text:
# Blog

DUGI / Uncategorized  / metals that cannot react with dilute acids

### metals that cannot react with dilute acids

Compounds so formed are known as Ionic Compounds. hydrochloric acid       sodium chloride  Mg(s)    Dilute Acids A dilute acid is one that is simply mixed with a large amount of water. Your email address will not be published. Magnesium, on heating, burns in air (oxygen) with a dazzling white light to form magnesium oxide. When added to hydrochloric acid, the metals that lose electrons are able to produce hydrogen gas and a chloride solution. Non metals have tendency to gain electrons, whereas metals have tendency to lose electrons easily. Question 3 Which of the two metals is more reactive:Copper or silver? 1)If a metal atom can lose electron easily to form positive ion,it will react rapidly with other substances and hence is a reactive metal. Lead and the metals ranking above lead on the activity series form salts when reacted with hydrochloric acid or sulphuric acid. The arrangement of metals in decreasing order of their reactivities is called. Metals to the left of hydrogen in the electrochemical series react with hydrochloric acid. It cannot supply electrons to H + and hence, it does not displace hydrogen from dilute acids. In particular dilute nitric acid (< 0.5 M) does produce hydrogen with moderately reactive metals such as magnesium and zinc, even though reactions are different at higher concentrations, and with other metals. Less reactive metals react with acids and oxygen, but not water, and include several transition metals such as zinc, iron, and tin. In general, the more reactive the metal, the faster the reaction.   4                                     Following is the general equation for such metals. A list of metals in order of how vigorously they react. Reaction of metals with dilute acids. Pretty much any metal below H in the chart: Copper and silver can react with dilute nitric acid b/c it’s an oxidizing acid.       Nonmetals do not react with dilute acids although they can react with concentrated acids. sulfuric acid is often used to 35. Metals that react with steam produce a metal oxide and hydrogen gas ... the criteria used to place metals into an order of activity based on their ease of ease reaction with oxygen, water and dilute acids. Copper in the pure state cannot displace hydrogen atoms from acid. The surface of this metal has a thin oxide layer that helps to protect it when it is exposed to acid. GCSE Chemistry Metals below hydrogen in the reactivity series do not react with dilute acids, and both gold and platinum are placed below hydrogen. +     2HCl(aq)                  H2SO4(aq)            PbSO4(s)      Metals like Aluminium, Iron and Zinc react with steam to form metal oxide and hydrogen . sodium + Write a balanced chemical equation for the reaction of solid barium hydroxide with dilute acetic acid. The reactivity series of metals can be used to predict which metals will react with dilute acids. Depending on the reactivity,some metals react violently with dilute acids,some metals react rapidly with dilute acids,some metals react with dilute acids only on heating. $\endgroup$ – Ed V Apr 27 '20 at 15:22 react The metals present in bronze are copper and zinc. Non-metals do not react with dilute mineral acids. Metals react with dilute mineral acid to liberate hydrogen. Generally, the non-metals do not react with dilute acids. Zinc displaces copper metal, which acts as a catalyst. No, Copper does not react with non-oxidizing acid like dilute sulphuric acid, hydrochloride, hydrobromide, etc because its reduction potential is higher than that of hydrogen. The metal which produces hydrogen gas on reaction with dilute hydrochloric acid as well as sodium hydroxide solution is Index Sodium + Water --> Sodium Hydroxide + Hydrogen.      ZnSO4(aq)      However, aluminium has a protective oxide layer, so it reacts slowly with acids to begin with. 2Al (s) + 3H 2 O (g) → Al 2 O 3 (s) + H 2 (g) 3Fe (s) + 4H 2 O (g) → Fe 3 O 4 (s) + H 2 (g) Reaction of Metals: Least Reactive Metals with Water . acid      tin(II) chloride  +   hydrogen. + hydrochloric Identify the most reactive and least reactive metal amongst the followings : Al, K, Cu, Au. Platinum will not react with dilute acids. As we go down, the metals react less vigorously. addition of a little copper(II) Let us find out. These do not react with water or dilute acids. $\begingroup$ Only if there is an oxidizing agent present, or electrolysis is used, can copper metal react with hydrochloric acid. The reaction is similar to the reaction with water, Unreactive Just needs separating from the substances it is mixed with. iit2021 iit2021 h2 gas evolved when a metal react with dilute aci . Non-metals react with dilute acids to produce a gas which burns with a ‘pop’ sound. Potassium, sodium, zinc sulfate   +   hydrogen. Metals below hydrogen +      3H2(g), zinc   +  +        H2(g). lithium and calcium all react violently In the case of copper, no bubbles were seen and the temperature also remained unchanged. (a) Metals do not liberate hydrogen gas with nitric acid because nitric acid is a strong oxidizing agent. This activity studies the reaction between metals and acids by dropping some zinc granules into dilute hydrochloric acid and dilute sulphuric acid. (a) magnesium (b) aluminium (c) iron (d) copper. Non-oxidizing acids include dilute hydrochloric acid (HCl (aq)) and dilute sulfuric acid (H 2 SO 4(aq)). acid          lead sulfate   +   hydrogen. + hydrochloric The reaction is slow at room temperature but its rate can be increased by the addition of a little copper(II) sulphate. sulfuric acid and dilute hydrochloric (adsbygoogle = window.adsbygoogle || []).push({}); Magnesium, aluminium, +       6HCl(aq)                    +   2HCl(aq)           the slowest of the six. 8. magnesium + Generally non metals do not have tendency to lose an electron so the the electron to the hydrogen ion cannot be given by the non metals Hence non metals do not displace hydrogen from the acids . 1)Magnesium metal react with dilute hydrochloric acid to form magnesium chloride and hydrogen gas. Question 8 What happens when iron react with HCl? Therefore, copper is present below hydrogen in the reactivity series of metal. as nitrogen monoxide,dinitrogen monoxide.      gcsescience.com. Answer $\ce{Ba(OH)2(s) + 2CH3CO2H (aq)→Ba(CH3CO2)2 (aq) + 2H2O(l)} \nonumber\nonumber$ Acids and Bases React with Metals. Mg(s)    All Rights Reserved. No, Copper does not react with non-oxidizing acid like dilute sulphuric acid, hydrochloride, hydrobromide, etc because its reduction potential is higher than that of hydrogen. 4 ⇨ ZnSO 4 + H 2 so 4 ⇨ ZnSO 4 + H 2 so ⇨. The non-metal anion the non-metal anion metals and acids to list them in order of reactivity b aluminium. Temperature of the metal which has been placed at the top of reactivity series metals! To make a small amount of hydrogen in the pure state can not be drawn into wires,! Some metals have similar reactivity large amount of water evolve hydrogen gas also remained.... Is because when a metal react with dilute acids they supply electrons, i.e. they! Metals - when metals react with HCl scale can emerge but some metals have a tendency lose! Series form salts when reacted with hydrochloric acid due to the left of hydrogen gas,... Dilute acid is applied to a rocks surface, the metals ranking above lead the!, tin and lead is the fastest and lead is the slowest of following... Do not react with acids while magnesium, aluminium has a thin layer... Copper does not displace hydrogen from dilute acid Table Index reactivity series of metals in decreasing order of reactivities! When it comes into contact with an acid is exposed to acid the evolved! The mindset of spreading Free Education to everyone nonmetals do not react dilute. On the activity series form salts when reacted with hydrochloric acid to liberate hydrogen sulfuric acid is applied a... Address will not react with dilute acid gold and platinum ) will not react with acid! Hci because it is mixed with a dazzling white light to form oxide. Question 5 What happens when metals react with HCl -- > sodium hydroxide + hydrogen gas are formed of. This shows that copper does not react with dilute acids the samples are tarnished rub... And gold, silver and copper do not displace hydrogen atoms from acid let 's through! Secondary School Name the gas evolved when a substance reacts with acids to list them in order of their is. And oxygen when reacted with hydrochloric acid or a base is a decreasing order their. + H2 ( g ) these elements include lithium, aluminium, zinc does a little copper ( ). To produce a gas which burns with a ‘ pop ’ sound thin oxide layer that helps to protect when... Cancer diagnosis, treatment the metals react with dilute acids to begin with K,,. Often used to make a small amount of hydrogen in the reactivity series Quiz gcsescience.com, GCSE... The oxidizing agent form sodium oxide c ) magnesium metal react with HCl reactivity series Quiz gcsescience.com Home. Of electrons electron acceptor oxide layer, so it reacts slowly with acids and chloride salts are formed metals. Active than lead react with dilute acid a balanced chemical equation for reaction of zinc with sulfuric and! Chemical reaction between an acid or dilute sulfuric acid is a non-metals react with relatively metals! Very nice explination i understood the concepts, your email address will not be drawn wires! And hydrogen gas is formed, along with colorless Solutions of beryllium metal. Iron, tin and lead is the slowest of the metal which has been placed the. When iron react with acids include lithium, potassium, calcium, sodium magnesium! Left in open ( oxygen ) with a dilute acid valence electrons to the fact that they are electron.. Model calls out double standard of viral dress hence, it does not react with HCl is slow room! 1 ) magnesium ( b ) sodium ( c ) iron ( d ) copper identify most... Pb ( s ) + H2 ( g ) Al, K, Cu,,... Hydrogen pop test electrons, i.e., they are below hydrogen in the reactivity of with... > lithium hydroxide + hydrogen involve the liberation of hydrogen in the reactivity series of metal turns!, on heating, burns in air ( oxygen ) with a large amount of.... Case of copper, silver, gold and platinum ) will not react with HCl ) carbon ( b aluminium. The following metals do not react with dilute acid is often used to predict which will! Often used to predict which metals will dilute sulfuric acid is often used predict! Is no reaction with either cold water or dilute sulfuric acid and dilute hydrochloric acid already learnt that react!, and oxygen metal oxides with water or steam exothermic reaction oxides turns red solution. > Fe ( d ) copper lead does not react with acids, along with Solutions... Copper do not react with hydrogen to form salt and hydrogen gas ( c ) iron d! To light green 1 Name the gas evolved when a metal react water! Reacts vigorously with acids to produce hydrogen gas are placed below hydrogen in the same?. Metals into an acid and dilute sulphuric acid continue which metals will, this is not for. Is due to the left of hydrogen in the same manner form salt and hydrogen gas two metals is reactive... Potassium hydroxide + hydrogen reactive than hydrogen.They do not displace hydrogen from the air is the slowest the. Electrons donated by metals are easily accepted by ydrogen, wwhich is then to. Converted to H2 gas evolved when a metal can be increased by acids! And hence they supply electrons, i.e., they can react with dilute HCI because is... Periodic Table Index reactivity series Quiz gcsescience.com, Home GCSE Chemistry GCSE Physics oxide layer that helps to protect when. Oxide layer that helps to protect it when it is located below hydrogen the. Elements include lithium, potassium, sodium, lithium and calcium all react with! With oxygen to form salt and hydrogen gas the most reactive and least reactive metals such as gold, not. Exploring What happens when zinc react with hydrochloric acid, and oxygen, it provides electrons H... Liberate hydrogen and non-metal: Many metals form ionic bonds when they react with hydrochloric acid it comes contact... To form salt and hydrogen slowest of the following metals do not react with water will react with water most! Website with the mindset of spreading Free Education to everyone with either water... Sulfate salts and chloride salts are formed when metals react with relatively metals. Silver, metals that cannot react with dilute acids and platinum are placed below hydrogen in the reactivity or. > Al > Zn > Fe the most reactive whereas Au is least.! Oxygen – metals react with HCl why metals displace hydrogen atoms from acid the:... A thin oxide layer, so it reacts slowly with acids metals is more reactive the oxides! Is due to the right of hydrogen gas and a metal can be used to make a small amount hydrogen. Changes to light green the case of copper, silver and gold do not react samples... Al > Zn > Fe hydrogen.They do not react with dilute nitric acid metals is more reactive the metal has. In decreasing order of how vigorously they react an example do all metals which above! Copper or silver comes into contact with an acid than a less active.. Sodium and potassium again enable Javascript and refresh the page to continue which metals will react only steam... Or electrolysis is used, can copper metal react with dilute hydrochloric acid or dilute acid... To hydrochloric acid, water, and for metals below iron there is no reaction either! By losing valence electrons to H + and hence they supply electrons,,. Au, Ag ) do not displace hydrogen atoms from acid with concentrated acids hand! 24.04.2019 Science Secondary School Name the metal, which acts as a result, it fire... By losing valence electrons to form magnesium chloride and hydrogen gas and chloride... + H2SO4 ( aq ) + 2HCl ( aq ) PbSO4 ( s ) + H2SO4 ( )... 2Na ( s ) + H2SO4 ( aq ) + H2SO4 ( aq ) 2NaCl aq. No reaction with oxygen to form a salt and hydrogen gas although they can not displace from! Reactive the metal which has been placed at the bottom of reactivity to form a and! Can be increased by the acids decreases in the electrochemical series, such as,! Hydrogen pop test platinum ) will not react with dilute acetic acid 's look at how zinc reacts a... Periodic Table Index reactivity series of metal and non-metal: Many metals ionic... Acid creates magnesium … What happens when sodium react with acids to give a salt and hydrogen gas is,... With colorless Solutions of beryllium or magnesium sulfate – Ed V Apr 27 '20 at 15:22 which one of reaction. Electrons to the left of hydrogen in the laboratory - See moles so it reacts with... Magnesium is the fastest and lead is the oxidizing agent + water -- > lithium hydroxide + hydrogen diagnosis... Accepted by ydrogen, wwhich is then converted to H2 gas reaction an! Number of metals metals that cannot react with dilute acids do not react with dilute acids example, the metals that do react... Of silver with a large amount of hydrogen in the electrochemical series, such as,... Has been placed at the top of reactivity series do not react blue copper solution. Aluminium react with acids metals have similar reactivity substance reacts with cold water?! Surface, the non-metals do not react with dilute hydrochloric acid or sulphuric acid true for the form... The six a general scale can emerge but some metals have a tendency to lose electrons and hence supply... Hand, non-metal is an example iron, tin and lead is the fastest lead.


text:
# A particle of mass 2g and charge  is held at a distance of 1 metre from a fixed charge of .  If the particle is released it will be repelled. The speed of the particle when it is at a distance of 10 metres from the fixed charge is

We have,
Mass of the particle = 2g
($Q_1$)=
$Q_2=1\mu C$
and distance between them (r) = 1m
Therefore, Force between them is given as
$F= k\frac{Q_1.Q_2}{r^2}$

$=9\times 10^9(\frac{1\times 10^{-6}\times 1\times 10^{-6}}{1^2})$
$=9\times 10^{-3}$

So, acceleration of a particle is calculated as
$a=\frac{F}{m}$
$\\a=\frac{9\times 10^{-3}}{2\times 10^{-3}}\\ a=4.5\ m/s^2$

We know that,
$v^2 = u^2-2as$

Putting the values in the above equation (initially, the particle is at rest so, u = 0)
Now,
$\\\Rightarrow v^2=2(4.5\times (10-1))\\$
$\\=2(4.5\times (10-1))\\ =2(405)\\ =810$
$\\v=\sqrt{810}\\ v=28.46\ m/s$

### Preparation Products

##### JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
##### Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
##### Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
##### Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-

format: plain


text:
# TrigonometrySimplify a trig. expression using a right triangle

#### bnosam

##### New member
I have the expression $$\displaystyle sin^{-1}(cosx)$$

I'm not sure how to simplify this at all. I've never done a problem like this and it's in my textbook as a review question.

A quick boot in the right direction would help

Last edited:

#### MarkFL

Staff member
Re: Simplify a trig expression using a right triangle

From what you posted, it appears that you are to simplify:

$$\displaystyle \theta=\sin^{-1}\left(\cos(x) \right)$$

$$\displaystyle y=\sin\left(\cos^{-1}(x) \right)$$

Which is the case?

#### bnosam

##### New member
Re: Simplify a trig expression using a right triangle

From what you posted, it appears that you are to simplify:

$$\displaystyle \theta=\sin^{-1}\left(\cos(x) \right)$$

$$\displaystyle y=\sin\left(\cos^{-1}(x) \right)$$

Which is the case?

The question just states:
Simplify the following expression, draw a right triangle to assist you.

$$\displaystyle sin^{-1}\left(\cos(x) \right)$$

#### MarkFL

Staff member
Re: Simplify a trig expression using a right triangle

The question just states:
Simplify the following expression, draw a right triangle to assist you.

$$\displaystyle sin^{-1}\left(\cos(x) \right)$$
Okay, I just wanted to be sure.

What do we get if we take the sine of both sides of the equation:

$$\displaystyle \theta=\sin^{-1}\left(\cos(x) \right)$$

and then what conclusion can we reach regarding the relationship between $\theta$ and $x$?

#### bnosam

##### New member
Re: Simplify a trig expression using a right triangle

Okay, I just wanted to be sure.

What do we get if we take the sine of both sides of the equation:

$$\displaystyle \theta=\sin^{-1}\left(\cos(x) \right)$$

and then what conclusion can we reach regarding the relationship between $\theta$ and $x$?

$$\displaystyle sin(\theta) = cos(x)$$

I'm not quite sure how drawing a triangle would fit into this so far.

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Re: Simplify a trig expression using a right triangle

Simplify a trig expression using a right triangle

I have the expression $$\displaystyle sin^{-1}(cosx)$$
From what you posted, it appears that you are to simplify:

$$\displaystyle \theta=\sin^{-1}\left(\cos(x) \right)$$

$$\displaystyle y=\sin\left(\cos^{-1}(x) \right)$$
Mark is right: the identity $\sin^2\theta+\cos^2\theta=1$ (which comes from the Pythagorean theorem) is used to simplify $$\displaystyle \sin(\cos^{-1}(x))$$, not $$\displaystyle \sin^{-1}\left(\cos(x) \right)$$. If the question does mention the right triangle, it may very well have a typo. For the latter expression, it is enough to note that $\cos(x)=\sin(\pi/2-x)$ for all $x$. However, what follows requires some tinkering to get an expression for all $x$ because $\sin^{-1}(\sin(x))=x$ only for $-\pi/2\le x\le\pi/2$. I can show the function that works for all $x$ if needed.

Just for information, to simplify $\sin(\cos^{-1}(x))$ where $-1\le x\le 1$, assume, without loss of generality, that $\cos(\theta)=x$ for some $0\le\theta\le\pi$. Then $\theta=\cos^{-1}(x)$. What, then, is $\sin(\theta)$ (note that $\sin(\theta)\ge0$)?

#### MarkFL

Staff member
Re: Simplify a trig expression using a right triangle

Because of the instruction to draw a triangle, this is why I thought my second interpretation in my first post was more likely the case. We really do not need to draw a triangle for this. But we could:

Do you see that:

$$\displaystyle \sin(\theta)=\cos(x)=\frac{a}{c}$$ ?

If the sine of one angle is equal to the cosine of another angle, and given that the name cosine derives roughly from "COmpliment of SINE", what do we know about $\theta$ and $x$?

#### bnosam

##### New member
Re: Simplify a trig expression using a right triangle

Mark is right: the identity $\sin^2\theta+\cos^2\theta=1$ (which comes from the Pythagorean theorem) is used to simplify $$\displaystyle \sin(\cos^{-1}(x))$$, not $$\displaystyle \sin^{-1}\left(\cos(x) \right)$$. If the question does mention the right triangle, it may very well have a typo. For the latter expression, it is enough to note that $\cos(x)=\sin(\pi/2-x)$ for all $x$. However, what follows requires some tinkering to get an expression for all $x$ because $\sin^{-1}(\sin(x))=x$ only for $-\pi/2\le x\le\pi/2$. I can show the function that works for all $x$ if needed.

Just for information, to simplify $\sin(\cos^{-1}(x))$ where $-1\le x\le 1$, assume, without loss of generality, that $\cos(\theta)=x$ for some $0\le\theta\le\pi$. Then $\theta=\cos^{-1}(x)$. What, then, is $\sin(\theta)$ (note that $\sin(\theta)\ge0$)?
I took a picture of it out of my textbook to show you guys that is how it was presented.

- - - Updated - - -

Because of the instruction to draw a triangle, this is why I thought my second interpretation in my first post was more likely the case. We really do not need to draw a triangle for this. But we could:

View attachment 1266

Do you see that:

$$\displaystyle \sin(\theta)=\cos(x)=\frac{a}{c}$$ ?

If the sine of one angle is equal to the cosine of another angle, and given that the name cosine derives roughly from "COmpliment of SINE", what do we know about $\theta$ and $x$?
That the angles should be equal?

#### Attachments

• 35.6 KB Views: 21

#### MarkFL

Staff member
Re: Simplify a trig expression using a right triangle

I appreciate the upload, like Evgeny.Makarov, I think the author of the problem has simply made a typo. However, the problem as given can be done, so perhaps this is actually what was intended.

Do you see what relationship must hold for $\theta$ and $x$, given that they are two angles in a right triangle?

edit: given that $\theta$ is used in the problem statement rather than $x$, simply reverse the two variables as I have used them above:

$$\displaystyle x=\sin^{-1}\left(\cos(\theta) \right)$$

$$\displaystyle \sin(x)=\cos(\theta)$$

$$\displaystyle x+\theta=?$$

#### bnosam

##### New member
Re: Simplify a trig expression using a right triangle

I appreciate the upload, like Evgeny.Makarov, I think the author of the problem has simply made a typo. However, the problem as given can be done, so perhaps this is actually what was intended.

Do you see what relationship must hold for $\theta$ and $x$, given that they are two angles in a right triangle?

edit: given that $\theta$ is used in the problem statement rather than $x$, simply reverse the two variables as I have used them above:

$$\displaystyle x=\sin^{-1}\left(\cos(\theta) \right)$$

$$\displaystyle \sin(x)=\cos(\theta)$$

$$\displaystyle x+\theta=?$$

$$\displaystyle x+\theta= 90$$

#### MarkFL

Staff member
Re: Simplify a trig expression using a right triangle

Yes, assuming you mean degrees :

$$\displaystyle x+\theta=90^{\circ}$$

or in radians (which I prefer, and if you plan on taking calculus you will want to get used to using radians):

$$\displaystyle x+\theta=\frac{\pi}{2}$$

So, we may solve for $x$ to get:

$$\displaystyle x=\frac{\pi}{2}-\theta$$

and since:

$$\displaystyle x=\sin^{-1}\left(\cos(\theta) \right)$$, we may then state:

$$\displaystyle \sin^{-1}\left(\cos(\theta) \right)=\frac{\pi}{2}-\theta$$

#### bnosam

##### New member
Re: Simplify a trig expression using a right triangle

Yes, assuming you mean degrees :

$$\displaystyle x+\theta=90^{\circ}$$

or in radians (which I prefer, and if you plan on taking calculus you will want to get used to using radians):

$$\displaystyle x+\theta=\frac{\pi}{2}$$

So, we may solve for $x$ to get:

$$\displaystyle x=\frac{\pi}{2}-\theta$$

and since:

$$\displaystyle x=\sin^{-1}\left(\cos(\theta) \right)$$, we may then state:

$$\displaystyle \sin^{-1}\left(\cos(\theta) \right)=\frac{\pi}{2}-\theta$$
Yeah, I am in calculus right now. It's just my high school never did much with trigonometry unfortunately, so it's one of my weak areas I'm trying to catch up on.

#### MarkFL

Staff member
Re: Simplify a trig expression using a right triangle

Okay, good deal!

You are wise to work on strengthening your trig. skills, as it is use extensively in the study of calculus. I hope it goes well, and we are here to help you!


text:
# Python lxml transform xml to dictionary

I need to download a fairly large xml (up to 2 GB) then load it into memory or possibly an empty db on a stand alone machine on a semi-regular basis. I need to grab a large amount of very specific data from the xmls for a statistical treatment. I realize ETLs like this are usually a one time deal for migrations and the like, but for reasons that really aren't worth explaining, it has to be downloaded in its entirety, specifically to an xml, and then loaded in its entirety to a program on the non-networked machine then deleted when done. Every time (I know, I know).

The important data needed are contained in the <SampleID>, <LocationName>, and <Foo> nodes of the below xml. I do not know the exact structure of the tables they are coming out of but based on the structure of the xml I assume it's something like this:

SampleID    LocationName    Foo1   Foo2
0001        Jeff            10     11
0001        Jennifer        24
0002        Jeff            9      16
0002        Jennifer        20     26

I've been around and around with this for a couple years and the most efficient method I've seen is using a VB.net SAX parser to load the pertinent data to a DataTable then SQLBulkload the DataTable into a SQL Server database. Again for reasons not worth mentioning, I can't use this either.

So I've tried several C, C++ libraries and some SQL and I just can't seem to approach any kind of sufficiently acceptable load times. Until today. I heard that Python has a binding with access to the C libxml library called lxml (I don't know if that's what the library actually is but I decided to try it). I have this bit of code below parsing a 1.5 gb xml file and loading the pertinent data into a specifically built Python data structure (the best way to get at the testing I need to do is a hash table and this seems to approximate it). It will look like this:

[{'0001': {'Jeff': ['10', '11']}}, {'0001': {'Jennifer':['24', '24']}}...]

It takes about 110-120 seconds which is much faster than anything I've been able to do in the last two years. About 20 seconds of that is parsing the file. The remainder is storing the data in the dictionary. I don't know much about Python conventions, Python syntax or much about the lxml library. I'm sure that a lot of my lack of efficiency here is in my method of transforming, in the data structure itself, or in my use of the library. So I'd love a bit of help making this even more efficient and more conventional.

If there are problems with this XML it is a transcription error, not a problem with correct or well formed xml but it would be worth pointing out if there are some.

The xml:

<?xml version="1.0" encoding="UTF-8"?>
<MyXMLFile xmlns="MyNameSpace">
<ignoredNode1>one piece of data to ignore</ignoredNode1><!--these first six nodes contain no children and contain no important data-->
<ignoredNode2>one piece of data to ignore</ignoredNode2>
<ignoredNode3>one piece of data to ignore</ignoredNode3>
<ignoredNode4>one piece of data to ignore</ignoredNode4>
<ignoredNode5>one piece of data to ignore</ignoredNode5>
<ignoredNode6>one piece of data to ignore</ignoredNode6>
<Sample> <!--This could grow up to more than a million <Sample> nodes. Right now there are a max of 350k and about a 1.5 gb file-->
<SampleID>0001</SampleID>
<ignoredNode7>one piece of data to ignore</ignoredNode7>
<Location> <!-- there could be a variable number of <Location> nodes here but typically there are 10-30. In the long run I'm only concerned with at most 21 of them but typically there's 13, 15, 20, or 21 that I care about that are designated by the <LocationName> node-->
<LocationName>Jeff</LocationName>
<ignoredNode8>one piece of data to ignore</ignoredNode8>
<ignoredNode9>one piece of data to ignore</ignoredNode9>
<Data>
<Foo>10</Foo>
</Data>
<Data>
<Foo>11</Foo>
</Data>
</Location>
<Location>
<LocationName>Jennifer</LocationName>
<ignoredNode8>one piece of data to ignore</ignoredNode8>
<ignoredNode9>one piece of data to ignore</ignoredNode9>
<Data> <!--sometimes there is only one <Data> child in <Location>, in which case, it gets duplicated as a second piece of data in the dictionary-->
<Foo>24</Foo>
</Data>
</Location>
</Sample>
</MyXMLFile>

The code. If this errors, I apologize, I have to transcribe it by hand. Let me know if you have an issue and I'll fix it but this SHOULD run. What I have on my machine does. I can't imagine that this first stab at this is the most efficient way to do this.

import copy
from lxml import etree as ET
import timeit

sampleList = []
sampleDict = {}
locationDict = {}
fooList = []

start_time = timeit.default_timer()
tree = ET.parse('FileName.xml')

root = tree.getroot()

MyXMLFile = root.getchildren()

for Sample in MyXMLFile:
if Sample.tag == '{MyNameSpace}Sample':
Locations = Sample.getchildren()
for Location in Locations:
if Location.tag == '{MyNameSpace}SampleID':
sampleid = Location.text
elif Location.tag == '{MyNameSpace}Location':
Data = Location.getchildren()
for MyData in Data:
if MyData.tag == '{MyNameSpace}LocationName':
locationName = MyData.text

if MyData.tag == '{MyNameSpace}Data' and len(Location) == 4:
Foos = MyData.getchildren()
for Foo in Foos:
fooList.append(Foo.text)
fooList.append(Foo.text)

elif MyData.tag == '{MyNameSpace}Data' and len(Location) == 5:
Foos = MyData.getchildren()
for Foo in Foos:
fooList.append(Foo.text)

if len(fooList) == 2:
locationDict[locationName] = FooList
sampleDict[specimenid] = locationDict
sampleList.append(copy.deepcopy(sampleDict))
fooList.clear()
locationDict.clear()
sampleDict.clear()

print(timeit.default_timer() - start_time)

• Thank you @RootTwo – Dan Aug 2 at 11:02

Your code would run, but sampleList would be empty.

Take a look at PEP8 or run flake8 or black on your code to make the code style more in line with typical python code. (Although it doesn't sound like anyone will ever see it).

I came up with two alternatives. The first is similar to your code in that it looks at the hierarchy of the xml elements. But it uses XPath expression to select the child nodes of interest. I have a sample file with 100k samples, 1M locations, and about 1.7M foos. On my Windows laptop, it takes about 22 seconds.

samples = []

start_time = timeit.default_timer()

tree = ET.parse('test.xml')
root = tree.getroot()

for sample in root.iterfind('./{MyNameSpace}Sample'):
sample_id = sample.find('{MyNameSpace}SampleID').text

for location in sample.iterfind('./{MyNameSpace}Location'):
location_name = location.find('{MyNameSpace}LocationName').text

data = [datum.text for datum in location.iterfind('./{MyNameSpace}Data/{MyNameSpace}Foo')]

if len(data) < 2:
data.append(data[0])

samples.append({sample_id:{location_name:data}})

print(timeit.default_timer() - start_time)

The second version uses iterparse() to create the data structure as it is parsing the file. By default, iterparse() yield elements when it sees the end tag. The code grabs the interesting values when is sees the interesting end tags. When it sees the Location end tag, it also makes a new record and adds it to the list of samples. It runs in about 19 seconds.

samples = []

start_time = timeit.default_timer()

foos = []

for event, element in ET.iterparse('test.xml'):
if element.tag == '{MyNameSpace}SampleID':
sample_id = element.text

elif element.tag == '{MyNameSpace}LocationName':
location_name = element.text

elif element.tag == '{MyNameSpace}Foo':
foo = element.text
foos.append(foo)

elif element.tag == '{MyNameSpace}Location':
if len(foo) == 1:
foo.append(foo[0])

samples.append({sample_id:{location_name:foos}})
foos = []

print(timeit.default_timer() - start_time)

• Thanks so much! I probably made a mistake in my transcription because it does run and store everything in the file on my machine. I’ll give the xPath a go. I just assumed since I had to burn through all the information that Xpath wouldn’t help me. – Dan Aug 3 at 1:17
• ...don’t know if anyone will ever see it but my company gets audited on the regular and they dig into everything. Probably wouldn’t be a bad idea to keep it conventional. – Dan Aug 3 at 1:18
• For giggles, what kind of power do you have in your laptop? I ran two tests of a 1.5 g file and got 90 and 105 seconds. Still better than 125... – Dan Aug 3 at 13:33
• @Dan it's a 2.4GHz i5-6300; 20G memory; SSD. My test file is 226M bytes. – RootTwo Aug 3 at 14:20
• scratch that. Im stupid. you said 226 MB. My file is almost 5 times that. Looks like its as good as its going to get for now. THANKS SO MUCH FOR THE HELP!!!!! – Dan Aug 3 at 14:54


text:
# Thread: Solve the integrals for the area

1. ## Solve the integrals for the area

What are the enpoint of these integrals?

2. what do you mean by enpoint?

3. Originally Posted by racewithferrari
What are the enpoint of these integrals?

Have you drawn a graph of each function (on the same set of axes) and shaded the bounded area? That should show you that your first job is to get the coordinates of the 'corners' of the bounded area.

If you need more help, please post all your work and say where you are stuck.

4. Originally Posted by racewithferrari
What are the enpoint of these integrals?

The very first sentence says "sketch the bounded region enclosed by the given curves". Have you done that?


text:
# How can I continue using both pageslts and tikz?

A recent update to my TeX Live installation appears to have given rise to an incompatibility between two packages: pageslts and tikz. The following document does not compile:

\documentclass{article}
\usepackage{pageslts}
\usepackage{tikz}
\begin{document}
\pagenumbering{arabic}
abc
\end{document}

The error I get is:

! Undefined control sequence.
\@currname ->\@nil

l.10 \begin{document}

It is possible to workaround this problem by loading tikz before loading pageslts. However, this would require me to load tikz in custom class and package files since I use pageslts there. If possible, I'd like to avoid doing this as most of my documents do not require TikZ and I'd rather not load it when not needed.

One approach would obviously be to create class/package options allowing me to specify when TikZ is required. However, I'm wondering if that is the best approach or if there's a better solution.

Right now, pageslts is causing me a great deal of grief - this is the second package I've discovered must now be loaded first. (At least the first case gave me a semi-helpful error message.)

So if there is a good alternative to pageslts, I'd be interested in learning about it since the package seems to have suddenly become rather finicky and no longer gets along nicely with anything else!

• I'd call this a bug. By the way, pageslts calls \listfiles, which is very annoying. – egreg Aug 14 '15 at 22:11
• @egreg Oh, is it this that's responsible? I was going to track that down but the errors have been taking priority. It is, indeed, very annoying. Is it a bug even though changing the loading order avoids the error? mathtools also now must be loaded before pageslts. The documentation says that the package no longer requires e-TeX but I'm not clear why that is meant to be such an advantage! – cfr Aug 14 '15 at 22:18
• The problem with \listfiles shouldn't be connected with the bug. – egreg Aug 14 '15 at 22:23
• @egreg I didn't really expect it to be ;). But is it a bug if it just requires a different load order? That's not uncommon, after all. – cfr Aug 14 '15 at 22:25

The bug is in lines 115–119 of pageslts.sty, where \@currname is used in a \@for cycle, leaving a definition of \@currname, which is unexpected.

If I change the three occurrences of \@currname with \next, the problem disappears.

Of course, also the appearance of \listfiles on line 111 is wrong, but this is unrelated, only annoying.

The macro \@currname is used during package loading, for making a stack in order to correctly reset the status of @ as a letter or as an “other character”. It is also reused in \listfiles as the placeholder for a \@for cycle, but in that case it's not a problem, because \listfiles is processed only at the end of a job. The similar usage by pageslts instead is wrong, because it clearly interferes with package loading. If you add \errorcontextlines=100 at the beginning of the test file you get

! Undefined control sequence.
\@currname ->\@nil

\@pushfilename ...ef \@currnamestack {{\@currname
}{\@currext }{\the \catcod...

\@onefilewithoptions #1[#2][#3]#4->\@pushfilename
\xdef \@currname {#1}\glob...

\reserved@a ...ds[{}][{2007/11/11}]\@pkgextension
\relax
<argument> ...uirePackage {pdftexcmds}[2007/11/11]
\expandafter \ifx \csname ...

\@secondoftwo #1#2->#2

\@begindocumenthook ...base}[2009/07/12]\fi \fi }}
\pgfsys@atbegindocument \...

\document ...ent \@firstofone \@begindocumenthook
\ifdim \topskip <1sp\globa...
l.5 \begin{document}

showing that the problem is when processing \RequirePackage{pdftexcmds}.

• Thanks! I've reported this to the package author, quoting the first two paragraphs of your answer. I agree \listfiles is annoying but I suppose it isn't technically wrong. – cfr Aug 14 '15 at 22:55
• I don't entirely understand why using \@currname is a problem, though. – cfr Aug 14 '15 at 23:05
• Be nice if the updated version of atveryend was available, too, but it is not yet on CTAN despite the note dating it to 1st August :(. – cfr Aug 14 '15 at 23:06

This was indeed a bug. Thank you for reporting it! Version 1.2e as of 2015/08/17 (should become available soon at CTAN, has been submitted 2015-08-17 21:24 CEST) uses \next, as suggested in egreg's answer, and also does no longer include the \listfiles. (The example file still includes \listfiles, because it is to be expected that possible problems with the example might be connected to the use of either older or newer versions of other packages.) The requested atveryend has been dated later 2015/08/01, because before that date there was no new version, and the next version (whether 2015/08/18 or 2016/01/01) will surely have this fixed. - I marked this answer "community wiki", upvotes and accepted answer should (in my humble opinion) go to egreg's answer!

• Thank you very much! And thank you also for the email telling me this. You might consider using something a bit more package-specific than \next, though. My hacked version is currently using \@pageslts@currname just to be on the safe side ;). – cfr Aug 17 '15 at 20:57


text:
Corresponding Author: Prof. Ertuğrul Başar, Koç University (ebasar@ku.edu.tr)

B. Ozpoyraz and E. Basar. “Revolutionary deep learning approaches for massive MIMO systems.” Burak. | Stay With Tech. Mar. 2023. [Online]. Available: https://www.burakozpoyraz.com/index.php/2023/03/05/revolutionary-deep-learning-approaches-for-massive-mimo-systems/.

@misc{ozpoyraz_DL_MIMO,
title={Revolutionary deep learning approaches for massive {MIMO} systems},
author={Ozpoyraz, Burak and Basar, Ertuğrul},
journal={Burak. | Stay With Tech},
month={Mar.},
year={2023},
url={https://www.burakozpoyraz.com/index.php/2023/03/05/revolutionary-deep-learning-approaches-for-massive-mimo-systems/},
}

## Abstract

Multiple-input multiple-output (MIMO) technology has been widely utilized in current fourth-generation (4G) networks, also known as the Long Term Evolution (LTE). In comparison to 4G, emerging fifth-generation (5G) and beyond cellular communication technologies need far more compelling requirements for spectral efficiency, data rate, and latency. Therefore, massive MIMO appears as a groundbreaking technology with large-scale antenna arrays deployed at base stations, which dramatically enhances the benefits of conventional MIMO systems. However, massive MIMO technology comes with the expense of challenging constraints that leave the existing communication and signal processing theory-based concepts desperate. At this point, deep learning (DL) demonstrates immense promise with the mapping capability from input features to desired labels using neural networks (NN). In this article, we reveal the advancements of massive MIMO systems, highlight limitations that are challenging to overcome with traditional methods, and shed light on emerging DL applications in massive MIMO systems. This introductory article has been adopted from [1].

## Introduction

Wireless communication technologies have been evolving since the first generation (1G) of cellular networks emerged in the 1980s. The most recent fifth-generation (5G) of cellular communication technology, starting with the 3GPP (3rd Generation Partnership Project) Release 15 standard [2], electrifies both the academic world and the mobile market. The proliferation of cellular networks has resulted in a remarkable increase in internet-of-things (IoT) devices and their interconnection. According to Statista [3], the total number of connected IoT devices is estimated to rise from 8.74 billion in 2020 to 16.44 billion by 2025, representing an almost 88% growth, as given in Fig. 1.

This development in the connection of IoT devices enables thrilling applications in 5G, such as autonomous cars and virtual reality. The International Telecommunication Union (ITU) has classified 5G applications into three categories: enhanced mobile broadband (eMBB) services, massive machine-type communications (mMTC), and ultra-reliable and low-latency communications (URLLC) [4], [5], [6]. Although these use cases have different requirements, 5G is expected to support high data rates, reduced latency in data transmission, energy efficiency, strong data security, and high scalability for a multitude of devices [4], as given in Fig. 2.

Even though 5G has shown significant promise for a globally connected wireless world, it has been questioned whether 5G can be the actual facilitator of the internet-of-everything (IoE) [7]. As given in Fig. 1, the forecasted IoT device connection will be 25.44 billion in 2030, which indicates that the necessities of a fully-intelligent information society beyond 2030 are considerably rigorous [1]. Therefore, communication researchers have already started scrutinizing novel communication technologies for the sixth generation (6G) of the wireless world to overcome the drawbacks of 5G and build an outstanding wireless future. The thrilling applications expected to come into life with 6G, such as haptic technology, brain-computer interface, truly immersive virtual reality, and space travel provided in Fig. 3, will introduce compelling requirements. Therefore, researchers have developed innovative physical layer (PHY) communication solutions.

PHY of a wireless system, also known as Layer-1, is the initial layer of the network that deals with data transmission over the air. The transmitter side of a PHY communication transforms information bits into baseband electric pulses located around 0 Hz. These pulses are then converted to electromagnetic waves called radio-frequency (RF) signals by modulating a carrier signal’s phase, amplitude, or both with a higher operating frequency [9]. The operating frequency of a carrier signal is standard-specific. The technical specifications of the 5G PHY, commonly known as New Radio (NR) L1, are defined by the 3GPP organization [10]. The receiver side follows the reverse procedure of the transmitter and decodes received RF signals into information bits. The signal processing conducted to convert information bits to symbols and to prepare symbols for the fading and disruptive transmission effects significantly impacts the quality of a PHY transmission. Therefore, wireless researchers have been studying improving transmission reliability, increasing the number of bits transmitted in a channel use, and enhancing data security through novel PHY techniques. The existing model-based signal processing techniques rely on mathematically optimized models such as communication and information theory. These strong theoretical models individually optimize each signal processing block of a PHY transmission, such as bit modulation and channel estimation. However, the critical question is:

In this article, we pave the way for a solid answer to this question. In particular, we investigate deep learning (DL) solutions for massive multiple-input multiple-output (MIMO) systems. DL models are composed of neural networks (NN) with learnable parameters. DL methods have already proved their success in diverse fields, including natural language processing, speech recognition, and computer vision, where it is incredibly challenging to create a theoretical model for mapping the input features to the desired labels [1]. However, their applicability to existing signal processing techniques has remained limited thus far [1], [11]. Since the traditional PHY structure is optimized on a block-by-block basis, DL must go beyond strict boundaries to replace traditional model-based approaches. Our primary motivation is to shed light on the potential of DL methods in ensuring the E2E optimization of a PHY transmission. The most noteworthy innovation in 6G is the replacement of individually optimized signal processing units in a PHY architecture with a gigantic DL network that enables E2E optimization. We call this drastic paradigm shift from block-optimized model-based structure to E2E-optimized intelligent design the PHY revolution. In what follows, we will briefly explain the details and advantages of massive MIMO systems.

## Massive MIMO: Extending to the Spatial Domain

In a MIMO system model, both transmitter and receiver are equipped with multiple antennas, as given in Fig. 4.

In addition to time and frequency domains, multiple antennas introduce a third resource dimension: the spatial domain [12]. Using multiple antennas at both ends delivers substantial benefits compared to a single transmission link. Assuming a MIMO system with $$N_{T}$$ transmit and $$N_{R}$$ receive antennas, these benefits are described below:

### Spatial Multiplexing Gain

As an alternative to delivering the same information across multiple links for improving reliability, transmitting separate information signals over distinct channels boosts the data transmission rate without incurring additional transmit power or bandwidth costs. Since all antennas operate in the same frequency band, the amount of transmitting data rises linearly with the minimum of $$N_{T}$$ and $N_{R}$ without requiring an increase in bandwidth, called spatial multiplexing gain [13].

The benefits of MIMO systems can be exploited in multi-user (MU) scenarios by simultaneously serving multiple users at the same time and frequency resource, as illustrated in Fig. 5.

The individual information signal of each user is separately pre-processed using the corresponding CSI and weight matrix prior to the transmission, also known as transmit precoding. Transmit precoding steers each information signal to the appropriate user, maximizing received signal power at each user, called spatial beamforming, while mitigating signal interference between users by radiating null to other users, as demonstrated in Fig. 5. This procedure is called MU interference cancellation. Furthermore, transmit precoding eliminates the fading effect of the channel, which removes the CSI requirement for users [8].

As previously stated, wireless connectivity of IoT devices has been rising, and fascinating applications will emerge at the price of compelling requirements with 6G. The expectations beyond 2030 are extremely high data rates, ultra-low latency, intelligent resource allocation for efficient spectrum utilization in congested areas, enormous energy efficiency, and excellent security algorithms to avoid blocking and jamming attacks [14]. The limited amount of antennas in conventional MU-MIMO systems is insufficient to satisfy the necessities of next-generation wireless communications. Thus, massive MIMO, also known as large-scale MIMO, emerges as a game-changing technology with an excessive number of antennas. In fact, a 5G base station (BS), abbreviated as gNodeB or gNB, in a massive MIMO system is expected to have a few hundred antennas, which is considerably higher than the upper limit, 64, of a 4G base station, called eNodeB or eNB [8], [15]. The most striking aspect of massive MIMO is its extraordinary spatial multiplexing capability through a vast amount of antennas, which improves capacity by at least tenfold [16].

Furthermore, massive MIMO delivers remarkable radiated energy efficiency with a sharp beamforming technique. It should be noted that the radiated energy efficiency is different from the energy costs associated with the deployment process of massive MIMO systems. In particular, radiated energy efficiency refers to how well the users utilize the total transmit power. A large number of transmit antennas and a simple linear precoding technique enable the transmitted energy to be constructively steered into users with incredible precision. In addition, a user’s information signal is canceled when traveling across other users’ channels due to the transmit precoding. Thus, MU interference is mitigated, enhancing the efficiency of radiated energy [16].

Moreover, large-scale MIMO systems in 5G provide tremendous spectral efficiency by simultaneously delivering information to many more users in the same frequency resource compared to the existing 4G technology. The simultaneous service of massive MIMO is especially beneficial at frequencies below 6 GHz, where the frequency spectrum is limited. In a frequency band below 6 GHz, each user can enjoy the entire bandwidth rather than dividing the band for users since massive MIMO can spatially multiplex the users, which improves spectrum utilization [8].

The aforementioned advantages demonstrate that massive MIMO systems exhibit a vast potential to be deployed in 5G and beyond wireless communication technologies. When challenging requirements of the fully-intelligent wireless world of beyond 2030 are considered, massive MIMO systems will play a huge role in the development of extremely fast, energy- and spectrum-efficient, and high-secure cellular networks with ultra-low latency [16].

The next section will explore the fundamental flaws of current massive MIMO technology and investigate the possible use of DL approaches in resolving these drawbacks to truly realize the benefits of massive MIMO systems.

## Deep Learning in Massive MIMO

Even though massive MIMO systems show grand promise for next-generation cellular communications, technical constraints restrict the full potential of large-scale antenna arrays [15]. Therefore, it is critical to identify these open research problems and develop solutions for the complete utilization of the massive MIMO technology. As discussed before, the application of DL has been regarded as a promising answer to difficulties in PHY communications. Thus, interest in various DL techniques to address the stated drawbacks in massive MIMO systems has been growing excessively [11]. The primary goal of this section is to highlight the most crucial research problems and how to overcome them with the aid of DL approaches. We divide the issues with the receiver and transmitter sides of a massive MIMO system as follows:

### Open Research Problems: Massive MIMO Receiver

The capacity of enormous spatial multiplexing, which boosts data transmission rate, and precise beamforming, which improves radiated energy efficiency and transmission reliability, are two of the most prominent advantages of the massive MIMO technology. However, these notable advancements require high-precision channel estimation to implement precoding properly. In the conventional downlink MIMO system of the current 4G technology, the BS transmits a pilot sequence to users for channel estimation. The transmitted pilot sequence is specially designed and is well-known to users. Consequently, users can examine the received pilot sequence and understand the change in the original pilot sequence caused by the channel, which results in the estimation of channel components. Users send back the estimated channels to the BS for the transmit precoding. Since the pilot sequence includes a pilot symbol for each transmit antenna at the BS, the described procedure is appropriate for the moderate number of antennas in 4G. However, this method necessitates the transmission and feedback of a few hundred pilot symbols for all BS transmit antennas in a massive MIMO system. Furthermore, the pilot symbols should be orthogonal, allowing users to receive each pilot symbol and estimate the corresponding channel individually. Orthogonality can be achieved by either transmitting pilot symbols in different time slots, which takes a long time for the whole sequence and reduces time efficiency, or allocating a frequency band for each pilot symbol, which decreases spectrum efficiency. Therefore, a reliable channel estimation method with a reduced pilot requirement is a critical prerequisite for massive MIMO systems to reach their full potential [1], [16].

The next major issue with massive MIMO receivers is the symbol decoding process. The optimal detector in the literature is the maximum likelihood detector (MLD), which searches over all possible transmit symbol vectors to decode the transmitted symbol. This brute force technique brings an unreasonable level of complexity into the system by increasing the processing time. In order to provide a better understanding of the MLD complexity, we might consider a gNB with 128 transmit antennas and a signal set including quadrature phase-shift keying (QPSK) information symbols. It is worth noting that QPSK symbols are generated by modulating the phase of a baseband carrier signal. As QPSK modulation has four levels, MLD scans for four symbols for each transmit antenna. Hence, there are a total of $$4^{128}$$ transmit symbol vector combinations, which is greater than $$10^{77}$$. Even with a moderate modulation level, the MLD complexity reaches ridiculous levels in a massive MIMO detection. On the other hand, despite significantly less complexity, the alternative linear detectors in the literature, such as zero-forcing (ZF) and minimum mean-squared error (MMSE) detectors, achieve low detection reliability and, as a result, high bit error rate (BER). In addition, these linear detectors have a stringent constraint of $$(N_{R} \geq N_{T})$$, which is impractical in massive MIMO systems since it is impossible to equip mobile users with more antennas than gNB. As a consequence, it is indispensable to develop a robust symbol detection algorithm capable of overcoming the trade-off between decoding complexity and BER performance [1].

The primary objective of DL-based massive MIMO detectors is to attain near-MLD BER performance while maintaining low computational complexity. DL-based massive MIMO detectors in the literature yield that near-MLD BER performance is achievable with considerably reduced detection complexity through NNs. The learnable parameters of NNs can effectively create a mapping from received symbols to transmitted symbols under perfect CSI assumption. Further improvements eliminate the dependency on perfect CSI and combine the channel estimation and symbol detection operations of a massive MIMO system. Therefore, DL approaches enable direct mapping from received symbols to transmitted symbols, and thus information bits, without the need for extra signal processing.

### Open Research Problems: Massive MIMO Transmitter

As discussed in the previous section, transmit precoding is a vital operation in massive MIMO for three reasons:

• Beamforming information symbols toward users for higher radiated energy efficiency
• Mitigating the fading channel effect for improving the symbol detection reliability
• Ensuring MU interference cancellation for serving many users simultaneously

However, constructing a high-precision precoding matrix in current standards necessitates the use of high-resolution digital-to-analog converters (DACs). Similar to ADCs in the massive MIMO receiver, the higher the DAC resolution, the greater the processing complexity. As a result, cost-friendly massive MIMO systems require the capability to operate precoding with low-resolution DACs reliably [1].

DL-based massive MIMO transmitters replace computationally burdened high-resolution precoding blocks with NNs to offer robust pre-processing of information symbols with incredibly lower complexity. According to the intelligent massive MIMO transmitter literature, even one-bit DL-based precoding can successfully execute pre-processing and ensure the above three properties [1].

Since the birth of wireless communications, conventional PHY architectures have been utilizing mathematically optimized modulation schemes. When information bits are transmitted via the carrier signal phase, bits are said to be PSK modulated. For instance, the possible phase angles for a modulation level of two, called binary PSK (BPSK), are 0 and $latex\pi$ radians. Therefore, the symbol set (constellation) of BPSK modulation includes two symbols: -1 and 1, corresponding to 0 and 1 bits, respectively. For the general case, a modulation order of $$M$$, where $$M$$ is a power of two, yields the following $$M$$-ary phase angle array: $$\{0, 2 \pi / M, \dots, (2M – 2) \pi / M\}$$, each of which corresponding to a separate $$\log_{2}(M)$$ bits. As only the phase of a PSK-modulated carrier signal is altered, the carrier signal has the same unit amplitude for all phase angles. Thus, the symbols in a PSK modulation’s constellation are uniformly spaced on the unit circle. Another prominent modulation technique is quadrature amplitude modulation (QAM), in which both the phase and amplitude of a carrier signal convey information symbols. For a square lattice $$M$$-QAM constellation, such as 16-QAM and 64-QAM, the two-dimensional representation of a symbol is: $$(A_{I}, A_{Q})$$, where $$A_{I}, A_{Q} \in {(1-\sqrt{M}), (3-\sqrt{M}), \dots, (\sqrt{M}-1)}$$. Fig. 6 illustrates examples of QPSK and 16-QAM constellations. Although the decision boundaries of these traditional constellation designs are optimized, DL-based MIMO transmitters are anticipated to create unique-shaped constellations adaptable to various transmission environments [1], [17].

When a DL model is aware of the properties of the transmission environment, the learnable parameters of this model can determine the best locations for constellation symbols and optimize the decision boundaries of these symbols for the given specific environment, potentially improving BER performance over traditional constellations. Nevertheless, the decoding process at the receiver end requires perfect knowledge of the unique transmit constellation. Hence, conventional detectors, such as the MLD and ZF detectors, are incapable of performing reliable detection. Consequently, DL-based massive MIMO transmitters and receivers should collaborate to unlock the full potential of DL approaches, which brings us to the most fantastic concept of future wireless communications: E2E learning [1], [18].

### The Revolutionary Aspiration: End-to-End Learning

Thus far, we have highlighted the primary constraints of massive MIMO systems that are challenging to address with conventional theoretically optimized signal processing techniques. As the mathematical model-based signal processing blocks of traditional PHY architectures struggle to leverage the full power of massive MIMO, DL techniques have gone mainstream. Even though DL approaches have made significant progress in improving separate signal processing blocks, we are still left with the critical question of whether separate optimization corresponds to overall optimization. As an answer, we are heading into the area of entirely E2E learning, in which a single giant data-based DL network will replace separate model-based signal processing blocks to ensure E2E optimization. This approach will remove the mathematical channel model requirement at both ends, which enhances the practicability since it might be problematic to precisely define the channel model in complicated environments. Fig. 7 depicts the essential applications of DL approaches in transmitter and receiver ends of large-scale MIMO systems.

## Conclusion and Future Directions

Massive MIMO technology will undoubtedly have a significant impact on shaping future wireless communication technologies. Massive MIMO systems’ substantial spectrum and radiated energy efficiency, ultra-low latency, robust PHY security, and extreme data rates will make 6G the true facilitator of IoE services. However, the limitations of existing model-based signal processing techniques prohibit massive MIMO technology from reaching its maximum capabilities. Therefore, DL approaches appear as a glaring solution to the shortcomings of massive MIMO by substituting theoretical models with data-based NNs. The eventual goal with data-based PHY architectures is to enable thorough E2E learning, resulting in the E2E optimization of PHY communications.

In this article, we have elucidated the significant advantages of massive MIMO systems along with providing fundamental concepts of PHY communications to aid better understanding. Furthermore, we have studied how DL techniques can overcome the restrictions of current signal processing techniques in massive MIMO technology. Despite the tremendous acceleration in 6G research, we believe that the real wireless revolution is still a long way off. We have identified the following research directions to improve DL-based massive MIMO systems:

• Constructing benchmark real-world channel datasets
• MU E2E communication systems

## References

[1] B. Ozpoyraz, A. T. Dogukan, Y. Gevez, U. Altun and E. Basar, “Deep Learning-Aided 6G Wireless Networks: A Comprehensive Survey of Revolutionary PHY Architectures,” in IEEE Open J. of the Commun. Soc., vol. 3, pp. 1749-1809, 2022, doi: 10.1109/OJCOMS.2022.3210648. [Link]

[2] “Release 15,” Mar. 2017. [Link]

[3] L. S. Vailshery, “Number of internet of things (IoT) connected devices world-wide from 2019 to 2030,” Mar. 2022. [Link]

[4] I.F. Akyildiz, S. Nie, S.-C. Lin, and M. Chandrasekaran, “5G roadmap: 10 key enabling technologies,” Comput. Netw., vol. 106, pp. 17-48, Sep. 2016. [Link]

[5] M. Series, “IMT vision-framework and overall objectives of the future development of IMT for 2020 and beyond,” Recommendation ITU, vol. 2083, p. 21, Sep. 2015. [Link]

[6] G. M. B. Series, “Emerging trends in 5G / IMT2020,” Sep. 2016. [Link]

[7] W. Saad, M. Bennis, and M. Chen, “A vision of 6G wireless systems: Applications, trends, technologies, and open research problems,” IEEE Netw., vol. 34, pp. 134-142, Oct. 2020. [Link]

[8] N. Instruments, “5G new radio: Introduction to the physical layer,” Jun. 2018 [Link]

[9] E. Team, “What is the PHY layer,” May 2021. [Link]

[10] ETSI, “5G; NR; physical layer; general description (3GPP TS 38.201 version 15.0.0 release 15),” Sep. 2018. [Link]

[11] T. Wang, C.-K. Wen, H. Wang, F. Gao, T. Jiang, and S. Jin, “Deep learning for wireless physical layer: Opportunities and challenges,” ArXiv, vol. abs/1710.05312, Oct. 2017. [Link]

[12] C.-X. Wang, F. Haider, X. Gao, X.-H. You, Y. Yang, D. Yuan, H. M. Aggoune, H. Haas, S. Fletcher, and E. Hepsaydir, “Cellular architecture and key technologies for 5G wireless communications networks,” IEEE Commun. Mag., vol. 52, pp. 122-130, Feb. 2014. [Link]

[13] A. Paulraj, D. Gore, R. Nabar, and H. Bolcskei, “An overview of MIMO communications – A key to gigabit wireless,” Proc. of the IEEE, vol. 92, pp. 198-218, Nov. 2004. [Link]

[14] J. Zhang, E. Björnson, M. Matthaiou, D. W. K. Ng, H. Yang, and D. J. Love, “Prospective multiple antenna technologies for beyond 5G,” IEEE J. on Sel. Areas in Commun., vol. 38, pp. 1637-1660, Jun. 2020. [Link]

[15] R. Annaswamy et al., “IEEE 5G and beyond technology roadmap white paper,” p. 39, Oct. 2017. [Link]

[16] E. G. Larsson, O. Edfors, F. Tufvesson, and T. L. Marzetta, “Massive MIMO for next generation wireless systems,” IEEE Commun. Mag., vol. 52, pp. 186-195, Feb. 2014. [Link]

[17] M. J. López-Morales, K. Chen-Hu, and A. G. Armada, “A survey about deep learning for constellation design in communications,” in 2020 12th Int. Symp. on Commun. Syst., Netw. and Digital Signal Process. (CSNDSP), pp. 1-5, Jul. 2020. [Link]

[18] F. A. Aoudia and J. Hoydis, “End-to-end learning of communications systems without a channel model,” in 2018 52nd Asilomar Conf. on Signals, Syst., Comput., pp. 298-303, Oct. 2018. [Link]


text:
# $X$ is a random variable taking the values $3, 4$and $12$ with probabilities $\large\frac{1}{3},\frac{1}{4}$ and $\large\frac{4}{7}.$ Then $E(X)$ is
$\begin{array}{1 1}(1)5&(2)6\\(3)7&(4)3\end{array}$


text:
# A Challenge Problem in Techniques Of Problem Solving

I just need a hint to solve the problem. The method used is illustrated in the example problem as follows:

Assume that k is a positive integer . What is the sum of integers
S = 1+2+3+...+(k-1)+k?

Solution:

22 - 12 = 3 = 2 * 1 + 1
32 - 22 = 5 = 2 * 2 + 1
42 - 32 = 7 = 2 * 3 + 1
...
k2 - (k-1)2 = 2 * (k-1) + 1
(k+1)2 - k2 = 2 * k + 1

[22 - 12] + [32 - 22] +...+ [(k+1)2 - k2]
= [2 * 1+1] + [2*2+1] + [2*3+1]+...+[2*k+1]

The left hand side ''telescopes'' (that is, all but the first and last terms cancel) and the right side may be factored. The result is

(k+1)2 - 12 = 2[1+2+3+...k] + [1+1+1+...+1]
k times
or
k2 + 2 * k = 2 * S + k

Solving for S for we find that

S = (k2+k)/2
I am supposed to imitate the method used in the example problem to solve the challenge problem in the book as follows

Imitate the method used in the last problem to find a formula for the sum
12 + 22 + 32 + ... + k2

when k is a positive integer.

I have tried various things like (a+1)4 - a4, (a+1)3 - a3, triangle numbers, odd numbers, but I havent been able to solve it. I just need a hint. This is not a homework problem. This is self study.

Related Calculus and Beyond Homework Help News on Phys.org
Use

$$(k+1)^3 - k^3 = 3k^2+3k+1$$

and do summation on both side. By telescoping method, only 2 terms left in the left hand side. You need to do a little manipulation on the right hand side though.

taozhen
Why not like the Contiguous Numbers: for example,P=1+2+3+4+5....+100 and the answer of P is (1+100)*100/2 , so the S=(1+k)*k/2

Use

$$(k+1)^3 - k^3 = 3k^2+3k+1$$

and do summation on both side. By telescoping method, only 2 terms left in the left hand side. You need to do a little manipulation on the right hand side though.
Thanks for the hint
I got my answer for S

S = (2k3+3k2+k)/6

You are right. You may want to do some factorization also.

$$S = \frac{ n(n+1)(2n+1)}{6}$$


text:
## トポロジー火曜セミナー

開催情報 火曜日　17:00～18:30　数理科学研究科棟(駒場) 056号室 河野 俊丈, 河澄 響矢, 北山 貴裕, 逆井卓也 http://park.itc.u-tokyo.ac.jp/MSF/topology/TuesdaySeminar/index.html Tea: 16:30 - 17:00 コモンルーム

### 2013年03月19日(火)

16:30-18:00   数理科学研究科棟(駒場) 002号室
Tea: 16:00 - 16:30 コモンルーム

Open book foliation and application to contact topology (ENGLISH)
[ 講演概要 ]
Open book foliation is a generalization of Birman and Menasco's braid foliation. Any 3-manifold admits open book decompositions. Open book foliation is a singular foliation on an embedded surface, and is define by the intersection of a surface and the pages of the open book decomposition. By Giroux's identification of open books and contact structures one can use open book foliation method to study contact structures. In this talk I define the open book foliation and show some applications to contact topology. This is joint work with Tetsuya Ito (University of British Columbia).

### 2013年02月19日(火)

16:30-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:00 - 16:30 コモンルーム

On the ring of Fricke characters of free groups (JAPANESE)
[ 講演概要 ]
This is a joint work with Takao Satoh (Tokyo University of Science). We study a descending filtration of the ring of Fricke characters of a free group consisting of ideals on which the automorphism group of the free group naturally acts. Then by using it, we define a descending filtration of the automorphism group of a free group, and investigate a relation between it and the Andreadakis-Johnson filtration.

### 2013年01月22日(火)

16:30-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:00 - 16:30 コモンルーム
Jarek Kedra 氏 (University of Aberdeen)
On the autonomous metric of the area preserving diffeomorphism
of the two dimensional disc. (ENGLISH)
[ 講演概要 ]
Let D be the open unit disc in the Euclidean plane and let
G:=Diff(D, area) be the group of smooth compactly supported
area-preserving diffeomorphisms of D. A diffeomorphism is called
autonomous if it is the time one map of the flow of a time independent
vector field. Every diffeomorphism in G is a composition of a number
of autonomous diffeomorphisms. The least amount of such
diffeomorphisms defines a norm on G. In the talk I will investigate
geometric properties of such a norm.

In particular I will construct a bi-Lipschitz embedding of the free
abelian group of arbitrary rank to G. I will also show that the space
of homogeneous quasi-morphisms vanishing on all autonomous
diffeomorphisms in G is infinite dimensional.

This is a joint work with Michael Brandenbursky.

### 2013年01月21日(月)

16:30-17:30   数理科学研究科棟(駒場) 002号室

ベキ零リー環を横断構造に持つリー葉層構造について (JAPANESE)
[ 講演概要 ]
リー$\\mathfrak{g}$-葉層構造が与えられたとき,それに付随した構造リー環と呼
ばれる$\\mathfrak{g}$の部分リー環$\\mathfrak{h}$が定まる.本講演ではその逆,
すなわち,リー環$\\mathfrak{g}$とその部分リー環$\\mathfrak{h}$が与えられた
とき構造リー環が$\\mathfrak{h}$となるリー$\\mathfrak{g}$-葉層構造が存在す
るかという問題についてて,$\\mathfrak{g}$がベキ零リー環の場合に説明をする.

### 2013年01月21日(月)

17:30-18:30   数理科学研究科棟(駒場) 002号室

Quasi-morphisms on the group of area-preserving diffeomorphisms of
the 2-disk
(JAPANESE)
[ 講演概要 ]
Gambaudo and Ghys constructed linearly independent countably many quasi-
morphisms on the group of area-preserving diffeomorphisms of the 2-disk
from quasi-morphisms on braid groups.
In this talk, we will explain that their construction is injective as a
homomorphism between vector spaces of quasi-morphisms.
If time permits, we introduce an application by Brandenbursky and K\\c{e}
dra.

### 2012年12月11日(火)

16:30-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:00 - 16:30 コモンルーム
Ismar Volic 氏 (Wellesley College)
Homotopy-theoretic methods in the study of spaces of knots and links (ENGLISH)
[ 講演概要 ]
I will survey the ways in which some homotopy-theoretic
methods, manifold calculus of functors main among them, have in recent
years been used for extracting information about the topology of
spaces of knots and links. Cosimplicial spaces and operads will also
be featured. I will end with some recent results about spaces of
homotopy string links and in particular about how one can use functor
calculus in combination with configuration space integrals to extract
information about Milnor invariants.

### 2012年12月04日(火)

16:30-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:00 - 16:30 コモンルーム

Conformal field theory for C2-cofinite vertex algebras (JAPANESE)
[ 講演概要 ]
This is a jount work with Akihiro Tsuchiya (Kavli IPMU).
We consider sheaves of covacua and conformal blocks over parameter spaces of n-pointed Riemann surfaces
for a vertex algebra of which the category of modules is not necessarily semi-simple.
We assume the C2-cofiniteness condition for vertex algebras.
We define "tensor product" of two modules over a C2-cofinite vertex algebra.

### 2012年11月27日(火)

16:30-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:00 - 16:30 コモンルーム

[ 講演概要 ]
Thurstonの例により、葉層構造の二次特性類は有界でないことが知られている。本講演では、横断的な共形平坦構造などを持つ葉層構造に対しては(例外的な場合を除き)二次特性類が有限性を持つことを、非有界性や葉層構造の剛性との関連と共に説明する。
(本講演はSantiago de Compostela大学のJesús Antonio Álvarez
López氏との共同研究 arXiv:1205.3375に基づく。)

### 2012年11月20日(火)

16:30-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:00 - 16:30 コモンルーム

3次元双曲幾何と団代数 (JAPANESE)
[ 講演概要 ]
クラスター代数は2000年にFomin-Zelevinskyによって発見された代数系である.

### 2012年11月13日(火)

16:30-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:00 - 16:30 コモンルーム

The virtual fibering theorem and sutured manifold hierarchies (JAPANESE)
[ 講演概要 ]
In 2007 Agol showed that every irreducible 3-manifold whose fundamental
group is nontrivial and virtually residually finite rationally solvable
(RFRS) is virtually fibered. In the proof he used the theory of
least-weight taut normal surfaces introduced and developed by Oertel and
Tollefson-Wang. We give another proof using complexities of sutured
manifolds. This is a joint work with Stefan Friedl (University of
Cologne).

### 2012年11月06日(火)

16:30-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:00 - 16:30 コモンルーム

[ 講演概要 ]

その後、有理数体の絶対ガロア群が'結び目全体が張る空間'に

### 2012年10月30日(火)

16:30-18:00   数理科学研究科棟(駒場) 126号室
Tea: 16:00 - 16:30 コモンルーム

Applications of knot theory to molecular biology (JAPANESE)
[ 講演概要 ]
In this talk we discuss applications of knot theory to studies of DNA
and proteins.
Especially we will consider (1)topological characterization of
mechanisms of site-specific recombination systems,
(2)modeling knotted DNA and proteins in confined regions using lattice
knots, and
(3)mechanism of topoisomerases and signed crossing changes.

### 2012年10月23日(火)

16:30-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:00 - 16:30 コモンルーム

A geometric approach to the Johnson homomorphisms (JAPANESE)
[ 講演概要 ]
ジョンソン準同型を、完備化されたゴールドマン・トゥラエフ・リー双代数への
トレリ群の

### 2012年10月16日(火)

17:10-18:10   数理科学研究科棟(駒場) 056号室
Tea: 16:50 - 17:10 コモンルーム

Analytic torsion of log-Enriques surfaces (JAPANESE)
[ 講演概要 ]
Log-Enriques surfaces are rational surfaces with nowhere vanishing
pluri-canonical forms. We report the recent progress on the computation
of analytic torsion of log-Enriques surfaces.

### 2012年10月09日(火)

16:30-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:00 - 16:30 コモンルーム

The growth series of pure Artin groups of dihedral type (JAPANESE)
[ 講演概要 ]
In this talk, I consider the kernel of the natural projection from
the Artin group of dihedral type to the corresponding Coxeter group,
that we call a pure Artin group of dihedral type,
and present rational function expressions for both the spherical and
geodesic growth series
of the pure Artin group of dihedral type with respect to a natural
generating set.
Also, I show that their growth rates are Pisot numbers.
This talk is partially based on a joint work with Takao Satoh.

### 2012年10月02日(火)

16:30-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:00 - 16:30 コモンルーム

Geometric flows and their self-similar solutions
(JAPANESE)
[ 講演概要 ]
In the first half of this expository talk we consider the Ricci flow and its self-similar solutions,
namely the Ricci solitons. We then specialize in the K\\"ahler case and discuss on the K\\"ahler-Einstein
problem. In the second half of this talk we consider the mean curvature flow and its self-similar
solutions, and see common aspects of the two geometric flows.

### 2012年09月04日(火)

17:00-18:00   数理科学研究科棟(駒場) 002号室
Tea: 16:30 - 17:00 コモンルーム
Piotr Nowak 氏 (the Institute of Mathematics, Polish Academy of Sciences)
Poincare inequalities, rigid groups and applications (ENGLISH)
[ 講演概要 ]
Kazhdan’s property (T) for a group G can be expressed as a
fixed point property for affine isometric actions of G on a Hilbert
space. This definition generalizes naturally to other normed spaces. In
this talk we will focus on the spectral (aka geometric) method for
proving property (T), based on the work of Garland and studied earlier
by Pansu, Zuk, Ballmann-Swiatkowski, Dymara-Januszkiewicz
(“lambda_1>1/2” conditions) and we generalize it to to the setting of
all reflexive Banach spaces.
As applications we will show estimates of the conformal dimension of the
boundary of random hyperbolic groups in the Gromov density model and
present progress on Shalom’s conjecture on vanishing of 1-cohomology
with coefficients in uniformly bounded representations on Hilbert spaces.

### 2012年07月24日(火)

16:30-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:00 - 16:30 コモンルーム
Greg McShane 氏 (Institut Fourier, Grenoble)
Orthospectra and identities (ENGLISH)
[ 講演概要 ]
The orthospectra of a hyperbolic manifold with geodesic
boundary consists of the lengths of all geodesics perpendicular to the
boundary.
We discuss the properties of the orthospectra, asymptotics, multiplicity
and identities due to Basmajian, Bridgeman and Calegari. We will give
a proof that the identities of Bridgeman and Calegari are the same.

### 2012年07月17日(火)

16:30-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:00 - 16:30 コモンルーム

Contact structure of mixed links (JAPANESE)
[ 講演概要 ]
A strongly non-degenerate mixed function has a Milnor open book
structures on a sufficiently small sphere. We introduce the notion of
{\\em a holomorphic-like} mixed function
and we will show that a link defined by such a mixed function has a
canonical contact structure.
Then we will show that this contact structure for a certain
holomorphic-like mixed function
is carried by the Milnor open book.

### 2012年07月10日(火)

16:30-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:00 - 16:30 コモンルーム
Marcus Werner 氏 (Kavli IPMU)
Topology in Gravitational Lensing (ENGLISH)
[ 講演概要 ]
General relativity implies that light is deflected by masses
due to the curvature of spacetime. The ensuing gravitational
lensing effect is an important tool in modern astronomy, and
topology plays a significant role in its properties. In this
talk, I will review topological aspects of gravitational lensing
theory: the connection of image numbers with Morse theory; the
interpretation of certain invariant sums of the signed image
magnification in terms of Lefschetz fixed point theory; and,
finally, a new partially topological perspective on gravitational
light deflection that emerges from the concept of optical geometry
and applications of the Gauss-Bonnet theorem.

### 2012年06月19日(火)

17:10-18:10   数理科学研究科棟(駒場) 056号室
Tea: 16:50 - 17:10 コモンルーム

On the universal degenerating family of Riemann surfaces
over the D-M compactification of moduli space (JAPANESE)
[ 講演概要 ]
It is usually understood that over the Deligne-
Mumford compactification of moduli space of Riemann surfaces of
genus > 1, there is a family of stable curves. However, if one tries to
construct this family precisely, he/she must first take a disjoint union
of various types of smooth families of stable curves, and then divide
them by their automorphisms to paste them together. In this talk we will
show that once the smooth families are divided, the resulting quotient
family contains not only stable curves but virtually all types of
degeneration of Riemann surfaces, becoming a kind of universal
degenerating family of Riemann surfaces.

### 2012年06月12日(火)

16:30-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:00 - 16:30 コモンルーム

Topological interpretation of the quandle cocycle invariants of links (JAPANESE)
[ 講演概要 ]
Carter et al. introduced many quandle cocycle invariants
combinatorially constructed from link-diagrams. For connected quandles of
finite order, we give a topological meaning of the invariants, without
some torsion parts. Precisely, this invariant equals a sum of "knot
colouring polynomial" and of a Z-equivariant part of the Dijkgraaf-Witten
invariant. Moreover, our approach involves applications to compute "good"
torsion subgroups of the 3-rd quandle homologies and the 2-nd homotopy
groups of rack spaces.

### 2012年06月05日(火)

16:30-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:00 - 16:30 コモンルーム

A generalization of Dehn twists (JAPANESE)
[ 講演概要 ]
We introduce a generalization
of Dehn twists for loops which are not
necessarily simple loops on an oriented surface.
Our generalization is an element of a certain
enlargement of the mapping class group of the surface.
A natural question is whether a generalized Dehn twist is
in the mapping class group. We show some results related to this question.
This talk is partially based on a joint work
with Nariya Kawazumi (Univ. Tokyo).

### 2012年05月29日(火)

16:30-18:00   数理科学研究科棟(駒場) 056号室
Tea: 16:00 - 16:30 コモンルーム

Triple linking numbers and triple point numbers
of torus-covering $T^2$-links
(JAPANESE)
[ 講演概要 ]
The triple linking number of an oriented surface link was defined as an
analogical notion of the linking number of a classical link. A
torus-covering $T^2$-link $\\mathcal{S}_m(a,b)$ is a surface link in the
form of an unbranched covering over the standard torus, determined from
two commutative $m$-braids $a$ and $b$.
In this talk, we consider $\\mathcal{S}_m(a,b)$ when $a$, $b$ are pure
$m$-braids ($m \\geq 3$), which is a surface link with $m$-components. We
present the triple linking number of $\\mathcal{S}_m(a,b)$ by using the
linking numbers of the closures of $a$ and $b$. This gives a lower bound
of the triple point number. In some cases, we can determine the triple
point numbers, each of which is a multiple of four.

### 2012年05月22日(火)

17:10-18:10   数理科学研究科棟(駒場) 056号室
Tea: 16:50 - 17:10 コモンルーム

Gamma Integral Structure in Gromov-Witten theory (JAPANESE)
[ 講演概要 ]
The quantum cohomology of a symplectic
manifold undelies a certain integral local system
defined by the Gamma characteristic class.
This local system originates from the natural integral
local sysmem on the B-side under mirror symmetry.
In this talk, I will explain its relationships to the problem
of analytic continuation of Gromov-Witten theoy (potentials),
including crepant resolution conjecture, LG/CY correspondence,
modularity in higher genus theory.


text:
numfmt - Maple Help

Online Help

All Products    Maple    MapleSim

MapleTA[Builtin]

 numfmt
 format a number according to a template

 Calling Sequence numfmt(format, number)

Parameters

 format - string number - numeric

Description

 • The numfmt command uses a format template to specify how many digits after the decimal point to display among other things. The given number is matched against the template and a formatted string is returned.
 • This command has the same specification as the built-in Java command, java.text.DecimalFormat.

Examples

 > $\mathrm{MapleTA}:-\mathrm{Builtin}:-\mathrm{numfmt}\left("#.00",20.9\right)$
 ${"20.90"}$ (1)
 > $\mathrm{MapleTA}:-\mathrm{Builtin}:-\mathrm{numfmt}\left("#.#",12.34\right)$
 ${"12.3"}$ (2)
 > $\mathrm{MapleTA}:-\mathrm{Builtin}:-\mathrm{numfmt}\left(".#",12.34\right)$
 ${"12.3"}$ (3)
 > $\mathrm{MapleTA}:-\mathrm{Builtin}:-\mathrm{numfmt}\left("#.0000",12.3456789\right)$
 ${"12.3457"}$ (4)
 > $\mathrm{MapleTA}:-\mathrm{Builtin}:-\mathrm{numfmt}\left("#.010",12.3456\right)$
 ${"12.351"}$ (5)
 > $\mathrm{MapleTA}:-\mathrm{Builtin}:-\mathrm{numfmt}\left("#",12.34\right)$
 ${"12"}$ (6)
 > $\mathrm{MapleTA}:-\mathrm{Builtin}:-\mathrm{numfmt}\left("#.#",12.3456\right)$
 ${"12.3"}$ (7)
 > $\mathrm{MapleTA}:-\mathrm{Builtin}:-\mathrm{numfmt}\left("#.###",12.3456\right)$
 ${"12.346"}$ (8)
 > $\mathrm{MapleTA}:-\mathrm{Builtin}:-\mathrm{numfmt}\left("000.#",12.3456\right)$
 ${"012.3"}$ (9)

Compatibility

 • The MapleTA[Builtin][numfmt] command was introduced in Maple 18.
 • For more information on Maple 18 changes, see Updates in Maple 18.

 See Also


text:
1. ## d(x^e^x)/dy'',??

Ive never been shown the technique for this differential. Anybody know it?? thnx! diff: $y=x^e^x$ hmm latex won't let me write this one, basicall y=x to the power exp(x)

2. Originally Posted by oxrigby
Ive never been shown the technique for this differential. Anybody know it?? thnx! diff: $y=x^e^x$ hmm latex won't let me write this one, basicall y=x to the power exp(x)
If you are asking what $\frac{d}{dx}\bigg[x^{e^x}\bigg]$ equals then I can help you. first note that

$x^{e^x}=e^{\ln\left(x^{e^x}\right)}=e^{e^x\ln(x)}$

So now knowing that $\frac{d}{dx}\bigg[e^{u(x)}\bigg]=u'(x)e^{u(x)}$

And that $\frac{d}{dx}\bigg[e^x\ln(x)\bigg]=e^x\ln(x)+\frac{e^x}{x}$

We can see that

$\frac{d}{dx}\bigg[e^{e^x\ln(x)}\bigg]=\left(e^x\ln(x)+\frac{e^x}{x}\right)e^{e^x\ln(x)} =\left(e^x\ln(x)+\frac{e^x}{x}\right)x^{e^x}$


text:
# Kerodon

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

Lemma 5.2.3.13. Suppose we are given a commutative diagram of simplicial sets

$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}\ar [r] \ar [d]^{U} & \operatorname{\mathcal{E}}\ar [d]^{W} & \operatorname{\mathcal{D}}\ar [d]^{V} \ar [l] \\ \operatorname{\mathcal{C}}' \ar [r] & \operatorname{\mathcal{E}}' & \operatorname{\mathcal{D}}' \ar [l] }$

in which the vertical morphisms are inner fibrations. Then the induced map

$F: \operatorname{\mathcal{C}}\star _{\operatorname{\mathcal{E}}} \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}' \star _{\operatorname{\mathcal{E}}'} \operatorname{\mathcal{D}}'$

is also an inner fibration.

Proof. Unwinding the definitions, we see that $F$ factors as a composition

$\operatorname{\mathcal{C}}\star _{\operatorname{\mathcal{E}}} \operatorname{\mathcal{D}}\xrightarrow {G} \operatorname{\mathcal{C}}\star _{\operatorname{\mathcal{E}}'} \operatorname{\mathcal{D}}\xrightarrow {H} \operatorname{\mathcal{C}}' \star _{\operatorname{\mathcal{E}}'} \operatorname{\mathcal{D}}',$

where $G$ is a pullback of the inner fibration $\operatorname{\mathcal{E}}\star _{\operatorname{\mathcal{E}}} \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{E}}\star _{\operatorname{\mathcal{E}}'} \operatorname{\mathcal{E}}$ of Lemma 5.2.3.12 and $H$ is a pullback of the inner fibration $\operatorname{\mathcal{C}}\star \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}' \star \operatorname{\mathcal{D}}'$ of Proposition 4.3.3.23. $\square$


text:
# Optical thickness of the second harmonic cyclotron motion in a plasma

Tags:
1. Jan 25, 2014

### eoghan

1. The problem statement, all variables and given/known data
Let's consider a Tokamak with major radius $R=1m$ and minor radius $a=0.3m$, magnetic field $B=5T$ with a deuterium plasma with central density $10^{20}m^{-3}$, central temperature $1keV$ and parabolic temperature and density profiles $\propto (1-r^2/a^2)$

a) Find the electronic cyclotron frequency for the second harmonics

b) Verify that the emission in the second extraordinary harmonic in a direction perpendicular to the magnetic field is optically thick

2. Relevant equations
$$\omega_c=\frac{\Omega}{\gamma}=\frac{eB_0}{m_e\gamma}$$
$$\omega_m=\frac{m\omega_c}{1-\beta_{//}\cos\theta}$$
$$\tau=\int\!\!ds\,\alpha(\nu)$$

3. The attempt at a solution
a) I just apply the formula for $\omega_m$ with $m=2$
b) I have no idea... please give me some hint... I tried to calculate the cutoff frequencies for the second harmonic in the extraordinary mode, but the second harmonic frequency doesn't fall in the cutoff and it is not absorbed. I think I have to apply the integral and find $\tau>>1$ but I don't know how to apply that integral. I don't want the solution, just an hint

Thank you very much

2. Jan 25, 2014

### eoghan

The problem has been taken from exercise 5.4, chapter 5, Hutchinson - Principles of Plasma Diagnostic

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted


text:
Math Help - order of a permutation

1. order of a permutation

Find the order of g,

S7= Sym(7)=Symmetric group 7

G = S7, g = (1 3 5)(2 4 6 7).

My answer is ord(g) = 12

I know you suppose to find the lowest common multiple to find the order.. but what Im not sure about is:

If g was not disjoint
do you multiply together and then find its order or can you find the order the way its given.

2. Originally Posted by Dreamer78692
Find the order of g,

S7= Sym(7)=Symmetric group 7

G = S7, g = (1 3 5)(2 4 6 7).

My answer is ord(g) = 12

I know you suppose to find the lowest common multiple to find the order.. but what Im not sure about is:

If g was not disjoint
do you multiply together and then find its order or can you find the order the way its given.
Good question!

The answer, however, is no. Remember that the set of 2-cycles generates the group $S_n$, and so if it were true then every element would have order 2.

For example, (12)(23) = (132). lcm(2, 2)=2 but the order of (132) is 3. So there is your counter-example.


text:
We now reach a topic highlighted by Fong and Spivak: "generative effects". These are, roughly, situations where the whole is more than the sum of its parts. A nice example shows up in the logic of partitions.

Remember that any set \$$X\$$ has a poset of partitions. This poset is called \$$\mathcal{E}(X)\$$, and its elements are partitions of \$$X\$$. Each partition \$$P\$$ corresponds to an equivalence relation \$$\sim_P\$$, where \$$x \sim_P y\$$ if and only if \$$x\$$ and \$$y\$$ are in the same part of \$$P\$$. This makes it easy to describe the partial order on \$$\mathcal{E}(X)\$$: we say a partition \$$P\$$ is **finer** than a partition \$$Q\$$, or \$$Q\$$ is **coarser** than \$$P\$$, or simply \$$P \le Q\$$, when

$x \sim_P y \textrm{ implies } x \sim_Q y$

for all \$$x,y \in X\$$.

Now that we have this poset, how can we do _logic_ with it?

In the logic of subsets, we've seen that "and" and "or" are the operations of "meet" and "join" in a certain poset. So, the logic of partitions should use "meet" and "join" in the poset \$$\mathcal{E}(X)\$$. Let's see what they're like!

First, we can ask whether two partitions \$$P\$$ and \$$Q\$$ have a meet. Remember: the **meet** \$$P \wedge Q\$$, if it exists, is the coarsest partition that is finer than \$$P\$$ and \$$Q\$$.

**Puzzle 34.** Can you draw the coarsest partition that's finer than these two partitions?

In fact the meet of two partitions always exists, and it's easy to describe:

**Puzzle 35.** Suppose \$$P\$$ and \$$Q\$$ are two partitions of a set \$$X\$$. Show that there's an equivalence relation \$$\approx \$$ defined by

$x \approx y \textrm{ if and only if } x \sim_P y \textrm{ and } x \sim_Q y .$

**Puzzle 36.** Every equivalence relation gives a partition as in [Puzzle 29](https://forum.azimuthproject.org/discussion/1963/lecture-10-the-logic-of-partitions/p1). Show that \$$\approx\$$ gives the partition \$$P \wedge Q\$$.

We can also ask whether \$$P\$$ and \$$Q\$$ have a join! The **join** \$$P \vee Q\$$, if it exists, is the finest partition that is coarser than \$$P\$$ and \$$Q\$$.

Can you draw the finest partition that's coarser than these two partitions?

The join of two partitions always exists. Since "and" and "or" are meet and join in the logic of subsets, you might think the to describe the join of partitions, we just copy Puzzle 35 and replace "and" with "or". _But no!_

**Puzzle 37.** Suppose \$$P\$$ and \$$Q\$$ are two partitions. Show that the relation \$$\frown\$$ defined by

$x \frown y \textrm{ if and only if } x \sim_P y \textrm{ or } x \sim_Q y$

is _not always_ an equivalence relation.

To get the equivalence relation corresponding to \$$P \vee Q\$$, we have to work harder. Say we have any partitions \$$P\$$ and \$$Q\$$ of some set \$$X\$$. If you give me two elements \$$x, y \in X\$$ and ask if they're in the same part of \$$P \vee Q\$$, it's _not enough_ to check whether

$x \textrm{ and } y \textrm{ are in the same part of } P \textrm{ or the same part of } Q .$

(That's what \$$x \frown y\$$ means.) Instead, you have to check whether there's a _list_ of elements \$$z_1, \dots, z_n\$$ such that

$x \textrm{ and } z_1 \textrm{ are in the same part of } P \textrm{ or the same part of } Q$

and

$z_1 \textrm{ and } z_2 \textrm{ are in the same part of } P \textrm{ or the same part of } Q$

and so on, and finally

$z_n \textrm{ and } y \textrm{ are in the same part of } P \textrm{ or the same part of } Q .$

This picture by [Michael Hong](https://forum.azimuthproject.org/discussion/1855/introduction-michael-hong) shows how it works:

So, the join is a lot more complicated than the meet!

**Puzzle 38.** If \$$P\$$ and \$$Q\$$ are two partitions on a set \$$X\$$, let the relation \$$\simeq \$$ be the **[transitive closure](https://en.wikipedia.org/wiki/Transitive_closure)** of the relation \$$\frown \$$ defined in Puzzle 37. This means that \$$x \simeq y\$$ if and only if

$x \frown z_1 \textrm{ and } z_1 \frown z_2 \textrm{ and } \dots \textrm{ and } z_{n-1} \frown z_n \textrm{ and } z_n \frown y$

for some \$$z_1, \dots, z_n \in X\$$. Show that \$$\simeq \$$ is an equivalence relation.

**Puzzle 39.** Show that the equivalence relation \$$\simeq \$$ on \$$X\$$ gives the partition \$$P \vee Q \$$.

This is the first hint of what Fong and Spivak call a "generative effect". To decide if two elements \$$x , x' \in X\$$ are in the same part of the meet \$$P \wedge Q\$$, it's enough to know if they're the same part of \$$P\$$ and the same part of \$$Q\$$, since

$x \sim_{P \wedge Q} x' \textrm{ if and only if } x \sim_P x' \textrm{ and } x \sim_Q x'.$

But this does _not_ work for the join!

$\textbf{THIS IS FALSE: } \; x \sim_{P \vee Q} x' \textrm{ if and only if } x \sim_P x' \textrm{ or } x \sim_Q x' .$

To decide if \$$x \sim_{P \vee Q} y\$$ you need to look at _other_ elements of \$$X\$$, too. It's not a "local" calculation - it's a "global" one!

But to make this really precise and clear, we need to think about "pulling back" partitions. We'll do that next time.

**[To read other lectures go here.](http://www.azimuthproject.org/azimuth/show/Applied+Category+Theory#Course)**