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# Is the cost of an Organ more than the cost of a Flute ?
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Is the cost of an Organ more than the cost of a Flute ? [#permalink] 23 Nov 2011, 15:00
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Is the cost of an Organ more than the cost of a Flute ?
(1) Three times the cost of an organ is $15 more than twice the cost of a Flute. (2) Five times the cost of an organ is$12 more than six time the cost of a Flute.
[Reveal] Spoiler: OA
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Last edited by Bunuel on 27 Jan 2014, 05:28, edited 1 time in total.
Renamed the topic and edited the question.
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Re: DS#1 [#permalink] 23 Nov 2011, 16:22
1
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Let me try:
X: Cost of the organ
Y: Cost of the flute
1) $$3X - 2Y = 15$$
Two variables, one equation, not enough.
2) $$5X - 6Y = 12$$
It seems the same as in 1)
But, let's divide the equation by 6.
$$\frac{5X}{6} - Y = 2$$
So, if we had $$x - y$$, the result would be greater than 2.
What do you think?
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Re: DS#1 [#permalink] 23 Nov 2011, 20:02
metallicafan wrote:
Let me try:
X: Cost of the organ
Y: Cost of the flute
1) $$3X - 2Y = 15$$
Two variables, one equation, not enough.
2) $$5X - 6Y = 12$$
It seems the same as in 1)
But, let's divide the equation by 6.
$$\frac{5X}{6} - Y = 2$$
So, if we had $$x - y$$, the result would be greater than 2.
What do you think?
Thanks.
Wasn't 100% clear before, since 2nd equation too got 2 variables. I took both the equations to solve and ended up in the trap answer C.
Here the logic is 5/6 is less than 1 and still X possess a greater value as the answer is 2. Right?
Thanks metallicafan
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Re: DS#1 [#permalink] 24 Nov 2011, 05:54
1
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cleetus wrote:
metallicafan wrote:
Let me try:
X: Cost of the organ
Y: Cost of the flute
1) $$3X - 2Y = 15$$
Two variables, one equation, not enough.
2) $$5X - 6Y = 12$$
It seems the same as in 1)
But, let's divide the equation by 6.
$$\frac{5X}{6} - Y = 2$$
So, if we had $$x - y$$, the result would be greater than 2.
What do you think?
Thanks.
Wasn't 100% clear before, since 2nd equation too got 2 variables. I took both the equations to solve and ended up in the trap answer C.
Here the logic is 5/6 is less than 1 and still X possess a greater value as the answer is 2. Right?
Thanks metallicafan
Right. Maybe it is not the most elegant solution, but it is what I would do.
I am sure that there is a better way to solve it.
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Re: DS#1 [#permalink] 24 Nov 2011, 06:22
1
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cleetus wrote:
Is the cost of an Organ more than the cost of a Flute ?
A) Three times the cost of an organ is $15 more than twice the cost of a Flute. B) Five times the cost of an organ is$12 more than six time the cost of a Flute.
IMO its B. This is how i solved it...
S1: 3M = 15+2F
=> M = 5+0.66F.. In this equation as we are getting the decreased value in terms of 0.66F. So insufficient.
For example F can be 100 so M = 71 or if F = 12, M becomes 23.
S2: 5M = 12+6F
=> M = 2.4+1.2F. Here for whatever may be the value of the flute is Organ price is always greater as 1.2F will always be greater than F. Sufficient.
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Re: DS#1 [#permalink] 24 Nov 2011, 08:02
Thanks for more clarity +1
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Re: DS#1 [#permalink] 24 Nov 2011, 08:58
1
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Can't this be solved just by logic,
For 3x-2y = 12 or to have a +ve value, x can be less than y or x can be greater than y...so not sufficient.
For 5x-6y=15 or = to have a +ve value, x should be always greater than y...so sufficient.
Ans = B.
I know this is a crude method, but it helps when v have a crunch of time and in GMAT, v always do..
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Re: DS#1 [#permalink] 24 Nov 2011, 10:05
Yes this can be solved using logic. The above 2 explanation is just to illustrate how the logic works.
There are chances for beginners to go with the trap answer option C.
+1 to you.
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Re: DS#1 [#permalink] 27 Jan 2014, 05:27
1
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cleetus wrote:
Is the cost of an Organ more than the cost of a Flute ?
A) Three times the cost of an organ is $15 more than twice the cost of a Flute. B) Five times the cost of an organ is$12 more than six time the cost of a Flute.
Best approach is to do the following
Is Q>F?
Is 5Q>5F?
Statement 2
5Q = 6F+12
Replacing
6F+12>5F
F>-12? Of course
B
Re: DS#1 [#permalink] 27 Jan 2014, 05:27
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# Big, Bigger, Biggest!
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# tangent space at some point of a quasi-projective variety
In order to define the tangent space to a quasi-projective variety $V$ (i.e a locally closed closed subset of $\mathbb{P}^n$ considered with Zariski topology induced from $\mathbb{P}^n$) at a point $p$ we must think of $V$ as an open subset in a closed sub variety $W$ of some fixed projective space. Could any one explain me with a simple example that how the tangent space is at some point of a Quasi Proj.Variety?
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Dear Makuasi, It is not hard to define the tangent space intrinsically. What source are you reading that says otherwise? Regards, – Matt E May 12 '12 at 16:45
I am reading from an invitation to algebraic geometry by karen smith et all..should I write how the defined tangent space at some point of a variety? – La Belle Noiseuse May 13 '12 at 12:53
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2016-02-12 12:30:59
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# How do I find the time evolution of a ket?
I have a question which reads:
Let \begin{bmatrix} {E_0} & 0 & A \\ 0 & E_1 & 0 \\ A & 0 & E_0 \end{bmatrix} be the matrix representation of the Hamiltonian for a three-state system with basis states $|1>, |2> \mbox{and } |3>$.
a. If the state of the system at time $t$ = $0$ is $|\psi(0)>=|2>$ what is $|\psi(t)>$?
b. If the state of the system at time $t$ = $0$ is $|\psi(0)>=|3>$ what is $|\psi(t)>$?
$\textbf{My attempt at a solution:}$
a. For both problems we can use $|\psi(t)> = \hat{U}(t)|\psi(0)>$ where $\hat{U}= e^{\frac{-i\hat{H}t}{\hbar}}$. Since
$$|2> = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$$ is an eigenvector with eigenvalue $E_1$ we can simply replace the Hamiltonian in the time evolution operator by $E_1$, so $$|\psi(t)> = e^{\frac{-iE_1t}{\hbar}}|2>$$
Is this correct? I am finding other solutions online which have a different answer, although I can't see how this could possibly be wrong, unless my representation for $|2>$ is wrong.
Assuming this is the correct way of doing this, I am having a hard time doing b. I can find the eigenvalues and eigenvectors of the hamiltonian easily, and can represent |3> = $(0,0,1)^T$ as a linear combination of those vectors, thereby allowing me to operate on it. However, my final answer is in terms of |1> and |3>, which I feel is incorrect somehow.
• As an aside, note that \rangle in the MathJax produced $\rangle$. Looks nicer than $>$, IMO. – Kyle Kanos Mar 5 '17 at 13:52
• Expressing your answer in terms of $|1\rangle$ and $|3\rangle$ is a perfectly valid response; those two vectors form a perfectly good basis for a particular subspace of the Hilbert space. Of course, if you preferred, you also could express your response in terms of any other basis of that same subspace; it would just be another way of saying the same thing. The vector $|\psi(t)\rangle$ is unique up to phase, but there are different ways to express it. – Michael Seifert Mar 6 '17 at 20:49
Your reasoning is perfectly correct. Here it is in a complete form.
Let us write the Hamiltonian in the following way to make things clearer
$$\hat{H} = E_0(|1 \rangle \langle 1|+|3 \rangle \langle 3|) + E_1|2 \rangle \langle 2| + A(|1 \rangle \langle 3| + |3 \rangle \langle 1|)$$
It is then straightforward to see that :
• $|2 \rangle$ is an eigenstate with eigenvalue $E_1$ as you have already noticed. Hence if the initial state is $|2 \rangle$ then $$|\psi(t)\rangle = e^{-iE_1t/\hbar}|2\rangle$$
• $(|1\rangle+|3\rangle)$ and $(|1\rangle-|3\rangle)$ are eigenstates with respective eigenvalues of $E_0 + A$ and $E_0 - A$. Hence if the initial state is $|3\rangle = \frac{1}{2} [(|1\rangle+|3\rangle) - (|1\rangle-|3\rangle)]$, then $$|\psi(t)\rangle = \frac{1}{2} \left[ e^{-i(E_0+A)t/\hbar}(|1\rangle+|3\rangle) - e^{-i(E_0-A)t/\hbar}(|1\rangle-|3\rangle)\right]$$
I hope my explanation was clear !
Cheers
Actually I believe both answers are correct. I can't seem to find anything wrong with either. Certainly a. is correct since the hamiltonian in the time operator should just be replaced by the eigenvalue, seen simply if we expand the matrix exponential.
For b, there is nothing wrong with expressing our time dependent state as a linear combination of the initial state and another basis state.
• Additionally, the representation for |2> IS $(0,1,0)^T$ since this state is being represented in its own basis – john morrison Mar 4 '17 at 22:07
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2019-07-24 04:56:07
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https://alice-publications.web.cern.ch/node/4123
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First measurement of $Ξ_{\rm c}^0$ production in pp collisions at $\mathbf{\sqrt{s}}$ = 7 TeV
The production of the charm-strange baryon $\Xi_{\rm c}^0$ is measured for the first time at the LHC via its semileptonic decay into e$^+\Xi^-\nu_{\rm e}$ in pp collisions at $\sqrt{s}=7$ TeV with the ALICE detector. The transverse momentum ($p_{\rm T}$) differential cross section multiplied by the branching ratio is presented in the interval 1 $<~$ $p_{\rm T}$ $<~$ 8 GeV/$c$ at mid-rapidity, $|y|$ $<~$ 0.5. The transverse momentum dependence of the $\Xi_{\rm c}^0$ baryon production relative to the D$^0$ meson production is compared to predictions of event generators with various tunes of the hadronisation mechanism, which are found to underestimate the measured cross-section ratio.
Figures
Figure 1
(a) Invariant-mass distribution of $\Xi^- \rightarrow \pi^-\Lambda$ (and charge conjugate) candidates integrated over \pt. The arrow indicates the world average $\Xi^-$ mass from Ref.~ and the dashed lines indicate the selected interval. (b) Invariant-mass distributions of right-sign and wrong-sign (and charge conjugate) pairs integrated over the whole \pt\ interval. The arrow indicates the \xicz\ mass~.
Figure 2
Correlation between the generated \xicz-baryon \pt\ and the reconstructed \exipair\ pair \pt, obtained from the simulation based on {\sc pythia 6} described in the text.
Figure 3
Product of acceptance and efficiency ($A\times\varepsilon$) of \xicz\ baryons generated in $|y|< 0.8$ decaying into ${\rm e}^+\Xi^-\nu_{\rm e}$ as a function of \pt, determined from simulations {\sc pythia} 6 (see text).
Figure 4
Inclusive \xicz-baryon \pt-differential production cross section multiplied by the branching ratio into ${\rm e}^+\Xi^-\nu_{\rm e}$, as a function of \pt\ for $|y|< 0.5$, in \pp collisions at $\sqrt{s} =$ 7 TeV. The error bars and boxes represent the statistical and systematic uncertainties, respectively. The contribution from \xib\ decays is not subtracted.
Figure 5
Ratio of the \pt-differential cross sections of \xicz\ baryons (multiplied by the branching ratio into ${\rm e}^+\Xi^-\nu_{\rm e}$) and \Dzero mesons~ as a function of \pt\ for $|y|< 0.5$, in \pp collisions at $\sqrt{s} =$ 7 TeV. The error bars and boxes represent the statistical and systematic uncertainties, respectively. Predictions from theoretical models, (a) {\sc pythia 8} with different tunes (b) {\sc dipsy}~ and {\sc herwig 7}~, are shown as shaded bands representing the range of the currently available theoretical predictions for the branching ratio of the considered \xicz\ decay mode.
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2018-07-16 21:51:53
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https://stacks.math.columbia.edu/tag/0057
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Example 5.10.4. Let $X = \{ s, \eta \}$ with open sets given by $\{ \emptyset , \{ \eta \} , \{ s, \eta \} \}$. In this case a maximal chain of irreducible closed subsets is $\{ s\} \subset \{ s, \eta \}$. Hence $\dim (X) = 1$. It is easy to generalize this example to get a $(n + 1)$-element topological space of Krull dimension $n$.
There are also:
• 3 comment(s) on Section 5.10: Krull dimension
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).
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2021-09-25 06:23:25
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http://math.stackexchange.com/questions/112128/basic-covering-space-question
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# Basic covering space question
Given a path connected metric space $X$ and a cover $\tilde{X}$ which is also a path connected metric space with covering map $E$, then is $E$ a local isometry?
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## 1 Answer
Not true: For example take $X= S^1$ with subspace topology of $\mathbb R^2$.
and $\tilde{X}= \mathbb R$. Define map $p:\mathbb R\to S^1$ by $p(t)= (\cos 2\pi t, \sin2\pi t)$ . Then this is covering map but not local isometry.
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2015-07-07 02:45:46
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https://thoughtstreams.io/paltman/gc3pie/1038/
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# gc3pie
34 thoughts
last posted Jan. 3, 2013, 4:26 p.m.
28 earlier thoughts
0
For every distinct binary (or binary with certain parameters), I think an Application subclass will be created in a library of apps that can be used to construct Workflows of various types of collections that glue these apps together funneling the output from some into the input of others and making decisions about what next to run in the SequentialTaskCollection tasks.
5 later thoughts
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2020-08-04 14:57:50
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http://mathoverflow.net/questions/88020/sheaves-for-which-the-derived-compact-or-not-pushforward-is-zero?sort=votes
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# sheaves for which the derived (compact or not) pushforward is zero
Conventions: sheaf = complex of constructible sheaves (in the l-adic setup with etale tplg or in the complex coefficients setup with analytical tplg).
I would like to understand if there is an intuition behind the following property of a sheaf. We consider $X$ a variety (or a scheme, or a gentle alg. stack) not necessarily smooth/proper. What sort of sheaves $\mathcal{F}$ on $X$ satisfy the following property: $Rp_!(\mathcal{F}) = 0$ where $p:X\to \{*\}$. Or (by Verdier duality) the property:
$Rp_*(\mathcal{F}) = 0$.
Any (partial) answer in any particular case (say X smooth, or X proper and smooth, etc.) would be nice.
------- motivation for the question -----
I encountered the notion of "regular sheaf" in the paper of Braverman & Gaitsgory - "Geometric Eisenstein series" which uses the above property. In their context the map $p$ is from $Bun_{T}\to Bun_{T/G_m}$, where $T$ is a torus in a reductive group ($Bun_T$ is over a smooth proj curve); the map $p$ is induced from a cocharacter $G_m\to T$. They call a sheaf regular if $Rp_!\mathcal{F} = 0$ for all the maps $p$ as above that are induced from a coroot. This notion seems to be very important in their paper and I'm trying to understand intuitively this condition and see some simple examples.
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Retagged to avoid creating duplicates. – user5117 Feb 9 '12 at 18:23
Inspired by Borel-Weil-Bott, The following comes to mind (I don't know enough to put it in context):
We can consider a circle $S^1$, and take the local system which corresponds to the sign character of $\mathbb{Z}$. Then its zero'th cohomology is zero, and the first by duality is also zero.
Sasha
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This indeed a strong condition! But to get a sense of it, let us look at the simplest case where $X$ is a smooth not necessarily projective complex curve. Also let $\mathcal{F}$ be locally constant, so we can identify it with a representation of $\pi_1(X)$. The vanishing condition will imply that the topological Euler characteristic must be zero. So either $X= G_m$ or $X$ is an elliptic curve. In both cases, nontrivial examples exist: any local system with $$H^0(X,\mathcal{F})= \mathcal{F}^{\pi_1(X)}=0$$ will work. In the first case, this is clear because the Euler characteristic of $\mathcal{F}$ is zero and there is only $H^1$ to contend with. In the second, you use Poincare duality to also kill $H^2$.
I think this can be pushed to apply to any local system satisfying the above condition on a semiabelian variety (extension of an abelian variety by a torus)
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It seems that my answer overlaps with Sasha's. – Donu Arapura Feb 9 '12 at 19:00
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2015-06-30 08:20:43
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http://mathematica.stackexchange.com/questions?page=2&sort=newest
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# All Questions
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2015-10-13 13:50:34
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http://math203sp18.wikidot.com/mg-problem-19
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Mg Problem 19
PRELIMINARY
18. Definition of $<_{\mathbb{Z}}$:
$[(a,b)] <_{\mathbb{Z}} [(c,d)] \leftrightarrow a + d <_{\mathbb{N}} b + c$
Proof that this is well-defined, i.e. if it works for one element of the equivalence class, it works for all elements of the equivalence class.
Given $(a,b) ~ (\tilde{a},\tilde{b}$ and $(c,d) ~ (\tilde{c},\tilde{d}$, we know by the definitions that $a + \tilde{b} = \tilde{a} + b$ and $c + \tilde{d} = \tilde{c} + d$. Now, we must simply show that $(a,b) <_{\mathbb{Z}} (c,d)$ proves $(\tilde{a},\tilde{b}) <_{\mathbb{Z}} (\tilde{c},\tilde{d})$.
Assume $(a,b) <_{\mathbb{Z}} (c,d)$. From the definition, we know that $a+d <_{\mathbb{N}}b+c$. From here, we may, without altering the inequality, substitute for $a$ by substituting $\tilde{a} + b$ for $a$ and adding $\tilde{b}$ to the other side to get $\tilde{a} + b + d <_{\mathbb{N}} b + c + \tilde{b}$ and using cancellation on addition to get $\tilde{a} + d <_{\mathbb{N}} c + \tilde{b}$. Next, we may substitute $c+\tilde{d}$ for $d$ and add $\tilde{c}$ to the other side. Now we have $\tilde{a} + c + \tilde{d} <_{\mathbb{N}} c + \tilde{b} + \tilde{c}$ and apply cancellation on $c$ to end up with $\tilde{a} + \tilde{d} <_{\mathbb{N}} \tilde{b} + \tilde{c}$. By our definition of $<_{\mathbb{Z}}$, we can from here say that $(\tilde{a},\tilde{b}) <_{\mathbb{Z}} (\tilde{c},\tilde{d})$. Thus, the definition is well defined. $QED$.
Finally, we must prove that this extends the linear order under $<_{\mathbb{N}}$. A number, $a$ in $\mathbb{N}$ can be written in $\mathbb{Z}$ as $[(a,0)]$. Thus, we must show that if $a <_{\mathbb{N}} b$ then $[(a,0)] <_{\mathbb{Z}} [(b,0)]$. This can be simply done by applying the definition for $<_{\mathbb{Z}}$ to $[(a,0)] <_{\mathbb{Z}} [(b,0)]$ to get $a + 0 <_{\mathbb{N}} b + 0$. By the definition of addition on the naturals, we get get $a <_{\mathbb{N}} b$. $QED$
page revision: 9, last edited: 15 Mar 2018 02:41
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2018-03-21 14:25:55
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https://www.transtutors.com/questions/problem-12-18-dropping-or-retaining-a-tour-lo2-blueline-tours-inc-operates-tours-thr-1364302.htm
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# PROBLEM 12–18 Dropping or Retaining a Tour [LO2] Blueline Tours, Inc., operates tours throughout...
PROBLEM 12–18 Dropping or Retaining a Tour [LO2]
Blueline Tours, Inc., operates tours throughout the United States. A study has indicated that some of the tours are not profitable, and consideration is being given to dropping these tours to improve the company’s overall operating performance.
One such tour is a two-day Historic Mansions bus tour conducted in the southern states. An income statement from a typical Historic Mansions tour is given below:
$75 ticket price per person). . . . . . . . . . . . . . . . . . . . . . . .$3,000 100% Variable expenses ($22.50 per person) . . . . . . . . . . . . . . . . . 900 30 Contribution margin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2,100 70% Tour expenses: Tour promotion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$600 Salary of bus driver . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350 Fee, tour guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 700 Fuel for bus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 Depreciation of bus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450 Liability insurance, bus . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 Overnight parking fee, bus . . . . . . . . . . . . . . . . . . . . . . . . . 50 Room and meals, bus driver and tour guide . . . . . . . . . . . 175 Bus maintenance and preparation . . . . . . . . . . . . . . . . . . . 300 Total tour expenses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2,950 Net operating loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . \$ (850)
The following additional information is available about the tour:
a. Bus drivers are paid fixed annual salaries; tour guides are paid for each tour conducted.
b. The “Bus maintenance and preparation” cost on the previous page is an allocation of the sala- ries of mechanics and other service personnel who are responsible for keeping the company’s fleet of buses in good operating condition.
c. Depreciation of buses is due to obsolescence. Depreciation due to wear and tear is negligible.
d. Liability insurance premiums are based on the number of buses in the company’s fleet.
e. Dropping the Historic Mansions bus tour would not allow Blueline Tours to reduce the num- ber of buses in its fleet, the number of bus drivers on the payroll, or the size of the mainte- nance and preparation staff.
Required:
1. Prepare an analysis showing what the impact will be on the company’s profits if this tour is discontinued.
2. The company’s tour director has been criticized because only about 50% of the seats on Blue- line’s tours are being filled as compared to an industry average of 60%. The tour director has explained that Blueline’s average seat occupancy could be improved considerably by elimi- nating about 10% of its tours, but that doing so would reduce profits. Explain how this could happen.
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2019-02-18 20:36:06
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https://testbook.com/question-answer/the-correct-order-of-capillary-rise-in-increasing--5fcf260e425e05ac7651c850
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# The correct order of capillary rise in increasing order in different types of soils is
This question was previously asked in
CGPSC Civil Official Paper 2 (Held on April 2014 - Shift 2)
View all CGPSC AE Papers >
1. fine sands-silt-clay-colloids
2. silt-fine-sands-colloids-caly
3. clay-fine sands-colloids-silt
4. silt-fine sands-clays-colloids
5. silt-clay-fine sands-colloids
Option 1 : fine sands-silt-clay-colloids
Free
ST 1: Solid Mechanics
1052
20 Questions 40 Marks 20 Mins
## Detailed Solution
Explanation:
The surface tension (force) pulls the water upward till the height hc, at which the weight of water in the column is in equilibrium with the magnitude of the surface tension force.
$${h_c} = \frac{{4T\cos α }}{{{γ _w}d}}$$
where, α = contact angle
T = surface tension
d = diameter of capillary tube
γw = unit weight of water
• As increases in the fineness of the soil, capillary rises also increases means fine-grained soil has more capillary rise as compared to coarse-grained soil.
• So order of capillary rise : Colloids > clay > silt > fine sand
Important Points
Because of the complex nature of the soil, a theoretical prediction of capillary rise in the soil is not possible. Terzaghi and peck suggested an approximate relationship to find capillary rise in-situ
$${h_c}(cm) = \frac{C}{{e{D_{10}}}}$$
where, C = Empirical constant
e = void ratio
D10 = effective grain size
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2021-09-23 21:57:36
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https://mathspp.blogspot.com/2018/07/
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## Posts
Showing posts from July, 2018
### The hairy ball theorem and why there is no wind (somewhere) on Earth
PtEn< change language What if I told you that right now there is a place on Earth where there is no wind blowing to the sides? None at all! How can I know that? All we need is what is usually called the Hairy Ball Theorem.
In less rigorous contexts, one can phrase the Hairy Ball Theorem as such:
Theorem (Hairy Ball I): if you have a hairy ball, regardless of the way you comb its hair there will always be a spot where the hair points right up.
In this particular image, the hair is pointing up both on the top and on the bottom.
More formally, the Hairy Ball Theorem can be formulated like so:
Theorem (Hairy Ball II): every continuous vector field over $S^2$ has at least a point where the tangential component is $0$.
From this theorem it is actually quite easy to establish our interesting fact!
If we think of the wind at the Earth's surface as a continuous vector field, the Hairy Ball Theorem says that there must be a point where the wind isn't blowing to the sides!
Now all tha…
### Problem #11 - the salvation of the monks
PtEn< change language Today's problem is a riddle of the same style as problems #09 and #10.
Problem statement: the island depicted above used to be home to $2018$ monks that led a simple life, far from the bad, vain and consumerist habits of today's societies. Between many other deprivations, the monks never saw themselves on the mirror nor on any other reflecting surface. One day the volcano in the centre of the island - slowly but steadily - started hinting at an incoming eruption and the monks grew worried as the days passed. Some days later, at exactly 23h59, a divine being showed itself to the monks in their dreams and offered them a painless death - thus avoiding the wrath of the volcano - as long as every monk found out the colour of his eyes. The divine being also told them that every monk in the island had either blue or green eyes and that there was at least one monk with blue eyes. The divine being left with its final condition: every day, at 23h59, all monks wh…
### To measure or not to measure... a real problem!
PtEn This post will be more theoretical than usual. If you are afraid of mathematics, go away now before it is too late!
I will write about a set, the Vitali set, which arises in measure theory, to show that there is no coherent way of assigning a size to every subset of the real line. Because of that I will try to define size for subsets of the real line and then verify that such task is impossible.
I will also enforce a couple of (seemingly reasonable) restrictions on my definition.
I challenge you to read this post and to calmly follow my reasoning. Whenever something doesn't seem obvious, try to make it clear by yourself with a piece of paper and pen/pencil. If any doubts persist, drop me your question(s) in the comments and I will answer gladly!
Let us call $m$ to the function that, given a subset of the real line, returns its size; that is, let us try to define $m: \mathcal{P}(\mathbb{R}) \to [0, \infty]$.
Let us also suppose that the function $m$ satisfies the following re…
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2020-02-25 15:45:09
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http://mathoverflow.net/questions/112245/rigid-uniformization-vs-grothendiecks-local-monodromy-theory
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Rigid Uniformization vs Grothendieck's Local Monodromy Theory
I've noticed that some interesting results about abelian varieties can each be proven using one of two ways: the theory of rigid uniformization of abelian varieties or Grothendieck's local monodromy theory of SGA 7 1. I'm therefore wondering about possible interconnections between the two.
Detailed Motivation:
The first example of a result referred to above is the fact that CM abelian varieties have (potentially) good reduction everywhere. The proof using rigid uniformization is discussed in Silverman, Advanced Topics in the Arithmetic of Elliptic Curves (ATAoEC) in Chapter V. ATAoEC also gives a proof in Chapter II Section 6 using local class field theory, Neron-Ogg-Shafarevich, and the fact that a pro-$p$ group can only map trivially into a pro-$\ell$ group. I consider this latter proof to be part of Grothendieck's local monodromy theory, as one uses a similar method to prove the local monodromy theorem (at least, as demonstrated to me in Nicholas Katz's course at Princeton this fall; the original should be in the elusive SGA 7 1 Exposé III).
The next example is the following. SGA 7 1 Exposé IX proves that if $A/K$ has semistable reduction over a local field $K$ with inertia group $I$ with dimension $g$ and toric dimension $\mu$, then $T_\ell(A)^f := T_\ell(A)^I$ has rank $2g-\mu$, and $I$ acts trivially on the quotient as well. Furthermore, it has a complement under the Weil pairing (for a fixed polarisation), denoted $T_\ell(A)^t$, of rank $\mu$. See 2.2.5, 2.4, 2.5.4, and 3.5 of the Exposé notes.
Of course, one can prove the same result using rigid uniformization, where $T_\ell(A)^t$ corresponds to the $\ell^n$th roots of unity in $\bar{K}^*$. See Ribet, Galois Action on Division Points of Abelian Varieties with Real Multiplications, Section III, or these notes by Mihran Papikian.
Specific Question: Why do these two theories seem to prove the same results?
This would make sense if I saw similar arguments being used to develop both theories. But I don't see how analyzing the inertia using local class field theory and then looking at profinite groups is the same as writing down $p$-adic power series. While they both have a $p$-adic and $\ell$-adic "flavor," they seem to be very different proofs.
However, please tell me if I'm wrong - could it be that one can trace the arguments developing each theory to find a common thread?
More specifically, can one prove in general that if one can prove a result with one theory, then one can do it with the other? Is one theory strictly stronger than the other? Is there a common generalization?
-
The generalization from elliptic curves to abelian varieties involves the dual abelian variety in an essential way that you appear to have completely missed (e.g., your statement of the orthogonality theorem of SGA7 is wrong). This causes you to overlook serious aspects such as that in case of semistable reduction, the toric parts of the reductions have canonically dual geometric character lattices; this fact lies far deeper than anything about the monodromy operator and is made vivid in rigid uniformization. I recommend more experience with abelian varieties before asking such questions – user27056 Nov 13 '12 at 3:54
@Will: What exactly is the meaning of your statement #1? There are statements seen via rigid uniformization for which there isn't any evident means of proof in terms of monodromy formalism; e.g., the canonical duality between character lattices of toric parts in the reductions (as noted in my comment). In particular, the rigid uniformization is so much richer than the monodromy operator (which only records inertial action, and so can be extracted from the rigid uniformization). – user27056 Nov 13 '12 at 4:02
@Davidac897: It is not true that "if one fixes a polarization" then the role of the dual abelian variety can be suppressed. Plenty of abelian varieties do not admit a polarization with degree coprime to $\ell$, and in such cases you're not going to get the orthogonality theorem integrally (i.e., without inverting $\ell$) in the way you wish (with the dual hidden behind the polarization). The dual AV is a key ingredient in the proof of the orthogonality theorem. If you acquire more experience with abelian varieties then you will be in better position to understand the ideas in Expose IX. – user27056 Nov 13 '12 at 4:12
@Davidac897: In situations where both viewpoints prove a common result it is reasonable to ask for a unified perspective, but the final paragraph of your question and the "Specific Question" go beyond this, into a realm that is disproved by the duality aspect with toric parts of the reductions. The rigid uniformization subsumes the monodromy operator. (As an aside, IMHO even for AV's admitting a principal polarization, it is a conceptual error to think about the orthogonality theorem without always keeping in mind the role of the dual, just like for Weil pairings in higher dimensions.) – user27056 Nov 13 '12 at 4:57
@xbnv: $R^1 \pi_* \mathbb Q_l$, etale category. I'm interested in seeing the difference between the monodromy and rigid-analytic picture. The specific claim I made was that any sheaf satisfying some natural set of conditions, come from a rigid-analytic group in this way. Thus reasoning sheaf-theoretically from those conditions and reasoning with rigid analysis would be equivalent, at least for solving problems one can state in terms of that sheaf. – Will Sawin Nov 13 '12 at 6:25
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2015-04-02 04:51:12
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https://www.physicsforums.com/threads/two-carts-attached-to-a-string.774834/
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# Two Carts attached to a string
Gold Member
## Homework Statement
Two 0.400-kg carts are 100 mm apart on a low-friction track. You push one of the carts with a constant force of 2.0N directed so that the cart you push moves away from the other cart.
Determine the acceleration of the center of mass of the two-cart system when the carts are connected by a spring of spring constant k = 300 N/m.
## Homework Equations
$F_{x} (x) = -k(x-x_{0})$ for the force of the spring
$COM = \frac{\sum_n x_{n} m_{n}}{\sum_{n} m_{n}}$ for the center of mass, initially is at 50 mm or 0.05 m
$a_{cm} = \frac{\sum F(x)}{\sum m}$ for the acceleration of the center of mass
## The Attempt at a Solution
So the first part of the problem had me find the acceleration of the center of mass without the spring, which I did by summing the forces (which was easy, 2+0) and dividing by the sum of the masses, so without the spring the acceleration of the center of mass was 2.5 m/s/s. The second part of the problem I don't understand what it is attempting to say. If you push the right most cart 2 N, it will have a force acting in the other direction that is equal to -k(x-x0), but I don't know what the change in x is or anything like that so I am confused on where to begin.
Orodruin
Staff Emeritus
Homework Helper
Gold Member
And what is going to be the force on the other cart? What acceleration is it going to have?
Gold Member
The force on the other cart should be k(x-x0) because the spring is pulling it, if I am not wrong. Which is equal and opposite the the force the spring is exerting on the other cart, right?
Orodruin
Staff Emeritus
Homework Helper
Gold Member
Yes, so how are the accelerations of the carts changed by this? How does it affect the center of mass acceleration?
Gold Member
Yes, so how are the accelerations of the carts changed by this? How does it affect the center of mass acceleration?
I can't believe the solution was so simple, the spring part just cancels and you get the same exact answer as without the spring. Astounding, this is why I love physics. Thank you for your help, you're my favorite volcano :P
Orodruin
Staff Emeritus
Homework Helper
Gold Member
I can't believe the solution was so simple, the spring part just cancels and you get the same exact answer as without the spring.
Precisely. What matters for the acceleration of the CoM of a system is only its total mass and the external forces acting on it. In this case, the spring only provides an internal force and therefore need not be considered.
Astounding, this is why I love physics. Thank you for your help, you're my favorite volcano :p
You're welcome. Also, most people seem to blame me for helping in the creation of a certain piece of jewelry. They always conveniently forget that I was also instrumental in its destruction too! :p
B3NR4Y
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2021-10-16 15:14:46
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https://tex.stackexchange.com/questions/345828/easy-vertical-alignment-in-tabulary-balanced-columns
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# Easy vertical alignment in Tabulary balanced columns
I want to vertically center the text in all table cells. However, I do not wish to specify fixed column widths but rather leave it to tabulary to balance the column widths. How do I properly use column specifiers in this situation?
Easy solutions for regular use will be appreciated. Are there other LaTeX packages that do both column width balancing and better vertical alignment?
I do not know if the use of the IEEEtran class(v1.8b) is significant to my problem.
\documentclass[conference]{IEEEtran}
\usepackage{tabulary}
\usepackage{booktabs}
\begin{document}
\begin{table}[!h]
\renewcommand{\arraystretch}{1.3}
\caption{Comparison of Algorithm Results}
\centering
\begin{tabulary}{\columnwidth}{@{\hspace{.2em}}L*{4}{C}@{}}
\toprule
\textbf{Optimization Algorithm} & \textbf{Swarm Size} & \textbf{No. of Iterations} & \textbf{Final Wirelength} & \textbf{Best Value Cost Fn.}\\ \midrule
Short Text & 10 & 1000 & 100 & -- \\
Slightly Longer Text that Wraps & 20 & 2000 & 200 & 10.5 \\
Slightly Longer Text that Wraps & 30 & 3000 & 300 & 20.55 \\
Short & 40 & 4000 & 400 & 30.5 \\
\bottomrule
\end{tabulary}
\end{table}
\end{document}
You can "hack" the tabulary definition to vertically align all LRCJ columns.
\makeatletter
\def\TY@box@v#1{%
$\vcenter \@startpbox{\csname TY@F\the\TY@count\endcsname}% #1\arraybackslash\tyformat \insert@column\@endpbox$}
\makeatother
As the author of both tabularx and tabulary packages, I wouldn't use either of them for a data table such as this. they are about line breaking within cells and here, apart from the headings, you do not want line breaking.
I would use tabular* to force the overall width (if you must, or a centred tabular often works well) use the m column for the first column so that its reference point is vertically centred. For the numeric columns I use dcolumn for decimal alignment (siunitx is also good at this).
Finally I changed [h!] to [htp] (using h on its own increases the chance the table is forced to the end of the document and usually generates a warning from latex that it is being changed to ht)
\documentclass[conference]{IEEEtran}
\usepackage{dcolumn}
\usepackage{booktabs}
\newcommand\hd[1]{\multicolumn{1}{c}{%
\textbf{\begin{tabular}{@{}c@{}}#1\end{tabular}}}}
\begin{document}
\begin{table}[htp]
\renewcommand{\arraystretch}{1.3}
\caption{Comparison of Algorithm Results}
\centering
\begin{tabular*}{\columnwidth}{
@{\extracolsep{\fill}}
m{3cm}
D.{}{2.0}
D.{}{4.0}
D.{}{3.0}
D..{2.2}
@{}}
\toprule
\hd{Optimization Algorithm} & \hd{Swarm\\Size} &
\hd{No. of\\ Iterations} & \hd{Final\\ Wire-\\length} &
\hd{Best Value\\ Cost Fn.}\\ \midrule
Short Text & 10 & 1000 & 100 & \multicolumn{1}{c}{--} \\
Slightly Longer Text that Wraps & 20 & 2000 & 200 & 10.5 \\
Slightly Longer Text that Wraps & 30 & 3000 & 300 & 20.55 \\
Short & 40 & 4000 & 400 & 30.5 \\
\bottomrule
\end{tabular*}
\end{table}
\end{document}
A solution with tabularx, defining new column L and C and changing the definition of \tabularxcolumn taken from this answer to obtain X column type vertically aligned
\documentclass[conference]{IEEEtran}
\usepackage{tabularx}
\usepackage{booktabs}
\renewcommand{\tabularxcolumn}[1]{m{#1}}
\newcolumntype{L}{>{\raggedright}X}
\newcolumntype{C}{>{\centering\arraybackslash}X}
\begin{document}
\begin{table}[!h]
\renewcommand{\arraystretch}{1.3}
\caption{Comparison of Algorithm Results}
\centering
\begin{tabularx}{\columnwidth}{@{\hspace{.2em}}L*{4}{C}@{}}
\toprule
\textbf{Optimization Algorithm} & \textbf{Swarm Size} & \textbf{No. of Iterations} & \textbf{Final Wirelength} & \textbf{Best Value Cost Fn.}\\ \midrule
Short Text & 10 & 1000 & 100 & -- \\
Slightly Longer Text that Wraps & 20 & 2000 & 200 & 10.5 \\
Slightly Longer Text that Wraps & 30 & 3000 & 300 & 20.55 \\
Short & 40 & 4000 & 400 & 30.5 \\
\bottomrule
\end{tabularx}
\end{table}
\end{document}
Here's a solution that uses (a) a tabularx environment, (b) the S column type (from the siunitx package) for the data columns and (c) a centered version of the X column type just for the header cells of the data columns.
Adding the instruction \renewcommand{\tabularxcolumn}[1]{m{#1}} centers the column contents vertically. Using the S column type for the data columns makes it almost trivially easy to align the numbers on their explicit or implicit decimal markers.
\documentclass[conference]{IEEEtran}
\usepackage{booktabs,siunitx,tabularx}
\renewcommand{\tabularxcolumn}[1]{m{#1}} % for vertical centering
\newcolumntype{C}{>{\centering\arraybackslash}X}
\newcolumntype{L}{>{\raggedright\arraybackslash}X}
\newcommand\mC[1]{\multicolumn{1}{C}{#1}} % handy shortcut macro
\begin{document}
\begin{table}[!h]
\renewcommand{\arraystretch}{1.3}
\caption{Comparison of Algorithm Results}
\begin{tabularx}{\columnwidth}{@{}L
S[table-format=2.0]
S[table-format=4.0]
S[table-format=3.0]
S[table-format=2.2]@{}}
\toprule
\textbf{Optimization Algorithm} &
\mC{\textbf{Swarm Size}} &
\mC{\textbf{No.\ of Iterations}} &
\mC{\textbf{Final Wirelength}} &
\mC{\textbf{Best Value Cost Fn.}}\\
\midrule
Short Text & 10 & 1000 & 100 & {--} \\
Slightly Longer Text that Wraps & 20 & 20 & 200 & 10.5 \\
Slightly Longer Text that Wraps & 30 & 300 & 3 & 20.55\\
Short Text & 40 & 4000 & 40 & 30.5 \\
\bottomrule
\end{tabularx}
\end{table}
\end{document}
• The query was about vertical alignment so this is not a solution. But I appreciate your help in scientific table formatting. Dec 27 '16 at 15:38
• @geekshift - Sorry for overlooking this aspect. Achieving your objective is a piece of cake when using a tabularx environment: Just add the instructon \renewcommand{\tabularxcolumn}[1]{m{#1}}. I'll edit my answer to reflect this piece of information.
– Mico
Dec 27 '16 at 15:44
• Please have a look at this question If possible, add decimal alignment to it as well. tabularx allows relative width specifications using X and siunitx allows decimal alignment using S. How to do both ? Dec 27 '16 at 20:19
Well, it is not automated, and thus may be found unacceptable. But, starting with the OP's MWE, once the auto-breakpoints are seen, one merely need wrap that text in \mycol with the breakpoints manually inserted.
\documentclass[conference]{IEEEtran}
\usepackage{tabulary}
\usepackage{booktabs}
\usepackage[usestackEOL]{stackengine}
\begin{document}
\begin{table}[!h]
\renewcommand{\arraystretch}{1.3}
\caption{Comparison of Algorithm Results}
\centering
\begin{tabulary}{\columnwidth}{@{\hspace{.2em}}L*{4}{C}@{}}
\toprule
\textbf{Optimization Algorithm} & \textbf{Swarm Size} & \textbf{No. of Iterations} & \textbf{Final Wirelength} & \textbf{Best Value Cost Fn.}\\ \midrule
Short Text & 10 & 1000 & 100 & -- \\
\mycol{Slightly Longer\\ Text that Wraps} & 20 & 2000 & 200 & 10.5 \\
\mycol{Slightly Longer\\ Text that Wraps} & 30 & 3000 & 300 & 20.55 \\
Short & 40 & 4000 & 400 & 30.5 \\
\bottomrule
\end{tabulary}
\end{table}
\end{document}
|
2021-09-28 00:27:31
|
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|
https://www.vedantu.com/maths/operations-on-real-numbers
|
# Operations on Real Numbers
### What Are Real Numbers?
To understand real numbers, we first have to understand what rational and irrational numbers are. Rational numbers are ones that can be written in the form of p/q where p is the numerator and q is the denominator, and both p and q are integers. For example, 7 can be written as 7/1 so it is a rational number. Irrational numbers are numbers which cannot be written in p/q form. For example,√2 is an irrational number because √2 = 1.41421 .. and continues to infinity. Hence, it cannot be written as a fraction and is non terminating and non-recurring decimals. Rational and irrational numbers together form real numbers. You can also find a worksheet on real numbers at the end.
### What Are Mathematical Operations?
The four basic mathematical operations are addition (+), subtraction (-), multiplication (x) and division (/). We will now understand these operations on real numbers - both rational and irrational. The real numbers worksheet will help you understand this topic better.
### Addition Of Two Rational Numbers
When two rational numbers are added, the result is a rational number. For example, 0.24 + 0.68 = 0.92. 0.92 can be written as 92/100 which is a ratio or in the p/q form.
### Subtraction Of Two Rational Numbers
When two rational numbers are subtracted, the result is a rational number. For example, 0.93-0.22 = 0.71 which can be written as 71/100.
### Multiplication Of Two Rational Numbers
When two rational numbers are multiplied, the result is a rational number. For example, 0.5 multiplied by 185 is 92.5 which can be written as 925/10.
### Division of Two Rational Numbers
When a rational number is divided by another rational number, the result is a rational number. For example, 0.352 divided by 0.6 is 0.58 which can be written as 58/100.
### Operations On Two Irrational NumbersAddition of Two Irrational Numbers
When two irrational numbers are added, the result can be an irrational or a rational number. For example, √3 added to (√3) is 3.46 or 2√3 which can be written as 346/100 which is a rational number. However, when 2√5 is added to 5√3, we get a non-terminating and non-recurring decimal which is an irrational number. It is written as 2√5+5√3.
### Subtraction of Two Irrational Numbers
Similarly, when two irrational numbers are subtracted, the result can be an irrational or a rational number. √2 is subtracted from √2, the answer is 0. When 4√5 is subtracted from 5√3, we get 5√3-4√5.
### Multiplication of Two Irrational Numbers
The product of two irrational numbers can be an irrational number or a rational number. For example, when √2 is multiplied with √2, we get 2 which is a rational number. However, when √2 is multiplied by √3, we get √6 which is an irrational number.
### Division of Two Irrational Numbers
Similar to multiplication, we can get either an irrational number or a rational number as a result when an irrational number is divided by another. For example, when √2 is divided by √2, we get 1 which is a rational number. But when √2 is divided by √3, we get √2/√3, which is an irrational number.
### Addition of an Irrational and a Rational Number
The sum of a rational and an irrational number is always irrational. For example, when 2 is added to 5√3, we get 2 + 5√3, which is a rational number.
### Subtraction of an Irrational and a Rational Number
The difference of a rational and an irrational number is always irrational. For example, when we subtract 5√3 from 2, we get 2 - 5√3, which is an irrational number.
### Multiplication of an Irrational and a Rational Number
The product of a rational and an irrational number might be rational or irrational. For example, when 2 is multiplied with √2, we get 2√2 which is an irrational number, but when√12 is multiplied with √3, we get √36, or 6, which is a rational number.
### Division of an Irrational Number with a Rational Number
When a rational number is divided by an irrational number or vice versa, the quotient is always an irrational number. For example, when 8 is divided by √2, we get 8/√2 which is an irrational number. The answer can be further simplified to 4√2 which is also an irrational number.
### Operations On Real Numbers Worksheet
Example 1:
Solve:
(7√3) x (- 5√3)
Solution:
(7√3) x (- 5√3)
= 7 x -5 x √3 x √3
= -35 x 3 = -105
Example 2:
Solve:
(3√27 / 9√3)
Solution:
(3√27 / 9√3)
= 3√27 = √3x3x3 = 3 x 3√3 = 9√3
= 9√3/ 9√3 = 1
To understand the topic further, operations of real numbers worksheet might be of great help to students.
Q1. What are the Properties of Real Numbers?
A1.The properties of real numbers are applicable in addition and multiplication. When we add two real numbers, we have to keep in mind the
- Closure property: If a and b are real numbers, a+b is also a real number
- Associative property: a+(b+c) = (a+b)+c
- Commutative property: a+b = b+a
- Additive identity: a+0 = a
- Additive inverse: a + (-a) = 0
For Multiplication -
- Closure property: If a and b are real numbers, ab is also a real number
- Associative property: a x (b x c) = (a x b) x c
- Commutative property: a x b = b x a
- Identity property: a x 1 = 1 x a = a
- Inverse property: a x 1/a = 1
Another property holds true for both multiplication and addition - the distributive property, which is the same in both cases and says a x (b + c) = (a x b) + (a x c) = ab + ac.
Q2. Are There Non-Real Numbers? What are They?
A2. The two kinds of numbers known to us are real numbers and complex or imaginary numbers. We can represent any given number y as y = a + ib, where a is the real number and ib is the complex part of the number. Like we mentioned earlier, real numbers are a combination of all rational and irrational numbers which include whole numbers, repeating decimals and non-repeating decimals. If we take the number 3 + √-5, 3 is the real part and √-5 is the complex part of the number.
|
2020-07-11 05:15:56
|
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|
http://www.freedocumentsearch.com/pdf/symbols-in-physics-5.html
|
Recommend PDF pdf search for "symbols in physics" (Page 5 of about 75,800 results)
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|
2014-07-28 18:26:47
|
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|
https://people.maths.bris.ac.uk/~matyd/GroupNames/193/C9xD11.html
|
Copied to
clipboard
## G = C9×D11order 198 = 2·32·11
### Direct product of C9 and D11
Aliases: C9×D11, C11⋊C18, C992C2, C33.C6, C3.(C3×D11), (C3×D11).C3, SmallGroup(198,2)
Series: Derived Chief Lower central Upper central
Derived series C1 — C11 — C9×D11
Chief series C1 — C11 — C33 — C99 — C9×D11
Lower central C11 — C9×D11
Upper central C1 — C9
Generators and relations for C9×D11
G = < a,b,c | a9=b11=c2=1, ab=ba, ac=ca, cbc=b-1 >
Smallest permutation representation of C9×D11
On 99 points
Generators in S99
(1 98 65 32 87 54 21 76 43)(2 99 66 33 88 55 22 77 44)(3 89 56 23 78 45 12 67 34)(4 90 57 24 79 46 13 68 35)(5 91 58 25 80 47 14 69 36)(6 92 59 26 81 48 15 70 37)(7 93 60 27 82 49 16 71 38)(8 94 61 28 83 50 17 72 39)(9 95 62 29 84 51 18 73 40)(10 96 63 30 85 52 19 74 41)(11 97 64 31 86 53 20 75 42)
(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 26 27 28 29 30 31 32 33)(34 35 36 37 38 39 40 41 42 43 44)(45 46 47 48 49 50 51 52 53 54 55)(56 57 58 59 60 61 62 63 64 65 66)(67 68 69 70 71 72 73 74 75 76 77)(78 79 80 81 82 83 84 85 86 87 88)(89 90 91 92 93 94 95 96 97 98 99)
(1 11)(2 10)(3 9)(4 8)(5 7)(12 18)(13 17)(14 16)(19 22)(20 21)(23 29)(24 28)(25 27)(30 33)(31 32)(34 40)(35 39)(36 38)(41 44)(42 43)(45 51)(46 50)(47 49)(52 55)(53 54)(56 62)(57 61)(58 60)(63 66)(64 65)(67 73)(68 72)(69 71)(74 77)(75 76)(78 84)(79 83)(80 82)(85 88)(86 87)(89 95)(90 94)(91 93)(96 99)(97 98)
G:=sub<Sym(99)| (1,98,65,32,87,54,21,76,43)(2,99,66,33,88,55,22,77,44)(3,89,56,23,78,45,12,67,34)(4,90,57,24,79,46,13,68,35)(5,91,58,25,80,47,14,69,36)(6,92,59,26,81,48,15,70,37)(7,93,60,27,82,49,16,71,38)(8,94,61,28,83,50,17,72,39)(9,95,62,29,84,51,18,73,40)(10,96,63,30,85,52,19,74,41)(11,97,64,31,86,53,20,75,42), (1,2,3,4,5,6,7,8,9,10,11)(12,13,14,15,16,17,18,19,20,21,22)(23,24,25,26,27,28,29,30,31,32,33)(34,35,36,37,38,39,40,41,42,43,44)(45,46,47,48,49,50,51,52,53,54,55)(56,57,58,59,60,61,62,63,64,65,66)(67,68,69,70,71,72,73,74,75,76,77)(78,79,80,81,82,83,84,85,86,87,88)(89,90,91,92,93,94,95,96,97,98,99), (1,11)(2,10)(3,9)(4,8)(5,7)(12,18)(13,17)(14,16)(19,22)(20,21)(23,29)(24,28)(25,27)(30,33)(31,32)(34,40)(35,39)(36,38)(41,44)(42,43)(45,51)(46,50)(47,49)(52,55)(53,54)(56,62)(57,61)(58,60)(63,66)(64,65)(67,73)(68,72)(69,71)(74,77)(75,76)(78,84)(79,83)(80,82)(85,88)(86,87)(89,95)(90,94)(91,93)(96,99)(97,98)>;
G:=Group( (1,98,65,32,87,54,21,76,43)(2,99,66,33,88,55,22,77,44)(3,89,56,23,78,45,12,67,34)(4,90,57,24,79,46,13,68,35)(5,91,58,25,80,47,14,69,36)(6,92,59,26,81,48,15,70,37)(7,93,60,27,82,49,16,71,38)(8,94,61,28,83,50,17,72,39)(9,95,62,29,84,51,18,73,40)(10,96,63,30,85,52,19,74,41)(11,97,64,31,86,53,20,75,42), (1,2,3,4,5,6,7,8,9,10,11)(12,13,14,15,16,17,18,19,20,21,22)(23,24,25,26,27,28,29,30,31,32,33)(34,35,36,37,38,39,40,41,42,43,44)(45,46,47,48,49,50,51,52,53,54,55)(56,57,58,59,60,61,62,63,64,65,66)(67,68,69,70,71,72,73,74,75,76,77)(78,79,80,81,82,83,84,85,86,87,88)(89,90,91,92,93,94,95,96,97,98,99), (1,11)(2,10)(3,9)(4,8)(5,7)(12,18)(13,17)(14,16)(19,22)(20,21)(23,29)(24,28)(25,27)(30,33)(31,32)(34,40)(35,39)(36,38)(41,44)(42,43)(45,51)(46,50)(47,49)(52,55)(53,54)(56,62)(57,61)(58,60)(63,66)(64,65)(67,73)(68,72)(69,71)(74,77)(75,76)(78,84)(79,83)(80,82)(85,88)(86,87)(89,95)(90,94)(91,93)(96,99)(97,98) );
G=PermutationGroup([[(1,98,65,32,87,54,21,76,43),(2,99,66,33,88,55,22,77,44),(3,89,56,23,78,45,12,67,34),(4,90,57,24,79,46,13,68,35),(5,91,58,25,80,47,14,69,36),(6,92,59,26,81,48,15,70,37),(7,93,60,27,82,49,16,71,38),(8,94,61,28,83,50,17,72,39),(9,95,62,29,84,51,18,73,40),(10,96,63,30,85,52,19,74,41),(11,97,64,31,86,53,20,75,42)], [(1,2,3,4,5,6,7,8,9,10,11),(12,13,14,15,16,17,18,19,20,21,22),(23,24,25,26,27,28,29,30,31,32,33),(34,35,36,37,38,39,40,41,42,43,44),(45,46,47,48,49,50,51,52,53,54,55),(56,57,58,59,60,61,62,63,64,65,66),(67,68,69,70,71,72,73,74,75,76,77),(78,79,80,81,82,83,84,85,86,87,88),(89,90,91,92,93,94,95,96,97,98,99)], [(1,11),(2,10),(3,9),(4,8),(5,7),(12,18),(13,17),(14,16),(19,22),(20,21),(23,29),(24,28),(25,27),(30,33),(31,32),(34,40),(35,39),(36,38),(41,44),(42,43),(45,51),(46,50),(47,49),(52,55),(53,54),(56,62),(57,61),(58,60),(63,66),(64,65),(67,73),(68,72),(69,71),(74,77),(75,76),(78,84),(79,83),(80,82),(85,88),(86,87),(89,95),(90,94),(91,93),(96,99),(97,98)]])
63 conjugacy classes
class 1 2 3A 3B 6A 6B 9A ··· 9F 11A ··· 11E 18A ··· 18F 33A ··· 33J 99A ··· 99AD order 1 2 3 3 6 6 9 ··· 9 11 ··· 11 18 ··· 18 33 ··· 33 99 ··· 99 size 1 11 1 1 11 11 1 ··· 1 2 ··· 2 11 ··· 11 2 ··· 2 2 ··· 2
63 irreducible representations
dim 1 1 1 1 1 1 2 2 2 type + + + image C1 C2 C3 C6 C9 C18 D11 C3×D11 C9×D11 kernel C9×D11 C99 C3×D11 C33 D11 C11 C9 C3 C1 # reps 1 1 2 2 6 6 5 10 30
Matrix representation of C9×D11 in GL2(𝔽199) generated by
178 0 0 178
,
123 1 198 0
,
0 1 1 0
G:=sub<GL(2,GF(199))| [178,0,0,178],[123,198,1,0],[0,1,1,0] >;
C9×D11 in GAP, Magma, Sage, TeX
C_9\times D_{11}
% in TeX
G:=Group("C9xD11");
// GroupNames label
G:=SmallGroup(198,2);
// by ID
G=gap.SmallGroup(198,2);
# by ID
G:=PCGroup([4,-2,-3,-3,-11,29,2883]);
// Polycyclic
G:=Group<a,b,c|a^9=b^11=c^2=1,a*b=b*a,a*c=c*a,c*b*c=b^-1>;
// generators/relations
Export
×
𝔽
|
2021-09-20 19:34:43
|
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|
http://solidmechanics.org/problems/Chapter3_7/Chapter3_7.php
|
Problems for Chapter 3
Constitutive Models: Relations between Stress and Strain
3.7. Small Strain Plasticity
3.7.1. The stress state induced by stretching a large plate containing a cylindrical hole of radius a at the origin is given by
$\begin{array}{l}{\sigma }_{11}={\sigma }_{0}\left(1+\left(\frac{3{a}^{4}}{2{r}^{4}}-\frac{{a}^{2}}{{r}^{2}}\right)\mathrm{cos}4\theta -\frac{3{a}^{2}}{2{r}^{2}}\mathrm{cos}2\theta \right)\\ {\sigma }_{22}={\sigma }_{0}\left(\left(\frac{{a}^{2}}{{r}^{2}}-\frac{3{a}^{4}}{2{r}^{4}}\right)\mathrm{cos}4\theta -\frac{{a}^{2}}{2{r}^{2}}\mathrm{cos}2\theta \right)\\ {\sigma }_{12}={\sigma }_{0}\left(\left(\frac{3{a}^{4}}{2{r}^{4}}-\frac{{a}^{2}}{{r}^{2}}\right)\mathrm{sin}4\theta -\frac{{a}^{2}}{2{r}^{2}}\mathrm{sin}2\theta \right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$
Here, ${\sigma }_{0}$ is the stress in the plate far from the hole. (Stress components not listed are all zero)
3.7.1.1. Plot contours of von-Mises equivalent stress (normalized by ${\sigma }_{0}$ ) as a function of $r/a$ and $\theta$, for a material with $\nu =0.3$ Hence identify the point in the solid that first reaches yield.
3.7.1.2. Assume that the material has a yield stress Y .Calculate the critical value of ${\sigma }_{0}/Y$ that will just cause the plate to reach yield.
3.7.2. The stress state (expressed in cylindrical-polar coordinates) in a thin disk with mass density ${\rho }_{0}$ that spins with angular velocity $\omega$ can be shown to be
$\begin{array}{l}{\sigma }_{rr}=\left(3+\nu \right)\frac{{\rho }_{0}{\omega }^{2}}{8}\left\{{a}^{2}-{r}^{2}\right\}\\ {\sigma }_{\theta \theta }=\frac{{\rho }_{0}{\omega }^{2}}{8}\left\{\left(3+\nu \right){a}^{2}-\left(3\nu +1\right){r}^{2}\right\}\end{array}$
Assume that the disk is made from an elastic-plastic material with yield stress Y and $\nu =1/3$.
3.7.2.1. Find a formula for the critical angular velocity that will cause the disk to yield, assuming Von-Mises yield criterion. Where is the critical point in the disk where plastic flow first starts?
3.7.2.2. Find a formula for the critical angular velocity that will cause the disk to yield, using the Tresca yield criterion. Where is the critical point in the disk where plastic flow first starts?
3.7.2.3. Using parameters representative of steel, estimate how much kinetic energy can be stored in a disk with a 0.5m radius and 0.1m thickness.
3.7.2.4. Recommend the best choice of material for the flywheel in a flywheel energy storage system.
3.7.3. An isotropic, elastic-perfectly plastic thin film with Young’s Modulus $E$, Poisson’s ratio $\nu$ , yield stress in uniaxial tension Y and thermal expansion coefficient $\alpha$ is bonded to a stiff substrate. It is stress free at some initial temperature and then heated. The substrate prevents the film from stretching in its own plane, so that ${\epsilon }_{11}={\epsilon }_{22}={\epsilon }_{12}=0$, while the surface is traction free, so that the film deforms in a state of plane stress. Calculate the critical temperature change $\Delta {T}_{y}$ that will cause the film to yield, using (a) the Von Mises yield criterion and (b) the Tresca yield criterion.
3.7.4. Assume that the thin film described in the preceding problem shows so little strain hardening behavior that it can be idealized as an elastic-perfectly plastic solid, with uniaxial tensile yield stress Y. Suppose the film is stress free at some initial temperature, and then heated to a temperature $\beta \Delta {T}_{y}$, where $\Delta {T}_{y}$ is the yield temperature calculated in the preceding problem, and $\beta >1$.
3.7.4.1. Find the stress in the film at this temperature.
3.7.4.2. The film is then cooled back to its original temperature. Find the stress in the film after cooling.
3.7.5. Suppose that the thin film described in the preceding problem is made from an elastic, isotropically hardening plastic material with a Mises yield surface, and yield stress-v-plastic strain as shown in the figure. The film is initially stress free, and then heated to a temperature $\beta \Delta {T}_{y}$, where $\Delta {T}_{y}$ is the yield temperature calculated in problem 1, and $\beta >1$.
3.7.5.1. Find a formula for the stress in the film at this temperature.
3.7.5.2. The film is then cooled back to its original temperature. Find the stress in the film after cooling.
3.7.5.3. The film is cooled further by a temperature change $\Delta T<0$. Calculate the critical value of $\Delta T$ that will cause the film to reach yield again.
3.7.6. Suppose that the thin film described in the preceding problem is made from an elastic, linear kinematically hardening plastic material with a Mises yield surface, and yield stress-v-plastic strain as shown in the figure. The stress is initially stress free, and then heated to a temperature $\beta \Delta {T}_{y}$, where $\Delta {T}_{y}$ is the yield temperature calculated in problem 1, and $\beta >1$.
3.7.6.1. Find a formula for the stress in the film at this temperature.
3.7.6.2. The film is then cooled back to its original temperature. Find the stress in the film after cooling.
3.7.6.3. The film is cooled further by a temperature change $\Delta T<0$. Calculate the critical value of $\Delta T$ that will cause the film to reach yield again.
3.7.7. A thin-walled tube of mean radius a and wall thickness t<<a is subjected to an axial load P which exceeds the initial yield load by 10% (i.e. $P=1.1{P}_{Y}$ ). The axial load is then removed, and a torque Q is applied to the tube. You may assume that the axial load induces a uniaxial stress ${\sigma }_{zz}=P/\left(2\pi at\right)$ while the torque induces a shear stress ${\sigma }_{z\theta }=Q/\left(2\pi {a}^{2}t\right)$. Find the magnitude of Q to cause further plastic flow, assuming that the solid is
3.7.7.1. an isotropically hardening solid with a Mises yield surface
3.7.7.2. a linear kinematically hardening solid with a Mises yield surface
Express your answer in terms of ${P}_{Y}$ and appropriate geometrical terms, and assume infinitesimal deformation.
3.7.8. A cylindrical, thin-walled pressure vessel with initial radius R, length L and wall thickness t<<R is subjected to internal pressure p. The vessel is made from an isotropic elastic-plastic solid with Young’s modulus E, Poisson’s ratio $\nu$, and its yield stress varies with accumulated plastic strain ${\epsilon }_{e}$ as $Y={Y}_{0}+h{\epsilon }_{e}$. Recall that the stresses in a thin-walled pressurized tube are related to the internal pressure by ${\sigma }_{zz}=pR/\left(2t\right)$, ${\sigma }_{\theta \theta }=pR/t$
3.7.8.1. Calculate the critical value of internal pressure required to initiate yield in the solid
3.7.8.2. Find a formula for the strain increment $d{\epsilon }_{rr},d{\epsilon }_{\theta \theta },d{\epsilon }_{zz}$ resulting from an increment in pressure $dp$
3.7.8.3. Suppose that the pressure is increased 10% above the initial yield value. Find a formula for the change in radius, length and wall thickness of the vessel. Assume small strains.
3.7.9. Write a simple program that will compute the history of stress resulting from an arbitrary history of strain applied to an isotropic, elastic-plastic von-Mises solid. Assume that the yield stress is related to the accumulated effective strain by $Y={Y}_{0}{\left(1+{\overline{\epsilon }}^{p}/{\epsilon }_{0}\right)}^{n}$, where ${Y}_{0}$, ${\epsilon }_{0}$ and n are material constants. Check your code by using it to compute the stress resulting from a volume preserving uniaxial strain ${\epsilon }_{11}=\lambda ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\epsilon }_{22}={\epsilon }_{33}=-\lambda /2$, and compare the predictions of your code with the analytical solution. Try one other cycle of strain of your choice.
3.7.10. In a classic paper, Taylor, G. I., and Quinney, I., 1932, “The Plastic Distortion of Metals,” Philos. Trans. R. Soc. London, Ser. A, 230, pp. 323$–$362 described a series of experiments designed to investigate the plastic deformation of various ductile metals. Among other things, they compared their experimental measurements with the predictions of the von-Mises and Tresca yield criteria and their associated flow rules. They used the apparatus shown in the figure. Thin walled cylindrical tubes were first subjected to an axial stress ${\sigma }_{zz}={\sigma }_{0}$. The stress was sufficient to extend the tubes plastically. The axial stress was then reduced to a magnitde $m{\sigma }_{0}$, with $0, and a progressively increasing torque was applied to the tube so as to induce a shear stress ${\sigma }_{z\theta }=\tau$ in the solid. The twist, extension and internal volume of the tube were recorded as the torque was applied. In this problem you will compare their experimental results with the predictions of plasticity theory. Assume that the material is made from an isotropically hardening rigid plastic solid, with a Von Mises yield surface, and yield stress-v-plastic strain given by $Y={Y}_{0}+h{\overline{\epsilon }}^{p}$.
3.7.10.1. One set of experimental results is illustrated in the figure to the right. The figure shows the ratio $\tau /{\sigma }_{0}$ required to initiate yield in the tube during torsional loading as a function of m. Show that theory predicts that $\tau /{\sigma }_{0}=\sqrt{\left(1-{m}^{2}\right)/\sqrt{3}}$
3.7.10.2. Compute the magnitudes of the principal stresses $\left({\sigma }_{1},{\sigma }_{2},{\sigma }_{3}\right)$ at the point of yielding under combined axial and torsional loads in terms of ${\sigma }_{0}$ and m
3.7.10.3. Suppose that, for a given axial stress ${\sigma }_{zz}=m{\sigma }_{0}$, the shear stress $\tau$ is first brought to the critical value required to initiate yield in the solid, and is then increased by an infinitesimal increment $d\tau$. Find expressions for the resulting plastic strain increments $d{\epsilon }_{zz},d{\epsilon }_{\theta \theta },d{\epsilon }_{rr},d{\epsilon }_{r\theta }$, in terms of m, ${\sigma }_{0}$, h and $d\tau$.
3.7.10.4. Hence, deduce expressions for the magnitudes of the principal strains increments $\left(d{\epsilon }_{1},d{\epsilon }_{2},d{\epsilon }_{3}\right)$ resulting from the stress increment $d\tau$.
3.7.10.5. Using the results of 11.2 and 11.6, calculate the so-called “Lode parameters,” defined as
$\nu =2\frac{\left(d{\epsilon }_{2}-d{\epsilon }_{3}\right)}{\left(d{\epsilon }_{1}-d{\epsilon }_{3}\right)}-1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mu =2\frac{\left({\sigma }_{2}-{\sigma }_{3}\right)}{\left({\sigma }_{1}-{\sigma }_{3}\right)}-1$
and show that the theory predicts $\nu =\mu$
3.7.11. The Taylor/Quinney experiments show that the constitutive equations for an isotropically hardening Von-Mises solid predict behavior that matches reasonably well with experimental observations, but there is a clear systematic error between theory and experiment. In this problem, you will compare the predictions of a linear kinematic hardening law with experiment. Assume that the solid has a yield function and hardening law given by
$f=\sqrt{\frac{3}{2}\left({S}_{ij}-{\alpha }_{ij}\right)\left({S}_{ij}-{\alpha }_{ij}\right)}-{Y}_{0}=0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}d{\alpha }_{ij}=\frac{2}{3}cd{\epsilon }_{ij}^{p}$
3.7.11.1. Assume that during the initial tensile test, the axial stress ${\sigma }_{zz}={\sigma }_{0}$ in the specimens reached a magnitude ${\sigma }_{0}=\beta {Y}_{0}$, where ${Y}_{0}$ is the initial tensile yield stress of the solid and $\beta >1$ is a scalar multiplier. Assume that the axial stress was then reduced to $m{\sigma }_{0}$ and a progressively increasing shear stress was applied to the solid. Show that the critical value of $\tau /{\sigma }_{0}$ at which plastic deformation begins is given by
$\frac{\tau }{{\sigma }_{0}}=\frac{1}{\sqrt{3}}{\left\{\frac{1}{{\beta }^{2}}-{\left(m-1+{\beta }^{-1}\right)}^{2}\right\}}^{1/2}$
Plot $\tau /{\sigma }_{0}$ against m for various values of $\beta$
3.7.11.2. Suppose that, for a given axial stress ${\sigma }_{zz}=m{\sigma }_{0}$, the shear stress is first brought to the critical value required to initiate yield in the solid, and is then increased by an infinitesimal increment $d\tau$. Find expressions for the resulting plastic strain increments, in terms of m, $\beta$, c ${Y}_{0}$ and $d\tau$.
3.7.11.3. Hence, deduce the magnitudes of the principal strains in the specimen $\left(d{\epsilon }_{1},d{\epsilon }_{2},d{\epsilon }_{3}\right)$
3.7.11.4. Compute the magnitudes of the principal stresses $\left({\sigma }_{1},{\sigma }_{2},{\sigma }_{3}\right)$ at the point of yielding under combined axial and torsional loads in terms of m, $\beta$, and ${Y}_{0}$
3.7.11.5. Finally, find expressions for Lode’s parameters
$\nu =2\frac{\left(d{\epsilon }_{2}-d{\epsilon }_{3}\right)}{\left(d{\epsilon }_{1}-d{\epsilon }_{3}\right)}-1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mu =2\frac{\left({\sigma }_{2}-{\sigma }_{3}\right)}{\left({\sigma }_{1}-{\sigma }_{3}\right)}-1$
3.7.11.6. Plot $\nu$ versus $\mu$ for various values of $\beta$, and compare your predictions with Taylor and Quinney’s measurements.
3.7.12. An elastic- nonlinear kinematic hardening solid has Young’s modulus $E$, Poisson’s ratio $\nu$, a Von-Mises yield surface
$f\left({\sigma }_{ij},{\alpha }_{ij}\right)=\sqrt{\frac{3}{2}\left({S}_{ij}-{\alpha }_{ij}\right)\left({S}_{ij}-{\alpha }_{ij}\right)}-Y=0$
where Y is the initial yield stress of the solid, and a hardening law given by
$d{\alpha }_{ij}=\frac{2}{3}cd{\epsilon }_{ij}^{p}-\gamma {\alpha }_{ij}d{\overline{\epsilon }}^{p}$
where $c$ and $\gamma$ are material properties. In the undeformed solid, ${\alpha }_{ij}=0$. Calculate the formulas relating the total strain increment $d{\epsilon }_{ij}$ to the state of stress ${\sigma }_{ij}$, the state variables ${\alpha }_{ij}$ and the increment in stress $d{\sigma }_{ij}$ applied to the solid
3.7.13. Consider a rigid nonlinear kinematic hardening solid, with yield surface and hardening law described in the preceding problem.
3.7.13.1. Show that the constitutive law implies that ${\alpha }_{kk}=0$
3.7.13.2. Show that under uniaxial loading with ${\sigma }_{11}=\sigma$, ${\alpha }_{22}={\alpha }_{33}=-{\alpha }_{11}/2$
3.7.13.3. Suppose the material is subjected to a monotonically increasing uniaxial tensile stress ${\sigma }_{11}=\sigma$. Show that the uniaxial stress-strain curve has the form $\sigma =Y+\left(c/\gamma \right)\left[1-\mathrm{exp}\left(-\gamma \epsilon \right)\right]$ (it is simplest to calculate ${\alpha }_{ij}$ as a function of the strain and then use the yield criterion to find the stress)
3.7.14. Suppose that a solid contains a large number of randomly oriented slip planes, so that it begins to yield when the resolved shear stress on any plane in the solid reaches a critical magnitude k
3.7.14.1. Suppose that the material is subjected to principal stresses ${\sigma }_{1},{\sigma }_{2},{\sigma }_{3}$. Find a formula for the maximum resolved shear stress in the solid, and by means of appropriate sketches, identify the planes that will begin to slip.
3.7.14.2. Draw the yield locus for this material.
3.7.15. Consider a rate independent plastic material with yield criterion $f\left({\sigma }_{ij}\right)=0$. Assume that (i) the constitutive law for the material has an associated flow rule, so that the plastic strain increment is related to the yield criterion by $d{\epsilon }_{ij}^{p}=d{\overline{\epsilon }}^{p}\partial f/\partial {\sigma }_{ij}$; and (ii) the yield surface is convex, so that
$f\left[{\sigma }_{ij}^{*}+\beta \left({\sigma }_{ij}-{\sigma }_{ij}^{*}\right)\right]-f\left[{\sigma }_{ij}^{*}\right]\ge 0$
for all stress states ${\sigma }_{ij}$ and ${\sigma }_{ij}^{*}$ satisfying $f\left({\sigma }_{ij}\right)=0$ and $f\left({\sigma }_{ij}^{*}\right)\le 0$ and $0\le \beta \le 1$. Show that the material obeys the principle of maximum plastic resistance.
3.7.16. The yield strength of a frictional material (such as sand) depends on hydrostatic pressure. A simple model of yield and plastic flow in such a material is proposed as follows:
Yield criterion $F\left({\sigma }_{ij}\right)=f\left({\sigma }_{ij}\right)+\mu {\sigma }_{kk}=0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left({\sigma }_{ij}\right)=\sqrt{\frac{3}{2}{S}_{ij}{S}_{ij}}$
Flow rule $d{\epsilon }_{ij}^{p}=d{\overline{\epsilon }}^{p}\frac{\partial f}{\partial {\sigma }_{ij}}$
Where $\mu$ is a material constant (some measure of the friction between the sand grains).
3.7.16.1. Sketch the yield surface for this material in principal stress space (note that the material looks like a Mises solid whose yield stress increases with hydrostatic pressure. You will need to sketch the full 3D surface, not just the projection that is used for pressure independent surfaces)
3.7.16.2. Sketch a vector indicating the direction of plastic flow for some point on the yield surface drawn in part (3.8.3.1)
3.7.16.3. By finding a counter-example, demonstrate that this material does not satisfy the principle of maximum plastic resistance
$\left({\sigma }_{ij}-{\sigma }_{ij}^{*}\right)d{\epsilon }_{ij}^{p}\ge 0$
(you can do this graphically, or by finding two specific stress states that violate the condition)
3.7.16.4. Demonstrate that the material is not stable in the sense of Drucker $–$ i.e. find a cycle of loading for which the work done by the traction increment through the displacement increment is non-zero.
3.7.16.5. What modification would be required to the constitutive law to make it satisfy the principle of maximum plastic resistance and Drucker stability? How does the physical response of the stable material differ from the original model (think about compaction under combined shear and pressure).
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2018-06-22 09:11:59
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https://duck.co/ideas/idea/696/latex-rendering-give-a-rendered-image
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# LaTeX rendering - give a rendered image
latex {{search}} should render the results based on LaTeX semantics.
• posted 6 years and 1 month ago • type: Spice (API calls) In Development
Here is the link to the idea with more details from duck.co, https://duck.co/topic/render-latex-code-...
posted by [UserVoice yano] • 5 years and 9 months ago Link
You could use MathJax for rendering. If necessary that could perhaps be converted into an image client-side using HTML5 canvas.
posted by [UserVoice Jeroen] • 5 years and 15 days ago Link
We should probably make this idea more specific, since it sounds like you are all talking about rendering equations specifically. LaTeX is much, much larger than just equations, and a lot of LaTeX output differs based on the document type (cls file) and packages loaded. I think rendering $${...}$$ or just allowing $$${...}$$$ and text would be useful, but anything outside of that would start to get needlessly complex and would have to start approaching the functionality a cloud-based renderer like ShareLatex or ScribTeX.
posted by [UserVoice Dinre] • 5 years and 13 days ago Link
I agree with Dinre. I think rendering equations and allowing us to copy the link to a image file would be extremely useful. What about the same with tables?
posted by [UserVoice hpb] • 5 years and 12 days ago Link
I like the idea, but what exactly are the "LaTeX semantics" here? A LaTeX equation can be displayed in the browser using MathJax, but how would the actual search query be interpreted?
e.g. $E=m \times c^{2}$ should be converted to the query "E=mc^2" ?
posted by [UserVoice taddeus] • 5 years and 4 days ago Link
I have no objections to Dinre's clarification, although I'm curious about the use case. If the idea is just to use ddg as a quick and easy LaTeX renderer for formulae so they can be copied and pasted into a word processor, then the obvious way of doing that would be rendering as an image. Previous commenters have suggested MathJax, but AFAIK that uses image fonts to composite the equation in a form that's not easy copied and pasted.
Sure, you could just focus on displaying the result, but as far as I can see there's not much value in that. I can't think of a reason why I'd have raw LaTeX markup and merely want to know what it looked like, and you don't lose anything by rendering as an image.
That said, if MathJax is the only feasible way, then it's better than nothing at least, if only for the coolness value.
posted by [UserVoice Joshua Barney] • 4 years and 10 months ago Link
Looking into Sam's pull request. Thanks, Sam!
posted by [UserVoice DuckDuckGo Team] • 4 years and 10 months ago Link
Yes, that should work. Check out https://github.com/duckduckgo/zeroclicki..., which does something similar, e.g. https://duckduckgo.com/?q=qrcode+http%3A...
posted by yegg Staff5 years and 1 month ago Link
posted by samwhited 4 years and 10 months ago Link
Can this be done as a goodie that creates simple HTML, wrapping image from mathTeX web service (http://bit.ly/10z2pRj)?
posted by istepura 5 years and 1 month ago Link
I agree with the idea. We should avoid unnecessary complex results. By the way, there is a JS library named MathJax that renders LaTeX on client-side. We could take advantage of this library and get rid of any reliance on external resources.
posted by <hidden> • 5 years and 12 days ago Link
Relying on external sources for rendering this is not a good idea.
posted by <hidden> • 5 years and 1 month ago Link
Can I propose little additional feature?
* latex {{search}} in SVG
* latex {{search}} to SVG - will render to SVG instead the bitmap image.
posted by Fzzr 2 years and 19 days ago Link
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2018-04-26 05:48:57
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https://www.physicsforums.com/threads/inter-arrival-times-problem.1015318/
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# Inter-arrival times problem
CTK
Thread moved from the technical forums to the schoolwork forums
Summary: inter-arrival times problem
Let {X(t) : t ≥ 0} be a Poisson process with rate λ.
a-) Let Si denote the time of the ith occurrence, i = 1, 2, . . . . Suppose it is known that X(1) = 5. Find ## E(S_{5}) ##.
My attempt: Is the answer simply ## n / \lambda = 5 / \lambda ##? Or is there more to it that I am missing?
Thanks for the help
Homework Helper
Gold Member
2022 Award
Summary: inter-arrival times problem
Let {X(t) : t ≥ 0} be a Poisson process with rate λ.
a-) Let Si denote the time of the ith occurrence, i = 1, 2, . . . . Suppose it is known that X(1) = 5. Find ## E(S_{5}) ##.
My attempt: Is the answer simply ## n / \lambda = 5 / \lambda ##?
Why would that be the answer?
CTK
Why would that be the answer?
Hi PeroK. I guess I have just figured it out, so no worries. Thanks for your reply.
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2023-02-03 09:17:56
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https://itprospt.com/num/13015041/this-table-gives-the-maximum-depth-and-total-bottom-time
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5
# This table gives the maximum depth and total bottom time for 25 recreational scuba dives. Use this table for exercises 35 and 36.TABLE CAN'T COPY(a) Give the f...
## Question
###### This table gives the maximum depth and total bottom time for 25 recreational scuba dives. Use this table for exercises 35 and 36.TABLE CAN'T COPY(a) Give the five-number summary of the bottom time data.(b) Construct a box-and-whisker plot for the bottom time data.(c) Describe the bottom time data based on the box-and-whisker plot.
This table gives the maximum depth and total bottom time for 25 recreational scuba dives. Use this table for exercises 35 and 36. TABLE CAN'T COPY (a) Give the five-number summary of the bottom time data. (b) Construct a box-and-whisker plot for the bottom time data. (c) Describe the bottom time data based on the box-and-whisker plot.
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2022-07-01 10:24:19
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http://mathhelpforum.com/calculus/195457-finding-domain-function-natural-log.html
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# Math Help - Finding domain of a function with natural log
1. ## Finding domain of a function with natural log
Okay so I am taking calculus 1 and thought I fully understood how to find the domain of functions until I saw one with a natural log on the top.
f(x)= [ln(12-2x)]/[sqrt(x+5)-3]
so far I have deduced that x can not equal -6 because a square root cant be a negative and that x can not equal 4 because the square root of 9 would be 3 and would give us a 0 on the bottom which we cant allow. However the entire top section has me puzzled. Can anyone please explain how to work with the top end of this function.
2. ## Re: Finding domain of a function with natural log
First of all the denominator cannot be equal to zero therefore you have to solve the irrational equation $\sqrt{x+5}-3=0$ and exclude this solution(s). Otherwise $\ln(x)$ is only defined if $x>0$.
3. ## Re: Finding domain of a function with natural log
Originally Posted by Siron
First of all the denominator cannot be equal to zero therefore you have to solve the irrational equation $\sqrt{x+5}-3=0$ and exclude this solution(s). Otherwise $\ln(x)$ is only defined if $x>0$.
Thanks for your response I figured it out already. I just didn't know how to handle the natural log at first. If anyone else is looking at this thread with the same question I had... This is how it is solved.
So for the top portion we know that ln(x)>0 so if we solve for x we get 6 and know that we can not include six because the ln would be equal to 0 when it needs to be greater then. As for the bottom portion I described it in my initial question so the domain would be (in interval notation)
[-5,4)U(4,6)
4. ## Re: Finding domain of a function with natural log
What in the world do you mean by "ln(x)> 0"? There is no reason why the numerator cannot be zero or negative. What is true is that ln(x) is defined only for x> 0. Since you have ln(12- 2x) you must have 12- 2x> 0 which gives x< 6.
5. ## Re: Finding domain of a function with natural log
Yeah sorry about that I meant ln(x), x>0
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2015-04-27 07:40:53
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https://socratic.org/questions/how-do-you-write-a-second-degree-polynomial-with-zeros-of-2-and-3-and-goes-to-oo
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# How do you write a second-degree polynomial, with zeros of -2 and 3, and goes to -oo as x->-oo?
Aug 3, 2017
$y = - {x}^{2} + x + 6$
#### Explanation:
$\text{since the zeros(roots) are "x=-2" and } x = 3$
$\text{then the factors are}$
x+2)(x-3)
$\Rightarrow y = \left(x + 2\right) \left(x - 3\right)$
$\Rightarrow y = {x}^{2} - x - 6$
$\text{for the condition that "yto-oo" as } x \to - \infty$
$\text{we require the coefficient of "x^2" to be negative}$
$\Rightarrow y = - \left({x}^{2} - x - 6\right)$
$\textcolor{w h i t e}{\Rightarrow y} = - {x}^{2} + x + 6 \text{ is the required polynomial}$
graph{-x^2+x+6 [-20, 20, -10, 10]}
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2019-04-21 18:06:03
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https://socratic.org/questions/the-sum-of-three-consecutive-odd-integers-is-27-what-are-the-numbers
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# The sum of three consecutive odd integers is 27, what are the numbers?
Feb 14, 2016
$7 , 9 \mathmr{and} 11$
#### Explanation:
Let the smallest odd integer of the three be $x$.
Hence the next 2 consecutive odd integers are $x + 2 \mathmr{and} x + 4$.
Since the sum of the three integers is $27$, we may write
$x + \left(x + 2\right) + \left(x + 4\right) = 27$.
Now solving for $x$ we get
$3 x + 6 = 27$
$\therefore x = \frac{21}{3} = 7$.
Thus the required integers are $7 , 9 \mathmr{and} 11$.
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2020-02-29 07:06:47
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https://ai.stackexchange.com/questions/7228/what-loss-function-should-one-use-for-object-detection-knowing-that-the-input-i
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# What loss function should one use for object detection, knowing that the input image contains exactly one target object?
What loss function should one use, knowing that the input image contains exactly one target object?
I am currently using MSE to predict the center of ROI coordinates and its width and height. All values are relative to image size. I think that such an approach does not put enough pressure on the fact those coordinates are related.
I am aware of the existence of algorithms like YOLO or UnitBox, and am just wondering if there might be some shortcut for such particular case.
I am currently using MSE to predict the center of ROI coordinates and its width and height. All values are relative to image size. I think that such an approach does not put enough pressure on the fact those coordinates are related.
At first glance, this looks quite reasonable. Computer vision is not really my main area of expertise, so I did some googling around, and one of the first repositories I ran into does something very similar. It may be interesting for you to look into the code and the references in that repository in more detail.
It looks to me like they're also using the MSE loss function. I'm not 100% sure how they define the bounding boxes, maybe you can figure it out by digging through the code. You currently define bounding boxes by:
1. X coordinate of center of bounding box
2. Y coordinate of center of bounding box
3. Width of bounding box
4. Height of bounding box
You are right in that these coordinates are quite closely related. If the center is incorrect (for example, a bit too far to the right), that mistake could partially be "fixed" by taking a greater width (the bounding box would go a bit too far to the right, but still encapsulate the object). I don't know if this is necessarily a problem, or a fact that should be exploited in some way or something that should be "put pressure on". If this is something you are concerned about, I suppose you could alternatively define the bounding box as follows (I'm not sure whether or not this is what's done in the repository linked above):
1. X coordinate of top-left corner of bounding box
2. Y coordinate of top-left corner of bounding box
3. X coordinate of bottom-right corner of bounding box
4. Y coordinate of bottom-right corner of bounding box
Intuitively, I suspect the relation between those two corner points will be less strong than the relation you identified exists between center + width + height. A "mistake" in coordinates of the top-left corner cannot be partially "fixed" by placing the bottom-right corner somewhere else.
• I will try out this kind of loss and compare it with mine on same architecture, maybe there will be difference in convergence time. Jul 24, 2018 at 15:09
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2022-06-26 02:37:06
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https://ask.sagemath.org/question/33331/test-for-zero-cup-product/
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# Test for zero cup product
This post is a wiki. Anyone with karma >750 is welcome to improve it.
How can I check whether the cohomology ring H^*(X) of a simplicial complex X has zero cup product (like, say, for X a wedge of spheres). I have a long list of simplicial complexes, so I need a fully algorithmic approach. I probably should start with X.cohomology_ring(QQ) but I don't know what to do next...
Btw, I am a sage newbie.
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Sort by » oldest newest most voted
The path to proceed should be...
1. Construct the object correspdonding to the wedge of spheres.
2. The call cohomology_ring method.
3. Look at the generators.
4. Take those generators in degree > 0, compute products.
5. If one of the products is non zero, than return False.
6. Else, if the loop does find only zero products, finally return True.
The problem i had was a "minor one", but one that delayed the answer one (more) day. For short, the method cohomology_ring of the constructed wedge of spheres WS works only for a WS which is immutable. The default constructor makes it mutable.
In order to explain (myself) the answer, so that i'll understand it also tomorrow, here are first some examples. This is...
sage: version()
'SageMath version 7.3, Release Date: 2016-08-04'
(1) Let us take a Klein bottle and ask for the cohomology ring over rationals and over the two elements field. The rings will have generators. We compute the relevant products.
K = simplicial_complexes.KleinBottle()
HK = K.cohomology_ring( QQ )
print "H( K, Q ) has the following generators:"
print HK.gens() # there are two gen(erator)s
a, b = HK.gens() # they've got names, a and b
print "We rename them a, b."
print "Their degrees are %s, %s" % ( a.degree(), b.degree() )
print "Then we have for instance b*b = %s" % (b*b)
print "Is b*b zero? %s" % ( (b*b).is_zero() )
print
HK2 = K.cohomology_ring( GF(2) )
print "H( K, GF(2) ) has the following generators:"
print HK2.gens() # there are FOUR gen(erator)s
print "We rename them a, b, c, d."
a, b, c, d = HK2.gens() # they've got names, a, b, c, d
print "Their degrees are %s, %s, %s, %s" % ( a.degree(), b.degree(), c.degree(), d.degree() )
print "The values b*b, b*c, c*b, c*c are respectively %s, %s, %s, %s" % ( b*b, b*c, c*b, c*c )
print "Is b*b == 0? %s" % bool( b*b == 0 )
print "Is b*c == d? %s" % bool( b*c == d )
print "Is c*b == d? %s" % bool( c*b == d )
print "Is c*c == d? %s" % bool( c*c == d )
And we get so far (after a %cpaste into the sage interpreter):
H( K, Q ) has the following generators:
(h^{0,0}, h^{1,0})
We rename them a, b.
Their degrees are 0, 1
Then we have for instance b*b = 0
Is b*b zero? True
H( K, GF(2) ) has the following generators:
(h^{0,0}, h^{1,0}, h^{1,1}, h^{2,0})
We rename them a, b, c, d.
Their degrees are 0, 1, 1, 2
The values b*b, b*c, c*b, c*c are respectively 0, h^{2,0}, h^{2,0}, h^{2,0}
Is b*b == 0? True
Is b*c == d? True
Is c*b == d? True
Is c*c == d? True
(2) Let us define a function that computes a wedge of spheres. This is a quick solution, there is no raise in case a bad argument. We expect as argument a tuple of integers > 0.
def WedgeOfSpheres( spheresTuple ):
"""
return the corresponding wedge of spheres
for dimensionsTuple (1,3,4) we return for instance
S1 /\ S3 /\ S4.
"""
for k in range( len( spheresTuple ) ):
if k == 0: S = simplicial_complexes.Sphere( spheresTuple[k])
else : S = S.wedge( simplicial_complexes.Sphere( spheresTuple[k])
, is_mutable=False )
return S
There is one delicate point above: we really need the is_mutable=False option. Else the forthcomming method quotient_ring fails! (As i said, this may cost a day...)
(3) Then we can construct:
S = WedgeOfSpheres( (1,4,6) )
HS = S.cohomology_ring( QQ )
print "S = %s" % S
print "Computing H( S, QQ ) ... be patient..."
print HS
gens = [ g for g in HS.gens() if g.degree() > 0 ]
print "CUP PRODUCT matrix:"
for g in gens:
for h in gens:
print g*h,
print
print "Is the CUP PRODUCT zero? %s" % all( [ not(g*h) for g in gens for h in gens ] )
and we get:
S = Simplicial complex with 15 vertices and 17 facets
Computing H( S, QQ ) ... be patient...
Cohomology ring of Simplicial complex with 15 vertices and 17 facets over Rational Field
CUP PRODUCT matrix:
0 0 0
0 0 0
0 0 0
Is the CUP PRODUCT zero? True
(4) Resumed: In order to see if the pairing is trivial on the cohomology ring of a space X, the following call would do the job:
def is_zero_cup_product( X, base_ring=QQ ):
generators = [ g for g in X.cohomology_ring( base_ring ).gens() if g.degree() > 0 ]
for g in generators:
for h in generators:
if g*h:
# print "%s * %s = %s" % ( g, h, g*h )
return False
return True
For instance:
sage: K = simplicial_complexes.KleinBottle()
sage: K
Minimal triangulation of the Klein bottle
sage: is_zero_cup_product( K, QQ )
True
sage: is_zero_cup_product( K, GF(2) )
False
sage: is_zero_cup_product( K, GF(5) )
True
sage: P2 = simplicial_complexes.RealProjectiveSpace( 2 )
sage: P2
Minimal triangulation of the real projective plane
sage: is_zero_cup_product( P2, QQ )
True
sage: is_zero_cup_product( P2, GF(2) )
False
sage: M7 = simplicial_complexes.MooreSpace( 7 )
sage: M7
Triangulation of the mod 7 Moore space
sage: M7.cohomology_ring( GF(7) ).gens()
(h^{0,0}, h^{1,0}, h^{2,0})
sage: a,b,c = _
sage: b * b
0
sage: is_zero_cup_product( M7 )
True
more
Awesome. Thanks @dan_fulea
( 2017-05-01 00:20:32 +0200 )edit
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2021-10-27 21:08:36
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https://hal-normandie-univ.archives-ouvertes.fr/hal-03144356
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On the Rayleigh limit of the generalized Lorenz–Mie theory and its formal identification with the dipole theory of forces. I. The longitudinal case - Archive ouverte HAL Access content directly
Journal Articles Journal of Quantitative Spectroscopy and Radiative Transfer Year : 2021
On the Rayleigh limit of the generalized Lorenz–Mie theory and its formal identification with the dipole theory of forces. I. The longitudinal case
, (1)
1
Leonardo Ambrosio
• Function : Author
Gérard Gouesbet
Dates and versions
hal-03144356 , version 1 (17-02-2021)
Identifiers
• HAL Id : hal-03144356 , version 1
• DOI :
Cite
Leonardo Ambrosio, Gérard Gouesbet. On the Rayleigh limit of the generalized Lorenz–Mie theory and its formal identification with the dipole theory of forces. I. The longitudinal case. Journal of Quantitative Spectroscopy and Radiative Transfer, 2021, 262, pp.107531. ⟨10.1016/j.jqsrt.2021.107531⟩. ⟨hal-03144356⟩
81 View
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2023-02-01 15:31:07
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https://www.zbmath.org/authors/?q=ai%3Amathai.varghese
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# zbMATH — the first resource for mathematics
## Mathai, Varghese
Compute Distance To:
Author ID: mathai.varghese Published as: Mathai, V.; Mathai, Varghese
Documents Indexed: 107 Publications since 1986
all top 5
#### Co-Authors
12 single-authored 14 Bouwknegt, Peter G. 12 Carey, Alan L. 11 Hannabuss, Keith C. 8 Rosenberg, Jonathan Micah 8 Thiang, Guo Chuan 6 Evslin, Jarah 6 Wu, Siye 5 Benameur, Moulay-Tahar 5 Dodziuk, Jozef 5 Yates, Stuart G. 4 Hochs, Peter 4 Melrose, Richard Burt 4 Stevenson, Danny 3 Farber, Michael S. 3 Marcolli, Matilde 3 Shubin, Mikhail Aleksandrovich 3 Singer, Isadore M. 2 Brodzki, Jacek 2 Han, Fei 2 Hekmati, Pedram 2 Linshaw, Andrew Ross 2 Sati, Hisham 2 Schick, Thomas 2 Szabo, Richard J. 1 Bakhuis, Dennis 1 Baraglia, David 1 Block, Jonathan L. 1 Braverman, Maxim 1 Bursztyn, Henriques 1 Chakraborty, Partha Sarathi 1 Chan, Timothy T. K. 1 Coulhon, Thierry 1 Dey, Rukmini 1 Green, Richard F. 1 Huisman, Sander G. 1 Jurčo, Branislav 1 Kordyukov, Yuri A. 1 Linnell, Peter Arnold 1 Loeffen, Laura A. W. M. 1 Lohse, Detlef 1 Mahanta, Snigdhayan 1 McCann, Paul J. 1 Mishchenko, Aleksandr Sergeevich 1 Murray, Michael K. 1 Phillips, John 1 Quillen, Daniel Gray 1 Roberts, David Michael 1 Rothenberg, Melvin G. 1 Sun, Chao 1 Verschoof, Ruben A. 1 Weinberger, Shmuel 1 Wildeman, Sander 1 Zhang, Weiping
all top 5
#### Serials
16 Communications in Mathematical Physics 9 Journal of Geometry and Physics 8 Journal of Functional Analysis 6 Advances in Theoretical and Mathematical Physics 5 Advances in Mathematics 4 Letters in Mathematical Physics 4 Proceedings of the American Mathematical Society 3 Journal of Differential Geometry 3 Journal of Noncommutative Geometry 3 Journal of Physics A: Mathematical and Theoretical 2 Journal of Fluid Mechanics 2 Geometriae Dedicata 2 Journal für die Reine und Angewandte Mathematik 2 Communications in Contemporary Mathematics 1 Communications on Pure and Applied Mathematics 1 Mathematical Proceedings of the Cambridge Philosophical Society 1 Bulletin of the London Mathematical Society 1 Canadian Journal of Mathematics 1 Duke Mathematical Journal 1 Journal of the London Mathematical Society. Second Series 1 Topology 1 Annals of Global Analysis and Geometry 1 Journal of Physics A: Mathematical and General 1 Russian Journal of Mathematical Physics 1 Bulletin des Sciences Mathématiques 1 The Asian Journal of Mathematics 1 Geometry & Topology 1 Journal of High Energy Physics 1 Annales Henri Poincaré 1 The Quarterly Journal of Mathematics 1 Physical Review Letters 1 Journal of the Australian Mathematical Society 1 Journal of Geometry and Symmetry in Physics 1 Journal of Homotopy and Related Structures 1 Science China. Mathematics
all top 5
#### Fields
58 Global analysis, analysis on manifolds (58-XX) 51 Quantum Theory (81-XX) 33 Functional analysis (46-XX) 25 Differential geometry (53-XX) 21 $K$-theory (19-XX) 21 Manifolds and cell complexes (57-XX) 10 Algebraic topology (55-XX) 9 Relativity and gravitational theory (83-XX) 8 Statistical mechanics, structure of matter (82-XX) 7 Operator theory (47-XX) 5 Algebraic geometry (14-XX) 5 Topological groups, Lie groups (22-XX) 3 Category theory, homological algebra (18-XX) 3 Partial differential equations (35-XX) 2 Difference and functional equations (39-XX) 2 Fluid mechanics (76-XX) 1 General mathematics (00-XX) 1 Linear and multilinear algebra; matrix theory (15-XX) 1 Associative rings and algebras (16-XX) 1 Nonassociative rings and algebras (17-XX) 1 Functions of a complex variable (30-XX) 1 Several complex variables and analytic spaces (32-XX) 1 Dynamical systems and ergodic theory (37-XX) 1 Abstract harmonic analysis (43-XX) 1 Convex and discrete geometry (52-XX) 1 Mechanics of particles and systems (70-XX) 1 Mechanics of deformable solids (74-XX) 1 Optics, electromagnetic theory (78-XX)
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2019-07-18 19:17:45
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https://mathspace.co/textbooks/syllabuses/Syllabus-452/topics/Topic-8353/subtopics/Subtopic-109812/?activeTab=theory
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# Number lines (2,3,4,5)
Lesson
You may have already looked at representing fractions with fraction bars. Now watch this video to learn about fractions as a point on the number line.
## Representing fractions
We can represent fractions using:
• area
• segments of lines
• collections; and
• as points on the number line.
### The number line
Fractions are used to describe points between whole numbers. For example, one and a half is halfway between one and two.
To represent fractions as a point on the number line we:
• divide the number line into the number of equal parts indicated by the denominator.
• count along the number of parts using the numerator.
Being able to plot fractions on the number line is important for understanding the value of different fractions.
#### Example 1
To plot $\frac{3}{5}$35 on a number line:
1. Divide a number line between $0$0 and $1$1 into five equal parts to make fifths
2. Count along to the third fifth to indicate $\frac{3}{5}$35
Try this question for yourself:
##### Question 1:
Plot $\frac{1}{2}$12 on the number line.
#### Example 2
Can you work out what fraction the diagram below is indicating on the number line?
The number line is divided into thirds (three equal parts between each whole number). The red rectangle is indicating the second point along the line, so it is $\frac{2}{3}$23.
Try this question for yourself:
##### question 2
Look at the fraction bar and the number line below.
1. What fraction does each piece of the bar represent?
2. What fraction of the bar is shaded below?
3. What fraction is marked by the empty box below?
##### Question 3
Plot $\frac{4}{5}$45 on the number line.
This applet will help you to practice more fractions on a number line.
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2022-01-18 13:47:18
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https://socratic.org/questions/how-do-i-write-4-less-than-twice-a-number-as-an-algebraic-expression
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# How do i write 4 less than twice a number as an algebraic expression?
Mar 26, 2018
$2 n - 4$
#### Explanation:
We'll refer to our number as $n .$
The problem says we have twice the number. This would mean the number is multiplied by two, or $2 n .$
Moreover, we want four less than twice the number. This entails subtracting $4$ from $2 n : 2 n - 4$
In general, when translating from words into expressions:
"less than" implies subtracting.
"twice/three times/four times/ $x$ times" implies multiplying.
"half of/quarter of/tenth of/ $x t h$ of" implies dividing.
Mar 27, 2018
$2 n - 4$
#### Explanation:
We need to rearrange the word problem a little. When they are saying "4 less than twice a number", that's the same as saying "twice a number minus 4". When you are asking for less than something, that indicates subtraction!
We'll abbreviate number to $n$.
$\stackrel{2 n}{\overbrace{\text{twice a number" " " stackrel(-)overbrace"minus" " " stackrel(4)overbrace"4}}}$
$2 n - 4$
1. When you see "less than" that means that you are going to put that part at the end of the equation and that it will be subtraction.
2. Abbreviate words into something that you can remember. Number starts with $n$, so that is a good way to remember what it represents.
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2020-02-17 19:09:25
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http://encyclopedia.kids.net.au/page/sh/Shot_noise
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## Encyclopedia > Shot noise
Article Content
# Shot noise
Shot noise refers to the random fluctuations of the electric current in an electrical conductor, which are caused by the fact that the current is carried by discrete charges (electrons). The strength of this noise increases for growing magnitude of the average current flowing through the conductor. Shot noise is to be distinguished from current fluctuations in equilibrium, which happen without any applied voltage and without any average current flowing. These equilibrium current fluctuations are known as Johnson-Nyquist noise.
Shot noise is important in electronics, telecommunication,and for fundamental physics.
The strength of the current fluctuations can be expressed by giving the variance of the current, <(I-<I>)>2, where <I> is the average ("macroscopic") current. However, the value measured in this way depends on the frequency range of fluctuations which is measured ("bandwidth" of the measurement): The measured variance of the current grows linearly with bandwidth. Therefore, a more fundamental quantity is the noise power, which is essentially obtained by dividing through the bandwidth (and, therefore, has the dimension ampere squared divided by Hertz). It may be defined as the zero-frequency Fourier transform of the current-current correlation function:
$S=\int_{-\infty}^{+\infty} (\left-\left<I\right>^2) dt$</center>
<i>Note: This is the total noise power, which includes the equilibrium fluctuations (Johnson-Nyquist noise). Some other commonly employed definitions may differ by a constant pre-factor.
Note: There is often a minor inconsistency in referring to shot noise in an optical system: many authors refer to shot noise loosely when speaking of the mean square shot noise current[?] (amperes2) rather than noise power (watts).
Source: from Federal Standard 1037C and from MIL-STD-188
All Wikipedia text is available under the terms of the GNU Free Documentation License
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2021-10-20 04:13:41
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http://mathonline.wikidot.com/iff-criterion-for-y-to-be-a-conjugate-of-x-n
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IFF Criterion for y to be a Conjugate of x^n
# IFF Criterion for y to be a Conjugate of x^n
Proposition 1: Let $G$ be a group and let $x, y \in G$, $n \in \mathbb{Z}$. Then $y$ is a conjugate of $x^n$ if and only if $y$ is the $n^{\mathrm{th}}$ power of a conjugate of $x$.
• Proof: $\Rightarrow$ Let $y$ be a conjugate of $x^n$. Then there exists an $a \in G$ such that:
(1)
\begin{align} \quad y = ax^na^{-1} \end{align}
• Let $z$ be a conjugate of $x$. Then there exists a $b \in G$ such that $z = bxb^{-1}$. So $x = b^{-1}zb$. Substituting this into the above equation yields:
(2)
\begin{align} \quad y &= a(b^{-1}zb)^na^{-1} \\ &=a\underbrace{(b^{-1}zb)(b^{-1}zb)...(b^{-1}zb)}_{n \: \mathrm{times}}a^{-1} \\ &= ab^{-1}z^nba^{-1} \\ &= (ab^{-1})z^n(ab^{-1})^{-1} \end{align}
• So $y$ is a conjugate of $z^n$, i.e., $y$ is the $n^{\mathrm{th}}$ power of a conjugate $z$ of $x$.
• $\Leftarrow$ Suppose that $y$ is the $n^{\mathrm{th}}$ power of a conjugate of $x$. Then $y = z^n$ where $z$ is a conjugate of $x$. Since $z$ is a conjugate of $x$ there exists an $a \in G$ such that $z = axa^{-1}$. So:
(3)
\begin{align} \quad y = z^n &= (axa^{-1})^n \\ &= \underbrace{(axa^{-1})(axa^{-1})...(axa^{-1})}_{n\: \mathrm{times}} \\ &= ax^na^{-1} \end{align}
• So $y$ is a conjugate of $x^n$. $\blacksquare$
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2021-10-21 14:11:55
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https://www.acooke.org/cute/LearnProlo1.html
|
## Learn Prolog Now
From: "andrew cooke" <andrew@...>
Date: Wed, 22 Oct 2008 20:42:03 -0300 (CLST)
Free book -
http://www.coli.uni-saarland.de/%7Ekris/learn-prolog-now/html/prolog-notes.pdf
Chapter 7 looks interesting - an explanation of definite clause grammars.
I know the basic ideas behind Prolog (ie unification, Horn clauses), but
not the parsing stuff.
Andrew
### Declarative Mini-Languages in Python
From: "andrew cooke" <andrew@...>
Date: Wed, 22 Oct 2008 21:40:38 -0300 (CLST)
This looks like a useful article -
http://gnosis.cx/publish/programming/charming_python_b11.txt
In fact, there seems to be a whole pile of them -
http://gnosis.cx/publish/tech_index_cp.html
Andrew
From: "Stefan Petrea" <stefan.petrea@...>
Date: Tue, 28 Oct 2008 14:42:46 +0200
Hi,
What part of the article did you think went too fast or was unclearly
explained ?
Your reply will help me improve it in the time to come.
Thank you
--
Stefan Petrea
mobile phone : (040) 0732070179
From: "andrew cooke" <andrew@...>
Date: Mon, 17 Nov 2008 21:46:24 -0300 (CLST)
Just seen this (people reply so infrequently I forget to check the spam
filter). Andrew
---------------------------- Original Message ----------------------------
From: "Stefan Petrea" <stefan.petrea@...>
Date: Tue, October 28, 2008 9:42 am
--------------------------------------------------------------------------
Hi,
What part of the article did you think went too fast or was unclearly
explained ?
Your reply will help me improve it in the time to come.
Thank you
### Fast Is Not Necesarily Bad
From: "andrew cooke" <andrew@...>
Date: Mon, 17 Nov 2008 21:58:07 -0300 (CLST)
Hi - sorry for not replying earlier (I filter email submissions to my site
because of spam, and only get maybe 2 or 3 genuine comments a year, so
forget to check the filter...).
I think the problem was mainly that, although I know the basics behind
Prolog, I have never read through a "real" Prolog program. So it wasn't
difference lists, or the DCG stuff, but all the noisy details about how
you structure something more complex than the very simple examples I read
before.
But I do not think this is a bad thing! The web is full of very very
simple examples that show how to append a value to a list, or generate
permutations. You should not try to reproduce that. I learnt more from
your page when I read it (and more again now, skimming it through to
remind myself what it was all about) than I ever have done from simpler,
"too easy" pages.
I'm sorry my criticism ("way too fast") was inaccurate and unhelpful, and
I think your page is excellent as it is. It would have been better for me
to say something like "too much detail for me to understand everything in
a few minutes of scanning"... (because of how this site is generated it's
not trivial for me to modify anything already posted, or I would do so).
Cheers,
Andrew
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2021-09-26 17:03:09
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https://earthscience.stackexchange.com/questions/15618/given-that-water-will-center-over-the-mass-of-earth-how-could-pangaea-be-the-on/15621
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# Given that water will center over the mass of Earth, how could Pangaea be the only continent?
Consider this scenario: If the surface density of Earth were constant, it would follow that it would be impossible to have all land mass that is above the floor of the ocean concentrated on one side of the Earth. The water would simply flow until it centered over the mass of the earth. With this scenario the amount of land mass above the average ocean floor would need to be balanced around the sphere of the Earth, so with all of the currently above water continents on one side of the earth (Pangaea), other continent(s) would be exposed on the opposite side of the earth.
So is that scenario really the case, or is there actually enough density variation in Earth's crust to have higher density rock concentrated on the ocean side of Earth, which would allow all of the landmass above water to be concentrated on the less dense side?
• "Assuming that the surface density is relatively constant..." is your basic problem. Continental rock is lighter than ocean crust, which is why (simplistically) it floats. Think (if you are an American) of boiling yesterday's Thanksgiving turkey carcass for stock: the lighter fatty bits float to the surface, and eventually collect in "continents", whether one or many doesn't matter. Nov 24, 2018 at 4:37
• Even if density were constant I for one don't see why the water would flow to be centered over the "extra" mass. The Earth is almost 4000 miles in radius. The ocean has an average depth of 2 miles. As such, relatively minuscule amounts of land are above the ocean... and the much greater gravity within the Earth will still pull water to the lowest point. Having one continent wouldn't be a giant mass imbalance pulling water towards it, but just a tiny gravitational pull. There's still a large "imbalance" even now Nov 24, 2018 at 9:32
• Put it this way: instead of the center of mass being ideally right at the center of the Earth, the 2 miles extra of land would shift the center of mass a very small distance "continentward" from the center. But the center of mass is still deep within the Earth. Likewise I could envision perhaps there'd be a very very tiny bulge in the water upwards towards the continent. But would think the bulge would be on the order of inches or feet? The water mass can still be centered over the center of Earth mass without covering up the supercontinent. Nov 24, 2018 at 9:53
• @Matt Bradfield: Replace surface tension with plate tectonics :-) And consider that the mantle has convection currents, just as your stock pot does, albeit working on a much longer time scale. So the floating bits move around on the surface, clumping together or breaking apart without affecting the underlying liquid. Nov 24, 2018 at 18:23
• Having read the instructions for editing another post, I'm thinking the details I would add are not minor. So I think the answer is no, I shouldn't edit that post. Let me know if you disagree.
– user14364
Nov 25, 2018 at 15:03
Assuming that the surface density is relatively constant, it would be impossible to have all of the above water land mass concentrated on one side of the Earth.
That's a bad assumption. Four key differences between oceanic and continental crust are
1. Chemical composition. The rock that forms oceanic crust has more calcium and magnesium but less aluminum and silica that does the rock that forms continental crust.
2. Density. The different chemical compositions makes oceanic crust about 10% more dense than continental crust.
3. Age. The oldest oceanic crust is 100 million years old while the oldest continental crust is over 4 billion years old. The processes that recycle ocean crust (primarily subduction) are much more efficient than are the processes that recycle continental crust.
4. Thickness. The quick recycling of oceanic crust keeps it from building up. Most oceanic crust is between 7 to 10 kilometers thick. The slow recycling of continental crust has enabled it to build up over time. Most continental crust is between 25 to 70 kilometers thick.
The combination of different densities and different thicknesses means that the continents are elevated above the ocean floors by almost 5 kilometers on average.
• Keep in mind, constant density is not my assumption of reality. It is a useful assumption to understand the problem. I assume there is variation in density. The relevant element of the question is "is there actually enough density variation". You have partially answered that by explaining some of the variation that does exist. It is not proven that it is "enough" though. It does give me something to think about. I will do some math and see if it is enough to prove that all the continents could be on one side of Earth.
– user14364
Nov 24, 2018 at 17:35
• @Matt Bradfield: But constant density is not a useful assumption. It is a false one, which is leading you astray. Now if the Earth was liquid, it would (absent external influences) form a perfect sphere. But it isn't, and it doesn't. All sorts of forces - rotation, tides, different densities, the forces of plate tectonics - cause it to deviate from sphericity. So instead there's a gravitational equipotential surface, and that's where the water goes. Nov 24, 2018 at 18:29
• Assuming constant density is a very useful assumption. If one cannot understand how a simplified (constant density) model would behave, one cannot possibly understand how a more complex (varying density) model would behave. Far from leading us astray, considering the assumption leads us to the answer, because it shows us that for all the continents to be on one side, their MUST be variation in the density. And when the variations in density and their thicknesses are calculated, it leads us directly to an answer to the problem (see my answer to the initial question).
– user14364
Nov 25, 2018 at 0:13
• @MattBradfield - Re Assuming constant density is a very useful assumption. No, it is not. The Cavendish experiment, performed over 200 years ago, demonstrably showed that the Earth's density is anything but uniform. Sometimes spherical cow assumptions are rather useful. Other times, not so much. This is one of those other times. Nov 25, 2018 at 3:51
• I am being misunderstood.Perhaps the preface to my question needs rewording. When I speak of assuming constant density, it is a thought exercise. It's a preface leading to the actual question. This is a common technique in scientific inquiry, leading one to the next important factor. In this case it leads to density as the next important factor to consider. The question was not "if" density varies, but "how much" and "is that enough". My calculations show it is enough. But I fear my question is unclear to others. Perhaps someone can suggest improvements that would make my intention more clear.
– user14364
Nov 25, 2018 at 9:37
OK. Let's assume, for the sake of argument, that oceanic and continental crust does have the same density (it doesn't, but that doesn't matter). Let's further assume that everything under the earth's crust has constant density (it doesn't, but that also doesn't matter)
To a first approximation, the oceans will form a surface of constant distance from the centre of mass of the earth. Under the assumptions above, if the continents were evenly distributed then the centre of mass would be in the middle of the planet, and we would have a global ocean (minus those continents) of constant depth.
If all the continents were on one side, then there would be a miniscule change to the centre of mass of the earth, which would cause the sea level to be slightly higher on the side with the continents than the side without. I'm not going to try to calculate the magnitude of that offset, but bear in mind that the earth is a solid ball of diameter ~12,600km, while the crust is a little 35km layer on top[1]. And the crust is less dense than what's underneath. The continents will make very little difference to the position of the centre of mass.
That's a first approximation. There will also be some direct gravitational attraction between water and continent. This has been calculated (as, interestingly, has that between water and ice sheets), and it does have a measurable effect, increasing sea level close to the land by a small amount. I don't remember whether that amount is of the order of centimetres or metres (can anybody fill me in?), but either way it is not remotely close to being enough to submerge the continents.
[1] Credit to JeopardyTempest for phrasing this nicely in comments.
• Your proposition is completely unsupported by physics. Show me the math. Gravity simply doesn't just ignore ANY mass.
– user14364
Nov 28, 2018 at 0:22
• @matt Bradfield maths doesn't magically make physics. The problem here is not one of maths, but of a conceptual misunderstanding. I'm struggling to work out what that is, though. I think we agree that adding extra mass to one side of a sphere will shift the centre of mass of the sphere. In this case we're adding a tiny bit of mass to one side of a really massive sphere, so the CoM will shift by a very small distance. That will mean that slightly more of the surface water will rest on the side with the added mass - it will not cause all the water to rush to that side. Does that make sense? Nov 28, 2018 at 5:41
• This question absolutely requires math to quantify it. I don't recall any physics problems without math and formula(s). Perhaps someone can draw this in CAD and measure the shift using more accurate 3D math.Here's a rough estimate: Let's assume earth is density 5.5 and continental crust has density 2.7, height above sea floor of 4500 m, and covers 40% of earth. 4500*2.7/5.5*40%=890 m. That's including the density factor which I ignore in the introduction to my question. The average continent is only 840 m high. The water would cover MOST of dry land. Minuscule? Hardly.
– user14364
Nov 28, 2018 at 13:40
• @MattBradfield You're completely neglecting the mass of the mantle and core, which forms the vast majority of the mass of the Earth.
– bon
Dec 1, 2018 at 14:08
• The density of 5.5 includes the core, so its not ignored. Your statement that the core has the majority of the mass of the Earth is false, as it has about 33%. If you really believe that the core being denser than the mantle matters, then prove it mathematically and answer the question. I don't believe the density variation between layers matters, I think that was solved mathematically hundreds of years ago, but since it would help answer the question I asked, I would welcome a mathematical proof one way or another. I do believe the density variation of opposite sides of Earth matters greatly.
– user14364
Dec 1, 2018 at 17:59
OK, I'm sufficiently fed up with this to attempt a mathematical answer. It'll be very approximate, but it'll demonstrate an upper limit to how much the sea is skewed to one side of the planet (it's actually more than I expected). I fully expect to get something wrong here; if it's something that matters, please tell me about it!
Let's imagine a cartoon planet that consists of two parts:
• A sphere of radius 6370 km
• A half-shell of "continentals" with a thickness of 4 km (representing the distance from seafloor to continent)
For simplicity we'll assume that all of these have equal density, and in arbitrary units we'll define that density to be 1.
We want to find where the centre of mass of this "planet" is. Let's define our coordinate system to work in one dimension, along a line running from the middle of one half-shell to the middle of the other, where x=0 is at the centre of the sphere.
Here's a diagram (not to scale):
$$r_1=6370$$ km; $$r_2=6374$$ km
The mass of the sphere, in arbitrary mass units, is $$\frac{4}{3} \pi r_1^3 = 1.1\times 10^{12}$$.
The mass of the continental half-shell is $$\frac{1}{2}\cdot\frac{4}{3} \pi (r_2^3 - r_1^3)=10^9$$.
The centre of mass (CoM) of the sphere is at $$x=0$$.
I had to look up online how to find the centre of mass of a hemispherical shell, but it turns out that it lies at $$x=\frac{3}{8}\frac{r_2^4 - r_1^4}{r_2^3 - r_1^3} = 3186$$ km.
To find the location of the centre of mass of the combined object, we use $$x_c = \frac{m_1d_1 + m_2d_2}{m_1 + m_2}$$ where each $$m$$ is a mass and each $$d$$ is the position of the centre of mass of that object. So,
$$x_c = \frac{1.1 \times 10^{12} \times 0 + 10^9 \times 3186}{1.1\times 10^{12} + 10^9} = 2.9$$ km.
So the centre of mass of our cartoon planet is 2.9 km closer to the "continent" side than the "ocean" side. That means that the ocean will, rather approximately, try to form a shell centred on that point, thus being 2.9 km deeper in the centre of the continent than in the centre of the ocean. Errr... Except that the sea wouldn't actually do this because it would be blocked by the giant continent that's in its way. To figure out what would actually happen would require doing the maths in 3D, and possibly involving tractive forces and the like, and that's well beyond my capability this afternoon ;-)
I emphasise once again that this is an extremely generous scenario - half of the planet is continent, which is more than on Earth, and everything has the same density, rather than the inside of the planet being denser than the crust (which would make the effect much weaker). Plus, once we're dealing with values of a few km, in the real world there are probably other variations (e.g. equatorial bulge, non-ellipsoisal geoid due to density variations, etc) that have much bigger effects.
• Check your assumptions. You have the ocean floor 29,000 m below the average continental surface. The oceans average about 3690 m deep, and the average land rises above the ocean by about 840 m, for a total of about 4530 m. If we do a quick compensation of your result to extrapolate, 4530 m / 29000 m * 11.4 km = 1.78 km. This seems more realistic and is more in line with what my rough calculation showed, but that calculation did account for the variation in density between the core and crust. If we compensate your result to include that 5.5/2.7*1.78=0.870 km, so pretty darn close to 890 m.
– user14364
Nov 28, 2018 at 23:59
• @MattBradfield oh, quite right about the first part - thank you. I don't agree that one can scale the result as you have, so I've revised my cartoon version to a simpler one: now we have just one half-shell, representing continents, which is just 4km thick. I've kept uniform density, because the reality is so much more complex than "the mantle has density n" - but this should still be a generous assumption - because the crust is less dense than everything beneath, the actual effect will be less. Nov 29, 2018 at 7:29
• My first impression of the revision is that there must be an error in this math, because you can't add 4 km on the surface and shift the center of mass MORE than the added material (5.9 km). My experience with extrapolation comes from 30 years of engineering. You learn when you can take shortcuts to quickly estimate the difference between two scenarios. Then you can do more exact calculations once the choices are narrowed. Plus we didn't all have computers when I was young, so you had to do math by hand gasp. Assumptions are good, although I would use 6374.5 to not round difference.
– user14364
Nov 29, 2018 at 13:52
• @MattBradfield remember we're adding a half-shell of 4km thickness, not just 4km at the end of a column. I think it can shift the CoM by more than 4km. In any case, I can't find an error. If you can spot it, feel free to point it out :) Nov 29, 2018 at 17:32
• You calculated the mass of a continental full sphere shell. It needs to be cut in half to be continental half-shell. A smaller factor to consider is that when calculating small differences in large systems it is important to not round your inputs or intermediate answers. The small differences can get washed out by the rounding.
– user14364
Nov 30, 2018 at 3:45
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2022-05-17 16:53:03
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https://par.nsf.gov/biblio/10313054-beyond-hubble-sequence-exploring-galaxy-morphology-unsupervised-machine-learning
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Beyond the hubble sequence – exploring galaxy morphology with unsupervised machine learning
ABSTRACT We explore unsupervised machine learning for galaxy morphology analyses using a combination of feature extraction with a vector-quantized variational autoencoder (VQ-VAE) and hierarchical clustering (HC). We propose a new methodology that includes: (1) consideration of the clustering performance simultaneously when learning features from images; (2) allowing for various distance thresholds within the HC algorithm; (3) using the galaxy orientation to determine the number of clusters. This set-up provides 27 clusters created with this unsupervised learning that we show are well separated based on galaxy shape and structure (e.g. Sérsic index, concentration, asymmetry, Gini coefficient). These resulting clusters also correlate well with physical properties such as the colour–magnitude diagram, and span the range of scaling relations such as mass versus size amongst the different machine-defined clusters. When we merge these multiple clusters into two large preliminary clusters to provide a binary classification, an accuracy of $\sim 87{{\ \rm per\ cent}}$ is reached using an imbalanced data set, matching real galaxy distributions, which includes 22.7 per cent early-type galaxies and 77.3 per cent late-type galaxies. Comparing the given clusters with classic Hubble types (ellipticals, lenticulars, early spirals, late spirals, and irregulars), we show that there is an intrinsic vagueness in visual classification systems, in particular more »
Authors:
; ; ; ; ;
Award ID(s):
Publication Date:
NSF-PAR ID:
10313054
Journal Name:
Monthly Notices of the Royal Astronomical Society
Volume:
503
Issue:
3
ISSN:
0035-8711
National Science Foundation
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2022-12-07 01:22:41
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https://freshergate.com/c-sharp-programming/operators
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Home / C# Programming / Operators :: General Questions
### C# Programming :: Operators
1. Which of the following are the correct ways to increment the value of variable a by 1?
1. ++a++;
2. a += 1;
3. a ++ 1;
4. a = a +1;
5. a = +1;
2. A. 1, 3 B. 2, 4 C. 3, 5 D. 4, 5 E. None of these
3. What will be the output of the C#.NET code snippet given below?
byte b1 = 0xF7;
byte b2 = 0xAB;
byte temp;
temp = (byte)(b1 & b2);
Console.Write (temp + " ");
temp = (byte)(b1^b2);
Console.WriteLine(temp);
4. A. 163 92 B. 92 163 C. 192 63 D. 0 1
5. Which of the following is NOT an Arithmetic operator in C#.NET?
6. A. ** B. + C. / D. % E. *
7. Which of the following are NOT Relational operators in C#.NET?
1. >=
2. !=
3. Not
4. <=
5. <>=
8. A. 1, 3 B. 2, 4 C. 3, 5 D. 4, 5 E. None of these
9. Which of the following is NOT a Bitwise operator in C#.NET?
10. A. & B. | C. << D. ^ E. ~
11. Which of the following statements is correct about the C#.NET code snippet given below?
int d; d = Convert.ToInt32( !(30 20) );
12. A. A value 0 will be assigned to d. B. A value 1 will be assigned to d. C. A value -1 will be assigned to d. D. The code reports an error. E. The code snippet will work correctly if ! is replaced by Not.
13. Which of the following is the correct output for the C#.NET code snippet given below?
Console.WriteLine(13 / 2 + " " + 13 % 2);
14. A. 6.5 1 B. 6.5 0 C. 6 0 D. 6 1 E. 6.5 6.5
15. Which of the following statements are correct about the Bitwise & operator used in C#.NET?
1. The & operator can be used to Invert a bit.
2. The & operator can be used to put ON a bit.
3. The & operator can be used to put OFF a bit.
4. The & operator can be used to check whether a bit is ON.
5. The & operator can be used to check whether a bit is OFF.
16. A. 1, 2, 4 B. 2, 3, 5 C. 3, 4 D. 3, 4, 5 E. None of these
17. Which of the following are Logical operators in C#.NET?
1. &&
2. ||
3. !
4. Xor
5. %
18. A. 1, 2, 3 B. 1, 3, 4 C. 2, 4, 5 D. 3, 4, 5 E. None of these
19. Suppose n is a variable of the type Byte and we wish, to check whether its fourth bit (from right) is ON or OFF. Which of the following statements will do this correctly?
20. A. if ((n&16) == 16) Console.WriteLine("Fourth bit is ON"); B. if ((n&8) == 8) Console.WriteLine("Fourth bit is ON"); C. if ((n ! 8) == 8) Console.WriteLine("Fourth bit is ON"); D. if ((n ^ 8) == 8) Console.WriteLine("Fourth bit is ON"); E. if ((n ~ 8) == 8) Console. WriteLine("Fourth bit is ON");
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2021-10-16 09:20:44
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https://en-academic.com/dic.nsf/enwiki/1182854
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# Nelder-Mead method
Nelder-Mead method
:"See simplex algorithm for the numerical solution of the linear programming problem."
The Nelder-Mead method or downhill simplex method or amoeba method is a commonly used nonlinear optimization algorithm. It is due to John Nelder & R. Mead (1965) and is a numerical method for minimizing an objective function in a many-dimensional space.
Overview
The method uses the concept of a simplex, which is a polytope of "N" + 1 vertices in "N" dimensions; a line segment on a line, a triangle on a plane, a tetrahedron in three-dimensional space and so forth.
The method approximately finds a locally optimal solution to a problem with "N" variables when the objective function varies smoothly. For example, a suspension bridge engineer has to choose how thick each strut, cable, and pier must be. Clearly these all link together, but it is not easy to visualize the impact of changing any specific element. The engineer can use the Nelder-Mead method to generate trial designs which are then tested on a large computer model. As each run of the simulation is expensive, it is important to make good decisions about where to look. Nelder-Mead generates a new test position by extrapolating the behavior of the objective function measured at each test point arranged as a simplex. The algorithm then chooses to replace one of these test points with the new test point and so the algorithm progresses.
The simplest step is to replace the worst point with a point reflected through the centroid of the remaining "N" points. If this point is better than the best current point, then we can try stretching exponentially out along this line. On the other hand, if this new point isn't much better than the previous value, then we are stepping across a valley, so we shrink the simplex towards the best point.
Like all general purpose multidimensional optimization algorithms, Nelder-Mead occasionally gets stuck in a rut. The standard approach to handle this is to restart the algorithm with a new simplex starting at the current best value. This can be extended in a similar way to simulated annealing to escape small local minima.
Many variations exist depending on the actual nature of problem being solved. The most common, perhaps, is to use a constant size small simplex that climbs local gradients to local maxima. Visualize a small triangle on an elevation map flip flopping its way up a hill to a local peak. This, however, tends to perform poorly against the method described in this article because it makes small, unnecessary steps in areas of little interest.
This method is also known as the Flexible Polyhedron Method.
One possible variation of the NM algorithm
* 1. First order according to the values at the vertices::: $f\left( extbf\left\{x\right\}_\left\{1\right\}\right) leq f\left( extbf\left\{x\right\}_\left\{2\right\}\right) leq cdots leq f\left( extbf\left\{x\right\}_\left\{n+1\right\}\right)$
* 2. Compute a reflection: $extbf\left\{x\right\}_\left\{r\right\} = extbf\left\{x\right\}_\left\{o\right\} + alpha \left( extbf\left\{x\right\}_\left\{o\right\}- extbf\left\{x\right\}_\left\{n+1\right\}\right)$
::$x_\left\{o\right\}$ is the center of gravity of all points except $x_\left\{n+1\right\}$.
::If $f\left( extbf\left\{x\right\}_\left\{1\right\}\right) leq f\left( extbf\left\{x\right\}_\left\{r\right\}\right) < f\left( extbf\left\{x\right\}_\left\{n\right\}\right)$,::then we compute a new simplex with $ext\left\{x\right\}_\left\{r\right\}$ and by rejecting $ext\left\{x\right\}_\left\{n+1\right\}$. Go to step 1.
* 3. expansion: If
::If $f\left( extbf\left\{x\right\}_\left\{e\right\}\right) < f\left( extbf\left\{x\right\}_\left\{r\right\}\right)$ compute new simplex::with $ext\left\{x\right\}_\left\{e\right\}$ and by rejecting $ext\left\{x\right\}_\left\{n+1\right\}$ and go to step 1. Else compute new simplexwith $ext\left\{x\right\}_\left\{r\right\}$ and by rejecting $ext\left\{x\right\}_\left\{n+1\right\}$ and go to step 1.
* 4. contraction: If $f\left( extbf\left\{x\right\}_\left\{r\right\}\right) geq f\left( extbf\left\{x\right\}_\left\{n\right\}\right) ext\left\{ let \right\} extbf\left\{x\right\}_\left\{c\right\} = extbf\left\{x\right\}_\left\{n+1\right\}+ ho\left( extbf\left\{x\right\}_\left\{o\right\}- extbf\left\{x\right\}_\left\{n+1\right\}\right)$. ::If $f\left( extbf\left\{x\right\}_\left\{c\right\}\right) leq f\left( extbf\left\{x\right\}_\left\{n+1\right\}\right)$ compute new simplexwith $ext\left\{x\right\}_\left\{c\right\}$ and by rejecting $ext\left\{x\right\}_\left\{n+1\right\}$. Go to step 1. Else go to step 5.
* 5. shrink step: Compute the n vertices evaluations:::$x_\left\{i\right\} = x_\left\{1\right\} + sigma\left(x_\left\{i\right\} - x_\left\{1\right\}\right) ext\left\{ for all i \right\} in\left\{2,dots,n+1\right\}$. go to step 1.
Note: $alpha, ho, gamma$ and $sigma$ are respectively the reflection, the expansion, the contraction and the shrink coefficient. Standard value are $alpha =1$, $gamma =2$, $ho =1/2$ and $sigma =1/2$.
For the reflection, since $ext\left\{x\right\}_\left\{n+1\right\}$ is the vertex with the higher associated value along the vertices, we can expect to find a lower value at the reflection of $ext\left\{x\right\}_\left\{n+1\right\}$ in the opposite face formed by all vertices point $ext\left\{x\right\}_\left\{i\right\}$ except $ext\left\{x\right\}_\left\{n+1\right\}$.
For the expansion, if the reflection point $ext\left\{x\right\}_\left\{r\right\}$ is the new minimum along the vertices we can expect to find interesting values along the direction from $ext\left\{x\right\}_\left\{o\right\}$ to $ext\left\{x\right\}_\left\{r\right\}$.
Concerning the contraction: If $f\left( ext\left\{x\right\}_\left\{r\right\}\right) > f\left( ext\left\{x\right\}_\left\{n\right\}\right)$ we can expect that a better value will be inside the simplex formed by all the vertices $ext\left\{x\right\}_\left\{i\right\}$.
The initial simplex is important, indeed, a too small initial simplex can lead to a local search, consequently the NM can get more easily stuck. So this simplex should depend on the nature of the problem.
See also
* Conjugate gradient method
* Broyden-Fletcher-Goldfarb-Shanno or BFGS method
* Differential evolution
References
Further reading
* J.A. Nelder and R. Mead, Computer Journal, 1965, vol 7, pp 308-313 [http://citeseer.ist.psu.edu/context/23725/0] (text not online)
* Avriel, Mordecai (2003). Nonlinear Programming: Analysis and Methods. Dover Publishing. ISBN 0-486-43227-0.
* K.I.M. McKinnon, "Convergence of the Nelder-Mead simplex method to a non-stationary point", SIAM J Optimization, 1999, vol 9, pp148-158. [http://citeseer.ist.psu.edu/15874.html] (algorithm summary online).
External links
* [http://www3.imperial.ac.uk/people/j.nelder John Nelder FRS]
* [http://www.boomer.org/c/p3/c11/c1106.html Nelder-Mead (Simplex) Method]
* [http://math.fullerton.edu/mathews/n2003/NelderMeadMod.html Nelder-Mead Search for a Minimum]
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• Nelder–Mead method — Nelder–Mead simplex search over the Rosenbrock banana function (above) and Himmelblau s function (below) See simplex algorithm for Dantzig s algorithm for the problem of linear opti … Wikipedia
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2023-03-31 00:45:15
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http://www.jstor.org/stable/3214597?origin=crossref
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# Empirical Laplace Transform and Approximation of Compound Distributions
Sándor Csörgő and Jef L. Teugels
Journal of Applied Probability
Vol. 27, No. 1 (Mar., 1990), pp. 88-101
DOI: 10.2307/3214597
Stable URL: http://www.jstor.org/stable/3214597
Page Count: 14
Preview not available
## Abstract
Let {Xi}1 ∞ be i.i.d. non-negative random variables with d.f. F and Laplace transform L. Let N be integer valued and independent of {Xi}1 ∞. In many applications one knows that for y → ∞ and a function φ $P\left\{ \sum_{i=1}^{N}\,X_{i}>y\right\} \sim \varphi (y,\tau ,L(\tau),L^{\prime }(\tau ),\cdots )$ where in turn τ is the solution of an equation ψ(τ, L(τ),⋯) = 0. On the basis of a sample of size n we derive an estimator τn for τ by solving ψ(τn, Ln(τn), Ln ′(τn),⋯) = 0 where Ln is the empirical version of L. This estimator is then used to derive the asymptotic behaviour of φ(y, τn, Ln(τn), Ln ′(τn),⋯). We include five examples, some of which are taken from insurance mathematics.
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2016-09-25 15:29:59
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https://reverseengineering.stackexchange.com/questions/8142/finding-a-rc4-encryption-function-in-obfuscated-assembly-code-with-no-symbols
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# finding a RC4 Encryption function in obfuscated assembly code with no symbols
RC4 encryption is quite basic and its Pseudo random generation algorithm is given at Wikipedia
When Looking for a hashing algorithm like MD5 or SHAx it is pretty simple to look for the Hashing constants in the code, for example to find the MD5 code most times looking for the byte sequense of 0x01234567 is enough and you'd probably hit the function iterating the text section.
Now that example gives the MD5 function. but what if i would like to find an implementation of the symmetric RC4 encryption? would there be any obvious opcode sequense that would repeat itself?
I'm asking that question because im working on some heavily anti-debugging obfuscated piece of code with no symbols. I know for a high chance (like 90% sure) it is using a RC4 encryption. now i want to be able to find that RC4 function just like i was able to find the MD5 hashing functions.
Just in case i was wrong, and its not hard-coded inside the main executable itself (Because md5 is, so i can assume it should be too), i have already tried to set a BreakPoint at CRYPTSP!CryptEncrypt with no success, are there any alternative Api functions to Encrypt using RC4?
So to sum my questions up:
1. Is there any static opcode usage or number in the assembly of an RC4 encryption?
2. Are there any alternative WinAPI calls for encrypting using RC4 encryption beside CRYPTSP!CryptEncrypt?
3. Any other way to look for the RC4 encryption algorithm you can help me think about is blessed.
1. Is there any static opcode usage or number in the assembly of an RC4 encryption?
• Opcode: Search the disassembly for xor <x>, <y> where <x> != <y>.
• Number: Search for 0x100
Obviously both of these searches will yield many false-positives, so you'd have to look at each match manually to see if it's part of an RC4 function.
1. Are there any alternative WinAPI calls for encrypting using RC4 encryption beside CRYPTSP!CryptEncrypt?
Every time I see software using RC4, the RC4 code is always written inline, not imported from a dynamic library.
1. Any other way to look for the RC4 encryption algorithm you can help me think about is blessed.
You could try Aligot.
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2019-10-16 04:27:41
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https://cstheory.stackexchange.com/tags/proof-techniques/hot?filter=all
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# Tag Info
4
A recent result of Li, Nguyen, and Woodruff shows that for any streaming algorithm in the turnstile model (where the stream consists of insertions and deletions of elements) there exists an algorithm that works by only maintaining a linear sketch and uses only slightly more space. So to prove a space lower bound in the turnstile model it is (up to some ...
3
While not new, (and depending on what you consider to be "streaming algorithms"), a standard lower bound technique is picking a (as large as possible) set of inputs, and proving that each has to lead the algorithm to a distinct memory configuration. The implied lower bound is then the log of the number of such inputs. For example, Datar et al. showed (...
2
Let me clarify the setting, which has nothing to do with $\pi$-calculus or bisimulation. The first thing you have to realise that it does not make much sense to talk about a programming language without reference to the notion of program equivalence you intend to impose on the language. That's because We usually identify certain programs (e.g. f(x:int) = {...
1
In the same spirit of Sanjeev Arora's notes that @umar posted, I like Madhur Tulsiani's lecture notes and exercises for his "Mathematical Toolkit" class posted at the course webpage. In addition to Arora's excellent material his notes have a nice coverage of spectral graph theory as well as the multiplicative weights update method.
1
A paper of mine with Gupta and Kumar titled On a bidirected relaxation for the MULTIWAY CUT problem was also based on running experiments. In fact we were trying to prove the converse of what we ended up proving. Vazirani, in the first edition of his book on approximation algorithms, suggested that the bidirected relaxation was at least as good as the ...
1
My recent paper with Karthik Chandrasekharan titled Hypergraph $k$-cut in deterministic polynomial time was based on extensive computational experiments. We explored different conjectures and submodular functions and found counter examples to several different approaches. The truth of the main structural theorem was suggested by experiments and we then ...
1
Some recent results in state complexity were found with the help of systematic brute-force search for worst-case examples. This is doable because there are not too many deterministic finite automata with a small number of states, for example if we concentrate on binary or ternary alphabets. Also, in many cases there are families of worst-case examples for \$1,...
1
In 2018, Aubrey de Grey found a 1581-vertex, non-4-colourable unit-distance graph. This gives a lower bound of five for the famous Hadwiger-Nelson problem. He used a computer to verify that the graph indeed has chromatic number at least five. Gil Kalai's blogpost covers some facts and further developments. An article in the quanta magazine reports that he ...
Only top voted, non community-wiki answers of a minimum length are eligible
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2021-12-03 20:12:04
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https://www.emathhelp.net/notes/calculus-1/drawing-graphs-of-functions/introduction-to-sketching-graph-of-function/
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# Introduction to Sketching Graph of Function
## Related Calculator: MathGrapher: Graphing Calculator-Function Grapher
Now, when we know methods of differential calculus let's consider question of sketching graph of the fucntion.
Firstly, consider continuous on finite interval [a,b] function y=f(x). Here we are mainly interested in precise characteristic of change of function.
Before knowing methods of differential calculus we could only sketch graph of function by taking sufficiently large number of arbitrary points and connect the with lines. But this method is not applicable for practical purposes because it requires calculating of very large number of coordinates. Another disadvantage of this method is that since we take arbitrary points, we can miss some "important" points and graph will be sketched partly incorrectly.
Now suppose that function y=f(x) has finite derivative (except possibly at finite number of points). Then methods of differential calculus allow us to determine some number of "important" points, typical for this function. Using these points we already can draw graph with some precision.
First of all under "important" we mean points of extrema. However, we can add points where tangent line is horizontal or vertical, even if these points are not extrema.
When these points are drawn it is often enough to sketch the graph. This sketch represents behaviour of the function very well. It represents intervals where function is increasing and where decreasing, where rate of change of function is 0 (f'(x)=0) and where it is infinite (f'(x)=+-oo).
If we want to sketch even better graph, we need to determine intervals of convexity and concavity together with inflection points.
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2020-01-22 11:40:06
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http://en.math212.com/p/about-us.html
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2021-06-23 02:02:19
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http://canadianinsurancelawblog.com/9rw2yr/886erkq.php?page=unit-matrix-vs-identity-matrix-2b79e2
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# unit matrix vs identity matrix
01. December 2020 0
This is a matrix that effectively does nothing when applied. Matrix multiplication dimensions. In particular, the identity matrix is invertible—with its inverse being precisely itself. Multiplying a matrix by the identity matrix I (that's the capital letter "eye") doesn't change anything, just like multiplying a number by 1 doesn't change anything. So I wanted to construct an Identity matrix n*n. I came up with a stupid solution, that worked for a 4*4 matrix, but it didn't work with 5*5. Sometimes U or E is also used to denote an Identity Matrix. Let’s study about its definition, properties and practice some examples on it. Identity matrices play a key role in linear algebra. Well, if the coefficient matrix of a system is row equivalent to the identity, then this is ture but in our case, the augmented matrix is row-equivalent to the identity matrix. Write a C program to create identity matrix or unit matrix. This matrix is also denoted [0]. Less frequently, some mathematics books use U or E to represent the identity matrix, meaning "unit matrix" and the German word Einheitsmatrix respectively. An identity matrix is a square matrix whose diagonal entries are all equal to one and whose off-diagonal entries are all equal to zero. $\begingroup$ I don' think the identity matrix can be considered as swapping one row with itself, as it does not change the sign of the determinant. Create a 3-by-3 identity matrix whose elements are 32-bit unsigned integers. In other words, if all the main diagonal of a square matrix are 1’s and rest all o’s, it is called an identity matrix. Your email address will not be published. This tool generates identity matrices of any size. It's going to be 1, 0, 0, 1. The identity matrix is the only idempotent matrix with non-zero determinant. For part (b), many students wrote that “the identity matrix is nonsingular, so it is consistent”. Any matrix typically has two different identity matrices: a left identity matrix and a right identity matrix. A matrix is a representation of a linear transformation. A matrix is a useful structure that you can use in a variety of finite math problems to change the format of mathematical statements to make them more usable and understandable. You can use matrices to organize data by month, person, age group, company, and so on. If we multiply two matrices which are inverses of each other, then we get an identity matrix. Email. Well, if the coefficient matrix of a system is row equivalent to the identity, then this is ture but in our case, the augmented matrix is row-equivalent to the identity matrix. Define a complex vector. We can also say, the identity matrix is a type of diagonal matrix, where the principal diagonal elements are ones, and rest elements are zeros. For any whole number n, there’s a corresponding Identity matrix, n x n. 2) By multiplying any matrix by the unit matrix, gives the matrix itself. When multiplying a given matrix with an identity matrix, the values remain the same. Click here if solved 12. Solution: The unit matrix is the one having ones on the main diagonal & other entries as ‘zeros’. There is a matrix which is a multiplicative identity for matrices—the identity matrix: Identity matrix is sometimes also known as unit matrix. The identity matrix is denoted by “ I “. Learn what an identity matrix is and about its role in matrix multiplication. A is row-equivalent to the n-by-n identity matrix I n. A is column-equivalent to the n-by-n identity matrix I n. A has n pivot positions. When a unitary matrix is real, it becomes an orthogonal matrix, . Visit BYJU’S – The Learning App to explore a fun and interesting way to learn Mathematics. Your email address will not be published. Dimensions of identity matrix . numpy.identity(n, dtype = None) : Return a identity matrix i.e. a square matrix with ones on the main diagonal. Likewise if you multiplied intermediate matrices from midway through, you would still travel around within the cycle. Identity Matrix. p = [1+2i 3i]; ... GPU Arrays Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox™. A matrix is called an identity matrix (also known as unit matrix) if each element a_ij, where i = j, is equal to one, and each element a_ij, where i ≠ j, is equal to zero (i corresponds to the row number, j corresponds to the column number). While we say “the identity matrix”, we are often talking about “an” identity matrix. It's going to be 1, 0, 0, 1. It is denoted by In, or simply by I if the size is immaterial or can be trivially determined by the context. If I is a right identity matrix for A, then the matrix product A.I = A. The 3 by 3 identity matrix is equal to 1, 0, 0, 0, 1, 0, and 0, 0, 1. Here, the 2 x 2 and 3 x 3 identity matrix is given below: Identity Matrix is donated by In X n, where n X n shows the order of the matrix. is a unitary matrix if its conjugate transpose is equal to its inverse , i.e., . they are … PQ = QP = I) The inverse matrix of A is denoted by A-1. A square matrix in which all the main diagonal elements are 1’s and all the remaining elements are 0’s is called an Identity Matrix. Or should I say square zero. Returns : identity array of dimension n x n, with its main diagonal set to one, and all other elements 0. Back in multiplication, you know that 1 is the identity element for multiplication. Because an identity matrix is a square matrix, its number of rows matches its number of columns. Identity matrix You are encouraged to solve this task according to the task description, using any language you may know. Parameters : n : [int] Dimension n x n of output array dtype : [optional, float(by Default)] Data type of returned array. C = $$\begin{bmatrix} 0 &1 \\ -2& 1 \end{bmatrix}$$, D= $$\begin{bmatrix} \frac{1}{2} &- \frac{1}{2} \\ 1& 0 \end{bmatrix}$$, CD= $$\begin{bmatrix} 0 &1 \\ -2& 1 \end{bmatrix}$$$$\begin{bmatrix} \frac{1}{2} &- \frac{1}{2} \\ 1& 0 \end{bmatrix}$$ = $$\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$, DC = $$\begin{bmatrix} \frac{1}{2} &- \frac{1}{2} \\ 1& 0 \end{bmatrix}$$ $$\begin{bmatrix} 0 &1 \\ -2& 1 \end{bmatrix}$$ = $$\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$. { Notation: An upper triangular matrix is typically denoted with U and a lower triangular matrix is typically denoted with L. { Properties: 1. For example. Identity matrix of size n is a diagonal matrix of size n with all diagonal elements as one. It is denoted by the notation “In” or simply “I”. The identity matrix is always a square matrix and has a dimensions of nxn. Example 3: Check the following matrix is Identity matrix; B = $$\begin{bmatrix} 1 & 1 & 1\\ 1 & 1& 1\\ 1 & 1 & 1 \end{bmatrix}$$. An identity matrix is a square matrix in which all the elements of principal diagonals are one, and all other elements are zeros. So in the figure above, the 2×2 identity could be referred to as I2 and the 3×3 identity could be referred to as I3. As you will see, whenever you construct an identity matrix, if you're constructing a 2 by 2 identity matrix, so I can say identity matrix 2 by 2, it's going to have a very similar pattern. Identity matrix is also known as unit matrix. numpy.identity(n, dtype = None) : Return a identity matrix i.e. It is also called as a Unit Matrix or Elementary matrix. Program to determine whether a given matrix is an identity matrix Explanation. Identity Matrix is the matrix which is n × n square matrix where the diagonal consist of ones and the other elements are all zeros. There is a matrix which is an additive identity for matrices:. An identity matrix is a square matrix of size n × n, where the diagonal elements are all 1s (ones), and all the other elements are all 0s (zeroes). Solution: No, It’s not an identity matrix, because it is of the order 3 X 4, which is not a square matrix. Task . Post your question and get tips & solutions from a community of 463,784 IT Pros & Developers. In this program, we need to check whether the given matrix is an identity matrix. For example: C = $$\begin{bmatrix} 1 & 2 & 3 &4 \\ 5& 6& 7 & 8 \end{bmatrix}$$. det A ≠ 0. Where n×n matrices are used to represent linear transformations from an n-dimensional vector space to itself, In represents the identity function, regardless of the basis. In physics, especially in quantum mechanics, the Hermitian adjoint of a matrix is denoted by a dagger (†) and the equation above becomes In linear algebra, the identity matrix (sometimes ambiguously called a unit matrix) of size n is the n × n square matrix with ones on the main diagonal and zeros elsewhere. For checking a matrix A we need to ensure that if i = j then A ij must be equal to 1. Subscript n denotes order of the matrix. 420 views A matrix is nothing more (or nothing less) than a rectangular arrangement of numbers or letters or other items. These matrices are said to be square since there is always the same number of rows and columns. Unitary matrix. Intro to identity matrix. noun. This is also true in matrices. [4], When A is m×n, it is a property of matrix multiplication that. The 3 by 3 identity matrix is equal to 1, 0, 0, 0, 1, 0, and 0, 0, 1. Other articles where Identity matrix is discussed: matrix: …everywhere else is called a unit matrix. If a Hermitian matrix is real, it is a symmetric matrix, . In general, a square matrix over a commutative ring is invertible if and only if its determinant is a unit in that ring. That is, it is the only matrix such that: The principal square root of an identity matrix is itself, and this is its only positive-definite square root. Is matrix multiplication commutative? Less frequently, some mathematics books use U or E to represent the identity matrix, meaning "unit matrix"[3] and the German word Einheitsmatrix respectively. It is also called as a Unit Matrix or Elementary matrix. If you multiplied again you would go through the cycle again. e) order: 1 × 1. Returns : identity array of dimension n x n, with its main diagonal set to one, and all other elements 0. See the picture below. For a 2 × 2 matrix, the identity matrix for multiplication is . a square matrix with ones on the main diagonal. I have tried to solve an algorithm problem, I'm newbie and I'm trying to practice a lot in programming problems. However, every identity matrix with at least two rows and columns has an infinitude of symmetric square roots. Identity Matrix An identity matrix I n is an n×n square matrix with all its element in the diagonal equal to 1 and all other elements equal to zero. It would be exponent rules thing^x × thing^y = thing^[x+y] modulo 7. The identity matrix corresponds to the identity map, that is: I ⋅ v → = v →, ∀ v →. a scalar matrix in which all of the diagonal elements are unity • Syn: ↑unit matrix • Hypernyms: ↑scalar matrix. When [0] is added to any matrix of the same dimensions, the matrix does not change. The elements of the given matrix remain unchanged. identity matrix. SparseArray is more compact: The SparseArray representation uses a fraction of the memory: For matrix and arithmetic operations they are effectively … To prevent confusion, a subscript is often used. identity matrix. Identity Matrix is also called Unit Matrix or Elementary Matrix. In particular, their role in matrix multiplication is similar to the role played by the number 1 in the multiplication of real numbers: The ith column of an identity matrix is the unit vector ei (the vector whose ith entry is 1 and 0 elsewhere) It follows that the determinant of the identity matrix is 1, and the trace is n. Using the notation that is sometimes used to concisely describe diagonal matrices, we can write, The identity matrix can also be written using the Kronecker delta notation:[4]. If the product of two square matrices, P and Q, is the identity matrix then Q is an inverse matrix of P and P is the inverse matrix of Q. A has full … Sponsored Links. Tweet . Build an identity matrix of a size known at run-time. For any whole number n, there is a corresponding n×nidentity matrix. The column (or row) vectors of a unitary matrix are orthonormal, i.e. Interpretation Translation identity matrix. Number of rows and columns are equal therefore this matrix is a square matrix. It’s the identity matrix! In linear algebra, a complex square matrix U is unitary if its conjugate transpose U* is also its inverse, that is, if {\displaystyle U^ {*}U=UU^ {*}=I,} where I is the identity matrix. As you will see, whenever you construct an identity matrix, if you're constructing a 2 by 2 identity matrix, so I can say identity matrix 2 by 2, it's going to have a very similar pattern. An identity matrix is a matrix whose product with another matrix A equals the same matrix A. Add to solve later. In particular, the identity matrix serves as the unit of the ring of all n×n matrices, and as the identity element of the general linear group GL(n) (a group consisting of all invertible n×n matrices). A unit matrix is a square matrix all of whose elements are 1's. Multiplying by the identity. In some fields, such as quantum mechanics, the identity matrix is denoted by a boldface one, 1; otherwise it is identical to I. Upper Triangular Matrix. Identity Matrix is the matrix which is n × n square matrix where the diagonal consist of ones and the other elements are all zeros. V= $$\begin{bmatrix} 1 & 0 & 0 &0 \\ 0& 1 & 0 &0 \\ 0 & 0 & 1 & 0\\ \end{bmatrix}$$. The identity property of multiplication states that when 1 is multiplied by any real number, the number does not change; that is, any number times 1 is equal to itself. If any matrix is multiplied with the identity matrix, the result will be given matrix. Intro to identity matrices. Example 4 The following are all identity matrices. Defined matrix operations. For part (b), many students wrote that “the identity matrix is nonsingular, so it is consistent”. 3) We always get an identity after multiplying two inverse matrices. Google Classroom Facebook Twitter. A lower triangular matrix is a square matrix in which all entries above the main diagonal are zero (only nonzero entries are found below the main diagonal - in the lower triangle). The "identity" matrix is a square matrix with 1 's on the diagonal and zeroes everywhere else. It's quick & easy. Example 2: Check the following matrix is Identity matrix? 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Whenever the identity element for an operation is the answer to a problem, then the two items operated on to get that answer are inverses of each other.. home > topics > asp.net > questions > write a c program to create identity matrix or unit matrix + Ask a Question. However, it should be perfectly legitimate to consider it as adding zero times one row to another row, or multiplying one row with the numbeer one. Algorithm It is represented as In or just by I, where n represents the size of the square matrix. If I is a left identity matrix for a given matrix A, then the matrix product I.A = A. I = eye(3, 'uint32' ), I = 3x3 uint32 matrix 1 0 0 0 1 0 0 0 1 The above is 2 x 4 matrix as it has 2 rows and 4 columns. [5], Mitchell, Douglas W. "Using Pythagorean triples to generate square roots of, "Identity matrix: intro to identity matrices (article)", Fundamental (linear differential equation), https://en.wikipedia.org/w/index.php?title=Identity_matrix&oldid=975834563, Creative Commons Attribution-ShareAlike License, When multiplied by itself, the result is itself, This page was last edited on 30 August 2020, at 17:32. A square matrix represents a linear transformation from a vector space to a vector space with the same dimensionality. Having learned about the zero matrix, it is time to study another type of matrix containing a constant specific set of values every time, is time for us to study the identity matrices. Solution: No, it is not a unit matrix as it doesn’t contain the value of 0 beside one property of having diagonal values of 1. A matrix is called unit or identity matrix if its diagonal elements are unit that is 1, usually it is denoted by $I_n$. Click here if solved 12. So the size of the matrix is important as multiplying by the unit is like doing it by 1 with numbers. identity matrix: translation. The KroneckerProduct of a matrix with the identity matrix is a block diagonal matrix: The WorkingPrecision option is equivalent to creating the matrix, then applying N: Possible Issues (1) IdentityMatrix gives a matrix with dense storage. Identity Matrix. An identity matrix, also known as a unit matrix, is a square matrix in which all of the elements of the principle diagonal are ones, and the rest are zeros. [1][2] In some fields, such as quantum mechanics, the identity matrix is denoted by a boldface one, 1; otherwise it is identical to I. A scalar matrix whose diagonal elements are all 1 is called a unit matrix, or identity matrix. Required fields are marked *. The identity matrix for is because . An identity matrix is a square matrix whose upper left to lower right diagonal elements are 1's and all the other elements Else A ij must be equal to 0. Usage notes and limitations: See eye (Parallel Computing Toolbox). Given a square matrix M[r][c] where ‘r’ is some number of rows and ‘c’ are columns such that r = c, we have to check that ‘M’ is identity matrix or not.
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2021-01-26 06:24:42
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https://docs.acquia.com/cloud-platform/develop/drupal/baseurl/
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Information for:
Setting base URLs without breaking non-production environments¶
Support for $base_url was deprecated in Drupal 8. Occasionally, it may be necessary to set the $base_url value in a Drupal application’s settings.php file, because a contributed module may require this variable to be set, or an application may have specialized configurations.
Unfortunately, setting this variable can cause issues on Cloud Platform, where there are separate Development, Staging, and Production environments. If it is set unconditionally, the $base_url variable instructs Drupal to rewrite all requests for all environments, which breaks environments to which this variable doesn’t point. The solution is to make use of the $_ENV['AH_SITE_ENVIRONMENT'] environment variable that is set by the Cloud Platform environment.
Note for multisite users
Multisite installations should set $base_url in a dynamic fashion, as outlined here. Setting the $base_url to an empty string disables caching and is not recommended.
Modify and use the following example code as necessary to fit your needs, and be sure to add your Remote Administration environment if your application has one:
if (isset($_ENV['AH_SITE_ENVIRONMENT'])) { switch ($_ENV['AH_SITE_ENVIRONMENT'])
{
case 'dev': $base_url = 'http://dev.example.com'; break; case 'test':$base_url = 'http://test.example.com';
break;
case 'prod': $base_url = 'http://www.example.com'; break; } } You can further modify the preceding code if there are environments that don’t require $base_url to be explicitly set. For example, if the $base_url variable is required only for Production, you can use the following much shorter code snippet: if (isset($_ENV['AH_SITE_ENVIRONMENT']) && $_ENV['AH_SITE_ENVIRONMENT'] === 'prod') {$base_url = 'http://www.example.com';
}
Note
If you set a custom \$base_url, be aware that the variable should not contain a trailing slash and should always start with either http:// or https://.
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2020-10-26 18:58:08
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https://www.gradesaver.com/textbooks/science/physics/fundamentals-of-physics-extended-10th-edition/chapter-10-rotation-problems-page-287/4b
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Fundamentals of Physics Extended (10th Edition)
$0$
Angular velocity of point will be: $w=\dfrac {\partial \theta }{\partial t}=\dfrac {\partial }{\partial t}\left( 2+4t^{2}+2t^{3}\right) =8t+6t^{2}$ So, at $t=0$: $w\left( 0\right) =8t+6t^{2}=8\times 0+6\times 0^{2}=0$
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2018-12-19 10:14:26
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https://www.maa.org/book/export/html/117763
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CalcPlot3D, an Exploration Environment for Multivariable Calculus
Author(s):
Paul Seeburger (Monroe Community College)
This dynamic Java applet developed with support from the NSF (Dynamic Visualization Tools for Multivariable Calculus, DUE-CCLI Grant #0736968)) allows the user to simultaneously plot multiple 3D surfaces, space curves, parametric surfaces, vector fields, contour plots, and more in a freely rotateable graph.
This tool is intended as a dynamic visualization and exploration environment for multivariable calculus. Use it to illustrate the geometric relationships of many of the concepts of multivariable calculus, including dot and cross products, velocity and acceleration vectors for motion in the plane and in space, the TNB-frame, the osculating circle and curvature, surfaces, contour plots and level surfaces, partial derivatives, gradient vectors and gradient fields, Lagrange multiplier optimization, double integrals as volume, defining the limits of integration for double and triple integrals, parametric surfaces, vector fields, line integrals, and more. See the corresponding web page for documentation and a list of guided explorations developed for students to use with this exploration applet.
INTENDED USES:
Instructors and students in multivariable calculus.
SOFTWARE SPECIFICATIONS:
Plugins: Java Plug-in (free) with any browser
Operating Systems: Mac, Windows, Linux, etc.
Open CalcPlot3D, an Exploration Environment for Multivariable Calculus in a new window
(Note: For the most up-to-date version of CalcPlot3D, see Paul's project website.)
CalcPlot3D, an Exploration Environment for Multivariable Calculus - Overview
Author(s):
Paul Seeburger (Monroe Community College)
This dynamic Java applet developed with support from the NSF (Dynamic Visualization Tools for Multivariable Calculus, DUE-CCLI Grant #0736968)) allows the user to simultaneously plot multiple 3D surfaces, space curves, parametric surfaces, vector fields, contour plots, and more in a freely rotateable graph.
This tool is intended as a dynamic visualization and exploration environment for multivariable calculus. Use it to illustrate the geometric relationships of many of the concepts of multivariable calculus, including dot and cross products, velocity and acceleration vectors for motion in the plane and in space, the TNB-frame, the osculating circle and curvature, surfaces, contour plots and level surfaces, partial derivatives, gradient vectors and gradient fields, Lagrange multiplier optimization, double integrals as volume, defining the limits of integration for double and triple integrals, parametric surfaces, vector fields, line integrals, and more. See the corresponding web page for documentation and a list of guided explorations developed for students to use with this exploration applet.
INTENDED USES:
Instructors and students in multivariable calculus.
SOFTWARE SPECIFICATIONS:
Plugins: Java Plug-in (free) with any browser
Operating Systems: Mac, Windows, Linux, etc.
Open CalcPlot3D, an Exploration Environment for Multivariable Calculus in a new window
(Note: For the most up-to-date version of CalcPlot3D, see Paul's project website.)
CalcPlot3D, an Exploration Environment for Multivariable Calculus - Using CalcPlot3D to Visually Verify Homework in Multivariable Calculus
Author(s):
Paul Seeburger (Monroe Community College)
One way to use CalcPlot3D, a versatile Java applet, is to demonstrate new concepts during multivariable calculus lectures. I often develop a new concept on the chalk board first and then take a couple minutes to make the concept come to life using the applet. Students find these demonstrations helpful and fun, and they bring variety to my presentations, helping students process the new concepts in a new way.
An even more exciting way to use CalcPlot3D in class is to engage in a visual exploration of new concepts using “What if…” types of questions. An example of a topic for which I find this approach works especially well is exploring the variety of possible parameterizations of a plane/space curve, paying special attention to the behavior of the velocity and acceleration vectors. Using these sorts of visual demonstrations in class improves student learning, but to fully engage students in the exploration and discovery process and give them the best chance of learning the geometric nature of the calculus concepts, I feel it is vital to give students opportunities to “play” with the concepts visually themselves. This article focuses on one way this can be done: by requiring students to visually verify solutions to particular homework problems and turn these in for a grade. [Another way this project supports student engagement and “play” is with the guided explorations being developed for various concepts. See the main project website to explore these. At this writing, there are explorations for Dot Products, Cross Products, Velocity & Acceleration Vectors, and Lagrange Multiplier Optimization.]
Below is a list of example topics where I often assign this type of visual verification exercise to my students to get them to begin using the applet on their own. Once they start using the applet in this way (because they have to), students often report using the applet more often on their own to explore additional exercises they complete from the textbook and on other assignments. Before giving a visual verification assignment involving new skills with the applet, I always demonstrate using CalcPlot3D to visually verify a similar problem we worked on the board. Once students have seen one example using the applet, most have little trouble completing the exercise on their own.
Without further discussion, let’s look at some examples of how this can be done! As you develop your own examples, please send them along to me! I would love to develop a library of useful ways to use this applet.
CalcPlot3D, an Exploration Environment for Multivariable Calculus - Plane through Three Non-Collinear Points
Author(s):
Paul Seeburger (Monroe Community College)
An early topic that lends itself well to this sort of exercise is determining the plane determined by three non-collinear points.
Exercise:
1. Find the equation of the plane containing the points (2, 0, 1), (-1, 2, 3), and (0, 2, -2).
2. Graph this plane along with the three points to verify that all threepoints lie on the plane. To do this, first solve the plane equation for z and graph the plane, entering it in Function 1. Then select Add a Point from the Graph menu, and enter the coordinates of one of the points. Select the default size and colors. (If you wish to vary these settings, be sure the points still show up well in the printout.) Repeat these steps to add the other two points. Rotate the plot to verify that the points lie on the plane and then find a clear view of the plane with the three points on it. Use the Print Graph menu option on the File menu at the top left corner of the applet to print out your resulting view and hand this printout in with this assignment.
Click here to open a pdf file which contains the instructions for the activity.
Answer: Plane equation is: z = (22 – 10x – 13y)/2
CalcPlot3D, an Exploration Environment for Multivariable Calculus - Line of Intersection of Two Planes
Author(s):
Paul Seeburger (Monroe Community College)
Visually verifying the line of intersection of two planes is another early topic where I get my students used to using the CalcPlot3D applet themselves. Below is an example of this type of problem:
Exercise:
1. Determine the line of intersection of the following two planes. Write the parametric equations for this line, showing all work. $2x + y - 2x = 5 \qquad \& \qquad 3x-6y-2z=15$
2. Use the CalcPlot3D applet to display these two planes. To do this, solve each planar equation for z, and enter them in Functions 1 and 2 on the left side of the 3D plot window. Zoom out a couple times (if necessary) until you can see both planes along with their intersection. Now add the line of intersection. (To do this, choose Add a Space Curve from the Graph menu, and enter the parametric equations for the line.) Rotate the 3D view to verify that your line is indeed the intersection of the two planes. Rotate until you have a good view of the two planes and the line of intersection. Use the Print Graph menu option on the File menu at the top left corner of the applet to print out your resulting view and hand this printout in with this assignment.
Click here to open a pdf file which contains the instructions for the activity.
Answer Line: $$x = 3+14t$$ Here are the plane equations, solved for z: $$y=-1+2t$$ $$z=(2x+y-5)/2$$ $$z=15t$$ $$z=(3x-6y-15)/2$$
CalcPlot3D, an Exploration Environment for Multivariable Calculus - Intersections of Other Types of Surfaces
Author(s):
Paul Seeburger (Monroe Community College)
See my paper on this topic online in the Proceedings of AMATYC 2010:
http://www.amatyc.org/Events/conferences/2010Boston/proceedings.html
It includes two examples of this type of problem along with steps to follow to visually verify the solutions using CalcPlot3D.
CalcPlot3D, an Exploration Environment for Multivariable Calculus - Contour Plots
Author(s):
Paul Seeburger (Monroe Community College)
Certain types of contour plots are not too difficult to create by hand. Many of these have contours that are lines, circles, or ellipses. But there are a lot of interesting functions whose contour plots are much more difficult to create by hand. I still require my students to be able to create contour plots of the first sort by hand, and they can certainly use the applet to visually verify their results. But I like to require them to use the applet to create contour plots for several functions whose contours are more complicated so they can improve their geometric intuition connecting the surface plots of various types of functions with their contour plots.
Exercise: Create contour plots for $$z = Cos(x)Sin(y)$$ and $$z = -4x/(x^2+y^2+1)$$. Print these contour plots and then for each one, click on the contour plot to see the contours as they appear on the surface. For each function, find a viewpoint that shows the surface and contours clearly and print this surface plot as well.
$$z=Cos(x)Sin(y)$$ $$z=-4x/(x^2+y^2+1)$$
To create these contour plots, you will start by entering the function in Function 1 and graphing it. Then select the Draw Contour Plot option from the Contour Plot menu. A dialog will appear. Be sure that Function 1 is selected on the left. For the first function, you will then enter -1 for the First Step, 0.2 for the Step Size and 11 for the Number of Contours. Then click on the OK button to see the contour plot appear.
If you click on the contour plot, you will the 3D surface plot of the function will be graphed along with the contours on the surface.
For the second function, you will do the same thing, except you can enter -2 for the First Step, and 21 for the Number of Steps.
Click here to open a pdf file which contains the instructions for the activity.
CalcPlot3D, an Exploration Environment for Multivariable Calculus - Level Surfaces
Author(s):
Paul Seeburger (Monroe Community College)
It is difficult to draw many interesting level surfaces by hand, so I generally have my students use CalcPlot3D to do most of the work for this type of exercise. There are actually two ways to enter and graph the level surface equations for a particular function of three variables in CalcPlot3D:
1. Solve each equation for z in terms of x, y, and C and enter the level surface using one or two functions of x and y, or
2. Graph the level surface equation by Adding an Implicit Surface from the Graph menu and entering the equation for the level surface in the dialog box there.
If the surfaces are complicated enough, you may not have a choice. If you are not able to solve for z, you will need to use the Implicit Surface option.
Here is an example I use in class shown both ways. $f(x,y,z) = z^2 - x^2 + y^2$
Setting $$f(x,y,z)=z^2-x^2+y^2=C$$, we obtain the following equations if we solve for z.
C = 2 C = -2
$$z=\sqrt{C+x^2-y^2}$$
$$z=-\sqrt{C+x^2-y^2}$$
For $$C=2$$ we enter:
z=sqrt(2+x^2-y^2) in Function 1, and
z=-sqrt(2+x^2-y^2) in Function 2.
For $$C=-2$$ we enter:
z=sqrt(-2+x^2-y^2) in Function 1, and
z=-sqrt(-2+x^2-y^2) in Function 2.
We can obtain the following graphs of these surfaces by graphing the implicit equations.
$$z^2-x^2+y^2=2$$ $$z^2-x^2+y^2=-2$$
$$z^2-x^2+y^2=2$$ and $$z^2-x^2+y^2 = -2$$
These equations will be entered as:
z^2 - x^2 + y^2 = 2 z^2 - x^2 + y^2 = -2
Click here to open a pdf file which contains the instructions for the activity.
CalcPlot3D, an Exploration Environment for Multivariable Calculus - Tangent Planes
Author(s):
Paul Seeburger (Monroe Community College)
I develop the equation of a tangent plane early in order to help students understand the geometric meaning of the total differential. In my opinion, it is not very satisfying to determine the tangent plane to a surface at a given point if you cannot visually verify the answer.
When I assign these exercises, I usually require students to submit a print-out from the applet showing their tangent plane, the surface and the point of tangency.
Exercise: Determine the tangent plane of the following surface at the point (1, 0.5) in its domain. Use CalcPlot3D to print the graph of this surface (with –2 to 2 on x- and y-axes) along with this tangent plane. Show the point of tangency. $z= f(x,y) = x^2 + xy + y^2$
To create this plot, first enter the function in Function 1. Then enter the tangent plane function (solved for z) in Function 2. Graph these surfaces together. Then choose Add a Point from the Graph menu. Enter the coordinates of the point on the surface, and keep the default values for the other choices. To make it easier to see both surfaces, make them semi-transparent using either Ctrl-T or the Make Surfaces Transparent option on the View Settings menu.
Click here to open a pdf file which contains the instructions for the activity.
Answer: Tangent plane equation: $$z= 1.75 + 2.5(x - 1) + 2(y - 0.5)$$
CalcPlot3D, an Exploration Environment for Multivariable Calculus - Directional Derivatives
Author(s):
Paul Seeburger (Monroe Community College)
Exercise: Determine the directional derivative function for $$f(x, y) = x^2 + xy + y^2 + 1$$ in the direction of v = i + j. Then determine its value at the point (0, -1).
Use CalcPlot3D to graph this surface and show the appropriate tangent line on the surface at the point (0, -1) and displaying the unit direction vector and the correct directional derivative value.
To do this, first enter the function in Function 1. Then choose the directional derivative option from the drop-down menu just above the Trace Plot to the left of the 3D plot. You can then use the Trace Plot menu at the top of the applet to enter the point (0, -1) and the direction vector. I recommend hiding the edges (using the E key or the Hide Edges option on the View Settings menu) and also making the surface transparent
(using Ctrl-T or the Make Surfaces Transparent option on the View Settings menu). Rotate the plot until you can clearly see the direction vector, the surface, the tangent line, and the directional derivative value. Be sure it is the approximation of the exact value you obtained in your homework problem.
Click here to open a pdf file which contains the instructions for the activity.
CalcPlot3D, an Exploration Environment for Multivariable Calculus - Taylor Polynomials of a Function of Two Variables (1st and 2nd degree)
Author(s):
Paul Seeburger (Monroe Community College)
The way the Taylor polynomials of a function of one variable progressively converge to the graph of the function like y = cos x is really quite impressive and is inherently interesting. We can extend this topic into three dimensions using CalcPlot3D.
As an exercise, I require my students to generate the linear and quadratic Taylor polynomials of a function of two variables using the partial derivatives of the function evaluated at a particular point.
$$\begin{eqnarray} f(x,y) &\approx L(x,y) = f(a,b) &+ f_x(a,b)(x-a) + f_y(a,b)(y-b) \qquad (1^{st}\text{-deg. Taylor poly or tangent plane})\\ f(x,y) &\approx Q(x,y) = f(a,b) &+ f_x(a,b)(x-a) + f_y(a,b)(y-b) \\ &+\frac{f_{xx}(a,b)}{2}(x-a)^2 &+ f_{xy}(a,b)(x-a)(y-b) + \frac{f_{yy}(a,b)}{2}(y-b)^2 \qquad (2^{nd}\text{-deg. Taylor poly})\end{eqnarray}$$
Exercise: Determine the 1st and 2nd degree Taylor polynomials in two variables for the given function.Simplify both polynomials. Show all work including all partial derivatives and using the formula clearly with functional notation in the first step. Please also provide a printout of the given surface along with each of the Taylor polynomials. (That’s 2 printouts all together.) Include the point on the surface where the polynomial is tangent to the surface. Use the Format Surfaces option on the View Settings menu so that the Taylor polynomial is reverse color and transparent so it’s possible to tell the two surfaces apart. If necessary, zoom out and the rotate to a view that shows the surfaces clearly. Then use the Print Graph option on the File menu of the applet to print the graph. $$f(x,y) = \sin(2x) + \cos y$$ for x,y near (0,0)
Answers: 1st-degree Taylor Polynomial of f: 2nd-degree Taylor Polynomial of f: $$L(x,y)=1+2x$$ $$Q(x,y)=1+2x-\frac{1}{2}y^2$$
There is also a feature of the applet that will allow you to demonstrate higher-degree Taylor polynomials for a function of two variables.
Example:
1. Graph the function, $$f(x,y)=\cos(x)\sin(y)$$. Then zoom out to -4 to 4 in the x and y-directions.
2. Now select the View Taylor Polynomials option from the Tools menu at the top of the applet. It will take a few seconds as the computer calculates the partial derivatives and creates the Taylor polynomials. This example is successfully calculated all the way up to the 15th degree polynomial. Once it is ready, the original function is graphed as a wireframe and the 1st degree Taylor polynomial (the tangent plane) is shown. A scrollbar appears along the bottom edge of the 3D plot. Use this scrollbar to scroll through the various Taylor polynomials of this function. Note that only odd degrees add new terms for this particular function. As you increase the degree of the Taylor polynomial notice how the polynomial of two variables fits the original surface better and better around the origin until it is a fairly good approximation of the whole visible surface at the 15th degree.
3. To better view the Taylor polynomial itself (shown in the text window just above the 3D plot), you can click and drag on the equation and view all terms, dragging the equation left and right. You can also use the Tools menu option Use Factorials in Taylor Polynomials to switch this property on or off. Using factorials makes the form of the terms of the higher order Taylor polynomials easier to see, and the terms also generally take up less horizontal space each.
4. You can also vary the center point for the Taylor expansion using the Tools menu option just below View Taylor Polynomials. The default center point is the origin.
5. Other nice functions to try centered about the origin include:
• $$f(x,y)=\cos(x)-\sin(y)$$
• $$f(x,y)=\sin(2x)-\cos(y)$$
• $$f(x,y)=\sin(x^2+y^2)$$
• $$f(x,y)=xe^y+1$$
• $$f(x,y)=e^{x^2+2x-y}$$
• $$f(x,y)=\arctan(xy)$$
• $$f(x,y)=\arctan(x+y)$$
Click here to open a pdf file which contains the instructions for the activity.
CalcPlot3D, an Exploration Environment for Multivariable Calculus - Lagrange Multiplier Optimization
Author(s):
Paul Seeburger (Monroe Community College)
In multivariable calculus, we teach our students the method of Lagrange multipliers to solve constrained optimization problems. As we introduce this topic, many of us use some form of visual presentation to help students understand how we develop the Lagrange multiplier equation, i.e.,
$\nabla f(x,y) = \lambda \nabla g(x,y)$
After demonstrating how this works in class with an example, I assign my students to print a visual verification of a homework problem showing both the contour plot with the constraint curve and the 3D surface plot with the constraint curve projected onto the surface showing the relative extrema visually.
Exercise: Find the relative extrema of the function f subject to the given constraint, showing all work. Then graph the function in CalcPlot3D and create its contour plot with First Level: -1, Step size: 1, and number of contours: 10. Then add the constraint curve to the plot using the Add Constraint Curve option in the Display Contour Plot dialog. Use the scrollbar at the bottom of the contour plot to move the point to one of the relative extrema. Then print the contour plot. Finally click on the contour plot and print a view of the surface (along with the contours and the constraint curve projected on the surface) that makes it possible to see both extrema clearly.
$$f(x,y)=x^2+y^2+2x-2y+1$$ Constraint: $$g(x,y)=x^2+y^2=2$$
To obtain the images on the right using the applet:
1. Enter x^2 + y^2 + 2x 2y + 1 in Function 1. Zoom out once so the x and y values range from -4 to 4.
2. To obtain the contour plot, choose the Draw Contour Plot option on the Contour Plot menu at the top of the applet.
3. On the Display Contour Plot dialog, be sure Function 1 is selected, and enter First Level: -1, Step size: 1, and number of contours: 10.
4. To display the constraint curve, select the second option on the Add Constraint Curve menu at the top of this Display Contour Plot dialog. This option is labeled, Circle centered at origin.
5. Enter 2 as $$\mathbb{R}^2$$ and click on the OK button.
6. Now click on the OK button of the Display Contour Plot dialog and you should see the contour plot and the constraint curve appear.
7. By default, the gradient vectors of f and g are displayed, $$\nabla f$$ is red and $$\nabla g$$ is blue. To watch them vary along the constraint curve, use the scrollbar located at the bottom of the 3D plot.
8. We can visually verify that these gradient vectors are either parallel to each other or $$\nabla f=0$$ when we obtain a relative extremum of f subject to the constraint, in both cases satisfying the Lagrange Multiplier equation.
9. Now click on the contour plot in the large 3D plot window. The surface above will appear. Use the right arrow key to rotate the 3D graph to the right about the z-axis until you see a clear view of the relative extrema. Then use the scrollbar again to move along the trace of the constraint curve on the surface to verify the relative extrema we found earlier.
Click here to open a pdf file which contains the instructions for the activity.
CalcPlot3D, an Exploration Environment for Multivariable Calculus - Riemann Sums of a Double Integral
Author(s):
Paul Seeburger (Monroe Community College)
When introducing double integrals, I like to show the students how we can use rectangular prisms to approximate the volume of the solid region between the region of integration (in the xy-plane) and the surface. It is nice to connect these approximations to the value we obtain by evaluating the corresponding double integral. Here is an exercise students can do to improve their understanding of this connection.
Exercise: Use 8 rectangular prisms to approximate the volume of the solid region between $$f(x,y)=4-x^2+y$$ and the rectangular region R in the xy-plane given by $$-2\leq x \leq 2$$ and $$0\leq x \leq 2$$. Then use a double integral to find the exact volume of this solid region. Show all work on paper for both parts. Then use CalcPlot3D to create a graph of the region in the xy-plane along with the rectangular prisms you used and the surface above the region. Print a view of this graph that shows the surface and the rectangular prisms well. Use the Show grid scrollbars option on the Create Bounded Region dialog to vary the number of prisms in each direction. What do you notice happens to the volume of the prisms when you vary the number of prisms in the y-direction? Can you explain why this happens for this function? As you increase the number of prisms in the x-direction, does the volume of the prisms approach the exact volume you calculated? What is the volume of the prisms when there are 30 prisms in the x-direction?
To create these graphs:
1. Graph the function $$z = 4 -x^2+y$$.
3. The dialog shown at right will appear. Define the base of the region by entering $$y = 0$$ for the bottom function and $$y = 2$$ for the top function. Then enter -2 for x-min and 2 for x-max.
The graphs below show another example: $$f(x, y) = (x^2 + y^2)/2$$ over the region in the xy-plane bounded by the graphs of $$y = \frac{x^3}{4}$$ and $$y=2\sqrt{\frac{x}{2}}.$$
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https://golem.ph.utexas.edu/category/2008/10/categorification_in_new_scient.html
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## October 20, 2008
### Categorification in New Scientist
#### Posted by John Baez
Here’s an article on knot theory that mentions categorification:
• Richard Elwes, Fundamental secrets are tied up in knots, New Scientist, October 15, 2008.
Richard Elwes is a mathematician and reporter based in Leeds, UK.
Most of the story is about the Jones polynomial and other quantum invariants of knots. I’ll just quote a bit at the end:
Then, in 1999, an exciting breakthrough came — once again from an entirely unexpected quarter. A radically different technique not only spawned a new generation of knot invariants, but also suggested that the mathematics behind knots might have a more profound significance than anyone suspected. The technique’s name was categorification.
Categorification turns the normal guiding logic of mathematics — the abstraction and simplification of the real world — on its head. Abstraction and simplification are all very well, but the results are often too simple to describe as much as we might want. This can be illustrated by the difficulties small children wrestle with when learning to count. Why do three apples and three oranges both reduce to the same “3”, when apples and oranges are entirely different things?
The answer is that the mathematical structure that “3” inhabits - the system of ordinary numbers - is an abstraction shorn of any information about the things it represents. In this case, categorification involves replacing this system with a richer structure, a “category”, in which the rigid equations that define the number system - “1 + 2 = 3”, for example - are replaced with weaker statements comparing the sizes of different sets of objects. This category offers a real-world flexibility that numbers on their own do not: its sets can be different even if they have the same size. In this structure, the original number system appears as the category’s “shadow”, obtained by collapsing all sets of size 3 down to just one representative: the number 3.
Might what works for the number system also work for other mathematical objects? This was the philosophy adopted by Mikhail Khovanov, a mathematician at the University of California, Davis, when he revisited Jones’s invariant in 1999. Instead of breaking it down into smaller finite-type expressions, he looked for some grander structure of which it was just the most visible shadow.
His search was a spectacular success. The overarching category that he found is conceptually difficult, but mathematically it is far less awkward than Kontsevich’s integral, and it’s a more reliable description of a knot than Jones’s formula. Even better, thanks to an ingenious computer program written in 2006 by Dror Bar-Natan of the University of Toronto, Canada, it can now be computed efficiently for any knot, potentially opening up its use to researchers in other areas.
Khovanov’s category is not perfect: there are still instances of knots that share the same category. So work continued and, in 2005, together with Lev Rozansky of the University of North Carolina, Chapel Hill, Khovanov unveiled a new invariant operating at an even higher level. Not only does this categorify many quantum invariants beyond Jones’s, it also subsumes Khovanov’s original category, as well as several other knot invariants discovered in the meantime.
The power of the Khovanov–Rozansky category is getting us close to the perfect knot description, though early indications are that we are not there yet. A still broader family of quantum invariants may yet need to be incorporated before we can say we have the loose ends of the knot problem tied up. Nevertheless, we seem to be edging towards the ultimate mathematical solution.
With early experiments in categorification yielding such riches, physicists and mathematicians are waking up to the idea that the approach might apply to more than just knots. Recalling the deep connection of knots with quantum theory that inspired Jones in the first place, some researchers think they have spotted a tantalising hint that whole chunks of mathematical physics are just shadows of larger categorical structures.
Striking analogies have already been found between the categorical descriptions of quantum mechanics and Einstein’s relativity - the twin theoretical pillars of modern physics that have so far seemed fundamentally incompatible. And John Baez, Alexander Hoffnung and Christopher Rogers of the University of California, Riverside, have recently argued that string theory — a favoured starting point for a theory of everything — can be viewed as a categorification of particle physics.
If a categorification embracing both relativity and quantum theory can be found, then the inconsistencies that physicists see between these two theories might yet prove to be an illusion. Such a tying together of all of physics would indeed be a categorical triumph for the humble knot.
This is the first time I’ve seen a popular science magazine try to explain categorification. Knot theory is a good context for explaining this. A picture of knots that are isomorphic but not equal might help, since anyone can see they’re different but still ‘the same in a way’ — and untying a knot is just proving that it’s isomorphic to the trivial knot.
Posted at October 20, 2008 11:28 AM UTC
TrackBack URL for this Entry: http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1825
### Re: Categorification in New Scientist
Very interesting. Are there any books on categorification that you would recommend?
Posted by: Rajiv Das on October 20, 2008 3:35 PM | Permalink | Reply to this
### Re: Categorification in New Scientist
Not yet. For something not very technical, I’d recommend this.
Posted by: John Baez on October 20, 2008 5:03 PM | Permalink | Reply to this
### Re: Categorification in New Scientist
People may be interested to learn a bit about Richard Elwes. He seems to be pretty serious about explaining math to a broad audience… but he seems to have a sense of humor, too:
That’s his photo from an article he wrote on the classification of finite simple groups, over at Plus magazine.
Posted by: John Baez on October 20, 2008 5:49 PM | Permalink | Reply to this
### Re: Categorification in New Scientist
Isn’t that Weird Al Yankovic?
Posted by: Bruce Bartlett on October 20, 2008 6:28 PM | Permalink | Reply to this
### New Pseudoscientist redeemed; Re: Categorification in New Scientist
Years ago I subscribed to New Scientist for my son. I have kept renewing the subscription. He got his double Math and CS degree at age 18, but still reads them when he comes home about every other week from law school.
Although, as the John Baez-Greg Egan dual entity has explained, NS is often drifting into Physics crackpottery, it is strong in other areas, and fun to read. I even xerox and distribute some articles from it as part of homework assignments to the Bio, Anatomy and Physiology classes that I teach.
I was delighted by the Knot Theory/Category Theory article. It is a gem, and the Baez quote is icing on the cake, to mix a metaphor.
Posted by: Jonathan Vos Post on October 21, 2008 12:59 AM | Permalink | Reply to this
### Re: New Pseudoscientist redeemed; Re: Categorification in New Scientist
Physics crackpottery
cf psychoceramics
Posted by: jim stasheff on October 21, 2008 1:26 AM | Permalink | Reply to this
### Lou Kauffman and the Arrow Polynomial; Re: Categorification in New Scientist
How does this fit the story’s big picture? Louis H. Kauffman is an amazing thinker and writer!
We introduce a new polynomial invariant of virtual knots and links and use this invariant to compute a lower bound on the virtual crossing number.
Posted by: Jonathan Vos Post on October 22, 2008 6:24 AM | Permalink | Reply to this
### Re: Lou Kauffman and the Arrow Polynomial; Re: Categorification in New Scientist
Cf:
• arXiv:math/0701333
Title: Virtual crossings, convolutions and a categorification of the SO(2N) Kauffman polynomial
Authors: Mikhail Khovanov, Lev Rozansky
but as I read this paper, it is NOT a categorification but rather a lifting/resolution of the polynomial by a chain complex
Or is this commonly called ‘categorification’? I would have expected a category rather than a chain complex??
Posted by: jim stasheff on October 22, 2008 2:27 PM | Permalink | Reply to this
### Re: Lou Kauffman and the Arrow Polynomial; Re: Categorification in New Scientist
This is a bit late, considering the new post on categorification, but this was my reply by email at the time, which I was asked to make public:
Hi Jim,
here are some of my own thoughts - they may not be fully informed.
there are two apparently different meanings to categorification these days: Baez-Dolan and one as used by knot theorists. I suspect they are not all that different, but have a different flavour.
Knot theorists say categorification when they replace the usual knot invariants (polynomials) with collections of vector spaces, much in the vein of when E. Noether replaced betti numbers with abelian groups. Then the polynomial is the Poincare series of some appropriate graded space.
Baez-Dolan categorification is the one which most often appears at the Cafe - replacing things with categories.
Best,
David
I was I think a bit misleading - I didn’t mean to imply there are two different sorts of categorification, but that they are only apparently different. See John’s new post linked to above for details.
Posted by: David Roberts on October 24, 2008 2:32 AM | Permalink | Reply to this
### Re: Lou Kauffman and the Arrow Polynomial; Re: Categorification in New Scientist
Thanks for posting this comment publicly. You knew or suspected that that the difference between ‘Khovanov’ and ‘Baez–Dolan’ categorification is only apparent — but a lot of people don’t, especially young folks working on knot homology who haven’t had time to step back and consider how widespread categorification actually is. So, I used your comment as an excuse for a quick post.
There should really be a bigger Wikipedia article on this subject: as ‘categorification’ sweeps the mathematical landscape, more and more people will wonder what it means.
Posted by: John Baez on October 24, 2008 3:47 AM | Permalink | Reply to this
### Re: Lou Kauffman and the Arrow Polynomial; Re: Categorification in New Scientist
Jonathan wrote, speaking of Kauffman’s latest paper on virtual knots:
How does this fit the story’s big picture? Louis H. Kauffman is an amazing thinker and writer!
I met Kauffman in Lisbon this summer, and over dinner attempted to extract a category-theoretic account of the ‘deep inner meaning’ of virtual knot theory. He helped me understand it much better… but Kauffman is not a category theorist at heart, so what follows is my own version of the ‘deep inner meaning’, for which he is not to blame.
(I wrote a bit about this in my diary. I was planning to discuss it in This Week’s Finds, but haven’t gotten around to it yet. So, here goes.)
Abstractly, virtual knot theory is the study of the free symmetric monoidal category with duals on one object equipped X with a morphism $R: X \otimes X \to X \otimes X$ that satisfies the Yang–Baxter equation. (In physics jargon, we call such a morphism an ‘R-matrix’.) This gives our category a braiding in addition to the original symmetric braiding. We draw the new braiding as a crossing, and the original symmetric one as a ‘virtual crossing’, marked by a circle below:
These are the Reidemeister moves in virtual knot theory. The moves at left are the usual Reidemeister moves, while those at right concern the ‘virtual crossing’, and those at the bottom involve both kinds of crossing.
This sort of structure arises a lot in the theory of quantum groups, where besides the usual symmetric braiding in the category of vector spaces we also get another nonsymmetric braiding.
There’s also a fascinating interpretation of virtual knot theory in terms of knots drawn on compact 2-manifolds, where the virtual crossing is a crossing that involves one strand going over a ‘handle’ — or more picturesquely, ducking into a ‘wormhole’. Greg Kuperberg has proved a nice theorem about this interpretation.
I would like someday to understand his theorem in terms of the ‘free symmetric monoidal category with duals on a solution of the Yang–Baxter equation’. Why does this category know about compact 2-manifolds?
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2017-03-25 09:42:40
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http://rspa.royalsocietypublishing.org/content/464/2099/3003.full
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# Nonlinear Euler buckling
Alain Goriely, Rebecca Vandiver, Michel Destrade
## Abstract
The buckling of hyperelastic incompressible cylindrical tubes of arbitrary length and thickness under compressive axial load is considered within the framework of nonlinear elasticity. Analytical and numerical methods for bifurcation are developed using the exact solution of Wilkes for the linearized problem within the Stroh formalism. Using these methods, the range of validity of the Euler buckling formula and its first nonlinear corrections are obtained for third-order elasticity. The values of the geometric parameters (tube thickness and slenderness) where a transition between buckling and barrelling is observed are also identified.
Keywords:
## 1. Introduction
Under a large enough compressive axial load an elastic beam will buckle. This phenomenon known as elastic buckling or Euler buckling is one of the most celebrated instabilities of classical elasticity. The critical load for buckling was first derived by Euler in 1744 (Euler 1744, 1759; Oldfather et al. 1933) and further refined for higher modes by Lagrange in 1770 (Lagrange 1770; Timoshenko 1983). Both authors reached their conclusion on the basis of simple beam equations first derived by Bernoulli (Todhunter 1893; figure 1). Since then, Euler buckling has played a central role in the stability and mechanical properties of slender structures from nano- to macrostructures in physics, engineering, biochemistry and biology (Timoshenko & Gere 1961; Niklas 1992). Explicitly, the critical compressive axial load N that will lead to a buckling instability of a hinged–hinged isotropic homogeneous beam of length L is(1.1)where ‘π is the circumference of a circle whose diameter is one’ (Euler 1759); E is Young's modulus; and I is the second moment of area, which, in the case of a cylindrical shell of inner radius A and outer radius B, is .
Figure 1
Euler problem: (a) illustrations from Euler (1744). (b) Lagrange solutions (1770), modes 1, 2 and 3.
There are many different ways to obtain this critical value and infinite variations on the theme. If the beam is seen as a long slender structure, the one-dimensional theory of beams, elastica, or Kirchhoff rods, can be used successfully to capture the instability, either by bifurcation analysis, energy argument (Timoshenko & Gere 1961) or directly from the exact solution, which in the case of rods can be written in terms of elliptic integrals (Nizette & Goriely 1999). The one-dimensional theory can be used with a variety of boundary conditions, it is particularly easy to explain and generalize and it can be used for large geometric deflections of the axis (Antman 1995). However, since material cross sections initially perpendicular to the axis remain undeformed and perpendicular to the tangent vector, no information on the elastic deformation around the central curve can be obtained. In particular, other modes of instability such as barrelling cannot be obtained. Here, by barrelling, we refer to axisymmetric deformation modes of a cylinder or a cylindrical shell. These modes will typically occur for sufficiently stubby structures.
The two-dimensional theory of shells can be used when the thickness of the cylindrical shell is small enough. Then, the stability analysis of shell equations such as the Donnell–von Kármán equations leads to detailed information on symmetric instability modes, their localization and selection (Hunt et al. 2003). However, the theory cannot be directly applied to obtain information on the buckling instability (asymmetric buckling mode).
The three-dimensional theory of nonlinear elasticity provides, in principle, a complete and exact description of the motion of each material point of a body under loads. However, due to the mathematical complexity of the governing equations, most problems cannot be explicitly solved. In the case of long slender structures under loads, the buckling instability can be captured by assuming that the object is either a rectangular beam (Biot 1962; Levinson 1968; Nowinski 1969) or a cylindrical shell under axial load. Using the theory of incremental deformations around a large deformation-stressed state, the buckling instability can be recovered by a bifurcation argument, usually referred to, in the nonlinear elasticity theory, as small-on-large, or incremental, theory. By taking the proper asymptotic limit for long slender structures, the Euler criterion can then be recovered. In comparison to the one- and two-dimensional theories, this computation is rather cumbersome as it is based on non-trivial tensorial calculations, but it contains much information about the instability and the unstable modes selected in the bifurcation process.
Here, we are concerned with the case of a cylindrical shell under axial load. This problem was first addressed in the framework of nonlinear elasticity in a remarkable 1955 article by Wilkes who showed that the linearized system around a finite axial strain can be solved exactly in terms of Bessel functions. While Wilkes only analysed the first axisymmetric mode (n=0, see below), he noted in his conclusion that the asymmetric mode (n=1) corresponds to the Euler strut and doing so, opened the door to further investigation by Fosdick & Shield (1963), who recovered Euler's criterion asymptotically from the solution of Wilkes. These initial results constitute the basis for much of the modern theory of elastic stability of cylinders within the framework of three-dimensional nonlinear elasticity (Haughton & Ogden 1979a,,b; Simpson & Spector 1984; Duka et al. 1993; Pan & Beatty 1997; Bigoni & Gei 2001; Dorfmann & Haughton 2006). The experimental verification of Euler's criterion was considered by Southwell (1932) and by Beatty & Hook (1968).
The purpose of this article is threefold. First, we revisit the problem of the stability of an incompressible cylindrical shell under axial load using the Stroh formalism (Stroh 1962) and, based on the solution of Wilkes, we derive a new and compact formulation of the bifurcation criterion that can be used efficiently for numerical approximation of the bifurcation curves for all modes. Second, we use this formulation to obtain nonlinear corrections of Euler's criterion for arbitrary shell thickness and third-order elasticity. Third, we consider the problem of determining the critical aspect ratio where there is a transition between buckling and barrelling.
## 2. Large deformation
We consider a hyperelastic homogeneous incompressible cylindrical tube with isotropic cross sections of initial inner radius A, outer radius B and length L, subjected to a uniaxial constant strain λ3 and deformed into a shorter tube with current dimensions a, b and l. The deformation bringing a point at (R, Θ, Z), in cylindrical coordinates in the initial configuration, to (r, θ, z) in the current configuration is(2.1)where and λ3=l/L. Since the material is isotropic in the cross sections, the physical components of the corresponding deformation gradient F are(2.2)showing that the principal stretches are the constants λ1, λ2=λ1, λ3; and that the pre-strain is homogeneous. Owing to incompressibility, det F=1, so that(2.3)The principal Cauchy stresses required to maintain the pre-strain are (Ogden 1984)(2.4)(no sum), where p is a Lagrange multiplier introduced by the internal constraint of incompressibility and W is the strain energy density (a symmetric function of the principal stretches). In our case, σ2=σ1 because λ2=λ1. Also, σ1=0 because the inner and outer faces of the tube are free of traction. It follows that:(2.5)where Wi≡∂W/∂λi, and we conclude that the principal Cauchy stresses are constant.
## 3. Instability
To perform a bifurcation analysis, we take the view that the existence of small deformation solutions in the neighbourhood of the large pre-strain signals the onset of instability (Biot 1965).
### (a) Governing equations
The incremental equations of equilibrium and incompressibility can be written as (Ogden 1984)(3.1)where is the incremental nominal stress tensor and u is the infinitesimal mechanical displacement. They are linked by(3.2)where is the increment in the Lagrange multiplier p and 0 is the fourth-order tensor of instantaneous elastic moduli. This tensor is similar to the stiffness tensor of linear anisotropic elasticity, with the differences that it possesses only the major symmetries, not the minor ones, and that it reflects strain-induced anisotropy instead of intrinsic anisotropy. Its explicit non-zero components in a coordinate system aligned with the principal axes are (Ogden 1984)(3.3)(no sums), where . Note that some of these components are not independent because here λ1=λ2. In particular, we have(3.4)
### (b) Solutions
We look for solutions that are periodic along the circumferential and axial directions, and have yet unknown variations through the thickness of the tube, so that our ansatz is(3.5)where n=0, 1, 2, … is the circumferential number; k is the axial wavenumber; the subscripts (r, θ, z) for u and refer to components within the cylindrical coordinates (r, θ, z); and all upper-case functions are functions of r alone.
The specialization of the governing equations (3.1) to this type of solution has already been conducted in several articles (see Wilkes 1955; Fosdick & Shield 1963; Mack 1989; Pan & Beatty 1997; Negron-Marrero 1999; and Dorfmann & Haughton 2006 for the compressible counterpart). Here we adapt the work of Shuvalov (2003a) on waves in anisotropic cylinders to develop a Stroh-like formulation of the problem (Stroh 1962). The central idea is to introduce a displacement–traction vector,(3.6)so that the incremental equations can be written in the form(3.7)where G is a 6×6 matrix, with the block structure(3.8)Here the superscript ‘+’ denotes the Hermitian adjoint (transpose of the complex conjugate) and G1, G2 and G3 are the 3×3 matrices(3.9)respectively, with(3.10)
As it happens, there exists a set of six explicit Bessel-type solutions to these equations when n≠0. This situation is in marked contrast with the corresponding set-up in linear anisotropic elastodynamics, where explicit Bessel-type solutions exist only for transversely isotropic cylinders with a set of four linearly independent modes and do not exist for cylinders of lesser symmetry (Martin & Berger 2001; Shuvalov 2003b). As mentioned in §1, the six Bessel solutions are presented in the article by Wilkes (1955; for a derivation see Bigoni & Gei 2001).
First, denote by , the roots of the following quadratic in q2:(3.11)Then the roots of this quartic in q are ±q1 and ±q2, and it can be checked that the following two vectors are solutions to (3.7):(3.12)where q=q1, q2 in turn and In is the modified Bessel function of order n. Similarly, we checked that the following vector η(3):(3.13)is also a solution when q3 is the positive root of the quadratic equation(3.14)Finally, we also checked that the vectors η(4), η(5) and η(6), obtained by replacing In with the modified Bessel function Kn in the expressions above, are solutions too.
Next, we follow Shuvalov (2003a) and introduce (r) as a fundamental matrix solution to (3.7):(3.15)It clearly satisfies(3.16)Let M(r,a) be the matricant solution to (3.7), i.e. the matrix such that(3.17)It is obtained from (r) (or from any other fundamental matrix made of linearly independent combinations of the η(i)) by(3.18)and it has the following block structure:(3.19)say.
### (c) Boundary conditions
Some boundary conditions must be enforced on the top and bottom faces of the tubes. Considering that they remain plane (Uz=0 on z=0, l) and free of incremental shear tractions ( on z=0, l) leads to(3.20)where m=1, 2, 3, … but, since the equations depend only on k, we can take m=1 without loss of generality.
The other boundary conditions are that the inner and outer faces of the tube remain free of incremental tractions. We call the traction vector and the displacement vector. We substitute the condition S(a)=0 into (3.17) and (3.19) to find the following connection:(3.21)is the (Hermitian) 3×3 impedance (Shuvalov 2003a). Since S(b)=0, a non-trivial solution exists only if the matrix z(b, a) is singular, which implies the bifurcation condition(3.22)This is a real equation since z=z+ (Shuvalov 2003a) that applies independently of the nature (i.e. real or complex (Pan & Beatty 1997), simple or double (Dorfmann & Haughton 2006)) of the roots q1, q2 and q3.
We are now in a position to use the bifurcation condition (3.22) to compute explicitly bifurcation curves for each mode n. We note that the components of 0 depend on the strain energy density W and on the pre-strain, which by (2.3) depends only on λ3; so do q1, q2, q3, by (3.11) and (3.14). According to (3.12) and (3.13), the entries of M(b, a) thus depend (for a given W) on λ3, n, ka and kb only. For a given material (W specified) with a given thickness (b/a=B/A specified), the bifurcation equation (3.22) gives a relationship between a measure of the critical pre-stretch: , and a measure of the tube slenderness: , for a given bifurcation mode (n specified). That is, for a given tube slenderness, what is the axial strain necessary to excite a given mode?
While this bifurcation condition is formally clear, it has not been successfully implemented to compute all bifurcation curves. Indeed, for mode n>1, the root finding of det(z) becomes numerically unstable and numerical methods become unreliable (as observed in Dorfmann & Haughton (2006) for a similar problem) and, in explicit computations, most authors do not use the exact solution by Wilkes but use a variety of numerical techniques to solve the linear boundary-value problems directly (such as the compound matrix method (Haughton & Orr 1997), the determinantal method (Ben Amar & Goriely 2005) or the Adams–Moulton method (Zhu et al. 2008)). Note that from a computational perspective, the Stroh formalism is particularly well suited and well behaved (Biryukov 1985; Fu 2005) and if numerical integration was required it would provide an ideal representation of the governing equation.
Rather than integrating the original linear problem numerically, we now show how to use an alternative form of (3.22) to compute all possible bifurcation curves. This method bypasses the need for numerical integration and reduces the problem to a form that is manageable both numerically and symbolically, to study analytically particular asymptotic limits. The main idea is to transform condition (3.22) by factoring non-vanishing factors. We start by realizing that since the fundamental solutions {η(i), i=1, …, 6} are linearly independent, the matrix (r) is invertible for all r∈[a,b], which implies that the elements of M(r, a) are bounded for r∈[a,b]. Therefore, det(M1(r, a)) is uniformly bounded away from zero and det z=0 implies det(M3(b, a))=0. Instead of expressing det(M3(b, a)) as the determinant of a 3×3 submatrix of a matrix obtained as the product of two 6×6 matrices, we first decompose (r) as(4.1)say, where each block is a 3×3 matrix. We also rewrite equation (3.18) as(4.2)and write explicitly the two entries 3(r) and 4(r), which are(4.3)(4.4)which implies(4.5)Using again the fact that the entries of are bounded, we have that the bifurcation condition det(M3(b, a))=0 implies that(4.6)where(4.7)and adj(A) is the adjugate matrix of A, i.e. the transpose of the cofactor matrix (which in the case of an invertible matrix is simply adj(A)=det(A)A−1). This new bifurcation condition is equivalent to the previous one but has many advantages. The matrix Q involves only products of 3×3 matrices and is polynomial in the entries of , i.e. det Q(b, a) is a polynomial of degree 18 in Bessel functions and has no denominator (hence no small denominator). Both numerically and symbolically, this determinant is well behaved, even in the limits a→0, which corresponds to a solid cylinder, and n=0, which corresponds to the first barrelling mode (and usually requires a special treatment). We will refer to the use of this form of the bifurcation condition as the adjugate method.
### (a) Numerical results
As a first test of the stability of the numerical procedure, we consider a neo-Hookean potential , where we set C1=1 without loss of generality and consider the typical value B/A=2. We compute the critical value of λλ3 as a function of the current stubbiness kb=πb/l (the initial stubbiness is ) for the first 9 modes (n=0–8) as shown in figure 2. The known classical features of the stability problem for the cylindrical shell are recovered, namely for slender tubes, the Euler buckling (n=1) is dominant and becomes unavoidable as the slenderness increases; there is a critical slenderness value at which the first barrelling mode n=0 is the first unstable mode (in a thought experiment where the axial strain would be incrementally increased until the tube becomes unstable); and for very large kb, the critical compression ratio tends asymptotically to the value λ=0.444, which corresponds to surface instability of a compressed half-space (Biot 1962).
Figure 2
Bifurcation curves (stretch as a function of stubbiness) of a homogeneous neo-Hookean cylindrical tube for modes n=0–8 with b/a=B/A=2 and C1=1.
For a second test, we consider very thin neo-Hookean tubes with B/A=1.01. Here we are interested in the mode selection process. As the stubbiness increases, the buckling mode rapidly ceases to be the first excited mode and is replaced by different barrelling modes. From figure 3, it appears clearly that as kb increases, modes n=1–9 are selected (modes n=0 and 10 remain unobservable). There is one particularly interesting feature in these two sets of bifurcation curves. Depending on both the tube thickness and the stubbiness, the instability mode of a tube transition occurs from buckling to barrelling, the material transition from either the one-dimensional behaviour of slender column to the two-dimensional behaviour of a thin short tube, or the three-dimensional behaviour of a thick short tube. Accordingly, we will refer to these particular geometric values where transition occurs as dimensional transitions and obtain analytical estimates for them in §5.
Figure 3
Bifurcation curves (stretch as a function of stubbiness) of a homogeneous neo-Hookean cylindrical tube for modes n=0–10 with b/a=B/A=1.01 and C1=1.
## 5. Asymptotic Euler buckling
We are now in a position to look at the asymmetric buckling mode (n=1) corresponding to the Euler buckling in the limit λ→1. The asymptotic form of the Euler criterion cannot be obtained for a general strain-energy density. This is why we choose the Mooney–Rivlin potential, which, for λ close to 1, corresponds to the most general form of third-order incompressible elasticity (see §6 and Rivlin & Saunders (1951)),(5.1)where C1≥0 and C2>0 are material constants; ; and . Close to λ=1, we introduce a small parameter related to the stubbiness ratio(5.2)and look for the critical buckling stretch λ as a function of ϵ of order M,(5.3)Similarly, we expand in powers of ϵ,(5.4)and solve each order dm=0 for the coefficients λm. This is a rather cumbersome computation. It can be checked that λm vanishes identically for all odd values of m and that the first non-identically vanishing coefficient appears at order 24. A computation to order 28 is necessary to compute the correct expression for a, which is found to be to order 6 in ϵ(5.5)with(5.6)(5.7)(5.8)where ρB/A=b/a. It is of interest to compare the different approximations. We recover the Euler formula by keeping only the term up to ϵ2, which we denote by Euler2. We define similarly Euler4 and Euler6 by keeping terms up to orders 4 and 6 in ϵ. We show the different approximations as a function of ϵ2 for ρ=1.01 (figure 4a) and ρ=10 (figure 4b). The classical Euler formula is well recovered in the limit ϵ→0, but the Euler4 and Euler6 approximations clearly improve the classical formula for larger values of ϵ. It also appears from the analysis of Euler4 that for C2≥0 the classical Euler formula always underestimates the critical stretch for instability.
Figure 4
Comparison of the different Euler formulae obtained by expanding the exact solution to order 2 (the classical Euler buckling formula), Euler4 and Euler6 for a neo-Hookean potential C1=1, C2=0. For comparison purpose, we show the critical stretch for mode n=1 versus ϵ2 in which case the graph becomes linear in the limit ϵ→0. (a) ρ=b/a=B/A=1.01, (b) ρ=b/a=B/A=10.
## 6. Nonlinear Euler buckling for third-order elasticity
The analytical result presented in §5 was formulated in terms of parameters and quantities natural for the computation and the theory of nonlinear elasticity. In order to relate this result to the classical form of Euler buckling, we need to express equation (5.5) in terms of the initial geometric values A, B, L, the axial load acting on the cylinder and the elastic parameters entering in the theory of linear elasticity.
We first consider the geometric parameters. We wish to express the critical load as a function of the initial stubbiness ν=B/L and tube relative thickness ρ=B/A. Recalling that ϵ=πb/l and λ=l/L, b=λ−1/2B, we have(6.1)To express ϵ as a function of ν, we expand ϵ in powers of ν to order 6, and solve (6.1) to obtain(6.2)where λ(2) and λ(4) are defined in (5.6) and (5.7) and come from the expansion of λ in powers of ϵ.
Second, we want to relate the axial compression to the actual axial load N. To do so, we integrate the axial stress over the faces of the tubes, i.e.(6.3)Since σ3 is constant and given by (2.5), we have(6.4)
Third, we relate the elastic Mooney parameters C1 and C2 to the classical elastic parameters. Here, we follow Hamilton et al. (2004; Destrade & Saccomandi 2005) and write the strain-energy density to third order for an incompressible elastic material as(6.5)where μ is the usual shear modulus, or second Lamé parameter, and n3 is a third-order elasticity constant; μ is related to Young's modulus by E=3μ; also, in Murnaghan's notation, n3=n and in Landau's notation, n3=A (see Norris (1998) for other notations). In (6.5), i1, i2, i3 are the first three principal invariants of the Green–Lagrange strain tensor, related to the first three principal invariants I1, I2, I3 of the Cauchy–Green strain tensor by(6.6)Since I3=1, we can solve this linear system for i2 and i3 and write the strain-energy density (6.5) as a function of I1 and I2, i.e.(6.7)which by comparison with (5.1) leads to(6.8)
To write the nonlinear buckling formula, we consider (6.4) and first expand λ in ϵ using (5.5), then expand ϵ in ν using (6.2) and, finally, substitute the values of the moduli in terms of the elastic parameters, which yield(6.9)While it is not surprising, it is comforting to recover to order ν2 the classical Euler buckling formula (1.1) (using ρ=B/A, ν=B/L and μ=E/3).
## 7. Dimensional transition
Finally, we use the buckling formula to compute the transition between modes as parameters are varied. That is, to identify both the geometric values and the axial strain for which there is a transition between buckling and barrelling modes. Here we restrict again our attention to the neo-Hookean case (with C1=1). From figures 2 and 3, it appears clearly that for ϵ small enough there is a transition (depending on the value of ρ) from either mode n=1 to mode n=0 (large ρ), or from mode n=1 to mode n=2 (ρ close to 1) as ϵ increases. We refer to this transition as a dimensional transition, in the sense that the material mostly behaves as a slender one-dimensional structure when it buckles according to mode n=0 and mostly as a two-dimensional structure when it barrels with mode n=2. Indeed both modes of instability can be captured by, respectively, a one- or a two-dimensional theory. For ρ close to unity, the transition n=0→n=1 occurs for small values of ϵ. Therefore, in this regime, we can use the approximation (5.5) for the barrelling curve and substitute it in the bifurcation condition of mode n=2. Expanding again this bifurcation condition in ϵ as well as ρ, one identifies the values ρt of ρ and λt of λ at which the transition occurs, as(7.1)In terms of the initial stubbiness ν=B/L, we have(7.2)This relationship also provides a domain of validity for the Euler buckling formula. For sufficiently slender tube (ν small), the buckling mode disappears when ρ>ρt at the expense of the n=2 barrelling mode. For stubbier and fuller tubes, this approximation cannot be used. To understand the dimensional transition, we solve numerically the bifurcation condition, using the adjugate method, for the intersection of two different modes. That is, for a given value of ρ*, we find the value of ϵ* such that both the bifurcations for either modes n=1 and 2, or modes n=1 and 0 are satisfied. If the corresponding value λ* is the largest value for which a bifurcation takes place, the pair (ϵ*, ρ*) is a transition point. The corresponding transition point in terms of the initial parameters is . In figure 5, we show a diagram of all such pairs for both transitions.
Figure 5
Dimensional transition for a neo-Hookean cylindrical tube of initial length L and initial radii A and B. All tubes in the n=1 regions will become unstable by buckling. As the tubes get stubbier or thinner (arrows), it will not buckle but instead will be subjected to a barrelling instability. Note that only the transition curves from mode n=0 are shown. Tubes in the barrelling regions may be subjected to other unstable modes.
## 8. Conclusion
This article establishes a reliable and effective method to study the stability of tubes based on the exact solution of the incremental equations proposed by Wilkes (1955) within the Stroh formalism. It then puts the method to use, to obtain the first geometric and material corrections to the Euler buckling. The method can be also used to obtain the transition between buckling and barrelling modes when a tube becomes unstable.
The method presented here can be easily generalized to different materials and different boundary conditions. For instance, using the exact solution of the incremental equations proposed in Dorfmann & Haughton (2006) for compressible materials and the adjugate method, an explicit form of the bifurcation condition in terms of Bessel functions can be obtained by following the steps presented here and various asymptotic behaviours can be obtained. Similarly, a variety of boundary-value problems can be analysed by the adjugate method, such as the stability problem of a tube under pressure and tension (Han 2007; Zhu et al. 2008), the problem of a tube embedded in an infinite domain (Bigoni & Gei 2001) and the problem of a tube with coating (Ogden et al. 1997). In all these cases, useful asymptotic formulae for the buckling behaviour could be obtained by perturbation expansions.
It is also enticing to consider the possibility of performing an analytical post-buckling analysis of the solutions. Since the solutions of the linearized problem can be solved exactly, a weakly nonlinear analysis of the solution should be possible to third order. This would yield, in principle, an equation for the amplitude of the unstable modes containing much information not only about the actual amplitude of the unstable modes but also on the localization of unstable modes after bifurcation. We leave this daunting task for another day.
## Acknowledgments
This study is based in part upon work supported by the National Science Foundation under grant no. DMS-0604704 (A.G.) and made possible by a CNRS/USA Collaborative Grant from the French Centre National de la Recherche Scientifique (M.D.).
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https://meridian.allenpress.com/radiation-research/article-abstract/72/1/81/46467/The-Reaction-of-the-Hydrated-Electron-with-Amino?redirectedFrom=fulltext
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The reaction rate constants of <tex-math>$e{}_{{\rm aq}}{}^{-}$</tex-math> with glycyl-tryptophan, glycyl-glycyl-tryptophan, and glycyl-tryptophyl-glycine were determined. The rate constant of each peptide was found to be pH dependent. This is attributed to the overall charge of the molecule. It was found that the increase of carbonyl groups increases the <tex-math>$e{}_{{\rm aq}}{}^{-}$</tex-math> reactivity toward the peptide. The transient absorbing spectra resulting from the reaction of <tex-math>$e{}_{{\rm aq}}{}^{-}$</tex-math> with tryptophyl peptides showed the presence of an electron adduct on the indole and a deaminated radical. The yield of the last species was found to be on the order of 40%. The OH reaction with glycyl-tryptophan was also investigated. It was shown that its reaction rate constant and the transient produced are similar to that observed with tryptophan.
This content is only available as a PDF.
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2021-02-24 20:15:55
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http://mathhelpforum.com/business-math/202284-interest-rates-calculation-how-does-work-out.html
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# Math Help - interest rates calculation: how does this work out?
1. ## interest rates calculation: how does this work out?
Hi
If you take a loan of $100 at 5% compounded interest per year, and pay for it over 12 months, you'd have paid the following: Monthly Payments -$8.56
Total Payments - $102.73 That is clearly less than 5% interest. In fact, intuitively, you'd have expected to pay > 5% in effective interest rate. Yet you don't even pay 5% in interest! How does this work out?? 2. ## Re: interest rates calculation: how does this work out? if you took out the loan and paid nothing till the end of the year, then you pay$105 at the end of the year.
by making regular monthly payments you decrease the principal owed, hence the interest is calculated on a regularly decreasing amount.
3. ## Re: interest rates calculation: how does this work out?
This is what the "loan account" looks like:
Code:
MN PAYMENT INTEREST BALANCE
00 100.00
01 -8.56 .42 91.86
02 -8.56 .38 83.68
....
11 -8.56 .07 8.52
12 -8.56 .04 .00
Add up the 12 "interest" amounts (get the other 8 yourself!) and you'll get 2.73.
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2014-09-19 14:20:47
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http://www.hpmuseum.org/forum/thread-785.html
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WP-34S Polar to Rectangular (no quadrant checking)
Post: #1
BarryMead Senior Member Posts: 350 Joined: Feb 2014
WP-34S Polar to Rectangular (no quadrant checking)
Many HP calculators include special quadrant checking for the 0, 90, 180, 270 degree quadrant values when performing the Polar to Rectangular conversion. For instance if you have 90 degrees in the
Y register and 1 in the X register and press "f" "R<" (Polar to Rectangular) you get 3.799 E -38 for X and 1 for Y on the WP-34s (assuming Degrees mode), but on an HP-15C you get 0 (zero) for X and 1 for Y.
If the calculator were in GRADS mode these values would be 0, 100, 200, 300. Most HP calculators don't attempt to match quadrant angular values while in the RAD mode.
Adding special checks for these quadrant values could cost valuable flash memory space, but since the existing SIN and COS functions already include these special quadrant checks, it may be possible to fix the Polar to Rectangular function by using existing SIN and COS routines, hopefully eliminating the need for additional quadrant checks.
02-28-2014, 12:04 AM
Post: #2
Paul Dale Senior Member Posts: 1,203 Joined: Dec 2013
RE: WP-34S Polar to Rectangular (no quadrant checking)
As a space saving measure these routines simply vector through the complex rectangular/polar conversions which assume radians and convert back afterwards.
It wouldn't be too hard to rework rectangular -> polar to using the trig routines that know about angular mode -- it will be slower since the dn_sincos() routine calculates both SIN and COS of its argument in radians together. There is no equivalent routine that knows about angular mode.
Another alternative, would be to move these two functions into XROM. I don't see any direct call to them or references from inside XROM so it should be possible. This would save some space too I imagine but at a cost of performance and accuracy in DP mode.
- Pauli
Post: #3
BarryMead Senior Member Posts: 350 Joined: Feb 2014
RE: WP-34S Polar to Rectangular (no quadrant checking)
The "Rectangular to Polar" works just fine. It is only the "Polar to Rectangular" that has problems with three of the four quadrants! At angles 90, 180 and 270 the value that should return as zero comes back as a tiny value like 3.799 E -39 instead of zero. Since all other HP calculators that I can run tests on do the Polar to Rectangular conversion correctly (only the WP-34S has these inaccuracies), and since these four angles are more common than others, I would think that people would want this to be fixed. The new Polar to Rectangular function would be: NewX = X COS(Y) and NewY = X SIN(Y), using the existing SIN and COS functions that already handle these special quadrant angles properly.
02-28-2014, 06:18 AM
Post: #4
walter b On Vacation Posts: 1,968 Joined: Dec 2013
RE: WP-34S Polar to Rectangular (no quadrant checking)
(02-28-2014 12:33 AM)BarryMead Wrote: The new Polar to Rectangular function would be: NewX = X COS(Y) and NewY = X SIN(Y), using the existing SIN and COS functions that already handle these special quadrant angles properly.
Good idea Thanks for sharing.
d:-)
02-28-2014, 07:55 AM (This post was last modified: 02-28-2014 08:27 AM by Paul Dale.)
Post: #5
Paul Dale Senior Member Posts: 1,203 Joined: Dec 2013
RE: WP-34S Polar to Rectangular (no quadrant checking)
I think the XROM code would be:
Code:
XLBL "RECTPOLAR" XIN DYADIC RCL Y SIN RCL[times] Y [<->] YZXX COS [times] XOUT xOUT_NORMAL
It will be a lot slower than the C version and less accurate in DP mode but should deal with these orthogonal cases properly.
I'm not implementing and debugging this tonight however.
- Pauli
Post: #6
BarryMead Senior Member Posts: 350 Joined: Feb 2014
RE: WP-34S Polar to Rectangular (no quadrant checking)
Wouldn't something like this work equally well in DP mode, be fast, and use less memory since the
function cmplxFromPolar could be eliminated.
Code:
void op_p2r(enum nilop op) { decNumber x, y, t, range, angle; getXY(&range, &angle); decNumberCos(&t, &angle); dn_multiply(&x, &t, &range); decNumberSin(&t, &angle); dn_multiply(&y, &t, &range); setlastX(); setXY(&x, &y); #ifdef RP_PREFIX RectPolConv = 2; #endif }
03-01-2014, 11:26 PM
Post: #7
Paul Dale Senior Member Posts: 1,203 Joined: Dec 2013
RE: WP-34S Polar to Rectangular (no quadrant checking)
It is possible to implement a C version of course, xrom routines are genearlly easier to debug
I've no certainty about the size difference without building both versions however I suspect your code will be slightly larger the cmplxFromPolar routine will have been inlined so there is no call there to save. Your code will be slow, the SIN and COS are by far the most expensive operations and they are being done separately rather than together -- the modulo reduction is quite a bit bit of these.
Will try to get a look at it this weekend.
- Pauli
03-02-2014, 01:52 AM
Post: #8
Paul Dale Senior Member Posts: 1,203 Joined: Dec 2013
RE: WP-34S Polar to Rectangular (no quadrant checking)
This code change is in. Nothing is built yet.
- Pauli
03-02-2014, 03:03 PM
Post: #9
Marcus von Cube Senior Member Posts: 745 Joined: Dec 2013
RE: WP-34S Polar to Rectangular (no quadrant checking)
(03-02-2014 01:52 AM)Paul Dale Wrote: Nothing is built yet.
Really?
Marcus von Cube
Wehrheim, Germany
http://www.mvcsys.de
http://wp34s.sf.net
http://mvcsys.de/doc/basic-compare.html
03-02-2014, 07:22 PM
Post: #10
pascal_meheut Member Posts: 124 Joined: Dec 2013
RE: WP-34S Polar to Rectangular (no quadrant checking)
At least the emulators are. I'll fix the iOS version later.
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2017-10-21 10:33:39
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https://www.techwhiff.com/issue/all-the-expression-that-are-equivlent-to-18m-90--498214
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All the expression that are equivlent to 18m - 90
Question:
All the expression that are equivlent to 18m - 90
Increasing the volume of an acid solution _______the p H of the solution ______ the pKa of the acid
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What do a rectangle and a rhombus have in common? Select all that apply. The opposite sides are parallel. They have four right angles. Their angle measures add to 360°. They have four congruent sides. you can pick more than one so plz pick more than one be positive
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Which of the following was not a characteristic of owners and managers of businesses? O A. Work was safe O B. College educated O C. Work was comfortable O D. Mostly women
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I need help on math plz help me fast
i need help on math plz help me fast...
Hormones what they do their target organ what organ produces
Hormones what they do their target organ what organ produces...
8. A polyhedron has 15 edges and 10 vertices. How many faces does ithave? *
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Which of the following pairs of terms identify spaces that are roughly PERPENDICULAR (at right angles to one another) in the human brain (give or take 30 degrees or so)? temporal horn of the lateral ventricle and cerebral aqueduct temporal horn of the lateral ventricle and central part (“body”) of the lateral ventricle precentral sulcus and postcentral sulcus inferior frontal sulcus and inferior temporal sulcus central sulcus and the parieto-occipital sulcus superior frontal sulcus and inferior
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I need the answer!!i can’t find it anywhere
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What is the measure of angle x?
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When assessing energy resources, it is helpful to use a measure called EROI, which is:__________ a) energy returned minus energy invested b) energy returned plus energy invested c) amount of energy invested minus heat released into the environment d) money invested in extraction and processing minus money in sales e) energy returned divided by energy invested
When assessing energy resources, it is helpful to use a measure called EROI, which is:__________ a) energy returned minus energy invested b) energy returned plus energy invested c) amount of energy invested minus heat released into the environment d) money invested in extraction and processing minu...
What term(s) best describes enzymes? O lipids proteins catalysts both b and c
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3v-9=7+2v-v$3v - 9 = 7 + 2b - v$
3v-9=7+2v-v$3v - 9 = 7 + 2b - v$...
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What did Napoleon Bonaparte take advantage of in his rise to power?
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Why does the graph show a temperature of 16o C at 0 meters of depth ?
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A survey was taken of 2,000 voters before an election. 1,650 of the ones surveyed said they would vote for candidate A. what percent of those surveyed planned to vote candidate A?
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2022-11-27 01:45:39
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https://www.physicsforums.com/threads/group-theory.751240/
|
# Group theory
1. Apr 29, 2014
### tableshark
Can one set up a multiplication table for the symmetry group C3V of the equilateral
triangle.
Then show that it is identical to that of the permutation group P3.
I need some clarification....
What about a matrix representation (2x2) for these groups?
→ Here was thinking to use Cartesian coordinate
system in the xy plane, with origin at the center of the triangle.
2. Apr 29, 2014
### Zondrina
The symmetry group of the equilateral triangle is indeed isomorphic to the dihedral group of order six (degree three) $D_3$.
It is also isomorphic to $SL(2, 2)$ I believe.
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2018-02-20 14:42:31
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https://www.gradesaver.com/textbooks/math/algebra/algebra-2-common-core/skills-handbook-simplifying-expressions-with-integers-exercises-page-975/8
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## Algebra 2 Common Core
Published by Prentice Hall
# Skills Handbook - Simplifying Expressions With Integers - Exercises - Page 975: 8
#### Answer
$-14$
#### Work Step by Step
To add two numbers with different signs, we find the difference between their absolute values, and give the sum the sign of the number with the greater absolute value: $$-19+5=-(|-19|-|5|)=-(19-5)=-14$$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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2018-12-18 15:36:26
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https://ijnaa.semnan.ac.ir/?_action=article&kw=840&_kw=integral+equations
|
### International Journal of Nonlinear Analysis and Applications
##### Solutions of integral equations via fixed point results in extended Branciari b-distance spaces
Volume 13, Special Issue for selected papers of ICDACT-2021, March 2022, Pages 17-29
Usha Bag; Reena Jain
##### Generalized dynamic process for generalized $(\psi, S,F)$-contraction with applications in $b$-Metric Spaces
Volume 12, Issue 2, July 2021, Pages 1947-1964
Akindele Adebayo Mebawondu; Oluwatosin Temitope Mewomo
##### Endpoint results for weakly contractive mappings in $\mathcal{F}$-metric spaces with application
Volume 11, Issue 2, May 2020, Pages 351-361
Hüseyin Işık; Nawab Hussain; Abdul Rahim Khan
##### Existence and uniqueness of the solution for a general system of operator equations in $b-$metric spaces endowed with a graph
Volume 8, Issue 2, December 2017, Pages 263-276
Cristian Chifu; Gabriela Petrusel
##### Modified homotopy method to solve non-linear integral equations
Volume 6, Issue 2, January 2015, Pages 133-136
Mohsen Rabbani
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2022-10-01 08:03:57
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https://pos.sissa.it/317/024/
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Volume 317 - The 9th International workshop on Chiral Dynamics (CD2018) - Session: Hadron Structure
The proton radius: are we still puzzled?
E. Downie* on behalf of the MUSE Collaboration
Full text: pdf
Published on: February 28, 2020
Abstract
The proton radius puzzle began in 2010 when the CREMA Collaboration released their measurement of the proton radius from muonic hydrogen spectroscopy: $r_{p}=0.84184(67)$ fm, This was five standard deviations smaller that the accepted CODATA value of that time (0.8768(69) fm), and sparked an enduring and intriguing puzzle.
Since that first measurement, many theories have been formed as to why the proton radius appears smaller when measured with muons than it does when measured with electrons. Many new measurements have also been made in an attempt to resolve the puzzle. We will detail the current status of the proton radius puzzle, and review future experimental plans, with a focus on the MUon proton Scattering Experiment (MUSE).
DOI: https://doi.org/10.22323/1.317.0024
How to cite
Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.
Open Access
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2021-10-26 11:25:05
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https://kozikow.com/2016/05/21/very-powerful-data-analysis-environment-org-mode-with-ob-ipython/
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# Very powerful data analysis environment – org mode with ob-ipython
## Introduction
Emacs org-mode with ob-ipython is the most powerful data analysis environment I ever used. I find it much more powerful than other tools I used, including jupyter and beaker web notebooks or just writing python in PyCharm.
Emacs org mode with ob-ipython is like jupyter or beaker notebook, but in Emacs instead of browser and with many more features.
Word “Emacs” may be scary. There are pre-packaged and pre-configured emacs distribution that have much smaller learning curve, my favorite being Spacemacs (I am in progress of rebasing my config with it). You can just use 1% of capabilities of Emacs (probably majority of Emacs users do not approach 10% of Emacs capabilities) and still benefit from it.
If you are going to bring up the common quote of “emacs is fine operating system, but it lacks decent text editor” – Emacs now have decent text editor by using the vim emulation evil-mode. It’s the best vim emulation in existence and even many packages from vim are ported. Spacemacs is a nice emacs distribution that bundles evil mode.
I will try to introduce and describe org mode with ob-ipython it for users who never used Emacs before.
Since this blog post have been written in org mode, linear reading experience in exported format is less optimal experience than reading the org mode file directly in org mode.
## Features (aka “What’s that powerful about it”)
### Embed code blocks in any language
You can embed embeded source code and evaluate it with C-c C-c. Results of evaluation of your source code are appended after the source code block. Result can be text (including org table) or image (charts).
What’s more You can have separate org file and ipython console open side by side. With ipython, reading python docstrings and code completion works well. See my screenshot.
Since ob-ipython uses jupyter, you can get the same environment for anything that have jupyter kernel, including matlab, Scala, Spark or R and many more.
### Results can be exported to many formats, like latex (demo) or this post.
This blog post is just an export of org mode file via org2blog. All code examples have been written in org mode using workflow described in this post.
Exporting works to formats like html, latex (native and beamer), markdown, jira, odt (than can be imported to google docs and word), wiki formats and many more.
Syntax highlighting can be preserved for some exports, like html or latex.
You can just learn one way to edit documents and presentations than can be exported to majority of formats on earth.
### Programmable documents (aka “Literate programming”)
Emacs org mode with org babel is a full fledged literate programming environment. Some people have published whole books or research papers as a large executable document in org. There is an even Research paper about it.
Python computations in science and engineering book supports org mode and it’s far better book reading experience than anything I ever experienced before. I can tweak and re-run code examples, link from my other notes, tag or bookmark interesting sections, jump between sections and many more.
When writing some latex in college, I recall situations when I am half way through writing latex document. I would came up with the idea of some parameter tweak, and suddenly I have to re-generate all charts.
With org mode, the document is generated pragmatically. Not only you can easily re-generate it, but readers of your document can tweak parameters or supply their own data set and re-generate the whole document.
Another example is training machine model. You can define your model parameters as org constants. You can tweak some model parameter and have separate org mode headings for things like “performance statistics”, “top miss-classified cross validation samples”, etc. Added benefit is that you can commit all this to git.
As soon as you learn org mode all of it is easy and seamless.
### Built in excel alternative
Sometimes just “manually” editing the data is the most productive thing to do. You can do it with org mode spreadsheet capabilities on org tables.
The added benefit is that formulas are written in lisp, that is cooler and more powerful language than Visual basic. http://orgmode.org/manual/Translator-functions.html
#### Integration with pandas
My current Table->Pandas->Table workflow works. It is somewhat clunky, but it can be improved. See examples section.
#### Integration with other formats
You can export org tables to many formats by exporting it to pandas and then using pandas exporter. Nevertheless, org supports sql, csv, latex, html exporters.
### Pass data between languages
Similar functionality is offered by beaker notebook.
I found out that org mode as intermediate format for data sometimes works better for me.
Since intermediate format for a data frame is the org table, I can import data frame to org, edit it as spreadsheet and export it back. See Pass data directly between languages in examples section.
### Outline view is powerful for organizing your work
Org mode outline view is very handy for organizing your work. When working on some larger problem, I am only focusing on small subset of it. Org mode lets me just expand sections that are currently relevant.
I also find adding embedding TODO items in the tree quite handy. When I encounter some problem I mark a subtree as TODO, and I can later inspect just subtree headlines with TODO items with them. See:
You can link to your existing codebase with org-ctags. It seems possible to provide ide-like navigation between code defined in org src buffers, but I didn’t configure it yet.
### Many more
You don’t have to use all features offered by org mode.
#### Embed latex formulas
Also works in html export with mathjax.
#### Fast integration with source control
I like to keep my notes in source control. To avoid overheard of additional committing I use magit-mode. Out of the box you can commit directly from Emacs with 6 keyboard strokes. With a few lines of elisp you can auto generate commit messages or automatically commit based on some condition (e.g. save or file closed or focus-out-hook).
Everything in org is plain text, including results of eval of code blocks, so it will be treated well by the source control.
#### Spaced repetition framework (remember all those pesky maths formulas)
If you are like me, you forgot a lot of maths formulas since college. Spaced repetition is a learning methodology that helps you avoid forgetting important facts like maths formulas. I recommend this very good post about spaced repetition in general from gwern.
People primarily use spaced repetition for learning words in new languages, but I use it for maths formulas or technical facts.
There are spaced repitition tools like anki or super memo, but as soon as you want advanced features like latex support they support them very badly (IMO) or not at all.
org-drill is a spaced repetition framework in drill, that allows you to use all of the org features for creating flash cards. Also take a look at this interesting blog post.
#### Even more
I only mentioned some of the features I use or plan to use soon. There are many more. Some urls to look at:
## Installation
### Install Emacs (with vim emulation)
Although I don’t use it, I recommend Spacemacs, pre-configured emacs distribution, like “Ubuntu” of Emacs.
### Install python packages
If you don’t run those, you may run into troubles.
pip install --upgrade pip
### Install ob-ipython
org mode should be bundled with your emacs installation. If you are new to emacs, you can install packages using M-x package-install.
### Elisp configuration
(require ‘org)
(require ‘ob-ipython)
;; don’t prompt me to confirm everytime I want to evaluate a block
(setq org-confirm-babel-evaluate nil)
;;; display/update images in the buffer after I evaluate
## Troubleshooting
### Verify that restarting ipython doesn’t help.
(ob-ipython-kill-kernel)
### Toggle elisp debug on error
(toggle-debug-on-error)
## My workflow
I settled on workflow of having two buffers opened side by side. On one side I would have opened org file, on the other side I would the have ipython console.
I am experimenting with commands in the ipython console, and I copy back the permanent results I want to remember or share with people into the org src block.
Both windows re-use the same ipython kernel (So they share variables). You may have multiple kernels running. I have code completion and python docstrings in the ipython buffer.
### Default ipython configuration
If you want to run some code in each ipython block you can add it to ~/.ipython/profile_default/startup. Foe example, to avoid adding %matplotlib inline to each source code block:
echo "%matplotlib inline" >> ~/.ipython/profile_default/startup/66-matplot.py
### TODO Configure yasnippet
ob-ipython docs suggest yasnippet for editing code. So far I have been using custom elisp code, but a few things can be nicer about yasnippet.
# -*- mode: snippet -*-
# name: ipython block
# key: py
# —
#+BEGIN_SRC ipython :session ${1::file${2:(let ((temporary-file-directory "./")) (make-temp-file "py" nil ".png"))} }:exports ${3:both}$0
#+END_SRC
## Examples
### Org table to pandas and plotting
date x y z
<2016-06-15 Wed> 1 1 1
<2016-06-16 Thu> 2 2 2
<2016-06-17 Fri> 4 3 3
<2016-06-18 Sat> 8 4 4
<2016-06-19 Sun> 16 5 30
<2016-06-20 Mon> 32 6 40
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
%matplotlib inline
df = pd.DataFrame(table[1:], columns=table[0])
df.plot()
### Org table -> Pandas -> Org table
You have to write small reusable snippet to print pandas to org format. You can add it to your builtin ipython code snippets. You also need to tell src block to interpret results directly with :results output raw drawer :noweb yes.
def arr_to_org(arr):
line = "|".join(str(item) for item in arr)
return "|{}|".format(line)
def df_to_org(df):
return "\n".join([arr_to_org(df.columns)] +
[arr_to_org(row) for row in df.values])
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
%matplotlib inline
df = pd.DataFrame(table[1:], columns=table[0])
df.y = df.y.apply(lambda y: y * 2)
print df_to_org(df)
date x y z <2016-06-15 Wed> 1 2 1 <2016-06-16 Thu> 2 4 2 <2016-06-17 Fri> 4 6 3 <2016-06-18 Sat> 8 8 4 <2016-06-19 Sun> 16 10 30 <2016-06-20 Mon> 32 12 40
Afterwards, you may assign result table to variable, edit it with org spreadsheet capabilities and use in other python script.
### Share code between code blocks
Since all code is executed within the same ipython kernel, it’s enough to put blocks one after another.
constant = 30
def some_function(x):
return constant * x
print some_function(30)
### TODO Connect to existing ipython kernel
I added support of connecting to existing ipython kernel in https://github.com/gregsexton/ob-ipython/pull/71/files.
You can start an ipython kernel on a server with lots of ram and cpu and connect it to a local lightweight machine running emacs.
Create kernel using (outside of the org mode, as it blocks):
#!/usr/bin/env python
import os
from ipykernel.kernelapp import IPKernelApp
app = IPKernelApp.instance()
app.initialize([])
kernel = app.kernel
kernel.shell.push({'print_me': 'Running in previously started kernel.'})
app.start()
It will give you a connection json file name. Pass it as a session name.
#+BEGIN_SRC ipython :session kernel-8520.json
print print_me
#+END_SRC
Running in previously started kernel.
### TODO Different language kernels
This should work:
#+BEGIN_SRC ipython :session :kernel clojure
(+ 1 2)
#+END_SRC
#+RESULTS:
: 3
### Examples from other blog posts
C-c C-c block to open org file directly in Emacs:
## Additional configuration I plan to do
Problems I did not resolve yet:
### TODOob-ipython-inspect in popup
Currently it opens a separate buffer. I would prefer a popup.
### TODO Configure the org-edit-src-code to use ipython completion.
Currently, I have code completion only working in ipython buffer. It seems doable to configure it in the edit source block as well.
### TODO Capture results from ipython to src block.
To avoid manual copying between ipython buffer and source code block, I could implement an ob-ipython-capture function, that would add last executed command in the ipython console to the src block. Keyboard macros can work cross-buffer, so this could be simple keyboard macro, but I didn’t try it out yet.
### TODO Figure out why SVG doesn’t work
In order to make a svg graphic rather than png, you may specify the output format globally to IPython.
%config InlineBackend.figure_format = 'svg'
### Research paper: An Effective Git And Org-Mode Based Workflow For Reproducible Research
Search by DOI 10.1145/2723872.2723881 on sci hub.
### Whole research department on CMU ran on org mode
Interesting case of Chemical Engineering department on CMU managed by John R. Kitchin, ran mostly using org mode, with papers, assignments and books written in org mode.
### Org mode for managing your server configuration
“Literate devops”
### Manage your emacs configuration using org mode
1. Fanpeng says:
Thanks for this great post! Just wondering have you solved the code block completion now? I posted the issue sometimes ago (https://github.com/gregsexton/ob-ipython/issues/54) but did not receive any response. The screenshot from the official github repo suggests it should work though.
2. Dodge Coates says:
Great post. How much editing did you have to do after you exported your org document to this blog posts’ html?
1. Generally all my posts within last year are org mode exports. I think I didn’t do anything special for this post. My config is in https://gist.github.com/anonymous/a161656a8064a3bcfb7df96ed076fe3c .
The only caveat is that I have to sometimes use different source code to get proper highlighting in the bog. For example, highlighting for Emacs Lisp does not work, but I get OK highlighting if I specify clojure.
1. Dodge Coates says:
Yeah, I’m facing this problem, too. I installed htmlize, which helped. Thanks for the gist. Very cool of you.
3. Dodge Coates says:
Heya, this bit of code simplifies the printing of tables a bit:
from tabulate import tabulate
“””
Pretty print DataFrame in an org table. Org tables are good.
They also export nicely.
“””
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2017-05-24 07:48:22
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https://math.stackexchange.com/questions/3177478/how-to-improve-this-bound
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# How to improve this bound?
As everyone reading this should very well know, $$F_0 = 0$$, $$F_1 = 1$$ and $$F_n = F_{n - 2} + F_{n - 1}$$ for all integers $$n > 1$$. The choice of uppercase F for the Fibonacci numbers seems to be fairly standard.
I'm not sure what the standard notation is for the binary weight function, so I'll use $$wt_2(n)$$. For example, $$wt_2(14) = 3$$ since $$14$$ in binary is $$1110$$ and that's three $$1$$s; $$wt_2(15) = 4$$ since $$15$$ in binary is $$1111$$ and that's three $$1$$s. For now, I'm unconcerned about negative integers.
Now, what is $$wt_2(F_n)$$? It's at most $$wt_2(F_{n - 2}) + wt_2(F_{n - 1})$$. But, except for $$F_3 = 2$$, that seems like overkill. Can this be improved for $$n > 3$$?
EDIT: As Robert pointed out, $$n = 10$$ is another example. But I've gone up to $$n = 2500$$ and it looks to me like $$wt_2(F_{n - 2}) + wt_2(F_{n - 1})$$ is a vast overestimate for $$wt_2(F_n)$$.
• for weight of 15 you wrote "and that's three ones" but correctly said weight is 4. – coffeemath Apr 6 at 22:33
• Are you interested in other upper bounds (not using weights)? – coffeemath Apr 6 at 22:35
• @coffeemath In your shoes I would have gone ahead and put in that correction, since it seems to be that David copied and pasted and neglected to make all the necessary changes. What I would get on his case about is $F_{10} = 55$, 0b110111, which follows 0b10101 and 0b100010. – Robert Soupe Apr 7 at 2:01
• @coffeemath I'm gonna have to read up on weights, I don't even know what the term means in this context. – David R. Apr 11 at 20:53
Well using induction on $$n$$, it can be shown that
$$F_n\lt 2^n$$ So the weight of $$F_n$$ is less than $$n$$.
• Indeed, this looks like a good bound (and with some further effort might be improved to something like $n-c$ for some constant $c$), so I don't understand why the OP offered a bounty instead of accepting this answer. – Alex M. Apr 29 at 7:25
By Binet’s formula, $$F_n=\tfrac 1{\sqrt{5}}(\varphi^n+(-\varphi)^{-n})$$ for each $$n$$, where $$\varphi=\frac {\sqrt{5}+1}2$$ is the golden ratio. This fact provides an upper bound for $$\operatorname{wt}_2 F_n$$ about $$n\log_2\varphi\simeq 0.694 n.$$ My computer calculations for small values of $$n$$ and a guess that approximately a half of binary digits of $$F_n$$ are $$1$$’s suggest a conjecture that $$\operatorname{wt}_2 F_n=n\frac {\log_2\varphi}2+o(n).$$
Using your data, you can try to look what happens up to $$n=2500$$.
• I don't have the time to check the details, but I think your conjecture can be proved using the formula $F_{n+d}=F_{d-1}F_{n-d+1}+F_{d-2}F_{n-d}$. You deduce an upper and a lower bound for $\frac{F_{n+d}}{F_n}$, and when $\frac{F_{n+d}}{F_n}-\phi^d$ is small, $\frac{wt_2(F_n)}{n}-\frac{\log_2(\phi)}{2}$ should be small also. – Ewan Delanoy Apr 30 at 16:38
• better than $o(n)$ I conjecture $O(\sqrt{n})$. – Somos May 3 at 12:40
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2019-12-06 16:08:52
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https://python-charts.com/distribution/violin-plot-plotly/
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# Violin plots in plotly
## Violin charts in plotly with violin
Violin plots are distribution charts similar to box plots that allow visualizing the underlying distribution of the data through a mirrored kernel density line of that data. With the violin function from the plotly express module you can create violin plots in Python. You will need to input a numerical variable to y or specify the name of the column of a data frame with the desired variable in order to create a vertical violin plot.
import plotly.express as px
import numpy as np
# Sample data
np.random.seed(5)
var = np.random.normal(0, 1, 500)
fig = px.violin(y = var)
# Alternative:
# import pandas as pd
# df = pd.DataFrame({'var': np.random.normal(0, 1, 500)})
# fig = px.violin(df, y = 'var')
fig.show()
Horizontal violin plot
If you pass the variable to x instead of to y you will create a horizontal violin plot.
import plotly.express as px
import numpy as np
# Sample data
np.random.seed(5)
var = np.random.normal(0, 1, 500)
fig = px.violin(x = var)
fig.show()
Violin plot with box plot inside
Recall that you can draw boxes inside the violin setting the box argument to True.
import plotly.express as px
import numpy as np
# Sample data
np.random.seed(5)
var = np.random.normal(0, 1, 500)
fig = px.violin(y = var, box = True)
fig.show()
Color customization
The default blue color of the violin plot can be customized passing an array with a single color to color_discrete_sequence. This will change the border and fill color for all the elements.
import plotly.express as px
import numpy as np
# Sample data
np.random.seed(5)
var = np.random.normal(0, 1, 500)
fig = px.violin(y = var, color_discrete_sequence = ['green'])
fig.show()
With the violin traces you will be able to customize everything of the plot, such as colors, line styles, etc. In the following example we highlight some of the most used arguments, whose names are self-explanatory, but recall to read the original documentation for the full list of arguments clicking on the previous link.
import plotly.express as px
import numpy as np
# Sample data
np.random.seed(5)
var = np.random.normal(0, 1, 500)
fig = px.violin(y = var, box = True)
fig.update_traces(fillcolor = 'green',
line_color = 'blue',
marker_line_outliercolor= 'black',
box_fillcolor = 'red',
opacity = 0.5)
fig.show()
Points
By default, the violin plot shows the outliers, if any. However, you can also remove them setting points = False or highlight the suspected outliers with points = 'suspectedoutliers'. The last alternative is to add all the observations with points = 'all', as in the example below.
import plotly.express as px
import numpy as np
# Sample data
np.random.seed(5)
var = np.random.normal(0, 1, 500)
fig = px.violin(y = var, points = 'all')
fig.show()
Violin plot side
The side argument of update_traces controls the side of the violin to be displayed. The argument defaults to 'both' but can also be set to 'positive' or 'negative'. This argument is specially useful to create splited violin plots.
import plotly.express as px
import numpy as np
# Sample data
np.random.seed(5)
var = np.random.normal(0, 1, 500)
fig = px.violin(y = var)
fig.update_traces(side = 'negative')
fig.show()
## Violin plots by group
If a categorical variable representing groups is passed to color as input a violin chart by group will be created.
import plotly.express as px
import numpy as np
import pandas as pd
import random; random.seed(2)
# Sample data
np.random.seed(5)
df = pd.DataFrame({'var': np.random.normal(0, 1, 500),
'group': random.choices(["G1", "G2", "G3"], k = 500)})
fig = px.violin(df, y = 'var', color = 'group')
fig.show()
The arguments used in the first section of this tutorial can also be used to customize the chart by group. For instance, you can add points for each group with points = 'all' or add box plots with box = True.
import plotly.express as px
import numpy as np
import pandas as pd
import random; random.seed(2)
# Sample data
np.random.seed(5)
df = pd.DataFrame({'var': np.random.normal(0, 1, 500),
'group': random.choices(["G1", "G2", "G3"], k = 500)})
fig = px.violin(df, y = 'var', color = 'group', points = 'all', box = True)
fig.show()
Overlayed violin plots
You can set the violinmode argument to 'overlay' in order to overlay the violins instead of display them side by side.
import plotly.express as px
import numpy as np
import pandas as pd
import random; random.seed(2)
# Sample data
np.random.seed(5)
df = pd.DataFrame({'var': np.random.normal(0, 1, 500),
'group': random.choices(["G1", "G2", "G3"], k = 500)})
fig = px.violin(df, y = 'var', color = 'group', violinmode = 'overlay')
fig.show()
Custom color for each group
The color for each group can be customized passing an ordered array of colors to color_discrete_sequence or a dict to color_discrete_map, as in the following block of code.
import plotly.express as px
import numpy as np
import pandas as pd
import random; random.seed(2)
# Sample data
np.random.seed(5)
df = pd.DataFrame({'var': np.random.normal(0, 1, 500),
'group': random.choices(["G1", "G2", "G3"], k = 500)})
fig = px.violin(df, y = 'var', color = 'group', box = True,
color_discrete_map = {'G1': '#66C2A5', 'G2': '#FC8D62', 'G3': '#8DA0CB'})
fig.show()
Grouped violin plot
Finally, if your data set contains groups and subgroups you can also create a grouped violin chart. For that purpose you will need to input your categorical data to x and to color, as shown below.
import plotly.express as px
import numpy as np
import pandas as pd
import random; random.seed(2)
# Sample data
np.random.seed(5)
df = pd.DataFrame({'var': np.random.normal(0, 1, 500),
'cat_var': random.choices(["A", "B"], k = 500),
'group': random.choices(["G1", "G2", "G3"], k = 500)})
fig = px.violin(df, y = 'var', x = 'cat_var', color = 'group', box = True,
color_discrete_map = {'G1': '#66C2A5', 'G2': '#FC8D62', 'G3': '#8DA0CB'})
fig.show()
A Primer on Making Informative and Compelling Figures
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2023-04-01 11:28:38
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http://martindemaine.org/papers/Jigsaw1xn_JCDCGGG2016/
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# Paper by Martin L. Demaine
Reference:
Jeffrey Bosboom, Erik D. Demaine, Martin L. Demaine, Adam Hesterberg, Pasin Manurangsi, and Anak Yodpinyanee, “Even $1 \times n$ Edge Matching and Jigsaw Puzzles are Really Hard”, in Abstracts from the 19th Japan Conference on Discrete and Computational Geometry, Graphs, and Games (JCDCGGG 2016), Tokyo, Japan, September 2–4, 2016, to appear.
Length:
The abstract is 2 pages.
Availability:
Currently unavailable. If you are in a rush for copies, contact me.
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2018-11-15 00:53:39
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http://www.mapleprimes.com/tags/optimization
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# Items tagged with optimizationoptimization Tagged Items Feed
### Error of minimization...
October 09 2014
0 2
Hi all
The aim of following program is minimization but it is unable to produce it. where is the mistake?
Taylor2.mws
thanks a lot.
Ph.D Candidate
Applied Mathematics Department
September 23 2014
1 2
### Prime sum question ...
September 11 2014
2 8
According to this site,"It is known that every even number can be written as a sum of at most six primes".
http://www.theage.com.au/national/education/christians-goldbachs-magic-sum-20140903-3es2t.html
i wanted to test this using maple.
restart:
> PF := proc (a::integer)
> local cst,obj,res;
> cst := add(x[i], i = 1 .. numtheory:-pi(prevprime(a))) <= 6;
> obj := add(x[i]*ithprime(i), i = 1 .. numtheory:-pi(prevprime(a)))-a;
> res := Optimization:-LPSolve(obj, {cst ,obj>=0}, assume={nonnegative,integer}); end proc:
> PF(30);
[0, [x[1] = 0, x[2] = 0, x[3] = 6, x[4] = 0, x[5] = 0, x[6] = 0,x[7] = 0, x[8] = 0, x[9] = 0, x[10] = 0]]
the third prime is 5 and 6 of them make 30. as an aside, it would be nice to know how to get maple to output "30 = 6x5".
this is obviously pretty limited, because 30 can be written as the sum of two primes (7+23 and 11+19) [GOLDBACH], but using DS's GlobalSearch for all solutions takes a long time to compute. also I have to nominate the highest prime.
any suggestions?
### Bilevel optimization...
July 24 2014
0 3
Dear all
is it possible to solve bilevel optimization problems in maple?
min F(x,y)
s.t. min G(x,y)
s.t. k(x,y)<=0
### Dear Friends Help a Maple Dude :...
July 22 2014
1 7
As I am trying to solve this integration:
restart; with(linalg); with(stats); with(plots); with(Statistics); with(LinearAlgebra); with(Optimization);
lambda0 := proc (t) options operator, arrow; gamma0+gamma1*t+gamma2*t^2 end proc;
lambda := lambda0(t)*exp(beta*s);
t1 := 145; t3 := 250; t2 := (t1+t3)*(1/2);
s := 1/(273.16+50); s1 := 1/(273.16+t1); s3 := 1/(273.16+t3); s2 := 1/(273.16+t2); gamma0 := 0.1e-3; gamma1 := .5; gamma2 := 0; beta := -3800;
c := 300; n := 200;
Theta := solve(1-exp(-(gamma0*tau1+(1/2)*gamma1*tau1^2+(1/3)*gamma2*tau1^3)*exp(beta*s1)) = 1-exp(-(gamma0*a+(1/2)*gamma1*a^2+(1/3)*gamma2*a^3)*exp(beta*s2)), a);
a := Theta[1];
Delta := solve(1-exp(-(gamma0*(a+tau2-tau1)+(1/2)*gamma1*(a+tau2-tau1)^2+(1/3)*gamma2*(a+tau2-tau1)^3)*exp(beta*s2)) = 1-exp(-(gamma0*b+(1/2)*gamma1*b^2+(1/3)*gamma2*b^3)*exp(beta*s3)), b);
b := Delta[1];
A1 := assuming([unapply(int(exp(beta*s1)*exp(-(gamma0*t+(1/2)*gamma1*t^2+(1/3)*gamma2*t^3)*exp(beta*s1))/(gamma0+gamma1*t+gamma2*t^2), t = N .. M), N, M)], [N > 0, M > 0]);
A2 := unapply(int(exp(beta*s2)*exp(-(gamma0*(a+t-tau1)+(1/2)*gamma1*(a+t-tau1)^2+(1/3)*gamma2*(a+t-tau1)^3)*exp(beta*s2))/(gamma0+gamma1*(a+t-tau1)+gamma2*(a+t-tau1)^2), t = N .. M), N, M);
A3 := unapply(int(exp(beta*s3)*exp(-(gamma0*(b+t-tau2)+(1/2)*gamma1*(b+t-tau2)^2+(1/3)*gamma2*(b+t-tau2)^3)*exp(beta*s3))/(gamma0+gamma1*(b+t-tau2)+gamma2*(b+t-tau2)^2), t = N .. M), N, M);
B1 := assuming([unapply(int(t^2*exp(beta*s1)*exp(-(gamma0*t+(1/2)*gamma1*t^2+(1/3)*gamma2*t^3)*exp(beta*s1))/(gamma2*t^2+gamma1*t+gamma0), t = N .. M), N, M)], [N > 0, M > 0]);
B2 := unapply(int((a+t-tau1)^2*exp(beta*s2)*exp(-(gamma0*(a+t-tau1)+(1/2)*gamma1*(a+t-tau1)^2+(1/3)*gamma2*(a+t-tau1)^3)*exp(beta*s2))/(gamma0+gamma1*(a+t-tau1)+gamma2*(a+t-tau1)^2), t = N .. M), N, M);
B3 := unapply(int((b+t-tau2)^2*exp(beta*s3)*exp(-(gamma0*(b+t-tau2)+(1/2)*gamma1*(b+t-tau2)^2+(1/3)*gamma2*(b+t-tau2)^3)*exp(beta*s3))/(gamma0+gamma1*(b+t-tau2)+gamma2*(b+t-tau2)^2), t = N .. M), N, M);
F0 := A1(0, tau1)+A2(tau1, tau2)+A3(tau2, c);
F1 := B1(0, tau1)+B2(tau1, tau2)+B3(tau2, c);
NLPSolve(1/(n^3*(F0*F1-F1)), tau1 = 115 .. 201, tau2 = 237 .. 273);
I need to have tau1 tau2 as varibles to get there optimal values ..
But this error keeps coming :
Error, (in Optimization:-NLPSolve) integration range or variable must be specified in the second argument, got HFloat(1.0) = HFloat(158.0) .. HFloat(255.0)
### Dear Friends Help a Maple Dude :...
July 18 2014
1 4
As am trying to solve this integration:
A(c,n,m):=evalf(int(1/(y)*exp(-c*y^(2)),y=n..m))
B(c,n,m):=evalf(int(exp(-c*y^(2)),y=n..m))
C(c,n,m):=evalf(int(y*exp(-c*y^(2)),y=n..m))
d(c,n,m):=evalf(int(y^(2)*exp(-c*y^(2)),y=n..m))
E(c,n,m):=evalf(int(y^(3)*exp(-c*y^(2)),y=n..m))
F0 := exp(beta*s1)*exp(gamma0^2*exp(beta*s1)/(2*gamma1))*A((1/2)*gamma1*exp(beta*s1), gamma0/gamma1,gamma0/gamma1+tau1)/gamma1+exp(beta*s2)*exp(gamma0^2*exp(beta*s2)/(2*gamma1))*A((1/2)*gamma1*exp(beta*s2), gamma0/gamma1+a, gamma0/gamma1+tau2-tau1+a)/gamma1+exp(beta*s3)*exp(gamma0^2*exp(beta*s3)/(2*gamma1))*A((1/2)*gamma1*exp(beta*s3), gamma0/gamma1+b, gamma0/gamma1+c-tau2+b)/gamma1
F1 := exp(beta*s1)*exp(gamma0^2*exp(beta*s1)/(2*gamma1))*(gamma0^2*A((1/2)*gamma1*exp(beta*s1), gamma0/gamma1, gamma0/gamma1+tau1)/gamma1^2-2*gamma0*B((1/2)*gamma1*exp(beta*s1), gamma0/gamma1, gamma0/gamma1+tau1)/gamma1+C((1/2)*gamma1*exp(beta*s1), gamma0/gamma1, gamma0/gamma1+tau1))/gamma1+exp(beta*s2)*exp(gamma0^2*exp(beta*s2)/(2*gamma1))*(gamma0^2*A((1/2)*gamma1*exp(beta*s2), gamma0/gamma1+a, gamma0/gamma1+tau2-tau1+a)/gamma1^2-2*gamma0*B((1/2)*gamma1*exp(beta*s2), gamma0/gamma1+a, gamma0/gamma1+tau2-tau1+a)/gamma1+C((1/2)*gamma1*exp(beta*s2), gamma0/gamma1+a, gamma0/gamma1+tau2-tau1+a))/gamma1+exp(beta*s3)*exp(gamma0^2*exp(beta*s3)/(2*gamma1))*(gamma0^2*A((1/2)*gamma1*exp(beta*s3), gamma0/gamma1+b, gamma0/gamma1+c-tau2+b)/gamma1^2-2*gamma0*B((1/2)*gamma1*exp(beta*s3), gamma0/gamma1+b, gamma0/gamma1+c-tau2+b)/gamma1+C((1/2)*gamma1*exp(beta*s3), gamma0/gamma1+b, gamma0/gamma1+c-tau2+b))/gamma1
F01 := exp(beta*s1)*exp(gamma0^2*exp(beta*s1)/(2*gamma1))*(B((1/2)*gamma1*exp(beta*s1), gamma0/gamma1, gamma0/gamma1+tau1)-gamma0*A((1/2)*gamma1*exp(beta*s1), gamma0/gamma1, gamma0/gamma1+tau1)/gamma1)/gamma1+exp(beta*s2)*exp(gamma0^2*exp(beta*s2)/(2*gamma1))*(B((1/2)*gamma1*exp(beta*s2), gamma0/gamma1+a, gamma0/gamma1+tau2-tau1+a)-gamma0*A((1/2)*gamma1*exp(beta*s2), gamma0/gamma1+a, gamma0/gamma1+tau2-tau1+a)/gamma1)/gamma1+exp(beta*s3)*exp(gamma0^2*exp(beta*s3)/(2*gamma1))*(B((1/2)*gamma1*exp(beta*s3), gamma0/gamma1+b, gamma0/gamma1+c-tau2+b)-gamma0*A((1/2)*gamma1*exp(beta*s3), gamma0/gamma1+b, gamma0/gamma1+c-tau2+b)/gamma1)/gamma1
Fβ := int((s1^2*(gamma0*t+(1/2)*gamma1*t^2)*exp(beta*s1)*(gamma1*t+gamma0)*exp(beta*s1))*exp(-(gamma0*t+(1/2)*gamma1*t^2)*exp(beta*s1)), t = 0 .. tau1)+int((s2^2*(gamma0*(a+t-tau1)+(1/2)*gamma1*(a+t-tau1)^2)*exp(beta*s2)*(gamma0+gamma1*(a+t-tau1))*exp(beta*s2))*exp(-(gamma0*(a+t-tau1)+(1/2)*gamma1*(a+t-tau1)^2)*exp(beta*s2)), t = tau1 .. tau2)+int((s3^2*(gamma0*(b+t-tau2)+(1/2)*gamma1*(b+t-tau2)^2)*exp(beta*s3)*(gamma0+gamma1*(b+t-tau2))*exp(beta*s3))*exp(-(gamma0*(b+t-tau2)+(1/2)*gamma1*(b+t-tau2)^2)*exp(beta*s3)), t = tau2 .. c)+int((s3^2*(gamma0*(b+t-tau2)+(1/2)*gamma1*(b+t-tau2)^2)*exp(beta*s3)*(gamma0+gamma1*(b+t-tau2))*exp(beta*s3))*exp(-(gamma0*(b+t-tau2)+(1/2)*gamma1*(b+t-tau2)^2)*exp(beta*s3)), t = c .. infinity)
I need to have tau2 as varibles to get there optimal values ..
Minimize(1/((F0*F1-F01^2)*n^3*Fβ), tau2 = 237..273})
But this error keeps coming :
Error, (in Optimization:-NLPSolve) integration range or variable must be specified in the second argument, got HFloat(1.0) = 121.0828419 .. HFloat(193.0828419)
### How do I do optimization...
July 18 2014
0 1
Hi all
I am trying to maximize a function f(x,y,z,w) in terms of x. (Only x is treated as a variable, and the others are treated as parameters).
However, all I know is that y,z,w they are parameters and they are non-negative. I have already tried with the "optmization help page" from maplesoft's website, and it looks like it will search the range of x,y,z,w, and it will return numerical values at which this function is maximized.
However, what I want is instead a close-form solution of x=g(y,z,w) that will maximize the function. In other words, I would like to keep the parameter in symbolic forms.
Can Maple do that?
### Maple and Matlab - Global Optimization toolbox...
July 17 2014
1 23
Hi,
On page 32 （PDF）
Two different results were obtained using the Global optimization.
Log likelihood does not differ much. BUT the estimates vary a lot, such as mu[p].
tmp.mw
tmp.pdf
When I tried to use one of the answer from a particular run, I get the HFLOATING error, see picture.
So how reliable is this? Could there be a better way to optimize this ?
Thanks!
As an additional note, if I have Matlab R2014a, could I use Matlab to optimize the target function? DO I need to purchase a seperate addon?
### founding the error in program...
July 13 2014
0 2
hi all.
I have wrore the following program for optimization with bernstein and block pulse hybrid functions.
the program have some errors which i can't understand.
Bernestien1.mws
> restart:
> alias(C=binomial): with(LinearAlgebra): macro(LA= LinearAlgebra):
> HybrFunc:=proc(N, M, tj) # N=Number of subintervals, M=Number of functions in subintervals local B, n, m; global b; for n from 1 to N do for m from 0 to M-1 do B := (i,m,t) -> C(m,i)*(1-t)^(m-i)*t^i: b[n,m]:=unapply(piecewise(t>=(n-1)*tj/N and tb[n,m](t)): #convert(%,vector); end proc: HybrFunc(3, 3, 1); # End Of Definition g2(t):=t; #*exp(t-1): # Any other function can be replaced here g1(t):=add(add(c[n,m]*b[n,m](t), m=0..2), n=1..3); Optimization[Minimize](sqrt(int((g2(t)-g1(t))^2, t=0.. 1))); assign(op(%[2])); plot([g2(t),g1(t)], t=0..1, 0..5, color=[blue,red],thickness=[1,3],discont, scaling=constrained);
Error, (in Optimization:-NLPSolve) complex value encountered Error, invalid left hand side in assignment (1)
I'll be so grateful if any one can help me.
Ph.D Candidate
Applied Mathematics Department
### how to minimize this penalty function which has ma...
June 26 2014
0 22
f1 := -2*x1 - x2;
f2 := -x1 - 4*x2;
g1 := 2*x1 + 3*x2 - 6;
g2 := -x1;
g3 := -x2;
penalty := lambda1*Max(f1-M,0)+lambda2*Max(f2-M,0)+(M^2)*(max(g1,0) + max(g2,0) + max(g3,0));
M := 1;
k := 1;
s := 1;
lambda1 := 0.5;
lambda2 := 0.5;
with(Optimization):
Minimize(subs(lambda2=0.5, subs(lambda1=0.5, subs(M=1, penalty))));
correct answer for these parameters should be below, however, Minimize got error
when lambda1 = 0.5, lambda2 = 0.5
(x1, x2) = 1.551123, 0.965918
f(x1, x2) = (–4.068164, –5.414795);
### how to do optimization with partial differential e...
June 05 2014
0 0
1. how to do optimization with partial differential equation as constraints in maple
2. how to do optimization with partial differential equation as objective function in order to make output obey this model
### nonlinear programming...
April 26 2014
1 1
Hi, I would like to solve this nonlinear problem :
$\left\{ \begin{array}{lrcll} & max \ \ \ \sum \limits_{i=1}^{n}c_i & \\ s.t. \\ & x_i & \ge & R & \forall i=1..n \\ & x_i & \le & L - R & \forall i=1..n \\ & y_i & \ge & R & \forall i=1..n \\ & y_i & \le & H - R & \forall i=1..n \\ & (x_i - x_j)^2 + (y_i - y_j)^2 - 4c_ic_jR^2 & \ge & 0 & \forall i,j=1..n, i \ne j \\ \\ & c_i & \in & \left\{0, 1 \right\} & \forall i=1..n \\ & x_i,y_i & \ge & 0 & \forall i=1..n \end{array} \right.$
with : $\left\{ \begin{array}{lcl} L & \ge & 2R \\ H & \ge & 2R \\ R & > & 0 \\ \end{array}\right.$
I use the NLPSolve command and i get this error : Error, (in Optimization:-NLPSolve) integer variables are not accepted.
Can you help me ?
### Solve can't find analytically solution... but I ca...
April 24 2014
2 2
Hi Maple-Prime-ers!
I have a system of equations, containing 18 variables and 13 equations, making this a 5 degree of freedom (DOF) system. I would like to analytically solve each of the equations in terms of each of these DOFs. Normally I would use solve(system, dof_variables) to accomplish this, but it doesn't return anything. Not even [].
I can solve this system by hand. I've included a hand-solution involving isolate() and subs() in the attached worksheet. I'm looking to incorporate this in an optimization algorithm with varying system, so I would like an automated way of doing this.
Does anybody have any suggestions to get solve to work as intended?
3driversys_FD_BRAKE_ICE_GEN.mw
Here is the system I am talking about:
The free variables are: {FD_T, FD_W, ICE_T, EM2_T, BRAKE_T}
I'm looking for a solution in this form:
### Dear Friends Help a Maple Dude :...
April 20 2014
3 12
As am trying to solve this integration:
int(B*eta^(-B)*t^(B-1)*exp(-(t/eta)^B)*(t-n*h), t = n .. (n+1)*h)
where,
ETA:=1000
B:=2.5
But this error keeps coming :
Error, (in Optimization:-NLPSolve) integration range or variable must be specified in the second argument, got 1. = 1. .. 2.
### Optimization [Minimize] and [Maximize] with Pi...
April 20 2014
2 3
what is the wrong with Pi set ::: in this function ::: Warning, no iterations performed as initial point satisfies first-order conditions
Optimization[Minimize](x^2 + y^2 + 25*(sin(x)^2+sin(y)^2), x=-2*Pi .. 2*Pi , y= -2*Pi .. 2*Pi);
Warning, no iterations performed as initial point satisfies first-order conditions
[0., [x = HFloat(0.0), y = HFloat(0.0)]]
Optimization[Maximize](x^2 + y^2 + 25*(sin(x)^2+sin(y)^2), x=-2*Pi .. 2*Pi , y= -2*Pi .. 2*Pi);
Warning, no iterations performed as initial point satisfies first-order conditions
[-0., [x = HFloat(0.0), y = HFloat(0.0)]]
--------------------------------
I got my good result when I apply it with this function :
f:= (x,y)->cos(x)*sin(y) -(x/(y^2+1));
Optimization[Maximize](f(x,y), x = -1 .. 2, y = -1 .. 1);
[0.994945017202501170,[x = HFloat(-0.6362676080636113), y = HFloat(1.0)]]
Optimization[Minimize](f(x,y), x = -1 .. 2, y = -1 .. 1);
[-2.02180678335978703,[x = HFloat(2.0), y = HFloat(0.10578346945175972)]]
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2014-10-21 21:31:25
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http://persistentstruggle.blogspot.com/2012_10_28_archive.html
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## Sunday, 28 October 2012
### Lender fees - at arms length or a transfer? Plus fee-or-no-fee?
I just want to make two points about the lender fee I've introduced recently in my discussion on understanding rates of interest. First, the world is a diverse and unpredictable place. There's nothing to stop a lender giving a loan at seemingly crazy rates (either very high seeming, or very now seeming). The other way of coming at this is to say that, for any given rate of return, there's some possible market environment which make this rate of return understandable. And even if it isn't understandable, human beings are free to agree whatever rates they want between them (absent any usury laws which may be in place, of course). If a lender lends money because they don't need to charge for it, or because they're favourably disposed towards the borrower, then the transaction is more like a transfer and less like an arms length business contract.
The second point I want to make is that you know nothing about how much of a fee, if any, is being charged if all you know about the loan is the rate of interest. This seems like such a strange thing to say that I'll pause a while before explaining more. I'm saying that if I told you that a man walked into a bar and asked for a one day lend of £100 and agreed a rate $r$ with a lender, then you the listener have no clear idea of how much of a fee the lender charged, if indeed he did charge anything at all. I will explain this more in a later post. Looking at a rate is often a great place to work out the fee charged by the lender, but as I will show, this can quite easily be thwarted by adverse economic conditions.
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2017-08-22 03:35:38
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https://www.borgholt.dk/lsplhld/coordinate-calculator-graph-b8318d
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Flight Club Ny, Residency Interviews 2020 Reddit, Yoshi Game Music, Italian Stanley Street, Gum Facts For School, Cashing Out 401k After Leaving Job, American Ballet Theatre, " />
This grid is widely a good choice for a range of subject areas. Graph equations by plotting points. 5.12 / Interpret Line Plots with Up to 5 Data Points. The east west axis is called as the x axis and north south axis is called as y axis. the one coordinate for x and the other one for f(x). Let’s first convert from polar to rectangular form; to do this we use the following formulas, as we can see this from the graph:. It is a line of slope and vertical axis intercept at . K-8 Math. General Math. The slope calculator updates the graph and the equation automatically when you enter new values for the points. In a coordinate graph having x, y axes. In Cartesian coordinate system a point can be defined with 3 real numbers : x, y, z. Check out which quadrant of the graph … Find more Mathematics widgets in Wolfram|Alpha. If resetting the app didn't help, you might reinstall Calculator to deal with the problem. Use the distance formula. Distance between two points. The next step is to determine the angle of rotation, theta. To graph in the polar coordinate system we construct a table of $$\theta$$ and $$r$$ values. Ask an Expert . 5.106 / Graph Points on a Coordinate Plane. Distance Between Coordinates. Get the free "Polar Graphs" widget for your website, blog, Wordpress, Blogger, or iGoogle. Math is about vocabulary. This page will help you to do that. The slope calculator determines the slope or gradient between two points in the Cartesian coordinate system. The most basic plotting skill it to be able to plot x,y points. Students find answers in different ways. Use the midpoint formula. If needed, Free graph paper is available. The grid allows you to determine a range of capabilities in a manner in which is proportionally appropriate. Grade 6. 5.13 / Create Line Plots. For example, students can use either Graph, Equation, or Matrix function to solve the simultaneous equations below. Upgrade . Office Tools downloads - Coordinate Calculator by Victor Nekrasov and many more programs are available for instant and free download. Use rectangular, polar, cylindrical, or spherical coordinates. Area in Polar Coordinates Calculator Added Apr 12, 2013 by stevencarlson84 in Mathematics Calculate the area of a polar function by inputting the polar function for "r" and selecting an interval. This will work for triangles, regular and irregular polygons, convex or concave polygons. If you are not sure what the gps coordinates are, you can use the coordinates converter to convert an address into latlong format or vice versa. Through these tutorials, you will be able to explore the properties of points and quadrants in rectangular coordinate systems. If you take a look on the function graphs, you see that intersects the y-axis at intersects the y-axis at . For this example we will say that point is (6,8). And of course, you can use this calculator to calculate vector difference as well, that is, the result of subtracting one vector from another. The first step is finding or determining the original coordinates. Home. Y0. How to calculate the equation of the line from a point and the slope? Math for Everyone. Download free on Amazon. Tom Lucas, Bristol. Tracie set out from Elmhurst, IL, to go to Franklin Park. Solve equations numerically, graphically, or symbolically. Finite Math. It is necessary to follow the next steps: Enter coordinates (xA x A, yA y A) and (xB x B, yB y B) of two points A and B in the box A and B in the box. X0. Plots & Geometry. Missing Coordinate Calculator. Polar-Rectangular Point Conversions. X1. Graph Individual (x,y) Points - powered by WebMath. Hence, the coordinates of a point on the y-axis are (0, y). Precalculus. We offer a large amount of good reference tutorials on subject areas starting from basic algebra to synthetic division Find the missing value using the slope method by entering any three values. Distance Calculator is use to calculate the distance between coordinates and distance between cities. What is the x coordinate, y coordinate of the following ordered pairs? Select the distance method and enter any three values for finding the missing coordinate value. This is because the vector difference is a vector sum with the second vector reversed, according to: To get reversed or opposite vector in cartesian form, you simply negate the coordinates. Monday, July 22, 2019 " Would be great if we could adjust the graph via grabbing it and placing it where we want too. Emmitt, Wesley College. On the way, she made a few stops to do errands. Solve the coordinate graph questions on your own and verify with the answers provided and test your preparation level. {y = x y = -x + 2. In the below distance on a coordinate plane calculator, enter the values for two set of x and y coordinates ie., X0, Y0 and X1, Y1 and click calculate to know the distance between 2 points in 2-dimensional space. We enter values of $$\theta$$ into a polar equation and calculate $$r$$. Example. Polar coordinate graph. Using Graph. Before we can use the calculator it is probably worth learning how to find the slope using the slope formula. The position of points and elements can be found by Coordinate system. Cartesian coordinate system. An online graphing calculator to graph points in rectangular coordinate system. Graph inequalities, contour plots, density plots and vector fields. Calculus. Graph Individual (x,y) Points. Slope Calculator Helps To Do: This calculator allows you to perform calculations corresponding to the slope and different other parameters: You can readily find $$(m)$$ or gradient of a line that passes through 2 points; It assists you to solve a coordinate for a given point, $$(m)$$ or angle $$(θ)°$$, and distance from a … Count 7 units to the upwards of the origin along the y-axis. I am sure your x y coordinate calculator can be solved faster here. To graph in the rectangular coordinate system we construct a table of $$x$$ and $$y$$ values. Pre-Algebra. Download free on Google Play. If you don't know your location, use the where am I right now to find out. Here is a simple online coordinate distance calculator to calculate the distance between two points on a coordinate plane. Solve the simultaneous equations. The graph of the linear equation is a straight line in rectangular coordinates. Go through them. You will probably be asked to convert coordinates between polar form and rectangular form.. Free graphing calculator instantly graphs your math problems. It uses the same method as in Area of a polygon but does the arithmetic for you. Go to the coordinate graph having lines X’OX, Y’OY. The vector calculator allows the calculation of the coordinates of a vector from the coordinates of two points online. For the sake of your comfort, we even listed the step by step solutions to all the sample problems provided. Approximately 5,000 predefined geographic and projected coordinate systems can be selected from the list, or quickly located using the handy search tool. Learn mathematics more deeply using Explore feature. By inspection, we quickly see that the points with rectangular coordinates and are on the graph. Calculate the coordinates of a vector from 2 points in a system of any dimension; The vector calculator is used according to the same principle for any dimension of systems. How to Find Coordinates of a Point? & Calculus. Cartesian, cylindrical and spherical coordinate systems. Convert a point in the Cartesian plane to its equal polar coordinates with this polar coordinate calculator. Icon display. Chemistry. This graph is the same as the graph of the function . Help With Your Math Homework. 5.111 / Graph Linear Functions. Download coordinate calculator .exe for free. Below given are the steps that are helpful to find the coordinates of a point. This calculator is intended for coordinates transformation from / to the following 3d coordinate systems: Cartesian; Cylindrical; Spherical. Visit Mathway on the web. so the point lies in the second quadrant. 6. Desmos supports an assortment of functions. Also, if you need to clear (reset) the calculator, you can use to clear the RAM in the calculator. 6.132 / Maps with Decimal Distances. Graphing Calculator for Macintosh, Windows, & iOS. Graphing. This is typically given, but can be calculated if needed. 5.121 / Relative Coordinates. This conversion is pretty straight-forward: Download free in Windows Store. Coordinate Graphs with Decimals and Negative Numbers. 5.112 / Coordinate Graphs as Maps. Graph equations with a graphing utility. coordinate inverse calculator, It's an online Geometry tool requires coordinates of 2 points in the two-dimensional Cartesian coordinate plane. Male or Female ? As the name says, it says where the function cuts the y-axis. Converting from Polar to Rectangular Coordinates. If ever you actually will need help with algebra and in particular with algebra calculator find holes in a graph or solution come visit us at Graph-inequality.com. Polar coordinates also take place in the x-y plane but are represented by a radius and angle as shown in the diagram below. You can plug in the questions and this product will go through it with you step by step so you can understand easily as you solve them. Graphing. There are some demos available so you can also get to know how incredibly helpful the program is. Statistics. With graph paper, you is capable of doing math equations or write research info with exact precision. Input Equations. Coordinate Calculator The TatukGIS Coordinate Calculator interactively converts the coordinates of any point between thousands of coordinate systems. … Basic Math. Algebra. Visit Cosmeo for explanations and help with your homework problems! Plot ordered pairs in a Cartesian coordinate system. Y1. The x coordinate of the point is negative and the y coordinate is positive. Note: To make your TI-84 C calculator graph a little faster, hit , scroll down to “Detect Asymptote” and turn this Off:. Learn and revise how to plot coordinates and create straight line graphs to show the relationship between two variables with GCSE Bitesize Edexcel Maths. get Go. Trigonometry. Wednesday, February 21, 2018 " It would be nice to be able to draw lines between the table points in the Graph Plotter rather than just the points. Download free on iTunes. Algebra. 5.110 / Convert Graphs to Input/Output Tables. Note: If your graphing calculator ever gets “stuck” and doesn’t seem to be graphing, you can hit to turn it off, and then to turn it back on again. X Y Coordinate Graph Paper Printable – Graph paper is a form of writing paper that accompany a prearranged grid. A graph of the line is drawn on a coordinate plane, along with the slope intercept equation. thus adjusting the coordinates and the equation. Other Stuff. Polygon area calculator The calculator below will find the area of any polygon if you know the coordinates of each vertex. Mark the point as D (-2, 7). Example 2: Draw the graph of the (nonlinear) equation . The solution is x = 1 y = 1. Find $$x$$-intercepts and $$y$$-intercepts. Thoroughly talk about the services that you need with potential payroll providers. To improve this 'New coordinates by rotation of axes Calculator', please fill in questionnaire. Trig. Linear Algebra. Graph functions in two and three dimensions, explicit, implicit, or parametric. Count 2 units from the left of the origin along the x-axis. 1. The following example is a step by step guide on using those equations to calculate the new coordinate points. These values must be real numbers or parameters; Mathway. 1. The slope is basically the amount of slant a line has, and can have a positive, negative, zero or undefined value. Inverse Calculator Reviews & Tips Inverse Calculator Ideas . You have to insert the point into the equation, i.e. 7 units to the upwards of the following 3d coordinate systems is x = 1 between... And verify with the slope calculator determines the slope intercept equation work for triangles, regular and polygons. Online graphing calculator for Macintosh, Windows, & iOS { y = x coordinate! Widely a good choice for a range of subject areas the original coordinates shown in the diagram below coordinates! To insert the point as D ( -2, 7 ), explicit, implicit or... Powered by WebMath hence, the coordinates of a point and the other one for (. The vector calculator allows the calculation of the origin along the y-axis at intersects the y-axis can have a,. For explanations and help with your homework problems area calculator the calculator, you probably! Coordinates and distance between cities system we construct coordinate calculator graph table of \ r\! Same as the x coordinate of the linear equation is a line of slope and vertical axis intercept at the. Lines x ’ OX, y points intersects the y-axis are ( 0 y! Of slope and vertical axis intercept at also get to know how incredibly helpful the program is we can the! And distance between two points on a coordinate graph having lines x ’,... Few stops to do errands will find the area of a point in the polar coordinate.! Also take place in the polar coordinate system we construct a table of \ ( \theta\ ) into polar! Is probably worth learning how to find out intersects the y-axis at intersects the y-axis at payroll... Or Matrix function to solve the coordinate graph having lines x ’ OX, y.. Faster here south axis is called as the x axis and north south axis is called as axis!, negative, zero or undefined value zero or undefined value few stops to do errands you need to (... Elements can be found by coordinate system coordinate calculator can be selected from the of. ( 6,8 ) demos available so you can also get to know how incredibly helpful program! The handy search tool ', please fill in questionnaire will work for triangles, regular irregular! On a coordinate plane, along with the answers provided and test preparation. A straight line in rectangular coordinate system units to the coordinate graph paper –... Polar graphs '' widget for your website, blog, Wordpress,,! Even listed the step by step solutions to all the sample problems provided you do know... Basic plotting skill it to be able to explore the properties of points and can., z many more programs are available for instant and free download questions on your own and verify with answers! The upwards of the function cuts the y-axis polar coordinate system we construct a table of \ ( y\ values. To all the sample problems provided the vector calculator allows the calculation of the equation. Verify with the problem thoroughly talk about the services that you need with potential payroll providers of two in. Negative and the equation automatically when you enter new values for the sake your. Rotation of axes calculator ', please fill in questionnaire y\ ) -intercepts coordinates transformation from / to the of. Accompany a prearranged grid located using the slope method by entering any three values step guide on those. Polar form and rectangular form accompany a prearranged grid coordinate is positive do n't know location... Intended for coordinates transformation from / to the coordinate graph questions on own... Nonlinear ) equation can have a positive, negative, zero or undefined value be solved faster here we... Inequalities, contour plots, density plots and vector fields you take a look on the y-axis at the! For explanations and help with your homework problems a manner in which is proportionally appropriate step is determine! To all the sample problems provided Victor Nekrasov and many more programs available. Shown in the Cartesian plane to its equal polar coordinates with this polar calculator... Of capabilities in a coordinate plane, along with the answers provided test... This polar coordinate system a point and the equation, or parametric know how incredibly helpful the is... Calculate the distance between two points online the problem plots and vector fields first step is or. Coordinate plane, along with the problem is widely a good choice for a range of subject.... Explore the properties of points and quadrants in rectangular coordinate system a point or iGoogle and test preparation... Of writing paper that accompany a prearranged grid the grid allows you to determine range! Be calculated if needed line has, and can have a positive, negative, zero or value! Coordinate systems, implicit, or Spherical coordinates paper that accompany a prearranged grid Cartesian system! A radius and angle as shown in the diagram below pretty straight-forward: Inverse Reviews.
Kategorier: Ikke-kategoriseret
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2021-04-14 17:24:27
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http://physics.stackexchange.com/tags/weak-interaction/new
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# Tag Info
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Short answer: mostly no. Slightly longer answer: the interaction of neutrinos with the molecules of your body take several forms, but they all come down to ionizing dose. However, there are many larger sources of ionizing dose in your like. Things like the Potasium-40 in the food you eat, radon and Carbon-14 in the air you breath, cosmic rays, and on and ...
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We need to distinguish between the concept of Dark Matter and the potential candidates that fill it's role. Dark matter, as originally posited, was just non-luminous matter that could be inferred only through gravity. As time went on, additional research shed more light on what dark matter was. Both the growth of large scale structure of the universe and ...
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pfnuesel's answer is absolutely correct and very satisfying to understand, and you should read it before this one. There is, however, a route which would permit $\pi\to e+\nu$ even if the electron were massless. The conserved quantity which suppresses that decay is not spin, but total angular momentum. There exists an electron+neutrino wavefunction with ...
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Since the spin of the charged $\pi$ is $0$, the spins of the daughter particles need to add up to $0$ as well, i.e., their spins need to be anti-parallel. That's nothing else than the conservation of angular momentum. Assuming the anti-neutrino to be massless, it is always right-handed. Right-handed means that the momentum vector and the spin vector are ...
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This is mostly to make an explicit connection with natural units - the unit system in which $\hbar$ and $c$ are both set to 1, which is the natural set of units for relativistic quantum theory. Because you've adimensionalized two units and you had three physical dimensions to start with (mass, length and time), natural units retain a single dimensional ...
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I am going to take a stab at answering my own questions with what I have been able to find on the subject. (Still very sparse.) 1) In this model the SU(2) confinement occurs at much higher energies than SU(3) confinement, so that the weak-partons confine first (giving us quarks and leptons) and then quarks confine into hadrons second. This explains why ...
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2016-02-10 01:01:50
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https://lucatrevisan.wordpress.com/2017/10/16/beyond-worst-case-analysis-lecture-9/
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# Beyond Worst-Case Analysis: Lecture 9
Scribed by Chinmay Nirkhe
In which we explore the Stochastic Block Model.
1. The ${G_{n,p,q}}$ problem
The Stochastic Block Model is a generic model for graphs generated by some parameters. The simplest model and one we will consider today is the ${G_{n,p,q}}$ problem.
Definition 1 (${G_{n,p,q}}$ graph distribution) The ${G_{n,p,q}}$ distribution is a distribution on graphs of ${n}$ vertices where ${V}$ is partitioned into two 2 subsets of equal size: ${V = V_1 \sqcup V_2}$. Then for ${\{i,j\}}$ pair of vertices in the same subset, ${\Pr( (i,j) \in E) = p}$ and otherwise ${\Pr( (i,j) \in E) = q}$.
We will only consider the regime under which ${p > q}$. If we want to find the partition ${V = V_1 \sqcup V_2}$, it is intuitive to look at the problem of finding the minimum balanced cut. The cut ${(V_1, V_2)}$ has expected size ${q n^2 / 4}$ and any other cut will have greater expected size.
Our intuition should be that as ${p \rightarrow q}$, the problem only gets harder. And for fixed ratio ${p/q}$, as ${p,q \rightarrow 1}$, the problem only gets easier. This can be stated rigorously as follows: If we can solve the problem for ${p,q}$ then we can also solve it for ${cp, cq}$ where ${c > 1}$, by keeping only ${1/c}$ edges and reducing to the case we can solve.
Recall that for the ${k}$-planted clique problem, we found the eigenvector ${{\bf x}}$ corresponding to the largest eigenvalue of ${A - \frac{1}{2} J}$. We then defined ${S}$ as the vertices ${i}$ with the ${k}$ largest values of ${|x_i|}$ and cleaned up ${S}$ a little to get our guess for the planted clique.
In the Stochastic Block Model we are going to follow a similar approach, but we are instead going to find the largest eigenvalue of ${A - \left( \frac{p + q}{2} \right) J}$. Note this is intuitive as the average degree of the graph is ${p(n/2 - 1) + q(n/2) \approx \frac{n}{2} (p+q)}$. The idea is simple: Solve ${{\bf x}}$ the largest eigenvector corresponding to the largest eigenvalue and define
$\displaystyle V_1 = \{ i : x_i > 0\}, \qquad V_2 = \{ i : x_i \leq 0 \} \ \ \ \ \ (1)$
As we proceed to the analysis of this procedure, we fix ${V_1, V_2}$. Prior to fixing, the adjacency matrix ${A}$ was ${\left( \frac{p+q}{2} \right) J}$.\footnote{The diagonal should be zeroes, but this is close enough.} Upon fixing ${V_1, V_2}$, the average adjacency matrix ${R}$ looks different. For ease of notation, if we write a bold constant ${{\bf c}}$ for a matrix, we mean the matrix ${cJ}$. It will be clear from context.
$\displaystyle R = \left( \begin{array}{c | c} {\bf p} & {\bf q} \\ {\bf q} & {\bf p} \end{array} \right) \ \ \ \ \ (2)$
Here we have broken up ${R}$ into blocks according to the partition ${V_1, V_2}$.
Theorem 2 If ${p, q > \log n / n}$ then with high probability, ${\lVert A - R \rVert < O \left(\sqrt{n(p+q)} \right)}$.
Proof: Define the graph ${G_1}$ as the union of a ${G_{n/2, p}}$ graph on ${V_1}$ and ${G_{n/2, p}}$ graph on ${V_2}$. Define the graph ${G_2}$ a a ${G_{n, q}}$ graph. Note that the graph ${G}$ is distributed according to picking a ${G_1}$ and ${G_2}$ graph and adding the partition crossing edges of ${G_2}$ to ${G_1}$. Let ${A_1}$ and ${A_2}$ be the respective adjacency matrices and define the follow submatrices:
$\displaystyle A_1 = \left( \begin{array}{c | c} A_1' & \\ & A_1'' \end{array} \right), \qquad A_2 = \left( \begin{array}{c | c} A_2' & A_2''' \\ {A_2'''}^\dagger & A_2'' \end{array} \right). \ \ \ \ \ (3)$
Then the adjacency matrix ${A}$ is defined by
$\displaystyle A = A_1 + A_2 - \left( \begin{array}{c | c} A_2' & \\ & A_2'' \end{array} \right) \ \ \ \ \ (4)$
Similarly, we can generate a decomposition for ${R}$:
$\displaystyle R = \left( \begin{array}{c | c} {\bf p} & \\ & {\bf p} \end{array} \right) + \bigg ( {\bf q} \bigg ) - \left( \begin{array}{c | c} {\bf q} & \\ & {\bf q} \end{array} \right). \ \ \ \ \ (5)$
Then using triangle inequality we can bound ${\lVert A - R \rVert}$ by bounding the difference in the various terms.
\displaystyle \begin{aligned} \lVert A - R \rVert &\leq \left \lVert A_1 - \left(\begin{array}{c | c} {\bf p} & \\ & {\bf p} \end{array} \right) \right \rVert + \left \lVert A_2 - ({\bf q}) \right \rVert + \left \lVert \left( \begin{array}{c | c} A_2' & \\ & A_2'' \end{array} \right) - \left( \begin{array}{c | c} {\bf q} & \\ & {\bf q} \end{array} \right) \right \rVert \\ &\leq O(\sqrt{np}) + O(\sqrt{nq}) + O(\sqrt{nq}) \\ &= O \left( \sqrt{n (p+q)} \right) \end{aligned} \ \ \ \ \ (6)
The last line follows as the submatrices are adjacency matrices of ${G_{n,p}}$ graphs and we can apply the results we proved in that regime for ${p, q > \log n / n}$. $\Box$
But the difficulty is that we don’t know ${R}$ as ${R = R(V_1, V_2)}$. If we knew ${R}$, then we would know the partition. What we can compute is ${\left \lVert A - \left(\frac{p+q}{2} \right) J \right \rVert}$.\footnote{The rest of this proof actually doesn’t even rely on knowing ${p}$ or ${q}$. We can estimate ${p + q}$ by calculating the average vertex degree.} We can rewrite ${R}$ as
$\displaystyle R = \left( \frac{p+q}{2}\right) J + \frac{p-q}{2} \left( \begin{array}{c | c} {\bf 1} & - {\bf 1} \\ - {\bf 1} & {\bf 1} \end{array} \right) \ \ \ \ \ (7)$
Call the matrix on the right ${C}$. It is clearly rank-one as it has decomposition ${n \chi \chi^\dagger}$ where ${\chi = \frac{1}{\sqrt{n}}\left( \begin{array}{c} {\bf 1} \\ -{\bf 1} \end{array} \right)}$. Therefore
$\displaystyle \left \lVert \left( A - \left( \frac{p+q}{2} \right) J \right) - \left( \frac{p-q}{2} \right) C \right \rVert = \left \lVert A - R \right \rVert \leq O \left( \sqrt{n (p+q)} \right). \ \ \ \ \ (8)$
Then ${A - \left( \frac{p+q}{2} \right) J}$ is close (in operator norm) to the rank 1 matrix ${\left( \frac{p-q}{2} \right) C}$. Then their largest eigenvalues are close. But since ${\left( \frac{p-q}{2} \right) C}$ has only one non-zero eigenvalue ${\chi}$, finding the corresponding eigenvector to the largest eigenvalue of ${A - \left( \frac{p+q}{2} \right) J}$ will be close to the ideal partition as ${C}$ describes the ideal partition. This can be formalized with the Davis-Kaham Theorem.
Theorem 3 (Davis-Kahan) Given matrices ${M, M'}$ with ${\lVert M - M' \rVert \leq \varepsilon}$ where ${M}$ has eigenvalues ${\lambda_1 \leq \ldots \leq \lambda_n}$ and corresponding eigenvectors ${{\bf v}_1, \ldots, {\bf v}_n}$ and ${M'}$ has eigenvalues ${\lambda_1' \leq \ldots \leq \lambda_n'}$ and corresponding eigenvectors ${{\bf v}_1', \ldots, {\bf v}_n'}$, then
$\displaystyle \sin \left( \mathrm{angle} \left( \mathrm{span}({\bf v}_1), \mathrm{span}({\bf v}_1') \right) \right) \leq \frac{\varepsilon}{|\lambda_1' - \lambda_2|} \leq \frac{\varepsilon}{|\lambda_1 - \lambda _2 - \varepsilon|}. \ \ \ \ \ (9)$
Equivalently,
$\displaystyle \min \left \{ \lVert {\bf v}_1 \pm {\bf v}_1' \rVert \right \} \leq \frac{\sqrt{2} \varepsilon}{\lambda_1 - \lambda_2 - \varepsilon}. \ \ \ \ \ (10)$
The Davis Kahan Theorem with ${M' = A - \left( \frac{p+q}{2} \right) J, M = \left( \frac{p-q}{2} \right) C}$, and ${\varepsilon = O \left( \sqrt{n (p+q)} \right)}$ states that
$\displaystyle \min \left \{ \lVert {\bf v}' \pm \chi \rVert \right \} \leq O \left( \frac{\sqrt{a + b}}{a - b - O \left( \sqrt{a + b} \right)} \right) \ \ \ \ \ (11)$
where ${{\bf v}'}$, the eigenvector associated with the largest eigenvalue of ${A - \left( \frac{p+q}{2} \right) J}$ and ${a = pn /2, b = qn/2}$, the expected degrees of the two parts of the graph. Choose between ${\pm {\bf v}'}$ for the one closer to ${\chi}$. Then
$\displaystyle \lVert {\bf v}' - \chi \rVert^2 \leq O \left( \left( \frac{\sqrt{a + b}}{a - b - O\left(\sqrt{a + b}\right)} \right)^2 \right). \ \ \ \ \ (12)$
Recall that ${\sum_i (v_i' - \chi_i)^2 = \lVert {\bf v}' - \chi \rVert^2}$. If ${v_i'}$ and ${\chi_i}$ disagree in sign, then this contributes at least ${1/n}$ to the value of ${\lVert {\bf v}' - \chi \rVert^2}$. Equivalently, ${n \cdot \lVert {\bf v}' - \chi \rVert^2}$ is at least the number of misclassified vertices. It is simple to see from here that if ${a - b \geq c_\varepsilon \sqrt{a + b}}$ then we can bound the number of misclassified vertices by ${\varepsilon n}$. This completes the proof that the proposed algorithm does well in calculating the partition of the Stochastic Block Model.
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2017-11-20 09:44:10
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https://indico.cern.ch/event/129980/contributions/1351191/
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# DPF 2011
8-13 August 2011
Rhode Island Convention Center
US/Eastern timezone
### Indico Support
12 Aug 2011, 10:30
15m
556 (Rhode Island Convention Center)
### 556
#### Rhode Island Convention Center
Parallel contribution Detector Technology and R&D
### Speaker
Prof. Neeti Parashar (Purdue University Calumet)
### Description
The present Compact Muon Solenoid silicon pixel tracking system has been designed for a peak luminosity of $10^{34}~\textrm{cm}^{-2} \textrm{s}^{-1}$ and total dose corresponding to two years of the Large Hadron Collider (LHC) operation. With the steady increase of the luminosity expected at the LHC, a new pixel detector with four barrel layers and three endcap disks is being designed. We will present the key points of the design: the new geometry, which minimizes the material budget and increases the tracking points, and the development of a fast digital readout architecture, which ensures readout efficiency even at high rate. The expected performances for tracking and vertexing of the new pixel detector are also addressed.
### Primary author
Prof. Neeti Parashar (Purdue University Calumet)
Slides
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2021-05-17 11:07:12
|
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|
http://math.stackexchange.com/questions/75220/finding-minimum-of-multidimensional-function
|
# Finding minimum of multidimensional function
My calculus knowledge is pretty limited, but unfortunately I need to solve a problem of the following kind:
I'm given a 2 dimensional function $f(x,y)$ from $\mathbb{R}^2$ to $\mathbb{R}$ and I want to know, where it attains its minimum value over $\mathbb{R}\times(a,b)$.
Put differently I want to find an $x$ value and a $y\in(a,b)$ such that $f(x,y) \leq f(x',y')$ for all x' in $\mathbb{R}$ and all $y \in (a,b)$.
I'll have to take the partial derivative of $f$ w.r.t $x$, but I don't understand how y will come into play.
-
There may be no minimum; you can try to locate the local minima using the Second partial derivatives test, and then comparing. – Arturo Magidin Oct 23 '11 at 23:13
Wait: do you really want to find an $x$ such that for each $y\in (a,b)$, $f(x,y)\leq f(x',y)$ for all $x;\in\mathbb{R}$, or do you want to find an $x\in\mathbb{R}$ and a $y\in (a,b)$ such that $f(x,y)\leq f(x',y')$ for all $x'\in\mathbb{R}$ and all $y'\in(a,b)$? That's what "the minimum for arbitrary $x$ but for $y$ in some interval $(a,b)$" means. – Arturo Magidin Oct 23 '11 at 23:20
oh yes i was ambiguous, I meant the second version, I edited it. – stefan Oct 23 '11 at 23:25
$f(x,y)$ has a critical point at $(x,y)$ if the gradient $\left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$ is the zero vector at that point.
So the procedure you'll want to follow is the following. I assume that $f$ does in fact attain its minium on $\mathbb{R} \times [a,b]$; if it's not obvious for your particular $f$, it's something you'll need to check.
1. Find the points where the gradient of $f$ vanishes. Throw out critical points with $y$ not in $(a,b)$.
2. Evaluate $f$ at these points to find the minimum on the interior of your region. (If there are no critical points in the region, skip this step.)
3. Find the minimum of the one-dimensional functions $f(x,a)$ and $f(x,b)$. This will give you the minima at the boundary of your region.
4. Evaluate $f$ at the three points from steps 2 and 3. Whichever gives the smallest $f$ is the global minimum.
|
2014-12-20 19:34:29
|
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|
https://shtools.oca.eu/shtools/python-examples.html
|
## Notebooks
Notebook name Description
Introduction 1 Grids and spherical harmonic coefficients.
Introduction 2 Localization windows and spectral analysis.
Introduction 3 Gravity and Magnetic fields.
Tutorial 1 Simple spherical harmonic analyses.
Tutorial 2 Localized spectral analysis on the sphere.
Tutorial 3 The SHTOOLS class interface.
Tutorial 4 Spherical harmonic normalizations and Parseval’s theorem.
Tutorial 5 Multitaper spectral analysis class interface.
Tutorial 6 3D plots of gridded data.
## Test programs
A variety of test programs can be found in the folders in examples/python.
Folder directory Description
ClassInterface/ Test the python class interfaces.
TestLegendre/ Test and plot the Legendre functions.
IOStorageConversions/ Read coefficients from a file and test conversions between real and complex coefficients.
GlobalSpectralAnalysis/ Test functions to compute different power spectra from real and complex coefficients.
LocalizedSpectralAnalysis/ Test the coupling matrix, localized spectral analysis, and bias routines.
GravMag/ Test the gravity and magnetics routines, and compute the crustal thickness of Mars.
TimingAccuracy/ Perform timing and accuracy tests using real and complex coefficients, with Driscoll and Healy (1994) and Gauss-Legendre quadrature grids.
Other/ Test a variety of other routines.
|
2019-03-19 18:41:20
|
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|
https://solvedlib.com/n/part-draw-short-segment-of-the-polymer-containing-two-repeating,14248830
|
# Part |Draw short segment of the polymer containing two repeating units and obtained from the following monomers. HcNCOHOCH,CH,OHOCN
###### Question:
Part | Draw short segment of the polymer containing two repeating units and obtained from the following monomers. Hc NCO HOCH,CH,OH OCN
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2023-04-01 23:18:47
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|
http://mathhelpforum.com/geometry/114020-coordinate-geometry.html
|
1. ## Coordinate Geometry
I had this big assignment to do, and I did most of the questions except for these few (or I did them and I'm not sure if that's how I go about working them out)
1. State the y_intercept of the line: 3x+4y = -3 with equations.
I did: 4y = 3x+8 / y_int = -8?
2. Show that (2, 1) lies on the same line with the equation y=2x+3
I started with: 2 x 2 + -3x1 = -3
3. Parallel to 3x-4y=2 passing through (1, 2)
I did: 3x-4y =3 x 1 - 4 x 1
=3x-4y=-1
4. Perpendicular to 3x-4y = 2 passing through (1, 2)
Absolutely no idea how to do this.
5. Two lines have slopes k+2=-1. Find k given that:
a. The lines are parallel
b. The lines are perpendicular
I got:
a. k+2 = -1
b. -1/2k
2. Originally Posted by suckatmaths
...
1. State the y_intercept of the line: 3x+4y = -3 with equations.
I did: 4y = 3x+8 / y_int = -8?
Where does the +8 com from?
You have to plug in x = 0 to calculate the y-intercept.
2. Show that (2, 1) lies on the same line with the equation y=2x+3
I started with: 2 x 2 + -3x1 = -3
Where did you find this -3 ? Plug in the coordinates of the point and check if you get a true statement. In your case:
$1 = 2 \cdot 2 +3~\implies~1 \neq 7$ that means the point doesn't belong to the line.
3. Parallel to 3x-4y=2 passing through (1, 2)
I did: 3x-4y =3 x 1 - 4 x 1
=3x-4y=-1
Why do you use the x-value in the place of the y-variable? Your method is OK but you should have gotten:
$3 \cdot 1 - 4 \cdot 2 = -5$ and therefore the equation of the parallel line is $3x - 4y = -5$
4. Perpendicular to 3x-4y = 2 passing through (1, 2)
If a straight line has the equation
$ax +by = c$ then the perpendicular line has the equation
$bx - ay = d$
So you know now the LHS of the equation. Find the value of d using the same way as in #3
Absolutely no idea how to do this.
5. Two lines have slopes k+2=-1. Find k given that:
a. The lines are parallel
b. The lines are perpendicular
I got:
a. k+2 = -1
b. -1/2k
...
to a): Two lines are parallel if the slopes $m_1$ and $m_2$ are equal. Thus $m_1=m_2~\implies~k+2=-1~\implies~k=-3$
to b) Two lines are perpendicular to each other if the slopes satisfy the equation
$m_1 \cdot m_2 = -1$
I'll leave the rest for you.
3. 1. State the y_intercept of the line: 3x+4y = -3 with equations.
I did: 4y = 3x+8 / y_int = -8?
rewrite the equation with y as subject and its coefficient always = +1 ,
hence y = mx + c, where c is the y-intercept, and m is the slope of line.
3x + 4y = -3
4y = -3x -3
y = $-\frac{3}{4}x - \frac{3}{4}$
so y-intercept = $-\frac{3}{4}$
2. Show that (2, 1) lies on the same line with the equation y=2x+3
I started with: 2 x 2 + -3x1 = -3
substitute (2,1) into y = 2x + 3 :
left hand side = 1
right hand side = 2(2) + 3 = 7
since left hand side is not equal to right hand side, the point (2,1) does not lie on y = 3x + 3 ........check your question again
3. Parallel to 3x-4y=2 passing through (1, 2)
I did: 3x-4y =3 x 1 - 4 x 1
=3x-4y=-1
What you presented is wrong. (1,2) lies on a line that is parallel to another line 3x - 4y = 2. it does not mean (1,2) simply lies on this 3x - 4y = 2.
Being parallel means the two lines share the same slope, m.
3x - 4y = 2
3x - 2 = 4y
y = $\frac{3}{4}x - \frac{1}{2}$
hence its slope = $\frac{3}{4}$
Therefore, the equation of the line parallel to 3x - 4y = 2 and passes through (1,2) is:
$\frac{y - 2}{x - 1} = \frac{3}{4}$
In the end you get 7x - 4y + 1 = 0
4. Perpendicular to 3x-4y = 2 passing through (1, 2)
Absolutely no idea how to do this.
Similar to question 3 but here the slope of the line you are working on is -\frac{4}{3} because the product of the slopes of two lines, perpendicular to each other, is equal to -1.
5. Two lines have slopes k+2=-1. Find k given that:
a. The lines are parallel
b. The lines are perpendicular
I got:
a. k+2 = -1
b. -1/2k
Don't understand what does k = -3 stand for?...
4. Originally Posted by ukorov
...
3. Parallel to 3x-4y=2 passing through (1, 2)
I did: 3x-4y =3 x 1 - 4 x 1
=3x-4y=-1
What you presented is wrong. (1,2) lies on a line that is parallel to another line 3x - 4y = 2. it does not mean (1,2) simply lies on this 3x - 4y = 2.
Being parallel means the two lines share the same slope, m.
3x - 4y = 2
3x - 2 = 4y
y = $\frac{3}{4}x - \frac{1}{2}$
hence its slope = $\frac{3}{4}$
Therefore, the equation of the line parallel to 3x - 4y = 2 and passes through (1,2) is:
$\frac{y - 2}{x - 1} = \frac{3}{4}$
In the end you get 7x - 4y + 1 = 0
...
I don't want to pick at you but you must have made a small mistake:
$\frac{y - 2}{x - 1} = \frac{3}{4}~\implies~4(y-2)=3(x-1) ~\implies~ 4y-8=3x-3~\implies~-3x+4y-5=0$
5. Originally Posted by earboth
I don't want to pick at you but you must have made a small mistake:
$\frac{y - 2}{x - 1} = \frac{3}{4}~\implies~4(y-2)=3(x-1) ~\implies~ 4y-8=3x-3~\implies~-3x+4y-5=0$
...foolish mistake
6. Originally Posted by ukorov
...foolish mistake
It is customary among the regulars to use the thanks button when someone spots an error in one of your posts.
CB
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2013-12-10 09:33:09
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https://socratic.org/questions/part-a-determine-the-empirical-formula-when-the-elemental-composition-by-mass-sh
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# Part A: Determine the empirical formula when the elemental composition by mass shows: 40.0% C, 6.71% H, 53.3% 0. For that I got CH2010 simplified to CHO5 Part B: Determine the molecular formula when the mole mass of the compound is 90.08 g/mol. (Help?)
Sep 29, 2015
${\text{C"_3"H"_6"O}}_{3}$
#### Explanation:
To get the molecular formula of the compound, you need to use its empirical formula and its molar mass.
The empirical formula tells you what the smallest integer ratio of atoms is for the elements that make up your compound.
Now, from what I can tell, you didn't get the correct empirical formula. When you are given the percent composition by mass of the compound, you can take a 100-g sample to make the calculations easier.
If you sample has a mass of 100 g, then you know that it contains
• 40.0 grams of carbon
• 6.71 g of hydrogen
• 53.3 grams of oxygen
Use the molar masses of the three elements to find how many moles of each you get in that sample
$\text{For C: " (40.0color(red)(cancel(color(black)("g"))))/(12.011color(red)(cancel(color(black)("g")))/"mol") = "3.33 moles C}$
$\text{For H: " (6.71color(red)(cancel(color(black)("g"))))/(1.01color(red)(cancel(color(black)("g")))/"mol") = "6.64 moles H}$
$\text{For O: " (53.3color(red)(cancel(color(black)("g"))))/(16.0color(red)(cancel(color(black)("g")))/"mol") = "3.33 moles O}$
Divide all the values by the smallest one to get
"For C: " (3.33color(red)(cancel(color(black)("moles"))))/(3.33color(red)(cancel(color(black)("moles")))) = 1
"For H: " (6.64color(red)(cancel(color(black)("moles"))))/(3.33 color(red)(cancel(color(black)("moles")))) = 1.994 ~~ 2
"For O: " (3.33color(red)(cancel(color(black)("moles"))))/(3.33color(red)(cancel(color(black)("moles")))) = 1
The empirical formula will thus be
$\text{C"_1"H"_2"O"_1" }$, or $\text{ " "CH"_2"O}$
So, you know that the smallest integer ratio of atoms in your compound is $1 : 2 : 1$, but you don't know exactly how many atoms of each element are needed to form one molecule.
This is where the molar mass of the compound comes into play. You know that one mole of molecules has a mass of $\text{90.08 g}$, and that you ahve a minimum of one mole of carbon, two moles of hydrogen, and one mole of oxygen for every mole of the compound.
This means that you can write
(1 xx M_"M carbon" + 2 xx M_"M hydrogen" + 1 xx M_"M oxygen") * color(blue)(n) = "90.08 g"
(1 xx "12.011 g" + 2 xx "1.01 g" + 1 xx "16.0 g") * color(blue)(n) = "90.08 g"
This will get you
$\textcolor{b l u e}{n} = \left(90.08 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))))/(30.03color(red)(cancel(color(black)("g}}}}\right) = 2.9997 \approx 3$
The molecular formula will thus be
("CH"_2"O")_3 = color(green)("C"_3"H"_6"O"_3
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2021-12-06 20:06:25
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http://openstudy.com/updates/5240dec2e4b090015417fef6
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## anonymous 3 years ago Evaluate the limit, if it exists lim h-0 ((1/(x+h)^2 - 1/x^2 ))/ h
1. myininaya
Try combining top fractions. And get rid of the compound fraction. Unless you already know derivatives then we can actually skip this part.
2. anonymous
so a common denominator for the top fractions? do i expand the (x+h)^2
3. myininaya
You will need to.
4. anonymous
so x^2 + 2xh + h^2, what would be the common denominator
5. myininaya
x^2(x+h)^2
6. myininaya
|dw:1379983226588:dw|
7. myininaya
|dw:1379983241382:dw|
8. anonymous
2xh + h^2/ x^2 (x+h)^2 x (1/h)
9. anonymous
can i factor out a h from the numerator?
10. anonymous
so i factored out an h from the numerator, than cancelled the h from 1/h
11. myininaya
What happen to your negative in from 2xh?
12. myininaya
h/h=1 yep yep
13. anonymous
i have lim h--0 2x+h/ x^2(x+h)^2
14. anonymous
so now i can "sub" in 0 into h, but what value goes for x?
15. myininaya
Well yeah but still you are missing a negative on top
16. myininaya
only h is going somewhere x stays
17. myininaya
It said what happens as h goes to zero (it said nothing for x)
18. myininaya
Did you figure out what negative you are missing?
19. anonymous
ok i think i see my mistake so i went thru the stages again and right now I have lim h--0 -2x-h/x^2 (x+h)^2
20. anonymous
because I factored out an h, but im confused as to my next step
21. myininaya
h goes to 0
22. anonymous
is my answer suppose to be just a number? or will it have a value and x
23. myininaya
24. anonymous
-2x-0/x^2 (x+0) ^2 ?
25. myininaya
(-2x-0)/(x^2(x+0)^2) yes Simplify.
26. anonymous
-2/x?
27. anonymous
not rlly syre how to simplify the denominator
28. mathslover
$$\cfrac{(-2x-0)}{(x^2)(x+0)^2}$$ (Just for ease - wrote the LaTeX)
29. myininaya
-2x-0=-2x x^2(x+0)^2=x^2(x)^2 (since x+0=x) Use law of exponents.
30. myininaya
Recall one of the laws of exponents a^m * a^n=a^(m+n)
31. anonymous
ok so my answer is -2/x^3
32. myininaya
yep
33. anonymous
thank you :)
34. mathslover
But the denominator should be x^4 @myin
35. myininaya
-2x/x^4=-2/x^3
36. anonymous
I have another question: find the limit if it exists limx-- -2 2-|x|/ 2+x
37. anonymous
thats lim x approaches -2
38. anonymous
can I just sub in -2 for x? so that makes it 0/0
39. mathslover
Oh sorry @myininaya - didn't saw that :) Thanks !
40. myininaya
$\lim_{x \rightarrow -2} \frac{2-|x|}{2+x}$ ?
41. anonymous
yes
42. myininaya
So we can't just simply plug in -2 because our denominator will be 0. Is there someway to make the function continuous at x=-2 so we can just plug in Well |x|=x if x>=0 and -x if x<0 Since x approaches -2 then x is negative since we are looking at what surrounds a negative number so we have |x|=-x
43. myininaya
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2016-12-04 22:22:53
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https://www.projecteuclid.org/euclid.rmjm/1414760960
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## Rocky Mountain Journal of Mathematics
### The existence of three solutions for $p$-Laplacian problems with critical and supercritical growth
#### Abstract
In this paper we deal with the existence and multiplicity of solutions for the $p$-Laplacian problems involving critical and supercritical Sobolev exponent via variational arguments. By means of the truncation combining with the Moser iteration, we extend the result obtained by Ricceri \cite{14} to the critical and supercritical case.
#### Article information
Source
Rocky Mountain J. Math., Volume 44, Number 4 (2014), 1383-1397.
Dates
First available in Project Euclid: 31 October 2014
https://projecteuclid.org/euclid.rmjm/1414760960
Digital Object Identifier
doi:10.1216/RMJ-2014-44-4-1383
Mathematical Reviews number (MathSciNet)
MR3274355
Zentralblatt MATH identifier
1350.35083
#### Citation
Zhao, Lin; Zhao, Peihao. The existence of three solutions for $p$-Laplacian problems with critical and supercritical growth. Rocky Mountain J. Math. 44 (2014), no. 4, 1383--1397. doi:10.1216/RMJ-2014-44-4-1383. https://projecteuclid.org/euclid.rmjm/1414760960
#### References
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• H. Brezis and L. Nirenberg, Positive solutions of nonlinear elliptic equations involving critical Sobolev exponent, Comm. Pure Appl. Math. 36 (1983), 437–477.
• J. Chabrowski and J. Yang, Existence theorems for elliptic equations involving supercritical Sobolev exponent, Adv. Differ Equat. 2 (1997), 231–256.
• F.J.S.A. Corrêa and G.M. Figueiredo, On an elliptic equation of $p$-Kirchhoff type via variational methods, Bull. Austral. Math. Soc. 74 (2006), 263–277.
• G.M. Figueiredo and M.F. Furtado, Positive solutions for some quasilinear equations with critical and supercritical growth, Nonlinear Anal. TMA 66 (2007), 1600–1616.
• J. Gao, P. Zhao and Y. Zhang, Compact Sobolev embedding theorems involving symmetry and its application, NoDEA Nonlin. Differ. Equat. Appl. 17 (2010), 161–180.
• J. García Azorero and I. Peral Alonso, Multiplicity of solutions for elliptic problems with critical exponent or with a nonsymmetric term, Trans. Amer. Math. Soc. 323 (1991), 877–895.
• J.P. García Azorero, I. Peral Alonso and J.J. Manfredi, Sobolev versus Hölder local minimizers and global multiplicity for some quasilinear elliptic equations, Comm. Cont. Math. 2 (2000), 385–404.
• M. Guedda and L. Véron, Quasilinear elliptic equations involving critical Sobolev exponents, Nonlin. Anal. 13 (1989), 879–902.
• P.L. Lions, The concentration-compactness principle in the calculus of variations. The limit case, Part 1, Rev. Mat. Iber. 1 (1985), 145–201.
• J. Moser, A new proof of De Giorgi's theorem concerning the regularity problem for elliptic differential equations, Comm. Pure Appl. Math. 13 (1960), 457–468.
• P.H. Rabinowitz, Variational methods for nonlinear elliptic eigenvalue problems, Ind. Univ. Math. J. 23 (1974), 729–754.
• B. Ricceri, A three critical points theorem revisited, Nonlin. Anal. TMA 70 (2009), 3084–3089.
• ––––, A further three critical points theorem, Nonlin. Anal. TMA 71 (2009), 4151–4157.
• E.A.B. Silva and M.S. Xavier, Multiplicity of solutions for quasilinear elliptic problems involving critical Sobolev exponents, Ann. Inst. Poincaré Anal. Non Linéaire 20 (2003), 341–358.
• G. Tarantello, On nonhomogeneous elliptic equations involving critical Sobolev exponent, Ann. Inst. Poincaré Anal. Non Linéaire 9 (1992), 281–304.
• W. Wang, Sobolev embeddings involving symmetry, Bull. Sci. Math. 130 (2006), 269–278.
• Z. Wei and X. Wu, A multiplicity result for quasilinear elliptic equations involving critical Sobolev exponents, Nonlin. Anal. TMA 18 (1992), 559–567.
• P. Zhao and C. Zhong, On the infinitely many positive solutions of a supercritical elliptic problem, Nonlin. Anal. TMA 44 (2001), 123–139.
• ––––, Positive solutions of elliptic equations involving both supercritical and sublinear growth, Houst. J. Math. 28 (2002), 649–663.
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2019-10-19 06:06:51
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https://zbmath.org/?q=an:0863.05063&format=complete
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# zbMATH — the first resource for mathematics
On 2-extendability of generalized Petersen graphs. (English) Zbl 0863.05063
Summary: Let $$\text{GP}(n,k)$$ be a generalized Petersen graph with $$(n,k)=1$$, $$n>k\geq4$$. Then every pair of parallel edges of $$\text{GP}(n,k)$$ is contained in a 1-factor of $$\text{GP}(n,k)$$. This partially answers a question posed by Larry Cammack and Gerald Schrag [Problem 101, Discrete Math. 73, No. 3, 311-312 (1989)].
##### MSC:
05C70 Edge subsets with special properties (factorization, matching, partitioning, covering and packing, etc.)
##### Keywords:
generalized Petersen graph; 2-extendable
Full Text:
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2021-03-05 16:43:55
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http://mathematica.stackexchange.com/tags/parametric-equations/hot
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# Tag Info
## Hot answers tagged parametric-equations
44
Consider this: ParametricPlot3D[ RotationTransform[a, {0, 1, 0}][{0, 0, Sin[3 a] + 5/4}], {a, 0, 2 Pi}, Evaluated -> True] Now rotate this around a circle, while rotating it at the same time around its' origin: ParametricPlot3D[ RotationTransform[b, {0, 0, 1}][{6, 0, 0} + RotationTransform[a + 3 b, {0, 1, 0}][{0, 0, Sin[3 a] + 5/4}]], ...
29
You asked for alternative approaches to what you did, so here is one: A completely different approach to the one-dimensional time-independent Schrödinger equation would be to use matrix techniques. The idea is to eliminate the need for NDSolve entirely. For bound-state problems, you can do this by choosing a basis satisfying the condition of vanishing wave ...
22
Here is another approach. It could be improved (I am sure) to properly determined the principle axes and translation (if I get time I will aim to update): lin = {#1^2, #1, #2, 2 #1 #2, #2^2} & @@@ points; lm = LinearModelFit[lin, {1, a, b, c, d}, {a, b, c, d}] Exploring model: lm["ParameterTable"] Determining quadric formula: pa = ...
18
We couldn't be really pleased if we didn't exploit existing Mathematica functionality to get exact solutions. Here we provide them with Reduce rewriting the given system to an exact one and using a trick by adding another variable x because one can see that any solutions are described by two different arguments t and t + x. Now we can realize that one can ...
15
I'm adding this answer to put on record an answer to the second part the question, "what is the parametric equation?". The parametric equation is implicit in Kirma's RotationTransform expression. To extract it, one need simply write something like Clear[a, b] quoit[a_, b_] := Evaluate @ RotationTransform[b, {0, 0, 1}][{6, 0, 0} + ...
15
In analytic geometry, the ellipse is defined as the set of points (X,Y) of the Cartesian plane that, in non-degenerate cases, satisfy the implicit equation with and where Lets fit points with second-order curve (which include ellipse). elipse = a11*x^2 + a22*y^2 + 2*a12*x*y + 2*a13*x + 2*a23*y + a33; coeff = {a11, a22, a12, a13, a23, a33}; ...
14
Here is a standard direct way to get the principal exes and other transformation data. Find the mean of the points, subtract it to center them, and take the singular value decomposition. The third and second components thereof give the rotation and scaling data necessary to form a circle on which the first component, viewed as a point set, roughly lies. The ...
13
You can get the curve in polynomial implicit form as below. poly = GroebnerBasis[{x^2 - ct, y^2 - st, ct^2 + st^2 - 1}, {x, y}, {ct, st}][[1]] (* Out[290]= -1 + x^4 + y^4 *) To get the area, integrate the characteristic function for the interior of the region. That that's where the polynomial is nonpositive (just notice that it is negative at the ...
13
If the potential is $(\tanh (x)+1) (\tanh (x)-1)$ you can obtain the analytic solution using Mathematica as follows: [I have omitted some of the detail - e.g. the asymptotic expansions - because the details are analogous to the simple harmonic oscillator case in my previous answer (see above).] Define the potential. u[x_] = (1 + Tanh[x]) (-1 + Tanh[x]) ...
11
My defined function next find {nextpoint, nextdirection} value from {startpoint, startdirection} using NSolve. next[{sp_, sd_}][δ_] := Module[{φ, sol, fp, fd}, sol = NSolve[{{x[φ, δ], y[φ, δ]} == sp + t sd, Abs[t] > 10^(-9), 0 <= φ < 2 π}, {t, φ}, Reals]// Quiet; sol = If[Length[sol] > 0, sol[[1]]]; fp = {x[φ, δ], y[φ, δ]} /. sol; ...
10
Can be done as follows. (1) Find implicit form from parametric. (2) Solve for pts (x,y) where implicit eqn and gradient simultaneously vanish. (3) Discard those solutions that correspond to cusps. The rest correspond to crossings. The code below handles steps (1) and (2). I found it useful to rationalize because we eventually get an overdetermined system ...
10
As Jens mentioned, the spatially discretize the equation is another alternative for bound state problem. Here is my very simple implementation of this approach. The basic idea is express the equation on a grid. The differentials can be expressed as finite differences. For example, the second order derivative can be expressed as \frac{d^2\psi}{d ...
10
I guess I know what you've encountered. The model doesn't print correctly though it looks OK in Mathematica, right? Specifically speaking, nothing seems to be wrong when you plot the object in Mathematica: ParametricPlot3D[{Cos[v]*(3 - u) + .25*Sin[4 u], Sin[v]*(3 - u) + .25 Sin[4 u], u}, {u, 6, 13}, {v, 6, 13}, PlotStyle -> {Thickness[.3], ...
10
Concerning the comment about creating the surfaces, sure: Mathematica is one of the best tools available for that. Here's the Klein bottle, for example. ParametricPlot3D[{ (3 + Cos[v/2]*Sin[u] - Sin[v/2]*Sin[2 u])*Cos[v], (3 + Cos[v/2]*Sin[u] - Sin[v/2]*Sin[2 u])*Sin[v], Sin[v/2]*Sin[u] + Cos[v/2]*Sin[2 u]}, {u, -Pi, Pi}, {v, 0, 2 Pi}, Axes -> ...
10
The approach mentioned in my comment using a polar representation: r[ a_, b_, theta_, t_] := a b/Sqrt[b^2 Cos[theta] Sin[2 t] - a^2 Sin[2 t] Sin[theta]^2 + (a^2 + b^2) Sin[t]^2 Sin[theta]^2 + Cos[t]^2 (b^2 Cot[theta]^2 + a^2 Sin[theta]^2)]; some random sample data: data = With[ {a = 1, b = 1/2, theta = Pi/3 , x0 = 1, y0 = 1} , ...
9
Here is a symbolic solution of your eigenvalue problem. Define the differential equation (setting $\hbar = \omega = m_0 = 1$). diffeq = -(1/2) \[Psi]''[x] + 1/2 x^2 \[Psi][x] == e \[Psi][x] Symbolically solve the differential equation. soln = DSolve[diffeq, \[Psi], x][[1, 1]] (* \[Psi] -> Function[{x}, C[2] ParabolicCylinderD[1/2 (-1 - 2 e), I ...
9
I upvoted the response by @J.M. and was tempted to leave it at that. This is similar but automates the process a bit further by explicitly implicitizing (is that an oxymoron?) the tori. Somehow I think that step deserves mention since it can be a useful thing in its own right. We start with code to take the trig parametrized tori and find algebraic implicit ...
9
For this, you could use the answer by J.M. to the question "Extruding along a path". The question here isn't a duplicate because it makes use of features in J.M.'s excellent answer that go beyond what the linked question actually asked for. In particular, that answer can deal with self-intersecting cross sectional curves, which is what you need for this ...
9
Finding the whole solution set Inspired by @Szabolcs idea, you can let Reduce solve this problem with the help of existential quantifiers: Reduce[ ForAll[x, 5 x^3 + z + x z + x^2 (1 + z) == 0 \[And] Im[z] == 0, Re[x] < 0], z] It immediately affirms your conjecture: (* z > 4 *) This problem formulation via ForAll can be read as: Reduce the ...
9
Since it seems to have not been mentioned yet: yet another way to obtain an approximation of the area of your Lamé curve is to use the shoelace method for computing the area. Here's a Mathematica demonstration: pts = First[Cases[ ParametricPlot[{Sqrt[Abs[Cos[t]]] Sign[Cos[t]], Sqrt[Abs[Sin[t]]] Sign[Sin[t]]}, {t, 0, 2 π}, ...
8
If you're happy with an approximate solution, you can use NSolve[]. As I mentioned in an answer to an earlier question of yours, GroebnerBasis[] can be used for parameter elimination. Let's do that for your three "tori": t1 = First @ GroebnerBasis[Thread[{x, y, z} == Rationalize[torus1[a, b]]] ~Join~ {Cos[a]^2 + Sin[a]^2 == 1, ...
8
Clear["Global`*"] f[x_, \[Theta]_] = RotationTransform[\[Theta], {1, 0, 1}, {5 Pi, 0, 5 Pi}][{x, 0, -((10 Pi)/6) Sin[x] + 5 Pi}][[{1, 3}]]; p1 = ParametricPlot[{x, x}, {x, -10 Pi, 10 Pi}, PlotRange -> {{-10 Pi, 10 Pi}, {-10 Pi, 10 Pi}, {-10 Pi, 10 Pi}}, ImageSize -> 300, Axes -> True]; n = 7; g[a_] := Evaluate[ t^(1/n) (5 a ...
8
I just finished blog post about the creation of nice graphics from Mathematica Graphics3D using the Blender render framework: http://wolfig-techblog.blogspot.de/2015/04/blender-as-shader-for-mathematica.html Maybe you can find some inspiration there for your own graphics. I managed to generate a reasonable Klein bottle with glass shading: Note: the ...
7
Third solution A slightly simpler and more geometric approach leads to a third form for the solution (including Artes' and my second) -- don't you just love trigonometric functions! By symmetry, two points starting from a vertex of the hypocycloid (star) and going in opposite directions at the same speed will meet at one of the desired crossings. If the ...
7
Similar to @ybeltukov, you can extract the lines from the plot. But to get a proper polygon, you need to reverse one of the lines. plot = ParametricPlot[{{u + Sin[u], -Cos[u]}, {u + Sin[u + Pi], Cos[u + Pi]}}, {u, 0, Pi}, Axes -> True]; {line1, line2} = Cases[plot, l_Line :> First@l, Infinity]; Graphics[ {Opacity[0.4], Darker@Blue, ...
7
Complex roots come in pairs therefore the equation will have a single real solution only if it has a real double root and two complex root or if it has a quadruple root. So we require poly = -4 l^2 + 4 l^2 x + (5 - l^2) x^2 - 4 x^3 + x^4 Solve[{poly == 0, D[poly, x] == 0}, {x, l}, Reals] This gives us three solutions for l: {0, -((2 + ...
7
This is an old (and possibly abandoned) question; however, its solution involves a number of issues related to NonlinearModelFit and ParametricNDSolveValue that show up on this site and it's perhaps constructive to look at this problem as a case study. Rescaling the fit data We wish to fit the following data: fitdata = {{1962, 0}, {1963, 0.9}, {1964, ...
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--- edit --- I forgot to address an issue which I now see was raised in a comment by @Michael E2. This setup is only going to give a tangent circle. If it osculates it is largely by accident. (Outright snogging, now that might be intentional.) --- end edit --- The issue is that these curves need not intersect at the same value of the parameter. So you can ...
7
You need to define partial sums (I will use Simon's definition in the comments which is faster than my older one): Manzoni[n_, y_] := Transpose@{Re@#, Im@#} &@Accumulate[Range[n]^-(.5 + y I)] The thing worth attention here is 1/k^(.5 + I y) - meaning 0.5 which makes it automatically numeric before the sum is taken. If you would keep 1/2 and wrap ...
7
Just for fun: f[a_, t_] := a {t - Sin[t], 1 - Cos[t]} Manipulate[ ParametricPlot[{f[1, 4 t], f[2, 2 t], f[4, t]}, {t, 0, 4 Pi}, PlotStyle -> {Red, Green, Blue}, Epilog -> {{Orange, Circle[{4 p, 1}, 1], Black, PointSize[0.015], Point[f[1, 4 p]]}, {Orange, Circle[{4 p, 2}, 2], Black, PointSize[0.015], Point[f[2, 2 p]]}, {Orange, ...
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2015-05-04 05:04:05
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Matt and Peter can do together a piece of work in 20 days
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Matt and Peter can do together a piece of work in 20 days. After they have worked together for 12 days Matt stops and Peter completes the remaining work in 10 days. In how many days Peter complete the work separately.
A. 26 days
B. 27 days
C. 23 days
D. 25 days
E. 24 days
[Reveal] Spoiler: OA
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Last edited by Bunuel on 23 Oct 2012, 05:22, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Time & Work [#permalink]
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Suppose rate of work for per day Matt = M & Rate of work per day for Peter = P
Together in a Day they can finish = P + M units
Total Work Done in 20 Days = 20(P+M)
Total Work Done in 12 Days = 12(P+M)
Work Done by Peter in 10 Days = 10P
Since total work is same we can say that 20(P+M) = 12(P+M) + 10P => 8M = 2P => P = 4M
So total work done by them together in 20 Days = 20(4M+M) = 100M
Since Pete does 4M unit of work per day, it will take him 25 Days (100M/4M) to finish up the work.
Answer D.
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Re: Time & Work [#permalink]
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together, they can do a piece in 20 days, i.e. 1/m + 1/p = 1/20
In 12 days, they can finish 12*(1/20) = 3/5 of the piece. After Matt leaves, 2/5 still needs to be done by Peter, which he does in 10 days.
10/(2/5) = 25
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Re: Time & Work [#permalink]
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30 Mar 2009, 15:36
Isn't this the same question? time-work-61136.html#p442431 But why do the answers differ...
pmenon wrote:
together, they can do a piece in 20 days, i.e. 1/m + 1/p = 1/20
In 12 days, they can finish 12*(1/20) = 3/5 of the piece. After Matt leaves, 2/5 still needs to be done by Peter, which he does in 10 days.
10/(2/5) = 25
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Re: Time & Work [#permalink]
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01 Apr 2009, 00:06
bigfernhead wrote:
Isn't this the same question? time-work-61136.html#p442431 But why do the answers differ...
Because Matt and Peter have some issues working together ..Just kidding.
Note that this problem is asking the the time taken by thr guy who did NOT stop working after 12 days while the question in link is asking the time taken by the guy who stopped working after 12 days.
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Re: Time & Work [#permalink]
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01 Apr 2009, 00:49
Best way is this.
1) Get the unit of quantity of work. Make it a number which you find LCM of the given digits.
2) Calculate the rate of work for each person.
And then calculate what is asked.
So as per above,
Lets assume that 20*12=240 unit of work is there.
Assume rate of work per day for Matt is m, and for Peter is p;
so..
Matt and Peter can do together a piece of work in 20 days. implies..
(m+p)*20= 240
m+p=12
Now, peter works for 22 day while, matt works for 12 days.
so, 22p+12m=240
Solving the equations, we find the value of m and p as 2.4, and 9.6 unit of work/day respectively.
So peter will take, 240/9.6=25 days.
prasannar wrote:
Matt and Peter can do together a piece of work in 20 days. After they have worked together for 12 days Matt stops and Peter completes the remaining work in 10 days. In how many days Peter complete the work separately.
26days
27days
23days
25days
24 days
What is the best way to solve these problems?
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Re: Time & Work [#permalink]
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11 Apr 2009, 22:10
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Another easy way is ..
M&P complete 60% of the work in 12 days (since they complete 100% in 20 days)
P completes the remaining 40% in 10 days .. => to complete 100% he would need 25 days.
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Re: Time & Work [#permalink]
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12 Apr 2009, 04:36
Work done by M&P in 12 days = 12/20 = 3/5
Remaining 2/5 is done by P alone in 10 days.
So P alone can do the entire work in (5/2) X 10 = 25 days
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Re: Time & Work [#permalink]
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18 Aug 2009, 02:46
24.Matt and Peter can do together a piece of work in 20 days. After they have worked together for 12 days Matt stops and Peter completes the remaining work in 10 days. In how many days Peter complete the work separately.
26days
27days
23days
25days
24 days
Rate Together * # of days working together + Rate of Peter * # of days working alone = 1 completed job
let P = 3 of hours Peter can complete one job alone
(1/20)*12 + (1/P)*10 = 1
P = 25
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Re: Time & Work [#permalink]
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05 Feb 2010, 16:51
Together they complete the job in 20 days means they complete 12/20 of the job after 12 days.
Peter completes the remaining (8/20) of the job in 10 days which means that the whole job(1) can be completed in X days.
<=> 8/20->10 <=> X=10/(8/20)= 25 Thus the answer is D.
1 -> X
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Re: Matt and Peter can do together a piece of work in 20 days. [#permalink]
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To me, the most intuitive approach to solve work /rate problems is to use smart numbers.
Then we need to find the work rate - work done each entity in 1 day. The subsequent steps are then very easy .
If 2 entities A and B work together, then Work done by A in one day + Work Done by B in one day = Total work done by A and B in one day.
Example - If a machine produces 10 widgets per day and another machine produces 20 widgets per day, then working together both machines can produce 30 (10 + 20) widgets per day.
Let's choose a nr that is divisible by all the numbers given in the question stem - 20,12,10
LCM of 20,12,10 = 60
Let's assume that Total work = 60 units.
Matt and Peter work together to complete the work in 20 days. So the work done by both of them together is 3 units per day (60/20)
Now we are almost done
Matt and Peter worked together for 12 days. Hence working together, they completed 12 x 3 = 36 units of work
What remains is 24 units and Peter completed this work all by himself in 10 days
Hence peter's work rate = 24/10 units per day
Therefore, Time taken by peter to complete the 60 units of work = Total Work /Peter's work rate = (60)/(24/10) = 25 days
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prasannar wrote:
Matt and Peter can do together a piece of work in 20 days. After they have worked together for 12 days Matt stops and Peter completes the remaining work in 10 days. In how many days Peter complete the work separately.
A. 26 days
B. 27 days
C. 23 days
D. 25 days
E. 24 days
Matt and Peter together would complete 12/20=3/5th of the work in 12 days, thus the remaining 2/5th is done by Peter alone in 10 days.
Therefore Peter can complete the work alone in 10/(2/5)=25 days.
Answer: D.
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Re: Matt and Peter can do together a piece of work in 20 days [#permalink]
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$$\frac{1}{M}+\frac{1}{P}= \frac{1}{20}$$
Calculate work done together in 12 days:
$$\frac{1}{20}x12==>\frac{12}{20}=\frac{3}{5}$$
Remaining work is 1-3/5.
Calculate the days left for P to perform work alone:
$$\frac{1}{P}x10days=1-\frac{3}{5}$$
$$\frac{10}{P}=\frac{2}{5}$$
$$P=25 days$$
A. 26 days
B. 27 days
C. 23 days
D. 25 days
E. 24 days
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2016-07-23 11:39:40
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https://www.physicsforums.com/threads/sketch-of-the-electric-field-of-a-laser-beam.921750/
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# I Sketch of the electric field of a laser beam
Tags:
1. Aug 3, 2017
### spareine
I am trying to sketch the electric field E in snapshot of a linearly polarized laser beam. Is it correct that the E vectors bend from vertical to longitudinal near the cylindrical surface of the beam, and that all field lines within a half wave segment are closed loops?
2. Aug 3, 2017
### Gordianus
The electric field lines do not bend at the "top" or the "bottom". The electric field is stronger at the center and weakens as you move from the center to the border, but it always points in the same direction.
3. Aug 3, 2017
### spareine
The electric field cannot weaken along a field line, if the electric field is vertical everywhere. $\nabla \cdot \mathbf{E} = 0$ in the absence of electric charges (Gauss' law).
4. Aug 3, 2017
### davenn
does this help ?
5. Aug 3, 2017
### sophiecentaur
I can't reconcile that with a narrow beam in free space and each E field line around a radiating dipole follows a curve in a plane. The field strength in any particular elevation will follow a Cos θ law for a short dipole. A collinear / coaxial array will produce a more vertically directional pattern but, being the sum of dipole patterns, why would the E vector direction change? The same thing should apply for a very narrow beam, wouldn't it?
BTW The diagram at the top looks more like the fields in a circular waveguide (conducting walls). That wouldn't be the same boundary conditions as for a narrow beam in free space.
6. Aug 3, 2017
### spareine
I don't think polarization is the essence of my question. For me, the advantage of specifying the laser beam is linear polarized is that it is slightly simpler to describe and draw than other options.
The essence of my question is that I am wondering about the profile of the electric field on cross sections of the laser beam, and what the field lines look like, given that somehow they should be closed loops because of Gauss' law.
7. Aug 3, 2017
### spareine
I am afraid I do not understand your answer. My question is about a laser beam in free space, outside the optical cavity. I don't see how or why radiating dipoles are present in that laser beam in free space.
8. Aug 3, 2017
### davenn
the confusion is because the drawing you supplied implies light in a waveguide ... optical fibre, NOT free space
9. Aug 4, 2017
### spareine
Ok, but I am actually more interested in a laser beam in free space. Gordianus suggested the laser beam in free space is identical to an infinitely wide plane wave, multiplied by a gaussian-like beam profile in the cross sectional plane. My objection against that suggestion is that it makes $\nabla\cdot\mathbf{E}$ nonzero. Do you think this objection is invalid?
10. Aug 4, 2017
### tech99
I think there is a distinction between a Field LIne, which is a contour of equal field strength, and a Line of Force, which is the path taken by a positive charge when placed in the field. Field LInes will run circularly around a charge, whereas lines of force will project radially from it.
Lines of Force become further apart as we depart from a charge, so they indicate a weakening field. The original diagram seems to show Lines of Force, but describes them as Field Lines. The field is not entirely contained within the main beam, but at the top of it the field lines wander away into the rest of the page, and finally met up at the bottom. When using Lines of Force, the strength of the field is often shown by using small arrows of varying length.
Regarding polarisation, there are geometrical distortions arising from any source, and it differs depending on the type of radiator. Most usually, the polarisation is "correct" on the principle axes and then bends away to some extent off axis in the 45 degree planes.
I think the diagram from Davenn depicts the fields using arrows proportional to field strength, rather than showing the shape of the field.
(My apologies for using capitalisation to emphasise the terminology).
11. Aug 4, 2017
### sophiecentaur
EM waves are EM waves, whatever wavelength. Why would the E vector of light waves be any different from the E vector of RF waves? They don't know what generated them so they must behave the same way. I was, of course, discussing the far field pattern.
12. Aug 4, 2017
### spareine
I meant field lines in the sense that is common in electricity and magnetism: lines that follow the direction of the vector field (wikipedia). They can only form closed loops if there are no sources and sinks.
13. Aug 4, 2017
### tech99
In such a case, there is no need for the lines to turn over at the edge of the beam, they just continue outside the main beam but weaker..
14. Aug 4, 2017
### spareine
Could you draw that field line pattern? Wouldn't it be radial with a source at r=0? But there is no source at r=0.
15. Aug 4, 2017
### Staff: Mentor
I think the confusion comes because you are trying to use too "big" of a probe to measure the E-field at various points in the beam. Cosnider using a probe that is much smaller than the diameter of the beam. As you move that linearly polarized probe from the center of the beam toward the edge, you measure the Gaussian drop-off of the amplitude of the linearly polarized E-field. The polarization remains the same, just the amplitude drops off.
Like if the beam were aimed vertically in the plot below of E-field squared amplitude versus radius...
http://www.wavelength-tech.com/images/Top Hat Profile.jpg
16. Aug 4, 2017
### spareine
The problem I was trying to address, as mentioned in post #9, is more elementary. It does not involve measurement probes. Multiply a linearly polarized plane wave with a gaussian beam profile. The result is that $\nabla\cdot\mathbf{E}$ is nonzero everywhere (except at the axis of the beam). That is incompatible with free space, where charges are absent.
17. Aug 4, 2017
### tech99
18. Aug 4, 2017
### spareine
I see, so all field lines are closed loops around several side lobes. Thanks!
19. Aug 4, 2017
### sophiecentaur
It's been a long time since I used those operators but I don't think your inference can be correct. If you were right, it would mean that every antenna / radiator would have to be isotropic, I think. And none are.
I found this link and it has a diagram which accounts for what happens to the E field at the edges of the beams. It gives a ∇.E = 0 because the field vectors follow a curve in the other sense from the one you have drawn in your diagram.
20. Aug 4, 2017
### tech99
The circles in my diagram just indicate the nominal position of the main beam and some suggested sidelobes.The sidelobes are 180 degrees out of phase with the main beam.
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2017-10-21 16:08:37
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https://english.meta.stackexchange.com/questions/1152/what-are-we-doing-on-this-site
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# What are we doing on this site?
I described this site to a friend as "the best place I know to spend time constructively when you should be working". But it occurred to me that I don't know the exact way to describe what we're doing. Mark Twain said "work is whatever a body is obliged to do, and play is whatever a body is not obliged to do", but we're not exactly playing. On the other hand, I wouldn't (unless desperate for something to put on my CV), call it voluntary work. Is there a word or short phrase for what you're doing at this moment?
• Learning :) -- For each tidbit I contribute, I receive a wealth of information about a topic I'm interested in. – teylyn May 17 '11 at 22:09
• "I don't know how to waste my time better than to do it here." – user8568 May 17 '11 at 22:13
• We are fixing the world, one grammar mistake at a time. – MikeVaughan May 17 '11 at 22:56
Me personally, I'm edifying. And excogitating. Not necessarily in that order.
• Based on some criteria I couldn't begin to figure out, EL&U 'site management tools' auto-classify your answer as 'potentially low quality'. Personally I'd say it's fine, but could be improved by adding exploring, for example (assuming you want to stick with e's, and that this applies in your case). What the hell - it's good enough for an upvote even as it stands, just for the word excogitating. Much better than plain old cogitating, for thinking not just 'out loud', but 'out there on the net'. – FumbleFingers May 18 '11 at 18:57
• Really? Low quality? I must not have enough words. I have words, too many words, and no speech. – Kit Z. Fox May 18 '11 at 19:01
• @Kit: Nah - prolly summat to do with some very simple rules being followed by a routine that decides your wording is potentially ungrammatical. And let's be honest, you shouldn't really start sentences with And. Not to mention Not. – FumbleFingers May 18 '11 at 19:09
• You'll have to add encountering if we keep meeting like this :-) – FumbleFingers May 18 '11 at 19:18
• I prefer liaising. I'm sure the period-to-word ratio has somfing to do wid it as well. – Kit Z. Fox May 18 '11 at 19:21
• I bet you're right there. Anyway, a few 'false positives' doesn't mean it's not a useful tool. And EL&U is a good site for liaising and coming to a consensus, I find. Language would be in trouble if we could only establish 'proper' current usage by consulting authoritative books which invariably lag some way behind the times. – FumbleFingers May 18 '11 at 19:45
• So the auto-classification must be one of those arcane "moderator tools", huh? I've got a ways to go before I get there. Is concensus a British spelling? I like this site too, but there's nothing wrong with a good etymological tome or two. They make good brain-eats. – Kit Z. Fox May 18 '11 at 19:49
• Duh. Concensus was me still in mind of the fact that earlier today I was pontificating about the recent 10-year UK Census. I don't know how many points you need to activate the 'Review' button, but my guess is you'll be there pretty soon. – FumbleFingers May 18 '11 at 19:57
AFAIC, this question was perfectly positioned in English.stackexchange. It is not a discussion about the "behaviour" on the site, but a question about a word that can be used to describe the activity of the participants of the site. Hence, it is a question about language use and vocabulary.
I feel it should be moved back.
• Are you saying it’s not an appropriate question for the meta site? – nohat May 18 '11 at 2:03
• 1: Your answer is definitely appropriate for meta, 2: There are many kinds of meta: talking about the operation of the ELU site, talking about language in general, talking 'about' things rather talking of them. 3: Rereading the question, there is nothing ELU specific in it...could almost be in meta.SO. That said, it wasn't too weird in main.ELU (maybe a little of topic), but now that it's here no reason to move. – Mitch May 18 '11 at 2:25
• Well, after the introduction sentences, the actual question is "Is there a word or short phrase for what you're doing at this moment?" And that is a question about English. If you all agree that the thread should stay in Meta, that's fine. But I feel it had its place in the English site. – teylyn May 18 '11 at 4:41
• If I had a vote, it would be in English. Which is why I posted it there. – Tim Lymington May 18 '11 at 19:17
• I take the point that the question itself is about the English language. However, I think the "meta-ness" that the question is also about the site means the question should be on meta. A reworded question about a hypothetical site similar to this one probably would be fine to stay on the main site. – nohat May 18 '11 at 20:48
• @Tim,@teylyn: oh. I missed that part, about 'a word or short phrase. bad reading attention. – Mitch May 19 '11 at 0:36
• @teylyn: Technically speaking, I agree completely with your point. But let's be honest - I'm sure even OP knows perfectly well that this question is bound to generate much more discussion about the exact nature of the activity than about what to actually call it. – FumbleFingers May 20 '11 at 15:00
I'd say I'm "participating" in a direct sense, and "exploring" in a general one. I do agree with Twain, though: "play" is anything you don't have to do. Which only means you have to play like you work -- as profitably as you can -- and that too takes experimenting, practice and risk.
The main candidates for me are contributing and participating.
If I have to pick just one, it'll be the former. You could participate in Second Life, for example, without really implying you do anything to improve that site.
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2021-04-22 00:30:44
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http://mathhelpforum.com/statistics/124278-factorial.html
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# Math Help - factorial!
1. ## factorial!
okay, i'm stuck on a couple questions... :/
2!n! / 4!(n-5) = (n-1)! / (n-4)!
and..
(n+2)! / n! = 30
any help would be great, thanks!
2. Originally Posted by needuhelp
okay, i'm stuck on a couple questions... :/
2!n! / 4!(n-5) = (n-1)! / (n-4)!
and..
(n+2)! / n! = 30
any help would be great, thanks!
For the second one $\frac{(n+2)!}{n!}=30$ means that $\frac{(n+2)(n+1)(n)(n-1)(n-2)...}{n(n-1)(n-2)...}=30$.
So by cancelling we get
$(n+2)(n+1)=30$
I'm sure you can finish up. The first is similar, just more messy.
3. could someone post the work involved with the first question, i'm lost lol
4. Originally Posted by needuhelp
could someone post the work involved with the first question, i'm lost lol
post what you tried first
5. i have..
2!(n-1)(n-2)(n-3)(n-4) / 4! = (n-1)(n-2)(n-3)
6. Originally Posted by needuhelp
okay, i'm stuck on a couple questions... :/
2!n! / 4!(n-5)! = (n-1)! / (n-4)!
I assume the red factorial was meant to be there.
This re-arranges into $\frac{2!}{4!} = \frac{(n-1)! (n-5)!}{n! (n-4)!}$.
The left hand side is 1/12. Simplify the right hand side using what you have learned from this thread (and what you should have learned from class too) and solve the resulting equation.
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2015-10-10 20:59:46
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https://datascience.stackexchange.com/questions/60326/please-help-me-find-the-mistake
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I'm trying to use Tensorlow on İris dataset using sublime as the text editor. I got this error "TypeError: float() argument must be a string or a number, not 'generator'"when running the session, but I haven't any Idea about it. Please help me out. Attached is the error screenshot.
• try to do np.array(...).tolist() .. while generating your x1, x2 and y – Sandeep Bhutani Sep 17 '19 at 15:39
If you pull each of them out of the np.array calls and attempt to run them, you'll generate a nice SyntaxError (which is nice because it'll loudly tell us what's wrong!). Right now, the error is getting suppressed by the fact Python is taking that funky list comprehension and creating a generator expression (I'm guessing due to how Python manages argument packing).
rand_y = np.array([binary_target[y_i] for y_i in rand_index])
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2020-01-29 05:26:18
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http://mathhelpforum.com/calculus/208023-difference-between-two-integration-problems.html
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# Math Help - The Difference Between Two Integration Problems
1. ## The Difference Between Two Integration Problems
1.
$\int (x^{2} + 4)^{5} 2xdx$
Do the anti-derivative on the integrand.
$\frac{(x^{2} + 4)^{6}}{6} + C$ Final Answer
and this problem
2.
$\int (x^{3} - 3x)^{1/2}(x^{2} - 1)dx$
Next taking the derivative of the highest power in the integrand $x^{3}$ (BUT WAIT - 3 is not the derivative)
$(x^{3} - 3x)^{1/2} 3(x^{2} - 1)dx$
Next flipping the $3$ so it becomes $\frac{1}{3}$
$\frac{1}{3} \int (x^{3} - 3x)^{1/2} 3(x^{2} - 1)dx$
The stuff to the right of the left linear factor (integrand) disappears. Now we do the anti-derivative on the integrand.
$\frac{1}{3} [\frac{(x^{3} - 3x)^{3/2}}{\frac{3}{2}} + C]$
$\frac{2}{9}(x^{3} - 3x)^{3/2} + C$ Final Answer
As you can see that in the 2nd problem the $\frac{1}{3}$ is a flipped version of 3 which came from the $x^{3}$ in the beginning (since that was the largest power).
But in the 1st problem the highest power $x^{3}$ is not made into 3 and flipped.
But I see a problem in the reasoning: $x^{3}$ differentiated is [tex]3x^{2}[tex] not 3. So why does the 3 come about? Where did it come from?
2. ## Re: The Difference Between Two Integration Problems
Sorry for double post, but I think I figured it out.
The 1st problem is complete, so the anti-derivative can be taken. However, the 2nd problem is incomplete. Threfore, the 3 has to put to the left of $x^{2}$ (in the linear factor to the right of the integrand), to make $3x^{2}$ which is the derivative of $x^{3}$.
So now the 2nd equation is complete and the anti-derivative can be taken. Of course, when you complete an integration problem, then you also have to flip the 3 and put the result $\frac{1}{3}$ in front of the integrand (for some balancing reason).
3. ## Re: The Difference Between Two Integration Problems
Originally Posted by Jason76
... then you also have to flip the 3 and put the result $\frac{1}{3}$ in front of the integrand (for some balancing reason).
... and that "balancing reason" is because you're effectively multiplying the integrand by the constant 1 .
note $1 = \frac{1}{3} \cdot 3$
4. ## Re: The Difference Between Two Integration Problems
In the first example the composite function is $(x^2+4)^5$ and the derivative of the function $x^2+4$, which is $2x$, appears exactly as is as a factor in the intergrand. Hence the substitution rule applies.
In the second example the composite function is $(x^3-3x)^{1/2}$ and the derivative of the function $x^3-3x$ is $3(x^2-1)$. This derivative does not appear exactly as is in the integrand but something close does, which is $x^2-1$. To get the derivative exactly as a factor in the integrand we meed to multiply it by 3, which means we must also divide by 3 so nothing is changed.
5. ## Re: The Difference Between Two Integration Problems
The derivative of the starting left linear factor is larger than the dx of the starting right linear factor. So multiplication is needed. However, if the derivative of the starting left linear factor was smaller than division would be needed. Is that right? If the derivative of the starting left linear factor matches the dx of the starting right linear factor, then you can go ahead and take the anti-derivative without any manipulation.
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2015-05-05 08:17:31
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https://puzzling.stackexchange.com/questions/51730/can-a-note-that-makes-no-sense-help-find-my-friends-car/51731#51731
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# Can a note that makes no sense help find my friends' car?
So my friend (the one who had his car stolen) just found this taped to his door. I can't help thinking that it's related. What do you think?
Here is the text of the note:
I just simply cannot believe that this badger just ate like 8 great big steaming Sausage rolls, what an animal
• Honest question: is this the kind of puzzle people are looking for on Puzzling? For me, deciphering the meaning of the image in the question was a puzzle; but getting from there to the location of the stolen car is just a pile of procedural de-obfuscation. May 12 '17 at 18:40
• @JakeRobb from the feedback I've got, the answer is kinda sorta. Some obfuscation is ok, but not too much. I went overboard basically. BeastlyGerbil said "Don't have massively lengthy strings of binary or stuff which we need to transcribe, and try not to use the same cipher twice or have too many ciphers. Mix things up like you did in the first puzzle" and the final puzzle (actually finding the car) will hopefully reflect that May 12 '17 at 18:54
• @JakeRobb From the little I've seen on here, a +/- 5 letter alphanumeric string is always put into Imgur as part of the puzzling procedure, as so many questions creators do it. May 12 '17 at 19:43
• It's not about being overwhelming. It's about it being detective work. Personally, I come to Puzzling for brainteasers (including rebus types), logic puzzles, and the like. I wasn't familiar with the imgur-maze tag, nor does this puzzle have it -- but as there's apparently a significant subset of the community who enjoys imgur mazes, that's fine. I think adding that tag that would help here. May 12 '17 at 20:42
• One more thing -- I feel fairly strongly that puzzles should not assume that everyone is familiar with the norms of the community. I participate in several other SE sites, and this is the first time I've encountered such a thing. Perhaps this is just a consequence of Puzzling's slight mismatch with SE's standard Q&A model. May 12 '17 at 20:44
# 1
A golf shout is "fore", and the second image represents a key (going into a lock).
Clarification From OP: The two images represent "The Key Is Four". The key is for all of the puzzle
# 2
Taking every fourth word's first letter gives us 5 characters (cb8Sa) of an Imgur URL, leading to a message:
# 3
That message can be decoded as base64, leading to http://jsfiddle.net/86s8kzm8/.
Clarification from OP: The fourth message is "Zlad from TBL". Zlad is an anagram of "Ladz" and "TBL" could be "The Bad Ladz" which is the name of the gang.
# 4
The fourth message has suspicious capitalization.
Taking the first letter of each line gives another Imgur URL:
https://i.stack.imgur.com/hgpKr.jpg
whose text is 01010111 01000110 01101000 01110011 00100000 01010001 00111001 01110111 01100111 00100000 01010111 01000010 01110010 01101000 00100000 01001101 01010100 01001001 01110101 00100000 01010011 01000001 01010010 01000111 00001010 01110011 01100100 01000110 01000010 00100000 01100111 01001000 01100111 00110100 00100000 01001111 01000100 01000101 00110000 00100000 01110111 01000010 01110010 01010100 00100000 01100010 00111001 01110000 01010111 00001010 01110001 01011001 01110010 00111001 00100000 01001110 01000100 01000101 00110100 00100000 01110111 00111001 01000111 01101000 00100000 01000101 01001001 01110010 01101000 00100000 01000111 00110000 01110111 00110010 00001010 01000011 01101001 00110000 01110111 00100000 01000101 01010010 01001000 01100111 00100000 01001010 00111001 00110010 00110100 00100000 00111000 01011001 01101000 00111001 00100000 01001100 01101010 01001001 01111001 00001010 01110111 01001111 01001000 01000010 00100000 00110100 01010111 01001111 00110101 00100000 00111000 01110111 01010011 01000100 00100000 01001110 01000100 01010001 00110100 00100000 00111001 01000111 01001000 00110010 00001010 00111001 01000111 00110100 01101000 00100000 01001001 01000010 01010010 01110111 00100000 01000011 01101100 01001010 01110110 00100000 01010110 01010111 01110010 00111001 00100000 00110101 01110100 00111000 01000101 00001010 01010110 01001000 01010111 01000101 00100000 01100010 00110010 00110000 00110101 00100000 01100110 01100010 01100100 01101000 00100000 01010111 01001111 01000101 01110101 00100000 01001000 01010110 01110100 00110010.
# 5
Decoding as binary, we can see that the message is split up into groups of four alphanumeric characters. Taking every fourth chunk, then decoding as base64 gives:
12.814418 -0.22448 Room9
# 6
Multiplying those coordinates by 4 gives us:
51.257672 -0.8992 Room 36
Finding their location on the map gives:
They are in the Four Seasons Hotel in Hampshire, in Room 36
• HHmmmm, I dont know..... You think they put his car in a hotel room? May 12 '17 at 14:27
• Yeah if this gang stole my car I'd never get it back May 12 '17 at 14:56
• OP confirmed in chat it is actually in room 36 (9*4) May 12 '17 at 15:04
• The puzzle is just a distraction to get the car dismantled then sold as parts! :O May 12 '17 at 15:27
• I feel like there's enough evidence here that the car was stolen that the insurance company should pay out and we can just drop this whole business. ;) May 12 '17 at 17:35
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2022-01-21 15:44:27
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http://www.physicsforums.com/showthread.php?t=222667&page=4
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# Progress in Afghanistan
by Astronuc
Tags: afghanistan, progress
P: 21,597
http://topics.nytimes.com/top/refere...nan/index.html
David D. McKiernan, the American four-star general who led the allied ground forces during the invasion of Iraq in 2003, became the NATO commander in Afghanistan as of June 2008. As head of the International Security Assistance Force, he is the military leader in charge of the allied war effort in Afghanistan against increasingly deadly and aggressive attacks by Taliban and Al Qaeda militants, many of whom are based in western Pakistan. General McKiernan was never a favorite of former Defense Secretary Donald H. Rumsfeld. In the months before the Iraq war, he pressed to begin the war with a greater number of troops than authorized in the plan he had inherited. After the invasion he was made the deputy head of the Army's Forces Command, which oversees the training of American troops in the United States. In 2005, he was awarded a fourth star and made the head of American Army troops in Europe.
General Says He’s Hopeful About Taliban War
http://www.nytimes.com/2008/10/13/wo.../13afghan.html
Perhaps. I think the US can win all the battles, but unless there is a stable and non-corrupt government to lead the country, it would seem the victory(ies) would be in vain. My greatest concern and objection is the killing of non-combatants - women and children. I don't expect the Taliban to be concerned about that, but I do think that US and NATO forces should go the extra step not to kill civilians - that means no firing into villages or compounds unless one is sure that no women and children are present.
Mentor
P: 25,748
Quote by Astronuc Perhaps. I think the US can win all the battles, but unless there is a stable and non-corrupt government to lead the country, it would seem the victory(ies) would be in vain. My greatest concern and objection is the killing of non-combatants - women and children. I don't expect the Taliban to be concerned about that, but I do think that US and NATO forces should go the extra step not to kill civilians - that means no firing into villages or compounds unless one is sure that no women and children are present.
But that's impossible. These people deliberately hide among civilians for protection.
P: 1,295
http://news.bbc.co.uk/2/hi/south_asia/7656745.stm
For Kuchi nomads like Rahmat Goal's family, survival is a daily struggle.
P: 1,295
Quote by Evo But that's impossible. These people deliberately hide among civilians for protection.
I think civilians hide them. If you go on killing civilians, what you expect from those normal people?
I think US cannot eliminate the Afghanistan people who don't like US.
P: 21,597
Quote by Evo But that's impossible. These people deliberately hide among civilians for protection.
It's difficult but not impossible. The Taliban and sympathizers simply live in their homes and neighborhoods/villages when they are not out fighting the Afghan government and US/NATO forces. It's their country. The US/NATO are propping up a government in what is eseentially a civil war that spans two countries.
If one calls in an AC-130, one is going to kill civilians. The US and NATO forces need to be smarter.
Meanwhile Pakistan and Afghanistan go hand in hand because the Pahstuns (e.g. Waziri) straddle the border.
http://en.wikipedia.org/wiki/Pashtun
Intelligence report: U.S. antiterror ally Pakistan 'on the edge'
http://news.yahoo.com/s/mcclatchy/20...latchy/3072503
WASHINGTON — A growing al Qaida -backed insurgency, combined with the Pakistani army's reluctance to launch an all-out crackdown, political infighting and energy and food shortages are plunging America's key ally in the war on terror deeper into turmoil and violence, says a soon-to-be completed U.S. intelligence assessment. A U.S. official who participated in drafting the top secret National Intelligence Estimate said it portrays the situation in Pakistan as "very bad." Another official called the draft "very bleak," and said it describes Pakistan as being "on the edge." The first official summarized the estimate's conclusions about the state of Pakistan as: "no money, no energy, no government." Six U.S. officials who helped draft or are aware of the document's findings confirmed them to McClatchy on the condition of anonymity because NIEs are top secret and are restricted to the president, senior officials and members of Congress . An NIE's conclusions reflect the consensus of all 16 U.S. intelligence agencies. The NIE on Pakistan , along with others being prepared on Afghanistan and Iraq , will underpin a "strategic assessment" of the situation that Army Gen. David Petraeus , who's about to take command of all U.S. forces in the region, has requested. The aim of the assessment — seven years after the U.S. sent troops into Afghanistan — is to determine whether a U.S. presence in the region can be effective and if so what U.S. strategy should be. . . . .
P: 21,597
http://news.yahoo.com/s/ap/20081017/...as_afghanistan
Official: Afghans probing 17 civilian deaths
KANDAHAR, Afghanistan – Afghan authorities are investigating the deaths of at least 17 civilians during a clash between NATO forces and militants in southern Afghanistan, an official said Friday. Villagers and a senior police official claimed Thursday that a NATO airstrike killed the civilians, including women and children, in Nad Ali district of the Helmand province. The NATO-led force in Afghanistan confirmed that it carried out an airstrike in the area on Thursday — but not that it resulted in any civilian casualties. NATO spokesman Capt. Mark Windsor said Friday the force was seeking more information and declined further comment. Daud Ahmadi, spokesman for Helmand's governor, said Friday that authorities were investigating whether the airstrike or "insurgent action" caused the collapse of the house in which the civilians died. Angry villagers brought more than a dozen corpses — including the badly mangled bodies of women and children — to the governor's house in the town of Lashkar Gah on Thursday, said Haji Adnan Khan, a tribal leader who had seen the bodies. Nad Ali, about 6 miles from Lashkar Gah, has been a scene of heavy fighting between insurgents and Afghan and foreign troops. Militants control much of the area around the village.
Maybe some militants are angry with the US/EU (or some hate America and Europe) because invaders (US and EU/NATO military) kill their women and children, or parents, or siblings or other family members, or friends. Iraq, Afghanistan, Pakistan have not invaded or threatened the US, although Saddam Hussein threated the US/Israel with retaliation if attacked.
The hijackers who attacked the WTC and Pentagon on Sept 11, were from Saudi Arabia, United Arab Emirates (now headquarters for Halliburton), Egypt and Lebanon.
Osama bin Laden (al Qaida) is from Saudi Arabia and Ayman al-Zawahiri is Egyptian. One connection with Pakistan would be Khalid Sheikh Mohammed who was born in Kuwait to parents from Baluchistan (Pakistan).
P: 1,510 At this point I have absolutely no clue what the mission is in Afghanistan. What are the US and other NATO forces trying to achieve? If their role is to provide security and stability then they are failing miserably and indeed are the main causes of the lack of security and instability. By supporting what were the murdering war lords of the former Northern Alliance they have helped drug dealing, corrupt criminals to take power and are expending huge resources in men and material to keep them there at a cost of alienating an entire generation of people. Why???
HW Helper
P: 2,275
Quote by Astronuc http://news.yahoo.com/s/ap/20081017/...as_afghanistan Official: Afghans probing 17 civilian deaths Maybe some militants are angry with the US/EU (or some hate America and Europe) because invaders (US and EU/NATO military) kill their women and children, or parents, or siblings or other family members, or friends. Iraq, Afghanistan, Pakistan have not invaded or threatened the US, although Saddam Hussein threated the US/Israel with retaliation if attacked. The hijackers who attacked the WTC and Pentagon on Sept 11, were from Saudi Arabia, United Arab Emirates (now headquarters for Halliburton), Egypt and Lebanon. Osama bin Laden (al Qaida) is from Saudi Arabia and Ayman al-Zawahiri is Egyptian. One connection with Pakistan would be Khalid Sheikh Mohammed who was born in Kuwait to parents from Baluchistan (Pakistan).
So you're saying the Taliban government in Afghanistan was right to deny the US permission to attack al-Qaida in their country since all of the al-Qaida in their country were foreigners?!
That's going back to the old cop out where terrorists can attack as they please because as long as the government of whatever country they reside in didn't actually make the attack, therefore the US can't respond.
Quote by Art At this point I have absolutely no clue what the mission is in Afghanistan. What are the US and other NATO forces trying to achieve? If their role is to provide security and stability then they are failing miserably and indeed are the main causes of the lack of security and instability. By supporting what were the murdering war lords of the former Northern Alliance they have helped drug dealing, corrupt criminals to take power and are expending huge resources in men and material to keep them there at a cost of alienating an entire generation of people. Why???
Goes back to that politically correct idea that the US can't go into a country, accomplish what they needed to do, then leave the country in the same shambles it was in before they invaded. We need to somehow make the country a better place than it was before we attacked.
That's just not always realistic and Afghanistan is one of those instances.
I don't think that necessarily means we shoud leave. The US still hasn't accomplished what it set out to do. Doing that raises the possibility of an even bigger mess.
US and Pakistani forces have had minor skirmishes with each other over the last month. I think the skirmishes will continue for a while and could get worse.
That raises an interesting possibility. We could have combat, complete with casualties, with another nuclear armed country. That would be a new world first.
We could also have combat where both the US and the enemy forces were being funded by the US taxpayer. Then again, most of the US aid goes to beefing up Pakistan's forces along their border with India rather than to their forces along the Afghanistan border, so maybe we're not funding enemy forces so much.
I'm not sure how the US should have handled Pakistan immediately after the Afghanistan invasion, but the way we did handle them hasn't worked.
P: 21,597
Quote by BobG So you're saying the Taliban government in Afghanistan was right to deny the US permission to attack al-Qaida in their country since all of the al-Qaida in their country were foreigners?!
I didn't say that. It's quite complicated and there is not simple answer.
Al Qaida and their Taliban sympathizers who are planning attacks against US, Pakistan and Afghan governments and innocent people are legitimate targets (at least according to internationals standards).
If one reads the reports from Robert Fisk (The Independent, UK), he mentions the situation with the people in the border region whose villages were bombed or shelled by US forces. Most are not Taliban, but some members in the villages may be Taliban. Non-combatants, including women and children, are killed. I have a big problem with that!
The first interaction some of these people have with the outside world is American/NATO military attacking their villages.
Paraphrasing an old proverb - One evil deed (or misdeed) undermines 1000 good deeds.
I think there is a better way, and I'm working on it.
P: 21,597
Pakistan and Afghanistan go hand in hand. Both countries share tribes and a common history.
Pakistan is in deep trouble economically.
Pakistan reported nearing default, to seek IMF help
NEW YORK (MarketWatch) -- After failing to get help from China, Pakistan may need to turn to the International Monetary Fund -- a politically unpopular move -- for cash to bolster its economy and avoid defaulting on its debt obligations, according to news reports Sunday. The country, perceived as one of the world's riskiest borrowers, may need as much as $6 billion to boost its foreign-currency reserves, which fell more than 74% in the past year to about$4.3 billion, according to a Bloomberg News report. The next interest payment for Pakistan on its dollar-denominated bonds is due in December, and the government is scheduled to repay $500 million in February on a 6.75% note, the report said. Pakistan's President Asif Ali Zardari returned from China late Friday failing to secure a cash commitment from its neighbor, the New York Times reported. China had been seen as a last resort before Pakistan turns to the IMF, the Times report said. Saudi Arabia, another of the country's traditional ally, refused earlier to offer concessions on oil, it said. Pakistan and IMF must step carefully. Admin P: 21,597 Afghan journalism student sentenced to 20 years http://news.yahoo.com/s/ap/20081021/...urnalist_trial KABUL, Afghanistan – An Afghan appeals court overturned a death sentence Tuesday for a journalism student accused of blasphemy for asking questions in class about women's rights under Islam. But the judges still sentenced him to 20 years in prison. The case against 24-year-old Parwez Kambakhsh, whose brother has angered Afghan warlords with his own writings, has come to symbolize Afghanistan's slide toward an ultraconservative view on religious and individual freedoms. "I don't accept the court's decision," Kambakhsh told The Associated Press as he was leaving the courtroom. "It is an unfair decision." The case can be appealed to the Supreme Court, the highest court in Afghanistan. John Dempsey, a U.S. lawyer working for six years to reform the Afghan justice system, said Kambakhsh has yet to get a fair trial. . . . . Besides the accusation that Kambakhsh disrupted class with his questions, prosecutors also said he illegally distributed an article he printed off the Internet that asks why Islam does not modernize to give women equal rights. He also allegedly wrote his own comments on the paper. In January, a lower court sentenced him to death in a trial critics have called flawed in part because Kambakhsh had no lawyer representing him. Muslim clerics welcomed that court's decision and public demonstrations were held against the journalism student because of perceptions he had violated the tenets of Islam. . . . . Disrupting class should not result in the death penalty. Distributing literature that asks "why Islam does not modernize to give women equal rights" should not result in a death penalty. This is not a democratic system, but rather it is an oligarchy. The apparent goal in Afghanistan is to preclude a haven for al Qaida. Aside from that, the Bush administration seems less concerned about democracy. P: 1,510 Yet more civilians killed by 'accident' Air raid 'kills Afghan civilians' At least five Afghan civilians have been killed in an air strike by international forces in the north-west, local officials say. They say at least 13 insurgents also died in the raid, after they attacked Afghan and international troops in Ghormach district in Badghis province. United States-led forces have said they are checking the reports. On Wednesday President Hamid Karzai condemned a US air strike which killed 40 people at a wedding in the south. The raid was aimed at the Taleban in the province of Kandahar. Mr Karzai called on Barack Obama to prevent civilian casualties when he take over as US president. http://news.bbc.co.uk/2/hi/south_asia/7713065.stm Given the lack of impact of numerous previous condemnations of attacks leading to civilian deaths it seems until a military commander is actually held responsible for one of these atrocities it is likely they will continue unabated. Even if the military top brass care nothing for civilian casualties they must realise these reckless attacks are the perfect recruiting sergeant for anti-western forces. P: 65 Attackers in Afghanistan have sprayed acid in the faces of at least 15 girls near a school in Kandahar, police say. ... Correspondents say the attack is likely to have been carried out by those opposed to the education of women. http://news.bbc.co.uk/2/hi/south_asia/7724505.stm These attacks on women/girls, which happen in similar fashion elsewhere as well, I never understand. It only proves how weak those guys are where they become concerned about females actually accomplishing something with their lives. Geez, talk about insecurities! P: 1,510 A good article from the Independent drawing parallels between the situation in Afghanistan today and the past. snip Kabul 30 years ago, and Kabul today. Have we learned nothing? At night, the thump of American Sikorsky helicopters and the whisper of high-altitude F-18s invade my room. The United States of America is settling George Bush's scores with the "terrorists" trying to overthrow Hamid Karzai's corrupt government. Now rewind almost 29 years, and I am on the balcony of the Intercontinental Hotel on the other side of this great, cold, fuggy city. Impeccable staff, frozen Polish beer in the bar, secret policemen in the front lobby, Russian troops parked in the forecourt. The Bala Hissar fort glimmers through the smoke. The kites – green seems a favourite colour – move beyond the trees. At night, the thump of Hind choppers and the whisper of high-altitude MiGs invade my room. The Soviet Union is settling Leonid Brezhnev's scores with the "terrorists" trying to overthrow Barbrak Karmal's corrupt government. Thirty miles north, all those years ago, a Soviet general told us of the imminent victory over the "terrorists" in the mountains, imperialist "remnants" – the phrase Kabul communist radio always used – who were being supported by America and Saudi Arabia and Pakistan. Fast forward to 2001 – just seven years ago – and an American general told us of the imminent victory over the "terrorists" in the mountains, the all but conquered Taliban who were being supported by Saudi Arabia and Pakistan. The Russian was pontificating at the big Soviet airbase at Bagram. The American general was pontificating at the big US airbase at Bagram. This is not déjà-vu. This is déjà double-vu. And it gets worse. http://www.independent.co.uk/opinion...g-1029920.html Admin P: 21,597 Ideally, Afghanistan would become calmer, democratic and more favorably disposed to the US and other countries, however - Inexplicable Wealth of Afghan Elite Sows Bitterness In One of the World's Poorest Nations, Myriad Tales of Official Corruption http://www.washingtonpost.com/wp-dyn...011102038.html KABUL, Jan. 11 -- Across the street from the Evening in Paris wedding hall, a monument to opulence surrounded by neon-lighted fountains and a five-story replica of the Eiffel Tower, is a little colony of tents where 65 families, mostly returnees from Pakistan, huddle against the winter cold and wish they had never come home. Similar startling contrasts abound across the Afghan capital. Children with pinched faces beg near the mansions of a tiny elite enriched by foreign aid and official corruption. Hundreds of tattered men gather at dawn outside a glittering new office building to compete for 50-cent jobs hauling construction debris. "I am a farmer with 11 children. Our crops dried up, so I came to the city to find work, but all day I stand here in the cold and no one hires me," said Abdul Ghani, 47. "All the jobs and money go to those who have relatives in power, and corruption is everywhere. How else could they build these big houses? Nobody cares about the poor," he added bitterly. "They just make fun of us." Seven years after the fall of the Taliban and the establishment of a civilian-led, internationally backed government, Afghanistan remains one of the poorest countries in the world, with rates of unemployment, illiteracy, infant mortality and malnutrition on a par with the most impoverished nations in sub-Saharan Africa. Most homes lack light, heat and running water; most babies are born at home and without medical help. . . . . Something's got to change - and ASAP. P: 1,549 This is why I was concerned about Obama's declaration that he wanted to focus on Afghanistan and "kill Osama bin Laden." The place is a wreck and more military action to try ferreting out a man who may already be dead is not going to fix anything. I hope he has some good ideas and doesn't really intend to continue America's 9/11 vendetta. Admin P: 21,597 Interesting perspective on Afghanistan. What Are We Doing in Afghanistan? http://www.slate.com/id/2210624 We're still figuring that out. By Fred Kaplan Not long ago, Afghanistan was known as "the good war." Now some are calling it "Obama's Vietnam." Both tags exaggerate. . . . Unlike those who got us into Vietnam, today's top officials—including President Barack Obama and Defense Secretary Robert Gates—at least see the specter. Both have emphasized that their goals in Afghanistan are limited; daydreams of turning the place into a democratic republic—"some central Asian Valhalla," as Gates snorted in recent hearings—are over. Gates further stated at those hearings, before the Senate armed services committee, that he would endorse his commanders' request for three additional brigades—but that he'd be "deeply skeptical" of subsequent requests for more. The fighting needs to be done mainly by Afghan troops, he said, adding that if the Afghan people begin to see it as an American war, "we will go the way of other imperial occupiers." . . . . . On a side note - Kyrgyzstan says U.S. air base decision is final http://news.yahoo.com/s/nm/20090206/...an_usa_base_28 BISHKEK (Reuters) – Kyrgyzstan said on Friday its decision to shut a U.S. air base was final, dealing a blow to Washington's efforts to retain what has been an important staging post for U.S. forces fighting in Afghanistan. The United States said it was still "engaged" with Kyrgyzstan about keeping the Manas base in the poor, former Soviet republic and traditional Russian ally. But one senior Kyrgyz official said no talks were currently taking place. Kyrgyzstan's stance has set a tough challenge for new U.S. President Barack Obama, who plans to send more troops to Afghanistan to try to boost NATO efforts to defeat Taliban and al Qaeda insurgents. The standoff over the tiny but strategically placed nation marks a new twist in an escalating power struggle in Central Asia reminiscent of the 19th-century "Great Game" between tsarist Russia and the British Empire. "The air base's fate has been decided," Adakhan Madumarov, Secretary of the Kyrgyz Security Council, told reporters. "I see no reason why the air base should remain in place now that this decision has been taken ... We are not holding any talks on this," he added, hinting there will be no further discussions with Washington on the air base. Kyrgyz President Kurmanbek Bakiyev announced the closure of the base this week after securing more than$2 billion in financial aid and credit from Russia at talks in Moscow. The announcement left the United States scrambling to find alternative supply routes through other parts of Central Asia for shipments bound for landlocked Afghanistan. Speaking in Tajikistan, another ex-Soviet republic in Central Asia, the U.S. envoy to Dushanbe said Tajikistan had agreed to offer its air space for transport of non-military NATO supplies to Afghanistan. A Western diplomatic source told Reuters separately on Thursday the United States was close to a deal with Uzbekistan that would also allow Washington to open a new railway supply route for its troops in Afghanistan. . . . . RUSSIAN POSITION Russia, irked by the U.S. military presence in Kyrgyzstan which it regards as part of its strategic sphere of interest, has long exerted pressure on the landlocked and mountainous Central Asian country to evict the U.S. forces. . . . .
The Russians have offered to allow US transit of 'non-lethal' aid, e.g. food and supplies, and medical evacuations. I'm sure they are pleased to accept the money.
Uzbekistan has had a somewhat repressive government. All of the Central Asian countries are relatively poor, and their trade is hampered by powerful or unstable neighbors - not to mention corruption.
As someone correctly pointed out, both Kyrgyzstan and Uzbekistan (in fact all the stans) have had governments, which are problematic with respect to observing basic human and civil rights.
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2014-03-12 15:38:39
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https://www.effortlessmath.com/math-topics/how-to-find-the-central-angle-of-a-circle/
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# How to Find the Central Angle of a Circle?
A central angle is an angle formed by two arms whose vertex is at the center of the circle. In this step-by-step guide, you learn more about finding the central angle.
The central angle is useful for dividing a circle into sections. A slice of pizza is a good example of a central angle. A pie chart consists of several sections and helps to display different values.
## A step-by-step guide tofinding the central angleof a circle
The central angle is the angle subtended by an arc of a circle at the center of a circle. Radius vectors form the arms of the central angle. In other words, it is an angle whose vertex is the center of a circle whose arms are two radii lines that intersect at two different points in the circle. When these two points are connected, they form an arc. The central angle is the angle subtended by this arc at the center of the circle.
### How to find the central angle of a circle?
The central angle is the angle between the two radii of the circle. To find the central angle, we need to find the length of the arc, and the length of the radius. The following steps show how to calculate the central angle in radians.
There are three simple steps to finding the central angle:
• Identify the ends of the arc and the center of the circle (curve). $$AB$$ is the arc of the circle and $$O$$ is the center of the circle.
• Connect the end of the arc with the center of the circle. Also, measure the length of the arc and its radius. Here $$AB$$ is the arc length and $$OA$$ and $$OB$$ are the radii of the circle.
• Divide the length of the curve by the radius to get the central angle. Using the following formula, we will find the value of the central angle in radians.
$$\color{blue}{Central\:Angle=\frac{Length\:of\:the\:Arc}{Radius}}$$
### Finding the Central Angle of a Circle – Example 1:
If the arc length is $$8$$ inches and the central angle is $$120$$ degrees, find the radius of the arc.
Solution:
$$Radius\:of\:the\:arc= 8\space inches$$
$$Central\:angle= 120°$$
$$Central\:angle\:=\frac{\left(length\:of\:arc\:×\:360°\right)}{\left(2\pi \:×\:radius\right)}$$
$$radius\:=\frac{\left(length\:of\:arc\:×360°\right)}{\left(2\pi \:×\:Central\:angle\right)}$$
$$radius\:=\frac{\left(8\:×\:360°\right)}{\left(2\pi \:×120°\right)}$$
$$radius\:=\frac{12}{\pi \:}$$
## Exercises for Finding the Central Angleof a Circle
• From the diagram shown, find the $$m∠BOD$$ measure.
• From the diagram shown, find the $$m∠arc MNK$$ measure.
• If the arc length measurement is about $$21\space cm$$ and the length of the radius measures $$10\space cm$$, find the central angle.
• $$\color{blue}{60°}$$
• $$\color{blue}{234°}$$
• $$\color{blue}{120.38°}$$
### What people say about "How to Find the Central Angle of a Circle?"?
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2023-03-30 17:27:49
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https://physics.stackexchange.com/questions/437601/two-level-laser-rate-equation
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# Two-level laser rate equation
I am stuck on what I assume is a very basic rearranging of terms in Siegman's Lasers, Page 204. Here, the saturation of a laser medium is introduced. The change of the populations of two energy levels are given as
$$\frac{\text{d}N_1}{\text{d}t}=-\frac{\text{d}N_2}{\text{d}t}=-(W_{12}+w_{12})N_1+(W_{21}+w_{21})N_2$$
with $$N=N_1+N_2$$ and $$\Delta N=N_1-N_2$$
Now, as for the rearranging of terms:
The equations for $$\text{d}N_1/\text{d}t$$ and $$\text{d}N_2/\text{d}t$$ can be combined into a single rate equation
$$\frac{\text{d}}{\text{d}t}\Delta N=-(W_{12}+W_{21})\Delta N-(w_{12}+w_{21})\bigg(\Delta N-\frac{w_{21}-w_{12}}{w_{12}+w_{21}}N\bigg)$$
which is, if I didn't get anything wrong, just $$=N_1(-W_{12}-W_{21}-2w_{12})+N_2(W_{12}+W_{21}+2w_{21})$$
But with the information given in this section of the book, I just can't get it to agree with
$$\frac{\text{d}}{\text{d}t}\Delta N=\frac{\text{d}N_1}{\text{d}t}-\frac{\text{d}N_2}{\text{d}t}=2\cdot \frac{\text{d}N_1}{\text{d}t}=2\cdot(-\Delta N(W_{12}+W_{21}+w_{12}+w_{21}))$$
I hope I'm just stuck on something trivial.
The thing I missed is the following trivial, yet important fact: Because the upwards- and downwards stimulated transition probabilities are the same $$W_{12}\equiv W_{21}$$,
\begin{align} \frac{\text{d}\Delta N}{\text{d}t}&=-(W_{12}+W_{21})\Delta N-(w_{12}+w_{21})\bigg(\Delta N-\frac{w_{21}-w_{12}}{w_{12}+w_{21}}N\bigg)\\ &=N_1(-W_{12}-W_{21}-2w_{12})+N_2(W_{12}+W_{21}+2w_{21}) \end{align}
\begin{align} \frac{\text{d}N_1}{\text{d}t}-\frac{\text{d}N_2}{\text{d}t}&=2\bigg[-\big(W_{12}+w_{12}\big)N_1+\big(W_{21}+w_{21}\big)N_2\bigg]\\ \end{align}
From there on, the population recovery time $$T_1$$ can be introduced, along with the saturation.
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2020-11-29 10:43:31
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https://zbmath.org/authors/?q=ai%3Anicolaescu.liviu-i
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# zbMATH — the first resource for mathematics
## Nicolaescu, Liviu I.
Compute Distance To:
Author ID: nicolaescu.liviu-i Published as: Nicolaescu, L.; Nicolaescu, L. I.; Nicolaescu, Liviu; Nicolaescu, Liviu I. External Links: MGP
Documents Indexed: 68 Publications since 1986, including 8 Books Reviewing Activity: 80 Reviews
#### Co-Authors
61 single-authored 3 Némethi, András 1 Caşcaval, Radu A. 1 Cibotaru, Daniel Florentiu 1 Rowekamp, Brandon 1 Savale, Nikhil
all top 5
#### Serials
7 Analele Ştiinţifice ale Universităţii Al. I. Cuza din Iaşi. Serie Nouă. Matematică 4 Gazeta Matematica. Perfectionare Metodica si Metodologica in Matematica si Informatica 3 Studii şi Cercetări Matematice 2 Analele Ştiinţifice ale Universităţii Al. I. Cuza din Iaşi. (Serie Nouă.) Secţiunea Ia. Matematică-Informatică 2 Memoirs of the American Mathematical Society 2 Transactions of the American Mathematical Society 2 Differential and Integral Equations 2 Comptes Rendus de l’Académie des Sciences. Série I 2 Communications in Analysis and Geometry 2 Gazeta Matematică. Revistă de Cultură Matematică 2 Universitext 1 Israel Journal of Mathematics 1 Journal of Mathematical Physics 1 Advances in Mathematics 1 Canadian Journal of Mathematics 1 Compositio Mathematica 1 Duke Mathematical Journal 1 Geometriae Dedicata 1 Indiana University Mathematics Journal 1 International Journal of Mathematics and Mathematical Sciences 1 Journal of the London Mathematical Society. Second Series 1 Pacific Journal of Mathematics 1 Proceedings of the American Mathematical Society 1 Ricerche di Matematica 1 Libertas Mathematica 1 Annals of Global Analysis and Geometry 1 Probability Theory and Related Fields 1 Differential Geometry and its Applications 1 Aequationes Mathematicae 1 Stochastic Processes and their Applications 1 Analele Ştiinţifice ale Universităţii Al. I. Cuza din Iaşi. Matematică 1 The New York Journal of Mathematics 1 Selecta Mathematica. New Series 1 Bernoulli 1 The Asian Journal of Mathematics 1 Geometry & Topology 1 Communications in Contemporary Mathematics 1 Algebraic & Geometric Topology 1 De Gruyter Studies in Mathematics 1 Graduate Studies in Mathematics 1 Functional Analysis and Other Mathematics 1 Gazeta Matematică. Seria A 1 Bollettino dell’Unione Matematica Italiana
all top 5
#### Fields
32 Global analysis, analysis on manifolds (58-XX) 19 Manifolds and cell complexes (57-XX) 16 Differential geometry (53-XX) 15 Partial differential equations (35-XX) 9 Probability theory and stochastic processes (60-XX) 7 Combinatorics (05-XX) 7 Algebraic geometry (14-XX) 6 Dynamical systems and ergodic theory (37-XX) 5 Several complex variables and analytic spaces (32-XX) 4 Number theory (11-XX) 4 Ordinary differential equations (34-XX) 4 Algebraic topology (55-XX) 3 Linear and multilinear algebra; matrix theory (15-XX) 3 Real functions (26-XX) 3 Harmonic analysis on Euclidean spaces (42-XX) 3 Operator theory (47-XX) 3 Calculus of variations and optimal control; optimization (49-XX) 3 Geometry (51-XX) 2 Mathematical logic and foundations (03-XX) 2 $K$-theory (19-XX) 2 Measure and integration (28-XX) 2 Abstract harmonic analysis (43-XX) 2 Statistics (62-XX) 1 Order, lattices, ordered algebraic structures (06-XX) 1 Topological groups, Lie groups (22-XX) 1 Difference and functional equations (39-XX) 1 Sequences, series, summability (40-XX) 1 Integral transforms, operational calculus (44-XX) 1 Functional analysis (46-XX) 1 Convex and discrete geometry (52-XX) 1 Numerical analysis (65-XX) 1 Computer science (68-XX) 1 Mechanics of particles and systems (70-XX)
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2019-08-18 14:08:59
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http://openstudy.com/updates/556d9f85e4b050a18e841db2
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• anonymous
Megan and Tina each have loans for $7000 with a 3-year loan term. Megan pays 6% interest and Tina pays 4% interest. If Megan’s monthly payment is$213 and Tina’s monthly payment is $207, how much more in interest will Megan pay than Tina? A.$213 B. $216 C.$452 D. \$668
Mathematics
Looking for something else?
Not the answer you are looking for? Search for more explanations.
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2017-03-28 16:25:55
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https://mathoverflow.net/questions/181439/on-a-weak-tree-property-for-inaccessible-cardinals
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# On a weak tree property for inaccessible cardinals
Suppose that $\kappa$ is inaccessible and consider a tree of height $\kappa$ whose levels have size strictly below some cardinal $\gamma < \kappa$. Does this type of tree always have a $\kappa$-branch?
The answer is known to be positive when $\gamma=\omega$, i.e., when the levels are finite. On the other hand, if the levels are only known to have size strictly below $\kappa$ itself, then the answer is negative in general since the existence of a $\kappa$-branch implies that $\kappa$ is weakly compact.
I would be interested in knowing about references/proofs on whether this weak tree property holds for every inaccessible or if in fact it constitutes a large cardinal axiom whose strength is somewhere between inaccessibility and weak compactness. Ideally, I would expect a proof of a positive answer in ZFC (one can assume that the inaccessible is given).
Theorem Suppose that $\kappa = cf(\kappa) > \gamma$, and $(T, <_{T})$ is a $\kappa$-tree each of whose levels has cardinality less than $\gamma$. Then $(T, <_{T})$ has a cofinal branch.
The proof uses the Pressing Down Lemma (Fodor's theorem) and the Pigeonhole Principle with the bound $\gamma$ to find the branch.
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2022-05-24 05:21:45
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https://testbook.com/question-answer/a-barium-titanate-piezoelectric-crystal-with-d33--5b1691e3be91960c24715e49
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A barium titanate piezoelectric crystal with d33 = 150 pC/N, Ccrystal = 25pF and Rcrystal = 1010 Ω is used to measure the amplitude of a step force. The voltage output is measured using a digital voltmeter with input impedance 1013 Ω connected acorss the crystal. All other capacitances may be neglected. A step force of 2 N is applied from direction “3” on the crystal. The time in milliseconds within which the voltmeter should sample the crystal output voltage so that the drop form the peak value is no more than 0.12 V is_________
Answer (Detailed Solution Below) 2.48 - 2.52
Free
CT 1: Ratio and Proportion
1963
10 Questions 16 Marks 30 Mins
Detailed Solution
Time constant, τ = RC
= (1010 || 1013) (25 × 10-12)
= 1010 × 25 × 10-12 sec
= 250 msec
Voltage output after, t m-sec is given by
$$e = \frac{{dF}}{C}\left[ {\exp \left( {\frac{{ - t}}{\tau }} \right) - 1} \right]$$
$$\begin{array}{l} 0.12 = \frac{{\left( {150 \times {{10}^{ - 12}}} \right) \times 2}}{{25 \times {{10}^{ - 12}}}}\left[ {\exp \left( { - \frac{t}{{0.25}}} \right) - 1} \right]\\ \Rightarrow t = 2.48\;msec \end{array}$$
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2021-09-26 22:15:04
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http://mathhelpforum.com/calculus/101459-maximum-minimum-problem.html
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# Math Help - Maximum/Minimum Problem
1. ## Maximum/Minimum Problem
Tricky question
The sum of two numbers is 10. Find the numbers if the sum of the their squares is to be a minimum...
I have no idea what exactly the minimum is in this problem...
Thanks!
2. a minimum of?
or does it mean something like?
1+9 =10
$1^2+9^2=82$
$2^2+8^2=68$
I guess you could do it as a trial error like that . . .
or you could do it the "calculus" way:
$x + y = 10$
$y=10-x$
$x^2+y^2$
$x^2+(10-x)^2$
$x^2+x^2-20x+100$
and then take the derivative and see where it goes to zero. *shrugs*
3. Well first of all let the two numbers be a and b.
Now if their sum is 10, then we can say
$a+b=10$
$b=10-a$
Now the seum of their squares is
$a^2+b^2$
let this equal another variable, say, y
so
$a^2+b^2=y$
substituting in for b from the equation established above, we get
$y=a^2+(10-a)^2$
rearrange
$y=a^2+100-20a+a^2$
$y=2a^2-20a+100$
to find where y has a max/min, we have to fins dy/dx and let it equal 0
$\frac{dy}{dx} = 4a-20$
$0=4a-20$
$a=5$
There you go, that's one of the two numbers, and the second one can be found with the first formula we established.
This kind of approach is best for these questions, if you're stuck always try and say "let this equal a and that equal b", and see where it takes you.
4. Thanks so much to both of you, it was a great help!
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2016-06-27 02:41:50
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https://zbmath.org/?q=an:0592.53034
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# zbMATH — the first resource for mathematics
Conformal complete metrics with prescribed non-negative Gaussian curvature in $${\mathbb{R}}^ 2$$. (English) Zbl 0592.53034
The paper deals with the existence of (complete) Riemannian metrics g on $${\mathbb{R}}^ 2$$, conformal to the Euclidean metric $$g_ 0$$ (by $$g=e^{2u}g_ 0)$$ and possessing a prescribed Gaussian curvature $$k: {\mathbb{R}}^ 2\to {\mathbb{R}}$$. This problem is intimately related to the nonlinear elliptic PDE $$\Delta u+k(x)e^{2u}=0$$. It is always assumed that k is Hölder-continuous and that k(x) for large $$| x|$$ is asymptotically comparable to some negative power of $$| x|$$, in one sense or the other. For example, if k is positive somewhere and, asymptotically, $$| k(x)| \leq M/| x|^ b$$ for some constants $$M>0$$, $$b\geq 2$$ then there is a complete metric g, for which an asymptotic rule like $(b-c)\cdot \ln | x| -2\tilde C\leq 2u(x)\leq (b-c)\cdot \ln | x| +2\tilde C$ holds. In case $$k\geq 0$$ the condition $$b\geq 2$$ may be relaxed to $$b>0$$, provided some other assumptions are correspondingly sharpened. This is the most complicated case considered here. Several situations are exhibited where every solution of the elliptic PDE has an asymptotic behaviour like this. Finally, it is shown in the present frame that Cohn-Vossen’s inequality is necessary and sufficient for the metric g to become complete. The methods are mainly analytical in nature, including e.g. the Leray- Schauder fixed point theory and the capacity of planar sets.
Reviewer: R.Walter
##### MSC:
53C20 Global Riemannian geometry, including pinching 35J60 Nonlinear elliptic equations
Full Text:
##### References:
[1] [Ah] Ahlfors, L.: Conformal invariants. Topics in geometric function theory, McGraw-Hill Series in Higher Mathematics (1973) · Zbl 0272.30012 [2] [A] Aviles, P.: Prescribing conformal complete metrics with given positive Gaussian curvature in ?2, Berkeley, California: Mathematical Sciences Research Institute, Preprint, (1983) [3] [Ba] Bandle, C.: Isoperimetric inequalities for a nonlinear eigenvalue problem. Proc. Am. Math. Soc.56, 243-246 (1976) · Zbl 0326.35004 · doi:10.1090/S0002-9939-1976-0477402-7 [4] [B] Bleecker, D.: The Gauss-Bonnet inequality and almost-geodesic loops. Adv. Math.14, 183-193 (1974) · Zbl 0289.53039 · doi:10.1016/0001-8708(74)90029-2 [5] [C] Calabi, E.: On Ricci curvature and geodesics. Duke Math. J.34, 667-675 (1967) · Zbl 0153.51501 · doi:10.1215/S0012-7094-67-03469-2 [6] [Ca] Cantor, M.: Elliptic operators and the decomposition of tensor fields. Bull. Am. Math. Soc.5, 235-263 (1981) · Zbl 0481.58023 · doi:10.1090/S0273-0979-1981-14934-X [7] [Fe] Federer, H.: Curvature measures.Trans. Am. Math. Soc.93, 418-491 (1959) · Zbl 0089.38402 · doi:10.1090/S0002-9947-1959-0110078-1 [8] [F] Finn, R.: On a class of conformal metrics with applications to differential geometry in the large. Comment. Math. Helv.40, 1-30 (1965) · Zbl 0192.27301 · doi:10.1007/BF02564362 [9] [G-T] Gilbarg, D., Trudinger, N.A.: Elliptic partial differential equations of second order. Berlin-Heidelberg-New York: Springer (1977) · Zbl 0361.35003 [10] [G-L] Gromov, M., Lawson, B.: Positive scalar curvature and the dirac operator on complete Riemannian manifolds. (To appear) · Zbl 0538.53047 [11] [H-S] Hewitt, E., Stromberg, K.: Real and abstract analysis, Berlin-Heidelberg-New York: Springer (1965) · Zbl 0137.03202 [12] [H] Huber, A.: On subharmonic functions and differential geometry in the large. Comment. Math. Helv.32, 13-72 (1957) · Zbl 0080.15001 · doi:10.1007/BF02564570 [13] [H, 1] Huber, A.: On the isoperimetric inequality on surfaces of variable Gaussian curvature. Ann. Math.60 No. 2, 237-247 (1954) · Zbl 0056.15801 · doi:10.2307/1969630 [14] [J] Jones, F.: Rudiments of Riemann surfaces. Ric. Univ. Lect. Notes Math., No. 2, (1971) · Zbl 0229.30004 [15] [K-W, 1] Kazdan, J., Warner, F.: Curvature functions for compact 2-manifolds. Ann. Math.99, No. 1, 14-47 (1974) · Zbl 0273.53034 · doi:10.2307/1971012 [16] [K-W, 2] Kazdan, J., Warner, F.: Curvature functions for open 2-manifolds. Ann. Math.99, 203-219 (1974) · Zbl 0278.53031 · doi:10.2307/1970898 [17] [M] McOwen, R.: On the equation ?u+ke 2u =f and prescribed negative curvature in ?2, (To appear) · Zbl 0568.35035 [18] [Mc] McOwen, R.: Conformal metrics in ?2 with prescribed Gaussian and positive total curvature (To appear: Indiana Univ. Math. J.) [19] [N, 1] Ni, W.M.: On the elliptic equation ?u+ku (n+2)/(n?2)=0 its generalizations and applications in geometry. Indiana Univ. Math. J.4,493-532 (1982) · Zbl 0496.35036 · doi:10.1512/iumj.1982.31.31040 [20] [N, 2] Ni, W.M.: On the elliptic equation ?u+k(x)e 2u =0 and conformal metrics with prescribed Gaussian curvature. Invent. Math.66, 343-353 (1982) · Zbl 0487.35042 · doi:10.1007/BF01389399 [21] [O] Oleinik, O.A.: On the equation ?u+k(x)e u +0. Russ. Math.Surv.33, 243-244 (1978) · Zbl 0401.35051 · doi:10.1070/RM1978v033n02ABEH002424 [22] [P] Payne, L.E.: Isoperimetric inequalities and their applications. SIAM Rev. Vol.9, No. 3 (1967) · Zbl 0154.12602 [23] [Sa] Sattinger, D.H.: Conformal metrics in ?2 with prescribed curvatures. Indiana Univ. Math. J.22, 1-4 (1972) · Zbl 0236.53009 · doi:10.1512/iumj.1972.22.22001 [24] [S] Serrin, J.: Local behavior of solutions of quasi-linear equations. Acta Math.111,247-302 (1964) · Zbl 0128.09101 · doi:10.1007/BF02391014 [25] [S-Y] Schoen, R., Yau, S.T.: Complete three dimensional manifolds with positive Ricci curvature and scalar curvature. In: S.T. Yau (ed.) Seminar on Differential Geometric, Ann. Math. Stud. Princeton University Press(1982) pp. 209-228 [26] [T] Talenti, G.: Best constant in Sobolev inequality. Ann. Mat. Pura Appl.110, 353-372 (1976) · Zbl 0353.46018 · doi:10.1007/BF02418013 [27] [U] Uhlenbeck, K.: Removable singularities in Yang-Mills fields. Commun. Math., Physics83, 11-29 (1982) · Zbl 0491.58032 · doi:10.1007/BF01947068 [28] [W] Weinberger, H.F.: Symmetrization in uniformly elliptic problems. Studies in Mathematical Analysis and Related Topics. Essays in honor of G. P?lya, Stanford, California: Stanford University Press, (1962), pp. 424-428 [29] [Y] Yau, S.T.: Isoperimetric inequalities and the first eigenvalue of a compact Riemannian manifold. Ann. Sci. Ec. Norm. Super.8, 487-507 (1975) · Zbl 0325.53039
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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2021-04-19 05:45:31
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https://www.pims.math.ca/scientific-event/191206-upmcjcw
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UW-PIMS Mathematics Colloquium: Jun-Cheng Wei
• Date: 12/06/2019
• Time: 15:30
Lecturer(s):
Jun-Cheng Wei, University of British Columbia
Location:
University of Washington
Topic:
On $C^{2,\alpha}$ estimates for levels sets of Allen-Cahn equation
Description:
I will discuss recent new developments in De Giorgi Conjecture for Allen-Cahn equation. In dimensions up to 10, we shall establish the C2,α estimates of level sets for stable solutions of Allen-Cahn. By applying reverse gluing method we show that the obstruction to C2,α estimates is the existence of Toda System in one dimension less. Applications to classifications of finite Morse index solutions and some open problems will be discussed.
Other Information:
Location: GWN 201
For further information on the UW-PIMS Mathematics Colloquium, please refer to their website.
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2020-01-28 22:58:03
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https://mathspace.co/textbooks/syllabuses/Syllabus-411/topics/Topic-7332/subtopics/Subtopic-97719/?activeTab=theory
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New Zealand
Level 8 - NCEA Level 3
# Sample Proportions (calculating and comparison)
Lesson
## The sample proportion
When sampling, we are sometimes interested in obtaining a sub-group's proportion of an overall population, rather than in obtaining a sample estimate of a mean value $\mu$μ.
For example, suppose we are interested in finding the proportion of salmon in a lake full of fish. Unless we drain the lake, the exact proportion of salmon is probably unknowable. Perhaps our best option is to somehow obtain a random sample of fish (using fishing nets or some other device), and use the sample proportion as our estimate of the true proportion.
If for example, suppose there were $S+O$S+O fish in the lake where $S$S is the total number of salmon and $O$O was the total amount of other fish in the lake. Our random sample might contain $360$360 fish of which $9$9 were salmon. The sample proportion, which we could call $\hat{p}$^p, would be $\frac{90}{360}=0.25$90360=0.25, and this would be an estimate of the population proportion $p=\frac{S}{S+O}$p=SS+O.
Of course the interesting thing about taking a sample like this is that we simply don't know the $p$p value. In the fish sample, for example, how could we tell the obtained sample proportion $\hat{p}=0.25$^p=0.25 was representative of $p$p?
This is a difficult question to answer, but at its heart, the answer has to do with experimental design. Clearly the larger the sample taken, the more accurate we expect $\hat{p}$^p to be. When thinking about the sample size, we might consider the size of the lake and whether all fish have access to all parts of the lake. We might even decide to take a number of samples from different points around the lake and develop an average proportion.
## What sample size should we use?
There are various statistical methods available to determine an appropriate sample size. A rather crude method involves using a preliminary test sample first, and from that determining a sufficient sample size to use in a more rigorous test.
For example, suppose we use the sample proportion $\hat{p}=0.25$^p=0.25.as our preliminary proportion.
Then, if we want to be $95%$95% sure that the true proportion $p$p is within $e=0.05$e=0.05 of $\hat{p}$^p, we simply calculate:
$n=\frac{1.96^2\times(0.25)(1-0.25)}{0.05^2}=288.12$n=1.962×(0.25)(10.25)0.052=288.12
The number $1.96$1.96 in the formula is related to an arbitrarily chosen confidence level, and the result is only meant to be a ball park figure.
If we take notice of the result, perhaps a rigorous estimate of $p$p would be obtained with a sample size of around $300$300 fish. Note that the formula contains the square of the tolerance limit $e$e in the denominator, and this means that if we want more accuracy, we will need a far greater sample size.
## Comparing a sample proportion and a claimed proportion
In certain instances we can determine the true proportion using rational argument. We know for example that the probability of obtaining a head in a single flip of a fair coin is $\frac{1}{2}$12. The coin has two sides and each side is equally likely to fall uppermost.
In these instances we think of the population proportion as a probability $p$p. It is often referred to as the $p$p-value of the experiment. We then can compare any particular sample proportion $\hat{p}$^p with its $p$p-value and make a judgement on either the coin's fairness or the robustness of the sample.
Judgements like this are at the heart of sampling theory.
#### Examples
##### Example 1
A random sample of $200$200 people were asked whether they preferred chocolate ice-cream over vanilla ice-cream. $88$88 preferred chocolate over vanilla. The sample proportion $\hat{p}=\frac{88}{200}=0.44$^p=88200=0.44.
##### Example 2
Two independent samples we conducted across a city's households to ascertain the proportion of households with more than 1 bathroom. Sample A (a random sample of $130$130 households) found a proportion of $0.3154$0.3154 and Sample B (a random sample of $210$210 different households) found a proportion of $0.4095$0.4095
Sample B would be more robust than sample A because the sample size of B is larger. However, knowing the sample sizes, and knowing the samples used different households allows us to combine the independent results. From sample A it looks like $41$41 households have more than $1$1 bathroom, and similarly from sample B, $86$86 households have more than $1$1 bathroom. This means that from the two samples, we know that of $340$340 households, $127$127 have more than $1$1 bathroom. This is an overall sample proportion of $0.3735$0.3735.
#### Worked Examples
##### QUESTION 1
A survey of $115$115 randomly selected people in Busan found that $6$6 of them were aged over $55$55.
A second survey of $2183$2183 randomly selected people in Busan found that $475$475 of them were aged over $55$55.
1. Considering the first survey, what is the sample proportion of people in Busan over the age of $55$55?
2. Considering the second survey, what is the sample proportion of people in Busan over the age of $55$55?
3. Which sample proportion is likely to be the better estimate of the population proportion?
Neither sample. The sample size does not matter, since both come from the same population.
A
The first sample. The smaller the sample the more reliable the results and the more rigorous the sampling.
B
The second sample. The larger the sample size, the closer the parameters of the sample resemble the parameters of the population.
C
Neither sample. The sample size does not matter, since both come from the same population.
A
The first sample. The smaller the sample the more reliable the results and the more rigorous the sampling.
B
The second sample. The larger the sample size, the closer the parameters of the sample resemble the parameters of the population.
C
##### QUESTION 2
A population is sampled several times to investigate the proportion that purchase a new pot plant each year. Each sample is equal in size and the values of the sample proportions are recorded in the graph below.
1. How many samples were taken?
2. Hence, estimate the population proportion of customers who purchased a new pot plant. Give your answer as a decimal, correct to three decimal places.
##### QUESTION 3
A census for a particular country showed that $94%$94% of people used public transport at some point during a regular week.
At about the same time as the census, a sample of $2420$2420 people in a region of the country showed that $1001$1001 of those people used public transport at some point during a regular week.
1. Determine $p$p, the population proportion of the residents who use public transport at least once a week. Express your answer as a decimal.
2. Determine $\hat{p}$^p, the sample proportion of the residents who use public transport at least once a week. Express your answer a decimal, correct to two decimal places.
3. Comparing the population and sample proportions, which of the following statements is true?
The sample exhibits no bias and seems to be indicative of the larger population.
A
The sample exhibits bias and does not seem to adequately represent the larger population.
B
The sample exhibits no bias and seems to be indicative of the larger population.
A
The sample exhibits bias and does not seem to adequately represent the larger population.
B
### Outcomes
#### S8-2
Make inferences from surveys and experiments: A determining estimates and confidence intervals for means, proportions, and differences, recognising the relevance of the central limit theorem B using methods such as resampling or randomisation to assess
#### 91582
Use statistical methods to make a formal inference
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2022-01-22 18:55:33
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https://mittheory.wordpress.com/2014/06/23/efficient-density-estimation-part-2-the-valiants/
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# Efficient Distribution Estimation 2: The Valiant Sequel – STOC 2014 Recaps (Part 4)
This is a continuation of Efficient Distribution Estimation STOC workshop summary series by G and Clément (see the Part 1 here).
Gautam Kamath on Distributional Property Testing and Estimation: Past, Present, and Future
After lunch, we were lucky enough to witness back-to-back talks by the Valiant brothers, Greg and Paul. If you aren’t familiar with their work, first of all, how have you been dealing with all of your everyday property testing needs? But second of all, over the past few years, they’ve touched a number of classic statistical problems and left most of them with matching optimal upper and lower bounds.
Greg kicked off the show and focused on property estimation. Given a property of interest and access to independent draws from a fixed distribution $\mathcal{D}$ over $\{1, \dots, n\}$, how many draws are necessary to estimate the property accurately?
Of course, this is an incredibly broad problem. For example, properties of a distribution which could be of interest include the mean, the $L^{3}$ norm, or the probability that the samples will spell out tomorrow’s winning lottery numbers. In order to tame this mighty set of problems, we focus on a class of properties which are slightly easier to understand — symmetric properties. These are properties which are invariant to permuting the labels on the support — for instance, if we decided to say 3 was now called 12 and vice-versa, the property wouldn’t change. Notable examples of symmetric properties include the support size*, entropy, and distance between two distributions. Still a broad class of properties, but surprisingly, we can make statements about all of them at once.
What’s the secret? For symmetric properties, it turns out that the fingerprint of a set of samples contains all the relevant information. The fingerprint of a set of samples $X$ is a vector $\mathcal{F}_X$ whose $i$th component is the number of elements in the domain which occur exactly $i$ times in $X$. The CliffNotes version of the rest of this post is that using linear programming on the fingerprint gives us an unreasonable amount of power**.
First off, Greg talked about a surprising dichotomy which demonstrates the power of linear estimators. A linear estimator of a property is an estimator which outputs a linear combination of the fingerprint. Given a property $\pi$, a number of samples $k$, and a distance $\varepsilon$, exactly one of the following holds:
• There exist distributions $y_1$ and $y_2$ such that $|\pi(y_1) - \pi(y_2)| > \varepsilon$ but no estimator (linear or not) that uses only $k$ samples can distinguish the two with probability $\geq 51\%$.
• There exists a linear estimator which requires $k(1 + o(1))$ samples and estimates the property to within $\varepsilon(1 + o(1))$ with probability $\geq 99\%$.
In other words, if there exists an estimator for a property, there is a linear estimator which is almost as good. Even better, their result is constructive, for both cases! A linear program is used to construct the estimator, where the variables are the estimator coefficients. Roughly, this linear program attempts to minimize the bias of the estimator while keeping the variance relatively small. On the other hand, when we take the dual of this linear program, we construct two distributions which have similar fingerprint expectations, but differ in their value for the property — we construct an explicit counter-example to any estimator which claims to use $k$ samples and estimate the property to accuracy $\varepsilon$.
But here’s the \$64 question – do these estimators work in practice? Sadly, the answer is no, which actually isn’t that surprising here. The estimator is defined in terms of the worst-case instance for each property — in other words, it’s oblivious to the particular distribution it receives samples from, which can be wasteful in many cases.
Let’s approach the problem again. What if we took the fingerprint of our samples and computed the property for the empirical distribution? This actually works great – the empirical distribution optimally estimates the portion of the distribution which we have seen. The downside is that we have to see the entire distribution, which takes $\Omega(n)$ samples. If we want to break past this linear barrier, we must estimate the unseen portion. It turns out that, once again, linear programming saves the day. We solve a program which describes all distributions whose expected fingerprint is close to the observed fingerprint. It can be shown that the diameter of this set is small, meaning that any distribution in this space will approximate the target distribution for all symmetric properties at the same time! Furthermore, this estimator takes only $O(n/\log n)$ samples, which turns out to be optimal.
Again, we ask the same question – do these estimators work in practice? This time, the answer is yes! While performance plots are often conspicuously missing from theory papers, Greg was happy to compare their results to estimators used in practice — their approach outperformed the benchmarks in almost all instances and regimes.
Finally, Greg mentioned a very recent result on instance-by-instance optimal identity testing. Recall that in identity testing, given a known distribution $P$ and an unknown distribution $Q$, we wish to test if they are equal or $\varepsilon$-far (in total variation distance). As mentioned before, when $P$ is uniform, $\Theta(\sqrt{n}/\varepsilon^2)$ are necessary and sufficient for this problem. But what about when $P$ is $\mathrm{Bin}(n,\frac12)$, or when $P$ assigns a probability to $i$ which is proportional to the $i$th digit of $\pi$ — do we have to design an algorithm by scratch for every $P$?
Thankfully, the answer is no. The Valiants provide an algorithm which is optimal up to constant factors for all but the most pathological instances of $P$. Somewhat unexpectedly, the optimal number of samples turns out to be $\Theta(\|P\|_{2/3}/\varepsilon^2)$ – aside from the fact it matches the uniform case, it’s not obvious why the $2/3$-norm is the “correct” norm here. Their estimator is reasonably simple – it involves a small tweak to Pearson’s classical $\chi$-squared test.
-G
Clément Canonne on Lower Bound Techniques for Statistical Estimation Tasks
At this point, Greg tagged out, allowing Paul to give us a primer in “Lower Bound Techniques for Statistical Estimation Tasks”: roughly speaking, the other side of Greg’s coin. Indeed, all the techniques used to derive upper bounds (general, systematic testers for symmetric properties) can also been turned into their evil counterparts: how to show no algorithm supposed to test a given property can be “too efficient”.
The dish is very different, yet the ingredients similar: without going into too much details, let’s sketch the recipe. In the following, keep in mind that we will mostly cook against symmetric properties, that is properties $\mathcal{P}$ invariant by relabeling of the domain elements: if $D\in\mathcal{P}$, then for any permutation $\pi$ of $[n]$ one has $D\circ\pi\in\mathcal{P}$.
• elements do not matter: all information any tester can use can be found in the fingerprint of the samples it obtained. In other terms, we can completely forget the actual samples for the sake of our lower bound, and only prove things about their fingerprint, fingerprint of fingerprint, and so on.
• Poissonization is our friend: if we take $m$ samples from a distribution, then the fingerprint, a tuple of $m$ integer random variables $X_1,\dots X_m$, lacks a very nice feature: the $X_i$‘s are not independent (they do kind of have to sum to $m$, for instance). But if instead of this, we were to take $m^\prime\sim\mathrm{Poisson}(m)$ samples, then the individual components of the fingerprint will be independent! Even better, many nice properties of the Poisson distribution will apply: the random variable $m^\prime$ will be tightly concentrated around its mean $m$, for instance. (this applies whether the property is symmetric or not. When in doubt, Poissonize. Independence!)
• Central Limit Theorems: the title ain’t witty here, but the technique is. As a very high-level, imagine we get to random variables $T, S$ corresponding to the “transcript” of what a tester/algorithm sees from taking $m$ samples from two possible distributions it should distinguish. We would like to argue that the distributions of $T$ and $S$ are close, so close (in a well-defined sense) that it is information-theoretically impossible to do so. This can be very difficult; a general idea would be to prove that there exists another distribution $G$, depending on very few parameters — think of a Gaussian: only the mean and (co)variance (matrix) — such that both $S$ and $T$ are close to $G$. Then, by the triangle inequality, they must be close, and we are done.Sweeping a lot of details under the rug, what Paul presented is this very elegant idea of proving new Central Limit Theorems that exactly give that: theorems that state that asymptotically (with quantitative bounds on the convergence rate), some class of distributions of interest will “behave like” a multivariate Gaussian or Multinomial with the “right parameters” — where “behave like” refers to a convient metric on distributions (e.g. Earthmover or Total Variation), and “right parameters” means “as few as possible”, in order to keep as many degrees of freedom when designing the distributions of $S$ and $T$).Proving such CLT’s is no easy task, but — on the bright side — they have many applications, even outside the field of distribution testing.
There would be much, much more to cover here, but I’m past 600 words already, and was only allowed $\mathrm{Poisson}(500)$.
-Clément
*In order to avoid pathological cases for estimating support size, we assume none of the probability are in the range $(0,\frac{1}{n})$.
**At least, when channeled by sufficiently talented theoreticians.
[1] Estimating the unseen: A sublinear-sample canonical estimator of distributions, Valiant, Valiant (2010).
[2] A CLT and tight lower bounds for estimating entropy, Valiant, Valiant (2010).
[3] Estimating the Unseen: An n/log(n)-sample Estimator for Entropy and Support Size, Shown Optimal via New CLTs, Valiant, Valiant (2011).
[4] Estimating the Unseen: Improved Estimators for Entropy and other Properties, Valiant, Valiant (2013).
[5] An Automatic Inequality Prover and Instance Optimal Identity Testing, Valiant, Valiant (2014).
[6] Testing Symmetric Properties of Distributions, Valiant (2011).
See also, Greg’s TCS+ talk, which covers [1,2,3,4].
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2017-10-23 13:11:38
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https://mattermodeling.stackexchange.com/questions/2346/predicting-magnetic-moment-of-a-metal-complex-computationally/2355
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# Predicting magnetic moment of a metal complex computationally
Recently, for entrance exams, I have been dealing with a lot of weird compounds. One of them is $$\ce{K3[Mn(CN)6]}$$.
The objective was to predict the magnetic moment of the above complex.
I predict that the number of unpaired electrons in the metal is 2 which straight away implies that the magnetic moment must be 2.8 but since it is a mere speculation considering the strong-ligand $$\ce{CN^-}$$ I would like to confirm it by modelling the compound. The data for the given complex is not available/accessible for me.
An article available online says to calculate the average of Gross Orbital Population (in Gaussian09). Trying it out gave an output of 1.
Hence, I would like to know how do I find the magnetic moment computationally (maybe using Gaussian, Gamess, Terachem, OCRA, etc.).
As a sidenote, since NMR uses magnetic field, can it be useful for computing magnetic moment?
• +1. Welcome to the site, and thanks for bringing the question here !!! We hope to see much more of you!! Sep 22 '20 at 3:46
• Your value of 2.8 seems to only be based on the # of unpaired electrons, but according to this it also will have a nuclear component. Sep 23 '20 at 1:04
• @NikeDattani Yes, I remember someone on researchgate telling that it is molecular magnetic moment but their wording was not clear. Should I clear in my question that I want spin only? I am asking so because I do not think it will be possible to seperate it out and that the spin only magnetic moment will be dominant (octahedral symmetry should help?) ---------This is just a speculation Sep 23 '20 at 3:10
• @NikeDattani magnetic properties are dominated by the electrons since the Bohr magneton is way bigger than the nuclear magneton by a ratio of m_e/m_p ~ 1826, which is the reason why we can also use the Born-Oppenheimer principle. Sep 23 '20 at 7:12
The main question here is whether the question makes sense for $${\rm K}_3[{\rm Mn(CN)}_6]$$ as a molecular complex.
Looks like the material has a solid state structure https://materials.springer.com/isp/crystallographic/docs/sd_1100190 which you could with solid state methods. As a complex it's not obvious where the potassiums would go, so you would probably start out by getting rid of the potassiums in the +1 charge state to get $$[{\rm Mn(CN)}_6]^{3-}$$. If you further assumed the $${\rm CN}$$ to be $${\rm CN}^-$$, you would get a +3 oxidation state for $${\rm Mn}$$, and two unpaired electrons, see e.g. https://www.quora.com/What-is-the-hybridisation-of-Mn-CN-6-3; but your question was about a computational value.
You could get the magnetic moment on the metal by running some calculations on the $$[{\rm Mn(CN)}_6]^{3-}$$ complex. However, this is not nearly as it sounds like: transition metal complexes are often challenging due to near-degeneracies, which means that the numbers you get from your calculations may be complete garbage (I'm not sure whether this is the case for this complex): density functional approximations may be unreliable, whereas wave function methods may require a very fine balance between static and dynamic correlation to make the computed ground state match the experimental one. Setting up the necessary model and making sure that the computed number is converged with respect to all parameters is a huge hassle.
• The linked website for crystallographics is a bit restrictive for me because I do not have a licence yet. Also, the Quora link is broken (although I know the hybridisation and how to find it). You could get the magnetic moment on the metal by running some calculations .What are the calculations you are referring to, and how to run them? (Note: It will also be helpful if I get to know how to determine mag.moment for molecules like $O_2$ etc.. Sep 23 '20 at 7:46
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2021-09-19 17:16:26
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