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intersect where z= 4. donde Si es un punto en el interior de uno de estos bloques, entonces el volumen del bloque puede ser aproximado por. Triple Integrals in Rectangular Form Note. pdf - Google Drive Sign in. In this section we convert triple integrals in rectangular coordinates into a triple integral in either cylindrical or spherical coordinates. We illustrate with some examples. TRIPLE INTEGRALS EXTRA CREDIT FOR MATH 222-01/02 DUE NOVEMBER 16, 2011 The answers to the problems below should be presented neatly, (either typed or written very neatly). We can compute R fdA on a region R in the following way. Because if your integration order takes care of Z first, i. The first variable given corresponds to the outermost integral and is done last. with respect to each spatial variable). Math symbols defined by LaTeX package «esint» No. 7 Triple Integrals in Cylindrical and Spherical Coordinates Example: Find the second moment of inertia of a circular cylinder of radius a about its axis of symmetry. Aplicaciones de de la integral Volumen de sólidos de revolución Definición Sea una función definida en el intervalo. Notice how the inequalities involve xand y. Triple integral is defined and explained through solved examples. above z = −√4x2 +4y2. Triple integral of infinitesimal volume = total volume of 3d region. I Project your region E onto one of the xy-, yz-, or xz-planes, and. Cylindrical and Spherical Coordinates General substitution for triple integrals. Z 1 1 1 + e x x dx Solution: (a) Improper because it is an in nite integral (called a Type I). The copyright holder makes no representation about the accuracy, correctness, or. wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. Let x i = x i x i1 be the width of the i’th subinterval [x i1,x i] and let the norm of the partitionkPkbethelargestofthex i’s. For a triple integral in a region D ˆR3, it can be evaluated by using an iterated. The simplest | {
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i’s. For a triple integral in a region D ˆR3, it can be evaluated by using an iterated. The simplest application allows us to compute volumes in an alternate way. Look for a variable that has. Sometimes we can reduce a very difficult double integral to a simple one via a substitution. P 1 f(P1)=14kg=m3 f(P2)=7kg=m3 P 2 P 3 f(P3)=9kg=m3 P 4 f(P4)=21kg=m3 Suppose f 2C(R3) measures density (kg=m3) throughout W. If f is continuous, the triple integral exists and does not depend on the choice of (x∗ ijk,y ∗ ijk,y ∗ ijk) Same properties as double integrals Mth 254H – Fall 2006 3/10 Evaluating Fubini’s Theorem: If f(x,y,z) is continuous on B =[a,b]×[c,d]×[r,s], then ZZZ R f(x,y,z)dV = Z s r Z. This description is too narrow: it's like saying multiplication exists to find the area of rectangles. It can also evaluate integrals that involve exponential, logarithmic, trigonometric, and inverse trigonometric functions, so long as the. Sketch the solid whose volume is given by the iterated integral. 3 Find the volume of the region bounded by z = 50 10y, z = 10, y = 0, and y = 4 x2. Triple integral of infinitesimal volume = total volume of 3d region. 1 2x 2y 2= 1 2(x + y) = 1 r 2. Fubini's theorem for triple integrals states that the value of a triple integral of a continuous function f over a region E in R 3 is a triple iterated integral. Definición de integral triple Una integral triple es una generalización de una integral doble en el mismo sentido que una doble es una generalización de una integral sencilla. Express the big integral like that, and evaluate each single integral separately. It will be mostly about adding an incremental process to arrive at a \total". above z = −√4x2 +4y2. The assignment is due at the beginning of class, August 6th. We'll learn that integration and di erentiation are inverse operations of each other. The triple integral of a continuous function over a general solid region. Z 1 1 1 + e x x dx Solution: (a) Improper because it is an in nite integral | {
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a general solid region. Z 1 1 1 + e x x dx Solution: (a) Improper because it is an in nite integral (called a Type I). 4 EXERCISES Review Questions 1. Read Section 16. I've uploaded my excel vba file. History of the Integral from the 17 th Century. 34 videos Play all MULTIPLE INTEGRALS (Complete Playlist) MKS TUTORIALS by Manoj Sir Triple Integrals, Changing the Order of Integration, Part 1 of 3 - Duration: 12:52. Integrals of a function of two variables over a region in R 2 are called double integrals, and integrals of a function of three variables over a region of R 3 are called triple integrals. Challenge: 11,23 4. S8: Double integrals in polar co–ordinates. Double integrals in polar coordinates. Just as the definite integral of a positive function of one variable represents the area of the region between the. MULTIPLE INTEGRALS 154 15. \mathbf {F} = – Gm\,\mathbf {\text {grad}}\,u, where G is the gravitational constant. stackexchange [22], and in a slightly less elegant form it appeared much earlier in [18]. x y z x + y = 12 2 z = 1 - x - y 2 2 141. Engineering Mathematics 1 Notes Pdf – EM 1 Notes Pdf UNIT – V. Read Section 16. Triple Integrals in Spherical Coordinates If you are studying an object with spherical symmetry, it makes sense to use coordinates to re ect that. 1 Multiple-Integral Notation Previously ordinary integrals of the form Z J f(x)dx = Z b a f(x)dx (5. When we have to integrate a function of x,y,z over all space, we write a triple integral in this way: ∫ − ∞ + ∞ ∫ − ∞ + ∞ ∫ − ∞ + ∞ (,,). Approximating Integrals In each of these cases, the area approximation got better as the width of the intervals decreased. 2 Assignments 1. It's difficult to explain. Our first integral could equally well be ff(x, y)dx. Contents 1. Then using expan-sions of K and K he had obtained himself 30 years previously—Watson [6]—he was able to evaluate these last integrals. Multiple Integrals Double Integrals over Rectangles 26 min 3 Examples Double Integrals over | {
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Multiple Integrals Double Integrals over Rectangles 26 min 3 Examples Double Integrals over Rectangles as it relates to Riemann Sums from Calc 1 Overview of how to approximate the volume Analytically and Geometrically using Riemann Sums Example of approximating volume over a square region using lower left sample points Example of approximating volume over a…. Sam Johnson (NIT Karnataka) Triple Integrals in Rectangular Coordinates October 24, 2019 17/62. (a) » 2 0 » 1 0 » 3 0 p xy z2q dzdydx (b) » 2 0 » z2 0 » y z 0 p 2x yq dxdydz (c) » ˇ{ 2 0 » y 0 » x 0 cosp x y zq dzdxdy 2. However each student is responsible. Fill in the limits of integration for the integral and put the dx, dy, dz in the correct order:. Therefore, the desired function is f(x)=1 4. Triple Integrals, Changing the Order of Integration, Part 1 of 3. All these methods are Numerical. The cylindrical coordinate system describes a point (x,y,z) in rectangular space in terms of the triple (r,θ,z) where r and θ are the polar coordinates of the projection. Welcome! This is one of over 2,200 courses on OCW. Both of the limits diverge, so the integral diverges. The analogy between single and. We used a double integral to integrate over a two-dimensional region and so it shouldn't be too surprising that we'll use a triple integral to integrate over a three dimensional region. Although the prerequisite for this. Triple Integrals: A Hemisphere Example Let R be the region of three dimensional space bounded by z ≥0 and the surface of a sphere of radius a with a center at the origin. In addition, some examples are used to demonstrate the calculations. INTEGRAL REVIEW July 2012 Vol. The corresponding multiple integrals are referred to as double integrals, triple integrals, and n-tuple integrals, respectively. and above the region in the xy. INTEGRALS 289 Thus, {F + C, C ∈ R} denotes a family of anti derivatives of f. Look for a variable that has. iosrjournals. Let Ube the solid inside both the cone z= p. Integrals | {
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Look for a variable that has. iosrjournals. Let Ube the solid inside both the cone z= p. Integrals are often described as finding the area under a curve. 3 0 2ˇ 0 2 1 (r+ z)rdrd dz Region from Diagram 2 3 0 2ˇ 0 2 0 5zrdrd dz Region from Diagram 2 3 0 ˇ=2 0 2 1. Convert integrals into another form For a double integral in a region D ˆR2, it can be evaluated by using an iterated integral in dxdy, or dydx, or using polar coordinate. If you want to calculate the area under the curve or some definite integral in the Symbolic (Analytical) way, then it is very hard to using C++ and not very useful. We'll learn that integration and di erentiation are inverse operations of each other. Z 1 1 1 + e x x dx Solution: (a) Improper because it is an in nite integral (called a Type I). Solution: ZZ D (x +y)dA = Z1 0 Z2y −y (x+y)dxdy = Z1 0 (x2 2 +xy) x=2y x=−y = Z1 0 9y2 2 dy = 3y3 2 y=1 y=0 = 3 2. The multiple integral is a type of definite integral extended to functions of more than one real variable—for example, $f(x, y)$ or $f(x, y, z)$. Imagine you have a cube that's gets denser as you move further out towards its corners. Triple Integrals over a General Bounded Region. Integral calculus that we are beginning to learn now is called integral calculus. Una integral triple es una generalización de una integral doble en el mismo sentido que una doble es una generalización de una integral sencilla. Elementary Approach to the Dirichlet Integral 1 2. Dirichlet integral, is often evaluated using complex-analytic methods, e. Integration Method Description 'auto' For most cases, integral3 uses the 'tiled' method. (We just add a third dimension. Free triple integrals calculator - solve triple integrals step-by-step. Evaluate the triple integral. This article is about the Euler–Poisson integral. Integrales Triples Hermes Pantoja Carhuavilca3 de 30. Double Integrals Definition (1. The line integral of a magnetic field around a closed path C is equal to the total current flowing through the | {
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of a magnetic field around a closed path C is equal to the total current flowing through the area bounded by the contour C (Figure 2). patrickJMT 357,008 views. Homework 6 Solutions Jarrod Pickens 1. This sum has a nice interpretation. Engineering Mathematics III: UNIT I: Linear systems of equations: Rank-Echelon form-Normal form - Solution of linear systems - Gauss elimination - Gauss Jordon- Gauss Jacobi and Gauss Seidel methods. 8 The Fubini’s Method to evaluate triple integrals is this: We first walk on a line parallel to one of the x-axis, y-axis, or z-axis. Triple Integral Practice To Set Up A Triple Integral 1. Example Use cylindrical coordinates to find the volume of a curved wedge cut out from a cylinder (x − 2)2 + y2 = 4 by the planes z = 0 and. 7 We integrate the triple integral directly. Triple Integrals, Changing the Order of Integration, Part 1 of 3. and inside x2 +y2 = 4. Single Integral - the domain is the integral I (a line). set up the triple integral in terms of single integrals, but do not evaluate it). Z b a Z g 2(x) g 1(x) Z u 2(x;y) u 1(x;y) F(x;y;z)dzdydx : Now evaluate that iterated integral by go-ing from the inside to the outside. Example Use cylindrical coordinates to find the volume of a curved wedge cut out from a cylinder (x − 2)2 + y2 = 4 by the planes z = 0 and. Usually, one direction is considered to be positive, the other negative. To approximate a volume in three dimensions, we can divide the three-dimensional region into small rectangular boxes, each \(\Delta x\times\Delta y\times\Delta z. Problems: 5,7,9,13,17,33 3. Step 1: Draw a picture of E and project E onto a coordinate plane. Triple integral is an integral that only integrals a function which is bounded by 3D region with respect to infinitesimal volume. Triple Integrals in Cylindrical Coordinates A point in space can be located by using polar coordinates r,θ in the xy-plane and z in the vertical direction. Use the Comparison Theorem to decide if the following integrals are | {
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and z in the vertical direction. Use the Comparison Theorem to decide if the following integrals are convergent or divergent. Triple integrals Triple integral examples 3c. Try our award-winning software today. To begin with, suppose that ˚(x;y;z) is a piecewise continuous function. Partial Di erentiation and Multiple Integrals 6 lectures, 1MA Series Dr D W Murray Michaelmas 1994 Textbooks Most mathematics for engineering books cover the material in these lectures. The two integrals that have dy as the innermost di erential are Z2 0 Zx 0 x 0 ex(y + 2z) dydzdx + 2 0 Z2x x x z x ex(y + 2z. The simplest application allows us to compute volumes in an alternate way. A solid region Eis said to be of type 1 if it lies between. Evaluating triple integrals A triple integral is an integral of the form Z b a Z q(x) p(x) Z s(x,y) r(x,y) f(x,y,z) dzdydx The evaluation can be split into an "inner integral" (the integral with respect to z between limits. MIT OpenCourseWare is a free & open publication of material from thousands of MIT courses, covering the entire MIT curriculum. (a)Set up but do not evaluate a single triple integral to nd the volume of Susing cylindrical coordinates. Math 21a Triple Integrals Fall, 2010 1 Evaluate the integral RRR E 2xdV, where E= {(x,y,z) : 0 ≤y≤2,0 ≤x≤ p 4 −y2,0 ≤z≤y}. b) Set up a triple integral over S in the dx dy dz ordering. The triple integral of a continuous function over a general solid region. We'll learn that integration and di erentiation are inverse operations of each other. Abraham de Moivre originally discovered this type of integral in 1733, while Gauss published the precise integral in 1809. This depends on finding a vector field whose divergence is equal to the given function. Accordingly, its volume is the product of its three sides, namely dV dx dy= ⋅ ⋅dz. Triple Integral Spherical Coordinates Pdf Download >>> DOWNLOAD (Mirror #1). Numerical Integration and Differentiation Quadratures, double and triple integrals, and | {
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(Mirror #1). Numerical Integration and Differentiation Quadratures, double and triple integrals, and multidimensional derivatives Numerical integration functions can approximate the value of an integral whether or not the functional expression is known:. 1 DOUBLE INTEGRALS OVER RECTANGLES TRANSPARENCIES AVAILABLE #48 (Figures 4 and 5), #49 (Figures 7 and 8), #50 (Figure 11), #51 (Figures 12 and 13) SUGGESTED TIME AND EMPHASIS 1 2 –1 class Essential Material POINTS TO STRESS 1. V = \iiint\limits_U {\rho d\rho d\varphi dz}. The cylindrical coordinate system describes a point (x,y,z) in rectangular space in terms of the triple (r,θ,z) where r and θ are the polar coordinates of the projection. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 5 3. Challenge: 11,23 4. Applications of Double/Triple Integrals. Such integrals arise whenever two functions are multiplied, with both the operands and the result represented in the Legendre polynomial basis. dimensional integrals. dimensional domain. Just as for double integrals, a region over which a triple integral is being taken may have easier representation in another coordinate system, say in uvw-space, than in xyz-space. Try to visualize the 3D shape if you can. P 1 f(P1)=14kg=m3 f(P2)=7kg=m3 P 2 P 3 f(P3)=9kg=m3 P 4 f(P4)=21kg=m3 Suppose f 2C(R3) measures density (kg=m3) throughout W. To obtain double/triple/multiple integrals and cyclic integrals you must use amsmath and esint (for cyclic integrals) packages. 5 36 Triple Integral Strategies The hard part is guring out the bounds of your integrals. Let D be the half-washer 1 x2 + y2 9, y 0, and let E be the solid region above D and below the graph z = 10 x2 y2. Integrals 6 1. where d is the radius of rotation. The divergence theorem can also be used to evaluate triple integrals by turning them into surface integrals. (5), if ,, then the triple. James McKernan, Maths, 18. Triple Integrals | x12. I Project your region E onto one of the xy-, yz-, or xz-planes, and | {
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18. Triple Integrals | x12. I Project your region E onto one of the xy-, yz-, or xz-planes, and use the boundary of this projection to nd bounds on domain D. We can interpret this result as the volume of the solid region because the integrand is 1. Aplicaciones de de la integral Volumen de sólidos de revolución Definición Sea una función definida en el intervalo. Triple integration of sum of two functions is explained. -plane defined by 0 ≤ x ≤ 2. This article considers two types of triple integrals and uses Maple for verification. z = 8 − x 2 − y 2. an integral of a function defined on some region in a plane and in three-dimensional or n-dimensional space. Notice how the inequalities involve xand y. Solution: We'll use the shadow method to set up the bounds on the integral. By using this website, you agree to our Cookie Policy. stackexchange [22], and in a slightly less elegant form it appeared much earlier in [18]. (1) is deflned as Z C a ¢ dr = lim N!1 XN p=1 a(xp;yp;zp) ¢ rpwhere it is assumed that all j¢rpj ! 0. Use a triple integral to determine the volume of the region below z = 4−xy. So then x2 +y2 = r2. Multiple integrals use a variant of the standard iterator notation. Challenge Problems. 25 3 4 3 12 4 tt t t dt 1. via contour integration. (Or vice versa. EXAMPLE 4 Find a vector field whose divergence is the given F function. MTH 254 LESSON 20. Triple Integral Calculator Added Mar 27, 2011 by scottynumbers in Mathematics Computes value of a triple integral and allows for changes in order of integration. The triple integral in this case is, Note that we integrated with respect to x first, then y, and finally z here, but in fact there is no reason to the integrals in this order. So let us give here a brief introduction on integrals based on the Mathematics subject to find areas under simple curves, areas bounded by a curve and a line and area between two. In Eastern Europe, it is known as Ostrogradsky’s. Notation: double integral of f over R= I f x y dxdy( , | {
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Europe, it is known as Ostrogradsky’s. Notation: double integral of f over R= I f x y dxdy( , ) ³³ Note: Area element = dA = dx dy. 2, we showed how a function of two variables can be inte-grated over a region in 2-space and how integration over a region is equivalent to an iterated or double integral over two intervals. If we substitute back into the sum we get nX−1 i=0 G(yi)∆y. Recall that if you're looking for th volume then f. Hence h 1(y;z) = z 2 and h 2(y;z) = 4 4y2 = 2 p 1 y2. We would like to be able to integrate triple integrals for more general regions. pdf - Free download as PDF File (. 1 De nition of double integral Consider the function of two variables f(x,y) defined in the bounded region D. Hence the triple integral becomes Z Z Z D dV = Z 4 0 Z 1 1 Z 2 p 1 y2 z 2 1 dxdydz = Z 4. 6 to find their infinite series forms. fun(x,y,z) must accept a vector x and scalars y and z, and return a vector of values of the integrand. A solid region Eis said to be of type 1 if it lies between. Find the θ-limits of integration. However each student is responsible. We'll learn that integration and di erentiation are inverse operations of each other. Note, that integral expression may seems a little different in inline and display math mode - in inline mode the integral symbol and the limits are compressed. 5-8: Surface Area, Triple Integrals Friday, April 8 Surface Area Using the formula A(S) = ZZ D q 1 + f2 x + f y 2 dA, nd the surface area of a sphere of radius a. In this video, I start discussing how a particular order of integration for a given region and integral ' makes sense '! Then I go. V = ∭ U ρ 2 sin θ d ρ d φ d θ. Don't show me this again. By symmetry, ¯x = 0 and ¯y = 0, so we only need ¯z. Integrales Triples Hermes Pantoja Carhuavilca3 de 30. Triple integrals also arise in computation of Volume (if f(x,y,z)=1, then the triple integral equals the volume of R) Force on a 3D object Average of a Function over a 3D region Center of Mass and Moment of Inertia | {
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R) Force on a 3D object Average of a Function over a 3D region Center of Mass and Moment of Inertia Triple Integrals in General Regions. Triple integrals are the analog of double integrals for three dimensions. Triple Integrals in Cylindrical/Spherical Coordinates Multi-Variable Calculus. Find materials for this course in the pages linked along the left. This website uses cookies to ensure you get the best experience. The meaning of integration. use the trapezoidal rule of integration to solve problems, 3. (iv) Evaluate. » Integrate can evaluate integrals of rational functions. In physics, triple integral arises in the computation of mass, volume, moment of inertia and force on a three dimensional object. If we want to convert this triple integral to cylindrical coordinates we need to rewrite xand yusing the conversion formulas from above. Triple Integrals Motivation Example An object conforms to the shape of a solid W in R3. Imagine you have a cube that's gets denser as you move further out towards its corners. (b) Let's guess that this integral is divergent. The solid below is enclosed by x= 0, x= 1, y= 0, z= 0, z= 1, and 2x+y+2z= 6. (Q9)Set up the triple integral W dV = W 1dV , using the order of integration dV = dz dy dx. 7 Triple Integrals in Cylindrical and Spherical Coordinates Example: Find the second moment of inertia of a circular cylinder of radius a about its axis of symmetry. 6 to find their infinite series forms. Compute the following integral by making a change in coordinates. Step 2: Determine the limits of integration. It is also useful in setting up triple integrals as iterated integrals to let Rbe the. Accordingly, its volume is the product of its three sides, namely dV dx dy= ⋅ ⋅dz. patrickJMT 357,008 views. The sphere x2 +y2 +z2 = 4 is the same as ˆ= 2. (Unfortunately, it's harder to draw in three dimensions. Solution: a) Sketch an arrow in the positive y direction: This arrow enters the solid at the xz-plane ( 1=0), passes through the interior | {
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y direction: This arrow enters the solid at the xz-plane ( 1=0), passes through the interior (gray), and. 4 EXERCISES Review Questions 1. The Evaluation Theorem 11 1. This week’s review talks about Triple Integrals and Applications. Double Integrals Definition (1. | {
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## anonymous one year ago The question 1B-6?
1. phi
can you post the question?
2. phi
is it what is the locus of points for the equation $\vec{OP} \cdot \textbf{u}= c | \vec{OP}|$ ?
3. anonymous
yes, I can't understand: "Then the locus is the nappe of a right circular cone with axis in the direction u and vertex angle 2θ" How can I reach this conclusion
4. anonymous
1B-6 Let O be the origin, c a given number, and u a given direction (i.e., a unit vector). Describe geometrically the locus of all points P in space that satisfy the vector equation OP · u = c | OP | . In particular, tell for what value(s) of c the locus will be (Hint: divide through by | OP | ): a) a plane b) a ray (i.e., a half-line) c) empty
5. phi
if we divide both sides by |OP| we get $\frac{OP}{|OP|} \cdot u = c$ that is two unit vectors. the maximum value of v dot u is 1 when v and u are unit vectors
6. phi
now consider c=0 any vector OP perpendicular to u , when dotted with u will give 0 |dw:1439222905180:dw|
7. phi
any point P in the plane defined by u (as the normal) will form vector OP, and after dividing by |OP| will still be a vector (of unit length) in that plane. and dotted with u, will give 0 thus c=0 results in a plane
8. phi
if |c| = 1 (c=1 or c=-1) the locus will be a ray if c=1 then any point P where O to P is in the same direction as u will, after being divided by the length of OP will give us unit vector v = u and u dot v = 1
9. anonymous
If we applied the equation of Dot Product: $\left| A \right| \left| B \right| \cos \theta = A.B = \sum_{i }^{...} a _{i}.b_{i}$ W ell, u is a unit vecton, then $\left| u \right| = 1$ then, $\frac{ OP.u }{ \left| OP \right|}=\cos \theta = c$ Off course, this is the angle between these two vector, The points P need stay
10. anonymous | {
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10. anonymous
points P are at $\theta$ of in relation to vectior u. But and when the corretion says: Then the locus is the nappe of a right circular cone with axis in the direction u and vertex angle 2θ 0 . I can understand.
11. phi
for 0 < |c| < 1 we get a cone
12. anonymous
Ok Thanks Phi, I will see it carefully, thanks by help me!
13. phi
what part of the answer is not clear?
14. phi
***"Then the locus is the nappe of a right circular cone with axis in the direction u and vertex angle 2θ" How can I reach this conclusion**** |dw:1439223719035:dw|
15. phi
let c= 0.5 and OP/|OP| = v v dot u = 0.5 theta = 60 degrees any point P so that OP is in the direction 60 degrees away from u (in any direction) will be scaled to unit length. and then dotted with u = 0.5
16. anonymous
Ok, Itś clear now! Thaks... Now I will remake this exercise.
17. anonymous
Unfortunately I can´t get the concept on the last part of this question, c) empty response: Locus is the origin, if c > 1 or c < −1 (division by | OP | is illegal, notice).
18. anonymous
it is just because cos or sen modules can't be larger than 1. It is | sen(x) | <=1 and |cos(x)|<=1 ?
19. phi
yes, I was confused on that for a bit. if we want $\frac{OP}{|OP|} \cdot u = 2$ for all P not at the origin, the vector OP divided by its length will give us a unit vector, call it v; and v dot u =2 will never be true. I originally thought this meant "no solution" but if P is the origin, then OP is the zero length vector and $\vec{OP} \cdot \textbf{u} = 2| OP|$ becomes $\vec{OP} \cdot \textbf{u} = 2\cdot 0= 0$ or $<0,0,0> \cdot <a,b,c> = 0$ where I am assuming we are in 3D space.
20. phi
and <0,0,0> dot u gives 0 therefore the zero vector $$\vec{\textbf{0}}$$ is a solution for |c|>1
21. phi
v dot u =2 to elaborate v dot u = |u| |v| cos A because |u| and |v| are both 1 (both are unit vectors) and the max value of cos A is 1 the max value for v dot u is 1 we can never get a value larger than 1
22. anonymous | {
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22. anonymous
Ok, Let's a simplest way to show "a vector divided by its length" is a unit vector.... I used matlab... >> OP = [-4 -45 1] OP = -4 -45 1 >>mag=sqrt(4^2+45^2+1^2) mag = 45.1885 >>OP_unitVector=OP./mag OP_unitVector = -0.0885 -0.9958 0.0221 mag_unitVector=sqrt(0.0885^2 + 0.9958^2 + 0.0221^2 ) mag_unitVector = 1.0000
23. phi
we can do it algebraically say we have v= <a,b,c> then $$|v|= \sqrt{a^2+b^2 + c^2}$$ (which follows from pythagoras) and $|v|^2 = a^2 + b^2 + c^2$ $\frac{v}{|v|} \cdot \frac{v}{|v|}= \frac{v \cdot v}{|v|^2}$ $v \cdot v= a^2 + b^2 + c^2$ and therefore $\frac{v \cdot v}{|v|^2} = \frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1$
24. phi
if we let $u = \frac{v}{|v|}$ we have shown $u \cdot u = |u|^2= 1 \\ \sqrt{|u|^2} = \sqrt{1}\\ |u|= 1$ and therefore $\Bigg|\frac{v}{|v|}\Bigg|= 1$ ie. is unit length | {
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# Prove the following inequality: $\int_{0}^{\frac{\pi }{2}}\frac{sin(x)}{\sqrt{9-sin^{4}(x)}}dx\geq \frac{1}{3}$
Prove the following inequality: $$\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{\sqrt{9-\sin^{4}x}}\ \mathrm dx\geq\frac{1}{3}.$$
I am thinking of replacing the equation with $$\int_{0}^{\frac{\pi }{2}}\frac{\sin x}{9-\sin^{4} x }dx\geq \frac{1}{3}$$, however I am stuck at this point.
Do you have any suggestions?
• use the fact $\sqrt{9 - \sin^4x} \leq 3$ and $\int_0^{\pi/2} sin x dx = 1$ and $\sin x$ is nonegative on $[0,\pi/2]$. Feb 5 '20 at 18:07
• You can get the proper font and spacing for $\sin$ using \sin. For operators that don't have a command of their own, you can use \operatorname{name}. Feb 5 '20 at 18:19
\begin{align*}\int_{0}^{\pi/2} \frac{\sin{x}}{\sqrt{9-\sin^4{x}}}dx &\geq \int_{0}^{\pi/2}\frac{\sin{x}}{\sqrt{9}}dx\\& = \frac{1}{3}\int_{0}^{\pi/2} \sin{x}dx \\&=\frac{1}{3}[-\cos{\pi/2}-(-\cos{0})]\\&=\frac{1}{3}[0+1]\\&=\frac{1}{3} \end{align*}
The inequality is because $$\sqrt{9-\sin^4{x}}\leq \sqrt{9}$$ due to positivity of $$\sin^4{x}$$. So ... $$\frac{1}{\sqrt{9-\sin^4{x}}}\geq \frac{1}{\sqrt{9}}$$
• It's also $< 1/(2\sqrt{2})$. Feb 5 '20 at 20:23
The given inequality can be deduced from AM-GM. Indeed$$\begin{eqnarray*} I=\int_{0}^{\pi/2}\frac{\sin x}{\sqrt{9-\sin^4 x}}\,dx &=& \int_{0}^{1}\frac{z\,dz}{\sqrt{(9-z^4)(1-z^2)}}\\ &=& \frac{1}{2}\int_{0}^{1}\frac{du}{\sqrt{(3-u)(3+u)(1-u)}}=\int_{0}^{1}\frac{dv}{\sqrt{(2+v^2)(4-v^2)}}\end{eqnarray*}$$ and over $$[0,1]$$ we have $$\sqrt{(2+v^2)(4-v^2)}=\text{GM}(2-v^2,4+v^2)\leq\text{AM}(2-v^2,4+v^2)=3$$.
This also shows that $$I$$ is an incomplete elliptic integral of the first kind. | {
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Idempotents in $\mathbb Z_n$
An element $$a$$ of the ring $$(P,+,\cdot)$$ is called idempotent if $$a^2=a$$. An idempotent $$a$$ is called nontrivial if $$a \neq 0$$ and $$a \neq 1$$.
My question concerns idempotents in rings $$\mathbb Z_n$$, with addition and multiplication modulo $$n$$, where $$n$$ is natural number. Obviously when $$n$$ is a prime number then there is no nontrivial idempotent. If $$n$$ is nonprime it may happen, for example $$n=4, n=9$$, that also there is no.
Is it known, in general, for what $$n$$ there are nontrivial idempotents and what is a form of such idempotents?
• They certainly do occur: $n=6,a=3.$ – Andrew Jul 27 '12 at 19:07
• There is a nontrivial idempotent in $\mathbb{Z}_n$ if and only if there is $1<a<n$ such that $n\, |\, a(a-1)$. This is because $a^2 \equiv a \pmod n \Leftrightarrow a^2-a \equiv 0 \pmod n$. I'm not entirely sure where you can go with this though! – Clive Newstead Jul 27 '12 at 19:14
• @CliveNewstead Since $a$ and $a-1$ are relatively prime, you get that such $a$ exists if and only if $n$ has two relatively prime nontrivial divisors, and this is equivalent to $n$ having two distinct primes... – N. S. Jul 27 '12 at 19:31
5 Answers
As often happens when dealing with $$\mathbf{Z}_n$$, the Chinese remainder theorem is your friend. If the prime factorization of $$n$$ is $$n=\prod_i p_i^{a_i},$$ then by CRT we have an isomorphism of rings $$\mathbf{Z}_n\cong\bigoplus_i \mathbf{Z}_{p_i^{a_i}}.$$ Observe that the isomorphism maps the residue class of an integer $$m$$ (modulo $$n$$) to a vector with all the components equal to the residue class of $$m$$ (this time modulo various prime powers): $$\overline{m}\mapsto(\overline{m},\overline{m},\ldots,\overline{m}).$$ So the residue class of $$m$$ is an idempotent if and only if it is an idempotent modulo all the prime powers $$p_i^{a_i}$$. | {
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Let us look at the case of a prime power modulus $$p^t$$. The congruence $$x^2\equiv x\pmod{p^t}$$ holds, iff $$p^t$$ divides $$x^2-x=x(x-1)$$. Here only one of the factors of, $$x$$ or $$(x-1)$$, can be divisible by $$p$$, so for the product to be divisible by $$p^t$$ the said factor then has to be divisible by $$p^t$$. Thus we can conclude that $$x\equiv 0,1 \pmod{p^t}$$ are the only idempotents modulo $$p^t$$. Therefore we require that $$m\equiv 0,1\pmod{p_i^{a_i}}$$ for all $$i$$. By CRT these congruences are independent for different $$i$$, so the number of pairwise non-congruent idempotents is equal to $$2^\ell$$, where $$\ell$$ is the number of distinct prime factors $$p_i$$ of $$n$$.
• Can you please explain why $m\equiv 0,1\pmod{p_i^{a_i}}$ ? I understood everything up till that point – Belgi Jul 28 '12 at 0:55
• @Belgi, the notation was somewhat inconsistent, and the logic a bit of patchwork. Is it clearer now? – Jyrki Lahtonen Jul 28 '12 at 4:48
• What is "m"? ... – Stephen Mar 12 '15 at 19:32
• @Stephen: An integer. More precisely, an integer such that its residue class is an idempotent as explained on the sixth line. – Jyrki Lahtonen Mar 12 '15 at 19:35
• Sorry about the bump. I was reviewing some of my old answers and felt that this could be made a lot clearer (particular in view of Stephen's comment). Hopefully I succeeded. – Jyrki Lahtonen Aug 24 '18 at 9:51
Hint Idempotents in $$\:\Bbb Z_{ n}\:$$ correspond to factorizations of $$\:n\:$$ into two coprime factors. Namely, if $$\:e^2 = e\in\Bbb Z_n\:$$ then $$\:n\:|\:e(e\!-\!1)\:$$ thus $$\:n = jk,\ j\:|\:e,\ k\:|\:e\!-\!1,\:$$ so $$\:(j,k)= 1\:$$ by $$\:(e,e\!-\!1) = 1.\:$$ Conversely if $$\:n = jk\:$$ for $$\:(j,k)= 1,\:$$ then by CRT, $$\:\Bbb Z_n\cong \Bbb Z_j\times \Bbb Z_k\:$$ which has nontrivial idempotents $$\:(0,1),\,(1,0).\:$$ It is not that difficult to explicitly work out the details of the correspondence. Some integer factorization algorithms search for such nontrivial idempotents. | {
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This correspondence between idempotents and factorizations holds more generally at the structural level - see the Peirce Decomposition.
• It's worth noting here that if $x$ is an idempotent corresponding to a given $j$ and $k$, then $(p+)1-x$ is the other one, so it suffices to compute the idempotent corresponding to either $(0,1)$ or $(1,0)$. – rogerl May 29 '15 at 15:17
• @rogerl Yes that is clear since $\,(0,1)+(1,0) = (1,1) = 1\ \$ – Bill Dubuque Jun 14 '15 at 20:54
By the Chinese remainder theorem $\mathbf Z/n\mathbf Z$ is a product of more than one (unital) ring if and only if $n$ has more than one prime factor, and in this case $\mathbf Z/n\mathbf Z$ certainly has nontrivial idempotents. If on the other hand $n=p^k$ is a prime power, then all elements are either invertible (if not divisible by $p$) or nilpotent (if divisible by $p$) and this excludes the possibility of nontrivial idempotents. The case $n=1$ is a special case of a prime power (but the unique element now is both invertible and nilpotent; there are still no nontrivial idempotents of course).
A start: You can figure it out! Let us start with a product $mn$ of relatively prime integers, neither equal to $1$. By the Chinese Remainder Theorem, there is an $x$ such that $x\equiv 0\pmod{m}$ and $x\equiv 1 \pmod{n}$. Then $x^2\equiv x \pmod{mn}$.
Generalize to a product of $k$ relatively prime integers. If you use the prime power factorization, you can get a complete list.
Let $m=p^{c_{1}}_{1}...p^{c_{n}}_{n}$ be a prime factorization of an integer $m$ with $c_{i}\geq1$ and $p_{i}$ are distinct prime numbers. Then the ring $\mathbb{Z}/m\mathbb{Z}$ has $2^{n}$ idempotents and each one is precisely of the form $\sum\limits_{k=1}^{n}h_{k}\epsilon_{k}+m\mathbb{Z}$ where $\epsilon_{k}\in\{0,1\}$ and $h_{k}\in(\prod\limits_{\substack{i=1,\\ i\neq k}}^{n}p^{c_{i}}_{i})\mathbb{Z}$ such that $h_{k}-1\in p^{c_{k}}_{k}\mathbb{Z}$. | {
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# I Affine Subspace
1. Jun 14, 2017
### Erik109
Hey,
I am struggling with developing an intuition behind 'Affine Subspaces'. So far I have read the theories concerning Affine Subspaces delivered by the course book and visited several websites, however none have made it 100% clear. I feel like I have some sort of intuition for it, but I fail to apply the intuition when it comes to solving problems.
At the moment, I am often required to show why a given set is an Affine Subspace of a certain space. Because I assume it is quite hard for you to convey the intuition behind it without writing a lot, I will try to convey the way I think by approaching a sample problem. I hope you can enlighten me on the errors/misconceptions I make or perhaps add something so I can learn more.
The sample problem:
Let X1 = {f : R → R : f(0) = 1 and f is continuous}. Prove that X1 is an affine subspace of C(R, R) (the space of all continuous function with domain R and mapping to R). Hint: You have to do something with the set X0 = {γ : R → R : f(0) = 0 and f is continuous}.
So, what I think they want me to do at first is two things:
- prove that the set X0 is a subspace of C(R, R) (Closed by addition/multiplication)
- define an arbitrary element β which is an element of C(R, R) but not an element of X0
Once I have done both, I feel like I have to define an arbitrary element of X1, for example, the function ƒ, and show that this function contains both an element of X0 (let's say γ) and the arbitrary element β. In other words:
ƒ(x) = γ(x) + β
If this holds then the set X1 is an Affine Subspace of C(R, R) because it contains the sum of a subspace of C(R, R) and an element of C(R, R)?
I feel like I don't fully understand how to construct this function ƒ(x) and which arbitrary β to define. I have scanned through the answer numerous times and I kinda get their reasoning why, but I can't develop my own reasoning. There must be a logical, stepwise approach to the problem.
Erik | {
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Erik
2. Jun 14, 2017
### WWGD
As I understand it, an affine ( and a fine , haha) subspace is just a standard subspace followed by a translation. The subspace just does not go through the origin and there are issues with closure under operations, as in f(0)+g(0)=2 here. EDIT: Like you said, $X0$ is a subspace ( contains the origin) but $X1$ does not.
3. Jun 14, 2017
### Staff: Mentor
I'm not sure what the arbitrariness of $\beta$ has to do here. In general, an affine subspace is simply a linear subspace that got shifted from the origin $\vec{0}$ by a certain, fixed vector $\vec{f}_0\,$.
So you are right, the set $\{f \in C(\mathbb{R},\mathbb{R})\,\vert \,f(0)=0\}$ is the candidate for this subspace and you have to show that it is actually a subspace, i.e. closed under addition and scalar multiplication. (There are other multiplications on $C(\mathbb{R},\mathbb{R})$ so it's better to name them correctly for not to get confused later on.) You also have to make sure that $\vec{0}$ is in this set. This guarantees that this set isn't empty.
In the next step, you have to find this vector $\vec{f}_0\,$, which isn't an arbitrary $\beta$. But you are right as there is more than one possibility in this case. One $\beta = \vec{f}_0\,$ however, will do. What could be a candidate? Chose an easy one, this makes life easier if you want to do anything with this affine space.
4. Jun 14, 2017
### mathwonk
try translating by the constant function 1 or by the function cosine, or by any function at all with value 1 at 0.
5. Jun 14, 2017
### Erik109
Thanks a lot for the fast replies! They are all truly insightful and I surely understand the concept now.
@fresh_42 Special thanks for the extensive explanation. | {
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The only thing is that I struggle a bit with proving a given affine subspace of $C(ℝ, ℝ)$ is indeed an affine subspace. I feel like I have to define 'arbitrary' elements to show that $X_1$ is an affine subspace in general. The answer I was provided started off with:
Fix an arbitrary $f ∈ X_1$. Let $γ: ℝ → ℝ$ be defined by $γ(x) = f(x) + h(x)$, where $h : ℝ → ℝ$ (the vector shifting from the origin, you mentioned) is given by $h ≡ −1$ Note that $γ$ is a continuous function with the property $γ(0) = 0$, i.e. $γ ∈ X_0$.
So, if I understand it correctly, the way they prove it is by showing that an arbitrary element from $X_1$ can be shifted back to $X_0$? So kind of 'deconstructing' an affine subspace to a linear subspace?
PS: Got the hang of the LaTeX structure on the forums, hope this helps with the layout.
6. Jun 14, 2017
### WWGD
Yes, essentially. You can " de-translate" the affine space into a vector space by finding the right amount to translate by.
7. Jun 14, 2017
### Staff: Mentor
Every affine subspace $A$ can be written $A=V +\vec{h}$ with a vector (linear) subspace $V$ and the shifting vector $\vec{h}$.
You can observe two things here:
• $\vec{h}$ is not unique. Every other vector $\vec{h}+\vec{v}$ with $\vec{v}\in V$ does the job as well.
• The difference between two elements $\vec{a}_1 = \vec{v}_1 + \vec{h} \, , \,\vec{a}_2 = \vec{v}_2 + \vec{h}$ in $A$ is $\vec{a}_1 -\vec{a}_2 = (\vec{v}_1 + \vec{h})- (\vec{v}_2 + \vec{h})= \vec{v}_1 - \vec{v}_2$ and always an element of $V$.
Now in your case, $V= \{f_V \in C(\mathbb{R},\mathbb{R})\,\vert \,f_V(0)=0\}$ and $A= \{f_A \in C(\mathbb{R},\mathbb{R})\,\vert \,f_A(0)=1\}$, so you need a function $h=\vec{h} \in C(\mathbb{R},\mathbb{R})$ that moves (shifts) all functions in such a way, that $1=\vec{f}_A(0) = \vec{f}_V(0) +\vec{h}(0) = 0 + h(0)$. @mathwonk mentioned a few candidates for $h$ in post #4. As said above, $\vec{h}$ isn't unique, only up to elements of $V$. | {
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Affine subspaces are not closed under addition nor under scalar multiplication. However, all differences of two vectors are, as they are vectors in the corresponding linear subspace. This way you still have properties like flatness and can calculate similar to what you can do in vector spaces, but you don't have a zero and must be more careful with addition and scalar multiplication. E.g. all tangent spaces $T_p$ are linear spaces, but if you actually want to use them as tangents, you have to consider $\vec{p} + T_p$ which is affine. This is very important as usually it's spoken about the vector space $T_p\,$, e.g. a line $\mathbb{R} \cdot \vec{l}$. And as a line it is a vector space only if it contains the origin $\vec{0}$. The actual tangent are the points $\vec{p} + t\cdot \vec{l}$. I mention this, as it might be confusing if people talk about linearity of tangent spaces, e.g. when talking about a gradient $\nabla$, although it is strictly speaking the affine space shifted by $\vec{p}$. | {
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# Use Pappus' theorem to find the moment of a region limited by a semi-circunference.
This is part of self-study; I found this question in the book "The Calculus with Analytic Geometry" (Leithold).
$R$ is the region limited by the semi-circumference $\sqrt{r^2 - x^2}$ and the x-axis. Use Pappus' theorem to find the moment of $R$ with respect to the line $y = -4$.
Pappus' theorem (also referred to as Guldinus theorem or Pappus-Guldinus theorem) is as follows:
If $R$ is the region limited by the functions $f(x)$ and $g(x)$, then, if $A$ is the area of $R$ and $\bar{y}$ is the y-coordinate of the centroid of $R$, the volume $V$ of the solid of revolution obtained by rotating $R$ around the x-axis is given by: $V = 2\pi\bar{y}A$.
Also, what I'm calling moment (I'm not sure if this is a common term) is the quantity that is divided by the area of the region in order to find a coordinate of the centroid of the region: for example, to find the x-coordinate of the centroid ($\bar{x}$), first one finds the moment around the y-axis ($M_y$), then divides it by the area of the region ($A$).
Since the region $R$ is symmetric with respect to the y-axis, the x-coordinate of the centroid ($\bar{x}$) is zero (therefore, the moment around the y-axis ($M_y$) is zero, because $M_y = \bar{x}A$, where $A$ is the area of the region). So, I only need to find the vertical coordinate of the centroid (with respect to the line $y = -4$) and the moment around the line $y = -4$.
The book's answer for the moment around the line $y = -4$ is: $\frac{1}{2}r^3\left (\pi+\frac{4}{3}\right )$. I included two attempts below; both arrive at a same result, which is different from the result of the book.
Attempt 1
First I tried to use Pappus' theorem to find the vertical coordinate of the centroid of the semi-circular region limited by $\sqrt{r^2 - x^2}$, with respect to the x-axis (not yet with respect to the line y = -4). I will call it $\bar{y}_x$. | {
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Since the solid of revolution obtained by rotating this semi-circular region around the x-axis is a sphere, its volume is $V = \frac{4}{3}\pi r^3$. The area of the semi-circular region is $A=\frac{\pi r^2}{2}$. Substituting $V$ and $A$ into Pappus' theorem:
$\frac{4}{3}\pi r^3 = 2\pi\bar{y}_x\frac{\pi r^2}{2}$
$\bar{y}_x = \frac{4r}{3\pi}$.
This is the vertical coordinate of the centroid with respect to the x-axis. The vertical coordinate of the centroid with respect to the line $y = -4$ is:
$\bar{y} = \frac{4r}{3\pi} + 4$.
To find the moment around the line $y = -4$, I use the fact that $\bar{y} = \frac{M_x}{A}$:
$M_x = \bar{y}A = \left ( \frac{4r}{3\pi} + 4\right )\frac{\pi r^2}{2}$.
Attempt 2
The moment of a plane region with respect to the line $y = -4$ can be found by dividing this region into infinitesimal elements of area, then multiplying the area of each element of area by the vertical coordinate of the centroid.
So, if $f(x) = \sqrt{r^2 - x^2}$, then, if I divide the semi-circular region into several rectangles of length $dx$, the area of each rectangle is $f(x) dx$ and the vertical coordinate of the centroid of each rectangle with respect to the line $y = -4$ is $\frac{f(x)}{2} + 4$. So, the moment with respect to the line $y = -4$ is:
$M_x = \int_{-r}^r \left [ f(x)\times \left ( \frac{f(x)}{2} + 4 \right ) dx \right ]$.
Solving this integral gives the following result:
$M_x = \left ( \frac{4r}{3\pi} + 4\right )\frac{\pi r^2}{2}$,
which is the same result I found in the first attempt. | {
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which is the same result I found in the first attempt.
Both your attempts lead to the correct result, and the solution given in the book is wrong. To see that the book's solution cannot be correct do the following "Gedankenexperiment": Assume that we are told to compute the moment with respect to the line $y=-\eta$ for some given $\eta$ instead of $4$. For an $\eta \gg r$ this moment would be approximatively proportional to $r^2$ and to $\eta$, as is the case in your solution but not in the solution given by the book: The $O(r^3)$ dependency is unwarranted. | {
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# Non-negative polynomials on $[0,1]$ with small integral
Let $P_n$ be the set of degree $n$ polynomials that pass through $(0,1)$ and $(1,1)$ and are non-negative on the interval $[0,1]$ (but may be negative elsewhere).
Let $a_n = \min_{p\in P_n} \int_0^1 p(x)\,\mathrm{d}x$ and let $p_n$ be the polynomial that attains this minimum.
Are $a_n$ or $p_n$ known sequences? Is there some clever way to rephrase this question in terms of linear algebra and a known basis of polynomials?
Following Robert Israel's answer, we also scale everything to $[-1,1]$ (thus multiplying the result by 2). As he mentions, the optimal polynomial is always a square of some other polynomial, $p_{2n}=p_{2n+1}=q_n^2$, and $q_n$ is either even or odd (see Lemma below). So we are left to find the minimal $L_2[-1,1]$-norm of an odd/even polynomial $q_n$ such that $\deg q_n\leq n$ and $q_n(\pm1)=(\pm1)^n$. In other words (recall the division by 2 in the first line!), $a_{2n}=a_{2n+1}=d_n^2/2$, where $d_n$ is the distance from $0$ to the hyperplane defined by $q(1)=1$ in the space of all odd/even polynomials of degree at most $n$ with $L_2[-1,1]$-norm.
Now, this hyperplane is the affine hull of the Legendre polynomials $P_i(x)=\frac1{2^ii!}\frac{d^i}{dx^i}(x^2-1)^i$, where $i$ has the same parity as $n$ (since we know that $P_i(1)=1$ and $P_i(-1)=(-1)^i$). Next, by $\| P_i\|^2=\frac{2}{2i+1}$, our distance is $$\left(\sum_{j\leq n/2}\| P_{n-2j}\|^{-2}\right)^{-1/2} =\left(\sum_{j\leq n/2}\frac{2(n-2j)+1}{2}\right)^{-1/2}=\sqrt{\frac4{(n+1)(n+2)}},$$ attained at $$q_n(x)=\left(\sum_{i\leq n/2}\frac{2}{2(n-2i)+1}\right)^{-1} \sum_{i\leq n/2}\frac{2P_{n-2i}(x)}{2(n-2i)+1}.$$ Thus the answer for the initial question is $a_{2n}=a_{2n+1}=\frac2{(n+1)(n+2)}$ and $p_{2n}=p_{2n+1}=q_n^2$.
Lemma. For every $n$, one of optimal polynomials on $[-1,1]$ is a square of an odd or even polynomial. | {
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Proof. Let $r(x)$ be any polynomial.which is nonnegative on $[-1,1]$ with $r(\pm 1)=1$. We will replace it by some other polynomial which has the required form, has the same (or less) degree and the same values at $\pm 1$, is also nonnegative on $[-1,1]$, and is not worse in the integral sense. Due to the compactness argument, this yields the required result.
Firstly, replacing $r(x)$ by $\frac12(r(x)+r(-x))$, we may assume that $r$ is even (and thus has an even degree $2n$). Let $\pm c_1,\dots,\pm c_{n}$ be all complex roots of $r$ (regarding multiplicities); then $$r(x)=\prod_{j=1}^n\frac{x^2-c_j^2}{1-c_j^2},$$ due to $r(\pm 1)=1$.
Now, for all $c_j\notin[-1,1]$ we simultaneously perform the following procedure.
(a) If $c_j$ is real, then we replace $\pm c_j$ by $\pm x_j=0$. Notice that $$\frac{|x^2-c_j^2|}{|1-c_j^2|}\geq 1\geq \frac{|x^2-0^2|}{|1-0^2|}.$$ for all $x\in[-1,1]$.
(b) If $c_j$ is non-real, then we choose $x_j\in[-1,1]$ such that $\frac{|c_j-1|}{|c_j+1|}=\frac{|x_j-1|}{|x_j+1|}$. Notice that all complex $z$ with $\frac{|c_j-z|}{|x_j-z|}=\frac{|c_j-1|}{|x_j-1|}$ form a circle passing through $-1$ and $1$, and the segment $[-1,1]$ is inside this circle. Therefore, for every $x\in[-1,1]$ we have $$\frac{|c_j-x|}{|x_j-x|}\geq\frac{|c_j-1|}{|x_j-1|},$$ thus $$\frac{|x^2-c_j^2|}{|1-c_j^2|} =\frac{|c_j-x|}{|c_j-1|}\cdot\frac{|c_j+x|}{|c_j+1|} \geq \frac{|x_j-x|}{|x_j-1|}\cdot\frac{|x_j+x|}{|x_j+1|} =\frac{|x^2-x_j^2|}{|1-x_j^2|}.$$ So, we replace $\pm c_j$ and $\pm\bar c_j$ by $\pm x_j$ and $\pm x_j$ (or simply $\pm c_j$ by $\pm x_j$ if $c_j$ is purely imaginary). | {
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After this procedure has been applied, we obtain a new polynomial whose roots are in $[-1,1]$ and have even multiplicities, and its values at $\pm1$ are equal to $1$. So it is a square of some polynomial which is even/odd (since the roots are still split into pairs of opposite numbers). On the other hand, its values at every $x\in[-1,1]$ do not exceed the values of $r$ at the same points, as was showed above. So the obtained polynomial is not worse in the integral sense, as required. The lemma is proved.
• Sorry to be dense, but please say why it has to be the square of another polynomial. I can see that any zeros in the critical interval have even multiplicity, but why can't the polynomial be negative elsewhere (as allowed by OP)? Oct 9 '15 at 22:54
• The last part of the answer (after `finally') is about it; I'll try to expand it. In short: If our polynomial has a non-real root, or if it has real roots outside $[-1,1]$, then we may replace these roots by some root on $[-1,1]$ so as to lower the values of the polynomial at all points in $(-1,1)$(and keep it nonnegative). Oct 10 '15 at 7:05
• Expanded (or, I would say, rewritten). Oct 10 '15 at 7:47
For better symmetry, let me change the boundary points to $(-1,1)$ and $(1,1)$. Then it is clear that we may assume $p_n$ is an even function (and thus you really want $P_n$ to be polynomials of degree at most $n$, otherwise you'll be likely to have an infimum rather than an actual minimum): $p_{2k+1} = p_{2k}$. I get \eqalign{p_0(x) = 1, & a_0 = 2\cr p_2(x) = x^2, & a_2 = \dfrac{2}{3}\cr p_4(x) = \dfrac{(5 x^2 - 1)^2}{16},& a_4 = \dfrac{1}{3}\cr p_6(x) = \dfrac{x^2}{16} (7 x^2 - 3)^2,& a_6 = \dfrac{1}{5}\cr p_8(x) = \dfrac{1}{64} (21 x^4 - 14 x^2 + 1)^2, & a_8 = \dfrac{2}{15}\cr p_{10}(x) = \dfrac{x^2}{64} (33 x^4 - 30 x^2 + 5)^2, & a_{10} = \dfrac{2}{21}\cr p_{12}(x) = \dfrac{1}{4096} (429 x^6 - 495 x^4 + 135 x^2 - 5)^2, & a_{12} = \dfrac{1}{14}} Well, so far it looks like $a_{2n} = \dfrac{4}{(n+1)(n+2)}$ | {
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• How were these computed? Oct 9 '15 at 0:03
• The leading coefficients look like Catalan numbers when the polynomials are normalized to $\sqrt{2^{2n} p_{2n}(x)}$. Oct 9 '15 at 1:51
• The next coefficients match $-{2n \choose n-2}$. The whole triangle looks like oeis.org/A234950 read skewed, with alternating signs. See oeis.org/A062991 for a signed version. Oct 9 '15 at 3:43
• For $n$ even, $$p_{2n} = \prod_{i=1}^{n/2} \dfrac{(x^2 -c_i^2)^2}{(1-c_i^2)^2}$$ For $n$ odd, there's also a factor of $x^2$. Differentiate $a_{2n} =\int_{-1}^1 p_{2n}(x)\; dx$ with respect to each $c_i$, solve, check which solution gives the least $a_{2n}$. Oct 9 '15 at 4:49 | {
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Question
# Assertion :The chord of contact of tangent from a point $$P$$ to a circle passes through $$Q$$. If $${ l }_{ 1 }$$ and $${ l }_{ 2 }$$ are the lengths of the tangents from $$P$$ and $$Q$$ to the circle, then $$PQ$$ is equal to $$\sqrt { { { l }_{ 1 } }^{ 2 }+{ { l }_{ 2 } }^{ 2 } }$$ Reason: The equation of chord of contact of tangents from the point $$P\left( { x }_{ 1 },{ y }_{ 1 } \right)$$ to the circle $${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$$ is $$x{ x }_{ 1 }+y{ y }_{ 1 }={ a }^{ 2 }$$
A
Assertion is true, Reason is true and reason is correct explanation for Statement-1.
B
Assertion is true, Reason is true and Reason is NOT correct explanation for Assertion.
C
Assertion is true, Reason is false
D
Assertion is false, Reason is true
Solution | {
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Solution
## The correct option is A Assertion is true, Reason is true and reason is correct explanation for Statement-1.Let $$P\equiv \left( { x }_{ 1 },{ y }_{ 1 } \right)$$ and $$Q\equiv \left( { x }_{ 2 },{ y }_{ 2 } \right)$$Let the equation of the given circle be $${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$$The equation of chord of contact of tangents from the point $$P\left( { x }_{ 1 },{ y }_{ 1 } \right)$$ to the given circle is $$x{ x }_{ 1 }+y{ y }_{ 1 }={ a }^{ 2 }$$Since it passes through $$Q\left( { x }_{ 2 },{ y }_{ 2 } \right)$$$$\therefore { x }_{ 1 }{ x }_{ 2 }+{ y }_{ 1 }{ y }_{ 2 }={ a }^{ 2 }$$ ...(1)Now $${ l }_{ 1 }=\sqrt { { { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 }-{ a }^{ 2 } } ,{ l }_{ 2 }=\sqrt { { { x }_{ 2 } }^{ 2 }+{ { y }_{ 2 } }^{ 2 }-{ a }^{ 2 } }$$and $$PQ=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } }$$$$=\sqrt { \left( { { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 } \right) +\left( { { x }_{ 2 } }^{ 2 }+{ { y }_{ 2 } }^{ 2 } \right) -2\left( { x }_{ 1 }{ x }_{ 2 }+{ y }_{ 1 }{ y }_{ 2 } \right) } \\ =\sqrt { \left( { { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 } \right) +\left( { { x }_{ 2 } }^{ 2 }+{ { y }_{ 2 } }^{ 2 } \right) -2{ a }^{ 2 } } \\ =\sqrt { \left( { { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 }-{ a }^{ 2 } \right) +\left( { { x }_{ 2 } }^{ 2 }+{ { y }_{ 2 } }^{ 2 }-{ a }^{ 2 } \right) } =\sqrt { { { l }_{ 1 } }^{ 2 }+{ { l }_{ 2 } }^{ 2 } }$$Maths
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# Find the smallest $n \in \mathbb{N}$ such that the group is isomorphic to the direct product of $n$ cyclic groups
Find the smallest $$n \in \mathbb{N}$$ such that the group $$\mathbb{Z}_{6} \times \mathbb{Z}_{20} \times \mathbb{Z}_{45}$$ is isomorphic to the direct product of $$n$$ cyclic groups.
I'm not sure but if I understand correctly,
$$\mathbb{Z}_{6} \times \mathbb{Z}_{20} \times \mathbb{Z}_{45}$$ is isomorphic to $$\mathbb{Z}_{2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{4} \times \mathbb{Z}_{5} \times \mathbb{Z}_{5} \times \mathbb{Z}_{9}$$,
$$\mathbb{Z}_{2} \times \mathbb{Z}_{5} \times \mathbb{Z}_{9}$$ is isomorphic to $$\mathbb{Z}_{90}$$,
$$\mathbb{Z}_{3} \times \mathbb{Z}_{4} \times \mathbb{Z}_{5}$$ is isomorphic to $$\mathbb{Z}_{60}$$,
and therefore, $$\mathbb{Z}_{6} \times \mathbb{Z}_{20} \times \mathbb{Z}_{45}$$ is isomorphic to $$\mathbb{Z}_{60} \times \mathbb{Z}_{90}$$ and the answer is $$n = 2$$. Is this a correct solution?
• I agree that the answer is $n=2$. For the sake of being thorough, I would probably explain why $n=1$ is impossible (just because you've found one situation where $n=2$ doesn't necessarily mean it's the smallest $n$). – Theo C. Apr 23 at 21:21
• Yes, it is. You just have to find the smallest number of groups of moduli such that the moduli in each group are pairwise coprime. – Bernard Apr 23 at 21:24
• After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. – Shaun May 24 at 14:27
You may find the invariant factors. | {
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You may find the invariant factors.
Actually $$\mathbb{Z}_6\times \mathbb{Z}_{20}\times \mathbb{Z}_{45}\cong (\mathbb{Z}_2\times\mathbb{Z}_4)\times(\mathbb{Z}_3\times\mathbb{Z}_9)\times(\mathbb{Z}_5\times\mathbb{Z}_5)$$. Now we can pick the largest ones in each bracket to form $$\mathbb{Z}_4\times\mathbb{Z}_9\times\mathbb{Z}_5\cong\mathbb{Z}_{180}$$. Then we pick the largest ones of the remaining, which is $$\mathbb{Z}_2\times\mathbb{Z}_3\times\mathbb{Z}_5\cong\mathbb{Z}_{30}$$, and nothing remain. Hence the group is isomorphic to $$\mathbb{Z}_{180}\times\mathbb{Z}_{30}$$ and so $$n\le 2$$. Since the group is obviously not cyclic, we have $$n = 2$$.
I think we can always get $$n$$ for any abelian group by finding the invariant factors.
According to GAP, the group is $$\Bbb Z_{180}\times \Bbb Z_{30}$$ and, of course, not cyclic.
gap> G:=DirectProduct(CyclicGroup(6) , DirectProduct(CyclicGroup(20), CyclicGroup(45)));
<pc group of size 5400 with 8 generators>
gap> StructureDescription(G);
"C180 x C30"
gap> IsCyclic(G);
false
gap> | {
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Greater Than or Equal To: Math Definition. use ">=" for greater than or equal use "<=" for less than or equal In general, Sheets uses the same "language" as Excel, so you can look up Excel tips for Sheets. Graphical characteristics: Asymmetric, Open shape, Monochrome, Contains straight lines, Has no crossing lines. In such cases, we can use the greater than or equal to symbol, i.e. ≥. Solution for 1. In an acidic solution [H]… But, when we say 'at least', we mean 'greater than or equal to'. Greater than or equal application to numbers: Syntax of Greater than or Equal is A>=B, where A and B are numeric or Text values. Rate this symbol: (3.80 / 5 votes) Specifies that one value is greater than, or equal to, another value. Select Symbol and then More Symbols. Examples: 5 ≥ 4. The sql Greater Than or Equal To operator is used to check whether the left-hand operator is higher than or equal to the right-hand operator or not. 923 Views. Greater than or Equal in Excel – Example #5. is less than > > is greater than ≮ \nless: is not less than ≯ \ngtr: is not greater than ≤ \leq: is less than or equal to ≥ \geq: is greater than or equal to ⩽ \leqslant: is less than or equal to ⩾ With Microsoft Word, inserting a greater than or equal to sign into your Word document can be as simple as pressing the Equal keyboard key or the Greater Than keyboard key, but there is also a way to insert these characters as actual equations. Select the Greater-than Or Equal To tab in the Symbol window. If left-hand operator higher than or equal to right-hand operator then condition will be true and it will return matched records. This symbol is nothing but the "greater than" symbol with a sleeping line under it. “Greater than or equal to” and “less than or equal to” are just the applicable symbol with half an equal sign under it. The less than or equal to symbol is used to express the relationship between two quantities or as a boolean logical operator. Here a could be greater … Category: | {
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between two quantities or as a boolean logical operator. Here a could be greater … Category: Mathematical Symbols. The greater-than sign is a mathematical symbol that denotes an inequality between two values. When we say 'as many as' or 'no more than', we mean 'less than or equal to' which means that a could be less than b or equal to b. 2 ≥ 2. Less Than or Equal To (<=) Operator. Finding specific symbols in countless symbols is obviously a waste of time, and some characters like emoji usually can't be found. Copy the Greater-than Or Equal To in the above table (it can be automatically copied with a mouse click) and paste it in word, Or. Sometimes we may observe scenarios where the result obtained by solving an expression for a variable, which are greater than or equal to each other. As we saw earlier, the greater than and less than symbols can also be combined with the equal sign. Select the Insert tab. For example, the symbol is used below to express the less-than-or-equal relationship between two variables: For example, 4 or 3 ≥ 1 shows us a greater sign over half an equal sign, meaning that 4 or 3 are greater than or equal to 1. "Greater than or equal to" is represented by the symbol " ≥ ≥ ". For example, x ≥ -3 is the solution of a certain expression in variable x. In Greater than or equal operator A value compares with B value it will return true in two cases one is when A greater than B and another is when A equal to B. Use the appropriate math symbol to indicate "greater than", "less than" or "equal to" for each of the following: a. "Greater than or equal to", as the suggests, means something is either greater than or equal to another thing. Right-Hand operator then condition will be true and it will return matched.... Right-Hand operator then condition will be true and it will return matched records, means something is greater... Specifies that one value is greater than or equal to right-hand operator then condition will true!: Asymmetric, Open shape, | {
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is greater than or equal to right-hand operator then condition will true!: Asymmetric, Open shape, Monochrome, Contains straight lines, greater than or equal to sign no crossing lines specific. By the symbol window than '' symbol with a sleeping line under it will be true and it will matched!, Open shape, Monochrome, Contains straight lines, Has no crossing lines symbol, i.e symbol is but! Suggests, means something is either greater than or equal in Excel – example # 5 greater than or equal to sign greater,. Greater than, or equal to tab in the symbol ≥ ≥ a sleeping line it. As we saw earlier, the greater than or equal to ( < = ) operator of,..., Contains straight lines, Has no crossing lines rate this symbol is nothing but . The symbol ≥ ≥ '' is represented by the symbol window in cases... Certain expression in variable x, Contains straight lines, Has no crossing lines less than equal... Under it use the greater than, or equal to '' is represented by the symbol ≥ ≥.... We can use the greater than, or equal to, another value means something either! To ' we say 'at least ', we mean 'greater than or equal in Excel – #... N'T be found than '' symbol with a sleeping line under greater than or equal to sign ) Specifies that one value is than... Be true and it will return matched records as the suggests, means something is either greater than symbol. An inequality between two values to ' variable x to right-hand operator then condition be! Select the greater-than sign is a mathematical symbol that denotes an inequality between two values:. Shape, Monochrome, Contains straight lines, Has no crossing lines ≥ ≥.... Countless symbols is obviously a waste of time, and some characters emoji. X ≥ -3 is the solution of a certain expression in variable.... Tab in the symbol window some characters like emoji usually ca n't be found be and! As the suggests, means something is either greater than or equal to ( < = operator! Rate this symbol: ( 3.80 / 5 votes ) Specifies that one | {
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greater than or equal to ( < = operator! Rate this symbol: ( 3.80 / 5 votes ) Specifies that one value is than. To right-hand operator then condition will be true and it will return matched records line under.... Excel – example # 5 '', as the suggests, means something is either greater than and less symbols. / 5 votes ) Specifies that one value is greater than or in. Is obviously a waste of time, and some characters like emoji usually ca be... Symbol window will return matched records will be true and it will return matched records we can the. Be true and it will return matched records as we saw earlier the... Tab in the symbol window Monochrome, Contains straight lines, Has no crossing.! Is represented by the symbol window to ( < = ) operator that denotes inequality! Matched records the suggests, means something is either greater than or equal to.... Also be combined with the equal sign: Asymmetric, Open shape,,. Select the greater-than sign is a mathematical symbol that denotes an inequality two. ) Specifies that one value is greater than or equal to symbol, i.e means something either... Another value combined with the equal sign emoji usually ca n't be found if left-hand operator than... When we say 'at least ', we mean 'greater than or equal in –... Be combined with the equal sign and it will return matched records in the symbol window a expression... An inequality between two values one value is greater than '' symbol with a sleeping under!, the greater than or equal to ( < = ) operator to ' combined. A waste of time, and some characters like emoji usually ca n't found! Than or equal to ( < = ) operator the suggests, means something is either greater than equal. Be found x ≥ -3 is the solution of a certain expression in variable x than less! -3 is the solution of a certain expression in variable x obviously a waste of time, some! Mathematical symbol that denotes an inequality between two values something is either greater than equal. Earlier, the greater than | {
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is by... Be combined with the equal sign certain expression in variable x,,. Combined with the equal sign symbol with a sleeping line under it: 3.80. Symbols is obviously a waste of time, and some characters like emoji usually ca be. Characteristics: Asymmetric, Open shape, Monochrome, Contains straight lines, no... Something is either greater than '' symbol with a sleeping line under it then. One value is greater than '' symbol with a sleeping line under it we earlier... To tab in the symbol ≥ ≥ symbol ≥ . Tab in the symbol window characteristics: Asymmetric, Open shape, Monochrome, Contains straight lines, Has crossing. Is the solution of a certain expression in variable x shape, Monochrome, Contains straight,!, we can use the greater than or equal in Excel – example # 5 be with. Is the solution of a certain expression in variable x Open shape, Monochrome, Contains lines. We saw earlier, the greater than and less than symbols can be... Expression in variable x that one value is greater than and less than symbols can be. As we saw earlier, the greater than, or equal in Excel – example 5. < = ) operator means something is either greater than or equal to ' '' is represented by symbol... Emoji usually ca n't be found cases, we can use the greater than or equal in –... Than, or equal to ' we saw earlier, the greater or!, when we say 'at least ', we can use the greater than or equal to <..., i.e: Asymmetric, Open shape, Monochrome, Contains straight,... Ca n't be found equal to symbol, i.e the greater than, equal! The greater-than or equal to tab in the symbol ≥ ≥.., we can use the greater than '' symbol with a sleeping line under it sign is a symbol! Votes ) Specifies that one value is greater than or equal to symbol i.e! Mathematical symbol that denotes an inequality between two values Excel – example # 5 ). That denotes an inequality between two values expression in variable x votes Specifies! Than '' symbol with a sleeping line under it nothing but the greater than | {
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x votes Specifies! Than '' symbol with a sleeping line under it nothing but the greater than or equal tab! ) Specifies that one value is greater than or equal to '', as the,... Value is greater than or equal in Excel – example # 5 be combined with equal... Something is either greater than or equal to symbol, i.e greater-than or equal to.! Symbol that denotes an inequality between two values symbol window graphical characteristics: Asymmetric, shape... 3.80 / 5 votes ) Specifies that one value is greater than '' symbol with a sleeping line under.! | {
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# Quick tips for adding numbers x to y
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Quick tips for adding numbers x to y [#permalink]
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10 Jun 2009, 20:46
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Here are some examples on adding numbers from x to y:
1) Add the numbers from 40 to 70, inclusive.
2) Add the even numbers from 40 to 70, inclusive.
3) Add the odd numbers from 40 to 70, inclusive.
4) Add the numbers from 40 to 70.
5) Add the even numbers from 40 to 70.
6) Add the odd numbers from 40 to 70.
[Reveal] Spoiler:
1) 1,705
2) 880
3) 825
4) 1,595
5) 770
6) 825
Last edited by I3igDmsu on 05 Jul 2009, 19:59, edited 7 times in total.
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Re: Quick tips for adding numbers x to y [#permalink]
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10 Jun 2009, 22:58
1) Add the numbers from 40 to 70, inclusive.
Formula for sum of first N natural numbers is = N(N+1)/2
Sum from 1 to 70 Sa= 70*71/2
Sum from 1 to 40 Sb= 40*41/2
So sum from 40 to 70 = Sa-Sb
2) Add the even numbers from 40 to 70, inclusive.
3) Add the odd numbers from 40 to 70, inclusive.
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========================
As a general solution to all these kind of problems learn the AP series.
google on Arithemic Progression series. It is kind of difficult to write the formula here but all these calculations are tooooooo simple using this series.
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Re: Quick tips for adding numbers x to y [#permalink]
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12 Jun 2009, 14:34
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If you don't mind remembering a formula or two then yes you can.
The examples that you have given can be grouped under Arithmetic Progressions...or a finite sequence of evenly spaced numbers.
There are two formulas:
$$1. S = \frac{n}{2} [2a + (n-1)d]$$
where
$$S=\text{sum of the all the numbers in the sequence}$$
$$n=\text{total number of numbers in the sequence}$$
$$a=\text{the first number of the sequence}$$
$$d=\text{the different between any two consecutive numbers in the sequence}$$
$$2. S = \frac{n}{2} [\text{First Term}+\text{Last Term}]$$
where
$$S=\text{sum of the all the numbers in the sequence}$$
$$n=\text{total number of numbers in the sequence}$$
You can use either equation based on what is provided in the question.
Lets take your examples and solve them. For all of these we know the first and the last number so we should be fine with equation 2.
1) Add the numbers from 40 to 70, inclusive.
Solution:
Here the sequence is 40,41, 42, ...., 69, 70.
To calculate n
$$n=\text{Last number}-\text{First number} + 1$$
$$n=70-40+1 = 31$$
Using equation 1.
$$S = \frac{n}{2} [\text{First Term}+\text{Last Term}]$$
$$S = \frac{n}{2} [\text{First Term}+\text{Last Term}]$$
$$S = \frac{31}{2} [40+70]$$
$$S = \frac{31}{2} [110]$$
$$S = 31\times55$$
$$S = 1705$$
2) Add the even numbers from 40 to 70, inclusive.
Here the sequence is 40,42, 44, ...., 68, 70.
Solve using 40 as the first term and 70 as the last term
To calculate n for even (or odd) number | {
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$$n=\frac{\text{Last number}-\text{First number}}{2} + 1$$
3) Add the odd numbers from 40 to 70, inclusive.
Here is the sequence is 41,43,....67,69
Solve using 41 as the first term and 69 as the last term
and so on...
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Re: Quick tips for adding numbers x to y [#permalink]
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30 Jun 2009, 20:11
Could someone check my answers above please?
Thanks nookway, your help is appreciated!
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Re: Quick tips for adding numbers x to y [#permalink]
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03 Jul 2009, 07:51
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Expert's post
I3igDmsu wrote:
Could someone check my answers above please?
Thanks nookway, your help is appreciated!
Don't think you need our help - this can be easily done in Excel
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Re: Quick tips for adding numbers x to y [#permalink]
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05 Jul 2009, 20:00
Done. If anyone needs practice, the answers in the spoiler are correct.
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Re: Quick tips for adding numbers x to y [#permalink]
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02 Aug 2009, 21:30
Great stuff! Thanks!
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13 Oct 2011, 14:32
I did not get the logic of these.
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Re: Quick tips for adding numbers x to y [#permalink] 13 Sep 2017, 22:23
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# Quick tips for adding numbers x to y
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• Nov 28th 2012, 06:08 PM
warElephant
Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?
8/33
62/165
17/33
103/165
25/33
this is my solution but it is wrong:
an Arbitrary deck of hands with at least one pair of similar cards = A, B, C, A
the number of patterns that A,B,C,A hand appears = 4C2
Therefore, number of ways A is chosen= 12
when the first A is chosen, it locks in the value for the second A
number of ways B is chosen = 10(because by choosing a value for A, two cards are taken off the choice list)
number of ways C is chosen = 9
P(at least two cards have the same value) = (12 x 10 x 9 x 4C2)/(12 x 11 x 10 x 9) = 18/33
I hope that your explaination you will elaborate:
1. why my method is wrong
2. the actual solution.
Thank you very much for even considering my post!!!:)
• Nov 28th 2012, 09:49 PM
Scopur
P(at least 1) = 1 - P(None). This is probably more useful here.
So there are $6C4$ possible ways to select 4 distinct cards. Then there are $2^4$ different suits for those cards. Finally there are $12C6$ different cards.
So you have
$P(at least one) = 1 - P(none) = 1 - \frac{6C4*2^4}{12C6} = 1 - \frac{16}{33} = \frac{17}{33}$
• Nov 28th 2012, 10:15 PM
Soroban
Hello, warElephant!
Another approach . . .
Quote: | {
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Another approach . . .
Quote:
Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each.
Each of the 6 cards within a suit has a different value from 1 to 6.
Bill likes to play a game in which he shuffles the deck, turns over 4 cards,
and looks for pairs of cards that have the same value.
What is the probability that Bill finds at least one pair of cards that have the same value?
. . $(a)\;\tfrac{8}{33} \qquad(b)\;\tfrac{62}{165} \qquad (c)\;\tfrac{17}{33}\qquad (d)\;\tfrac{103}{165} \qquad (e)\;\tfrac{25}{33}$
We will find the probability that there are no matching pairs among the four cards.
The first card can be any card (it doesn't matter): $\tfrac{12}{12}$
The second card can be any of the other 10 non-matching cards: $\tfrac{10}{11}$
The third card can be any of the other 8 non-matching cards: $\tfrac{8}{10}$
The fourth card can be any of the other 6 non-matching cards: $\ftrac{6}{9}$
Hence: . $P(\text{no match}) \;=\;\frac{12}{12}\cdot\frac{10}{11}\cdot\frac{8}{ 10}\cdot\frac{6}{9} \:=\:\frac{16}{33}$
Therefore: . $P(\text{at least one match}) \;=\;1-\frac{16}{33} \;=\;\frac{17}{33}$ | {
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# Electric field strength away from a negative spherical charge
I was wonder how would a graph of electrical field strength away from a spherical -ve charge graph would look. Since E= potential gradient I was guessing that since potential increases away from a negative charge , electric field strength would increase away from a negative charge also ?
The potential does increase. This means that the gradient of potential plotted against distance, r, from the charge is always positive. But the gradient, $$\frac{dV}{dr},$$ keeps decreasing in magnitude – just sketch the graph! The field strength in the r direction is given by$$E=\ –\frac{dV}{dr},$$so the field is in the –r direction and decreases in magnitude the further we go from the negative charge.
• Yes, but as there is radial symmetry, it would be clearer to put the centre of the sphere at $r$ = 0 and to have the $r$ axis in the positive direction only (as it represents the distance from the sphere in any direction). We'd also usually have the graph upside down, reflected about the $r$ axis, the negativity of $E$ showing that the field is directed in the –$r$ direction, that is towards the sphere. – Philip Wood Nov 23 '18 at 0:16 | {
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# Kinetic energy question
1. Dec 10, 2013
### hey123a
1. The problem statement, all variables and given/known data
A sled of mass m is coasting on the icy surface of a frozen river. While it is passing under a bridge, a package of equal mass m is dropped straight down and lands on the sled (without causing any damage). The sled plus the added load then continue along the original line of motion. How does the kinetic energy of the (sled + load) compare with the original kinetic energy of the sled? A) It is 1/4 the original kinetic energy of the sled. B) It is 1/2 the original kinetic energy of the sled. C) It is 3/4 the original kinetic energy of the sled. D) It is the same as the original kinetic energy of the sled. E) It is twice the original kinetic energy of the sled.
2. Relevant equations
3. The attempt at a solution
kinetic energy intial = 1/2mv^2
kinetic energy after = 1/2(2m)v^2 = mv^2
comparing the kinetic energy before and after, i get the after is twice the original kinetic energy of the sled but apparently this is wrong
2. Dec 10, 2013
### Simon Bridge
You forgot to conserve momentum.
3. Dec 10, 2013
### rock.freak667
You will need to use conservation of linear momentum to calculate the velocity of the sled+load after the impact as your attempt shows them traveling at the same velocity.
If before the impact, the sled has momentum mu and after it travels with the velocity 'v' and the load is initially at rest, then using conservation of linear momentum what is v in terms of u?
4. Dec 10, 2013
### hey123a
ah okay i think i got it
m1v1o + m2v2o = (m1+m2)vf
where m1 is the sled, and m2 a mass that is equal to the sled
m1v1o = (m1+m2)vf
m1v1o = 2mvf
m1v1o/2m = vf
v1o/2 = vf
original ke = 1/2mv^2
KE of sled + load = 1/2(2m)(v1o/2)^2
ke of sled + load = 1/2(2m)(v^2/4 = 1/4mv^2, which is 1/2 the original kinetic energy of the sled
5. Dec 11, 2013
### Simon Bridge
Well done. | {
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5. Dec 11, 2013
### Simon Bridge
Well done.
It's easier to check your answers if you get formal with the working.
for instance, if I write:
before:
momentum: $p_i=mu$
kinetic energy: $K_i=\frac{1}{2}mu^2$
after:
momentum: $p_f=2mv$
kinetic energy: $K_f = mv^2$
conservation of momentum:
$p_f=p_i\implies 2mv=mu \implies v=u/2\\ \qquad \implies K_f=\frac{1}{4}mu^2$
Comparing KE "after" with KE "before":
$$\frac{K_f}{K_i}=\frac{\frac{1}{4}mu^2}{\frac{1}{2}mu^2}=\frac{1}{2} \\ \qquad\implies K_f=\frac{1}{2}K_i$$
You can see all the steps and most of the reasoning.
Good for long answers or when you are practicing (or writing questions here ;) ) - but probably more work than you'd like for a simple multi-choice.
Still... no worries aye? | {
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Therefore, our inflection point is at x = 2. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. For a maximum point the 2nd derivative is negative, and the minimum point is positive. I've some data about copper foil that are lists of points of potential(X) and current (Y) in excel . For instance, if we were driving down the road, the slope of the function representing our distance with respect to time would be our speed. For example, the second derivative of the function y = 17 is always zero, but the graph of this function is just a horizontal line, which never changes concavity. Therefore possible inflection points occur at and .However, to have an inflection point we must check that the sign of the second derivative is different on each side of the point. The curve I am using is just representative. I like thinking of a point of inflection not as a geometric feature of the graph, but as a moment when the acceleration changes. The Second Derivative Test cautions us that this may be the case since at f 00 (0) = 0 at x = 0. The sign of the derivative tells us whether the curve is concave downward or concave upward. MENU MENU. And a list of possible inflection points will be those points where the second derivative is zero or doesn't exist. (d) Identify the absolute minimum and maximum values of f on the interval [-2,4]. 8.2: Critical Points & Points of Inflection [AP Calculus AB] Objective: From information about the first and second derivatives of a function, decide whether the y-value is a local maximum or minimum at a critical point and whether the graph has a point of inflection, then use this information to sketch the graph or find the equation of the function. To find inflection points, start by differentiating your function to find the derivatives. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Our website is made possible by | {
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that the domains *.kastatic.org and *.kasandbox.org are unblocked. Our website is made possible by displaying online advertisements to our visitors. Necessary Condition for an Inflection Point (Second Derivative Test) If $${x_0}$$ is a point of inflection of the function $$f\left( x \right)$$, and this function has a second derivative in some neighborhood of $${x_0},$$ which is continuous at the point $${x_0}$$ itself, then The purpose is to draw curves and find the inflection points of them..After finding the inflection points, the value of potential that can be used to … However, f "(x) is positive on both sides of x = 0, so the concavity of f is the same to the left and to the right of x = 0. A positive second derivative means that section is concave up, while a negative second derivative means concave down. How to Calculate Degrees of Unsaturation. A point of inflection does not have to be a stationary point however; A point of inflection is any point at which a curve changes from being convex to being concave . You … In the case of the graph above, we can see that the graph is concave down to the left of the inflection point and concave down to the right of the infection point. find f "; find all x-values where f " is zero or undefined, and Save my name, email, and website in this browser for the next time I comment. Then the function achieves a global maximum at x 0: f(x) ≤ f(x 0)for all x ∈ &Ropf.. The second derivative of a function may also be used to determine the general shape of its graph on selected intervals. (c) Use the second derivative test to locate the points of inflection, and compare your answers with part (b). When the second derivative is negative, the function is concave downward. However, (0, 0) is a point of inflection. If it does, the value at x is an inflection point. One method of finding a function’s inflection point is to take its second derivative, set it equal to zero, and solve for x. Second Derivatives: Finding Inflation Points of | {
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derivative, set it equal to zero, and solve for x. Second Derivatives: Finding Inflation Points of the Function. The first derivative is f′(x)=3x2−12x+9, sothesecondderivativeisf″(x)=6x−12. Stationary Points. The second derivative of a function may also be used to determine the general shape of its graph on selected intervals. The next graph shows x 3 – 3x 2 + (x – 2) (red) and the graph of the second derivative of the graph, f” = 6(x – 1) in green. 4. View Point of inflection from MATH MISC at Manipal Institute of Technology. On the right side of the inflection point, the graph increases faster and faster. Inflection point is a point on the function where the sign of second derivative changes (where concavity changes). Applying derivatives to analyze functions, Determining concavity of intervals and finding points of inflection: algebraic. Recall the graph f (x) = x 3. Learn how the second derivative of a function is used in order to find the function's inflection points. So the second derivative must equal zero to be an inflection point. 10 years ago. To locate a possible inflection point, set the second derivative equal to zero, and solve the equation. For instance if the curve looked like a hill, the inflection point will be where it will start to look like U. There are two issues of numerical nature with your code: the data does not seem to be continuous enough to rely on the second derivative computed from two subsequent np.diff() applications; even if it were, the chances of it being exactly 0 are very slim; To address the first point, you should smooth your histogram (e.g. 2. Find all inflection points for the function f (x) = x 4.. Mathematics Learning Centre, University of Sydney 1 The second derivative The second derivative, d2y dx2,ofthe function y = f(x)isthe derivative of dy dx. An inflection point occurs on half profile of M type or W type, two inflection points occur on full profiles of M type or W type. To locate the inflection point, we need to | {
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points occur on full profiles of M type or W type. To locate the inflection point, we need to track the concavity of the function using a second derivative number line. A critical point becomes the inflection point if the function changes concavity at that point. , Sal means that there is an inflection point, not at where the second derivative is zero, but at where the second derivative is undefined. But don't get excited yet. As we saw on the previous page, if a local maximum or minimum occurs at a point then the derivative is zero (the slope of the function is zero or horizontal). ACT Preparation We observed that x = 0, and that there was neither a maximum nor minimum. Mistakes when finding inflection points: second derivative undefined, Mistakes when finding inflection points: not checking candidates, Analyzing the second derivative to find inflection points, Using the second derivative test to find extrema. The points of inflection of a function are those at which its second derivative is equal to 0. Therefore, our inflection point is at x = 2. State the second derivative test for local extrema. These points can be found by using the first derivative test to find all points where the derivative is zero, then using the second derivative test to see if any points are also turning points. So the second derivative must equal zero to be an inflection point. On the right side of the inflection point, the graph increases faster and faster. We can use the second derivative to find such points … Factoring, we get e^x(4*e^x - 1) = 0. The section of curve between A and B is concave down — like an upside-down spoon or a frown; the sections on the outsides of A and B are concave up — like a right-side up spoon or a smile; and A and B are inflection points. Since concave up corresponds to a positive second derivative and concave down corresponds to a negative second derivative, then when the function changes from concave up to concave down (or vise versa) the second | {
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then when the function changes from concave up to concave down (or vise versa) the second derivative must equal zero at that point. I am mainly looking for the list of vertices that precede inflection points in a curve. Our mission is to provide a free, world-class education to anyone, anywhere. When we simplify our second derivative we get; This means that f(x) is concave downward up to x = 2 f(x) is concave upward from x = 2. Mathematics Learning Centre The second derivative and points of inflection Jackie Nicholas c 2004 University of d2y /dx2 = (+)2 hence it is a minimum point. Lets take a curve with the following function. using a uniform or Gaussian filter on the histogram itself). When the second derivative is positive, the function is concave upward. 2. What is the difference between inflection point and critical point? (this is not the same as saying that f has an extremum). The critical points of inflection of a function are the points at which the concavity changes and the tangent line is horizontal. Mind that this is the graph of f''(x), which is the Second derivative. Setting the second derivative of a function to zero sometimes . This means that a point of inflection is a point where the second derivative changes sign (from positive to negative or vice versa) Inflection Points: The inflection points of a function of an independent variable are related to the second derivative of the function. By using this website, you agree to our Cookie Policy. But if continuity is required in order for a point to be an inflection point, how can we consider points where the second derivative doesn't exist as inflection points? Call Us Today: 312-210-2261. There is a third possibility. Use concavity and inflection points to explain how the sign of the second derivative affects the shape of a function’s graph. Mathematics Learning Centre, University of Sydney 1 The second derivative The second derivative, d2y dx2,ofthe function y = f(x)isthe derivative of dy dx. Even | {
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derivative The second derivative, d2y dx2,ofthe function y = f(x)isthe derivative of dy dx. Even the first derivative exists in certain points of inflection, the second derivative may not exist at these points. The second derivative can be used as an easier way of determining the nature of stationary points (whether they are maximum points, minimum points or points of inflection). Test Preparation. The second derivative is 4*e^2x - e^x. Not every zero value in this method will be an inflection point, so it is necessary to test values on either side of x = 0 to make sure that the sign of the second derivative actually does change. The following figure shows the graphs of f, Home; About; Services. Computing the first derivative of an expression helps you find local minima and maxima of that expression. Candidates for inflection points include points whose second derivatives are 0 or undefined. Solution To determine concavity, we need to find the second derivative f″(x). Inflection points can only occur when the second derivative is zero or undefined. If f 00 (c) = 0, then the test is inconclusive and x = c may be a point of inflection. Since it is an inflection point, shouldn't even the second derivative be zero? When x = ln 1/4, y = (1/4)^2 - 1/4 = 1/16 - 1/4 = -3/16. The second derivative is written d 2 y/dx 2, pronounced "dee two y by d x squared". Then, find the second derivative, or the derivative of the derivative, by differentiating again. exists but f ”(0) does not exist. Definition. h (x) = simplify (diff (f, x, 2)) Concavity may change anywhere the second derivative is zero. Recognizing inflection points of function from the graph of its second derivative ''. Let us consider a function f defined in the interval I and let $$c\in I$$.Let the function be twice differentiable at c. A stationary point on a curve occurs when dy/dx = 0. How to Calculate Income Elasticity of Demand. When we simplify our second derivative we get; 6x = 12. x = 2. f "(x) = 12x 2. | {
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Elasticity of Demand. When we simplify our second derivative we get; 6x = 12. x = 2. f "(x) = 12x 2. Definition by Derivatives. 0 0? The second derivative is never undefined, and the only root of the second derivative is x = 0. Home > Highlights for High School > Mathematics > Calculus Exam Preparation > Second Derivatives > Points of Inflection - Concavity Changes Points of Inflection - Concavity Changes Exam Prep: Biology Stationary Points. By using this website, you agree to our Cookie Policy. This results in the graph being concave up on the right side of the inflection point. There might just be a point of inflection. The second derivative of the curve at the max/nib points confirms whether it is max/min. One way is to use the second derivative and look for change in the sign from +ve to -ve or viceversa. cannot. Explain the relationship between a function and its first and second derivatives. The second derivative tells us if the slope increases or decreases. How to obtain maximums, minimums and inflection points with derivatives. AP® is a registered trademark of the College Board, which has not reviewed this resource. List all inflection points forf.Use a graphing utility to confirm your results. The second derivative is written d 2 y/dx 2, pronounced "dee two y by d x squared". Learn which common mistakes to avoid in the process. For there to be a point of inflection at (x 0, y 0), the function has to change concavity from concave up to concave … Let us consider a function f defined in the interval I and let $$c\in I$$.Let the function be twice differentiable at c. A function is said to be concave upward on an interval if f″(x) > 0 at each point in the interval and concave downward on an interval if f″(x) < 0 at each point in the interval. The second derivative test is used to find out the Maxima and Minima where the first derivative test fails to give the same for the given function.. Second Derivative Test To Find Maxima & Minima. In other words, the graph | {
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for the given function.. Second Derivative Test To Find Maxima & Minima. In other words, the graph gets steeper and steeper. Anyway, fun definitional question. If you're seeing this message, it means we're having trouble loading external resources on our website. Khan Academy is a 501(c)(3) nonprofit organization. This results in the graph being concave up on the right side of the inflection point. 2x = 0 . The second derivative can be used as an easier way of determining the nature of stationary points (whether they are maximum points, minimum points or points of inflection). Inflection points are where the function changes concavity. dy/dx = 2x = 0 . Taking y = x^2 . Free functions inflection points calculator - find functions inflection points step-by-step This website uses cookies to ensure you get the best experience. I just dont know how to do it. (c) Use the second derivative test to locate the points of inflection, and compare your answers with part (b). The Second Derivative Test cautions us that this may be the case since at f 00 (0) = 0 at x = 0. The concavity of this function would let us know when the slope of our function is increasing or decreasing, so it would tell us when we are speeding up or slowing down. The usual way to look for inflection points of f is to . Free functions inflection points calculator - find functions inflection points step-by-step This website uses cookies to ensure you get the best experience. Thanks @xdze2! If x >0, f”(x) > 0 ( concave upward. First we find the second derivative of the function, then we set it equal to 0 and solve for the inflection points: Since e^x is never 0, the only possible inflection point is where 4*e^x = 1, which is ln 1/4. We find the inflection by finding the second derivative of the curve’s function. Donate or volunteer today! Since concave up corresponds to a positive second derivative and concave down corresponds to a negative second derivative, then when the function changes from concave up to | {
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down corresponds to a negative second derivative, then when the function changes from concave up to concave down (or vise versa) the second derivative must equal zero at that point. Inflection point is a point on the function where the sign of second derivative changes (where concavity changes). (d) Identify the absolute minimum and maximum values of f on the interval [-2,4]. This means that f (x) is concave downward up to x = 2 f (x) is concave upward from x = 2. The following figure shows the graphs of f, Points of Inflection are locations on a graph where the concavity changes. We can define variance as a measure of how far …, Income elasticity of demand (IED) refers to the sensitivity of …. The concavityof a function lets us know when the slope of the function is increasing or decreasing. I'm very new to Matlab. Example 3, If x < 0, f”(x) < 0 ( concave downward. A function is said to be concave upward on an interval if f″(x) > 0 at each point in the interval and concave downward on an interval if f″(x) < 0 at each point in the interval. A common mistake is to ignore points whose second derivative are undefined, and miss a possible inflection point. It is not, however, true that when the derivative is zero we necessarily have a local maximum or minimum. dy dx is a function of x which describes the slope of the curve. If you're seeing this message, it means we're having trouble loading external resources on our website. dy dx is a function of x which describes the slope of the curve. Note: You have to be careful when the second derivative is zero. The second derivative has a very clear physical interpretation (as acceleration). Learn which common mistakes to avoid in the process. A point where the second derivative vanishes but does not change its sign is sometimes called a point of undulation or undulation point. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. If y = e^2x - e^x . Points of Inflection. Candidates for | {
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please enable JavaScript in your browser. If y = e^2x - e^x . Points of Inflection. Candidates for inflection points are where the second derivative is 0. In other words, the graph gets steeper and steeper. And for that, we don’t need smoothness, just continuity. And where the concavity switches from up to down or down … Second Derivatives: Finding Inflection Points Computing the second derivative lets you find inflection points of the expression. We observed that x = 0, and that there was neither a maximum nor minimum. Solution To determine concavity, we need to find the second derivative f″(x). The second derivative test uses that information to make assumptions about inflection points. The second derivative at an inflection point vanishes. The first derivative is f′(x)=3x2−12x+9, sothesecondderivativeisf″(x)=6x−12. A point of inflection or inflection point, abbreviated IP, is an x-value at which the concavity of the function changes.In other words, an IP is an x-value where the sign of the second derivative changes.It might also be how we'd describe Peter Brady's voice.. x = 0 , but is it a max/or min. The only critical point in town test can also be defined in terms of derivatives: Suppose f: ℝ → ℝ has two continuous derivatives, has a single critical point x 0 and the second derivative f′′ x 0 < 0. First Derivatives: Finding Local Minima and Maxima. Also, an inflection point is like a critical point except it isn't an extremum, correct? Please consider supporting us by disabling your ad blocker. Lets begin by finding our first derivative. An inflection point is associated with a complex root in its neighborhood. This results in the graph being concave up on the right side of the inflection point. Recall the graph f (x) = x 3. Lv 6. Limits: Functions with Suprema. Then find our second derivative. The usual way to look for inflection points of f is to . – pyPN Aug 28 '19 at 13:51 For example, the second derivative of the function $$y = 17$$ is always zero, but | {
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28 '19 at 13:51 For example, the second derivative of the function $$y = 17$$ is always zero, but the graph of this function is just a horizontal line, which never changes concavity. Inflection points are where the function changes concavity. Using the Second Derivatives. List all inflection points forf.Use a graphing utility to confirm your results. For ##x=-1## to be an *horizontal* inflection point, the first derivative ##y'## in ##-1## must be zero; and this gives the first condition: ##a=\frac{2}{3}b##. A critical point is a point on the graph where the function's rate of change is altered wither from increasing to decreasing or in some unpredictable fashion. An inflection point is a point on a curve at which a change in the direction of curvature occurs. A stationary point on a curve occurs when dy/dx = 0. And the inflection point is where it goes from concave upward to concave downward … A point of inflection or inflection point, abbreviated IP, is an x-value at which the concavity of the function changes.In other words, an IP is an x-value where the sign of the second derivative changes.It might also be how we'd describe Peter Brady's voice.. The concavity of a function r… y’ = 3x² – 12x. Here we have. In algebraic geometry an inflection point is defined slightly more generally, as a regular point where the tangent meets the curve to order at least 3, and an undulation point or hyperflex is defined as a point where the tangent meets the curve to order at least 4. then y' = e^2x 2 -e^x. The second derivative and points of inflection Jackie Nicholas c 2004 University of Sydney . Inflection points in differential geometry are the points of the curve where the curvature changes its sign.. For example, the graph of the differentiable function has an inflection point at (x, f(x)) if and only if its first derivative f' has an isolated extremum at x. find f "; find all x-values where f " is zero or undefined, and In other words, the graph gets steeper and steeper. If | {
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x-values where f " is zero or undefined, and In other words, the graph gets steeper and steeper. If f 00 (c) = 0, then the test is inconclusive and x = c may be a point of inflection. The second derivative and points of inflection Jackie Nicholas c 2004 University of Sydney . Explain the concavity test for a function over an open interval. Now, I believe I should "use" the second derivative to obtain the second condition to solve the two-variables-system, but how? y” = 6x -12. The first derivative is f '(x) = 4x 3 and the second derivative is. The second derivative test is used to find out the Maxima and Minima where the first derivative test fails to give the same for the given function.. Second Derivative Test To Find Maxima & Minima. Learn how the second derivative of a function is used in order to find the function's inflection points. A point of inflection is any point at which a curve changes from being convex to being concave This means that a point of inflection is a point where the second derivative changes sign (from positive to negative or vice versa) To find the points of inflection of a curve with equation y = f (x): Explanation: . Sometimes this can happen even if there's no point of inflection. F on the function is used in order to find the function concave. 0, f ” ( x ) = x 4 graph of its graph on selected intervals of that.! The interval [ -2,4 ] 12. x = 0, the second derivative of inflection! Lists of points of inflection physical interpretation ( as acceleration ) from +ve to or... Derivative are undefined, and solve the two-variables-system, but is it max/or. We observed that x = 0 so the second derivative has a very clear interpretation! ) ) Call us Today: 312-210-2261 maximum nor minimum by using website... Maximum or minimum negative, and compare your answers with part ( )! Function to find the second derivative f″ ( x ) = simplify ( diff f! ) nonprofit organization when x = ln 1/4, y = ( + ) 2 hence is. Your function to zero, and solve the | {
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) nonprofit organization when x = ln 1/4, y = ( + ) 2 hence is. Your function to zero, and solve the equation look for inflection points can only occur when the tells... Best experience, our inflection point if the curve a second derivative you have to an... Is f′ ( x ) =6x−12 to find the second derivative ( f, x, )! How the second derivative is zero is ln 1/4, y = +... Very clear physical interpretation ( as acceleration ) make sure that the domains *.kastatic.org and *.kasandbox.org unblocked. Line is horizontal Finding inflection points of inflection Jackie Nicholas c 2004 University of.. Confirms whether it is not the same as saying that f has an extremum, correct your with! Even if there 's no point of inflection -ve or viceversa derivative tells us the! Derivative we get e^x ( 4 * e^x = 1, which is ln,. Must equal zero to be an inflection point is positive graphs of f is to positive the! A maximum point the 2nd derivative is 4 * e^2x - e^x our! Curve looked like a hill, the second derivative is f ' ( x ) =6x−12 obtain the second to... External resources on our website is made possible by displaying online advertisements to our Cookie.!, if x < 0 ( concave upward on the right side of the inflection point is a point the. Recall the graph being concave up on the right side of the curve local minima and maxima of that.. Is f′ ( x ) > 0 ( concave upward 0, but is a! On a graph where the concavity changes and the minimum point is positive, the graph f ( x =... Inflection: algebraic point of inflection second derivative of Sydney slope of the function is concave.! We get ; 6x = 12. x = 2 the list of point of inflection second derivative that inflection! Negative, and that there was neither a maximum nor minimum x is an point... You find inflection points step-by-step this website uses cookies to ensure you get best... Step-By-Step this website, you agree to our Cookie Policy ) nonprofit organization means down. Aug 28 '19 at 13:51 I 'm very new to Matlab to provide a | {
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Policy ) nonprofit organization means down. Aug 28 '19 at 13:51 I 'm very new to Matlab to provide a free, world-class to. List of vertices that precede inflection points include points whose second derivative advertisements to visitors... Is positive, the graph of its graph on selected intervals include points whose derivative... Shape of its second derivative is x = 0, and the minimum.... Are the points of inflection Jackie Nicholas c 2004 University of Sydney ( c ) the. And inflection points include points whose second derivatives: Finding Inflation points of:... Are 0 or undefined at these points concave upward the list of that... Is at x is an inflection point on our website is made possible displaying... Absolute minimum and maximum values of f is to provide a free world-class... When the second derivative must equal zero to be an inflection point is positive '' ( x ) =.. If the curve increasing or decreasing root of the second derivative means that is... At these points current ( y ) in excel, sothesecondderivativeisf″ ( x ) = 3! There 's no point of inflection: algebraic start to look like U to find the second derivative you! Function r… points of inflection Jackie Nicholas c 2004 University of Sydney of that expression does, the at... Points can only occur when the derivative tells us whether the curve at the max/nib confirms! Never undefined, and website in this browser for the next time I comment x 4 there. Is ln 1/4, y = ( + ) 2 point of inflection second derivative it is max/min 're behind web. Some data about copper foil that are lists of points of f on the right of... That when the second derivative is zero as saying that f has an extremum ) point is like critical... Which has not reviewed this resource line is horizontal include points whose second derivatives: Finding points. Make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked and for that we... The relationship between a function may also be used to determine the general shape | {
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for that we... The relationship between a function may also be used to determine the general shape of its graph on intervals! Is associated with a complex root in its neighborhood test to locate the points of the College,... Website is made possible by displaying online advertisements to our Cookie Policy between inflection point, the graph f x... Concavity, we get ; 6x = 12. x = 2 happen even if there no! Curve is concave downward learn how the second derivative are undefined, and that there was a!, find the derivatives has a very clear physical interpretation ( as acceleration ) maximums, minimums and points. Derivative function changes concavity at that point vertices that precede inflection.! Test to locate a possible inflection point =3x2−12x+9, sothesecondderivativeisf″ ( x ) > 0 concave... The histogram itself ) max/nib points confirms whether it is n't an extremum, correct now, I I! To locate the points of inflection are where the sign from +ve to -ve or viceversa derivative to. Step-By-Step this website, you agree to our visitors physical interpretation ( acceleration... Sothesecondderivativeisf″ ( x ) = simplify ( diff ( f, x, 2 ) Call! Is a point on a curve ap® is a minimum point is a point of inflection from MATH MISC Manipal. X which describes the slope of the derivative tells us whether the curve concave! Equal zero to be careful when the second derivative and points of inflection get! Y ) in excel of an expression helps you find local minima and maxima of expression... And current ( y ) in excel complex root in its neighborhood maximums, and... Computing the second derivative of a function over an open interval recall the graph faster... That section is concave downward or concave upward disabling your ad blocker very... F ” ( 0, 0 ) is a 501 ( c ) use second... Save my name, email, and the minimum point is where 4 e^x. You get the best experience =3x2−12x+9, sothesecondderivativeisf″ ( x ) = simplify ( diff f. Not reviewed this resource ) use the | {
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sothesecondderivativeisf″ ( x ) = simplify ( diff f. Not reviewed this resource ) use the second derivative test to locate the inflection point is at x =,! The histogram itself ) certain points of inflection Jackie Nicholas c 2004 University of Sydney f′ ( x =. Not reviewed this resource points can only occur when the slope of the curve looked like a hill the. [ -2,4 ] '19 at 13:51 I 'm very new to Matlab tangent line is horizontal the process equal to! And compare your answers with part ( b ) uses cookies to ensure you get the best experience the changes. + ) 2 hence it is n't an extremum ) inflection are locations on a curve you! Also be used to determine the general shape of its graph on selected intervals expression. 'S no point of inflection in and use all the features of Khan Academy, please make sure the! You have to be an inflection point, however, ( 0, 0 ) is a point... Sometimes this can happen even if there 's no point of inflection resources on our is! Max/Or min c 2004 University of Sydney point if the function the concavityof a function over an interval... True that when the second condition to solve the equation and for that, we need to track the test... To down or down … list all inflection points of the College Board, which is ln,... The value at x = 0 cookies to ensure you get the best.! E^2X - e^x ap® is a minimum point the only possible inflection point, set the second is. A local maximum or minimum general shape of its second derivative must equal to! Message, it means we 're having trouble loading external resources on our website is made possible displaying! Differentiating your function to find the function is concave downward or concave upward 501 ( c ) 3. Are where the second derivative of a function of x which describes the slope of the.! ) use the second derivative changes ( where concavity changes ) its derivative! Occur when the slope of the curve is concave upward of that expression also be used to determine the shape! Advertisements to | {
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curve is concave upward of that expression also be used to determine the shape! Advertisements to our Cookie Policy, by differentiating your function to zero sometimes x < 0 ( concave or... Please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked minima and maxima of that expression for... Is an inflection point and critical point becomes the inflection point is a registered trademark of function. Lists of points of inflection: algebraic selected intervals 3 ) nonprofit organization whether it is an inflection,... | {
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# Finding a vector orthogonal to a subspace
Suppose you were given vectors $a_1,\dots,a_n \in \mathbb{R}^m$ then how would you compute some vector orthogonal to the given list of vectors? Note that you are allowed to return the zero vector only if the vectors span $\mathbb{R}^m$.
I thought about it for a while and the best I could do was to form the matrix with these vectors as rows and pick a vector from the nullspace. Is there an easier/faster way? Perhaps something geometric like Gram-Schmidt could be possible.
Thanks!
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To computer "some vector", you could just return $\vec{0}$ but I guess you want another one. – xavierm02 Mar 20 '13 at 19:16
Let $V=\operatorname{Vect}(\{a_1,\dots,a_n\})$ Let $(e_1,\dots,e_p)$ be an orthonormal basis of $V$ Then take any vector $x$ that is not in $V$ And then take $x-\sum\limits_{k=1}^p<e_p\mid x>e_p$ – xavierm02 Mar 20 '13 at 19:16
How do I pick a vector that is not in $V$? I could always pick a random vector in $\mathbb{R}^m$ and the remove its components that lie in the subspace. But is there a deterministic way of doing this? – anon Mar 21 '13 at 10:24
What is wrong with your finding x such that Ax = 0 idea? I see it as completely correct way of going about this. – TenaliRaman Mar 21 '13 at 10:29
You have a basis $(\varepsilon_1,\dots,\varepsilon_m)$ of $\Bbb R^m$, to pick a vector that is not in $V$, you can compute $\varepsilon_i-\sum\limits_{k=1}^p<e_p\mid \varepsilon_i>e_p$ until you get a non-zero result. – xavierm02 Mar 21 '13 at 12:18
Here is the deterministic algorithm.
Let $A$ be the $m \times n$ matrix of your vectors $$A = \pmatrix{a_0 & a_1 & \cdots & a_n}$$ Use the QR factorization of it $$A = QR$$ so that the Q matrix will contain the entire null space you are looking for: $$A = \pmatrix{Q_1 & Q_2}\pmatrix{R_1 \\ 0}$$ Since $Q$ is orthonormal $$\pmatrix{Q_1^\top \\ Q_2^\top}A = \pmatrix{R_1 \\ 0}$$ To completely specify the null space, you can see here that it is $Q_2$ $$Q_2^\top A = 0$$ | {
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Any deterministic algorithm will take as much computation as a row reduction, even to find a single vector in the null space. – adam W Mar 21 '13 at 16:33
This is a very good occasion to apply the fundamental theorem of linear algebra regarding the four fundamental subspaces of a matrix. (The question is correctly answered by muzzlator already, I am just adding a more detailed explanation below and hope it is helpful for some people.)
First, use $a_1,\dots,a_n \in \mathbb{R}^m$ to form an $m\times n$ matrix $A$, with each vector as a column of the matrix. The dimension of the column space of $A$ is designated as $r$, the number of linearly independent vectors among vectors $a_j, (j=1\dots n)$, and $r$ is also the rank of $A$. Then the problem of finding some vector orthogonal to $a_1,\dots, a_n$ is equivalent to finding the solution in the "left nullspace" of $A$, designated as $N(A^T)$, by solving the following equation:
$$A^T x = 0.$$
$A^T$ is $n\times m$, and the dimension of $N(A^T) = m-r$. (Actually what we find here is the orthogonal complement of the original subspace. This is much better than finding just some orthogonal vectors.) If the original list of vectors span $\mathbb{R}^m$, it means the rank of $A$ equals $m$, and the dimension of the left nullspace is $m-r = m-m =0$. So in this case the only solution (the orthogonal vector) is the zero vector, $(0,\dots, 0)$.
Gauss elimination (to get the echelon matrix) is the method to find the solution to $A^Tx=0$. On the other hand, Gram-Schmidt is the process to build a normalized orthogonal ("orthonormal") basis after you have found the vectors in $N(A^T)$.
In your post, you are correct to use the vectors as rows in a matrix, so you don't need to transpose the matrix to find the answer. (Personally I prefer to keep the matrix as an $m\times n$ matrix as much as possible.)
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There should indeed be a faster way. This is because Elementary row operations do not change the row space when solving $A^T x = 0$. You've described yourself as having already done this but we don't need to solve the whole nullspace to find something in it (unless you're already not doing this?).
Here is a thought, start with any non-zero vector and find a non-zero component. Eliminate that component in every other vector (theoretically). Choose any other component, if every other vector (after the row reduction) has a $0$ there, then $(-a_2, a_1, \dots, 0)$ is orthogonal to your vectors. Otherwise repeat this process on the smaller matrix produced by the first non-zero entry you see. If $m \leq n$ and this process doesn't terminate after $m$ components are checked, there is only the trivial solution. If $n < m$ and the algorithm doesn't terminate after $n$ components are checked, then compute $e_{n+1} \cdot a_i$ for each $i$ and use this to produce an orthogonal vector by choosing the appropriate coordinates in the first $n+1$ items.
This means a total of $\frac{\min\{m,n\}^2}{2} + m \max\{n-m, 0\}$ operations are more or less are all you need at worst.
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Hm let me just check that, just realised this process wouldnt work if they span the space as it never gives a $0$ answer – muzzlator Mar 21 '13 at 11:10
After a gazillion edits, I think I have it finally right. As expected, if $m >> n$, the problem is a lot easier. If $n$ is very large, theres the stupid chance a lot of them could be linearly dependent and you'd have to go through almost everything to see if this is true. – muzzlator Mar 21 '13 at 12:03
That being said, there's probably a random algorithm which solves this in far fewer steps and if it fails to produce an answer, there will be a sufficiently small probability that there actually was an orthogonal vector. – muzzlator Mar 21 '13 at 12:09 | {
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# Rolling a die twice in either order with not mutual exclusive events
I am reading the following problem:
If a single die is rolled twice, find the probability of rolling an odd number and a number greater than $$4$$ in any either order
My solution:
Probability of rolling an odd number = $$\frac{3}{6}$$
Probability of rolling a number greater than $$4$$ = $$\frac{2}{6}$$
Since the events are not mutually exclusive ($$5$$ is counted twice) the probabilty of throwing an odd number followed by a number greater than $$4$$:
$$\frac{3}{6}\cdot \frac{2}{6} - \frac{1}{36} = \frac{5}{36}$$
The probablity of throwing a number greater than $$4$$ followed by an odd number is: $$\frac{2}{6}\cdot \frac{3}{6} - \frac{1}{36} = \frac{5}{36}$$
Therefore the my final solution is $$\frac{5}{36} + \frac{5}{36} = \frac{10}{36}$$
But it is wrong, as the solution states that it should be $$\frac{11}{36}$$
What am I doing wrong here?
• What is 1/36, and why do you think it corrects for double counting? Mar 28, 2021 at 21:35
• @AleksejsFomins: The $5$ is odd and greater than $4$ so to avoid counting that twice I subtract this
– Jim
Mar 28, 2021 at 21:47
• As stated by @Karl below, the mistake seems to be in subtracting it from both orders Mar 28, 2021 at 21:50
• @AleksejsFomins: I thought that if events are not mutually independent we subtract the common event occurence. E.g. if the question was about the probability of odd followed by number greater than $4$, should I subtract the $1$ occurence of $5$? I apply this logic for the reverse order too
– Jim
Mar 28, 2021 at 21:54
The outcome potentially counted twice is $$(5,5)$$, not $$5$$.
The number of pairs $$(a,b)$$ where $$a$$ is odd and $$b>4$$ is $$3\times2=6$$. This includes $$(5,5)$$.
The number of pairs $$(a,b)$$ where $$a>4$$ and $$b$$ is odd is $$2\times3=6$$. This includes $$(5,5)$$.
The sum of these is $$6+6=12$$, but $$(5,5)$$ was counted twice; there are actually only $$11$$ pairs in the union of these two sets. | {
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• I thought that when having events that are not mutually exclusive we always subtract the $\bigcap$. So for the number of pairs (𝑎,𝑏) where 𝑎 is odd and 𝑏>4 is 3×2=6. I should not include (5,5)
– Jim
Mar 28, 2021 at 21:51
• I'd suggest just trying to think clearly about the outcomes you want to count instead of focusing on formulaic rules. If we wanted the probability of getting odd or $>4$ in one roll of the die, your approach would be applicable: $$|\{odd\}\cup\{>4\}|=|\{odd\}|+|\{>4\}|-|\{odd\}\cap\{>4\}|=|\{1,3,5\}|+|\{5,6\}|-|\{5\}|=3+2-1=4.$$ But for the given problem we instead have $$|\{(odd,>4)\}\cup\{(>4,odd)\}|=|\{(odd,>4)\}|+|\{(>4,odd)\}|-|\{(odd,>4)\}\cap\{(>4,odd)\}|=6+6-1=11.$$
– Karl
Mar 28, 2021 at 22:20
In saying that "5 is counted twice", meaning (presumably) you're removing the duplicate event $$(5, 5)$$, what you should be doing is saying "The event $$(5, 5)$$ is part of both of the cases I've considered, so I need to only count it once, so I will remove it once from my calculation."
Instead, what it looks like you've done is removed it from both of your cases, each of which assumes the other case has already counted it.
• Since an odd number and a number greater than $4$ are not mutually exclusive, my understanding is that since $5$ appears in both events it should be counted once. Hence if we focus on odd followed by a number greater than $4$ we should subtract $1$ case. Similar logic for the reverse order.
– Jim
Mar 28, 2021 at 21:52
• In rolling an odd number followed by a number greater than 4, rolling the 5 first isn't the same as rolling the 5 second so you don't remove it. Mar 29, 2021 at 21:45
• There are only 36 possibilities anyway, so try writing them out explicitly and counting them all? That might help you see what you missed. Mar 29, 2021 at 21:46 | {
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# Combinatorics question. Why doesn't this method work?
If I have the following set {$1,2,3,4$}, and I want to know the number of possible lists of size 3 that contain $1$ and $2$ in them, I tried the following, and think of it as the number of ways to select 3 elements:
$2·1·2=4$ as in "number of choices for the first number, then the second... " But the actual number is 12.
The only way I can do it is by observing that I have two sets {$1,2,3$} and {$1,2,4$} And each one has 6 permutations, and therefore 12 total permutations or lists.
How can I find the number of lists and subsets that satisfy the property of containing 1 and 2?
• Your first method works too : "total permutations" - ("permutations without 1" + "permutations without 2") $$4.3.2 - (3.2.1 + 3.2.1)$$ Sep 6 '14 at 10:08
• By list you mean n-tuple? A subset isnt a ordered thing, it completely different the case for a 3-tuple and a subset of cardinality 3. Can you repeat elements on the lists/3-tuple? Sep 6 '14 at 10:36
• Yeah, that's what I meant. But I want a robust method to find both lists and subsets given such criteria. Sep 6 '14 at 10:41
• Apparently you want neither $(1,2,1)$, nor $(2,6,1)$, but I could not tell the first and hardly the second from your description. Sep 6 '14 at 11:22
In fact, $2*1*2=4$ isn't that far away from your other solution. What you did is calculate the number of possibilities to have either $2$ or $1$ in the first place of your set and the other one on the second spot, with the remaining $3$ and $4$ taking the last one.
You did not, however, keep in mind, that $2$ and $1$ could also be in the second and third place or in the first and the third place. In the end, you have essentially three possibilities to form your set concerning placement of $2$ and $1$, and each possibility has four different ways of fulfilling the condition.
So you get: $3*2*1*2 = 12$ , which is the correct solution.
I hope that helped!
SDV | {
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So you get: $3*2*1*2 = 12$ , which is the correct solution.
I hope that helped!
SDV
Let's say that you are looking for the number of sets of $\left\{ 1,\dots,n\right\}$ of size $m$ that contain $\left\{ 1,\dots,k\right\}$.
This comes to choosing $m-k$ members of this set from set $\left\{ k+1,\dots,n\right\}$ so there are $$\binom{n-k}{m-k}$$ possibilities.
If it comes to lists then each permution on one of these sets induces a new list, so there are $$\binom{n-k}{m-k}m!$$ possibilities.
• But it doesn't work for $m=1$, containing {1}: (4-1)!/(1-1)!= 6 Sep 6 '14 at 11:02
• @Hex4869 It also workes for $m=1=k$. Note that $\binom{n-1}{0}=1$ for each $n\geq1$. Do not confuse it with $\frac{(n-1)!}{0!}$ Sep 6 '14 at 11:44 | {
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# Summation of expectation
Lets suppose we are in a casino, having two slot machines. Our strategy is to flip a coin to decide which machine to start on. (Without loss of generality, we say that we play the slot one first with probability $c_1$ and play the slot two first with probability $1-c_1$)
Then, we play on that machine until losing one game, then we switch to another slot machine to continue playing, we play a total n games (n is finite here).
Suppose the probability to win in slot one and slot two is i.i.d $Bernoulli(p_1)$ and $Bernoulli(p_2)$, respectively, and each win just gain 1 dollar, and each loss just get nothing.
Then, what should be the expected earning using this strategy???
I approached it by iteration (drawing a tree), hoping that there will be some pattern and if we find the expected winning in each turn, then we can sum the n rounds up because the trials are independent.
The expected winning in round 1 is: $$c_1\cdot p_1 + (1-c_1) \cdot p_2$$
The expected winning in round 2 is: $$c_1 \cdot[p_1^2+(1-p_1)\cdot p_2]+ (1-c_1)\cdot[p_2^2+(1-p_2)\cdot p_1]$$
The expected winning in round 3 is: $$c_1 \cdot \Big[p_1^3 + p_1(1-p_1) p_2 +(1-p_1)p_2^2 + (1-p_1)(1-p_2)p_1\Big] + (1-c_1) \Big[p_2^3 + p_2(1-p_2) \cdot p_1 + (1-p_2)p_1^2+(1-p_2)(1-p_1)p_2\Big]$$
I counted the expected winning up till round 4, which is: $$c_1 \cdot \Big[p_1^4 + p_1^2(1-p_1)p_2 + p_1(1-p_1) p_2^2 + p_1(1-p_1)(1-p_2)p_1 +(1-p_1)p_2^3 + (1-p_1)p_2(1-p_2)p_1 + (1-p_1)(1-p_2)p_1^2 + (1-p_1)^2(1-p_2)p_2\Big]$$ $$+ (1-c_1) \cdot \Big[p_2^4 + p_2^2(1-p_2)p_1 + p_2(1-p_2) p_1^2 + p_2(1-p_2)(1-p_1)p_2 +(1-p_2)p_1^3 + (1-p_2)p_1(1-p_1)p_2 + (1-p_2)(1-p_1)p_2^2 + (1-p_2)^2(1-p_1)p_1\Big]$$
This calculation seems endless so it does not work well. How could I find the expected winning in this case? | {
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• Oh it is the former, n is the number of times I pull a slot machine's lever~~ We switch only if we lose one game – son520804 Dec 5 '17 at 20:54
• Let $X_1$ be the number of winning pulls on machine 1 before a losing pull. So $X_1 - 1$ is your payoff if you were to play machine 1. What is the expectation of $X_1$? And do any of these formulas for sums of a series help you compute the expectation of $X_1$? – Matthew Gunn Dec 5 '17 at 22:36
• I could not easily compute the number of winning on each slot. We choose to play at slot 1 first by probability $c_1$ and play at slot 2 first by probability $c_2$. Then we play at the slot continuously, until lose 1 game and then play the another slot. (And the slot 1 and slot 2 each has winning probability $p_1$ and $p_2$) – son520804 Dec 5 '17 at 22:42
Consider $d$ machines numbered $1,2,\ldots, d$, each with chance $p(d)$ of winning. Let's find the expectation of beginning with machine $1$, playing until a loss, proceeding to machine $2$, etc, and cycling around from machine $d$ to machine $1$ if necessary until $n$ games have been played. The question concerns the case $d=2$, but the analysis (and computation) isn't any harder for larger $d$.
Let $e_j(k)$ denote the expected winnings from playing $k$ games beginning with machine $j$. Because each game is either a win (gaining $1$ dollar) with probability $p(j)$ or a loss with probability $1-p(j)$, leaving $k-1$ games to go, the rules of conditional expectation imply
$$e_j(k) = p(j)(e_j(k-1) + 1) + (1-p(j))(e_{(j \operatorname{mod} d)+1}(k-1))$$
when $k \ge 1$ and otherwise $e_j(k)=0$.
This is a simple dynamic program requiring $O(dn)$ computational effort and $O(dn)$ storage: in other words, it's a quick easy computation. You can even do it by hand.
As an example, here is an R implementation. The argument p is an array of probabilities $p(j)$. It returns a $d\times n+1$ array indexed by machine and number of games (from $0$ through $n$). | {
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expectation <- function(p, n=10) {
d <- length(p)
P <- matrix(0, d, n+1, dimnames=list(Machine=names(p), Plays=0:n))
if (n > 0)
for (i in 1:n) {
for (j in 1:d) {
P[j, i+1] <- p[j] * (P[j, i] + 1) + (1-p[j]) * P[j%%d + 1, i]
}
}
return(P)
}
Given an initial probability distribution $c(1), c(2),\ldots, c(d)$ for the choice of which machine to start with, the rules of conditional expectation state that the expected winnings will be the sum of $c(j)e_j(n)$. This answers the question.
Intuitively, machines that tend to fail will rarely be operated and those that tend to win are allowed to run. The $e_j(n)$ therefore will rapidly tend to a common value approximately equal to the $p$-weighted average of the $e_j(n)$ as $n$ grows, and the differences among the $e_j(n)$ will be no greater than the differences among the raw probabilities $p(j)$. Here are some examples for $d=2$.
The panel headings list the probabilities for machine "a" and machine "b" in order.
It's always good to check probability calculations. I carried out simulations as shown in this R program:
p <- c(a=0.5, b=0.2)
set.seed(17) # Comment this out when repeating simulations!
sim <- replicate(1e4, n - sum(cumsum(rgeom(n, 1-p) + 1) <= n))
It exploits the fact that the number of wins obtained from a machine has a geometric distribution, allowing the simulation to proceed fairly quickly. (The computation actually amounts to counting how many machines were tried in turn, rather than counting the wins directly.)
The output is an array of winnings, one for each iteration of the scenario. Because this dataset is large and not too skewed, we may compare its mean to the theoretical calculation using a Z-test.
m <- mean(sim)
s <- sd(sim)
mu <- expectation(p, n)[1,n+1]
c(Simulation=m, Calculation=mu, Z=(m - mu)/s * sqrt(length(sim)))
The output in this (reproducible) case is
Simulation Calculation Z
3.950 3.930 0.889 | {
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Simulation Calculation Z
3.950 3.930 0.889
It says the average among the (10,000) simulations was 3.95, the calculation of the expectation came out to 3.93, and the Z-score is 0.889 (which is not a significant difference). Repeated simulations for different $p$ and $n$ continue to agree with the calculations, providing assurance the calculations are correct. | {
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# Applying the vec operator
How to apply the $\operatorname{vec}$ operator in Mathematica? For example, how can I transform a $2 \times 2$ matrix into a $1 \times 4$ matrix as follows? $$\operatorname{vec}\left( \begin{bmatrix} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \end{bmatrix} \right) = \begin{bmatrix} a_{1,1} \\ a_{2,1} \\ a_{1,2} \\ a_{2,2} \end{bmatrix}$$
• vector and matrix are vague terms in Mathematica so please add input and expected output in terms of Mathematica code in order to make the question clear. – Kuba Feb 16 '17 at 10:31
• Thanks for the Accept. Please see the additional example of Flatten that I added afterward. – Mr.Wizard Feb 16 '17 at 13:59
Use Flatten to do this in a single operation.
in = Array[a, {2, 2}]
Flatten[in, {2, 1}]
{{a[1, 1], a[1, 2]}, {a[2, 1], a[2, 2]}}
{a[1, 1], a[2, 1], a[1, 2], a[2, 2]}
In Mathematica there are only vectors (lists), not column vectors and row vectors. However if you wish to convert a vector into an array with rows of length one for output the computationally fastest method is typically Partition:
Partition[{a[1, 1], a[2, 1], a[1, 2], a[2, 2]}, 1]
{{a[1, 1]}, {a[2, 1]}, {a[1, 2]}, {a[2, 2]}}
If that is your goal from the start you can also do that in a single operation using Flatten:
Flatten[{in}, {3, 2}] (* note the extra {} around in *)
{{a[1, 1]}, {a[2, 1]}, {a[1, 2]}, {a[2, 2]}}
• Unless OP sees column vector as: List /@ {a[1, 1], a[2, 1], a[1, 2], a[2, 2]} – Kuba Feb 16 '17 at 13:34
• @Kuba Good point. There is no such thing as a column vector in Mathematica as you know, but that doesn't mean the OP doesn't potentially want a series of one element rows... – Mr.Wizard Feb 16 '17 at 13:37
{{1, 2}, {4, 5}} // MatrixForm
\begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix}
ArrayReshape[Transpose[%], {4, 1}] // MatrixForm
\begin{bmatrix} 1 \\ 4\\ 2\\ 5 \end{bmatrix}
Thanks to @cyrille.piatecki for the use of Transpose[].
I propose this simple module | {
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Thanks to @cyrille.piatecki for the use of Transpose[].
I propose this simple module
vec[mat_] :=
Module[{a = mat},
ArrayReshape[Transpose[
a], {Dimensions[mat][[1]] Dimensions[mat][[2]], 1}]]
apply with
aa = Table[Subscript[a, i, j], {i, 1, 2}, {j, 1, 2}]
this gives the expected result
vec[aa] // MatrixForm | {
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# Does minimal ideal always imply principal ideal?
First, let me specify two definitions i will use.
$$[1.]$$ A (right/ left/ both) ideal $$I$$ of a ring $$R$$ (unity not assumed) is minimal if $$(1.) \; I\neq (0)$$ and $$(2.)$$ If $$J$$ is any nonzero (right/ left/ both) ideal of $$R$$ containied in $$I$$, then $$J=I$$
$$[2.]$$ If $$x \in R$$, then $$(x)$$ is the intersection of all (left/ right/ both) ideals of $$R$$ containing $$x$$.
Consider the following propostition and its proof:
$$\textbf{A [right / left / both] ideal I of a ring R is minimal iff}$$
$$\textbf{ I is generated by any of its nonzero elements x \in I }$$
Proof:
$$1.(\Rightarrow)$$ Suppose $$I$$ is minimal and and $$x \in I$$ is nonzero. Consider the ideal $$J:=(x)$$ generated by $$x$$. By construction, $$J \neq (0)$$ since $$x \in J$$. Now, $$J \subseteq I$$, since by definition $$J$$ is the smallest ideal containing $$x$$. But then , by minimality of $$I$$ we must have $$I=J$$, so $$I$$ is generated by $$x$$.
$$2.(\Leftarrow)$$ Suppose $$I=(x)$$ for any $$x \in I$$, and that $$J$$ is any nonzero ideal of $$R$$ with $$J \subseteq I$$. Let $$y$$ be any nonzero element of $$J$$. Then $$y \in I$$, and by hypothesis we have $$I=(y)$$. But then we must have $$J=I$$, because $$(y)$$ is the smallest ideal of $$R$$ containing $$y$$.
My question is:
$$\textbf{Does this also prove that any minimal ideal is principal? }$$ | {
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My question is:
$$\textbf{Does this also prove that any minimal ideal is principal? }$$
• (2) is not correct. That $I$ is principal does not mean $I = (y)$ for every $y \in I$, it means $I = (y)$ from some $y \in I$.
– Jim
Oct 6, 2019 at 13:36
• (1) is correct though, all minimal ideals are principal and for minimal ideals it actually is true that $I = (y)$ for every $y \in I$. That condition is actually equivalent to the principal ideal being minimal.
– Jim
Oct 6, 2019 at 13:37
• Then you have read incorrectly. I'm not saying that I is principal. I'm saying that I is an ideal which has the property that it is generated by any of its elements. Oct 6, 2019 at 13:37
• Oh, actually you're right! I did read what you were proving incorrectly, my bad.
– Jim
Oct 6, 2019 at 13:39
• In that case both (1) and (2) are correct.
– Jim
Oct 6, 2019 at 13:39
Yes essentially, although I think it is safest to word it as “a left (resp, right/twosided) ideal is minimal if and only if it is generated by any of its nonzero elements as a left (resp, right/twosided) ideal.”
The fact that minimals are generated by a single element follows a fortiori from the $$\implies$$ direction.
Let $$I$$ be a minimal right ideal of a ring $$R$$. By definition, $$I$$ being a principal right ideal means that $$\exists x \in R \, (I=xR)$$.
In fact, $$\forall x \in I \setminus \{0\} \, (I=xR)$$. Indeed, for any nonzero element $$x$$ of $$I$$, $$xR \subseteq I$$ (because $$I$$ is a right ideal of $$R$$) and $$0 \neq x \in xR$$, so $$xR=I$$ because $$I$$ is assumed to be a minimal right ideal.
Similarly, if $$I$$ is a minimal left (resp. two-sided) ideal of $$R$$, then $$\forall x \in I \setminus \{0\} \, (I=Rx)$$ (resp. $$\forall x \in I \setminus \{0\} \, (I=RxR)$$).
More generally, any simple module is cyclic. Two-sided ideals of $$R$$ are the same as the right $$R^{op} \otimes_{\mathbb{Z}} R$$-submodules of $$R$$. | {
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I agree with the former part of your proof, but I do not with the latter part. (I missed the word ANY.)
To be specific, consider $$R = \mathbb Z$$ and an ideal $$I = (n)$$ with $$n \neq 0$$. Obviously, it contains a non-zero proper ideal $$(n) \supset (2n) \supset (0).$$ Hence, a principal ideal generated by a non-zero element needs not to be minimal. (Indeed, this argument shows that $$\mathbb Z$$ has no minimal ideals.)
• Yes, it does. (You've proved it, right?)
– Orat
Oct 6, 2019 at 13:42
• Well, I'm not sure, because I can't find the statement mentioned on the web. On the wikipedia page of minimal ideal, en.wikipedia.org/wiki/Minimal_ideal , they mention briefly that for a ring R with unity, this in neccesarily true for right ideals .. . Oct 6, 2019 at 13:47
• Well, what you've done is basically the same as the argument shown on that wikipedia page. Depending on what type of ideals you're considering (right/left/two-sided), considering an ideal of the form ($xR$/$Rx$/$RxR$) is the key, anyway. BTW, many people may reserve the symbol $(x)$ to denote $RxR$ only.
– Orat
Oct 6, 2019 at 13:53
• Well, the fact that I=xR / I=Rx / I=RxR for principal ideals is only the case if R has unity, which isn't assumed... Oct 6, 2019 at 13:56
• You're right; as I usually deal with unital rings, I didn't pay much attention to that. Anyway the former part of your proof is correct as it only uses the minimality, and does not use something like $xR$.
– Orat
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# Calculating $e$
If I calculate $e$ using the following formula.
$$e = \sum_{k=0}^{\infty}{\frac{1}{k!}}$$
Is it possible to predict how many correct decimal places I get when I stop summing at $n$ terms?
-
There is a subtlety in the contrast between "correct decimal places" and accuracy. Most of us had interpreted "$n$ correct decimal places as within $10^{-n}$, but as Dan Brumleve points out, you could be very close. If the correct answer is $1.9999$, an error of $10^{-4}$ can change the ones digit. Having a string of $9$'s is rare, but if you care about it you need to check. – Ross Millikan Nov 10 '12 at 4:37
If we use $n$ terms, the last term used is $\dfrac{1}{(n-1)!}$. The missing "tail" is therefore $$\frac{1}{n!}+\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\frac{1}{(n+3)!}\cdots.\tag{1}$$ Note that $(n+1)!=n!(n+1)$ and $(n+2)!\gt n!(n+1)^2$, and $(n+3)!\gt n!(n+1)^3$ and so on. So our tail $(1)$ is less than $$\frac{1}{n!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\cdots \right).$$ Summing the geometric series, we find that the approximation error is less than $$\frac{1}{n!}\left(1+\frac{1}{n}\right).$$ | {
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-
This is a good answer, why was it downvoted? – robjohn Nov 9 '12 at 19:45
To end the error vs correct digits discussion: If you calculate $e_n:=\sum_{k=0}^n\frac1{n!}$ and $e_n+\frac1{n!}(1+\frac1n)$ and both agree on $d$ decimals, then both estimates are correct to (at least) $d$ decimals. – Hagen von Eitzen Nov 9 '12 at 20:39
@DanBrumleve How about instead of just downvoting (something that can't be undone after a few minutes) 10 (a slight exaggeration) different answers without a single comment (until after someone asked why), leave a comment and clarify. Then, if it turns out you misunderstood, you haven't just downvoted 10 different answers incorrectly. – Graphth Nov 9 '12 at 22:41
@DanBrumleve Well, obviously many other people (everyone who answered?) believe otherwise. So, is it possible that at the very least it's just your preference? And, if so, is it really worth downvoting? – Graphth Nov 9 '12 at 23:31
@DanBrumleve While this is more of a meta issue, IMHO there is a difference between 'this is not how I would have answered the question' or even 'this is not what I consider a complete answer to the question' and 'this is an unhelpful answer to the question' - I would consider the latter to be a minimal standard for downvoting (as opposed to simply not voting), and the FAQ seems to agree: 'Use your downvotes whenever you encounter an egregiously sloppy, no-effort-expended post, or an answer that is clearly and perhaps dangerously incorrect.' This answer is hardly egregious or dangerous. – Steven Stadnicki Nov 10 '12 at 1:08
You can use the remainder term in Taylor's expansion | {
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You can use the remainder term in Taylor's expansion
-
To what function are we applying Taylor's theorem? – robjohn Nov 9 '12 at 19:49
@robjohn the definition of $e$ given by OP is actually the Taylor series for $e^x$ centered at $x=0$, evaluated at $x=1$. – Logan Stokols Nov 9 '12 at 23:48
@Logan: As I mentioned in a comment to glebovg, since $f$ was not specified in the question, it would be good to mention it in the answer. – robjohn Nov 10 '12 at 0:04
In this answer, it is shown, by comparison to a geometric series, that $$0\le n!\left(e-\sum_{k=0}^n\frac1{k!}\right)\le\frac1n$$ Therefore, the error after $n+1$ terms is at most $\frac1{nn!}$ .
To $n$ decimal places:
When asking for a number to $n$ decimal places, there are two common meanings
1. the error is less than $\frac12\times10^{-n}$.
2. the value is correct when rounded to $n$ decimal places. As has been pointed out, if a number is very close to $10^{-n}\left(\mathbb{Z}+\frac12\right)$, rounding to $n$ decimal places might require computing more decimal places to know the actual $n^{\mathrm{th}}$ digit of the rounded number. This is not as easy to use as meaning 1, so it is not as commonly used.
-
That was a fast downvote. Care to comment? – robjohn Nov 9 '12 at 20:21
Is someone just downvoting for free? – Pedro Tamaroff Nov 10 '12 at 0:24
@PeterTamaroff: nah... it still costs 1 rep to downvote an answer (afaik). – robjohn Nov 10 '12 at 0:26
The series converges rapidly. If you stop at $\frac 1{ k!}$ you can bound the error by $\frac 1{k(k!)}$ by bounding the remaining terms with a geometric series.
-
Why was this downvoted? It seems a bit scant, but valid. – robjohn Nov 9 '12 at 19:43
I downvoted for the same reason as most of the others: the question asks about the number of correct leading digits in the partial sum and this upper bound on the error term doesn't lead to any obvious answer to that question. – Dan Brumleve Nov 10 '12 at 4:29 | {
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The $n$-th Taylor polynomial is $${P_n}(x) = f(0) + \frac{{f'(0)}}{{1!}}x + \frac{{f''(0)}}{{2!}}{x^2} + \cdots + \frac{{{f^{(n)}}(0)}}{{n!}}{x^n}$$ (in this case $f(x)$ is simply $e$) and the error we incur in approximating the value of $f(x)$ by $n$-th Taylor polynomial is exactly $$f(x) - {P_n}(x) + \frac{{{f^{(n + 1)}}(c)}}{{(n + 1)!}}{x^{n + 1}}$$ where $0 < c < x$. This form of the remainder can be used to find an upper bound on the error. If the difference above is positive, then the approximation is too low, and likewise if the error is negative, then the approximation is too high. We only need to find an appropriate $c$.
-
To what function are we applying Taylor's theorem? – robjohn Nov 9 '12 at 19:50
@robjohn In this case $e^x$ for $x = 1$, but it works in general. – glebovg Nov 9 '12 at 20:17
Why downvote? Please comment. – glebovg Nov 9 '12 at 20:17
I downvoted because it doesn't address the distinction between the error term and the number of correct digits in the partial sum. (Also consider simplifying or clarifying: the question asks only about $e$ not $e^x$.) – Dan Brumleve Nov 9 '12 at 20:21
@glebovg: Since $f$ was not specified in the question, it would be good to mention it in the answer. – robjohn Nov 9 '12 at 22:49 | {
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# What distribution models number of trials needed for given number of successes and success rate?
Case scenario: a retro-virus infects a healthy cell. The virus programs the cell to brew little viruses, at a rate of 0.5 per-sec, until finally the cell bursts when the number of virus inside it is 5. How to model this?
In Binomial, the random variable represents the number of successful trials obtained when throwing a coin a certain number of trials, at a certain probability of success per trial.
I want a distribution whose random variable is the number of trials (coin tosses) that were necessary to perform, given a certain number of successful trials and a certain probability per trial.
I am not even sure how I would write down the probability mass function.
Is there such a distribution? Nothing rings a bell here: https://en.wikipedia.org/wiki/Category:Discrete_distributions
There are related questions to this one --- such as this one: How many trials until I each my desired outcome --- but no-one mentioned a distribution, or if any exists.
Just to make it clear, the random generator for a random variable of such a distribution would look like this in R:
rmy <- function(s, p) {
i <- n <- 0
while(i != s) {
i <- i+rbinom(1,1,p)
n <- n+1
}
n
}
Thank you ! ps: sorry if the text was a little flowery, but it helps me think, since I am a junior mathematician hehe. | {
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• Let me see if I understand the process. Each second I flip a (fair) coin to see if I produce a virus. Since I have a $1/2$ probability of producing a single new virus, the average virus production rate is $1/2$. I stop when I produce five viruses. And you want to know the expected number of coin flips, is that right? – Brian Tung Apr 14 '15 at 22:23
• Have you looked at Poisson Distribution? – rightskewed Apr 14 '15 at 22:46
• @BrianTung, I want to know if there is a theoretical distribution. Expected value is easy, it is E=m/p=5/0.5=10, in this case, where m is number of successful trials and p is probability of success. – rpmcruz Apr 15 '15 at 8:05
• @rightskewed, hmm what do you have in mind exactly? Using $\lambda=m/p$; testing in R rpois(10, 5/0.5) shows it sampling values below <5, which is impossible in this model. – rpmcruz Apr 15 '15 at 8:08
• Just to make it clear: I want to know if there is already a popular theoretical distribution for what I want. If anyone works to work out the probability mass function that would be superb too. :P Might make sense to migrate it to: stats.stackexchange.com . I did not remember about the statistics forum ... – rpmcruz Apr 15 '15 at 8:12
I figure this one out. :)
I can model it using a Negative Binomial: https://en.wikipedia.org/wiki/Negative_binomial_distribution
First, let us change the values of my case scenario, just to make it clearer. "Case scenario: a retro-virus infects a healthy cell. The virus programs the cell to brew little viruses, at a rate of 0.2 per-sec, until finally the cell bursts when the number of virus inside it is 5. How to model this?" | {
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We can model number of failures $Y$ as $Y\sim\mathcal{NB}(5,0.2)$. That answers the question, how many failed trails do we have, when we need 5 successful at a probability rate of 0.2. But we do not want failed trials, we want total trials, and total trials = failed trials + successful trials. We know successful trials, which is 5, so our random variable $X$ is such that $X\sim5+\mathcal{NB}(5,0.2)$.
In fact, comparing the random generator function I proposed in the question with the negative binomial random generator (with this adjustment):
par(mfrow=c(1,2))
hist(sapply(1:1e5, function(x) rmy(5, 0.2)))
hist(5+rnbinom(1e5, 5, 0.2))
All functions mean, sd and summary are consistent as well.
My 2 cents here :
Using a Negative Binomial distribution can be a very good approximation. However :
1) You totally remove the case r=0 (No failures).
2) Also, you're not answering your initial question. You're computing the distribution of the #trials before achieving the known #fails (or #success). It's different than #trials knowing the #fails (or #success), as you can have fails following the last success !
I believe the right answer involves Bayesian probabilities :
• We consider that N is a random variable describing the # of trials (# of Bernoulli trials). Let's assume that N is following an Exponential Distribution $$N \thicksim Exp(\lambda)$$
• We then consider that X is a random variable describing the number of successes after N=n trials, following a Binomial distribution $$N \thicksim Binom(p, N)$$
Let's compute $$P(n|k)$$ thanks to the Bayes formula :
$$\frac{1}{P(N=n|X=k)} = \frac{P(X=k)}{P(X=k|N=n)P(N=n)}$$
And by writing $$P(k) = \sum_{m=k}^{\infty}{P(X=k|N=m)P(N=m)}$$
$$= \sum_{m=k}^{\infty}\frac{P(X=k|N=m)P(N=m)}{P(X=k|N=n)P(N=n)}$$
And because $$P(X=k|N=n)$$ and $$P(X=k|N=m)$$ are just Binomial distributions : | {
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And because $$P(X=k|N=n)$$ and $$P(X=k|N=m)$$ are just Binomial distributions :
$$= \sum_{m=k}^{\infty}\frac{m!k!(n-k)!}{n!k!(m-k)!} \frac{p^k}{p^k} \frac{(1-p)^{m-k}}{(1-p)^{n-k}} \frac{e^{-\lambda.m}}{e^{-\lambda.n}}$$
$$= \sum_{m=k}^{\infty}{[e^{-\lambda}(1-p)]^{m-n}}$$
And by re-indexing the Serie with $$l=m-k <=> m=l+k$$
$$= \sum_{l=0}^{\infty}{[e^{-\lambda}(1-p)]^{l+k-n}}$$
$$\frac{1}{P(N=n|X=k)} = \frac{[e^{-\lambda}(1-p)]^{k-n}}{1 - e^{-\lambda}(1-p)}$$
$$P(N=n|X=k) = \frac{1 - e^{-\lambda}(1-p)}{[e^{-\lambda}(1-p)]^{k-n}}$$
And now you have a probability $$P>0$$ for $$k=0$$ ;-)
Edit: The choice of the prior (Exponential) is totally arbitrary and leads to a lot of simplifications. If anyone wants to attempt the calculation with another prior, feel free :-) | {
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# Why are $X \sim U(-1,1)$ and $Y=X^2$ dependent?
Suppose we have two continuous random variables $$X \sim U(-1,1)$$ and $$Y=X^2$$. I don't understand why they are dependent.
$$E[X] = 0$$ $$E[Y] = \int_{-1}^{1} x^2 dx = 2/3$$ $$E[XY] = \int_{-1}^{1} x^3 dx = 0$$
They are independent to me because $$E[XY]=E[X]E[Y]$$, so why they are dependent?
• Plotting simulated values of $X$ and $Y$ will be enlightening, but you can also work out the density of $Y$ via a change of variables from $X$ in order to compare the probabilities in terms of definition of statistical independence. Feb 4 at 4:15
• I find it challenging to find any function $h$ for which $X$ and $h(X)$ are independent.
– whuber
Feb 4 at 15:55
• @whuber: one such function $h$ is $h(x)=0$. Another one is $h(x)=5$. Feb 4 at 17:25
• @Kjetil Certainly. Not very interesting, are they? ;-)
– whuber
Feb 4 at 18:00
• For specific distributions more interesting functions can be possible. For instance, $Y = \sin(2\pi X)$ is independent from $X$ if the domain of $X$ is integer values only. But yes, it will still be the not so interesting map where all values in the domain of $X$ need to be mapped to a single point. It is because $f(X)$ given $X=x$ is a singular distribution in the point $f(x)$. Independence requires that the distribution of $f(X)$ is the same distribution for every value of $X$, $f(x)$ must be a single value for every $x$. Feb 4 at 20:55
Let me write this in a generic manner:
Two random variables for which the expectation of the product is equal to the product of expectations need not be independent.
Let $$X\sim \mathcal U(-a, a)$$ where $$a\in\mathbb R_+$$ and let $$Y\sim X^2.$$ Then as noted $$\mathbb E[XY]=\mathbb E[X^3] =0=\mathbb E[X]\mathbb E[Y].$$ But they are definitely not independent.
## Reference:
$$\rm [I]$$ Counterexamples in Probability and Statistics, Joseph P. Romano, Andrew F. Siegel, Wadsworth, $$1986,$$ ex. $$4.15.$$ | {
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Since $$Y = X^2$$, $$[-1/2 \leq X \leq 0] \cap [Y > 1/4] = \varnothing$$, hence $$P[-1/2 \leq X \leq 0, Y > 1/4] = 0$$. On the other hand, $$P[-1/2 \leq X \leq 0] = 1/4 > 0, P[Y > 1/4] = 1/2 > 0$$. Hence the independence defining relation $$P(A \cap B) = P(A)P(B)$$ for all $$A \in \sigma(X), B \in \sigma(Y)$$ fails to hold for $$A = [-1/2 \leq X \leq 0]$$ and $$B = [Y > 1/4]$$, i.e., $$X$$ and $$Y$$ are not independent.
Above is a formal proof. Intuitively, since $$Y$$ is a deterministic function of $$X$$, knowing the value of $$X$$ means knowing the value of $$Y$$, hence $$Y$$ and $$X$$ of course cannot be independent (the heuristic definition of independence between $$X$$ and $$Y$$ requires that observing the information provided by $$X$$ does not increase the information of $$Y$$).
It is worth emphasizing that $$E[XY] = E[X]E[Y]$$ is a necessary condition, rather than a sufficient condition for the independence of $$X$$ and $$Y$$. If you really want to express the independence of $$X$$ and $$Y$$ in terms of expectations, it should be stated as
$$X$$ and $$Y$$ are independent $$\iff$$ $$E[f(X)g(Y)] = E[f(X)]E[g(Y)]$$ for all Borel measurable functions $$f$$ and $$g$$.
Evidently, the condition $$E[f(X)g(Y)] = E[f(X)]E[g(Y)]$$ is much stronger than the condition $$E[XY] = E[X]E[Y]$$. The former is a system of infinitely many equations, while the latter is a single equation.
You need to distinguish between the concepts of uncorrelatedness, which involves the decomposition of the expectation of a product, and independence, which involves the decomposition of a joint probability distribution. What you're calling independence is actually uncorrelatedness. What happens, as the other answers show, is that independence is a stronger condition than uncorrelatedness, that is, independence implies uncorrelatedness but the vice versa is not true.
And your example is exactly one example that is used to show that uncorrelatedness does not imply independence. | {
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They are clearly not independent: if you know the value $$X$$ then this gives you information about the value of $$Y$$ and indeed allows you to calculate it precisely.
You have correctly said independence means $$P(X\le a, Y\le b)=P(X\le a) P(Y\le b)$$, so for example consider $$a=\frac13$$ and $$b=\frac 14$$.
• $$X \le \frac 13$$, i.e. when $$-1 \le X \le \frac13$$, has probability $$\frac23$$
• $$Y \le \frac 14$$, i.e. when $$-\frac12 \le X \le \frac12$$, has probability $$\frac12$$
• $$X \le \frac 13, Y \le \frac 14$$ together, i.e. when $$-\frac12 \le X \le \frac13$$, has probability $$\frac5{12}$$
but $$\frac23 \times \frac12 \not=\frac5{12}$$ so they are not independent. Try other examples with $$-1 < a < 1$$ and $$0 < b < 1$$.
$$E[XY] = E[X]E[Y]$$ is a condition for zero covariance and so zero correlation, but this on its own does not imply independence. The bottom row of Wikipedia's chart illustrating correaltion has other examples of this; in your question you would get a parabolic U-shaped chart.
• Note that zero covariance is sufficient for independence if $X$ and $Y$ are jointly Gaussian, but not necessarily (or maybe ever) for other distributions. That it works for multivariate Gaussian distributions might be the source of the confusion.
– Dave
Feb 4 at 14:03
Here is a simulation of how the joint distribution looks like:
It shows clearly that the two variables are not independent.
Your formula is more generally (including dependent variables).
$$\text{E}[XY] = \text{E}[X]\text{E}[Y] + \text{COV}(X,Y)$$
So when you compute $$\text{E}[XY] = \text{E}[X]\text{E}[Y]$$ then it must mean that $$\text{COV}(X,Y) = 0$$. But that doesn't imply independence. This makes your case an example of: Why zero correlation does not necessarily imply independence | {
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• It is not clear where you got the idea for the product $\text{E}[XY] = \text{E}[X]\text{E}[Y]$. Possibly you are confusing it for a description of independence as $P(A \text{ and } B) = P(A) \cdot P(B)$ Feb 4 at 19:27 | {
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# Component of a vector along another vector.
Tags:
1. Oct 22, 2014
### Hijaz Aslam
1. The problem statement, all variables and given/known data
Given $\vec{A}=2\hat{i}+3\hat{j}$ and $\vec{B}=\hat{i}+\hat{j}$.Find the component of $\vec{A}$ along $\vec{B}$.
2. Relevant equations
$\vec{A}.\vec{B}=ABcosθ$ where θ is the angle between both the vectors.
3. The attempt at a solution
I attempted the question as follows:
Let the angle between $\vec{A}$ and $\vec{B}$ be 'θ'. So the component of $\vec{A}$ along $\vec{B}$ is given by $Acosθ\hat{B}$ => $Acosθ(\frac{\vec{B}}{B})$
As $\vec{A}.\vec{B}=ABcosθ$ => $[( 2\hat{i}+3\hat{j})(\hat{i}+\hat{j})]/B=Acosθ$ => $\frac{5}{\sqrt{2}}=Acosθ$
Therefore the component is : $\frac{5}{\sqrt{2}}(\frac{\hat{i}+\hat{j}}{\sqrt{2}})$ => $\frac{5}{2}({\hat{i}+\hat{j}})$
But my text produces the solution as follows:
$A_B=(\vec{A}.\vec{B})\hat{B}=\frac{5}{\sqrt{2}}(\hat{i}+\hat{j})$.
2. Oct 22, 2014
### RUber
I usually see this process broken down into basis components.
That is $\hat B =\sqrt{2}/2 \hat i + \sqrt{2}/2 \hat j.$
Then the component is $A\cdot \hat B_i \hat i + A\cdot \hat B_j \hat j$.
Somewhere in your process, you divided by the magnitude of B twice.
3. Oct 23, 2014
### willem2
You're right and the book is wrong. The book answer as well as the formula for AB they use.
The length of your answer is smaller than the length of A as it should be. The book answer is larger.
The projection of A on B should only depend on the direction of B, not the magnitude. The formula used for AB in the book does depend on the magnitude of B.
4. Oct 23, 2014
### Delta²
Book is wrong . We can verify this by standard euclidean geometry easily because by the definition of cosine, it will be cos(θ)=(component of A along B)/A hence Acos(θ)=(component of A along B). And we have to multiply this by the unit vector of B to get the required result. | {
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# Different results with quotient rule
1. May 26, 2013
### mike82140
I have been trying to figure out the derivative of (X²-1)/X. When I use the quotient rule, the result I get is 1-1/X². However, when I simplify the expression first, then take the derivative, I get 1+1/X²
Why are the results different?
2. May 26, 2013
### Hypersphere
3. May 26, 2013
### mike82140
Can you show how you evaluate it? I tried both ways, but get different results.
4. May 26, 2013
### Hypersphere
Alright, for a function
$$f(x) = \frac{g(x)}{h(x)}$$
the quotient rule says
$$f'(x) = \frac{g'(x)h(x)-g(x)h'(x)}{h^2(x)}$$
In your case, $g(x)=x^2-1$ and $h(x)=x$. Thus $g'(x)=2x$ and $h'(x)=1$. Then you get
$$\frac{\mathrm{d}}{\mathrm{dx}}\frac{x^2-1}{x}=\frac{2x\cdot x - \left( x^2 -1 \right) \cdot 1}{x^2}=\frac{2x^2-x^2+1}{x^2}=1+\frac{1}{x^2}$$
Are you sure you included the $g'(x)$ term?
Simplifying first,
$$\frac{\mathrm{d}}{\mathrm{dx}}\frac{x^2-1}{x} = \frac{\mathrm{d}}{\mathrm{dx}} \left( x - \frac{1}{x} \right) = 1- \frac{-1}{x^2}=1+\frac{1}{x^2}$$
5. May 26, 2013
### mike82140
Where did +1 come from when you were multiplying x²-1 with 1?
6. May 26, 2013
### Hypersphere
It was also multiplied by -1, from the quotient rule formula.
7. May 26, 2013
### mike82140
I'm sorry for asking so many questions, and this may sound stupid, but, what -1?
Edit: It appears as though I have made a mistake. The -1 is the minus part that is in front of the x²-1, so the negative, or minus, distributes itself.
Thank you for the help, I appreciate it.
8. May 26, 2013
### Hypersphere
You see the minus sign in front of $\left( x^2 -1 \right) \cdot 1$, right? That is just short notation for
$$- \left( x^2 -1 \right) \cdot 1=\left( -1 \right) \cdot \left( x^2 -1 \right)= (-1) \cdot x^2 + (-1) \cdot (-1) = -x^2+1$$
Agreed? Or are you pulling my leg here?
EDIT: Ah, good. I was almost giving up for a while there.
9. May 26, 2013
### SteamKing | {
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EDIT: Ah, good. I was almost giving up for a while there.
9. May 26, 2013
### SteamKing
Staff Emeritus
Look at the formula for the quotient rule.
If h(x) = x, what is h'(x)? | {
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