces with a simple closed-form representation. In the first case, the trader buys the option and hedges with the underlying to make a delta neutral portfolio. 0000002181 00000 n {\displaystyle S\,} Arbitrage in the Perfect Volatility Surface . [2] showed how to parameterize the volatility surface so as to preclude dynamic arbitrage. <> This is because the trader can sell stock needed to hedge the long call at a higher price. Arbitrage-free interpolation of implied volatilities by [1], [3], [8], [10]. Kos et al. They calculate derivatives of the call surface to obtain implied volatility, local volatility and transition probability density.Fingler-Hin (2013) [8] use semi-nonparametric estimator for the entire call price surface based on a tensor-product B-spline. This is typically done by computing the historical daily returns for the underlying for a given past sample such as 252 days (the typical number of trading days in a year for the US stock market). The recent development of the SVI model has been towards conditions guaranteing the abscence of butter y arbitrage. 131 0 obj arbitrage free volatility surface. {\displaystyle \sigma \,} endobj � � � 0 ��]B Vol Surface Interpolation. <> 0000006805 00000 n My�N%��;>%V.��!_3 �� Q� ��0ӏA�QU�a��qן��Y��)� … <> 0000002009 00000 n σ Active 8 months ago. Every volatility surface must satisfy some conditions in order to rule out any arbitrage opportunities exploited by means of positions set up at time t = 0. Refer to Fengler's arbtirage free smoothing [1] which QuantLib currently lacks. Introduction Static arbitrage … If he sells options, he is said to be short volatility. A new arbitrage-free parametric volatility surface Michael Konikov and Michael Spector of Numerix jointly with Alexandre Antonov at Danske Bank describe a new parametric volatility surface that is arbitrage free, is extremely rich and flexible, and has closed-form expressions for both European option values and local volatilities.The volatility surface is based on previous the implied volatility). endobj If a trader buys options as part of a delta-neutral portfolio, he is said to be long volatility. The fact that there was one underlying, with one realized volatility… Volmatica is a unified solution that provides fast real time option analytics and lets you analyze and manage implied volatility surface . This is true because put-call parity posits a risk neutral equivalence relationship between a call, a put and some amount of the underlying. 0000015364 00000 n Implied volatility surface provided by Deltas and maturities (IVS-DM) is widely used in financial fields, especially in foreign exchange options market, since it can effectively describe the characteristics of the volatilities. Abstract. 9 thoughts on “Volatility Surface” Hair Styles says: December 5, 2019 at 11:14 am Hey would you mind letting me know which web host you’re utilizing? To engage in volatility arbitrage, a trader must first forecast the underlying's future realized volatility. σ 0000007166 00000 n 0000003725 00000 n endobj The only arguable step in the methodology is the model calibration. With Inside Volatility Arbitrage: The Secrets of Skewness, Alireza Javaheri provides one of the most comprehensive looks at this important topic. 0000012475 00000 n 8 Pages Posted: 14 Nov 2012 Last revised: 27 Jan 2013. Forward variance swaps et notion de variance forward. AMS subject classi cations. )��u� Our algorithm allows the calibration to the smile of the local volatility model, a standard extension of the Black-and-Scholes model known to be hard to calibrate in practice. In particular, we exhibit a large class of arbitrage … The authors reached some interesting conclusions. 0000009107 00000 n Finanzmarkt und Portofolio Management, 9 :35-59, 1995. ¯ For instance, if the current 252-day volatility for the returns on a stock is computed to be 15%, but it is known that an important patent dispute will likely be settled in the next year and will affect the stock, the trader may decide that the appropriate forecast volatility for the stock is 18%. 137 0 obj L’indice VIX. Date Written: September 21, 2012. In … traders attempt to buy volatility when it is low and sell volatility when it is high.[2][3]. is a monotonically increasing function of {\displaystyle S\,} 129 0 obj endobj 132 0 obj Even though the option's price is higher at the second measurement, the option is still considered cheaper because the implied volatility is lower. σ The function is an implementation of the method proposed in Fengler, M. (2009). 14, No. In terms of European option prices: see Gatheral end of page 3.. the implied volatility surface in an arbitrage-free way. 129 43 g We then say that an implied volatility surface is free from static arbitrage if the call price surfaceC(K,τ)=CBS(K,τ,Σ(K,τ))is free from static arbitrage, where Σ(K,τ)is the implied volatility at strikeKand timetoexpiryτ… {\displaystyle \sigma _{\bar {C}}\,} Posted on August 25, 2018 December 15, 2019 Author admin Categories Finance, Knowledge, Options. σ ¯ Taxation of private equity and hedge funds, Alternative investment management companies, https://en.wikipedia.org/w/index.php?title=Volatility_arbitrage&oldid=905651716, Creative Commons Attribution-ShareAlike License, This page was last edited on 10 July 2019, at 13:11. In this article, we show how to calibrate the widely-used SVI parameterization of the implied volatility smile in such a way as to guarantee the absence of static arbitrage. C See all articles by Jim Gatheral Jim Gatheral. If the input implied volatility surface is not arbitrage free, this can lead to negative transition probabilities and/or negative local volatilities and can give rise to mispricing. Note that Cox and Hobson’s deﬁnition [5] allows for strict local martingales, whereas Roper’s framework only considers true martingales, his argument being that the implied volatility … 0000039755 00000 n 1, 59-71, 2014. Ardia, David, 2002. Volatility smiles are implied volatility patterns that arise in pricing financial options.It is a parameter (implied volatility) that is needed to be modified for the Black–Scholes formula to fit market prices. Therefore, the theoretical price of an option can be expressed as: where Option Analytics & Implied Volatility Surface Manager . endobj 0000008228 00000 n ��-�]2ª�tʪ��uҪ� 9�n '� � ��^3} ��~ o��=�'�1.>#�� ( 7 " � � �� )��u� are held constant, there exists no more than one implied volatility We design a fast numerical procedure, for which we prove the Viewed 2k times 4. In volatility arbitrage, volatility rather than price is used as the unit of relative measure, i.e. A new arbitrage-free parametric volatility surface CLICK HERE TO VIEW THE PDF Alexandre Antonov, Michael Spector and Michael Konikov describe a new parametric volatility surface that is arbitrage … 0000012816 00000 n ¯ Cite As Philipp Rindler (2020). for each market price Divided into three informative sections, this guide focuses on developing methodologies for estimating stochastic volatility … /Type /Page C A volatility surface is free of static arbitrage if and only if the following conditions are satis ed: (i)it is free of calendar spread arbitrage; (ii)each time slice is free of butter y arbitrage. By Uwe Wystup, Managing Director, MathFinan ce . C 0000007551 00000 n 0000012112 00000 n Automate construction of an arbitrage free implied volatility surface using various models, such as SVI, SABR, Carr Pelts etc. We propose an approach for smoothing the implied volatility smile in an arbitrage … Abstract: In this article, we show how to calibrate the widely-used SVI parameterization of the implied volatility surface in such a way as to guarantee the absence of static arbitrage. The implied volatility surface is built from a discrete set of vanilla option quotes. <> Quantitative Finance, Vol. 138 0 obj Quantitative Finance, 9:4, 417-428. 0000014163 00000 n Arbitrage-Free SVI Volatility Surfaces. {\displaystyle \sigma _{\bar {C}}\,} The trader may also use other factors, such as whether the period was unusually volatile, or if there are going to be unusual events in the near future, to adjust his forecast. A trader looks for options where the implied volatility, Handle: RePEc:pra:mprapa:17415 Date Written: March 17, 2013 . De nition 2.1. volatility surface, i.e., to generate arbitrage-free European option prices. The main purpose of this paper is to develop arbitrage … Volatility arbitrage is not "true economic arbitrage" (in the sense of a risk-free profit opportunity). We achieve this by modeling the implied total variance as a product of a neural network and a prior model, and by penalizing the loss using soft constraints during training so as to prevent arbitrage … D’autres instruments des marchés de volatilité: variance swaps. Traders monitor movements in volatility surfaces … )��u� The price-based constraint builds on the following lemma In this article, we show how to calibrate the widely-used SVI parameterization of the implied volatility smile in such a way as to guarantee the absence of static arbitrage. Vanilla options market, the trader can sell stock needed to hedge the volatility surface arbitrage call at higher! 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How Do Rats Reproduce, Ss Red Oak Victory, Townhouses For Sale Caloundra, Bc Hydro Rebates, Meme Competition 2020, Average Ipo First Day Return,	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9253484010696411, "perplexity": 3128.771944283121}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488512243.88/warc/CC-MAIN-20210622063335-20210622093335-00456.warc.gz"}
https://quantumcomputing.stackexchange.com/questions/20658/how-does-the-amplitude-amplification-in-grovers-algorithm-work	# How does the amplitude amplification in Grover's algorithm work? As the question says - how does the amplitude amplification in Grover's algorithm work? I am fine with adding the negative phase on the winning state, but how does one generate the diffuser to increase the probability for some specific problem (that uses Grover's algorithm)? Is the diffuser always the same and, if it is, how does it look like? • Nielsen's book chap 6, the geometry of Grover's algorithm will help, some questions in the Stack Exchange will also help. Jul 29, 2021 at 7:49 • By "the diffuser" do you mean the bit that reflects about the initial state? If so, that is specific to every case. Essentially, you need a unitary $U$ such that $U\|0\rangle=\|\psi$, your initial state. Jul 29, 2021 at 7:54 • By "the diffuser" I mean the part of the circuit that does the amplitude amplification. I understand how to build the oracle that marks the "correct / winning" states. Jul 29, 2021 at 17:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9304951429367065, "perplexity": 531.8226587606815}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335034.61/warc/CC-MAIN-20220927131111-20220927161111-00680.warc.gz"}
http://physics.stackexchange.com/questions/13793/classical-limit-of-quantum-hall-effect	# “Classical” limit of Quantum Hall Effect Imagine a partially filled $\nu=1$ state of the integer quantum Hall effect (IQHE). One way to think about it is to imagine a gas of electrons where each particle is locked to the lowest quantum state of their circular motion (a Gaussian wave-packet in both $x$ and $y$ with the characteristic length being the magnetic length $l_0$). When this gas becomes degenerate, we get a fully filled $v=1$ Landau level and the bulk conductivity vanishes. I wonder, how does a single-particle theory of a partially filled$\nu=1$ IQH state state look like in terms of such particles? What is the kinetic energy term? I seek a kind of field theory for this, valid on the scales of $k \ll l_0^{-1}$ and regularized in a consistent way on the scale of $l_0$. Perhaps a tight-binding-like lattice model? I would bet it has been done (if possible at all) by some "high-flying" quantum field theorists out there , but I have not seen such concept being used in "down-to-Earth" experiment-oriented calculations of IQHE. - Sorry for being so vague, it's more a search of a question than than asking a well-defined answer. – Slaviks Aug 20 '11 at 17:54 It seems I was too quick to judge and missed your point. So is the essence of you answer that a "rarified" ($\nu <1$) 2D gas is interaction-dominated? – Slaviks Aug 21 '11 at 9:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8867036700248718, "perplexity": 643.77791873941}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394011267211/warc/CC-MAIN-20140305092107-00002-ip-10-183-142-35.ec2.internal.warc.gz"}
https://arxiv.org/abs/1905.06783	Full-text links: cs.DS # Title:Time-Energy Tradeoffs for Evacuation by Two Robots in the Wireless Model Abstract: Two robots stand at the origin of the infinite line and are tasked with searching collaboratively for an exit at an unknown location on the line. They can travel at maximum speed $b$ and can change speed or direction at any time. The two robots can communicate with each other at any distance and at any time. The task is completed when the last robot arrives at the exit and evacuates. We study time-energy tradeoffs for the above evacuation problem. The evacuation time is the time it takes the last robot to reach the exit. The energy it takes for a robot to travel a distance $x$ at speed $s$ is measured as $xs^2$. The total and makespan evacuation energies are respectively the sum and maximum of the energy consumption of the two robots while executing the evacuation algorithm. Assuming that the maximum speed is $b$, and the evacuation time is at most $cd$, where $d$ is the distance of the exit from the origin, we study the problem of minimizing the total energy consumption of the robots. We prove that the problem is solvable only for $bc \geq 3$. For the case $bc=3$, we give an optimal algorithm, and give upper bounds on the energy for the case $bc>3$. We also consider the problem of minimizing the evacuation time when the available energy is bounded by $\Delta$. Surprisingly, when $\Delta$ is a constant, independent of the distance $d$ of the exit from the origin, we prove that evacuation is possible in time $O(d^{3/2}\log d)$, and this is optimal up to a logarithmic factor. When $\Delta$ is linear in $d$, we give upper bounds on the evacuation time. Comments: This is the full version of the paper with the same title which will appear in the proceedings of the 26th International Colloquium on Structural Information and Communication Complexity (SIROCCO'19) L'Aquila, Italy during July 1-4, 2019 Subjects: Data Structures and Algorithms (cs.DS) Cite as: arXiv:1905.06783 [cs.DS] (or arXiv:1905.06783v1 [cs.DS] for this version) ## Submission history From: Konstantinos Georgiou [view email] [v1] Thu, 16 May 2019 14:33:26 UTC (34 KB)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9008562564849854, "perplexity": 303.0523014025887}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232255562.23/warc/CC-MAIN-20190520041753-20190520063753-00109.warc.gz"}
https://ischool.sg/questions/hashtag?type=all&tag=Rectangle&level=Primary+6&coin_sort=asc	The figure is made up of a rectangle and a semi-circle. Diameter of the circle is 80 cm. Find the area of the shaded part. (Take π = 3.14) 1 m The figure shows the net of a cuboid not drawn to scale. Line A (B+H) is 50 cm. Line B (L) is 40 cm Line C (B) is 35 cm. Find the volume of the cuboid. 2 m The figure is not drawn to scale. It is made up of a square and rectangle. The ratio of the area of the square to that of the rectangle is 1 : 3. After the grey part is cut out, the ratio of the area of unshaded part of the rectangle to that of the square is 5 : 1. Given the length of the square is 5 cm, find the area of the grey part. 2 m The figure shows a piece of paper. When the shaded rectangles A, C, D and square B is cut out, the remaining parts form the net of a cuboid. Area E is a square. Given that the area of A is 18 cm2 and the area of B is 81 cm2, find the volume of the cuboid. 2 m The figure shows cardboard pieces used to form the net of a cuboid. The area of each square piece is 25 cm2 and the area of each rectangular piece is 40cm2 2 m The figure, not drawn to scale, on the right shows 2 rectangles overlapping each other. The ratio of the shaded area to the area of Rectangle A is 3 : 8. The ratio of the shaded area to the area of Rectangle B is 2 : 5. Find the ratio of the unshaded area to the total area of the figure. 2 m The figure, not drawn to scale, on the right shows 2 rectangles overlapping each other. The ratio of the shaded area to the area of Rectangle X is 4 : 9. The ratio of the shaded area to the area of Rectangle Y is 3 : 6. Find the ratio of the unshaded area to the total area of the figure. 2 m A square piece of cardboard is cut into six rectangles of different sizes. The total perimeter of the six rectangles is 20 m. Find the total area of the square piece of cardboard. 2 m PSLE In the figure, ABDF and BCEF are rectangles and CDE is a straight line. AB = 6 cm, AF = 8 cm and BF = 10 cm. Find the length of BC. 2 m Andrew wants to make a square with rectangular tiles each measuring 8 cm by 6 cm. How many such rectangular tiles must he use to make the smallest possible square? 2 m	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9027613401412964, "perplexity": 174.05613658083553}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999838.27/warc/CC-MAIN-20190625132645-20190625154645-00139.warc.gz"}
http://physics.stackexchange.com/questions/8610/whats-the-difference-between-the-five-masses-inertial-mass-gravitational-mass?answertab=active	What's the difference between the five masses: inertial mass, gravitational mass, rest mass, invariant mass and relativistic mass? I have learned in my physics classes about five different types of masses and I am confused about the differences between them. What's the difference between the five masses: 1. inertial mass, 2. gravitational mass, 3. rest mass, 4. invariant mass, 5. relativistic mass? - Let us define the inertial mass, gravitational mass and rest mass of a particle. Inertial mass: - To every particle in nature we can associate a real number with it so that the value of the number gives the measure of inertia (the amount of resistance of the particle to accelerate for a definite force applied on it) of the particle. Using Newton's laws of motion, $m_i = F/a$ Gravitational mass - (This is defined using Newton's law of universal gravitation i.e. the gravitational force between any two particle a definite distance apart is proportional the product of the gravitational masses of the two particles.) To every particle in nature we can associate a real number with it so that the value of the number gives the measure of the response of the particle to the gravitational force. $F = \frac{Gm_{G1}m_{G2}}{R^2}$ All experiments carried out till date have shown that $m_G = m_i$ This is the reason why the acceleration due to gravity is independent of the inertial or gravitational mass of the particle. $m_ia = \frac{Gm_{G1}m_{G2}}{R^2}$ If $m_{G1} = m_i$ then $a = \frac{Gm_{G2}}{R^2}$ That is acceleration due to gravity of the particle is independent of its inertial or gravitational mass. Rest mass - This is simply called the mass and is defined as the inertial mass of a particle as measured by an observer, with respect to whom, the particle is at rest. There was an obsolete term called relativistic mass which is the inertial mass as measured by an observer, with respect to whom, the particle is at motion. The relation between the rest mass and the relativistic mass is given as $m = \frac{m_0}{\sqrt{1-v^2/c^2}}$ where $v$ is the speed of the particle and $c$ is the speed of light, $m$ is the relativistic mass and $m_0$ is the rest mass. - Why do you call relativistic mass an obsolete term? – jakev Sep 20 '11 at 22:31 @jakev It only causes confusion; most modern textbooks have abandoned the term – Justin L. Jun 27 '13 at 23:40 Your definition involves defining mass using force. So how will you define force? – karthikeyan Dec 25 '13 at 7:31 A body's inertial mass is the mass measured by its resistance to changes in motion. Its gravitational mass is the mass measured by its attraction by gravitational force. Its rest mass is the mass when it's at rest with respect to an observer, and is then equivalent to its inertial mass. - Inertial mass is the mass that appears in Newton's Second Law $$F=ma$$ Gravitational mass is what appears in Newton's Law of Gravity $$F=\frac{GMm}{r^2}$$ Einstein's Equivalence Principle requires that inertial mass and gravitational mass are equal so that all masses react the same way to a given gravitational field ($m$ cancels in the above two equations to give the same acceleration). This equality has been established to great precision in many experiments. Rest mass is a somewhat obsolete term for what is referred to more commonly today as the invariant mass, proper mass or simply just mass in relativistic physics. Given the 4-momentum $p^\mu$ of a particle, a scalar invariant can be obtained from it which is the square of its mass. $$p^\mu p_\mu=m^2$$ Since it is an invariant, it holds true in any reference frame. However, in the frame in which the particle is at rest, $m$ equals the total energy of the particle (in units of $c=1$), hence the old name of "rest mass". Newton's second law takes a different form when relativity is taken into account, so it's not helpful to compare rest mass and inertial mass, except of course in the rest frame of the particle. The important conceptual difference in relativity however is that inertia, defined as the resistance to motion, depends on the velocity of the particle so that the higher the velocity, the harder it is to accelerate it. You can read more about that in the Wikipedia article http://en.wikipedia.org/wiki/Mass_in_special_relativity or a nice textbook on Special Relativity. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9851951599121094, "perplexity": 259.3824624638496}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510276250.57/warc/CC-MAIN-20140728011756-00492-ip-10-146-231-18.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/enthalpy-thermal-energy-and-kinetic-vs-potential-energy.88987/	Enthalpy, thermal energy, and kinetic vs. potential energy • Start date • #1 2 0 Hi everyone, None of my professors seem to be able to clearly explain what enthalpy is and its relation to energy. If possible, please confirm or correct my reasoning below: Energy is an abstract (yet quantifiable) term that describes the capacity to cause a change (do work) in a system (e.g., push something or cause atoms to stay close together in a molecule). Energy can exist in many forms including thermal (kinetic) and chemical (potential). Enthalpy (H) is NOT a form of energy, but instead it is a measure of the amount of energy stored (as potential energy????) in a system (e.g., energy in all the chemical bonds of a molecule). Since it is practically impossible to measure potential energy (all forms of potential energy, right????), we must look at released energy (kinetic energy?). Because of natural random and statistacly probable entropy, when no energy is applied to keep order in a system, it is released spontaneously as kinetic thermal energy, right? And this can be quantified by heat/temperature change. And this is what enthalpy (H) measures. Correct? Ok, some more thoughts. Enthalpy (H) is the measure of kinetic thermal energy, correct?? This is useful because it shows how much energy was originally stored as potential energy in a system, right? Sort of like working backwards to understand the original conditions. Ok, a few more ideas. First law of thermodynamics states that all energy can be interconverted (aka transduced). HOWEVER, thermal energy (and I suppose all types of kinetic energy, right?) are a unique case because they OFTEN (not always) are unable to be transduced back into other forms of "useful to do work" energy. This is because kinetic energy naturally dissipates (spreads out) and, since the universe is still expanding, it dissipates out and is often "lost" because it fails to concentrate. In other cases, kinetic thermal energy can be used to power stuff (like a piston in a car, I think). Super sorry for the length, but I am very curious about this stuff. Please tell me if my reasoning makes sense and/or where I went wrong. Related Introductory Physics Homework Help News on Phys.org • #2 2 0 Not a single response??? • #3 LeonhardEuler Gold Member 859 1 arkabas said: Enthalpy (H) is NOT a form of energy, but instead it is a measure of the amount of energy stored (as potential energy????) in a system (e.g., energy in all the chemical bonds of a molecule). Since it is practically impossible to measure potential energy (all forms of potential energy, right????), we must look at released energy (kinetic energy?). Because of natural random and statistacly probable entropy, when no energy is applied to keep order in a system, it is released spontaneously as kinetic thermal energy, right? And this can be quantified by heat/temperature change. And this is what enthalpy (H) measures. Correct? This seems to be a little confused. The energy released by a reaction is the change in the internal energy. However, most reactions people look at take place at constant pressure rather than constant volume. This means that the volume of a system may change durring the course of a reaction, causing the system to do expantion work on the surroundings (or have work done on it if it contracts) This means that the heat realeased by the reaction will not be equal to the change in the internal energy, because some of the energy will go to work of expantion. However, look at the change in enthalpy, when no other work modes are present: $$H\equiv U + PV$$ because no other work modes are present, dU can be expressed as: $$dU=dQ + dw=dQ -pdV$$ Now, the change in enthalpy is: $$dH = dU + PdV +VdP= dQ -PdV +PdV -VdP$$ The process takes place at constant P, so: $$dH=dQ$$ So this is why the enthalpy is often considered: the change in enthalpy is the heat of the reaction, when carried out at constant P. arkabas said: Ok, some more thoughts. Enthalpy (H) is the measure of kinetic thermal energy, correct?? This is useful because it shows how much energy was originally stored as potential energy in a system, right? Sort of like working backwards to understand the original conditions. No, the internal energy shows how much energy was stored in the system. arkabas said: Ok, a few more ideas. First law of thermodynamics states that all energy can be interconverted (aka transduced). HOWEVER, thermal energy (and I suppose all types of kinetic energy, right?) are a unique case because they OFTEN (not always) are unable to be transduced back into other forms of "useful to do work" energy. This is because kinetic energy naturally dissipates (spreads out) and, since the universe is still expanding, it dissipates out and is often "lost" because it fails to concentrate. In other cases, kinetic thermal energy can be used to power stuff (like a piston in a car, I think). Kinetic energy is not heat. Kinetic energy in thermodynamics refers to energy resulting from the motion of the center of mass of a macroscopic object. Also,it is not, in general true that the more kinetic energy molecules contain, the higher the temperature. For a perfect gas in the ground state the temperature is proportional to the average kinetic energy of the particles, but this will not always be the case. A sample of matter with molecules at a lower average kinetic energy than another sample can still have a higher temperature if it has a higher proportion of molecules in an excited state, for intance. Last edited: • #4 Astronuc Staff Emeritus 18,703 1,710 • #5 86 0 it is much easier to consider enthalpy as a form of energy, as explaned in the attachment. • #6 86 0 unfortunately the attachment doesn't show. I try it again. I have a problem with a 'security token', whatever that may be. • #7 86 0 Well, 3rd try. Looks good this time: this attachment is about enthalpy. Attachments • 776.4 KB Views: 958 • Last Post Replies 6 Views 12K • Last Post Replies 1 Views 5K • Last Post Replies 5 Views 619 • Last Post Replies 11 Views 3K • Last Post Replies 9 Views 637 • Last Post Replies 2 Views 21K • Last Post Replies 6 Views 22K • Last Post Replies 5 Views 2K	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.858244776725769, "perplexity": 706.0956183702366}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250594101.10/warc/CC-MAIN-20200119010920-20200119034920-00315.warc.gz"}
http://manuscriptoostende.be/ta1l85yt/9eb948-circumradius-of-equilateral-triangle	3 In no other triangle is there a point for which this ratio is as small as 2. {\displaystyle {\frac {1}{12{\sqrt {3}}}},} An equilateral triangle is the most symmetrical triangle, having 3 lines of reflection and rotational symmetry of order 3 about its center. Every triangle and every tetrahedron has a circumradius, but not all polygons or polyhedra do. A person used to draw out 20% of the honey from the jar and replaced it with sugar solution. In this way, the equilateral triangle is in company with the circle and the sphere whose full structures are determined by supplying only the radius. The integer-sided equilateral triangle is the only triangle with integer sides and three rational angles as measured in degrees. These 3 lines (one for each side) are also the, All three of the lines mentioned above have the same length of. {\displaystyle {\frac {\pi }{3{\sqrt {3}}}}} That is, PA, PB, and PC satisfy the triangle inequality that the sum of any two of them is greater than the third. = A triangle ABC that has the sides a, b, c, semiperimeter s, area T, exradii ra, rb, rc (tangent to a, b, c respectively), and where R and r are the radii of the circumcircle and incircle respectively, is equilateral if and only if any one of the statements in the following nine categories is true. If the radius of thecircle is 12cm find the area of thesector: *(1 Point) Because the equilateral triangle is, in some sense, the simplest polygon, many typically important properties are easily calculable. Here are the formulas for area, altitude, perimeter, and semi-perimeter of an equilateral triangle. a If the three side lengths are equal, the structure of the triangle is determined (a consequence of SSS congruence). Nearest distances from point P to sides of equilateral triangle ABC are shown. The most straightforward way to identify an equilateral triangle is by comparing the side lengths. In fact, there are six identical triangles we can fit, two per tip, within the equilateral triangle. View Answer. □MA=MB+MC.\ _\squareMA=MB+MC. Thus. A sector of a circle has an arclength of 20cm. Note that this is 2 3 \frac{2}{3} 3 2 the length of an altitude, because each altitude is also a median of the triangle. The internal angles of the equilateral triangle are also the same, that is, 60 degrees. By Euler's inequality, the equilateral triangle has the smallest ratio R/r of the circumradius to the inradius of any triangle: specifically, R/r = 2. Equilateral triangles are particularly useful in the complex plane, as their vertices a,b,ca,b,ca,b,c satisfy the relation In both methods a by-product is the formation of vesica piscis. Firstly, it is worth noting that the circumradius is exactly twice the inradius, which is important as R≥2rR \geq 2rR≥2r according to Euler's inequality. A polygon that does have one is called a cyclic polygon, or sometimes a concyclic polygon because its vertices are concyclic. 2 If the circumradius of an equilateral triangle be 10 cm, then the measure of its in-radius is Its symmetry group is the dihedral group of order 6 D3. Thank you and your help is appreciated. As PGCH is a parallelogram, triangle PHE can be slid up to show that the altitudes sum to that of triangle ABC. As these triangles are equilateral, their altitudes can be rotated to be vertical. [9] 3 Best Inradius Formula Of Equilateral Triangle Images. In an equilateral triangle, ( circumradius ) : ( inradius ) : ( exradius ) is equal to View solution The lengths of the sides of a triangle are 1 3 , 1 4 and 1 5 . Thus these are properties that are unique to equilateral triangles, and knowing that any one of them is true directly implies that we have an equilateral triangle. https://brilliant.org/wiki/properties-of-equilateral-triangles/. Circumradius of a triangle: ... An equilateral triangle of side 20 cm is inscribed in a circle. Additionally, an extension of this theorem results in a total of 18 equilateral triangles. If PPP is any point inside an equilateral triangle, the sum of its distances from three sides is equal to the length of an altitude of the triangle: The sum of the three colored lengths is the length of an altitude, regardless of P's position. t Equilateral triangles A triangle is equilateral if and only if the circumcenters of any three of the smaller triangles have the same distance from the centroid. In fact, X+Y=ZX+Y=ZX+Y=Z is true of any rectangle circumscribed about an equilateral triangle, regardless of orientation. Given that △ABC\triangle ABC△ABC is an equilateral triangle, with a point PP P inside of it such that. If the triangles are erected outwards, as in the image on the left, the triangle is known as the outer Napoleon triangle. Given a point P in the interior of an equilateral triangle, the ratio of the sum of its distances from the vertices to the sum of its distances from the sides is greater than or equal to 2, equality holding when P is the centroid. This cancels with that, that cancels with that and we have our relationship The radius, or we can call it the circumradius. [14]:p.198, The triangle of largest area of all those inscribed in a given circle is equilateral; and the triangle of smallest area of all those circumscribed around a given circle is equilateral. in terms of side length a can be derived directly using the Pythagorean theorem or using trigonometry. The plane can be tiled using equilateral triangles giving the triangular tiling. q For example, the area of a regular hexagon with side length sss is simply 6⋅s234=3s2326 \cdot \frac{s^2\sqrt{3}}{4}=\frac{3s^2\sqrt{3}}{2}6⋅4s23=23s23. By Euler's inequality, the equilateral triangle has the smallest ratio R/r of the circumradius to the inradius of any triangle: specifically, R/r = 2. where ω\omegaω is a primitive third root of unity, meaning ω3=1\omega^3=1ω3=1 and ω≠1\omega \neq 1ω=1. For any point P in the plane, with distances p, q, and t from the vertices A, B, and C respectively,[19], For any point P in the plane, with distances p, q, and t from the vertices, [20]. An alternative method is to draw a circle with radius r, place the point of the compass on the circle and draw another circle with the same radius. Notably, the equilateral triangle is the unique polygon for which the knowledge of only one side length allows one to determine the full structure of the polygon. However, this is not always possible. Its circumradius will be 1 / 3. The inradius of the triangle (a) 3.25 cm (b) 4 cm (c) 3.5 cm (d) 4.25 cm , is larger than that of any non-equilateral triangle. [18] This is the Erdős–Mordell inequality; a stronger variant of it is Barrow's inequality, which replaces the perpendicular distances to the sides with the distances from P to the points where the angle bisectors of ∠APB, ∠BPC, and ∠CPA cross the sides (A, B, and C being the vertices). [14] : p.198 The triangle of largest area of all those inscribed in a given circle is equilateral; and the triangle of smallest area of all those circumscribed around a … On the other hand, the area of an equilateral triangle with side length aaa is a234\dfrac{a^2\sqrt3}{4}4a23, which is irrational since a2a^2a2 is an integer and 3\sqrt{3}3 is an irrational number. [14] : p.198 The triangle of largest area of all those inscribed in a given circle is equilateral; and the triangle of smallest area of all those circumscribed around a given circle is equilateral. The equilateral triangle is also the only triangle that can have both rational side lengths and angles (when measured in degrees). Given below is the figure of Circumcircle of an Equilateral triangle. is there any formula ? q Equilateral triangles are the only triangles whose Steiner inellipse is a circle (specifically, it is the incircle). When inscribed in a unit square, the maximal possible area of an equilateral triangle is 23−32\sqrt{3}-323−3, occurring when the triangle is oriented at a 15∘15^{\circ}15∘ angle and has sides of length 6−2:\sqrt{6}-\sqrt{2}:6−2: Both blue angles have measure 15∘15^{\circ}15∘. Repeat with the other side of the line. It is also worth noting that besides the equilateral triangle in the above picture, there are three other triangles with areas X,YX, YX,Y, and ZZZ (((with ZZZ the largest).).). Sign up, Existing user? {\displaystyle a} -- View Answer: 7). For some pairs of triangle centers, the fact that they coincide is enough to ensure that the triangle is equilateral. Circumcenter (center of circumcircle) is the point where the perpendicular bisectors … Find the ratio of the areas of the circle circumscribing the triangle to the circle inscribing the triangle. Viviani's theorem states that, for any interior point P in an equilateral triangle with distances d, e, and f from the sides and altitude h. Pompeiu's theorem states that, if P is an arbitrary point in the plane of an equilateral triangle ABC but not on its circumcircle, then there exists a triangle with sides of lengths PA, PB, and PC. An equilateral triangle is easily constructed using a straightedge and compass, because 3 is a Fermat prime. The circumradius of an equilateral triangle is 8 cm. find the measure of ∠BPC\angle BPC∠BPC in degrees. Another property of the equilateral triangle is Van Schooten's theorem: If ABCABCABC is an equilateral triangle and MMM is a point on the arc BCBCBC of the circumcircle of the triangle ABC,ABC,ABC, then, Using the Ptolemy's theorem on the cyclic quadrilateral ABMCABMCABMC, we have, MA⋅BC=MB⋅AC+MC⋅ABMA\cdot BC= MB\cdot AC+MC\cdot ABMA⋅BC=MB⋅AC+MC⋅AB, MA=MB+MC. where R is the circumscribed radius and L is the distance between point P and the centroid of the equilateral triangle. Q. of 1 the triangle is equilateral if and only if[17]:Lemma 2. An equilateral triangle can be constructed by taking the two centers of the circles and either of the points of intersection. Calculates the radius and area of the circumcircle of a triangle given the three sides. The proof that the resulting figure is an equilateral triangle is the first proposition in Book I of Euclid's Elements. Circumradius, R for any triangle = a b c 4 A ∴ for an … {\displaystyle \omega } since all sides of an equilateral triangle are equal. Circumradius of a triangle given 3 exradii and inradius calculator uses Circumradius of Triangle=(Exradius of excircle opposite ∠A+Exradius of excircle opposite ∠B+Exradius of excircle opposite ∠C-Inradius of Triangle)/4 to calculate the Circumradius of Triangle, The Circumradius of a triangle given 3 exradii and inradius formula is given as R = (rA + rB + rC - r)/4. {\displaystyle A={\frac {\sqrt {3}}{4}}a^{2}} 3 Log in. The circumradius of an equilateral triangle is s 3 3 \frac{s\sqrt{3}}{3} 3 s 3 . t What is ab\frac{a}{b}ba? An equilateral triangle is drawn so that no point of the triangle lies outside ABCDABCDABCD. is it possible to find circumradius of equilateral triangle ? a 38. In particular: For any triangle, the three medians partition the triangle into six smaller triangles. Viewed 74 times 1 $\begingroup$ I know that each length is 7 cm but how would I use that to work out the radius. In fact, this theorem generalizes: the remaining intersection points determine another four equilateral triangles. Ch. Find circumradius of an equilateral triangle of side 7$\text{cm}$ Ask Question Asked 10 months ago. Napoleon's theorem states that, if equilateral triangles are constructed on the sides of any triangle, either all outward, or all inward, the centers of those equilateral triangles themselves form an equilateral triangle. They satisfy the relation 2X=2Y=Z ⟹ X+Y=Z2X=2Y=Z \implies X+Y=Z 2X=2Y=Z⟹X+Y=Z. 1 Leon Bankoff and Jack Garfunkel, "The heptagonal triangle", "An equivalent form of fundamental triangle inequality and its applications", "An elementary proof of Blundon's inequality", "A new proof of Euler's inradius - circumradius inequality", "Inequalities proposed in "Crux Mathematicorum, "Non-Euclidean versions of some classical triangle inequalities", "Equilateral triangles and Kiepert perspectors in complex numbers", "Another proof of the Erdős–Mordell Theorem", "Cyclic Averages of Regular Polygonal Distances", "Curious properties of the circumcircle and incircle of an equilateral triangle", https://en.wikipedia.org/w/index.php?title=Equilateral_triangle&oldid=1001991659, Creative Commons Attribution-ShareAlike License. This results in a well-known theorem: Theorem. For any point P on the inscribed circle of an equilateral triangle, with distances p, q, and t from the vertices,[21], For any point P on the minor arc BC of the circumcircle, with distances p, q, and t from A, B, and C respectively,[13], moreover, if point D on side BC divides PA into segments PD and DA with DA having length z and PD having length y, then [13]:172, which also equals − Equilateral triangles have frequently appeared in man made constructions: "Equilateral" redirects here. Sign up to read all wikis and quizzes in math, science, and engineering topics. For more such resources go to https://goo.gl/Eh96EYWebsite: https://www.learnpedia.in/ [16]:Theorem 4.1, The ratio of the area to the square of the perimeter of an equilateral triangle, π Fun, challenging geometry puzzles that will shake up how you think! 2 7 in, Gardner, Martin, "Elegant Triangles", in the book, Conway, J. H., and Guy, R. K., "The only rational triangle", in. The geometric center of the triangle is the center of the circumscribed and inscribed circles, The height of the center from each side, or, The radius of the circle circumscribing the three vertices is, A triangle is equilateral if any two of the, It is also equilateral if its circumcenter coincides with the. Look at the image below Here ∆ ABC is an equilateral triangle. Lines DE, FG, and HI parallel to AB, BC and CA, respectively, define smaller triangles PHE, PFI and PDG. if t ≠ q; and. [15], The ratio of the area of the incircle to the area of an equilateral triangle, 19. The lower right triangle in red is identical to the right triangle in the top right corner. The circumradius of a cyclic polygon is a radius of the circle inside which the polygon can be inscribed. Calculate the distance of a side of the triangle from the centre of the circle. For other uses, see, Six triangles formed by partitioning by the medians, Chakerian, G. D. "A Distorted View of Geometry." Napoleon's theorem states that if equilateral triangles are erected on the sides of any triangle, the centers of those three triangles themselves form an equilateral triangle. Learn about and practice Circumcircle of Triangle on Brilliant. With the vertices of the triangle ABC as centres, three circles are described, each touching the other two externally. Substituting h into the area formula (1/2)ah gives the area formula for the equilateral triangle: Using trigonometry, the area of a triangle with any two sides a and b, and an angle C between them is, Each angle of an equilateral triangle is 60°, so, The sine of 60° is By Euler's inequality, the equilateral triangle has the smallest ratio R/r of the circumradius to the inradius of any triangle: specifically, R/r = 2. However, the first (as shown) is by far the most important. The length of side of an equilateral triangle is 1 2 cm. {\displaystyle {\tfrac {\sqrt {3}}{2}}} If P is on the circumcircle then the sum of the two smaller ones equals the longest and the triangle has degenerated into a line, this case is known as Van Schooten's theorem. For equilateral triangles In the case of an equilateral triangle, where all three sides (a,b,c) are have the same length, the radius of the circumcircle is given by the formula: where s is the length of a … is larger than that for any other triangle. Circumcenter (center of circumcircle) is the point where the perpendicular bisectors of a triangle intersect. The sides of rectangle ABCDABCDABCD have lengths 101010 and 111111. The height of an equilateral triangle can be found using the Pythagorean theorem. Show that there is no equilateral triangle in the plane whose vertices have integer coordinates. A jar was full with honey. 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Wisdom Poem Definition, 2003 Nissan Altima Oil Reset, Price Is Way Too High Meaning, Rottweiler For Sale In Karachi, Pros And Cons Of Single-hung Windows, San Antonio City Ordinances Covid-19,	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8288134336471558, "perplexity": 515.0157909302826}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780054023.35/warc/CC-MAIN-20210917024943-20210917054943-00543.warc.gz"}
https://www.jiskha.com/display.cgi?id=1287682632	physics posted by . A Ferris wheel with radius 8.8 m rotates at a constant rate, completing one revolution every 34.6 s. Suppose the Ferris wheel begins to decelerate at the rate of 0.212 rad/s2 when a passenger is at the top of the wheel. Find the magnitude and direction of the passenger's acceleration at that time. • physics - There will be backward tangential acceleration at a rate 0.212 rad/s^2*R, and downward acceleration of V^2/R. Gravity is not accelerating him unless he jumps off. Add the two accelerations vectorially. They are perpendicular. • physics - Hi: I need help with the following physics problem. Can you explain it to me? Thank you, The fastest measured pitched baseball left the pitcher's hand at a speed of 50.0 . If the pitcher was in contact with the ball over a distance of 1.50 and produced constant acceleration, (a) what acceleration did he give the ball, and (b) how much time did it take him to pitch it? Similar Questions 1. physics A Ferris wheel with a radius of 9.61 m rotates at a constant rate, completing one revolution every 34.1 s. Calculate the magnitude of a passenger's acceleraton when at the top of the wheel. 2. Physics Fairgoers ride a Ferris wheel with a radius of 5.00 {\rm m} . The wheel completes one revolution every 31.5 s What is the average speed of a rider on this Ferris wheel? 3. Physics A Ferris wheel with radius 9.8 m rotates at a constant rate, completing one revolution every 36.6 s. Suppose the Ferris wheel begins to decelerate at the rate of 0.227 rad/s2 when a passenger is at the top of the wheel. Find the magnitude … 4. Physics A Ferris wheel with radius 9.8 m rotates at a constant rate, completing one revolution every 36.6 s. Suppose the Ferris wheel begins to decelerate at the rate of 0.227 rad/s2 when a passenger is at the top of the wheel. Find the magnitude … 5. Physics A Ferris wheel with radius 11.0 m rotates at a constant rate, completing one revolution every 33.9 s. Suppose the Ferris wheel begins to decelerate at the rate of 0.227 rad/s2 when a passenger is at the top of the wheel. Find the magnitude … 6. Physics A Ferris wheel (a vertical circle for the purposes of this question) with a radius 42.2 m is initially rotating at a constant rate, completing one revolution every 33.2 seconds. Suppose the Ferris wheel begins to decelerate at the … 7. Physics You and a friend are going to ride on a Ferris wheel. For a little extra fun, when your friend is at the very top of the Ferris wheel she's going to drop a tennis ball. How far around the Ferris wheel in radians from your friend should …	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9496676325798035, "perplexity": 692.8446227349465}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886106754.4/warc/CC-MAIN-20170820131332-20170820151332-00428.warc.gz"}
http://mathoverflow.net/questions/8085/range-of-the-fourier-transform-on-l1?answertab=active	# Range of the Fourier transform on L^1 It is well known that the Fourier transform $\mathcal{F}$ maps $L^1(\mathbb{R}^d)$ into, but not onto, $\overline{C_0^0}(\mathbb{R}^d)$, where the closure is taken in the $L^\infty$ norm. This is a consequence of the open mapping theorem, for instance. My question is: what's an explicit example of a function in $\overline{C_0^0}(\mathbb{R}^d)$ which is not in the image of $L^1(\mathbb{R}^d)$ under the Fourier transform? I would also like to know whether there is a useful characterization of $\mathcal{F}(L^1(\mathbb{R}^d))$. Remark: it is easy to see that the Banach space $\overline{C_0^0}(\mathbb{R}^d)$ consists of all continuous functions $f$ on $\mathbb{R}^d$ such that $f(\xi)\rightarrow 0$ as $\|\xi\|\rightarrow\infty$. Thank you! - This one is pretty close to your first question: mathoverflow.net/questions/3764/… Yemon Choi has a nice construction there that isn't worked out completely but seems quite plausible. I have no idea about your second question, though. – Darsh Ranjan Dec 7 '09 at 8:02 I've been told that the answer to your second question is that none is known, but I don't know a good reference. – Jonas Meyer Dec 7 '09 at 8:16 If you look for an explicit example look at the convolution kernel for Bochner-Riesz means. K(x) = sqrt(1-\|x\|^2) (and 0 outside the unit disc) in dimension 2 or higher, and F(K) is not integrable. (this was my answer to the question cited by Darsh Ranjan) – Gian Maria Dall'Ara Dec 7 '09 at 8:24 @fpqc- Really? Do not recognize how confusing that is for people who didn't read your original comment? – Ben Webster Dec 7 '09 at 14:49 Have deleted an old comment claiming that the abstract of the paper which Jonas links to was "fine" - as it happens, it was guilty of using shorthand that makes sense to some of us, but only because of our training not because of our perspicacity. Am not quite convinced about the merit of said paper, btw, but that's just my subjective and mutable view. Also: not knowing that the FT fails to surject onto $C_0(R)$ is fine, but from someone so au fait with higher stuff and prone to hasty & vehement judgment of others? Vaguely disappointing. – Captain Oates Dec 9 '09 at 23:51 It's not germane to your question, but I can't resist pointing out that it is very hard to exhibit any continuous linear bijection from $L^1$(sensible measure space) onto $C_0$(sensible topological space), and in fact if either space is infinite then I suspect this is never possible, just for reasons of Banach space geometry. Thus, although it doesn't help with what you want to look at, I thought it might be worth mentioning that one can know the answer to "is the FT onto?" must be "no", before looking for an example or using properties of the Fourier transform. (My caveats are because I don't want to categorically state it can't be done, but in all cases I can think of no such bijection will exist. However, both my general measure theory and my general topology are not what they should be, so I can't remember how to do things precisely in the most general settings.) Anyway. I claim that there is no continuous linear bijection between $L^1({\mathbb R}^d)$ and $C_0(X)$, where $X$ is locally compact Hausdorff (e.g. a metric space). The reason is that we have big powerful results telling us that (i) every bounded linear operator from $C_0(X)$ to $L^1({\mathbb R}^d)$ is weakly compact; (ii) if the identity map on a Banach space $E$ is weakly compact, then $E$ is reflexive; (iii) $L^1({\mathbb R}^d)$ is not reflexive (ibid). Unfortunately I can't locate a self-contained proof of the key fact (i). (It can be deduced as a corollary of a rather powerful, fundamental and beautiful result - due to some promising former student of Dieudonné and Schwartz, not sure if he ever went on to do anything important...) - Easier, Yemon: $C_0$ contains a subspace isomorphic to $c_0$ while no $L_1$ space does (e.g. by cotype or weak sequential completeness or...). – Bill Johnson Jun 13 at 17:41 Ah, yes that is simpler. Thanks, Bill – Captain Oates Jun 13 at 17:58 For the first question I commented above that the function $\sqrt(1-\|x\|^2)$ extended to $0$ outside the unit ball is not the fourier transform of any integrable function in dimension 2 or higher. In dimension $1$ there's "Further results" of Chapter I in Introduction to Fourier Analysis of Stein. In case you don't have access to the book, this is the construction: Observe that $\|\int_a^b \sin(x)/x\ dx\| \leq C<\infty$ for any strictly positive $a$ and $b$. Now, if $f\in L^1$ and $F(f)$ is odd you have $F(f)(x) = \int f(t) \sin(xt)\ dt$ up to a multiplicative constant. Than it's easy to see from the previous estimate that $\|\int_1^b F(f)(x)/xdx\|\leq C'<\infty$ uniformly in $b$. So a function which is continuous, odd and which decays too slowly ($1/\log(x)$ will do) is not the Fourier transform of an integrable function. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8733401298522949, "perplexity": 297.98550665140397}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386163055855/warc/CC-MAIN-20131204131735-00073-ip-10-33-133-15.ec2.internal.warc.gz"}
http://mathoverflow.net/questions/100323/whats-the-name-for-the-analogue-of-divided-power-algebras-for-xi-i/100586	What's the name for the analogue of divided power algebras for x^i/i? I recently came across divided power algebras here: http://amathew.wordpress.com/2012/05/27/lazards-theorem-ii/ It interests me because the free divided power algebra on one variable $x$, where $x^{(i)}$ models $x^i/i!$, seems like a good way of handling exponential and exponential-type series. (Divided power algebras that aren't generated by a single variable are more complicated to axiomatize, so I will stick to the one-variable case). Quick summary: A system of divided powers for an element $x$ in an associative unital ring $R$ associates to each nonnegative integer $n$ a ring element $x^{(n)}$ with $x^{(0)} = 1$ and $x^{(1)} = x$ and satisfying the following condition for all $i,j \ge 0$: $$x^{(i)}x^{(j)} = \binom{i + j}{i} x^{(i + j)}$$ As we can see, if $R$ is an algebra over the rationals, the obvious (and unique!) choice is to set $x^{(i)} = x^i/i!$ for all $i$. Suppose I am interested in finding some $x^{[i]}$ that models $x^i/i$ instead of $x^i/i!$, i.e., I am interested in the kind of terms that appear in the expansions of logarithmic and inverse trig series. My best guess for the right analogue to consider is this: $x^{[i]}$ is defined for all positive integers $i$ with $x^{[1]} = x$ and satisfy the following condition. For each $i,j > 0$, write $ij/(i + j)$ as a reduced fraction $r/s$. Then, we must have: $$rx^{[i]}x^{[j]} = sx^{[i + j]}$$ I'd like to know whether this structure has been studied in the past, and/or what names have been given to the structure in question. Also, I'd be interested if anybody has opinions on whether the above is a reasonable way of trying to model $x^i/i$. - First of all, let me remark that one normally speaks of a divided power structure on an ideal $I\subset R$ (rather than on the whole $R$), by asking the existence of divided powers only for elements in the ideal; the requirement is then that all divided powers but the $0$-th also lie in the ideal. I think that your divided power structure would not be very useful – let me call it logarithmic divided power as Liz suggested. If you have the "usual" one on your ideal $I$ (let me denote it by $\gamma$ as in Berthelot&Ogus' book) you can immediately deduce a logarithmic one (call it $\lambda$) by setting $$\lambda_i(x)=(i-1)!\gamma_i(x)$$ for all $x\in I$. Since $(i-1)!\in\mathbb{Z}$, this defines a unique logarithmic divided power structure on $I$. But unfortunately, your logarithmic powers are weaker than usual ones: for instance, we know (see Berthelot&Ogus' book, page 3.2, Lemma 3.3) that if $V$ is a discrete valuation ring of mixed characteristic $(0,p)$ and uniformizer $(\pi)$, we can put a divided power structure on $(\pi^k)$ if and only if the absolute ramification index $e$ of $V$ verifies $e\leq k(p-1)$ (only the case $k=1$ is treated in the book, but computations are analogous for $k\geq 1$). If you want to stress that your logarithmic structure exist, you only need to insist that $v_\pi[\lambda_i(\pi^k)]\geq 1$ for all $i\geq 1$. Computing explicitly, this corresponds to $$e\leq \frac{ik-1}{v_p(i)}$$ which, by monotonicity of $i/v_p(i)$, is satisfied as soon as $e\leq kp-1$. Since this is (in general) bigger than $k(p-1)$ you see that there are rings admitting one structure but not the other; since in crystalline cohomology and related topics one really needs exponentials (so, dividing out by $n!$), yours are too weak to be useful. You started your question saying "Suppose I am interested in finding...": are you really? - Filippo, thanks a lot for your answer. I don't know anything about crystalline cohomology, but I think I understand the rest of your answer. For the application that I had in mind, it isn't a disadvantage if the "logarithmic divided powers" are defined in a more diverse array of situations than the usual divided powers; that might even be an advantage. But I suspect that you may be right that the situations where logarithmic divided powers exist, and the usual ones don't, are not very interesting. – Vipul Naik Jun 26 '12 at 16:07 Vipul, I think Laurent's answer is very enlightening, but as you say that you are not very familiar with crystalline cohomology I dare commenting a bit. The ring $B_{cris}$ you might have heard of is the divided power envelope of some ''universal'' creature; what Laurent observes is that one can, analogously, apply a whole bestiary of different divided power constructions to the universal creature to find many more rings which all look natural in $p$-adic Hodge theory, like $B_{cris}$. Your ''logarithmic divided power'' then corresponds to a (useful!) ring of functions with a growth condition. – Filippo Alberto Edoardo Jun 27 '12 at 12:23 Vipul, I don't have an answer to your question, but I thought I would remind you why divided power algebras are so very natural (in contrast ...?), since that makes understanding the many variable case easy, at least if you are willing to have your algebras be commutative. Working over a field (not necessarily of characteristic zero), look at a polynomial algebra and give it a bialgebra structure by letting its generators be primitive. The dual algebra to the resulting coalgebra is the divided polynomial algebra. This makes sense and gives the right (commutative) definition even if you have infinitely many generators. Of course, you have to be careful about saying the dual is a bialgebra, but it is in zillions of graded situations in algebraic topology where there are finitely many generators of positive degree (even if the characteristic isn't $2$). - Thank you for the insight, Professor May! – Vipul Naik Jun 27 '12 at 18:47 One possible way to generalize divided powers is to model sequences $x^n/a_n$ where $a_{n+m}/(a_na_m)$ is integral (an integer, or at least integral in some sense). You can model these in the obvious way, like divided powers. Choosing $a_n = n!$ gives you divided powers, and choosing $a_n=p^n$ gives you a sequence which, from the point of view of $p$-adic analysis, is much more "regular" in its growth. You can come up with other examples. As you may know, one uses divided powers to construct Fontaine's ring of periods $B_{cris}$ and using $a_n=p^n$ instead of $a_n=n!$ gives a ring $B_{max}$ that is similar in nature to $B_{cris}$ but much better behaved (see III.2 of Colmez' 1998 paper "Théorie d'Iwasawa des représentations de de Rham d'un corps local"). More generally, tweaking divided powers to get various convergence conditions is something that happens often in $p$-adic Hodge theory. See for instance 5.2.3 of Fontaine's "Le corps des périodes $p$-adiques" for one of many examples of custom made convergence conditions. In the $p$-adic setting, I would say that your log series belongs to the "space of holomorphic functions on the $p$-adic open unit disk, having order of growth $\leq 1$". Using precise analytic conditions rather than "divided powers" seems the right way to study the analytic series that interest you. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9524193406105042, "perplexity": 276.67470993126676}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609535745.0/warc/CC-MAIN-20140416005215-00516-ip-10-147-4-33.ec2.internal.warc.gz"}
http://mathhelpforum.com/number-theory/62811-help.html	# Math Help - help 1. ## help can anyone help me with these 2 super hard questions... 1. If n is positive integer what is the number of solutions (x,y) (with x and y postive integers) to the equation 1/x +1/y =1/n 2. let p be an odd prime, and i be an integer, 0<= i <= p-2. show that 1^i +2^i +3^i +.....+ (p-1)^i = 0 (mod p) 2. Originally Posted by felixmcgrady can anyone help me with these 2 super hard questions... 1. If n is positive integer what is the number of solutions (x,y) (with x and y positive integers) to the equation 1/x +1/y =1/n 2. let p be an odd prime, and i be an integer, 0<= i <= p-2. show that 1^i +2^i +3^i +.....+ (p-1)^i = 0 (mod p) 1) is not too hard. If you multiply out the fractions, then the equation $\tfrac1x+\tfrac1y = \tfrac1n$ can be written $(x-n)(y-n)=n^2$. So the number of solutions is the same as the number of ways of factorising $n^2$ as a product of two factors (which I'll leave you to think about). 2) seems to be harder (unless I'm missing a simple solution). First, the question as stated is wrong, since the result is clearly false when i=0. So we want to prove it for $1\leqslant i \leqslant p-2$. The only way I can see to do this is to quote the result that the multiplicative group $\mathbb{Z}_p^$ of nonzero elements of $\mathbb{Z}_p$ is cyclic. Let r be a generator of this group. Then $\sum_{k=1}^{p-1}k^i = \sum_{j=1}^{p-1}r^{ji}$, because the elements $r,r^2,\ldots,r^{p-1}$ are just 1,2,...,p-1 in some order. But if $1\leqslant i \leqslant p-2$ then the map consisting of multiplication by i is a bijection of $\mathbb{Z}_p^$. So the terms $r^i,r^{2i},\ldots,r^{(p-1)i}$ are also just 1,2,...,p-1 in some order. That reduces the problem to showing that $\sum_{k=1}^{p-1}k\equiv0\!\!\!\pmod p$. But that is obviously true because $\sum_{k=1}^{p-1}k = \tfrac12p(p-1)$. 3. Originally Posted by Opalg (unless I'm missing a simple solution). That is the standard (simple) way of proving it. Except I just write out $1^i + ... + (p-1)^i \equiv r^i + r^{2i} + ... + r^{(p-1)i}$ And now apply geometric series. We from there that $1^i + ... + (p-1)^i \equiv 0$ if $i \not \| (p-1)$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 16, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9575210809707642, "perplexity": 344.2028774191348}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644064919.24/warc/CC-MAIN-20150827025424-00250-ip-10-171-96-226.ec2.internal.warc.gz"}
https://groupprops.subwiki.org/wiki/Intermediate_isomorph-conjugacy_is_normalizing_join-closed	# Intermediate isomorph-conjugacy is normalizing join-closed This article gives the statement, and possibly proof, of a subgroup property (i.e., intermediately isomorph-conjugate subgroup) satisfying a subgroup metaproperty (i.e., normalizing join-closed subgroup property) View all subgroup metaproperty satisfactions \| View all subgroup metaproperty dissatisfactions \|Get help on looking up metaproperty (dis)satisfactions for subgroup properties Get more facts about intermediately isomorph-conjugate subgroup \|Get facts that use property satisfaction of intermediately isomorph-conjugate subgroup \| Get facts that use property satisfaction of intermediately isomorph-conjugate subgroup\|Get more facts about normalizing join-closed subgroup property ## Statement Suppose $H, K$ are intermediately isomorph-conjugate subgroups of a group $G$, such that $K \le N_G(H)$. Then, the join $HK$ (which is also equal to the product of subgroups) is also intermediately isomorph-conjugate. ## Proof CONVENTION WARNING: This article/section uses the left-action convention. The left and right action conventions are equally powerful and statements/reasoning here can be converted to the alternative convention (the main reason being that every group is naturally isomorphic to its opposite group via the inverse map). For more on the action conventions and switching between them, refer to switching between the left and right action conventions. This proof follows the left-action convention, though it can be stated equally well using the right-action convention (a similar proof for pronormality uses the right-action convention -- the proofs use essentially the same idea). We denote by $c_g$ the operation $x \mapsto gxg^{-1}$ of conjugation by $g$. Given: A group $G$, intermediately isomorph-conjugate subgroups $H,K \le G$ such that $K$ normalizes $H$. $\sigma$ is an isomorphism from $HK$ to some subgroup of $G$. To prove: There exists $l \in \langle HK, \sigma(HK) \rangle$ such that $c_l(HK) = \sigma(HK)$. Proof: 1. (Given data used: $H$ is intermediately isomorph-conjugate in $G$): $H$ and $\sigma(H)$ are isomorphic subgroups, so, since $H$ is intermediately isomorph-conjugate, there exists a $g \in \langle H, \sigma(H) \rangle$ such that $c_g(H) := gHg^{-1} = \sigma(H)$. Note that $g \in \langle H, \sigma(H) \rangle \le \langle HK, \sigma(HK) \rangle$. 2. (No given data used): Define $\sigma' = c_g^{-1} \circ \sigma$ as a map from $HK$ to $G$. Now, $\sigma'(HK) \le \langle HK, \sigma(HK) \rangle$, because it involves composing $\sigma$ with conjugation by an element inside $\langle H, \sigma(H) \rangle \le \langle HK, \sigma(HK) \rangle$. Thus, $\langle HK, \sigma'(HK) \rangle \le \langle HK, \sigma(HK) \rangle$. 3. (Given data used: $K$ normalizes $H$): We have $K \le N_G(H)$. Now, $\sigma'(H) = H$ by construction. Thus, $\sigma'(K) \le N_G(\sigma'(H)) = N_G(H)$. Thus, $\langle K, \sigma'(K) \rangle \le N_G(H)$. 4. (Given data used: $K$ is intermediately isomorph-conjugate in $G$): Since $K$ is intermediately isomorph-conjugate in $G$, the conclusion of step (3) tells us that there exists $k \in \langle K, \sigma'(K) \rangle$ such that $c_k(K) = \sigma'(K)$. In particular, $k \in N_G(H)$, so $c_k(H) = H$. Thus, $c_k(HK) = c_k(H)c_k(K) = H\sigma'(K) = \sigma'(H)\sigma'(K) = \sigma'(HK)$, with $k \in \langle K, \sigma'(K) \le \langle HK, \sigma'(HK)\rangle$. By the conclusion of step (2), this yields $k \in \langle HK, \sigma(HK) \rangle$. 5. Now, consider the product $l = gk$ We claim that this works (No problem data used): Clearly, since both $g$ and $k$ are elements of $\langle HK, \sigma(HK) \rangle$, so is $l$. Further, $c_l = c_g \circ c_k$. In particular, $c_l(HK) = c_g(c_k(HK)) = c_g(\sigma'(HK)) = \sigma(HK)$. Thus, $HK$ and $\sigma(HK)$ are conjugate in the subgroup they generate.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 57, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9520895481109619, "perplexity": 960.4022966871878}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178381230.99/warc/CC-MAIN-20210307231028-20210308021028-00148.warc.gz"}
https://edurev.in/course/quiz/attempt/2433_Olympiad-Test-Chemical-Effects-Of-Electric-Current/8f24c32e-c124-4577-899d-27875895180f	Courses # Olympiad Test: Chemical Effects Of Electric Current ## 20 Questions MCQ Test Science Class 8 \| Olympiad Test: Chemical Effects Of Electric Current Description This mock test of Olympiad Test: Chemical Effects Of Electric Current for Class 8 helps you for every Class 8 entrance exam. This contains 20 Multiple Choice Questions for Class 8 Olympiad Test: Chemical Effects Of Electric Current (mcq) to study with solutions a complete question bank. The solved questions answers in this Olympiad Test: Chemical Effects Of Electric Current quiz give you a good mix of easy questions and tough questions. Class 8 students definitely take this Olympiad Test: Chemical Effects Of Electric Current exercise for a better result in the exam. You can find other Olympiad Test: Chemical Effects Of Electric Current extra questions, long questions & short questions for Class 8 on EduRev as well by searching above. QUESTION: 1 ### Plastic wire is Solution: Plastic wire is an insulator because it doesn't conduct heat. QUESTION: 2 Solution: QUESTION: 3 ### Adding common salt to distilled water makes it: Solution: QUESTION: 4 An electrolyte is: Solution: QUESTION: 5 Copper wire is a: Solution: QUESTION: 6 Poor conductors are Solution: QUESTION: 7 Distilled water is Solution: Distilled water is a poor conductor of electricity because it doesn't contain any dissolved salts in it which can provide it ions to conduct electricity. QUESTION: 8 Flow of electrons is called Solution: Electricity is a term used to describe the energy produced (usually to perform work) when electrons are caused to directional (not randomly) flow from atom to atom. In fact, the day-to-day products that we all benefit from, rely on the movement of electrons. This movement of electrons between atoms is called electrical current. QUESTION: 9 Electroplating prevents Solution: QUESTION: 10 An electric lamp glows due to Solution: When electric current passes through a bulb, the filament of the bulb heats up to a high temperature. This is called heating effect of current. The temperature is so high that it starts glowing. However, if the current through a circuit is weak, the filament does not get heated up sufficiently and it does not glow. QUESTION: 11 Plastics are the good conductor of electricity. Solution: QUESTION: 12 Flow of electrons is called electric current. Solution: QUESTION: 13 Distilled water also conducts the electric current. Solution: Distilled water which is a pure form of water, is neither acidic nor basic in nature. So distilled water does not dissociate into ions. Since, conduction of electricity requires free ions so, distilled water does not conduct electricity. QUESTION: 14 Circuit is the closed path through which current flows. Solution: The correct option is A. circuit A path through which electric charges flow. A continuous, unbroken path through which electrons can flow is a closed circuit. Charges, or current, can flow only through a closed circuit. A break or opening in a circuit creates an open circuit. QUESTION: 15 Electric bulb glows due to chemical effect of electricity. Solution: The electric bulb has a filament called tungsten when electricity passes through this filament, it heats up and glows. This heat is generated due to the passage of electric current and the drift of electrons due to the current and the resistance it causes. Hence, the glowing of electric bulb is based on the heating effect of electric current. QUESTION: 16 Copper wire is poor conductor of electricity. Solution: QUESTION: 17 LED is an electric bulb which is used in a tester. Solution: QUESTION: 18 Electricity shows magnetic effect Solution: QUESTION: 19 When salt is dissolved in distilled water, it does not conduct electricity. Solution: QUESTION: 20 Electroplating is based on the heating effect of electricity. Solution: The process of depositing a thin layer of a desired metal over a metal object with the help of electric current is called electroplating. So, Electroplating is based on chemical effect of electricity.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8526304960250854, "perplexity": 2446.128489489427}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780060908.47/warc/CC-MAIN-20210928214438-20210929004438-00663.warc.gz"}
http://www2.math.binghamton.edu/p/seminars/stat/171109	### Sidebar seminars:stat:171109 Statistics Seminar Department of Mathematical Sciences DATE: Thursday, November 9, 2017 1:15pm – 2:15pm WH 100E Lin Yao, Binghamton University Optimal Model Averaging Estimation for Generalized Linear Models and Generalized Linear Mixed-Effects Models Abstract Considering model averaging estimation in generalized linear models, we propose a weight choice crite- rion based on the Kullback–Leibler (KL) loss with a penalty term. This criterion is different from that for con- tinuous observations in principle, but reduces to the Mallows criterion in the situation. We prove that the corresponding model averaging estimator is asymptotically optimal under certain assumptions. We further extend our concern to the generalized linear mixed-effects model framework and establish associated the- ory. Numerical experiments illustrate that the proposed method is promising. seminars/stat/171109.txt · Last modified: 2017/11/16 09:26 by qyu	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9256013035774231, "perplexity": 2279.582029201848}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703521987.71/warc/CC-MAIN-20210120182259-20210120212259-00320.warc.gz"}
https://www.physicsforums.com/threads/optimization-why-is-there-a-constant-term-in-the-time-equation.720216/	# Optimization - Why is there a constant term in the time equation? 1. Nov 1, 2013 ### Qube Optimization - Why is there a constant term in the time equation?? Mod note: The OP apparently answered his own question. 1. The problem statement, all variables and given/known data 1) A man can run at 8 km/hr and swim at 4 km/hr. He is currently 6 km from the shore of a lake, which is due east from him. He wants to get to a point 10 km south on the shore of his current position. How should he proceed? 2. Relevant equations 1) Pythagorean Theorem; speed is distance / time; time is distance / speed. 3. The attempt at a solution http://i.minus.com/jb1uFVsmbSfNtE.jpg [Broken] My question is: intuitively, if the man ran 0 km on the shore and swam all the way, he would have swam the square root of 136 km - two legs of the triangle are 6 and 10. sqrt136 km divided by a rate of 4 km/hr yields a time of 2.91 hours. Distance over speed = time. However, T(0) yields a different time of 3.25 hours. How come? I see that when x = 0 in the equation T(x) the first term - the (6-x)/8 term doesn't go to zero. Did I set up the problem incorrectly? ----- Wait, never mind. I think I see why now. If x = 0, that wouldn't be that he ran 0 km. That would actually imply the opposite just looking at the diagram I setup. It would imply he ran 6 km and then swam 10 km. And time would correctly be 6 km / 8 plus 10/4. Last edited by a moderator: May 6, 2017	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8074281811714172, "perplexity": 678.0235040519336}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647612.53/warc/CC-MAIN-20180321102234-20180321122234-00034.warc.gz"}
http://mathhelpforum.com/algebra/125105-t.html	# Math Help - T = ? 1. ## T = ? I have T=1 for the following equation. Is this correct? Thanks! (3t-5)/(t-1) = 2 + (2t)/(2-t) 2. Originally Posted by jay1 I have T=1 for the following equation. Is this correct? Thanks! (3t-5)/(t-1) = 2 + (2t)/(2-t) Your solution should have two values (one may be extraneous) . And by the way, how can 1 be a solution if its not in the domain?	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8913494944572449, "perplexity": 1164.2990392472157}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394678692841/warc/CC-MAIN-20140313024452-00096-ip-10-183-142-35.ec2.internal.warc.gz"}
https://biz.libretexts.org/Sandboxes/[email protected]/MGT_235/04%3A_The_Normal_Distribution/4.02%3A_The_Standard_Normal_Distribution	# 4.2: The Standard Normal Distribution $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ The standard normal distribution is a normal distribution of standardized values called z-scores. A z-score is measured in units of the standard deviation. The mean for the standard normal distribution is zero, and the standard deviation is one. What this does is dramatically simplify the mathematical calculation of probabilities. Take a moment and substitute zero and one in the appropriate places in the above formula and you can see that the equation collapses into one that can be much more easily solved using integral calculus. The transformation $$z=\frac{x-\mu}{\sigma}$$ produces the distribution $$Z \sim N(0, 1)$$. The value $$x$$ in the given equation comes from a known normal distribution with known mean $$\mu$$ and known standard deviation $$\sigma$$. The z-score tells how many standard deviations a particular $$x$$ is away from the mean. ## Z-Scores If $$X$$ is a normally distributed random variable and $$X \sim N(\mu, \sigma)$$, then the z-score for a particular $$x$$ is: $z=\frac{x-\mu}{\sigma}\nonumber$ The z-score tells you how many standard deviations the value $$\bf{x}$$ is above (to the right of) or below (to the left of) the mean, $$\bf{\mu}$$.Values of $$x$$ that are larger than the mean have positive z-scores, and values of $$x$$ that are smaller than the mean have negative z-scores. If x equals the mean, then x has a z-score of zero. Example $$\PageIndex{1}$$ Suppose $$X \sim N(5, 6)$$. This says that $$X$$ is a normally distributed random variable with mean $$\mu = 5$$ and standard deviation $$\sigma = 6$$. Suppose $$x = 17$$. Then: $z=\frac{x-\mu}{\sigma}=\frac{17-5}{6}=2\nonumber$ This means that $$x = 17$$ is two standard deviations $$(2\sigma)$$ above or to the right of the mean $$\mu = 5$$. Now suppose $$x = 1$$. Then: $$z=\frac{x-\mu}{\sigma}=\frac{1-5}{6}=-0.67$$ (rounded to two decimal places) This means that $$\bf{x = 1}$$ is 0.67 standard deviations $$\bf{(–0.67\sigma)}$$ below or to the left of the mean $$\bf{\mu = 5}$$. ## The Empirical Rule If $$X$$ is a random variable and has a normal distribution with mean $$\mu$$ and standard deviation $$\sigma$$, then the Empirical Rule states the following: • About 68% of the $$x$$ values lie between $$–1\sigma$$ and $$+1\sigma$$ of the mean $$\mu$$ (within one standard deviation of the mean). • About 95% of the $$x$$ values lie between $$–2\sigma$$ and $$+2\sigma$$ of the mean $$\mu$$ (within two standard deviations of the mean). • About 99.7% of the $$x$$ values lie between $$–3\sigma$$ and $$+3\sigma$$ of the mean $$\mu$$ (within three standard deviations of the mean). Notice that almost all the x values lie within three standard deviations of the mean. • The z-scores for $$+1\sigma$$ and $$–1\sigma$$ are $$+1$$ and $$–1$$, respectively. • The z-scores for $$+2\sigma$$ and $$–2\sigma$$ are $$+2$$ and $$–2$$, respectively. • The z-scores for $$+3\sigma$$ and $$–3\sigma$$ are $$+3$$ and $$–3$$ respectively. Figure $$\PageIndex{1}$$ Example $$\PageIndex{1}$$ Suppose $$x$$ has a normal distribution with mean 50 and standard deviation 6. • About 68% of the $$x$$ values lie within one standard deviation of the mean. Therefore, about 68% of the $$x$$ values lie between $$–1\sigma = (–1)(6) = –6$$ and $$1\sigma = (1)(6) = 6$$ of the mean 50. The values $$50 – 6 = 44$$ and $$50 + 6 = 56$$ are within one standard deviation from the mean 50. The z-scores are –1 and +1 for 44 and 56, respectively. • About 95% of the $$x$$ values lie within two standard deviations of the mean. Therefore, about 95% of the $$x$$ values lie between $$–2\sigma = (–2)(6) = –12$$ and $$2\sigma = (2)(6) = 12$$. The values $$50 – 12 = 38$$ and $$50 + 12 = 62$$ are within two standard deviations from the mean 50. The z-scores are –2 and +2 for 38 and 62, respectively. • About 99.7% of the $$x$$ values lie within three standard deviations of the mean. Therefore, about 99.7% of the $$x$$ values lie between $$–3\sigma = (–3)(6) = –18$$ and $$3\sigma = (3)(6) = 18$$ of the mean 50. The values $$50 – 18 = 32$$ and $$50 + 18 = 68$$ are within three standard deviations from the mean 50. The z-scores are –3 and +3 for 32 and 68, respectively. This page titled 4.2: The Standard Normal Distribution is shared under a CC BY license and was authored, remixed, and/or curated by .	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9842690825462341, "perplexity": 130.92812295608033}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030338213.55/warc/CC-MAIN-20221007143842-20221007173842-00271.warc.gz"}
https://link.springer.com/article/10.1007/s10590-021-09282-0	# Dual contextual module for neural machine translation ## Abstract Self-attention-based encoder-decoder frameworks have drawn increasing attention in recent years. The self-attention mechanism generates contextual representations by attending to all tokens in the sentence. Despite improvements in performance, recent research argues that the self-attention mechanism tends to concentrate more on the global context with less emphasis on the contextual information available within the local neighbourhood of tokens. This work presents the Dual Contextual (DC) module, an extension of the conventional self-attention unit, to effectively leverage both the local and global contextual information. The goal is to further improve the sentence representation ability of the encoder and decoder subnetworks, thus enhancing the overall performance of the translation model. Experimental results on WMT’14 English-German (En$$\rightarrow$$De) and eight IWSLT translation tasks show that the DC module can further improve the translation performance of the Transformer model. ## Introduction Self-Attention Networks (SAN) (Parikh et al. 2016; Lin et al. 2017) have been shown to significantly improve the performance of neural models for natural language processing (NLP) tasks including Neural Machine Translation (NMT) (Gehring et al. 2017; Vaswani et al. 2017; Dou et al. 2019), acoustic modeling ( Sperber et al. 2018), document summarization (Al-Sabahi et al. 2018; Wang and Ren 2018) and reading comprehension (Yu et al. 2018). The state-of-the-art NMT model, the Transformer network (Vaswani et al. 2017), is a prominent non-RNN model based entirely on self-attention. One of the strengths of SANs is their ability to capture long-range dependencies between tokens within a given sentence. SANs achieve this by attending to all tokens present irrespective of their distance apart ( Sperber et al. 2018; Yang et al. 2019a). That is, to generate the contextual representation for a token, self-attention tends to consider all tokens in the sentence. The implication of this approach is there is a higher probability that the self-attention mechanism may overlook important short-range dependencies between neighbouring tokens (Sperber et al. 2018; Yang et al. 2019a). Recently, there has been a growing amount of research dedicated to enhancing the performance of SANs at capturing both the long- and short-range dependencies between the sentence tokens. For example, Sperber et al. (2018) applied a locality restriction approach to limit the attention scope/span of the attention mechanism to the neighbouring elements to further enhance performance on the task of acoustic modeling. Similarly, Yang et al. (2018) employed a learnable Gaussian bias to model localness by revising the attention weight distribution to focus more on a dynamically varying window of tokens. In a different direction, other work ( Wu et al. 2019; Yang et al. 2019b) explored convolutional concepts to restricting the attention span to a fixed size window. Even though these approaches enhance the sentence representation ability of the SAN, Xu et al. (2019) argue that restricting the attention scope to some extent limits the ability of the self-attention mechanism to efficiently learn global contextual information. An interesting question here is the following: how can we effectively exploit local contextual information together with global contextual information without restricting the attention scope of the self-attention mechanism? To this end, this work proposes the Dual Contextual (DC) module, an extension to the self-attention unit to leverage both the local and global contextual information to further enhance translation quality. The DC module shown in Fig. 1 comprises two sub-modules, namely the Local Contextual Unit and the Context Interaction Unit. The Local Contextual Unit is a CNN-based network employed to capture the local contextual information respective to a neighbourhood window determined by the convolution filter size. The generated sentence representation from the Local Contextual Unit is passed to the Context Interaction Unit which employs two multi-head attention-based units and an aggregation unit to model the interaction between the global and local contextual information. The proposed approach imposes no restrictions on the attention scope, allowing the self-attention to fully capture any long-range dependencies. The learning of the short-range dependency information is assigned to the Local Contextual Unit. One of the multi-head attention units in the Context Interaction Unit is employed to perform the self-attention computation across the input representation. The DC module is incorporated into the Transformer network. Experimental results on nine translation tasks show that the proposed DC module can further improve the translation quality of the Transformer model. Furthermore, analysis based on ten linguistic probing tasks (Conneau et al. 2018) demonstrates that the leveraging both global and local contextual information further enhances the model’s ability to effectively capture the necessary linguistic properties required for the source translation task. The contributions of this work are the following: 1. 1. Proposing the Dual Contextual module to leverage both the local and global contextual information to further improve translation performance. 2. 2. Demonstrating consistent improvement over the strong Transformer baselines across nine translation tasks. 3. 3. Providing analysis on the performance impact of limiting the application of the DC module to the layers of either the encoder subnetwork or the decoding subnetwork. 4. 4. Providing an ablation study on the impact of limiting the contextual modeling via the DC module to only some combinations of encoding layers. The remainder of the paper is organized as follows: Sect. 2 briefly reviews the related literature and Sect. 3 provides a background to NMT. The proposed Dual Contextual module is presented in Sect. 4. The experiments conducted are presented in Sect. 5, and the results are compared and discussed in Sect. 6. Section 7 presents a detailed analysis on the impact of the DC module on translation performance of the Transformer network. Finally, the conclusions are presented in Sect. 8, together with avenues for future work. ## Related work Leveraging useful word-to-word contextual information has been shown to be essential to achieve higher translation quality in statistical machine translation (SMT) models (Gimpel and Smith 2008; He et al. 2008; Marton and Resnik 2008). One of the earliest works exploiting both local and global contextual information to improve NMT performance was Luong et al. (2015). Specifically, to utilize both local and global information of the source tokens, the authors employ two attention mechanisms, namely the global and local attention units. During the decoding step t, the global attention considers all the source tokens, whereas the local attention computation is performed across a subset of the source tokens. The improvements in performance highlight the importance of capturing both global and local contextual information. Motivated by this, recent work ( Sperber et al. 2018; Yang et al. 2018; Xu et al. 2019) has investigated strategies to learn both long-distance and short-range dependencies between tokens to further enhance the sentence representational ability of the self-attention mechanism. To model localness, Yang et al. (2018) proposed a modification of the self-attention mechanism with a learnable Gaussian bias which specifies the central position and window of tokens neighbourhood to which more attention should be paid. Similarly, Sperber et al. (2018) explored two masking techniques for controlling the contextual range of self-attention for the task of acoustic modeling. Other work ( Wu et al. 2019; Yang et al. 2019b) has explored convolutional concepts to restrict the attention scope/span to a window of neighbouring tokens. The ability of self-attention to efficiently capture global contextual information has been identified as one of its salient strengths, improving the performance of downstream NLP tasks such as semantic modeling (Yang et al. 2019a) and constituency parsing (Kitaev and Klein 2018). However, the approaches by Wu et al. (2019) and Yang et al. (2019b) restricting the attention scope to some degree can result in loss of important global and long-distance dependencies (Xu et al. 2019). The work of Song et al. (2018b) and Xu et al. (2019) exploit a hybrid attention mechanism to learn local and global contextual information. These approaches augment the self-attention mechanism with masking designed purposely to learn the local contextual patterns without restricting the self-attention mechanism’s ability to efficiently model the global and long-distance dependencies. Another form of hybrid attention mechanism is the QANet proposed by Yu et al. (2018), which employs a multi-layer depth-wise separable CNN to initially model the local contextual representation before the application of the self-attention unit. The resulting model significantly outperformed RNN-based models for the task of machine comprehension. In a similar direction, our work proposes the Dual Contextual Module to leverage both local and global information to further improve the performance of NMT. However, unlike Shen et al. (2018); Song et al. (2018b); Yu et al. (2018) and Xu et al. (2019), the DC module employs two separate attention computation units. Furthermore, the modeling of the local and global contextual information is applied to all layers within the encoder and decoder subnetworks. This work argues that the decoder subnetwork requires rich contextual source representation to achieve higher translation quality. This implies that the output of the encoder subnetwork has to ‘fully’ encapsulate both the local and global contextual information of the source tokens. ## Background A typical NMT system learns the conditional probability $${\mathsf {P}}\left( Y\mid X;\,\theta \right)$$ mapping from a given source sequence $$X=\left( x_1,x_2,\cdots ,x_M\right)$$ to a target language sentence $$Y=\left( y_1,y_2,\cdots ,y_N\right)$$ of length N, where $$x_i$$ and $$y_t$$ denote the $${i}^{th}$$ and $${t}^{th}$$ tokens of X and Y respectively. The model parameter $$\theta$$ is learned by maximizing the likelihood, as in (1): \begin{aligned} {\mathsf {P}}\left( Y\mid X;\theta \right) = \prod ^{M}_{t=1}{{{\mathsf {P}}\left( y_t\mid y_{<t} , X ;\theta \right) }} \end{aligned} (1) where $$y_{<t}=y_1,\cdots ,y_{t-1}$$ is the partial target sequence generated. NMT models usually consist of two subnetworks, namely the Encoder and Decoder. The Encoder generates the hidden representation $$H^e=[h^e_1,h^e_2,\cdots ,h^e_M]$$ based on the source embedding $$E_x=[e_1,e_2,\cdots ,e_M]$$, where $$e_i\in {\mathbb {R}}^d$$ is the embedding vector for the $${i}^{th}$$ source token (i.e. $$x_i$$). During the decoding step t, the target token $$y_t$$ is generated by the decoder subnetwork based on the partial target sequence $$y_{<t}$$ and source semantic representation $$H^e$$ from the encoding subnetwork. ### The transformer model Similar to NMT models based on RNNs (LSTM/GRU) (Cho et al. 2014; Bahdanau et al. 2015) and CNN (Gehring et al. 2017), the Transformer network (Vaswani et al. 2017) employs the Encoder-Decoder architectural structure. The encoder and decoder subnetworks consist of a stack of L-identical layers. For the encoding subnetwork, each layer consists of two sublayers, a self-attention (SA) module followed by a position-wise feed-forward network. A decoding layer is composed of a similar network structure to that of an encoding layer, but it employs an encoder-decoder attention unit in between the SA module and position-wise feed-forward network. To enhance the flow of gradient information, both the encoder and decoder subnetworks employ residual connection (He et al. 2016) and layer normalization (Ba et al. 2019) around each sublayer. A linear transformation layer with a softmax activation is employed to convert the decoder’s output representations $$H^{L}_d$$ into output probabilities over the target vocabulary. To further improve the model’s performance, recent work (Inan et al. 2016; Pappas et al. 2018) has proposed a linear transformation layer sharing the same weights with the word embedding layers of the decoder and encoder subnetworks. Furthermore, this strategy reduces the size of the model in terms of the number of trainable parameters. ### Self-attention mechanism An attention mechanism aims at modelling the direct relations between tokens of a given pair of sequence representations. Formally, the attention mechanism generates the token-to-token contextual representation c in (2): \begin{aligned} \begin{aligned} c&= {\text {ATT}}\left( Q, K, V\right) \\ {\text {ATT}}\left( Q, K, V\right)&= \alpha {\cdot V} \\ \alpha&= {{\,\rm {softmax}\,}}\left( {{\,\rm {score}\,}}(Q,K)\right) \\ \end{aligned} \end{aligned} (2) where $$Q \in {\mathbb {R}}^{Z\times d_{model}}$$, $$K \in {\mathbb {R}}^{J\times d_{model}}$$, and $$V \in {\mathbb {R}}^{J\times d_{model}}$$ are the query, key and value vectors, respectively. $$\alpha$$ is the attention weight distribution computed across the tokens represented by K. Z and J are the number of tokens in the given sequence representations Q and K respectively. $$d_{model}$$ denotes dimension of the hidden representation. Finally, $${{\,\rm {score}\,}}(\cdot )$$ is a scaled dot product function defined as in (3): \begin{aligned} {{\,\rm {score}\,}}(Q,K) = \frac{Q \times K^\intercal }{\sqrt{d_k}} \end{aligned} (3) where $$\sqrt{d_k}$$ is a scaling factor employed to stabilise the attention computation. Self-attention, a variant of the attention mechanism, performs the attention computation across a single sentence representation. Specifically, self-attention models the long-range dependencies between the tokens within the input sequence. Recent work including Gehring et al. (2017); Vaswani et al. (2017); Shaw et al. (2018) and Yu et al. (2018) has shown the potential performance gain of the self-attention mechanism over RNN by capturing long-range contextual information and dependencies between the pairs of tokens within the given sequence. For a layerFootnote 1l, the self-attention unit computes the sentence representation $$h^l$$ by attending to the hidden representation $$H^{l-1}$$ from the preceding layerFootnote 2 ($$l-1$$). The first stage of the self-attention unit is the computation of the query (Q), value (V) and key (K) based on three separate projections of the $$H^{l-1}$$, as in (4): \begin{aligned} \begin{aligned} Q&= H^{l-1}{W^Q} \in {\mathbb {R}}^{J\times d_{model}} \\ V&= H^{l-1}{W^V} \in {\mathbb {R}}^{J\times d_{model}} \\ K&= H^{l-1}{W^K} \in {\mathbb {R}}^{J\times d_{model}} \\ \end{aligned} \end{aligned} (4) where $$\{ W^V, W^Q, W^K\} \in {\mathbb {R}}^{d_{model}\times d_{model}}$$ are the projection weights employed to generate the value, query and key vectors respectively. The self-attention unit employed by the Transformer model is based on the multi-head attention ($${\text {MHA}}$$) mechanism. Specifically, the $${\text {MHA}}$$ defines $$N_h$$ attention heads where each $${head}_i$$ generates a separate attention weight distribution $$\alpha ^i$$. The attention head $${head}_i$$ attends to tokens at different positions based on the $$\alpha ^i$$ when generating the contextual representation $$c^i \in {\mathbb {R}}^{J\times \frac{d_{model}}{N_h}}$$. The $${\text {MHA}}$$ is formulated as in (5): \begin{aligned} \begin{aligned} h^l&= {\text {MHA}}\left( Q, K, V\right) \\ {\text {MHA}}\left( Q, K, V\right)&= O_c{W_o} \\ O_c&={{\,\rm {Concat}\,}}(c^1,c^2,\cdots ,c^{N_h}) \\ c^i&= {\text {ATT}}(Q^i,K^i,V^i) \end{aligned} \end{aligned} (5) where $$W_o \in {\mathbb {R}}^{d_{model}\times d_{model}}$$ is a trainable weight parameter employed to generate the output representation $$h^l \in {\mathbb {R}}^{J\times d_{model}}$$ based on concatenation of the contextual representations from all the attention heads $$O_c$$. $$\{ Q^{i} ,K^{i} ,V^{i} \} \in \mathbb{R}^{{J \times {{d_{{\bmod el}} } \mathord{\left/ {\vphantom {{d_{{\bmod el}} } {N_{h} }}} \right. \kern-\nulldelimiterspace} {N_{h} }}}}$$ are the query, key and value vectors with respect to the $${i}^{th}$$ attention head, respectively. ## Dual contextual module Our approach seeks to further improve translation performance by leveraging both the local and global contextual information to enhance the sentence representation ability of the encoding and decoding subnetworks. To this end, for a given layer (of either encoder or decoder subnetwork), the self-attention unit is replaced with a Dual Contextual Module as illustrated in Fig. 2. As shown in Fig. 1, the DC module consists of two sub-modules, namely the Local Contextual Unit and the Context Interaction Unit. To enhance the flow of gradient information, residual connections and layer normalization are applied across the output of each sub-module. For an encoder layer l, the input sentence representation to the DC module (r) denotes the output of the $$({l-1})^{th}$$ encoder layer $$H^{l-1}_e$$. Similarly, for a decoding layer l, r is the output of the preceding layer $$l-1$$ (i.e. $$H^{l-1}_d$$). ### Local context unit To capture local contextual information, this unit employs a single layer one-dimensional (1-D) convolution network with Gated Linear Unit (GLU) activation (Dauphin et al. 2017). The 1-D convolution can learn local dependencies between the source tokens within a neighbourhood of width determined by the kernel size of the filter (Song et al. 2018a; Yu et al. 2018). The local contextual representation $$l_c$$ is generated as in (6): \begin{aligned} \begin{aligned} l_c&={{\,\rm {LayerNorm}\,}}({\hat{r}} + r)\\ {\hat{r}}&= {{\,\rm {Concat}\,}}({\hat{r}}_1, {\hat{r}}_2,\cdots ,{\hat{r}}_J)\\ {\hat{r}}_t&= {\text {g}}\left( [r_{t-f/2},\cdots ,r_{t+f/2}]W^r +b^r \right) \end{aligned} \end{aligned} (6) where $$W^r \in {\mathbb {R}}^{{f\cdot d_{model}}\times {2\cdot d_{model}}}$$ and f are the convolution filter and kernel size, respectively. $$b^r$$ is the bias and $${\text {g}}(\cdot )$$ the GLU activation. $${\hat{r}}_t$$ is the local contextual representation of the $$t$$th token in the sequence. As shown in (6), $${\hat{r}}$$ is generated from the concatenation of the local contextual representations with respect to all input tokens. ### Context interaction unit This is the core of the DC module computing the context-rich representation $$z_f$$ based on the $$l_c$$ from the Local Contextual Unit and r. Given $$l_c$$ and r, the Context Interaction Unit generates the hidden representations $$h_l$$ and $$h_g$$ via two Feature Interaction ($${\text {FI}}$$) units as shown in Fig. 1. Each $${\text {FI}}$$ employs multi-head attention mechanism to model the interactions between the unit’s inputs, as in (7): \begin{aligned} \begin{aligned} h_l&={\text {FI}}(r,l_c)\\ h_g&={\text {FI}}(r,r)\\ \text {where}\\ {\text {FI}}(A,B)&= {\text {concat}}({M}_1,{M}_2,\cdots ,{M}_{n_h})\\ {M}_i&={\text {ATT}}(AW^q_i,BW^k_i,BW^v_i)\\ \end{aligned} \end{aligned} (7) where A and B represent the inputs to the $${\text {FI}}$$ units. $${AW^q_i,BW^k_i,BW^v_i} \in {\mathbb {R}}^{J\times \frac{d_{model}}{N_h}}$$ are the projections of the query (A), key (B) and value (B) vectors with respect to the attention head $${head}_i$$ sub-space, respectively. As shown in Fig. 1, $$h_g$$ is the global contextual representation obtained from a self-attention operation across the r. The $$h_l$$ is sentence representation generated from the attention operation between r and the $$l_c$$ (from the Local Contextual Unit). A feature Aggregation unit (AGG) is employed to generate the hidden representation $${\hat{z}}$$ from the combination the outputs of the FI units, as in (8): \begin{aligned} {\hat{z}} = [h_l;h_g]W_z + b_z \end{aligned} (8) where $$W_z \in {\mathbb {R}}^{2\cdot d_{model}\times d_{model}}$$ and $$b_z \in {\mathbb {R}}^{d_{model}}$$ are trainable model parameters. $$[\cdot ;\cdot ]$$ denotes the concatenation operation. Formally, the Context Interaction Unit computes the $$z_f$$ as in (9): \begin{aligned} \begin{aligned} z_f&= {{\,\rm {LayerNorm}\,}}\left( {\hat{z}}+r\right) \end{aligned} \end{aligned} (9) As shown in Fig. 2, the generated $$z_f \in {\mathbb {R}}^{J\times d_{model}}$$ is passed to the subsequent sublayers for further processing. For the encoder subnetwork, it is fed into the position-wise feed-forward network. However, in the case of the decoder subnetwork, it is one of the inputs to the encoder-decoder attention unit. ## Experimental setup ### Datasets The effectiveness of the proposed approach is evaluated on WMT’14 English-German (En$$\rightarrow$$De), and IWSLT tasks: French-English (Fr$$\leftrightarrow$$En), Spanish-English (Es$$\leftrightarrow$$En), Romanian-English (Ro$$\leftrightarrow$$En) and English-Vietnamese (En$$\leftrightarrow$$Vi) translation tasks. The dataset for the En$$\rightarrow$$De translation task consists of about 4.5M sentence pairs for training. Consistent with existing works (Vaswani et al. 2017; Yang et al. 2018; Xu et al. 2019), the newstest2013 and newstest2014 sets are employed as the validation and test sets, respectively. For the Es$$\leftrightarrow$$En and Ro$$\leftrightarrow$$En tasks, the datasets employed are respectively from the Spanish-to-English and Romanian-to-English translation tracks of the IWSLT 2014 evaluation campaign (Cettolo et al. 2014).Footnote 3 The training sets for these tasks consist of 183k and 182k sentence pairs, respectively. The test set is the tst2014 split and the validation set is created by concatenating the tst2010, tst2011, tst2012 and dev2010 sets. The En$$\leftrightarrow$$Vi translation task is performed on the English to Vietnamese track of IWSLT 2015 (Cettolo et al. 2015). The dataset consists of 133k training sentence pairs. The tst2013 (consisting of about 1.2k sentences pairs) and tst2012 (consisting of 1553 sentences pairs) are employed as the test and validation data, respectively. For the Fr$$\leftrightarrow$$En task, the dataset consisting of 207k training sentence pairs is from the IWSLT 2015 campaign. The test set is from the combination of the tst2014 and tst2015 splits. The validation set is the tst2013 set. For a given translation task, the size of the vocabulary is limited to a few numbers of words to reduce the complexity of the model. However, limiting the vocabulary size usually results in the out-of-vocabulary problem, which can be mitigated by learning a shared vocabulary via byte-pair-encodingFootnote 4 (Sennrich et al. 2016) for both the encoding and decoding subnetworks. The resulting subword-based shared vocabulary is employed to encode the sentence pairs (source and target sentences).Footnote 5 The vocabularies for the En$$\rightarrow$$De, Es$$\leftrightarrow$$En, En$$\leftrightarrow$$Vi, Ro$$\leftrightarrow$$En and Fr$$\leftrightarrow$$En translation tasks consist of 32k, 34k, 21k, 32k and 31k subword tokens, respectively. The translation quality of the proposed model is reported based on the BLEU metric (Papineni et al. 2002). Specifically, the 4-gram case-sensitive NIST BLEU metric is employed as the evaluation metric for the En$$\rightarrow$$De task. The translation quality for En$$\leftrightarrow$$Vi is reported based on the case-sensitive BLEU score computed with sacreBLEU.Footnote 6 Finally, for the other IWSLT translation tasks, the case-sensitive BLEU metric evaluated with multi-bleu.plFootnote 7 is employed. The statistical significance of the BLEU scores between the baseline and our DC based models is evaluated with paired bootstrap resampling (Koehn 2004) using the compare-mtFootnote 8 (Neubig et al. 2019) library with 1000 resamples. For simplicity, the statistical significance test is performed on the WMT’14 En$$\rightarrow$$De task mainly due to the size of the test dataset. ### Model settings The proposed Dual Contextual Module is integrated into the Transformer network (Vaswani et al. 2017). The parameter settings such as the number of encoder and decoder layers, number of attention heads and hidden size employed for each translation task are chosen based on the number of sentence-pairs within the training set. For the low-resource (IWSLT) tasks, the model consists of 4-layer encoder and 4-layer decoder subnetworks each with hidden size and number of attention heads set as 256 and 4, respectively. However, for En$$\rightarrow$$De, the hidden size, number of attention heads and number of layers are 512, 8 and, 6, respectively. Based on an analysis performed on the En$$\rightarrow$$De task (see Sect. 7.2), the filter size of the CNN unit of the Local Context Unit is set to 2. ### Training and inference For the IWSLT tasks, the models are trained with a batch size of 2048 tokens and the number of training iterations is 200k. The batch size and the number of iterations employed to train the model on the En$$\rightarrow$$De are 4960 tokens and 160k iterations, respectively. Adam (Kingma and Ba 2014) (with $$\beta _1 = 0.9, \beta _2 = 0.98, \epsilon = 10^{9}$$) is used as the optimizer to train the models. Unlike Vaswani et al. (2017), the learning rate scheduling algorithm employed in this work is the single-cosine-cycle with warm-up (So et al. 2019). During inference, the target translations are generated using the beam search algorithm. For the En$$\rightarrow$$De task, a beam-size of 4 and a length penalty of 0.6 is employed. Finally, for the IWSLT tasks, the beam size and length penalty are 6 and 1.1, respectively. The proposed DC module can be applied to both subnetworks of the Transformer model. Furthermore, it can also be limited to either the encoder or decoder subnetwork. For simplicity, the Transformer model trained with the DC module applied to both the encoder and decoder subnetworks is denoted as Full-DC. The model with the DC module applied to only the encoder is denoted as Enc-DC. Finally, Dec-DC is the Transformer model trained with the DC module employed within only the decoder subnetwork. ## Results This section presents the evaluation performance of the proposed Dual Contextual model on the translation tasks under consideration. Table 1 shows translation performance on the WMT’14 En$$\rightarrow$$De dataset, with the value in parentheses denoting the progressive gain over the Transformer baseline. Similarly for the IWSLT tasks, the results are summarized in Table 2. Finally, Table 3 summarizes the performance of existing models on the IWSLT En$$\rightarrow$$Vi and Es$$\rightarrow$$En translation tasks. On the En$$\rightarrow$$De translation task, the DC unit significantly improves the performance of the Transformer model by $$+0.89$$ BLEU in the case of the Enc-DC model and $$+0.49$$ BLEU with respect to the Dec-DC model. The performance of the Dec-DC model is improved by $$+0.25$$ BLEU when the DC module is applied to both subnetworks. However, the performance achieved by the Full-DC model is lower than that achieved by the Enc-DC model. Besides, the translation performance obtained via the DCM unit is higher than all the existing models further demonstrating the superiority of the proposed approach. The languages under consideration for the IWSLT tasks belong to different language families. The overall performance of an NMT model is affected by the linguistic and syntactic properties or structures of the language pairs under consideration. The translation quality achieved demonstrates the effectiveness of our proposed DC-based models at translating between languages of different families as shown in Table 2. On average, the Enc-DC model consistently achieves a higher performance gain over the Transformer baseline across all the IWSLT translation tasks. Compared to existing work, the baseline Transformer model consistently outperforms the models of Luong and Manning (2015) and Huang et al. (2018), with a performance improvement of $$+2.51$$ BLEU score on the En$$\rightarrow$$Vi translation task as summarized in Tables 2 and 3. The proposed Enc-DC and Dec-DC models achieved BLEU scores of 31.28 and 31.24, respectively. This represents a further performance gain up to $$+0.7$$. In contrast, a lower gain in translation quality of $$+0.52$$ BLEU is achieved when DC is applied to both subnetworks (i.e. Full-DC). Similarly, on the Es$$\rightarrow$$En task, all our models outperform the Transformer baseline. A marginal performance gain of $$+0.16$$ BLEU is achieved when the DC module is applied to only the decoder subnetwork (i.e. Dec-DC). On this dataset, the best performance is achieved by the Enc-DC and Full-DC models enhancing the translation quality by $$+0.38$$ BLEU and $$+0.35$$ BLEU, respectively. However, the translation performance of the Enc-DC, Dec-DC and Full-DC models is lower than that achieved by the models in He et al. (2018) and Xia et al. (2019), outperforming only the model of Cettolo et al. (2014). Overall, the performance achieved across all translation tasks indicates the benefits of leveraging both the local and global contextual information. The Enc-DC model consistently outperforms the Dec-DC approach on all datasets. This could be attributed to the bias/mask employed by self-attention units within the decoding layers. During the decoding step t, the attention bias limits the decoder’s self-attention to only consider the target sub-sequence (i.e. $$y_{<t}$$) generated so far. This implies that the decoder layer is able to exploit the local information within the neighbourhood of $$y_{<t}$$. Therefore unlike the encoder subnetwork, the CNN unit generating $$l_c$$ has a limited impact on the performance of the decoding layers and the decoder subnetwork. This is consistent with the observations made by Zhang et al. (2018). Furthermore, applying the DC module to both subnetworks (Full-DC) in most cases only improves performance from the Dec-DC model’s perspective. Besides, the Full-DC achieved a higher performance gain over the Enc-DC model only on the Vi$$\rightarrow$$En task. On a number of the translation tasks under consideration, there is no significant difference in the performance of the Full-DC and Enc-DC models. As shown in Table 1, the DC module introduces new parameters due to the additional attention computation as well as the LC units. Applying the DC module to all the subnetworks (Full-DC model) results in about a 25M increase compared to the 12.7M by the Enc-DC and Dec-DC models. The computational speed of a neural model is affected by quantities such as the optimizer, number of network parameters and the other computations that directly modify the formulation of the network (Popel and Bojar 2018). Accordingly, our DC-based models have lower training speeds compared to the baseline. Compared to the Full-DC model, the Enc-DC and Dec-DC models are shown to have the least decrease in training speed. Based on the translation quality achieved and the computation speeds, this work suggests limiting the explicit modeling of the local contextual information via the DC module to only the encoding subnetwork. ## Analysis This section presents analyses performed to better understand the performance improvement introduced by the Dual Contextual Module. These analyses are performed on the En$$\rightarrow$$De dataset. As shown in Sect. 6, the best performance is obtained when the DC module is applied across the encoder subnetwork, so the analyses presented in Sects. 7.1, 7.2 and 7.3 are performed only on the Enc-DC model. As shown in Tables 1 and 2, the Enc-DC model further improves the performance of the Transformer baseline model across all the translation tasks under consideration. However, little is know about the linguistic perspectives or properties improved by the proposed module. Following Conneau et al. (2018), Li et al. (2019), ten classification tasks are conducted to study the linguistic properties enhanced by the DC module. The ten classification tasks are divided into three categories: • Surface (Surf) focuses on the surface properties captured by the sentence representation or embedding. This category consists of the Sentence Length (SentLen) and Word Content (WC) tasks. Under the SentLen task, the goal is to predict the length of the input sentence based on the number of its tokens. The WC task tests the possibility of recovering information about the original word in a sentence given its embedding. • Syntactic (Sync) evaluates the ability of the encoder subnetwork to learn/capture syntactic information. The syntactic tasks under consideration are: Tree Depth (TDep), Bigram Shift (BShift) and Top Constituent (ToCo). The TDep task tests the capability of the encoder to infer the hierarchical structure of the input sentence. For the BShift, the sensitivity of the encoder to the legal word order is evaluated. Specifically, the goal is to predict if two consecutive tokens within the sentence have been inverted or not. Finally, under the ToCo task, the sentence is classified in terms of the sequence of top constituents immediately below the sentence node. • Semantic (Sem) evaluates the capabilities of the encoder to understand the denotation of a given sentence. To achieve higher performance on this task, the sentence embedding should encapsulate the syntactic structure (Conneau et al. 2018). Here, there are five tasks under consideration. The first semantic probing task is the Tense classification task which evaluates the tense of the main clause verb (whether it is in the past or present tense). The next is the Subject Number (SubjN) which focuses on predicting the number of the subject in the main clause. In contrast, the Object Number (ObjN) predicts the number of the direct object of the main clause. Under the Semantic odd man out (SoMo) task, the sentences are modified by randomly replacing a noun or verb with another noun or verb. Here the task is to predict whether a sentence has been modified or not. Finally for the Coordination Inversion (CoIn), the sentences are divided into two coordinate clauses. In half of the sentences, the order of the clauses is inverted and the task here is to check whether a sentence is modified or is left intact. The above tasks are performed based on the sentence representations generated by the encoding subnetwork. Specifically, the decoding subnetwork of the pre-trained NMT model is replaced with a two-layer classifier. L2 regularization of $$\lambda =1e^{-4}$$ is applied to the hidden layer of the classifier. For each classification task, the input representation to the classifier is the mean of the output representation from the top-level (last) encoding layer. The resulting classification models are trained and evaluated on the dataset presented by Conneau et al. (2018).Footnote 9 The training corpus for each task comprises 100k sentences, with 10k sentences for validation and 10k sentences for testing. During training, only the parameters of the classifier are updated. The classifiers are trained for 100 epochs with the RMProp optimizer using the learning rate of $$2.5e^{-4}$$ and mini-batch size of 64. An early stopping criterion is applied during training based on the accuracy score on the validation set. The parameters of the pre-trained encoding subnetwork are fixed to quantify the linguistic properties and information captured by the pre-trained encoder subnetwork of the Transformer baseline and our Enc-DC model. #### Results The results of the probing tasks are summarized in Table 4. Despite the DC-based encoding subnetwork of the Enc-DC model achieving best performance on three of the five Semantic tasks, on average it achieved identical performance as the baseline. The actual performance gain introduced by the DC module can be seen across the Surface and Syntactic tasks. Across these tasks, the Enc-DC model significantly outperformed the Transformer baseline. Tasks such as WC and CoIn require global contextual information, whereas local contextual information is generally required to achieve higher performance on tasks such as ToCo, BShift and SentLen. Overall, the performance on the Syntactic, Surface and Semantic probing tasks demonstrates the effectiveness of the DC module in learning both the global and local contextual information required to achieve higher translation performance. The DC module allows the encoder subnetwork to effectively learn the surface and syntactic information of the source sentence with minimal impact on its ability to encode the deeper linguistic features. ### Effect of CNN kernel size To analyse the impact of the CNN kernel size (employed by the Local Context unit) on translation quality, different Enc-DC models were trained with the kernel size $$f \in [2, 3, 4, 5, 6, 7, 8]$$. The translation performance of each Enc-DC model is summarized in Fig. 3. As shown, the Enc-DC models achieved almost identical BLEU scores when trained with the filter size $$5\le f\le 7$$. This is also true for the values of $$f\in {[3,4]}$$. The worse performance is obtained with $$f=8$$, producing a marginal performance gain of about $$+0.10$$ BLEU over the Transformer baseline. However, the best performance is achieved with $$f \in [2,3,4]$$. Overall, the performance of the Enc-DC model generally decreases as the filter size increases. One possible reason for this is that for larger filter sizes there is a higher possibility of information overlap between $$l_c$$ and $$h_g$$. As a result, the FI unit generating the $$h_l$$ (based on $$l_c$$ and $$H^{l-1}_e$$) has only a limited impact on the performance of the DC unit at learning the contextual sentence representation. This is because the generation of the $$h_g$$ and $$h_l$$ will be more meaningful if each FI unit captures diverse information. The above hypothesis is investigated by analyzing the relationship between the attention weight distributions $$\alpha _g$$ and $$\alpha _l$$ associated with the generation of the hidden contextual representations $$h_g$$ and $$h_l$$, respectively. As mentioned in Sect. 4.2, each FI unit is a multi-head attention unit employing $$n_h$$ attention heads. For simplicity, the attention weights for each encoding layer are represented by the attention head with the maximum weight distribution among all the $$n_h$$ heads employed by the FI units, as in (10): \begin{aligned} \begin{aligned} \alpha _g&= \max \left( [\alpha _g^1,\alpha _g^2,\cdots ,\alpha _g^{n_h}]\right) \\ \alpha _l&= \max \left( [\alpha _l^1,\alpha _l^2,\cdots ,\alpha _l^{n_h}]\right) \\ \end{aligned} \end{aligned} (10) where $$\alpha ^i_g$$ and $$\alpha ^i_l$$ are the weight distribution employed by the $${i}^{th}$$ attention head, respectively. To this end, Jensen-Shannon divergence, JS P Q, is employed as a distance metric to measure how far the $$\alpha _g$$ and $$\alpha _l$$ are from each other. The distance is measured using JS P Q because it is symmetric and bounded. Given the multi-dimensional attention weight distributions $$\alpha _l$$ and $$\alpha _g$$, the divergence with respect to the encoding layer l, $${dv}^l$$ is computed as in (11): \begin{aligned} \begin{aligned} {dv}^l&= JS{\alpha _g}{\alpha _l} \\ JS{\alpha _g}{\alpha _l}&= \frac{1}{2}\times KL{\alpha _g}{m} + \frac{1}{2}\times KL{\alpha _l}{m} \\ m&=\frac{1}{2}\times \left( \alpha _g + \alpha _l\right) \end{aligned} \end{aligned} (11) where KL P Q is the Kullback-Leibler divergence formulated as in (12): \begin{aligned} KL{P}{Q}= \sum _x P(x) \log {P(x)} - \sum _x P(x) \log {Q(x)} \end{aligned} (12) The overall divergence ($$E_d$$) for the L-layer encoding subnetwork is computed as the average of the divergence ($${dv}^l$$) computed for each encoding layer, as in (13): \begin{aligned} E_d= \frac{1}{L} \sum ^{L}_{l=1} {{dv}^l} \end{aligned} (13) Larger values of $$E_d$$ indicate that, on average across the multiple encoding layers, the two FI units (of the DC module) can capture more diverse information using the attention weights $$\alpha _l$$ and $$\alpha _g$$. Fig. 4 illustrates the divergence between $$\alpha _l$$ and $$\alpha _g$$ for the different filter kernel sizes. As displayed in Fig. 4a, variation in the divergence across the encoding layers is dependent on the value of f employed to train the Enc-DC model. In all cases, the divergence is larger across the top-4 encoding layers. Interestingly, the $${dv}^l$$ is closer to zero for values of $$f\ge 5$$ across the second encoding layer compared to that of the models trained with $$f\in [2,3,4]$$. This implies that for these models with $$f\ge 5$$, the two FI units capture almost identical contextual information. Overall the models trained with $$f\in [2,3,4]$$ have the lowest divergence across the top three layers among all the values of f under consideration. In contrast, the models with $$f\ge 5$$ have the lowest divergence between the FI units across the first three layers. Fig. 4b shows that, on average, the divergence across the entire encoding subnetwork is higher for smaller values of f. Specifically, the $$E_d$$ for the models trained with $$f \in [2,3,4]$$ is greater than that of $$f\ge 5$$. This implies that each FI unit within the encoding layers for the models with $$f \in [2,3,4]$$ can capture more diverse contextual information when generating $$h_g$$ and $$h_l$$. However, there is less diversity with $$f\ge 5$$. Overall, the variation of the divergence as shown in Fig. 4 and the corresponding translation quality displayed in Fig. 3 confirm our hypothesis that smaller values of f allow the DC module to effectively capture more diverse contextual information to further improve the performance of the translation model. ### Layers to consider Recent work (Belinkov et al. 2017; Peters et al. 2018; Raganato and Tiedemann 2018) has revealed that each encoder layer captures different levels of abstraction of the source sequence. Therefore, these layers tend to have different requirements for the local contextual information. Based on these findings, Yang et al. (2018), Xu et al. (2019) and Shen et al. (2018) argue in favour of limiting the explicit localness modeling to the first few encoding layers, in order to allow the top-level layers to focus more on capturing the global contextual information. In contrast, this work argues that higher performance can be achieved when the local and global information modeling is applied across all encoder layers. To investigate this further, an ablation study was performed where the DC module is applied to different combinations of the encoding layers. The translation performance achieved for the layer combinations is summarized in Table 5. As shown, all combinations of the encoder layers consistently outperform the Transformer baseline, further confirming the importance of learning both global and local contextual source information. For example, training the model with the DC module across only the lower 2 encoder layers (Row 3) produced a performance gain of about $$+0.61$$ BLEU over the baseline. Among the different combinations of layers, limiting the DC module to only lower-level layers (Rows 2-6) achieved higher translation quality compared to when employed across only the top-level layers (Rows 6, 7 and 8). The worst performance (28.47 BLEU) is obtained when the DC module is incorporated into only the top-2 encoding layers [5-6] (Row 7). Besides, the different combinations across the lower-level layers (Rows 2-6) produced fairly identical results with [1-1] (Row 2) and [1-4] (Row 5) achieving the best performance. This is consistent with the observations made by Shen et al. (2018), Yang et al. (2018) and Xu et al. (2019). However, our NMT model achieves the best translation performance when the DC module is applied to all the encoding layers. ### Length Previous work (Luong et al. 2015; Dou et al. 2018) argued that one of the major weaknesses of NMT models is the translation of long sentences. To achieve higher performance on source sentences of any arbitrary length, the encoder-decoder model should be able to effectively capture both the long-distance and short-range dependencies between the tokens (Dou et al. 2018). Following Luong et al. (2015), sentences of identical or equal length are grouped and the translation quality of the outputs from the models for each group is calculated. The comparison presented here is based on the following sentence length groups: <10, 10-20, 20-30, 30-40, 40-50, and >50. For each length group, the translation quality is evaluated for outputs from the models under consideration. Fig. 5 summarizes the impact of the length of the source sentence. As shown, both the Transformer baseline and our DC-based models (Enc-DC, Dec-DC and Full-DC) display identical variation in the translation quality across the different source sentence lengths. This is true especially for sentences with length greater than or equal to 20 subword tokens. However, our DC-based models consistently outperform the baseline across the different sentence groups with greater than 10 subword tokens. The performance gain achieved by our models is attributed to the addition of the DC module which allows both the encoder (in the case of Enc-DC), the decoder (in the case of Dec-DC) or both (in the case of Full-DC) to effectively exploit both the local and global contextual information required to improve the generation of the target translation. For shorter sentences with fewer than ten subword tokens, the global contextual model of the self-attention module is shown to be effective enough for the target generation. In contrast, for longer source sentences, further improvement in translation quality is achieved when both global and local contextual modelling are employed as shown across the sentence groups 10-20, 20-30, 30-40, 40-50 and>50. Overall, the best performance across the sentence groups (greater than 10 tokens) is achieved when the DC module is applied to only the encoding subnetwork (i.e. Enc-DC), which supports our recommendation to limit the global and local modelling to only the encoder subnetwork. ## Conclusion This work proposes the Dual Contextual (DC) module to leverage local and global contextual information to improve translation performance of the Transformer model. Three possible applications of the DC module, namely Enc-DC, Dec-DC and Full-DC, were presented. The experimental results on the nine translation tasks demonstrate the effectiveness of the proposed DC module. Furthermore, the analyses performed suggests that: • exploiting both the global and local contextual information is beneficial to the overall performance of the translation model. • the best performance is achieved when the DC module is applied to only the encoding layers (i.e. Enc-DC). The decoding subnetwork employing the DC module (in the case of Dec-DC and Full-DC) produces lower performance gains over the baseline. • in contrast to the findings of recent work (Shen et al. 2018; Yang et al. 2018; Xu et al. 2019), applying the DC module across all encoding layers results in higher performance gains compared to limiting its application to only the first few lower-level encoding layers (e.g. from layer 1 to 3). Future work includes validating the performance of the proposed DC module on other NLP tasks such as document summarization, sequence tagging, and machine reading comprehension. Another interesting direction will consider investigating further the performance–model complexity trade-off of the application of the DC module. To further improve performance, we plan to extend the DC module to exploit the self-attention techniques presented in Song et al. (2018b), Yang et al. (2018) and Xu et al. (2019). ## Notes 1. 1. A layer of either the encoder or decoding subnetwork. 2. 2. For $$l=0$$, $$H^{l-1}$$ is the output of the embedding layer. 3. 3. 4. 4. 5. 5. The original casing for the tokens in each sentence is preserved. 6. 6. 7. 7. 8. 8. 9. 9. ## References 1. Al-Sabahi K, Zuping Z, Nadher M (2018) A hierarchical structured self-attentive model for extractive document summarization (HSSAS). IEEE Access 6:24205–24212 2. Ba JL, Kiros JR, Hinton GE (2019) Layer normalization. arXiv: 1607.06450, 2016 3. Bahdanau D, Cho K, Bengio Y (2015). Neural machine translation by jointly learning to align and translate. 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Melbourne, pp 1789–1798 ## Author information Authors ### Corresponding author Correspondence to Isaac Kojo Essel Ampomah. ### Publisher's Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. ## Rights and permissions Reprints and Permissions Ampomah, I.K.E., McClean, S. & Hawe, G. Dual contextual module for neural machine translation. Machine Translation (2021). https://doi.org/10.1007/s10590-021-09282-0	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8693976998329163, "perplexity": 2222.246612878198}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363791.16/warc/CC-MAIN-20211209091917-20211209121917-00262.warc.gz"}
https://www.fachschaft.informatik.tu-darmstadt.de/forum/viewtopic.php?f=474&t=17892	## Ex. 08, Homework Moderator: Post Quantum Cryptography Patr0rc Sonntagsinformatiker Beiträge: 260 Registriert: 8. Feb 2008 11:43 ### Ex. 08, Homework Hi everybody. There is another thing that was not considered in the task before, but it was told to me by Markus Rückert: The equation in task a) was wrong. Now it is corrected. A new sheet is online. The cession date is extended until January 15th in 2010 since there were many mistakes in the exercise sheet. I'm very sorry for this. Best regards. Zuletzt geändert von Patr0rc am 17. Dez 2009 18:31, insgesamt 2-mal geändert. rueckert Mausschubser Beiträge: 57 Registriert: 9. Apr 2008 09:25 ### Re: Ex. 08, Homework Actually, the point is that you have to show that the equation holds over the integers without the reduction mod q. In other words, reducing mod q must not change anything.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8436396718025208, "perplexity": 3573.9002308249023}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540519149.79/warc/CC-MAIN-20191209145254-20191209173254-00033.warc.gz"}
http://blog.nietofineart.com/docs/parallel-vector-calculator-766d41	Walk through homework problems step-by-step from beginning to end. For example, 'A' can paint a room in 5 … Select the vectors dimension and the vectors form of representation; Press the button "Find vector projection" and you will have a detailed step-by-step solution. Enter values into Magnitude and Angle ... or X and Y. You can add, subtract, find length, find dot and cross product, Two vectors A and B may be added to obtain their resultant or sum A + B, where the two vectors are the two legs of the parallelogram. Solution to Question 5 A vector V is represented in three dimentional space in terms of the sum of its three mutually perpendicular components. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. Free vector dot product calculator - Find vector dot product step-by-step This website uses cookies to ensure you get the best experience. This calculator can be used for work problems. are given by their components as follows: by 4 in the first equation to obtain a new equation. A = B = --- Enter \|A\| and \|B\| vectors or Vectors Video. a) the equation of the line through point A(1 , 1) and parallel to vector U. Find the equation of the tangents through the point D(2 , 4) to the circle of center C(0 , 0) and radius 2. Expand and simplify to obtain the equation of the circle (2 - x)(4 - x) + (-2 - y)(-3 - y) = 0 Menu. From the definition of the cross product the following relations between the vectors are apparent: The cross product obey the following laws: Triple scalar product is defined as the determinant: The derivative of a vector P according to a scalar variable t is: The derivative of the sum of two vectors is: The derivative of the product of a vector P and a scalar u(t)according to t is: The derivative of two vectors dot product: Gradient If Ï is a scalar function defined by Ï=f(x,y,z),we define the gradient of The subtraction of a vector is the same as the addition of a negative vector. Welcome to OnlineMSchool. Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. 5 x + 2 y = 3 Given vector U = (2 , -5), find. The dot or scalar product of two vectors A and B is defined as: Two vectors are perpendicular when their dot product is: The cross or vector product of two vectors A and B is defined as: n - unit vector whose direction is perpendicular to vectors A and B. vector-unit-calculator. ,$\color{blue}{\text{ 2r3 } = 2\sqrt{3}}$ To calculate the unit vector in the same direction, you have to follow these steps: Write down the x, y and z components of the vector. In this case, x₁ = 8, y₁ = -3 and z₁ = 5. The following diagram shows several vectors that are parallel. Practice online or make a printable study sheet. Addition and subtraction of two vectors, Online calculator. As soos as, scalar triple product of the vectors can be the negative number, and the volume of geometric body is not, one needs to take the magnitude of the result of the scalar triple product of the vectors when calculating the volume of the parallelepiped: Vectors addition (A ± B) This web site owner is mathematician Miloš Petrović. You can input only integer numbers, decimals or fractions in this online calculator (-2.4, 5/7, ...). Direction cosines of a vector, Online calculator. By using this website, you agree to our Cookie Policy. If u and v are two non-zero vectors and u = c v, then u and v are parallel. The point is just any point on the line (therefore you got infinitely many possibilities which vector to take.) All rights reserved. Solution to Question 5 a) A point M(x , y) is on the line through point A(1 , 1) and parallel to vector U = (2 , -5) if and only if the vectors AM and U are parallel. Given vector $v_1 = (8, -4)$, calculate the the magnitude. BM = (x - (-2) , y - (-3)) = (x + 2 , y + 3) Now that you know the magnitude of the vector u, you probab Length of a vector, magnitude of a vector in space. Volume of the parallelepiped a) the equation of the line through point A(1 , 1) and parallel to vector U. b) the equation of the line through point B(-2 , -3) and perpendicular to vector U. a) A point M(x , y) is on the line through point A(1 , 1) and parallel to vector U = (2 , -5) if and only if the vectors AM and U are parallel. Explore anything with the first computational knowledge engine. For every operation, calculator will generate a detailed explanation. Six operations with two dimensional vectors + steps. Component form of a vector with initial point and terminal point in space, Exercises.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.820723831653595, "perplexity": 421.90018707621897}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363336.93/warc/CC-MAIN-20211207045002-20211207075002-00378.warc.gz"}
http://mathhelpforum.com/algebra/130847-root-equation.html	# Math Help - root of the equation 1. ## root of the equation (p^2)(x^2)-12x+p+7=0 has the root 3/2..find the value of p. 3. To explain HallsOfIvy's solution, remember that the roots of an equation are the values of $x$ when $y = 0$. Here, you are given an equation in the form $y = f(x)$ with $y = 0$ and $f(x) = (p^2)(x^2)-12x+p+7$ for some $p$. Since you are given that one root is $\frac{3}{2}$, you know that when $y = 0$ (which is our case), $x = \frac{3}{2}$. So you can just substitute $x = \frac{3}{2}$ into $f(x)$ and solve for $p$, to find which values of $p$ (there may be more than one) satisfy the equation $f(x) = 0$ when $x = \frac{3}{2}$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 15, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9349023103713989, "perplexity": 118.98089646159376}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398451648.66/warc/CC-MAIN-20151124205411-00329-ip-10-71-132-137.ec2.internal.warc.gz"}
https://www.global-sci.org/intro/article_detail/jcm/8715.html	Volume 25, Issue 5 Fast Parallelizable Methods for Computing Invariant Subspaces of Hermitian Matrices DOI: J. Comp. Math., 25 (2007), pp. 583-594 Published online: 2007-10 Preview Full PDF 138 1742 Export citation Cited by • Abstract We propose a {\it quadratically} convergent algorithm for computing the invariant subspaces of an Hermitian matrix. Each iteration of the algorithm consists of {\it one} matrix-matrix multiplication and {\it one} QR decomposition. We present an accurate convergence analysis of the algorithm without using the big $O$ notation. We also propose a general framework based on implicit rational transformations which allows us to make connections with several existing algorithms and to derive classes of extensions to our basic algorithm with faster convergence rates. Several numerical examples are given which compare some aspects of the existing algorithms and the new algorithms. • Keywords Eigenvalue Invariant subspace Hermitian matrix QR method Parallelizable method • AMS Subject Headings 15A18 65F05 65F35. @Article{JCM-25-583, author = {}, title = {Fast Parallelizable Methods for Computing Invariant Subspaces of Hermitian Matrices}, journal = {Journal of Computational Mathematics}, year = {2007}, volume = {25}, number = {5}, pages = {583--594}, abstract = { We propose a {\it quadratically} convergent algorithm for computing the invariant subspaces of an Hermitian matrix. Each iteration of the algorithm consists of {\it one} matrix-matrix multiplication and {\it one} QR decomposition. We present an accurate convergence analysis of the algorithm without using the big $O$ notation. We also propose a general framework based on implicit rational transformations which allows us to make connections with several existing algorithms and to derive classes of extensions to our basic algorithm with faster convergence rates. Several numerical examples are given which compare some aspects of the existing algorithms and the new algorithms.}, issn = {1991-7139}, doi = {https://doi.org/}, url = {http://global-sci.org/intro/article_detail/jcm/8715.html} } TY - JOUR T1 - Fast Parallelizable Methods for Computing Invariant Subspaces of Hermitian Matrices JO - Journal of Computational Mathematics VL - 5 SP - 583 EP - 594 PY - 2007 DA - 2007/10 SN - 25 DO - http://doi.org/ UR - https://global-sci.org/intro/article_detail/jcm/8715.html KW - Eigenvalue KW - Invariant subspace KW - Hermitian matrix KW - QR method KW - Parallelizable method AB - We propose a {\it quadratically} convergent algorithm for computing the invariant subspaces of an Hermitian matrix. Each iteration of the algorithm consists of {\it one} matrix-matrix multiplication and {\it one} QR decomposition. We present an accurate convergence analysis of the algorithm without using the big $O$ notation. We also propose a general framework based on implicit rational transformations which allows us to make connections with several existing algorithms and to derive classes of extensions to our basic algorithm with faster convergence rates. Several numerical examples are given which compare some aspects of the existing algorithms and the new algorithms.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8249889016151428, "perplexity": 647.5714969343879}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178350717.8/warc/CC-MAIN-20210225041034-20210225071034-00524.warc.gz"}
http://export.arxiv.org/abs/math/0409006	math (what is this?) # Title: Differential Algebra Structures on Familes of Trees Abstract: It is known that the vector space spanned by labeled rooted trees forms a Hopf algebra. Let k be a field and let R be a commutative k-algebra. Let H denote the Hopf algebra of rooted trees labeled using derivations D in Der(R). In this paper, we introduce a construction which gives R a H-module algebra structure and show this induces a differential algebra structure of H acting on R. The work here extends the notion of a R/k-bialgebra introduced by Nichols and Weisfeiler. Comments: 31 pages, 8 figures Subjects: Quantum Algebra (math.QA) Cite as: arXiv:math/0409006 [math.QA] (or arXiv:math/0409006v1 [math.QA] for this version) ## Submission history From: Robert Grossman [view email] [v1] Wed, 1 Sep 2004 02:27:47 GMT (147kb) Link back to: arXiv, form interface, contact.	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8626676797866821, "perplexity": 2594.105876497232}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496669813.71/warc/CC-MAIN-20191118182116-20191118210116-00557.warc.gz"}
https://physics.stackexchange.com/questions/528638/is-net-force-on-a-geostationary-satellite-when-observer-is-standing-just-below	Is net force on a geostationary satellite, (when observer is standing just below it on earth) equal to '0'? Since a geo stationary satellite appears to be at rest form earth, should not the net force acting on the satellite equal to $$0$$. Now since earth is a non-inertial frame we will have to apply a pseudo force on satellite to make Newtons laws of motion valid. So for the observer standing on earth, just below the geostationary satellite, two forces are acting on satellite, 1) gravitational force, $$\frac{GMm}{(R+h)^2}$$, and 2) pseudo force, $$m\omega^2R$$. (Here $$G$$ = gravitational constant, $$M$$ = mass of earth, $$m$$ = mass of satellite, $$R$$ = radius of earth, $$h$$ = height of satelite above earth surface and $$\omega$$ = angular velocity of earth.) Now for earth to be at rest these two forces should be equal in magnitude, since they are opposite in direction. Now here since we know the value of all the quantities invoved in the two forces, when I equated them they were not equal. Please guide me where I am wrong. • Please make use of this MathJax tutorial to typeset equations to make them more readable. – Sam Feb 2 at 7:02 Taking an inertial frame of reference, the satellite experiences a centripetal force towards earth which is equal to $$F=\frac {GMm}{(R+h)^2}$$. This force keeps the satellite in a fixed circular orbit around Earth. As the satellite has a fixed speed, the net force on the satellite is downwards. Now when we consider the geostationary satellite with respect to Earth's rotating frame of reference, we need to add the respective pseudo-force. As the Earth used to rotate the satellite with the force $$F$$, the add the pseudo-force $$F'$$ in the opposite direction. This centrifugal force in the opposite direction has the exact same magnitude as $$F$$ and results in net force being exactly zero. • @Sameernilkan The acceleration of the observer with respect to the satellite is $\omega^2(R+h)$ – Sam Feb 2 at 15:22 The mistake is that you should have used $$m\omega^2(R+h)$$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9577318429946899, "perplexity": 311.87284221471646}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590348513230.90/warc/CC-MAIN-20200606093706-20200606123706-00038.warc.gz"}
https://www.physicsforums.com/threads/what-formula-or-principle-governs-this-observed-phenomena.711833/	# What formula or principle governs this observed phenomena? 1. Sep 22, 2013 ### VMey Hello, Just discovered this forum as I'm so intensely curious about this question I sought out just such a place! I'm currently designing a circular patio using 6-in square blocks in concentric circles. While using a combination of Excel and Adobe Illustrator, I uncovered something unexpected. Each new circle required a consistent number of additional blocks, when rounded up to the value of a whole block. 7 to be exact. It is my goal to understand how and why this works, and hopefully to learn a formula that will let me repeat it and change things like block size. I began with a 36-inch circle which would be a firepit. I determined I would need 18.85 6-inch blocks. I'm not planning to cut them, so I rounded up to 19 blocks and backed into the circumference that would accommodate it (114 inches). I went to the next ring and more or less repeated the same process: approximated circumference, determined a fractional number of blocks, rounded up to nearest whole block, backed into precise circumference. Rinse, repeat. After about six rings, I wanted to know how many rings I'd need to get to a 16-foot diameter, so I figured I'd average the diameter increase between each consecutive ring, hoping to extrapolate an estimate. I was surprised to find each circle was the exact same increase from its adjacent circle (1.114085 ft). And that's when I noticed the consistency in the blocks. If I was aiming to have circumferences that accommodated only whole blocks, I learned that each new circle required exactly 7 additional blocks. How is this so precise and predictable? I'm assuming it has something to do with effectively turning the circle into a polygon by using blocks, since they are flat. So the first "circle" is really a 19-sided polygon, the next a 26-sided polygon, etc. But I don't know the first thing about complex polygons. Furthermore, I'd love learn HOW this works. If I had a formula of some kind, I could change the size of the blocks at will and speed up my estimates. Thanks for any help! Attached an image for reference #### Attached Files: • ###### Patio plan.png File size: 25.9 KB Views: 59 2. Sep 22, 2013 ### Simon Bridge I went a bit further ... If you have blocks which are x wide and y long, and lay them in concentric circles with the long-axis radial, then each circle will have an inner diameter 2y longer than the last. If we start, as you did, with the inner circumference rigged so a natural number, $b$, of blocks fits comfortably around it, then the inner circumference is $C_0=bx$; so the inner diameter is $D_0=bx/\pi$. The next circle out has an inner diameter of $D_1=bx/\pi +2y$ so the next circumference is $C_1=(bx/\pi +2y)\pi=bx+2\pi y$ ... See the pattern? The circumference for the nth circle will be $C_n=bx+2n\pi y$ and each circumference differs from the last by $\Delta C = 2\pi y$ which is a difference of $2\pi y/x$ blocks. If y=x then that number will be $2\pi$ which is about 6.3 ... which gives an overlap of 2/3 of a block. You could round down and have biggish gaps between bricks or round up to the next circumference, which is 7 blocks around. Does that sound like what you did? For square bricks this will leave a larger gap between circles that you may like - which can be fixed by making y slightly longer than x. You can reverse the above calculation to figure out what shape blocks to use for a circular pattern with a snug fit. 3. Sep 24, 2013 ### Erland For every new circle of blocks you add, the radius of the circle which encloses all the blocks increases with 1 unit (= the side of the square block). But then, the circumference of the circle increases with 2π units, since the circumference is 2πr. This means that you can fit 2π≈6.28 more blocks in than the previous time. Rounding of upwards gives 7. Similar Discussions: What formula or principle governs this observed phenomena?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8316053152084351, "perplexity": 910.7877622526937}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948513478.11/warc/CC-MAIN-20171211105804-20171211125804-00155.warc.gz"}
https://nalinkpithwa.com/2020/11/09/exercises-1-alfred-tarski-introduction-to-logic/	# Exercises 1: Alfred Tarski, Introduction to Logic I. Which among the following expressions are sentential functions, and which are designatory functions: a) x is divisible by 3 b) the sum of the numbers x and 2 c) $x^{2}-z^{2}$ d) $y^{2}=z^{2}$ e) $x+2< y+3$ f) $(x+3) - (y+5)$ g) the mother of x and z h) x is the mother of z? Problem 2: Give examples of sentential and designatory functions from the field of geometry. Problem 3: The sentential functions which are encountered in arithmetic and which contain only one variable (which may, however, occur at several different places in the given sentential function) can be divided into three categories : (1) functions satisfied by every number; (ii) functions not satisfied by any number; (iii) functions satisfied by some numbers, and not satisfied by others. To which of these categories do the following sentential functions belong: (a) $x+2=5+x$ (b) $x^{2}=49$ (c) $(y+2).(y-2) (d) $y+24>36$ (e) $z=0$ or $z<0$ or $z>0$ (f) $z+24>z+36$? Problem 4:Give examples of universal, absolutely existential and conditionally existential theorems from the fields of arithmetic and geometry. Problem 5: By writing quantifiers containing the variables “x” and “y” in front of the sentential function: $x>y$ it is possible to obtain various sentences from it, for instance: for any numbers x and y, $x>y$; for any number x, there exists a number y such that $x>y$; there is a number y such that, for any number x, $x>y$. Formulate them all (there are six altogether) and determine which of them are true. Problem 6: Do the same as in problem 5 for the following sentential functions: $x+y^{2}>1$ and ” x is the father of y.” (assuming that the variables x and y in the latter stand for names of human beings.) Problem 7: State a sentence of every day language that has the meaning as: For every x, if x is a dog, then x has a good sense of smell. And, your sentence must not contain any quantifier or variables. Problem 8: Replace the following sentence: “some snakes are poisonous” by one which has the same meaning but is formulated with the help of quantifiers and variables. Problem 9: Differentiate, in the following expressions, between the free and bound variables: (a) x is divisible by y. (b) for any x, $x-y = x +(-y)$ (c) if $x, then there is a number z such that $ and $y; (d) for any number y, if $y>0$, then there is a number z such that $x=y.z$ (e) if $x=y^{2}$ and $y>0$, then for any number z, $x>-z^{2}$; (f) if there exists a number y such that $x>y^{2}$, then, for any number z, $x>-z^{2}$. Formulate the above expressions by replacing the quantifiers by the symbols introduced in Section 4. Problem 10: If, in the sentential function, (e) of the preceding exercise, we replace the variable “z” in both places by “y”, we obtain an expression in which “y” occurs in some places as a free and in others as a bound variable; in what places and y? (In view of some difficulties in operating with expressions in which the same variable occurs both bound and free, some logicians prefer to avoid the use of such expressions altogether and not to treat them as sentential functions.) Problem 11: Try to state quite generally under which conditions a variable occurs at a certain place of a given sentential function as a free or as a bound variable. Problem 12: Which numbers satisfy the sentential function: there is a number y such that $x=y^{2}$, and which satisfy: there is a number y such that $x.y=1$? Cheers, Nalin Pithwa This site uses Akismet to reduce spam. Learn how your comment data is processed.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 30, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.899151623249054, "perplexity": 791.5717002521808}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038072366.31/warc/CC-MAIN-20210413122252-20210413152252-00042.warc.gz"}
https://gmatclub.com/forum/we-define-f-n-as-the-highest-non-negative-integer-power-of-7-that-105582.html	GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 24 Jun 2018, 05:33 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # We define f(n) as the highest non-negative integer power of 7 that new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Retired Moderator Joined: 02 Sep 2010 Posts: 775 Location: London We define f(n) as the highest non-negative integer power of 7 that [#permalink] ### Show Tags 01 Dec 2010, 00:30 8 38 00:00 Difficulty: 95% (hard) Question Stats: 27% (01:46) correct 73% (01:52) wrong based on 478 sessions ### HideShow timer Statistics We define f(n) as the highest non-negative integer power of 7 that divides n. What is the minimum value of k such that $$f(1)+f(2)+....+f(k)$$ is a positive multiple of 7 ? (A) 7 (B) 49 (C) 56 (D) 91 (E) 98 _________________ Manager Joined: 30 Aug 2010 Posts: 89 Location: Bangalore, India Re: We define f(n) as the highest non-negative integer power of 7 that [#permalink] ### Show Tags 01 Dec 2010, 02:03 i go with B i.e.49. As 7 is a prime #, 7 can devide multiples of 7 only but can not devide the multiples of any other # for n = 1 thru 6 the highest power of 7 should be 0 as 7^0 = 1 that evenly devides 1 thru 6 for n = 7 the highest power of 7 should be 1 as 7^1 = 7 again for n = 8 thru 13 the highest power of 7 shuld be 0 as 7^0 = 1 hence, every n that is multiple of 7, can be devided by 7 hence in f(0)+f(1)....f(k) , K has to be a multiple of 7 However it is also given that f(0)+f(1)....f(k) is a +ve multiple of 7 k when = 7 gives the SUM = 1 that is NOT a mupltiple of 7 k when 72=14 gives the SUM = 2 that is NOT a mupltiple of 7 . . . when k = 76 give the SUM = 6 that is NOT a multiple of 7 hence K should be atleast 7^2 = 49 gives the SUM = 8 and for K>49 (precisely when K = 70) we can arrive for the SUM (precisely the SUM = 14)that is a multiple of 7 min(K)=49 ASNWER "B" WHat is the OA. Regards, Murali. Kudos? Manager Joined: 01 Nov 2010 Posts: 149 Location: Zürich, Switzerland Re: We define f(n) as the highest non-negative integer power of 7 that [#permalink] ### Show Tags 01 Dec 2010, 11:06 2 4 All function f(x) values will be divisible by power of 7 as 0 except for those where x is divisible by 7. Thus, f(1)+........ +f(7) gives sum as 1 f(8)+.........+f(14) gives sum as 1+1=2 ...till ... + f(42)............ gives total sum as 6 however at f(49), 49 is 7 power 2 to give value as 2 total = 6(1)+ 2 = 8.......... going by this way 1 keeps on adding at places 56, 63, 70, 77, 84 and 91 at f(91) we reach across total sum as 6 +2 + 6 =14 which is divisible by 7 Manager Joined: 19 Aug 2010 Posts: 69 Re: We define f(n) as the highest non-negative integer power of 7 that [#permalink] ### Show Tags Updated on: 02 Dec 2010, 04:16 2 shrouded1 wrote: We define $$f(n)$$ as the highest power of 7 that divides n. What is the minimum value of k such that $$f(1)+f(2)+....+f(k)$$ is a positive multiple of 7 ? (a) 7 (b) 49 (c) 56 (d) 91 (e) 98 Everything from $$f(1)$$ until $$f(6)$$ ist $$0$$. $$f(7)$$ is the first that equals $$1$$. Therefore multiples of $$7$$ will all equal $$1$$ until we reach $$7^2=49$$ which eqquals $$2$$ . For easier understanding: 7-->1 14-->1 21-->1 28-->1 35-->1 42-->1 49-->2 and the sum of all so far is 8 $$(16+2)$$ Meaning we need 6 more multiples of 7 to reach a sum that will be divisible by 7, in the case 14 $$(14-8=6)$$ 6 more multiples of 7 means that the number we are looking for is $$49+67=91$$ Originally posted by medanova on 01 Dec 2010, 14:15. Last edited by medanova on 02 Dec 2010, 04:16, edited 1 time in total. Retired Moderator Joined: 02 Sep 2010 Posts: 775 Location: London Re: We define f(n) as the highest non-negative integer power of 7 that [#permalink] ### Show Tags 02 Dec 2010, 00:07 Correct answer is indeed (d) F(n) will be 0 for all numbers that are not multiples of 7. So the sum till k=7 is 1, k=14 is 2 and so on ... When you get to k=49, the sum will jump from 6 to 8, as 49 is divisible by 7^2 So we need another 76 or 42 numbers to get to 8+6=14 on the sum .... So answer is 49+42=91 _________________ Manager Joined: 25 Jan 2010 Posts: 106 Location: Calicut, India Re: We define f(n) as the highest non-negative integer power of 7 that [#permalink] ### Show Tags 02 Dec 2010, 04:11 Is functions a topic for GMAT? i thought GMAT never asks questions from Funvtions _________________ If u think this post is useful plz feed me with a kudo Manager Joined: 01 Nov 2010 Posts: 149 Location: Zürich, Switzerland Re: We define f(n) as the highest non-negative integer power of 7 that [#permalink] ### Show Tags 02 Dec 2010, 10:04 Quote: Is functions a topic for GMAT? i thought GMAT never asks questions from Funvtions Yes. Functions are included in the official syllabus of GMAT. Manager Joined: 30 Apr 2017 Posts: 74 Re: We define f(n) as the highest non-negative integer power of 7 that [#permalink] ### Show Tags 02 Nov 2017, 11:04 can someone please explain the question? I think I'm missing something but I don't know what! "f(n) as the highest power of 7 that divides n." means highest power of 7 which can divide on N, when it says the highest power it could be anything am I wrong? Math Expert Joined: 02 Sep 2009 Posts: 46305 Re: We define f(n) as the highest power of 7 that divides n. Wha [#permalink] ### Show Tags 02 Nov 2017, 23:57 1 1 soodia wrote: can someone please explain the question? I think I'm missing something but I don't know what! "f(n) as the highest power of 7 that divides n." means highest power of 7 which can divide on N, when it says the highest power it could be anything am I wrong? We define f(n) as the highest non-negative integer power of 7 that divides n. What is the minimum value of k such that $$f(1)+f(2)+....+f(k)$$ is a positive multiple of 7 ? (A) 7 (B) 49 (C) 56 (D) 91 (E) 98 f(n) is the highest non-negative integer power of 7 that divides n. This means that f(n) is the highest value of x (where x is a non-negative integer) for which n/7^x is an integer. Or simply put f(n) is the value of the power of 7 in prime factorisation of n. For example: If n = 14, then f(14) = 1 because the highest power of 7 that divides 14 is 1: 14 = 27^1; If n = 15, then f(14) = 0 because the highest power of 7 that divides 14 is 0: 14 = 357^0; ... You can notice that if n is NOT a multiple of 7, then f(n) = 0. The question asks to find the value of k such that $$f(1)+f(2)+....+f(k)$$ is a positive multiple of 7. As discussed all values which are not multiples of 7 will give the value of f as 0. So, we should concentrate on the multiple of 7. f(7) = 1; f(14) = 1; f(21) = 1; f(28) = 1; f(35) = 1; f(42) = 1; f(49) = 2 (because 49 = 7^2); At this point the sum is $$f(1)+f(2)+....+f(49) = 1+1+1+1+1+1+2=8$$, which is not a multiple of 7. f(56) = 1 (sum = 9); f(63) = 1 (sum = 10); f(70) = 1 (sum = 11); f(77) = 1 (sum = 12); f(84) = 1 (sum = 13); f(91) = 1 (sum = 14 = multiple of 7). So, basically we need thirteen multiples of 7, out of which twelve give the value of f as 1 and one gives the value of f as 2 (f(49) = 2): $$f(1)+f(2)+....+f(91) = 1+1+1+1+1+1+2+1+1+1+1+1+1=14$$. Hope it's clear. _________________ Manager Joined: 30 Apr 2017 Posts: 74 Re: We define f(n) as the highest non-negative integer power of 7 that [#permalink] ### Show Tags 03 Nov 2017, 14:58 Bunuel wrote: soodia wrote: can someone please explain the question? I think I'm missing something but I don't know what! "f(n) as the highest power of 7 that divides n." means highest power of 7 which can divide on N, when it says the highest power it could be anything am I wrong? We define f(n) as the highest non-negative integer power of 7 that divides n. What is the minimum value of k such that $$f(1)+f(2)+....+f(k)$$ is a positive multiple of 7 ? (A) 7 (B) 49 (C) 56 (D) 91 (E) 98 f(n) is the highest non-negative integer power of 7 that divides n. This means that f(n) is the highest value of x (where x is a non-negative integer) for which n/7^x is an integer. Or simply put f(n) is the value of the power of 7 in prime factorisation of n. For example: If n = 14, then f(14) = 1 because the highest power of 7 that divides 14 is 1: 14 = 27^1; If n = 15, then f(14) = 0 because the highest power of 7 that divides 14 is 0: 14 = 35*7^0; ... You can notice that if n is NOT a multiple of 7, then f(n) = 0. The question asks to find the value of k such that $$f(1)+f(2)+....+f(k)$$ is a positive multiple of 7. As discussed all values which are not multiples of 7 will give the value of f as 0. So, we should concentrate on the multiple of 7. f(7) = 1; f(14) = 1; f(21) = 1; f(28) = 1; f(35) = 1; f(42) = 1; f(49) = 2 (because 49 = 7^2); At this point the sum is $$f(1)+f(2)+....+f(49) = 1+1+1+1+1+1+2=8$$, which is not a multiple of 7. f(56) = 1 (sum = 9); f(63) = 1 (sum = 10); f(70) = 1 (sum = 11); f(77) = 1 (sum = 12); f(84) = 1 (sum = 13); f(91) = 1 (sum = 14 = multiple of 7). So, basically we need thirteen multiples of 7, out of which twelve give the value of f as 1 and one gives the value of f as 2 (f(49) = 2): $$f(1)+f(2)+....+f(91) = 1+1+1+1+1+1+2+1+1+1+1+1+1=14$$. Hope it's clear. wow! I don't know how should I say thank you Bunuel it was the amazing explanation I am really thank you Re: We define f(n) as the highest non-negative integer power of 7 that [#permalink] 03 Nov 2017, 14:58 Display posts from previous: Sort by # We define f(n) as the highest non-negative integer power of 7 that new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group \| Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8795398473739624, "perplexity": 1066.9767446917294}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267866937.79/warc/CC-MAIN-20180624121927-20180624141927-00339.warc.gz"}
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2231360/?tool=pubmed	• We are sorry, but NCBI web applications do not support your browser and may not function properly. More information Theor Biol Med Model. 2007; 4: 38. Published online Sep 26, 2007. PMCID: PMC2231360 # Optimization of biotechnological systems through geometric programming ## Abstract ### Background In the past, tasks of model based yield optimization in metabolic engineering were either approached with stoichiometric models or with structured nonlinear models such as S-systems or linear-logarithmic representations. These models stand out among most others, because they allow the optimization task to be converted into a linear program, for which efficient solution methods are widely available. For pathway models not in one of these formats, an Indirect Optimization Method (IOM) was developed where the original model is sequentially represented as an S-system model, optimized in this format with linear programming methods, reinterpreted in the initial model form, and further optimized as necessary. ### Results A new method is proposed for this task. We show here that the model format of a Generalized Mass Action (GMA) system may be optimized very efficiently with techniques of geometric programming. We briefly review the basics of GMA systems and of geometric programming, demonstrate how the latter may be applied to the former, and illustrate the combined method with a didactic problem and two examples based on models of real systems. The first is a relatively small yet representative model of the anaerobic fermentation pathway in S. cerevisiae, while the second describes the dynamics of the tryptophan operon in E. coli. Both models have previously been used for benchmarking purposes, thus facilitating comparisons with the proposed new method. In these comparisons, the geometric programming method was found to be equal or better than the earlier methods in terms of successful identification of optima and efficiency. ### Conclusion GMA systems are of importance, because they contain stoichiometric, mass action and S-systems as special cases, along with many other models. Furthermore, it was previously shown that algebraic equivalence transformations of variables are sufficient to convert virtually any types of dynamical models into the GMA form. Thus, efficient methods for optimizing GMA systems have multifold appeal. ## Background Model based optimization of biotechnological processes is a key step towards the establishment of rational strategies for yield improvement, be it through genetic engineering, refined setting of operating conditions or both. As such, it is a key element in the rapidly emerging field of metabolic engineering [1,2]. Optimization tasks involving living organisms are notoriously difficult, because they almost always involve large numbers of variables, representing biological components that dominate cell operation, and must account for multitudinous and complex nonlinear interactions among them [3]. The steady increase in the ready availability of computing power has somewhat alleviated the challenge, but it has also, together with other technological breakthroughs, been raising the level of expectation. Specifically, modelers are more and more expected to account for complex biological details and to include variables of diverse types and origins (metabolites, RNA, proteins...). This trend is to be welcomed, because it promises improved model predictions, yet it easily compensates for the computer technological advances and often overwhelms available hardware and software methods. As a remedy, effort has been expanded to develop computationally efficient algorithms that scale well with the growing number of variables in typical optimization tasks. The most straightforward attempts toward improved efficiency have been based, in one form or another, on the reduction of the originally nonlinear task to linearity, because linear optimization tasks are rather easily solved, even if they involve thousands of variables. One variant of this approach is the optimization of stoichiometric flux distribution models [4]. The two great advantages of this method are that the models are linear and that minimal information is needed to implement them, namely flux rates, and potentially numerical values characterizing metabolic or physico-chemical constraints. The significant disadvantage is that no regulation can be considered in these models. An alternative is the use of S-system models within the modeling framework of Biochemical Systems Theory [5-7]. These models are highly nonlinear, thus allowing suitable representations of regulatory features, but have linear steady-state equations, so that optimization under steady-state conditions again becomes a matter of linear programming [8]. The disadvantages here are that much more (kinetic) information is needed to set up numerical models and that S-systems are based on approximations that are not always accepted as valid. Linear-logarithmic models [9] similarly have the advantage of linearity at steady state and the disadvantage of being a local approximation. An extension of these linear approaches is the Indirect Optimization Method [10]. In this method, any type of kinetic model is locally represented as an S-system. This S-system is optimized with linear methods, and the resulting optimized parameter settings are translated back into the original model. If necessary, this linearized optimization may be executed in sequential steps. An alternative to using S-system models is the General Mass Action (GMA) representation within BST. GMA systems are very interesting for several reasons. First, they contain both stoichiometric and S-system models as direct special cases, which would allow the optimization of combinations of the two. Second, mass action systems are special cases of GMA models, so that, in some sense, Michaelis-Menten functions and other kinetic rate laws are special cases, if they are expressed in their elemental, non-approximated form. Third, it was shown that virtually any system of differential equations may be represented exactly as a GMA system, upon equivalence transformations of some of the functions in the original system. Thus, GMA systems, as a mathematical representation, are capable of capturing any differentiable nonlinearity that one might encounter in biological systems. We show here that GMA systems, while highly nonlinear, are structured enough to permit the application of efficient optimization methods based on geometric programming. ### Formulation of the optimization task Pertinent optimization problems in metabolic engineering can be stated as the targeted manipulation of a system in the following way: max or min f0(X) subject to: metabolic and physico-chemical constraints cell viability In this generic representation, (1) usually targets a flux or a yield. The optimization must occur under several constraints. The first set (2) ensures that the system will operate under steady-state conditions. Other constraints (3) are imposed to retain the system within a physically and chemically feasible state and so that the total protein or metabolite levels do not impede cell growth. Yet other constraints (4) guarantee that no metabolites are depleted below minimal required levels or accumulate to toxic concentrations. These sets of constraints are designed to allow sustained operation of the system. ### Biochemical Systems Theory (BST) Biological processes are usually modeled as systems of differential equations in which the variation in metabolites X is represented as: $dXdt=N⋅v MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaadaWcaaqaaiabdsgaKHqabiab=HfaybqaaiabdsgaKjabdsha0baacqGH9aqpcqWGobGtcqGHflY1cqWF2bGDaaa@37F4@$ The elements ni,j of the stoichiometric matrix N are constant. The vector v contains reaction rates, which are in general functions of the variables and parameters of the system. This structure is usually associated with metabolic systems, but it is similarly valid for models describing gene expression, bioreactors, and a wide variety of other processes in biotechnology. In typical stoichiometric analyses, the reaction rates are considered constant. Furthermore, the analysis is restricted to steady-state operation, with the consequence that (5) is set equal to 0 and thereby becomes a set of linear algebraic equations, which are amenable to a huge repertoire of analyses. In analyses accounting for regulation, the reaction rates become functions that depend on system variables and outside influences. Even at steady state, these may be very complex, thereby rendering direct analysis of the system a formidable task [11]. As a remedy, BST suggests to represent these rate functions with power laws: $vi=γi∏j=1n+mXjfi,j MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaacqWG2bGDdaWgaaWcbaGaemyAaKgabeaakiabg2da9GGaciab=n7aNnaaBaaaleaacqWGPbqAaeqaaOWaaebCaeaacqWGybawdaqhaaWcbaGaemOAaOgabaGaemOzay2aaSbaaWqaaiabdMgaPjabcYcaSiabdQgaQbqabaaaaaWcbaGaemOAaOMaeyypa0JaeGymaedabaGaemOBa4Maey4kaSIaemyBa0ganiabg+Givdaaaa@4502@$ In analogy with chemical kinetics, γi is called the rate constant and fi,j are kinetic orders, which may be any real numbers. Positive kinetic orders indicate augmentation, whereas negative values are indicative of inhibition. Kinetic orders of 0 result in automatic removal of the corresponding variable from the term. In the notation of BST, the first n variables are often considered the dependent variables, which change dynamically under the action of the system, while the remaining variables Xi for i = n + 1 ... m + n are considered independent variables and typically remain constant throughout any given simulation study. Thus, metabolites, enzymes, membrane potentials or other system components can easily be made dependent or independent by the modeler without requiring alterations in the structure of the equations. BST is very compact and explicitly distinguishes variables from parameters. Because we will later introduce concepts of geometric programming, it is noted that the power-law term in Eq. 6 is also called a monomial. If this monomial is an approximation of reaction rate V, its parameters can be directly related to V, by virtue of the fact that the monomial is in fact a Taylor linearization in logarithmic space [12]. Thus, choosing an operating point with index 0, one obtains: $ln⁡vi=ln⁡V0+\|∂ln⁡V∂ln⁡X1\|0(ln⁡X1−\|ln⁡X1\|0)+⋯+\|∂ln⁡V∂ln⁡Xm+n\|0(ln⁡Xm+n−\|ln⁡Xm+n\|0) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=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@87EF@$ Thus, it follows directly from 7 that the parameters of a power-law (monomial) term can be computed as $γi=\|vi\|0\|∏j=1n+mXjfi,j\|0 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaaiiGacqWFZoWzdaWgaaWcbaGaemyAaKgabeaakiabg2da9maalaaabaWaaqWaaeaacqWG2bGDdaWgaaWcbaGaemyAaKgabeaaaOGaay5bSlaawIa7amaaBaaaleaacqaIWaamaeqaaaGcbaWaaqWaaeaadaqeWaqaaiabdIfaynaaDaaaleaacqWGQbGAaeaacqWGMbGzdaWgaaadbaGaemyAaKMaeiilaWIaemOAaOgabeaaaaaaleaacqWGQbGAcqGH9aqpcqaIXaqmaeaacqWGUbGBcqGHRaWkcqWGTbqBa0Gaey4dIunaaOGaay5bSlaawIa7amaaBaaaleaacqaIWaamaeqaaaaaaaa@4D5E@$ $fi,j=\|∂ln⁡vi∂ln⁡Xj\|0=\|∂vi∂XjXjvi\|0 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=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@5927@$ System equations in BST may be designed in slightly different ways. For the GMA form, each reaction is represented by its own monomial, and the result is therefore $dXidt=∑j=1pni,jγj∏k=1n+mXkfj,ki=1...n MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=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@5C92@$ Note that this is actually a spelled-out version of Eq. 5, where the reaction rates are monomials as in Eq. 6. As an alternative to the GMA format, one may, for each dependent variable, collect all incoming reactions in one term $Vi+ MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaacqWGwbGvdaqhaaWcbaGaemyAaKgabaGaey4kaScaaaaa@304B@$ and do the same with all outgoing fluxes, which are collectively called $Vi− MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaacqWGwbGvdaqhaaWcbaGaemyAaKgabaGaeyOeI0caaaaa@3056@$. These aggregated terms are now represented as monomials, and the result is $dXidt=Vi+−Vi−=αi∏j=1n+mXjgi,j−βi∏j=1n+mXjhi,j MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaadaWcaaqaaiabdsgaKjabdIfaynaaBaaaleaacqWGPbqAaeqaaaGcbaGaemizaqMaemiDaqhaaiabg2da9iabdAfawnaaDaaaleaacqWGPbqAaeaacqGHRaWkaaGccqGHsislcqWGwbGvdaqhaaWcbaGaemyAaKgabaGaeyOeI0caaOGaeyypa0dcciGae8xSde2aaSbaaSqaaiabdMgaPbqabaGcdaqeWbqaaiabdIfaynaaDaaaleaacqWGQbGAaeaacqWGNbWzdaWgaaadbaGaemyAaKMaeiilaWIaemOAaOgabeaaaaaaleaacqWGQbGAcqGH9aqpcqaIXaqmaeaacqWGUbGBcqGHRaWkcqWGTbqBa0Gaey4dIunakiabgkHiTiab=j7aInaaBaaaleaacqWGPbqAaeqaaOWaaebCaeaacqWGybawdaqhaaWcbaGaemOAaOgabaGaemiAaG2aaSbaaWqaaiabdMgaPjabcYcaSiabdQgaQbqabaaaaaWcbaGaemOAaOMaeyypa0JaeGymaedabaGaemOBa4Maey4kaSIaemyBa0ganiabg+Givdaaaa@6766@$ Thus, there are at most one positive and one negative term in each S-system equation. The conversion of a GMA into an S-system will become important later. It is achieved by collecting the aggregated fluxes into vectors $V+=N+vV−=N−v MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaafaqabeGabaaabaacbeGae8Nvay1aaWbaaSqabeaacqGHRaWkaaGccqGH9aqpcqWGobGtdaahaaWcbeqaaiabgUcaRaaakiab=zha2bqaaiab=zfawnaaCaaaleqabaGaeyOeI0caaOGaeyypa0JaemOta40aaWbaaSqabeaacqGHsislaaGccqWF2bGDaaaaaa@3AD7@$ where N+ and N- are matrices containing respectively the positive and negative coefficients of N such that N = N+ - N-. With these definitions, we can derive the matrices of kinetic orders of S-systems from those of the corresponding GMA representation. Namely, $G=(V+)−1N+VFH=(V−)−1N−VF MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaafaqadeGabaaabaGaem4raCKaeyypa0JaeiikaGIaemOvay1aaWbaaSqabeaacqGHRaWkaaGccqGGPaqkdaahaaWcbeqaaiabgkHiTiabigdaXaaakiabd6eaonaaCaaaleqabaGaey4kaScaaGqabOGae8NvayLaemOrayeabaGaemisaGKaeyypa0JaeiikaGIaemOvay1aaWbaaSqabeaacqGHsislaaGccqGGPaqkdaahaaWcbeqaaiabgkHiTiabigdaXaaakiabd6eaonaaCaaaleqabaGaeyOeI0caaOGae8NvayLaemOrayeaaaaa@4647@$ where V, V+ and V- are square matrices of zeros having the corresponding vectors as their main diagonals. G and H contain the kinetic orders of the S-system while F contains those of the GMA [13]. GMA systems may be constructed in three manners [11]. First, given a pathway diagram, each reaction rate is represented by a monomial, and equations are assembled from all reaction rates involved. Second, it is possible (though not often actually done) to dissect enzyme catalyzed reactions into their underlying mass action kinetics, without evoking the typical quasi-steady-state assumption. The result is directly the special case of a GMA system where most kinetic orders are zero, one, or in some cases 2. Third, it has been shown that virtually any nonlinearity can be represented equivalently as a GMA system [14]. As an example for this recasting technique, consider a simple equation where production and degradation are formulated as traditional Michaelis-Menten rate laws: where X0 is a dependent or independent variable describing the substrate for the generation of X1. To effect the transformation into a GMA equation, define auxiliary variables as X2 = KM,2 + X1 and X3 = KM,1 + X0. The equation then becomes $dX1dt=Vmax,1X0X3−1−Vmax,2X1X2−1 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaadaWcaaqaaiabdsgaKjabdIfaynaaBaaaleaacqaIXaqmaeqaaaGcbaGaemizaqMaemiDaqhaaiabg2da9iabdAfawnaaBaaaleaacqWGTbqBcqWGHbqycqWG4baEcqGGSaalcqaIXaqmaeqaaOGaemiwaG1aaSbaaSqaaiabicdaWaqabaGccqWGybawdaqhaaWcbaGaeG4mamdabaGaeyOeI0IaeGymaedaaOGaeyOeI0IaemOvay1aaSbaaSqaaiabd2gaTjabdggaHjabdIha4jabcYcaSiabikdaYaqabaGccqWGybawdaWgaaWcbaGaeGymaedabeaakiabdIfaynaaDaaaleaacqaIYaGmaeaacqGHsislcqaIXaqmaaaaaa@5119@$ For simplicity of discussion, suppose that X0 is a constant, independent variable. Thus, X3 is also constant and does not need its own equation. By contrast, X2 is a new dependent variable and from its definition we can calculate its initial value and see that its derivative must be equal to that of X1. Therefore the equations: $dX1dt=Vmax,1X0X3−1−Vmax,2X1X2−1dX2dt=Vmax,1X0X3−1−Vmax,2X1X2−1X1(t0)=X10X2(t0)=KM,2+X10 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=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@90C7@$ form a system that is an exact equivalent of the original system but in GMA format. Recasting can be useful with equations that are difficult to handle otherwise or for purposes of streamlining a model structure and its analysis. One must note though that often the number of variables increases significantly. In the case shown, the number of equations rises from one to two if X0 is independent or to three if it is a dependent variable. ### Current optimization methods based on BST The overall task is to reset some of the independent variables so that some objective is optimized. The independent variables in question are typically enzyme activities, which are experimentally manipulated through genetic means, such as the application of customized promoters or plasmids. The objective is usually the maximization of a metabolite concentration or a flux. Three approaches have been proposed in the literature. #### Pure S-systems Among a number of convenient properties, the steady states of an S-system can be computed analytically by solving a system of algebraic linear equation [6]. Equating Eq. 11 to zero and rearranging one obtains: $αi∏j=1nXjgi,jβi∏j=1nXjhi,j=1 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaadaWcaaqaaGGaciab=f7aHnaaBaaaleaacqWGPbqAaeqaaOWaaebmaeaacqWGybawdaqhaaWcbaGaemOAaOgabaGaem4zaC2aaSbaaWqaaiabdMgaPjabcYcaSiabdQgaQbqabaaaaaWcbaGaemOAaOMaeyypa0JaeGymaedabaGaemOBa4ganiabg+GivdaakeaacqWFYoGydaWgaaWcbaGaemyAaKgabeaakmaaradabaGaemiwaG1aa0baaSqaaiabdQgaQbqaaiabdIgaOnaaBaaameaacqWGPbqAcqGGSaalcqWGQbGAaeqaaaaaaSqaaiabdQgaQjabg2da9iabigdaXaqaaiabd6gaUbqdcqGHpis1aaaakiabg2da9iabigdaXaaa@523C@$ which is a monomial of the form $αiβi∏j=1nXjgi,j−hi,j=1. MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaadaWcaaqaaGGaciab=f7aHnaaBaaaleaacqWGPbqAaeqaaaGcbaGae8NSdi2aaSbaaSqaaiabdMgaPbqabaaaaOWaaebCaeaacqWGybawdaqhaaWcbaGaemOAaOgabaGaem4zaC2aaSbaaWqaaiabdMgaPjabcYcaSiabdQgaQbqabaWccqGHsislcqWGObaAdaWgaaadbaGaemyAaKMaeiilaWIaemOAaOgabeaaaaaaleaacqWGQbGAcqGH9aqpcqaIXaqmaeaacqWGUbGBa0Gaey4dIunakiabg2da9iabigdaXiabc6caUaaa@4AE2@$ Monomial equations become linear by taking logarithms on both sides thus reducing the steady-state computation to a linear task: y = b where Ai,j = gi,j - hi,j yi = In Xi $bi=ln⁡βiαi MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaacqWGIbGydaWgaaWcbaGaemyAaKgabeaakiabg2da9iGbcYgaSjabc6gaUnaalaaabaacciGae8NSdi2aaSbaaSqaaiabdMgaPbqabaaakeaacqWFXoqydaWgaaWcbaGaemyAaKgabeaaaaaaaa@39C0@$ Monomial objective functions become linear by taking logarithms and so holds for many constraints on metabolites or fluxes. Therefore, constrained optimization of pathways modeled as S-systems becomes a straightforward linear program [8]. Any other relevant constraint or objective function that is not a power law can also be approximated using the abovementioned methods. Then logarithms can be taken and Eqns 1–4 can be rewritten as: max or min F(y) Subject to: A·y = b B·y = d C·y e yL y yU Where F is the logarithm of the flux or variable to be optimized, and superscripts L and U refer to lower and upper bounds. Eq. 20 assures operation at steady state. Matrix B and vector d account for additional equality constraints and C and e are analogous constraints for additional inequalities, which could, for instance, limit the magnitude of a metabolite concentration or flux, and improve the chances of viability. Optimization problems of this type are called linear programs (LPs) and can be solved very efficiently for large numbers of variables and constraints [15]. The advantage of the pure S-system approach is its great speed combined with the fact that S-system models have proven to be excellent representations of many pathways. The disadvantage is that the optimization process, by design, moves the system away from the chosen operating point, so that questions arise as to how accurate the S-system representation is at the steady state suggested by the optimization. #### Indirect Optimization Method If the pathway is not modeled as an S-system, the reduction of the optimization task to linearity is jeopardized. A compromise solution that has turned out to be quite effective is the Indirect Optimization Method (IOM) [10]. The first step of IOM is approximation of the alleged model with an S-system. This S-system is optimized as shown above. The solution is then translated back into the original system in order to confirm that it constitutes a stable steady state and is really an improvement from the basal state of the original model. The S-system solution typically differs somewhat from a direct optimization result with the original model, but since it is obtained so fast, it is possible to execute IOM in several steps with relatively tight bounds, every time choosing a new operating point and not deviating too much from this point in the next iteration [16]. The speed of the process is slower than in the pure S-system case, but still reasonable. Variations on IOM are to search for subsets of independent variables to be manipulated for optimal yield at lower cost and for multi-objective optimization tasks [17,18]. #### Global GMA optimization A global optimization method for GMA systems [19] has been recently proposed based on branch-and-reduce methods combined with convexification. These methods are interesting because of the variety of roles that GMA models can play (see above). The disadvantage of the global method is that it quickly leads to very large systems that are non-convex, even though they allow relatively efficient solutions. ### Geometric programming Geometric programming (GP) [20] addresses a class of problems that include linear programming (LP) and other tasks within the broader category of convex optimization problems. Convex problems are among the few nonlinear tasks where, thanks to powerful interior point methods, the efficient determination of global optima is feasible even for large scale systems. For example, a geometric program of 1,000 variables and 10,000 constraints can be solved in less than a minute on a desktop computer [21]; the solution is even faster for sparse problems as they are found in metabolic engineering. Furthermore, easy to use solvers are starting to become available [22,23]. GP addresses optimization programs where the objective function and the constraints are sums of monomials, i.e., power-law terms as shown in Eq. 6. Because of their importance in GP, sums of monomials, all with positive sign, are called posynomials. If some of the monomials enter the sum with negative signs, the collection is called a signomial. The peculiarities of convexity and GP methods render the difference between posynomials and signomials crucial. A GP problem has the generic form: min P0(x) Subject to: Pi(x) ≤ 1 i = 1...n Mi(x) = 1 i = 1...p where Pi(x) and Mi(x) must fulfill strict conditions. Every function Mi(x) must be a monomial, while the objective function P0(x) and the functions Pi(x) involved in inequalities must be posynomials. Signomials are not permitted, and optimization problems involving them require additional effort. The equivalence between monomials and power laws immediately suggests the potential use of GP for optimization problems formulated within BST. In the next sections, several methods will be proposed to develop such potential. ## Results and discussion It is easy to see that steady-state equations of S-systems are readily arranged as monomials as shown in Eq 18 and that optimization tasks for S-systems directly adhere to the format of a GP, except that GP mandates minimization. However, this is easily remedied for maximization tasks by minimizing the inverse of the objective, which again is a monomial. By contrast, steady-state GMA equations as shown in Eq. 10 do not automatically fall within the GP structure, because GMA systems usually include negative terms, thus making them signomials. Furthermore, inversion of an objective that contains more than one monomial is not equivalent to a monomial. When the objective or some restriction falls outside the GMA formalism, it can be recast into proper form as has been discussed above and will be shown in one of the case studies. ### Two strategies The proposed solutions for adapting GP solvers to treat GMA systems rely on condensation [24], but they do it in different ways. Condensation is a standard procedure in GP which is exactly equivalent to aggregation in BST. Namely, the sum of monomials is approximated by a single monomial. In the terminology of GP, the condensation $C^() MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaacuWGdbWqgaqcaiabcIcaOiabcMcaPaaa@2F7D@$ is generically denoted as $C^(P(x)=C^(M1(x)+⋯+Mn(x))=M0(x) MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaacuWGdbWqgaqcaiabcIcaOiabdcfaqjabcIcaOGqabiab=Hha4jabcMcaPiabg2da9iqbdoeadzaajaGaeiikaGIaemyta00aaSbaaSqaaiabigdaXaqabaGccqGGOaakcqWF4baEcqGGPaqkcqGHRaWkcqWIVlctcqGHRaWkcqWGnbqtdaWgaaWcbaGaemOBa4gabeaakiabcIcaOiab=Hha4jabcMcaPiabcMcaPiabg2da9iabd2eannaaBaaaleaacqaIWaamaeqaaOGaeiikaGIae8hEaGNaeiykaKcaaa@4C4F@$ and, in the terminology of Eqs. 10 and 11, defined as: $C^(∑j=1kni,jγj∏k=1n+mXkfj,k)=αi∏j=1nXjgi,j MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaacuWGdbWqgaqcamaabmaabaWaaabCaeaacqWGUbGBdaWgaaWcbaGaemyAaKMaeiilaWIaemOAaOgabeaaiiGakiab=n7aNnaaBaaaleaacqWGQbGAaeqaaaqaaiabdQgaQjabg2da9iabigdaXaqaaiabdUgaRbqdcqGHris5aOWaaebCaeaacqWGybawdaqhaaWcbaGaem4AaSgabaGaemOzay2aaSbaaWqaaiabdQgaQjabcYcaSiabdUgaRbqabaaaaaWcbaGaem4AaSMaeyypa0JaeGymaedabaGaemOBa4Maey4kaSIaemyBa0ganiabg+GivdaakiaawIcacaGLPaaacqGH9aqpcqWFXoqydaWgaaWcbaGaemyAaKgabeaakmaarahabaGaemiwaG1aa0baaSqaaiabdQgaQbqaaiabdEgaNnaaBaaameaacqWGPbqAcqGGSaalcqWGQbGAaeqaaaaaaSqaaiabdQgaQjabg2da9iabigdaXaqaaiabd6gaUbqdcqGHpis1aaaa@62C7@$ where αi and gi,j are chosen such that equality holds at a chosen operating point; thus, the result is equivalent to the Taylor linearization that is fundamental in BST as was shown in eqn. 7 [5,7,12]. As in the Taylor series, the condensed form is equal to the original equation at the operating point. For any other point, as it can be shown that the left and right hand side of eqn. 29 are equivalent to those of the Arithmetic-Geometric inequality: $∑i=1nai≥∏i=1n(aiwi)wi MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaadaaeWbqaaiabdggaHnaaBaaaleaacqWGPbqAaeqaaOGaeyyzIm7aaebCaeaadaqadaqaamaalaaabaGaemyyae2aaSbaaSqaaiabdMgaPbqabaaakeaacqWG3bWDdaWgaaWcbaGaemyAaKgabeaaaaaakiaawIcacaGLPaaadaahaaWcbeqaaiabdEha3naaBaaameaacqWGPbqAaeqaaaaaaSqaaiabdMgaPjabg2da9iabigdaXaqaaiabd6gaUbqdcqGHpis1aaWcbaGaemyAaKMaeyypa0JaeGymaedabaGaemOBa4ganiabggHiLdaaaa@49D6@$ and therefore, the condensed form is an understimation of the original. Objective functions can only be minimized in GP, this is seldom a problem given that the functions to maximize are often monomials that can be inverted: a variable, a reaction rate or a flux ratio. Posynomial objectives are usually entitled for minimization, like the sum of certain variables. Nonetheless, it is also relevant in metabolic engineering to consider the maximization of posynomials, such as the sum of variables or fluxes. In such cases, condensation or recasting can be used. For en extensive introduction on GP modelling see [25]. #### A local approach: Controlled Error Method The steady-state equation of a GMA system may be written as the single difference of two posynomials: P(x) - Q(x) = 0 If both posynomials are condensed, every equation will be reduced to the standard form for monomial equations: $C^(P(x))C^(Q(x))=1 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaadaWcaaqaaiqbdoeadzaajaGaeiikaGIaemiuaaLaeiikaGccbeGae8hEaGNaeiykaKIaeiykaKcabaGafm4qamKbaKaacqGGOaakcqWGrbqucqGGOaakcqWF4baEcqGGPaqkcqGGPaqkaaGaeyypa0JaeGymaedaaa@3D00@$ Because the division of a monomial by another is itself a monomial. Since the steady state equations of the GMA have been condensed to those of an s-system, this method could be regarded as a direct generalization of classical IOM methods. One of the advantages of this approach is the possibility of keeping posynomial inequalities and objectives as they are and therefore reduce the amount of condensation (approximation) needed, but there is another interesting possibility. When a posynomial is approximated by condensation, the A-G inequality, Eq. 30, guarantees that the monomial is an underestimation of the constraint. Furthermore, the posynomial structure is not altered when divided by a monomial so the quotient between a posynomial and its condensed form is always greater than or equal to 1 and provides the exact error as a posynomial function. Therefore the problem can be constrained to allow a maximum error per condensed constraint: $∑jδj∏kXkbj,kC^(∑jδj∏kXkbj,k)≤1+ε MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=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@57AB@$ So the original problem is solved as a series of GPs in which the GMA equations are successively condensed using the previous solution as the reference point. To assure validity an extra set of constraints is added to ensure that every iteration will only explore the neighborhood of the feasible region in which error due to condensation remains below an arbitrary tolerance set by the user. #### A global approach: Penalty Treatment A similar yet distinct strategy that minimizes the use of condensation is an extension of the penalty treatment method [26], a classic algorithm for signomial programming. In this method, a signomial constraint such as P(x) - Q(x) = 0 where P and Q are posynomials, is replaced by two posynomial equalities through the creation of an ancilliary variable t: $P(x)=tQ(x)=t MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaafaqabeGabaaabaGaemiuaaLaeiikaGccbeGae8hEaGNaeiykaKIaeyypa0JaemiDaqhabaGaemyuaeLaeiikaGIae8hEaGNaeiykaKIaeyypa0JaemiDaqhaaaaa@3A53@$ These are not valid GP constraints, so the following relaxed version is used: $P(x)≤tQ(x)≤t MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=vr0dc8meaabaqaciaacaGaaeqabaqabeGadaaakeaafaqabeGabaaabaGaemiuaaLaeiikaGccbeGae8hEaGNaeiykaKIaeyizImQaemiDaqhabaGaemyuaeLaeiikaGIae8hEaGNaeiykaKIaeyizImQaemiDaqhaaaaa@3BB1@$ Upon dividing by t, the feasible area of the original problem is contained in the feasible area of the new relaxed version and aproximation by condensation is not needed. In order to force these inequalities to be tight in the final solution, the objective function is augmented with penalty terms that grow with the slackness of the constraints, namely the inverses of the condensation of the relaxed constraints. The result of this procedure is a legal GP: Where the condensed terms are calculated at the basal steady state. If the obtained solution falls within the feasible area of the original problem, it is taken as a solution, if it does not (any of the relaxed inequalities is below 1, the solution is used as the next reference point: condensations are calculated again, the weights of the violated constraints are increased and the new problem is solved. This procedure is repeated until a satisfactory solution is obtained. The original method used 1 as the initial value of the weights and increased them all in every iteration, some modifications are useful for our purposes: • The initial weights are selected such that the overall penalty terms are just a fraction of the total objective in the initial point. In the case studies explored in this paper, such fraction was 10%. • The weights are only increased if their corresponding constraint was violated in the last iteration. In such cases, the weight would be multiplied times a fixed value. For the case studies considered here, the choice in the value of such multiplier didn't have a significant impact in the performance of the method. These variations on the original method serve to prevent the penalty terms from dominating the objective function and pushing the relaxed problem towards the boundaries of the feasible region from the very beginning. ### Case studies In order to illustrate the combination of GP with BST, some optimization tasks were explored. The first example demonstrates the procedure with a very simple two variable GMA system. The second example is a model of the anaerobic fermentation pathway in Saccharomyces cerevisiae. The third example revisits an earlier case study concerned with the tryptophan operon in E. coli. These systems were optimized using the Matlab based solver ggplab [23] running on an ordinary laptop (1.6 GHz Pentium centrino, 512 Mb RAM). Matlab scripts were written in order to perform all the transformations required by the two methods described. For comparison, the models were also optimized using IOM [10] as well as Matlab's optimization toolbox. The function used in this toolbox, fmincon(), is based on an iterative algorithm called Sequential Quadratic Programming, which uses the BGFS formula to update the estimated Hessian matrix during every iteration [27,28]. #### A seemingly simple problem A very distinctive difference between the alternative methodsfor GMA optimization can be ilustrated by a problem modified from [24], which presents the simplest possible fragmented feasible region (see Fig. Fig.11). Feasible area of the first example. The lines show the nullclines of each of the two equations of the system. They intersect at two (unconnected) points, which constitute the only feasible solutions. The feasible area of the relaxed problem in the penalty ... The feasible region of this problem consists of two points (1.178,2.178) and (3.823,4.823), of which clearly the first solution is superior, because X1 is to be minimized. As these points are not connected, local methods are not able to find one solution using the other as a starting point. The problem was solved using IOM, controlled error and penalty treatment methods. The initial point was set to be (3.823,4.823), which is disconnected from the true optimal solution. While both IOM and the Controlled-Error method reported the initial point as the solution, the penalty treatment algorithm found the global optimum at (1.178,2.178). In this case, most methods failed to find the optimal solution because the approximated s-system had the operating point as the only feasible solution while the relaxed problem for the penalty treatment algorithm had a feasible area (shadowed in Fig. Fig.1)1) that included and connected both feasible solutions. #### Anaerobic fermentation in S. cerevisiae This GMA model [29] (see also appendix) is derived from a previous version [30] formulated with traditional Michaelis Mentem kinetics to explain experimental data, and has been used to illustrate other optimization methods [10,17,19]. It has the following structure (see Fig. Fig.22): Anaerobic fermentation in S. cerevisiae. $X˙1=vin−vHKX˙2=vHK−vPFK=vPOLX˙3=vPFK−vGAPD−12vGOLX˙4=2⋅vGAPD−vPKX˙5=2⋅vGAPD+vPK−vHK−vPFK−vPOL−vATP MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=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@9E11@$ The model was already formulated [29] as a GMA system, so that all its fluxes are monomials: $vin=0.8122X2−0.2344X6vHK=2.8632X10.7464X50.0243X7vPFK=0.5232X20.7318X5−0.3941X8vGAPD=0.011X30.6159X40.1308X9X14−0.6088vPK=0.0945X30.05X40.533X5−0.0822X10vPOL=0.0009X28.6107X11vGOL=0.0945X30.05X40.533X5−0.0822X12vATP=X5X13 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=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@1401@$ The objective is (constrained) maximization of the ethanol production rate, vPK. Together with the upper and lower bounds of the variables, two extra constraints will be studied. The first is an upper limit to the total amount of protein. This is especially important for pathways of the central carbon metabolism as they represent a significant fraction of the total amount of cell protein and increasing the expression of its enzymes by large amounts might compromise cell viability. As a first example, we assume that the activity to protein ratio is the same for every enzyme and set an arbitrary limit of four times the amount of enzymes in the basal state. As an alternative, we explore the effect of limiting the total substrate pool. This constraint will later be subject to tradeoff analysis in order to see its influence in the optimum steady state (see Fig Fig3).3). Being posynomial functions, the constraints will be supported by GP without any transformation. The Appendix contains a complete formulation of the optimization problem. Tradeoff curve for the anaerobic fermentation pathway if the total substrate pools are kept fixed. No upper limit for total enzyme was used in this case. The results are sumarized in Table Table1.1. Both GP methods and the SQP found the same solution, although GP finished in 0.5 s while SQP was significantly slower, taking 1.5 s for the calculation. The IOM method was as fast as GP but it's solution violated one constraint. Optimization results for the GMA glycolitic model in S. cerevisiae. Constraint violations are shown in boldface. GP column stands for both methods #### Tryptophan operon The third example addresses the tryptophan operon in E. coli, as illustrated in Fig. Fig.4.4. This is an appealing benchmark system, because it has already been optimized with other methods [16,31]. A model of the tryptophan operon. Adapted from [32]. A model of the system was recently presented by [32] and includes transcription, translation, chemical reactions and tryptophan consumption for growth. It is thus more than a simple pathway model and demonstrates that GP and BST are applicable in more complex contexts. Finally, this model doesn't follow the structure of any standard formalism so it will be a good example on how recasting widens the applicability of the method to a higher degree of generality. The model takes the form $X˙1=v1−v2X˙2=v3−v4X˙3=v5−v6−v7−v8 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=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@50B4@$ Here X1, X2 and X3 are dimensionless quantities representing mRNA, enzyme levels and the tryptophan concentration, respectively. The rate equations are: $v1=X3+11+(1+X5)X3v2=(0.9+X4)X1v3=X1v4=(0.02+X4)X2v5=X2X62X62+X32v6=X3X4v7=0.0022X3X51+X3v8=(1−7.5X4)X4X3X7X3+0.005 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=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@A8C5@$ The GMA format is obtained by defining the following ancillary variables: $X8=1+X5X9=X3+1X10=1+X8X3X11=0.9+X4X12=0.02+X4X13=X62+X32X14=X3+0.005X15=1−7.5X4 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=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@8070@$ which turns the rates into power laws: $v1=X9X10−1v2=X11X1v3=X1v4=X12X2v5=X2X62X13−1v6=X3X4v7=0.0022X3X5X9−1v8=X15X3X7X14−1 MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfKttLearuWrP9MDH5MBPbIqV92AaeXatLxBI9gBaebbnrfifHhDYfgasaacH8akY=wiFfYdH8Gipec8Eeeu0xXdbba9frFj0=OqFfea0dXdd9vqai=hGuQ8kuc9pgc9s8qqaq=dirpe0xb9q8qiLsFr0=vr0=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@8B0D@$ The objective function consists simply of v8, which may be regarded as an aggregate term for growth and tryptophan excretion. A recurrent feature of previously found IOM solutions was the noticeable violation of a constraint retaining a minimum tryptophan concentration. This discrepancy is a feature for comparisons between methods beyond computational efficiency. The Appendix contains a complete formulation of the optimization problem. In order to test the effectiveness of the controlled error approach, two variants were used in this model: • Fixed tolerance. The standard method in which every iteration is limited to a maximum condensation error of 10% by constraints described in Eq. 33. • Fixed step. No limit on the condensation error. The variation of the variables in every iteration is limited to 10% distance from the reference state. When the constraints were absent (fixed step), the variation of the variables was restricted to a fraction of the total range in every iteration, in order to prevent them from moving too far from the operating point. Fig. Fig.55 shows the evolution of the objective function and condensation errors through iterations, both for fixed step and fixed tolerance. Though both methods find the same solution, the fixed tolerance method is much faster and keeps the error within a limit specified a priori. The fixed step method remains within a lower margin of error in this case due to the good quality of the condensed approximation but this margin is not under direct control and will depend on the size of the subintervals and on the model in an unforeseeable way. When the error tolerance was lowered to match the values observed for the fixed step method, both performed very similarly with a slight advantage of the fixed tolerance. Effect of the error constraints in the optimization algorithm. Results of optimizing the model of the tryptophan operon using fixed step and fixed tolerance. Both the controlled error and penalty treatment methods yielded the same results while SQP returned a solution that was feasible but yielded a lower flux. As can be seen in Table Table22 no constraint violations occurred with GP. When the lower bound was extended to include the levels reached by other methods, all previous results were reproduced. The tradeoff curve resulting from solving the problem for different tryptophan lower bounds is depicted as Fig Fig6.6. SQP and error controlled method took about 1 s to find the solution while the penalty tratment took 0.3 s. Comparison of results obtained for the tryptophan model with different methods. All the results that violate the lower bound for X3 were reproduced with GP by relaxing such bound. Constraint violations are shown in boldface. Tradeoff analysis for tryptophan model showing flux against lower bound for tryptophan. ## Conclusion The main challenge of non-linear optimization is dealing with non-convexities. In some cases, like GP, there is an elegant transformation that convexifies the problem without adding undue complexity. But this is seldom the case and dealing with non-convexities usually implies developing ad hoc tricks such as subdividng the system in many subsystems, finding convex relaxations of the constraints, adding extra variables or a combination of several of these strategies. Geometric programming provides a simple and efficient tool for the optimization of biotechnological systems that takes advantage of the structural regularity and flexibility of GMA systems. In this work we have presented two different strategies to do so, of which the penalty treatment seems to be the most promising. The methods are quite general, as this treatment of GP and recasting can be applied to any rational function, which in fact include almost all rate functions used in representations of metabolic processes. The use of geometric programming also provides a solution for the problem of constraint violations in the two strategies considered. The possibility of keeping an arbitrarily small approximation error in every iteration prevents the buildup of discrepancies in the Controlled Error Method which results in a "safer" condensation while the Penalty treatment doesn't rely on condensation to define the feasible area. It has been shown elsewhere [21] that GP can deal with big systems, and the sparse nature of the problems in metabolic engineering improves the capabilities of the approach. It is therefore reasonable to expect both strategies considered here to scale well for big problems but it is yet to be seen which one of the two behaves better in such cases. Geometric programming is a relatively recent and active area in operations research, which implies that further improvements and refinements for the optimization of GMA systems are to be expected. But even with existing methods, the optimization of this large class of systems, which is further expanded by the technique of recasting, has become feasible for execution of moderately sized tasks even on simple desktop computers. ## A Optimization problems A.1 Anaerobic fermentation by error controlled method A.2 Anaerobic fermentation by penalty treatment A.3 Tryptophan by error controlled method A.4 Tryptophan penalty approach ## Competing interests The author(s) declare that they have no competing interests. ## Acknowledgements This work was supported by a research grant from the Spanish Ministry of Science and Education ref. BIO2005-08898-C02-02. ## References • Stephanopoulos G, Aristidou A, Nielsen J. Metabolic Engineering: Principles and Methodologies. Academic Press; 1998. • Torres N, Voit E. Pathway Analysis and Optimization in Metabolic Engineering. Cambridge University Press; 2002. • Mendes P, Kell D. 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