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# Saturation current on chip inductors

Inductors are really not my strong point, and I have been getting a bit confused on the rated current vs saturation current.

I know the rated current is how much current the inductor can handle before the temperature increases too much, and all I know about saturation current is when that threshold is approached and exceeded, the core is filled with magnetic flux and the inductance starts to drop.

I have come across a chip inductor HERE which I am planning on using in a boost regulator circuit. It is cheaper and smaller than the current inductor I am using, with the same rated current. However, whereas my old one had a saturation current of 900mA, this one is very low at 250mA.

Why is there such a difference in saturation current in chip inductors? My boost regulator chip has recommended inductor values of 2.2uH to 4.7uH, so if my current peaks at 300mA for a few ms and the inductance drops somewhat, will this still be ok to use? And would this change in inductance affect the output stability in any way?

As much as I would love to save money and PCB space, I don't know enough about these parameters to make an informed decision. Any advice on how to proceed is appreciated.

• You are going to have to either calculate or use design software to find the Peak inductor current under the conditions you are operating on. Be sure to check the extreme edges of your operating ranges. For a boost converter, check peak current with maximum load current and minimum input voltage. – Dwayne Reid Feb 22 '19 at 11:34

Looking at the datasheet of the "chip inductor", page 2: Inductance over current, I conclude that only up to about 100 mA this inductor behaves "properly".

The other inductor has a similar plot, also on page 2, from which I would say that it is usable up to 600 mA.

That's a 6x difference!

Only if you are 100% sure that your application keeps the current through the inductor always below 100mA would I consider the chip inductor.

Note that depending if we're dealing with a buck or boost converter and what the input/output voltage ratios are, the momentary current through the inductor can be many factors larger than the (average) currents going into and coming out of the converter. In extreme cases the current through the inductor can be 10x as high or even more.

The chip inductor is physically smaller and has a different construction so it simply is less capable at storing magnetic energy.

• It is a boost converter, with a single AAA battery input (1.6V - 0.95V) and an output of 3V. The overall current (measured by an Agilent power analyser) is an average of 400uA, but during operation, spikes of 50mA for about 10-20ms occur, and potential peaks of 200-220mA for about 1-2ms. – MCG Feb 22 '19 at 11:48
• Thank you for the response, this has been the most useful answer to the question so far! – MCG Feb 22 '19 at 11:49
• What would be the biggest side effect if the saturation current was exceeded momentarily? I'm not too sure what happens to the actual boost regulator once the inductor saturates. I was going to experiment with my dev board, but thought it best to ask before I go potentially smoking components! – MCG Feb 22 '19 at 11:57
• @MCG When the coil saturates the inductance part of the coil basically becomes a short so you're left with the series resistance of the coil. That causes the current to increase even more and could potentially destroy the switching transistors. Also once saturated, the coil cannot store more energy so even if the converter keeps trying to charge the coil more that energy is lost. I would stay away from saturation. You can investigate it on a dev. board but use controlled circumstances for example to prevent currents from becoming too large. – Bimpelrekkie Feb 22 '19 at 13:15
• Lovely, that's what I was after, along with the points in the answer. Thanks for the help! – MCG Feb 22 '19 at 13:35

The general equation:

$$\I_{Sat} = \frac{B_{max} \cdot N \cdot A_q}{L}\$$

$$\B_{max}\$$ is given by core material. If you want higher $$\I_{Sat}\$$, but don't want to reduce inductance $$\L\$$, then you need more windings and/or core cross-section $$\A_q\$$, both resulting in a larger and more expensive inductance.

Dependent on core-material the indcutance might drop dramatically after saturation, and you might get very high current peaks at each switching. This will heat up your semiconductors and reduce their lifetime or immediately destroy them.

• That's good for general use, but what about with these particular components linked in the question? And in particular why do these chip inductors have a much less saturation current, but an equal rated current? Also, note i mentioned the peaks would only be for a few ms, so was interested in the output stability – MCG Feb 22 '19 at 11:21
• Rated current is related to thermal issues (wire properties), saturation current is related to core properties (core geometry, gap, turn number). Both are input parameters for an inductor design. – UweD Feb 22 '19 at 11:30
• There could be instabiltiy in the current control loop, depends on the controller design. – UweD Feb 22 '19 at 11:33

You are going to have to either calculate or use design software to find the Peak inductor current under the conditions you are operating on. Be sure to check the extreme edges of your operating ranges.

For a boost converter, check peak current with maximum load current and minimum input voltage. If the peak current does not exceed the saturation current of the smaller inductor, feel free to use it.


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Algebra Through Puzzles

# Scale and Lever Logic

How much does a single red square weigh?

This Exploration investigates how the logic you intuitively apply when solving puzzles mirrors fundamental algebraic strategies.

To do so, we will teach you new mathematical techniques and push you to recognize and generalize the algebraic techniques that you already intuitively use as you problem-solve.

• A full jar of honey weighs 1000 grams.
• With half of the honey, the total weight is 600 grams.

What is the weight of the empty jar in grams?

What weight will the fourth scale display?

Think carefully! There's a quick and clever shortcut for solving this.

At the end of this Exploration, you'll have a thorough toolbox of algebraic techniques that feel natural to use and empower you to master complex problems and puzzles!

How many green squares should be placed on the right side of the bottom scale in order to make it balance when released?

We’ll also delve into some applications of algebra in physics!

Which way will the balance above tilt if released?

How to Solve Leverage Puzzles:

Archimedes once said:

“Give me a place to stand and with a lever I will move the whole world.”

His claim was in reference to the fact that the amount of force exerted on a lever arm is equal to the weight of the object sitting on it multiplied by the distance of that object from the fulcrum of the lever: $\text{Force} = \text{Weight} \times \text{Distance from Fulcrum}.$ This equation explains why the lever system below is balanced, even though there’s a rock on one side and only a small bird on the other: $$120 \times 1 = 40 \times 3.$$

And, at least in theory, if there were a force pulling the earth and Archimedes both "down" and something fixing the position of the fulcrum in space...

(Yes, Archimedes would also need a space suit.)

×

format: plain


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## A note on the Bramson-Kalikow process.(English)Zbl 1321.60071

Summary: We consider discrete-time stationary processes with long-range dependencies, $$X_{n}\in\{\pm1\}$$, $${n\in\mathbb{Z}}$$, specified by a regular attractive $$g$$-function, similar to those considered by M. Bramson and S. Kalikow [Isr. J. Math. 84, No. 1–2, 153–160 (1993; Zbl 0786.60043)]. We give an explicit set of conditions that imply the existence of at least two distinct processes specified by the same $$g$$-function, and consider a few examples that emphasize the role played by the smoothness of the majority rule at the origin.

### MSC:

 60G10 Stationary stochastic processes

Zbl 0786.60043
Full Text:

### References:

 [1] Berger, N., Hoffman, C. and Sidoravicius, V. (2003). Non-uniqueness for specifications in $$\ell^{2+\varepsilon}$$. Available at . · Zbl 1388.37039 [2] Bramson, M. and Kalikow, S. (1993). Non-uniqueness in $$g$$-functions. Israel J. Math. 84 , 153-160. · Zbl 0786.60043 [3] Dias, J. C. A. and Friedli, S. (2015). Uniqueness vs. non-uniqueness in complete connections with modified majority rules. Probab. Theory Related Fields . · Zbl 1336.60142 [4] Doeblin, W. and Fortet, R. (1937). Sur les chaînes à liaisons complètes. Bull. Soc. Math. France 65 , 132-148. · Zbl 0018.03303 [5] Hulse, P. (1991). Uniqueness and ergodic properties of attractive $$g$$-measures. Ergodic Theory Dynam. Systems 11 , 65-77. · Zbl 0723.28008 [6] Iosifescu, M. (1992). A coupling method in the theory of dependence with complete connections according to Doeblin. Rev. Roumaine Math. Pures Appl. 37 , 59-66. · Zbl 0757.60110 [7] Johansson, A. and Öberg, A. (2003). Square summability of variations of $$g$$-functions and uniqueness of $$g$$-measures. Math. Res. Lett. 10 , 587-601. · Zbl 1051.28008 [8] Samson, P.-M. (2000). Concentration of measure inequalities for Markov chains and $$\phi$$-mixing processes. Ann. Probab. 28 , 416-461. · Zbl 1044.60061
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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# GATE Mechanical 2021 Set 2 | GA Question: 10

1 vote

The world is going through the worst pandemic in the past hundred years. The air travel industry is facing a crisis, as the resulting quarantine requirement for travelers led to weak demand.

In relation to the first sentence above, what does the second sentence do?

1. Restates an idea from the first sentence
2. Second sentence entirely contradicts the first sentence
3. The two statements are unrelated
4. States an effect of the first sentence

recategorized

The air travel industry is facing a crisis, as the resulting quarantine requirement

The usage of “resulting” shows that the second statement is a result/effect of the first one.

Correct option: D

5.7k points 3 6 14

## Related questions

1 vote
Consider the following sentences: The number of candidates who appear for the $\text{GATE}$ examination is staggering. A number of candidates from my class are appearing for the $\text{GATE}$ examination. The number of candidates who appear for the $\text{GATE}$ examination are staggering. A number of candidates ... $\text{(i) and (iii)}$ $\text{(ii) and (iii)}$ $\text{(ii) and (iv)}$
$X$ is an online media provider. By offering unlimited and exclusive online content at attractive prices for a loyalty membership, $X$ is almost forcing its customers towards its loyalty membership. If its royalty membership continues to grow at its current rate, ... cable television Cable television operators don't subscribe to $X's$ loyalty members The $X$ is cancelling accounts of non-members
In forced convective heat transfer, Stanton number $\text{(St)}$, Nusselt number $\text{(Nu)}$, Reynolds number $\text{(Re)}$ and Prandtl number $\text{(Pr)}$ are related as $St=\frac{Nu}{Re\:Pr}$ $St=\frac{Nu\:Pr}{Re}$ $\text{St = Nu Pr Re}$ $St=\frac{Nu\:Re}{Pr}$


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## 48.29 The fundamental class of an lci morphism

In this section we will use the computations made in Section 48.15. Thus our result will suffer from the same kind of non-uniqueness as we have in that section.

Lemma 48.29.1. Let $X$ be a locally ringed space. Let

$\mathcal{E}_1 \xrightarrow {\alpha } \mathcal{E}_0 \to \mathcal{F} \to 0$

be a short exact sequence of $\mathcal{O}_ X$-modules. Assume $\mathcal{E}_1$ and $\mathcal{E}_0$ are locally free of ranks $r_1, r_0$. Then there is a canonical map

$\wedge ^{r_0 - r_1}\mathcal{F} \longrightarrow \wedge ^{r_1}(\mathcal{E}_1^\vee ) \otimes \wedge ^{r_0}\mathcal{E}_0$

which is an isomorphism on the stalk at $x \in X$ if and only if $\mathcal{F}$ is locally free of rank $r_0 - r_1$ in an open neighbourhood of $x$.

Proof. If $r_1 > r_0$ then $\wedge ^{r_0 - r_1}\mathcal{F} = 0$ by convention and the unique map cannot be an isomorphism. Thus we may assume $r = r_0 - r_1 \geq 0$. Define the map by the formula

$s_1 \wedge \ldots \wedge s_ r \mapsto t_1^\vee \wedge \ldots \wedge t_{r_1}^\vee \otimes \alpha (t_1) \wedge \ldots \wedge \alpha (t_{r_1}) \wedge \tilde s_1 \wedge \ldots \wedge \tilde s_ r$

where $t_1, \ldots , t_{r_1}$ is a local basis for $\mathcal{E}_1$, correspondingly $t_1^\vee , \ldots , t_{r_1}^\vee$ is the dual basis for $\mathcal{E}_1^\vee$, and $s'_ i$ is a local lift of $s_ i$ to a section of $\mathcal{E}_0$. We omit the proof that this is well defined.

If $\mathcal{F}$ is locally free of rank $r$, then it is straightforward to verify that the map is an isomorphism. Conversely, assume the map is an isomorphism on stalks at $x$. Then $\wedge ^ r\mathcal{F}_ x$ is invertible. This implies that $\mathcal{F}_ x$ is generated by at most $r$ elements. This can only happen if $\alpha$ has rank $r$ modulo $\mathfrak m_ x$, i.e., $\alpha$ has maximal rank modulo $\mathfrak m_ x$. This implies that $\alpha$ has maximal rank in a neighbourhood of $x$ and hence $\mathcal{F}$ is locally free of rank $r$ in a neighbourhood as desired. $\square$

Lemma 48.29.2. Let $Y$ be a Noetherian scheme. Let $f : X \to Y$ be a local complete intersection morphism which factors as an immersion $X \to P$ followed by a proper smooth morphism $P \to Y$. Let $r$ be the locally constant function on $X$ such that $\omega _{Y/X} = H^{-r}(f^!\mathcal{O}_ Y)$ is the unique nonzero cohomology sheaf of $f^!\mathcal{O}_ Y$, see Lemma 48.17.11. Then there is a map

$\wedge ^ r\Omega _{X/Y} \longrightarrow \omega _{Y/X}$

which is an isomorphism on the stalk at a point $x$ if and only if $f$ is smooth at $x$.

Proof. The assumption implies that $X$ is compactifyable over $Y$ hence $f^!$ is defined, see Section 48.16. Let $j : W \to P$ be an open subscheme such that $X \to P$ factors through a closed immersion $i : X \to W$. Moreover, we have $f^! = i^! \circ j^! \circ g^!$ where $g : P \to Y$ is the given morphism. We have $g^!\mathcal{O}_ Y = \wedge ^ d\Omega _{P/Y}[d]$ by Lemma 48.15.7 where $d$ is the locally constant function giving the relative dimension of $P$ over $Y$. We have $j^! = j^*$. We have $i^!\mathcal{O}_ W = \wedge ^ c\mathcal{N}[-c]$ where $c$ is the codimension of $X$ in $W$ (a locally constant function on $X$) and where $\mathcal{N}$ is the normal sheaf of the Koszul-regular immersion $i$, see Lemma 48.15.6. Combining the above we find

$f^!\mathcal{O}_ Y = \left(\wedge ^ c\mathcal{N} \otimes _{\mathcal{O}_ X} \wedge ^ d\Omega _{P/Y}|_ X\right)[d - c]$

where we have also used Lemma 48.17.9. Thus $r = d|_ X - c$ as locally constant functions on $X$. The conormal sheaf of $X \to P$ is the module $\mathcal{I}/\mathcal{I}^2$ where $\mathcal{I} \subset \mathcal{O}_ W$ is the ideal sheaf of $i$, see Morphisms, Section 29.31. Consider the canonical exact sequence

$\mathcal{I}/\mathcal{I}^2 \to \Omega _{P/Y}|_ X \to \Omega _{X/Y} \to 0$

of Morphisms, Lemma 29.32.15. We obtain our map by an application of Lemma 48.29.1.

If $f$ is smooth at $x$, then the map is an isomorphism by an application of Lemma 48.29.1 and the fact that $\Omega _{X/Y}$ is locally free at $x$ of rank $r$. Conversely, assume that our map is an isomorphism on stalks at $x$. Then the lemma shows that $\Omega _{X/Y}$ is free of rank $r$ after replacing $X$ by an open neighbourhood of $x$. On the other hand, we may also assume that $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(R)$ where $A = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ and where $f_1, \ldots , f_ m$ is a Koszul regular sequence (this follows from the definition of local complete intersection morphisms). Clearly this implies $r = n - m$. We conclude that the rank of the matrix of partials $\partial f_ j/\partial x_ i$ in the residue field at $x$ is $m$. Thus after reordering the variables we may assume the determinant of $(\partial f_ j/\partial x_ i)_{1 \leq i, j \leq m}$ is invertible in an open neighbourhood of $x$. It follows that $R \to A$ is smooth at this point, see for example Algebra, Example 10.137.8. $\square$

Lemma 48.29.3. Let $f : X \to Y$ be a morphism of schemes. Let $r \geq 0$. Assume

1. $Y$ is Cohen-Macaulay (Properties, Definition 28.8.1),

2. $f$ factors as $X \to P \to Y$ where the first morphism is an immersion and the second is smooth and proper,

3. if $x \in X$ and $\dim (\mathcal{O}_{X, x}) \leq 1$, then $f$ is Koszul at $x$ (More on Morphisms, Definition 37.59.2), and

4. if $\xi$ is a generic point of an irreducible component of $X$, then we have $\text{trdeg}_{\kappa (f(\xi ))} \kappa (\xi ) = r$.

Then with $\omega _{Y/X} = H^{-r}(f^!\mathcal{O}_ Y)$ there is a map

$\wedge ^ r\Omega _{X/Y} \longrightarrow \omega _{Y/X}$

which is an isomorphism on the locus where $f$ is smooth.

Proof. Let $U \subset X$ be the open subscheme over which $f$ is a local complete intersection morphism. Since $f$ has relative dimension $r$ at all generic points by assumption (4) we see that the locally constant function of Lemma 48.29.2 is constant with value $r$ and we obtain a map

$\wedge ^ r\Omega _{X/Y}|_ U = \wedge ^ r \Omega _{U/Y} \longrightarrow \omega _{U/Y} = \omega _{X/Y}|_ U$

which is an isomorphism in the smooth points of $f$ (this locus is contained in $U$ because a smooth morphism is a local complete intersection morphism). By Lemma 48.21.5 and the assumption that $Y$ is Cohen-Macaulay the module $\omega _{X/Y}$ is $(S_2)$. Since $U$ contains all the points of codimension $1$ by condition (3) and using Divisors, Lemma 31.5.11 we see that $j_*\omega _{U/Y} = \omega _{X/Y}$. Hence the map over $U$ extends to $X$ and the proof is complete. $\square$

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# LMIs in Control/Applications/An LMI for Multi-Robot Systems/Consensus for Multi-Agent Systems

LMIs in Control/Applications/An LMI for Multi-Robot Systems/Consensus for Multi-Agent Systems

This application gives an instance of the use of LMIs in achieving consensus for multi-agent systems. More focus is placed on the flocking algorithm, which is one of the consensus algorithms extensively used in multi-robot systems.

## Introduction

A robot swarm is a network of multiple robots working together as a unit to achieve an objective. In order for robots in a swarm to work together, they need to come to consensus/agreement on a number of parameters. A number of algorithms called consensus algorithms exist to serve that purpose. Consensus Algorithms are algorithms that allow a set of agents to reach an agreement. Various areas such as graph theory, control theory and matrix theory overlap when it comes to designing consensus algorithms for multi-agent systems.

## Flocking Algorithm

The flocking algorithm is one of the consensus algorithms used in trajectory planning for multi-agent networks. The algorithm helps in maintaining separation, so that the agents do not collide, alignment, so that the agents have the same headings, and cohesion, so that the agents do not wonder far away from each other. The flocking algorithm can be represented in a graph network form, where the nodes in the graph represent the agents or individual robots, and the edges/links represent the nature of interaction between the nodes. Consequently. the interactions between the nodes can be Below is the general equation for a flocking algorithm.

${\displaystyle {\dot {v_{i}}}-\sum _{i=N_{i}}^{n}i=(v_{i}-v_{j})+\sum _{i=N_{i}}^{n}i\Delta _{r_{i}}V_{ij}(r_{ij})}$, where

${\displaystyle V_{ij}}$ is the artificial potential function

${\displaystyle n}$ is the number of robots

${\displaystyle {\dot {v_{i}}}}$ is the input/acceleration control

${\displaystyle r_{i}}$ is the position vector of robot ${\displaystyle i}$

${\displaystyle r_{ij}=r_{i}-r_{j}}$, which is the distance between two neighboring robots

## Mathematical Model:

Before we begin, we need to obtain a state space representation for our system. Based on observation, it can be deduced that the general form of a consensus algorithm bears some resemblance to the state space equation. The general equation for a consensus algorithm in terms of the Laplacian matrix is:

${\displaystyle {\dot {x}}=-Lx}$, where ${\displaystyle L}$ represents the Laplacian matrix.

Comparing the above equation to the state space equation, ${\displaystyle {\dot {x}}=Ax}$, we notice that the ${\displaystyle A}$ has been substituted with a ${\displaystyle -L}$.

The A, B and C matrices are represented as diagonal blocks because of the decentralized nature of the control framework. Each block in the A, B and C matrices is associated with its corresponding state, input, or output vector.

{\displaystyle {\begin{aligned}\ A&={\begin{bmatrix}\ a_{11}&a_{1n}\\\ a_{n1}&a_{nn}\\\end{bmatrix}}&\ B={\begin{bmatrix}\ b_{11}&b_{1n}\\\ b_{n1}&b_{nn}\\\end{bmatrix}}&\ &C={\begin{bmatrix}\ c_{11}&c_{1n}\\\ c_{n1}&c_{nn}\\\end{bmatrix}}\end{aligned}}}

x, u and y represent the state, input and output vectors respectively.

{\displaystyle {\begin{aligned}\ x&={\begin{bmatrix}\ x_{1}&...&x_{n}\\\end{bmatrix}}^{T}&\ y&={\begin{bmatrix}\ y_{1}&...&y_{n}\\\end{bmatrix}}^{T}&\ u&={\begin{bmatrix}\ u_{1}&...&u_{n}\\\end{bmatrix}}^{T}&\end{aligned}}}

## Further Discussion:

In Elham Semsar-Kazerooni and Khashayar's paper, "Optimal Consensus Seeking in a Network of Multi-Agent Systems", further work was done through a series of variable substitutions and other mathematical operations to obtain the an LMI as seen below. For more information on the LMI formulation procedure, and the details of the variables in the above LMI, please visit follow the link in the reference section to review the paper.

Objective: ${\displaystyle \min trace(P)}$.

Subject to::

{\displaystyle {\begin{aligned}\ {\begin{bmatrix}\ \gamma &{\bar {S}}^{*}ZQ^{1/2}&{\bar {S}}^{*}W^{*}R^{1/2}\\\ Q^{1/2}Z{\bar {S}}&-I&0\\\ R^{1/2}W{\bar {S}}&0&-I\\\end{bmatrix}}\ <0,\end{aligned}}}


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# zbMATH — the first resource for mathematics

A lumped mass finite-element method with quadrature for a nonlinear parabolic problem. (English) Zbl 0591.65079
The authors consider the nonlinear uniformly parabolic initial boundary value problem $u_ t-\nabla (a(x,t,u)\nabla u)=f(x,t,u)\quad in\quad \Omega \times [0,T],$ $u=0\quad on\quad \partial \Omega \times [0,T],\quad u(.,0)=v\quad in\quad \Omega,$ where $$\Omega$$ is a two- dimensional domain with smooth boundary $$\partial \Omega$$ and the functions a(x,t,u) and f(x,t,u) are sufficiently smooth. By quasi-uniform triangularization of $$\Omega$$ and use of corresponding pyramidal basis functions they formulate the standard semi-discrete Galerkin method in form of a system of ordinary differential equations with time t as independent variable. They simplify this system by approximating inner products by quadrature with equal weights for the vertices of each triangle (separately for each triangle). Denoting by h the maximal diameter of a triangle they show that the approximate solution has error of order $$h^ 2$$ in $$L_ 2(\Omega)$$-norm, $$h^ 2 \log (1/h)$$ in $$L_{\infty}(\Omega)$$-norm, whereas the gradient has error of order h in the $$L_ 2(\Omega)$$-norm. Then they discretize time t also, using k as time step, and investigate the forward Euler scheme (requiring the usual restriction of time step) and linearized versions of the backward Euler scheme and the Crank-Nicolson scheme, and again errors come out as expected, for the approximate solution in the latter two schemes of orders $$h^ 2+k$$ or $$h^ 2+k^ 2$$, respectively, in the $$L_ 2(\Omega)$$-norm, with a factor log(1/h) to be adjoined if the $$L_{\infty}$$-norm is used. Finally, monotonicity properties of the discrete solution are discussed.
Reviewer: R.Gorenflo

##### MSC:
 65N30 Finite element, Rayleigh-Ritz and Galerkin methods for boundary value problems involving PDEs 65N15 Error bounds for boundary value problems involving PDEs 65N40 Method of lines for boundary value problems involving PDEs 35K60 Nonlinear initial, boundary and initial-boundary value problems for linear parabolic equations
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# Question #41f77

Nov 15, 2015

The most common oxide of aluminum is $A {l}_{2} {O}_{3}$.

#### Explanation:

An oxide is a binary compound with oxygen and another element (in this case, aluminum).

Oxygen typically has a charge of $- 2$ since it likes to gain $2$ electrons to fill its valence shell. Aluminum commonly has a charge of $+ 3$ since it likes to lose $3$ electrons to reach noble gas configuration.

In order for the compound to form, there needs to be the same amount of electrons gained and lost, so there needs to be a common multiple of electrons either gained or lost on each side: $6$.

Since oxygen has a $- 2$ charge, it will be denoted as ${O}_{3}$, showing that there are $3$ oxygen atoms total, which will gain $6$ electrons in all. Aluminum's $+ 3$ charge means it will be in the compound in the form $A {l}_{2}$, since the $2$ aluminum atoms will donate $3$ electrons each, $6$ total.

The two elements in the compound will be written together, starting with the more metallic element, or the element further left on the periodic table: $A {l}_{2} {O}_{3}$.

Because $A l$ sometimes forms $+ 1$ ions so that it will still have a full $2 s$ subshell, and for other exceptions, aluminum($I$) oxide ($A l O$) and aluminum($I I$) oxide ($A {l}_{2} O$) also form, but they only do so very rarely.

Nov 15, 2015

Al2O3

#### Explanation:

Al has oxidation state +3 and Oxygen has -2 i.e $A {l}^{+} 3 , {O}^{-} 2$. By combining the oxidation state are cross multiplied.


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# Can the eigenvalue for a quantum mechanical operator be zero? [closed]

Can the eigenvalue for a quantum mechanical operator be zero?

## closed as too broad by Tyberius, Mithoron, paracetamol, Pritt Balagopal, ronAug 24 '17 at 15:04

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• yes it could be, here you could find a satisfying answer link – Jaafar Mehrez Aug 24 '17 at 11:31
• The question is way broader than you seem to think; you might want to be more specific. As it stands now, the answer is "yes", but it is a really useless "yes", about the same level as if you were asking "can a difference of two numbers be zero". Sure it can; so what? – Ivan Neretin Aug 24 '17 at 11:31

$S_z$ (spin along the z-axis) is a QM operator, and it frequently is zero - such as in closed-shell molecules.


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# Edexcel A Levels Core Maths 4 (C4)

Unit C4: Core Mathematics 4

A2 compulsory unit for GCE Mathematics and GCE Pure Mathematics

Prerequisites

A knowledge of the specifications for C1, C2 and C3 and their preambles, prerequisites and associated formulae, is assumed and may be tested.

Module 1 Algebra and functions - Edexcel Core Maths 4 (39.08 mins)

Rational functions. Partial fractions (denominators not more complicated than repeated linear terms).

Partial fractions to include denominators such as

(ax + b)(cx + d)(ex + f)

and (ax + b)(cx + d)2.

The degree of the numerator may equal or exceed the degree of the denominator. Applications to integration, differentiation and series expansions.

Quadratic factors in the denominator such as (x2 + a), a > 0, are not required.

Unit 1 Algebra and functions - Edexcel Core Maths 4 PDF
Unit 2 Algebra and functions - Edexcel Core Maths 4 Video
Module 2 Coordinate geometry - Edexcel Core Maths 4 (57.83 mins)

Parametric equations of curves and conversion between Cartesian and parametric forms.

Students should be able to find the area under a curve given its parametric equations. Students will not be expected to sketch a curve from its parametric equations.

Unit 1 Coordinate geometry - Edexcel Core Maths 4 PDF
Unit 2 Coordinate geometry - Edexcel Core Maths 4 Video
Module 3 Sequences and series - Edexcel Core Maths 4 (71.34 mins)

Binomial series for any rational n.

For $$\vert x \vert < \dfrac{b}{a}$$ , students should be able to obtain the expansion of $$(ax + b)^n$$, and the expansion of rational functions by decomposition into partial fractions.

Unit 1 Sequences and series - Edexcel Core Maths 4 PDF
Unit 2 Sequences and series - Edexcel Core Maths 4 Video
Module 4 Differentiation - Edexcel Core Maths 4 (81.55 mins)

Differentiation of simple functions defined implicitly or parametrically.

The finding of equations of tangents and normals to curves given parametrically or implicitly is required.

Exponential growth and decay.

Knowledge and use of the result $$\dfrac{d}{dx}(a^x) = a^x \ln a$$ is expected.

Formation of simple differential equations.

Questions involving connected rates of change may be set.

Unit 1 Differentiation - Edexcel Core Maths 4 PDF
Unit 2 Differentiation - Edexcel Core Maths 4 Video
Module 5 Integration - Edexcel Core Maths 4 (420.38 mins)

Integration of $$e^x, \dfrac{1}{x} , \sin x, \cos x$$.

To include integration of standard functions such as $$\sin 3x, \sec^2 2x, \tan x, e^{5x}, \dfrac{1}{2x}$$.

Students should recognise integrals of the form $$\displaystyle\int \dfrac{f'(x)}{f(x)} dx = \ln f(x) + c$$.

Students are expected to be able to use trigonometric identities to integrate, for example, $$\sin^2 x, \tan^2 x, \cos^2 3x$$.

Evaluation of volume of revolution.

$$\pi \displaystyle\int y^2 dx$$ is required, but not $$\pi \displaystyle\int x^2 dy$$. Students should be able to find a volume of revolution, given parametric equations.

Simple cases of integration by substitution and integration by parts. These methods as the reverse processes of the chain and product rules respectively.

Except in the simplest of cases the substitution will be given.

The integral $$\displaystyle\int \ln x dx$$ is required.

More than one application of integration by parts may be required, for example $$\displaystyle\int x^2 e^x dx$$.

Simple cases of integration using partial fractions.

Integration of rational expressions such as those arising from partial fractions, eg $$\dfrac{2}{3x + 5}, \dfrac{3}{(x −1)^2}$$.

Note that the integration of other rational expressions, such as $$\dfrac{x}{x^2 + 5}$$ and $$\dfrac{2}{(2x −1)^4}$$ is also required (see above paragraphs).

Analytical solution of simple first order differential equations with separable variables.

General and particular solutions will be required.

Numerical integration of functions.

Application of the trapezium rule to functions covered in C3 and C4. Use of increasing number of trapezia to improve accuracy and estimate error will be required. Questions will not require more than three iterations.

Simpson’s Rule is not required.

Unit 1 Integration - Edexcel Core Maths 4 PDF
Unit 2 Integration - Edexcel Core Maths 4 Video
Module 6 Vectors - Edexcel Core Maths 4 (239.12 mins)

Vectors in two and three dimensions.

Magnitude of a vector.

Students should be able to find a unit vector in the direction of $$a$$, and be familiar with $$\vert a \vert$$.

Algebraic operations of vector addition and multiplication by scalars, and their geometrical interpretations.

Position vectors.

$$\overrightarrow{OB} - \overrightarrow{OA} = \overrightarrow{AB} = b − a$$ .

The distance between two points.

The distance $$d$$ between two points $$(x_1 , y_1 , z_1)$$ and $$(x_2 , y_2 , z_2)$$ is given by $$d^2 = (x_1 – x_2)^2 + (y_1 – y_2)^2 + (z_1 – z_2)^2$$ .

Vector equations of lines.

To include the forms r = a + tb and r = c + t(dc).

Intersection, or otherwise, of two lines.

The scalar product. Its use for calculating the angle between two lines.

Students should know that for

$$\overrightarrow{OA} = a = a_1i + a_2j + a_3k$$ and

$$\overrightarrow{OB} = b = b_1i + b_2j + b_3k$$ then

$$a . b = a_1b_1 + a_2b_2 + a_3b_3$$ and

$$\cos \angle AOB = \dfrac{a.b}{\vert a \vert \vert b \vert}$$ .

Students should know that if $$a . b = 0$$, and that a and b are non-zero vectors, then a and b are perpendicular.

Unit 1 Vectors - Edexcel Core Maths 4 PDF
Unit 2 Vectors - Edexcel Core Maths 4 Video

** Our syllabus is current and updated to 2018


text:
# Properties

 Label 882.3.b.f Level $882$ Weight $3$ Character orbit 882.b Analytic conductor $24.033$ Analytic rank $0$ Dimension $4$ CM no Inner twists $2$

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$882 = 2 \cdot 3^{2} \cdot 7^{2}$$ Weight: $$k$$ $$=$$ $$3$$ Character orbit: $$[\chi]$$ $$=$$ 882.b (of order $$2$$, degree $$1$$, minimal)

## Newform invariants

 Self dual: no Analytic conductor: $$24.0327593166$$ Analytic rank: $$0$$ Dimension: $$4$$ Coefficient field: $$\Q(\sqrt{-2}, \sqrt{7})$$ Defining polynomial: $$x^{4} + 8 x^{2} + 9$$ Coefficient ring: $$\Z[a_1, \ldots, a_{5}]$$ Coefficient ring index: $$2^{4}$$ Twist minimal: no (minimal twist has level 126) Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$

## $q$-expansion

Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\beta_2,\beta_3$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.

 $$f(q)$$ $$=$$ $$q + \beta_{1} q^{2} -2 q^{4} + ( \beta_{1} - \beta_{2} ) q^{5} -2 \beta_{1} q^{8} +O(q^{10})$$ $$q + \beta_{1} q^{2} -2 q^{4} + ( \beta_{1} - \beta_{2} ) q^{5} -2 \beta_{1} q^{8} + ( -2 + \beta_{3} ) q^{10} + ( -2 \beta_{1} + 2 \beta_{2} ) q^{11} + ( 8 + \beta_{3} ) q^{13} + 4 q^{16} + ( -13 \beta_{1} + \beta_{2} ) q^{17} -20 q^{19} + ( -2 \beta_{1} + 2 \beta_{2} ) q^{20} + ( 4 - 2 \beta_{3} ) q^{22} + ( -2 \beta_{1} + 2 \beta_{2} ) q^{23} + ( -33 + 2 \beta_{3} ) q^{25} + ( 8 \beta_{1} + 2 \beta_{2} ) q^{26} + ( -19 \beta_{1} - 2 \beta_{2} ) q^{29} + ( -4 - 2 \beta_{3} ) q^{31} + 4 \beta_{1} q^{32} + ( 26 - \beta_{3} ) q^{34} + 38 q^{37} -20 \beta_{1} q^{38} + ( 4 - 2 \beta_{3} ) q^{40} + ( -27 \beta_{1} - 3 \beta_{2} ) q^{41} + ( 20 + 6 \beta_{3} ) q^{43} + ( 4 \beta_{1} - 4 \beta_{2} ) q^{44} + ( 4 - 2 \beta_{3} ) q^{46} -12 \beta_{1} q^{47} + ( -33 \beta_{1} + 4 \beta_{2} ) q^{50} + ( -16 - 2 \beta_{3} ) q^{52} + ( -3 \beta_{1} - 12 \beta_{2} ) q^{53} + ( 116 - 4 \beta_{3} ) q^{55} + ( 38 + 2 \beta_{3} ) q^{58} + ( -20 \beta_{1} - 4 \beta_{2} ) q^{59} + ( -58 + 4 \beta_{3} ) q^{61} + ( -4 \beta_{1} - 4 \beta_{2} ) q^{62} -8 q^{64} + ( -48 \beta_{1} - 6 \beta_{2} ) q^{65} + ( -48 - 8 \beta_{3} ) q^{67} + ( 26 \beta_{1} - 2 \beta_{2} ) q^{68} + ( -2 \beta_{1} + 2 \beta_{2} ) q^{71} + ( 24 + 5 \beta_{3} ) q^{73} + 38 \beta_{1} q^{74} + 40 q^{76} + ( 76 - 4 \beta_{3} ) q^{79} + ( 4 \beta_{1} - 4 \beta_{2} ) q^{80} + ( 54 + 3 \beta_{3} ) q^{82} + ( -64 \beta_{1} + 4 \beta_{2} ) q^{83} + ( 82 - 14 \beta_{3} ) q^{85} + ( 20 \beta_{1} + 12 \beta_{2} ) q^{86} + ( -8 + 4 \beta_{3} ) q^{88} + ( -51 \beta_{1} + 9 \beta_{2} ) q^{89} + ( 4 \beta_{1} - 4 \beta_{2} ) q^{92} + 24 q^{94} + ( -20 \beta_{1} + 20 \beta_{2} ) q^{95} + ( 72 + 11 \beta_{3} ) q^{97} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4q - 8q^{4} + O(q^{10})$$ $$4q - 8q^{4} - 8q^{10} + 32q^{13} + 16q^{16} - 80q^{19} + 16q^{22} - 132q^{25} - 16q^{31} + 104q^{34} + 152q^{37} + 16q^{40} + 80q^{43} + 16q^{46} - 64q^{52} + 464q^{55} + 152q^{58} - 232q^{61} - 32q^{64} - 192q^{67} + 96q^{73} + 160q^{76} + 304q^{79} + 216q^{82} + 328q^{85} - 32q^{88} + 96q^{94} + 288q^{97} + O(q^{100})$$

Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{4} + 8 x^{2} + 9$$:

 $$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$($$$$\nu^{3} + 5 \nu$$$$)/3$$ $$\beta_{2}$$ $$=$$ $$($$$$2 \nu^{3} + 22 \nu$$$$)/3$$ $$\beta_{3}$$ $$=$$ $$4 \nu^{2} + 16$$
 $$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$($$$$\beta_{2} - 2 \beta_{1}$$$$)/4$$ $$\nu^{2}$$ $$=$$ $$($$$$\beta_{3} - 16$$$$)/4$$ $$\nu^{3}$$ $$=$$ $$($$$$-5 \beta_{2} + 22 \beta_{1}$$$$)/4$$

## Character values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/882\mathbb{Z}\right)^\times$$.

 $$n$$ $$199$$ $$785$$ $$\chi(n)$$ $$1$$ $$-1$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
197.1
 2.57794i − 1.16372i 1.16372i − 2.57794i
1.41421i 0 −2.00000 8.89753i 0 0 2.82843i 0 −12.5830
197.2 1.41421i 0 −2.00000 6.06910i 0 0 2.82843i 0 8.58301
197.3 1.41421i 0 −2.00000 6.06910i 0 0 2.82843i 0 8.58301
197.4 1.41421i 0 −2.00000 8.89753i 0 0 2.82843i 0 −12.5830
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Inner twists

Char Parity Ord Mult Type
1.a even 1 1 trivial
3.b odd 2 1 inner

## Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 882.3.b.f 4
3.b odd 2 1 inner 882.3.b.f 4
7.b odd 2 1 126.3.b.a 4
7.c even 3 2 882.3.s.i 8
7.d odd 6 2 882.3.s.e 8
21.c even 2 1 126.3.b.a 4
21.g even 6 2 882.3.s.e 8
21.h odd 6 2 882.3.s.i 8
28.d even 2 1 1008.3.d.a 4
35.c odd 2 1 3150.3.e.e 4
35.f even 4 2 3150.3.c.b 8
56.e even 2 1 4032.3.d.j 4
56.h odd 2 1 4032.3.d.i 4
63.l odd 6 2 1134.3.q.c 8
63.o even 6 2 1134.3.q.c 8
84.h odd 2 1 1008.3.d.a 4
105.g even 2 1 3150.3.e.e 4
105.k odd 4 2 3150.3.c.b 8
168.e odd 2 1 4032.3.d.j 4
168.i even 2 1 4032.3.d.i 4

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
126.3.b.a 4 7.b odd 2 1
126.3.b.a 4 21.c even 2 1
882.3.b.f 4 1.a even 1 1 trivial
882.3.b.f 4 3.b odd 2 1 inner
882.3.s.e 8 7.d odd 6 2
882.3.s.e 8 21.g even 6 2
882.3.s.i 8 7.c even 3 2
882.3.s.i 8 21.h odd 6 2
1008.3.d.a 4 28.d even 2 1
1008.3.d.a 4 84.h odd 2 1
1134.3.q.c 8 63.l odd 6 2
1134.3.q.c 8 63.o even 6 2
3150.3.c.b 8 35.f even 4 2
3150.3.c.b 8 105.k odd 4 2
3150.3.e.e 4 35.c odd 2 1
3150.3.e.e 4 105.g even 2 1
4032.3.d.i 4 56.h odd 2 1
4032.3.d.i 4 168.i even 2 1
4032.3.d.j 4 56.e even 2 1
4032.3.d.j 4 168.e odd 2 1

## Hecke kernels

This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{3}^{\mathrm{new}}(882, [\chi])$$:

 $$T_{5}^{4} + 116 T_{5}^{2} + 2916$$ $$T_{11}^{4} + 464 T_{11}^{2} + 46656$$ $$T_{13}^{2} - 16 T_{13} - 48$$

## Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$( 2 + T^{2} )^{2}$$
$3$ $$T^{4}$$
$5$ $$2916 + 116 T^{2} + T^{4}$$
$7$ $$T^{4}$$
$11$ $$46656 + 464 T^{2} + T^{4}$$
$13$ $$( -48 - 16 T + T^{2} )^{2}$$
$17$ $$79524 + 788 T^{2} + T^{4}$$
$19$ $$( 20 + T )^{4}$$
$23$ $$46656 + 464 T^{2} + T^{4}$$
$29$ $$248004 + 1892 T^{2} + T^{4}$$
$31$ $$( -432 + 8 T + T^{2} )^{2}$$
$37$ $$( -38 + T )^{4}$$
$41$ $$910116 + 3924 T^{2} + T^{4}$$
$43$ $$( -3632 - 40 T + T^{2} )^{2}$$
$47$ $$( 288 + T^{2} )^{2}$$
$53$ $$64738116 + 16164 T^{2} + T^{4}$$
$59$ $$9216 + 3392 T^{2} + T^{4}$$
$61$ $$( 1572 + 116 T + T^{2} )^{2}$$
$67$ $$( -4864 + 96 T + T^{2} )^{2}$$
$71$ $$46656 + 464 T^{2} + T^{4}$$
$73$ $$( -2224 - 48 T + T^{2} )^{2}$$
$79$ $$( 3984 - 152 T + T^{2} )^{2}$$
$83$ $$53231616 + 18176 T^{2} + T^{4}$$
$89$ $$443556 + 19476 T^{2} + T^{4}$$
$97$ $$( -8368 - 144 T + T^{2} )^{2}$$


text:
# Pattern color differs from normal color?

I have the following code:

\documentclass{minimal}
\usepackage{tikz}
\usetikzlibrary{scopes,patterns,intersections,calc}
\begin{document}
\begin{tikzpicture}[thick]
\path (-2,0) coordinate (t1) (2,0) coordinate (t2);
\path (0,-2) coordinate (s1) (0,2) coordinate (s2);
{ [color=blue!50!red!50,pattern color=blue!50!red!50]
\draw (s1) -- (s2);
\path [pattern=north west lines] (s1) +(-0.2,0) rectangle (s2);
}
{ [color=magenta,pattern color=magenta]
\draw (t1) -- (t2);
\path [pattern=north east lines] (t1) +(0,-0.2) rectangle (t2);
}
\end{tikzpicture}
\end{document}

As in the first (vertical) scope I would expect that the pattern has the same color as the line in both scopes. For me, I get three different colors; the pattern and the main line of the horizontal scope differ quite much (looks nearly midway between colors of hor. and vert. scope).

I am using pdfLaTeX to compile the file. Different viewers don't change anything.

Can you explain me this behavior?

-
Could you upload a screenshot, and say which version of TikZ/PGF that you are using? When I compile your code then I get the same colour for the lines as their respective patterns. –  Andrew Stacey Jan 12 '12 at 9:25
I stand corrected! The effect is a little more subtle than I was expecting and I had to zoom in quite far to see it. (A screenshot would still be useful) –  Andrew Stacey Jan 12 '12 at 9:26
I don't know why this is happening, but it seems to be the line colour that is wrong. I find I can fix it by loading xcolor with the svgnames package and using the colours defined by that, whence Magenta instead of magenta. –  Andrew Stacey Jan 12 '12 at 9:31
That did the trick. Thanks –  Christian Wolf Jan 12 '12 at 9:56

I'm not a great expert with color and latex but I think magenta is defined somewhere with the rgb system by (1,0,1) . With color=magenta magenta is not defined by rgb(1,0,1) but something like rgb (0.79216,0.12156,0.48236). Then when you use the pattern, pgf/tikz uses extra tools to convert the color in a rgb color and you get finally in all cases rgb(1,0,1). A possible solution is to define yourself magenta with \definecolor{magenta}{rgb}{1,0,1} or with \definecolor{magenta}{rgb}{0.79216,0.12156,0.48236} or something like that. In the first case the color for the line is rgb (1,0,1)and it's the same for the pattern and in the second case the conversion with \convertcolorspec gives your definition rgb(0.79216,0.12156,0.48236).

\documentclass{minimal}
%\usepackage[dvipsnames]{xcolor}
\usepackage{tikz}
\usetikzlibrary{scopes,patterns,intersections,calc}
\begin{document}

%\definecolor{magenta}{rgb}{1,0,1}
\definecolor{magenta}{rgb}{0.79216,0.12156,0.48236}

\begin{tikzpicture}[thick]
\path (-2,0) coordinate (t1) (2,0) coordinate (t2);
\path (0,-2) coordinate (s1) (0,2) coordinate (s2);
{ [color=blue!50!red!50,pattern color=blue!50!red!50]
\draw (s1) -- (s2);
\path [pattern=north west lines] (s1) +(-0.2,0) rectangle (s2);
}
{ [color=magenta,pattern color=magenta]
\draw (t1) -- (t2);
\path [pattern=north east lines] (t1) +(0,-0.2) rectangle (t2);
}
\end{tikzpicture}
\end{document}

-
Not sure if anyone else has noticed this, but I found that tinting a colour leads to a different colour in TikZ (in pgfplots, at least). E.g. if I paint something black, it comes out black, but if I paint something black!100!black it comes out as a dark grey. Not sure why, I guess it's a bug.
This issue can also be resolved by explicitly loading the xcolor package with the rgb option before loading the tikz package (the loading order is important as tikz implicitly loads xcolor and that leads to an option clash when xcolor is loaded after tikz). But this will have the side effect that all colors in your document are converted to the rgb model. As long as you're fine with that this should solve your problem.


text:
# Every closed subset $E\subseteq \mathbb{R}^n$ is the zero point set of a smooth function

In Walter Rudin's Principles of mathematical analysis Exercise 5.21, it is proved that for any closed subset $E\subseteq \mathbb{R}$, there exists a smooth function $f$ on $\mathbb{R}$ such that $E=\{x\in \mathbb{R}\mid f(x)=0\}$.

Since $E^c$ is open in $\mathbb{R}$, $E^c$ is a countable disjoint union of open intervals $(a_i,b_i)$. On an interval $(a,b)$, if we let $f(x)=\exp(\frac{1}{(x-a)(x-b)})$, $x\in(a,b)$; $f(x)=0$, $x\notin(a,b)$, ($f(x)=\exp(\frac{1}{(x-a)})$ or $f(x)=\exp(\frac{1}{(x-b)})$ on $(a,\infty)$ and $(-\infty,b)$ resp.), then $f$ is smooth. Hence we can get a smooth function on $\mathbb{R}$ such that $E$ is the zero point sets of $f$.

(1). How about a closed subset $E$ of $\mathbb{R}^n$? is it true that for any closed subset $E\subseteq \mathbb{R}^n$, there exists a smooth function $f$ on $\mathbb{R}^n$ such that $E=\{x\in \mathbb{R}^n\mid f(x)=0\}$?

An open set in $\mathbb{R}^n$ cannot be written as a disjoint union of countable open balls hence the proof above is not valid. The partition of unity only claims that for an open set $U$, there exists open $V$ in $U$ such that $\bar V\subseteq U$ and a smooth function $f$ on $\mathbb{R}^n$ such that $supp f\subseteq U$, $f|_V=1$.

(2). The taylor series of $f(x)=\exp(\frac{1}{(x-a)(x-b)})$ does not converge around points $a,b$ hence $f$ is not analytic at $a,b$. Is it true that for any closed subset $E\subseteq \mathbb{R}$, there exists an analytic function $f$ on $\mathbb{R}$ such that $E=\{x\in \mathbb{R}\mid f(x)=0\}$?

1) Let $\{B_{r_j}(a_j)\}$ be a countable collection of open balls (radius $r_j$, centre $a_j$) whose union is the complement of $E$. Take $f(x) = \sum_j c_j g(\|x - a_j\|/r_j)$ for some smooth function $g$ which is nonzero on $[0,1)$ and $0$ on $[1,\infty)$ and a suitable sequence of positive numbers $c_j$.
• You don't need it to be locally finite: just choose the $c_j$ decaying rapidly enough that, with $f_j (x) = g(\|x - a_j\|/r_j)$, $|c_j | \|f_j^{(k)}\|_\infty < 2^{-j}$ for $k \le j$. – Robert Israel Apr 27 '15 at 5:19
• Don't we need the function composed by $g$ to be smooth for $f$ to be smooth ? For that we can squared the function, right ? – Subhadip Majumder Mar 13 '17 at 15:44
• @RobertIsrael why do you demand that $|c_j|||f_j^{(k)}||_\infty<2^{-j}$? What is that for? – rmdmc89 Feb 1 '19 at 22:13


text:
# zbMATH — the first resource for mathematics

On the positivity step size threshold of Runge–Kutta methods. (English) Zbl 1073.65077
The author considers the initial value problem (IVP) $U'(t)- f(t,U(t)),\quad t\geq t_0,\quad U(t_0)= u_0,$ where $$t_0\in\mathbb{R}$$, $$u_0\in\mathbb{R}^n$$, $$f:\mathbb{R}\times \mathbb{R}^n\to \mathbb{R}^n$$.
IVP is called positive if $$U(t)\geq 0$$ for $$t\in [t_0, t^*]$$ and $$f\in P$$. For the numerical solution of the IVP the general Runge-Kutta methods (RK) with $$s$$ stages are used, defined by a matrix $$A= (a_{ij})$$ and vectors $$b= (b_i)$$, $$c= (c_i)$$, $$i,j= 1,2,\dots, s$$.
$$H$$ is called the step size threshold of positivity, if for step $$h_{im}\leq H$$ the solutions $$u_m$$ obtained by this method, are positive for any $$m$$ finite from $$\mathbb{R}$$.
For an irreducible non-confluent RK-method the threshold $$H$$ is equivalent with the positivity radius $$R(A,b)$$, introduced and studied thoroughly in numerical analysis literature.
In the second part the positive invariant cone for a linear problem and for a quasilinear problem is constructed.

##### MSC:
 65L50 Mesh generation, refinement, and adaptive methods for ordinary differential equations 65L06 Multistep, Runge-Kutta and extrapolation methods for ordinary differential equations 34C14 Symmetries, invariants of ordinary differential equations 65L05 Numerical methods for initial value problems involving ordinary differential equations 34A34 Nonlinear ordinary differential equations and systems
Full Text:
##### References:
 [1] Bermann, A.; Plemmons, R.J., Nonnegative matrices in the mathematical sciences, (1979), Academic Press New York [2] Berzins, M., Modified mass matrices and positivity preservation for hyperbolic and parabolic pdes, Commun. numer. methods engrg., 17, 9, 659-666, (2001) · Zbl 0986.65091 [3] Bolley, C.; Crouzeix, M., Conservation de la positivité lors de la discrétisation des problèmes d’évolution paraboliques, RAIRO anal. numér., 12, 3, 237-245, (1978) · Zbl 0392.65042 [4] Butcher, J.C., The numerical analysis of ordinary differential equations—runge – kutta and general linear methods, (1987), Wiley Chichester · Zbl 0616.65072 [5] Chawla, M.M.; Al-Zanaidi, M.A.; Evans, D.J., Generalized trapezoidal formulas for parabolic equations, Internat. J. comput. math., 70, 3, 429-443, (1999) · Zbl 0926.65081 [6] Dekker, K.; Kraaijevanger, J.F.B.M.; Schneid, J., On the relation between algebraic stability and B-convergence for runge – kutta methods, Numer. math., 57, 249-262, (1990) · Zbl 0702.65079 [7] Faragó, I.; Horváth, R.; Korotov, S., Discrete maximum principle for linear parabolic problems solved on hybrid meshes, Appl. numer. math., 53, 2-4, 249-264, (2005), (this issue) · Zbl 1070.65094 [8] L. Ferracina, M.N. Spijker, Stepsize restrictions for the total-variation-diminishing property in general Runge-Kutta methods, Report MI 2002-21, University of Leiden · Zbl 1080.65087 [9] Gerisch, A.; Griffiths, D.F.; Weiner, R.; Chaplain, M.A.J., A positive splitting method for mixed hyperbolic – parabolic systems, Numer. methods partial differential equations, 17, 2, 152-168, (2001) · Zbl 0981.65112 [10] Gerisch, A.; Weiner, R., The positivity of low order explicit runge – kutta schemes applied in splitting methods, Comput. math. appl., 45, 53-67, (2003) · Zbl 1036.65061 [11] Hairer, E.; Wanner, G., Solving ordinary differential equations II, (1991), Springer Berlin · Zbl 0729.65051 [12] Horváth, Z., Positivity of runge – kutta and diagonally split runge – kutta methods, Appl. numer. math., 28, 309-326, (1998) · Zbl 0926.65073 [13] Horváth, Z., On the positivity of matrix – vector products, Linear algebra appl., 393, 253-258, (2004) · Zbl 1070.15018 [14] Hundsdorfer, W., Numerical solution of advection – diffusion – reaction equations, Lecture notes for ph.D. course, Thomas Stieltjes institute, note NM-N9603, (1996), CWI Amsterdam [15] Hundsdorfer, W.; Verwer, J., Numerical solution of time-dependent advection – diffusion – reaction equations, Springer series in computational mathematics, vol. 33, (2003), Springer Berlin · Zbl 1030.65100 [16] Kraaijevanger, J.F.B.M., Contractivity of runge – kutta methods, Bit, 31, 482-528, (1991) · Zbl 0763.65059 [17] Shu, C.W.; Osher, S., Efficient implementation of essentially non-oscillatory shock capturing schemes, J. comput. phys., 77, 439-471, (1988) · Zbl 0653.65072 [18] Smith, H.L., Monotone dynamical systems: an introduction to the theory of competitive and cooperative systems, (1995), American Mathematical Society Providence, RI · Zbl 0821.34003 [19] Stoyan, G.; Mihálykó, C.; Ulbert, Z., Convergence and nonnegativity of numerical methods for an integrodifferential equation describing batch grinding, Comput. math. appl., 35, 12, 69-81, (1998) · Zbl 0999.65151 [20] Stuart, A., Perturbation theory for infinite dimensional dynamical systems, (), 181-290 · Zbl 0847.34070 [21] Vejchodský, T., On the nonnegativity conservation in semidiscrete parabolic problems, () · Zbl 1073.65096
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text:
Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Example 4.6.1.4. Let $X$ be a topological space containing a pair of points $x$ and $y$, which we regard as objects of the $\infty$-category $\operatorname{Sing}_{\bullet }(X)$. Then we have a canonical isomorphism of Kan complexes

$\operatorname{Hom}_{ \operatorname{Sing}_{\bullet }(X) }(x, y) \simeq \operatorname{Sing}_{\bullet }( P_{x,y} ),$

where $P_{x,y}$ denotes the topological space of continuous paths $p: [0,1] \rightarrow X$ satisfying $p(0) = x$ and $p(1) = y$ (equipped with the compact-open topology). See Example 3.4.0.5.


text:
# The Parameterization of Algebraic Structures and Applications I

 ANALYTIC AND GEOMETRIC NUMBER THEORY MINI-COURSE Topic: The Parameterization of Algebraic Structures and Applications I Speaker: Manjul Bhargava Affiliation: Princeton University Date: Wednesday, March 3 Time/Room: 2:00pm - 3:30pm/West Bldg. Lecture Hall

We describe how various fundamental algebraic structures (involving, for example, number fields, class groups, and algebraic curves) can be parameterized via the orbits of appropriate group representations. By developing techniques to count such orbits, we can then use these parameterizations to prove analytic density theorems about the corresponding algebraic objects. In this series of talks, we discuss several recent such results (as well as the techniques behind them), and in particular, we lead up to a proof that the average rank of all elliptic curves (when ordered by their heights) is bounded. [In fact, it is bounded by 1.5.]


text:
# How should $\lambda x. \lambda y. f z x y \stackrel{\eta}{=} \lambda x. f z x$ be read?

I'm currently learning untyped $\lambda$-calculus and especially the $\eta$-reduction.

The professor hat the following in his slides:

$$\lambda x. \lambda y. f~z~x~y \stackrel{\eta}{=} \lambda x. f~z~x$$

I don't know how to read and understand this. Could you please explain it to me?

## My try

On the left side and the right side is a function and they are $\eta$-equivalent.

It is a $\lambda$ function with one bound parameter $x$ and two unbound parameters $f$ and $z$. But here start my problems: What does $f~z~x$ mean? Is this a triple? Is it a function $f$ and a function $z$ and a function $x$?

I think in $\lambda$-calculus everything can be viewed as a $\lambda$-function, so $f$, $z$ and $x$ are functions. So the right side is the same as

$$\lambda(x) = f(z(x))$$

correct?

Applying the same logic on the left side, I get:

\begin{align} \lambda x. \lambda y. f~z~x~y &= \lambda(x)\\ &= f(z(x(y))) \end{align}

As $y$ is not in $\lambda(x)$ it is not bound. So we can simply remove it. Correct?

I think this could work. But I still don't know how to read it. I guess it will be something like:

"The lambda-function with the parameter $x$ that is a lambda function with a parameter y is f applied to z applied to x applied to y is eta-equivalent to the lambda function with the paramter x that is f applied to z applied to x".

• Much of what you write does not make sense to me. What is "λ(x)" supposed to be? Is that a legitimate syntax for a λ-expression? Whatever the theory, if you write something that is not syntactically correct, it does not mean anything, both in a formal and in an informal sense. - - - Also what do you mean by "As y is not in λ(x) it is not bound."? See answer. – babou Mar 3 '14 at 7:10

Applications in Lambda calculus are assumed to be left associative, this means that $f~z~x~y$ is the same as $((f~(z))~(x))~(y)$.

With this is mind eta-conversion makes more sense. Wikipedia says:

Eta-conversion expresses the idea of extensionality, which in this context is that two functions are the same if and only if they give the same result for all arguments. Eta-conversion converts between λx.(f x) and f whenever x does not appear free in f.

This means that $\lambda x. \lambda y. f~z~x~y \stackrel{\eta}{=} \lambda x. f~z~x$ is assumed to be true because $y$ does not change the result of the application $f~z~x$, its value is applied after $f~z~x$ is already evaluated.

You can find out more about eta-conversion by playing with it in the Haskell REPL:

Prelude> let z = 1
Prelude> let f x y = (+) ((+) z x) y
Prelude> :t f
f :: Integer -> Integer -> Integer
Prelude> f 1 1
3
Prelude> let f x = (+) ((+) z x)
Prelude> :t f
f :: Integer -> Integer -> Integer
Prelude> f 1 1
3

As you can se, it is easily possible to leave out a the parameter y without changing the meaning of your program.

You do not have to try to interpret each side of the equality. This may be far too complicated in some cases, and $\lambda$-calculus is a formal calculus to begin with. It is better to know the calculus rules and apply them.

This said, you still have to read the expressions correctly:

$\lambda x.fzx$ is to be read as $\lambda x.((fz)x)$, that is a function that takes an argument $x$ and then returns the result from applying the function $f$ to $z$, and then applying the resulting function to $x$.

By definition of $\eta$-conversion you have: $\lambda x.fx \stackrel{\eta}{=} f$ whenever $x$ does not occur free in $f$

Though $\lambda$-calculus is a formal calculus, which can be analyzed without interpretation, you can try to understand intuitively the intent of such a rule. Essentially it says that if you have a function $f$ and you build from it a function $g=\lambda x.fx$ which takes one argument (here called $x$) and applies $f$ to it, then $g$ is equal to $f$. The idea is that $f$ and $g$ will both give the same result when applied to any argument.

Of course, in this definition, $f$ stands for any lambda expression.

There is however the requirement that no free variable named $x$ appear in $f$. The reason is that a free variable names something in the outer environment where $f$ is used. But, if $g$ replaces $f$, such an occurence of $x$ in $g$ would refer to the "parameter" $x$ in the definition of $g$, hence no longer to whatever it named in the outer environment.

But much of this explanation is for your intuition. The lambda game is formal, even though it does have interpretations.

Hence, if you apply this definition to the $\lambda$-expression $fzx$ , you get: $\lambda y.fzxy \stackrel{\eta}{=} fzx$ since the variable y does not appear free in $fzx$

Being equal, they can be substituted for each other in any context (precisely because of the restriction on free variables).

Hence $\lambda x.\lambda y.fzxy \stackrel{\eta}{=} \lambda x.fzx$

All this is of course remindful of scope rules in many programming languages, and that is not due to chance.


text:
# Missing the first eigenVector/eigenvalue

Hello !
I’m currectly using FreeFEM++ to compute electronic structures.
I wrote a FreeFREM++ code supposed to compute the lowest eigenvalues and eigenfunctions of the Schrödinger equation in 2D, that is to solve the elliptic eigenvalue problem

• \frac 12 \Delta u(x) + V(x) u(x) = E u(x) on a square with periodic boundary, where V is a given function.

The eigenvalues (E_n) are ranked in nondecreasing order : E_1 \le E_2 \le E_3 …

Following the example of acoustics I found in Acoustics , I used the following code lines to solve the resulting generalized eigenvalue problem:

matrix K= rigidite(Vh,Vh,solver=Crout,factorize=1);
matrix M= masse(Vh,Vh,solver=CG,eps=1e-20);

// Computation of the 5 lowest eigenvalues

int nev=5;
real[int] ev(nev);
Vh[int] eV1 (nev);
real sigma= 0.0;
real scale=10;
int k=EigenValue(K,M,sym=true, sigma=sigma, value=ev,vector=eV1,tol=1e-10,maxit=0,ncv=0);

cout << "First eigenvalue : " << ev[0] << endl;
cout << "Second eigenvalue : " << ev[1] << endl;
cout << "Third eigenvalue : " << ev[2]<< endl;
cout << "Fourth eigenvalue : " << ev[3] << endl;
cout << "Fifth eigenvalue : " << ev[4] << endl;

Surprisingly, it seems that the code misses the lowest eigenvalue (ev[0] \simeq E_2).

Does someone know how to fix this problem?

Grégoire

I’d recommend that you use SLEPc instead of ARPACK, see, e.g., FreeFem-sources/schrodinger-square-well-1d-SLEPc.edp at develop · FreeFem/FreeFem-sources · GitHub or FreeFem-sources/schrodinger-harmonic-oscillator-2d-SLEPc.edp at develop · FreeFem/FreeFem-sources · GitHub. It’s much easier to debug things there.

I tried to use your example but couldn’t make it work because of an unsuccessful installation of PETSc & SLEPc on my virtual machine.
So I tried to make it work on the cluster of my workplace, where FreeFEM++ is compiled without SLEPc/PETSc in /3rdparty folder. (I need to sudo make install some files, but don’t have rights to sudo.
I tried to load my own installation of SLEPc/PETSc : the library seems to be loaded but I can’t call a single function of it.
Example given :

If I try to call for example the function EPSSolve, I’m told that the identifier EPSSolve doesn’t exist.
Maybe isn’t it possible to load a library this way ?

You don’t need to use sudo if you build FreeFEM with a prefix. You must do that when on a cluster. Please configure and install FreeFEM correctly, see, e.g., README.md, and then try again.


text:
# List reference by author

I am trying to use this code to order my references by author. I am on TeXworks and I am using Biber at the moment.I want to list all refernces with the author Weber in it.

\documentclass{article}
\usepackage[T1]{fontenc}
\usepackage[french]{babel}
\usepackage[style=alphabetic,maxnames=6,natbib=true,backend=biber]{biblatex}
%\bibliography{biblatex-examples}

\begin{document}
\DeclareSourcemap{
\maps[datatype=bibtex]{
\map{
\step[fieldsource=author,
match=Weber,
final]
\step[fieldset=keywords, fieldvalue=weber]
}
}
}

\printbibliography[keyword=weber]
\printbibliography[notkeyword=weber]
\end{document}

This is the error I get:

INFO - This is Biber 2.7 INFO - Logfile is 'biberpractice.blg' ERROR - Cannot find control file 'biberpractice.bcf'! - did you pass the "backend=biber" option to BibLaTeX? INFO - ERRORS: 1

• What bibliography package and style are you using? As far as I am aware this is only supported by biblatex with Biber (maybe also with BibTeX as backend, but Biber is preferred). We need to see an MWE/MWEB. Can you explain in more detail why you need this feature? I don't see what ids has to do with citing co-authors. – moewe Oct 12 '17 at 15:02
• Im using \usepackage{cite} and the \bibliographystyle{ieeetr}. Sorry I'm new to using LaTeX and BibTeX. I am trying to find a command in LaTeX that will go through my bib file (which is full of articles and publications) and list all of which that are referenced by the specified author. I tried to make IDS for each author trying to act as a type of keyword instead of the cite key. – Michael Oct 13 '17 at 12:47
• Sorry, I still don't quite understand what you want (but that might just be my limited grasp of the English language). You can edit your question to include the additional information and to make it more clear. – moewe Oct 13 '17 at 13:27
• I edited by question. I am trying find a command that will filter through my bib file (full of a bunch of references with numerous authors) and list all references by the author specified. – Michael Oct 16 '17 at 13:20
• I have added an MWE that works and filters out citations by 'Knuth'. I have a hunch that you might have more basic problems with Biber at the moment. Have a look at Biblatex with Biber: Configuring my editor to avoid undefined citations and Troubleshooting for biber if my MWE does not work for you (you should not alter the MWE in any way it will work out of the box, the file biblatex-examples.bib is automatically found if biblatex is installed properly). – moewe Oct 16 '17 at 13:32

Your MWE is almost there. \DeclareSourcemap should go into the preamble.

Here we filter out citations authored by Knuth and Sigfridsson. Note that you have to \cite or \nocite{*} all the entries you want to consider for filtering.

\documentclass{article}
\usepackage[T1]{fontenc}
\usepackage[french]{babel}
\usepackage[style=alphabetic, maxnames=6, natbib=true, backend=biber]{biblatex}

\DeclareSourcemap{
\maps[datatype=bibtex]{
\map{
\step[fieldsource=author,
match=Knuth,
final]
\step[fieldset=keywords, fieldvalue=knuth]
}
\map{
\step[fieldsource=author,
match=Sigfridsson,
final]
\step[fieldset=keywords, fieldvalue=sigfridsson]
}
}
}

\begin{document}
\nocite{sigfridsson,knuth:ct:a,companion,worman,knuth:ct:c}
\printbibliography[keyword=knuth]
\printbibliography[keyword=sigfridsson]
\printbibliography[notkeyword=knuth,notkeyword=sigfridsson]
\end{document}

• When Im using TeXworks and after I hold ctrl "T" the pdfLaTeX only shows "Hi, I'm Kant {knuth:ct,knuth:ct:a,companion}" Is the software not up to date? – Michael Oct 16 '17 at 14:31
• @Michael Did you compile my MWE? It does not contain anything related to Kant. Please have a look at the two links I gave in my comment to your question. – moewe Oct 16 '17 at 14:56
• Sorry I got my examples I was working with mixed. I tried to compile your MWE but it gives me ERROR - Cannot find control file 'knuth.bcf'! - did you pass the "backend=biber" option to BibLaTeX? – Michael Oct 16 '17 at 18:48
• @Michael Check your set up with tex.stackexchange.com/q/154751/35864. Did you set your editor up to use a 'build folder'? Or did you tell it to move auxiliary files to a different directory? Can you show a picture of your TeXworks configuration. – moewe Oct 16 '17 at 18:51
• I dont know if I should be using Biber or not, I saw a solution that required it so thats why I am using it. I already added biber to my TeXworks and im using UTF-8 under my preferences. – Michael Oct 17 '17 at 13:01