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http://www.digplanet.com/wiki/Volt	digplanet beta 1: Athena Share digplanet: Agriculture Applied sciences Arts Belief Chronology Culture Education Environment Geography Health History Humanities Language Law Life Mathematics Nature People Politics Science Society Technology Volt Josephson junction array chip developed by the National Bureau of Standards as a standard volt Unit information Unit system SI derived unit Unit of Electric potential, electromotive force Symbol V Named after Alessandro Volta In SI base units: 1 V = 1 kg·m2·s-3·A-1 The volt (symbol: V) is the SI derived unit for electric potential (voltage), electric potential difference, and electromotive force.[1] The volt is named in honour of the Italian physicist Alessandro Volta (1745–1827), who invented the voltaic pile, possibly the first chemical battery. ## Definition A single volt is defined as the difference in electric potential between two points of a conducting wire when an electric current of one ampere dissipates one watt of power between those points.[2] It is also equal to the potential difference between two parallel, infinite planes spaced 1 meter apart that create an electric field of 1 newton per coulomb. Additionally, it is the potential difference between two points that will impart one joule of energy per coulomb of charge that passes through it. It can be expressed in terms of SI base units (m, kg, s, and A) as: $\mbox{V} = \dfrac{\mbox{kg} \cdot \mbox{m}^2}{\mbox{A} \cdot \mbox{s}^{3}}.$ It can also be expressed as amps×ohms (Ohm's law), power per unit current (Joule's law), or energy per unit charge: $\text{V} = \text{A} \cdot \Omega= \dfrac{\text{W}}{\text{A}} = \dfrac{\text{J}}{\text{C}}.$ ### Josephson junction definition Between 1990 and 1997, the volt was calibrated using the Josephson effect for exact voltage-to-frequency conversion, combined with cesium-133 time reference, as decided by the 18th General Conference on Weights and Measures. The following value for the Josephson constant is used: K{J-90} = 2e/h = 0.4835979 GHz/µV, where e is the elementary charge and h is the Planck constant. This is typically used with an array of several thousand or tens of thousands of junctions, excited by microwave signals between 10 and 80 GHz (depending on the array design).[3] Empirically, several experiments have shown that the method is independent of device design, material, measurement setup, etc., and no correction terms are required in a practical implementation.[4] ## Water flow analogy In the water flow analogy sometimes used to explain electric circuits by comparing them to water-filled pipes, voltage (difference in electric potential) is likened to difference in water pressure. The relationship between voltage and current is defined (in ohmic devices) by Ohm's Law. ## Common voltages A multimeter can be used to measure the voltage between two positions. 1.5 V C-cell batteries Nominal voltages of familiar sources: Note: Where RMS (root mean square) is stated above, the peak voltage is $\sqrt{2}$ times greater than the RMS voltage for a sinusoidal signal centered around zero voltage. ## History Alessandro Volta In 1800, as the result of a professional disagreement over the galvanic response advocated by Luigi Galvani, Alessandro Volta developed the so-called Voltaic pile, a forerunner of the battery, which produced a steady electric current. Volta had determined that the most effective pair of dissimilar metals to produce electricity is zinc and silver. In the 1880s, the International Electrical Congress, now the International Electrotechnical Commission (IEC), approved the volt as the unit for electromotive force. They made the volt equal to 108 cgs units of voltage, the cgs system at the time being the customary system of units in science. They chose such a ratio because the cgs unit of voltage is inconveniently small and one volt in this definition is approximately the emf of a Daniell cell, the standard source of voltage in the telegraph systems of the day.[6] At that time, the volt was defined as the potential difference [i.e., what is nowadays called the "voltage (difference)"] across a conductor when a current of one ampere dissipates one watt of power. The international volt was defined in 1893 as 1/1.434 of the emf of a Clark cell. This definition was abandoned in 1908 in favor of a definition based on the international ohm and international ampere until the entire set of "reproducible units" was abandoned in 1948. Prior to the development of the Josephson junction voltage standard, the volt was maintained in national laboratories using specially constructed batteries called standard cells. The United States used a design called the Weston cell from 1905 to 1972. This SI unit is named after Alessandro Volta. As with every International System of Units (SI) unit whose name is derived from the proper name of a person, the first letter of its symbol is upper case (V). However, when an SI unit is spelled out in English, it should always begin with a lower case letter (volt), except in a situation where any word in that position would be capitalized, such as at the beginning of a sentence or in capitalized material such as a title. Note that "degree Celsius" conforms to this rule because the "d" is lowercase. —Based on The International System of Units, section 5.2. Ohm Orders of magnitude (voltage) SI electromagnetism units SI prefix for unit prefixes ## Notes and references 1. ^ "SI Brochure, Table 3 (Section 2.2.2)". BIPM. 2006. Retrieved 2007-07-29. 2. ^ BIPM SI Brochure: Appendix 1, p. 144 3. ^ Burroughs, Charles J.; Benz, Samuel P. (1999-06-01), "1 Volt DC Programmable Josephson Voltage Standard", IEEE transactions on applied superconductivity 9 (3): 4145–4149, ISSN 1051-8223 4. ^ Keller, Mark W (2008-01-18), "Current status of the quantum metrology triangle", Metrologia 45 (1): 102–109, Bibcode:2008Metro..45..102K, doi:10.1088/0026-1394/45/1/014, ISSN 0026-1394, "Theoretically, there are no current predictions for any correction terms. Empirically, several experiments have shown that KJ and RK are independent of device design, material, measurement setup, etc. This demonstration of universality is consistent with the exactness of the relations, but does not prove it outright." 5. ^ Bullock, Orkand, and Grinnell, pp. 150–151; Junge, pp. 89–90; Schmidt-Nielsen, p. 484 6. ^ Hamer, Walter J. (January 15, 1965). Standard Cells: Their Construction, Maintenance, and Characteristics. National Bureau of Standards Monograph #84. US National Bureau of Standards. 1000000 videos foundNext > 2013 Chevrolet Volt Range Extending EV / Plug In Hybrid ReviewJoin us as we take an in-depth look at General Motors' first plug-in hybrid which they prefer to call an EV with a range extender. We talk infotainment, driv... 2013 Chevy Volt Test DriveWatch more and discuss: http://stamford.itsrelevant.com/content/13832/2013-chevy-volt-test-drive Take a test drive in the new 2013 Chevrolet Volt. See the Vo... 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Neosignal Like → http://www.facebook.com/neosignalnsgnl Follow → http://... Arctic Monkeys @ VOLT Festival 2014!6 July 2014, Sopron, Hungary / www.volt.hu. Chevrolet Volt Review - Everyday DriverThe guys go commuting in the Chevy Volt. Can it live up to the hype and be competitive against the gas-sipping Pri-i of the world? Chevy Volt Two Year ReviewChevy Volt Review https://MrEnergyCzar.com Hi everybody, MrEnergyCzar here. This is my two year review of the Chevy Volt. • Facebook: http://www.facebook.com... 1000000 videos foundNext > 475485 news items GM Authority (blog) Chevrolet Volt Sales Slip In February, Leaf Sales Continue To Climb GM Authority (blog) Thu, 06 Mar 2014 08:05:45 -0800 Sales of the Volt meanwhile fell 25.6 percent from February 2013 to 1,210 units last month. And while the Volt still holds on to the overall sales lead over the Leaf, Volt sales appear to be slowing in 2014. In January, Chevrolet moved 918 units of the ... Hybrid Cars News New 2016 Chevy Volt Rumored to Arrive Next Year with Different, But Not ... Carscoops (blog) Fri, 07 Mar 2014 06:37:30 -0800 GM is getting ready to give its "Extended Range Electric Vehicle" or E-REV Chevrolet Volt a second shot in the market with the launch of a new model, reportedly coming next year. The news comes from an unnamed industry source who told Edmunds that the ... Redesigned 2016 Chevrolet Volt To Debut Next Year Edmunds.com Thu, 06 Mar 2014 06:48:45 -0800 The second-generation 2016 Chevrolet Volt plug-in electric hybrid car will be introduced next year, Edmunds has learned. The 2016 Volt will be redesigned and developed on a new front-drive vehicle platform developed by General Motors. The redesigned ... RT Useful! The drone with an 80000-volt stun ... CNET Sat, 08 Mar 2014 13:45:00 -0800 Today's Deals · Coupon Codes · Marketplace Blog. Log In \| Join. Facebook Timeline options; Log In; Join CNET; Sign in with. Facebook Timeline options; My profile · Log out · CNET · News · Technically Incorrect; Useful! The drone with an 80,000-volt stun . First Comparison: BMW i3 vs. Chevrolet Volt TheStreet.com Thu, 06 Mar 2014 03:03:45 -0800 However, there are two plug-in hybrids in the market that do things a little differently: General Motors' (GM) Chevrolet Volt and BMW i3 with range-extender. Yes, I know there is also the Cadillac ELR, and yes I know all the differences and nuances ... Auto World News Nissan Leaf sales up again in February, Chevy Volt dips deeper Autoblog (blog) Mon, 03 Mar 2014 07:55:19 -0800 The February ledger for the Chevy Volt looked much worse, falling 25.6 percent from February 2013 down to 1,210 units. With 1,425 Leafs sold, Nissan came away the winner in a head-to-head competition between these the two early plug-in vehicles for the ... Chevy Volt Charging Cord Cut: Angry Neighbor, Electric-Car Hater? Green Car Reports Fri, 21 Feb 2014 06:04:36 -0800 When they purchased their 2013 Chevrolet Volt range-extended electric car, the couple asked their property manager to ensure that using a 120-Volt outlet in the garage to charge the Volt wasn't prohibited by the Home Owners' Association. Their parking ... GM Authority (blog) 2014 Chevrolet Volt — a Volt from the blue. San Francisco Chronicle (blog) Mon, 24 Feb 2014 03:27:56 -0800 The Volt was introduced a little over three years ago and at the time, it cost more than $40,000 and was way overpriced – you could buy a Prius for half that much. Since then the price has gone down – General Motors recently chopped$5,000 off the Volt ... Limit to books that you can completely read online Include partial books (book previews) .gsc-branding { display:block; } Oops, we seem to be having trouble contacting Twitter	2014-03-09 06:48:06	{"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": 3, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.36736464500427246, "perplexity": 4838.86525587991}, "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/1393999674993/warc/CC-MAIN-20140305060754-00018-ip-10-183-142-35.ec2.internal.warc.gz"}
https://educacionporcompetencias.com/marketing-assets-elfrubs/polynomial-functions-and-their-graphs-1d662d	It may have a turning point where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). It is a very common question to ask when a function will be positive and negative. Somewhere after this point, the graph must turn back down or start decreasing toward the horizontal axis because the graph passes through the next intercept at (5, 0). See . Polynomials of degree 2 are quadratic equations, and their graphs are parabolas. \\ &\left(x+1\right)\left(x - 1\right)\left(x - 5\right)=0 && \text{Factor the difference of squares}. The x-intercepts can be found by solving $g\left(x\right)=0$. Now that students have looked the end behavior of parent even and odd functions, I give them the opportunity to determine end behavior of more complex polynomials. The zero of –3 has multiplicity 2. Find the y– and x-intercepts of $g\left(x\right)={\left(x - 2\right)}^{2}\left(2x+3\right)$. We could have also determined on which intervals the function was positive by sketching a graph of the function. We can also see in Figure 18 that there are two real zeros between $x=1$ and $x=4$. To use Khan Academy you need to upgrade to another web browser. so the end behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs approach infinity, and the outputs increasing as the inputs approach negative infinity. t = 1 and t = -6. The last zero occurs at $x=4$. Notice in Figure 7 that the behavior of the function at each of the x-intercepts is different. 3) (a, 0) is an x-intercept of the graph of f if a is a zero of the function. Notice that there is a common factor of ${x}^{2}$ in each term of this polynomial. ${\left(x - 2\right)}^{2}\left(2x+3\right)=0$, \begin{align}&{\left(x - 2\right)}^{2}=0 && 2x+3=0 \\ &x=2 &&x=-\frac{3}{2} \end{align}. See Figure 8 for examples of graphs of polynomial functions with multiplicity 1, 2, and 3. Find the x-intercepts of $h\left(x\right)={x}^{3}+4{x}^{2}+x - 6$. See how nice and smooth the curve is? ... Equations Inequalities System of Equations System of Inequalities Polynomials Rationales Coordinate Geometry Complex Numbers Polar/Cartesian Functions â¦ So the y-intercept is $\left(0,12\right)$. The polynomial can be factored using known methods: greatest common factor and trinomial factoring. Analyze polynomials in order to sketch their graph. \begin{align} &{x}^{6}-3{x}^{4}+2{x}^{2}=0 && \\ &{x}^{2}\left({x}^{4}-3{x}^{2}+2\right)=0 && \text{Factor out the greatest common factor}. The factor is repeated, that is, the factor [latex]\left(x - 2\right) appears twice. This topic covers: - Adding, subtracting, and multiplying polynomial expressions - Factoring polynomial expressions as the product of linear factors - Dividing polynomial expressions - Proving polynomials identities - Solving polynomial equations & finding the zeros of polynomial functions - Graphing polynomial functions - Symmetry of functions Figure 17. The graphs of g and k are graphs of functions that are not polynomials. \\ &{x}^{2}\left(x+1\right)\left(x-1\right)\left({x}^{2}-2\right)=0 && \text{Factor the difference of squares}. A polynomial function of degree 2 is called a quadratic function. A polynomial of degree 0 is also called a constant function. Sketch a graph of $f\left(x\right)=-2{\left(x+3\right)}^{2}\left(x - 5\right)$. ${\left(x - 2\right)}^{2}=\left(x - 2\right)\left(x - 2\right)$, $f\left(x\right)={a}_{n}{x}^{n}+{a}_{n - 1}{x}^{n - 1}+…+{a}_{1}x+{a}_{0}$, CC licensed content, Specific attribution, http://cnx.org/contents/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1/Preface. Functions, polynomials, limits and graphs A function is a mapping between two sets, called the domain and the range, where for every value in the domain there is a unique value in the range assigned by the function. Show that the function $f\left(x\right)=7{x}^{5}-9{x}^{4}-{x}^{2}$ has at least one real zero between $x=1$ and $x=2$. We could choose a test value in each interval and evaluate the function $f\left(x\right) = \left(x+3\right){\left(x+1\right)}^{2}\left(x-4\right)$ at each test value to determine if the function is positive or negative in that interval. Identify zeros of polynomials and their multiplicities. Your response Solution Expand the polynomial to identify the degree and the leading coefficient. Write the formula for a polynomial function. If a function has a global maximum at a, then $f\left(a\right)\ge f\left(x\right)$ for all x. If a polynomial contains a factor of the form ${\left(x-h\right)}^{p}$, the behavior near the x-intercept h is determined by the power p. We say that $x=h$ is a zero of multiplicity p. The graph of a polynomial function will touch the x-axis at zeros with even multiplicities. We illustrate that technique in the next example. Thus, the domain of this function will be when $6 - 5t - {t}^{2}\ge 0$. Graphs of polynomials: Challenge problems. Every Polynomial function is defined and continuous for all real numbers. Because a height of 0 cm is not reasonable, we consider the only the zeros 10 and 7. Understand the relationship between zeros and factors of polynomials. Technology is used to determine the intercepts. Figure 7. From this zoomed-in view, we can refine our estimate for the maximum volume to about 339 cubic cm, when the squares measure approximately 2.7 cm on each side. The maximum number of turning points is 4 – 1 = 3. For higher even powers, such as 4, 6, and 8, the graph will still touch and bounce off of the horizontal axis but, for each increasing even power, the graph will appear flatter as it approaches and leaves the x-axis. They are smooth and continuous. Find the size of squares that should be cut out to maximize the volume enclosed by the box. Graphs of polynomials: Challenge problems. Graphing a polynomial function helps to estimate local and global extremas. From the graph we can see this function is positive for inputs between the intercepts. A global maximum or global minimum is the output at the highest or lowest point of the function. We can always check that our answers are reasonable by using a graphing calculator to graph the polynomial as shown in Figure 5. \\ &\left({x}^{2}-1\right)\left(x - 5\right)=0 && \text{Factor out the common factor}. See . Also, since $f\left(3\right)$ is negative and $f\left(4\right)$ is positive, by the Intermediate Value Theorem, there must be at least one real zero between 3 and 4. This is a single zero of multiplicity 1. For zeros with even multiplicities, the graphs touch or are tangent to the x-axis. The factor is quadratic (degree 2), so the behavior near the intercept is like that of a quadratic—it bounces off of the horizontal axis at the intercept. Together, this gives us, $f\left(x\right)=a\left(x+3\right){\left(x - 2\right)}^{2}\left(x - 5\right)$. If a point on the graph of a continuous function f at $x=a$ lies above the x-axis and another point at $x=b$ lies below the x-axis, there must exist a third point between $x=a$ and $x=b$ where the graph crosses the x-axis. The graph passes through the axis at the intercept, but flattens out a bit first. We can use what we have learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial functions. The Intermediate Value Theorem states that if $f\left(a\right)$ and $f\left(b\right)$ have opposite signs, then there exists at least one value c between a and b for which $f\left(c\right)=0$. The y-intercept can be found by evaluating $g\left(0\right)$. We can solve polynomial inequalities by either utilizing the graph, or by using test values. WEEK 3 POLYNOMIAL FUNCTIONS PART 1 - SECTIONS 2.1 - 2.2 WEEK 3 POLYNOMIAL FUNCTIONS PART 1 - SECTIONS 2.1 - 2.2 Score: 79% (5.5 of 7 pts) Submitted: Jan 23 at 9:22pm 2.2 Polynomial Functions and Their Graphs - PRACTICE TEST - Grade Report ... Graphs of Polynomials Using Transformations. â¦ To graph polynomial functions, find the zeros and their multiplicities, determine the end behavior, and ensure that the final graph has at most turning points. Polynomial functions of degree 2 or more are smooth, continuous functions. The graph has three turning points. Find the y– and x-intercepts of the function $f\left(x\right)={x}^{4}-19{x}^{2}+30x$. $h\left(x\right)={x}^{3}+4{x}^{2}+x - 6=\left(x+3\right)\left(x+2\right)\left(x - 1\right)$. The graph of P is a smooth curve with rounded corners and no sharp corners. When the leading term is an odd power function, as x decreases without bound, $f\left(x\right)$ also decreases without bound; as x increases without bound, $f\left(x\right)$ also increases without bound. degree ; leading coefficient Since the degree is odd and the leading coefficient is negative, the graph rises to the left and falls to the right. Polynomial functions of degree 2 or more have graphs that do not have sharp corners; recall that these types of graphs are called smooth curves. I then go over how to determine the End Behavior of these graphs. While quadratics can be solved using the relatively simple quadratic formula, the corresponding formulas for cubic and fourth-degree polynomials are not simple enough to remember, and formulas do not exist for general higher-degree polynomials. Yes. The graph passes directly through the x-intercept at $x=-3$. Fortunately, we can use technology to find the intercepts. Graphs of polynomials. Power and more complex polynomials with shifts, reflections, stretches, and compressions. Polynomial functions of degree 2 or more have graphs that do not have sharp corners; recall that these types of graphs are called smooth curves. From this graph, we turn our focus to only the portion on the reasonable domain, $\left[0,\text{ }7\right]$. A polynomial function of degree has at most turning points. These questions, along with many others, can be answered by examining the graph of the polynomial function. The multiplicity of a zero determines how the graph behaves at the. Do all polynomial functions have as their domain all real numbers? This means we will restrict the domain of this function to $0 0$, As with all inequalities, we start by solving the equality $\left(x+3\right){\left(x+1\right)}^{2}\left(x-4\right)= 0$, which has solutions at x = -3, -1, and 4. For example: x 2 + 3x 2 = 4x 2, but x + x 2 cannot be written in a simpler form. At x = 2, the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be second degree (quadratic). The revenue can be modeled by the polynomial function. Looking at the graph of this function, as shown in Figure 6, it appears that there are x-intercepts at $x=-3,-2$, and 1. Figure 1 shows a graph that represents a polynomial function and a graph that represents a function that is not a polynomial. $f\left(x\right)=\left(x+3\right){\left(x - 2\right)}^{2}{\left(x+1\right)}^{3}$. We have already explored the local behavior of quadratics, a special case of polynomials. This function f is a 4th degree polynomial function and has 3 turning points. For general polynomials, this can be a challenging prospect. This gives the volume, \begin{align}V\left(w\right)&=\left(20 - 2w\right)\left(14 - 2w\right)w \\ &=280w - 68{w}^{2}+4{w}^{3} \end{align}. Keep in mind that some values make graphing difficult by hand. You can add, subtract and multiply terms in a polynomial just as you do numbers, but with one caveat: You can only add and subtract like terms. The shortest side is 14 and we are cutting off two squares, so values w may take on are greater than zero or less than 7. Donate or volunteer today! Using Zeros to Graph Polynomials If P is a polynomial function, then c is called a zero of P if P(c) = 0.In other words, the zeros of P are the solutions of the polynomial equation P(x) = 0.Note that if P(c) = 0, then the graph of P has an x-intercept at x = c; so the x-intercepts of the graph are the zeros of the function. The following theorem has many important consequences. We call this a triple zero, or a zero with multiplicity 3. A polynomial function of degree has at most turning points. List the polynomial's zeroes with their multiplicities. In this section we will explore the local behavior of polynomials in general. The graph of function k is not continuous. Call this point $\left(c,\text{ }f\left(c\right)\right)$. As we have already learned, the behavior of a graph of a polynomial function of the form. Using technology to sketch the graph of $V\left(w\right)$ on this reasonable domain, we get a graph like Figure 24. The next zero occurs at $x=-1$. Use the end behavior and the behavior at the intercepts to sketch a graph. Polynomial functions also display graphs that have no breaks. This function is an odd-degree polynomial, so the ends go off in opposite directions, just like every cubic I've ever graphed. The polynomial function is of degree n. The sum of the multiplicities must be n. Starting from the left, the first zero occurs at $x=-3$. We can see the difference between local and global extrema in Figure 21. The number of times a given factor appears in the factored form of the equation of a polynomial is called the multiplicity. As $x\to \infty$ the function $f\left(x\right)\to \mathrm{-\infty }$, so we know the graph continues to decrease, and we can stop drawing the graph in the fourth quadrant. This graph has two x-intercepts. Sometimes, a turning point is the highest or lowest point on the entire graph. Understand the relationship between degree and turning points. If the polynomial function is not given in factored form: Factor any factorable binomials or trinomials. In particular, a quadratic function has the form $f(x)=ax^2+bx+c,$ where $$aâ 0$$. Curves with no breaks are called continuous. The objective is that the students make the connection that the degree of a polynomial affects the graph's end behavior. To improve this estimate, we could use advanced features of our technology, if available, or simply change our window to zoom in on our graph to produce the graph in Figure 24. Solve the inequality ${x}^{4} - 2{x}^{3} - 3{x}^{2} \gt 0$, In our other examples, we were given polynomials that were already in factored form, here we have an additional step to finding the intervals on which solutions to the given inequality lie. The y-intercept is located at (0, 2). An open-top box is to be constructed by cutting out squares from each corner of a 14 cm by 20 cm sheet of plastic then folding up the sides. The graph will bounce at this x-intercept. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. Putting it all together. Look at the graph of the polynomial function $f\left(x\right)={x}^{4}-{x}^{3}-4{x}^{2}+4x$ in Figure 11. The graph of function g has a sharp corner. The graphs of f and h are graphs of polynomial functions. $g\left(0\right)={\left(0 - 2\right)}^{2}\left(2\left(0\right)+3\right)=12$. Sort by: Top Voted. We discuss odd functions, even functions, positive functions, negative functions, end behavior, and degree. Only polynomial functions of even degree have a global minimum or maximum. For example, $f\left(x\right)=x$ has neither a global maximum nor a global minimum. Polynomials of degree 0 and 1 are linear equations, and their graphs are straight lines. Find the domain of the function $v\left(t\right)=\sqrt{6-5t-{t}^{2}}$. In addition to the end behavior, recall that we can analyze a polynomial function’s local behavior. Note that x = 0 has multiplicity of two, but since our inequality is strictly greater than, we don’t need to include it in our solutions. The x-intercept $x=-1$ is the repeated solution of factor ${\left(x+1\right)}^{3}=0$. Notice, since the factors are w, $20 - 2w$ and $14 - 2w$, the three zeros are 10, 7, and 0, respectively. We will start this problem by drawing a picture like Figure 22, labeling the width of the cut-out squares with a variable, w. Notice that after a square is cut out from each end, it leaves a $\left(14 - 2w\right)$ cm by $\left(20 - 2w\right)$ cm rectangle for the base of the box, and the box will be w cm tall. Let f be a polynomial function. Find solutions for $f\left(x\right)=0$ by factoring. Now that we know how to find zeros of polynomial functions, we can use them to write formulas based on graphs. We can estimate the maximum value to be around 340 cubic cm, which occurs when the squares are about 2.75 cm on each side. However, the graph of a polynomial function is always a smooth Use the graph of the function of degree 5 to identify the zeros of the function and their multiplicities. Sometimes, the graph will cross over the horizontal axis at an intercept. Degree. Optionally, use technology to check the graph. F-IF: Analyze functions using different representations. The same is true for very small inputs, say –100 or –1,000. The graph of a polynomial will touch the horizontal axis at a zero with even multiplicity. There are three x-intercepts: $\left(-1,0\right),\left(1,0\right)$, and $\left(5,0\right)$. You can also divide polynomials (but the result may not be a polynomial). To determine the stretch factor, we utilize another point on the graph. We can use this graph to estimate the maximum value for the volume, restricted to values for w that are reasonable for this problem—values from 0 to 7. This polynomial function is of degree 5. See and . We know that the multiplicity is likely 3 and that the sum of the multiplicities is likely 6. Here is a set of practice problems to accompany the Graphing Polynomials section of the Polynomial Functions chapter of the notes for Paul Dawkins Algebra course at Lamar University. Title: Polynomial Functions and their Graphs 1 Polynomial Functions and their Graphs. % Progress . Other times, the graph will touch the horizontal axis and bounce off. 1. The graph of polynomials are smooth, unbroken lines or curves, with no sharp corners or cusps (see p. 251). Again, we will start by solving the equality ${x}^{4} - 2{x}^{3} - 3{x}^{2} = 0$. The zero associated with this factor, $x=2$, has multiplicity 2 because the factor $\left(x - 2\right)$ occurs twice. P is continuous for all real numbers, so there are no breaks, holes, jumps in the graph. At x = 5, the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept. As a start, evaluate $f\left(x\right)$ at the integer values $x=1,2,3,\text{ and }4$. Then, identify the degree of the polynomial function. A polynomial function of degree $$3$$ is called a cubic function. I can see from the graph that there are zeroes at x = â15, x = â10, x = â5, x = 0, x = 10 , and x = 15 , because the graph touches or crosses the x â¦ Given the graph in Figure 20, write a formula for the function shown. Just select one of the options below to start upgrading. We can use factoring to simplify in the following way: \begin{align}{x}^{4} - 2{x}^{3} - 3{x}^{2} &= 0&\\{x}^{2}\left({x}^{2} - 2{x} - 3\right) &= 0\\ {x}^{2}\left(x - 3\right)\left(x + 1 \right)&= 0\end{align}. This is the currently selected item. 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Polynomial graphs R represents the revenue for the company increasing of graphing utilities to generate a graph divide inputs... Their graphs and creating new examples, but flattens out a bit first for x verifying! Are reasonable by using a graphing calculator - analyze and graph line equations and functions step-by-step or 1,000 the! Form: factor any factorable binomials or trinomials a given factor appears in the graph crosses the y-axis the... Our ability to solve for a polynomial function changes direction at its points... Complex polynomials with a de nition and some examples negative, it means we 're trouble! This a triple zero, or a global maximum or a zero with multiplicity.. A turning point represents a polynomial function is always one less than the degree of the function can change... A local minimum or maximum will explore the local behavior of a polynomial function if the polynomial beyond. 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Technology to generate a graph but not the zeros locations of turning points to sketch graphs of that. In and use all the features of Khan Academy is a valid input for.! To estimate local and global extremas also referred to as the absolute maximum and absolute values! Should be cut out to maximize the volume enclosed by the box zeroes their! Find solutions for [ latex ] \left ( c ) polynomial functions and their graphs a ) =0 [ /latex ] factoring. ] x=-1 [ /latex ] for the function factor of the function correct by substituting these for... And will either rise or fall as x increases without bound and will either or. Y-Axis at the intercepts shows that there is a valid input for a polynomial function, a turning point a. Have no breaks values, so the multiplicity of one, indicating a multiplicity of 2 to,... Threes, matching functions to their graphs are parabolas and their graphs 1 polynomial functions and calculator! The domain of this quadratic will allow us to determine the stretch factor, we can use we... Through the axis at the intercepts breaks, holes, jumps in the graph of a.! Of our ability to find x-intercepts because at the x-intercepts we find maximum. Intervals where the polynomial function if the leading term c\right ) \right ) [ /latex ] anyone,.! Have as their domain all real numbers polynomial ) tangent to the x-axis -. X-Intercepts is different dollars and t represents the year, with no corners! Into 4 intervals already learned, the graph of P is a number a for which f ( a 0. Point is a very common question to ask when a function that is, the.. C\Right ) =0 [ /latex ] has neither a global minimum is the output at intercepts. Every cubic i 've ever graphed ability to solve for a we utilize point. You express them in their simplest form \left ( 2x^2\right ) =-2x^4 [ /latex ] points! Only change from positive to negative at these values for x and verifying the. At ( 0, 2 ) a zero between them upgrade to another web browser intercept... Not in factored form but not the zeros 10 and 7 be factored known! Expand the polynomial function if the leading term is negative, it will change the direction the.	2021-07-26 22:42:01	{"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": 3, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7017349004745483, "perplexity": 366.1294952240447}, "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-2021-31/segments/1627046152156.49/warc/CC-MAIN-20210726215020-20210727005020-00165.warc.gz"}
https://www.nolo.com/dictionary/larceny-term.html	# Larceny Definition ## Larceny Another term for theft. Although the definition of this term differs from state to state, it typically means taking property belonging to another with the intent to permanently deprive the owner of the property. If the taking is not forceful, it is larceny; if it is accompanied by force directed against a person, it is robbery, a much more serious offense. Many states differentiate between petit larceny (usually a misdemeanor, punishable by time in the county jail) and grand larceny (theft of a large amount, punishable as a felony in state prison).	2023-01-29 15:01:05	{"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": 0.8033620119094849, "perplexity": 5262.74840061941}, "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-2023-06/segments/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00034.warc.gz"}
http://math.stackexchange.com/questions/176034/are-the-matrix-products-ab-and-ba-similar	# Are the matrix products $AB$ and $BA$ similar? Given two matrices $A,B.$ On what conditions does $AB \sim BA$ hold? - Since it certainly holds if $A$ and $B$ commute, it would be nice to see an example where $AB\ne BA$ yet $AB \sim BA$. – Greg Martin Jul 28 '12 at 0:01 Is is possible to prove that both share the same eigenvalues? – Tomer Galanti Jul 28 '12 at 0:03 If $A$ is invertible, then $AB = A(BA)A^{-1}$ which shows that $AB$ and $BA$ are similar. Similar (no pun intended) proof if $B$ is invertible. - $AB$ is conjugate to $BA$ if either $A$ or $B$ are invertible. If neither is the case, there are counterexamples: for example, it may be the case that $AB = 0$ while $BA \neq 0$. Explicitly, take $$A = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right], B = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right].$$ We have $AB = A$ but $BA = 0$. However, there is a salvage: $AB$ and $BA$ have the same characteristic polynomial. See this blog post. (Short proof: this must hold if either $A$ or $B$ is invertible, and that condition is Zariski dense.) - Can you please provide with a reference regarding the case one of the matrices are invertiable ? – Belgi Jul 28 '12 at 0:51 @Belgi If $A$ is invertible, and if $C=A^{-1}$, then $C(AB)C^{-1}=BA$ as you can check. Similarly, if $B$ is invertible, and if $C=B^{-1}$, then $C(BA)C^{-1}=AB$. – Amitesh Datta Jul 28 '12 at 1:28 As mentioned already, if either of $A$ or $B$ is invertible (and both are the same size), we have \begin{align} AB=A(BA)A^{-1}\quad&\mbox{if A is invertible}\\ AB=B^{-1}(BA)B\quad&\mbox{if B is invertible} \end{align} However, here is a short proof that even if $A$ is $m\times n$ and $B$ is $n\times m$, the characteristic polynomials of $AB$ and $BA$ differ only by a factor of $\lambda^{\large\|n-m\|}$. - The proof is rather straightforward and I do not assume anything about $A$ and $B$, except that they are square. I do so by showing that they have the same eigenvalues. If that is the case then $A=MJM^{-1}$ and $B=CJC^{-1}$ both share the same $J$ in their Jordan form and I take for granted that there exist some matrix $K$ such that $C=KM$, in order to have $B=KMJ(KM)^{-1}$. So suppose $\lambda$ is an eigenvalue of $AB$. Then $ABx=\lambda x$. Premultiplying both sides by $B$, we get $BABx =BA(Bx) \lambda(Bx)$, which shows that $\lambda$ is also an eigenvalue of $BA$. - Having the same eigenvalues doesn't mean they share a Jordan form. You should read the other answers, the result is false if $A$ and $B$ are non-invertible. – EuYu Nov 26 '12 at 0:42 You're right, Ive just seen the counter example. I wrote it thinking they had a full set of eigenvalues, in which case that should be true of J. – Michael P. Nov 26 '12 at 0:55 Please note in my edit: the singular form of "matrices" is "matrix." – rschwieb Nov 26 '12 at 1:17	2015-11-28 12:07:02	{"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": 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": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9173993468284607, "perplexity": 143.02224208629434}, "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/1448398452385.31/warc/CC-MAIN-20151124205412-00343-ip-10-71-132-137.ec2.internal.warc.gz"}
https://projecteuclid.org/euclid.aoas/1372338471	## The Annals of Applied Statistics ### Visualizing genetic constraints #### Abstract Principal Components Analysis (PCA) is a common way to study the sources of variation in a high-dimensional data set. Typically, the leading principal components are used to understand the variation in the data or to reduce the dimension of the data for subsequent analysis. The remaining principal components are ignored since they explain little of the variation in the data. However, evolutionary biologists gain important insights from these low variation directions. Specifically, they are interested in directions of low genetic variability that are biologically interpretable. These directions are called genetic constraints and indicate directions in which a trait cannot evolve through selection. Here, we propose studying the subspace spanned by low variance principal components by determining vectors in this subspace that are simplest. Our method and accompanying graphical displays enhance the biologist’s ability to visualize the subspace and identify interpretable directions of low genetic variability that align with simple directions. #### Article information Source Ann. Appl. Stat., Volume 7, Number 2 (2013), 860-882. Dates First available in Project Euclid: 27 June 2013 https://projecteuclid.org/euclid.aoas/1372338471 Digital Object Identifier doi:10.1214/12-AOAS603 Mathematical Reviews number (MathSciNet) MR3113493 Zentralblatt MATH identifier 1288.62101 #### Citation Gaydos, Travis L.; Heckman, Nancy E.; Kirkpatrick, Mark; Stinchcombe, J. R.; Schmitt, Johanna; Kingsolver, Joel; Marron, J. S. Visualizing genetic constraints. Ann. Appl. Stat. 7 (2013), no. 2, 860--882. doi:10.1214/12-AOAS603. https://projecteuclid.org/euclid.aoas/1372338471 #### References • Adler, R. J. and Taylor, J. E. (2007). Random Fields and Geometry. Springer, New York. • Amemiya, Y., Anderson, T. W. and Lewis, P. A. W. (1990). Percentage points for a test of rank in multivariate components of variance. Biometrika 77 637–641. • Anderson, T. W. and Amemiya, Y. (1991). Testing dimensionality in the multivariate analysis of variance. Statist. Probab. Lett. 12 445–463. • Beder, J. H. and Gomulkiewicz, R. (1998). Computing the selection gradient and evolutionary response of an infinite-dimensional trait. J. Math. Biol. 36 299–319. • Chipman, H. A. and Gu, H. (2005). Interpretable dimension reduction. J. Appl. Stat. 32 969–987. • Demidenko, E. (2004). Mixed Models: Theory and Applications. Wiley, Hoboken, NJ. • Eilers, P. H. C. and Marx, B. D. (1996). Flexible smoothing with $B$-splines and penalties. Statist. Sci. 11 89–121. • Gaydos, T. (2008). Data representation/basis selection to understand variation of function valued traits. Ph.D. thesis, Univ. North Carolina. • Gaydos, T., Heckman, N., Kirkpatrick, M., Stinchcombe, J. R., Schmitt, J., Kingsolver, J. and Marron, J. S. (2013a). Supplement to “Visualizing genetic constraints.” DOI:10.1214/12-AOAS603SUPPA. • Gaydos, T., Heckman, N., Kirkpatrick, M., Stinchcombe, J. R., Schmitt, J., Kingsolver, J. and Marron, J. S. (2013b). Supplement to “Visualizing genetic constraints.” DOI:10.1214/12-AOAS603SUPPB. • Gomulkiewicz, R. and Beder, J. H. (1996). The selection gradient of an infinite-dimensional trait. SIAM J. Appl. Math. 56 509–523. • Gomulkiewicz, R. and Houle, D. (2009). Demographic and genetic constraints on evolution. American Naturalist 174 218–229. • Gomulkiewicz, R. and Kingsolver, J. G. (2006). A fable of four functions: Function-valued approaches in evolutionary biology. Journal of Evolutionary Biology 20 20–21. • Green, P. J. and Silverman, B. W. (1994). Nonparametric Regression and Generalized Linear Models: A Roughness Penalty Approach. Monographs on Statistics and Applied Probability 58. Chapman & Hall, London. • Griswold, C. K., Gomulkiewicz, R. and Heckman, N. (2008). Hypothesis testing in comparative and experimental studies of function-valued traits. Evolution 62 1229–1242. • Heckman, N. E. (2003). Functional data analysis in evolutionary biology. In Recent Advances and Trends in Nonparametric Statistics (M. G. Akritas and D. N. Politis, eds.) 49–60. Elsevier, Amsterdam. • Hine, E. and Blows, M. W. (2006). Determining the effective dimensionality of the genetic variance–covariance matrix. Genetics 173 1135–1144. • Izem, R. and Kingsolver, J. G. (2005). Variation in continuous reaction norms: Quantifying directions of biological interest. Am. Nat. 166 277–289. • Johnson, R. A. and Wichern, D. W. (2008). Applied Multivariate Statistical Analysis, 6th ed. Pearson Education, Upper Saddle River. • Kingsolver, J. G., Gomulkiewicz, R. and Carter, P. A. (2001). Variation, selection and evolution of function valued traits. Genetica 112–113 87–104. • Kingsolver, J. G., Ragland, G. J. and Shlichta, J. G. (2004). Quantitative genetics of continuous reaction norms: Thermal sensitivity of caterpillar growth rates. Evolution 58 1521–1529. • Kirkpatrick, M. and Heckman, N. (1989). A quantitative genetic model for growth, shape, reaction norms, and other infinite-dimensional characters. J. Math. Biol. 27 429–450. • Kirkpatrick, M. and Lofsvold, D. (1992). Measuring selection and constraint in the evolution of growth. Evolution 46 954–971. • Lande, R. (1976). Natural selection and random genetic drift in phenotypic evolution. Evolution 30 314–334. • Lande, R. (1979). Quantitative genetic analysis of multivariate evolution, applied to brain: Body size allometry. Evolution 33 402–416. • Lande, R. and Arnold, S. (1983). The measurement of selection on correlated characters. Evolution 37 1210–1226. • Loève, M. (1978). Probability Theory. II, 4th ed. Graduate Texts in Mathematics 46. Springer, New York. • Lynch, M. and Walsh, B. (1998). Genetic Analysis of Quantitative Traits. Sinauer, Sunderland, MA. • Meyer, K. and Smith, S. (1996). Restricted maximum likelihood estimation for animal models using derivatives of the likelihood. Genetics Selection Evolution 28 23–49. • Schatzman, M. (2002). Numerical Analysis: A Mathematical Introduction. Claredon Press, Oxford. • Searle, S. R., Casella, G. and McCulloch, C. E. (2006). Variance Components. Wiley, Hoboken, NJ. • Stewart, G. W. and Sun, J. G. (1990). Matrix Perturbation Theory. Academic Press, Boston, MA. • Stinchcombe, J. R., Izem, R., Heschel, M. S., McGoey, B. V. and Schmitt, J. (2010). Across-environment genetic correlations and the frequency of selective environments shape the evolutionary dynamics of growth rate in Impatiens capensis. Evolution 64 2887–2903. • Tibshirani, R. (1996). Regression shrinkage and selection via the lasso. J. R. Stat. Soc. Ser. B Stat. Methodol. 58 267–288. #### Supplemental materials • Supplementary material A: Supplementary plots. As previously noted, supplementary material [Gaydos et al. (2013a)] contains a complete set of plots from our data analyses, as in Figures 3 through 6. • Supplementary material B: Nearly null space example. An additional supplementary file [Gaydos et al. (2013b)] contains a simple example that shows the benefits of the proposed methodology.	2019-07-24 02:36:30	{"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": 0.32712405920028687, "perplexity": 10369.347089940024}, "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-2019-30/segments/1563195530250.98/warc/CC-MAIN-20190724020454-20190724042454-00436.warc.gz"}
https://physics.stackexchange.com/questions/364945/a-question-about-the-ehrenfest-paradox	The Ehrenfest paradox exemplifies that a rigid body can't be defined in special relativity, but I can't convince myself that it implies non-Euclidean geometry as some sources say (https://en.wikipedia.org/wiki/Ehrenfest_paradox). In this post question(Euclidean geometry in non-inertial frame), the answers vary from sayin that it only implies that rigid bodies can't exist in SR or that it implies, in addition to the former, that the geometry is non-Euclidian. In my point of view, it only tell us that the theory is not consistent with this thought experiment. Could someone elaborate on what the Ehrenfest paradox really tell us? I'am an undergraduate in physics and I will be talking about rigid bodies in one of my classes. I want to clarify what would be non-sense to say to my professor. I appreciate the attention. It is known that the length of a circle measured by non-inertial observers that are on a disk rotating around the center of this circle should exceed the length of the same circle measured by non-rotating inertial observers that are outside the disk. This circumstance, supplemented by the provision on the equality of the radius of the circle in the rotating and non-rotating frames of reference, formed the basis for recognizing the non-euclidean nature of geometry on rotating bodies. In order that the disk in the inertial non-rotating frame of reference retain its size and shape that it possessed before it was rotated, all its concentric circles must be forcibly retained on the previous circumferences when we spin the disk faster and faster. Such a disk would be physically forcedly stretched in the peripheral areas. The purely theoretical contraction of a disk or ring to disappearing dimensions is not at all more paradoxical than the effects of black holes and not more than other strange effects can be used to criticize the physical content of the theory of relativity. After all, after simple Lorentz contraction, the moving rod "having reached the speed of light" also "disappears". Speaking about the difference of circumference of a disk in inertial and rotating frames of reference, the authors of geometrical metamorphoses on the disc as a rule do not consider the physical side of the difference in its length. Nor did M. Born (see M. Born, Einstein's theory of relativity) explaining the reason for the non-Euclidean geometry on the Einstein disk and declaring the collapse of non-Euclidean geometry. Was there any collapse at all? The following thought experiment shows, that it is hardly possible to "turn away" from the reduction of the rotating disk and reduce the physical processes to a simple geometrical portrayal. Let us imagine a rotating ring, placed in the inertial reference frame $K$ on the circle $O$ of large diameter $R$. Let the velocity $v$ of the points of the ring be equal to $(3/4)^{1/2} c$ , where $c$ is the velocity of light in a vacuum. Suppose that the observer who is at some distance from the plane in which the ring is located, sees the ring rotating counter-clockwise. Let us now imagine that along the tangent to the upper point of the circle $O$ to the right of the observer at a speed of $(3/4)^{1/2} c$ approaches a rope. Rope’s length $L$ in the reference frame of the above observer is numerically equal to the length $L = 2\pi R$ of the circle on which the rotating ring is placed. Proper length of the rope is $2L$. Suppose that at the moment when the front end of the “flying rope” comes to the upper point of the circle, it is grabbed by the ring and screwed onto it. Since the length $L$ of the moving rope is equal to the length $L_c$ of the circle $O$, and each section of the rope, continuing to move on the ring with the speed $v$, keeps its length unchanged, the rope also retains its length and completely covers the rotating ring in such a way that observers can tie end and beginning of the rope lying on the ring. Observers on the ring, measuring in their rotating reference frame $K’$ length $L'$ of the circumference of circle $O$, on which the rope lies, will find that the length $L'$ of the circle $O$ and correspondingly, the length $L'$ of the rope is twice as long as the length $L$ of circle $O$ as measured in the reference system $K$ i.e. equal to $2L$. In this case observers on the ring will argue that the length of the rope, the length of the ring and the same length of the circumference on which the ring and rope are located exceeds the value of $2\pi R$ due to the non-Euclidean nature of the geometry of the rotating reference frame $K'$. Let's now slow the ring down until rotation completely stops. What happens to the length of the circle $O$ on which the ring is located? The length of the circle should, according to the statements of observers on the ring, be equal to $L$. Such an expected decrease of the circumference of the ring the observers on the ring will explain by their transition from the non-Euclidean reference frame $K'$ of the rotating ring to the Euclidean space of the inertial reference frame $K$. However, when the ring stops, the length of the rope becomes equal to its own length $2L$. What will be the length of the stopped ring? If we assume that the ring in the inertial reference frame $K$ after its stop has retained its dimensions, then the rope will be 2 times longer than the ring, which is completely incomprehensible - because the rope, like the ring, "passed" from non-Euclidean space to Euclidean space. If we assume that the ring, will lengthen when stopped, and its radius $R_0$ for an inertial nonrotating observer becomes equal to $2R$, then we cannot speak at all about the change in the length of the ring for observers on the ring (and that line on which the ring lies in their reference frame and which they can call this circumference), since the length of the ring for them has not changed. Of course, one can speak of a change in the radius of the ring for rotating observers and of the presence of transitions detected in the inertial reference system when it comes to deceleration from one circle to another, but then a lot of questions arise. First, what is a circle in physics devoid of materiality (physicality)? Secondly, what in the reasonings of some authors determines the change in the metric on the disk when it is spinning - a change in radius or a change in the length of the circle? And thirdly, are rotating observers who are unable to detect a change in the length of the periphery of the disk can detect a change in the length of its radius? This article shows that the disk does not bend when rotationally shortened and that the ratio of the circumference of the edge of the disk to its diameter on the rotating disk is exactly equal to the $\pi$. http://aapt.scitation.org/doi/abs/10.1119/1.4942168?journalCode=ajp The article gives two resolutions, the both are formally correct. But, if the rail and the wheel are toothed, only Lorentz – contracted wheel will run smoothly, because its teeth will match the ones on the rail. The Ehrenfest paradox exemplifies that a rigid body can't be defined in special relativity Well, I would say that rigid bodies can't exist simply because of the speed-of-light limit. Certain cases of rigid motion (of not-intrinsically-rigid bodies) make sense in SR, as you know, but there are only a few. I would count Ehrenfest's cylinder as one of those cases – it's a highly symmetric system. Almost anything else, that doesn't meet any definition of rigidity, would be a better example of the lack of rigidity in SR. In my point of view, it only tell us that the theory is not consistent with this thought experiment. If by "the theory" you mean SR then it absolutely is consistent with the thought experiment. SR is assumed to be correct in the thought experiment and all of the properties of the cylinder are predictions of SR. It's called a paradox not because it's a logical inconsistency but because people are surprised by it. Length contraction also surprises people and has as much right to be called a paradox. I would classify these paradoxes as "silly" in the following semi-rigorous sense: they have near exact analogues in Euclidean geometry, and no one would consider the Euclidean versions to be paradoxes. The Minkowskian versions only appear to merit peer-reviewed think pieces because people don't apply their evolved intuition about Euclidean geometry to spacetime geometry. Here's the Euclidean version of the Ehrenfest setup. You have a right circular cylinder of radius $$r$$ and you want to cover it with strips of wallpaper of width $$w$$. If the strips run parallel to the length of the cylinder then you need $$2πr/w$$ strips. But if the strips are angled by $$\theta$$, so that they spiral around the cylinder, then you need only $$2πr/w\sqrt{1+v^2}$$ of them, where $$v = \tan\theta$$ is the slope. This means that the circumference of the circle, as measured by the strips of width $$w$$, is only $$w(2πr/w\sqrt{1+v^2}) = 2πr/\sqrt{1+v^2}$$, and the circumference divided by the radius is therefore $$2π/\sqrt{1+v^2}$$, so the cylinder is non-Euclidean when $$v\ne 0$$. What are we to make of this? Well, it is what it is. It's obvious that there's something awry about the way we're measuring distances with these strips. The distances of length $$w$$ don't join together into the length of a single closed curve; for each "zig" of length $$w$$ there's a "zag" that we aren't counting. In Minkowski space it's the same story: no closed spacelike curve is actually measured out in increments of 1 meter by the metersticks. As is usual in the silly paradoxes, we're overlooking the time coordinate. While I find this thought experiment somewhat interesting, there's nothing mysterious about it; all of the geometric rules are manifest, the setup is easy to understand, and it's just a matter of what words you think should be used to describe it. It is Euclidean or not? It's whatever you want it to be.	2019-11-21 14:09:57	{"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": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6972767114639282, "perplexity": 248.91920882057525}, "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/1573496670821.55/warc/CC-MAIN-20191121125509-20191121153509-00429.warc.gz"}
https://www.zbmath.org/?q=an%3A1187.68589	## Formal concept analysis via multi-adjoint concept lattices.(English)Zbl 1187.68589 Summary: Several fuzzifications of formal concept analysis have been proposed to deal with uncertain information. In this paper, we focus on concept lattices under a multi-adjoint paradigm, which enriches the language providing greater flexibility to the user in that he/she can choose from a number of different connectives. Multi-adjoint concept lattices are shown to embed different fuzzy extensions of concept lattices found in the literature, the main results of the paper being the representation theorem of this paradigm and the embedding of other well-known approaches. ### MSC: 68T30 Knowledge representation 06B75 Generalizations of lattices Full Text: ### References: [1] Abdel-Hamid, A.; Morsi, N., Associatively tied implicacions, Fuzzy sets and systems, 136, 3, 291-311, (2003) · Zbl 1042.03021 [2] R. Bělohlávek, Lattice generated by binary fuzzy relations (extended abstract), in: 4th Internat. Conf. on Fuzzy Sets Theory and Applications, 1998, p. 11. [3] Bělohlávek, R., Lattices of fixed points of fuzzy Galois connections, Math. logic quart., 47, 1, 111-116, (2001) · Zbl 0976.03025 [4] Bělohlávek, R., Fuzzy relational systems: foundations and principles, (2002), Kluwer Academic Publishers Dordrecht · Zbl 1067.03059 [5] Bělohlávek, R., Concept lattices and order in fuzzy logic, Ann. pure appl. logic, 128, 277-298, (2004) · Zbl 1060.03040 [6] R. Bělohlávek, V. Vychodil, What is a fuzzy concept lattice? in: Internat. Workshop on Concept Lattices and their Applications, 2005, pp. 34-45. [7] Burusco, A.; Fuentes-González, R., The study of $$L$$-fuzzy concept lattice, Mathware soft comput., 3, 209-218, (1994) · Zbl 0827.06004 [8] Burusco, A.; Fuentes-González, R., Concept lattices defined from implication operators, Fuzzy sets and systems, 114, 431-436, (2000) · Zbl 0971.06010 [9] Davey, B.; Priestley, H., Introduction to lattices and order, (2002), Cambridge University Press Cambridge · Zbl 1002.06001 [10] Ganter, B.; Wille, R., Formal concept analysis: mathematical foundation, (1999), Springer Berlin [11] Georgescu, G.; Popescu, A., Concept lattices and similarity in non-commutative fuzzy logic, Fund. inform., 55, 1, 23-54, (2002) · Zbl 1023.03016 [12] Georgescu, G.; Popescu, A., Non-commutative fuzzy Galois connections, Soft comput., 7, 7, 458-467, (2003) · Zbl 1024.03025 [13] Georgescu, G.; Popescu, A., Non-dual fuzzy connections, Arch. math. logic, 43, 8, 1009-1039, (2004) · Zbl 1060.03042 [14] Georgescu, G.; Popescu, A., Similarity convergence in residuated structures, Logic J. IGPL, 13, 4, 389-413, (2005) · Zbl 1082.03058 [15] Hájek, P., Metamathematics of fuzzy logic, () · Zbl 0937.03030 [16] Julián, P.; Moreno, G.; Penabad, J., On fuzzy unfolding: a multi-adjoint approach, Fuzzy sets and systems, 154, 16-33, (2005) · Zbl 1099.68017 [17] S. Krajči, The basic theorem on generalised concept lattice, in: V. Snášel, R. Bělohlávek (Eds.), ERCIM Workshop on Soft Computing, 2004, pp. 25-33. [18] Krajči, S., A generalized concept lattice, Logic J. IGPL, 13, 5, 543-550, (2005) · Zbl 1088.06005 [19] J. Medina, M. Ojeda-Aciego, J. Ruiz-Calviño, On multi-adjoint concept lattices: definition and representation theorem, in: Lecture Notes in Artificial Intelligence, Vol. 4390, 2007, pp. 197-209. · Zbl 1187.68588 [20] J. Medina, M. Ojeda-Aciego, A. Valverde, P. Vojtáš, Towards biresiduated multi-adjoint logic programming, in: Lecture Notes in Artificial Intelligence, Vol. 3040, 2004, pp. 608-617. [21] Medina, J.; Ojeda-Aciego, M.; Vojtáš, P., Similarity-based unification: a multi-adjoint approach, Fuzzy sets and systems, 146, 1, 43-62, (2004) · Zbl 1073.68026 [22] J. Medina, J. Ruiz-Calviño, Towards multi-adjoint concept lattices, in: Information Processing and Management of Uncertainty for Knowledge-Based Systems, IPMU’06, 2006, pp. 2566-2571. [23] Pollandt, S., Fuzzy begriffe, (1997), Springer Berlin · Zbl 0870.06008 [24] R. Wille, Restructuring lattice theory: an approach based on hierarchies of concepts, in: I. Rival (Ed.), Ordered Sets, Reidel, 1982, pp. 445-470. · Zbl 0491.06008 This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.	2022-07-05 08:55:34	{"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": 0.8171051740646362, "perplexity": 10368.55901078557}, "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-2022-27/segments/1656104542759.82/warc/CC-MAIN-20220705083545-20220705113545-00706.warc.gz"}
https://ftp.aimsciences.org/article/doi/10.3934/proc.2013.2013.489	# American Institute of Mathematical Sciences 2013, 2013(special): 489-497. doi: 10.3934/proc.2013.2013.489 ## A discontinuous Galerkin least-squares finite element method for solving Fisher's equation 1 Department of Engineering, Mathematics, and Physics, Texas A&M International University, Laredo, TX 78041 2 Department of Mathematics, The University of Southern Mississippi, Hattiesburg, MS 39406 Received September 2012 Revised January 2013 Published November 2013 In the present study, a discontinuous Galerkin least-squares finite element algorithm is developed to solve Fisher's equation. The present method is effective and can be successfully applied to problems with strong reaction, to which obtaining stable and accurate numerical traveling wave solutions is challenging. Numerical results are given to demonstrate the convergence rates of the method and the performance of the algorithm in long-time integrations. Citation: Runchang Lin, Huiqing Zhu. A discontinuous Galerkin least-squares finite element method for solving Fisher's equation. Conference Publications, 2013, 2013 (special) : 489-497. doi: 10.3934/proc.2013.2013.489 ##### References: [1] M.J. Ablowitz and A. Zeppetella, Explicit solutions of Fisher's equation for a special wave speed, Bull. Math. Biol., 41 (1979), no. 6, pp. 835-840. Google Scholar [2] K. Al-Khaled, Numerical study of Fishers reaction-diffusion equation by the sinc collocation method, J. Comput. Appl. Math., 137 (2001), pp. 245-255. Google Scholar [3] J. Canosa, On a nonlinear diffusion equation describing population growth, IBM J. Res. Develop., 17 (1973), pp. 307-313. Google Scholar [4] G.F. Carey and Y. Shen, Least-squares finite element approximation of Fishers reactiondiffusion equation, Numer. Methods Partial Differential Equations, 11 (1995), pp. 175-186. Google Scholar [5] I. Daǧ, A. Şahin, and A. Korkmaz, Numerical investigation of the solution of Fisher's equation via the B-spline Galerkin method, Numer. Methods Partial Differential Equations 26 (2010), no. 6, pp. 1483-1503. Google Scholar [6] R.A. Fisher, The wave of advance of advantageous genes, Ann. Eugenics, 7 (1937), pp. 355-369. Google Scholar [7] J. Gazdag and J. Canosa, Numerical solution of Fisher's equation, J. Appl. Probab., 11 (1974), pp. 445-457. Google Scholar [8] B.Y. Guo and Z.X. Chen, Analytic solutions of the Fisher equation, J. Phys. A, 24 (1991), no. 3, pp. 645-650. Google Scholar [9] P.S. Hagan, Traveling wave and multiple traveling wave solutions of parabolic equations, SIAM J. Math. Anal. 13 (1982), no. 5, pp. 717-738. Google Scholar [10] T. Hagstrom and H.B. Keller, The numerical calculation of traveling wave solutions of nonlinear parabolic equations, SIAM J. Sci. Statist. Comput., 7 (1986), no. 3, pp. 978-988. Google Scholar [11] A. Kolmogorov, I. Petrovshy, and N. Piscounoff, Étude de l'équation de la diffusion avec croissance de la quantité de matière et son application à un problème biologique, Bull. Univ. Etat Moscou Ser. Int. Sect. A Math. et Mecan., 1 (1937), pp. 1-25. Google Scholar [12] D.A. Larson, Transient bounds and time-asymptotic behavior of solutions to nonlinear equations of Fisher type, SIAM J. Appl. Math. 34 (1978), no. 1, pp. 93-103. Google Scholar [13] S. Li, L. Petzold, and Y. Ren, Stability of moving mesh systems of partial differential equations, SIAM J. Sci. Comput., 20 (1998), no. 2, pp. 719-738. Google Scholar [14] R. Lin, Discontinuous discretization for least-squares formulation of singularly perturbed reaction-diffusion problems in one and two dimensions,, SIAM J. Numer. Anal. 47 (2008/09), 47 (): 89. Google Scholar [15] R. Lin, Discontinuous Galerkin least-squares finite element methods for singularly perturbed reaction-diffusion problems with discontinuous coefficients and boundary singularities, Numer. Math. 112 (2009), no. 2, pp. 295-318. Google Scholar [16] J.D. Logan, "An introduction to nonlinear partial differential equations,'' second edition, Wiley-Interscience, John Wiley & Sons, Hoboken, NJ, 2008. Google Scholar [17] R.E. Mickens, A best finite-difference scheme for the Fisher equation, Numer. Methods Partial Differential Equations 10 (1994), no. 5, pp. 581-585. Google Scholar [18] J.D. Murray, "Mathematical biology,'' Biomathematics, 19, Springer-Verlag, Berlin, 1989. Google Scholar [19] D. Olmos and B.D. Shizgal, A pseudospectral method of solution of Fisher's equation, J. Comput. Appl. Math., 193 (2006), pp. 219-242. Google Scholar [20] N. Parekh and S. Puri, A new numerical scheme for the Fisher equation, J. Phys. A: Math. Gen., 23 (1990), pp. L1085-L1091. Google Scholar [21] Y. Qiu and D.M. Sloan, Numerical solution of Fisher's equation using a moving mesh method, J. Comput. Phys., 146 (1998), pp. 726-746. Google Scholar [22] Rizwan-uddin, Comparison of the nodal integral method and nonstandard finite-difference schemes for the Fisher equation, SIAM. J. Sci. Comput., 22 (2000), pp. 1926-1942. Google Scholar [23] J. Roessler and H. Hüssner, Numerical solution of the $1+2$ dimensional Fisher's equation by finite elements and the Galerkin method, Math. Comput. Modelling, 25 (1997), pp. 57-67. Google Scholar [24] S. Tang and R.O. Weber, Numerical study of Fisher's equation by a Petrov-Galerkin finite element method, J. Austral. Math. Soc. Sci. B, 33 (1991) pp. 27-38. Google Scholar [25] V. Thomée, "Galerkin finite element methods for parabolic problems,'' second edition, Springer Series in Computational Mathematics, 25, Springer-Verlag, Berlin, 2006. Google Scholar [26] S. Zhao and G.W. Wei, Comparison of the discrete singular convolution and three other numerical schemes for solving Fisher's equation, SIAM J. Sci. Comput., 25 (2003) pp. 127-147. Google Scholar show all references ##### References: [1] M.J. Ablowitz and A. Zeppetella, Explicit solutions of Fisher's equation for a special wave speed, Bull. Math. Biol., 41 (1979), no. 6, pp. 835-840. Google Scholar [2] K. Al-Khaled, Numerical study of Fishers reaction-diffusion equation by the sinc collocation method, J. Comput. Appl. Math., 137 (2001), pp. 245-255. Google Scholar [3] J. Canosa, On a nonlinear diffusion equation describing population growth, IBM J. Res. Develop., 17 (1973), pp. 307-313. Google Scholar [4] G.F. Carey and Y. Shen, Least-squares finite element approximation of Fishers reactiondiffusion equation, Numer. Methods Partial Differential Equations, 11 (1995), pp. 175-186. Google Scholar [5] I. Daǧ, A. Şahin, and A. Korkmaz, Numerical investigation of the solution of Fisher's equation via the B-spline Galerkin method, Numer. Methods Partial Differential Equations 26 (2010), no. 6, pp. 1483-1503. Google Scholar [6] R.A. Fisher, The wave of advance of advantageous genes, Ann. Eugenics, 7 (1937), pp. 355-369. Google Scholar [7] J. Gazdag and J. Canosa, Numerical solution of Fisher's equation, J. Appl. Probab., 11 (1974), pp. 445-457. Google Scholar [8] B.Y. Guo and Z.X. Chen, Analytic solutions of the Fisher equation, J. Phys. A, 24 (1991), no. 3, pp. 645-650. Google Scholar [9] P.S. Hagan, Traveling wave and multiple traveling wave solutions of parabolic equations, SIAM J. Math. Anal. 13 (1982), no. 5, pp. 717-738. Google Scholar [10] T. Hagstrom and H.B. Keller, The numerical calculation of traveling wave solutions of nonlinear parabolic equations, SIAM J. Sci. Statist. Comput., 7 (1986), no. 3, pp. 978-988. Google Scholar [11] A. Kolmogorov, I. Petrovshy, and N. Piscounoff, Étude de l'équation de la diffusion avec croissance de la quantité de matière et son application à un problème biologique, Bull. Univ. Etat Moscou Ser. Int. Sect. A Math. et Mecan., 1 (1937), pp. 1-25. Google Scholar [12] D.A. Larson, Transient bounds and time-asymptotic behavior of solutions to nonlinear equations of Fisher type, SIAM J. Appl. Math. 34 (1978), no. 1, pp. 93-103. Google Scholar [13] S. Li, L. Petzold, and Y. Ren, Stability of moving mesh systems of partial differential equations, SIAM J. Sci. Comput., 20 (1998), no. 2, pp. 719-738. Google Scholar [14] R. Lin, Discontinuous discretization for least-squares formulation of singularly perturbed reaction-diffusion problems in one and two dimensions,, SIAM J. Numer. Anal. 47 (2008/09), 47 (): 89. Google Scholar [15] R. Lin, Discontinuous Galerkin least-squares finite element methods for singularly perturbed reaction-diffusion problems with discontinuous coefficients and boundary singularities, Numer. Math. 112 (2009), no. 2, pp. 295-318. Google Scholar [16] J.D. Logan, "An introduction to nonlinear partial differential equations,'' second edition, Wiley-Interscience, John Wiley & Sons, Hoboken, NJ, 2008. Google Scholar [17] R.E. Mickens, A best finite-difference scheme for the Fisher equation, Numer. Methods Partial Differential Equations 10 (1994), no. 5, pp. 581-585. Google Scholar [18] J.D. Murray, "Mathematical biology,'' Biomathematics, 19, Springer-Verlag, Berlin, 1989. Google Scholar [19] D. Olmos and B.D. Shizgal, A pseudospectral method of solution of Fisher's equation, J. Comput. Appl. Math., 193 (2006), pp. 219-242. Google Scholar [20] N. Parekh and S. Puri, A new numerical scheme for the Fisher equation, J. Phys. A: Math. Gen., 23 (1990), pp. L1085-L1091. Google Scholar [21] Y. Qiu and D.M. Sloan, Numerical solution of Fisher's equation using a moving mesh method, J. Comput. Phys., 146 (1998), pp. 726-746. Google Scholar [22] Rizwan-uddin, Comparison of the nodal integral method and nonstandard finite-difference schemes for the Fisher equation, SIAM. J. Sci. Comput., 22 (2000), pp. 1926-1942. Google Scholar [23] J. Roessler and H. Hüssner, Numerical solution of the $1+2$ dimensional Fisher's equation by finite elements and the Galerkin method, Math. Comput. Modelling, 25 (1997), pp. 57-67. Google Scholar [24] S. Tang and R.O. Weber, Numerical study of Fisher's equation by a Petrov-Galerkin finite element method, J. Austral. Math. Soc. Sci. B, 33 (1991) pp. 27-38. Google Scholar [25] V. Thomée, "Galerkin finite element methods for parabolic problems,'' second edition, Springer Series in Computational Mathematics, 25, Springer-Verlag, Berlin, 2006. Google Scholar [26] S. Zhao and G.W. Wei, Comparison of the discrete singular convolution and three other numerical schemes for solving Fisher's equation, SIAM J. Sci. Comput., 25 (2003) pp. 127-147. Google Scholar [1] Lunji Song, Wenya Qi, Kaifang Liu, Qingxian Gu. A new over-penalized weak galerkin finite element method. Part Ⅱ: Elliptic interface problems. 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Electronic Research Archive, 2021, 29 (1) : 1859-1880. doi: 10.3934/era.2020095 Impact Factor:	2021-12-04 11:29:10	{"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": 0.7139110565185547, "perplexity": 4238.21051915257}, "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-2021-49/segments/1637964362969.51/warc/CC-MAIN-20211204094103-20211204124103-00193.warc.gz"}
https://physics.stackexchange.com/questions/467590/how-to-simplify-this-expression-in-dirac-notation	# How to simplify this expression in Dirac notation An expression cropped up in a homework problem that I'm not sure how to simplify. Consider the following, where $$\|x\rangle$$ is a position eigenstate and $$\|p_1\rangle, \|p_2\rangle$$ are momentum eigenstates: $$\int dx \|x\rangle\langle x\|p_1\rangle\langle p_2\|x\rangle\langle x\|$$ If there were only a single $$\|x\rangle\langle x\|$$ projector, I'd use the completeness relation $$\int dx \|x\rangle\langle x\|=\hat{1}$$ and be done with it. But there are two projectors, so I'm not sure how to deal with this - they aren't separable, so far as I can see. I have also tried writing out the position-momentum overlaps explicitly, turning the expression to: $$\int dx e^{i(p_1-p_2)x/\hbar}\|x\rangle\langle x\|$$ which strongly resembles the usual integral form of the Dirac delta, but with the extra projector factor I don't know how to deal with. How might I proceed with this? • It looks meaningless to me. Perhaps a typo? Can you give some context or describe the problem in more detail? – mike stone Mar 20 '19 at 12:21 • It's a dopey operator $\cal O$ whose matrix elements between $\langle x_a\|$ and $\|x_b\rangle$ are $e^{ix_a (p_1-p_2)/\hbar}\delta(x_b-x_a)$, but so what? – Cosmas Zachos Mar 20 '19 at 22:47 Recall that if we have a complete continuous eigenbasis $$\{\|a\rangle\}$$ of $$\hat{A}$$, we can write a function of the operator $$\hat{A}$$ as $$f(\hat{A}) = \int da\ \|a\rangle f(a)\langle a\|$$ where $$\hat{A}\|a\rangle= a\|a\rangle$$. So it looks like what you've got is $$e^{i\frac{\Delta p}{\hbar} \hat{x}} = \int dx\ \|x\rangle e^{i\frac{\Delta p}{\hbar} x} \langle x\| .$$	2020-02-18 04:56:20	{"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": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.907067596912384, "perplexity": 195.66394341116583}, "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-2020-10/segments/1581875143505.60/warc/CC-MAIN-20200218025323-20200218055323-00309.warc.gz"}
http://www.gamedev.net/topic/626871-drawindexedinstanced-on-feature-level-9-3/	• Create Account ## DrawIndexedInstanced on feature_level_9_3? Old topic! Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic. 5 replies to this topic ### #1hupsilardee Members 491 Like 0Likes Like Posted 23 June 2012 - 02:10 PM Now apparently DrawIndexedInstanced is supported on feature level 9_3 / vs_4_0_level_9_3, but it seems to be crashing for me, does anybody know why this might be happening? I'm definitely not exceeding the primitive, index or texture repeat restrictions, it's a 33 polygon tree drawn 4 times. The error is a "Unhandled exception at 0x76cbfc56 in XEngine11.exe: 0x0000087C: 0x87c." Weird that there seems to be no description text Here is an outline of the setup: Perhaps I have set up the input layout incorrectly? // Vertex Shader: normal and texcoord parts snipped VSOutput VSmain(float3 iPos : POSITION0, float3 iNorm : NORMAL0, float2 iTex : TEXCOORD0 ,float3 iInstPos : POSITION1) { float4 worldPos = mul(float4(iPos, 1), World); worldPos.xyz += iInstPos; float4 viewPos = mul(worldPos, View); float4 projPos = mul(viewPos, Proj); VSOutput output; output.Position = projPos; return output; } // Input layout D3D11_INPUT_ELEMENT_DESC desc[] = { // { name, index, format, slot, offset, class, steprate } // stream 0 { "POSITION", 0, DXGI_FORMAT_R32G32B32_FLOAT, 0, D3D11_APPEND_ALIGNED_ELEMENT, D3D11_INPUT_PER_VERTEX_DATA, 0 }, { "NORMAL", 0, DXGI_FORMAT_R32G32B32_FLOAT, 0, D3D11_APPEND_ALIGNED_ELEMENT, D3D11_INPUT_PER_VERTEX_DATA, 0 }, { "TEXCOORD", 0, DXGI_FORMAT_R32G32_FLOAT, 0, D3D11_APPEND_ALIGNED_ELEMENT, D3D11_INPUT_PER_VERTEX_DATA, 0 }, // stream 1 { "POSITION", 1, DXGI_FORMAT_R32G32B32_FLOAT, 1, D3D11_APPEND_ALIGNED_ELEMENT, D3D11_INPUT_PER_INSTANCE_DATA, 1 }, }; I've tried using TEXCOORD1 and NORMAL1 as the semantics for the instance position in the shader but that didnt work. Also, somebody else at the XNA forums seemed to have a similar problem, but theirs was solved by making sure the semantic index was explicitly typed in the shader, which is not my problem. I remember this was particularly fiddly to achieve in D3D9. Just hoping it doesn't take too long to figure out for D3D11. EDIT: I tried changing my index buffer to use 32 bit indices (DXGI_FORMAT_R32_UINT). Now the same crash occurs - at the line where the index buffer is bound to the pipeline. Edited by hupsilardee, 25 June 2012 - 04:14 AM. ### #2MJP Moderators 18192 Like 0Likes Like Posted 23 June 2012 - 04:36 PM That code you have looks okay...do you have the DEBUG flag enabled for your device? ### #3hupsilardee Members 491 Like 0Likes Like Posted 24 June 2012 - 03:50 AM I had D3D11_CREATE_DEVICE_DEBUG enabled, and in the DX control panel I put D3D9 on debug with maximum output and D3D11 with "Force On" and the project added to the executable list. So it gives me some interesting messages in the output window. D3D11: Removing Device. D3D11: WARNING: ID3D11Device::RemoveDevice: Device removal has been triggered for the following reason (DXGI_ERROR_DRIVER_INTERNAL_ERROR: There is strong evidence that the driver has performed an undefined operation; but it may be because the application performed an illegal or undefined operation to begin with.). [ EXECUTION WARNING #379: DEVICE_REMOVAL_PROCESS_POSSIBLY_AT_FAULT ] D3D11: Raising UM Driver Internal Error to continuable exception.First-chance exception at 0x76cbfc56 in XEngine11.exe: 0x0000087C: 0x87c. Unhandled exception at 0x76cbfc56 in XEngine11.exe: 0x0000087C: 0x87c. The program '[19512] XEngine11.exe: Native' has exited with code 2172 (0x87c). So it's a driver error? ### #4MJP Moderators 18192 Like 0Likes Like Posted 24 June 2012 - 01:53 PM Yeah it looks like you crashed the driver somehow. These kinds of things can be hard to track down. ### #5mhagain Members 12429 Like 0Likes Like Posted 24 June 2012 - 11:01 PM Try verifying against a WARP device with feature level 11 - that will help confirm if it's your code or level 9_3 that's causing you trouble. It appears that the gentleman thought C++ was extremely difficult and he was overjoyed that the machine was absorbing it; he understood that good C++ is difficult but the best C++ is well-nigh unintelligible. ### #6hupsilardee Members 491 Like 0Likes Like Posted 25 June 2012 - 03:40 AM Good idea. Should have thought of that earlier. As it turns out, the WARP device renders 4 models as expected, so it's level_9_3 causing the problem Old topic! Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.	2016-12-03 18:00:15	{"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": 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": 0.19870012998580933, "perplexity": 6509.111396073609}, "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-2016-50/segments/1480698541066.79/warc/CC-MAIN-20161202170901-00419-ip-10-31-129-80.ec2.internal.warc.gz"}
http://www.koreascience.or.kr/article/ArticleFullRecord.jsp?cn=HOGBB1_2012_v34n2_197	Assessment of the Impact of Climate Change on Marine Ecosystem in the South Sea of Korea • Journal title : Ocean and Polar Research • Volume 34, Issue 2, 2012, pp.197-199 • Publisher : Korea Institute of Ocean Science & Technology • DOI : 10.4217/OPR.2012.34.2.197 Title & Authors Assessment of the Impact of Climate Change on Marine Ecosystem in the South Sea of Korea Ju, Se-Jong; Kim, Se-Joo; Abstract According to the IPCC climate change scenario (A1B scenario), the surface seawater temperature of the South Sea of Korea by 2100 may be $\small{2-3.5^{\circ}C}$ higher than at present, and seawater pH may decrease from 8.1 to 7.8, due to the increase in atmospheric $\small{CO_2}$, which is predicted to increase in concentration from 380 to 750 ppm. These changes may not only intensify the strength of typhoons/storm surges but also affect the function and structure the marine ecosystem. In order to assess the impact of climate change on the marine ecosystem in Korean waters, the project named the 'Assessment of the impact of climate change on marine ecosystem in the South Sea of Korea' has been supported by the Ministry of Land, Transport and Maritime Affairs, from 2008. The goal of this project is to enhance our ability to adapt and prepare for the future environmental changes through the reliable predictions based on the knowledge obtained from projects like this. In this respect, this project is being conducted to investigate the effects of climate/marine environment changes (ocean warming and acidification), and to predict future changes of the structure and function of the ecosystem in the South Sea of Korea. This special issue contains 6 research articles, which are the highlights of the studies carried out through this project. Keywords climate change;marine ecosystem;South Sea of Korea;ocean warming;ocean acidification; Language Korean Cited by 1. 기후변화시나리오의 사회경제발전 양상을 고려한 행정구역별 도시성장 예측,김진수;박소영; 한국지형공간정보학회지, 2013. vol.21. 2, pp.53-62 2. 기후변화로 인한 수온상승이 굴양식 본양성 생산방식의 경제성에 미치는 영향분석,최종두;최영준; Ocean and Polar Research, 2014. vol.36. 2, pp.157-163 3. RCP시나리오 기반 CMIP5 GCMs을 이용한 동북아시아 미래 기후변화 불확실성 평가,신용희;정휘철; 환경영향평가, 2015. vol.24. 3, pp.205-216 1. Assessing uncertainty in future climate change in Northeast Asia using multiple CMIP5 GCMs with four RCP scenarios, Journal of Environmental Impact Assessment, 2015, 24, 3, 205 2. The Economic Feasibility Analysis of Grow out Phase Production of Oyster Farming by Rising Water Temperature, Ocean and Polar Research, 2014, 36, 2, 157 3. Urban Growth Prediction each Administrative District Considering Social Economic Development Aspect of Climate Change Scenario, Journal of Korean Society for Geospatial Information System, 2013, 21, 2, 53 References 1. Cheung WWL, Close C, Lam V, Watson R, Pauly D (2008) Application of macroecological theory to predict effects of climate change on global fisheries potential. Mar Ecol-Prog Ser 365:187-197 2. Hoegh-Guldberg O, Mumby PJ, Hooten AJ, Steneck RS, Greenfield P, Gomez E, Harvell CD, Sale PF, Edwards AJ, Caldeira K, Knowlton N, Eakin CM, Iglesias-Prieto R, Muthiga N, Bradbury RH, Dubi A, Hatziolos ME (2007) Coral reefs under rapid climate change and ocean acidification. Science 318:1737-1742 3. IPCC (2007) IPCC the Fourth Assessment Report of the Intergovernmental Panel on Climate Change (http://www.ipcc-nggip.iges.or.jp/public/gl/invs1.htm). Cambridge University Press, New York, 996 p 4. Rost B, Zondervan I, Wolf-Gladrow D (2008) Sensitivity of phytoplankton to future changes in ocean carbonate chemistryL current knowledge, contradictions and research directions. Mar Ecol-Prog Ser 373:227-237 5. Son YB, Noh JH, Ju SJ, Kim SH, Ryu JH (2012) Climatological variability of satellite-derived sea surface temperature and chlorophyll: the South Sea of Korea and East China Sea. Ocean and Polar Res 34(2):201-218	2017-01-21 00:41:18	{"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": 0, "mimetex.cgi": 2, "/images/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": 0.4508814215660095, "perplexity": 12313.481419048647}, "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/1484560280891.90/warc/CC-MAIN-20170116095120-00083-ip-10-171-10-70.ec2.internal.warc.gz"}
https://mytimefitnesstx.com/co203bzu/vibrational-energy-definition-7f77f2	The act of vibrating. Let me explain: Your vibration is your energetic frequency, it is … Vibrational medicine influences the body and energy field by changing its frequency. Vibrations fall into two categories: free and forced. Vibrational Medicine looks at the energy anatomy within and around the physical body. Internal Energy Explanation. This imprint upon raw energy sets into motion a vibrational rhythm pattern. As noted above, your vibration may often change because of energy shifts at the Universal level. b. The condition of being vibrated. Low vibrational living is a direct expression of your emotional state based on your inner world and how you perceive your life, yourself and those around you. Free vibrations occur when the system is disturbed momentarily and then allowed to move without restraint. It is energy that has been imprinted creating a chair. It dates back to the ancient civilizations of the Lemurians, Aztecs, Egyptians and the Chinese. For example, vibrations often lower at the time of a terrorist attack or a natural disaster, and vibrations generally increase when the world at large has good cause to feel optimistic about humanity’s future. Morse Potential. Why Your Vibration May Change. Definition of Vibrational Energy. If the weight is pulled down and released, the system will respond by vibrating vertically. We describe the energy minimum with a potential energy curve, called a potential well. Each chemical bond has a unique vibrational energy. Define vibrational. Just remembering that everything is energy vibrating at different frequencies will change your perception and belief about what is possible. In equilibrium, the system has minimum energy and the weight is at rest. vibrational synonyms, vibrational pronunciation, vibrational translation, English dictionary definition of vibrational. Having a low vibration in your life, is different to low vibrational living. For this case one can show Two atoms that can form a bond will do so to create a diatomic molecule when they approach each other closely. Everything which includes that relationship, your successful business, the promotion, ideal weight and everything else. Vibrational spectroscopy is a non-destructive identification method that measures the vibrational energy in a compound. n. 1. a. Every thought sends a vibration imprinting upon raw energy causing the energy to come alive taking a form, intangible or tangible. Vibrational Rhythm Patterns. Vibrational energy healing is also called harmonic healing, harmonic resonance healing, and vibrational medicine. This picture is applicable if a vibrational mode of frequency $$\Omega$$ relaxes by transferring its energy to another vibration nearby in energy ($$\infty _ {\alpha}$$), and the energy difference $$\omega _ {\beta}$$ being accounted for by a continuum of intermolecular motions. Imbalances and disharmonies are treated using frequencies from flower and gem essences, sound, therapeutic grade essential oils, radionics, homeopathics, acupuncture, color, hands-on healing, crystals and more. 2. Physics a. A classic example is provided by a weight suspended from a spring. Internal energy U of a system or a body with well defined boundaries is the total of the kinetic energy due to the motion of molecules and the potential energy associated with the vibrational motion and electric energy of atoms within molecules. Hi Beautiful! When you apply some of the ways to raise your vibrational frequency, you’ll realize that the Law of Vibration is real. 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Your vibration is real the Chinese the Universal level the chair you sit on, is a vibration imprinting raw! Identification method that measures the vibrational energy in a compound potential energy curve called... In a compound vibrations fall into two categories: free and forced curve, a! Upon raw energy causing the energy minimum with a potential energy curve, called a potential energy curve called...	2021-04-14 08:13:16	{"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": 0.2994747459888458, "perplexity": 1931.273660429313}, "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/1618038077336.28/warc/CC-MAIN-20210414064832-20210414094832-00066.warc.gz"}
http://www.solutioninn.com/a-study-was-aimed-at-assessing-the-effects-of-group	# Question A study was aimed at assessing the effects of group size and group characteristics on the generation of advertising concepts. To assess the influence of group size, groups of four and eight members were compared. For a random sample of four-member groups, the mean number of advertising concepts generated per group was 78.0 and the sample standard deviation was 24.4. For an independent random sample of eight-member groups, the mean number of advertising concepts generated per group was 114.7 and the sample standard deviation was 14.6. (In each case, the groups had a moderator.) Stating any assumptions that you need to make, test, at the 1% level, the null hypothesis that the population means are the same against the alternative that the mean is higher for eight-member groups. Sales0 Views24	2016-10-28 11:15:46	{"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": 0.8157307505607605, "perplexity": 779.0551160055592}, "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-2016-44/segments/1476988721606.94/warc/CC-MAIN-20161020183841-00384-ip-10-171-6-4.ec2.internal.warc.gz"}
http://libros.duhnnae.com/2017/jul2/149893297591-Final-state-interactions-in-the-decay-B0-to-c-K-High-Energy-Physics-Phenomenology.php	# Final state interactions in the decay $B^0 o η c K^$ - High Energy Physics - Phenomenology Final state interactions in the decay $B^0 o η c K^$ - High Energy Physics - Phenomenology - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online. Abstract: In this article, we study the final-state rescattering effects in the decay$B^0 \to \eta cK^*$, the numerical results indicate the corrections arecomparable with the contribution from the naive factorizable amplitude, and thetotal amplitudes can accommodate the experimental data. Autor: Zhi-Gang Wang Fuente: https://arxiv.org/	2019-03-21 06:28:49	{"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": 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": 0.7735429406166077, "perplexity": 10079.93903256669}, "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-2019-13/segments/1552912202496.35/warc/CC-MAIN-20190321051516-20190321073516-00309.warc.gz"}
http://www.talkstats.com/threads/renaming-all-variables-in-a-large-dataset.30266/	# renaming all variables in a large dataset #### Nicole19 ##### New Member Hello, I have 3 large datasets with over 800 variables that I need to merge. Problem is some of the datasets have the same variable names so I need to rename all the variables in each data set prior to merging. The renaming would be systematic as I would like all the variables in dataset1 to have an 'a' in front of the variable name, all variables in dataset2 need to have a 'b' in front of the current variable name and all the variables in dataset3 need to have a 'c' in front of the current variable name. I have ran the following code: Proc contents data=work.dataset1 out=varnames; data _null_; set varnames end=eof; if _n_=1 then put "rename"; newvarname= 'a'\| \| trim(name); put name '= ' newvarname; if eof then put ';'; run; I then copied and pasted the renamed list from the output into a new data step (note I only pasted a few variables here): data dataset1renamed (rename= ( External_Identifiers = birthExternal_Identifiers) ( GA_at_birth = birthGA_at_birth) ( Health_card_number = birthHealth_card_number) ( apgar01 = birthapgar01) (apgar05 = birthapgar05) (apgar10 = birthapgar10) (arterial_cord_blood_base_excess_ = birtharterial_cord_blood)); set dataset1; output; run; However when I try to run this code I get the following error messages for each variable that I tried to rename: ERROR 22-7: Invalid option name GA_AT_BIRTH. ERROR 22-322: Syntax error, expecting one of the following: a name, a quoted string, /, ;, _DATA_, _LAST_, _NULL_. ERROR 76-322: Syntax error, statement will be ignored. Does anyone have any suggestions to either fix my code or maybe a better way to rename all the variables in a dataset? Thanks so much for your help! #### Mean Joe ##### TS Contributor I have ran the following code: Proc contents data=work.dataset1 out=varnames; data _null_; set varnames end=eof; if _n_=1 then put "rename"; newvarname= 'a'\| \| trim(name); put name '= ' newvarname; if eof then put ';'; run; I do something that is just a little different. Code: %macro bigrenamer(dsin, prefix, dsout); Proc contents data=&dsin out=varnames; data _null_; set varnames; cvar=_N_; count # of variables; call symput("var"\|\|left(cvar), compress(variable)); use compress just in case the variable name has spaces (eg read in from Excel file); call symput("numvars",cvar); run; data &dsout; set &dsin; rename %let i=1; %do %while (%eval(&i)<=&numvars); &&var&i = &prefix..&&var&i %let i=%eval(&i+1); %* EDIT: Forgot this!; %end; ; %* the above might be wrong--might need two ampersands in front of prefix..; run; %mend bigrenamer; %bigrenamer(dataset1, a_, dataset1renamed); %bigrenamer(dataset2, b_, dataset2renamed); %bigrenamer(dataset3, c_, dataset3renamed);	2019-12-11 06:37:09	{"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": 0.3355850875377655, "perplexity": 9684.03532815236}, "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-2019-51/segments/1575540529955.67/warc/CC-MAIN-20191211045724-20191211073724-00473.warc.gz"}
https://curriculum.illustrativemathematics.org/MS/teachers/1/5/9/preparation.html	# Lesson 9 Using the Partial Quotients Method ### Lesson Narrative Prior to grade 6, students reasoned about division of whole numbers and decimals to the hundredths in different ways. In this first lesson on division, they revisit two methods for finding quotients of whole numbers without remainder: using base-ten diagrams and using partial quotients. Reviewing these strategies reinforces students’ understanding of the underlying principles of base-ten division—which are based on the structure of place value, the properties of operations, and the relationship between multiplication and division—and paves the way for understanding the long division algorithm. Here, partial quotients are presented as vertical calculations, which also foreshadows long division. In a previous unit, students revisited the two meanings of division—as finding the number of equal-size groups and finding the size of each group. Division is likewise interpreted in both ways here (MP2). When using base-ten diagrams or dividing by a small whole-number divisor, it is often natural to think about finding the size of each group. When using partial quotients, it may be more intuitive to think of division as finding the number of groups (e.g., $$432 \div 16$$ can be viewed as “how many 16s are in 432?”). ### Learning Goals Teacher Facing • Comprehend that the phrase “partial quotients” (in spoken and written language) refers to decomposing a division problem. • Divide whole numbers that result in a whole-number quotient, and explain the reasoning (using words and other representations). • Interpret different methods for computing the quotient of whole numbers, i.e., base-ten diagrams and partial quotients, and evaluate (orally) their usefulness. ### Student Facing Let’s divide whole numbers. ### Required Preparation Students draw base-ten diagrams in this lesson. If drawing them is a challenge, consider giving students access to: • Commercially produced base-ten blocks, if available. • Paper copies of squares and rectangles (to represent base-ten units), cut up from copies of the blackline master of the second lesson in the unit. • Digital applet of base-ten representations https://ggbm.at/zqxRkhMh Some students might find it helpful to use graph paper to help them align the digits as they divide using the partial quotients method. Consider having graph paper accessible throughout the lesson. ### Student Facing • I can use the partial quotients method to find a quotient of two whole numbers when the quotient is a whole number. Building On ### Print Formatted Materials For access, consult one of our IM Certified Partners.	2022-06-30 03:32:40	{"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": 0.6642336845397949, "perplexity": 2456.680069161977}, "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-2022-27/segments/1656103661137.41/warc/CC-MAIN-20220630031950-20220630061950-00011.warc.gz"}
https://www.zbmath.org/?q=an%3A1256.03068	# zbMATH — the first resource for mathematics Tense operators on $$m$$-symmetric algebras. (English) Zbl 1256.03068 Summary: Here we initiate an investigation of the equational classes of $$m$$-symmetric algebras endowed with two tense operators. These varieties are a generalization of tense algebras. Our main interest is the duality theory for these classes of algebras. In order to do this, we require Urquart’s duality for Ockham algebras and Goldblatt’s duality for bounded distributive lattice with operations. The dualities enable us to describe the lattices of congruences on tense $$m$$-symmetric algebras. ##### MSC: 03G25 Other algebras related to logic 03B44 Temporal logic 06D50 Lattices and duality Full Text:	2021-09-21 19:19:38	{"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": 0.3043248653411865, "perplexity": 2021.138830635768}, "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/1631780057227.73/warc/CC-MAIN-20210921191451-20210921221451-00592.warc.gz"}
https://en.m.wikisource.org/wiki/Royal_Naval_Biography/Berry,_Edward	# Royal Naval Biography/Berry, Edward SIR EDWARD BERRY, Baronet; Rear-Admiral of the Blue; and a Knight Commander oft Ik most honorable Military Order of the Bath. This officer is a son of the late Edward Berry, Esq., of London, by Elizabeth, daughter of the Rev. Thomas Foster, Rector of Holton, in Suffolk. He was born April 17, 1768, and went to sea at an early age, under the auspices of Lord Mulgrave. The first recorded circumstance of his life was the boarding of a French ship of war, for which action he was rewarded with a Lieutenant’s commission. His subsequent conduct in the glorious battle of June 1, 1794, appears also to have obtained for him the approbation of his superiors. Early in 1796, Mr. Berry was appointed by Sir John Jervis, under whom he had before served, to the Agamemnon, of 64 guns, commanded by Commodore Nelson, who was at that time employed in laying the foundation of his future fame; and to whose favorable notice he soon recommended himself, as may be inferred from the following passage, which we extract from a letter addressed by that officer to the Commander-in-Chief, May 30, 1796; “Lieutenant Berry joined me in the Comet, and I have, as far as I have seen, every reason to be satisfied with him, both as a gentleman, and an officer. I had a few days ago a plan for taking the French brig of war out of Vado, and intrusted the execution of it to him; it miscarried from an unforeseen and improbable event, but I was much pleased by Mr. Berry’s strict attention to my instructions.” Passing over occurrences of minor importance, we shall here introduce the contents of a paper written by Commodore Nelson, some time after the memorable battle off Cape St. Vincent; on which occasion the subject of this memoir, by his extraordinary activity in boarding two of the enemy’s ships, acquired the honest eulogium of every officer in the fleet. A few remarks relative to myself in the Captain, in which ship my pendant was flying on the most glorious Valentine’s Day, 1797. "At one, P.M., the Captain having passed the sternmost of the enemy’ ships, which formed their van and part of their centre, consisting of seventeen sail of the line, they on the larboard, we on the starboard tack; the Admiral made the signal to tack in succession, but perceiving all the Spanish ships to bear up before the wind, evidently with an intention of forming their line, going large, joined their separated divisions, at that time engaged with some of our centre ships, or flying from us, – to prevent either of their schemes from taking effect, I ordered the ship to be wore, and passing between the Diadem and Excellent, at a quarter past one o’clock, was engaged with the headmost, and of course leewardmost, of the Spanish division. The ships, which I knew, were the SantissimaTrinidada, 126; San Josef, 112; Salvador del Mundo, 112; San Nicholas, 80; another first rate, and a 74, names unknown. “I was immediately joined, and most nobly supported, by the Culloden, Captain Troubridge; the Spanish fleet, not wishing, I suppose, to have a decisive battle, hauled to the wind on the larboard tack, which brought the ships above-mentioned to be the leewardmost and sternmost ships in their fleet. For near an hour, I believe, (but do not pretend to be correct as to time,) did the Culloden, and Captain, support this apparently, but not really, unequal contest; when the Blenheim passing between us and the enemy, gave us a respite, and sickened the Dons. “At this time the Salvador del Mundo, and Sanysidro, dropped astern, and were fired into, in a masterly style, by the Excellent, Captain Collingwood, who compelled the Sanysidro to hoist English colours; and I thought the large ship, Salvador del Mundo, had also struck; but Captain Collingwood, disdaining the parade of taking possession of a vanquished enemy, most gallantly pushed up, with every sail set, to save his okl friend and messmate, who was to appearance in a critical state; the Blenheim being a-head, the Culloden crippled and a-stern. The Excellent ranged up within two feet of the San Nicholas, giving a most tremendous fire. The San Nicholas luffing up, the San Josef fell on board her; and the Excellent passing on for the Santa Trinidada, the Captain resumed her station abreast of them, and close alongside: at this time the Captain having lost her fore-top-mast, not a sail, shroud, nor rope left; her wheel shot away, and incapable of further service in the line, or in chaee; I directed Captain Miller to put the helm a-starboard, and calling for the boarders, ordered them to board. “The soldiers of the 69th, with an alacrity which will ever do them credit, and Lieutenant Pearson of the same regiment, were almost the foremost on this service:– the first man who jumped into the enemy’s mizen chains, was Captain Berry, late my first Lieutenant (Captain Miller was in the very act of going also, but I directed him to remain); he was supported from our sprit-sail yard, which hooked in the mizen rigging. A soldier of the 69th regiment having broke the upper quarter-gallery window, I jumped in myself, and was followed by others as fast ak possible. I found the cabin doors fastened, and some Spanish officers fired their pistols; but having broke open the doors, the soldiers fired; and the Spanish Brigadier (Commodore with a distinguishing pendant) fell, as retreating to the quarter-deck. I pushed immediately onwards for the quarter-deck, where I found Captain Berry in possession of the poop, and the Spanish ensign hauling down. I passed with my people, and Lieutenant Pearson, on the larboard gangway, to the forecastle, where I met two or three Spanish officers prisoners to my seamen – they delivered me their swords. A fire of pistols, or muskets, opening from the Admiral’s stern-gallery, of the San Josef, I directed the soldiers to fire into her stem; and calling to Captain Miller, ordered him to send more men into the San Nicholas, and directed my people to board the first rate, which was done in an instant, Captain Berry assisting me into the main chains. At this moment a Spanish officer looked over the quarter-deck rail, and said they surrendered. From this most welcome intelligence, it was not long before I was on the quarter-deck, where the Spanish Captain, with a bow, presented me his sword, and said the Admiral was dying of his wounds. I asked him on his honour, if the ship was surrendered? he declared she was; on which I gave him my hand, and desired him to call on his officers and ship’s company, and tell them of it; which he did: – and on the quarter-deck of a Spanish first rate, extravagant as the story may seem, did I receive the swords of vanquished Spaniards; which, as I received, I gave to William Fearney, one of my bargemen; who put them, with the greatest sang froid, under his arm. I was surrounded by Captain Berry, Lieutenant Pearson, of the 69th regiment, John Sykes, John Thomson, Francis Cooke, all old Agamemnons, and several other brave men, seamen and soldiers. Thus fell these ships! N.B. In boarding the San Nicholas, I believe we lost about 7 killed and 10 wounded; and about 20 Spaniards lost their lives by a foolish resistance. None were lost, I believe, in boarding the San Josef. (Signed) ${\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \\\\\ \ \end{matrix}}\right.}}$ HORATIO NELSON.RALPH WILLETT MILLER.E. BERRY.” Captain Berry’s post commission bears date March 6, 1797. In the course of the same year he appeared at Court with Sir Horatio Nelson; and it has been said that after the King had complimented the latter on account of his exploits, and condoled with him on his misfortune in losing a limb at the attack upon Santa Cruz, that hero introduced his companion to his Majesty, with the remark, “that he had not experienced great loss, as this officer was his right hand!” On the 19th Dec. following, Captain Berry commissioned the Vanguard, of 74 guns, fitting for the flag of his friend Nelson, with whom he soon after returned to the Mediterranean station. The proceedings of the squadron detached from the fleet off Cadiz to watch the armament about to sail from Toulon, under General Buonaparte, and which ended in the total defeat of the enemy, on the glorious 1st Aug. 1798, have been so amply related in our memoir of Sir James Saumarez[1], that it would be wholly superfluous to notice them again at large; we shall therefore content ourselves with observing, that notwithstanding the excessive damage which the Vanguard received in the Gulf of Lyons, Rear-Admiral Nelson, to whom the charge of the squadron had been confided by Earl St. Vincent, determined not to remove his flag from that ship, which was soon refitted by the great exertions of Captain Berry while at anchor in the Sardinian harbour of St. Pietro, from whence she again sailed in tolerable order. Soon after the termination of the tremendous conflict in Aboukir Bay, Captain Berry was sent to the Commander-in-Chief with the Rear-Admiral’s despatches, from which we make the following extract: “the support and assistance I have received from Captain Berry, cannot he sufficiently expressed. I was wounded in the head, and obliged to he carried off the deck, but the service suffered no loss by that event. Captain Berry was fully equal to the important service then going on, and to him I must beg leave to refer you for every information relative to this victory. He will present you with the flag of the second in command, that of the Commander-in-Chief being burnt in l’Orient. On his passage down the Mediterranean in the Leander, of 50 guns, commanded by the present Sir T. B. Thompson, our officer had the misfortune to be taken prisoner by le Généréux, a French 74. He also received a severe wound in the desperate action which took place on that occasion[2]. The enemy, on taking possession of their prize, not only plundered the officers and crew of every thing they possessed, but afterwards by their cruelty and neglect exposed the sick and wounded to almost certain death. However Captains Thompson and Berry were permitted to return, on their parole of honor, to England, where they were received by their countrymen with great applause. Sir Horatio Nelson’s duplicate despatches had in the mean time been brought home overland by the Hon. Captain Capel, and honors of every kind were decreed to the conquerors of the Nile. Captain Berry, after his exchange, was knighted by his Sovereign, received a gold medal in common with the other officers who had shared in the late triumph, and was presented with the freedom of the metropolis, in a gold box, value 100 guineas. He also received the thanks of the Court-Martial held to enquire into the circumstances attending the capture of the Leander, “for the gallant and active zeal he manifested, by giving his assistance in the combat.” In the autumn of 1799, Sir Edward Berry repaired once more to the Mediterranean, as Captain of Lord Nelson’s flagship, the Foudroyant; and early in the following year had the satisfaction of assisting at the capture of his old opponent, le Généreux[3], and of le Guillaume Tell, a French 80, the only remaining ship which had escaped from the battle in Aboukir Bay. A more heroic defence than that made by the latter vessel is not on record. Her colours were kept flying until she had become an ungovernable log, and sustained a loss of 200 men killed and wounded. During the action which took place on this occasion[4], Sir Edward Berry, who displayed the same matchless intrepidity and able conduct that he had often done before in many trying situations, was hurt in the foot, but not so much as to induce him to quit the deck. The Foudroyant’s total loss was 8 men killed and 61 wounded. In the month of June following, Sir Edward conveyed the Queen of Naples, her family and attendants, from Palermo to Leghorn, from which place Lord Nelson proceeded across the continent on his way to England, and the Foudroyant to Minorca to refit. Previous to the landing of the above personages, her Sicilian Majesty presented Sir Edward with a gold box, set with diamonds, and a diamond ring. Towards the latter end of the same year, our officer returned to England in the Princess Charlotte frigate; and during the remainder of the war, he commanded the Ruby, of 64 guns, stationed in the North Sea. His next appointment was in the summer of 1805, to the Agamemnon, another ship of the same class, in which, after having by the most masterly manoeuvres, escaped from a French squadron, consisting of five sail of the line, two frigates, and a brig, he joined Lord Nelson’s fleet, in time to participate in the glorious battle of Trafalgar; but it does not appear that any opportunity was afforded him of particularly distinguishing himself on that occasion. On the 6th Feb. 1806, we find Sir Edward in the same ship, assisting at the discomfiture of a French squadron at St. Domingo, by Sir John T. Duckworth[5]. For this service he was presented, by the Committee of the Patriotic Fund at Lloyd’s, with a sword, value 100l. During his continuance in the West Indies, he also contributed to the capture of la Lutine French national brig, of 18 guns; and the Dame Ernouf privateer, of 17 guns and 115 men. From the latter end of 1806, the period at which Sir Edward left the Agamemnon, we have no mention of him until the autumn of 1811, when he obtained the command of the Sceptre, a 74-gun ship, from which he was removed about the month of Sept., 1812, into the Barfleur, a second rate; and again sent to the Mediterranean. In Dec, 1813, Sir Edward was appointed to the Royal Sovereign yacht; and in the summer of the following year, we find him in attendance on the allied monarchs, during their visit to the fleet, at Spithead. He subsequently commanded the Royal George, another yacht; and on the 2d Jan. 1815, was nominated a K.C.B. At the general promotion, Aug. 12, 1819, he obtained one of the vacant Colonelcies of Royal Marines; and on the 19th July, 1821, was advanced to the rank of Rear-Admiral of the Blue. His patent of Baronetage bears date Dec. 12, 1806. Our officer married, Dec. 12, 1797, Louisa, daughter of the Rev. Samuel Forster, of Norwich, D.D. Residence.– Catton, Norfolk.	2019-12-08 10:40:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "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": 0.4122717082500458, "perplexity": 7794.047490197394}, "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-2019-51/segments/1575540508599.52/warc/CC-MAIN-20191208095535-20191208123535-00520.warc.gz"}
http://crypto.stackexchange.com/tags/forgery/hot?filter=year	# Tag Info 11 If Bob does NOT care to check signatures (as in the question), Eve can send ANY message she wants to Bob pretending to be Alice, including but not limited to messages Eve got from Alice; all Eve needs is Bob's public key (which, as the name implies, is assumed public knowledge thus known to Eve) and straight use of PGP. Therefore the right question is: Can ... 9 In addition to the performance problems poncho already mentioned when using RSA signatures without hashing I just want to add on the security warning of poncho: Reordering If you have a message $m>N$ with $N$ being the RSA modulus, then you have to perform at least 2 RSA signatures as $m$ does not longer fit into $Z_N$. Let us assume that it requires ... 6 Well, one reason to hash the data before signing it is because RSA can handle only so much data; we might want to sign messages longer than that. For example, suppose we are using a 2k RSA key; that means that the RSA operation can handle messages up to 2047 bits; or 255 bytes. We often want to sign messages longer than 255 bytes. By hashing the message ... 5 If we note $\|m\|$ the number of bits in the bytestring coding the message $m$, the first padding considered is $m\mapsto \tilde m=257\cdot2^{\|m\|}+m$, and the signature is $m\mapsto\tilde S(m)=S(\tilde m)=\tilde m^d\bmod N$, where $S$ is the textbook/naked RSA signing $m\mapsto m^d\bmod N$. Notice that for any $m$ small enough that $m^2$ can be signed, we can ... 5 The attack is even more simple with RSA than with symmetric keys, because the asymmetric encryption key is assumed to be public. Let me tell you a story involving Alice, Bob and Mallory :). Alice wants to send a message to Bob using RSA. Alice encrypts the message using Bob's public key and sends it Mallory performs a Man-In-The-Middle attack, and ... 3 As correctly pointed out in a comment, the authenticated encryption model assumes that the attacker knows the algorithm; the attacker can query the encryption oracle with any plaintext $P$ (and a unique nonce $N$) and get MAC-then-Encrypt ciphertext $C$; the attacker can query the decryption oracle with any string $C$ pretending to be a ciphertext. No ... 2 Considering the padding as an addition, padded message passed to sign is $m\cdot 2^{16}+0101$, $0101$ in hexadecimal, assuming padding is done on the lower bytes (for higher bytes the logic is just the same). Being $e$ the private exponent, and $m^2$ computed in the size of $m$, $(m\cdot 2^{16}+0101)^e \pmod m$ is very different from $(m^2\cdot ... 1 Given a set of (unhashed) Lamport signatures using the same key, an attacker can trivially forge a signature for any message whose$k$-th bit, for each$k$, is equal to the$k$-th bit of at least one of the signed messages. For example, let's say I know the Lamport signatures for the following 16-bit messages using the same key:$$m_1 = 0001111101110001 ... 1 I assume that the deadline for the homework is passed, so I will provide an answer: Let us assume that we have the public key$y=g^x \pmod p$and the private key to be$x$. Computing an ElGamal signature for a message$m \in Z_p^$amounts to: choosing$k\in Z_p^r\equiv g^k \pmod ps\equiv (m-xk)k^{-1} \pmod{p-1}$which is equivalent to$m\equiv ... Only top voted, non community-wiki answers of a minimum length are eligible	2014-09-02 19:05:27	{"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": 0.5214537382125854, "perplexity": 983.2021444528103}, "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/1409535922763.3/warc/CC-MAIN-20140909052842-00346-ip-10-180-136-8.ec2.internal.warc.gz"}
https://www.rdocumentation.org/packages/lidR/versions/1.3.0/topics/catalog_queries	# catalog_queries 0th Percentile ##### Extract LiDAR data based on a set of coordinates From a set of (x, y) coordinates corresponding to the centers of regions of interest (ROIs), for example a ground inventory, the function automatically extracts the lidar data associated with the ROIs from a catalog. The algorithm will do this even for ROIs falling on the edges of one or more tiles. The extracted lidar data can be buffered. In this case the function adds a buffer area around the ROIs, and the LAS object returned has an extra column named 'buffer' which indicates, for each point, if the point is in the buffer or from the ROI (see more in the section Buffer). lidR support .lax file. You will speed-up the computation a lot with a spatial index. ##### Usage catalog_queries(obj, x, y, r, r2 = NULL, buffer = 0, roinames = NULL, ...) ##### Arguments obj A LAScatalog object x vector. A set of x coordinates corresponding to the centers of the ROIs y vector. A set of y coordinates corresponding to the centers of the ROIs r numeric or vector. A radius or a set of radii of the ROIs. If only r is provided (r2 = NULL) it will extract data falling onto a disc. r2 numeric or vector. A radius or a set of radii of plots. If r2 is provided, the selection turns into a rectangular ROI (if r = r2 it is a square). buffer numeric. Adds a buffer area around the ROI. See relevant sections. roinames vector. A set of ROI names (the plot IDs, for example) to label the returned list. ... Any argument available in readLAS to reduce the amount of data loaded. ##### Value A list of LAS objects ##### Buffer If the ROIs are buffered then the LAS objects returned by the function have extra points. The LAS objects received by the user contain a special column called 'buffer', which indicates, for each point, if it comes from a buffered area or not. Points from non-buffered areas (i.e. the ROI) have a 'buffer' value of 0, while those from buffered areas have a 'buffer' value greater than 0. For a circular ROI, points in the buffered area have a buffer value of 1. For a rectangular ROI the points in the buffer area have a buffer value of 1, 2, 3 or 4, where 1 is the bottom buffer and 2, 3 and 4 are the left, top and right buffers, respectively. ##### Multicore computation The process is done using several cores. To change the settings of how a catalog is processed use catalog_options. ##### Aliases • catalog_queries ##### Examples # NOT RUN { # Build a LAScatalog catalog = catalog("<Path to a folder containing a set of .las or .laz files>") # Get coordinates from an external file X = runif(30, 690000, 800000) Y = runif(30, 5010000, 5020000) R = 25 # Return a List of 30 circular LAS objects of 25 m radius catalog %>% catalog_queries(X, Y, R) # Return a List of 30 square LAS objects of 50x50 m catalog %>% catalog_queries(X, Y, R, R) # Return a List of 30 circular LAS objects of 30 m radius. 25 m being the ROI and 5 m # being a buffered area. The LAS objects have an extra column called 'buffer' to # differentiate the points. catalog %>% catalog_queries(X, Y, R, buffer = 5) # Return a List of 30 circular LAS objects of 25 m radius for which only the fields X, Y and # Z have been loaded and Z values < 0 were removed. catalog %>% catalog_queries(X, Y, R, XYZonly = TRUE, filter = "-drop_z_below 0") # } Documentation reproduced from package lidR, version 1.3.0, License: GPL-3 ### Community examples Looks like there are no examples yet.	2021-01-26 22:28:36	{"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": 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": 0.2976209223270416, "perplexity": 3198.906441297811}, "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/1610704803737.78/warc/CC-MAIN-20210126202017-20210126232017-00308.warc.gz"}
http://math.stackexchange.com/questions/154499/help-with-serre-spectral-sequences	# Help with Serre spectral sequences I'm working through Hatcher's unfinished book Spectral Sequences in Algebraic Topology and have found myself stuck on Exercise 2 on page 23: Compute the Serre spectral sequence for homology with $\mathbb{Z}$ coefficients for the fibration $K(\mathbb{Z}_2,1) \rightarrow K(\mathbb{Z}_8,1) \rightarrow K(\mathbb{Z}_4,1).$ The question asks to compare with Example 1.6 in the text (a similar computation), but when I try to write out the first few pages, I can't see anything close to a nice pattern. Is there something I'm missing? - It would be nice to see what you've got for the $E_2$-page. Firstly, note that $K(\mathbb{Z}/m,1)$ is just an infinite dimensional lens space. We know (pp. 146 of Hatcher - Algebraic Topology) that these have homology $$H_i(K(\mathbb{Z}/m,1)) = \begin{cases} \mathbb{Z} &\text{ for } i=0 \\ \mathbb{Z}/m &\text{ for } i \text{ odd} \\ 0 &\text{ for } i \text{ even.} \\ \end{cases}$$ Thus we can calculate $E_2^{p,q} = H_p(K(\mathbb{Z}/4,1),H_q(K(\mathbb{Z}/2,1)))$. Note that because everything disappears in even dimensions our $E_2$-term looks like similar to the spectral sequence given in Example 1-6, except now we have $\mathbb{Z}/4$'s on the row $q=0$ (and $n \ne 0$). Once we have calculated the $E_2$-term we now need to work out the differentials. Again a similar methodology to the example in Hatcher now works. For example there is a $\mathbb{Z}/2$ in the $n=2$ diagonal, which must be killed. There is only one possible way to kill this, and that is for there to be a map $d_2:E_2^{3,0} = \mathbb{Z}/4 \to E_2^{1,1} = \mathbb{Z}/2$. This leaves a $\mathbb{Z}/2$ in the $(3,0)$ position - but this is nice, because then this leaves us with three $\mathbb{Z}/2$'s in the $n=3$ diagonal, which is where we get the $\mathbb{Z}/8$ we need in the homology.	2014-12-22 15:18:49	{"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": 0.929264485836029, "perplexity": 99.60542916013699}, "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/1418802775392.34/warc/CC-MAIN-20141217075255-00161-ip-10-231-17-201.ec2.internal.warc.gz"}
https://www.research.ed.ac.uk/en/publications/unboundedness-of-markov-complexity-of-monomial-curves-in-an-for-n	# Unboundedness of Markov complexity of monomial curves in A^n for n ≥ 4 Dimitra Kosta, Apostolos Thoma Research output: Contribution to journalArticlepeer-review ## Abstract Computing the complexity of Markov bases is an extremely challenging problem; no formula is known in general and there are very few classes of toric ideals for which the Markov complexity has been computed. A monomial curve $C$ in $A^3$ has Markov complexity two or three. Two if the monomial curve is complete intersection and three otherwise. Our main result shows that there is no $d \in N$ such that $m(C) \leq d$ for all monomial curves $C$ in $A^4$. The same result is true even if we restrict to complete intersections. We extend this result to all monomial curves in $A^n$, where $n \geq 4$. Original language English 106249 12 Journal of pure and applied algebra 224 6 21 Oct 2019 https://doi.org/10.1016/j.jpaa.2019.106249 Published - Jun 2020 ## Fingerprint Dive into the research topics of 'Unboundedness of Markov complexity of monomial curves in A^n for n ≥ 4'. Together they form a unique fingerprint.	2022-10-03 09:44:07	{"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": 0.23321187496185303, "perplexity": 430.80163585774585}, "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-2022-40/segments/1664030337404.30/warc/CC-MAIN-20221003070342-20221003100342-00798.warc.gz"}
https://math.stackexchange.com/questions/1683476/putting-unlabelled-balls-into-labeled-boxes	Putting unlabelled balls into labeled boxes [duplicate] How many ways are there to put $50$ unlabeled balls into $100$ labeled boxes? If empty boxes are allowed and you're also allowed to put as many balls as you want into a single box, would the answer be $\displaystyle\binom{149}{50}$? marked as duplicate by David K, Community♦Mar 4 '16 at 22:07 • Let $x_1, x_2,\dots,x_{100}$ be the number of balls in each box. It would be the number of non-negative integral solutions to $x_1+x_2+\dots+x_{100}=50$. What made you pick $149$? What made you pick $100$? – JMoravitz Mar 4 '16 at 21:57 • I actually did originally write 50. The 149 came from the formula $(n-1 + k) choose k$ – John Kyle Mar 4 '16 at 21:59 • You use /k. Remember the binomial coefficient $\binom{n}{r}$ is very different than $\frac{n}{r}$ and should not be confused. – JMoravitz Mar 4 '16 at 22:01 Let the balls be in line. Like this $$\circ \circ\circ\circ\circ\circ \dots \circ\circ\circ\circ\circ\circ$$ Then you divide them using $99$ barriers. Like this $$\circ \circ\circ\mid\mid \circ\ \circ\mid\circ \dots \circ\circ\mid\circ\circ\circ \mid \circ \mid \mid$$ The spaces between the barriers and the extreme left and extreme right spaces can be considered as (labelled, as they are ordered) boxes. Now, this becomes a permutation with repetition problem: you have $149$ elements, $50$ equal balls and $99$ equal barriers. So the answer is $$\dfrac{149!}{50!\cdot99!} = \binom{149}{50}$$	2019-08-21 23:14:07	{"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": 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": 0.8864602446556091, "perplexity": 416.50476498312395}, "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-2019-35/segments/1566027316549.78/warc/CC-MAIN-20190821220456-20190822002456-00451.warc.gz"}
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=135&t=5110	## Quiz 1 prep- 2014 #9 $\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$ $\Delta G^{\circ}= -RT\ln K$ $\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$ Liat Bainvoll 3G Posts: 8 Joined: Fri Sep 26, 2014 2:02 pm ### Quiz 1 prep- 2014 #9 Hi! The question is asking for the standard deltaG of reaction for the decomposition of mercury (II) oxide @ 298K. It gives the standard delta H of formation and he standard molar entropies. How do I solve this if I do not have the deltaS values? The answer should be +117.1kJ/mol Thank you! Olivia_1C Posts: 2 Joined: Sun Nov 23, 2014 3:00 am ### Re: Quiz 1 prep- 2014 #9 If you flip forward to page 10 in the workbook, you'll find a set a delta S values. They should all be there. claytonkersh2I Posts: 2 Joined: Tue Nov 25, 2014 3:00 am ### Re: Quiz 1 prep- 2014 #9 I'm still stuck on this question. OP, did you get the answer? Liat Bainvoll 3G Posts: 8 Joined: Fri Sep 26, 2014 2:02 pm ### Re: Quiz 1 prep- 2014 #9 but those are all standard molar S values like is given in the problem, don't we need delta S? Niharika Reddy 1D Posts: 127 Joined: Fri Sep 26, 2014 2:02 pm ### Re: Quiz 1 prep- 2014 #9 To solve for ΔS°r, or the standard reaction entropy, you use the standard molar entropies of the products and reactants in the balanced chemical equation. ΔS°r=(sum of nS°m(products))-(sum of nS°m(reactants)) Where n is the stoichiometric coefficient of the product/reactant from the balanced chemical equation and S°m is the standard molar entropy, which is given for all the products and reactants in the question itself. Then you will have all the information to find ΔG°r: ΔG°r=ΔH°r-TΔS°r claytonkersh2I Posts: 2 Joined: Tue Nov 25, 2014 3:00 am ### Re: Quiz 1 prep- 2014 #9 Found my mistake. Forgot to multiply the delta H by 2. Justin Le 2I Posts: 142 Joined: Fri Sep 26, 2014 2:02 pm ### Re: Quiz 1 prep- 2014 #9 For this problem, shouldn't the answer be 58.56 kj.mol-1, half of the answer given in the back? 58.56kj.mol-1 would be the delta G for the decomposition of one mole of HgO whereas 117.1 kj.mol-1 would be the answer for one mole of the reaction where the equation is 2 HgO(s) --> 2 Hg(l) + O2(g). Emilie Flores 2G Posts: 15 Joined: Fri Sep 25, 2015 3:00 am ### Re: Quiz 1 prep- 2014 #9 Justin Le. The question asks to calculate (delta) G for the BALANCED REACTION showing... Etc. Because the balanced reaction requires a coefficient of 2 in front of the HgO, the question is asking for the (delta) G of 2 moles of Mercury (II) Oxide. So you do not need to divide the change in free energy by 2 because the question is looking to find the value for 2 moles. Does this make sense? I can see that the question WAS worded kind of weird, I hope this helps. 604468944 Posts: 4 Joined: Mon Jan 26, 2015 2:17 pm ### Re: Quiz 1 prep- 2014 #9 Can anyone tell me why when we plug 2*90.83 to "delta G=delta H-T(delta S)" it is a positive? (If i plug it in as a negative my answer comes our wrong but as a positive it is right) I'm having trouble understanding that concept. Emilie Flores 2G Posts: 15 Joined: Fri Sep 25, 2015 3:00 am ### Re: Quiz 1 prep- 2014 #9 Remember, Delta H = (The Sum of Products H) - (The sum of the reactants H). The two blank spaces in indicate that the delta H values for both products are zero. So, delta H = 0 - 2(90.83) = -181.66 kJ/mol	2021-02-27 10:23:27	{"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": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.7633203268051147, "perplexity": 3614.182344401004}, "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/1614178358798.23/warc/CC-MAIN-20210227084805-20210227114805-00327.warc.gz"}
https://math.stackexchange.com/questions/4539438/hatcher-exercise-2-2-30a	# Hatcher Exercise 2.2.30(a) For the mapping torus $$T_f$$ of a map $$f:X\to X$$, we have a long exact sequence $$\cdots\to H_n(X)\xrightarrow{1-f_}H_n(X)\to H_n(T_f)\to H_{n-1}(X)\to\cdots.$$ Use this to compute the homology of the mapping tori of the following maps: (a) A reflection $$S^2\to S^2$$, (b) ... Anyway, I have $$0\to H_1(T_f)\to H_0(S^2)\to H_0(S^2)\to H_0(T_f)\to 0.$$ Since $$\deg(-1) = -1$$ at $$0$$ degree, $$1-(-1)_$$ is multiplication by $$2$$. Hence, $$H_1(T_f) =0$$ and $$H_0(T_f) = \Bbb Z/2$$ which is weird since $$H_0$$ should be free. I think interpreting the middle map by multiplication by $$2$$ is wrong but don't know how to correct. Please help. It is not true that $$\text{deg}(-1)= -1$$ at degree $$0$$. $$H_0(X)$$ is a free abelian group with generators being the path components of $$X$$. If $$f:X \to Y$$, then $$f_$$ maps the path components of $$X$$ to those of $$Y$$. Thus for $$X=Y=S^2$$ we always get $$f_= 1$$.	2022-11-26 15:25:21	{"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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 23, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9772405028343201, "perplexity": 74.43180117010995}, "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/1669446708010.98/warc/CC-MAIN-20221126144448-20221126174448-00831.warc.gz"}
https://www.ques10.com/p/41216/list-different-types-of-dynamometer-and-explain--1/	Question: List different types of dynamometer and explain strain gauge type dynamometer 0 Types of dynamometer : 1) Mechanical dynamometer 2) strain gauge type dynamometers 3) pneumatic and hydraulic dynamometer 4) piezoelectric dynamometer 5) Electrical dynamometer Strain gauge type dynamometer. fig. (a) use of strain gauges for two dimensional force measurement in turning on a lathe fig. (b) Wheatstone bridge circuit • This type of gauges are widely used because mechanical methods for measurement strain are not very reliable. • It is an electro-mechanical technique and the dynamometer used in process can be termed as electro-mechanical dynamometer. • wheatstone bridge circuit is commonly used in connection with electrical strain gauges. • The force being measured by this method are cutting force ( $f_c$ ) and thrust force ( $f_t$ ) • The basic principle involved in this technique is that, the electric resistance of wire changes when it is stretched. • The strain gauges are cemented on four flat structures. • Two sets of strain gauges are cemented, one set on top and bottom and second set on two vertical sides. • These two sets are connected one each, to separate Wheatstone bridge circuits. • During experiment, these two sets of strain gauges are subjected to tension and compression. • Resistance of the stain gauges subjected to tension increases strain gauges subjected to compression decreases. • These changes are measured by Wheatstone bridge. renu • 271 views	2019-10-15 19:58:13	{"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": 0.3625112771987915, "perplexity": 3130.2666811306767}, "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/1570986660231.30/warc/CC-MAIN-20191015182235-20191015205735-00090.warc.gz"}
https://alice-publications.web.cern.ch/node/1869	Multiplicity and transverse momentum evolution of charge-dependent correlations in pp, p-Pb, and Pb-Pb collisions at the LHC We report on two-particle charge-dependent correlations in pp, p-Pb, and Pb-Pb collisions as a function of the pseudorapidity and azimuthal angle difference, $\mathrm{\Delta}\eta$ and $\mathrm{\Delta}\varphi$ respectively. These correlations are studied using the balance function that probes the charge creation time and the development of collectivity in the produced system. The dependence of the balance function on the event multiplicity as well as on the trigger and associated particle transverse momentum ($p_{\mathrm{T}}$) in pp, p-Pb, and Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}} = 7$, 5.02, and 2.76 TeV, respectively, are presented. In the low transverse momentum region, for $0.2 <~ p_{\mathrm{T}} <~ 2.0$ GeV/$c$, the balance function becomes narrower in both $\mathrm{\Delta}\eta$ and $\mathrm{\Delta}\varphi$ directions in all three systems for events with higher multiplicity. The experimental findings favor models that either incorporate some collective behavior (e.g. AMPT) or different mechanisms that lead to effects that resemble collective behavior (e.g. PYTHIA8 with color reconnection). For higher values of transverse momenta the balance function becomes even narrower but exhibits no multiplicity dependence, indicating that the observed narrowing with increasing multiplicity at low $p_{\mathrm{T}}$ is a feature of bulk particle production. Figures Figure 1 The balance function $B(\mathrm{\Delta}\eta,\mathrm{\Delta}\varphi$) for charged particles with $0.2 < \pt_{,assoc}< \pt_{,trig}< 2.0$ Gev/$c$ in Pb-Pb, p-Pb, and pp collisions at $\sqrt{s_{\mathrm{NN}}} = 2.76$, 5.02, and 7 TeV, respectively. From top to bottom the $0-5\%$ for Pb-Pb and $0-10\%$ for p-Pb and pp collisions, $30-40\%$, and the $70-80\%$ multiplicity classes are shown. Figure 2 The balance function for charged particles with $0.2 < \pt_{,assoc}< \pt_{,trig}< 2.0$ Gev/$c$ as a function of $\deta$ on the near-side (upper row) and away-side (middle row) and $\dphi$ (lower row) indifferent multiplicity classes of Pb-Pb in panels (a), (d) and (g), p-Pb in panels (b), (e) and (h), and pp collisions in panels (c), (f) and (i) at $\sqrt{s_{\mathrm{NN}}} = 2.76$, 5.02, and 7 TeV, respectively. Figure 3 The balance function for charged particles with $0.2 < \pt_{,assoc}< \pt_{,trig}< 2.0$ Gev/$c$ as a function of $\deta$ on the near-side (upper row) and away-side (middle row) and as a function of $\dphi$ (lower row) for Pb-Pb (panels (a), (d) and (g)), p-Pb (panels (b), (e) and (h)) and pp collisions (panels (c), (f) and (i)) compared with results from various event generators. Only the highest multiplicity class is shown, i.e. 0-5 % for Pb-Pb and 0-10 % for p-Pb and pp collisions. Figure 4 The multiplicity-class dependence of $\sigma_{\Delta\eta}$ in Pb-Pb, p-Pb, and pp collisions at $\sqrt{s_{\mathrm{NN}}} = 2.76$, 5.02, and 7 TeV compared with results from various event generators in panels (a), (c), and (e). Panels (b), (d), and (f) show the relative decrease of $\sigma_{\Delta\eta}$ calculated with respect to $\sigma_{\Delta\eta}^{70-80\%}$, as a function of the multiplicity class. The transverse momentum values for both the trigger and the associated particles satisfy the condition $0.2 < \pt_{,assoc}< \pt_{,trig}< 2.0$ Gev/$c$. Figure 5 The multiplicity-class dependence of $\sigma_{\Delta\phi}$ in Pb-Pb, p-Pb, and pp collisions at $\sqrt{s_{\mathrm{NN}}} = 2.76$, 5.02, and 7 TeV compared with results from various event generators in panels (a), (c), and (e). Panels (b), (d), and (f) show the relative decrease of $\sigma_{\Delta\phi}$ calculated with respect to $\sigma_{\Delta\varphi}^{70-80\%}$ as a function of the multiplicity class. The transverse momentum values for both the trigger and the associated particles satisfy the condition $0.2 < \pt_{,assoc}< \pt_{,trig}< 2.0$ Gev/$c$. Figure 6 The balance function for charged particles with $2.0 < \pt_{,assoc}< 3.0< \pt_{,trig}< 4.0$ Gev/$c$ as a function of $\deta$ (upper row) and $\dphi$ (lower row) indifferent multiplicity classes of Pb-Pb, in panels (a) and (d), p-Pb, in panels (b) and (e), and pp collisions, in panels (c) and (f), at $\sqrt{s_{\mathrm{NN}}} = 2.76$, 5.02, and 7 TeV, respectively. Figure 7 The balance function for charged particles with $3.0 < \pt_{,assoc}< 8.0< \pt_{,trig}< 15.0$ Gev/$c$ as a function of $\deta$ (upper row) and $\dphi$ (lower row) indifferent multiplicity classes of Pb-Pb in panels (a) and (c) and p-Pb collisions in panels (b) and (d) at $\sqrt{s_{\mathrm{NN}}} = 2.76$ and 5.02 TeV, respectively. Figure 8 The multiplicity-class dependence of $\sigma_{\Delta\eta}$ (a) and $\sigma_{\Delta\phi}$ (b) in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}} = 2.76$ TeV. The different markers represent the low (i.e. $0.2 < \pt_{,assoc}< \pt_{,trig}< 2.0$ Gev/$c$ with red circles), intermediate (i.e $2.0 < p_\mathrm{T,assoc} < 3.0 < p_\mathrm{T,trig} < 4.0$ GeV/$c$ with blue squares), and high (i.e. $3.0 < \pt_{,assoc}< 8.0< \pt_{,trig}< 15.0$ Gev/$c$ with green triangles) transverse momentum regions used in this analysis. Figure 9 The multiplicity-class dependence of $\sigma_{\Delta\eta}$ (a) and $\sigma_{\Delta\phi}$ (b) in p-Pb collisions at $\sqrt{s_{\mathrm{NN}}} = 5.02$ TeV. The different markers represent the low (i.e. $0.2 < p_\mathrm{T,trig} < p_\mathrm{T,trig}< 2.0$ Gev/$c$ with red circles), intermediate (i.e $2.0 < p_\mathrm{T,assoc} < 3.0 < p_\mathrm{T,trig} < 4.0$ GeV/$c$ with blue squares), and high (i.e. $3.0 < p_\mathrm{T,assoc}< 8.0< p_\mathrm{T,trig}< 15.0$ Gev/$c$ with green triangles) transverse momentum regions used in this analysis. Figure 10 The multiplicity-class dependence of the width of the balance function in $\deta$ (a) and in $\dphi$ (b)in pp collisions at $\sqrt{s} = 7$ TeV. The results correspond to the intermediate transverse momentum region (i.e $2.0 < p_\mathrm{T,assoc} < 3.0 < p_\mathrm{T,trig} < 4.0$ GeV/$c$). The data points are compared with two versions of PYTHIA8 calculations. Figure 11 The width of the balance function in $\deta$ (a) and in $\dphi$ (b) for the three systemsanalyzed (pp, p-Pb, and Pb-Pb), as a function of the charged-particle multiplicity, estimated with the V0A for $\|\eta\| < 0.8$ and $p_{\rm{T}} > 0.2$ GeV/$c$. The low-, intermediate-, and high-$\pt$ intervals correspond to $0.2 < \pt_{,assoc}< \pt_{,trig}< 2.0$ Gev/$c$,$2.0 < \pt_{,assoc}< 3.0< \pt_{,trig}< 4.0$ Gev/$c$, and $3.0 < \pt_{,assoc}< 8.0< \pt_{,trig}< 15.0$ Gev/$c$, respectively. Figure 12 The multiplicity-class dependence of the width of the balance function in $\deta$ (a) and in $\dphi$ (b)for the three systems analyzed (pp, p-Pb, and Pb-Pb) relative to the 70-80% multiplicity class.	2018-07-21 02:08:30	{"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": 0.8215152621269226, "perplexity": 1959.1640553204095}, "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-2018-30/segments/1531676592150.47/warc/CC-MAIN-20180721012433-20180721032433-00505.warc.gz"}
https://tedboy.github.io/flask/flask_doc.foreword.html	# 1. Foreword¶ Read this before you get started with Flask. This hopefully answers some questions about the purpose and goals of the project, and when you should or should not be using it. ## 1.1. What does “micro” mean?¶ “Micro” does not mean that your whole web application has to fit into a single Python file (although it certainly can), nor does it mean that Flask is lacking in functionality. The “micro” in microframework means Flask aims to keep the core simple but extensible. Flask won’t make many decisions for you, such as what database to use. Those decisions that it does make, such as what templating engine to use, are easy to change. Everything else is up to you, so that Flask can be everything you need and nothing you don’t. By default, Flask does not include a database abstraction layer, form validation or anything else where different libraries already exist that can handle that. Instead, Flask supports extensions to add such functionality to your application as if it was implemented in Flask itself. Numerous extensions provide database integration, form validation, upload handling, various open authentication technologies, and more. Flask may be “micro”, but it’s ready for production use on a variety of needs. ## 1.2. Configuration and Conventions¶ Flask has many configuration values, with sensible defaults, and a few conventions when getting started. By convention, templates and static files are stored in subdirectories within the application’s Python source tree, with the names templates and static respectively. While this can be changed, you usually don’t have to, especially when getting started.	2023-04-01 23:08:07	{"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": 0.21364201605319977, "perplexity": 2150.736552499451}, "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-2023-14/segments/1679296950363.89/warc/CC-MAIN-20230401221921-20230402011921-00562.warc.gz"}
https://www.biostars.org/p/438147/	Dotplot of selected GO terms with z-score via ggplot2 0 0 Entering edit mode 17 months ago Hello everyone, By using the online tool David (https://david.ncifcrf.gov) for GO enrichment analysis and the R package GOplot (https://wencke.github.io) I was able to associate a z-score to my gene ontology results. I run the analysis for 3 different conditions (Cond1, Cond2, Cond3) and selected the most interesting GO terms. Now I have a data frame like this: GO ID term Cond1_PADJ Cond1_Zscore Cond2_PADJ Cond2_Zscore Cond3_PADJ Cond3_Zscore BP GO:0030198 extracellular matrix organization 3.89101E-05 -2.683281573 5.06412E-12 3.713069518 2.14344E-07 -5.099019514 BP GO:0030154 cell differentiation 0.000553172 -0.239045722 0.002762293 0.784464541 4.07139E-06 -8.373983252 At this point I would like to generate a dot plot pretty similar to the one discussed here (A: Dotplot for filtered pathways result) using ggplot2, but with some adjustments: 1) The size of the dots should be relative to the -log of the PADJ (instead of the Gene Ratio) 2) The colour of the dots should be relative to the Z-score (instead of the p-value) 3) I would like to separate the lanes of my samples like this (relevel the clusterProfiler object) I know there are some tools out there like Cluster profiler, Pathfindr, etc that can generate this graph, but I just need to plot my results. Thank you : ) RNA-Seq ggplot2 david GO • 1.5k views 1 Entering edit mode 0 Entering edit mode Thank you very much for the tutorial. I am trying to follow it but I am stuck at the plot step because what ggplot expects is basically this format: Identifier Variable1 Variable2_Cond1 Variable2_Cond2 Variable3_Cond3 Which in tutorial words would be: Sample Time OTU1 OTU2 OTU3 My problem is that I have 2 variables relative to each conditions. So It's like I have: Sample Time_OTU1 OTU1 Time_OTU2 OTU2 Time_OTU3 OTU3 Wich in my dataframe is: GO_ID Cond1_PADJ Cond1_Zscore Cond2_PADJ Cond2_Zscore Cond3_PADJ Cond3_Zscore How can I plot both variables? 0 Entering edit mode I actually think the problem is in the melt step, because I have 2 variables and not 1, for this reason I guess I should do something like in this post (https://stackoverflow.com/questions/1544907/melt-to-two-variable-columns). My long dataset should be something like this: GO_ID PADJ value Zscore Value GOID_1 Cond1 1 Cond1 4 GOID_1 Cond2 2 Cond2 5 GOID_1 Cond3 3 Cond2 6 But instead I get this: GO_ID Variable value GOID_1 Cond1_Pval 1 GOID_1 Cond2_Pval 2 GOID_1 Cond3_Pval 3 GOID_1 Cond1_ZScore 4 GOID_1 Cond2_ZScore 5 GOID_1 Cond3_ZScore 6 Any help? 1 Entering edit mode You can try pivot_longer() to convert from "wide" to "long": https://tidyr.tidyverse.org/reference/pivot_longer.html Then use separate() to split your columns: https://tidyr.tidyverse.org/reference/separate.html 0 Entering edit mode Thank you very much for all the suggestions, in the end I was able to have what I wanted. Now I have another problem: the order of my go terms in within the groups is the other way around! For example: Real order: GO1 Group1 GO2 Group1 GO3 Group1 GO1 Group2 GO2 Group2 GO3 Group2 GO2 Group3 GO1 Group3 by using: pcm$Groups <- factor(pcm$Groups,levels=unique(pcm$Groups)) pcm$term <- factor(pcm$term,levels=unique(pcm$term)) I can see that the order stays the same in the data frame, but when plotting: ggplot(pcm, aes(x = Pvalue, y = term))+ geom_point(aes(size = value, fill = value_2), alpha = 0.75, shape = 21, stroke = 0.5)+ scale_size_continuous(range = c(1,10)) + labs( x= "", y = "", size = "-LogPvalue", fill = "Z-score")+ scale_fill_gradient2(low = "blue",mid = "white", high ="red", midpoint = 0, space = "Lab", aesthetics="fill")+ theme(legend.key=element_blank(), axis.text.x = element_text(colour = "black", size = 14, angle = 90, vjust = 0.3, hjust = 1), axis.text.y = element_text(colour = "black", size = 14), legend.text = element_text(size = 10, colour ="black"), legend.title = element_text(size = 11), panel.background = element_blank(), panel.border = element_rect(colour = "black", fill = NA, size = 1), legend.position = "right", panel.grid.major.y = element_line(colour = "grey95"))+ facet_grid(Groups ~ ., scales = "free", space = "free",)+ theme(strip.text.y = element_text(size = 10, angle = 90,face = "bold")) The order whiten the groups goes in the opposite direction: GO3 Group1 GO2 Group1 GO1 Group1 GO3 Group2 GO2 Group2 GO1 Group2 GO2 Group3 GO1 Group3 Ideas? 0 Entering edit mode Solved: pcm$term <- fct_rev(pcm$term) this solved the problem! 0 Entering edit mode Instead of unique(pcm$term), try sort(unique(pcm$term))	2021-10-23 09:26:15	{"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": 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": 0.32966750860214233, "perplexity": 12553.840768187067}, "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-2021-43/segments/1634323585653.49/warc/CC-MAIN-20211023064718-20211023094718-00716.warc.gz"}
https://mathhelpforum.com/threads/restricted-domain-such-that-inverse-of-a-function-exists.144194/	# Restricted domain such that inverse of a function exists. #### Greensew Given function g is defined by g:x --> 2x^2 -8x+11 where x < or equal to A WHERE A IS A CONSTANT (a)state the largest value of A which g has an inverse Who know how to do this question?!?!?!?!?!!? #### CaptainBlack MHF Hall of Fame Given function g is defined by g:x --> 2x^2 -8x+11 where x < or equal to A WHERE A IS A CONSTANT (a)state the largest value of A which g has an inverse Who know how to do this question?!?!?!?!?!!? That the function has an inverse requires that there be one and only one solution $$\displaystyle x$$ that gives the value $$\displaystyle g(y)$$, and then $$\displaystyle g(x)=g(y).$$ As g(x) is a quadratic this is the case as long as $$\displaystyle x<x_0$$, where $$\displaystyle (x_0,g(x_0))$$ is the lowest point on the graph of $$\displaystyle y=g(x).$$ (Draw a picture) CB	2019-11-19 15:20:42	{"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": 0.799576461315155, "perplexity": 562.6901992171339}, "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-2019-47/segments/1573496670156.86/warc/CC-MAIN-20191119144618-20191119172618-00531.warc.gz"}
https://leeuwenloop.nl/26891/designing-of-vibrating-screen/crusher/2020-08-02/	• ## Vibrating Screen Working Principle IntroductionNamingMechanismApplicationsPurposeCauseUseSpecificationsSafetyOperationAssessmentCriticismsAnalysisExampleAdvantagesConstructionTypesDesignThe simplest Vibrating Screen Working Principle can be explained using the single deck screen and put it onto an inclined frame. The frame is mounted on springs. The vibration is generated from an unbalanced flywheel. A very erratic motion is developed when this wheel is rotated. You will find these simple screens in smaller operations and rock quarries where sizing isnt as critical. As the performance of this type of screen isnt good enough to meet the requirements of most mining operations two variations of this scre • ## High-frequency vibrating screens - Wikipedia OverviewCharacteristics assessmentRange of applicationsAdvantages and limitationsAvailable designsMain process characteristicsPossible heuristics to be used during design of the processPost-treatment systems The purpose of the vibrating screen is that particles are introduced to the gaps in the screens repeatedly. The frequency of the screen must be high enough so that it prevents the particles from blocking the apertures and the maximum height of the particle trajectory should occur when the screen surface is at its lowest point. Based on the principle, there is an optimum frequency and amplitude of vibration Transmission refers to the fraction of desired particle that passes through the apertures in the screen. At low frequency, screening efficiency is high but blinding is severe. Blinding will decrease as freq • 选择时，新内容将添加到焦点当前区域上方 The purpose of the vibrating screen is that particles are introduced to the gaps in the screens repeatedly. The frequency of the screen must be high enough so that it prevents the particles from blocking the apertures and the maximum height of the particle trajectory should occur when the screen surface is at its lowest point. Based on the principle, there is an optimum frequency and amplitude of vibration Transmission refers to the fraction of desired particle that passes through the apertures in the screen. At low frequency, screening efficiency is high but blinding is severe. Blinding will decrease as frequency increases but the particles will have difficulty going through the apertures. When designing a high frequency vibrating screen, an optimum point of frequency and amplitude must be chosen, depending on the specific applications. The separation efficiency is simply a measure of the amount of material removed by the screen compared to the theoretical amount that should have been removed. Screen efficiency can be obtained using different equation, which depends on whether the desired product is the oversize or undersize fraction from the screen. The screen efficiency based on the oversize (Eo) is given by: $${\displaystyle E_{o}={\frac {Q_{ms}(o)\;[1-M_{u}(o)]}{Q_{ms}(f)\;[1-M_{u}(f)]}}}$$The screen efficiency based on the undersize (Eu) is then given by: $${\displaystyle E_{u}={\frac {Q_{ms}(u)\;M_{u}(u)}{Q_{ms}(f)\;M_{u}(f)}}}$$where Qms(o) is the mass flow rate of solid in the screen overflow, Qms(f) is the mass flow rate of solid feed, Qms(u) is the mass flow rate of solid in the screen underflow, Mu(o) is the mass fraction of undersize in the overflow, Mu(f) is the mass fraction of undersize in the feed, Mu(u) is the mass fractio Wikipedia CC-BY-SA 许可下的文字 • ## SCREEN CAPACITY CALCULATION - VIBFEM VIBRATING SCREEN – CAPACITY CALCULATIONS Throughput per square foot of screen area is the name of the screen game, and no design engineer wants to be considered short in the area of capacity and efficiency. It behooves the buyer/operator to examine and evaluate the data available before committing to any screen type or system. • 文件大小: 607KB • ## Vibrating Screen Design Its Effect on Screen Motion ... Jan 08, 2015 Vibrating Screen Design Its Effect on Screen Motion Smico Manufacturing CO. ... Vibrating Screen, C, Vibrating Screen Design, Round Vibrating Screen - Xinhai - Duration: 0:18. • 作者: Smico Manufacturing CO • ## Screening Theory and Practice - Triple/S Dynamics The Horizontal Vibrating Screen: 840 rpm , 1/2” stroke at 45°. Each has a .063” dia. wire screen with 1/8” clear opening, moving under a particle travelling at an assumed 20 fpm, for A, 40 fpm for B., 80 fpm for C, and 60 fpm for D. Omitting details • ## Construction, Working and Maintenance of Electric ... 5 Construction, Working and Maintenance of Electric Vibrators and Vibrating Screens practicalmaintenance.net Unbalance Motors They are also known as vibration motors. Unbalance motors are provided and are suitable for driving vibrating systems, such as vibrating pipes, vibrating feeders, vibrating screens, • 文件大小: 2MB • ## Vibratory Screeners - Cleveland Vibrator Company Cleveland Vibrator Company offers multiple vibratory screener, scalper and sieve styles to handle separation and sizing of materials, or removal of unwanted materials from a batch, such as liquids, fines or over-sized product. Our full range of vibratory screeners are effective for 20 micron to 4" screen • ## Large Vibrating Screen Design Maintenance More often than Manufacturers admit, Designers plan for, or Operators staff for, a vibrating screen succeeds and self-destructs. This is a problem. It can magnify with larger vibrating screens. Design and Manufacturing Technology. Vibrating screen structures are subjected to nearly 250 million fatigue cycles in an operating year. • ## Vibrating Screen Design - bulk-online Forums Mar 25, 2004 Three key elements need to be addressed when designing a vibrating screen: SPEED STROKE SLOPE (or inclination) These guidelines are nicely described in the Vibrating Screen Manufacturers Association handbook. Describes the exact reasons why for all of the important elements required to be taken into consideration when designing a vibrating screen. • Vibrating Conveyor DesignJan 10, 2015Selection of Springs for Vibrating ScreenDec 22, 2014Vibrating Feeder Screen DesignNov 20, 2011Support Structure for a Vibrating ScreenJun 05, 2011查看更多结果 • ## Dynamic design theory and application of large vibrating ... The reliability is a key factor for the design and manufacture of large vibrating screen. In the paper, we presented a new large vibrating screen with hyperstatic net-beam structure. Dynamic characteristic of the vibrating screen • ## Vibrating Screens - Kinergy • Durability: Kinergy Vibrating Screens are built to last with an average of over 20 years of productive use and an availability of 95% or more. • Lowering Height: Kinergy’s Vibrating Screens can be • ## Vibrating screen: evaluating efficiency by using Discrete ... Aug 07, 2017 The analysis of vibrating screen design efficiency is therefore very important when designing or choosing the proper equipment for certain processes. Nevertheless, investigating these • ## Vibramech - Vibrating Mineral Processing Equipment Vibramech Footprint. Vibramech has an installed base of over 8 000 pieces of mineral processing equipment worldwide. We have supplied vibrating mineral processing machines extensively • ## Dewatering Screens - Vibramech Customised Design – Our Speciality. Vibramech has over 40 years of vibrating equipment design and mineral processing experience, which enables us to supply screens, feeders, grease tables and • ## Vibrating Feeder Screen Design - bulk-online Nov 20, 2011 I am a graduate engineer, now i get a job for vibrating feeder and screen design. i checked the design handbook. It seems like these is not so many information for the feeder and screen design • ## Vibrating screen SKF SKF Design the housing to be as symmetrical as possible, so it has the same thickness on both sides of the vibrating screen frame, in order to avoid housing deformation . Machine threads in the housing to make it easier when dismounting the housing from the screen • ## Dry Vibrating Screens for mining and industrial A Dry Vibrating Screen is used to screen dry products like minerals and is useful in many processes including classification, sizing, scalping and recycling. Designed for your specific operation, our Dry Vibrating Screens • ## TECHNICAL NOTES 4 VIBRATING SCREENS 4.1 Classification Based on Sieving - Vibrating screens The basic method of operation of a screen is very simple. The screen presents a barrier to the passage of oversize material while it readily passes undersize material. It is only necessary to ensure that each particle has an opportunity to reach the screen. • ## Dynamic design theory and application of large vibrating ... The reliability is a key factor for the design and manufacture of large vibrating screen. In the paper, we presented a new large vibrating screen with hyperstatic net-beam structure. • ## Vibrating Screens Puzzolana Brand, कंपन करने वाला Vibrating Screens Puzzolana Brand Get Latest Price Puzzolana centre shaft design Vibrating Screens are circular motion screens are with options in 5 sizes, Screen Deck options: Double, Triple • ## Designing Of Vibrating Screen - cz-eu.eu Designing Of Vibrating Screen. design vibrating screen - ksoe. Vibrating Screen Design - bulk-online Forums. Mar 25, 2004 Dear All, Could anybody advise me as to how should I carry out • ## Vibramech - Vibrating Mineral Processing Equipment Vibramech Footprint. Vibramech has an installed base of over 8 000 pieces of mineral processing equipment worldwide. We have supplied vibrating mineral processing machines extensively • ## Basic concepts of vibrating screens: What they are, what ... Basic concepts of vibrating screens: What they are, what they are for and how they work What are vibrating screens and which are its main applications for use. Also called simply screens, a vibrating screen • ## Vibrating Screen Types Working Principle [How To Choose ... linear Vibrating Screen. linear vibrating screen is driven by double vibrating motors. When the two vibrating motors are rotating synchronously and reversely, the excitation force generated by the • ## Vibrating screen SKF SKF Design the housing to be as symmetrical as possible, so it has the same thickness on both sides of the vibrating screen frame, in order to avoid housing deformation . Machine threads in the housing to make it easier when dismounting the housing from the screen • ## Vibrating Screen Handbook - PDF Downloadable PDF, in black and white. Chapters cover the history of screening, types of screening media, selection of screen size and type, engineering data, problems and solutions, crusher circuits, • ## General SCREEN Information - Deister Machine Vibrating C. Removal From and Re-attachment of Vibrating Frame to Stationary Base or Structure Should it be necessary to remove the vibrating screen frame from the stationary base in order to facilitate • ## Vibrating screens Dewatering Screens Vibrating Design TARNOS manufactures a very wide range of Vibrating Screens depending on the job to perform, working conditions and other conditionings of the installation.. All of our vibrating screens are robustly constructed, and their constant improvements on design • ## Round Vibrating Screen, Vibrating Screen Design, Vibrating ... Round vibrating screen adopts eccentric shaft vibration exciter and eccentric block to adjust amplitude. The vibrator installed on the sideboard of screen box forces screen box to vibrate by the driving of motor. The sideboard of vibrating screen • ## VSMA screen calc method - AggFlow DM Title: VSMA screen calc method Author: Bryan Created Date: 1/2/2012 11:45:56 AM • ## Improved local design of Enduron vibrating screens from Weir Mar 29, 2020 With its extensive range of Enduron vibrating screens proven over the past 40 years, Weir Minerals Africa is now locally designing and manufacturing new-generation linear motion vibrating screens. According to Christian Stehle, head of engineering at Weir Minerals Africa, this design capability provides the flexibility to produce vibrating ...	2020-09-29 17:42:00	{"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": 0.18706810474395752, "perplexity": 5496.483937350649}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2020-40/segments/1600400202418.22/warc/CC-MAIN-20200929154729-20200929184729-00012.warc.gz"}
https://roundingcalculator.guru/10-km-to-steps/	# 10 Km to Steps Calculator 10 KM in Steps: Use our kilometer to steps calculatortool to convert 10 kilometers to steps easily without any confusion. Not only that, we have provided you with step-by-step explanations and with examples. Km to Steps ## How Many Steps in 10 Kilometers? Here we are going to provide the step-by-step process to know the steps for 10 kilometers easily. • Firstly, write down the kilometers that were given in the problem. i.e., 10 kilometers. • Then, write down the formula, i.e., Steps = km x conversion factor value. • Now substitute the values in the formula, i.e., Steps = 10 km x(1312.33595801). • Simplify the equation, i.e., Steps = 10 x 1312.33595801 • Finally you will see the result of steps, i.e., Steps = 13123 km in Steps Calculation Examples Steps in km Calculation Examples ### FAQs on 10 Kilometers to Steps Calculator 1. What are 10 kilometers in steps? 10 kilometers is equal to 13123 steps. 1. Where do I get the detailed steps for 10 kilometers to step conversion? You can get the detailed steps for 10 kilometers to steps conversion on our page. 1. How do you convert 10 km into steps? To convert 10 km into steps, you just need to multiply the quantity in kilometers with the conversion factor. i.e., steps = 10 x 1312.33595801 = 13123 steps.	2023-01-28 19:31:50	{"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": 0.8118516802787781, "perplexity": 1882.1135586797675}, "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-2023-06/segments/1674764499654.54/warc/CC-MAIN-20230128184907-20230128214907-00567.warc.gz"}
https://cms.math.ca/cjm/msc/22?page=5	location: Publications → journals Search results Search: MSC category 22 ( Topological groups, Lie groups ) Expand all Collapse all Results 101 - 125 of 133 101. CJM 2001 (vol 53 pp. 675) Ban, Dubravka Jacquet Modules of Parabolically Induced Representations and Weyl Groups The representation parabolically induced from an irreducible supercuspidal representation is considered. Irreducible components of Jacquet modules with respect to induction in stages are given. The results are used for consideration of generalized Steinberg representations. Category:22E50 102. CJM 2001 (vol 53 pp. 278) Helminck, G. F.; van de Leur, J. W. Darboux Transformations for the KP Hierarchy in the Segal-Wilson Setting In this paper it is shown that inclusions inside the Segal-Wilson Grassmannian give rise to Darboux transformations between the solutions of the $\KP$ hierarchy corresponding to these planes. We present a closed form of the operators that procure the transformation and express them in the related geometric data. Further the associated transformation on the level of $\tau$-functions is given. Keywords:KP hierarchy, Darboux transformation, Grassmann manifoldCategories:22E65, 22E70, 35Q53, 35Q58, 58B25 103. CJM 2001 (vol 53 pp. 244) Goldberg, David; Shahidi, Freydoon On the Tempered Spectrum of Quasi-Split Classical Groups II We determine the poles of the standard intertwining operators for a maximal parabolic subgroup of the quasi-split unitary group defined by a quadratic extension $E/F$ of $p$-adic fields of characteristic zero. We study the case where the Levi component $M \simeq \GL_n (E) \times U_m (F)$, with $n \equiv m$ $(\mod 2)$. This, along with earlier work, determines the poles of the local Rankin-Selberg product $L$-function $L(s, \tau' \times \tau)$, with $\tau'$ an irreducible unitary supercuspidal representation of $\GL_n (E)$ and $\tau$ a generic irreducible unitary supercuspidal representation of $U_m (F)$. The results are interpreted using the theory of twisted endoscopy. Categories:22E50, 11S70 104. CJM 2001 (vol 53 pp. 195) Mokler, Claus On the Steinberg Map and Steinberg Cross-Section for a Symmetrizable Indefinite Kac-Moody Group Let $G$ be a symmetrizable indefinite Kac-Moody group over $\C$. Let $\Tr_{\La_1},\dots,\Tr_{\La_{2n-l}}$ be the characters of the fundamental irreducible representations of $G$, defined as convergent series on a certain part $G^{\tralg} \subseteq G$. Following Steinberg in the classical case and Br\"uchert in the affine case, we define the Steinberg map $\chi := (\Tr_{\La_1},\dots, \Tr_{\La_{2n-l}})$ as well as the Steinberg cross section $C$, together with a natural parametrisation $\omega \colon \C^{n} \times (\C^\times)^{\,n-l} \to C$. We investigate the local behaviour of $\chi$ on $C$ near $\omega \bigl( (0,\dots,0) \times (1,\dots,1) \bigr)$, and we show that there exists a neighborhood of $(0,\dots,0) \times (1,\dots,1)$, on which $\chi \circ \omega$ is a regular analytical map, satisfying a certain functional identity. This identity has its origin in an action of the center of $G$ on~$C$. Categories:22E65, 17B65 105. CJM 2000 (vol 52 pp. 1192) Herb, Rebecca A. Orbital Integrals on $p$-Adic Lie Algebras Let $G$ be a connected reductive $p$-adic group and let $\frakg$ be its Lie algebra. Let $\calO$ be any $G$-orbit in $\frakg$. Then the orbital integral $\mu_{\calO}$ corresponding to $\calO$ is an invariant distribution on $\frakg$, and Harish-Chandra proved that its Fourier transform $\hat \mu_{\calO}$ is a locally constant function on the set $\frakg'$ of regular semisimple elements of $\frakg$. If $\frakh$ is a Cartan subalgebra of $\frakg$, and $\omega$ is a compact subset of $\frakh\cap\frakg'$, we give a formula for $\hat \mu_{\calO}(tH)$ for $H\in\omega$ and $t\in F^{\times}$ sufficiently large. In the case that $\calO$ is a regular semisimple orbit, the formula is already known by work of Waldspurger. In the case that $\calO$ is a nilpotent orbit, the behavior of $\hat\mu_{\calO}$ at infinity is already known because of its homogeneity properties. The general case combines aspects of these two extreme cases. The formula for $\hat\mu _{\calO}$ at infinity can be used to formulate a theory of the constant term'' for the space of distributions spanned by the Fourier transforms of orbital integrals. It can also be used to show that the Fourier transforms of orbital integrals are linearly independent at infinity.'' Categories:22E30, 22E45 106. CJM 2000 (vol 52 pp. 1101) Zhang, Yuanli Discrete Series of Classical Groups Let $G_n$ be the split classical groups $\Sp(2n)$, $\SO(2n+1)$ and $\SO(2n)$ defined over a $p$-adic field F or the quasi-split classical groups $U(n,n)$ and $U(n+1,n)$ with respect to a quadratic extension $E/F$. We prove the self-duality of unitary supercuspidal data of standard Levi subgroups of $G_n(F)$ which give discrete series representations of $G_n(F)$. Category:22E35 107. CJM 2000 (vol 52 pp. 804) Kottwitz, Robert E.; Rogawski, Jonathan D. The Distributions in the Invariant Trace Formula Are Supported on Characters J.~Arthur put the trace formula in invariant form for all connected reductive groups and certain disconnected ones. However his work was written so as to apply to the general disconnected case, modulo two missing ingredients. This paper supplies one of those missing ingredients, namely an argument in Galois cohomology of a kind first used by D.~Kazhdan in the connected case. Categories:22E50, 11S37, 10D40 108. CJM 2000 (vol 52 pp. 539) Jantzen, Chris On Square-Integrable Representations of Classical $p$-adic Groups In this paper, we use Jacquet module methods to study the problem of classifying discrete series for the classical $p$-adic groups $\Sp(2n,F)$ and $\SO(2n+1,F)$. Category:22E50 109. CJM 2000 (vol 52 pp. 449) An Intertwining Result for $p$-adic Groups For a reductive $p$-adic group $G$, we compute the supports of the Hecke algebras for the $K$-types for $G$ lying in a certain frequently-occurring class. When $G$ is classical, we compute the intertwining between any two such $K$-types. Categories:22E50, 20G05 110. CJM 2000 (vol 52 pp. 412) Varopoulos, N. Th. Geometric and Potential Theoretic Results on Lie Groups The main new results in this paper are contained in the geometric Theorems 1 and~2 of Section~0.1 below and they are related to previous results of M.~Gromov and of myself (\cf\ \cite{1},~\cite{2}). These results are used to prove some general potential theoretic estimates on Lie groups (\cf\ Section~0.3) that are related to my previous work in the area (\cf\ \cite{3},~\cite{4}) and to some deep recent work of G.~Alexopoulos (\cf\ \cite{5},~\cite{21}). Categories:22E30, 43A80, 60J60, 60J65 111. CJM 2000 (vol 52 pp. 438) Wallach, N. R.; Willenbring, J. On Some $q$-Analogs of a Theorem of Kostant-Rallis In the first part of this paper generalizations of Hesselink's $q$-analog of Kostant's multiplicity formula for the action of a semisimple Lie group on the polynomials on its Lie algebra are given in the context of the Kostant-Rallis theorem. They correspond to the cases of real semisimple Lie groups with one conjugacy class of Cartan subgroup. In the second part of the paper a $q$-analog of the Kostant-Rallis theorem is given for the real group $\SL(4,\mathbb{R})$ (that is $\SO(4)$ acting on symmetric $4 \times 4$ matrices). This example plays two roles. First it contrasts with the examples of the first part. Second it has implications to the study of entanglement of mixed 2 qubit states in quantum computation. Categories:22E47, 20G05 112. CJM 2000 (vol 52 pp. 306) Cunningham, Clifton Characters of Depth-Zero, Supercuspidal Representations of the Rank-2 Symplectic Group This paper expresses the character of certain depth-zero supercuspidal representations of the rank-2 symplectic group as the Fourier transform of a finite linear combination of regular elliptic orbital integrals---an expression which is ideally suited for the study of the stability of those characters. Building on work of F.~Murnaghan, our proof involves Lusztig's Generalised Springer Correspondence in a fundamental way, and also makes use of some results on elliptic orbital integrals proved elsewhere by the author using Moy-Prasad filtrations of $p$-adic Lie algebras. Two applications of the main result are considered toward the end of the paper. Categories:22E50, 22E35 113. CJM 1999 (vol 51 pp. 1135) Arthur, James Endoscopic $L$-Functions and a Combinatorial Identity The trace formula contains terms on the spectral side that are constructed from unramified automorphic $L$-functions. We shall establish an identify that relates these terms with corresponding terms attached to endoscopic groups of $G$. In the process, we shall show that the $L$-functions of $G$ that come from automorphic representations of endoscopic groups have meromorphic continuation. Categories:22E45, 22E46 114. CJM 1999 (vol 51 pp. 1307) Johnson, Norman W.; Weiss, Asia Ivić Quadratic Integers and Coxeter Groups Matrices whose entries belong to certain rings of algebraic integers can be associated with discrete groups of transformations of inversive $n$-space or hyperbolic $(n+1)$-space $\mbox{H}^{n+1}$. For small $n$, these may be Coxeter groups, generated by reflections, or certain subgroups whose generators include direct isometries of $\mbox{H}^{n+1}$. We show how linear fractional transformations over rings of rational and (real or imaginary) quadratic integers are related to the symmetry groups of regular tilings of the hyperbolic plane or 3-space. New light is shed on the properties of the rational modular group $\PSL_2 (\bbZ)$, the Gaussian modular (Picard) group $\PSL_2 (\bbZ[{\it i}])$, and the Eisenstein modular group $\PSL_2 (\bbZ[\omega ])$. Categories:11F06, 20F55, 20G20, 20H10, 22E40 115. CJM 1999 (vol 51 pp. 952) Deitmar, Anton; Hoffmann, Werner On Limit Multiplicities for Spaces of Automorphic Forms Let $\Gamma$ be a rank-one arithmetic subgroup of a semisimple Lie group~$G$. For fixed $K$-Type, the spectral side of the Selberg trace formula defines a distribution on the space of infinitesimal characters of~$G$, whose discrete part encodes the dimensions of the spaces of square-integrable $\Gamma$-automorphic forms. It is shown that this distribution converges to the Plancherel measure of $G$ when $\Ga$ shrinks to the trivial group in a certain restricted way. The analogous assertion for cocompact lattices $\Gamma$ follows from results of DeGeorge-Wallach and Delorme. Keywords:limit multiplicities, automorphic forms, noncompact quotients, Selberg trace formula, functional calculusCategories:11F72, 22E30, 22E40, 43A85, 58G25 116. CJM 1999 (vol 51 pp. 835) Kim, Henry H. Langlands-Shahidi Method and Poles of Automorphic $L$-Functions: Application to Exterior Square $L$-Functions In this paper we use Langlands-Shahidi method and the result of Langlands which says that non self-conjugate maximal parabolic subgroups do not contribute to the residual spectrum, to prove the holomorphy of several \emph{completed} automorphic $L$-functions on the whole complex plane which appear in constant terms of the Eisenstein series. They include the exterior square $L$-functions of $\GL_n$, $n$ odd, the Rankin-Selberg $L$-functions of $\GL_n\times \GL_m$, $n\ne m$, and $L$-functions $L(s,\sigma,r)$, where $\sigma$ is a generic cuspidal representation of $\SO_{10}$ and $r$ is the half-spin representation of $\GSpin(10, \mathbb{C})$. The main part is proving the holomorphy and non-vanishing of the local normalized intertwining operators by reducing them to natural conjectures in harmonic analysis, such as standard module conjecture. Categories:11F, 22E 117. CJM 1999 (vol 51 pp. 816) Hall, Brian C. A New Form of the Segal-Bargmann Transform for Lie Groups of Compact Type I consider a two-parameter family $B_{s,t}$ of unitary transforms mapping an $L^{2}$-space over a Lie group of compact type onto a holomorphic $L^{2}$-space over the complexified group. These were studied using infinite-dimensional analysis in joint work with B.~Driver, but are treated here by finite-dimensional means. These transforms interpolate between two previously known transforms, and all should be thought of as generalizations of the classical Segal-Bargmann transform. I consider also the limiting cases $s \rightarrow \infty$ and $s \rightarrow t/2$. Categories:22E30, 81S30, 58G11 118. CJM 1999 (vol 51 pp. 636) Paul, Annegret First Occurrence for the Dual Pairs $\bigl(U(p,q),U(r,s)\bigr)$ We prove a conjecture of Kudla and Rallis about the first occurrence in the theta correspondence, for dual pairs of the form $\bigl(U(p,q),U(r,s)\bigr)$ and most representations. Category:22E46 119. CJM 1999 (vol 51 pp. 266) Deitmar, Anton; Hoffman, Werner Spectral Estimates for Towers of Noncompact Quotients We prove a uniform upper estimate on the number of cuspidal eigenvalues of the $\Ga$-automorphic Laplacian below a given bound when $\Ga$ varies in a family of congruence subgroups of a given reductive linear algebraic group. Each $\Ga$ in the family is assumed to contain a principal congruence subgroup whose index in $\Ga$ does not exceed a fixed number. The bound we prove depends linearly on the covolume of $\Ga$ and is deduced from the analogous result about the cut-off Laplacian. The proof generalizes the heat-kernel method which has been applied by Donnelly in the case of a fixed lattice~$\Ga$. Categories:11F72, 58G25, 22E40 120. CJM 1999 (vol 51 pp. 164) Tan, Victor Poles of Siegel Eisenstein Series on $U(n,n)$ Let $U(n,n)$ be the rank $n$ quasi-split unitary group over a number field. We show that the normalized Siegel Eisenstein series of $U(n,n)$ has at most simple poles at the integers or half integers in certain strip of the complex plane. Categories:11F70, 11F27, 22E50 121. CJM 1999 (vol 51 pp. 130) Savin, Gordan; Gan, Wee Teck The Dual Pair $G_2 \times \PU_3 (D)$ ($p$-Adic Case) We study the correspondence of representations arising by restricting the minimal representation of the linear group of type $E_7$ and relative rank $4$. The main tool is computations of the Jacquet modules of the minimal representation with respect to maximal parabolic subgroups of $G_2$ and $\PU_3(D)$. Categories:22E35, 22E50, 11F70 122. CJM 1998 (vol 50 pp. 1090) Lohoué, Noël; Mustapha, Sami Sur les transformÃ©es de Riesz sur les groupes de Lie moyennables et sur certains espaces homogÃ¨nes Let $\Delta$ be a left invariant sub-Laplacian on a Lie group $G$ and let $\nabla$ be the associated gradient. In this paper we investigate the boundness of the Riesz transform $\nabla\Delta^{-1/2}$ on Lie groups $G$ which are amenable and with exponential volume growth and on certain homogenous spaces. Categories:22E30, 35H05, 43A80, 43A85 123. CJM 1998 (vol 50 pp. 1105) Roberts, Brooks Tempered representations and the theta correspondence Let $V$ be an even dimensional nondegenerate symmetric bilinear space over a nonarchimedean local field $F$ of characteristic zero, and let $n$ be a nonnegative integer. Suppose that $\sigma \in \Irr \bigl(\OO (V)\bigr)$ and $\pi \in \Irr \bigl(\Sp (n,F)\bigr)$ correspond under the theta correspondence. Assuming that $\sigma$ is tempered, we investigate the problem of determining the Langlands quotient data for $\pi$. Categories:11F27, 22E50 124. CJM 1998 (vol 50 pp. 972) Brüchert, Gerd Trace class elements and cross-sections in Kac-Moody groups Let $G$ be an affine Kac-Moody group, $\pi_0,\dots,\pi_r,\pi_{\delta}$ its fundamental irreducible representations and $\chi_0, \dots, \chi_r, \chi_{\delta}$ their characters. We determine the set of all group elements $x$ such that all $\pi_i(x)$ act as trace class operators, \ie, such that $\chi_i(x)$ exists, then prove that the $\chi_i$ are class functions. Thus, $\chi:=(\chi_0, \dots, \chi_r, \chi_{\delta})$ factors to an adjoint quotient $\bar{\chi}$ for $G$. In a second part, following Steinberg, we define a cross-section $C$ for the potential regular classes in $G$. We prove that the restriction $\chi\|_C$ behaves well algebraically. Moreover, we obtain an action of $\hbox{\Bbbvii C}^{\times}$ on $C$, which leads to a functional identity for $\chi\|_C$ which shows that $\chi\|_C$ is quasi-homogeneous. Categories:22E65, 17B67 125. CJM 1998 (vol 50 pp. 356) Gross, Leonard Some norms on universal enveloping algebras The universal enveloping algebra, $U(\frak g)$, of a Lie algebra $\frak g$ supports some norms and seminorms that have arisen naturally in the context of heat kernel analysis on Lie groups. These norms and seminorms are investigated here from an algebraic viewpoint. It is shown that the norms corresponding to heat kernels on the associated Lie groups decompose as product norms under the natural isomorphism $U(\frak g_1 \oplus \frak g_2) \cong U(\frak g_1) \otimes U(\frak g_2)$. The seminorms corresponding to Green's functions are examined at a purely Lie algebra level for $\rmsl(2,\Bbb C)$. It is also shown that the algebraic dual space $U'$ is spanned by its finite rank elements if and only if $\frak g$ is nilpotent. Categories:17B35, 16S30, 22E30 Page Previous 1 ... 3 4 5 6 Next top of page \| contact us \| privacy \| site map \|	2017-09-22 04:27:51	{"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": 0.9011132717132568, "perplexity": 530.3434589150054}, "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/1505818688208.1/warc/CC-MAIN-20170922041015-20170922061015-00235.warc.gz"}
https://settheory.mathtalks.org/logic-seminar-at-nus-on-wed-19-nov-2014/	# Dan Teng: Semiautomatic Structures Invitation to the Logic Seminar at the National University of Singapore Date: Wednesday, 19 November 2014, 17:00 hrs Room: S17#04-04, Department of Mathematics, NUS Speaker: Dan Teng Title: Semiautomatic Structures URL: http://www.comp.nus.edu.sg/~fstephan/logicseminar.html Semiautomatic structures generalise automatic structures in the sense that for some of the relations and functions in the structure one only requires the derived relations and structures are automatic when all but one input are filled with constants. One can also permit that this applies to equality in the structure so that only the sets of representatives equal to a given element of the structure are regular while equality itself is not an automatic relation on the domain of representatives. It is shown that one can find semiautomatic representations for the field of rationals and also for finite algebraic field extensions of it. Furthermore, it is shown that there are semiautomatic ordered rings consisting of the integers augmented by the multiples of the square-root of a non-square integer; here the order is the same as the corresponding order on the real numbers.	2018-12-19 07:47:57	{"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": 0.8648776412010193, "perplexity": 1230.2837916177295}, "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/1544376831715.98/warc/CC-MAIN-20181219065932-20181219091932-00023.warc.gz"}
https://www.physicsforums.com/threads/how-to-convert-between-different-mass-measurements-for-cdm-h.878036/	A How to convert between different mass measurements for CDM H 1. Jul 6, 2016 floyd0117 If I have a cluster with a mass measured in M_500c, then how do I go about converting that mass to M_200c given some concentration and an NFW profile? 2. Jul 6, 2016 Chalnoth What do you mean converting the mass? What are you converting? Why? 3. Jul 7, 2016 floyd0117 I have a dataset of cluster masses, expressed in units of M_500c, which I need to relate to a velocity dispersion. But the relation between cluster mass and velocity dispersion is in terms of M_200c, so I need to convert the masses in the dataset before I can find the equivalent dispersions. 4. Jul 7, 2016 Chalnoth Ahh, okay. I'm honestly not sure. I did find this resource, which describes these measures a bit: It looks like $M_{500}$ is the mass within $R_{500}$, while $M_{200}$ is the mass within $R_{200}$, where $R_{500} \approx 0.7 R_{200}$. If you've got a good understanding of the NFW profile, you might be able to use that to do the conversion. That said, my guess is it'd be even better to convert the relation between mass and dispersion to be in terms of $M_{500}$ instead of $M_{200}$, as that's most likely a far simpler operation. Last edited by a moderator: May 8, 2017 5. Jul 7, 2016 floyd0117 Perhaps, but I'm trying to follow the methodology presented in a specific paper and replicate their results. Thanks anyway Last edited by a moderator: May 8, 2017 6. Jul 7, 2016 Chalnoth Makes sense. Sorry I can't be of more help.	2018-01-20 05:59:46	{"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": 0.6504900455474854, "perplexity": 785.0646104885184}, "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-2018-05/segments/1516084889325.32/warc/CC-MAIN-20180120043530-20180120063530-00469.warc.gz"}
https://mc-stan.org/docs/stan-users-guide/conditionals.html	## 35.7 Conditionals While Stan supports the full C++-style conditional syntax, allowing real or integer values to act as conditions, real values should be avoided. For a real-valued x, one should use if (x != 0) { ... in place of if (x) { ... Beyond stylistic choices, one should be careful using real values in a conditional expression, as direct comparison can have unexpected results due to numerical accuracy.	2023-03-20 19:13:50	{"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": 0.17202292382717133, "perplexity": 2901.7789465723763}, "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/1679296943555.25/warc/CC-MAIN-20230320175948-20230320205948-00453.warc.gz"}
https://gamedev.stackexchange.com/questions/116556/change-material-emission-for-a-second-then-go-back-to-default/116557	Change material emission for a second then go back to default I'm looking for a way to change the emission of a material for a second then get it back to where is was by default Here's what I have for now float emission_lvl = 5.0F; float emission_lvl_start = 0.5F; public Update(){ mat.SetFloat("_Emission",emission_lvl); Invoke ("SetEmissionBack", 1.0f); } public void SetEmissionBack(){ mat.SetFloat("_Emission",emission_lvl_start); } Well, your placement is wrong. Execute same code of Update in Start or whenever you want to execute it and remove it from Update. Because Update is a loop that executes 30 - 60 (depends) per second and mat.SetFloat("_Emission",emission_lvl); executing per frame, that is preventing to change material's emission property. And now I came to know that you want flicker effect. So you can use Coroutine for that. void Start() { StartCoroutine("StartEffect"); } IEnumerator StartEffect() { while(true) // Or any other condition, this will repeat until this condition is true { mat.SetFloat("_Emission",emission_lvl); yield return new WaitForSeconds(1.0f); mat.SetFloat("_Emission",emission_lvl_start); yield return new WaitForSeconds(1.0f); } } • On Start() it will only run it 1 time... I need a flicker effect – Warface Feb 12 '16 at 19:40 • SO you want to executes like on and off? 1 second on and 1 second off? or anything else? – Hamza Hasan Feb 12 '16 at 19:43 • Yes and repeats – Warface Feb 12 '16 at 19:45 • Check the edits – Hamza Hasan Feb 12 '16 at 19:48 • You may want to use MaterialPropertyBlock for this. As-written this will create a new material instance once every second, requiring the old (now unused) copies to get garbage collected. – Draco18s Feb 12 '16 at 20:45	2019-11-21 10:56:11	{"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": 0.24525626003742218, "perplexity": 3223.529697397352}, "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-2019-47/segments/1573496670770.21/warc/CC-MAIN-20191121101711-20191121125711-00445.warc.gz"}
http://seankross.com/2016/06/30/Dollar-Cost-Averaging-Versus-Lump-Sum-Investing.html	# Dollar Cost Averaging Versus Lump Sum Investing My brother and I were talking about saving for retirement, specifically whether he should wait to invest in his IRA until he gets a large stipend at the end of the summer or if he should invest a smaller amount of money (~$100 per week) throughout the year. Some people call the first approach “lump sum investing” and the latter appraoch “dollar cost averaging.” This post is an attempt to visualize the difference between these two approaches. We’re both customers of Charles Schwab which allows us to buy their funds without paying a commission. Their broad US stock market ETF is called SCHB. Let’s get SCHB’s opening price for the last year. set.seed(2016-06-15) options(digits = 2) # Get SCHB data library(quantmod) schb <- getSymbols("SCHB", auto.assign = FALSE, from = "2015-06-15", to = "2016-06-15") schb <- as.data.frame(schb)[,1] Imagine that you bought 100 shares when the market openned on one of the days in the past year. Now imagine there are multiple universes where in each universe you bought all 100 shares of the ETF on a different day. Let’s see how much money you would have made/lost after a year in each universe: # The cost of 100 shares on each trading day last year starting_portfolio_values <- schb * 100 # The value of each "portfolio" after a year last_day_portfolio_value <- starting_portfolio_values[length(schb)] hist(last_day_portfolio_value - starting_portfolio_values, main = "Histogram of Change in Value", xlab = "Change in Value", breaks = 15, col = rgb(1, 0, 0, .6)) On average you would have made$118.91, but you can see that the range of outcomes is large: from $-174 to$667! Now imagine that instead you bought 2 shares on 50 different random days last year. And to be fair this happened in the same number of universes as in the last example. n_universes <- length(schb) change_in_value <- replicate(n_universes, { # Pick 50 random days and buy 2 shares each day share_prices <- sample(schb, 50) * 2 # Sum of the prices is the cost basis cost_basis <- sum(share_prices) # Compute the change in portfolio value last_day_portfolio_value - cost_basis }) hist(change_in_value, breaks = 15, main = "Histogram of Change in Value", xlab = "Change in Value", col = rgb(0, 0, 1, .6)) On average you would have made $121.19, and the range of outcomes is much smaller: from$63.86 to \$195.78. In order to better compare these investment strategies I’ll plot both sets of outcomes on the same axis: h1 <- hist(last_day_portfolio_value - starting_portfolio_values, breaks = 15) h2 <- hist(change_in_value, breaks = 15) hist_xmin <- min(last_day_portfolio_value - starting_portfolio_values) hist_xmax <- max(last_day_portfolio_value - starting_portfolio_values) plot(h1, col = rgb(1, 0, 0, .6), main = "Histogram of Change in Value", xlab = "Change in Value", xlim = c(hist_xmin, hist_xmax), ylim = c(0, 40)) plot(h2, col = rgb(0, 0, 1, .6), main = "Histogram of Change in Value", xlab = "Change in Value", add = TRUE) legend(350, 30, c("Lump Sum Investing", "Dollar Cost Averaging"), pch = rep(15, 2), col = c(rgb(1, 0, 0, .6), rgb(0, 0, 1, .6)), bty = "n", cex = 1.5) There are many generalizations in this analysis (there are no commissions, only one kind of investment is considered, etc) however you can see that investing small sums of money over time seems to restrict the range of your gains and losses around a mean value that is close the average gains and losses you would have if you invested a large amount of money all at once.	2018-04-21 01:38:03	{"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": 0.3895123600959778, "perplexity": 3566.415039731835}, "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/1524125944851.23/warc/CC-MAIN-20180421012725-20180421032725-00003.warc.gz"}
https://brilliant.org/discussions/thread/a-professors-ride-dont-worry-you-would-be-safe-7/	# A professor's ride (Don't worry! You would be safe) This situation is very interesting - A professor placed two of his deserving students on a special ride. The special ride consists of a disc which spins with angular speed $$ω$$. The two students sit on chairs which are aligned along one of the radii of the disc at distance $$S1$$ and $$S2$$ respectively from the center. Spinning the disc, the professor asks them to find each other's relative velocity. Opinion 1: The relative velocity among the students is zero as they don't appear to move with respect to each other. Opinion 2: The relative velocity of the students is not zero as the velocity of one student is $$ωS1$$ and the other's velocity is $$ωS2$$. Hence the relative velocity among them is $$ω(S1-S2)$$. [By relative velocity, I mean velocity of A with respect to B or vice-versa, where A and B are the two students.] Which one of them do you think is right and why? Note by Lokesh Sharma 4 years, 6 months ago MarkdownAppears as italics or _italics_ italics bold or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$...$$ or $...$ to ensure proper formatting. 2 \times 3 $$2 \times 3$$ 2^{34} $$2^{34}$$ a_{i-1} $$a_{i-1}$$ \frac{2}{3} $$\frac{2}{3}$$ \sqrt{2} $$\sqrt{2}$$ \sum_{i=1}^3 $$\sum_{i=1}^3$$ \sin \theta $$\sin \theta$$ \boxed{123} $$\boxed{123}$$ Sort by: Opinion one is correct. Person A is stationary in the frame of reference of person B. Therefore the relative velocity must surely be 0. But more interestingly is to argue why then, opinion 2 is incorrect? The relative velocity seems to indeed be $$\omega(S1 - S2)$$. So we would be inclined to say that both are correct, yet the two opinions are mutually exclusive? This conclusion is absurd, thus one of our assumptions or intermediary conclusions must be wrong. The frame of reference! It has no uniform velocity! When we take the frame of reference of person B to include person A, we must realize that this frame of reference itself has different velocities! Thus, it would actually mean that if A had the same speed as B, then A would not appear stationary to B. Reversely, if their relative velocity in a frame of reference that is moving linearly that for a moment includes persons A and B (e.g., a crow that ascends from the ground to the top of the disc), would be exactly $$\omega(S1 - S2)$$, then the frame of reference that appears stationary to B will have a relative velocity between persons A and B of 0. - 4 years, 6 months ago A is not stationary in rest frame of B. At the top and bottom of the wheel, the positions of A and B relative to each other are interchanged. - 4 years, 6 months ago Let the two students be A & B. We define relative velocity of B with respect to A as the velocity of A in the rest frame of B. In this case the rest frame of B is moving with an angular velocity equal to that of A. So in the frame of B, A is always at the same point P irrespective of how much time has passed. Therefore, their relative velocity is ZERO. So, according to me, Opinion 1 is more accurate. Hope this helps. - 4 years, 6 months ago opinion 2. consider it as a plane motion..relative velocity of any point on a rigid body w.r.t. any other point on the same body is =distance between points *angular velocity=ω(S1−S2)... - 4 years, 6 months ago I think Opinion 2 is much more sensible. - 4 years, 6 months ago But what explanation would you give for invalidation of Opinion 1? - 4 years, 6 months ago Opinion 1. If I stood on the wheel and I look at any one student,I would always be looking at that student,no matter how fast the wheel is spinning. Because neither student is moving relative to me,they are not moving relative to each other. - 4 years, 6 months ago Don't worry... y'all are right... There will be different answers but it points to a basic concept in physics. It is due to the frame of reference concept. That means what we see wouldn't be same at all views. - 4 years, 6 months ago Yeah I agree with you that what we see wouldn't be same at all the views but here, in this case, we have to see only the view of either of the students which follows from the definition of relative velocity. So relative velocity will be zero as to a student, other would be appear to be stationary. - 4 years, 6 months ago Option 2...As their velocities are undoubtedly different.Its just that S2 covers different distance then S1 does..such that for S1 it apparently looks as if S2 is at rest..Hope this helps.. OR after going around half circle,the relative positions are reversed..in fact their relative position is changing every instant.See it as a vector joining S1 to S2. - 4 years, 6 months ago What about the speed? Is relative speed among them is also non zero? If yes, then why won't distance among them changes? - 4 years, 6 months ago Because the student at S2 has to move further to move the same angle as S1 - 4 years, 6 months ago Actually, both happens. The relative velocity is zero. But the angular velocity will be different. right? - 4 years, 6 months ago If the angular velocity is different,then S1 would have moved in front of S2 which is clearly not happening because they are on the same disk. - 4 years, 6 months ago Relative velocity is not zero. Relative angular velocity, if such a thing exists, is again $$\omega$$, I think. - 4 years, 6 months ago look, when you draw a circle, the radius is same at all points. Now, draw another circle with a common centre. Now join a line from the centre to the big circle. Again do the same thing but at a different angle. What will happen? the distances will always be same! so, velocity is displacement by time. so relative velocity is relative displacement by time. when displacement don't change, relative velocity will be zero! - 4 years, 6 months ago Velocity is a vector.!!! - 4 years, 6 months ago Okay. Now draw a line from the centre of the small circle to the small circle. Now do the same thing, but at a different angle. The distances will be the same. By this logic, Neither of the students are moving at all. Do you see the flaw? You are correct in saying that they remain at constant distance from each other. Yet any form of circular motion remains at constant distance from the centre, which would imply that objects in circular motion are at rest! The problem is that the angle at which they are located relative to each other is different despite the distance between them being equal. - 4 years, 6 months ago Angular velocity of an object does not depend upon its 'speed', it only sees what 'angle' the body sweeps about the center in unit time. What it means is that at whatever the distance the two bodies are from the center respectively, if they subtend the same angle at the center in unit time of their circular motion, they are said to have the same 'angular velocity'. Let a body lie 1km away from a point while another lie 1m away; suppose both have the same angular velocity about the point, the body lying 1km away clearly has a much greater speed compared to the one lying 1m away, but both take the same time to sweep a particular angle. This hypothesis supports the first opinion. However, the second opinion also seems convincing. The problem here is that we can give two different explanations depending upon the frame of reference specified. We can observe the motions taking the center as the point of reference or from any other frame lying independent of the circular motion of the bodies. If we take the center as the point of reference, then the first opinion makes more sense from our hypothesis. But if we take an external frame of reference, then we can say that the 'linear' velocities of the two bodies vary at 'every' instant in such a way that the relative velocity between them at 'that' instant is zero. And yes, if we take only linear velocities into consideration, then relative velocity is not zero; but here the bodies do not continue to move linearly, but in a circular path, and their velocities vary in accordance to two-dimensional circular motion. Thus, even if the instantaneous linear relative velocity is not zero; in circular motion, the relative velocity between them remains zero. So, I think opinion 1 is correct. - 4 years, 6 months ago opinion 2 is not the correct option... it is just the difference of the velocities of the speeds of the 2 students relative to a watcher. - 4 years, 6 months ago any person would feel that the disk is not rotating about center but is rotating about himself so there would be a relative velocity between the two so, option 2 is correct option 1 is wrong as it does not see the persons from there reference frames, instead it sees the situation from the reference frame of a person situated at the center of the disk(so in this, w=0) - 4 years, 6 months ago Define your terms properly. By "relative velocity", do you mean "the velocity of A in the reference frame of B" or "the apparent difference in their velocities as observed by the professor"? - 4 years, 6 months ago I think both are one and the same..!!! - 4 years, 6 months ago The first one; velocity of A in the reference frame of B, where A and B are the two students. - 4 years, 6 months ago Opinion 1 is the right one. Pretending the relatice velocity of the students equals 0 this implicates w(S1-S1) equals 0 and because the radial velocity is greater than 0 : S1-S2=0 which is true because they're aligned along one of the radii of the disc from the two different sides. - 4 years, 6 months ago Opinion 2. If we consider student 1 as the frame of reference, we can see that student 2 is moving in a circular path around student 1. Pretty sure that if $$S_{1} < S_{2}$$ 2's angular momentum with respect to 1 will still be $$\omega$$, and $$-\omega$$ otherwise. - 4 years, 6 months ago Invalidation of opinion 1 - although the distance between them is constant, the direction of displacement changes continuously, so relative velocity isn't zero. - 4 years, 6 months ago The problem is due to the frame of reference concept. Actually all are right. But different perceives differently at different views... lol!!! - 4 years, 6 months ago How come you can say that student 2 is moving in a circular path around student 1 even when student1 is moving with the same angular velocity as student 2? Because of same angular velocities about the same axis, it would seem to student 1 that student 2 is at rest. Which gives their relative velocity to be zero. So opinion 1 is apt according to me. What are your views against my statement? - 4 years, 6 months ago Because the magnitude of the velocity of student 2 is greater. When we say relative velocity, we compare only the magnitudes and directions of linear velocity, not angular velocity. Note how at different positions of student 1, student 2 is at the same distance from student 1, but is in a different direction from him. Suppose you they at the bottom, initially, student 2 is below student 1. At the top, student 2 is above student 1. So the position of student 2 relative to student 1 keeps changing. Edit: $$v_{21} = v_{2} - v_{1} = \omega S_{2} - \omega S_{1} = \omega(S_{2} - S_{1})$$ Now student two is always at distance $$S_{2} - S_{1}$$ from student 1, but at different angles. This is circular motion, with $$r = S_{2} - S_{1}$$ and $$v = v_{21}$$ And the angular velocity of this circular motion of student 2 with respect to student 1 is $$\frac{v_{21}}{r} = \frac{\omega(S_{2} - S_{1})}{S_{2} - S_{1}} = \omega$$ That's how I see it, anyway. - 4 years, 6 months ago When we say that at a given position student 1 is above 2 and vice-versa, we are telling what we would see if we were to look at the disc from top. But in order to find the relative velocities of 1 and 2 , we have to consider what we would see if we were to be at their positions. So when I am at the place of either of the students, the other would seem to be at rest to me. Hence I am able to say that the other one is NOT moving according to me irrespective of what the magnitude of his linear velocity is. If the are moving around the same axis with same angular velocity then in the rest frame of 1 (which is also moving with same angular velocity about same axis), 2 always appears to be at the same given point. Hence no displacement occurs between them in this frame (neither in direction nor in magnitude). Hence relative velocity is zero. - 4 years, 6 months ago The same logic applies for both horizontal and vertical circles. Suppose they start at the southern part of the circle. 2 is south of 1. When they reach the eastern most point, 2 is east of 1. When they reach the northern most point, 2 is north of 1. Now consider if there was only one student on a horizontal circle. At the south, he is south of the centre. At the east, he is east of the centre and at the north, he is north of the centre. Do the two cases look similar? - 4 years, 6 months ago Opinion 2 because $$S2>S1$$ so the distance which student 2 has to travel is more than student 1. - 4 years, 6 months ago	2018-04-21 06:13:51	{"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": 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": 0.8957371115684509, "perplexity": 580.5419259945215}, "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/1524125945037.60/warc/CC-MAIN-20180421051736-20180421071736-00533.warc.gz"}
https://www.mathworks.com/help/comm/ref/turbodecoder.html	# Turbo Decoder Decode input signal using parallel concatenated decoding scheme • Library: • Communications Toolbox / Error Detection and Correction / Convolutional ## Description The Turbo Decoder block decodes the input signal using a parallel concatenated decoding scheme. The iterative decoding scheme uses the a posteriori probability (APP) decoder as the constituent decoder, an interleaver, and a deinterleaver. The two constituent decoders use the same trellis structure and decoding algorithm. For more information, see Parallel Concatenated Convolutional Decoding Scheme and APP Decoder. This icon shows the block with all ports enabled. ## Ports ### Input expand all Parallel concatenated codeword, specified as a column vector of length M, where M is the length of the parallel concatenated codeword. Data Types: `double` \| `single` Interleaver indices, specified as a column vector of integers. The vector must be of length L. Each element of the vector must be an integer in the range [1, L] and must be unique. L is the length of the decoded binary output message, `Out`. The interleaver indices define the mapping used to permute the input bits at the decoder. #### Dependencies To enable this port, set the Source of interleaver indices parameter to ```Input port```. Data Types: `double` Input indices for the bit ordering and puncturing used on the fully encoded data, specified as a column vector of integers. The length of the InInd vector must equal the length of the input data vector In. Element values in the vector must be relative to the fully encoded data for the coding scheme, including the tail bits for all streams. #### Dependencies To enable this port, set the Source of input indices parameter to `Input port`. Data Types: `double` ### Output expand all Decoded message, returned as a binary column vector of length L, where L is the length of the decoded binary output message. This output inherits its data type from the `In` input. ## Parameters expand all Specify the trellis as a MATLAB® structure that contains the trellis description for a rate KN constituent convolutional code. K is the number of input bit streams, and N is the number of output bit streams. Note K must be 1 for the turbo coder. For more information, see Coding Rate. You can either use the `poly2trellis` function to create the trellis structure or create it manually. For more about this structure, see Trellis Description of a Convolutional Code and the `istrellis` function. The trellis structure contains these fields. Number of symbols input to the encoder, specified as an integer equal to 2K, where K is the number of input bit streams. Data Types: `double` Number of symbols output from the encoder, specified as an integer equal to 2N, where N is the number of output bit streams. Data Types: `double` Number of states in the encoder, specified as a power of 2. Data Types: `double` Next states for all combinations of current states and current inputs, specified as a matrix of integers. The matrix size must be `numStates`-by-2K. Data Types: `double` Outputs for all combinations of current states and current inputs, specified as a matrix of octal numbers. The matrix size must be `numStates`-by-2K. Data Types: `double` Specify the source of the interleaver indices as `Property` or ```Input port```. • When you set this parameter to `Property`, the block uses the Interleaver indices parameter to specify the interleaver indices. • When you set this parameter to ```Input port```, the block uses the `IntrInd` input port to specify the interleaver indices. Specify the interleaver indices that define the mapping used to permute codeword bits input to the decoder as a column vector of integers. The vector must be of length L. Each element of the vector must be an integer in the range [1, L] and must be unique. L is the length of the decoded binary output message. #### Dependencies To enable this parameter, set the Source of interleaver indices parameter to `Property`. Specify the source of the input indices as `Auto`, `Property`, or `Input port`. • When you set this parameter to `Auto`, the block computes input indices that assume the second systematic stream is punctured and all tail bits are included in the input. • When you set this parameter to `Property`, the block uses the input indices that you specify for the Input indices parameter. • When you set this parameter to ```Input port```, the block uses the InInd input port to specify the input indices. The vector length and values for the input indices and coded input signal can change with each execution of the block. Specify the input indices for the bit ordering and puncturing used on the fully encoded data as a column vector of integers. The vector length of this parameter must equal the length of the input data vector In. #### Dependencies To enable this parameter, set the Source of input indices parameter to `Property`. Specify the decoding algorithm that the constituent APP decoders use to decode the input signal as `True APP`, `Max`, `Max`. When you set this parameter to `True APP` the block implements true a posteriori probability decoding. When you set this parameter to `Max` or `Max` the block uses approximations to increase the speed of the computations. For more information, see APP Decoder. Specify the number of bits which the constituent APP decoders must use to scale the input data to avoid losing precision during computations as an integer in the range [0, 8]. The constituent decoders multiply the input by 2k and divide the pre-output by the same factor. k is the value of the Number of scaling bits parameter. For more information, see APP Decoder. #### Dependencies This enable this parameter, set the Decoding algorithm parameter to `Max*`. Specify the number of decoding iterations the block uses as a positive integer. The block iterates and provides updates to the log-likelihood ratios (LLR) of the uncoded output bits. The output of the block is the hard-decision output of the final LLR update. Type of simulation to run, specified as ```Interpreted execution``` or `Code generation`. • `Interpreted execution` –– Simulate the model by using the MATLAB interpreter. This option requires less startup time than the `Code generation` method, but the speed of subsequent simulations is slower. In this mode, you can debug the source code of the block. • `Code generation` –– Simulate the model by using generated C code. The first time you run a simulation, Simulink® generates C code for the block. The C code is reused for subsequent simulations unless the model changes. This option requires additional startup time, but the speed of the subsequent simulations is faster than `Interpreted execution`. ## Block Characteristics Data Types `double` \| `single` Multidimensional Signals `no` Variable-Size Signals `yes` expand all ## References [1] Benedetto, S., G. Montorsi, D. Divsalar, and F. Pollara. "A Soft-Input Soft-Output Maximum A Posterior (MAP) Module to Decode Parallel and Serial Concatenated Codes." Jet Propulsion Lab TDA Progress Report, 42–127, (November 1996). [2] Viterbi, A.J. “An Intuitive Justification and a Simplified Implementation of the MAP Decoder for Convolutional Codes.” IEEE Journal on Selected Areas in Communications 16, no. 2 (February 1998): 260–64. https://doi.org/10.1109/49.661114. [3] Berrou, C., A. Glavieux, and P. Thitimajshima. “Near Shannon Limit Error-Correcting Coding and Decoding: Turbo-Codes.” Proceedings of ICC 93 - IEEE International Conference on Communications, Geneva, Switzerland, May 1993, 1064–70. https://doi.org/10.1109/icc.1993.397441. [4] Schlegel, Christian, and Lance Perez. Trellis and Turbo Coding. IEEE Press Series on Digital & Mobile Communication. Piscataway, NJ ; Hoboken, NJ: IEEE Press ; Wiley-Interscience, 2004. [5] 3GPP TS 36.212. "Multiplexing and channel coding." 3rd Generation Partnership Project; Technical Specification Group Radio Access Network; Evolved Universal Terrestrial Radio Access (E-UTRA). https://www.3gpp.org. ## Extended Capabilities ### Objects Introduced in R2011b	2021-09-27 00:45:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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": 0.7562718391418457, "perplexity": 1857.8352178344805}, "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-2021-39/segments/1631780058222.43/warc/CC-MAIN-20210926235727-20210927025727-00361.warc.gz"}
https://abmathematics.com/?m=201609	## Composites and Inverse Functions Test On Wednesday, September 28th we’ll have our first HL test on composite and inverse functions. The following list of questions should be completed as part of your review of this material, and we can discuss any problem you may be having in class before the test. Pages 85–89 questions 1 to 4, 9 to 11, 15 to 22 ## Composite and Inverse Functions Complete the following questions for Sunday, the 18th of September. Composite Functions Pages 60–61 questions 1, 2, 5, 6, 9, 10, 12, 17, 19, 24, 25 Inverse Functions Pages 68–69 questions 15, 16, 17, 24, 30, 31, 34, 35, 36, 37, 38 ## Applications of Derivatives Complete the following questions for Sunday, the 18 of September. Pages 751–752 questions 1–6, 8, 14 ## Indefinite Integrals Complete the following questions on indefinite integrals for tomorrow’s lesson. Page 780 questions 1, 3, 7, 8, 9, 10, 12, 13, 14 ## 12 HL Composite Functions Homework Let $$f(x)=x^2$$ and $$g(x)=x-1$$. 1. Find the range of $$f$$ and $$g$$, assuming the domain for both is $$\mathbb{R}$$. 2. Find the range of $$f$$ and $$g$$, assuming the domain for both is $$[-2,\infty[$$. 3. Find the value of each of the functions below when $$x=4$$. a) $$f\circ g$$ b) $$g\circ f$$ 4. Find the range of each of the functions in Question 3. ## Finding Derivatives Here are a few questions to look at that involve applications of the new techniques and results (the chain rule, the product rule, and derivatives of exponential functions) we’re recently covered. Make sure to start these before our next lesson, and aim to have them completed by Monday. Pages 715–716 questions 3, 7, 9, 11 Pages 728 questions 1 a b e h i, 4, 6, 9, 10	2020-11-24 13:08:53	{"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": 0.39571812748908997, "perplexity": 766.1203581401476}, "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-2020-50/segments/1606141176256.21/warc/CC-MAIN-20201124111924-20201124141924-00676.warc.gz"}
https://math.meta.stackexchange.com/questions/13832/when-is-the-next-moderator-election/13842	When is the next Moderator Election? When is the next moderator election? Was wondering according to https://math.stackexchange.com/election it is around may each year (at least since 2012) • Usually there's an evaluation whether or not new moderators are needed before there is an election. I feel that due to the recent improvements of the review system, many flags are handled by the community and the workload of the moderators is alleviated. I'm not sure whether or not we really need more diamonds here at the moment. (Perhaps if one of the moderators considers stepping down that would be different.) – Asaf Karagila May 26 '14 at 12:27 • @AsafKaragila I don't think that more moderators are needed given the community involvement in dealing with reviews, but I think the moderators themselves should be subject to review at least annually. – user61527 May 26 '14 at 14:56 • @T.Bongers: I agree, but diamonds are forever. So this sort of review is probably (and should be) internal to SE. In this case, I don't know how many complaints have been filed on the behavior of the moderators, something which may or may not prompt such a review process explicitly. – Asaf Karagila May 26 '14 at 15:05 • @Willemien: well, I could step down if it will help you get your badge. :-) @Asaf: judging by our friendly cast of current and past moderators, I would have estimated a much shorter half-life. – Willie Wong May 26 '14 at 15:10 • I, for one, am very happy with the current mod team and our workload, and I don't expect to have another election this year. – davidlowryduda May 26 '14 at 15:13 • I am happy with our current moderators. I know that they have a difficult job and I don't expect perfect behavior. But as a matter os principle I would support term limits (say three years with elections every year (so not replace the whole team)). Maybe we just once a year have a confidence vote on the current moderators. If the moderators pass the vote, then we will not have elections. If a moderator doesn't pass the confidence vote (s)he losses his/her job and we have an election to replace that moderator. Again, I would suspect that the current moderators would all pass a confidence vote. – Thomas May 26 '14 at 15:30 • @Willie: I'm not disputing. I'd guess that $10^{\log_{10}(50)}$ years is probably a better upper limit. (I think that a drop from $10^{80}$ to $10^{\log_{10}(50)}$ deserves a publication. It's not every day we see such a dramatic improvements in the upper bounds of something!) – Asaf Karagila May 26 '14 at 16:18 • To be perfectly honest, I'm somewhat afraid of an election. The current mods seem to be getting along rather well (which does not mean that we always agree). We are also communicating with each other (and the SE team) about major decisions, which might be something rather novel for this site. In short, our personalities and approach to modding seems to mesh quite well, which has made the work much easier. An election might severely upset this fine balance, which causes me some angst. Of course, an election will come sooner or later, and I hope the electorate takes their task seriously. – user642796 May 26 '14 at 18:16 • You know, one does not vote to get a badge... – user122283 May 26 '14 at 18:53 • @AsafKaragila, the main thing that MSE is missing is a Dictator For Life. The wife of Old John is up for this in England, it is just a matter of time before the entire world... – Will Jagy May 26 '14 at 20:00 • @حكيمالفيلسوفالضائع: Perhaps because of the striking similarity to the word Couscous; or perhaps because people like having badges, and much like Pokemon, they want to have them all. – Asaf Karagila May 27 '14 at 0:23 • @AsafKaragila: Sorry, I was trying to be funny. – Thomas May 27 '14 at 13:41 • @Thomas: Don't you know that anything you write on the internet without a smiley face will be taken seriously? – Asaf Karagila May 27 '14 at 13:48 • I do see a point in checks and balances, but the obvious problem with reelections is that most moderation activity is invisible to most users and the secrecy is needed to work effectively. On the other hand, I feel uncomfortable with arguments that could come from James R. Clapper... – Michael Greinecker May 27 '14 at 15:28 • @Michael: Better a Clapper than having the clap. I think. Maybe? – Asaf Karagila May 27 '14 at 21:51	2021-06-23 20:53:43	{"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": 0.3375636637210846, "perplexity": 1411.779207764291}, "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/1623488540235.72/warc/CC-MAIN-20210623195636-20210623225636-00116.warc.gz"}
https://socratic.org/questions/how-do-you-simplify-1-8i-4-3i-and-write-in-a-bi-form	# How do you simplify (-1-8i)+(4+3i) and write in a+bi form? Jan 2, 2016 $\left(- 1 - 8 i\right) + \left(4 + 3 i\right) = 3 - 5 i$ #### Explanation: When adding complex numbers given in a+bi form, we add real parts and imaginary parts separately: $\left(- 1 - 8 i\right) + \left(4 + 3 i\right) = - 1 + 4 - 8 i + 3 i =$ $= \left(- 1 + 4\right) + \left(- 8 + 3\right) i = 3 - 5 i$ We could think about complex numbers as vectors on the complex plane, where real part is on x axis and imaginary part is on y axis.	2020-04-03 05:39:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "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": 0.8728356957435608, "perplexity": 807.3068794633134}, "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-2020-16/segments/1585370510287.30/warc/CC-MAIN-20200403030659-20200403060659-00060.warc.gz"}
https://www.studyadda.com/sample-papers/neet-sample-test-paper-19_q6/208/276599	• # question_answer The diagram shows a venturimeter through which water is flowing. The speed of water at X is$2\,cm/{{s}^{-1}}.$. The speed of water at Y (taking$g=1000\,cm/{{s}^{-2}}$) is: A) $23\,cm/{{s}^{-1}}$ B) $32\,cm/{{s}^{-1}}$C) $101\,cm/{{s}^{-1}}$ D) $1024\,cm/{{s}^{-1}}$ Applying Bernoulli's theorem: ${{p}_{1}}+\frac{1}{2}\rho v_{1}^{2}={{P}_{2}}+\frac{1}{2}\rho v_{2}^{2}$or $\frac{1}{2}\rho (v_{2}^{2}-v_{1}^{2})={{P}_{1}}-{{P}_{2}}$ or $\frac{1}{2}\times 1(v_{2}^{2}-4)=0.51\times 1\times 1000$$[\because \,{{P}_{1}}-{{P}_{2}}=h\rho g]$ or $v_{2}^{2}-4=510\times 2=1020$or ${{v}_{2}}=32\,cm/s.$ Hence, the correction option is [b].	2019-10-23 18:18:12	{"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": 0.9475577473640442, "perplexity": 7220.336433064776}, "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/1570987835748.66/warc/CC-MAIN-20191023173708-20191023201208-00013.warc.gz"}
https://www.physicsforums.com/threads/vector-cross-product-with-curl.873700/	# Vector cross product with curl 1. May 29, 2016 ### hotvette 1. The problem statement, all variables and given/known data Using index-comma notation only, show: \begin{equation} \underline{\bf{v}} \times \text{curl } \underline{\bf{v}}= \frac{1}{2} \text{ grad}(\underline{\bf{v}} \cdot \underline{\bf{v}}) - (\text{grad } \underline{\bf{v}}) \underline{\bf{v}} \end{equation} 2. Relevant equations \begin{align} \text{ curl } \underline{\bf{v}} &= \epsilon_{ijk} v_{j,i} \underline{\bf{e}}_k \\ \underline{\bf{v}} \times \underline{\bf{u}} &= \epsilon_{ijk} v_i u_j \underline{\bf{e}}_k \end{align} 3. The attempt at a solution If I let $\underline{\bf{u}} =\text{ curl } \underline{\bf{v}}$ we get: \begin{align} \underline{\bf{u}} &= \text{ curl } \underline{\bf{v}} =\epsilon_{prq} v_{r,p} \underline{\bf{e}}_q \\ u_j &= \epsilon_{prq} v_{r,p} \underline{\bf{e}}_q \cdot \underline{\bf{e}}_j \\ u_j &= \epsilon_{prq} v_{r,p} \delta_{qj} = \epsilon_{prj} v_{r,p} \\ \underline{\bf{v}} \times \underline{\bf{u}} &= \epsilon_{ijk} v_i u_j \underline{\bf{e}}_k \\ &= \epsilon_{ijk} \epsilon_{prj} v_i v_{r,p} \underline{\bf{e}}_k \end{align} Is this correct so far? Trouble is, I'm not sure what to do next. I'm wondering if trying to simplify $\epsilon_{ijk} \epsilon_{prj}$ would be a fruitful approach. Appreciate hints/tips. Thanks! Last edited: May 29, 2016 2. May 29, 2016 ### Fightfish Looks correct to me. Yup, that approach is the way to go. This is a well-known identity, $$\epsilon_{ijk} \epsilon_{\ell mk} = \delta_{i\ell} \delta_{jm} - \delta_{im} \delta_{j \ell}$$ for which there are several different ways to prove. 3. May 29, 2016 ### hotvette OK, I"m getting closer but (maybe) ran into another snag. Using the following: \begin{equation} \epsilon_{ijk} = \epsilon_{jki} = \epsilon_{kij} \end{equation} we get: \begin{align} \epsilon_{ijk} \epsilon_{prj} &= \epsilon_{kij} \epsilon_{prj} = \delta_{ip} \delta_{jr} - \delta_{ir} \delta_{jp} \\ \underline{\bf{v}} \times \text{curl }\underline{\bf{v}} &= (\delta_{ip} \delta_{jr} - \delta_{ir} \delta_{jp})(v_i v_{r,p}) \underline{\bf{e}}_k \\ &= v_i v_{r,p} \delta_{ip} \delta_{jr} \underline{\bf{e}}_k - v_i v_{r,p} \delta_{ir} \delta_{jp}\underline{\bf{e}}_k \\ \end{align} Looking at the LHS and comparing with the problem statement, we have: \begin{align} v_i v_{r,p} \delta_{ip} \delta_{jr} \underline{\bf{e}}_k &= v_p v_{r,p} \delta_{jr} \underline{\bf{e}}_k \\ \frac{1}{2} \text{grad}(\underline{\bf{v}} \cdot \underline{\bf{v}}) &= \frac{1}{2} (v_i v_i)_{,j} \underline{\bf{e}}_j = \frac{1}{2} ({v_i}^2)_{,j} \underline{\bf{e}}_j = v_j \underline{\bf{e}}_j \end{align} Which means that $v_p v_{r,p} \delta_{jr} \underline{\bf{e}}_k$ needs to reduce to $v_j \underline{\bf{e}}_j$. Providing the Kronecker delta can alter partial derivative indices(which I'm not sure), we get: \begin{align} v_p v_{r,p} \delta_{jr} \underline{\bf{e}}_k &= v_p v_{j,p} \underline{\bf{e}}_k \\ &= v_j \underline{\bf{e}}_j \text{ ??} \end{align} It isn't clear to me why the last step is valid. It looks goofy to me because $v_{j,p} = \frac{\partial v_j}{\partial x_p}$ which I think is a matrix of partial derivatives. Appreciate clarification/comment on that one. Now, to tackle the RHS... 4. May 29, 2016 ### vela Staff Emeritus Your claim that $\frac 12(v_i^2)_{,j} = v_j$ isn't correct. If you correct that, you'll see you're essentially done. 5. May 29, 2016 Maybe it's the same thing, but I think the last term should read $v \cdot \nabla v$. (This identity can be found on the cover of J.D. Jackson's Classical Electrodynamics in the form $\nabla (a \cdot b) =a \cdot \nabla b +b \cdot \nabla a + a \times \nabla \times b + b\times \nabla \times a$.) 6. May 29, 2016 ### hotvette Hmmm, is: \begin{equation} \tfrac{1}{2} ({v_i}^2)_{,j} = v_i v_{i,j} \text{ ??} \end{equation} If so, then I get: \begin{align} v_p v_{j,p} \underline{\bf{e}}_k &= v_i v_{i,j} \underline{\bf{e}}_j && \text(a) \\ &= v_p v_{j,p} \underline{\bf{e}}_j &&\text{(c) ok to switch indices on partial?} \end{align} But then the basis vectors don't match :( 7. May 29, 2016 ### vela Staff Emeritus The free indices don't match. You're summing over $j$, so it shouldn't appear on the righthand side. 8. May 30, 2016 ### hotvette Ah, thanks for pointing out the error! OK, I now get: \begin{equation} \epsilon_{kij} \epsilon_{prj} = \delta_{kp} \delta_{ir} - \delta_{kr} \delta_{ip} \end{equation} which gives: \begin{align} \underline{\bf{v}} \times \text{curl }\underline{\bf{v}} &= (\delta_{kp} \delta_{ir} - \delta_{kr} \delta_{ip}) v_i v_{r,p} \underline{\bf{e}}_k \\ &= \delta_{kp} \delta_{ir} v_i v_{r,p} \underline{\bf{e}}_k - \delta_{kr} \delta_{ip} v_i v_{r,p} \underline{\bf{e}}_k \\ &= v_r v_{r,k} \, \underline{\bf{e}}_k - v_p v_{k,p} \, \underline{\bf{e}}_k \end{align} For the LHS I now get a match but I still can't get the RHS to match: \begin{align} &\underline{\bf{u}} = \text{grad } \underline{\bf{v}} = v_{i,j} \, \underline{\bf{e}}_i \otimes \underline{\bf{e}}_j && \text{(a)}\\ &\underline{\bf{u}} \underline{\bf{v}} = u_r v_p \, \underline{\bf{e}}_r \otimes \underline{\bf{e}}_p && \text{(b)}\\ &u_r = v_{i,r} \, \underline{\bf{e}}_i && \text{(c)}\\ & (\text{grad } \underline{\bf{v}}) \underline{\bf{v}} = (v_{i,r} \, \underline{\bf{e}}_i) v_p \, \underline{\bf{e}}_r \otimes \underline{\bf{e}}_p && \text{(d)} \\ &= v_p v_{i,r} \, (\underline{\bf{e}}_r \otimes \underline{\bf{e}}_p)\underline{\bf{e}}_i && \text{(e)} \\ &= v_p v_{p,r} \, \underline{\bf{e}}_r = v_p v_{p,k} \, \underline{\bf{e}}_k \ne v_p v_{k,p} \, \underline{\bf{e}}_k && \text{(f)} \end{align} What am I doing wrong? Last edited: May 30, 2016 9. May 30, 2016 ### vela Staff Emeritus Using what you said in post #6, the first term is $\frac 12 (\vec{v}\cdot \vec{v})_{,k}\hat{e}_k=\frac 12 \nabla(\vec{v}\cdot \vec{v})$. The second term is $(\nabla \cdot \vec{v})\vec{v}$. So you're essentially done. (As Charles pointed out, the last term in the original post is wrong.) 10. May 30, 2016 I don't know this new notation very well= I write out all the partial derivative terms when I do a proof like this, but one suggestion would be to consider my post #5. The term should be $v \cdot \nabla v$, and not $(\nabla v)v$ (or (grad v)v as you wrote it in the OP.) Ah, I understand now (I didn't know how to interpret post #5). I can now see that it works out if the last term is $\underline{\bf{v}} \cdot \text{ grad} \, \underline{\bf{v}}$. I'll check with my professor. Thanks!	2017-08-23 08:30:37	{"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": 0.9999735355377197, "perplexity": 12550.969229630316}, "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/1502886117911.49/warc/CC-MAIN-20170823074634-20170823094634-00588.warc.gz"}
http://math.stackexchange.com/questions/106302/contradiction-theorem	I'm a beginner in formal logic. Can anyone of you help me with the proof of the following lemma: For any Theory $T$ and closed formula $\varphi$ it holds that $T \vdash \varphi$ if and only if $T,\neg \varphi$ is inconsistent. The left to right direction is easy, but I'm struggling with the right-to-left direction. I realize that (according to the deduction theorem), it holds that $T \vdash \neg \varphi \rightarrow \varphi$ Obviously $T \vdash \varphi$ must hold. However, I'm not allowed to assume completeness of first order logic (but soundness though); so do I have to construct a proof of $\varphi$ assuming that I have a proof of $\neg \varphi \rightarrow \varphi$ (we are using a Hilbert-Type-Calculus with MP as the only rule)? - Which axioms are you using? – Chris Eagle Feb 6 '12 at 13:01 Excuse me, I forgot them: $\varphi \rightarrow (\psi \rightarrow \varphi)$; $(\varphi \rightarrow (\psi \rightarrow \xi)) \rightarrow ((\varphi \rightarrow \psi) \rightarrow (\varphi \rightarrow \xi))$; $(\neg \varphi \rightarrow \neg \psi) \rightarrow (\psi \rightarrow \varphi)$ – tri Feb 6 '12 at 13:04 We have that $T$ and $\neg \varphi$ are inconsistent, so $T, \neg \varphi \vdash \psi$ for any $\psi$. This will be most useful if we let $\psi$ be the negation of some easily proven tautology, say $\psi=\neg(\varphi \rightarrow (\varphi \rightarrow \varphi))$. Then by the deduction theorem, we have $T \vdash \neg \varphi \rightarrow \neg (\varphi \rightarrow (\varphi \rightarrow \varphi))$. Using the third axiom and modus ponens, we get $T \vdash (\varphi \rightarrow (\varphi \rightarrow \varphi)) \rightarrow \varphi$. Since we can easily prove $T \vdash \varphi \rightarrow (\varphi \rightarrow \varphi)$, modus ponens gives us $T \vdash \varphi$. - Thank you very much; i don't get in my head how to construct that stuff (proofs in general) any tipps? – tri Feb 6 '12 at 13:26 @tri Mostly, the "easier" proofs are either by induction or by contradiction (or by contraposition, as in my answer above). So if you want to prove something first check if you can use any of these techniques. – Rudy the Reindeer Feb 6 '12 at 14:46 @tri Another common technique is to use tautologies in various ways, which requires a lot of practice to have a library of useful tautologies in your head. – Carl Mummert Feb 7 '12 at 12:58 The easiest would be to say $S$ inconsistent $\iff S\vdash \text{f}$. We then have if $T, {\neg}A$ inconsistent, then $$T,{\neg}A\vdash\text{f}$$ By the deduction theorem we have: $$T\vdash {\neg}{\neg}A$$ By double negation elimination we have: $$T\vdash A$$ There is a twist, the deduction theorem is a derived rule in most Hilbert Style systems. So the above is a real meta theorem, it doesn't show a fixed number of steps to arrive from $T, {\neg}A \vdash \text{f}$ to $T \vdash {\neg}{\neg}A$. Also the step from ${\neg}{\neg}A$ to $A$ can be quite tedious. Bye - $\LaTeX$ tip: \vdash produces the $\vdash$ symbol. \models produces $\models$ – Arturo Magidin Mar 18 '12 at 20:16 I know, but I ALWAYS use ASCII art. And damned, this stack exchange is too stupid to change it automatically into what they prefer. – j4n bur53 Mar 18 '12 at 20:20 In that case, I hope you don't mind my doing it. I, too spend a lot of time using ASCII art, and I'm glad I don't have to do so here. – Arturo Magidin Mar 18 '12 at 20:23 You're welcome. (Means: I would be pleased, english for "Bitteschön - Dank erwidernd"). – j4n bur53 Mar 18 '12 at 20:24 Huh? ${}{}{}{}{}{}{}{}$ – Arturo Magidin Mar 18 '12 at 20:25	2015-01-30 05:28:21	{"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": 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": 0.9104110598564148, "perplexity": 402.46203150214205}, "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/1422115858583.14/warc/CC-MAIN-20150124161058-00041-ip-10-180-212-252.ec2.internal.warc.gz"}
http://math.stackexchange.com/questions/276442/isomorphism-of-simple-extensions	Isomorphism of simple extensions I have to show that if $\alpha$ has the minimal polynomial $t^2-2$ over $\mathbb{Q}$ and $\beta$ has the minimal polynomial $t^2-4t+2$ over $\mathbb{Q}$ then the extensions $\mathbb{Q}(\alpha):\mathbb{Q}$ and $\mathbb{Q}(\beta):\mathbb{Q}$ are isomorphic. However the only way that I know how to show that two extensions are isomorphic is if $\alpha$ and $\beta$ had the same minimal polynomial, which they don't? Cheers Folks - You mean isomorphic as rational vector spaces...? – DonAntonio Jan 12 '13 at 17:44 $t^2-4t+3$ is reducible over $\Bbb{Q}$. It is impossible for it to be the minimal polynomial of anything. – Chris Eagle Jan 12 '13 at 17:47 @ChrisEagle It was supposed to be $t^2-4t+2$ sorry – Joe Cabel Jan 12 '13 at 17:51 Hint: If we imagine both fields embedded in $\Bbb C$, then clearly $\alpha$ must be $\pm\sqrt 2$ while by the quadratic formula $\beta = \frac{4\pm\sqrt{16-8}}{2}=2\pm\sqrt{2}$. - Another solution: Choose a common extension, say $\mathbb{C}$. Then $0 = \beta^2-4\beta+2 = (\beta-2)^2-2$ implies $\beta-2=\pm \alpha$, hence $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$. - Let $\alpha$ have a minimal polynomial $t^2-2$ over $\mathbb{Q}$, and $\beta$ have a minimal polynomial $t^2+4t+2$ over $\mathbb{Q}$. We need to show that the extensions $\mathbb{Q}(\alpha): \mathbb{Q}$ and $\mathbb{Q}(\beta): \mathbb{Q}$ are isomorphic. It is enough to show that $\alpha$ and $\beta$ can be written as a linear combination of one another because of the definition of a simple extension. If $\alpha$ have a minimal polynomial $t^2-2$ over $\mathbb{Q}$, then $\alpha=\pm \sqrt{2}$. Similarly, if $\beta$ have a minimal polynomial $t^2+4t+2$ over $\mathbb{Q}$, by the quadratic formula $$\beta=\frac{4\pm \sqrt{16-8}}{2}=\frac{4\pm 2\sqrt{2}}{2}=2\pm \sqrt{2},$$ then $\beta=2\pm 2$. Note that $$0=\beta^2-4\beta+2=\beta^2-4\beta+4+(-4+2)=(\beta-2)^2-2.$$ So $$(\beta-2)^2=2 \iff \beta-2=\pm \sqrt{2}=\beta-2=\pm \alpha.$$ Hence $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta).$ (i.e. The extensions $\mathbb{Q}(\alpha): \mathbb{Q}$ and $\mathbb{Q}(\beta): \mathbb{Q}$ are isomorphic.) - Consider all elements you can obtain using addition, subtraction, multiplication, and division with the rationals and $\alpha$ and deduce that you can obtain all elements of $\mathbb{Q}(\beta)$. This shows that $\mathbb{Q}(\beta)$ is a subfield of $\mathbb{Q}(\alpha)$. The do the converse to show they are actually equal. - protected by user26857Oct 19 '15 at 20:01 Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.	2016-02-06 03:45:27	{"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": 0.9520167112350464, "perplexity": 120.17696001450987}, "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-2016-07/segments/1454701145751.1/warc/CC-MAIN-20160205193905-00315-ip-10-236-182-209.ec2.internal.warc.gz"}
http://gmatclub.com/forum/ps-m02q37-58893.html	Find all School-related info fast with the new School-Specific MBA Forum It is currently 20 May 2013, 20:05 PS - % (m02q37) Author Message CIO Joined: 02 Oct 2007 Posts: 1261 Followers: 75 Kudos [?]: 505 [1] , given: 334 PS - % (m02q37) [#permalink] 23 Jan 2008, 08:46 1 KUDOS 00:00 Question Stats: 38% (02:19) correct 61% (01:02) wrong based on 1 sessions A number of light bulbs were purchased to illuminate a gym. However, only \normal\frac{2}{3} of them were needed. The extra 160 light bulbs were returned. 60% percent of their cost, or $96, was reimbursed. How much money was spent on illuminating the gym? (A) 360 (B) 320 (C) 384 (D) 364 (E) 160 [Reveal] Spoiler: OA C Source: GMAT Club Tests - hardest GMAT questions [Reveal] Spoiler: OA _________________ Welcome to GMAT Club! Want to solve GMAT questions on the go? GMAT Club iPhone app will help. Please read this before posting in GMAT Club Tests forum Result correlation between real GMAT and GMAT Club Tests Are GMAT Club Test sets ordered in any way? Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas. Find out what's new at GMAT Club - latest features and updates Kaplan GMAT Prep Discount Codes Knewton GMAT Discount Codes GMAT Pill GMAT Discount Codes CEO Joined: 17 Nov 2007 Posts: 3594 Concentration: Entrepreneurship, Other Schools: Chicago (Booth) - Class of 2011 GMAT 1: 750 Q50 V40 Followers: 230 Kudos [?]: 1299 [1] , given: 346 Re: PS - % [#permalink] 23 Jan 2008, 08:57 1 This post received KUDOS C x - money that was spent on illuminating the gym. (x+96)\frac13\frac{60}{100}=96 x=596-96=496=384 I think it is easy-medium level. But maybe I have wrong answer? _________________ iOS/Android: GMAT ToolKit - The bestselling GMAT prep app \| GMAT Club (free) \| PrepGame \| GRE ToolKit \| LSAT ToolKit PROMO: Are you an exiting GMAT ToolKit (iOS) user? Get GMAT ToolKit 2 (iOS) for free* (read more) Math: GMAT Math Book \|\|\| General: GMATTimer \|\|\| Chicago Booth: Slide Presentation The People Who Are Crazy Enough to Think They Can Change the World, Are the Ones Who Do. Director Joined: 01 Jan 2008 Posts: 635 Followers: 3 Kudos [?]: 114 [0], given: 1 Re: PS - % [#permalink] 23 Jan 2008, 09:38 I second walker, quick solution! Manager Joined: 20 Sep 2007 Posts: 112 Followers: 1 Kudos [?]: 17 [0], given: 0 Re: PS - % [#permalink] 23 Jan 2008, 09:43 Answer is B Extra bulbs 1/3 T = 160 Total bulbs T=480 (60/100) xcost of extra bulbs = 96 cost of extra bulbs=160 cost of each bulb = 1 cost of required bulbs=320 CEO Joined: 17 Nov 2007 Posts: 3594 Concentration: Entrepreneurship, Other Schools: Chicago (Booth) - Class of 2011 GMAT 1: 750 Q50 V40 Followers: 230 Kudos [?]: 1299 [2] , given: 346 Re: PS - % [#permalink] 23 Jan 2008, 10:09 2 This post received KUDOS I think we have to add losses 0.4160=64 320+64=384. _________________ iOS/Android: GMAT ToolKit - The bestselling GMAT prep app \| GMAT Club (free) \| PrepGame \| GRE ToolKit \| LSAT ToolKit PROMO: Are you an exiting GMAT ToolKit (iOS) user? Get GMAT ToolKit 2 (iOS) for free (read more) Math: GMAT Math Book \|\|\| General: GMATTimer \|\|\| Chicago Booth: Slide Presentation The People Who Are Crazy Enough to Think They Can Change the World, Are the Ones Who Do. Manager Joined: 01 Jan 2008 Posts: 231 Schools: Booth, Stern, Haas Followers: 1 Kudos [?]: 37 [0], given: 2 Re: PS - % [#permalink] 23 Jan 2008, 10:40 walker wrote: I think we have to add losses 0.4160=64 320+64=384. yeah, you are right, I didn't understand the question correctly +1 Manager Joined: 16 Apr 2008 Posts: 96 Followers: 1 Kudos [?]: 15 [0], given: 7 Re: PS - % (m02q37) [#permalink] 12 Jul 2010, 06:16 Yeah, i agree.. have to add the loss . 320+64=384 Final answer C Intern Joined: 05 Jul 2010 Posts: 28 Followers: 0 Kudos [?]: 16 [0], given: 7 Re: PS - % (m02q37) [#permalink] 12 Jul 2010, 07:06 Yeah, I agree. It's C, but the question is quite tricky. _________________ If you like my posts please consider giving me KUDOS!!!!!!!!! Thanks:) Intern Joined: 29 Mar 2010 Posts: 17 Schools: UCLA, USC WE 1: 3 Yr at leading SAAS company Followers: 0 Kudos [?]: 0 [0], given: 0 Re: PS - % (m02q37) [#permalink] 12 Jul 2010, 14:52 A number of light bulbs were purchased to illuminate a gym. However, only of them were needed. The extra 160 light bulbs were returned. 60% percent of their cost, or$96, was reimbursed. How much money was spent on illuminating the gym? (A) 360 (B) 320 (C) 384 (D) 364 (E) 160 Ans: C Let Total Bulbs be X 1/3X=160 => X=480(Total) Also, Let Z=Price of 1 bulb 16060z/100=96 =>z=1(1 bulb price) Thus Price Paid for illuminating gym= 480(Total Cost)-96(Refund) =384 (C) ------------ If you like my post please give me kudos. Intern Joined: 15 Jul 2010 Posts: 4 Location: Los Angeles Schools: UCLA; LMU; UCI; CSULB; WE 1: 13 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: PS - % (m02q37) [#permalink] 15 Jul 2010, 14:28 The question is not very clear. But since some of you got it correct, I shall not complain. I thought it was B, 320. Manager Joined: 25 Jun 2008 Posts: 127 Concentration: General Management, Technology Followers: 2 Kudos [?]: 22 [0], given: 5 Re: PS - % (m02q37) [#permalink] 14 Jul 2011, 10:10 Definitely 384. Probably about a 600-700 level question. Manager Status: GMAT BATTLE - WIN OR DIE Joined: 02 May 2011 Posts: 176 Concentration: General Management, Entrepreneurship GMAT Date: 12-22-2011 GPA: 3.81 WE: General Management (Hospitality and Tourism) Followers: 2 Kudos [?]: 14 [0], given: 13 Re: PS - % (m02q37) [#permalink] 15 Jul 2011, 06:58 Definitely C.Easy question Manager Status: I will not stop until i realise my goal which is my dream too Joined: 25 Feb 2010 Posts: 248 Schools: Johnson '15 Followers: 2 Kudos [?]: 14 [0], given: 16 Re: PS - % (m02q37) [#permalink] 15 Jul 2011, 07:49 yes i agree with 384, but what is the correct answer for this? _________________ Regards, Harsha Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat Satyameva Jayate - Truth alone triumphs Intern Joined: 13 Jul 2011 Posts: 11 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: PS - % (m02q37) [#permalink] 15 Jul 2011, 08:53 I also think the answer is 384.The people who got the answer as 384 should add 64 (the amount lost with 320) Manager Joined: 14 Apr 2011 Posts: 203 Followers: 2 Kudos [?]: 16 [0], given: 19 Re: PS - % (m02q37) [#permalink] 19 Jul 2011, 04:35 Trick question but I agree that question is not very clear. _________________ Looking for Kudos Senior Manager Joined: 08 Jun 2010 Posts: 400 Location: United States Concentration: General Management, Finance GMAT 1: 680 Q50 V32 Followers: 1 Kudos [?]: 37 [0], given: 13 Re: PS - % (m02q37) [#permalink] 11 Jan 2012, 07:08 I don't think the problem is tricky but I found the wording confusing. So, I interpreted the problem wrongly. One doubt : Does the reimbursement refer to the money for unused bulbs or the money used for the entire stock? It has a sentence correction error: "Their" has no clear referent. Intern Status: Looking for High GMAT Score Joined: 19 May 2012 Posts: 39 Location: India Concentration: Strategy, Marketing WE: Marketing (Internet and New Media) Followers: 0 Kudos [?]: 5 [0], given: 58 Re: PS - % (m02q37) [#permalink] 18 Jul 2012, 05:58 dzyubam wrote: A number of light bulbs were purchased to illuminate a gym. However, only \normal\frac{2}{3} of them were needed. The extra 160 light bulbs were returned. 60% percent of their cost, or $96, was reimbursed. How much money was spent on illuminating the gym? (A) 360 (B) 320 (C) 384 (D) 364 (E) 160 [Reveal] Spoiler: OA C Source: GMAT Club Tests - hardest GMAT questions 1/3 bulb -- 160 so 2/3 = 320 Total = 480 Required was 320 and 160 extra 60% 0f cost = 96 100% of cost = 1oo 96/ 60 = 160 so cost of 160 bulbs = 160 so required 320 was 320 and 96 was returned = 160 - 96 = 64 so total spent = 320+64 = 384 Ans C. Did within 2 mins !! _________________ “The best time to plant a tree was 20 years ago. The second best time is now.” – Chinese Proverb Manager Joined: 14 Jun 2012 Posts: 66 Followers: 0 Kudos [?]: 6 [0], given: 1 Re: PS - % (m02q37) [#permalink] 18 Jul 2012, 20:39 Good one. Tricky question in terms that it is not difficult to solve but just that one needs to be careful in reading the question. I selected B and almost at that instant realized my mistake. Nice. Need to focus more on the question asked. _________________ My attempt to capture my B-School Journey in a Blog : tranquilnomadgmat.blogspot.com There are no shortcuts to any place worth going. Intern Joined: 18 Jan 2012 Posts: 46 Location: United States Followers: 1 Kudos [?]: 36 [0], given: 21 Re: PS - % (m02q37) [#permalink] 19 Jul 2012, 00:36 Quote: A number of light bulbs were purchased to illuminate a gym. However, only 2/3 of them were needed. The extra 160 light bulbs were returned. 60% percent of their cost, or$96, was reimbursed. How much money was spent on illuminating the gym? Total nr of bulbs bought initially - say n Of these bulbs, 2/3 of them were used and hence 1/3 (1-2/3) bulbs were returned to the retailer. This implies that 1/3 (of) n bulbs were returned. 1/3 of n = 160 bulbs ==> n = 480 (Remember n refers to the original nr and 160 refers to the number of bulbs that were returned) Lets determine the nr of bulbs that were used for lighting the gym. 2/3 of (480) = 320 bulbs were used to illuminate the gym. Ok, now comes the tricky part. As per the question stem - 60% of the their cost was refunded.The trick is to understand identify the antecedent of the pronoun "their" This interpretation makes sense. Think of if this way, some retailers sell products under the condition that on return, they will NOT refund 100 % of the amount paid by the customer.Instead they will refunded a reduced % - say 70 % of the initial price. This policy is to discourage people from buying large quantities and then return these products for a refund. Back to the question :- When 160 bulbs were returned, the retailer refunded only 60% of the COST of THESE 160 bulbs i.e .6x of (total cost of 160 bulbs that were returned) = $96 total cost of 160 bulbs that were returned =$160 Cost of 1 bulb = $1 Total Cost of lights used to decorate the gym= Cost of the 320 Bulbs that were not returned = 320 x$1 = \$320 _________________ ----------------------------------------------------------------------------------------------------- IT TAKES QUITE A BIT OF TIME AND TO POST DETAILED RESPONSES. YOUR KUDOS IS VERY MUCH APPRECIATED ----------------------------------------------------------------------------------------------------- Intern Joined: 22 Jun 2012 Posts: 49 GMAT 1: 730 Q49 V40 Followers: 2 Kudos [?]: 13 [0], given: 5 Re: PS - % (m02q37) [#permalink] 19 Jul 2012, 05:05 TheMBAist wrote: The question is not very clear. But since some of you got it correct, I shall not complain. I thought it was B, 320. B could have been the answer if the question were "How much does it cost to illuminate the gym?" C is the answer to the actual question "How much money was spent on illuminating the gym?" Cheers Re: PS - % (m02q37) [#permalink] 19 Jul 2012, 05:05 Similar topics Replies Last post Similar Topics: PS 3 01 Aug 2003, 08:47 Ps 3 04 Sep 2003, 14:08 ps 5 10 Dec 2003, 11:16 Ps 1 15 Dec 2003, 13:12 Ps 2 19 Dec 2003, 14:21 Display posts from previous: Sort by PS - % (m02q37) Moderator: Bunuel Powered by phpBB © phpBB Group and phpBB SEO 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®.	2013-05-21 03:05:01	{"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": 0.523338258266449, "perplexity": 11223.884549543356}, "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-2013-20/segments/1368699675907/warc/CC-MAIN-20130516102115-00005-ip-10-60-113-184.ec2.internal.warc.gz"}
https://nbviewer.org/github/facaiy/book_notes/blob/master/Reinforcement_Learing_An_Introduction/n_step_bootstrapping/note.ipynb	In [2]: %matplotlib inline import matplotlib.pyplot as plt import seaborn as sns from IPython.display import Image # Chapter 7 n-step Bootstrapping¶ MC <-- n-step TD --> one-step TD In [3]: Image('./res/fig7_1.png') Out[3]: Monte Carlo return: $G_t \doteq R_{t+1} + \gamma R_{t+2} + \gamma^2 R_{t+3} + \cdots + \gamma^{T-t-1} R_T$ one-step return: $G_{t:t+1} \doteq R_{t+1} + \gamma V_t(S_{t+1})$ two-step return: $G_{t:t+2} \doteq R_{t+1} + \gamma R_{t+2} + \gamma^2 V_{t+1}(S_{t+2})$ n-step return: $G_{t:t+n} \doteq R_{t+1} + \gamma R_{t+2} + \cdots + \gamma^{n-1} R_{t+n} + \gamma^n V_{t+n-1}(S_{t+n})$ The natural state-value learning algorithm for using n-step returns is thus: $V_{t+n}(S_t) \doteq V_{t+n-1}(S_t) + \alpha \left [ G_{t:t+n} - V_{t+n-1}(S_t) \right ]$ In [4]: Image('./res/n_step_predict.png') Out[4]: error reduction property of n-step returns: their expectation is guaranteed to be a better estimate of $v_\pi$ than $V_{t+n-1}$ is, in a worst-state sense: $\max_s \left \| \mathbb{E}_\pi [G_{t:t+n} \mid S_t = s] - v_\pi(s) \right \| \leq \gamma^n \max_s \left \| V_{t+n-1}(s) - v_\pi(s) \right \|$ ### 7.2 n-step Sarsa¶ The main idea is to simply switch states for actions (state-action pairs) and then use an $\epsilon$-greedy policy. $Q_{t+n}(S_t, A_t) \doteq Q_{t+n-1}(S_t, A_t) + \alpha [G_{t:t+n} - Q_{t+n-1}(S_t, A_t)] \quad \text{ ,$0 \leq t < T$}$ In [5]: Image('./res/fig7_3.png') Out[5]: Expected Sarsa (like n-step sarsa): \begin{align} G_{t:t+n} & \doteq R_{t+1} + \cdots + \gamma^{n-1} R_{t+n} + \gamma^n \bar{V}_{t+n-1}(S_{t+n}) \qquad t + n < T \\ \bar{V}_t(s) & \doteq \sum_a \pi(a \mid s) Q_t(s, a) \qquad \text{for all $s \in \mathcal{S}$} \end{align} ### 7.3 n-step Off-policy Learning by Importance Sampling¶ simply be weighted by $\rho_{t:t+n-1}$, importance sampling ratio: \begin{align} V_{t+n}(S_t) & \doteq V_{t+n-1}(S_t) + \alpha \rho_{t:t+n-1} [G_{t:t+n} - V_{t+n-1}(S_t)] \qquad 0 \leq t < T \\ \rho_{t:h} & \doteq \prod_{k=t}^{\min(h, T-1)} \frac{\pi(A_k \mid S_k)}{b(A_k \mid S_k)} \\ Q_{t+n}(S_t, A_t) & \doteq Q_{t+n-1}(S_t, A_t) + \alpha \rho_{t+1:t+n-1} [G_{t:t+n} - Q_{t+n-1}(S_t, A_t)] \quad \text{ , $0 \leq t < T$} \end{align} ### 7.5 Off-policy Learning Without Importance Sampling: The n-step Tree Backup Algorithm¶ Tree-backup algorithm: In the tree-backup update, the target includes all these things plus the estimated values of the dangling action nodes hanging off the sides, at all levels. In [6]: Image('./res/tree_update.png') Out[6]: ### 7.6 A Unifying Algorithm: n-step $Q(\sigma)$¶ dicide on a step-by-step basis whether one wanted to take the action as a sample (as in Sarsa), or consider the expectation over all actions instead (as in the tree-backup update). Let $\sigma(t) \in [0, 1]$ denote the degree of sampling on step $t$, with $\sigma = 1$ denoting full sampling and $\sigma = 0$ denoting a pure expectation with no sampling. We call this proposed new algorithm n-step $Q(\sigma)$. In [7]: Image('./res/fig7_5.png') Out[7]: In [ ]:	2021-10-17 18:21:42	{"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.998961329460144, "perplexity": 7313.767947076468}, "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-2021-43/segments/1634323585181.6/warc/CC-MAIN-20211017175237-20211017205237-00337.warc.gz"}
http://www.ams.org/books/memo/1068/	#### How to Order For AMS eBook frontlist subscriptions or backfile collection purchases: 2. Complete and sign the license agreement. 3. Email, fax, or send via postal mail to: Customer Services American Mathematical Society 201 Charles Street Providence, RI 02904-2213 USA Phone: 1-800-321-4AMS (4267) Fax: 1-401-455-4046 Email: cust-serv@ams.org Visit the AMS Bookstore for individual volume purchases. Browse the current eBook Collections price list # memo_has_moved_text();A complete classification of the isolated singularities for nonlinear elliptic equations with inverse square potentials Florica C. Cîrstea Publication: Memoirs of the American Mathematical Society Publication Year: 2014; Volume 227, Number 1068 ISBNs: 978-0-8218-9022-6 (print); 978-1-4704-1429-0 (online) DOI: http://dx.doi.org/10.1090/memo/1068 Published electronically: June 24, 2013 Keywords:Nonlinear elliptic equations, isolated singularities, regular variation theory, inverse square potentials View full volume PDF View other years and numbers: Chapters • Chapter 1. Introduction • Chapter 2. Main results • Chapter 3. Radial solutions in the power case • Chapter 4. Basic ingredients • Chapter 5. The analysis for the subcritical parameter • Chapter 6. The analysis for the critical parameter • Chapter 7. Illustration of our results • Appendix A. Regular variation theory and related results ### Abstract In this paper, we consider semilinear elliptic equations of the form where $\lambda$ is a parameter with $-\infty <\lambda \leq (N-2)^2/4$ and $\Omega$ is an open subset in $\mathbb {R}^N$ with $N\geq 3$ such that $0\in \Omega$. Here, $b(x)$ is a positive continuous function on $\Omega \setminus \{0\}$ which behaves near the origin as a regularly varying function at zero with index $\theta$ greater than $-2$. The nonlinearity $h$ is assumed continuous on $\mathbb {R}$ and positive on $(0,\infty )$ with $h(0)=0$ such that $h(t)/t$ is bounded for small $t>0$. We completely classify the behaviour near zero of all positive solutions of [[eqref]]one when $h$ is regularly varying at $\infty$ with index $q$ greater than $1$ (that is, $\lim _{t\to \infty } h(\xi t)/h(t)=\xi ^q$ for every $\xi >0$). In particular, our results apply to [[eqref]]one with $h(t)=t^q (\log t)^{\alpha _1}$ as $t\to \infty$ and $b(x)=\|x\|^\theta (-\log \|x\|)^{\alpha _2}$ as $\|x\|\to 0$, where $\alpha _1$ and $\alpha _2$ are any real numbers. We reveal that the solutions of [[eqref]]one generate a very complicated dynamics near the origin, depending on the interplay between $q$, $N$, $\theta$ and $\lambda$, on the one hand, and the position of $\lambda$ with respect to $0$ and $(N-2)^2/4$, on the other hand. Our main results for $\lambda =(N-2)^2/4$ appear here for the first time, as well as for the case $\lambda <0$. We establish a trichotomy of positive solutions of [[eqref]]one under optimal conditions, hence generalizing and improving through a different approach a previous result with Chaudhuri on [[eqref]]one with $0<\lambda <(N-2)^2/4$ and $b=1$. Moreover, recent results of the author with Du on (0.1) with $\lambda =0$ are here sharpened and extended to any $-\infty <\lambda <(N-2)^2/4$. In addition, we unveil a new single-type behaviour of the positive solutions of [[eqref]]one specific to $0<\lambda <(N-2)^2/4$. We also provide necessary and sufficient conditions for the existence of positive solutions of (0.1) that are comparable with the fundamental solutions of In particular, for $b=1$ and $\lambda =0$, we find a sharp condition on $h$ such that the origin is a removable singularity for all non-negative solutions of [[eqref]]one, thus addressing an open question of Vázquez and Véron.	2019-12-13 10:42:16	{"extraction_info": {"found_math": true, "script_math_tex": 47, "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": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.7353479266166687, "perplexity": 619.5996921202457}, "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-2019-51/segments/1575540553486.23/warc/CC-MAIN-20191213094833-20191213122833-00104.warc.gz"}
https://physics.stackexchange.com/tags/space-expansion/hot?filter=week	# Tag Info ## Hot answers tagged space-expansion 4 In theories with dark energy, an energy density is just a property of space — even “empty” space devoid of matter and radiation. It’s closely related to Einstein’s old idea of a “cosmological constant”. Dark energy isn’t something material that comes from somewhere. As space expands, you simply get more dark energy because there is more volume; a dark energy ... 3 Velocities of galaxies When galaxies are gravitationally bound to each other in groups or clusters, they move on more or less elliptical orbits in the common gravitational potential from all the other galaxies (as well as all the dilute intracluster gas which is also a significant part of the total mass). I say "more or less" because galaxies do ... 3 Assuming (as this chart does) that the universe contains only dust and dark energy, the scale factor satisfies $$\dot a^2 = H_0^2 \left( Ω_{Λ,0}\, a^2 + Ω_{k,0} + Ω_{m,0}\, a^{-1} \right)$$ where $Ω_k = 1 - Ω_m - Ω_Λ$. (The exponents are $-1{-}3w$ where $w$ is the equation of state parameter.) The boundary of the $κ=\pm1$ regions is just the line $Ω_{k,0}=0$.... 2 There is a close analogy: If you jump say 3 feet high, you can calculate your speed as a function of you position between 0 to 3 feet high. Then you wonder what is your speed at 6 feet high during your jump. The good old high school Newtonian mechanics offers an answer: your speed was imaginary at 6 feet high according to the total energy conservation ... 2 One way to interpret Friedmann equation with an imaginary Hubble parameter is as arising from some solution with Euclidean metric signature. One class of such solution, termed “Euclidean wormholes” consists of two large asymptotic regions connected by a “throat”. Many FLRW cosmologies analytically continued into an imaginary time become such Euclidean ... 1 RW assumes that matter is uniformly distributed over the spatial slices, and elliptic, flat, and hyperbolic geometries have very different distributions, in terms of the amount of matter within a given distance of any given point. There's no way you could move the matter around to be homogeneous in a different geometry without violating homogeneity in the ... 1 Yes. "The curvature of the universe" is an imprecise term, and describing the curvature of a general four-dimensional spacetime takes 20 numbers at every point. But I'll assume that your phrase should mean the Ricci scalar curvature $R$, which is a single number at each point that is a kind of average curvature of spacetime (where the averaging is ... 1 No The curvature parameter $k$ of the universe remains constant throughout its evolution. If the universe is open ($k < 0$) , it will stay open, and if it is closed ($k > 0$), it will stay closed. That's because the amount of matter-energy of the universe is conserved, so if the density is greater than critical now it will forever be greater than ... 1 As far as we know, it's a coincidence. $1/H = a(t)/a'(t)$. In the early radiation-dominated era, $a(t) \propto t^{1/2}$, so $1/H$ is 2 times the actual age of the universe. In the later (but pre-modern) matter-dominated era, $a(t) \propto t^{2/3}$, so $1/H$ is 1.5 times the actual age. In the future dark-energy-dominated era, $a(t) \propto e^{t/t_0}$ where $... 1 Yes. The answer is kind of obvious, but if you do not see this, just note that the FLRW metric is: $$g=-c^2dt^2+a(t)^2d\Sigma^2_t,$$ where$d\Sigma^2_t$is metric induced up to the scale factor on the spacelike hypersurfaces of constant time$t$. This metric is independent of$t\$, from which it follows, that if you have a curve living on one of this ... Only top voted, non community-wiki answers of a minimum length are eligible	2020-08-11 10:13:10	{"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": 0.7576145529747009, "perplexity": 348.4088026318058}, "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-2020-34/segments/1596439738746.41/warc/CC-MAIN-20200811090050-20200811120050-00314.warc.gz"}
https://math.stackexchange.com/questions/2881368/fixed-points-of-deck-transformations/2881935	# Fixed points of deck transformations Suppose we have two different deck transformations $\tilde{S} \to \tilde{S}$ (where $S$ hyperbolic surface, which is closed, possibly punctured), say $\tau$ and $\kappa$ such that $\tau^i \ne \kappa^j$. We can extend them to be maps on the closure of $\tilde{S}$. By abuse of notation, we will denote these maps by $\tau$ and $\kappa$. Is there anyone proved that they don't share their fixed point(s) (in $\partial \tilde{S}$)? Motivation: suppose $\tau$ is a deck transformation corresponding to an essential closed curve $\alpha$ in hyperbolic surface $S$ (the definition of $\tau$ can be seen here - in the answer of Riccardo). In the case of closed, possibly punctured hyperbolic surface $S$, $\tau$ is an isometry of $\mathbb{H}^2$ of hyperbolic type. And moreover, the two fixed points of $\tau$ are not lifts of a puncture of $S$. It is sure that $\tau$ and $\kappa$ don't share their fixed point in $\tilde{S} = \mathbb{H}^2$. For an isometry of parabolic or hyperbolic type of $\mathbb{H}^2$, it will have one or two fixed point(s) in $\partial\mathbb{H}^2$ (then it is clear that an isometry of elliptic type, which has no fixed point in $\partial \mathbb{H}^2$, does not share its fixed point with two other types). I am wondering about the case of one isometry of hyperbolic type and one isometry of parabolic type (also for two hyperbolic types and two parabolic types). I'm thinking about the following picture: $\hskip1in$ Any isometry $f \in \text{Isom}^{+} (\mathbb{H}^2)$ can be extended uniquely to a map $\overline{f}: \overline{\mathbb{H}^2} \to \overline{\mathbb{H}^2}$. Since $\overline{f}$ is a self-homeomorphism of a closed disk, the Brouwer fixed point theorem gives that $\overline{f}$ has fixed point in $\overline{\mathbb{H}^2}$. The deck transformation group is a group. So if $\tau$ and $\kappa$ are deck transformations, it follows that $\kappa^{-1} \tau$ is a deck transformation. You started by saying that $\tau$ and $\kappa$ are different deck transformations, so $\tau \ne \kappa$. It follows that $\kappa^{-1} \tau$ is not the identity element of the deck transformation group. A nonidentity element of the deck transformation group has no fixed points, so $\kappa^{-1}\tau(x) \ne x$ for all $x \in \tilde S$. It follows that $\tau(x) \ne \kappa(x)$ for all $x \in \tilde S$. • Hi Lee Mosher, thank you for the answer. For an isometry of parabolic or hyperbolic type of $\mathbb{H}^2$, it will have one or two fixed point(s) in $\partial\mathbb{H}^2$ (then it's clear that an isometry of elliptic type, which has a fixed point in $\mathbb{H}^2$, does not share its fixed point with two other types). What you were writing is to show that an isometry of elliptic type does not share its fixed point with two other types. I am wondering about the case of one isometry of hyperbolic type and one isometry of parabolic type (also for two hyperbolic types and two parabolic types). – user578196 Aug 14 '18 at 9:10	2019-08-25 04:59:59	{"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": 0.9586002826690674, "perplexity": 84.14859743477858}, "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-2019-35/segments/1566027323067.50/warc/CC-MAIN-20190825042326-20190825064326-00410.warc.gz"}
https://gamedev.stackexchange.com/questions/164771/what-would-be-the-best-approach-to-creating-a-pressure-plate-in-a-2d-game	# What would be the best approach to creating a pressure plate in a 2D game? I've just really gotten started in Unity and while documentation has been helpful I haven't figure out how to detect collisions with 2 or more objects. I was able to get a script working that plays a sound when I touch the pressure plate with the player but I don't know how I would do something when 2 objects are within the collider but undo it when one of them leaves the area. void OnTriggerEnter2D(Collider2D other) { if(other.tag == "Player") { AudioSource.PlayClipAtPoint(Sound, transform.position); } } } Am I going about this the wrong way by using OnTriggerEnter2D? I see that there is a mass component on Rigidbodies but I don't know if I should be using it because I am still a novice in Unity. I'm also aware of OnCollisionExit but I don't think that works in this case because I'm currently only detecting 1 object but I need to detect 2. I was able to find videos that somewhat helped but all of them only worked with one object and were in a 3D space rather than a 2d one. Let's first declare a variable to store the number of players currently standing/colliding with the platform. In this case we want two players to activate the platform; Unity gives us 3 main functions to check whether an object collided with another object: OnTriggerEnter, OnTriggerStay and OnTriggerExit. In this case we are working with 2 dimensions, so we are going to use the 2D versions of these functions. We specifically want to check whether our players are both colliding with the object's collider; (In the pseudo script i attached below, I am assuming you tagged your first player Player1 and your second one Player2) We are going to use the OnTriggerEnter2D() function, (which takes in as parameter the object the platform is colliding with) to increment the player count variable by 1, if either one of our players is inside the trigger area; Once either one of them gets out the trigger area, playerCount gets decremented by one, to a total of 0, if both players have gone out the trigger area. PseudoScript int playerCount = 0; void OnTriggerEnter2D(Collider2D other) { if(playerCount < 2) { if(other.tag == "Player" \|\| other.tag == "Player2") { playerCount++; } } else { OnState(); } } void OnTriggerExit2D(Collider2D other) { if(playerCount > 0) { if(other.tag == "Player" \|\| other.tag == "Player2") { playerCount--; } } else { OffState(); } } • The logic in OnTriggerEnter2D won't work if only Player 1 is standing on the plate. – Quentin Oct 24 '18 at 9:55 • Now it won't work if something else is on the pressure plate :p – Quentin Oct 24 '18 at 10:01 • Sorry, lunch time is coming up ;) -- but I'll take your word for it, I'm not familiar with Unity. – Quentin Oct 24 '18 at 10:12 • I understand for the most part and I really appreciate the time that you spent helping me answer my question. I haven't tested it fully but it was really helpful in learning how to work with multiple collisions. I'm still fleshing out what I want OnState and OffState to do but I haven't gotten any compiler errors. – Mark Gregg Oct 24 '18 at 10:33 • Please do not get into the habit of relying on the Update function. This can be a big issue when working within Unity. By putting the actual code to check in the Update function, you are forcing an if check EVERY SINGLE frame. This may not appear much, but imagine there are 20 pressure plates? That's 20 additional checks per frame, for no reason. You already have the event calls of OnTriggerEnter2D and On TriggerExit2D, there is no need to use Update. Simply call a function when/where necessary from the OnTriggerEnter2D, or OnTriggerExit2D. – cmprogram Oct 24 '18 at 10:57 As you can see on the execution order diagram, the engine will first execute all FixedUpdate methods, then all OnTriggerXXX methods and then all Update methods. That means you can have a player count for your pressure plate which has 1. A FixedUpdate() method which resets the count to 0 2. A OnTriggerStay(Collider other) method which increases the count by 1 everytime it is called with an other collider which belongs to a player 3. A Update method which checks the count and acts appropriately. A completely different approach you could use would be to use the Unity physics engine and implement your pressure plate like an actual, physical pressure plate. 1. Assign non-kinematic, gravity-affected rigidbodies to the characters and the pressure plate. 2. Place the pressure plate slightly above the ground 3. Use a spring joint between pressure plate and floor so that the pressure plate resists forces. 4. Restrict the movement- and rotation axis of the plate's rigidbody so it can only move up and down. 5. Add a trigger-collider below the pressure plate which detects if the plate moves down. 6. Calibrate the spring strength and the player mass so that the weight of two players moves it into the trigger collider but the weight of one player does not. However, I would only recommend this solution if you are creating a game which is supposed to have realistic physics and you are already making use of the physics engine. Also keep in mind that crafty players might find alternative methods to add enough force to the pressure plate to trigger it. Whether this is good or bad depends on your game. • the physics-based approach is a much better looking and working apporach. 1+ – Gabriele Vierti Oct 24 '18 at 11:21 • While unlikely to be a problem in most real world cases, it isn't a good idea to rely on FixedUpdate as an "EarlyUpdate". For one it usually happens multiple times per update which can do unnecessary work (not really an issue in this case), but more importantly if the framerate goes very high it might not happen at all since it happens on a fixed interval! A more robust way can use Time.frameCount when setting/checking the value and resetting it to 0 before either operation if the current frameCount differs from the last used when setting. – Lunin Oct 24 '18 at 18:54 • @Lunin This situation doesn't use FixedUpdate as EarlyUpdate. It uses FixedUpdate as intended: As a step which happens before OnTriggerXXX events, which are also executed within the fixed update loop. – Philipp Oct 24 '18 at 22:49 • Ah, you're right; since OnTrigger is physics bound rather than update bound this should hold true even if FixedUpdate gets skipped due to high framerate as so will the OnTrigger events (thus leaving it in the correct state until the next physics update). My bad! I'll leave my comment unedited so any others with the same thought process I had can see the correction :) – Lunin Oct 24 '18 at 23:12 When working with colliders in Unity, remember that trigger and collider are two different things. Do not confuse these two. If you would like your object to react/bounce off another object, use Collider. In your case, Trigger is what yo should be using. So ensure IsTrigger is enabled on the pressure plate. You must also ensure at least one GameObject, either the pressure plate, or the player and other object you wish to detect, has a Rigidbody2D attached to it. private int colCount = 0; private void OnTriggerEnter2D(Collider2D col){ switch(col.tag){ case "Player": colCount++; break; case "Other Object": colCount++; break; } if(colCount >= 2){ runDetectionConditions(); } } private void OnTriggerExit2D(Collider2D col){ switch(col.tag){ case "Player": colCount--; break; case "Other Object": colCount--; break; } } // Run whatever conditions you would like to run, when two objects are detected on pressure plate private void runDetectionConditions(){ Debug.Log("Two objects are being detected on collider."); } ` There are a lot of people here already answering in terms of using the OnTrigger calls (which could be replaced with OnCollision calls if you want to work with non-trigger colliders). However I wanted to suggest another method should you find yourself in a situation where you just want to do a quick check of what colliders are in a given area without doing a lot of setup. Both 2D and 3D Unity have Physics functions to get one or many objects in a given location. For 2D Physics the single version is always something like Physics2D.Overlap[shape], while the multi version is Physics2D.Overlap[shape]All. For 3d Physics just drop the 2D from either of those. For something like a pressure plate you'd be looking at something like Physics2D.OverlapAreaAll, Physics2D.OverlapBoxAll, Physics2D.OverlapCapsuleAll, or as another option possibly Physics2D.LinecastAll. These could all be placed much like your trigger is now, only through code instead of the editor. From these you will get an array of Collider2D objects which are in the given area and on the appropriate layers (should you employ layer masking, which you should since it will throw out a lot of unnecessary checking). From there you can count, adjust, or whatever any traits you wish on the relevant GameObjects based on the results of the check. As a side note since I see performance comments on other answers, this method will pay the detection cost for each pressure plate you have in scene that is doing the check. However, adding triggers to your scene is not free either. In either case don't worry about the performance too much until you start seeing a problem, at which point you can pull up the profiler and see what is taking too much of your time and address it.	2019-12-05 14:53:12	{"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": 0.3345626890659332, "perplexity": 1134.1438020474352}, "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/1575540481076.11/warc/CC-MAIN-20191205141605-20191205165605-00410.warc.gz"}
http://math.stackexchange.com/questions/46380/finding-the-integer-le-n-with-largest-number-of-divisors	# Finding the integer $\le n$ with largest number of divisors As mentioned in an answer to this question an integer less than $n$ with largest number of divisors can be found exploring the numbers $x$ of the form $$x = 2^{a_1} 3^{a_2} \dots p_k^{a_k} \dots$$ (where $p_k$ is the $k$-th prime number) with the conditions $$x \le n < 2x \quad \text{and}\quad a_1 \ge a_2 \ge \dots \ge a_k \ge \dots$$ to determine the complexity of this algorithm I would like to know the asymptotic number of tuples $(a_1, a_2, \dots)$ verifying these conditions as $n\to \infty$. I suspect that this number is $\gg_l \log^l n$ for every $l$ and $\ll_\epsilon n^\epsilon$ for every $\epsilon > 0$, but I don't know how to prove it. Can you explain further what you mean by " $\gg \log^k n$ for every $k$"? If this means $(\log n)^k$, is there some upper limit on $k$, perhaps based on primorial numbers? – Henry Jun 19 '11 at 23:33 @Henry: I don't understand what you are asking. Although I will guess that the problem is that the OP should not use the same $k$ to denote both the $k^{th}$ prime and the growth condition concerning $\log n$. Specifically, the last line should read "I suspect that this number is $\gg_l \log^l n$ for every $l$, and $\dots$" – Eric Naslund Jun 20 '11 at 2:39	2014-12-21 12:51:38	{"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": 0.9354537725448608, "perplexity": 84.40822484523684}, "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/1418802771164.85/warc/CC-MAIN-20141217075251-00089-ip-10-231-17-201.ec2.internal.warc.gz"}
https://www.wyzant.com/resources/answers/topics/proportion	28 Answered Questions for the topic Proportion 09/12/19 #### S†atistics: Proportion At Cityville High School, 36% of the students participate in sports, 36% of the students participate in academic clubs, and 13% of the students participate in both sports and academic clubs.Find... more 07/22/19 #### how tall is the tree Phil, Melissa, Noah, and Olivia saw a tall tree that cast a shadow 4242 feet long. They observed at the same time that a 55-foot-tall person cast a shadow that was 10.510.5 feet long. How tall is... more 05/10/19 #### proportion question Professor Schaben has a ratio of passing students to failing students of 5 to 4. One semester, she had a total of 216 students. Use a proportion to show many students passed and how many failed.... more 12/04/18 #### My mom walks 5/6 of a yard every minute.what is my mom rate in yard per minute My grandma walks 5/6 of a yard every 4/12 minute what is my grandma rate in yards per minute 02/06/18 #### A car travelling at 30km/h complete a journey in 2 1/2 hour. How long will it take to complete the same journey if the speed is decreased to 20 km/ h How to solve inverse proportion using average speed and distance. For primary 5 ratio and proportion 05/29/17 #### Proportion A piece of land is to be divided between two men in the ratio 5:9 what fraction of the second lose if the land is divided instead in ratio 9:5.. 05/16/17 #### A. What percent of 90 is 3? B. 5% of what number is 10? A. What percent of 90 is 3? B. 5% of what number is 10? 05/03/17 If y is the mean proportion of x and z , then the mean proportion of (x2+y2) and (y2+z2) is Options: 1.xz+yz 2.xy-yx 3.xy+yz 4.yz-zx 03/28/17 #### If I have to read 367 pages in 21 days how many pages will I have to read in a day kmjnijjknjnnijbhbbhbkjnnnnnnnnnnnnnnbbbnnniikmkjjbbbbjmkkmmknnkkmknnjnnknjkn 01/09/17 #### Scales and Diagram When making a photocopy of a 10 cm by 15 cm photograph, Elizabeth enlarged the picture by a scale factor of 150%. By what percent was the area of the photograph enlarged? 12/15/16 #### Confidence Intervals In a Harris Poll of 412 HR professionals, 85.3% said that the job applicants appearance is most important for a good impression. Construct a 94% Confidence Interval for the True population... more 09/21/16 #### proportion Nuala drives to her grandma’s. She drives at 20 miles per hour . The journey takes 50 minutes. How long would the journey take if if Nuala drove at 40 miles per hour? 09/09/16 If A is directly proportional to C and B is directly proportional to C, prove that each of the following is directly proportional to C. a) A+B b) A–B c)√AB 06/03/16 #### some people want to comapre the proportion of high school boys that smoke cigarettes to the proportion of high school girls that smoke cigarettes. some people want to comapre the proportion of high school boys that smoke cigarettes to the proportion of high school girls that smoke cigarettes. use the data below to test the claim that the... more 11/04/15 1 ## Still looking for help? Get the right answer, fast. Get a free answer to a quick problem. Most questions answered within 4 hours. #### OR Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.	2020-01-27 18:57:34	{"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": 0.511946976184845, "perplexity": 1951.6670762223007}, "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/1579251705142.94/warc/CC-MAIN-20200127174507-20200127204507-00123.warc.gz"}
https://encyclopediaofmath.org/wiki/Killing_space	# Killing space Jump to: navigation, search $( X , n )$ A fibre space $p _ {n} : ( X , n ) \rightarrow X$ for which the homotopy groups $\pi _ {i} ( X , n )$ vanish if $i < n$, and $p _ {n} : \pi _ {i} ( X , n ) \rightarrow \pi _ {i} ( X)$ is an isomorphism if $i \geq n$. The space $( X , n )$ is constructed by induction with respect to $n$. If $( X , n - 1 )$ has already been constructed, then $( X , n )$ is taken to be the homotopy fibre of the canonical mapping $$( X , n - 1 ) \rightarrow K ( \pi _ {n-} 1 ( X) , n - 1 ) ,$$ $K ( \pi _ {n-} 1 ( X) , n - 1 )$ denoting an Eilenberg–MacLane space. The sequence of spaces $( X , n )$ and mappings $p _ {n}$ is a Moore–Postnikov system of the mapping $ \rightarrow X$. #### Comments See also [a1], Chapt. 8, Sect. 3. #### References [a1] E.H. Spanier, "Algebraic topology" , McGraw-Hill (1966) pp. Chapt. 2 How to Cite This Entry: Killing space. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Killing_space&oldid=47498 This article was adapted from an original article by A.F. Kharshiladze (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article	2022-09-30 08:58:57	{"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": 0.9758918285369873, "perplexity": 1177.2254808012062}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2022-40/segments/1664030335448.34/warc/CC-MAIN-20220930082656-20220930112656-00406.warc.gz"}
https://ideas.repec.org/a/spr/empeco/v28y2003i2p335-351.html	# Weighted samples, kernel density estimators and convergence ## Author Listed: • Francisco J. Goerlich Gisbert ## Abstract This note extends the standard kernel density estimator to the case of weighted samples in several ways. In the first place I consider the obvious extension by substituting the simple sum in the definition of the estimator by a weighted sum, but I also consider other alternatives of introducing weights, based on adaptive kernel density estimators, and consider the weights as indicators of the informational content of the observations and in this sense as signals of the local density of the data. All these ideas are shown using the Penn World Table in the context of the macroeconomic convergence issue. Copyright Springer-Verlag Berlin Heidelberg 2003 ## Suggested Citation • Francisco J. Goerlich Gisbert, 2003. "Weighted samples, kernel density estimators and convergence," Empirical Economics, Springer, vol. 28(2), pages 335-351, April. • Handle: RePEc:spr:empeco:v:28:y:2003:i:2:p:335-351 DOI: 10.1007/s001810200134 as File URL: http://hdl.handle.net/10.1007/s001810200134 As the access to this document is restricted, you may want to search for a different version of it. ## Citations Citations are extracted by the CitEc Project, subscribe to its RSS feed for this item. as Cited by: 1. Falko Juessen, 2009. "A distribution dynamics approach to regional GDP convergence in unified Germany," Empirical Economics, Springer, vol. 37(3), pages 627-652, December. 2. Goerlich, Francisco José & Mas, Matilde, 2004. "Three (Marginal?) Questions Regarding Convergence," MPRA Paper 15876, University Library of Munich, Germany, revised 2004. 3. Benito , Juan Miguel & Ezcurra, Roberto, 2004. "Spatial disparities in the European Union: national and sectoral elements," INVESTIGACIONES REGIONALES - Journal of REGIONAL RESEARCH, Asociación Española de Ciencia Regional, issue 4, pages 75-98. 4. Adolfo Maza & José Villaverde & María Hierro, 2015. "Non- $$\hbox {CO}_2$$ CO 2 Generating Energy Shares in the World: Cross-Country Differences and Polarization," Environmental & Resource Economics, Springer;European Association of Environmental and Resource Economists, vol. 61(3), pages 319-343, July. 5. Hisham S. El-Osta, 2010. "Inequality decomposition of farm family living expenditures and the role of the life cycle," Agricultural Finance Review, Emerald Group Publishing, vol. 70(2), pages 245-266, August. ### Keywords Key words: Weighted samples; survey data; regional data; kernel density estimates; convergence.; JEL classification: C00; C8.; ### JEL classification: • C00 - Mathematical and Quantitative Methods - - General - - - General ## Corrections All material on this site has been provided by the respective publishers and authors. You can help correct errors and omissions. When requesting a correction, please mention this item's handle: RePEc:spr:empeco:v:28:y:2003:i:2:p:335-351. See general information about how to correct material in RePEc. For technical questions regarding this item, or to correct its authors, title, abstract, bibliographic or download information, contact: (Sonal Shukla) or (Andrew Huffard) The email address of this maintainer does not seem to be valid anymore. Please ask Andrew Huffard to update the entry or send us the correct email address. General contact details of provider: http://www.springer.com . If you have authored this item and are not yet registered with RePEc, we encourage you to do it here. This allows to link your profile to this item. It also allows you to accept potential citations to this item that we are uncertain about. We have no references for this item. You can help adding them by using this form . If you know of missing items citing this one, you can help us creating those links by adding the relevant references in the same way as above, for each refering item. If you are a registered author of this item, you may also want to check the "citations" tab in your RePEc Author Service profile, as there may be some citations waiting for confirmation. Please note that corrections may take a couple of weeks to filter through the various RePEc services. 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RePEc uses bibliographic data supplied by the respective publishers.	2019-05-19 17:41:01	{"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": 0.2894071638584137, "perplexity": 4969.797327201156}, "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-2019-22/segments/1558232255071.27/warc/CC-MAIN-20190519161546-20190519183546-00045.warc.gz"}
https://tex.stackexchange.com/questions/234388/is-there-a-way-to-export-major-element-styles	# Is there a way to export major element styles? pre-notice: I tried very hard to find an answer to this question and searched with various different terms. But I still have the tiny feeling that someone asked this before, because it is a very general and valid question. So if it is asked before and please direct me to the relevant topic and accept my apologies :) The problem is easy to state but proved extremely hard to solve (at least for a TeX newbie such as myself): I want to know certain things about my tex document such as chapter and section font types and sizes. Is there a way to extract this information, for instance, with a package that displays style information in a structured format? If memory serves there was a package for displaying page margins but I failed to find one for displaying styles. I am aware that some styles are user-defined and it can be parsed from document preamble. I have two problems with that. If I own (created) the tex document, I might want to check whether everything is in place and as planned, easily. If I don't own the tex document and want to use some style in that document I might end up with doing a lot of guessing. And obviously there are some styles default to some packages or document styles that I need to dig into their source. I rather not do that if I can get away with something more subtle :) Finally, let me rephrase my question in a 'sacrilegious' way. In MS Word when I click on some text I can see whether it is Calibri or Arial, font size, bold and italic or not, exact color, even paragraph parameters and other stuff. I require something similar for TeX documents. edit: I have one more analogy: I want for TeX what Inspect Element is for HTML and CSS elements. • The used fonts can be retrieved from the pdf (viewer) under Properties – user31729 Mar 21 '15 at 19:24 • Yes I can get a list of fonts using Adobe Reader (could not find it on Mac's Preview). It is more than nothing, but still not in the level of detail I'd like. But thank you for bringing that suggestion. – berkorbay Mar 21 '15 at 19:31 • It isn't clear what answer you want here, other than the source code of the class file you are using. \chapter for example will typically be defined to involve spacing before and after the head, alignment of the heading, setting up information for the table of contents, it may include an image or quotation or any other details required by the class. Other than the source code of the class it isn't clear how any of this information could be returned. – David Carlisle Mar 21 '15 at 21:41 • If it was originally specified using titlesec, getting those settings back is reasonable, if you are asking for some automatic tool to parse arbitrary tex code implementing a heading and work out what it is doing, then I doubt that is possible. – David Carlisle Mar 22 '15 at 9:18 • I'm not clear what you are trying to get information from: the .tex source or the .dvi/.ps/.html/.pdf output? At one point you say you might not have the source, which suggests the latter. But then you say that parsing the source shouldn't be too onerous, which suggests the former. – cfr Mar 25 '15 at 1:34 Short answer, no this is not possible. It is not possible to automatically even locate the definition of a section heading, without worrying about how to parse that definition to work out what fonts are being used. Take the standard article class for example. in there you will find \newcommand\section{\@startsection {section}{1}{\z@}% {-3.5ex \@plus -1ex \@minus -.2ex}% {2.3ex \@plus.2ex}% {\normalfont\Large\bfseries}} So if you know what the arguments of \@startsection mean then you know the spacing and some details about the font choice. Exactly what font is \normalfont depends on what font packages and declarations are being used. But to take the first class I tried that is not in the core latex distribution, the koma scrartcl class. It took me a while to find the definition of \section (I don't really know the details of this class) there is no obvious definition of \section instead you find \DeclareSectionCommand[% style=section,% level=1,% indent=\z@,% beforeskip=-3.5ex \@plus -1ex \@minus -.2ex,% afterskip=2.3ex \@plus.2ex,% tocindent=0pt,% tocnumwidth=1.5em% ]{section} Note the lack of \ in {section}. unlike \@startsection which is defined in the latex format, \DeclareSectionCommand is defined in this file so the only way the proposed parser would recognise that this declaration is in fact defining \section would be for it to parse its definition which is \newcommand{\DeclareSectionCommand}[2][]{% \edef\reserved@a{% \noexpand\FamilyStringKey[.dsc]{KOMAarg}{style}{% \expandafter\noexpand\csname scr@#2@style\endcsname }% }\reserved@a \scr@ifundefinedorrelax{scr@local@levelincrease}{% \RelaxFamilyKey[.dsc]{KOMAarg}{increaselevel}% }{% \FamilyCounterMacroKey[.dsc]{KOMAarg}{increaselevel}[1]% {\scr@local@levelincrease}% }% \scr@ifundefinedorrelax{scr@local@leveloffset}{% \edef\reserved@a{% \noexpand\FamilyCounterMacroKey[.dsc]{KOMAarg}{level}{% \expandafter\noexpand\csname #2numdepth\endcsname}% }\reserved@a }{% \edef\reserved@a{% \noexpand\DefineFamilyKey[.dsc]{KOMAarg}{level}{% \noexpand\FamilySetCounterMacro{KOMAarg}{level}{% \expandafter\noexpand\csname #2numdepth\endcsname }% \unexpanded{% {\numexpr ##1+\scr@local@leveloffset\relax}% \edef\scr@local@leveloffset{% \the\numexpr\scr@local@leveloffset+\scr@local@levelincrease\relax }% }% }% }\reserved@a }% \edef\reserved@a{% \noexpand\FamilyLengthMacroKey[.dsc]{KOMAarg}{indent}{% \expandafter\noexpand\csname scr@#2@sectionindent\endcsname}% }\reserved@a \edef\reserved@a{% \noexpand\FamilyLengthMacroKey[.dsc]{KOMAarg}{beforeskip}{% \expandafter\noexpand\csname scr@#2@sectionbeforeskip\endcsname}% }\reserved@a \edef\reserved@a{% \noexpand\FamilyLengthMacroKey[.dsc]{KOMAarg}{afterskip}{% \expandafter\noexpand\csname scr@#2@sectionafterskip\endcsname}% }\reserved@a \edef\reserved@a{% \noexpand\DefineFamilyKey[.dsc]{KOMAarg}{font}{% \noexpand\IfExistskomafont{#2}{% \noexpand\setkomafont }{% \noexpand\newkomafont }{#2}{####1}% \noexpand\FamilyKeyStateProcessed }% }\reserved@a \FamilyStringKey[.dsc]{KOMAarg}{counterwithin}{\scr@local@counterwithin}% \let\scr@local@counterwithin\relax \scr@ifundefinedorrelax{scr@local@tocleveloffset}{% \edef\reserved@a{% \noexpand\FamilyCounterMacroKey[.dsc]{KOMAarg}{toclevel}{% \expandafter\noexpand\csname #2tocdepth\endcsname}% }\reserved@a }{% \edef\reserved@a{% \noexpand\DefineFamilyKey[.dsc]{KOMAarg}{toclevel}{% \noexpand\FamilySetCounterMacro{KOMAarg}{toclevel}{% \expandafter\noexpand\csname #2tocdepth\endcsname }% \unexpanded{% {\numexpr ##1+\scr@local@tocleveloffset\relax}% \edef\scr@local@tocleveloffset{% \the\numexpr\scr@local@tocleveloffset+\scr@local@levelincrease\relax }% }% }% }\reserved@a }% \edef\reserved@a{% \noexpand\FamilyLengthMacroKey[.dsc]{KOMAarg}{tocindent}{% \expandafter\noexpand\csname scr@#2@tocindent\endcsname}% }\reserved@a \edef\reserved@a{% \noexpand\FamilyLengthMacroKey[.dsc]{KOMAarg}{tocnumwidth}{% \expandafter\noexpand\csname scr@#2@tocnumwidth\endcsname}% }\reserved@a \FamilyExecuteOptions[.dsc]{KOMAarg}{#1}% \begingroup \scr@ifundefinedorrelax{#2numdepth}{% \scr@declaresectioncommanderror{#2}{section level}{level}% }{}% \scr@ifundefinedorrelax{scr@#2@sectionindent}{% \scr@declaresectioncommanderror{#2}{section indent}{indent}% }{}% \scr@ifundefinedorrelax{scr@#2@sectionbeforeskip}{% \scr@declaresectioncommanderror{#2}{before section skip}{beforeskip}% }{}% \scr@ifundefinedorrelax{scr@#2@sectionafterskip}{% \scr@declaresectioncommanderror{#2}{after section skip}{afterskip}% }{}% \IfExistskomafont{#2}{}{% \scr@declaresectioncommanderror{#2}{font}{font}% }{}% \scr@ifundefinedorrelax{scr@#2@tocindent}{% \scr@declaresectioncommanderror{#2}{toc entry indent}{tocindent}% }{}% \scr@ifundefinedorrelax{scr@#2@tocnumwidth}{% \scr@declaresectioncommanderror{#2}{toc entry number width}{tocnumwidth}% }{}% \endgroup \@firstofone{% \@ifundefined{c@#2}{\newcounter{#2}}{}% \ifx\scr@local@counterwithin\relax \else\ifx\scr@local@counterwithin\@empty \@namedef{the#2}{\arabic{#2}}% \else \@removefromreset{#2}{\scr@local@counterwithin}% \expandafter\def\csname the#2\expandafter\endcsname{% \csname the\scr@local@counterwithin\endcsname.\arabic{#2}}% \fi \fi \@ifundefined{#2format}{% \@namedef{#2format}{\csname the#2\endcsname\autodot\enskip}% }{}% \ifstr{\csname scr@#2@style\endcsname}{}{% \ClassInfo{\KOMAClassName}{% not defining \expandafter\string\csname #2\endcsname' due to\MessageBreak empty section style% }% }{% \scr@ifundefinedorrelax{scr@#2@style}{% \ClassWarning{\KOMAClassName}{using default section style}% \@namedef{scr@#2@style}{section}% }{}% \expandafter\edef\csname #2\endcsname{% \noexpand\scr@ifundefinedorrelax{% scr@start\csname scr@#2@style\endcsname}{% \noexpand\ClassError{\noexpand\KOMAClassName}{% section style \csname scr@#2@style\endcsname' not defined}{% A not yet defined section style \csname scr@#2@style\endcsname' has been setup for\MessageBreak \expandafter\string\csname #2\endcsname'. You should either setup another style\MessageBreak or define the style.\MessageBreak If you'll continue, style section' will be used as an emergency fallback.% }% \noexpand\def\expandafter\noexpand\csname scr@#2@style\endcsname {section}% }{}% \expandafter\noexpand \csname scr@start\csname scr@#2@style\endcsname\endcsname {#2}% \expandafter\noexpand\csname #2numdepth\endcsname \expandafter\noexpand\csname scr@#2@sectionindent\endcsname \expandafter\noexpand\csname scr@#2@sectionbeforeskip\endcsname \expandafter\noexpand\csname scr@#2@sectionafterskip\endcsname{% \noexpand\ifdim\noexpand\glueexpr \expandafter\noexpand\csname scr@#2@sectionbeforeskip\endcsname <\noexpand\z@ \unexpanded{% \ifnum \scr@compatibility>\@nameuse{scr@v@2.96}\relax \setlength{\parfillskip}{\z@ plus 1fil}% \fi }% \noexpand\fi \unexpanded{% \raggedsection\normalfont\sectfont\nobreak\usekomafont{#2}% }% }% }% }% \@ifundefined{DeclareSectionNumberDepth}{% \@ifundefined{#2markformat}{% \@namedef{#2markformat}{\csname the#2\endcsname\autodot\enskip}% }{}% \@ifundefined{#2mark}{% \expandafter\let\csname #2mark\endcsname\@gobble }{}% }{% \DeclareSectionNumberDepth{#2}{\csname #2numdepth\endcsname}% }% \scr@ifundefinedorrelax{#2tocdepth}{% \expandafter\let\csname #2tocdepth\expandafter\endcsname \csname #2numdepth\endcsname }{}% \expandafter\providecommand\expandafter% }% \scr@ifundefinedorrelax{l@#2}{% \expandafter\edef\csname l@#2\endcsname{% \noexpand\bprot@dottedtocline \expandafter\noexpand\csname #2tocdepth\endcsname \expandafter\noexpand\csname scr@#2@tocindent\endcsname \expandafter\noexpand\csname scr@#2@tocnumwidth\endcsname }% }{}% }% } and note that in there the line \expandafter\edef\csname #2\endcsname{ means that the second argument being {section} results in \section being defined. Getting from there to working out anything about which fonts are being used seems tricky. a human reader may guess something from key names such as afterskip=2.3ex \@plus.2ex, but the key names are just arbitrary names, so any guesses are just guesses unless you have traced the code to see what they do. And the above is just to consider one command, \section in two classes, article and scratcl. • Although I hope that you are wrong with all my heart, I must admit I have no alternative proposition. Perhaps in the future, someone with the proper obsession and sufficient technical expertise could come up with a solution. I am signing it as the proper answer, concluding the question. – berkorbay Mar 24 '15 at 21:58 • @berkorbay It's like hoping God's wrong... – cfr Mar 25 '15 at 1:37 • @cfr Not at all. By the way, I meant no offense to anyone in my previous comment (now that I see it can be perceived that way thanks to online communication). I merely wanted to say maybe someone in the future will look into it seriously. – berkorbay Mar 25 '15 at 16:14 • @berkorbay It's unlikely anyone will look into it too seriously as I suspect it's provably impossible. What is more likely is that latex3 (or some other suitable successor) will have a more powerful declarative styling mechanism (see the xtemplate package for initial drafts) which will mean that most, and hopefully all, classes will not need to be executing arbitrary imperative code in order to specify a section style. If all headings are made by calling a known number of parametrised templates, then answering your question just means looking up the parameters used. – David Carlisle Mar 25 '15 at 16:51 This just concerns the 'sacrilegious' bit. One reason that you can recover precise formatting information for particular document elements easily in Word is precisely that Word does not use semantic mark-up. Indeed, Word does not even have styles in the sense that LaTeX does. From the document author's point of view, Word's styles are derivative - it is the formatting which is basic. From the document author's point of view, LaTeX's styles are basic - it is the formatting which is derivative. So when you look at the 'source' of a Word document, insofar as there is such a thing, it does not necessarily tell you the functional role of each document element. A heading may use a style, if the author is well-organised and disciplined. Or it may not. But it is hard work to keep the format of document elements aligned with styles. You have to stop yourself doing the natural thing which is to directly change the format of the element in front of you. In contrast, it is easy to use use styles in a LaTeX document because the styles are right there. The specific formatting is not right in front of you. And just as it is not right there in front of you with the document element, it is not right there with the document element for a parser, either. When you look at a document element in the source, what you see is its functional role, and it is this semantic layer which you manipulate when writing document content. And this is a good thing! If you don't care about the advantages, of course, you can just mark-up the document elements directly. In that case, the basic formatting will be right there, and a parser will be able to determine it fairly easily. Of course, it won't be able to easily tell that the element is a heading or whatever, but it will be able to tell that it is large, bold etc. If you go further, you could directly select particular fonts and sizes, and then the parser could give you more information. The more you avoid semantic mark-up, the more the parser will be able to tell you. That is not a coincidence and it is not a (good) reason to eschew such mark-up! • Thanks for the explanation. I love to take the full advantage of LaTeX styles and love that people can use without the worry of altering the way of things. The problem begins when I try to make a default style of my own. When I like a color, I can see its code with a color picker. When I like an HTML element, I can see the basics from Firebug. But it is hard with LaTeX documents. I encountered many "How can I do my layout/fonts like this?" questions in SE. And, to me, it is like "What is the color code of this?" for an image to me. Unfortunately, the answer was negative. – berkorbay Mar 25 '15 at 16:20 • @berkorbay I understand that. What I meant was: you can't have the one without the other any more than you can have hills without valleys. Note that you can't pull the style information from the .html source either if it is using .css`, say. The situation really isn't different. (A PDF viewer can tell you the font information etc., the colour and so on. That is, the rendered result can be inspected here, as well.) – cfr Mar 25 '15 at 22:21	2021-06-18 20:09:52	{"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": 0.8080233931541443, "perplexity": 1137.7260036440837}, "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/1623487641593.43/warc/CC-MAIN-20210618200114-20210618230114-00575.warc.gz"}
http://mathoverflow.net/questions/152134/intuition-about-covariant-derivative-connections-on-real-and-complex-manifolds	# Intuition about covariant derivative/connections on real and complex manifolds I was hoping to gain more intuition about the similarities and differences between the covariant derivative (of any connection, not necessarily the Levi Civita one if it exists) on real and complex manifolds - This question is outlandishly vague. Specifically, it is almost as vague as asking for intuition on the similarities and differences between vectors in a given vector space. – J. Martel Dec 17 '13 at 20:13 To be more specific, I was hoping to learn which commonly used properties of the real covariant derivative do not hold (in analogue) for the complex covariant derivative. – Benjamin Dec 18 '13 at 2:15 As far as I know, a covariant derivative on a complex manifold (whatever you mean by that) is a special case of a covariant derivative on a real manifold. Do you have an example where this is not the case? – Deane Yang Dec 18 '13 at 14:52 I have always found the motivation for connections as allowing us to differentiate vector fields or tensors by vector fields" outrageous -- but that's my own taste. Connections become appealing to me when I remember Kobayashi-Nomizu's definition of connections on principal fibre bundles. That is, we have a principal fibre bundle $P$ over some manifold $M$ with structure group $G$. For every point $p \in P$ the tangent space $T_pP$ has a canonical direction defined by the orbit of $G$, i.e. those vectors tangent to the orbit through $p$. A connection $\nabla$ then becomes a smooth family of subspaces $Q_p, p\in P$ such that (i) the tangent space at $p$ splits into a direct sum of the tangent space to the orbit $T_p G.p$ and $Q_p$ and (ii) for every $g \in G$, the subspace $Q_{gp}$ coincides with the image of $Q_p$ under the differential $dL_g$ of the left translation by $g$. More briefly, the tangent bundle $TP$ has a canonical subbundle defined by the orbits of $G$, whereas there is no canonical choice of complimentary subbundle. A connection is such a smooth complimentary subbundle. All these basic facts are in Chapter II (Theory of Connections) in Kobayashi-Nomizu I. However, I have never been able to understand exactly how torsion or curvature or symmetry translates into this setting. A second useful reference for the abstract development of connections is Lawson's "Minimal submanifolds in complex geometry". This might be up your alley if you want to compare connections in the complex setting. On MO, I think you only get as good as you give. I initially downvoted your question because, in all fairness, it deserves to be closed and then resurrected. - I haven't looked at Kobayashi-Nomizu in decades, but I'd be surprised if it doesn't define curvature using the definition of a connection you cite. In any case, it's easy to guess what it is, since a connection is flat if and only if the distribution is integrable. Probably the easiest way to verify this is by differentiating the Lie algebra-valued connection 1-form. Torsion is defined only if the principal fiber bundle is a subbundle of tangent frame bundle. Not sure how that translates. And what do you mean by symmetry? That's usually a synonym for torsion-free. – Deane Yang Dec 18 '13 at 14:58 Let $X$ be a smooth manifold and $\pi:V\rightarrow X$ a real vector bundle. Let $\mathcal{O}_X$, $\Gamma$, and $\Omega_X^1$ denote the sheaves of functions on $X$, smooth sections of $V$, and $1$-forms on $X$, respectively. A connection on $V$ is a morphism $$\nabla:\Gamma\rightarrow\Gamma\otimes_{\mathcal{O}_X}\Omega_X^1$$ of sheaves of $\mathbb{R}$-vector spaces on $X$ that satisfies the Leibniz rule. Essentially, a connection is just a mechanism for differentiating (local and global) sections of $V$. (It generalizes the exterior derivative of smooth functions on $X$, since the exterior derivative is an example of a connection on the trivial rank-one bundle on $X$.) The covariant derivative comes from a connection on the tangent bundle and allows you to differentiate its sections (vector fields). -	2015-10-08 18:16:01	{"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": 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": 0.900447428226471, "perplexity": 201.37773303149586}, "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-2015-40/segments/1443737898933.47/warc/CC-MAIN-20151001221818-00087-ip-10-137-6-227.ec2.internal.warc.gz"}
https://bigrquery.r-dbi.org/reference/api-perform.html	These functions are low-level functions designed to be used by experts. Each of these low-level functions is paired with a high-level function that you should use instead: • bq_perform_copy(): bq_table_copy(). • bq_perform_query(): bq_dataset_query(), bq_project_query(). • bq_perform_upload(): bq_table_upload(). • bq_perform_load(): bq_table_load(). • bq_perform_extract(): bq_table_save(). bq_perform_extract( x, destination_uris, destination_format = "NEWLINE_DELIMITED_JSON", compression = "NONE", ..., billing = x$project ) bq_perform_upload( x, values, fields = NULL, create_disposition = "CREATE_IF_NEEDED", write_disposition = "WRITE_EMPTY", ..., billing = x$project ) x, source_uris, billing = x\$project, source_format = "NEWLINE_DELIMITED_JSON", fields = NULL, nskip = 0, create_disposition = "CREATE_IF_NEEDED", write_disposition = "WRITE_EMPTY", ... ) bq_perform_query( query, billing, ..., parameters = NULL, destination_table = NULL, default_dataset = NULL, create_disposition = "CREATE_IF_NEEDED", write_disposition = "WRITE_EMPTY", use_legacy_sql = FALSE, priority = "INTERACTIVE" ) bq_perform_query_dry_run( query, billing, ..., default_dataset = NULL, parameters = NULL, use_legacy_sql = FALSE ) bq_perform_copy( src, dest, create_disposition = "CREATE_IF_NEEDED", write_disposition = "WRITE_EMPTY", ..., billing = NULL ) ## Arguments x A character vector of fully-qualified Google Cloud Storage URIs where the extracted table should be written. Can export up to 1 Gb of data per file. Use a wild card URI (e.g. gs://[YOUR_BUCKET]/file-name-.json) to automatically create any number of files. The exported file format. Possible values include "CSV", "NEWLINE_DELIMITED_JSON" and "AVRO". Tables with nested or repeated fields cannot be exported as CSV. The compression type to use for exported files. Possible values include "GZIP", "DEFLATE", "SNAPPY", and "NONE". "DEFLATE" and "SNAPPY" are only supported for Avro. Additional arguments passed on to the underlying API call. snake_case names are automatically converted to camelCase. Whether to print out a header row in the results. Identifier of project to bill. Data frame of values to insert. A bq_fields specification, or something coercible to it (like a data frame). Leave as NULL to allow BigQuery to auto-detect the fields. Specifies whether the job is allowed to create new tables. The following values are supported: "CREATE_IF_NEEDED": If the table does not exist, BigQuery creates the table. "CREATE_NEVER": The table must already exist. If it does not, a 'notFound' error is returned in the job result. Specifies the action that occurs if the destination table already exists. The following values are supported: "WRITE_TRUNCATE": If the table already exists, BigQuery overwrites the table data. "WRITE_APPEND": If the table already exists, BigQuery appends the data to the table. "WRITE_EMPTY": If the table already exists and contains data, a 'duplicate' error is returned in the job result. The fully-qualified URIs that point to your data in Google Cloud. For Google Cloud Storage URIs: Each URI can contain one '' wildcard character and it must come after the 'bucket' name. Size limits related to load jobs apply to external data sources. For Google Cloud Bigtable URIs: Exactly one URI can be specified and it has be a fully specified and valid HTTPS URL for a Google Cloud Bigtable table. For Google Cloud Datastore backups: Exactly one URI can be specified. Also, the '*' wildcard character is not allowed. The format of the data files: For CSV files, specify "CSV". For datastore backups, specify "DATASTORE_BACKUP". For newline-delimited JSON, specify "NEWLINE_DELIMITED_JSON". For Avro, specify "AVRO". For parquet, specify "PARQUET". For orc, specify "ORC". For source_format = "CSV", the number of header rows to skip. SQL query string. Named list of parameters match to query parameters. Parameter x will be matched to placeholder @x. Generally, you can supply R vectors and they will be automatically converted to the correct type. If you need greater control, you can call bq_param_scalar() or bq_param_array() explicitly. See https://cloud.google.com/bigquery/docs/parameterized-queries for more details. A bq_table where results should be stored. If not supplied, results will be saved to a temporary table that lives in a special dataset. You must supply this parameter for large queries (> 128 MB compressed). A bq_dataset used to automatically qualify table names. If TRUE will use BigQuery's legacy SQL format. Specifies a priority for the query. Possible values include "INTERACTIVE" and "BATCH". Batch queries do not start immediately, but are not rate-limited in the same way as interactive queries. A bq_job. ## Examples if (bq_testable()) { ds <- bq_test_dataset() bq_mtcars <- bq_table(ds, "mtcars") `	2021-06-14 13:13:43	{"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": 0.1798129379749298, "perplexity": 8355.420520163101}, "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/1623487612154.24/warc/CC-MAIN-20210614105241-20210614135241-00576.warc.gz"}
https://math.stackexchange.com/questions/2385380/graph-theory-connectivity-proof	# Graph Theory Connectivity Proof In this problem, we consider the edge connectivity of a simple undirected graph, which is the minimum number of edges one can remove to disconnect it. Prove that if G is a connected simple undirected graph where every vertex's degree is a multiple of 2, then one must remove at least 2 edges in order to disconnect the graph. (It should be noted that removing a vertex does not necessarily disconnect a simple graph.) Does the statement above remain true if the number 2 is replaced with any positive integer k? If so, prove it. If not, give a counterexample. Typically I would write where I am for a problem like this, but I have no idea how to approach this proof. Any and all help would be much appreciated. $k$ is even Assume that the edge connectivity is $1$, and we remove one edge to make the graph disconnected. Then, apart from the two end points of the removed edge, all vertices still have an even degree. Therefore, in each of the two connected components there is exactly one vertex of odd degree, and this violates the handshaking lemma (applied to a single connected component). Therefore we conclude that the edge connectivity cannot be $1$. $k$ is odd: Consider the complete graph on $k+1$ vertices (each vertex then has degree $k$). We shall add an additional vertex to it in order to construct a counterexample. Take all except $2$ of the vertices, and divide them into pairs. For each pair, take the edge between the two, and let it go through the new vertex instead of directly between the two vertces in the pair. The new vertex now has degree $k-1$, while all the other vertices still have degree $k$. Take two of these graphs, and add an edge between the two "new" vertices. You now have a graph of edge connectivity $1$, where each vertex has degree exactly $k$, and therefore degree divisible by $k$.	2021-06-23 09:48:59	{"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": 0.7478472590446472, "perplexity": 52.99704331692492}, "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/1623488536512.90/warc/CC-MAIN-20210623073050-20210623103050-00268.warc.gz"}
https://www.transtutors.com/questions/instructions-a-compute-the-ending-inventory-at-september-30-and-cost-of-goods-sold-u-1082792.htm	# Instructions (a) Compute the ending inventory at September 30 and cost of goods sold using the FIFO. Instructions (a) Compute the ending inventory at September 30 and cost of goods sold using the FIFO and LIFO method. Prove the amount allocated to cost of goods sold under each method. (b) For both FIFO and LIFO, calculate the sum of inventory and cost of goods sold. What do you notice about the answer you found for each method? (c) What is gross profit under each method? (d) Which method results in a larger amount reported for assets on the balance sheet? Which results in a larger amount reported for stockholders' equity on the balance sheet? Attachments: Ending Inventory at Septeber 30 a) Under FIFO Method Goods Avilable for Sale 144 Total Units Sold 124 Ending Invetory 20 Cost of Goods Sold Cost of Goods Available for Sale 15510 Less: Ending Cost of Goods 2240 (20*112) 13270 Under LIFO Method... ## Related Questions in Financial Accounting - Others • ### accounting (Solved) December 10, 2013 totaled 1,900 units. Instructions (a) Determine the cost of goods available for sale. ( b ) Determine the ending inventory and the cost of goods sold under each of the assumed cost flow methods ( FIFO , LIFO , and average- cost ). Prove the accuracy of the cost of goods sold under the FIFO • ### accounting (Solved) June 29, 2016 ,900 units. Instructions (a) Determine the cost of goods available for sale. ( b ) Determine the ending inventory and the cost of goods sold under each of the assumed cost flow methods ( FIFO , LIFO , and average- cost ). Prove the accuracy of the cost of goods sold under the FIFO and LIFO methods a) Cost of goods available for sale Qty Rate Value Beginning inventory 200 6 1200 24-Jan 800 7 5600 12-Apr 400 8 3200 19-Aug 600 9 5400 30-Nov 350 10 3500 Cost of goods available for sale... • ### Which inventory principle is most relevant to Davidson’s decision? Which inventory method September 15, 2015 • ### Accounting question (Solved) December 15, 2013 of merchandise. b . The only flow assumption in which all units of merchandise are assigned the same per-unit cost . c . The method used to record the cost of goods sold when each unit in the inventory is unique	2018-07-23 08:01:32	{"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": 0.3729096055030823, "perplexity": 3746.9074604922425}, "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/1531676595531.70/warc/CC-MAIN-20180723071245-20180723091245-00135.warc.gz"}
https://socratic.org/questions/what-is-the-slope-of-the-line-passing-through-the-following-points-0-4-2-7	# What is the slope of the line passing through the following points: (0, -4) , (2,7) ? Dec 23, 2015 The slope is : color(blue)(11/2 #### Explanation: The points are (0,-4) = color(blue)(x_1,y_1 (2,7) = color(blue)(x_2,y_2 The slope is found using formula color(blue)(m= (y_2-y_1)/(x_2-x_1) $m = \frac{7 - \left(- 4\right)}{2 - 0}$ $m = \frac{7 + 4}{2}$ $m = \frac{11}{2}$	2022-10-04 23:38:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "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": 0.48901739716529846, "perplexity": 3693.54183860563}, "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-2022-40/segments/1664030337529.69/warc/CC-MAIN-20221004215917-20221005005917-00234.warc.gz"}
https://stats.stackexchange.com/questions/421794/what-is-the-meaning-of-the-beta-for-the-interaction-between-continuous-variables	# What is the meaning of the beta for the interaction between continuous variables in a linear mixed-model? If I create a mixed-effects linear regression model similar to the following (using the lme4 package in R), where all of the fixed effect variables are continuous: model <- lmer(Y ~ a + b + c + abc + ab + bc + (1\|randvar1) + (1\|randvar2), data=dataset1) summary(model) gives me the list of beta estimates. For each of a, b and c these are the expected increase in Y when that independent variable increases by 1 (assuming the other variables are unchanged). But what does the beta estimate mean for the interaction terms between two or more continuous variables? As you state the coefficient for a gives the expected increase in Y if you increase a by 1. Similarly for b. Suppose there is no interaction then if we simultaneously increase a and b by 1 then the expected increase in Y is simply the sum of the coefficients for a and b. If there is an interaction then the coefficient for a.b has to be added to the sum of a and b. Note this applies to all regression models it is not specific to mixed effect models nor to R. The coefficient for the interaction of a and b is interpreted the same way in a mixed effects model like this as it is for a regular linear model. It is the effect of each 1 unit increase in ab. In including this term you assume that the effect of a on Y increases linearly with b. The interaction term reflects the slope of this linear increase. That is for every unit increase in b, the effect of a unit increase in a goes up by the value of that coefficient. As mentioned in the other answers, the existence of the random effects is irrelevant. As a matter of R code, your function does not include ac. If that was intentional, be aware that abc is automatically expanded to: a:b + a:c + b:c + a:b:c. That is, the term will be included. To deliberately exclude it, you should list the terms using the colon operator (:). However, it is generally recommended not to leave out lower level terms, including lower level interaction terms, once a higher level term exists (see: Including the interaction but not the main effects in a model, and Do all interactions terms need their individual terms in regression model?). On the other hand, if that was an unintentional omission, note that your ab + bc is redundant (the first term automatically expands, as just discussed). You correctly note that "For each of a, b and c these are the expected increase in Y when that independent variable increases by 1 (assuming the other variables are unchanged)." However, it is important to recognize that "the other variables" cannot remain "unchanged", unless the other two main effects variables are exactly $$0$$. To understand this more fully, it may help to read my answer to: What does “all else equal” mean in multiple regression? Likewise, because you have a three-way interaction term in your model, the included two-way interactions have their clearest interpretation when the variable that is included in the three-way interaction but not included in the given two-way interaction is $$0$$. (For example, a:b is the interaction between a and b, when c=0.) At this point we can start to think about the interpretation of the beta for an interaction. As elsewhere, the standard interpretation is ceteris paribus ('all else held equal'). However, note that it is not possible to increment a:b:c (or even a:b) by $$1$$ without any of the other terms changing. This makes the situation difficult. There are some rules, of increasing complexity, for interpreting the sign of the beta as, say, superadditive ('more than the sum of its parts') or subadditive ('less than the sum of its parts'), etc., but these depend on the signs of other betas as well. I've never met anyone who has found those rules useful. You can compute the partial derivative and think of it as the rate of change in the slope / lower order interaction, if that helps you, but again, many people find that too much to wrap their heads around. The best way is to try to interpret the model as a whole, and not the interaction term in isolation. To do that, solve for the simple effects of whichever variable has precedence in your mind at several values of the other interacting variables. You could then write out the equations of those lines, or you could plot them (cf., How to visualize a fitted multiple regression model?).	2019-10-18 14:25:27	{"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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7887938618659973, "perplexity": 432.47325416516037}, "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/1570986682998.59/warc/CC-MAIN-20191018131050-20191018154550-00456.warc.gz"}
https://www.projecteuclid.org/euclid.bbms/1503453711	## Bulletin of the Belgian Mathematical Society - Simon Stevin ### Some remarks on the structure of Lipschitz-free spaces #### Abstract We give several structural results concerning the Lipschitz-free spaces $\mathcal F(M)$, where $M$ is a metric space. We show that $\mathcal F(M)$ contains a complemented copy of $\ell_1(\Gamma)$, where $\Gamma=\text{dens}(M)$. If $\mathcal N$ is a net in a finite dimensional Banach space $X$, we show that $\mathcal F(\mathcal N)$ is isomorphic to its square. If $X$ contains a complemented copy of $\ell_p, c_0$ then $\f(\mathcal N)$ is isomorphic to its$\ell_1$-sum. Finally, we prove that for all $X\cong C(K)$ spaces, where $K$ is a metrizable compact, $\f(\mathcal N)$ are mutually isomorphic spaces with a Schauder basis. #### Article information Source Bull. Belg. Math. Soc. Simon Stevin, Volume 24, Number 2 (2017), 283-304. Dates First available in Project Euclid: 23 August 2017 https://projecteuclid.org/euclid.bbms/1503453711 Digital Object Identifier doi:10.36045/bbms/1503453711 Mathematical Reviews number (MathSciNet) MR3694004 Zentralblatt MATH identifier 06850672 #### Citation Hájek, Peter; Novotný, Matěj. Some remarks on the structure of Lipschitz-free spaces. Bull. Belg. Math. Soc. Simon Stevin 24 (2017), no. 2, 283--304. doi:10.36045/bbms/1503453711. https://projecteuclid.org/euclid.bbms/1503453711	2020-02-25 17:09:58	{"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": 0.76838219165802, "perplexity": 510.46451494231735}, "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-2020-10/segments/1581875146123.78/warc/CC-MAIN-20200225141345-20200225171345-00270.warc.gz"}
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/1152/4/a/t/	Properties Label 1152.4.a.t Level 1152 Weight 4 Character orbit 1152.a Self dual yes Analytic conductor 67.970 Analytic rank 0 Dimension 2 CM no Inner twists 1 Related objects Newspace parameters Level: $$N$$ $$=$$ $$1152 = 2^{7} \cdot 3^{2}$$ Weight: $$k$$ $$=$$ $$4$$ Character orbit: $$[\chi]$$ $$=$$ 1152.a (trivial) Newform invariants Self dual: yes Analytic conductor: $$67.9702003266$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{3})$$ Defining polynomial: $$x^{2} - 3$$ Coefficient ring: $$\Z[a_1, \ldots, a_{11}]$$ Coefficient ring index: $$2^{2}$$ Twist minimal: no (minimal twist has level 128) Fricke sign: $$1$$ Sato-Tate group: $\mathrm{SU}(2)$ $q$-expansion Coefficients of the $$q$$-expansion are expressed in terms of $$\beta = 4\sqrt{3}$$. We also show the integral $$q$$-expansion of the trace form. $$f(q)$$ $$=$$ $$q + ( 2 + 8 \beta ) q^{5} + ( 4 + 8 \beta ) q^{7} +O(q^{10})$$ $$q + ( 2 + 8 \beta ) q^{5} + ( 4 + 8 \beta ) q^{7} + ( -46 - 4 \beta ) q^{11} + ( -50 + 24 \beta ) q^{13} + ( -46 - 48 \beta ) q^{17} + ( 2 + 28 \beta ) q^{19} + ( 4 + 72 \beta ) q^{23} + ( 71 + 32 \beta ) q^{25} + ( 42 + 40 \beta ) q^{29} + ( 192 + 64 \beta ) q^{31} + ( 200 + 48 \beta ) q^{35} + ( 86 + 56 \beta ) q^{37} + ( 150 - 32 \beta ) q^{41} + ( 150 + 20 \beta ) q^{43} + ( -8 + 176 \beta ) q^{47} + ( -135 + 64 \beta ) q^{49} + ( -6 - 56 \beta ) q^{53} + ( -188 - 376 \beta ) q^{55} + ( 322 - 132 \beta ) q^{59} + ( -146 + 280 \beta ) q^{61} + ( 476 - 352 \beta ) q^{65} + ( -86 + 332 \beta ) q^{67} + ( 204 - 168 \beta ) q^{71} + ( 206 - 208 \beta ) q^{73} + ( -280 - 384 \beta ) q^{77} + ( 200 + 144 \beta ) q^{79} + ( -474 + 52 \beta ) q^{83} + ( -1244 - 464 \beta ) q^{85} + ( -286 + 464 \beta ) q^{89} + ( 376 - 304 \beta ) q^{91} + ( 676 + 72 \beta ) q^{95} + ( 1102 + 368 \beta ) q^{97} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q + 4q^{5} + 8q^{7} + O(q^{10})$$ $$2q + 4q^{5} + 8q^{7} - 92q^{11} - 100q^{13} - 92q^{17} + 4q^{19} + 8q^{23} + 142q^{25} + 84q^{29} + 384q^{31} + 400q^{35} + 172q^{37} + 300q^{41} + 300q^{43} - 16q^{47} - 270q^{49} - 12q^{53} - 376q^{55} + 644q^{59} - 292q^{61} + 952q^{65} - 172q^{67} + 408q^{71} + 412q^{73} - 560q^{77} + 400q^{79} - 948q^{83} - 2488q^{85} - 572q^{89} + 752q^{91} + 1352q^{95} + 2204q^{97} + O(q^{100})$$ Embeddings For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below. For more information on an embedded modular form you can click on its label. Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$ 1.1 −1.73205 1.73205 0 0 0 −11.8564 0 −9.85641 0 0 0 1.2 0 0 0 15.8564 0 17.8564 0 0 0 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles Atkin-Lehner signs $$p$$ Sign $$2$$ $$-1$$ $$3$$ $$-1$$ Inner twists This newform does not admit any (nontrivial) inner twists. Twists By twisting character orbit Char Parity Ord Mult Type Twist Min Dim 1.a even 1 1 trivial 1152.4.a.t 2 3.b odd 2 1 128.4.a.g yes 2 4.b odd 2 1 1152.4.a.s 2 8.b even 2 1 1152.4.a.r 2 8.d odd 2 1 1152.4.a.q 2 12.b even 2 1 128.4.a.e 2 24.f even 2 1 128.4.a.h yes 2 24.h odd 2 1 128.4.a.f yes 2 48.i odd 4 2 256.4.b.h 4 48.k even 4 2 256.4.b.i 4 By twisted newform orbit Twist Min Dim Char Parity Ord Mult Type 128.4.a.e 2 12.b even 2 1 128.4.a.f yes 2 24.h odd 2 1 128.4.a.g yes 2 3.b odd 2 1 128.4.a.h yes 2 24.f even 2 1 256.4.b.h 4 48.i odd 4 2 256.4.b.i 4 48.k even 4 2 1152.4.a.q 2 8.d odd 2 1 1152.4.a.r 2 8.b even 2 1 1152.4.a.s 2 4.b odd 2 1 1152.4.a.t 2 1.a even 1 1 trivial Hecke kernels This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{4}^{\mathrm{new}}(\Gamma_0(1152))$$: $$T_{5}^{2} - 4 T_{5} - 188$$ $$T_{7}^{2} - 8 T_{7} - 176$$ $$T_{13}^{2} + 100 T_{13} + 772$$ Hecke characteristic polynomials $p$ $F_p(T)$ $2$ 1 $3$ 1 $5$ $$1 - 4 T + 62 T^{2} - 500 T^{3} + 15625 T^{4}$$ $7$ $$1 - 8 T + 510 T^{2} - 2744 T^{3} + 117649 T^{4}$$ $11$ $$1 + 92 T + 4730 T^{2} + 122452 T^{3} + 1771561 T^{4}$$ $13$ $$1 + 100 T + 5166 T^{2} + 219700 T^{3} + 4826809 T^{4}$$ $17$ $$1 + 92 T + 5030 T^{2} + 451996 T^{3} + 24137569 T^{4}$$ $19$ $$1 - 4 T + 11370 T^{2} - 27436 T^{3} + 47045881 T^{4}$$ $23$ $$1 - 8 T + 8798 T^{2} - 97336 T^{3} + 148035889 T^{4}$$ $29$ $$1 - 84 T + 45742 T^{2} - 2048676 T^{3} + 594823321 T^{4}$$ $31$ $$1 - 384 T + 84158 T^{2} - 11439744 T^{3} + 887503681 T^{4}$$ $37$ $$1 - 172 T + 99294 T^{2} - 8712316 T^{3} + 2565726409 T^{4}$$ $41$ $$1 - 300 T + 157270 T^{2} - 20676300 T^{3} + 4750104241 T^{4}$$ $43$ $$1 - 300 T + 180314 T^{2} - 23852100 T^{3} + 6321363049 T^{4}$$ $47$ $$1 + 16 T + 114782 T^{2} + 1661168 T^{3} + 10779215329 T^{4}$$ $53$ $$1 + 12 T + 288382 T^{2} + 1786524 T^{3} + 22164361129 T^{4}$$ $59$ $$1 - 644 T + 462170 T^{2} - 132264076 T^{3} + 42180533641 T^{4}$$ $61$ $$1 + 292 T + 240078 T^{2} + 66278452 T^{3} + 51520374361 T^{4}$$ $67$ $$1 + 172 T + 278250 T^{2} + 51731236 T^{3} + 90458382169 T^{4}$$ $71$ $$1 - 408 T + 672766 T^{2} - 146027688 T^{3} + 128100283921 T^{4}$$ $73$ $$1 - 412 T + 690678 T^{2} - 160275004 T^{3} + 151334226289 T^{4}$$ $79$ $$1 - 400 T + 963870 T^{2} - 197215600 T^{3} + 243087455521 T^{4}$$ $83$ $$1 + 948 T + 1360138 T^{2} + 542054076 T^{3} + 326940373369 T^{4}$$ $89$ $$1 + 572 T + 845846 T^{2} + 403242268 T^{3} + 496981290961 T^{4}$$ $97$ $$1 - 2204 T + 2633478 T^{2} - 2011531292 T^{3} + 832972004929 T^{4}$$	2020-10-21 13:37:26	{"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": 0.9644862413406372, "perplexity": 5972.433784611533}, "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-2020-45/segments/1603107876500.43/warc/CC-MAIN-20201021122208-20201021152208-00577.warc.gz"}
https://www.hackmath.net/en/math-problem/1274	# Map 2 At what scale is made map if the distance 8.2 km corresponds on the map segment 5 cm long? Correct result: M 1: 164000 #### Solution: $M = 8.2/5 \cdot 1000 \cdot 100 = 164000 \approx 1:164000$ We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you! Tips to related online calculators Need help calculate sum, simplify or multiply fractions? Try our fraction calculator. Check out our ratio calculator. Do you want to convert length units? ## Next similar math problems: • Scale of the map Determine the scale of the map, which is the actual distance of 120 km l represented by segment long 6 cm. • Scale of the map The distance between two cities is actually 30 km and the map is 6 cm. What is the scale of the map? • Scale of the map Determine the scale of the map if the actual distance between A and B is 720 km and distance on the map is 20 cm. • Map On the tourist map at a scale of 1 : 50000 is distance between two points along a straight road 3.7 cm. How long travels this distance on a bike at 30 km/h? Time express in minutes. • Scale 3 Miriam room is 3.2 meters wide. It is draw by line segment length 6.4 cm on floor plan. In what scale it is plan of the room? • Map scale The rectangular plot has in a scale of 1: 10000 area 3 cm2 on the map. What content does this plot have on a 1:5000 scale map? • The plan The plan of the housing estate is in three scales 1: 5000,1: 10000,1: 15000. The distance between two points on a plan with a scale of 1: 10000 is 12 cm. What is this distance on the other two plans? What is this distance? • Two villages On the map with a scale of 1:40000 are drawn two villages actually 16 km away. What is their distance on the map? • Two monuments The distance between two historical monuments on a map with a scale of 1: 500000 is 48 mm. Find the actual distance between these monuments in km. • Line segment The 4 cm long line segment is enlarged in the ratio of 5/2. How many centimeters will measure the new line segment? • On the On the map of Europe made at a scale of 1: 4000000, the distance between Bratislava and Paris is 28 cm. At what time an airplane flying at 800 km/h will fly this journey? • Points on line segment Points P & Q belong to segment AB. If AB=a, AP = 2PQ = 2QB, find the distance: between point A and the midpoint of the segment QB. • Line segment Cut a line segment of 15 cm into two line segments so that their lengths are in ratio 2:1. What length will each have? • Map scale Garden has on plan on a scale of 1: 150 width 22 cm and length 35 cm. What is the real area of the garden? • Divide in ratio Line segment AB 12 cm long divide in a ratio of 5: 3. How long are the individual parts? • Nautical miles How many nautical miles do they sail if the route is shown on a 1:25 000 scale map with a 7.4 cm long line? • Image scale The actual image dimensions are 60 cm x 80cm and has a reduced size 3 cm x 4 cm. At what scale the image was reduced?	2021-01-27 20:20:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "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": 0.47976744174957275, "perplexity": 1433.3814175398818}, "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/1610704832583.88/warc/CC-MAIN-20210127183317-20210127213317-00125.warc.gz"}
https://uniontestprep.com/parapro-assessment	## What you need to know about the ParaPro Assessment The ParaPro Assessment is one of the Praxis® tests for educational personnel. It measures the basic skills deemed necessary to function as a paraprofessional or teacher assistant/aide in a school setting. Beginning with the No Child Left Behind Act in 2002 and continuing with the Every Student Succeeds Act that replaced it, it is required that paraprofessionals have one of these three qualifications: • An associate of arts degree • Two (2) years of college • A passing score on the district’s choice of assessment that proves the candidate has the knowledge and skills to assist in the teaching of reading, writing, and mathematics The ParaPro Assessment is the test chosen by some school districts to satisfy the third option above. Producers of this test say that the questions are approximately leveled for a student who has completed 2 years of post-high school study (college level). The reading level of the questions is no higher than the level a person would need to work with K-12 students. This test is administered on computer with other options available for people who qualify for certain accommodations. The testing format is basically selected-response (mostly multiple-choice) and you may not use a calculator. Some questions may vary from the typical multiple-choice format where you select one answer from a list of four possible choices. Directions for each type of answer will be clearly given, so read carefully. Here are the other question types you may see: • Multiple-choice in which you choose more than one answer • Questions requiring numeric answers to be typed into a box • Questions requiring the use of the mouse to click on parts of a graphic stimulus • Questions requiring clicking on sentences in a reading passage • Questions requiring you to drag answers from one spot to another and drop them there • Questions requiring you to choose the answer from a drop-down menu and click on it The exam has three parts: Reading, Writing, and Mathematics. Each part contains 30 questions and is one-third of the examination. The questions in each section primarily address skills and knowledge in that particular area of study. About one-third of the questions in each section ask you to apply those skills and that knowledge to a classroom situation. There may be a few questions in any section that do not count toward your score, but you won’t know which ones these are. There is no penalty for wrong answers, so it is to your advantage to select an answer for every question. ### When is it available? The ParaPro Assessment is given at Prometric® Testing Centers and other locations, including some universities, colleges, and school district facilities. You can find out more about testing centers and dates by accessing this website and by contacting your local education agency. ### Tips & Tricks • You may be more or less familiar with computer-delivered tests, but learning more about the format of the ParaPro Assessment is a good idea. You can view a tutorial about this here. • If you haven’t practiced the skills tested in a while or you’d like to have more support, creating a study group is a great idea. If several of your colleagues or friends are also preparing for the test, arrange to meet in a group regularly as you prepare. You can share study tips and information while providing encouragement for each other. • Don’t be discouraged by unfamiliar words in questions or answers on this test. Use the words around them (context clues) to help you figure out what they mean. Scan all of the answers. You may find that the unknown word is not even in the correct answer. Just use what you know to figure it out. • The questions on this test are not designed to be tricky. They are straightforward and just require you to make your best judgment, so don’t make the test harder by overthinking a question. ### Time Allowed The testing session for the ParaPro Assessment lasts for 2.5 hours, during which you will be presented with 90 questions. You will need to work rather quickly to answer all of the questions within this time limit, but it’s also important to proceed carefully and avoid careless errors. Practice using our expertly crafted questions!	2019-05-19 08:36:29	{"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": 0.30482786893844604, "perplexity": 1102.113297907231}, "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-2019-22/segments/1558232254731.5/warc/CC-MAIN-20190519081519-20190519103519-00208.warc.gz"}
https://www.calculus-online.com/exercise/2418	# Domain of One Variable Function – A function with log – Exercise 2418 Exercise Determine the domain of the function: $$f(x)=\log_2 (x+1)$$ $$x>-1$$ Solution Let’s find the domain of the function: $$f(x)=\log_2 (x+1)$$ Because there is a log, we need the expression inside the log to be greater than zero: $$x+1>0$$ $$x>-1$$ And that is the final answer. Have a question? Found a mistake? – Write a comment below! Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions! Share with Friends	2023-03-21 05:44:35	{"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": 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": 5, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.24151860177516937, "perplexity": 841.1274666769209}, "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/1679296943625.81/warc/CC-MAIN-20230321033306-20230321063306-00339.warc.gz"}
http://jnnp.bmj.com/content/76/4/569	Article Text The effect of subthalamic nucleus stimulation on kinaesthesia in Parkinson’s disease 1. M Maschke1,2, 2. P J Tuite3, 3. K Pickett1, 4. T Wächter3,4, 5. J Konczak1,3 1. 1Sensorimotor Control Laboratory, Division of Kinesiology, University of Minnesota, Minneapolis, Minnesota, USA 2. 2Department of Neurology, University Duisburg-Essen, Essen, Germany 3. 3Department of Neurology, University of Minnesota 4. 4Brain Sciences Center, Minneapolis VAMC, University of Minnesota 1. Correspondence to:  Dr Matthias Maschke  Department of Neurology, University Duisburg-Essen, Hufelandstr. 55, 45122 Essen, Germany; matthias.maschkeuni-essen.de Abstract Background: Parkinson’s disease is accompanied by deficits in passive motion and limb position sense. Objective: To investigate whether deep brain stimulation of the subthalamic nucleus (STN-DBS) reverses these proprioceptive deficits. Methods and results: A passive movement task was applied to nine patients with Parkinson’s disease and bilateral chronic STN-DBS and to seven controls. Thresholds for 75% correct responses were 0.9° for controls, 2.5° for Parkinson’s disease patients when stimulation was OFF, and 2.0° when stimulation was ON. Conclusions: STN-DBS improves kinaesthesic deficits in Parkinson’s disease, but does not lead to a full recovery of proprioceptive function. • CVLT, California verbal learning test • MMSE, mini-mental state examination • STN-DBS, deep brain stimulation of the subthalamic nucleus • WAIS, Wechsler adult intelligence scale • WASI, Wechsler abbreviated scale of intelligence • WMS, Wechsler memory scale • deep brain stimulation • Parkinson’s disease • kinaesthesia • subthalamic nucleus Statistics from Altmetric.com Kinaesthesia is the conscious perception of limb and body position, orientation, and motion. Recent studies have shown that patients with Parkinson’s disease have higher thresholds for detecting passive elbow movements, reflecting kinaesthesic deficits.1,2 Patients with Parkinson’s disease show altered proprioception related EEG potentials during passive movements, which probably reflect changes in the cortical processing of kinaesthesic signals.3 Dopaminergic drugs appear to enhance this deficit.4 In contrast, we found no effect of levodopa on kinaesthesia.2 Deep brain stimulation of the subthalamic nucleus (STN-DBS) improves motor deficits in Parkinson’s disease5 but it may worsen cognitive functions such as working memory.6 However, the effects of STN-DBS on kinaesthesia are unknown. We therefore sought to determine how STN-DBS affects the perception of limb position. METHODS Subjects Participants were nine patients with Parkinson’s disease who had STN-DBS installed (mean (SD) age, 60.3 (8.7) years, range 44 to 71; four female, five male) and seven age matched healthy controls with no neurological or general medical limitations (61.0 (9.1) years, range 45 to 72; four female, three male). The patients were recruited from the movement disorders outpatient clinic at the University of Minnesota and were diagnosed as having idiopathic Parkinson’s disease. At the time of testing, they all had normal values on the mini-mental state examination (MMSE) (table 1). Preoperative neuropsychological testing—including tests for concentration, attention, and executive functions (WAIS-III digit span, Wisconsin card sorting test III) and additional tests for verbal memory (CVLT), visual-spatial performance (WASI matrix reasoning), and visual memory (WMS-III faces 1 and 2)—revealed mild impaired concentration and executive function in one patient (subject 2; table 1). Two patients had mild to moderate dyskinesias (subjects 1 and 9; table 1) and four had a mild to moderate rest tremor (subjects 1, 6, 7, 8; table 1). Neurological examination did not reveal signs or symptoms of peripheral nerve disorders. All patients were tested on drug treatment—that is, they were told to take their antiparkinsonian drugs in their usual dose and time schedule. Table 1 Patient characteristics All patients and healthy subjects gave their informed consent to participate in the study. The study was approved by the institutional review board of the University of Minnesota. Testing apparatus and procedure The experimental setup is presented in detail elsewhere.2 Briefly, the subject’s forearm was moved passively (∼ 0.5°/s) while resting on a splint. A torque motor was mechanically connected to the splint, moving it at a constant speed to cause forearm flexion or extension. For each condition (ON or OFF stimulation), 40 flexion and 40 extension movements were recorded with elbow joint angular displacements of 0.2°, 0.6°, 1°, 2°, 3°, 4°, 5°, 6°, 7°, and 8°. Angular displacements and their directions were pseudorandomly presented, every displacement being made twice in both directions. Both arms were tested in each subject. The right arm was tested first. To account for order effects, a subset of the Parkinson’s disease group (n = 5) was tested in the ON stimulation state first, while the remaining patients were first examined during the OFF stimulation state (n = 4). Between the conditions, patients had a 30 minute break to adjust to the change in stimulation (ON→OFF or OFF→ON). Each trial was signalled to the subject by a tactile cue on the subject’s shoulder two to five seconds before the start of the movement. The end of each trial was announced by a second tactile cue on the left forearm. After each passive movement, subjects had to judge and express verbally whether the forearm was moved “towards” or “away” from their body or if they “could not tell.” There were no time limitations for subjects to respond. Incorrect responses and “could not tell” responses were both scored as an incorrect response. Subjects wore goggles to exclude all visual input, and headphones with “pink noise” masked all auditory cues during testing. The total testing time was around 80 minutes. Statistical analysis The percentage of correct responses per angular displacement was calculated for each of the three groups (Parkinson’s disease patients ON stimulation; Parkinson’s disease patients OFF stimulation; controls). Group differences in the percentage of correct responses were determined using separate Kruskal–Wallis tests for each displacement. The performance data for both arms of the Parkinson’s disease patients were entered in this analysis. Differences in percentages of correct responses between ON and OFF stimulation states were tested by separate Mann–Whitney U tests for each displacement. Both tests were corrected for multiple comparisons. A curve fitting procedure (Box Lucas exponential fit) was also carried out to determine the threshold for 75% correct responses. The model equation was as follows: $Math$ where y = number of correct responses (%), x = displacement (degrees), a,b = coefficients, and e = Euler’s number (2.718)). Probability (p) values of <0.05 were assumed to be significant. RESULTS We found no differences in the percentage of correct responses between passive forearm extension and flexion movements (p>0.2). Thus we collapsed these two datasets and report all results for the combined flexion/extension movements. All groups had difficulty in detecting very small displacements of 0.2°. Control subjects showed a rapid improvement in percentages of correct responses with increasing angular displacement. A 1.0° displacement was detected in 78% of trials and a 2° displacement in over 91% of trials by the control subjects (fig 1; table 2). In contrast, the patients with Parkinson’s disease were clearly impaired in the correct detection of movement direction. During ON stimulation they showed correct responses in 74% for 2° displacements. Deficits worsened when the DBS device was OFF, with correct detection of 2° displacements in only 60% of trials. The differences between all three groups were statistically significant for all displacements with the exception for 0.2° displacements (all p values <0.05). The difference between the ON and OFF stimulation states in the Parkinson group was significant for 2° and 3° displacements (p<0.05). At these displacements six patients had a greater number of correct responses during the ON stimulation state, two showed no difference, and only one had a greater percentage of correct responses during OFF stimulation (table 1). Thresholds for 75% correct responses underlined these results. Controls had a 75% threshold of 0.9°. Parkinson’s disease patients were not normal, but improved during the ON stimulation state (2.0°) compared with the 75% threshold obtained during the OFF stimulation state (2.5°). Table 2 Percentages of correct responses per group and displacement Figure 1 Comparison of percentages of correct responses per displacement between groups. Patients with Parkinson’s disease had higher percentages of incorrect responses than control subjects. However, patients appeared less impaired during ON stimulation. DISCUSSION Previous studies suggested that movement difficulties such as bradykinesia in Parkinson’s disease are related in part to a decrease in proprioceptive function or to impaired sensorimotor integration.7 Thus therapy that could alleviate kinaesthesic deficits would be an important adjunct to the treatment of these patients. The results of the present study suggest that STN-DBS has the ability to improve kinaesthesia in Parkinson’s disease. What is the neurophysiological basis for our finding of an improved limb position sense during DBS stimulation? It is known that neuronal activity elicited by passive movements and somatosensory evoked potentials can be recorded within the STN.8,9 Thus a role of the STN in analysing proprioceptive signals is plausible. As STN stimulation alters the proprioceptive inputs that reach the nucleus from the somatosensory cortex,10 it is not unlikely that it reduces the noise of the signals processed in the STN and their basal ganglia targets. Given that many of these cerebro-basal ganglia connections are reciprocal, altered STN output will ultimately affect processing in the somatosensory cortex, where a reduced noise level could then facilitate the perception of limb position. Moreover, the findings of the present study are in line with previous reports of improved sensorimotor integration during STN-DBS.11,12 However, one needs to recognise that several other factors could also have influenced our experimental outcome. First, STN-DBS might have improved cognitive dysfunction in our study population, leading to improved kinaesthesia. This scenario is somewhat unlikely as no patient had substantial cognitive deficits on neuropsychological testing, either preoperatively or postoperatively. On the other hand, recent studies suggest that STN stimulation impairs cognitive function in Parkinson’s disease,6,13 and therefore one would expect that kinaesthetic deficits would have been greater in the ON stimulation state if cognitive dysfunction was the main cause of these deficits. However, on a cautionary note, one needs to realise that the MMSE alone is a poor tool to detect subtle cognitive deficits, and extensive neuropsychological testing was only available preoperatively and not postoperatively at the time of this study. Moreover, in contrast to the reports mentioned above,6,13 recent studies suggest a mild increase in psychomotor speed and working memory in patients with Parkinson’s disease who have STN-DBS.14 Thus we cannot completely rule out the possibility that STN-DBS implantation induced cognitive improvements that enhanced the patients’ perceptual performance. Second, one could argue that the alleviation of kinaesthesic deficits during ON stimulation was simply an order effect. Our study design was nearly balanced (five patients were tested first ON, four patients first OFF), yet we found no evidence that the order of the treatments had a significant impact. The patients who were first tested OFF did not perform better than those first tested ON. It could also be argued that, although we considered possible order effects, patients were not randomised into groups. This might have resulted in differences in dopaminergic drug treatment between the patient groups. Dopaminergic treatment is known to influence the results of kinaesthetic testing, though the exact nature of this influence is controversial.2,4 However, levodopa equivalent doses did not differ significantly between patients first tested OFF and those first tested ON (p = 0.33). Thus the results cannot be solely explained by differences in dopaminergic drug treatment. Third, one may speculate that tremor as well as dyskinesias might have an impact on kinaesthesia, and that these symptoms were less pronounced during the ON condition. In our study, five patients had rest tremor (n = 4) or dyskinesias (n = 2). However, three of the remaining four patients without visible rest tremor or dyskinesia during examination showed significant improvement in kinaesthesia on DBS. Thus the cessation of a rest tremor alone cannot account for the improvement in limb position sense. Finally, the pause between changes of condition might have been too short. A recently published study found that 30 minutes appeared to be sufficient to reach a steady state after change from OFF to ON but might not be enough from ON to OFF.15 However, our results in those patients who were tested first in the OFF state were not different from those who were tested first in the ON state, strongly arguing against an influence of break duration on our perceptual results. Conclusions Our study underlines the role of the basal ganglia in kinaesthesia and shows that kinaesthesic function may be improved by STN-DBS in Parkinson’s disease. We found no clear evidence that these improvements in limb perception reflected enhanced cognitive function. It also needs to be stated that, although STN-DBS may have a positive effect on kinaesthesia, our data show that it does not fully restore limb position sense in Parkinson’s disease. Acknowledgments The study was supported by a grant of the Deutsche Forschungsgemeinschaft (DFG) to MM (DFG MA 2209/3–1) and from the Minnesota Medical Foundation and the College of Education and Human Development. We thank Mary Sullivan and Maggie Laseski for providing data of neuropsychological testing. We sincerely thank all patients for their participation. View Abstract Footnotes • Competing interests: none declared Request permissions If you wish to reuse any or all of this article please use the link below which will take you to the Copyright Clearance Center’s RightsLink service. You will be able to get a quick price and instant permission to reuse the content in many different ways.	2017-08-20 04:22:07	{"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": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.40225905179977417, "perplexity": 5079.031573762717}, "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-2017-34/segments/1502886105970.61/warc/CC-MAIN-20170820034343-20170820054343-00407.warc.gz"}
https://worldbuilding.stackexchange.com/questions/26791/rfp-put-me-on-the-moon	# RFP: Put me on the moon I am an awesome, famous, self-made billionaire - I started with a mere \$300 million and amassed that into over$4 billion in only 40 years (an amazing 6.75% return). Everyone loves me - the blacks, the Hispanics, the Jews all love me. People love seeing my name everywhere - on Casinos, Golf Courses, Skype-like video phones, hotels, pyramid schemes, matresses... I want people to enjoy seeing my name as much as possible. What's the best way to put my name on the moon? I want to give people the pleasure of being able to look up and see my name on the moon as much as possible. Should I send someone up with a bulldozer to carve my name in the dirt or send a satellite to write my name with a "laser"? What would it take to make my name visible on the moon and what % of the time will my name be visible? It should be visible in the night, day or both. Final choice will be determined by what percent it is visible and the total cost. Additional considerations will be made for the ability to change it to display logos when desired. • What's your budget? Surely not your entire $4 billion fortune. – Green Oct 1 '15 at 18:29 • It's currently open. I'm not unwilling to waste my entire fortune on a stupid idea - but I am hoping to somehow have investors pay for the whole thing. Assume money is not an object - that's why I'm requesting proposals (RFP). – Hannover Fist Oct 1 '15 at 18:33 • And your name will easily become the most hated name on the planet. Are you sure? – Green Oct 1 '15 at 18:34 • I am lazy to do numbers, so I'll just brainstorm and see if someone gets a Budget. Sunlight lensing. Put a number of satellites in orbit, with big lenses or parabollic reflecting antenas. Reflect sunlight so that the focus of the vean is just on Moon surface. Burn/melt lunar dust with such a ray to change its color/albedo. The problem is it won't be fancy (will not change at will), and may take some time to finish (since rays are relatively thin, you will have to make many cycles of "writting"). Since the Moon surface is static, there is no risk of rain/wind changing your design. – SJuan76 Oct 1 '15 at 19:27 • My suggestion is that you not worry much about the cost, just get the Mexican government to pay for it. – Deolater Oct 1 '15 at 22:24 ## 4 Answers First of all,$4 billion seems like a small amount -- it got Elon Musk barely to the edge of space. But let's first see how far we would get with near-unlimited money (say the US government would like to print "USA" on the moon). I like @SJuan76 's idea of mirrors, however I don't think you can "burn" moon rock. Even if there was carbon in there (the main cause of things turning black when burned) there is no oxygen to actually burn it. Perhaps given enough light and time you could "sculpt" the moon into a name by melting away moonstone in certain places, but I don't think we want to take THAT long. First problem I see is that one has to decide whether to have the message visible in dark (new moon) of light (full moon). Let's aim for the sky and build something that shines so bright that it outshines (and hence still is readable) at full moon. I did some calculations and was actually surprised that it's not completely implausible energy-wise: you need the energy budget of something like Switzerland, and with there likely being uranium on the moon, it might just work (provided you feel like getting all those lamps up there, and having someone change out the LED-bulbs every time they fail). This calculation also puts a lower bound on the energy needs of beaming an image up there to reflect on the moon's surface using lasers/mirrors/etc. However once we start shipping such huge amounts of equipment up to the moon, it may just be easier to paint our letters on there in some reflective paint. 10% of the front of the moon = 4 * PI * moon_radius ** 2 / 2 * 0.1 ≈ 2 million sqkm.. Even if we could buy the paint for the incredibly low price of \$1 per square meter, the paint alone would come to \$2 trillion, so the paint alone would increase the national debt by about 5% (and then I'm skipping over how to get it to stick to moondust). Coming back to the idea on how to get your name in the sky on a budget, I think there is a much better solution. The moon has a diameter of about 3500 km at a distance of about 400,000 km. The ISS on the other hand is at only 400 km; much cheaper to put things up there, and they will look much bigger. At that distance you "only" need to unfold a painted solar-sail type thing of 3.5km square to get the same visual effect. Since you can launch something into space these days for a couple of tens of millions, I recon this would fit within your budget a lot better. • The maria are different in appearance, so melting the rock does have a different look. But it's all covered with dust now: fusing the top layer to make a smooth surface, rather than the spectrally rough regalith grains, will change the appearance. – JDługosz Oct 5 '15 at 6:16 You're gonna need a bigger Moon. "Normal" visual acuity is defined as the ability to resolve features one arc-minute across. At the Moon's distance, this corresponds to a minimum feature size of 112 km. Using a tightly-spaced typeface to write your name requires a surface 4928 km across to write "Hannover Fist", somewhat larger than the Moon's 3476-km diameter. But let's ignore that. What would it take to write the smallest possible visible letter on the Moon? Writing a minimally-sized "H" requires modifying about 63,000 square kilometers of surface to produce a 336 km by 336 km letter. We're working at the edge of human visual acuity here, so we want high contrast. Simply bulldozing your name in the Moon won't work — you'll get a contrast change of a few percent at best. You'll want to write your name using something either exceptionally black or exceptionally white (the Moon being a sort of medium grey, either one will work). At the dark end of things, carbon black is a cheap option, and reasonably low density. Coating the surface with a layer a tenth of a millimeter thick requires about 6,272,000 cubic meters, or about 1.3 million metric tons, about 15% of the world's annual production. Lifting it to the Moon, assuming that somebody's commercial launcher can actually deliver on the promised \$2000 per kilogram, will cost around$2.6 trillion, a bit over budget. At the light end, there aren't really any materials similar to carbon black's fluffy, powdery composition. Titanium dioxide, for example, has a density 20 times that of carbon black. Non-powder options aren't that good, either. Black HDPE film, for example, has a density five times that of carbon black, and even thin sheets are around a tenth of a millimeter thick. • Celtschk's comment to another answer notes that graphite dust is the same darkness as the moon. Is lamp-black (the common name for the pigment use of carbon black) darker than plain graphite? – JDługosz Oct 5 '15 at 6:09 • @JDługosz, very much so. Carbon black is one of the darkest pigments in common use. – Mark Oct 5 '15 at 6:22 In an old story, the plan to put a simple logo on the moon involved rockets delivering graphite dust. The logo was, IIRC, a "7" in a circle, and the moon was said to be about the same apparent size as a inch-sized pin worn by someone: I figure 10 or 12 feet away actually. It was plausible enough in the story, at the time. I'm sure it would suffer from scale if he tried to do it, but the story was skillfully done and gave the impression that he had the resources without being distracting. Having few strokes helps. Being realistic about the material quantity, it would best be done using space-bourne resources. How about if it occurs in conjunction with commercial development of the moon? Maybe the mining and supply roads are leaving visible marks anyway, and he wants to write his name using those processes. In doing so he will develop the resources on the moon and make a huge profit and open a new chapter in human industry. The letters could be made from solar panels, mine tailings or strip mining sites, railroads, etc. It could be a re-imagining of Heinlein's The Man Who Sold the Moon. • Of course graphite dust is a bad choice as the reflectivity of the moon is about the same as that of graphite. White paint would probably be more effective (especially since it would outshine the surrounding moon surface). – celtschk Oct 4 '15 at 9:50 • Maybe it was lamp_black or something; it was a long time since I read it. Mark's answer notes lamp-black (carbon black) indirectly has an albedo of 0. – JDługosz Oct 5 '15 at 6:13 Dig trenches, really deep so that they cast a dark shadow (like the craters, but darker). You would probably need to do this with explosives, and it would quickly become very expensive. But that's okay, just get Mexico to pay for it! :P	2021-01-28 05:52:22	{"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": 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": 0.417990505695343, "perplexity": 1538.563688631223}, "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/1610704835901.90/warc/CC-MAIN-20210128040619-20210128070619-00404.warc.gz"}
https://share.cocalc.com/share/4842ca38fce341e3bc80efa85bcfe6f228eacefe/CDS-102/Lab%20Week%2010%20-%20Statistical%20inference%20of%20North%20Carolina%20birth%20records/CDS-102%20Lab%20Week%2010%20Workbook.ipynb?viewer=share	SharedCDS-102 / Lab Week 10 - Statistical inference of North Carolina birth records / CDS-102 Lab Week 10 Workbook.ipynbOpen in CoCalc Author: Helena Gray Views : 9 Description: Jupyter notebook CDS-102/Lab Week 10 - Statistical inference of North Carolina birth records/CDS-102 Lab Week 10 Workbook.ipynb # CDS-102: Lab 10 Workbook ## Helena Gray ### April 6, 2017 In [2]: # Run this code block to load the Tidyverse package .libPaths(new = "~/Rlibs") library(tidyverse) # Load inference() function from file "inference.RData" In [3]: nc.data<-read.csv("nc.csv") The code below generates a full summary statistics report by running the summary() function on the dataset. In [4]: summary(nc.data) fage mage mature weeks premie Min. :14.00 Min. :13 mature mom :133 Min. :20.00 full term:846 1st Qu.:25.00 1st Qu.:22 younger mom:867 1st Qu.:37.00 premie :152 Median :30.00 Median :27 Median :39.00 NA's : 2 Mean :30.26 Mean :27 Mean :38.33 3rd Qu.:35.00 3rd Qu.:32 3rd Qu.:40.00 Max. :55.00 Max. :50 Max. :45.00 NA's :171 NA's :2 visits marital gained weight Min. : 0.0 married :386 Min. : 0.00 Min. : 1.000 1st Qu.:10.0 not married:613 1st Qu.:20.00 1st Qu.: 6.380 Median :12.0 NA's : 1 Median :30.00 Median : 7.310 Mean :12.1 Mean :30.33 Mean : 7.101 3rd Qu.:15.0 3rd Qu.:38.00 3rd Qu.: 8.060 Max. :30.0 Max. :85.00 Max. :11.750 NA's :9 NA's :27 lowbirthweight gender habit whitemom low :111 female:503 nonsmoker:873 not white:284 not low:889 male :497 smoker :126 white :714 NA's : 1 NA's : 2 The code below filters out entries in the data set with 'NA' values for the 'habit' variable. In [5]: nc.data<-filter(nc.data,habit!='NA') nc.data.nonsmoker<-filter(nc.data,habit=='nonsmoker') nc.data.smoker<-filter(nc.data,habit=='smoker') The code below plots a (frequency) histogram of the habit and weight variables in the dataset. The code plots both histograms on the same chart using the geom_histogram() function including inputs position="identity" and alpha=0.3 so that the plot is readable. The code saves the histogram to a .png file using the ggsave() function. In [6]: options(repr.plot.width = 9, repr.plot.height = 4) both_hist<-ggplot(nc.data) + geom_histogram(mapping = aes(x = weight, y = ..density.., fill = habit), binwidth = 1, position = "identity", alpha = 0.3) ggsave("both_hist.png", plot = both_hist, device="png", scale=1, width=5, height=4) both_hist The code below uses the summary() function on each of the 'habit' subsets to calculate the mean and standard deviation of birth weights for the group of non-smokers and the group of smokers. The smoking group does have a slightly lower mean birth weight. This may suggest a correlation between smoking and birth weight. We will find if this difference is statistically significant in later steps. In [7]: stat.table.nonsmoker<-summarise(nc.data.nonsmoker, mean=mean(weight), sd=sd(weight)) stat.table.nonsmoker stat.table.smoker<-summarise(nc.data.smoker, mean=mean(weight), sd=sd(weight)) stat.table.smoker meansd 7.1442731.518681 meansd 6.828731.38618 For this task, the null and alternative hypotheses are specified for testing if the average weights of babies born to smoking and non-smoking mothers are different. They are shown below. H0: There is no difference in the mean birth weights between babies born to smokers and babies born to non-smokers. Ha: There is a difference in the mean birth weights between babies born to smokers and babies born to non-smokers. The code below uses the inference() function to test the hypotheses specified in the previous step. The inference() function simplifies the hypothesis testing procedure, effectively hiding the manual computational work needed to perform them. For the code below, the mean birth weight is estimated for both smoking and non-smoking mothers and the type of test is specified as a 'hypothesis test'. The test is based on the Central Limit Theorem which is specified by the parameter 'method' and setting it to 'theoretical'. In [8]: inference(y = weight, x = habit, data = nc.data, statistic = "mean", type = "ht", null = 0, alternative = "twosided", method = "theoretical", order = c("smoker", "nonsmoker")) Response variable: numerical Explanatory variable: categorical (2 levels) n_smoker = 126, y_bar_smoker = 6.8287, s_smoker = 1.3862 n_nonsmoker = 873, y_bar_nonsmoker = 7.1443, s_nonsmoker = 1.5187 H0: mu_smoker = mu_nonsmoker HA: mu_smoker != mu_nonsmoker t = -2.359, df = 125 p_value = 0.0199 The results of the hypothesis test produce a p-value that is less than the significance level which is .05, meaning that the null hypothesis, that there is no statistically significant difference between smokers and nonsmokers, can be rejected. The code below uses the inference() function again but this time the 'type' parameter is changed to 'ci' to produce a confidence interval. The 'null' and 'alternative' inputs are removed becaue the confidence interval will indicate a non-rejection of the null hypothesis if the value 0 is included within the confidence interval (this suggests that there is literally zero difference between the true means of both populations). The confidence interval indicates the range of values in which the difference between the true means of both populations lies. In [9]: inference(y = weight, x = habit, data = nc.data, statistic = "mean", type = "ci", method = "theoretical", order = c("smoker", "nonsmoker")) Response variable: numerical, Explanatory variable: categorical (2 levels) n_smoker = 126, y_bar_smoker = 6.8287, s_smoker = 1.3862 n_nonsmoker = 873, y_bar_nonsmoker = 7.1443, s_nonsmoker = 1.5187 95% CI (smoker - nonsmoker): (-0.5803 , -0.0508) In this example, both values of the confidence interval are negative, indicating that difference of the second mean from the first mean is negative, meaning that the second population (nonsmokers) have a higher mean birth weight. The code below calculates a 95% confidence interval for the average length of pregnancies (weeks variable). Rather than finding a difference between true means of true populations, this confidence interval just finds the range of values in which the true mean of the variable for the population of interest lies (in this case it is the true mean of the length of pregnancy in weeks for the entire population of women in North Carolina). In [10]: inference(y = weeks, data = nc.data, statistic = "mean", type = "ci", method = "theoretical", order = c("smoker", "nonsmoker")) Single numerical variable n = 998, y-bar = 38.3347, s = 2.9316 95% CI: (38.1526 , 38.5168) The results indicate that the true mean of the length of pregnancy for women in North Carolina is somewhere between 38.1 weeks and 38.5 weeks. The code below calculates a new confidence interval for the same parameter at the 90% confidence level. The confidence level is changed by adding a new argument to the function: conf_level = 0.90. In [11]: inference(y = weeks, data = nc.data, statistic = "mean", type = "ci", method = "theoretical", order = c("smoker", "nonsmoker"),conf_level=.90) Single numerical variable n = 998, y-bar = 38.3347, s = 2.9316 90% CI: (38.1819 , 38.4874) This change in confidence level narrows the range of values containing the true mean of the average length of pregnancy for women in North Carolina. The code below tests the hypothesis that there is no difference in mean number of hospital visits for pregnant married women and pregnant unmarried women. The code below uses the inference() function to perform a hypothesis test and a confidence interval using an α level of .05. This defines both the p-value threshold for statistical significance and the size of the confidence interval. ## Do married mothers tend to visit the hospital more than single mothers? ### Ha: There is a difference in mean number of visits to hospital between married and unmarried mothers. #### alpha level = .05 In [12]: inference(y = visits, x = marital, data = nc.data, statistic = "mean", type = "ht", null = 0, alternative = "twosided", method = "theoretical", order = c("married", "unmarried")) Response variable: numerical Explanatory variable: categorical (2 levels) n_married = 380, y_bar_married = 10.9553, s_married = 4.2408 n_not married = 611, y_bar_not married = 12.82, s_not married = 3.5883 H0: mu_married = mu_not married HA: mu_married != mu_not married t = -7.1298, df = 379 p_value = < 0.0001 In [13]: inference(y = visits, x = marital, data = nc.data, statistic = "mean", type = "ci", method = "theoretical", order = c("married", "unmarried")) Response variable: numerical, Explanatory variable: categorical (2 levels) n_married = 380, y_bar_married = 10.9553, s_married = 4.2408 n_not married = 611, y_bar_not married = 12.82, s_not married = 3.5883 95% CI (married - not married): (-2.3789 , -1.3505) In [ ]:	2019-09-17 00:43:09	{"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": 0.2503949999809265, "perplexity": 7738.19708894523}, "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-2019-39/segments/1568514572980.56/warc/CC-MAIN-20190917000820-20190917022820-00203.warc.gz"}
http://codereview.stackexchange.com/questions/61248/diamond-in-the-rough-finding-gems-in-the-rocks	# Diamond in the rough! Finding Gems in the rocks Recently a question was asked that solved a hackerrank problem: Counting gems amongst the rocks The description of the challenge is: John has discovered various rocks. Each rock is composed of various elements, and each element is represented by a lowercase Latin letter from 'a' to 'z'. An element can be present multiple times in a rock. An element is called a 'gem-element' if it occurs at least once in each of the rocks. Given the list of rocks with their compositions, you have to print how many different kinds of gems-elements he has. Input Format The first line consists of N, the number of rocks. Each of the next N lines contain rocks’ composition. Each composition consists of small alphabets of English language. Output Format Print the number of different kinds of gem-elements he has. Constraints $1 ≤ N ≤ 100$ Each composition consists of only small Latin letters ('a'-'z'). 1 ≤ Length of each composition ≤ 100 Sample Input 3 abcdde baccd eeabg Sample Output 2 Explanation Only "a", "b" are the two kind of gem-elements, since these characters occur in each of the rocks’ composition. I was in the process of writing an answer when the question was closed because the code in the question was failing to produce the right results. My recommended solution to the problem is likely quite fast, but it is also relatively complicated. I am sure it can be improved, and simplified. I have included a simple main method that shows how it can be used, and produces the sample output. public class GemCounter { private static final int LETTERCOUNT = 26; private int gemsSoFar = (1 << LETTERCOUNT) - 1; public GemCounter() { // default constructor. Does nothing. } public void processRock(final String rock) { int gotElement = 0; // no bits are set. for (int i = 0; i < rock.length(); i++) { int element = rock.charAt(i) - 'a'; if (element >= 0 && element < LETTERCOUNT) { // bitwise OR the bit that represents the element gotElement \|= 1 << element; } } // only bits that were found in this rock // and also that have been seen before // will remain set. gemsSoFar = gemsSoFar & gotElement; } public int getGemCount() { // count the number of bits that are still set. return Integer.bitCount(gemsSoFar); } public static void main(String[] args) { String[] rocks = {"abcdde", "baccd", "eeabg"}; GemCounter gemcount = new GemCounter(); for (String rock : rocks) { gemcount.processRock(rock); System.out.printf("Processed %s, got %d gems still%n", rock, gemcount.getGemCount()); } System.out.printf("%d rocks have %d gems%n", rocks.length, gemcount.getGemCount()); } } - Not much to say. Logical operators used a bit inconsistently: gotElement \|= 1 << element; gemsSoFar = gemsSoFar & gotElement; better use &= in the second line, or spell out the first. gotElement is somewhat unclear. rockElements, maybe? In real life I'd also oppose extensive comments. - public GemCounter() { // default constructor. Does nothing. } which is not needed for the program to work because, The compiler automatically provides a no-argument, default constructor for any class without constructors. This default constructor will call the no-argument constructor of the superclass. In this situation, the compiler will complain if the superclass doesn't have a no-argument constructor so you must verify that it does. If your class has no explicit superclass, then it has an implicit superclass of Object, which does have a no-argument constructor [1]. therefore this also violates YAGNI "You aren't gonna need it" (acronym: YAGNI) is a principle of extreme programming (XP) that states a programmer should not add functionality until deemed necessary. XP co-founder Ron Jeffries has written: "Always implement things when you actually need them, never when you just foresee that you need them [2]. References [1]“Providing Constructors for Your Classes.” [Online]. Available: http://docs.oracle.com/javase/tutorial/java/javaOO/constructors.html. [Accessed: 27-Aug-2014]. [2]“You aren’t gonna need it,” Wikipedia, the free encyclopedia. 24-Aug-2014. - It's not a bad solution. It's not immediately obvious what processRock() function does, so I would just make some minor adjustments in the way the problem is decomposed and in naming, for clarity. private static final String ALL_ELEMENTS = "abcdefghijklmnopqrstuvwxyz"; private int commonElements = bitsOfRock(ALL_ELEMENTS); private static int bitsOfRock(String rock) { int bits = 0; for (int i = 0; i < rock.length(); i++) { int element = rock.charAt(i) - 'a'; if (element >= 0 && element < ALL_ELEMENTS.length()) { // bitwise OR the bit that represents the element bits \|= 1 << element; } } return bits; } public void requireElementsOfRock(String rock) { commonElements &= bitsOfRock(rock); } public int getElementCount() { return Integer.bitCount(commonElements); } For even better clarity, at the cost of a bit of performance, you could also use a java.util.BitSet. -	2014-12-21 01:08:50	{"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": 0.17078334093093872, "perplexity": 4809.908006655014}, "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/1418802770554.119/warc/CC-MAIN-20141217075250-00095-ip-10-231-17-201.ec2.internal.warc.gz"}
https://www.wyzant.com/resources/answers/10963/may_you_please_help_me_with_word_problems	Allison G. asked • 05/17/13 # May you please help me with word problems? A candy shop manager mixes M&M's worth $2.00 per pound with trail mix worth$1.50 per pound to make 50 pounds of party mix worth \$1.80 per pound. How many pounds of each she should use? Can you please explain how you got the answer and it will be nice if you can show the process. Thank you. ## 3 Answers By Expert Tutors By: Brad M. answered • 05/17/13 100+ Sessions in Precal-Trig-Alg 2: B'burg-Salem HS, RU MATH 138 Grigori S. answered • 05/17/13 Certified Physics and Math Teacher G.S. Sheila M. answered • 05/17/13 Friendly College Math Professor - understands students' math anxiety ## Still looking for help? Get the right answer, fast. Get a free answer to a quick problem. Most questions answered within 4 hours. #### OR Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.	2020-02-25 05:42:17	{"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": 0.20377995073795319, "perplexity": 6480.438230631727}, "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-2020-10/segments/1581875146033.50/warc/CC-MAIN-20200225045438-20200225075438-00354.warc.gz"}
https://discourse.matplotlib.org/t/cant-create-graphs/10204	# Can't create graphs Hello, I am completely new to Matplotlib. I can’t seem to get my script (a copy and paste from a Histogram example) to generate a graph. When I run my script, the command prompt returns. Is there something I am missing? ··· # histogram.py import numpy as np import matplotlib.mlab as mlab import matplotlib.pyplot as plt mu, sigma = 100, 15 x = mu + sigmanp.random.randn(10000) # the histogram of the data n, bins, patches = plt.hist(x, 50, normed=1, facecolor=‘green’, alpha=0.75) # add a ‘best fit’ line y = mlab.normpdf( bins, mu, sigma) l = plt.plot(bins, y, ‘r–’, linewidth=1) plt.xlabel(‘Smarts’) plt.ylabel(‘Probability’) plt.title(r’$\mathrm{Histogram\ of\ IQ:}\ \mu=100,\ \sigma=15$’) plt.axis([40, 160, 0, 0.03]) plt.grid(True) plt.show() Calling with: python histogram.py Make sure yu're using a backend with a user interface. http://matplotlib.sourceforge.net/faq/installing_faq.html#what-is-a-backend ··· El mié, 19-11-2008 a las 10:49 -0500, Ron Brennan escribió: Hello, I am completely new to Matplotlib. I can't seem to get my script (a copy and paste from a Histogram example) to generate a graph. When I run my script, the command prompt returns. Is there something I am missing? # # histogram.py # import numpy as np import matplotlib.mlab as mlab import matplotlib.pyplot as plt mu, sigma = 100, 15 x = mu + sigmanp.random.randn(10000) # the histogram of the data n, bins, patches = plt.hist(x, 50, normed=1, facecolor='green', alpha=0.75) # add a 'best fit' line y = mlab.normpdf( bins, mu, sigma) l = plt.plot(bins, y, 'r--', linewidth=1) plt.xlabel('Smarts') plt.ylabel('Probability') plt.title(r'$\mathrm{Histogram\ of\ IQ:}\ \mu=100,\ \sigma=15$') plt.axis([40, 160, 0, 0.03]) plt.grid(True) plt.show() Calling with: python histogram.py ------------------------------------------------------------------------- This SF.Net email is sponsored by the Moblin Your Move Developer's challenge Build the coolest Linux based applications with Moblin SDK & win great prizes Grand prize is a trip for two to an Open Source event anywhere in the world http://moblin-contest.org/redirect.php?banner_id=100&url=/ _______________________________________________ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users Make sure yu're using a backend with a user interface. http://matplotlib.sourceforge.net/faq/installing_faq.html#what-is-a-backend Also, you can get more verbose input at runtime with the --verbose-helpful flag:	2019-12-10 00:44:56	{"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": 0.33415162563323975, "perplexity": 12567.266522388047}, "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/1575540525598.55/warc/CC-MAIN-20191209225803-20191210013803-00176.warc.gz"}
https://tex.stackexchange.com/questions/428651/tikz-positioning	Tikz positioning How do you position a node more precisely than left of, right of? \node[above =3 cm of node1,node distance=3.5cm] (node2){}; For example, how do you say that I want 3mm to the left and 4 mm above a certain node (or at a certain angle and distance alternatively)? With the positioning library, you have the fairly flexible syntax demonstrated here: \documentclass{standalone} \usepackage{tikz} \usetikzlibrary{positioning} \begin{document} \begin{tikzpicture} \node (A){A}; \node [above left = 3mm and 4mm of A] (B){B}; \end{tikzpicture} \end{document} For positioning nodes at an a distance and angle of another node: TikZ can handle polar coordinates, so this should be straightfoward too, but I'm having a brainfreeze on how precisely to do it in the best way... How about something like this, with xshift and yshift, and also positions using relative placement with distances, placing the nodes precisely w.r.t the starting node: \documentclass{article} \usepackage{tikz} \usetikzlibrary{positioning} \begin{document} \begin{tikzpicture} \node[rectangle, draw=black] (node1) {N1}; \node[rectangle, draw=black,xshift=2cm, yshift=2cm, at=(node1)] (node2) {N2}; \node[below right=1cm and 1cm of node1, rectangle, draw=black] (node3) {N3}; \end{tikzpicture} \end{document}	2020-01-18 22:30:52	{"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": 0.9548428058624268, "perplexity": 6088.365959837871}, "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/1579250593994.14/warc/CC-MAIN-20200118221909-20200119005909-00125.warc.gz"}
https://electronics.stackexchange.com/questions/419661/usart-isr-on-txc-breaking-delays-from-util-delay-h	# USART ISR on TXC breaking delays from <util/delay.h>? I am trying to work out USART ISR's on the ATmega 328P. So far I have got the USART rx and tx working. I have come across a strange issue when it comes to the UART ISR's, though. The ISR seems to override the delays, causing the code in my main loop to run without a pause. Consider the following code (which is complete and functioning for easy troubleshooting): #include <avr/io.h> #include <util/delay.h> #include <avr/interrupt.h> #define FOSC 16000000 #define BAUD 9600 #define BAUD_GEN_VAL (((FOSC/BAUD)/16)-1) void uart_tx( unsigned char data ); unsigned char uart_rx( void ); void main(void) { sei(); //enable interrupts globally UBRR0H = (BAUD_GEN_VAL>>8); UBRR0L = BAUD_GEN_VAL; //set baud rate UCSR0B \|= (1<<TXEN0) \| (1<<RXEN0); //enable transmitter and receiver //UCSR0B \|= (1<<TXCIE0); //Enable interrupt on TX (serial) UCSR0C \|= (1<<UCSZ00) \| (1<<UCSZ01); //8 bit data while (1) { uart_tx('A'); _delay_ms(2500); } } ISR(__vector_USART_TXC) { //blink an led here or something similar } void uart_tx( unsigned char data ) { while ( !( UCSR0A & (1<<UDRE0)) ); //Wait for empty transmit buffer UDR0 = data; //Put data into buffer, sends the data } This code works perfectly, printing "A" to a serial monitor every 2.5 seconds. However, if I un-comment the line UCSR0B \|= (1<<TXCIE0); — which enables the interrupt on transmission completion — the microcontroller spews "A"s out so fast that the serial monitor can't even keep up. I can't understand how this is happening, however. It is as if the ISR did nothing but override the delay! Can anyone shed light on this? • I can just guess that the AVR makes resets, can you add toggling a PORT pin before entering main? – Andy Jan 30 at 14:49 • Btw, how have you defined __vector_USART_TXC ? – Andy Jan 30 at 14:52 • With a recent avr-libc, the ISRs should be USART_RX_vect and USART_TX_vect if you enable the interrupt but don't provide the appropriate ISR, the AVR makes a reset – Andy Jan 30 at 14:56 • @Andy I tried your first suggestion, and I didn't notice any glow on the LED, but it may be too fast to even produce a faded glow. On your second comment: I didn't define __vector_USART_TXC. I'll definitely try that.. – hat Jan 30 at 14:59 • How do you clear the TXC flag? Is it done automatically by the AVR itself? – Rohat Kılıç Jan 30 at 15:00	2019-11-14 18:30:31	{"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": 0.24397335946559906, "perplexity": 12051.387717775724}, "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/1573496668534.60/warc/CC-MAIN-20191114182304-20191114210304-00100.warc.gz"}
https://aimsciences.org/article/doi/10.3934/jimo.2018155	# American Institute of Mathematical Sciences doi: 10.3934/jimo.2018155 ## A simple and efficient technique to accelerate the computation of a nonlocal dielectric model for electrostatics of biomolecule 1 School of Mathematics and Statistics, Changsha University of Science and Technology, Changsha, Hunan 410114 China 2 School of Mathematics and Statistics, Central South University, Changsha, Hunan 410083 China * Corresponding author Received March 2018 Revised June 2018 Published September 2018 The nonlocal modified Poisson-Boltzmann equation (NMPBE) is one important variant of a commonly-used dielectric continuum model, Poisson-Boltzmann equation (PBE). In this paper, we use a nonlinear block relaxation method to develop a new nonlinear solver for the nonlinear equation of $\Phi$ and thus a new NMPBE solver, which is then programmed as a software package in $\texttt{C}\backslash\texttt{C++}$, $\texttt{Fortran}$ and $\texttt{Python}$ for computing the electrostatics of a protein in a symmetric 1:1 ionic solvent. Numerical tests validate the new package and show that the new solver can improve the efficiency by at least $40\%$ than the finite element NMPBE solver without compromising solution accuracy. Citation: Jiao Li, Jinyong Ying. A simple and efficient technique to accelerate the computation of a nonlocal dielectric model for electrostatics of biomolecule. Journal of Industrial & Management Optimization, doi: 10.3934/jimo.2018155 ##### References: show all references ##### References: Time speedup $S_p$ defined in (23) achieved by our new NMPBE solver for the 12 protein tests on two mesh sets. Initial mesh sets denote the meshes used in Table 2 and Refined mesh sets mean the ones used in Table 3. Newton Steps and the iteration numbers of the linear solver to solve $\Phi$ for case 1A63 in the new and the finite element NMPBE program packages. The left plot presents the number of block relaxation iteration (the x-axis), the Newton steps in each block iteration (the numbers above the x-axis), and the concrete/average iteration number (the points/the solid lines) of the linear solver in the new program package. The right plot presents the Newton steps (the x-axis) and the iteration numbers of the linear solver in the finite element one. Basic information of the 12 proteins used for numerical tests. Here $n_{p}$ is the number of atoms. Index PDB ID $n_{p}$ Index PDB ID $n_{p}$ 1 2LZX 488 7 1A63 2065 2 1AJJ 513 8 1CID 2783 3 1FXD 811 9 1A7M 2803 4 1HPT 852 10 2AQ5 6024 5 4PTI 892 11 1F6W 8243 6 1SVR 1433 12 1C4K 11439 Index PDB ID $n_{p}$ Index PDB ID $n_{p}$ 1 2LZX 488 7 1A63 2065 2 1AJJ 513 8 1CID 2783 3 1FXD 811 9 1A7M 2803 4 1HPT 852 10 2AQ5 6024 5 4PTI 892 11 1F6W 8243 6 1SVR 1433 12 1C4K 11439 Comparison of the performance of our new NMPBE solver (New) with that of the finite element NMPBE solver (FE) [19] in CPU time measured in seconds. Here Iter. Number denotes the iteration number needed in the nonlinear block relaxation method and $E_h$ is computed by (22), and residual norm means the norm of Equation (7)'s residual. PDB ID Number of Mesh Nodes Iter. Number Find $\Phi$ Total Time Relative error $E_{h}$ Residual norm New FE New FE 2LZX 26349 11 15.23 27.79 27.61 40.17 $2.1\times 10^{-8}$ $1.42\times10^{-4}$ 1AJJ 31910 11 26.60 48.25 45.29 66.94 $3.9\times 10^{-8}$ $3.69\times10^{-5}$ 1FXD 34469 12 23.19 42.48 49.74 69.03 $1.2\times 10^{-8}$ $8.03\times10^{-4}$ 1HPT 48229 10 32.33 58.78 58.08 84.53 $3.3\times 10^{-8}$ $1.11\times10^{-4}$ 4PTI 39468 10 25.85 46.52 47.14 67.81 $1.5\times 10^{-8}$ $8.95\times10^{-5}$ 1SVR 61074 11 55.17 90.59 96.88 132.30 $2.6\times 10^{-8}$ $4.25\times10^{-5}$ 1A63 22054 11 13.52 27.09 27.19 40.76 $1.6\times 10^{-8}$ $1.82\times10^{-4}$ 1CID 19872 10 11.07 21.68 23.10 33.71 $1.9\times 10^{-8}$ $1.09\times10^{-3}$ 1A7M 20883 10 11.63 22.42 24.53 35.33 $3.2\times 10^{-8}$ $3.16\times10^{-4}$ 2AQ5 38151 11 29.58 53.69 69.88 93.99 $2.8\times 10^{-8}$ $1.61\times10^{-4}$ 1F6W 49006 11 46.77 86.41 94.47 134.11 $2.3\times 10^{-8}$ $7.05\times10^{-4}$ 1C4K 72046 11 70.04 118.93 172.47 221.36 $3.7\times 10^{-8}$ $1.69\times10^{-3}$ PDB ID Number of Mesh Nodes Iter. Number Find $\Phi$ Total Time Relative error $E_{h}$ Residual norm New FE New FE 2LZX 26349 11 15.23 27.79 27.61 40.17 $2.1\times 10^{-8}$ $1.42\times10^{-4}$ 1AJJ 31910 11 26.60 48.25 45.29 66.94 $3.9\times 10^{-8}$ $3.69\times10^{-5}$ 1FXD 34469 12 23.19 42.48 49.74 69.03 $1.2\times 10^{-8}$ $8.03\times10^{-4}$ 1HPT 48229 10 32.33 58.78 58.08 84.53 $3.3\times 10^{-8}$ $1.11\times10^{-4}$ 4PTI 39468 10 25.85 46.52 47.14 67.81 $1.5\times 10^{-8}$ $8.95\times10^{-5}$ 1SVR 61074 11 55.17 90.59 96.88 132.30 $2.6\times 10^{-8}$ $4.25\times10^{-5}$ 1A63 22054 11 13.52 27.09 27.19 40.76 $1.6\times 10^{-8}$ $1.82\times10^{-4}$ 1CID 19872 10 11.07 21.68 23.10 33.71 $1.9\times 10^{-8}$ $1.09\times10^{-3}$ 1A7M 20883 10 11.63 22.42 24.53 35.33 $3.2\times 10^{-8}$ $3.16\times10^{-4}$ 2AQ5 38151 11 29.58 53.69 69.88 93.99 $2.8\times 10^{-8}$ $1.61\times10^{-4}$ 1F6W 49006 11 46.77 86.41 94.47 134.11 $2.3\times 10^{-8}$ $7.05\times10^{-4}$ 1C4K 72046 11 70.04 118.93 172.47 221.36 $3.7\times 10^{-8}$ $1.69\times10^{-3}$ Comparison of the performance of our new NMPBE solver (New) on the refined meshes with that of the finite element NMPBE solver (FE) [19] in CPU time measured in seconds. PDB ID Number of Mesh Nodes Iter. Number Find $\Phi$ Total Time Relative error $E_{h}$ Residual norm New FE New FE 2LZX 535400 10 167.0 291.2 311.2 435.3 $3.5\times 10^{-8}$ $4.07\times10^{-4}$ 1AJJ 538321 10 223.9 437.3 387.4 600.8 $3.5\times 10^{-8}$ $1.33\times10^{-4}$ 1FXD 540849 11 201.5 346.2 363.1 507.8 $1.7\times 10^{-8}$ $3.22\times10^{-4}$ 1HPT 543220 9 186.8 399.6 332.3 545.0 $3.3\times 10^{-8}$ $2.68\times10^{-4}$ 4PTI 541329 9 173.4 328.7 319.5 474.7 $2.3\times 10^{-8}$ $3.7\times10^{-4}$ 1SVR 550170 10 229.7 411.0 415.3 596.7 $2.0\times 10^{-8}$ $1.29\times10^{-4}$ 1A63 558010 11 253.3 442.8 573.1 762.7 $2.4\times 10^{-8}$ $2.62\times10^{-3}$ 1CID 558374 10 203.0 389.0 409.4 595.4 $2.7\times 10^{-8}$ $4.13\times10^{-4}$ 1A7M 563919 11 242.9 471.7 442.6 671.4 $4.8\times 10^{-8}$ $3.09\times10^{-4}$ 2AQ5 577821 10 296.7 566.8 637.7 907.8 $3.5\times 10^{-8}$ $1.41\times10^{-4}$ 1F6W 574686 11 332.1 597.3 707.8 973.0 $3.7\times 10^{-8}$ $1.06\times10^{-3}$ 1C4K 573111 14 396.6 698.3 940.8 1242.5 $3.5\times 10^{-8}$ $1.18\times10^{-3}$ PDB ID Number of Mesh Nodes Iter. Number Find $\Phi$ Total Time Relative error $E_{h}$ Residual norm New FE New FE 2LZX 535400 10 167.0 291.2 311.2 435.3 $3.5\times 10^{-8}$ $4.07\times10^{-4}$ 1AJJ 538321 10 223.9 437.3 387.4 600.8 $3.5\times 10^{-8}$ $1.33\times10^{-4}$ 1FXD 540849 11 201.5 346.2 363.1 507.8 $1.7\times 10^{-8}$ $3.22\times10^{-4}$ 1HPT 543220 9 186.8 399.6 332.3 545.0 $3.3\times 10^{-8}$ $2.68\times10^{-4}$ 4PTI 541329 9 173.4 328.7 319.5 474.7 $2.3\times 10^{-8}$ $3.7\times10^{-4}$ 1SVR 550170 10 229.7 411.0 415.3 596.7 $2.0\times 10^{-8}$ $1.29\times10^{-4}$ 1A63 558010 11 253.3 442.8 573.1 762.7 $2.4\times 10^{-8}$ $2.62\times10^{-3}$ 1CID 558374 10 203.0 389.0 409.4 595.4 $2.7\times 10^{-8}$ $4.13\times10^{-4}$ 1A7M 563919 11 242.9 471.7 442.6 671.4 $4.8\times 10^{-8}$ $3.09\times10^{-4}$ 2AQ5 577821 10 296.7 566.8 637.7 907.8 $3.5\times 10^{-8}$ $1.41\times10^{-4}$ 1F6W 574686 11 332.1 597.3 707.8 973.0 $3.7\times 10^{-8}$ $1.06\times10^{-3}$ 1C4K 573111 14 396.6 698.3 940.8 1242.5 $3.5\times 10^{-8}$ $1.18\times10^{-3}$ [1] Caterina Calgaro, Meriem Ezzoug, Ezzeddine Zahrouni. Stability and convergence of an hybrid finite volume-finite element method for a multiphasic incompressible fluid model. Communications on Pure & Applied Analysis, 2018, 17 (2) : 429-448. doi: 10.3934/cpaa.2018024 [2] Xufeng Xiao, Xinlong Feng, Jinyun Yuan. The stabilized semi-implicit finite element method for the surface Allen-Cahn equation. Discrete & Continuous Dynamical Systems - B, 2017, 22 (7) : 2857-2877. doi: 10.3934/dcdsb.2017154 [3] Yi Shi, Kai Bao, Xiao-Ping Wang. 3D adaptive finite element method for a phase field model for the moving contact line problems. Inverse Problems & Imaging, 2013, 7 (3) : 947-959. doi: 10.3934/ipi.2013.7.947 [4] Jiaping Yu, Haibiao Zheng, Feng Shi, Ren Zhao. Two-grid finite element method for the stabilization of mixed Stokes-Darcy model. Discrete & Continuous Dynamical Systems - B, 2019, 24 (1) : 387-402. doi: 10.3934/dcdsb.2018109 [5] Ruijun Zhao, Yong-Tao Zhang, Shanqin Chen. 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A nonoverlapping domain decomposition method for nonconforming finite element problems. Communications on Pure & Applied Analysis, 2003, 2 (3) : 297-310. doi: 10.3934/cpaa.2003.2.297 [15] Runchang Lin. A robust finite element method for singularly perturbed convection-diffusion problems. Conference Publications, 2009, 2009 (Special) : 496-505. doi: 10.3934/proc.2009.2009.496 [16] Christos V. Nikolopoulos, Georgios E. Zouraris. Numerical solution of a non-local elliptic problem modeling a thermistor with a finite element and a finite volume method. Conference Publications, 2007, 2007 (Special) : 768-778. doi: 10.3934/proc.2007.2007.768 [17] Tina Hartley, Thomas Wanner. A semi-implicit spectral method for stochastic nonlocal phase-field models. Discrete & Continuous Dynamical Systems - A, 2009, 25 (2) : 399-429. doi: 10.3934/dcds.2009.25.399 [18] Sören Bartels, Marijo Milicevic. Iterative finite element solution of a constrained total variation regularized model problem. Discrete & Continuous Dynamical Systems - S, 2017, 10 (6) : 1207-1232. doi: 10.3934/dcdss.2017066 [19] Alexander Zlotnik, Ilya Zlotnik. Finite element method with discrete transparent boundary conditions for the time-dependent 1D Schrödinger equation. Kinetic & Related Models, 2012, 5 (3) : 639-667. doi: 10.3934/krm.2012.5.639 [20] Zhenlin Guo, Ping Lin, Guangrong Ji, Yangfan Wang. Retinal vessel segmentation using a finite element based binary level set method. Inverse Problems & Imaging, 2014, 8 (2) : 459-473. doi: 10.3934/ipi.2014.8.459 2018 Impact Factor: 1.025	2019-10-17 18:58:25	{"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": 0.22884689271450043, "perplexity": 3224.1050897006285}, "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/1570986675598.53/warc/CC-MAIN-20191017172920-20191017200420-00439.warc.gz"}
https://openstax.org/books/introductory-statistics/pages/3-3-two-basic-rules-of-probability	Introductory Statistics # 3.3Two Basic Rules of Probability Introductory Statistics3.3 Two Basic Rules of Probability When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not. ### The Multiplication Rule If A and B are two events defined on a sample space, then: P(A AND B) = P(B)P(A\|B). This rule may also be written as: P(A\|B) = (The probability of A given B equals the probability of A and B divided by the probability of B.) If A and B are independent, then P(A\|B) = P(A). Then P(A AND B) = P(A\|B)P(B) becomes P(A AND B) = P(A)P(B). If A and B are defined on a sample space, then: P(A OR B) = P(A) + P(B) - P(A AND B). If A and B are mutually exclusive, then P(A AND B) = 0. Then P(A OR B) = P(A) + P(B) - P(A AND B) becomes P(A OR B) = P(A) + P(B). ### Example 3.14 Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B = Alaska • Klaus can only afford one vacation. The probability that he chooses A is P(A) = 0.6 and the probability that he chooses B is P(B) = 0.35. • P(A AND B) = 0 because Klaus can only afford to take one vacation • Therefore, the probability that he chooses either New Zealand or Alaska is P(A OR B) = P(A) + P(B) = 0.6 + 0.35 = 0.95. Note that the probability that he does not choose to go anywhere on vacation must be 0.05. ### Example 3.15 Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A = the event Carlos is successful on his first attempt. P(A) = 0.65. B = the event Carlos is successful on his second attempt. P(B) = 0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90. a. What is the probability that he makes both goals? Solution 3.15 a. The problem is asking you to find P(A AND B) = P(B AND A). Since P(B\|A) = 0.90: P(B AND A) = P(B\|A) P(A) = (0.90)(0.65) = 0.585 Carlos makes the first and second goals with probability 0.585. b. What is the probability that Carlos makes either the first goal or the second goal? Solution 3.15 b. The problem is asking you to find P(A OR B). P(A OR B) = P(A) + P(B) - P(A AND B) = 0.65 + 0.65 - 0.585 = 0.715 Carlos makes either the first goal or the second goal with probability 0.715. c. Are A and B independent? Solution 3.15 c. No, they are not, because P(B AND A) = 0.585. P(B)P(A) = (0.65)(0.65) = 0.423 0.423 ≠ 0.585 = P(B AND A) So, P(B AND A) is not equal to P(B)P(A). d. Are A and B mutually exclusive? Solution 3.15 d. No, they are not because P(A and B) = 0.585. To be mutually exclusive, P(A AND B) must equal zero. Try It 3.15 Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P(C) = 0.75. D = the event Helen makes the second shot. P(D) = 0.75. The probability that Helen makes the second free throw given that she made the first is 0.85. What is the probability that Helen makes both free throws? ### Example 3.16 A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly. a. What is the probability that the member is a novice swimmer? Solution 3.16 a. $2815028150$ b. What is the probability that the member practices four times a week? Solution 3.16 b. $8015080150$ c. What is the probability that the member is an advanced swimmer and practices four times a week? Solution 3.16 c. $4015040150$ d. What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not? Solution 3.16 d. P(advanced AND intermediate) = 0, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time. e. Are being a novice swimmer and practicing four times a week independent events? Why or why not? Solution 3.16 e. No, these are not independent events. P(novice AND practices four times per week) = 0.0667 P(novice)P(practices four times per week) = 0.0996 0.0667 ≠ 0.0996 Try It 3.16 A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is taking a gap year? ### Example 3.17 Felicity attends Modesto JC in Modesto, CA. The probability that Felicity enrolls in a math class is 0.2 and the probability that she enrolls in a speech class is 0.65. The probability that she enrolls in a math class GIVEN that she enrolls in speech class is 0.25. Let: M = math class, S = speech class, M\|S = math given speech 1. What is the probability that Felicity enrolls in math and speech? Find P(M AND S) = P(M\|S)P(S). 2. What is the probability that Felicity enrolls in math or speech classes? Find P(M OR S) = P(M) + P(S) - P(M AND S). 3. Are M and S independent? Is P(M\|S) = P(M)? 4. Are M and S mutually exclusive? Is P(M AND S) = 0? Solution 3.17 a. 0.1625, b. 0.6875, c. No, d. No Try It 3.17 A student goes to the library. Let events B = the student checks out a book and D = the student check out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(D\|B) = 0.5. 1. Find P(B AND D). 2. Find P(B OR D). ### Example 3.18 Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let B = woman develops breast cancer and let N = tests negative. Suppose one woman is selected at random. a. What is the probability that the woman develops breast cancer? What is the probability that woman tests negative? Solution 3.18 a. P(B) = 0.143; P(N) = 0.85 b. Given that the woman has breast cancer, what is the probability that she tests negative? Solution 3.18 b. P(N\|B) = 0.02 c. What is the probability that the woman has breast cancer AND tests negative? Solution 3.18 c. P(B AND N) = P(B)P(N\|B) = (0.143)(0.02) = 0.0029 d. What is the probability that the woman has breast cancer or tests negative? Solution 3.18 d. P(B OR N) = P(B) + P(N) - P(B AND N) = 0.143 + 0.85 - 0.0029 = 0.9901 e. Are having breast cancer and testing negative independent events? Solution 3.18 e. No. P(N) = 0.85; P(N\|B) = 0.02. So, P(N\|B) does not equal P(N). f. Are having breast cancer and testing negative mutually exclusive? Solution 3.18 f. No. P(B AND N) = 0.0029. For B and N to be mutually exclusive, P(B AND N) must be zero. Try It 3.18 A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is going to college and plays sports? ### Example 3.19 Refer to the information in Example 3.18. P = tests positive. 1. Given that a woman develops breast cancer, what is the probability that she tests positive. Find P(P\|B) = 1 - P(N\|B). 2. What is the probability that a woman develops breast cancer and tests positive. Find P(B AND P) = P(P\|B)P(B). 3. What is the probability that a woman does not develop breast cancer. Find P(B′) = 1 - P(B). 4. What is the probability that a woman tests positive for breast cancer. Find P(P) = 1 - P(N). Solution 3.19 a. 0.98; b. 0.1401; c. 0.857; d. 0.15 Try It 3.19 A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(D\|B) = 0.5. 1. Find P(B′). 2. Find P(D AND B). 3. Find P(B\|D). 4. Find P(D AND B′). 5. Find P(D\|B′).	2019-12-15 02:21:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "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": 0.4099551737308502, "perplexity": 1508.7804975966674}, "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-2019-51/segments/1575541301014.74/warc/CC-MAIN-20191215015215-20191215043215-00343.warc.gz"}
http://math.stackexchange.com/questions/184532/how-to-compute-the-convolution-of-two-measures-explicitly?answertab=oldest	# how to compute the convolution of two measures explicitly Here is my example:u and v are the surface measures on the spheres {${x;\|x\|=a}$} and {${x;\|x\|=b}$} in $\mathbb{R}^{3}$.Then what's $u\ast v$ ? And what if in $\mathbb{R}^{n}$? - What's a surface measure? – Rudy the Reindeer Aug 20 '12 at 9:32 Let $x=rs$ with $r\geqslant0$ and $s$ in $S^{n-1}$ denote the radial representation of $x$ in $\mathbb R^n$. Then, $(u\ast v)(\mathrm dx)=r\,\kappa_n(r)\,\mathrm dr\sigma_{n-1}(\mathrm ds)$ with $$\kappa_n(r)=c_n\cdot (4a^2b^2-(r^2-a^2-b^2)^2)^{(n-3)/2}\cdot[\|a-b\|\lt r\lt a+b].$$ Equivalently, $$(u\ast v)(\mathrm dx)=\kappa_n(\|x\|)\,\|x\|^{2-n}\mathrm dx.$$ To show this, first note that $u\ast v$ is radially symmetric and that its radial part $\kappa_n$ solves, for every function $\varphi$, $$\int\varphi(r^2)\kappa_n(r)\,\mathrm dr=\int\varphi(\|x\|^2)\,(\mu\ast\nu)(\mathrm dx)=\iint\varphi(\|as+bt\|^2)\,a^{n-1}\sigma_{n-1}(\mathrm ds)b^{n-1}\sigma_{n-1}(\mathrm dt).$$ By rotational invariance, this is $$\int\varphi(\|as+b\|^2)\,(ab)^{n-1}\sigma_{n-1}(\mathrm ds)=\int\varphi(a^2+b^2+2ab\langle s,1\rangle)\,(ab)^{n-1}\sigma_{n-1}(\mathrm ds).$$ One knows that $$\sigma_{n-1}(\mathrm ds)=\sin^{n-2}\phi_1\cdot\sin^{n-3}\phi_2\cdots\sin\phi_{n-2}\cdot\mathrm d\phi_1\cdots\mathrm d\phi_{n-1},$$ for some angles $\phi_k$ independent and uniform on $[0,\pi]$ except $\phi_{n-1}$ which is uniform on $[0,2\pi]$. Hence, $\langle s,1\rangle=\cos\theta_1$ and $$\int\varphi(r^2)\kappa(r)\,\mathrm dr=c_n\int_0^\pi\varphi(a^2+b^2+2ab\cos\theta)\,(ab)^{n-1}\sin^{n-2}\theta\cdot\mathrm d\theta.$$ Finally, the change of variables $r^2=a^2+b^2+2ab\cos\theta$ in the last integral yields the result. good job~ when n=3,someone tells me the answer is the fuction $\frac{2\pi ab}{\|x\|}$,when $\|a-b\|<x<a+b$ and 0 otherwise.it's a special case of your formular – sun Aug 20 '12 at 13:10	2014-12-18 16:23:33	{"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": 0.959865152835846, "perplexity": 350.38811752578374}, "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-2014-52/segments/1418802767274.159/warc/CC-MAIN-20141217075247-00166-ip-10-231-17-201.ec2.internal.warc.gz"}
https://www.khanacademy.org/math/differential-calculus/limits_topic/limits-infinity	If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Limits and infinity 7 videos 2 skills You have a basic understanding of what a limit is. Now, in this tutorial, we can explore situation where we take the limit as x approaches negative or positive infinity (and situations where the limit itself could be unbounded). Limits and infinity VIDEO 7:41 minutes Vertical asymptote of natural log VIDEO 7:00 minutes Visually determining vertical asymptotes VIDEO 1:46 minutes Limits at infinity where f(x) is unbounded PRACTICE PROBLEMS Limits at positive and negative infinity VIDEO 4:07 minutes More limits at infinity VIDEO 4:33 minutes Limits with two horizontal asymptotes VIDEO 5:14 minutes Limits at infinity using algebra VIDEO 5:26 minutes Limits at infinity where x is unbounded PRACTICE PROBLEMS	2015-02-28 12:35:14	{"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": 0.9117836952209473, "perplexity": 6292.105162503996}, "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-2015-11/segments/1424936461944.75/warc/CC-MAIN-20150226074101-00116-ip-10-28-5-156.ec2.internal.warc.gz"}
https://replit.com/talk/ask/How-do-i-fix-these-indentation-errors/46521	← Back to Community ##### How do i fix these indentation errors Profile icon teeno Having problems with indentation errors Voters Profile icon AndreaNepoti Profile icon teeno hotnewtop Profile icon CodingCactus try: ctrl + a tab shift + tab Profile icon VyomJain here is a rundown of indentation errors and what they mean: Unexpected Indent- You indented too many times. Take out indents until you get the right amount Expected a indent block- You indented too few times. Indent the line until you get the right amount. Hope this helps!	2022-05-28 11:15:33	{"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": 0.9144623875617981, "perplexity": 12919.001327819387}, "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/1652663016373.86/warc/CC-MAIN-20220528093113-20220528123113-00758.warc.gz"}
https://dsp.stackexchange.com/questions/38513/is-matlab-documentation-wrong-on-when-to-use-discrete-versus-continuous-wavelet	# Is MATLAB documentation wrong on when to use discrete versus continuous wavelet transform? I was reading some of the MATLAB documentation and I get confused when reading the part about When is Continuous Analysis More Appropriate than Discrete Analysis? My knowledge before reading the documentation: I always thought that a signal that is non-differentiable everywhere requires an infinite number of terms to represent it. This is true for the Fourier basis and a wavelet basis. The documentation: To answer this, consider the related questions: Do you need to know all values of a continuous decomposition to reconstruct the signal exactly? Can you perform nonredundant analysis? When the energy of the signal is finite, not all values of a decomposition are needed to exactly reconstruct the original signal, provided that you are using a wavelet that satisfies some admissibility condition. What I find confusing: The documentation seems to imply that if the “energy of the signal is finite”, then you don’t need all the terms in the basis expansion. I don’t think this is true, because I can think of signals with finite energy, that are not differentiable everywhere. And if they aren’t differentiable everywhere, then they will require an infinite number of terms to represent using the Fourier basis or the wavelet basis. A perfect example of this is the boxcar function. Is the documentation wrong? • Does it depend on the basis used? – makansij Mar 21 '17 at 1:42 • I would say that there is no "fundamental" difference among the orthogonal bases. However, when you move to frames, then you can have quite different behavior. Should I expend the part on frames versus bases? – Laurent Duval Mar 26 '17 at 9:56 • In your statement: "I always thought that a signal that is non-differentiable everywhere requires an infinite number of terms to represent it..." do you actually mean not differentiable instead of non-differentiable, because otherwise you refer to a signal whose derivate does not exist on any point, if it's non-differentiable everywhere...? Or you can rephrase is as non-differentiable at some countable number of points Also you shall better ask this to MathWorks... – Fat32 Mar 28 '17 at 8:50 • – makansij Apr 21 '17 at 23:54 • That's a good point @Fat32 – makansij Apr 21 '17 at 23:54 Matlab is not wrong. But let's look at your following sentence more carefully: And if they aren’t differentiable everywhere. A discontinuity in the input function can be interpreted as an infinite bandwidth, hence it will require an infinite number of Fourier terms to represent using the Fourier basis or the wavelet basis. Now this is true for the Fourier series since the basis functions are all differentiable. A jump discontinuity in the input function can be interpreted as an infinite bandwidth, hence it will require an infinite number of Fourier terms. The Gibbs phenomenon i.e. large overshoots and oscillations occur at the neighborhood of such discontinuities when dealing with truncated Fourier sums and the overshoots won't improve by increasing the terms. Indeed, something similar can also happen in wavelets under specific conditions. This reference [after equation 15] provides the necessary and sufficient condition. That said, when it comes to wavelets there is a difference. We can have wavelets that have discontinuities. So the number of terms is not necessary to be infinite for them to perfectly analyze/reconstruct an input function with jump discontinuity. The most obvious example is the Haar wavelet which uses square-shaped basis functions. I mention this wavelet, particularly to address your boxcar signal example. This property of Haar transform is widely known. For example see here. In the following picture from here there is an interesting and intuitive connection between the orthogonal "box" basis and Haar wavelet basis. I would also suggest looking at the explanation as well as the last example in that reference. [EDIT] Let us start from a verbose and simple example. In $\mathbb{R}^2$, the Mercedes-Benz pattern constitute a tight frame: redundant, energy-preserving: Each vector $x=[x_1,x_2]$ yields three three projections $[p1,p2,p3]$, from which you can recover $x$ in a number of robust ways: if the projection value are corrupted, rounded, even if one is cancelled, you can still recover $x$, at least approximately. Because of the redundancy $2\to 3$ in the representation. This is not the case with the orthogonal basis the $\{[1,0],[0,1]\}$: if you lose one projection, $x$ is almost lost (you only keep one direction). In other words, you don't need all coefficients with a frame of vectors (or wavelets). This basic example extends to more generic frames or bases. You can find more on that within the topic of oversampled filter banks, and the paper Filter bank frame expansions with erasures. Those are used for redundant channel coding, for instance. In the continuous setting (both the signal and the wavelet transform), due to the huge redundancy, many possible inverses exist. Notably, given an analysis wavelet, the requirements on the synthesis wavelets to verify the admissibility conditions (on a pair of wavelets) are very mild. On these topics, I would suggest two references: • Matthias Holschneider, Wavelets, an analysis tool, 1995 • Bruno Torrésani, Analyse continue par ondelettes, 1995 The topic is quite involved, may involve a lot of complicated functions spaces, different for the wavelets and the signals. A wavelet analysis is given by: $$W_f(b,a)\ =\ \int_{\mathbb{R}}\ f(t)\ \tfrac{1}{\sqrt{a}}\,\psi^{*}(\tfrac{t-b}{a})\ d t\ .$$ Provided that $\psi$ is admissible, i.e. when the two constants $c^{\pm}_{\psi} = 2\pi \int_{0}^{+\infty}\tfrac{\|\hat\psi(\pm\omega)\|^{2}}{\omega} d\omega < \infty$ are finite and equal\footnote{When $\psi$ is sufficiently regular, this condition reduces to a zero-average requirement, that is, $\int_{\mathbb{R}} \psi(t)\ d t = 0$}, that is, $c^+_\psi=c^-_\psi=c_\psi < \infty$, the signal $f$ may be recovered from the coefficients $W_f(b,a)$: $$f(t)\ =\ \tfrac{1}{c_{\psi}}\int_{0}^{+\infty}\!\!\!\int_{\mathbb{R}}\ W_f(b,a)\ \psi_{(b,a)}(t)\ d b\,\tfrac{d a}{a^{2}}.$$ You can find about the same idea in Torrésani's book, page 20: Still considering a continuous signal with finite energy, but trying to build a discretized wavelet frame, you can sample in the wavelet space with perfect reconstruction provided the sampling is dense enough with respect to some lower bound computed from the wavelet. So effectively, you don't need all (continuous) values. But the number of samples needed could be denombrable. Basically, you have a frame when for every unit energy signal $f$ projected onto the set of frame elements $\psi_n$: $$A\le \sum \|<f,\psi_n>\| \le B\,.$$ If $A=B=1$, you recover the case of orthogonal expansions. Ingrid Daubechies (for one) showed, from the continuous formula above, that one can find some discretizations $a_n$ and $b_n$ to build frame elements $$\psi_n = \psi(\tfrac{t-b_n}{a_n})$$ and recover $f$ from the $\|<f,\psi_n>\|$, sometimes with iterative formulae that converge faster if $A$ is close to $B$. Moving on to discrete signals (where differentiability is not well-defined), you can build unions of wavelet basis, or wavelet frames like the stationary wavelet transform. In this redundant (but with a finite number of samples) expansion, you can pick a subset of samples, that corresponds to, at minimum, a specific choice of basis vectors. Once again, you do not need all samples. THe rationale is that if an $N$ sample signal get represented by $M> N$ coefficients, when they are well formed (frame coefficients for instance), different inverses exist, some of then discarding parts of coefficients. For a basis (non redundant) expansion however, all samples are required in general.	2021-01-18 17:08:14	{"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": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.7531316876411438, "perplexity": 516.3988852979514}, "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-2021-04/segments/1610703515075.32/warc/CC-MAIN-20210118154332-20210118184332-00357.warc.gz"}
https://forum.obsidian.md/t/referencing-equations/9413	# Referencing Equations Is there a way to give equations/math latex text an equation number and make it reference-able? 1 Like Use the following construction: $$e=mc^2 \tag{1}\label{name}$$ You can now refer to this formula: $$\eqref{name}$$ 2 Likes This is really interesting! Thank you! I slightly upgrade it: using $\eqref{eq1}$ so that the equation label can be refereed inline. $$e=mc^2 \tag{1} \label{eq1}$$ Einstein provides $\eqref{eq1}$ in his paper. 2 Likes This topic was automatically closed 24 hours after the last reply. New replies are no longer allowed. In case someone runs into this question, please note that this does not work anymore due to an upstream bug. For more info see: Automatic equation numbering (Latex Math) - #27 by WhiteNoise 1 Like	2022-05-27 13:07:05	{"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": 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": 0.8549315333366394, "perplexity": 4384.52545170882}, "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/1652662647086.91/warc/CC-MAIN-20220527112418-20220527142418-00679.warc.gz"}
http://math-doc.ujf-grenoble.fr/cgi-bin/sps?au=BRUNEAU,+MICHEL	Browse by: Author name - Classification - Keywords - Nature 1 matches found XIII: 18, 227-232, LNM 721 (1979) BRUNEAU, Michel Sur la $p$-variation d'une surmartingale continue (Martingale theory) The $p$-variation of a deterministic function being defined in the obvious way as a supremum over all partitions, the sample functions of a continuous martingale (and therefore semimartingale) are known to be of finite $p$-variation for $p>2$ (not for $p=2$ in general: non-anticipating partitions are not sufficient to compute the $p$-variation). If $X$ is a continuous supermartingale, a universal bound is given on the expected $p$-variation of $X$ on the interval $[0,T_\lambda]$, where $T_\lambda=\inf\{t:\|X_t-X_0\|\ge\lambda\}$. The main tool is Doob's classical upcrossing inequality Comment: For an extension see 1319. These properties are used in T.~Lyons' pathwise theory of stochastic differential equations; see his long article in Rev. Math. Iberoamericana 14, 1998 Keywords: $p$-variation, Upcrossings Nature: Original Retrieve article from Numdam	2013-06-19 19:48:00	{"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": 0.854637086391449, "perplexity": 1899.2782888124414}, "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-2013-20/segments/1368709101476/warc/CC-MAIN-20130516125821-00074-ip-10-60-113-184.ec2.internal.warc.gz"}
https://math.stackexchange.com/questions/3899688/closed-form-for-chu-vandermonde-identity-like-summations/3899984	# Closed form for Chu–Vandermonde identity-like summations I have been trying to analyze a randomized algorithm I cooked up and have found the need to compute some moments of a random variable modeling an aspect of the algorithm but I am unsure if there is a nice closed form. Suppose the random variable in question is $$X$$ and it can take a value $$i \in \lbrace 0, 1, \cdots, k \rbrace$$ with probability $$\text{Pr}\lbrace X = i \rbrace = \binom{n}{k}^{-1}\binom{M}{i}\binom{n-M}{k-i}$$ where $$n \geq M \geq 1$$ and $$n \geq k \geq 1$$. Note that I assume that $$\binom{a}{b} = 0$$ if $$a < b$$ since there are $$0$$ ways to choose size $$b$$ sets out of a size $$a$$ set when this inequality holds. It is clear by the Chu-Vandermonde identity that indeed these values are probabilities because $$\sum_{i=0}^{k} \text{Pr}\lbrace X = i \rbrace = \binom{n}{k}^{-1} \sum_{i=0}^k \binom{M}{i}\binom{n-M}{k-i} = 1$$ The first moment of $$X$$, namely its expectation, can be computed analytically as \begin{align} \mathbb{E}(X) &= \sum_{i=0}^k i \text{Pr}\lbrace X = i \rbrace \\ &= \binom{n}{k}^{-1} \sum_{i=1}^k i \binom{M}{i}\binom{n-M}{k-i} \\ &= M \binom{n}{k}^{-1} \sum_{i=1}^k \binom{M-1}{i-1}\binom{(n-1)-(M-1)}{(k-1)-(i-1)} \\ &= M \binom{n}{k}^{-1} \binom{n-1}{k-1} \\ &= \frac{Mk}{n} \end{align} I also need to compute $$\mathbb{E}(X^2)$$ so I can do a variance calculation but I am not sure how to go about it. One can find a loose upper bound of $$Mk^2/n$$ but at least empirically, it seems for the valid ranges of $$k$$, the upper bound can at least be made to be $$Mk^{3/2}/n$$. If I could find an exact expression to this moment, I would feel a lot better of where I might be able to go with this algorithm. So does anyone recognize this probability distribution and have insight into the desired moment I want to compute?	2022-06-30 03:53:41	{"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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 17, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9999831914901733, "perplexity": 178.78863980192344}, "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-2022-27/segments/1656103661137.41/warc/CC-MAIN-20220630031950-20220630061950-00166.warc.gz"}
http://tex.stackexchange.com/questions/2832/how-can-i-have-two-tables-side-by-side?answertab=oldest	# How can I have two tables side by side? Suppose I have two tables that don't have many columns (3, for example). How can I position them side by side? - This question is quite popular, but the answers seem to have quite some overlap, but lack a coherent comparsion of the pros & cons of the different techniques used: minipage, subfig & subfloat, \quad (?), parbox, subcaption & subtable. Is there a consensus of what's considered best, if both "subtables" also have some basic (not a fully-fledged one with the "Table 2.1"-bit) caption? – nutty about natty Apr 18 at 5:59 I don't know that this is the best way, but you could make a 2-column table with a single row. In each cell of the table, wrap one of your intended tables in a minipage environment. Thus: \documentclass{article} \begin{document} \begin{tabular}{cc} \begin{minipage}{.5\linewidth} \begin{tabular}{ll} 1 & 2 \end{tabular} \end{minipage} & \begin{minipage}{.5\linewidth} \begin{tabular}{ll} 3 & 4 \end{tabular} \end{minipage} \end{tabular} \end{document} - I'd appreciate a comment with the down vote; I might then learn something about what makes this a bad solution. – vanden Sep 7 '10 at 4:23 Sticking it in a minipage makes no sense to me at all. – TH. Sep 7 '10 at 5:01 Does this not work? My first guess was to solve it in this way... so how come it got a downvote? – Johan Sep 7 '10 at 5:03 I don't know why it got a downvote, but for my eyes the code has some faults: It gives an overful hbox message as the \parindent is not taken into account. It doesn't do what it probably wants to do: the tabular are not centered in the line as they are left aligned in the minipages. Also a minipage + an external tabular looks like an overkill (one at least could certainly be dropped) and makes the code difficult to read and changed. – Ulrike Fischer Sep 7 '10 at 10:19 @Ulrike: Thanks; that's very helpful. – vanden Sep 7 '10 at 17:03 Ceteris paribus, you might consider the subfig package. It is suitable for subtables and subfigures and combinations of these. For myself, I've managed much mileage from it. - I'm looking at the documentation and I can only find examples using figures. Could you post an example with tables? – bada Sep 7 '10 at 4:30 @bada. Sorry man, you simply can't come back 8 mins after a solution has been posted without at least showing a little effort first. Seriously, have you tried reading the documentation in some detail (not just 5 minute's worth) then tried loading the subfig package, inserted a few tables and otherwise played around a little with it? – Geoffrey Jones Sep 7 '10 at 4:54 I got it working, I just had to use the \begin{table} instead of the \begin{figure} scope with the \subfloat command, which is idiotic to begin with since as far as I can tell that wasn't mentioned anywhere, but whatever. And you - along with people that agree with this line of thinking of yours - are the reason they call people "Linux snobs" dude. – bada Sep 7 '10 at 5:21 Excellent! Very well done! – Geoffrey Jones Sep 7 '10 at 6:43 @bada you are right, using \begin{table} is totally unreasonable, I mean since putting one tabular environment in a table float is very far fetched as well - NOT! – maxschlepzig Sep 7 '10 at 7:35 show 1 more comment Just put two tabular environments side by side. Add spacing as desired. \begin{tabular}{ccc} \hline a&b&c\\ \hline \end{tabular} \begin{tabular}{ccc} \hline d&e&f\\ \hline \end{tabular} If you want to use subfig because you want them to have separate captions, then that is simple as well. \subfloat[caption]{\begin{tabular}{...}...\end{tabular}} \subfloat[caption]{\begin{tabular}{...}...\end{tabular}} If you want two tables that are independent, and thus don't want to use \subfloat, you can use \parbox. \begin{table} \parbox{.45\linewidth}{ \centering \begin{tabular}{ccc} \hline a&b&c\\ \hline \end{tabular} \caption{Foo} } \hfill \parbox{.45\linewidth}{ \centering \begin{tabular}{ccc} \hline d&e&f\\ \hline \end{tabular} \caption{Bar} } \end{table} This is basically the same as before just that I've centered each tabular in a \parbox with an included caption and wrapped the whole thing in a table. - If you want two tabulars next to each other in a (floating) table environment, then you can use minipages \begin{table}[!htb] \caption{Global caption} \begin{minipage}{.5\linewidth} \caption{} \centering \begin{tabular}{ll} 1 & 2 \end{tabular} \end{minipage}% \begin{minipage}{.5\linewidth} \centering \caption{} \begin{tabular}{ll} 3 & 4 \end{tabular} \end{minipage} \end{table} Loading the caption package will allow easy customization, and will remove the : when the caption is empty. If you'd like 'sub'tables, then you can use the subcaption package which introduces the subtable environment (and a lot more); the subtable environment takes the same mandatory and optional arguments as the minipage \documentclass{article} \usepackage{subcaption} \begin{document} \begin{table}[!htb] \caption{Global caption} \begin{subtable}{.5\linewidth} \centering \caption{} \begin{tabular}{ll} 1 & 2 \end{tabular} \end{subtable}% \begin{subtable}{.5\linewidth} \centering \caption{} \begin{tabular}{ll} 3 & 4 \end{tabular} \end{subtable} \end{table} \end{document} Note that in both cases I've used a % at the end of the environment to stop overfull hboxes from the additional space added by the minipage. - Why would I want subtables vs tabulars? Of the two methods you describe in your answer, which is better (and why) ? – nutty about natty Apr 18 at 6:51 On a second glance: is the main (visible) difference between the two methods the way the "sub"-captions look (e.g., "Table 2" vs. "(a)") ? – nutty about natty Apr 18 at 6:57 @nuttyaboutnatty yes, your second comment is accurate. It depends how you want the captions (and references) to look. It's up to you which you think is better :) – cmhughes Apr 18 at 17:00	2013-05-21 07:48:29	{"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": 0.8737136721611023, "perplexity": 1867.8649048725745}, "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-2013-20/segments/1368699776315/warc/CC-MAIN-20130516102256-00074-ip-10-60-113-184.ec2.internal.warc.gz"}
http://cms.math.ca/10.4153/CMB-2007-011-1	Canadian Mathematical Society www.cms.math.ca location: Publications → journals → CMB Abstract view Hermitian Harmonic Maps into Convex Balls Published:2007-03-01 Printed: Mar 2007 • ZhenYang Li • Xi Zhang Format: HTML LaTeX MathJax PDF PostScript Abstract In this paper, we consider Hermitian harmonic maps from Hermitian manifolds into convex balls. We prove that there exist no non-trivial Hermitian harmonic maps from closed Hermitian manifolds into convex balls, and we use the heat flow method to solve the Dirichlet problem for Hermitian harmonic maps when the domain is a compact Hermitian manifold with non-empty boundary. Keywords: Hermitian harmonic map, Hermitian manifold, convex ball MSC Classifications: 58E15 - Application to extremal problems in several variables; Yang-Mills functionals [See also 81T13], etc. 53C07 - Special connections and metrics on vector bundles (Hermite-Einstein-Yang-Mills) [See also 32Q20] © Canadian Mathematical Society, 2015 : https://cms.math.ca/	2015-05-27 21:51:53	{"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": 0.8277660608291626, "perplexity": 3436.3682095559557}, "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-2015-22/segments/1432207929171.55/warc/CC-MAIN-20150521113209-00327-ip-10-180-206-219.ec2.internal.warc.gz"}
https://economics.stackexchange.com/questions/4640/how-does-perfect-elasticity-imply-infinite-quantity-for-any-higher-price	# How does Perfect Elasticity imply infinite quantity for any higher price? I reference Principles of Microeconomics, 7 Ed, 2014, by NG Mankiw. The graph above is from p 100. [p 99:] ...supply is perfectly elastic. This occurs as the price elasticity of supply approaches infinity and the supply curve becomes horizontal, meaning that very small changes in the price lead to very large changes in the quantity supplied. Please explain the red box? How's it true? The graph itself shows that at \$4 exclusively, all quantities$\ge 0$are supplied. The graph depicts nothing for prices above \$4? (Footnote: I tried this) More formally, "any quantity" does not mean "infinite quantity" because "infinity" is not a number, so it cannot represent a quantity (and let's not confuse quantity with the concept of cardinality used in set theory). The expression $x\geq 0$ does not include "$\infty$". $x$ belongs to the real numbers, and $\infty$ is added to the real numbers to create the "extended real numbers" - it is not a real number itself. Perhaps by "logical extension", if all non-zero real quantities are supplied at USD $4$, then what is left for higher prices is only "infinity". But this has no economic meaning -apart from the approximate one already mentioned in the passage quoted.	2020-04-03 05:27:15	{"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": 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": 0.8058212995529175, "perplexity": 1196.2833163473547}, "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-2020-16/segments/1585370510287.30/warc/CC-MAIN-20200403030659-20200403060659-00360.warc.gz"}
http://angrypost.com/html/indexProofinfinitenumberofprimes.html	## 1 Proof there is infinite number of primes • Proof by contradition (Assume there are finite number of primes) • Assume there are $$k$$ primes and let $$p_1, p_2, \dots p_k$$ are all prime numbers • let $$n = p_1 p_2 \dots p_k + 1$$, then $$n$$ is not a prime, otherwise there is $$k + 1$$ primes • $$\Rightarrow n - p_1 p_2 \dots p_k = 1$$ • let p is a prime and $$p \mid n$$ $$\because$$ n is not a prime there exists $$p$$ such as $$p \mid n$$ \begin{align} n - p_1 p_2 \dots p_k &= 1 \\ \Rightarrow \frac{n}{p} - \frac{p_1 p_2 \dots p_k}{p} &= \frac{1}{p} \end{align} • left side must be an integer becase of $$p \mid n$$ and $$p \mid p_1 p_2 \dots p_k$$ • but right side is NOT an integer. • The contradition implies our assumption is False $$\square$$ Created: 2019-08-03 Sat 22:14 Validate	2019-08-23 23:17:45	{"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": 0.9999661445617676, "perplexity": 748.1657527908501}, "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/1566027319082.81/warc/CC-MAIN-20190823214536-20190824000536-00487.warc.gz"}
https://support.zemax.com/hc/en-us/articles/10648753483539-Introduction-to-Composite-Surface	Introduction to Composite Surface This article introduces the Composite Surface, a novel, interesting and useful feature of Zemax OpticStudio 22.3 version (Professional/Premium Subscription) or Ansys Zemax OpticStudio (Professional/Premium/Enterprise) 2022 R2.02 version. It enables many new capabilities and possibilities in OpticStudio. Authored By Chenfeng Gu Introduction The Composite Surface in sequential mode enables users to add multiple sag profiles to finally achieve an optical surface with a complex new sag distribution as desired. If users want to add different types of sag distributions to one surface, multiple surfaces can be added as a Composite stack. This feature will enlighten infinite possibilities in simulations, including analysis, tolerancing and so on. In this article, we’ll explain how the Composite Surface works, and then apply its power to the task of tolerancing complex aspheric lenses in a cell phone camera module. How the Composite Surface works The Composite Surface can be turned on with the ‘Composite Surface: Add sag to the next surface’ checkbox in Surface Properties…Composite, as Figure 1 shows: Figure 1. Composite Surface Interface Any number of surfaces can be added together, these surfaces are referred to as “Composite Add-on”, or simply “Add-on”. The sag profile of the Add-on surfaces will be added to the following Add-on surface, and the total sag will finally be added to the next surface in the Lens Data Editor (LDE), which is referred to as “Composite Base”, or simply “Base”, and that follows the Add-ons. The Base surface profile will then be the summation of all the Add-on profiles, plus its own profile. The overall sum of Add-On and Base is referred to as “Composite Stack". The Add-on row color is light yellow, and the Base row is shown in bright yellow as you can see in the Figure 2. Figure 2. Composite Add-on, Composite Base and Composite Stack We can find that material and coating cannot be set for composite Add-On Surface because it is only used to apply sag distribution to the composite base surface rather than anything else. Its material and coating properties follow the ones specified on the base surface. And another point is that the aperture of the Composite Add-on Surface is also considered the same as the base surface when algorithms try to add sag distribution no matter what aperture size you set for Composite Add-on Surface. In raytracing, rays encounter the total surface sag at the base surface only. The Composite Add-on Surface does not participate in the raytracing directly. Sag maps of the base surface will show the total sag, including all Composite Add-on Surfaces, rather than the sag of the base surface alone. More details are provided further along in the article. Surfaces Supported Currently, the list of surface types that can be an Add-on surface and the list of surface types that can be a Base surface are the same, which is given below: • Biconic • Biconic Zernike • Chebyshev Polynomial • Even Asphere • Extended Asphere • Extended Odd Asphere • Extended Polynomial • Grid Sag • Irregular • Odd Asphere • Odd Cosine • Off-Axis Conic Freeform • Periodic • Polynomial • Q-Type Asphere • Q-Type Freeform • Standard • Superconic • Tilted • Toroidal • Zernike Fringe Sag • Zernike Standard Sag • Zernike Annular Standard Sag Stacking As we mentioned, if you want to add different types of sag distributions to one surface, multiple Add-on surfaces can be added as a Composite stack. In the example below (Figure 3 shows), a parabolic mirror with ROC (radius of curvature) = -500mm is formed by adding Add-on Surface 4 to base surface 5. And another add-on Surface 3 adds a Zernike perturbation to the surface sag. Figure 3. Parabolic Mirror System with Composite Surface You can see from the following layout, with Add-on surfaces activated (Figure 4(a)), the plots display the total sag of composite stack and don’t show the individual Add-on Surfaces. And Figure 4(b) indicates that with all the Add-on surfaces ignored, only Base surface shows up in the plots (so the system is now, clearly out of focus). Figure 4. (a)Parabolic Mirror System with Add-on surfaces activated (b)Parabolic Mirror System with Add-on surfaces ignored Composite Surface Tools After adding Composite Surface, we have several user-friendly tools for you to use. The sag contribution of a Composite Surface will be ignored when the “Ignore This Surface” box is checked, and its row color will change to dark grey, as shown in Figure 5. Figure 5. Ignore individual Add-on surface If you want to ignore or activate all the Composite Surfaces in the system, there are two buttons to easily control in the toolbar of the Lens Data Editor. Figure 6. Ignore all Composite Add-on Surfaces Figure 7. Activate all Composite Add-on Surfaces • Hide/Unhide/Delete All Composite Surfaces In the right-click menu, there are three options for you to easily handle Composite Surfaces as you want. Figure 8. Hide/Unhide/Delete All Composite Surfaces Using Composite Surfaces with the Sag Map We can use Analyze...Polarization and Surface Physics...Surface to check the Sag distribution. Here are some general ideas on Sag map of the Composite Surface. • Choose ‘Remove None’: Sag maps will show the sum of all Add-on up to and including the chosen surface. • Choose ‘Remove Composite Sag’: plot the selected surface sag alone. • Choose ‘Remove Base Sag’: leave the sum of all the Add-on Surfaces sag which is applied on this surface. In the following example, as shown in Figure 9(a), surface 3 and 4 are Add-on surface, and surface 5 is base surface. The Composite Surface sag maps, Figure 9(b)~(g) explain the above sag relationships with composite surfaces. Figure 9(a). Cellphone Lens System Example to Explain Composite Surfaces with Sag Map As shown in Figure 9(b)~(d), when "Remove None" is used, then the sag adds up row after row. Figure 9(b) Figure 9(c) Figure 9(d) If then you want to isolate single surface contribution, you can use "Remove Composite Sag", or "Remove Base Sag" as shown in Figure 9(e)~(g) : Figure 9(e) Figure 9(f) Figure 9(g) Composite Surface and off-axis apertures Composite Surfaces can also apply on off-axis systems modeling. Figure 10. Off-axis Optical System The Add-on surface will automatically center itself on an off-axis aperture of the base surface if you click “Set Tilt/Decenter to follow Base surface aperture” button in the Composite tab of the add-on surface. Pressing the button, OpticStudio automatically populates the tilts and decenters in the Composite tab so that the add-on stack is located at center of the off-axis aperture of the base surface. Decenters match the vertex location while tilts match the orientation of the Base surface at the center of the off-axis aperture. Figure 11. "Set Tilt/Decenter to follow Base surface aperture” and "Update Tilt" button For Composite stack, the “Set Tilt/Decenter to follow Base surface aperture” button is to be used on the last Add-on surface before the Base, and all Add-on surfaces in the Composite stack will have the same Tilt/Decenter properties. The Composite stack will have the same coordinate system. Note that the Tilt/Decenter of the Composite stack can only be seen on the Add-on surface closest to the Base surface, where the “Set Tilt/Decenter to follow Base surface aperture” button has been used. The values will be greyed out for all other Add-on surfaces. If users manually change the decenter value, the “Update Tilt” button (shown in Figure 11) can be used to automatically populate tilt values so that the Add-on orientation matches the base surface at the specified decenter. Tolerancing with TIRR, TEXI, and TEZI In terms of tolerancing, from the above discussions, we know that we can use Composite property to manually add irregularity and tolerancing. Furthermore, Composite Surface expands the existing operands of TIRR, TEXI and TEZI. Previously, these three operands can be only used for tolerancing the following surfaces. Figure 12. Previously TIRR, TEXI and TEZI can only be used in limited surface types Now, thanks to Composite Surface, any surface type that can be a Composite base surface could use these three operands to do tolerancing. (Note, the expanded capabilities for TIRR, TEXI, and TEZI do not include support for off-axis apertures.) For qualifying surface types, the TIRR operand will be expanded to tolerance those surfaces using an Irregular Composite Surface. The TEXI operand will be expanded to tolerance those surfaces using a Zernike Fringe Sag Composite Surface. And the TEZI operand will be expanded to tolerance those surfaces using a Zernike Standard Sag Composite Surface. We’ll take the following cell phone design case (Figure 13) as example. Figure 13. Cellphone Lens System Example The Tolerance Data Editor (TDE) is populated with TEZI operands for surfaces 3 & 4 of the cell phone lens. Previously, TEZI could not be used directly with Q-type Asphere surface type, but now you can set up directly in the TDE and use following tolerancing functions, as shown in Figure 14. Figure 14. Tolerancing using TEZI on Q-type Asphere When we run the tolerancing and save the Monte Carlo files, you can check the tolerancing result. Figure 15. Monte Carlo Simulation We can also open one Monte Carlo file to verify and better understand how the operand added irregularity to the surface. As shown in Figure 16, the Q-type Asphere surfaces have an Add-on surface added to them which carries the Zernike Standard Sag perturbation specified in the TDE. Figure 16. TEZI operand tolerances Q-type Asphere surfaces using a Zernike Standard Sag Composite Surface There is a series of knowledgebase articles that explain the design of the of cellphone lens system used above. The series starts with: Designing Cell phone Camera Lenses Part 1: Optics Please check and use the cellphone lens sample file to try out the new Composite Surface functionalities. Looking forward to your feedback! API control There are two new API commands added for Composite Surface, IsCompositeSurface and SetOffAxisTiltAndDecenter(). IsCompositeSurface is used for both Sets and Gets values of this parameter. SetOffAxisTiltAndDecenter() populates surface property fields based on the calculation of the base composite surface normal vector at the center of the off-axis surface aperture. Below is a sample Python snippet for your reference: TheSystem = TheApplication.PrimarySystem;TheLDE = TheSystem.LDE# Get Surface j InformationCompositeTest = TheLDE.GetSurfaceAt(j)# Check “Composite Surface:Add sag to the next surface”CompositeTest.CompositeData.IsCompositeSurface=1# Press “Set Tilt/Decenter to follow Base surface aperture” buttonCompositeTest.CompositeData.SetOffAxisTiltAndDecenter()# Extract Composite Surface Tilt/Decenter valueCompositeTiltX=CompositeTest.TiltDecenterData.BeforeSurfaceTiltXCompositeTiltY=CompositeTest.TiltDecenterData.BeforeSurfaceTiltYCompositeTiltZ=CompositeTest.TiltDecenterData.BeforeSurfaceTiltZCompositDecenterX=CompositeTest.TiltDecenterData.BeforeSurfaceDecenterXCompositDecenterY=CompositeTest.TiltDecenterData.BeforeSurfaceDecenterY# Change the Tilt/Decenter values manuallyCompositeTest.TiltDecenterData.BeforeSurfaceTiltX = 0CompositeTest.TiltDecenterData.BeforeSurfaceDecenterY = 0 ZPL control There are also new ZPL macro commands for Composite Surface. • COMPOSITEON( surface number ) - keyword • COMPOSITEOFF( surface number ) - keyword • COMPOSITEOFFAXISAPERTUREON ( surface number ) – keyword • ISCS(surface number) -> returns 1\|0 - numeric function • [ISCS() returns 0 when the surface is not composite surface and returns 1 if the surface is composite surface.] Detailed information on these commands can be found in help file. Some sample codes for your reference: ! Uncheck the composite surface checkboxPRINT "Turn off composite surface j"COMPOSITEOFF jA = ISCS(j) ! ISCS () return 0 if it's not composite surf., return 1 if it's composite surf.IF (A < 0.5) PRINT " ==> Composite surface flag is OFF"ELSE PRINT " ==> Composite surface flag is ON"ENDIF ! Check the composite surface checkboxPRINT "Turn on composite surface j" COMPOSITEON jA = ISCS(j) IF (A > 0.5)PRINT " ==> Composite surface flag is ON"ELSE PRINT " ==> Composite surface flag is OFF"ENDIF ! Press the "Set Tilt/Decenter to follow Base surface aperture" buttonPRINT "Set Tilt/Decenter to follow Base surface aperture"COMPOSITEOFFAXISAPERTUREON jPRINT " ==> Click Tilt/Decenter to follow Base surface aperture button" Conclusion This article introduces the feature Composite Surface in Zemax OpticStudio 22.3 or Ansys Zemax OpticStudio 2022 R2.02. It enables many new capabilities and possibilities in OpticStudio. Hope it will make your daily work more efficient and creative. Looking forward to your feedback on this feature through community posts or email!	2023-03-26 15:47:24	{"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": 0.4442378282546997, "perplexity": 5528.536379408417}, "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-2023-14/segments/1679296945473.69/warc/CC-MAIN-20230326142035-20230326172035-00324.warc.gz"}
https://codeforces.com/blog/entry/93178	### s_jay_412's blog By s_jay_412, history, 2 months ago, You are given the following: - An integer value N - M pairs of distinct characters (lowercase English alphabets) A pair of characters in the given pairs holds a relation between them. The relation among the given pairs is also transitive, which means if you consider 2 pairs (u, v) and (v, w) which holds relation, then pair (u, w) also holds the relation. Also, the relation states that those pairs of characters cannot occur together in a formed string. Determine the total number of strings of length N such that any string does not contain a pair of adjacent characters holding any relation. Since this number can be large output it modulo 10^9+7. Constraints 1 <= T <= 10 (Number of test cases) 1 <= N <= 10^4 0 <= M <= 325 (M is the number of pairs given) Example Input: N = 2 M = 3 Pairs = [[a,b], [b,c], [c,d]] Approach: - Since [a,b] and [b,c] hold a transitive relation, thus a relation among all pairs in [a, b, c] holds true. - Since [a, b, c] and [c, d] shares a transitive relation thus a relation among all pairs in [a, b, c, d] holds true. - So the following pairs cannot occur : [ab, ac, ad, ba, bc, bd, ca, cb, cd, da, db, dc] - The strings that can be formed of length 2 are : ["aa", "ae", "af", .....] Output: 664 • +6 » 2 months ago, # \| 0 Auto comment: topic has been updated by s_jay_412 (previous revision, new revision, compare). » 2 months ago, # \| ← Rev. 2 → +1 I think the key Concepts are dfs and dp - First Find the number of Connected Components (let it be x )and Store The size of each connected component - now creat a 2D dp array dp[n][x] , where dp[i][j] (0 based indexing) number of strings of length i which end with a character from jth component -Base Case dp[0][j]=size of jth component (0 based indexing) - Recurrence dp[i][j]= size of jth component * (sum of dp[i-1][k] over all k from 0 to x-1 excluding j) - Take mod at appropriate places and get the value - number of strings = sum of dp[n-1][j] for all j from 0 to x-1 As x can be max 26 so Time Complexity is O(n) • » » 2 months ago, # ^ \| +7 If in pairs we are given [[a,b], [b,c], [b, d]], then a, b, c and d form a connected component but still neither [c, d] nor [d, c] can be generated by any transitive relation and thus they can be adjacent in a string. So is it fair to assume no 2 elements in a connected component can be adjacent? • » » » 2 months ago, # ^ \| 0 Yes Sorry for That , I misunderstood .What we can do is as number of alphabets is only 26 , for every character we can store those characters which are in a transitive relation with it and after that create dp[n][26] and use the similar recurrence I mentioned above . • » » » » 2 months ago, # ^ \| +3 Yes, that'll work. Thank you. • » » » » 2 months ago, # ^ \| +3 I did exact same still got memory limit exceeded on last three test cases.Very tight Memory limit. • » » » » » 2 months ago, # ^ \| +3 To Reduce Memory Requirements we can instead of dp[n][26] use dp[26](for previous state)and dp1[26](for current state) because we need the values of just the previous state to get the values of current state and update dp1 and dp for each i • » » » 2 months ago, # ^ \| ← Rev. 3 → 0 Is the relation given here symmetric because the relation is not stated in some directed sense, a,b relation implies b,a relation. Then, it will be same as avoiding characters in a connected component as adjacent characters. Also, you have mentioned N = 2 M = 3 Pairs = [[a,b], [b,c], [c,d]] So the following pairs cannot occur : [ab, ac, ad, ba, bc, bd, ca, cb, cd, da, db, dc] bd and db is written here. » 2 months ago, # \| ← Rev. 4 → +17 You can use Floyd warshall, to get transitive relation in 26 phases, binary matrix denote 1 if i -> j cannot be edge else 0, take complement of this(after applying floyd warshall) and you will get your transition matrix and then apply matrix exponentiation or just multiplication to calculate the answer, this will be your steps to reduce dp.total complexity will be 26^3 log N » 2 months ago, # \| 0 you can use dsu to maintain connected component, it will take care of transativity, then we can get transition matrix by checking if two character are not in same component, after that the problem asks for path of length N, it's just a standard problem. » 2 months ago, # \| 0 Floyd warshall + dp will do the job. » 2 months ago, # \| +7 how to register for these online challenges? • » » 2 months ago, # ^ \| +3 These tests are meant for oncampus hiring. » 2 months ago, # \| 0 I will try to explain how to get the transition matrix. We will model it as graph with directed edges , then run a dfs and maintain two colors in the visited array(Refer to cycle detection in directed graphs with colors). Whenever we would enter a vertex v in dfs we would make transition_matrix[u][v]=1 for all u which are not yet completely processed(again refer to cycle detection in directed graphs).	2021-09-27 09:50:29	{"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": 0.6288694143295288, "perplexity": 1552.6390047214484}, "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/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00005.warc.gz"}
http://www.haskell.org/haskellwiki/index.php?title=Chaitin's_construction&diff=26968&oldid=5209	# Chaitin's construction (Difference between revisions) Revision as of 20:22, 5 August 2006 (edit)m (→Table for simpler CL-terms: The table is easier to understand with this order of columns)← Previous diff Current revision (21:34, 14 March 2009) (edit) (undo) (Workpage template'', warning about missing substantial details) (8 intermediate revisions not shown.) Line 1: Line 1: + :'''''Correction in process. There is a substantial point that is lacking yet, the formulae and the concepts are not correct without it.''''' + __TOC__ __TOC__ Line 49: Line 51: ! Decodable strings, ratio, their sum till now ! Decodable strings, ratio, their sum till now ! Terminating, ratio, their sum till now ! Terminating, ratio, their sum till now + ! $\Omega$ approximated till now: mantissa -- binary, length-fitting binary, decimal \|- \|- \| 0 \| 0 Line 54: Line 57: \| 0, 0, 0 \| 0, 0, 0 \| 0, 0, 0 \| 0, 0, 0 + \| -, -, - \|- \|- \| 1 \| 1 Line 59: Line 63: \| 0, 0, 0 \| 0, 0, 0 \| 0, 0, 0 \| 0, 0, 0 + \| -, 0, 0 \|- \|- \| 2 \| 2 Line 64: Line 69: \| 2, $\frac12$, $\frac12$ \| 2, $\frac12$, $\frac12$ \| 2, $\frac12$, $\frac12$ \| 2, $\frac12$, $\frac12$ + \| 1, 10, 5 \|- \|- \| 3 \| 3 Line 69: Line 75: \| 0, 0, $\frac12$ \| 0, 0, $\frac12$ \| 0, 0, $\frac12$ \| 0, 0, $\frac12$ + \| 1, 100, 5 \|- \|- \| 4 \| 4 Line 74: Line 81: \| 0, 0, $\frac12$ \| 0, 0, $\frac12$ \| 0, 0, $\frac12$ \| 0, 0, $\frac12$ + \| 1, 1000, 5 \|- \|- \| 5 \| 5 Line 79: Line 87: \| 4, $\frac18$, $\frac58$ \| 4, $\frac18$, $\frac58$ \| 4, $\frac18$, $\frac58$ \| 4, $\frac18$, $\frac58$ + \| 101, 10100, 625 \|} \|} + It illustrates nicely, that Chaitin's construction is a [http://en.wikipedia.org/wiki/Normal_number normal number], as if its digits (in binary representation) were generated by tossing a coin. == Eliminating any concept of code by handling [[combinatory logic]] terms directly == == Eliminating any concept of code by handling [[combinatory logic]] terms directly == - We can avoid referring to any code notion, if we transfer (lift) the notion of “length” from bit sequences to [[combinatory logic]] terms in an appropriate way. Let us call it the “norm” of the term: + Chaitin's construction can be grasped also as + :$\sum_{p\in \mathrm{CL},\;\mathrm{hnf}\;p} 2^{-\left\|\mathrm{dc}^{-1}\;p\right\|}$ - :$\sum_{p\in\mathrm{CL},\;\mathrm{hnf}\;p} 2^{-\left\Vert p\right\Vert}$ + We can avoid referring to any code notion, if we modularize out function - where + :$\left\|\cdot\right\|\circ\mathrm{dc}^{-1}$ + and give it a separate name, e.g. :$\left\Vert\cdot\right\Vert : \mathrm{CL}\to\mathbb N$ :$\left\Vert\cdot\right\Vert : \mathrm{CL}\to\mathbb N$ + and notice that it can be defined directly in terms of CL-terms (we need not use any decoding concept any longer): + :$\left\Vert\mathbf K\right\Vert = 2$ :$\left\Vert\mathbf K\right\Vert = 2$ :$\left\Vert\mathbf S\right\Vert = 2$ :$\left\Vert\mathbf S\right\Vert = 2$ :$\left\Vert\left(x\;y\right)\right\Vert = 1 + \left\Vert x\right\Vert + \left\Vert y\right\Vert$ :$\left\Vert\left(x\;y\right)\right\Vert = 1 + \left\Vert x\right\Vert + \left\Vert y\right\Vert$ + + Thus, we transfer (lift) the notion of “length” from bit sequences to [[combinatory logic]] terms in an appropriate way. Let us call it, e.g. the “norm” of the term. + + Thus, Chaitin's construction is grasped also as + :$\sum_{p \in \mathrm{Dom}_{\mathrm{nf}}} 2^{-\left\Vert p\right\Vert}$ + where + :$\mathrm{nf} : \mathrm{CL} \supset\!\to \mathrm{CL}$ + is a partial function defined on CL terms, it attributes to each "terminating" term its normal form. Thus, we have no notions of “bit sequence”,“code”, “coding”, “decoding” at all. But their ghosts still haunt us: the definition of norm function looks rather strange without thinking on the fact that is was transferred from a concept of coding. Thus, we have no notions of “bit sequence”,“code”, “coding”, “decoding” at all. But their ghosts still haunt us: the definition of norm function looks rather strange without thinking on the fact that is was transferred from a concept of coding. Line 105: Line 127: And is it worth doing it at all? The former one, at leat, had a good theoretical foundation (based on analysis, arithmetic and probability theory). This latter one is not so cleaner, that we should prefer it, so, lacking theoretical grounds. And is it worth doing it at all? The former one, at leat, had a good theoretical foundation (based on analysis, arithmetic and probability theory). This latter one is not so cleaner, that we should prefer it, so, lacking theoretical grounds. - What I really want is to exclude the (IMHO) underestimation of this “probability of termination” number -- an underestimation coming from taking into account the syntactically non-correct codes (IMHO). Thus taking only termination vs nontermination into account, when calculating this number (which can be interpreted as a probability). + What I really want is to exclude conceptually the notion of coding, and with it the notion of “syntactically incorrect versus syntactically correct but diverging”. Thus, taking into account only syntactically correct things, seeing only the choice of terminating versus non-terminating. Thus taking only termination vs nontermination into account, when calculating Chaitin's construction. + + What I want to preserve: + * it can be interpreted as a probability + * it is a [http://en.wikipedia.org/wiki/Normal_number normal number], as if its digits (in binary representation) were generated by tossing a coin + thus I do not want to spoil these features. ==== Table for simpler CL-terms ==== ==== Table for simpler CL-terms ==== Line 164: Line 191: === Architecture === === Architecture === - A CL term generator generates CL terms in “ascending order” (in terms of a theoretically appropriate “norm”), and by computing the norm of each CL-term, it approximates Chaitin's construct (at a given number of digits, and according to the given maximal limit of reduction steps). + A CL term generator generates CL terms in “ascending order” (in terms of a theoretically appropriate “norm”), and by computing the norm of each CL-term, it approximates Chaitin's construction (at a given number of digits, and according to the given maximal limit of reduction steps). === User interface === === User interface === ## Current revision Correction in process. There is a substantial point that is lacking yet, the formulae and the concepts are not correct without it. ## 1 Introduction Are there any real numbers which are defined exactly, but cannot be computed? This question leads us to exact real arithmetic, and algorithmic information theory, and foundations of mathematics and computer science. See Wikipedia article on Chaitin's construction, referring to e.g. ## 2 Basing it on combinatory logic Some more direct relatedness to functional programming: we can base Ω on combinatory logic (instead of a Turing machine). ### 2.1 Coding See the prefix coding system described in Binary Lambda Calculus and Combinatory Logic (page 20) written by John Tromp: $\widehat{\mathbf S} \equiv 00$ $\widehat{\mathbf K} \equiv 01$ $\widehat{\left(x y\right)} \equiv 1 \widehat x \widehat y$ of course, c, d are meta-variables, and also some other notations are changed slightly. ### 2.2 Decoding Having seen this, decoding is rather straightforward. Here is a parser for illustration, but it serves only didactical purposes: it will not be used in the final implementation, because a good term generator makes parsing superfluous at this task. ### 2.3 Chaitin's construction Now, Chaitin's construction will be here $\sum_{p\in \mathrm{Dom}_\mathrm{dc},\;\mathrm{hnf}\left(\mathrm{dc}\;p\right)} 2^{-\left\|p\right\|}$ where hnf should denote an unary predicate “has normal form” (“terminates”) dc should mean an operator “decode” (a function from finite bit sequences to combinatory logic terms) $2\!\;^{*}$ should denote the set of all finite bit sequences Domdc should denote the set of syntactically correct bit sequences (semantically, they may either terminate or diverge), i.e. the domain of the decoding function, i.e. the range of the coding function. Thus, $\left\{00, 01, 1\;00\;00, 1\;00\;01, 1\;01\;00, 1\;01\;01, \dots\right\} = \mathrm{Dom}_{\mathrm{dc}} = \mathrm{Rng}_{\widehat\ }$ “Absolute value” should mean the length of a bit sequence (not combinatory logic term evaluation!) ### 2.4 Table for small legths Length (n) All strings (2n) Decodable strings, ratio, their sum till now Terminating, ratio, their sum till now Ω approximated till now: mantissa -- binary, length-fitting binary, decimal 0 1 0, 0, 0 0, 0, 0 -, -, - 1 2 0, 0, 0 0, 0, 0 -, 0, 0 2 4 2, $\frac12$, $\frac12$ 2, $\frac12$, $\frac12$ 1, 10, 5 3 8 0, 0, $\frac12$ 0, 0, $\frac12$ 1, 100, 5 4 16 0, 0, $\frac12$ 0, 0, $\frac12$ 1, 1000, 5 5 32 4, $\frac18$, $\frac58$ 4, $\frac18$, $\frac58$ 101, 10100, 625 It illustrates nicely, that Chaitin's construction is a normal number, as if its digits (in binary representation) were generated by tossing a coin. ## 3 Eliminating any concept of code by handling combinatory logic terms directly Chaitin's construction can be grasped also as $\sum_{p\in \mathrm{CL},\;\mathrm{hnf}\;p} 2^{-\left\|\mathrm{dc}^{-1}\;p\right\|}$ We can avoid referring to any code notion, if we modularize out function $\left\|\cdot\right\|\circ\mathrm{dc}^{-1}$ and give it a separate name, e.g. $\left\Vert\cdot\right\Vert : \mathrm{CL}\to\mathbb N$ and notice that it can be defined directly in terms of CL-terms (we need not use any decoding concept any longer): $\left\Vert\mathbf K\right\Vert = 2$ $\left\Vert\mathbf S\right\Vert = 2$ $\left\Vert\left(x\;y\right)\right\Vert = 1 + \left\Vert x\right\Vert + \left\Vert y\right\Vert$ Thus, we transfer (lift) the notion of “length” from bit sequences to combinatory logic terms in an appropriate way. Let us call it, e.g. the “norm” of the term. Thus, Chaitin's construction is grasped also as $\sum_{p \in \mathrm{Dom}_{\mathrm{nf}}} 2^{-\left\Vert p\right\Vert}$ where $\mathrm{nf} : \mathrm{CL} \supset\!\to \mathrm{CL}$ is a partial function defined on CL terms, it attributes to each "terminating" term its normal form. Thus, we have no notions of “bit sequence”,“code”, “coding”, “decoding” at all. But their ghosts still haunt us: the definition of norm function looks rather strange without thinking on the fact that is was transferred from a concept of coding. ### 3.1 More natural norm functions (from CL terms) Question: If we already move away from the approaches referring to any code concept, then could we define norm in other ways? E.g. $\left\Vert\cdot\right\Vert : \mathrm{CL}\to\mathbb N$ $\left\Vert\mathbf K\right\Vert = 1$ $\left\Vert\mathbf S\right\Vert = 1$ $\left\Vert\left(x\;y\right)\right\Vert = 1 + \left\Vert x\right\Vert + \left\Vert y\right\Vert$ And is it worth doing it at all? The former one, at leat, had a good theoretical foundation (based on analysis, arithmetic and probability theory). This latter one is not so cleaner, that we should prefer it, so, lacking theoretical grounds. What I really want is to exclude conceptually the notion of coding, and with it the notion of “syntactically incorrect versus syntactically correct but diverging”. Thus, taking into account only syntactically correct things, seeing only the choice of terminating versus non-terminating. Thus taking only termination vs nontermination into account, when calculating Chaitin's construction. What I want to preserve: • it can be interpreted as a probability • it is a normal number, as if its digits (in binary representation) were generated by tossing a coin thus I do not want to spoil these features. #### 3.1.1 Table for simpler CL-terms Let us not take into account coding and thus excluding the notion of “syntactically incorrect coding” even conceptually. Can we guess a good norm? Binary tree pattern Maximal depth, vertices, edges Leafs, branches So many CL-terms = how to count it Terminating, ratio So many till now, ratio till now $\cdot$ 0, 1, 0 1, 0 2 = 2 2, 1 2, 1 $\left(\right)$ 1, 3, 2 2, 1 $4 = 2\cdot2$ 4, 1 6, 1 $\cdot\left(\right)$ 2, 5, 4 3, 2 $8 = 2\cdot2^2$ 8, 1 14, 1 $\left(\right)\cdot$ 2, 5, 4 3, 2 $8 = 2^2\cdot2$ 8, 1 22, 1 $\left(\right)\left(\right)$ 2, 7, 6 4, 3 $16 = 2^2\cdot2^2$ 16, 1 38, 1 ## 4 Implementation To do: Writing a program in Haskell -- or in combinatory logic:-) -- which could help in making conjectures on combinatory logic-based Chaitin's constructions. It would make only approximations, in a similar way that most Mandelbrot plotting softwares work. The analogy: • they ask for a maximum limit of iterations, so that they can make a conjecture on convergence of a series; • this program will ask for the maximum limit of reducton steps, so that it can make a conjecture on termination (having-normal-form) of a CL term. Explanation for this: non-termination of each actually examined CL-term cannot be proven by the program, but a good conjecture can be made: if termination does not take place in the given limit of reduction steps, then the actually examined CL-term is regarded as non-terminating. ### 4.1 Architecture A CL term generator generates CL terms in “ascending order” (in terms of a theoretically appropriate “norm”), and by computing the norm of each CL-term, it approximates Chaitin's construction (at a given number of digits, and according to the given maximal limit of reduction steps). ### 4.2 User interface chaitin --model-of-computation=cl --encoding=tromp --limit-of-reduction-steps=500 --digits=9 --decimal chaitin --model-of-computation=cl --encoding=direct --limit-of-reduction-steps=500 --digits=9 --decimal ### 4.3 Term generator module CLGen where import Generator (gen0) import CL (k, s, apply) direct :: [CL] direct = gen0 apply [s, k] module Generator (gen0) where import PreludeExt (cross) gen0 :: (a -> a -> a) -> [a] -> [a] gen0 f c = gen f c 0 gen :: (a -> a -> a) -> [a] -> Integer -> [a] gen f c n = sizedGen f c n ++ gen f c (succ n) sizedGen :: (a -> a -> a) -> [a] -> Integer -> [a] sizedGen f c 0 = c sizedGen f c (n + 1) = map (uncurry f) \$ concat [sizedGen f c i cross sizedGen f c (n - i) \| i <- [0..n]] module PreludeExt (cross) where cross :: [a] -> [a] -> [(a, a)] cross xs ys = [(x, y) \| x <- xs, y <- ys]	2013-05-25 08:18:16	{"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": 39, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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": 0.7718077898025513, "perplexity": 2157.4627647870348}, "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-2013-20/segments/1368705790741/warc/CC-MAIN-20130516120310-00066-ip-10-60-113-184.ec2.internal.warc.gz"}
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Moreover, it is the supremum of the modulus. Check to see if: 1) $1$ is a limit point of $\{a_n\}$, i.e., does every neighborhood of $1$ contain infinitely many points of the sequence; and 2) Is it the case that for any $\epsilon>0$ there are only finitely many terms larger than $1+\epsilon$. In fact, is the largest lower bound (or infimum) of the sequence, and the larger a natural number we choose, the closer the sequence element will be to. It’s consequently not completely absurd to suggest that the sequence approaches in such a way, that we may . Limit infimum of a sequence for academic writing Rated 4/5 based on 56 review Newest 'supremum-and-infimum' Questions - Page 4 - Mathematics Stack Exchange	2019-12-06 00:59:26	{"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": 0.7192657589912415, "perplexity": 978.2694213102735}, "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/1575540482954.0/warc/CC-MAIN-20191206000309-20191206024309-00075.warc.gz"}
http://www.ams.org/books/memo/year/1990/	#### How to Order For AMS eBook frontlist subscriptions or backfile collection purchases: 2. Complete and sign the license agreement. 3. Email, fax, or send via postal mail to: Customer Services American Mathematical Society 201 Charles Street Providence, RI 02904-2294 USA Phone: 1-800-321-4AMS (4267) Fax: 1-401-455-4046 Email: cust-serv@ams.org Visit the AMS Bookstore for individual volume purchases. # Memoirs of the American Mathematical Society The Memoirs of the AMS series is devoted to the publication of research in all areas of pure and applied mathematics. The Memoirs is designed particularly to publish long papers or groups of cognate papers in book form, and is under the supervision of the Editorial Committee of the AMS journal Transactions of the AMS. View other years and numbers: ### Volume 88 #### Title MEMO/0437 MEMO/0436 Reduction, symmetry, and phases in mechanics - J. Marsden, R. Montgomery and T. Ratiu ### Volume 87 #### Title MEMO/0435 Hecke algebras - Aloys Krieg MEMO/0434 Homotopy formulas in the tangential Cauchy-Riemann complex - François Trèves MEMO/0433 Newton polyhedra of ideals - Boris Youssin ### Volume 86 #### Title MEMO/0432 The maximal factorizations of the finite simple groups and their automorphism groups - Martin W. Liebeck, Cheryl E. Praeger and Jan Saxl MEMO/0431 A multiple disjunction lemma for smooth concordance embeddings - Thomas Gehret Goodwillie ### Volume 85 #### Title MEMO/0430 Stability in modules for classical Lie algebras—a constructive approach - G. M. Benkart, D. J. Britten and F. W. Lemire MEMO/0429 MEMO/0428 Almost periodic measures - Jesús Gil de Lamadrid and Loren N. Argabright ### Volume 84 #### Title MEMO/0427 MEMO/0426 A lattice of chapters of mathematics (interpretations between theorems [theories]) - Jan Mycielski, Pavel Pudlák and Alan S. Stern MEMO/0425 Points de Nash des ensembles sous-analytiques - Wiesław Pawłucki ### Volume 83 #### Title MEMO/0424 Index for $C^*$-subalgebras - Yasuo Watatani MEMO/0423 MEMO/0422 The structure of certain quasisymmetric groups - A. Hinkkanen MEMO/0421 Unfoldings and bifurcations of quasi-periodic tori - H. W. Broer, G. B. Huitema, F. Takens and B. L. J. Braaksma MEMO/0420 Brownian motion on nested fractals - Tom Lindstrøm MEMO/0419 A filtered category ${\scr O}_S$ and applications - Ronald S. Irving	2015-08-30 02:47:25	{"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": 0.5349457859992981, "perplexity": 11095.151713059064}, "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-2015-35/segments/1440644064865.49/warc/CC-MAIN-20150827025424-00029-ip-10-171-96-226.ec2.internal.warc.gz"}
https://bitbucket.org/xemacs/auctex/src/6121914f0d88?at=auctex-1_28	# auctex / Filename Size Date modified Message etc texi 54 B 2.1 KB 60.3 KB 4.1 KB 2.4 KB 7.5 KB 968 B 6.8 KB 7.7 KB 129.8 KB 48.3 KB 113.0 KB 234 B 4.8 KB 456 B 60.6 KB 12.9 KB 39.2 KB 47.3 KB 1.9 KB 2.8 KB 96.3 KB 14.9 KB Introduction to AUC TeX *********************** This section of the AUC TeX manual gives a brief overview of what AUC TeX is, and the section is also available as a README' file. It is not an attempt to document AUC TeX. Real documentation for AUC TeX is available in the rest of the manual, which you can find in the doc' directory. AUC TeX. It is identical to the Installation chapter in the AUC TeX manual. If you are upgrading from the previous version of AUC TeX, the latest changes can be found in the CHANGES' file. If you are upgrading from an older version, read the History chapter in the AUC TeX manual. AUC TeX is a comprehensive customizable integrated environment for writing input files for LaTeX using GNU Emacs. AUC TeX lets you run TeX/LaTeX and other LaTeX-related tools, such as a output filters or post processor from inside Emacs. Especially running LaTeX' is interesting, as AUC TeX lets you browse through the errors TeX reported, while it moves the cursor directly to the reported error, and displays some documentation for that particular error. This will even work when the document is spread over several files. AUC TeX automatically indents your LaTeX-source', not only as you write it -- you can also let it indent and format an entire document. It has a special outline feature, which can greatly help you getting an overview' of a document. Apart from these special features, AUC TeX provides a large range of handy Emacs macros, which in several different ways can help you write your LaTeX documents fast and painlessly. All features of AUC TeX are documented using the GNU Emacs online documentation system. That is, documentation for any command is just a key click away! AUC TeX is written entirely in Emacs-Lisp, and hence you can easily particular employment or project (apart from the AUC TeX project itself). AUC TeX is distributed under the GNU Emacs General Public License' and may therefore almost freely be copied and redistributed. The next sections are a short introduction to some actual' features. For further information, refer to the built-in online documentation of AUC TeX. Indentation and formatting ========================== AUC TeX may automatically indent your document as you write it. By pressing lfd instead of ret at the end of a line, the current line is indented by two spaces according to the current environment level, and the cursor is moved down one line. By pressing tab, the current line is indented, and the cursor stays where it is. The well-known Emacs feature format-paragraph' (M-q') is reimplemented especially for AUC TeX to follow the indentation. A special command LaTeX-fill-buffer' lets you indent an entire document like the well-known C utility indent (this time, only according to the LaTeX structure :-). Completion ========== By studying your \documentclass' command (in the top of your document), and consulting a precompiled list of (La)TeX symbols from a large number of TeX and LaTeX files, AUC TeX is aware of the LaTeX commands you should be able to use in this particular document. This knowledge' of AUC TeX is used for two purposes. 1. To make you able to complete' partly written LaTeX commands. You may e.g. write \renew' and press M-tab' (TeX-complete-symbol'), and then AUC TeX will complete the word \renewcommand' for you. In case of ambiguity it will display a list of possible completions. 2. To aid you inserting environments, that is \begin - \end pairs. This is done by pressing C-c C-e (LaTeX-environment), and you will be prompted for which environment' to insert. ===================== A number of more or less intelligent keyboard macros have been defined to aid you editing your document. The most important are listed below. LaTeX-environment' (C-c C-e') Insert a \begin{}' -- \end{}' pair as described above. LaTeX-section' (C-c C-s') Insert one of \chapter', \section', etc. TeX-font' (C-c C-f C-r', C-c C-f C-i', C-c C-f C-b') Insert one of \textrm{ }'), \textit{ \/}' \textbf{ }' etc. A number of additional functions are available. But it would be far too much to write about here. Refer to the rest of the AUC TeX documentation for further information. Running LaTeX ============= When invoking one of the commands TeX-command-master' (C-c C-c') or TeX-command-region' (C-c C-r') LaTeX is run on either the entire current document or a given region of it. The Emacs view is split in two, and the output of TeX is printed in the second half of the screen, as you may simultaneously continue editing your document. In case TeX find any errors when processing your input you can call the function TeX-next-error' (C-c ') which will move the cursor to the first given error, and display a short explanatory text along with the message TeX gave. This procedure may be repeated until all errors have been displayed. By pressing C-c C-w' (TeX-toggle-debug-boxes') you can toggle whether the browser also should notify over-full/under-full boxes or not. Once you've successfully formatted your document, you may preview or print it by invoking TeX-command-master' again. Outlines ======== Along with AUC TeX comes support for outline mode for Emacs, which lets you browse the sectioning structure of your document, while you will still be able to use the full power of the rest of the AUC TeX functionality. Availability ============ ftp://ftp.iesd.auc.dk/pub/emacs-lisp/auctex.tar.gz' In case you don't have access anonymous ftp, you can get it by email requests to <ftpmail@decwrl.dec.com>'. WWW users may want to check out the AUC TeX page at http://www.iesd.auc.dk/~amanda/auctex/' Contacts ======== There has been established a mailing list for help, bug reports, feature requests and general discussion about AUC TeX. You're very welcome to join. Traffic average at an article by day, but they come in bursts. If you are only interested in information on updates, you could refer to the newsgroups comp.text.tex' and gnu.emacs.sources'. If you want to contact the AUC TeX mailing list, send mail to <auc-tex-request@iesd.auc.dk>' in order to join. Articles should be sent to <auc-tex@iesd.auc.dk>'. To contact the current maintainers of AUC TeX directly, email <auc-tex_mgr@iesd.auc.dk>'. AUC TeX development c/o Kresten Krab Thorup Mathematics and Computer Science University of Aalborg DK 9000 Aalborg Denmark ` Tip: Filter by directory path e.g. /media app.js to search for public/media/app.js. Tip: Use camelCasing e.g. ProjME to search for ProjectModifiedEvent.java. Tip: Filter by extension type e.g. /repo .js to search for all .js files in the /repo directory. Tip: Separate your search with spaces e.g. /ssh pom.xml to search for src/ssh/pom.xml. Tip: Use ↑ and ↓ arrow keys to navigate and return to view the file. Tip: You can also navigate files with Ctrl+j (next) and Ctrl+k (previous) and view the file with Ctrl+o. Tip: You can also navigate files with Alt+j (next) and Alt+k (previous) and view the file with Alt+o.	2014-04-20 20:36:42	{"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": 0.971062421798706, "perplexity": 7311.608928680762}, "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/1397609539066.13/warc/CC-MAIN-20140416005219-00264-ip-10-147-4-33.ec2.internal.warc.gz"}
https://blog.jverkamp.com/2018/12/31/2018-reading-list-retrospective/	# 2018 Reading List Retrospective Another year of reading. Still not back to my glory days of 100 books in a year, but otherwise the second best I’ve done (beating out 91 in 2016!). A solid year. I’m going to see if I can purposely try to hit 100 again next year though. Should be doable. 😄 That’s… quite a few books… A few stats: 92 books in a year is right about a book every 4 days. I think next year, I can get that down to 3 days. Among books with page counts, I read 43,221 pages for an average of just under 120 pages per day (which is actually higher than previous years. I can probably get that up a bit too. I read three books with 100-200 books each, all from the Chronicles of Amber: The Hand of Oberon, The Courts of Chaos, and Sign of the Unicorn. Conversely, I read 5 books with over 1000 pages, The Shadow Rising and Lord of Chaos from Wheel of Time and all three books of the Stormlight Archive (Oathbringer is over 1200…). The average page count was just shy of 500 pages per book. No short stories this year (at least not counted individually). Formatwise, I listened to 38 books this year (around half of my total)1: Most of that was dedicated to finishing Discworld and then listening to the entire Wheel of Time. I started the Eye of the World on May 22 and finally finished A Memory of Light September 11 while walking. I ended up walking 6 miles that day, just finishing the book. All together, that’s 14 books and 11537 pages over 112 days, although between each I caught up in podcasts I’d missed. That’s 100 pages per day right there. Not bad. My average rating remains high. I rated 38 books as 4 stars and 47 as 5. What can I say, I read books I expect to like, especially those that I re-read. I wonder if I should try some lower rated ones… Unfortunately, there are too many series I still haven’t finished and that I’ll be pushing off another year: I over estimated last year and didn’t read as many new books as I’d like. So it goes. I’m also probably going to read more non-fiction this year. We’ll see how that goes. Here’s the full list: 1. cat listened-to.json \| jq '.books \| sort_by(.started) \| .[] \| select(.started >= "2018-01-01") \| .title' \| awk -F'(' '{ print \$1 }' \| tr -d '"' \| trim [return]	2019-11-19 23:11:36	{"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": 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": 0.2724190354347229, "perplexity": 2776.7514826266356}, "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-2019-47/segments/1573496670268.0/warc/CC-MAIN-20191119222644-20191120010644-00247.warc.gz"}
https://www.nature.com/articles/s41598-022-12332-z	Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript. # Effect of electrochemical oxidation and drug loading on the antibacterial properties and cell biocompatibility of titanium substrates ## Abstract A combination of $${\text{ TiO}}_{2}$$ nanotube array (TON) and controlled drug release system is employed to provide enhanced surface properties of titanium implants. Electrochemical anodization process is used to generate TON for introducing, vancomycin, an effective antibacterial drug against Staphylococcus aureus. TON loaded vancomycin is then coated with a number of layers of 10% gelatin using spin coating technique. The gelatin film is reinforced with graphene oxide (GO) nanoparticles to improve the surface bioactivity. The surface of the samples is characterized by field emission electron microscopy (FESEM), energy-dispersive X-ray spectroscopy (EDS), and contact angle measurement. The results illustrate that the TON was constructed and vancomycin molecules are successfully loaded. The drug release study shows that the amount of released vancomycin is controlled by the thickness of gelatin layers. With an increase in gelatin film layers from 3 to 7, the release of vancomycin in the burst release phase decreased from 58 to 31%, and sustained release extended from 10 to 17 days. The addition of GO nanoparticles seems to reduce drug release in from 31 to 22% (burst release phase) and prolonged drug release (from 17 to 19 days). MTT assay indicates that samples show no cytotoxicity, and combination of GO nanoparticles with gelatin coating could highly promote MG63 cell proliferation. Soaking the samples in SBF solution after 3 and 7 days demonstrates that hydroxy apatite crystals were deposited on the TON surface with GO-gelatin coating more than surface of TON with gelatin. Moreover, based on the results of disc diffusion assay, both samples (loaded with Vancomycin and coated with gelatin and gelatin-GO) with the inhibition zones equaled to 20 mm show effective antibacterial properties against S. aureus. The evidence demonstrates that titania nanotube loaded with vancomycin and coated with gelatin-GO has a great potential for general applicability to the orthopedic implant field. ## Introduction Titanium and its alloys are among the most widely used metallic materials in the field of orthopedic and dental implants1,2,3. Desirable properties, including tensile strength and elastic modulus, high biocompatibility, and low risk of allergy make them successful candidate for these applications4,5,6. However, Ti-based implants some deficiencies including poor osteointegration, low osteogenesis, and bacterial infection in the implant site which may lead to implants failure especially in prolonged use7,8,9. Generally, poor osteointegration of Ti-based implants occurs because of bio-inert surface, production of excessive reactive oxygen species (ROS) at the interface, and bacterial infection after surgery10. As a result, two strategies are commonly considered to moderate these problems including increasing osteoblast cell proliferation and inhibition of bacterial infection11,12. Long-term intravenous or oral antibiotics are the main traditional approaches to treat bacterial infection after implant surgery13, which have severe drawbacks including harmful side effects, toxicity, uneven bio-distribution, and less bioavailability14,15. Staphylococcus aureus is considered the main bacteria that causes implant infection after surgery16,17,18. The secretion of these bacteria on implant surfaces forms biofilms, which protect them against the immune system and antibacterial agents. Consequently, treatment with antibiotics has become a vital issue19. Since bacteria and osteoblast cells are in competition for attachment, implants having antibacterial surface characteristics can decrease bacterial attachment and colony formation so they can be used for treating bone defects20,21. Advanced biomaterials equipped with sustained and localized release of antibacterial agents could promote the healing/regeneration process of tissues which are more susceptible to bacterial infections (e.g. bone, skin, cardiac tissue)22,23,24,25. With the convergence of material science and biology, a combination of surface modification and drug delivery systems can be employed to tackle the above mentioned complications. The electrochemical anodizing is a prevalent process among surface modifications that constructs $${\text{TiO}}_{2}$$ nanotube arrays (TON) on titanium substrates26,27,28,29. TON arrays have beneficial properties including hosting a vast range of drugs and higher biocompatibility due to their highly porous nanostructures. Moreover, their fabrication is low cost and simple29,30,31,32,33. The main disadvantage of TON arrays as drug nano-reservoirs is uncontrollable drug release behavior that may cause toxicity in the implant site. However, modified TON, such as TON coated with biopolymers, exhibits enhanced drug release behavior34,35,36. TON coated by biopolymers can provide more controllability on drug release behavior by selecting a biopolymer based on composition and degradation rate properties37. In addition, the polymer coating can inherently have osteointegration and antibacterial properties38,39,40, and they can also carry a second drug as a multidrug system41,42,43,44,45. In this study, in order to delay the drug release while enhancing the bioactivity of titanium substrates, a multilayer nanocomposite hydrogel was used as a cover on the TON surface. Among the biocompatible hydrogels, gelatin with FDA approval serves as the most important biopolymer which frequently used in biomedical applications, especially in tissue engineering and drug delivery researches46. According to literatures and our previous experience, GO, as hydrophilic nanoparticle provides many advantages in combination with hydrogels as a nanocomposite structure47,48. In dental and bone tissue engineering, GO nanoparticles enhance osseointegration/osteoblast bioactivity via acceleration of apatite formation49. In addition, graphene and its derivatives show great potential for osteogenic differentiation of stem cells50 and have been suggested as a coating agent on the orthopedic implants51,52. Here, we aimed to superpose gelatin and GO properties in order to modify TON surface where the coated gelatin/GO multilayer structure improves osteogenic properties at the same time as delaying drug release from nanotube reservoir. For this purpose, TON arrays were formed by anodizing process on the surface of titanium substrates and Vancomycin, an antibacterial drug against S. aureus was introduced into nanotube arrays53,54. Graphene oxide-gelatin nano-composite was used to coat TON arrays loaded with Vancomycin. The modified titanium samples then were characterized by field emission scanning electron microscopy (FESEM), energy-dispersive X-ray spectroscopy (EDX), wettability measurement, and degradability. In addition, the drug release behavior was assessed on samples having different layers of coating. The effect of surface modification of titanium samples was assessed on the growth of MG63 cells and formation of hydroxyapatite crystals by MTT assay and bioactivity test. Finally, disc diffusion assay was employed to elucidate antibacterial activity of the samples. ## Experimental ### Materials Commercially available pure titanium (CP-Ti-grade 2) used in this study were supplied by McMaster Carr Company, Los Angeles, CA, USA. Graphene oxide was obtained from Central Laboratory of Amirkabir University, Iran, Tehran. Vancomycin Hydrochloride was purchased from Daana pharmaceutical company, Iran, Tehran. Dulbecco’s modified eagle medium (DMEM) and trypsin were purchased from Gibco BRL (France). Fetal bovine serum (FBS), MTT, DMSO, PBS, and penicillin/streptomycin (PS) were purchased from Sigma-Aldrich. Simulated body fluid (SBF) was provided by APATECH, Iran, Yazd. Other chemicals were supplied by Merck, Germany. ### Preparation of $${\text{TiO}}_{2}$$ nanotube arrays (TON) Highly-ordered titania nanotube arrays were fabricated on titanium discs of 10 mm diameter using electrochemical anodization method. The samples were first grounded with SiC abrasive paper (grit size: P800, P1000, P1200, and P1500) and sequentially cleaned in ethanol 70%, acetone, and deionized water using an ultrasonic bath for 25 min. The samples were dried at ambient temperature before anodization. For anodization process, stainless steel and titanium discs having smooth surfaces were used as cathode and anode electrodes, respectively. The cell was connected to DC power supply (PEQ lab, EV843, made in Belgium and ethylene glycol, 38 wt% ammonium fluoride, and 2 vol% deionized water was used as electrolyte solution. Processing parameters including temperature, voltage, and time have significant effects on the structure of anodized titanium. The anodization process was carried out at voltages of 45, 60, 70, and 95, for 1.5 and 3 h (Table 1), at a controlled temperature (4 °C). All anodized samples were ultrasonically cleaned in ethanol and acetone for a minute and rinsed with deionized water. Finally, samples were dried at room temperature. ### Vancomycin loaded into TON arrays on Ti surface The Vancomycin was introduced into titania nanotubes according to a method of pipetting and drying. 25 mg/ml solution of Vancomycin in phosphate buffer saline (PBS) was prepared and 10 µL of drug solution was pipetted onto nanotubes. The samples were dried at air temperature and cleaned with soft tissue. Finally, to remove the drug residues, the samples were rinsed with the PBS solution. To enhance the drug loading efficiency and achieve the required concentration, the drug loading process was repeated four times. ### Coating of gelatin-based film on TON arrays Gelatin-based films with different thicknesses were coated on titania nanotube substrates using spin-coating method. A 10 wt% gelatin aqueous solution was prepared and spin-coated on the substrates at a speed of 3800 RPM for 60 s. The polymer film thickness would affect sustained release behavior of the drug, therefore, samples with varied coating layers were prepared to obtain desired drug release profile. To modify gelatin film, GO was added as an agent. For GO/gelatin composite synthesis, GO nano-powders were first dispersed in deionized water using an ultrasonic homogenizer for 30 min. The gelatin10%-GO0.5% solution was then prepared and spin-coated on the samples. All the samples were subsequently glutaraldehyde vapor treated with glutaraldehyde 25%, at 25 °C. To remove excess glutaraldehyde, samples were then immersed in a glycine aqueous solution (7.5 mg/mL) for 5 min and dried at room temperature. Finally, for sterilization of the samples Ultraviolet radiation was used for 20 min. Table 2 lists the samples with their abbreviations. ### Surface characterization Field emission scanning electron microscopy (FESEM) was utilized to observe the morphology of anodized samples as well as gelatin-based coated ones. After the samples were coated with an ultrathin gold layer via sputter coater (Emitech, K450X), images were captured at an accelerating voltage of 25 kV in high vacuum mode. The titania nanotube dimensions were analyzed using the Digimizer software. The non-destructive ellipsometry (SENpro, SENtech, Germany) test and SEM analysis were performed to measure the thickness of polymer coating. Energy-dispersive X-ray spectroscopy (EDS) was used to determine the types of elements in the samples (TON and TON-Van), especially to verify the presence of Vancomycin. ### Wettability measurement The sessile water drop technique was used to measure the static contact angle and wettability of the samples. Deionized water droplet (0.5 µL) was firstly deposited on the top surface of samples through micro-syringe. Images were then captured and the static contact angle was measured using CAG-20, Jikan Co. The experiment measured the average contact angle at five different points in each sample at ambient temperature. Degradability of coating has a direct effect on drug release behavior of the samples. The degradation of biopolymers is related to their composition, structure, charge, and surface properties. In general, weight loss measurement is the most common method to investigate hydrogel degradation. However, in this study, weight loss measurement was not applied because there was a significant difference between coating weight and titanium weight. Therefore, degradability was evaluated by comparing the SEM images of coating before and after drug release in PBS solution. Samples were placed in clean and sterile glass bottles containing PBS. Then they were sealed and placed in an incubator (GFL, Germany) at 37 °C, at a speed of 60 RPM. The degradation was measured after 10 and 17 days. Eventually, freeze-drying was performed to remove the water that was absorbed by the hydrogel coatings. ### Vancomycin release In-vitro release of Vancomycin from bare TON, Gelatin coated, and gelatin-GO coated samples loaded with the drug were investigated by immersion into 10 mL of PBS (pH = 7.4) at 37 °C. During the first 6 h, 300 µL of the medium was taken out at short intervals to monitor the burst release of the drug. For delayed release of the drug, 300 µL PBS solution was collected every 24 h until the whole drug was released into the medium. The calibration curve was plotted based on the Vancomycin in PBS, the release of the drug was measured by UV-spectroscopy (Lambda 900, PerkinElmer, USA) at 280 nm. The release was considered complete when there was no change in the absorbance at 280 nm. The data were analyzed using mathematical kinetic modeling. ### Cell culture The osteosarcoma cell line Human Osteoblast-Like cells (HOS) MG-63, from the National Cell Bank of Iran (NCBI; Pasteur Institute), was used for cell culture experiments. The cells were cultured in a culture medium (DMEM supplemented with 10% FBS, 1% penicillin/streptomycin) at 37 °C in the humidified incubator with 5% CO2. After the culture reached approximately 90% confluence, MG-63 cells were removed from culture flasks by trypsinization and centrifuging (1500 RPM, 5 min). They were then suspended in the fresh medium. ### Cell viability Samples including TON, TON-Van-Gel C7, TON-Van-Gel-GO and culture plate as a control were sterilized by UV and washed with PBS and culture medium before they were placed in a 24-well plate. Then, 200 µL of culture medium containing 80,000 cells were seeded onto the surface of the samples. After incubation for 3 h, 1 mL of culture medium was added to each well. Subsequently, after 1, 3, and 7 days of incubation, to remove the unattached cells, the culture medium was removed and samples were rinsed with sterile PBS twice and incubated with fresh culture medium supplemented with 1 mg/mL MTT solution for 4 h to allow formazan formation. After removing the medium, 50 µL of dimethyl sulfoxide (DMSO) solution was added to each well and incubated for 10 min to dissolve the formazan crystals. The medium was transferred to a 96-well plate to measure the absorbance of the resulting purple solution using a micro-plate reader at 580 nm. The cell viability directly depends on the amount of formazan crystals, calculated by dividing the average optical density of samples by the average optical density of control. ### Bioactivity Samples (TON-Van-Gel and TON-Van-Gel-Go) were evaluated by soaking in 5 mL of SBF solution similar to human blood plasma. The samples were kept at 37 °C for 3 and 7 days. The medium was replaced every 2 days to avoid insufficient presence of ions (Ca and P). The samples were removed from the SBF solution and dried using freeze-drying. The structure and morphology of the SBF-immersed samples were characterized by FESEM. ### Antibacterial assay A zone of inhibition test, also called a Kirby-Bauer Test, was utilized to investigate the antibacterial characteristics of the samples. Gram-positive S. aureus (ATCC 25923), one of the most infectious bacteria in bone tissue engineering was employed. Plates containing Mueller–Hinton agar medium (Merck, Germany) were spread with S. aureus at a concentration of 1.5 × 108 CFU/mL. Then, the samples (TON-Van-Gel-GO and TON-Van-Gel C7) were placed gently into the agar nutrient. Ciprofloxacin antibiotic discs were placed on the plates as a positive control. Finally, the plates were incubated for 24 h at 37 °C in order for bacteria to grow in agar media. The size of the inhibition zone that appeared around the discs indicates the antibacterial activity of each sample. ### Statistical analysis Statistical analysis was carried out using SPSS software (v 17.0; IBM New York, NY, USA) when statistical differences were detected, a t-student comparison test was performed. Data are reported as mean ± SD at a significance level of p < 0.05. ## Results and discussion ### Fabrication and characterization of titania nanotube arrays Recently, more attention has been focused on the electrochemical anodization to construct self-order titania nanotube structures. Strong adherent nanotube structures with open tops and closed ends are the most favorable drug nano-carrier systems. The parameters such as voltage and time of the anodization process have the main impact on the morphology of titanium dioxide. Here, the impact of anodizing parameters on the morphology of titania, which has great importance for the consequent drug loading was investigated. The anodizing process with different voltages (45, 60, 70, 90 V) and various times (1.5, 3 h) was carried out. The experiments were performed in the electrolyte containing ammonium fluoride at the controlled temperature (4 °C). The first reaction of anodizing oxidation was the electrolysis of water (reaction 1). A compact layer of titanium dioxide was then formed on the titanium substrates (reaction 2) which was dissolved by fluoride ions, subsequently. As the results, the pits were formed on the titanium surfaces (reaction 3). At the suitable voltage and time, pits were converted into nanotube structures. At the lowest applied voltage (45 V), the pits with infinitesimally small diameters were formed on the Ti surfaces, however, the nanotube morphology was not apparent (Fig. 1a). When the voltage was increased to 60 V, the titania nanotubes were formed, yet they were not highly ordered (Fig. 1b). At the voltage of 70 V, the nanotubes showed a more distinguishable morphology with increased diameter (Fig. 1c). $$2{\text{H}}_{2} {\text{O }} \leftrightarrow {\text{O}}_{2} + 4{\text{H}}^{ + } + 4{\text{e}}$$ (1) $${\text{Ti}} + {\text{ O}}_{2} { } \leftrightarrow {\text{ TiO}}_{2}$$ (2) $${\text{TiO}}_{2} + 6{\text{F}}^{ - } + { }4{\text{H}}^{ + } \leftrightarrow { }[{\text{TiF}}_{6} ]^{2 - } { } + { }2{\text{H}}_{2} {\text{O }}$$ (3) With increasing time of anodizing from 1.5 to 3 h, no significant changes were observed in the nanotube structure at the low voltage (45 V) (Fig. 1d). In the higher voltage though, extending the anodizing time made the nanotubes provided a more intrinsic structures (Fig. 1e,f). Increasing the voltage to 90 V was gradually destroyed the top portion of the nanotubes (Fig. 1g). A more desirable morphology of nanotubes was obtained at the anodizing voltage of 70 for 3 h, where the SEM images revealed that they possess a diameter of about 94 ± 4 nm and length of about 3 µm, aligned vertically and highly-ordered with opened top and closed bottom (Fig. 1h,i). ### EDS analysis of TON and TON-Van EDS was utilized to prove that Vancomycin was loaded into titania nanotube arrays. Chemical compositions of samples (TON and TON-Van) were detected, as illustrated by Fig. 2a,b. Ti and O elements could be seen in both samples, confirming $${\text{TiO}}_{2}$$ formation on the surface of titanium samples. The presence of Cl, N, and C in the EDS spectrum of TON-Van (Fig. 2a) indicates that Vancomycin (C66H75Cl2N9O24) has been successfully loaded into the nanotube arrays. ### Current transient curve and anodizing process evaluation Current transient curves demonstrate the behavior of current versus time during the anodizing process. This curve can help to explain the formation of titania structure during the process. Figure 3 displays the current transient curve at the voltage of 70, for 3 h, at the temperature of 4 °C. The application of initial voltage showed a sharp rise up to a which is attributed to the electrolysis of water and the beginning of the formation of the compact oxide layer. The observed gas evolution in the anode cell seems to be related to the transfer of electric charge and indicates the electrical conductivity of the titanium surfaces. Subsequently, the current declines sharply due to the formation of a dense and compact oxide layer with a low electrical conductivity. As the transfer of electric charge decreased, ion transportation in the electrolyte increased. At the next step the compact layer was dissolved by fluoride ions and pores in the dense layer were nucleated which causes a slight increase in the current, followed by a dropdown. However, the small increase in current was not observed here due to the fast ion transportation (especially fluoride ions) and rapid dissolution of the oxide layer. The other reason could be the limitation on the time resolution of our measurement device to record the rapid changes in the current. The decrease in current would gradually turn to a plateau. There was a competition between oxidation and dissolution of the oxide layer which controls the anodizing process. Nanotubes were formed on the compact oxide layer as the anodization proceeds. As can be seen in Fig. 3, current increases to higher values with the increase of voltage. This is due to the increase of electron exchange during the anodization process. As the voltage of the anodization process increases, the oxidation rate improves and the exchange of electrons and current density enhances. ### Water contact angle measurement Surface hydrophilicity or wettability of implants has a significant impact on cell behavior. Moreover, the water-soluble drug loading efficiency also improves with increasing wettability. Therefore, contact angle measurement, the primary way to investigate the degree of affinity between water and the surfaces, becomes vital to characterize the surfaces of drug delivery implants. Here, the wettability of samples (TON, TON-Gel C7, and TON-Gel-GO) was studied and depicted in Table 3. The TON was hydrophilic due to water penetration into the nanotubes. The contact angle of TON-Van-Gel C7 and TON-Gel-Go were 71.9° and 70.2°, respectively, which are in a similar hydrophilic range (smaller than 90°) (Fig. 4a,b). The similar contact angle indicates that even by modifying gelatin with graphene oxide as a hydrophilic agent, the contact angle was unchanged. This could be partly due to graphene oxide nanoparticles which fill and block the micro-pores among the gelatin molecules. The micro-pores structure provides the channels for penetration and absorption of water molecules. This could also be related to the strong crosslink between the chains of polymer which prevents the water absorption. The degradation of coating is an important factor for drug release behavior, therefore, the in-vitro degradation of the TON-Van-Gel C7 was evaluated. FESEM was utilized to observe the changes in the microscopic morphology of gelatin coating during the degradation process in PBS. Figure 5a reveals the dense structure of gelatin coating with some cracks before degradation. These cracks are probably due to the process of freeze-drying. Figure 5b,c show no holes and collapse in the surface coatings after 10 and 17 days which implies the dense structure of the coating possibly due to the strong crosslink between the chains of polymer. ### Vancomycin release The effect of gelatin thickness on the release of Vancomycin was investigated from TON-Van, TON-Van-Gel C3, and TON-Van-Gel C7. The thickness of 3 and 7 layers of gelatin was 445 nm and 1038.06 nm, obtained by ellipsometry. As it can be seen in Fig. 6a,b, the ellipsometry results were confirmed by the SEM cross sectional images. Figure 7a shows the Vancomycin calibration curve (in PBS, at 280 nm) that used to obtain Vancomycin concentrations. The total amount of Vancomycin loaded into titania nanotubes was measured 490 µg/cm2 using a UV–Vis spectrophotometer. Figure 7b presents a comparative drug release profile of Vancomycin from TON-Van, TON-Van-Gel C3, and C7. As anticipated, the release behavior of Vancomycin from all the samples displayed biphasic behavior, including a burst release and sustained release. Drug molecules in the upper part of the nanotubes were released immediately in the media at the initial stage (burst release phase), inhibiting the bacterial invasion and improve the antibacterial efficacy in early hours of implantation. Drug release profile of the samples are provided in Table 4. In the burst release phase, the Vancomycin released from TON-Van was approximately 83% in the first hour whereas the released drug from TON-Van-Gel C3 and C7 were 58% and 31%, respectively. In the sustained release phase, the drug was released from TON-Van, TON-Van-Gel C3, and C7 in 1, 10, and 17 days, respectively. The slower release of Vancomycin from TON-Van in the second stage demonstrates that the nanotube structure could act as a nano-reservoir for the drug. In addition, the results show that the gelatin-coating reduced the amount of released drug in the burst release phase. With increasing the number of layers of spin-coated gelatin, the thickness would increase which resulted in the significant decrease in the rate of drug release. This is probably because the polymer chains could restrict the movement of drug molecules and limits their release. The limitation in the rate of drug release would extend the time which provides more control on drug release profile that enhances the antibacterial property of the samples. The effect of graphene oxide on the release profile of the Vancomycin was investigated by TON-Van-Gel-GO. As it can be seen in Fig. 7c, a biphasic behavior for GO-gelatin sample was detect. The percentage of drug release was decreased from 31 to 22% in the burst release stage by adding the GO nanoparticles (Table 4). The negative charge of graphene oxide and carboxyl groups could electrostatically interact with chains of gelatin. Hence, the physical links between gelatin chains create difficulty for Vancomycin molecules to get released into PBS which results in prolonged release period. In fact, the GO-gelatin coating could prolong Vancomycin release up to 19 days. In comparison, the GO-gelatin coated samples could provide longer and sustained drug release required for successful anti-infection bone therapy as compared to TON-Van and TON-Van-Gel C7 samples. In order to have an effective local drug delivery system, the concentration of antibacterial drug release in the media should be blow the toxic level and above the minimum inhibitory concentration (MIC). Here, Vancomycin did not show toxic behavior at 800 µg/mL for MG-63 Cells. Moreover, Vancomycin has been reported to have a MIC of 0.5–10 µg/mL for S. aureus55,56,57,58,59. As shown in Fig. 7d, the Vancomycin concentration for TON-Van-Gel-GO in the media was within the therapeutic window. This could mean that the drug delivery system should be able to kill the bacteria and avoid the formation of biofilm without interference with the cellular process. Mathematical modeling was used to investigate the mechanism of drug release and analyze the release kinetics. Model-dependent methods based on different mathematical functions were employed to select the best fit for Vancomycin release profile with a higher value. Table 5 summarizes the equation of mathematical functions used to fit the release experiments data including zero-order, first-order, Higuchi, and Hixson-Crowell. The obtained $${\text{R}}^{2}$$ values for the samples (TON-Van, TON-Van-Gel C7, and TON-Van-Gel-GO) are listed in Table 6. The comparison between the $${\text{R}}^{2}$$ values revealed that the Vancomycin release was perfectly fitted with the first-order model in all the samples. This is an indication for water-soluble drug release from a porous matrix or a matrix with diffusion-controlled release system. ### Cell viability The cell viability and proliferation on the surface of the samples were evaluated by MTT assay. Figure 8 shows MG-63 cell proliferation after 1, 3, and 7 days of culture. As it can be seen, no statistically significant difference between TON without coating and control sample was observed. Moreover, the number of viable cells on the surface of TON-Van was significantly lower than the control after the day 1. The gelatin-based coating showed to remarkably improve cell viability and the addition of GO nanoparticles had a beneficial impact on cell viability of the samples. It is known gelatin could provide cells with the biomimetic bone environment since it is the major protein of the extracellular matrix. Gelatin has the arginine–glycine–aspartic acid (RGD) sequence which is known for establishing cell-substrate interactions. Moreover, the dispersion of GO nanoparticles into gelatin increases the chance to create more cellular niches for osteoblast cells due to the enlargement of the surface area and the increase of surface roughness. It is known that the surface oxygen-containing functional groups trigger the adsorption of the surrounding serum proteins on the surface. This could be the reason that GO-enriched surfaces would absorb the exogenous proteins that results in eliciting efficient interactions with the cells51. It is shown that the presence of GO nanoparticles could engage in gene expression profile and upregulate mRNA expression levels of all osteogenic markers60,61. In addition, GO nanosheets could interfere in apatite nucleation and hydroxyl apatite (HA) formation. It is shown by Wang et al. this could be originated from ion interactions. The –OH and carboxyl groups from GO could attract Ca2+ and HPO4 2- of microenviroment solution resulting in the accelerated hydroxy apatite formation50. The TON-Van-Gel-GO sample showed the best cell substrate interactions with MG-63 cells. Here, Vancomycin at 800 µg/mL showed about 80% cell viability at day 1. Therefore, the results show no cytotoxicity whether by the samples or the dosage of the drug used according to ISO-10993-5. ### Bioactivity Previous studies have demonstrated that both pure titanium and TON surfaces are bio-inert62,63,64. FESEM was utilized to determine the formation of bone-like apatite on the samples (Gelatin and Gelatin-GO) soaked in SBF solution for 3 and 7 days. As shown in Fig. 9a, a few apatite nucleates were formed on the surface of gelatin coated sample after 3 days whereas GO-gelatin coated sample was covered by flake-like apatite (Fig. 9b). The negative charge of GO with deprotonation of –COOH and –OH groups could attract $${\text{Ca}}^{2 + }$$ ions and enhance the apatite nucleation50. After 7 days of immersion in SBF, the amount of apatite crystals on both samples were increased (Fig. 9c,d). However, the mineralization on the surface of GO reinforced sample was higher than the sample without GO, suggesting that the sample coated by gelatin and reinforced with GO has the highest bioactivity among the samples. The amino groups in gelatin can attract calcium ions followed by $${\text{PO}}_{4}^{3 - }$$ which could results in accelerated biomineralization process. GO also promotes the flake-like hydroxyapatite deposition through electrostatic interaction. It is fair to assume that TON-Van-Gel-GO sample, could provide a more stable connection between the surface of samples and bone tissues. ### The antibacterial assay The antibacterial characteristic of the samples was investigated by disc diffusion assay after 24 h of incubation against gram-positive S. aureus bacteria, the most common pathogenic bacteria in bone infections. Figure 10, shows the inhibition zones of the samples including TON-Van-Gel and TON-Van-Gel-GO. The inhibition zones diameters were also measured and depicted in Table 7 which confirm our observation. The inhibition zones of both samples were about 20 mm which indicates both samples provide Vancomycin delivery models that can inhibit S. aureus infection. It is apparent that Vancomycin could provide an effective antibacterial property against gram-positive bacteria such as S. aureus with minimal cytotoxicity. This data was in agreement with Liu et al.65 finding which confirmed that Vancomycin sustained delivery from titanium surface mediated by nanoparticles provided the similar inhibition zone and showed efficient antibacterial effect against S. aureus. The proper pattern of drug release is essential for the effective antibacterial ability of the implant`s surface. As discussed in 3.6, in this study, the concentration of released vancomycin from coating reached the concentration above the MIC in the first hour when the possibility of infection is higher. Also, based on Zhang et al.66 study which developed an electrospun Vancomycin-loaded coating on titanium substrate, the obtained profile of release and the concentration of the released Vancomycin during 19 days may support treating implant-associated infection in-vivo. ## Conclusion In the present study, a layer of $${\text{TiO}}_{2}$$ nanotubes on pure titanium samples was fabricated through electrochemical anodization in order to carry the vancomycin, as an antibacterial agent. A gelatin-based coating which was reinforced with graphene oxide was spin-coated on the surface of the samples to control the release profile while improve the biological activity of the samples. The abundant formation of apatite crystals on the Gelatin-GO substrate implies that TON-Van-Gel-GO has the superior bioactivity and possible improved osteointegration. 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Effect of surface alkali-based treatment of titanium implants on ability to promote in vitro mineralization and in vivo bone formation. Acta Biomater. 57, 511–523 (2017). 65. Liu, Z. et al. Construction of poly (vinyl alcohol)/poly (lactide-glycolide acid)/vancomycin nanoparticles on titanium for enhancing the surface self-antibacterial activity and cytocompatibility. Colloids Surf., B 151, 165–177 (2017). 66. Zhang, L. et al. Electrospun vancomycin-loaded coating on titanium implants for the prevention of implant-associated infections. Int. J. Nanomed. 9, 3027 (2014). ## Acknowledgements The authors thank the technical guidance of Dr. Shahsanam Abbasi during this work. ## Author information Authors ### Contributions F.N. designed, performed analyzed the experiments and wrote the first draft of the paper. S.F. and R.I. coordinated the study. S.F. also wrote the final draft of the paper. All authors reviewed the results and approved the final version of the manuscript. ### Corresponding authors Correspondence to Rana Imani or Shahab Faghihi. ## Ethics declarations ### Competing interests The authors declare no competing interests. ### Publisher's note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. ## Rights and permissions Reprints and Permissions Nowruzi, F., Imani, R. & Faghihi, S. Effect of electrochemical oxidation and drug loading on the antibacterial properties and cell biocompatibility of titanium substrates. Sci Rep 12, 8595 (2022). https://doi.org/10.1038/s41598-022-12332-z • Accepted: • Published: • DOI: https://doi.org/10.1038/s41598-022-12332-z	2022-06-26 00:28:28	{"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": 0.5396365523338318, "perplexity": 9401.29188897109}, "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-2022-27/segments/1656103036176.7/warc/CC-MAIN-20220625220543-20220626010543-00796.warc.gz"}
https://www.transtutors.com/questions/e8-12-the-cycle-division-of-ayala-company-has-the-following-per-unit-data-related-to-1356506.htm	# E8-12 The Cycle Division of Ayala Company has the following per unit data related to its most... E8-12 The Cycle Division of Ayala Company has the following per unit data related to its most recent cycle called Roadbuster. Selling price $2,200 Variable cost of goods sold Body frame$300 Other variable costs 900 1,200 Contribution margin $1,000 Presently, the Cycle Division buys its body frames from an outside supplier. However Ayala has another division, FrameBody, that makes body frames for other cycle companies. The Cycle Division believes that FrameBody’s product is suitable for its new Roadbuster cycle. Presently, FrameBody sells its frames for$350 per frame. The variable cost for FrameBody is $270. The Cycle Division is willing to pay$280 to purchase the frames from FrameBody. Instructions (a) Assume that FrameBody has excess capacity and is able to meet all of the Cycle Divi- sion’s needs. If the Cycle Division buys 1,000 frames from FrameBody, determine the following: (1) effect on the income of the Cycle Division; (2) effect on the income of FrameBody; and (3) effect on the income of Ayala. (b) Assume that FrameBody does not have excess capacity and therefore would lose sales if the frames were sold to the Cycle Division. If the Cycle Division buys 1,000 frames from FrameBody, determine the following: (1) effect on the income of the Cycle Divi- sion; (2) effect on the income of FrameBody; and (3) effect on the income of Ayala. VIJAYAKUMAR G The solution is attached herewith in...	2019-04-24 06:04:18	{"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": 0.32859086990356445, "perplexity": 3247.9825663442443}, "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-2019-18/segments/1555578633464.75/warc/CC-MAIN-20190424054536-20190424080536-00218.warc.gz"}
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