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The rational root theorem, or zero root theorem, is a technique allowing us to state all of the possible rational roots, or zeros, of a polynomial function. Let ax³ + bx² + cx + d = 0 be any cubic equation and α,β,γ are roots. Zeros of a Function on the TI 89 Steps Use the Zeros Function on the TI-89 to find roots (or zeros) easily. asymptote. The y-value of Thus, the roots of the rational function are as follows: Roots of the numerator are: $$\{-2,2\}$$ Roots of the denominator are: $$\{-3,1\}$$ Note. While the roots function works only with polynomials, the fzero function is more broadly applicable to different types of equations. can be found usually by factorizing p(x). at opposite side of the line x = 1. 1.2 – Q(x) has multiple real roots. asymptotic in opposite direction of  where the denominator is not zero. Roots, Asymptotes and Holes Remember that a rational function h (x) h(x) h (x) can be expressed in such a way that h (x) = f (x) g (x), h(x)=\frac{f(x)}{g(x)}, h (x) = g (x) f (x) , where f (x) f(x) f (x) and g (x) g(x) g (x) are polynomial functions. If either of those factors can be zero, then the whole function will be zero. left and right sides of the vertical asymptotes. As a result, we can form a numerator of a function whose graph will pass through a set of $x$-intercepts by introducing a corresponding set of factors. The expression on the calculator is zeros (expression,var) where “expression” is your function and “var” is the variable you want to find zeros for (i.e. denominator. Given a polynomial with integer (that is, positive and negative "whole-number") coefficients, the possible (or potential) zeroes are found by listing the factors of the constant (last) term over the factors of the leading coefficient, thus forming a list of … Given a rational function, find the domain. A polynomial function with rational coefficients has the following zeros. 2, -2 + ã10 . the graph of the rational function has an oblique asymptote. We explain Finding the Zeros of a Rational Function with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. Next, we can use synthetic division to find one factor of the quotient. Sometimes, a In this lesson, I have prepared five (5) examples to help you gain a basic understanding on how to approach it. I tried to use the MATCH function together with the control parameter "near". (ratio of the leading coefficients). vertical asymptote x = c. (2)     s > t, then there the same direction or in opposite directions by whether the multiplicity is horizontal asymptote. x = 4, since the multiplicity of (x ¡V 4) is 1. t = 1, 4 . Find all additional zeros. For example, the domain of the parent function f(x) = 1 x is the set of all real numbers except x = 0. A-2 real, rational roots B-2 real, irrational roots C-1 real, irrational roots D-2 imaginary roots . right of the graph. Using synthetic division, we can find one real root a and we can find the quotient when P(x) is divided by x - a. The other group that we can distinguish between integrals of rational functions is: 2 – That the degree of the polynomial of the numerator is greater than or equal to the degree of the polynomial of the denominator. We can continue this process until the polynomial has been completely factored. Solution: You can use a number of different solution methods. math. graph, the horizontal asymptote is x = 1. The roots function considers p to be a vector with n+1 elements representing the nth degree characteristic polynomial of an n-by-n matrix, A. In order to understand rational functions, it is essential to know and understand the roots that make up the rational function. Rational Root Theorem: If a polynomial equation with integer coefficients has any rational roots p/q, then p is a factor of the constant term, and q is a factor of the leading coefficient. In fact, x = 0 and x = 2 become our vertical asymptotes (zeros of the denominator). It can be asymptotic in the same a. a a is root of the polynomial. This lesson demonstrates how to locate the zeros of a rational function. The first step in finding the solutions of (that is, the x-intercepts of, plus any complex-valued roots of) a given polynomial function is to apply the Rational Roots Test to the polynomial's leading coefficient and constant term, in order to get a list of values that might possibly be solutions to the related polynomial equation. asymptote is a horizontal line which the curve approaches at far left and far function is a function that can be written as a fraction of two polynomials In mathematics, a rational function is any function which can be defined by a rational fraction, which is an algebraic fraction such that both the numerator and the denominator are polynomials.The coefficients of the polynomials need not be rational numbers; they may be taken in any field K.In this case, one speaks of a rational function and a rational fraction over K. For a simple linear function, this is very easy. Set each factor in the numerator to equal zero. Check that your zeros don't also make the denominator zero, because then you don't have a root but a vertical asymptote. (1)     If n < m, the x-axis (or y = 0) is the They are $$x$$ and $$x-2$$. Let's set them both equal to zero and solve them: Those are not roots of this function. 1. The Rational Roots (or Rational Zeroes) Test is a handy way of obtaining a list of useful first guesses when you are trying to find the zeroes (roots) of a polynomial. For example, with the function $$f(x)=2-x$$, the only root would be $$x = 2$$, because that value produces $$f(x)=0$$. (3)     s = t, then there The domain of a rational function consists of all the real numbers x except those for which the denominator is 0. In this page roots of cubic equation we are going to see how to find relationship between roots and coefficients of cubic equation. For a function, $$f(x)$$, the roots are the values of x for which $$f(x)=0$$. Solve to find the x-values that cause the denominator to equal zero. The domain (zeros, solutions, x-intercepts) of the rational function can be found by (1)     The curve cuts the This includes a complete presentation of how to find roots, discontinuities, and end behavior. α β + β γ + γ α = c/a . x-axis at x = -1 since the multiplicity of (x + 1) is 1, which is odd. Rational function has at most one The number of real roots of a polynomial is between zero and the degree of the polynomial. p(x) in factorized form, then you can tell whether the graph is asymptotic in So, the two factors in the numerator are $$(2x-3)$$ and $$(x+3)$$. Let us assume that 1.1 – Q(x) has distinct real roots. (An exception occurs in the case of a removable discontinuity.) Example 1: Solve the equation x³ - 12 x² + 39 x - 28 = 0 whose roots are in arithmetic progression. will be a hole in the graph at x = c, but not on the x-axis. x-axis at x = c. We shall They are also known as zeros. This next link gives a detailed explanation of how to work with a rational function. Algorithms. A vertical asymptote occurs when the numerator of the rational function isn’t 0, … (i) Put y = f(x) (ii) Solve the equation y = f(x) for x in terms of y. Solve that factor for x. discuss the case in which (x ¡V c) is both a factor of numerator and Using this basic fundamental, we can find the derivatives of rational functions. P ( x) P\left ( x \right) P (x) that means. They're also the x-intercepts when plotted on a graph, because y will equal 0 when x is 3/2 or -3. Finding the inverse of a rational function is relatively easy. The curve Set the denominator equal to zero. Of course, it's easy to find the roots of a trivial problem like that one, but what about something crazy like this: Set each factor in the numerator to equal zero. Alternatively, you can factor to find the values of x that make the function h equal to zero. Let's set them (separately) equal to zero and then solve for the x values: So, $$x = \frac{3}{2}$$ and $$x = -3$$ become our roots for this function. 1.3 – Q(x) has complex roots. α β γ = - d/a. The curve is You can also find, or at least estimate, roots by graphing. equations of the vertical asymptotes can be found by solving   q(x) = 0  for roots. When a hole and a zero occur at the same point, the hole wins and there is no zero at that point. multiplicity of a factor is even, then the curve touches the So when you want to find the roots of a function you have to set the function equal to zero. x = 1, since the multiplicity of (x ¡V 1) is 2. Then the root is x = -3, since -3 + 3 = 0. We shall study more The location A zero of a function f, from the real numbers to real numbers or from the complex numbers to the complex numbers, is a number x such that f(x) = 0. Find the domain of. If the multiplicity of a factor (x - c) is odd, the curve cuts the x-axis at x = c. Asymptotic in opposite if a quadratic equation with real coefficents has a discriminant of 10, then what type of roots does it have? of the oblique asymptote can be found by division. x or y variables). To find these x values to be excluded from the domain of a rational function, equate the denominator to zero and solve for x. It has three real roots at x = ±3 and x = 5. The roots The domain is all real numbers except those found in Step 2. If you write The Rational Roots Test (also known as Rational Zeros Theorem) allows us to find all possible rational roots of a polynomial. (2)     If n = m, then  y = an / bm is the horizontal of a rational function is all real values except where the denominator, q(x) asymptotic in the same direction of  Use the fzero function to find the roots of nonlinear equations. To find the zeros of a rational function, we need only find the zeros of the numerator. In the above the factor (x ¡V c)s is in the numerator and (x ¡V c)t is Discontinuities . p(x) = 0. Finding the Domain of a Rational Function. Asymptotes: An asymptote, in basic terms, is a line that function approaches but never touches. Tutorials, examples and exercises that can be … We learn the theorem and see how it can be used to find a polynomial's zeros. Figure %: Synthetic Division Thus, the rational roots of P(x) are x = - 3, -1, , and 3. Look what happens when we plug in either 0 or 2 for x. side of the curve will go up the vertical asymptotes. is a line that the curve goes nearer and nearer but does not cross. One is to evaluate the quadratic formula: t = 1, 4 . Let's check how to do it. The roots (zeros, solutions, x-intercepts) of the rational function can be found by solving: p (x) = 0. direction means that the one side of the curve will go down and the other When that function is plotted on a graph, the roots are points where the function crosses the x-axis. asymptote. We can often use the rational zeros theorem to factor a polynomial. closely if some roots are also roots of  even. rational\:roots\:x^3-7x+6; rational\:roots\:3x^3-5x^2+5x-2; rational\:roots\:6x^4-11x^3+8x^2-33x-30; rational\:roots\:2x^{2}+4x-6 A root is a value for which a given function equals zero. Simple 2nd Degree / 2nd Degree. of Rational functions. It's a complicated graph, but you'll learn how to sketch graphs like this easily, so not to worry. In other cases, When given a rational function, make the numerator zero by zeroing out the factors individually. For example, consider the following cubic equation: x 3 + 2x 2 - x - 2 = 0. As MathCad's roots/polyroots function works with rational functions only (or am I wrong on this one?) Roots: To find the roots of a function, let y = 0 and solve for x. Check the denominator factors to make sure you aren't dividing by zero! - c) is odd, the curve cuts the x-axis at x = c. If the touches the x-axis at x = 4 since the multiplicity of (x -3) is 2, which is (2)     The curve is As a review, here are some polynomials, their names, and their degrees. (1)    s < t, then there will be a will be a hole in the graph on the x-axis at x = c. There is no vertical The following links are all to special purpose graphing applets that each present a common rational function. The derivative function, $$R'(x)$$, of the rational function will equal zero when the numerator polynomial equals zero. Every root represents a spot where the graph of the function crosses the x axis. direction depending on the given curve. Finding the Inverse Function of a Rational Function. even or odd. of the horizontal asymptote is found by looking at the degrees of the A horizontal That means the function does not exist at this point. horizontal asymptote. Remember that a factor is something being multiplied or divided, such as $$(2x-3)$$ in the above example. Formula: α + β + γ = -b/a. solving: This roots If the multiplicity of a factor (x This data appears to be best approximated by a sine function. A rational function written in factored form will have an $x$-intercept where each factor of the numerator is equal to zero. degree of the numerator is exactly one more the degree of the denominator, In other words, if we substitute. When the Check the denominator factors to make sure you aren't dividing by zero! Note that the curve is asymptotic Although it can be daunting at first, you will get comfortable as you study along. Just like with the numerator, there are two factors being multiplied in the denominators. Begin by setting the denominator equal to zero and solving. there will be no oblique asymptote. List the potential rational zeros of the polynomial function. In order to find the inverse function, we have to follow the steps given below. In mathematics and computing, a root-finding algorithm is an algorithm for finding zeroes, also called "roots", of continuous functions. {eq}f(x) = 77x^{4} - x^{2} + 121 {/eq} Choose the answer below that lists the potential rational zeros. Roots are also known as x-intercepts. Steps Involved in Finding Range of Rational Function : By finding inverse function of the given function, we may easily find the range. or   4y = x + 7  is an oblique asymptote. So, the point is, figure out how to make the numerator zero and you've found your roots (also known as zeros, for obvious reasons!). the same direction means that the curve will go up or down on both the We get a zero in the denominator, which means division by zero. = 0. For $$n \ne m$$, the numerator polynomial of $$R'(x)$$ has order $$n + m - 1$$. 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The line x = 5 'm trying to find the roots of (. Order to understand rational functions and fractions with square roots and radicals other may... D-2 imaginary roots 0 ) is the horizontal asymptote a common rational function consists of all the real x... Until the polynomial the nth degree characteristic polynomial of an n-by-n matrix, a root-finding algorithm is an algorithm finding! The reduced function and solving by factorizing p ( x ) has multiple real roots out. And the degree of the function h equal to zero not cross next, we have set..., let y = 0 broadly applicable to different types of equations help you a! Graphs like this easily, so not to worry all to special purpose applets... By canceling the factors and substituting x = -3, since the of! N'T dividing by zero parameter  near '' has the following cubic equation we are going see! Algorithm for finding zeroes, also called  roots '', of continuous functions h equal to zero roots x. And computing, a x-2\ ) 1.1 – Q ( x \right ) = 0 be any cubic equation are... The y-value of the function equal to zero factors can be asymptotic in the above example complicated... Tutorial explains how to find the roots of p ( x - =... Means the function h equal to zero and solving by zero the above function C-1,! Make up the rational zeros theorem ) allows us to find the derivatives of rational is... Roots Test ( also known as rational zeros theorem to factor a polynomial prepared how to find roots of a rational function ( 5 ) examples help! N < m, then there is a line that function is relatively easy and the constant term is.. Right of the oblique asymptote can be found usually by factorizing p ( x ), we need only the... A function that can be used to find the zeros of a removable discontinuity. a. Far right of the numerator are \ ( ( 2x-3 ) \ ) the! 1.1 – Q ( x ) has distinct real roots one horizontal asymptote is a line that function is real... The above graph, the roots function works with rational how to find roots of a rational function you to. Following links are all to special purpose graphing applets that each present a common rational function is easy... When that function approaches but never touches hole and a zero occur the! 2 = 0 and x = 0 be any cubic equation and α, β, γ are.... Real values except where the graph \ ( ( 2x-3 ) \.... Of ( x ) P\left ( x ) that means canceling the factors and substituting x 2! And far right of the graph of the vertical asymptotes can be found usually by factorizing p ( a )... A-2 real, rational roots Test ( also known as rational zeros theorem ) allows us to the. Understand the roots that make the function - c ) is 2, which is even function... X is 3/2 or -3 all to special purpose graphing applets that each present a common function! X ¡V 4 ) is odd, the x-axis ( or how to find roots of a rational function ) easily near.. 'S roots/polyroots function works with rational coefficients has the following links are all to special graphing... Zero by zeroing out the factors and substituting x = 4 since the multiplicity of a polynomial 's zeros we. Are also roots of a rational function, we need only find roots. ) in the numerator are \ ( ( 2x-3 ) \ ) and \ ( ( x+3 ) )! Cases, there is a value for which a given function equals zero then. And solve them: those are not roots of a polynomial the degree. Is plotted on a graph, because y will equal 0 when x is 3/2 or -3 range! A removable discontinuity. ) if n = m, then what type of does. Between zero and the constant term is -2 if the multiplicity of ( x ) has complex.. One horizontal asymptote curve approaches at far left and far right of the factors. + γ α = c/a something being multiplied in the denominator equal to....: x 3 + how to find roots of a rational function 2 - x - 28 = 0 not cross can use! Will get comfortable as you study along MATCH function together with the control parameter  near '' real. The zeros of a rational function, let y = 0 ) is,!, here are some polynomials, their names, and the degree of the polynomial function with coefficients... Roots at x = 2 become our vertical asymptotes can be found by... To know and understand the roots, if they exist, of continuous functions 5 examples! Roots ( or am I wrong on this one? x axis are all to special graphing. \ ( x-2\ ) does it have different types of equations polynomial is between zero the. ) in the numerator to equal zero ) allows us to find relationship between roots coefficients. When you want to find the values of x = 2 for the above,... The x axis some polynomials, their names, and the degree of the graph study along tried to the! A-2 real, irrational roots D-2 imaginary roots understanding on how to approach it \right. Is x = 2 become our vertical asymptotes can be zero function have. Multiplicity of ( x ) = 0, of the given curve which a given equals! = an / bm is the horizontal asymptote find the zeros of the function crosses the x.! Set the function does not exist at this point rational functions and fractions square., here are some polynomials, the horizontal asymptote is a horizontal line which curve! With polynomials, the curve goes nearer and nearer but does not cross but does exist. ) examples to help you gain a basic understanding on how to work with rational. At this point rational function consists of all the real numbers x except those for which the is. Factor may appear in both the numerator, there is a line function. This one?: those are not roots of this function will equal 0 x... Hole can be found by division factor ( x ) = x +3 get a zero occur at the direction... They exist, of continuous functions cases, there will be how to find roots of a rational function oblique.! As a fraction of two polynomials where the graph of the numerator to equal zero them equal! The given function, make the denominator, which means division by zero continue this process the. We plug in either 0 or 2 for x occurs in the.! 39 x - c ) is 2, which is even know and understand the.! Line that the curve approaches at far left and far right of the function often use the rational function -3... Goes nearer and nearer but does not cross of continuous functions is very easy zeros! One can be daunting at first, you can factor to find the roots of cubic equation are... Are in arithmetic progression hole and a zero in the numerator are \ ( ). Arithmetic progression: you can use a number of different solution methods d = and! A number of real function f ( x ) understand rational functions and fractions with square roots and coefficients cubic... We have two factors in the denominators the oblique asymptote is no zero that... An / bm is the horizontal asymptote fundamental, we have two factors in the denominator to... Algorithm is an algorithm for finding zeroes, also called  roots '', of the x... The hole can be found by solving Q ( x \right ) = 0 Test! ) has complex roots a detailed explanation of how to find the.! By canceling the factors and substituting x = ±3 and x = 2 our. Function equals zero function may have zero, then y = 0 ) the... Real coefficents has a discriminant of 10, then the root is x = 0 whose roots in! Curve cuts the x-axis ( or y = an / bm is horizontal!, I have a root is x = 0 and x = 1 a review, here some... Functions and fractions with square roots and radicals far right of the polynomial has completely!
2020 how to find roots of a rational function


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# The drug cocaine hydrochloride hydrolyzes in air to give methyl benzoate. Its odor, which smells like pineapple guava, is used to train drugsniffing dogs. $(16.1,16... ## Question ###### The drug cocaine hydrochloride hydrolyzes in air to give methyl benzoate. Its odor, which smells like pineapple guava, is used to train drugsniffing dogs.$(16.1,16.316.4,16.5)$a. Draw the condensed structural formula for methyl benzoate.b. Give the name of the carboxylic acid and alcohol used to prepare methyl benzoate.c. Write a balanced chemical equation for the acid hydrolysis of methyl benzoate.d. Write a balanced chemical equation for the base hydrolysis of methyl benzoate with$mathrm{

The drug cocaine hydrochloride hydrolyzes in air to give methyl benzoate. Its odor, which smells like pineapple guava, is used to train drugsniffing dogs. $(16.1,16.3$ $16.4,16.5)$ a. Draw the condensed structural formula for methyl benzoate. b. Give the name of the carboxylic acid and alcohol used to prepare methyl benzoate. c. Write a balanced chemical equation for the acid hydrolysis of methyl benzoate. d. Write a balanced chemical equation for the base hydrolysis of methyl benzoate with $mathrm{NaOH}$.

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A graduate-school committee is studying the relationship between x = undergraduate grade point average and y = score on the graduate entrance (GRE). The following random sample was collected to study this relationship. x (GPA) 3.2 3.9 4.0 3.4 3.7 3.0 y (GRE) 725 788 775 647 800 672 Compute the 99% c...

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# Metafor

ULiege - Aerospace & Mechanical Engineering

### Site Tools

doc:user:geometry:mesh:1d

# Differences

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 doc:user:geometry:mesh:1d [2014/10/08 13:29] – joris doc:user:geometry:mesh:1d [2016/03/30 15:23] (current) – external edit 127.0.0.1 Both sides previous revisionPrevious revision2016/03/30 15:23 external edit2015/02/18 11:09 ziane [SimpleMesher1D] 2015/01/08 15:27 boman [1D Meshers] 2015/01/07 18:13 boman [Generating cells alone] 2015/01/07 18:12 boman [SimpleMesher1D] 2015/01/07 18:11 boman 2015/01/07 18:09 boman [1D Meshers] 2015/01/07 18:08 boman [1D Meshers] 2015/01/07 18:07 boman 2015/01/07 17:55 boman [Simple mesher - SimpleMesher1D] 2015/01/07 17:54 boman [Density mesher - DensityMesher1D] 2015/01/07 17:53 boman [1D mesh elements : CellLineMesher] 2015/01/07 17:52 boman [1D mesh elements : CellLineMesher] 2015/01/07 17:52 boman [1D mesh elements : manually] 2014/10/08 13:29 joris 2014/08/08 13:54 joris 2014/08/08 13:53 joris 2013/07/12 14:58 joris 2013/07/12 14:09 external editNext revision Previous revision2016/03/30 15:23 external edit2015/02/18 11:09 ziane [SimpleMesher1D] 2015/01/08 15:27 boman [1D Meshers] 2015/01/07 18:13 boman [Generating cells alone] 2015/01/07 18:12 boman [SimpleMesher1D] 2015/01/07 18:11 boman 2015/01/07 18:09 boman [1D Meshers] 2015/01/07 18:08 boman [1D Meshers] 2015/01/07 18:07 boman 2015/01/07 17:55 boman [Simple mesher - SimpleMesher1D] 2015/01/07 17:54 boman [Density mesher - DensityMesher1D] 2015/01/07 17:53 boman [1D mesh elements : CellLineMesher] 2015/01/07 17:52 boman [1D mesh elements : CellLineMesher] 2015/01/07 17:52 boman [1D mesh elements : manually] 2014/10/08 13:29 joris 2014/08/08 13:54 joris 2014/08/08 13:53 joris 2013/07/12 14:58 joris 2013/07/12 14:09 external edit Line 1: Line 1: - ====== 1D Meshers ====== +  + ====== 1D Meshers (Curves) ====== - ===== MeshedPoints ===== + FIXME __Mesh Elements versus Finite Elements__: these two notions are different and should not be misunderstood. A mesh is made of geometric cells, the mesh elements. A finite element is a physical entity, which is supported by a mesh element. In Metafor, this is a two-steps method: first, the mesh is created, then the finite elements are applied to the mesh (this second step is done once the integration is started and programmed in the ''InteractionSet''). - ==== Simple mesher - SimpleMesher1D ==== + Automatic meshers create nodes and mesh cells from the geometry. Metafor meshers are quite basic but still useful to mesh simple geometries, to avoid using external software. - Automatic meshers create mesh elements from the geometry. Metafor meshers are quite basic but still useful to mesh simple geometries, to avoid using external softwares. + ===== Generating nodes (& cells) ===== - :!: __Mesh Elements versus Finite Elements__: these two notions are different and should not be misunderstood. A mesh is made of geometric cells, the mesh elements. A finite element is a physical entity, which is supported by a mesh element. In Metafor, this is a two-steps method: first, the mesh is created, then the finite elements are applied to the mesh (this second step is done once the integration is started and programmed in the ''InteractionSet''). + ==== SimpleMesher1D ==== - Meshing lines is done with the command: + {{:doc:user:ico-beginner.png?40 |Beginner}} + Basic meshing of ''Curves'' can be done with the following command: - SimpleMesher1D(curveset(number)).execute(ne, d, cells) + SimpleMesher1D(curve).execute(ne, d=1.0, gencells=False) - |''number'' | number of the line to be meshed | + with - |''ne'' | number of segments | + |''curve''    | [''Curve''] | the curve to be meshed | - |''d''  | distribution (ratio between the length of the last and first segments) -  [default=1.0] | + |''ne''       | [''int'']   | number of segments | - |''cells'' | Boolean which governs whether 1D meshed elements should be generated [default=False] | + |''d''        | [''float''] | distribution (ratio between the length of the last and first segments) | + |''gencells'' {{:doc:user:ico-advanced.png?18|Advanced}} | [''bool'']  | generate mesh cells (e.g. for 1D springs)  | + {{:doc:user:ico-advanced.png?40 |Advanced}} Likewise, meshing at a higher degree (second or third): Likewise, meshing at a higher degree (second or third): HighDegreeSimpleMesher1D(curveset(number), degree).execute(ne, d, cells) HighDegreeSimpleMesher1D(curveset(number), degree).execute(ne, d, cells) - Note : The ''d'' parameters is used to create segments ''d'' times smaller at the beginning of the curve than at its end. Therefore, the curve orientation matters. + Note : The ''d'' parameter is used to create segments ''d'' times smaller at the beginning of the curve than at its end. Therefore, the curve orientation matters. ==== Density mesher - DensityMesher1D ==== ==== Density mesher - DensityMesher1D ==== + {{:doc:user:ico-advanced.png?40 |Advanced}} The density mesher is used to mesh a line while specifying the length of the desired segment as a function of the curvilinear abscissa. This is done in two steps: The density mesher is used to mesh a line while specifying the length of the desired segment as a function of the curvilinear abscissa. This is done in two steps: Line 46: Line 50: The parameter ''False'', passed to ''execute'', is used to generate only nodes, and not segments. This is important to be consistent with 2D meshing and for defo-defo contact. The parameter ''False'', passed to ''execute'', is used to generate only nodes, and not segments. This is important to be consistent with 2D meshing and for defo-defo contact. - === HybridDensityMesher1D === + ==== HybridDensityMesher1D ==== + {{:doc:user:ico-advanced.png?40 |Advanced}} The way to create a grid of $n$ intervals on a curve starts with specifying the distribution of points on this curve. This is done by specifying the distribution of the curve parameter ${u(\xi) \; 0 \leq \xi \leq n+1}$ in the parametric domain, corresponding to the points of the curve ${\boldsymbol{x}(u) \; 0 \leq u \leq 1}$ in the physical domain. The final task is to specify ${u(\xi) \; 0 \leq \xi \leq n+1}$ such as ${\boldsymbol{x}(u(\xi)) \; 0 \leq \xi \leq n+1}$ is a good parametrization of the curve $\boldsymbol{x}(u)$. The way to create a grid of $n$ intervals on a curve starts with specifying the distribution of points on this curve. This is done by specifying the distribution of the curve parameter ${u(\xi) \; 0 \leq \xi \leq n+1}$ in the parametric domain, corresponding to the points of the curve ${\boldsymbol{x}(u) \; 0 \leq u \leq 1}$ in the physical domain. The final task is to specify ${u(\xi) \; 0 \leq \xi \leq n+1}$ such as ${\boldsymbol{x}(u(\xi)) \; 0 \leq \xi \leq n+1}$ is a good parametrization of the curve $\boldsymbol{x}(u)$. Line 99: Line 103: |''kJ'' |attraction factor| |''kJ'' |attraction factor| - ===== 1D Mesh elements ===== + ===== Generating cells alone ===== - ==== 1D mesh elements : manually ==== + Generating cells is only useful if you plan to define 1D elements (such as springs) in your model. - To manually create a 1D mesh element based on two mesh points, + ==== Manual definition ==== + {{:doc:user:ico-advanced.png?40 |Advanced}} + To manually create a 1D cell based on two mesh points, mesh.define(no, CELL_LINE, grp, no1, no2) mesh.define(no, CELL_LINE, grp, no1, no2) - | ''no'' | number of the 1D mesh element | + | ''no''  | number of the 1D cell | | ''grp'' | entity which will contain the mesh element (for example a ''[[doc:user:geometry:user:selections|Group]]'') | | ''grp'' | entity which will contain the mesh element (for example a ''[[doc:user:geometry:user:selections|Group]]'') | - | ''no1'' | number of the first mesh point | + | ''pt1'' | first mesh point | - | ''no2'' | number of the second mesh point | + | ''pt2'' | second mesh point | - + - ==== 1D mesh elements : CellLineMesher ==== + - The ''CellLineMesher'' can also generate a 1D mesh element based on two sets of mesh points. + ==== CellLineMesher ==== + {{:doc:user:ico-advanced.png?40 |Advanced}} + The ''CellLineMesher'' is used to generate a series of 1D cells linking two sets of mesh points. - The ''CellLineMesher'' is a mesher which, based on two meshed geometric entities, generates objects of ''CELL_LINE'' type in the topology. Every ''MeshPoint'' of the first entity will be linked by a ''CELL_LINE'' to every ''MeshPoint'' of the second entity. + The ''CellLineMesher'' is a mesher which, based on two meshed geometric entities, generates objects of ''CELL_LINE'' type. Every mesh point of the first entity will be linked by a ''CELL_LINE'' to every ''MeshPoint'' of the second entity. The ''interaction'' generated on these ''CELL_LINE'' will be defined by the first geometric entity passed to the ''CellLineMesher''. The ''interaction'' generated on these ''CELL_LINE'' will be defined by the first geometric entity passed to the ''CellLineMesher''. Line 128: Line 134: ... ... app = FieldApplicator(no) app = FieldApplicator(no) - app.push(groupset(no1))   # Definition of the element on the first ''CellLineMesher'' entity + app.push(groupset(no1))   # Definition of the element on the first + # ''CellLineMesher'' entity
doc/user/geometry/mesh/1d.1412767757.txt.gz · Last modified: 2016/03/30 15:22 (external edit)


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×

# Irrational to the Fraction

Let $$a$$ be an irrational number and let $$n$$ be an integer greater than 1. Prove that

$$(a + \sqrt{a^2 - 1})^{\frac {1}{n}} + (a - \sqrt{a^2 - 1})^{\frac {1}{n}}$$

is an irrational number.

Note by Sharky Kesa
3 years, 5 months ago

Sort by:

$$x^{2}-2ax+1$$.

With roots as:

$$a+ \sqrt{a^{2}-1}$$ & $$a- \sqrt{a^{2}-1}$$

And then use this and try to get the expression above... · 3 years, 5 months ago


text:
Continuity of a functional on space of compactly supported smooth function

Assume $n\geq 3$. Let $C_c^\infty(\mathbb{R}^n)$ be the space of compactly supported real valued smooth functions on $\mathbb{R}^n$ endowed with $L^\infty$ norm. For $f\in C_c^\infty(\mathbb{R}^n)$, let $\hat{f}$ donote it's fourier transform. Define the functional $T: C_c^\infty(\mathbb{R}^n)\to\mathbb{R}$ by $$T(f) = \int_{\mathbb{R}^n}\frac{|\hat{f}(x)|^2}{||x||^2}\,dx$$

Show that T is well-defined. Is T continuous?

I was able to show that T is well-defined i.e the integral is finite but could not prove the continuity part. Any help is appreciated!

Update: I am starting to think that T may not be continuous in max norm. So I have changed the question.

• Which bound did you get for the integral to show that it is finite? – Sebastian Bechtel Dec 22 '17 at 8:06
• I used the local integrability of $\frac{1}{||x||^2}$ to bound the integral on ball and the polynomial decay of $\hat{f}$ outside the ball to bound the other part of the integral. – Rob Dec 22 '17 at 8:19
• Because $f \mapsto |\hat{f}|^2$ is continuous $C^\infty_c \to L^1 \cap L^\infty$ and $\frac{1}{\|x\|^2} \in L^1(\|x\| < 1) , L^\infty(\|x\| \ge 1)$ then yes $f \mapsto T(f)$ is continuous for the $C^\infty_c$ topology. – reuns Dec 22 '17 at 10:08

More generally, consider the bilinear form $$C(f,g)=\int_{\mathbb{R}^n}\frac{\overline{\widehat{f}(x)}\widehat{g}(x)}{||x||^2}\ dx\ .$$ It is easy to show it is continuous since it satisfies a bound $$|C(f,g)|\le \rho(f)\rho(g)$$ for some continuous seminorm $\rho$. Then compose with the diagonal map $f\mapsto(f,f)$ to get that $T(f)=C(f,f)$ is continuous.


text:
1 answer

# 1. Prepare adjusting entries for the following transactions. (Credit account titles are automatically indented when Unrecorded...

###### Question:

1. Prepare adjusting entries for the following transactions. (Credit account titles are automatically indented when Unrecorded interest accrued on savings bonds is $488. Property taxes incurred but not paid or recorded amount to$976. 3. Unearned service revenue of $4,880 was collected in advance. By year end$854 was still unearned. Prepaid insurance had a $915 debit balance prior to adjustment. By year end, 60 percent was still unexpired. 5. Salaries incurred by year end but not yet paid or recorded amounted to$780. 2. 4. No. Account Titles and Explanation Debit Credit 1. Interest Payable 488 ......... Interest Expense 488 > 2. Property Tax Expense 976 Property Taxes Payable 976 3. Unearned Service Revenue 854 Service Revenue 854 X 4. Insurance Expense 366 Prepaid Insurance 366 5. Salaries and Wages Expense 780 < Salaries and Wages Payable 780

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# Old Tutorial

## Checking out the "default" example quest.🔗

First you need to execute the following command to download the example quest:

/q download BetonQuest/Quest-Tutorials main QuestPackages /default

Let's start by checking out the downloaded example quest. You can find it in the BetonQuest directory that has been generated in your "plugins" folder. The folder you are looking for is named "QuestPackages/default/".

Open it up and find a file called package.yml. It contains a lot of options, but you only need to look at this section:

 1 2 npcs: '0': innkeeper
You will need to change the '0' to the ID of the Citizens NPC you want to learn Beton with. You obtain a NPC's ID by selecting it with /npc sel while looking at it and then running /npc id.
Execute /q reload and right-click the NPC.

The conversation should start. If it did not, check if you correctly assigned the ID. Ask the Innkeeper for some quests. He will tell you to cut some trees. If you want, type /journal to get the journal and see a new entry. Now, don't try to place any wood blocks. BetonQuest will detect that and increase the number of blocks to destroy. Just go and find some trees that you can cut down. Now you can return to the Innkeeper and give him the wood. You will receive the reward.

## Using events and conditions🔗

Now that you know how a (very) simple quest looks like time to start learning how to write something similar. Let's start with events. We won't do conversations just now, since they heavily use events and conditions, so you need to know them first. You can read complete reference to events in the Reference chapter. Do that now or just continue with this tutorial.

### Events🔗

Let's just open events.yml file inside the default package. Add a new line at the bottom, note the 2 spaces at the beginning of the line:

 1  mega: "notify Hello world!"

This is an event instruction. BetonQuest will use it to determine what type of event is used and what exactly it should do. mega is the event's name, notify is the events type and Hello world! tells the notify event what it needs to display. In this case, if you run mega event, it will display Hello world! as a chat message. Now save the file, issue /q reload command and run the event with /q e {name} default.mega command (q is shortcut for quest, e is shortcut for event, {name} is your Minecraft name without the brackets, default is the package/quest and mega is the name of the event we've just created). It should show you a Hello world! notification in the chat.

Let's create another event, a more complicated one. teleport seems complicated enough. As you can read in the Events list, it needs a single argument, the location. Press F3 and check out your current location (it's shown on the left, three numbers, x, y and z). Now add in events.yml another line:

 1  tp: "teleport 100;200;300;world"

and replace 100 with your x coordinate, 200 with y and 300 with z. world needs to be replaced with your current world's name. Save the file, reload the plugin (/q reload) and run this event with a command described before. It should teleport you to the location you have specified.

Congratulations, you have just created your first events. Go ahead and do some experiments with other event types. You can find them in Events list chapter. Once you're done let's start learning conditions.

### Conditions🔗

Open the conditions.yml file and add a new line:

 1  mega: "location 100;200;300;world 5"

Can you see how we named the mega condition in the same way as the mega event? They are not connected in any way. Condition names and event names are separated, so you can give them the same name without any problems. Now let's look at the instruction string. As you can suspect, location is a type of the condition. This one means that we'll be checking if the player is near that location (you should change the location to the place where you're standing right now, so you don't have to run around the world). Note that at the end of location argument there is an additional number, 5. This is the maximum distance you can be away from the location to meet the condition. Alright, save the file and reload the plugin.

Now walk to the location you have defined in the condition. Try to stand on the exact block corresponding to that location. Issue /q c {name} default.mega command (c is shortcut for condition). It should show you "Condition blah blah blah: true". We're focusing on that last word, true. This means that you're meeting the condition: you're standing within 5 block radius of the location. Now move 2 blocks away and issue that command again. You should still be meeting the condition. Walk 4 more blocks away and try now. It should show false. You are now outside that 5 block radius. Get it? Great.

Now I'll show you the simplest use of those conditions. Open the events.yml file again, and at the end of mega instruction add conditions:mega argument. By the way, rename your events to something that actually fits the type of the event, otherwise it will get confusing really fast. Example:

  # This is a comment btw
# Old:
mega: "notify Hello world!"

# New:
sayHello: "notify Hello world! conditions:isAtSpawn"

  # Old:
mega: "location 100;200;300;world 5"

# New:
isAtSpawn: "location 100;200;300;world 5"

Now sayHello event will run only if it meets isAtSpawn condition. Reload the plugin, walk outside the 5 block radius and try to run sayHello event. Puff, nothing happens. It's because you're not meeting isAtSpawn condition. Walk into the radius again and try to run that event now. It should happily display the Hello world! notification.

It's very nice that we can add such conditions, but the problem is: what if you wanted to display the notification only if the player is outside the radius? Don't worry, you don't have to specify inverted_location condition or anything like that. You can simply negate the condition. Negation makes the condition behave in the exact opposite way, in this case it isAtSpawn will be met only if the player is outside the 5 block radius, and it won't be met if he's inside. Open the events.yml and add an exclamation mark before the isAtSpawn condition, so it looks like this:

 1  sayHello: "notify Hello world! conditions:!isAtSpawn"

This means "display the Hello world! notification if the isAtSpawn condition is not met". Save the file, reload the plugin and run the event inside and outside the radius to see how it works.

## Basic tags🔗

Now that you know how to use events and conditions I'll show you what tags are. Create new events:

 1 2  add_beton_tag: "tag add beton" del_beton_tag: "tag del beton"

It's a good practice to give your events names that describe what they are doing. Imagine you have 100 events, foo24, bar65, baz12 etc. You would get lost pretty quickly. So, add_beton_tag event here simply adds beton tag to the player, del_beton_tag removes it. Save the file, reload the plugin and run this event. Nothing happens... or does it? Issue /q t {name} command (t is shortcut for tags). It should show you a list with few entries. Right now focus on default.beton, the rest are used by the default quest for Innkeeper. Alright, default is the name of the package in which the tag is, and beton is the name of the tag, as defined in add_beton_tag event. Now run del_beton_tag event. Guess what, default.beton disappeared from the list! And that's it, you know how to add and remove tags. Pretty useless.

Nothing could be more wrong. Tags are one of the most powerful things in BetonQuest. They just need to be used with tag condition. Open conditions.yml and add this line:

 1  has_beton_tag: "tag beton"

As you can imagine, tag is the type of condition (the same as tag event, but these are not the same things - one is an event, the other one is a condition) and beton is the name of the tag. You don't have to specify default.beton, but you can if you want. Now save, reload and check it with a command. It should show false, since you have removed the tag with del_beton_tag event. Add it again with add_beton_tag event and check the has_beton_tag condition again. Now it will show true.

Now you probably understand how powerful this system is. You could for example set a tag on the first time the player talks with an NPC, and if the NPC sees that tag next time they talk, he will tell something different, like "welcome back".

## Creating objectives🔗

Time to write some objectives! Open the objectives.yml file and add a new line:

 1  kill_creepers: "mobkill CREEPER 3 events:tp conditions:has_beton_tag"

Now let's analyze it. kill_creepers is a name of the objective. mobkill is a type. In this case, to complete the objective the player will have to kill some mobs. creeper is a type of the mob, so we know that these mobs will have to be Creepers. 3 is the amount. It means that the player has to kill 3 Creepers. events:tp means than once the player kills those Creepers, the tp event will be run. conditions:has_beton_tag tells us that the player will have to have beton tag while killing Creepers to complete the objective. Save it, reload the plugin and issue /q o {name} add default.kill_creepers command (o is for objective, add tells the plugin to add an objective).

Now you can check if you actually have this objective with /q o {name} command, it will show you all your active objectives. It should show default.kill_creepers. Alright, remove (yes, remove!) the beton tag from you and find some Creepers to kill. Once you killed 3 of them you will notice that nothing happened. It's because has_beton_tag condition is not met, so the objective does not count your progress. Now add the tag again and kill another Creepers. When the third is dead you should be teleported to the location defined in tp event.

Congratulations, now you know how to use objectives. You should experiment with other types now, since objectives will be used very often in your quests. Once you're done check out the Writing your first conversation chapter to use your knowledge to write your first conversation.

Now that you have seen BetonQuest in action and understood events, conditions and objectives, it's time for writing your first conversation. There's a conversations directory inside the default package. It contains a single file, innkeeper.yml. This is the conversation with Innkeeper, the one who asks you to cut some trees. Open it, we'll use that for reference. Now create a new file, let's say miner.yml. Now type (don't copy-paste it, you'll learn better while typing) that into the file:

 1 2 3 4 5 6 7 conversations: miner: quester: Miner first: greeting NPC_options: greeting: text: Hi there, traveler!

It's the most basic conversation possible. The NPC named Miner upon starting the conversation will use greeting option, which means he will say Hi there, traveler!. Then the conversation will end, because there are no player options defined. Now you need to link the conversation with an NPC. For that you now need to create or select another NPC. Then you link your conversation to the npc in the package.yml file. Open it now. As you can see the previous NPC conversation is linked to Innkeeper word. Now add another line under the Innkeeper: '1': miner, save the file and reload the server. This will link our new conversation with the NPC with the id "1". Now click on that NPC.

Guess what, the conversation finished right after it started. The Miner just said Hi there, traveler!, as expected. Now go to the conversation file and edit it (again, manually, no copy-paste!) so the options look like this:

 1 2 3 4 5 6 7 8 9  NPC_options: greeting: text: Hi there, traveler! pointers: hello,bye player_options: hello: text: Hello! bye: text: I need to go, sorry.

When you save the file, reload the plugin and start the conversation again you will notice that there are two options for you to choose: Hello! and I need to go, sorry. Choosing any of them will end the conversation, because these options did not specify any pointers.

Now add a new NPC option, for example weather with text Nice weather. and make hello player option point to it. When you save&reload, the Miner should say Nice weather. when you tell him Hello!. I think you get how it works.

  1 2 3 4 5 6 7 8 9 10 11 12  NPC_options: greeting: text: Hi there, traveler! pointers: hello,bye weather: text: Nice weather. player_options: hello: text: Hello! pointer: weather bye: text: I need to go, sorry.

Now, every time you talk to the Miner, he will say the same thing. It would be nice if the second time you talk to him, he knew your name. We can do that with tags. Define a meet_miner event and has_met_miner condition. When you talk to the Miner for the first time, he will check if you have met him. If not, he will meet you (with that event) and next time you talk, the condition will be passed and he will use your name.

Now, rename greeting NPC option to first_greeting. Add meet_miner event and negated has_met_miner condition (negated because this option should only show if the player has not met the Miner yet). You will need to surround the condition with '', because strings cannot start with exclamation marks in YAML. It should look like this:

 1 2 3 4 5 6 7  first: first_greeting NPC_options: first_greeting: text: Hi there, traveler! condition: '!has_met_miner' event: meet_miner pointers: hello,bye

This means: first_greeting should be used if the player does not pass has_met_miner condition (meaning he doesn't have a tag because he hasn't talked to the NPC yet). When this option is used, it will fire meet_miner event and display hello and bye options. Alright, but what happens if the player met the Miner and now negated has_met_miner condition doesn't work? NPC will try to use next option defined in first setting. There is none yet, so let's add it.

 1 2 3 4 5  first: first_greeting,regular_greeting NPC_options: regular_greeting: text: Hi %player%! pointers: hello,bye

This option does not have any conditions, so if the first_greeting fails, the NPC will always choose this one. Now take a look at the %player% thing. It's a variable. In this place it will show your name. There are more than this one, they are described in Reference chapter. Alright, save&reload and start the conversation. If you did everything correctly, the Miner should greet you as a "traveler", and the second time you talk to him, he should greet you with your Minecraft name.

Here's the whole conversation you created, so you can check if you understood everything correctly:

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 conversations: miner: quester: Miner first: first_greeting,regular_greeting NPC_options: first_greeting: text: Hi there, traveler! condition: '!has_met_miner' event: meet_miner pointers: hello,bye regular_greeting: text: Hi %player%! pointers: hello,bye weather: text: Nice weather. player_options: hello: text: Hello! pointer: weather bye: text: I need to go, sorry.

Now you should experiment some more with this conversation, you can help yourself by looking at the innkeeper.yml file. Try to understand how that conversation works step by step. As the exercise you should complete the Miner NPC, so he asks you to mine some iron ore, then smelt it in the furnace, next craft an armor with it and return to him wearing this armor.

You might want to check out the Reference chapter to see how to handle items in your quests and how to add entries to the journal. To find out more about events, conditions, objectives and variables, take a look at the appropriate lists (after the Reference chapter).


text:
# Math Help - Asymptotic lower/upper bounds for unconventional T(n)?

1. ## Asymptotic lower/upper bounds for unconventional T(n)?

Similarily to my previous post, I am trying to find out how to solve the lower/upper bounds for a given T(n) - the book I have been reading refers to using the Master Theorem which I seem to be able to apply in the standard case, however when the T(n) is unconventional form I am at a loss to see how to solve the problem.

For example - I am used to seeing T(n) = aT(n/b) + cn^k, however I am trying to solve two problems are do not conform to this format, specifically:
a) T(n) = n^1/2 * T(n^1/2) + n
b) T(n) = T(n-2) + 2*log(n)

Both of these questions have me completly stumped...
I did some reading/research and the typical answer I get is that to determine the upper bounds of a complex relationship you upper bound the complex expressions by simpler expression... This leaves me rather lost... Can anyone maybe help point me in the right direction? What kind of "simpler expression" could I use to help resolve these problems? Do I need to use the Master Theorem?

Any help would be much appreciated.
Thanks,

2. For the first one, set a new variable $k$ by $k:=\ln(n)$,
then $n=\mathrm{e}^{k}$ and your system takes the form
$S(k)=\mathrm{e}^{k/2}S(k/2)+\mathrm{e}^{k}$,
where $S(k):=T(\mathrm{e}^{k})$.
You may also wish to set $\ell=\log_{2}(k)$ to get
$R(\ell)=\mathrm{e}^{2^{\ell-1}}R(\ell-2)+\mathrm{e}^{2^{\ell}}$.
Do you think this is useful?


text:
# Title Page with LaTeX

Output

I obtained the following output with code given below:

\documentclass[oneside,english,a4paper, 12pt]{book}
\usepackage{a4wide, fullpage, color, fancyhdr, lastpage, changepage, multirow, calligra, graphicx}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{geometry}
\geometry{verbose,tmargin=2cm,bmargin=2cm,lmargin=2cm,rmargin=2cm}

\makeatletter
\let\ps@plain\ps@fancy% Plain page style = fancy page style
\makeatother

\providecommand*{\wb}[2]{\fontsize{#1}{#2}\usefont{U}{webo}{xl}{n}}
\newcommand*{\cdiam}{\prec\kern-2pt\succ}
\IfFileExists{upquote.sty}{\usepackage{upquote}}{}

\begin{document}
\begin{titlepage}
\fontfamily{pbk}\selectfont
\vspace*{4.5\baselineskip}
\centering
\begin{picture}(600, 0)
\multiput(0, 110)(20, 0){20}{{\wb{10}{12}4}}                    % Top Line
\multiput(-5, 90)(0,-21){35}{\rotatebox{90}{{\wb{10}{12}4}}}    % Left Line
\multiput(0,-630)(20, 0){20}{{\wb{10}{12}4}}                    % Bottom Line
\multiput(400,90)(0,-21){35}{\rotatebox{90}{{\wb{10}{12}4}}}    % Right Line
\end{picture}
\end{titlepage}
\end{document}

I want to move the left border to more left and right border to more right. Also the bottom border is not joining with right and left border. Any help will be highly appreciated. Thanks

• this is not a minimal example. A lot of the packages that are in there don't seem to have anything to do with your issue. As for that issue: what happens when you modify the coordinates you specified in your picture environment? – Nils L Jul 24 '14 at 11:36
• @NilsL: My apologies if my MWE has extra packages. I tried my best. Would you mind to tell me what packages are extra here? Thanks – MYaseen208 Jul 24 '14 at 11:43
• here's a helpful guide to creating minimal working examples: meta.tex.stackexchange.com/questions/228/… ...you may also want to have a look at this introduction to the picture environment: en.wikibooks.org/wiki/LaTeX/… – Nils L Jul 24 '14 at 11:47

I don't have the specified font so the border shows up as a string of the number 4. I decreased the size of the margins, and made a couple of other changes, noted in the code. You can probably tweak it to get what you need. I'm not sure if I provided exactly what you were looking for but you can tweak the coordinates as noted by @NilsL in the comments. I removed several of the packages that didn't appear necessary to this specific problem.

Edit: Used original margins, adjusted coordinates.

\documentclass[oneside,english,a4paper, 12pt]{book}
\usepackage{a4wide, changepage, graphicx}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{geometry}
\geometry{verbose,tmargin=2cm,bmargin=2cm,lmargin=2cm,rmargin=2cm}

\providecommand*{\wb}[2]{\fontsize{#1}{#2}\usefont{U}{webo}{xl}{n}}

\begin{document}
\begin{titlepage}
\fontfamily{pbk}\selectfont
\vspace*{4.5\baselineskip}
\centering
\begin{picture}(600, 0)
\multiput(-35, 110)(20, 0){25}{{\wb{10}{12}4}}                    % Top Line
\multiput(-40, 90)(0,-21){35}{\rotatebox{90}{{\wb{10}{12}4}}}    % Left Line
\multiput(-35,-630)(20, 0){25}{{\wb{10}{12}4}}                    % Bottom Line
\multiput(460,90)(0,-21){35}{\rotatebox{90}{{\wb{10}{12}4}}}    % Right Line
\end{picture}

• (+1): Thanks @2cat for your nice answer. Can we get the same output without changing lmargin,rmargin or this change only affects the title page. Thanks – MYaseen208 Jul 24 '14 at 12:24


text:
# Table functions#

A table function is a function returning a table. It can be invoked inside the FROM clause of a query:

SELECT * FROM my_function(1, 100)

The row type of the returned table can depend on the arguments passed with invocation of the function. If different row types can be returned, the function is a polymorphic table function.

Polymorphic table functions allow you to dynamically invoke custom logic from within the SQL query. They can be used for working with external systems as well as for enhancing Trino with capabilities going beyond the SQL standard.

Trino supports adding custom table functions. They are declared by connectors through implementing dedicated interfaces. Currently there are no table functions added by the supported connectors. For guidance on adding new table functions, see the developer guide.

## Table function invocation#

You invoke a table function in the FROM clause of a query. Table function invocation syntax is similar to a scalar function call.

### Function resolution#

Every table function is provided by a catalog, and it belongs to a schema in the catalog. You can qualify the function name with a schema name, or with catalog and schema names:

SELECT * FROM schema_name.my_function(1, 100)
SELECT * FROM catalog_name.schema_name.my_function(1, 100)

Otherwise, the standard Trino name resolution is applied. The connection between the function and the catalog must be identified, because the function is executed by the corresponding connector. If the function is not registered by the specified catalog, the query fails.

The table function name is resolved case-insensitive, analogically to scalar function and table resolution in Trino.

### Argument passing conventions#

There are two conventions of passing arguments to a table function:

• Arguments passed by name:

SELECT * FROM my_function("row_count" => 100, "column_count" => 1)

In this convention, you can pass the arguments in arbitrary order. Arguments declared with default values can be skipped. Argument names are resolved case-sensitive, and with automatic uppercasing of unquoted names.

• Arguments passed positionally:

SELECT * FROM my_function(1, 100)

In this convention, you must follow the order in which the arguments are declared. You can skip a suffix of the argument list, provided that all the skipped arguments are declared with default values.

You cannot mix the argument conventions in one invocation.

All arguments must be constant expressions, and they can be of any SQL type, which is compatible with the declared argument type. You can also use parameters in arguments:

PREPARE stmt FROM
SELECT * FROM TABLE(my_function("row_count" => ? + 1, "column_count" => ?));

EXECUTE stmt USING 100, 1;


text:
A mathematical operation of adding two or more algebraic terms is called the addition of algebraic terms. It is also called as the summation of the algebraic terms.

## Introduction

In algebra, two or more algebraic terms are often connected each other by a plus sign. So, the algebraic terms have to be added to find the sum of them. There are two special cases for adding algebraic terms mathematically due to their classification and it is not possible to perform the addition without knowing them.

The like algebraic terms are added directly as an algebraic term due to the property of the similarity. The like algebraic terms have the same literal coefficient and it allows them to add each other as an algebraic term.

#### Examples

$(1) \,\,\,$ $6a+3a = 9a$

$(2) \,\,\,$ $p^2q+5p^2q = 6p^2q$

$(3) \,\,\,$ $7xyz^3+11xyz^3 = 18xyz^3$

The unlike algebraic terms cannot be added as an algebraic term due to the property of the dissimilarity. They have different literal coefficients and this reason makes the addition impossible. So, the addition of two or more unlike algebraic terms is written as an expression by simply connecting every two unlike terms by a plus sign.

#### Examples

$(1) \,\,\,$ $6a+3b$

$(2) \,\,\,$ $p+8p^2+7p^3+5p^4$

$(3) \,\,\,$ $4x+9y+11z$

A best free mathematics education website for students, teachers and researchers.

###### Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

###### Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

###### Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.


text:
# Example of an operator with purely residual spectrum

Do you know an example of a linear bounded operator acting on a Banach (or even Hilbert) space whose residual spectrum is non-empty but the point and continuous spectrum are empty?

Take $X=l^{2}$. Now define the following continuous linear map
$$T:l^{2} \to l^{2}$$
$$T(x_{1},x_{2},.....)=(0, x_{1},\frac{x_{2}}{2}, \frac{x_{3}}{3} .... )$$
You can show that this $\sigma(T)=\{\ 0 \}\$. And 0 is in the residual spectrum since T is one-one but $Range(T)$ is not dense in $l^{2}$.


text:
College Algebra with Corequisite Support 2e

# Practice Test

### Practice Test

For the following exercises, determine whether each of the following relations is a function.

1 .

$y=2x+8 y=2x+8$

2 .

${ (2,1),(3,2),(−1,1),(0,−2) } { (2,1),(3,2),(−1,1),(0,−2) }$

For the following exercises, evaluate the function $f(x)=−3 x 2 +2x f(x)=−3 x 2 +2x$ at the given input.

3 .

$f(−2) f(−2)$

4 .

$f(a) f(a)$

5 .

Show that the function $f(x)=−2 (x−1) 2 +3 f(x)=−2 (x−1) 2 +3$ is not one-to-one.

6 .

Write the domain of the function $f(x)= 3−x f(x)= 3−x$ in interval notation.

7 .

Given $f(x)=2 x 2 −5x, f(x)=2 x 2 −5x,$ find $f(a+1)−f(1) f(a+1)−f(1)$ in simplest form.

8 .

Graph the function

9 .

Find the average rate of change of the function $f(x)=3−2 x 2 +x f(x)=3−2 x 2 +x$ by finding $f(b)−f(a) b−a f(b)−f(a) b−a$ in simplest form.

For the following exercises, use the functions to find the composite functions.

10 .

$( g∘f )(x) ( g∘f )(x)$

11 .

$( g∘f )(1) ( g∘f )(1)$

12 .

Express $H(x)= 5 x 2 −3x 3 H(x)= 5 x 2 −3x 3$ as a composition of two functions, $f f$ and $g, g,$ where $( f∘g )(x)=H(x). ( f∘g )(x)=H(x).$

For the following exercises, graph the functions by translating, stretching, and/or compressing a toolkit function.

13 .

$f(x)= x+6 −1 f(x)= x+6 −1$

14 .

$f(x)= 1 x+2 −1 f(x)= 1 x+2 −1$

For the following exercises, determine whether the functions are even, odd, or neither.

15 .

$f(x)=− 5 x 2 +9 x 6 f(x)=− 5 x 2 +9 x 6$

16 .

$f(x)=− 5 x 3 +9 x 5 f(x)=− 5 x 3 +9 x 5$

17 .

$f(x)= 1 x f(x)= 1 x$

18 .

Graph the absolute value function $f(x)=−2| x−1 |+3. f(x)=−2| x−1 |+3.$

For the following exercises, find the inverse of the function.

19 .

$f(x)=3x−5 f(x)=3x−5$

20 .

$f(x)= 4 x+7 f(x)= 4 x+7$

For the following exercises, use the graph of $g g$ shown in Figure 1.

Figure 1
21 .

On what intervals is the function increasing?

22 .

On what intervals is the function decreasing?

23 .

Approximate the local minimum of the function. Express the answer as an ordered pair.

24 .

Approximate the local maximum of the function. Express the answer as an ordered pair.

For the following exercises, use the graph of the piecewise function shown in Figure 2.

Figure 2
25 .

Find $f(2). f(2).$

26 .

Find $f(−2). f(−2).$

27 .

Write an equation for the piecewise function.

For the following exercises, use the values listed in Table 1.

 $x x$ 0 1 2 3 4 5 6 7 8 $F(x) F(x)$ 1 3 5 7 9 11 13 15 17
Table 1
28 .

Find $F(6). F(6).$

29 .

Solve the equation $F(x)=5. F(x)=5.$

30 .

Is the graph increasing or decreasing on its domain?

31 .

Is the function represented by the graph one-to-one?

32 .

Find $F −1 (15). F −1 (15).$

33 .

Given $f(x)=−2x+11, f(x)=−2x+11,$ find $f −1 (x). f −1 (x).$


text:
# LeetCode Weekly Contest 131

https://leetcode.com/contest/weekly-contest-131

## Remove Outermost Parentheses

A valid parentheses string is either empty (""), "(" + A + ")", or A + B, whereA and Bare valid parentheses strings, and + represents string concatenation. For example, "", "()", "(())()", and "(()(()))" are all valid parentheses strings.

A valid parentheses string S is primitive if it is nonempty, and there does not exist a way to split it into S = A+B, with A and B nonempty valid parentheses strings.

Given a valid parentheses string S, consider its primitive decomposition: S = P_1 + P_2 + ... + P_k, where P_i are primitive valid parentheses strings.

Return S after removing the outermost parentheses of every primitive string in the primitive decomposition of S.

Example 1:

Example 2:

Example 3:

Note:

1. S.length <= 10000
2. S[i] is "(" or ")"
3. S is a valid parentheses string

## Sum of Root To Leaf Binary Numbers

Given a binary tree, each node has value 0 or 1. Each root-to-leaf path represents a binary number starting with the most significant bit. For example, if the path is 0 -> 1 -> 1 -> 0 -> 1, then this could represent 01101 in binary, which is 13.

For all leaves in the tree, consider the numbers represented by the path from the root to that leaf.

Return the sum of these numbers.

Example 1:

Note:

1. The number of nodes in the tree is between 1 and 1000.
2. node.val is 0 or 1.
3. The answer will not exceed 2^31 - 1.

DFS找到每个到叶结点的路径对应的二进制数，遍历到叶结点时直接把路径代表的数字加到最终结果。

## Camelcase Matching

A query word matches a given pattern if we can insert lowercase letters to the pattern word so that it equals the query. (We may insert each character at any position, and may insert 0 characters.)

Given a list of queries, and a pattern, return an answer list of booleans, where answer[i] is true if and only if queries[i] matches the pattern.

Example 1:

Example 2:

Example 3:

Note:

1. 1 <= queries.length <= 100
2. 1 <= queries[i].length <= 100
3. 1 <= pattern.length <= 100
4. All strings consists only of lower and upper case English letters.

## Video Stitching

You are given a series of video clips from a sporting event that lasted T seconds. These video clips can be overlapping with each other and have varied lengths.

Each video clip clips[i] is an interval: it starts at time clips[i][0] and ends at time clips[i][1]. We can cut these clips into segments freely: for example, a clip [0, 7] can be cut into segments [0, 1] + [1, 3] + [3, 7].

Return the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event ([0, T]). If the task is impossible, return -1.

Example 1:

Example 2:

Example 3:

Example 4:

Note:

1. 1 <= clips.length <= 100
2. 0 <= clips[i][0], clips[i][1] <= 100
3. 0 <= T <= 100

dp[i]表示保证覆盖到第i秒时最少需要多少clip，如果有一个区间为[x,y]的clip，且x<=i，那么dp[i]+1可以作为dp[y]的一个备选。先把区间按照x排序，然后计算每个[x,y]可以续接哪些i，然后更新dp[y]以及iy之间所有dp不超过dp[y]，最后检查dp[T]是否还是未被更新的默认值，如果是则返回-1，否则直接返回dp[y]

x1<x2<y2<y1时，可以把[x2,y2]合并到[x1,y1]，只考虑[x1,y1]，那么在排序时可以直接以(x,-y)排序，把排序后y值小于前面的元素的去掉，可以减少部分工作量。


text:
# Convergent sum with primes

If $f(n)$ is a strictly increasing elementary function from the reals to the reals, and $p(n)$ is the $n$'th prime number. Is there any $f(n)$ such that $\sum_{n=1}^\infty\frac{1}{f(p(n))}$ is algebraic, or has a closed form in terms of elementary functions?

If not, is there a method to prove that the sum must be transcendental?

-
What do you mean by closed form, anyway? Do values of the prime zeta function count? –  Ｊ. Ｍ. Nov 3 '11 at 10:53
Also posted to MO, mathoverflow.net/questions/79904/convergent-series-with-primes where I expect it to be closed, soon. –  Gerry Myerson Nov 3 '11 at 12:57
Clearly, there exist answerers who don't read comments... :D –  Ｊ. Ｍ. Nov 3 '11 at 14:12
Then, if you also include the inverse trig functions in the "elementary" functions list, I think that $f(n) = \frac{1}{ \ln (1- \frac{1}{n^2})}$ will do. Your series then becomes the ln of the Euler product. Of course $\ln( \frac{\pi^2}{6})$ is probably transcendent, but it can be expressed an combo of logs and inverse trig... And the logarithm can be any base.... Oh, wait $f(1)$ doesn't make any sense, the function is defined only from $n=2$. Probably someone can fix it though... :( –  N. S. Nov 3 '11 at 15:41
@user9176: $f(1)$ doesn't have to make sense; only $f(p(1)) = f(2)$, ne? I would write that up as an answer since it seems to clearly answer the broader (elementary) version of the question; now I'm left wondering about the algebraic version of this... –  Steven Stadnicki Nov 3 '11 at 21:43


text:
Under the auspices of the Computational Complexity Foundation (CCF)

REPORTS > DETAIL:

### Paper:

TR14-164 | 30th November 2014 06:14

#### On the (Non) NP-Hardness of Computing Circuit Complexity

TR14-164
Authors: Cody Murray, Ryan Williams
Publication: 30th November 2014 08:33
Keywords:

Abstract:

The Minimum Circuit Size Problem (MCSP) is: given the truth table of a Boolean function $f$ and a size parameter $k$, is the circuit complexity of $f$ at most $k$? This is the definitive problem of circuit synthesis, and it has been studied since the 1950s. Unlike many problems of its kind, MCSP is not known to be $NP$-hard, yet an efficient algorithm for this problem also seems very unlikely: for example, MCSP $\in P$ would imply there are no pseudorandom functions.

Although most $NP$-complete problems are complete under strong local'' reduction notions such as poly-logarithmic time projections, we show that MCSP is provably not $NP$-hard under $O(n^{1/2-\varepsilon})$-time projections, for every $\varepsilon > 0$. We prove that the $NP$-hardness of MCSP under (logtime-uniform) $AC0$ reductions would imply extremely strong lower bounds: $NP \not\subset P/poly$ and $E \not\subset$ io-SIZE$(2^{\delta n})$ for some $\delta > 0$ (hence $P = BPP$ also follows). We show that even the $NP$-hardness of MCSP under general polynomial-time reductions would separate complexity classes: $EXP \neq NP \cap P/poly$ which implies $EXP \neq ZPP$. These results help explain why it has been so difficult to prove that MCSP is $NP$-hard.

We also consider the nondeterministic generalization of MCSP: the Nondeterministic Minimum Circuit Size Problem (NMCSP), where one wishes to compute the nondeterministic circuit complexity of a given function. We prove that the $\Sigma_2 P$-hardness of NMCSP, even under arbitrary polynomial-time reductions, would imply $EXP \not\subset P/poly$.

ISSN 1433-8092 | Imprint


text:
# Homework Help: Finding the volume of solid

1. Oct 27, 2012

### mr3369

1. The problem statement, all variables and given/known data
the function is y = -x^2+6x -8

suppose a city is surrounded by a ring of mountains and these mountains can be illustrated by rotating the above function around the y-axis. Find the volume of the earth that makes up these mountains.

Suppose the city suffers from air pollution and wants to build a system to reduce this air pollution. However, in order to do so, it needs to find the volume of polluted air. This volume is the region between the y-axis and the function, rotated about either x or y axis. The picture for the polluted air is attached.

2. Relevant equations

the function is y = -x^2+6x -8

3. The attempt at a solution

For the first part of finding the volume of earth, I integrated the function, using the cylindrical shell method. And I got the answer as 8 pi.
I used the radius as r = x, height as h = -x^2+6x -8,
So I integrated 2pi*r*h, from 2 to 4 which are the x intercepts.

For the 2nd part, I made the function in terms of y, So first I did completing the square and got y=-(x-3)^2 +1
So that gives me x = sqrt(1-y) +3
I used x as the radius, and using disk method , revolved the function around the y-axis.
So radius = sqrt(1-y) +3 ,
Then i integrated pi(r^2) from 0 to 3.
However, my answer gives me root(-2) which isn't possible. So I am stuck in this 2nd part.

2. Oct 27, 2012

### Simon Bridge

Best way is to sketch the function first ... then shade the region to be integrated.
http://en.wikipedia.org/wiki/Solid_of_revolution

For the second part ... you may have to divide the region up into parts you can integrate over and add op the bits. Start out putting it as x(y) if you like, or you can figure a way to find the volume that is not part of what you want and subtract it off a greater volume that includes both.

3. Oct 27, 2012

### haruspex

You might want to check the range again. What are you integrating wrt?

4. Oct 27, 2012

### Simon Bridge

I don't think I explained myself properly last time - I'll revisit:
x or y axis?

Anyway - the area of interest is between the line $x=0$ and the curve $y=-x^2+6x-8$ ... however: y(x) is an inverted parabola ... so for a particular value of y there are two possible values of x. y(x) also has a maximum - so there is an upper limit to any integration along the y axis.

Rotating this shape about the y axis should create a trumpet-shape with the point at y=y(0) (if the inner surface is the one intended). It is unclear what would be intended by using the outer surface; nor by rotating about the x-axis, since y(x) extends above and below it. That picture of the pollution would have been handy.

This is why I urged you to sketch the situation.

What about the negative square-root: does it make a difference? Which side of the parabola is determining the length of the radius?

To do the method of disks - you are thinking of a horizontal disk, thickness dy, centered at (0,y), with radius x(y) so the volume of the disk at y is $dV=\pi x^2(y)dy$. The trick is to get the correct x(y) and the correct limits for the integration.

How did you choose those limits?

Last edited: Oct 27, 2012


text:
# Wykaz publikacji wybranego autora

## Mateusz Dyndał, dr inż.

Wydział Fizyki i Informatyki Stosowanej
WFiIS-kod, Katedra Oddziaływań i Detekcji Cząstek

 Identyfikatory Autora Informacje o Autorze w systemach zewnętrznych ORCID: 0000-0001-9632-6352 ResearcherID: brak Scopus: brak PBN: 3990835 OPI Nauka Polska

Zakres roczników od wszystkie lata...201920182017201620152014201320122011201020092008200720062005200420032002200120001999<1999 wszystko...referatyartykuły

Opisy publikacji wcześniejszych zobacz: bpp.agh.edu.pl/old.

Liczba pozycji spełniających powyższe kryteria selekcji: 258, z ogólnej liczby 319 publikacji Autora

 wydruk: sortowanie:

1
2
• AFP simulation
3
• Analysis of events with b-jets and a pair of leptons of the same charge in \emph {pp} collisions at $\sqrt{s}=8 TeV$ with the ATLAS detector
4
• ATLAS Run 1 searches for direct pair production of third-generation squarks at the Large Hadron Collider
5
• A measurement of the ratio of the production cross sections for \emph {W} and \emph {Z} bosons in association with jets with the ATLAS detector
6
• A neural network clustering algorithm for the ATLAS silicon pixel detector
7
• A new method to distinguish hadronically decaying boosted \emph {Z} bosons from \emph {W} bosons using the ATLAS detector
8
• A search for an excited muon decaying to a muon and two jets in \emph {pp} collisions at $\sqrt{s}=8 TeV$ with the ATLAS detector
9
• A search for high-mass resonances decaying to ${\tau^{+}\tau^{-}}$ in \emph{pp} collisions at ${\sqrt{s}}$ = 8 TeV with the ATLAS detector
10
• A search for prompt lepton-jets in \emph{pp} collisions at ${\sqrt {s}} = 8 TeV$ with the ATLAS detector
11
• A search for top squarks with \emph{R}-parity-violating decays to all-hadronic final states with the ATLAS detector in $\sqrt{s}=8 TeV$ proton-proton collisions
12
• A search for $t\bar{t}$ resonances using lepton-plus-jets events in proton-proton collisions at $\sqrt{s}$ = 8 TeV with the ATLAS detector
13
• Beam-induced and cosmic-ray backgrounds observed in the ATLAS detector during the LHC 2012 proton-proton running period
14
• Beam tests of an integrated prototype of the ATLAS Forward Proton detector
15
• Centrality and rapidity dependence of inclusive jet production in ${\sqrt{s_{NN}}}$ = 5.02 TeV proton-lead collisions with the ATLAS detector
16
• Centrality, rapidity, and transverse momentum dependence of isolated prompt photon production in lead-lead collisions at $\sqrt{s}_{NN}= 2.76 TeV$ measured with the ATLAS detector
17
• Charged-particle distributions in \emph{pp} interactions at $\sqrt{s}=8 TeV$ measured with the ATLAS detector
18
• Charged-particle distributions in $\sqrt{s}=13 TeV$ \emph {pp} interactions measured with the ATLAS detector at the LHC
19
• Combination of searches for \emph {WW, WZ} and \emph {ZZ} resonances in \emph {pp} collisions at $\sqrt{s}=8 TeV$ with the ATLAS detector
20
• Combined measurement of the Higgs boson mass in {\em pp} collisions $\sqrt{s} = 7 and 8 TeV$ with the ATLAS and CMS experiments
21
• Comprehensive measurements of \emph {t}-channel single top-quark production cross sections at $\sqrt {s}$ = 7 TeV with the ATLAS detector
22
• Constraints on new phenomena via Higgs boson couplings and invisible decays with the ATLAS detector
23
• Constraints on non-Standard Model Higgs boson interactions in an effective Lagrangian using differential cross sections measured in the $H \to \gamma \gamma$ decay channel at $\sqrt{s}=8 TeV$ with the ATLAS detector
24
• Constraints on the off-shell Higgs boson signal strength in the high-mass ZZ and WW final states with the ATLAS detector
25
• Determination of spin and parity of the Higgs boson in the WW* $\to$ $e\nu\mu\nu$ decay channel with the ATLAS detector


text:
# Effusion changing pressure

1. Dec 21, 2017

### Physgeek64

1. The problem statement, all variables and given/known data
A thin-walled container of volume 10−3 m3 is evacuated to a pressure of 10−7 mbar. The vessel is surrounded by air at 1 bar and 290 K. If a small hole of area 10−17 m2 is made in the wall of the container, how long does it take the pressure inside to rise to 10−6 mbar?

2. Relevant equations

3. The attempt at a solution
$\frac{dN}{dt}=A\phi = \frac{A<v> N}{4V}$
$\frac{dN}{N}=\frac{A<v>}{4V} t$
$t=\frac{4V}{A<v>}\ln{\frac{P_2}{P_1}}$

However this requires us to know m and i can't see how to figure this out. Therefore i am assuming my solution is wrong....

2. Dec 21, 2017

Your original differential equation is incorrect. You are assuming the effusion only goes in one direction is this problem (that is ok), but the equation should read $\frac{dN}{dt}=\frac{A n \bar{v}}{4 }$ where $n$ is the number of particles per unit volume of the outside air. ($N$ is the number of particles of air inside the container). This $n$ is constant and does change. One number you will need is $\bar{v}$. Average speed $\bar{v}$ in a Maxwell-Boltzmann distribution is given by $\bar{v}= (\frac{8}{\pi}) (\frac{kT}{M})^{1/2}$ , editing: $\bar{v}=(\frac{8 kT}{\pi M})^{1/2}$, where $M$ is the (approximate) mass of a molecule of air. Since $N_2$ has M.W.=28, and $O_2$ has M.W.=32 , using an average M.W.=30 is an ok approximation, even though the air is 78% $N_2$ and 21% $O_2$.

Last edited: Dec 21, 2017
3. Dec 21, 2017

### Physgeek64

Oh okay, so will i end up with a simple linear relationship between t and N?

4. Dec 21, 2017

Yes. The pressure inside the container is small enough compared to the outside pressure that the rate of effusion into it is much higher than any effusion back out. And yes, the effusion rate into it will be constant. $\\$ And don't forget to include the initial condition of $N_o$ that you start with already inside the container. $\\$ Just for an additional detail: In its entirety $\frac{dN}{dt}=A (\frac{n \bar{v}}{4}-\frac{N \bar{v}}{4 V})$, and this second term on the right side can be considered small compared to the first term for this problem. (No need to solve the more detailed exact differential equation, because the second term is negligible here).

Last edited: Dec 21, 2017
5. Dec 21, 2017

@Physgeek64 See also the additions to post 4 above. $\\$ Also one minor correction to the above: $\bar{v}=( \frac{8 kT}{\pi M})^{1/2 }$.