# Triple Integral Limit Calculator

Definite Integral Calculator computes definite integral of a function over an interval utilizing numerical integration. The integral is equal to the area of. Functions like sin and sqrt work, as do special constants like pi and e. 2 Double Integrals over General Regions In this section we define and evaluate double integrals over bounded regions in the plane which are more general than rectangles. In double integrals we saw how to calculate volume by integrating a three dimensional function using limits of intgration of a bounded AREA. Triple integrals 3. For example, if f ( x) is positive for. To get a viewing window containing a and b, these values must be between Xmin. Fast and easy to use. Write a triple integral, including limits of integration, that gives the volume between 3x+y+z=0 and 4x+4y+z=0, and above x+y<=1, x>=0, y>=0. TRIPLE INTEGRALS 3 5B-2 Place the solid hemisphere D so that its central axis lies along the positive z-axis and its base is in the xy-plane. Notes on Triple Integration Dr. Convert this triple integral into cylindrical coordinates and evaluate $\int_{-1}^{1}\int_{0}^{\sqrt{1-x^2}}\int_{0}^{y}x^2dz\; dy\; dx onumber$ Solution. A volume integral is a specific type of triple integral. The sphere x2 +y2 +z2 = 4 is the same as ˆ= 2. Second Derivative. Triple Integrals in Cartesian Coordinates: Over rectangular. Includes derivatives, integration, volume, quadratic equation and trig identities. Elliott Jacobs On Wednesday, March 4, you saw how to set up triple and calculate triple integrals. Now is the lower limit and is. For example, if f ( x) is positive for. pro [email protected] It is the height of a thin stick as in Section 14. The key idea is to replace a double integral by two ordinary "single" integrals. Let's do limit comparison to 1/t3: lim t→∞ 1/t3 1/t3−t =lim t→∞ t3−t t3 =lim t→∞ t3−t t3. $$\int_0^4 \int_{0}^{1-x} \int x^2 + y^2 dzdydx$$ I think those two limits may be correct but I don't know how to get the third, I understand how to find limits form a shape like a tetrahedron but a bit confused and my notes don't help me with this shape. We calculate as follows:. dzdxdy, then once you find your Z limits in the first integral, then you are done with Z altogether, the next step to solve the triple integral is to project into the remaining variables' plane, in other words, project into the x-y plane. So we should just calculate that limit, for arbitrary a, and then let a!1. V = \iiint\limits_U { {\rho ^2}\sin \theta d\rho d\varphi d\theta }. Changing the order of integration of a triple integral blackpenredpen. Third Derivative. A similar situation occurs with triple integrals, but here we need to distinguish between cylindrical symmetry and spherical symmetry. restart: Setting limits of integration and evaluating. • Evaluate double integrals over general regions. When there are limits, and we need to use U-Substitution, there are a few things we need to keep in mind:. Summary : The integral function calculates online the integral of a function between two values. Recently Asked Math Questions. Arc Length Cartesian & Polar Coordinates, 2D & 3D Parametric Curves. In python we use numerical quadrature to achieve this with the scipy. To approximate a volume in three dimensions, we can divide the three-dimensional region into small rectangular boxes, each $\Delta x\times\Delta y\times\Delta z. Set up (but do not evaluate) iterated triple integrals, with appropriate limits, for nd the volume of the solid bounded by z= x2 + y2 and z= 8 x2 y2 in: (a)rectangular coordinates (b)cylindrical coordinates 9. The purpose of these notes is to present a number of triple integral examples and discuss how to set up the limits of integration. Notes on Triple Integration Dr. The limits of integration for this graph are (0,2). You can also use it to solve differential and integral equations. Multiple Integral Calculator - eMathHelp This site contains an online calculator that finds multiple integrals (double or triple integrals). The concept of integrals is fundamental in calculus, and has broad application in all engineering disciplines. V = \iiint\limits_U {\rho d\rho d\varphi dz}. Watch if the algorithm is converging. Let's return to the previous visualization of triple integrals as masses given a function of density. The int function can be used for definite integration by passing the limits over which you want to calculate the integral. I am trying to solve below mentioned (image) equation in R for double integral over an area. 0 (R14), you will need to use a function file or. Simplify a calculation by changing the order of integration of a triple integral. This is somewhat subtle in the physical interpretation but can be summarized as "generality". An integral is a mathematical result that represents the area between a function and a plane (e. , 26 MB] Improper integrals - part 2 - integrals with integrand undefined at an endpoint [video; 21 min. Sometimes you need to change the order of integration to get a tractable integral. For example, if you tried to evaluate ∫1 0∫1 xey2dydx directly, you would run into trouble. Triple Integral Cylindrical Coordinates. To set up a double. Create the worksheets you need with Infinite Calculus. Calculate the average value of a function of three variables. Integral bounds , also called limits of integration, define the area that you’ll be integrating. set up triple integrals to compute the volume of a solid. Math 6B is the second quarter of a two quarter sequence in vector calculus and infinite series. 6 Implicit Differentiation. So, if possible, please write the logic for the solution. 973 #55,56) Calculate mass by integrating density. Enter all requested information on the top of this page, and put your initials on the top of every page, in case the pages become separated. Evaluating Double Integrals by Repeated Integrals 5. This calculator evaluates derivatives using analytical differentiation. Never runs out of questions. ranges here in the interval 0 \le x \le 1, and the variable y. ) Finding the Limits. Calculate the volume of the solid. You can also easily calculate multiple integrals as well as use mathematical constants such as. The limit of the Riemann sum is the triple integral of f over D and f is continuous. For example: 𝑟 𝑟 𝜃 3 −3 2 0 2π 0 is the triple integral used to calculate the volume of a cylinder of height 6 and radius 2. For a sequence indexed on the natural number set , the limit is said to exist if, as , the value of the elements of get arbitrarily close to. ( Click here for an explanation) [ ti-83/ti-84 ] Approximate Integration. Evaluating double integrals is similar to evaluating nested functions: You work from the inside out. Absolute Minimum. Triple integral - volume of sphere. Evaluate a triple integral by expressing it as an iterated integral. ranges in the interval 0 \le y \le 2 – 2x. Triple Integral Calculator. In this section we convert triple integrals in rectangular coordinates into a triple integral in either cylindrical or spherical coordinates. (1 point) Write a triple integral including limits of integration that gives the volume of the cap of the solid sphere x2 + y +2 s 25 cut off by the plane z 3 and restricted to the first octant. Triple Integral Spherical Coordinates. We start from the simplest case when the region of integration $$U$$ is a rectangular box $$\left[ {a,b} \right] \times \left[ {c,d} \right]$$ $$\times \left[ {p,q} \right]$$ (Figure $$1$$). If you have a function f(x), there are several ways to mark the derivative of f when it comes to x. sets, logic, proofs. Fill in the integrand, the limits, and the integrating variable for. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. Analysis & calculus symbols table - limit, epsilon, derivative, integral, interval, imaginary unit, convolution, laplace transform, fourier transform. the triple integral of a continuous function $$f(x,y,z)$$ over a rectangular solid box $$B$$ is the limit of a Riemann sum for a function of three variables, if this limit exists Contributors Gilbert Strang (MIT) and Edwin "Jed" Herman (Harvey Mudd) with many contributing authors. 7 Triple Integrals Be able to evaluate a given triple integral. Evaluate a Triple Integral Using Cylindrical Coordinates - Triple Integral of e^z Evaluate a Triple Integral Using Spherical Coordinates - Triple Integral of 1/(x^2+y^2+z^2) Find the Moment of Inertia about the z-axis of a Solid Using Triple Integrals Find the Center of Mass of a Solid Using Triple Integrals. • Calculate the Jacobian of a transformation of two and three variables. Calculate the volume of the solid. Create the worksheets you need with Infinite Calculus. I The average value of a function in a region in space. Derivative Calculator. I Triple integrals in arbitrary domains. Let’s do limit comparison to 1/t3: lim t→∞ 1/t3 1/t3−t =lim t→∞ t3−t t3 =lim t→∞ t3−t t3. They will make you ♥ Physics. The Integral Calculator supports definite and indefinite integrals (antiderivatives) as well as integrating functions with many variables. What does mean where is a region in the plane. This is a topic that takes some practice. V = \iiint\limits_U { {\rho ^2}\sin \theta d\rho d\varphi d\theta }. Calculate the average value of a function of three variables. So I'll start with the triple integral. To find those limits on the z integral, follow a line in the z direction. Further, it is possible to show that the limit. BYJU’S online triple integral calculator tool makes the calculation faster, and it displays the integrated value in a fraction of seconds. Absolutely free, online math solution. Limit Calculator calculates an established limit of the function with respect to a variable in a specific point. is given by where R(xyz) is the region of integration in xyz space, R(uvw) is the corresponding region of integration in uvw space, and the. Next, I'll substitute the limits. io development by creating an account on GitHub. One of the ways in which definite integrals can be improper is when one or both of the limits of integration are infinite. Define the indefinite integral of a vector-valued function. Integrals of a function of two variables over a region in R 2 are called double integrals, and integrals of a function of three variables over a region of R 3 are called triple integrals. The limits on z run from 0 to h. quad command. Triple Integral Spherical Coordinates. In this section we convert triple integrals in rectangular coordinates into a triple integral in either cylindrical or spherical coordinates. Math terminology from differential and integral calculus for functions of a single variable. sets, logic, proofs. Online Triple & Double Integral Calculator With Steps. And then finish with dx to mean the slices go in the x direction (and approach zero in width). index: subject areas. Fraction calculator. What Everybody Dislikes About Double Integral Calculator and Why. The general steps required to perform the above integral. Since the plane ABC. Wolfram|Alpha is a great tool for calculating indefinite and definite triple integrals. Figure 12-10 shows an example. In this example, since the limits of integration are constants, the order of integration can be changed. In many cases, it is convenient to represent the location of in an alternate set of coordinates, an example of which are the so-called polar coordinates. ha πha Outer integral: 2π =. In Calculus, the four important concepts are limits, continuity, derivatives and integrals. Second Derivative. Integrals involving radicals for instance, we want to get rid of the square root. How are triple integrals in rectangular coordinates evaluated? How are the limits of integration determined? Give an example. You can calculate integrals numerically using techniques such as the Simpson quadrature, Lobatto quadrature, and Gauss. It enters the prism at z = 0 and exits at the sloping face y + 32 = 3. We start from the simplest case when the region of integration $$U$$ is a rectangular box $$\left[ {a,b} \right] \times \left[ {c,d} \right]$$ $$\times \left[ {p,q} \right]$$ (Figure $$1$$). Reversing the order of integration. Create the worksheets you need with Infinite Calculus. More than just an online integral solver. If you're behind a web filter, please make sure that the domains *. Enter the series to calculate its sum: This calculator for to calculating the sum of a series is taken from Wolfram Alpha LLC. Integral Calculator If you were looking for a way to calculate the Integral value of a set of mumbers, then the Integral calculator is exactly what you need. To something small like 0. Partial Derivative. This integral is improper at infinity only, and for large t we know that t3 is the dominant part. This tutorial demonstrates how to evaluate integrals using the TI-89, TI-92+, or Voyage 200 graphing calculators. Expression Calculator evaluates an expression in a given context. The integral is equal to the area of. Write a triple integral, including limits of integration, that gives the volume between 3x+y+z=0 and 4x+4y+z=0, and above x+y<=1, x>=0, y>=0. It is easier to calculate triple integrals in spherical coordinates when the region of integration U is a ball (or some portion of it) and/or when the integrand is a kind of f\left ( { {x^2} + {y^2} + {z^2}} \right). Muliple Integration Section 1: DOUBLE INTEGRALS PROBLEM: Consider the solid E in 3-space bounded above by the surface z = 40 − 2xy and bounded below by the rectangular region D in the xy-plane (z = 0) deﬁned by the set D = {(x,y) : 1 ≤ x ≤ 3, 2 ≤ y ≤ 4}. Next, I’ll substitute the limits. Integration over surfaces, properties, and applications of integrals. Hold ˚and xed, and let ˆincrease. My work so far: Since it's a paraboloid, where each cross section parallel to the plane x = 5 is a circle, cylindrical polars is what I used, so my bounds are 5y 2 +5z 2 ≤x ≤ 5 -----> 5r2 ≤ x ≤ 5, since each cross-section is a full circle 0 ≤ θ ≤ 2π. Explore the solid defining the boundaries of the region for a triple integral. Changes of variable can be made using Jacobians in much the same way as for double integrals. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Using (19) and (22), we calculate ∂(x,y) ∂(u. Multiple-version printing. How to Use Definite Integral Calculator Integration can. sets, logic, proofs. Try int(x^2,x=t+2. The Integral Calculator Trap The last result is provided by taking the very first limit from the second. It is possible to calculate the limit at a of a function where a represents a real : If the limit exists and that the calculator is able to calculate, it returned. Among other things, they let us calculate the volume under a surface. So it will be. In polar coordinates, the point is located uniquely by specifying the distance of the point from the origin of a given coordinate system and the angle of the vector from the origin to the point from the positive -axis. An example shows how to set them up and how to evaluate them. Integral online. Enter a valid algebraic expression to find the derivative. Scalar line integrals can be used to calculate the mass of a wire; vector line integrals can be used to calculate the work done on a particle traveling through a field. Triple integrals 3. First of all, I’ll integrate with respect to. As a final example, we see how to compute the length of a curve given by parametric equations. Triple Integrals in Cartesian Coordinates: Over rectangular. Limits at Jump Discontinuities and Kinks. Evaluating Double Integrals by Repeated Integrals 5. Triple Integrals, Changing the Order of Integration, Part 1 of 3. The Riemann Sum formula is as follows: Below are the steps for approximating an integral using six rectangles: Increase the number of rectangles (n) to create a better approximation: Simplify this formula by factoring out w […]. Watch if the algorithm is converging. Just as with double integrals, the only trick is determining the limits on the iterated integrals. The syntax in dblquad is a bit more complicated than in Matlab. There are three steps that must be done in order to properly convert a triple integral into cylindrical coordinates. What Everybody Dislikes About Double Integral Calculator and Why. Try this handy Limit calc right now!. The key idea is to replace a double integral by two ordinary "single" integrals. Double and triple integrals. (Very simple problem). Apply, evaluate, and understand integrals of multi-variable scalar-valued functions. You can also check your answers! Interactive graphs/plots help visualize and better understand the functions. Mathispower4u 6,729 views. Don't forget to use the magnify/demagnify controls on the y-axis to adjust the scale. Please Explain Reasoning Behind Limits. Double Integrals over General Regions Type I and Type II regions Examples 1-4 Examples 5-7 Swapping the Order of Integration Area and Volume Revisited Double integrals in polar coordinates dA = r dr (d theta) Examples Multiple integrals in physics Double integrals in physics Triple integrals in physics Integrals in Probability and Statistics. Calculate Partial derivatives with incredible ease! Use this handy partial derivative calculator with a step-by-step solution and graph. Higher Order Derivatives. Finding volume for triple integrals using spherical coordinates. , when integrate is called as integrate (expr, x, a, b). I Triple integrals in arbitrary domains. Definite Integral Calculator computes definite integral of a function over an interval using numerical integration. index: subject areas. 9 Best Free Integral Calculator Software For Windows Here is a list of best free Integral Calculator Software to solve integrations. using Type I region using Type region calculate one fo the above integrals. In spherical coordinates, the volume of a solid is expressed as. advanced topics. The integral is the line integral of a continuous real-valued function or , i. Automatic spacing. Any time we have an iterated integral (and we can go beyond triple integrals) we can have the limits be non-constant functions. MTH 229 projects. It enters the prism at z = 0 and exits at the sloping face y + 32 = 3. Also recall the chapter prelude, which showed the opera house l’Hemisphèric in Valencia, Spain. 5b: Integrals in probability and statistics:. Series Expansions Number & Power Series, Fourier / Taylor / Laurent / Puiseux Series. Homework Statement Find the centroid of the solid: Limit of a function as n approaches infinity Finding the eigenfunctions and eigenvalues associated with an operator. As with most definite integrals, you should ignore the bounds (0 and 2) at first and focus on how to find an antiderivative of the function inside the integral. Use the upper right corner as your sample point of each rectangle to approximate the double integral. The Integral Calculator supports definite and indefinite integrals (antiderivatives) as well as integrating functions with many variables. So it will be. To begin, enter the limit expression in graphing or "y =" mode, in Table Setup, set Tbl to start to the arrow-number, and then set. I Triple integrals in arbitrary domains. Online Derivative Calculator. You are allowed to take one. Functions 3D Plotter is an application to drawing functions of several variables and surface in the space R3 and to calculate indefinite integrals or definite integrals. Solve limits step-by-step. See exercise 27. Instead of integrating a function of two variables over an area, we are integrating a function of three variables over a volume. De nition: The triple integral of fover Bis ZZZ B f(x;y;z)dV = lim jjPjj!0 Xl i=1 m j=1 Xn k=1 f(x ijk;y ijk;z ijk) V ijk provided that the limit exists. The innermost integral is the one that is integrated over first. In all of our examples above, the integrals have been indefinite integrals - in other words, integrals without limits of integration (the "a" and "b" in the statement "the integral from a to b"). Here is a list of best free Integral Calculator Software to solve integrations. Let's return to the previous visualization of triple integrals as masses given a function of density. In many cases, it is convenient to represent the location of in an alternate set of coordinates, an example of which are the so-called polar coordinates. Also recall the chapter prelude, which showed the opera house l’Hemisphèric in Valencia, Spain. Changing the order of integration in triple integrals Triple Integrals, Changing the Order of Integration,. numbers & symbols. Improper integrals - part 1 - introduction; integrals with an infinite limit of integration [video; 21 min. The mechanics for double and triple integration have been wrapped up into the functions dblquad and tplquad. A triple integral can be represented as , where f(x,y,z) is the integrated function defined over the three-dimensional shape E , and E is the region of integration in the (x,y,z) three. Includes derivatives, integration, volume, quadratic equation and trig identities. In triple integrals, the integral will be taken of a four-dimensional function using limits of integration of a bounded VOLUME. Type in the triple integral problem to solve To get started, type in a value of the triple integral and click «Submit» button. Since is defined as a limit of Riemann sums, the continuity of is enough to guarantee the existence of the limit, just as the integral exists if g is continuous over. Triple Integral Calculator at a Glance The 5-Minute Rule for Triple Integral Calculator. The text is Vector Calculus by M. Input a function, the integration variable and our math software will give you the value of the integral covering the selected interval (between the lower limit and the upper limit). By this, I mean you can take the volume of any three dimensional object with a triple integral, but you are somewhat limited with a double integral. Integration over surfaces, properties, and applications of integrals. If expr is a constant, then the default integration variable is x. We can solve for z to determine the upper limit of integration. }\) Activity 11. This differential volume can be expressed in six possible ways. The x and y coordinates lie in a disk of radius a, so 0 ≤ r ≤ a and 0 <θ ≤ 2π. ) is written as y = 2 – 2x. Define the indefinite integral of a vector-valued function. Observe that this function is symmetric across all eight octants, so we. Change the camera position and the direction of view in three dimensions. surface integral (1) is deﬁned to be this limit. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 9 Setting up the volume as a triple integral in spherical coordinates, we have: ZZZ S dV = Z 0 Z 2ˇ 0 Z R 0 ˆ2 sin˚dˆd d˚ = Z 0 Z 2ˇ 0 [1 3 ˆ 3]ˆ=R ˆ=0 sin˚d d˚ = 1 3 R 3(2ˇ)[ cos˚]˚= ˚=0 = 2 3 ˇR 3(1 cos ): In the special case = ˇ, we recover the well-known formula that. To calculate. We calculate as follows:. Triple integrals in Cartesian coordinates (Sect. Figure 1 In order for the double integral to exist, it is sufficient that, for example, the region D be a closed (Jordan) measurable region and that the function f(x, y ) be continuous throughout D. See more ideas about Vector calculus, Calculus and Calculus 2. Math terminology from differential and integral calculus for functions of a single variable. The integral is the line integral of a continuous real-valued function or , i. (B)A graphing calculator is required. Wolfram|Alpha is a great tool for calculating antiderivatives and definite integrals, double and triple integrals, and improper integrals. Introduction; Finding the area under a curve is a useful tool in a large number of problems in many areas of science, engineering, and business. 0 a 4 Middle integral: hr dr3 =. Handout 8: Plots of three dimensional regions associated with triple integrals. int uses the default integration variable determined by symvar ( expr,1 ). The limit calculator finds if it exists the limit at any point, at the limit at 0, the limit at +oo and the limit at -oo of a function. If expr is a constant, then the default integration variable is x. The limits on z run from 0 to h. Definite Integrals Calculator. Online Triple & Double Integral Calculator With Steps. How to Calculate Multiple Integrals. index: subject areas. No need to calculate any antiderivatives. The user enters a function of two or three variables and corresponding limits of integration and the tool evaluates the integral. Calculate the volume of the solid. Choose "Find the Derivative" from the menu and click to see the result!. These integral calculator can be used to calculate and solve definite integrals and indefinite integrals. My work so far: Since it's a paraboloid, where each cross section parallel to the plane x = 5 is a circle, cylindrical polars is what I used, so my bounds are 5y 2 +5z 2 ≤x ≤ 5 -----> 5r2 ≤ x ≤ 5, since each cross-section is a full circle 0 ≤ θ ≤ 2π. Triple integrals also arise in computation of Volume (if f(x,y,z)=1, then the triple integral equals the volume of R) Force on a 3D object Average of a Function over a 3D region Center of Mass and Moment of Inertia Triple Integrals in General Regions. Equation solver can find both numerical and parametric solutions of equations. Fubini’s Theorem for double integrals over rectangles; iterated integrals. I am using the following code by help of the following post: Triple integral in R (how to specifying the domain) Not sure, if I am missing anything as if I change value of A, it still gives me the same results. Summary : The integral function calculates online the integral of a function between two values. Triple integral limits over pyramid. This is a topic that takes some practice. (L points) et C be the half cylinder bounded by y = 0, z = 0, z = 2 and x2 +y2 = 1 for y. zip: 1k: 03-03-08: Ultimate Calculus Program Program includes Rolles, Trapezoidal, Mean Value, Rieman Sum, and Integral Equations. #N#Index for Calculus. numbers & symbols. The every single and general integration techniques and even unique, important functions being provided. Triple Integral Calculator at a Glance The 5-Minute Rule for Triple Integral Calculator. 7 Triple Integrals Be able to evaluate a given triple integral. Tech- II Subject: Engineering Mathematics II Unit-3 RAI UNIVERSITY, AHMEDABAD. BYJU’S online integral calculator tool makes the calculations faster, showing the integral value for the given function in a fraction of seconds. Step by step calculus inside your TI-89 & Titanium calculator. The cone z = p. By using this website, you agree to our Cookie Policy. 1-Definitions. If you're seeing this message, it means we're having trouble loading external resources on our website. We start from the simplest case when the region of integration $$U$$ is a rectangular box $$\left[ {a,b} \right] \times \left[ {c,d} \right]$$ $$\times \left[ {p,q} \right]$$ (Figure $$1$$). 3 Triple Integrals Question Find the prism volume in the order dz dy dx (six orders are possible). and convert it to cylindrical coordinates. use triple integral to prove volume of ice cream cone c is : pib^(3) Calculus: Apr 8, 2013: Cone volume with triple integral: Calculus: Dec 18, 2012: Cone volume-triple integral: Calculus: Jun 12, 2012: Mass of Ice Cream Cone using Triple Integrals: Calculus: Mar 22, 2012. • Apply the Change of Variables in Multiple Integrals. How to use the Indefinite Integral Calculator. This gives us a ray going out from the origin. The x and y coordinates lie in a disk of radius a, so 0 ≤ r ≤ a and 0 <θ ≤ 2π. I Examples: Changing the order of integration. advanced algebra. Be able to find the limits of integration for an integral given the solid region of integration. In a moment you will receive the calculation result. In polar coordinates, the point is located uniquely by specifying the distance of the point from the origin of a given coordinate system and the angle of the vector from the origin to the point from the positive -axis. I Examples: Changing the order of integration. Absolute Convergence. Step by step calculus inside your TI-89 & Titanium calculator. There is nothing that says that triple integrals set up as this is must only have constants as limits! So, here is the $$x$$ integration. This differential volume can be expressed in six possible ways. Triple integrals are the analog of double integrals for three dimensions. Double integral calculator provides you the facility of step by step solution of the problem which means that you can get a solution like your teachers to solve it on a white board. The limits of all inner integrals need to be defined as functions. Physical Applications of Triple Integrals : volume of sphere. In StandardForm, Integrate [ f, x] is output as ∫ f x. Evaluate the triple integral ∫∫∫E 5x dV, where E is bounded by the paraboloid x = 5y 2 + 5z 2 and the plane x = 5. Free Fourier Series calculator - Find the Fourier series of functions step-by-step This website uses cookies to ensure you get the best experience. To calculate the limits for an iterated integral. Of course, you can have Maple calculate multiple integrals. Requires the ti-83 plus or a ti-84 model. Theorem 13. We need an antiderivative of √(4. What is a cross product? A cross product, also known as a vector product, is a mathematical operation in which the result of the cross product between 2 vectors is a new vector that is perpendicular to both vectors. It is the height of a thin stick as in Section 14. Triple Integrals and Volume - Part 2 - Duration: 8:06. In general integrals in spherical coordinates will have limits that depend on the 1 or 2 of the variables. The definite integral of from to , denoted , is defined to be the signed area between and the axis, from to. The limits of all inner integrals need to be defined as functions. For a curve produced by a function, you may be able to integrate the function from a to b and calculate the area under the curve. When you see the table, you will mostly see the y values getting closer to the limit answer as homes in on arrow-number. xmin lower limit of outer integral. By using this website, you agree to our Cookie Policy. 2 A Catalog of Essential Functions 1. Let’s do limit comparison to 1/t3: lim t→∞ 1/t3 1/t3−t =lim t→∞ t3−t t3 =lim t→∞ t3−t t3. It is possible to calculate the limit at a of a function where a represents a real : If the limit exists and that the calculator is able to calculate, it returned. You solve this type of improper integral by turning it into a limit problem where c approaches infinity or negative infinity. The usual cautions about numerical methods apply, particularly when the function is not well behaved. It also shows plots, alternate forms, and other relevant information to enhance your mathematical intuition. Type in the indefinite integral problem to solve To get started, type in a value of the indefinite integral and click «Submit» button. One special case of the product rule is the constant multiple rule, which states that if c is a number and f(x) is a differential function, then cf(x) is also differential, and its derivative is (cf)'(x)=cf'(x). Example 4 Find volume of the tetrahedron bounded by the coordinate planes and the plane through$(2,0,0)$,$(0,3,0)$, and$(0,0,1)$. Get more help from Chegg Get 1:1 help now from expert Calculus tutors Solve it with our calculus problem solver and calculator. It also shows plots, alternate forms, and other relevant information to enhance your mathematical. The notation for the general triple integrals is, $\iiint\limits_{E}{{f\left( {x,y,z} \right)\,dV}}$ Let’s start simple by integrating over the box,. Triple integrals in spherical coordinates Our mission is to provide a free, world-class education to anyone, anywhere. coordinates, change of variables in multiple integrals. Classical integration theorems of vector calculus Math 6B. Step 2: Click the blue arrow to submit. 0, 1e5 or an expression that evaluates to a float, such as exp(-0. Double integrals over general regions. In cylindrical coordinates, the volume of a solid is defined by the formula. The Integral Calculator Trap The last result is provided by taking the very first limit from the second. The multiple integral is a definite integral of a function of more than one real variable, for example, f(x, y) or f(x, y, z). using Type I region using Type region calculate one fo the above integrals. This page shows you two ways to compute a definite integral with numeric limits, and how to plot an accumulation function. The magnitude of this new vector is equal to the area of a para. Definite Integral. Now is the lower limit and is. Suppose we have a double integral in x and y, which we would like to change to a double integral in two new variables, u, and v, where u = 3x + 4y and v = x - 2y. That gives the upper limit z = (3 -y)/3. Multiple-version printing. int uses the default integration variable determined by symvar ( expr,1 ). Second Derivative. Now here the solid is enclosed by the planes and the surface. TI-84 Plus and TI-83 Plus graphing calculator program for AP calculus students. First Derivative. If you took integral calculus, you probably learned that there are many functions whose integrals don't show up in any integral tables, simply because they are unknown to mathematics. For each fixed x we integ- rate with respect to y. If it's not clear what the y. Do a change of variables on the integral R dA, using x = u y =3v, where R be the region bounded by 9x 2+y = 36. Two key concepts expressed in terms of line integrals are flux and circulation. integrate (expr, x) is an indefinite integral, while integrate (expr, x, a, b) is a definite integral, with limits of integration a and b. You can solve double integrals in two steps: First evaluate the inner integral, and then plug this solution into the outer integral and solve that. Please Explain Reasoning Behind Limits. 3 Triple Integrals Question Find the prism volume in the order dz dy dx (six orders are possible). 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Multiple Integrals -- First Example: Degenerate Double Integral | # Following Examples are Variations of Examples from Math 210 Calculus III | # Textbook Multivariable Calculus, Third Edition, by James Stewart | # from Section 13. There are examples of valid and invalid expressions at the bottom of the page. Although we define triple integrals using a Riemann sum, we usually evaluate triple integrals by turning them into iterated integrals involving three single integrals. Volume integrals are especially important in physics for many applications, for example, to calculate flux densities. index: click on a letter. Integral Calculator If you were looking for a way to calculate the Integral value of a set of mumbers, then the Integral calculator is exactly what you need. Logo, Boxshot & ScreenShot. Note, that integral expression may seems a little different in inline and display math mode - in inline mode the integral symbol and the limits are compressed. 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# Expressing $\Bbb N$ as an infinite union of disjoint infinite subsets.

The title says it. I thought of the following: we want $$\Bbb N = \dot {\bigcup_{n \geq 1} }A_n$$
We pick multiples of primes. I’ll add $1$ in the first subset. For each set, we take multiples of some prime, that hasn’t appeared in any other set before. Then \begin{align} A_1 &= \{1, 2, 4, 6, 8, \cdots \} \\ A_2 &= \{3, 9, 15, 21, 27, \cdots \} \\ A_3 &= \{5, 25, 35, 55, \cdots \} \\ A_4 &= \{7, 49, 77, \cdots \} \\ &\vdots \end{align}

I’m heavily using the fact that there are infinite primes. I think these sets will do the job. Can someone check if this is really ok? Also, it would be nice to know how I could express my idea better, instead of that hand-waving. Alternate solutions are also welcome. Thank you!

Edit: the subsets must be also infinite.

#### Solutions Collecting From Web of "Expressing $\Bbb N$ as an infinite union of disjoint infinite subsets."

Here is another way to do it.
Let $A_{i}$ consist of all the numbers of the form $2^im$ where $2\nmid m$. That is, $A_i$ consists of all the numbers that have exactly a factor of $2^i$ in them. So
\begin{align} A_0 &= \{1,3,5,7,9,11, \dots\}\\ A_1 &= \{2, 6 =2^1\cdot 3, 10 = 2^1\cdot 5, 14 = 2^1\cdot 7, \dots\}\\ A_2 &= \{4 = 2^2, 12=2^2\cdot 3, 20=2^2\cdot 5, \dots\}\\ A_3 &= \{8=2^3, 24=2^3\cdot 3, 40=2^3\cdot 5, \dots \}\\ &\vdots \end{align}
In general $A_i = \{2^im: p\nmid m\}$. You can of course pick any other prime instead of $2$.

I like @Thomas’s answer best, but I would have enumerated $\mathbb N\times\mathbb N$ and then taken the inverse images of the separate columns for the subsets.

This is an alternate solution.

Let $\alpha$ and $\beta$ be any pair of irrational numbers such that
$$1 < \alpha < 2 < \beta\quad\text{ and }\quad \frac{1}{\alpha} + \frac{1}{\beta} = 1.$$
The two sequences
$\displaystyle\;\left\lfloor\alpha k\right\rfloor\;$ and
$\displaystyle\;\left\lfloor\beta k\right\rfloor\;$, $k \in \mathbb{Z}_{+}$
are called Beatty sequence and it is known they form a partition of $\mathbb{Z}_{+}$.
Define two functions $\tilde{\alpha}, \tilde{\beta} : \mathbb{Z}_{+} \to \mathbb{Z}_{+}$ by
$$\tilde{\alpha}(k) = \left\lfloor\alpha k\right\rfloor \quad\text{ and }\quad \tilde{\beta}(k) = \left\lfloor\beta k\right\rfloor$$
We have
$$\mathbb{Z}_{+} = \tilde{\alpha}(\mathbb{Z}_{+}) \uplus \tilde{\beta}(\mathbb{Z}_{+})$$ where $\uplus$ stands for disjoint union.
Replace the rightmost $\mathbb{Z}_{+}$ recursively by this relation, we get

\begin{align} \mathbb{Z}_{+} &= \tilde{\alpha}(\mathbb{Z}_{+}) \uplus \tilde{\beta}(\mathbb{Z}_{+})\\ &= \tilde{\alpha}(\mathbb{Z}_{+}) \uplus \tilde{\beta}(\tilde{\alpha}(\mathbb{Z}_{+})) \uplus \tilde{\beta}(\tilde{\beta}(\mathbb{Z}_{+})))\\ &= \tilde{\alpha}(\mathbb{Z}_{+}) \uplus \tilde{\beta}(\tilde{\alpha}(\mathbb{Z}_{+})) \uplus \tilde{\beta}(\tilde{\beta}(\tilde{\alpha}(\mathbb{Z}_{+}))) \uplus \tilde{\beta}(\tilde{\beta}(\tilde{\beta}(\mathbb{Z}_{+}))))\\ &\;\vdots \end{align}
As a consequence, if one define a sequence of subsets $A_1, A_2, \ldots \subset \mathbb{Z}_{+}$ recursively by
$$A_1 = \tilde{\alpha}(\mathbb{Z}_{+}) \quad\text{ and }\quad A_n = \tilde{\beta}(A_{n-1}), \quad\text{ for } n > 1,$$
these subsets will be pairwise disjoint. It is clear all these $A_n$ are infinite sets.
Since $\beta > 2$, we have

$$\tilde{\beta}(k) = \left\lfloor \beta k \right\rfloor > \beta k – 1 \ge (\beta – 1) k\quad\text{ for all } k \in \mathbb{Z}_{+}$$
This implies
$$\tilde{\beta}^{\circ\,\ell}(k) = \underbrace{\tilde{\beta}(\tilde{\beta}( \cdots \tilde{\beta}(k)))}_{\ell \text{ times}} > (\beta-1)^\ell k \ge (\beta-1)^\ell \quad\text{ for all } k, \ell \in \mathbb{Z}_{+}$$

As a result,

$$\bigcap_{\ell=1}^\infty \tilde{\beta}^{\circ\,\ell}(\mathbb{Z}_{+}) = \emptyset \quad\implies\quad \mathbb{Z_{+}} = \biguplus_{k = 1}^\infty A_k$$
i.e. $\mathbb{Z}_{+}$ is an infinite disjoint union of infinite sets $A_k$. Since there are uncountable choices for $\alpha$, there are uncountable ways of such infinite disjoint unions.

For a concrete example, let $\alpha = \phi, \beta = \phi^2$ where $\phi$ is the golden mean, we get something like

$$\begin{array}{rll} \mathbb{Z}_{+} = & \{\; 1,3,4,6,8,9,11,12,14,16,17,19,\ldots\;\}\\ \uplus & \{\; 2,7,10,15,20,23,28,31,36,41,\ldots\;\}\\ \uplus & \{\; 5,18,26,39,52,60,73,81,94,\ldots\;\}\\ \uplus &\{\; 13,47,68,\ldots\;\}\\ \vdots\; & \end{array}$$

If you follow the rule that you only take the multiples that didn’t show up already, then you’re fine, since by construction you’ll be making all the subsets disjoint and by the Fundamental Theorem of Arithmetic every element will be in some $A_i$.

An example of simple infinite disjoint union would be $A_i=\{i\}$ and then $\mathbb{N}=\bigcup_{i=0}^\infty{A_i}$.

With edit: A simple way to split up $\mathbb{N}$ into a disjoint union of infinite subsets is to start with $A_0=\{0,2,4,\ldots\}$, and then let $A_1$ be every other element of $\mathbb{N}\setminus A_0$ (i.e. $\{1,5,9,13,\ldots\}$). In general, let $A_n$ be the set containing “every other element” of the set $\mathbb{N}\setminus \bigcup_{i=0}^{n-1}{A_n}$.


# A natural proof of the Cauchy-Schwarz inequality

Most of the proofs of the Cauchy-Schwarz inequality on a pre-Hilbert space use a fact that if a quadratic polynomial with real coefficients takes positive values everywhere on the real line, then its discriminant is negative(e.g. Conway: A course in functional analysis). I think this is somewhat tricky. Moreover I often forget its proof when the pre-Hilbert space is defined over the field of complex numbers. Is there a more natural proof (hence it's easy to remember) which is based on a completely different idea?

• To someone who voted to close, would you please explain the reason for the vote? – Makoto Kato Jul 5 '13 at 5:59
• You might want to browse the first chapter of The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities by J. Michael Steele. – Did Jul 5 '13 at 6:26
• I'm not sure about what could qualify as "natural" or "more natural" proof for you, but in this case I can't think, off the top of my head, of anything *simpler", basic and straightforward as working with a quadratic's discriminant: this is Junior High School stuff! – DonAntonio Jul 5 '13 at 8:45
• @DonAntonio As I wrote, I don't think the complex pre-Hilbert space case is not so straightforward. – Makoto Kato Jul 5 '13 at 9:01
• A question which has 3 upvotes, 2 favorites and a 6 upvoted answer should not be closed. Please reset the close votes. – Makoto Kato Jul 5 '13 at 19:24

Recall the Pythagorean theorem: If $u_1, \cdots u_n$ are pairwise orthogonal, then $$\| u_1 + \cdots u_n \|^2 = \|u_1\|^2 + \cdots + \| u_n \|^2.$$

I want to use this to tell us something about two non-zero vectors $u$ and $v,$ but they aren't necessarily orthogonal. So consider the projection of $u$ onto the plane of vectors orthogonal to $v:$ $$w = u - \frac{ \langle u,v \rangle}{\|v\|^2} v.$$ This is certainly orthogonal to $v,$ and the Pythagorean theorem applied to $w$ and $v$ gives the Cauchy-Schwarz inequality.

• This is excellent. Thanks. By the way, I think $<v, u>$ is a typo($<v, v>$). – Makoto Kato Jul 5 '13 at 9:19
• @MakotoKato Thank you for spotting that. Also, I find using \langle , \rangle for inner products is more pleasing to the eye than $<$ and $>.$ – Ragib Zaman Jul 5 '13 at 14:14
• @RagibZaman You're welcome. I just didn't know how to write $\langle, \rangle$ instead of < and >. By the way again, I think $\|v||$ should be squared or replaced by $\langle v, v \rangle$. Regards. – Makoto Kato Jul 5 '13 at 19:19
• I fail to see how you obtain the CS inequality in your last paragraph. Would you explain that? – Martin Argerami Jul 5 '13 at 20:42
• @MartinArgerami Let $u_1 = w$, $u_2 = \frac{ \langle u,v \rangle}{\langle v, v \rangle} v$. Then apply the Pythagorean formula $\| u_1 + u_2\|^2 = \|u_1\|^2 + \| u_2 \|^2 \ge \|u_2\|^2$. Regards. – Makoto Kato Jul 6 '13 at 0:41

There is also an approach by "amplification" which is really cool. Also the exact same trick works to prove Hölder's inequality and is generally a very important principle for improving inequalities.

It goes like this: We start out with $$\langle a-b,a-b\rangle\ge 0$$ for $a,b$ in your inner product space, and $a\not=0$, $b\not=0$. This implies $$2\langle a,b\rangle\le \langle a, a\rangle + \langle b, b\rangle$$ Now notice that the left hand side is invariant under the scaling $a\mapsto \lambda a$, $b\mapsto \lambda^{-1}b$ for $\lambda>0$. This gives $$2\langle a,b\rangle \le \lambda^2 \langle a,a\rangle + \lambda^{-2}\langle b, b\rangle$$ Now look at the right hand side as a function of the real variable $\lambda$ and find the optimal value for $\lambda$ using calculus (set the derivative to $0$):

$$\lambda^2=\sqrt{\frac{\langle b,b\rangle}{\langle a,a\rangle}}$$

Plugging this value in, we obtain

$$2\langle a,b\rangle\le \sqrt{\langle a,a\rangle}\sqrt{\langle b,b\rangle}+\sqrt{\langle a,a\rangle}\sqrt{\langle b,b\rangle}$$

i.e.

$$\langle a,b\rangle\le\sqrt{\langle a,a\rangle}\sqrt{\langle b,b\rangle}$$

Notice how we took a trivial observation and "optimized" the expression by exploiting scaling invariance.

• As explained (and expanded) by Terence Tao on his blog. – Did Jul 5 '13 at 13:08
• Could your proof be modified to show that the equality holds iff $a$ is proportional to $b$? I've been thinking a while on it, but I haven't figured anything out. – jinawee Oct 1 '14 at 19:54
• @jinawee: Yes, the only inequality in this proof is $\langle \mu a-b,\mu a-b\rangle\ge 0$ for an optimal constant $\mu$ depending on $a,b$. Equality implies $b=\mu a$. – J.R. Oct 2 '14 at 7:17

The inequality $| \langle a, b \rangle | \leq \| a \| \| b \|$ can be rewritten as $$| \langle a, \frac{b}{\|b\|} \rangle | \leq \| a \|$$ (assuming $b \neq 0$).

On the left we see the component of $a$ on the unit vector $u = \frac{b}{\| b \|}$. On the right we have the norm of $a$. Of course the norm of $a$ is larger than the component of $a$ on $u$, this is very intuitive. To make this into a rigorous proof, we merely have to write $a = \langle a, u \rangle u + v$, note that $v \perp u$, and use the pythagorean theorem.

Another approach is to start with the inequality $$0 \leq \| a - \text{proj}_b(a) \|^2.$$

Just expand the right hand side and Cauchy-Schwarz pops out.

This is not exactly an answer but an explanation for the idea behind the Ragib Zaman's answer.

Let $K$ be the field of real numbers or the field of complex numbers. Let $E$ be a pre-Hilbert space over $K$.

Let $x, y$ be elements of $E$ such that $\langle x, y \rangle = 0$. Then $\|x + y\|^2 = \langle x+y, x+ y\rangle = \langle x, x\rangle + \langle x, y \rangle + \langle y, x\rangle + \langle y,y\rangle = \|x\|^2 + \|y\|^2$.

Let $x, y$ be elements of $E$ such that $\langle x - y, y \rangle = 0$. Then, by the above formula, $\|x\|^2 = \|x - y\|^2 + \|y\|^2$. Hence $\|x\| \ge \|y\|$.

Finally let $u, v$ be non-zero elements of $E$. Let $t$ be an element of $K$ such that $\langle u - tv, v \rangle = 0$. $t$ must be $\frac{ \langle u,v \rangle}{\langle v, v\rangle}$ since $v \neq 0$. Then $\|u\| \ge \|tv\|$ by the previous inequality. This leads to the Cauchy-Schwarz inequality immediately.

Here's my favorite proof, mainly because it's nicely symmetric, easy to remember and not impossible to come up with (the main trick is that $2=1+1$):

We want to prove $$|\langle x,y\rangle|\le\|x\|\|y\|$$ This is linear in $x$ or $y$ and obviously holds for $x=0$ or $y=0$. Therefore without loss of generality we can suppose that $\|x\|=\|y\|=1$: $$|\langle x,y\rangle|\le\|x\|\|y\|\Longleftrightarrow|\langle x,y\rangle|\le1\Longleftrightarrow2|\langle x,y\rangle|\le2\Longleftrightarrow2|\langle x,y\rangle|=\|x\|^2+\|y\|^2$$

Let $u\in\mathbb C$ be a complex unit (i.e. $|u|=1$), then \begin{align}0\le\|x-uy\|^2&=\langle x-uy,x-uy\rangle=\langle x,x\rangle-u\langle x,y\rangle-\overline{u\langle x,y\rangle}+\langle y,y\rangle\\&=\|x\|^2+\|y\|^2-2\,\text{Re}(u\langle x,y\rangle)\end{align} Now just set $u$ so that $\text{Re}(u\langle x,y\rangle)=|\langle x,y\rangle|$ and you are done.

Of course in the real case you can just expand $0\le\|x\pm y\|^2$.


# Is this a valid approach to solving the inequality $\frac{1}{x} < x < 1$?

I have been given the inequality $\frac{1}{x} < x < 1$ and have been told to find the values of $x$ which satisfy this inequality, and I have also been told to find these values using a case-by-case approach. I'd like to know whether my reasoning is valid.

Case 1: $x=0$

In this case we find that the value of $\frac1x$ is undefined, so we know that $x \neq 0$.

Case 2: $0<x<1$

Given the inequality $\frac{1}{x} < x < 1$ and that $x$ lies in the range $0 < x < 1$, we find that $1<x^2$. Therefore, $x>1$ or $x<-1$. However, $x\not>1$ and in this case $x\not<-1$ since we are looking at the case $0<x<1$. Thus we conclude that no values of $x$ in the range $0<x<1$ satisfy the original inequality.

Case 3: $x<0$

From the original equality and the fact that we know that $x<1$ we find:

$$\frac1x < x$$

$$\implies 1 > x^2$$ (since $x<0$) $$\implies -1<x<1$$ but we know that $x$ cannot lie in the range $0<x<1$ so we have that $-1<x$ and that $x<0$ and combining these inequalities we have that $-1<x<0$.

This argument seems to be supported by looking at the graph, but I am unsure whether all the steps I have made are valid and whether this is what is meant by a 'case analysis'.

• This is absolutely fine and there is no part of the solution that is wrong. – Prakhar Nagpal Jul 12 '18 at 16:00
• I think the argument is good! ${}{}{}{}{}{}{}$ – Andres Mejia Jul 12 '18 at 16:01
• Thank you for the feedback. – Benjamin Jul 12 '18 at 16:02

Overall, I would say that the argument is spot on! Just a couple comments you might want to consider. First of all, to be super rigorous in your answer, you might mention that clearly we know $x \ngeq 1$ so we need not consider that case. Another thing, I felt that your argument via contradiction for Case 2 was a bit lengthy there at the end. Once you reached the point $1 < x^2$ you could have just said that since we have $0 < x < 1$ (via assumption) and no such value squared is greater than 1, we have a contradiction and are done. The whole considering $x > 1$ or $x < -1$ is just overkill.
Also, that is exactly what it means by a case analysis, or case-by-case approach. Use cases to consider every possible value of $x$, and see what happens. Using cases is actually a very common proof technique!


# A set of elements in a reduced unity ring

Let $$(A,+,\cdot)$$ be a unity ring with the property that if $$x \in A$$ and $$x^2=0$$ then $$x=0$$. Consider the set $$M=\{a\in A | a^3=a\}$$. Prove that:

a) $$2a\in Z(A)$$, $$\forall a\in M$$, where $$Z(A)$$ denotes the centre of the ring $$A$$;
b) $$ab=ba$$, $$\forall a,b\in M$$.

My attempts revolved around the fact that an idempotent element in a reduced ring is central.
So, since for $$a\in M$$ we have that $$(a^2)^2=a^2$$, it follows that $$a^2\in Z(A)$$, $$\forall a\in M$$.
The next thing I wanted to use in order to solve a) was that $$Z(A)$$ is a subring of $$A$$, so if I had proved that $$(a+1)^2 \in Z(A)$$, $$\forall a\in M$$, then we would have reached the desired conclusion. However, I couldn't prove this and I honestly doubt that it is true.
Another idea that I had was to prove that $$M$$ is a subring of $$A$$. Of course, this didn't work out because I cannot even prove that $$M$$ is closed under addition. Again, I don't know if this is true and it most likely isn't.
As for b), I think that a) should be of use, but I don't know how. It is a well-known problem that a ring with $$x^3=x$$ for any $$x$$ in that ring is commutative, but since $$(M,+,\cdot)$$ is almost definitely not a ring, this doesn't help.
EDIT: Is there any chance that this question is simply wrong? I tended to believe this before asking it here too, but since nobody has made any progress on it until now I am even more inclined to think so.

• If you honestly doubt that $(a+1)^2 \in Z(a)$ for all $a \in M$, then you honestly doubt that part (a) is true, since if part (a) is true (and since you've already noted $a^2 \in Z(a)$ for all $a \in M$), you get $(a+1)^2 = a^2+2a+1 \in Z(a)$ for all $a \in M$. – mathworker21 Nov 9 at 18:50
• @mathworker21 I agree with you. But since I have spent hours and hours trying to prove that $(a+1)^2 \in Z(A)$ for all $a\in M$ and nothing worked out, then I am inclined to believe that this statement may actually be wrong. Yet, since I cannot provide a counterexample, here I am, still hoping that someone better than me at rings will solve it.... – Math Guy Nov 9 at 21:34
• I don't know. I don't think the proof that any ring with $x^3 = x$ for all $x$ implies the ring is commutative is that easy... I could imagine spending hours failing to find that proof. By the way, where did you find this problem? – mathworker21 Nov 9 at 21:35
• @mathworker21 I know the "classical" one where $x^3=x$ for all $x$ holds in a ring, but since here we do not have a ring it doesn't really help I think. The problem is from a magazine from my country. – Math Guy Nov 9 at 21:37
• you missed my point. my point was that working on these kinds of problems for hours and failing doesn't mean they are false. (Of course that is always true, but I find the statement more meaningful here). The evidence/example I gave was the $x^3=x$ problem. For that problem, I could imagine working on it for several hours without finding the solution. And of course that problem is true. – mathworker21 Nov 9 at 21:54

## 1 Answer

We show $$M \subseteq Z(A)$$. This immediately gives (a) and (b).

Lemma 1: $$yx = 0 \implies xzy = 0$$ for any $$z$$.

Proof: $$(xzy)(xzy) = xz(yx)zy = 0$$.

Lemma 2: $$x^2 = x$$ implies $$x \in Z(A)$$.

Proof: For any $$y \in A$$, a short computation shows $$(xy-xyx)(xy-xyx) = 0 = (yx-xyx)(yx-xyx)$$.

Lemma 3: $$a \in M \implies a^2 \in Z(A)$$.

Proof: $$(a^2)^2 = a^4 = a^2$$, so use Lemma 2.

Claim: For any $$a \in M$$, $$a \in Z(A)$$.

Proof: Since $$(a-1)[a(a+1)]=0$$, Lemma 1 implies that for any $$b \in A$$, $$0 = a(a+1)b(a-1) = (a^2b+ab)(a-1).$$ Also, $$a(a+1)(a-1)=0$$ implies $$0 = ba(a+1)(a-1) = (ba^2+ba)(a-1) = (a^2b+ba)(a-1),$$ where the last equality used Lemma 3. Subtracting gives: (1) $$0 = (ba-ab)(a-1)$$. The exact same argument shows: (2) $$0 = (a-1)(ba-ab)$$. (1) immediately implies $$0 = (ba-ab)(a-1)b = (ba-ab)(ab-b)$$, and (2) with Lemma 1 implies $$0 = (ba-ab)b(a-1) = (ba-ab)(ba-b)$$. Subtracting the two results gives $$(ba-ab)^2 = 0$$.

• Thank you ! Could you tell me how you came up with this solution? – Math Guy Nov 10 at 11:23
• @MathGuy well, I played around with it for a while, so I exhausted many different approaches. Then I eventually stumbled upon something like Lemma $1$, first in the form of $0 = a(a+1)b(a-1)$ and realized it started giving equations I hadn't seen/derived before. So I knew I had something good. Then it was just a matter of finishing up, which was pretty easy (the proof of the claim is rather short). – mathworker21 Nov 10 at 13:02


# Numerical approximation of $\pi$

Points are randomly scattered inside the unit square, some fall within the unit circle with probability $P=\pi/4$.

so $P$ is approximated by the fraction $$P\approx \frac{\text{Number of red points}}{\text{Number of all points}}$$ this leads $$\pi \approx 4\frac{\text{Number of red points}}{\text{Number of all points}}$$ (see following image)

There is a code for this:

tinyColor[color_, point_] := {PointSize[Small], color, Point[point]}
colorChoose[point_] :=
If[Norm[point] <= 1, tinyColor[Red, point], tinyColor[Blue, point]]
darts = RandomReal[{0, 1}, {40000, 2}];
coloredDarts =ParallelMap[colorChoose, darts];
insides = Map[Boole[Norm[#] <= 1] &, darts];
piapprox = Accumulate[insides]/Range[Length[darts]]
inner = Select[darts, Norm[#] <= 1 &];
outer = Select[darts, Norm[#] > 1 &];

Show[Plot[Sqrt[1 - x^2], {x, 0, 1}, Filling -> Axis, AspectRatio -> 1,
PlotLabel -> n == Length[darts] TildeTilde[π, 4.0*piapprox[[-1]]]],
ListPlot[{inner, outer},
PlotStyle -> {{PointSize[Tiny], Red}, {PointSize[Tiny], Blue}},
ImageSize -> {500, 500}]]

I tried to simplify this problem:

pts = RandomPoint[Rectangle[], 40000];

ListPlot[pts, AspectRatio -> 1, PlotStyle -> Blue]

The problem is following:

How can I split set of points pts into two parts, "inside the circle " and "outside the circle"?

• You should also investigate RegionMember, e.g., rf = RegionMember[Disk[]]; rf[darts] – Carl Woll Feb 10 '17 at 22:26

You can also use Select:

ptsin = Select[pts, Norm[#] < 1 &];

N[Length[ptsin]/Length[pts]]*4
(* 3.1496 *)

How can I split set of points pts into two parts, "inside the circle " and "outside the circle"?

{in, out} = SortBy[GatherBy[pts, Norm[#] < 1 &], Norm[#[[1, 1]]] &];

ListPlot[{in, out}, AspectRatio -> 1, PlotStyle -> {Red, Blue}]

• What's the purpose of SortBy? – anderstood Feb 4 '17 at 18:27
• The approximation of $\pi$ can be recovered with {in, out} = GatherBy[pts, Norm[#] < 1 &]; 4*Length[in]/(Length[out] + Length[in]) // N. – anderstood Feb 4 '17 at 18:30
• @anderstood, depending on the Norm of the first element in lst, the the first list in the output produced by GatherBy may be the "inside sublist" or the "outside sublist". Sorting the output of GatherBy makes the first list the "inside" one. – kglr Feb 4 '17 at 18:32
• Comment to my own comment: it should be {in, out} = SortBy[GatherBy[pts, Norm[#] < 1 &], Norm[#[[1, 1]]] &], cf kglr's comment above. – anderstood Feb 4 '17 at 18:55

Playing with Norm as shown in other answers:

pts = RandomReal[1, {40000, 2}];

4 True/(True + False) /. CountsBy[pts, Norm[#] < 1 &]

3.1474

That doesn't help with drawing the graphic however.

You can compute the norm of the point and verify if it is inside the circle.

At = 4;
d = 2;
totalpoints = 40000;
pts = RandomReal[{-1, 1}, {totalpoints, 2}];
pointsinsidecircle = Select[pts, Norm[#] < 1 &];
counter = Length[pointsinsidecircle];
approxpi = (4. At counter/totalpoints)/d^2;
Print["approx \[Pi] = ", approxpi]
ListPlot[{pts, pointsinsidecircle}, AspectRatio -> 1,
PlotStyle -> {Blue, Red}]

(*approx \[Pi] = 3.1328*)

• You could you Norm directly, and avoid AppendTo which is very slow. Using Select is probably a much faster option. – anderstood Feb 4 '17 at 18:24
• @anderstood i have changed the answer. Thank you. – Diogo Feb 4 '17 at 18:30
• Tip: Don't save graphics as JPEGs, they become fuzzy and lose color fidelity. Use PNGs. – Rahul Feb 10 '17 at 21:19

A slightly different approach:

inside = Pick[pts, Map[# ∈ Disk[] &, pts], True];
outside = Complement[pts, inside];

Also as pointed in a comment above:

inside = Pick[#, RegionMember[Disk[]][#], True] &@pts
outside = Complement[pts, inside];
`


Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.

Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.

I spent a lot of time trying to solve this and, having consulted some books, I came to this:
$$2x^2+2y^2+2z^2 \ge 2xy + 2xz + 2yz$$ $$2xy+2yz+2xz = 1-(x^2+y^2+z^2)$$ $$2x^2+2y^2+2z^2 \ge 1 - x^2 -y^2 - z^2$$ $$x^2+y^2+z^2 \ge \frac{1}{3}$$ But this method is very unintuitive to me and I don't think this is the best way to solve this. Any remarks and hints will be most appreciated.

• This does seem like one of the best ways to tackle this problem. Why do you find this method counterintuitive? – vrugtehagel Apr 23 '17 at 12:56
• @vrugtehagel It entails using a property which is not directly derived from the problem. – ILoveChess Apr 23 '17 at 12:58
• Are you comfortable with calculus? Here's a completely different approach. $z = 1 - x - y$ so $x^2 + y^2 + z^2 = x^2 + y^2 +(1 - x - y)^2$. Now use some calculus to find when that has a maximum. It will be when $x = y = \frac{1}{3}$ hence also $z = \frac{1}{3}$. – badjohn Apr 23 '17 at 13:03
• – Martin Sleziak Apr 23 '17 at 21:29

Cauchy- Schwarz works: $$x^2+y^2+z^2=\frac{1}{3}(1^2+1^2+1^2)(x^2+y^2+z^2)\geq\frac{1}{3}(x+y+z)^2=\frac{1}{3}$$

$x^2+y^2+z^2$ only depends on the squared distance of $(x,y,z)$ from the origin and the constraint $x+y+z=1$ tells us that $(x,y,z)$ lies in a affine plane. The problem is solved by finding the distance between such plane and the origin: since the plane is orthogonal to the line $x=y=z$,

$$\min_{x+y+z=1}x^2+y^2+z^2 = \left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2 = \frac{1}{3}$$ and we are done.

• clear: the sphere from the origin is tangent to the plane in x=y=z=1/3 – G Cab Apr 23 '17 at 14:49

This is not a proof in itself, but if you've studied statistics, then you've seen a proof that

$$0\le V(X)=E(X^2)-E(X)^2$$

If we now consider a random variable $X$ with three equally likely values, $X=x,y$, and $z$, then we have

$$E(X)={x+y+z\over3}\qquad\text{and}\qquad E(X^2)={x^2+y^2+z^2\over3}$$

If, in addition, we assume $x+y+z=1$, then we have $E(X)={1\over3}$, which implies $E(X^2)\ge\left(1\over3\right)^2={1\over9}$, or $x^2+y^2+z^2\ge{1\over3}$.

Excellent algebraic have been given (I voted for them). Intuition can here be obtained through visual proofs.

The equation $x+y+z=1$ defines a plane. $x^2+y^2+z^2=1/3$ defines a sphere. The following visualization depicts the plane in blue, the sphere in red.

From that, you can imagine that the question could be rephrased as (in a mundane way): for any point in the plane, its distance from the $(0,0,0)$ origin is higher than $1/\sqrt{3}$? So the sphere should remain "below" the plane. Except when they meet. The symmetry of the problem tells you that the tangency point has equal coordinates $(1/\sqrt{3},1/\sqrt{3},1/\sqrt{3})$, which is one of the motivations behind the equation $(3x-1)^2+(3y-1)^2+(3z-1)^2$.

Would the sphere be bigger (a radius higher than $1/\sqrt{3}$), it would intersect the plane in more than a single tangency point.

All in all, this resorts to finding the distance of the plane to the origin, which is exactly where the sphere and the plane meet. So what you are looking at is the distance of the plane to the origin.

If the plane is given by $ax+by+cz+d$, the signed distance of a point $(x_0,y_0,z_0)$ to it is (Point-Plane Distance):

$$D = \frac{ax_0+by_0+cz_0+d}{\sqrt{a^2+b^2+c^2}}$$

which in your case gives exactly $1/\sqrt{3}$. Any point in the plane is farther to $(0,0,0)$ than $|D|$.

Hint:

Expand $(3x-1)^2+(3y-1)^2+(3z-1)^2\geqslant 0$ And simplify.

If you rewrite your proof as:

$3 ( x^2+y^2+z^2 ) \ge ( x^2+y^2+z^2 ) + ( 2xy+2yz+2zx ) = (x+y+z)^2 = 1$.

You would find that it is not so unintuitive after all.

Since you know that $x+y+z = 1$, a natural thing to do to the original equation is to homogenize it; namely make all terms have the same degree. This gives us "$3 ( x^2+y^2+z^2 ) \ge (x+y+z)^2$, and expanding out immediately tells us the solution.

In general, a technique that works for many cyclic polynomial inequalities is to try to 'smooth' the terms out. Terms that are just a power of one variable are the 'biggest', and using inequalities like "$x^2+y^2 \ge 2xy$" will 'mix' powers and thereby 'reduce' them.

It is enough to prove the result for $x,y,z$ positive since $$\frac{|x|+|y|+|z|}{3} \geq \frac{x+y+z}{3}$$

One can use the generalized AM-GM inequality: If $$M_p = \left(\frac{x^p+y^p+z^p}{3}\right)^{\frac{1}{p}}$$ then for $p < q$, $M_p \leq M_q$ with equality holding if and only if $x=y=z$. Here, using $M_2 \geq M_1$, we get $$\left(\frac{x^2+y^2+z^2}{3}\right)^{\frac{1}{2}} \geq \frac{|x|+|y|+|z|}{3} \geq \frac{x+y+z}{3} = \frac{1}{3}$$

We can minimize $x^2 + y^2 + z^2$ subject to the constraint $x+y+z = 1$ using Lagrange multipliers, we then find that $x = y = z = \frac{1}{3}$, therefore $x^2 + y^2 + z^2\geq\frac{1}{3}$.

• This only shows that this is a local minimum, right? – Carsten S Apr 24 '17 at 14:05

One more way to prove it is by substituting out for $z$:

$$x^2+y^2+z^2 = x^2+y^2+(1-x-y)^2=2x^2+2y^2+2xy-2x-2y+1$$ Now, substitute $x=\hat x+a$ and $y=\hat y+b$. We get $$2\hat x^2+2\hat y^2+2\hat x\hat y+(4a+2b-2)\hat x+(2a+4b-2)\hat y+(2(a^2+ab+b^2-a-b)+1)$$ We can eliminate the order 1 terms by letting $4a+2b-2=2a+4b-2=0$, which gives $a=b=\frac13$. With this substitution, we have $$2(\hat x^2+\hat x\hat y+\hat y^2)+\frac13$$ We can express the bracketed term as a sum of squares, giving us $$3(\hat x+\hat y)^2+(\hat x-\hat y)^2 + \frac13$$ and we can see that the smallest value this can take is $\frac13$. Indeed, it takes this value when $\hat x=\hat y=0$ - that is, when $x=y=\frac13$.

Ok, you want an intuitive proof, not a better proof. That's fine.

First note that for $x=y=z$, we have $x^2+y^2+z^2 = 3\left(\frac13\right)^2= \frac13$. Next let us show that $x^2+y+2+z^2$ is not minimal if $x\ne y$ or $y\ne z$. Indeed if $a\ne b$ then $$a^2 + b^2 - \left(\left(\frac{a+b}2\right)^2+\left(\frac{a+b}2\right)^2\right) = \frac12\left( a^2 + b^2 -2ab \right)=\frac12(a-b)^2>0,$$ so we can replace two of the numbers be their averages and lower the sum of squares.

Unfortunately we not yet quite done. We are done if we can show that there is a global minimum of $x^2+y^2+z^2$, given that $x+y+z=1$. Because then that global minimum must also be a local minimum, but for that we cannot have $x\ne y$ or $y\ne z$, so $x=y=z$ must be the global minimum, and its value is $\frac13$.

For that we can first argue that we can restrict ourselves to $x,y,z\ge0$ (e.g. if $x<0$ replace $(x,y,z)$ by $\frac1{-x+y+z}(-x,y,z)$, which yields a smaller value since $-x+y+z = 1 - 2x > 1$) and then appeal to the compactness of $\{(x,y,z)\colon \text{$x+y+z=1$,$x,y,z\ge0$}\}$.

I did not say that this would be pretty, but it is pretty intuitive to me :)

Here's another way to approach this. It's easy to see that the value of $\frac13$ is obtained when each of $x, y, z$ is $\frac13$. We want to show that as the variables deviate from this point (with their sum still being 1) the value cannot decrease.

So we look at the deviations from $\frac13$: $x=\frac13+\epsilon_1$, $y=\frac13+\epsilon_2$, $z=\frac13+\epsilon_3$ with $\epsilon_1+\epsilon_2+\epsilon_3=0$. you have

$x^2+y^2+z^2=\\ (\frac13+\epsilon_1)^2+(\frac13+\epsilon_2)^2+(\frac13+\epsilon_3)^2=\\\left(\frac19+\frac23\epsilon_1+\epsilon_1^2\right)+\left(\frac19+\frac23\epsilon_2+\epsilon_2^2\right)+\left(\frac19+\frac23\epsilon_3+\epsilon_3^2\right)=\\ \left(\frac19+\frac19+\frac19\right)+\frac23(\epsilon_1+\epsilon_2+\epsilon_3)+(\epsilon_1^2+\epsilon_2^2+\epsilon_3^2)=\\ \frac13+(\epsilon_1^2+\epsilon_2^2+\epsilon_3^2) \ge \frac13$

The desired inequality follows by noting that from the convexity of $f(t) = t^{2},$ we have $f\left(\dfrac{x+y+z}{3}\right) \leq \dfrac{f(x)+f(y)+f(z)}{3}$


Combinatorics and Probability Problem

The problem I am working on is:

An ATM personal identification number (PIN) consists of four digits, each a 0, 1, 2, . . . 8, or 9, in succession.

a.How many different possible PINs are there if there are no restrictions on the choice of digits?

b.According to a representative at the author’s local branch of Chase Bank, there are in fact restrictions on the choice of digits. The following choices are prohibited: (i) all four digits identical (ii) sequences of consecutive ascending or descending digits, such as 6543 (iii) any sequence start-ing with 19 (birth years are too easy to guess). So if one of the PINs in (a) is randomly selected, what is the prob-ability that it will be a legitimate PIN (that is, not be one of the prohibited sequences)?

c. Someone has stolen an ATM card and knows that the first and last digits of the PIN are 8 and 1, respectively. He has three tries before the card is retained by the ATM (but does not realize that). So he randomly selects the $2nd$ and $3^{rd}$ digits for the first try, then randomly selects a different pair of digits for the second try, and yet another randomly selected pair of digits for the third try (the individual knows about the restrictions described in (b) so selects only from the legitimate possibilities). What is the probability that the individual gains access to the account?

d.Recalculate the probability in (c) if the first and last digits are 1 and 1, respectively.

---------------------------------------------

For part a): The total number of pins without restrictions is $10,000$

For part b): The number of pins in either ascending or descending order is $10 \cdot 1 \cdot 1 \cdot 1$, because once the first digit is known, then the three other spots containing digits are already spoken for. The number of pins where each slot contains the same digit is $10 \cdot 1 \cdot 1 \cdot 1$, because once the first digit is known there is only one option left to the rest of the slots. The number of pins that have their first and second slot occupied by 1 and 9, respectively, is $1 \cdot 1 \cdot 10 \cdot 10 \cdot$. So, if R is the set that contains these restricted pins, then $|R| = 130$; and if N is the set that contains the non-restricted ones, meaning R and N are complementary sets, then $|N| = 10,000 - 130$. Hence, the probability is then $P(N) = 9780/10000 = 0.9870.$ However, the answer is $0.9876$. What did I do wrong?

For part c): The sample space, containing all of the outcomes of the experiment that will take place, is $|N|=9870$. When it says that the thief won't use the same pair of digits in each try, does that not allow him trying the pin 8 5 2 1 in one try and the pin 8 2 5 1 in another try?

-
Ascending or descending is $14$, not $20$. That takes care of the disagreement. –  André Nicolas Feb 3 '13 at 20:44
@AndréNicolas I didn't get 14 or 20, I got 10. Your saying the answer is 14, how did you get that? –  Mack Feb 3 '13 at 20:56
@Eli: You got $2\cdot10=20$ for ascending and descending together, and André is saying that it should be $2\cdot7=14$ for both together. –  joriki Feb 3 '13 at 21:04
Oh, I see. The 2 corresponds to the two choices (ascending or descending); and the 7 corresponds to the fact that you can't start the 4-digit pin with 7,8, or 9 when ascending, and you can't start a 4-digit pin with 0,1, or 2 when descending. –  Mack Feb 3 '13 at 21:14

For b): Which is the descending sequence starting with $1$ that you counted?

For c): Good question; the problem is badly worded in that regard. Taking it literally, I'd tend to interpret it as referring to unordered pairs, but since it makes little sense to couple two different PINs in this manner, I suspect that they actually mean ordered pairs. However, note that the answer doesn't depend on this.

I understand neither why the question says that the thief knows the restrictions, nor why you say that the sample space has size $9870$. The thief knows that the first and last digits are $8$ and $1$, respectively; that's not compatible with any of the sequences excluded by the restrictions, and it doesn't allow for $9870$ possibilities.

-
I’d interpret is as referring to ordered pairs, partly from the language, and partly from the overall level of difficulty of the exercise. –  Brian M. Scott Feb 3 '13 at 20:41
@joriki I don't really understand what you are asking when you say, "Which is the descending sequence starting with 1 that you counted?" –  Mack Feb 3 '13 at 21:00
@Eli: Exactly. And $10-3=7$. –  joriki Feb 3 '13 at 21:08
@Eli: I found another version of the book online with the answers included. It has $.0337$ for d., which is also wrong, but can be interpreted as a rounded version of the correct result $3/89$ (see Metin's answer). So it seems they just round results without indicating it (which is rather bad style). So it seems likely that $0.0333$ is a rounded version of $3/90$. That leaves the question how they arrive at $90$ options. Perhaps they meant to say $1$ and $8$ instead of $8$ and $1$; then the birth year rule would lead to a count of $90$. –  joriki Feb 4 '13 at 8:35
@Eli: That would also explain why part c already mentions that the thief knows the restrictions, even though that's irrelevant for the question as posed. –  joriki Feb 4 '13 at 8:37

For d): We have the case: $1$ * * $1$.

But the thief knows, by prohibition (i), it can not be $1111$. Thus he eleminates $1$ possibility.

Also he knows, by (iii), the second digit can not be $9$. There are exactly $10$ number of the form $19$ * $1$, namely, $1900$, $1910$, $1920$... So, at this stage he eleminates $10$ possibilities.

All in all, if he had no restrictions, there were $100$ choices for the form $1$ * * $1$. But he excluded $10 + 1 = 11$ of them and get 89 possible choices. Since he has 3 chances, the resulting possibility is $3/89$.

-

Just my two cents,

10^4 possibilities.

There are 14 ascending and descending groups of 4.

Keyspace 9,876

If the badguy knows two of the four spaces, he only has to guess through entropy 10^2. (None of the restrictions meet up with the range 8xx1)

3 Tries in 100

:)

-
(0123)(1234)(2345)(3456)(4567)(5678)(6789) And then the reverse of those... –  Ben Sep 9 '13 at 5:33

Regarding part [c], I do not think that there is something wrong with the book. As @Mack said in one of his comments, thief has to guess only two second numbers. Since the given restrictions actually do not apply on this order of numbers, he has to guess among 100 (10*10) possible combinations of the numbers.

Since he has 3 tries, the probability that he will gain desired access is 3/100.

According to Wolfram Alpha, 3/100 is exactly 0.03, which is the same as written solution from the book.

Correct me if I am wrong.

-


# Condition for equilibrium of a rigid body on a horizontal surface

The following question is from a past paper from Further Mechanics and it has been bothering me immensely I have spent hours cracking at a way to make sense of it, the question is in two parts

Part 1:

An object consists of a uniform solid circular cone, of vertical height 4r and radius 3r, and a uniform solid cylinder, of height 4r and radius 3r. The circular base of the cone and one of the circular faces of the cylinder are joined together so that they coincide. The cone and the cylinder are made of the same material.

Find the distance of the centre of mass of the object from the end of the cylinder that is not attached to the cone.

I first roughly sketched the object as such

Then finding the $$\bar{x}$$ of each part seperately:

For the cone I used the standard result of the centre of mass being $$\frac{1}{4}r$$ away from the base:

$$\bar{x}_{Cone}=\frac{1}{4}.4r + 4r=5r$$

For the cylinder I derived $$\bar{x}$$ via integration:

$$y=3r$$

$$\bar{x}_{cylinder}=\frac{\int_0^{4r} xdV}{\int_0^{4r}dV}$$

$$\because dV=y^2 \pi dx$$

$$\implies \bar{x}_{cylinder}=\frac{\int_0^{4r} 9r^2 \pi x dx}{\int_0^{4r}9r^2 \pi dx}$$

$$\implies \bar{x}_{cylinder}=\frac{9r^2 \pi \int_0^{4r} x dx}{9r^2 \pi\int_0^{4r} 1 dx}$$

$$\implies \bar{x}_{cylinder}=\frac{\frac{1}{2}\left[x^2 \right]^{4r}_0}{\left[x \right]^{4r}_0}$$

$$\therefore \bar{x}_{cylinder}=2r$$

Then by taking the weighted average of both objects I arrived at the correct value for the distance of the centre of gravity from the base of the cylinder which was

$$\bar{x}=\frac{11}{4}r$$

However the next part has had me at a complete loss for the better part of the day, it states that:

Show that the object can rest in equilibrium with the curved surface of the cone in contact with a horizontal surface.

I tried coming up with a rough sketch for this also

However I do not understand how to tackle this question at all, all I know is that for a body to be at equilibrium on such a surface, the centre of mass must pass through the point of suspension, however clearly this does not give an insight on how to answer this.

The condition for this question given in the marking scheme is that

Can someone explain what this means and why this is the case?

Given the geometry of the object, the center of gravity is very close to the surface between the cylinder and the cone. More specifically the distance between the center of gravity and the base of the cone is $$\frac{1}{4}r$$.

Basically what you need to prove is that the angle of the cone $$\phi$$ is such, so that when the object is tilted the weight crosses over the last $$\frac{1}{4}r$$.

So you need to prove that the angle $$\theta$$ is smaller that angle $$\phi$$.

Angle $$\theta$$ is calculated as: $$\tan\theta = \frac{\text{distance to be covered}}{\text{cylinder radius}}= \frac{\frac{11}{4}r}{3r}=\frac{11}{12} \Rightarrow \theta =42.5 [deg]$$

while Angle $$\phi$$ is calculated as: $$\tan\phi= \frac{\text{height of cone}}{\text{cone radius}}= \frac{4r}{3r}=\frac{4}{3} \Rightarrow \phi =53.1[deg]$$

Therefore, since $$\phi> \theta$$, the object becomes vertical enough so that the weight passes through the coned surface.

– NMech
Mar 23, 2021 at 9:52
• Reminds me of a crafty problem my undergrad QM prof posed. I'll adapt it to this particular object: using Heisenberg's Uncertainty Principle, how long will this object remain stable resting on the cone? It's not intended as a "real-world" problem but rather to see how you might use paired-parameters to guesstimate the stability. Mar 24, 2021 at 12:43

The CG is found by assuming the mass of the cone as 1/3m and the cylinder m, then

$$\bar{X}= \frac{2r*m+5r*m/3}{4/3m} =r*11/4$$

The cone side length (like the sharp tip of a pencil) is $$5r$$, the side of a 3,4,5 triangle, and the interior half tip angle is 36.87 degrees.

The CG is $$5.25r$$ from the tip of the cone and at rotation to rest on the side of the cone will be at $$5.25r*cos36.87= 4.19r<5r$$

I let you prove the condition.

It is well within the footprint of the pencil shape so it is stable.


# Transport equation $u_t + xu_x + u = 0$ with $u(x_0, 0) = \cos(x_0)$

I have been studying PDEs using Peter Olver's textbook. I have learnt how to solve equations such as $u_t + 2u_x = \sin(x)$ subject to an initial condition such as $u(0,x) = \sin x$. Letting $\epsilon = x - 2t$ and $u(t,x) = v(t,\epsilon)$, I then plug this into the transport equation.

However, I am not sure how to define a characteristic to solve the following equation

$$u_t + xu_x + u = 0, \qquad u(x_0, 0) = \cos(x_0)$$

because it has a variable 'speed' term $x$ and it is also not homogenous because of the term $u$.

A solution would be very helpful so I can see how to approach these problems.

• If you change coordinates to $x = e^y$ then $u_y = xu_x$ so your PDE in these new coordinates simplifies to $u_t + 2u_y + u = 0$. You can even simplify it further by taking $v = e^{t} u$ then $v_t + 2v_y = 0$ which is the equation you say you already know how to solve. – Winther May 9 '18 at 17:48

$$u_t + xu_x = -u$$ $\frac{dt}{1}=\frac{dx}{x}=\frac{du}{-u}$

First characteristics, from $\quad \frac{dt}{1}=\frac{dx}{x}$ :

$$x\,e^{-t}=c_1$$

Second characteristics, from $\quad\frac{dx}{x}=\frac{du}{-u}$ : $$x\,u=c_2$$ General solution of the PDE : $\quad x\,u=F(x\,e^{-t})$

$$u(x,t)=\frac{1}{x}F(x\,e^{-t})$$ $F$ is an arbitrary function, to be determined according to the boundary condition.

Condition : $\quad u(x_0, 0) = \cos(x_0)=\frac{1}{x_0}F(x_0\,e^{0})$

$F(x_0)=x_0\cos(x_0)$. Now the function $F$ is determined, i.e.: $F(X)=X\cos(X)$.

We put it into the above general solution , where $X=x\,e^{-t}$ , thus $F(x\,e^{-t})=(x\,e^{-t})\cos(x\,e^{-t})$ :

$$u(x,t)=\frac{1}{x}(x\,e^{-t})\cos(x\,e^{-t})=e^{-t}\cos(x\,e^{-t})$$

• Hi, thank you. Can you explain how you obtain your general solution of the PDE from the two characteristics? – PhysicsMathsLove May 11 '18 at 10:37
• The general solution can be expressed on various forms. For example on the form of implicit equation : $\Phi(c_1,c_2)=0$, or $c_1=f(c_2)$ or $c_2=F(c_1)$ or other forms. All functions introduced are arbitrary, but they are related one to the other. Doesn't matter the form chosen, the final result is the same after applying the boundary condition. – JJacquelin May 11 '18 at 10:51
• Ok I am still not sure how you obtain $xu = F(xe^{-t})$ from your characteristics, I.e. how it depends on $xe^{-t}$? – PhysicsMathsLove May 11 '18 at 10:57
• $c_2=F(c_1)$ with $c_2=xu$ and $c_1=xe^{-t}$ gives $xu=F(xe^{-t})$. – JJacquelin May 11 '18 at 11:02
• Why do you know one constant is a function of the other? – PhysicsMathsLove May 11 '18 at 11:12

Let us apply the method of characteristics. We get the characteristic equations $$\frac{\text{d} t}{\text{d} s} = 1 \, , \qquad \frac{\text{d} x}{\text{d} s} = x \, , \qquad \frac{\text{d} u}{\text{d} s} = -u \, .$$ Letting $t(0) = 0$, we know $t=s$. Letting $x(0) = x_0$, we get $x(t) = x_0\, e^t$. Since $u(0) = \cos(x_0)$, we have $u(t) = \cos(x_0)\, e^{-t}$. Finally, $$u(x,t) = \cos(x\, e^{-t})\, e^{-t} \, .$$

Note that \begin{align} \frac{\rm d}{{\rm d}t}u(e^t,t+t_0)&=\frac{\partial u}{\partial t}(e^t,t+t_0)+e^t\frac{\partial u}{\partial x}(e^t,t+t_0)\\ &=\left(\frac{\partial u}{\partial t}+x\frac{\partial u}{\partial x}\right)(e^t,t+t_0)\\ &=-u(e^t,t+t_0) \end{align} holds for all $t$ and $t_0$. Thus $$\frac{\rm d}{{\rm d}t}u(e^t,t+t_0)+u(e^t,t+t_0)=0,$$ or equivalently, $$\frac{\rm d}{{\rm d}t}\left(e^tu(e^t,t+t_0)\right)=0.$$ Therefore, $$e^tu(e^t,t+t_0)=e^{-t_0}u(e^{-t_0},-t_0+t_0)=e^{-t_0}u(e^{-t_0},0)=e^{-t_0}\cos e^{-t_0},$$ or equivalently, $$u(e^t,t+t_0)=e^{-t-t_0}\cos e^{-t_0}.$$ Finally, let $x=e^t>0$ and $\tau=t+t_0$. This gives $t=\log x$ and $t_0=\tau-\log x$. Hence $$u(x,\tau)=e^{-\log x-\tau+\log x}\cos e^{-\tau+\log x}=e^{-\tau}\cos\left(e^{-\tau}x\right).$$


# count the pairs in a set of Data

I have a list of Data, e.g.,

list1={0, 1, 1, 1, 1, 2, 2, 3, 3}

list2={0, 1, 1, 1, 1, 2, 2, 3, 3, 0}

and I want to get as an output True or False if I have only even number of Data of similar or not. in the above example I like to write a code that returns False for list1 and True for list2, since in list1 I have only one 0, but in the list2 there are even number of each number.

Thanks!

## 4 Answers

Using Counts

f = And @@ EvenQ[Values[Counts[#]]] &
{f[list1], f[list2]}

{False, True}

• +1. Or f = Counts /* AllTrue[EvenQ] – WReach Nov 1 '17 at 14:27
• Yours is much concise. I think you should post it as a separate answer. – Anjan Kumar Nov 1 '17 at 14:30
• I did, but when I saw yours, I deleted it :) I shall undelete. – WReach Nov 1 '17 at 14:36

Given:

f = Counts /* AllTrue[EvenQ];

Then:

f[list1]
(* False *)

f[list2]
(* True *)

You can use the Tally function:

Tally[list]
tallies the elements in list, listing all distinct elements together with their multiplicities.

So as

And@@EvenQ[Tally[list1][[;; , 2]]]

or

And@@EvenQ[Tally[list2][[;; , 2]]]

check[list_] := If[
Cases[EvenQ[Count[list, #] & /@ Union[list]], False] == {},
True, False]

check[list2]

True