text:
Question

# If a and b are unit vectors inclined at an angle $\alpha ,\alpha \in \left[ {0,\pi } \right]$ to each other and $\left| {a + b} \right| < 1$ . Then $\alpha$ belongs to (a) $\left( {\dfrac{\pi }{3},\dfrac{{2\pi }}{3}} \right)$ (b) $\left( {\dfrac{{2\pi }}{3},\pi } \right)$ (c) $\left( {0,\dfrac{\pi }{3}} \right)$ (d) $\left( {\dfrac{\pi }{4},\dfrac{{3\pi }}{4}} \right)$

Hint: For this question we have to know that the $\left| {a + b} \right| = \sqrt {{{\left| a \right|}^2} + {{\left| b \right|}^2} + 2\left| a \right|\left| b \right|\cos \alpha }$ on squaring on both side we get ${\left| {a + b} \right|^2} = {\left| a \right|^2} + {\left| b \right|^2} + 2\left| a \right|\left| b \right|\cos \alpha$ and use of $\left| a \right| = 1$ , $\left| b \right| = 1$ and $\left| {a + b} \right| < 1$ we will find the range of the $\alpha$.

It is given that in the question that $\left| {a + b} \right| < 1$ and if we square on both side we get ;
${\left| {a + b} \right|^2} < 1$
Now we know that from the vectors property $\left| {a + b} \right| = \sqrt {{{\left| a \right|}^2} + {{\left| b \right|}^2} + 2\left| a \right|\left| b \right|\cos \alpha }$
Where $a$ and $b$ are the unit vectors and $\alpha$ is the angle between them .
Now that the for the unit vectors $\left| a \right| = 1$ , $\left| b \right| = 1$
On Squaring the above equation became : ${\left| {a + b} \right|^2} = {\left| a \right|^2} + {\left| b \right|^2} + 2\left| a \right|\left| b \right|\cos \alpha$
Put the values $\left| a \right| = 1$ $\left| b \right| = 1$
${\left| {a + b} \right|^2} = {1^2} + {1^2} + 2\cos \alpha$
From above it is proved that
${\left| {a + b} \right|^2} < 1$
$\left| {a + b} \right| < 1$
hence ${\left| {a + b} \right|^2} = {1^2} + {1^2} + 2\cos \alpha$
$2 + 2\cos \alpha < 1$
$2\cos \alpha < - 1$
$\cos \alpha < - \dfrac{1}{2}$
Hence the
$\alpha \in \left[ {\dfrac{{2\pi }}{3},\pi } \right]$
It is given that the
$\alpha \in \left[ {0,\pi } \right]$
Now we know that the cosine is negative in the second quadrant and by taking the intersection from the given value we will find out that $\alpha \in \left[ {\dfrac{{2\pi }}{3},\pi } \right]$

So, the correct answer is “Option B”.

Note: Always be careful on solving the trigonometric inequalities . If you have any problem in solving it then try to make the graph of the given trigonometric function . So it will become easier to solve .
The dot product tells you what amount of one vector goes in the direction of another .
Dot product of the perpendicular vector is always zero .
Cross product always gives the vector which is perpendicular to the both vectors and its direction would be determined by Right hand rule.


text:
# Find:11.2 Multiplication of 0.15 - Mathematics

Find:11.2 × 0.15

#### Solution

11.2 × 0.15 = 112/10 xx 15/100 = 1680/10000 = 1.680 = 1.68

Concept: Multiplication of Decimal Numbers by 10, 100 and 1000
Is there an error in this question or solution?

#### APPEARS IN

NCERT Class 7 Maths
Chapter 2 Fractions and Decimals
Exercise 2.6 | Q 5.06 | Page 52

format: plain


text:
Article Contents
Article Contents

# On the oscillation behavior of solutions to the one-dimensional heat equation

• We study the oscillation behavior of solutions to the one-dimensional heat equation and give some interesting examples. We also demonstrate a simple ODE method to find explicit solutions of the heat equation with certain particular initial conditions.

Mathematics Subject Classification: 35K05, 35K15.

 Citation:

•  [1] P. Collet and J. -P. Eckmann, Space-time behavior in problems of hydrodynamic type: A case study, Nonlinearity, 5 (1992), 1265-1302.  doi: 10.1088/0951-7715/5/6/004. [2] S. D. Eidel'man, Parabolic System, North-Holland, Amsterdam, 1969. [3] S. Kamin, On stabilization of solutions of the Cauchy problem for parabolic equations, Proc. Roy. Soc. Edinburgh Sect. A, 76/77 (1976), 43-53.  doi: 10.1017/S0308210500019478. [4] M. Nara and M. Taniguchi, The condition on the stability of stationary lines in a curvature flow in the whole plane, J. Diff. Eq., 237 (2007), 61-76.  doi: 10.1016/j.jde.2007.02.012. [5] W.-M. Ni, The Mathematics of Diffusion, CBMS-NSF Regional Conference Series in Applied Mathematics, v. 82, SIAM, 2011. doi: 10.1137/1.9781611971972. [6] V. D. Repnikov and S. D. Eidel'man, A new proof of the theorem on the stabilization of the solution of the Cauchy problem for the heat equation, Math. USSR Sb., 2 (1967), 135-139.  doi: 10.1070/SM1967v002n01ABEH002328.


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# Need explanation of Heap's Algorithm

Okay I am trying to understand the Heap's Algorithm from the original research paper published by B.R Heap.

The most confusing aspect I found that while Heap says that the example described on the first page takes N=12 ie. example with 12 objects to provide a permutations for them.

But yet he says that the 12 objects are the letters A to Q. Now unless I have forgotten my counting totally, A to Q is 17 objects not 12. So am I missing something obvious ?

Secondly, can someone please explain the Table 1 given in the bottom of the first page ?

Thanks.

• We expect references to fulfill the minimal scholarly requirements and be as robust over time as possible. Please take some time to improve your post in this regard. We have collected some advice here. – Raphael Jul 18 '16 at 10:32
• Please make your question self-contained by quoting the relevant passages and definitions. – Raphael Jul 18 '16 at 10:32

Pay no attention to the letter $Q$. $Q$ is just a "short-hand" for "the $N$th letter". Think of $Q$ as a kind of variable, rather than a letter itself.


text:
IFC 4.3.0.1 (IFC4X3_TC1)

# 4.7.1.2.10 Sine Spiral Transition Segment

The sine spiral transition segment is a special case of a spiral where the curvature rate of change is a sine function and the terms are dependent on the length L measured from the inflection point. The parameter value is defined as the deflection i.e. bearing angle &Theta.

The values of its individual terms are identical and equal to the length of the segment L: SineTerm = L LinearTerm = L

The following diagram shows the generic classes and relationships used when applying this concept.


text:
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# Half-integer level shift of vortex bound states in an iron-based superconductor

### Subjects

An Author Correction to this article was published on 13 March 2020

An Author Correction to this article was published on 13 March 2020

## Abstract

Vortices in topological superconductors may host Majorana zero modes (MZMs), which have been proposed as the building blocks of fault-tolerant topological quantum computers. Recently, a new single-material platform with the potential for realizing MZMs has been discovered in iron-based superconductors, without involving hybrid semiconductor–superconductor structures. Here, we report a detailed scanning tunnelling spectroscopy study of a FeTe0.55Se0.45 single crystal and show that this material hosts two distinct classes of vortex. These differ by a half-integer level shift in the energy spectra of the vortex bound states. This level shift is directly tied to the presence or absence of a zero-bias conductance peak and also alters the ratios of higher energy levels from integer to half-odd-integer. Our model calculations fully reproduce the spectra of these two types of vortex bound state, suggesting the presence of regions with and without topological surface states, which coexist within the same crystal. Our findings provide strong evidence for the presence of MZMs in FeTe0.55Se0.45 and establish it as an excellent platform for further studies.

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## Data availability

The data that support the findings of this study are available from the corresponding authors on reasonable request.

## Change history

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## References

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31. Yin, Z. P., Haule, K. & Kotliar, G. Spin dynamics and orbital–antiphase pairing symmetry in iron-based superconductors. Nat. Phys. 10, 845–850 (2014).

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47. Noguchi, R. et al. A weak topological insulator state in quasi-one-dimensional bismuth iodide. Nature 566, 518–522 (2019).

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49. Shi, X. et al. FeTe1 - xSex monolayer films: towards the realization of high-temperature connate topological superconductivity. Sci. Bull. 62, 503–507 (2017).

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## Acknowledgements

We thank N.F. Yuan, C.-K. Chiu, C. Schrade, S.-S. Qin and R.-X. Zhang for helpful discussions and F.-Z. Yang, G.-J. Qian for technical assistance. The work at IOP is supported by grants from the Ministry of Science and Technology of China (2015CB921000, 2015CB921300 and 2016YFA0202300), the National Natural Science Foundation of China (11234014, 11574371 and 61390501), and the Chinese Academy of Sciences (XDB28000000 and XDB07000000). The work at MIT is supported by DOE Office of Basic Energy Sciences, Division of Materials Sciences and Engineering under award no. DE-SC0019275. G.D.G. is supported by US DOE DE-SC0012704. J.S. and R.D.Z. are supported by the Center for Emergent Superconductivity, an EFRC funded by the US DOE.

## Author information

Authors

### Contributions

H.D. and H.-J.G. designed the experiments. S.Z., L.C., H.C. and Y.X. performed the STM experiments with assistance from L.K., W.L., D.W., P.F. and S.D. M.P., H.I. and L.F. provided theoretical models and simulations. J.S., R.Z. and G.D.G. provided samples. L.K., S.Z. and H.D. analysed experiment data with input from all other authors. L.K., M.P. and S.Z. plotted figures with input from all other authors. All authors participated in writing the manuscript. H.D., H.-J.G. and L.F. supervised the project.

### Corresponding authors

Correspondence to Liang Fu, Hong-Jun Gao or Hong Ding.

## Ethics declarations

### Competing interests

The authors declare no competing interests.

Peer review information: Nature Physics thanks Peter Wahl and the other, anonymous, reviewer(s) for their contribution to the peer review of this work.

Publisher’s note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

## Supplementary information

### Supplementary Information

Additional theoretical and experimental details, Supplementary Figs. 1–11 and refs. 1–41.

## Rights and permissions

Reprints and Permissions

Kong, L., Zhu, S., Papaj, M. et al. Half-integer level shift of vortex bound states in an iron-based superconductor. Nat. Phys. 15, 1181–1187 (2019). https://doi.org/10.1038/s41567-019-0630-5

• Accepted:

• Published:

• Issue Date:

• DOI: https://doi.org/10.1038/s41567-019-0630-5

• ### Topological superconducting vortex from trivial electronic bands

• Lun-Hui Hu
• Rui-Xing Zhang

Nature Communications (2023)

• ### Two-dimensional superconductors with intrinsic p-wave pairing or nontrivial band topology

• Wei Qin
• Jiaqing Gao
• Zhenyu Zhang

Science China Physics, Mechanics & Astronomy (2023)

• ### Ordered and tunable Majorana-zero-mode lattice in naturally strained LiFeAs

• Meng Li
• Geng Li
• Hong-Jun Gao

Nature (2022)

• ### Dual topological states in the layered titanium-based oxypnictide superconductor BaTi2Sb2O

• Z. Huang
• W. L. Liu
• D. W. Shen

npj Quantum Materials (2022)

• ### Iron pnictides and chalcogenides: a new paradigm for superconductivity

• Rafael M. Fernandes
• Amalia I. Coldea
• Gabriel Kotliar

Nature (2022)


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# Doom, the editor that I love

## Workspaces

Wow this just works so well. SPC TAB is the root menu for this great feature. From there you have a tons of options but really I have only used the following:

1. SPC TAB BackTick - Flips between last workspace
2. SPC TAB TAB - Shows the workspace bar for quick selection by number
3. SPC TAB . - Switches to workspace by typing a name or letters matching the name

## Python and testing

I have found that you need the following combination of settings to make sure that this all works:

Open your .doom.d/init.el file and add this:

(doom! )
(python
+lsp
+poetry
+pyenv)
)

### Preparing a python version

First, select the version you want to use and get it installed:

pyenv install 3.9.14

Then you need to install the following packages to on a global scale to get it prepared:

pyenv global 3.9.14
pip install python-lsp-server
pip install flake8
pip install poetry
pip install pylint

### Making a Python project

mkdir project_name
pyenv local 3.9.14
poetry init
# and then start coding!
# make sure to add pytest of at least version 1.2.0

### Running things with poetry

Basic running of your code inside poetry is done with:

poetry shell

or

poetry run the_command --with-args

## Projects

### Add a project in Doom

SPC p a

then find the folder you want to add. Best that it be a git directory!

### Closing a project

Reasons for closing a project are to re-evaluate the settings picked up by major or minor modes:

SPC p k

This will kill all the buffers under the project and hope you have everything saved.

## Editing

### Mass file changes

Below are the key commands in order to make mass edits across multiple files along with a description of what the key command is trying to do. Follow them in order and you can do a massive search and replace of text.

• SPC s p Search across the project for the text pattern to be modified
• C-; This implements a “right click” menu where you want to get options based on the search you are running
• E Invokes the embark option to export the search for modification
• C-c C-p This makes the buffer be able to be updated
• Now you can make your edits all you want. I like to sometimes use :%s/pattern/replace/g or other times, I just made edits where I see the need to do so.
• Z Z Save all the changes to every file updated.


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# GATE2017 CE-1: GA-4

1 vote
234 views

If the radius of a right circular cone is increased by $50\%$ its volume increases by

1. $75\%$
2. $100\%$
3. $125\%$
4. $237.5\%$

recategorized
Migrated from GO Civil 1 year ago by Arjun

Volume of  the cone $V=\frac{1}{3}\pi r^{2}h$

Here only radius change so we can write $V\propto r^{2}$

$r\rightarrow r+\frac{50}{100}r \implies r\rightarrow 1.5r$

New volume ${V}'\propto (1.5r)^{2} \propto 2.25r^{2}$

Percentage changes in volume $=\frac{2.25r^{2}-r^{2}}{r^{2}}\times 100=1.25\times 100=125\%$

So, correct option is $(C)$

selected by
0
$V = \dfrac{1}{3}\pi r^{2}h$

$\implies V \propto r^{2}$

$\implies V = (1.5)^{2} = 2.25 = (2.25-1)\times 100\% = 1.25\times 100\% = 125\%\Big\Uparrow$

## Related questions

1 vote
1
353 views
The following sequence of numbers is arranged in increasing order: $1,x,x,x,y,y,9,16,18$. Given that the mean and median are equal, and are also equal to twice the mode, the value of $y$ is $5$ $6$ $7$ $8$
Two machine $M1$ and $M2$ are able to execute any of four jobs $P,Q,R$ and $S$. The machines can perform one job on one object at a time. Jobs $P,Q,R$ and $S$ take $30$ minutes, $20$ minutes, $60$ minutes and $15$ minutes each respectively. There are $10$ objects each ... Job $S$ on $4$ objects. What is the minimum time needed to complete all the jobs? $2$ hours $2.5$ hours $3$ hours $3.5$ hours
The bar graph below shows the output of five carpenters over one month. each of whom made different items of furniture: chairs, tables, and beds. Consider the following statements. The number of beds made by carpenter $C2$ is exactly the same as the number of tables made by ... than the total number of tables. Which one of the following is true? Only i Only ii Both i and ii Neither i nor ii
The radius as well as the height of a circular cone increases by $10\%.$ The percentage increase in its volume is ________. $17.1$ $21.0$ $33.1$ $72.8$


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# Fraktal funksjoner

For our first example of recursion, let's look at how to compute the factorial function. We indicate the factorial of n by $n!$. It's just the product of the integers 1 through n. For example, 5! equals 1, dot, 2, dot, 3, dot, 4, dot, 5, or 120. (Note: Wherever we're talking about the factorial function, all exclamation points refer to the factorial function and are not for emphasis.)
You might wonder why we would possibly care about the factorial function. It's very useful for when we're trying to count how many different orders there are for things or how many different ways we can combine things. For example, how many different ways can we arrange n things? We have n choices for the first thing. For each of these n choices, we are left with n, minus, 1 choices for the second thing, so that we have n, dot, left parenthesis, n, minus, 1, right parenthesis choices for the first two things, in order. Now, for each of these first two choices, we have n, minus, 2 choices for the third thing, giving us n, dot, left parenthesis, n, minus, 1, right parenthesis, dot, left parenthesis, n, minus, 2, right parenthesis choices for the first three things, in order. And so on, until we get down to just two things remaining, and then just one thing remaining. Altogether, we have $n \cdot (n-1) \cdot (n-2) \cdots 2 \cdot 1$ ways that we can order n things. And that product is just $n!$ (n factorial), but with the product written going from n down to 1 rather than from 1 up to n.
Another use for the factorial function is to count how many ways you can choose things from a collection of things. For example, suppose you are going on a trip and you want to choose which T-shirts to take. Let's say that you own n T-shirts but you have room to pack only k of them. How many different ways can you choose k T-shirts from a collection of n T-shirts? The answer (which we won't try to justify here) turns out to be $n! / (k! \cdot (n-k)!)$. So the factorial function can be pretty useful. You can learn more about permutations and combinations here, but you don't need to understand them to implement a factorial algorithm.
The factorial function is defined for all positive integers, along with 0. What value should 0! have? It's the product of all integers greater than or equal to 1 and less than or equal to 0. But there are no such integers. Therefore, we define 0! to equal the identity for multiplication, which is 1. (Defining 0! = 1 meshes nicely with the formula for choosing k things out of n things. Suppose that we want to know how many ways there are to choose n things out of n things. That's easy, because there is only one way: choose all n things. So now we know that, using our formula, $n! / (n! \cdot (n-n)!)$ should equal 1. But $(n-n)!$ is 0!, so now we know that $n! / (n! \cdot 0!)$ should equal 1. If we cancel out the $n!$ in both the numerator and denominator, we see that $1/(0!)$ should equal 1, and it does because 0! equals 1.)
So now we have a way to think about $n!$. It equals 1 when n, equals, 0, and it equals $1 \cdot 2 \cdots (n-1) \cdot n$ when n is positive.

This content is a collaboration of Dartmouth Computer Science professors Thomas Cormen and Devin Balkcom, plus the Khan Academy computing curriculum team. The content is licensed CC-BY-NC-SA.


text:
# Magnetic braking

The magnetic field lines rotates together with the Sun as a solid object. Ionized material carried along the field lines will at some point rc escape the magnetic field lines and thus take away the Sun's angular momentum

Magnetic braking is a theory explaining the loss of solar angular momentum due to material getting captured by the solar magnetic field and thrown out at great distance from the surface of the Sun.

## The problem

The currently accepted theory of the solar system's evolution states that the Solar System originates from a contracting gas cloud. As the cloud contracts the angular momentum L must be conserved, this is in accordance with the angular momentum of the major planets, but when the planets' angular momentum are compared to the Sun's own, the Sun has less than 1% of its supposed angular momentum. In other words, the Sun has slowed down its spin while the planets have not.

## The idea behind magnetic braking

Even though material falling in towards the Sun has its angular momentum conserved and will thus add to the spin, The ionized material captured by the magnetic field lines will rotate with the Sun as a solid body. If material escapes from the Sun due to solar wind, i.e., driven out by thermal pressure, the highly ionized material will get captured by the field lines and rotate with the same angular velocity as the Sun even though its carried far away from the Sun's surface and should in principle have its rotation governed by "normal" celestial mechanics. This effect of carrying mass into a great distance and throwing it away robs the Sun of its spin. A likewise analog effect is used in slowing the spin of a rotating satellite, here two wires spool out weights into a distance slowing the satellites spin, then the wires are cut, letting the weights escape into space and permanently robbing the spacecraft of its angular momentum.

## Theory behind magnetic braking

As ionized material follows the Sun's magnetic field lines, due to the effect of the field lines being frozen in the plasma, the charged particles feel a force $\mathbf{F}$ of the magnitude:

$\mathbf{F}=qv \times \mathbf{B}$

where q is the charge, v is the velocity and $\mathbf{B}$ is the magnetic field vector. This bending action forces the particles to "corkscrew" around the magnetic field lines while held in place by an "magnetic pressure"PB, or "energy density", while rotating together with the Sun as a solid body:

$P_B=\frac{B^2}{2 \mu_0}$

Since magnetic field strength decreases with the cube of the distance there will be a place where the kinetic gas pressure Pg of the ionized gas is great enough to break away from the field lines:

Pg = nmv2

where n is the number of particles, m is the mass of the individual particle and v is the radial velocity away form the Sun, or the speed of the solar wind.

The magnetic field strength as a function of radius can be approximated as:

$B(r)=\frac{D_\odot}{r^3}$

The critical distance where the material will break away from the filed lines can then be calculated as the distance where the kinetic pressure and the magnetic pressure are equal, i.e.:

$P_B=P_g \Rightarrow$
$\frac{B(rc)^2}{2\mu_0}=n m v^2\Rightarrow$
$\frac{D_{\odot}^2}{2\mu_0 r_c^6}=n m v^2$

If the solar mass loss is omni-directional the mass loss then $n m= \frac{dM/dt}{4\pi r^2 v}$, plugging this into the above equation an isolating the critical radius it follows that:

$r_c=\sqrt[4]{\frac{2\pi D_\cdot ^2}{\mu_0 v (dM/dt)}}$

### Present day value

Currently it is estimated that:

• The mass loss of the Sun is about $dM/dt=2 \cdot 10^9 kg/s$
• The solar wind speed is $v=5\cdot10^5m/s$
• The dipole field is $D_{\odot}=8 \cdot 10^22 T m^3$

This leads to a critical radius of $r_c =3.4 R_\odot$. This means that the ionized plasma will rotate together with the Sun as a solid body until it reaches an distance of nearly 3½ the radius of the Sun, from there the material will break off and stop affecting the Sun.

The amount of solar mass needed to be thrown out along the field lines to make the Sun completely stop the rotation can then be calculated using the specific angular momentum:

$\frac{j_\odot}{j_c}=(\frac{R_\odot}{r_c})^2 \approx 10%$

About 1/10 of the solar mass should thus have be lost by now due to this effect, but if the present mass loss rate was stable throughout the Sun's "life" only about 0.01% of the Sun's original mass would have been lost. This indicates that either the Sun's mass loss must have been higher in the past or that the Sun's magnetic field has been much stronger.

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text:
I have a question which relate adjoint space. That is: ” Is $C^*[0,1]$ linear isomorphism with $C[0,1]$?” $C[0,1]$ is space of continous function in $[0,1]$. If we choose inner product of $C[0,1]$ is $<,>: C[0,1] \longrightarrow\Bbb{R}$, by $=\int_0^1 f(x)g(x)dx$ then $C^*[0,1]$ $=$ $C[0,1]$ (linear isomorphism)?

Problem of division boy and girl set

TST Lam Dong 2010

Make $n$ boys and $n$ girls a line. Divide the line by two parts such that number of boys and number of girls are equivalent. Let $A$ be number of cases which cannot divide, let $B$ be number of cases which can divide by one way. Prove that $B = 2A$.

Solution. One way division the line is choosing $k$ boys and $k$ girls. The line has two part, left part and right part. And two ways division are equivalent if their left part has same of number boys and girls.

With the only way division, (suppose $k$ boys and $k$ girls in left part), we change a boy in left part with a girl in right part then we attain a impossible way. And we change a girl in left part with a boy in right part then we attain a impossible way too.

So we have correspondence of one-one: the one way division and two impossible way division, this implies $B = 2A$. QEA

Bài tập về khử dạng vô định

Khi tính giới hạn hàm số, ta sẽ gặp những bài toán có dạng $\infty - \infty, \dfrac{\infty}{\infty}, \dfrac{0}{0}, 0.\infty, 1^{\infty}$. Đó là các dạng vô định. Để làm các kiểu bài tập này, các em phải biết cách khử chúng.

Yêu cầu: Nắm vững phần giới hạn hàm số, giới hạn một phía + các hằng đẳng thức.

Bài 1.

1. $\underset{x\to 1}{\lim}\dfrac{-2{{x}^{2}}+x+1}{{{x}^{2}}-4x+3}$.
2. $\underset{x\to 2}{\lim}\dfrac{\sqrt{2x-1}-\sqrt{x+1}}{{{x}^{2}}-3x+2}$.
3. $\underset{x\to 0}{\lim }\dfrac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x}$.
4. $\underset{x\to 1}{\lim }\dfrac{\sqrt[3]{x+7}-\sqrt{x+3}}{{{x}^{2}}-3x+2}$.

Bài 2.

1. $\underset{x\to 2}{\lim }\dfrac{\sqrt{x+7}-3}{{{x}^{2}}-4}$.
2. $\underset{x\to -1}{ \lim }\dfrac{\sqrt{{{x}^{2}}+3}-2}{\sqrt[3]{x}+1}$.
3. $\underset{x\to -\infty }{\lim }\dfrac{\sqrt{{{x}^{2}}+2x+3}+2x}{\sqrt{4{{x}^{2}}+2}-x+1}$.

Một số dạng vô định có chứa lượng giác, ta chú ý giới hạn sau:

$\lim\limits_{x \to 0}\dfrac{\sin x}{x}=1$.

Bài 3.

1. $\lim\limits_{x \to o}\dfrac{1 - \cos x}{x^2}$.
2. $\lim\limits_{x \to 0}\dfrac{\sin 2x}{\sin 5x}$.
3. $\lim\limits_{x \to 0}\dfrac{\sqrt{1+x^2} - \cos x}{x^2}$.


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# Convergence of a matrix

Let $A_j$ be a sequence in $\mathbb {C}^{n\times n}$. Show that $\displaystyle \sum_{j=0}^\infty A_j$ converges if $\displaystyle \sum_{j=0}^\infty ||A_j||$ does.

Note that $\displaystyle \Vert A\Vert=\sup_{|x|=1} \vert Ax \vert$. I know that since $\displaystyle \sum_{j=0}^\infty ||A_j||$ does converge, then $\displaystyle \Vert A_j \Vert$ eventually converges to zero. Where do I go from here?

-
I'm confused by your use of $j$. Sometimes it is a subscript, sometimes it seems like a vector or scalar multiple. –  Arkamis Nov 4 '12 at 4:51
@ Ed sorry about that, I fixed it. –  Klara Nov 4 '12 at 4:59
Notice that $\| \cdot \|$ as you give defines a norm and adapt the proof that an absolutely convergent series of real numbers converges. –  Jose27 Nov 4 '12 at 6:22

Hint: $\mathbb{C}^{n \times n}$ is complete, so we need only show that the sequence $S_n = \sum_{j=0}^{n} A_j$ is Cauchy.
Suppose without loss of generality that $n>m$. Then: $\|S_n-S_m\| =\|\sum_{j=m+1}^n A_j\| \leq \sum_{j=m+1}^n \|A_j\|= |\sum_{j=0}^n \|A_j\| - \sum_{j=0}^m \|A_j\||$ Since $\sum_{j=0}^\infty \|A_j\|$ is convergent, it is Cauchy, hence it follows that $S_n$ is also Cauchy.


text:
Synopsis

Get the Hydrogen Out

Physics 9, s95
Using clean helium made with a two-step purification process could prevent damaging blockages in cryostats.

Liquid helium is the unsung partner in numerous low-temperature experiments. Unfortunately, the cryogen is expensive. So to preserve it, a growing number of labs have liquefiers, which collect evaporated helium and cool it back into liquid. Conrado Rillo, of the University of Zaragoza, Spain, and the Spanish National Research Council, and his colleagues have now developed a “clean helium” recipe for liquifiers that should prevent a costly problem for cryogenic experiments: blockages caused by solidified hydrogen molecules.

For certain low-temperature applications, researchers need a cryogen that’s colder than standard liquid helium (4.2 K). Pushing helium through a narrow capillary such that it cools upon expanding at the exit can chill the liquid to 3 K and below. But at these temperatures, molecular hydrogen ( ${\text{H}}_{2}$)—a common impurity in helium that is tough to remove—almost completely solidifies. Rillo and his co-workers thought that ${\text{H}}_{2}$ collecting on the capillary walls might explain numerous reported failures in cryogenic experiments. To test their hypothesis, they pumped helium through a pinched capillary. Within 10 hours the thin tube had a clog, which only cleared when the tube was warmed above ${\text{H}}_{2}$’s melting point.

The researchers estimate that, to prevent blockages, the ${\text{H}}_{2}$ concentration has to be roughly ten million times lower than that of commercial grade helium. To reach this purity, they propose a two-step solution: a pre-freeze that condenses out the ${\text{H}}_{2}$ (and other impurities) and the use of a getter purifier, which absorbs the remaining ${\text{H}}_{2}$. They produced this clean helium at the University of Zaragoza’s helium liquefaction plant, and using it kept their cryogenic instruments blockage-free for more than a year.

This research is published in Physical Review Applied.

–Jessica Thomas

Jessica Thomas is the Editor of Physics.

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text:
# Abstract

We will present our study of the flux and spectral variability of the Crab above 100 MeV on different timescales ranging from days to weeks. In addition to the four main intense and day-long flares detected by AGILE and Fermi-LAT between Sept. 2007 and Sept. 2012, we find evidence for week-long and less intense episodes of enhanced gamma-ray emission that we call "waves". Statistically significant "waves" show timescales of 1-2 weeks, and can occur by themselves or in association with shorter flares. The Sept. - Oct. 2007 gamma-ray enhancement episode detected by AGILE shows both "wave" and flaring behavior.

We extend our analysis to the publicly available Fermi-LAT dataset and show that several additional "wave" episodes can be identified. We discuss the spectral properties of the September 2007 "wave"/flare event and show that the physical properties of the "waves" are intermediate between steady and flaring states. Plasma instabilities inducing "waves" appear to involve spatial distances $l \sim 10^{16} \,$cm and enhanced magnetic fields $B \sim (0.5 - 1)\,$}mG.

Day-long flares are characterized by smaller distances and larger local magnetic fields. Typically, the deduced total energy associated with the "wave" phenomenon ($E_w \sim 10^{42} \, \rm erg$, where $E_w$ is the kinetic energy of the emitting particles) is comparable with that associated to the flares, and can reach a few percent of the total available pulsar spindown energy. Most likely, flares and waves are the product of the same class of plasma instabilities that we show acting on different timescales and radiation intensities.


text:
# Properties of Definite Integrals: Definite Integral Definition (For CBSE, ICSE, IAS, NET, NRA 2022)

Doorsteptutor material for competitive exams is prepared by world's top subject experts: get questions, notes, tests, video lectures and more- for all subjects of your exam.

We will be exploring some of the important properties of definite integrals and their proofs in this article to get a better understanding. Integration is the estimation of an integral. It is just the opposite process of differentiation. Integrals in maths are used to find many useful quantities such as areas, volumes, displacement, etc. There are two types of Integrals namely, definite integral and indefinite integral. Here, we will learn about definite integrals and its properties, which will help to solve integration problems based on them.

• Definite Integral Definition
• Properties
• Proofs
• Example

## Definite Integral Definition

If an integral has upper and lower limits, it is called a Definite Integral. There are many definite integral formulas and properties. Definite Integral is the difference between the values of the integral at the specified upper and lower limit of the independent variable. It is represented as;

## Definite Integral Properties

Following is the list of definite integrals in the tabular form which is easy to read and understand.

 Properties Description Property 1 Property 2 Also Property 3 Property 4 Property 5 Property 6 Property 7 2 parts Property 8 2 parts or it՚s an even function or it՚s an odd function

## Properties of Definite Integrals Proofs

Property 1:

This is the simplest property as only a is to be substituted by t, and the desired result is obtained.

Property 2:

, also

Suppose

If is the anti-derivative of then use the second fundamental theorem of calculus, to get Also, if then

Hence, .

Property 3: If is the anti-derivative of f, then use the second fundamental theorem of calculus, to get;

Let՚s add equations (2) and (3) , to get

Property 4:

Let, so that

Also, note that when and when . So, wil be replaced by when we replace a by t. Therefore,

from equation (4)

From property 2, we know that . Use this property, to get

Now use property 1 to get

Property 5:

Let, so that

Also, observe that when and when So, will be replaced by when we replace a by t. Therefore,

from equation (5)

From Property 2, we know that Using this property, we get

Next, using Property 1, we get

Property 6:

From property 3, we know that

Therefore,

Where, and

Let, so that

Also, note that when , and when Hence, when we replace a by t. Therefore,

from equation (7)

From Property 2, we know that . Using this property, we get

Next, using Property 1, we get

Replacing the value of I2 in equation (6) , we get

Property 7:

we know that

Now, if then equation (8) becomes

And, if then equation (8) becomes

Property 8: or it is an even function and or it is an odd function.

Using Property 3, we have

Where,

Consider

Let,

Also, observe that when . Hence, will be replaced by when we replace a by t. Therefore,

From Property 2, we know that , use this property to get,

Next, using Property 1, we get

Replacing the value of in equation (9) , we get

Now, if ‘f’ is an even function, then Therefore, equation (11) becomes

And, if ‘f’ is an odd function, then Therefore, equation (11) becomes

Now, let us evaluate Definite Integral through a problem sum.

### Example

Question: Evaluate

Solution: Observe that,

Hence, using Property 3, we can write

Solving the integrals, we get

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