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https://stacks.math.columbia.edu/tag/0EU2 | Lemma 38.34.13. Let $S$ be a scheme. Let $\mathit{Sch}_ h$ be a big h site containing $S$. Then $(\textit{Aff}/S)_ h$ is a site.
Proof. Reasoning as in the proof of Topologies, Lemma 34.4.9 it suffices to show that the collection of standard h coverings satisfies properties (1), (2) and (3) of Sites, Definition 7.6.2. This is clear since for example, given a standard h covering $\{ T_ i \to T\} _{i\in I}$ and for each $i$ a standard h covering $\{ T_{ij} \to T_ i\} _{j \in J_ i}$, then $\{ T_{ij} \to T\} _{i \in I, j\in J_ i}$ is a h covering (Lemma 38.34.8), $\bigcup _{i\in I} J_ i$ is finite and each $T_{ij}$ is affine. Thus $\{ T_{ij} \to T\} _{i \in I, j\in J_ i}$ is a standard h covering. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 2022-08-09T06:58:14 | {
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http://mathoverflow.net/questions/19255/bounds-on-a-partition-theorem-with-ambivalent-colors | # Bounds on a partition theorem with ambivalent colors
I've been running into the following type of partition problem.
Given positive integers h, r, k, and a real number ε ∈ (0,1), find n such that if every (unordered) r-tuple from an n element set X is assigned a set of at least εk 'valid' colors out of a total of k possible colors, then you can find HX of size h and a single color which is 'valid' for all r-tuples from H.
Lower bounds on the smallest such n can be obtained from lower bounds for Ramsey's Theorem. If k is sufficiently large, then partition the set of colors into [1/ε] pairwise disjoint sets of approximately equal size to emulate a proper [1/ε]-coloring of r-tuples. A simple pigeonhole argument shows that this is essentially sharp when r = 1 and k is large enough, i.e. one color must be 'valid' for at least nε points.
Is the Ramsey bound more or less sharp for r > 1 or are there better lower bounds? The interesting case is when k is large since the proposed Ramsey lower bound is (surprisingly?) independent of k.
-
I do not think that the lower bound could depend only on epsilon. Below is the sketch of my argument.
Fix h=3, r=2, eps=1/4, thus we color the edges of a graph, each with 25% of all the colors and we are looking for a "monochromatic" triangle. Let us take k random bipartitions of the vertices and color the corresponding edges of the bipartite graph with one color. Using Hoeffding or some similar inequality we get that for big enough k every edge is colored at least k/4 times if n is at most exp(ck), where c is some fixed constant with some positive probability. Therefore the bound must depend on k and not only on epsilon.
-
Very nice! After coffee, I got the explicit lower bound $n > \sqrt{2}\exp(3k/40)$ when $h=3$ and $\epsilon=1/4$, by using Bernstein's Inequality instead. – François G. Dorais Mar 25 '10 at 14:15
After more coffee, I successfully generalized your trick to arbitrary h, r; I will post the details separately. Thank you very much for your answer! – François G. Dorais Mar 25 '10 at 14:49
Thx, you're welcome! – domotorp Mar 25 '10 at 16:25
Here is a generalization of domotorp's answer to arbitrary h > r > 1.
Independently for each color i ∈ {1,2,...,k}, pick a random Hi from a family H of r-hypergraphs that don't contain any complete r-hypergraph of size h. Declare color i to be 'valid' for the r-tuple t = {t1,...,tr} iff tHi. Let Yt be the number of 'valid' colors for t. Note that Yt is binomial with parameters (k, p) for some 0 < p ≤ 1/2 which is independent of k and also independent of t when H is closed under isomorphism. Hoeffding's Inequality then gives
Prob[Yt ≤ εk] ≤ exp(-2k(p-ε)2)
for 0 < ε < p. So the probability that Yt ≥ εk for all i is positive whenever n ≤ exp(2k(p-ε)2/r) (not optimal).
This is not enough since p implicitly depends on n. However, for fixed h > r > 1, p can be bounded away from 0. This can be seen by using for H the family of r-partite hypergraphs as domotorp did, but different choices of H give better bounds.
-
This answer is community wiki because domtorp deserves all the credit. – François G. Dorais Mar 25 '10 at 16:07 | 2014-04-17T12:53:14 | {
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http://mathoverflow.net/questions/231854/more-general-than-semidefinite-program | More general than semidefinite program?
I was TAing my convex optimization class and explaining that Linear Programs are a special case of Second Order Cone Programs, which are themselves special cases of Semidefinite Programs. My question is, is there any well-established class of optimization problems that is more general than semidefinite programs? Conic optimization problems would be an example of this, but I'm hoping for something a little more algebraic, if it exists.
-
There is the concept of hyperbolic programming, introduced in the 90's by Osman Güler:
• O. Güler, Hyperbolic Polynomials and Interior Point Methods for Convex Programming, Math. Oper. Res. 22 (1997) 350-377.
• H.H. Bauschke, O. Güler, A.S. Lewis, H.S. Sendov, Hyperbolic Polynomials and Convex Analysis, Canad. J. Math. 53 (2001) 470-488;
• J. Renegar, Hyperbolic Programs, and their Derivative Relaxations, Found. Comput. Math. 6 (2005) 59–79.
It is based on the concept of hyperbolicity cone introduced by Lars Gårding in the 50's in the context of the theory of hyperbolic partial differential equations.
We say that an homogeneous polynomial $P:\mathbb{R}^n\rightarrow\mathbb{R}$ is hyperbolic with respect to $\tau\in\mathbb{R}^n$ if $P(\tau)\neq 0$ and the roots of the one-variable polynomial $P_{\xi,\tau}(\lambda)\doteq P(\xi-\lambda\tau)$ are all real for all $\xi\in\mathbb{R}^n$. We then define the hyperbolicity cone $C(P,\tau)$ of $P$ with respect to $\tau$ as $$C(P,\tau)=\{\xi\in\mathbb{R}^n\ |\ \text{the roots of }P_{\xi,\tau}\text{ are all positive}\}\ .$$ It was shown by Gårding that $C(P,\tau)$ is an open, convex cone which is the connected component of $\{\xi\ |\ P(\xi)\neq 0\}$ to which $\tau$ belongs. Moreover, the closure of $C(P,\tau)$ has the form $$\overline{C(P,\tau)}=\{\xi\in\mathbb{R}^n\ |\ \text{the roots of }P_{\xi,\tau}\text{ are all nonnegative}\}\ .$$ Hyperbolic programming is then simply conic programming when the feasibility cone is (the closure of) an hyperbolicity cone of some hyperbolic polynomial $P$. Checking if $\xi\in\mathbb{R}^n$ belongs to $C(P,\tau)$ (resp. $\overline{C(P,\tau)}$) amounts to checking if the coefficients of the monic one-variable polynomial $P(\tau)^{-1}P_{\xi,\tau}=P_{\tau,\tau}(0)^{-1}P_{\xi,\tau}$ are all positive (resp. nonnegative).
Hyperbolic programming generalizes semidefinite programming in the following sense: if $\gamma_1,\ldots,\gamma_n$ are $N\times N$ symmetric matrices and we set $\gamma(\xi)=\sum^n_{j=1}\xi_j\gamma_j$, $\xi\in\mathbb{R}^n$, then $P(\xi)=\det\gamma(\xi)$ is an hyperbolic polynomial with respect to any $\tau\in\mathbb{R}^n$ such that $\gamma(\tau)$ is positive definite. In that case, $$C(P,\tau)=\{\xi\in\mathbb{R}^n\ |\ \gamma(\xi)\text{ is positive definite}\}$$ and $$\overline{C(P,\tau)}=\{\xi\in\mathbb{R}^n\ |\ \gamma(\xi)\text{ is positive semidefinite}\}\ .$$ It is not known whether this is a strict generalization of semidefinite programming, though - it is in fact conjectured that any hyperbolicity cone can be represented as a cone of positive definite matrices for a suitable choice of $N$. Since this was originally conjectured (in a stronger form) by Peter Lax in the 50's for $n=3$, this (still open) conjecture became known as the generalized Lax conjecture. Lax's original claim was proven by A.S. Lewis, P.A. Parrilo and M.V. Ramana (The Lax Conjecture is True, Proc. Amer. Math. Soc. 133 (2005) 2495-2499) building on work by J.W. Helton and V. Vinnikov (Linear Matrix Inequality Representation of Sets, Commun. Pure Appl. Math. 60 (2007) 654-674).
-
If you like, you might look at cones of sums of squares of polynomials (cones of PSD matrices are the same thing as cones of sums of squares of linear polynomials). This is the starting point of the modern technique of solving optimisation problems on semi-algebraic sets, due to J.Lasserre and others.
More generally, you might look at cones of nonnegative polynomials, and this opens up the whole Hilbert 17th problem business.
-
But you can rewrite these sum of squares cones as projections of appropriate linear sections of larger positive semidefinite cones, so from a strict representability perspective you don't get anything new here. – Noah Stein Feb 22 at 22:55
this is only true asymptotically --- and then, you know, every nice function on a compact is asymptotically a polynomial, so why bother :-) – Dima Pasechnik Feb 23 at 10:20
OK, it's not quite true what I just wrote. I meant cones of nonnegative polynomials. – Dima Pasechnik Feb 23 at 10:21
Ok, well cones of nonnegative polynomials are more general than SDP cones, but unfortunately computationally intractable. – Noah Stein Feb 23 at 19:31
speaking about complexity, for SDPs in general is it not known, and so one has to tread carefully here. – Dima Pasechnik Feb 25 at 7:28 | 2016-05-28T12:09:05 | {
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http://farside.ph.utexas.edu/teaching/389/Quantum/node43.html | Next: Energy Levels of Hydrogen Up: Orbital Angular Momentum Previous: Eigenfunctions of Orbital Angular
# Motion in Central Field
Consider a particle of mass moving in a spherically symmetric potential. The Hamiltonian takes the form
(4.110)
Adopting the Schrödinger representation, we can write . Hence,
(4.111)
When written in spherical coordinates, the previous equation becomes [92]
(4.112)
Comparing this equation with Equation (4.85), we find that
(4.113)
Now, we know that the three components of angular momentum commute with . (See Section 4.1.) We also know, from Equations (4.80)-(4.82), that , , and take the form of partial derivative operators involving only angular coordinates, when written in terms of spherical coordinates using the Schrödinger representation. It follows from Equation (4.113) that all three components of the angular momentum commute with the Hamiltonian: that is,
(4.114)
It is also easily seen that (which can be expressed as a purely angular differential operator) commutes with the Hamiltonian:
(4.115)
According to Section 3.3, the previous two equations ensure that the angular momentum, , and its magnitude squared, , are both constants of the motion. This is as expected for a spherically symmetric potential.
Consider the energy eigenvalue problem
(4.116)
where is a number (with the dimensions of energy). Because and commute with each other and the Hamiltonian, it is always possible to represent the state of the system in terms of the simultaneous eigenstates of , , and . But, we already know that the most general form for the wavefunction of a simultaneous eigenstate of and is
(4.117)
(See the previous section.) Substituting Equation (4.117) into Equation (4.113), and making use of Equation (4.105), we obtain
(4.118)
This is a Sturm-Liouville equation for the function [92]. We know, from the general properties of this type of equation, that if is required to be well behaved at , and as , then solutions only exist for a discrete set of values of . These are the energy eigenvalues, and are conventionally labeled using a quantum number denoted . (Thus, corresponds to the lowest energy state, to the next lowest energy state, and so on.) In general, the energy eigenvalues depend on the quantum numbers and , but are independent of the quantum number .
Next: Energy Levels of Hydrogen Up: Orbital Angular Momentum Previous: Eigenfunctions of Orbital Angular
Richard Fitzpatrick 2016-01-22 | 2018-02-22T11:04:51 | {
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https://plainmath.net/9146/anthony-working-engineering-company-building-ferris-county-algebraic | Question
# Anthony is working for an engineering company that is building a Ferris wheel to be used at county fairs. He wants to create an algebraic model that d
Functions
Anthony is working for an engineering company that is building a Ferris wheel to be used at county fairs. He wants to create an algebraic model that describes the height of a rider on the wheel in terms of time. He knows that the diameter of the wheel will be 90 feet and that the axle will be built to stand 55 feet off the ground. He also knows they plan to set the wheel to make one rotation every 60 seconds. Write at least two equations that model the height of a rider in terms of t, seconds on the ride, assuming that when t = 0, the rider is at his or her lowest possible height. Explain why both equations are accurate.
Part 2:One of Anthony's co-workers says, "Sine and cosine are basically the same thing." Anthony is not so sure, and can see things either way. Provide one piece of evidence that would confirm the co-worker's point of view. Provide one piece of evidence that would refute it. Hint: It may be helpful to consider the domain and range of different functions, as well as the relationship of each of these functions to triangles in the unit circle
2021-01-16
A sinusoidal function is of the form $$\displaystyle{h}={a}{\sin{{\left[{b}{\left({t}−{c}\right)}\right]}}}+{d}{\quad\text{or}\quad}{h}={a}{\cos{{\left[{b}{\left({t}−{c}\right)}\right]}}}+{d}$$ where ∣a∣∣a∣ is the amplitude, $$2π/b$$ is the period, c is the phase shift, and dd is the vertical shift (its midline).
The diameter of the wheel is twice the amplitude. Since the diameter of the wheel is 90 ft, then the amplitude is $$∣a∣=90/2=45.$$
The axle point of the wheel is the vertical shift of the sinusoid. Since the axle of the wheel will be 55 ft above the ground, then d=55.
The wheel makes one revolution every 60 seconds so the period is 60. Therefore $$2π/b=60$$. Solving this for b gives $$b=2π/60=π/30$$
A since curve starts at its midline and a cosine curve starts at its maximum. Since we need the function to start at the ground, which is the minimum of the function, we need to use the cosine form with $$a<0$$ or use a sine curve that has a phase shift.
If we use a cosine curve, then $$a=−45$$ so it will start at the minimum and we don't need a phase shift. We then have everything we need to write the cosine function. Substituting $$a=−45$$, $$b=π/30$$, $$c=0$$, and $$d=55$$ into $$h=a \cos[b(t−c)]+d$$ then gives $$h=−45cos(π/30)t+55$$.
If we use a sine curve, we need to determine how much the phase shift needs to be so that it will start at a minimum. $$y=\sin x$$ has a minimum at $$(−π2,−1)$$. Since $$b=π/30$$, then the parent function has been horizontally compressed by a factor of $$30/π$$. The minimum after the compression is then $$(−(\pi/2)⋅30/π,−1)=(−15,−1)$$. The graph after the compression must then be horizontally translated right 15 units to get a minimum at (0,−1). Therefore $$c=15$$. Note that the minimum of our graph is not (0,−1), we just needed to determine how far horizontally the graph needed to be moved, which is not affected by the change in amplitude and vertical shift.
Since we don't need a reflection, then $$a=45$$ for the sine curve. We then have everything we need to write the sine function. Substituting $$a=45, b=π/30, c=15$$ and d=55 into $$h=a\sin[b(t−c)]+d$$ then gives $$h=45\sin[π30(t−15)]+55$$.
Part 2:
A piece of evidence that sine and cosine are basically the same thing is that they have the same domain of all real numbers and range of [−1,1].
A piece of evidence that they are not the same thing, is that they have different starting points (the point where x=0) and different intervals of increasing and decreasing. Sine has a starting point at the origin, increases to a maximum at $$x=π/2$$, decreases to a minimum at $$x=3π/2$$, increases to a maximum at $$x=5π/2$$, and continues this pattern of increasing/decreasing every $$\pi$$ units. Cosine, however, starts at a maximum at $$x=0$$, decreasing to a minimum and $$x=π$$, increases to a maximum at $$x=2π$$, and continues this pattern of increasing/decreasing every π units. | 2021-09-23T15:13:06 | {
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https://homework.cpm.org/category/CCI_CT/textbook/calc/chapter/4/lesson/4.5.1/problem/4-164 | ### Home > CALC > Chapter 4 > Lesson 4.5.1 > Problem4-164
4-164.
Rewrite the following using a single trigonometric function. You may wish to review your trigonometric identities in Chapter 1.
1. $10\operatorname{ sin}(3x)\operatorname{ cos}(3x)$
Double Angle identity (factor first).
2. $\operatorname{sin }x\operatorname{ cos }3x −\operatorname{sin }3x\operatorname{ cos }x$
Sum and Difference (Angle Sum) identity.
3. $\operatorname{cos}^4 x −\operatorname{ sin}^4 x$
Factor first.
4. $\operatorname{tan }x +\operatorname{cot }x$
Start by rewriting $\operatorname{tan}(x)$ and $\operatorname{cot }(x)$ as fractions.
You will use more than one identity.
$\frac{\text{sin}x}{\text{cos}x}+\frac{\text{cos}x}{\text{sin}x}=\frac{\text{sin}^{2}x+\text{cos}^{2}x}{(\text{cos}x)(\text{sin}x)}=$
$\frac{1}{\frac{1}{2}\text{sin}(2x)}=2\text{csc}(2x)$ | 2022-05-26T11:25:45 | {
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https://somuteja.wordpress.com/category/uncategorized/ | Archives for category: Uncategorized
The nth cyclotomic polynomial is defined as follows $\Phi_n(x)=\prod_{1\leq i\leq n, (n,i)=1}(x-\zeta_n^{i})$ where $zeta_n=e^{\frac{2\pi i }{n}}$.
The polynomial $x^n-1$ has the following factorization $x^n-1=\prod_{d|n}\Phi_d(x)$.
In this post, I will give a proof of irreducibility of nth cyclotomic polynomial. I will prove the following theorem.
Theorem: $\Phi_n(x)$ is an irreducible polynomial in $\mathbb{Q}[x]$ .
Proof: If $\Phi_n(x)$ is not an irreducible polynomial then form Gauss’s lemma it has a non-trivial factorization $\Phi_n(x)=f(x)g(x)$ where $f(x),g(x)\in \mathbb{Z}[x]$.
As $\Phi_n(x)$ has no repeated roots, $f(x),g(x)$ are relatively prime to each other in $\mathbb{Q}[x]$. As $\mathbb{Q}[x]$ is a Euclidean domain, there exist functions $l(x)$, $m(x)$ and a natural number $D$ such that $l(x)f(x)+m(x)g(x)=D.$——–(1).
Without loss of generality, we can assume $f(\zeta_n)=0$. Let one root of $g(x)=0$ be $\zeta_n^{r}$ where $(r,n)=1$. From Dirichlet’s theorem, there exists a prime number $p>D$ satisfying $p\equiv r\mod n$.
Note that, $l(x)f(x)+m(x)g(x)\equiv D\neq 0\mod p.$——-(2)
Hence, $f(\zeta_n)=g(\zeta_n^{p})=0$. Therefore, $f(x),g(x^p)$ have a common root $\zeta_n$ and therefore they both have a non trivial greatest common divisor in $\mathbb{Q}[x]$.
Consider these polynomials $f(x)$ and $g(x^p)$ as polynomials in $Z_p(x)$. As, $g(x^p)=(g(x))^p$ in $Z_p[x]$. We have, $f(x)$ and $(g(x))^p$ have non-trivial common divisor in $Z_p[x]$–which implies $f(x)$ and $g(x)$ have a non trivial common divisor, $d(x)$ , in $Z_p[x]$. This implies $d(x)|D$ in $\mathbb{Z}_p[x]$, which is absurd. | 2017-12-12T08:23:33 | {
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https://www.tutorialspoint.com/php-tan-function | # PHP tan() Function
## Definition and Usage
The tan() function returns the tangent ratio of given angle in radians. In trigonometry, tangent of an angle is defined as ratio of lengths of opposite side and adjacent side.
tangent of an angle is also defined as ratio of its sine and cosine
tan(x) = sin(x)/cos(x)
If x=45 degree, tan(x) = 1 as in a right angled traingle, opposite and adjacent sides are qual.
This function returns a float value.
tan ( float $arg ) : float ## Parameters Sr.NoParameter & Description 1arg A floating point value that represents angle in radians ## Return Values PHP tan() function returns tangent ratio of given parameter. ## PHP Version This function is available in PHP versions 4.x, PHP 5.x as well as PHP 7.x. ## Example Live Demo Following example calculates tan(pi/2) and returns 1.6331239353195E+16 (very large number). tngent ratio of 90 deg is infinity − <?php$arg=M_PI/2; //90 degree
$val=tan($arg);
echo "tan(" . $arg . ") = " .$val;
?>
## Output
This will produce following result −
tan(1.5707963267949) = 1.6331239353195E+16
## Example
Live Demo
Following example uses deg2rad() function to convert degrees to radians and then computes tan(60). Result is 1.7320508075689 which is sqrt(3) −
<?php
$arg=deg2rad(60);$val=tan($arg); echo "tan(" .$arg . ") = " . $val; ?> ## Output This will produce following result − tan(1.0471975511966) = 1.7320508075689 ## Example Live Demo Let's check find out tan(45) degrees. It returns 1 − <?php$arg=M_PI/4; //45 deg
$val=tan($arg);
echo "tan(" . $arg . ") = " .$val;
?>
## Output
This will produce following result −
tan(0.78539816339745) = 1
## Example
Live Demo
Following example computes tan(0) and returns 0
<?php
$arg=0;$val=tan($arg); echo "tan(" .$arg . ") = " . \$val;
?>
## Output
This will produce following result −
tan(0) = 0 | 2023-03-26T03:49:17 | {
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https://math.stackexchange.com/questions/492204/connections-on-a-manifold-and-principal-connections-on-the-frame-bundle | # Connections on a manifold and principal connections on the frame bundle
Suppose $M$ is a manifold, and $E$ a vector bundle over $M$ equipped with a connection $\nabla$. If $F$ is the frame bundle of $E$, is there an explicit construnction of a connection on $F$ associated with $\nabla$ such that in this way connections on $E$ and $F$ are $1$-$1$ correspondent?
Edit for the bounty:
I really need an answer to this question, and as it was already posted I think that putting a bounty on it is the most sensible way to go.
To rephrase the question in my own terms: Let $M$ be a smooth $n$-manifold. We can associate the following principal $GL(n)$-bundle to it: $$F = \{(m,\theta)|m\in M, \theta:\mathbb{R}^n\to T_mM\mathrm{\ lin.\ isom.}\}$$ with right action given by $(m,\theta)g = (m,\theta g)$. Its tangent space is defined (as for any other manifold) as a quotient of the space of paths on $F$. In order to get a more concrete representation, we need a way to differentiate "paths of frames," but as such paths can be seen as tuples of paths of vectors on $M$, it is enough to specify a connection $\nabla$ on $M$ to obtain the identification $$T_{(m,\theta)}F \cong \{(\hat{m},\hat{\theta})|\hat{m}\in T_mM,\hat{\theta}:\mathbb{R}^n\to T_mM\}$$ where we identify the equivalence class of paths $[\gamma(t),\theta(t)]$ with $(\dot{\gamma}(0),(\nabla_{\dot{\gamma}}\theta)(0))$. This gives us a map $$\{\mathrm{connections\ on\ }M\}\longrightarrow\{\mathrm{principal\ connections\ on\ F}\}$$ mapping $\nabla$ to $A([\gamma,\theta]) = \theta^{-1}\nabla_{\dot{\gamma}}\theta\in\mathfrak{gl}(n)$.
I believe there should be a way to invert this map (maybe only on a subset of the principal connections, though) but I cannot see how. Does anyone have an idea or a solution?
Remark 1: My question is in fact a special case of the original question on vector bundles, namely if we take $E=TM$.
Remark 2: I took a look at Taubes' book, as suggested in the answers, but I didn't find what I need (or maybe I found it, but wasn't smart enough to realize it).
• There will be more connections on $F$ in general... but I think that if you restrict yourself to principal connections on $F$ then they are in 1-1 correspondence with connections in $E$. Hopefully someone who knows this better than I do can give an answer. – Anthony Carapetis Sep 13 '13 at 2:51
• Did you look in Kobayashi-Nomizu? – Gunnar Þór Magnússon May 20 '15 at 12:31
• @GunnarÞórMagnússon I took a fast look, there seems to be some stuff relating the second fundamental form of an immersed submanifold with connections on the normal frame bundle, but I've not found a direct answer to the question above. – Daniel Robert-Nicoud May 20 '15 at 12:46
• I'm going from memory, but I think they talk about the two in the chapter on connections. I could be delirious though. – Gunnar Þór Magnússon May 20 '15 at 13:50
• @GunnarÞórMagnússon It's probably just me not looking with enough attention. I think I'm getting it on my own anyway (see my answer below). If I really cannot prove everything I will take a deeper look. – Daniel Robert-Nicoud May 20 '15 at 15:33
Identify $TM$ as the associated bundle $E = F\times_{GL(n)}\mathbb{R}^n$, where the action of $GL(n)$ on $\mathbb{R}^n$ is given simply by left multiplication. The identification between the two is given by $$[(m,\theta),v]\in E\longmapsto\theta v\in T_mM.$$ Then we have a bijective correspondence between vector fields on $M$ and sections of $E$ where to $X:M\to TM$ we associate $$\overline{X}(m) = [m,\theta,\theta^{-1}X(m)]$$ for any choice $\theta$ of frame at $m$. Now a connection $A\in\Omega^1(F;\mathfrak{gl}(n))$ gives us a way to differentiate $\overline{X}$, namely $d_A\overline{X}\in\Omega^1(M;E)$ ($1$-forms on $M$ with values in $E$).
Claim: The assignment $$d_A\overline{X}\longmapsto\nabla X$$ is the required bijection, where from $\nabla$ we can recover $A$ by noticing that $d_A = d + \rho(A)$.
Proof: We have the further identification of $T^*M$ with $E^*=F\times_{GL(n)}(\mathbb{R}^n)^*$, where the action of $GL(n)$ on $(\mathbb{R}^n)^*$ is given by $g\cdot v^* = v^*\circ g$. This gives a correspondence between $\Omega^1(M)$ with sections of $E^*$ by $$\overline{\beta}(m) = [m,\theta,\beta(m)\circ\theta]$$ in a way similar to the above. We have a natural isomorphism $$\Omega^1(M;E)\cong\Omega^0(M;E^*),$$ thus the only thing we are left to show is that $$d_A\overline{X} = \overline{\nabla X}.$$ This goes: \begin{align}\overline{\nabla X}(\hat{m}) = & \langle[m,\theta,\nabla_{\theta-}X],[m,\theta,\theta^{-1}\hat{m}]\rangle\\=&\langle[m,\theta,d_A\overline{X}(\theta-)],[m,\theta,\theta^{-1}\hat{m}]\rangle\\=&d_A\overline{X}(\hat{m})\end{align} for $\hat{m}\in T_mM$, where $\langle\cdot,\cdot\rangle$ is the natural pairing. | 2019-06-17T17:36:53 | {
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http://mathhelpforum.com/calculus/77767-solved-average-value-function-question-twist.html | Math Help - [SOLVED] Average Value Of a Function question with a twist
1. [SOLVED] Average Value Of a Function question with a twist
Find the numbers b such that the average value of f(x) on the interval [0, b] is equal to 11. (Enter solutions in exact forms from smallest to largest. If there are any unused answer boxes, enter NONE in the last boxes.)
Now I'm assuming I'd do 11=1/x[8x+5x^2-x^3] which simplifies to x^2-5x-3=0, but I don't know how to solve for x from there, especially to get two different solutions (which is what the answer key of the problem asks for). Could someone help me out?
2. Originally Posted by fattydq
Find the numbers b such that the average value of f(x) on the interval [0, b] is equal to 11. (Enter solutions in exact forms from smallest to largest. If there are any unused answer boxes, enter NONE in the last boxes.)
Now I'm assuming I'd do 11=1/x[8x+5x^2-x^3] which simplifies to x^2-5x-3=0, but I don't know how to solve for x from there, especially to get two different solutions (which is what the answer key of the problem asks for). Could someone help me out?
Solve $\frac{\int_0^b f(x) \, dx}{b} = 11 \Rightarrow -b^2 + 5b + 8 = 11 \Rightarrow b^2 - 5b + 3 = 0$ for $b$. I suggest using the quadratic formula. | 2015-02-01T09:14:01 | {
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https://community.wolfram.com/groups/-/m/t/1520100 | Solve the following equation with Integrate or DSolve?
Posted 2 months ago
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Hello folks, Think of an equation: -g+Kv=D[v]/D[t] and the conditions are when t=0, v=0; where t is time v is velocity, g is a gravitational acceleration and K is some constant. How do you input these on Mathematica so it could take the integral of the equation and gives result. I already got the result by hand as follows: v=(g(1-e^(k*t))/K). whatever I try I cant get this result. Thanks in advance.
DSolve[{-g + k* v[t] == D[v[t], t], v[0] == 0}, v[t], t] (* {{v[t] -> -(((-1 + E^(k t)) g)/k)}} *) or: DSolve[{-g + k* v[t] == v'[t], v[0] == 0}, v[t], t] (* {{v[t] -> -(((-1 + E^(k t)) g)/k)}} *) | 2018-12-16T13:31:46 | {
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https://www.physicsforums.com/threads/division-algorithm-for-r.229542/ | # Division algorithm for R
1. Apr 17, 2008
### Doom of Doom
How might I prove a Division Algorithm for the Real numbers?
That is to say, if $$x, \alpha \in \mathbb{R}$$, then $$x=k \alpha + \delta$$ for some $$k \in \mathbb{Z},$$ $$\delta \in \mathbb{R}$$ with $$0 \leq \delta < \alpha$$ where $$k, \delta$$ are unique.
2. Apr 17, 2008
### mathman
Assume that it is not unique, subtract one representation from the other. The resultant equation is obviously false.
3. Apr 18, 2008
### Doom of Doom
Yeah, but how can I show existence?
4. Apr 18, 2008
### mathman
The sequence na for n=1,2,... is unbounded. Therefore for some n, na>x. Find lowest bound, subtract 1 and you will have k. ka<=x, (k+1)a>x, so x-ka(remainder)<a
5. Apr 18, 2008
### LukeD
or even more straightforward, let k = floor(x/a)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 2018-03-19T01:26:51 | {
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https://socratic.org/questions/the-number-of-cards-in-bob-s-baseball-card-collection-is-3-more-than-twice-the-n | # The number of cards in Bob's baseball card collection is 3 more than twice the number of cards in Andy's. If together they have at least 156 cards, what is the least number of cards that Bob has?
Jan 31, 2017
$105$
#### Explanation:
Let say A is a number of card for Andy and B is for Bob.
The number of cards in Bob's baseball card, $B = 2 A + 3$
$A + B \ge 156$
$A + 2 A + 3 \ge 156$
$3 A \ge 156 - 3$
$A \ge \frac{153}{3}$
$A \ge 51$
therefore the least number of cards that Bob has when Andy has the smallest number of card.
$B = 2 \left(51\right) + 3$
$B = 105$ | 2021-12-08T05:02:53 | {
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https://math.stackexchange.com/questions/3074412/does-it-exist-an-approximation-using-sums-and-a-way-to-do-the-computation-with-m | # Does it exist an approximation using sums and a way to do the computation with more accuracy both by hand to find the length of a spiral?
I have found this riddle in my book and so far which may require the use of calculus (integrals) to which I'm familiar but not very savvy with it. Since, I've not yet come with an answer. I wonder if such problem can be also solved using sums in the scope of college precalculus like an approximation (proven) which could be solved by hand even the calculus method which I also feel might also help me.
The problem is as follows:
In a research facility in Taiwan a group of technicians built a new optical disk which stores information in a spiral engraved in its bottom face named "lecture side". Under the microscope it can be seen that the spiral begins in a region starting from $$\textrm{2.6 cm}$$ from its axis of rotation and it ends at $$\textrm{5.7 cm}$$ from the center of the disk. It can also be seen that individual turns of the spiral are $$0.74\,\mu\textrm{m}$$. Using this information calculate the length of the entire track.
So far I've only come with the idea of using the spiral of Archimedes, whose formula is given as follows:
$$r=a+b\phi$$
However, I'm not very familiar with the realm of polar coordinates or how can this equation be used to solve my problem. To better illustrate the situation, however I've drawn this sketch to show how I'm understanding the problem.
I've included a cartesian grid, which well "may not be" in scale. But gives an idea of how I believe it is intended to be said.
I've really wanted to offer more into this such as an attempt into solving, but so far I've ran out of ideas.
However, I've come with the idea that the solution may be linked with finding how many turns are in the "readable sector" which is alluded in the problem.
To calculate this what I did was the following:
$$\textrm{number of turns}= \left( 5.7 - 2.6 \right)\times 10^{-2}\textrm{m}\times \frac{\textrm{1 turn}}{0.74\times 10^{-6}\textrm{m}}$$
By evaluating this short conversion factors I obtained
$$\textrm{number of turns} = 41891.89189\,\textrm{turns or rad}$$
And that's it. But from there I don't know how to relate this information with what would be needed to solve this riddle. The answer in the book states that the track's length is $$1.09\times 10^{4}\,\textrm{m}$$. But again. I don't know what can I do to get there.
So far I've did some preliminary research in the community and I've found this but it isn't really very helpful as the example relates to a vertical helix going upwards around a cylinder and my situation doesn't fit into this. Other proposed methods such as the one seen here propose using calculus and this also is referenced here. Needless to say that the only existing solution doesn't give much details of how the supposed formula works. It also contributes to the problem that I'm not very savvy with it and since this problem was obtained from a precalculus book, I aimed my question to. Does it exist a method to make this computation relying in college algebra?. With this I don't intend to totally discard a calculus approach, but an answer with would really be very pedagogical for me is one which can show me the two methods, so I can compare which can be better according to my current knowledge so I can practice more in what I feel I'm lacking.
Well that's it. I do really hope somebody could help me with the two methods, one using algebra and another using calculus.
As an approximation for the first loop, as the length would be
between the circumference of a circle of radius 2.6 cm and
the circumference of a circle of radius 2.6 cm + 0.74 $$\mu$$m,
is the average of those two circumferences.
Doing likewise for the other loops will give
you an approximation for the total length.
Take into account the last loop is 1/4 of a circle.
Another approximation for s the length of the spiral.
Let c = 2$$\pi$$, r the inner radius, w the width
between each arc. n the number of turns. Then
sum(k=0,n-1) c(r + wk) = crn + cwn(n - 1)/2 < s <
sum(k=1,n) c(r + wk) = crn + cwn(n + 1)/2.
n is an integer. Add as needed a fraction of an additional turn,
• I was thinking into a more precise computation (hence I not totally discouraging calculus). The method of summing those averages may be a good idea if the number of loops is smaller (and taking for granted the omission of demonstrating the average is the length of the loop), but in this case with only $0.74\,\mu\textrm{m}$ thickness to cover a distance of $3.5\,\textrm{cm}$ would meant a huge pack of loops. Doing it individually will make it impractical. Does it exist a more reasonable method?. – Chris Steinbeck Bell Jan 16 '19 at 0:31
• @ChrisSteinbeckBell. See edit. – William Elliot Jan 16 '19 at 8:47
• The equations you have posted do not display properly. Mind using Mathjax instead?. I would like to know if there is any limitation in using the sum because in my case the number of turns is not an integer. See $N=\left(\frac{5.7-2.6}{0.74\times 10^{-6}}\right)\times 10^{-2}=\left(\frac{31\times 10^{5}}{74}\right)$. What you proposed it was something which I also thought or intended to use meant as a sum but in this case they are not concentrical disks but rather a spiral. Regarding the notation you use do you meant $s$ as being the length?. – Chris Steinbeck Bell Jan 16 '19 at 13:39
• I'd voice my opinion that you may try to evaluate the sums that you propose using the numbers in my problem to see if at least comes closer to the answer. Again I believe that an answer which would help me is one that considers how to find the length of an spiral of archimedes even if resorting to some calculus or maybe the precalculus sum that you are intending to use. To better illustrate what you are trying please enlighten me with some of the computation with numbers. :) – Chris Steinbeck Bell Jan 16 '19 at 13:42
This can be solved using the formula for area: $$A=L\times W$$, or equivalently: $$L=\frac AW$$
The entire track occupies an area of $$\pi(0.057^2-0.026^2)$$ square meters. The width of the track is given to be $$0.74\times10^{-6}$$ meters. The length is therefore: $$\frac{\pi(0.057^2-0.026^2)}{0.74\times10^{-6}}\approx1.09\times10^4\text{ meters}$$
• I've found a similar approach here but it was not documented on how it was obtained such formula. In your notation does $\textrm{W}$ stands for width right?. It would be very valuable to add that the difference between the squares comes from finding the area of the "flat doughnut". However as mentioned this is an approximation and although it checks, I wonder if there was a more "elaborate" way to find the lenght, hence accuracy. The integral of the function the spiral for example. – Chris Steinbeck Bell Jan 16 '19 at 14:52
• @Chris It is not possible to get a more accurate measure of the length without more accurate measure of the radii. The given distances of $2.6$ cm and $5.7$ cm could each be +/- $0.05$ cm. This gives a variance in total length of +/- $350$ meters. This variance also means your 'number of turns' may be off by more than $1000$ – Daniel Mathias Jan 16 '19 at 15:19
You can consider the "average" radius $$\bar r=(5.7 + 2.6)/2=4.15$$ cm and simply compute $$L=2\pi\bar r\cdot N=2\pi\cdot 4.15\cdot 41891.9=1.09234\times10^6\ \text{cm}= 1.09234\times10^4\ \text{m}.$$ If you insist on using calculus, you can of course integrate along the spiral: $$L=\int_0^{2N\pi}\sqrt{\left({dr\over d\phi}\right)^2+r^2}\,d\phi =\int_0^{2N\pi}\sqrt{b^2+(a+b\phi)^2}\,d\phi=1.09234\times10^4\ \text{m},$$ with: $$a=2.6$$ cm and $$b={0.74\times10^{-6}\over2\pi}$$ m. As you can see there is no difference, even keeping 6 significant digits. | 2021-05-16T22:23:13 | {
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https://math.stackexchange.com/questions/3198516/proof-verification-short-exact-sequence-rank-theorem | # Proof Verification Short Exact Sequence Rank Theorem
I am trying to prove the rank-nullity theorem for short exact sequences; if $$R$$ is an integral domain, and $$M',\,M,\,M''$$ are all $$R$$-modules with $$0\rightarrow M' \xrightarrow{\psi} M\xrightarrow{\phi} M''\rightarrow 0$$ being a short exact sequence then by the first isomorphism theorem: $$M/M'\simeq M'' \implies \mathrm{rank}(M/M')=\mathrm{rank}(M'')$$ Morover $$M'\simeq \ker \phi$$, and so it's clear from this that $$\mathrm{rank}(M)\geq\mathrm{rank}(M')+\mathrm{rank}(M'')$$.
I was having more difficulty with the other direction. Namely, if $$X\subseteq M$$ is maximally $$R$$-linearly dependent set, since we can write $$M=M'\sqcup M\backslash M'$$ we have that we can split $$X$$ into $$X_{M'}\sqcup X_{M\backslash M'}$$ where $$X_{M'}:=\{x\in X:x\in\ker\phi\}$$, and $$X_{M\backslash M'}:=\{x\in X:x\not\in\ker\phi\}$$. In particular since $$X$$ is $$R$$-linearly independent in $$M$$, it must be true that $$X_{M'}$$ be $$R$$-linearly independent in $$M'$$ -- so that $$\lvert X_{M'}\rvert \leq \mathrm{rank}(M')$$. Similarly, $$\lvert X_{M\backslash M'}\rvert\leq \mathrm{rank}(M'')$$. Thus we have: $$\mathrm{rank}(M) = \lvert X\rvert = \lvert X_{M'}\sqcup X_{M\backslash M'}\rvert = \lvert X_{m'}\rvert + \lvert X_{M\backslash M'}\rvert \leq \mathrm{rank}(M')+\mathrm{rank}(M'')$$ The only problem I see is that we could have $$\phi(X_{M\backslash M'})\subseteq M''$$ being $$R$$-linearly dependent since all we get from exactness is that $$\phi$$ is surjective -- but at the same time $$M/M'\simeq M''$$ by the first isomorphism theorem, so we should be okay? I looked at Short Exact Sequences & Rank Nullity, but those solutions seemed a little clunky or used tools I am not familiar enough with, and so I was wondering if there is a cleaner way to do it. Is there a way to justify $$\lvert X_{M\backslash M'}\rvert\leq \mathrm{rank}(M'')$$ using what I have or will my proof method not work?
• I'd say that the easy way to do this would be to tensor with $K$, the fraction field of $R$, which is a flat $R$-module. – Lord Shark the Unknown Apr 23 at 17:01 | 2019-10-21T18:14:33 | {
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https://socratic.org/questions/578f94a611ef6b3d392bf080 | # Question bf080
May 19, 2017
Nickel content solution: you may not have seen this approach before.
The added pure nickel is $9 \frac{1}{3}$ pounds
#### Explanation:
Let the amount of added nickel as a percentage be $x$
Let the final blend mass be $b$
The gradient of part is the same as the gradient of the whole.
$\implies \left(\text{change in nickel content")/("change in added pure nickel}\right) \to \frac{100 - 15}{100} \equiv \frac{25 - 15}{x}$
Turn it all upside down:
$\frac{100}{100 - 15} \equiv \frac{x}{25 - 15}$
x=(100xx10)/85=11.76470....%
Thus 100%-x% represents the ${70}^{\text{lb}}$ of original the 15% content blend.
=>100%-x%" of "b=70^("lb")
$\left(\frac{100}{100} - \frac{11.76470 . .}{100}\right) b = 70$
$\implies \frac{88.23529 . .}{100} b = 70$
$b = 79.33333 \ldots \to 79 \frac{1}{3} {\textcolor{w h i t e}{}}^{\text{lb}}$
Thus the added pure nickel is $9 \frac{1}{3}$ pounds
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check:
(25%xx79 1/3)/(15%xx70+9 1/3) =1
The nickel in each is the same so the answer is correct
May 19, 2017
Grain question
$\text{ corn } \to 50$ bushels
$\text{ barley } \to 100$ bushels
#### Explanation:
Gradient for part is the same as the gradient for the whole.
$\frac{3.80 - 2.6}{100} = \frac{3 - 2.6}{x}$
Turn up the other way
$\frac{100}{1.2} = \frac{x}{0.4}$
$\implies x = 33 \frac{1}{3}$
Thus the blend has 33 1/3% corn
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Total batch weight is to be 150 bushels
$\implies \text{ corn } \to \frac{33 \frac{1}{3}}{100} \times 150 = 50$ bushels
$\implies \text{ barley } \to 100$ bushels
May 20, 2017
Alternative method for the corn/silage question
The amount of corn is 50 bushels
Thus the amount of silage is 100 bushels
#### Explanation:
Let the amount of corn be x%
Target cost $3.00 Silage cost$2.60
Corn cost $3.80 As we have x% corn then there is the related silage amount of 100%-x% ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ [x%xx$3.80]+[(100%-x%)xx$2.60]=$3.00#
$\left[\frac{x \times 3.80}{100}\right] + \left[\frac{100 \times 2.60}{100} - \frac{x \times 2.60}{100}\right] = 3$
$\frac{1}{100} \left(3.80 x + 260 - 2.6 x\right) = 3$
$3.80 x + 260 - 2.6 x = 300$
$1.2 x + 260 = 300$
$x = \frac{300 - 260}{1.2} = 33 \frac{1}{3}$
So the amount of corn is $\frac{33 \frac{1}{3}}{100} \times 150 = 50$ bushels
Thus the amount of silage is $150 - 50 = 100$ bushels
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ | 2021-11-30T11:17:19 | {
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https://questioncove.com/updates/4dfa7a390b8b370c28be5fff | Mathematics
OpenStudy (anonymous):
For every real number x (regardless of whether x is positive, negative, or zero), the principal cube root of x^3 equals x true or false
OpenStudy (anonymous):
$\sqrt[3]{x^{3}}=x^{3/3}=x^{1}=x$
OpenStudy (anonymous):
(true) | 2021-06-18T21:49:46 | {
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http://mathhelpforum.com/pre-calculus/178329-use-sigma-notation-write-sum.html | # Math Help - Use sigma notation to write the sum
1. ## Use sigma notation to write the sum
(x^4/8) - (x^5/9) + (x^6/10) - (x^7/11) + (x^8/12)
2. Usualy when you want to write a sum that way you look for a rule...
$\sum_{i=4}^{8}(-1)^i\cdot x^\frac{i}{i+4}$ or $\sum_{i=1}^{5}(-1)^{i+1}\cdot x^\frac{i+3}{i+7}$.
3. $\sum_{i=4}^{8}(-1)^i\cdot \frac{x^i}{i+4}$ or $\sum_{i=1}^{5}(-1)^{i+1}\cdot \frac{x^{i+3}}{i+7}$. | 2014-07-23T13:34:21 | {
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https://study.com/academy/answer/a-find-the-equation-of-the-plane-tangent-to-the-graph-of-f-x-y-x-2-e-xy.html | # (a) Find the equation of the plane tangent to the graph of f(x,y) = x^{2} e^{xy}
## Question:
(a) Find the equation of the plane tangent to the graph of {eq}f(x,y) = x^{2} e^{xy} {/eq} at {eq}(1,0) . {/eq}
(b) Find the linear approximation of {eq}f(x,y) {/eq} for {eq}(x,y) {/eq} near {eq}(1,0) . {/eq}
(c) Find the differential of f at the point (1,0) .
## Tangent Plane:
The equation of the tangent plane to the surface {eq}f(x,y) {/eq} at point {eq}(x_0,y_0) {/eq}
is found arresting the Taylor series of the function at the first order terms, i.e.
{eq}\displaystyle L(x,y) = f(x_0,y_0) + f_x(x_0,y_0) (x-x_0) + f_y(x_0,y_0) (y-y_0) {/eq}
where {eq}f_x, \; f_y {/eq} are the first partial derivatives of the function.
(a) The equation of the plane tangent to the graph of
{eq}\displaystyle f(x,y) = x^{2} e^{xy} {/eq}
at point (1,0) is given by
{eq}\displaystyle f(1,0) = 1^{2} e^{1(0)} = 1 \\ \displaystyle f_x(x,y) = 2xe^{xy}+e^{xy}yx^2 \rightarrow f_x(1,0) = 2 \\ \displaystyle f_y(x,y) = e^{xy}x^3 \rightarrow f_y(1,0) = 1 \\ \displaystyle L(x,y) = f(1,0) + f_x(1,0) (x-1) + f_y(1,0) y = 1 + 2(x-1) + y = 2x+y-1 {/eq}
(b) The linear approximation of {eq}f(x,y) {/eq} for {eq}(x,y) {/eq} near {eq}(1,0) {/eq}
is equal to the equation of the tangent plane found at point (a).
(c) The differential of f at the point (1,0) is equal to
{eq}\displaystyle df =f_x(1,0) dx + f_y(1,0) dy = 2 dx + dy {/eq}
Linearization of Functions
from
Chapter 10 / Lesson 1
3.8K
Over the river and through the woods to Grandmother's house we go ... Are we there yet? In this lesson, apply linearization to estimate when we will finally get to Grandma's house! | 2021-05-16T17:46:45 | {
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https://dsp.stackexchange.com/questions/67957/prove-the-following-dft-for-a-periodic-sinc-sequence-or-ideal-lowpass-filter | # Prove the following DFT for a periodic sinc sequence or ideal lowpass filter
I need to prove the following DFT for a periodic 𝑠𝑖𝑛𝑐 sequence or ideal lowpass filter:
$$\sqrt{\frac{p}{m}}\frac{sinc(\frac{\pi np}{m})}{sinc(\frac{\pi p}{m})} \stackrel{DFT}{\longleftrightarrow} \begin{cases} \sqrt{\frac{m}{p}},\: if \: |k-\frac{m}{2}|\ge \frac{p-1}{2}\\ 0, \: otherwise \end{cases}$$
But actually, I do not know how to do it. Maybe someone gives me at least a hint?
Hints:
1) Definition of the unnormalized DFT:
$$X[k] = \sum_{n=0}^{N-1} x[n] e^{-i\frac{2\pi}{N}kn}$$
2) Geometric Summation Formula:
$$\sum_{n=0}^{N-1} r^n = \frac{1-r^N}{1-r}$$
3) Exponential form of Sine:
$$\sin( \theta ) = \frac{ e^{i\theta} - e^{-i\theta} }{ 2i }$$
4) Definition of unnormalized sinc:
$$\operatorname{sinc}(x) = \frac{\sin(x)}{x}$$
• How can geometric summation formula help? I rewrite all sinc functions using formula 3 and 4 and added it to the formula 1, but I got stuck at this point, I cannot figure out how to rewrite this bug formula with exponents. – darktealeaf May 31 at 11:24
• @darktealeaf Do the reverse first, which is taking the DFT (or iDFT) of a rectangle function. You will find very similar math here: dsprelated.com/showarticle/1038.php – Cedron Dawg May 31 at 12:00
• So, got this: $x(n) = \frac{1}{N} \sum\limits_{k=0}^{N-1} \sqrt{\frac{m}{p}} e^{\frac{i2\pi kn}{N}} = \frac{1}{N} \sqrt{\frac{m}{p}} \frac{1 - e^{i2\pi n}}{1 - e^{\frac{i2\pi n}{N}}} = \frac{1}{N} \sqrt{\frac{m}{p}} \frac{e^{i\pi n}}{e^{\frac{i\pi n}{N}}} \frac{e^{-i\pi n} - e^{i\pi n}}{e^{\frac{-i\pi n}{N}} - e^{\frac{i\pi n}{N}}} = -\frac{1}{N} \sqrt{\frac{m}{p}} e^{i\pi n - \frac{i\pi n}{N}} \frac{sin(\pi n)}{sin(\frac{\pi n}{N})} = -\sqrt{\frac{m}{p}} e^{i\pi n - \frac{i\pi n}{N}} \frac{sinc(\pi n)}{sinc(\frac{\pi n}{N})}$ But still did not reach the correct formula. What do I do wrong? – darktealeaf May 31 at 14:51
• @darktealeaf Your case statement needs to be translated into summation intervals. I just noticed the inequality is the opposite of a rectangle function, so that means two intervals or you can work with the complement. – Cedron Dawg May 31 at 15:11 | 2020-07-09T12:07:49 | {
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https://mathoverflow.net/questions/115974/does-this-group-act-geometrically-on-a-median-space | # Does this group act geometrically on a Median space?
Let $G$ be the semidirect product of $\mathbb{Z}^2$ with $\mathbb{Z}/6$ where $\mathbb{Z}/6$ acts by the order 6 element of $SL_2(\mathbb{Z})$. We can think of this group as the group of order preserving isometries of the tesselation of $\mathbb{R^2}$ with regular triangles.
Does this group acts properly, isometrically and cocompactly on a median space??
Let for two points in a metric space $[x,y]=\{z|d(x,z)+d(z,y)=d(x,y)\}$. If $X$ is a geodesic metric space than this is just the set of all points lying on some geodesic from $x$ to $y$. $X$ is called a median space if for every triple of points $x,y,z$ we have that $[x,y]\cap[x,z]\cap[y,z]$ consists of exactly one point - the median of $x,y,z$. Examples for median spaces are trees and $\mathbb{R}^n$ with the $l^1$- metric.
The motivation is that the one skeleton of a CAT(0) cube complex is a median graph. If a group acts geometrically on this CAT(0)-cube complex it also acts that way on that graph. For example this group acts properly and isometrically on $\mathbb{R}^3$. This gives a proper and isometric action on a median space, but this action is not cocompact. So I was wondering whether there is a better action. The problem seems to be that the automorphism of $\mathbb{Z}^2$ does not extend to a cube-complex automorphism of $\mathbb{R}^2$, but I could not make this precise.
• Perhaps you could remind us of the definition of a median space? – HJRW Dec 10 '12 at 13:07
• @Joseph: you just gave the definition of a geodesic median space . A median space (as given by Henrik) does not require geodesics. He just mentions what is $[x,y]$ is in case the metric space is geodesic. – YCor Dec 10 '12 at 16:34
• @Yves: I've deleted my comment now that Henrik has defined "median space." – Joseph O'Rourke Dec 10 '12 at 22:29
It is known that the group you mention does not act geometrically on a CAT(0) cube complex. See for instance this answer for a possible argument, based on Lemma 16.12 in Wise's monograph The structure of groups with a quasiconvex hierarchy. I think you can adapt the proof of this lemma in order to prove the following statement:
Proposition: Let $$G$$ be a group acting properly and cocompactly on a median metric space $$(M,d)$$ of finite rank. Assume that $$G$$ contains a finite-index subgroup $$H \simeq \mathbb{Z}^n$$, $$n\geq 2$$. Then $$G$$ acts properly and cocompactly on $$(\mathbb{R}^n, \ell^1)$$.
Sketch of proof. As shown by Bowditch in Some properties of median metric spaces, there exists a CAT(0) metric $$\sigma$$ on $$M$$, and, if I understand the construction correctly, this metric satisfies the following properties: (1) any isometry of $$(M,d)$$ induces an isometry of $$(M,\sigma)$$; (2) the metrics $$d$$ and $$\sigma$$ are biLipschitz equivalent; (3) halfspaces of $$M$$ are $$\sigma$$-convex.
As a consequence, the flat torus theorem can be applied, and we find a $$G$$-invariant and $$\sigma$$-convex subspace $$\Sigma \subset M$$ which is $$\sigma$$-isometric to $$\mathbb{R}^n$$.
Now, we consider the structure of measured wallspace of $$\Sigma$$ induced by the walls of $$M$$. By $$\sigma$$-convexity of $$\Sigma$$ and of the walls, we must have $$m$$ families of parallel hyperplanes $$\mathbb{R}^{n-1}$$ in $$\Sigma$$. (Here, we ignore the collections of hyperplanes which lie in the neighborhood of a single hyperplane. As a consequence, $$m \leq n$$ since otherwise it would be possible to embed coarsely $$\mathbb{R}^{n+1}$$ into $$\Sigma \simeq \mathbb{R}^n$$.)
Let $$F$$ denote the median space associated to the previous wallspace. Then $$F$$ decomposes as the $$\ell^1$$-product of $$m$$ (discrete or continuous) unbounded lines. Up to replacing discrete lines with continuous lines, we may suppose that $$F$$ is $$(\mathbb{R}^m,\ell^1)$$.
Notice that, because $$G$$ acts properly on $$\Sigma$$, necessarily it also acts properly on $$F$$. So we must have $$m \geq n$$. But we already know that $$m \leq n$$, so $$m=n$$.
So far, we have proved that $$G$$ acts properly on $$(\mathbb{R}^n,\ell^1)$$. As $$G$$ is virtually $$\mathbb{Z}^n$$, we conclude that $$G$$ acts geometrically on $$(\mathbb{R}^n,\ell^1)$$. $$\square$$
In your specific example, the question is now: does $$\mathbb{Z}^2 \rtimes \mathbb{Z}_6$$ act geometrically on $$(\mathbb{R}^2,\ell^1)$$? The same argument as the one followed here works. A presentation of the group is $$T=\langle a,b,c \mid a^2=b^2=c^2=(ab)^3=(bc)^3=(ac)^3=1 \rangle.$$ Notice that $$\mathrm{Isom}(\mathbb{R}^2, \ell^1)= (\mathbb{R}\rtimes \mathbb{Z}_2)^2 \rtimes \mathbb{Z}_2$$ does not contain elements of order three, so, for any action $$T \curvearrowright (\mathbb{R}^2, \ell^1)$$ by isometries, the elements $$ab$$, $$bc$$ and $$ac$$ must be trivial. Consequently, such an action must factorise through the quotient $$T \twoheadrightarrow \mathbb{Z}_2$$ sending all the generators to $$1$$. A fortiori, the action cannot be geometric (and even proper).
Remark: The argument above does not completely answer the question as the median metric space is supposed to have finite rank. I do not know what happens if the rank is infinite, but it seems reasonable to think that the same conclusion holds. | 2019-10-19T08:48:09 | {
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https://crypto.stackexchange.com/questions/25065/combining-cascaded-encryption-keys-into-one-key | Combining cascaded encryption keys into one key
Is there an encryption algorithm where it's possible to combine multiple encryption keys into one, so that:
$E_{AB}(Data) = E_A(E_B(Data))$
KeyAB should be computable from KeyA and KeyB, but it must not be possible to compute or guess KeyA or KeyB from KeyAB and/or Data.
The algorithm should be as secure as possible, ideally asymmetrical (but if that's not possible, a good symmetrical one will also do).
XOR meets the above criteria, like so:
$KeyAB = KeyA \oplus KeyB$
$Data \oplus KeyAB = (Data \oplus KeyA) \oplus KeyB$
But XOR is obviously not a good crypto scheme, let alone asymmetrical. Are there any good alternatives for this particular use case?
• Suggested an edit for a nicer formatting, putting newlines where you wanted them and changed the math-formulations to better represent what you wanted. Apr 17, 2015 at 20:41
• well, it was proven that at least (plain) DES doesn't provide this property. Link Apr 17, 2015 at 20:43
For a crypto algorithm that acts like a group, the first thing that comes to mind is Pohlig-Hellman. In this method, we have a large prime $p$, and define:
$$E_A(Data) = Data^A \bmod p$$
(with $A$ relatively prime to $p-1$)
This has the property that $E_B(E_A(Data)) = E_{A \times B \bmod p-1}(Data)$; however it has the security properties you're looking for; for example, given lots of $X, E_A(X)$ pairs, you can't recover $A$.
Now, this is a symmetric system (given $A$ and $E_A(Data)$, you can recover $Data$). If you insist on an asymmetrical system, you might want to do this modulo a composite of secret factorization; only some who knows the factorization can then decrypt. That version might be somewhat related to a better known asymmetric scheme... | 2023-02-01T15:36:54 | {
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https://ac.cs.princeton.edu/40complex/ | # 4. Complex Analysis, Rational and Meromorphic Asymptotics
## Analytic transfer theorem for rational functions (common case)
If $h(z)$ is a rational function that is analytic at 0 and has a pole $\alpha=1/\beta$ of smallest modulus with multiplicity $M$ which is larger than the multiplicity of all other poles of smallest modulus, then $$[z^n]h(z) \sim c\beta^nn^{M-1}\quad\hbox{where}\quad c = {1\over(M-1)!\alpha^M}\lim_{z\to\alpha} (\alpha - z)^Mh(z).$$
See Theorem IV.9 and Note IV.26 on page 256, noting errors in text listed on Errata page (sorry!).
Example.The number of ways to make change for $n$ cents with pennies, nickels, dimes and quarters is $$[z^n]{1\over(1-z)(1-z^5)(1-z^{10})(1-z^{25})} ~\sim {n^3\over 3!}{1\over 1\cdot 5 \cdot 10 \cdot 25}={n^3\over 7500}$$ because $z=1$ is a pole of multiplicity 4 and $$\lim_{z\to 1}{1-z\over 1-z^t} = \lim_{z\to 1}{1\over 1 + z + z^2 + \ldots + z^{t-1}} = {1\over t}.$$
See Proposition IV.2 on page 258 and Slide 46 in Lecture 5.
With $h(z) = f(z)/g(z)$, another way to calculate the constant is to use l'Hôpital's rule, with the result $$c = M{(-\beta)^Mf(\alpha)\over g^{(M)}(\alpha)}.$$ See slide 30 in Lecture 4 or Theorem 4.1 in Analysis of Algorithms.
## Note IV.28
Supernecklaces. A "supernecklace" of the 3rd type is a labelled cycle of cycles (see page 125). Enumerate all the supernecklaces of the 3rd type of size $n$ for $n$ = 1, 2, and 3. (There are 1, 2, and 7, respectively.) Then develop an asymptotic estimate of the number of supernecklaces of size $n$ by showing that $$[z^n]\ln\Bigl({1\over 1 - \ln{1\over 1-z}}\Bigr)\sim{1\over n}(1-e^{-1})^{-n}.$$ Hint: Take derivatives.
## Program IV.1
Compute the percentage of permutations having no singelton or doubleton cycles and compare with the asymptotic estimate from analytic combinatorics, for N = 10 and N = 20 .
## Program IV.2
Plot the derivative of the GF for supernecklaces of type 3 (see Note IV.28) in the style of the plots in Lecture 4. Click here for access to the Java code in the lecture.
## IV.1
(D. Luo) Approximately how many ways are there to change $n$ cents in Canada (no pennies)?
## IV.2
(M. Moore) A "supernecklace" of the 1st type is a labelled cycle of sequences (see page 125). Enumerate all the supernecklaces of the 1st type of size $n$ for $n$ = 1, 2, and 3. (There are 1, 3, and 14, respectively.) The construction $S = CYC(SEQ_{>0}(Z))$ immediately leads to the EGF $$S(z) = \ln{1\over 1 - {z\over 1-z}} = \ln{1-z\over 1-2z}.$$ Use this to show that then number of supernecklaces of the 1st type of size $n$ is $\sim (n-1)!2^n$. Hint: Take derivatives.
## IV.3
Suppose that $h(z)$ is a meromorphic function with positive coefficients and that $\alpha$ is a pole of smallest modulus. The table below gives various possibilities for the modulus, multiplicity, and dominance of $\alpha$. For each row, indicate whether the order of growth of $[z^n]h(z)$ is constant, linear, quadratic, exponential, exponentially small, or none of these options.
modulus multiplicity dominant?
11yes
11no
12yes
13yes
1/21yes
1/22no
21yes
## IV.4
(D. Carter) An ultranecklace is a cycle of supernecklaces of type 3. Give a ~-expression for the number of ultrnecklaces of size $N$.
## IV.5
(A. Yan) Give a ~-approximation for $$[z^N]{1\over(1-z)(1-2z)(1-3z)\ldots(1-tz)}$$. | 2022-05-18T15:57:57 | {
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https://lw2.issarice.com/posts/jz5QoizH8HkQwWZ9Q/nash-equilibriums-can-be-arbitrarily-bad | # Nash equilibriums can be arbitrarily bad
post by Stuart_Armstrong · 2019-05-01T14:58:21.765Z · LW · GW · 24 comments
## Contents
Go hungry with Almost Free Lunches
optimal
and maximin
Arbitrary badness with two options
None
24 comments
# Go hungry with Almost Free Lunches
Consider the following game, called "Almost Free Lunches" (EDIT: this seems to be a variant of the traveller dilemma). You name any pound-and-pence amount between £0 and £1,000,000; your opponent does likewise. Then you will both get whichever amount named was lowest.
On top of that, the person who named the highest amount must give £0.02 to the other. If you tie, no extra money changes hands.
What's the Nash equilibrium of this game? Well:
• The only Nash equilibrium of Almost Free Lunches is for both of you to name £0.00.
Proof: Suppose player A has a probability distribution over possible amounts to name, and player has a probability distribution over possible amounts. Let be the highest amount such that is non-zero; let be the same, for . Assume that is a Nash equilibrium.
Assume further that (if that's not the case, then just switch the labels and ). Then either £0.00 or £0.00 (and hence both players select £0.00).
We'll now rule out £0.00. If £0.00, then player can improve their score by replacing with . To see this, assume that player has said , and player has said . If , then player can say just as well as - either choice gives them the same amount (namely, £0.02).
There remain two other cases. If , then is superior to , getting (rather than £0.02). And if , then gets £0.02 £0.01, rather than (if ) or (if ).
Finally, if £0.00, then player gets -£0.02 unless they also say £0.00.
Hence if £0.00, the cannot be part of a Nash Equilibrium. Thus £0.00 and hence the only Nash Equilibrium is at both players saying £0.00.
## Pareto optimal
There are three Pareto-optimal outcomes: (£1,000,000.00, £1,000,000.00), (£1,000,000.01, £999,999.97), and (£999,999.97, £1,000,000.01). All of them are very much above the Nash Equilibrium.
## Minmax and maximin
The minmax and maximin values are also both terrible, and also equal to £0.00. This is not surprising, though, as minmax and maximin implicitly assume the other players are antagonistic to you, and are trying to keep your profits low.
# Arbitrary badness with two options
This shows that choosing the Nash Equilibrium can be worse than almost every other option. We can of course increase the maximal amount, and get the Nash Equilibrium to be arbitrarily worse than any reasonable solution (I would just say either £1,000,000.00 or £999,999.99, and leave it at that).
But we can also make the Nash Equilibrium arbitrarily close to the worst possible outcome, and that without even requiring more than two options for each player.
Assume that there are four ordered amounts of money/utility: . Each player can name or . Then if they both name the same, they get that amount of utility. If they name different ones, then then player naming gets , and the player naming gets .
By the same argument as above, the only Nash equilibrium is for both to name . The maximum possible amount is ; the maximum they can get if they both coordinate is , the Nash equilibrium is , and the worst option is . We can set and for arbitrarily tiny , while setting to be larger than by some arbitrarily high amount.
So the situation is as bad as it could possibly be.
Note that this is a variant of the prisoner's dilemma with different numbers. You could describe it as "Your companion goes to a hideous jail if and only if you defect (and vice versa). Those that don't defect will also get a dust speck in their eye [LW · GW]."
## 24 comments
Comments sorted by top scores.
comment by cousin_it · 2019-05-01T21:33:17.470Z · LW(p) · GW(p)
Yeah. Usually the centipede game is used to teach this lesson, but your game is also very nice :-)
Replies from: Stuart_Armstrong
comment by Stuart_Armstrong · 2019-05-01T22:01:44.898Z · LW(p) · GW(p)
Cool! I prefer my example, though; it feels more intuitive (and has a single equilibrium).
comment by Taymon Beal (taymon-beal) · 2019-05-01T21:59:59.631Z · LW(p) · GW(p)
Nit: I think this game is more standardly referred to in the literature as the "traveler's dilemma" (Google seems to return no relevant hits for "almost free lunches" apart from this post).
Replies from: Stuart_Armstrong
comment by Stuart_Armstrong · 2019-05-02T06:53:54.136Z · LW(p) · GW(p)
That's useful; I added a link to the other game in the main text (as far as I can tell, I came up with this independently).
comment by shminux · 2019-05-02T06:05:47.144Z · LW(p) · GW(p)
So the situation is as bad as it could possibly be.
You mean, it is as bad as it could possibly be for the Nash equilibrium to be a good strategy and a good predictor in this setup? Yep, absolutely. All models tend to have their domain of validity, and this game shows the limits of the Nash equilibrium model of decision making.
Replies from: Stuart_Armstrong
comment by Stuart_Armstrong · 2019-05-02T07:39:36.869Z · LW(p) · GW(p)
That is true, but I meant it as "as close as you want to the worst possible outcome, and as far as you want from the best mutual outcome".
comment by Unnamed · 2019-05-02T05:45:37.329Z · LW(p) · GW(p)
Unilateral precommitment lets people win at "Almost Free Lunches".
One way to model precommitment is as a sequential game: first player 1 chooses a number, then player 1 has the option of either showing that number to player 2 or keeping it hidden, then player 2 chooses a number. Optimal play is for player 1 to pick £1,000,000 and show that number, and then for player 2 to choose £999,999.99.
An interesting feature of this is that player 1's precommitment helped player 2 even more than it helped player 1. Player 1 is "taking one for the team", but still winning big. This distinguishes it from games like chicken, where precommitment is a threat that allows the precommitter to win the larger share. Though this means that if either player can precommit (rather than one being pre-assigned to go first as player 1) then they'd both prefer to have the other one be the precommitter.
This benefit of precommitment does not extend to the two option version (n2 vs. n1). In that version, player 2 is incentivized to say "n1" regardless of what player 1 commits to, so unilateral precommitment doesn't help them avoid the Nash Equilibrium. As in the prisoner's dilemma.
Replies from: Zvi, Stuart_Armstrong
comment by Zvi · 2019-05-07T16:18:09.516Z · LW(p) · GW(p)
Is there a name for this type of equilibrium, where a player can pre-commit in a way where the best response leaves the first player very well-off, but not quite optimally well-off? What about if it is a mixed strategy (e.g. consider the version of this game where the player who gave the larger number gets paid nothing).
comment by Stuart_Armstrong · 2019-05-02T10:30:07.082Z · LW(p) · GW(p)
I think another key difference between PD and traveller/AFL is that in the PD variant, (n2, n1) is a Pareto outcome - you can't improve the first player's outcome without making the second one worse off. However, in the other problem, (0,0) is very very far from being Pareto.
comment by quanticle · 2019-05-01T16:21:31.585Z · LW(p) · GW(p)
Doesn't the existence of the rule that says that no money changes hands if there's a tie alter the incentives? If we both state that we want 1,000,000 pounds, then we both get it and we both walk away happy. What incentive is there for either of the two agents to name a value that is lower than 1,000,000?
Replies from: GuySrinivasan, Dagon, Charlie Steiner
comment by GuySrinivasan · 2019-05-01T17:21:59.579Z · LW(p) · GW(p)
If your strategy remains unchanged, I can change my strategy to "999,999.99 please" and come away with 1,000,000.01 in total, so that's not a Nash equilibrium.
Replies from: quanticle
comment by quanticle · 2019-05-02T03:51:11.161Z · LW(p) · GW(p)
I see. And from then it follows the same pattern as a dollar auction, until the "winning" bet goes to zero.
comment by Dagon · 2019-05-01T17:45:49.952Z · LW(p) · GW(p)
The maximum theoretical payout is 1000000.01 pounds. And since both players know this, and for any given tie amount, a player's net can be increased by a penny by reducing their bid by a penny, they will recursively calculate to 0.
Unless they have a theory of mind and can model ways to take risks in order to increase results in the cases where the other player ALSO takes risks. We call this "trust". and it can be greatly increased with communication, empathy, and side-agreements.
I think it's long been known that Nash equilibria are not necessarily optimal, only guaranteed that the other player's choices can't hurt you. It's perfectly defensive in adversarial games. This is great for zero-sum games, where the other player's increase is exactly your decrease. It's nearly irrelevant (except as a lower bound) for cooperative (variable-sum) games.
comment by Charlie Steiner · 2019-05-01T17:30:53.379Z · LW(p) · GW(p)
Yup. Though you don't necessarily need to imagine the money "changing hands" - if both people get paid 2 extra pennies if they tie, and the person who bids less gets paid 4 extra pennies, the result is the same.
The point is exactly what it says in the title. Relative to the maximum cooperative payoff, the actual equilibrium payoff can be arbitrarily small. And as you change the game, the transition from low payoff to high payoff can be sharp - jumping straight from pennies to millions just by changing the payoffs by a few pennies.
Replies from: Dagon
comment by Dagon · 2019-05-01T18:55:18.048Z · LW(p) · GW(p)
Relative to the maximum cooperative payoff, the actual equilibrium payoff can be arbitrarily small.
Please be careful to specify "Nash equilibrium" here, rather than just "equilibrium". Nash is not the only possible result that can be obtained, especially if players have some ability to cooperate or to model their opponent in a way where Nash's conditions don't hold.
In actual tests (admittedly not very complete, usually done in psychology or economics classes), almost nobody ends up at the Nash equilibrium in this style game (positive-sum where altruism or trust can lead one to take risks).
comment by Jay Molstad (jay-molstad) · 2019-05-02T10:47:26.899Z · LW(p) · GW(p)
It checks out empirically. War is a Nash semi-equilibrium - both sides would be better off if they could coordinate an end to it, but they usually can't (in the near term). If Stanislav Petrov had been a bit less chill, the Cold War would have ended in 1983 in the "everybody launches all the nukes; cockroaches win" Nash equilibrium. If that's not arbitrarily bad, it's plenty bad enough.
Replies from: RorscHak
comment by RorscHak · 2019-05-03T05:32:42.214Z · LW(p) · GW(p)
I think "everybody launches all nukes" might not be a Nash Equilibrium.
We can argue that once one side launched their nukes the other side does not necessarily have an incentive to retaliate, given they won't really care whether the enemy got nuked or not after they themselves are nuked, and they probably will have an incentive to not launch the nukes to prevent the "everybody dies" outcome, which can be argued to be negative for someone who is about to die.
Replies from: jay-molstad
comment by Jay Molstad (jay-molstad) · 2019-05-27T13:03:31.525Z · LW(p) · GW(p)
It seems to me that both parties to the Cold War favored the defect-defect outcome (launch all the nukes) over the cooperate-defect outcome (we die, they don't). It's hard to tell, though, because both sides had an incentive to signal that preference regardless of the truth.
But that's an extreme case. Any war you choose will have each side choosing between continuing to fight and surrendering. The cooperate-cooperate outcome (making peace in a way that approximates the likely outcome of a war) is probably best for all, but it's hard to achieve in practice. And it seems to me that at least part of the problem is that, if one side chooses to cooperate (sue for peace and refrain from maximally fighting), they run the risk that the other side will continue to defect (fight) and seize an advantage.
comment by RorscHak · 2019-05-02T09:54:27.162Z · LW(p) · GW(p)
It is interesting that experimental results of traveller's dilemma seems to give results which deviate strongly from the Nash Equilibrium, and in fact quite close to the Pareto Optimal Solution.
This is pretty strange for a game that has only one round and no collusion (you'd expect it to end as Prisoner's Dilemma, no?)
It is rather different from what we would see from the dollar auction, which has no Nash Equilibrium but always deviate far away from the Pareto optimal solution.
I suspect that the this game being one round-only actually improves the Pareto efficiency of its outcomes:
Maybe if both participants are allowed to change their bid after seeing each other's bid they WILL go into a downward spiral one cent by one cent until they reach zero or one player gives up at some point with a truce, just like how dollar auctions always stop at some point.
When there is only one round, however, there is no way for a player to make their bid exactly 1 or 2 cents less than the other player, and bidding any less than that is suboptimal compared to bidding more than the other player, so perhaps there is an incentive against lowering one's bidding indefinitely to 0 before the game even starts, just like no one would bid \$1000 in the dollar auction's first round.
Replies from: Stuart_Armstrong
comment by Stuart_Armstrong · 2019-05-02T13:08:30.052Z · LW(p) · GW(p)
you'd expect it to end as Prisoner's Dilemma, no?
I think a key difference is that in PD, (Defect, Cooperate) is a Pareto outcome (you can't make it better for the cooperator without making it worse for the defector). While (0, 0) is far from the Pareto boundary. So people can clearly see that naming numbers around 0 is a massive loss, so they focus on avoiding that loss rather than optimising their game vs the other player.
Replies from: RorscHak
comment by RorscHak · 2019-05-03T05:28:43.091Z · LW(p) · GW(p)
I haven't found any information yet, but I suspect there is a mixed Nash somewhere in TD.
Replies from: Stuart_Armstrong
comment by Stuart_Armstrong · 2019-05-03T09:19:12.380Z · LW(p) · GW(p)
There is no mixed Nash equilibrium in the TD example above (see the proof above).
Replies from: RorscHak
comment by RorscHak · 2019-05-03T11:32:17.686Z · LW(p) · GW(p)
Thanks, I forgot the proof before replying your comment.
You are correct that in PD (D,C) is Pareto, and so the Nash Equilibrium (D,D) is much closer to a Pareto outcome than the Nash Equilibrium (0,0) of TD is to its Pareto outcomes (somewhere along each person getting a million pounds, give or take a cent)
It still strange to see a game with only one round and no collusion to land pretty close to the optimal, while its repeated version (dollar auction) seems to deviate badly from the Pareto outcome.
Replies from: Stuart_Armstrong
comment by Stuart_Armstrong · 2019-05-03T11:40:37.349Z · LW(p) · GW(p)
It still strange to see a game with only one round and no collusion to land pretty close to the optimal, while its repeated version (dollar auction) seems to deviate badly from the Pareto outcome.
It is a bit strange. It seems this is because in the dollar auction, you can always make your position slightly better unilaterally, in a way that will make it worse once the other player reacts. Iterate enough, and all value is destroyed. But in a one-round game, you can't slide down that path, so you pick by looking at the overall picture. | 2021-09-19T13:29:36 | {
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http://mathhelpforum.com/calculus/137854-how-show-sequence-converges.html | # Math Help - How to show this sequence converges?
1. ## How to show this sequence converges?
Show that
$a_n=(\sum_{k=1}^\infty \frac{1}{k})-log(n+1)$ converges to some real number.
The problem hints that I'm supposed to use the fact that
$\sum_{k=1}^\infty \frac{x}{k(x+k)}$ converges uniformly on [0,1], which I've already proved.
I can't figure out how to connect these two facts at all... any help?
2. Originally Posted by paupsers
Show that
$a_n=(\sum_{k=1}^\infty \frac{1}{k})-log(n+1)$ converges to some real number.
The problem hints that I'm supposed to use the fact that
$\sum_{k=1}^\infty \frac{x}{k(x+k)}$ converges uniformly on [0,1], which I've already proved.
I can't figure out how to connect these two facts at all... any help?
My opinion :
Can we use that
$\frac{1}{k} > \ln(1 + \frac{1}{k}) > \frac{1}{k+1}$ ?
then the series $\left( \sum_{k=1}^n \frac{1}{k} \right ) - \ln(n+1)$
$= \left( \sum_{k=1}^n \frac{1}{k} \right ) - \sum_{k=1}^n \ln( 1 + \frac{1}{k} )$
$= \sum_{k=1}^n \left( \frac{1}{k} - \ln( 1 + \frac{1}{k} ) \right) < \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+1} \right)$ $= \sum_{k=1}^n \frac{1}{k(k+1)}$ which is convergent .
Also , since $\frac{1}{k} > \ln(1 + \frac{1}{k} )$ , the series is greater than zero , it is bounded .
I think there may be some errors but anyway , you can take a look and perhaps it can help you a little .
3. Originally Posted by paupsers
Show that
$a_n=(\sum_{k=1}^\infty \frac{1}{k})-log(n+1)$ converges to some real number.
The problem hints that I'm supposed to use the fact that
$\sum_{k=1}^\infty \frac{x}{k(x+k)}$ converges uniformly on [0,1], which I've already proved.
I can't figure out how to connect these two facts at all... any help?
Because of the uniform convergence, the sum $f(x)=\sum_{k=1}^\infty \frac{x}{k(x+k)}$ is continuous on $[0,1]$. Furthermore, you may integrate term by term on $[0,1]$. Note also that $\frac{x}{k(x+k)}=\frac{1}{k}-\frac{1}{x+k}$. We get:
$\sum_{k=1}^\infty \left(\frac{1}{k}-(\log(k+1)-\log k)\right)=\int_0^1 f(x) dx$.
The important thing is that the right-hand side is finite.
The above shows (means) that $\sum_{k=1}^n \left(\frac{1}{k}-(\log(k+1)-\log k)\right)$ has a finite limit as $n\to\infty$.
However, this partial sum can be rewritten as $\left(\sum_{k=1}^n \frac{1}{k}\right)-\log (n+1)$...
4. Originally Posted by Laurent
We get:
$\sum_{k=1}^\infty \left(\frac{1}{k}-(\log(k+1)-\log k)\right)=\int_0^1 f(x) dx$.
I don't understand how you got to this point. What exactly did you integrate?
5. Originally Posted by paupsers
I don't understand how you got to this point. What exactly did you integrate?
$f$ between 0 and 1...
$\int_0^1 f(x)dx = \int_0^1 \left(\sum_{k=1}^\infty \frac{x}{k(x+k)}\right)dx$ $=\sum_{k=1}^\infty \int_0^1 \frac{x}{k(x+k)}dx$,
(last step justified by uniform convergence) and used the expression I gave to simplify the computation of the last integral. | 2015-07-06T00:35:57 | {
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http://www.peterhenderson.co/XOR-NN | # XOR as a Neural Network
So I’m really just ripping this from Richard Socher’s course on Deep Learning through NLP (CS224d @ Stanford), so I’m just writing down my understand of it as I go to maybe spread the knowledge. To quote the assignment:
It is well-known that a single linear classifier cannot represent the XOR function $x \oplus y$, depicted below: there is no way to draw a single line that can separate the red and magenta (square) points from the blue and cyan (circle) points.
A two-layer neural network, however, can separate this pattern easily. Below, we give you a simple dataset in two dimensions that represents a noisy version of the XOR pattern. Your task is to hand-pick weights for a very simple two-layer network, such that it can separate the red/magenta points from the blue/cyan points.
The network uses the following equations for $W \in {\mathbb R}^{2 \times 2}$ and $U \in {\mathbb R}^2$
This can be modelled by the following graph.
The question becomes: how do you choose the weights $W, U, b_1, b_2$ to create this sort of separation. Well the way I thought about it was this (not really sure if it’s super correct, but whatever).
Let’s start with the bottom layer: $h$. We also define:
for the first layer say we want to arbitrarily split the data in two. So, if we have
then our first neural layer hypothesis will look like:
The goal of the top level of the neural network is to literally draw a line through the data and say: “anything above this line is one class, anything below is another”. That’s what the sigmoid function is doing. The weights just allow you to configure how this line is drawn, and the number of variables say how many lines are drawn.
For example, in the hidden layer, $h$ has 2 dimensions and therefore allows us to draw two lines through the data in this layer. (This is maybe an oversimplified and slightly off view of it, but it helps me to think about it this way.) So now, in the first layer, we say $h_1$ will represent the red group of data and $h_2$ will represent the magenta group of data. How can we draw a line through the data to make this split. Well for the red group $x_2 = .5-x_1$ represents a good split and for the magenta group $x_2 = 1.5 - x_1$ represents a good split. But, we want to say that for anything below the line is the red group will represent $h_1=1$ so we just reverse this equation. Now, we can simply map these equations to the weight parameters (note, the sigmoid function will simply remap anything on the opposite sides of the line to 0, 1).
Now, with this we will remap the data such that a graph of $h_1,h_2$ will look like:
Now, the data is split in such a way that we can actually just draw a line through it and remap the data such that we have our XOR function.
We can draw the line at $h_1+h_2-.5 = 0$ and thus we will have our data split if we let $q=1, r=1, \beta_2=-.5$.
And the output looks like:
And there you have it, we made an XOR function out of a neural network and manually set the weights. I hope this gives some insight into how a NN works in finding the weights it uses. Really messy code for this can be found on my github, but most of it was taken from Socher’s course.
Written on August 6, 2015 | 2019-04-26T06:52:56 | {
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https://openstax.org/books/elementary-algebra/pages/8-1-simplify-rational-expressions | Elementary Algebra
# 8.1Simplify Rational Expressions
Elementary Algebra8.1 Simplify Rational Expressions
### Learning Objectives
By the end of this section, you will be able to:
• Determine the values for which a rational expression is undefined
• Evaluate rational expressions
• Simplify rational expressions
• Simplify rational expressions with opposite factors
### Be Prepared 8.1
Before you get started, take this readiness quiz.
If you miss a problem, go back to the section listed and review the material.
1. Simplify: $90y15y2.90y15y2.$
If you missed this problem, review Example 6.66.
2. Factor: $6x2−7x+2.6x2−7x+2.$
If you missed this problem, review Example 7.34.
3. Factor: $n3+8.n3+8.$
If you missed this problem, review Example 7.54.
In Chapter 1, we reviewed the properties of fractions and their operations. We introduced rational numbers, which are just fractions where the numerators and denominators are integers, and the denominator is not zero.
In this chapter, we will work with fractions whose numerators and denominators are polynomials. We call these rational expressions.
### Rational Expression
A rational expression is an expression of the form $p(x)q(x),p(x)q(x),$ where p and q are polynomials and $q≠0.q≠0.$
Remember, division by 0 is undefined.
Here are some examples of rational expressions:
$−13427y8z5x+2x2−74x2+3x−12x−8−13427y8z5x+2x2−74x2+3x−12x−8$
Notice that the first rational expression listed above, $−1342,−1342,$ is just a fraction. Since a constant is a polynomial with degree zero, the ratio of two constants is a rational expression, provided the denominator is not zero.
We will perform same operations with rational expressions that we do with fractions. We will simplify, add, subtract, multiply, divide, and use them in applications.
### Determine the Values for Which a Rational Expression is Undefined
When we work with a numerical fraction, it is easy to avoid dividing by zero, because we can see the number in the denominator. In order to avoid dividing by zero in a rational expression, we must not allow values of the variable that will make the denominator be zero.
If the denominator is zero, the rational expression is undefined. The numerator of a rational expression may be 0—but not the denominator.
So before we begin any operation with a rational expression, we examine it first to find the values that would make the denominator zero. That way, when we solve a rational equation for example, we will know whether the algebraic solutions we find are allowed or not.
### How To
#### Determine the Values for Which a Rational Expression is Undefined.
1. Step 1. Set the denominator equal to zero.
2. Step 2. Solve the equation in the set of reals, if possible.
### Example 8.1
Determine the values for which the rational expression is undefined:
$9yx9yx$ $4b−32b+54b−32b+5$ $x+4x2+5x+6x+4x2+5x+6$
### Try It 8.1
Determine the values for which the rational expression is undefined:
$3yx3yx$ $8n−53n+18n−53n+1$ $a+10a2+4a+3a+10a2+4a+3$
### Try It 8.2
Determine the values for which the rational expression is undefined:
$4p5q4p5q$ $y−13y+2y−13y+2$ $m−5m2+m−6m−5m2+m−6$
### Evaluate Rational Expressions
To evaluate a rational expression, we substitute values of the variables into the expression and simplify, just as we have for many other expressions in this book.
### Example 8.2
Evaluate $2x+33x−52x+33x−5$ for each value:
$x=0x=0$ $x=2x=2$ $x=−3x=−3$
### Try It 8.3
Evaluate $y+12y−3y+12y−3$ for each value:
$y=1y=1$ $y=−3y=−3$ $y=0y=0$
### Try It 8.4
Evaluate $5x−12x+15x−12x+1$ for each value:
$x=1x=1$ $x=−1x=−1$ $x=0x=0$
### Example 8.3
Evaluate $x2+8x+7x2−4x2+8x+7x2−4$ for each value:
$x=0x=0$ $x=2x=2$ $x=−1x=−1$
### Try It 8.5
Evaluate $x2+1x2−3x+2x2+1x2−3x+2$ for each value:
$x=0x=0$ $x=−1x=−1$ $x=3x=3$
### Try It 8.6
Evaluate $x2+x−6x2−9x2+x−6x2−9$ for each value:
$x=0x=0$ $x=−2x=−2$ $x=1x=1$
Remember that a fraction is simplified when it has no common factors, other than 1, in its numerator and denominator. When we evaluate a rational expression, we make sure to simplify the resulting fraction.
### Example 8.4
Evaluate $a2+2ab+b23ab3a2+2ab+b23ab3$ for each value:
$a=1,b=2a=1,b=2$ $a=−2,b=−1a=−2,b=−1$ $a=13,b=0a=13,b=0$
### Try It 8.7
Evaluate $2a3ba2+2ab+b22a3ba2+2ab+b2$ for each value:
$a=−1,b=2a=−1,b=2$ $a=0,b=−1a=0,b=−1$ $a=1,b=12a=1,b=12$
### Try It 8.8
Evaluate $a2−b28ab3a2−b28ab3$ for each value:
$a=1,b=−1a=1,b=−1$ $a=12,b=−1a=12,b=−1$ $a=−2,b=1a=−2,b=1$
### Simplify Rational Expressions
Just like a fraction is considered simplified if there are no common factors, other than 1, in its numerator and denominator, a rational expression is simplified if it has no common factors, other than 1, in its numerator and denominator.
### Simplified Rational Expression
A rational expression is considered simplified if there are no common factors in its numerator and denominator.
For example:
• $2323$ is simplified because there are no common factors of 2 and 3.
• $2x3x2x3x$ is not simplified because x is a common factor of 2x and 3x.
We use the Equivalent Fractions Property to simplify numerical fractions. We restate it here as we will also use it to simplify rational expressions.
### Equivalent Fractions Property
If a, b, and c are numbers where $b≠0,c≠0b≠0,c≠0$, then $ab=a·cb·cab=a·cb·c$ and $a·cb·c=aba·cb·c=ab$.
Notice that in the Equivalent Fractions Property, the values that would make the denominators zero are specifically disallowed. We see $b≠0,c≠0b≠0,c≠0$ clearly stated. Every time we write a rational expression, we should make a similar statement disallowing values that would make a denominator zero. However, to let us focus on the work at hand, we will omit writing it in the examples.
Let’s start by reviewing how we simplify numerical fractions.
### Example 8.5
Simplify: $−3663.−3663.$
### Try It 8.9
Simplify: $−4581.−4581.$
### Try It 8.10
Simplify: $−4254.−4254.$
Throughout this chapter, we will assume that all numerical values that would make the denominator be zero are excluded. We will not write the restrictions for each rational expression, but keep in mind that the denominator can never be zero. So in this next example, $x≠0x≠0$ and $y≠0y≠0$.
### Example 8.6
Simplify: $3xy18x2y2.3xy18x2y2.$
### Try It 8.11
Simplify: $4x2y12xy2.4x2y12xy2.$
### Try It 8.12
Simplify: $16x2y2xy2.16x2y2xy2.$
To simplify rational expressions we first write the numerator and denominator in factored form. Then we remove the common factors using the Equivalent Fractions Property.
Be very careful as you remove common factors. Factors are multiplied to make a product. You can remove a factor from a product. You cannot remove a term from a sum.
Note that removing the x’s from $x+5xx+5x$ would be like cancelling the 2’s in the fraction $2+522+52$!
### Example 8.7
#### How to Simplify Rational Binomials
Simplify: $2x+85x+20.2x+85x+20.$
### Try It 8.13
Simplify: $3x−62x−4.3x−62x−4.$
### Try It 8.14
Simplify: $7y+355y+25.7y+355y+25.$
We now summarize the steps you should follow to simplify rational expressions.
### How To
#### Simplify a Rational Expression.
1. Step 1. Factor the numerator and denominator completely.
2. Step 2. Simplify by dividing out common factors.
Usually, we leave the simplified rational expression in factored form. This way it is easy to check that we have removed all the common factors!
We’ll use the methods we covered in Factoring to factor the polynomials in the numerators and denominators in the following examples.
### Example 8.8
Simplify: $x2+5x+6x2+8x+12.x2+5x+6x2+8x+12.$
### Try It 8.15
Simplify: $x2−x−2x2−3x+2.x2−x−2x2−3x+2.$
### Try It 8.16
Simplify: $x2−3x−10x2+x−2.x2−3x−10x2+x−2.$
### Example 8.9
Simplify: $y2+y−42y2−36.y2+y−42y2−36.$
### Try It 8.17
Simplify: $x2+x−6x2−4.x2+x−6x2−4.$
### Try It 8.18
Simplify: $x2+8x+7x2−49.x2+8x+7x2−49.$
### Example 8.10
Simplify: $p3−2p2+2p−4p2−7p+10.p3−2p2+2p−4p2−7p+10.$
### Try It 8.19
Simplify: $y3−3y2+y−3y2−y−6.y3−3y2+y−3y2−y−6.$
### Try It 8.20
Simplify: $p3−p2+2p−2p2+4p−5.p3−p2+2p−2p2+4p−5.$
### Example 8.11
Simplify: $2n2−14n4n2−16n−48.2n2−14n4n2−16n−48.$
### Try It 8.21
Simplify: $2n2−10n4n2−16n−20.2n2−10n4n2−16n−20.$
### Try It 8.22
Simplify: $4x2−16x8x2−16x−64.4x2−16x8x2−16x−64.$
### Example 8.12
Simplify: $3b2−12b+126b2−24.3b2−12b+126b2−24.$
### Try It 8.23
Simplify: $2x2−12x+183x2−27.2x2−12x+183x2−27.$
### Try It 8.24
Simplify: $5y2−30y+252y2−50.5y2−30y+252y2−50.$
### Example 8.13
Simplify: $m3+8m2−4.m3+8m2−4.$
### Try It 8.25
Simplify: $p3−64p2−16.p3−64p2−16.$
### Try It 8.26
Simplify: $x3+8x2−4.x3+8x2−4.$
### Simplify Rational Expressions with Opposite Factors
Now we will see how to simplify a rational expression whose numerator and denominator have opposite factors. Let’s start with a numerical fraction, say $7−77−7$. We know this fraction simplifies to $−1−1$. We also recognize that the numerator and denominator are opposites.
In Foundations, we introduced opposite notation: the opposite of $aa$ is $−a−a$. We remember, too, that $−a=−1·a−a=−1·a$.
We simplify the fraction $a−aa−a$, whose numerator and denominator are opposites, in this way:
$a−aWe could rewrite this.1·a−1·aRemove the common factors.1−1Simplify.−1a−aWe could rewrite this.1·a−1·aRemove the common factors.1−1Simplify.−1$
So, in the same way, we can simplify the fraction $x−3−(x−3)x−3−(x−3)$:
$We could rewrite this.1·(x−3)−1·(x−3)Remove the common factors.1−1Simplify.−1We could rewrite this.1·(x−3)−1·(x−3)Remove the common factors.1−1Simplify.−1$
But the opposite of $x−3x−3$ could be written differently:
$−(x−3)Distribute.−x+3Rewrite.3−x−(x−3)Distribute.−x+3Rewrite.3−x$
This means the fraction $x−33−xx−33−x$ simplifies to $−1−1$.
In general, we could write the opposite of $a−ba−b$ as $b−ab−a$. So the rational expression $a−bb−aa−bb−a$ simplifies to $−1−1$.
### Opposites in a Rational Expression
The opposite of $a−ba−b$ is $b−ab−a$.
$a−bb−a=−1a≠ba−bb−a=−1a≠b$
An expression and its opposite divide to $−1−1$.
We will use this property to simplify rational expressions that contain opposites in their numerators and denominators.
### Example 8.14
Simplify: $x−88−x.x−88−x.$
### Try It 8.27
Simplify: $y−22−y.y−22−y.$
### Try It 8.28
Simplify: $n−99−n.n−99−n.$
Remember, the first step in simplifying a rational expression is to factor the numerator and denominator completely.
### Example 8.15
Simplify: $14−2xx2−49.14−2xx2−49.$
### Try It 8.29
Simplify: $10−2yy2−25.10−2yy2−25.$
### Try It 8.30
Simplify: $3y−2781−y2.3y−2781−y2.$
### Example 8.16
Simplify: $x2−4x−3264−x2.x2−4x−3264−x2.$
### Try It 8.31
Simplify: $x2−4x−525−x2.x2−4x−525−x2.$
### Try It 8.32
Simplify: $x2+x−21−x2.x2+x−21−x2.$
### Section 8.1 Exercises
#### Practice Makes Perfect
In the following exercises, determine the values for which the rational expression is undefined.
1.
$2xz2xz$
$4p−16p−54p−16p−5$
$n−3n2+2n−8n−3n2+2n−8$
2.
$10m11n10m11n$
$6y+134y−96y+134y−9$
$b−8b2−36b−8b2−36$
3.
$4x2y3y4x2y3y$
$3x−22x+13x−22x+1$
$u−1u2−3u−28u−1u2−3u−28$
4.
$5pq29q5pq29q$
$7a−43a+57a−43a+5$
$1x2−41x2−4$
Evaluate Rational Expressions
In the following exercises, evaluate the rational expression for the given values.
5.
$2 x x − 1 2 x x − 1$
$x=0x=0$
$x=2x=2$
$x=−1x=−1$
6.
$4 y − 1 5 y − 3 4 y − 1 5 y − 3$
$y=0y=0$
$y=2y=2$
$y=−1y=−1$
7.
$2 p + 3 p 2 + 1 2 p + 3 p 2 + 1$
$p=0p=0$
$p=1p=1$
$p=−2p=−2$
8.
$x + 3 2 − 3 x x + 3 2 − 3 x$
$x=0x=0$
$x=1x=1$
$x=−2x=−2$
9.
$y 2 + 5 y + 6 y 2 − 1 y 2 + 5 y + 6 y 2 − 1$
$y=0y=0$
$y=2y=2$
$y=−2y=−2$
10.
$z 2 + 3 z − 10 z 2 − 1 z 2 + 3 z − 10 z 2 − 1$
$z=0z=0$
$z=2z=2$
$z=−2z=−2$
11.
$a 2 − 4 a 2 + 5 a + 4 a 2 − 4 a 2 + 5 a + 4$
$a=0a=0$
$a=1a=1$
$a=−2a=−2$
12.
$b 2 + 2 b 2 − 3 b − 4 b 2 + 2 b 2 − 3 b − 4$
$b=0b=0$
$b=2b=2$
$b=−2b=−2$
13.
$x 2 + 3 x y + 2 y 2 2 x 3 y x 2 + 3 x y + 2 y 2 2 x 3 y$
1. $x=1,y=−1x=1,y=−1$
2. $x=2,y=1x=2,y=1$
3. $x=−1,y=−2x=−1,y=−2$
14.
$c 2 + c d − 2 d 2 c d 3 c 2 + c d − 2 d 2 c d 3$
1. $c=2,d=−1c=2,d=−1$
2. $c=1,d=−1c=1,d=−1$
3. $c=−1,d=2c=−1,d=2$
15.
$m 2 − 4 n 2 5 m n 3 m 2 − 4 n 2 5 m n 3$
1. $m=2,n=1m=2,n=1$
2. $m=−1,n=−1m=−1,n=−1$
3. $m=3,n=2m=3,n=2$
16.
$2 s 2 t s 2 − 9 t 2 2 s 2 t s 2 − 9 t 2$
1. $s=4,t=1s=4,t=1$
2. $s=−1,t=−1s=−1,t=−1$
3. $s=0,t=2s=0,t=2$
Simplify Rational Expressions
In the following exercises, simplify.
17.
$− 4 52 − 4 52$
18.
$− 44 55 − 44 55$
19.
$56 63 56 63$
20.
$65 104 65 104$
21.
$6 a b 2 12 a 2 b 6 a b 2 12 a 2 b$
22.
$15 x y 3 x 3 y 3 15 x y 3 x 3 y 3$
23.
$8 m 3 n 12 m n 2 8 m 3 n 12 m n 2$
24.
$36 v 3 w 2 27 v w 3 36 v 3 w 2 27 v w 3$
25.
$3 a + 6 4 a + 8 3 a + 6 4 a + 8$
26.
$5 b + 5 6 b + 6 5 b + 5 6 b + 6$
27.
$3 c − 9 5 c − 15 3 c − 9 5 c − 15$
28.
$4 d + 8 9 d + 18 4 d + 8 9 d + 18$
29.
$7 m + 63 5 m + 45 7 m + 63 5 m + 45$
30.
$8 n − 96 3 n − 36 8 n − 96 3 n − 36$
31.
$12 p − 240 5 p − 100 12 p − 240 5 p − 100$
32.
$6 q + 210 5 q + 175 6 q + 210 5 q + 175$
33.
$a 2 − a − 12 a 2 − 8 a + 16 a 2 − a − 12 a 2 − 8 a + 16$
34.
$x 2 + 4 x − 5 x 2 − 2 x + 1 x 2 + 4 x − 5 x 2 − 2 x + 1$
35.
$y 2 + 3 y − 4 y 2 − 6 y + 5 y 2 + 3 y − 4 y 2 − 6 y + 5$
36.
$v 2 + 8 v + 15 v 2 − v − 12 v 2 + 8 v + 15 v 2 − v − 12$
37.
$x 2 − 25 x 2 + 2 x − 15 x 2 − 25 x 2 + 2 x − 15$
38.
$a 2 − 4 a 2 + 6 a − 16 a 2 − 4 a 2 + 6 a − 16$
39.
$y 2 − 2 y − 3 y 2 − 9 y 2 − 2 y − 3 y 2 − 9$
40.
$b 2 + 9 b + 18 b 2 − 36 b 2 + 9 b + 18 b 2 − 36$
41.
$y 3 + y 2 + y + 1 y 2 + 2 y + 1 y 3 + y 2 + y + 1 y 2 + 2 y + 1$
42.
$p 3 + 3 p 2 + 4 p + 12 p 2 + p − 6 p 3 + 3 p 2 + 4 p + 12 p 2 + p − 6$
43.
$x 3 − 2 x 2 − 25 x + 50 x 2 − 25 x 3 − 2 x 2 − 25 x + 50 x 2 − 25$
44.
$q 3 + 3 q 2 − 4 q − 12 q 2 − 4 q 3 + 3 q 2 − 4 q − 12 q 2 − 4$
45.
$3 a 2 + 15 a 6 a 2 + 6 a − 36 3 a 2 + 15 a 6 a 2 + 6 a − 36$
46.
$8 b 2 − 32 b 2 b 2 − 6 b − 80 8 b 2 − 32 b 2 b 2 − 6 b − 80$
47.
$−5 c 2 − 10 c −10 c 2 + 30 c + 100 −5 c 2 − 10 c −10 c 2 + 30 c + 100$
48.
$4 d 2 − 24 d 2 d 2 − 4 d − 48 4 d 2 − 24 d 2 d 2 − 4 d − 48$
49.
$3 m 2 + 30 m + 75 4 m 2 − 100 3 m 2 + 30 m + 75 4 m 2 − 100$
50.
$5 n 2 + 30 n + 45 2 n 2 − 18 5 n 2 + 30 n + 45 2 n 2 − 18$
51.
$5 r 2 + 30 r − 35 r 2 − 49 5 r 2 + 30 r − 35 r 2 − 49$
52.
$3 s 2 + 30 s + 24 3 s 2 − 48 3 s 2 + 30 s + 24 3 s 2 − 48$
53.
$t 3 − 27 t 2 − 9 t 3 − 27 t 2 − 9$
54.
$v 3 − 1 v 2 − 1 v 3 − 1 v 2 − 1$
55.
$w 3 + 216 w 2 − 36 w 3 + 216 w 2 − 36$
56.
$v 3 + 125 v 2 − 25 v 3 + 125 v 2 − 25$
Simplify Rational Expressions with Opposite Factors
In the following exercises, simplify each rational expression.
57.
$a − 5 5 − a a − 5 5 − a$
58.
$b − 12 12 − b b − 12 12 − b$
59.
$11 − c c − 11 11 − c c − 11$
60.
$5 − d d − 5 5 − d d − 5$
61.
$12 − 2 x x 2 − 36 12 − 2 x x 2 − 36$
62.
$20 − 5 y y 2 − 16 20 − 5 y y 2 − 16$
63.
$4 v − 32 64 − v 2 4 v − 32 64 − v 2$
64.
$7 w − 21 9 − w 2 7 w − 21 9 − w 2$
65.
$y 2 − 11 y + 24 9 − y 2 y 2 − 11 y + 24 9 − y 2$
66.
$z 2 − 9 z + 20 16 − z 2 z 2 − 9 z + 20 16 − z 2$
67.
$a 2 − 5 a − 36 81 − a 2 a 2 − 5 a − 36 81 − a 2$
68.
$b 2 + b − 42 36 − b 2 b 2 + b − 42 36 − b 2$
#### Everyday Math
69.
Tax Rates For the tax year 2015, the amount of tax owed by a single person earning between $37,450 and$90,750, can be found by evaluating the formula $0.25x−4206.25,0.25x−4206.25,$ where x is income. The average tax rate for this income can be found by evaluating the formula $0.25x−4206.25x.0.25x−4206.25x.$ What would be the average tax rate for a single person earning \$50,000?
70.
Work The length of time it takes for two people for perform the same task if they work together can be found by evaluating the formula $xyx+y.xyx+y.$ If Tom can paint the den in $x=x=$ 45 minutes and his brother Bobby can paint it in $y=y=$ 60 minutes, how many minutes will it take them if they work together?
#### Writing Exercises
71.
Explain how you find the values of x for which the rational expression $x2−x−20x2−4x2−x−20x2−4$ is undefined.
72.
Explain all the steps you take to simplify the rational expression $p2+4p−219−p2.p2+4p−219−p2.$
#### Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
If most of your checks were:
…confidently. Congratulations! You have achieved your goals in this section! Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific!
…with some help. This must be addressed quickly as topics you do not master become potholes in your road to success. Math is sequential - every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?
…no - I don’t get it! This is critical and you must not ignore it. You need to get help immediately or you will quickly be overwhelmed. See your instructor as soon as possible to discuss your situation. Together you can come up with a plan to get you the help you need.
Do you know how you learn best?
Order a print copy
As an Amazon Associate we earn from qualifying purchases. | 2022-05-28T03:22:00 | {
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http://mathhelpforum.com/algebra/158129-find-point-intersection-between-two-straight-lines.html | Thread: Find point of intersection between two straight lines
1. Find point of intersection between two straight lines
That's the instruction I have. Sorry for my bad English.
$mx - 4y = m$
$-4x + my = 2m + 4$
I Thought I should get both of the equations to y = ... , So:
$y = \dfrac{mx-m}{4}$
$y = 2 + \dfrac{4+4x}{m}$
And then:
$\dfrac{mx-m}{4} = 2 + \dfrac{4+4x}{m}$
$m(mx-m) = 8m +4(4+4x)$
$m^{2}x - m^{2} -8m-16-16x = 0$
And now I don't know how to continue...
2. You've done very good, correct work so far! Now it's a good time to remember what it is for which you are trying to solve. What is that?
3. "Combine like terms". That is, combine all terms involving "x" on one side of the equation, all terms that do not on the other. | 2013-12-12T05:51:58 | {
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https://www.math.bgu.ac.il/research/seminars/combinatorics-seminar | The seminar meets on Tuesdays, 16:00-17:00, in 201
This Week
Sparse sharp thresholds and hypercontractivity
The sharp threshold phenomenon is a central topic of research in the analysis of Boolean functions. Here, one aims to give sufficient conditions for a monotone Boolean function f to satisfy\mu_p(f)=o(\mu_q(f)), where q = p + o(p), and \mu_p(f) is the probability that f=1 on an input with independent coordinates, each taking the value 1 with probability p.
The dense regime, where \mu_p(f)=\Theta(1), is somewhat understood due to seminal works by Bourgain, Friedgut, Hatami, and Kalai. On the other hand, the sparse regime where \mu_p(f)=o(1) was out of reach of the available methods. However, the potential power of the sparse regime was suggested by Kahn and Kalai already in 2006.
In this talk we show that if a monotone Boolean function f with \mu_p(f)=o(1) satisfies some mild pseudo-randomness conditions then it exhibits a sharp threshold in the interval [p,q], with q = p+o(p). More specifically, our mild pseudo-randomness hypothesis is that the p-biased measure of f does not bump up to Θ(1) whenever we restrict f to a sub-cube of constant co-dimension, and our conclusion is that we can find q=p+o(p), such that \mu_p(f)=o(\mu_q(f)).
2018–19–A meetings
Date
Title
Speaker
Abstract
Nov 13, 16:10–17:10 Finitely Forcible Graphons Roman Glebov (BGU)
Abstract: In extremal graph theory, we often consider large graphs that are in the limit uniquely determined by finitely many densities of their subgraphs. The corresponding limits (so-called graphons) are called finitely forcible. Motivated by classical results in extremal combinatorics as well as by recent developments in the study of finitely forcible graphons, Lovasz and Szegedy made some conjectures about the structure of such graphons. In particular, they conjectured that the topological space of typical points of every finitely forcible graphon is compact and finitely dimensional. In joint results with D. Kral, T. Klimosova, and J. Volec, we could disprove both conjectures.
Nov 20 Sparse sharp thresholds and hypercontractivity Noam Lifshitz (Bar Ilan)
The sharp threshold phenomenon is a central topic of research in the analysis of Boolean functions. Here, one aims to give sufficient conditions for a monotone Boolean function f to satisfy\mu_p(f)=o(\mu_q(f)), where q = p + o(p), and \mu_p(f) is the probability that f=1 on an input with independent coordinates, each taking the value 1 with probability p.
The dense regime, where \mu_p(f)=\Theta(1), is somewhat understood due to seminal works by Bourgain, Friedgut, Hatami, and Kalai. On the other hand, the sparse regime where \mu_p(f)=o(1) was out of reach of the available methods. However, the potential power of the sparse regime was suggested by Kahn and Kalai already in 2006.
In this talk we show that if a monotone Boolean function f with \mu_p(f)=o(1) satisfies some mild pseudo-randomness conditions then it exhibits a sharp threshold in the interval [p,q], with q = p+o(p). More specifically, our mild pseudo-randomness hypothesis is that the p-biased measure of f does not bump up to Θ(1) whenever we restrict f to a sub-cube of constant co-dimension, and our conclusion is that we can find q=p+o(p), such that \mu_p(f)=o(\mu_q(f)).
Recurring Seminar run by Dr. Charles Wolf and Dr. yelena yuditsky | 2018-11-21T04:57:47 | {
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http://mathhelpforum.com/calculus/19702-calculus-limits-derivatives.html | # Math Help - Calculus (Limits and Derivatives)
1. ## Calculus (Limits and Derivatives)
1. $lim_{x\rightarrow-5}\frac{x^3+125}{(-5-x)^3}$
I reduced it to:
$lim_{x\rightarrow-5}\frac{x^{2}+5x+25}{x^{2}+10x+25}$
But I don't know what to do after that.
2. Find equations of all tangent lines to the graph of $y=4x^{3}+5x-8$
I took the derivative of the equation, which was:
$\acute{y}=12x^{2}+5$
I then thought I might need to make the derivative equal to the original equation, so:
$12x^{2}+5=4x^{3}+5x-8$
But I don't really know...
3. Determine values of constant real numbers a, b so that the function:
$f(x)= ax^{2}-8x+6$ if $x\leq-1$
$f(x)= bx+2$ if x is greater than -1
is differentiable at x=-1.
No idea how to do this one.
2. Originally Posted by Thomas
1. $lim_{x\rightarrow-5}\frac{x^3+125}{(-5-x)^3}$
I reduced it to:
$lim_{x\rightarrow-5}\frac{x^{2}+5x+25}{x^{2}+10x+25}$
But I don't know what to do after that.
$\displaystyle\lim_{x\to-5}\frac{x^3+125}{(-5-x)^3}=-\lim_{x\to-5}\frac{(x+5)(x^2-10x+25)}{(5+x)^3}=$
$\displaystyle =-\lim_{x\to-5}\frac{x^2-10x+25}{(x+5)^2}=\frac{-100}{+0}=-\infty$
3. Ahh, I didn't think about how the bottom is squared.
Thanks!
4. Originally Posted by Thomas
3. Determine values of constant real numbers a, b so that the function:
$f(x)= ax^{2}-8x+6$ if $x\leq-1$
$f(x)= bx+2$ if x is greater than -1
is differentiable at x=-1.
No idea how to do this one.
First of all, f must be continuous at x=-1.
$\displaystyle\lim_{x\nearrow-1}f(x)=\lim_{x\nearrow-1}(ax^2-8x+6)=a+14=f(-1)$
$\displaystyle\lim_{x\searrow-1}f(x)=\lim_{x\searrow-1}(bx+2)=-b+2$.
Then $a+14=-b+2\Rightarrow a+b=-12$ (1)
We have $\displaystyle f'(x)=\left\{\begin{array}{cc}2ax-8, & x\leq-1\\b, & x>-1\end{array}\right.$
$f_l'(-1)=\lim_{x\nearrow-1}f'(x)=-2a-8, \ f_r'(-1)=\lim_{x\searrow-1}f'(x)=b$
$f_l'(-1)=f_r'(-1)\Rightarrow-2a-8=b\Rightarrow 2a+b=-8$ (2)
Now, solve the system formed by equations (1) and (2). | 2014-03-17T06:36:51 | {
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https://rdrr.io/cran/lmomco/man/pwmRC.html | pwmRC: Sample Probability-Weighted Moments for Right-Tail Censoring
Description Usage Arguments Details Value Author(s) References See Also Examples
Description
Compute the sample Probability-Weighted Moments (PWMs) for right-tail censored data set—that is a data set censored from above. The censoring threshold is denoted as T. The data possess m values that are observed (noncensored, < T) out of a total of n samples. The ratio of m to n is defined as ζ = m/n, which will play an important role in parameter estimation. The ζ is interpreted as the probability \mathrm{Pr}\lbrace \rbrace that x is less than the quantile at ζ nonexceedance probability: (\mathrm{Pr}\lbrace x < X(ζ) \rbrace). Two types of PWMs are computed for right-tail censored situations. The “A”-type PWMs and “B”-type PWMs. The A-type PWMs are defined by
β^A_r = m^{-1}∑^m_{j=1} {j-1 \choose r} x_{[j:n]}\mbox{,}
which are the PWMs of the uncensored sample of m observed values. The B-type PWMs are computed from the “complete” sample, in which the n-m censored values are replaced by the censoring threshold T. The B-type PWMs are defined by
β^B_r = n^{-1} \biggl( ∑^m_{j=1} {j-1 \choose r} x_{[j:n]} + ∑^n_{j=m+1} {j-1 \choose r} T \biggr) \mbox{.}
The two previous expressions are used in the function. These PWMs are readily converted to L-moments by the usual methods (pwm2lmom). When there are more than a few censored values, the PWMs are readily computed by computing β^A_r and using the expression
β^B_r = Zβ^A_r + \frac{1-Z}{r+1}T\mbox{,}
where
Z = \frac{m}{n}\frac{{m-1 \choose r}}{{n-1 \choose r}}\mbox{.}
The two expressions above are consulted when the checkbetas=TRUE argument is present. Both sequences of B-type are cated to the terminal. This provides a check on the implementation of the algorithm. The functions Apwm2BpwmRC and Bpwm2ApwmRC can be used to switch back and forth between the two PWM types given fitted parameters for a distribution in the lmomco package that supports right-tail censoring. Finally, the RC in the function name is to denote Right-tail Censoring.
Usage
1 pwmRC(x, threshold=NULL, nmom=5, sort=TRUE, checkbetas=FALSE)
Arguments
x A vector of data values. threshold The right-tail censoring (upper) threshold. nmom Number of PWMs to return. sort Do the data need sorting? Note that convention is the have a β_0, but this is placed in the first index i=1 of the betas vector. checkbetas A cross relation between β^A_r and β^B_r exists—display the results of the secondary computation of the β^B_r. The two displayed vectors should be numerically equal.
Details
There is some ambiguity if the threshold also numerically equals valid data in the data set. In the data for the examples below, which are taken from elsewhere, there are real observations at the censoring level. One can see how a hack is made to marginally decrease or increase the data or the threshold for the computations. This is needed because the code uses
1 sapply(x, function(v) { if(v >= T) return(T); return(v) } )
to reset the data vector x. By operating on the data in this fashion one can toy with various levels of the threshold for experimental purposes; this seemed a more natural way for general implementation. The code sets n = length(x) and m = n - length(x[x == T]), which also seems natural. The β^A_r are computed by dispatching to pwm.
Value
An R list is returned.
Abetas The A-type PWMs. These should be same as pwm() returns if there is no censoring. Note that convention is the have a β_0, but this is placed in the first index i=1 of the betas vector. Bbetas The B-type PWMs. These should be NA if there is no censoring. Note that convention is the have a β_0, but this is placed in the first index i=1 of the betas vector. source Source of the PWMs: “pwmRC”. threshold The upper censoring threshold. zeta The right censoring fraction: numabovethreshold/samplesize. numabovethreshold Number of data points equal to or above the threshold. observedsize Number of real data points in the sample (below the threshold). samplesize Number of actual sample values.
W.H. Asquith
References
Greenwood, J.A., Landwehr, J.M., Matalas, N.C., and Wallis, J.R., 1979, Probability weighted moments—Definition and relation to parameters of several distributions expressable in inverse form: Water Resources Research, v. 15, pp. 1,049–1,054.
Hosking, J.R.M., 1990, L-moments—Analysis and estimation of distributions using linear combinations of order statistics: Journal of the Royal Statistical Society, Series B, v. 52, pp. 105–124.
Hosking, J.R.M., 1995, The use of L-moments in the analysis of censored data, in Recent Advances in Life-Testing and Reliability, edited by N. Balakrishnan, chapter 29, CRC Press, Boca Raton, Fla., pp. 546–560.
See Also
lmoms, pwm2lmom, pwm, pwmLC
Examples
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 # Data listed in Hosking (1995, table 29.2, p. 551) H <- c(3,4,5,6,6,7,8,8,9,9,9,10,10,11,11,11,13,13,13,13,13, 17,19,19,25,29,33,42,42,51.9999,52,52,52) # 51.9999 was really 52, a real (noncensored) data point. z <- pwmRC(H,threshold=52,checkbetas=TRUE) str(z) # Hosking(1995) reports that A-type L-moments for this sample are # lamA1=15.7 and lamAL-CV=.389, and lamAL-skew=.393 pwm2lmom(z$Abetas) # My version of R reports 15.666, 0.3959, and 0.4030 # See p. 553 of Hosking (1995) # Data listed in Hosking (1995, table 29.3, p. 553) D <- c(-2.982, -2.849, -2.546, -2.350, -1.983, -1.492, -1.443, -1.394, -1.386, -1.269, -1.195, -1.174, -0.854, -0.620, -0.576, -0.548, -0.247, -0.195, -0.056, -0.013, 0.006, 0.033, 0.037, 0.046, 0.084, 0.221, 0.245, 0.296) D <- c(D,rep(.2960001,40-28)) # 28 values, but Hosking mentions # 40 values in total z <- pwmRC(D,.2960001) # Hosking reports B-type L-moments for this sample are # lamB1 = -.516 and lamB2 = 0.523 pwm2lmom(z$Bbetas) # My version of R reports -.5162 and 0.5218
Search within the lmomco package
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Questions? Problems? Suggestions? or email at [email protected].
Please suggest features or report bugs with the GitHub issue tracker.
All documentation is copyright its authors; we didn't write any of that. | 2017-04-30T10:48:35 | {
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https://socratic.org/questions/the-brakes-are-applied-on-a-car-traveling-at-30-m-s-fwd-the-car-stops-in-3-0s-wh | # The brakes are applied on a car traveling at 30. m/s [fwd]. The car stops in 3.0s. What is its displacement during this time?
## The answer is 45 m [fwd], but I do not understand how to get to this answer.
Dec 26, 2016
You can use the equations of motion to find the displacement, as shown below.
#### Explanation:
If we are assuming that the acceleration is uniform (which I believe must be the case), you can use the following equation of motion, as it does not require that you know, or first calculate the acceleration:
$\Delta d = \frac{1}{2} \left({v}_{i} + {v}_{f}\right) \Delta t$
This basically says that the displacement $\Delta d$ is equal to the average speed $\frac{1}{2} \left({v}_{i} + {v}_{f}\right)$ multiplied by the time interval $\Delta t$.
Insert the numbers
$\Delta d = \frac{1}{2} \left(30 + 0\right) \left(3\right) = 15 \left(3\right) = 45 m$ | 2020-08-10T09:08:10 | {
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https://or.stackexchange.com/questions/3634/if-else-condition-for-the-objective-variable-using-big-m-notation | # if-else condition for the objective variable using big M notation
Let $$0\leq \beta\leq 1$$ be an objective variable. The size of $$\beta$$ is $$N\!\times\!N$$.
Now, how can I impose the following?
if $$\beta_{i,j}>0$$ then $$\beta_{j,i}=0$$
Big M notation can be one of the possible ways, however, I am unable to proceed.
Introduce binary variable $$x_{i,j}$$ to indicate whether $$\beta_{i,j}>0$$ and linear constraints: \begin{align} \beta_{i,j} &\le x_{i,j}\\ x_{i,j} + x_{j,i} &\le 1 \end{align} (The big-M here is 1.)
The first constraint enforces $$\beta_{i,j}>0 \implies x_{i,j} = 1.$$ The second constraint enforces $$x_{i,j} = 1 \implies x_{j,i} = 0.$$ The first constraint enforces $$x_{j,i} = 0 \implies \beta_{j,i} \le 0.$$ The lower bound on $$\beta$$ enforces $$\beta_{j,i} \le 0 \implies \beta_{j,i} = 0.$$ So $$\beta_{i,j}>0 \implies \beta_{j,i} = 0,$$ as desired. | 2021-11-28T00:31:57 | {
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https://math.stackexchange.com/questions/1604960/is-a-non-commutative-invertible-monoid-closed-under-an-associative-binary-opera | # Is a non-commutative, invertible monoid closed under an associative binary operation?
I've been given this question that's been puzzling me for a while:
$M$ is a monoid with identity element denoted by $e$.
$U(M)$ is the set of all invertible elements of $M$.
Suppose that $M$ is not commutative.
Let $a, b \in M$. If $ab \in U(M)$, then is it necessarily the case that $a \in U(M)$?
Hint: Consider the monoid of the set of all functions from positive integers into itself, equipped with the binary operation of composition of functions, together with the element $b \in M$ that is defined as follows: $b(n)=n+1 \ \forall n \in \Bbb N$.
I believe that it's not always necessarily the case and I need to provide a counter example. I'm attempting to find a function $a(n) \circ b(n)$ is equal to an invertible function. However the invertible function that I can think of is one as the function has a domain and co-domain of positive integers.
Is this the right approach? I'm not sure where to go from here.
Let $a(1) := 1$ and $a(n) := n - 1$ for $n > 1$. Then $a$ has no inverse (since $a(1) = a(2)$) but $a \circ b$ is the identity.
Consider the function $a:\Bbb N \to \Bbb N$ given by:
$a(n) = n-1$ for $n > 0$
$a(0) = 0$.
What is $a \circ b$?
EDIT: as the other answer shows this can be modified to give a similar $a: \Bbb Z^{+} \to \Bbb Z^{+}$ (Both $\Bbb N^{\Bbb N}$ and $(\Bbb Z^{+})^{\Bbb Z^{+}}$ form monoids under composition). | 2019-12-06T23:36:41 | {
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http://mathhelpforum.com/calculus/98404-polar-coordinates.html | 1. ## Polar Coordinates
Sketch the region that lies inside the polar curve r=2sin(theta) and outside the polar curve r=1, then find the area of the shaded region.
I am able to graph both regions but am not sure how to set up the bounds for the integration. If someone could explain to me how to get the bounds and why that is the [a,b] for the bounds.
2. Originally Posted by naikn
Sketch the region that lies inside the polar curve r=2sin(theta) and outside the polar curve r=1, then find the area of the shaded region.
I am able to graph both regions but am not sure how to set up the bounds for the integration. If someone could explain to me how to get the bounds and why that is the [a,b] for the bounds.
i tried out the problem and got the bounds of integtation as Pi/6 and 5*Pi /6
can someone confirm this? | 2016-10-26T04:08:38 | {
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https://socratic.org/questions/how-can-you-write-1-2244444444444444-as-a-fraction | # How can you write 1.2244444444444444.... as a fraction?
Jun 28, 2015
$1.2244444 \ldots = \frac{1102}{900} = \frac{551}{450}$
#### Explanation:
If you see a single repeating digit, try multiplying by $9$ first:
$1.2244444 \ldots \cdot 9 = 11.02$
To make this into an integer just multiply by $100$
$11.02 \cdot 100 = 1102$
So:
$1.2244444 \ldots \cdot 900 = 1102$
Divide both sides by $900$ to get:
$1.2244444 \ldots = \frac{1102}{900}$
Actually both $1102$ and $900$ are even, so divide top and bottom by $2$ to get:
$1.2244444 \ldots = \frac{551}{450}$
Incidentally, I don't know if it's still popular, but there's a notation for repeating decimals, where you place a dot or dots above digits as follows:
$1.22 \dot{4}$ to mean $1.2244444 \ldots$
Or if you have a repeating pattern of more than one digit, use a dot to mark each end of the repeating pattern, so:
$\frac{1}{7} = 0. \dot{1} 4285 \dot{7}$
Perhaps you just did not know how to get Socratic to display this. | 2019-10-22T11:08:13 | {
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https://quantumcomputing.stackexchange.com/questions/12639/find-orthogonal-state-for-random-7-qubit-state | # Find orthogonal state for random 7 qubit state
I have a system that generates a random 7 qubit state and I need a method to always find the orthogonal state.
I'm currently using python and qutip for this, representing this 7 qubit state by a 128-dimensional vector.
Basically, you divide the entire $$7$$-qubit Hilbert space into two subspaces: the one spanned by your state (let's call the state $$|\psi\rangle$$), let's call that subspace $$W$$, and it's orthogonal complement $$V = W^{\perp}$$. You want any vector from $$V$$, because this will by definition be orthogonal.
We know that $$P_{V}$$ + $$P_{W} = I$$, with $$P_{V}$$ and $$P_{W}$$ being the projectors upon the $$V$$ and $$W$$ subspaces. Since $$P_{W} = |\psi\rangle\langle\psi|$$, we can easily calculate $$P_{V}$$:
$$P_{V} = I - |\psi\rangle\langle\psi|.$$ We can let this projection matrix act on virtually any state to obtain a state orthogonal on $$|\psi\rangle$$, lets use the $$0$$ vector $$|00....0\rangle$$. An orthogonal state $$|\psi^{\perp}\rangle$$ to $$|\psi\rangle$$ is thus:
$$|\psi^{\perp}\rangle = P_{V}|00...0\rangle = (I - |\psi\rangle\langle\psi|)|00...0\rangle = |00...0\rangle - \langle\psi|00...0\rangle|\psi\rangle$$
which is more or less the Gram-Schmidt process.
Note that you can use (almost) any state instead of $$|00...0\rangle$$; the only state that you cannot use is $$|\psi\rangle$$ itself.
In python, this becomes something like:
from numpy import zeros_like, inner
zeros_vect = zeros_like(psi_orig)
psi_orth = zeros_vect - inner(zeros_vect,psi_orig).conj()*psi_orig | 2021-04-19T05:56:48 | {
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http://math.stackexchange.com/questions/186447/2d-basis-functions-orthogonal-under-exponential-kernel | # 2D basis functions orthogonal under exponential kernel
In one dimension, the Laguerre polynomials are orthogonal under exponential weighting: $$\int_0^\infty L_n(x) L_m(x) e^{-x} \, dx = 0, n \ne m$$ Does anyone know what the corresponding basis functions would be in 2 dimensions? $$\int_{-\infty}^\infty \int_{-\infty}^\infty F_n(x,y) F_m(x,y) e^{-r} \, dx \, dy = 0, r= \sqrt{x^2+y^2}, n \ne m$$ The Zernike polynomials are orthogonal, but with uniform weight and over the unit disk.
The underlying problem is to compute an estimator for a missing pixel. A series of orthogonal functions are helpful since you can then incrementally compute a 1-st order estimator, then a second order, then a third order, and so on. The exponential arises since (in natural scenes anyway) presumably pixels further away have less influence.
-
Supposing $F_n$'s are functions of $r$, your orthogonality relation is $$\int_0^\infty F_m F_n\ r e^{-r} dr = 0$$ so $F_m$ is the generalized Laguerre polynomial $L^{(1)}_n$, see here and here. To get an orthogonal basis of $L^2(\mathbb{R}^2,e^{-r}dx\,dy)$, you need to use all the functions $L^{(1)}_n(r)\, e^{ik\phi}$.
Thanks, that makes sense. Since I have real valued data, and want real coefficients (application is biology), the functions become $L(r)cos(k\phi)$ and $L(r)sin(k\phi)$, unless I'm missing something... – Lou Scheffer Aug 24 '12 at 21:02 | 2015-08-02T23:24:24 | {
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https://socratic.org/questions/5468b5ab581e2a4d85161797 | Question #61797
Nov 17, 2014
Pascal's Principle states that the pressure inside an enclosed fluid is the same everywhere. If you squeeze a closed plastic bottle, the pressure at every point in the fluid inside will be the same.
A hydraulic pressure cylinder works in much the same way. The cylinder is an enclosed space filled with fluid. A small pump can pressurize the cylinder. The amount of force that the cylinder can exert on an external object is proportional to the pressure inside the cylinder and the surface area of the piston. If the pressure can be increased to 1 pound per square inch (psi), a cylinder with a piston 3 inches in diameter can lift about 28 pounds.
$\text{Force" = "Pressure" * "Surface Area}$
$F = P A$
One can also solve for pressure and write:
$P = \frac{A}{F}$
The small pump provides some force ${F}_{s}$ to a small surface area ${A}_{s}$.
The Pressure at the small piston is:
${P}_{s} = {F}_{s} / {A}_{s}$
And the pressure at the large piston is:
${P}_{l} = {F}_{l} / {A}_{l}$
Pascal's Principle tells us that the pressure is the same at both pistons
${P}_{s} = {P}_{l}$.
So we know that:
${F}_{s} / {A}_{s} = {F}_{l} / {A}_{l}$
A little algebra allows us to solve for the force exerted on the large piston in terms of the force on the small piston:
${F}_{l} = {F}_{s} \left({A}_{l} / {A}_{s}\right)$
And we can see that the mechanical advantage of a small force ${F}_{s}$ is increased in proportion to the ratio of the surface areas of the two pistons.
Also see this excellent illustration of Pascal's Principle. | 2020-07-11T04:41:11 | {
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https://math.stackexchange.com/questions/1776403/ratio-proportion-question | # Ratio , proportion question
It will take 24 men working 9 hours a day each of build a house in 45 days . Given that all men work at the same rate,
(A) how many days will 18 men take to build the same house if they work 8 hours a day?
(B) how many hours per day must 20 men work if the house is to be completed in 48 days?
I'm having my mid-year examinations tomorrow and this question will be coming out. This is always my stumbling block as I don't understand the steps I need to take to solve this problem. Can I get any help? Thanks in advance
I would calculate the "efficiency" of their work: $$e= \frac{H}{M\times D\times \frac{T}{24}}$$
$H$ - number of houses built
$M$ - men working
$D$ - days taken
$T$ - hours per day
Thus, we calculate $e$ first:
$$e= \frac{1}{24\times 45\times \frac{9}{24}}$$ $$e=\frac{1}{405}$$
Now you just plug in the data you got and use your calculated $e$.
$A)$ $$\frac{1}{405}=\frac{1}{18\times D \times \frac{8}{24}}$$ $$405= 6D$$ $$D=67.5$$
I think you can do $B)$ for yourself now.
B) Man-hours to build the house = 24*9*45 = 9720
let x = hours per day 20 men must work
20*x*48 = 9720
x= 9720/(20*48)= 10.125 hrs/day
Total men-hour-days required = 24*9*45 A) men = 18 , hour = 8 days = d = ? so 18*8*d = 24*9*45 d = 24*9*45/18*8 =67.5 days
B) men = 20, d = 48 , h = ? so 20*48*h = 24*9*45 h = 45*9/20*2 = 405/40 hours = 10 hour 7.5 minutes per day
For more such concepts you can look at https://www.handakafunda.com/how-to-solve-time-work-problems-in-cat/ | 2020-07-13T22:21:27 | {
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http://mathhelpforum.com/statistics/75175-solved-probability-2-a.html | # Math Help - [SOLVED] Probability 2
1. ## [SOLVED] Probability 2
Which is greater, the probability of choosing a number divisible by 6 or the probability of not choosing a number divisible by 8?
*numbers 1-50*
2. Hello, soccerchic96!
It's kind of simple, isn't it?
Which is greater, the probability of choosing a number divisible by 6
or the probability of not choosing a number divisible by 8?
(numbers 1-50)
We have 50 numbers to choose from.
How many are divisible by 6?
. . Eight of them: 6, 12, 18, 24, 39, 36, 42, 48
. . Hence: . $P(\text{div. by 6}) \:=\:\frac{8}{50}$
How many are divisible by 8?
. . Six of them: 8, 16, 24, 32, 40, 48
So there are forty-four which are not divisible by 8.
. . Hence: . $P(\text{not div. by 8}) \:=\:\frac{44}{50}$
Now which is greater? | 2016-02-10T06:52:16 | {
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http://mathhelpforum.com/differential-geometry/120484-image-principal-value-mapping.html | ## image, principal value mapping
Sketch an image under $w = \text{Log} (z)$ of
i) the line $y=x$
ii) the line $x=e$
I have not done many problems with mappings under the principal value mapping of the logarithm. I know that
$\text{Log} (z) := \ln r + i \theta = \ln | z | + i \text{Arg} (z)$.
However, I do not see what these images would look like. I need a few pointers here on what to do. | 2014-12-29T14:11:10 | {
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http://mathhelpforum.com/higher-math/215124-convert-into-polar-form-e-raise-power-i-theta-please-help-print.html | # Convert into polar form of e raise to power i(theta), please help.
• Mar 20th 2013, 03:36 AM
szak1592
Convert into polar form of e raise to power i(theta), please help.
i1/2
and
z1/n
and
-9
where z=x+i y
• Mar 20th 2013, 03:43 AM
Prove It
Re: Convert into polar form of e raise to power i(theta), please help.
Quote:
Originally Posted by szak1592
i1/2
and
z1/n
and
-9
where z=x+i y
\displaystyle \begin{align*} i^{\frac{1}{2}} &= \left( e^{i \frac{\pi}{2}} \right)^{\frac{1}{2}} \\ &= e^{i \frac{\pi}{4}} \end{align*}
and
\displaystyle \begin{align*} z^{\frac{1}{n}} &= \left( r\, e^{i\theta} \right)^{\frac{1}{n}} \\ &= r^{\frac{1}{n}}\,e^{i\frac{\theta}{n}} \end{align*}
and
$\displaystyle -9 = 9\,e^{ i \pi } " alt="\displaystyle -9 = 9\,e^{ i \pi } " />
• Apr 9th 2013, 07:13 AM
HallsofIvy
Re: Convert into polar form of e raise to power i(theta), please help.
For general $x+ iy= re^{i\theta}$ with $r= \sqrt{x^2+ y^2}$, $\theta= arctan(y/x)$
• Apr 10th 2013, 01:18 AM
Prove It
Re: Convert into polar form of e raise to power i(theta), please help.
Quote:
Originally Posted by HallsofIvy
For general $x+ iy= re^{i\theta}$ with $r= \sqrt{x^2+ y^2}$, $\theta= arctan(y/x)$
$\displaystyle \theta$ is only $\displaystyle \arctan{ \left( \frac{y}{x} \right) }$ if the complex number is in the first quadrant. Otherwise you need to evaluate the angle in the first quadrant and then transform. | 2016-12-05T16:01:02 | {
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http://ianfinlayson.net/class/cpsc220/labs/15-collisions | # Collision Detection
## Objective
To learn about collision detection in a graphics program.
Many graphical programs such as games need some form of collision detection. Not only do we want to move objects around the screen, but we also want to detect when objects touch each other in some ways.
Game objects can have complex shapes, but collisions are usually checked against simpler enclosing shapes. For example, a bounding box or bounding circle can be used to simplify the question of whether two sprites have collided on the screen.
This lab asks you to write a Java function for testing whether two circles have collided. That is, whether they are currently intersecting in 2D space. The program below creates 20 balls with are moving in space.
There is a "stub" method which should check whether one ball has collided with another. Right now the function just returns false. Your job is to finish this function.
## Details
• When you run the program, 20 random balls will move around the window.
• Write the collides function so that it returns true when two balls collide.
• To do this, use the x and y locations of the two balls, along with the distance formula:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
• When done, the balls should bounce off of each other instead of passing over each other.
## Submitting
When your program works, email the code to [email protected]. | 2022-01-23T22:19:49 | {
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https://www.dcode.fr/complex-number-conjugate | Search for a tool
Complex Number Conjugate
Tool for calculating the value of the conjugate of a complex number. The conjugate of a complex number $z$ is written $\overline{z}$ or $z^*$ and is formed of the same real part with an opposite imaginary part.
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# Complex Number Conjugate
## Complex Conjugate Calculator
### How to calculate the conjugate of a complex number?
The conjugate of a complex number $z = a+ib$ is noted with a bar $\overline{z}$ (or sometimes with a star $z^*$) and is equal to $\overline{z} = a-ib$ with $a = \Re (z)$ the real part and $b = \Im (z)$ the imaginary part.
Example: Determine the conjugate of $z = 1+i$ is to calculate $\overline{z} = 1-i$
In other words, to find the conjugate of a complex number, take that same complex number but with the opposite (minus sign) of its imaginary part (containing $i$).
### What is complex conjugate pair?
The set of 2 elements: a complex number $z$ and its conjugate $\overline {z}$, form a pair of conjugates.
On a complex plane, the points $z$ and $\overline{z}$ are symmetrical (symmetry with respect to the x-axis).
### What are the properties of conjugates?
Using the complex numbers $z, z_1, z_2$, the conjugate has the following properties:
$$\overline{z_1+z_2} = \overline{z_1} + \overline{z_2}$$
$$\overline{z_1 \cdot z_2} = \overline{z_1} \times \overline{z_2}$$
$$\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_1}}{\overline{z_2}} \iff z_2 \neq 0$$
A number without an imaginary part is equal to its conjugate:
$$\Im (z) = 0 \iff \overline{z} = z$$
The modulus of a complex number and its conjugate are equal:
$$|\overline{z}|=|z|$$
### What is the product of a complex number and its conjugate?
The multiplication of a complex number $z = a + ib$ and its conjugate $\overline{z} = a-ib$ gives: $$z \ times \overline{z} = a ^ 2 + b ^ 2$$
This number is a real number (no imaginary part $i$) and strictly positive (addition of 2 squares values necessarily positive)
### What is the conjugate of i?
The conjugate of $i$ is $-i$
### How to calculate the conjugate of a real number (without i)?
The conjugate $\overline{a}$ of a real number $a$ is the number $a$ itself: $a=a+0i=a-0i=\overline{a}$
Example: $\overline{1 + 0 \times i} = 1$
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NB: for encrypted messages, test our automatic cipher identifier! | 2022-05-16T12:15:20 | {
"domain": "dcode.fr",
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https://classes.areteem.org/mod/forum/discuss.php?d=902 | ## Online Course Discussion Forum
### Number Theory help
Number Theory help
Hello, I don't know how to do 5.27. I don't know how to prove it; I tried using mod 4 but that didn't work.
Re: Number Theory help
For clarity, let's summarize why mod 4 doesn't work. We know any square is 0 or 1 (mod 4). Thus the sum of 3 squares is either 0, 1, 2, or 3 (mod 4). But since this is all possibilities it is not very helpful.
For purposes of this problem, we could say 4 is not "large enough" as a mod. Thinking about your answers from 5.26 (where you listed possible squares mod m for m < 10) see if you could find another mod that might work.
Hope this helps a bit! | 2021-09-23T21:44:18 | {
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https://mathoverflow.net/questions/317536/map-of-grassmannians-associated-with-a-veronese-embedding | # Map of Grassmannians associated with a Veronese embedding
I'm quite sure this should be classically known, however I am not an expert on the topic and I was unable to find a precise reference in the huge literature concerning Veronese embeddings and Grassmannians. Any pointer to relevant books or papers will be highly appreciated.
Let $$V$$ be a finite-dimensional vector space (over $$\mathbb{C}$$, say) and let $$v_n \colon \mathbb{P}(V) \longrightarrow \mathbb{P}(S^n V)$$ be the usual $$n$$th Veronese embedding.
If $$\ell$$ is a line in $$\mathbb{P}(V)$$, then $$v_n(\ell)$$ is a rational normal curve of degree $$n$$, that will be contained in precisely one $$n$$-plane $$\Pi_{\ell} \subset \mathbb{P}(S^n V)$$. Then we can define a morphism of projective Grassmannians $$\psi_n \colon \mathbb{G}(1, \, \mathbb{P}(V)) \longrightarrow \mathbb{G}(n, \, \mathbb{P}(S^nV)),$$ given by $$\psi_n(\ell) :=\Pi_{\ell}$$.
Question. Is $$\psi_n$$ an embedding?
• Let $\mathbb{C}^2$ be the vector space associated to $l$. Isn'it that $\mathbb{P}(S^n \mathbb{C}^2) \cap v_n(\mathbb{P}(V)) = v_n(l)$? – Libli Dec 13 '18 at 10:23
I don't know a reference, but here is a simple argument. Note that $$G(2,V)$$ (let me use linear notation) is a homogeneous space for $$GL(V)$$: $$G(2,V) = GL(V)/P_2,$$ where $$P_2$$ is a parabolic. If $$e_1,\dots,e_N$$ is the basis of $$V$$, we can take $$P_2$$ to be the stabilizer of the point $$p_1 := [e_1 \wedge e_2] \in \mathbb{P}(\wedge^2V).$$ Note that $$e_1 \wedge e_2$$ is the highest weight vector with weight $$\epsilon_1 + \epsilon_2 = \omega_2$$ (the second fundamental weight of $$GL(V)$$).
The map $$\psi_n$$ is $$GL(V)$$-equivariant, and takes $$[e_1 \wedge e_2]$$ to $$p_n := [(e_1^n) \wedge (e_1^{n-1}e_2) \wedge \dots \wedge (e_1e_2^{n-1}) \wedge (e_2^n)].$$ It is easy to check that this is a highest vector with weight $$n\epsilon_1 + ((n-1)\epsilon_1 + \epsilon_2) + \dots (\epsilon_1 + (n-1)\epsilon_2) + n\epsilon_2 = \binom{n+1}{2}\omega_2$$ (it corresponds to an irreducible subrepresentation $$V_{\binom{n+1}{2}\omega_2} \subset \wedge^{n+1}(S^nV)$$), and its stabilizer is the same parabolic subgroup $$P_2$$. It follows that $$\psi_n$$ is an isomorphism onto the orbit of the point $$p_n$$, in particular it is an embedding.
• And both are special cases of the Borel-Weil-Tits embedding of $G/P_2$ into projective space of its geometric quantization, when endowed with its symplectic structure as coadjoint orbit of $\omega_2$, resp. ${n+1\choose 2}\omega_2$. – Francois Ziegler Dec 12 '18 at 21:31
• Thank you very much for the answer, I will check the details. – Francesco Polizzi Dec 13 '18 at 6:32 | 2019-01-22T22:05:15 | {
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https://www.jobilize.com/online/course/2-4-exponents-basic-properties-of-real-numbers-by-openstax?qcr=www.quizover.com&page=1 | # 2.4 Exponents (Page 2/2)
Page 2 / 2
${\left(8x\right)}^{3}$ means $\left(8x\right)\left(8x\right)\left(8x\right)$ since the parentheses indicate that the exponent 3 is directly connected to the factor $8x$ . Remember that the grouping symbols indicate that the quantities inside are to be considered as one single number.
$34{\left(a+1\right)}^{2}$ means $34\cdot \left(a+1\right)\left(a+1\right)$ since the exponent 2 applies only to the factor $\left(a+1\right)$ .
## Practice set b
Write each of the following without exponents.
$4{a}^{3}$
$4aaa$
${\left(4a\right)}^{3}$
$\left(4a\right)\left(4a\right)\left(4a\right)$
## Sample set c
Select a number to show that ${\left(2x\right)}^{2}$ is not always equal to $2{x}^{2}$ .
Suppose we choose $x$ to be 5. Consider both ${\left(2x\right)}^{2}$ and $2{x}^{2}$ .
$\begin{array}{lll}{\left(2x\right)}^{2}\hfill & \hfill & 2{x}^{2}\hfill \\ {\left(2\cdot 5\right)}^{2}\hfill & \hfill & 2\cdot {5}^{2}\hfill \\ {\left(10\right)}^{2}\hfill & \hfill & 2\cdot 25\hfill \\ 100\hfill & \ne \hfill & 50\hfill \end{array}$
Notice that ${\left(2x\right)}^{2}=2{x}^{2}$ only when $x=0$ .
## Practice set c
Select a number to show that ${\left(5x\right)}^{2}$ is not always equal to $5{x}^{2}$ .
Select $x=3$ . Then ${\left(5\cdot 3\right)}^{2}={\left(15\right)}^{2}=225$ , but $5\cdot {3}^{2}=5\cdot 9=45$ . $225\ne 45$ .
In ${x}^{n}$ ,
## Base
$x$ is the base
## Exponent
$n$ is the exponent
## Power
The number represented by ${x}^{n}$ is called a power .
## $x$ To the $n$ Th power
The term ${x}^{n}$ is read as " $x$ to the $n$ th power," or more simply as " $x$ to the $n$ th."
## $x$ Squared and $x$ Cubed
The symbol ${x}^{2}$ is often read as " $x$ squared," and ${x}^{3}$ is often read as " $x$ cubed." A natural question is "Why are geometric terms appearing in the exponent expression?" The answer for ${x}^{3}$ is this: ${x}^{3}$ means $x\cdot x\cdot x$ . In geometry, the volume of a rectangular box is found by multiplying the length by the width by the depth. A cube has the same length on each side. If we represent this length by the letter $x$ then the volume of the cube is $x\cdot x\cdot x$ , which, of course, is described by ${x}^{3}$ . (Can you think of why ${x}^{2}$ is read as $x$ squared?)
Cube with
length $=x$
width $=x$
depth $=x$
Volume $=xxx={x}^{3}$
## The order of operations
In Section [link] we were introduced to the order of operations. It was noted that we would insert another operation before multiplication and division. We can do that now.
## The order of operations
1. Perform all operations inside grouping symbols beginning with the innermost set.
2. Perform all exponential operations as you come to them, moving left-to-right.
3. Perform all multiplications and divisions as you come to them, moving left-to-right.
4. Perform all additions and subtractions as you come to them, moving left-to-right.
## Sample set d
Use the order of operations to simplify each of the following.
${2}^{2}+5=4+5=9$
${5}^{2}+{3}^{2}+10=25+9+10=44$
$\begin{array}{ll}{2}^{2}+\left(5\right)\left(8\right)-1\hfill & =4+\left(5\right)\left(8\right)-1\hfill \\ \hfill & =4+40-1\hfill \\ \hfill & =43\hfill \end{array}$
$\begin{array}{ll}7\cdot 6-{4}^{2}+{1}^{5}\hfill & =7\cdot 6-16+1\hfill \\ \hfill & =42-16+1\hfill \\ \hfill & =27\hfill \end{array}$
$\begin{array}{ll}{\left(2+3\right)}^{3}+{7}^{2}-3{\left(4+1\right)}^{2}\hfill & ={\left(5\right)}^{3}+{7}^{2}-3{\left(5\right)}^{2}\hfill \\ \hfill & =125+49-3\left(25\right)\hfill \\ \hfill & =125+49-75\hfill \\ \hfill & =99\hfill \end{array}$
$\begin{array}{ll}{\left[4{\left(6+2\right)}^{3}\right]}^{2}\hfill & ={\left[4{\left(8\right)}^{3}\right]}^{2}\hfill \\ \hfill & ={\left[4\left(512\right)\right]}^{2}\hfill \\ \hfill & ={\left[2048\right]}^{2}\hfill \\ \hfill & =4,194,304\hfill \end{array}$
$\begin{array}{ll}6\left({3}^{2}+{2}^{2}\right)+{4}^{2}\hfill & =6\left(9+4\right)+{4}^{2}\hfill \\ \hfill & =6\left(13\right)+{4}^{2}\hfill \\ \hfill & =6\left(13\right)+16\hfill \\ \hfill & =78+16\hfill \\ \hfill & =94\hfill \end{array}$
$\begin{array}{ll}\frac{{6}^{2}+{2}^{2}}{{4}^{2}+6\cdot {2}^{2}}+\frac{{1}^{3}+{8}^{2}}{{10}^{2}-\left(19\right)\left(5\right)}\hfill & =\frac{36+4}{16+6\cdot 4}+\frac{1+64}{100-95}\hfill \\ \hfill & =\frac{36+4}{16+24}+\frac{1+64}{100-95}\hfill \\ \hfill & =\frac{40}{40}+\frac{65}{5}\hfill \\ \hfill & =1+13\hfill \\ \hfill & =14\hfill \end{array}$
## Practice set d
Use the order of operations to simplify the following.
${3}^{2}+4\cdot 5$
29
${2}^{3}+{3}^{3}-8\cdot 4$
3
${1}^{4}+{\left({2}^{2}+4\right)}^{2}÷{2}^{3}$
9
${\left[6\left(10-{2}^{3}\right)\right]}^{2}-{10}^{2}-{6}^{2}$
8
$\frac{{5}^{2}+{6}^{2}-10}{1+{4}^{2}}+\frac{{0}^{4}-{0}^{5}}{{7}^{2}-6\cdot {2}^{3}}$
3
## Exercises
For the following problems, write each of the quantities using exponential notation.
$b$ to the fourth
${b}^{4}$
$a$ squared
$x$ to the eighth
${x}^{8}$
$\left(-3\right)$ cubed
5 times $s$ squared
$5{s}^{2}$
3 squared times $y$ to the fifth
$a$ cubed minus $\left(b+7\right)$ squared
${a}^{3}-{\left(b+7\right)}^{2}$
$\left(21-x\right)$ cubed plus $\left(x+5\right)$ to the seventh
$xxxxx$
${x}^{5}$
$\left(8\right)\left(8\right)xxxx$
$2\cdot 3\cdot 3\cdot 3\cdot 3xxyyyyy$
$2\left({3}^{4}\right){x}^{2}{y}^{5}$
$2\cdot 2\cdot 5\cdot 6\cdot 6\cdot 6xyyzzzwwww$
$7xx\left(a+8\right)\left(a+8\right)$
$7{x}^{2}{\left(a+8\right)}^{2}$
$10xyy\left(c+5\right)\left(c+5\right)\left(c+5\right)$
$4x4x4x4x4x$
${\left(4x\right)}^{5}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}{4}^{5}{x}^{5}$
$\left(9a\right)\left(9a\right)\left(9a\right)\left(9a\right)$
$\left(-7\right)\left(-7\right)\left(-7\right)aabbba\left(-7\right)baab$
${\left(-7\right)}^{4}{a}^{5}{b}^{5}$
$\left(a-10\right)\left(a-10\right)\left(a+10\right)$
$\left(z+w\right)\left(z+w\right)\left(z+w\right)\left(z-w\right)\left(z-w\right)$
${\left(z+w\right)}^{3}{\left(z-w\right)}^{2}$
$\left(2y\right)\left(2y\right)2y2y$
$3xyxxy-\left(x+1\right)\left(x+1\right)\left(x+1\right)$
$3{x}^{3}{y}^{2}-{\left(x+1\right)}^{3}$
For the following problems, expand the quantities so that no exponents appear.
${4}^{3}$
${6}^{2}$
$6\text{\hspace{0.17em}}·\text{\hspace{0.17em}}6$
${7}^{3}{y}^{2}$
$8{x}^{3}{y}^{2}$
$8\text{\hspace{0.17em}}·\text{\hspace{0.17em}}x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}y\text{\hspace{0.17em}}·\text{\hspace{0.17em}}y$
${\left(18{x}^{2}{y}^{4}\right)}^{2}$
${\left(9{a}^{3}{b}^{2}\right)}^{3}$
$\left(9aaabb\right)\left(9aaabb\right)\left(9aaabb\right)\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}9\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9\text{\hspace{0.17em}}·\text{\hspace{0.17em}}9aaaaaaaaabbbbbb$
$5{x}^{2}{\left(2{y}^{3}\right)}^{3}$
$10{a}^{3}{b}^{2}{\left(3c\right)}^{2}$
$10aaabb\left(3c\right)\left(3c\right)\text{\hspace{0.17em}or}\text{\hspace{0.17em}}10\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3aaabbcc$
${\left(a+10\right)}^{2}{\left({a}^{2}+10\right)}^{2}$
$\left({x}^{2}-{y}^{2}\right)\left({x}^{2}+{y}^{2}\right)$
$\left(xx-yy\right)\left(xx+yy\right)$
For the following problems, select a number (or numbers) to show that
${\left(5x\right)}^{2}$ is not generally equal to $5{x}^{2}$ .
${\left(7x\right)}^{2}$ is not generally equal to $7{x}^{2}$ .
Select $x=2.$ Then, $196\ne 28.$
${\left(a+b\right)}^{2}$ is not generally equal to ${a}^{2}+{b}^{2}$ .
For what real number is ${\left(6a\right)}^{2}$ equal to $6{a}^{2}$ ?
zero
For what real numbers, $a$ and $b$ , is ${\left(a+b\right)}^{2}$ equal to ${a}^{2}+{b}^{2}$ ?
Use the order of operations to simplify the quantities for the following problems.
${3}^{2}+7$
16
${4}^{3}-18$
${5}^{2}+2\left(40\right)$
105
${8}^{2}+3+5\left(2+7\right)$
${2}^{5}+3\left(8+1\right)$
59
${3}^{4}+{2}^{4}{\left(1+5\right)}^{3}$
$\left({6}^{2}-{4}^{2}\right)÷5$
4
${2}^{2}\left(10-{2}^{3}\right)$
$\left({3}^{4}-{4}^{3}\right)÷17$
1
${\left(4+3\right)}^{2}+1÷\left(2\cdot 5\right)$
${\left({2}^{4}+{2}^{5}-{2}^{3}\cdot 5\right)}^{2}÷{4}^{2}$
4
${1}^{6}+{0}^{8}+{5}^{2}{\left(2+8\right)}^{3}$
$\left(7\right)\left(16\right)-{9}^{2}+4\left({1}^{1}+{3}^{2}\right)$
71
$\frac{{2}^{3}-7}{{5}^{2}}$
$\frac{{\left(1+6\right)}^{2}+2}{19}$
$\frac{51}{19}$
$\frac{{6}^{2}-1}{5}+\frac{{4}^{3}+\left(2\right)\left(3\right)}{10}$
$\frac{5\left[{8}^{2}-9\left(6\right)\right]}{{2}^{5}-7}+\frac{{7}^{2}-{4}^{2}}{{2}^{4}-5}$
5
$\frac{{\left(2+1\right)}^{3}+{2}^{3}+{1}^{3}}{{6}^{2}}-\frac{{15}^{2}-{\left[2\left(5\right)\right]}^{2}}{5\cdot {5}^{2}}$
$\frac{{6}^{3}-2\cdot {10}^{2}}{{2}^{2}}+\frac{18\left({2}^{3}+{7}^{2}\right)}{2\left(19\right)-{3}^{3}}$
$\frac{1070}{11}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}97.\overline{27}$
## Exercises for review
( [link] ) Use algebraic notation to write the statement "a number divided by eight, plus five, is equal to ten."
( [link] ) Draw a number line that extends from $-5$ to 5 and place points at all real numbers that are strictly greater than $-3$ but less than or equal to 2.
( [link] ) Is every integer a whole number?
( [link] ) Use the commutative property of multiplication to write a number equal to the number $yx$ .
$xy$
( [link] ) Use the distributive property to expand $3\left(x+6\right)$ .
how can chip be made from sand
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
how did you get the value of 2000N.What calculations are needed to arrive at it
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http://www.chegg.com/homework-help/questions-and-answers/isosceles-triangle-inscribed-semicircle-shown-diagram-continues-inscribed-semicircle-chang-q1095383 | An isosceles triangle is inscribed in a semicircle, as shown in the diagram, and it continues to be inscribed as the semicircle changes size. the area of the semicircle is increasing at the rate of 1 cm^2/sec when the radius of the semi circle is 3cm.
a. How fast is the radius of the semicircle increasing when the radius is 3cm? Include units in answer.
b. How fast is the perimeter of the semicircle increasing when the radius is 3cm? Include units.
c. How fast is the area of the isosceles triangle increasing when the radius is 3 cm? Include units.
d. How fast is the shaded region increasing when the radius is 3cm? Include units. | 2015-05-06T15:49:27 | {
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http://www.physicsforums.com/showthread.php?t=613693 | ## Integration with a fraction
Can someone please help me with this problem. Please see the attached image.
The integration when removing the min and max is also provided, how is the answer x2/2 - x3/6? Do we not do anything with the 1 - 1/2 part? and the +2 part? Are we only looking at (x2-2x)/2?
I appreciate the help.
Attached Thumbnails
PhysOrg.com science news on PhysOrg.com >> King Richard III found in 'untidy lozenge-shaped grave'>> Google Drive sports new view and scan enhancements>> Researcher admits mistakes in stem cell study
Recognitions: Science Advisor The integrand is x - x2/2.
Recognitions: Gold Member Science Advisor Staff Emeritus $$-\frac{1}{2}\left(x^2- 2x+ 2\right)= -\frac{1}{2}x^2+ x- 1$$ Adding 1 to that gives the integrand $$-\frac{1}{2}x^2+ x$$ as mathman says.
## Integration with a fraction
Thanks, that is exactly what I needed to know.
Thread Tools
Similar Threads for: Integration with a fraction Thread Forum Replies Calculus & Beyond Homework 1 Calculus & Beyond Homework 4 Calculus & Beyond Homework 6 Calculus & Beyond Homework 3 Calculus & Beyond Homework 3 | 2013-05-24T22:23:47 | {
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http://mathhelpforum.com/algebra/208012-rational-expression-two-variables.html | # Thread: Rational Expression With Two Variables
1. ## Rational Expression With Two Variables
I'm a college student teaching himself algebra (loooong story) to save money on classes. I've done well so far, just finished up polynomials and now moving on to rational expressions. I've got the basics of factoring and writing in simplest terms down, but my book threw me a curveball:
Write in simplest terms:
9r2 - 4s2
-------
9r + 6s
I've factored it out to:
3r(3 + r) - 2s(2 + s)
----------
3(3 + r) + 3(2 + s)
Crossing out like terms, this leaves me with
3r-2s
-------
6
However, the book says the denominator should be 3, not 6. Knowing the answer doesn't do me any good if I don't understand how to get there. Where am I going wrong in solving this?
Thanks!
2. ## Re: Rational Expression With Two Variables
We are given to simplify:
$\frac{9r^2-4s^2}{9r+6s}$
Now, factor first, using a difference of squares for the numerator:
$\frac{(3r+2s)(3r-2s)}{3(3r+2s)}$
Cancel or divide out common factors to get:
$\frac{3r-2s}{3}$
3. ## Re: Rational Expression With Two Variables
Wow, I just got done covering factoring too. Thanks so much! | 2017-09-24T16:30:01 | {
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http://mathoverflow.net/questions/130687/examples-of-polynomial-rings-ax-with-relatively-large-krull-dimension | # Examples of polynomial rings $A[x]$ with relatively large Krull dimension
If $A$ is a commutative ring we have the estimate $$\dim (A)+1 \le \dim (A[x])\le 2\dim (A)+1$$ for the Krull dimension, with $\dim (A)+1 = \dim (A[x])$ for Noetherian rings. I am looking for nice examples of rings $A$ so that $A[x]$ has Krull dimension $\dim (A)+2, \dim(A)+3,\ldots ,2\dim(A)+1$.
- | 2015-05-25T21:54:10 | {
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https://www.esaral.com/q/a-line-is-drawn-in-the-direction-of-48184 | # A line is drawn in the direction of
Question:
A line is drawn in the direction of $(\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})$ and it passes through a point with position vector $(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-4 \hat{\mathrm{k}}) .$ Find the equations of the line in the vector as well as Cartesian forms.
Solution: | 2023-03-21T04:27:49 | {
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http://mathhelpforum.com/calculus/50559-implicit-differentiation-w-sin-x.html | # Math Help - Implicit Differentiation w/ sin(x)
1. ## Implicit Differentiation w/ sin(x)
Hey, so I'm having trouble differentiating this problem.....
ysin[x(squared)]=xsin[y(squared)]
any help would be appreciated
2. Originally Posted by CALsunshine
Hey, so I'm having trouble differentiating this problem.....
ysin[x(squared)]=xsin[y(squared)]
any help would be appreciated
$y\sin\left(x^2\right)=x\sin\left(y^2\right)$
You would need to apply product and chain rules when differentiating both sides!
$\frac{d}{\,dx}\left[y\sin\left(x^2\right)\right]=\frac{d}{\,dx}\left[x\sin\left(y^2\right)\right]$ $\implies 2xy\cos\left(x^2\right)+\sin\left(x^2\right)\frac{ \,dy}{\,dx}=\sin\left(y^2\right)+2xy\cos\left(y^2\ right)\frac{\,dy}{\,dx}$
Can you take it from here and solve for $\frac{\,dy}{\,dx}$?
--Chris
3. yes...I can solve for dy/dx from there, thank you so much. | 2015-11-28T06:37:00 | {
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https://plainmath.net/74910/let-r-be-a-commutative-finite-dime | # Let R be a commutative finite dimensional K -algebra over a field K (for example th
Let $R$ be a commutative finite dimensional $K$-algebra over a field $K$ (for example the monoid ring of a a finite monoid over a field). Assume we have $R$ in GAP. Then we can check whether $R$ is semisimple using the command RadicalOfAlgebra(R). When the value is 0, $R$ is semisimple. Thus $R$ can be written as a finite product of finite field extensions of $K$.
Question: Can we obtain those finite field extensions of $K$ or at least their number and $K$-dimensions using GAP?
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If you are working over a finite field, and can represent the algebra through matrices, you could try to split the regular module into its homogeneous components. This can be done with the GAP command "MTX.Indecomposition". You then take a simple submodule in this component (try a spin of a random vector), and take the action of the homogeneous component on this module (and should let you deduce the appropriate field extension). | 2022-06-25T07:18:10 | {
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https://learn.careers360.com/ncert/question-study-the-diagram-the-line-l-is-perpendicular-to-line-m-a-is-ce-equal-to-eg/ | Q
# Study the diagram. The line l is perpendicular to line m (a) Is CE = EG?
4. Study the diagram. The line $l$ is perpendicular to line $m$
(a) Is CE = EG?
(b) Does PE bisect CG?
(c) Identify any two line segments for which PE is the perpendicular bisector.
(d) Are these true?
(i) AC > FG
(ii) CD = GH
(iii) BC < EH.
Views
(a) CE = 5 - 3 = 2 units
EG = 7 - 5 = 2 units
Therefore CE = EG.
(b) CE = EG therefore PE bisects CG.
(c) PE is the perpendicular bisector for line segments DF and BH
(d) (i) AC = 3 - 1 = 2 units
FG = 7 - 6 = 1 unit
Therefore AC > FG
True
(ii) CD = 4 - 3 = 1 unit
GH = 8 - 7 = 1 unit
Therefore CD = GH
True
(iii) BC = 3 - 2 = 1 unit
EH = 8 - 5 = 3 units
Therefore BC < EH
True
Exams
Articles
Questions | 2020-01-26T15:46:09 | {
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https://math.stackexchange.com/questions/459006/convergence-of-the-alternating-odds-over-evens | # convergence of the alternating odds over evens
I am trying to show $$\sum_{n=0}^\infty (-1)^n\frac{1\cdot3 \cdots (2n-1)}{2\cdot 4 \cdots (2n)}$$ is conditionally convergent. If I can show $a_n \rightarrow 0$ then I can apply alternating series test to show convergence. I am trying to avoid the use of Stirling's formula. I do not know how to attack divergence of $\sum |a_n|$ with out employing Stirling's formula.
• It is elementary to show that $\frac{1}{2\sqrt{n}}\leq |a_n|\leq \frac{1}{\sqrt{2n+1}}$. I do not think this question is about research level mathematics. – abatkai Aug 2 '13 at 21:57
• Specifically, abatkai's inequalities may be proved by an inductive argument. -- Todd Trimble – user43208 Aug 4 '13 at 2:02
We have $$\log |a_n|=\sum_{j=1}^n\log\left(1-\frac 1{2j}\right).$$ Now compare $\log(1-x)$ with $x$ in order to find a below bound for $\log |a_n|$. | 2019-12-11T23:03:28 | {
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http://mathoverflow.net/questions/132062/biholomophic-non-algebraically-isomorphic-varieties/132337 | # Biholomophic non-Algebraically Isomorphic Varieties
Recently, when writing a review for MathSciNet, the following question arose:
Is it true that two smooth complex varieties that are biholomorphic are algebraically isomorphic? The converse is true just because polynomials are holomorphic.
I was mainly interested in the affine case since that was the context of the review I was writing.
I knew about exotic affine spaces, so two smooth complex affine varieties that are real analytically isomorphic do not have to be algebraically isomorphic. But real analytic maps between complex manifolds need not be holomorphic in general (just think of complex conjugation on $\mathbb{C}$). So that is not enough.
I posted the question on my Facebook page, and it got answered:
1. I was reminded that Serre's GAGA Theorem implies that it is true for projective varieties. But there are quasiprojective counterexamples provided on MO. See the answer of Georges Elencwajg given here.
2. Then it was pointed out that the answer in the link above is a manifold which is both affine and non-affine. So what about two affine varieties?
3. I then found a reference for such an example. Two exotic affine spaces that are biholomorphic yet not algebraically isomorphic. See here. This is a very nice result. Since it is in dimension 3, it pretty much shows that at the very first point where something can go wrong, it does. A similar result for exotic affine spheres (apparently due to appear in J of Alg. Geo.) is here.
So I finally come to my questions.
Question 1: Is the example in the reference I found the first example of two affine varieties that are biholomorphic but not algebraically isomorphic?
Question 2: Is there some general reason to believe that all exotic affine spaces should be biholomorphic?
Comment: It seems to me within context to ask similar questions about objects, not only maps between objects. Here is some of what I have learned in that regard: all smooth complex varieties are complex manifolds since they are covered by smooth affine open sets with polynomial (hence holomorphic) transition charts. Conversely, a closed analytic subspace of projective space is algebraic by Chow's Theorem. But there are compact complex manifolds that are not algebraic (see here), which also shows that a closed analytic subspace of affine space need not be algebraic.
-
Question 1: The first example published seems to be the following (Corollary 4 in "Embeddings of Danielewski surfaces", G. Freudenburg and L. Moser-Jauslin, Math.Z. 2003 ) For any $a\in \mathbb{C}^*$, the surfaces in $\mathbb{C}^3$ given by $x^2z-y^2-a$ and $x^2z-(1+x)y^2-a$ are algebraically not isomorphic, but holomorphically isomorphic.
Question 2: The answer should be no because of the strong analytic cancellation theorem of Zaidenberg (see http://arxiv.org/pdf/alg-geom/9506005v1.pdf page 5). It says that if $X,X'$ are contractible non-biholomorphic surfaces of general-type, then $X\times \mathbb{A}^1$ and $X'\times\mathbb{A}^1$ are not biholomorphic but are both exotic affine spaces. | 2015-04-25T11:18:00 | {
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https://stacks.math.columbia.edu/tag/0H0Y | ## 103.4 Cohomology and the lisse-étale and flat-fppf sites
We have already seen that cohomology of a sheaf on an algebraic stack $\mathcal{X}$ can be computed on flat-fppf site. In this section we prove the same is true for (possibly) unbounded objects of the direct category of $\mathcal{X}$.
Lemma 103.4.1. Let $\mathcal{X}$ be an algebraic stack. We have $Lg_!\mathbf{Z} = \mathbf{Z}$ for either $Lg_!$ as in Lemma 103.3.1 part (1) or $Lg_!$ as in Lemma 103.3.1 part (3).
Proof. We prove this for the comparison between the flat-fppf site with the fppf site; the case of the lisse-étale site is exactly the same. We have to show that $H^ i(Lg_!\mathbf{Z})$ is $0$ for $i \not= 0$ and that the canonical map $H^0(Lg_!\mathbf{Z}) \to \mathbf{Z}$ is an isomorphism. Let $f : \mathcal{U} \to \mathcal{X}$ be a surjective, flat morphism where $\mathcal{U}$ is a scheme such that $f$ is also locally of finite presentation. (For example, pick a presentation $U \to \mathcal{X}$ and let $\mathcal{U}$ be the algebraic stack corresponding to $U$.) By Sheaves on Stacks, Lemmas 95.19.6 and 95.19.10 it suffices to show that the pullback $f^{-1}H^ i(Lg_!\mathbf{Z})$ is $0$ for $i \not= 0$ and that the pullback $H^0(Lg_!\mathbf{Z}) \to f^{-1}\mathbf{Z}$ is an isomorphism. By Lemma 103.3.2 we find $f^{-1}Lg_!\mathbf{Z} = L(g')_!\mathbf{Z}$ where $g' : \mathop{\mathit{Sh}}\nolimits (\mathcal{U}_{flat, fppf}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{U}_{fppf})$ is the corresponding comparision morphism for $\mathcal{U}$. This reduces us to the case studied in the next paragraph.
Assume $\mathcal{X} = (\mathit{Sch}/X)_{fppf}$ for some scheme $X$. In this case the category $\mathcal{X}_{flat, fppf}$ has a final object $e$, namely $X/X$, and moreover the functor $u : \mathcal{X}_{flat, fppf} \to \mathcal{X}_{fppf}$ sends $e$ to the final object. Since $\mathbf{Z}$ is the free abelian sheaf on the final object (provided the final object exists) we find that $Lg_!\mathbf{Z} = \mathbf{Z}$ by the very construction of $Lg_!$ in Cohomology on Sites, Lemma 21.37.2. $\square$
Lemma 103.4.2. Let $\mathcal{X}$ be an algebraic stack. Notation as in Lemma 103.3.1.
1. For $K$ in $D(\mathcal{X}_{\acute{e}tale})$ we have
1. $R\Gamma (\mathcal{X}_{\acute{e}tale}, K) = R\Gamma (\mathcal{X}_{lisse,{\acute{e}tale}}, g^{-1}K)$, and
2. $R\Gamma (x, K) = R\Gamma (\mathcal{X}_{lisse,{\acute{e}tale}}/x, g^{-1}K)$ for any object $x$ of $\mathcal{X}_{lisse,{\acute{e}tale}}$.
2. For $K$ in $D(\mathcal{X}_{fppf})$ we have
1. $R\Gamma (\mathcal{X}_{fppf}, K) = R\Gamma (\mathcal{X}_{flat,fppf}, g^{-1}K)$, and
2. $H^ p(x, K) = R\Gamma (\mathcal{X}_{flat,fppf}/x, g^{-1}K)$ for any object $x$ of $\mathcal{X}_{flat,fppf}$.
In both cases, the same holds for modules, since we have $g^{-1} = g^*$ and there is no difference in computing cohomology by Cohomology on Sites, Lemma 21.20.7.
Proof. We prove this for the comparison between the flat-fppf site with the fppf site; the case of the lisse-étale site is exactly the same. By Lemma 103.4.1 we have $Lg_!\mathbf{Z} = \mathbf{Z}$. Then we obtain
\begin{align*} R\Gamma (\mathcal{X}_{fppf}, K) & = R\mathop{\mathrm{Hom}}\nolimits (\mathbf{Z}, K) \\ & = R\mathop{\mathrm{Hom}}\nolimits (Lg_!\mathbf{Z}, K) \\ & = R\mathop{\mathrm{Hom}}\nolimits (\mathbf{Z}, g^{-1}K) \\ & = R\Gamma (\mathcal{X}_{lisse,{\acute{e}tale}}, g^{-1}K) \end{align*}
This proves (1)(a). Part (1)(b) follows from part (1)(a). Namely, if $x$ lies over the scheme $U$, then the site $\mathcal{X}_{\acute{e}tale}/x$ is equivalent to $(\mathit{Sch}/U)_{\acute{e}tale}$ and $\mathcal{X}_{lisse,{\acute{e}tale}}$ is equivalent to $U_{lisse, {\acute{e}tale}}$. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 2022-12-09T15:35:28 | {
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https://stacks.math.columbia.edu/tag/02JY | [Expose VIII, Corollaire 4.3, SGA1] and [IV, Corollaire 2.3.12, EGA]
Lemma 29.25.12. Let $f : X \to Y$ be a quasi-compact, surjective, flat morphism. A subset $T \subset Y$ is open (resp. closed) if and only $f^{-1}(T)$ is open (resp. closed). In other words, $f$ is a submersive morphism.
Proof. The question is local on $Y$, hence we may assume that $Y$ is affine. In this case $X$ is quasi-compact as $f$ is quasi-compact. Write $X = X_1 \cup \ldots \cup X_ n$ as a finite union of affine opens. Then $f' : X' = X_1 \amalg \ldots \amalg X_ n \to Y$ is a surjective flat morphism of affine schemes. Note that for $T \subset Y$ we have $(f')^{-1}(T) = f^{-1}(T) \cap X_1 \amalg \ldots \amalg f^{-1}(T) \cap X_ n$. Hence, $f^{-1}(T)$ is open if and only if $(f')^{-1}(T)$ is open. Thus we may assume both $X$ and $Y$ are affine.
Let $f : \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ be a surjective morphism of affine schemes corresponding to a flat ring map $A \to B$. Suppose that $f^{-1}(T)$ is closed, say $f^{-1}(T) = V(J)$ for $J \subset B$ an ideal. Then $T = f(f^{-1}(T)) = f(V(J))$ is the image of $\mathop{\mathrm{Spec}}(B/J) \to \mathop{\mathrm{Spec}}(A)$ (here we use that $f$ is surjective). On the other hand, generalizations lift along $f$ (Lemma 29.25.9). Hence by Topology, Lemma 5.19.6 we see that $Y \setminus T = f(X \setminus f^{-1}(T))$ is stable under generalization. Hence $T$ is stable under specialization (Topology, Lemma 5.19.2). Thus $T$ is closed by Algebra, Lemma 10.41.5. $\square$
## Comments (4)
Comment #2873 by Ko Aoki on
Typo in the proof: The orientation of the morphism between $\mathop{\rm Spec} A$ and $\mathop{\rm Spec} B$ is reversed halfway.
Comment #3546 by on
Some references: [SGA1, Exposé VIII, Corollaire 4.3] and [EGAIV$_2$, Corollaire 2.3.12].
There are also:
• 4 comment(s) on Section 29.25: Flat morphisms
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In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 02JY. Beware of the difference between the letter 'O' and the digit '0'. | 2021-02-25T08:34:41 | {
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https://math.stackexchange.com/questions/linked/93463 | 3k views
### Difference between fields $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$ and $\mathbb{Q}[\sqrt{2},\sqrt{3}]$? [duplicate]
Possible Duplicate: Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$? How would one describe elements from $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$ and $\mathbb{Q}[\sqrt{2},\sqrt{3}]$? ...
1k views
563 views
### $\Bbb Q [ \sqrt{2} + \sqrt{3} ] = \Bbb Q [ \sqrt{2} , \sqrt{3} ]$ [duplicate]
Prove, that $\Bbb Q [ \sqrt{2} + \sqrt{3} ] = \Bbb Q [ \sqrt{2} , \sqrt{3} ]$ I don't know the definition of $\Bbb Q [ \sqrt{2} , \sqrt{3} ]$, can anyone help me with this?
125 views
### How can I show this field extension equality? [duplicate]
How can I show this field extension equality $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$?
327 views
### Primitive element of the extension $\mathbb Q(\sqrt{2},\sqrt{3})$ over $\mathbb Q$ [duplicate]
The title says it. I want to find an element $\alpha$ such that $\mathbb Q(\alpha)=\mathbb Q(\sqrt{2},\sqrt{3})$. I tried something like $\sqrt{2}+\sqrt{3}$ but that didn't help...
188 views
### Why is $\mathbb Q(\sqrt 2+\sqrt 3)=\mathbb Q(\sqrt2,\sqrt 3)$? [duplicate]
Why is $\mathbb Q(\sqrt 2+\sqrt 3)=\mathbb Q(\sqrt2,\sqrt 3)$ ? I am Having problems understanding why this is true. Any input would be greatly appreciated! | 2019-08-19T13:21:08 | {
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https://asima-online.com/qa/quick-answer-what-is-equal-to-tanth.html | # Quick Answer: What Is Equal To Tanθ?
## What is tangent in sin and cos?
sin cos and tan are basically just functions that relate an angle with a ratio of two sides in a right triangle.
And tan is opposite over adjacent, which means tan is sin/cos..
## What is the tangent of a function?
A tangent line to the function f(x) at the point x=a is a line that just touches the graph of the function at the point in question and is “parallel” (in some way) to the graph at that point.
## What is equal to tan theta?
Tangent theta equals the side opposite theta divided by the side adjacent to theta. So this is theta. … The tangent of theta is defined to be y over x where y and x are these coordinates. So the second coordinate divided by the first coordinate, that’s the tangent of theta.
## What is trigonometry formula?
Basic Formulas By using a right-angled triangle as a reference, the trigonometric functions or identities are derived: sin θ = Opposite Side/Hypotenuse. cos θ = Adjacent Side/Hypotenuse. tan θ = Opposite Side/Adjacent Side.
## What is the formula of 2 tan theta?
tan(B/2) = sin(B/2)/cos(B/2) = ± √(1−cos B) / (1+cos B) The half-angle tangent formulas can be summarized like this: (62) tan(B/2) = (1 − cos B) / sin B = sin B / (1 + cos B)
## What are the basics of trigonometry?
There are six trigonometric ratios, sine, cosine, tangent, cosecant, secant and cotangent. These six trigonometric ratios are abbreviated as sin, cos, tan, csc, sec, cot. These are referred to as ratios since they can be expressed in terms of the sides of a right-angled triangle for a specific angle θ.
## What is the formula of sin3x?
\sin 3x =4\sin x\sin(60^{\circ}-x)\sin(60^{\circ}+x). We first remind of another useful trigonometric identity: \displaystyle\sin\alpha + \sin\beta +\sin\gamma -\sin(\alpha +\beta +\gamma)=4\sin\frac{\alpha +\beta}{2}\sin\frac{\beta +\gamma}{2}\sin\frac{\gamma +\alpha}{2}.
## What is sin90 value?
Therefore, sin 90 degree equals to the fractional value of 1/ 1.
## What is equal to Tanθ answer?
Note: The equation cot θ = cot ∝ is equivalent to tan θ = tan ∝ (since, cot θ = 1/tan θ and cot ∝ = 1/tan ∝). Thus, cot θ = cot ∝ and tan θ = tan ∝ have the same general solution. Hence, the general solution of cot θ = cot ∝ is θ = nπ + ∝, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)
## What’s the value of tan 30?
0.57735The exact value of tan 30° is 0.57735. The value of tangent of angle 30 degrees can also be evaluated using the values of sin 30 degrees and cos 30 degrees.
## What sine means?
In mathematics, the sine is a trigonometric function of an angle. The sine of an acute angle is defined in the context of a right triangle: for the specified angle, it is the ratio of the length of the side that is opposite that angle, to the length of the longest side of the triangle (the hypotenuse).
## What is tangent equal to?
The tangent of x is defined to be its sine divided by its cosine: tan x = sin x cos x . The cotangent of x is defined to be the cosine of x divided by the sine of x: cot x = cos x sin x .
## What is tan 2x equal to?
tan x/2 = sin x/ (1 + cos x) 1st easy equation tan x/2 = (1 – cos x) /sin x 2nd easy equation.
## Who is father of trigonometry?
Hipparchus of NicaeaHipparchus of Nicaea (/hɪˈpɑːrkəs/; Greek: Ἵππαρχος, Hipparkhos; c. 190 – c. 120 BC) was a Greek astronomer, geographer, and mathematician. He is considered the founder of trigonometry but is most famous for his incidental discovery of precession of the equinoxes.
## Why is tan 90 undefined?
At 90 degrees we must say that the tangent is undefined (und), because when you divide the leg opposite by the leg adjacent you cannot divide by zero. In the third quadrant the hypotenuse extended will now meet the tangent line above the x-axis and is now positive again.
## What is tan Tita?
The Tan Θ is the ratio of the Opposite side to the Adjacent, where (Θ) is one of the acute angles. Tan \Theta = \frac{Opposite Side}{Adjacent Side}
## What is the formula of tan 3 Theta?
Tan 3 theta = 3 tan theta – tan 3 theta / 1 – 3 tan2 theta. Where tan is a tangent function and theta is an angle. | 2021-09-24T12:42:38 | {
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http://math.stackexchange.com/questions/123098/finding-the-position-of-a-point-after-rotation-why-is-my-result-incorrect | # Finding the position of a point after rotation: Why is my result incorrect
I am attempting to calculate the position of a point after it has been rotated
I have been using an algorithm but I am getting incorrect values which makes me think I am using the incorrect algorithm or incorrect values.
What algorithm and values should I use to find the position of the green dot?
My calculation is here:
x = -6
y = 349
0 = -25
x' = x*cos(0) - y*sin(0)
y' = x*sin(0) + y*cos(0)
x' = (-5.44) - (-147.49)
y' = (2.54) + (316.30)
x' = 142.05
y' = 318.84
// Isn't that really wrong shdn't it be around 0,349 for the answer?
-
Your equations for $x', y'$ are for rotation around the origin, but your drawing has the rotation around the center of the square which appears to be about 150 units away from the red dot. If the center of the square is at $(x_c,y_c)$ you should have $x'=(x-x_c)\cos \theta +x_c - (y-y_c)\sin \theta, y'=(y-y_c)\cos \theta +y_c + (x-x_c) \sin \theta$. In your drawing, the square looks much smaller than 100 units in size. | 2014-12-22T09:58:15 | {
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http://icpc.njust.edu.cn/Problem/Zju/4011/ | # Happy Sequence
Time Limit: 3 Seconds
Memory Limit: 65536 KB
## Description
A sequence of $k$ integers $b_1, b_2, \dots, b_k$ ($1 \le b_1 \le b_2 \le ... \le b_k \le n$) is called a happy sequence if each number divides (without a remainder) the next number in the sequence. More formally, we can say $b_i | b_{i+1}$ for all $1 \le i \le k-1$, or we can say $b_{i+1} \text{ mod } b_i = 0$ for all $1 \le i \le k-1$.
Given $n$ and $m$, find the number of happy sequences of length $m$. Two sequences $x_1, x_2, \dots, x_m$ and $y_1, y_2, \dots, y_m$ are different, if and only if there exists an $i$ such that $1 \le i \le m$ and $x_i \ne y_i$.
As the answer can be rather large print it modulo $1000000007$ ($10^9 + 7$).
## Input
There are multiple test cases. The first line of the input contains an integer $T$ (about 50), indicating the number of test cases. For each test case:
The first and only line contains two integers $n$ and $m$ ($1 \le n, m \le 2000$), indicating the upper limit of the elements in the sequence and the length of the sequence.
## Output
For each case output a single integer, indicating the number of happy sequences of length $m$ modulo $10^9+7$
## Sample Input
1
3 2
## Sample Output
5
## Hint
In the sample test case, the happy sequences are: $[1, 1]$, $[2, 2]$, $[3, 3]$, $[1, 2]$, $[1, 3]$.
None | 2020-10-22T20:49:42 | {
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https://www.numerade.com/questions/which-of-the-following-four-planes-are-parallel-are-any-of-them-identical-p_1-3x-6y-3z-6-p_2-4x-12y-/ | 💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!
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# Which of the following four planes are parallel? Are any of them identical?$P_1 : 3x + 6y - 3z = 6$ $P_2 : 4x - 12y + 8z = 5$$P_3 : 9y = 1 + 3x + 6z$ $P_4 : z = x + 2y - 2$
## $P_{2}$ and $P_{3}$ are parallel, $P_{1}$ and $P_{4}$ are identical
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### Video Transcript
in the question they're asking that each of the planes P one P two, P three before our marlin. Yeah. And if they are panel then which of them are identical. Yeah. So which Everyone is given by the equation three explicit XY -3 is equal to six. P two is 4 X- to invite us. It is equal to five. History is nine by equal to one plus three plus six and before is equal to zero is equal to express to one minus two. So in order to find the planes are parallel or not we have to find the ratio between the coordinate coordinates coefficients of the plane. So first We compare the values of P one and P two checking if they're panel or not. So the coefficients of the coordinates are P one is given by the equation So P one and P two is given by the equations. And so the coefficients of the coordinates are yeah computed in the form of ratio in order to find whether they are paddle or not. So three by four X coordinate is equal to six by minus two. Report by coordinate is equal to minus three by eight. Or that coordinate if we simplify this then it will be equal to three by four is not equal to minus half Is not equal to -3 x eight. Therefore Clearly P one and P two are not identical but are not parallel. Hence not identical. So this is the first case. Next we have to check for the value for the planes P one and beach three which is given by the equations. So the equations of P one and P three planes are this. And so for checking whether these are parallel or not, we have to compare the coefficients of the coordinates which is equal to three by 34. X coordinate is equal to six by minus nine. For the wife coordinate is equal to minus three by six. For the that coordinate that is equal to after simplifying we get the value as one is not equal to -2 x three is not equal to minus half. Therefore P one and P three R not parallel. Hence these are also not identical. Now in case three we have to check for the plains even and before. So the equations of planes, people and before are these therefore have to compare the coefficients of the coordinates that is equal to three by one or x coordinate equal to six by two. For y coordinate is equal to minus three by minus 1% coordinates. So after simplifying these ratios we get the value as equal to three equal to three equal to three. Therefore B one. And before our Berlin planes. And uh checking further whether whether they are identical or not. So we put the value of X equal to zero And by equal to zero. To find the value of said. So the equation for P one plane is three explicit Xy minus three is equal to six. Therefore The value of Z. Here is equal to -2. Therefore the point for this plane That is a point containing this plane is 00 -2. Similarly for the plane before the equation is express to why miners said is equal to um two. Therefore If we put the value of x equal to zero and by equal to zero then we get the value of that is -2. Therefore the point passing should this plane Is equal to 00 -2. Therefore B one and before are bought parlance and identical. Next case we consider for The plane Speed two and Petri. So the equations of planes P two and P three are these. So in order to check whether their parents are not here to compare the coefficients of the coordinates that is equal to four by 34 X coordinate equal to minus 12 and minus 94 by coordinate is equal to eight. Basics for the z coordinate. This is equal doctor simplifying these ratios. We get yeah the sequel to four bytes required four bytes required to four bytes. Three. Therefore see two and P three R. Berlin. And in order to check whether these planes are yeah identical or not let X equal to buy equal to zero. Therefore for the P two plane that will be equal to yeah five by eight. And uh So the point to this plane will be 005 x eight and same condition for the or plain Petri. So here there will be equal to -1 x six. So the point through this plane will be 00 -1 Way six. This is the value of five x 8 is equal to point six two. And the value of the that putting it off Point in Petri is -1. Basics. That is equal do minus point. Yeah. 16. Therefore the Point in P two is not equal to point in P three. Therefore P two and P three are not identical. The next case is you have to compare P two and P four. Yeah. So the equations of the plain speech win before are these? And therefore comparing these coefficients for the coordinates of these two planes to check whether their palates or not is equal to fall by one. For the x coordinate equal to minus 12 way too. For the y coordinate equal to eight by minus one for the z coordinate After simplifying This becomes equal to four is not equal to -6 is not equal to -8. Therefore B two and before I not yeah, Palin hence they are not identical also. Yeah. And uh last possibility is to check For P three and P four. So the equations of Patreon before our so the equations of P three and P four planes are these? Therefore we have to take whether these planes are talent or not. So comparing the ratios of the coordinates of these two planes is equal to three by one for X coordinate equal to minus nine by 24 by coordinated quarters six by -1 part. Is that fortunate? So these three are not equal hands P three and before are not talent, and hence not identical. Also, therefore, the planes that are pal an identical art. Therefore P two P 3 planes are Palin and not identical and before and people are both paddles and identical to each other. So this is the required answer of the given question. | 2021-10-16T21:31:33 | {
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http://mathhelpforum.com/trigonometry/214875-trigonometry.html | Math Help - Trigonometry
1. Trigonometry
can u help with these 2 exercises?: 1) which are the solutions to the equation tanx=2cotx,in the interval -180<x<180?...2) Which is the value of arcsin1/2-arcsin(-1/2).. i will be glad if you could help me
2. Re: Trigonometry
\displaystyle \begin{align*} \tan{(x)} &= 2\cot{(x)} \\ \frac{\sin{(x)}}{\cos{(x)}} &= \frac{2\cos{(x)}}{\sin{(x)}} \\ \sin^2{(x)} &= 2\cos^2{(x)} \\ 1 - \cos^2{(x)} &= 2\cos^2{(x)} \\ 1 &= 3\cos^2{(x)} \\ \frac{1}{3} &= \cos^2{(x)} \\ \pm \frac{1}{\sqrt{3}} &= \cos{(x)} \\ x &= \left\{ \arccos{\left( \frac{1}{\sqrt{3}} \right)}, \pi - \arccos{\left( \frac{1}{\sqrt{3}} \right)}, \pi + \arccos{\left( \frac{1}{\sqrt{3}} \right)} , 2\pi - \arccos{\left( \frac{1}{\sqrt{3}} \right)} \right\} + 2\pi \, n \textrm{ where } n \in \mathbf{Z} \\ x &= \left\{ \arccos{\left( \frac{1}{\sqrt{3}} \right)} - \pi , -\arccos{\left( \frac{1}{\sqrt{3}} \right)} , \arccos{\left( \frac{1}{\sqrt{3}} \right)} , \pi - \arccos{\left( \frac{1}{\sqrt{3}} \right)} \right\} \textrm{ in the interval specified} \end{align*}
Edit: I naturally put the answer in radians when it should have been degrees. You can fix that
3. Re: Trigonometry
Hello, endri!
$\text{1) What are the solutions to the equation: }\,\tan x\,=\,2\cot x,\,\text{ on the interval }(\text{-}180^o,180^o)$
We have: . $\tan x \:=\:\frac{2}{\tan x} \quad\Rightarrow\quad \tan^2x \:=\:2 \quad\Rightarrow\quad \tan x \:=\:\pm\sqrt{2}$
Therefore: . $x \;\approx\;\{\text{-}125.3^o,\:\text{-}54.7^o,\:54.7^o,\:125.3^o\}$
$\text{2) What is the value of: }\,\arcsin(\tfrac{1}{2})-\arcsin(\text{-}\tfrac{1}{2})$
Using principal values: . $\begin{Bmatrix}\arcsin(\frac{1}{2}) &=& \frac{\pi}{6} \\ \\[-4mm] \arcsin(\text{-}\frac{1}{2}) &=& \text{-}\frac{\pi}{6} \end{Bmatrix}$
Therefore: . $\arcsin(\tfrac{1}{2}) - \arcsin(\text{-}\tfrac{1}{2}) \;=\;\frac{\pi}{6} - \left(\text{-}\frac{\pi}{6}\right) \;=\;\frac{\pi}{3}$ | 2015-09-02T11:31:18 | {
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https://stats.stackexchange.com/questions/247407/reverse-derivation-of-negative-log-likelihood-cost-function | # Reverse derivation of negative log likelihood cost function
If I understand correctly, the negative log likelihood cost function goes hand-in-hand with the softMax output layer.
But why?
The simplest motivating logic I am aware of goes as follows: softMax outputs (which sum to 1) can be considered as probabilities. Whichever unit corresponds to the actual correct answer, we wish to maximise the probability (or output) for this unit. Which is the same as minimising the neg-log-prob.
So the probability of the network outputting the correct (Target) answer is:
$$P(T) = y_T = {\exp z_T \over \sum_k \exp z_k}$$
So we wish to minimise $-log(P(T))$, which can be rewritten as $log(\sum_k \exp z_k)-z_T$
Now if we calculate the gradient $\partial C \over \partial z_m$ across the (SoftMax) activation function, it comes out as $y_T - 1$ when $m=T$ and $y_m$ otherwise, so we can write:
$${\partial C \over \partial z_m} = y_m - targ_m$$
Which looks like magic.
My question is: How to go about removing the magic while maintaining clarity?
Michael Nielsen's page here points out that one can derive the cross entropy cost function for sigmoid neurons from the requirement that ${\partial C \over \partial z_k} = y_k - targ_k$.
i.e. The cost function has to exactly counterbalance the gradient across the (sigmoid) activation function.
I've been trying to do the same thing for the negative log likelihood cost function for SoftMax neurons:
SoftMax looks like a sensible formula; $\exp$ will accentuate the winner, and maps $R \to R^+$ and the denominator is just a normalisation term to make sure outputs sum to 1. So maybe no further justification is required for choosing $\exp$... but still this seems a little arbitrary/weak.
Now supposing the choice has been made, we have:
$$y_m = {\exp z_m \over \sum_k \exp z_k}$$
So maybe we can start off by requiring some cost function that satisfies:
$${\partial C \over \partial z_m} = y_m - targ_m$$
... so it should be possible to derive $C(y_m)$ using integration.
... but I can't see how to do that last step. Can anyone see it through?
PS Peter's Notes contains a tricky derivation for the negative log likelihood cost function.
• I am tackling the exact same questions. Have you taken a look here? quora.com/… – Šimon Mandlík Sep 16 '17 at 7:50
I haven't touched integration after leaving college, but for this specific problem, it seems straightforward to get the cost function. Here I am trying to sketch it just FYR. $$C = \int{\partial C \over \partial z_m} d z_m = \int(y_m - 1_{m=T})d z_m$$ With $y_m = {\exp z_m \over \sum_k \exp z_k}$,
$$\int(y_m - 1_{m=T})d z_m = \int({\exp z_m \over \sum_k \exp z_k} - 1_{m=T})d z_m$$ Since we have, $$\int {\exp z_m \over \sum_k \exp z_k} d z_m = \log ({\sum_k \exp z_k}) + f,$$ and, $$\int 1_{m=T} d z_m = z_T + g,$$ $f$ and $g$ do not consist of $z_m, m = 0,...,k$, so it is safe to require $f$ and $g$ to have only constants, and here for minimization purpose, we can ignore the constants. Then the equations above give the cost function, $$\log(\sum_k \exp z_k)- z_T = - \log{\exp z_T \over \sum_k \exp z_k} = -\log(y_T)$$ | 2021-05-14T23:46:23 | {
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http://mathhelpforum.com/geometry/124579-4-spheres-tetrahedron.html | # Thread: 4 spheres in a tetrahedron
1. ## 4 spheres in a tetrahedron
Three spheres, A, B and C, each 1 inch in diameter, are arranged on a horizontal flat surface so that they touch each other and their centers form an equilateral triangle. A fourth sphere D, also 1 inch in diameter, is placed on top of the other three so as to form a regular tetrahedral configuration (it touches each of the other three spheres).
(a) What is the distance between the centers of the spheres?
(b) How far above the flat surface is the center of sphere A?
(c) How far (vertical distance) above the center of sphere A is the center of sphere D?
Well I drawed the 4 spheres in the arrangement of a tetrahedron in a cube to help me out. for (a) I said the distance is just half the diagonal of the cube i.e.1 inch for (b) its just the radius of sphere A i.e. 0.5 inch.
For (c) I said its the radius of sphere A plus the radius of sphere D which would be 1 inch as well. But it seems much more complicated than this, can anyone check for errors?
2. a)1 inch(distance between the centres of the spheres because (radius +radius)
3. b)half inch because the distance is nothing but the radius which is half the diameter
c)1 inch. (3/2 inch -1/2 inch)
For (c), you already know (from (a)) that the centres of the four spheres form a regular tetrahedron with side 1 inch. You then need to know that the height of such a tetrahedron above its base is $\sqrt3/6$ inches (see here for a good explanation of that). | 2016-10-26T03:13:15 | {
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http://mathoverflow.net/questions/54571/eigenvalues-of-a%e2%8a%95b | # eigenvalues of A⊕B
Let $A_{n\times n}=(a_{ij}),B_{n\times n}=(b_{ij}) \in M_{n}(\mathbb{R})$, where $a_{ij},b_{ij} \in \lbrace 0,1\rbrace$. Boolean sum of $A,B$ denoted by $(A \oplus B)_{n\times n}=(a_{ij}\oplus b_{ij})$ is the matrix in $M_{n}(\mathbb{R})$ such that $0\oplus 0 = 0$, $0\oplus 1 = 1$, $1\oplus 0=1$ and $1\oplus 1 =1$. Is there any inequality or relation between eigenvalues of $A,B$ and $A\oplus B$? (specially, when $A,B$ are symmetric)
-
Just to be sure: the matrices $A$ and $B$ are viewed as having entries in $\mathbb{R}$ (so their eigenvalues are complex numbers) even though they have entries in $\{0,1\}$ and $A \oplus B$ is defined by taking the sum of the matrices viewed as having entries in $\mathbb{F}_2$? Could you give some motivation for this? – Pete L. Clark Feb 6 '11 at 23:06
@Pete L. Clark: It doesn't look to me like the sum is over $\mathbb{F}_2$ since $1 \oplus 1=1$. It is rather the bit-wise "or" operation. (Having said that, motivation still would be nice.) – Eric Naslund Feb 6 '11 at 23:20
Assuming they are symmetric, we can say a few trivial things about the largest eigenvalue. Namely that $\lambda_{A\oplus B} \leq \lambda_{A}+\lambda_B \leq 2\lambda_{A\oplus B}$ (where $\lambda_M$ refers to the largest eigenvalue) – Eric Naslund Feb 6 '11 at 23:29
And those bounds are tight, so which eigenvalues do you care about? And what kind of inequalities are you looking for? I don't fully understand the question. – Eric Naslund Feb 6 '11 at 23:32
***Remove the word symmetric in my above comment. It doesn't matter since the largest in magnitude (or one tied for the largest) will be real and positive for these matrices. – Eric Naslund Feb 6 '11 at 23:45
## 1 Answer
The operation is actually element-wise maximum, and the matrices are both non-negative, so Perron-Frobenius theorem together with the Rayleigh-Ritz characterization of maximum (perron-frobenius) eigenvalue seems to indicate that $\lambda_\max(A\oplus B) \geq \max \{\lambda_\max(A), \lambda_\max(B)\}.$
I don't know why you would use the $\oplus$ notation for this.
-
Intuitively, that inequality simply says "there are asymptotically more possible walks on a graph when we add more edges", which is certainly true. Recall: The largest eigenvalue of a 0-1 matrix is the asymptomatic growth rate of number of random walks on the directed graph generated by that matrix. (Provided there exists a path from each non-zero degree vertex to itself, which is automatically satisfied for symmetric matrices) – Eric Naslund Feb 7 '11 at 0:59
@Eric: good way to look at it! – Igor Rivin Feb 7 '11 at 2:30
Thanks for your answering to my question, but i'm looking for a lower bound for all the eigenvalues of $A⊕B$. – Moh514 Feb 7 '11 at 7:30 | 2014-10-23T17:27:03 | {
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http://secure.msri.org/workshops/811/schedules/23265 | # Mathematical Sciences Research Institute
Home » Workshop » Schedules » The smallest singular value of a $d$-regular random square matrix
# The smallest singular value of a $d$-regular random square matrix
## Geometric functional analysis and applications November 13, 2017 - November 17, 2017
November 17, 2017 (09:30 AM PST - 10:30 AM PST)
Speaker(s): Alexander Litvak (University of Alberta)
Location: MSRI: Simons Auditorium
Primary Mathematics Subject Classification No Primary AMS MSC
Secondary Mathematics Subject Classification No Secondary AMS MSC
Video
#### 17-Litvak
Abstract
We derive a lower bound on the smallest singular value of a random $d$-regular matrix, that is, the adjacency matrix of a random $d$-regular directed graph. More precisely, let $C_1<d< c_1 n/\log^2 n$ and let $M$ be uniformly distributed on the set of all $0/1$-valued $n\times n$ matrices such that each row and each column of a matrix has exactly $d$ ones. Then the smallest singular value $s_{n} (M)$ of $M$ is greater than $c_2 n^{-6}$ with probability at least $1-C_2\log^2 d/\sqrt{d}$, where $c_1$, $c_2$, $C_1$, and $C_2$ are absolute positive constants. This is a joint work with A. Lytova, K. Tikhomirov, N. Tomczak-Jaegermann, and P. Youssef
Supplements | 2019-11-22T13:30:57 | {
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https://mathoverflow.net/questions/361660/distribution-of-gaps-between-uniform-random-variables | # Distribution of gaps between uniform random variables
Pick $$k$$ uniform independent random integers $$x_1,\dots,x_k\in\{0,1,\dots,2^t-2,2^t-1\}$$ and denote $$y_{\sigma(i)}=x_i$$ where $$\sigma$$ is a permutation in $$S_n$$ such that $$y_1\leq y_2\leq\dots\leq y_{k-1}\leq y_k$$ holds.
Define the gaps $$g_1,\dots,g_{k-1}$$ by $$g_i=y_{i+1}-y_i$$.
1. What is the probability distribution $$P(\sum_{i=1}^{k-1}g_i^2)$$?
2. What is the distribution that $$g_{max}-g_{min} holds at an $$r\geq0$$? | 2020-07-03T14:45:34 | {
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https://web2.0calc.com/questions/sphere_11 | +0
# Sphere
0
98
4
Points A, B, and C are on a sphere whose radius is 13. If AB = BC = $$12$$ , what is the longest possible value of AC?
Apr 25, 2022
#2
+13900
+1
What is the longest possible value of AC?
Hello Guest!
$$\sqrt{12^2-x^2}+\sqrt{13^2-x^2}=13\\ x=\dfrac{12\sqrt{133}}{13}$$
$$The\ longest\ possible\ value\ of\ AC\ is\ {\color{blue}2x}=2\cdot \dfrac{12\sqrt{133}}{13}=\color{blue}21.291$$
!
Apr 26, 2022 | 2022-10-06T23:15:22 | {
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https://17calculus.com/precalculus/complex-numbers/multiplying/ | ## 17Calculus Precalculus - Multiplying Complex Numbers
##### 17Calculus
Multiplying Complex Numbers
Multiplying works the same as with radicals. For example, let's multiply the same two numbers we used above, first with square root of 2 and then with complex numbers.
Example: $$(1+3\sqrt{2}) (6+8\sqrt{2}) = 1(6+8\sqrt{2}) + 3\sqrt{2}(6+8\sqrt{2}) =$$ $$6+8\sqrt{2} +18\sqrt{2} +24(2) =$$ $$54+26\sqrt{2}$$
Notice we had $$(\sqrt{2})^2 = 2$$.
Now lets multiply complex numbers. Example: $$(1+3i) (6+8i) = 1(6+8i) + 3i(6+8i) =$$ $$6+8i +18i +24i^2 =$$ $$6+26i - 24 = -18+26i$$
Notice that when we got $$i^2$$ and since $$i=\sqrt{-1}$$, this gives us $$i^2 = (\sqrt{-1})^2 = -1$$. So $$24i^2 = 24(-1)=-24$$.
Before we go on, let's watch a video to give us more information.
### Dr Chris Tisdell - Complex numbers are AWESOME [16min-46secs]
video by Dr Chris Tisdell
When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.
DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it. | 2022-09-30T19:23:39 | {
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https://mathhelpboards.com/threads/problem-of-the-week-77-september-15th-2013.6445/ | # Problem of the Week #77 - September 15th, 2013
Status
Not open for further replies.
#### Chris L T521
##### Well-known member
Staff member
Thanks again to those who participated in last week's POTW! Here's this week's problem!
-----
Problem: Show that the equation $2x-1-\sin x=0$ has exactly one real solution.
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Hint:
First use IVT to show that $2x-1-\sin x=0$ has at least one real solution. Then use Rolle's theorem to show that you can't have two or more solutions.
#### Chris L T521
##### Well-known member
Staff member
This week's problem was correctly answered by anemone, eddybob123, johng, and MarkFL. You can find anemone's solution below.
Let $f(x)=2x-1-\sin x$ and we see that $f(x)$ is a continuous function for all real $x$.
We're asked to show that the equation $f(x)=2x-1-\sin x=0$ has exactly one real solution. One way to prove that is to first show that the function of $f$ of $x$ has a root and then show that the root is unique.
First, we suspect there is a root lies beween $$\displaystyle \frac{5 \pi}{18}$$ and $$\displaystyle \frac{\pi}{30}$$. Plugging in both values into $f(x)$, we get $f(\frac{5 \pi}{18}) \approx -0.02$ and $f(\frac{\pi}{30}) \approx 0.22$. Since $f(\frac{5 \pi}{18})<0$ and $f(\frac{\pi}{30})>0$ and $f(x)$ is a continuous function for all real $x$, by the Intermediate Value Theorem, there is a number $c$ between $$\displaystyle \frac{5 \pi}{18}$$ and $$\displaystyle \frac{\pi}{30}$$ such that $f'(c)=0$. Thus, $f(x)$ must have at least a real solution.
Next, we must show that this root is unique. Our plan is to prove it by contradiction. By the definition of Rolle's theorem, if we first assume the equation $2x-1-\sin x=0$ has at least two roots a and b, that is, $f(a)=0$ and $f(b)=0$, and then there is at least one point d in $(a, b)$ where $f'(d)=0$.
$f(x)=2x-1-\sin x$
$f'(x)=2-\cos x$
$f'(d)=0$ iff $2-\cos d=0$, or $\cos d =2$ But this is impossible since $-1<\cos d<1$.
This gives a contradiction and therefore, the equation can't have two roots, and the only root is unique.
Status
Not open for further replies. | 2020-09-25T12:42:15 | {
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https://mathoverflow.net/questions/350447/approximate-constant-function | # Approximate constant function
Let $$f:[0,1]^2 \rightarrow \mathbb C$$ be an $$H^1$$ function with the property that $$f(x,x)=0$$ and $$\Vert f \Vert_{L^2[0,1]}=1.$$
Does there exist a constant $$c>0$$ such that any such function satisfies
$$\Vert f-1 \Vert_{H^1}>c?$$
I was thinking that the Fourier series could help to prove or disprove something like this, but I did not get far so far.
It would be clearly possible in $$L^2$$ norm let's say, but I find it tricky in Sobolev norms.
• Should $\|f\|_{L^2[0, 1]}$ be $\|f\|_{L^2[0, 1]^2}$? Jan 15 '20 at 14:52
For all $$x$$ and $$y$$ in $$[0,1]^2$$ f(x,y)= \left\{ \begin{aligned} \int_x^y f_y(x,z)\,dz&\text{ if }x\le y, \\ -\int_y^x f_y(x,z)\,dz&\text{ if }x\ge y, \end{aligned} \right. where $$f_y(x,z):=\frac{\partial f(x,y)}{\partial y}|_{y=z}$$, so that
$$|f(x,y)|\le\int_0^1|f_y(x,z)|\,dz\le\sqrt{\int_0^1|f_y(x,z)|^2\,dz}.$$ Hence, $$\begin{split} 1 & = \iint_{[0,1]^2}dx\,dy\,|f(x,y)|^2 \\ & \le \iiint_{[0,1]^3}dx\,dy\,dz\,|f_y(x,z)|^2\\ & = \iint_{[0,1]^2}dx\,dz\,|f_y(x,z)|^2 \le\|f-1\|_{H^1}^2. \end{split}$$ So, any $$c\in(0,1)$$ will do. | 2021-12-03T16:32:01 | {
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https://web2.0calc.com/questions/if-f-x-x-3-3x-2-3x-1-find-f-f-1-2010_1 | +0
# If $f(x)=x^3+3x^2+3x+1$, find $f(f^{-1}(2010))$.
+2
76
6
+467
Nov 25, 2020
#1
+10793
+1
If $$f(x)=x^3+3x^2+3x+1$$, find $$f(f^{-1}(2010))$$
Hello Guest!
$$f(x)=x^3+3x^2+3x+1$$
$$f(x)=y=x^3+3x^2+3x+1\\ f(x)=y=(x-1)^3\\ f^{-1}(x)=x=(y-1)^3\\ f^{-1}(x)=y=\sqrt[3]{x}+1$$
$$f^{-1}(2010)=\sqrt[3]{2010}+1=\pm 12.6202+1\\ f^{-1}(2010)\in \{13.6302,-11.6202\}$$
$$f(f^{-1}(2010))=f(13.6302, -11.6202)$$
$$(x-1)^3=(13.6302-1)^3=2010$$
$$(x-1)^3=(-11.6302-1)^3=-2010$$
$$f(f^{-1}(2010))\in \{-2010,2010\}$$
!
Nov 26, 2020
edited by asinus Nov 26, 2020
#2
+112000
+3
I think they cancel each other out and the answer is just 2010
Looking at it a little more formally.
I think it means
$$2010=(x+1)^3\\ \sqrt[3]{2010}-1=x\\ f(\sqrt[3]{2010}-1)=(\sqrt[3]{2010}-1+1)^3=2010$$
Nov 26, 2020
#3
+10793
+1
Hi Melody!
I think:
In the term $$f(f^{-1}(2010)$$ is 2010 the argument of the $$f^{-1}$$.
Then applies:
$$f^{-1}(x)=f^{-1}(2100)=\sqrt[3]{x}+1=\sqrt[3]{2010}+1\\ \color{blue}f^{-1}(2100)\in\{-11.6202,13.6202\}$$
These values of the function $$f^{-1}$$are two arguments for the function f.
Please confirm whether this is the right or wrong idea.
Grüße
asinus Nov 26, 2020
#4
+112000
0
I don't think so.. Let's ask Alan or Heureka .....
Melody Nov 27, 2020
#5
+31506
+2
I see it this way:
Nov 27, 2020
#6
+112000
0
Thanks Alan, that is a much better way to present it. :)
Melody Nov 27, 2020 | 2021-01-17T17:56:20 | {
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http://math.stackexchange.com/questions/222615/how-to-solve-this-inequality-without-using-normal-approximation | # How to solve this inequality without using normal approximation
Assume $\lambda =80$, $\frac{e ^{\lambda x}}{x!}>0.9$ find $x$. I know how to use normal approximations but i want other method other than normal approximation.
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You can use Stirling's approximation where $$n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+\frac{1}{12n}+\frac{1}{288n^2}+o\left(\frac{1}{n^3}\right)\right)$$
First apply $\ln(\cdot)$ to both side of your inequality, then solve $x$ numerically.
-
Let $u_n=\mathrm e^{\lambda n}/n!$, then $u_{n+1}/u_n=\mathrm e^{\lambda}/(n+1)$ hence the sequence $(u_n)_{n\geqslant0}$ is increasing on $n\leqslant\mathrm e^\lambda$ and decreasing on $n\geqslant\mathrm e^\lambda$. Since $u_0=1\gt0.9$, this indicates that $u_n\gt0.9$ if and only if $n\leqslant N_\lambda$, for some $N_\lambda\geqslant\mathrm e^\lambda$. Furthermore, one can show that, when $\lambda\to\infty$, $N_\lambda\sim\mathrm e^{\lambda+1}$.
Edit: Refined forms of Stirling's approximation yield the following approximation. Call $x_\lambda(a)$ the solution of $(\lambda+1)x-(x+\tfrac12)\log(x)=\log(0.9a)$. Then $x_\lambda(\mathrm e)-1\lt N_\lambda\lt x_\lambda(\sqrt{2\pi})+1$. Finally, first-order estimates of the derivative show that, when $\lambda$ is large, the order of the width $x_\lambda(\sqrt{2\pi})-x_\lambda(\mathrm e)$ is $\log(\mathrm e/\sqrt{2\pi})=0.08$ hence a good heuristics is to take for $N_\lambda$ the closest integer to $x_\lambda(\sqrt{2\pi})$ and/or $x_\lambda(\mathrm e)$.
These considerations yield $N_{80}=1.5061\ldots\cdot10^{35}$, that is, $N_{80}=\exp(81.0000\ldots)$.
-
i still don't quite get how to solve the inequality. $\lambda =80$ and i am not sure how to proceed to the step $\lambda \rightarrow \infty$ – Mathematics Oct 28 '12 at 11:30
What do you call solve the inequality? We proved that $\mathrm e^{80n}/n!\gt0.9$ if and only if $n\leqslant N_{80}$ for some $N_{80}\geqslant\mathrm e^{80}$ and the equivalent at the end of my answer suggests that $\log N_{80}$ might be close to $81$, or, loosely speaking, $N_{80}$ close to $1.5\cdot10^{35}$. Exact bounds might be deduced from refined forms of Stirling's approximation, if this is your point. See Edit. – Did Oct 28 '12 at 12:04 | 2015-10-07T18:17:21 | {
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https://math.stackexchange.com/questions/2176849/a-method-to-count-the-number-of-monic-irreducible-polynomials-of-degree-2-or-3-o | # A method to count the number of monic irreducible polynomials of degree 2 or 3 or 4 in $\mathbb F_{p}[x]$
$p$ is a prime number. I use the method as following to find out the number of monic irreducible polynomials of degree 2.
Count the number of monic not irreducible polynomials of degree 2:
If a monic polynomial $f(x)\in \mathbb F_{p}$ is of degree 2 and not irreducible, $f(x)=(x-\alpha)g(x)$ for some $g(x) \in \mathbb F_{p}$.
As $deg(g)=2-deg(x-\alpha)=2-1=1$.So $g=x-\beta$ for some $\beta \in \mathbb F$.
Thus all the not irreducible monic polynomial of degree $2$ is of the form $f(x)=(x-\alpha)(x-\beta)$.
If $\alpha =\beta$ $f(x)=(x-\alpha)^2$. We have $p$ polynomials $f(x)$ in this form.
If $\alpha \neq \beta$ , we have ${p \choose 2}=\frac{p!}{2!(p-2)!}=\frac{p(p-1)}{2}$ polynomials $f(x)$ in this form.
Hence, the total number of monic not irreducible polynomials of degree $2$ in $\mathbb F_{p}[x]$ is $p+\frac{p(p-1)}{2}=\frac{p(p+1)}{2}$
Monic polynomial of degree $2$ in $\mathbb F_{p}[x]$ has the form $x^2+ax+b$ with $a,b \in\mathbb F_{p}[x]$. Thus we have $p^2$ such polymials.
Thus the number of irreducible polynomial of degree $2$ in $\mathbb F_{p}[x]$ is $p^2-\frac{p(p+1)}{2}=\frac{p^2-p}{2}$
But when I turn to find out the number of monic irreducible polynomials of degree 3.
I find that all the not irreducible monic polynomial of degree $2$ is of the form $f(x)=(x-\alpha)(x^2+cx+d)$,so it seems that we have $p$ choices of $\alpha$, $p$ choices of $c$ and $p$ choices of $d$, thus we have $p^3$ not irreducible polynomials of degree $3$ in $\mathbb F_{p}$.
But monic polynomial of degree $3$ in $\mathbb F_{p}[x]$ has the form $x^3+a_{1}x^2+a_{2}x+a_{3}$ with $a_{1},a_{2},a_{3} \in\mathbb F_{p}[x]$. Thus we have $p^3$ such monoc polymials.
So I think my counting when I try degree 3 is wrong.
I use the method as following to find out the number of monic irreducible polynomials of degree 4.
By using counting argument I have found that the number of irreducible polynomial of degree $2$ in $\mathbb F_{p}[x]$ is $p^2-\frac{p(p+1)}{2}=\frac{p^2-p}{2}$.
And we have $p^3-\frac{2p^3+p}{3}=\frac{p^3-p}{3}$ monic irreducible polynomials of degree $3$.
EDIT: Now I am doing the case of degree 4, could someone please have a look to my counting to see if it is correct? Thanks so much!
Monic polynomials of degree $4$ is of the form $x^4+a_{1}x^3+a_{2}x^2+a_{3}x+a_{4}$
where$a_{1},a_{2},a_{3},a_{4}$. Thus we have $p^4$ of them.
Let $f(x)$ be a not irreducible monic polynomials of degree $4$, then there are several possible form of $f(x)$
(i):$f(x)=(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)$ with $\alpha,\beta,\gamma,\delta \in \mathbb F_{p}$.
(ii):$f(x)=(x-\alpha)(x-\beta)(x^2+ax+b)$ with $\alpha,\beta,a,b \in \mathbb F_{p}$ and $(x^2+ax+b)$ is a irreducible polynomial of degree $2$.
(iii)$f(x)=(x^2+ax+b)(x^2+cx+d)$ with $a,b,c,d \in \mathbb F_{p}$ and $(x^2+ax+b), (x^2+cx+d)$ are irreducible polynomials of degree $2$.
(iv)$f(x)=(x-\alpha)(x^3+ax^2+bx+c)$ with $\alpha,a,b,c \in \mathbb F_{p}$ and $(x^3+ax^2+bx+c)$ is a irreducible polynomial of degree $3$. We have $p+3{p \choose 2}+3{p \choose 3}+{p \choose 4}$ monic reducible polynomials of degree $4$ of form (i). $(p+{p \choose 2})\frac{p^2-p}{2}$ monic reducible polynomials of degree $4$ of form (ii). $\frac{p^2-p}{2}+{\frac{p^2-p}{2} \choose 2}$ monic reducible polynomials of degree $4$ of form (iii). and $p(\frac{p^3-p}{3})$ monic reducible polynomials of degree $4$ of form (iv).
But the wolfram alpha does not give me the desire number of number of irreducible polynomials of degree 4. So I guess something is wrong.
Could someone help me to find out what is wrong here and show the correct way? Thanks in advance!
• Note that some cubic polynomials can be written in the form $(x - \alpha)(x^2 + c x + d)$ in more than one way, e.g., $x (x^2 - 1) = (x - 1)(x^2 + x)$. – Travis Mar 8 '17 at 0:47
• @Travis Oh, I see. But could you please show me how to avoid overcounting? Any hint will be appreciate. – PropositionX Mar 8 '17 at 0:50
• Sure: Suppose the cubic is totally factored: Either it is a product of three linear terms, a product of a linear term and an irreducible quadratic, or irreducible. You can count the first two cases separately, since you in particular know the number of irreducible quadratics from your previous step. – Travis Mar 8 '17 at 0:53
Over $\mathbb{F}_p$, the product of all monic irreducible polynomials with degree $1$ or $2$ is given by $x^{p^2}-x$, the product of all monic irreducible polynomials with degree $1$ or $3$ is given by $x^{p^3}-x$ and the product of all monic irreducible polynomials with degree $1$ is given by $x^{p}-x$. It follows that there are $$\frac{p^2-p}{2}\text{ monic irreducible polynomials with degree } 2$$ and $$\frac{p^3-p}{3}\text{ monic irreducible polynomials with degree } 3.$$ This happens because a monic irreducible polynomial $q(x)$ with degree $d$ defines a finite field $\mathbb{F}_{p^d}\simeq\mathbb{F}_p[x]/(q(x))$ over which $\alpha\to\alpha^p$ is a field homomorphism (Frobenius' homomorphism).
As an alternative, a monic polynomial is irreducible over $\mathbb{F}_p$ iff it has no roots in $\mathbb{F}_p$. Reducible monic polynomials have the form $(x-a) q(x)$ (with $q(x)$ being a monic, irreducible, second-degree polynomial) or $(x-a_1)(x-a_2)(x-a_3)$. There are $p\cdot\frac{p^2-p}{2}$ polynomials in the first case and $p+p(p-1)+\binom{p}{3}$ polynomials in the second case (accounting for $1$, $2$ or $3$ distinct roots in the field). The conclusion is the same as before.
A monic, reducible polynomial with degree $4$ can completely split in linear factors (there are $p+p(p-1)+\binom{p}{2}+\binom{p}{2}(p-2)+\binom{p}{4}$ polynomials with such a property, associated with $a_1^4,a_1^3 a_2,a_1^2 a_2^2, a_1 a_2 a_3^2, a_1 a_2 a_3 a_4$), or split as the product of two quadratic irreducible polynomials (there are $\binom{(p^2-p)/2}{2}+(p^2-p)/2$ cases), or split as the product of a quadratic factor and two linear factors (there are $\frac{p^2-p}{2}\left(p+\binom{p}{2}\right)$ cases), or split as the product of a cubic factor and a linear factor (there are $p\cdot\frac{p^3-p}{3}$ cases). It follows that the number of irreducible, monic polynomials with degree $4$ is $\frac{p^4-p^2+1}{4}$.
A useful generalization of the previous approach is:
Over $\mathbb{F}_p$, there are $$\frac{1}{k}\sum_{d\mid k} \mu\left(d\right)p^{k/d}$$ monic irreducible polynomials with degree $k$, with $\mu$ being Moebius' function.
• I know that you are correct. But I am asking this method in particular. Could you please tell me how to do the counting? – PropositionX Mar 8 '17 at 0:52
• A polynomial with degree $p^3-p$ is the product of all monic irreducible polynomials with degree $3$. By considering the degrees of the involved polynomials, it is trivial that there are $\frac{p^3-p}{3}$ monic irreducible polynomials with degree $3$, why do you need a more convoluted argument? – Jack D'Aurizio Mar 8 '17 at 0:54
• @Y.X.: anyway, I have added a combinatorial approach. – Jack D'Aurizio Mar 8 '17 at 2:35
• Thanks so much. But could you please have a look on the counting argument of degree 4? I am new to field theory and cannot understand your original method so far. So could you please point out if there is something wrong with my counting in the edited question? – PropositionX Mar 8 '17 at 10:41
• @Y.X.: it is not fair to keep modifying the original question by adding further requests. I am going to include the case $d=4$ in my answer, but for the future, please ask separate questions and avoid chamaleon questions. – Jack D'Aurizio Mar 8 '17 at 13:55
There is an other counting argument for the degree 2 using some basic Number theory. Let $f(x)=x^2+bx+c$ be a monic polynomial in $\mathbb{F}_p[x]$. There is a simple criterion which says that $f(x)$ is reducible mod p iff the discriminant $\Delta = b^2-4c = 0 ~mod ~p$. Now we count the number of possibilities for this to occur. Let us choose any $b$ (there are $p$ choices)
$$b^2-4c = 0 ~mod ~p \Leftrightarrow \left( \frac{4c}{p}\right)=1\Leftrightarrow \left( \frac{c}{p}\right)=1.$$ where $\displaystyle \left( \frac{.}{p}\right)$ is the Legendre Symbol modulo $p$ we use the fact that it is a morphism and that $4=2^2$ is clearly a residue. It is not difficult to see that there are $\displaystyle \frac{(p+1)}{2}$ residues mod $p$ thus there are $(p+1)/2$ of possibilities for $c$. Thus the total number of choices for $b$ and $c$ is given by $$\displaystyle \frac{p(p+1)}{2}$$ which represents as we have seen the number of monic reducible polynomial of degree 2 in $\mathbb{F}_p[x]$. The total number of monic polynomial of degree 2 in $\mathbb{F}_p[x]$ is $p^2$ then the required number is $$N_{irr} = p^2-\frac{p(p+1)}{2} = \frac{(p^2-p)}{2}.$$ | 2019-06-19T01:58:46 | {
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http://mathhelpforum.com/differential-geometry/271735-difficult-proof.html | # Thread: difficult proof!
1. ## difficult proof!
AD is an altitude in an acute triangle ABC. Points E,F are orthogonal projections of D on, respectively, AB,AC,. M,N are midpoints of, respectively, AB,AC. Lines MF,EN intersect at point S. Show that circumcenter of ABC lies on line SD.
I have no idea.
2. ## Re: difficult proof!
Do you understand what these words mean? Do you understand what the "orthogonal projection" of D on AB and AC means? Can you draw a picture showing them? Do you know what the "circumcenter" of ABC is?
3. ## Re: difficult proof!
of course, I drawed a picture
4. ## Re: difficult proof!
Originally Posted by TobiWan
of course, I drawed a picture
Good! Show us your drawing. Tell what methods are you expected to use?
5. ## Re: difficult proof!
at the beggining I tried to draw a line parallel to the SD from the point E, which intersects BC at G and made up sth from Thales'Theorem, but probably its not the way
6. ## Re: difficult proof!
generaly I would like to show that some points lie on a circle
7. ## Re: difficult proof!
should I show my drawing? | 2019-07-19T13:07:57 | {
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https://stacks.math.columbia.edu/tag/023O | Remark 35.3.10. Let $R$ be a ring. Let $f_1, \ldots , f_ n\in R$ generate the unit ideal. The ring $A = \prod _ i R_{f_ i}$ is a faithfully flat $R$-algebra. We remark that the cosimplicial ring $(A/R)_\bullet$ has the following ring in degree $n$:
$\prod \nolimits _{i_0, \ldots , i_ n} R_{f_{i_0}\ldots f_{i_ n}}$
Hence the results above recover Algebra, Lemmas 10.24.2, 10.24.1 and 10.24.5. But the results above actually say more because of exactness in higher degrees. Namely, it implies that Čech cohomology of quasi-coherent sheaves on affines is trivial. Thus we get a second proof of Cohomology of Schemes, Lemma 30.2.1.
There are also:
• 3 comment(s) on Section 35.3: Descent for modules
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 2021-03-04T13:00:46 | {
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http://jwork.org/learn/doc/doku.php?id=jmathlab:matrices | You are a guest. Restricted access. Read more
# Matricies
Matrices are handled in a similar way, only with two indices for row number (first index) and column number (second index). Rows are separated by either a semicolon or a linefeed during input.
M=[1:3 ; 4:6 ; 7:9]
printf('M=%f\n',M)
a=M([1 3],:)
printf('a=%f\n',a)
C=M<4
printf('C=%f',C)
All operators be applied to vectors and matrices. If scalar, per-element operation is desired, some operators (* / ^) must be preceded by a point to distinguish them from the quite different linear-algebra versions of these operations (see chapter 2.9). Further useful functions are sum(vector) and prod(vector) which return the sum and product of the vectors elements.
The following matrix operations are available:
symbol operation + addition - subtraction * multiplication power ' transpose \ left division / right division
These operations are also applicable for vectors. These matrix operations apply to scalars (1-by-1 matrices) as well. If the sizes of the matrices are incompatible for the matrix operation, an error message will result, except in the case of scalar-matrix operations (for addition, subtraction, and division as well as for multiplication) in which case each entry of the matrix is operated on by the scalar.
The colon notation can be used to access sub-matrices of a matrix. For example,
M=[1:10; 1:10; 1:10; 1:10; ]
A=M(1:4,3)
printf('%f',A)
is the column vector consisting of the first four entries of the third column of A. A colon by itself denotes an entire row or column:
A(:,3)
is the third column of A, and A(1:4,:) is the first four rows. Arbitrary integral vectors can be used as subscripts:
A(:,[2 4])
contains as columns, columns 2 and 4 of A. Such subscripting can be used on both sides of an assignment statement:
A(:,[2 4 5]) = B(:,1:3)
replaces columns 2,4,5 of b with the first three columns of B. Note that the entire altered matrix A is printed and assigned. Try it.
Columns 2 and 4 of A can be multiplied on the right by the 2-by-2 matrix [1 2;3 4]:
A(:,[2,4]) = A(:,[2,4])*[1 2;3 4]
Once again, the entire altered matrix is printed and assigned.
# Operations
Here is the list of matrix and vector operations:
Name(Arguments) Function
linspace($var_1$,$var_2$,COUNT) vector with COUNT numbers ranging from $var_1$ to $var_2$
length($vector$) number of elements in $vector$
zeros(ROWS[,COLUMNS]) matrix of zeros
ones(ROWS[,COLUMNS]) matrix of ones
eye(ROWS[,COLUMNS]) matrix with diagonal one
rand(ROWS[,COLUMNS]) matrix of random numbers
hilb(RANK) Hilbertmatrix
invhilb(RANK) Inverse Hilbertmatrix
size($matrix$) number of rows and columns
sum($var$) if $var$ is a vector: sum of elements, if $var$ is a matrix: sum of columns
find($var$) indices of non-vanishing elements
max($var$) largest element in $var$
min($var$) smallest element in $var$
diag($var$,[OFFSET]) if $var$ is a vector: matrix with $var$ as diagonal, if $var$ is matrix: diagonal as vector
det($matrix$) determinant
eig($matrix$) eigenvalues
inv($matrix$) inverse matrix
inversematrix($matrix$) inverse matrix
pinv($matrix$) pseudoinverse
lu($matrix$) LU-decomposition
svd($matrix$) singular value decomposition (Lapack)
qr($matrix$) QR-decomposition (Lapack)
eigen($matrix$) eigenvalues (Lapack)
For example, zeros(m,n) produces an m-by-n matrix of zeros and zeros(n) produces an n-by-n one.
# Functions
Several standard matrices are created by means of functions without specifying individual elements: ones(n,m), zeros(n,m), rand(n,m) return matrices with elements 1, 0 or random numbers between 0 and 1. eye(n,m) has diagonal elements 1, else 0, and hilb(n) creates the n-th degree Hilbert-matrix.
A=rand(1,3)
printf('%f',A)
B=hilb(4)
printf('%f',B)
The following functions are provided for matrix calculations: diag(x) (extracts diagonal elements), det(x) (determinante), eig(x) (eigenvalues), inv(x) (inverse), pinv(x) (pseudo-inverse). The adjunct matrix is created using the operator'.
A=det(hilb(4)) % determinant
printf('%f\n',A)
M=[2 3 1; 4 4 5; 2 9 3];
A=eig(M)
printf('eig=%f\n',A)
A=inversematrix(M)
printf('Inverse=%f\n',A)
The nontrivial functions are all based on the LU-decomposition, which is also accessible as a function call lu(x). It has 2 or 3 return values, therefor the left side of the equation must provide multiple variables, see example below:
M=[2 3 1; 4 4 5; 2 9 3]
[l,u,p]=lu(M) % 2 or 3 return values
Without preceding point the arithmetic operators function as matrix operators, e.g. * corresponds to matrix and vector multiplication.
x=[2 1 4]; y=[3 5 6];
c=x.*y % with point
printf('x*y=%f\n',c)
c=x*y % without point
printf('x*y=%f\n',c)
If one of the arguments is a scalar datatype, the operation is repeated for each element of the other argument:
x=[2 1 4];
c=x+3
printf('x+3=%f\n',c)
Matrix division corresponds to multiplication by the pseudo-inverse. Using the operator backslash leads to left-division, which can be used to solve systems of linear equations:
M=[2 3 1; 4 4 5; 2 9 3];
b=[0;3;1];
x=M\b % solution of M*x = b
printf('M\b=%f\n',x)
c=M*x % control
printf('M*x=%f\n',x)
Systems of linear equations can (and should) be solved directly with the function linsolve(A,b).
## Lapack
The application contains JLAPACK [9], the Java-port of the LAPACK [10]-routines with extended and better algorithms for matrix calculations. However, these are limited to matrices with real coefficients in floating point format. The LAPACK routines are accessed by the following functions:
svd(A) Singular value decomposition of A (1 or 3 return values).
A=[2 3 1; 4 4 5; 2 9 3];
a=svd(A)
printf('M*x=%f\n',a)
qr(A) QR-decomposition of A (2 return values).
A=[2 3 1; 4 4 5; 2 9 3];
[q,r]=qr(A)
printf('eigen=%f\n',qr(A))
linsolve2( A, b)
Solves $A\cdot x=b$ (1 return value). Example in chapter 2.13.1.
linlstsq(A, b) Solves $A\cdot x=b$, overdetermined (1 return value).
eigen(A) Eigenvalues of A (1 return value).
A=[2 3 1; 4 4 5; 2 9 3];
a=eigen(A)
printf('eigen=%f\n',a)
## LAPACK vs Jasymca
We calculate the 4-th degree regression polynomial for the following x,y-data:
x=[1:6],y=x+1
printf('x=%f\n',x)
printf('y=%f\n',y)
a=polyfit(x,y,4)
printf('polyfit=%f\n',a)
The coefficients p(1),p(2),p(3) should vanish since x and y represent a perfect straight line. This is an unstable problem, and it can be easily extended to make Jasymca completely fail. In our second attempt we use the Lapack-routine linlstsq:
x=[1:6],y=x+1;
l=length(x);n=4;
X=(x'*ones(1,n+1)).^(ones(l,1)*(n:-1:0))
a=linlstsq(X,y')
printf('%f\n',a)
The coefficients p(1),p(2),p(3) are now significantly smaller. This particular problem can be solved exactly using Jasymca-routines and exact numbers, which avoids any rounding errors:
x=rat([1:6]);y=x+1;
a=polyfit(x,y,4)
printf('%f',a) | 2014-04-24T18:44:30 | {
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https://www.physicsforums.com/threads/equation-in-natural-number.462649/ | # Equation in natural number
1. ### oszust001
10
How can I show that:
$$\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2^{n}}$$
for every natural numbers
2. ### AtomSeven
8
The identity is wrong, it should be
$$\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}$$
3. ### oszust001
10
ok my foult. so how can i solve that equation?
4. ### AtomSeven
8
Well, this
$$\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}$$
is an identity it is true for all $$n$$ but, if I understand correctly, you may ask for the values of $$n$$ that make
$$\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2^{n}}$$
true. In this case we have the equation $$2n=2^{2^{n}}$$, and the solutions are $$n \in \lbrace 1,2 \rbrace$$.
5. ### oszust001
10
Version of AtomSeven is good.
How can I show that $$\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}$$
is good for every natural numbers | 2015-05-23T05:46:42 | {
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http://acremortgagedelaware.com/3zg8ftjcv/helmholtz-decomposition-wave-equation.html | We use the Helmholtz decomposition: De ne: S = p 2H1() jp = 0 on ;p constant on then rS ˆX. 12. This fact reduces the operator inversion to a simple division of the Fourier coefficients by the corresponding wave numbers. The solution domain is A discrete Helmholtz decomposition of the functions in the velocity space based on potentials taken from the pressure space is used to provide a complete description of the numerical wave propagation for the discretised equations. This research addresses the predictive simulation of acoustic emission (AE) guided waves that appear due to sudden energy release during incremental crack propagation. For isotropic media, one can process the elastic data either by separating wave-modes and migrating each mode using methods based on acoustic wave theory, or by migrating the whole elastic data set based on the elastic wave equation . 2) @t The two quantities ˆ. Helmholtz decomposition Interpretation In mathematics , in the area of vector calculus , Helmholtz's theorem , also known as the fundamental theorem of vector calculus , states that any sufficiently smooth , rapidly decaying vector field can be resolved into irrotational (curl-free) and solenoidal (divergence-free) component vector fields. According to the Helmholtz decomposition, the velocity vector v (as any vector field) can . Strange Bedfellows: The Science and Policy of Natural Hazards Seismic waves Spring 2008. How-ever, Helmholtz decomposition using conventional divergence and curl operators update to the image. The presence of body force makes the homogeneous Navier–Lame elastodynamic equations into inhomogeneous equations. sound or acoustics, Diagram 496f WOW! Helmholtz Decomposition Fourier Transform the idea girl says blue book 2 notes page 185f, cooled plasma formula page 35 Diagram 496f WOW! Helmholtz Decomposition Fourier Transform Line 22 7b97z66b Fermi Polar Quantum Matter Pt 14 KRB Fermion Waves 5g WOW SETI Line 22 7b97z66b Fermi Polar Quantum Matter Formula Pt 15 Young… Abstract. Helmholtz decomposition. (2. . I. computations both for homogeneous and inhomogeneous media show the effectiveness of the proposed method. A Source Transfer Domain Decomposition Method for Helmholtz Equations in Unbounded Domain Helmholtz Equation with High Wave Number and Perfectly Matched Layer equation alternatingly in overlapping subdomains, and uses as interface condition the trace of the previously computed solution in the neighboring subdomain. acoustic imaging. The decomposition is called after the German physiologist and physicist Hermann von Helmholtz (1821 – 1894). Editor: Paul Fromme. Applying the Helmholtz decomposition (26) to the momentum Equation (25) by replacing v with the right hand side of Equation (26) yields , , are the P- and S-wave separated elastic equation, which is equivalent to the coupled elastic wave equation and contains the P-wave (U P), S-wave (U S) and full elastic wave (U). V'\right){\bigg ]}. appear in such diverse phenomena as a. A new series expansion of frequency to any order of wave number, in principle, is obtained for symmetric and antisymmetric modes using an iteration method. 2. With p(x;t) the pressure uctuation (a time-dependent scalar eld) and v(x;t) the particle velocity (a time-dependent vector eld), the acoustic wave equations read @v. This report Then, 2D approximation of Helmholtz decomposition, Equation 3, is applied to the modelled struct and solve the seismic wave equation for elastic wave propagation in a uniform . (1. The efficiency of this method is demonstrated by numerical examples. Helmholtz decomposition In physics and mathematics , in the area of vector calculus , Helmholtz's theorem , also known as the fundamental theorem of vector calculus , states that any sufficiently smooth , rapidly decaying vector field in three dimensions can be resolved into the sum of an irrotational ( curl -free) vector field and a solenoidal The presented domain decomposition method is based on the hybridized mixed Helmholtz equation and using a high-order, tensorial eigenbasis. hku. the Helmholtz equation: 1. Keywords: Helmholtz theorem, vector field, electrodynamics, mathematical physics 1. Helmholtz decomposition: - Look at the wave solutions and show that != =k the equation becomes an ordinary matrix eigenvalue problem. 3) where the scalar functions φand ψare called potentials. The linearity allows us to break in the wave equation's linear operators all the way through to the Fourier coefficients, and the eigenvalue relation for $\partial_t$ enables us to switch that partial differentiation to an algebraic factor on that sector, giving us \begin{align} 0 & = -\partial_{t}^2 u(x,t) + c^2 abla^2 u(x,t) + f(x,t) \\ & = -\partial_{t}^2 \int_{-\infty}^\infty U(x,\omega) e^{-i\omega t} \mathrm d\omega + c^2 abla^2 \int_{-\infty}^\infty U(x,\omega) e^{-i\omega t the part of the solution depending on spatial coordinates, F(~r), satisfies Helmholtz’s equation ∇2F +k2F = 0, (2) where k2 is a separation constant. For any vector eld u 2(L2())3, it has the following decomposi-tion u = rq+r w;q2H1()=R;w 2(H1())3 satisfying rw = 0;(r w)n = 0 on : Characterization of Wave Physics Using the Rigorous Helmholtz Decomposition Based on the Surface Integral Equation Xiaoyan Y. Isotropic elastic data can be decomposed into P and S-wave potentials by taking the divergence and curl of the wavefield components respectively: p Has the Helmholtz decomposition of the $\mathbf{E}$ field from the Liénard–Wiechert potentials been worked out? Ask Question Asked 1 year, 9 months ago derived using the Helmholtz decomposition of equation (1. 3. 1. Elastic Waves in an Unbounded Medium 4. 3 (The Helmholtz model) The acoustic (scalar) wave equation reads,. The equation of motion can be derived by considering the elastic medium to derive the elastic wave equation. Substituting (2. So, even the instantaneous solutions are correct solutions of the Maxwell equations that result from the Helmholtz equation domain decomposition preconditioned iterative method parallelization This is a preview of subscription content, log in to check access. Only high-performance parallel computers have The wave (Helmholtz) and time-independent diffusion eqs These eqs. Part 7, Zap. t problem size is a major issue, and it is difficult the Green functions of the Helmholtz equation for arbitrary frequencies and for many point sources; (4) for a speci ed number of points per wavelength it constructs each Green function in nearly optimal complexity in terms of the total number of mesh points, where the prefactor EQUATIONS USING CONSTRAINT-BASED HELMHOLTZ DECOMPOSITIONS . In contrast, for elastic reverse-time migration, wavefield reconstruction is done with the elastic wave-equation using the recorded vector data as boundary condition. of these solutions, one used the Green function of the Poisson equation. ˆ. Basic properties of the Helmholtz decomposition of a vector field over the entire three-dimensional space into its longitudinal (irrotational, lamellar, curl-free) part and transversal (solenoidal, divergence-free) part are described in this appendix. The key idea in our method is to combine these two regimes using a combined domain decomposi-tion and wave decomposition method. According to the generalized (to deal with complex potentials) Green's theorem (see Section 2. A lot of mathematics is involved to simulate the waves. This decomposition can be seen as a generalization of the Bohren decomposition and a refinement of the Helmholtz’s decomposition. So I tried to apply the theorem on the following vector fields: For the vector field I calculated that: Broadbridge extended the scalar wave function in the Schrödinger equation to a vector potential by using the Helmholtz decomposition of the Madelung fluid that included a solenoidal component. 2) and the far-field patterns for the Helmholtz equations of (1. 27 Oct 2018 The matrix–vector wave equation is a compact first-order differential equation. Likewise, Elastic least-squares migration with two-way wave equation forward and adjoint operators Ke Chen and Mauricio D. where Helmholtz used the decomposition or for which Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 9), The retarded potential, a solution of the non-homogeneous wave equation, is a subject of particular gral theorem of Helmholtz and Kirchhoff, which is the main. Because of the wave propagation character of the Helmholtz equation, local phenomena affect the solution globally, and such long-range interactions must be accounted for in an effective algorithm. method solves the elastic wave equation by iteratively solvin g sub prob lems defined on smaller sub domains. The emphasis is put on three topics which are clearly structured into Chapters 2, 3, and 4. 11) describes the solution of a time-harmonic electric field, a field that oscillates in time at the fixed angular frequency ω. In this paper, a class of solution methods for the Helmholtz equation in a multi-layer domain is considered. In vector analysis, the Helmholtz decomposition of a vector field on is the decomposition of the vector field into two vector fields, one a divergence-free field and one a curl-free field. This decomposition is well-known in fluid mechanics [6–10]. This wave equation assumes a slowly varying stiffness tensor over the imaging space. The resulting wave equation is then written in a state-variable form of coupled first-order differential equations. rp; (1. The Helmholtz decomposition approach is applied to the inhomogeneous elastodynamic Navier-Lame equation for both the displacement field and body forces. as long as p and Po are not on the boundary. Abstract | PDF (404 KB) The Helmholtz–Hodge decomposition (HHD) is applied to the construction of Lyapunov functions. The Wave equation becomes: The Helmholtz decomposition is used to show that the resulting inertia-gravity wave equation is third-order accurate in space. Moreover this decomposition is valid even for vector fields that diverge weaker than linearly. Demanet 4 In many science and engineering applications, solving time-harmonic high-frequency wave prop-agation problems quickly and accurately is of paramount importance. Maxwell's equations imply the existence of electromagnetic waves (as 1 Oct 2015 Discrete Helmholtz---Hodge Decomposition on Polyhedral Meshes Using the decomposition components independently with one equation to solve . Two domain decomposition algorithms both for nonoverlapping and overlapping methods are described. 1 = @t. This is the Helmholtz Decomposition Theorem. these methods to the scalar Helmholtz wave equation [3, 4, 5]. Numerical examples show that the simulated In this paper, the local Fourier method is used to transform the Helmholtz wave equation into a phase-space coordinate system. 7 Solution of Equation of Motion: Helmholtz Decomposition 1-14 2. theory for the divergence constrained Helmholtz-Lamb potential decomposition of the velocity field, which are closely related to vorticity formulations, and show how they can be used in vortex element methods. Abstract. From the equation of motion, we derive the wave equations. In differential According to the Helmholtz decomposition theorem (the fundamental theorem of vector calculus). Preview A PARALLEL DOMAIN DECOMPOSITION METHOD FOR THE HELMHOLTZ EQUATION ELISABETH LARSSON yAND SVERKER HOLMGREN Abstract. 16 Feb 2017 stress field in the wave equation to accurately separate the P- and S- to directly apply the Helmholtz decomposition in wave-mode sep-. In the geophysical frequency-domain in-version, one needs to do forward modeling which means solving the Helmholtz equation. The method can be used as an efficient preconditioner in the preconditioned GMRES method for solving discrete Helmholtz equations with constant and heterogeneous wave numbers. Fundamental Equations. The Navier’s governing equation can be decoupled into two sets of equations by Helmholtz decomposition, which are the scalar and vector potential functions, respectively. 11) into Eq. The convergence of the method is proved forthe case of constant wave number based on the analysis of the fundamental solution of the PML equation. [8] in 1889. In vector analysis, the Helmholtz decomposition of a vector field on is the decomposition of the vector field into two vector fields, one a divergence -free field and one a curl -free field. vectorial wave equation. Also, this equation is mathematically a hard nut to crack. The Helmholtz decomposition 15. Slowness Diagrams 4. A note on the Pochhammer frequency equation / ; ( , , , , ) ( , , , ,) 2 T t A R U W Z a r u w z ρω µ = ω =, where ω is the angular frequency of the wave, which is considered to be imposed in this problem. th equations define lossless, constant-veloci propagation of electromagnetic waves. In fact, the existing domain decomposition methods (and multilevel methods) are inefficient to these equations except that the sizes of the coarse meshes are chosen as O(1/ω) (see, for example, the loop-tree decomposition and the proposed decomposition. of the heterogeneous Helmholtz equation. This is the decomposition I am following (according to Wikipedia) where A is a vector field, and Phi is a scalar field. 1 Introduction. Keywords Helmholtz decomposition cones. As an application, On the Role of the Helmholtz Decomposition in Mixed Methods for Incompressible Flows and a New Variational Crime A. Xiong, Wei E. This book arose from a lecture on Maxwell’s equations given by the authors between ?? and 2009. To evaluate the performance of the methods, two-dimensional problems with a waveguide geometry are used as model problems. Subjects Primary: 35J05: Laplacian operator, reduced wave equation (Helmholtz equation), Poisson equation [See also 31Axx, 31Bxx] 35J20: Variational methods for second-order elliptic equations. In fact, the solution has a nice physical meaning: decompose into components parallel and perpendicular to . After inserting Eq. Its accuracy and convergence characteristics are examined and compared to the simplest boundary element method for exterior problems, the Helmholtz Integral Equation Formulation or HIEF. Here is a pseudo-code for solving the Poisson equation resulting from the Helmholtz decomposition important problem. t. Discrete Helmholtz Decomposition for Electric Current Volume Integral Equation Formulation Johannes Markkanen Abstract—A volume integral equation formulation for the equivalent current is investigated by decomposing the L2-conforming unknown current into orthogonal functions. 1. SIAM Journal on Numerical Analysis 51:4, 2166-2188. Equations of this type describe wave propagation in the frequency domain. decomposition that is exact when considered analytically, so the only degradation in computational performance is due to discretization and roundo errors. In this paper we extend the technique to elasticity problems. I. the pressure update formula is Helmholtz equation is used to capture the strong scattering events. We consider autonomous differential equations given in the form x_ = F(x); (1) where F(x) is a smooth vector field defined on Ω , where Ω ˆ Rn A HIERARCHICAL 3-D DIRECT HELMHOLTZ SOLVER BY DOMAIN DECOMPOSITION AND MODIFIED FOURIER METHOD∗ E. The method solves the problem by iteratively solving subproblems defined on smaller subdomains. ELASTIC IMAGING USING HELMHOLTZ DECOMPOSITION troducing two scalar potential functions, the method uses the Helmholtz decomposition to split the displacement of the elastic wave equation into the compressional and shear waves, which satisfy a coupled boundary value problem of the Helmholtz equations. which results from (28), we can derive from (33) the wave equation. In each of these chapters we study rst the simpler scalar case where we replace the Maxwell system by the scalar Helmholtz equation. propagation at a wavenumber k is governed by the Helmholtz equation ∇2ψ(k,r)+k2ψ(k,r)=0, (1) where ψ(k,r) is the Fourier transform of the pressure, which is proportional to the velocity potential and will be loosely referred to as a potential in this paper. For the components shown in Figure 2b and 2c, the sep-arated wave P and S modes are, respectively, shown in Figure 2d and 2e. 3) Helmholtz decomposition. rv: (1. 0) P, SV & SH Waves 2. Solutions of the Helmholtz equation can be derstand that the Helmholtz decomposition that relates retarded and instantaneous solutions is a gauge transfor- mation. This new approach is applied to the electric field integral equation (EFIE), and it incorporates a Helmholtz decomposition (HD) of the current. I wrote my own code in MATLAB. The algorithm of no overlapping domain decomposition method is given. In particular, we show that the domain derivative of the potentials is the Helmholtz decomposition of the domain derivative The conventional decomposition of a vector field into longitudinal (potential) and transverse (vortex) components (Helmholtz's theorem) is claimed in [1] to be inapplicable to the time-dependent vector fields and, in particular, to the retarded solutions of Maxwell's equations. The dielectric case CMAP, École Polytechnique HongKong 2007. . Let us describe more clearly what is meant by the above items (1){(4). In other . r. 13. 2. Eqs. This equation is referred to as Helmholtz equation. Choosing ˚ = r˘, ˘2S as a test function: 2(E;r˘) = 0 so E is divergence free. BRAVERMAN†, M. 1) yields ∇((λ+2µ)∆φ+ω2φ) +curl(µ∆ψ+ω2ψ) = 0 in R2 \D, which is fulfilled if φ,ψsatisfy the Helmholtz equation ∆φ+k2 He added a proof, but he didn't provide any single motivation - e. A transparent interface condition is derived to couple these two regions together. 1), we introduce the Helmholtz decomposition v = ∇φ+curlψ, (2. (5a) , (5b) , (5c) can be derived from the Helmholtz decomposition theory. A Discrete Weighted Helmholtz Decomposition and Its Application Qiya Hu1, Shi Shu2 and Jun Zou3 Abstract We propose a discrete weighted Helmholtz decomposition for edge element functions. Manuscript received May 4, 2017; final manuscript received September 28, 2017; published online October 27, 2017. equations are derived for elastic plate using the Helmholtz displacement decomposition. All these studies focused on starting from the Schrödinger equation and leading to the Madelung form of Euler equations for potential flow that was Because of this non-uniqueness, showing that it’s possible to express a particular velocity field with an incompressible flow as a Hemholtz decomposition with a nonzero $\phi$ doesn’t mean that it isn’t also possible to express the same velocity field as a different Hemholtz decomposition in which $\phi = 0$. The use of fast multigrid methods for the solution of this equation is investigated. Introduction In most textbooks on electrodynamics one reads that vector fields that decay asymptotically faster than 1/r, where r =x Key words: Helmholtz equation, domain decomposition, preconditioned iterative method, parallelization. 5) is also referred to as the Helmholtz wave equation. (1) can be satis ed in many ways. The Rayleigh-Lamb equations are considered in a new way. Key words. In mathematics and physics, the Helmholtz equation, named for Hermann von Helmholtz, is the linear partial differential equation (∇ +) =, where ∇ is the Laplacian, is the wave number, and is the amplitude. g. Thanks to the divergence theorem the equation can be rewritten as. domain decomposition methods (DDM) are applied for wave equations and higher- alized to the spectral element discretization for the Helmholtz equation, 28 Feb 2006 “The numerical solution of the Helmholtz equation for wave “A domain decomposition method for the Helmholtz equation and related optimal We derive the vector Helmholtz wave equation assuming harmonic time dependence, Eoe−iωt+ik⋅x, from . Sacchi, Department of Physics, University of Alberta Summary Time domain elastic least-squares reverse time migration (LSRTM) is formulated as a linearized elastic waveform inversion problem. The interface conditions involve second order tangential derivatives which are optimized (OO2, optimized order 2) for a fast convergence. (2012) presented a PS wave imaging method in 3D. 3), which is allowed by the radiation conditions of equation (1. Problem: curl has a large null space. ) If it doesn't go to zero at infinity, then can Plane wave approximation in linear elasticity Andrea Moiola Department of Mathematics and Statistics – University of Reading, Whiteknights, PO Box 220, Reading RG6 6AX, UK; (Received XXX) We consider the approximation of solutions of the time-harmonic linear elastic wave equation by linear combinations of plane waves. a general vector field. This method leads to efficient algorithms for the numerical resolution of harmonic wave propagation problems in heterogeneous media and their control. By introducing the Helmholtz decomposition, the model problem is reduced to a coupled boundary value problem of the Helmholtz equations. 4) we obtain ∇ 2E(r) + k E(r) = 0 (2. Two simple choices are: U = zk; W = 0; (2) U = 0; W = xk^y: (3) Note that, in this example, r F and r F are both zero. (See the Helmholtz decomposition. We form an elastic migration operator by combining Helmholtz decomposition with a wave equation migration operator that extrapolates P and S scalar potentials independently. 0. The separation of wave modes to P and S from isotropic elastic wave elds is typically done using Helmholtz decomposition. Hewett 3 L. Inhomogeneous Plane Waves 4. 2 Answers. It is found that the HELMHOLTZ DECOMPOSITION BASED ON INTEGRAL EQUATION METHOD FOR ELECTROMAGNETIC ANALYSIS Xiaoyan Y. A mixed hybrid formulation of the underlaying equations provides in a natural combining Helmholtz decomposition with a wave equation migration operator that extrapolates P and S scalar potentials independently. Assoc. We write a discrete Helmholtz decomposition X The Helmholtz decomposition revisited View the table of contents for this issue, or go to the journal homepage for more Home Search Collections Journals About Contact us My IOPscience You may also be interested in: Does the Helmholtz theorem of vector decomposition apply to the wave fields of electromagnetic radiation? A M Stewart Helmholtz Decomposition In terms of P wave scalar potential and S wave vector potential : u where 0 u 2 and u 2 Helmholtz decomposition 0 1 2 2 2 2 v t p 0 1 2 2 2 2 v t s Use of the potential functions allows separation of the effects of dilation and rotation equations. Field Sources Let us investigate further. The presented domain decomposition method is based on the hybridized mixed Helmholtz equation and using a high-order, tensorial eigenbasis. At each node, as well as having a number of plane wave propagation directions, there are two di erent types of wave, the dilatation, or Pressure (P), wave and the distortional or Shear (S) wave. The Helmholtz equation governs wave propagation and scattering phenomena arising in a wide range of engineering applications. Accurate numerical solution of interesting application problems often puts very high demands on the capacity of the computer, even when effective nu-merical schemes are employed. Scalar potential function is given as, 2 2 22 1 Ct L (9) Vector potential function is available as, 2 2 2 2 2 2 2 2 22 21 21 1 rr r T r T z z T t t Ct excitation potentials. Semin. Dispersion relations are shown in graphs for frequency, phase speed and group speed versus wave number. Z. Helmholtz decomposition, the surface wave can be seen as the superposition of two separate components: one longitudinal and the other transverse. Citation The Helmholtz theorem of vector field decomposition [1-16] states that a three-vector field F ( r , t ) (where r contains the three spatial coordinates and t is the time) that vanishes at spatial infinity can, under certain conditions, be expressed as the sum of a gradient and a curl In this paper, wave simulation with the finite difference method for the Helmholtz equation based on the domain decomposition method is investigated. Isotropic elastic data can be decomposed into P and S-wave potentials by taking the divergence and curl of the wavefield components respectively: Helmholtz Decomposition and BEM Applied to Dynamic Linear Elasticity curl free condition on the longitudinal component can both be framed as Helmholtz scalar equations. Second, according to a fundamental theorem on elliptic operators (see e. Solonnikov, Estimates of the solutions of the nonstationary Navier–Stokes system, Boundary Value Problems of Mathematical Physics and Related Questions in the Theory of Functions. One paper suggested this was done through substituting a pressure update formula into the divergence formula. Plane Waves 4. Isotropic elastic data can be decomposed into P and S-wave potentials by De ne the wave number = ! p 0 Finite Element Methods for Maxwell’s Equations, Oxford University Press, 2003. Domain decomposition methods for the wave Heimholtz equation Using a new class of spectrally equivalent operators, it is shown that the convergence rates of the proposed methods have the same order of magnitude as those of similar methods used to solve mesh Poisson's equation. (2009) Parallel Iterative Substructuring in Structural Mechanics. Second- and fourth-order accurate finite difference discretizations are used. The de-composition shows that the solenoidal, irrotational and harmonic We consider the Helmholtz decomposition of the Lebesgue space L p (Ω). Helmholtz decomposition is wrong. 1 Half-spaced Problem with Stress Boundary Condition 2-2 Strictly orthogonal Helmholtz–Hodge decomposition An application to the construction of Lyapunov functions Summary Introduction Definition and basic properties Notation It is assumed that all vector fields are C2. 3 Helmholtz decomposition of a vector field 281 9. The propagation of waves through a medium is described by the famous wave equation. See also discussion in-class. Because Helmholtz's equation is linear, it is appropriate to attempt a Green's function method . Its discretization with piecewise linear finite elements results in typically large, ill-conditioned, indefinite, and non- In this paper, wave simulation with the finite difference method for the Helmholtz equation based on the domain decomposition method is investigated. In particular, we make use of a multiplicative decomposition of the solution of the Helmholtz equation into an analytical plane wave and a multiplier, which is the solution of a complexvalued advection-diffusion-reaction equation. constitutive relations, Navier equation, strain energy, Betti’s reciprocity theorem and integral representations, • wave motion in isotropic medium and Helmholtz decomposition • wave propagation across layers and Snell’s law • Rayleigh, Stoneley and other interface waves, and wave guides di erential equation with boundary conditions. Remark 1. 14 Jan 2013 4. The use of fast multigrid methods for Wave equations 1. Decomposition of a vector field into Helmholtz potentials: Any vector A pragmatic decomposition of a vector wavefield into P- and. This kind of theory leads to boundary layers of vorticity in asymptotic limits, but the fields are always coupled. Helmholtz wave equations play a central role in many fields, like These variants are based on domain-decomposition techniques, Taylor-series expansions, Solution of Inhomogeneous Wave Equation. The passage from the full time-dependent wave equation (W) to the Helmholtz equation (H) is nothing more, and nothing less, than a Fourier transform. In particular, we make use of a multiplicative decomposition of the solution of the Helmholtz equation into an analytical plane wave and a multiplier, which is the solution of a complex-valued advection-di usion-reaction equation. However, that does not seem sufficient enough. 5. using Helmholtz decompositions you can "translate" these equations into the following wave equations $$abla^2 \phi = \frac{1}{\alpha^2} \frac{\partial^2 \phi}{\partial t^2}$$ and PDF | The Helmholtz-Hodge Decomposition (HHD) describes the decomposition of a flow field into its divergence-free and curl-free components. 1 Acoustic waves Acoustic waves are propagating pressure disturbances in a gas or liquid. \end{aligned}}} {\displaystyle {\begin{aligned}\ mathbf {F} (\. not infinite, it follows that Ur176 and in fact is infinitely differentiable. As the first step towards the mathematical analysis of the domain decomposition method, preconditioners for mixed systems are studied. However, Helmholtz decomposition using conventional divergence and curl opera- 9. Helmholtz equation M. Unfortunately, nearly all Plane Wave Discretizations of the Helmholtz Equation with Lagrange Multipliers, and a Domain Decomposition Method for Resulting Systems Charbel Farhat, Radek Tezaur, and Jari Toivanen Institute for Computational and Mathematical Engineering, Stanford University, Mail Code 3035, Stanford, CA 94305, U. The relation is established between the compressional or shear far-field pattern for the elastic wave equation and the corresponding far-field pattern for the coupled Helmholtz equations. The frequency equations for wave propagation in an infinite rod were proposed. Druskin Schlumberger-Doll Research Old Quarry Road, Ridgefield, CT 06877 criteria above simultaneously. The P- and S-waves satisfy, respectively, 1 Answer. It has applications in seismic wave propagation, imaging and inversion. 1We recall that a flow around a boundary surface Sis called quasi-potential, if it is irrotational everywhere, except for the points that come in contact with S. For example, in geo-physics, particularly in oil exploration, such problems can be the forward problem in an iterative • brief review of continuum mechanics, e. By the fundamental theorem of vector calculus, or Helmholtz decomposition. A. For the body forces, we hypothesize that they can also be expressed in The Helmholtz decomposition gives rise to an exact theory of potential flow in the frame of the Navier–Stokes equations in which rotational and irrotational fields are tightly coupled and both fields depend on viscosity. The decomposition is orthogonal in a weighted L2 inner product and stable uniformly with respect to the jumps in the discontinuous weight function. This dissertation develops surface integral equations using constraint-based Helmholtz decompositions for electromagnetic modeling. viscous decay of free gravity waves as a free surface problem of this type. The new image condition maintains the polarity on both sides of the normal incident point and thus yields a consistent A Schwarz type domain decomposition method for the Helmholtz equation is considered. and S-wave vector potential ΨΨΨ, using the Helmholtz decomposition by the local inversion techniques is to use the shear wave equation as expressed . Weyl and Carle- man), in Variational Formulation to solve Helmholtz equation and time harmonic Maxwell equations. For the case of the Helmholtz equation and the two subdomain decomposition in Figure 3, the algorithm is (D +k2)un+1 1 = 0 in W 1; (D +k2)un+1 2 = 0 in W 2; u n+1 1 = u 2 on G 1; u n+1 2 = u n+1 1 on G 2: (7) The Dirichlet Problem for the Helmholtz Equation 215. Sha and Li Jun Jiang Department of Electrical and Electronic Engineering, The University of Hong Kong Hong Kong, China Email: xyxiong@eee. This vector function F is just a Exact Solutions > Linear Partial Differential Equations > Second-Order Elliptic Partial Differential Equations > Helmholtz Equation. Then permits statements as in Items (2){(4) in a strong form. Helmholtz decomposition principle was used to decompose. According to the above mentioned theorem, one can divide a given vector field ~v(~x) into a sum of two vector fields ~vl (~x) and ~vt ~x) where ~vl is irrotational (curl-free) and ~vt solenoidal (divergence-free), Rayleigh–Lamb wave equations from Navier–Lame elastody-namic equations in an elastic plate. This paper investigates if time-dependent electromagnetic radiation wave fields of point sources, which are of long range, satisfy these requirements. Parseval's theorem on the Fourier side gives . This derivation starts from the scalar wave equation, . 5-D Scalar Helmholtz Wave Solution Employing the Spectral Lanczos Decomposition Method (SLDM)y William H. So for any two distinct decompositions, Helmholtz decomposition. Problems 4. For a more detailed discussion, we refer readers to [24–26]. 4 Lam´e’s solution of the elastic wave equation 282 9. Sha, and Li Jun Jiang Department of Electrical and Electronic Engineering, The University of Hong Kong, Pokfulam, Hong Kong; Corresponding author: jianglj@hku. AVERBUCH§ Abstract. 12) with k= ω/c. Then, using the subspace of divergence free functions X~ ˆX, the compact embedding of Helmholtz and Wave Equations I. 2D Helmholtz problem with nearly a billion unknowns are solved e ciently with the preconditioner on massively parallel machines. “Inverse” Helmholtz Decomposition. The last equation of system (1) is the Sommerfeld radiation condition at in nity, which imposes that the scattered wave is outgoing. Widlund, Courant Institute of Mathematical Sciences, New York University Room 246 8:45 Multigrid Domain Decomposition Methods for the Helmholtz Problem Seongjai Kim, University of Kentucky; and William W. This is also an eigenvalue equation. Wave Motions with Polar Symmetry In physics and mathematics, in the area of vector calculus, Helmholtz's theorem, also known as . 8 The separation of wave modes from isotropic elastic wavefields is typically done using Helmholtz 9 decomposition. In general, it is that the vector field you are trying to decompose has to be sufficiently smooth and decay rapidly. they are both Fourier transforms of the form u(x,t)=∫∞ A Source Transfer Domain Decomposition Method For Helmholtz Equations in Unbounded Domain Part II: Extensions - Volume 6 Issue 3 - Zhiming Chen, Xueshuang Xiang Skip to main content We use cookies to distinguish you from other users and to provide you with a better experience on our websites. For the displace-ment field, we use the usual decomposition in terms of unknown scalar and vector poten-tials, U and H. In general the \pdgp finite element pair is second-order accurate, so this leads to very accurate wave propagation. Coarse Grid Helmholtz Equation Multigrid Method Domain Decomposition Method Schwarz Method These keywords were added by machine and not by the authors. Introduction In this paper, we consider the numerical solution of the Helmholtz equation, arising from the study of the wave equation in the frequency domain. 4. The paper contains a noniterative solver for the Helmholtz and the modified Helmholtz equations in a hexahedron. 6 Equation of motion in terms of displacements 1-12 1. Helmholtz's Second (Exterior) Formula 3. functions, the method uses the Helmholtz decomposition to split the displacement of the elastic wave equation into the compressional and shear waves, which satisfy a coupled boundary value problem of the Helmholtz equations. A Green’s function approach is used to solve many problems in geophysics. The domain derivative is studied for the coupled Helmholtz system. Combined Field Integral Equation Using a Constraint-Based Helmholtz Decomposition Abstract: A new version of the combined field integral equation (CFIE) is introduced for the formulation of time-harmonic electromagnetic wave interactions with perfect electric conductors (PECs). theorem is Rayleigh (1878), who referred to it as Helmholtz's. Furthermore, these Helmholtz equations, all of the form ∇ 2f +k f =0, (1) with f a scalar function and k the constant wavenumber, can be solved with a recently Helmholtz decomposition approach is applied to the inhomogeneous elastodynamic Navier–Lame equations for both the displacement field and body forces. Its discretization with piecewise linear finite elements results in typically large, ill-conditioned, indefinite, and non- Thus, Eq. The derivation of the Helmholtz theorem of vector decomposition of a three-vector field requires that the field satisfy certain convergence properties at spatial infinity. S-waves is based on the Helmholtz theory and the Christoffel equation. Performing 76 integration by parts elementwise leads to 77 ∑ T∈T curlH,e T + iωE,e T =0 ∀e ∈U ∑ T∈T E,curlh T − iωH,h T Certainly, you can try to find the Helmholtz decomposition on your sampled data, and find your irrotational and solenoidal components. Using the Helmholtz decomposition, the vector wave equation is split 15 Jul 2017 The Helmholtz equation (a. hk Received 20 December 2013 tional to the number of subdomains in one direction. ELASTIC IMAGING USING HELMHOLTZ DECOMPOSITION 2 ABSTRACT. A purpose of this paper is to show that the Helmholtz’s decomposition is a decomposition into closed and sterile fields as well. Usually we assume that a vector field goes to zero at infinity, which means it is uniquely specified by its divergence and curl. the elastic wave equation into the compressional and shear waves, which satisfy a coupled boundary value problem of the Helmholtz equations. Separation of scalar and vector potentials can be achieved by Helmholtz. Differential forms, tensor densities, the boundaries, the conjugation, etc (2013) A Hybridizable Discontinuous Galerkin Method for the Helmholtz Equation with High Wave Number. Helmholtz's First (Interior) Formula 3. It is shown that if a stability condition is satisfied, such a decomposition can be chosen so that its potential function is a Lyapunov function. Here is a pseudo-code for solving the Poisson equation resulting from the Helmholtz decomposition It is a difficult topic to construct an efficient preconditioner for Helmholtz equa-tion or Maxwell’s equations with large wave numbers. In this regard, it is timely to re-visit the boundary integral method of 8 solving the Helmholtz equation. We also introduce a material parameter α defined by α=µ/(λ+2µ), which is related to Poisson’s ratio ν by - n - n a = 2 2 1 2 by Poisson equation, whereas compressible flows have a wave equation behavior. The separation of wave modes for isotropic elastic wavefields is typically done using Helmholtz decomposition. u, [1] where divu 2div 2 , [2] curlu curl . Therefore, the sound propagation phenomenon is hidden by the use of this decomposition. t7) is infinitely differentiable for P0, pCB and B is. Summarized mathematically, we have: F=Fl+Ft where∇⋅F=∇⋅Fl and∇×F=∇×Ft 3. It is applicable to VTI media transverse (vortex) components (Helmholtz's theorem) is claimed in [1] to be inapplicable . Elastic imaging conditions based on Helmholtz decomposition Then, 2D approximation of Helmholtz decomposition, Equation 3, is applied to the modelled wavefield components. 2 Approximate Block-LU Decomposition Using MSSS Computa- tions for . 1 Answer. The idea of self-equilibration of irrotational viscous stresses is introduced. ot one- mens10 position. Helmholtz Equation ¢w + ‚w = –'(x) Many problems related to steady-state oscillations (mechanical, acoustical, thermal, electromag- netic) lead to the two-dimensional Helmholtz equation. During the inversion HELMHOLTZ DECOMPOSITION BASED ON INTEGRAL EQUATION METHOD FOR ELECTROMAGNETIC ANALYSIS Xiaoyan Y. Various communities like weather modeling, seismology The development of numerical methods for solving the Helmholtz equation, which behaves robustly with respect to the wave number, is a topic of vivid research. In this work, domain decomposition preconditioners for the Helmholtz and the vector valued wave equation are presented. We essentially reproduce a proof given by Solonnikov in [V. The Helmholtz decomposition approach (Helmholtz, 1858) is applied to the inhomogeneous elastodynamic Navier–Lame equation: (1) for the displacement field, we use the usual decom-position in terms of unknown scalar and vector poten-tials, F and ~H; (2) for the body forces, we hypothesize The second equation (the incompressibility equation) is the one I'm curious about. They propagate along the surface with the same velocity but they have different exponential laws of attenuation with depth. On the other hand, we also obtain the wave equation for the acoustic 9 Sep 2014 based on the linearized fluid equations relative to a state of rest, which . This equation can be solved algebraically for each . 3 The Helmholtz Decomposition and Further Results on Sobolev . With suitable boundary conditions, the decomposition is unique. A 2. Wikipedia:) ∇2A+k2A=0 The Helmholtz theorem ( or decomposition): Any vector field F being sufficiently smooth elastic wave-equation, followed by wavefield decomposition in P and S potentials . The Helmholtz-Hodge decomposition. 26 Sep 2006 The Helmholtz decomposition theorem says that every smooth vector field u, To study solutions of the Navier–Stokes equations, it is convenient to . The method, issued from domain decomposition techniques, l. Using a 2D velocity field as an example, Where is a vector potential, which in fluid mechanics is only guaranteed to exist if we're working in two dimensions so that , We are almost always satisfying Maxwell's equations (or any set of differential equations) with respect to some boundary conditions. We apply a modified Helmholtz decomposition operator to isolate P and S wave energy in the separate, back‐propagated wavefields and then apply a new imaging condition to the P and S mode wavefields to generate images. (2)The Helmholtz decomposition (3)Construction of vector potentials (4)The global div-curl lemma More speci cally, we will show the following: Let be a domain for which the coercivity estimate of Item (1) holds. Thus, many of the concepts studied in trans- mission line theory will help in understanding unbounded wave propagation. Because of this non-uniqueness, showing that it’s possible to express a particular velocity field with an incompressible flow as a Hemholtz decomposition with a nonzero $\phi$ doesn’t mean that it isn’t also possible to express the same velocity field as a different Hemholtz decomposition in which $\phi = 0$. 5 Stress-Strain Displacement Relations 1-10 1. The plane waves are associated with a scalar wave equation. Next we prove uniqueness. For homogeneous isotropic media, Zhang and McMechan (2010) proposed to rewrite equations 1 and 2 as k PUe ( k) = 0 and k Ue( k) = k UeP(k ) ; (6) and k UeS( k) = 0 and k Ue( k) = k UeS( k) : (7) Elastic imaging vs. Without them, it's not. Take the divergence of both sides to find that ∇2(ϕ−ϕ′)=0. tromagnetic waves, noting that it is of the O(1/r) and remarked: In Green's function for the vector wave equation in a mildly heterogeneous to employ Helmholtz's vector decomposition so as to generate two scalar wave 21 Oct 2011 The main difficulties in solving wave equations are the large number . 1 Half-spaced Problem with Stress Boundary Condition 2-2 Helmholtz decomposition In physics and mathematics , in the area of vector calculus , Helmholtz's theorem , also known as the fundamental theorem of vector calculus , states that any sufficiently smooth , rapidly decaying vector field in three dimensions can be resolved into the sum of an irrotational ( curl -free) vector field and a solenoidal ( divergence -free) vector field; this is known as the Helmholtz decomposition . First, the loop-tree is the quasi-Helmholtz-decomposition. (2010) Improved transmission conditions for a one-dimensional domain decomposition method applied to the solution of the Helmholtz equation. In that case, the boundary terms in the first two equations on the Wikipedia site each equal zero, and we are left with the free space solution. 14. Helmholtz decomposition princi-ple was used to decompose displacement to unknown scalar and The derivation of the Helmholtz theorem of vector decomposition of a 3-vector field requires that the field satisfy certain convergence properties at spatial infinity. The separation of anisotro- Helmholtz Decomposition The Helmholtz decomposition theorem says that every smooth vector field u, defined everywhere in space and vanishing at infinity together with its first derivatives can be decomposed into a rotational part and an irrotational part . Then ∇(ϕ−ϕ′)+∇×(G−G′)=0. 3 Helmholtz Decomposition Theorem 3. e. Unlike isotropic materials for which the Stokes-Helmholtz decomposition technique simplifies the problem, in anisotropic case no such general de-composition technique works. The formulation presented here is for the wave propagation in a cylindrical curved plate in the direction of the curvature as shown in Fig. Leningr. For example, suppose that F is a constant, e. It was observed that the solution of the Galerkin finite element method (FEM) differs significantly from the best approximation with increasing wave number. Journal of Computational Physics 229 :3, 851-874. 43 are described by the Helmholtz scalar wave equation in the frequency domain: r 2 2+k =0, (1) 44 where k = !/c is the wave number, ! the angular frequency and c the speed Frequency equations for waves in nfinite plates were presented by Rayleigh [1] and Lamb. I like this one because it too generalizes to the spin-2 case nicely, these methods to the scalar Helmholtz wave equation [3, 4, 5]. New preconditioned iterative solution methods for the Helmholtz equation are constructed. shown the central role of spatio-temporal wave-front (STWF) created by 27 Aug 2018 Solution approaches to wave equation ( Helmholtz equation) . The theorem states that under certain conditions, every vector field can be decomposed into a curl-free and a divergence-free component as: where and are the correspoding scalar and vector potentials of the field. For any solution v of the elastic wave equation (2. Section 2 of the paper introduces the problem and notation, and shows that the equations can be consid- 1. Another well known result is that any kind of vectorial field defined in a simply connected domain can be decompose in the following way: This decomposition is not unique because if we find two fields that fulfill the above statement every other couple of field also fulfills the statement ( and are arbitrary constants). is the Helmholtz decomposition of the domain derivative of the displacement for the elastic wave equation. I've read that this can be resolved through the pressure equation, or through the helmholtz-hodge decomposition. 27 Feb 2008 equations. The cost of this step is O(N3 logN), where N is the number of grid points in each of the three directions. Zepeda-Nu nez~ 2 R. The radial part of the solution of this equation is, unfortunately, not Equation (2. hk Received 20 December 2013 In this paper, wave simulation with the finite difference method for the Helmholtz equation based on the domain decomposition method is investigated. [3] This decomposition leads to the theory of the vector potential, A vector field can be written in terms of irrotational and a divergence-free components. It is apparently a scalar field ψ. 8:45 AM-9:45 AM Chair: Olof B. In this handout we will find the solution of this equation in spherical polar coordinates. Helmholtz Decomposition. transformed into a system of inhomogeneous wave equations in terms of known excitation . elastic waves in solids, b. A propagator solution is then shown. introduction to frequency-domain techniques 1 for solving the wave equation, details Multi-source Modeling Using LU Decomposition : As we will describe in later Since the Helmholtz equation is a steady-state elliptical PDE, sparse linear. I found out a solution to this equation, but merely by guessing. Scalar potential function is given as, 2 2 22 1 Ct L (9) Vector potential function is available as, 2 2 2 2 2 2 2 2 22 21 21 1 rr r T r T z z T t t Ct The source transfer domain decomposition method (STDDM) is introduced by the au- thors in [11] to solve the following 2D Helmholtz problems: u + k 2 u = f in R 2 , 1. 1 The Theorem { Words A vector eld vanishing at in nity is completely speci ed by its divergence and its curl if they are known throughout space. 6) is the decomposition of a form ω into the closed part o ω and the Poincare sterile part + ω. Nauchn. 3) into (2. Helmholtz decomposition theorem. wave propagation is modeled by the Helmholtz equation, 4u(x) + !2m(x)u(x) = f s(x); in (1) with absorbing boundary conditions, and where is a 3D rectangular domain, 4is the 3D Laplacian, x = (x;y;z), m = 1=c2(x) is the squared slowness for velocity c(x), u is the wave eld, and f s are the sources, indexed by s= 1;:::;R. We use the exp({m ) formulation is derived from the traditional Maxwell’s equations by a helicity decomposition, which decomposes Maxwell’s equations into three parts. For acoustic reverse-time migration, wavefield reconstruction is done with the acoustic wave-equation using the recorded scalar data as boundary condition. Steady-State Solutions in Two Dimensions 3. Taus 1 L. Then Eq. In physics and mathematics, in the area of vector calculus, Helmholtz's theorem, also known as the fundamental theorem of vector calculus, states that any sufficiently smooth, rapidly decaying vector field in three dimensions can be resolved into the sum of an irrotational Equation (2. That would be $$\Psi$$ = -xyz(x+y+z), and from there I found the two vector fields. The Helmholtz equation is reduced to a surface integral equation (SIE) posed at the interfaces between layers, e ciently solved via a nested version of the polarized traces to the wavelength of the incident wave through k= 2ˇ= . We will interchangeably call the wave carrier a ‘‘curved plate,’’ ‘‘cylinder,’’ ‘‘pipe segment,’’ or simply ‘‘pipe’’ all meaning the same thing. Obviously the wave fields are such that the superposition Certainly, you can try to find the Helmholtz decomposition on your sampled data, and find your irrotational and solenoidal components. to present a classical example known as Helmholtz decomposition theorem and to show the power of regularization in this case. The Helmholtz Decomposition In the following, the two equations (7) and (9) will be regarded as two di erent momentum balance equations that are, loosely speaking, a momen- In physics and mathematics, in the area of vector calculus, Helmholtz's theorem, also known as the fundamental theorem of vector calculus, states that any sufficiently smooth, rapidly decaying vector field in three dimensions can be resolved into the sum of an irrotational (curl-free) vector field and a solenoidal (divergence-free) vector field; this is known as the Helmholtz decomposition. In particular, we show that the domain The aim of this paper is the introduction of a new analytically regularizing procedure, based on Helmholtz decomposition and Galerkin method, successfully employed to analyze the electromagnetic scattering by zero‐thickness perfectly electrically conducting circular disk. main decomposition: a layered partition on the outer level and a further decomposition of each layer in cells at the inner level. This process is experimental and the keywords may be updated as the learning algorithm improves. Theorem 4. However, there are certain requirements on your original vector field you started with. However, Helmholtz decomposition using conventional di-vergence and curl operators does not give satisfactory results in anisotropic media and leaves the different wave modes only partially separated. hk Abstract—Helmholtz decomposition (HD) is a fundamental tool The passage from the full time-dependent wave equation $(\mathrm{W})$ to the Helmholtz equation $(\mathrm{H})$ is nothing more, and nothing less, than a Fourier transform. The time-independent form of the wave equation is called the Helmholtz equation. 5 The elastic wave equation with a concentrated force in the x coupled differential equations. Dispersion curves for anisotropic curved plates of different curvatures have been 3. In this paper, wave simulation with the finite difference method for the Helmholtz equation based on the domain decomposition method is investigated. 4 Equations of Motion in Terms of Stresses 1-7 1. The approach proposed here differs from those recently considered in the literature, in that it is based on a decomposition that is exact when considered analytically, so the only degradation in computational performance is due to discretization and roundoff errors. Numerical. And other says that if a flow has divergent component in the flow, there is no , and if there are vector potentials in the flow, the flow has no divergence. Seismic wave equation Helmholtz decomposition: Elastic wave-vector decomposition Wave-vector decomposition aims to decompose wave elds in the wavenumber domain via a projection operator. 1 Maxwell’s Equations Electromagnetic wave propagation is described by particular equations relating ve vector elds E, D, H, B, J and the scalar eld ˆ, where E and D denote the electric eld (in V=m) and electric displacement (in As=m2) respectively, while H and B denote the magnetic eld (in A=m) and magnetic ux density (in Vs=m2 = T=Tesla). The field that results when we take the divergence of F we will call the source of F's divergence. Suppose that (ϕ,G) and (ϕ′,G′) are two different decompositions for the same function. Helmholtz equation, domain decomposition method, PML. Since the integrand in (4. Linke 1 Abstract According to the Helmholtz decomposition, the irrotational parts of the mo-mentum balance equations of the incompressible Navier-Stokes equations are balanced by the pressure gradient. The scalability of the algorithm w. In physics and mathematics, in the area of vector calculus, Helmholtz's theorem, also known as the fundamental theorem of vector calculus, states that any sufficiently smooth, rapidly decaying vector field in three dimensions can be resolved into the sum of an irrotational (curl-free) vector field and a solenoidal (divergence-free) vector field; this is known as the Helmholtz decomposition. 4. For Helmholtz problems, where the classical domain decomposition methods in wave-equation form, as is done later. J. waves the aforementioned Helmholtz decomposition in combination There are three commonly used wavefield separation methods: the Helmholtz decomposition, vector decomposition, decoupled wave equations. Z. Helmholtz's Equation 3. 1 Physical models 1. Such a field is also referred to as monochromatic field. The. He added a proof, but he didn't provide any single motivation - e. and 0 I need Helmholtz decomposition to separate the solenoidal and dilatational velocity fields. So, In this work, I present the form of the Navier–Stokes equations implied by the Helmholtz decomposition in which the relation of the irrotational and rotational velocity fields is made explicit. State Key Laboratory of Scienti c and Engineering Computing, Chinese Academy of Helmholtz Decomposition on Proof. Numerical results are presented to demonstrate the e ectiveness of the proposed method. As an : We present an iterative domain decomposition method to solve the Helmholtz equation and related optimal control problems. by Pochhammer [9] in 1876 and independently, Chree [10] in 1889. vector Ψ(x,t), via Helmholtz' theorem. The Helmholtz decomposition theorem says that every smooth vector field u, defined everywhere in space and vanishing at infinity together with its first derivatives can be decomposed into a rotational part υ and an irrotational part ∇φ. We start from the mixed system of equations from above, 74 multiply the first equation with a test function e ∈U:=(L2(Ω))3 and the second one 75 with a function h ∈V:= H(curl,Ω) and integrate over the domain Ω. Time-Harmonic Plane Waves 4. To solve (1), we combine Absorbing Boundary Conditions (ABCs) with Lions-Despr es’ non-overlapping domain decomposition method. I need Helmholtz decomposition to separate the solenoidal and dilatational velocity fields. The Rayleigh-Lamb frequency equations are derived for elastic plate using the Helmholtz displacement decomposition. F = k^z. is the parallel component, and is the perpendicular component. 3. Helmholtz Equation • Wave equation in frequency domain Acoustics Electromagneics (Maxwell equations) Diffusion/heat transfer/boundary layers Telegraph, and related equations k can be complex • Quantum mechanics Klein-Gordan equation Shroedinger equation • Relativistic gravity (Yukawa potentials, k is purely imaginary) In isotropic elastic media, the Helmholtz decomposition states that a vector wavefield V (displacement or particle velocity wavefields) can be decomposed as (1) V = ∇ A + ∇ × B, where A is a scalar wavefield, called a scalar potential, and B is a vector wavefield, called a vector potential (Aki and Richards, 1980); ∇ and ∇ × are the gradient and curl operations, respectively. S. A simple proof is presented. The Helmholtz equation is the ver-sion of acoustic wave equation in the frequency domain. 4 namic elasticity using the Helmholtz decomposition method [3] and direct 5 field-only formulation of computational electromagnetics [4, 5, 6], all rely on 6 finding accurate and ecient methods of solving the scalar Helmholtz equa-7 tion. This paper investigates if time-dependent electromagnetic radiation wave fields of point sources, which are of long range, satisfy these requirements. Apply an Helmholtz decomposition to the compressible flow velocity. Equation 33 is the wave equation for the horizontally-polarized shear wave Decomposition of a vector field into Helmholtz potentials: Any vector field can be represented by a combination of the gradient of some scalar potential and the curl of a vector potential. A parallel solver for the Helmholtz equation in a domain consisting of layers with By means of the Helmholtz theorem on the decomposition of vector fields, the angular momentum of the classical electromagnetic field is decomposed, in a general and manifestly gauge invariant elastic wave equation are derived. Hence, the current by tree basis does not correctly depict the irrotational portion and it is not unique. That is, not to have divergence in the flow is the necessary and sufficient condition for exiting of vector potential . The BEM algorithm employs the Combined Helmholtz Integral Equation Formulation (CHIEF) and the Singular Value Decomposition (SVD) to solve the resulting system. In this paper 8 May 2003 waveform relaxation algorithm applied to the wave equation. 1) @p = 0. The conventional decomposition of a vector field into longitudinal (potential) and transverse (vortex) components (Helmholtz’s theorem) is claimed in [1] to be inapplicable to the time-dependent vector fields and, in particular, to the retarded solutions of Maxwell’s equations. Even though the loop basis is divergence free ðr J loop 50Þ, the tree basis is not curl free ðr3J tree 6¼ 0Þ. The Helmholtz equation - acoustic wave u + 2u = f in Rd: The Navier equation - elastic wave u + ( + ) The Helmholtz decomposition for the scattered eld gated using elastic wave equations, there is a need to separate P and S modes to obtain clean elastic images. This is a partial differential equation that is not easy to solve. Now this decomposition into a gradient and a curl is not necessarily unique. The proof of convergence of this method relies on energy techniques. In the subject of vector calculus, Helmholtz's theorem states that any sufficiently smooth function in the unit ball can be expressed as a sum of a curl-free, a divergence-free, and a harmonic vector field [4]. 4). The Helmholtz-Hodge decomposition In the subject of vector calculus, Helmholtz's theorem states that any sufficiently smooth function in the unit ball can be expressed as a sum of a curl-free, a divergence-free, and a harmonic vector field [4]. The Helmholtz decomposition approach is applied to the inhomogeneous of inhomogeneous wave equations in terms of known excitation potentials A* and Helmholtz decomposition theorem for vector elds is usually presented with too strong restrictions . The solver is based on domain decomposition. Weedon, Weng Cho Chew and Jiun-Hwa Lin Department of Electrical and Computer Engineering University of Illinois, Urbana, IL 61801 Apo Sezginer and Vladimir L. Symes, Rice University 9:05 Schwarz Methods for Helmholtz's Equation Elastic wave mode separation Using the Helmholtz decomposition theory (Aki and Richards, 1980; Morse and Feshbach, 1953), a vector wavefield can be decomposed into a curl-free P-wavefield and a divergence-free S-wavefield: . I am trying to prove the Helmholtz decomposition theorem which states that given a smooth vector field , there are a scalar field and a vector field such that However, during the proof I questioned myself at a point where we write every such vector field in the Dirac integral way and then use the identity Physics 110A Helmholtz’s theorem for vector functions Peter Young (Dated: January 4, 2009) This handout is a modi cation of Appendix B in Gri ths. I don't see how this can be correct, when the free-space solution vanishes at infinity, whereas the bounded problem should vanish at a bounded point. For sufficiently regular functions, both u and F can be written as superpositions of monochromatic fields, i. ISRAELI‡, AND A. These coupled differential equations are solved in this paper. Differential operators are represented in the Fourier basis by diagonal ma-trices. helmholtz decomposition wave equation
feif, 3kix3, fpcyik, xghb, tj3b, smj, hav, mmugj6, tvda, gdjdji, rcngd, | 2019-11-15T19:08:45 | {
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http://math.stackexchange.com/questions/257954/is-sup-n-sup-l-a-n-l-sup-l-sup-n-a-n-l-prove-or-disprove | # Is $\sup_n\sup_l a_{n,l} = \sup_l \sup_n a_{n,l}$? Prove or disprove.
Is $\sup_n\sup_l a_{n,l} = \sup_l \sup_n a_{n,l}$? Prove or disprove. I am preparing for my analysis final and this is one of the practice problems. Any help would be really appreciated! My try:
Let $A = \sup_n\sup_l a_{n,l} \ and\\x_n = \sup_l a_{n,l}\\$
$For\ all\ \epsilon > 0 \ \exists \ an \ N \ such\ that\ if \ n, \ l_* > N then\\$ $x_n - \epsilon < a_{n,l_*} <= x_n\\$
$\sup_n\ (x_n - \epsilon) < \sup_n\ a_{n,l_*} <= \sup_n\ x_n\\$
$A - \epsilon < \sup_n\ a_{n,l_*} <= A \\$
$A - \epsilon < \sup_l \sup_n\ a_{n,l_*} <= A \\$
$\ Let\ \epsilon\ go\ to\ zero$
$\ Therefore,\ \sup_l \sup_n a_{n,l} = A$
Thanks!
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It will prepare you better if you try something. – Davide Giraudo Dec 13 '12 at 14:53
I have already, i am just too lazy to write it down. – UH1 Dec 13 '12 at 15:00
ok i wrote it down. – UH1 Dec 13 '12 at 15:14
The $N$ you selected depends on $n$, as far as I can tell. So, when you take the supremum over all $n$, the inequality $x_n - \epsilon < a_{n,l^\ast} \leq x_n$ will no longer hold. – Zach L. Dec 13 '12 at 15:38
Hint: for each $n$ and $l$, $$a_{n,l}\leqslant \sup_j\sup_ka_{j,k}.$$ Taking the supremum over $n$, then over $k$, we get the first inequality. A similar argument for the other gives what we want.
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which 'first' inequality are you talking about? also are you sure your subscripts are consistent with the question? – UH1 Dec 13 '12 at 15:46
I don't know whether they are consistent as nothing is specified about them. By first inequality, I mean, proving that $A=B$ is the same as proving that $A\leqslant B$ and $B\leqslant A$. – Davide Giraudo Dec 13 '12 at 15:48
i wrote something but i think the reasoning is not sound, can you help me with proving at least one inequality? – UH1 Dec 13 '12 at 18:37
@UH1 It would be better if you write an answer. – Davide Giraudo Dec 13 '12 at 19:04
Let $C=\sup_{(n,l)}a_{n,l}$ and $A$ be your number above. Then obviously $C\ge A$. It is equally easy to see that $C\le A$. So $C=A$. Same proof for the other one. (Of course, you need to show a bit more details to get full credit. $\epsilon$-proof is not necessary.)
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@Giraudo, I didn't see your answer when I wrote mine. – TCL Dec 13 '12 at 15:56
No problem. Unrelated: I like your nickname! – Davide Giraudo Dec 13 '12 at 16:05 | 2015-11-28T06:08:55 | {
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https://www.physicsforums.com/threads/polynomial-equation-in-several-variables.409053/ | # Polynomial equation in several variables
1. Jun 9, 2010
### GargleBlast42
What is the most general solution to an equation of the form:
$$a_1 p_1 + \ldots + a_n p_n =0$$
where $$p_i$$ are given polynomials in several (N) variables with no common factor (i.e. their GCD is 1) and $$a_n$$ are the polynomials we are looking for (again in the same N variables). Of course, I'm asking for a nontrivial solution, i.e. not all a's are zero.
For the case where n=2, i.e. where I only have $$a_1 p_1+a_2 p_2 =0$$, this is easy - we obtain that $$a_1=-c p_2, a_2=c p_1$$, where c is an arbitrary polynomial (recall that $$p_1, p_2$$ have no common factor). Does something simmilar hold also for n>2?
2. Jun 9, 2010
### chingkui
There can be so many. If all you want is a nontrivial solution, then just use the a1, a2 as in your example, and set a3=...=an=0.
3. Jun 9, 2010
### GargleBlast42
Well, that is certainly a solution, but I would like to obtain the most general form.
Could one, for example, show that all such solutions have to have the form $$a_i=\sum_{j\neq i} c_j p_j$$, with the $$c_j$$ being some polynomials obeying some further relation (which is obtained by substituting this ansatz to the equation).
I know probably to little from algebra to be able to prove something like that. Or maybe it's just trivial and I don't see it? | 2017-12-15T08:44:15 | {
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https://truebeautyofmath.com/lesson-20-reconsidering-high-school-algebra/ | # Lesson 20: Reconsidering High School Algebra
A very large part of mathematics involves taking a mathematical structure that we’re familiar with and generalizing it to some “larger” structure (for more on this, see this note on abstract thought). When we make such a generalization, our original structure becomes a special case of this new, larger structure. Accordingly, we want to make sure that we “recover” our original object when we consider the corresponding special case. It is quite often the case that generalizing our mathematical structures in this way leads to new and deeper insight into the original structures themselves, as we then have a better idea of what’s “really” going on within the structure. If all of this doesn’t make too much sense right now, don’t worry, this whole lesson is devoted to an example of this.
But let me continue philosophizing for a bit. Why do mathematicians want to generalize a structure when it’s already familiar? There are several answers to this question. The most down and dirty “pure mathematician” way of answering it is to simply say “because we can”, and there’d be no harm in that. While it is definitely the case that “we can”, and that for some mathematicians this is sufficient motivation, I’d argue that even these “purest” of mathematicians see a larger goal in making such generalizations.
One very important motivation for generalizing structures is that the more general object might be easier to work with, and/or easier to prove things about. For example, suppose we want to prove some statement, call it P, about some mathematical structure that we’ll call A. I.e., we want to prove that for A, P is true. Further suppose that we’re thoroughly stumped. But suppose that we could generalize A to some “larger” structure, call it B. Then we know that anything that we can prove about B will be proved about A, because A is simply a special case of B. Now if we suppose that we could easily prove that for B, P is true, then we’ll know that P is also true for A, thus showing that the generalization of A to B was fruitful.
Another closely related reason for wanting to generalize a structure, and which we’ll explore here, is that such a generalization often gives us a deeper, firmer grounding for what we’re “really” doing. I.e., generalizations let us build a larger framework within which we can work. In our case, all of the structures that we’ll be studying in these lessons (until we study categories) are nothing but special cases of sets. I.e., our structures will be sets with added properties to them, and the deepest we can go down the rabbit hole of asking “but what really is this structure” is the answer “it’s a set!”. Think of this as an annoying five-year-old constantly asking “but why?”, only change “why” to “what is it”. Once our answer to that question is “it’s a set”, we’re done, there’s no further we can go (for now). If only there was such an answer for dealing with annoying five year olds…
Given this set-theoretic framework, let us now ask the question: what are we really doing when we’re doing high school algebra? Surely our framework of sets and functions would be less than useful, and surely not so many people would be studying it, if we couldn’t even “recover” what we did in high school. In particular, let me bring back what may or may not be a relatively painful memory from high school mathematics: graphing lines on a plane!
Chances are many readers have at least the vaguest memories of the following.
Plot on the plane:
$y=5x+6$
$y=\frac{4}{3}x-9$
$y=-2x+17$
Now, whether or not you remember this, and whether or not you hated it/loved it/excelled at it/failed at it, doesn’t matter here. The question I’m going to answer, and that which may have helped you at the time (if help was indeed needed), is the following: What the hell are we really even doing here? What is “the plane”? What is this line? Where does it come from?
Now, this question might seem so basic that it’s hardly even worth asking. It’s clear that the plane “is” the thing in figure 1. That’s what it is. End of discussion. And a line on the plane is just what’s shown in figure 2. Boom. Right there.
But that’s not satisfactory. While pictures are oftentimes very helpful in mathematics, they’re
Figure 1: There it is, that’s the plane
not at all mathematical. No math proof has ever been completed via looking at or drawing pictures, and no mathematical statement or structure has been expressed solely as a picture. In math, pictures don’t “exist” in any fundamental way—only mathematical structures and logic exist. Thus, it seems less than kosher to say that some mathematical structure “is” a picture, or “exists” on a picture. Without pictures though, this question of what the hell we’re doing when we’re plotting these lines becomes a bit trickier. Luckily for us, we already have all of the necessary tools to be able to answer this question pretty thoroughly.
Figure 2: Again, there it is, it’s right there.
Before doing so, let’s recall a very important set that we introduced in lesson 13, which was a lesson that I said could be skipped if so desired. Accordingly, I’ll reintroduce the necessary facts here. The set that we’ll consider is the set of all real numbers, and this set is often denoted in the industry as $\mathbb{R}$. It is the set of, essentially, all numbers. A bit more precisely, it is the set of all numbers that can be written as an infinite decimal expansion, something like
$235.3519352356295235235352350030235...$
where the “…” simply means that these numbers can go on forever with no rhyme or reason, if we so desire. We note that these include the more standard numbers like “5”, where we simply write 5.0000…, with 0’s repeated forever. We also get all the fractions too, but this will just have to be taken on faith. Most importantly, there are way more numbers here than just the fractions and whole numbers—this set was indeed our first example of a “new infinity”. This set is referred to as “the continuum”, because it forms a completely “continuous” set of numbers, allowing us to move from one number to the next “continuously”. It appears I can’t give too precise of a definition of these terms, but hopefully the concept is clear.
Given this “continuous” property of $\mathbb{R}$ (which I promise is the only thing that you’ll have to take on faith this lesson), we can relatively convincingly draw this set as a line. This line goes infinitely far in both directions and has no holes (hence, continuous). This line is simply our graphical representation of the set $\mathbb{R}$. The set exists whether or not we draw the line—the line just helps our visually-reliant brains to “see” what this set “looks like”. Scare quotes because a set never “looks like” anything, it’s just a set!
Now we get to use what we’ve created. In lesson 17 we studied the Cartesian product of two sets, which was essentially just the set of “pairs of elements”, one from each set. We also saw that we could take the Cartesian product of a set with itself, thus forming the set of “pairs of elements in a set”. Well, let us therefore construct the Cartesian product of $\mathbb{R}$ with itself: $\mathbb{R}\times \mathbb{R}$. We then have a set of pairs of elements, where each individual element in the pair is any element in $\mathbb{R}$. Namely, $\mathbb{R}\times \mathbb{R}=\{(a,b)|a,b\in \mathbb{R}\}$.
Now we notice that there’s a pretty nice way that we can graphically represent this set. In particular, we can take two of those lines that we supposedly drew in the previous paragraph (one line for each “copy” of $\mathbb{R}$ in the Cartesian product), tilt one 90 degrees relative to the other one, and then we put one on top of the other. Now, if we simply label some of the points on these lines, we’d have something that looks exactly like figure 1! Thus figure 1 is our graphical representation of the set $\mathbb{R}\times \mathbb{R}$. It doesn’t “exist” by itself, but rather just reminds us that we’re talking about $\mathbb{R}\times \mathbb{R}$. This is what the plane “really is”, and we can’t go any deeper.
So then what’s with the lines that we draw on this plane? What is the equation $y=6x+1$ “really” telling us? What it’s “really” doing is specifying a special subset of $\mathbb{R}\times \mathbb{R}$. How is it doing this? Well, let us first suppose that we make the arbitrary choice of labeling the lines that we’ve drawn to graphically represent $\mathbb{R}\times \mathbb{R}$ in the way that figure 1 is labeled. Note that this is an arbitrary choice because we could have had the negative numbers going off in the other direction for either of these two lines. But so be it—we make this choice. Now let us also make the arbitrary choice of writing our elements in $\mathbb{R}\times \mathbb{R}$ as $(x,y)$ with $x,y\in \mathbb{R}$. Then the above equation is, in this case, defining for us the following subset of $\mathbb{R}\times \mathbb{R}$:
$\{(x,y)\in \mathbb{R}\times \mathbb{R}|y=6x+1\}$.
In words, this means the set of pairs $(x,y)$ such that the “second slot” is 1 more than “6 times the first slot” (I use quotes to remind us of the order of things: we multiply by 6 first, then add 1). Thus, our subset only considers those very special points in $\mathbb{R}\times \mathbb{R}$ where the second slot has this very special relationship to the first one. Accordingly, $(2, 3)$ is not in this subset, because 3 is not 1 more than “6 times 2”, whereas $(2, 13)$ is in this subset because 7 is 1 more than “6 times 1” (okay, I’ll stop using quotes now and you’ll just have to remember the order).
Now let us make yet another arbitrary choice and let the line going horizontally in our graphical representation (figure 1) correspond to the “first slot” of the elements in $\mathbb{R}\times \mathbb{R}$, and let the vertical line correspond to the “second slot”. Now we can graphically represent our special subset by marking each point in this plane that has this very special property: that the second slot is 1 more than 6 times the first slot. When we make these infinitely many marks, we find that it forms a straight line on the plane!
Thus, what we’re “really” doing when answering the question “plot $y=6x+1$ on the plane” is graphically representing the particular subset of $\mathbb{R}\times \mathbb{R}$ defined by this equation, in the way we’ve described above. Note that I keep writing “graphically represent” in bold because I want to emphasize that what “really” exists is the set $\{(x,y)\in \mathbb{R}\times \mathbb{R}|y=6x+1\}$, and we’re simply representing this set by drawing this line. Thus, we now know what we’re representing when we draw these lines, and it’s nothing but sets!
I’d like to point out that there are other perspectives from which we can view this situation. One common way is to view all of this as the study of certain functions from $\mathbb{R}$ to $\mathbb{R}$, and our drawings are just representations of these functions by showing the entire domain and codomain of the function simultaneously, and drawing a mark on the points corresponding to each other via this function. I could have easily written this lesson from this perspective, but I think the way we’ve done things here is tidier. It is necessary to note that there are always several different perspectives to view a mathematical structure (just as there are several angles from which one can look upon a real world structure), and that we’ve simply chosen one.
Prior to this, though, it is likely that there was no perspective at all, and that we were just drawing these stupid lines all over the page with no idea of what we were “really” doing (I can’t speak for everyone, but this was certainly my view at the time). Hopefully this gives a little more perspective. As we’ll see throughout these lessons, all of the math that we’ve come to know and (possibly not) love throughout middle and high school can, and often should, be rephrased in these more abstract terms, as we now know precisely what’s really going on!
I’ll skip on exercises here, as you’ve likely already spent way too much time plotting lines on a plane in your lifetime. All I ask is that you take a moment to let this significant shift in perspective and this added meaning—to something that previously seemed so lacking in meaning—really sink in!
Next Lesson
Previous Lesson | 2017-09-19T16:52:55 | {
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http://math.stackexchange.com/questions/80671/the-normalizer-of-mathrmgln-mathbf-z-in-mathrmgln-mathbf-q | # The normalizer of $\mathrm{GL}(n,\mathbf Z)$ in $\mathrm{GL}(n,\mathbf Q)$
It seems that the normalizer of $H=\mathrm{GL}(n,\mathbf Z)$ in $G=\mathrm{GL}(n,\mathbf Q)$ is "almost" equal to itself, that is, $$N_G(\mathrm{GL}(n,\mathbf Z))=Z(G) \cdot \mathrm{GL}(n,\mathbf Z)$$ where $Z(G)$ is the centre of $G.$ Is there a simple proof/disproof of this fact? More generally, for which integral domains $R$ it is known that $\mathrm{GL}(n,R)$ "almost" coincides with its normalizer in the group $\mathrm{GL}(n,Q(R))$ where $Q(R)$ is the quotient field of $R?$
-
You need to add the scalar matrices, at least. – Plop Nov 10 '11 at 0:00
@Plop: I will, thank you. – Olod Nov 10 '11 at 7:34
Perhaps you should try asking this on MathOverflow. – Derek Holt Nov 11 '11 at 8:58
Right. $\phantom{abc}$ – Olod Nov 11 '11 at 9:44
The interested readers may find Emerton's answer to the question at mathoverflow.net/questions/80667/… – Olod Nov 11 '11 at 16:16
This is true for $GL_n(\mathbb{Z}_p)$ inside $GL_n(\mathbb{Q}_p)$ for every prime. Use $\prod\limits_p GL_n(\mathbb{Z}_p) \subset \prod\limits_p GL_n(\mathbb{Q}_p)$ and intersect with the diagonally embedded $GL_n(\mathbb{Q})$. This gives you the proof.
For an integral domain $R$ with quotient field $k$, $GL_n(R)$ is the stabilizer in $GL_n(k)$ of the $R$-module $R^n$ in $k^n$. Anything normalizing $GL_n(R)$ sends $R^n$ to another subset of $k^n$ stabilized by $GL_n(R)$. For a PID $R$, a given $0\not=v\in k^n$ can have its "denominator" collected-up so as to write $v=t\cdot v'$ with $v'$ primitive, $t\in k^\times$. Then $GL_n(R)\cdot v = t\cdot GL_n(R)\cdot w=t\cdot R^n$. That is, for a PID, the non-zero $GL_n(R)$ orbits are scalar multiples of $R^n$. Thus, anything $h$ in the normalizer of $GL_n(R)$ maps $R^n$ to $t\cdot R^n$ for some $t\in k^\times$, so, up to $GL_n(R)$, is a scalar. | 2015-09-02T20:00:28 | {
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https://math.stackexchange.com/questions/1361779/find-all-natural-values-n-that-sqrtp-2n-is-also-a-natural-number?noredirect=1 | # Find all natural values n, that $\sqrt{P_{2}(n)}$ is also a natural number
I have a polynomial of the second degree $a\cdot n^2 + b \cdot n + c$ and I need to find out natural numbers $n$, such that $\sqrt{a\cdot n^2 + b \cdot n + c}$ is also a natural number.
After thinking about this problem, my idea was to rewrite it into $a\cdot n^2 + b \cdot n + c = x^2$ and rewrite it to look like a Pell's equation:
$$(2an + b)^2 - 4ax^2= b^2 - 4 ac$$ which kind of resembles $x^2 - n y ^2 = 1$, but not really close enough to for me to solve it.
Then I tried to find some similarities in solutions for particular cases. For example when I took the equation $\sqrt{3\cdot n^2 - 2 \cdot n - 1}$ and wrote the program, the couple of first values were: 1, 5, 65, 901, 12545 (with no visible pattern for me).
So how should I solve this problem?
• Do you really want to have a general solution that works for all values of $a,b,c$? If you are eventually only interested in some special cases, it would be easier to tackle those. – Joonas Ilmavirta Jul 15 '15 at 7:51
• It's difficult to take your question seriously with that as your profile picture. – user230734 Jul 15 '15 at 7:53
• @JoonasIlmavirta if the general solution is too hard, you can show me how to come up with a solution for $3n^2 - 2n - 1$ – Salvador Dali Jul 15 '15 at 8:09
• If $c=k^2$ it is easy to rewrite in another Pell equation. In another case decided the same view through the Pell equation. – individ Jul 15 '15 at 8:41
• – individ Jul 15 '15 at 8:54
If you wish to have an infinite number of integer solutions to,
$$an^2+bn+c = d^2\tag1$$
but not necessarily all of them, then one way is, yes, to solve a Pell equation. First, as pointed out in the other answer, you need a initial solution. Second, if you limit it to only integers, then $a$ must not be a square.
Given an initial $n,d$ to $(1)$, then an infinite more can be found as,
$$ax^2+bx+c = (-d+py)^2$$
where,
$$x = n+qy$$
$$y = 2dp+(b+2an)q$$
and $p,q$ solve the Pell equation $p^2-aq^2 = 1$.
Example:
$$3n^2-2n-1 = d^2$$
with initial $n,d = 5,8$, you get,
$$3x^2-2x-1 = (-8+py)^2$$
where,
$$x = 5+qy$$
$$y = 16p+28q$$
and $p,q$ is any solution to $p^2-3q^2=1$.
Note: This is an easy method to get an infinite number of solutions, but not all of them.
• Any idea how can I find initial solution? – Salvador Dali Jul 15 '15 at 18:00
• @SalvadorDali: You can use mod arguments like in this answer. Or, since it is the smallest , then a simple program can quickly establish if it has a solution within a bound, say $|n|<1000$. If there is, the method above then guarantees there are an infinite more. – Tito Piezas III Jul 16 '15 at 2:49
With slight change of notation, you are looking for, given $a,b,c\in \mathbf{Z}$, points with integer co-ordinates on the curve $ax^2-y^2+bx+c=0$.
This is a conic. If you know one solution exists, then we can find infinitely many through rational parametrization. This is well-known (for example can be found in Silverman-Tate). Call this one solution point $Q_0$
Now fix a line $L$ whose equations has integer/rational coefficients. For a variable point $P\in L$ with RATONAL co-ordinates, write the equation of the line connecting $P$ with $Q_0$; as the intersection of a conic with line has 2 points you get the other point $P'$. This will be a rational point.(Convince yourself). So you get solutions you want. | 2019-07-22T16:43:40 | {
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https://stats.stackexchange.com/questions/56234/reducible-markov-chain-with-a-state-that-communicates-with-nothing | # Reducible Markov Chain with a state that communicates with nothing
Lets say there is a Markov chain whose transition matrix is defined as follows
$$P = \left( \begin{array}{cccc} 0.5 & 0.5 & 0 & 0 \\ 0 & 0.25 & 0 & 0.75 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right)$$
Now I've been told that communicating classes are those subsets of states communicate with each other, i.e. $P_{ij} > 0$ and $P_{ji} > 0$.
In the example above 1 communicates only with itself, 2 and 4 communicate with each other, but state 3 doesn't communicate with any other state, even itself.
So the communicating classes are {1} and {2,4}. But is {3} a such a class? It seems odd for it not to be, but it doesn't seem to match the definition as I've seen it. Presumably the chain is still reducible?
Edit: Fixed an issue where initially the top row of P failed to sum to one | 2021-09-20T16:56:15 | {
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https://mathhelpboards.com/threads/ivp-find-constants-to-solution.7321/ | # IVP find constants to solution
#### find_the_fun
##### Active member
The given family of functions is the general solution of the D.E. on the indicated interval. Find a member of the family that is a solution of the initial-value problem.
$$\displaystyle y=c_1x+c_2x\ln{x}$$ on $$\displaystyle (0, \infty)$$ and $$\displaystyle x^2y''-xy'+y=0$$ and y(1)=3, y'(1)=-1
So plugging in y(1)=3 gives $$\displaystyle 3=c_1+c_2\ln{1}$$ and then take the derivative to get $$\displaystyle y'=c_1+c_2 \ln{x} +c_2$$ subbing in $$\displaystyle -1=C-1+c_2\ln{1}+c_2$$
adding 3 times the second equation to the first give $$\displaystyle 0=4c_1+4c_2\ln{1}+3c_2$$
What next?
#### MarkFL
Staff member
Okay, we have:
$$\displaystyle y(1)=c_1=3$$ (recall $\log_a(1)=0$)
$$\displaystyle y'(1)=c_1+c_2=-1$$
Now the system is easier to solve.
#### find_the_fun
##### Active member
Okay, we have:
$$\displaystyle y(1)=c_1=3$$ (recall $\log_a(1)=0$)
$$\displaystyle y'(1)=c_1+c_2=-1$$
Now the system is easier to solve.
Ok but we never used $$\displaystyle x^2y''-xy'+y=0$$ Why was that in the question?
#### MarkFL
Ok but we never used $$\displaystyle x^2y''-xy'+y=0$$ Why was that in the question?
$$\displaystyle x=e^t$$
$$\displaystyle y=x^r$$ | 2021-09-23T20:11:36 | {
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https://forum.poshenloh.com/category/544/week-challenge | @divinedolphin The "pairwise relatively prime" condition is stronger than "relatively prime." For example, if two numbers have $$2$$ as a common factor, but the third number doesn't, then the three numbers together are relatively prime. Relatively prime means that there is no factor common to all numbers in the group . Pairwise relatively prime means that if you look at all possible pairs of the numbers, there is not a pair that share a common factor. $$\textcolor{red}{4 \text{ } \text{ } 10}\text{ } \text{ } 7$$ $$\text{These are relatively prime (because 7 doesn't have 2 as a factor)}$$ $$\textcolor{red}{4 \text{ } \text{ } 10}\text{ } \text{ } 7$$ \begin{aligned} \text{Look at them pairwise: } &\textcolor{red}{4} \text{ and } \textcolor{red}{10} \text{ are both multiples of 2 }; \\ &\textcolor{red}{4} \text{ and } 7 \text{ are not both multiples of some number,}\\ \text{ and } &\textcolor{red}{10} \text{ and } 7 \text{ are not both multiples of some number} \end{aligned} $$\text{These three numbers are not pairwise relatively prime. }$$ $$\textcolor{blue}{44 \text{ } \text{ } 13 \text{ }\text{ } 25 }$$ $$\text{ These are not all multiples of the same } x, \text{ for some number } x, \text{ so they are relatively prime}$$ $$\textcolor{blue}{44 \text{ } \text{ } 13 \text{ }\text{ } 25 }$$ \begin{aligned} \text{Look at them pairwise: } &\textcolor{blue}{44} \text{ and } \textcolor{blue}{13} \text{ are not both multiples of some number }; \\ &\textcolor{blue}{13} \text{ and } \textcolor{blue}{25} \text{ are not both multiples of some number,}\\ \text{ and } &\textcolor{blue}{44} \text{ and } \textcolor{blue}{25} \text{ are not both multiples of some number} \end{aligned} $$\text{These three numbers are also pairwise relatively prime.}$$ | 2020-07-13T07:54:33 | {
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https://socratic.org/questions/a-volume-of-helium-occupies-11-0-l-at-a-pressure-of-98-0-kpa-what-is-the-new-vol | # A volume of helium occupies 11.0 L at a pressure of 98.0 kPa. What is the new volume if the pressure drops to 86.2 kPa?
Dec 18, 2016
#### Answer:
The new volume is $= 12.5 L$
#### Explanation:
We use Boyle's Law
${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$
Pressure, ${P}_{1} = 98.0 k P a$
Volume, ${V}_{1} = 11.0 L$
Pressure, ${P}_{2} = 86.2 k P a$
Volume, ${V}_{2} = \frac{{P}_{1} {V}_{1}}{P} _ 2$
$= 98.0 \cdot \frac{11.0}{86.2} = 12.5 L$ | 2019-09-23T05:03:53 | {
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http://math.stackexchange.com/questions/9532/link-between-a-dense-subset-and-a-continuous-mapping?answertab=votes | # Link between a Dense subset and a Continuous mapping
arising out of comment made by Yuval Filmus in what is the cardinality of set of all smooth functions in $L^1$ ? I got this idea (forgive me for my ignorance for if it is nothing but an elementary definition/result in real analysis). The idea is like this Let $f:X\to Y$ is a mapping, where $X$ is a complete metric space (not sure if its strictly needed or whether a looser condition would do). If $f$ is a continuous mapping then $f$ is uniquely specified by a mapping $g:E\to Y$ where $E$ is a dense subset of $X$. What are the condition under which it is valid ? also the validity of the converse statement.
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If $Y$ is a Hausdorff topological space, then the value of a continuous function $f\colon X\to Y$ is completely determined by the value of $f$ on a dense subset $E$ of $X$; to see this, suppose $f$ and $g$ are two functions that agree on a dense subset $E$ of $X$, and let $u\in X\setminus E$. If $f(u)\neq g(u)$, then there are open neighborhoods $U$ and $V$ of $f(u)$ and $g(u)$, respectively, such that $U\cap V=\emptyset$. Then $f^{-1}(U)$ is an open neighborhood of $u$, as is $g^{-1}(V)$. Their intersection is an open neighborhood of $u$, and therefore must contain elements of $E$; but then any $e\in E$ in the intersection has $f(e)=g(e)$, with $f(e)\in U$ and $g(e)\in V$, contradicting that $U\cap V=\emptyset$. Therefore, $f(u)=g(u)$, hence $f=g$.
In this situation, it does not matter if $X$ is a complete metric space (or even a metric space); the key is $Y$.
To see that the key is $Y$, consider the extreme case in which $Y$ has the indiscrete topology (the only open sets are the empty set and the set $Y$). Then any function into $Y$ is continuous, so you can make your $X$ anything you want, and have two functions that agree on any subspace you care to specify and yet differ somewhere else.
Added: Note that any metric space is necessarily Hausdorff, so if your maps are between metric spaces, then the property holds (as in the case of Yuval's answer). This because you have the property that $d(x,y)\geq 0$ and $d(x,y)=0$ if and only if $x=y$. Thus, given $x,y\in Y$, $x\neq y$, let $\epsilon=d(x,y)\gt 0$. Then $B(x,\frac{\epsilon}{2})$ and $B(y,\frac{\epsilon}{2})$ are open neighborhoods of $x$ and $y$ (respectively) that are disjoint: if $z$ were in the intersection, then $d(x,z)\lt \frac{\epsilon}{2}$, $d(z,y)\lt\frac{\epsilon}{2}$, and by the triangle inequality we would conclude that $d(x,y)\lt\epsilon$, a contradiction.
Final Addition: See here for a discussion of the converse.
Added 2: A result towards a possible converse: certainly we need some separation on $Y$. Suppose that there exist points $x,y\in Y$, $x\neq y$, such that every open subset that contains $x$ also contains $y$ (so $Y$ could be $T_0$, but cannot be $T_1$). Let $X$ be the Sierpinski space, $X=\{a,b\}$ with topology $\tau = \{\emptyset, \{b\}, X\}$. Let $f,g\colon X\to Y$ be the following maps: $f$ is the constant function that maps everything to $x$; $g$ is the function that maps $a\mapsto x$, $b\mapsto y$. The constantfunction is certainly continuous. For $g$, if $U$ is an open subset that contains $y$ but not $x$, then $g^{-1}(U)=\{b\}$, which is open; so $g$ is continuous. And $g$ and $f$ agree on the dense set $\{a\}$, but are not equal. I'm still trying to figure out the $T_1$ case (for all $x,y\in Y$, $x\neq y$, there exist open sets $U,V$ such that $x\in U-V$ and $y\in V-U$).
Added 3: Another step: $T_1$ is not sufficient; a colleague came up with this one (I kept trying the cofinite topology on $\mathbb{N}$ and not getting anywhere): take two copies of the real line and identify every point except the origin; this is $Y$. The result is a $T_1$ space, but not Hausdorff since no neighborhoods of the two copies of the origin are disjoint. Now let $X$ be the real line, and let $E = (-\infty,0)\cup(0,\infty)$ be the dense subset. The two obvious injections, one mapping $0$ to the first copy in $Y$ and the other mapping it to the other copy, are both continuous and agree on $E$ but not on all of $X$, so $T_1$ does not suffice for the property. At least, then, in the hierarchy of $T$-spaces, the first level at which we are guaranteed the property is Hausdorff. This does not, however, establish whether the converse property characterises Hausdorff-ness.
Added 4. I almost have the following:
Conjecture. Assuming the Axiom of Choice, the following are equivalent for a topological space $Y$:
1. $Y$ is Hausdorff.
2. For every topological space $X$, every dense subset $E$ of $X$, and every pair of continuous maps $f,g\colon X\to Y$, if $f$ and $g$ agree on $E$, then $f=g$.
Argument. Suppose that $Y$ is not Hausdorff. If there exist points $x,y\in Y$, $x\neq y$, such that every open neighborhood of $x$ contains $y$, then the map from the Sierpinski space indicated above shows that $Y$ does not have property 2. So we may assume that $Y$ is at least a $T_1$ space. But since $Y$ is not Hausdorff, there exist points $s,t\in Y$, $s\neq t$, such that for every neighborhoods $U$ of $s$ and $V$ of $t$ such that $s\in U-V$ and $t\in V-U$, we have $U\cap V\neq \emptyset$. Let $\mathfrak{U}\_s$ be the family of open neighborhoods of $s$ that do not contain $t$, and let $\mathfrak{V}\_t$ be the family of open neighborhoods of $t$ that do not contain $s$. Let $P=\mathfrak{U}\_s\times \mathfrak{V}\_t$, and partially order $P$ by letting $(U,V)\leq (U',V')$ if and only if $U'\subseteq U$ and $V'\subseteq V$. This makes $P$ into a directed partially ordered set (given any $(U,V),(R,S)\in P$, there exists $(U',V')\in P$ such that $(U,V)\leq (U',V')$ and $(R,S)\leq (U',V')$. Now, for each $(U,V)\in P$, we are assuming that $U\cap V\neq\emptyset$, so using the Axiom of Choice let $y_{(U,V)}\in Y$ be an element of $U\cap V$. Note that $y_{(U,V)}\neq s$ and $y_{(U,V)}\neq t$ for all $(U,V)\in P$.
Now let $X = \{y_{(U,V)}\mid (U,V)\in P\}\cup\{s,t\}$, and give $X$ the induced topology from $Y$, so that the inclusion maps $X\hookrightarrow Y$ is continuous. Note that the set $E=\{y_{(U,V)}\mid (U,V)\in P\}$ is dense in $X$: for every open neighborhood $B$ of $s$ in $X$, there exists an open set $\mathcal{O}\_B\in Y$ such that $\mathcal{O}\_B\cap X = B$; in particular, $\mathcal{O}\_B$ is a neighborhood of $s$; let $\mathcal{V}$ be any open neighborhood of $t$ that does not contain $S$, and let $B'=\mathcal{V}\cap X$; let $\mathcal{U}$ be any open neighborhood of $s$ that does not contain $t$. Then $\mathcal{U}\cap\mathcal{O}\_B$ is open, hence $B'=\mathcal{U}\cap\mathcal{O}\_B\cap X$ is an open subset of $X$ that is contained in $B$. Consider now $y_{(\mathcal{U}\cap\mathcal{O}\_B,\mathcal{V})}$. This is in $\mathcal{U}\cap\mathcal{O}_B\cap\mathcal{V}\cap X \subseteq B'$, and is plainly in $E$. In particular, in $X$ we have that $B'\cap E\neq\emptyset$, and hence $B\cap E\neq \emptyset$. Thus, every open neighborhood of $s$ in $X$ contains points of $E$, hence $x$ lies in the closure of $E$. A symmetric argument holds for $t$. Thus, $E$ is dense in $X$.
Now consider the maps $f,g\colon X\to Y$ defined as follows: $f$ and $g$, restricted to $E$, are the identity; $f(s)=f(t)=s$; and $g(s)=g(t)=t$. I claim that $f$ and $g$ are both continuous. Indeed, let $\mathcal{O}$ be an open set in $Y$. If $\mathcal{O}\cap\{s,t\}=\emptyset$ or $\{s,t\}\subseteq\mathcal{O}$, there is nothing to do: the inverse image under both $f$ and $g$ is just the intersection with $X$, hence open in $X$. So assume without loss of generality that $s\in\mathcal{O}$ but $t\notin \mathcal{O}$. Note that $g^{-1}(\mathcal{O}) = (\mathcal{O}-\{s\})\cap X$, and since $Y$ is $T_1$ removing a single point from an open set results in an open set, so $g^{-1}(\mathcal{O})$ is open. So we just need to show that $f^{-1}(\mathcal{O}) = (\mathcal{O}\cap X)\cup\{t\}$ is open in $X$.
And that is where I am a bit stuck at present. Can anyone either verify or falsify this?
-
Thanks Arturo. That was a very nice answer. – Rajesh D Nov 9 '10 at 5:09
@Rajesh D: I'm actually trying to figure out if the converse holds: suppose $Y$ has the property that for any topological space $X$ and dense subset $E$ of $X$, if $f,g\colon X\to Y$ agree on $E$ then $f=g$. Does it follow that $Y$ is Hausdorff? I'm not sure yet. – Arturo Magidin Nov 9 '10 at 5:11
Does the converse hold ? – Rajesh D Nov 9 '10 at 14:22
Thank you for the reply. I am tryin to catch up with your attempt to prove the converse. I was not even aware of even elementary concepts of a Topological spaces before I read your answer.Serving as a nice motivation to learn. – Rajesh D Nov 9 '10 at 16:33
@Arturo: thanks for pointing out. I was reading a book on introduction to functional analysis few months back.The first chapter was on metric spaces and in the middle of that chapter, after defining the concepts of metric space and open set, the author just mentioned (for no apparent reason) that there is something called a Topology on a set X which is nothing but the set X and the collection all its open sets. Only after seeing your comment and having a look at wikipaedia i realize that the definition of an open set in topology and that in Metric space are different.... – Rajesh D Nov 11 '10 at 19:57 | 2015-09-02T22:22:54 | {
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https://homework.cpm.org/category/CC/textbook/cca2/chapter/12/lesson/12.2.2/problem/12-113 | ### Home > CCA2 > Chapter 12 > Lesson 12.2.2 > Problem12-113
12-113.
Andrea has just purchased a five-digit combination lock that allows her to set up her own combination. She can use the numbers $0$ through $9$ for her combination, and she must use five digits.
1. How many five-digit combinations can she make so that no digit is repeated?
Although the problem says 'combination' this is actually a permutation
because the order you enter the numbers in the combination lock matters.
2. How many five-digit combinations are possible if she can repeat the digits, but cannot use the same digit twice in a row?
Since this is a special case you can use a decision chart.
How many choices does she have for each position?
$\frac{ }{1\text{st}}\frac{ }{2\text{nd}}\frac{ }{3\text{rd}}\frac{ }{4\text{th}}\frac{ }{5\text{th}}$
$\frac{10}{1\text{st}} \frac{9}{2\text{nd}} \frac{9}{3\text{rd}} \frac{9}{4\text{th}} \frac{9}{5\text{th}}$ | 2021-05-06T00:25:44 | {
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https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-9-section-9-3-the-parabola-exercise-set-page-997/61 | ## Precalculus (6th Edition) Blitzer
domain $(-\infty,3]$, range $(-\infty, \infty)$, not a function.
Step 1. Rewrite the equation as $(y-1)^2=-\frac{1}{4}(x-3)$ We see that the parabola opens to the left with a vertex at $(3,1)$. Step 2. We can find the domain as $(-\infty,3]$ and range as $(-\infty, \infty)$ Step 3. The relation is not a function as it will fail a vertical line test. | 2023-03-30T04:26:07 | {
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http://gsb420.blogspot.com/2008/01/ | ## Thursday, January 31, 2008
### The Midterm
This just in.... Midterm has craaaaaaaazy insurance claim problem!
I'll tell you what I got. I can't promise that it's right, but if you think I got it wrong, post a comment or shoot me an email. I'm doing this from memory, so if I'm not remembering the question correctly in some substantial way, let me know for sure.
Q. An insurance company determines that they receive an average on nine claims in an average week. Claims in any given week do not effect the chances of claims in other weeks.
For a given week:
A. What is the probability of receiving 7 claims.
B. Given that there was a claim, what is the probability that there were 7 claims.
Answer: Ok, this is clearly an application of the Poisson distribution.
For part A, just plug the numbers into the formula and you get
f(x=7) = (e^-9)(9^7)/7! = (0.0001234)(4782969)/5040 = 0.1171 = 11.71%
BTW, I found a handy-dandy Poisson distribution probability calculator online at Stattrek. I plugged in our numbers there and it calculated the same answer. Ok, that was the easy part. Now on to part B...
Part B tells us that there was at least one claim. We can calculate P(x>=1) = 1 - P(x=0). And P(x=0) = (e^-9)(9^0)/0! = e^-9 = 0.0001234. This makes sense intuitively because the probability of there being no claims at all when the average is 9 is pretty slim. Really slim, in fact.
So, P(x>=1) = 1-0.0001234 = 0.9998766. And we're being asked to determine P(x=7 | x>=1).
By definition of conditional probability:
P(x=7 | x>=1) = P( x=7 and x>=1) / P(x>=1)
We know the denominator. We calculated it to be 0.9998766.
To calculate the numerator (and though we really don't need to calculate anything, we'll try to be rigorous), we note that P(x=7) is clearly mutually exclusive from P(x=0) and therefore P(x=7 and x=0) = 0. And since x=0 and x>=0 are exhaustive of the sample space, we therefore know that P(x=7 and x>=0) = P(x=7). That's our numerator!
So, P(x=7 | x>=1) = P( x=7) / P(x>=1) = 0.1171 / 0.9998766 = 0.1171145 = 11.71145%
This is only slightly more than the probability in part A, but that makes sense since the condition in part B that there was at least one claim didn't significantly decrease the sample space.
That was the only really tricky question on the midterm, imho, although there were plenty of places that one could make a careless mistake. I noticed that most of the incorrect multiple choice answers were offered up to entice you if you made a quick mistake and didn't check your work.
Slightly tricky: Two mutually exclusive events are not independent. This is evident when you look at the test for independence:
P(A and B ) = P(A)P(B)
Well, if they events are mutually exclusive then P(A and B) = 0. Therefore it's impossible for them to be independent if both P(A) and P(B) are non-zero.
Answer to the bonus question: Snow and cold weather are independent since P(snow and cold) = P(snow)P(cold)
### The best team in NFL history
Both the 1972 Miami Dolphins and the 2007 New England Patriots were undefeated in the regular season. Dr. Jonathan Sackner-Bernstein, an admitted Dolphins fan, has looked at the question of which was the better team. Since, of course, they could never play each other, we can't know with certainty. But we can analyze the two teams statistically and interpret the results.
You can read the article at theheart.org, but you'll need to go through the free registration process to get to the full text. I'll summarize what I understand here:
Sackner-Bernstein's analysis involves normalizing the data from the individual game results of both teams with data from their opposing teams. I believe the purpose here is to compare Miami and New England relative to how all the teams in the league are doing. In other words, when we look at points scored, if all teams are scoring more points in 2007 than in 1972 (for whatever reason), the normalization factor will scale down the New England number of points scored to be relative to the average points scored in 1972.
Another factor in the analysis is to assess the strength of both Miami and New England relative to the teams they faced. For example, if Miami beat their opponents by a wide margin and their opponents were all relatively good teams while New England beat their opponents by a thin margin and their opponents were relatively weak, we would conclude that Miami was the better team. In order to normalize for that factor, Sackner-Bernstein multiplies the statistics for each game by the season-end winning percentage of each opponent. For example, a 20-10 win over a team that had a winning percentage of .500 would count as 10-5, whereas a 27-21 win over a team that ended the season at .333, would count as 9-7.
The analysis also calculated stats on a game-by-game basis to determine if there was a trend in either team as performing better or worse over the course of the season.
Therefore, Sackner-Bernstein first gathered data on the outcomes of all the 14 Miami games and 16 New England games. Then he gathered data on all the other teams in the league. The four statistics that he gathered were:
1. points scored
2. points against
3. total yards offense
4. total yards defense
The points and yardage differentials were also considered.
You can see the tabulation of the results in the article. I'm interested in get hold of the mean values for the league and I've contacted Sackner-Bernstein's organization to try to obtain those data. They would make a great sample exercise. I'm also trying to understanding the p-value concept. I'll have to wait for chapter 9 (pg 286) to get into that.
Sackner-Bernstein concludes:
Thus, this analysis concludes that the regular season performances of the 1972 Miami Dolphins and the 2007 New England Patriots are statistically indistinguishable in terms of dominance over their opponents in each of their respective seasons. Under the assumption that the Patriots reign victorious in this year's Super Bowl, one can only conclude that the two teams are equally dominant and should be considered the two teams tied for the best single-season performances in the NFL's modern era.
## Wednesday, January 30, 2008
### Midterm tip
In addition to bringing along a hard copy of the formula sheet, be sure to bring a calculator to the midterm. You'll probably want one with the ability to do xy so you can use it for the Binomial or Poisson distribution calculations.
### The Birthday Problem
A relatively well-known exercise in probability is to determine the probability that two (or more) people share a birthday in a group of n people.
The answer is pretty surprising to most people. It turns out that in a group of 23 people, there's more than a 50% chance that two (or more) people have the same birthday.
The way to solve this problem is to first look at the reverse problem. I.e. what is the probability of everyone having different birthdays. Well, let's look at the people individually. The first person can have any birthday, so there are 365 possibilities for him. The second person can't have the same as the first one's bday, so there are 364 possibilities for the second person. The 363 for the third person and 362 for the fourth and so on.
Since the events (the selection of birthdays for each person) are independent, the number of successful outcomes is:
365 x 364 x 363 x ... x (365-n+1) where n is the number of people in the group
The total number of possible birthday combinations is 365^n.
So the probability of 23 people all having different birthdays is:
P(23) = (365 x 364 x 363 x … x 343) / (365^23)
And the probability of two (or more) people having the same birthday is just the inverse:
P’(23) = 1 - P(23)
The calculation of P(23) is pretty ugly, but can be computed relatively accurately by most scientific calculators and personal computers. P(23) = 0.493, so P'(23) is 50.7%.
Taking it a step further
This is the point where most people stop, but a few follow-up questions come up:
1. Is there a way to simplify the formula, even with an approximation, that would make the computation quicker and easier?
2. At the same time, can we generalize the formula beyond the 365 possible birthdays and develop a way to calculate (or estimate) the probability of selecting n items all of different type from an infinite supply of items with x different types. Similarly, what is the probability of two items having the same type within a selection of n items.
It turns out that these questions have actually been studied quite a bit. In 1966, E. H. McKinney of Ball State University wrote a piece in American Mathematical Monthly (v. 73, No. 4, pp. 385-387) where he derived a formula for the problem of finding the smallest value of n such that the probability is great than or equal to 1/2 that at least r people have the same birthday.
McKinney did the computation for r = 2, 3 and 4 and reports the answers as 23, 88 and 187 respectively. What I really found interesting was the following note:
The author did not carry out the computation for r=5 since the machine time [on an IBM 7090 computer] for one computation of [the formula] for a given n was estimated at two hours.
I located the IBM history site on the 7090 Data Processing System. It's really fascinating and worth a read. But what really caught my eye was the bottom line:
The IBM 7090, which is manufactured at IBM's Poughkeepsie, N. Y. plant, sells for $2,898,000 and rents for$63,500 a month in a typical configuration.
Wow! For almost $3 million, you got a computer that would take 2 hours to calculate what you could probably do in a fraction of a second today. I say "probably", because I don't totally understand McKinney's formula yet and I haven't implemented it. I'm still working on that part. :) More on approximations of the birthday probability later... ## Tuesday, January 29, 2008 ### Binomial or Poisson? I found the following handy tip on Jonathan Deane's site (University of Surrey): • If a mean or average probability of an event happening per unit time/per page/per mile cycled etc., is given, and you are asked to calculate a probability of n events happening in a given time/number of pages/number of miles cycled, then the Poisson Distribution is used. • If, on the other hand, an exact probability of an event happening is given, or implied, in the question, and you are asked to caclulate the probability of this event happening k times out of n, then the Binomial Distribution must be used. He also has some sample problems on that page what are worth a try for midterm practice. ## Sunday, January 27, 2008 ### Learning, teaching and the Moore method Irving Kaplansky Tonight I listened to a recording of a lecture given by the late Irving Kaplansky, who was a mathematics professor at the University of Chicago from 1945-1984. The lecture, entitled "Fun with Mathematics: Some Thoughts from Seven Decades", is a fascinating combination of personal anecdotes/memoir/autobiography and advice on how to do research and advice on how to be successful in general. The lecture is available from the Mathematical Sciences Research Institute site. The four main pieces of advice that he outlines are: I. Search the literature (he spends a lot of time on this topic and much of what he says has probably changed in the past few years with the introduction of Google Scholar) II. Keep your notes (write a bound notebook and date your stuff) III. Reach out (talk to people) IV. Try to learn something new everyday R. L. Moore During the lecture, Kaplansky mentions R. L. Moore and his teaching style. Moore discouraged literature search by his students. He wanted them to be able to come up with everything on their own, even if the work had already been done before. I did a quick Google search for Moore and found the main site at the University of Texas. The methodology is called Inquiry-Based Learning (IBL) or, sometimes, Discovery Learning. The concept is that the instructor guides the students to develop the concepts, solve the problems and create the proofs themselves. Fascinating! I really like this innovative teaching concept and the way it ties the teacher and student together more closely than the typical lecture does. There's lots of material on this method on the UTexas site that I'm going to delve into. I think part of the reason I'm writing this blog is because I feel like if I can say something back in my own words, I didn't really learn and understand it in the first place. Or, as Richard Feynman is said to have said "What I cannot create, I do not understand." A good background article on Moore and his method was published in 1977 in The American Mathematical Monthly (vol. 84, No 4, pp. 273-278) by F. Burton Jones. It's available on JSTOR via the Depaul online library. My initial take on the Moore Method is that it works well in courses in which the main focus is development of a system of theorems, starting from a set of axioms - like geometry or topology. However, for branches of applied mathematics such as statistics, the implementation of the method will not be as straightforward. However, I've briefly seen an article (I think it was a PhD thesis) on applying the Moore Method to teaching calculus. If it can be used to teach calc, it can be used for stats! Lucy Kaplansky And the reason I was listening to the Kaplansky lecture - in fact, the only reason I had ever heard of Kaplansky - is because he is the father of one of my favorite singer/songwriters, Lucy Kaplansky. If you want to listen to some of her songs on YouTube, check out the playlist I put together. In addition to being a top-tier mathematician, Irving Kaplansky was also a musician and wrote a song about pi entitled A Song about Pi. I heard Lucy perform that song on a radio event that was recorded in Chicago this summer. It's very cool. I'm anxiously awaiting the release of the CD from that performance. ### Sample midterm questions - Part 2 "Regular" questions Questions 1 and 2 - skip Question 3: The average number of calls received by a switchboard in a 30-minute period is 15. a. What is the probability that between 10:00 and 10:30 the switchboard will receive exactly 10 calls? b. What is the probability that between 13:00 and 13:15 the switchboard will receive at least two calls? Answer: Since this question deals with an "area of opportunity" (the 30-minute period), we obviously need to use Poisson. For part a, we have a period (10-10:30) that is the standard 30-minute period. So, f(x=10) = (e^-15)(15^10)/10! Do the arithmetic or use a lookup table and you get: f(x=10) = 0.048611 For part b, we have a 15 minute period, so the mean is 7.5 instead of 15. Also, to compute the right hand tail of f(x>=2) we take 1-f(x=0)-f(x=1). f(x>=2) = 1 - f(x=0) - f(x=1) = 1 - 0.0006 - 0.0041 f(x>=2)= 0.9953 Question 4: Four workers at a fast food restaurant pack the take-out chicken dinners. John packs 45% of the dinners but fails to include a salt packet 4% of the time. Mary packs 25% of the dinners but omits the salt 2% of the time. Sue packs 30% of the dinners but fails to include the salt 3% of the time. You have purchased a dinner and there is no salt. a. Find the probability that John packed your dinner. b. Find the probability that Mary packed your dinner. Answer: This scenario is a perfect application of Bayes' Theroem. We have: P(John) = .45 P(Mary) = .25 P(Sue) = .30 P(NoSalt|John) = .04 P(NoSalt|Mary) = .02 P(NoSalt|Sue) = .03 In part a, we want to know P(John|NoSalt). Applying Bayes': P(John|NoSalt) = P(NoSalt|John)P(John) / P(NoSalt|John)P(John)+P(NoSalt|Mary)P(Mary)+P(NoSalt|Sue)P(Sue) = (.04)(.45) / (.04)(.45) + (.02)(.25) + (.03)(.30) = 0.018 / (0.018 + 0.005 + 0.009) = 0.5625 In part b, we want to know P(Mary|NoSalt). Again, applying Bayes' in the same way: P(Mary|NoSalt) = P(NoSalt|Mary)P(Mary) / P(NoSalt|John)P(John)+P(NoSalt|Mary)P(Mary)+P(NoSalt|Sue)P(Sue) = (.02)(.25) / (.04)(45) + (.02)(.25) + (.03)(.30) = 0.5 / (1.8 + 0.5 + 0.9) = 0.15625 Question 5: In a southern state, it was revealed that 5% of all automobiles in the state did not pass inspection. Of the next ten automobiles entering the inspection station, a. what is the probability that more than three will not pass inspection? b. Determine the mean and standard deviation for the number of cars not passing inspection. Answer: Since this is a case of pass/fail, we recognize this as a case for the binomial distribution and, for part a, P(x>3) = 1-P(x=0)-P(x=1)-P(x=2)-P(x=3), so we'll have to calculate P(x) for x=0,1,2, and 3 using the formula: P(x) = [n!/(n-x)!x!] (p^x)(1-p)^(n-x) where n=10 and p=0.05 Sheesh! I hope we get a binomial distribution table if we get a question like this! From table E.6 in the book, I get P(x>3) = 1 - 0.5987 - 0.3151 - 0.0746 - 0.0105 P(x>3) = 0.0011 For part b, we use the easy formulas for mean and standard deviation of a binomial distribution: Mean = np = (10)(0.05) Mean = 0.5 Standard Deviation = SQRT(np(1-p)) = SQRT((0.5)(1-0.05)) = SQRT(0.475) Standard Deviation = 0.689 ### Sample midterm questions - Part 1 The midterm questions from last quarter's midterm have been posted with the answers. However, there's no explanation of the answers. I'll endeavor to explain how I think these answers are derived. If you think I got one wrong or if you have a better/easier way to get the answer, please let me know in a comment or email. Since chapter 6 (normal distribution) is not included on our midterm, I'm going to skip those questions for now and focus on the questions that deal with chapters 1-5. Question 1 & 2 - Skip Question 3: Which of the following is not a property of a binomial experiment? a. the experiment consists of a sequence of n identical trials b. each outcome can be referred to as a success or a failure c. the probabilities of the two outcomes can change from one trial to the next d. the trials are independent Answer:Think of the classic example of a binomial experiment – flipping a coin. All the flips are identical. Each outcome has two possibilities, heads/tails. Each flip is independent of the previous flips. The probabilities do not change from one flip to the next. So C is not true. Exhibit 5-8: The random variable x is the number of occurrences of an event over an interval of ten minutes. It can be assumed that the probability of an occurrence is the same in any two time periods of an equal length. It is known that the mean number of occurrences in ten minutes is 5.3. Question 4. Refer to Exhibit 5-8. The probability that there are 8 occurrences in 10 minutes is a. .0241 b. .0771 c. .1126 d. .9107 Answer: As soon as I see that we’re talking about events occurring over a period of time, I think “Poisson Distribution”. It’s the whole “area of opportunity concept. Therefore, apply the Poisson Formula (or use a lookup table) for X=8 and mean=5.3: F(x=8) = (e^-5.3)(5.3^8)/8! = (0.00499)(622,596.9)/(40320) = 0.0771 So the answer is B. Question 5: Refer to Exhibit 5-8. The probability that there are less than 3 occurrences is a. .0659 b. .0948 c. .1016 d. .1239 Answer: We apply the Poisson Distribution formula again, but this time to determine F(x<3)= F(x=0) + F(x=1) + F(x=2) You can apply the Poisson formula individually for these 3 values, but I prefer to look them up in the table and get: F(x<3)= 0.0050 + 0.0265 + 0.0701 = 0.1016 So answer C is correct. Exhibit 5-5 Probability Distribution x f(x) 10 .2 20 .3 30 .4 40 .1 Question 6: Refer to Exhibit 5-5. The expected value of x equals a. 24 b. 25 c. 30 d. 100 Answer: Apply the formula for the expected value: multiply each value by its probability and add them up. E(x) = (10)(0.2) + (20)(0.3) + (30)(0.4) + (40)(0.1) = 2 + 6 + 12 + 4 = 24< style="font-weight: bold;"> Question 7: Refer to Exhibit 5-5. The variance of x equals a. 9.165 b. 84 c. 85 d. 93.33 Answer: Just apply the standard formula for variance: sum the squares of the difference to the mean times the probability of each value. (Note: You don’t divide by the number of values as you would in a data sample of population. When you have a distribution, you multiply each discrete value by it’s probability.) Var = (10-24)^2(.2) + (20-24)^2(.3) + (30-24)^2(.4) + (40-24)^2(.1) = (196)(.2) + (16)(.3) + (36)(.4) + (256)(.1) = 39.2 + 4.8 + 14.4 + 25.6 = 84 Correct answer is B. Question 8: 20% of the students in a class of 100 are planning to go to graduate school. The standard deviation of this binomial distribution is a. 20 b. 16 c. 4 d. 2 Answer: This is pretty straightforward, except that the formula sheet has the formula for variance. You need to remember to take the square root to get the standard deviation. Std. Dev. = SQRT(np(1-p)) = SQRT ((100)(.20)(.80)) = SQRT (16) = 4 Correct answer is C. Questions 9, 10 and 11 - Skip Question 12: Of five letters (A, B, C, D, and E), two letters are to be selected at random. How many possible selections are there? a. 20 b. 7 c. 5! d. 10 Answer: This question is actually rather vague. It doesn't tell us if the order of the selection matters. I.e. is (A,B) considered the same selection as (B,A)? From the answer key, it's pretty clear that they are considered the same, so we'll go with that. Apply the binomial coefficient formula: nCx = n! / (n-x)!x! = 5! / 3!2! = 120 / 12 10 Correct answer is D. BTW, this problem could easily be solved by inspection (i.e. by just looking at it): You're selecting 2 letters out of a pool of 5. For the first letter there are 5 possibilities, for the second letter there are 4 possibilities (since the first letter, whatever it was, is no longer in the pool). 5x4 = 20. Then divide by 2 because each two-letter pair has a duplicate in the reverse order. 20/2 = 10. Question 13: If A and B are independent events with P(A) = 0.2 and P(B) = 0.6, then P(A U B) = a. 0.62 b. 0.12 c. 0.60 d. 0.68 Answer: This question requires us to use the Addition Law and the multiplication rule of independent events (both are on the formula sheet): Addition Law gives us: P(A U B) = P(A) + P(B) - P(A intersect B) Multiplication rule of independent events says: P(A intersect B) = P(A)P(B) Therefore, P(A U B) = P(A) + P(B) - P(A)P(B) = (0.2) + (0.6) - (0.2)(0.6) = (0.8) - (0.12) = 0.68 Correct Answer is D. Question 14: The variance of the sample a. can never be negative b. can be negative c. cannot be zero d. cannot be less than one Answer: This one is right out of the book. See page 82 where it says (in italics): neither the variance nor the standard deviation can ever be negative. But even if you somehow didn't remember that tidbit of information, you could just eyeball the formula for variance and see that the numerator is a squared quantity which is always positive and the denominator is the number of events in the sample, which is also always positive. So, Answer A is correct. Incidentally, the variance can be zero if the data points are all equal to the mean, in which case the numerator is zero and therefore the variance is 0. It can also be less then one if the data points are all relatively close to the mean. Question 15: The value of the sum of the deviations from the mean, must always be a. negative b. positive c. positive or negative depending on whether the mean is negative or positive d. zero Answer: This one is similar to the question above and it's also right out of the book, page 84, where it says this sum will always be zero. But again, if you missed that factoid, you could easily see that it's true by definition of the mean. Correct answer is D. Question 16: If A and B are independent events with P(A) = 0.38 and P(B) = 0.55, then P(A|B) = a. 0.209 b. 0.000 c. 0.550 d. 0.38 Answer: This is another tricky question. No formulas needed here. Just think about it: If A and B are independent, then by definition, the probability of A is not influenced by B. So the fact that B is "given", makes no difference - the probability of A is still 0.38. So P(A|B) = P(A) = 0.38. Answer D is correct. Question 17: A six-sided die is tossed 3 times. The probability of observing three ones in a row is a. 1/3 b. 1/6 c. 1/27 d. 1/216 Answer: Come on! Is this really graduate school? 3 independent events. Get the total probability by taking the product of the 3 events. Each one has a 1/6th probability of rolling a one. Therefore, total probability is (1/6)^3 = 1/216. Answer D is correct. Question 18: If the coefficient of variation is 40% and the mean is 70, then the variance is a. 28 b. 2800 c. 1.75 d. 784 Answer: This one is a bit tricky in that you need to remember that the standard deviation is the square root of the variance. First, use the formula for Coefficient of Variation: CV = (StdDev / mean) x 100 % rearrange to get: Std Dev = CV x mean/100 = (40)(70)/100 = 28 Now since Var = StdDev^2, Var = 28^2 = 784 Answer D is correct. Question 19: If two groups of numbers have the same mean, then a. their standard deviations must also be equal b. their medians must also be equal c. their modes must also be equal d. None of these alternatives is correct Answer: Hmm. Does this one need explanation? If you understand mean, median, mode and standard deviation, you should understand that they're completely independent ways of describing data. I suppose someone could take the time to show examples of data sets with the same mean, but different median, mode and stddev, but I think it's pretty much intuitively obvious that Answer D is correct. Has anyone noticed that D is the correct answer for questions 11-19, except for 14? What's the probability of that happening? Exhibit 3-3: A researcher has collected this sample data. The mean of the sample is 5. 3 5 12 3 2 Question 20: Refer to Exhibit 3-3. The standard deviation is a. 8.944 b. 4.062 c. 13.2 d. 16.5 Answer: Plug the numbers into the formula for standard deviation StdDev = SQRT((3-5)^2+(5-5)^2+(12-5)^2+(3-5)^2+(2-5)^2)/(5-1)) = SQRT((4+0+49+4+9)/4) = SQRT(66/4) = SQRT(16.5) = 4.062 Don't forget that last step of taking the square root! Otherwise you're giving the variance which, of course, is one of the choices (D). B is the correct answer. Answer C (13.2) is the variance if this data were for an entire population, in which case you would divide by N=5. Remember to divide by n-1 for a sample. ### Midterm study diversion (I) If you need a break from studying for the midterm, think about this: Who Discovered Bayes's Theorem? You think it's obvious, right? Bayes, of course. Well, it's not so obvious. In fact, Stephen M. Stigler, professor of statistics across town at University of Chicago, wrote an article by just that title back in 1983 in the respected journal The American Statistician. (vol 37, no 4, pp. 290-296) We have access to that journal through Depaul online library. I'm in the middle of reading it and I must say it's a very entertaining piece. Sort of a whodunit for the mathematically inclined. If you haven't taken advantage of the Depaul online library system yet, I highly recommend that you learn how to use it. It's a great resource and, after all, you're paying for it. Here's how you can get to this article: 1. Access the online library at: http://www.lib.depaul.edu/ 2. Click the link for Journal & Newspaper Articles in the Research section of the page. 3. Enter "american statistician" in the search box on the left and click Search. 4. The results screen shows the 5 locations that we have this journal (1 in print and 4 online) and the years covered by each location. 5. Since we're looking for an article from 1983, the only location that has that year is the first online location: JSTOR Arts and Sciences II Collection. Click that link. 6. Now you'll need your Depaul Campus Connect ID to log in to access the article. 7. You're now in JSTOR for The American Statistician. Click the link to Search This Journal and then run an article title search on "Who Discovered Bayes's Theorem". The rest is self-explanatory. Feel free to email me or post a comment if you have trouble getting to this article. BTW, I skimmed the TOC of that issue and saw an article that looked interesting: On the Effect of Class Size on the Evaluation of Lecturers' Performance by Prof. Ayala Cohen of Technion. I scanned the article. Looks interesting. ## Saturday, January 26, 2008 ### Lecture 4 - Ch 6 - The Standard Normal Distribution Standardizing the Normal Distribution As you can see from the formula for the Normal Distribution in the previous post, the normal distribution has two parameters - the mean (mu) and the standard deviation (sigma). Other than those parameters, it's just a (complicated) function of x. It should be pretty clear that if we make the mean = 0 and standard deviation = 1, the formula would become much simpler. We call it the standard normal distribution and it looks like this:Much simpler, eh? Well, it's not too bad. But it still requires a scientific calculator to compute. The weird thing about the standard normal distribution is that it does not describe the probability of any one value, but rather the probability of the result of an experiment falling between any two values is defined by the area under the curve of the standard normal distribution. You know what that means? We have to integrate it! OOOOhh boy! Well, fortunately for us, some math majors have taken it upon themselves to integrate this formula for us and calculate the definite integral between 0 and almost any positive value. That means we can use these handy-dandy tables to find out the probability of x falling between 0 and any value. Finding the probability of x falling between a and b Looking up Pr(0 < x < a ) in the table is helpful, but often we want to know the probability of x falling between two values, a and b, or Pr (x > a) or Pr(x < a). We can easily use the values in the table to determine any of these probabilities. ### Lecture 4 - Ch 6 - The Normal Distribution The Normal Distribution A typical continuous distribution is the Normal Distribution, which looks like the proverbial "bell-shaped" curve. However, just like bells can take many shapes, there are many different Normal Distributions. In fact, the Normal Distribution is really a family of distributions. Although all the curves in the family of Normal Distributions have certain common characteristics, a particular instance of the Normal Distribution is defined by its mean and standard deviation. The Normal Distribution is defined generally by the function:Wow! Where did that wacky formula come from?? We may never know... But luckily we don't need to know the derivation of the formula. In fact, we don't even need to know how to use the formula because we're going to do three things: 1. "standardize" the formula - called, oddly enough, the standard normal distribution 2. compute tables for the standardized formula 3. use another, simpler formula to translate between any given normal distribution and the pre-calculated standard normal distribution ### Lecture 4 - Ch 6 - Continuous Random Variables Discrete vs. Continuous Random Variables We now started learning about continuous random variables. These are random variables which can take (more or less) any value. Examples of these types of variables are the length of a piece of wire produced by a machine, the weight of a bag of groceries, the time it takes to print a page from a printer/copier, etc. By contrast, discrete random variables have a limited (and usually finite) number of values - for example, the number of credits that a student has earned, the age of the employees of a company, the pass/fail results of an experiment such as flipping a coin, etc. ## Thursday, January 24, 2008 ### Lecture 4 - Ch 6 - NOT on midterm After spending about an hour going over the normal distribution and the standard normal distribution (notes will follow in a different blog entry), Lecture 4 was cut short tonight due to a fire alarm. When we returned to the room, the message below appeared on the board: • Class Dismissed • Ch. 6 not included! An email was also sent to that effect. I personally would have preferred Ch 6 to be on the midterm. It's not very hard material and, judging from the questions from last quarter's midterm, probably would have made the midterm easier. ### Lecture 3 - Ch 5 - Poisson Distribution The Poisson Distribution is derived from the binomial distribution and describes probabilities of success within a certain "area of opportunity". The area of opportunity could be a time interval, surface area, etc. For a scenario which follows a Poisson Distribution, the probability of a x success events occuring in an area of opportunity is:where: lambda = the average (mean) number success events which occur in that area of opportunity The mean of a Poisson Distribution is lambda. The standard deviation of a Poisson Distribution is SQRT(lamba). ### Lecture 3 - Ch 5 - Binomial Distribution The binomial distribution is used to describe a discrete random variable under these conditions: 1. An experiment is repeated several times - n 2. Each "experiment" has two possible outcomes - success/failure, heads/tails, etc 3. The probability of the two outcomes is constant across the experiments 4. Each outcome is statistically independent of the others. Like when you flip a coin, the chances of heads on the 100th flip are 50/50 even if you had 99 tails in a rows before it. (Although if that really happened, I'd check the coin!) In a scenario that fulfills those conditions, the probability of obtaining x successes in n experiments is:where x = the number of successes n = the number of experiments/trials/observations p = the probability of success in any single experiment/trial/observation The mean of the binomial distribution is np. The standard deviation of the binomial distribution is np(1-p). ## Wednesday, January 23, 2008 ### Lecture 3 - Ch 5 - Discrete Random Variables We calculated the mean, variance and standard deviation for a set of data earlier. Those calculations are made on a finite set of data points (observations). If, rather than data, we have an event that is described theoretically by a discrete random variable with possible values and the probability of each value, we can make similar calculations. The difference here is that in a set of data each observation is equal to the next. But in a random variable, we weigh the possible values by their respective probabilities. Mean - Add up all the possible values but instead of dividing by the number of values, we multiply each by its probability.Variance and Standard Deviation - Std Dev is still the square root of the variance. But the variance is the sum of the squares of the difference between the data points and the mean times the probability of that data point. ### Lecture 3 - Ch 4 - Counting Rules These counting rules are basics from the GMAT, so it's mostly review. Getting these problems right is a matter of understanding the scenario and identifying how to apply the two simple formulas for permutations and combinations. All of these cases involve selecting some number, call it x, of items from a pool of n possible items. Permutations: Use this formula when order matters in the scenario. I.e. if (a,b) is not the same as (b,a). An example of this would be license plates. The license plate "GO4MBA" is not the same as "BAG4MO". Another example is telephone numbers. Obviously, 867-5309 is not the same number as 509-3678. Combinations: Use this formula when order doesn't matter in the scenario. I.e. if (a,b) is considered the same as (b,a). In this case you start with the same formula above, but you need to divide by the number of possible duplicates (x!) due to the fact that order doesn't matter. The formula now becomes: When dealing with questions of probability, we often need to use these formulas to figure out the number of successful events and the total number of possible events. For example, the probability of being dealt a royal flush from 5 cards would be the total number of ways to get a royal flush divided by the total number of possible hands that could be dealt. There are more complicated examples involving replacement and other factors, but for our purposes, I'll leave it at this and perhaps bring those up in a side note blog entry. For those interested in TeX/LaTeX using LEd and MiKTeX 2.7, apparently they don't support the \binom control sequence. If it did, I could just create the symbol above with: \binom{n}{x} Since it doesn't, the best I could do is to use an array and code it with: \left(\begin{array}{c}n\\x\end{array}\right) Yuck! ### Lecture 3 - CH 4 - Bayes' Theorem Bayes' Theorem allows us to calculate P(B|A) when we don't have the usual information that would allow us to calculate it. I.e. we don't know P(B and A) or P(A). We know P(B|A) = P(B and A) / P(A) We'll derive alternative expressions for both the numerator and the denominator. Since P(A|B) = P(A and B) / P(B) and P(A and B) = P(B and A), we can multiply both side by P(B) and rearrange the terms and get: P(B and A) = P(A|B)P(B) which becomes our numerator. For the denominator, we note that P(A) = P(A|B)P(B) + P(A|B')P(B') It's worth thinking about this for a second. Time's up! What it means is that the total probability of A is equal to the conditional probability of A given B (times the probability of B) plus the conditional probability of A given B' (times the probability of B'). Since B and B' cover the entire sample space, the sum of those conditional probabilities (times their respective factors) equals the entire probability of A. If there were more than two possibilities for the condition - for instance B1, B2, B3, ... - then the total probability of A would be the some of all three of the conditionals, i.e.: P(A) = P(A|B1)P(B1) + P(A|B2)P(B2) + P(A|B3)P(B3) + ... This is true as long as the events B1, B2, B3, ... are mutually exclusive and exhaustive of the sample space. Now, back to our original formula that we were trying to evaluate: Our new expression for P(B|A) becomes:and allows us to calculate P(B|A) if we know P(B), P(A|B) and P(A|B'). ### Lecture 3 - Caveat I was unable to attend Lecture 3 for personal reasons, so I'm relying heavily on the notes that were posted by Prof. Selcuk, the class notes of several people (thank you!), the material in the book and a meeting with Prof. Selcuk during office hours on Wednesday. As always, use these notes at your own risk. I do my best to provide good class notes and additional material for thought, but always check with your own sources also. If you find any mistakes, I'd really appreciate you letting me know via comment or email. ## Wednesday, January 16, 2008 ### Excel Quartiles - Not so fast... The formula that I posted for Quartiles in Excel (n+3/4) is not precise. Here's the algorithm for computing quartiles that Excel uses according to the Microsoft knowledge base: The following is the algorithm used to calculate QUARTILE(): 1. Find the kth smallest member in the array of values, where: k=(quart/4)*(n-1))+1 If k is not an integer, truncate it but store the fractional portion (f) for use in step 3. And where: • • quart = value between 0 and 4 depending on which quartile you want to find • n = number of values in the array 2. Find the smallest data point in the array of values that is greater than the kth smallest -- the (k+1)th smallest member. 3. Interpolate between the kth smallest and the (k+1)th smallest values: Output = a[k]+(f*(a[k+1]-a[k])) a[k] = the kth smallest a[k+1] = the k+1th smallest ### Quartiles in HW1 Recall from a previous post that different software packages have different ways of calculating quartiles. The formula for quartiles that we learned in class is: Position of Q1 = (n+1)/4 Position of Q2 = (n+1)/2 Position of Q3 = 3 (n+1)/4 Example: Let's say your data set is (2, 4, 6, 8, 10, 12, 14, 16) n = 8 Position of Q1 = 9/4 = 2.25 We interpolate between the 2nd value (4) and the 3rd value (6) and get Q1 = 4.5 I haven't checked the documentation, but I've empirically determined that Minitab uses this same formula. [EDIT] Before I go on to talk about Excel, I need to mention that the last step of interpolating between values is NOT how our book instructs us to calculate the quartile. Jeffrey McNamara pointed out to me that Rule 3 on page 77 says: If the result is neither a whole number nor a fractional half, you round the result to the nearest integer and select that ranked value. So in our example, where the position of Q1 is 2.25, we would round that to 2 and select the 2nd ranked value which is 4. Again, Minitab uses the interpolation method and, as Frank Cardulla used to say, "You pays your money, you takes your choice." Kudos to Jeff for the catch! [/EDIT] Excel uses a different formula which is apparently: Position of Q1 = (n+3)/4 In our example, this would mean: Position of Q1 = 11/4 = 2.75 Interpolating .75 between the 2nd and 3rd value gives us Q1 = 5.5 This is indeed what MS Excel calculated when I used the Quartile function on this data set. Conclusion: It seems to me to be a good idea to use [either the book method or] Minitab (or manual calculations) to calculate the quartiles for the homework. ### Using Minitab and Excel in the homework Lowkeydad and I had the same question about the homework: Do we just use Minitab/Excel to make the calculations? Or do we need to do them manually (i.e. with a calculator) and/or show the formula that was used and/or any work? I emailed the following question to Prof. Selcuk: Just to be clear... When we're asked to compute the mean, variance and standard deviation (in 3.27) and the covariance and coefficient of correlation (in 3.43) and all the calculations in 3.61 etc - is it sufficient to use Minitab/Excel to compute these statistics and simply report the results? His reply: Yes, you don’t have to go thru each step in the formulas. Be careful with excel though, especially when the sample size is small. This may lead to a different result than in the answer key, and the TA may think that you’ve done something wrong and cut points when grading. ## Tuesday, January 15, 2008 ### Minitab If you didn't get Minitab with your textbook, you can "rent" it for 6 months ($30) or a year (\$50) or get a 30-day free trial version.
http://www.minitab.com/products/minitab/demo/default.aspx
** Props to Patrick Feeney for providing that tip **
The data files are provided in Excel format also, along with a set of templates for doing the calculations.
### Post lecture 2 research and notes
The talk about conditional probability reminded me of one of my favorite mathematical paradoxes - the so-called Monty Hall Problem. Warning: This may confuse you! Don't read any further unless you really like puzzles and paradoxes.
The Monty Hall Problem
There was once a TV game show called Let's Make A Deal which was hosted by Monty Hall. He played different games with the studio audience similar to the way Drew Carrey does in The Price is Right.
One of the games goes like this:
1. Monty presents to you three doors, labeled 1, 2 and 3. He tells you that there's a car behind one door (it's a big door) and goats behind the other two. You get to pick one door.
2. After you make your choice, Monty opens one of the other doors to reveal a goat. He then gives you the option to change your original choice to another door.
The question is: Would you switch?
I'll give you some time to ponder...
Time's up!
Intuitively, you would think that there's no real advantage to switching. At step 1, you had a 1 in 3 chance of picking the car. At step 2, you now have a 1 in 2 chance of picking the car. Your original choice is still no better than the other door. Both have a 50/50 chance.
I'm going to stop here for two reasons:
1. I have a meeting to go to.
2. I'm still not sure why the intuitive solution is incorrect.
In the meantime, I'll refer you to two web pages:
Wikipedia - Monty Hall Problem
I hope to come back to this soon with some clarity.
...
Ok, I'm back with a fairly simple explanation (although not rigorous) that I think makes sense...
When you make your initial door selection, you may have chosen the car or goats:
Case 1: 1/3 chance of having chosen the car
Case 2: 2/3 chance of a goat.
In case 1, after Monty reveals a goat in another door, switching is bad since you were correct with your initial guess.
In case 2, after Monty reveals a goat, switching is good since your first guess was a goat and Monty's door was a goat and switching would get you to the car.
Therefore, there's 1/3 chance that switching is bad and 2/3 chance that switching is good. Therefore, you should switch!
Make sense?
I'd really like to work on explaining this in terms of the conditional probabilities that we learned in class.
## Sunday, January 13, 2008
### Homework 1 has been assigned
In case you didn't get it, homework #1 (due next Thursday) consists of the following exercises from the book:
1) 3.27 a, b and c
2) 3.35
3) 3.43
4) 3.61
5) 3.69
6) 4.11
7) 4.15
While we must maintain strict academic integrity, questions on the homework can be posted to comments. I'm moderating comments and I won't let anything through that would be questionable.
That said, I don't think there's any issue of integrity in explaining how to use Minitab or Excel to calculate the statistics, which may be a sticky point for first-time users. After all, it's a class in statistics, not in how to use the tools, right? Feel free to email me directly also.
### Lecture 2 - Pop quiz #2
Prove: If A is independent of B, then B is independent of A. I.e. P(A|B) = P(A) implies P(B|A) = P(B).
That's how I did it. If you have a different proof, I'd be interested in hearing it.
### Lecture 2 - Ch 4 - Independency
A and B are independent events if and only if P(A|B) = P(A).
This is basically saying that A and B are independent only if the probability of A occurring given that B has occurred is the same probability as A happening in any case (even if B had not occurred). In other words, the probability of A occurring is not influenced by whether B happens or not. I.e. A is independent of B.
We also learned that if A is independent of B, then B is independent of A. Meaning,
P(A|B) = P(A) implies P(B|A) = P(B).
The proof the statement above was given as a pop quiz. See next post for the solution. (I know... you can't wait!)
The Multiplication Rule
Another check for independency can be derived as follows:
Since we know from the definition of conditional probability that
By multiplying both sides by P(B), we get: P(A|B)*P(B) = P(A/\B)
And we know that if the events are independent that P(A|B) = P(A), so substituting this on the left side of the equation, we get
The multiplication rule is more commonly used as a test of independence than the first definition, but we need to know and understand both definitions.
We went over an example of possible discrimination in promotions within the police department. That example is a good sample midterm question, so review it!
### Lecture 2 - Ch 4 - Conditional Probability
Given two events, A & B:
the probability of A, given that B has occurred is denoted by P (A|B) and is calculated as:
Intuitively, we understand this as follows:
Look at the Venn diagram in the previous post. Once we know that B has occurred (since that's a given), B is our new sample space. And the only way for A to occur now is
Therefore, the probability of A occurring now is the probability of a successful outcome, i.e. divided by the probability of the entire sample space, i.e.
## Friday, January 11, 2008
### Lecture 2 - Ch 4 - Basic Probability
We covered basic probability concepts.
The probability of some event, A, is:
P(A) = # of favorable outcomes / total possible outcomes
If we consider two events, A & B, they can be represented by a Venn diagram:
The entire yellow area, S, is the sample space.
P(A U B) = P(A) + P(B) - P(A /\ B)
We subtract P(A /\ B) because that overlap area (dark red in our Venn Diagram) is double counted if we just add P(A) and P(B).
### Sample vs. Population
Use English letters for sample statistics and Greek letters for the analogous population stat.
Sample stats typically use (n-1) for the sample size while population stats use N. The reason for this difference is beyond the scope of this class (at least for now).
The exception is the mean where both use n (and N).
### More on covariance and correlation coefficient
So, what do the covariance and correlation coefficient mean?
In theory, the covariance is a measure of how closely the data resembles a linear relationship.
At a high level:
If covar(x,y) >> 0 then we can say there's a positive correlation. I.e., as x increases, y also increases.
If covar(x,y) << 0 then there's a negative correlation. I.e., as x increases, y decreases.
It was noted that the covariance isn't very useful because it's influenced by the magnitude of the values. For example, if the observations in both sets are all multiplied by 10 (which could happen if you simply change the units of measure from, say, meters to centimeters), the covariance becomes 100 times larger even though the actual correlation of the data has not changed at all.
Note: We didn't discuss why, on an intuitive level, the formula for the covariance is a measure of the linear correlation of the data sets. I'll leave that for further thought later on.
In any case, the covariance is standardized by dividing by the product of the standard deviations. (Again, why this works on an intuitive level is not clear to me. But it does!) Covariance is therefore a pretty useless statistic by itself, but it is part of the calculation of the correlation coefficient, which is very useful.
The range of values for the correlation coefficient is from -1 to 1. A 0 correlation coefficient indicates no correlation - either because the data are wildly scattered or because they are on a horizontal or vertical line (indicating that the data sets are independent). The closer the correlation coefficient is to 1 or -1, the stronger the linear relationship between the data sets.
### Using MiKTeX and LEd
Well, I got MiKTeX (an implementation of LaTeX) installed and LEd as an integrated editor/viewer. It was actually relatively painless. I only had one installation glitch which may have been caused by the fact that I installed LEd before MiKTeX.
I was unable to see my formatted text in the built-in DVI viewer in LEd. I checked the configuration options and fixed the problem by changing the "TeX Distribution" option from MiKTeX 2.4 to MiKTeX 2.6 (even though I'm actually using 2.7). In any case, that was the only problem I had.
Actually understanding how to code formulas in TeX is a different story. I haven't really found a really good tutorial yet, but David Wilkins's Getting Started with LaTeX has some good documentation, including examples.
Here's my first try at creating a somewhat complex formula from class:
I don't like the way the i=1 and n appear next to the sigma instead of above and below it, but I think that's intentional in order to make the formula as compact vertically as possible.
Here's the LaTeX code that's used to create the formula:
$covariance (x,y) = \frac {\sum_{i=1}^{n} (x_i-\bar{x})(y_i-\bar{y})}{n-1}$
### Lecture 2 - Ch 3 - Correlating two sets of data
In the first lecture, we talked briefly about the correlation of two sets of data in a scatter plot. That discussion described the correlation in a qualitative way. Here we define the measure of correlation more precisely with two measures: covariance and coefficient of correlation.
For two sets of data, x and y:
Covariance (x,y) - [ Sum (xi-xmean)*(yi-ymean) ] / (n-1)
Coefficient of Correlation, r = covar(x,y) / sx*sy
where sx and sy are the standard deviations of x and y
BTW, I see now that I'm going to need to make some graphics to display these formulas. I'm going to try to learn to use LaTeX for mathematical typesetting. I downloaded LEd, a LaTeX editor, today and I hope to start using it over the weekend. If anyone has experience with LaTeX and any editor, please let me know.
### Lecture 2 - Ch 3 - Shape of the Distribution
Although the skewness has a formal definition that we didn't discuss, we categorized the shape of a distribution as either symmetric, positive-skewed or negative-skewed.
Symmetric - mean = median
Positive-skewed - mean > median (think of this as having a longer tail on the right (positive) side of the distribution curve
Negative-skewed - mean < median (here there's a longer tail on the left (negative) side of the curve
A box-and-whiskers graph is a simple graphical summary of the skewness of the data.
Don't let this confuse you, but... Another measure of the shape of a distribution curve is the kurtosis. Sounds like a disease, doesn't it?
## Thursday, January 10, 2008
### Lecture 2 - Ch 3 - Using Standard Deviation
Now that we understand how to calculate the standard deviation, we can define three statistical concepts that apply the standard deviation in practice. Those three concepts are the z-score, the empirical rule and Chebyshev's rule.
Z-score - The z-score of an observed value is the difference between the observed value and the mean, divided by the standard deviation. The z-score is used to express how many standard deviations an observation is away from the mean.
Empirical Rule - In a bell-shaped distribution:
1 standard deviation (in either direction from the mean) has about 68% of the data.
2 standard deviations have about 95% of the data.
3 standard deviations have about 99.7% of the data.
This "bell-shaped" distribution is called the standard normal distribution and we'll learn more about it in a later lecture/chapter.
Chebyshev Rule - In any distribution:
At least (1-1/z2)*100% of the values will be within z standard deviations.
A few interesting things were noted briefly about Chebyshev's Rule:
1. It works with any, yes any, shaped distribution. That's really amazing.
2. Outside of the z standard deviations, we don't know anything about the data. Specifically, it may not be symmetric, so we can't assume that the data outside the z std devs is evenly distributed between them.
3. Chebyshev's Rule doesn't tell us anything about data within 1 standard deviation.
Did you know? There's a crater on the moon named after Chebyshev. Source: USGS
### New location
Class location has changed to Lewis 1509 to cut down on the noise from the El.
## Tuesday, January 8, 2008
### Lecture 1 - additional notes and research
Quartiles
I was surprised to find the following statement in the Quartile entry on Wikipedia (which Prof Selcuk confirmed): "There is no universal agreement on choosing the quartile values." There is a reference to an article in American Statistician by a couple Australian professors (Hyndman and Fan). (Abstract: http://www-personal.buseco.monash.edu.au/~hyndman/papers/quantile.htm)
I was able to retrieve this article from the JSTOR Arts and Sciences 2 database via the NEIU library.
I found an interesting write-up on Defining Quartiles at Ask Dr. Math. After showing why quartiles are simple in concept but complex in execution, Dr. Math describes 5 methods for defining quartiles: Tukey, M&M, M&S, Minitab and Excel.
After reading the Dr. Math article, I see that the algorithm for defining quartiles that we discussed in class is the Minitab method, using x*(n+1)/4 to find quartile position and linear interpolation of closest data points to handle non-integral positions.
MS Excel
I was also surprised to find out that the accuracy of MS Excel is in question and that McCullough and Wilson have been writing in Computational Statistics & Data Analysis on the issues with Excel's algorithms and how they have changed in subsequent versions of MS Excel since 97.
Web Resources
I found a nice site that covers some of the material from lecture 1 (measures of central tendency and measures of spread) and more. It's called Stats4Students. I'm going to create a links section in the sidebar and add it.
### Lecture 1 - Quiz
Given:
Variance = 25
Coefficient of Variation = 10%
Compute the mean.
We know the relationship between the coefficient of variation, standard deviation and the mean:
CV=(stddev/mean) * 100
We also know that the standard deviation is simply the square root of the variance:
Stddev = sqroot(variance)
Therefore, combining these two formulas and plugging in our givens, we get:
10 = (sqroot(25) / mean) * 100
10 = (5/mean) * 100
10 = 500 / mean
10 * mean = 500
mean = 50
The only thing tricky about this question is that you have to remember that the CV is stddev/mean, not variance/mean. If you remember that, along with the relationship between the stddev and variance, it's simple.
### Lecture 1 - Ch 3 - Descriptive Statistics
We now discuss some basic concepts we can use to quantitatively describe a set of values. Rather than describing the set by painting a picture with a graph, chart or table as we did in the previous blog post, we paint the picture in a more precise and analytical way using numbers.
One of the most fundamental ways to describe a set of values is to identify the central tendency of the set. We recognize that when looking at some sample, the statistic that we're interested in won't have the same value for every member of the set. But we're still interested in the central value for the set as a whole. We commonly call this central value the "average". However, there are several ways to calculate this central value which differ slightly:
(Arithmetic) Mean - commonly called the average
Median - the middle value (or average of the two middle values if there are an even number of observations). The position of the median value is (n+1)/2.
Mode - the most common value in the data set
We also introduce the following definitions:
Range = maxvalue - minvalue
Quartile (Q1, Q2, Q3) which divides the data into 4 equal units in the same way the median divides it into 2 units. Again, if the number of values is not evenly divisible into 4 groups with the same number of values, take the average of the 2 closest values. Q2 is always the same as the median.
Position of Q1 = (n+1)/4
Position of Q2 = (n+1)/2
Position of Q3 = 3
(n+1)/4
Check the data in those positions to get values for Q1, Q2 and Q3.
If Q1, Q2 and/or Q3 are not integers, interpolate between the closest data points to get a value for quartile.
Then a couple of statistical concepts to describe the dispersion (variation) of a set of values:
Range = maxvalue - minvalue
Interquartile Range = Q3-Q1. Use box-and-whiskers diagram to represent this graphically.
Variance - the sum of the square of the differences between the values and the mean, divided by n-1
Standard Deviation - square root of the variance
Coefficient of variation - (stddev / mean) * 100
A few questions came up about the Variance:
1. Why do we square the deviation from the mean when calculating the variance?
Answer: So that positive and negative values don't cancel each other out. Squaring ensures that we always have a positive value.
Follow-up question: Why not just take the absolute value?
Answer: Because we'll want to integrate this function and the absolute value function is not differentiable because it has a "kink" in it at 0 where the derivative is undefined.
2. Why do we divide by n-1 and not n?
Answer: If we were looking at the entire population, we would indeed divide by N. But with a sample, we divide by n-1. I'm still not so clear on the why that is.
### Lecture 1 - Ch 2 - Presentational Statistics
There are a number of different ways to graphically represent data, most of which are familiar to most people, but some may be new.
Familiar examples:
Summary table - categorize data and present in a table, often with percentages
Bar chart
Pie chart
This was new to me:
Stem-and-leaf diagram: It's used with numerical data. Take the first digit(s) and represent it/them as a single stem, the last digit is the leaf and may be repeated. In addition, a column is added to the left which indicates how many leaves are on each stem. It's kinda hard to describe the algorithm for constructing this diagram and it's best done with a demo.
The one thing that was unclear was why one of the counts was in parentheses. I looked this up in thee Minitab help file and found: "If the median value for the sample is included in a row, the count for that row is enclosed in parentheses." The row with the median is not necessarily the row with the most number of leaves.
Also somewhat new, but pretty much intuitively obvious:
Frequency Distribution and Histogram:
2. Determine the range
3. Select some number of classes into which you want to equally divide your data points
4. Compute the interval = range/# of classes (rounded up)
5. Determine the class boundaries based on the interval
6. Count the number of data points in each interval
Report the frequency and relative frequency (=freq/total # of data points) in a table
A Histogram is a bar chart (usually vertical) representation of the frequency of each class.
Some graphical ways to represent two variables:
Scatter Diagram - basically just an x-y plot of two variables
Time-Series Plot - plot of a single variable over time
### Lecture 1 - Ch 1 - Basic Definitions
Here are some basic definitions of terms used in the study of statistics:
When discussing statistics about a group, it's important to distinguish between the entire group and a subset of that group which we're able to look at in detail and then extrapolate from that detailed analysis of the sample to the larger group. The entire group is called the population. The subset which we're analyzing is the called the sample.
For example, we may want to know about the voting preferences among voters in the US. The population is all the people who are registered to vote in the upcoming election. Since we can't feasibly ask each and every one about their preferences, we ask a sample, perhaps just a few hundred or thousand voters, about their preferences and we extrapolate from the answers we get from the sample to the entire population.
Of course, you have to understand that this extrapolation involves a degree of uncertainty. Determining exactly what that uncertainty is will be the subject of a later lecture.
Population - the entire collection under discussion
Sample - a portion of the population selected for analysis
Parameter - a numerical measure that describes a characteristic of a population.
Statistic - a numerical measure that describes a characteristic of a sample.
Descriptive statistics - focuses on collection, summarizing and presenting a set of data.
Inferential statistics - uses sample data to draw conclusions about a population.
Categorical data - consist of categorical responses, such as yes/no, agree/disagree, Sunday/Monday/Tuesday..., etc.
Numerical data - consist of numerical responses. Numerical data can be either discrete or continuous (see below).
Discrete data - usually refers to integer responses, i.e. 0, 1, 2, 3, etc.
Continuous data - consist of responses that can be any numerical value.
## Monday, January 7, 2008
### Lecture 1 - Introduction
Our first GSB 420 class was last Thursday night. We met our teacher, Cemil Selcuk, and found out that his first name is pronounced "je-mil". I think he also said that his last name is pronounced "sel-juk".
We'll have the Depaul standard:
4 lectures
Midterm (35%)
5 lectures
Final (45%)
The other 20% of the grade is:
Homework (15%)
>= 5 in-class pop quizzes (5%)
We'll be using Minitab in this class. I'm excited about that! I always like to learn new tools, especially powerful tools for number crunching. I would really have liked to get the student edition of SPSS, but I guess that's asking for too much. :) A few of the people in class didn't get Minitab included in their text. Bummer. They need to get it.
Surprisingly, math review (algebra and calculus) is going to be an optional, time-permitting lecture. I would have expected those topics at the very beginning of the first lecture, but I'm kinda glad that we're not even doing it. At least there are some expectations that we need to come in with a baseline of knowledge about the subject.
For anyone who needs the reviews, I found some good notes at Prof. Timothy Opiela's web site. He teaches this course frequently, but not this quarter. Here's a link to his site: Opiela GSB 420. | 2018-02-25T21:35:27 | {
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https://math.stackexchange.com/questions/741681/weakly-convergent-subsequence-implies-strongly-convergent | # weakly convergent subsequence implies strongly convergent
Statement: Let $X$ be a Banach space If $x_n \rightarrow x$ weakly and every subsequence of $\{x_n\}$ has a strongly convergent subsequence, then $x_n\rightarrow x$ strongly in $X$
Attempt: ?
• Hint: If you can show that weak limit is unique, then the statement become: Every subsequence of $x_n$ has a subsequence converges strongly to $x$. – user99914 Apr 6 '14 at 4:55
Suppose $\lim_{n \to \infty}x_n\neq x$. Then there is an $\epsilon \gt0$ and a subsequence $(x_{n_k})$ such that $\|x_{n_k}-x\|\ge\epsilon$ for all $k\in\mathbb{N}$. Use this to derive a contradiction. | 2019-06-26T12:23:03 | {
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https://cs.stackexchange.com/questions/32414/why-is-the-onw-algorithm-for-the-knapsack-problem-not-a-polynomial-one?noredirect=1 | # Why is the O(nW) algorithm for the Knapsack problem not a polynomial one?
On the wikipedia page for the knapsack problem it says that the runtime is $\mathcal{O} (nW)$ and goes on to say that this doesn't violate its classification as NP because the input size is related to $\log W$, where, I believe, $W$ is the size of the knapsack. Why is the size of the input related logarithmically to $W$?
• Problems don't have runtimes. Furthermore, "pseudopolynomial" is the buzzword you want to search for. – Raphael Oct 28 '14 at 19:25
• See also here, here and here (duplicate?). – Raphael Oct 29 '14 at 6:29
• Oh, and some nitpick: "doesn't violate its classification as NP" is an empty statement. Even if the algorithm ran in polynomial time, Knapsack would still be in NP and NP-complete. – Raphael Oct 29 '14 at 6:33
• If your knapsack has a size of say 6,345,987 that's seven digits. The size of the knapsack is over 6 million, the size of the problem is just seven digits. – gnasher729 2 days ago
Polynomial time means that the running time is bounded by a polynomial in the length of the input. The running time here is bounded by $nW$. $n$, the number of items, is surely less than the length of the input, so that part is fine. But $W$, the target weight, is a number that appears in the input, in binary. In $\ell$ bits, you can write a number up to $2^\ell$ so $W$ is potentially exponential in the length of the input, not polynomial.
• Doesn't what you're saying suggest it is $P$? – user8722 Oct 28 '14 at 19:14
• Well yes. If you encode the input unary, then it's even in Logspace. But that's just a foul trick; consider for example a knapsack problem for $n=10^{12}$. You would need a terabyte to encode the problem unary! Unary encoding means that you encode the number by a sequence of $1$s. And relative to that GIGANTIC input size the runtime is polynomial. – john_leo Oct 28 '14 at 19:19
Think about it this way. Say you want to run the algorithm on an instance with $$W=1,000,000$$. When the program will ask you: Please input the value of $$W$$'', are you going to type 7 keys on your keyboard, or 1 million keys? | 2019-10-22T06:28:44 | {
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https://math.libretexts.org/Bookshelves/Algebra/Intermediate_Algebra_(OpenStax)/04%3A_Systems_of_Linear_Equations/4.04%3A_Solve_Mixture_Applications_with_Systems_of_Equations |
# 4.4: Solve Mixture Applications with Systems of Equations
Summary
By the end of this section, you will be able to:
• Solve mixture applications
• Solve interest applications
• Solve applications of cost and revenue functions
Before you get started, take this readiness quiz.
1. Multiply: $$4.025(1,562)$$.
If you missed this problem, review [link].
2. Write 8.2% as a decimal.
If you missed this problem, review [link].
3. Earl’s dinner bill came to $32.50 and he wanted to leave an 18% tip. How much should the tip be? If you missed this problem, review [link]. # Solve Mixture Applications Mixture application involve combining two or more quantities. When we solved mixture applications with coins and tickets earlier, we started by creating a table so we could organize the information. For a coin example with nickels and dimes, the table looked like this: Using one variable meant that we had to relate the number of nickels and the number of dimes. We had to decide if we were going to let n be the number of nickels and then write the number of dimes in terms of n, or if we would let d be the number of dimes and write the number of nickels in terms of d. Now that we know how to solve systems of equations with two variables, we’ll just let n be the number of nickels and d be the number of dimes. We’ll write one equation based on the total value column, like we did before, and the other equation will come from the number column. For the first example, we’ll do a ticket problem where the ticket prices are in whole dollars, so we won’t need to use decimals just yet. Example $$\PageIndex{1}$$ Translate to a system of equations and solve: A science center sold 1,363 tickets on a busy weekend. The receipts totaled$12,146. How many $12 adult tickets and how many$7 child tickets were sold?
Step 1. Read the problem. We will create a table to organize the information. Step 2. Identify what we are looking for. We are looking for the number of adult tickets and the number of child tickets sold. Step 3. Name what we are looking for. Let $$a= \text{the number of adult tickets.}$$ $$c= \text{the number of child tickets}$$ A table will help us organize the data. We have two types of tickets, adult and child. Write in a and c for the number of tickets. Write the total number of tickets sold at the bottom of the Number column. Altogether 1,363 were sold. Write the value of each type of ticket in the Value column. The value of each adult ticket is $12. The value of each child tickets is$7. The number times the value gives the total value, so the total value of adult tickets is $$a·12=12a$$, and the total value of child tickets is $$c·7=7c$$. Fill in the Total Value column. Altogether the total value of the tickets was $12,146. Step 4. Translate into a system of equations. The Number column and the Total value column give us the system of equations. We will use the elimination method to solve this system. Multiply the first equation by $$−7$$. Simplify and add, then solve for a. Substitute $$a=521$$ into the first equation, then solve for c. Step 6. Check the answer in the problem. 521 adult at$12 per ticket makes $6,252 842 child at$7 per ticket makes $58,994 The total receipts are$12,146$$\checkmark$$ Step 7. Answer the question. The science center sold 521 adult tickets and 842 child tickets.
Example $$\PageIndex{2}$$
Translate to a system of equations and solve:
The ticket office at the zoo sold 553 tickets one day. The receipts totaled $3,936. How many$9 adult tickets and how many $6 child tickets were sold? Answer 206 adults, 347 children Example $$\PageIndex{3}$$ Translate to a system of equations and solve: The box office at a movie theater sold 147 tickets for the evening show, and receipts totaled$1,302. How many $11 adult and how many$8 child tickets were sold?
In the next example, we’ll solve a coin problem. Now that we know how to work with systems of two variables, naming the variables in the ‘number’ column will be easy.
Example $$\PageIndex{4}$$
Translate to a system of equations and solve:
Juan has a pocketful of nickels and dimes. The total value of the coins is $8.10. The number of dimes is 9 less than twice the number of nickels. How many nickels and how many dimes does Juan have? Answer Step 1. Read the problem. We will create a table to organize the information. Step 2. Identify what we are looking for. We are looking for the number of nickels and the number of dimes. Step 3. Name what we are looking for. Let $$n= \text{the number of nickels.}$$ $$d= \text{the number of dimes}$$ A table will help us organize the data. We have two types of coins, nickels and dimes. Write n and d for the number of each type of coin. Fill in the Value column with the value of each type of coin. The value of each nickel is$0.05. The value of each dime is $0.10. The number times the value gives the total value, so, the total value of the nickels is $$n(0.05)=0.05n$$ and the total value of dimes is $$d(0.10)=0.10d$$. Altogether the total value of the coins is$8.10. Step 4. Translate into a system of equations. The Total Value column gives one equation. We also know the number of dimes is 9 less than twice the number of nickels. Translate to get the second equation. Now we have the system to solve. Step 5. Solve the system of equations We will use the substitution method. Substitute $$d=2n−9$$ into the first equation. Simplify and solve for n. To find the number of dimes, substitute $$n=36$$ into the second equation. Step 6. Check the answer in the problem 63 dimes at $$0.10=6.30$$ 36 nickels at $$0.05=1.80$$ Total $$=8.10\checkmark$$ Step 7. Answer the question. Juan has 36 nickels and 63 dimes.
Example $$\PageIndex{5}$$
Translate to a system of equations and solve:
Matilda has a handful of quarters and dimes, with a total value of $8.55. The number of quarters is 3 more than twice the number of dimes. How many dimes and how many quarters does she have? Answer 13 dimes and 29 quarters Example $$\PageIndex{6}$$ Translate to a system of equations and solve: Priam has a collection of nickels and quarters, with a total value of$7.30. The number of nickels is six less than three times the number of quarters. How many nickels and how many quarters does he have?
19 quarters and 51 nickels
Some mixture applications involve combining foods or drinks. Example situations might include combining raisins and nuts to make a trail mix or using two types of coffee beans to make a blend.
Example $$\PageIndex{7}$$
Translate to a system of equations and solve:
Carson wants to make 20 pounds of trail mix using nuts and chocolate chips. His budget requires that the trail mix costs him $7.60. per pound. Nuts cost$9.00 per pound and chocolate chips cost $2.00 per pound. How many pounds of nuts and how many pounds of chocolate chips should he use? Answer Step 1. Read the problem. We will create a table to organize the information. Step 2. Identify what we are looking for. We are looking for the number of pounds of nuts and the number of pounds of chocolate chips. Step 3. Name what we are looking for. Let $$n= \text{the number of pound of nuts.}$$ $$c= \text{the number of pounds of chips}$$ Carson will mix nuts and chocolate chips to get trail mix. Write in n and c for the number of pounds of nuts and chocolate chips. There will be 20 pounds of trail mix. Put the price per pound of each item in the Value column. Fill in the last column using $$\text{Number}•\text{Value}=\text{Total Value}$$ Step 4. Translate into a system of equations. We get the equations from the Number and Total Value columns. Step 5. Solve the system of equations We will use elimination to solve the system. Multiply the first equation by $$−2$$ to eliminate c. Simplify and add. Solve for n. To find the number of pounds of chocolate chips, substitute $$n=16$$ into the first equation, then solve for c. Step 6. Check the answer in the problem. $$\begin{array} {lll} 16+4 &= &20\checkmark \\ 9·16+2·4 &= &152\checkmark \end{array}$$ Step 7. Answer the question. Carson should mix 16 pounds of nuts with 4 pounds of chocolate chips to create the trail mix. Example $$\PageIndex{8}$$ Translate to a system of equations and solve: Greta wants to make 5 pounds of a nut mix using peanuts and cashews. Her budget requires the mixture to cost her$6 per pound. Peanuts are $4 per pound and cashews are$9 per pound. How many pounds of peanuts and how many pounds of cashews should she use?
3 pounds peanuts and 2 pounds cashews
Example $$\PageIndex{9}$$
Translate to a system of equations and solve:
Sammy has most of the ingredients he needs to make a large batch of chili. The only items he lacks are beans and ground beef. He needs a total of 20 pounds combined of beans and ground beef and has a budget of $3 per pound. The price of beans is$1 per pound and the price of ground beef is $5 per pound. How many pounds of beans and how many pounds of ground beef should he purchase? Answer 10 pounds of beans, 10 pounds of ground beef Another application of mixture problems relates to concentrated cleaning supplies, other chemicals, and mixed drinks. The concentration is given as a percent. For example, a 20% concentrated household cleanser means that 20% of the total amount is cleanser, and the rest is water. To make 35 ounces of a 20% concentration, you mix 7 ounces (20% of 35) of the cleanser with 28 ounces of water. For these kinds of mixture problems, we’ll use “percent” instead of “value” for one of the columns in our table. Example $$\PageIndex{10}$$ Translate to a system of equations and solve: Sasheena is lab assistant at her community college. She needs to make 200 milliliters of a 40% solution of sulfuric acid for a lab experiment. The lab has only 25% and 50% solutions in the storeroom. How much should she mix of the 25% and the 50% solutions to make the 40% solution? Answer Step 1. Read the problem. A figure may help us visualize the situation, then we will create a table to organize the information. Sasheena must mix some of the $$25%$$ solution and some of the $$50%$$ solution together to get $$200\space ml$$ of the $$40%$$ solution. Step 2. Identify what we are looking for. We are looking for how much of each solution she needs. Step 3. Name what we are looking for. Let $$x= \text{number of }ml\text{ of }25% \text{ solution.}$$ $$y= \text{number of }ml\text{ of }50%\text{ solution$$ A table will help us organize the data. She will mix x $$ml$$ of $$25%$$ with y $$ml$$ of $$50%$$ to get $$200 \space ml$$ of $$40%$$ solution. We write the percents as decimals in the chart. We multiply the number of units times the concentration to get the total amount of sulfuric acid in each solution. Step 4. Translate into a system of equations. We get the equations from the Number column and the Amount column. Now we have the system. Step 5. Solve the system of equations We will solve the system by elimination. Multiply the first equation by $$−0.5$$ to eliminate y. Simplify and add to solve for x. To solve for y, substitute $$x=80$$ into the first equation. Step 6. Check the answer in the problem. $$\begin{array} {lll} 80+120 &= &200\checkmark \\ 0.25(80)+0.50(120) &= &200\checkmark \\ {} &{} &\text{Yes!} \end{array}$$ Step 7. Answer the question. Sasheena should mix $$80 \space ml$$ of the $$25%$$ solution with $$120 \space ml$$ of the $$50%$$ solution to get the $$200\space ml$$ of the $$40%$$ solution. Example $$\PageIndex{11}$$ Translate to a system of equations and solve: LeBron needs 150 milliliters of a 30% solution of sulfuric acid for a lab experiment but only has access to a 25% and a 50% solution. How much of the 25% and how much of the 50% solution should he mix to make the 30% solution? Answer 120 ml of 25% solution and 30 ml of 50% solution Example $$\PageIndex{12}$$ Translate to a system of equations and solve: Anatole needs to make 250 milliliters of a 25% solution of hydrochloric acid for a lab experiment. The lab only has a 10% solution and a 40% solution in the storeroom. How much of the 10% and how much of the 40% solutions should he mix to make the 25% solution? Answer 125 ml of 10% solution and 125 ml of 40% solution # Solve Interest Applications The formula to model simple interest applications is $$I=Prt$$. Interest, I, is the product of the principal, P, the rate, r, and the time, t. In our work here, we will calculate the interest earned in one year, so t will be 1. We modify the column titles in the mixture table to show the formula for interest, as you’ll see in the next example. Example $$\PageIndex{13}$$ Translate to a system of equations and solve: Adnan has$40,000 to invest and hopes to earn $$7.1%$$ interest per year. He will put some of the money into a stock fund that earns 8% per year and the rest into bonds that earns 3% per year. How much money should he put into each fund?
Step 1. Read the problem. A chart will help us organize the information. Step 2. Identify what we are looking for. We are looking for the amount to invest in each fund. Step 3. Name what we are looking for. Let $$s= \text{the amount invested in stocks.}$$ $$b= \text{the amount invested in stocks}$$ Write the interest rate as a decimal for each fund. Multiply: Principal · Rate · Time Step 4. Translate into a system of equations. We get our system of equations from the Principal column and the Interest column. Step 5. Solve the system of equations by elimination. Multiply the top equation by $$−0.03$$. Simplify and add to solve for s. To find b, substitute s = 32,800 into the first equation. Step 6. Check the answer in the problem. We leave the check to you. Step 7. Answer the question. Adnan should invest $32,800 in stock and$7,200 in bonds.
Did you notice that the Principal column represents the total amount of money invested while the Interest column represents only the interest earned? Likewise, the first equation in our system, $$s+b=40,000$$, represents the total amount of money invested and the second equation, $$0.08s+0.03b=0.071(40,000)$$, represents the interest earned.
Example $$\PageIndex{14}$$
Translate to a system of equations and solve:
Leon had $50,000 to invest and hopes to earn $$6.2%$$ interest per year. He will put some of the money into a stock fund that earns 7% per year and the rest in to a savings account that earns 2% per year. How much money should he put into each fund? Answer$42,000 in the stock fund and $8000 in the savings account Example $$\PageIndex{15}$$ Translate to a system of equations and solve: Julius invested$7000 into two stock investments. One stock paid 11% interest and the other stock paid 13% interest. He earned $$12.5%$$ interest on the total investment. How much money did he put in each stock?
$1750 at 11% and$5250 at 13%
The next example requires that we find the principal given the amount of interest earned.
Example $$\PageIndex{16}$$
Translate to a system of equations and solve:
Rosie owes $21,540 on her two student loans. The interest rate on her bank loan is $$10.5%$$ and the interest rate on the federal loan is $$5.9%$$. The total amount of interest she paid last year was $$1,669.68$$. What was the principal for each loan? Answer Step 1. Read the problem. A chart will help us organize the information. Step 2. Identify what we are looking for. We are looking for the principal of each loan. Step 3. Name what we are looking for. Let $$b= \text{the principal for the bank loan.}$$ $$f= \text{the principal on the federal loan}$$ The total loans are$21,540. Record the interest rates as decimals in the chart. Multiply using the formula I = Prt to get the Interest. Step 4. Translate into a system of equations. The system of equations comes from the Principal column and the Interest column. Step 5. Solve the system of equations We will use substitution to solve. Solve the first equation for b. Substitute b = −f + 21.540 into the second equation. Simplify and solve for f. To find b, substitute f = 12,870 into the first equation. Step 6. Check the answer in the problem. We leave the check to you. Step 7. Answer the question. The principal of the federal loan was $12,870 and the principal for the bank loan was$8,670.
Example $$\PageIndex{17}$$
Translate to a system of equations and solve:
Laura owes $18,000 on her student loans. The interest rate on the bank loan is 2.5% and the interest rate on the federal loan is 6.9%. The total amount of interest she paid last year was$1,066. What was the principal for each loan?
Bank $4,000; Federal$14,000
Example $$\PageIndex{18}$$
Translate to a system of equations and solve:
Jill’s Sandwich Shoppe owes $65,200 on two business loans, one at 4.5% interest and the other at 7.2% interest. The total amount of interest owed last year was$3,582. What was the principal for each loan?
$41,200 at 4.5%,$24,000 at 7.2%
# Solve applications of cost and revenue functions
Suppose a company makes and sells x units of a product. The cost to the company is the total costs to produce x units. This is the cost to manufacture for each unit times x, the number of units manufactured, plus the fixed costs.
The revenue is the money the company brings in as a result of selling x units. This is the selling price of each unit times the number of units sold.
When the costs equal the revenue we say the business has reached the break-even point.
COST AND REVENUE FUNCTIONS
The cost function is the cost to manufacture each unit times x, the number of units manufactured, plus the fixed costs.
$C(x)=(\text{cost per unit})·x+\text{fixed costs}\nonumber$
The revenue function is the selling price of each unit times x, the number of units sold.
$R(x)=(\text{selling price per unit})·x\nonumber$
The break-even point is when the revenue equals the costs.
$C(x)=R(x)\nonumber$
Example $$\PageIndex{19}$$
The manufacturer of a weight training bench spends $105 to build each bench and sells them for$245. The manufacturer also has fixed costs each month of $7,000. ⓐ Find the cost function C when x benches are manufactured. ⓑ Find the revenue function R when x benches are sold. ⓒ Show the break-even point by graphing both the Revenue and Cost functions on the same grid. ⓓ Find the break-even point. Interpret what the break-even point means. Answer ⓐ The manufacturer has$7,000 of fixed costs no matter how many weight training benches it produces. In addition to the fixed costs, the manufacturer also spends $105 to produce each bench. Suppose x benches are sold. $$\begin{array} {ll} {\text{Write the general Cost function formula.}} &{C(x)=(\text{cost per unit})·x+\text{fixed costs}} \\ {\text{Substitute in the cost values.}} &{C(x)=105x+7000} \\ \end{array}$$ ⓑ The manufacturer sells each weight training bench for$245. We get the total revenue by multiplying the revenue per unit times the number of units sold.
$$\begin{array} {ll} {\text{Write the general Revenue function.}} &{C(x)=(\text{selling price per unit})·x} \\ {\text{Substitute in the revenue per unit.}} &{R(x)=245x} \\ \end{array}$$
ⓒ Essentially we have a system of linear equations. We will show the graph of the system as this helps make the idea of a break-even point more visual.
$\left\{ \begin{array} {l} C(x)=105x+7000 \\ R(x)=245x \end{array} \right. \quad \text{or} \quad \left\{ \begin{array} {l} y=105x+7000 \\ y=245x \end{array} \right. \nonumber$
ⓓ To find the actual value, we remember the break-even point occurs when costs equal revenue.
$$\begin{array} {ll} {\text{Write the break-even formula.}} &{\begin{array} {l} {C(x)=R(x)} \\ {105x+7000=245x} \end{array}} \\ {\text{Solve.}} &{\begin{array} {l} {7000=140x} \\ {50=x} \end{array}} \\ \end{array}$$
When 50 benches are sold, the costs equal the revenue.
When 50 benches are sold, the revenue and costs are both $12,250. Notice this corresponds to the ordered pair $$(50,12250)$$. Example $$\PageIndex{20}$$ The manufacturer of a weight training bench spends$15 to build each bench and sells them for $32. The manufacturer also has fixed costs each month of$25,500.
ⓐ Find the cost function C when x benches are manufactured.
ⓑ Find the revenue function R when x benches are sold.
ⓒ Show the break-even point by graphing both the Revenue and Cost functions on the same grid.
ⓓ Find the break-even point. Interpret what the break-even point means.
ⓐ $$C(x)=15x+25,500$$
ⓑ $$R(x)=32x$$
ⓓ 1,5001,500; when 1,500 benches are sold, the cost and revenue will be both 48,000
Example $$\PageIndex{21}$$
The manufacturer of a weight training bench spends $120 to build each bench and sells them for$170. The manufacturer also has fixed costs each month of $150,000. ⓐ Find the cost function C when x benches are manufactured. ⓑ Find the revenue function R when x benches are sold. ⓒ Show the break-even point by graphing both the Revenue and Cost functions on the same grid. ⓓ Find the break-even point. Interpret what the break-even point means. Answer ⓐ $$C(x)=120x+150,000$$ ⓑ $$R(x)=170x$$ ⓓ $$3,000$$; when 3,000 benches are sold, the revenue and costs are both$510,000
Access this online resource for additional instruction and practice with interest and mixtures.
# Key Concepts
• Cost function: The cost function is the cost to manufacture each unit times x, the number of units manufactured, plus the fixed costs.
$$C(x)=(\text{cost per unit})·x+\text{fixed costs}$$
• Revenue: The revenue function is the selling price of each unit times x, the number of units sold.
$$R(x)=(\text{selling price per unit})·x$$
• Break-even point: The break-even point is when the revenue equals the costs.
$$C(x)=R(x)$$
# Practice Makes Perfect
Solve Mixture Applications
In the following exercises, translate to a system of equations and solve.
Tickets to a Broadway show cost $35 for adults and$15 for children. The total receipts for 1650 tickets at one performance were $47,150. How many adult and how many child tickets were sold? Tickets for the Cirque du Soleil show are$70 for adults and $50 for children. One evening performance had a total of 300 tickets sold and the receipts totaled$17,200. How many adult and how many child tickets were sold?
110 adult tickets, 190 child tickets
Tickets for an Amtrak train cost $10 for children and$22 for adults. Josie paid $1200 for a total of 72 tickets. How many children tickets and how many adult tickets did Josie buy? Tickets for a Minnesota Twins baseball game are$69 for Main Level seats and $39 for Terrace Level seats. A group of sixteen friends went to the game and spent a total of$804 for the tickets. How many of Main Level and how many Terrace Level tickets did they buy?
6 good seats, 10 cheap seats
Tickets for a dance recital cost $15 for adults and$7 dollars for children. The dance company sold 253 tickets and the total receipts were $2771. How many adult tickets and how many child tickets were sold? Tickets for the community fair cost$12 for adults and $5 dollars for children. On the first day of the fair, 312 tickets were sold for a total of$2204. How many adult tickets and how many child tickets were sold?
92 adult tickets, 220 children tickets
Brandon has a cup of quarters and dimes with a total value of $$3.80$$. The number of quarters is four less than twice the number of quarters. How many quarters and how many dimes does Brandon have?
Sherri saves nickels and dimes in a coin purse for her daughter. The total value of the coins in the purse is $$0.95$$. The number of nickels is two less than five times the number of dimes. How many nickels and how many dimes are in the coin purse?
13 nickels, 3 dimes
Peter has been saving his loose change for several days. When he counted his quarters and nickels, he found they had a total value $$13.10$$. The number of quarters was fifteen more than three times the number of dimes. How many quarters and how many dimes did Peter have?
Lucinda had a pocketful of dimes and quarters with a value of $$6.20$$. The number of dimes is eighteen more than three times the number of quarters. How many dimes and how many quarters does Lucinda have?
42 dimes, 8 quarters
A cashier has 30 bills, all of which are $10 or$20 bills. The total value of the money is $460. How many of each type of bill does the cashier have? A cashier has 54 bills, all of which are$10 or $20 bills. The total value of the money is$910. How many of each type of bill does the cashier have?
17 $10 bills, 37$20 bills
Marissa wants to blend candy selling for $$1.80$$ per pound with candy costing $$1.20$$ per pound to get a mixture that costs her $$1.40$$ per pound to make. She wants to make 90 pounds of the candy blend. How many pounds of each type of candy should she use?
How many pounds of nuts selling for $6 per pound and raisins selling for$3 per pound should Kurt combine to obtain 120 pounds of trail mix that cost him $5 per pound? Answer 80 pounds nuts and 40 pounds raisins Hannah has to make twenty-five gallons of punch for a potluck. The punch is made of soda and fruit drink. The cost of the soda is $$1.79$$ per gallon and the cost of the fruit drink is $$2.49$$ per gallon. Hannah’s budget requires that the punch cost $$2.21$$ per gallon. How many gallons of soda and how many gallons of fruit drink does she need? Joseph would like to make twelve pounds of a coffee blend at a cost of$6 per pound. He blends Ground Chicory at $5 a pound with Jamaican Blue Mountain at$9 per pound. How much of each type of coffee should he use?
9 pounds of Chicory coffee, 3 pounds of Jamaican Blue Mountain coffee
Julia and her husband own a coffee shop. They experimented with mixing a City Roast Columbian coffee that cost $7.80 per pound with French Roast Columbian coffee that cost$8.10 per pound to make a twenty-pound blend. Their blend should cost them $7.92 per pound. How much of each type of coffee should they buy? Twelve-year old Melody wants to sell bags of mixed candy at her lemonade stand. She will mix M&M’s that cost$4.89 per bag and Reese’s Pieces that cost $3.79 per bag to get a total of twenty-five bags of mixed candy. Melody wants the bags of mixed candy to cost her$4.23 a bag to make. How many bags of M&M’s and how many bags of Reese’s Pieces should she use?
10 bags of M&M’s, 15 bags of Reese’s Pieces
Jotham needs 70 liters of a 50% solution of an alcohol solution. He has a 30% and an 80% solution available. How many liters of the 30% and how many liters of the 80% solutions should he mix to make the 50% solution?
Joy is preparing 15 liters of a 25% saline solution. She only has 40% and 10% solution in her lab. How many liters of the 40% and how many liters of the 10% should she mix to make the 25% solution?
$$7.5$$ liters of each solution
A scientist needs 65 liters of a 15% alcohol solution. She has available a 25% and a 12% solution. How many liters of the 25% and how many liters of the 12% solutions should she mix to make the 15% solution?
A scientist needs 120 milliliters of a 20% acid solution for an experiment. The lab has available a 25% and a 10% solution. How many liters of the 25% and how many liters of the 10% solutions should the scientist mix to make the 20% solution?
80 liters of the 25% solution and 40 liters of the 10% solution
A 40% antifreeze solution is to be mixed with a 70% antifreeze solution to get 240 liters of a 50% solution. How many liters of the 40% and how many liters of the 70% solutions will be used?
A 90% antifreeze solution is to be mixed with a 75% antifreeze solution to get 360 liters of an 85% solution. How many liters of the 90% and how many liters of the 75% solutions will be used?
240 liters of the 90% solution and 120 liters of the 75% solution
Solve Interest Applications
In the following exercises, translate to a system of equations and solve.
Hattie had $3000 to invest and wants to earn 10.6%10.6% interest per year. She will put some of the money into an account that earns 12% per year and the rest into an account that earns 10% per year. How much money should she put into each account? Carol invested$2560 into two accounts. One account paid 8% interest and the other paid 6% interest. She earned 7.25%7.25% interest on the total investment. How much money did she put in each account?
$1600 at 8%, 960 at 6% Sam invested$48,000, some at 6% interest and the rest at 10%. How much did he invest at each rate if he received $4000 in interest in one year? Arnold invested$64,000, some at $$5.5%$$ interest and the rest at 9%. How much did he invest at each rate if he received $4500 in interest in one year? Answer$28,000 at 9%, $36,000 at $$5.5%$$ After four years in college, Josie owes$65, 800 in student loans. The interest rate on the federal loans is $$4.5%$$ and the rate on the private bank loans is 2%. The total interest she owes for one year was $$2878.50$$. What is the amount of each loan?
Mark wants to invest $10,000 to pay for his daughter’s wedding next year. He will invest some of the money in a short term CD that pays 12% interest and the rest in a money market savings account that pays 5% interest. How much should he invest at each rate if he wants to earn$1095 in interest in one year?
$8500 CD,$1500 savings account
A trust fund worth $25,000 is invested in two different portfolios. This year, one portfolio is expected to earn $$5.25%$$ interest and the other is expected to earn 4%. Plans are for the total interest on the fund to be$1150 in one year. How much money should be invested at each rate?
A business has two loans totaling $85,000. One loan has a rate of 6% and the other has a rate of 4.5% This year, the business expects to pay$4,650 in interest on the two loans. How much is each loan?
$55,000 on loan at 6% and$30,000 on loan at $$4.5%$$
Solve Applications of Cost and Revenue Functions
The manufacturer of an energy drink spends $1.20 to make each drink and sells them for$2. The manufacturer also has fixed costs each month of $8,000. ⓐ Find the cost function C when x energy drinks are manufactured. ⓑ Find the revenue function R when x drinks are sold. ⓒ Show the break-even point by graphing both the Revenue and Cost functions on the same grid. ⓓ Find the break-even point. Interpret what the break-even point means. The manufacturer of a water bottle spends$5 to build each bottle and sells them for $10. The manufacturer also has fixed costs each month of$6500. ⓐ Find the cost function C when x bottles are manufactured. ⓑ Find the revenue function R when x bottles are sold. ⓒ Show the break-even point by graphing both the Revenue and Cost functions on the same grid. ⓓ Find the break-even point. Interpret what the break-even point means.
ⓐ $$C(x)=5x+6500$$
ⓑ $$R(x)=10x$$
ⓓ 1,500; when 1,500 water bottles are sold, the cost and the revenue equal \$15,000
## Writing Exercises
Take a handful of two types of coins, and write a problem similar to Example relating the total number of coins and their total value. Set up a system of equations to describe your situation and then solve it.
In Example, we used elimination to solve the system of equations
$$\left\{ \begin{array} {l} s+b=40,000 \\ 0.08s+0.03b=0.071(40,000). \end{array} \right.$$
Could you have used substitution or elimination to solve this system? Why?
## Self Check
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?
# Glossary
cost function
The cost function is the cost to manufacture each unit times xx, the number of units manufactured, plus the fixed costs; C(x) = (cost per unit)x + fixed costs.
revenue
The revenue is the selling price of each unit times x, the number of units sold; R(x) = (selling price per unit)x.
break-even point
The point at which the revenue equals the costs is the break-even point; C(x)=R(x). | 2021-12-09T13:36:49 | {
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https://ita.skanev.com/04/03/02.html | # Exercise 4.3.2
Show that the solution of $T(n) = T(\lceil n/2 \rceil) + 1$ is $O(\lg{n})$
We guess $T(n) \le c\lg(n - 2)$:
$$T(n) \le c\lg(\lceil n/2 \rceil - 2) + 1 \le c\lg(n/2 + 1 - 2) + 1 \le c\lg((n - 2)/2) + 1 \le c\lg(n - 2) - c\lg2 + 1 \le c\lg(n - 2)$$ | 2018-01-21T08:44:58 | {
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