text: Calculate the coordinates of the centroid of ABC (the intersection of the medians). Calculations at a simple polygon. Online calculator to calculate the area and perimeter of a triangle given the coordinates of its vertices. . Given the length of the rhombus' diagonals, use the formula in solving for the area. The units are in place to give an indication of the order of the calculated results such as ft, ft 2 or ft 3. Calculate its area and its interior angles. to find the length of the diagonal if we know the side length of the square. The online calculator below calculates the area of a rectangle, given coordinates of its vertices. A polygon consists of straight edges and at least three vertices. Completing the square is another way to find the vertex of a quadratic equation. Alternate solution using slope: https://www.youtube.com/watch?v=Z0jWwzxhFSI For example, if a farmer needs to distribute 100g of fertilizer per square meter of a field, they can use the calculator to calculate the area of the field. Ratio iso triangle area and perimeter of a diagonals Calculator: The diagonals of the square are the segments that connect the opposite vertices. Final Answer: The area of the rhombus is 320 square inches. How many vertices in a tetrahedron? The calculations will be very difficult! Problem 2: Square Inscribed in a Semicircle Because, the area of the quadrilateral is never negative. Calculate the perimeter and area of the trapezoid. Triangle in circle d = a√2. Free Ellipse Vertices calculator - Calculate ellipse vertices given equation step-by-step This website uses cookies to ensure you get the best experience. The Regular Polyhedrons Calculator. In the figure above, click 'reset'. Write the equations of the parallels to the given line passing through the given point. The shape has four equal sides and four 90-degree angles; thus, it is called a regular quadrilateral. Visual Studio Languages , .NET Framework > Visual Basic. Vertices of a right triangle Show that the points D(2,1), E(4,0), F(5,7) are vertices of a right triangle. How many vertices are there in triangular prism? MHF Helper. So, for example, if the square side is equal to 5 in, then the diagonal is 5√2 in ≈ 7.071 in. Area: To calculate the area of a square, we multiply the lengths of 2 consecutive sides. Good luck! A vertex is a corner. They are the same length and intersect perpendicular to the center of the square. I thought I was onto something with this post: Given coordinates of hypotenuse, how can I calculate coordinates of other vertex? In order to verify if the given points are vertices of right triangle, we have to follow the steps given below. A cuboid has six rectangular faces. Number Of Vertices calculator uses Number Of Vertices=2+Number Of Edges-Number Of Faces to calculate the Number Of Vertices, The Number Of Vertices formula is defined as the number of vertices in the given two dimensional figure on a Euclidean or Cartesian Plane. of these triangles. View a scaled diagram of the resulting triangle, or explore many other math calculators, as well as hundreds of other calculators addressing finance, health, fitness, and more. Pyramid has different types such as triangular pyramid, square pyramid, pentagonal pyramid and so on. It's easy enough if the rectangle has not been rotated, I figured that part out. Welcome to OnlineMSchool. The vertices of the square are placed on a circle around M with $$\displaystyle r = \frac12 \cdot \sqrt{p^2+q^2}$$ 4. The algorithm of this right triangle calculator uses the Pythagorean theorem to calculate the hypotenuse or one of the other two sides, as well as the Heron formula to find the area, and the standard triangle perimeter formula as described below. A vertex (vertices) is a point where two or more lines meet. Pythagoras' Theorem significant figures. In a poll 37% of the people polled answered yes to … Alternatively, if you need to buy some tiles or new carpet for a room, the tool will tell you how much material you need to buy. once i've done that, i can read off the information i need from the. This Pentagon Has 5 Edges . Formulas, explanations, and graphs for each calculation. Another way to calculate the exterior angle of a triangle is to subtract the angle of the vertex of interest from 180°. In real life measurements, like in construction, engineering, landscaping, etc. Find out the length of square diagonal with our diagonal of a square calculator. This is the hypotenuse of a right angle triangle so you can calculate the lengths of the sides. Polygon Calculator. Face is a flat surface that forms part of the boundary of a solid object. Use vector addition to find all vertices. Given here two sample problems of calculating the vertices of square pyramid and triangular pyramid. Coordinates of a centroind Let’s A = [3, 2, 0], B = [1, -2, 4] and C = [1, 1, 1] be 3 points in space. Explore many other math calculators like the area and surface area calculators, as well as hundreds of other calculators related to finance, health, fitness, and more. Calculate the total volume (in … Squares are geometric figures with 4 equal sides, 4 vertices and 4 right angles. Given that the vertices of a square are A (6, -13), B (-2,2), C (13, 10), and D (21,-5), i. hypotenuse Show That the Points are the Vertices of a Right Triangle : Here we are going to see, how to show that the given points are vertices of right triangle. Figure out how to transform the vertices from local/object/model space (the ones I figured out below) to world space. The calculator builds convex hull, using Jarvis march, then calculates four sides of the hull and one diagonal between first and third points of the hull. Since a cube has all equal length, the formula is written as V=Length³. Units: Note that units of length are shown for convenience. A cuboid has 8 vertices. Use the calculator above to calculate the properties of a square. It is simple when the edges don't intersect, so if the polygon isn't crossed. 1. Get more help from Chegg. Quadrilateral 2 Show that the quadrilateral with vertices P1(0,1), P2(4,2) P3(3,6) P4(-5,4) has two right triangles. Find the value. Perimeter of a square formula. They do not affect the calculations. It has 6 equal square surfaces and 8 equal vertices. Calculator Technique. Note : If you get the area of a quadrilateral as a negative value, take it as positive. And this pentagon has 5 vertices: Edges. This calculator can compute area of the triangle, altitudes of a triangle, medians of a triangle, centroid, circumcenter and orthocenter. calculating vertices of a square. The shape has four equal sides and four 90-degree angles; thus, it is called a regular quadrilateral. Apr 2005 20,249 7,909. The number of vertices is the total number of sides + 1 (apex point). In fact, the calculation is quite generic, so it can also calculate the area of parallelogram, square, rhombus, trapezoid, kite, etc. A square is a polygon with four vertices. Stonehenge Leaflet Ks2, I Love You Forever In Roman Numerals, Component form of a vector with initial point and terminal point, Online calculator. State the center, foci, vertices, and co vertices of the ellipse with equation 25x 2 4y 2 100x – 40y 100 = 0. also state the lengths of the two axes. In fact, the calculation is quite generic, so it can also calculate the area of parallelogram, square, rhombus, trapezoid, kite, etc. For this method, when you get to the end, you'll be able to find your x and y coordinates right away, instead of plugging the x coordinate back in to the original equation. number of faces (4, 6, 8, 12 or 20) length of edge = units: surface area = square units: volume = cubic units: in-diameter = units: circum-diameter = units: Remember: Appropriate units need to be attached. It is not possible for a triangle to have more than one vertex with internal angle greater than or equal to 90°, or it would no longer be a triangle. Area of a triangle (Heron's formula - given lengths of the three sides) Area of a triangle (By formula, given coordinates of vertices) Area of a triangle (Box method, given coordinates of vertices) Limitations The calculator will produce the wrong answer for crossed polygons, … How many vertices are there in triangular prism? This online calculator calculates area of convex quadrilateral (rectangles too) given coordinates of four vertices (corners). Given here two sample problems of calculating the vertices of square pyramid and triangular pyramid. Pyramid has different types such as triangular pyramid, square pyramid, pentagonal pyramid and so on. Calculate the vector torque on the square loop (with vertices at (-2, 2, 0), (-2, -2, 0), (2, -2, 0), (2, 2, 0)) about: a. I … If a square cannot be formed using these two vertices, print -1. So, area of the given quadrilateral is 28 square units. that is, the area of any convex quadrilateral. Let us say it as M. Step 2) Calculate the distance between the midpoint M and anyone vertice of the square. a. The online calculator below calculates the area of a rectangle, given coordinates of its vertices. (i) Find the length of each side. We can use Final Answer: The area of the rhombus is 320 square inches. A polygon consists of straight edges and at least three vertices. 1000 square feet >> Related Questions . Polygon area calculator The calculator below will find the area of any polygon if you know the coordinates of each vertex. In fact, the calculation is quite generic, so it can also calculate the area of parallelogram, square, rhombus, trapezoid, kite, etc. The number of vertices is the total number of sides + 1 (apex point). Parallelogram Area Definition. Similar calculators • Area of rectangle by coordinates • Area of a convex quadrilateral. Uses Heron's formula and trigonometric functions to calculate area and other properties of a given triangle. line segment All I could do now is to calculate the width of rectangle from those two points. The diagonals of the square are the segments that connect the opposite vertices. For example, enter the side length and the diagonal will be calculated. The diagonals have the following properties: In the figure above, click 'reset'. Its diameter is given here by the in-diameter. The formula for the area of a square is side 2, as seen in the figure below:. You then multiply it by itself to get to the area, so the formula used in this area of a square calculator is just as simple. Vertices, Edges and Faces. This tetrahedron has 4 vertices. The diagonal of the square is the A vertex is defined in geometry as the common endpoint of two or more rays or line segments. It is simple when the edges don't intersect, so if the polygon isn't crossed.Here the edge lengths as well as the perimeter and area of the polygon can be … Similarly, if you enter the area, the side length needed to get that area will be calculated. Example of creating a calculator. 2. This will work for triangles, regular and irregular polygons, convex or concave polygons. This calculator can compute area of the triangle, altitudes of a triangle, medians of a triangle, centroid, circumcenter and orthocenter. Parallelogram inscribed in a quadrilateral, Perimeter of a polygon (regular and irregular), Each diagonal divides the square into two. A vertex is a corner. Cube, Prism, Square Dragging the slider will split the solid open to help you to count faces, edges and vertices...be sure not to double-count any of the edges or vertices! Check out below how to find the diagonal of a square formula or simply give our tool a try - you won't be disappointed. The sides meet in four corners, which are called vertices. H. HallsofIvy. Find the coordinate of the center of the square. Circumcenter of the triangle with vertices (0,0),(3,1),(1,3). Diagonal is a line segment that joins two non-neighboring vertices. which seems to be in the right ballpark, but I am stuck at the step where you need to subtract two linear equations. An edge is a line segment joining two vertex. Calculate the area of the square (the side is 2*apothem), Calculate the sides of the right triangles either by using the 45 45 90 triangle calculator or the fact that they are right isosceles triangles, and use the special right triangles calculator. Also, explore the surface area or volume calculators, as well as hundreds of other math, finance, fitness, and health calculators. Given two opposite corners you can calculate the length of the diagonal. The perimeter: To calculate the perimeter of a square, add the length of the 4 sides. if two opposite vertices of a square are (5 ,4) and (1, -6) find other two vertices? Get more help from Chegg. Since it is a sqaure so it is somewhat easier. A vertex (plural: vertices) is a point where two or more line segments meet. Here, for example, quadrilateral is split by diagonal to two triangles, and the area of each triangle then calculated using Hero's formula. As you can see, a diagonal of a square divides it into two right triangles,BCD and DAB. Given the length of the rhombus' diagonals, use the formula in solving for the area. Triangle area calculator by points. The solution to the equation is straightforward multiplication and this is the formula used in our area of a square online calculator. The distance formula is used to find the distances between vertices then these distances are used to find the perimeter and area of the triangle. How to find the diagonal of a square - formula. First is the circum-sphere This is the sphere which fits around the outside of the polyhedron so as to touch all its vertices (or corners). . i first have to rearrange this equation into conics form by completing the square and dividing through to get "=1". The perimeter: To calculate the perimeter of a square, add the length of the 4 sides. It uses the same method as in Area of a polygon but does the arithmetic for you. Please try again using a different payment method. Thanks Peter. Each quadrilateral has two diagonals, a square too. that is, the area of any convex quadrilateral. vertices (corners) of the square. d1 = 32 inches d2 = 20 inches Area of rhombus = 1 / 2 (d1) (d2) Area of rhombus = 1 / 2 (32) (20) Area of rhombus = 320 square inches. Calculate the perimeter and area of the quadrilateral formed by the points (0,0) and (5,10) and (10,15) and (5,5) Free Square Area & Perimeter Calculator - calculate area & perimeter of a square step by step This website uses cookies to ensure you get the best experience. Subtract the area of a right triangle four times from the area of the square, Enjoy success! Click hereto get an answer to your question  The two opposite vertices of a square are ( - 1, 2) and (3, 2) . The diagonal of the square is the hypotenuseof these triangles.We can use Pythagoras' Theoremto find the length of the diagonal if we know the side length of the square. This free triangle calculator computes the edges, angles, area, height, perimeter, median, as well as other values of a triangle. For example: The given midline has two points: A (10.767008,106.665884) B (10.767715,106.667151) Broadth is 4-metre long. Here the edge lengths as well as the perimeter and area of the polygon can be calculated from the cartesian coordinates. drawn between the opposite ... How can I calculate area of the polygon and throw exception if it's greater than 400 square miles? Practice Questions. Polygon Calculator. 0. An easy to use, free perimeter calculator you can use to calculate the perimeter of shapes like square, rectangle, triangle, circle, parallelogram, trapezoid, ellipse, octagon, and sector of a circle. The interior angles of a triangle always add up to 180° while the exterior angles of a triangle are equal to the sum of the two interior angles that are not adjacent to it. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. If the rectangle has been rotated, I thought of two possible ways to calculate the vertices. a. By intersecting the non-parallel lines, you will find other vertices. d1 = 32 inches d2 = 20 inches Area of rhombus = 1 / 2 (d1) (d2) Area of rhombus = 1 / 2 (32) (20) Area of rhombus = 320 square inches. Triangle Plane coordinates of vertices: K[11, -10] L[10, 12] M[1, 3] give Triangle KLM. Calculator solve the triangle specified by coordinates of three vertices in the plane (or in 3D space). A vertex is defined in geometry as the common endpoint of two or more rays or line segments. It is a Corner. It is a Corner. Square Pyramid; Triangular Prism; Triangular Pyramid; 3D Shape – Faces, Edges and Vertices. Below is the algorithm you can use :) Let us say the 2 vertices are A and C (i.e diagonally opposite) Step 1) Calculate the midpoint between A and C using midpoint theorem. Find coordinates of vertices C and D, C 'and D'. 1) How many FACES on the cube? Calculator Technique. Point (2,1) and (6,7) are opposite vertices of a square inscribed in a circle, calculate the centre and the equation if the circle. As you can see, a diagonal of a square divides it into two Everyone who receives the link will be able to view this calculation, Copyright © PlanetCalc Version: How to use the calculator Enter the x and y coordinates of the three vertices A, B and C of the triangle and press "calculate". Find the area of the each quadrilateral whose vertices are In the first scenario, these two points are opposite vertices of a square, and I need to work out the coordinates of the other two vertices. 1 Answers. An edge is a line segment between faces. Find the coordinate of the center of the square. That is, we always take the area of quadrilateral as positive. Calculate the total volume (i n m 3) of a mole of hydrogen atom?Give 1 A = 1 0 − 1 0 m The sides meet in four corners, which are called vertices. Perimeter of a triangle calculation using all different rules: SSS, ASA, SAS, SSA, etc. Diagonal of square. Supports many imperial and metric units: mm, cm, meters, km, inches, feet, yards, miles, and more. Given the coordinates of any two vertices of a square (X 1, Y 1) and (X 2, Y 2), the task is to find the coordinates of the other two vertices. How many vertices in a tetrahedron? Click hereto get an answer to your question  Find the potential energy of a system of four particles placed at the vertices square of side 'a' .Also obtain the potential at the center of the square(i) The size of hydrogen atom is about 0.4 A. Type that value into the diagonal of a square calculator to check it yourself! Ask Question Asked 3 years, 5 months ago. Vertices in square prism? A square is a polygon with four vertices. BCD and DAB. Enter any one value and the other three will be calculated. Result: Coordinates of 4 vertices of rectangle. To calculate the diagonal of a square, multiply the length of the side by the square root of 2:. This free area calculator determines the area of a number of common shapes using both metric units and US customary units of length, including rectangle, triangle, trapezoid, circle, sector, ellipse, and parallelogram. Given that the vertices of a square are A (6, -13), B (-2,2), C (13, 10), and D (21,-5), i. By … To find the volume of a cube , we need to calculate its area of square base first which is A=length×length, and then multiply by its length. Let us look more closely at each of those: Vertices. Vertices of RT Show that the points P1 (5,0), P2 (2,1) & P3 (4,7) are the vertices of a right triangle. A square has two diagonals. Any other base unit can be substituted. Vertices; Square prism; DeepAMarudhachalaM 2015-06-23 12:48:22. Calculate the coordinates of the points of intersection of the circle and the line AC. Enter the set of points into the calculator below - one point per line, with x and y coordinates separated by semicolon. Calculations at a simple polygon. This free volume calculator can compute the volumes of common shapes, including that of a sphere, cone, cube, cylinder, capsule, cap, conical frustum, ellipsoid, and square pyramid. Then the calculation is trivial. A square calculator is a special case of the rectangle where the lengths of a and b are equal. Second is the in-sphere This is the sphere which fits inside the polyhedron so as to touch all its faces. They are the same length and intersect perpendicular to the center of the square.


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I have decided to rescind this proposal.  Stars in a galaxy travel at one velocity, light at the speed c.  There is only one velocity involved in applying the product rule to the definition of a force as a change in momentum.  Note that is not a trivial application of the product rule either.  Newton's second law is invariant under Galilean transformation.  The dm/dt dR/dt is not invariant.

The Problem with Velocity Profiles

The velocity profile of thin disk galaxies is a riddle worth pondering. Studies have shown that the velocity of stars in a galaxy rise to a peak as one leaves the center, and then stays flat as one looks further out. Newtonian gravitational theory can predict the peak correctly, but there must always be a decline.

From Galaxy rotation curve, with A is the Newtonian prediction, B is the data.

There are two parts to any force law: the cause and the effect. Consider the elliptical motion of the Moon around the Earth.  The cause is the gravitational field of the Earth acting like a point source.  The effect is the motion of the Moon around the Earth.

$-\frac{G M_{Earth}}{R^2}m_{Moon}=-\frac{V^2}{R}m_{Moon}$

The rules of algebra allow one to cancel the mass of the Moon on both sides.  Einstein appreciated the depth of that step, the equivalence of gravitational and inertial masses.  One can spot a core message of general relativity in this classical context. How much the Moon weighs is not relevant to its own gravitational motion. Gravity demands all masses orbit this way due to the Earth's gravitational field.

One can solve this equation to get the circular velocity:

$V_c = \pm \sqrt{\frac{G M_{Earth}}{R}}$

Both G and the mass of the Earth are fixed numbers. Go out a larger distance and the velocity must drop. While the equation for a potential is quite different for a disk galaxy, the fact remains that the circular velocity must fall with distance.

There are two proposals to fix this issue that have received much research attention. The first is Modification Of Newtonian Dynamics, or MOND.  When gravitational fields are weak, some ten orders of magnitude smaller than the acceleration of gravity on the surface of the Earth, then the form of the force law changes from a one over distance squared to one over distance. The proposal works great for the rotation profile of galaxies, closely matching over a hundred cases. MOND also does well with the Tully-Fisher relation.  For even larger scales such as clusters of galaxies, MOND is not as successful.  See The MOND Pages for a balanced assessment of the technical issues.

Dark matter is another proposal that tries to solve the riddle. Here one modifies the cause to include enough dark matter to produce the velocity curves that are seen. The proposal does not work so well for the rotation profile of galaxies because there are few constraints on the distribution of dark matter in space.  Dark matter is viewed as being a useful hypothesis for cosmology. A significant problem is that many dark matter candidates have been considered and rejected. Dark matter appears as the only player that can explain the weak gravitational lensing happening as a bullet cluster passes through another. In MOND, the gravitational and inertial masses must be exactly on top of each other. Yet the data says otherwise, the matter we see is in a different place from the gravitational potential. That is possible with dark matter.

The Light Matter Proposal

The light matter proposal starts from the simplest observation: galaxies emit light. They do so by converting mass into photons. We need to account for this mass change somewhere in our velocity profile equation. While the amount of mass that gets consumed by fusion is exceptionally small, the change in the form of the velocity profile equation needs to be investigated.

Now consider the Earth in an elliptical orbit around the Sun. The numbers will change, but much of the math remains the same. A difference is that the Sun undergoes fusion, where two hydrogen atoms fuse to create one helium atom. Because the rest mass of the two hydrogen atoms is more than one helium atom, the mass decrease according to Einstein's most famous equation - E = m c2 - releases an enormous amount of energy. About 4 billion kilograms of mass are converted to various forms of energy per second. While that may sound huge, in the context of the Sun such a mass is not.  The mass of the Sun is more than twenty orders of magnitude larger.

The gravitational potential of the Sun can be split into two components: the bulk which does not change in time, and the mass that does change in time due to fusion. Because these two masses enter as a sum into the potential, we can treat the Sun's mass as a constant. We cannot measure rotational speeds of planets to over ten significant digits, so the detail about fusion gets lost.

Finally, consider the rotation of stars in a thin disk galaxy. A number of things change. First, the source can no longer be viewed as a point. The potential function is no longer just the mass over the distance. The math problem was tackled in the 1960s by Alar Toomre using elliptical integrals. Later techniques used modified Bessel functions. With modern symbolic math programs, the rotation profile of a disk galaxy can be calculated by a computer in less time than it takes to type in the expression. Even with the fancier potential, a decline in circular velocity as the distance increases from the center of the galaxy remains unavoidable, yet inconsistent with observation.

Another change is that fusion happens for the masses effected by the force of gravity. All forces are a change in momentum, the product of mass times velocity. By the product rule of calculus:

$F=\frac{d (m V)}{d t}=m \frac{d V}{d t}+V\frac{d m}{d t}$

The first term is the well-known mass times acceleration. The second term is the stuff of rocket science. Rockets are launched in a particular direction, and that is the direction of this term. How would stars undergoing fusion contribute to this term? The photons go out in every direction, so when everything is even, there is no direction. A star is not a rocket.

Let's review why circular motion leads to the centripetal force that points radially. The velocity is perpendicular to the radius (see figure 2a).

Momentum is the product of mass times velocity, so at a time t, the momentum is perpendicular to the radius. A short time later, the star has moved along an elliptical path, retaining its momentum perpendicular to the radius. It is the direction of momentum that has changed. The only way to change momentum is through a force. The re-directional force points along the radius. The centripetal force of moving along an elliptical path is balanced by the gravitational force from other stars in the galaxy.

For a star undergoing fusion in an elliptical orbit, the mass being held in orbit decreases (see figure 2b). As a star loses mass, it takes less force to keep it in place. Would that translate into the star going faster? Consider a speed skater with several weights going around a corner. The skater is doing everything he can to stay on the path. He lets go of weights in the turn. He works just as hard as before, but can now go faster because he weighs less.

The next step in solving the riddle is to determine what function one needs to use for the change in mass per unit of time. Start by defining a value for the fusion rate at the center of the galaxy:

$\frac{d m_{R=0}}{d t}=F_0$

Go out some arbitrary distance. Can we say something about that rate of fusion based on what happens at the center of the galaxy?

The change in mass as a function of distance from the center of the galaxy can be determined from observation.  One uses the luminosity profile which has the form of an exponential decay along with assumptions about the mass to light ratio.  For thin disk galaxies, the rate of change in the mass per unit area is

$\frac{M(R)}{A}=\Sigma_0 e^{-R/R_d}$

where Ris a length scale for the decay in the density, often on the order of 2-6 kpc.

I propose a strict proportionality between the amount of mass and the amount of fusion going on: the more mass, the more fusion, the less mass, the less fusion.  At the center of the galaxy:

$\gamma \frac{m(t)_{R=0}}{A}=\Sigma_0$

Gamma is a coupling factor. Consider it a free parameter in the proposal.

For this to work at different places away from the center, the form of m(t) must match that of the luminosity profile:

$\gamma \frac{m(t)}{A}=\gamma \frac{m_0}{A}e^{-ct/R_d}=\Sigma_0 e^{-R/R_d}$

After eliminating the unit area, take the derivative of this function with respect to time:

$\frac{d m}{dt}=-\gamma \frac{c m_0}{R_d} e^{-ct/R_d}=-\gamma \frac{c}{R_d} m$

Substitute this into the force equation:

$m \frac{d \phi}{dR} = - m \frac{V^2}{R} - \gamma m \frac{c}{R_d} V$

Notice something miraculous: we again get to cancel out the small m mass.  This time it feels fortuitous because it happened based on the exponential form of the luminosity profile.

$V^2 + \gamma c \frac{R}{R_d} V - r \frac{d \phi}{dR}=0$

The solution for the circular velocity comes  out from the binomial equation:

$V_c=-\frac{1}{2} \left(\gamma c\frac{R}{R_d}\pm \sqrt{\left(\gamma c\frac{R}{R_d}\right)^2+4 R\frac{d \phi}{dR}}\right)$

For the Earth/Moon system, gamma equals zero and we can see the simple square root result. When R is much smaller than the length scale Rd, the velocity will be similar to that for Newtonian theory. For large enough values of the ratio, fusion will cease, making gamma equal to zero and reverting to Newtonian theory. Between those two limits, for an exponentially decaying disk galaxy, there will be a linear contribution to the velocity profile happening because stars undergo fusion. A linear increase in the circular velocity from the fusion term with distance sounds quite odd, but it is the way the math cards happen to be for this proposal.

There are two real solutions, a positive and negative circular velocity.  For circular motion, there are two directions, clockwise and counterclockwise. With the signs both negative, the magnitude of the resulting negative velocity is greater than the Newtonian solution. With the signs different, the positive rotational velocity is smaller.  Because a decrease in mass should lead to faster speeds, I interpret the same sign solution as the physical one.

Deriving the Velocity Profile Equation

What potential should one use for a thin, exponentially decaying disk galaxy? Answering this question will require a digression because the math is not trivial. Feel free to skip this section and go directly to the plots, they are more fun :-)

If you are interested in the details of the technical approach, get a hold of "Galactic Dynamics", by James Binney and Scott Tremaine, and focus in on Section 2.6, "The Potentials of Disks".  I will provide an impression of how to get to the velocity profile, but a number of the details are beyond my reach.  If you have questions on this part of the blog, please consult the source.

Rewrite the Laplace equation in cylindrical coordinates, so

$\nabla^2\phi=0$

becomes:

$\frac{1}{R} \left( \frac{d}{dR} \left( R \frac{d \Phi}{dR} \right) \right) + \frac{1}{R}\frac{d^2 \Phi }{d \phi ^2}+\frac{d^2 \Phi}{d z^2}=0$

The radius is R, phi is an angle, and z is a complex number.

One wants to separate the variables so there are three functions:

$\Phi = J(R) F(\phi) Z(z)$

The magic of separation of variables is used to get two darn similar equations made up of these parts linked by a constant (k2) :

$\frac{1}{R J} \left( \frac{d}{dR} \left( R \frac{d J}{dR} \right) \right) + \frac{1}{R^2 F}\frac{d^2 F }{d \phi ^2}=-\frac{1}{Z}\frac{d^2 Z}{d z^2}=-k^2$

Put these separate equations on separate lines and rearrange slightly:

$0=\frac{d^2 Z}{d z^2} - k^2 Z$

$0=\frac{d^2 F }{d \phi ^2} + m^2 F$

$0=R\left(\frac{d}{dR}\left( R \frac{d J}{dR}\right)\right)+(k^2R^2-m^2)J$

The first two equations can be solved with exponential functions. The third equation requires Bessel functions:

$J = c_1 J_0(k R) + c_2 Y_0(k R)$

Only the Bessel J function is used.  While not discussed in Galactic Dynamics, I think it has to do with the behaviour at the origin, where Bessel J is equal to unity, but Bessel Y is not defined (small values of R evaluate to large negative numbers). Put the solutions all back together to define a potential:

$\Phi_{\pm}=e^{i(m \phi \pm k z)} J_m (k R)$

This is where the math starts going beyond anything I will understand. They talk about a generalized surface function, Σ(R).  Using the miracle powers of Hankel transformations which are similar to Fourier transformations, they define the inverse of this as an "S(k)" function.. for a an exponential disk (eq. 2-158, first edition) is:

$S(k)=-2 \pi G \int_0^{\infty}dR ~R ~J_0(k R) ~\Sigma(R)$

$S(k)=-\frac{2 \pi G \Sigma_0 R_d}{(1+k^2 R^2_d)^{3/2}}$

Plug this into an integral with a Bessel function to get the potential.

$\Phi(R,z)=-2 \pi G \Sigma_0 R^2_d \int_0^{\infty} dk \frac{J_0(k R) e^{-k|z|}}{(1+(k R_d)^2)^{3/2}}$

If one consults two of those old technical books on integrals, Gradshtey and Ryzhik, and Abramowitz and Stegun, then together you can evaluate the integral:

$\Phi(R,0)=-\pi G \Sigma_0 R \left(I_0\left(\frac{R}{2 R_d}\right)K_1\left(\frac{R}{2 R_d}\right)-I_1\left(\frac{R}{2 R_d}\right)K_0\left(\frac{R}{2 R_d}\right)\right)$

[Math sidebar] I know this blog is math heavy.  That is mildly ironic because this is purely classical stuff, nothing so difficult as quantum mechanics. I tried to "follow along" with Mathematica. The program of course had the Bessel functions, and in the equation above, the modified Bessel functions. Yet it did not find these compact results, choosing instead to write the integrals as MeijerG functions, the proverbial granddaddy of special functions. Special functions can be defined in a few different ways, so they are playing it safe. I thought it was cool that this math was too hard for Mathematica to express in a compact fashion.
[end math sidebar]

Recall the circular velocity expression:

$V_c = \pm \sqrt{R \frac{d \Phi}{dR}}$

One needs to take the derivative of the potential.  That requires a dive into the appendix to justify:

$\frac{d J_0}{d R} = -J_1$

One does that magic with finding a compact expression for an integral with Bessel functions, and the circular velocity for Newtonain gravitational theory is:

$\frac{V_c}{\sqrt{G M/R_d}} = \pm \sqrt{\frac{R^2}{4 R_d}\left(I_0\left(\frac{R}{2 R_d}\right)K_0\left(\frac{R}{2 R_d}\right)-I_1\left(\frac{R}{2 R_d}\right)K_1\left(\frac{R}{2 R_d}\right)\right)}$

This is the equation needed to make the velocity profile of a disk galaxy (with no bulge).

Velocity Profile Plots

Plot the dimensionless velocity profile for a disk galaxy:

This is just like Galactic Dynamics figure 2.17 (omitting the negative velocities), with a peak around 2 R/Rd of a bit over 0.6.   I am more interested in seeing the real velocities.  I used the mass for the disk of the Milky Way galaxy (1.2 x 1041 kg) and scale factor 2.5 kpc.

The curve has the same shape.  The peak velocity is a touch over 220 km/s, which is similar to what happens in the Milky Way.

The same potential will be used to create a new velocity plot taking into account fusion happening within the stars in elliptical motion. There is one free parameter, gamma, set to 0.00009:

The peak velocity is slightly higher. The most important difference is the shape of the curve which is relatively flat from a few kpc out to more than 15 kpc.  Note that the curve does start to increase at even greater distances. I don't think there is any information content in the negative sign, but graphed it that way to be honest.

The graph is sensitive to the value of gamma.

Increase it by an order of magnitude and one gets a straighter line. Decrease it by an order of magnitude and the graph looks like Newton's theory unaltered.

Discussion

A big open question for this proposal is why gamma should be about 10-4? Certainly Nature is not playing with a parameter until the velocity profile is flat.  Here is one possibility.  Fusion in stars happens near the core with the temperature on the order of 10 million degrees. How much of the energy of the fusion stays in the stars in order to maintain its state? The star uses most of the energy of fusion to have a 10 million degree core and resist gravitational collapse.  The amount of energy that leaves might be the fraction between the temperature at the surface - thousands of degrees - versus the core, a difference of four orders of magnitude.

The bulge that appears in many galaxies including our own was ignored in this modeling. It is unclear how the bulge would change the resulting velocity profile.

A concern with using potential theory in this case is that this system is not conservative. While energy is conserved for the Universe, mass is leaving the galaxy as light energy in the three term model used here.

MOND is a daring hypothesis: gravity changes its form when the acceleration gets very small. Dark matter is a daring hypothesis: we don't know what it is, but there is even more of it than the stuff we see.

The light matter proposal in this initial phase is also bold. The fraction of mass involved in fusion is orders of magnitude beyond trivial. It is easy to appreciate those skilled in approximation to ignore the weight loss stars undergo to make light. The fraction of fusion mass is eliminated by a "lucky" cancellation. One is left with how the rate changes. The change in fusion rate should mimic the change in bulk mass.  It is that algebraic trick based on observation that may allow the small amount of fusion to have a large effect.

If we can see something, then there is fusion going on.  We may have to account for fusion in much cosmology work.  Changing the form of equations has a huge impact.

Death to dark matter.  Long live light matter.

A Mathematica notebook is available as a pdf or notebook.

COMMENT NOTE: I will delete comments from anyone who does not log into the site, whether positive or negative, it is simple to enforce. I have found such people are often impolite and make comments that are not worth reading.  Let's stick to the technical issues (toss in a "personal suggestion" and I will delete those too).

format: plain


text:
# Airplane for the Common Denominator

### Help Support Homebuilt Aircraft & Kit Plane Forum:

#### Toobuilder

##### Well-Known Member
Hiperbipe is indeed a 2 place.

#### Bigshu

##### Well-Known Member
So at least for my purposes, it's not aerobatic. The g limits might be, but it looks like the handling for enjoyable aerobatics and formation aren't there. But I've also been spoiled by doing those things in an RV-6.
Gotcha. Sounds like you found a design niche!

#### Pops

##### Well-Known Member
My Falconar F-12 , 2 or 3 place, is rated at +9 and -6 at 1600 lbs with a GW of 1800 lbs. Can be built taildragger, tri-gear, or RG. My F-12 was a taildragger. Could do anything that you can do with a RV-6. Fun airplane and good for traveling.

#### challenger_II

##### Well-Known Member
And didn't rattle and oilcan when manouvering! Gotta LOVE wood!

#### Toobuilder

##### Well-Known Member
20,000+ RV kits sold and 10,000+ completed and flying....

(Thats the mic drop moment)

#### raytol

##### Well-Known Member
Fuji-200 is a cross country capable aerobatic 4 seat aircraft if you can find one or a YAK-55 if you can feed the round pots!

##### Well-Known Member
Zenith 601/701. Small footprint for a T-hangar or even a simple shed or at worst, a carport.

#### Victor Bravo

##### Well-Known Member
The 4 seat Yak ?!? This has gone from the aircraft for the "Common Denominator" to the aircraft for the "Comrade Proletariat"!

The Antonov An-2 can carry ten or twelve common denominators to Burning Man as well....

#### BJC

##### Well-Known Member
The 4 seat Yak ?!?
Bud and Ross Granley fly (flew?) a nice airshow in a Yak 55 and a Yak 18T, the four seat version of the 18.

BJC

#### Tom DM

##### Well-Known Member

Hey!! On an AN-2 the oil is not consumed. OK: it is spit out through the exhaust but technically that is not consumed .
If it were consumed, the oil droplets would not land on the laundry hanging out to dry on the approach to Final

Found memories of a blue AN-2 in the South of Poland , sat at its controls, parked N2501A under its wings, its top wing is ondulated too.

1000 HP on AvGas, a flywheel starter the co-pilot has to wind to exhaustion, a bi-plane with a toilet and cargo for either 5 VIPs, 7 paratroopers or a lot of chemicals. Go anywhere, land anywhere, drink gasoline faster than a cow water... what is not like?

#### Tiger Tim

##### Well-Known Member
To clarify, the idea of a common denominator was to try and home in on a no-more-excuses, ‘shut up and build’ type of airplane. While it slides in at a bunch more expensive than I was picturing, Toobuilder has a pretty valid point on nominating the RV-8 to fit the bill. It has twice the seats and twice the power of what I was imagining but I t’s pretty okay at a bunch of different stuff, can be had a little at a time, has excellent support and the things routinely get finished and fly lots. Those last two points are pretty telling of the whole design, IMO. As an added bonus as near as I can tell they actually have decent resale value too.

BJC

#### challenger_II

##### Well-Known Member
Then I suppose the Common Denominator is One that can afford a $60,000 to$100,000 airplane.
Most of us would be considered less-than common, then.

#### challenger_II

##### Well-Known Member
Flying has always been an expensive endeavor. Making it manageable to One's financial position has been a challenge for over a hundred years, now.
Homebuilt airplanes have been a hedge on the expense since the 1920's: just peruse through the old Flying and Glider manuals for a reference.
Back in the late 1920's and through the 1930's, those Big Plane boys flying the WACO's and Travel Airs looked down their collective noses at the less financially capable folks for piddling with those tiny Aeronca C-2's and such.
Today, it is much the same: the RV drivers and the like think that they own the world, and those tinkering with something slower, made of wood and fabric and such are beneath contempt. These attitudes are what squashes innovation, and positive progress towards having an airplane for that "Common Denominator".

In the morning I am gonna go drag out my 1932 vintage Longster II and go scare the rabbits. Might not go very fast, but the ride is good, the view is excellent, and I WILL BE in the air.

Nuff said.

#### BJC

##### Well-Known Member
Today, it is much the same: the RV drivers and the like think that they own the world, and those tinkering with something slower, made of wood and fabric and such are beneath contempt. These attitudes are what squashes innovation, and positive progress towards having an airplane for that "Common Denominator".
There was a time, not very long ago, when owners / drivers of TC aircrsft looked down on all E-AB aircraft. Today, they look with envy at the performance, configuration, and maintenance cost advantage of E-AB.

People who let other peoples’ attitudes “squash innovation” never will own an airplane.

BJC

#### TFF

##### Well-Known Member
While many a RV are 100k, most think of it as all the money right then and fly tomorrow. While not cheap, a kit, a used engine with fixed pitch prop and basic instruments will add up, but it will take at least ten years to build.

That puts it up at $350 a month. Still while not cheap, there are plenty out there paying that much a month for car insurance for something pretty and shiny, and they still have the car note for that pretty. Counteract with used car, but on still funneling money into a different hobby they think is more important. Homebuilt, it’s supposed to take time to build. It’s supposed to be a primary hobby not a back hobby. It’s not instant gratification. It is way cheaper to buy something that flys. Build a scratch built Cub with no legacy parts. Skybolt from plans? T-18 or Mustang II? Bearhawk? EZ? You won’t build those for less than the RV. Not really. It might feel like$3K of tubing is a cheap fuselage, but after fittings, controls and everything a fuselage needs, is it cheaper?

Can you build a cheaper plane? Sure. You do start paring back with some sort of performance. Finishing someone’s project is the cheapest way. Let them take the depreciation hit.

While my interests are in different places, it’s hard to deny the truth of the RVs.

#### Tom DM

##### Well-Known Member
Today, it is much the same: the RV drivers and the like think that they own the world, and those tinkering with something slower, made of wood and fabric and such are beneath contempt. These attitudes are what squashes innovation, and positive progress towards having an airplane for that "Common Denominator".

Don't share the view about the RV-jockeys. RV's are a reaction to the ruthless exploitation in General Aviation (that being aluminium build airplanes) RV's are not modern yet are in some way a proof that even when Innovation is buried in concrete, It will find a way out.

Wood is yesteryear: no contempt to it , just a material gone out of grace because it was outperformed and there was better. And as such no new craftsmen were schooled while the old ones slowly take off West. Now is the time of aluminum yet its successor is already here.

The discussion about "common denominator" circles always around the same hot potato: money. However there is a deeper question: is there for society a more-value to more people flying? Of that I am not certain.

#### BJC

##### Well-Known Member
However there is a deeper question: is there for society a more-value to more people flying? Of that I am not certain
I consider my flying to be the greatest expression of personal freedom, so your “deeper question”, for me, may be restated and generalized as “Does society benefit by more people exercising and enjoying personal freedom?”

My answer is “Yes”. I wish that other areas of the world cherished personal freedom more. It would make for a better world.

This reminds me of the discussion here Flight Helmets?

BJC

#### Tom DM

##### Well-Known Member
I consider my flying to be the greatest expression of personal freedom, so your “deeper question”, for me, may be restated and generalized as “Does society benefit by more people exercising and enjoying personal freedom?”

My answer is “Yes”. I wish that other areas of the world cherished personal freedom more. It would make for a better world.

This reminds me of the discussion here Flight Helmets?

BJC

I am inclined to the affirmative answer because of the word Freedom. Freedom comes with doing it your way, gaining skills which can be applied in others domains, making something out of nothing and thus generating wealth / contributing the society.

However Freedom has its limits as there is an oscillation between Freedom- Security. The more cynical in me says the most can not bear the responsibility of Freedom and are but too happy to exchange it for Security, for not having to Think. I bump my head regularly to that.

There is also another "little" thing: keeping things in perspective in a kind of moral way.
Is one right to sell something for 100-1000 times its intrinsic value just because the circumstances allow to get away with it? Stupid Greed will not hesitate to do so, Smart Greed however... Smart Greed transforms easily into Stupid Greed, I still have to witness the opposite.

As to other parts of the World having to cherish personal freedom more, that comes from an "american" view on how an better World is supposed to look. The USA and the World have both found out the hard way that not everybody subscribes to that view.

Will check Flight Helmets

Blue Skies


text:
# 11  Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.

The equation of the plane through the intersection of the given two planes, $x+y+z =1$  and  $2x+3y+4z =5$ is given in Cartesian form as;

$(x+y+z-1) +\lambda(2x+3y+4z -5) = 0$

or $(1+2\lambda)x(1+3\lambda)y+(1+4\lambda)z-(1+5\lambda) = 0$             ..................(1)

So, the direction ratios of (1) plane are $a_{1},b_{1},c_{1}$ which are $(1+2\lambda),(1+3\lambda),\ and\ (1+4\lambda)$.

Then, the plane in equation (1) is perpendicular to $x-y+z= 0$ whose direction ratios $a_{2},b_{2},c_{2}$ are $1,-1,\ and\ 1$.

As planes are perpendicular then,

$a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0$

we get,

$(1+2\lambda) -(1+3\lambda)+(1+4\lambda) = 0$

or  $1+3\lambda = 0$

or  $\lambda = -\frac{1}{3}$

Then we will substitute the values of $\lambda$ in the equation (1), we get

$\frac{1}{3}x-\frac{1}{3}z+\frac{2}{3} = 0$

or   $x-z+2=0$

This is the required equation of the plane.

Exams
Articles
Questions


text:
# How can I insert a bounding box or polygon in a PostGIS geometry field?

I have a field in my table called 'geometry' that was created thus:

CREATE TABLE "public"."sites" (
"id" int8 DEFAULT nextval('sites_id_seq' :: regclass) NOT NULL,
#... A few MORE fields
"geometry" "public"."geometry",
CONSTRAINT "sites_pkey" PRIMARY KEY ("id")
) WITH (OIDS = FALSE);
CREATE INDEX "index_sites_on_geometry" ON "public"."sites" USING gist (geometry);

I can create a point with the command:

UPDATE sites
SET geometry = ST_SetSRID (
ST_MakePoint (44.28, - 96.68, 495),
4326
),
WHERE
ID = 12;

I would like to create a bounding box, but being new to the syntax have not been able to get the correct syntax.

My best attempt:

UPDATE sites
SET geometry = ST_SetSRID (
ST_3DMakeBox (
ST_MakePoint (44.2, - 96.6, 400),
ST_MakePoint (44.3, - 97.0, 400)
),
4326
)
WHERE
ID = 1123;

Gives the error:

Column has Z dimension but geometry does not

I get this error even after trying

1. wrapping the ST_SetSRID(...) in a select statement as some online examples show (though the select statement returns a 2D result)
2. using ST_Envelope(ST_Makeline as suggested by @mattmakesmaps

My apologies that I am new to this. I have been looking through the documentation and googling postgis insert bouding box etc, but at my level, this isn't clear.

You can use St_GeomFromText to add geometries in WKT format

UPDATE sites
SET  geometry = st_geomfromtext('POLYGON((0 0, 1 1, 2 2, 3 3, 0 0))')
WHERE ID = 1123

• when I wrap your example in select st_geomfromtext('POLYGON((0 0, 1 1, 2 2, 3 3))') I get " ERROR: geometry contains non-closed rings" – David LeBauer Aug 21 '14 at 0:56
• Ooops sorry. Coordinates should end with the first point. I will edit – amenadiel Aug 21 '14 at 1:02

You may want to use the ST_Envelope() function.

Specifically, see the 2nd example using a LINESTRING geometry. Given a LINESTRING composed of two vertices that represent your points, you should be able to get back a Polygon geometry which might be more useful to you.

• I still get the error "Column has Z dimension but geometry does not" with update sites set geometry = ( select ST_SetSRID(ST_Envelope(ST_Makeline(ST_MakePoint(42.794448, -76.116081, 415),ST_MakePoint(42.79231, -76.116679, 415))),4326)) where id = 1123 – David LeBauer Aug 20 '14 at 23:09
• Is the column you're trying to load these data into a 3d column? If so you can use the ST_Force_3D function, as seen in this previous question – mattmakesmaps Aug 20 '14 at 23:31
• Based on your above example, the following SQL Statement outputs a Polygon Z geometry: SELECT ST_AsText(ST_Force_3D(ST_SetSRID(ST_Envelope(ST_Makeline(ST_MakePoint(42.794448, -76.116081, 415),ST_MakePoint(42.79231, -76.116679, 415))),4326))) AS geometry – mattmakesmaps Aug 20 '14 at 23:32
• The output is as follows: geometry ------------------------------------------------------------------------------------------------------------------------------- POLYGON Z ((42.79231 -76.116679 0,42.79231 -76.116081 0,42.794448 -76.116081 0,42.794448 -76.116679 0,42.79231 -76.116679 0)) (1 row)  – mattmakesmaps Aug 20 '14 at 23:33
• ST_Envelope drops the Z dimension, making any geometry a 2D polygon. – Mike T Aug 21 '14 at 1:32

Perhaps you may need to think through the data type for sites. What is it for? It seems that geometry is defined as geometry(PolygonZ,4326), which I think is a bit silly. A polygon could be expressed as a 3D object if all the linear rings are coplanar (like a triangle or a lake). However, if it were, for example, the border of Nepal, why keep elevations with these coordinates? I'd argue it's nonsense, since you can't visualize this kind of PolygonZ.

If you want to keep the Z dimension, use ST_Force3D (or ST_Force_3D for older versions), which will assign a default Z of 0. Otherwise, you may want to change the geometry type to geometry(Polygon,4326). Lastly, you could always use a box3d type, if you want to only store 3D boxes.

Lastly, your axis order is reversed. It is always (long, lat) for PostGIS.

• We will want to keep elevation information, primarily because it informs the physical properties of the site (i.e., meteorology, hydrology). – David LeBauer Aug 22 '14 at 17:07
• That's not what I said. I was specifically singling out the meaning of a PolygonZ geometry. On the other hand, I think PointZ and LineStringZ are fantastic geometry types for features with elevation. – Mike T Aug 23 '14 at 1:18


text:
# Unemployment rate vs deficit spending

#### Orion1

I attempted the following plot on Wolfram Alpha:
Code:
plot United States unemployment rate,(United States federal deficit/United States GNP)
The resulting graph is as attachment.

According to my understanding of this graph, the derivative of the United States unemployment rate is a direct result of deficit spending by the United States Federal Government.

The greater the amount of deficit spending the exponentially higher the unemployment rate.

Are there any economists here that can improve upon my calculation?

Reference:
http://www.wolframalpha.com/input/?i=plot+United+States+unemployment+rate,(United+States+federal+deficit/United+States+GNP)"

#### Attachments

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Related General Discussion News on Phys.org

#### Gokul43201

Staff Emeritus
Gold Member
What is it that you are trying to calculate? If anything, you can extract a correlation between unemployment and deficits. Without a proposal for a mechanism, you have provided no basis for a causation.

Moreover, a naive inspection of your plot shows a lag in the deficit relative to unemployment. And it's hard to not notice that rising unemployment causes governments to go into spending overdrive in an attempt to stave off job losses, stimulate economic activity, or provide more unemployment benefits. So there's clearly an argument for the causation to be primarily in the opposite direction to your claim.

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#### Ivan Seeking

Staff Emeritus
Gold Member
Moreover, a naive inspection of your plot shows a lag in the deficit relative to unemployment. And it's hard to not notice that rising unemployment causes governments to go into spending overdrive in an attempt to stave off job losses, stimulate economic activity, or provide more unemployment benefits. So there's clearly an argument for the causation to be primarily in the opposite direction to your claim.
What's more, if we assume causation, then the graph suggests that the government succeeds in reducing unemployment through spending, but it takes a few years for the full effect to be seen.

#### imiyakawa

We must understand OPs position, from 1990 onwards there seems to be identical correlation, it doesn't seem to be a lagging indicator, which doesn't help with the hypothesis that unemployment is causing deficit spending.

I suggest everyone thinks about the actual events that are happening in line with the changes in the gradients.

Our limits in infering given this data are:
-One cannot conclude that unemployment is causing deficit spending.
-One cannot conclude that deficit spending is causing unemployment.
-One CAN conclude that global events are causing both unemployment and deficit spending.

(1) beginning of 2000, Stock market crashes -- Fiscal Stimulus AS WELL AS inevitable unemployment. Both have a root cause, (A) isn't causing (B), (B) isn't causing (A), (C) is causing (A) and (B).

(2) 1990s, general domestic economic prosperity within USA (call this (C)). This caused (A) reducing unemployment, (B) reducing deficit. Again, (C) caused (A) and (B).

(3) 2010, SubPrime mortgage crisis, 383 mortgage-related financial institutions bust! (ml-implode.com), (A) is not causing (B), (B) is not causing (A), (C - collapsing financial system), is causing but (A - unemployment) and (B - a responsive deficit).

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#### Ivan Seeking

Staff Emeritus
Gold Member
We must understand OPs position, from 1990 onwards there seems to be identical correlation, it doesn't seem to be a lagging indicator, which doesn't help with the hypothesis that unemployment is causing deficit spending.
Really? I clearly see the spending [the deficit] lagging the unemployment rate before and after 1990. Where on the graph [approximate date] do you see the opposite?

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#### imiyakawa

Really? I clearly see the spending [the deficit] lagging the unemployment rate before and after 1990. Where on the graph [approximate date] do you see the opposite?
Just the small patch between 1993.5 to 2002.5 :p

There looks like identical correlation.

ALthough you of course have grounds to argue against that patch. 2001 you could argue that the gradient of the deficit increased markedly not only in response to the stock market crash but in response to quickly growing unemployment.

#### russ_watters

Mentor
Moreover, a naive inspection of your plot shows a lag in the deficit relative to unemployment. And it's hard to not notice that rising unemployment causes governments to go into spending overdrive in an attempt to stave off job losses, stimulate economic activity, or provide more unemployment benefits. So there's clearly an argument for the causation to be primarily in the opposite direction to your claim.
Also, high unemployment lags behind a drop in GDP and a drop in GDP has a direct impact on tax revenue.

#### russ_watters

Mentor
What's more, if we assume causation, then the graph suggests that the government succeeds in reducing unemployment through spending, but it takes a few years for the full effect to be seen.
Right: you can't assume causation, you have to demonstrate it with logic. The was the problem with the OP!

#### Ivan Seeking

Staff Emeritus
Gold Member
Right: you can't assume causation, you have to demonstrate it with logic. The was the problem with the OP!
Indeed.

#### Matterwave

Gold Member
Perhaps it's best if you also included a regression on the two variables. Instead of graphing them over time, plot the data points for each year on a graph with deficit spending on the x-axis and unemployment on the y-axis. This should give you some better indication if there is correlation (on an instantaneous basis). You can then expand your analysis by plotting say, the unemployment rate at time t and the deficit spending at time t+3 or something to see if there is supposedly lagged correlation.

This is all correlation...it's very hard to come up with causation arguments without first correcting for every omitted variable...omitted variable bias is incredibly hard to get rid of in something as complicated as this haha.

#### hamster143

Are there any economists here that can improve upon my calculation?
Not an economist but I can surely improve on that calculation:

plot United States unemployment rate, (United States federal receipts/GNP), (United States federal outlays/GNP)

2001 and 2008 recessions are clearly accompanied by sharp drops in government revenues. Government spending did not increase substantially during the 2001 recession, and it only started to rise with considerable delay during the 2008 recession.

#### imiyakawa

Perhaps it's best if you also included a regression on the two variables. Instead of graphing them over time, plot the data points for each year on a graph with deficit spending on the x-axis and unemployment on the y-axis. This should give you some better indication if there is correlation (on an instantaneous basis). You can then expand your analysis by plotting say, the unemployment rate at time t and the deficit spending at time t+3 or something to see if there is supposedly lagged correlation.

This is all correlation...it's very hard to come up with causation arguments without first correcting for every omitted variable...omitted variable bias is incredibly hard to get rid of in something as complicated as this haha.
THIS. .. THIS. THIS. THIS!!!!!

Yes!!!!!!

Although such a regression will be basically nonsense. As matterwave said, omitted variable bias. This means there's an X2 that's causing both X1 and Y1, and this X2 isn't in your model. This makes it look like you have very neat correlation, but you actually don't! (E.G., when comparing student test scores to class size, an example of X2, X3, X4 would be income, geographic area, and whether there are computers in the classroom). An example of that X2 in what we're talking about are real world events, such as recessions, etc. Recessions are causing correlated directional changes in both GDP and unemployment, but it's not in you rmodel, and so it makes unemployment look like it's having a huge causal effect when it's not!

I dunno how you're going to account for this, frankly.
----------------------------------------------------
There is some reason to expect this kind of negative correlation after all!
Inflation and employment are negatively correlated. It's not unwise to suspect that the government would risk a deficit during times of non-inflation. This will make it seem like a deficit is causing unemployment. Another omitted variable!

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#### Orion1

unemployment rate vs government spending...

Code:
plot United States unemployment rate, (United States federal receipts/GNP), (United States federal outlays/GNP)
Given that causation has not been established and there is an omitted variable bias, it still appears that the government spending rate drives the unemployment rate.

Reference:
http://www.wolframalpha.com/input/?i=plot+United+States+unemployment+rate,+(United+States+federal+receipts/GNP),+(United+States+federal+outlays/GNP)+"

#### Attachments

• 33.8 KB Views: 261
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#### WhoWee

You might want to consider using GDP.

#### hamster143

Given that causation has not been established and there is an omitted variable bias, it still appears that the government spending rate drives the unemployment rate.

Federal spending comes in two major categories, discretionary and mandatory.

Discretionary spending (e.g. deciding to spend extra $5 billion on the space shuttle program, or to save$5 billion by shutting down Tevatron) does not vary much (except when the government decides to start a new war). And it is a small part of the total bill. In the last pre-recession budget, non-defense discretionary spending accounted for less than 20% of all spending.

Big fluctuations you see in the chart come from mandatory spending, they reflect recession-time increases in safety-net spending on Medicaid, unemployment insurance, welfare, and food stamps. These all increase automatically as a consequence of rising unemployment and poverty.

Likewise, fluctuations in federal receipts are causally linked to unemployment (fewer people working mean fewer taxes collected).

#### mheslep

Gold Member
The OP is attempting to show, I believe, how spending or the deficit drives unemployment, not the other way around.

#### hamster143

The OP is attempting to show, I believe, how spending or the deficit drives unemployment, not the other way around.
And we're trying to explain to him that he's wrong.

#### mheslep

Gold Member
And we're trying to explain to him that he's wrong.
??? Not in this thread. The only arrows thrown are ones casting doubt on the causality, not that he's wrong.

#### hamster143

??? Not in this thread. The only arrows thrown are ones casting doubt on the causality, not that he's wrong.
He is wrong because he posits causality in one direction, when in actuality the causality goes in the opposite direction.

#### mheslep

Gold Member
He is wrong because he posits causality in one direction, when in actuality the causality goes in the opposite direction.
That does not mean his thesis is wrong. Causality might go both ways.

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#### russ_watters

Mentor
Given that causation has not been established and there is an omitted variable bias, it still appears that the government spending rate drives the unemployment rate.
No, it really doesn't. The recent data patterns appear pretty much exactly superimposed to me!

IIRC, people are eligible for unemployment compensation a week after they become unemployed. So when someone becomes unemployed, there is an immediate drop in the taxes they pay and a week later a rise in the benefits they receive. I don't see how there could be any question that the major year-to-year fluctuations in deficit are caused by unemployment fluctuations and not the other way around.

#### russ_watters

Mentor
He is wrong because he posits causality in one direction, when in actuality the causality goes in the opposite direction.
Yes, the long-term fluctuation in deficit could result in long-term fluctuation in unemployment.... but the OP didn't attempt that argument!

#### John Creighto

I attempted the following plot on Wolfram Alpha:
Code:
plot United States unemployment rate,(United States federal deficit/United States GNP)
The resulting graph is as attachment.

According to my understanding of this graph, the derivative of the United States unemployment rate is a direct result of deficit spending by the United States Federal Government.

The greater the amount of deficit spending the exponentially higher the unemployment rate.

Are there any economists here that can improve upon my calculation?

Reference:
http://www.wolframalpha.com/input/?i=plot+United+States+unemployment+rate,(United+States+federal+deficit/United+States+GNP)"
Aside from what the causation actually is it is a striking relationship in the latter part of the graph. I do not know how these documents work so I don’t know what to infer about the methodology. Is the computation more complex then shown or is it exactly as shown? Is there any way to see the source code?

What I wonder is if the tighter relationship in the latter part of the graph is not a result of how the graph is drawn, then what quantities are held constant in the latter period of the graph that weren’t held constant in the earlier period of the graph? For instance how did inflation and the interest rate change over these time periods, how did government spending change, and how did tax revenue change? Additionally, how did the demographics of government tax revenue change thought these periods. Reducing taxes on the supper rich would have likely made the deficit more sensitive to unemployment as the national unemployment rate is not likely as correlated with the unemployment rate of the wealthiest Americans.

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#### Attachments

• 33.5 KB Views: 293
Last edited by a moderator:

#### John Creighto

Code:
plot United States unemployment rate, (United States federal receipts/GDP), (United States federal outlays/GDP)
Done. At the end of the graph, the recession begins in late 2007.

Reference:
http://www.wolframalpha.com/input/?i=plot+United+States+unemployment+rate,+(United+States+federal+receipts/GDP),+(United+States+federal+outlays/GDP)"
That's interesting that the correlation is stronger when deviding by GNP then GDP. I suppose this makes sence.

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### Physics Forums Values

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text:
Hardcover | $35.00 Short | £24.95 | ISBN: 9780262013444 | 200 pp. | 6 x 9 in | 19 figures, 8 tables| September 2009 Ebook |$24.95 Short | ISBN: 9780262260091 | 200 pp. | 19 figures, 8 tables| September 2009

## Overview

Information retrieval in the age of Internet search engines has become part of ordinary discourse and everyday practice: “Google” is a verb in common usage. Thus far, more attention has been given to practical understanding of information retrieval than to a full theoretical account. In Human Information Retrieval, Julian Warner offers a comprehensive overview of information retrieval, synthesizing theories from different disciplines (information and computer science, librarianship and indexing, and information society discourse) and incorporating such disparate systems as WorldCat and Google into a single, robust theoretical framework. There is a need for such a theoretical treatment, he argues, one that reveals the structure and underlying patterns of this complex field while remaining congruent with everyday practice. Warner presents a labor theoretic approach to information retrieval, building on his previously formulated distinction between semantic and syntactic mental labor, arguing that the description and search labor of information retrieval can be understood as both semantic and syntactic in character. Warner’s information science approach is rooted in the humanities and the social sciences but informed by an understanding of information technology and information theory. The chapters offer a progressive exposition of the topic, with illustrative examples to explain the concepts presented. Neither narrowly practical nor largely speculative, Human Information Retrieval meets the contemporary need for a broader treatment of information and information systems.


text:
### antontrygubO_o's blog

By antontrygubO_o, 6 months ago, translation,

It’s finally time for one of the biggest events in the competitive programming world. The SnackDown ‘21 Grand Finale is here.

The entire Grand Finale event is scheduled to be completed in two days — January 8 (Saturday), and January 9 (Sunday).

### For the Finalists

All the information has been shared over email, but summarising — The Grand Finale will be spread over two days and will be a closed event accessible only to the finalists, and a few invited guests. The SnackDown 2021 Final contest will be held on Jan 9th between 6:30 — 10:30 PM (IST).

For the last time for this SnackDown, meet the panelists!

### For the Community

For those who would like to have a taste of the SnackDown Final problems, we also have the SnackDown Final Parallel Unrated Contest scheduled for January 9, 2022, 6:30 PM — 10:30 PM (IST). The problems will be the same as the Finale. Since the main Finals will be a restricted session, the rank list can be viewed here. Note that the rank list will be frozen for the last hour, and will be resolved in the post-contest ceremony on YouTube.

### The Grand Finale Closing & Crowning Ceremony

The SnackDown ‘21 Finale closing ceremony will start on Jan 9th at 10.30 PM IST. It will be hosted by Kamil Debowski and Ashish Gupta, who will be analyzing the contest problems and the strategies and performances of the participants. This is also where you can witness the moment of truth, the SnackDown ‘21 Grand Finale winner announcement! Everyone is cordially invited to the event, and we request you to set your reminders here.

• +203

 » 6 months ago, # |   +207 as a tester, i overslept the elimination round
 » 6 months ago, # |   0 as a setter, i have tried to make interesting and beautiful problems and recommend to give this contest
 » 6 months ago, # |   +31 Up, the contest starts in 30 minutes.
 » 6 months ago, # | ← Rev. 2 →   -30 .
 » 6 months ago, # |   0 WHere can I find the live leaderboard??
•  » » 6 months ago, # ^ |   0
 » 6 months ago, # | ← Rev. 2 →   0 i think codechef should keep this leaderboard for regular standing too. it looks quite nice and able to see all problems at once makes a big difference . i think it will be much better than normal standing page in codechef rounds
•  » » 6 months ago, # ^ |   +15 You mean displaying all the problems in the leaderboard without needing to scroll, right?
•  » » » 6 months ago, # ^ |   0 yes
•  » » » » 6 months ago, # ^ |   0 Sure. Thank you for the feedback. We are already revamping the rankings page, will try to accommodate this too. :)
•  » » » » » 6 months ago, # ^ |   +13 Would it be difficult to also show timestamps of submissions for each problem on the leaderboard? This information is useful during contests, because it allows to roughly guess how much time people (from someone's friend list) typically spend on each problem.And after a contest, being able to sort submissions by execution time would be also very useful. I'm sometimes interested in checking the most performance optimized submission, but the CodeChef UI doesn't seem to provide an easy way to find it.
 » 6 months ago, # | ← Rev. 2 →   +95 The contest has finished, we hope you liked the problems!. We enjoyed seeing the finalists compete fiercely up to the very last minute!If you are interested, there is a wonderful final ceremony (with some comments on the contest by none other than antontrygubO_o and Um_nik, and with the revelation of the final ranking) hosted by Errichto and Ashishgup.
•  » » 6 months ago, # ^ |   +26 Ashish Gupta cohosted it
 » 6 months ago, # |   +39 Congratulations slime!, tourist! and Benq!
 » 6 months ago, # |   +82 I really enjoyed the problems. It was the best onsite final problemset in 2021. Thank you!dario2994 could you mind sharing some tricky cases for Climbing?
•  » » 6 months ago, # ^ |   +64 I am very happy to read that.Here is a list of 1000 small tests (they are not tricky, but they are many): in and out.Achtung: The editorial of CLIMBINGRANK will appear only after the first AC.
•  » » » 6 months ago, # ^ | ← Rev. 2 →   -84 So that you can use that AC solution as an editorial lmao
•  » » » 6 months ago, # ^ |   +106 Congrats to Benq who is the first solver of CLIMBINGRANK!
 » 6 months ago, # |   +181 Thank you for the contest! I enjoyed thinking about every problem. ONESGUESSA good easy problem. ANDOFMAXESI like the $O(n \log A)$ solution, and I actually almost implemented it (but I used a set instead of maintaining stack maxima). It's fine that solutions with an extra $\log$ factor passed too. PERMRANGESThe first problem that made me think I was missing something obvious, given how quickly many people were solving it. Turned out I failed to recognize Stirling numbers :) The problem is far from obvious anyway, though. EQPRFMAXSPLTA fun story: my first impression was "I must have solved this problem at AtCoder". I tried searching for it, but only found AGC052 D. Given that the problem is similar (but is asking for "equal LIS" instead of "equal number of records"), and the writer of both problems is antontrygubO_o, I thought "ok, I guess this is the problem I am looking for, it just looks similar, but it's a different problem" and stopped searching. Turns out I should have dug deeper to find AGC028 E. ROBBERYPLANThe second problem that made me think I was missing something obvious. I spent a lot of time thinking about various applications of Minkowski sums, but none of them helped to solve the problem even for $k \le 10^3$, let alone $k \le 10^6$. In the last hour, I figured out the reordering trick: we can reorder the stolen items so that the sum on every prefix of length $i$ is close to $w \cdot \frac{i}{k}$. It was easy enough from there. BAKERYI came up with a slow DP solution pretty early into the contest (before solving my 3rd problem), but only figured out how to reduce its complexity during the last hour. It's fun to have a problem asking for a floating-point answer (which actually allows to speed up the solution too) instead of the exact answer modulo whatever, at least once in a while. CLIMBINGRANKGiven the contest situation, I basically discarded this problem, I only came up with a simple $O(n^4)$ solution. Looks fun to think about, though.
•  » » 6 months ago, # ^ |   +26 For ANDOFMAXES we originally wanted only $O(n \log C)$, but closer to the contest it turned out that we can't separate it well enough from $O(n \log n \log C)$ but we wanted to have a second problem before "hard part" so we decided to lower the limitations to let $O(n \log n \log C)$ to pass freely.
•  » » 6 months ago, # ^ | ← Rev. 3 →   +8 Thanks for the contest. I really enjoyed solving the first 4 problems. In the contest, I implemented $O(n * (logn + logC))$ soln for ANDOFMAXES and it's a bit different from the soln given in the editorial. Probably others would have implemented it as well but since solns aren't public I will write it here for the sake of completeness. SpoilerGiven a fixed mask $M$ and let's define $dp[L][R][freeLeft][freeRight]$ as maximum no of subarrays we can divide subarray $A[L....R]$ such that AND of all maximums is a supermask of $m$, we are free to leave prefix if $freeLeft$ is true and we are free to leave a suffix if $freeRight$ is true.The main idea is we query for the maximum in that range and if it's the subset of $M$ then it splits the range into 2 parts and we are allowed to skip prefix in one part and a suffix in another part, which would correspond to the subarray to which this maximum belongs.If it isn't the supermask of $M$ then we will have to include this index in prefix or suffix which we are going to skip.We want to know if $dp[1][n][false][false] \geq k$ for fixed mask $M$.We can find range maximums using RMQ in O(1). SpoilerTo find the value of particular $dp[L][R][freeLeft][freeRight]$ we first query for maximum in $A[L...R]$ let the index of maximum value be $I$.If $A_{I}$ is supermask of $M$ then $dp[L][R][freeLeft][freeRight] = dp[L][I-1][freeLeft][true] + 1 + dp[I+1][R][true][freeRight]$Otherwise,$dp[L][R][freeLeft][freeRight]$ is -$max(dp[L][I-1][freeLeft][freeRight],dp[I+1][R][freeLeft][freeRight])$ if both $freeLeft$ and $freeRight$ are true, $dp[L][I-1][freeLeft][freeRight]$ if $freeRight$ is true,$dp[I+1][R][freeLeft][freeRight]$ if $freeLeft$ is true,$-INF$ otherwise.It can be seen that these have $O(n)$ states if implemented recursively.There would be $O(logC)$ candidature $M$ and checking for each one takes $O(n)$ time. Overall $O(n logC)$. Building RMQ would take $O(n log n)$.My submission
•  » » » 6 months ago, # ^ |   +2 "solns aren't public"Sorry, missed that. All submissions should be public now.
•  » » » » 6 months ago, # ^ |   -7 What is your take on this :: i think codechef should keep this leaderboard for regular standing too. it looks quite nice and able to see all problems at once makes a big difference . i think it will be much better than normal standing page in codechef rounds
 » 6 months ago, # |   0 Congratulations to the winners!Since, snackdown is over, can I know when can I receive my Amazon-coupon and T-shirt? I received a mail back on 16th November regarding the coupon and T-shirt. I had also filled the form, but I didn't listen anything back from codechef since. I won them through the referral leaderboard system.Thank You. CodeChef_admin suraj_sharma


text:
algebra - QuestionCove
OpenStudy (anonymous):

algebra

1 year ago
OpenStudy (anonymous):

1 year ago
OpenStudy (mathmale):

I'm surprised at you! I gave you guidance towards finding the slope formula, whereupon you simply started a new post in the hope that someone else would do your work for you. This is not acceptable. Please go back to our previous discussion, look up the "slope of a straight line" formula, and then find the slope of the line thru those two points.

1 year ago
OpenStudy (mathmale):

By deleting our previous conversation, you wasted my time as well as yours.

1 year ago
OpenStudy (mathmale):

Look up the slope formula and type it in here, please!

1 year ago
OpenStudy (anonymous):

4,2 -2-4

1 year ago
OpenStudy (mathmale):

http://cs.selu.edu/~rbyrd/math/slope/ I'm asking you to type in the general slope formula. Only after doing that, you apply this formula to the 2 points in question, to find the slope of the line passing through both points. Slope formula, please?

1 year ago
OpenStudy (anonymous):

y=mx+b

1 year ago
OpenStudy (anonymous):

how can i put those number on that equation?

1 year ago
OpenStudy (mathmale):

Show me the slope equation first, please. It's all there in black and white if you'd refer to the links I've given you.

1 year ago
OpenStudy (anonymous):

dude i dont even know what is a slope

1 year ago
OpenStudy (mathmale):

1 year ago
OpenStudy (mathmale):

This example not only defines the slope, but gives you an example of how to find it.

1 year ago
OpenStudy (anonymous):

damm giv me a site where i just plug the numbers

1 year ago
OpenStudy (mathmale):

Life's not like that. If you're not willing to make the slightest effort to look up and learn what you need to know, such as the slope of a straight line, then I'm not willing to continue investing time in you until that attitude changes. http://cs.selu.edu/~rbyrd/math/slope/

1 year ago
OpenStudy (anonymous):

bro im still lost

1 year ago
OpenStudy (mathmale):

Have you gone to the site whose URL I've given you, read the definition of "slope," and copied it down here on Open Study? If not, why are you so resistant to doing so?

1 year ago
OpenStudy (anonymous):

y=-4?

1 year ago
OpenStudy (anonymous):

x=2?

1 year ago
OpenStudy (mathmale):

The slope formula looks like:$m=slope=\frac{ y _{2}-y _{1} }{ x _{2}-x _{1} }$

1 year ago
OpenStudy (mathmale):

The two given points are .... what? type them out, enclosing the coordinates of each one in parentheses.

1 year ago
OpenStudy (anonymous):

still lost

1 year ago
OpenStudy (mathmale):

Tell my what you yourself have done so far. Where is the slope formula? Type in the two points given in this problem.

1 year ago


text:
# The Idea Behind Taylor Series

I understand that they are viewed as approximations, but was that Taylor's original hope? Assuming that a function can be written as a power series seems to me to be a wild assumption, without some kind of formal argument first being made. Was there such an argument?

-

In the best of worlds all functions should be polynomials.
Since this fails, one uses Taylor's formula to replace (some) functions by polynomials which reasonably approximate them.
Mathematicians who can't face these harsh realities become algebraic geometers and live in a dream world where indeed all functions are polynomials.

-
Why should all functions be polynomials? This doesn't help me, I don't see why one should think that, intuitively, sine should be presentable as polynomials. –  Anthony Sep 12 '13 at 0:07
@Tony: which functions could you compute exactly and easily without machines or tables except polynomials (and rational functions). How would you compute $\sin(7)$ or $\log (5)$ by hand? –  Georges Elencwajg Sep 12 '13 at 0:20
This makes sense then-But still, it doesn't seem to me that all functions should have some form in the polynomials. –  Anthony Sep 12 '13 at 6:45

One possible motivation comes from calculus.

A derivative is nothing more than a local approximation of a functions behaviour by a linear map.

In particular, we have:

$$f(x)=f(a)+f'(x)(x-a)+o(|x-a|)$$

for differentiable functions from $\mathbb{R}^n$ to $\mathbb{R}$.

So a differentiable function is one that can be "well-approximated" by a linear polynomial in the above sense everywhere.

A natural generalisation is to expand our class of approximating functions (linear polynomials) to higher order polynomials in the hope that this looser view will strengthen the bounds we can put on our error term. As it happens this is exactly what happens if we have higher regularity, this is Taylor's theorem.

It is a miracle that for some functions, the Taylor series we construct can not only give us good local approximations by truncating to a finite number of terms, but converges EXACTLY to the original function $f$ in a small neighbourhood of $a$. These are the analytic functions, whose importance becomes especially apparent if one studies complex analysis.

-

Taylor himself argued for his "Taylor series" on the basis of Newton's forward-difference formula. It may be summarised as follows. An infinite polynomial A+Bx+Cx^2+Dx^3+... has infinite degrees of freedom. Therefore we expect to be able to construct an infinite polynomial passing through an infinite number of given points, just as a parabola of the form y=Ax^2+Bx+C can be constructed going through essentially any three points, but not any four, owing to its having three coefficients (similarly you can make y=Ax+B go through any two points etc.). Newton's forward-difference formula constructs such a polynomial, namely a polynomial which takes the same values as a given function at the x-values 0,b,2b,3b,.... Taylor's derivation of his series consists in letting b go to zero is this formula.

-

I think Taylor's formula is already used by Newton. Taylor was working as Newton's assistant or something during that period. But this is "unofficial" mathematical history I learned from a class without proper citations.

However if you take this as believable, then Taylor's formula looks rather natural as Newton proved the binomial theorem for arbitrally powers. So it is not difficult to assume he would consider how to attack the problem in general. Then the passage from $(x+y)^{n}$ to $f(x+y)$ becomes rather natural, as around that time people must already know the Taylor expansion of $\log[1+x],e^{x},\sin[x]$, etc. So all one need is to assume $f$ is already a polynomial of infinite degree and solve the problem in reversed order.

-


text:
• # question_answer A metallic wire of density d is lying horizontal on the surface of water. The maximum length of wire so that it may not sink will be A)  $\sqrt{\frac{2Tg}{\pi d}}$                    B)  $\sqrt{\frac{2\pi T}{dg}}$C)  $\sqrt{\frac{2T}{\pi dg}}$               D)  any length

$mg=2TL\Rightarrow \pi {{r}^{2}}Ldg=2TL\Rightarrow \pi {{r}^{2}}dg=2T.$ This relation is independent of L.


text:
# Gaussian distribution!

In my course textbook it is written that "approximately 68% of the measurements from a normally distributed set lie within +-1 standard deviation of the mean value".
What do they mean by standard deviation of the mean value? They give a definition for "the mean"(of a set of measurements(data)) right before talking about gaussian distribution.
Also when they say "68% of the measurements" do they use the word "measurement" as in the meaning of data(e.g. lenght of an object) or set of data(e.g. lenghts of objects)?

You can think of the standard deviation as "the average distance from the average."

Homework Helper
In my course textbook it is written that "approximately 68% of the measurements from a normally distributed set lie within +-1 standard deviation of the mean value".
What do they mean by standard deviation of the mean value? They give a definition for "the mean"(of a set of measurements(data)) right before talking about gaussian distribution.
Here they are talking in terms of distance: roughly 68% of the measurements are within a distance of one standard deviation of the mean. If you think of a sketch of a bell curve, then when you locate
the two values $\mu - \sigma$ and $\mu + \sigma$, you can say that roughly 68% of the values from that distribution is between those two values

Also when they say "68% of the measurements" do they use the word "measurement" as in the meaning of data(e.g. lenght of an object) or set of data(e.g. lenghts of objects)?


text:
# Underbracing on a set of points in a plot with pgfplots

I have two sets of points graphed in the same plot. I want to mark them grouped with an under- or overbrazing with a large curly bracket, as I have illustrated by hand here:

I am looking for the same type of brazing as in mathematical formatting with $$\underbrace{a+b+c}_n$$:

\documentclass{article}

\usepackage{pgfplots}
\usepackage{xcolor}

\begin{document}

\begin{tikzpicture}
\pgfplotsset{height=7cm, width=8.5cm}

\begin{axis}[
axis lines = left,
hide x axis,
xmin=0, xmax=8,
ymin=600,ymax=1000
]
(1,700)
(2,700)
(3,700)
(4,700)
(5,700)
};
\end{axis}

\begin{axis}[
hide x axis,
hide y axis,
ymin=600,ymax=1000,
xmin=0, xmax=8
]
(6,900)
(7,900)
};
\node at (axis cs:1, 725) {1};
\node at (axis cs:2, 725) {2};
\node at (axis cs:3, 725) {3};
\node at (axis cs:4, 725) {4};
\node at (axis cs:5, 725) {5};
\node at (axis cs:6, 925) {6};
\node at (axis cs:7, 925) {7};

\end{axis}
\end{tikzpicture}

\end{document}

You can use decorations={brace} to create a brace with tikz between two points.

I defined the required nodes with A, B etc. to reuse the positions.

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{decorations.pathreplacing}

\usepackage{pgfplots}
\usepackage{xcolor}

\begin{document}

\begin{tikzpicture}
\pgfplotsset{height=7cm, width=8.5cm}

\begin{axis}[
axis lines = left,
hide x axis,
xmin=0, xmax=8,
ymin=600,ymax=1000
]
(1,700)
(2,700)
(3,700)
(4,700)
(5,700)
};
\end{axis}

\begin{axis}[
hide x axis,
hide y axis,
ymin=600,ymax=1000,
xmin=0, xmax=8
]
(6,900)
(7,900)
};
\node (A) at (axis cs:1, 725) {1};
\node at (axis cs:2, 725) {2};
\node at (axis cs:3, 725) {3};
\node at (axis cs:4, 725) {4};
\node (B) at (axis cs:5, 725) {5};
\node (C) at (axis cs:6, 925) {6};
\node (D) at (axis cs:7, 925) {7};

\draw [decorate,decoration={brace,amplitude=10pt,mirror,raise=6mm}]
(A.west) -- (B.east) node [black,midway, below, yshift=-10mm] {a};

\draw [decorate,decoration={brace,amplitude=5pt,mirror,raise=6mm}]
(C.west) -- (D.east) node [black,midway, below, yshift=-10mm] {b};

\end{axis}
\end{tikzpicture}

\end{document}


text:
# Help manipulating this equation

I have this equation:

$$A(f)=\frac{1}{\sqrt{\left(1-\left(\frac{f}{F}\right)^2\right)^2+c^2\left(\frac{f}{F}\right)^2}}$$

and I am told this:

Note that for a fixed amplitude $A$, the equation relating amplitude and $f$ can be converted to a 4th degree polynomial in $f$ by squaring both sides (to eliminate the square root), taking the reciprocal of both sides, and then subtracting $A$. Do this algebraic manipulation "by hand."

I got it down to a quadratic:

$$f^2 = \frac{F^4-F^4y^2}{y^2(F^2C^2-2F^2+1)}$$

Please help :)

-
Joe, I hope you don't mind, I rolled back and then incorporated your edits into my edit. –  Zev Chonoles Mar 6 '12 at 22:47
@ZevChonoles: Don't mind at all. –  Joe Johnson 126 Mar 6 '12 at 22:48
Somehow, when you "got it down to a quadratic," the symbol $A$ disappeared altogether. How'd you do that? –  Gerry Myerson Mar 6 '12 at 22:49

## 1 Answer

$$A=\frac{1}{\sqrt{\left(1-\left(\frac{f}{F}\right)^2\right)^2+c^2\left(\frac{f}{F}\right)^2}}$$ $$A^2 = \frac{1}{(1-(\frac{f}{F})^2)^2+c^2(\frac{f}{F})^2}$$ $$\frac{1}{A^2} = (1-(\frac{f}{F})^2)^2+c^2(\frac{f}{F})^2$$ $$\frac{1}{A^2} = 1+\frac{f^4}{F^4} - \frac{2f^2}{F^2} + \frac{c^2f^2}{F^2}$$ $$\frac{1}{A^2} = \frac{1}{F^4} f^4 + \left( \frac{c^2-2}{F^2}\right)f^2+1$$

where the last line gives a 4 degree polynomial in $f$. (Not sure where to subtract $A$!)

-
Surely the "subtract $A$" was a mistake. –  Gerry Myerson Mar 6 '12 at 23:10
I assume so, but perhaps I am missing some clever algebraic trick? –  Aru Ray Mar 7 '12 at 3:30
I don't think you're missing anything. The problem statement is very sloppy, asking to convert an equation to a polynomial - strictly speaking, that makes no sense. It's not surprising that someone who could be that sloppy could also be sloppy about "subtract $A$". –  Gerry Myerson Mar 7 '12 at 6:23


text:
# If the impedance of each phase of a balanced star-connected load is (3 + 2j) Ω, what will be the impedance of the equivalent delta-connected load?

This question was previously asked in
DSSSB JE EE 2019 Official Paper (Held on 25 Oct 2019)
View all DSSSB JE Papers >
1. (3 + j2) Ω
2. (9 + j6) Ω
3. (3 + j6) Ω
4. (2 + j3) Ω

Option 2 : (9 + j6) Ω

## Detailed Solution

Concept:

1) Delta to Star conversion:

$${R_A} = \dfrac{{{R_{AB}} \times {R_{AC}}}}{{{R_{AB}} + {R_{AC}} + {R_{BC}}}}$$

$${R_B} = \dfrac{{{R_{AB}} \times {R_{BC}}}}{{{R_{AB}} + {R_{AC}} + {R_{BC}}}}$$

$${R_C} = \dfrac{{{R_{BC}} \times {R_{AC}}}}{{{R_{AB}} + {R_{AC}} + {R_{BC}}}}$$

2) Star to Delta conversion:

$${R_{AB}} = \dfrac{{({R_A} \times {R_C}) + ({R_A} \times {R_B})+({R_B} \times {R_C})}}{{{R_C}}}$$

$${R_{AC}} = \dfrac{{({R_A} \times {R_C}) + ({R_A} \times {R_B})+({R_B} \times {R_C})}}{{{R_B}}}$$

$${R_{BC}} = \dfrac{{({R_A} \times {R_C}) + ({R_A} \times {R_B})+({R_B} \times {R_C})}}{{{R_A}}}$$

This concept is valid not only for Resistances but also for any

impedance.

If all the branch Impedances of the star network are equal to Z,

then each branch in its equivalent delta network is equal to 3 Z.

Application:

Given: Zan = Zbn = Zcn = (3 + j 2) Ω

According to the question, by converting the given Star to Delta, we get

$$Z_{AC}=Z_{AB}=Z_{BC}=3Z_{an}$$

= 3 (3 + j 2) Ω

= (9 + j 6) Ω