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# If the height of a vertical pole is $\sqrt{3}$ times the length of its shadow on the ground, then the angle of elevation of the sun at that time is

A. 30° B. 60° C. 45° D. 75° Answer: Option B
Let AB be a vertical pole and let its shadow be BC Let BC = x m, then length of pole $=\sqrt{3} x$ $\theta$ be the angle of elevation $\begin{array}{l} \therefore \tan \theta=\frac{A B}{B C}=\frac{\sqrt{3} x}{x}=\sqrt{3} \\ =\tan 60^{\circ} \\ \therefore \theta=60^{\circ} \end{array}$


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Stochastic programming 메모

Multistage Stochastic Optimisation Problems 관련 논문 이해를 위해

Stochastic programming?

In the field of mathematical optimization, stochastic programming

is a framework for modeling optimization problems that involves uncertainty.

The goal here is to find some policy that is feasible for all (or almost all) the possible data instances and maximizes the expectation of some function of the decisions and the random variables.

Two-stage problems

The basic idea of two-stage stochastic programming is that (optimal) decisions should be based on data available at the time the decisions are made and cannot depend on future observations.

where $Q(x, \xi)$ is the optimal value of the second-stage problem

The classical two-stage linear stochastic programming problems can be formulated as

where $Q(x, \xi)$ is the optimal value of the second-stage problem

$x\in \Bbb R^n$ is the first-stage decision variable vector.
$y\in \Bbb R^m$ is the second-stage decision variable vector.
$\xi(q, T, W, h)$ contains the data of the second-stage problem.

at the first stage we have to make a “here-and-now” decision $x$ before the realization of the uncertain data $\xi$, viewed as a random vector, is known. At the second stage, after a realization of $\xi$ becomes available, we optimize our behavior by solving an appropriate optimization problem.

Distributional assumption(분포 가정?)

The formulation of the above two-stage problem assumes that the second-stage data $\xi$ can be modeled as a random vector with a known probability distribution (not just uncertain).

Discretization

To solve the two-stage stochastic problem numerically, one often needs to assume that the random vector $\xi$ has a finite number of possible realizations, called scenarios, say $\xi_1,...,\xi_K$, with respective probability masses $p_1,...,p_K$. Then the expectation in the first-stage problem’s objective function can be written as the summation:

and, moreover, the two-stage problem can be formulated as one large linear programming problem (this is called the deterministic equivalent of the original problem).

format: plain


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# Computing class group of $\mathbb Q(\sqrt{6})$

I am calculating the class group of $\mathbb Q(\sqrt 6)$. My working is as follows:

The Minkowski bound is $\lambda(6)=\sqrt 6<3$ so we only need to look at prime ideals of norm $2$. $2$ divides the discriminant, $24$, of $\mathbb Q(\sqrt 6)$, so $2$ ramifies, and so for any prime ideal $\mathfrak p$ of norm $2$, $\mathfrak p^2$ is principal. But $(2, \sqrt 6)^2=(2)$, so $\mathfrak p=(2, \sqrt 6)$ has norm $2$. If I can show $\mathfrak p$ is not principal, then I can conclude that $Cl(\mathbb Q(\sqrt 6))\cong C_2.$ But if there was an element of $\mathcal O_{\mathbb Q(\sqrt 6)}$ of norm 2, this would correspond to a solution of the diophantine equation $x^2-6y^2=2$. Clearly $x$ is even, so writing $x=2n$ we get $2n^2-3y^2=1$. Now reducing modulo $3$, we get $2n^2\equiv 1 \pmod 3$ which has no soultions. Hence $\mathfrak p$ is not principal and $Cl (\mathbb Q(\sqrt 6)) \cong C_2.$

However, this link says that the class group is trivial. Where have I gone wrong? What am I misunderstanding? Is $(2, \sqrt 6)$ principal?

There are indeed no elements of norm $2$, but there are elements of norm $-2$.
• $(2+\sqrt 6)(\sqrt 6-2)=2 \in (2+\sqrt 6)$ and $2+\sqrt 6 - 2=\sqrt 6 \in (2+\sqrt 6)$. So $(2+\sqrt 6)=(2,\sqrt 6)$ – James Apr 28 '15 at 10:46
Where you went wrong is where you overlooked the Diophantine equation $x^2 - 6y^2 = -2$. The fundamental unit of this ring is $5 + 2 \sqrt 6$, which has a norm of 1, not $-1$. This means that if $x^2 - 6y^2 = n$ has no solutions, it does not rule out that $x^2 - 6y^2 = -n$ might.
And indeed $x^2 - 6y^2 = -2$ does have infinitely many solutions, with the most obvious one being $x = 2$, $y = 1$, which you yourself later discovered. Then $(2 + \sqrt 6)(5 + 2 \sqrt 6)$ gives another, and from there you can figure out how to get as many as you want.
More importantly to your query, however, $(2 - \sqrt 6)(2 + \sqrt 6) = -2$, and since $-2 \in \langle 2 \rangle$, combined with the fact that $(2 + \sqrt 6)(3 - \sqrt 6) = \sqrt 6$, we find that both $\langle 2 \rangle$ and $\langle \sqrt 6 \rangle$ are properly contained in $\langle 2 + \sqrt 6 \rangle$, which is not only a principal ideal but a prime ideal after all.
You were right that 2 ramifies, as we have $\langle 2 \rangle = \langle 2 + \sqrt 6 \rangle^2$. In general, if $\mathbb Q(\sqrt{pq})$ (with $p$ and $q$ distinct primes) is a principal ideal domain, you can find factorizations for $\pm p$, $\pm q$ in that domain, and consequently that $\sqrt{pq}$ is reducible.


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# Formally proving no algorithm exists [closed]

Are there standard techniques to show that no algorithms exist for given complexity constraints?

For example, consider the following problem. The input is a list of items with exactly one duplicate, and you wish to find the duplicate item.

How do you formally prove that there cannot be an algorithm that does this in constant space and linear time?

• Mar 12, 2017 at 4:26
• I don't see how to solve this problem in constant space with any amount of time, since you need superconstant space just to write down the answer.... But anyway one answer to your question is information theory, also communication complexity which is closely related.
– usul
Mar 13, 2017 at 5:40

See this lower bound for sorting, for example: http://planetmath.org/lowerboundforsorting

You typically need to assume something about the algorithm's access to the input data. In this case, the assumption is that the algorithm sorts via pairwise comparisons. Any such algorithm must make at least $\Omega(n\log n)$ such comparisons. This is because the pairwise comparisons effectively induce a binary tree on $n!$ leaves -- one leaf for each possible input ordering. Such a tree must have depth $\Theta(\log(n!))=\Theta(n\log n)$.

Note that in a different data access model, algorithms can achieve linear-time sorting: https://en.wikipedia.org/wiki/Radix_sort

Obiously, faster than linear time is impossible, since you have to at least read all of the inputs.

• This is not entirely correct for the example. You do not need to sort the elements in order to find the single duplicate. You can do it in linear time easily, but you need at least two pointers, each of size of log n bits. Mar 12, 2017 at 17:11
• Ah, ok -- was answering a different part of the question, before the "For example". I think the answer does offer some insight for that part, but it's a judgement call. Mar 12, 2017 at 17:25

What you are asking for is methods in proving lower bounds on the computational complexity (measured in space, time, etc) of given computational problems, and the answer is mostly that we have made very little progress on general methods for this and that is essentially the entire subject of complexity theory.

That being said, if you're more interested in estimates that are almost certainly going to be useful, you can use the tool of the reduction which will essentially place your problem in a partial order representing the difficulty of problems, allowing you to then think about how hard it is relative to the other problems we study, given some granularity of reduction (in the link, reductions in FP).


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# SOLUTION: still not sure of this answer. graph (-2,5) slope of -1 I know the formula is y=mx+b so would that be y=-1x+5 so then -1x+5-1x+5 x= -5 or am i way off??

Algebra ->  -> SOLUTION: still not sure of this answer. graph (-2,5) slope of -1 I know the formula is y=mx+b so would that be y=-1x+5 so then -1x+5-1x+5 x= -5 or am i way off??      Log On

 Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Mathway solves algebra homework problems with step-by-step help!


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# Weitere Informationen zum ausgewählten Vortrag

## Algebraic cycles and refined unramified cohomology

### Stefan Schreieder (Hannover)

#### Freitag, 15. Januar 2021 10:30Uhr

We introduce refined unramified cohomology groups. This notion allows us to give in arbitrary degree a cohomological interpretation of the failure of integral Hodge- or Tate-type conjectures, of l-adic Griffiths groups, and of the subgroup of the Griffiths group that consists of torsion classes with trivial transcendental Abel--Jacobi invariant. Our approach simplifies and generalizes to cycles of arbitrary codimension previous results of Bloch--Ogus, Colliot-Thélène--Voisin, Voisin, and Ma that concerned cycles of codimension two or three. As an application, we give for any i>2 the first example of a uniruled smooth complex projective variety for which the integral Hodge conjecture fails for codimension i-cycles in a way that cannot be explained by the failure on any lower-dimensional variety.


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# Exporting images on the command line

Recent builds from git will bring you a new executable, “darktable-cli”. With this tool you can export images using the processing power of darktable on the command line.

The simplest way to call the utility is

darktable-cli <input> <output>

This will take the input image, look for the XMP file associated with it, process it at maximal resolution and write the output to output, trying to guess the output format from the output filename. You can also explicitly give a XMP file by running

darktable-cli <input> <xmp> <output>

When a scaled down output is needed you can overwrite the default 0x0 size by adding one or both of

--width <max x size> --height <max y size>

These settings work the same as in the darktable gui. There is also a setting

--bpp <max bitdepth>

which is not working at the time of writing this, so currently the highest possible bit depth for the chosen output format is used.

All of this is subject to change so expect more options to be added in the future. To be on the safe side darktable-cli will not overwrite files but just terminate when output already exists.

As always, play with it, use it, break it, and report back. Preferably on the mailing lists or in IRC.

## 2 thoughts on “Exporting images on the command line”

1. Hey there, would feedback in form of tickets not be a good thing?

2. Pingback: 1.1 release candidate 1 | darktable


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# Persistent Mounts with or etc or fstab

The chief difference between defining a mount in fstab versus autofs is that the kernel dedicates resources to keep fstab mounts in place all the time. Conversely, autofs mounts are “on-demand” and are better suited for NFS or DVD drives.

## Understanding the fstab

Here’s a sample fstab:

# Sample fstab
# default = rw, suid, dev, exec, auto, nouser, and async

# Device                  Mount Point   FS Type  Options         Dump?   fsck

/dev/VolGroup00/LogVol00  /             ext3     defaults        1       1
LABEL=/boot               /boot         ext3     defaults        1       2
tmpfs                     /dev/shm      tmpfs    defaults        0       0
devpts                    /dev/pts      devpts   gid=5,mode=620  0       0
sysfs                     /sys          sysfs    defaults        0       0
proc                      /proc         proc     defaults        0       0
/dev/VolGroup00/LogVol01  swap          swap     defaults        0       0

Column What it is
Device Name of the block device (like /dev/sdc1, etc.) You can also use LABEL= or UUID=
Mount Point Where to mount the device
FS (FileSystem) Type Can be ext2, ext3, reiserfs, vfat and swap (many others depending upon kernel support.) The interesting one is auto, whereby the kernel attempts to detect the filesystem automagically. This is good for CD/DVD-ROMs and (gasp!) floppies. Use an actual FStype for all others
Options For a much more comprehensive list of options, see man mount. Here are some common ones:
auto/noauto - Mount/do not mount at boot
user/nouser - Should normal users be able to mount? If so, use user. Else, it’s nouser
exec/noexec - Execute binaries off the device? noexec off a Linux root partition might be unpleasant…
ro - Mount as read-only
rw - Mount as read/write
sync/nosync - Changes to device are done synchronously with sync (written at the same time the command is issued). Good for USB sticks and (gasp!) floppies. nosync may be preferable for HDD performance.
Dump This is a small backup utility. Set to 0 almost all the time.
fstab Determine the order in which to check filesystems on devices. 0 will cause it to ignore it.

## Using better device identifiers

Given that the device names might change due to myriad factors (e.g. plug HDD into a different USB or SATA port), it’s a better idea to use the device label or, even better, the device’s UUID. I’m going to deal with UUIDs in this example.

### Figure out the device’s UUID

UNIX-like simplicity itself.

[user@example ~]# ls -l /dev/disk/by-uuid/
total 0
lrwxrwxrwx 1 user user 10 Jun 25 11:00 43b64bd4-ee2b-4e4d-b2af-e868cb47d596 -> ../../sda1
lrwxrwxrwx 1 user user 10 Jun 25 11:00 7c488bee-a567-4432-bdbe-c30e9d33a927 -> ../../sdb5
lrwxrwxrwx 1 user user 10 Jun 25 11:00 c1e9157e-6572-4ec1-b1d0-8adb97052a59 -> ../../sdd1
lrwxrwxrwx 1 user user 10 Jun 28 10:25 edabbb0e-d87a-4c5b-a9aa-492e553c98d1 -> ../../sdc5

You can see what each device is using fdisk -l /dev/sdc5 or, since you know the UUID:

[user@example ~]# fdisk -l /dev/disk/by-uuid/edabbb0e-d87a-4c5b-a9aa-492e553c98d1

Disk /dev/disk/by-uuid/edabbb0e-d87a-4c5b-a9aa-492e553c98d1: 1000.2 GB, 1000202208768 bytes
255 heads, 63 sectors/track, 121600 cylinders
Units = cylinders of 16065 * 512 = 8225280 bytes

### Create the fstab entry

UUID=edabbb0e-d87a-4c5b-a9aa-492e553c98d1 /media/usbdrive ext3  defaults  0 0

Et voila!

### Problems with this approach

You might see a “special device UUID=edabbb0e-d87a-4c5b-a9aa-492e553c98d1 does not exist” when you reboot your system, especially when dealing with external drives. Some people report that specifying the full path helps (e.g. UUID=/dev/disk/by-uuid/...) but I haven’t had any luck with this.

Essentially, and to keep things simple, you could still keep the fstab entry and issue a mount -a to mount all the devices you’ve specified after booting up. While there’s really no ‘fix’ to this AFAIK (even using LABEL= doesn’t help), you could alternately add a quick line to /etc/rc.local.

echo -e "Waiting for external drives to mount..." && sleep 30 && echo -e "Done waiting."

This has worked for me, even though I see the “special device” error at startup.

## Miscellaneous

• blkid is another way to figure out UUIDs.

[user@example ~]# blkid
/dev/mapper/VolGroup00-LogVol01: TYPE="swap"
/dev/mapper/VolGroup00-LogVol00: UUID="5aafa32b-d5c0-40a7-9a6a-417b8a9c7832" TYPE="ext3"
/dev/sda1: LABEL="/boot" UUID="43b64bd4-ee2b-4e4d-b2af-e868cb47d596" TYPE="ext3"
/dev/hda: LABEL="CentOS_5.3_Final" TYPE="iso9660"
/dev/VolGroup00/LogVol00: UUID="5aafa32b-d5c0-40a7-9a6a-417b8a9c7832" TYPE="ext3"
/dev/VolGroup00/LogVol01: TYPE="swap"
/dev/sdc5: UUID="edabbb0e-d87a-4c5b-a9aa-492e553c98d1" TYPE="ext3" SEC_TYPE="ext2"
/dev/sdb5: LABEL="/rsync_drive_3" UUID="7c488bee-a567-4432-bdbe-c30e9d33a927" TYPE="ext3"

• To see the changes you’ve made to fstab without rebooting your system, use mount -a
• Samba shares with spaces in them can get a little tricky. I’ve tried single, double quotes with no success. Octal representation of a space (\040) seemed to work :) Here’s an example:
//19.67.18.160/Lab\040Documents /media/secretary cifs rw,credentials=/media/secretary.credentials 0 0

I’ve stored the credentials for the connection in the file /media/secretary.credentials. It has 600 permissions and looks like this:
username=nikhil


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## College Algebra (11th Edition)

$\dfrac{1}{2}\log_m5+\dfrac{3}{2}\log_mr-\dfrac{5}{2}\log_mz$
$\bf{\text{Solution Outline:}}$ Use the properties of radicals and logarithms to rewrite the given expression, $\log_m\sqrt{\dfrac{5r^3}{z^5}} .$ $\bf{\text{Solution Details:}}$ Using the Quotient Rule of radicals which is given by $\sqrt[n]{\dfrac{x}{y}}=\dfrac{\sqrt[n]{x}}{\sqrt[n]{y}}{},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \log_m\dfrac{\sqrt{5r^3}}{\sqrt{z^5}} .\end{array} Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the expression above is equivalent \begin{array}{l}\require{cancel} \log_m\sqrt{5r^3}-\log_m\sqrt{z^5} .\end{array} Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \log_m\sqrt[2]{5r^3}-\log_m\sqrt[2]{z^5} \\\\= \log_m(5r^3)^{1/2}-\log_mz^{5/2} .\end{array} Using the extended Power Rule of the laws of exponents which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \log_m\left(5^{1/2}r^{3\cdot{\frac{1}{2}}}\right)-\log_mz^{5/2} \\\\= \log_m\left(5^{\frac{1}{2}}r^{{\frac{3}{2}}}\right)-\log_mz^{5/2} .\end{array} Using the Product Rule of Logarithms, which is given by $\log_b (xy)=\log_bx+\log_by,$ the expression above is equivalent \begin{array}{l}\require{cancel} \log_m5^{\frac{1}{2}}+\log_mr^{{\frac{3}{2}}}-\log_mz^{5/2} .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the expression above is equivalent \begin{array}{l}\require{cancel} \dfrac{1}{2}\log_m5+\dfrac{3}{2}\log_mr-\dfrac{5}{2}\log_mz .\end{array}


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A wire of diameter 1mm  breaks under a tension of 1000 N. Another wire, of same material as that of the first one, but of diameter 2 mm breaks under a tension of

(a)   500 N                                (b)   1000 N

(c)   10000 N                            (d)   4000 N

High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

There is no change in the volume of a wire due to change in its length on stretching. The Poisson's ratio of the material of the wire is

(a)   + 0.50                                (b)   – 0.50

(c)   0.25                                   (d)   – 0.25

High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

The force required to stretch a steel wire of cross-section  to 1.1 times its length would be

(a) $2×{10}^{6}N$                               (b) $2×{10}^{3}N$

(c) $2×{10}^{-6}N$                             (d) $2×{10}^{-7}N$

High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

When a weight of 10 kg is suspended from a copper wire of length 3 metres and diameter 0.4 mm, its length increases by 2.4 cm. If the diameter of the wire is doubled, then the extension in its length will be

(a) 9.6 cm                               (b) 4.8 cm

(c) 1.2 cm                               (d) 0.6 cm

Concept Questions :-

Hooke's law
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

A fixed volume of iron is drawn into a wire of length L. The extension x produced in this wire by a constant force F is proportional to

(a) $\frac{1}{{L}^{2}}$                                              (b) $\frac{1}{L}$

(c) ${L}^{2}$                                                (d) L

Concept Questions :-

Hooke's law
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

On applying a stress of $20×{10}^{8}$ N/${m}^{2}$ the length of a perfectly elastic wire is doubled. Its Young’s modulus will be

(a) $40×{10}^{8}N/{m}^{2}$                         (b) $20×{10}^{8}N/{m}^{2}$

(c) $10×{10}^{8}N/{m}^{2}$                         (d) $5×{10}^{8}N/{m}^{2}$

Concept Questions :-

Hooke's law
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

The length of an elastic string is a metre when the longitudinal tension is 4 N and b metre when the longitudinal tension is 5 N. The length of the string in metre when the longitudinal tension is 9 N is

(a) a - b                                             (b) 5b - 4a

(c) 2b -$\frac{1}{4}a$                                        (d) 4a - 3b

Concept Questions :-

Hooke's law
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

How much force is required to produce an increase of 0.2% in the length of a brass wire of diameter 0.6 mm ?

(Young’s modulus for brass = $0.9×{10}^{11}N/{m}^{2}$)

(a) Nearly 17 N                        (b) Nearly 34 N

(c) Nearly 51                        (d) Nearly 68 N

Concept Questions :-

Hooke's law
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

A 5 m long aluminium wire $\left(Y=7×{10}^{10}N/{m}^{2}\right)$ of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire $\left(Y=12×{10}^{10}N/{m}^{2}\right)$ of the same length under the same weight, the diameter should now be, in mm

(a)   1.75                                   (b)   1.5

(c)   2.5                                     (d)   5.0

Concept Questions :-

Hooke's law
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

A steel wire of 1 m long and  cross section area  is hang from rigid end. When mass of 1kg is hung from it then change in length will be (given $Y=2×{10}^{11}N/{m}^{2}$)

(a)   0.5 mm                              (b)   0.25 mm

(c)   0.05 mm                            (d)   5 mm

Concept Questions :-

Hooke's law


text:
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Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp01h989r543m
 Title: Zonal Flows and Turbulence in Fluids and Plasmas Authors: Parker, Jeffrey Advisors: Krommes, John A Contributors: Plasma Physics Department Keywords: pattern formationturbulencezonal flow Subjects: Plasma physicsAtmospheric sciences Issue Date: 2014 Publisher: Princeton, NJ : Princeton University Abstract: In geophysical and plasma contexts, zonal flows are well known to arise out of turbulence. We elucidate the transition from statistically homogeneous turbulence without zonal flows to statistically inhomogeneous turbulence with steady zonal flows. Starting from the Hasegawa--Mima equation, we employ both the quasilinear approximation and a statistical average, which retains a great deal of the qualitative behavior of the full system. Within the resulting framework known as CE2, we extend recent understanding of the symmetry-breaking `zonostrophic instability'. Zonostrophic instability can be understood in a very general way as the instability of some turbulent background spectrum to a zonally symmetric coherent mode. As a special case, the background spectrum can consist of only a single mode. We find that in this case the dispersion relation of zonostrophic instability from the CE2 formalism reduces exactly to that of the 4-mode truncation of generalized modulational instability. We then show that zonal flows constitute pattern formation amid a turbulent bath. Zonostrophic instability is an example of a Type Is instability of pattern-forming systems. The broken symmetry is statistical homogeneity. Near the bifurcation point, the slow dynamics of CE2 are governed by a well-known amplitude equation, the real Ginzburg-Landau equation. The important features of this amplitude equation, and therefore of the CE2 system, are multiple. First, the zonal flow wavelength is not unique. In an idealized, infinite system, there is a continuous band of zonal flow wavelengths that allow a nonlinear equilibrium. Second, of these wavelengths, only those within a smaller subband are stable. Unstable wavelengths must evolve to reach a stable wavelength; this process manifests as merging jets. These behaviors are shown numerically to hold in the CE2 system, and we calculate a stability diagram. The stability diagram is in agreement with direct numerical simulations of the quasilinear system. The use of statistically-averaged equations and the pattern formation methodology provide a path forward for further systematic investigations of zonal flows and their interactions with turbulence. URI: http://arks.princeton.edu/ark:/88435/dsp01h989r543m Alternate format: The Mudd Manuscript Library retains one bound copy of each dissertation. Search for these copies in the library's main catalog Type of Material: Academic dissertations (Ph.D.) Language: en Appears in Collections: Plasma Physics

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text:
# How to add a forced line break inside a table cell

I have some text in a table and I want to add a forced line break. I want to insert a forced line break without having to specify the column width, i.e. something like the following:

\begin{tabular}{|c|c|c|}
\hline
Foo bar & Foo <forced line break here> bar & Foo bar \\
\hline
\end{tabular}

I know that \\ inserts a line break in most cases, but here it starts a new table row instead.

A similar question was asked before: How to break a line in a table

-

It's a quite old question, but I'll add my answer anyway, as the method I suggest didn't appear in the others

\begin{tabular}{|c|c|c|}
\hline
Foo bar & \begin{tabular}[x]{@{}c@{}}Foo\\bar\end{tabular} & Foo bar \\
\hline
\end{tabular}

where x is either t, c, or b to force the desired vertical alignment.

In case this is needed in more than a couple of places, it's better to define a command

\newcommand{\specialcell}[2][c]{%
\begin{tabular}[#1]{@{}c@{}}#2\end{tabular}}

so the table line before can be one of

Foo bar & \specialcell{Foo\\bar} & Foo bar \\    % vertically centered
Foo bar & \specialcell[t]{Foo\\bar} & Foo bar \\ % aligned with top rule
Foo bar & \specialcell[b]{Foo\\bar} & Foo bar \\ % aligned with bottom rule

More variations are possible, for instance specifying also the horizontal alignment in the special cell.

Notice the @{} to suppress added space before and after the cell text.

-
For those wanting to control the horizontal alignment, change c@ to l@ or r@ (or make it another parameter like the vertical alignment?). Thanks egreg –  Dolan Antenucci Apr 19 '12 at 13:31
@dolan I'd go with another parameter: #1 vertical alignment (optional), #2 horizontal alignment, #3 text. –  egreg Apr 19 '12 at 14:46
\parbox will do the same, but you have set a width: \parbox[t]{5cm}{foo\\bar} –  Born2Smile Dec 8 '12 at 14:11
@Born2Smile There's the varwidth environment (package of the same name), but it's overkill. –  egreg Dec 8 '12 at 14:15
This wins the prize for the most incomprehensible answer. –  Ricky Robinson Jan 13 at 13:00

It really is no wonder why LaTeX is said to be complicated! Just look at your answers to such an easy question! How about an easy solution to an every day problem?

\usepackage{pbox}

\begin{tabular}{|l|l|} \hline
\pbox{20cm}{This is the first \\ cell} & second \\ \hline
3rd & and the last cell \\ \hline
\end{tabular}

which looks like:

Note that the width supplied to \pbox is a maximum width. If the content is shorter the length of the longest line is taken.

-
Once again the simplest solution is the best. Thanks! –  Connor Doyle Mar 15 '12 at 20:39
+1 Nice and easy...love it! –  WAF Aug 26 '13 at 10:40
This may be an obvious question. What is the 20cm for, and is there a way to remove it? (automatically choose the correct value) –  Annan Dec 2 '13 at 4:38
The \pbox solution is great. Alignment might be a bit tricky. If you run into trouble using this solution, see here tex.stackexchange.com/a/55861/13450 –  Christian Apr 8 at 17:23

You can switch your cell layout to paragraph to use the \newline command.

\begin{tabular}{|p{2cm}|p{2cm}|}
\hline
Test & foo \newline bar \\
...

-
I did. I tried both \newline and \linebreak, but they only work in paragraph mode. cims.nyu.edu/cgi-comment/info2html?(latex)%5Cnewline –  Denilson Sá Aug 28 '10 at 0:56
Hmm. I thought it did. I could have sworn I've done this before. The only thing I can think of right now is to put a single column tabular within the other tabular. E.g., Foo bar & \begin{tabular}{c} Foo \\ bar \end{tabular} & foo bar \\ –  frabjous Aug 28 '10 at 2:59
@frabjous: Could you clarify the status of this answer? Does it work, and if so, how? (A complete MWE would be good.) –  doncherry Jun 9 '12 at 15:06
@doncherry: This answer works, iff your column type is p{width} or X (in a tabularx environment). It does not work with c, l or r columns, as requested by the asker, though. –  Fritz Aug 28 '12 at 13:11
+1 This solution (unlike some others I've tried) also works well with \rowcolors –  Dana Mar 12 at 17:26

Use the tabularx environment instead of tabular, and then use \newline where you want line breaks within a cell.

\usepackage{tabularx}
\begin{tabularx}{\textwidth}{lX}
Section:   &  This is my     \newline
long paragraph \\
\end{tabularx}

The tabularx environment has a special column type, X, in addition to the usual ones, and its first argument is the desired width of the table. The X column will have the necessary width in order to make the whole table the desired width.

Details on tabularx can be found here.

-
Nice one, I hadn't realised tabularx supported this, thanks for mentioning it. –  sharky Feb 23 '12 at 4:02
By far the easiest solution! –  mtsz Oct 9 '12 at 21:33
\begin{tabular}{lll}
a&\vbox{\hbox{\strut ASDF}\hbox{\strut ASDF}\hbox{\strut This is my
really long line}}&c
\end{tabular}

The \strut is essential for spacing.

-
This looks like good method. Maybe too much to ask, but is there any easy way to adjust vertical justification in the cells? E.g., so single line cells are even with top of multi-line cell? Or so all are vertically centered? –  Herbert Sitz Aug 28 '10 at 15:33
Replace \vbox with \vtop to get alignment at the top. –  TH. Aug 29 '10 at 0:49
Great answer. How do you center horizontally? –  denilw Oct 21 '10 at 20:47
@denilw: Oops. I misread and wrote a previous comment (now deleted). If you want to center the lines, the easiest way is to compute the width of the widest \hbox and use that as the width of the other \hboxes. Something like \vbox{\setbox0\hbox{\strut This is the widest one.}\hbox to\wd0{\hss\strut ASDF\hss}\copy0\hbox to\wd0{\hss\strut asdf\hss}} –  TH. Oct 21 '10 at 21:05
@TH, works well thanks –  denilw Oct 21 '10 at 21:20

Here's a very simple way to do it, using Plain TeX commands within the tabular environment:

\begin{tabular}{|c|c|c|}
\hline
here&\vtop{\hbox{\strut top line}\hbox{\strut botline}}&more\\
\hline
x&y&z\\
\hline
\end{tabular}

By using hboxes within the vtop we've stayed in vertical mode and therefore the width of the text in the hboxes determines the width of the vtop. This way we don't need to know the width of the text in advance. \strut will maintain the right space above and below the text in the hbox.

-
Hi Amy, and welcome to TeX.sx. You might want to read Welcome to TeX.SX! and note that signing your posts is not necessary since your name appears at the bottom automatically. See also Rules about linking to external pages. With respect to this particular answer, how is it different from TH's answer? –  Alan Munn Sep 10 '13 at 20:08
Thank you! This works when you are in a multicell but they are left aligned (even through I defined it to be in the centre). Is there a way around this? –  user2822693 May 14 at 16:41

You can also just fake it:

\begin{tabular}{|c|c|c|}
\hline
Foo bar & Foo & Foo bar \\
~       & bar &  ~      \\
\hline
\end{tabular}

-

How about using \parbox in a custom command

\documentclass{article}
\newsavebox\mybox
\newlength\mylength
\newcommand\boxup[2]{%
\savebox\mybox{#1}%
\setlength\mylength{\wd\mybox}%
\parbox{\mylength}{#1 \\ #2}%
}
\begin{document}
\begin{tabular}{|c|c|c|}
Foo bar & \boxup{Foo}{bar} & Foo bar \\
\end{tabular}
\end{document}

This takes two arguments with the assumption that the first line is longer than the second. It would be possible to refine the code to work through an arbitrary number of lines and find the longest. If that's of interest I'll write something, probably using expl3 for the looping.

-
Gives me issues with too little vertical spacing. \vspace only helps for the bottom margin. –  Aleksandr Levchuk Oct 23 '11 at 22:09

This is a really old question, but since this was linked from a recent question on separating content and presentation, I'll add a ConTeXt answer for comparision.

In ConTeXt, \crlf adds a forced line-break, so achieving a forced line-break in a table is as simple as just adding \crlf in the appropriate place. Here is the complete example:

% Setup for rules at the top and bottom
\startsetups table:rules
\setupTABLE[frame=off]
\setupTABLE[row][first][topframe=on]
\setupTABLE[row][last][bottomframe=on]
\stopsetups

% Setups for middle alignment
\startsetups table:middle
\setupTABLE[align=middle]
\stopsetups

\starttext

\startTABLE[setups={table:rules, table:middle}]
\NC Foo bar \NC Foo \crlf Bar \NC Foo bar \NC \NR
\NC Foo bar \NC Foo \crlf Bar \NC Foo bar \NC \NR
\stopTABLE

\stoptext

which gives:

-
your source code does not match what the illustration shows. –  Kurt Pfeifle Nov 17 '13 at 13:28
@KurtPfeifle: It does for me! Which version of context are you using? –  Aditya Nov 17 '13 at 20:34
Sorry, didn't see your question earlier. My context is v. 0.60 (2013.04.20 01:15). However, now re-testing your code, I can confirm that your code works. I have to withdraw my previous statement. Can't remember now what caused me to make it. –  Kurt Pfeifle Feb 11 at 19:30

I believe I have the simplest answer here:

If you are using a paragraph column in a table, you can put text in an alignment environment and the table does not pick up the \\ as a new table row, so you can use it normally. For example:

\begin{tabular}{p{2in}p{2in}}
\begin{flushleft}
Some text over here \\
newline! \\
\bigskip all the way down
\end{flushleft}
&
\begin{flushleft}
Lorem ipsum dolor sit amet, consectetur
tempor incididunt ut labore et dolore
magna aliqua. Ut enim ad minim veniam,
quis nostrud exercitation ullamco laboris
nisi ut aliquip ex ea commodo\\ \bigskip
consequat. Duis aute irure dolor in
reprehenderit in voluptate velit esse
cillum dolore eu fugiat nulla pariatur.
Excepteur sint occaecat cupidatat non
proident, sunt in culpa qui officia
deserunt mollit anim id est laborum.\\
\bigskip \bigskip lorem ipsum\\
\end{flushleft}\\
\end{tabular}\\

Will give you a table with line breaks using the \\ command.

-
I think the explanation is wrong - I just tried {flushleft} in a l column, and it fails; I think the reason why this example works is because it has p{2in} column type, which sets the width in advance.. –  sdaau Jul 10 at 1:08

I think multirow is a simple elegant solution, at least for simple tables.

For example, I tried a few of the suggestions above but I found this worked the best:

\usepackage{multirow}
\begin{table}[ht]
\caption{RESTful Resources}
\centering
\begin{tabular}{l l l}
\hline
Resource & Methods & Description \\
\hline
\multirow{2}{*}{Resource1} & \multirow{2}{*}{POST, PUT}
& This resource contains ...\\
& & \emph{Media types: text/plain, application/json}\\
\multirow{2}{*}{Resource2} & \multirow{2}{*}{POST, PUT}
& This resource contains ...\\
& & \emph{Media types: text/plain, application/json} \\
\hline
\end{tabular}
\label{table:resources}
\end{table}

-

Strangely, no answer (unless I've misread them) mentions a package that is dedicated to this precise question: makecell, which allows for common formatting of certain cells, thanks to its \thead and \makecell commands, and for line breaks in those cells. The horizontal and vertical alignments can chosen independently from those of the table they're included in. So here is a demo:

\documentclass[12pt]{article}
\usepackage[utf8]{inputenc}
\usepackage{fourier}
\usepackage{array}
\usepackage{makecell}

\renewcommand\cellgape{\Gape[4pt]}

\begin{document}

\begin{center}
\begin{tabular}{ | c | c | c |}
\hline
\hline
Some text &  \makecell{Some really \\ longer text}  & Text text text  \\
\hline
\end{tabular}
\end{center}

\end{document}

-

I don't have direct help on how to add a linebreak, but using the multirow package may provide different route to same desired end. You can read a tutorial here:

http://andrewjpage.com/index.php?/archives/43-Multirow-and-multicolumn-spanning-with-latex-tables.html

But I have a feeling the \parbox answer in the similar question you linked is what you're looking for. I think using a box in the cell is going to be simplest and best way, was there something about that that wasn't working for you?

If you want to use parbox without having to hardcode in a width, here's one way that works for me. Not perfect, since it requires you to know beforehand the text of the longest line in the multi-line column, but it does work:

% define new length and set to length of longest line
\newlength{\longline}
\settowidth{\longline}{longest line in table}\\

\begin{tabular}{lll}

one line & \parbox[t]{\longline}{longest line in table \\ and another line} & more text \\
second line here& and what & more text
\end{tabular}

-
\parbox works, but it requires me to type the desired column width. Thus, I lose the auto-column-size that happens otherwise. I bet \multirow will work, but the source-code will be a lot messy. –  Denilson Sá Aug 28 '10 at 1:39
Gotcha. Added code sample to my question that uses column autosizing with parbox and is probably cleaner than using multirow package. I think there must be a better way, but at least it works. . . –  Herbert Sitz Aug 28 '10 at 4:38

You can use minipage:

\begin{minipage}[t]{0.2\columnwidth}%
This is the first line
\newline
second line
\newline
...
\end{minipage}

-
You have to specify width in advance, though... –  sdaau Jul 10 at 1:06

This approach isn't so different from some of the others already given, except that it "externalises" all of the design decisions from the table definition itself (that is, leaving it cleaner to write and read each table should you want to include multiple tables like this throughout your doc or docs).

\usepackage{array} % needed if you're going to use \\ together with centering,
% raggedright, raggedleft in your column specifier (see
% manuals on \arraybackslash)
\newlength{\mywidestcolwidth}
\newcommand*\mycellformat{\protect\footnotesize}
\newcommand*\mytableparskip{\setlength\parskip{4pt}}
\newcommand*\setmywidestcolwidth[1]{
\settowidth{\mywidestcolwidth}{\mycellformat #1}
}

Usage:

\setmywidestcolwidth{foo baz bar}
\begin{tabular}{|>{\mycellformat}c
|>{\mytableparskip\mycellformat\centering\arraybackslash}p{\mywidestcolwidth}
|>{\mycellformat}c|}
\hline
foo bar baz&
foo

foo baz bar

baz&
foo

foo baz bar

baz\\ \hline
\end{tabular}

Skip the \mycellformat and \parskip if you have no need for them (although not necessary in your case, I included \parskip since it's useful to visually distinguish paragraphs from one another since paragraph indentation is automatically switched off in tables, and added \mycellformat to allow for uniform document-wide table design settings).

-

changed suggestion above to accomplish different problem with second line being longer...

\mybox\usepackage{ifthen}

\newsavebox{\mybox}%
\newsavebox{\myboxx}%
\newlength{\mylengthA}%
\newlength{\mylengthB}%
\newlength{\mylength}%
\newcommand{\boxup}[2]{%
\savebox{\mybox}{#1}%
\setlength\mylengthA{\wd\mybox}%
\savebox{\myboxx}{#2}%
\setlength\mylengthB{\wd\myboxx}%
\ifthenelse{\lengthtest{\mylengthA>\mylengthB}}
{\setlength{\mylength}{\mylengthA}}
{\setlength{\mylength}{\mylengthB}}
\parbox{\mylength}{\centering #1 \\ #2}%
}

-

Good afternoon, when I was younger I was typesetting a tabular environment inside a tabular environment. Since we have TikZ I use tikzpicture environment inside tabulars, or even tikzpicture inside tikzpicture. I prefer putting simple TikZ nodes next and below each other, but it depends on actual task. I enclose an MWE with fast text height correction.

\documentclass{article}
\usepackage{tikz}\pagestyle{empty}
\begin{document}
\tikzset{inner sep=0pt}
\begin{tabular}{|l|l|} \hline
\begin{tikzpicture}
\node[align=left, text height=4.5ex]{This is the\\[3pt]first cell};
\end{tikzpicture}
& second \\ \hline
3rd & and the last cell \\ \hline
\end{tabular}
\end{document}

-

## protected by toheczMar 17 at 11:24

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.


text:
# Channel Estimation in OFDM Systems

Channel estimation in single-carrier systems has been described in a previous article. In OFDM systems, each subcarrier acts as an independent channel as long as there is no Inter-Carrier Interference (ICI) left in the synchronized signal. The options of both a training sequence and individual pilots are available for channel estimation and the choice between the two depends on time variation rate of the channel as well as the computational complexity. Many systems acquire the channel through the preamble while employ the pilots for channel tracking. The discussion in this article is mostly based on Ref. [1].

For a simplified OFDM system model, the signal after the FFT at the receiver can be written as a product between the data symbols $a[k]$ and channel frequency response $H[k]$ at each subcarrier $k$.
\begin{equation*}
\begin{aligned}
Z[k] = a[k]\cdot H[k]~~& \\
&\hspace{-.5in}\text{for each }k = -N/2,\cdots,N/2-1
\end{aligned}
\end{equation*}

Based on this model, there are different estimation schemes as follows.

## Training Based Estimation

In some systems, training OFDM symbols are periodically sent in the Rx signal so that the channel estimate can be updated at the same rate. In such a scenario, all subcarriers of an OFDM symbol are occupied by the training as shown in the figure below where OFDM symbols for $m=0$ and $m=4$ are the training symbols. After estimating the channel for the initial training sequence, it is common to assume a time-invariant channel (and hence valid estimates for subsequent OFDM data symbols) until the next training sequence arrives. Then, the channel estimate is updated and used again for subsequent OFDM symbols and this process is known as piecewise constant interpolation. For example, training based channel estimation is adopted in IEEE 802.11a/b/g and fixed WiMAX systems.

Since the symbols $a[k]$ are known in this context, we can write
\label{eqOFDMLS}
\begin{aligned}
\hat H[k] = \frac{Z[k]}{a[k]}~~& \\
&\hspace{-.5in}\text{for each }k = -N/2,\cdots,N/2-1
\end{aligned}

This is commonly known as Least Squares (LS) channel estimation which in case of OFDM subcarriers is also Maximum Likelihood (ML) solution. Depending on the channel characteristics and the system requirements, this can suffice alone or can be computed as a first estimate before refining it through a more advanced technique.

From the above figure, it is evident that frequency selectivity of the channel is not a problem due to its division into underlying flat fading subcarriers. Even if the channel is drastically different from one subcarrier to the next, it can satisfactorily be estimated by probing on all frequencies.

On the other hand, it is the time domain where the gaps are left and hence susceptible to time selectivity of the channel. Ideally, the training sequences need to be as far apart as possible to minimize the overhead and maximize the data rate. To find out the maximum required spacing between them, we refer to the sampling theorem. The period between two training sequences needs to be at least as frequent as the coherence time of the channel which is the inverse of the Doppler spread (normalized by subcarrier spacing).

Since the OFDM symbols between the training are used to deliver the data symbols, a decision-directed approach can also be employed to smooth the channel tracking between the training sequences.

## Pilot Based Estimation

Another option is to insert regularly spaced pilots in some subcarriers that are kept occupied for each OFDM symbol. In this way, the pilots are multiplexed with the data and this arrangement is shown in the figure below. For example, the subcarriers $k=-2$ and $k=1$ are taken by the pilots here. Then, a similar equation as in training case can also be used for pilot-aided channel estimation for values of $k$ corresponding to the pilot symbols.

In this case, tracking rapid time variations of the channel is not a problem since the pilots are present for each OFDM symbol. On the other hand, there is a gap between them in the frequency domain and hence they are susceptible to frequency selectivity of the channel. For this purpose, the maximum gap between two pilots is governed by the coherence bandwidth.

As before, the data symbols between the pilots can be exploited in a decision-directed manner to improve on the channel estimates.

### Optimal pilot locations

This is determined by the channel fading characteristics in time and frequency domains. Our objective is to insert a minimum number of pilots in the OFDM time-frequency grid. A larger number is a waste of power while a smaller number is a performance loss due to insufficient channel sampling. The tradeoffs involved are accuracy (the more, the better) and spectral efficiency (the lesser, the better). Once a certain number of pilots is determined, it has been reported that the minimum mean square error is obtained when the pilots are equispaced with maximum distance from each other.

### Power and modulation

Many systems transmit more power at the pilot subcarriers than the data subcarriers. Also, a lower-order modulation scheme, such as BPSK, is well suited for pilot assignment since a higher data rate is not an objective and maximum power can be allocated to BPSK symbols instead of dividing it into $M$ PSK symbols for normalization, thus implying a longer range.

### Interpolation

Interpolation between the pilot subcarriers can be carried out in both time and frequency domains. A piecewise constant interpolation technique that implies a constant channel between two training sequences can be employed. In the case of pilot subcarriers, linear interpolation offers better performance without adding much complexity to the block.

Assume that two pilot measurements are available at subcarriers $k=k_1$ and $k=k_2$ where
\begin{equation*}
k_2 – k_1 = Q_P
\end{equation*}

Then, we can find the channel estimate for any other $k$ by linear interpolation.
\begin{equation*}
\hat H(k) = \mu_k \hat H[k_1] + (1-\mu_k)\hat H[k_2], \qquad k_1 < k < k_2 \end{equation*} where $\mu_k$ is given by \begin{equation*} \mu_k = \frac{k}{Q_P} \end{equation*} It can be viewed as a $2$ tap filter applied to the channel estimate sequence after upsampling the channel estimates at pilot locations by $Q_P$. This is shown in the figure below. Better interpolation techniques such as Wiener, Gaussian, cubic or spline are also frequently applied for this purpose. Finally, taking an iFFT/FFT pair also accomplishes pilot interpolation if the FFT size is divisible by the pilot spacing.

Another option is to arrange the pilots on adjacent subcarriers in adjacent OFDM symbols such that they cover the time-frequency grid in a diagonal fashion. This helps in sampling the channel at all times and all subcarriers in a periodic manner.

## Transform Domain Techniques

There are scenarios in which the number of taps in the channel impulse response is much less than the Cyclic Prefix (CP) length. This extra information helps in refining an initial channel estimate (such as the Least Squares (LS) estimate) by concentrating on the channel taps within the delay spread and nullifying the effect of the noise outside the farthest channel tap.

Consider an OFDM system with the CP length $N_{CP}$ and the number of subcarriers $N$ operating in a channel with length $N_{Tap}+1$. Let $\hat H[k]$ for each subcarrier $k$ denote the Least Squares estimate for the corresponding channel gain. Next, we take the iFFT of $\hat H[k]$ and denote it by $\hat h[n]$ that represents an estimate of the channel impulse response $h[n]$. This sequence in time domain is given by
\begin{equation*}
\hat h[n] = h[n] + \text{noise}, \qquad n = 0,1,\cdots,N-1
\end{equation*}

Since the actual channel length is $N_{Tap}+1$, all the taps after $N_{Tap}$ consist of noise only. Here, we form a new channel estimate $\hat h_{FFT}[n]$ by removing all the taps from $\hat h[n]$ beyond $N_{Tap}$.
\begin{equation*}
\hat h_{FFT}[n] =
\begin{cases}
h[n] + \text{noise} & n = 0,1, \cdots, N_{Tap} \\
0 & n = N_{Tap}+1,\cdots,N-1
\end{cases}
\end{equation*}

Now when we perform an FFT of the above sequence to go back to the frequency domain, the resultant channel estimates are free of noise beyond $n=N_{Tap}$ and consequently yield better performance.
\begin{equation*}
\hat H_{FFT}[k] = \text{FFT} \Big\{\hat h_{FFT}[n]\Big\}
\end{equation*}

A block diagram for the implementation of such an approach is drawn in the figure below.

Although a similar concept can be applied to some other transforms, the FFT is particularly attractive due to its efficient implementation.

## Decision-Directed Channel Estimation

Due to its widespread use in single-carrier systems, decision-directed channel estimation was one of the earliest methods explored for OFDM systems. It is particularly useful because after the initial estimates, the channel coefficients can be updated with the help of decisions from data symbols. The basic idea is to employ the channel estimates from the previous OFDM symbol to detect the data in the current OFDM symbol.

• Assume that the channel is slowly varying from one OFDM symbol to the next. Let $Z_m[k]$ be the $m^{th}$ Rx signal after the FFT and $\hat H_{m-1}[k]$ be the channel estimate for $(m-1)^{th}$ OFDM symbol. Then, it can be employed to detect the current data symbols as
• \begin{equation*}
\hat a_m[k] = \frac{Z_m[k]}{\hat H_{m-1}[k]}
\end{equation*}

• Next, the symbol candidate $\hat a[_m[k]$ can be used for estimating the channel $\hat H_m[k]$ at time $m$ as
\begin{equation*}
\hat H_m[k] = \frac{Z_m[k]}{\hat a_m[k]}
\end{equation*}

As another option, the data symbol decision $\text{sign}$ $\{\hat a_m[k]\}$ can also be used for this purpose to reduce the effect of noise.

• \begin{equation*}
\hat H_m[k] = \frac{Z_m[k]}{\text{sign}\{\hat a_m[k]\}}
\end{equation*}

There are two main problems with this kind of decision-directed approach.

• Channel estimate expiry: For a quickly varying channel on a time scale of an OFDM symbol, the channel estimates at time $m-1$ are no longer valid at time $m$. Consequently, the detected symbols $\hat a_m[k]$ increasingly suffer from errors.
• Error propagation: Wrong data symbols fed back to estimate the current channel gives rise to error propagation which renders the system dysfunctional. This is a problem associated with all decision-directed techniques whether in the context of equalization or synchronization.

To resolve such issues, significant improvement over the decision-directed channel estimation as well as training and pilot based methods described above can be gained through involving a powerful error correcting code for the channel estimation process. Such codes operate in a recursive manner and improve the reliability of symbol decisions in each iteration. Inserting the channel estimation process within such a feedback loop improves the quality of the channel estimates which in turn refines the reliability of the decisions.

### References

M. Ozdemir and H. Arslan. Channel estimation for wireless ofdm systems. IEEE Communications Surveys & Tutorials, 9(2), 2007.


text:
# A unity negative feedback system has the open – loop transfer function ${\rm{G}}\left( {\rm{s}} \right) = \frac{{\rm{k}}}{{{\rm{s}}\left( {{\rm{s}} + 0 votes 119 views closed A unity negative feedback system has the open – loop transfer function \({\rm{G}}\left( {\rm{s}} \right) = \frac{{\rm{k}}}{{{\rm{s}}\left( {{\rm{s}} + 1} \right)\left( {{\rm{s}} + 3} \right)}}$. The value of the gain ${\rm{K}}( > 0)$ at which the root locus crosses the imaginary axis is _________.

by (37.3k points)
selected

The characteristic equation is $1 + {\rm{G}}\left( {\rm{s}} \right){\rm{H}}\left( {\rm{s}} \right) = 0$

$\begin{array}{l} \Rightarrow 1 + \frac{{\rm{k}}}{{{\rm{s}}\left( {{\rm{s}} + 1} \right)\left( {{\rm{s}} + 3} \right)}} = 0\\ \Rightarrow {{\rm{s}}^3} + 4{{\rm{s}}^2} + 3{\rm{s}} + {\rm{k}} = 0 \end{array}$

Solving the Routh’s table,

 ${{\rm{s}}^3}$ 1 3 ${{\rm{s}}^2}$ 4 k ${{\rm{s}}^1}$ $\frac{{12 - {\rm{k}}}}{4}$ ${{\rm{s}}^0}$ k

For finding imaginary axis poles, we set $\frac{{12 - {\rm{k}}}}{4} = 0$

Thus, ${\rm{k}} = 12$


text:
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How many positive integers less than 5,000 are evenly divisi  [#permalink]

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12 Jul 2011, 21:20
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How many positive integers less than 5,000 are evenly divisible by neither 15 nor 21?

A. 4,514
B. 4,475
C. 4,521
D. 4,428
E. 4,349
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Re: Integers less than 5,000 are not divisible by 15 or 21 ?  [#permalink]

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12 Jul 2011, 22:00
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integers less than 5000 divisible by 15 5000/15 =333.something , so 333

integers less than 5000 divisible by 21 5000/21 = 238.## , so 238

we have double counted some, so take LCM of 15 and 21 =105 and divide by 5000, we get 47. so all numbers divisible by 15 and 21 = 333 +238 -47 =524

now subtract that from 4999. 4999- 524 = 4475 answer B.
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Re: Integers less than 5,000 are not divisible by 15 or 21 ?  [#permalink]

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04 Apr 2014, 05:10
Aj85 wrote:
integers less than 5000 divisible by 15 5000/15 =333.something , so 333

integers less than 5000 divisible by 21 5000/21 = 238.## , so 238

we have double counted some, so take LCM of 15 and 21 =105 and divide by 5000, we get 47. so all numbers divisible by 15 and 21 = 333 +238 -47 =524

now subtract that from 4999. 4999- 524 = 4475 answer B.

Question here, why don't you multiply 47*2

I thought that 105 was contained in both 15 and 21? Is there any reason why you don't do this?
Thanks
Cheers
J
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Posts: 52161
Re: Integers less than 5,000 are not divisible by 15 or 21 ?  [#permalink]

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04 Apr 2014, 07:12
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jlgdr wrote:
Aj85 wrote:
integers less than 5000 divisible by 15 5000/15 =333.something , so 333

integers less than 5000 divisible by 21 5000/21 = 238.## , so 238

we have double counted some, so take LCM of 15 and 21 =105 and divide by 5000, we get 47. so all numbers divisible by 15 and 21 = 333 +238 -47 =524

now subtract that from 4999. 4999- 524 = 4475 answer B.

Question here, why don't you multiply 47*2

I thought that 105 was contained in both 15 and 21? Is there any reason why you don't do this?
Thanks
Cheers
J

47 numbers are counted TWICE, we need them to be counted ONCE, so that's why are subtracting 47 once.

Similar questions to practice:
what-is-the-number-of-integers-from-1-to-1000-inclusive-126153.html
what-is-the-total-number-of-positive-integers-that-are-less-128104.html
how-many-integers-from-1-to-200-inclusive-are-divisib-109333.html
how-many-even-integers-n-where-100-n-200-are-divisib-103779.html

Hope it helps.
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Re: Integers less than 5,000 are not divisible by 15 or 21 ?  [#permalink]

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04 Apr 2014, 20:57
Bunuel wrote:
jlgdr wrote:
Aj85 wrote:
integers less than 5000 divisible by 15 5000/15 =333.something , so 333

integers less than 5000 divisible by 21 5000/21 = 238.## , so 238

we have double counted some, so take LCM of 15 and 21 =105 and divide by 5000, we get 47. so all numbers divisible by 15 and 21 = 333 +238 -47 =524

now subtract that from 4999. 4999- 524 = 4475 answer B.

Question here, why don't you multiply 47*2

I thought that 105 was contained in both 15 and 21? Is there any reason why you don't do this?
Thanks
Cheers
J

47 numbers are counted TWICE, we need them to be counted ONCE, so that's why are subtracting 47 once.

Similar questions to practice:
what-is-the-number-of-integers-from-1-to-1000-inclusive-126153.html
what-is-the-total-number-of-positive-integers-that-are-less-128104.html
how-many-integers-from-1-to-200-inclusive-are-divisib-109333.html
how-many-even-integers-n-where-100-n-200-are-divisib-103779.html

Hope it helps.

Hi Bunuel,
Is there a shorter way around this question?This approach took me 5 minutes!
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Re: How many positive integers less than 5,000 are evenly divisi  [#permalink]

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04 Apr 2014, 20:59
Option B.(Took 5 minutes to solve with the following approach!)
All nos. from $$1$$ to $$5000$$ since we're given POSITIVE integers.$$=4999$$
Multiples of $$15$$ in this range$$=4995/15=333$$
Multiples of $$21=4998/21=238$$
Multiples of both $$15$$ and $$21$$=multiples of $$105=47$$
Now $$answer=4999-333-238+47=4475$$
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Re: How many positive integers less than 5,000 are evenly divisi  [#permalink]

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16 Apr 2014, 22:53
2
AKG1593 wrote:
Option B.(Took 5 minutes to solve with the following approach!)
All nos. from $$1$$ to $$5000$$ since we're given POSITIVE integers.$$=4999$$
Multiples of $$15$$ in this range$$=4995/15=333$$
Multiples of $$21=4998/21=238$$
Multiples of both $$15$$ and $$21$$=multiples of $$105=47$$
Now $$answer=4999-333-238+47=4475$$

This seems to be one of those killer problems that aren't necessarily conceptually difficult, but unless you have (what for me is) very high-powered arithmetic (long division) skills you're going to be in a time crunch here and in danger of compromising your quant section on test day. Practice, practice, practice those fundamentals!

One trick, because the ACs are so close together, is that you want to be careful to account for the fact that "less than 5000" accounts for only 4,999 integers, not including 5000. (Turns out on this problem it's not of such great concern, but on many other problems, it definitely is!) For this question, huge waste of about 20-30 seconds was the initial rephrase: "How many integers.. divisible by NEITHER 15 or 21." Don't know why but that tripped me up. Do they mean all of the numbers not divisible by 105 (Neither 15 AND 21, I saw the language trick there and ruled this out but not after wasting some time on math). What were are therefore looking for is Opposite: 4999 (minus) all the numbers that 15 or 21 go into (minus) overlap between multiples of 15 and 21. Let the long division begin!
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Re: How many positive integers less than 5,000 are evenly divisi  [#permalink]

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21 Nov 2014, 18:11
hey Bunuel,
What does evenly divisible number means? I thought the quotient for such division should be multiple of 2.

Thanks,
Amit

Bunuel wrote:
jlgdr wrote:
Aj85 wrote:
integers less than 5000 divisible by 15 5000/15 =333.something , so 333

integers less than 5000 divisible by 21 5000/21 = 238.## , so 238

we have double counted some, so take LCM of 15 and 21 =105 and divide by 5000, we get 47. so all numbers divisible by 15 and 21 = 333 +238 -47 =524

now subtract that from 4999. 4999- 524 = 4475 answer B.

Question here, why don't you multiply 47*2

I thought that 105 was contained in both 15 and 21? Is there any reason why you don't do this?
Thanks
Cheers
J

47 numbers are counted TWICE, we need them to be counted ONCE, so that's why are subtracting 47 once.

Similar questions to practice:
what-is-the-number-of-integers-from-1-to-1000-inclusive-126153.html
what-is-the-total-number-of-positive-integers-that-are-less-128104.html
how-many-integers-from-1-to-200-inclusive-are-divisib-109333.html
how-many-even-integers-n-where-100-n-200-are-divisib-103779.html

Hope it helps.
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Posts: 52161
Re: How many positive integers less than 5,000 are evenly divisi  [#permalink]

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22 Nov 2014, 05:36
Amit0507 wrote:
hey Bunuel,
What does evenly divisible number means? I thought the quotient for such division should be multiple of 2.

Thanks,
Amit

Nope.

Evenly divisible means divisible without a remainder, so simply divisible.
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Re: How many positive integers less than 5,000 are evenly divisi  [#permalink]

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21 Feb 2017, 12:49
Bunuel wrote:
jlgdr wrote:
Aj85 wrote:
integers less than 5000 divisible by 15 5000/15 =333.something , so 333

integers less than 5000 divisible by 21 5000/21 = 238.## , so 238

we have double counted some, so take LCM of 15 and 21 =105 and divide by 5000, we get 47. so all numbers divisible by 15 and 21 = 333 +238 -47 =524

now subtract that from 4999. 4999- 524 = 4475 answer B.

Question here, why don't you multiply 47*2

I thought that 105 was contained in both 15 and 21? Is there any reason why you don't do this?
Thanks
Cheers
J

47 numbers are counted TWICE, we need them to be counted ONCE, so that's why are subtracting 47 once.

Similar questions to practice:
http://gmatclub.com/forum/what-is-the-n ... 26153.html
http://gmatclub.com/forum/what-is-the-t ... 28104.html
http://gmatclub.com/forum/how-many-inte ... 09333.html
http://gmatclub.com/forum/how-many-even ... 03779.html

Hope it helps.

In questions like these do we always have to take an LCM ?
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Re: How many positive integers less than 5,000 are evenly divisi  [#permalink]

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02 Apr 2018, 23:38
I’m sorry for my ignorance, but why do we take the LCM of 15 and 21 to remove double count? I multiplied 15*21= 315 and divided 5000/315= 15 times to get the overlap. I know it is wrong now, but can someone explain it to me please?

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Posts: 52161
Re: How many positive integers less than 5,000 are evenly divisi  [#permalink]

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03 Apr 2018, 02:41
1
Juliaz wrote:
I’m sorry for my ignorance, but why do we take the LCM of 15 and 21 to remove double count? I multiplied 15*21= 315 and divided 5000/315= 15 times to get the overlap. I know it is wrong now, but can someone explain it to me please?

Posted from my mobile device

The numbers which are divisible by both 15 and 21 are numbers which are divisible by the least common multiple of 15 and 21, which is 105: 105, 210, 315, 420, ... If you simply multiply you are loosing many numbers, for example, 105, 210, 420, ...
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Re: How many positive integers less than 5,000 are evenly divisi  [#permalink]

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18 Dec 2018, 22:30
We can find the no of integers divisible by 15 using the formula of A.P. with first term as 15 and common difference also 15.
Since, the largest term smaller than 5000 divisible by 15 is 4995.
4995 = 15 + (n-1)15. We get, n = 333.
Similarly, we can find no of terms divisible by 21.
So, no of integers from 0 to 5000 divisible by 21 = 238.
Now, we need to find the no of integers divisible by 15 x 21 because these integers are counted twice as the factors of 15 as well as the factors of 21.
No of factors of LCM of 15 &21 till 5000 = 47.
So, total no of positive integers till 5000 divisible by either 15 or 21 = 333 + 238 – 47 = 524.
Hence, no of integers not divisible by 15 or 21 = 4999 – 524 = 4475.
Re: How many positive integers less than 5,000 are evenly divisi &nbs [#permalink] 18 Dec 2018, 22:30
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# Math Help - example of sequence of functions

1. ## example of sequence of functions

Can anyone think of an example of a sequence of functions $f_n:[0,1]\to\mathbb{R}$ that converges uniformly on $(0,1)$ but not on $[0,1]$?

2. I assume the functions need to be continuous, in which case no such sequence exists. The behavior of a continuous function is determined by its behavior on any dense subset.

3. Originally Posted by retep
Can anyone think of an example of a sequence of functions $f_n:[0,1]\to\mathbb{R}$ that converges uniformly on $(0,1)$ but not on $[0,1]$?

If we don't assume $f_n$ need to be continuous a trivial example is:

$f_n(x)=\begin{Bmatrix} 0 & \mbox{ if }& x\in [0,1)\\(-1)^n & \mbox{if}& x=1\end{matrix}$

Fernando Revilla

4. can someone help me , i now what does mean the sequence of real number , but i confused what does it mean by sequence of vector space? for example vector space of polynomials ? and n*m matrices ?how would be their sentences of the related sequences? and why the convergent of these sequences depends on the norm that we define for these spaces ?