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https://math.stackexchange.com/questions/617407/evaluate-the-limit-lim-limits-n-to-infty-frac11n2-frac22n2 | # Evaluate the limit $\lim\limits_{n \to \infty} \frac{1}{1+n^2} +\frac{2}{2+n^2}+ \ldots +\frac{n}{n+n^2}$
Evaluate the limit
$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$
My approach :
If I divide numerator and denominator by $n^2$ I get :
$$\lim_{ n \to \infty} \dfrac{\frac{1}{n^2}}{\frac{1}{n^2} +1} +\dfrac{\frac{2}{n^2}}{\frac{2}{n^2} +1} + \ldots+ \dfrac{\frac{1}{n}}{\frac{1}{n} + 1}=0$$
but the answer is $\dfrac{1}{2}$ please suggest how to solve this.. thanks.
• Do you familiar with Riemann's sums? – Salech Rubenstein Dec 24 '13 at 16:15
• And what is $\lim_{n\to\infty}\frac{1}{n}+\frac{1}{n}+\cdots+ \frac{1}{n}$, where $\frac{1}{n}$ appears $n$ times? – Salech Rubenstein Dec 24 '13 at 16:17
For each $1\leq i\leq n$, $\frac{1}{n^2+i}\leq \frac{1}{n^2}$ and $\frac{1}{n^2+i}\geq \frac{1}{n^2+n}$, and so we may bound the sum from above and below by $$\sum_{i=1}^{n}\frac{i}{n+n^{2}}\leq\sum_{i=1}^{n}\frac{i}{i+n^{2}}\leq\sum_{i=1}^{n}\frac{i}{n^{2}}.$$ Since $\sum_{i=1}^{n}i=\frac{n(n+1)}{2},$ this becomes $$\frac{1}{2}=\frac{n(n+1)}{2n(n+1)}\leq\sum_{i=1}^{n}\frac{i}{i+n^{2}}\leq\frac{n(n+1)}{2n^{2}}=\frac{1}{2}\left(1+\frac{1}{n}\right),$$ and so it follows from the squeeze theorem that the limit is $\frac{1}{2}$.
• brilliant! (all that follows is extra text inserted to satisfy the software censor's rule that "comments MUST BE at least 15 characters in length". I'm sure this excellent, and absolutely non-arbitrary rule must keep a lot of footling short comments from wasting certain peoples' very valuable time). I only wish there were far more such rules applied to comments, to encourage a more structured approach to the process of commenting. – David Holden Dec 24 '13 at 19:20
• Short and sweet! – ireallydonknow Dec 27 '13 at 2:13
• @David, I believe the rule is to avoid myriad comments such as "me too" that will typically appear in any public forum, adding little or nothing to the information content and obscuring potentially more valuable contributions – holdenweb Oct 19 '16 at 13:02
$$S_n=\sum_{k=1}^n \frac{k}{k+n^2} = \sum_{k=1}^n \left( 1 -\frac{n^2}{k+n^2} \right) \\ = n - \sum_{k=1}^n \left(\frac{n^2}{k+n^2} \right)$$ but $$\sum_{k=1}^n \left(\frac{n^2}{k+n^2} \right) = \sum_{k=1}^n \frac1{1+\frac{k}{n^2}}$$
$$= \sum_{k=1}^n \sum_{j=0}^{\infty} \left(\frac{-k}{n^2} \right)^j \\ = n -\frac1{n^2}\frac{n(n+1)}{2} +O\left(\frac1{n}\right) = n - \frac12 +O\left(\frac1{n}\right)$$ so $$\lim_{n \rightarrow \infty} S_n = n - (n - \frac12) = \frac12$$
Since $$1+n^2 \le k+n^2\le n+n^2 \quad \forall k \in \{1,2,\ldots,n\},$$ it follows that $$\frac{n(n+1)}{2(n+n^2)}=\sum_{k=1}^n\frac{k}{n+n^2}\le \sum_{k=1}^n\frac{k}{k+n^2}\le\sum_{k=1}^n\frac{k}{1+n^2}=\frac{n(n+1)}{2(1+n^2)} \quad \forall n \ge 1.$$ Thus $$\lim_n\sum_{k=1}^n\frac{k}{k+n^2}=\frac12.$$
• But, $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$ – lab bhattacharjee Dec 24 '13 at 16:24
• This is just Eric's answer. – Andrés E. Caicedo Dec 24 '13 at 18:23
• @lab (+1) thank you for that succinct statement. I have previously had to go through the angst of detail every time i have applied this line of reasoning. now I can just apply this result. – David Holden Dec 24 '13 at 19:37
• @labbhattacharjee This is not a Riemann sum. So, one would need to be careful in applying such a framework. – Mark Viola Feb 17 '17 at 20:15
Use Riemann sums to show that $S\in\Big[\tfrac{\ln2}2,\tfrac12\Big]$ :
$$\sum_{k=1}^n\frac k{k^2+n^2}<\sum_{k=1}^n\frac k{k+n^2}<\sum_{k=1}^n\frac k{n^2}\quad\iff\quad\int_0^1\frac x{1+x^2}dx<S<\int_0^1xdx$$
• This provides bounds, but does not yield the limit. Apology in advance if I've missed something here. -Mark – Mark Viola Feb 17 '17 at 20:17
• @Dr.MV: I was able to determine $\tfrac12$ as the upper limit, but was unable, at the time, to provide a better lower limit. Meanwhile, other answers, showing that the latter is also $\tfrac12,$ have already been posted. – Lucian Feb 17 '17 at 22:03
• No worry. Your answers are often bedazzling. So, I thought I might have missed something. -Mark – Mark Viola Feb 17 '17 at 22:26
$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$
$$\lim_{n \to \infty} \dfrac{1}{n^2}(\dfrac{1}{1/n^2 + 1} +\dfrac{2}{2/n^2+1}+ \ldots+\dfrac{n}{n/n^2+1})$$
$$\lim_{n \to \infty} \dfrac{1}{n^2}\dfrac{n*(n+1)}{2}$$
$$\lim_{n \to \infty}\dfrac{1}{2}*(1+\dfrac{1}{n})$$
$$=\dfrac {1}{2}$$
• What happened from line 2 to 3? – Andrés E. Caicedo Dec 24 '13 at 16:43
• maybe he or she use limit first as $1/n^2\to 0$, big mistake! – mathlove Dec 24 '13 at 16:48
• Can I not use product of limits? – Satish Ramanathan Dec 24 '13 at 16:51
• Not the way you are doing it here. – Andrés E. Caicedo Dec 24 '13 at 17:02
• Understood, David's method is what I intended but applied limits improperly – Satish Ramanathan Dec 24 '13 at 17:07 | 2019-06-26T09:45:34 | {
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https://math.stackexchange.com/questions/3568751/a-problem-regarding-the-in-equality-of-complex-numbers | # A problem regarding the in-equality of complex numbers .
$$\mathbf {The \ Problem \ is}:$$ Let, $$z_1,z_2 \cdots z_n$$ be such that the real and imaginary parts of each $$z_i$$ are non-negative . Show that $$\bigg|\sum_{i=1}^n z_i\bigg| \geq \frac{1}{\sqrt2} \sum_{i=1}^n |z_i|.$$
$$\mathbf {My \ approach} :$$ Actually, at a very first glance, it seems that it can be proved using induction ,but I tried by using the triangular in-equality, though it didn't work .
I think there are some tricks related to it, a small hint is warmly appreciated .
• A straightforward approach should work. Establish that $|z_1 + z_2| \geq \frac{|z_1|+|z_2|}{\sqrt2}$. You can then use induction to obtain the result. Mar 4, 2020 at 7:02
Hint: in conjunction with the two given expressions, also consider the expressions $$\bigg| \sum_{j=1}^n x_j + i \sum_{j=1}^n y_j \bigg| \quad\text{and}\quad \frac1{\sqrt2} \bigg( \sum_{j=1}^n x_j + \sum_{j=1}^n y_j \bigg).$$
• great hint: I couldn't solve the original task, but was able to solve it given your answer. but wondering how to come up to the hint itself Mar 4, 2020 at 7:13
• I found one part of the hint by realizing that if the $n$ numbers $z_1,\dots,z_n$ were replaced by the $2n$ numbers $x_1,\dots,x_n,iy_i,\dots,iy_n$, then the left-hand side wouldn't change but the right-hand side would increase. Thus we might as well consider this case (where all numbers are either real or imaginary), since it really implies the full case. The other part of the hint came from trying to understand where the $\sqrt2$ might come from, by thinking about the case $n=1$ (which is trivial in the original formulation but slightly nontrivial in the new one). Mar 4, 2020 at 17:31
Since it was somewhere tagged as duplicate here my answer:
You may use
• $$\sqrt{a+b}\leq \sqrt a + \sqrt b$$ for $$a,b \geq 0$$ and
• $$(x+y)^2 \leq 2(x^2+y^2)$$ (This is $$2$$-dimensional Cauchy-Schwarz inequality)
Let $$z_k = x_k + iy_k$$ for $$k=1, \ldots , n$$ where $$x_k,y_k\geq 0$$.
$$\left(\sum_{k=1}^n \lvert z_k \rvert\right)^2 =\left(\sum_{k=1}^n \sqrt{x_k^2 + y_k^2}\right)^2 \leq \left(\sum_{k=1}^n \left(x_k + y_k\right)\right)^2$$ $$= \left(\sum_{k=1}^n x_k + \sum_{k=1}^n y_k\right)^2$$ $$\leq 2\left(\left(\sum_{k=1}^n x_k\right)^2 + \left(\sum_{k=1}^n y_k\right)^2\right) = 2 \lvert \sum_{k=1}^n z_k \rvert^2$$
Let us define $$\hat{m}=\frac{1}{\sqrt{2}}+j\ \frac{1}{\sqrt{2}}$$ and $$\hat{n}= -\frac{1}{\sqrt{2}}+j\ \frac{1}{\sqrt{2}}$$ and $$z_{i}=a_{i}\hat{m}+b_{i}\hat{n}$$.
Since $$z_{i}$$ has non - negative real and imaginary part,
$$\left(b_{i}\right)^{2}\leq\left(a_{i}\right)^{2}\rightarrow\sum{\frac{1}{\sqrt{2}}\sqrt{\left(a_{i}^{2}+b_{i}^{2}\right)}}\leq\sum{a_{i}=\sqrt{\left(\sum{a_{i}}\right)^{2}}}\leq\sqrt{\left(\sum{a_{i}}\right)^{2}+ \left(\sum{b_{i}}\right)^{2}}$$
aka
$$\frac{1}{\sqrt{2}}\sum{\left|z_{i}\right|}\leq\left|\sum{z_{i}}\right|$$
The $$\sqrt{2}$$ is a hint for me as it reminds me of $$\sin{\left(\frac{\pi}{4}\right)}$$ and $$\cos{\left(\frac{\pi}{4}\right)}$$. BTW could You share Your own answer too?
I'll provide a different approach to the ones mentioned so far.
First let $$z_k=x_k+iy_k$$, then $$\vert z_1+\cdots+z_k\vert=\sqrt{(x_1+\cdots+x_k)^2+(y_1+\cdots+y_k)^2}$$ but $$h(t):=\sqrt{t},\, t\in\mathbb R\,$$ is a concave function, thus $$\sqrt{\frac{(x_1+\cdots+x_k)^2+(y_1+\cdots+y_k)^2}2}\stackrel{\color{red}{(*)}}{\geq} \frac{\sqrt{(x_1+\cdots+x_k)^2}+\sqrt{(y_1+\cdots+y_k)^2}}{2}$$ which implies that \begin{align*}\sqrt{(x_1+\cdots+x_k)^2+(y_1+\cdots+y_k)^2}&\geq\frac{1}{\sqrt2}\left[(x_1+\cdots+x_k)+(y_1+\cdots+y_k)\right]\\ &=\frac{1}{\sqrt2}\left(\sqrt{(x_1+y_1)^2}+\cdots+\sqrt{(x_k+y_k)^2}\right)\\ &=\frac{1}{\sqrt2}\left(\sqrt{(x_1^2+2x_1y_1+y_1^2)}+\cdots+\sqrt{x_k^2+2x_ky_k+y_k^2}\right) \end{align*} If $$\Re\left(z_k\right)>0$$ and $$\Im\left(z_k\right)>0$$ for each $$k$$, then $$\sqrt{x_k^2+2x_ky_k+y_k^2}\geq\sqrt{x_k^2+y_k^2}$$ so \begin{align*}\sqrt{(x_1+\cdots+x_k)^2+(y_1+\cdots+y_k)^2}&\geq\frac{1}{\sqrt2}\left(\sqrt{x_1^2+y_1^2}+\cdots+\sqrt{x_k^2+y_k^2}\right)\\ &=\frac{1}{\sqrt2}\left(|z_1|+\cdots+|z_k|\right). \end{align*}
As mentioned $$\color{red}{(*)}$$ follows from concavity: a real function is concave if and only if it is midpoint concave. Also, this is a particular case of Jensen's Inequality, specifically the finite form one, with $$\varphi(x)=\sqrt{x}$$. | 2022-06-30T13:08:24 | {
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https://iul.riesenloeffel.de/news/laplacian-in-curvilinear-coordinates.html | lottery algorithm formula pdf amazon basics wide ruled 85
# Laplacian in curvilinear coordinates
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A curvilinear coordinate system is one where at least one of the coordinate surfaces is curved, e.g. in cylindrical coordinates the line between and is a circle. If the coordinate surfaces are mutually perpendicular, it is an orthogonal system, which is generally desirable. A useful attribute of a coordinate system is its line element , which .... In geometry, curvilinear coordinates are a coordinate system for Euclidean space in which the coordinate lines may be curved. ... and Laplacian) can be transformed from one coordinate system to another, according to transformation rules for scalars, vectors, and tensors. Such expressions then become valid for any curvilinear coordinate system. 3. Know how to choose coordinates and basis vectors wisely 4. Know your limits 5. Do the integrals 5 Divergence Theorem and Stokes’ Theorem 1. Know integral forms of grad, div, curl and laplacian 2. Divergence theorem 3. Stokes theorem (Green’s theorem as special case) 4. Know geometrical interpretations of div and curl 5. Physical examples.
Figure 2: Volume element in curvilinear coordinates. The sides of the small parallelepiped are given by the components of dr in equation (5). Vector v is decomposed into its u-, v- and w-components. 3 Divergence and laplacian in curvilinear coordinates Consider a volume element around a point P with curvilinear coordinates (u;v;w). The diver-. Laplacian operators in curvilinear coordinates can all be expressed in terms of these coefficients. This allows us to do the computations once and only once for every orthogonal curvilinear coordi- nate system, or more generally any curvilinear coordinate system.. Laplacian — From the formulas ... In maple, a vector entry in curvilinear coordinates that has not been designated as a vector field consists of the values of the three coordinates of its head when its tail is placed at the origin, with the entries permanently identified as being in the coordinate system that was active when the vector was. Section 4: The Laplacian and Vector Fields 11 4. The Laplacian and Vector Fields If the scalar Laplacian operator is applied to a vector field, it acts on each component in turn and generates a vector field. Example 3 The Laplacian of F(x,y,z) = 3z2i+xyzj +x 2z k is: ∇2F(x,y,z) = ∇2(3z2)i+∇2(xyz)j +∇2(x2z2)k. Let x,, x-t be the two components of a vector x with respect to the orthonormal basis {fi, }- In the sys- tem of Cartesian coordinates (,Xi, x^), the contravar- iant metric tensor is the two-dimensional unit tensor Iz and the Laplacian is simply .^,_6^ 6xf 6xj' The conventional method of computing the Lapla- cian in curvilinear coordinates ^(^i, ^2) and 2 (xi, ^2) requires evaluation. The Laplacian appears in several partial differential equations used to model wave propagation. Summation-by-parts--simultaneous approximation term (SBP-SAT) finite difference methods are often used for such equations, as they combine computational efficiency with provable stability on curvilinear multiblock grids. Hi if you have a simple ring, then you can define a cylindrical coordinate system at the ring centre (Model - Definition Coordinate systems - Cylindrical coordinates ) if this is then sys2, you have access to sys2.r (=sqrt((x-x0)^2+(y-yo)^2) if the axis is along Z, and x0,y0 are the centre offset) and to sys2.phi (=atan2(y,x) the angular variable hypotheses as before) and to sys2.z. Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 11 1Summary: Grad, Div, Laplacianand Curl in Curvilinear Coordi-nates 1. Gradient Operator in Cylindrical snd Spherical Coordinates 2. General Expressions for Div, Lapla- cian and Curl 3.
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Orthogonal Curvilinear Coordinates Often it is convenient to use coordinate systems other than rectangular ones. You are familiar, for example, with polar coordinates (r, θ) in the plane. One can transform from polar to regular (Cartesian) coordinates by way of the transformation equations x = r cos θ, y = r sin θ. Video created by 香港科技大学（The Hong Kong University of Science and Technology） for the course "Vector Calculus for Engineers". Integration can be extended to functions of several variables. We learn how to perform double and triple integrals.. Vectors Tensors 16 Curvilinear Coordinates. By Daniel Nguyen. Differential Geometry in Physics. By Waliyudin Anwar. A Student's Guide to Vectors and Tensors. By Víctor Ayala. Fleisch Vectors And Tensors. By Aayush Mani. VECTOR AND TENSOR CALCULUS. By angel barragan. Download PDF. About; Press; Blog; People; Papers; Job Board. Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 11 1Summary: Grad, Div, Laplacianand Curl in Curvilinear Coordi-nates 1. Gradient Operator in Cylindrical snd Spherical Coordinates 2. General Expressions for Div, Lapla- cian and Curl 3. 4. Divergence, Curl and Laplacian in Curvilinear Coordinates (1) Spherical coordinate s Fig. 9 shows the relationship between the spherical coordinates and the Cartesian coordinates. The spherical coordinates (r = = = θ.
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nonorthogonal curvilinear coordinates. Outline: 1. Cartesian coordinates 2. Orthogonal curvilinear coordinate systems 3. Differential operators in orthogonal curvilinear coordinate systems 4. Derivatives of the unit vectors in orthogonal curvilinear coordinate systems 5. Incompressible N-S equations in orthogonal curvilinear coordinate systems 6. general curvilinear coordinates, defined in (2-22a,b) f(m) ... Laplacian of a scalar, defined in (A-14) gJ' gi covariant and contravariant general metric tensors, defined in (2-29c, 2-34) g(ij) physical components of gii g determinant of gtj G turbulence generation term. Applying boundary conditions and Up: Appendix I: Scaling expansion Previous: Expansion of the dilation Curvilinear boundary coordinates We express for in the form where is the outward normal coordinate, and is a (periodic) tangential location on the boundary , increasing in an anti-clockwise fashion (see Fig. I.1). The Laplacian in curvilinear coordinates Substitute the components of gradU into the expression for diva 6.19 Much grindng gives the folowing expression for the Laplacian in general or- thogonal co-ordinates: Laplacian in curvi coords is: v2U h h ôU ôV ôw h h ôU hw ÔW ôu h h ôU hu ôU ôv Grad, etc, the 3D polar coordinate 6.20. Curvilinear coordinates, line, surface, and volume elements; grad, div, curl and the Laplacian in curvilinear coordinates. More examples. Syllabus The Contents section of this document is the course syllabus! Books The course will not use any particular textbook. The.
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32,455 recent views. This course covers both the basic theory and applications of Vector Calculus. In the first week we learn about scalar and vector fields, in the second week about differentiating fields, in the third week about multidimensional integration and curvilinear coordinate systems. The fourth week covers line and surface integrals. Mathematical Sciences - Mellon College of Science at CMU - Mathematical. A) (2), (3) and (4) both give (5) or (6) the Laplacian in spherical polar coordinates. B) (4) is equivalent to (3) the general 3-dimensional expression. C) (4) is equivalent to (2) the general 𝑛-dimensional expression. A was quite easy. B follows immediately from the formula for the covariant derivative. But in 2020 I could not prove C. Curvilinear Coordinates Outline: 1. Orthogonal curvilinear coordinate systems 2. Differential operators in orthogonal curvilinear coordinate systems 3. ... F for cylindrical coordinates 2.4 Laplacian acting on a scalar 2 23 31 12 123 1 11 2 2 2 3 33 1 hh hh hh f hhh x hx x h x x hx. This is because spherical coordinates are curvilinear, so the basis vectors are not the same at all points. For small variations, however, they are very similar. Because of the mathematical complexity that arises in curvilinear coordinate systems, you might wonder why we would want to use anything other than the Cartesian coordinate system. But, as we’ll highlight here, there are some applications where curvilinear coordinate systems are particularly useful. When to Use Curvilinear Coordinate Systems.
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pq = 0 for p 6= q then the system of coordinates in orthogonal. We denote by G the matrix with elements g ik and g ik = (G 1) ik (the element (i;k) of the inverse of G ) and g = detjG j. The Laplacian in curvilinear coordinates is = 1 p g X ik=1;N @ @u i p gg ik @ @u k The quantization: a simple two-dimensional case: One starts from the Hamilton. In addition find the laplacian ; Question: 3. In curvilinear coordinates the gradient, divergence, curl, and the lamarckian are give by: 1 au 5 % 24 a a 1 a (Vihah) hihahs agi Vzhzhs) + əqs (Vahiha) 91 42 1 ☺ x V հլիշht:3| да да2 дуз hivi h2V2 h3V3 Apply these three for cylindrical and spherical coordinates, check your answers. which is the ordinary Laplacian. In curvilinear coordinates, such as spherical or cylindrical coordinates, one obtains alternative expressions. ... is the D'Alembertian. Spherical Laplacian. The spherical Laplacian is the Laplace–Beltrami operator on the (n − 1)-sphere with its canonical metric of constant sectional curvature 1. Indeed, many textbooks opt to present the Laplacian in spherical polar coordinates without bothering to derive it from its Cartesian form (e.g. [5–7]). Specialising from the general expression for the Laplacian in curvilinear coordinates is sometimes recommended (e.g. [ 8 ] citing [ 9 ], and [ 10 ]), which is a viable option for students who have had a rigorous course in. View CurvilinearCoordinates.pdf from MATH MISC at Ying Wa College. Orthogonal Curvilinear Coordinates: Div, Grad, Curl, and the Laplacian The most common way that the gradient of a. Problem 2.6a. Show that the wave equation (2.5a) can be written in cylindrical coordinates (see Figure 2.6a) as.
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which is the ordinary Laplacian. In curvilinear coordinates, such as spherical or cylindrical coordinates, one obtains alternative expressions. ... is the D'Alembertian. Spherical Laplacian. The spherical Laplacian is the Laplace–Beltrami operator on the (n − 1)-sphere with its canonical metric of constant sectional curvature 1. math 2443–008 calculus iv spring 2014 orthogonal curvilinear coordinates in 3–dimensions consider coordinate system in r3 defined by r(u1 u2, u3 hx(u1 u2 u3 y. 1.4: Curvilinear Coordinates 1.4.1: Spherical Coordinates. You can label a point P by its Cartesian coordinates (x, y, z), but sometimes it is more convenient to use spherical coordinates ; is the distance from the origin (the magnitude of the position vector r), (the angle down from the z axis) is called the polar angle, and (the angle around from the x axis) is the azimuthal angle.. View Homework Help - Appendix.pdf from PHYSICS 2020 at National University of Singapore. Gradient, Divergence, Curl, and Laplacian in Curvilinear Coordinate Systems if f. Prolate Spheroidal Coordinates. A system of Curvilinear Coordinates in which two sets of coordinate surfaces are obtained by revolving the curves of the Elliptic Cylindrical Coordinates about the x -Axis, which is relabeled the z -Axis. The third set of coordinates consists of planes passing through this axis. where , , and. According to Mathworld, in three dimensions there are 13 coordinate systems in which Laplace's equation is separable, and 11 for the Helmholtz equation.I've read the relevant chapters of the book by Morse & Feshbach. Apart from a recent paper by Phil Lucht, almost nothing has been written about this since the sixties.The situation is actually more complicated. Problem 2.6a. Show that the wave equation (2.5a) can be written in cylindrical coordinates (see Figure 2.6a) as.
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Feb 27, 2016 · I'm looking for a simple expression for the vector Laplacian $abla^2\mathbf{A}$ in orthogonal curvilinear coordinates. Actually, I don't require the whole thing, just the part of \$\mathbf{u}_i\.... Orthogonal Curvilinear Coordinates: Arc length, surface area and volume element in curvilinear coordinate system. Gradient, divergence, curl and Laplacian in curvilinear coordinate system. Cylindrical Co-ordinate system, spherical polar coordinate system. Applications. Books Recommended: 1. Vector Calculus, S. J. Colley. (Pearson) 2. The vector Laplacian has to be handled with some care. In Cartesian components you simply have $$\Delta A^j=\partial_k \partial^k A^j=\delta^{ik} \partial_i \partial_k A^j.$$ It's not as easy in general coordinates (even not in general orthogonal curvilinear coordinates). In one dimension the Laplacian of u is simply the second derivative of u and so we look at the limit of the second order incremental quotient. Recall that: u00(x) = lim h!0 u(x+h)u(x) h u(x)u(xh) h h = lim h!0 u(x + h) 2u(x) + u(x h) h2. List of illustrations Prologue Modelling solids 1.1 Introduction 1.2 Hookes law 1.3 Lagrangian and Eulerian coordinates 1.4 Strain 1.5 Stress 1.6 Conservation of momentum 1.7 Linear elasticity 1.8 The incompressibility approximation 1.9 Energy 1.10 Boundary conditions and well-posedness 1.11 Coordinate systems Exercises Linear elastostatics 2.1.
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Vector identities; the Laplacian. Curvilinear coordinate system. Application: Maxwell equations and boundary conditions. Gauss' Law, Poisson and Laplace equations Electric field from point charge and charge distribution; Dirac delta function. Derivation of Gauss's Law from Coulomb's law. Scalar potential; Poisson and Laplace equations. Fourier's Law in radial coordinates r dT q kA dr Substituting the area of a sphere The one-dimensional heat conduction equations based on the dual-phase-lag theory are derived in a unified form which can be used for Cartesian, cylindrical , and spherical coordinates Heat Transfer, Vol Only linear brick, first-order axisymmetric, and second-order modified tetrahedrons are available for modeling. Apr 13, 2020 · Vector laplacian in Curvilinear coordinate systems. ( ∇ 2 A) r = ∇ 2 A r − A r r 2 − 2 r 2 ∂ A φ ∂ φ ( ∇ 2 A) φ = ∇ 2 A φ − A φ r 2 + 2 r 2 ∂ A r ∂ φ ( ∇ 2 A) z = ∇ 2 A z. but didn't give any derivation. So I wonder how to derive this and also the expression of vector laplacian in spherical coordinates. Also, can this be derived in genral form in any orthogonal curvilinear coordinates, via the Lamé coefficients?. A number of coordinates are used for analyzing problems such as the colliding phenomena, magnetofluid dynamic equilibrium and stability of plasma or the analysis of magnetic field configuration, depending on each problem. Therefore, consideration on the vector anal. View Notes - MATH 2443 Orthogonal Curvilinear Coordinates Notes from MATH 2443 at The University of Oklahoma. MATH 2443008 Calculus IV Spring 2014 Orthogonal Curvilinear Coordinates in 3Dimensions 1.
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pq = 0 for p 6= q then the system of coordinates in orthogonal. We denote by G the matrix with elements g ik and g ik = (G 1) ik (the element (i;k) of the inverse of G ) and g = detjG j. The Laplacian in curvilinear coordinates is = 1 p g X ik=1;N @ @u i p gg ik @ @u k The quantization: a simple two-dimensional case: One starts from the Hamilton.
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Laplacian — From the formulas ... In maple, a vector entry in curvilinear coordinates that has not been designated as a vector field consists of the values of the three coordinates of its head when its tail is placed at the origin, with the entries permanently identified as being in the coordinate system that was active when the vector was. The paper presents a new combination of two methods for tool path generation for five-axis machining proposed earlier by the authors. The first method is based on the grid generation technologies whereas the second method exploits the space-filling. Nov 07, 2016 · The gradient of a vector is. As expected, we see that the gradient splits nicely into a dot and curl. where the cylindrical representation of the divergence is seen to be. and the cylindrical representation of the curl is. Should we want to, it is now possible to evaluate the Laplacian of using. , which will have the following components.. Curvilinear coordinates, namely polar coordinates in two dimensions, and cylindrical and spherical coordinates in three dimensions, are used to simplify problems with circular, cylindrical or spherical symmetry. We learn how to write differential operators in curvilinear coordinates and how to change variables in multidimensional integrals ....
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Problem 2.6a. Show that the wave equation (2.5a) can be written in cylindrical coordinates (see Figure 2.6a) as. In cylindrical (R,Phi,Z) coordinates, for example, the radial component of the Laplacian of a vector V is. DEL2(Vr) - Vr/R^2 - 2*DPHI(Vphi)/R^2. The extra 1/R^2 terms have arisen from the differentiation of the unit vectors. FlexPDE performs the correct expansion of the differential operators in all supported coordinate systems.
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Curvilinear co-ordinates- Scale factors, Base vectors, Cylindrical-polar coordinates, Spherical-polar coordinates - Transformationsbetween Cartesian and curvilinear systems , Orthogonality. Elements of arc, area and volume in curvilinear system, Gradient, Divergence, Curl and Laplacian in orthogonal curvilinear coordinates. Unit –IV 09 Hrs. A curvilinear coordinate system is one where at least one of the coordinate surfaces is curved, e.g. in cylindrical coordinates the line between and is a circle. If the coordinate surfaces are mutually perpendicular, it is an orthogonal system, which is generally desirable. A useful attribute of a coordinate system is its line element , which .... Curvilinear Coordinates and Vector Calculus 1 1. Orthogonal Curvilinear Coordinates Let the rectangular coordinates (x, y, z) of any point be expressed as functions of (u1, u2, u 3) so that ... Divergence, Curl and Laplacian in Curvilinear Coordinates (1) Spherical coordinate s.
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2. Electric Field in Curvilinear Coordinates Using the Generalized Functions Method The derivation presented in this section is similar to the one shown by Walsh and Donnelly (1987a) for the two-body scattering. Here, however, the system of equations is derived purely by using vector and dyadic calculus identities, independent of a coordinate. as I know, it works for all curvilinear coordinate systems and the Laplacian. Don't have a derivation though for all coordinate systems. At various times, I have thought about teaching my B-spline codes about the metric tensor, so I could easily do any coordinate system I want. The only thing that is a bit difficult is that the variable. The most general coordinate system for fluid flow problems are nonorthogonal curvilinear coordinates. A special case of these are orthogonal curvilinear coordinates. Here we shall derive the appropriate relations for the latter using vector technique. It should be recognized that the derivation can also be accomplished using tensor analysis 1. Laplacian (r:r ) r:r = @ @x @ @x + @ @y @ @y + @ @z @ @z Introduction to Curvilinear Co-ordinate System The Curvilinear co-ordinates are the common name of di erent sets of co-ordinates other than Cartesian coordinates. In many problems of physics and applied mathematics it is usually necessary to write vector equations.
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The problem is considered in the curvilinear coordinate system (3.123).As it turns out, at b / a < ~ 2, the cross section of the surface φ = const by the planes z = const is almost exactly determined by the ellipse with the semi-axes given by curves 1 and 2 in Figure 38.(X is the distance along the normal to the beam boundary.)Curves 1 correspond to η = 0; curves 2 correspond to η = π/2. Mathematical Sciences - Mellon College of Science at CMU - Mathematical. as I know, it works for all curvilinear coordinate systems and the Laplacian. Don't have a derivation though for all coordinate systems. At various times, I have thought about teaching my B-spline codes about the metric tensor, so I could easily do any coordinate system I want. The only thing that is a bit difficult is that the variable. Section 4: The Laplacian and Vector Fields 11 4. The Laplacian and Vector Fields If the scalar Laplacian operator is applied to a vector field, it acts on each component in turn and generates a vector field. Example 3 The Laplacian of F(x,y,z) = 3z2i+xyzj +x 2z k is: ∇2F(x,y,z) = ∇2(3z2)i+∇2(xyz)j +∇2(x2z2)k. What happens in curvilinear coordinates? Naive pattern matching with (4.11.1) might lead you to believe that the position vector in spherical coordinates is given by: →r = r^r +θ^θ +ϕ ^ϕ (incorrect). (4.11.2) (4.11.2) r → = r r ^ + θ θ ^ + ϕ ϕ ^ (incorrect). 🔗. However, if you try to follow this equation as a literal set of. Section 14.18 The Laplacian. One second derivative, the divergence of the gradient, occurs so often it has its own name and notation. It is called the Laplacian of the function $$V\text{,}$$ and is written in any of the forms. In curvilinear coordinates we have: h1dq 1 , h2dq 2 , and h3dq 3 . Note that each has dimension of length. Our gradient in curvilinear coordinates is then ... Show our Laplacian in curvilinear coordinates, which is ∇2 = 1 ∂ ( h2h3 ∂ )+ ∂ ( h1h3 ∂ )+ ∂ ( h1h2.
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Equation of Poisson and cylindrical coordinates, Uncharged conducting Laplace, applications of Laplace's equation to and dielectric sphere in uniform electric field, problems (Conductors and dielectrics) having spherical cylindrical and Cartesian symmetry, Electrostatic Images, Point charge near an electrical images (conductors and dielectrics). We realize that the gradient operator in curvilinear coordinates can in general be written as ~Ñf = 3 å j=1 ~e j 1 h j ¶f ¶a j (23) where h j = ¶~x ¶aj are scaling factors in the respective coordinate system (for example in cylindrical coordinates they are given in Eq. (9)). This is also readily verified in cartesian coordinates. 1.2.4. Let Ω ′ be the parallelogram domain depicted in Fig. 1 a. By varying the angle φ, we obtain a family of parallelograms whose base and height are of unit length and whose bottom and top boundaries are horizontal. The grid lines are chosen to be equidistant and parallel to the domain boundaries. For φ = π / 2, Ω ′ is square and the grid is Cartesian. The Laplacian in curvilinear coordinates Substitute the components of gradU into the expression for diva 6.19 Much grindng gives the folowing expression for the Laplacian in general or- thogonal co-ordinates: Laplacian in curvi coords is: v2U h h ôU ôV ôw h h ôU hw ÔW ôu h h ôU hu ôU ôv Grad, etc, the 3D polar coordinate 6.20. v. t. e. In mathematics, the Laplace operator or Laplacian is a differential operator given by the divergence of the gradient of a scalar function on Euclidean space. It is usually denoted by the symbols ∇ ⋅ ∇, ∇ 2 (where ∇ is the nabla operator ), or Δ. In a Cartesian coordinate system, the Laplacian is given by the sum of second. Reset your password. If you have a user account, you will need to reset your password the next time you login. You will only need to do this once.
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A number of coordinates are used for analyzing problems such as the colliding phenomena, magnetofluid dynamic equilibrium and stability of plasma or the analysis of magnetic field configuration, depending on each problem. Therefore, consideration on the vector anal. 32,455 recent views. This course covers both the basic theory and applications of Vector Calculus. In the first week we learn about scalar and vector fields, in the second week about differentiating fields, in the third week about multidimensional integration and curvilinear coordinate systems. The fourth week covers line and surface integrals. | 2022-11-26T09:50:15 | {
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https://mathhelpboards.com/threads/solving-equation-ii.9395/ | # Solving equation II
#### anemone
##### MHB POTW Director
Staff member
Solve the equation $\left\lfloor \dfrac{25x-2}{4} \right\rfloor=\dfrac{13x+4}{3}$.
##### Well-known member
Solve the equation $\left\lfloor \dfrac{25x-2}{4} \right\rfloor=\dfrac{13x+4}{3}$.
Subtract 1 from both sides
$\lfloor\frac{25x-6}{4}\rfloor = \frac{13x+1}{3}$
So $\frac{13x + 1}{3}$ = integer say n
So x = $\frac{3n-1}{13}$
So we get
$\lfloor\frac{25\frac{3n-1}{13}-6}{4}\rfloor = n$
Or
$\lfloor\frac{75n-103}{52}\rfloor = n$
Or 23n-103= > 0 and < 52
Or 103 <= 23n < 155
Or $\frac{103}{23} <=n < \frac{155}{23}$
So n = 5 or 6
Hence x = $\frac{14}{13}$ or $\frac{17}{13}$
#### anemone
##### MHB POTW Director
Staff member
But, if you don't mind me asking, I see that we could make the substitution right from the start, why would you do so only after subtracting both sides of the equation by 1? | 2021-08-01T17:51:47 | {
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https://in.mathworks.com/help/optim/ug/sudoku-puzzles-problem-based.html | # Solve Sudoku Puzzles Via Integer Programming: Problem-Based
This example shows how to solve a Sudoku puzzle using binary integer programming. For the solver-based approach, see Solve Sudoku Puzzles Via Integer Programming: Solver-Based.
You probably have seen Sudoku puzzles. A puzzle is to fill a 9-by-9 grid with integers from 1 through 9 so that each integer appears only once in each row, column, and major 3-by-3 square. The grid is partially populated with clues, and your task is to fill in the rest of the grid.
### Initial Puzzle
Here is a data matrix B of clues. The first row, B(1,2,2), means row 1, column 2 has a clue 2. The second row, B(1,5,3), means row 1, column 5 has a clue 3. Here is the entire matrix B.
B = [1,2,2;
1,5,3;
1,8,4;
2,1,6;
2,9,3;
3,3,4;
3,7,5;
4,4,8;
4,6,6;
5,1,8;
5,5,1;
5,9,6;
6,4,7;
6,6,5;
7,3,7;
7,7,6;
8,1,4;
8,9,8;
9,2,3;
9,5,4;
9,8,2];
drawSudoku(B) % For the listing of this program, see the end of this example.
This puzzle, and an alternative MATLAB® solution technique, was featured in Cleve's Corner in 2009.
There are many approaches to solving Sudoku puzzles manually, as well as many programmatic approaches. This example shows a straightforward approach using binary integer programming.
This approach is particularly simple because you do not give a solution algorithm. Just express the rules of Sudoku, express the clues as constraints on the solution, and then MATLAB produces the solution.
### Binary Integer Programming Approach
The key idea is to transform a puzzle from a square 9-by-9 grid to a cubic 9-by-9-by-9 array of binary values (0 or 1). Think of the cubic array as being 9 square grids stacked on top of each other, where each layer corresponds to an integer from 1 through 9. The top grid, a square layer of the array, has a 1 wherever the solution or clue has a 1. The second layer has a 1 wherever the solution or clue has a 2. The ninth layer has a 1 wherever the solution or clue has a 9.
This formulation is precisely suited for binary integer programming.
The objective function is not needed here, and might as well be a constant term 0. The problem is really just to find a feasible solution, meaning one that satisfies all the constraints. However, for tie breaking in the internals of the integer programming solver, giving increased solution speed, use a nonconstant objective function.
### Express the Rules for Sudoku as Constraints
Suppose a solution $x$ is represented in a 9-by-9-by-9 binary array. What properties does $x$ have? First, each square in the 2-D grid (i,j) has exactly one value, so there is exactly one nonzero element among the 3-D array entries $x\left(i,j,1\right),...,x\left(i,j,9\right)$. In other words, for every $i$ and $j$,
$\sum _{k=1}^{9}x\left(i,j,k\right)=1.$
Similarly, in each row $i$ of the 2-D grid, there is exactly one value out of each of the digits from 1 to 9. In other words, for each $i$ and $k$,
$\sum _{j=1}^{9}x\left(i,j,k\right)=1.$
And each column $j$ in the 2-D grid has the same property: for each $j$ and $k$,
$\sum _{i=1}^{9}x\left(i,j,k\right)=1.$
The major 3-by-3 grids have a similar constraint. For the grid elements $1\le i\le 3$ and $1\le j\le 3$, and for each $1\le k\le 9$,
$\sum _{i=1}^{3}\sum _{j=1}^{3}x\left(i,j,k\right)=1.$
To represent all nine major grids, just add 3 or 6 to each $i$ and $j$ index:
$\sum _{i=1}^{3}\sum _{j=1}^{3}x\left(i+U,j+V,k\right)=1,$ where $U,V\phantom{\rule{0.5em}{0ex}}ϵ\phantom{\rule{0.5em}{0ex}}\left\{0,3,6\right\}.$
### Express Clues
Each initial value (clue) can be expressed as a constraint. Suppose that the $\left(i,j\right)$ clue is $m$ for some $1\le m\le 9$. Then $x\left(i,j,m\right)=1$. The constraint $\sum _{k=1}^{9}x\left(i,j,k\right)=1$ ensures that all other $x\left(i,j,k\right)=0$ for $k\ne m$.
### Sudoku in Optimization Problem Form
Create an optimization variable x that is binary and of size 9-by-9-by-9.
x = optimvar('x',9,9,9,'Type','integer','LowerBound',0,'UpperBound',1);
Create an optimization problem with a rather arbitrary objective function. The objective function can help the solver by destroying the inherent symmetry of the problem.
sudpuzzle = optimproblem;
mul = ones(1,1,9);
mul = cumsum(mul,3);
sudpuzzle.Objective = sum(sum(sum(x,1),2).*mul);
Represent the constraints that the sums of x in each coordinate direction are one.
sudpuzzle.Constraints.consx = sum(x,1) == 1;
sudpuzzle.Constraints.consy = sum(x,2) == 1;
sudpuzzle.Constraints.consz = sum(x,3) == 1;
Create the constraints that the sums of the major grids sum to one as well.
majorg = optimconstr(3,3,9);
for u = 1:3
for v = 1:3
arr = x(3*(u-1)+1:3*(u-1)+3,3*(v-1)+1:3*(v-1)+3,:);
majorg(u,v,:) = sum(sum(arr,1),2) == ones(1,1,9);
end
end
sudpuzzle.Constraints.majorg = majorg;
Include the initial clues by setting lower bounds of 1 at the clue entries. This setting fixes the value of the corresponding entry to be 1, and so sets the solution at each clued value to be the clue entry.
for u = 1:size(B,1)
x.LowerBound(B(u,1),B(u,2),B(u,3)) = 1;
end
Solve the Sudoku puzzle.
sudsoln = solve(sudpuzzle)
Solving problem using intlinprog.
LP: Optimal objective value is 405.000000.
Optimal solution found.
Intlinprog stopped at the root node because the objective value is within a gap
tolerance of the optimal value, options.AbsoluteGapTolerance = 0 (the default
value). The intcon variables are integer within tolerance,
options.IntegerTolerance = 1e-05 (the default value).
sudsoln = struct with fields:
x: [9x9x9 double]
Round the solution to ensure that all entries are integers, and display the solution.
sudsoln.x = round(sudsoln.x);
y = ones(size(sudsoln.x));
for k = 2:9
y(:,:,k) = k; % multiplier for each depth k
end
S = sudsoln.x.*y; % multiply each entry by its depth
S = sum(S,3); % S is 9-by-9 and holds the solved puzzle
drawSudoku(S)
You can easily check that the solution is correct.
### Function to Draw the Sudoku Puzzle
type drawSudoku
function drawSudoku(B)
% Function for drawing the Sudoku board
% Copyright 2014 The MathWorks, Inc.
figure;hold on;axis off;axis equal % prepare to draw
rectangle('Position',[0 0 9 9],'LineWidth',3,'Clipping','off') % outside border
rectangle('Position',[3,0,3,9],'LineWidth',2) % heavy vertical lines
rectangle('Position',[0,3,9,3],'LineWidth',2) % heavy horizontal lines
rectangle('Position',[0,1,9,1],'LineWidth',1) % minor horizontal lines
rectangle('Position',[0,4,9,1],'LineWidth',1)
rectangle('Position',[0,7,9,1],'LineWidth',1)
rectangle('Position',[1,0,1,9],'LineWidth',1) % minor vertical lines
rectangle('Position',[4,0,1,9],'LineWidth',1)
rectangle('Position',[7,0,1,9],'LineWidth',1)
% Fill in the clues
%
% The rows of B are of the form (i,j,k) where i is the row counting from
% the top, j is the column, and k is the clue. To place the entries in the
% boxes, j is the horizontal distance, 10-i is the vertical distance, and
% we subtract 0.5 to center the clue in the box.
%
% If B is a 9-by-9 matrix, convert it to 3 columns first
if size(B,2) == 9 % 9 columns
[SM,SN] = meshgrid(1:9); % make i,j entries
B = [SN(:),SM(:),B(:)]; % i,j,k rows
end
for ii = 1:size(B,1)
text(B(ii,2)-0.5,9.5-B(ii,1),num2str(B(ii,3)))
end
hold off
end | 2021-06-21T07:55:12 | {
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http://mathhelpforum.com/calculus/90133-volume-integral.html | # Math Help - Volume integral
1. ## Volume integral
Find the volume of the region bounded by the cone $x^2 + y^2 = z^2~ (z \ge 0)$ and the paraboloid $z =2x^2 + 2y^2$.
------
I'm having a hard time even visualising the integration region. The following is my attempt to plot the paraboloid (in blue) and the cone (in brown) from x=-1..1 and y=-1..1.
Would I do this in rect-coords or cylindrical or what?
2. Originally Posted by scorpion007
Find the volume of the region bounded by the cone $x^2 + y^2 = z^2~ (z \ge 0)$ and the paraboloid $z =2x^2 + 2y^2$.
------
I'm having a hard time even visualising the integration region. The following is my attempt to plot the paraboloid (in blue) and the cone (in brown) from x=-1..1 and y=-1..1.
Would I do this in rect-coords or cylindrical or what?
two surfaces intersect at $z=0, \frac{1}{2}.$ for $0 \leq z \leq \frac{1}{2}$ the cone is the top surface. so the volume is: $V= \int \int_R [\sqrt{x^2+y^2} - 2(x^2+y^2)] dA,$ where $R=\{(x,y): \ x^2+y^2 \leq \frac{1}{4} \}.$
using polar coordinates we get: $V=\int_0^{2 \pi} \int_0^{\frac{1}{2}}(r^2-2r^3) \ dr d \theta = \frac{\pi}{48}.$
3. Thanks, but I have some questions:
How did you find their intersection? Did you solve the quartic polynomial:
$x^2+y^2 = 4x^4+8x^2y^2+4y^4$ ?
Don't we need a triple integral here? Actually, I guess not since the integrand would be a constant "1" for volume. So you somehow used a double over R...
How did you determine that the integrand should be the difference between the surface functions, I.e. $z_1 - z_2$?
4. Originally Posted by scorpion007
Thanks, but I have some questions:
How did you find their intersection? Did you solve the quartic polynomial:
$x^2+y^2 = 4x^4+8x^2y^2+4y^4$ ?
Don't we need a triple integral here? Actually, I guess not since the integrand would be a constant "1" for volume. So you somehow used a double over R...
How did you determine that the integrand should be the difference between the surface functions, I.e. $z_1 - z_2$?
to find the intersection: $z^2=x^2+y^2=\frac{z}{2}.$ thus $2z^2=z$ and hence $z=0, \frac{1}{2}.$ to answer your second question, recall that the volume of the region $D$ bounded by the two surfaces:
$z=f(x,y), \ z=g(x,y),$ with $f(x,y) \leq g(x,y), \ \forall (x,y) \in R,$ is $\int \int_R (g(x,y) - f(x,y)) dA,$ where $R$ is the image of $D$ on the xy plane.
5. Ah yes! Thank you very much! | 2014-12-25T12:13:56 | {
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https://math.stackexchange.com/questions/209950/an-inequality-on-trace-of-product-of-two-matrices/209991 | # An inequality on trace of product of two matrices
Suppose we have two $n \times n$ positive semidefinite matrices, $A$ and $B$, such that $\mbox{tr}(A), \mbox{tr}(B) \le 1$.
Can we say anything about $\mbox{tr}(AB)$? Is $\mbox{tr}(AB) \le 1$ too?
In the space of positive semi-definite matrices, trace is a proper inner-product (it is easy to show that), i.e. it obeys the Cauchy-Schwarz inequality: $\langle x,y \rangle \leq \sqrt{ \langle x,x \rangle \langle y,y\rangle}$. So
$$\mbox{tr}\{AB\}\leq \sqrt{\mbox{tr}\{A^2\} \mbox{tr}\{B^2\}}$$
Now, since $A$ is positive semidefinite, $\mbox{tr}\{A^2\} \leq \mbox{tr}\{A\}^2$, i.e., the eigenvalues of $A^2$ are squared eigenvalues of $A$, and since they are positive
$$\mbox{tr} \{A^2\} = \sum_{i=1}^{N}\lambda_{i}^{2}\leq \left( \sum_{i=1}^{N}\lambda_{i} \right)^{2} = \mbox{tr}\{A\}^2 \leq 1$$
A similar argument for B proves $\mbox{tr}\{B^2\}\leq 1$ . So $\mbox{tr}\{AB\}\leq 1$. Hope this answers your question.
First, note that $(A-B)^{2}$ is positive semi-definite, so we have:
$$0\leq\mathrm{Tr}(A-B)^{2}=\mathrm{Tr}(A^{2})+\mathrm{Tr}(B^{2})-2\mathrm{Tr}(AB)$$
$$\mathrm{Tr}(AB)\leq\frac{1}{2}(\mathrm{Tr}(A^{2})+\mathrm{Tr}(B^{2}))$$
Second, for $A$ positive semi-definite, suppose that all of eigenvalues are $\lambda_{1}$, $\lambda_{2}$, $\cdots$, $\lambda_{n}$, then $\lambda_{i}\geq0$ and $\mathrm{Tr}A=\sum_{i=1}^{n}\lambda_{i}\leq1$, so $\mathrm{Tr}(A^{2})=\sum_{i=1}^{n}\lambda_{i}^{2}\leq\sum_{i=1}^{n}\lambda_{i}\leq1$.
Similarly, $\mathrm{Tr}(B^{2})\leq1$, so $\mathrm{Tr}(AB)\leq1$.
Remark. More generally, we can conclude that the range of $\mathrm{Tr}(AB)$ is $[0,1]$.
As $A$ and $B$ are positive semi-definite, so there exist $C$ and $D$ such that $A=C^{T}C$ and $B=D^{T}D$, so $\mathrm{Tr}(AB)=\mathrm{Tr}(C^{T}CD^{T}D)=\mathrm{Tr}(CD^{T}DC^{T})=\mathrm{Tr}[CD^{T}(CD^{T})^{T}]\geq0$.
Set $A=diag[1,0,0,\cdots,0]$ and $A=diag[0,1,0,\cdots,0]$, then $\mathrm{Tr}(AB)=0$.
Set $A=B=diag[1,0,0,\cdots,0]$, then $\mathrm{Tr}(AB)=1$.
Accoading to above, we can conclude that $\text{Range}(\mathrm{Tr}(AB))=[0,1]$.
• When does equality holds for $\text{Tr}(AB)\leq\frac{1}{2}[\text{Tr}(A^2)+\text{Tr}(B^2)]$?
– Tan
Jan 5 at 17:39
An extension to @dineshdileep 's answer can also be found here, in which it shows that: $$tr(A^TB)\le \sqrt{tr(A^TA)tr(B^TB)}$$
• Hello, when does equality hold?
– Tan
Jan 5 at 17:39 | 2022-05-26T02:00:16 | {
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https://kr.mathworks.com/help/matlab/ref/lsqminnorm.html?lang=en | # lsqminnorm
Minimum norm least-squares solution to linear equation
## Description
example
X = lsqminnorm(A,B) returns an array X that solves the linear equation AX = B and minimizes the value of norm(A*X-B). If several solutions exist to this problem, then lsqminnorm returns the solution that minimizes norm(X).
example
X = lsqminnorm(A,B,tol) additionally specifies the tolerance that lsqminnorm uses to determine the rank of A.
example
X = lsqminnorm(___,rankWarn) specifies an optional flag to display a warning if A has low rank. You can use any of the input argument combinations in previous syntaxes. rankWarn can be 'nowarn' (default) or 'warn'.
## Examples
collapse all
Solve a linear system that has infinitely many solutions with backslash (\) and lsqminnorm. Compare the results using the 2-norms of the solutions.
When infinite solutions exist to $\mathrm{Ax}=\mathit{b}$, each of them minimizes $‖\mathrm{Ax}-\mathit{b}‖$. The backslash command (\) computes one such solution, but this solution typically does not minimize $‖\mathit{x}‖$. The solution computed by lsqminnorm minimizes not only norm(A*x-b), but also norm(x).
Consider a simple linear system with one equation and two unknowns, $2{\mathit{x}}_{1}+3{\mathit{x}}_{2}=8$. This system is underdetermined since there are fewer equations than unknowns. Solve the equation using both backslash and lsqminnorm.
A = [2 3];
b = 8;
x_a = A\b
x_a = 2×1
0
2.6667
x_b = lsqminnorm(A,b)
x_b = 2×1
1.2308
1.8462
The two methods obtain different solutions because backslash only aims to minimize norm(A*x-b), whereas lsqminnorm also aims to minimize norm(x). Calculate these norms and put the results in a table for easy comparison.
s1 = {'Backslash'; 'lsqminnorm'};
s2 = {'norm_Ax_minus_b','norm_x'};
T = table([norm(A*x_a-b); norm(A*x_b-b)],[norm(x_a); norm(x_b)],'RowNames',s1,'VariableNames',s2)
T=2×2 table
norm_Ax_minus_b norm_x
_______________ ______
Backslash 0 2.6667
lsqminnorm 8.8818e-16 2.2188
This figure illustrates the situation and shows which solutions each of the methods return. The blue line represents the infinite number of solutions to the equation ${\mathit{x}}_{2}=-\frac{2}{3}{\mathit{x}}_{1}+\frac{8}{3}$. The orange circle represents the minimum distance from the origin to the line of solutions, and the solution returned by lsqminnorm lies exactly at the tangent point between the line and circle, indicating it is the solution that is closest to the origin.
Show how specifying a tolerance for the rank computation in lsqminnorm can help define the scale of the problem so that random noise does not corrupt the solution.
Create a low-rank matrix of rank 5 and a right-hand side vector b.
rng default % for reproducibility
U = randn(200,5);
V = randn(100,5);
A = U*V';
b = U*randn(5,1) + 1e-4*randn(200,1);
Solve the linear system $\mathrm{Ax}=\mathit{b}$ using lsqminnorm. Compute the norms of A*x-b and x to check the quality of the solution.
x = lsqminnorm(A,b);
norm(A*x-b)
ans = 0.0014
norm(x)
ans = 0.1741
Now add a small amount of noise to the matrix A and solve the linear system again. The noise affects the solution vector x of the linear system disproportionately.
Anoise = A + 1e-12*randn(200,100);
xnoise = lsqminnorm(Anoise,b);
norm(Anoise*xnoise - b)
ans = 0.0010
norm(xnoise)
ans = 1.1215e+08
The reason for the big difference in the solutions is that the noise affects the low-rank approximation of A. In other words, lsqminnorm is treating small values on the diagonal of the R matrix in the QR decomposition of A as being more important than they are. Ideally, these small values on the diagonal of R should be treated as zeros.
Plot the diagonal elements of the R matrix in the QR decomposition of Anoise. A large number of the diagonal elements are on the order of 1e-10.
[Q,R,p] = qr(Anoise,0);
semilogy(abs(diag(R)),'o')
The solution to this issue is to increase the tolerance used by lsqminnorm so that a low-rank approximation of Anoise with error less than 1e-8 is used in the calculation. This makes the result much less susceptible to the noise. The solution using a tolerance is very close to the original solution x.
xnoise = lsqminnorm(Anoise, b, 1e-8);
norm(Anoise*xnoise - b)
ans = 0.0014
norm(xnoise)
ans = 0.1741
norm(x - xnoise)
ans = 1.0811e-14
Solve a linear system involving a low-rank coefficient matrix with warnings turned on.
Create a 3-by-3 matrix that is of rank 2. In this matrix, you can obtain the third column by adding together the first two columns.
A = [1 2 3; 4 5 9; 6 7 13]
A = 3×3
1 2 3
4 5 9
6 7 13
Find the minimum norm least-squares solution to the problem $\mathrm{Ax}=\mathit{b}$, where $\mathit{b}$ is equal to the second column in $\mathit{A}$. Specify the 'warn' flag for lsqminnorm to display a warning if it detects that A is of low rank.
b = A(:,2);
x = lsqminnorm(A,b,'warn')
Warning: Rank deficient, rank = 2, tol = 1.072041e-14.
x = 3×1
-0.3333
0.6667
0.3333
## Input Arguments
collapse all
Coefficient matrix. The coefficient matrix appears in the system of linear equations on the left as Ax = B. The coefficient matrix can be full or sparse.
Data Types: single | double
Complex Number Support: Yes
Input array, specified as a vector or matrix. B appears in the system of linear equations on the right as Ax = B. If B is a matrix, then each column in the matrix represents a different vector for the right-hand side.
Data Types: single | double
Complex Number Support: Yes
Rank tolerance, specified as a nonnegative scalar. Specifying the tolerance can help prevent the solution from being susceptible to random noise in the coefficient matrix. By default, lsqminnorm computes tol based on the QR decomposition of A.
lsqminnorm computes the rank of A as the number of diagonal elements in the R matrix of the QR decomposition [Q,R,p] = qr(A,0) with absolute value larger than tol. If the rank of A is k, then the function forms a low-rank approximation of A by multiplying the first k columns of Q by the first k rows of R. Changing the tolerance affects this low-rank approximation of A.
Example: X = lsqminnorm(A,B,1e-2)
Data Types: double
Warning toggle for low-rank matrices, specified as either 'nowarn' or 'warn'. Specify 'warn' to indicate that lsqminnorm should produce warnings if the coefficient matrix A is rank deficient.
Example: X = lsqminnorm(A,B,'warn')
## Tips
• The minimum-norm solution computed by lsqminnorm is of particular interest when several solutions exist. The equation Ax = b has many solutions whenever A is underdetermined (fewer rows than columns) or of low rank.
• lsqminnorm(A,B,tol) is typically more efficient than pinv(A,tol)*B for computing minimum norm least-squares solutions to linear systems. lsqminnorm uses the complete orthogonal decomposition (COD) to find a low-rank approximation of A, while pinv uses the singular value decomposition (SVD). Therefore, the results of pinv and lsqminnorm do not match exactly.
• For sparse matrices, lsqminnorm uses a different algorithm than for dense matrices, and therefore can produce different results.
## Version History
Introduced in R2017b | 2022-08-19T05:18:52 | {
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https://stats.stackexchange.com/questions/199463/mle-for-independent-exponentials?noredirect=1 | MLE for Independent Exponentials
Here is the problem statement:
Let $(X_i,Y_i)$ be a random sample from a distribution with pdf $$f(x,y;\theta)= > \frac{1}{\theta^3}\exp\left(\frac{-x}{\theta}-\frac{y}{\theta^2}\right)\qquad > 0<x, 0<y.$$ (a) Find the MLE for $\theta$.
(b) Give the asymptotic distribution of $\sqrt{n}(\hat{\theta}-\theta).$
Here are my thoughts: We basically have two random samples, $X\sim \exp(\theta)$ and $Y\sim \exp(\theta^2)$. This pdf is the joint pdf for $X$ and $Y$. The parameter has no impact on the support of the variables $X$ and $Y$, so calculus should suffice for this problem. $$L(x,y;\theta) = \prod_{i=1}^n f(x,y;\theta) = \frac{1}{\theta^{3n}}\exp\left(\frac{-1}{\theta^2}\sum_{i=1}^n \theta x_i+y_i\right),$$ so $$l(x,y;\theta) = -3n\ln(\theta)-\frac{1}{\theta^2}\sum_{i=1}^n \theta x_i + y_i,$$ which has partial $\theta$ derivative $$\frac{-3n\theta^2+\sum_{i=1}^n \theta x_i +2y_i}{\theta^3}.$$ Equating to zero and solving for $\theta$ requires solving a quadratic in $\theta$.
The second $\theta$ derivative is $$\frac{3n\theta^2 - \left(\sum_{i=1}^n 2\theta x_i + 6y_i\right)}{\theta^4}.$$ The solutions to the quadratic are $$\theta = \frac{\sum_{i=1}^n x_i\pm \sqrt{\left(\sum_{i=1}^nx_i\right)^2+24n\sum_{i=1}^n y_i}}{6n}.$$ Is there no easier way to do the problem? Am I forced to make this substitution to find the maximum?
• Some brackets inside your SUMS would make your notation clearer. Your derivation so far is correct. You need to break the sum of $(x_i+y_i)$ into separate terms for x and y. Then you can solve the quadratic in $\theta$. – wolfies Mar 2 '16 at 4:10
• The roots are not convenient to write and the second derivative is not obviously nonpositive, so I feel I am still stuck on deciding if my roots are minima. Substitution seems nightmarish. – Hamilton Mar 2 '16 at 4:56
• It's just occurred to me: Is the MLE of a product the product of the MLE's? That would save quite a headache with algebra. – Hamilton Mar 2 '16 at 5:10
• Consider $\bar{x}=1$ and $\bar{y}=2$ and draw the log-likelihood(with $n=4$ if you need it, though $n$ is actually immaterial, you can pull a factor of $n$ out of every term). [You may find it easier to identify what's going on if you plot log-likelihood vs 1/theta but YMMV.] – Glen_b Mar 2 '16 at 5:21
• If you think about what's going on with the two solutions, you should be able to write a single solution down (the other you can eliminate for two reasons, one trivial, one still fairly simple, either of which is sufficient to rule it out). You should be able to then argue why the remaining turning point must indeed be a maximum. You can also simplify the formula a little though it's not necessary. – Glen_b Mar 2 '16 at 5:29
I hesitated to 'answer' the question before, because it was not clear if this was a homework assignment, ... but as you have now answered it yourself (and neatly so), I thought I might briefly reply to the question posed:
$$\text{"Is there no easier way to do the problem?"}$$
In exactly this vein, my co-author, Murray Smith, and I sought to automate these sorts of MLE problems, symbolically and exactly, in precisely the way you have proceeded - step by step - but getting the computer to do the grunt work. See, for instance:
• Rose, C and Smith, M.D. (2000), Symbolic maximum likelihood estimation with Mathematica, Journal of the Royal Statistical Society, Series D: The Statistician, 49(2), 2000, 229-240. Download available: here
A more sophisticated version of same was then built into the mathStatica software package which we later developed.
GIVEN: $$(X,Y)$$ have a bivarate Exponential distribution, with parameter $$\theta >0$$, with joint pdf $$f(x,y)$$:
(source: tri.org.au)
Activate the special SuperLog function:
(source: tri.org.au)
For a random sample of size $$n$$ drawn on $$(X,Y)$$, the log-likelihood function for parameter $$\theta$$ is:
(source: tri.org.au)
The score function is the gradient of the log-likelihood with respect to $$\theta$$:
(source: tri.org.au)
Setting the score to zero and solving for $$\theta$$ corresponds to the first-order condition:
(source: tri.org.au)
We require the second (positive) solution. As for second-order conditions, the Hessian, evaluated at the optimal solution, is:
(source: tri.org.au)
... which is strictly negative, given that $$X>0$$ and $$Y>0$$.
• I was thinking more about it today and I realized that the first derivative is sufficient. Once I have the solutions (which correspond to roots of a concave parabola), I can analyze the behavior near a zero to classify one solution as a maximum and one as a minimum. The second derivative IS unnecessary. Thanks for this code. – Hamilton Mar 2 '16 at 22:49 | 2021-04-13T05:04:56 | {
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http://math.stackexchange.com/questions/63176/is-there-a-weighted-left-inverse-of-a-matrix | # Is there a weighted left-inverse of a matrix?
We can find a left inverse $A^{L} = (A^T A)^{-1}A^T$
In a situation $Ax=b$ using this left inverse I can obtain $x=A^{L}b$
This provides with "best-fit" solution for $x$, if I were to re-compute $b' = Ax$ I will get $b \ne b'$
This "best-fit" notion is based on a root-mean-square deviation. ${1 \over n} \sqrt{\sum{(b_i-b_i')^2}}$
Question: if I wanted certain values within $b$ to contribute less than other values, I would want some "weight" associated with it. However, I do not want to simply evaluate a weighted RMSD value. I want to compute a "weighted" left inverse. -- How can this be done?
-
I think you mean $A^L = (A^T A)^{-1} A^T$ (assuming of course that $A^T A$ is invertible). – Robert Israel Sep 9 '11 at 19:20
that's what I meant! Fixed – Mikhail Sep 9 '11 at 19:21
So we're picturing $A$ having many more rows than columns. Then $M = (A^T A)^{-1}A^T$ is a left inverse of $A$ in the sense that $MA=\text{a small identity matrix}$, where "small" means having only as many rows and columns as $A$ has columns. Instead of speaking of minimizing the root-mean-square deviation, why not keep it simple and speak of minimizing the sum of the squares of the deviations? Anyway, "weighted" least squares is done similarly with certain formulas involving the weights on the diagonal. More on this later, unless someone beats me to it..... – Michael Hardy Sep 9 '11 at 19:26
There's also "generalized" least squares, where one uses some other inner product than the usual one and its matrix is not necessarily diagonal. – Michael Hardy Sep 9 '11 at 19:27
If you have a non-standard inner product (e.g., a weighted inner product), you want to consider $A$ as a linear transformation on the space, but compute its coordinate matrix relative to an orthonormal basis with respect to that inner product; then use the pseudo-inverse of that matrix to obtain a best-fit solution (or a minimal solution if the system is consistent) in terms of the orthonormal basis. Finally, translate back to standard coordinates. – Arturo Magidin Sep 9 '11 at 19:27
Consider the problem of minimizing $f(x) = \sum_i w_i (A x - b)_i^2$ where $w_i$ are positive weights. You can write this as $f(x) = (A x - b)^T W (A x - b)$ where $W$ is the diagonal matrix with diagonal entries $w_i$. Assuming $A^T W A$ is invertible, the solution is $x = (A^T W A)^{-1} A^T W b$.
-
It's not very intuitive (I'm not a math person) but I'll give it a try. Is there a way to perform this weighted derivation with SVD too? – Mikhail Sep 9 '11 at 20:33
Basically this is the same as the unweighted version if you replace $A$ by $W^{1/2} A$ and $b$ by $W^{1/2} b$. So if you know how to do unweighted least squares using SVD, you can use that with this adjusted matrix and vector. – Robert Israel Sep 9 '11 at 21:37
Here is a reformulation of the previous answers and comments which I hope will be somewhat helpful to the OP.
A. The problem you are interested in is the following: given an inner product $\langle \cdot, \cdot \rangle$ find $x$ such that $$\langle b - Ax, b - Ax \rangle$$ is minimized.
When $\langle \cdot, \cdot \rangle$ is the ordinary inner product, this is the ordinary least squares solution. When $\langle x, y \rangle = x^T W y$ where $W$ is some positive diagonal matrix, this the weighted case you are interested in.
B. The solution will satisfy the following optimality criterion: the error must be orthogonal to the column space of $A$.
Formally, let $a_1, \ldots, a_n$ be the columns of $A$. Then the optimal $x^*$ will satisfy $$\langle a_i, b-Ax^* \rangle = 0$$ for all $i$.
Why? Because if the error could be orthogonally decomposed as
$$b- Ax = x_{R(A)} + x_{R(A)^\perp}$$ where $x_{R(A)} \neq 0$ is the projection onto the range of $A$, and $x_{R(A)^\perp}$ is a projection onto its a complement, then we could pick a different $x$ to get a smaller error. Indeed, $$\langle b - Ax, b-Ax \rangle = \langle x_{R(A)}, x_{R(A)} \rangle + \langle x_{R(A)^\perp}, x_{R(A)^\perp} \rangle$$ by the Pythagorean theorem. Now if $x_{R(A)} = Ay$, then $$\langle b-A(x+y), b-A(x+y) \rangle = \langle x_{R(A)^\perp}, x_{R(A)^\perp} \rangle$$ which is smaller.
C. For the case of the ordinary inner product, the above optimality principle can be restated as $$A^T (b-Ax^*) = 0$$ which immediately gives you your least-squares solution; and for the case of the weighted inner product, it can be restated as $$A^T W (b-Ax^*)=0$$ which immediately gives you the weighted solution.
- | 2014-07-25T13:57:09 | {
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http://mathhelpforum.com/math-topics/97997-last-two-digit.html | 1. ## Last two digit
Find the last digit two digit of the expression?
(201 * 202 * 203 *204 * 246 * 247 * 248 * 249)^2.
Is there any shortcut to find the last two digit of the above expression.
Any help would be appreciated.
Thanks,
Ashish
2. Do you know the chinese remainder theorem? If so let me know, and I will show you a much easier way. I got 76.
If not, you can just multiply and reduce modulo 100 at each stage, and the worst you will ever have to multiply is some 2 digit numbers. It will take a while, but you will get the correct answer. I suggest multiplying in a smart fashion to reduce some of the work like do
2*49=98=-2 (mod 100)
4*48=2*2*48=2*-4=-8 (mod 100)
etc
Can you please let me know how we can do it with Chinese remainder theorem?
Thanks,
4. ## Chinese Remainder Theorem in Action
Sure thing mate. So the goal is to figure out what this beast is mod 100. Fortunately, we have $\displaystyle 100=2^2\cdot 5^2$. Now we notice that gcd(5^2,2^2)=1, which means we are in business for using the chinese remainder theorem. So this means we can just figure what this thing is mod 4 and mod 25 separately and the chinese remainder theorem will give us the unique solution mod $\displaystyle 100=25\cdot 4$.
mod 4 is really easy because one of the numbers is 0 mod 4, so the product will be 0.
We are not quite as fortunate when we look mod 25.
This is the product when each term has been reduced.
$\displaystyle (1\cdot 2\cdot 3\cdot 4\cdot -4 \cdot -3 \cdot -2 \cdot -1)^2=(24\cdot 24)^2=(-1\cdot -1)^2=1^2=1$ (mod 25).
So there is precisely 1 number between 1 and 100 which is both 0 mod 4 and 1 mod 25. By inspection it is pretty clear it is 76, so this will be the last two digits of this product.
Just to prove to you that the CRT works in general, you have numbers that are relatively prime, that is gcd(a,b)=1. Then say we are looking to solve this congruence
$\displaystyle n\equiv x$ (mod a)
$\displaystyle n\equiv y$ (mod b)
Then there is 1 number mod ab that will satisfy this and this is how you figure out in general what it is.
Use the Euclidean Algorithm to find integers s,t, such that $\displaystyle as+bt=1$. Now take the number yas+xbt notice the sort of cross multiplying aspect of choosing this number.
now we consider $\displaystyle yas+xbt$ (mod ab)
$\displaystyle yas+xbt \equiv xbt \equiv x(1-as) \equiv x-xas \equiv x$ (mod a)
Similarly
$\displaystyle yas+xbt \equiv yas \equiv y(1-bt) \equiv y-xbt \equiv y$ (mod b)
And presto, there you have it in general.
In our case we notice that $\displaystyle -6\cdot 4 + 1\cdot 25=1$, so we should choose $\displaystyle 1\cdot (-6)\cdot 4 + 0\cdot 1\cdot 25=-24=76$ (mod 100) as claimed. There ya have it.
5. Thanks a lot Gamma,it is very informative | 2018-06-19T07:15:44 | {
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https://byjus.com/question-answer/there-are-two-circles-whose-equations-are-x-2-y-2-9-and-x-2-1/ | Question
# There are two circles whose equations are $$x^{2}+y^{2}=9$$ and $$x^{2}+y^{2}-8x-6y+n^{2}=0,\ n\in Z$$. If the two circles have exactly two common tangents, then the number of possible values of $${n}$$ is
A
2
B
8
C
9
D
5
Solution
## The correct option is D $$9$$$$S_{1}:x^{2}+y^{2}=9$$$$C_{1}\equiv (0,0)$$ and $$r_{1}=3$$$$S_{2}:x^{2}+y^{2}-8x-6y+n^{2}=0$$$$C_{2}\equiv (4,3)$$ and $$r_{2}=\sqrt{25-n^{2}}$$ $$r_{2}=\sqrt{25-n^{2}}\gt0$$ so $$-5\leq n\leq 5$$ $$\therefore$$ Given $$S_{1}$$ and $$S_{2}$$ have two common tangents, $$S_{1}$$ and $$S_{2}$$ are intersecting each other.$$\therefore r_{1}+r_{2}>C_{1}C_{2} > |r_1-r_2|$$$$C_1C_2 > |r_1-r_2|$$ is true for all $$n$$.$$3+\sqrt{25-n^{2}}>5$$$$\sqrt{25-n^{2}}>2$$$$25-n^{2}>4$$$$n^{2}<21$$$$(n-\sqrt{21})(n+\sqrt{21})<0$$$$n\in (-\sqrt{21},\sqrt{21})$$ ...(1) and $$-5\leq n\leq 5$$ ...(2)From (1) & (2) $$n\in (-\sqrt{21},\sqrt{21}),n\in Z.$$$$n=-4,-3,-2,-1,0,1,2,3,4$$$$\Rightarrow$$ Number of values of $$n$$ are $$9$$Mathematics
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View More | 2022-01-20T18:21:22 | {
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https://math.stackexchange.com/questions/491633/eigenvalues-and-the-characteristic-equation | Eigenvalues and the Characteristic Equation
Given the following matrix,
$$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$$
assuming eigenvectors exist for $A$, they can be found by first solving for $\lambda$ (i.e. the roots of the equation) in the characteristic equation:
$$\text{det}(A-\lambda I) = 0$$
I know that if the determinant of a matrix is equal to zero, the matrix is non-invertible; also, I know that, for a given a matrix $A$, eigenvector $x$ and eigenvalue $\lambda$, $Ax = \lambda x$; hence, with respect to $x$, $A$ is somewhat "equivalent" to $\lambda$, but I'm not entirely sure why solving for the characteristic equation provides the eigenvalues for the matrix $A$.
Given this, my question is: could somebody provide some logic on why the above works?
Also, as an aside, assuming the correct eigenvalues have been found, solving for the system,
$$\begin{bmatrix} a_{11}-\lambda & a_{12} \\ a_{21} & a_{22}-\lambda \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \vec{0}$$
will provide the associated eigenvector for a given $\lambda$.
Is it correct to assume the reasoning behind this is because of the following.
Firstly,
$$\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \lambda \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix}$$
This implies,
$$\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \left( \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} \right) \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \vec{0}$$
Thus,
$$\begin{bmatrix} a_{11} - \lambda & a_{12} \\ a_{21} & a_{22} - \lambda \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \vec{0}$$
• It's really confusing what you want to ask – Mathematician Sep 12 '13 at 12:53
You're nearly there.
A square matrix $B$ is non-invertible if and only if there exists a non-zero vector $v$ such that $Bv=0$. For example, necessity follows because $Bv=0$ implies that the function $f(v)=Bv$ is not injective (or "one-to-one") and hence not bijective (which is necessary for $f$ to have an inverse -- see the link). It is easy to see that $B$ is not injective since $B(2v)=2Bv=2\cdot0=0=Bv$, that is, $f$ maps both $v$ and $2v$ onto the same point $0$.
So, if $\lambda$ is an eigenvalue of $A$, and $x$ is its corresponding eigenvector,
$$Ax=\lambda x\Leftrightarrow Ax-\lambda x=0\Leftrightarrow (A-I\lambda)x=0.$$
Hence, $\lambda$ must be such that $B=A-I\lambda$ is non-invertible. Thus $\lambda$ is an eigenvalue of $A$ if and only if it satisfies the characteristic equation $\det(A-I\lambda)=0$.
Aside: If $\lambda$ is real, $x$ is simply a vector that function $f(v)=Av$ maps onto "itself" just stretches it and/or reflects it across the origin. For example, if $\lambda=2$, $f(x)$ simply "stretches" $x$ by two, and if $\lambda=-1$, $f(x)$ reflects $x$ across the origin (rotates it by $180$ degrees).
Edit: You might find these cam-casts of interest.
• Additionally, if some $\lambda_i = 0$, then $A$ is not invertible, and the eigenspace corresponding to $\lambda_i$ is the null space of $A$. – rcollyer Sep 12 '13 at 15:05
Your reasoning is correct; $\lambda$ is an eigenvalue of square matrix $A$ by definition iff there exists a nonzero vector $x$ s.t. $(A-\lambda I)x = 0$. This is equivalent to $(A-\lambda I)$ being singular (non-invertible), and to $\det(A-\lambda I) = 0$.
So if you solve for the polynomial roots of the "characteristic equation" $\det(A-\lambda I) = 0$, those values will be exactly the eigenvalues of $A$.
Finding an eigenpair $(\lambda, x)$ involves finding the eigenvalue $\lambda$ and its eigenvector $x$, such that \begin{gather} Ax = \lambda x. \end{gather}
This is equivalent to \begin{gather} Ax- \lambda x = 0. \end{gather}
This system can only have a nonzero solution $x$, if the matrix $(A-\lambda I)$ ($I$ being the identity) is singular. And one criterion for being singular is that its determinant is zero.
you want to find solutions ,$\lambda$,such that : $Ax=\lambda x$ where $x\ne0$ or equivalently $Ax-\lambda x=(A-\lambda I)x=0$ in order to have solutions for this equation the expression $|A-\lambda I|$ must equal zero because if the $A-\lambda I$ was invertable then we have only one solution which is zero vector and by the definition of the problem this not valid so the matrix $A-\lambda I$ must be non-invertable and it is non invertable if and only if $|A-\lambda I|=0$ . so the problem of finding eigenvalues is reducible to the problem of finding roots of polynomial . then we want to find the span of all vectors that satisfies the equations , so you must find a set of independent vectors that for all $\lambda$ the set is a space of all possible solutions $x$ | 2019-07-21T19:45:44 | {
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https://math.stackexchange.com/questions/1354828/how-to-solve-this-recurrence-kn-2kn-1-kn-2c/1354874 | # How to solve this recurrence $K(n)=2K(n-1)-K(n-2)+C$?
The recurrence is $K(n)=2K(n-1)-K(n-2)+C$ where $C$ is a constant. What I have tried is substituting $2K(n-1)$ as we do in fibonnacical recurrences. It didn't gave me a fruitful expression! Can someone help in solving it? Not a homework problem.
$$\begin{bmatrix} K(n+1) \\ K(n) \end{bmatrix}=A \begin{bmatrix} K(n) \\ K(n-1) \end{bmatrix}+ \begin{bmatrix} C \\ 0 \end{bmatrix}$$
where
$$A= \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}.$$
Therefore,
$$\begin{bmatrix} K(n+1) \\ K(n) \end{bmatrix}=A^n \begin{bmatrix} K(1) \\ K(0) \end{bmatrix}+ \left(\sum_{k=1}^{n-1}A^k+I\right) \begin{bmatrix} C \\ 0 \end{bmatrix}$$
and
$$K(n)=nK(1)-(n-1)K(0)+\frac{1}{2}Cn(n-1)$$
by noticing that
$$A^k= \begin{bmatrix} k+1 & -k \\ k & k-1 \end{bmatrix}.$$
### The other answers are way too complicated for this particular problem.
They're useful in more general cases, but they're completely overkill here.
\begin{align*} K(n) &= 2 K(n - 1) - K(n - 2) + C \\ K(n) - K(n - 1) &= K(n - 1) - K(n - 2) + C \end{align*}
Just look at this equation for a few seconds.
It's literally telling you that the difference between successive elements increases by $C$ every step.
So... just go ahead and count how many times you add $C$ to the difference $K(1) - K(0)$:
$$K(n) = K(0) + \sum_{k=1}^{n} K(1) - K(0) + (k - 1) C$$
Notice you don't need any linear algebra, eigenvectors, or other higher-level math for this problem. It's just algebraic manipulation.
I'll leave the last step of simplifying the summation to you.
## Edit:
or I'll just do it for you myself, since you seem to think it leads to another recurrence...
\begin{align*} K(n) &= K(0) + n(K(1) - K(0)) + C\sum_{k=1}^{n} (k-1) \\ &= K(0) + n(K(1) - K(0)) + C \frac{n(n-1)}{2} \end{align*}
• indeed Beautiful ! – Shubham Sharma Jul 9 '15 at 11:09
• Thanks! I'll admit that it wasn't obvious to me -- at first I was trying to be "creative" by turning this into a continuous equation, via converting differences into derivatives and seeing if I knew the solution to the (delay-?)differential equation. But as soon as I subtracted $K(n-1)$ from both sides to turn the differences into derivatives, I saw there was a much easier way to solve this, hence this answer. :) – Mehrdad Jul 9 '15 at 11:12
• @MichaelGaluza: Dude, my solution is quadratic... – Mehrdad Jul 9 '15 at 15:42
• @ShubhamSharma: Michael is totally wrong, but do you really need me to do the last step for you? No, this shouldn't lead you to another recurrence. It just simplifies to $K(0) + n(K(1) - K(0)) - n + C\sum_{k=1}^{n} k$ and $\sum_{k=1}^{n} k$ is just $n(n+1)/2$. This is the right answer to your question. – Mehrdad Jul 9 '15 at 15:46
• @ShubhamSharma: I just realized, you might want to look up the phrase linear constant-coefficient difference equation. – Mehrdad Jul 9 '15 at 17:08
To keep things general, suppose $k_0=A$ and $k_1=B$. Then the next few terms are: $$k_2=2A-B+C\\ k_3=4A-3B+3C\\ k_4=6A-5B+6C\\ k_5=8A-7B+10C\\ k_6=10A-9B+15C$$
The pattern seems to be (for $n\ge2$) that 2 more $A$'s are added, 2 more $B$'s are subtracted, and $n-1$ more $C$'s are added, $$k_n=(2n-2)A-(2n-3)B+\frac{(n-1)(n)}{2}C$$
A proof by strong induction along with some messy algebra will give you your answer
More standard way. Rewrite your equation: $$K(n)-2K(n-1)+K(n-2)=C \tag{1}\label{1}$$ Solution of this is $K=K_0+K_{part}$, where $$K_0(n)-2K_0(n-1)+K_0(n-2)=0\tag{2}\label{2}$$ and $K_{part}$ is any solution of $\eqref{1}$.
Now we're finding solution of homogenuous equation $\eqref{2}$ in a form $K(n)=\mathrm{const}\cdot \lambda^n$ and get $$\lambda^{n}-2\lambda^{n-1}+\lambda^{n-2}=0\Longrightarrow \lambda^2-2\lambda+1=0;$$ $\lambda_1=\lambda_2=1$, and $$K_0(n)=A+Bn.\tag{3}\label{3}$$
$K_{part}$ we'll find in a form $K_{part}=\alpha n^2$ (polynom of degree $0$ and $1$ we used yet). Substitute it in $\eqref{1}$ and get $2\alpha=C$.
So, solution is $$K(n)=A+Bn+\frac{Cn^2}{2}.$$
If you prefer, we can take $K(0)=A$ and $K(1)=A+B+C/2$; hence, $$K(n)=K_0+(K_1-K_0)n+\frac{Cn(n-1)}{2}$$
• might be nice to explain how you just came up with the form $c \lambda^n$, because it looks pretty magical to someone who doesn't already know that's the right thing to do. – Mehrdad Jul 9 '15 at 10:02
• @Mehrdad, I can say just "it works". But, if you want perfect rigor, read whatever contains "recurrence relation". – Michael Galuza Jul 9 '15 at 11:35 | 2019-05-19T23:02:32 | {
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https://math.stackexchange.com/questions/465103/simplify-sqrt-5-sqrt6-sqrt7%E2%88%92-sqrt5-sqrt6-sqrt7-sqrt5-%E2%88%92-s | # Simplify : $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7)$
The question is to simplify $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7)$ without using a calculator .
My friend has given me this challenge . I solved it by expanding $$(a+b+c)(-a+b+c)(a-b+c)(a+b-c) = -a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4$$ and then substituting $a,b,c=\sqrt 5 , \sqrt6 , \sqrt7$ respectively to get the answer $104$ .
But I suppose there is a more elegant and easy way to solve this problem .
Can anyone find it ?
• Which contest's question did your friend get it from?Is it from the BdMO 2006 Nationals? – rah4927 Aug 11 '13 at 16:46
• @rahul I don't know but it probably isn't from bdMO . – A Googler Aug 11 '13 at 16:57
• Hmm. . .But I am sure this was the 6th question in the Junior section of the Bdmo 2006 nationals. – rah4927 Aug 11 '13 at 17:03
• @rahul really ? then it may be. Can you share the link to it ? – A Googler Aug 11 '13 at 17:32
• You can find the question from BdMO's official website in the Questions section. – rah4927 Aug 11 '13 at 17:36
As $$(a+b+c)(-a+b+c)=\{(b+c)+a\}\{(b+c)-a\}=(b+c)^2-a^2,$$
$$(\sqrt5 +\sqrt6 +\sqrt7)(−\sqrt5+\sqrt6+\sqrt7)$$ $$=(\sqrt6+\sqrt7)^2-(\sqrt5)^2=6+7+2\sqrt7\cdot\sqrt6-5=8+2\sqrt{42}=2\sqrt{42}+8$$
Again as $$(a-b+c)(a+b-c)=\{a+(b-c)\}\{a-(b-c)\}=a^2-(b-c)^2,$$ $$(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7)=\{\sqrt5 − (\sqrt6-\sqrt7)\}\{\sqrt5 + (\sqrt6-\sqrt7)\}$$ $$=(\sqrt5)^2-(\sqrt6 − \sqrt7)^2=5-(6+7-\sqrt7\cdot\sqrt6)=2\sqrt{42}-8$$
So, the product $$=(2\sqrt{42}-8)(2\sqrt{42}+8)=(2\sqrt{42})^2-8^2=4\cdot42-64=104$$
Consider Heron's formula: the area of a triangle with sides $a, b, \text{and } c$ is
$$\sqrt{s(s-a)(s-b)(s-c)}$$
where $s$ is the semi-perimeter $\frac12 (a + b + c)$.
Let $a, b, \text{and } c$ be $\sqrt{5}, \sqrt{6}, \text{and } \sqrt{7}$. Then the area is the square root of your expression divided by $4$. So, what is the area of this triangle? Use the law of cosines to find the cosine of the angle $C$ opposite $c$:
\begin{align} 7 &= 5 + 6 - 2 \sqrt{5}\sqrt{6}\cos{C}\\ 2\sqrt{30}\cos{C} &= 4\\ \cos{C} &= \frac{2}{\sqrt{30}} \end{align}
But the area of the triangle is $\frac12 ab\sin{C}$.
$$\frac12 ab\sin{C} = \frac12 \sqrt{30} \frac{\sqrt{26}}{\sqrt{30}} = \frac12\sqrt{26}.$$
Your expression is therefore the square of $2\sqrt{26}$, which is $104$.
• Delightful! ${}$ – Jyrki Lahtonen Aug 12 '13 at 6:47
• The poster wanted a solution without needing a calculator or using complex algebra, and the expression looked like the one in Heron's formula. – Eric Jablow Aug 12 '13 at 12:51
• Wow. I'm speechless. – A Googler Aug 12 '13 at 16:47
Use $$(a+b)(a-b)=a^2-b^2.$$
With $a=\sqrt6+\sqrt7$, $b=\sqrt5$ you see that the product of first two numbers is $(\sqrt6+\sqrt7)^2-5=8+2\sqrt{42}$. With $a=\sqrt5$, $b=\sqrt7-\sqrt6$ you get that the product of last two is $5-(\sqrt7-\sqrt6)^2=-8+2\sqrt{42}$. One last application of this rule tells you that the answer is $$(2\sqrt{42})^2-8^2=168-64=104.$$
Here's a nice way to get the expansion the OP used:
The expression $$P(a,b,c)=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)$$ is clearly a homogeneous polynomial of degree $4$, symmetric in its three variables. It's also clear that the coefficient of $a^4$ (hence also $b^4$ and $c^4$) is $-1$. Moreover, $$P(-a,b,c)=P(a,-b,c)=P(a,b,-c)=P(a,b,c)$$ which implies $P$ has no terms with any variable taken to an odd degree. Therefore $P$ must be of the form
$$P(a,b,c)=r(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)$$
for some coefficient $r$. It's easy to determine $r$ by letting $a=b=c=1$, for which we have
$$3=3\cdot1\cdot1\cdot1=P(1,1,1)=r(1+1+1)-(1+1+1)=3r-3$$
so $r=2$. The rest of the answer follows what the OP did:
$$P(\sqrt5,\sqrt6,\sqrt7)=2(5\cdot6+6\cdot7+7\cdot5)-(5^2+6^2+7^2)=104$$
Added 8/12/13: Eric Jablow's invocation of Heron's formula inspires one more approach:
Consider the triangle formed by the origin and two vectors $x$ and $y$, with $|x|=a$, $|y|=b$, and $|x-y|=c$. There are two formulas for the area of the triangle: Heron's formula
$$\sqrt{s(s-a)(s-b)(s-c)}$$
where $s=(a+b+c)/2$, and the dot-product formula
$${1\over2}\sqrt{|x|^2|y|^2-(x\cdot y)^2}$$
Putting these together, we have
$$(a+b+c)(-a+b+c)(a-b+c)(a+b-c)=4((a^2b^2-(x\cdot y)^2)$$
What's nice is that this holds for vectors $x$ and $y$ in any dimension. So now let $x=(2,1,0,0)$ and $y=(0,2,1,1)$, so that $x-y=(2,-1,-1,-1)$. We have $a=\sqrt5$, $b=\sqrt6$, $c=\sqrt7$, and $x\cdot y=2$, and thus the OP's product of square roots simplifies to
$$4(5\cdot6-2^2) = 104$$
If you want to know why we went into the fourth dimension for the vectors $x$ and $y$, it's because $7$, like all positive integers, can be written as the sum of four squares, but not as the sum of three. Another possibility, which makes it clear one can handle arbitrary $a$, $b$, and $c$ (as long as the triangle inequality is satisfied, at least) is to let $x=(1,1,1,1,1,0,0,0,0)$ and $y=(0,0,0,1,1,1,1,1,1)$, so that $x-y=(1,1,1,0,0,-1,-1,-1,-1)$. (Edit: Actually, what I just said only works when the triangle inequality holds and $a+b+c$ is even.)
• Sorry , but I don't understand why P has no terms with odd degree . Can you please explain it ? – A Googler Aug 11 '13 at 17:35
• @AGoogler, $P(a,b,c)=(P(a,b,c)+P(-a,b,c))/2$ rules out the possibility for $P$ to have any terms with $a$ of odd degree, and similarly for $b$ and $c$. – Barry Cipra Aug 11 '13 at 19:06
Given that your original numbers are square roots and your expanded equation involves only even powers of your original numbers, you could say:
$$A=a^2=5, B=b^2=6, C=c^2=7$$
$$2(AB+AC+BC) - (A^2+B^2+C^2) = 2(30+35+42) - (25+36+49) = 214-110 = 104$$ | 2021-07-30T00:40:16 | {
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http://www.mathworks.com/help/matlab/ref/rat.html?nocookie=true | # rat
Rational fraction approximation
## Syntax
• ```[N,D] = rat(___)``` example
## Description
example
````R = rat(X)` returns the rational fraction approximation of `X` to within the default tolerance, `1e-6*norm(X(:),1)`. The approximation is a string containing the truncated continued fractional expansion.```
example
````R = rat(X,tol)` approximates `X` to within the tolerance, `tol`.```
example
``````[N,D] = rat(___)``` returns two arrays, `N` and `D`, such that `N./D` approximates `X`, using any of the above syntaxes.```
## Examples
collapse all
### Approximate Value of π
Approximate the value of π using a rational representation of the quantity `pi`.
The mathematical quantity π is not a rational number, but the quantity `pi` that approximates it is a rational number since all floating-point numbers are rational.
Find the rational representation of `pi`.
```format rat pi ```
```ans = 355/113 ```
The resulting expression is a string. You also can use `rats(pi)` to get the same answer.
Use `rat` to see the continued fractional expansion of `pi`.
`R = rat(pi)`
```R = 3 + 1/(7 + 1/(16)) ```
The resulting string is an approximation by continued fractional expansion. If you consider the first two terms of the expansion, you get the approximation $3+\frac{1}{7}=\frac{22}{7}$, which only agrees with `pi` to 2 decimals.
However, if you consider all three terms printed by `rat`, you can recover the value `355/113`, which agrees with `pi` to 6 decimals.
$3+\frac{1}{7+\frac{1}{16}}=\frac{355}{113}\text{\hspace{0.17em}}.$
Specify a tolerance for additional accuracy in the approximation.
```R = rat(pi,1e-7) ```
```R = 3 + 1/(7 + 1/(16 + 1/(-294))) ```
The resulting approximation, `104348/33215`, agrees with `pi` to 9 decimals.
### Express Array Elements as Ratios
Create a 4-by-4 matrix.
```format short; X = hilb(4)```
```X = 1.0000 0.5000 0.3333 0.2500 0.5000 0.3333 0.2500 0.2000 0.3333 0.2500 0.2000 0.1667 0.2500 0.2000 0.1667 0.1429```
Express the elements of `X` as ratios of small integers using `rat`.
`[N,D] = rat(X)`
```N = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 D = 1 2 3 4 2 3 4 5 3 4 5 6 4 5 6 7```
The two matrices, `N` and `D`, approximate `X` with `N./D`.
View the elements of `X` as ratios using ```format rat```.
```format rat X```
```X = 1 1/2 1/3 1/4 1/2 1/3 1/4 1/5 1/3 1/4 1/5 1/6 1/4 1/5 1/6 1/7 ```
In this form, it is clear that `N` contains the numerators of each fraction and `D` contains the denominators.
## Input Arguments
collapse all
### `X` — Input arraynumeric array
Input array, specified as a numeric array of class `single` or `double`.
Data Types: `single` | `double`
Complex Number Support: Yes
### `tol` — Tolerancescalar
Tolerance, specified as a scalar. `N` and `D` approximate `X`, such that `N./D - X < tol`. The default tolerance is `1e-6*norm(X(:),1)`.
## Output Arguments
collapse all
### `R` — Continued fractionstring
Continued fraction, returned as a string. The accuracy of the rational approximation via continued fractions increases with the number of terms.
### `N` — Numeratornumeric array
Numerator, returned as a numeric array. `N./D` approximates `X`.
### `D` — Denominatornumeric array
Denominator, returned as a numeric array. `N./D` approximates `X`.
collapse all
### Algorithms
Even though all floating-point numbers are rational numbers, it is sometimes desirable to approximate them by simple rational numbers, which are fractions whose numerator and denominator are small integers. Rational approximations are generated by truncating continued fraction expansions.
The `rat` function approximates each element of `X` by a continued fraction of the form
$\frac{N}{D}={D}_{1}+\frac{1}{{D}_{2}+\frac{1}{\left({D}_{3}+...+\frac{1}{{D}_{k}}\right)}}\text{\hspace{0.17em}}.$
The Ds are obtained by repeatedly picking off the integer part and then taking the reciprocal of the fractional part. The accuracy of the approximation increases exponentially with the number of terms and is worst when `X = sqrt(2)`. For `X = sqrt(2)` , the error with `k` terms is about `2.68*(.173)^k`, so each additional term increases the accuracy by less than one decimal digit. It takes 21 terms to get full floating-point accuracy. | 2015-05-05T21:50:44 | {
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https://math.stackexchange.com/questions/238997/prove-the-divergence-of-the-sequence-left-sinn-right-n-1-infty?noredirect=1 | # Prove the divergence of the sequence $\left\{ \sin(n) \right\}_{n=1}^{\infty}$.
I am looking for nice ways of proving the divergence of the sequence $\left\{x_n\right\}_{n=1}^{\infty}$ defined by $$x_n=\sin{(n)}.$$ One (not so nice) way is to construct two subsequences: one where the indexes are picked such that they lie in the intervals $$I_k=\left(\dfrac{\pi}{6}+2\pi(k-1),\dfrac{5\pi}{6}+2\pi(k-1)\right)$$ and one where they lie in $$J_k=\left(\dfrac{7\pi}{6}+2\pi(k-1),\dfrac{11\pi}{6}+2\pi(k-1)\right).$$ If ${x_n}$ converges, then all its subsequences must converge to the same limit, but here the first subsequence has all its values in the interval $\left[ \frac{1}{2}, 1\right]$ while the second has its values in $\left[-1,-\frac{1}{2}\right]$. Contradiction.
Filling the details of this proof is rather tedious and not very elegant. Anyone has a better idea?
• Hi, Sos440 Is it possible to extend this method for a proof of divergence of $sin(n!)$? – M.R. Yegan Oct 1 '14 at 4:43
Equidistribution argument is very elegant, but it becomes a sledgehammer method when it comes to a question of mere convergence.
A simple argument can reveal the divergence of $(\sin n, n \geq 1)$. Let
$$(x_n, y_n) = (\cos n, \sin n)$$
(I changed the notation for the sake of consistency of notation.) Then the application of the addition formula for trigonometric function or the rotation matrix gives
\begin{align*} x_{n+1} &= x_n \cos 1 - y_n \sin 1 \\ y_{n+1} &= x_n \sin 1 + y_n \cos 1. \end{align*}
Now assume $(y_n)$ converges. Then since $\sin 1 \neq 0$, we have
$$x_{n+1} = (y_{n+1} - y_n \cos 1) \cot 1 - y_n \sin 1$$
and hence $(x_n)$ also converges. Now let $(x_n, y_n) \to (\alpha, \beta)$. Then taking limit to the recursive formula we have
\begin{align*} \alpha &= \alpha \cos 1 - \beta \sin 1 \\ \beta &= \alpha \sin 1 + \beta \cos 1. \end{align*}
Solving this system of linear equations give $(\alpha, \beta) = (0, 0)$. On the other hand, since
$$x_n^2 + y_n^2 = 1,$$
we must have
$$\alpha^2 + \beta^2 = 1,$$
a contradiction! Therefore $(y_n)$ cannot converge. ////
Of course, we can say much more on $(y_n)$. For example, we can show that the set of limit points of $(y_n)$ is exactly $[-1, 1]$, and the Cesaro mean of $(y_n)$ is 0 from Weyl's criterion.
(This proof is from the book Problems in Real Analysis Advanced Calculus on the Real Axis.)
• This is very nice! – Jason DeVito Nov 17 '12 at 2:42
• Very nice indeed! Thanks! – Spenser Nov 17 '12 at 2:45
• Thanks! Anyway I forgot to comment that this is not my original idea, but the idea that I learned from the book Problems in Real Analysis Advanced Calculus on the Real Axis. – Sangchul Lee Nov 17 '12 at 2:49
• Very nice argument! So this also proves that $(x_n)=(\cos n)$ cannot converge. – Paul Apr 16 '13 at 21:21
• Another quick way to see that, if $\lim_{n \to \infty} \sin(n)$ and $\lim_{n \to \infty} \cos(n)$ exist, then both limits are zero is to note that $\sum_{n=0} \sin(n)$ and $\sum_{n=0}^\infty \cos(n)$ have bounded partial sums. Indeed, $\sum_{k=0}^n \sin(k)$ and $\sum_{k=0}^n \cos(k)$ are the real and imaginary parts of $\sum_{k=0}^n e^{ik} = \frac{e^{i(n+1)} - 1}{e^i - 1}$ which is bounded. – Mike F May 26 '14 at 8:34
If $\alpha$ is an irrational number, the numbers $n\alpha$, considered mod $1$, are dense in $[0,1]$. (A stronger result is that they are equidistributed.) Thus the numbers $2n/\pi$ are dense in $[0,1]$, and therefore the integers are dense mod $\pi/2$. It follows that $\sin n$ diverges.
Imagine marching around the circumference of the unit circle in steps of arc length $1$. Then $\sin(n)$ is the $y$ coordinate at the $n$th step. Since $\pi\gt1$, the $y$ coordinate will be positive infinitely often and negative infinitely often. Consequently the limit of $\sin(n)$, if it exists, would have to be $0$. But for the same reason ($\pi\gt1$), $\sin(n)\ge\sin({\pi-1\over2})=\sin(\pi-{\pi-1\over2})$ for infinitely many $n$, and hence the limit, if it exists, would have to be at least $\sin({\pi-1\over2})$, which is greater than $0$. These contradictory requirements show that the limit does not exist.
Remark: This proof does not rely on $\pi$ being an irrational number; nor does it use any trig identities beyond the simplest symmetries of the sine function.
This is a simplified version of the proof posted by Sangchul Lee.
If the limit of $\sin{n}$ is $s$, from $$\sin{(n+1)}= \sin{n}\cos{1}-\cos{n}\sin{1}$$ it follows that $\cos{n}$ is also convergent to a limit $c$. Taking limits in the equality above and the similar one for $\cos{(n+1)}$, we get that $$s=s\cdot\cos{1} + c\cdot\sin{1}$$ and $$c=c\cdot\cos{1} - s\cdot\sin{1}.$$ The simplification is that we don't need to solve the system, just add these two equalities after multiplying them by $s$ and $c$, respectively, to obtain $1=\cos{1}$, a contradiction.
• Welcome to math.SE. The answer is difficult to read. You could use MathJax to format your formula. – Yujie Zha May 21 '17 at 16:41
• For completeness, you would also have to rule out $s=c=0$, since you are dividing by $s^2 + c^2$ to conclude that $1 = \cos 1$. – Bungo May 21 '17 at 17:57
• There is no need for that, the fact that $s^2+c^2=1$ follows from taking limits in $\sin^2{n}+\cos^2{n}=1$. – Serban May 21 '17 at 18:25
This question was posed today (April 16, 2013) as https://mathoverflow.net/questions/127726/integer-multiples-of-a-irrational-dense-in-r-z and Douglas Zare gave a nice trig-identity-based proof in comments there.
• You might wants to give a quick paraphrase/quote the comment text here. Even though we may not expect the MO like to rot, but comments are ephemeral. (Perhaps, equally important, this s a link-only answer). – The Long Night Nov 3 '18 at 12:24
Essentially, every point in the interval $[-1, 1]$ is a limit point for the sequence $\left\{ {\sin n} \right\}$. Since there is more than one limit point, the sequence diverges.
Definition:
[x] is a rounding function if [x] is the biggest integer such that for a real number x, ⌊x⌋
Theorem 1:
Define x_n=[〖10〗^n x]/〖10〗^n for a real number x. Then lim(n→∞)〖x_n=x〗. Proof:
|x_n-x|=|([〖10〗^n x]-〖10〗^n x)/〖10〗^n |≤1/〖10〗^n
Theorem 2:
|[〖10〗^n π]-〖10〗^n π| diverges.
Proof:
Suppose |[〖10〗^n π]-〖10〗^n π| converges to a real number. This implies that there are integers k between 0 and 10 and K such that for n≥K, the nth decimal place of π is k. This is absurd since π is irrational.
Theorem 3:
sin(n) diverges.
Proof:
If sin[〖10〗^n x] has a limit s, then for any ϵ>0 there exist an integer K such that for n≥K, |s-sin[〖10〗^n x]|< ϵ. However, this implies that |[〖10〗^n π]-〖10〗^n π| converges which is proven to be false.
To see that the sequence $\left|\sin(n)\right|$ isn't Cauchy it is by the periodicity of $\sin$ enough to show the following: whenever $x_{0}\in[0,\pi-2], x_{1}=x_{0}+1, x_{2}=x_{0}+2$, we have $$\left|\sin(x_{0})-\sin(x_{1})\right|>\eta \text{ or } \left|\sin(x_{1})-\sin(x_{2})\right|>\eta.$$ But this is obvious because either $x_0,x_1\in[0,\pi/2]$ or $x_1,x_2\in[\pi/2,\pi]$, and $\sin$ is strictly monotone in these intervals. For example in the case $x_{0},x_{1}\in[0,\pi/2]$ we have $$\left|\sin(x_{0})-\sin(x_{1})\right|=\sin(x_1)-\sin(x_0)\leq\sin(\pi/2)-\sin(\pi/2-1)=:\eta,$$ where we estimated using the concavity of sin in the interval $[0,\pi/2]$.
A known resultis that if {x_n} and {y_n} are two sequences such that y_n is divergent and x_n\y_n converges to 0 then x_n also diverges. Take x_n=sin (n) and y_n=n then the result follows.
• This is false. (What happens if you take $x_n = 1$ and $y_n = n$ with the same argument?) – mrf May 26 '14 at 7:59 | 2019-12-10T08:14:26 | {
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http://math.stackexchange.com/questions/328468/find-q-and-r-with-0-leq-r-leq-b-such-that-a-qbr/328471 | # Find $q$ and $r$ with $0\leq r\leq |b|$, such that $a=qb+r$
Find $q$ and $r$, with $0\leq r\leq |b|$, such that $a=qb+r$ for
• $a=115,\ b=26$
• $a=400,\ b=-17$
• $a=-312,\ b=-64$
Sadly I missed the class where the prof went over this, so I have no idea what to do. Can somebody point me in the direction of the name of a process or theorem or something so that I can find out how to do it?
Thanks..
-
Hmmm. I understand how the Euclidean algorithm works for finding relative primality and the GCD of two numbers... Is this similar? It didn't appear to be when I first read it. – agent154 Mar 12 '13 at 14:46
Yes. Very similar. This is a single iteration in Euclid's algorithm for findsing the gcd. – Jyrki Lahtonen Mar 12 '13 at 15:15
This is the integer division algorithm with quotient $\rm\,q\,$ and remainder $\rm\,r.\:$ Yes, the division algorithm is the inductive step in the Euclidean algorithm for the gcd. – Math Gems Mar 12 '13 at 18:02
You should have dealt with the case when $a \ge 0$ and $b > 0$ in school.
When $a < 0$ and $b > 0$ divide first $-a$ by $b$ $$-a = b q + r, \qquad 0 \le r < b,$$ then consider $$a = b (-q) -r$$ If $r = 0$, then you're done. If $0 < r < b$, then $$a = b (-q -1) + b - r, \qquad 0 < b - r < b.$$ Finally, if $b < 0$, divide $a$ by $-b$ $$a = (-b) q + r = b (-q) + r, \qquad 0 \le r < -b = \vert b \rvert,$$ done.
-
Hint:
If $(q_0,r_0)$ is the solution to $a=qb+r$ then $(q_0-k,r_0)$ is the solution to $a-kb = qb+r$. In fact $q = \frac{a-r}{b}$ and because $0\leq r < |b|$ we have $0\leq\left|\frac{a}{b}-q\right|<1$.
Good luck ;-)
- | 2014-08-28T11:41:02 | {
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http://openstudy.com/updates/50494a19e4b063f50cff4b16 | ## eliassaab Group Title Let m and n be two positive integers. Show that (36 m+ n)(m+36 n) cannot be a power of 2. 2 years ago 2 years ago
1. sauravshakya
LET (36m+n)=2^x AND (m+36n)=2^y THEN, (36m+n)(m+36n)=2^(x+y) AlSO, x and y must me both integers
2. sauravshakya
Now, n=2^x - 36m THEN, m+36n=2^y m+36(2^x-36m)=2^y 36 * 2^x -1295m=2^y
3. sauravshakya
Now, I think I have to prove that y is never a integer
4. sauravshakya
2^y=36*2^x-1295m 2^y=2^x(36-1295m/2^x) Now, 2^y must be positive so, (36-1295m/2^x) must be positive... Now let 2^z=36-1295m/2^x HERE z also must be an positive integer..... So, 2^z can be 4,8,16,32 NOW,
5. sauravshakya
36-1295m/2^x= 4, 8, 16, 32 1295m/2^x=32, 28 ,16 ,4 2^x=1295m/32 , 1295m/28 , 1295m/16 , 1295m/4
6. sauravshakya
Now, 2^x=1295m/32 , 1295m/28 , 1295m/16 , 1295m/4 2^x=40.47 m , 46.25m , 80.94m ,323.75m Thus, for no positive integer value of m we will get x a integer value......
7. sauravshakya
Hence, (36 m+ n)(m+36 n) cannot be a power of 2.
8. sauravshakya
I have jumped some step......... I hope I made it clear
9. sauravshakya
@eliassaab
10. sauravshakya
@ganeshie8
11. mukushla
for $$m=n$$ the statement is true suppose $$m\ge n$$ in order for $$(36 m+ n)(m+36 n)$$ to be a power of $$2$$ $36m+n=2^a$$m+36n=2^b$$a> b$$\frac{36m+n}{m+36n}=2^{a-b}=2^c \ \ \ c\ge 1$$m+36n | 36m+n$$m+36n|36(m+36n)-(36m+n)=1295n$since $$\gcd(m+36n \ , \ n)=1$$ so $$m+36n$$ is a divisor of $$1295$$ so $$(36 m+ n)(m+36 n)$$ is not a power of 2
12. experimentX
looks like gcd is very important ... i never liked it.
13. eliassaab
Here is my proof. Suppose not, let m and n be the smallest such that this product is a power of 2. It is easy to see that m and n are divisible by 2. So m = 2 M and n = 2 N. Since ( 36 (2M) + 2 N)(36(2N)+2 M) is a power of 2,then ( 36 ( M) + N)(36( N+ M) is a power of 2 (contradiction).
14. mukushla
Short and Neat :) | 2014-09-22T14:23:48 | {
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https://trustreview.net/polaroid-png-voqvnq/b3325b-correlation-matrix-formula | The Pearson product-moment correlation coefficient (also referred to as Pearson’s r, or simply r) measures the strength of the linear association between two variables. That's fantastic !!! If one variable tends to increase as the other decreases, the correlation coefficient is negative. A correlation matrix is a table showing correlation coefficients between variables. Formula: 1) Sum of Squared Matrix . Correlation Matrix formula. Large values in this matrix indicate serious collinearity between the variables involved. ; If you would like a P-value so that you can test that each population correlation is 0, put a check mark in the box labeled Display p-values by clicking once on the box. Correlation is used to measure strength of the relationship between two variables. For variable 1 and variable 2, the syntax would be =CORREL(B3:B50, C3:C50). The values in the individual cells of the correlation matrix tell us the Pearson Correlation Coefficient between each pairwise combination of variables. The correlation coefficient r has a value of between −1 and 1. |||ly, Conversely, if the two variables tend to increase together the correlation coefficient is positive. ROWS($1:2) returns 2. Awesome, this saved me tons of time! However, the nonexistence of extreme correlations does not imply lack of collinearity. Pxy = SSxy / √(SSxx X SSyy). A correlation matrix is used as an input for other complex analyses such as exploratory factor analysis and structural equation models. If the correlation is 1, they move perfectly together and if the correlation is -1 then stock moves perfectly in opposite directions. Let’s take an example to understand the calculation of Covariance … Similarly, using the same data-matrix and the covariance matrix, let us define the correlation matrix (R): As we see here, the dimension of the correlation matrix is again p × p. Now, if we look at the individual elements of the correlation matrix, the main diagonal all comprises of 1. “Correlation” on the other hand measures both the strength and direction of the linear relationship between two variables. ⁄ Example 4.5.8 (Correlation-I) Let X have a uniform(0,1) distribution and Z have a uni-form(0,0.1) distribution. Referring to Figure 2 of Determining the Number of Factors, the reproduced correlation in Figure 1 is calculated by the array formula =MMULT(B44:E52,TRANSPOSE(B44:E52)) The greater is the absolute value the stronger the relationship tends to be. Additional Resources. Pearson correlation. If your data changes, you will need to rerun the data analysis to update the correlation matrix. Conclusions. So calculate Covariance.Mean is calculated as:Covariance is calculated using the formula given belowCov(x,y) = Σ ((xi – x) * (yi – y)) / (N – 1) 1. Daily Closing Prices of Two Stocks arranged as per returns. Pearson correlation measures a linear dependence between two variables (x and y). Correlation is used to measure strength of the relationship between two variables. In this post I show you how to calculate and visualize a correlation matrix using R. |r| > 0.7 strong correlation For example, r = -0.849 suggests a strong negative correlation. It’s also known as a parametric correlation test because it depends to the distribution of the data. In 5 X 5 matrix , paste down to 5 rows and right to 5 columns. To continue reading you need to turnoff adblocker and refresh the page. Consequently, each is necessarily a positive-semidefinite matrix. Several bivariate correlation coefficients can be calculated simultaneously and displayed as a correlation matrix. The plot of y = f(x) is named linear regression curve.. It looks like you are using an ad blocker! The correlation coefficient assumes a value between −1 and +1. The correlation coefficient may take on any value between +1 and -1. However, the nonexistence of extreme correlations does not imply lack of collinearity. J. Ferré, in Comprehensive Chemometrics, 2009. 3.02.3.5.3(i) Correlation matrix. The Correlation Matrix Definition Correlation Matrix from Data Matrix We can calculate the correlation matrix such as R = 1 n X0 sXs where Xs = CXD 1 with C = In n 11n10 n denoting a centering matrix D = diag(s1;:::;sp) denoting a diagonal scaling matrix Note that the standardized matrix Xs has the form Xs = 0 B B B B B @ (x11 x 1)=s1 (x12 The drawback of this method is the output is static. Moreover, the correlation matrix is strictly positive definite if no variable can have all its values exactly generated as a linear function of the values of the others. Calculate the matrix value of Correlation Matrix. What sets them apart is the fact that correlation values are standardized whereas, covariance values are not. Covariance Matrix is a measure of how much two random variables gets change together. The drawback of this method is the output is static. The pearson correlation formula is : $r = \frac{\sum{(x-m_x)(y-m_y)}}{\sqrt{\sum{(x-mx)^2}\sum{(y-my)^2}}}$ The correlation matrix is a (K × K) square and symmetrical matrix whose ij entry is the correlation between the columns i and j of X.Large values in this matrix indicate serious collinearity between the variables involved. Hence, ROWS($1:2)-1 returns 1, While I love having friends who agree, I only learn from those who don't. Covariance Matrix Formula. The correlation matrix of $${\displaystyle n}$$ random variables $${\displaystyle X_{1},\ldots ,X_{n}}$$ is the $${\displaystyle n\times n}$$ matrix whose $${\displaystyle (i,j)}$$ entry is $${\displaystyle \operatorname {corr} (X_{i},X_{j})}$$. 3.02.3.5.3 (i) Correlation matrix The correlation matrix is a (K × K) square and symmetrical matrix whose ij entry is the correlation between the columns i and j of X. A correlation matrix can be used as an input in other analyses. To estimate the market risk SCR, for example, six sub-risks (interest rate, equity, property, spread, currency and concentration risk) are aggregated using the market risk correlation matrix where the correlations between equity and the property and spread risks are 0.75 and all other correlations are 0.5. The correlation matrix in Excel is built using the Correlation tool from the Analysis ToolPak add-in. What is Correlation matrix ? Compute inverse matrix MINVERSE is the function which returns the inverse matrix stored in an array. All rights reserved © 2020 RSGB Business Consultant Pvt. Cov(x,y) = ((0.2 * (-1.02)) +((-0.1) * 0.78)+(0.5 * 0.98) +(0.… How to Create a Correlation Matrix in Excel A correlation matrix is a table showing correlation coefficients between variables. Measures the degree of linear relationship between two variables. A correlation matrix is used to summarize data, as an input into a more advanced analysis, and as a diagnostic for advanced analyses. The correlation coefficient may take on any value between +1 and -1. The variance covariance matrix of the b weights is: which is the variance of estimate (mean square residual) times the inverse of the SSCP matrix (the inverse of the deviation scores premultiplied by the transpose of the deviation scores). This allows you to I was asked two days ago how to compute a correlation matrix using an excel formula. Variance is … Pearson correlation measures a linear dependence between two variables (x and y). Measuring correlation in Google Sheets. The Spearman correlation is calculated by applying the Pearson correlation formula to the ranks of the data. To start, here is a template that you can apply in order to create a correlation matrix using pandas: df.corr() Next, I’ll show you an example with the steps to create a correlation matrix for a given dataset. Then select variables for analysis. Paste the formula below to N rows x N columns. First, let us calculate the matrix value for Sum of Squared Matrix. A correlation matrix is a table of correlation coefficients for a set of variables used to determine if a relationship exists between the variables. Correlation and Regression formulas list online. The greater is the absolute value the stronger the relationship tends to be. n = N x N Matrix Value |r| > 0.7 strong correlation For example, r = -0.849 suggests a strong negative correlation. Compute inverse matrix MINVERSE is the function which returns the inverse matrix stored in an array. Referring to Figure 2 of Determining the Number of Factors, the reproduced correlation in Figure 1 is calculated by the array formula =MMULT(B44:E52,TRANSPOSE(B44:E52)) ȳ = (6 + 5 + 4) / 3 = 5 z̄ = (9 + 5 + 1) / 3 = 5. The value in the ith row an jth column corresponds to the correlation between the variables $$X_i$$ and $$X_j$$. A matrix of differences can be displayed to compare the two types of correlation matrices . A correlation with many variables is pictured inside a correlation matrix. This applies both to the matrix of population correlations (in which case $${\displaystyle \sigma }$$ is the population standard deviation), and to the matrix of sample correlations (in which case $${\displaystyle \sigma }$$ denotes the sample standard deviation). Q. Minitab Procedure (v.16 & v.17) Select Stat >> Basic statistics >> Correlation...; In the box labeled Variables, specify the two (or more) variables for which you want the correlation coefficient(s) calculated. SSxx = ∑(xi - x̄)2 Each cell in the table shows the correlation between two variables. The correlation matrix is a table that shows the correlation coefficients between the variables at the intersection of the corresponding rows and columns. Definition: Correlation matrix is a type of matrix, which provides the correlation between whole pairs of data sets in a matrix. Figure 1 – Reproduced Correlation Matrix. SSxy = ∑(xi - x̄) X (yi - ȳ) Coefficients have a range of -1 to 1; -1 is the perfect negative correlation while +1 is the perfect positive correlation. “Covariance” indicates the direction of the linear relationship between variables. Correlation matrix analysis is very useful to study dependences or associations between variables. i. coefficient. Find out the correlation matrix from the given 3 X 3 matrix? Then select variables for analysis. The cor() function returns a correlation matrix. Correlation is a very useful statistic to determine if your data is related. Thank you for the step-by-step instructions. 1/ (n-1) SS xx: SS xy: SS xz: Correlation is a function of the covariance. Correlation formula is an important formula which tells the user the strength and the direction of a linear relationship between variable x and variable y. The coefficient indicates both the strength of the relationship as well as the direction (positive vs. negative correlations). x̄ = (1 + 4 + 7) / 3 = 4 Or if there is zero correlation then there is no relations exist between them. Pearson's correlation coefficient, when applied to a sample, is commonly represented by and may be referred to as the sample correlation coefficient or the sample Pearson correlation coefficient. In simple words, both the terms measure the relationship and the dependency between two variables.
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https://math.stackexchange.com/questions/1598893/near-integer-solutions-to-y-x12-sqrt2-given-by-sequence-why | # Near-integer solutions to $y=x(1+2\sqrt{2})$ given by sequence - why?
EDIT: I've asked the same basic question in its more progressed state. If that one gets answered, I'll probably accept the answer given below (although I'm uncertain of whether or not this is the community standard; if you know, please let me know).
I've found a sequence $x_{i+1}=\|x_i(2+k)\|$, where $k=1+2\sqrt{2}$ and where $||a||$ rounds $a$ to the nearest integer, that seems to minimize the distance $P_n$ to an integer solution (for $x$ and $y$) of the equation in the title. $P_n$ is more rigorously defined as the absolute value of $\|y_n\|-y_n$, where $y_n=nk$ for $n \in \mathbb{N}$.
Starting with $x_0=1$ the sequence becomes $x_1=\|2+k\|=\|5.83\ldots\|=6,x_2=\|6(2+k)\|=\|34.97\ldots\|=35,204,1189,6930,\ldots$ where $P_i$ very quickly becomes small.
In Fig. 1 below, I've plotted $P_n$ for $n$. In Fig. 2 only low values of $P$ is shown and it seems that the $P_i$'s from the sequence (in red) are the lowest value up until that $n$ (I've checked this up to $n=10^6$).
My main question is this: Why does this sequence give the solutions nearest integer-solutions? (Edit for further clarification:) And why are these the nearest up until that point (see Fig. 2)? Can it be proven, starting from the original equation, that the elements in this sequence will give the best approximations to integers up until that point, e.g. that it's error dies off faster than all other possible sequences?
Fig. 1
Fig. 2
Further questions: I've also "found" something that looks like an attractor, see Fig. 3 below. Can someone explain what is going on here? I haven't really studied dynamical systems, so if you could dumb it down a bit, I'd be grateful.
Also, as seen in Fig. 1, there's a high degree of regularity here, with all the seemingly straight lines. If I remove the absolute value-part of the def. of $P_n$ the cross-pattern in Fig. 1 becomes (seemingly) straight, parallel declining lines. Why do these lines form? Could it be explained via some modular arithmetic?
Fig. 3
Thanks!
EDIT: Changed "Bonus" to "Further", as I would also really like to hear answers to these questions. Should I post a new question with these, so I could accept an answer there that answers just those?
• Maybe you are interested in why $(3+2\sqrt{2})^n$ is nearly an integer. Imagine calculating $(3+2\sqrt{2})^n+(3-2\sqrt{2})^n$, say using the binomial theorem. The $\sqrt{2}$ stuff cancels, and we get an integer.. But $(3-2\sqrt{2})^n$ goes to $0$ fast, so for largish $n$ the number $(3+2\sqrt{2})^n$ is nearly an integer. – André Nicolas Jan 3 '16 at 22:11
• Very nice, thanks! But I'm still interested in knowing why this sequence is the best (quickest) at producing integers. – Bobson Dugnutt Jan 3 '16 at 22:16
• There are lots of others with similar properties, coming out of other Pell equations. For example $(2+\sqrt{3})^n$. – André Nicolas Jan 3 '16 at 22:22
$$\left( 3 + \sqrt 8 \right)^n + \left( 3 - \sqrt 8 \right)^n$$ is an integer, while $$3 - \sqrt 8 = \frac{1}{3 + \sqrt 8}$$ has absolute value smaller than one.
My sequence would be $$x_{n+2} = 6 x_{n+1} - x_n$$ with $x_0 = 2,$ $x_1 = 6,$ $x_2 = 34$
I think I see what you did. Instead of taking powers $\left( 3 + \sqrt 8 \right)^n,$ you took the nearest integer and multiplied by it again. So you get $1,6,35,204,$ but once the error is small enough you also settle into the necessary $x_{n+2} = 6 x_{n+1} - x_n .$ That is, you have $x_1 = 6,$ $x_2 = 35,$ $x_3 = 204,$ $x_4 = 1189,$ $x_5 = 6930,$ $x_6 = 40391.$ Start it with $x_0 = 1,$ because you have $$\left( \frac{8 + 3 \sqrt 8}{16} \right)\left( 3 + \sqrt 8 \right)^n + \left( \frac{8 - 3 \sqrt 8}{16} \right)\left( 3 - \sqrt 8 \right)^n$$ This is equal to $$\left( \frac{ \sqrt 8}{16} \right) \left( \left( 3 + \sqrt 8 \right)^{n+1} - \left( 3 - \sqrt 8 \right)^{n+1} \right)$$
• Thanks for the answer! I see why $(3+\sqrt{8})^n$ produces integers, but I'm still puzzled as to why exactly this factor is the one that finds the integers the fastest (see Fig. 2). I'm sorry if your sequence answers this (I didn't get it then), but to me it seems you've just reformulated it? What is the benefit of this? Cheers! – Bobson Dugnutt Jan 4 '16 at 10:47 | 2019-12-12T19:22:48 | {
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https://math.stackexchange.com/questions/2407095/what-is-meant-by-a-polynomial-that-is-irreducible-and-a-prime-polynomial | # What is meant by a polynomial that is “irreducible”? And a “prime” polynomial?
Let $P(\mathbb{F})$ be the set of all polynomials of one variable over a field $\mathbb{F}$. It is known that such a set, with usual polynomials operations, is a Euclidean domain, that is, a domain that has the Euclidean division algorithm. So we have the following
Theorem. Let $\deg: P(\mathbb{F})\setminus 0 \to\mathbb{N}$ the degree function. Then $(P(\mathbb{F}), \deg)$ is a Euclidean domain, that is:
a) $P(\mathbb{F})$ is a domain;
b)For all $f,g\in P(\mathbb{F}), g\ne 0$, exist unique polynomials $t,r\in P(\mathbb{F})$ such that $$f(x) = g(x)q(x) + r(x), \text{ with }\deg(r)<\deg(g) \text{ or }r = 0.$$
What I want to know: In such a domain, what is meant by a irreducible or prime polynomial? Moreover, if $p, m\in P(\mathbb{F})$ are such that $p\vert m$ and $p$ is irreducible (or prime) and $m$ is monic (coefficient of the higher degree term is $1$), why we have that $\deg\left(\dfrac mp\right)<\deg(m)$, that is, why $p$ can't have zero degree?
• Are you sure you don't mean irreducible? rather than irreductible? – jgon Aug 26 '17 at 22:13
• I don't know for sure, I'm using a text from my teacher that is incomplete and dont have such definition, besides uses it in propositions. But the text is not in english (neither I'm all proficient in that language). – AnalyticHarmony Aug 26 '17 at 22:14
• @jgon irréductible is French for irreducible. – André 3000 Aug 27 '17 at 16:05
## 3 Answers
I'll talk about the definitions in general domains, and it should be clear how they apply to the polynomial rings you're interested in.
In general, in any domain, $R$, a non-unit element, $p$, is said to be prime if $p\mid ab$ implies $p\mid a$ or $p\mid b$. A non-unit element, $r$, is said to be irreducible if $r=ab$ implies that one of $a$ or $b$ is a unit.
All primes, $p$, are irreducible.
Proof: If $p=ab$, then $p \mid ab$, so $p\mid a$ or $p\mid b$, without loss of generality, we may assume $p\mid a$. Then $a=pv$ for some $v\in R$, and $p=pvb$. Since $R$ is a domain, we may cancel to get $1=vb$, so $b$ is a unit. Hence $p$ is irreducible.
On the other hand, it isn't always the case that irreducible elements are prime. This is however a necessary condition for a ring to be a unique factorization domain, and hence it is in fact true in every Euclidean domain. For a proof, see the answers here, one of which gives a direct proof from the Euclidean property.
To sum up the answer to your first question, in a Euclidean domain, an element is prime ($p\mid ab \implies p\mid a\text{ or } p\mid b$) if and only if it is irreducible ($s=ab \implies a$ or $b$ is a unit).
As for your second question, note that polynomials over a field with degree zero are units, and hence aren't considered irreducible. Thus if $p\mid m$, with $p$ prime/irreducible, then $\deg(m/p) = \deg(m) - \deg(p) < \deg m$. (I've assumed you meant $m$ rather than $f$ in the fraction in your question.)
• What you wrote applies only to domains, not general commutative rings. In non domains basic notions such as irreducible, associate etc, bifurcate into a few inequivalent notions, e.g. see here. – Bill Dubuque Aug 26 '17 at 22:48
• @BillDubuque Ah yes, that's a fair point, that was careless. Edited. Primes still ought to be the right notion, but looking at the answer you linked to, it appears the definition I gave corresponds to the notion of strong associate, rather than associate. I'll record the definition of irreducible given in one of the papers in the linked answer for future curious people, because it's quite nice, and nicely generalizes the idea that it is weaker than prime. $s$ is irreducible in a commutative ring if $s=ab$ implies $s\mid a$ or $s\mid b$. – jgon Aug 26 '17 at 22:56
• Though I should record your note on that answer, that the terminology isn't entirely standardized. – jgon Aug 26 '17 at 23:01
• I often use that definition of irreducible since it makes the implication $\rm prime\Rightarrow irreducible$ completely obvious, e.g. see this answer. – Bill Dubuque Aug 26 '17 at 23:20
• Ah, sorry, yes, that is the unit element of the ring. A unit element, or more succinctly, a unit, is an invertible element, i.e. a $u$ such that there exists $v$ such that $uv=1$. – jgon Aug 26 '17 at 23:44
In $P(\mathbb{F})$, an irreducible polynomial $f$ is a a polynomial OF POSITIVE DEGREE that cannot be written as the product of two polynomials of smaller degree. This is the same as being prime (a prime in a ring is an element $p$ such that if $p | ab$ then $p | a$ or $p | b$) and this is a consequence of $P(\mathbb{F})$ being a unique factorization domain, (because it is a Euclidean domain). The note that Hungerford's Abstract Algebra gives on this matter is:
You could just as well all such a polynomial "prime," but "irreducible" is the customary term with polynomials.
As for your next question, I think you are then asking why an irreducible polynomial cannot be constant since you are assuming that $p$ is irreducible, then asking why it cannot be constant. Irreducible polynomials are nonconstant by definition. The reason is that in $P(\mathbb{F})$, ever constant polynomial (except $0$) is a unit, they have inverses since $\mathbb{F}$ is a field. So, when you talk about a polynomial being irreducible or prime, one runs into the same sort of reason that $1$ is not prime, because the notion of irreducibility is only a useful concept when you ignore the constants since you can always multiple or divide by a constant and nothing really changes much.
I am not sure what exactly is going on with $f$ and $m$, but I hoped I answered your questions.
• I would argue for being prime being stronger just because if you exclude non-constant polynomials/ scalars like wikipedia does for irreducible you might find $3x^2+2x+1$ is irreducible on the integers, but only a subset of these don't have a factor of 2 ( namely only when x is even, is the whole thing odd, in the x is even case it reduces to: $12y^2 + 4y + 1$) – user451844 Aug 27 '17 at 0:41
• I would note that $2$ dividing a polynomial does not mean that it divides the values of the polynomial when you plug in a value for $x$ (though the converse is true: if $2$ divides the polynomial, then whenever you plug in values for $x$ you get a multiple of $2$). The definition of a polynomial $a(x)$ dividing a polynomial $b(x)$ is that there is a polynomial $q(x)$ such that $a(x) = q(x)b(x)$, where the multiplication is done by using the rules of algebra. Note that $x$ in a polynomial is an indeterminate, not an integer/value of our field: – user357980 Aug 27 '17 at 4:20
• See: math.stackexchange.com/questions/495465/… for a description of an indeterminate. – user357980 Aug 27 '17 at 4:20
• I would also comment that what I said about polynomial division was for polynomials over a field $\mathbb{F}$. $\mathbb{Z}$ is not a field, but due to Gauss's lemma being irreducible over $\mathbb{Q}$ is the same as being irreducible over $\mathbb{Z}$, so irreducible is the same as being prime. – user357980 Aug 27 '17 at 4:23
$\underline {\text{Irreducible polynomial}} \text{: A non-constant polynomial that can't be factored within a given field or ring}$
so in the integers: $x^2+1$ is irreducible but $x^2-1$ isn't, this is because the latter factors as $(x+1)\cdot (x-1)$ both of which have coefficients that are integer.
• okay you got me, though quoting wikipedia it seems to also apply in rings: "In mathematics, an irreducible polynomial is, roughly speaking, a non-constant polynomial that cannot be factored into the product of two non-constant polynomials. The property of irreducibility depends on the field or ring to which the coefficients are considered to belong." – user451844 Aug 26 '17 at 22:26 | 2019-06-25T05:31:39 | {
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https://math.stackexchange.com/questions/108709/is-the-graph-g-f-x-fx-in-x-times-y-x-in-x-a-closed-subset-of-x | # Is the graph $G_f=\{(x,f(x)) \in X \times Y\ : x \in X \}$ a closed subset of $X \times Y$?
I'm thinking about Hausdorff spaces, and how mappings to Hausdorff spaces behave. Suppose I have an arbitrary (continuous) function $f:X \longrightarrow Y$, where $Y$ is a Hausdorff space (I think it is irrelevant for my question whether $X$ is Hausdorff or not, so I just consider it to be a topological space - if this is incorrect, please correct me!).
Can we say that the graph $$G_f=\{(x,f(x)) \in X \times Y\ : x \in X \}$$ is a closed subset of $X \times Y$? It seems quite obvious that it is the case, but I cannot see how to prove it. If anyone can offer a proof I'd be very interested. Regards.
EDIT 1
In response to Hennning Makholm:
I wasn't really aware of any variation in 'definition'; I guess I'm considering closed sets to be those with an open complement (though naturally this definition gives rise to other definitions, such as the subset equalling its closure etc.). For continuity of such a map, I would normally consider continuity to mean that $f^{-1}(V)$ is closed in $X$ whenever $V$ is closed in $Y$, though again definitions involving convergence of sequences and the notion that $f$ is continuous iff $f(\overline{A}) \subset \overline{f(A)}$ for every $A \subset X$ are also known to me.
• Yes, this is true (but only if $f$ is continuous). The details of proving it are sensitive to the exact definition of "closed" and "continuous" you're using. Please amend your question to contain the relevant definitions. Feb 13 '12 at 0:47
• @HenningMakholm: I have attempted to clarify, though I thought such ideas were reasonably standardised. Feb 13 '12 at 1:03
• Yes, this is true for arbitrary continuous functions, and for linear functions between Banach spaces, the converse (closed graph implies continuous) is also true: en.wikipedia.org/wiki/Closed_graph_theorem Feb 13 '12 at 1:07
Suppose that $\langle x,y\rangle\in (X\times Y)\setminus G_f$. Then $y\ne f(x)$, and $Y$ is Hausdorff, so there are disjoint open $U,V$ in $Y$ such that $y\in U$ and $f(x)\in V$. Since $f$ is continuous, there is an open nbhd $W$ of $x$ such that $f[W]\subseteq V$; clearly $W\times U$ is an open nbhd of $\langle x,y\rangle$ disjoint from $G_f$.
It is necessary to require that $Y$ be Hausdorff. For a simple example, let $X=\{0,1\}$ have the discrete topology, and let $Y=\{0,1\}$ with the Sierpiński topology, whose open sets are $\varnothing,\{0\}$, and $Y$ itself. Let $f:X\to Y$ be the identity function; $f$ is certainly continuous, since $X$ is discrete, but $\langle 0,1\rangle$ is in the closure of $G_f$, since every nbhd of $\langle 0,1\rangle$ contains $\langle 0,0\rangle$.
Added: The space $Y$ in that example is $T_0$ but not $T_1$; here’s an example in which $Y$ is $T_1$. Let $X=\mathbb{N}\cup\{p\}$, where $p\notin\mathbb{N}$, and let $Y=\mathbb{N}\cup\{p,q\}$, where $q\notin\mathbb{N}$ and $p\ne q$. In both $X$ and $Y$ the points of $\mathbb{N}$ are isolated, and in both $X$ and $Y$ a local base at $p$ consists of all sets of the form $\{p\}\cup(\mathbb{N}\setminus F)$ such that $F$ is a finite subset of $\mathbb{N}$. Finally, a local base at $q$ in $Y$ consists of all sets of the form $\{q\}\cup(\mathbb{N}\setminus F)$ such that $F$ is a finite subset of $\mathbb{N}$. The points $p$ and $q$ in $Y$ do not have disjoint open nbhds; they are the only pair of points in $Y$ that cannot be separated by disjoint open sets.
Let $f:X\to Y:x\mapsto x$ be the identity function; it’s easy to see that $f$ is not just continuous, but an embedding. The point $\langle p,q\rangle\in X\times Y$ is not in $G_f$, but you can check that if $U$ is an open nbhd of $\langle p,q\rangle$ in $X\times Y$, then there is an $m\in\mathbb{N}$ such that $\langle n,n\rangle\in U$ whenever $n\ge m$, so $U\cap G_f\ne\varnothing$. Thus, $\langle p,q\rangle$ is in the closure of $G_f$.
• I have a counterexample showing that $G_f$ is not closed in $X\times Y$ : suppose $y=x^3$ and $x\in [1,3]$ and suppose $U_a$ be any open set in standard topology on $\mathbb{R}^2$ such that $U_a \cap G_f=$∅. Because union of all $U_a$'s is open and equals $\mathbb{R}^2-G_f$, then $G_f$ is closed in $X\times Y=\mathbb{R}^2$. Am I right?
– L.G.
Jan 3 '15 at 9:25
• @Ali.E.: I can’t make any sense of this. First you say that $G_f$ is not closed, then you say that it is. Also, you write $X\setminus G_f$, as if $G_f$ were a subset of $X$, but then you say that $G_f$ is a subset of $X\times Y$. Finally, you seem to limit $x$ to the set $[1,3]$, in which case $\Bbb R^2$ is irrelevant: you should be asking whether $G_f$ is closed in $[1,3]\times\Bbb R$. Jan 3 '15 at 9:29
• Yes, sorry! My counterexample is not correct; in fact, I have given an example that $G_f$ is closed, as it must be. I edited $X\setminus G_f$ to $\mathbb{R}^2-G_f$.
– L.G.
Jan 3 '15 at 9:40
• @Ali.E.: The edited version is correct. However, it does take some work to show that the union of those sets $U_a$ actually is all of $\Bbb R^2\setminus G_f$. Jan 3 '15 at 9:44
• What if we consider an open interval $x\in (1,3)$. Then $f(x)$ would be a curve not including $f(1)$ and $f(3)$, which is homomorphic to an open interval on real line, say $U=(a,b)$, by just defining a bijective function between $G_f$ and $U$. Is is right that $G_f$ is open?
– L.G.
Jan 3 '15 at 10:17
Suppose $(x,y)\in \overline {G_f}$, then there is a net $(x_\alpha,f(x_\alpha))\ (\alpha\in \Lambda)$ so that $(x_\alpha,f(x_\alpha))\to (x,y)$ by the definition of the product topology $x_\alpha$ converges to $x$ and $f(x_\alpha)$ converges to $y$ since $f$ is continuous $f(x_\alpha)$ converges to $f(x)$. Since $Y$ is Hausdorff limits are unique hence $y=f(x)$. Thus, $(x,y)\in G_f$ so $G_f$ is closed. | 2021-09-20T15:03:18 | {
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https://www.physicsforums.com/threads/relative-velocity-boat-river-question.674426/ | # Relative velocity/Boat & River Question
## Homework Statement
280m wide river, destination 120m upstream, river current is 1.35 m/s downstream and the boat speed in still water is 2.70 m/s. What should the boat's heading angle be (relative to the shore)?
## Homework Equations
$V_x = Vcosθ$
$V_y = Vsinθ$
$x = V_0 t$ (starting points set to 0 for x and y, no acceleration involved)
$V_{BS} = V_{BW} + V_{WS}$
## The Attempt at a Solution
y component of velocity unaffected by current, so $V_{BS} = V_{BW} + V_{WS}$ will be relevent for x component only.
$V_x = Vcosθ = 2.7 cosθ$
$V_y = Vsinθ = 2.7 sinθ$
$V_{BS} = V_{BW} + V_{WS} = (2.7 cosθ) - 1.35$
$y = V_0 t = 280 = (2.7 sinθ)t$
$t = 280/(2.7 sin)$
$x = V_0 t = 120 = ((2.7 cosθ) - 1.35) t$
$t = 120/((2.7 cosθ)-1.35)$
time it takes to go across river (y component) equals time to travel upstream (x component)
$t = t = 280/(2.7 sin) = 120/((2.7 cosθ)-1.35)$
$280((2.7 cosθ) - 1.35) = 120 (2.7 sinθ)$
$(756 cosθ) - 378 = (324 sinθ)$
$(14 cosθ) - 7 = (6 sinθ)$
$7((2 cosθ) - 1) = (6 sinθ)$
$(2 cosθ) - 1 = (6/7) sinθ$
$((2 cosθ) - 1)/sinθ = 6/7$
$(2(cosθ / sinθ)) - (1/sinθ) = 6/7$
$2 cotθ - cscθ = 6/7$
I then plotted this using a graphing calculator and found where $y = 2 cot(x) - csc(x)$ crossed $y = 6/7$ and found θ = 39.5°. This answer does seem to check, but I have a feeling I'm doing something wrong, and that I should be able to solve for θ without using a graphing calculator. I'm not sure if the law of sines or law of cosines will come into play, or if I'm missing a certain trigonometric identity. Any insight would be greatly appreciated. Thanks.
Last edited:
## Answers and Replies
Related Introductory Physics Homework Help News on Phys.org
ehild
Homework Helper
Your equation 14cos(θ)-6sin(θ)=7 is correct.
You can proceed by substituting cosθ=sqrt(1-sin2θ) and solve a quadratic equation.
The other method is to rewrite the equation as A(cosβcosθ-sinβsinθ)=Acos(β+θ)=7, where Acosβ=14 and Asinβ=6: tanβ=3/7 and A2=142+62.
ehild
Thanks. I used the quadratic equation way and got the same answer as my previous attempt.
ehild
Homework Helper
Splendid.
ehild
I have a question for this. Doesn't going upstream mean going against the water flow?
ehild
Homework Helper
Yes, it is against the water flow.
ehild
Yes, it is against the water flow.
ehild
In the picture wouldn't the 120m distance be the whole horizontal distance? I am bit confused on that.
ehild
Homework Helper
120 m is the distance along the river where the boat is due.
ehild
120 m is the distance along the river where the boat is due.
ehild
Dalkiel had this: "280m wide river, destination 120m upstream, river current is 1.35 m/s downstream and the boat speed in still water is 2.70 m/s. What should the boat's heading angle be (relative to the shore)?"
In the bolded part wouldn't it make more sense if it did not say upstream and just " destination 120m" or something like that? Because the water is coming from the right side.
ehild
Homework Helper
120 m alone could mean both upstream and downstream, that is to the right from the starting point or to the left.
ehild
120 m alone could mean both upstream and downstream, that is to the right from the starting point or to the left.
ehild
Oh so even though the destination will be at an angle from starting point it still means upstream or downstream. Thanks!
ehild
Homework Helper
That 120 m is the distance on the other side of the river from the point opposite to the starting point.
ehild
That 120 m is the distance on the other side of the river from the point opposite to the starting point.
ehild
so 280 is horizontal distance then
so 280 is horizontal distance then
280 is the width of the river. if there was no current, and the boat wanted to go straight across the river, it would have to travel exactly 280 m. In my (horribly drawn) picture, 280 is the vertical distance, or the y component of the displacement.
The answer that was found (39.5°) is correct, but there was actually another way to solve this, using the law of sines. I'll try to draw another picture to describe it when I get home.
The answer that was found (39.5°) is correct, but there was actually another way to solve this, using the law of sines. I'll try to draw another picture to describe it when I get home.
Thanks! I was mostly confused with the wording of the question.
Sorry about the wording, I basically just paraphrased the problem to the basic components.
Here's the solution that was explained to me, using another horribly drawn ms paint picture as a guide.
The yellow line is straight across the river, green line is where we want to go, brown is where we have to aim, dark blue is the velocity of the water. Since we know the length of two sides of the right triangle, we can find the angles. tanβ = 120/280. so β=23.2° and γ=66.8°. Since we know γ, we can find σ= 113.2°. Then use law of sines to find 2.7/sin(113.2) = 1.35/sinα. α=27.4. 90-β-α=θ=39.4°. Which matches all previous attempts.
I personally didn't care for this method because I'm mixing angles involved with distance with those using velocity. My initial intuition would be to not mix the two, but this seemed to work out.
Last edited:
Sorry about the wording, I basically just paraphrased the problem to the basic components.
Here's the solution that was explained to me, using another horribly drawn ms paint picture as a guide.
The yellow line is straight across the river, green line is where we want to go, brown is where we have to aim, dark blue is the velocity of the water. Since we know the length of two sides of the right triangle, we can find the angles. tanβ = 120/280. so β=23.2° and γ=66.8°. Since we know γ, we can find σ= 113.2°. Then use law of sines to find 2.7/sin(113.2) = 1.35/sinα. α=27.4. 90-β-α=θ=39.4°. Which matches all previous attempts.
I personally didn't care for this method because I'm mixing angles involved with distance with those using velocity. My initial intuition would be to not mix the two, but this seemed to work out.
Oh I see and thanks for posting this up. Do you mind posting up the whole question? :shy:
Oh I see and thanks for posting this up. Do you mind posting up the whole question? :shy:
Sure, but it's actually a combination of two questions. One is direct from a textbook, and then and then there's my professor's part.
Original from text (problem #70): A boat, whose speed in still water is 2.70 m/s, must cross a 280-m-wide river and arrive at a point 120 m upstream from where it starts. To do so, the pilot must head the boat at a 45.0° upstream angle. What is the speed of the river's current?
Professor's question: In prob 70, suppose the river current is 1.35 m/s and the boat speed in still water is 2.70 m/s. What should the boat's heading angle be in that case, in order to reach the same destination? | 2020-08-12T01:54:22 | {
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https://math.stackexchange.com/questions/730888/microcontinuous-vs-continuous/732509 | # Microcontinuous vs Continuous
I've been studying infinitesimals and came upon the idea of uniformly microcontinuous functions. My question is: if a function $f^*: \mathbb{R}^* \to \mathbb{R}^*$ the natural extension of $f: \mathbb{R} \to \mathbb{R}$ is microcontinuous on $A\subset\mathbb{R}$ does that imply that it is continuous with the $\epsilon-\delta$ definition on $A$? If this is not true does the (I think)weaker statement hold that if the function is uniformly microcontinuous on all of $\mathbb{R^*}$ does that imply that it is continuous with the $\epsilon-\delta$ definition on $\mathbb{R}$?
Just a note that the definition I'm using for microcontinuous at $x$ is $\forall x' (x \approx x') \implies (f(x) \approx f(x'))$ where $a \approx b$ means that the standard part of $a-b$ is $0$. From this a function is micrcontinuous on some domain $S$ means that $\forall x\in S\, \forall x' (x \approx x') \implies (f(x) \approx f(x'))$
• @tomasz A microcontinuous function is a function that is microcontinuous for all points in its domain I believe. I'll edit that into the definition though. – ruler501 Mar 29 '14 at 3:04
• Yeah, I see that now. Sorry, I'm little tired. – tomasz Mar 29 '14 at 3:04
• Understandable. I should've made sure it was stated clearly since it doesn't seem to be a common concept for people to know. – ruler501 Mar 29 '14 at 3:06
• @ruler501: What book are you using? – Jose Antonio Mar 29 '14 at 3:11
• @JoseAntonio I have been reading from math.wisc.edu/~keisler/calc.html and math.wisc.edu/~keisler/foundations.html but I got the definition of microcontinuity from wikipedia. – ruler501 Mar 29 '14 at 3:14
Yes, it does. To see that, choose any (standard) real $x_0\in A$. Now for any $x$ such that $$\models \{x_0-1/n<x<x_0+1/n\mid n\in {\bf N}\}$$ we have $$\models \{ f(x_0)-1/m<f(x)<f(x_0)+1/m\mid m\in {\bf N}\}$$ This means that in particular, for each $m\in {\bf N}$, we have $$\{x_0-1/n<x<x_0+1/n\mid n\in {\bf N}\}\vdash f(x_0)-1/m<f(x)<f(x_0)+1/m$$ and by compactness there is some $n\in {\bf N}$ such that $$x_0-1/n<x<x_0+1/n\vdash f(x_0)-1/m<f(x)<f(x_0)+1/m$$ but then $f$ is clearly ($\varepsilon$-$\delta$) continuous at $x_0$ when considered as a function on real numbers, so it's also just continuous, since $x_0$ was arbitrary. Since $\varepsilon$-$\delta$ continuity is first order, it is also $\varepsilon$-$\delta$ continuous on any given nonstandard model.
Note that it only applies if $f$ is a standard real function (a symbol of the language) to begin with. Otherwise the part about restricting to reals does not make sense and neither does the compactness argument, and indeed we can find a counterexample: choose an infinitesimal $\varepsilon$ and let $f(x)=0$ for $\lvert x\rvert >\varepsilon$, $f(0)=0$, and $f(x)=\varepsilon$ for $x\in [-\varepsilon,0)\cup (0,\varepsilon]$. Then $f$ is microcontinuous but not continuous.
• I added to the question that it should be an $f^*$ that is the natural extension of an $f: \mathbb{R} \to \mathbb{R}$ so that the question makes more sense(addressing your note). – ruler501 Mar 29 '14 at 4:15
It is not clear what you mean exactly by "uniformly microcontinuous". If you simply mean "microcontinuous", then microcontinuity of $f^*$ at every point of a real set $A$ implies continuity of $f$ on $A$. If you assume that $f^*$ is microcontinuous on all of $\mathbb{R}^*$ then it follows that $f$ is not merely continuous on $\mathbb{R}$ but is actually uniformly continuous on $\mathbb{R}$. | 2021-07-27T18:58:00 | {
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https://www.naukovy.com.ua/p4c6x9/set-notation-symbols-1478d8 | The domains and ranges used in the discrete function examples were simplified versions of set notation. In this section, we will introduce the standard notation used to define sets, and give you a chance to practice writing sets in three ways, inequality notation, set-builder notation, and interval notation. In set-builder notation, the set is specified as a selection from a larger set, determined by a condition involving the elements. mathematical sets • A (finite) set can be thought of as a collection of zero or more . They are { } and { 1 }. Consider the set $\left\{x|10\le x<30\right\}$, which describes the behavior of $x$ in set-builder notation. S et theory is a branch of mathematics dedicated to the study of collections of objects, its properties, and the relationship between them. Because null set is not equal to A. Set theory is one of the foundational systems for mathematics, and it helped to develop our modern understanding of infinity and real numbers. Quiz & Worksheet Goals Set Theory Symbols Posted in engineering by Christopher R. Wirz on Wed Feb 08 2017. Google Classroom Facebook Twitter. Researchers and mathematicians have developed a language and system of notation around set theory. Lots symbols look similar but mean different things. Purplemath. On the Insert tab, in the Symbols group, click the arrow under Equation, and then click Insert New Equation. The Universal Set … If you … ALT Codes for Math Symbols: Set Membership & Empty Sets Read More » The Wolfram Language has the world's largest collection of consistent multifont mathematical notation characters\[LongDash]all fully integrated into both typesetting and symbolic expression construction . This carefully selected compilation of exam questions has fully-worked solutions designed for students to go through at home, saving valuable time in class. ... Set Language And Notation. There are many different symbols used in set notation, but only the most basic of structures will be provided here. Bringing the set operations together. Some notations for sets are: {1, 2, 3} = set of integers greater than 0 … Solution: Let P be the set of all members in the math This section is to introduce the notation to the reader and explain its usage. It is still a set, so we use the curly brackets with nothing inside: {} The Empty Set has no elements: {} Universal Set. After school they signed up and became members. Use set notation to describe: (a) the area shaded in blue (b) the area shaded in purple. Usually, you'll see it when you learn about solving inequalities, because for some reason saying "x < 3" isn't good enough, so instead they'll want you to phrase the answer as "the solution set is { x | x is a real number and x < 3 }".How this adds anything to the student's understanding, I don't know. MS Word Tricks: Typing Math Symbols 2015-05-14 Category: MS Office. Set Theory • A mathematical model that we will use often is that of . The default way of doing it is to use the Insert > Symbols > More Symbols dialog, where you can hunt for the symbol you want. The colon means such that.. For example: {x: x > 5}.This is read as x such that x is greater than > 5.. Admin Igcse Mathematics Revision Notes, O Level Mathematics Revision Notes 2 Comments 12,074 Views. This cheat sheet is extremely useful. The table below contains one example set… They wrote about it on the chalkboard using set notation: P = {Kyesha, Angie and Eduardo} When Angie's mother came to pick her up, she looked at the chalkboard and asked: What does that mean? CCSS.Math: HSS.CP.A.1. Because rarely used symbol may look very different on another computer. Inequalities can be shown using set notation: {x: inequality}where x: indicates the variable being described and inequality is written as an inequality, normally in its simplest form. Cardinality and ordinality Relative complement or difference between sets. Look at the venn diagram on the left. Set notation practice. When using set notation, we use inequality symbols to describe the domain and range as a set of values. Symbol Symbol Name Meaning / definition Example { } set: a collection of elements: A = {3,7,9,14}, This quiz and attached worksheet will help gauge your understanding of set notation. Below is the complete list of Windows ALT codes for Math Symbols: Set Membership & Empty Sets, their corresponding HTML entity numeric character references, and when available, their corresponding HTML entity named character references, and Unicode code points. Sets, in mathematics, are an organized collection of objects and can be represented in set-builder form or roster form.Usually, sets are represented in curly braces {}, for example, A = {1,2,3,4} is a set. Demo. For example, a set F can be specified as follows: = {∣ ≤ ≤}. Set theory starter. take the previous set S ∩ V ; then subtract T: This is the Intersection of Sets S and V minus Set T (S ∩ V) − T = {} Hey, there is nothing there! Let us discuss the next stuff on "Symbols used in set theory" If null set is a super set Email. Intersection and union of sets. elements . any. Hyperbolic functions The abbreviations arcsinh, arccosh, etc., are commonly used for inverse hyperbolic trigonometric functions (area hyperbolic functions), even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area. To fully embrace the world of professional Venn diagrams, you should have a basic understanding of the branch of mathematical logic called ‘set theory’ and its associated symbols and notation. Preview. Which is why the bulk of this follow-up piece covers the very basics of set theory notation, operations & visual representations extensively. Set Notation. Set notation and Venn diagrams questions. The guide you are now reading is a “legend” to how we notate drum and percussion parts when we engrave music at Audio Graffiti. Thankfully, there is a faster way. The following list documents some of the most notable symbols in set theory, along each symbol’s usage and meaning. Also, check the set symbols here.. Under Equation Tools , on the Design tab, in the Symbols group, click the More arrow. 8 February 2019 OSU CSE 1. Topics you will need to know in order to pass the quiz include sets, subsets, and elements. Probability and statistics symbols table and definitions - expectation, variance, standard deviation, distribution, probability function, conditional probability, covariance, correlation Sets. Set notation. The table below lists all of the necessary symbols for compact set notation. Basic set operations. Note that it's unnecessary to load amsmath if you load mathtools. The symbols shown in this lesson are very appropriate in the realm of mathematics and in mathematical logic. Set notation is an important convention in computer science. and symbols. Subset, strict subset, and superset. The following table gives a summary of the symbols use in sets. Example: Set-Builder Notation: Read as: Meaning: 1 {x : x > 0}the set of all x such that x is greater than 0. any value greater than 0: 2 {x : x ≠ 11}the set of all x such that x is any number except 11. any value except 11: 3 {x : x < 5}the set of all x such that x is any number less than 5. any value less than 5 Set notation is used to help define the elements of a set. Illustration: Use set notation to describe: (a) the area shaded in green (b) the area shaded in red : Look at the venn diagrams on the left. Sometimes the set is written with a bar instead of a colon: {x¦ x > 5}. Here null set is proper subset of A. Symbolab: equation search and math solver - solves algebra, trigonometry and calculus problems step by step Let’s kick off by introducing the two most basic symbols for notating a set & it’s corresponding elements. When picking a symbol, best to trust the symbol's unicode name for its meaning, not appearance. Basic set notation. Symbols Used in this Book; Glossary; While a comprehensive list of notation is included in the appendix, that is meant mostly as a reference tool to refresh the reader of what notation means. Problem 1: Mrs. Glosser asked Kyesha, Angie and Eduardo to join the new math club. You never know when set notation is going to pop up. Occasionally we will introduce a new symbol to cater for an unusual requirement of a client. A variant solution, also based on mathtools, with the cooperation of xparse allows for a syntax that's closer to mathematical writing: you just have to type something like\set{x\in E;P(x)} for the set-builder notation, or \set{x_i} for sets defined as lists. Shading task. State whether each … That is OK, it is just the "Empty Set". For example, let us consider the set A = { 1 } It has two subsets. If you want to get in on their secrets, you'll want to become familiar with these Venn diagram symbols. Click the arrow next to the name of the symbol set, and then select the symbol set that you want to display. Author: Created by Maths4Everyone. Universal set and absolute complement. Compact set notation is a useful tool to describe the properties of each element of a set, rather than writing out all elements of a set. A set is a collection of objects, things or symbols which are clearly identified.The individual objects in the set are called the elements or members of the set. Crow's foot notation, however, has an intuitive graphic format, making it the preferred ERD notation for Lucidchart. IGCSE 9-1 Exam Question Practice (Sets + Set Notation) 4.9 34 customer reviews. We attempt to follow the standards set out in Norman Weinberg’s Guide to Standardized Drumset Notation. Mathematical Set Notation. Created: Jan 19, 2018 | Updated: Feb 6, 2020. of . Null set is a proper subset for any set which contains at least one element. Basic Set Theory . 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https://codereview.stackexchange.com/questions/36655/count-possible-paths-through-a-maze | # Count possible paths through a maze
You have to find a path through which the rat move from the starting position (0,0) to the final position where cheese is (n,n). List the total no of possible paths which the rat can take to reach the cheese.
Sample Input:
7
0 0 1 0 0 1 0
1 0 1 1 0 0 0
0 0 0 0 1 0 1
1 0 1 0 0 0 0
1 0 1 1 0 1 0
1 0 0 0 0 1 0
1 1 1 1 0 0 0
Sample Output:
4
My code:
import java.util.*;
class maze
{
static int n;
static int[][] a;
static int path;
public static void main(String[] ar)
{
Scanner sc = new Scanner(System.in);
n = sc.nextInt();
a = new int[n][n];
for(int i = 0; i < n; i++)
{
for(int j = 0; j < n; j++)
{
a[i][j] = sc.nextInt();
}
}
search(0,0);
System.out.println(path);
}
public static void search(int i, int j)
{
if(!exist(i,j) || a[i][j] == 1)
return;
if(i == n-1 && j == n-1)
{
path++;
return;
}
a[i][j] = 1;
search(i+1,j);
search(i-1,j);
search(i,j+1);
search(i,j-1);
a[i][j] = 0;
}
public static boolean exist(int i, int j)
{
return i>=0 && j >=0 && i < n && j < n;
}
}
Can anybody recommend a better solution than this or if there are any, please point out the mistakes in this. I made it using simple logic and I don't know much about graph theory. Is this DFS or BFS?
• There's no input about where the cheese is. – CodyBugstein Jul 19 '16 at 19:05
There are not many things to criticize here ... the basic algorithm in general looks good.
The three items I would caution you on are:
1. you are keeping track of the path-count in the paths static variable. This is a pattern that, although works, is not very 'pretty', there's a better way... I'll explain.
2. you modify the source array. This can be OK, but, in general, when you want to modify the source data you should instead work off a copy of the data.
3. all your other variables (the maze itself and the size of the maze) are static.
With a slight shift in the way you think of your search method, instead of updating the number of paths, you should instead think 'how many paths from here?'
Also, lets fix the static variable issues too (we will need two methods for this):
public static final int search(int[][] data) {
int[] mymap = new int[data.length][];
for (int i = 0; i < data.length; i++) {
mymap[i] = Arrays.copyOf(data[i], data[i].length);
}
return search(mymap, 0, 0);
}
Now, the search(...) recursive method should return an int, and it takes the maze as input, not from a static variable.
To avoud the use of the 'n' variable, we use the basic information from the maze. The following is a copy/paste of your code with slight modifications:
1. it returns int
2. return-statements inside return a value now
3. the input includes the 'maze' (instead of being a static variable)
4. there are no references to n, just to the a array
5. it has an internal paths variable
6. it changes how it calls exist(...)
public static int search(int[][] a, int i, int j)
{
if(!exist(i,j) || a[i][j] == 1)
return 0; // no path here.
if(i == a.length - 1 && j == a[i].length - 1)
{
return 1; // 1 path here.
}
a[i][j] = 1; // mark that we have seen this spot here
int paths = 0; // introduce a counter...
paths += search(a, i+1,j); // add the additional paths as we find them
paths += search(a, i-1,j);
paths += search(a, i,j+1);
paths += search(a, i,j-1);
a[i][j] = 0;
return paths; // return the number of paths available from this point.
}
The exist() function will also need to change:
public static boolean exist(int[][] a, int i, int j)
{
return i>=0 && j >=0 && i < a.length && j < a[i].length;
}
Note that this no longer references the n static variable either.
Finally, with these changes (and, really, in the scheme of things they are 'small'), your main method simply becomes:
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[][] a = new int[n][n];
for(int i = 0; i < n; i++)
{
for(int j = 0; j < n; j++)
{
a[i][j] = sc.nextInt();
}
}
System.out.println(search(a));
For this, the a does not need to be static either.
If you code the process like I suggest then the methods become much more generic, and you have other advantages like:
1. the methods are all thread-safe now (you can find many paths in parallel using the same methods).
2. the data and the logic are both self-contained.
3. there are no static variables.
Finally, the variable names you have chosen are really short..... consider renaming the a variable at least.
Your recursive search is fine, as long as the paths through the maze are short enough not to overflow the stack. Since you have to find all paths, an exhaustive search is necessary, so I don't think you need to use any fancier algorithms. (Recursion generally results in depth-first search behaviour.)
I agree with @rolfl that static variables should be avoided, and that modifying the input matrix is naughty. However, I recommend a different remedy: make a Maze object. (By the way, class names should be UpperCaseLikeThis in Java.)
Once your Maze object has copied the maze definition into a private instance variable, it's free to do whatever it wants with the array, including modifying it with temporary roadblocks.
I've renamed search() to pathsFrom() to clarify the meaning of its parameters. I've also renamed exist() to isInBounds() for clarity.
The count of the number of paths is not part of the state of the maze, so it should not be an instance variable or a class variable. It should just be returned from the pathsFrom() method.
import java.util.*;
class Maze
{
private int n;
private int[][] a;
/**
* Array is a square matrix, whose elements are 0 for paths and 1 for walls.
*/
public Maze(int[][] array)
{
// Copy the array, assuming that it is a square matrix
n = array.length;
a = new int[n][];
for (int i = n - 1; i >= 0; i--)
{
a[i] = Arrays.copyOf(array[i], n);
}
}
public int pathsFrom(int i, int j)
{
if(!isInBounds(i,j) || a[i][j] == 1)
{
return 0;
}
if(i == n-1 && j == n-1)
{
return 1;
}
try
{
a[i][j] = 1;
return pathsFrom(i+1,j) +
pathsFrom(i-1,j) +
pathsFrom(i,j+1) +
pathsFrom(i,j-1);
}
finally
{
// Restore the state of the maze before returning
a[i][j] = 0;
}
}
private boolean isInBounds(int i, int j)
{
return i>=0 && j >=0 && i < n && j < n;
}
public static void main(String[] ar)
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[][] a = new int[n][n];
for(int i = 0; i < n; i++)
{
for(int j = 0; j < n; j++)
{
a[i][j] = sc.nextInt();
}
}
Maze m = new Maze(a);
System.out.println(m.pathsFrom(0, 0));
}
}
In the implementation of pathsFrom() above, I've used a slick language trick to sneak in a statement before returning. That code is equivalent to
a[i][j] = 1;
int paths = pathsFrom(i+1,j) +
pathsFrom(i-1,j) +
pathsFrom(i,j+1) +
pathsFrom(i,j-1);
a[i][j] = 0;
return paths;
… except that the version using finally restores the state of the matrix even if an exception is thrown. | 2020-02-23T23:04:37 | {
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https://mathematica.stackexchange.com/questions/98956/smallest-element-distance-between-two-lists/98967 | Smallest element distance between two lists
I have a problem, where I generate two lists of points, and I want to see which points are the closest. If I just want to find which ones are exactly the same, I can use Intersection[list1,list2]. I even know that I can use SameTest to define the test. As an example, I could use
testlist1 = Table[k, {k, 0, 25, 0.25}];
testlist2 = Table[k, {k, 0, 49, 0.49}];
Intersection[testlist1, testlist2]
and I get the result
{0., 12.25, 24.5}
I can use SameTest to define a closeness function that will test if the two elements are very close:
testlist1 = Table[k, {k, 0, 25, 0.25}];
testlist2 = Table[k, {k, 0, 49, 0.49}];
Intersection[testlist1, testlist2,
SameTest -> (Abs[#1 - #2] <= 0.01 &)]
and now I get the elements
{0., 0.5, 11.75, 12.25, 12.75, 24., 24.5}
For my problem, I have two lists that will always contain at least one pair of corresponding points between the list, although they may not always be equal, and the difference will not always be the same; and there may also be more than one set of corresponding points. I've generated the following code, which (mostly) works, but is slow, and depends on a new function introduced in 10.3 which is labeled as [[EXPERIMENTAL]], specifically, the function DistanceMatrix
testlist1 = Table[k, {k, 0, 25, 0.25}];
testlist2 = Table[k+.0057, {k, 0, 49, 0.49}];
Intersection[testlist1, testlist2,
SameTest -> (Abs[#1 - #2] <= Min[DistanceMatrix[testlist1,testlist2]] &)]
which returns
{0.5}
Which is (mostly) correct. See bullet 3 for why I'm unsatisfied with the answer.
Now my problems are these:
1. This seems inefficient, and runs quite slowly
2. This depends on an experimental function, which is only available in the latest version of Mathematica (10.3)
3. This returns the element in the first list called in Intersection corresponding to the SameTest returning true. Rather, I'd like it to return the matching elements from both lists. A dirty solution would be to call Intersection twice, with the lists swapped in the slots, but for a solution that already seems inefficient, I'd rather not call it multiple times.
So: is there a better way to do this? Even returning the current result would be ok, if not ideal. Ideally, a list of pairs would be returned, with each pair corresponding to the matching set.
DistanceMatrix is fast. It evaluates multiple times in the SameTest. If you rewrite your code, you can see the difference:
min=Min[DistanceMatrix[testlist1,testlist2]];//AbsoluteTiming
{0.000501174,Null}
Intersection[testlist1,testlist2,SameTest->(Abs[#1-#2]<=min&)]//AbsoluteTiming
{0.0236643,{0.5}}
VS original code
Intersection[testlist1,testlist2,SameTest->(Abs[#1-#2]<=Min[DistanceMatrix[testlist1,testlist2]]&)]//AbsoluteTiming
{5.39343,{0.5}}
DistanceMatrix will stay in the future versions of Mathematica. Also you can find DistanceMatrix in HierarchicalClustering package or use this code.
mat=DistanceMatrix[testlist1,testlist2];
pos=Position[mat,Min@mat]
{{3,2}}
testlist1[[pos[[;;,1]]]]
{0.5}
testlist2[[pos[[;;,2]]]]
{0.4957}
mind[l1_List, l2_List] :=
MinimalBy[Table[{k, First@Nearest[l1, k]}, {k, l2}], Norm]
RepeatedTiming[mind[testlist1, testlist2]]
(* {0.0061, {{0.0057, 0.}}} *)
For huge lists you may change Table for ParallelTable
This is based on the cool solution provided by @rhermans and it simply exploits the NearestFunction instead of Nearest, making his solution almost one order of magnitude faster. So, given the two lists you are working with:
testlist1 = Table[k, {k, 0, 25, 0.25}];
testlist2 = Table[k + .0057, {k, 0, 49, 0.49}];
first define the auxiliary NearestFunction for testlist1 and then define the actual function that computes what you need.
nf1 = Nearest@testlist1
mindFast[l2_List] := MinimalBy[{#, First@nf1@#} & /@ l2, Norm];
Let's compare it with @rhermans 's mind
RepeatedTiming[mind[testlist1, testlist2]]
{0.0048, {{0.0057, 0.}}}
RepeatedTiming[mindFast[testlist2]]
{0.000623, {{0.0057, 0.}}} | 2021-07-25T03:50:54 | {
"domain": "stackexchange.com",
"url": "https://mathematica.stackexchange.com/questions/98956/smallest-element-distance-between-two-lists/98967",
"openwebmath_score": 0.5606675744056702,
"openwebmath_perplexity": 1699.763763835051,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9546474168650673,
"lm_q2_score": 0.8824278571786139,
"lm_q1q2_score": 0.8424074744253403
} |
https://plainmath.net/1515/polynomial-polynomials-polynomial-polynomial-calculating-calculating | DISCOVER: Nested Form of a Polynomial Expand Q to prove that the polynomials P and Q ae the same P(x) = 3x^{4} - 5x^{3} + x^{2} - 3x +5 Q(x) = (((3x - 5)x + 1)x 3)x + 5 Try to evaluate P(2) and Q(2) in your head, using the forms given. Which is easier? Now write the polynomial R(x) =x^{5} - 2x^{4} + 3x^{3} - 2x^{2} + 3x + 4 in “nested” form, like the polynomial Q. Use the nested form to find R(3) in your head. Do you see how calculating with the nested form follows the same arithmetic steps as calculating the value ofa polynomial using synthetic division?
Question
Polynomial arithmetic
DISCOVER: Nested Form of a Polynomial Expand Q to prove that the polynomials P and Q ae the same $$P(x) = 3x^{4} - 5x^{3} + x^{2} - 3x +5$$
$$Q(x) = (((3x - 5)x + 1)x 3)x + 5$$
Try to evaluate P(2) and Q(2) in your head, using the forms given. Which is easier? Now write the polynomial
R(x) =x^{5} - 2x^{4} + 3x^{3} - 2x^{2} + 3x + 4\) in “nested” form, like the polynomial Q. Use the nested form to find R(3) in your head.
Do you see how calculating with the nested form follows the same arithmetic steps as calculating the value ofa polynomial using synthetic division?
2021-02-22
Step 1
Given $$P(x) = 3x^{4} - 5x^{3} + x^{2} - 3x + 5$$
$$Q(x) = (((3x - 5)x + 1)x-3)x+5$$
$$R(x) = x^{5} -2x^{4} + 3x^{3} - 2x^{2} + 3x + 4$$
Expand Q
$$Q(x) = (((3x - 5)x + 1)x-3)x + 5$$
$$=((3x^{2} - 5x + 1)x-3)x + 5$$
$$=(3x^{3} - 5x^{2} + x - 3)x + 5$$
$$= 3x^{4} - 5x^{3} + x^{2} - 3x + 5$$
$$\text{So}, P(x) = Q(x) = 3x^{4} -5x^{3} +x^{2} - 3x + 5$$
Hence proved
Step 2
Evaluate P(2) and Q(2)
$$P(x) = 3x^{4} - 5x^{3} + x^{2} - 3x + 5$$
$$P(2) = 3(2)^{4} - 5(2)^{3} + (2)^{2} - 3(2) + 5$$
$$= 48 - 40 + 4 - 6 + 5$$
$$=11$$
$$Q(2) = (((3(2) - 5)2+1)2- 3) 2 + 5$$
$$=((3(2) + 1)2 - 3)2 +5$$
$$=((3(2) - 3)2 + 5$$
$$= (3)2+5$$
$$= 11$$
Nested form of R(x)
$$R(x) = x^{5} - 2x^{4} + 3x^{3} - 2x^{2} + 3x +4$$
$$R(x) = (x^{4} - 2x^{3} + 3x^{2} - 2x + 3)x +4$$
$$= ((x^{3} - 2x^{2} + 3x - 2)x + 3)x +4$$
$$= (((x^{2} - 2x + 3)x - 2)x + 3)x +4$$
$$= ((((x - 2)x + 3)x - 2)x + 3)x +4$$
$$R(x) = ((((x - 2)x + 3)x - 2)x + 3)x +4$$
$$R(3) = ((((3 - 2)3 + 3)3 - 2)3 + 3)3 + 4$$
$$=167$$
Relevant Questions
Nested Form of a Polynomial Expand Q to prove that the polynomials P and Q ae the same $$\displaystyle{P}{\left({x}\right)}={3}{x}^{{4}}-{5}{x}^{{3}}+{x}^{{2}}-{3}{x}+{5}{N}{S}{K}{Q}{\left({x}\right)}={\left({\left({\left({3}{x}-{5}\right)}{x}+{1}\right)}{x}-{3}\right)}{x}+{5}$$ Try to evaluate P(2) and Q(2) in your head, using the forms given. Which is easier? Now write the polynomial $$\displaystyle{R}{\left({x}\right)}={x}^{{5}}—{2}{x}^{{4}}+{3}{x}^{{3}}—{2}{x}^{{3}}+{3}{x}+{4}$$ in “nested” form, like the polynomial Q. Use the nested form to find R(3) in your head. Do you see how calculating with the nested form follows the same arithmetic steps as calculating the value ofa polynomial using synthetic division?
Given the following function: $$\displaystyle{f{{\left({x}\right)}}}={1.01}{e}^{{{4}{x}}}-{4.62}{e}^{{{3}{x}}}-{3.11}{e}^{{{2}{x}}}+{12.2}{e}^{{{x}}}-{1.99}$$ a)Use three-digit rounding frithmetic, the assumption that $$\displaystyle{e}^{{{1.53}}}={4.62}$$, and the fact that $$\displaystyle{e}^{{{n}{x}}}={\left({e}^{{{x}}}\right)}^{{{n}}}$$ to evaluate $$\displaystyle{f{{\left({1.53}\right)}}}$$ b)Redo the same calculation by first rewriting the equation using the polynomial factoring technique c)Calculate the percentage relative errors in both part a) and b) to the true result $$\displaystyle{f{{\left({1.53}\right)}}}=-{7.60787}$$
Given the following function:
$$\displaystyle{f{{\left({x}\right)}}}={1.01}{e}^{{{4}{x}}}-{4.62}{e}^{{{3}{x}}}-{3.11}{e}^{{{2}{x}}}+{12.2}{e}^{{{x}}}$$
a) Use three-digit rounding frithmetic, the assumption that $$\displaystyle{e}^{{{1.53}}}={4.62}$$, and the fact that $$\displaystyle{e}^{{{n}{x}}}={\left({e}^{{{x}}}\right)}^{{{n}}}$$ to evaluate $$\displaystyle{f{{\left({1.53}\right)}}}$$
b) Redo the same calculation by first rewriting the equation using the polynomial factoring technique
c) Calculate the percentage relative errors in both part a) and b) to the true result $$\displaystyle{f{{\left({1.53}\right)}}}=-{7.60787}$$
Case: Dr. Jung’s Diamonds Selection
With Christmas coming, Dr. Jung became interested in buying diamonds for his wife. After perusing the Web, he learned about the “4Cs” of diamonds: cut, color, clarity, and carat. He knew his wife wanted round-cut earrings mounted in white gold settings, so he immediately narrowed his focus to evaluating color, clarity, and carat for that style earring.
After a bit of searching, Dr. Jung located a number of earring sets that he would consider purchasing. But he knew the pricing of diamonds varied considerably. To assist in his decision making, Dr. Jung decided to use regression analysis to develop a model to predict the retail price of different sets of round-cut earrings based on their color, clarity, and carat scores. He assembled the data in the file Diamonds.xls for this purpose. Use this data to answer the following questions for Dr. Jung.
1) Prepare scatter plots showing the relationship between the earring prices (Y) and each of the potential independent variables. What sort of relationship does each plot suggest?
2) Let X1, X2, and X3 represent diamond color, clarity, and carats, respectively. If Dr. Jung wanted to build a linear regression model to estimate earring prices using these variables, which variables would you recommend that he use? Why?
3) Suppose Dr. Jung decides to use clarity (X2) and carats (X3) as independent variables in a regression model to predict earring prices. What is the estimated regression equation? What is the value of the R2 and adjusted-R2 statistics?
4) Use the regression equation identified in the previous question to create estimated prices for each of the earring sets in Dr. Jung’s sample. Which sets of earrings appear to be overpriced and which appear to be bargains? Based on this analysis, which set of earrings would you suggest that Dr. Jung purchase?
5) Dr. Jung now remembers that it sometimes helps to perform a square root transformation on the dependent variable in a regression problem. Modify your spreadsheet to include a new dependent variable that is the square root on the earring prices (use Excel’s SQRT( ) function). If Dr. Jung wanted to build a linear regression model to estimate the square root of earring prices using the same independent variables as before, which variables would you recommend that he use? Why?
1
6) Suppose Dr. Jung decides to use clarity (X2) and carats (X3) as independent variables in a regression model to predict the square root of the earring prices. What is the estimated regression equation? What is the value of the R2 and adjusted-R2 statistics?
7) Use the regression equation identified in the previous question to create estimated prices for each of the earring sets in Dr. Jung’s sample. (Remember, your model estimates the square root of the earring prices. So you must actually square the model’s estimates to convert them to price estimates.) Which sets of earring appears to be overpriced and which appear to be bargains? Based on this analysis, which set of earrings would you suggest that Dr. Jung purchase?
8) Dr. Jung now also remembers that it sometimes helps to include interaction terms in a regression model—where you create a new independent variable as the product of two of the original variables. Modify your spreadsheet to include three new independent variables, X4, X5, and X6, representing interaction terms where: X4 = X1 × X2, X5 = X1 × X3, and X6 = X2 × X3. There are now six potential independent variables. If Dr. Jung wanted to build a linear regression model to estimate the square root of earring prices using the same independent variables as before, which variables would you recommend that he use? Why?
9) Suppose Dr. Jung decides to use color (X1), carats (X3) and the interaction terms X4 (color * clarity) and X5 (color * carats) as independent variables in a regression model to predict the square root of the earring prices. What is the estimated regression equation? What is the value of the R2 and adjusted-R2 statistics?
10) Use the regression equation identified in the previous question to create estimated prices for each of the earring sets in Dr. Jung’s sample. (Remember, your model estimates the square root of the earring prices. So you must square the model’s estimates to convert them to actual price estimates.) Which sets of earrings appear to be overpriced and which appear to be bargains? Based on this analysis, which set of earrings would you suggest that Dr. Jung purchase?
1. Explain with numerical examples what Real Numbers and Algebraic Expressions are. 2. Explain with numerical examples Factoring and finding LCMs (least common multiples). Explain factoring of a larger number. 3. Explain with numerical examples arithmetical operations (addition, subtraction, multiplication, division) with fractions 4, Explain with numerical examples arithmetical operations (addition, subtraction, multiplication, division) with percentages 5. Explain with numerical examples exponential notation 6. Explain with numerical examples order (precedence) of arithmetic operations 7. Explain with numerical examples the concept and how to find perimeter, area, volume, and circumference (use related formulas)
An experiment on the probability is carried out, in which the sample space of the experiment is
$$S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.$$
Let event $$E={2, 3, 4, 5, 6, 7}, event$$
$$F={5, 6, 7, 8, 9}, event G={9, 10, 11, 12}, and event H={2, 3, 4}$$.
Assume that each outcome is equally likely. List the outcome s in For G.
Now find P( For G) by counting the number of outcomes in For G.
Determine P (For G ) using the General Addition Rule.
The table below shows the number of people for three different race groups who were shot by police that were either armed or unarmed. These values are very close to the exact numbers. They have been changed slightly for each student to get a unique problem.
Suspect was Armed:
Black - 543
White - 1176
Hispanic - 378
Total - 2097
Suspect was unarmed:
Black - 60
White - 67
Hispanic - 38
Total - 165
Total:
Black - 603
White - 1243
Hispanic - 416
Total - 2262
Give your answer as a decimal to at least three decimal places.
a) What percent are Black?
b) What percent are Unarmed?
c) In order for two variables to be Independent of each other, the P $$(A and B) = P(A) \cdot P(B) P(A and B) = P(A) \cdot P(B).$$
This just means that the percentage of times that both things happen equals the individual percentages multiplied together (Only if they are Independent of each other).
Therefore, if a person's race is independent of whether they were killed being unarmed then the percentage of black people that are killed while being unarmed should equal the percentage of blacks times the percentage of Unarmed. Let's check this. Multiply your answer to part a (percentage of blacks) by your answer to part b (percentage of unarmed).
Remember, the previous answer is only correct if the variables are Independent.
d) Now let's get the real percent that are Black and Unarmed by using the table?
If answer c is "significantly different" than answer d, then that means that there could be a different percentage of unarmed people being shot based on race. We will check this out later in the course.
Let's compare the percentage of unarmed shot for each race.
e) What percent are White and Unarmed?
f) What percent are Hispanic and Unarmed?
If you compare answers d, e and f it shows the highest percentage of unarmed people being shot is most likely white.
Why is that?
This is because there are more white people in the United States than any other race and therefore there are likely to be more white people in the table. Since there are more white people in the table, there most likely would be more white and unarmed people shot by police than any other race. This pulls the percentage of white and unarmed up. In addition, there most likely would be more white and armed shot by police. All the percentages for white people would be higher, because there are more white people. For example, the table contains very few Hispanic people, and the percentage of people in the table that were Hispanic and unarmed is the lowest percentage.
Think of it this way. If you went to a college that was 90% female and 10% male, then females would most likely have the highest percentage of A grades. They would also most likely have the highest percentage of B, C, D and F grades
The correct way to compare is "conditional probability". Conditional probability is getting the probability of something happening, given we are dealing with just the people in a particular group.
g) What percent of blacks shot and killed by police were unarmed?
h) What percent of whites shot and killed by police were unarmed?
i) What percent of Hispanics shot and killed by police were unarmed?
You can see by the answers to part g and h, that the percentage of blacks that were unarmed and killed by police is approximately twice that of whites that were unarmed and killed by police.
j) Why do you believe this is happening?
Do a search on the internet for reasons why blacks are more likely to be killed by police. Read a few articles on the topic. Write your response using the articles as references. Give the websites used in your response. Your answer should be several sentences long with at least one website listed. This part of this problem will be graded after the due date.
Testing for a Linear Correlation. In Exercises 13–28, construct a scatterplot, and find the value of the linear correlation coefficient r. Also find the P-value or the critical values of r from Table A-6. Use a significance level of $$\alpha = 0.05$$. Determine whether there is sufficient evidence to support a claim of a linear correlation between the two variables. (Save your work because the same data sets will be used in Section 10-2 exercises.) Lemons and Car Crashes Listed below are annual data for various years. The data are weights (metric tons) of lemons imported from Mexico and U.S. car crash fatality rates per 100,000 population [based on data from “The Trouble with QSAR (or How I Learned to Stop Worrying and Embrace Fallacy),” by Stephen Johnson, Journal of Chemical Information and Modeling, Vol. 48, No. 1]. Is there sufficient evidence to conclude that there is a linear correlation between weights of lemon imports from Mexico and U.S. car fatality rates? Do the results suggest that imported lemons cause car fatalities? $$\begin{matrix} \text{Lemon Imports} & 230 & 265 & 358 & 480 & 530\\ \text{Crashe Fatality Rate} & 15.9 & 15.7 & 15.4 & 15.3 & 14.9\\ \end{matrix}$$ | 2021-05-14T01:08:48 | {
"domain": "plainmath.net",
"url": "https://plainmath.net/1515/polynomial-polynomials-polynomial-polynomial-calculating-calculating",
"openwebmath_score": 0.48904573917388916,
"openwebmath_perplexity": 1252.9991607719146,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9715639669551474,
"lm_q2_score": 0.8670357615200474,
"lm_q1q2_score": 0.8423807039543943
} |
https://docs.juliahub.com/SymPy/KzewI/1.1.4/Tutorial/solvers/ | # Solvers
From
>>> from sympy import *
>>> x, y, z = symbols("x y z")
>>> init_printing(use_unicode=True)
##### In Julia:
julia> using SymPy
julia> @syms x, y, z
(x, y, z)
Recall from the :ref:gotchas <tutorial_gotchas_equals> section of this tutorial that symbolic equations in SymPy are not represented by = or ==, but by Eq.
>>> Eq(x, y)
x = y
##### In Julia:
julia> Eq(x, y)
x = y
However, there is an even easier way. In SymPy, any expression not in an Eq is automatically assumed to equal 0 by the solving functions. Since a = b if and only if a - b = 0, this means that instead of using x == y, you can just use x - y. For example
>>> solveset(Eq(x**2, 1), x)
{-1, 1}
>>> solveset(Eq(x**2 - 1, 0), x)
{-1, 1}
>>> solveset(x**2 - 1, x)
{-1, 1}
##### In Julia:
julia> solveset(Eq(x^2, 1), x)
{-1, 1}
julia> solveset(Eq(x^2 - 1, 0), x)
{-1, 1}
julia> solveset(x^2 - 1, x)
{-1, 1}
This is particularly useful if the equation you wish to solve is already equal to 0. Instead of typing solveset(Eq(expr, 0), x), you can just use solveset(expr, x).
## Solving Equations Algebraically
The main function for solving algebraic equations is solveset. The syntax for solveset is solveset(equation, variable=None, domain=S.Complexes) Where equations may be in the form of Eq instances or expressions that are assumed to be equal to zero.
Please note that there is another function called solve which can also be used to solve equations. The syntax is solve(equations, variables) However, it is recommended to use solveset instead.
When solving a single equation, the output of solveset is a FiniteSet or an Interval or ImageSet of the solutions.
>>> solveset(x**2 - x, x)
{0, 1}
>>> solveset(x - x, x, domain=S.Reals)
ℝ
>>> solveset(sin(x) - 1, x, domain=S.Reals)
⎧ π ⎫
⎨2⋅n⋅π + ─ | n ∊ ℤ⎬
⎩ 2 ⎭
##### In Julia:
• S is not exported, as it is not a function, so we create an alias:
julia> const S = sympy.S
PyObject S
julia> solveset(x^2 - x, x)
{0, 1}
julia> solveset(x - x, x, domain=S.Reals)
ℝ
julia> solveset(sin(x) - 1, x, domain=S.Reals)
⎧ π │ ⎫
⎨2⋅n⋅π + ─ │ n ∊ ℤ⎬
⎩ 2 │ ⎭
If there are no solutions, an EmptySet is returned and if it is not able to find solutions then a ConditionSet is returned.
>>> solveset(exp(x), x) # No solution exists
∅
>>> solveset(cos(x) - x, x) # Not able to find solution
{x | x ∊ ℂ ∧ -x + cos(x) = 0}
##### In Julia:
julia> solveset(exp(x), x) # No solution exists
∅
julia> solveset(cos(x) - x, x) # Not able to find solution
{x │ x ∊ ℂ ∧ (-x + cos(x) = 0)}
In the solveset module, the linear system of equations is solved using linsolve. In future we would be able to use linsolve directly from solveset. Following is an example of the syntax of linsolve.
• List of Equations Form:
>>> linsolve([x + y + z - 1, x + y + 2*z - 3 ], (x, y, z))
##### In Julia:
Rather than a vector, we pass a tuple:
julia> linsolve((x + y + z - 1, x + y + 2*z - 3), (x, y, z))
{(-y - 1, y, 2)}
A tuple
Tuples
The linsolve function expects a list of equations, whereas PyCall is instructed to promote the syntax to produce a list in Python into a Array{Sym} object. As such, we pass the equations in a tuple above. Similar considerations are necessary at times for the sympy.Matrix constructor. It is suggested, as in the next example, to work around this by passing Julian arrays to the constructor or bypassing it altogether.
• Augmented
Matrix Form:
>>> M = Matrix(((1, 1, 1, 1), (1, 1, 2, 3)))
>>> system = A, b = M[:, :-1], M[:, -1]
>>> linsolve(system, x, y, z)
{(-y - 1, y, 2)}
##### In Julia:
We use Julian syntax for matrices:
julia> A = [1 1 1; 1 1 2]; b = [1,3]
2-element Vector{Int64}:
1
3
The augmented form is not available
julia> aug = [A b]
2×4 Array{Int64,2}:
1 1 1 1
1 1 2 3
julia> linsolve(sympy.Matrix(aug), (x,y,z)) # not {(-y - 1, y, 2)}!
∅
In lieu of using sympy.Matrix, the matrix can be created symbolically, as:
julia> A = Sym[1 1 1; 1 1 2]; b = [1,3]
2-element Vector{Int64}:
1
3
julia> aug = [A b]
2×4 Matrix{Sym}:
1 1 1 1
1 1 2 3
julia> linsolve(aug, (x,y,z)) # {(-y - 1, y, 2)};
Finally, linear equations are solved in Julia with the \ (backslash) operator:
A \ b
The variables are generated within \ in the sequence x1, x2, ...
• A*x = b Form
>>> M = Matrix(((1, 1, 1, 1), (1, 1, 2, 3)))
>>> system = A, b = M[:, :-1], M[:, -1]
>>> linsolve(system, x, y, z)
{(-y - 1, y, 2)}
##### In Julia:
We follow the syntax above to construct the matrix (tuple of tuples), but not the Julian matrix construtor would be recommended:
julia> M = sympy.Matrix(((1, 1, 1, 1), (1, 1, 2, 3)))
2×4 Matrix{Sym}:
1 1 1 1
1 1 2 3
julia> system = A, b = M[:, 1:end-1], M[:, end]
(Sym[1 1 1; 1 1 2], Sym[1, 3])
julia> linsolve(system, x, y, z)
{(-y - 1, y, 2)}
Note
The order of solution corresponds the order of given symbols.
In the solveset module, the non linear system of equations is solved using nonlinsolve. Following are examples of nonlinsolve.
1. When only real solution is present:
>>> a, b, c, d = symbols('a, b, c, d', real=True)
>>> nonlinsolve([a**2 + a, a - b], [a, b])
{(-1, -1), (0, 0)}
>>> nonlinsolve([x*y - 1, x - 2], x, y)
{(2, 1/2)}
##### In Julia:
• we pass [a,b] as either a, b or using a tuple, as in (a,b), but not as a vector, as this gets mapped into a vector of symbolic objects which causes issues with nonlinsolve:
julia> @syms a::real, b::real, c::real, d::real
(a, b, c, d)
julia> nonlinsolve([a^2 + a, a - b], a, b)
{(-1, -1), (0, 0)}
julia> nonlinsolve([x*y - 1, x - 2], x, y)
{(2, 1/2)}
1. When only complex solution is present:
>>> nonlinsolve([x**2 + 1, y**2 + 1], [x, y])
{(-ⅈ, -ⅈ), (-ⅈ, ⅈ), (ⅈ, -ⅈ), (ⅈ, ⅈ)}
##### In Julia:
julia> nonlinsolve([x^2 + 1, y^2 + 1], (x, y))
{(-ⅈ, -ⅈ), (-ⅈ, ⅈ), (ⅈ, -ⅈ), (ⅈ, ⅈ)}
1. When both real and complex solution is present:
>>> from sympy import sqrt
>>> system = [x**2 - 2*y**2 -2, x*y - 2]
>>> vars = [x, y]
>>> nonlinsolve(system, vars)
{(-2, -1), (2, 1), (-√2⋅ⅈ, √2⋅ⅈ), (√2⋅ⅈ, -√2⋅ⅈ)}
>>> n = Dummy('n')
>>> system = [exp(x) - sin(y), 1/y - 3]
>>> real_soln = (log(sin(S(1)/3)), S(1)/3)
>>> img_lamda = Lambda(n, 2*n*I*pi + Mod(log(sin(S(1)/3)), 2*I*pi))
>>> complex_soln = (ImageSet(img_lamda, S.Integers), S(1)/3)
>>> soln = FiniteSet(real_soln, complex_soln)
>>> nonlinsolve(system, [x, y]) == soln
True
##### In Julia:
• we must remove the spaces within []
• we must pass vars as a tuple:
julia> system = [x^2-2*y^2-2, x*y-2]
2-element Vector{Sym}:
x^2 - 2*y^2 - 2
x⋅y - 2
julia> vars = (x, y)
(x, y)
julia> nonlinsolve(system, vars)
{(-2, -1), (2, 1), (-√2⋅ⅈ, √2⋅ⅈ), (√2⋅ⅈ, -√2⋅ⅈ)}
However, the next bit requires some modifications to run:
• the system array definition must have extra spaces removed
• Dummy, Mod, ImageSet, FiniteSet aren't exported
• we need PI, not pi to have a symbolic value
• we compare manually
julia> n = sympy.Dummy("n")
n
julia> system = [exp(x)-sin(y), 1/y-3]
2-element Vector{Sym}:
exp(x) - sin(y)
-3 + 1/y
julia> real_soln = (log(sin(S(1)/3)), S(1)/3)
(log(sin(1/3)), 1/3)
julia> img_lamda = Lambda(n, 2*n*IM*PI + sympy.Mod(sin(S(1)/3), 2*IM*PI))
n ↦ 2⋅n⋅ⅈ⋅π + (sin(1/3) mod 2⋅ⅈ⋅π)
julia> complex_soln = (sympy.ImageSet(img_lamda, S.Integers), S(1)/3)
(ImageSet(Lambda(_n, 2*_n*I*pi + Mod(sin(1/3), 2*I*pi)), Integers), 1/3)
julia> soln = sympy.FiniteSet(real_soln, complex_soln)
{(log(sin(1/3)), 1/3), ({2⋅n⋅ⅈ⋅π + (sin(1/3) mod 2⋅ⅈ⋅π) │ n ∊ ℤ}, 1/3)}
julia> nonlinsolve(system, (x, y))
{({2⋅n⋅ⅈ⋅π + log(sin(1/3)) │ n ∊ ℤ}, 1/3)}
1. If non linear system of equations is Positive dimensional system (A system with
infinitely many solutions is said to be positive-dimensional):
>>> nonlinsolve([x*y, x*y - x], [x, y])
{(0, y)}
>>> system = [a**2 + a*c, a - b]
>>> nonlinsolve(system, [a, b])
{(0, 0), (-c, -c)}
##### In Julia:
• again, we use a tuple for the variables:
julia> nonlinsolve([x*y, x*y-x], (x, y))
{(0, y)}
julia> system = [a^2+a*c, a-b]
2-element Vector{Sym}:
a^2 + a*c
a - b
julia> nonlinsolve(system, (a, b))
{(0, 0), (-c, -c)}
Note:
1. The order of solution corresponds the order of given symbols.
2. Currently nonlinsolve doesn't return solution in form of LambertW (if there is solution present in the form of LambertW).
solve can be used for such cases:
>>> solve([x**2 - y**2/exp(x)], [x, y], dict=True)
⎡⎧ ⎛y⎞⎫⎤
⎢⎨x: 2⋅LambertW⎜─⎟⎬⎥
⎣⎩ ⎝2⎠⎭⎦
##### In Julia:
it is similar
julia> u = solve([x^2 - y^2/exp(x)], [x, y], dict=true)
2-element Vector{Dict{Any, Any}}:
Dict(y => -x*sqrt(exp(x)))
Dict(y => x*sqrt(exp(x)))
To get prettier output, the dict may be converted to have one with symbolic keys:
julia> convert(Dict{SymPy.Sym, Any}, first(u))
Dict{Sym, Any} with 1 entry:
y => -x*sqrt(exp(x))
1. Currently nonlinsolve is not properly capable of solving the system of equations
having trigonometric functions.
solve can be used for such cases(not all solution):
>>> solve([sin(x + y), cos(x - y)], [x, y])
⎡⎛-3⋅π 3⋅π⎞ ⎛-π π⎞ ⎛π 3⋅π⎞ ⎛3⋅π π⎞⎤
⎢⎜─────, ───⎟, ⎜───, ─⎟, ⎜─, ───⎟, ⎜───, ─⎟⎥
⎣⎝ 4 4 ⎠ ⎝ 4 4⎠ ⎝4 4 ⎠ ⎝ 4 4⎠⎦
##### In Julia:
julia> solve([sin(x + y), cos(x - y)], [x, y])
4-element Vector{Tuple{Sym, Sym}}:
(-3*pi/4, 3*pi/4)
(-pi/4, pi/4)
(pi/4, 3*pi/4)
(3*pi/4, pi/4)
solveset reports each solution only once. To get the solutions of a polynomial including multiplicity use roots.
>>> solveset(x**3 - 6*x**2 + 9*x, x)
{0, 3}
>>> roots(x**3 - 6*x**2 + 9*x, x)
{0: 1, 3: 2}
##### In Julia:
julia> solveset(x^3 - 6*x^2 + 9*x, x)
{0, 3}
julia> roots(x^3 - 6*x^2 + 9*x, x) |> d -> convert(Dict{Sym, Any}, d) # prettier priting
Dict{Sym, Any} with 2 entries:
3 => 2
0 => 1
The output {0: 1, 3: 2} of roots means that 0 is a root of multiplicity 1 and 3 is a root of multiplicity 2.
Note:
Currently solveset is not capable of solving the following types of equations:
• Equations solvable by LambertW (Transcendental equation solver).
solve can be used for such cases:
>>> solve(x*exp(x) - 1, x )
[LambertW(1)]
##### In Julia:
julia> solve(x*exp(x) - 1, x )
1-element Vector{Sym}:
W(1)
## Solving Differential Equations
To solve differential equations, use dsolve. First, create an undefined function by passing cls=Function to the symbols function.
>>> f, g = symbols('f g', cls=Function)
##### In Julia:
julia> @syms f() g()
(f, g)
f and g are now undefined functions. We can call f(x), and it will represent an unknown function.
>>> f(x)
f(x)
##### In Julia:
julia> f(x)
f(x)
Derivatives of f(x) are unevaluated.
>>> f(x).diff(x)
d
──(f(x))
dx
##### In Julia:
julia> f(x).diff(x)
d
──(f(x))
dx
(see the :ref:Derivatives <tutorial-derivatives> section for more on derivatives).
To represent the differential equation $f''(x) - 2f'(x) + f(x) = \sin(x)$, we would thus use
>>> diffeq = Eq(f(x).diff(x, x) - 2*f(x).diff(x) + f(x), sin(x))
>>> diffeq
2
d d
f(x) - 2⋅──(f(x)) + ───(f(x)) = sin(x)
dx 2
dx
##### In Julia:
julia> diffeq = Eq(f(x).diff(x, x) - 2*f(x).diff(x) + f(x), sin(x)); string(diffeq)
"Eq(f(x) - 2*Derivative(f(x), x) + Derivative(f(x), (x, 2)), sin(x))"
To solve the ODE, pass it and the function to solve for to dsolve.
>>> dsolve(diffeq, f(x))
x cos(x)
f(x) = (C₁ + C₂⋅x)⋅ℯ + ──────
2
##### In Julia:
• we use dsolve for initial value proplems
julia> dsolve(diffeq, f(x)) |> string
"Eq(f(x), (C1 + C2*x)*exp(x) + cos(x)/2)"
dsolve returns an instance of Eq. This is because in general, solutions to differential equations cannot be solved explicitly for the function.
>>> dsolve(f(x).diff(x)*(1 - sin(f(x))), f(x))
f(x) + cos(f(x)) = C₁
##### In Julia:
julia> dsolve(f(x).diff(x)*(1 - sin(f(x))), f(x))
f(x) = C₁
The arbitrary constants in the solutions from dsolve are symbols of the form C1, C2, C3, and so on.
## Julia alternative interface
SymPy.jl adds a SymFunction class, that makes it a bit easier to set up a differential equation, though not as general.
We use either the SymFunction constructor
julia> f = SymFunction("f")
f
or the @syms macro, as in @syms f() to define symbolic functions. The Differential function (who's functionality is lifted from ModelingToolkit). Can simplify things:
julia> D = Differential(x);
julia> diffeq = Eq(D(D(f))(x) - 2*D(f)(x) + f(x), sin(x)); string(diffeq)
"Eq(f(x) - 2*Derivative(f(x), x) + Derivative(f(x), (x, 2)), sin(x))"
julia> dsolve(diffeq, f(x)) |> string
"Eq(f(x), (C1 + C2*x)*exp(x) + cos(x)/2)"
Or:
julia> dsolve(D(f)(x)*(1 - sin(f(x))), f(x))
f(x) = C₁
Initial conditions can be specified using a dictionary.
For the initial condition f'(x0) = y0, this would be specified as Dict(D(f)(x0) => y0).
For example, to solve the exponential equation $f'(x) = f(x), f(0) = a$ we would have:
julia> @syms x, a, f()
(x, a, f)
julia> dsolve(D(f)(x) - f(x), f(x), ics = Dict(f(0) => a)) |> string
"Eq(f(x), a*exp(x))"
To solve the simple harmonic equation, where two initial conditions are specified, we combine the tuple for each within another tuple:
julia> ics = Dict(f(0) => 1, D(f)(0) => 2);
julia> dsolve(D(D(f))(x) - f(x), f(x), ics=ics) |> string
"Eq(f(x), 3*exp(x)/2 - exp(-x)/2)" | 2022-08-12T14:31:20 | {
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https://physics.stackexchange.com/questions/469617/comparing-measurements-of-a-2d-quantum-harmonic-oscillator-between-cartesian-and | # Comparing measurements of a 2D quantum harmonic oscillator between cartesian and rotated cartesian coordinates
I've come across an old quantum exam problem that's causing me a bit of confusion, and I'm hoping someone can offer some clarity:
There is a particle in a 2D harmonic oscillator potential such that it is described by the Hamiltonian (in unitless form):
$$H = -\frac{1}{2}(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}) + \frac{1}{2}(x^2+y^2)$$
Person A measures the system using x and y, while Person B uses a rotated coordinate system:
$$x' = x \: cos(\alpha) - y \: sin(\alpha)$$
$$y' = x \: sin(\alpha) + y \: cos(\alpha)$$
w/ $$\:\:0 < \alpha < \frac{\pi}{2}$$
The eigenstates for a 1D Harmonic Oscillator are:
$$\phi_0(x) = (\frac{1}{\pi})^{\frac{1}{4}}\: e^{-\frac{1}{2}x^2}$$
$$\phi_1(x) = (\frac{4}{\pi})^{\frac{1}{4}}\: x \: e^{-\frac{1}{2}x^2}$$
$$\phi_2(x) = (\frac{1}{4\pi})^{\frac{1}{4}}\: (2x^2-1) \: e^{-\frac{1}{2}x^2}$$
a) Is Person B's system still a two-dimensional harmonic oscillator?
To answer this, I inverted the expressions for x' and y' to get x and y in terms of x' and y' instead:
$$x = x' \: cos(\alpha) + y' \: sin(\alpha)$$
$$y = - x' \: sin(\alpha) + y' \: cos(\alpha)$$
And then replace the given Hamiltonian's x and y with these transformed versions:
$$H' = -\frac{1}{2} ([cos(\alpha)\frac{\partial}{\partial x'} + sin(\alpha)\frac{\partial}{\partial y'}]^2 +[-sin(\alpha)\frac{\partial}{\partial x'} + cos(\alpha)\frac{\partial}{\partial y'}]^2)$$
$$+ \frac{1}{2}([x' \: cos(\alpha) + y' \: sin(\alpha)]^2 +[- x' \: sin(\alpha) + y' \: cos(\alpha)]^2)$$
Doing some algebra, this yields:
$$H' = -\frac{1}{2}(\frac{\partial^2}{\partial x'^2}+\frac{\partial^2}{\partial y'^2}) + \frac{1}{2}(x'^2+y'^2)$$
(or we could just recognize that $$x^2 + y^2 = r^2$$, which is unchanged by rotation about the origin)
b) Person A puts the particle in state:
$$(n_x,n_y) = (1,0)$$ w/ $$E = 2\hbar\omega_0$$
Will Person B measure the same energy eigenvalue as Person A? Will Person B measure the same average energy as Person A?
First, I noted that $$E = 2\hbar\omega_0$$ is E = 2 in natural units. Using the given eigenstates, I computed:
$$\phi_{1,0}(x,y) = \phi_{1}(x)\phi_{0}(y) = (\frac{2}{\pi})^{\frac{1}{2}}\: x \: e^{-\frac{1}{2}(x^2+y^2)}$$
Then, to get this in terms of Person B's coordinates I again substituted x(x',y') and y(x',y'):
$$\phi'(x',y') = (\frac{2}{\pi})^{\frac{1}{2}}\: [x' \: cos(\alpha) + y' \: sin(\alpha)] \: e^{-\frac{1}{2}(x'^2+y'^2)}$$
Then,
$$H'\phi' = 2 \cdot (\frac{2}{\pi})^{\frac{1}{2}}\: [x' \: cos(\alpha) + y' \: sin(\alpha)] \: e^{-\frac{1}{2}(x'^2+y'^2)}$$ (after a moderate amount of math)
Clearly, this indicates that:
$$H'\phi' = 2 \cdot \phi'$$
And so, $$E = 2\hbar\omega_0$$ is also an eigenvalue in the rotated system of Person B (this makes intuitive sense, as we wouldn't expect that measuring a system at a different angle would change measured values of energy). As it is an eigenvalue for Person B's system, it should be a guaranteed measurement in the prepared state (thus also being the average value of energy, and the same as Person A's).
c) Will Person B say that the system's particle is in an eigenstate?
I think this is where I'm starting to get a bit lost in the language. If my conclusion from b) is correct, then this must be 'yes', right? However, subsequent part d makes me question that this is so:
d) Determine the probability that Person B will say that the particle is in each of the following states:
1) $$(n_x',n_y') = (1,0)$$
2) $$(n_x',n_y') = (0,1)$$
3) $$(n_x',n_y') = (0,0)$$
If we've just measured the particle in an eigenstate of Person B's basis, then the probability of a subsequent measurement of the system resulting in a system in a different eigenstate must be zero, right?
What am I misunderstanding here? part d was worth nearly half of the problem's points, so it seems odd to result in such a trivial solution. Much appreciated!
It's probably not appropriate to have written:
$$\phi'(x',y') = (\frac{2}{\pi})^{\frac{1}{2}}\: [x' \: cos(\alpha) + y' \: sin(\alpha)] \: e^{-\frac{1}{2}(x'^2+y'^2)}$$
Rather, since this is an eigenstate of the unprimed system but in terms of x' and y', it's probably better labeled just as $$\phi_{n_x = 1, n_y = 0}(x',y')$$
Since the primed system is also a harmonic oscillator, the degenerate eigenstates that produce E=2 would be:
$$\phi'_{1,0}(x',y') = (\frac{2}{\pi})^{\frac{1}{2}}\: x' \: e^{-\frac{1}{2}(x'^2+y'^2)}$$
$$\phi'_{0,1}(x',y') = (\frac{2}{\pi})^{\frac{1}{2}}\: y' \: e^{-\frac{1}{2}(x'^2+y'^2)}$$
So that the prepared state in terms of eigenstates of the primed basis is:
$$\phi_{1,0} = \cos\alpha \: \phi'_{1',0'} + \sin\alpha \: \phi'_{0',1'}$$
(The notation has become a bit convoluted, but hopefully you understand my meaning)
This result seems to make sense--- if $$\alpha = 0$$, we get back the original state, and if $$\alpha = \frac{\pi}{2}$$ we get the orthogonal (0',1') state instead.
Thank you again!
• Your answer to part (b) indeed tells you that Person B will measure the same energy as Person A. But Person B has two possible states that have that same energy. So it might be a superposition of the two! – octonion Mar 30 '19 at 21:59
The energy eigenvalue is known: it's 2, as you said, for B as well as for A. But energy eigenvalue $$E=2$$ has a 2D eigenspace, spanned by base vectors (1,0) and (0,1) both for B as for A. However these quantum numbers have different meaning for them: for A they refer to $$n_x$$, $$n_y$$ whereas for B they refer to $$n'_x$$, $$n'_y$$.
An observation of energy starting form state $$(n_x=1,n_y=0)$$ will certainly give an eigenvalue $$E=2$$ and the resulting state will be the projection of initial state in the subspace spanned by $$(n'_x=1,n'_y=0)$$ and $$(n'_x=0,n'_y=1)$$.
I leave for you to find that projection. (Hint: express $$n_x$$ as a linear combination of $$n'_x$$, $$n'_y$$.)
• Thank you! I've appended the new solution to my original question (as it was too long for a comment). My only follow-up, if you don't mind: do I remember correctly that, for degenerate eigenvalues, our resulting state is not either \phi'_{1',0'} or \phi'_{0',1'}, but rather exactly the linear combination of the two states --- $\phi_{1,0} = cos\alpha \: \phi'_{1',0'} + sin\alpha \: \phi'_{0',1'}$? – leo_africanus Apr 1 '19 at 15:22 | 2020-02-21T16:41:38 | {
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https://math.stackexchange.com/questions/1592198/what-is-the-intersection-of-these-two-cylinders | What is the intersection of these two cylinders?
$$0\le x^2 + z^2 \le 1$$
$$0 \le y^2 + z^2 \le 1$$
I want to compute the volume of the intersection.
Sketching it out on paper is sort of nice: I see cross-sections that are disks, the first cylinder, the y-coordinate is free to vary, and for the second cylinder, the x-coordinate is free to vary.
The intersection, I would guess, seems to be something spherical.
So how can I pin down the actual set of points?
Well, one thing I thought of was to try to manipulate both inequalities to make use of the equation of a sphere, so I try looking at these inequalities instead:
$$y^2\le x^2 + y^2 + z^2 \le 1 +y^2$$
$$x^2 \le x^2 +y^2 + z^2 \le 1+x^2$$
Am I heading in the right direction? Where can I go from here?
Thanks,
• It's definitely not a sphere, not sure you can say much more about it than it's the intersection of two perpendicular cylinders. – Gregory Grant Dec 28 '15 at 23:51
• Hi Professor Grant, thanks for your comment, and especially for noting the orthogonality relationship. I think I am almost there... – user301446 Dec 29 '15 at 0:13
• This comment is to link this post as one of the (abstract) duplicates to the current choice of mother/target post. – Lee David Chung Lin Jan 22 at 10:29
The intersection of two cylinders is called a Steinmetz solid. You can give a description of the edges of the solid by
$$x = \pm z, \quad y = \pm \sqrt{1 - z^2}$$
and use these to give corresponding inequalities.
• Hi @user, thanks so much for your quick response. Well, using the above inequalities, I do see these coming out: $x = +/- \sqrt{1-z^2}$ and $y = +/- \sqrt{1-z^2}$. How come you don't mention this guy: $x = +/- \sqrt{1-z^2}$? ... is it already like implied somewhere in your description? Thanks, – user301446 Dec 28 '15 at 23:59
• Ooh @user, I think I see what you are saying ... – user301446 Dec 29 '15 at 0:00
• So my above comment, I have basically found my integration limits for both x and y...is that correct, @user? And now, I just have to find the integration limits for $z$ ... – user301446 Dec 29 '15 at 0:01
• But, solving for $z$ in each inequality gives us two upper and lower limits of integration, which is strange, @user ... – user301446 Dec 29 '15 at 0:03
• Perhaps the limits for $z$ is simply +/- 1, since the cross-sections are disks of radius 1, @user ... what do you think? Thanks, – user301446 Dec 29 '15 at 0:06
I could not resist to model this in GeoGebra.
Zenith is $(0, 0, 1)$ and nadir $(0, 0, -1)$. A slice at height $z$ is a square with side length $$a = 2 \sqrt{1-z^2}$$ so $$dV = (2 \sqrt{1-z^2})^2 dz = 4(1-z^2) dz$$
• haha...@mvw, thanks so much. I am practicing for exams, so I think I had better pin down the algebra on paper for now. Hmm...so the intersection is indeed not spherical, as Professor Grant points out in his comment above... – user301446 Dec 29 '15 at 0:11
• I hoped its surface intersection operation could deal with this one, but that is either too much yet for GeoGebra or I am doing something wrong. – mvw Dec 29 '15 at 0:14 | 2019-06-26T08:50:27 | {
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http://math.stackexchange.com/questions/216000/a-logic-statement-or-in-abstract-algebra-groups | # A logic statement. “or” in Abstract algebra - groups
Let H be the subset of $M_2(\mathbb{R})$ consisting of all matrices of the form
$H^* = \left \{ \begin{pmatrix} a &-b \\ b&a \end{pmatrix} : a,b\in\mathbb{R} , a\neq 0 \; \text{or} \; b \neq 0\right\}$
Is (H*, .) a group under multiplication?
I said no because since $b \neq 0$, the identity matrix can't exist. But my prof briefly said "either a or b can't be 0, not a and b"
Am I right or his wrong?
-
Note that your last line is false (and syntactically incorrect). You're not right and he's not wrong. However you are wrong, and he is right. – Asaf Karagila Oct 18 '12 at 1:16
He’s right, and you’re wrong. The definition
$$H^* = \left \{ \begin{pmatrix} a &-b \\ b&a \end{pmatrix} : a,b\in\mathbb{R} , a\neq 0 \; \text{or} \; b \neq 0\right\}$$
makes $H^*$ the collection of all $2\times 2$ real matrices such that at least one of $a$ and $b$ is non-zero. The identity matrix is such a matrix: for it you have $a\ne 0$.
The condition $$a\ne 0\text{ or }b\ne 0$$ is equivalent to the condition $$(a\ne 0\text{ and }b\ne 0)\text{ or }(a\ne 0)\text{ and }b=0)\text{ or }(a=0\text{ and }b\ne 0)\;;$$ the only case that it excludes is $a=b=0$. In mathematics and logic, or is inclusive: ‘$a\text{ or }b$’ is ‘$a\text{ or }b\text{ or both}$’.
-
So what is an example of what i mean? That is when $b \neq 0$ ever? How do I say that? – Hawk Oct 21 '12 at 5:32
@jak: If the definition had been $$H^*=\left\{\pmatrix{a&-b\\b&a}:a,b\in\Bbb R,a\neq 0\text{ and }b\neq 0\right\}\;,$$ your objection would have been correct. That definition requires that $a$ and $b$ both be non-zero and therefore excludes the identity matrix. – Brian M. Scott Oct 21 '12 at 10:44
In mathematics the use of the word or is inclusive, not exclusive. It is possible that only $b\neq 0$, it is possible that only $a\neq 0$, and it is possible that both $a\neq0$ and $b\neq 0$.
It is only impossible that both $a=b=0$.
So the identity matrix is in $H^\ast$, since $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ has the wanted property.
- | 2016-04-29T13:02:24 | {
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http://yme.cefalugibilmanna.it/questions-on-domain-and-range-of-a-function-pdf.html | # Questions On Domain And Range Of A Function Pdf
List the interval(s) on which fis decreasing. If it is, use the graph to find its domain and range, the intercepts, if any, and any symmetry with respect to the x - axis, the y - axis, or the origin. Although the Fourier transform is a complicated mathematical function, it isn’t a complicated concept to understand and relate to your measured signals. Find (3 Marks) 2. When functions are expressed as "rules" (formulas), be sure to think about possible "problem" spots before. A function maps elements of its Domain to elements of its Range. Question Group #1: Directions and/or Common Information: Determine if each set of information given below defines a function or a relation. • Student Worksheet 1: Linear Equations in Real World Problems, two pages, guided questions for students to complete. The range of a function is the set of output values when all x-values in the domain are evaluated into the function, commonly known as the y-values. gebraic methods to computethe value of a limit of a function. (a) Interpret the domain and range. If there is a requirement that a y-value produced by a function. Every function is also a relation 2. † identified the domain and range of functions. A function is surjective if for every y in the codomain B there is at least one x in the domain. 2 Represent Functions and Relations A2. Return to Exercises. The graph of f is the graph of the equation y = f(x). 1 1 domain range and end behaviortebook end behavior as x as x —00 explain 1 identifying a function s domain range and end behavior from its graph recall that the domain of a function fis the set of input values x and the range is the set of output values f x the end behavior of a function describes what happens to the f x values as the x. An example { tangent to a parabola16 3. Algebra functions lessons with lots of worked examples and practice problems. Domain and Range Name: _____ State the domain and range for each graph and then tell if the graph is a function (write yes or no). Similarly, the domain of a particular solution to a differential equation can be restricted for reasons other than the function formula not being defined, and indeed, may be a subset of what the domain would be when the solution is considered only as a function. Suppose the function H(t) gives the heart rate of a runner at various points in time during a 20 mile run. The graph looks like the following: This equation is a function because you can draw a vertical line through any value of x in the domain and it will cross the graph of the equation only once. Functions have a domain and range and can be found in two ways. Step 6: Insert any identified “Hole(s)” from Step 1. 51: Domain and Range 1: Determine the domain and range of a function from its graph 1 The effect of pH on the action of a certain enzyme is shown on the accompanying graph. Domains and Ranges are sets. First, this tells us the basic shape of the graph (see Figure 30 on page 440). Domain of a function – this is the set of input values for the function. Introduction to the domain and range of a function. Nevertheless, if we restrict the domain of ln(x) to x values for which it is defined, then it is continuous on that domain. The tangent to a curve15 2. The line is the graph of f (x) = mx + n. Determine domains and ranges of particular functions. Evaluate lim x 1 fx(). A straight line and its slope. 5 Functions as Arrow Diagrams When the domain and codomain of a function are nite sets then one can represent the function by an arrow diagram. D The number of sponsors. Give examples of nonlinear functions. If not, explain. 8)-10 -5 5 10 x y 10 5-5-10 MULTIPLE CHOICE. Review function notation. One-to-One. Question 1, below, combines elements from more than one question from the 2019 exam so that you can practice working on a 25-minute question. consists of two real number lines that intersect at a right angle. Domains can be found algebraically; ranges are often found algebraically and graphically. State the domain and range of each relation. 1 Understand that a function from one set (the input, called the domain) to another set (the output, called the range) assigns to each element of the domain exactly one element of the range, i. State the domain and range of each relation. The Quadratic Function 1. Rowell 10/22/04 1. P 0 P 1 y 1-y 0 x 1-x 0 x 0 x 1 y 0 y 1 n 1 m Figure 4. Find (3 Marks) 4. Therefore, we need only consider points that are inside both the $$δ$$ disk and the domain of the function. Let’s get started! Trigonometry is the branch of mathematics that deals with the relationship between the sides and angles of triangles, as well as relations and functions. A function, f, is an assignment of exactly one element of set B to each element of set A. To be able to identify the range and domain of a function. De nition 1 A function whose domain is a sample space and whose range is some set of real numbers is called a random variable. In fact, I have been taking the lack of standards (or logical consistency) in the terminology as justification to use exactly this terminology when I teach my students about functions. Example: arccos(p rs) can only have 0 rs 1. [2 marks] 6b. y = x - 2x - 6 x y O x y O x y O Find the vertex, the equation of the axis of symmetry, and the y-intercept of the graph of each function. radical functions using tables of values or technology, or by transforming the base radical function, y= √ __ x. x y 4) Determine if the following relation is a function then state the domain and range. • Other parts might have been omitted, because the assessed content was taken from an excluded unit. The range of the absolute value function is the set of non-negative numbers. A straight line and its slope. parametric density function. Definition. Snyder 2014 FUNCTION NOTATION (DAY 3) Function Notation: For every x-value in the domain that you _____ into an equation there is a ____value in the range that is the OUTPUT. Created Date: 10/27/2011 2:20:07 PM. Inverse Functions Thinking of a function as a process like we did in Section. [Q] True or False. Domain and Range of Inverse Trigonometric Functions As we can see from the graph of the sine function, many different angles θ \theta θ map to the same value of sin ( θ ) \sin(\theta) sin ( θ ) :. 15 Questions Show answers. Example 1: A continuous graph with two endpoints. Write the Range | Function Rule - Level 1. In other words, it is the set of y-values that you get when you plug all of the possible x-values into the function. Definition: A function f is said to be one-to-one, or injective, if and only if f(x) = f(y) implies x = y for all x, y in the domain of f. By Theorem5. The sampling interval is denoted as T s and its reciprocal, the sampling-. Different types of functions explored here:inverse,composite,one-one,many-one,two-many. 2: Domain and Range 3b 1 The range of the function f(x) x 3 5 is 2 If f(x) x2 2, which interval describes the range of this function? 3 The range of the function f(x) x2 2x 8 is all real numbers. Deinition 1 A relation is a 78 Further Topics in Functions. y = 2-x2 + 3 3. f(x) = (2x+ 14; if x 3 16 2x; if x > 3 (a)Domain of f (b)Range of f (c. In this lesson you will learn how to determine the domain and range of a parabola by looking at the graph. if moved 4 units up b. 13) 15) If f (x) =. Limits and Continuous Functions21 1. Model Effective Teaching Practices from Principles to Actions 4. If you want to know how to find the range of a function, just follow these steps. Find (4 Marks). 2: Domain and Range 3b Name: _____ www. Determine whether a function is onto or not 5. The log function loga(x) is de ned to be the inverse of the exponential function ax. Include a description of the kinds of numbers in the domain as well as any limitations on their values. WARNING 2: Sometimes, the limit value lim x a fx() does not equal the function value fa(). Vocabulary quadratic function parabola vertex minimum maximum Why learn this? The height of a soccer ball after it is kicked into the air can be described by a quadratic function. What is the domain of this function? 1) 4 ≤x≤13 2) 4 ≤y≤13 3) x≥0. Directions: Identify the domain and. Free PDF Download of JEE Main Sets Relations and Functions Important Questions of key topics. Functions 1. Ask students the meaning of f (5) (or another value) in the context of the problem. • For each x in the domain of f, x is a member of the input of the function f, f. A function is nothing but a rule which is applied to the values inputted. We will deal with real-valued functions of real variables--that is, the variables and functions will only have values in the set of real numbers. –2 –1 0 1 2 5 xy 5. 1x + 1 Y 6 4 (3. From Ramanujan to calculus co-creator Gottfried Leibniz, many of the world's best and brightest mathematical minds have belonged to autodidacts. Questions for Critical Thinking can be used in the classroom to develop all levels of thinking within the cognitive domain. On the back, the student is g. Thanks to all of you who support me on Patreon. Understand that a function from one set (called the domain) to another set (called the range) assigns to each element of the domain exactly one element of the range. Define and use Piecewise functions in context 2. By Theorem5. 1 - Interpreting Graphs. Sine functions and cosine functions have a domain of all real numbers and a range of -1 ≤y≥ 1. It also guarantees that the graph has the y-axis as a vertical asymptote, and that the domain of loga(x) is(0;+1), the sameas the rangeof ax. Reconciling this with our definition of a relation, we see that 1. The range is all of the real numbers greater than (or equal to) zero, since if y = x 2, y cannot be negative. For more information on finding the domain of a function, read the tutorial on Defintion of Functions. function notation. Although f cannot be evaluated at because substituting for results in the undefined quantity 0x 0, can be calcu-lated at any number that is very x closeto 4. Find (4 Marks). See full list on analyzemath. Here, F is a function of the acceleration, a. A function is bijective if for every y in the codomain there is exactly one x in the domain. Functions • Onto Function • A function is onto if each element in the co-domain is an image of some pre-image • A function f: A→B is subjective (onto) if the image of f equals its range. Domain and Range of Inverse Trigonometric Functions As we can see from the graph of the sine function, many different angles θ \theta θ map to the same value of sin ( θ ) \sin(\theta) sin ( θ ) :. Remember that an arrow diagram represents a function if exactly one arrow must leave each element of the domain. State the domain and range of the inverse. After you finish the domain and range activity, begin your inverse functions exploration. For questions 5 – 7, the blanks provided are for the following information: Identify the domain and range for each. The domain of f consists of all x values at which the function is defined, and the range consists of all possible values f can have. Its domain is Z, its codomain is Z as well, but its range is f0;1;4;9;16;:::g, that is the set of squares in Z. It is the point where the graph intersects its axis of symmetry. After Learning Domain we will learn How to find Range of Function and we are going to learn various methods for finding Range. Why are linear functions useful in real-world settings? Why would you use multiple representations of linear equations and inequalities? How are systems of linear equations and inequalities useful in interpreting real world situations? Why is it important to consider slope, domain, and range in problem situations?. ? 2) Questions #2 and #3 have strictly numerical answers. Range = {–10, –8, –3, 4} d. Properties of Logarithms and Exponents* 13. Here, F is a function of the acceleration, a. Functions have a domain and range and can be found in two ways. Give the domain and range of fand the inverse function f 1. To see that, we observe that the natural domain of this function is [1,+∞) since we request that the expression from which we extract the square root is non. a) Express the area function for the three pens together, in terms of x, where x is the length of each pen. The inverse of a trigonometric function f may be indicated using the inverse function notation f -1 or with the prefix “arc” (e. Use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context. range: [—2, 2]; not a function 19. Domain And Range - Displaying top 8 worksheets found for this concept. To be able to identify the range and domain of a function. domain and range of graphs practice worksheet ANSWERS. Quadratic Functions-Worksheet Find the vertex and “a” and then use to sketch the graph of each function. Determine The Domain And The Range Of Each Of The Functions Graphed In Exercises 1-6. Every function is also a relation 2. Range and domain of composite functions. The following is an example of finding the composition of three functions. A quadratic function has selected values shown in the table below. R - {0} Another Way to Find Range of Rational Functions. Domain and Range Examples The domain of a function f is the set of all values x for which f(x) is defined. If expr includes a function, it can be either built-in or user-defined, but not another domain aggregate or SQL aggregate function. Vertical Asymptotes Horizontal Asymptote. f(1;5);(2;5)g 4. P 0 P 1 y 1-y 0 x 1-x 0 x 0 x 1 y 0 y 1 n 1 m Figure 4. The range of a function f(x) is the set of all output values (y-values) for the function. Graph the function. De nition 1 A function whose domain is a sample space and whose range is some set of real numbers is called a random variable. What is the range of F? 3. Definition Domain- Set of values D of which a function is defined. F or some rational functions, it is bit difficult to find inverse function. Find the pdf of Z. , in this section we seek another function. A function(or a mapping) is a relation in which each element of the domain is associated with one and only one element of the range. The domain of any and all polynomial(s) includes "all real numbers. Let's return to the subject of domains and ranges. Example 2: Use the mapping diagrams to. Domain and codomain of a function f is a set of all real numbers x for which f(x) is a real number. 380 Further Topics in Functions f. • Diagrams are NOT accurately drawn, unless otherwise indicated. Types of Circulation Color and Label. –2 –1 0 1 2 5 xy 5. This time we will tackle how to find the domain and range of more interesting functions, namely, radical functions and rational functions. domain of log(x) (x^2+1)/(x^2-1) domain; find the domain of 1/(e^(1/x)-1) function domain: square root of cos(x). Definition: A fractional function which in its denominator and numerator are polynomial is called rational function. Understand that a function from one set (called the domain) to another set (called the range) assigns to each element of the domain exactly one element of the range. each element of the domain exactly one element of the range. Its graph is a vertical one which opens up (we know that because a = 2 is positive). The graph of a function f. {The relation described by the set of points ( ) ( ) ( ( )}is NOT a function. This question requires the examinee to analyze rational, radical, absolute value, and piece-wise defined functions in terms of domain, range, and asymptotes. (d) The intervals of decrease. Plug in the values of x in the function rule to determine the range. Jacquie Hargreaves Janet Hunter. x y--7-5-2-4 1-3-2-1-0-x y O Graph the ordered pairs in the table and connect them with a smooth curve. Worked examples and illustrations. Therefore, we need only consider points that are inside both the $$δ$$ disk and the domain of the function. Plug in the values of x in the function rule to determine the range. Sample Graph – A rational function, , can be graphed by following a series of steps. In this case, the function encompasses all of the positive y-values if the parabola goes up, or all of the negative y-values if the parabola goes. The range of a function f consists of all values f(x)it assumes when x ranges over its domain. Define and use Piecewise functions in context 2. Another way to identify the domain and range of functions is by using graphs. The IIEF is a 15-question survey about the patients’ sexual experiences over the preceding 4 weeks. Vocabulary quadratic function parabola vertex minimum maximum Why learn this? The height of a soccer ball after it is kicked into the air can be described by a quadratic function. domain The range is −2, 0, 2, and 4. Since our answer to that question is yes, that means by definition, y is a function of x. Graph piecewise-defined functions. Functions are rules that take inputs from one set and produce outputs in another. In which function is the range equal to the domain? A. For exercises 7-12, use each table to determine whether the relation is a function. The range of the function is the set of all values that the function can take, in other words all of the possible values of y when y = f(x). Highly important questions on Domain & Range ( series : Domain Range #1) - Duration: 4:25. College Algebra Questions With Answers sample 5 : Domain and Range of Functions. Keep in mind. probability dual function. We can use set-builder notation to express the domain or range of a function. D The number of sponsors. 1 Understand that a function from one set (the input, called the domain) to another set (the output, called the range) assigns to each element of the domain exactly one element of the range, i. • The arrow on the right side indicates that the graph will continue without bound. 6(A) determine the domain and range of quadratic functions and represent the domain and range using inequalities A. The range (the y-values) is all real numbers greater than or equal to -3, which is the minimum. To see that, we observe that the natural domain of this function is [1,+∞) since we request that the expression from which we extract the square root is non. Functions 1. Type of blood Vessels EdPuzzle •Complete one before getting the next papers. You can think about finding the range by imagining horizontal lines and seeing at what y-values they do (and do not) intersect the graph. #1 - 11 Pg. HOLT RINEHART AND WINSTON Textbook online mode mean median range ; management 8th ed sample quiz questions pdf ; Functions, Statistics, and Trigonometry. State whether the matches form a function. Informal de nition of limits21 2. domain of the function), the values that these inputs are mapped to (called the range of the function). In pairs, determine the domain and range of the graphs and functions. D f −1 (x)= x 2 +4; no it is not a function ____ 29 Find the inverse of the function: f(x)= (x−2) 2 +3. This topic covers: - Evaluating functions - Domain & range of functions - Graphical features of functions - Average rate of change of functions - Function combination and composition - Function transformations (shift, reflect, stretch) - Piecewise functions - Inverse functions - Two-variable functions. domain and range of a linear function in mathematical problems; determine reasonable domain and range values for real‐world situations, both continuous and discrete; and represent domain and range using inequalities. You know the basic function is the sqrt(x) and you know the domain and range of the sqrt(x) are both [0,+infinity). Sometimes, a function is a set of points. WARNING 2: Sometimes, the limit value lim x a fx() does not equal the function value fa(). Determine the inverse function of f, which we write as f 1. 82 #9 (constant and intervals of increase. Nonlinear Functions I CAN… Determine if a relationship is linear or nonlinear from a table, graph, equation, or verbal description. GRAPH OF A FUNCTION Take the function The domain is all real values of x. List the interval(s) on which fis decreasing. y = F(x) + 5 8. 42 10 5 5 10 x y Find the domain and range of each function. State the domain and range for each graph and then tell if the graph is a function (write yes or no). If you want to know how to find the range of a function, just follow these steps. Question 4 Find the domain and range of the following function f (x) = p-x 2-3 x + 10. Function Notation Practice Worksheet. The graph of a function f. The range of the inverse of a function is identical with the domain of the original function. and the range is $(-\infty,0)$ and $(0,\infty)$ notice unbounded below and open notation, the parenthesis instead of brackets, and the open and unbounded above, again the. Ifx is positive or 0, then the absolute value of x is x itself. In function notation, questions are often asked like this: 3 2 1 f (x) x Find f(4): This is a fancy way of saying plug 4 in. The domain of f(x) = x 2 is all real numbers and the range is all nonnegative real numbers. Domain and Range Practice Quiz. Domain: 1 ≤ x ≤ 6 Range: −4 ≤ y ≤ 5 B. The line is the graph of f (x) = mx + n. In all questions of this form, you have to first: identify the domain and range, and second: write it in set-builder notation. Given the graph of a function, determine its domain or range. The dependent variable is F and the independent variable is a. f is not a 1-1 function If f* has domain (- , 0) f is one to one. 2: Domain and Range 3b Name: _____ www. Let fbe a function from Xto Y, X;Ytwo sets, and consider the subset SˆX. com Example 1 consider the linear graphs below, they have same equation f(x) =x Graph (a) ( -4) (4) Graph (b) For graph(a) For graph(b) Domain: x ε all real numbers Domain: -4 ≤ x ≤ 4 Range : y ε all real numbers Range : -4 ≤ x ≤ 4 REMEMBER -The domain and range states to what. The domain of this function is 1) all positive real numbers 2) all positive integers 3) x≥0 4)x≥−1 2 What is the domain of the function shown below? 1)−1≤x≤6 2)−1≤y≤6. Not every relation is a. Given: f x x( ) 9 2 , write a function 𝒈, a. Domain and Range of General Functions The domain of a function is the list of all possible inputs (x-values) to the function. Question: Write a linear function C(x) giving the total cost of producing x T-shirts. The sampling interval is denoted as T s and its reciprocal, the sampling-. Define and use Piecewise functions in context 2. consists of two real number lines that intersect at a right angle. is a function such that a. Find the ordered pairs for the funciton F G Let f(x) = x 2;g(x) = x2 + x, both with domain all real numbers. This is the contrapositive of the definition. The Quadratic Function 1. Improve your math knowledge with free questions in "Domain and range of relations" and thousands of other math skills. y 2= 2x - 8x + 6 5. Analyze each graph, write the minimum and maximum points for both domain and In this set of pdf worksheets, the function rule is expressed as a linear function and the domain is also provided in each problem. Express answers in interval notation. To denote different functions, you may often need to use g(x) or h(x). Ex 1) Determine the domain and range of f(x) = 3 (x 2)2 + 5 Note: When quadratic and linear functions model realworld situations, the domain and range are often restricted to values that _____ _____ for the situation. The range is all real numbers less than or equal to the maximum value, or {f(x) | f (x ` y = í3x2 + 6 x + 3 62/87,21 a. The range consists of the values of ƒ(x) where x is in the domain of ƒ. (See Exercise 51. radical functions using tables of values or technology, or by transforming the base radical function, y= √ __ x. The domain and range in a cubic graph is always real values. (d) The intervals of decrease. Copy and complete the input-output table for the function. Which graph illustrates a quadratic relation whose domain is all. Algebraically: There is no set way to find the range algebraically. Domain: Range: Transformations of g(x) = b x (c > 0): For Problems 3 – 14, graph each exponential function. y 2= 2x - 8x + 6 5. Domain: positive integers; range: positive integers; no, it is not a function. Range and domain of $$f$$. Answer the questions below to identify the range for various real-world functions. Find Domain Range and Function. Express answers in interval notation. A function is a unique mapping from the domain (the inputs) to the range (the outputs). In this set of pdf worksheets, the function rule is expressed as a linear function and the domain is also provided in each problem. 6(A) determine the domain and range of quadratic functions and represent the domain and range using inequalities A. linear function 3 2 1 EXAMPLE 4 graph dependent variable. Inverse functions and Implicit functions10 5. More examples on domain, co-domain and range of function: 1. From Ramanujan to calculus co-creator Gottfried Leibniz, many of the world's best and brightest mathematical minds have belonged to autodidacts. Its vertex is (-2, -3). Question 1. 2 Notes – Linear vs. If so, you have a function! Watch this tutorial to see how you can determine if a relation is a function. Range = {–10, –8, –3, 4} d. Questions for Critical Thinking can be used in the classroom to develop all levels of thinking within the cognitive domain. The IIEF is a 15-question survey about the patients’ sexual experiences over the preceding 4 weeks. Mathematics. A f: A B B A f: A B B. The domain is the set of all possible input values. The gamma distribution is also related to the normal distribution as will be discussed later. We substitute (“plug in”) x =1 and evaluate f ()1. The domain of the function f(x) is the set of all real numbers because f(x) is a polynomial. If there is a requirement that a y-value produced by a function. Example 4: Given three functions , , and , Find the domain and range of each of the three functions. Find the range of. State the domain and range for the elements matched in the diagram below. (The domain of. Explain why this graph represents a function. In pairs, determine the domain and range of the graphs and functions. NOTES : Functions – Domain & Range How to read domain and range from a graph: Domain: _____ Range: _____ In each of the following graphs, a) tell whether or not each graph is a function and, b) identify the domain and range of each relation. The domain of a function is the set of inputs allowed for the function, i. 5 Convexity = convexity along all lines Theorem 1. Open Domain Question Answering (QA) is evolving from complex pipelined systems to end-to-end deep neural networks. When we are solving real problems, we use meaningful letters like. all functions of this form. The domain of any and all polynomial(s) includes "all real numbers. Assignment: The Range of a Function. Also, students will identify the domain and range of a given relation/function. It can be a table name or a query name for a query that does not require a parameter. monotonically decreasing function. A function is said to be an injection if it is one-to-one. Make sure you look for minimum and maximum values of y. 2 Notes – Linear vs. Monitoring Progress Help in English and Spanish at BigIdeasMath. I wanted to know if I was doing the domain part right and needed help with the range. Add associations necessary to record the relationships that must be retained 4. † identified the domain and range of functions. The graph looks like the following: This equation is a function because you can draw a vertical line through any value of x in the domain and it will cross the graph of the equation only once. What are the domain and range of this function? Graph these ordered xy-211-11 0 -5 1 -7 2 -5 31 411 O x y 2 6 4 2 2 4 4 10 y 2x2 4x 5 range domain pairs and connect them with a smooth curve. This technique will be handy later, so remember it. A function is usually denoted by a lower case letter (e. Identify candidate conceptual classes 2. x y 2) Determine if the following relation is a function then state the domain and range. See how to find the domain and range of a function implied by a word problem with this free video math lesson. On the other. Include a description of the kinds of numbers in the domain as well as any limitations on their values. These high quality math worksheets are delivered in a PDF format and includes the answer keys. Essentially, it. • The unit step function u(t–τ) makes the integrand zero for τ > t, so the upper limit is t. Have your say about what you just read! Leave me a comment in the box below. The range is the set of all possible output values. 5 Functions as Arrow Diagrams When the domain and codomain of a function are nite sets then one can represent the function by an arrow diagram. same as functions? Not quite… Recall the following definition from past Lecture. Domain and Range of Inverse Trigonometric Functions As we can see from the graph of the sine function, many different angles θ \theta θ map to the same value of sin ( θ ) \sin(\theta) sin ( θ ) :. For example, the set given by, {x | x ≠ 0}, is in set-builder notation. If the random variable is denoted by Xand has the sample space = fo 1;o 2;:::;o ngas domain, then we write X(o k) for the value of Xat element o k. Determine whether the relation is a function. Be careful, the multivariable erms may limit the domain. Sample Questions for Relations and Functions 4. State the domain and the range. Determine whether a relationship is a function or not 2. Domain and Range of a Function. fx jx > 1g 13. Domain And Range - Displaying top 8 worksheets found for this concept. Whitman College. The range is all of the y values that lie on the function in the graph fro the lowest y to the highest y. Review function notation. To find the domain of a function, just plug the x-values into the quadratic formula to get the y-output. Identify the domain and range of a function from tables, equations, and 15. 2 Domain and Range Graphically. These high quality math worksheets are delivered in a PDF format and includes the answer keys. A function is a unique mapping from the domain (the inputs) to the range (the outputs). In order to find the domain of a function, if it isn’t stated to begin with, we need to look at the function definition to determine what values are not allowed. the equation as well domain and range. " The range is the set of possible outputs (y-values) and is (-3, infinity). Explain why. For example, x 2 ≥ 0, and √ x ≥ 0. odd and even functions, period and amplitude of a function, composite function, inverse function, trigonometric functions, sketching graphs. Evaluating functions using equations and graphs. , f is not defined at 2. Range (y) = Domain (y-1) Therefore, the range of y is. 2 Domain & Range (Graphs) 2 Write your questions and thoughts here! Increasing/Decreasing Functions: 1. For more information on finding the domain of a function, read the tutorial on Defintion of Functions. The simplest function of all, sometimes called the identity function, is the one that assigns as value the argument itself. Some Common Restrictions for Certain Functions In order to determine the domain and range for functions, particular attention has to be paid on functions involving square root Question 6 PDF. The line is the graph of f (x) = mx + n. To find the domain, I need to To find the range is a bit trickier than finding the domain. The ranges of (f g)(x) in (g) and (h) are hard to nd because the functions produce graphs you may not have covered previously. Th e set of all inputs for a function is known as the domain of the function. The relations in Example 3 are a special type of relation called a function. Alternative: A function is one-to-one if and only if f(x) f(y), whenever x y. Explain why. • Answer all questions. Opening – The teacher will define a piecewise function, and go over Key Idea p. Worked examples and illustrations. Express answers in interval notation. We will prove this later on using the moment generating function. If you're in the mood for a scary movie, you may want to check. 51: Domain and Range 1 Name: _____ www. • Diagrams are NOT accurately drawn, unless otherwise indicated. That is, it is the set of all y values for which there is an x value such that y = f(x): For many functions, the domain is easy to. For example, the set given by, {x | x ≠ 0}, is in set-builder notation. MathScore EduFighter is one of the best math games on the Internet today. Domain and Range Worksheet. (See Part C. Hence, the domain can be read on the horizontal axis, the range on the vertical. , f is not defined at 2. This means the identity function I 2 applies both to the domain of hand the domain of g. This indicates that each element in the domain corresponds to exactly one element in the range. Domain: Range: Transformations of g(x) = b x (c > 0): For Problems 3 – 14, graph each exponential function. When we are solving real problems, we use meaningful letters like. Remember that we always read the values o the graph from left to right, from bottom to top. 5we described a function as a process and de ned the notation necessary to work with functions algebraically. Range of a Real Function. Find (4 Marks). See , , and. Algebraic Test )– (Substitute −𝑥 in for 𝑥 everywhere in the function and analyze the results (of )𝑓−𝑥, )by comparing it to the original function 𝑓(𝑥. let me do this one for you and then give you another one the domain for this function is all values for x and all values of except the point (0,0), i think you know why. The graph in this question is a function based on the vertical line test. After Learning Domain we will learn How to find Range of Function and we are going to learn various methods for finding Range. Worked examples and illustrations. ? 2) Questions #2 and #3 have strictly numerical answers. f(1;4);(3;6);(4;2)g 3. Give domain and range of the relation. Questions for Critical Thinking can be used in the classroom to develop all levels of thinking within the cognitive domain. Function - Find Domain and Range(Questions) II फलन के डोमेन व रेंज ज्ञात करना(Lecture 5). Let be a function whose domain is a set X. Relation and Functions Worksheet (pdf) with Key. functional relationship? A The number of cookies ordered. 2: Domain and Range 1a 1 The function f(x) is graphed below. What is the range of F? 3. Domain and range could be limited to some discrete values, or they might incorporate all numbers everywhere, to infinity and beyond. State the domain and range for each graph and then tell if the graph is a function (write yes or no). is a function such that. Write the Range | Function Rule - Level 1. See full list on analyzemath. #1 - 5 Hw: pg 126 in textbook. Function? Function? Function? Explain: Explain: Explain:. Graphing Piecewise-Defined Functions. Domain definition, a field of action, thought, influence, etc. Range a function Domain —3 Range O Find the domain and Range of the graphs. You can use the basic cubic function, f(x) = x3, as the parent function for a family of cubic functions related through transformations of the graph of f(x) = x3. Domain and Range of Inverse Trigonometric Functions As we can see from the graph of the sine function, many different angles θ \theta θ map to the same value of sin ( θ ) \sin(\theta) sin ( θ ) :. (a) An a ne function (b) A quadratic function (c) The 1-norm Figure 2: Examples of multivariate convex functions 1. 1) Now that you are starting to learn about the domain and the range of a function, how is the answer for question #1 related to the domain. Domain and Range of Trigonometric Functions. Although the Fourier transform is a complicated mathematical function, it isn’t a complicated concept to understand and relate to your measured signals. Graph Radical Functions Using Tables of Values Use a table of values to sketch the graph of each function. In order to find the range and domain, we can start by studying the graph. Exam Questions – Domain and range Range of a function (example) : ExamSolutions Maths Revision - OCR C3 June 2013 Q7(i. † identified independent and dependent variables. In general, we determine the domain of each function by looking for those values of the independent variable Latest Functions and Graphs forum posts: Got questions about this chapter?. C The number of student participants. This question requires the examinee to analyze rational, radical, absolute value, and piece-wise defined functions in terms of domain, range, and asymptotes. The line is the graph of f (x) = mx + n. State whether the matches form a function. y = logx D. Compare the two relations on the below. Although you'll learn more advanced ways of finding the range later on, for now, the simplest way to find the range of this function is to apply the function to each element of the domain, and track your results. The inverse trigonometric functions: arcsin and arccos The arcsine function is the solution to the equation: z = sinw = eiw − e−iw 2i. If you're seeing this message, it means we're having trouble loading external resources on our website. x y 4) Determine if the following relation is a function then state the domain and range. Its Range is a sub-set of its Codomain. Determine whether a relation is a function. Finding Domains and Ranges of the Toolkit Functions. The structure of a function determines its domain and range. Informal de nition of limits21 2. yes a function is a mapping. We can use set-builder notation to express the domain or range of a function. Let T be the set of all triangles in a plane with R a relation in T given by R = {(T 1,T 2):T 1 ≅ T 2) show that R is an equivalence relation. Recall the domain is the set of values of the independent variable, the range the set of values of the dependent variable. Basically Range is subset of co- domain. Its graph is a vertical one which opens up (we know that because a = 2 is positive). (b)Give the domain and range of fand of f 1. How to Sketch the Graph of a Function. This is a double-sided practice page over Functions, Domain & Range and Function Notation. I can determine the appropriate domain and range of a quadratic equation or event. Inverse Functions If f is a one-to-one function with domain A and range B, we can de ne an inverse function f 1 (with domain B ) by the rule f 1(y) = x if and only if f(x) = y: This is a sound de nition of a function, precisely because each value of y in the domain of f 1 has exactly one x in A associated to it by the rule y = f(x). Find the domain and range of each relation, and determine whether it is a function. Type of blood Vessels EdPuzzle •Complete one before getting the next papers. y = logx D. Exercises13 Chapter 2. Lesson 5 – Introduction to Exponential Functions Mini-Lesson Page 179 Graph of a generic Exponential Growth Function f(x) = abx, b > 1 • Domain: All Real Numbers • Range: f(x) > 0 • Horizontal Intercept: None • Vertical Intercept: (0, a) • Horizontal Asymptote: y = 0 • Left to right behavior of the function: INCREASING. Use the graph to find: (a) Its domain and range (b) The x- and y- intercepts (c) The intervals of increase. The graph of a function f. A real life example of this is using a simple calculator to add two numbers together. Domain and Range of Inverse Trigonometric Functions As we can see from the graph of the sine function, many different angles θ \theta θ map to the same value of sin ( θ ) \sin(\theta) sin ( θ ) :. , the set of values that can be fed into the function to give a valid output. EXAMPLE 1 Finding Domain and Range from a Graph. Another way to identify the domain and range of functions is by using graphs. A very important fact that we have to know about the domain of a logarithm to any base is, "A logarithmic function is defined only for positive values of argument". The line is the graph of f (x) = mx + n. This article explains STEP-BY-STEP, how to find the Domain and range of a parabola with any orientation. ∗In our conventions, the real inverse tangent function, Arctan x, is a continuous single-valued function that varies smoothly from − 1 2π to +2π as x varies from −∞ to +∞. Remember that the independent variable must appear in the exponent for the function to be exponential. Here, F is a function of the acceleration, a. The vertex of a parabola is an extreme point of a quadratic function and in general, it is known as maximum or minimum of a parabola. / Exam Questions - Domain and range. the domain values in one oval are joined to the range values in the other oval using arrows. Some questions in Passport to Advanced Math will ask you to build a quadratic or exponential function or an equation that describes a context or to interpret the function, the graph of the function, or the solution to the equation in terms of the context. √ A) B) C) Power Function D) Powers Property of Equality What is the domain and range of the following chart: A. x y 4) Determine if the following relation is a function then state the domain and range. 1) Now that you are starting to learn about the domain and the range of a function, how is the answer for question #1 related to the domain. Individuals from a variety of disciplines, including mathematics, mathematics education, and cognitive psy- chology, are conducting this work. Identify candidate conceptual classes 2. Step 2: A relation is a function if each element in the domain is paired with one and only one element in the range. You might want to know what exactly is going on at this point. Pre-AP Algebra 2 Lesson 2 – Using inverse functions to find the range; Graphing the logarithmic function Objectives: Students will be able to find the range of f by finding the domain of f-1. is a function such that. Plug in the values of x in the function rule to determine the range. In addition, you’re going to see how it’s used to represent the domain and range of a function in a simplified and beautiful way. I can apply quadratic functions to model real-life situations, including quadratic. To see that, we observe that the natural domain of this function is [1,+∞) since we request that the expression from which we extract the square root is non. For the codomain, try and solve the equation $y=\dfrac{x^2}{1+x^2}$ This becomes. pdf: File Size: 375 kb: File Type: pdf. The range of a function f is the set of all values that f(x) takes on as x runs through the domain of f. domain of the function), the values that these inputs are mapped to (called the range of the function). What are the domain and range of each relation? 1. For questions 5 – 7, the blanks provided are for the following information: Identify the domain and range for each. 2) What is the domain of a function? ) For each question, decide if it is a function. Consider the relation attached to the function f(x) = x3: namely R= f(x;x3) jx2Rg: This relation is described by the equation y= x3; certainly it is a function, and its domain is R. A good example of a functional relation can be seen in the linear equation y = x + 1, graphed in Figure 3. This technique will be handy later, so remember it. Not every relation is a. {The relation described by the set of points ( ) ( ) ( ( )}is NOT a function. Just hearing the word is enough to send some students running for the hills. A relation is a function if there is exactly one arrow leading from each value in the domain. the equation as well domain and range. Find Domain Range and Function. The following table summarizes the domains and ranges of the inverse trig functions. Two Reasons to Restrict the Domain Consider the initial value problem dy. Values that don't make sense are excluded. range of f y = f (x) (x, f (x)) x domain of f Figure 3. ){ (−5,3),−2,1,(1,−1),(4,−3)} 2. 80 #6 (domain and range), also p. foo = function(){ console. See , , and. When a function, such as the line above, is both injective and surjective (when it is one-to-one and onto) it is said to be bijective. Revision of Solving Quadratic Equations In this lesson we investigate the standard and the turning point forms of the quadratic formula. We then look at the properties of these formulae and the graph by doing examples. *Each test may contain a variety of functions, including linear, polynomial (degree ≤ 5), rational, absolute value, power, exponential, logarithmic, and piecewise-defined. Inequalities Involving Rational Functions* 11. Worked examples and illustrations. For example, x 2 ≥ 0, and √ x ≥ 0. If expr includes a function, it can be either built-in or user-defined, but not another domain aggregate or SQL aggregate function. We learn the domain of a function is the set of possible x-values and the range is the resulting set of y-values. In this case, the domain is all the x− values and the range is all the y− values. Give the domain and range of each relation. f(1;5);(2;5)g 4. For the function {(0,1), (1,-3), (2,-4), (-4,1)}, write the domain and range. If f is a function and x is an element of its domain, then f(x) denotes the output of f corresponding to the input x. Let be a function whose domain is a set X. Section 3: Introduction to Functions Section 3 – Topic 1 Input and Output Values A function is a relationship between input and output. Domains can be found algebraically; ranges are often found algebraically and graphically. Step 3: From the mapping diagram, it can be observed that the given relation is not a function as '3' in the domain is paired with two elements - 1 and - 2 in the range and '6' is paired with - 1 and - 2. Sometimes, a function is a set of points. the equation as well domain and range. x y 4 2 0 6 0 2 4 6 STUDY TIP A relation. Graphing Piecewise-Defined Functions. In the example above, the domain of $$f\left( x \right)$$ is set A. Steps to create a Domain Model 1. Ø Range is the set of values of " calculated from the domain for the _____ of the function. To summarise, for domain look along x-axis, for range look along y-axis. This specifies what the output of the function is when is the input. * What is a function? -A Relation relates two sets of data Examples: Pounds of bananas and cost Ex2: Gallons of gas and miles traveled Ex3: Grade on test and time spent studying (underlined variables are the inputs). 12 Number and algebraic methods. relationship can be expressed by the function f(s) = 3x + 30, where s is the. Direct and Inverse Variation Solving eq. The range is all real values of y which is the same as all real values of f(x). The domain of f(x) = x 2 is all real numbers and the range is all nonnegative real numbers. If it is a function, respond "yes" and state the domain and range. For a function f defined by an expression with variable x, the implied domain of f is the set of all real numbers variable x can take such that the expression defining the function is real. Define a suitable restriction for f, f ∗, such that f ∗−1 exists. Thus, when speaking of continuity, it is very important to be cognizant of what the domain of the function is. The range is all of the real numbers greater than (or equal to) zero, since if y = x 2, y cannot be negative. Functions: Simplifying Difference Quotients* 6. Ling 310, adapted from UMass Ling 409, Partee lecture notes March 1, 2006 p. You can think about finding the range by imagining horizontal lines and seeing at what y-values they do (and do not) intersect the graph. Mastering Domain and Range Of Function is a course which will teach How to find Domain of various types of Expressions and there combinations. domain and range of graphs practice worksheet ANSWERS. Function - Find Domain and Range(Questions) II फलन के डोमेन व रेंज ज्ञात करना(Lecture 5). The graph in the figure below suggests that the function has no absolute maximum value and has an absolute minimum of 0, which occurs at x = 0. Keep in mind. Given a function, its domain is the set of all values taken on by the independent variable {eq}x, {/eq} and its range is the set of all values taken on by the dependent variable {eq}y. It is the point where the graph intersects its axis of symmetry. Graph functions by plotting points. y = F(3x) Graph one cycle of each function. Give the domain and range. Question: What is the domain of an exponential function f(x) = kb x? What is the range? Describe the shape of the graph for b > 1, and for b < 1. By definition, some of the points of the $$δ$$ disk are inside the domain and some are outside. Find the domain and range of each of the following functions. The graph in this question is a function based on the vertical line test. Sometimes, a function is a set of points. Give the domain and range of each relation. The range is the set of possible output values, which are shown on the $y$-axis. #3 – 15, 17. What is the definition of function? Q. Celina has a part-time job that pays \$8 per hour. • Diagrams are NOT accurately drawn, unless otherwise indicated. In other words, there is a lack of awareness of teachers in presenting the topic of domain and range function. Find the domain and range of each relation, and determine whether it is a function. You can think about finding the range by imagining horizontal lines and seeing at what y-values they do (and do not) intersect the graph. If you know an element in the domain of any polynomial function, you can find the corresponding value in the range. If the function is graphed, the domain and range can be identified from the picture, or the domain and range can be found algebraically with some reasoning. Properties of Logarithms and Exponents* 13. Type of blood Vessels EdPuzzle •Complete one before getting the next papers. An example { tangent to a parabola16 3. Reconciling this with our definition of a relation, we see that 1. find the domain, codomain and range of a function; (Functions) define the different types of functions such as injective function (one-to-one function), surjective function (onto function), bijective function, give examples of each kind of function, and solve problems based on them; (Functions). Many root functions have a range of (-∞, 0] or [0, +∞) because the vertex of the sideways parabola is on the horizontal, x-axis. Provide a reason for your answer. College Algebra Questions With Answers sample 5 : Domain and Range of Functions. So now it’s time to look at functions graphically again, only this time we’ll do so with the notation de ned in Section1. Sampling The first thing we have to do, is to obtain signal values from the continuous signal at regular time-intervals. A language function refers to the purpose for which speech or writing is being used.
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# In the coordinate plane a slope of the line K is 4 times the y-interce
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In the coordinate plane a slope of the line K is 4 times the y-interce [#permalink]
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In the coordinate plane a slope of the line K is 4 times the y-intercept of the line K. What is the x-intercept of the line K?
A. -4
B. 4
C. -1/4
D. 1/4
E. 2
*An answer will be posted in 2 days.
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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Intern Joined: 17 Jun 2013 Posts: 6 Re: In the coordinate plane a slope of the line K is 4 times the y-interce [#permalink] ### Show Tags 07 Jun 2016, 21:50 MathRevolution wrote: In the coordinate plane a slope of the line K is 4 times the y-intercept of the line K. What is the x-intercept of the line K? A. -4 B. 4 C. -1/4 D. 1/4 E. 2 *An answer will be posted in 2 days. $$y=mx+h$$, but $$m = 4h$$. So, we have $$y = (4h).x + h$$. Let´s find the X-intercept of the line K: In this case, $$y = 0$$. Thus, $$0 = (4h).x + h$$. Solving this equation we will find $$x = -\frac{1}{4}$$ The correctt answer is letter (C). _________________ Prof Marcelo Roseira Curso FDX GMAT Prep Online Brazil Twitter: @cursofdx Instagram: @cursofdx Skype: fdx.anpad Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 5876 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: In the coordinate plane a slope of the line K is 4 times the y-interce [#permalink] ### Show Tags 09 Jun 2016, 20:29 As Y=4mx+m, from 0=4mx+m we get x=-1/4. Hence, the correct answer choice is C. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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Re: In the coordinate plane a slope of the line K is 4 times the y-interce [#permalink]
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09 Jun 2016, 21:24
MathRevolution wrote:
In the coordinate plane a slope of the line K is 4 times the y-intercept of the line K. What is the x-intercept of the line K?
A. -4
B. 4
C. -1/4
D. 1/4
E. 2
*An answer will be posted in 2 days.
Line equation
y = mx+c (m is the slope, c is the y intercept)
Given values, m= 4c
equation becomes
y = (4c)x+c
X intercept would be (Put y =0)
x = -(c/4c) = -(1/4)
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Re: In the coordinate plane a slope of the line K is 4 times the y-interce [#permalink]
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22 Nov 2017, 12:22
MathRevolution wrote:
In the coordinate plane a slope of the line K is 4 times the y-intercept of the line K. What is the x-intercept of the line K?
A. -4
B. 4
C. -1/4
D. 1/4
E. 2
We can use the slope-intercept form of line, recalling that m is the slope and b is the y-intercept:
y = mx + b
Since m = 4b, we have:
y = 4bx + b
To find the x-intercept, we set y equal to zero:
0 = 4bx + b
0 = b(4x + 1)
0 = 4x + 1
-1 = 4x
-1/4 = x
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Re: In the coordinate plane a slope of the line K is 4 times the y-interce &nbs [#permalink] 22 Nov 2017, 12:22
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# In a corporation, 50 percent of the male employees and 40
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In a corporation, 50 percent of the male employees and 40 percent of the female employees are at least 35 years old. If 42 percent of all the employees are at least 35 years old, what fraction of the employees in the corporation are females?
A. 3/5
B. 2/3
C. 3/4
D. 4/5
E. 5/6
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udaymathapati wrote:
In a corporation, 50 percent of the male employees and 40 percent of the female
employees are at least 35 years old. If 42 percent of all the employees are at least 35
years old, what fraction of the employees in the corporation are females?
A. 3/5
B. 2/3
C. 3/4
D. 4/5
E. 5/6
Given weighted average employees who are at least 35 years old: $$\frac{0.5m+0.4f}{m+f}=0.42$$. Question: $$\frac{f}{m+f}=?$$
$$\frac{0.5m+0.4f}{m+f}=0.42$$ --> $$50m+40f=42m+42f$$ --> $$4m=f$$ --> $$\frac{f}{m}=4$$ --> $$\frac{f}{m+f}=\frac{4}{5}$$.
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26 Aug 2010, 08:46
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I have a different approach that may be useful for you:
Let's say that “x” is the % of male employees in the corporation. So, (1-x) is the % of female employees in the same company.
Therefore:
$$(50%)(x) + (40%)(1-x) = 42%$$
Expressed in fractions:
$$(1/2)(x) + (2/5)(1-x) = 21/50$$
We solve the equation and we have:
$$x = 1/5$$
$$(1-x) = 4/5$$
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Hi Buneul -
I understood your explanation until the last step. How do you get from f/m= 4 to the 4/5?
Thanks!
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Atrain13gm wrote:
Hi Buneul -
I understood your explanation until the last step. How do you get from f/m= 4 to the 4/5?
Thanks!
$$4m=f$$ --> $$\frac{f}{m+f}=\frac{4m}{m+4m}=\frac{4m}{5m}=\frac{4}{5}$$.
Hope it's clear.
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eybrj2 wrote:
In a corporation, 50 percent of the male employees and 40 percent of the female employeesa re at least 35 years old. If 42 percent of all the employees are at least 35 years old, what fraction of the employees in the corporation are females?
a) 3/5
b) 2/3
c) 3/4
d) 4/5
e) 5/6
You can use the weighted averages formula for a 10 sec solution.
No of females/No of males = (50 - 42)/(42 - 40 ) = 4/1
No of females as a fraction of total employees = 4/(4+1) = 4/5
For details of this method, check: http://www.veritasprep.com/blog/2011/03 ... -averages/
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Re: In a corporation, 50 percent of the male employees and 40 [#permalink]
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02 Jun 2012, 18:06
your post on veritas blog is amazing Karishma
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Re: In a corporation, 50 percent of the male employees and 40 [#permalink]
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28 Sep 2014, 23:20
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Let Male employees = x
Let Female employees = y
Total Employees = x+y
We require to find $$\frac{y}{x+y}$$
Setting up the equation:
$$\frac{50x}{100} + \frac{40y}{100} = \frac{42}{100} (x+y)$$
$$\frac{y}{x} = 4$$
$$\frac{y}{x+y} = \frac{4}{1+4} = \frac{4}{5}$$
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Re: In a corporation, 50 percent of the male employees and 40 [#permalink]
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05 Oct 2015, 20:42
2
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x- total number of employees
f - all females
x-f - all males
we need f/x
50(x-f)+40f=42x
50x-50f+40f=42x
10f=8x
f/x=8/10=4/5
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Re: In a corporation, 50 percent of the male employees and 40 [#permalink]
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22 Aug 2016, 06:08
.5m+.40f=.42
m+f=1
solve these two equations,
f=4/5
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Re: In a corporation, 50 percent of the male employees and 40 [#permalink]
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06 Nov 2016, 19:47
I used the 2 way chart to help create the expression :
(50/100)x + (40/100)y = 42/100(x+y) .... where x is number of males, y is number of females and x+y is total.
Now when you simplify it, you will get 4x = y. Means for every 4 females there is 1 male... No of females is 80% or 4/5
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Re: In a corporation, 50 percent of the male employees and 40 [#permalink]
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16 Dec 2017, 08:32
Males + Females =100 - Equation 1
50 * Males + 40 * Females = 42 [ Males + Females ] - Equation 2
Solve for Females.
You will get 80 Answer ( Which is 80%)
therefore = 80/100=4/5
Re: In a corporation, 50 percent of the male employees and 40 [#permalink] 16 Dec 2017, 08:32
Display posts from previous: Sort by | 2018-01-24T04:04:06 | {
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https://math.stackexchange.com/questions/2211965/how-to-efficiently-read-a-predicate-logic-formula-best-practices | # How to efficiently read a predicate logic formula (best practices)
This no question about how to understand a predicate logic proposition in general, it's about fast understanding such a proposition.
E.g. as a simple example, the convergence definition for $(a_n)_{n\in \mathbb{N}}$ with $a_n\in M$ (for all $n\in \mathbb{N}$ ; $M$ ordered ; e.g. $M=\mathbb{R}$ ) is
$\exists a\in M \forall \epsilon \in M_{>0} \exists N\in \mathbb{N} \forall n\in \mathbb{N}\colon n\ge N \implies |a_n -a|<\varepsilon$
When you read something like that (unknown to you): What is your approach?
So far, mine is the following:
Part 1: The part, that does include the quantifier declations. (e.g. "$\exists a\in M \forall \epsilon \in M_{>0} \exists N\in \mathbb{N} \forall n\in \mathbb{N}$"
Part 2: The part, that does not include the quantifier declrations. (e.g. "$n\ge N \implies |a_n -a|<\varepsilon$")
a) Directly read the formula as it is:
1. Look very shortliy at part 1 to get a short impression of the quantifiers (mostly to see what variables are used).
3. Look closely to the part 1 again and read it carfully from right to left. While doing that, i look from time to time to part 2 again, to see how the read variable is used in context of part 2.
b) Transform the formula:
If the formula appears to be too "chaotic" for me, i transform it to an equivalent predicate logic propositions which composes of a declaration part (part 1) and a coresponding part (part 2). After that i do a).
My questions now are:
1. What is your approach/procedure to read a (new to you) formula efficiently?
2. Do you occasionally transform the formula in a better readable formula? And if yes: How do you transform (e.g. do you also separate the declaration part of the other part; do you write the discure universum conditions alltogether in the declaration part [e.g. short $\forall n \in \mathbb{N}\ge N$])?
• It seems to me that it's quite opinion-based and rather related to experience/being used to syntax rather than "procedures". – Boris E. Mar 31 '17 at 22:32
• @BorisEng I think the way of understanding a proof often results in a common way of understanding cognitive sequence. That's also why many people write prose when writing proofs (there is one train of thought). My question relates to what you do, if you just have a formula and no written prose. What heuristic leads, in your experience, often to that train of thought, which helps you understanding the formula? – SearchSpace Apr 11 '17 at 7:49
The best solution to the problem is not to create the problem. Write formulas in a readable way. You might have to rewrite it yourself.
1) Don't use the variable "a" to refer to 2 different things. Avoid using $n/N/\mathbb N$ to refer to different things. Sometimes a convention helps, something like upper case for sets (and maybe sequences), lower case for primitive values.
2) Don't try to shove everything into 1 formula. Give different names to different concepts.
$$\exists l \in M~ \forall \varepsilon \in M_{>0} ~\exists k \in \mathbb{N} ~\forall j \in \mathbb{N} ~\colon~ j \ge k \implies |a_j - l|<\varepsilon$$
Break it up:
Eventuality: $E(a, P) = \exists k \in \mathbb N ~ \forall j \in \mathbb N ~:~ j \ge k \implies P(a_j)$ : "Eventually every value in a sequence has property P"
Boundedness : $B(a, l) = \forall \varepsilon \in M_{>0} ~:~ E(a, a_j \mapsto |a_j - l| < \varepsilon)$ : "For every deviation $\varepsilon$, (eventually every value in a sequence has property) that it is within $\varepsilon$ of L".
Existance : $F(a) = \exists l ~:~ B(a, l)$ : "There exists a limit L such that (for every deviation $\varepsilon$, (eventually every value in a sequence has property) that it is within $\varepsilon$ of L)"
You probably will have to start with the first part of the formula that makes sense, and go from there. It could be inside out, or left to right, or right to left.
• This seems to be a good idea when writing proofs. Question to the example: $E(a,a_j \implies |a_j-l|<\varepsilon )$ here you mean $E(a,|a_j-l|<\varepsilon )$? This probably also reflects your procedure when reading proofs, roughly speaking, you go from the inside out? – SearchSpace Apr 11 '17 at 7:43
• @SearchSpace Oh I was just writing an anonymous function. Instead of $f(x) = x^2$ I wrote $f = (x \mapsto x^2)$. There's a lot of conventions. – DanielV Apr 11 '17 at 16:38 | 2019-08-21T23:06:33 | {
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http://math.stackexchange.com/questions/250768/is-mathbb-r2-a-field | # Is $\mathbb R^2$ a field?
I'm new to this very interesting world of mathematics, and I'm trying to learn some linear algebra from Khan academy.
In the world of vector spaces and fields, I keep coming across the definition of $\mathbb R^2$ as a vector space ontop of the field $\mathbb R$.
This makes me think, Why can't $\mathbb R^2$ be a field of its own? Would that make $\mathbb R^2$ a field and a vector space?
Thanks
-
A field has multiplication. How would you define multiplication on $\mathbb R^2$ so that it is a field? (There is a way to do so, but it isn't "obvious" until you realize that the resulting field is the complex numbers...) – Thomas Andrews Dec 4 '12 at 16:07
$\mathbb{R}^2$ can be a field but with multiplication defined as follows: $(a,b)(c,d) = (ac - bd, ad + bc)$. Indeed, this is one way of defining the complex numbers. – Rankeya Dec 4 '12 at 16:08
But, if you want to try to do this for $\mathbb{R}^n$, for $n \geq 3$ such that $\mathbb{R}$ is naturally embedded in $\mathbb{R}^n$ as a subfield, then it is not possible to do so, and this is a harder fact to prove. – Rankeya Dec 4 '12 at 16:11
ok this just answered my follow-up question. Is there any proof of that being true? – vondip Dec 4 '12 at 16:11
Vondip - Perhaps this is at a slight tangent, but a significant difference between R and C is that R is an ordered field and C is not. e.g. 5 is larger than 3, but which is "larger", 4 + 7i or 6 + 5i ? (Answer: well, defining how "large" or "the length" a complex number is is not as obvious as for the reals. In fact, there are many different ways of defining the length of a complex number). Just something to think about. – Adam Rubinson Dec 4 '12 at 16:37
If you define:
$$(a,b)+(x,y):=(a+x,b+y)$$
$$(a,b)\cdot (x,y):=(ax-by,ay+bx)$$
then the set $\,\Bbb R^2=\Bbb R\times\Bbb R\,$ turns into a field, and a rather well known and important one. Can you identify it?
-
Spoiler: take a look at one of the comments given. $\^\smile\^$ – FrenzY DT. Dec 4 '12 at 16:10
complex numbers indeed! fantastic how it all connects! Would that make R^2 a field and a vector space? – vondip Dec 4 '12 at 16:11
Yes, it would, because addition in $\mathbb{R}^2$, as @DonAntonio defines it, is component wise (the usual way). Remember, the vector space structure depends only on the underlying abelian group. – Rankeya Dec 4 '12 at 16:15
What do you mean "vector space"? Any field is a vector space over any subfield, so the field $\,\Bbb R^2\cong\Bbb C\,$ is a vector field ove an infinite number of subfields, say $\,\Bbb C\,,\,\Bbb R\,,\,\Bbb Q\,,\,\Bbb Q(i)\ldots\,$ , etc. – DonAntonio Dec 4 '12 at 16:15
So would that mean that I any vector space could be defined when F -being the field, as : F^n ? – vondip Dec 4 '12 at 16:17
It is important to understand that a set on its own has no algebraic structure. By defining operators on $\mathbb{R}^2$ you could turn it into (almost) anything you like.
The natural operators on $\mathbb{R}^2$, namely $(x, y) + (a, b) \mapsto (x+a, y+b)$ and $(x, y) \cdot (a, b) \mapsto (x\cdot a, y\cdot b)$ do not define a field as $(0, 1)$ has no multiplicative inverse. | 2013-05-24T01:24:45 | {
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http://math.stackexchange.com/questions/123449/how-many-squares-does-a-line-between-two-points-pass-through/123513 | # How many squares does a line between two points pass through?
Suppose I have a square, let's say the sides have length 1. I will then partition the square into $N^2$ sub-squares, where $N \in \mathbb{N}$ and the sub-squares are all the same size. Now, suppose we place two points $A$ and $B$ randomly within the large square a distance of $D$ apart from each other, and then draw the line segment $AB$. The question is, what is the expected number of small squares that $AB$ will pass through?
I have no idea how to solve this analytically, and was thinking of attempting to get a sense using simulation, but was curious what I could learn about it here. (This isn't a homework question; it's related to a problem I'm working on in economics research.)
-
So basically you have $N$ copies of the partitioned square (one for each $x$) and you pick the same points $A$ and $B$ in each copy and add up the total squares intersected in the copies? Also, is each partition "uniform"? i.e. the sub-squares are equal sized, like a chessboard grid etc? – Aryabhata Mar 22 '12 at 22:40
All of the sub-squares are the same size. But we only pick one point $A$ and $B$ in the big square, not in each sub-square. Maybe it's clearer if you think about the points first, and then do the partition. Thanks. – itzy Mar 22 '12 at 23:48
It is still not clear. What does "partition into $x^2$ sub-squares, where $x \in \{1,2,\dots, N\}$" exactly mean? Perhaps you meant to say partition into $N^2$ sub-squares? Please clarify. – Aryabhata Mar 23 '12 at 0:01
I mean partition it into some prespecified number of squares. Now that I think about it, $x=1$ doesn't make any sense -- that wouldn't be a partition. But we could partition it into 4 squares, or 9 squares, etc. So I meant, choose an integer greater than one, square it, and that's how many sub-squares there will be. I could have specified the number of partitions, but I want the answer to depend on $x$. – itzy Mar 23 '12 at 0:05
If you just say $N^2$ sub-squares, the answer will depend on $N$ and that would be good enough, won't it? – Aryabhata Mar 23 '12 at 0:07
The number of of the cells touched by the line is (following Byron's answer) $N (|X_1-X_2| + |Y_1-Y_2| +1)$ . Calling $X =|X_1 - X_2|$ and $Y = |Y_1 - X_2|$ and leaving apart the $N$ factor, this is approximately equivalent, for large $N$, to $d_M=X+Y$, the Manhattan distance.
Hence, I'll attack the following problem: We throw two points at random on the unit square; we want to compute $$E(d_M | d)$$
where $d=\sqrt{X^2+Y^2}$ is the euclidean distance.
The expected number of touched cells, in this approximation, would be $N E(d_M | d) + 1$.
It's not difficult to see that $X$ (absolute value of the difference of two uniform v.a) has a triangular density : $f_X(x)=2(1-x)$ The same for $Y$, and both are independent, hence: $$f_{XY}(x y)=4(1-x)(1-y)$$ on the unit square.
UPDATE: Below I found a much simpler solution
--------------- begin ignore -----
Conditioning on a fixed value of $d$ implies that we must restrict to an arc, the intersection of $x^2+y^2=d^2$ and the unit square.
Let's assume first that $d<1$
Note that $d\le d_M \le \sqrt{2} d$
To compute $P(d_M \le \xi | d)$, we must integrate over two pieces of arcs begining on the axis (because of symmetry, we compute just one and multiply by two). (blue arcs in the figure) The first limit point is given by $$x_1=\frac{\xi - \sqrt{2 d^2-\xi^2}}{2}$$ Further to change the integration over the arc for an integral over $x$, we note that $ds=\sqrt{1+\frac{x^2}{y^2}}{dx} = \frac{d}{y}dx$. So, leaving out the constants (that disappear on the division), we get that the cumulative distribution function (conditioned on $d$) is given by
$$F_{d_M|d}(\xi;d)=P(d_M \le \xi | d) = \frac{2 \int_{0}^{x_1}\left( 1-x\right) \,\left( \frac{1}{\sqrt{{d}^{2}-{x}^{2}}}-1\right) dx}{\int_{0}^{d}\left( 1-x\right) \,\left( \frac{1}{\sqrt{{d}^{2}-{x}^{2}}}-1\right) dx}$$
Finally, we must integrate to get the expected value
$$E(d_M |d) = d+ \int_{d}^{\sqrt{2} d} \left[1- F_{d_M|d}(\xi;d) \right] \; d\xi$$
This is all.. but it seems pretty complicated to get in close form. Let's see Maxima:
assume(x1>0);assume(d>0);assume(d>x1);
integrate((1-x)*(1/sqrt(d^2-x^2)-1), x,0,x1);
$$\frac{2\,\mathrm{asin}\left( \frac{x1}{d}\right) +2\,\sqrt{{d}^{2}-{x1}^{2}}+{x1}^{2}-2\,x1-2\,d}{2}$$
ix2(x1,d):=(2*asin(x1/d)+2*sqrt(d^2-x1^2)+x1^2-2*x1-2*d)/2;
x1(dm,d):=(dm-sqrt(2*d^2-dm^2))/2;
assume(dm>d); assume(dm<sqrt(2)*d);
fdist(dm,d):=2*(ix2(xx(dm,d),d))/ix2(d,d);
This gets a little messy. Let's try at least some numerical values:
dd:0.8;
1.01798
Let's simulate to check: Matlab/Octave:
N=300000;
t=rand(N,4);
xy = [abs(t(:,1)-t(:,3)),abs(t(:,2)-t(:,4))];
t=[];
d2=xy(:,1).^2+xy(:,2).^2;
d=sqrt(d2);
dm=xy(:,1)+xy(:,2);
step=0.02;
dround=round(d/step)*step;
%size(dround(dround==0.8))
mean(dm(dround==0.8))
>>>ans = 1.0174
Not bad, I'd say. Some other values:
d Maxima Octave (simul)
0.9 1.15847 1.1569
0.8 1.01798 1.0174
0.7 0.88740 0.8863
0.6 0.75983 0.7579
0.5 0.63328 0.6331
0.4 0.50698 0.5063
0.3 0.38062 0.3808
The case $d>1$ should be solved in a similar way.
--------------- end ignore -----
A much simpler solution:
Let's change to polar coordinates: $x = r \cos(t)$, $y = r \sin(t)$. The joint density is
$$f_{r,t}(r,t) = 4 r (1- r \cos(t))(1- r \sin(t))$$
in the domain $0 \le r \le \sqrt{2}$, $0 \le t \le \pi/2$ for $r\le 1$, $arccos(1/r) \le t \le arcsin(1/r)$ for $r > 1$
The conditional density is then
$$f_{t|r}(t|r) = \frac{1}{g(r)} (1- r \cos(t))(1- r \sin(t))$$
where $g(r)$ is the normalization constant (indepent of $t$).
Assuming first that $r<1$, then
$$g(r) = \int_0^{\pi/2} (1- r \cos(t))(1- r \sin(t)) dt = \frac{{r}^{2}}{2}-2\,r+\frac{\pi }{2}$$
Further, $d_M= x + y= r (\cos(t)+\sin(t))$, so the conditional expectation is given by
$$E[d_M | r] = r \frac{1}{g(r)} \int_0^{\pi/2} (1- r \cos(t))(1- r \sin(t)) (\cos(t)+\sin(t)) dt$$
which gives
$$E[d_M | r] = r \frac{4\,{r}^{2}-3\,\left( \pi +2\right) \,r+12}{3\,{r}^{2}-12\,r+3\,\pi }$$
For $r>1$ it's more complicated. The result is
$$E[d_M | r] = r \frac{\sqrt{{r}^{2}-1}\,\left( 4\,{r}^{2}+8\right) +\left( 6\,\mathrm{asin}\left( \frac{1}{r}\right) -6\,\mathrm{acos}\left( \frac{1}{r}\right) -6\right) \,{r}^{2}-4}{3\,{r}^{3}-12\,r\,\sqrt{{r}^{2}-1}-\left( 6\,\mathrm{asin}\left( \frac{1}{r}\right) -6\,\mathrm{acos}\left( \frac{1}{r}\right) -6\right) \,r}$$
The figure shows $E[d_M | r] / r$, ie. the factor by which the expected Manhattan distance exceeds the euclidean distance (we already knew that this must be in the $[1,\sqrt{2}]$ range).
(BTW: sorry if the formatting is not optimal, but I got sick of my Chorme crashing on the edition, lots of times, sometimes losing changes - am I the only one?)
Added: Notice that for small distances ($r \to 0$) the expected number of squares results $r N \frac{4}{\pi} +1$, which agrees with Ronald's answer.
-
Imagine the subdivided square as an $N\times N$ chessboard of subsquares. Every point on the board belongs to one subsquare with coordinates $(r,c)$, where $r$ (the row index) ranges from $1$ to $N$ and similarly for the $c$ (the column index).
Now the number of subsquares that the line segment $\overline{AB}$ between two uniformly selected points $A$ and $B$ touches equals $d(A,B)=|A(r)-B(r)|+|A(c)-B(c)|+1$. Here $A(r)$ means row index for $A$, and similarly for the other notation. The points $A$ and $B$ are in "general position" because of the randomness, so the segment always passes through the maximum number of subsquares. The segment doesn't hit any corners.
Since the random variables $A(r),B(r),A(c),B(c)$ are independent and uniformly distributed on $\{1,\dots, N\}$, it is not hard to calculate that the expected number is $$\mathbb{E}(d(A,B))=1+{2(N^2-1)\over3N}.$$
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I think your answer is correct if we don't condition on $D$. But I meant for the answer to depend on $D$. So for example, if $N$=2 and $D$ is small enough, then I would think $E(d(a,B))$ would be close to 1. But if $D$ is large $E(\cdot)$ must be at least 2. Does that make sense? – itzy Mar 23 '12 at 13:35
@itzy Yes, I could only solve the simplest case of this problem. – Byron Schmuland Mar 23 '12 at 15:47
Suppose that you divide your square into $N^2$ sub-squares, where $N\in\mathbb{N}$. Then suppose that $A=(x_{1},y_{1})$ and $B=(x_{2},y_{2})$ where $x_{1},x_{2},y_{1}$ and $y_{2}$ are real numbers in $[0,1]$. Then the segment $AB$ must pass through at least $N|y_{2}-y_{1}|$ squares in the $y$ (or vertical) direction and at least $N|x_{2}-x_{1}|$ squares in the $x$ (or horizontal) direction. So, we would expect the segment $AB$ to pass through $N|x_{2}-x_{1}|+ N|y_{2}-y_{1}|$ squares.
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Isn't this double-counting some squares? A square that is passed through in the $x$ direction may be the same square that is passed through in the $y$ direction. – itzy Mar 23 '12 at 13:27
It ends up that this answer is an undercount by 1, because we should include the squares at the end and beginning of each direction (contrary to usual counting). The $x$ and $y$ directions are kind of independent, because you can either pass up into the next square, or left into the next square - there isn't really a risk of double-count. However, we are assuming that the line doesn't pass exactly through a corner, which would complicate things a bit. Which is fine, because it's almost sure that it won't pass through a corner. – Ronald Mar 31 '12 at 20:11
I am assuming that the size of a small square is 1, for simplicity, compared to a distance of D.
If the question is to place A and B "randomly" with a given distance of D apart, this may fall foul of Bertrand's paradox: that this concept of randomness is not clearly enough defined.
However, it seems to make sense to place A first, suggesting we should place B on an arc with centre A and radius D. The arc will be restricted depending on how close it is to the edge of the square.
Let's imagine that we choose the point A, and then the angle $\theta$ is chosen randomly. Each direction $\theta$ is equally likely (by assumption here and by symmetry of the square).
Then the number of squares passed through will be $D|\sin(\theta)|+D|\cos(\theta)|+1$ from Byron's work and using a right-angled triangle.
$$E[D|\sin(\theta)|+D|\cos(\theta)|+1]$$ $$= D \cdot E[|\sin(\theta)|+|\cos(\theta)|] + 1$$ $$= 4D/\pi + 1$$
Or, considering $(1/N)$ to be the length of a little square (according to the question as given) $$= 4ND/\pi + 1$$
In the case where $N = 2$ ($2$ little squares per side) and $D = 0.5$ ($D$ is half the length of a long side), we expect to pass through $2.27$ squares. Seems right. Sometimes it will pass through only $1$, but more often $3$.
This will break down badly when D gets long compared to the side of the big square, I imagine ($D>1$?), when the line no longer has the option of being horizontal or vertical.
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+1. I have tried to make your formulae a little prettier. When you say "Then the number of squares passed though", I think you mean "Then the expected number of squares passed though" – Henry Mar 31 '12 at 2:44
@Henry Thanks, and you're right. Any thoughts on the N=1 case? Have I made an error? – Ronald Mar 31 '12 at 2:46
I am not interested in the original finite square: I think the question would be better posed as an infinite integer grid. – Henry Mar 31 '12 at 2:51
agree; at least, it's much easier in an infinite grid :) – Ronald Mar 31 '12 at 9:50
+1 for mentioning Bertrand's paradox. In fact that is what I had in mind when I asked itzy to clarify how the points are selected. Alas, I might have been too terse, and that comment was ignored. Now we have a bounty! on a possibly ill-formed question. – Aryabhata Apr 1 '12 at 20:06 | 2016-07-28T20:48:37 | {
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http://oscarfalk.com/dc95nd/8e3611-differentiability-implies-continuity | Playing next. B The converse of this theorem is false Note : The converse of this theorem is false. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. The converse is not always true: continuous functions may not be differentiable… Obviously this implies which means that f(x) is continuous at x 0. Differentiability at a point: graphical. Differentiable Implies Continuous Differentiable Implies Continuous Theorem: If f is differentiable at x 0, then f is continuous at x 0. The best thing about differentiability is that the sum, difference, product and quotient of any two differentiable functions is always differentiable. f is differentiable at x0, which implies. Get NCERT Solutions of Class 12 Continuity and Differentiability, Chapter 5 of NCERT Book with solutions of all NCERT Questions.. Differentiability and continuity. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. differentiable on \RR . Applying the power rule. Continuously differentiable functions are sometimes said to be of class C 1. Calculus I - Differentiability and Continuity. But the vice-versa is not always true. Differentiation: definition and basic derivative rules, Connecting differentiability and continuity: determining when derivatives do and do not exist. Starting with \lim _{x\to a} \left (f(x) - f(a)\right ) we multiply and divide by (x-a) to get. A function is differentiable on an interval if f ' (a) exists for every value of a in the interval. Intermediate Value Theorem for Derivatives: Theorem 2: Intermediate Value Theorem for Derivatives. A function is differentiable if the limit of the difference quotient, as change in x approaches 0, exists. UNIFORM CONTINUITY AND DIFFERENTIABILITY PRESENTED BY PROF. BHUPINDER KAUR ASSOCIATE PROFESSOR GCG-11, CHANDIGARH . Differentiability Implies Continuity If is a differentiable function at , then is continuous at . infinity. Continuity. Then This follows from the difference-quotient definition of the derivative. Nevertheless, Darboux's theorem implies that the derivative of any function satisfies the conclusion of the intermediate value theorem. Note To understand this topic, you will need to be familiar with limits, as discussed in the chapter on derivatives in Calculus Applied to the Real World. Differential coefficient of a function y= f(x) is written as d/dx[f(x)] or f' (x) or f (1)(x) and is defined by f'(x)= limh→0(f(x+h)-f(x))/h f'(x) represents nothing but ratio by which f(x) changes for small change in x and can be understood as f'(x) = lim?x→0(? x) = dy/dx Then f'(x) represents the rate of change of y w.r.t. Nevertheless, Darboux's theorem implies that the derivative of any function satisfies the conclusion of the intermediate value theorem. Proof: Differentiability implies continuity. Hence, a function that is differentiable at $$x = a$$ will, up close, look more and more like its tangent line at $$( a , f ( a ) )$$, and thus we say that a function is differentiable at $$x = a$$ is locally linear. Follow. A … x) = dy/dx Then f'(x) represents the rate of change of y w.r.t. Theorem Differentiability Implies Continuity. constant to obtain that \lim _{x\to a}f(x) - f(a) = 0 . Continuity and Differentiability Differentiability implies continuity (but not necessarily vice versa) If a function is differentiable at a point (at every point on an interval), then it is continuous at that point (on that interval). function is differentiable at x=3. DIFFERENTIABILITY IMPLIES CONTINUITY AS.110.106 CALCULUS I (BIO & SOC SCI) PROFESSOR RICHARD BROWN Here is a theorem that we talked about in class, but never fully explored; the idea that any di erentiable function is automatically continuous. Get Free NCERT Solutions for Class 12 Maths Chapter 5 continuity and differentiability. However, continuity and … exist, for a different reason. Donate or volunteer today! It is perfectly possible for a line to be unbroken without also being smooth. Fractals , for instance, are quite “rugged” $($see first sentence of the third paragraph: “As mathematical equations, fractals are … Report. If you update to the most recent version of this activity, then your current progress on this activity will be erased. one-sided limits \lim _{x\to 3^{+}}\frac {f(x)-f(3)}{x-3}\\ and \lim _{x\to 3^{-}}\frac {f(x)-f(3)}{x-3},\\ since f(x) changes expression at x=3. continuous on \RR . So, if at the point a a function either has a ”jump” in the graph, or a corner, or what and thus f ' (0) don't exist. A function is differentiable if the limit of the difference quotient, as change in x approaches 0, exists. If is differentiable at , then is continuous at . Our mission is to provide a free, world-class education to anyone, anywhere. Checking continuity at a particular point,; and over the whole domain; Checking a function is continuous using Left Hand Limit and Right Hand Limit; Addition, Subtraction, Multiplication, Division of Continuous functions Proof: Suppose that f and g are continuously differentiable at a real number c, that , and that . Differentiability and continuity. You are about to erase your work on this activity. It is possible for a function to be continuous at x = c and not be differentiable at x = c. Continuity does not imply differentiability. continuity and differentiability Class 12 Maths NCERT Solutions were prepared according to CBSE … Differentiability Implies Continuity If f is a differentiable function at x = a, then f is continuous at x = a. 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http://math.stackexchange.com/questions/119253/interesting-square-of-log-sin-integral | interesting square of log sin integral
I ran across this challenging log sin integral and am wondering what may be a good approach.
$$\int_{0}^{\frac{\pi}{2}}x^{2}\ln^{2}(2\cos(x))dx=\frac{11{{\pi}^{5}}}{1440}$$
This looks like it may be able to be connected to the digamma or incomplete beta function somehow.
I tried using the identity $$\cos(x)=\frac{e^{ix}+e^{-ix}}{2},$$ but that square threw a wrench in my plans.
Maybe write it as $$(x\ln(2\cos(x)))^{2}=\ln^{2}(2)x^{2}+2\ln(2)x^{2}\ln(\cos(x))+x^{2}\ln^{2}(\cos(x))$$ and expand. I doubt if that does any good though.
Does anyone know of a clever way to approach this? It looks like a fun one if I knew a good starting point.
Is there an identity that goes with $$(2\cos(t))^{a}.$$
If $$t^{2}(2\cos(t))^{a}$$ were differentiated twice w.r.t a, then we would have
$$(2\cos(t))^{a}t^{2}\ln^{2}(2\cos(t))$$ Maybe the trig form of Beta which would give $$2^{a}\cdot B(1/2, \frac{a+1}{2}).$$ Then, differentiate this twice and it would somehow relate to the digamma? $$\frac{1}{2}B(1/2,\frac{a+1}{2})=\frac{\sqrt{\pi}\Gamma(\frac{a+1}{2})}{2\Gamma(\frac{a+2}{2})}.$$ Differentiate twice and obtain some sort of Psi relation.
Cheers Everyone.
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Well, you can rewrite $2\cos x$ as $(e^{2ix}+1)e^{-ix}$ and use the Taylor expansion of $\log(1+u)$, expand out the square, interchange summation and integration, evaluate some definite integrals of $x^3e^{imx}$ and $x^2e^{imx}$'s and then get some zeta functions and... Well maybe that's not the best route. – anon Mar 12 '12 at 13:02
Thanks. I actually tried that, but to no avail. I could get it if it weren't for that x^2 term. Maybe $$\zeta(4)$$ is in there somewhere?. $$\frac{11\pi}{16}\cdot \frac{{\pi}^{4}}{90}$$ – Cody Mar 12 '12 at 15:32
I spent some time yesterday trying to find an alternative solution to this problem using complex analysis. I haven't succeeded yet, but I've "reduced" the problem to computing either of the integrals $$\int_0^\infty x\log^2(1-e^{-2x})\,dx = -\int_0^\infty \frac{2x^2}{e^{2x} - 1}\log{(1-e^{-2x})}\,dx$$ or $$\int_0^{\pi/2} \log^4{(2\cos{x})}\,dx.$$ I've included some more detail in a recent answer to another question of @Cody. Anyways, if anyone has any ideas... – Nick Strehlke Mar 14 '12 at 2:09
As I remarked in a comment, I have spent a while trying to derive this identity by means of contour integration. Though I have not succeeded at that, (I take great pleasure in striking that out; see my other answer below!) I have found some related identities in my search which surprised me quite a bit, and I think they're worth sharing. For example, using the integral you asked about, I have shown that $$$$\sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{5}{4}\cdot \frac{\pi^4}{90} = \frac{5}{4} \zeta(4). \tag{1}$$$$ Here $$H_n = \sum_{k = 1}^n \frac{1}{k}$$ is the $n$th harmonic number. Here are a couple more examples. I will use the first to derive $(1)$.
1. With the method from this answer, one can show that $$\int_0^{\pi/2} x^2\log^2(2\cos{x})\,dx = \frac{1}{5}\left(\frac{\pi}{2}\right)^5 + \pi \int_0^\infty y \log^2(1-e^{-2y})\,dy,$$ and it then follows from the result of your question that $$\pi \int_0^\infty y \log^2(1- e^{-2y})\,dy = \frac{11}{45}\left(\frac{\pi}{2}\right)^5 - \frac{1}{5}\left(\frac{\pi}{2}\right)^5 = \frac{\pi}{8} \cdot \frac{\pi^4}{90}.$$
2. I also shared this example in the other answer, but for completeness, let me mention that $$\int_0^{\pi/2} x^2\log^2(2\cos{x})\,dx = \frac{1}{30}\left(\frac{\pi}{2}\right)^5 + \frac{1}{6}\int_0^{\pi/2} \log^4(2\cos{x})\,dx,$$ hence $$\int_0^{\pi/2} \log^4(2\cos{x})\,dx = \frac{19}{15}\left(\frac{\pi}{2}\right)^5.$$
More details about the derivations of 1. and 2. are contained in the answer I referenced above (near the bottom).
Let me now turn to deriving $(1)$. Taking $x = - e^{-2y}$ in the series expansion $$\log{(1+x)} = \sum_{n = 1}^\infty \frac{(-1)^{n+1}x^n}{n}$$ and inserting the result into the integral evaluated in 1. gives \begin{align} \frac{1}{8}\cdot\frac{\pi^4}{90} &= \int_0^\infty y \log^2(1 - e^{-2y})\,dy \\ & = \int_0^\infty y \left(\sum_{n = 1}^\infty \frac{e^{-2ny}}{n}\right)^2\,dy \\ & = \sum_{m = 1}^\infty\sum_{n = 1}^\infty\frac{1}{nm} \int_0^\infty y e^{-2(n + m)y}\,dy \\ & = \frac{1}{4}\sum_{m = 1}^\infty\sum_{n = 1}^\infty \frac{1}{nm(n+m)^2}. \tag{2} \end{align} Now put $r = m+n$ and $s=n$ in order to write \begin{align} \sum_{m = 1}^\infty\sum_{n = 1}^\infty \frac{1}{nm(n+m)^2} & = \sum_{r = 2}^\infty \frac{1}{r^2}\sum_{s = 1}^{r-1} \frac{1}{s(r-s)} = 2 \sum_{r=2}^\infty \frac{1}{r^3} \sum_{s=1}^{r-1}\frac{1}{s}, \tag{3} \end{align} with the last equation a consequence of the identity $\frac{1}{s(r-s)} = \frac{1}{r}\left(\frac{1}{s}+ \frac{1}{r-s}\right)$. Insert $(3)$ into $(2)$ and multiply both sides by $2$ to get $$\frac{1}{4} \cdot \frac{\pi^4}{90} = \sum_{r=2}^\infty \frac{1}{r^3} \sum_{s=1}^{r-1}\frac{1}{s} = \sum_{r=2}^\infty \frac{H_{r-1}}{r^3}.$$ Since $H_r - H_{r-1} = 1/r$, the identity $(1)$ now follows from the equations $$\frac{\pi^4}{90} = \sum_{r=1}^\infty \frac{1}{r^4} = \sum_{r = 1}^\infty \frac{H_r}{r^3} - \sum_{r=2}^\infty \frac{H_{r-1}}{r^3} = \sum_{r = 1}^\infty \frac{H_r}{r^3} - \frac{1}{4}\cdot\frac{\pi^4}{90}.$$
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Very good answer! – Daniel Montealegre Mar 22 '12 at 6:20
With a little time on my hands these past couple of days, I have finally succeeded in computing this integral through contour integration; the method is very similar in spirit to my answer here. The only background needed is knowledge of Cauchy's integral theorem and the fact that $\zeta(4) = \pi^4/90$. I've made a few remarks on generalization towards the bottom.
The strategy is as follows. We integrate the principal branch of $f(z) = z\log^3{(1+e^{2i z})}$ over an appropriately chosen contour in order to prove \begin{align} \int_{-\pi/2}^{\pi/2}x^2\log^2{(2\cos{x})}\,dx & = \int_{-\pi/2}^{\pi/2}x^4\,dx-\frac{\pi}{3}\int_0^\infty \log^3{(1-e^{-2y})}\,dy. \tag{1} \end{align} This leaves us in good shape; due to the fact that the integrand is even, the left-hand side is twice the integral we wish to evaluate. The first integral on the right is easily computed as $\pi^5/80$. To evaluate the second integral, put $e^{-v} = 1-e^{-2y}$; mild computations give $-(e^v - 1)^{-1}\,dv = dy$, and then \begin{align} -\frac{\pi}{3}\int_0^\infty \log^3{(1-e^{-2y})}\,dy & = \frac{\pi}{6}\int_0^\infty\frac{v^3}{e^v - 1}\,dv = \pi\zeta(4). \tag{2} \end{align} The last equation follows from the well-known identity $$\Gamma(s)\zeta(s) = \int_0^\infty \frac{v^{s-1}}{e^{v}-1}\,dv$$ which holds for $\text{Re}\,s > 1$ and can be derived easily by developing the integrand in a geometric series. Upon inserting $(2)$ into $(1)$ and simplifying we arrive at the desired result.
It remains to prove $(1)$; as mentioned, the argument is entirely analogous to one I gave in this answer to another of Cody's questions. For completeness, I include it here. Consider the region obtained from $\mathbb C$ by removing the half-lines on which $\text{Re}\,z$ is an integer multiple of $\pi/2$ and $\text{Im}\,z \leq 0$. On this region, we can define a branch of $\log{(1+e^{2iz})} = \log{(2e^{iz}\cos{z})}$; we choose the branch with imaginary part between $-\pi$ and $\pi$. Having done this, let $f(z) = z\log^3{(1+e^{2iz})}$. We wish to integrate $f(z)$ over the contour obtained by removing the corners from the rectangle determined by $-\pi/2$ and $\pi/2 + iR$, where $R > 0$. To complete the contour we replace the bottom corners with quarter-circles of radius $\delta >0$. The intention is to let $R \to \infty$ and $\delta \to 0$. ${}$
For fixed values of $\delta$ and $R$, Cauchy's theorem says that $f(z)$ integrates to zero over the contour. As $R \to \infty$, the contribution from the upper horizontal side tends to $0$ because $f(x+iR) \to 0$ uniformly for $-\pi/2 \leq x \leq \pi/2$. By writing $1+e^{2i z} = 1- e^{2i(z-\pi/2)}$ one sees that $1+e^{2i z} = O(z-\pi/2)$, and therefore that $\log{(1+e^{2iz})} = O(\log{|z-\pi/2|})$ as $z \to \pi/2$. It follows that the contribution from the left quarter-circle is $O(\delta^2\log^3{\delta})$, hence that it vanishes in the limit $\delta \to 0$. The same argument applies to the other quarter-circle, and we are left with the contributions from the vertical sides and the lower horizontal side.
After taking limits, the contribution from the vertical sides is \begin{align} i\int_0^\infty f(iy+\pi/2)\,dy -i\int_0^\infty f(iy - \pi/2)\,dy = i\pi\int_0^\infty\log^3{(1-e^{-2y})}\,dy. \tag{3} \end{align} Now for $x$ between $-\pi/2$ and $\pi/2$, the quantity $2\cos{x}$ is positive. This means that the unique value of $\arg{(2e^{ix}\cos{x})}$ which lies between $-\pi$ and $\pi$ is simply $x$. As we have chosen the principle branch, we obtain $\log{(2e^{ix}\cos{x})} = \log{(2\cos{x})}+ix$. Therefore the contribution from the lower horizontal side may be written $$\int_{-\pi/2}^{\pi/2} f(x)\,dx = \int_{-\pi/2}^{\pi/2}x\left(\log{(2\cos{x})} + ix\right)^3\,dx. \tag{4}$$ By the preceding analysis, $(3)$ and $(4)$ sum to zero. Since $(3)$ is purely imaginary, this means in particular that the imaginary part of $(4)$ is equal to the negative of $(3)$. The last statement is equivalent to $(1)$.
Because I have spent quite some time with this question, I would like to make a few remarks in the direction of a generalization. By taking $g(z) = p(z) \log^m(1+e^{2iz})$ in place of $f(z)$ above, where $p(z)$ is a polynomial and $m \in \mathbb N$, one finds by repeating the same arguments that \begin{align} \int_{-\pi/2}^{\pi/2} p(x)&\left(\log{(2\cos{x})} + ix\right)^m\,dx \\ &= -i \int_0^\infty \left(p(iy + \pi/2)-p(iy - \pi/2)\right)\log^m{(1-e^{-2y})}\,dy. \tag{5} \end{align} Clever choices of $p$ and $m$ then furnish a number of integral and series identities. Moreover, the identities $$\int_0^\infty y^n \log{(1-e^{-2y})}\,dy = -\frac{1}{2^{n+1}}\Gamma(n+1)\zeta(n+2) \tag{6}$$ and $$\int_0^\infty \log^m{(1-e^{-2y})}\,dy = \frac{(-1)^{m}}{2}\int_0^\infty \frac{y^m}{e^y - 1}\,dy = \frac{(-1)^m}{2}\Gamma(m+1)\zeta(m+1) \tag{7}$$ provide a connection between the integrals on the left-hand side of $(5)$ and the values taken by $\zeta$ at positive integers $n\geq2$. Here $(6)$ is derived by expanding $\log{(1-e^{-2y})}$ in powers of $e^{-2y}$ while $(7)$ is derived in the same way as $(2)$.
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This is truly great stuff. – Antonio Vargas Mar 31 '12 at 3:17
Thanks Antonio. – Nick Strehlke Mar 31 '12 at 8:45
Wow, Nick. Beautiful. Thank you. I am sorry I did not get back sooner. This was an older thread and I just noticed your wonderful response. – Cody Apr 6 '12 at 16:01
I know it has been a long time since you answered this, but I wanted to thank you again for the time you took to post all this. I have recently begun studying CA and your post has proven valuable. Absolute genius. – Cody May 30 '12 at 17:10
Thank you for the kind words Cody. I really enjoyed this problem, and I'm glad you found my answer useful! – Nick Strehlke Jun 2 '12 at 3:26
I found the solution to this problem. If anyone is interested, see here:
http://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale:_Form_R%28x,log,cos%29#0.2
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Very nice collection, in most cases with proofs. Thanks. – vesszabo Aug 29 '12 at 14:21 | 2014-08-31T05:07:39 | {
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https://www.physicsforums.com/threads/maclaurin-series-of-a-function.652629/ | # Maclaurin series of a function
1. Nov 15, 2012
### Mangoes
1. The problem statement, all variables and given/known data
Find the maclaurin series of:
$$f(x) = \int_{0}^{x}(e^{-t^2}-1) dt$$
3. The attempt at a solution
I know $$e^t = \sum_{n=0}^{∞} \frac{t^n}{n!}$$
Simple substitution gives me:
$$e^{-t^2} = \sum_{n=0}^{∞}\frac{(-t^2)^n}{n!}$$
Which I rewrote as
$$e^{-t^2} = \sum_{n=0}^∞\frac{(-1)^n(t^{2n})}{n!}$$
Since I notice that when n = 0, the term simplifies to 1,
$$e^{-t^2} - 1 = \sum_{n=1}^∞ \frac{(-1)^n(t^{2n})}{n!}$$
Going back to the original expression and substituting the infinite series
$$\int_0^x\sum_{n=1}^∞ \frac{(-1)^n(t^{2n})}{n!} dt$$
Now, I'm integrating with respect to t, so after using the power rule and applying the upper limit (lower limit simplifies to 0)
$$\sum_{n=1}^∞\frac{(-1)^nx^{2n+1}}{n!(2n+1)}$$
And this is what I would think is the maclaurin representation of the function, but it's wrong and I have no idea why. Would anyone please point to where I'm going wrong with this?
Last edited: Nov 15, 2012
2. Nov 15, 2012
### Dick
How did your lower limit n=0 appear in the appear in the answer? It was n=1 in the step before.
3. Nov 15, 2012
### Mangoes
Sorry about that, typo from copying and pasting.
4. Nov 15, 2012
### SammyS
Staff Emeritus
How do you know it's wrong?
You can change the sum to start at n=0, if that's the problem.
5. Nov 15, 2012
### Dick
Well, then I'm having trouble finding anything wrong with it. But does the answer you are checking against expect you to shift the lower limit to n=0?
6. Nov 15, 2012
### Mangoes
$$\sum_{n=0}^∞\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)(n+1)!}$$
They shifted the lower bound to start at 0 and compensated by replacing n by (n+1).
Pretty frustrating considering I've been hitting my head against the wall thinking I was doing something wrong all this time...
Thanks a lot for the help to both of you. Just learned about power series about a week ago so still not too used to working with them.
7. Nov 15, 2012
### rbj
there isn't anything wrong with it. this
$$f(x) = \int_{0}^{x}(e^{-t^2}-1) dt = \sum_{n=1}^\infty \frac{(-1)^n x^{2n+1}}{n!(2n+1)}$$
is in fact correct. | 2017-08-23T13:10:40 | {
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https://math.stackexchange.com/questions/2811476/cool-way-of-finding-cos-left-frac-pi5-right | Cool way of finding $\cos\left(\frac{\pi}{5}\right)$
while trying to solve a problem, I stumbled upon a way of finding $\cos\left(\frac{\pi}{5}\right)$ using identities and the cubic formula. Is it possible to find other values of sine or cosine in a similar way ?
Consider $$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right).$$ Using the difference of cosines identity, we have $$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right) = -2\sin\left(\frac{3\pi}{10}\right)\sin\left(-\frac{\pi}{10}\right).$$
Now we change the RHS using the identity $\sin(x) = \cos\left(\frac{\pi}{2}-x\right)$ and the fact that $\sin(x)$ is odd.
$$-2\sin\left(\frac{3\pi}{10}\right)\sin\left(-\frac{\pi}{10}\right) = 2\cos\left(\frac{\pi}{5}\right) \cos\left(\frac{2\pi}{5}\right)$$ So, $$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right) = 2\cos\left(\frac{\pi}{5}\right) \cos\left(\frac{2\pi}{5}\right)$$
Now we make use of the identity $\cos(2x)=2\cos^2(x)-1$. $$\cos\left(\frac{\pi}{5}\right) - 2\cos^2\left(\frac{\pi}{5}\right)+1 = 2\cos\left(\frac{\pi}{5}\right) \left(2\cos^2\left(\frac{\pi}{5}\right)-1\right)$$ Let $y=\cos\left(\frac{\pi}{5}\right)$ and we have $$y-y^2+1=2y(2y^2-1)$$ $$4y^3+2y^2-3y-1=0$$ which has the correct solution $$y=\frac{\sqrt{5}+1}{4} =\cos\left(\dfrac{\pi}{5}\right)$$ One of the roots is also $\sin\left(\dfrac{\pi}{10}\right)$ which I'm guessing is because you end up with the same cubic if you apply the above to sin too.
• There's no question – Jakobian Jun 7 '18 at 15:12
• I added the question at the end. – Adam Jun 7 '18 at 15:14
• I'm 100% sure there is a similar way to calculate arbitrary values of $\sin(\cdot)$ with your way. Another way I know would be to use the triple angle identity$$\sin 3\theta=3\sin\theta-4\sin^3\theta$$but you'll need to calculate what $\sin^3 3\theta$ is, which can be tedious. – Frank W. Jun 7 '18 at 15:17
• $\cos\frac{2\pi}{n}$ is an algebraic number over $\mathbb{Q}$ width degree $\frac{1}{2}\varphi(n)$. In particular finding an explicit form for $\cos\frac{\pi}{5}$ only requires the quadratic formula. – Jack D'Aurizio Jun 7 '18 at 15:19
• Is that the method with the 36-72-72 triangle ? – Adam Jun 7 '18 at 15:22
Using complex numbers we can derive $\sin(\frac{\pi}{5})$ and $\cos(\frac{\pi}{5})$ $$z^5=-1 \implies z^4-z^3+z^2-z+1=0 \implies z^2+\frac{1}{z^2}-z-\frac{1}{z}+1=0$$ By substitution $t=z+\frac{1}{z}$ we get: $$t^2-t-1=0 \implies t=\frac{1\pm\sqrt{5}}{2}$$ And because $\cos(\frac{\pi}{5})$ is positive, we may consider just $t=\frac{1+\sqrt{5}}{2}$
$$z^2-z\frac{1+\sqrt{5}}{2}+1=0 \implies z=\frac{1+\sqrt{5}}{4}\pm i\frac{\sqrt{10-2\sqrt{5}}}{4}$$ And we get that $\cos(\frac{\pi}{5}) = \frac{1+\sqrt{5}}{4}$ and $\sin(\frac{\pi}{5})= \frac{\sqrt{10-2\sqrt{5}}}{4}$
• that's very cool! – Adam Jun 7 '18 at 18:02
• I was quite confused how this arrived at $z^2-z\frac{1+\sqrt{5}}{2}+1=0$. But, for anyone confused, you're substituting the value of $t$ you've found into the equation for $z$, to solve for $z$. – Jam Jun 14 '18 at 12:10
There is a slightly shorter way:
$$\sin\frac{\pi}{5}=\sin\frac{4\pi}{5}=2\sin\frac{2\pi}{5}\cos\frac{2\pi}{5}=4\sin\frac{\pi}{5}\cos\frac{\pi}{5}\cos\frac{2\pi}{5}$$
Hence
$$\cos\frac{\pi}{5}\cos\frac{2\pi}{5}=\frac14$$
You can also write this
$$\cos\frac{\pi}{5}\cos\frac{3\pi}{5}=-\frac14$$
Then, using the formula $2\cos a\cos b=\cos(a+b)+\cos(a-b)$, you have:
$$\frac12=2\cos\frac{\pi}{5}\cos\frac{2\pi}{5}=\cos\frac{\pi}{5}+\cos\frac{3\pi}{5}$$
Hence $\cos\frac{\pi}{5}$ and $\cos\frac{3\pi}{5}$ are the roots of $t^2-\frac12t-\frac14$. The rest is easy, and the positive root is $\cos\frac{\pi}5$.
To answer your question: in general, it's not possible to find values of $\cos\frac{\pi}{n}$ only by radicals, even though they are algebraic numbers. Even in the case you obtain an irreducible cubic equation (which can be solved by radicals), it's the "trigonometric case" here, which has no expression with real radicals (and a complex cubic root needs the trigonometric functions anyway). For instance, $\cos1^\circ$ can't be computed with real radicals. But $\cos3^\circ$ can.
Even if in general it's not possible, there are a few tricks you can apply, for instance
$$\cos\frac{\pi}{12}=\cos\left(\frac{\pi}{3}-\frac{\pi}{4}\right)$$
Also, if you can compute $\cos x$ by radicals, then you can compute $\cos\dfrac{x}{2^n}$ too.
So, which angles of the form $\frac{\pi}{n}$ leads to trigonometric functions computable by radicals? The answer is given by the Gauss-Wantzel theorem. For instance, one can compute $\cos\dfrac{\pi}{17}$. However, the computation is not obvious, see http://mathworld.wolfram.com/TrigonometryAnglesPi17.html
Everything begins with finding $\sin\left(\frac{\pi}{10}\right)$
Okay... Lets say $x=\frac{\pi}{10}$
$$5x=\frac{\pi}{2}$$ $$2x=\frac{\pi}{2}-3x$$ $$\sin(2x)=\sin\left(\frac{\pi}{2}-3x\right)$$ $$\sin(2x)=\cos(3x)$$ $$2\sin(x)\cos(x)=4\cos^3(x)-3\cos(x)$$ $$2\sin(x)\cos(x)-4\cos^3(x)+3\cos(x)=0$$ $$\cos(x)\left(2\sin(x)-4\cos^2(x)+3\right)=0$$ $$2\sin(x)-4\cos^2(x)+3=0$$ $$2\sin(x)-4(1-sin^2(x))+3=0$$ $$2\sin(x)-4+4\sin^2(x)+3=0$$ $$4\sin^2(x)+2\sin(x)-1=0$$ Now use the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ $$a=4,b=2,c=-1$$ $$=\frac{-2\pm \sqrt{2^2-4(4)(-1)}}{2(4)}$$ $$=\frac{-2\pm \sqrt{20}}{8}$$ $$\sin\left(\frac{\pi}{10}\right)=\frac{-1\pm \sqrt{5}}{4}$$
One another way of finding the value of $\cos\left(\frac{\pi}{5}\right)$
$$\cos\left(\frac{\pi}{5}\right)=\cos\left(2\left(\frac{\pi}{10}\right)\right)$$ $$=1-2\left(\sin\frac{\pi}{10}\right)^2$$ Since $\sin\left(\frac{\pi}{10}\right)=\frac{\sqrt{5}-1}{4}$ $$=1-2\left(\frac{\sqrt{5}-1}{4}\right)^2$$ $$=1-\frac{(\sqrt{5}-1)^2}{8}$$ $$=\frac{8-5-1+2\sqrt{5}}{8}$$ $$=\frac{2+2\sqrt{5}}{8}$$ $$\cos\left(\frac{\pi}{5}\right)=\frac{1+\sqrt{5}}{4}$$ Now lets find $\sin\left(\frac{\pi}{5}\right)$ $$\sin\left(\frac{\pi}{5}\right)=\sqrt{1-\cos^2\left(\frac{\pi}{5}\right)}$$ $$=\sqrt{1-\left(\frac{1+\sqrt{5}}{4}\right)^2}$$ $$=\sqrt{1-\frac{1+5+2\sqrt{5}}{16}}$$ $$=\sqrt{\frac{10-2\sqrt{5}}{16}}$$ $$\sin\left(\frac{\pi}{5}\right)=\frac{\sqrt{10-2\sqrt{5}}}{4}$$
• But how would you find $\sin(\pi /5)$ ? – Adam Jun 7 '18 at 18:03
• How do you find $\sin(\pi/10)$? – Blue Jun 7 '18 at 18:24
• @Blue I added it in my answer. Finding $\sin(\pi/10)$ was really challenging and it took a bit to find out the way – tien lee Jun 7 '18 at 18:43
Let $\Delta ABC$ be a triangle with $\angle BAC=36^o$, $\angle ABC=\angle ACB=72^o.$ $BD$ bisects $\angle ABC$. $BE\perp AC$. You would find that $AD=BD=BC$ and $\Delta ABC \sim \Delta BDC.$ Thus, as the figure shows, we may obtain $$\frac{x}{2y}=\frac{x+2y}{x}.$$ Thus $$\left(\frac{x}{y}\right)^2-2\left(\frac{x}{y}\right)-4=0.$$
Solve it. We have $$\frac{x}{y}=1+\sqrt{5}.$$
Another negative root is not what we want.Thus, $$\cos 36^o=\frac{AE}{AB}=\frac{x+y}{x+2y}=\dfrac{\dfrac{x}{y}+1}{\dfrac{x}{y}+2}=\frac{1+\sqrt{5}}{4}.$$ | 2019-04-19T04:46:02 | {
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https://questioncove.com/updates/4ea8c0fee4b02eee39cd0c82 | Mathematics 83 Online
OpenStudy (anonymous):
Help with a probability function question! A woman has 7 keys on a keyring, one of which fits the door she wants to unlock. She randomly selects a key and tries it. If it does not unlock the door, she randomly selects another from these remaining and tries to unlock the door with it She continues in this manner until the door is unlocked. Let X be the number of keys she tries before unlocking the door, counting the key that actually worked. Find the probability function of X. Help with a probability function question! A woman has 7 keys on a keyring, one of which fits the door she wants to unlock. She randomly selects a key and tries it. If it does not unlock the door, she randomly selects another from these remaining and tries to unlock the door with it She continues in this manner until the door is unlocked. Let X be the number of keys she tries before unlocking the door, counting the key that actually worked. Find the probability function of X. @Mathematics
OpenStudy (amistre64):
$^7C_x\ (\frac{1}{7})^x(\frac{6}{7})^{1-x}$ maybe
OpenStudy (amistre64):
that wont do, you cant have 6 keys fit ...
OpenStudy (amistre64):
the chances are: P(1st) = 1/7 P(2nd) = 1/6 P(3rd) = 1/5 that aint it, is it
OpenStudy (amistre64):
P(y) = 1/7 P(n,y) = 6/7 * 1/6 P(n,n,y) = 6/7 * 5/6 * 1/5 feels better
OpenStudy (anonymous):
The probability of opening in the first key is 1/7 The probability of opening in the second key is the probability of NOT opening on the first one * the probability of opening if she started with 6 keys, so 6/7 * 1/6 Then, not opening on the first, not opening on the second and opening if she had 5 keys 6/7*5/6*1/5 and it goes on...
OpenStudy (anonymous):
I understand how to come up with that but Im not sure how I would come up with the function...that's the part where Im not sure how to approach...
OpenStudy (anonymous):
The probability of opening in the first key is 1/7 The probability of opening in the second key is the probability of NOT opening on the first one * the probability of opening if she started with 6 keys, so 6/7 * 1/6 Then, not opening on the first, not opening on the second and opening if she had 5 keys 6/7*5/6*1/5 and it goes on...
OpenStudy (anonymous):
1/7 = 1/7 6/7 * 1/6 = 1/7 6/7 * 5/6 * 1/5 = 1/7 ...
OpenStudy (amistre64):
$\frac{^NC_m\ ^?C_?}{^NC_x}$ cant recall the formula or even if its right
OpenStudy (anonymous):
That formula doesnt make sense haha...Im still trying to figure it out...thanks everyone for the help though...very useful!
OpenStudy (amistre64):
Zarkon is keen with these
OpenStudy (zarkon):
$f_X(x)=\left\{\begin{matrix}\frac{1}{7} &\text{if} &x\in\{1,2,3,4,5,6,7\} \\0 && \text{otherwise}\end{matrix}\right.$
OpenStudy (anonymous):
Zarkon, can you explain that please?
OpenStudy (anonymous):
And is vitor's explanation correct?
OpenStudy (zarkon):
what is there to explain. if x takes on any of the (integer) values 1 through 7 then P(X=x)=1/7 otherwise the answer is 0
OpenStudy (amistre64):
i was thinking about the hypergeometric probability distribution
OpenStudy (zarkon):
this is a discrete uniform distribution
OpenStudy (amistre64):
not that mine is correct, but this would explain what i was thinking better http://129.81.170.14/~slukens/Math114_fall2009/Hypergeometric.pdf
OpenStudy (anonymous):
How does Vitor's explanation differ from yours Zarkon?
OpenStudy (amistre64):
Vitors is not in function format ...
OpenStudy (anonymous):
Oh righttt nevermind
OpenStudy (anonymous):
Got it...thanks all for the help!
OpenStudy (amistre64):
you got the answer? or got the explanation :)
OpenStudy (amistre64):
casue id love to know the answer lol
OpenStudy (anonymous):
I understand how vitor's and zarkon's answers relate: The first key does have a probability of 1/7 The prob of not getting the key on first try and then getting it on the second try is: (6/7)*(1/6) = 1/7 The prob of not getting key on first try, not on second try, but on third try is: (6/7)*(5/6)*(1/5) = 1/7 And so on...all of them are equal to probability of 1/7. It is a 0 probability otherwise! | 2022-11-27T03:11:18 | {
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http://mathhelpforum.com/discrete-math/201219-power-set-n-intersection-finite-subsets-n-print.html | # Power Set of N-Intersection of finite subsets of N
• Jul 21st 2012, 07:26 AM
Neutriiino
Power Set of N-Intersection of infinitely many finite subsets of N
Hello everyone,
This is a problem I'm having a hard time solving:
Let $P(\mathbb{N})$= the power set of $\mathbb{N}$ and for every natural number, n, $P(\mathbb{N}_n)$ be the power set of $\mathbb{N}_n$ which is a finite subset of $\mathbb{N}: \mathbb{N}_n=\left \{ 1,2,3,...,n \right \}$.
Now, if for every n, $A_n=P(\mathbb{N})-P(\mathbb{N}_n)$,
what can be said about $\bigcap_{n=1}^{\infty }A_n?$
Is that an infinite subset of $\mathbb{N}$?
• Jul 21st 2012, 08:26 AM
Plato
Re: Power Set of N-Intersection of infinitely many finite subsets of N
Quote:
Originally Posted by Neutriiino
This is a problem I'm having a hard time solving:
Let $P(\mathbb{N})$= the power set of $\mathbb{N}$ and for every natural number, n, $P(\mathbb{N}_n)$ be the power set of $\mathbb{N}_n$ which is a finite subset of $\mathbb{N}: \mathbb{N}_n=\left \{ 1,2,3,...,n \right \}$.
Now, if for every n, $A_n=P(\mathbb{N})-P(\mathbb{N}_n)$,
what can be said about $\bigcap_{n=1}^{\infty }A_n?$
Is that an infinite subset of $\mathbb{N}$?
We realize that this is your first posting. Nonetheless, it is best if you show some work so that we know where to begin.
If $F$ is a finite subset of $\mathbb{N}$ is it possible for $F\in\bigcap_{n=1}^{\infty }A_n~?$
If $F$ is an infinite subset of $\mathbb{N}$ must for $F\in\bigcap_{n=1}^{\infty }A_n~?$
Tell us what you think?
• Jul 21st 2012, 08:29 AM
Deveno
Re: Power Set of N-Intersection of finite subsets of N
Note that N is an element of P(N), but cannot be an element of any P(Nn) (since these sets are all finite). Hence N is in every An, and thus in the intersection.
The same exact logic hold for N - S, where S is any finite subset of N. Thus any co-finite subset of N is in the infinite intersection. Since we obtain a co-finite subset of N from any finite set S, simply by taking N - S, there are at least as many co-finite subsets of N, as there are finite subsets of N. Hence the intersection of all the An is indeed infinite.
• Jul 21st 2012, 11:05 AM
Neutriiino
Re: Power Set of N-Intersection of finite subsets of N
Thanks Deveno, I got it!
Thank you too, Plato, for your rather Socratic method of instruction through questions. :)
well, here we go:
If F is a finite subset of $\mathbb{N}$, then $F\subseteq \mathbb{N}_n_m_a_x$, where nmax is the largest number in F. hence $F\notin P(\mathbb{N})-P(\mathbb{N}_n_m_a_x)$, $F\notin A_n$ (for any value of n)
If G is an infinite subset of N, for every n, $G\notin P(\mathbb{N}_n)$, hence $G\in P(\mathbb{N})-P(\mathbb{N}_n)$.
Right?
I found this in a previously held exam in our department, and now I'm wondering where I can find more problems like this.
Thanks, once again. | 2017-03-23T07:15:03 | {
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https://math.stackexchange.com/questions/2367841/what-is-the-best-way-to-solve-modular-arithmetic-equations-such-as-9x-equiv-33/2368266 | # What is the best way to solve modular arithmetic equations such as $9x \equiv 33 \pmod{43}$?
What is the best way to solve equations like the following:
$9x \equiv 33 \pmod{43}$
The only way I know would be to try all multiples of $43$ and $9$ and compare until I get $33$ for the remainder.
Is there a more efficient way ?
Help would be greatly appreciated!
• Divide both sides by $3$ and then multiply by $14$ – lab bhattacharjee Jul 22 '17 at 8:58
• @labbhattacharjee Thank you for your reply. Is there a more general definition for what you did? And what would be the solution? I know that the solution is $18$ but I don't see how you get that from what you wrote. – ViktorG Jul 22 '17 at 9:00
• See my answer here math.stackexchange.com/questions/407478/… – lab bhattacharjee Jul 22 '17 at 9:04
• Find $n$ such that $9n\equiv A \pmod {43}$ with $|A|<9.$ E.g let $9n<43<9(n+1)$ or $9(n-1)<43<9n.$ In this case take $n=5.$ Then $9x\equiv 33\iff 2x\equiv 45x\equiv (5)(33)=165 \equiv 36.$.... We could repeat this method, e.g. $2x\equiv 36 \iff x\equiv 44x\equiv (22)(2x) \equiv (22)(36),$ but the "common -divisor" short cut can be used when available : Since $\gcd(2,43)=1$ we have $2x\equiv 36 \iff (2)(x)\equiv (2)(18) \iff x\equiv 18.$..... Or we can use the "short-cut" at the start: $9x\equiv 33 \iff 3x\equiv 11 \iff -x\equiv 42x =(14)(3x)\equiv (14)(11)=154\equiv -18.$ – DanielWainfleet Jul 22 '17 at 19:51
How would we solve it in $$\mathbb{R}$$? Divide both sides by $$9$$ of course—or, in other words, multiply both sides by the multiplicative inverse of $$9$$. This setting is no different.
The challenge is knowing the multiplicative inverse of $$9$$ in $$\mathbb{Z}_{43}$$. What is key$$^\dagger$$ is that $$\gcd(9,43)=1$$, which guarantees integers $$n$$ and $$m$$ such that $$9n + 43m = 1$$. Modding out by $$43$$, we see that $$9n \equiv 1 \pmod{43}$$. Thus, multiplying both sides of $$9x \equiv 33 \pmod{43}$$ by $$n$$ gives us $$x$$.
The integers $$n$$ and $$m$$ can be found by using the extended Euclidean algorithm.
$$^\dagger$$ This coprimality condition is if-and-only-if. An integer $$x$$ will not have a multiplicative inverse $$(\text{mod} \ n)$$ if $$\gcd(x,n) \neq 1$$.
• Thank you very much! We just realized our mistake :D Have a great day ! – ViktorG Jul 22 '17 at 9:12
• @ViktorG What was your mistake? – Bill Dubuque Jul 22 '17 at 18:00
• @BillDubuque we already calculated the multiplicative inverse of 9 and didn't think about multiplying both sides with it to make it simpler – ViktorG Jul 22 '17 at 23:20
There no "best" way in general. The extended Euclidean algorithm is an efficient algorithmic way to compute modular inverses & fractions, but often there are quicker ways for small or special numbers.
We show $$5$$ ways to compute $$\ x\equiv 33(9^{-1})=: \dfrac{33^{\phantom{|}}\!}9\equiv\dfrac{-10}9\pmod{\!43} =$$ unique root of $$\, 9x\equiv 33$$
Cancel invertible factor $$3$$ then $$\rm\color{#c00}{twiddle}\,$$(add $$\,\pm 43j\,$$ to make division exact, cf. inverse reciprocity)
$$\dfrac{33}9\equiv \dfrac{\color{#c00}{11}}3 \equiv \dfrac{\color{#c00}{54}}3\equiv 18$$
Factor the fraction then $$\rm\color{#c00}{twiddle}$$ the top
$$\dfrac{-10}9\equiv \dfrac{\color{#c00}{-2}}9\ \dfrac{5}1\equiv\dfrac{\color{#c00}{-45}}9\ \dfrac{5}1\equiv -5\cdot 5\equiv 18$$
Gauss's algorithm
$$\dfrac{-10}9\equiv \dfrac{-50}{45}\equiv\dfrac{-50}2\equiv -25\equiv 18$$
Extended Euclidean algorithm in forward equational form, and associated fractional form
$$\begin{array}{rr} \bmod 43\!:\ \ \ \ \ \ \ \ [\![1]\!] &43\, x\,\equiv\ \ 0\ \\ [\![2]\!] &\ \color{#c00}{9\,x\, \equiv -10}\!\!\!\\ [\![1]\!]-5\,[\![2]\!] \rightarrow [\![3]\!] & \color{#0a0}{-2\,x\, \equiv\ \ 7}\ \\ [\![2]\!]+\color{orange}4\,[\![3]\!] \rightarrow [\![4]\!] & \color{#90f}{1\,x\, \equiv 18}\ \end{array}\qquad\qquad\qquad$$
\dfrac{0}{43}\ \overset{\large\frown}\equiv \underbrace{\color{#c00}{\dfrac{-10}{9}}\ \overset{\large\frown}\equiv \ \color{#0a0}{\dfrac{7}{-2}}\ \overset{\large\frown}\equiv\ \color{#90f}{\dfrac{18}{1}}} _{\!\!\!\Large \begin{align}\color{#c00}{-10}\ \ + \ \ &\!\color{orange}4\,(\color{#0a0}{\ \, 7\ \, }) \ \ \equiv \ \ \color{#90f}{18}\\ \color{#c00}{9}\ \ +\ \ &\!\color{orange}4\,(\color{#0a0}{-2} ) \ \ \equiv\ \ \ \color{#90f}{1}\end{align}}\quad
Beware $$\$$ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.
• See also this answer for a handful of methods applied to a similar problem. – Bill Dubuque Dec 1 '19 at 17:02 | 2020-10-31T16:32:37 | {
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https://math.stackexchange.com/questions/2497408/set-theory-universal-set | # Set Theory, Universal Set
If I have two sets A and B, given that $A := \{1, 4, 9, 16\}$ and $B := \{1, 8, 27\}$. Then is it correct to assume that the universal set is $U := \{1, 2, 3, ..., 27\}$?
Another thing, since the difference of two sets $A$ and $B$ is $A \cap B^c$, is $\mathcal{P}(A)-\mathcal{P}(B) = \mathcal{P}(A) \cap \mathcal{P}(B)^c$ or $\mathcal{P}(A) \cap \mathcal{P}(B^c)$? Thanks a lot!
This depends on the context. Usually the universal set is made clear beforehand. As for the second question, the former is true. Maybe it's a bit more clear if you set $C := \mathcal{P}(A)$ and $D := \mathcal{P}(B)$. Now apply the definition of $C - D$ and afterwards, just 'remember' what $C$ and $D$ where.
As a side note, I'd assume the universal set here is $\mathbb{N}_0$, but really there is no way to know without more information on what you're working with.
• I am given with just the two sets A and B. And I needed to work on the complement of A so I assumed I need a universal set. I also need to list down the elements of P(B-A) and P(B)-P(A). – frstrtdmthmtcn Oct 30 '17 at 23:55
• I see. If no context is provided, you will have to choose a universal set (and clearly state it before you proceed). Are you sure there's no contextual information that indicates a certain universal set? Sometimes this is assumed from the definition of the sets, e.g if I say "let $X,Y \subseteq \mathbb{R}$" with no additional information, you will assume $\mathbb{R}$ is the universal set here. – Guido A. Oct 31 '17 at 0:14
• The problem is "Given the sets A={1, 4, 9, 16} and B={1, 8, 27}. Determine the sets in P(B-A) and P(B)-P(A)." – frstrtdmthmtcn Oct 31 '17 at 0:27
• @frstrtdmthmtcn: Why do you need a universal set for that? – Asaf Karagila Oct 31 '17 at 0:27
• To get the complement of P(A). Is it wrong? – frstrtdmthmtcn Oct 31 '17 at 0:28
Let $A$, $B$, $U$, and $U^{'}$ be four sets satisfying
$\quad A \cup B \subset U$
and
$\quad A \cup B \subset U^{'}$
Then $A \cap B^c = A \cap B^{c^{'}}$, where $^c$ (resp. $^{c^{'}}$ is the complement in $U$ (resp. $U^{'}$).
So if you have only one definition for the set difference operation, and no indication for the universal set, simply set the context yourself and let $U = A \cup B$, defining the universal set to be the minimal set containing both $A$ and $B$.
"Then is it correct to assume that the universal set is U = {1, 2, 3, ..., 27}?"
The only thing you can know for certain is $\{1,4,8,9,16, 27\} \subseteq U$. But you don't need to know what the universal set is most of the time. If unstated it probably is simply anything that is conceivable.
.....
hmmm... $P(A) - P(B)$ are all the subset of $A$ that are not subsets of $B$.
That is $P(A) \cap P(B)^c$.
$P(B^c)$ and $P(B)^c$ are not the same. The former is the subsets of elements not in $B$ and the latter is everything that is not a subset of $B$. $P(B^c) \subset P(B)^c$ but $P(B)^c \not \subset P(B^c)$.
Example: if $A = \{1,2,3\}$ and $B= \{2,3,4\}$ and let the universal set $U = \{1,2,3,4,5\}$
$P(A) - P(B) = \{\{1\},\{1,2\},\{1,3\},\{1,2,3\}\}$
$P(B)^c =$ anything that isn't a subset of $B$ which if our universal set is the subsets of $U$ is everything that has $1$ or $5$ as an element.
$P(B^c) = P (\{1, 5\}) = \{\emptyset, \{1\},\{5\},\{1,5\}\}$
$P(A) \cap P(B)^c = P(A) - P(B)$ but $P(A) \cap P(B)^c =\{\emptyset, \{1\}\}$.
=======
In a comment in the other answer you wrote that the original problem is "The problem is "Given the sets A={1, 4, 9, 16} and B={1, 8, 27}. Determine the sets in P(B-A) and P(B)-P(A)"
Since $B- A = \{8,27\}$ the $P(B-A)$ are the subsets of $\{8,27\}$. i.e $\{\emptyset, \{8\},\{27\}, \{8,27\}\}$.
$P(B)$ are the $2^3 = 8$ subsets of $B$. $\{\emptyset, \{8\},\{27\}, \{8,27\}, \{1\},\{1,8\},\{1,27\}, \{1,8,27\}\}$
$P(A)$ are the subsets of $A$. I'm not going to list them. There are $2^{4} =16$ of them. But
$P(B) - P(A)$ are the subsets of $B$ that are not subsets of $A$. As $B$ and $A$ have only the element $1$ in common, the only subsets that have in common are $\emptyset$ and $\{1\}$. So those are removed from $P(B)$.
So $P(B) - P(A) =${ {8},{27}, {8,27},{1,8},{1,27}, {1,8,27}}\$
• What if I have sets A and B and the universal set U. If I get the complement of A x B, should I use U x U as basis? – frstrtdmthmtcn Oct 31 '17 at 1:43 | 2020-01-27T09:23:32 | {
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http://mathhelpforum.com/calculus/40950-finding-parametric-symmetric-equations-line.html | # Math Help - finding parametric & symmetric equations of line
1. ## finding parametric & symmetric equations of line
Have a question which is as above for line through points (0, 1/2, 1) and (2,1, -3)
They way I understand it this is what you do. Get a directional vector between the two points ie let (0, 1/2, 1) be point A and (2, 1, -3) be point B. Thus vector AtoB is < (2-0), (1-1/2), (-3-1)> giving < 2, 1/2, -4>
Now parametrics would be x= (0+2t) , y=( 1/2+1/2t) and z = (1+ (-4t))
giving symetric as being x/2 = (2y-1) = (z-1)/-4
But it looks like I'm wrong can someone please advise what I am doing wrong here.
2. Hello, Craka!
Find the equations for line through points $(0,\frac{1}{2}, 1)\text{ and }(2,1, -3)$
The way I understand it this is what you do.
Get a directional vector between the two points $A(0,\frac{1}{2}, 1)\text{ and }B(2, 1, -3)$
Thus: $\vec{v} \:=\:\overrightarrow{AB} \:=\:\langle 2-0,\:1-\frac{1}{2},\:-3-1\rangle \:=\:\langle 2,\:\frac{1}{2},\:-4\rangle$
The parametrics would be: . $\begin{Bmatrix} x&=& 0+2t \\ y& =& \frac{1}{2}+\frac{1}{2}t \\ z &=& 1-4t \end{Bmatrix}$
giving symmetric as: . $\frac{x}{2} \;=\; \frac{2y-1}{1} \;= \;\frac{z-1}{-4}$
But it looks like I'm wrong . . . . WHO said you're wrong?
There is an unavoidable problem with the equations of a line.
. . I wish textbook authors and math teachers would be own up to it.
There are a zillion possible forms of the equations of a line.
. . There is no "one right answer."
You used point $A(0,\frac{1}{2},1)$ and $\vec{v} = \langle 2,\frac{1}{2},1\rangle$ . . . and your answers are correct.
Suppose you used $B(2,1,-3)\text{ and }\vec{v} = \langle 2,\frac{1}{2},1\rangle$
You would have: . $\begin{Bmatrix} x &=& 2 + 2t \\ y &=& 1 + \frac{1}{2}t \\ z &=& \text{-}3 -4t \end{Bmatrix}\;\;\text{ and }\;\;\frac{x-2}{2} \:=\:\frac{2y-2}{1} \:=\:\frac{z+3}{-4}$ . . also correct.
$\text{Instead of }\,\vec{v} = \langle 2,\frac{1}{2},-4\rangle$, we can use a parallel vector: . $2\vec{v} = \langle 4,1,-8\rangle$
With point $B(2,1,-3)$, we would have: . $\begin{Bmatrix}x &=& 2 +4t \\ y &=& 1 + t \\ z&=& \text{-}3 -8t \end{Bmatrix}\;\;\text{ and }\;\;\frac{x-2}{4} \:=\:\frac{y-1}{1} \:=\:\frac{z+3}{-8}$
You see, the variations are endless . . .
3. Thankyou. That was causing me alot of anguish, as all the different sources I looked at seemed to have the method I was using, however the answer in the rear of my book was different to what I was getting. | 2014-08-22T09:52:08 | {
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http://openstudy.com/updates/5132a087e4b0034bc1d78d43 | ## appleduardo Group Title what is the integral of the following function? --> sec^4 (2x+1)dx I hope somebody can help me out. one year ago one year ago
1. appleduardo Group Title
$\int\limits_{}^{}\sec^4(2x+1)dx$
2. appleduardo Group Title
so far ive done this--> $\int\limits_{}^{}[1+\tan^2(2x)] [\sec^2(2x+1)]$
3. appleduardo Group Title
so then I get:$\int\limits_{}^{}\sec^2(2x+1) +\int\limits_{}^{}\tan^2(2x+1)\sec^2(2x+1)$
4. appleduardo Group Title
what is the next step? could somebody help me?
5. mathsmind Group Title
Use the reduction formula for m=4 $\int\limits \sec^m(x)dx=\frac{\sin(x)\sec^m-1(x)}{m-1}+\frac{m-2}{m-1}\int\limits \sec^{m-2}(x)dx$
6. mathsmind Group Title
now whatever answer u get multiply by 1/2 because $u =2x+1 \Longrightarrow du=2dx \longrightarrow dx=\frac{1}{2}du$
7. mathsmind Group Title
$\frac{1}{2}\int\limits\limits \sec^m(u)du=\frac{\sin(u)\sec^{m-1}(u)}{m-1}+\frac{m-2}{m-1}\int\limits\limits \sec^{m-2}(u)du$
8. mathsmind Group Title
now let m=4
9. mathsmind Group Title
$\frac{1}{2}\int\limits\limits\limits\limits \sec^4(u)du=\frac{\sin(u)\sec^{4-1}(u)}{4-1}+\frac{4-2}{4-1}\int\limits\limits\limits\limits \sec^{4-2}(u)du$
10. mathsmind Group Title
$=\frac{1}{2}[\frac{\sin(u)\sec^{3}(u)}{3}+\frac{2}{3}\int\limits\limits\limits\limits \sec^{2}(u)du]$
11. mathsmind Group Title
$=\frac{1}{6} \tan(u)\sec^2(x)+\frac{1}{3}\int\limits\limits \sec^2(u)du$
12. mathsmind Group Title
$=\frac{1}{6} \tan(u)\sec^2(u)+\frac{1}{3}\tan^2(u)+c$
13. mathsmind Group Title
u =2x+1
14. mathsmind Group Title
$=\frac{1}{6} \tan(2x+1)\sec^2(2x+1)+\frac{1}{3}\tan^2(2x+1)+c$
15. mathsmind Group Title
now using the reduction formula is useful because it reduces the amount of repetition of integration by parts, and simplifies the solution with less probability of errors...
16. mathsmind Group Title
An alternative way to solve this problem
17. mathsmind Group Title
there are many trig identities that can help in solving this problem one way is:
18. mathsmind Group Title
$\int\limits \sec^2(u)\sec^2(u)du=\frac{1}{2}\int\limits \sec^2(u)(1+\tan^2(u))du$
19. mathsmind Group Title
expand the function and distribute the integral
20. mathsmind Group Title
$=\frac{1}{2}\int\limits \sec^2(u)du + \frac{1}{2} \int\limits \sec^2(u)\tan^2(u)du$
21. mathsmind Group Title
$=\frac{1}{2}\tan(u) +\frac{1}{6} \tan^3(u)+c$
22. mathsmind Group Title
$=\frac{1}{2}\tan(2x+1) +\frac{1}{6} \tan^3(2x+1)+c$
23. mathsmind Group Title
if u think that the answers are different in the two methods, then that is not the case they are both the same u may use the trig identities to justify both solutions...
24. appleduardo Group Title
thank you very much mathsmind! i was kind of stressed, :/ but i have one queation:how did u get tan^3 ??in the final result.
25. mathsmind Group Title
ur welcome! i was expecting ur question i did that on purpose...
26. mathsmind Group Title
27. mathsmind Group Title
when u see a function with its derivative then u use the normal 1/(n+1tan^(n+1) like we do with 1/(n+1)x^(n+1)
28. mathsmind Group Title
the other way to visualize this concept is by assuming that tan(u) =v
29. mathsmind Group Title
dv = sec^2(u)du -> du=dv/sec^(u)
30. mathsmind Group Title
$\int\limits \tan^2(u)\sec^2(u)du=\int\limits v^2\sec^2(u)\frac{dv}{\sec^2(u)}$
31. mathsmind Group Title
now can u see the full picture...
32. mathsmind Group Title
(v^3)/3+c
33. mathsmind Group Title
but v=tan(u)
34. mathsmind Group Title
=(1/3)tan^3(u)+c, where u = 2x+1
35. mathsmind Group Title
hope this clarifies the point...
36. mathsmind Group Title
are u there?
37. appleduardo Group Title
yeep i am still here! i am trying to understand what you just typed! :P. uhmm i have another question, uhmm my teacher once told me that we can only use the reduction formula when "n" isnt a pair number, and in this case n=4, so uhmm is it right to solve it with that formula anyways?
38. mathsmind Group Title
Your teacher must have gone into a detailed answer, what happens is u need to consider if the function is an even one or an odd one, so because sec is an even function then u can use an even power...
39. mathsmind Group Title
So u can correct the information for ur teacher, if u think this is what she/he meant...
40. mathsmind Group Title
but anyways i mean if u r not familiar with the even and odd functions then use the 2nd method...
41. mathsmind Group Title
by the way this problem can be solved in many ways...
42. appleduardo Group Title
uhmm yeah i think i have to understand the second method first. i'll have to read it once again
43. mathsmind Group Title
just write it down on a piece of paper, you'll see how easy it is...
44. appleduardo Group Title
omG i just did and i understood it all! :D thank you very much mathsmind!, now id like to understand the reduction method, so uhmm what does "even" and "odd" fuctions mean ..?
45. mathsmind Group Title
if f(x) = (f-x) then that function is said to be even, e.g. x^2, x^4, x^6...etc or you may say x^n, where n = even number, also cos(x) is an even function, but u have to be careful u always need to perform a test and not make a mistake because for example (x+4)^2 is not an even function because $f(x) \neq f(-x)$
46. mathsmind Group Title
An odd function on the other hand is defined as -f(x)=f(-x), such as x^3, x^5, x^7, and sin(x) is also an odd function but x^3+1 is not an odd function...
47. mathsmind Group Title
one of the easiest to test if the function is odd or even is visualizing the graph, if the graph is symmetrical about the y-axis then its an even function, but if the graph is symmetrical about y=x then its an odd function, in other words symmetrical about the origin, however there are cases where the functions are neither odd or even...
48. appleduardo Group Title
:o cool!! :D but when i want to make use of the reduction formula.. is the same formula i have to use for an "odd" and an "even" function?
49. mathsmind Group Title
nope they are different and depends, it could be confusing if u don't have experience, i would say try solving the elementary way...
50. appleduardo Group Title
wow there's a lot of things to consider then and thank you so much for all your support. :P i have another question. When you solved this integral by the elementary way, you said tan^3 was the result of 1/(n+1)tan^(n+1) ... but why you did that? i mean, in order to get ...what??
51. mathsmind Group Title
that was the integration
52. mathsmind Group Title
the integral of tan^2(u)sec^(u)du
53. mathsmind Group Title
now when u differentiate tan u get sec
54. mathsmind Group Title
let me show u
55. mathsmind Group Title
$\frac{1}{2} \int\limits\limits \sec^2(u)\tan^2(u)du$
56. mathsmind Group Title
let $v = \tan(u) \therefore dv = \sec^2(u)du, so \space du=\frac{dv}{\sec^2(u)}$
57. mathsmind Group Title
$= \frac{1}{2} \int\limits\limits \sec^2(u)v^2\frac{dv}{\sec^2(u)}$
58. mathsmind Group Title
can u see how sec^2(u) is cancelled so u are left with
59. mathsmind Group Title
$\frac{1}{2} \int\limits v^2dv$
60. mathsmind Group Title
and u know the rest, or do u want me to finish it to the end...
61. appleduardo Group Title
yeep its alright! :D thank you! but if tan was originally tan^2 and then you took just tan to be your "u" function, it doesnt change the final result??
62. mathsmind Group Title
yes tan(u)=v therefore tan^2(u)=v^2
63. mathsmind Group Title
is that what u meant? or i misunderstood?
64. appleduardo Group Title
yeep! :D thanks, so if tan^2 were tan^3 then "V" would equal to v^3 in the final result?
65. mathsmind Group Title
well in the case of the question above v^4/4+c
66. mathsmind Group Title
tan^4(u)/4 +c
67. mathsmind Group Title
v^99
68. mathsmind Group Title
tan^100(u)/100+c
69. mathsmind Group Title
if u did not have the sec^2(u), then you can't do that...
70. mathsmind Group Title
that's what i meant of the function multiplied by its derivative
71. mathsmind Group Title
if i tell u to integrate sin(x)cos(x), what answer would i get?
72. appleduardo Group Title
i think i'd use -->:$\int\limits_{}^{}sen udu$ and since u=sen, and du=cos, then :$-\cos(x)+c$
73. appleduardo Group Title
is that correct??
74. mathsmind Group Title
u mean -(1/2)cos^2(x)+c
75. mathsmind Group Title
this is the whole point always when u integrate check if the function has its derivative...
76. mathsmind Group Title
what is the integral of sin(x)
77. appleduardo Group Title
its: -cos (u) +c :D
78. mathsmind Group Title
and what result did u get for sin(x)cos(x) again?
79. appleduardo Group Title
uhmm i thought the result was −cos(x)+c , but if i make use of $\frac{ 1 }{( n+1)(\cos ^{n+1})} =\frac{ 1 }{ 2\cos^2 }$ , then i'd get:$-\frac{ 1 }{ 2 }\cos^2(x)+c$
80. mathsmind Group Title
no never do that
81. mathsmind Group Title
a rcipical of cos never gives u a minus sign be careful never make such a mistake !
82. mathsmind Group Title
$\frac{1}{2}\frac{1}{\cos^2(x)}=\frac{1}{2}(\cos(x))^{-2}$
83. appleduardo Group Title
:o oohh, so is that the final result? +c
84. mathsmind Group Title
no that is the result of what you thought it was the result which is not the result in this case
85. mathsmind Group Title
$\int\limits \sin(x)\cos(x)dx = -\frac{1}{2}\cos^2(x)+c$
86. appleduardo Group Title
but how did u get the 1/2 and cos^2 ?
87. mathsmind Group Title
use the u sub and check for urself
88. appleduardo Group Title
yeep thanks! i'll try it :D :D | 2014-07-28T04:30:01 | {
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https://mathhelpboards.com/threads/show-that-k%E2%89%A51.27703/ | # [SOLVED]Show that K≥1
#### anemone
##### MHB POTW Director
Staff member
Let $a,\,b$ and $c$ be real numbers and let $K$ be the maximum of the function $y=|4x^3+ax^2+bx+c|$ in the interval $[-1,1]$. Show that $K\ge 1$. For which $a,\,b$ and $c$ is the equality occurs?
#### Opalg
##### MHB Oldtimer
Staff member
Let $p(x) = -ax^2 - bx - c$, so $p(x)$ is a quadratic (or lower degree) polynomial, and we want to find the smallest possible value of $K = \max\{|4x^3 - p(x)|:-1\leqslant x\leqslant 1\}.$
Notice that if $|4x^3 - p(x)| \leqslant K$ for $-1\leqslant x\leqslant 1$, then $|4(-x)^3 - p(-x)| = |4x^3 + p(-x)| \leqslant K$ for $-1\leqslant x\leqslant 1$. Therefore $|4x^3 - q(x)| \leqslant K$ for $-1\leqslant x\leqslant 1$, where $q(x) = \frac12(p(x) - p(-x))$. But $q(x)$ is an odd function. So to find the function that minimises $K$ we need only look at odd functions, and the only odd polynomials of degree at most $2$ are multiples of $x$.
Therefore in the polynomial $p(x) = -ax^2 - bx - c$ we should take $a=c=0$, and it is easy to see that the optimum value of $b$ is $-3$. Then (as in the diagram below) the graph of $y= 4x^3$ lies between the lines $y = 3x\pm1$ in the interval $[-1,1]$, and $1 = \max\{|4x^3 - 3x|:-1\leqslant x\leqslant 1\}$ is the smallest possible value for $K$.
\begin{tikzpicture}
[xscale=4,yscale=5]
\draw [step=0.25cm, help lines] (-1.1,-1.1) grid (1.1,1.1) ;
\draw (-1.1,0) -- (1.1,0) ;
\draw (0,-1.1) -- (0,1.1) ;
\draw[very thick, domain=-1.1:1.1] plot (\x,{\x^3});
\draw [thick] (-1.1, -0.825) -- (1.1,0.825) ;
\draw [dashed,thin] (-1.1, -1.075) -- (1.1,0.575) ;
\draw [dashed,thin] (-1.1, -0.575) -- (1.1,1.075) ;
\foreach \x in {-1,0.5,1} \draw (\x,-0.07) node [fill=black!8] {$\x$} ;
\foreach \y in {-4,-3,...,4} \draw (-0.13,\y/4) node [fill=black!8] {$\y$} ;
\end{tikzpicture}
Last edited: | 2020-07-05T18:03:36 | {
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https://math.stackexchange.com/questions/4026493/absolute-value-global-local-minima-and-maxima-piecewise-function/4026504 | absolute value global/local Minima and Maxima (piecewise function)
given the function $$f(x)= \begin{cases} |x-2|&\text {} \, 0 which of the following statements is true?(there can be more than one true or all false)
1. In the open interval $$(0,7)$$ there isn't a global maxima
2. In the open interval $$(4,5)$$ there is a global minima
3. The point $$x=3$$ is a local maxima or minima point in the interval $$(0,7)$$
What I tried:
First thing was to write it in a more comfortable way without the absolute value
$$f(x)= \begin{cases} -x+2&\text {} \, 0
then I checked if there is continuity at these points $$(x=2,x=3,x=5)$$ $$\lim \limits_{x \to 2^+}=0$$ $$\lim \limits_{x \to 2^-}=0$$ , $$\lim \limits_{x \to 3^+}=2$$ $$\lim \limits_{x \to 3^-}=1$$ (not continuous at x=3) and $$\lim \limits_{x \to 5^+}=0$$ $$\lim \limits_{x \to 5^-}=0$$.
then I calculated the derivative $$f'(x)= \begin{cases} -1&\text {} \, 0 checked the limits to see if it is differentiable at the points mentioned above and found that they are all critical points (sorry if it is not the right word but what I mean is "suspicious" points).
Did a small table to check if the points are minima or maxima. found out that $$x=2$$ is a minima , $$x=3$$ is a maxima and $$x=5$$ is also a minima but I cannot tell if local or global so I cannot tell if statements 1 or 2 are right since I cannot calculate $$f(2)$$ or $$f(5)$$ $$f(3)$$
Can anyone give me hints on how to continue from here? and if my way was correct? Thank you!
Edit: I just checked the limits of the interval but what can I learn from that? $$\lim \limits_{x \to 0^+}=2$$ $$\lim \limits_{x \to 7^-}=2$$
from what I personally know I can conclude from that if the limits are infinite
• You need also consider the end points $x=0, x=7$ and not the question focuses on *open *intervals – WA Don Feb 15 at 13:42
Actually, you can calculate $$f(2)$$ and $$f(5)$$. To check which of the two minima is the global one (both of them are local minima), you can see a bit more carefully the domains on which $$f$$ is defined. By doing this you should also discover that the function has an additional maximum, other than the one you found.
• Thank you for the help , I got that $f(7)=2$, $f(4)=1$ , $f(5)=0$ , $f(2)=0$ and $f(3)=2$ Does that mean $f(2)$ and $f(5)$ are local minimum and $f(7)$ and $f(3)$ are global maximum? if that is the case then statement 1 is false , statement 3 is false and the second statement is also false? or the second statement is true ? – Adamrk Feb 15 at 14:25
• $f(2)$ and $f(5)$ are local minima, while $f(3)$ and $f(7)$ are local maxima. The first question restricts you to the open interval $(0,7)$, where the only local maximum is $f(3)$. Thus, it is a global maximum and the first statement is false. In the open interval $(4,5)$ there is no local maximum or minimum, thus the second statement is false. Finally, as you said $f(3)$ is a local maximum in $(0,7)$ and the third statement is true. – Cyclops Feb 15 at 14:34 | 2021-06-25T01:46:48 | {
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http://math.stackexchange.com/questions/135723/integral-of-sin-x-cos-x-using-two-methods-differs-by-a-constant | # Integral of $\sin x \cos x$ using two methods differs by a constant?
$$\int \sin \theta\cos \theta~d \theta= \int \frac {1} 2 \sin 2\theta~ d \theta=-\frac {1} 4 \cos 2\theta$$ But, if I let $$u=\sin \theta , \text{ then }du=\cos \theta~d\theta$$ Then $$\int \sin \theta\cos \theta~d \theta= \int u ~ du =\frac { u^2 } 2 =\frac {1} 2 \sin^2 \theta$$ Since $$\sin^2 \theta =\frac {1} 2 - \frac {1} 2 \cos 2\theta$$ The above can be written as $$\int \sin \theta\cos \theta~d \theta= \frac {1} 2 \sin^2 \theta =\frac {1} 2 \left( \frac {1} 2 - \frac {1} 2 \cos 2\theta \right)=\frac {1} 4-\frac {1} 4 \cos 2\theta$$ Why are the two results differ by the constant $1/4$? Thank you.
-
Didn't your calculus instructor drum into your head to always write $+C$ when computing an indefinite integral? I guess your head could use a bit more drumming! – GEdgar Apr 23 '12 at 13:07
@Tony : Please notice my edits to your question. If you write 3\sin 5 in $\TeX$, the backslash on \sin not only prevents italicization, but also results in proper spacing before and after $\sin$, so you don't need to insert those spaces yourself. – Michael Hardy Apr 23 '12 at 14:57
@Michael Hardy, thank you very much for the useful info!! – Tony May 2 '12 at 2:43
The answer to the indefinite integration is the family of functions, which differ by a constant on every connected area of the domain.
That is, the correct way to write the answer to $\int f(x)dx$ (where $f$ is defined on a continuous area) is $g(x) + C$.
Note that $C-\dfrac{1}{4}\cos{2\theta}$ defines the same family of functions as $C+\dfrac{1}{4}-\dfrac{1}{4}\cos{2\theta}$.
-
A "series" of functions is a potentially confusing word choice, because "series" has a different technical meaning in analysis. Better to speak of a "family" of functions -- or just a "set" of them. – Henning Makholm Apr 23 '12 at 10:30
@HenningMakholm Of course you're right, it is just my knowledge of English which failed me... – penartur Apr 23 '12 at 10:35
@penartur, Thank you very much for your kind answers and comments!! – Tony Apr 23 '12 at 14:33
@Henning Makholm, Thank you very much for your comments!! – Tony Apr 23 '12 at 14:38
@penartur, Again thank you very much for your kind and detailed comments!! – Tony Apr 23 '12 at 14:43 | 2014-04-20T08:29:44 | {
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http://mathhelpforum.com/calculus/229750-integration.html | 1. ## Integration
Could any one possibly help me these questions. I have written the question and then my attempt at the solution next.
Find each of the following indefinite integrals, identifying any general rules
of calculus that you use.
[integral]xcos(1/3x)dx
Solving the equation by integration by parts. Let f(x)=x and g'(x)=cos(1/3x); then f'(x)=1 and g(x)=3 sin(1/3x) Substituting into the equation for integration by parts f(x)g(x)-∫f ' (x)g(x)dx. we obtain ∫xcos(1/3x)dx=3xsin(1/3x)-∫1(3 sin(1/3)x)=3xsin(1/3x)-3∫sin(1/3)x=3xsin(1/3 x)+9cos(1/3 x)+C
x/√1-x^4
Take u=1-x^4; thendu/dx=-4x^3. Hence∫x/√(1-x^4) dx=x/√u (-4x^3 )dx=x/√u du= -1/2 u^(-3/2). Plugging u back into the equation1/2 4x^3 ^(-3/2)
Explain why the graph of the function f(x) = (x2 + 1) ln(1/2x) lies
below the x-axis for 1 < x < 2 and above the x-axis for 2 < x < 3.
The graph of the function f(x)=(x^2+1) ln(1/2x) is positive above the x-axis and negative below the x-axis.
Use this fact to find the area enclosed by the graph and the x-axis
between x = 1 and x = 3, giving your answer to five decimal
places.
I have no idea how to answer this.
Find the volume of the solid of revolution obtained when the graph of
f(x)= sec x + tan x, from x=−[pie]/3 to x = [pie]/4
The volume is [pie]∫(-[pie]/3) ([pie]/4) secx+tanx=[pie](ln(tan(x)+sec(x) )+ln(sec(x)))=[pie]([(ln(tan(-[pie]/3)+sec(-[pie]/3)+ln(sec(-[pie]/3) ]-[(ln(tan([pie]/4)+sec([pie]/4)+ln(sec([pie]/4)]=[pie [ln(-√3+2)+ln(2)-[(ln(1+√2)+ln(√2)]=[pie][-0.63-1.2)=-1.83[pie]
2. ## Re: Integration
Hello, thomasthetankengine!
Thank you for showing your work!
$\int x\cos(\tfrac{x}{3})\,dx$
Solving the equation by integration by parts.
. . $\begin{Bmatrix}u &=& x && dv &=& \cos(\frac{x}{3}) \\ du &=&dx && v &=& 3\sin(\frac{x}{3})\end{Bmatrix}$
Then: . $\int x\cos(\tfrac{x}{3})\,dx \;=\;3x\sin(\tfrac{x}{3}) - 3\int\sin(\tfrac{x}{3})\,dx \;=\;3x\sin(\tfrac{x}{3}) + 9\cos(\tfrac{x}{3}) + C$
Excellent!
Good work!
$\int\frac{x\,dx}{\sqrt{1-x^4}}$
Let $u = x^2 \quad\Rightarrow\quad du = 2x\,dx \quad\Rightarrow\quad x\,dx \,=\,\tfrac{1}{2}du$
Substitute: . $\int \frac{\frac{1}{2}\,du}{\sqrt{1-u^2}} \;=\;\tfrac{1}{2}\int\frac{du}{\sqrt{1-u^2}} \;=\;\tfrac{1}{2}\arcsin u + C$
Back-substitute: . $\tfrac{1}{2}\arcsin(x^2) + C$
3. ## Re: Integration
Thanks. I can see where I have went wrong now. Is there any chance that you can help me with finding the volume of the solid of revolution obtained when the graph of secx+tanx, from x=-pie/3 to x=pie/4, is rotated about the x-axis. I understand that I need to compute the integral then set the limits of integration and then mulitply that by pie? The only problem that I'm having is computing the integral of secx+tanx. I have got secx+tanx=ln(tan(x)+sec(x)+ln(sec). I'm not sure if this is correct, I don't think it is. I'm not sure if the the integral of secx is ln(tan(x)+sec(x)) or ln(sec(x)+tan(x). Also the integral of tan(x) i assume that the integral of tan(x)=-lncos(x). I'm really not sure what i'm doing wrong. Anyhelp would be great. Thanks.
4. ## Re: Integration
Hello, thomasthetankengine!
You forgot to square the function.
$\text{Find the volume of the solid of revolution obtained when the graph}$
$\text{of }\,y \:=\:\sec x+\tan x\,\text{ from }x=\text{-}\tfrac{\pi}{3}\text{ to }x=\tfrac{\pi}{4}\,\text{ is rotated about the x-axis.}$
I'll omit the limits for now . . .
$V \;=\;\pi\int (\sec x + \tan x)^2\,dx \;=\;\pi \int\left(\sec^2\!x + 2\sec x\tan x + \tan^2\!x\right)\,dx$
. . $=\;\pi\left[\int\sec^2\!x\,dx + 2\int\sec x\tan x\,dx +\int\tan^2\!x\,dx\right]$
The first two integrals have standard formulas:
. . $\int \sec^2\!x\,dx \;=\;\tan x + C$
. . $\int\sec x\tan x\,dx \;=\;\sec x + C$
The third can handled like this:
. . $\int\tan^2\!x\,dx \;=\;\int(\sec^2\!x - 1)\,dx \;=\;\tan x - x+C$
5. ## Re: Integration
Originally Posted by thomasthetankengine
Thanks. I can see where I have went wrong now. Is there any chance that you can help me with finding the volume of the solid of revolution obtained when the graph of secx+tanx, from x=-pie/3 to x=pie/4, is rotated about the x-axis. I understand that I need to compute the integral then set the limits of integration and then mulitply that by pie? The only problem that I'm having is computing the integral of secx+tanx. I have got secx+tanx=ln(tan(x)+sec(x)+ln(sec). I'm not sure if this is correct, I don't think it is. I'm not sure if the the integral of secx is ln(tan(x)+sec(x)) or ln(sec(x)+tan(x). Also the integral of tan(x) i assume that the integral of tan(x)=-lncos(x). I'm really not sure what i'm doing wrong. Anyhelp would be great. Thanks.
Because I don't like using a lot of integration formulas, I tend to prefer converting things to sines and cosines as they're usually easy to integrate. I usually only remember \displaystyle \begin{align*} \int{\sin{(x)}\,\mathrm{d}x} = -\cos{(x)} + C, \int{\cos{(x)}\,\mathrm{d}x} = \sin{(x)} + C, \int{\sec^2{(x)}\,\mathrm{d}x} = \tan{(x)} + C \end{align*} and a few important identities (Pythagorean, Double Angle and Compound Angle).
Your volume will be calculated as:
\displaystyle \begin{align*} V &= \int_{-\frac{\pi}{3}}^{\frac{\pi}{4}}{ \pi \left[ \sec{(x)} + \tan{(x)} \right] ^2 \, \mathrm{d}x } \\ &= \pi \int_{-\frac{\pi}{3}}^{\frac{\pi}{4}}{ \sec^2{(x)} + 2\sec{(x)}\tan{(x)} + \tan^2{(x)}\,\mathrm{d}x } \\ &= \pi \int_{-\frac{\pi}{3}}^{\frac{\pi}{4}}{ \sec^2{(x)} + 2 \left[ \frac{1}{\cos{(x)}} \right] \left[ \frac{\sin{(x)}}{\cos{(x)}} \right] + \sec^2{(x)} - 1 \,\mathrm{d}x } \\ &= \pi \int_{-\frac{\pi}{3}}^{\frac{\pi}{4}}{ 2\sec^2{(x)} + \frac{2\sin{(x)}}{\cos^2{(x)}} - 1 \, \mathrm{d}x } \\ &= \pi \left\{ 2\int_{-\frac{\pi}{3}}^{\frac{\pi}{4}}{ \sec^2{(x)}\,\mathrm{d}x } - 2\int_{-\frac{\pi}{3}}^{\frac{\pi}{4}}{ \cos^{-2}{(x)} \left[ -\sin{(x)} \right]\,\mathrm{d}x } - \int_{-\frac{\pi}{3}}^{\frac{\pi}{4}}{1\,\mathrm{d}x} \right\} \\ &= \pi \left\{ 2\left[ \tan{(x)} \right]_{-\frac{\pi}{3}}^{\frac{\pi}{4}} - 2 \int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}}{u^{-2}\,\mathrm{d}u} - \left[ x \right]_{-\frac{\pi}{3}}^{\frac{\pi}{4}} \right\} \textrm{ after making the substitution } u = \cos{(x)} \implies \mathrm{d}u = -\sin{(x)}\,\mathrm{d}x \\ &= \pi \left\{ \left[ \tan{ \left( \frac{\pi}{4} \right) } - \tan{ \left( -\frac{\pi}{3} \right) } \right] - 2 \left[ -u^{-1} \right]_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} - \left[ \frac{\pi}{4} - \left( -\frac{\pi}{3} \right) \right] \right\} \\ &= \pi \left[ 1 + \sqrt{3} - 2 \left( -2 + \sqrt{2} \right) - \frac{7\pi}{12} \right] \\ &= \pi \left( 1 + \sqrt{3} + 4 - 2\sqrt{2} - \frac{7\pi}{12} \right) \\ &= \pi \left( 5 - 2\sqrt{2} + \sqrt{3} - \frac{7\pi}{12} \right) \end{align*}
P.S. I absolutely love your screen name
6. ## Re: Integration
May I ask why the function must be squared?
7. ## Re: Integration
I think I understand where I was going wrong now. I didn't square the function. The formula of the 'Volume of revolution'=[pie]int(f(x))^2dx.
8. ## Re: Integration
Just an aside: \displaystyle \begin{align*} \pi \end{align*} is the Greek letter "pi", not "pie". Interestingly, they use that symbol because it is their letter p, and this number is DEFINED as the Perimeter of a circle with a diameter of 1 unit.
9. ## Re: Integration
Here's what I got so far but i'm really not sure what next to do, or even if it's right.
The volume is π∫▒〖〖(sec〖x+tan〖x)〗 〗〗^2 dx=〗 π∫▒〖(sec^2〖x+2 sec〖x tan〖x+tan^2〖x)〗 〗 〗 〗=π(〗 ∫▒sec^2〖x dx+2∫▒sec〖x tan〖x dx+∫▒tan^2〖x dx=π(tan〖x+sec〖x+〗 〗 〗 〗 〗 〗 tan〖x+c=〗 〖〖(tan〗〖x+sec〖x+tan〖x-x〗)〗 〗〗_(-π/3)^(π/4)=π[〖tan π/4〗〖+sec〖π/4+tan〖π/4-π/4〗]-〖[tan〗〖-π/3+sec〖-π/3+tan〖-π/3∓π/3〗]=π[1+√2+1-π/4]-[+2〗 〗 〗 〗
10. ## Re: Integration
Originally Posted by thomasthetankengine
Here's what I got so far but i'm really not sure what next to do, or even if it's right.
The volume is π∫▒〖〖(sec〖x+tan〖x)〗 〗〗^2 dx=〗 π∫▒〖(sec^2〖x+2 sec〖x tan〖x+tan^2〖x)〗 〗 〗 〗=π(〗 ∫▒sec^2〖x dx+2∫▒sec〖x tan〖x dx+∫▒tan^2〖x dx=π(tan〖x+sec〖x+〗 〗 〗 〗 〗 〗 tan〖x+c=〗 〖〖(tan〗〖x+sec〖x+tan〖x-x〗)〗 〗〗_(-π/3)^(π/4)=π[〖tan π/4〗〖+sec〖π/4+tan〖π/4-π/4〗]-〖[tan〗〖-π/3+sec〖-π/3+tan〖-π/3∓π/3〗]=π[1+√2+1-π/4]-[+2〗 〗 〗 〗
Can I beg you to use Latex (see our help forum) or some other sort of math compiler? Or at least put it on several lines so it isn't bunched up like this. For example, I just tried to translate this out so it is more legible, but your errors in the parenthesis made this astoundingly painful until I scrolled back and found Prove It's post.
Here's the last line as best I can figure:
$= \pi \left [ tan \left ( \frac{\pi}{4} \right ) + sec \left ( \frac{\pi}{4} \right ) + tan \left ( \frac{\pi}{4} - \frac{\pi}{4} \right ) - \left ( \frac{\pi}{3} \right) + sec \left ( -\frac{\pi}{3} \right ) + tan \left ( -\frac{\pi}{3} \mp \frac{\pi}{3} \right ) \right ]$
$= \pi \left [ 1 + \sqrt{2} + 1 - \frac{\pi}{4} \right ] - (2)$
11. ## Re: Integration
Did you read my post? I put the complete solution... | 2017-04-29T12:23:26 | {
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https://physics.stackexchange.com/questions/383659/is-a-quadratic-majorana-hamiltonian-exactly-solvable | # Is a quadratic Majorana Hamiltonian exactly solvable?
Given $2N$ Majorana operators $\{a_i\}$ where $i=1,2,3,4,\cdots,2N$
The system Hamiltonian is the most general quadratic form:
$H=\sum A_{ij}a_i a_j$
where $\{a_i,a_j\}=2\delta_{ij} \quad a^\dagger_i=a_i$ is the Majorana operator comes from a set of fermion operators $c_i \ c^\dagger_i$ and $a_i=c^\dagger_i+c_i \quad a_{2i}=i(c^\dagger_i-c_i)$
$A_{ij}$ is chosen to guarantee that $H^\dagger=H$
So, such a Hamiltonian describes a lot of systems.
My question is:
(1)Since it's quadratic, is it exactly solvable? what is people's understanding of it?
By exactly solvable, my intuition is the analogy of the ordinary fermion quadratic problem $H=\sum h_{ij} c^\dagger_i c_j +h.c.$ , after linear transformation we can find a nice basis $H=\sum \epsilon_n v^\dagger_n v_n +...$ so, $v^\dagger_n v^\dagger_m \cdots |ground>$ and $<|c^\dagger_i c^\dagger_j \cdots|>$ can be trivially sovled via inverse linear transformation and expansion.
(2) let's choose $2L$ different Majorana operators, $a_{c_1},a_{c_2},a_{c_3},\cdots,a_{c_{2L}} \in \{a_i\}$
I want to calculate ground state average for the product of these operators.
$<G|\prod_{i=1}^{2L} a_{c_i}|G>=\text{complicated function of }\{c_{n};A_{ij}\}$
is this trivial or difficult ?
Is the "complicated function" linear function, polynomial, separable, involves determinant, pfaffian ?
I encounter this structure in my quantum Ising model research.
It would be helpful, if someone could show me similar problems as well.
If the problem is $H=\sum (B_{ij}c^\dagger_i c_j ＋h.c.)$, all the answer seems to be trivial.
• Sounds like you just have to 'diagonalize' the matrix $A_{ij}$. Consider to include your definition of 'Majorana operator' and 'exactly solvable'. – Qmechanic Feb 2 '18 at 9:57
• @Qmechanic The matrix $A_{ij}$ won't have eigenvectors that correspond to Majorana operators, unfortunately. A REAL linear combination of Majorana operators results in a new Majorana operator, but the eigenvectors will have complex entries which break the relation $a^\dagger=a$. – Jahan Claes Feb 2 '18 at 20:39
• Agreed. The word 'diagonalize' was meant as a shorthand for block diagonalize. – Qmechanic Feb 2 '18 at 21:18
• @Qmechanic Fair enough! – Jahan Claes Feb 2 '18 at 22:10
Let's start with your Hamiltonian $$i\sum_{jk} A_{jk} a_j a_k$$, where I've factored out an $$i$$ for convenience, and deduce some properties of the matrix $$A$$. Since the $${a}_{j}$$ anticommute, we can assume without loss of generality that $$A_{jk}$$ is antisymmetric. In fact, if $$A$$ is any matrix, than the antisymmetric part of $$A$$, $$\frac{1}{2}(A-A^T)$$, generates the same Hamiltonian as $$A$$ up to an irrelevant constant term. We also know $$H$$ is Hermitian, so that $$i\sum_{jk}A_{jk}a_ja_k=(i\sum_{jk}A_{jk}a_ja_k)^\dagger= -i\sum_{jk}A_{jk}^*a_ka_j=i\sum_{jk}A_{jk}^*a_ja_k$$. Thus, we conclude that the matrix $$A$$ must have all real entries. Therefore, $$A$$ is a real, antisymmetric matrix.
It is a theorem that real antisymmetric matrices can be put into a near-diagonal form by a real orthogonal transformation. To be precise,
$$A = O^T \left[\begin{matrix} 0 & \lambda_1 & &\cdots&&&0\\ -\lambda_1 & 0 & & &&&\vdots\\ & & 0 & \lambda_2 &&&\\ \vdots&&-\lambda_2&0\\ &&&&\ddots\\ &&&&&0&\lambda_n\\ 0&\cdots&&&&-\lambda_n&0 \end{matrix}\right]O$$
Using this decomposition, we can define new Majorana operators $$\bar a_i=\sum_j O_{ij}a_j$$. Using the fact that $$O$$ is real orthogonal, you can prove that these $$\bar a$$ are Majorana operators as well (they square to one, they are their own conjugate, and they anticommute with each other). In terms of these new operators, our Hamiltonian becomes very simple:
$$H= i\sum_n \lambda_n(\bar a_{2n}\bar a_{2n+1}-\bar a_{2n+1}\bar a_{2n})=2i\sum_n\lambda_n\bar a_{2n}\bar a_{2n+1}$$
We've thus taken a Hamiltonian that consisted of $$2n$$ coupled Majorana operators, and transformed it into a Hamiltonian that consists of $$n$$ decoupled systems, each with two Majorana operators.
We can then transform it back into a Fermion Hamiltonain by writing $$\bar{c}_n=\frac{\bar{a}_{2n}+i\bar{a}_{2n+1}}{2}$$ and noting that $$i\bar a_{2n}\bar a_{2n+1} = 2\bar{c}_n^\dagger\bar{c}_n+\text{constant}$$. Thus,
$$H = \sum_{n}4\lambda_n\bar{c}_n^\dagger\bar{c}_n+\text{constant}$$
Note that if you want to calculate the ground state expectation value of products of $$a$$ operators, it's not so bad. You can write your $$a$$ operators out in terms of the $$\bar{a}_i$$s, and you can figure out how the $$\bar{a}_i$$ act on the ground state.
• thank you. And by the way, is there a standard to classify matrix $A_{ij}$ represented in the site Majorana basis, thus classify the system? To me, it describes most quadratic solvable systems. All of them are linear transformation, but some are more "entangled" than others, when you mix creation and annihilation. – Jian Feb 5 '18 at 18:26
• Hey, I am wondering about the following: "Since the $a_i$ anticommute, we can assume without loss of generality that $A_{ij}$ is antisymmetric." What about the diagonal entries though? You wrote later the majoranas square to zero, but they actually square to one, so is it that we just neglect the constant off-set by the diagonal when we assume $A$ to be anti-symmetric? – Marsl Dec 17 '18 at 10:50
• @Marsi Hi, you're right that I should have written "square to one", that was a silly mistake! And yes, there could be a constant offset if $A$ has diagonal entries, but all that does is shift all the energies, so it can be neglected in the solution. – Jahan Claes Dec 17 '18 at 15:38
• @Marsl Actually, looking over the whole thing, I discovered a few different typos. Might want to look over my edited answer if this is something you're working on. Or just read Kitaev's paper that I linked! – Jahan Claes Dec 17 '18 at 15:49 | 2021-08-01T14:36:12 | {
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http://mathhelpforum.com/algebra/5345-evaluation-non-linear-series-print.html | Evaluation a non-linear series?
• Sep 5th 2006, 05:06 PM
chancey
Evaluation a non-linear series?
How do I evaluate $\sum_{n=1}^{20} (2^n - 1)$ without havng to add up each element?
Thanks
• Sep 5th 2006, 05:47 PM
ThePerfectHacker
Quote:
Originally Posted by chancey
How do I evaluate $\sum_{n=1}^{20} (2^n - 1)$ without havng to add up each element?
Thanks
This is a geometric series.
---
$(2^1+2^2+...+2^{20})-(1+1...+1)$
Thus,
$2(1+2+2^2+...+2^{19})-20$
Use geometric sum formula,
$2(2^{20}-1)-20$
Simplify,
$2^{21}-22$
• Sep 5th 2006, 06:50 PM
chancey
Quote:
Originally Posted by ThePerfectHacker
This is a geometric series.
---
$(2^1+2^2+...+2^{20})-(1+1...+1)$
Thus,
$2(1+2+2^2+...+2^{19})-20$
Use geometric sum formula,
$2(2^{20}-1)-20$
Simplify,
$2^{21}-22$
Right, but thats not the answer in the back of the book...
I did it this way:
$\sum_{n=1}^{20} (2^n - 1) = \sum_{n=1}^{20} (2^n) - 20$
$\sum_{n=1}^{20} (2^n) = \frac{1-2^{21}}{1-2} - 1 = 2097150$
$\because \sum_{n=0}^{k} (ar^n) = \frac{a(1-r^{k+1})}{1-r}$
$\therefore \sum_{n=1}^{20} (2^n) - 20 = 2097150 - 20 = 2097130$
Which is the same answer as yours, however the answer in the book says $2097170$. They must of added the 20 instead of subtracted it. OK thanks
• Sep 5th 2006, 06:53 PM
ThePerfectHacker
Quote:
Originally Posted by chancey
Right, but thats not the answer in the back of the book...
I did it this way:
$\sum_{n=1}^{20} (2^n - 1) = \sum_{n=1}^{20} (2^n) - 20$
$\sum_{n=1}^{20} (2^n) = \frac{1-2^{21}}{1-2} - 1 = 2097150$
$\because \sum_{n=0}^{k} (ar^n) = \frac{a(1-r^{k+1})}{1-r}$
$\therefore \sum_{n=1}^{20} (2^n) - 20 = 2097150 - 20 = 2097130$
Which is the same answer as yours, however the answer in the book says $2097170$. They must of added the 20 instead of subtracted it. OK thanks
Then the book is inncorect. | 2016-09-30T06:50:05 | {
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"lm_q1_score": 0.9777138131620183,
"lm_q2_score": 0.861538211208597,
"lm_q1q2_score": 0.8423378096655416
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http://cnx.org/content/m19840/1.3/ | # Connexions
You are here: Home » Content » Lab 3: Convolution and Its Applications
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# Lab 3: Convolution and Its Applications
Note: You are viewing an old version of this document. The latest version is available here.
This lab involves experimenting with the convolution of two continuous-time signals. The main mathematical part is written as a .m file, which is then used as a LabVIEW MathScript node within the LabVIEW programming environment to gain user interactivity. Due to the discrete-time nature of programming, an approximation of the convolution integral is needed. As an application of the convolution concept, echoes are removed from speech recordings using this concept.
## Numerical Approximation of Convolution
In this section, let us apply the LabVIEW MathScript function conv to compute the convolution of two signals. One can choose various values of the time interval ΔΔ size 12{Δ} {} to compute numerical approximations to the convolution integral.
### Convolution Example 1
In this example, use the function conv to compute the convolution of the signals x(t)=exp(at)u(t)x(t)=exp(at)u(t) size 12{x $$t$$ ="exp" $$- ital "at"$$ u $$t$$ } {} and h(t)=exp(bt)u(t)h(t)=exp(bt)u(t) size 12{h $$t$$ ="exp" $$- ital "bt"$$ u $$t$$ } {}with u(t)u(t) size 12{u $$t$$ } {}representing a step function starting at 0 for 0t80t8 size 12{0 <= t <= 8} {}. Consider the following values of the approximation pulse width or delta: Δ=0.5,0.1,0.05,0.01,0.005,0.001Δ=0.5,0.1,0.05,0.01,0.005,0.001 size 12{Δ=0 "." 5,0 "." 1,0 "." "05",0 "." "01",0 "." "005",0 "." "001"} {}. Mathematically, the convolution of h(t)h(t) size 12{h $$t$$ } {}and x(t)x(t) size 12{x $$t$$ } {}is given by
y(t)=1ab(ebteat)u(t)y(t)=1ab(ebteat)u(t) size 12{y $$t$$ = { {1} over {a - b} } $$e rSup { size 8{ - ital "bt"} } - e rSup { size 8{ - ital "at"} }$$ u $$t$$ } {}
(1)
Compare the approximation yˆ()yˆ() size 12{ { hat {y}} $$nΔ$$ } {}obtained via the function conv with the theoretical value y(t)y(t) size 12{y $$t$$ } {}given by Equation (1). To better see the difference between the approximated yˆ()yˆ() size 12{ { hat {y}} $$nΔ$$ } {}and the true yˆ()yˆ() size 12{ { hat {y}} $$nΔ$$ } {}values, display yˆ(t)yˆ(t) size 12{ { hat {y}} $$t$$ } {}and y(t)y(t) size 12{y $$t$$ } {} in the same graph.
Compute the mean squared error (MSE) between the true and approximated values using the following equation:
MSE=1Nn=1N(y()yˆ())2MSE=1Nn=1N(y()yˆ())2 size 12{ ital "MSE"= { {1} over {N} } Sum cSub { size 8{n=1} } cSup { size 8{N} } { $$y \( nΔ$$ - { hat {y}} $$nΔ$$ \) rSup { size 8{2} } } } {}
(2)
where N=TΔN=TΔ size 12{N= left lfloor { {T} over {Δ} } right rfloor } {}, T is an adjustable time duration expressed in seconds and the symbol .. size 12{ left lfloor "." right rfloor } {} denotes the nearest integer. To begin with, set T=8T=8 size 12{T=8} {}.
As you can see here, the main program is written as a .m file and placed inside LabVIEW as a LabVIEW MathScript node by invoking Functions Programming Structures MathScript. The .m file can be typed in or copied and pasted into the LabVIEW MathScript node. The inputs to this program consist of an approximation pulse width ΔΔ size 12{Δ} {}, input exponent powers aa size 12{a} {}and bb size 12{b} {} and a desired time duration TT size 12{T} {}. To add these inputs, right-click on the border of the LabVIEW MathScript node and click on the Add Input option as shown in Figure 1.
After adding these inputs, create controls to allow one to alter the inputs interactively via the front panel. By right-clicking on the border, add the outputs in a similar manner. An important consideration is the selection of the output data type. Set the outputs to consist of MSE, actual or true convolution output y_ac and approximated convolution output y. The first output is a scalar quantity while the other two are one-dimensional vectors. The output data types should be specified by right-clicking on the outputs and selecting the Choose Data Type option (see Figure 2).
Next write the following .m file textual code inside the LabVIEW MathScript node:
t=0:Delta:8;
Lt=length(t);
x1=exp(-a*t);
x2=exp(-b*t);
y=Delta*conv(x1,x2);
y_ac=1/(a-b)*(exp(-b*t)-exp(-a*t));
MSE=sum((y(1:Lt)-y_ac).^2)/Lt
With this code, a time vector t is generated by taking a time interval of Delta for 8 seconds. Convolve the two input signals, x1 and x2, using the function conv. Compute the actual output y_ac using Equation (1). Measure the length of the time vector and input vectors by using the command length(t). The convolution output vector y has a different size (if two input vectors m and n are convolved, the output vector size is m+n-1). Thus, to keep the size the same, use a portion of the output corresponding to y(1:Lt) during the error calculation.
Use a waveform graph to show the waveforms. With the function Build Waveform (Functions → Programming → Waveforms → Build Waveforms), one can show the waveforms across time. Connect the time interval Delta to the input dt of this function to display the waveforms along the time axis (in seconds).
Merge together and display the true and approximated outputs in the same graph using the function Merge Signal (Functions → Express → Signal Manipulation → Merge Signals). Configure the properties of the waveform graph as shown in Figure 3.
Figure 4 illustrates the completed block diagram of the numerical convolution.
Figure 5 shows the corresponding front panel, which can be used to change parameters. Adjust the input exponent powers and approximation pulse-width Delta to see the effect on the MSE.
### Convolution Example 2
Next, consider the convolution of the two signals x(t)=exp(2t)u(t)x(t)=exp(2t)u(t) size 12{x $$t$$ ="exp" $$- 2t$$ u $$t$$ } {}and h(t)=rect(t22)h(t)=rect(t22) size 12{h $$t$$ = ital "rect" $${ {t - 2} over {2} }$$ } {} for , where u(t)u(t) size 12{u $$t$$ } {}denotes a step function at time 0 and rect a rectangular function defined as
rect(t)={10.5t<0.50otherwiserect(t)={10.5t<0.50otherwise size 12{ ital "rect" $$t$$ = left lbrace matrix { 1 {} # - 0 "." 5 <= t<0 "." 5 {} ## 0 {} # ital "otherwise"{} } right none } {}
(3)
Let Δ=0.01Δ=0.01 size 12{Δ=0 "." "01"} {}. Figure 6 shows the block diagram for this second convolution example. Again, the .m file textual code is placed inside a LabVIEW MathScript node with the appropriate inputs and outputs.
Figure 7 illustrates the corresponding front panel where x(t)x(t) size 12{x $$t$$ } {}, h(t)h(t) size 12{h $$t$$ } {} and x(t)h(t)x(t)h(t) size 12{x $$t$$ * h $$t$$ } {} are plotted in different graphs. Convolution ()() size 12{ $$*$$ } {} and equal (=)(=) size 12{ $$=$$ } {}signs are placed between the graphs using the LabVIEW function Decorations.
### Convolution Example 3
In this third example, compute the convolution of the signals shown in Figure 8.
Figure 9 shows the block diagram for this third convolution example and Figure 10 the corresponding front panel. The signals x1(t)x1(t) size 12{x1 $$t$$ } {}, x2(t)x2(t) size 12{x2 $$t$$ } {} and x1(t)x2(t)x1(t)x2(t) size 12{x1 $$t$$ * x2 $$t$$ } {} are displayed in different graphs.
## Convolution Properties
In this part, examine the properties of convolution. Figure 11 shows the block diagram to examine the properties and Figure 12 and Figure 13 the corresponding front panel. Both sides of equations are plotted in this front panel to verify the convolution properties. To display different convolution properties within a limited screen area, use a Tab Control (Controls ModernContainersTab Control) in the front panel.
## Linear Circuit Analysis Using Convolution
In this part, let us consider an application of convolution in analyzing RLC circuits to gain a better understanding of the convolution concept. A linear circuit denotes a linear system, which can be represented with its impulse response h(t)h(t) size 12{h $$t$$ } {}, that is, its response to a unit impulse input. The input to such a system can be considered to be a voltage v(t)v(t) size 12{v $$t$$ } {}and the output to be the circuit current i(t)i(t) size 12{i $$t$$ } {}. See Figure 14.
For a simple RC series circuit shown in Figure 15, the impulse response is given by (Reference) ,
h(t)=1RCexp(1RCt)h(t)=1RCexp(1RCt) size 12{h $$t$$ = { {1} over {RC} } "exp" $$- { {1} over { ital "RC"} } t$$ } {}
(4)
which can be obtained for any specified values of R and C. When an input voltage v(t)v(t) size 12{v $$t$$ } {} (either DC or AC) is applied to the system, the circuit current i(t)i(t) size 12{i $$t$$ } {} can be obtained by simply convolving the system impulse response with the input voltage, that is
i(t)=h(t)v(t)i(t)=h(t)v(t) size 12{i $$t$$ =h $$t$$ * v $$t$$ } {}
(5)
Similarly, for the simple RL series circuit shown in Figure 16, the impulse response is given by (Reference) ,
h(t)=RLexp(RLt)h(t)=RLexp(RLt) size 12{h $$t$$ = { {1} over {L} } "exp" $$- { {R} over {L} } t$$ } {}
(6)
When an input voltage v(t)v(t) size 12{v $$t$$ } {} is applied to the system, the circuit current i(t)i(t) size 12{i $$t$$ } {} can be obtained by computing the convolution integral.
Figure 17 shows the block diagram of this linear system and Figure 18 the corresponding front panel. From the front panel, one can control the system type (RL or RC), input voltage type (DC or AC) and input voltage amplitude. One can also observe the system response by changing R, L and C values. Three graphs are used to display the input voltage v(t)v(t) size 12{v $$t$$ } {}, impulse response of the circuit h(t)h(t) size 12{h $$t$$ } {} and circuit current i(t)i(t) size 12{i $$t$$ } {}.
## Lab Exercises
### Exercise 1
Echo Cancellation
In this exercise, consider the problem of removing an echo from a recording of a speech signal. The LabVIEW MathScript function sound() or the function Play Waveform in LabVIEW can be used to play back the speech recording. To begin, load the .m file echo_1.wav provided on the book website by using the function wavread(‘filename’). This speech file was recorded at the sampling rate of 8 kHz, which can be played back through the computer speakers by typing
>> sound(y)
You should be able to hear the sound with an echo. If the LabVIEW function Play Waveform(Functions Programming Graphics & Sound Sound Output Play Waveform) is used to play the sound, you first need to build a waveform based on the loaded data and the time interval dt=1/8000dt=1/8000 size 12{ ital "dt"=1/"8000"} {} because this speech was recorded using an 8 kHz sampling rate. Connect the waveform to the function Play Waveform.
An echo is produced when the signal (speech, in this case) is reflected off a non-absorbing surface like a wall. What is heard is the original signal superimposed on the signal reflected off the wall (echo). Because the speech is partially absorbed by the wall, it decreases in amplitude. It is also delayed. The echoed signal can be modeled as ax(tτ)ax(tτ) size 12{ ital "ax" $$t - τ$$ } {}where a<1a<1 size 12{a<1} {} and ττ size 12{τ} {} denotes the echo delay. Thus, one can represent the speech signal plus the echoed signal as [7]
y(t)=x(t)+ax(tτ)y(t)=x(t)+ax(tτ) size 12{y $$t$$ =x $$t$$ + ital "ax" $$t - τ$$ } {}
(7)
What is heard is y(t)y(t) size 12{y $$t$$ } {}. In many applications, it is important to recover x(t)x(t) size 12{x $$t$$ } {} – the original, echo-free signal – from y(t)y(t) size 12{y $$t$$ } {}.
Method 1
In this method, remove the echo using deconvolution. Rewrite Equation (7) as follows [7]:
y[]=x[]+ax[(nN)Δ]=x[](δ[]+[nN]Δ)=x[]h[]y[]=x[]+ax[(nN)Δ]=x[](δ[]+[nN]Δ)=x[]h[] size 12{y $nΔ$ =x $nΔ$ + ital "ax" $$$n - N$$ Δ$ =x $nΔ$ * $$δ $nΔ$ +aδ $n - N$ Δ$$ =x $nΔ$ * h $nΔ$ } {}
(8)
The echoed signal is the convolution of the original signal x()x() size 12{x $$nΔ$$ } {} and the signal h()h() size 12{h $$nΔ$$ } {}. Use the LabVIEW MathScript function deconv(y,h) to recover the original signal.
Method 2
An alternative way of removing the echo is to run the echoed signal through the following system:
z[]=y[]az[(nN)Δ]z[]=y[]az[(nN)Δ] size 12{z $nΔ$ =y $nΔ$ - ital "az" $$$n - N$$ Δ$ } {}
(9)
Assume that z[]=0z[]=0 size 12{z $nΔ$ =0} {}for n<0n<0 size 12{n<0} {}. Implement the above system for different values of aa size 12{a} {} and NN size 12{N} {}.
Display and play back the echoed signal and the echo-free signal using both of the above methods. Specify the parameters aa size 12{a} {}and NN size 12{N} {}as controls. Try to measure the proper values of aa size 12{a} {}and NN size 12{N} {}by the autocorrelation method described below.
The autocorrelation of a signal can be described by the convolution of a signal with its mirror. That is,
Rxx[n]=x[n]x[n]Rxx[n]=x[n]x[n] size 12{R rSub { size 8{ ital "xx"} } $n$ =x $n$ * x $- n$ } {}
(10)
Use the autocorrelation of the output signal (echo-free signal) to estimate the delay time ( NN size 12{N} {}) and the amplitude of the echo ( aa size 12{a} {}). For different values of NN size 12{N} {}and aa size 12{a} {}, observe the autocorrelation output. To have an echo-free signal, the side lobes of the autocorrelation should be quite low, as shown in Figure 19.
Figure 20 shows a typical front panel for this exercise. It is not necessary to obtain the same front panel but there should be controls for aa size 12{a} {} and NN size 12{N} {} as well as graphs to observe the echoed signal, echo-free signal and autocorrelation function of the echo-free signal.
#### Solution
Insert Solution Text Here
### Exercise 2
Noise Reduction Using Mean Filtering
The idea of mean filtering is simply to replace each value in a signal with the mean (average) value of its neighbors. A mean filter is widely used for noise reduction.
Start by adding some random noise to a signal (use the file echo_1.wav or any other speech data file). Then, use mean filtering to reduce the introduced noise. More specifically, take the following steps:
1. Normalize the signal values in the range [0 1].
2. Add random noise to the signal by using the function randn. Set the noise level as a control.
3. Convolve the noise-added signal with a mean filter. This filter can be designed by taking an odd number of ones and dividing by the size. For example, a 1×31×3 size 12{1 times 3} {} size mean filter is given by [1/3 1/3 1/3] and a 1×51×5 size 12{1 times 5} {}size mean filter by [1/5 1/5 1/5 1/5 1/5]. Set the size of the mean filter as an odd number control (3, 5 or 7, for example).
#### Solution
Insert Solution Text Here
### Exercise 3
Impulse Noise Reduction Using Median Filtering
A median filter is a non-linear filter that replaces a data value with the median of the values within a neighboring window. For example, the median value for this data stream [2 5 3 11 4] is 4. This type of filter is often used to remove impulse noise. Use the file echo_1.wav or any other speech data file and take the following steps:
1. Normalize the signal values in the range [0 1].
2. Randomly add impulse noise to the signal by using the LabVIEW MathScript function randperm. Set the noise density as a control.
3. Find the median values of neighboring data using the function median` and replace the original value with the median value. Set the number of neighboring values as an odd number control (3, 5 or 7, for example).
#### Solution
Insert Solution Text Here
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| External bookmarks | 2013-06-19T13:47:52 | {
"domain": "cnx.org",
"url": "http://cnx.org/content/m19840/1.3/",
"openwebmath_score": 0.6266806721687317,
"openwebmath_perplexity": 3395.228428659996,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9777138151101525,
"lm_q2_score": 0.8615382076534743,
"lm_q1q2_score": 0.8423378078680412
} |
https://mathhelpboards.com/threads/degree-of-extension-invariant-upto-isomorphism.5314/ | # Degree of extension invariant upto isomorphism?
#### caffeinemachine
##### Well-known member
MHB Math Scholar
Let $K$ be a field and $F_1$ and $F_2$ be subfields of $K$. Assume that $F_1$ and $F_2$ are isomorphic as fields. Further assume that $[K:F_1]$ is finite and is equal to $n$.
Is it necessary that $[K:F_2]$ is finite and is equal to $n$??
___
I have not found this question in a book so I don't know the answer to the above question. I could not construct a counterexample.
If the isomorphism between $F_1$ and $F_2$ can be extended to an automorphism of $K$ (this probably required additional hypothesis) then the result can be proved in the affirmative.
#### PaulRS
##### Member
Here is an idea:
Let $F = {\mathbb Q}$ (to fix ideas). Consider $F_1 = F(x)$ and $F_2 = F(x^2)$, and finally $K = F(x)$. The field $F_1$ should be isomorphic to $F_2$ since $x^2$ is trascendental over $F$.
Clearly $K/F_1$ has degree 1. However .
#### caffeinemachine
##### Well-known member
MHB Math Scholar
Here is an idea:
Let $F = {\mathbb Q}$ (to fix ideas). Consider $F_1 = F(x)$ and $F_2 = F(x^2)$, and finally $K = F(x)$. The field $F_1$ should be isomorphic to $F_2$ since $x^2$ is trascendental over $F$.
Clearly $K/F_1$ has degree 1. However .
So $K/F_2$ doesn't have degree $1$ simply because $1$ and $x$ in $K$ are LI over $F_2$. Is that right?
#### PaulRS
##### Member
So $K/F_2$ doesn't have degree $1$ simply because $1$ and $x$ in $K$ are LI over $F_2$. Is that right?
Right. Otherwise $x\in F_2$ and so $x = f(x^2) / g(x^2)$ for some polynomials $f$ and $g$ ($g\neq 0$), which you can check is nonsense.
#### caffeinemachine
##### Well-known member
MHB Math Scholar
Right. Otherwise $x\in F_2$ and so $x = f(x^2) / g(x^2)$ for some polynomials $f$ and $g$ ($g\neq 0$), which you can check is nonsense.
That's great Paul.
I have one more question.
I think that $[K:F_2]$ is finite.
My Argument:
Clearly $K=F_2(x)$. Now define a polynomial $p(y)=x^2y^3-x^4y$ in $F_2[y]$. Clearly $p(x)=0$. Thus $[K:F_2]$ is finite.
Is this okay?
#### PaulRS
##### Member
That's great Paul.
I have one more question.
I think that $[K:F_2]$ is finite.
My Argument:
Clearly $K=F_2(x)$. Now define a polynomial $p(y)=x^2y^3-x^4y$ in $F_2[y]$. Clearly $p(x)=0$. Thus $[K:F_2]$ is finite.
Is this okay?
Correct.
There is a simpler polynomial $p(y)\in F_2[y]$ such that $p(x) = 0$, can you find it?
#### caffeinemachine
##### Well-known member
MHB Math Scholar
Correct.
There is a simpler polynomial $p(y)\in F_2[y]$ such that $p(x) = 0$, can you find it?
$p(y)=y^2-x^2$. That's just awesome!! | 2020-09-28T12:05:02 | {
"domain": "mathhelpboards.com",
"url": "https://mathhelpboards.com/threads/degree-of-extension-invariant-upto-isomorphism.5314/",
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"openwebmath_perplexity": 309.90783102563705,
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http://math.stackexchange.com/questions/41983/does-convexity-of-a-norm-imply-the-triangle-inequality | # Does convexity of a 'norm' imply the triangle inequality?
Given a vector space $V$ (for convenience, defined over $\mathbb{r}$), we call $d:V\rightarrow\mathbb{R}$ a norm for $V$ if $\forall \mathbf{u}, \mathbf{v} \in V$ and $\forall r \in \mathbb{R}$ we have:
1. $d(r \mathbf{v}) = |r|d(\mathbf{v})$,
2. $d(\mathbf{v})\ge 0$, with equality iff $\mathbf{v} = 0$, and
3. $d(\mathbf{u})+d(\mathbf{v}) \ge d(\mathbf{u}+\mathbf{v})$ (triangle inequality)
I've read in a few places that an important property of a norm is that it is convex; that is, given $\mathbf{u},\mathbf{v} \in V$, and $p \in (0,1)$, we have $d(p \mathbf{u} + (1-p) \mathbf{v}) \le p d(\mathbf{u}) + (1-p) d(\mathbf{v})$. This clearly follows from the triangle inequality.
My question is: Does the reverse also hold? i.e. does a function satisfying (1) and (2) above which is convex necessarily satisfy the triangle inequality? If not, what is an instructive counterexample?
Thanks! (btw: please feel free to suggest better tags / improvements to the question; I'm new to this!)
-
Set $p = \frac{1}{2}$ and use property 1. – Qiaochu Yuan May 29 '11 at 16:58
@Qiaochu Yuan: Thanks! Feeling a bit silly now! I even tried that at some point, and convinced myself it didn't work. :) – duncanm May 29 '11 at 17:09
It's cool to know that a convex-(1)-(2) function is equivalent to being a norm... I didn't know the triangle inequality was equivalent to convexity under (1) and (2). Good to know. Thumbs up! – Patrick Da Silva May 29 '11 at 17:14
@Patrick: Cheers! – duncanm May 29 '11 at 19:23
@Patrick: The homogeneity condition $d(r\mathbf{v}) = |r| d(\mathbf{v})$ is of course crucial here. Translation invariance and the triangle inequality are not enough to infer homogeneity. Some time ago I wrote an answer trying to elucidate the relations between scalar products, norms and metrics, maybe you find that interesting. – t.b. May 29 '11 at 21:06
add comment
## 1 Answer
Resolved in comments.
Setting $p=\frac12$ in the definition of convexity, we have $$d\Big( \frac{\mathbf u + \mathbf v}{2} \Big) \leqslant \frac12 d(\mathbf u) + \frac12 d(\mathbf v).$$ By the scaling or homogeneity, the left hand side is simply $\frac12 d(\mathbf u + \mathbf v)$; plugging in this and simplifying, we get the triangle inequality.
-
add comment | 2013-12-20T15:50:22 | {
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https://www.physicsforums.com/threads/probability-of-forming-an-increasing-geometric-progression.829466/ | # Probability of forming an increasing Geometric Progression
Tags:
1. Aug 26, 2015
### Titan97
1. The problem statement, all variables and given/known data
If three number are chosen randomly from the set ${1,3,3^2,......3^n}$ without replacement, then the probability that they form an increasing geometric progression is?
(a) 3/2n if n is odd
(b) 3/2n if n is even
(c)3n/2(n² -1) if n is even
(d) 3n/2(n² -1) if n is odd
2. Relevant equations
None
3. The attempt at a solution
This problem can be easily solved if I chose the end terms first.
Let $3^a$ be the first term and $3^b$ be the third term. Then the middle term has to be $3^{\frac{a+b}{2}}$.
Now, both $a$ and $b$ have to be even or both has to be odd.
let $n$ be odd. Then number of terms with even exponents=$\frac{n+1}{2}$
number of terms with odd exponents=$\frac{n+1}{2}$
Total number of ways of choosing 2 numbers $3^a$ and $3^b$ = $2\cdot ^{\frac{n+1}{2}}C_2$
Probability is $$P(n)=\frac{2\cdot ^{\frac{n+1}{2}}C_2}{^{n+1}C_3}$$
But I am not getting the answers given in the options.
2. Aug 26, 2015
### RUber
Start by checking n =2 and n = 3, those cases are pretty straightforward and should at least eliminate some of your options.
3. Aug 26, 2015
### Titan97
For n=3, the series is $1,3,9,27$
The possible GPs are : $1,3,9$ and $3,9,27$.
Probability is $2/4=1/2$.
I get the same answer if I substitute $n=3$ in $P(n)$
4. Aug 26, 2015
### RUber
Simplifying your equation for P(n) does give one of the answers for n odd.
Have you worked it out for n even?
$P(n) = 2 \frac{ (\frac{n+1}{2} )! }{ (\frac{n+1}{2}-2 )! 2!} \frac{ (n+1-3)!(3)!}{ (n+1)!}$
5. Aug 26, 2015
### Ray Vickson
Let's condition on the first chosen number, so $P(GP|i) =$ the (conditional) probability of a GP, given the first chosen number is $i$, for $i = 1, 3$.
Given $i = 1$ as the first number chosen, that leaves $3, 9, 27$. The chance of choosing $3$ next is $1/3$; that leaves $9, 27$, from which we must choose $9$. Thus, $P(GP|1) = (1/3)(1/2) = 1/6$.
Given $i = 3$ that leaves $1, 9, 27$, from which we must next choose $9$, then choose $27$ from the remaining set $1, 27$. Thus, $P(GP|3) = (1/3)(1/2) = 1/6$.
Altogether, we have $P(GP) = P(1) PGP|1) + P(3) P(GP|3) = (1/4)(1/6) + (1/4)(1/6) = 1/12$.
6. Aug 26, 2015
### RUber
I think the question implies that order doesn't matter. Simply that the 3 choices form a geometric progression.
That is to say, for n=2, P=1.
7. Aug 26, 2015
### Titan97
Yes. Only difference is that, number of ways of choosing $3^a$ and $3^b$ is $^{\frac{n}{2}+1}C_2+^{\frac{n}{2}}C_2$
8. Aug 26, 2015
### Titan97
@Ray Vickson , are you considering that the probability of choosing the second term changes after choosing the first term? I have considered that all three terms are picked at once.
9. Aug 26, 2015
### RUber
So, are A and C both correct?
10. Aug 26, 2015
### Titan97
Its a "more than one correct" type question.
11. Aug 26, 2015
### Ray Vickson
Well, they DID say increasing GP. I would assume that if they just meant "geometric progression (in any order)" they would have omitted the word "increasing", because it would be redundant.
However, I suppose this is a moot point.
Last edited: Aug 26, 2015
12. Aug 26, 2015
### Ray Vickson
Yes, because that is typically what is meant when speaking of choosing things without replacement. However, let's look carefully at the way you wanted to do it.
If order does matter (so {1,3,9} is acceptable, but not {3,1,9} or {9,1,3} ... ) then you need to use permutations, not combinations. The number of permutations of 3 things drawn from 4 distinct things is $_4P_3 = 4 \cdot 3 \cdot 2 = 24$. Among these 24 possibilities, only two of them (namely, {1,3,9} and {3,9,27}) are acceptable, so you end up with a probability of 2/24 = 1/12, exactly as obtained from the sequential, one-at-a-time argument.
If order does not matter we should look at combinations instead of permutations. There are $_4C_3 = 4$ possible "outcomes", among which two of them are "acceptable", so the probability you want would be 1/2.
Last edited: Aug 26, 2015
13. Aug 26, 2015
### RUber
I think the answer selection implied that order didn't matter much more convincingly than the question itself.
My initial sense was to interpret it the same way, but when n=2 plugged into the answer choices gave 3/4 or 1, and 1/6 was nowhere to be seen, I had to infer that order didn't matter. | 2017-08-23T08:45:45 | {
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https://math.stackexchange.com/questions/2400239/closed-balls-are-closed/2400280 | # Closed balls are closed
Let $E \subset X, p \in X$ and $(X,d)$ a metric space such that $E := \{q \in X\mid d(p,q) \leq r\}$. By definition, this is a closed ball with center $p$ and radius $r$. I now want to show that this set is closed (i.e. it contains all its limit points).
Proof
Let $k$ be a limit point of $E$. Let $\epsilon >0$ and define $N_{\epsilon}(k)$ as the open ball (neighborhood) with center $k$ and radius $\epsilon$. By definition of limit point, there exists $z \in E \cap N_{\epsilon}(k)$ such that $z \neq k$. Hence we have $d(p,z) \leq r$ and $d(k,z) < \epsilon$, so by triangle inequality, we have $d(p,k) < r + \epsilon$. Because $\epsilon >0$ is chosen arbitrarily, it follows that $d(p,k) \leq r$ (if $d(p,k) > r$, then $d(p,k) - r > 0$, implying that $d(p,k) < d(p,k)$, which is absurd). We deduce that $k \in E$ and hence $E$ is closed. $\quad \square$
Is my proof correct? Is there an easier proof?
• – Henry Aug 20 '17 at 14:34
• I know. I saw this post but that proof uses complements, where as mine is more direct. – user370967 Aug 20 '17 at 14:34
• The center of the closed ball $E$ – user370967 Aug 20 '17 at 14:42
• Your proof is good. Your proof uses only the def'n of a metric, so it can't really get any easier. E.g. if $X=\mathbb R$ and $d(x,y)=|x-y|,$ prove that $[-1,1]$ contains all its limits...BTW: An intro-course teacher might want you to show details of the use of the triangle inequality......BTW: In general, be careful to note that the $closure$ of the open ball $B_d(x,r)$ may fail to equal $\{y: d(y,x)\leq r\}.$ E.g. if $X$ has more than 1 point and $d(x,y)=1$ when $x\ne y,$ then $B_d(x,1)=Cl(B_d(x,1))=\{x\}$ but $\{y: d(y,x)\leq 1\}=X.$ – DanielWainfleet Aug 20 '17 at 17:40
• Thanks for your useful comment. Earlier this day, I conjectured that the closure of an open ball would be a closed ball, but as you showed yourself this isn't true. – user370967 Aug 20 '17 at 18:56
Your proof looks good to me. The key point, that $a<b+\epsilon$ for all $\epsilon>0$ implies $a\le b,$ is used explicitly, which is good expository style.
An alternative proof: the "closed" ball $E$ is the inverse image of the closed set $[0,r]$ under the continuous map $q\mapsto d(p,q).$ Hence it is closed. Or, it is the complement of the inverse image of the open set $(r,\infty)$, and thus closed.
• Sorry I'm not interested in alternative proofs. I just want to know whether my proof works (In my text, continuity is not yet discussed btw) – user370967 Aug 20 '17 at 15:14
• Then why does the last line of your question ask "Is there an easier proof?" – kimchi lover Aug 20 '17 at 15:16
• I first want to know whether my proof is correct, then we can talk about alternative proofs. I should have been clearer. Sorry. – user370967 Aug 20 '17 at 15:18
An alternate proof is to show that $X - E$ is open. To show that $x \in X - E$ is an interior point, consider the open ball centered at $x$ with radius $d(x, p) - 1$, and then argue that this is disjoint from $E$ using the triangle inequality.
Your proof is correct.You proved that the closure of $E$ is equal to $E$ thus $E$ is closed.
To answer also your second question,there is an easier proof for this:
Let $x_n \in E$ such that $x_n \rightarrow x$.
Then $d(x,p) \leq d(x,x_n)+d(x_n,p) \leq d(x,x_n)+r$
Taking limits to both sides of the inequality we have that
$$d(x,p) \leq r \Rightarrow x \in E$$
Thus $E$ is closed. | 2021-04-14T09:19:15 | {
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https://www.xarg.org/book/stochastics/monte-carlo-integration/ | Monte Carlo integration
Imagine we want to measure the area of a pond of arbitrary shape in the middle of a field with a known area $$A$$. If we throw $$N$$ stones randomly into the known area and count the number the stone falls into the pond $$N_{\text{pond}}$$, the area of the pond gets approximated better and better the more stones we throw with
$A_{\text{pond}} = \frac{N_{\text{pond}}}{N}A$
This procedure is a simple example of the Monte Carlo integration. But we can formalize this approach.
Given a real function $$f:[a, b]\to[0,\infty)$$ and a probability density function $$p(x)$$ such that $$\int\limits_a^bp(x)dx=1$$. The integral $$I(f)=\int\limits_a^bf(x)dx$$ can be approximated like this:
$\begin{array}{rl}I(f)=&\int\limits_a^bf(x)dx\\=&\int\limits_a^b\underbrace{\frac{f(x)}{p(x)}}_{g(x)}p(x)dx\\=&\mathbb{E}[g(x)]\\\overset{\text{estimate}}{=}&\lim\limits_{N\to\infty}\frac{1}{N}\sum\limits_{i=1}^Ng(x_i)\\=&\lim\limits_{N\to\infty}I_N(f)\end{array}$
It follows that for the case that $$x_i\sim \text{unif}(a, b)$$,
$I_n(f) = \frac{1}{N}\sum\limits_{i=1}^N\frac{f(x_i)}{p(x_i)} = \frac{b-a}{N}\sum\limits_{i=1}^Nf(x_i)$
Error analysis
The Monte Carlo method clearly yields an approximation. The accuracy depends on the number of samples $$N$$ that are used to average. A possible measure of this error is the variance
$\begin{array}{rl}Var[I_n(f)] &= \mathbb{E}[I_n(f)^2]-\mathbb{E}[I_n(f)]^2\\&= \mathbb{E}\left[\left(\frac{1}{N}\sum\limits_{i=1}^Ng(x_i)\right)^2\right]-\mathbb{E}\left[\frac{1}{N}\sum\limits_{i=1}^Ng(x_i)\right]^2\end{array}$
We should expect that the error decreases with the number of samples $$N$$, but $$Var[I_n(f)]$$ does not. A solution to this problem is performing the experiment $$I_n(f)$$ up to $$M$$ times. According to the central limit theorem, these values would be normally distributed around an expected value $$\mathbb{E}[I]$$. Suppose we have a set of $$M$$ of such $$I_n(f)$$, a convenient measure is the variance deviation of the means $$Var_M[I_n(f)]$$:
$\begin{array}{rl}Var_M[I_n(f)] &= \mathbb{E}[I^2]-\mathbb{E}[I]^2\\&= \frac{1}{M}\sum\limits_{n=1}^MI^2_n - \left(\frac{1}{M}\sum\limits_{n=1}^MI_n\right)^2\end{array}$
Although $$Var_M[I_n(f)]$$ gives an estimate of the actual error, making multiple measurements is not practical. But this is not needed, since it can be shown that
$Var_M[I_n(f)]\approx\sqrt{\frac{1}{N}Var[I_n(f)]}$
This relation becomes exact in the limit $$N$$. This expression implies, that the error decreases with the square root of the number of trials, meaning if we want to reduce the error by factor 10, we need 100 times more points.
Variance reduction
The quality of the approximation of the integral highly depends on the number of samples $$N$$. How can we reduce the noise with a moderate number of samples?
Importance sampling
With importance sampling, we define the probability density function $$p(x)$$ according to the function $$f(x)$$ so that large values of $$f(x)$$ are also large in $$p(x)$$. The optimal
$p_{\text{optimal}}(x)\propto f(x)$
For example the function $$f(x)=e^{-x^2}$$ in the interval $$[0,1]$$ will have most points in a region where the value of $$f(x)$$ is very small, which requires a large number of samples to achieve a decent accuracy. A reasonable choice for $$p$$ is $$p(x)=\frac{1}{z}e^{-x^2}$$ where $$z=\int_0^1 e^{-x^2} dx = \frac{1}{2} \sqrt\pi \text{erf}(1)\approx 0.746824$$
Examples
Constant function
Given a constant function $$f(x) = k$$ and a uniform probability density function. We can calculate the integral of the function analytically $$\int\limits_a^bf(x)dx = \int\limits_a^bkdx = k(b-a)$$ or with Monte Carlo:
$\int\limits_a^bf(x)dx = \int\limits_a^bkdx \approx \frac{1}{N}\sum\limits_{i=1}^Nk(b-a)=k(b-a)$
Approximation of Pi
Given a two dimensional function $$f:[0,1]^2\to [0,\infty)$$ and a uniform probability density function. We can approximate pi by defining the function $$f(x,y)$$ such that the return value is one if $$(x,y)$$ is within a circle and zero if it is outside. The area of the quarter circle is thus
$A=\int\limits_0^1\int\limits_0^1f(x,y)dxdy\approx\frac{1}{N}\sum\limits_{i=1}^Nf(x_i,y_i)$
From $$A=\frac{\pi}{4}$$ we can come up with
$\pi\approx\frac{4}{N}\sum\limits_{i=1}^Nf(x_i,y_i)$
Implementation
An implementation for the uniform distributed Monte Carlo algorithm can be implemented as follows
function integrate(f, a, b, N) {
let sum = 0
for (let i = 0; i < N; i++) {
sum+= f(a + Math.random() * (b - a));
}
return sum / N * (b - a);
} | 2020-01-23T18:28:21 | {
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http://judaicabennysart.com/rationale-meaning-voyoaxs/third-fundamental-theorem-of-calculus-c44a6e | third fundamental theorem of calculus
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פורסם בקטגוריה מאמרים בנושא יודאיקה. אפשר להגיע לכאן עם קישור ישיר. | 2021-05-15T04:25:43 | {
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http://gpuf.shantideva.it/find-an-equation-of-the-set-of-all-points-equidistant-from-the-points-calculator.html | One way we can define a parabola is that it is the locus of points that are equidistant from both a line called the directrix and a point called the focus. Click here 👆 to get an answer to your question ️ Find the equation of the set of points which are equidistant from the points (1,2,3) and (3,2,-1). if a point P is ‘equidistant’ from two points A and B, then the distance between P and A is the same as the distance between P and B, as illustrated here:. Derive the Equation of a Parabola (Vertex at Origin) Definition: A parabola is the set of points equidistant from a fixed line (the directrix) and a fixed point (the focus) not on the line. If a sphere is constructed that passes through these 2 points, then the center will be equidistant from both. Write an equation of the circle with center at (2, -5) and a radius of 7. The standard form of a parabola with vertex (0, 0) and the x-axis as its axis of symmetry can be used to graph the parabola. The center of a circle is defined as the point from which all points of the circumference are equidistant. 2 Translations and Shifts of Quadratic Functions ( discuss the effects of the symbol before the leading coefficient, the effect of the magnitude of the leading coefficient, the vertical shift of equation y = x2 c, the horizontal shift of y = (x - c)2. The sketch is shown in Figure 5. Finite Math. 1 Quadratic Function – give examples in standard form and demonstrate how to find the vertex and axis of symmetry. (ii) Construct the locus of points at a distance of 3. How to Find the Q-point of a Transistor Circuit. Basically, I need to be able to pass a function the number of points that I would like to generate and get back an array of points. If we place the center point on the origin point, the equation of a circle with center point (0, 0) and radius r is:. Find the equation of the set of all points the sum of whose distances from the points $(3, 0)$ and $(9, 0)$ is 12. Locus From Two Lines. Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1). The horizontal distance can be expressed as:. Equation Of Plane Equidistant Bewtween Two Points. 4 ) This unit shall be assessed by methods given in Volume 1, Part 3 'Assessment Guidelines'. Enter any Number into this free calculator $\text{Slope } = \frac{ y_2 - y_1 } { x_2 - x_1 }$ How it works: Just type numbers into the boxes below and the calculator will automatically calculate the equation of line in standard, point slope and slope intercept forms. For example, consider the line segment containing the end points A and B and midpoint P. The equation of a straight line can also be found from the graph by reading off two points and using them to find the equation as outlined above. The distance between any point of the circle and the centre is called the radius. Enjoy! eulers. The less they are aligned, the more the coefficient will get closer to zero. If we consider P(x, y) a point on the parabola then we can derive a parabola using the same ratio condition. Subtracting the second equation from the first equation gives (z-1) 2 - (z+1) 2 = 0 You can expand the squares and solve for z or factor the expression as a difference of squares and then solve for z. So the code would involve finding the equation of cubic polynomial connecting the two successive points. 6 units Since the sides all have different lengths, the triangle is scalene. Locus is a set of points that satisfy a given condition. A parabola 4 is the set of points in a plane equidistant from a given line, called the directrix, and a point not on the line, called the focus. The distance from the foci is just 2. Explanation of Solution Formula used:. Free Mathematics Tutorials. 1 Quadratic Function – give examples in standard form and demonstrate how to find the vertex and axis of symmetry. Definition 14. Use the distance formula to find the length of each side of the triangle. Objectives. So each point P on the parabola is the same distance from the focus as it is from the directrix as you can see in the animation below. Practice 2. Find locus of points where the potential a zero. Find coordinates for the point equidistant from (2,1) (2,-4) (-3,1) You don't need to do all that calculating the other tutor did. 2) The difference of the distance of P(x,y) from (-3,0) and (3,0) is equal to 2. Click Calculate to find standard deviation, variance, count of data points n, mean and sum of squares. Find the equation of an osculating circle? I need help with this question. So, just a few points start to look like a circle, but when we collect ALL the points we will actually have a circle. The standard form of a parabola that we are now going to use helps us to find the focus and the directrix. Notice that if we are given the equation of a plane in this form we can quickly get a normal vector for the plane. Question 1118795: Find an equation describing all points P(x,y) equidistant from Q(-3,4) and R(1,-3). Weight Watcher Points Calculator. Free line equation calculator - find the equation of a line given two points, a slope, or intercept step-by-step This website uses cookies to ensure you get the best experience. The calculator will generate a step-by-step explanation on how to obtain the result. asked Feb 18, 2015 in CALCULUS by anonymous plane-equation. When there are more than one independent variable it is called as multiple linear regression. Since it's equidistant, finding the midpoint should allow you to calculate the direction ratios of the normal to the plane. These online calculators find the equation of a line from 2 points. With three points A, B and C, three pairs of points can be formed, they are:. A is the set of all points P such that the difference of the distances from P to two fixed points, called the is constant. By using this website, you agree to our Cookie Policy. Forms of Linear Equations. Determine the equation of the trajectory of the midpoint of the ladder. (- 3, 1, 2), (6, - 2, 4) Find an equation of the plane that contains all the points that are equidistant from the given points. Find coordinates for all the points on a number line that are (a) six units from 0; (b) six units from four; (c) six units from −7; (d) six units from x. If a sphere is constructed that passes through these 2 points, then the center will be equidistant from both. YEAR 12 MATHEMATICS Coordinate Geometry QUESTION TWO A drain runs along a straight line equidistant between the points (1, -2) and (3, -4). Length of AB 34 5. , Find the normal to the these two vectors. 4 CHAPTER 2. 1 Quadratic Function – give examples in standard form and demonstrate how to find the vertex and axis of symmetry. Find the equation of this fence line. Equation of a line that passes halfway between two points in 2942 equidistant equation of a plane between two points find an equation of the plane consisting all points that are solved find the equation of plane each point which Equation Of A Line That Passes Halfway Between Two Points In 2942 Equidistant Equation Of A Plane Between…. Fractions should be entered with a forward slash such as '3/4. I think I have the general idea of this one. This theorem is a royal mouthful, so the best way to. Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions approximately, e. com and read and learn about rationalizing, multiplying and a variety of other math subject areas. So the part of the plane equidistant from points A and B are all points along the y-axis. Slope formula method to find that points are collinear. YEAR 12 MATHEMATICS 63 64 Coordinate Geometry. Given 2 points, we will find the equation of the line equidistant between them. Prove that the equation (k — 2)x2 + 2x —k = 0 has real roots for all values of When the expression x5 + 2x2 + ax + b is divided by x2 —4, the remainder is 3x +1. "A locus is a curve or other figure formed by all the points satisfying a particular equation. How to enter numbers: Enter any integer, decimal or fraction. The conic sections we will be studying all have B so we will not have a Bxyterm. Any point on that perpendicular line is equidistant from the two points. ) & perpendicular to the sgmt. New coordinates by rotation of axes. The user is asked to enter a set of x and y-axis data-points, and then each of these is joined by a cubic polynomial. Recall that the definition of locus is the set points that meet some given conditions. The term “opposite” refers to the additive inverse of a number, meaning the number that when added to a given number results in zero. I don't know how to solve this kind of system composed of quadratic equation with one variable and if the process which consist to fix n-1 dimension to determine the last one lead to an equidistant. Since every plane through the two points contains one of these lines, we can image the set of all possible points as a plane. What type of gure is described by this equation? We see from the description that we’re talking about the parabola with focus (1;1) and directrix the line y = x. This will simplify the calculations. Find an equation for the plane, each of whose points is equidistant from (-1, -4, -2) and (0, -2, 2). Let $\,(x,y)\,$ be a typical point on the parabola. He shows, using the distance formula, that indeed, we have an. A(?3, 4, 3) and B(6, 3, ?3). 2, 4 Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1). Newton interpolation with equidistant points. In the figure above click "show locus" and see that the green dotted line is the locus of all points that are equidistant from Q and R - a straight line. It can also be seen that Δx and Δy are line segments that form a right triangle with hypotenuse d, with d being the distance between the points (x 1, y 1) and (x 2, y 2). Tracing paper may be used. Formulas involving circles often contain a mathematical constant, pi, denoted as π; π ≈ 3. equation of the line that is an equal distance between points (3,10) and (13,4) a committeeis choosing a site for a county fair, The site needs to be located the same distance from the two main towns in the county. Answered by Penny Nom. 1] Given an equation, to find the locus. That sphere's center would be the minimal-distance equidistant point from the (defining) 3 points. [if its an upward opening parabola with vertical symmetry which yours is. , Substitute one of the points (A, B, or C) to get the specific plane required. Find an equation of the set of all points equidistant from the points A(−1, 6, 4) and B(5, 3, −3). pleasee help. Find a function to represent the set of all points equidistant from the point (-1,3) and the line y=7 2. Fundamentals Of 2d And 3d Graphs. (sin 9 + cosec + (cosØ+sec6)2 = 7+tan2 6 + cot2 6. Homework Statement Find an equation of the set of all points equidistant from the points A(-1,5,3) and B(6,2,-2). RELATIONS A set of ordered pairs is called a relation. Hope this is fine 2) I found most of them beyond UCL, indicating a bigger process shift. Connect all points in a circular motion. com and read and learn about rationalizing, multiplying and a variety of other math subject areas. Fundamentals Of 2d And 3d Graphs. (a) (b) Write down the equation of the ellipse Describe a sequence of transformations which, when applied to the circle with equation x2 +. So the code would involve finding the equation of cubic polynomial connecting the two successive points. Incenter of a Triangle. ? asked Mar 2, 2013 in GEOMETRY by Find an equation of the plane that contains all the points that are equidistant from the given points. The Parabola is defined as the set of all points P in a plane equidistant from a fixed line and a fixed point in the plane Parabola: Standard Equation. Conics: Hyperbola. 2 b = 10 → b = 5. pl confirm this is correct. Find an equation of the set of all points equidistant from the points. A circle is a two-dimensional geometric figure where all points are equidistant from the center. A set of points on a plain surface that forms a curve such that any point on that curve is equidistant from the focus is a parabola. There are five fundamental locus rules. Graph points and identify coordinates of points on the Cartesian coordinate plane (all four quadrants). For example, let's look at the point. [2] b Write the equation of the locus sketched in part a. Thank you in advance. To find out if a point is on a line with an equation, we just need to substitute in the point's and values and see if the equation holds true. Locus of point is defined as the set of point having a special. Applications The Circle: Wheels Gears The Chunnel Definition The circle is the locus (set) of all points that are equidistant from a given point called the center. On the Marks card, on the drop-down menu for Distance, select Dimension. From that alone, we can find its q-point. (sin 9 + cosec + (cosØ+sec6)2 = 7+tan2 6 + cot2 6. In this one, we were to find out the locus of a point such that it is equidistant from two fixed points, which was the perpendicular bisector of the line joining the points. 2 Educator Answers Find the equation of the locus of a point that it is equidistant from (-2,4) and the y-axis. Proven in North America and abroad, this classic text has earned a reputation for excellent accuracy and mathematical rigour. Think of the geometry. A plane is the set of all points in 3-D space equidistant from two points, A and B. thanks for your help. Find the equation of the set of all points P(x,y) that is equidistant from (-3,0) and (3,-5). University. I know that for all parabolas: distance from foci = 1/(4*abs(a)). When two points (x1, x2), (y1, y2) are given and the equation contains these two points, the first step is to find the slope of the line. the set of points in a plane such that the difference of the distances from (2,5) and (2,-5) is 8. This online calculator will find and plot the equation of the circle that passes through three given points. Sketch the graph of a cubic function that has x-intercepts at (3, 0), (5, 0), and (-6, 0) and has an end-behavior of down-up. The flxed distance is called the radius and the flxed point is called the center of the circle. positive constant. This task gives the important characterization of the perpendicular bisector of a line segment as the set of points equidistant from the endpoints of the segment. Because the distance between any. Michael is learning how to use his graphing calculator. ie, [tex](\vec{r} - \vec{a}). We need to find function with known type (linear, quadratic, etc. It is in the form: y=a(x-h)^2+k. P, Q, and R are three points in a plane, and R does not lie on line PQ. Should you require advice on a polynomial as well as systems of linear equations, Sofsource. So, a circle whose center is located at the origin (0, 0) with a radius of 4 has the equation: x² + y² = 16. describe the set. The distance formula can be used to obtain an equation of the circle whose center is the origin and whose radius is r. Using the y-intercept and one other point. I don't know how to solve this kind of system composed of quadratic equation with one variable and if the process which consist to fix n-1 dimension to determine the last one lead to an equidistant. [2] b Write the equation of the locus sketched in part a. What is equidistant in math? Create an account to start this course today. Then the directrix, being perpendicular to the axis, is a horizontal line and it must be p units away from V. The standard form of a parabola with vertex (0, 0) and the x-axis as its axis of symmetry can be used to graph the parabola. the center). Describe the set. Find The Parametric Equation Of A Line Segment Between Two Points. y + 4= (1/2)(x - 2)^2 b. Find a fully reduced equation for the set of all points in R3 (i. [For example, a problem such as: Find the equation of the locus of a point which is equidistant from the points (-2, 3) and (5, 8)]. In this article, I will show you solving equations in Excel. Finding the intersection point. Parametric. Defines the equation of a line equidistant from any two points. The distance formula is used to find the distance between two points in the coordinate plane. Find all of the points on the y-axis which are 5 units from the point (−5, 3). protractor, compasses, pen, HB pencil, eraser, calculator. 5 A g−1 in 1 M LiOH aqueous solution, much higher than that of Ni-MOF (306. • Both classes had the same third quartile score. What are the coordinates of the center of this circle and the length of its radius? (1) ( 2,5) and 16 (3) ( 2,5) and (2) (2, 5) and 16 (4) (2, 5) and 19 The equation of a line is y. Question 63. How to find an equation of the plane consisting of all points that are equidistant from (-5, -3, -3) and (-3, 4, -3). Think of the geometry. The parabola is the set of points equidistant from the focus and the directrix. com and master linear equations, adding and a large number of additional math topics. I think it is hard to visualize and understand without it:. About the Book Author Mark Ryan is the founder and owner of The Math Center in the Chicago area, where he provides tutoring in all math subjects as well as test preparation. 1] Given an equation, to find the locus. 1) P(x,y) is equidistant from the y axis and (4,0). Since the center point p is equidistant from a, b, and c, it lies at the intersection of these three plane equations. - 10873807. Linear Algebra. Using the y-intercept and a second point the equation can be found. the parabola has a downward opening. Let A (1, 2, 3) & B (3, 2, –1) Let point P be (x, y, z,) Let point P (x, y, z) be at equal distance from point A (1, 2, 3) & B (3, 2, – 1) i. [12] 35 In the diagram below, circles X and Y have two tangents drawn to them from external point T. Objectives. In three dimensions, the locus of points equidistant from two given points is a plane, and generalising further, in n. Find the equation of the circle whose center is (1,2) that contains the point (4,2). 1)Locus of points:The set of points in a plane equidistant from a fixed point called the center. A Rhombus is a 4-sided quadrilateral in which all the four sides are equal. The center is found by first finding the plane that is the bisector of any two points,say point 1 and point 2. Incenter of a Triangle. Vertex of a parabola is the coordinate from which it takes the sharpest turn whereas a is the straight line used to generate the curve. Describe the set. Equation Of Plane Equidistant Bewtween Two Points. Homework Equations None The Attempt at a Solution Um, I've drawn a graph of the two points and that's about as far as I could get. We can also find the solutions to a linear system by using determinants and Cramer's Rule. Coordinates of Point 1 (x 1 ,y 1 ): x=. Hypotenuse : The longest side of a right-angled triangle, always opposite to the right angle itself. Fill in 3 of the 6 fields, with at least one side, and press the 'Calculate' button. Show that an equation for the parabola with the focus (o, p) and directex y = -p is y = 1/4p x 2. All circles with centers located at the origin (0, 0) have the equation: x² + y² = r² Where r is the radius of the circle. Concept: Distance of a Point from a Plane. describe the set. Slope formula method to find that points are collinear. D) It is a line. Technically, a parabola is the set of points that are equidistant from a line (called the directrix) and another point not on that line (called the focus, or focal point). By default, points used to create each output line feature will be used in the order they are found. Requires the ti-83 plus or a ti-84 model. , Find the normal to the these two vectors. In this lesson you will learn how to write equations of circles and graphs of circles will be compared to their equations. Center (h,k) Radius P(x,y) Definition The standard form of the equation of a circle with radius r and center at (h,k) is: Example Write the equation of the circle with center at (4. If the sum of their reciprocals is. php(143) : runtime-created function(1) : eval()'d code(156. A parabola is the set of points (locus of points) that are equidistant from a given point and a given line in a plane. Technically, the line consists of the locus of all points that are equidistant from your two circle points; you can think of the line as a mirror where each of your two points appears as a reflection of the other. You can also see the work peformed for the calculation. Newton interpolation with equidistant points. In both the expression is common so we can cancel it. A parabola is the set of points (locus of points) that are equidistant from a given point and a given line in a plane. then be divided into red apples and green apples, but they are all still apples. Find locus of points where the potential a zero. Eliminating the parameter A curve traced out by a point on the circumfrence of a circle as the circle rolls along a straight line in a plane is called a __________. Since every plane through the two points contains one of these lines, we can image the set of all possible points as a plane. A point is said to be equidistant from a set of objects if the distances between that point and each object in the set are equal. If you know two points that a line passes through, this page will show you how to find the equation of the line. The user is asked to enter a set of x and y-axis data-points, and then each of these is joined by a cubic polynomial. Solving Systems of Linear Equations A system of linear equations is just a set of two or more linear equations. Finding the intersection point. 1) Graph of the equation - the set of all solution points of an equation. The form of these two equations makes them particularly nice to solve. The distance from the foci is just 2. Find The Parametric Equation Of A Line Segment Between Two Points. 3) Now there was situation,5 are within control limits and 4 are above. Incenter of a Triangle. x is the independent variable and y is the dependent variable. It is a perfectly round geometrical object that mathematically, is the set of points that are equidistant from a given point at its center, where the distance between the center and any point on the sphere is the radius r. net does not purport to have any relationship with Weight Watchers ® and has no intent to present Weight Watchers' ® product as its own. It only takes a minute to sign up. 2 Lines Intersection Calculator. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Warning: Unexpected character in input: '\' (ASCII=92) state=1 in /home1/grupojna/public_html/yu1woun/gbgd. 2, 5 Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (-4, 0, 0) is equal to 10. Center (h,k) Radius P(x,y) Definition The standard form of the equation of a circle with radius r and center at (h,k) is: Solution to Do Now h=-5 k=6 (-5,6) Try This Graph the circle. (sin 9 + cosec + (cosØ+sec6)2 = 7+tan2 6 + cot2 6. Midpoints and endpoints: 2000-02-15: From Jessica Sipes:. The fixed line is the ‘directrix’ and the fixed point is the ‘focus’ {Fig. t t Answer all questions. In either. Determine the equation of the trajectory of the midpoint of the ladder. The equation solver allows to solve equations with an unknown with calculation steps : linear equation, quadratic equation, logarithmic equation, differential equation. You know that for any two points there's a "mirror line" that goes halfway between them. pl confirm this is correct. Find an equation of the set of all points equidistant from the points A(−1, 6, 4) and B(5, 3, −3). This is the plane through the midpoint, at right angles to the segment (the spatial right bisector of the segment!). Because all the equations equal to 0 also the determinant of the coefficients should be equal to 0 and the value of the determinant will be as follows: Notice that we got the equation of a circle with the determinants equal to the coefficients A, B, C and D as follows; (x 2 + y 2) A + xB + yC + D = 0. Upenn Math 114 Finding The Equation Of A Plane Equidistant From Two. In the driving example above, the domain is the set of all possible speeds and the range is the set of resulting miles per gallon. Basically, I need to be able to pass a function the number of points that I would like to generate and get back an array of points. Substituting values of x and y in the equation for the coordinates of the points. Example 7 Find Vector Equation For Line 1 0 2 3 4 6. Two point form calculator This online calculator can find and plot the equation of a straight line passing through the two points. A trapezium is a quadrilateral that has one pair of parallel sides. 2 Translations and Shifts of Quadratic Functions ( discuss the effects of the symbol before the leading coefficient, the effect of the magnitude of the leading coefficient, the vertical shift of equation y = x2 c, the horizontal shift of y = (x - c)2. In , any equation involving , , and/or draws a surface. All circles with centers located at the origin (0, 0) have the equation: x² + y² = r² Where r is the radius of the circle. A is the set of all points P such that the difference of the distances from P to two fixed points, called the is constant. Very useful for all you saxon-math lovers: eulerformulageometry. All the points a given distance from a central point in a plane. For example, the locus of points that are 1cm from the origin is a circle of radius 1cm centred on the origin, since all points on this circle are 1cm from the origin. In the equation above, y 2 - y 1 = Δy, or vertical change, while x 2 - x 1 = Δx, or horizontal change, as shown in the graph provided. (h, k) = (2 + (− 4) 2, 1 + 1 2) = (− 2 2, 2 2) = (− 1, 1) Length of b: The minor axis is given as 10, which is equal to 2b. , Find the normal to the these two vectors. pl confirm this is correct. Download free on Amazon. Finding the control points of a Bezier Curve can be a difficult task. In this definition of a parabola, it is the shape created by the points that are the same distance from a given point (call the focus ) and a given line (called the. Concept: Distance of a Point from a Plane. The given point is called the focus, and the line is called the directrix. Find an equation of the set of all points equidistant from the points A(-1,5,3) and B(6,2,-2). this formula to nd an equation for the set of points in the plane that are equidistant from the line y = x and the point (1;1). A circle is the locus of all points equidistant from a central point. A(?3, 4, 3) and B(6, 3, ?3). 5 Example 1. Complex numbers represent a number system that combines both real and imaginary numbers. Describe the given set of points with a single equation or with a pair of equations. asked by Jim on July 6, 2016; Calc. A triangle is determined by 3 of the 6 free values, with at least one side. Here is a set of notes used by Paul Dawkins to teach his Calculus III course at Lamar University. Example: A Circle is: "the set of all points on a plane that are a fixed distance from a central point". Hence the required points are (-3,-3). Type your algebra problem into the text box. 1( Parabola): A parabola is the collection of all points in some plane that are equidistant from a given line and point in that plane. Find the distance between two points on horizontal or vertical number lines. Standard form equation of a Circle. The usual Euclidean definition of a Parabola is the locus of points equidistant from a line and a fixed point. The system of equation refers to the collection of two or more linear equation working together involving the same set of variables. A trapezium is a quadrilateral that has one pair of parallel sides. 7 Example 1. Not sure how to go about recalculating control limits, pl guide for below two possibillities a. YEAR 12 MATHEMATICS Coordinate Geometry QUESTION TWO A drain runs along a straight line equidistant between the points (1, -2) and (3, -4). 2) The difference of the distance of P(x,y) from (-3,0) and (3,0) is equal to 2. It also outputs slope and intercept parameters and displays line on a graph. In two-dimensional Euclidean geometry, the locus of points equidistant from two given (different) points is their perpendicular bisector. Once you know the equation of the new line, finding the intersection point between it and the first (given) line is a straightforward task. P, Q, and R are three points in a plane, and R does not lie on line PQ. A parabola is the set of all points in a plane and a given line. 2 b = 10 → b = 5. Find the equation of the set. This is the equation of a straight line with a slope of minus 1. you know this because the focal point is always inside the parabola]. (use the general equation of a line - input the numerical coefficient of each term of the equation below). I was pretty sure that all points of this last great circle must be equidistant to the initial locations but I was wrong. y + 2 = (1/8)(x - 2)^2 c. Describe the given set of points with a single equation or with a pair of equations. 2 units Length of CA 74 8. Find the equation of the set of all points equidistant from the line x = −3 and the point (−1,0). Identify the surface. Fundamentals Of 2d And 3d Graphs. points in the plane, (x1,y1) and (x2,y2), with x1 ̸= x2, determine a unique first-degree polynomial in x whose graph passes through the two points. asked by Jim on July 6, 2016; Calc. Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1). Which equation does the graph at right represent? Explain your answer. This is a parabola with focus (3,5) an directrix on the x-axis. Find an answer to your question Find an equation of the set of all points equidistant from the points a(−1, 6, 3) and b(4, 3, −2). To graph the parabola, it is also a good idea to determine the y-intercept by letting x = 0. Find an equation for the surface consisting of all points that are equidistant from the point (−1, 0, 0) and the plane x = 1. Very useful for all you saxon-math lovers: eulerformulageometry. Thus the focus is and the directrix is. Get an answer for 'Find the value of x so the distance between the points A(3,5) and B(x,8) is 5 units. The standard form of a parabola that we are now going to use helps us to find the focus and the directrix. Find coordinates for the point equidistant from (2,1) (2,-4) (-3,1) You don't need to do all that calculating the other tutor did. A system of equations is a set of two or more equations with the same variables. Each seed generates its own polygon called a Voronoi “cell,” and all of 2-dimensional (2D) space is associated with only one cell. Brock University. Other activities to help include hangman, crossword, word scramble, games, matching, quizes, and tests. asked by Justsimplelikeme on July 6, 2016; Analytic geometry. Find The Parametric Equation Of A Line Segment Between Two Points. 2, 5 Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (–4, 0, 0) is equal to 10. 6: Describe the equation x = 2 as it appears in 3. Not sure how to go about recalculating control limits, pl guide for below two possibillities a. A hyperbola has two branches and two asymptotes. Two point form calculator This online calculator can find and plot the equation of a straight line passing through the two points. A parabola is defined as the set of points the same distance from $$(6,2)$$ and the line $$y=4$$. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Method #1 (requires a TI-83+ or greater calculator): If there are n equations (and n variables), create a matrix that is n×n+1 (i. For example, a parabola can be defined as the set of points in a plane that are equidistant from a focus F and a directrix. The fixed line is the ‘directrix’ and the fixed point is the ‘focus’ {Fig. (This particular mapping can be inverted: x = ½(y - 1). Many geometric shapes are most naturally and easily described as loci. The calculator will generate a step-by-step explanation on how to obtain the result. In two-dimensional Euclidean geometry, the locus of points equidistant from two given (different) points is their perpendicular bisector. A calculator to find the equation of a line that is parallel to another line and passing through a point. the sum of the distances) just as a circle is the set of points which are equidistant from one point (i. Thus the focus is and the directrix is. A parabola is the set of all points (x, y) in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix. Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1). There are three possibilities: The lines intersect at zero points. He shows, using the distance formula, that indeed, we have an. Or as a function of 3 space coordinates (x,y,z), all the points satisfying the following lie on a sphere of radius r centered at the origin x 2 + y 2 + z 2 = r 2. MATH 11011 FINDING THE EQUATION KSU OF A CIRCLE Deflnitions: † Circle: is the set of all points in a plane that lie a flxed distance from a flxed point. (100 points) Find a fully reduced equation for the set of all points in R2 i. com and master linear equations, adding and a large number of additional math topics. Solutions: (1) (2) , (3) Set of all points in a plane equidistant from a fixed point. (iv) Mark 2 points X and Y which are at a distance of 3. Comparing equation (6) with the Fourier Series given in equation (1), it is clear that this is a form of the Fourier Series with non-integer frequency components. Find The Parametric Equation Of A Line Segment Between Two Points. A parabola is the set of points in the plane equidistant from a xed point Fcalled the focus and a xed line lcalled the directrix. Visit Mathway on the web. By signing up, you'll get thousands. Perhaps let the user be able to set a length for this line (the 500-mile-long mid-point line). The standard form of a parabola with vertex (0, 0) and the x-axis as its axis of symmetry can be used to graph the parabola. Standard Form of a. Find all of the points on the x-axis which are 2 units from the point (−1, 1). Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. Applications The Circle: WHEELS AND TABLE TOPS Definition The circle is the locus (set) of all points that are equidistant from a given point called the center. In this section we define ordinary and singular points for a differential equation. Similar to the "How far can I travel" tool. In other words, every point on the circumference of a circle is. Equations -. For example, a parabola can be defined as the set of points in a plane that are equidistant from a focus F and a directrix. Equation of a line that passes halfway between two points in 2942 equidistant equation of a plane between two points find an equation of the plane consisting all points that are solved find the equation of plane each point which Equation Of A Line That Passes Halfway Between Two Points In 2942 Equidistant Equation Of A Plane Between…. Find a point on the y-axis that is equidistant from the points (1, -7) and (5, -1). Hello, Q: Find equation of the set of all points equidistant from the point (0,0,4) and the xy-plane. Given 2 points, we will find the equation of the line equidistant between them. If he receives a dividend off. So, this point does fall on the line. 3) Now there was situation,5 are within control limits and 4 are above. The center of a circle is defined as the point from which all points of the circumference are equidistant. Table of contents. Detailed explanations are also provided. So, if you input 3 points, this will compute the circle's center point, radius and equation. "A locus is a curve or other figure formed by all the points satisfying a particular equation. Problem 8: Find a relationship between x and y so that the distance between the points (x, y) and (-2, 4) is equal to 5. A circle is defined as the set of all points in a plane that are equidistant from a fixed point called the center. Find the equation of the set of all points P(x,y) that satisfy the given conditions. 1 Quadratic Function – give examples in standard form and demonstrate how to find the vertex and axis of symmetry. a line perpendicular to A3 'Z' a cube With diagonal AB v a plane perpendicular to AB a sphere With diameter AB Find the unit vector, in the first quadrant, that is. Examples: 1. (This particular mapping can be inverted: x = ½(y - 1). Group Work Problems 1. 40 a Sketch the locus of points equidistant from circles x2 + y2 = 4 and x2 + y2 = 36. asked by Jim on July 6, 2016; Calc. Find the point on the y-axis that is equidistant from the points (2,-1,1) and (0,1,3). Example: A Circle is: "the set of all points on a plane that are a fixed distance from a central point". Given , solve for p when c =. Question Find an equation of the set of all points equidistant from the points A(−2, 5, 2) and B(5, 1, −3). The equation solver allows to solve equations with an unknown with calculation steps : linear equation, quadratic equation, logarithmic equation, differential equation. 15 with center 1h, k2 and radius r for r 7 0. The middle point is one that divides the line segment into two equal segments. = a(y - k). It only takes a minute to sign up. Of course, according to its fundamental definition, a circle is the set of all points equidistant from a fixed center. Vertex of a parabola is the coordinate from which it takes the sharpest turn whereas a is the straight line used to generate the curve. Parametric. Download free on Amazon. Answer the questions in the spaces provided – tthere may be more space than you need. The equation y = 2x + 1 maps x into y and the equation specifies how each value of x corresponds with y. The set of all points equidistant from two points in n-space is a linear set -- in 2-space, it's a line, in 3-space it's a plane, and in 4-space it's a hyperplane (kind of abstract), and so forth. Conics Parabolas Definition - a parabola is the set of all points equal distance from a point (call PowerPoint Presentation - Parabolas are shaped like a U or C. on a map these towns have coordinates (3,10) and (13, 4). - 10873807. The last column of the result are the answers. 6: Describe the equation x = 2 as it appears in 3. Example 12: Apply the Secant method to find a root of the equation 2x3 + 3x2 + 3x + 1 = 0. Parabolas wit Vertex h k. One way we can define a parabola is that it is the locus of points that are equidistant from both a line called the directrix and a point called the focus. yx =−+6 32 E. The symbol. Since the vertex is halfway between the focus and directrix, the directrix is the line $\,y = -2\,$. If the curve passing through the points you're putting on the table are perfectly aligned, you'll get a correlation coefficient of ±1 (r=±1). The equation of the hyperbola is in standard form if the two fixed points are on either the x-axis or the y-axis, and they are equidistant from the. This line is known as the locus of the point P. Describe the set. Linear Algebra. Find an equation of the set of all points equidistant from the points a(−3, 4, 3) and b(6, 1, −2). Take the initial approximations as xo = –0. First sketch the two given points Notice that the point (-1, -3) is equidistant from the two given points. Next, set CU equal to x. A sphere is the set of all points in space equidistant from a fixed point, As described earlier, vectors in three dimensions behave in the same way as vectors in a plane. • Both classes had the same third quartile score. If you are given an equation of a straight line and asked to draw its graph all you need to do is find two points whose coordinates satisfy the equation and plot the points. For example, a parabola can be defined as the set of points in a plane that are equidistant from a focus F and a directrix. 4 CHAPTER 2. positive constant. To create a movable point, use parameters instead of numerical coordinates, like this: (h, k). 6: Describe the equation x = 2 as it appears in 3. Equations -. A Rhombus is a 4-sided quadrilateral in which all the four sides are equal. asked by Becky on August 30, 2012; Analytic geometry. What this does here is giving you the equation of a REGRESSION LINE passing through points that you put in a table. from the line y = –6 to (12, 7). I think I have the general idea of this one. I know the. It is in the form: y=a(x-h)^2+k. From how to use my t89 calculator to adding and subtracting polynomials, we have got all the pieces included. - 10873807. If the area of the circle is given, the formula d = sqrt (4a/pi) can also be used to solve for the diameter, where "a" denotes the area. xml CUAT006-EVANS 32 August 3, 2005 13:22 Essential Mathematical Methods Units 1 & 2 Example 12 Find the equation of the line shown in the graph. But each point will cost 1 percent of your mortgage balance. In general: two distinct points will determine a straight. Points where the parabola intersects the $$x$$-axis are called the $$x$$-intercepts. In the diagram below, AC = DF and points A, C, D, and F are collinear on line ℓ. Then find the plane of the bisector of point 2 and point 3. This locus of points equidistant between the given points. You may use a cal-culator to do routine. Find a function to represent the set of all points equidistant from the point (-1,3) and the line y=7 2. The point estimate refers to the probability of getting one of the results. 17 How many points are 5 units from a line and also equidistant from two points on the line? (1) 1 (3) 3 (2) 2 (4) 0 18 The equation of a circle is (x 2)2 (y 5)2 32. Notice that if we are given the equation of a plane in this form we can quickly get a normal vector for the plane. Right-click the Seattle number and select Mark Label > Never Show. Beside performing different statistical, financial analysis we can solve equations in Excel. Describe the set. A parabola is a conic section that is described as: Definition of a parabola: The set of all points (x, y) in a plane that are the same distance from a fixed line, called. Because the distance between any. That sphere's center would be the minimal-distance equidistant point from the (defining) 3 points. It will move through all the points equidistant from the focus point and the directrix line. B) How to Sketch the Graph of an Equation by Point Plotting. Find the equation of the set of all points P(x,y) that satisfy the given conditions. So, just a few points start to look like a circle, but when we collect ALL the points we will actually have a circle. Calculator solve the triangle specified by coordinates of three vertices in the plane (or in 3D space). yx =3 2− D. E) It is a circle. Question Find an equation of the set of all points equidistant from the points A(−2, 5, 2) and B(5, 1, −3). A perpendicular bisector is a line that cuts a line segment connecting two points exactly in half at a 90 degree angle. The method illustrated in this section is useful in solving, or at least getting an approximation of the solution, differential equations with coefficients that are not constant. Find an equation of the set of all points equidistant from the points A(−1, 6, 4) and B(5, 3, −3). To improve this 'Plane equation given three points Calculator', please fill in questionnaire. A normal vector is,. y + 4= (1/2)(x - 2)^2 b. This is the plane through the midpoint, at right angles to the segment (the spatial right bisector of the segment!). If you want to know how to find the perpendicular bisector of two points, just follow these steps. Visit Mathway on the web. Select all points that are on this parabola. Circle: Set of all points equidistant from a given point, the center. If (x 1,y 1. Find the equation of the set of all points P(x,y) that is equidistant from (-3,0) and (3,-5). In other words, every point on the circumference of a circle is. = a(y - k). Vertex Point Calculator. Problem - Find the vertex, focus and directrix of a parabola when the coefficients of its equation are given. asked Feb 18, 2015 in CALCULUS by anonymous plane-equation. But each point will cost 1 percent of your mortgage balance. 4MB0/01 Instructions tt Use black ink or ball-point pen. (- 3, 1, 2), (6, - 2, 4) Find an equation of the plane that contains all the points that are equidistant from the given points. (Note: if more than 3 fields are filled, only a third used to determine the triangle, the others are (eventualy) overwritten. There are two commonly used methods to find two points. t t Answer all questions. Because all the equations equal to 0 also the determinant of the coefficients should be equal to 0 and the value of the determinant will be as follows: Notice that we got the equation of a circle with the determinants equal to the coefficients A, B, C and D as follows; (x 2 + y 2) A + xB + yC + D = 0. 1) P(x,y) is equidistant from the y axis and (4,0). It is in the form: y=a(x-h)^2+k. Question Find an equation of the set of all points equidistant from the points A(−2, 5, 2) and B(5, 1, −3). Homework Equations None The Attempt at a Solution Um, I've drawn a graph of the two points and that's about as far as I could get. 6: Describe the equation x = 2 as it appears in 3. the sum of the distances) just as a circle is the set of points which are equidistant from one point (i. Example : Find the stationary point(s) of the function z =xe −− x y 2 2. (iii) Construct the locus of points equidistant from AC and BC. -----in general, given 2 points, you can find the slope by using the equation: slope = -----in general, given the slope and one of the points on the line, you can find the y intercept by solving for b in the general equation of: y = m*x + b. The set of all points equidistant from two points in n-space is a linear set -- in 2-space, it's a line, in 3-space it's a plane, and in 4-space it's a hyperplane (kind of abstract), and so forth. Find the equation of the parabola with focus (1,4) and vertex (−3,4). The graph of a function f is the set of all points in the plane of the form (x, f(x)). Substitution into the equation gives us or , which is true. Find an equation for the surface consisting of all points that are equidistant from the point (−1, 0, 0) and the plane x = 1. One is slope formula method and the other is area of triangle method. This is a quadric surface. Basic Math. Determine the equation of the parabola with focus (0,2) and directrix y = −4. Question Find an equation of the set of all points equidistant from the points A(−2, 5, 2) and B(5, 1, −3). Distance formula vs. A circle is the set of all points in a plane equidistant from a fixed point. 4 ) This unit shall be assessed by methods given in Volume 1, Part 3 'Assessment Guidelines'. In future videos we'll try to think about, how do you relate these points, the focus and directrix, to the actual, to the actual equation, or the actual. You can find the center of that curve (which is, by definition, the exact same distance from each of the three points) with the following method:-Draw a triangle with your three points as the vertices-Find the perpendicular bisector of each side of the triangle. Section 6-2: Equations of Circles Definition of a Circle A circle is the set of all points in a plane equidistant from a fixed point called the center point. Fundamentals Of 2d And 3d Graphs. In space, the equation describes all points This equation defines the yz-plane. Before that, let us understand the basics of the different types of triangle. chapter 12 cbse class 11 ncert solutions , download ncert solutions for class 11 chapter 12,. In order to do this, all we have to do is DC analysis of the transistor circuit. Find the point not on the y-axis that is equidistant from the points (2,-1,1) and (0,1,3). y - 2 = (1/4)(x - 2)^2 d. Distance Formula Calculator Find the distance between two points from x and y coordinates with this distance formula calculator. This is called the scalar equation of plane. Determine a way to express the distance from a line and use that to write an equation for a parabola that can be graphed with Graphing Calculator 3. Equation cercle. SketchAndCalc™ is an irregular area calculator app for all manner of images containing irregular shapes. Generating sampling points - making a series of points in the horizontal centre of a long polygon, equidistant spacing on vertical axis I have a polygon showing the extent of a beach. Which equation does the graph at right represent? Explain your answer. Equation of a straight line - online calculator Below you can use a calculator prepared to find the equation of a straight line. Plane analytic geometry. Cartesian Equation: (Hint: Pythagoras) (This has been used as a Computer Science interview question at both Oxford and Cambridge) If we draw a line down from (x,y), we get a right-angled triangle. A circle is defined as the set of all points in a plane that are equidistant from a fixed point called the center. If you are given an equation of a straight line and asked to draw its graph all you need to do is find two points whose coordinates satisfy the equation and plot the points. on a map these towns have coordinates (3,10) and (13, 4). (no calculator) 35min Monday, Part 2 50min (M 4/30 is an A-day) Write the equation of a circle with. - 10873807. The line that joins two infinitely close points from a point on the circle is a Tangent. The standard form of a parabola with vertex (0, 0) and the x-axis as its axis of symmetry can be used to graph the parabola. There are three possibilities: The lines intersect at zero points. So that's what they are. How to Find the Q-point of a Transistor Circuit. Recall that the definition of locus is the set points that meet some given conditions. If it will help to see it, the set of all points in a plane that are equidistant from points A and B in the. What is the meaning of 12 and 37 in the context of this trip? 3. Visit Mathway on the web. Find the equation of the plane that contains all the points that are equidistant from the given points $(-9, 3, 3), (6, -2, 4)$ I think the plane described lies in the midpoint of these points, and it is perpendicular to the line connecting the two points. Get an answer for 'Find the value of x so the distance between the points A(3,5) and B(x,8) is 5 units. System of Equations Calculator. centroid triangle calculator - step by step calculation, formula & solved example to find the mid or center point of 3 given points of a triangle (x 1, y 1), (x 2, y 2) & (x 3, y 3) on the multi-dimensional coordinate system or plane. MATH 11011 FINDING THE EQUATION KSU OF A CIRCLE Deflnitions: † Circle: is the set of all points in a plane that lie a flxed distance from a flxed point. The flxed distance is called the radius and the flxed point is called the center of the circle. Target question: Is point R equidistant from points (-3,-3) and (1,-3)? This question is a great candidate for rephrasing the target question. (Here's an easy way to think about it: If you have two pairs of congruent segments, then there's a perpendicular bisector. (1,1), (2, 4), (3,9), etc. To improve this 'Plane equation given three points Calculator', please fill in questionnaire. | 2020-08-06T13:55:13 | {
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http://mathhelpforum.com/pre-calculus/32778-graph-transformations.html | 1. ## Graph transformations...
Hello! I am a little confused about graph transformations and I have some questions that I don't really know how to solve...Please help! Thank you!
1)
The transformations A,B and C are as follows:
A: A translation of 1 unit in the negative y direction
B: A stretching parallel to the y-axis (with x-axis invariant) with a scale factor of 2
C: A reflection about the x-axis
A curve undergoes in succession, the above transformations A,B and C and the equation of the resulting curve is y = -2(x^2) -6. Determine the equation of the curve before the transformations were effected. [Is it possible to view the equation of the resulting curve as y= -2 (x^2 -3)? But I think this will give a different original equation from if I use the given resulting equation??]
2)
The equation of a graph is y = (ax)/(bx-y), and y= c is the horizontal asymptote while x = d is the vertical asymtote of this graph. Given that c = d = 1/2, what are the values of a and b? [If I am not wrong, b should be 2, but I don't know how to find a....Is a = 1?]
3)
Consider the curve (5x + 10)^2 - (y-3)^2 = 25. Prove, using an algebraic method, that x cannot lie between two certain values (which are to be determined). [eartboth has solved this here but I don't understand the method...Can someone explain?]
4)
The curve of y = x^2 + (3/x) undergoes a reflection about the line y=4 and the point (2, 5.5) is mapped onto (2, 2.5). Hence find the equation of the new curve in the form y = f(x).
Thank you very very much for helping me!
2. Originally Posted by Tangera
...
3)
Consider the curve (5x + 10)^2 - (y-3)^2 = 25. Prove, using an algebraic method, that x cannot lie between two certain values (which are to be determined). ...
...
$\frac{(x+2)^2}{1^2} - \frac{(y-3)^2}{5^2} = 1$
which is the equation of a hyperbola with the center (-2, 3) and the semi-axes a = 1 and b = 5.
Then I calculated the coordinates of the vertices and got the intervall in which x cannot lie.
I've attached a drawing of the curve to show where you can find this interval.
3. Originally Posted by Tangera
Hello! I am a little confused about graph transformations and I have some questions that I don't really know how to solve...Please help! Thank you!
1)
The transformations A,B and C are as follows:
A: A translation of 1 unit in the negative y direction
B: A stretching parallel to the y-axis (with x-axis invariant) with a scale factor of 2
C: A reflection about the x-axis
A curve undergoes in succession, the above transformations A,B and C and the equation of the resulting curve is y = -2(x^2) -6. Determine the equation of the curve before the transformations were effected. [Is it possible to view the equation of the resulting curve as y= -2 (x^2 -3)? But I think this will give a different original equation from if I use the given resulting equation??]
...
You have:
$y = -2x^2-6 = -2(x^2+3)$
Reflection about the x-axis will yield: $y = 2(x^2+3)$
The stretching parallel to the y-axis (with x-axis invariant) with a scale factor of $\frac12$ will yield: $y = x^2+3$
The translation of 1 unit in the positive y direction will yield: $\boxed{y = x^2+4}$
I've attached a sketch of the complete process. You start with the blue graph and you end with the red one. | 2016-12-09T05:15:01 | {
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https://mathhelpforum.com/threads/integration-of-product-of-two-exponential-functions.145182/ | # Integration of product of two exponential functions
#### nortyrich
Hello,
I apologise in advance if this is a naive question (which it probably is) but my maths is very rusty and my old notes and text books don't seem to cover this particular problem.
What is the correct approach to integrating the product of two exponential functions of the form below?
I have current, i(t) = 10e^-5000t
voltage, v(t) = 50(1-e^-5000t).
Power, p(t) = i(t).v(t)
Energy for t is greater than equal to zero is therefore the integral of p(t) for t = 0 to infinity.
I know that the answer is 50 mJ (or 0.050 joules) from calculating p(t) in a spreadsheet for small increments of t until p(t) tends to zero but cannot demonstrate it mathematically.
I've tried integrating by parts but this does not seem to close the deal as I always end up with a integral of the same or similar form to evaluate.
What is the blindingly obvious thing that I am missing ? Am i on the right track with integrating by parts or is there an easier reduction of the problem given the similarities between the two functions?
Many thanks
#### tonio
Hello,
I apologise in advance if this is a naive question (which it probably is) but my maths is very rusty and my old notes and text books don't seem to cover this particular problem.
What is the correct approach to integrating the product of two exponential functions of the form below?
I have current, i(t) = 10e^-5000t
voltage, v(t) = 50(1-e^-5000t).
Power, p(t) = i(t).v(t)
Energy for t is greater than equal to zero is therefore the integral of p(t) for t = 0 to infinity.
I know that the answer is 50 mJ (or 0.050 joules) from calculating p(t) in a spreadsheet for small increments of t until p(t) tends to zero but cannot demonstrate it mathematically.
I've tried integrating by parts but this does not seem to close the deal as I always end up with a integral of the same or similar form to evaluate.
What is the blindingly obvious thing that I am missing ? Am i on the right track with integrating by parts or is there an easier reduction of the problem given the similarities between the two functions?
Many thanks
According to what you wrote, you have the improper integral:
$$\displaystyle \int\limits^\infty_010e^{-5,000t}\cdot 50(1-e^{-5,000t})\,dt=$$ $$\displaystyle 500\int\limits_0^\infty\left(e^{-5,000t}-e^{-10,000t}\right)\,dt=$$ $$\displaystyle 500\lim_{b\to\infty}\left[-\frac{1}{5,000}\,e^{-5,000t}+\frac{1}{10,000}\,e^{-10,000t}\right]^b_0=$$ $$\displaystyle \frac{1}{20}\lim_{b\to\infty}\left(-\frac{2}{e^{5,000b}}+2e^0+\frac{1}{e^{10,000b}}-e^0\right)=$$
$$\displaystyle =\frac{1}{20}=0.05$$
Tonio
#### pickslides
MHF Helper
Multiply the functions together before integrating, what do you get? | 2019-11-13T12:45:34 | {
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http://math.stackexchange.com/questions/14672/computing-a-rational-number-between-two-others-minimizing-numerator-and-denomin | # Computing a rational number between two others, minimizing numerator and denominator
Given two positive rational number $\frac{a_1}{b_1}$ and $\frac{a_3}{b_3}$ (written in lowest terms) such that $$\frac{a_1}{b_1} < \frac{a_3}{b_3},$$ I want to find a rational number $\frac{a_2}{b_2}$ such that $$\frac{a_1}{b_1} < \frac{a_2}{b_2} < \frac{a_3}{b_3}$$ and $a_2$ and $b_2$ are the smallest possible numbers
I've come with the solution $a_2 = a_1 + a_3$ and $b_2 = b_1 + b_3$ then dividing $a_2$ and $b_2$ by their greatest common divisor but there must be better solution.
-
I wouldn't be surprised if your recipe is already the optimal one, although I don't have time to think about it right now. This should have something to do with it: en.wikipedia.org/wiki/Stern-Brocot_tree – Hans Lundmark Dec 17 '10 at 13:59
Ah, so it wasn't optimal after all. (Which should have been quite obvious, considering the simple counterexample given by TonyK below.) – Hans Lundmark Dec 18 '10 at 12:35
See Wikipedia's article on the mediant, which explains when your solution is the best.
-
Just so you know, I fixed the broken link in your answer. – Mike Spivey Dec 17 '10 at 16:03
@Mike: thanks . – lhf Dec 17 '10 at 17:03
This is also the method used by Chuquet (in the Middle ages) to approximate $k$th roots of rational numbers. – A Walker Oct 25 '11 at 20:22
Your solution is not the best. (Take, for instance, 1/2 and 9/10.) One way to solve your problem is to calculate the continued fractions of the two numbers until they differ. Say the continued fractions start $[x_0;x_1,x_2,x_3]$ and $[x_0;x_1,x_2,y_3]$, with $x_3 \ne y_3$. Then you can just try all values $z_3$ between $x_3$ and $y_3$ inclusive to get your answer. There might be a cleverer way to find the best $z_3$, but this way should work: it relies on the fact that continued-fraction convergents are alternately greater and less than the value that they are approximating, so one of $[x_0;x_1,x_2,x_3]$ and $[x_0;x_1,x_2,y_3]$ will lie between $a1/b1$ and $a3/b3$.
There is a slight complication: suppose $x_3$ and $y_3$ differ by one, and one of the convergents is exact. Then you might not be able to find $z_3$ such that $[x_0;x_1,x_2,z_3]$ is strictly between $a1/b1$ and $a3/b3$. But in this case you can use your formula on the two convergents (instead of on $a1/b1$ and $a3/b3$) to get the answer, because the condition that $x_3$ and $y_3$ differ by one is sufficient to guarantee that your mediant is the best approximation.
For example, 1/2=[0;2] and 9/10=[0;1,9]. So they differ at $x_1$, which means we should try [0;1] = 1 and [0;2] = 1/2. Neither is satisfactory, so we take their mediant, which is 2/3.
-
HINT $\rm\displaystyle\quad \frac{A}B < \frac{X}Y < \frac{C}D\ \iff\ X = \frac{MA+NC}{BC-AD},\ Y = \frac{MB+ND}{BC-AD}\quad$ for some integers $\rm\:M,N>0$
This has a nice geometric interpretation in terms of plane lattices - which provides a vivid geometric interpretation of good approximations by Farey mediant searching - see my post here
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https://math.stackexchange.com/questions/860565/arctan-3-2-doesnt-give-expected-result | # $\arctan(-3/2)$ doesn't give expected result.
Let's say I want to find the angle measure (in degrees) such that $\tan(x) = -3/2$.
It turns out that $x \approx 123.7$, and when I compute $\tan(123.7)$, I get $\approx -3/2$; so far so good.
However, when I compute $\arctan(-3/2)$, I get $\approx -56.3$ where I expected $\approx 123.7$ since $\arctan$ is the inverse of $\tan$.
• How does it turn out that $x \approx 123.7$? How do you know?
Jul 8, 2014 at 22:06
• @Brad It was given in the text, I just don't know how that value was derived though…I thought arctan of the ratio would give the same value, but apparently not. Jul 8, 2014 at 22:10
By definition, $\arctan x$ is the angle (in radians) between $-\pi/2$ and $\pi/2$ whose tangent is $x$.
There are infinitely many numbers $y$ such that $\tan y=-3/2$.
• But then, how do you find the value of angle x in this case? Jul 8, 2014 at 22:01
• tan(x) is periodic with period $\pi$ or $180^o$ so you should get about 123 by adding 180 to the -56 you get
– Jam
Jul 8, 2014 at 22:02
• André's point was that you will only get an output of arctan that is within limits of about -1.6 and 1.6 ($-90^o$ and $90^o$)(though there are infinitely many other answers including yours)
– Jam
Jul 8, 2014 at 22:05
• @oliveeuler Thank you; would you know how that number (123.7) was derived? It's in the context of a rotation transformation in Linear Algebra. It says that the angle of rotation determined by $\tan(x) = -3/2$ is in the second quadrant, does that make a difference? Jul 8, 2014 at 22:12
• If it was given in the text then I have no idea without reading it :)
– Jam
Jul 8, 2014 at 22:25
I get $−56.3$ where I expected $123.7$ since $\arctan$ is the inverse of $\tan$.
Hint: How much is $180^\circ-56.3^\circ$ ?
The points on the circle corresponding to $-56.3^\circ$ and to $123.7^\circ$ are antipodal to each other, since $-56.3^\circ+ 180^\circ= 123.7^\circ$. The period of the tangent and cotangent functions is only a half circle, i.e. $180^\circ$, rather than a full circle as with the sine, cosine, secant, and cosecant. Thus $\tan(-56.3^\circ)=\tan123.7^\circ$.
So the question is which of these two, $-56.3^\circ$ or $123.7^\circ$, should we take to be the inverse-tangent of $-3/2$? The picture below suggests that taking it to be the negative one allows the inverse tangent function to be continuous, since the tangent does not go up to infinity as the angle stays between $\pm90^\circ$. | 2022-10-03T11:10:20 | {
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https://math.stackexchange.com/questions/620495/prove-that-the-multiplicative-groups-mathbbr-0-and-mathbbc-0?noredirect=1 | # Prove that the multiplicative groups $\mathbb{R} - \{0\}$ and $\mathbb{C} - \{0\}$ are not isomorphic.
Is my proof correct? I have made use of the fact isomorphism preserves order of elements, which I proved couple of exercises back. I am also interested in other ways of proving it. Is there a more explicit way or is this explicit enough?
Problem Prove that the multiplicative groups $\mathbb{R} - \{0\}$ and $\mathbb{C} - \{0\}$ are not isomorphic.
Solution Recall that isomorphism preserves order of elements and hence if there exists an isomorphism from $\phi: \mathbb{C}-\{0\} \mapsto \mathbb{R}-\{0\}$, then $x \in \mathbb{C} - \{0\}$, and $\phi(x) \in \mathbb{R} - \{0\}$, then $\vert x \vert = \vert \phi(x) \vert$. Now note that the element $i \in \mathbb{C} - \{0\}$ has order $4$. However, no element in $\mathbb{R}-\{0\}$ has order $4$. Hence, no isomorphism can exist. Hence, the multiplicative groups $\mathbb{R} - \{0\}$ and $\mathbb{C} - \{0\}$ are not isomorphic.
Thanks
• This would be probably best way... I guess there would be no easier way.... well done! – user87543 Dec 28 '13 at 15:16
• @DietrichBurde I am working on Dummit and Foote. So as I write down the solution, if I am unsure and have some questions, I ask for verification to make sure my understanding is correct. Is it not allowed on this website? – John Smith Dec 28 '13 at 16:49
• Your argument also works with the real numbers under addition. – CopyPasteIt May 23 '17 at 14:14
Another way would be. Let's take $i$ and see what happen:
a) $f(i^2)=f(-1)=-1$, using basic properties of a morphism.
b) on the other side, we have $f(i)^2=x^2$ that can't produce -1.
• How do you know that $f(-1)=-1$? $f$ is just an isomoprhism of multiplicative groups, so it's not clear how it interacts with an additive inverse. Of course, you can argue that $-1$ is the unique element of order $2$, but if you're going to do an order argument, you might as well give the argument in the OP. – Lukas Heger Oct 13 '19 at 10:59 | 2021-05-09T08:19:48 | {
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http://math.stackexchange.com/questions/72123/express-this-curve-in-the-rectangular-form | # Express this curve in the rectangular form
Express the curve $r = 9/(4+\sin \theta)$ in rectangular form. And what is the rectangular form?
if I get the expression in rectangular form, how am I able to convert it back to polar coordinate?
-
Clear the denominator, distribute, and notice that $y=r\sin(\theta)$ and $r=\sqrt{x^2+y^2}$. – Bill Cook Oct 12 '11 at 21:00
"Rectangular form" means that instead of polar coordinates with $\theta$ an angle and $r$ the distance to the center, you must describe the points of the curve in a standard Cartesian ($x$, $y$) coordinate system. – Henning Makholm Oct 12 '11 at 21:01
As @Bill mentioned, use $\sin\theta=\frac{y}{r}=\frac{y}{\sqrt{x^2+y^2}}$ and simplify. To get back the polar, undo (reverse) the process e.g., change $x=r\cos\theta$ and $y=r\sin\theta$. – Tapu Oct 12 '11 at 21:15
what is the rectangular form?
It is the $y=f(x)$ expression of the curve in the $x,y$ referential (see picture). It can also be the implicit form $F(x,y)=F(x,f(x))\equiv 0$.
Steps:
1) transformation of polar into rectangular coordinates (also known as Cartesian coordinates) (see picture)
$$x=r\cos \theta ,$$
$$y=r\sin \theta ;$$
2) from trigonometry and from 1) $r=\sqrt{x^2+y^2}$
$$\sin \theta =\frac{y}{r}=\frac{y}{\sqrt{ x^{2}+y^{2}}};$$
3) substitution in the given equation $$r=\frac{9}{4+\sin \theta }=\dfrac{9}{4+\dfrac{y}{\sqrt{x^{2}+y^{2}}}}=9\dfrac{\sqrt{x^{2}+y^{2}}}{4\sqrt{x^{2}+y^{2}}+y};$$
4) from 1) $r=\sqrt{x^2+y^2}$, equate
$$9\frac{\sqrt{x^{2}+y^{2}}}{4\sqrt{ x^{2}+y^{2}}+y}=\sqrt{x^{2}+y^{2}};$$
5) simplify to obtain the implicit equation
$$4\sqrt{x^{2}+y^{2}}+y-9=0;$$
6) Rewrite it as $$4\sqrt{x^{2}+y^{2}}=9-y,$$ square it (which may introduce extraneous solutions, also in this question), rearrange as
$$16y^{2}+18y+15x^{2}-81=0,$$
and solve for $y$
$$y=-\frac{3}{5}\pm \frac{4}{15}\sqrt{81-15x^{2}}.$$
7) Check for extraneous solutions.
if I get the expression in rectangular form, how am I able to convert it back to polar coordinate?
The transformation of rectangular to polar coordinates is
$$r=\sqrt{x^{2}+y^{2}}, \qquad \theta =\arctan \frac{y}{x}\qquad \text{in the first quadrant},$$
or rather $\theta =\arctan2(y,x)$ to take into account a location different from the first quadrant. (Wikipedia link).
As commented by J.M. the curve is an ellipse. Here is the plot I got using the equation $16y^{2}+18y+15x^{2}-81=0$.
-
As a check: the original polar equation is in fact that of a conic with focus at the origin. Rewriting the expression as $r=\dfrac{\frac94}{1+\frac14\sin\,\theta}$, we find that we have an ellipse with eccentricity $\frac14$. The same result can be seen if we use completing the square to convert the Cartesian equation to a standard form and compute the eccentricity from the values of the semimajor and smiminor axes... – J. M. Oct 13 '11 at 11:09
@J.M. Thanks! I added a picture. – Américo Tavares Oct 13 '11 at 11:29
$$4r + 4r\sin\theta = 9.$$ Therefore $$\sqrt{x^2+y^2} + 4y = 9.$$ $$\sqrt{x^2+y^2} = 9 - y.$$ Now square both sides and go on from there.
But remember that squaring both sides can lead to extraneous roots. For example $3^2=9$ and $(-3)^2=9$, so a "$\pm$" is introduced when you go back to the unsquared form. In other words, the graph you get after squaring may contain additional points beyond those you should get, and you have to check for those.
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https://math.stackexchange.com/questions/823970/is-i-irrational/823981 | # Is $i$ irrational?
On the one hand, $i(=\sqrt{-1})$ cannot be expressed as a ratio of integers, so, by definition, $i$ is not rational $\iff i$ is irrational.
However, the set of irrational numbers, $\mathbb{J}=\mathbb{R}\setminus\mathbb{Q}$ is defined to be the set of all real numbers that are not in $\mathbb{Q},$ but, clearly, $i \notin \mathbb{R}$, so it must be that $i \notin \mathbb{J}$, so $i$ is not irrational.
Clearly, the first two paragraphs seem to contradict each other, so I'm asking: is $i$ irrational or rational?
• It's an integer ;) (it's a zero of the monic polynomial $X^2 +1 \in \mathbb{Z}[X]$, so it's an algebraic integer) – Daniel Fischer Jun 7 '14 at 14:56
• If is it not real, it cannot be rational [no joke] gweigt.net – AlainD Jun 7 '14 at 14:56
• If you define irrationals to be $\mathbb{R}\setminus \mathbb{Q}$ then to be an irrational number one must be in $\mathbb{R}$. So if you stick to your definition then you must not count it as an irrational number. – user50948 Jun 7 '14 at 14:56
• In a sense it is like asking whether a tree is irrational, on the one hand it cannot be expressed as a ratio of integers, one the other hand... (i'm not trying to be smart, just illustrating the importance of having a clear idea of what one is talking about). – user50948 Jun 7 '14 at 14:58
• @user50948 No, it's not. It's a perfectly reasonable question to ask whether a number is rational- it's an unrelated and irrelevant question to ask whether a tree has the properties of a number. – beep-boop Jun 8 '14 at 0:56
This depends on convention. As you say, if the irrationals are defined as $\,\Bbb R\setminus \Bbb Q\,$ then $\,i\,$ is neither irrational nor rational. However, many authors use "irrational" to mean "not rational", i.e. $\,\not\in \Bbb Q,\,$ therefore $\,i\,$ is irrational. This is quite common usage in university-level algebra.
For example, in my 2006/3/8 sci.math post I remarked that if one searches books.google.com for "irrational algebraic" one finds such usage by many eminent mathematicians: e.g. John Conway, Gelfond, Manin, Ribenboim, Shafarevich, Waldschmidt (esp. in diophantine approximation, e.g. Thue-Siegel-Roth theorem, Gelfond-Schneider theorem, etc). See also other posts in that sci.math thread titled "Is $\,i\,$ irrational"?
Our of curiosity, I ran another Google Books search on "real irrational" numbers. Authors using such terminology presumably employ the more general definition of irrational numbers. Among such authors I found the following eminent mathematicians: Bombieri, Davenport, Dedekind, Euler, Hurwitz, Kronecker, Kirilov, Mahler, Lang, Ostrowski, Ribenboim, Weil.
However, it is not easy to find an explicit definition since higher-level textbooks assume the reader already knows basic terminology. I vaguely recall that Gerry Myerson once posted to sci.math some links to definitions which make it unquestionably clear that the author employs the more general definition of "irrational". Perhaps someone can dig those up, or locate others.
In any case, it is a matter of definition. In most cases one can quickly infer the intended denotation from the context, so there is little chance for confusion.
• Could this concept be wxtwbded to other integer rings? Say I am working with $Z[\sqrt{-2}]$ and I divide $1+\sqrt{-2}$ by $2+\sqrt{-2}$. The thing I get does not belong the the algebraic integer ring, but might it be called "rational" with respect to the ring? – Oscar Lanzi Dec 31 '16 at 23:28
• @OscarLanzi Yes, fractions over a domain $D$ are sometimes called $D$-rational - see my comment on Tom's answer. – Bill Dubuque Jan 3 '17 at 16:01
• One context in which it's not immediately clear which convention is being used is the Gelfond-Schneider Theorem - it pertains to values $a^b$ where $a$ and $b$ are algebraic and $b$ is irrational, but it's not clear without knowing the details of the proof whether $i$ counts as irrational or not (it does). Personally I'd want to call nonreal numbers "non-rational" or something to avoid confusion, and leave "irrational" to mean real and non-rational. – Ken Williams Feb 14 '17 at 4:27
It is neither. Just as it's neither positive nor negative. The real numbers are partitioned into rational and irrational numbers; but things which aren't real numbers don't have to be one or the other.
Your conundrum is similar to saying "A $2 \times 2$ matrix is not rational. So it must be irrational. But it's not irrational; a contradiction." Does it seem more clear why that's a false dichotomy?
If I tell you that for lunch you can either have pizza or pasta, and you see someone eating a burger, does that negate my statement? Or reality?
Context is not absolute in mathematics, and it's important to remember that. It is true that in the context of the real numbers we have rationals and irrationals, and whatever is not rational is irrational. But in the context of the real numbers we also don't have $\sqrt{-1}$.
In the context of $\Bbb C$ we often talk less about irrational numbers so the context can be taken to both direction. It is possible to declare "In the context of the complex numbers, the irrational numbers are $\Bbb{C\setminus Q}$" in which case $i$ is certainly irrational, but it is also possible to make other declarations, like "The irrational numbers are still $\Bbb{R\setminus Q}$", in which case $i$ is not irrational -- but it doesn't make it rational either.
It's a question of context and it's not true that there's always a conventional and agreed upon context. This very question is a good example for that.
$\sqrt{-1} \notin \mathbb{R}$ and $\mathbb{N} \subsetneq \mathbb{Z} \subsetneq \mathbb{Q} \subsetneq \mathbb{R} \subsetneq \mathbb{C}.$
Note that $i$ is a number, and for any number, the question to ask whether it is rational or not is makes sense. It is not rational, since it is not a ratio of two integers. Hence, it is irrational, as irrational numbers are the complement of the rational ones (complement depending on context, either reals or complex numbers).
Actual definition of irrational number says that an irrational number is any real number that cannot be expressed as a ratio of integers. So your first line is infact wrong.
• Not necessarily true - see my answer. – Bill Dubuque Jun 7 '14 at 15:07
• @BillDubuque see this.Also i can be wrong, just clarify it – RE60K Jun 7 '14 at 15:08
• Definitions are conventions. Wikipedia chose one (the most elementary one). As I mentioned in my answer, that convention is not universal. – Bill Dubuque Jun 7 '14 at 15:18
• @BillDubuque it is a good point to be noted – RE60K Jun 7 '14 at 15:19
Several previous comments and answers apply, but no one yet seems to have mentioned Gaussian Integers, (see http://en.wikipedia.org/wiki/Gaussian_integer) - complex numbers whose real and imaginary parts are both (real) integers. By this definition, i is a Gaussian Integer. the concept extends to Gaussian Rationals (see http://en.wikipedia.org/wiki/Gaussian_rational), so that i is also a Gaussian Rational.
• But this is employing a different convention, namely the convention of referring to the elements of the fraction field of a a domain $\,D\,$ as $\,D$-rational. By that convention, any number can be considered "rational" relative to an appropriate domain (e.g. often used to distinguish such rationals from higher degree algebraic numbers over $\,D).\ \$ – Bill Dubuque Jun 7 '14 at 16:03
• @BillDubuque Well the issue in the OP question is one of mixing conventions, i.e. using terms that normally describe real numbers to create a confusion about an imaginary number: my answer is simply pointing out that there is a convention which does address rationality in the complex numbers. – Tom Collinge Jun 7 '14 at 16:12 | 2019-10-17T18:53:12 | {
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https://aparande.gitbook.io/berkeley-notes/ee120-0/ee120-7 | Sampling
# Continuous Time
Sampling a continuous-time signal means representing it as a sequence of points measured at regular intervals
$T$
. Notice that if we were to take a signal
$x(t)$
and multiply it by an impulse train, then we would get a series of impulses equal to
$x(t)$
at the sampling points and
$0$
everywhere else. We can call this signal
$x_p(t)$
.
$p(t) = \sum_{k=-\infty}^{\infty}{\delta(t-kT)}$
$x_p(t) = x(t)p(t) = \sum_{k=-\infty}^{\infty}{x(t)\delta(t-kT)}$
In the Fourier Domain,
\begin{aligned} X_p(\omega) &= \frac{1}{2\pi}X(\omega)*P(\omega)\\ P(\omega) &= \frac{2\pi}{T}\sum_{k=-\infty}^{\infty}{\delta(\omega-k\omega_0)}\\ \therefore X_p(\omega) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}{X(\theta)P(\omega-\theta)d\theta} = \frac{1}{T}\sum_{k=-\infty}^{\infty}{X(\omega-k\omega_0)}\end{aligned}
What this tells us is that the Fourier Transform of our sampled signal is a series of copies of
$X(\omega)$
, each centered at
$k\omega_0$
where
$\omega_0 = \frac{2\pi}{T}$
For example, lets say that our original signal has the following Fourier Transform. Notice the signal is band-limited by
$\omega_M$
.
There are two major cases: if
$\omega_0 > 2\omega_m$
and
$\omega_0 < 2\omega_M$
. Case One:
$\omega_s > 2\omega_m$
When
$\omega_s > 2\omega_M$
, the shifted copies of the original
$X(\omega)$
(shown in blue) do not overlap with each other or which the original copy. If we wanted to recover the original signal, we could simply apply a low pass filter to isolate the unshifted copy of
$X(\omega)$
and then take the inverse Fourier Transform. Case Two:
$\omega_s < 2\omega_m$
Notice how in this case, the shifted copies overlap with the original
$X(\omega)$
. This means in our sampled signal, the higher frequency information is bleeding in with the lower frequency information. This phenomenon is known as aliasing. When aliasing occurs, we cannot simply apply a low pass filter to isolate the unshifted copy of
$X(\omega)$
.
When
$\omega_0 = 2\omega_M$
, then our ability to reconstruct the original signal depends on the shape of its Fourier Transform. As long as
$X_p(k\omega_m)$
are equal to
$X(\omega_m)$
and
$X(-\omega_m$
), then we can apply an LPF because we can isolate the original
$X(\omega)$
and take its inverse Fourier Transform.
Remember that an ideal low pass filter is a square wave in the frequency domain and a sinc in the time domain. Thus if we allow
$X_r(\omega) = X_p(\omega)\cdot \begin{cases} T & \text{if } |\omega| < \frac{\omega_s}{2}\\ 0 & \text{else } \end{cases}$
then our reconstructed signal will be
$x_r(t) = x_p(t)*\text{sinc}\left(\frac{t}{T}\right) = \sum_{n=-\infty}^{\infty}{X(nT)\text{sinc}\left(\frac{t-nT}{T}\right)}.$
This is why we call reconstructing a signal from its samples "sinc interpolation." This leads us to formulate the Nyquist Theorem.
## Theorem 18 (CT Nyquist Theorem)
Suppose a continuous signal
$x$
is bandlimited and we sample it at a rate of
$\omega_s > 2\omega_M$
, then the signal
$x_r(t)$
reconstructed by sinc interpolation is exactly
$x(t)$
# Discrete Time
Sampling in discrete time is very much the same as sampling in continuous time. Using a sampling period of
$N$
we construct a new signal by taking an impulse train and multiplying elementwise with the original signal.
\begin{aligned} p[n]=\sum_{n=-\infty}^{\infty}{\delta[n-kN]}\\ x_p[n] = x[n]p[n] = \sum_{n=-\infty}^{\infty}{x[kN]\delta[n-kN]}\\ X_p(\omega) = \frac{1}{N}\sum_{k=0}^{N-1}{X(\omega-k\omega_s)}\end{aligned}
Our indices only go from
$k$
to
$N-1$
in the Fourier Domain because we can only shift a particular number of times before we start to get repeated copies. This is the impulse train sampled signal. It has 0’s at the unsampled locations. If we want to, we could simply remove those zeros and get a downsampled signal
$x_p[n] = x[Nn]$
Like in continuous time, the reconstructed signal is recovered via sinc interpolation.
$x_r[n] = \sum_{k=-\infty}^{\infty}{x_p[n]\text{sinc}\left(\frac{n-kN}{N}\right)}$
The Nyquist Theorem in DT will tell us when this works.
## Theorem 19 (DT Nyquist Theorem)
Suppose a discrete signal
$x$
is bandlimited by
$\frac{\pi}{N}$
and we sample it at a rate of
$\omega_s > 2\omega_M$
, then the signal
$x_r[n]$
reconstructed by sinc interpolation is exactly
$x[n]$
Thus as long as the Nyquist Theorem holds, we can take a downsampled signal and upsample it (i.e reconstruct the missing pieces) by expanding
$y$
by a factor of
$N$
and putting
$0's$
for padding, and then applying sinc-interpolation to it.
# Sampling as a System
Notice that we have two ways of representing our sample signal. We can either write it as a discrete time signal
$x_d[n] = x(nT)$
or we can write it as an impulse train
$x_p(t)=\sum_{-\infty}^{\infty}{x(nT)\delta(t-nT)}$
. Based on their Fourier Transforms,
\begin{aligned} X_d(\Omega)=\sum_{n=-\infty}^{\infty}{x(nT)e^{-j\Omega n}}\\ X_p(\omega)=\sum_{n=-\infty}^{\infty}{x(nT)e^{-j\omega nT}}\end{aligned}
Thus if we let
$\Omega=\omega T$
, then we see that these two representations of a signal have the same Fourier Transforms and thus contain the same information. This means that for some continuous signals, convert them to discrete time via sampling, use a computer to apply an LTI system, and convert the result back to a CT output.
We must be careful though because as long as the DT system we apply is LTI, the overall CT system will be linear too, but it will not necessarily be time invariant because sampling inherently depends on the signal’s timing.
\begin{aligned} Y_d(\Omega) &= H_d(\Omega)X_d(\Omega) = H_d(\Omega)X_p\left(\frac{\Omega}{T}\right)\\ Y_p(\omega) &= Y_d(\omega T) = H_d(\omega T)X_p(\omega)\\ Y(\omega) &= \left\{ \begin{array}{cc} T & |\omega| < \frac{\omega_s}{2}\\ 0 & |\omega| \ge \frac{\omega_s}{2} \end{array} \right\} \cdot Y_p(\omega) = \left\{ \begin{array}{cc} TH_d(\omega T)X_p(\omega) & |\omega| < \frac{\omega_s}{2}\\ 0 & |\omega| \ge \frac{\omega_s}{2} \end{array} \right\}\end{aligned}
Assuming that the Nyquist theorem holds,
\begin{aligned} X_p(\omega) &= \frac{1}{T}X(\omega)\\ \therefore Y(\omega) &= \left\{ \begin{array}{cc} H_d(\omega T)X(\omega) & |\omega| < \frac{\omega_s}{2}\\ 0 & |\omega| \ge \frac{\omega_s}{2} \end{array} \right\}\\ \therefore H_{system} &= \left\{\begin{array}{cc} H_d(\omega T) & |\omega| < \frac{\omega_s}{2}\\ 0 & |\omega| \ge \frac{\omega_s}{2} \end{array} \right\}\end{aligned}
This shows us that as long as the Nyquist theorem holds, we can process continuous signals with a disrete time LTI system and still have the result be LTI. | 2022-07-01T13:15:08 | {
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https://math.stackexchange.com/questions/2797813/finding-an-unknown-multiplying-the-determinant-of-a-matrix-a-when-given-a-modifi | # Finding an unknown multiplying the determinant of a matrix A when given a modified matrix A
I need to solve the following problem:
$\det\begin{bmatrix} 7a_1 & 7a_2 & 7a_3\\4b_1+8c_1 & 4b_2+8c_2 & 4b_3+8c_3 \\ 6c_1 & 6c_2 & 6c_3\end{bmatrix}=k\det \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{bmatrix}$
I did the following: $\det\begin{bmatrix} 7a_1 & 7a_2 & 7a_3\\4b_1+8c_1 & 4b_2+8c_2 & 4b_3+8c_3 \\ 6c_1 & 6c_2 & 6c_3\end{bmatrix}=7\cdot 6\cdot \det\begin{bmatrix} a_1 & a_2 & a_3\\4b_1+8c_1 & 4b_2+8c_2 & 4b_3+8c_3 \\ c_1 & c_2 & c_3\end{bmatrix}$
I then recalled that if matrix A is obtained from matrix B by adding a multiple of one row to another, then $\det A=\det B$. So, $\det\begin{bmatrix} a_1 & a_2 & a_3\\4b_1+8c_1 & 4b_2+8c_2 & 4b_3+8c_3 \\ c_1 & c_2 & c_3\end{bmatrix} = \det\begin{bmatrix} a_1 & a_2 & a_3\\4b_1 & 4b_2 & 4b_3 \\ c_1 & c_2 & c_3\end{bmatrix} = 4\cdot \det \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{bmatrix}$
This makes $k=6\cdot 7\cdot 4=168$. Is my logic correct here?
• Yes, you're correct. – poyea May 27 '18 at 10:59
Yes, your logic and step by step simplification to get $$\det\begin{bmatrix} 7a_1 & 7a_2 & 7a_3\\4b_1+8c_1 & 4b_2+8c_2 & 4b_3+8c_3 \\ 6c_1 & 6c_2 & 6c_3\end{bmatrix}=168\det \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{bmatrix}$$ | 2019-08-26T04:59:42 | {
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https://math.stackexchange.com/questions/1223034/non-trivial-lower-bound-approximation-of-a-convex-function-using-the-second-deri | # Non-trivial lower bound approximation of a convex function using the second derivative at the minimum
Say that I am given an infinitely differentiable convex function $f: \mathbb{R}^n \rightarrow \mathbb{R}$.
I am wondering if I can construct a meaningful lower bound approximation of $f$ using it's minimum value $f(x^*)$ as well as the value of it's second derivative $\nabla^2 f(x^*)$ at the minimum.
Specifically, I am looking for a function $g: \mathbb{R}^n \rightarrow \mathbb{R}$ such that:
$$g(x) \leq f(x) \text{ for all } x \in \mathbb{R}^n$$
By meaningful, I mean any function $g(x)$ other than the constant function $g(x) = f(x^*)$.
EDIT: I have asked the question in a general form, but it if helps, the function that I am interested in is the $L_2$-regularized logistic loss function $$f(x) = \log(1+\exp(s^Tx)) + C\|x\|^2_2$$
where $s \in \mathbb{R}^n$ and $C > 0$.
The function is not only infinitely differentiable in $x$, but is also strictly convex in $x$ and with a unique minimum.
EDIT #2: To add to @Michael's answer, you can also find a non-trivial lower / upper bound for strictly convex functions in Section 9.1.2 of Stephen Boyd's Convex Optimization Book.
• Is the above supposed to be $C||x||^2$? – Michael Apr 6 '15 at 23:39
No. For the general case, you cannot get a lower bound that is better than the constant function $f(x)\geq f(x^*)$ for all $x$.
Claim: For any $a>0$, any $z>0$, and any $\epsilon>0$, we can construct a function $f:\mathbb{R}\rightarrow\mathbb{R}$ that is infinitely differentiable, convex, has minimum at $x=0$ given by $f(0)=0$, has $f''(0)=a$, and satisfies $0 \leq f(x)\leq \epsilon$ for all $x \in [0,z]$.
Proof: Fix $a>0$. Let $b>0$ be a given constant and define $f:\mathbb{R}\rightarrow\mathbb{R}$ by:
$$f(x) = \frac{a(bx+e^{-bx}-1)}{b^2}$$ Thus: \begin{align} f'(x) &= \frac{a(1-e^{-bx})}{b}\\ f''(x) &= a e^{-bx} \end{align}
Clearly $f''(x) \geq 0$ for all $x$, so the function is convex. The function is also infinitely differentiable and has minimum at $x=0$, given by $f(0)=0$. Further, regardless of the value of $b>0$, we get $f''(0)=a$. However, the value of $f(z)$ is:
$$f(z) = \frac{a(bz + e^{-bz} - 1)}{b^2}$$
which can be made smaller than $\epsilon$ by choosing a suitably large $b>0$. Since $f(x)$ is non-decreasing for $x \geq 0$, it follows that $0 \leq f(x) \leq \epsilon$ for all $x \in [0,z]$.
$\Box$
Note: I got this example by starting with $f''(x) = ae^{-bx}$ and integrating.
You might be able to get a non-trivial lower bound for your specific function.
If you have a uniform bound on $f''(x)$ for all $x \in \mathbb{R}$ then you can get what you want: Suppose there is a constant $c>0$ such that $f''(x) \geq c$ for all $x \in \mathbb{R}$. Then by the Taylor expansion there is a $z$ in between $x^*$ and $x$ such that: \begin{align} f(x) &= f(x^*) + (x-x^*)f'(x^*) + \frac{(x-x^*)^2}{2}f''(z) \\ &= f(x^*) + \frac{(x-x^*)^2}{2}f''(z)\\ &\geq f(x^*) + \frac{c(x-x^*)^2}{2} \end{align} where the second equality holds because $f'(x^*)=0$.
In the $n$-dimensional case you would want a matrix $C$ such that $\nabla^2 f(x) - C$ is positive semi-definite for all $x \in \mathbb{R}^n$. Then the lower bound would be $f(x) \geq f(x^*) + \frac{1}{2}(x-x^*)^TC(x-x^*)$ for all $x \in \mathbb{R}^n$.
• Thank you so much for a clear explanation + proof! I do actually have the uniform bound on $f''(x)$! One follow-up question: how did you use the Taylor expansion to justify that there was a $z$ between $x$ and $x^*$ such that your first equation holds? – Elements Apr 7 '15 at 2:03
• See "Mean value forms of the remainder" here: en.wikipedia.org/wiki/Taylor%27s_theorem – Michael Apr 7 '15 at 3:04 | 2021-03-07T01:56:29 | {
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https://math.stackexchange.com/questions/2465561/intersection-of-uncountable-set-and-countable-set | Intersection of uncountable set and countable set.
My friend told me that say you have an uncountable set $A$ and a countable set $B$, then the intersection of these two sets is the empty set. But wouldn't something like $A = [0,1]$ and $B = \{1, 2\}$ have the intersection of $\{1\}$? Also is there a way to prove this? Thanks!
• Friends do say some peculiar things.... – Lord Shark the Unknown Oct 10 '17 at 6:04
• Your counterexample is correct. – Qudit Oct 10 '17 at 6:05
• With friends like that, you will never need enemies – Mark Fischler Oct 10 '17 at 6:05
• Yeah I figured he was wrong so does this also mean that the intersection of countable and uncountable set is always countable? Seems like it should be. – aspookyghost20 Oct 10 '17 at 6:07
• Yes, you cannot generate an uncountable set from an intersection (subset) of a countable set. It will be countable or empty. Proof by contradiction – Daniel Oct 10 '17 at 6:43
$A \cap B \subseteq B$.
Hence $A \cap B$ is countable but as you have shown, it need not be empty.
Another counter example would be let $A = \mathbb{R}$ and $B = \mathbb{Q}$, then $A \cap B = \mathbb{Q} \neq \emptyset.$ | 2019-10-15T02:25:55 | {
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http://mathhelpforum.com/number-theory/122696-cardinality-finite-sets.html | # Math Help - cardinality of finite sets
1. ## cardinality of finite sets
Show that |P(X)| = 2 ^ (|X|) for all finite sets X
Any help would be greatly appreciated!! Thank you!
2. Originally Posted by pseudonym
Show that |P(X)| = 2 ^ (|X|) for all finite sets X
Any help would be greatly appreciated!! Thank you!
Prove it by induction on $|X|$ (hint: if A is any subset of a set $\{x_1,\ldots,x_{n-1}\}$ with n-1 elements, how many subsets of $\{x_1,\ldots,x_{n-1},x_n\}$ can be formed out of it ?)
Tonio
3. Originally Posted by pseudonym
Show that |P(X)| = 2 ^ (|X|) for all finite sets X
Any help would be greatly appreciated!! Thank you!
Let $f(n)$ denote the number of subsets of $\left\{1\cdots,n\right\}$. Consider the power set of $\left\{1,\cdots,n,n+1\right\}$. Partition set into two blocks $B_1=\left\{A\in\wp\left(\left\{1,\cdots,n+1\right\ }\right):n+1\in A\right\}$ and $\wp\left(\left\{1,\cdots,n+1\right\}\right)-B_1=B_2$. Clearly, we have that $\left|B_2\right|=f(n)$. And, with the exception of $\{n+1\}$ we may (and can) only put $\{n+1\}$ into each of the subsets of in $B_1$. Thus, we have that $f\left(n+1\right)=f(n)+f(n)+1$...almost...we double counted one thing (what is it??). So $f(n+1)=f(n)+f(n)+1-1=2f(n)$. Thus, it follows by induction that the recurrence relation $f(n+1)=2f(n)$ has solution $f(n)=C\cdot 2^n$ for some $C$. Noting though that $f(0)$ is the number of subsets of the empty set we may conclude that $f(0)=1=C\cdot 2^0=C$. The conclusion follows.
EDIT: I am sorry tonio. I didn't see your post!
4. Why so complicated? Just notice that if you want to construct a subset $S$ of $X$, you have two possibilities for every element of $X$: either you include it, or you exclude it. So in total you have $2^{|X|}$ possibilities.
Another proof : clearly the number of subsets of $X$ is
${n \choose 1} + {n \choose 2} + \dots + {n \choose n} = (1+1)^n = 2^n$
5. Originally Posted by Bruno J.
Why so complicated? Just notice that if you want to construct a subset $S$ of $X$, you have two possibilities for every element of $X$: either you include it, or you exclude it. So in total you have $2^{|X|}$ possibilities.
Another proof : clearly the number of subsets of $X$ is
${n \choose 1} + {n \choose 2} + \dots + {n \choose n} = (1+1)^n = 2^n$
I feel as though what I said is exactly what you said.
6. Which of the two proofs is exactly like yours?
7. Originally Posted by Bruno J.
Which of the two proofs is exactly like yours?
I didn't mean exactly exactly . I only meant that the first one is the exact point of my proof. I just did it in a slightly different way.
8. Your proof relies on induction whereas my proof relies on a combinatorial argument... I don't think they're similar at all!
9. Originally Posted by Bruno J.
Your proof relies on induction whereas my proof relies on a combinatorial argument... I don't think they're similar at all!
The induction is merely the formality of it. The point was that if you fix an element in $\left\{1,\cdots,n+1\right\}$ it only has two places to go. I really do think we are saying the exact same thing, just the means of conveyance is different.
10. I'm just teasing you at this point.
11. Originally Posted by Bruno J.
I'm just teasing you at this point.
12. Originally Posted by Bruno J.
Why so complicated? Just notice that if you want to construct a subset $S$ of $X$, you have two possibilities for every element of $X$: either you include it, or you exclude it. So in total you have $2^{|X|}$ possibilities.
Another proof : clearly the number of subsets of $X$ is
${n \choose 1} + {n \choose 2} + \dots + {n \choose n} = (1+1)^n = 2^n$
The "another proof" is not so unless you first prove that there are $\binom{n}{k}$ subsets with k elements out of a set with n elements, $k\le n$, and also you forgot the number $\binom{n}{0}$ in the sum and also, and perhaps most important, this another proof relies on "hidden induction": after all, we need to show that's true for all n, just like the formal proof of Newton's binomial theorem is usually, and as far as I am aware uniquely, done by induction.
Tonio | 2015-05-27T15:00:09 | {
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https://math.stackexchange.com/questions/710038/finding-an-orthonormal-basis-for-the-space-p-2-with-respect-to-a-given-inner-p | # Finding an orthonormal basis for the space $P_2$ with respect to a given inner product
I am so confused on what to do for this question.
The questions asks to find an orthonormal basis of $P_2$, the space of quadratic polynomials, with respect to the inner product $$\langle p, q\rangle = 2\int_{0}^{1} p(x)q(x)\, dx.$$ I don't know how to this the question with integrals. All I know is that it has something do with the Gram-Schmidt process.
Any help is appreciated thanks!
• I've TeX-ified your question and added a clarifying comment based on context. Readers will be grateful if you format your future posts. :) – Andrew D. Hwang Mar 12 '14 at 22:03
• For starters, do you understand how the Gram-Schmidt algorithm works in, say, $\mathbf{R}^3$? – Andrew D. Hwang Mar 12 '14 at 22:04
The standard method of orthogonalisation works like this:
1) take a basis $\{a_n\}$
2) take first vector $a_1$ , normalise it, call it the first vector in the new basis: $$b_1:=\frac{a_1}{\sqrt{\langle a_1,a_1\rangle}}.$$
3) take second vector $a_2$, find its component orthogonal to the first vector in the new basis: $$a'_2=a_2- \langle a_2,b_1\rangle b_1 ,$$ normalise it, then call it the second vector in the new basis: $$b_2:=\frac{a'_2}{\sqrt{\langle a'_2,a'_2\rangle}}.$$
4) repeat for all consecutive vectors - orthogonalisation comes for all previous vectors in new basis: $$a_k'=a_k- \sum_{j=1}^{k-1} \langle a_k,b_j\rangle b_j,\quad b_k= \frac{a_k'}{\sqrt{\langle a_k',a_k'\rangle}}.$$
So, let's take a canonical basis in $P_2[x]$: $\{1,x,x^2\}$.
First vector is $1$. It's norm with respect to our inner product is $2\int_0^11\cdot 1dx =2$, hence the first vector in the new basis is $b_1=\frac{1}{\sqrt 2}$.
Second vector is $x$. The orthogonalisation yields $$x- \langle x,1/\sqrt 2\rangle 1/\sqrt 2= x-\frac{1}{\sqrt 2}\cdot2\int_0^1\frac{x}{\sqrt 2}dx =x-\frac 12.$$
Can you now normalise it (i.e. find the second vector in the new basis) and then find the third vector in the new basis?
• for the second vector do i do the formula thats something like x2 X v1/v1 x v1? – user123204 Mar 12 '14 at 22:18
• just wondering did you forget to multiply the 2 in front of the integral to solve for the second vector? – user123204 Mar 12 '14 at 23:00
• @user123204 I edited the post for clarity. If you still have questions, feel free to ask. – TZakrevskiy Mar 13 '14 at 7:24 | 2021-05-13T08:59:53 | {
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https://www.coursehero.com/file/p75jo6/Consider-the-linear-transformation-L-P-3-7-P-3-defined-by-L-p-1-x-2-d-2-p-dx-2-2/ | # Consider the linear transformation l p 3 7 p 3
• Notes
• 13
• 100% (2) 2 out of 2 people found this document helpful
This preview shows page 6 - 9 out of 13 pages.
4. (20 points) Consider the linear transformation L : P 3 7→ P 3 defined by L ( p ) = (1 - x 2 ) d 2 p dx 2 - 2 x dp dx a) Find the matrix representation of L in the standard basis { 1 , x, x 2 , x 3 } b) Is L onto? one-to-one? c) Find the eigenvalues and eigenvectors of L . Your eigenvectors must be ele- ments of P 3 . d) Give a basis B of P 3 such that [ L ] B is diagonal. In your basis what is [ L ] B ? Solution.
c) Since the matrix is triangular the eigenvalues are the diagonal entries λ = 0 , - 2 , - 6 , - 12. We’ve actually already calculated the first two eigenvectors as well. Since L (1) = 0 = 0 · 1, and L ( x ) = - 2 x = - 2 · x we have that v 0 ( x ) = 1 and v - 2 ( x ) = x are the eigenvectors for λ = 0 and λ = - 2 respectively. To compute the other eigenvectors we proceed as usual: [ L ] S + 6 I = 6 0 2 0 0 4 0 6 0 0 0 0 0 0 0 - 6 3 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 v - 6 = 1 0 - 3 0 v - 6 ( x ) = 1 - 3 x 2 [ L ] S + 12 I = 12 0 2 0 0 10 0 6 0 0 6 0 0 0 0 0 1 0 0 0 0 5 0 3 0 0 1 0 0 0 0 0 v - 12 = 0 3 0 - 5 v - 12 ( x ) = 3 x - 5 x 3 d) We have now a basis of P 3 consisting of eigenvectors of L : B = { 1 , x, 1 - 3 x 2 , 3 x - 5 x 3 } . In this basis the matrix representation of L is diagonal: L B = 0 0 0 0 0 - 2 0 0 0 0 - 6 0 0 0 0 - 12 .
5. (15 points) For each of the following statements, decide whether it is true or false and give a brief proof or an explicit counterexample. a) If A, B , and P are n × n matrices which satisfy the equation AP = PB , then A and B are similar.
b) If A and B are n × n matrices and 0 is an eigenvalue of A , then 0 is an eigenvalue of AB . | 2021-12-05T08:31:00 | {
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https://kokecacao.me/page/Course/S22/15-259/Lecture_013.md | # Lecture 013
## Parameter Estimation
We want to estimate the probability $p$ of a coin by sampling. We took $n$ samples where each sample is $X_i$. Together, assume each $X_i$ are i.i.d., we have:
X = \sum_{i = 1}^n X_i
Therefore, we want to find out $\delta$:
\begin{align*} &Pr\{p \in [\frac{X}{n} - \delta, \frac{X}{n} + \delta]\} \geq 0.95\\ \iff& Pr\{|\frac{X}{n} - p| > \delta\} \leq 0.05\\ \iff& Pr\{|X - np| > n\delta\} \leq 0.05 \tag{multiply both side by $n$}\\ \iff& Pr\{|X - E[X]| > n\delta\} \leq 0.05 \tag{by $X \sim \text{Binomial}(n, p) \implies E[X] = np$}\\ \iff& 2e^{-\frac{2(n\delta)^2}{n}} \leq 0.05 \tag{by Pretty Chernoff Bound}\\ \iff& \delta \geq \sqrt{\frac{-\ln(0.025)}{2n}}\\ \iff& \delta \geq \sqrt{\frac{1.84}{n}}\\ \end{align*}
Since $n\delta \in \Theta(n)$, it makes sense to use Pretty Chernoff Bound for i.i.d. Binomial. Also notice that $\delta$ grows as $\frac{1}{\sqrt{n}}$.
So we conclude $[\frac{X}{n} - 0.043, \frac{X}{n} + 0.043]$ forms a 95\% confidence interval on the true $p$.
Of course there are many issues that come up in statistical parameter estimation. For example, it is not obvious how to get "independent", equally weighted samples.
## Balls and Bins
We randomly distribute $n$ balls in $n$ bins. Assuming $n$ is sufficiently large, we want to show with high probability, no bin will have more than $\frac{3\ln n}{\ln \ln n} - 1 \in O(\frac{\ln(n)}{\ln \ln n})$ balls.
• Sufficiently Large: $n \rightarrow \infty$
• With High Probability: $Pr \geq 1 - \frac{1}{n}$ with high $n$
Note that we chose $\frac{3\ln n}{\ln \ln n} - 1$ to simplify calculation. The reason we chose $k = \frac{3\ln n}{\ln \ln n} - 1$ because it is slower than $\ln(n)$
k = \frac{\ln n}{\ln \ln n} \sim \frac{3\ln n}{\ln \ln n} - 1 < \frac{\ln n}{10000} < \ln(n)
We define the total number of balls in bin $j$.
\begin{align*} B_j =& \text{Binomial}(n, \frac{1}{n})\\ =& \sum_{i = 1}^n X_i \tag{where $X_i = \begin{cases}1 & \text{if ith ball go to bin j}\\0&\text{otherwise}\end{cases}$}\\ \end{align*}
Want to show:
\begin{align*} &Pr\{\forall j, B_j < k\} \geq (1 - \frac{1}{n}) \tag{where $k = \frac{3\ln n}{\ln \ln n} - 1$}\\ \iff& Pr\{\exists j (B_j > k)\} \leq \frac{1}{n}\\ \iff& Pr\{B_1 > k \cup B_2 > k \cup ... \cup B_n > k\} \leq \frac{1}{n}\\ \iff& Pr\{B_1 > k\} + Pr\{B_2 > k\} + ... + Pr\{B_n > k\} \leq \frac{1}{n} \tag{Union Bound}\\ \iff& Pr\{B_j > k\} \leq \frac{1}{n^2} \tag{$\forall j$}\\ \iff& Pr\{B_j \geq 1+k\} \leq \frac{1}{n^2}\\ \iff& \left(\frac{e^k}{(1 + k)^{1 + k}}\right) \leq \frac{1}{n^2}\tag{Ugly Chernoff Bound}\\ \iff& k - (1 + k)\ln(1 + k) \leq -2\ln n \tag{$\ln$ both sides}\\ \iff& \frac{3\ln n}{\ln \ln n} - 1 - \frac{3 \ln n}{\ln \ln n} \cdot \ln \left(\frac{3 \ln n}{\ln \ln n}\right) \leq -2 \ln n \tag{subsitute $k = \frac{3\ln n}{\ln \ln n} - 1$}\\ \iff& \frac{3}{\ln \ln n} - \frac{1}{\ln n} - \frac{3 \ln 3}{\ln \ln n} - 3 + \frac{3 \ln \ln \ln n}{\ln \ln n} \leq -2 \tag{simplify}\\ \iff& o(1) + o(1) + o(1) - 3 + o(1) \leq -2 \tag{assume $n \rightarrow \infty$}\\ \end{align*}
Notice $o(1)$ is used with all positive sign. This is because $o(1) \rightarrow 0$ with high $n$ and cannot exceed $1$.
Table of Content | 2023-03-27T04:14:12 | {
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http://forum.math.toronto.edu/index.php?PHPSESSID=vr7acj5rroo41qop46p208c185&action=printpage;topic=831.0 | # Toronto Math Forum
## APM346-2016F => APM346--Tests => Q6 => Topic started by: XinYu Zheng on November 10, 2016, 08:54:19 PM
Title: Q6
Post by: XinYu Zheng on November 10, 2016, 08:54:19 PM
Solve
$$\begin{cases} u_{xx}+u_{yy}=0& r<a\\ u_r|_{r=a}=f(\theta) \end{cases}$$
Where
$$f(\theta)=\begin{cases} 1 & 0<\theta<\pi\\ -1 & \pi<\theta <2\pi \end{cases}$$
BONUS:
(a) What is a necessary and sufficient condition on $f(\theta)$ for solution to exist? Is this condition satisfied here?
(b) Is the solution unique?
Solution:
For Laplace's equation in the inner disk we know that the general solution takes the form
$$u(r,\theta)=\frac{A_0}{2}+\sum_{n\geq 1} r^n (A_n\sin(n\theta)+B_n\cos(n\theta))$$
Where we have dropped the logarithm and any terms with $r^{-n}$. Applying the boundary condition we have
$$f(\theta)=\sum_{n\geq 1}n a^{n-1}(A_n\sin(n\theta)+B_n\cos(n\theta))$$
At this point, the coefficients can be directly calculated:
$$A_n=\frac{a^{1-n}}{n\pi}\int_0^{2\pi}f(\theta)\sin(n\theta)\,d\theta=\frac{a^{1-n}}{n\pi}\left(\int_0^{\pi}\sin(n\theta)\,d\theta+\int_\pi^{2\pi}-\sin(n\theta)\,d\theta\right)=\frac{a^{1-n}}{n^2\pi}(\cos(n\theta)|_{\pi}^0+\cos(n\theta)|_\pi^{2\pi})=\frac{4a^{1-n}}{n^2\pi}\,\,\,\text{n odd, 0 otherwise}$$
Similarly,
$$B_n=\frac{a^{1-n}}{n\pi}\int_0^{2\pi}f(\theta)\cos(n\theta)\,d\theta=\frac{a^{1-n}}{n\pi}\left(\int_0^{\pi}\cos(n\theta)\,d\theta+\int_\pi^{2\pi}-\cos(n\theta)\,d\theta\right)=0$$
Where the integrals are 0 because we will be evaluating sine functions at integer multiples of $\pi$. Thus we have our solution:
$$u(r,\theta)=\frac{A_0}{2}+\sum_{n\geq 1, n\,\,odd}\frac{4a^{1-n}}{n^2\pi}r^n\sin(n\theta)$$
BONUS:
(a) Note that in the fourier expansion of $f(\theta)$ we have no constant term. Thus for solution to exist we must demand
$$\int_0^{2\pi}f(\theta)\,d\theta=0$$
Which is fulfilled here.
(b) No. In Neumann problems the solution is defined up to a constant, in this case $A_0/2$. | 2018-07-23T15:10:19 | {
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https://math.stackexchange.com/questions/3297028/finding-int-frac11x3dx-without-partial-fractions | # Finding $\int \frac{1}{1+x^3}dx$ without partial fractions
Find, without partial fractions $$\int\dfrac{1}{x^3+1}dx$$
My Attempt: I was able to do it via partial fractions by factoring the denominator as
$$(x+1)(x^2-x+1)$$
However, I then tried a different approach without using partial fractions. I added and subtracted $$+x^3$$ in the numerator and wrote the integrand as
$$1-\dfrac{x^3}{x^3+1}.$$
Then, as the first term is easily integrable, I took the second term and wrote it as
$$\dfrac{x^2\cdot x}{x^3+1}.$$
Using Integration by Parts, I integrated
$$\dfrac{x^2}{x^3+1}$$
and differentiated $$x$$. I ended up with a term and a new integral,
$$\dfrac{x\cdot \ln{(x^3+1)}}{3} + \int \dfrac{\ln{(x^3+1)}}{3}dx$$
To evaluate the second integral, I again used integration by parts wherein I integrated $$x$$ and differentiated
$$\ln{(x^3+1)}.$$
Finally, I got the original integral as one of the parts. However, when I undid all the integration by parts to substitute in the original integral, both sides had the same terms and I ended up with
$$0 = 0.$$
Is there any other way to solve this integral?
• See here for a tutorial about how to use MathJax. Jul 18, 2019 at 17:18
• the partial fractions approach here is certainly the easiest Jul 18, 2019 at 17:21
• Do you mean you did something like this? $\int f'g=fg-\int g'f=fg-\left(fg-\int f'g\right)$
– user239203
Jul 18, 2019 at 17:21
• @Gae.S. yeah I used by parts twice
– user671231
Jul 18, 2019 at 17:25
• @GeorgeDewhirst yes I know that. I just want to learn and explore other methods to expand my arsenal or integration skills
– user671231
Jul 18, 2019 at 17:26
Start off with the substitution: $$x=\frac{1-t}{1+t}\Rightarrow dx=-\frac{2}{(1+t)^2}dt$$ This substitution produces a nice cancelation in the denominator, since: $$(1+t)^3+(1-t)^3=1+3t +3t^2+t^3 +1-3t+3t^2 -t^3=2(1+3t^2)$$
$$\int \frac{1}{1+x^3}dx=-\int\frac{1}{\frac{(1+t)^3}{(1+t)^3}+\frac{(1-t)^3}{(1+t)^3}}\frac{2}{(1+t)^2}dt=-\int \frac{1+t}{1+3t^2}dt$$ $$=-\frac{1}{\sqrt 3}\arctan(\sqrt 3 t)-\frac16 \ln(1+3t^2)+C,\quad t=\frac{1-x}{1+x}$$ | 2022-07-04T00:58:39 | {
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http://mathhelpforum.com/advanced-statistics/53053-simple-probability-problem.html | # Math Help - Simple Probability Problem
1. ## Simple Probability Problem
In a particular town 60% of the population are women. 4% of the men and 1% of the women are taller than 180 cm.
i) What % of the town's population is taller than 180 cm?
ii) If a person is chosen at random and is taller than 180 cm, what is the probability that the person is a woman?
Ummm I'm not that good at probability so could someone please give me some help?
For:
i) I did:
Percentage = 100(0.04*0.4 + 0.01*0.6)
= 0.022*100
= 2.2%
ii) I did:
P = 0.022*0.2
= 11/2500
The answers don't look rite though...
2. You have the first part right, for the second one you need to use Baye's theorem.
$
P(woman|taller than 180)= \frac{P(taller than 180|woman)P(woman)}{P(taller than 180|woman)P(woman)+P(taller than 180|man)P(man)}
$
3. Originally Posted by xwrathbringerx
In a particular town 60% of the population are women. 4% of the men and 1% of the women are taller than 180 cm.
i) What % of the town's population is taller than 180 cm?
ii) If a person is chosen at random and is taller than 180 cm, what is the probability that the person is a woman?
Ummm I'm not that good at probability so could someone please give me some help?
For:
i) I did:
Percentage = 100(0.04*0.4 + 0.01*0.6)
= 0.022*100
= 2.2% Mr F says: Correct.
ii) I did:
P = 0.022*0.2 Mr F says: Wrong. Where has the 0.2 come from?
= 11/2500
The answers don't look rite though...
ii requires conditional probability.
Here is the magic recipe, the big secret, the great trick, the golden elixir. Read it carefully ........
Set up careful notation and definitions. Carefully define all known probabilities in terms of this notation.
Let T be the event height greater than 180 cm.
You have:
Pr(M) = 2/5, Pr(W) = 3/5,
Pr(T | M) = 1/25, Pr(T | W) = 1/100,
Pr(M and T) = Pr(T| M) Pr(M) = (1/25) (2/5) = 2/125,
Pr(W and T) = Pr(T| W) Pr(W) = (1/100) (3/5) = 3/500,
Pr(T) = Pr(M and T) + Pr(W and T) = 2/125 + 3/500 = 11/500 (= 0.022 as you found).
You require Pr(W | T).
$\Pr(W | T) = \frac{\Pr(W \cap T)}{\Pr(T)} = \frac{3/500}{11/500} = 3/11$.
4. You have the first part right, for the second one you need to use Baye's theorem.
To keep not that messy, Let
W=woman
M=Man
T=taller than 180
$
P(W|T)= \frac{P(T|W)P(W)}{P(T|W)P(W)+P(T|M)P(M)}=\frac{(0. 01)(0.60)}{(0.01)(0.60)+(0.04)(0.40)}
$
$
=\frac{0.006}{0.006+0.016}=\frac{0.006}{0.022}=\fr ac{3}{11}
$ | 2014-03-11T16:41:09 | {
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https://math.stackexchange.com/questions/1750660/conditional-probability-on-dice | Conditional Probability on dice
I have some trouble understanding the conditional probability of the following question.. Do pardon me even though it is a simple question..
Two fair dice are rolled and the sum on the faces is calculated. Find the probability : Pr(Sum of two dice is 8 | we know that the number on both dice is even)
This is my working
Let E be the event that the Sum of two dice is 8. $n(E) = {(2,6),(3,5),(4,4),(5,3),(6,2)}, P(E) = 5/36$
Let F be the event that the number on both dice is even $n(F) = {(2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)}, P(F) = 9/36$
$P(E|F) = (5/36)/(9/36) = 5/9$
However, in my textbook, while P(F) is the same answer as mine but P(E) is written as 3/36 and hence the overall answer is 1/3 I am unable to get my head around the part for P(E), other than 3/36 is obtained due to ${(2,6),(4,4),(6,2)}$ as these three are the even numbers..
Am I correct to assume as such?
• You enumerated all possible outcomes given both die are even...only three of the $9$, $\{(2, 6), (4, 4), (6, 2)\}$, result in a sum of $8$, hence it's $\frac{3}{9} = \frac{1}{3}$. – Jared Apr 20 '16 at 4:11
• @Jared okay.. I think I am having issues interpreting when given a conditional probability. This question is a good example of it though.. Many thanks – dissidia Apr 20 '16 at 4:34
The conditional probability $\mathbb{P}(E|F)$ is not equal to $\frac{\mathbb{P}(E)}{\mathbb{P}(F)}$, but rather to $\frac{\mathbb{P}(E\cap F)}{\mathbb{P}(F)}$.
There are three outcomes in $E\cap F$, namely $(2,6),(4,4)$, and $(6,2)$, hence the conditional probability is $$\frac{\frac{3}{36}}{\frac{9}{36}}=\frac{1}{3}$$
• My bad on the 'signs'. Will take note of that. I think I am still having issues understand when given a conditional prob. So suppose if now the question is stated something like Pr(Sum of two dice is 8 | we know that the number on both dice is <= 10), it will then be (5/36)/(33/36)? – dissidia Apr 20 '16 at 4:32
• I'm not sure I understand, aren't the numbers on standard dice always at most 6? – carmichael561 Apr 20 '16 at 4:36
• @carmichael561 Unless dissidia meant "... given that the sum is at most 10", which would indeed have a probability of $5/33$. – Graham Kemp Apr 20 '16 at 4:40
• Perhaps that is it. – carmichael561 Apr 20 '16 at 4:41
• @dissidia In this case you have $E\subset G$ so $E\cap G=E$ $$\Pr(E\mid G) = \dfrac{\Pr(E\cap G)}{\Pr(G)} = \dfrac{\Pr(E)}{\Pr(G)} = \dfrac{5}{33}$$ – Graham Kemp Apr 20 '16 at 4:45
You wish to know the probability that the sum is eight ($E$) given that the two dice are even ($F$). $\Pr(E\mid F)$
As you stated $E$ is the event set: $\{(2,6),(3,5),(4,4),(5,3),(6,2)\}$ and $F$ is the event set $\{(2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)\}$
However, we want only want the conditional probability of $E$ when $F$.
Now the event of both dice being even and the sum equaling eight is: $F\cap E = \{(2,6) (4,4), (6,2)\}$
Since all outcomes of this experiment are equally likely $\Pr(E\mid F) = \dfrac {\lvert F\cap E\rvert}{\lvert F\rvert} = \dfrac{3}{9}$ | 2019-09-18T05:04:35 | {
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https://math.stackexchange.com/questions/2778159/show-that-unit-circle-is-not-homeomorphic-to-the-real-line/2778164 | # Show that unit circle is not homeomorphic to the real line
Show that $S^1$ is not homeomorphic to either $\mathbb{R}^1$ or $\mathbb{R}^2$
$\mathbf{My \ solution}$:
So first we will show that $S^1$ is not homeomorphic to $\mathbb{R}^1$. To show that they are not homeomorphic we need to find a property that holds in $S^1$ but does not hold in $\mathbb{R}^1$ or vice-versa. $S^1$ is compact however $\mathbb{R}^1$ is not compact.
The set $\{1\}$ is closed, and the map $$f: \Bbb R^2 \longrightarrow \Bbb R,$$ $$(x, y) \mapsto x^2 + y^2$$ is continuous. Therefore the circle $$\{(x,y) \in \Bbb R^2 : x^2 + y^2 = 1\} = f^{-1}(\{1\})$$ is closed in $\Bbb R^2$.
Set $S^1$ is also bounded, since, for example, it is contained within the ball of radius $2$ centered at 0 of $\Bbb R^2$ (in the standard topology of $\Bbb R^2$).
Hence it is also compact.
However real line $\Bbb R^1$ is not because there is a cover of open intervals that does not have a finite subcover. For example, intervals (n−1, n+1) , where n takes all integer values in $\mathbb{Z}$, cover $\mathbb{R}$ but there is no finite subcover.
Hence $S^1$ can not be isomorphic to $\mathbb{R}^1$.
How to show now that $S^1$ is not homeomorphic to $\mathbb{R}^2$? Can i show it now in the same way? They can not be homeomorphic since $S^1$ is compact however $\mathbb{R}^2$ not. How to show that $\mathbb{R}^2$ is not compact?
• $\Bbb R^2$ is covered by all open balls around the origin, but not by finitely many of these. -- For an alternative proof ide: $S 1$ minus two point sis disconnected, but $\Bbb R^2$ minus two points is connected. – Hagen von Eitzen May 12 '18 at 16:41
• $\mathbb{R}^2$ is not compact for the same reason that $\mathbb{R}$ isn't. Consider growing balls of radius $n$, for example, to build an open cover without finite subcover. Another argument is that removing a single point from a circle leaves a connected space, while that is not true of $\mathbb{R}$. – user296602 May 12 '18 at 16:41
• Open balls of radius 2 centered at every point with integer coordinates would work. No finite collection of those covers $\Bbb R^2$. – G Tony Jacobs May 12 '18 at 16:49
• Not to mention, with the euclidean metric $\mathbb R^2$ is not closed and bounded. – fleablood May 12 '18 at 16:52
• A closed subspace of a compact Hausdorff space is compact. So if $\Bbb R^2$ were compact then $\Bbb R\times \{0\},$ which is closed in $\Bbb R^2,$ and is homeomorphic to $\Bbb R$, would also be compact.... Also no metric space with an unbounded metric can be compact. – DanielWainfleet May 13 '18 at 3:48
To prove that $$\mathbb R^2$$ is not compact: Assume that it is. The image of a compact space under a continuous map is compact. The mapping $$f:\mathbb R^2 \to \mathbb R, (x,y)\mapsto x$$ is continuous and has image $$\mathbb R$$. Hence $$\mathbb R$$ is compact. But you yourself showed that $$\mathbb R$$ is not compact. Contradiction.
To show that $$S^1$$ is not homeomorphic to $$\mathbb R^2$$: Observe that $$S^1$$ is compact but $$\mathbb R^2$$ isn't. Done.
You can certainly show these using compactness. The following proofs, however, I find simpler:
The removal of any one point from $\Bbb R$ results in a disconnected space, but if you remove one point from $S^1$, you still have a connected space.
The removal of any two points from $S^1$ results in a disconnected space, but if you remove two points from $\Bbb R^2$, you still have a connected space.
• Why do you say connectedness is "simpler" than compactness? – GEdgar May 12 '18 at 17:15
• I find it simpler, conceptually. YMMV. – G Tony Jacobs May 12 '18 at 17:19
Assume that $\mathbb{R}^2$ is homeomorphic to $S^1$. You already showed that $S^1$ is compact. Let $\mathcal{O}=\{O_i\}_{i\in I}$ be an open cover of $\mathbb{R}^2$ and $f: \mathbb{R}^2\to S^1$ a homeomorphism. Then $f(\mathcal{O})=\bigcup\limits_{i\in I}f(O_i)$ is an open cover for $S^1$, by compactness of $S^1$ it must contain a finite open subcover $\{f(O_{i_j})\}_{j=1}^k$. This gives,
$$\{f(O_{i_j})\}_{j=1}^k \supset S^1 \implies \{O_{i_j}\}_{j=1}^k\}\supset \mathbb{R}^2$$
which means that $\mathbb{R}^2$ is compact, contradiction.
If $f:S^{1} \rightarrow \mathbb{R}^{1}$ was a homeomorphism then, since $S^{1}$ is compact, $f(S^{1}) = \mathbb{R}^{1}$ would have to be compact. A contradiction.
The same argument works with $\mathbb{R}^{2}$.
• I think OP already knows this. Their question was how to show that $\mathbb R^2$ is not compact. – John Gowers May 12 '18 at 16:52
• I can't see anything there that isn't verbatim contained in the OP. And the OP contains the question "How to show that R2 is not compact?" which this answer completely ignores. – fleablood May 12 '18 at 16:54 | 2019-09-23T14:10:18 | {
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https://online.stat.psu.edu/stat414/book/export/html/666 | # 5.3 - Mutual Independence
5.3 - Mutual Independence
## Example 5-6
Consider a roulette wheel that has 36 numbers colored red ($$R$$) or black ($$B$$) according to the following pattern:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 R R R R R B B B B R R R R B B B B B 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19
and define the following three events:
• Let $$A$$ be the event that a spin of the wheel yields a RED number = $$\{1, 2, 3, 4, 5, 10, 11, 12, 13, 24, 25, 26, 27, 32, 33, 34, 35, 36\}$$.
• Let $$B$$ be the event that a spin of the wheel yields an EVEN number = $$\{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36\}$$.
• Let $$C$$ be the event that a spin of the wheel yields a number no greater than 18 = $$\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18\}$$.
Now consider the following two questions:
1. Are the events $$A$$, $$B$$, and $$C$$ "pairwise independent?" That is, is event $$A$$ independent of event $$B$$; event $$A$$ independent of event $$C$$; and $$B$$ independent of event $$C$$?
2. Does $$P(A\cap B\cap C)=P(A)\times P(B)\times P(C)$$?
Let's take a look:
So... this example illustrates that something seems to be lacking for the complete independence of events $$A, B, \text{ and }C$$. And, that's why the second condition exists in the following definition.
Mutually Independent Event
Three events $$A, B,\text{ and }C$$ are mutually independent if and only if the following two conditions hold:
1. The events are pairwise independent. That is,
• $$P(A\cap B)=P(A)\times P(B)$$ and...
• $$P(A\cap C)=P(A)\times P(C)$$ and...
• $$P(B\cap C)=P(B)\times P(C)$$
2. $$P(A\cap B\cap C)=P(A)\times P(B)\times P(C)$$
The idea of mutual independence can be extended to four or more events — each pair, triple, quartet, and so on, must satisfy the above type of multiplication rule.
## Example 5-7
One ball is drawn randomly from a bowl containing four balls numbered 1, 2, 3, and 4. Define the following three events:
• Let $$A$$ be the event that a 1 or 2 is drawn. That is, $$A=\{1, 2\}$$.
• Let $$B$$ be the event that a 1 or 3 is drawn. That is, $$B = \{1, 3\}$$.
• Let $$C$$ be the event that a 1 or 4 is drawn. That is, $$C = \{1, 4\}$$.
Are events $$A, B,\text{ and }C$$ pairwise independent? Are they mutually independent?
This example illustrates, as does the previous example, that pairwise independence among three events does not necessarily imply that the three events are mutually independent.
## Example 5-8
A pair of fair six-sided dice is tossed once yielding the following sample space:
$$S=\left\{\begin{array}{l}{(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)} \\ {(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)} \\ {(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)} \\ {(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)} \\ {(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)} \\ {(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)}\end{array}\right.$$
Define the following three events:
• Let $$A$$ be the event that the first die is a 1, 2, or 3.
• Let $$B$$ be the event that the first die is a 3, 4, or 5.
• Let $$C$$ be the event that the sum of the two faces equals 9. That is $$C = \{(3,6), (6,3), (4,5), (5,4)\}$$.
Are events $$A, B,\text{ and }C$$ pairwise independent? Are they mutually independent?
In solving that problem, I admit to being a little loosey-goosey with the definition of "mutual independence." That's why I said "a sort of mutual independence." Now that I haven't been perfectly clear so far, let me set the record straight by being perfectly clear. This example illustrates that the second condition of mutual independence among the three events $$A, B,\text{ and }C$$ (that is, the probability of the intersection of the three events equals the probabilities of the individual events multiplied together) does not necessarily imply that the first condition of mutual independence holds (that is, three events $$A, B,\text{ and }C$$ are pairwise independent). In order to check for mutual independence, you clearly need to check both the first and second conditions.
[1] Link ↥ Has Tooltip/Popover Toggleable Visibility | 2022-01-24T10:48:34 | {
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http://mathhelpforum.com/advanced-algebra/190031-determine-if-group.html | # Thread: Determine if this is a group
1. ## Determine if this is a group
Problem:
Determine if the set with the binary operation forms a group.
Set S = All real numbers except -1.
$\displaystyle a*b = ab+a+b$
****************
I know I need to check three things: associativity, if there is an identity, and if there is an inverse for every element in S.
I know how to do associativity since it's straightforward. For the identity, I just guess and checked and found that $\displaystyle e=0$ works. (Is there any way to find an identity without guessing/checking?)
For the inverse, I am having difficulty.
I know that this must be satisfied to have an inverse:
$\displaystyle a*a' = a'*a = e$
Since I know the identity element is 0,
$\displaystyle a*a' = a'*a = 0$
Expanding this out with the definition of the binary operation,
$\displaystyle aa'+a+a' = a'a+a'+a = 0$
I said that an inverse does not exist, because from the last equation, it is impossible to get $\displaystyle aa'+a+a'$ equal to 0 for all a. And therefore, this is not a group.
Am I correct?
2. ## Re: Determine if this is a group
let's try to find an identity in a more systematic way:
we want a*x = a.
(it should be obvious that a*b = b*a, so we just need to check for a one-sided identity).
now a*x = ax + a + x. set this equal to a, and try to solve for x:
ax + a + x = a
ax + x = 0
(a+1)x = 0. now, can we divide by a+1? yes we can, because a will not be -1.
x = 0/(a+1) = 0.
are you sure it is impossible to find an inverse?
again, we want b such that a*b = 0. we want to solve for b in terms of a, if we can.
ab + a + b = 0 | 2018-03-22T14:44:44 | {
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http://www.lofoya.com/Solved/2215/if-the-integers-m-and-n-are-chosen-at-random-from-integers-1-to | Difficult Probability Solved QuestionAptitude Discussion
Q. If the integers $m$ and $n$ are chosen at random from integers 1 to 100 with replacement, then the probability that a number of the form $7^{m}+7^{n}$ is divisible by 5 equals:
✔ A. $\dfrac{1}{4}$ ✖ B. $\dfrac{1}{7}$ ✖ C. $\dfrac{1}{8}$ ✖ D. $\dfrac{1}{49}$
Solution:
Option(A) is correct
Table below can be scrolled horizontally
Form of the exponent
$m$ $4x +1$ $4x+3$ $4x+2$ $4x$
$n$ $4y+3$ $4y+1$ $4y$ $4y+2$
last digit of
$7^m+7^n$
$0$ $0$ $0$ $0$
Number of
selections
$25 \times 25$ $25 \times 25$ $25 \times 25$ $25 \times 25$
If a number ends in a 0 then the number must be divisible by 5.
Hence required probability is,
$=\dfrac{625 \times 4}{100^2}$
$=\dfrac{1}{4}$
Edit: Thank you, Barry, for the very good explanation in the comments.
Edit 2: Thank you Vaibhav, corrected the typo now it's $25 \times 25$ and not $26 \times 25$.
Edit 3: For yet another approach of solving this question, check comment by Murugan.
(7) Comment(s)
Murugan
()
This sum is very simple. Power cycles of 7 are 7, 9, 3, 1
So totally 4 possibilities.
Total possibilities are $^4P_1 \times ^4P_1=16$
For selecting a number from 1 to 100 which are divisible by 5,
$m=9$, $n=1$ or $m=1$, $n=9$ or $m=7$, $n=3$ or $m=3$, $n=7$
i.e. 4 chances.
Therefore, probability is $\frac{4}{16}=\frac{1}{4}$
Vaibhav
()
Its should be $25*25$ not $26*25$
Deepak
()
Thank you Vaibhav, corrected typo in the solution.
Apurv
()
I did this: Let $m>n$ (Clearly m and n can't be equal because $5$ can't divide $2*7^{m}$).
Now $7^{m}+7^{n}=7^{n}(7^{m-n}+1)$. If $5$ has to divide this, it implies that 5 has to divide $(7^{m-n}+1)$ (because it cannot divide a power of $7$).
Since powers of $7$ have a cyclic order of $7,9,3,1$; $7^{m-n}$ has to therefore end in a $9$ and therefore $m-n$ can be $2,6,10,...,98$.
Hence the only set of values of $n$ are $(1,2,3,...,97),(1,2,3,...,93),(1,2,3,...,89),...,(1)$. Also fixing $n$ would fix $m$.
Therefore the number of favorable cases is $97+93+89+...+2=1224$. Which means that the required probability should be $\frac{1224}{^{100}C_2$ which turns out to be $\frac{68}{275}$, which is not matching with any of the options..
Where did I go wrong ??
Barry
()
You made three mistakes, or possibly four. One or two of them reside in your calculation $97+93+89+\cdots+2=1224$. It should be $97+93+89+\cdots+1=1225$. It's a little odd that the correction seemingly subtracts $1$ on one side and adds it to the other, so let me spell out the correct calculation:
\begin{align} 97+93+89+\cdots+1&=(4\cdot24+1)+(4\cdot23+1)+(4\cdot22+1)+\cdots+(4\cdot1+1)+1\\ &=4(24+23+22+\cdots+1)+25\\ &=4\left({24\cdot25\over2}\right)+25=1225 \end{align}
However, these aren't the correct numbers to be adding anyway. The correct calculation would have been
$$98+94+90+\cdots+2=1250$$
This is because the correct sets of values for $n$, corresponding to $m-n=2,6,\ldots,98$ are $(1,2,\ldots,98),(1,2,\ldots,94),\ldots,(1,2)$, since presumably you are allowing $m$ to take the value $100$ as well as $99$.
The final, most substantial error resides in what you did with this number after you got it, which was to divide it by ${100\choose2}=4950$. This computes the probability that of two different numbers the given form will be divisible by $5$. But the problem allows for the two numbers to be the same.
At this point you might be tempted to add the $100$ ways two numbers can be the same to the $4950$ ways they can be different and get $1225/5050=49/202$, but this would also be the wrong answer. The correct thing is to double the number $1225$ to get the total number of ways to choose $m$ and $n$ so that $7^m+7^n$ is divisible by $5$ regardless of which one is larger, and then divide by the total number of ways to choose two numbers between $1$ and $100$, which is simply $100^2=10000$. So the correct answer is $2500/10000=1/4$, as others have pointed out.
All that said, it's worthwhile understanding what you did wrong, and it's also worthwhile understanding (from the answers other people have given) how the problem could have been solved by taking a simpler approach.
Derek
()
$7\equiv 2 \bmod 5, 2^1 \equiv 2,2^2 \equiv 4, 2^3\equiv 3 ,2^4\equiv 1$. morever if $m\equiv a\bmod4$ then $2^m\equiv2^a$. (because $2^4\equiv 1 \bmod5)$
so if you want the sum to be divisible by 5 there are 4 ways for this to happen:
$m\equiv1$ and $n\equiv4 \bmod 4$
$m\equiv4$ and $n\equiv1 \bmod 4$
$m\equiv2$ and $n\equiv3 \bmod 4$
$m\equiv3$ and $n\equiv2 \bmod 4$
So if $m$ is any congruence class then the probability $n$ is the corresponding correct class is $\frac{1}{4}$
The numbers are probably intended to be independently chosen, with the possibility that $m=n$. The remainder of $7^m$ on division by $5$ is equally likely to be one of $2,4,3,1$, since $\varphi(5)$ divides $100$. So is the remainder of $7^n$.
Whatever the remainder of $7^m$ happens to be, there is a unique value of $7^n$ modulo $5$ that gives us sum $0$ modulo $5$. That value has probability $\frac{1}{4}$.
Remark: If we choose a pair of numbers (no replacement), then the answer changes. We can assume that the numbers are chosen in order, $m$ first. Whatever value of $m$ is chosen, there are $25$ choices for $n$ that will give sum $0$ modulo $7$, so the probability is $\frac{25}{99}$. | 2017-03-23T04:18:50 | {
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https://math.stackexchange.com/questions/2095057/amount-of-even-zeros-in-a-0-1-alphabet-n-length-word/2095073 | # Amount of even zeros in a 0,1 alphabet n length word
How many words at the length of n above {0,1} alphabet the amount of zeros is even.
I understand the answer is $2^{n-1}$. I'm trying to come up with a reasonable explanation for why that is.
If I could prove that the amount of even and uneven numbers is always the same, I could arrive to the conclusion that even numbers are $2^{n-1}$. Help?
• n מעל {0,1} מספר האפסים הוא זוגי? נסו לפתור במספר דרכים שונות, אך הגישו לבדיקה רק דרך אחת שלטעמכם היא הפשוטה והבטוחה ביותר. What's this? – Gyanshu Jan 12 '17 at 17:17
• @Gyanshu looks like the same question but in a different language? Could it be? – RGS Jan 12 '17 at 17:20
• The Hebrew question asks for the simplest (P'shutah) way to prove this. – Mark Fischler Jan 12 '17 at 17:27
• I think the induction/last digit argument is the simplest. Go us! – mdave16 Jan 12 '17 at 17:28
HINT: If you're given a word and you change the last letter, what happens to the number of zeros?
EDIT: This hint can give the answer without induction. Let $E_n$ be the number of words of length $n$ with an even number of $0$'s and $O_n$ the number of words of length $n$ with an odd number of $0$'s. Define $\phi:\{0,1\}^n \to \{0,1\}^n$ to flip the last letter of the word. Then $\phi^2 = \text{id}$ with $\phi(E_n) \subset O_n$ and $\phi(O_n) \subset E_n$. This immediately implies that $|E_n| = |O_n|$ Since there are $2^n$ total words, this implies $|E_n| = |O_n| = 2^{n-1}$.
A proof by induction could be done:
If $n=1, 2^n = 2^0 = 1$ and $1$ word of length $1$ has an even amount of $0$s, which is the word $1$.
Now imagine you have a sequence of length $k-1$. If it has an odd number of $0$s, then you can append a $0$ to the end of it. If it has an even number of $0$s, you can append a $1$ to it.
Assume then that it is true that, for words of length $k$, there are $2^{k-1}$ words with an even number of $0$s.
To those $2^{k-1}$ which have an even number of $0$s you add a $1$, making $2^{k-1}$ words of length $k+1$ that have an even amount of $0$s. To the other $2^{k-1}$ that have an odd number of $0$s, you add a $0$, making another $2^{k-1}$ words that have an even number of $0$s and length $k+1$. Therefore, you have a total of $2^{k-1} + 2^{k-1} = 2\cdot2^{k-1} = 2^k$ words, of length $k+1$, with an even number of $0$s, concluding your proof.
What you are considering are numbers $0000\dots0$ to $1111\dots1$ in binary. Indeed in total there are $2^n$ numbers.
Now, notice it is true for $n=1$. Assume it might be true for the number before your favourite natural number $k-1$.
Now, consider the possibilities for $k$. The last digit is either $0$ or $1$. So the total amount with even number when last is $0$, is all the $k-1$ length strings with odd amount, i.e. $2^{k-1}$. When last is $1$, we seek all the $k-1$ length strings with even amount, i.e. $2^{k-1}$.
Now, the total number would be $2^{k-1} + 2^{k-1} = 2^k$. So it's true for all the numbers you care about.
Two cases: If $n$ is odd, then any word of length $n$ containing an odd number of zeros is a word containing an even number of ones. So under the mapping that turns every one into a zero and every zero into a one, every word that contains an odd number of zeros corresponds to a word with an even number of zeros. That implies that half the words have an even number of zeros.
If $n$ is odd, then omit the last digit of the word. Note that half the words in that $n-1$ universe have an even number of zeros, and will pair with an last digit of zero to give an even number. And half the words in that $n-1$ universe have an odd number of zeros, and will pair with an last digit of oneto give an even number. So the fraction of words of length $n$ with an even number of zeros is $\frac12\cdot \frac12 + \frac12\cdot \frac12 = \frac12$. | 2019-05-22T23:35:02 | {
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https://physics.stackexchange.com/questions/314638/variational-derivative-of-function-with-respect-to-its-derivative | Variational derivative of function with respect to its derivative [closed]
What is $$\frac{\delta f(t)}{\delta \dot{f}(t)}~?$$
Where $\dot{f}(t) = df/dt$.
• could you provide some context please? – ZeroTheHero Feb 25 '17 at 3:32
• It is purely a math question ,so better to ask in mathematics stack exchange – Lapmid Feb 25 '17 at 4:00
• @SherlockHolmes yeah and all possible answers are purely math too. – ZeroTheHero Feb 25 '17 at 6:05
• Related question by OP: physics.stackexchange.com/q/263261/2451 – Qmechanic Feb 25 '17 at 7:36
• – Qmechanic Feb 25 '17 at 7:47
The definition of the functional derivative of a functional $I[g]$ is the distribution $\frac{\delta I}{\delta g}(\tau)$ such that $$\left\langle \frac{\delta I}{\delta g}, h\right\rangle := \frac{d}{d\alpha}\bigg\rvert_{\alpha=0} I[g+ \alpha h]$$ for every test function $h$. In our case, assuming to deal with functions which suitably vanish before reaching $\pm \infty$, $$I[g] = \int_{-\infty}^t g(x)dx$$ so that $$I[\dot{f}]= f(t)$$ as requested. Going on with the procedure $$\left\langle \frac{\delta I}{\delta g}, h\right\rangle = \frac{d}{d\alpha}|_{\alpha=0} \int_{-\infty}^t(g(\tau)+ \alpha h(\tau)) d\tau = \int_{-\infty}^t h(\tau) dx = \int_{-\infty}^{+\infty} \theta(t-\tau)h(\tau) d\tau$$ where $\theta(\tau)=1$ for $\tau\geq 0$ and $\theta(\tau)=0$ for $\tau<0$ and so $$\frac{\delta f(t)}{\delta \dot{f}}(\tau) = \frac{\delta I}{\delta g}(\tau)= \theta(t-\tau)$$
• Just a LaTeX tip: you can use \big, \bigg and so forth to have a larger vertical line, \rvert. (See edit.) – JamalS Feb 25 '17 at 9:54
• Thanks. I usually use $\left.$ $\right|$ but I did not exploit them here. – Valter Moretti Feb 25 '17 at 9:56
• Great explanation, thanks! I guess it makes physical sense too - varying at $\tau$ you only expect any effect at $t\geq\tau$ – smörkex Feb 26 '17 at 3:43
The important thing to keep in mind is that a functional derivative is more like a gradient than an ordinary derivative. The reason that this is an important consideration is because, practically, we always specify functions with (possibly infinite) lists of numbers, be they: Taylor series coefficients, continued fraction constants, a list of constant values (approximating with boxcars), a list of points (connect the dots), Fourier series coefficients, or etc.
The important part of this consideration is that the function's derivative doesn't carry any information about a constant vertical offset. Thus, because any function of the form $f(t) + c$ has the same derivative, $\dot{f}(t)$, the functional derivative in the question will not be defined in the "direction" that corresponds to the degree of freedom defined by $c$.
In equations, let \begin{align} g(t) &\equiv \dot{f}(t) \Rightarrow \\ f(t) - f(t_0) & = \int_{t_0}^t g(t') \operatorname{d} t'\end{align} From there: \begin{align} \frac{\delta f(t)}{ \delta \dot{f}(\tau)} - \frac{\delta f(t_0)}{ \delta \dot{f}(\tau)} & = \frac{\delta \int_{t_0}^t g(t') \operatorname{d}t'}{ \delta g(\tau)} \\ & = \int_{t_0}^t \delta(t' - \tau) \operatorname{d}t' \\ & = \Theta(t-\tau) \, \Theta(\tau - t_0) - \Theta(t_0 - \tau)\, \Theta(\tau - t). \end{align}
This now satisfies: $$\frac{\partial}{\partial t} \left(\frac{\delta f(t)}{\delta \dot{f}(\tau)}\right) = \delta(t - \tau),$$ as expected. Because $\dot{f}(t)$ doesn't carry any information about the vertical offset of $f(t)$, only differences of the functional derivative, like above, are well defined.
If the space of functions is limited to those that satisfy $\lim_{t\rightarrow -\infty} f(t) = 0$, then we can take $t_0\rightarrow -\infty$ to get the expression from Valter Moretti's answer.
• Since the choice of $t_0$ is arbitrary, your calculation seems to suggest that this variation $\frac{\delta f(t)}{\delta \dot{f}(t)}$ is not well defined. – taper Feb 25 '17 at 4:15
• @taper I have now addressed that, and you're right, only differences in that functional derivative are well defined. – Sean E. Lake Feb 25 '17 at 19:35
• Thanks for this insight. So combining with Valter Moretti's answer, the full solution is then $\delta f(t) / \delta \dot{f}(\tau) = \theta(t-\tau) + c(\tau)$. But what kinds of initial conditions would determine $c(\tau)$ - it seems like for this problem these should be context-independent of what $f$ actually is. One natural condition seems to be $\delta f(a) / \delta \dot{f}(b) = 0$ where $b>a$. Then $0 = \delta f(a) / \delta \dot{f}(b) = \theta(a-b) + c(b) = c(b)$, so the full solution is just $\delta f(t) / \delta \dot{f}(\tau) = \theta(t-\tau)$ - is this true? – smörkex Feb 26 '17 at 3:53
• I would disagree with that. Looking at your other questions, related to Euler-Lagrange equations, this isn't relevant anyway. There are two main ways to do that problem. First, variational derivative of the action w.r.t. $x(t)$ witch uses chain rules and $\frac{\delta x(t)}{\delta x(t')} = \delta(t-t')$. The second is to use partial derivatives in which $x$, $\dot{x}$, $\ddot{x}$, etc are all treated as independent variables. – Sean E. Lake Feb 26 '17 at 5:59
• @Kurt In either case, the result is: \begin{align} \frac{\delta S[x]}{\delta x(t)} &= \int \left(\frac{\partial L}{\partial x} \delta(t'-t) + \frac{\partial L}{\partial \dot{x}} \delta'(t'-t) + \frac{\partial L}{\partial \ddot{x}} \delta''(t'-t) + \ldots \right) \operatorname{d}t' \\ & = \sum_{n=0}^\infty (-1)^n \frac{\operatorname{d}^n}{\operatorname{d}t^n} \left(\frac{\partial L}{\partial \frac{\mathrm{d}^n\, x}{\mathrm{d}\, t^n}}\right) \end{align} – Sean E. Lake Feb 26 '17 at 6:03 | 2019-12-16T10:03:11 | {
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https://www.physicsforums.com/threads/how-to-differentiate-an-absolut-value-f-x-x-2-4.73747/ | # How to differentiate an absolut value, f(x)=│x^2-4│
1. Apr 30, 2005
### gillgill
how do u differentiate f(x)=│x^2-4│....?
i don't know how to do it with absolute values...
2. Apr 30, 2005
### cepheid
Staff Emeritus
If I remember right, define it in a piecewise fashion. Can you see that:
|x^2 - 4| = x^2 - 4 if x^2 - 4 > 0
|x^2 - 4| = -(x^2 - 4) if x^2 - 4 < 0
The first condition becomes: if x^2 > 4 ===> |x| > 2, i.e. if x > 2 OR x < -2
The second becomes: if x^2 < 4 ===> |x| < 2, i.e. -2 < x < 2
So you have two cases. For x > 2 and x < -2, the function is:
f(x) = x^2 - 4, and you can differentiate it.
For -2 < x < 2, the function is:
f(x) = -x^2 + 4, and you can differentiate it.
The function is differentiable at x = 2 and x = -2 if and only if the left and right hand derivatives exist at those points.
3. Apr 30, 2005
### arildno
It is quite instructive to use the DEFINITION of the derivative at the problem points 2 and -2.
I'll take the "2"-case:
In general, we have:
$$f'(2)=\lim_{\bigtriangleup{x}\to{0}}\frac{f(2+\bigtriangleup{x})-f(2)}{\bigtriangleup{x}}$$
if it exists.
In our case, $$f(x)=|x^{2}-4|$$
which implies $$f(2)=0,f(2+\bigtriangleup{x})=|(2+\bigtriangleup{x})^{2}-4|=|4\bigtriangleup{x}+(\bigtriangleup{x})^{2}|$$
Hence, we must have:
$$f'(2)=\lim_{\bigtriangleup{x}\to0}\frac{|\bigtriangleup{x}|}{\bigtriangleup{x}}|4+\bigtriangleup{x}|$$
if it exists.
Can it exist?
4. Apr 30, 2005
### gillgill
i found the first reply easier to understand...
f(x)=x^2-4
f'(x)=2x
f'(x) as x->2+ would equal 2(2)=4
f(x)=-x^2+4
f'(x)=-2x
f'(x) as x->2- would equal -2(2)=-4
so does not exist at x=2
f(x)=x^2-4
f'(x)=2x
f'(x) as x->-2- would equal 2(-2)=-4
f(x)=-x^2+4
f'(x)=-2x
f'(x) as x->-2+ would equal -2(-2)=4
so does not exist at x=-2
is that right?
5. Apr 30, 2005
### Pyrrhus
Yes that's right, because the limit does not exist therefore the derivative does not exist at those points.
6. Apr 30, 2005
### gillgill
okay...i see...thanks guys... | 2017-01-18T08:08:12 | {
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https://math.stackexchange.com/questions/3250988/calculation-of-a-series-using-series-definition/3251005 | # Calculation of a series using series definition
I have to calculate the series $$\sum_{k=2}^\infty \left(\frac 1k-\frac 1{k+2}\right)$$
Using the definition: $$L = \lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=0}^na_k$$ Obviously $$\lim_{n\to\infty} (\frac 1n-\frac 1{n+2})=0$$, but I don't think that this is the right way to calculate the value of the series.
• Have you ever heard the term "Telescoping" before used in a mathematical context? – JMoravitz Jun 4 at 17:42
• Welcome to Math Stack Exchange. Before taking the limit, first try to evaluate $S_n$ using telescoping – J. W. Tanner Jun 4 at 17:48
• I completely forgot about the telescopic series, I was too focused on using the definition – Diego Di Marzo Jun 4 at 18:09
As suggested from the comment by JMoravitz, you should write down the series explicitly and see if you can see something interesting:
$$\sum_{2}^\infty \big( \frac{1}{k} - \frac{1}{k+2} \big) = \bigg(\frac{1}{2} - \frac{1}{4} \bigg) + \bigg(\frac{1}{3} - \frac{1}{5} \bigg) + \bigg(\frac{1}{4} - \frac{1}{6} \bigg) + \bigg(\frac{1}{5} - \frac{1}{7} \bigg) + \bigg(\frac{1}{6} - \frac{1}{8} \bigg) + \cdots + \bigg(\frac{1}{n} - \frac{1}{n+2} \bigg) + \bigg(\frac{1}{n+1} - \frac{1}{n+3} \bigg) + \bigg(\frac{1}{n+2} - \frac{1}{n+4} \bigg) + \bigg(\frac{1}{n+3} - \frac{1}{n+5} \bigg) + \cdots$$
From here, you can see which terms are cancelling with one another and which one is left.... Essentially you should see that all the terms will cancel with each other except the $$\frac{1}{2}$$ and $$\frac{1}{3}$$. Therefore, the answer is $$\frac{5}{6}$$.
$$\textbf{Another, Different way}$$ that uses limit like you wanted to is to find the partial sum! You first compute, $$S_2 = \frac{1}{2} - \frac{1}{4}$$, $$S_3 = S_2 + \bigg( \frac{1}{3} - \frac{1}{5} \bigg)$$ and so on... What you will find (and you can prove this fact by induction) is that $$S_n = \dfrac{5n^2 + 3n -8}{6(n+1)(n+2)}$$ Now, $$\lim_{n \to \infty} \dfrac{5n^2 + 3n -8}{6n^2 + 18n + 12} = \frac{5}{6}$$
• Thank you so much, I was so focused on looking the limit that I forgot to write down the series. – Diego Di Marzo Jun 4 at 18:06
Let us use $$1/k=\int_{0}^1 t^{k-1} dt$$. Then $$S=\sum_{k=2}^{\infty} \left (\frac{1}{k}-\frac{1}{k+2} \right)= \int_{0}^{1} \sum_{k=2}^{\infty} ~[t^{k-1}- t^{k+2}]~dt =\int_{0}^{1} \frac{t-t^3}{1-t} dt=\int_{0}^{1} (t+t^2) dt=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}.$$
Hint:
Try writing out the first few partial sums and look to see if anything can cancel.
Even more explicitly, looking ahead to somewhere in the middle, you will have if you expand out the summation the following:
$$\dots +\left(\dfrac{1}{50}-\color{blue}{\dfrac{1}{52}}\right)+\left(\dfrac{1}{51}-\color{red}{\dfrac{1}{53}}\right)+\left(\color{blue}{\dfrac{1}{52}}-\dfrac{1}{54}\right)+\left(\color{red}{\dfrac{1}{53}}-\dfrac{1}{55}\right)+\dots$$
Now, notice the colors I used and think about why I put colors there. Where else in the series does something like this happen? What does this imply in the end will be left over if we take a partial sum? What does this imply in the end will be left over if we consider the limit of partial sums?
The more general name for this property is "Telescoping." We refer to this series as a "Telescoping Series."
It is $$\sum_{k=2}^\infty \frac1k-\frac1{k+2}=\lim_{n\to\infty}\sum_{k=2}^n \frac1k-\frac{1}{k+2}$$
As being said in the comments, this series is telescoping.
If you write down some terms for $$n=5$$, we would get:
$$\sum_{k=2}^5 \frac1k-\frac{1}{k+2}=(\frac12-\frac14)+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+(\frac{1}{5}-\frac{1}{7})$$
Here we see, what is called "the series is telescoping". There is a pattern in addition and subtraction of specific summands. First we have $$-\frac14$$. Later in the series, we have $$+\frac14$$. The same goes for $$\frac15$$. These summands cancel each other out.
So the long summation can be 'pushed together', like a telescop. If we would look at higher values of $$n$$, we would see this to more effect.
That being said, we have to formalize that.
It is $$\lim_{n\to\infty}\sum_{k=2}^n\frac1k-\frac{1}{k+2}=\lim_{n\to\infty}\sum_{k=2}^n\frac1k-\sum_{k=2}^n\frac{1}{k+2}$$
Note here, that we have to make this step for partial sums, and can not calculate like this for infinite series, because both sums would not converge.
Now we make an index shift:
We get $$\sum_{k=2}^n\frac{1}{k+2}\to\sum_{k=4}^{n+2}\frac{1}{k}$$.
Finally we arrive at:
$$\lim_{n\to\infty} \sum_{k=2}^n \frac{1}{k}-\sum_{k=4}^{n+2}\frac1k$$
Now we can see, what we observed above. Namely the partial equal terms in both sums, getting added and subtracted.
To be more formal:
$$\sum_{k=2}^n \frac{1}{k}-\sum_{k=4}^{n+2}\frac1k=\left(\frac12+\frac13+\sum_{k=4}^n\frac1k\right)-\left(\sum_{k=4}^n\frac{1}{k}+\frac1{n+1}+\frac{1}{n+2}\right)=\frac{1}{2}+\frac{1}{3}-\frac{1}{n+1}-\frac{1}{n+2}$$
So we end up with:
$$\lim_{n\to\infty}\sum_{k=2}^n \frac1k-\frac{1}{k+2}=\frac{1}{2}+\frac{1}{3}-\frac{1}{n+1}-\frac{1}{n+2}\to_{n\to\infty} \frac12+\frac13-0-0=\frac{5}{6}$$ | 2019-07-24T01:08:36 | {
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http://irmj.coroilcontrappunto.it/application-logarithms-real-life.html | # Application Logarithms Real Life
In the equation is referred to as the logarithm, is the base , and is the argument. The logarithm base 2 of y, denoted log2 y, is defined to be the number x such that 2x = y. MEASURING SOUND Standard unit: decibels Decibel scale: reflection of the logarithmic response of the human ear to changes in sound intensity Measured by first assigning an intensity I0 to a soft sound Equation: CHEMISTRY POPULATION GROWTH ECONOMICS Make linear/exponential. Get up-to-date Celebrity and Music News. Careers Find out what it's like to work at Life. Statisticians use logarithms to determine the possibilities in real life. An application of exponential functions is compound interest. The solution should also enable database administrators to monitor, track, instantly identify. That's really easy as well. Remove Ads. Angles: Real Life Applications of Trigonometry Classroom Video - Duration: 15:10. 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They'll give your presentations a professional, memorable appearance - the kind of sophisticated look that today's audiences expect. The inverse of y =ln (x +1) is y =ex º1. Modelling Exponential Decay - Using Logarithms. Just like PageRank, each 1-point increase is a 10x improvement in power. Lifecasting Smooth-On Alja-Safe® alginate and Body Double® silicone rubber have become Hollywood favorites and industry standards for capturing / reproducing detail from the human body. June 10, 2020, 9:06 a. Also the base a cannot be 1 or 0. If the problem has more than one logarithm on either side of the equal sign then the problem can be simplified. M11GM-Id-1 2. In real life, yes, logs are indispensible to. Logarithmic Functions Their Graphs And Applications¶ Rewriting exponentials into logarithms and logarithms into exponentials using common log, natural log, and logarithms of other bases ¶ Source : I made these up. Get your real rate today. Practical Applications of Mathematics in Everyday Life. y = log b x. it looks like meeting people in real life was actually working for them. This lesson builds on students' prior work with logarithmic functions. We explain Exponential Functions in the Real World with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. Explain how absolute value would be used to express that number in this situation. Jun 14, 2020 - Logarithms - Real Life Applications Quant Video | EduRev is made by best teachers of Quant. This video is highly rated by Quant students and has been viewed 326 times. I can use the logarithmic equation to formulate an educated guess of my future spending habits. Thermal Energy, or heat, is the vibration and movement of the atoms and molecules within substances. By continuing to use this site you consent to the use of cookies on your device as described in our cookie policy unless you have disabled them. Exponential & Logarithmic Applications Compound Interest In compound interest formulas, is the balance, is the principal, is the annual interest rate (in decimal form), and is the time in years. There are real life applications of logarithms but there is little that you would do every life, I mean technically the computer uses it but there rly isn't anything you as a person would do daily. F1 2020 Will Let You Race On The New Hanoi Circuit In Vietnam Before Its Real-Life Debut F1 2020 will travel to Vietnam and let players race along this iconic track on PC, PS4, Xbox One, and Stadia. Two common bases that you may encounter in problems and in real life applications are 10 and the number e (e ≈ 2. In fact, we can use the Exponential Growth and Decay Formula to find snow depth levels, the magnitude of a star, how temperature affects a body, or how a fast-food chain expands its business as Khan. But what are some other real life applications/examples of logarithms? Thank You. To forestall the chance that black jurors could save black defendants from conviction by refusing to join a guilty verdict and thus hanging. Introduction. logarithm base 2, that is the inverse of the exponential function 2x. Free Algebra 2 worksheets (pdfs) with answer keys-each includes visual aides, model problems, exploratory activities, practice problems, and an online component. In addition, Logarithmic scales are used in Another application of proportions in the real life is in movies screens, because in order to project. Our home of compelling and award-winning documentary content, Real Stories was launched in Autumn 2015 and has grown to include over 300 high quality documentary and factual titles, including over a dozen BAFTA and Emmy award winning programmes. But there's so much more that kids can learn with mobile apps, including some skills that may surprise you. Start your wellness journey and sign up today. Logarithms -- The Wonder of Logarithms A detailed inside to Logarithms. If you have a single logarithm on each side of the equation having the same base then you can set the Read more Solving Logarithmic Equations. The list below reviews some of the Azure Monitor options:. I like the example you chose to explain real-life situations of geometry. Tes Global Ltd is registered in England (Company No 02017289) with its registered office at 26 Red Lion Square London WC1R 4HQ. For instance, a student may say, “As the speed increases, your times gets closer and closer to 0 because you are going faster. Journaling helps you make sense of your life, helping you notice connections, gain insight, and see life patterns. Substituting these values into the formula. How to Be Mindful at a Wedding. Habitica can help you achieve your goals to become healthy and happy. 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Absolute Value Functions and Graphs - Word Docs & PowerPoints To gain access to our editable content Join the Algebra 2 Teacher Community! Here you will find hundreds of lessons, a community of teachers for support, and materials that are always up to date with the latest standards. If the meter charges the customer a rate of $1. Digital imaging is another real life application of this marvelous science. For each of the two applications, include the following as part of your response. Using logarithms to solve real world problems Interest Compounded Annually Suppose that$10,000 is invested at 6% interest compounded annually. Compound interest calculations often involve radicals. Save the application logs from the event viewer. The logarithm is the power in the exponential form. Welcome to a new era of Roche within Life Science We celebrate the students, the researchers, the managers who work tirelessly in the pursuit of discovery. Inverse relations. Asking someone to find a real life example of a number (integer, rational, etc) is the same as asking to find a real world example for a straight line. Some logs which should be collected are listed below: For Windows Operating System. Connections between exponential functions and logarithms. Expressed mathematically, x is the logarithm of n to the base b if b x = n, in which case one writes x = log b n. (Capital typically includes plant and equipment. For example, if you get a loan at a bank that has continuous interest (they all do), if you need to calculate how long it will take to pay it back, you need to use logarithms. Buy online. LOGARITHMS Mathematical functions using logarithms simplify computations with rates and areas that result from situations in physic, biology, medicine, and finance. Logarithm, the exponent or power to which a base must be raised to yield a given number. in response to Ryan D. Explore math with Desmos. I have developed group projects involving real-life applications of exponential and logarithmic functions for College Algebra courses. The components of message based software systems like Client-/Server-, Multitier- and Service-Oriented-Architectures (SOA) may be simulated as well as tested in isolation and in their supposed interaction. General Logarithms. Logarithm(log) of a number to given base is the power or exponent to which the base must be raised in order to produce that number. If you need a detailed discussion of index and log laws, then the Mathematics Learning Centre booklet: Introduction to Exponents and Logarithms is the place to start. I like the example you chose to explain real-life situations of geometry. We will see that when we translate this verbal statement into a differential equation, we arrive at a differential equation. contributed by Rob Terpilowski. You can find games, interactive lessons plans, and many other teacher educational resources to make real-life connections in the classroom to increase engagement online. LESSON 10: A Wrap on Logarithmic dB's and An Extension to Logarithmic Properties LESSON 11: Classroom Journey: Exponential and Logarithmic PropertiesLESSON 12: Laws of Logarithms and Real ApplicationsLESSON 13: Calculator Boot-Camp and Solving Exponential EquationsLESSON 14: Logs, Loans, and Life Lessons!LESSON 15: "Demystifying e" Day #1. Data is from MeasuringWorth. Erdinç Çakıroğlu April, 2004, 123 pages The purpose of the study was to investigate the effects of discovery and. The Basics of RISC-V and Its Real Life Applications – Nassif Jabbour – Sondos Osman Date: May 20, 2020. Ax or Saw – cut two logs about 8 feet in length; Rope – non-elastic is preferable for safety reasons. 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Spectroscopy Applications. e is the base of natural logarithms (2. Intuitive use of logarithms. The exponential decay is a model in which the exponential function plays a key role and is one very useful model that fits many real life application theories. Applications of Logarithms Logs are used in a variety of applications in sciences, some of the most common are: measuring loudness (decibels), measureing earthquake intensity (Richter scale), radioactive decay, and acidity (pH= -log 10 [H +]). I know right now you may think that real life is a lot more intimidating than sending a woman you like a message on an online dating site, but I’m going to share. It is very important in solving problems related to growth and decay. com - 2008-2020 You are free to copy, share and adapt any text in the article, as long as you give appropriate credit and provide a link/reference to this page. A logarithmic function can be used to transform an exponential function into a linear one. Applying Quadratics to Real-Life Situations By Mary Jane Sterling Quadratic equations lend themselves to modeling situations that happen in real life, such as the rise and fall of profits from selling goods, the decrease and increase in the amount of time it takes to run a mile based on your age, and so on. The 2 most common bases that we use are base 10 and base e, which we meet in Logs to base 10 and Natural Logs (base e) in later sections. ) In addition, Napier recognized the potential of the recent developments in mathematics, particularly those of prosthaphaeresis, decimal fractions, and symbolic index arithmetic, to tackle the issue of reducing computation. Tutorial on differentials. So, it is very important to understand the goal and collect appropriate logs. Any time you need to have guidance on final review or algebra exam, Solve-variable. There are real life applications of logarithms but there is little that you would do every life, I mean technically the computer uses it but there rly isn't anything you as a person would do daily that would make them more applicable especially if you are still in college or high school, maybe if you actually worked in electronics, engineering, or some sort of math field then you would. Three of the most common applications of exponential and logarithmic functions have to do with interest earned on an investment, population growth, and carbon dating. How long will it take to accumulate $20,000 in the account?. 81 Remove a respirator from behind. 142-143 DOI: 10. Just like PageRank, each 1-point increase is a 10x improvement in power. Logarithms: Logarithms are used in measuring many physical quantities. The components of message based software systems like Client-/Server-, Multitier- and Service-Oriented-Architectures (SOA) may be simulated as well as tested in isolation and in their supposed interaction. Domain and Range of Exponential and Logarithmic Functions Recall that the domain of a function is the set of input or x -values for which the function is defined, while the range is the set of all the output or y -values that the function takes. In fact, we can use the Exponential Growth and Decay Formula to find snow depth levels, the magnitude of a star, how temperature affects a body, or how a fast-food chain expands its business as Khan. An application of the logarithmic function. Absolute Value Functions and Graphs - Word Docs & PowerPoints To gain access to our editable content Join the Algebra 2 Teacher Community! Here you will find hundreds of lessons, a community of teachers for support, and materials that are always up to date with the latest standards. Radicals and Rational Exponents and Applications of Logarithms MTH133-0902B-32 College Algebra Set of data that can be modeled as a polynomial function series is a function with index as independent variable series is a function with index as independent variable series is a function with index as independent variable. Get up-to-date Celebrity and Music News. , 1997;Baker et al. They are essential in mathematics to solve certain exponential-type problems. in response to Ryan D. Interest is, generally, a fee charged for the borrowing of money. One way to manually evaluate logarithms is to use the relationship between the logarithm and exponential functions given above. Try the -LevelDisplayName and -ListLog parameters. V (x) = 4x3 482+144x (PS, Tito rex, its four and 3 squared and the other one is 48 squared) the polynomial function, and the real life problem would be to find the value of x that makes the volume a maximum. This course applies principles learned in my course “Introduction to Engineering Mechanics” to analyze real world engineering structures. Properties. Logarithms And. , 2000;Ubuz, 2007) and while. Discover how exponential functions can be used to model social, scientific, or personal finance situations. Exponential functions. I wanted to include this section of machine learning use cases that did not quite fit in our above categories. The list below reviews some of the Azure Monitor options:. Functions - Compound Interest Objective: Calculate final account balances using the formulas for com-pound and continuous interest. The first of these is the exponential function. In fact, we can use the Exponential Growth and Decay Formula to find snow depth levels, the magnitude of a star, how temperature affects a body, or how a fast-food chain expands its business as Khan. The natural logarithm is differentiable. Suppose a driver wants to know how many miles he has to drive to earn$100. Haven Life offers an easy way to buy high-quality, affordable term life insurance online. 0 on the Richter scale caused a massive amount of devastation. Get up-to-date Celebrity and Music News. But what are some other real life applications/examples of logarithms? Thank You. Application: In the real world if you were trying to figure out how much fruit punch juice your square punch container can hold, you could use a cube root function to figure it out. Since both k and t are exponents, we must use logarithms. Auditing database activities manually is a herculean task. Are you ready to spend some mulah? Or are you ready to save some dough? This section is all about using the equations for loans, investments, and present value which you can use when you are shopping for that car or house. For a source in an econometrics textbook saying that either form of logarithms could be used, see Gujarati, Essentials of Econometrics 3rd edition 2006 p 288. 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To estimate the data in logs obtained from magnitude scales for earthquakes. Formulas: Compounding times per Year Since the half-life is 60 days, when # , ,. There are millions of geocaches worldwide, just waiting for you to find them. Think of a real-life situation that can be represented by a logarithmic function, translate the situation to the function, and solve the function and represent it graphically. Join us for a dynamic showcase of live poetry readings and music with some of the world's most exciting and groundbreaking poets, featuring Yrsa Daley-Ward, Aja Monet, New York Times Best-Selling author, poet, and musician George Watsky, and more. ) In addition, Napier recognized the potential of the recent developments in mathematics, particularly those of prosthaphaeresis, decimal fractions, and symbolic index arithmetic, to tackle the issue of reducing computation. You can find games, interactive lessons plans, and many other teacher educational resources to make real-life connections in the classroom to increase engagement online. They are essential in mathematics to solve certain exponential-type problems. That is -$10 The balance is decreased by$10, so use a negative number. The number e is a mathematical constant approximately equal to 2. By condensing the logarithms, we can create an equation with only one log, and can use methods of exponentiation for solving a logarithmic equation with multiple logs. They are essential in mathematics to solve certain exponential-type problems. In particular, they are quite good for describing distance-speed-time questions, and modeling multi-person work problems. We recommend keeping it to 1-2 paragraphs. application note Authors Geert Vanden Poel DSM Resolve The Netherlands Vincent B. Its an example for modeling with Exponential and Logarithmic Equations: Use Newton's Lay of Cooling, T = C + (T 0 - C)e -kt , to solve this exercise. As an object is heated up, its atoms. This module will introduce logarithms and explore the properties of logarithmic functions of the form $$y=\log_c(x),c>0,c eq1$$. Real World Applications of Exponential and Logarithmic Functions Within this section, you will be captivated by the various uses of logarithms. Compound interest calculations often involve radicals. Also the base a cannot be 1 or 0. This function y = ln x can be viewed graphically:. It should be noted that no proofs are available which states the non existence of such algorithm. In real life application, isometric drawing is used in the design of the video games. Step 2 : Use the properties of logarithms to simplify the pro blem if needed. This is an online scientific calculator with double digit precision that support both button click and keyboard type. Sometimes the quantity which is to be solved for in an equation is in an exponential function. 0 on the Richter scale caused a massive amount of devastation. Maths plays an essential role in fighting COVID-19, which is why the pandemic has featured a lot on Plus. With thousands of events put on annually by student organizations, social sororities and fraternities and UF Student Government, there is an involvement opportunity for every Gator. In addition, it gives personal help, teaching notes, and sermon ideas that will address needs, answer questions, and provide insight for applying God’s Word to life today. In this function, a represents the starting value such as the starting population or the starting dosage level. 13 word problems with Exponential. Defines common log, log x, and natural log, ln x, and works through examples and problems using a calculator. Exponential functions. Erdinç Çakıroğlu April, 2004, 123 pages The purpose of the study was to investigate the effects of discovery and. Intervals, Exponents, Logarithms. Break through to improving results with Pearson's MyLab & Mastering. A logarithmic function can be used to transform an exponential function into a linear one. The mathematical model of exponential growth is used to describe real-world situations in population biology, finance and. World of Logs combat log analyzer allows gamers to save, share and analyze their raiding experiences conveniently and thoroughly in the Blizzard MMO World of Warcraft. This follows the rule that ⋅ = +. A process is likely to satisfy a logarithmic model if how it evolves is inversely proportional to its current state. Collect the logs according to your needs. So, it is very important to understand the goal and collect appropriate logs. Lifecasting Smooth-On Alja-Safe® alginate and Body Double® silicone rubber have become Hollywood favorites and industry standards for capturing / reproducing detail from the human body. "By shortening the labour, the invention of logarithms doubled the life of astronomers". This is a topic that I always struggle to come up with concrete “real world” applications of. The example is very easy to understand and is very detailed. Applications Involving Exponential Models Exponential functions are useful in modeling many physical phenomena, such as populations, interest rates, radioactive decay, and the amount of medicine in the bloodstream. When Do We Use Logarithms?. When the base a is equal to e, the logarithm has a special name: the natural logarithm, which we write as ln x. represents real-life situations using one-to one functions. Learn Applications in Engineering Mechanics from Georgia Institute of Technology. The properties of logarithms allow you to solve logarithmic and exponential equations that would be otherwise impossible. Logarithms in the Real World. The doubling time of a population exhibiting exponential growth is the time required for a population to double. For Assistance with Fifth Third Private Bank Rewards. The inversion z ↦ 1 z causes for the logarithmic spiral a reflexion against the imaginary axis and a rotation around the origin, but the image is congruent to the original one. Learn how to apply square root functions to predict and examine real world situations. If you need a detailed discussion of index and log laws, then the Mathematics Learning Centre booklet: Introduction to Exponents and Logarithms is the place to start. Kjetil Muri Skarstein/Shutterstock. certain functions, discuss the calculus of the exponential and logarithmic functions and give some useful applications of them. is distinction between exponential growth and decay. So, it is very important to understand the goal and collect appropriate logs. Solve log 2. One example is the use of air bags in automobiles. As an object is heated up, its atoms. SheLovesMath. We have already explored some basic applications of exponential and logarithmic functions. This website and its content is subject to our Terms and Conditions. The Real Life scenario of Logarithms is to measure the acidic, basic or neutral of a substance that describes a chemical property in terms of pH value. Search by what matters to you and find the one thats right for you. Quite simply, exponents tell you to multiply a number by itself using the superscript numeral to determine how many times you do this. Christopher Wanamaker. You can monitor many different applications and services with Azure Monitor. Auditing database activities manually is a herculean task. This first application is compounding interest and there are actually two separate formulas that we'll be looking at here. Fibonacci numbers and Phi are related to spiral growth in nature. Obama logs on amid the coronavirus pandemic The former president comes off the sidelines to share health information, say thanks and offer inspiration — but avoids criticism of Trump. These television coroners, as well as their real-life counterparts, use logarithms to make such determinations. These examples of chemical reactions from everyday life are a small sampling of the hundreds of thousands of reactions you experience as you go about your day. Skip navigation Sign in. In general, we also use properties and applications of logarithms in various geological circumstances: 1. Raygun is free to try, easy to set up, and works seamlessly across your. In the simplest case, the logarithm counts the number of occurrences of the same factor in repeated multiplication; e. In this section, we explore some important applications in more depth, including radioactive isotopes and …. ) A production function might be used to study tradeoffs in the use of. Now, instead of plotting length, what if we plot the logarithm of length? There will be as much space on the graph between 0. The golden ratio is referred to by many diverse terms, such as golden mean, golden section, medial section, divine proportion, golden cut, and extreme and mean ratio. In this article I will share with you 5 advantages of using very real life scenarios in eLearning. Six real word examples of exponential growth in a Powerpoint slide show (3. If I understand Logarithms correctly it is the orders of magnitude? Either exponentially growing or exponentially decaying. The first published use of the "ln" notation for the base-e logarithm was Stringham's, in his 1893 text "Uniplanar Algebra". The ball hits the ground when h = 0. Solving Logarithmic Equations Generally, there are two types of logarithmic equations. To estimate the data in logs obtained from magnitude scales for earthquakes. I've heard that exponential growth is the same as the growth of cancer cells. It can also be calculated as the sum of the infinite series = ∑ = ∞! = + + ⋅ + ⋅ ⋅ + ⋯. Summary This article will help fresher candidates to understand ASP. 7 real-life human cyborgs See how bionic technology has enhanced people's lives with everything from robotic limbs to electronic eyes. These formulas can arm you with ways to calculate how much money you are actually spending if you leave a balance on your credit card. Angles: Real Life Applications of Trigonometry Classroom Video - Duration: 15:10. The Asus ZenBook 13 UX333 is the best college laptop you can buy. Find log2,760 2. certain functions, discuss the calculus of the exponential and logarithmic functions and give some useful applications of them. Real-life money problems are used throughout this unit for consumers. This doubling time is illustrated in the following applet. We will see that when we translate this verbal statement into a differential equation, we arrive at a differential equation. You think of the whole scenario with rabbits multiplying rapidly, that's exponential growth. find interesting. In the real world, calculators may lose precision, so use a direct log base 2 function if possible. Logarithm base 2 Suppose y is positive number. Get your real rate today. Sunshine is radiant energy, which provides the fuel and warmth that make life on Earth possible. Applications of Exponential and Logarithmic Function. Monday - Friday, 9am to 9pm, Saturday 9 am to 7 pm, ET. If we follow this in software systems, we can create efficient software systems just like other efficient systems around you. Each time you use an app like Facebook, send an instant message, or check the weather on your phone, you're using an API. International dating expert Hayley Quinn, gives advice on how to meet women in real life to help give you the skills you need to be great at dating women, wherever you meet them. Generally, the simple logarithmic function has the following form, where a is the base of the logarithm (corresponding, not coincidentally, to the base of the exponential function). Practical Applications of Mathematics in Everyday Life. 24/7 help for fall, medical, shower, out of home emergencies. Search by what matters to you and find the one thats right for you. The notation is read "the logarithm (or log) base of. 5 Applications of Exponential and Logarithmic Functions As we mentioned in Section6. The University of Florida offers involvement opportunities for all students to maximize their university experience. The purpose of this lab is to familiarize you with some applications from real life involving exponential and logarithmic functions. Students continue an examination of logarithms in the Research and Revise stage by studying two types of logarithms—common logarithms and natural logarithm. The first published use of the "ln" notation for the base-e logarithm was Stringham's, in his 1893 text "Uniplanar Algebra". , when we concluded the most violent year in our history. In this section, we explore some important applications in more depth, including radioactive isotopes and …. (Consider, for instance, the inverse function: if y = ln -a, then e y = -a, but there is no real value of y that could make an exponential negative. Logarithms in the Real World. Author information: (1)Information Technologies Institute, Centre for Research and Technology Hellas, Thessaloniki, Greece. Smith (SHSU) Elementary Functions 2013 2 / 21 Applications of logarithms A worked example. How Are Logarithms Used In Real Life. 4 magnitude (reduced from an initial estimate of 5. Base 5 or base 7 have no real importance, other than providing practice in working with logarithms. Hypothetically in the same year, there is another earthquake recorded in South America that was eight times stronger. This includes such things as plant or population growth or decay such as a bouncing spring. 80 Encourage workers to wear face coverings. 1 5) or other such calculations. The two classic cases are (1) interest accrued as part of loan and (2) interest accrued in a savings or other account. 4 I think your example of real-life geometry is great. By condensing the logarithms, we can create an equation with only one log, and can use methods of exponentiation for solving a logarithmic equation with multiple logs. Signals will stand out from the noise of your life. Substitut-ing these values into our compound interest formula, we get A(t) = P ‡ 1+ r n ·nt A(7. In this lesson we show several Real Life uses of Exponents, as well as their impact on our understanding of the modern world around us. Applications of Pressure in Daily Life Some of the applications of pressure are given below. Newton's Law of Cooling Newton's Law makes a statement about an instantaneous rate of change of the temperature. Most of these real life patterns were evolved over a long period of time by brilliant people to have efficient systems in the society. Search MLS listings directly on your local Coldwell Banker ® office website to find the most up-to-date homes for sale. This includes such things as plant or population growth or decay such as a bouncing spring. In addition, Logarithmic scales are used in Another application of proportions in the real life is in movies screens, because in order to project. Personally I would have done 3 examples of good, bad and great straight from loopnet. In this section, we explore some important applications in more depth, including radioactive isotopes and …. M11GM-Id-1 2. y =ln (x +1) Write original function. LOGARITHMIC FUNCTIONS (Interest Rate Word Problems) 1. That means the logarithm of a given number x is the exponent to which another fixed number, the base b, must be raised, to produce that number x. Students, in turn, can take this valuable experience solving real world problems and building applications to stand out in the interview process as they look for new job opportunities. This course applies principles learned in my course “Introduction to Engineering Mechanics” to analyze real world engineering structures. † Range is (0;1). Carbon dating is based upon the decay of 14 C, a radioactive isotope of carbon with a relatively long half-life (5700 years). f (x) = ln(x). As Pierre-Simon Laplace, a scholar who worked in the fields of mathematics, statistics, physics and astronomy, remarked, “By. Trigonometry is a subject that has lots of practical applications. The Real-Life Possibilities of Black Panther's Wakanda, According to Urbanists and City Planners The design and infrastructure of Black Panther 's fictional East African nation, Wakanda, has. Seven strangers enter The Real World house ready to enjoy the single life, but they don't know that they'll be sharing the space with their exes. Common and Natural Logarithms We can use many bases for a logarithm, but the bases most typically used are the bases of the common logarithm and the natural logarithm. And since (it seems). Exercise 1a. Buy online. This builds towards students' understanding of. NET MVC 5 & real life example. Practical Applications of Mathematics in Everyday Life. Its an example for modeling with Exponential and Logarithmic Equations: Use Newton's Lay of Cooling, T = C + (T 0 - C)e -kt , to solve this exercise. On the other hand, exponential word problems tend to be much more involved, requiring, among other things, that the student first generate the exponential. Printable in convenient PDF format. Lecture 10 - How Science Is. In addition, Logarithmic scales are used in Another application of proportions in the real life is in movies screens, because in order to project. ) A production function might be used to study tradeoffs in the use of. The goal of the project is to determine whether including applications as group activities in College Algebra courses will improve student understanding of the material. Logarithms in the Real World. Applications Involving Exponential Models Exponential functions are useful in modeling many physical phenomena, such as populations, interest rates, radioactive decay, and the amount of medicine in the bloodstream. A common example of exponential decay is radioactive decay. Gaming is an obvious virtual reality application as are virtual worlds but there are a whole host of uses for virtual reality – some of which are more challenging or unusual than others. x Get email notifications whenever Second Life creates , updates or resolves an incident. You can find games, interactive lessons plans, and many other teacher educational resources to make real-life connections in the classroom to increase engagement online. Explain the relevance and application of logarithmic functions in real life situation - Answered by a verified Tutor We use cookies to give you the best possible experience on our website. With the advent of the use of logarithms, mulitplying large numbers became a simple matter, relatively speaking, of looking up numbers in a table. Millions of jobs. Creating one logarithm from a sum. This problem demonstrates the application of a classroom math problem to a real life situation. The area of the edge of a knife’s blade is extremely small. What is logarithmic function? The logarithmic function is the inverse of the exponential function. Implicit in this definition is the fact that, no matter when you start measuring, the population will always take the same amount of time to double. Logos Bible Software combines books, a search engine, and tools that empower anybody to understand the Bible. 1 Using the Rule of 72 we estimate that a 3% investment should double in approximately 72 3 = 24 years. This page is a resource for Second Life developers of all kinds. Lifecasting Smooth-On Alja-Safe® alginate and Body Double® silicone rubber have become Hollywood favorites and industry standards for capturing / reproducing detail from the human body. Exponential functions and their applications, page 3 PROBLEMS 1. Compound interest calculations often involve radicals. Let's take a look at each property individually. M11GM-Id-1 2. If 0 b 1 the function represents exponential decay. Any advice provided is general only and may not be right for you. Likewise, using the logarithm function, small values can be intensified and thereby it can be represented for proper evaluation. Consider for instance that the scale of the graph below ranges from 1,000 to. There may be various types of logs, which might not be useful for the incident under analysis. 1 that illustrates this point. The solution t = 4 tells you when the ball hits the ground. There are real life applications of logarithms but there is little that you would do every life, I mean technically the computer uses it but there rly isn't anything you as a person would do daily. If you sum the squares of any series of Fibonacci numbers, they will equal the last Fibonacci number used in the series times the next Fibonacci number. Applications of Pressure in Daily Life Some of the applications of pressure are given below. The inverse of y =ln (x +1) is y =ex º1. You can monitor many different applications and services with Azure Monitor. 12/12/2008в в· there are real life applications of logarithms but there is little that explain the relevance and application of logarithmic functions in real-life math 11011 applications of logarithmic functions ksu the logarithmic function with base a, denoted loga x, is deflned by its half-life is given by h = ln2 r. To estimate the data in logs obtained from magnitude scales for earthquakes. Free Algebra 2 worksheets created with Infinite Algebra 2. The inverse relationship between exponential and logarithmic functions is also useful for graphing logarithmic functions. On the page Definition of the Derivative, we have found the expression for the derivative of the natural logarithm function $$y = \ln x:$$ \[\left( {\ln x} \right)^\prime = \frac{1}{x}. Though very suc-cessful, the treatment of calculus in those days is not rigorous by nowadays mathematical standards. Data is from MeasuringWorth. y =ln (x +1) Write original function. He compared shared resources to a common grazing pasture; in this scenario, everyone with rights to the pasture grazes as many animals as possible, acting in self-interest for the greatest short-term …. Logarithm, the exponent or power to which a base must be raised to yield a given number. If I understand Logarithms correctly it is the orders of magnitude? Either exponentially growing or exponentially decaying. Applications of Logarithms Logs are used in a variety of applications in sciences, some of the most common are: measuring loudness (decibels), measureing earthquake intensity (Richter scale), radioactive decay, and acidity (pH= -log 10 [H +]). For news updates, please visit our blog. Now the logarithmic form of the statement xy = an+m is log a xy = n +m. But what are some other real life applications/examples of logarithms? Thank You. From determining whether it is better to take a low interest rate or cash back to forecasting the cost of living, learners get to use. If not, stop and use the Steps for Solving Logarithmic Equations Containing Terms without Logarithms. A prime location in southeastern Wisconsin, Racine County is located approximately 30 miles south of Milwaukee and 60 miles north of Chicago. With Lightrun, as the co-founders showed me in a demo, developers can easily add new logs and metrics to their code from their IDE and then receive real-time data from their real production or. , a coroner arrived at the home of a person who had died during the night. A bit about us Who We Are. (In some books, this topic is treated in a special chapter called "Related Rates", but since it is a simple application of the chain rule, it is hardly deserving of title that sets it apart. 20 Real-World Uses for Blockchain Technology Blockchain may also be able to put your end-of-life concerns to rest. You will have to cut those with your truck ax or takedown bucksaw. I am both a Mechanical and an Electrical engineer ( aka use math in real life every day) and I work every day with systems described by exponential or logarithmic functions. This module will introduce logarithms and explore the properties of logarithmic functions of the form $$y=\log_c(x),c>0,c eq1$$. These only work if the base a and the argument are positive. So, it is very important to understand the goal and collect appropriate logs. Exponential Decay Formula: Make a substitution for A and t since it is known that the half-life is 1690 years and : Solve for the decay rate k: Start by dividing both sides by the coefficient to isolate the exponential factor. When students have a solid foundation in logarithms, they are prepared for advanced science classes, and they can feel confident in any career choice. See episodes of your favorite MTV Shows. Now it will go to partial view of ViewForSnacs & see our URL goes to Partial View of ViewForSnacs. What are logarithms and why are they useful? Get the basics on these critical mathematical functions -- and discover why smart use of logarithms can determine whether your eyes turn red at the swimming pool this summer. Tools for Growth We want to equip you, your family, and your LifeGroup to grow as followers of Christ. Since $$2^4 = 16$$ , then $$\log_2(16) = 4$$. Rewards Service Center. Real Life Algorithms. It can't be said too often: a logarithm is nothing more than an exponent. Example 1: A $1,000 deposit is made at a bank that pays 12% compounded annually. Radioactive materials, and some other substances, decompose according to a formula for exponential decay. In this lesson, students will relate the concept of algorithms back to everyday real-life activities by planting an actual seed. Our home of compelling and award-winning documentary content, Real Stories was launched in Autumn 2015 and has grown to include over 300 high quality documentary and factual titles, including over a dozen BAFTA and Emmy award winning programmes. 6931 and ln3 1. from patricia. 20 Real-World Uses for Blockchain Technology Blockchain may also be able to put your end-of-life concerns to rest. Each time you use an app like Facebook, send an instant message, or check the weather on your phone, you're using an API. Graph Magics - an ultimate software for graph theory, having many very useful things, among which a strong graph generator and more than 15 different algorithms that one may apply to graphs (ex. Thus, the domain of the logarithm base b function is the range of the b x function (all positive numbers) and the range of the logarithm base b function is the domain of the b x function (all numbers). Real-Life Application of pH Scale To prove that the pH Scale formula is correct and to relate what we have learnt about logarithms into reality, we have made this small little experiment. We are going to discuss several types of word problems. 1 Using the Rule of 72 we estimate that a 3% investment should double in approximately 72 3 = 24 years. in response to Ryan D. We can measure this intensity on the decibel scale, which is a logarithmic scale. Exponential and Logarithmic Functions, Applications, and Models Exponential FunctionsIn this section we introduce two new types of func-tions. 6931 and ln3 1. By continuing to use this site you consent to the use of cookies on your device as described in our cookie policy unless you have disabled them. Real-life money problems are used throughout this unit for consumers. If$5;000 is invested at a rate of 8%, compounded weekly, flnd the value of the investment after 7 years. Likewise, using the logarithm function, small values can be intensified and thereby it can be represented for proper evaluation. At Real Life Dental, our patients are our highest priority. When students have a solid foundation in logarithms, they are prepared for advanced science classes, and they can feel confident in any career choice. On such shows, coroners often attempt to determine how long a body has been dead. Logarithmic inequalities are inequalities in which one (or both) sides involve a logarithm. I am both a Mechanical and an Electrical engineer ( aka use math in real life every day) and I work every day with systems described by exponential or logarithmic functions. certain functions, discuss the calculus of the exponential and logarithmic functions and give some useful applications of them. You will learn how to develop equations using square roots,a nd practice graphing your results to present your findings. The inversion z ↦ 1 z causes for the logarithmic spiral a reflexion against the imaginary axis and a rotation around the origin, but the image is congruent to the original one. Geometry Vocabulary. In t years an investment will grow to the amount expressed by the function, where t is time (in years). LOGARITHMIC FUNCTIONS (Interest Rate Word Problems) 1. In the same mild but devastatingly logical spirit, Gorsuch shot down a real-life embodiment of the “living Constitution” doctrine in his majority opinion for the Court in Ramos v. Well the career that i am pursuing is aerospace engineering so i will do the project on that. So, it is very important to understand the goal and collect appropriate logs. Jun 14, 2020 - Logarithms - Real Life Applications Quant Video | EduRev is made by best teachers of Quant. In 1996, the average cost of tuition was $2,975 per year. , 1997;Baker et al. They are important in measuring the magnitude of earthquakes, radioactive decay and population growth. The Asus ZenBook 13 UX333 is the best college laptop you can buy. This event log pertains to a loan application process of a Dutch financial institute. Ax or Saw – cut two logs about 8 feet in length; Rope – non-elastic is preferable for safety reasons. by TeenLife. The basic concept of logarithms can be. In this chapter, "Real Work for Real Audiences," Richardson envisions students creating work that is relevant. In this study, they take notes about the two special types of logarithms, why they are useful, and how to convert to these forms by using the change of base formula. You can monitor many different applications and services with Azure Monitor. The derivative of f(x) is:. I know right now you may think that real life is a lot more intimidating than sending a woman you like a message on an online dating site, but I’m going to share. Encouraging students to be active, mentally and physically, is the best way to make a real world connection. Real Life Mobile App. Amansmathsblogs. Personally I would have done 3 examples of good, bad and great straight from loopnet. Explore math with Desmos. Since equations like (*) need to be solved all the time in real-life applications such as engineering, complex numbers are needed. Intervals, Exponents, Logarithms. In the realm of medical instrumentation, a notable real-life application is Omron’s fuzzy- logic-based and widely used blood pressure meter. Apart from academic studies, logarithm has vast application in real life operations. RealLife English brings you a passionate team of native fluency coaches who are committed to guiding you beyond the classroom, so that you can learn, live, and speak English in the real world. Smith (SHSU) Elementary Functions 2013 2 / 21 Applications of logarithms A worked example. Excuse if it sucks) The graph illustrates the situation. Exponential growth means something is getting bigger. This is not the case for ez; we have. if n and a are positive real numbers, and a is not equal to 1, then If a x = n, then log a n = x. It is understood to be 10 10 1 0 10. Solving Radical Cube Roots. In general, we also use properties and applications of logarithms in various geological circumstances: 1. Students can learn the properties and rules of these functions and how to use them in real world applications through word problems such as those involving compound interest and. Since both k and t are exponents, we must use logarithms. Collect the logs according to your needs. Modern JavaScript Frameworks like Angular, React and Vue. In addition, Logarithmic scales are used in Another application of proportions in the real life is in movies screens, because in order to project. Exponentials and Natural Logs. Radicals and Rational Exponents and Applications of Logarithms MTH133-0902B-32 College Algebra Set of data that can be modeled as a polynomial function series is a function with index as independent variable series is a function with index as independent variable series is a function with index as independent variable. Once accepted, as long as you pay your premiums on time, we guarantee to renew your cover despite changes to your age, health and lifestyle. You often see logarithms in action on television crime shows, according to Michael Breen of the American Mathematical Society. Each study involved students in four college algebra sections in two two-year campuses. Are you ready to spend some mulah? Or are you ready to save some dough? This section is all about using the equations for loans, investments, and present value which you can use when you are shopping for that car or house. In DNA Interactive: Applications, investigate techniques of forensic analysis, how DNA science is applied to healthcare, & into mysteries of our human origins. In t years an investment will grow to the amount expressed by the function, where t is time (in years). Interactive questions, awards, and certificates keep kids motivated as they master skills. Solving Radical Cube Roots. Know and apply the properties of logarithms. x =ln (y + 1) Switch x and y. (The exact answer is slightly more. They are essential in mathematics to solve certain exponential-type problems. Detect, diagnose and resolve issues with greater speed and accuracy. It is the limit of (1 + 1/n) n as n approaches infinity, an expression that arises in the study of compound interest. Notice that the domain of the logarithm is limited to x > 0, since there are no real values defined for the logarithm when x is zero or less than zero. Sign up for an account to be matched to a local Coldwell Banker ® agent in your area, be the first to know about new MLS listings with email notifications, and customize your MLS search. Check out the newest additions to the Desmos calculator family. Types of Logarithmic Equations The first type looks like this. Most of these real life patterns were evolved over a long period of time by brilliant people to have efficient systems in the society. Light is one type of radiant energy. From determining whether it is better to take a low interest rate or cash back to forecasting the cost of living, learners get to use. We have already explored some basic applications of exponential and logarithmic functions. The various octaves of a given note, say $$C$$, sound similar to one another. These awesome 15 STEM summer online enrichment programs range between elementary, middle, and high school student offerings. Suppose a driver wants to know how many miles he has to drive to earn$100. We like to think these tests are accurate, yet, horror stories seem to abound. Solve log 2 3 x. what is an example of an application of logarithms in real life situations? the mechanical slide rule is actually an accurate "ruler" that shows numbers on its face, but they are spaced logarithmically. (b) Evaluate log2 1 32. certain functions, discuss the calculus of the exponential and logarithmic functions and give some useful applications of them. The Asus ZenBook 13 UX333 is the best college laptop you can buy. Out are the old school text based scenarios; in are graphic novels and comic strip panels using real employee and workplace photography. The diversity of the processes which are described by the natural exponential function appears amazing. In the same fashion, since 10 2 = 100, then 2 = log 10 100. Logarithm, the exponent or power to which a base must be raised to yield a given number. 12/12/2008в в· there are real life applications of logarithms but there is little that explain the relevance and application of logarithmic functions in real-life math 11011 applications of logarithmic functions ksu the logarithmic function with base a, denoted loga x, is deflned by its half-life is given by h = ln2 r. Exponential & Logarithmic Applications Compound Interest In compound interest formulas, is the balance, is the principal, is the annual interest rate (in decimal form), and is the time in years. What are logarithms, and what do they do? 1. Explain the relevance and application of exponential functions in real-life situations Explain the relevance and application of logarithmic functions in real-life situations. Collect the logs according to your needs. Tests used for detecting things like drug abuse, intoxication, disease, genetic and birth defects, etc. And since (it seems). Free Algebra 2 worksheets (pdfs) with answer keys-each includes visual aides, model problems, exploratory activities, practice problems, and an online component. the population doubles every 20 years). In this section, we explore some important applications in more depth, including radioactive isotopes and …. 5 Exponential and Logarithmic Models 259 Additional Example Radioactive iodine is a by-product of some types of nuclear reactors. They are essential in mathematics to solve certain exponential-type problems. They are used in science, business, medicine, and even more fields. Think of a real-life situation that can be represented by a logarithmic function, translate the situation to the function, and solve the function and represent it graphically. Applications of the Derivative 6. From determining whether it is better to take a low interest rate or cash back to forecasting the cost of living, learners get to use. LOGARITHMIC FUNCTIONS (Interest Rate Word Problems) 1. In 1996, the average cost of tuition was \$2,975 per year. Solving -16t 2 + 48t + 64 = 0, you factor to get -16(t - 4)(t + 1) = 0. Consider for instance the graph below. log of any number to base as itself is 1. You think of the whole scenario with rabbits multiplying rapidly, that's exponential growth. Money Math: Lessons for Life This free four-lesson collection of real-life examples from the world of finance includes a teacher's guide with lesson plans, activity pages, and teaching. It is very important in solving problems related to growth and decay. Logarithms graphs are well suited. An application of the logarithmic function. They learn the equation to find intensity, Beer's law, and how to use it. In order to work with these application problems you need to make sure you have a basic understanding of arithmetic sequences, arithmetic series, geometric sequences, and geometric series. Modelling Exponential Decay - Using Logarithms. The natural and common logarithm can be found throughout Algebra and Calculus. Consider the scenario described above and assume that this is a human population where a generation represents about 20 years. Intervals, Exponents, Logarithms. engineering, physics). I am both a Mechanical and an Electrical engineer ( aka use math in real life every day) and I work every day with systems described by exponential or logarithmic functions. Then they use the associated activity to investigate real world applications prior to completing a sheet of practice problems that use the Beer's law equation. 9 Real-Life Scenarios That Show How The Internet Of Things Could Transform Our Lives. Explain the relevance and application of logarithmic functions in real life situation - Answered by a verified Tutor We use cookies to give you the best possible experience on our website. JournalLife is the perfect place to hear yourself think, to listen to your own life being lived. When given a percentage of growth or decay, determined the growth/decay factor by adding or subtracting the percent, as a. Since the best predictor of the future is the past, there is a good chance that this downward trend will continue for some time into the. 5 into the formula, you get that h = 100 feet. The Real Life scenario of Logarithms is to measure the acidic, basic or neutral of a substance that describes a chemical property in terms of pH value. Real-Life Application of Logarithms in Measuring Sound Intensity. We will see that when we translate this verbal statement into a differential equation, we arrive at a differential equation. There are millions of geocaches worldwide, just waiting for you to find them. It is the limit of (1 + 1/n) n as n approaches infinity, an expression that arises in the study of compound interest. To solve an exponential or logarithmic word problems, convert the narrative to an equation and solve the equation. Search by what matters to you and find the one thats right for you. (2019, February 26). Find x such that. Get all the diagnostics you need to find and solve performance issues, fast. | 2020-11-24T20:00:28 | {
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https://math.stackexchange.com/questions/1739918/use-induction-to-prove-that-that-8n-4n-for-all-positive-integers-n | Use induction to prove that that $8^{n} | (4n)!$ for all positive integers $n$
Use induction to prove that that $8^{n} | (4n)!$ for all positive integers $n$
So far I have: Base case (n = 1) = $8^{1} | (4(1))!$
= $8 | 24$ which is true.
Induction Step:
$8^{n + 1} | (4(n + 1))!$
$8^{n + 1} | (4n + 4)!$
• A bit confused as to how to close this proof out, also wanted to make sure my current progress is correct as well. Any help is appreciated.
• Hint: $(4(n+1))! = (4n+4)! = (4n+4)(4n+3)(4n+2)(4n+1)(4n)! = 8(n+1)(4n+3)(2n+1)(4n+1) (4n)!$ – GohP.iHan Apr 12 '16 at 23:37
• You can also forego induction: Let $[x]$ denote the largest integer not exceeding $x.$ For prime $p$, the largest $k$ such that $p^k$ divides $n!$ is $k= \sum_{j=1}^n [n/p^j].$... Because $[n/p]$ counts how many multiples of $p$ are less or equal to $n$, and $[n/p^2]$ adds $1$ more for each multiple of $p^2$ not exceeding $n,$ and $[n/p^3]$ adds one more for......(etc). We have $\sum_{j=1}^n [4 n/2^j]\geq$ $[4 n/2^1+ 4 n/2^2]=$ $2 n +n=3 n.$ – DanielWainfleet Apr 13 '16 at 1:29
Well, you know that $(4n)! = 8^nk$ for some integer $k$.
$$\Rightarrow (4n + 4)! = (4n+4)(4n+3)(4n+2)(4n+1)(4n)!$$ $$= 8^nk(4n+4)(4n+3)(4n+2)(4n+1)$$
Factor out of the even terms: $$= 8^n 8 k (n+1)(4n+3)(2n+1)(4n+1)$$ $$= 8^{n+1} k'$$
For some integer $k'$. This means that $(4n+1)!$ is divisible by $8^{n+1}$.
Good so far, to finish up just note that $$(4(n + 1))! = (4n + 4)! = (4n + 4)(4n + 3)(4n + 2)(4n + 1)(4n)!.$$ Since $4$ divides $(4n + 4)$ and $2$ divides $(4n + 2$), we have that $8$ divides $(4n + 4)(4n + 3)(4n + 2)(4n + 1)$. By the induction hypothesis, $8^n$ divides $(4n)!$. Therefore $8^{n+1}$ divides $(4(n + 1))!$, completing the induction step.
You can also prove this statement without induction. Let $S = \{1,2\ldots,4n\}$. Note that there are $2n$ multiplies of $2$ in $S$ and $n$ multiplies of $4$ in $S$. Can you see why this implies that $2^{2n} \cdot 2^n = 2^{3n} = 8^n$ divides $(4n)!$?
Use the fact for any integer $n$, one of the numbers inside $n,n+1,n+2,n+3$ divisible by $4$ and one of them is divisible by $2$ but not $4$.
This easy fact says us any $4$ consecutive integers divisible by $8$.
We will use this fact in induction step. | 2021-03-08T12:40:40 | {
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https://tex.stackexchange.com/questions/375680/aligning-different-for-each-pair-of-equations-in-same-align-environment | # Aligning different for each pair of equations in same align-environment
I am using the following environment and want to have alignment for each the first two, the second two and the third two equations at the =-Symbol. In this moment everything aligns at the same point but this way the last equation for instance is too far on the left. What would you suggest? Splitting it up in three align-environments doesn't help as then the seperation between the equations would be too big.
\begin{align}
Cov\left(\tilde{c}_D,\tilde{r}_M\right) &= \sum_{i=1}^4f_i\left(c_{D,i}-E(\tilde{c}_D)\right)\left(r_{M,i}-E(\tilde{r}_M)\right),\\
Cov\left(\tilde{c}_E,\tilde{r}_M\right) &= \sum_{i=1}^4f_i\left(c_{E,i}-E(\tilde{c}_E)\right)\left(r_{M,i}-E(\tilde{r}_M)\right),\\
V_D &= \frac{E(\tilde{c}_D) - \left( E\left(\tilde{r}_M\right) - r_f\right)\frac{Cov\left(\tilde{c}_D,\tilde{r}_M\right)}{\sigma_{\tilde{r}_M}^2}}{1+r_f},\\
V_U &= \frac{E(\tilde{c}_E) - \left( E\left(\tilde{r}_M\right) - r_f\right)\frac{Cov\left(\tilde{c}_E,\tilde{r}_M\right)}{\sigma_{\tilde{r}_M}^2}}{1+r_f},\\
\mu_D &= \frac{E(\tilde{c}_D)}{V_D} - 1 \text{~und~}\\
\mu_E &= \frac{E(\tilde{c}_E)}{V_E} - 1.
\end{align}
• For future questions: You really should make your code minimal working examples (MWE). A MWE is compilable and creates the described problem. It starts at \documentclass and ends at \end{document} and includes all packages that are needed for your code (but not more). – Skillmon Jun 19 '17 at 11:18
ok, I found a solution by myself as I googled for another problem. This helps:
\begin{gather}
\begin{align}
Cov\left(\tilde{c}_D,\tilde{r}_M\right) &= \sum_{i=1}^4f_i\left(c_{D,i}-E(\tilde{c}_D)\right)\left(r_{M,i}-E(\tilde{r}_M)\right),\\
Cov\left(\tilde{c}_E,\tilde{r}_M\right) &= \sum_{i=1}^4f_i\left(c_{E,i}-E(\tilde{c}_E)\right)\left(r_{M,i}-E(\tilde{r}_M)\right),
\end{align}\\
\begin{align}
V_D &= \frac{E(\tilde{c}_D) - \left( E\left(\tilde{r}_M\right) - r_f\right)\frac{Cov\left(\tilde{c}_D,\tilde{r}_M\right)}{\sigma_{\tilde{r}_M}^2}}{1+r_f},\\
V_U &= \frac{E(\tilde{c}_E) - \left( E\left(\tilde{r}_M\right) - r_f\right)\frac{Cov\left(\tilde{c}_E,\tilde{r}_M\right)}{\sigma_{\tilde{r}_M}^2}}{1+r_f},
\end{align}\\
\begin{align}
\mu_D &= \frac{E(\tilde{c}_D)}{V_D} - 1 \text{~und~}\\
\mu_E &= \frac{E(\tilde{c}_E)}{V_E} - 1.
\end{align}
\end{gather}
• @DavidCarlisle It works: the amsmath documentation explains (well hidden in the last part about error messages) "the chief exception is that align and most of its variants can be used inside the gather environment". – campa Jun 19 '17 at 11:26
• @campa sigh I wonder who's maintaining amsmath these days:-) (thanks I'll delete the comment) – David Carlisle Jun 19 '17 at 11:29
• Add \DeclareMathOperator{\Cov}{Cov} to your preamble and use \Cov in the formulas. Each of the \left and \right you have here should be omitted. – egreg Jun 19 '17 at 12:22
I suggest some improvements: using subequations since the equations go in pairs,medium-sized fractions from nccmath for fractions in fractions, declare Cov as a math operator (for now, Cov looks like the product of three variables), and another placement of the word ‘und’ with the ArrowBetweenLines command from mathtools:
\documentclass{article}
\usepackage{mathtools, nccmath}
\DeclareMathOperator{\Cov}{Cov}
\begin{document}
\begin{gather}
\begin{subequations}
\begin{align}
\Cov\left(\tilde{c}_D,\tilde{r}_M\right) & = ∑_{i=1}⁴f_i\left(c_{D,i}-E(\tilde{c}_D)\right)\left(r_{M,i}-E(\tilde{r}_M)\right), \\
\Cov\left(\tilde{c}_E,\tilde{r}_M\right) & = ∑_{i=1}⁴f_i\left(c_{E,i}-E(\tilde{c}_E)\right)\left(r_{M,i}-E(\tilde{r}_M)\right),
\end{align}
\end{subequations}
\\
\begin{subequations}
\begin{align}
V_D & = \frac{E(\tilde{c}_D) - \left( E\left(\tilde{r}_M\right) - r_f\right)\mfrac{\Cov\left(\tilde{c}_D,\tilde{r}_M\right)}{\sigma_{\tilde{r}_M}²}}{1+r_f}, \\
V_U & = \frac{E(\tilde{c}_E) - \left( E\left(\tilde{r}_M\right) - r_f\right)\mfrac{\Cov\left(\tilde{c}_E,\tilde{r}_M\right)}{\sigma_{\tilde{r}_M}²}}{1+r_f},
\end{align}
\end{subequations}\\
\begin{subequations}
\begin{alignat}{2}
& & \mu_D & = \frac{E(\tilde{c}_D)}{V_D} - 1 \\
\ArrowBetweenLines[\text{und~}]
& & \mu_E & = \frac{E(\tilde{c}_E)}{V_E} - 1.
\end{alignat}
\end{subequations}
\end{gather}
\end{document} | 2019-07-24T01:01:00 | {
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https://math.stackexchange.com/questions/882877/produce-unique-number-given-two-integers/882961 | # Produce unique number given two integers
Given two integers, $a$ and $b$, I need an operation to produce a third number $c$. This number does not have to be an integer. The restrictions are as follows:
1. $c$ must be unique for the inputs (but it does not have to be reversible).
2. $a$ and $b$ must be interchangeable ($a$ & $b$ = $b$ & $a$)
Initially, the first thing I thought of was simply $a+b$, however naturally that does not fit restriction 1. Then I considered a hash function of some sort, but that doesn't fit 2.
Any thoughts?
• What do you mean that doesn't fit 1, there is only one unique number which is the sum of a and b. Jul 30 '14 at 18:08
• I believe that the OP wants an injective operation from $\mathbb Z \times \mathbb Z$ to $\mathbb Z$ as condition 1, and for the operation to be commutative for condition 2. Jul 30 '14 at 18:08
• @qaphla Almost, except it should be $2$-to-$1$, since $f(a, b) = f(b, a)$.
– user61527
Jul 30 '14 at 18:09
• It's impossible to have such an injection from ordered pairs, but an injection from two-element subsets is fine, since then the 'commutativity' condition is implied -- $f(\{a,b\}) = f(\{b,a\})$. Jul 30 '14 at 18:14
• I'd say that a commutative function on ordered pairs can be pretty clearly identified with a function on sets of size two, and the difference between the two isn't worth caring about. Jul 30 '14 at 18:31
If we restrict $a,b$ to be non-negative integers, we can try $f(a,b) = \dfrac{\max(a,b)(\max(a,b)+1)}{2}+\min(a,b)$.
This satisfies $f(a,b) = f(b,a)$ and grows quadratically with $\max(a,b)$. To help you see the pattern:
$f(0,0) = 0$,
$f(1,0) = 1$, $f(1,1) = 2$,
$f(2,0) = 3$, $f(2,1) = 4$, $f(2,2) = 5$,
$f(3,0) = 6$, $f(3,1) = 7$, $f(3,2) = 8$, $f(3,3) = 9$,
...
If you want to allow any integers $a,b$ then let $g$ be your favorite bijection from $\mathbb{Z}$ to $\mathbb{N}_0$, then let $f(a,b) = \dfrac{\max(g(a),g(b))(\max(g(a),g(b))+1)}{2}+\min(g(a),g(b))$.
One such bijection is $g(n) = \begin{cases} -2n & n \le 0 \\ 2n-1 & n \ge 1\end{cases}$.
• This is certainly the best solution so far from an implementation point of view. Jul 30 '14 at 20:18
• Great answer - I can see this being very useful. Jul 31 '14 at 4:01
• Additionally, if you wanted $f$ to take on negative values as well as positive values, you could define $$f(a, b) = g^{-1}\left(\dfrac{\max(g(a),g(b))(\max(g(a),g(b))+1)}{2}+\min(g(a),g(b))\right).$$
– user88319
Jul 31 '14 at 4:09
• I should look to see if there is a $g$ that maps my answer to yours.
– robjohn
Aug 1 '14 at 15:51
• this is brilliant, can you please describe how you found the solution / proved it? Oct 22 '20 at 16:46
How about just $2^a+2^b$? This represents the binary number with $1$s at exactly the $a$-th and $b$-th positions if $a\ne b$, and a single $1$ at the $(a+1)$-th position if $a=b$.
• This is a better answer than mine. Jul 30 '14 at 18:28
• reversible? Using this method we get a and b from c. Jul 30 '14 at 19:08
• @mridul Any method we use is theoretically reversible (up to specifying which input is $a$ and which is $b$) because we have a uniqueness requirement - meaning from $f(a,b)$ we can extract the information $\{a,b\}$. Jul 30 '14 at 19:18
• If we're talking about a computer, this becomes trickier as soon as $a$ or $b$ get bigger than $64$... not a great range. Jul 30 '14 at 20:30
• @JackM The goal of this answer was to give a nice mathematical solution to the problem regardless of practicalities. For a good way to actually solve the problem on a computer, please see the other answers - say TonyK's answer, or JimmyK4542's answer. Jul 30 '14 at 20:34
I think $a\circ b=(2^a+1)(2^b+1)$ works, if I understand correctly. $a \circ b = b \circ a$ and $a$ and $b$ can be found (up to permutation) from $(2^a+1)(2^b+1)$ (assuming $a$ and $b$ are integers.)
• Nice answer; especially since one can tell the original two numbers by glancing at the binary representation of the result. Jul 30 '14 at 18:19
• -1 Mathematically elegant but this isn't practical to implement into code for large $a$ and $b$. Jul 30 '14 at 18:21
• Who said anything about practicality for code? Jul 30 '14 at 18:23
• Finding a power of 2 modulo a prime is trivial. Jul 30 '14 at 18:35
• @ozo If that's what the OP wants, then he ought to ask a more specific question. This answer perfectly fits the requirements as stated. It's ridiculous to downvote someone for not reading the OP's mind or making unwarranted assumptions from the question. Jul 30 '14 at 19:05
Peter Woolfitt's suggestion ($2^a + 2^b$) is the simplest so far, but it becomes extremely large for quite small values of $a$ and $b$. For a more manageable function, I suggest interleaving the binary representations of $a$ and $b$. Then the result will be no larger than $\max(a,b)^2$.
To make it commutative ($f(a,b)=f(b,a)$), you will first have to swap $a$ and $b$ if $a< b$. Then it goes something like this:
f := 1
while a != 0 or b != 0
// Incorporate bottom bit of a
f := 2 * f
if a is odd then f := f + 1
a := a/2 // Discard bottom bit of a
// Incorporate bottom bit of b
f := 2 * f
if b is odd then f := f + 1
b := b/2 // Discard bottom bit of b
wend
• And what happens to the uppermost bits? It seems they get left out... Jul 30 '14 at 18:42
• @abiessu: No, every bit gets inspected and incorporated. That's why a and b get divided by 2. Jul 30 '14 at 18:44
• You're right, I mixed up my "negated input AND" logic and thought it would end prematurely. Jul 30 '14 at 18:48
• Oh, no! You're right! That 'and' should be an 'or'. I have fixed it now. Jul 30 '14 at 18:54
• @ozo: Yes, you are right. I have fixed that with f:=1 instead of f:=0; perhaps it works now... Jul 30 '14 at 19:19
Note: This answer is substantially the same as the one given by JimmyK4542. I am leaving it here in case some minor difference in wording helps someone understand the derivation.
If we can additionally assume that the integers are nonnegative, I believe that the following will satisfy the conditions given.
First note that commutativity can be guaranteed by sorting the elements, so without loss of generality we assume $a \geq b$. Call the given pair which has been sorted $(a_0, b_0)$. We can identify the following sequence which uniquely transforms a sorted pair $(a,b)$:
\begin{aligned} (0,0) &\rightarrow 0 \\ (1,0) &\rightarrow 1 \\ (1,1) &\rightarrow 2 \\ (2,0) &\rightarrow 3 \\ (2,1) &\rightarrow 4 \\ (2,2) &\rightarrow 5 \\ &\vdots \end{aligned}
From this it is clear that if we can compute the number of elements in this sequence which have $a < a_0$, and then add $b_0$, we have an answer that works. Let $$N(k) = \sum_{n=0}^{k}{n} = (k)(k+1)/2$$
Then we have a mapping $$(a \, \& \, b) \rightarrow N(max(a,b))+min(a,b)$$ which satisfies the two properties given.
As a benefit, this answer also scales only quadratically with the largest number in the pair.
• It appears that the answer by JimmyK4542 was posted while I was typing this up. What is the proper etiquette here (I'm new)? Should I delete my answer, being that it is basically the same? Jul 30 '14 at 19:58
• The rule of thumb is: if you think that your explanation is clearer/more concise than someone else's then keep it. Otherwise, it's good etiquette to delete it. Jul 30 '14 at 20:20
• @Hunter There is not really a consensus on this issue of etiquette. Please see here and here for discussions on this topic. Jul 30 '14 at 20:46
More programmerly than mathily, use a nondigit separator, like the convenient decimal point. In this case I concat a string together then cast it to a float.
c = parseFloat(max(a,b) + '.' + min(a,b))
c will be unique and reversible for all interchangeable combinations of a and b.
so for example,
myhash(124,24) = 124.24
myhash(24,124) = 124.24
myhash(11231,26611) = 26611.11231
I think some index systems like the Dewey Decimal system and some part numbering schemes use this sort of bin-based technique.
But oops, then there is a problem if the second number ends with trailing zeros, so stringwise reverse it to preserve them:
c = parseFloat(max(a,b) + '.' + reverse(min(a,b)))
then
myhash(123,456) = 123.654
myhash(12300,4560) = 12300.0654
• Whoops, this won't handle negative integers very well. Jul 30 '14 at 20:06
• Assuming $a$ and $b$ are of a similar size then this limits them each to about 3 significant figures (32-bit IEEE float) or 8 significant figures (64-bit IEEE float). Jul 30 '14 at 20:15
• You could encode the sign of min(a,b) in the first decimal digit (so for example myhash(2,1) = 2.01, myhash(2,-1) = 2.11).
– hvd
Jul 31 '14 at 12:24
• @ozo That assumes they'll be stored in an IEEE floating point variable. That's a problem for the programming aspect, but not for the maths aspect.
– hvd
Jul 31 '14 at 12:26
Sort a and b ,then apply hash function.
c=hash(sort(a,b))
Here,
1. c is unique for the inputs and not reversible.
2. a and b are interchangeable (a & b = b & a).
• All hash functions have collisions, multiple inputs with the same output. So this violates requirement 1. The probability of getting a collision for two randomly chosen inputs may be very low, and so not worth worrying about in practice, but it can theoretically happen. Jul 31 '14 at 0:10
Let $p$ and $q$ be distinct primes, and let
$$c=\min(p^aq^b,p^bq^a)$$
In fact this generalizes to a function for $n$ interchangeable variables $a_1,\ldots,a_n$, using distinct primes $p_1\ldots,p_n$:
$$c=\min_{\pi}\{\prod_{i=1}^np_i^{a_{\pi(i)}}\}$$
where the min is taken over all permutations of $\{1,\ldots,n\}$.
(Note, the integers $a$ and $b$ need not be positive. The Fundamental Theorem of Arithmetic still guarantees that different multisets correspond to different values of $c$.)
Well, there is a rather simple way to do this using string operations:
1. Let x = min(a,b) and y=max(a,b)
2. Treating x and y as strings and + as the concatenation operator, let z = x + "|" + y
3. Use your favorite string to binary/hex/decimal function to turn z into a number.
Let's look at an example. f(25,-36).
x = -36 and y = 25.
z = "-36|25".
Using a standard ASCII to Decimal converter, we get the number 4551541245053.
• I am trying to create combinations of a set of numbers in excel and thus realized that I need a function like the one mentioned in the main question of this thread. While I already tried 2^a + 2^b i realized it does not work for large a and b due to excel issues... but i Found some really great mathematical answers on this page and learnt a lot.. William Forcier from an excel point of view your answer is awesome. I made only 1 change...instead of converting the concatted text to a number I used a countifs and running countifs to weed out duplicates... Thank you very much for the base idea :) Feb 10 '19 at 5:56
Given $\max(|a|,|b|)$ and $a+b$, we can recover the unordered pair $(a,b)$ as $$\textstyle\left(\min(a+b,0)+\max(|a|,|b|),\max(a+b,0)-\max(|a|,|b|)\right)\tag{1}$$ There are $4n+1$ unordered pairs of integers so that $\max(|a|,|b|)=n$; their sums being $$\{-2n,-2n+1,\dots,0,\dots2n-1,2n\}\tag{2}$$ Since $$\sum_{k=1}^n(4k+1)=2n^2+3n\tag{3}$$ we will set $$f(a,b)=2\max(|a|,|b|)^2+\max(|a|,|b|)+a+b\tag{4}$$ Then the greatest $f(a,b)$ can be for $\max(|a|,|b|)=n$ is $$2n^2+n+2n=2n^2+3n\tag{5}$$ and the least it can be for $\max(|a|,|b|)=n+1$ is $$2(n+1)^2+(n+1)-2(n+1)=2n^2+3n+1\tag{6}$$ Therefore, the $f$ given in $(4)$ maps unordered pairs of integers to non-negative integers injectively (and incidentally, surjectively).
• Beautiful. Looks like the best solution so far to me, since the OP apparently wants to use it as a perfect hashing function or something like that. -- There are parenthesis missing around $4k+1$ in (3).
– user20796
Aug 1 '14 at 13:58
• @HansAdler: thanks. I can see your point about the parentheses, so I have added them. However, I usually write $1+\sum\limits_{k=1}^n4k$ when I intend the other association, just to be sure.
– robjohn
Aug 1 '14 at 15:59
If $P$ are all the primes, then:
$f(a,b) = P(a) P(b)$
f(0,0)=4
f(1,0)=6
f(1,1)=9
f(2,0)=10
f(2,1)=15
f(2,2)=25
f(3,0)=14
f(3,1)=21
f(3,2)=35
f(3,3)=49
$a$ & $b$ are interchangeable. It is reversible, but very hard for large numbers of c. I think this is also how asymmetric cryptography works in its most simple form with large numbers of '$a$' & '$b$'.
Example:
• 'P(a)' being the public-key on the server side
• 'P(b)' being the private-key on the client side
• 'c' being the token to be transmitted.
2. server transmits token to client and requests for a factor of it.
3. client divides token with the locally stored private-key and send the result to the server
4. Server compares stored public-key with given result, if same then grants access
• And it works also if you have more then 2 input variables. Dec 9 '14 at 12:23
Since your answer need to be a integer the following simple steps should work:
1. Sort the numbers a and b with the smallest first if a = b just put a first.
2. Now put the first number in front of the dot and the second number behind the decimal, then append one to the end (so that if the second number ended in 0 it will now end in 01 to keep it unique)
3. now you have a unique easily reversible number.
This method will also be reasonably memory efficient, and also ensure uniqueness (and you can easily find a given b and b given a).
• I see two difficulties with this approach: 1) The result for numbers with trailing zeroes won't be unique (1.50 = 1.500) - this can be resolved by reversing the number behind the decimal point. 2) the question states "integers", how can this function be improved to handle negative integers while still being injective? Jul 31 '14 at 8:51
• @waldrumpus, as TonyK observed when I asked much the same question of his answer, you could handle signs by encoding $a\le b$ as $ax.yr$, where $r$ is $b$ written in reverse (as you suggested) and $x,y\in\{0,1\}$ to indicate the signs of $a$ and $b$. Jul 31 '14 at 19:17
• added a simple trick just append 1 to the end and it will always be unique Jul 31 '14 at 19:45
• @BarryCipra Very nice! Since this answer was the first thing to pop into my mind, I'm glad to see it improved and perfectly acceptable. Aug 1 '14 at 7:51
Use complex numbers $u+vI$ with integer coefficients $u, v$. Let $c=a*b=(a+b)+abI$. It is easy to see that if we know $u=a+b$ and $v=ab$ then we know $a, b$. To see this consider the polynomial $x^2-ux+v$ then this has roots $a,b$.
If you want the answer to be real then you could use $(a+b)+(ab)\pi$
Yet another (and quite standard) solution is using the Cantor pairing function $\pi: \mathbb N\times\mathbb N\to \mathbb N$. It is a bijection. So now let's only find a bijection between unordered pairs ${\mathbb N}\choose{2}$ and ordered pairs $\mathbb N\times \mathbb N$. One such is $$\sigma:\{a,b\}\mapsto(\min\{a,b\},\max\{a,b\}-\min\{a,b\})$$ with the inverse $$\sigma^{-1}:(x,y)\mapsto \{x,x+y\}.$$
Therefore $\pi\circ\sigma$ satisfies your requirements, moreover, it is a bijection between unordered pairs of natural numbers and natural numbers, so it's efficient.
If you need integers instead of natural numbers, use your favourite bijection $\mathbb Z\to\mathbb N$, there's plenty of them.
Since some seem to be talking about programming and this is too long for a comment, I'll post this answer even though it's not really appropriate for a mathematics question. Given two 16 bit signed ints a and b, you can sort them and or them together. Meaning:
int uniqueCombo(int a, int b) { //assuming -2^15<=a,b<=2^15-1
if(a<b)
return (a&0xFFFF)|(b<<16);
else
return (b&0xFFFF)|(a<<16);
}
This, by the magic of signed and unsigned integers, unless I'm mistaken, will handle positives and negatives perfectly fine and is easy to invert (to get a as a 32 bit signed int, take out the first sixteen bits. If the biggest place digit is a one, pad the new 16 large place digits with ones. Else zeros.) | 2021-09-18T19:54:53 | {
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http://mathhelpforum.com/calculus/116956-indefinite-integral.html | # Math Help - Indefinite Integral
1. ## Indefinite Integral
Evaluate $\int{x^3\sqrt{x^2+36}} dx$
Stuck on this one.
2. Originally Posted by Em Yeu Anh
Evaluate $\int{x^3\sqrt{x^2+36}}\,dx$
Stuck on this one.
Apply the substitution $u=x^2+36\implies\tfrac{1}{2}\,du=x\,dx$.
Thus, $\int x^3\sqrt{x^2+36}\,dx\xrightarrow{u=x^2+36}{}\tfrac {1}{2}\int (u-36)\sqrt{u}\,du$.
Can you take it from here?
3. Originally Posted by Chris L T521
Apply the substitution $u=x^2+36\implies\tfrac{1}{2}\,du=\,dx$.
Thus, $\int x^3\sqrt{x^2+36}\,dx\xrightarrow{u=x^2+36}{}\tfrac {1}{2}\int (u-36)\sqrt{u}\,du$.
Can you take it from here?
I'm slightly confused on the last part that has (u-36); is that supposed to represent the x^3?
$u=x^2+36\implies\tfrac{1}{2}\,du=x\,dx$
$x^2=u-36$
$\int x^3\sqrt{x^2+36}\,dx=\int x^2\sqrt{x^2+36}\,xdx=\int (u-36)\sqrt{u}\,\tfrac{1}{2}du$ $=\tfrac{1}{2}\int (u-36)\sqrt{u}\,du$
5. Anyone mind checking the complete solution before I submit it?
$\frac{1}{2}\int(u-36)\sqrt{u} = \frac{1}{2}\int{u^{\frac{3}{2}}-36u^{\frac{1}{2}}}$
$= \frac{1}{2}(\frac{2}{5}u^{\frac{5}{2}}-24u^{\frac{3}{2}})+ C = \frac{2}{10}(x^2+36)^{\frac{5}{2}} - 12(x^2+36)^{\frac{3}{2}} + C$
6. Originally Posted by Em Yeu Anh
Anyone mind checking the complete solution before I submit it?
$\frac{1}{2}\int(u-36)\sqrt{u} = \frac{1}{2}\int{u^{\frac{3}{2}}-36u^{\frac{1}{2}}}$
$= \frac{1}{2}(\frac{2}{5}u^{\frac{5}{2}}-24u^{\frac{3}{2}})+ C = \frac{2}{10}(x^2+36)^{\frac{5}{2}} - 12(x^2+36)^{\frac{3}{2}} + C$ | 2016-06-27T09:59:12 | {
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https://proofwiki.org/wiki/Group_is_Normal_in_Itself | # Group is Normal in Itself
Jump to navigation Jump to search
## Theorem
Let $\struct {G, \circ}$ be a group.
Then $\struct {G, \circ}$ is a normal subgroup of itself.
## Proof
First we note that $\struct {G, \circ}$ is a subgroup of itself.
To show $\struct {G, \circ}$ is normal in $G$:
$\forall a, g \in G: a \circ g \circ a^{-1} \in G$
as $G$ is closed by definition.
Hence the result.
$\blacksquare$ | 2020-10-19T22:52:16 | {
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http://math.stackexchange.com/questions/26424/a-samurai-cuts-a-piece-of-bamboo | A samurai cuts a piece of bamboo
Suppose a samurai wants to try out his new sword and cuts a piece of bamboo twice, randomly, so now there are $3$ lenghts of bamboo. What is the probability of these 3 pieces being able to form a triangle?
I have never came across a continuous probability problem before, but I tried doing it anyway and got a result of 0.25 probability.
My solution: Let $L$ be the original lenght of the bamboo, $x$ be the place of the first cut and $y$ be the place of the second cut. Writing out all the 3 triangle inequalities, we come to the conclusion that no piece of bamboo can have more than $L/2$ lenght, then the probability we're looking for is: $$\frac{\int_{x=0}^{L/2}(\int_{y=L/2}^{x+L/2}(1)dy)dx}{\int_{x=0}^{L}(\int_{y=x}^L(1)dy)dx}=0.25$$
-
This question was asked on Math Overflow a while back. There are some good answers there: mathoverflow.net/questions/2014 – Mike Spivey Mar 11 '11 at 18:01
I like this question! – Asaf Karagila Mar 11 '11 at 18:09
Indeed, the accepted answer is very clever. Is my reasoning correct thought? I'm not sure, even thought I arrived at the same answer. – Leonardo Fontoura Mar 11 '11 at 18:11
Yes, I think your reasoning is correct. – joriki Mar 11 '11 at 18:27
@Asaf: I favorited this question solely because it features a samurai. – JCCyC Mar 13 '11 at 0:44
There is also a nice geometric-probability solution to the problem. For simplicity, let $L=1$, with $x$ and $y$ as you describe. The space of all possible values of $x$ and $y$ is the unit square $[0,1]\times[0,1]$, with each point being equally likely (as $x$ and $y$ are uniformly distributed). In order for the three pieces to form a triangle, each piece must have length less than $\frac{1}{2}$, so:
• if $x<y$: $x<\frac{1}{2}$, $y-x<\frac{1}{2}$, and $1-y<\frac{1}{2}$;
• or if $x>y$: $y<\frac{1}{2}$, $x-y<\frac{1}{2}$, and $1-x<\frac{1}{2}$.
Graphing these in the unit square gives the shaded region shown below.
This region is $\frac{1}{4}$ of the total area of the square, so the probability is $\frac{1}{4}$.
It's worth noting this similar but slightly different question, which arose from a mis-written Monte Carlo simulation of this problem.
-
Very intuitive answer, thank you! – Leonardo Fontoura Mar 12 '11 at 15:17
I've deleted my other comment; no need to have two links to the same question. Nice answer, by the way. – Mike Spivey Mar 12 '11 at 19:00
That works, assuming randomly means each point is uniformly distributed on $[0,L]$.
A similar approach is to note that the density of $x$ and $y$ is each $\frac{1}{L}$ so the probability of a triangle is $$\int_{x=0}^{L/2}\int_{y=L/2}^{x+L/2}\frac{1}{L^2} dy\,dx + \int_{x=L/2}^{1}\int_{y=x-L/2}^{L/2}\frac{1}{L^2}dy\,dx = \int_{x=0}^{L/2}\frac{x}{L^2}dx + \int_{x=L/2}^{1}\frac{L-x}{L^2}dx = \frac{1}{4}$$
- | 2016-02-13T09:04:02 | {
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https://math.stackexchange.com/questions/1811793/prove-a-b-cap-b-a-emptyset | # Prove $(A-B) \cap (B-A) = \emptyset$
My first instinct with this proof is to assume the opposite of the hypothesis, as in a proof by contradiction.
My work is as follows:
Suppose $(A-B) \cap (B-A) \neq \emptyset$.
Consider an $x \in (A-B) \cap (B-A)$.
If $x \in (A-B) \land x \in (B-A)$
$(x \in A \land x \notin B) \land (x \in B \land x \notin A)$
$(x \in A \land x\notin A) \land (x \in B \land x \notin B)$
These are contradictions. Hence such an element $x$ does not exist.
$\implies (A-B) \cap (B-A) = \emptyset$.
My issue is that it is apparent that assuming that the set is not empty is apparently not good practice? Is there a better way to approach this proof?
• What a problem do it directly? – BBVM Jun 4 '16 at 2:27
• I posted another approach to the problem and I personally don't think your solution is 'not a good practice'. – zxcvber Jun 4 '16 at 2:30
• Assuming a set is non empty is only a "bad habit" if you expect people to accept it un questioning. A proof by contradiction is the exact opposite. You assume it ONLY to tear it down. Nothing wrong with that! What is wrong is assuming it is non _ empty, come to a conclusion, and never verify that it was nonempty. – fleablood Jun 4 '16 at 2:39
• Well, I don't recommend this but $A \subset A-B$ and $B-A \subset A^$ so $A-B \cap B-A \subset A\capA^c=\emptyset$. But to answer your question there is nothing wrong with saying "let x \in A" if you are willing to accept that it might turn out that x doesn't exist. – fleablood Jun 4 '16 at 2:46
• "Assuming the set is not empty is not good practice" depends on what you're doing with the set. Here you're using a proof by contradiction to show that the set is empty. In this case, absolutely nothing wrong with assuming the set is not empty. In fact, it's required to do what you need to do. If it makes you feel better you can start by pointing out that there are only two possibilities: the set is empty or it isn't. If it's empty, we're done. If it's not, then contradiction. This covers all possible cases. – user307169 Jun 4 '16 at 3:04
The contradictory method works, although one of the other options available is to prove that
$$x \in (A-B) \implies x \notin (B-A).$$
This would implies that these two sets ($(A-B)$ and $(B-A)$) are disjoint and thus $$(A-B) \cap (B-A) = \emptyset.$$
If we see that $$x \in (A-B)$$ $$\implies x \in A \wedge x \notin B$$ $$\implies x \notin B$$ $$\implies x \notin B-A,$$
This claim concludes.
Just another solution;
Since $A-B=A\cap B^{C}$ (here, the $^{C}$ is for complementary set),
$$(A-B)\cap(B-A)=(A\cap B^{C})\cap(B\cap A^{C})$$
Since the operation $\cap$ is commutative, the equation above can be written as
$$A\cap B^{C}\cap B \cap A^{C}$$
The result is of course, a null set.
To do a proof by contradiction you assume to intersection isn't empty. Therefore we can pick x in the intersection.
And *that * we can show is imposible.
$x \in (A-B) \cap (B-A)$ so $x \in A-B$ and $x\in B-A$. So $x\in A$, $x\not \in B$, $x \in B$, and $x \not \in A$. Which is contradictory. | 2020-09-24T12:17:05 | {
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https://math.stackexchange.com/questions/3222581/prove-that-if-a-sequence-x-n-tends-to-infinity-then-frac1x-n-converges | # Prove that if a sequence $x_n$ tends to infinity, then $\frac{1}{x_n}$ converges to zero.
Question: $$(x_n)_{n=1}^\infty$$ is a sequence with $$x_n\neq0$$ for all $$n$$, also let $$x_n$$ tend to infinity. Let $$(y_n)_{n=1}^\infty$$ be defined by $$y_n=\frac{1}{x_n}$$, show that this converges to zero.
Definition for tending to infinity:$$\forall K \in \mathbb{R} \exists N\in \mathbb{N} \forall n \in \mathbb{N} ,n>N:x_n>K$$
Definition for convergence to zero:$$\forall \varepsilon > 0 \exists N\in \mathbb{N} \forall n \in \mathbb{N} ,n>N:∣x_n∣<\varepsilon$$
Idea: I know that $$x_n$$ tends to infinity since this is assumed. Then it must satisfy the defintion, therefore there must exist an $$N$$ which satisfies the defintion which i will call $$N_x$$ to prevent confusion.
To prove that $$y_n$$ converges to zero then I should choose $$N_y=\lceil\frac{1}{N_x}\rceil$$
Proof (attempt): Given $$\varepsilon>0$$ choose $$N_y=\lceil\frac{1}{N_x}\rceil$$. Given $$n \in \mathbb{N}$$, $$n>N$$ we have $$∣y_n∣=∣\frac{1}{x_n}∣\leq \frac{1}{N_x}\leq\varepsilon$$
Not entirely sure my selection for $$N_y$$ is correct, I understand why the statement is true however the proving put I am struggling.
• For any $\varepsilon >0$ there exists $M \in \Bbb{N}$ such that $1/M< \varepsilon$ solves this, as $1/n$ is obviously decreasing – B.Swan May 11 at 20:53
Take $$\varepsilon>0$$. Since $$\lim_{n\to\infty}x_n=\infty$$, there is some $$N\in\mathbb N$$ such that $$n\geqslant N\implies x_n>\frac1\varepsilon$$. But then $$n\geqslant N\implies 0<\frac1{x_n}<\varepsilon$$. In particular, $$\left\lvert\frac1{x_n}\right\rvert<\varepsilon$$.
• Hi thank you for your reply, would you mind clarifying how it implies that $x_n>\frac{1}{\varepsilon}$? – ViB May 11 at 21:17
• I am using the definition of $\lim_{n\to\infty}x_n=\infty$, taking $K=\frac1\varepsilon$. Is it clear now? – José Carlos Santos May 11 at 21:21
Your intuition is right: the definition of $$N_y$$ is a bit off.
When you choose $$N_x$$, this value is for a particular $$K$$, from the "For all $$K$$ there exists $$N\in\mathbb{N}$$" part of the first statement. As $$K$$ increases without bound, so will $$N_x$$. Here, I think you're choosing $$N_y$$ with the goal of showing that $$y_n$$ is sufficiently small for all $$n\geq N_y$$. But if you let $$K\rightarrow\infty$$, choose a corresponding $$N_x$$, and then let $$N_y=\lceil\tfrac{1}{N_x}\rceil$$, this will result in $$N_x$$ shooting off to $$\infty$$ but $$N_y$$ will end up being $$1$$ all the time. Your goal will then be proving that $$y_n$$ is sufficiently small for all $$n\geq N_y=1$$, which won't be true.
For this problem, you've been given a relationship between the values of $$\{x_n\}$$ and $$\{y_n\}$$, so you should use this relationship to somehow connect the formal statements of "$$x_n\rightarrow\infty$$" and "$$y_n\rightarrow 0$$". Let's restate your proof goal:
Choose any $$\epsilon>0$$ (this choice is out of your control, your proof must work for any choice). You want to produce an $$N\in\mathbb{N}$$ such that $$\{y_n\}$$ satisfies the definition.
First, pick some $$K>0$$ (you're allowed to do this since we know that $$x_n\rightarrow 0$$) such that if $$x_n>K$$ then $$y_n<\epsilon$$. This $$K$$ will be expressed algebraically in terms of $$\epsilon$$. (EDIT: I think you responded to another post very recently with this choice of $$K$$ - use that one.) Then, if you take some $$N_x$$ such that $$x_n>K$$ for all $$n\geq N_x$$, then you'll be able to say something about $$N_x$$ and $$\{y_n\}$$ similarly. | 2019-07-19T12:05:07 | {
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https://math.stackexchange.com/questions/1795276/does-this-fractal-have-a-name | # Does this Fractal Have a Name?
I was curious whether this fractal(?) is named/famous, or is it just another fractal?
I was playing with the idea of randomness with constraints and the fractal was generated as follows:
1. Draw a point at the center of a square.
2. Randomly choose any two corners of the square and calculate their center.
3. Calculate the center of the last drawn point and this center point of the corners.
4. Draw a new point at this location.
Not sure if this will help because I just made these rules up while in the shower, but sorry I do not have any more information or an equation.
Thank you.
• Where did it come from? Do you have any other information? Maybe an equation? – jdods May 22 '16 at 12:39
• How does one generate this fractal? Also I suppose the fractal isn't what the picture itself represents, but rather some "limit object" we approach when we keep on generating. – Wojowu May 22 '16 at 12:42
• where did you get the picture from? How was it generated? Without this information its impossible to say, although to me it looks like it was generated by some version of the "chaos game" – user140776 May 22 '16 at 12:42
• At first I thought that this is similar to the cayley graph of the free group in two generators, but there's a difference. Considering how easy it is to describe this fractal, I agree there should be a name, but I am not familiar with fractals. – blue May 22 '16 at 12:44
• I just made these rules up while in the shower That gets my upvote :) – 6005 May 23 '16 at 21:15
Your image can be generated using a weighted iterated function system or IFS. Specifically, let \begin{align} f_0(x,y) &= (x/2,y/2), \\ f_1(x,y) &= (x/2+1,y/2), \\ f_2(x,y) &= (x/2,y/2+1), \\ f_3(x,y) &= (x/2-1,y/2), \text{ and } \\ f_4(x,y) &= (x/2,y/2-1). \end{align} Let $(x_0,y_0)$ be the origin and define $(x_n,y_n)$ by a random, recursive procedure: $$(x_n,y_n) = f_i(x_{n-1},y_{n-1}),$$ where $i$ is chosen randomly from $(0,1,2,3,4)$ with probabilities $p_0=1/3$ and $p_i=1/6$ for $i=1,2,3,4$.
If we iterate the procedure $100,000$ times, we generate the following image:
This image is a solid square but the points are not uniformly distributed throughout that square. Technically, this illustrates a self-similar measure on the square.
To be a bit more clear, an invariant set of an IFS is a compact set $E\subset\mathbb R^2$ such that $$E = \bigcup_{i=0}^4 f_i(E).$$ It's pretty easy to see that the square with vertices at the points $(-2,0)$, $(0,-2)$, $(2,0)$, and $(0,2)$ is an invariant set for this IFS. It can be shown that an IFS of contractions always has a unique invariant set; thus, this square is the only invariant set for this IFS.
Let's call this square $E$, in honor of its status as an invariant set. We can get a deterministic understanding of the distribution of points on $E$ by thinking in terms of a mass distribution on the square (technically, a measure). Start with a uniform mass distribution throughout the square. Generate a second mass distribution on $E$ by distributing $1/3$ of the mass to $f_0(E)$ and $1/6$ of the mass to each of $f_i(E)$ for $i=1,2,3,4$. We can then iterate this procedure. The step from the original distribution to the next to the next might look like so:
The evolution of the first 8 steps looks like
• A "solid square"? Surely this process can only produce points with dyadic rational coordinates. – David Zhang May 22 '16 at 15:09
• @DavidZhang I agree, but there is a limiting object. – Mark McClure May 22 '16 at 15:10
• If you remove the first generator, you get the same square, just with a uniform distribution – John Dvorak May 22 '16 at 17:56
• @DavidZhang Surely, my 'limiting object' comment makes perfect sense? The square in question is just the closure of that set of dyadic points. Similarly, the closed unit interval can be described at the attractor of the IFS with functions $f_1(x)=x/2$ and $f_2(x)=x/2+1/2$. But, if we play the chaos game starting at the origin, we can generate only dyadic rationals. Similarly, the classic chaos game that generates the Sierpinski triangle approximates that uncountable set with only countably many points. This is pretty fundamental. – Mark McClure May 22 '16 at 20:04
• @MarkMcClure Oh yes, your first comment made perfect sense. I just hadn't gotten around to replying until now, thanks. – David Zhang May 23 '16 at 0:56
As others have noted, what you're describing is an iterated function system — specifically, a system of affine contraction maps — which is a common way of constructing self-similar fractals. In particular, it can be written as a system consisting of the following five maps:
\begin{aligned} (x,y) &\mapsto \tfrac12 (x,y) \\ (x,y) &\mapsto \tfrac12 (x,y) + \tfrac12(0,1)\\ (x,y) &\mapsto \tfrac12 (x,y) + \tfrac12(0,-1) \\ (x,y) &\mapsto \tfrac12 (x,y) + \tfrac12(1,0) \\ (x,y) &\mapsto \tfrac12 (x,y) + \tfrac12(-1,0) \\ \end{aligned}
where I've taken the center of your square to be at the origin, and its corners to be at $(\pm 1, \pm 1)$.
Written like this, you can indeed see that each of these maps represents taking the average of the current point $(x,y)$ and some fixed target point. The first map (where the target point is simply the origin) arises whenever the two corners you choose in step 2 are opposite, and is thus twice as likely to be chosen in your randomized iteration as the other four maps, each of which results from picking two corners that are adjacent to each other.
The first, obvious question then is what the fixed set of your iterated function system (i.e. the unique non-empty compact set $S \subset \mathbb R^2$ such that applying each of your affine maps to $S$ and taking the union of the results yields the same set $S$) is. This fixed set is the closure of the limit set obtained by infinitely iterating your function system from any starting point, and thus, in some sense, represents "the" limit shape obtained by iterating your system.
Alas, for your system, the answer is kind of boring: the fixed set is simply a (tilted) square with corners at the top, right, left and bottom of the outer square (i.e. at $(0,\pm1)$ and $(\pm1,0)$ in the parametrization I used above).
The easy way to see that is to observe that the last four maps in your system map this square into four smaller squares that precisely tile the original square (and the first map just produces a square that redundantly overlaps the other four). Another, more rigorous way is to show that any point within this square (and only within this square!) can be approached arbitrarily closely from any starting point by iterating the maps given above in a suitable order.
For this, it helps to rotate, scale and translate the coordinates so that the fixed square (with corners at $(0,\pm1)$ and $(\pm1,0)$) is mapped to the unit square (with corners at $x,y \in \{0,1\}$). With this coordinate transformation, the maps given above become:
\begin{aligned} (x,y) &\mapsto \tfrac12 (x,y) + \tfrac12(\tfrac12, \tfrac12)\\ (x,y) &\mapsto \tfrac12 (x,y) + \tfrac12(0,0)\\ (x,y) &\mapsto \tfrac12 (x,y) + \tfrac12(0,1) \\ (x,y) &\mapsto \tfrac12 (x,y) + \tfrac12(1,0) \\ (x,y) &\mapsto \tfrac12 (x,y) + \tfrac12(1,1) \\ \end{aligned}
We can then interpret the last four maps as shifting the binary representation of the coordinates $x$ and $y$ right by one place, and setting the first binary digits after the radix point to one of the four possible combinations. Thus, starting from the origin (which we can approach arbitrarily closely from any starting point just by iterating the second map above), we can reach any point with dyadic rational coordinates — i.e. coordinates whose binary representation terminates — inside the unit square with a finite number of steps. Since the dyadic rationals are dense in the unit square, any point within the square can be approached arbitrarily closely in this way.
(For completeness, we should also show that no point outside the unit square can be a limit of iterating these maps. A simple way to do that is to show that if the point $(x,y)$ lies outside the unit square — i.e. either of $x$ and $y$ is negative or greater than $1$ — then applying any of these maps will move it closer to the unit square.)
So, if the limit of iterating your system is simply a square, why are you seeing all that interesting fractal-looking structure, then?
The reason for that is the redundant first map (the one that just pulls each point closer to the center of the square), which causes some points in the square to be more likely that others to occur during the iteration. Thus, the invariant measure of your iterated function system will not be the uniform measure over the square (like you'd get if you removed the first map from the system). Rather, it ends up looking like this (rotated 45°, like in my transformed system above):
$\hspace{64px}$
The picture above is a discretized approximation of the invariant measure, with the darkness of each pixel being proportional to the probability of the randomly iterated point landing in that pixel. I obtained this picture simply by starting with the uniform measure on the unit square, and repeatedly adding together scaled and translated copies of it. Specifically, I used the following Python code to do this:
import numpy as np
import scipy.ndimage
import scipy.misc
d = 9 # log2(image size)
s = 2**d / 4 # number of pixels to shift first map by
a = np.ones((2**d, 2**d))
# iteration should converge in about d steps; run for 2*d to make sure
for i in range(2*d):
b = scipy.ndimage.zoom(a, 0.5, order=1) * (4/6.0)
a = np.tile(b, (2, 2)) # last four maps just tile the square
a[s:3*s, s:3*s] += 2*b # add first map (twice, since it has higher weight)
scipy.misc.imsave('measure.png', -a) # negate to invert colors
Of course, keep in mind that this is just an approximation of the actual invariant measure, which would seem to be singular.
(If anyone can characterize this measure more explicitly, I'd very much like to see it; the numerical approximation is suggestive, but doesn't really tell much about what really happens in the limit.)
BTW, there do exist actual fractals that resemble your iterated function system. For example, the following system:
\begin{aligned} (x,y) &\mapsto \tfrac13 (x,y) \\ (x,y) &\mapsto \tfrac13 (x,y) + \tfrac23(0,1)\\ (x,y) &\mapsto \tfrac13 (x,y) + \tfrac23(0,-1) \\ (x,y) &\mapsto \tfrac13 (x,y) + \tfrac23(1,0) \\ (x,y) &\mapsto \tfrac13 (x,y) + \tfrac23(-1,0) \\ \end{aligned}
which differs from your original system just by weighing the fixed corner points by twice as much in the average, generates the Vicsek fractal as its fixed set (again, shown rotated by 45° below):
$\hspace{200px}$
The Sierpiński space-filling curve also bears some resemblance to your system: it also has the entire square as its limit set, but the intermediate stages of the construction show a similar fractal-like structure.
• Wow! Thanks a lot for the explanation. – SilverSlash May 23 '16 at 13:34
• @IlmariKaronen The characterization of the measure is complicated by the fact that the IFS has overlaps (technically, the open set condition is not satisfied). This situation has been studied a lot over the last decade or so, though. Sze-Man Ngai has links to quite a few papers on his webpage. The singularity of the measure follows from theorem 1.1 of this paper. – Mark McClure May 23 '16 at 17:46
• By the way: The Vicsek fractal is the universal covering of a figure eight. – Christian Blatter May 31 '16 at 8:13
This is an "iterated function system" or IFS, with $5$ linear functions and their probabilities corresponding to the midpoints of the $6$ equally probable choices of pairs of corners of the square (the two pairs of diagonally opposite corners lead to the same function). The picture is of the attractor or invariant measure of that IFS. So the specific fractal picture probably does not have a name, but there is a name for the process by which it was generated.
The graph looks intriguingly similar to the picture of the free group on $2$ generators.
https://en.wikipedia.org/wiki/Cayley_graph#/media/File:Cayley_graph_of_F2.svg
• Yes, the image can be generated by an IFS. In fact, it can be generated by the specific IFS with specific probabilities presented in the accepted answer. So, I'm not sure what this answer contributes? The "fractal picture" (i.e., the attractor of the IFS) does have a name - it's called a "square". The invariant measure of the IFS with probabilities is concentrated on that looks something like a standard visualization of the free group but I see no logarithmic scale. – Mark McClure May 22 '16 at 17:48
• "I'm not sure what this answer contributes". Since you seem to insist on retaining that comment, here is the reply. (1) this answer is the one that first contributed the terms Iterated Function System and IFS to the thread. You edited that term into the accepted answer after criticizing its appearance in this one. Whatever redundancy exists between the two answers was created by your edit. (2) This answer is (more) explicit about the relation between IFS weights and the probabilities of pairs of corners, which is still not covered in accepted answer. (3) Relation to the free group. – zyx May 23 '16 at 18:24
To better describe your distribution, I will rotate and scale it so that the original square has corners $(0, \pm 2)$, $(\pm 2, 0)$. Let $x_n$ be the point of the $n$-th iteration ($x_0 = (0, 0)$). Then we have $x_{n+1} = \frac{1}{2}(x_n + a_n)$ where $a_n$ is randomly sampled from the multiset $\{(0, 0), (0, 0), (+1, +1), (+1, -1), (-1, +1), (-1, -1)\}$.
It is then a simple exercise of induction to show that $x_n$ is either $(0, 0)$ or of the form $\frac{1}{2^m}(a, b)$ where $m \leq n$, and $a$, $b$ are odd integers with absolute value below $2^m$. In other words, the coordinates are dyadic, with common minimal denominator, and inside the unit square.
A more interesting exercise is to show that this is exactly the support of our distribution.
Indeed, if we restrict ourselves to $a_n \neq (0, 0)$, each coordinate with common minimal denominator $2^m$ can be reached in exactly one way after $m$ iterations (which would give a uniform distribution on the support, converging to a uniform distribution on the unit square). Hint: Work backwards from $x_m$ to $x_0 = (0, 0)$.
Coming back to our original situation, after solving for $x_n$: $$x_n = \sum_{k=0}^{n-1} 2^{k-n} a_k,$$ we can easily compute, approximate and sketch its distribution.
Indeed, denoting by $\alpha$ the distribution of $a_k$ and by $\xi_n$ the distribution of $x_n$, the above equation means $$\xi_n = \sum_{k=1}^n 2^{-k}\alpha,$$ where sum of distributions represents the distribution of the sum of independent variables.
From that we arrive at the essential property $$\xi_{m+n} = \xi_m + 2^{-m}\xi_n.$$
This property is useful in several ways. For instance, we can iteratively compute/sketch $\xi_8$ from $\xi_4$. Also, it shows that $\xi_m$ approximates $\xi_n$ for arbitrarily large $n$ to precision $2^{-m}$.
Finally, it shows that the limiting distribution satisfies $$\xi = \alpha + \frac{1}{2}\xi,$$ which characterises it as self-similar with base pattern $\alpha$.
Note: Your original plot is not exactly the result of independently sampling from $\xi_n$, but the path to $x_N$ for some large $N$. However, the result is almost the same.
Indeed, after the first few samples, every point's distribution approximates $\xi$, and, from the equation for $x_n$, we can see that only the latest terms are relevant, which means that samples will be approximately independent unless they are very close. | 2019-06-25T22:07:20 | {
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