text:
Zipping Trees, Part 1
Lastmod: 2020-03-23

While trawling hackerrank in order to procrastinate doing "real" work, I came across this problem in the Functional Programming section. It has you perform various incremental operations on a rose tree, such as inserting and deleting nodes and visiting adjacent nodes in the tree. In the spirit of sticking with Simple Haskell, our first attempt at the problem will use only one benign language extension and two imports.

{-# LANGUAGE LambdaCase #-}
import Data.Tree

Maintaining a focus into a recursive data structure can be elegantly solved by zippers. We can adapt the content from this excellent introduction to zippers in Learn You a Haskell directly to suit the problem at hand. Using the Tree type from containers (essentially data Tree a = Node a [Tree a]) we can define a corresponding "Crumb" and Zipper type.

data Crumb
= Crumb
Int -- parent node's value
[Tree Int] -- left siblings
[Tree Int] -- right siblings
deriving (Show)
type Zipper = (Tree Int, [Crumb])

We could define them polymorphically, but the problem only uses Int s.

The first operation is to visit the $n^\mathrm{th}$ child of the current node. We accomplish this by splitting the current focus's children at the $n^\mathrm{th}$ position, making the $n^\mathrm{th}$ element our new focus, and appending a new crumb with its left and right siblings and parent value to the list of preceding crumbs.

visitChild :: Int -> Zipper -> Zipper
visitChild n (Node x children, crumbs) =
let (left, (focus : right)) = splitAt n children
in  (focus, Crumb x left right : crumbs)

Visiting the parent requires us to join the most recent crumb's left and right siblings with the current focus to construct a new focus.

visitParent :: Zipper -> Zipper
visitParent (focus, Crumb parent left right : cs) =
(Node parent (left ++ [focus] ++ right), cs)

Deleting the current node is almost identical to visiting the parent, except that we drop the focus instead of sandwiching it in between the left and right siblings.

delete :: Zipper -> Zipper
delete (_, Crumb parent left right : cs) =
(Node parent (left ++ right), cs)

To visit the right sibling we deconstruct the most recent crumb's list of right siblings, set its head to the new focus, and add the current focus to the end of the list of left siblings.

visitRight :: Zipper -> Zipper
visitRight (focus, Crumb parent left (r : rs) : cs) =
(r, Crumb parent (left ++ [focus]) rs : cs)

Visiting the left sibling is very similar, but due to the asymmetrical nature of lists, we can't use the same elegant pattern matching we used for the right case.

visitLeft :: Zipper -> Zipper
visitLeft (focus, Crumb parent left right : cs) =
(last left, Crumb parent (init left) (focus : right) : cs)

Inserting a new node to the left or right of the current node is as simple as constructing a new tree with no children and adding it to the appropriate place in the current crumb.

insertLeft :: Int -> Zipper -> Zipper
insertLeft x (focus, Crumb parent left right : cs) =
(focus, Crumb parent (left ++ [Node x []]) right : cs)

insertRight :: Int -> Zipper -> Zipper
insertRight x (focus, Crumb parent left right : cs) =
(focus, Crumb parent left (Node x [] : right) : cs)

Inserting a new child is stipulated to always happen at the leftmost position and only requires modification of the current focus.

insertChild :: Int -> Zipper -> Zipper
insertChild x (Node y children, crumbs) =
(Node y (Node x [] : children), crumbs)

The final operation, changing the value of the current focus, is the easiest, needing no explanation.

change :: Int -> Zipper -> Zipper
change x (Node _ forest, crumbs) = (Node x forest, crumbs)

With our logic written, we can write the main loop to read in the instructions and print the value of the current focus when we are told to do so. The input format, as is common in hackerrank problems, is an integer, $q$, on one line, denoting the number of instructions, followed by $q$ lines, each containing one instruction of the form "change 2", "print", "visit child 3", "insert left", etc. Although we could just throw away the first line and read in the rest of the contents line by line, it feels wasteful not to use information.

main :: IO ()
main = do
instructions <- replicateM q (words <$> getLine) foldM_ go (Node 0 [], []) instructions The foldM function from Control.Monad takes a function of type Monad m => (a -> b -> m a), an initial value of type a, a collection of type Foldable t => t b, and returns the final accumulation m a. foldM_ is identical to foldM except that it throws away the final result, returning m (), which is what we want since we just care about printing to STDOUT. Specializing the polymorphic type to our particular case, we have foldM_ :: (Zipper -> [String] -> IO Zipper) -> Zipper -> [[String]] -> IO (). The go function takes care of the actual logic of parsing each instruction and dispatching to the correct zipper function. go :: Zipper -> [String] -> IO Zipper go zipper = \case ["change", x] -> pure$ change (read x) zipper
["print"]     -> case zipper of
(Node x _, _) -> zipper <$print x ["visit", dir] -> let f = case dir of "left" -> visitLeft "right" -> visitRight "parent" -> visitParent _ -> error$ "Invalid direction: " <> dir
in  pure (f zipper)
-- The hackerrank problem counts children from 1.
-- Haskell lists are 0-indexed, though because
-- the Haskell language designers aren't monsters
["visit", "child", n] -> pure (visitChild (read n - 1) zipper)
["insert", dir, x] ->
let f = case dir of
"left"  -> insertLeft
"right" -> insertRight
"child" -> insertChild
_       -> error \$ "Invalid direction: " <> dir
in  pure (f (read x) zipper)
["delete"] -> pure (delete zipper)
other      -> error ("Invalid instruction: " <> unwords other)

This solution is correct and runs quickly enough to satisfy hackerrank's time constraints. It uses only Simple Haskell: a benign language extension, pattern matching, and basic algebraic datatypes. There are, however, some things it leaves to be desired.

1. Using unnamed tuples and product types with no record syntax can obscure the meaning of each field and requires some extra typing in each function for elements of the zipper that remain unchanged during a particular transformation.

2. The Tree type from containers uses lists, which is useful for infinite trees. However, our trees are guaranteed to be finite. Moreover, we have to append to the end of a list or drop its last element when visiting or inserting nodes, which takes linear time in the size of the list. We are told that any given node will have at most 10 children, so this isn't a huge issue, but it would be a serious performance problem with a larger branching factor.

3. All of the visit functions are partial and will crash if there is an invalid operation in the instruction list, such as visiting the child of a leaf node or the parent of the tree root. Hackerrank guarantees that all operations will be valid, so crashing on what should be unreachable cases is reasonable. We could return Maybe Zipper from all of our functions, but that would be more cumbersome to deal with. Perhaps the best solution would be to explicitly call error with an informative message rather than getting something about "irrefutable patterns."

In part 2, we will see how going slightly up the Haskell complexity chain can solve these problems, as well as adding a whole slew of new ones!


text:
# Synchronous Impedance Method (EMF Method) for Finding Voltage Regulation of Alternator

Electronics & ElectricalElectronDigital Electronics

The synchronous impedance method or EMF method is used to determine the voltage regulation of the larger alternators. The synchronous impedance method is based on the concept of replacing the effect of armature reaction by an imaginary reactance.

For an alternator,

$$\mathrm{𝑽 = 𝑬_{𝒂} − 𝑰_{𝒂}𝒁_{𝒔} = 𝑬_{𝒂} − 𝑰_{𝒂}(𝑅_{𝒂} + 𝑗𝑋_{𝑠}) … (𝟏)}$$

At first, the synchronous impedance (𝑍𝑠) is measured and then, the value of actual generated EMF (𝐸a) is calculated. Thus, from the values of (𝐸a) and V, the voltage regulation of the alternator can be calculated.

## Measurement of Synchronous Impedance

In order to determine the value of synchronous impedance, following tests are performed on an alternator −

• DC Resistance Test
• Open-Circuit Test
• Short-Circuit Test

## DC Resistance Test

The circuit diagram for the DC resistance test is shown in Figure-1.

Consider the alternator is star-connected with the field winding open-circuited. Now, measure the DC resistance between each pair of terminals either by using Wheatstone’s bridge or ammeter-voltmeter method. The average of three sets of resistance values Rt is taken. This value of Rt is divided by 2 to obtain the DC resistance per phase.

While performing the test, the alternator should be at rest, because the AC effective resistance is greater than DC resistance due to skin effect. The AC effective resistance per phase may be obtained by multiplying the DC resistance by a factor 1.20 to 1.75 depending upon the size of the alternator.

## Open-Circuit Test

To perform the open-circuit test, the load terminals are kept open and the alternator is run at rated synchronous speed. The circuit diagram of the open-circuit test is shown in Figure-2.

Initially, the field current is set to zero. Then, the field current is gradually increased in steps and the open-circuit terminal voltage Et is measured in each step. The field current may be increased to obtain 25 % more than rated voltage of the alternator.

A graph is plotted between the open-circuit phase voltage $((𝐸_{𝑝ℎ} = 𝐸_{𝑡}/ \sqrt{3}))$ and the field current (𝐼𝑓). The obtained characteristic curve is known as open-circuit characteristic (O.C.C) of the alternator (see Figure-3).

The shape of O.C.C. is same as a normal magnetisation curve. When the linear portion of the O.C.C. is extended, it given the air-gap line of the characteristic.

## Short-Circuit Test

For performing the short-circuit test, the armature terminals are short-circuited through three ammeters as shown in Figure-4.

Before starting the alternator, the field current should be decreased to zero. Each ammeter should have a range more than the rated full-load value. Now, the alternator is run at synchronous speed. Then, the field current is gradually increased in steps and the armature current is measured at each step. The field current may be increased to obtain the armature currents up to 150 % of the rated value.

The field current (𝐼𝑓) and the average of the three ammeter readings is taken at each step. A graph is plotted between the armature current (𝐼a) and the field current (𝐼𝑓). The obtained characteristic is known as short-circuit characteristic (S.C.C.) of the alternator and this characteristic is a straight line as shown in Figure-5.

## Calculation of Synchronous Impedance (𝒁𝒔)

In order to calculate the synchronous impedance of the alternator, the O.C.C. and the S.C.C. are drawn on the same curve sheet, as shown in Figure-6.

Then, determine the value of short-circuit current (𝐼𝑆𝐶) at the field current that gives the rated voltage per phase of the alternator. The synchronous impedance (𝑍𝑠) will then be equal to the ratio of the open-circuit voltage to the short-circuit current at the field current which gives the rated voltage per phase, i.e.,

$$\mathrm{𝑍_{𝑠} =\frac{open\:circuit\:voltage\:per \:phase}{short\:circuit\:armature\:current}… (2)}$$

From the figure-6, the synchronous impedance can be written as

$$\mathrm{\Rightarrow\: 𝑍_{𝑠} =\frac{𝐴𝐵\:(volts)}{𝐴𝐶\: (amperes)}… (3)}$$

Also, the synchronous reactance of the alternator is

$$\mathrm{𝑋_{𝑠} =\sqrt{𝑍^{2}_{𝑠}-𝑅^{2}_{𝑎}}… (4)}$$

Therefore, the percentage voltage regulation of the alternator will be,

$$\mathrm{Percentage\:voltage\:regulation =\frac{𝐸_{𝑎} − 𝑉}{𝑉}× 100 … (5)}$$

## Numerical Example

A 400 V, 35 kVA single-phase alternator has an effective armature resistance of 0.3Ω. An excitation current 15 A produces 250 A armature current on short-circuit and an EMF of 380 V on open-circuit. Calculate the synchronous impedance and synchronous reactance of the alternator.

Solution

Synchronous impedance of the given alternator is,

$$\mathrm{𝑍_{𝑠} =\frac{Open\:circuit\:voltage}{Short \:circuit\:armature\:current}=\frac{380}{250}= 1.52\:Ω}$$

The synchronous reactance of the alternator is,

$$\mathrm{𝑋_{𝑠} =\sqrt{𝑍^{2}_{𝑠}-𝑅^{2}_{𝑎}}=\sqrt{1.52^{2} − 0.3^{2}}= 1.49 Ω}$$

Published on 25-Sep-2021 07:09:27

format: plain


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Butterworth IIR Filter using Bilinear Transformation Method | MATLAB Tutorial

Bilinear Transformation Method:

1. Poles are transformed by using equation,

Where, Ts is Sampling Time

1. Mapping is one to one.
2. Aliasing effect is not present.
3. High pass Filter and band Reject Filter can be designed.
4. Poles as well as zeros can be mapped.
5. Frequency warping effect is present.

Butterworth approximation:

Many a times it is necessary to approximate the characteristics of analog filter. This approximation is required because the practical characteristics of a filter are not identical to the ideal characteristics. The Butterworth filter is a type of signal processing filter designed to have as flat a frequency response as possible in the passband. It is also referred to as a maximally flat magnitude filter. It was first described in 1930 by the British engineer and physicist Stephen Butterworth in his paper entitled “On the Theory of Filter Amplifiers”.

Design equations and Design steps:

Let

Ap= Attenuation in pass band.

As= Attenuation in Stop Band.

Ωp= Pass Band edge frequency.

Ωc= Cut-off Frequency.

Ωs= Stop Band edge Frequency.

Step 1: From the given specification of digital Filter, obtain equivalent analog filter as follows:

Here,

Ω= Frequency of analog Filter.

ω= Frequency of digital Filter.

Ts= Sampling Time.

Step 2: Calculate the order ‘N’ of Filter using the equation,

If the specifications are given in decibel (dB) then use,

Step 3: Calculation of Cut off Frequency (Ωc),

When ωc is not given then use the equation,

And if the specifications are in decibel (db) then,

Step 4: Calculate the poles using,

If the poles are complex conjugate then organize the poles (Pk) as complex conjugate pairs that means,

S1 and S1*,  S2 and S2*, etc.

Step 5: Calculate the system transfer function of analog filter using,

and if the poles are complex conjugate then,

Step 6: Design the digital filter by obtaining transfer function H(z) simply by substituting,

in the equation of H(s). We have to assume sampling time Ts= 1 if not given.

MATLAB CODE:

Consider a problem: Design a Butterworth IIR Filter using Bilinear Transformation by taking T= 1 sec and

Solution:

The Given Specifications are

Ap= 0.6                 omega_p= 0.35 π

As= 0.1                 omega_s= 0.7 π

clear all
clc
Ap=0.6;
As=0.1;
PEF_D=0.35*pi;
SEF_D=0.7*pi ;
T=1;
alpha_P=-20*log10(Ap);
alpha_S=-20*log10(As);
PEF_A=(2/T)*tan(PEF_D/2);
SEF_A=(2/T)*tan(SEF_D/2);
[N,la]=buttord(PEF_A,SEF_A,alpha_P,alpha_S,'s');
[Bn,An]=butter(N,1,'s');
display('Normalized transfer function is-');
Hsn=tf(Bn,An)
[B,A]=butter(N,la,'s')
display('Denormalized transfer function is-');
Hs=tf(B,A)
w=0:pi/16:pi
[num,den]=bilinear(B,A,1/T)
display('Digital Transfer function-');
Hz=tf(num,den,T)
display('Frequency response is-');
Hw=freqz(num,den,w)
display('Magnitude response is-');
Hw_mag=abs(Hw)
plot(w/pi,Hw_mag,'k');grid;
title('Magnitude response','fontweight','b');
xlabel('Normalized frequency,\omega/\pi','fontweight','b')
ylabel('Magnitude','fontweight','b')

Output:

Normalized transfer function is-

Hsn =

1

—————–

s^2 + 1.414 s + 1

Continuous-time transfer function.

Denormalized transfer function is-

Hs =

1.548

——————–

s^2 + 1.76 s + 1.548

Continuous-time transfer function.

Digital Transfer function-

Hz =

0.1708 z^2 + 0.3415 z + 0.1708

——————————

z^2 – 0.5407 z + 0.2237

Sample time: 1 seconds

Discrete-time transfer function.[/showhide]

What is Frequency Warping?

Because of the non-linear mapping, the amplitude response of digital IIR filter is expanded at lower frequencies and compressed at higher frequencies in comparison to the analog filter. This effect is called as frequency warping.

What is aliasing effect ?

Aliasing is an effect that causes different signals to become indistinguishable (or aliases of one another) when sampled. It also refers to the distortion or artifacts that result when the signal reconstructed from samples is different from the original continuous signal.


text:
## Quiver - take better notes

Developer: HappenApps, Inc.

Current Version: 3.1.1

Last Updated: 2 months ago

4.5

Description:

Quiver is a notebook built for programmers. It lets you easily mix text, code, Markdown and LaTeX within one note, edit code with an awesome code editor, live preview Markdown and LaTeX, and find any note instantly via the full-text search.

# Mix Text, Code, Markdown and LaTeX

A note in Quiver is comprised of cells --- snippets of text, code, Markdown, LaTeX (via MathJax) or diagrams (sequence diagram, flowchart). You can freely mix different cell types within one note. You can set different languages for different code cells, too.

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Review by Smanny1

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Found helpful by 3 out of 8 people

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5/5
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Review by _NM_

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text:
# 5.2 Infinite series  (Page 3/16)

 Page 3 / 16

Determine whether the series $\sum _{n=1}^{\infty }\left(n+1\right)\text{/}n$ converges or diverges.

The series diverges because the $k\text{th}$ partial sum ${S}_{k}>k.$

## The harmonic series

A useful series to know about is the harmonic series    . The harmonic series is defined as

$\sum _{n=1}^{\infty }\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\text{⋯}.$

This series is interesting because it diverges, but it diverges very slowly. By this we mean that the terms in the sequence of partial sums $\left\{{S}_{k}\right\}$ approach infinity, but do so very slowly. We will show that the series diverges, but first we illustrate the slow growth of the terms in the sequence $\left\{{S}_{k}\right\}$ in the following table.

 $k$ $10$ $100$ $1000$ $10,000$ $100,000$ $1,000,000$ ${S}_{k}$ $2.92897$ $5.18738$ $7.48547$ $9.78761$ $12.09015$ $14.39273$

Even after $1,000,000$ terms, the partial sum is still relatively small. From this table, it is not clear that this series actually diverges. However, we can show analytically that the sequence of partial sums diverges, and therefore the series diverges.

To show that the sequence of partial sums diverges, we show that the sequence of partial sums is unbounded. We begin by writing the first several partial sums:

$\begin{array}{l}{S}_{1}=1\hfill \\ {S}_{2}=1+\frac{1}{2}\hfill \\ {S}_{3}=1+\frac{1}{2}+\frac{1}{3}\hfill \\ {S}_{4}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}.\hfill \end{array}$

Notice that for the last two terms in ${S}_{4},$

$\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}.$

Therefore, we conclude that

${S}_{4}>1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)=1+\frac{1}{2}+\frac{1}{2}=1+2\left(\frac{1}{2}\right).$

Using the same idea for ${S}_{8},$ we see that

$\begin{array}{cc}\hfill {S}_{8}& =1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)\hfill \\ & =1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=1+3\left(\frac{1}{2}\right).\hfill \end{array}$

From this pattern, we see that ${S}_{1}=1,$ ${S}_{2}=1+1\text{/}2,$ ${S}_{4}>1+2\left(1\text{/}2\right),$ and ${S}_{8}>1+3\left(1\text{/}2\right).$ More generally, it can be shown that ${S}_{{2}^{j}}>1+j\left(1\text{/}2\right)$ for all $j>1.$ Since $1+j\left(1\text{/}2\right)\to \infty ,$ we conclude that the sequence $\left\{{S}_{k}\right\}$ is unbounded and therefore diverges. In the previous section, we stated that convergent sequences are bounded. Consequently, since $\left\{{S}_{k}\right\}$ is unbounded, it diverges. Thus, the harmonic series diverges.

## Algebraic properties of convergent series

Since the sum of a convergent infinite series is defined as a limit of a sequence, the algebraic properties for series listed below follow directly from the algebraic properties for sequences.

## Algebraic properties of convergent series

Let $\sum _{n=1}^{\infty }{a}_{n}$ and $\sum _{n=1}^{\infty }{b}_{n}$ be convergent series. Then the following algebraic properties hold.

1. The series $\sum _{n=1}^{\infty }\left({a}_{n}+{b}_{n}\right)$ converges and $\sum _{n=1}^{\infty }\left({a}_{n}+{b}_{n}\right)=\sum _{n=1}^{\infty }{a}_{n}+\sum _{n=1}^{\infty }{b}_{n}.$ (Sum Rule)
2. The series $\sum _{n=1}^{\infty }\left({a}_{n}-{b}_{n}\right)$ converges and $\sum _{n=1}^{\infty }\left({a}_{n}-{b}_{n}\right)=\sum _{n=1}^{\infty }{a}_{n}-\sum _{n=1}^{\infty }{b}_{n}.$ (Difference Rule)
3. For any real number $c,$ the series $\sum _{n=1}^{\infty }c{a}_{n}$ converges and $\sum _{n=1}^{\infty }c{a}_{n}=c\sum _{n=1}^{\infty }{a}_{n}.$ (Constant Multiple Rule)

## Using algebraic properties of convergent series

Evaluate

$\sum _{n=1}^{\infty }\left[\frac{3}{n\left(n+1\right)}+{\left(\frac{1}{2}\right)}^{n-2}\right].$

We showed earlier that

$\sum _{n=1}^{\infty }\frac{1}{n\left(n+1\right)}$

and

$\sum _{n=1}^{\infty }{\left(\frac{1}{2}\right)}^{n-1}=2.$

Since both of those series converge, we can apply the properties of [link] to evaluate

$\sum _{n=1}^{\infty }\left[\frac{3}{n\left(n+1\right)}+{\left(\frac{1}{2}\right)}^{n-2}\right].$

Using the sum rule, write

$\sum _{n=1}^{\infty }\left[\frac{3}{n\left(n+1\right)}+{\left(\frac{1}{2}\right)}^{n-2}\right]=\sum _{n=1}^{\infty }\frac{3}{n\left(n+1\right)}\underset{n=1}{\overset{\infty }{+\sum }}{\left(\frac{1}{2}\right)}^{n-2}.$

Then, using the constant multiple rule and the sums above, we can conclude that

$\begin{array}{ll}\sum _{n=1}^{\infty }\frac{3}{n\left(n+1\right)}+\sum _{n=1}^{\infty }{\left(\frac{1}{2}\right)}^{n-2}\hfill & =3\sum _{n=1}^{\infty }\frac{1}{n\left(n+1\right)}+{\left(\frac{1}{2}\right)}^{-1}\sum _{n=1}^{\infty }{\left(\frac{1}{2}\right)}^{n-1}\hfill \\ & =3\left(1\right)+{\left(\frac{1}{2}\right)}^{-1}\left(2\right)=3+2\left(2\right)=7.\hfill \end{array}$

Evaluate $\sum _{n=1}^{\infty }\frac{5}{{2}^{n-1}}.$

$10.$

## Geometric series

A geometric series    is any series that we can write in the form

$a+ar+a{r}^{2}+a{r}^{3}+\text{⋯}=\sum _{n=1}^{\infty }a{r}^{n-1}.$

Because the ratio of each term in this series to the previous term is r , the number r is called the ratio. We refer to a as the initial term because it is the first term in the series. For example, the series

how can chip be made from sand
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Leaves accumulate on the forest floor at a rate of 2 g/cm2/yr and also decompose at a rate of 90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?
You have a cup of coffee at temperature 70°C, which you let cool 10 minutes before you pour in the same amount of milk at 1°C as in the preceding problem. How does the temperature compare to the previous cup after 10 minutes?
Abdul


text:
### Find the integer $x$ that satisfies the equation $x^5 - 101 x^3 - 999 x^{ 2 } + 100900 = 0$.

$10$
1. We need to find the integer value of $x$ that satisfies the equation $x^5 - 101 x^3 - 999 x^{ 2 } + 100900 = 0.$
2. \begin{align} & x^5 - 101 x^3 - 999 x^{ 2 } + 100900 = 0 \\ \implies & x^5 - 101 x^3 - 999 x^{ 2 } + 100899 + 1 = 0 \\ \implies & x^3(x^{ 2 } - 101) - 999(x^{ 2 } - 101) + 1 = 0 \\ \implies & (x^{ 2 } - 101)(x^3 - 999) + 1 = 0 && \ldots (1) \\ \end{align}
3. We observe that the only integer that satisfies $(1)$ is $10.$
Hence, the value of $x$ is $10.$


text:
# THE ZEEMAN EFFECT IN THE OPTICAL SPECTRUM OF MANGANESE MONOHYDRIDE: MnH1

Please use this identifier to cite or link to this item: http://hdl.handle.net/1811/31396

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 Title: THE ZEEMAN EFFECT IN THE OPTICAL SPECTRUM OF MANGANESE MONOHYDRIDE: MnH1 Creators: Gengler, Jamie; Steimle, T. C.; Stoll, Michael Issue Date: 2007 Publisher: Ohio State University Abstract: The only successful demonstration of buffer-gas cooling and magnetic trapping of paramagnetic molecules was that performed by Doyle and co-workers in 1998} \textbf{395}, 148 (1998).}. It is anticipated that the relatively large rotational spacing and small spin-spin and spin-rotation interaction in the $X^7\Sigma^+$ (v = 0) state of MnH is a signature of a favorable ratio of elastic to inelastic collisions with helium atoms which is required for buffer-gas cooling and trapping. Furthermore, the large magnetic moment of the $N$ = 0, $X^7\Sigma^+$ (v = 0) state of MnH enables one to create a deep trap, and this allows a large enough trapping time to be able to isolate the sample from its environment by cryopumping the helium away. Here we report on the magnetic properties of MnH studied as a necessary preliminary to our use of it in buffer-gas cooling and magnetic trapping experiments. The Zeeman effect in the $P_{1}$(0) branch feature ($\nu$ = 17568.3536 cm$^{-1}$) of the $A^7\Pi - X^7\Sigma^+$ (0,0) band system of MnH was selected for Zeeman measurements. The field free spectra is complicated but has been thoroughly analyzed} \textbf{95}(3), 1563-1576 (1991).}'} \textbf{156}, 296-318 (1992).}'} \textbf{241}, 192 (2007).}. Description: Author Institution: Department of Chemistry and Biochemistry, Arizona State \\University, Tempe, AZ, USA 85287; Molecular Physics, Fritz-Haber-Institut der Max-Planck-Gesselschaft, Faradayweg 4-6, 14195 Berlin, Germany URI: http://hdl.handle.net/1811/31396 Other Identifiers: 2007-MF-11


text:
# Positive-real function

Positive-real functions, often abbreviated to PR function, are a kind of mathematical function that first arose in electrical network analysis. They are complex functions, Z(s), of a complex variable, s. A rational function is defined to have the PR property if it has a positive real part and is analytic in the right halfplane of the complex plane and takes on real values on the real axis.

In symbols the definition is,

{\displaystyle {\begin{aligned}&\Re [Z(s)]>0\quad {\text{if}}\quad \Re (s)>0\\&\Im [Z(s)]=0\quad {\text{if}}\quad \Im (s)=0\end{aligned}}}

In electrical network analysis, Z(s) represents an impedance expression and s is the complex frequency variable, often expressed as its real and imaginary parts;

${\displaystyle s=\sigma +i\omega \,\!}$

in which terms the PR condition can be stated;

{\displaystyle {\begin{aligned}&\Re [Z(s)]>0\quad {\text{if}}\quad \sigma >0\\&\Im [Z(s)]=0\quad {\text{if}}\quad \omega =0\end{aligned}}}

The importance to network analysis of the PR condition lies in the realisability condition. Z(s) is realisable as a one-port rational impedance if and only if it meets the PR condition. Realisable in this sense means that the impedance can be constructed from a finite (hence rational) number of discrete ideal passive linear elements (resistors, inductors and capacitors in electrical terminology).[1]

## Definition

The term positive-real function was originally defined by[1] Otto Brune to describe any function Z(s) which[2]

• is rational (the quotient of two polynomials),
• is real when s is real
• has positive real part when s has a positive real part

Many authors strictly adhere to this definition by explicitly requiring rationality,[3] or by restricting attention to rational functions, at least in the first instance.[4] However, a similar more general condition, not restricted to rational functions had earlier been considered by Cauer,[1] and some authors ascribe the term positive-real to this type of condition, while other consider it to be a generalization of the basic definition.[4]

## History

The condition was first proposed by Wilhelm Cauer (1926)[5] who determined that it was a necessary condition. Otto Brune (1931)[2][6] coined the term positive-real for the condition and proved that it was both necessary and sufficient for realisability.

## Properties

• The sum of two PR functions is PR.
• The composition of two PR functions is PR. In particular, if Z(s) is PR, then so are 1/Z(s) and Z(1/s).
• All the zeros and poles of a PR function are in the left half plane or on its boundary the imaginary axis.
• Any poles and zeroes on the imaginary axis are simple (have a multiplicity of one).
• Any poles on the imaginary axis have real strictly positive residues, and similarly at any zeroes on the imaginary axis, the function has a real strictly positive derivative.
• Over the right half plane, the minimum value of the real part of a PR function occurs on the imaginary axis (because the real part of an analytic function constitutes a harmonic function over the plane, and therefore satisfies the maximum principle).
• For a rational PR function, the number of poles and number of zeroes differ by at most one.

## Generalizations

A couple of generalizations are sometimes made, with intention of characterizing the immittance functions of a wider class of passive linear electrical networks.

### Irrational functions

The impedance Z(s) of a network consisting of an infinite number of components (such as a semi-infinite ladder), need not be a rational function of s, and in particular may have branch points on the negative real s-axis. To accommodate such functions in the definition of PR, it is therefore necessary to relax the condition that the function be real for all real s, and only require this when s is positive. Thus, a possibly irrational function Z(s) is PR if and only if

• Z(s) is analytic in the open right half s-plane (Re[s] > 0)
• Z(s) is real when s is positive and real
• Re[Z(s)] ≥ 0 when Re[s] ≥ 0

Some authors start from this more general definition, and then particularize it to the rational case.

### Matrix-valued functions

Linear electrical networks with more than one port may be described by impedance or admittance matrices. So by extending the definition of PR to matrix-valued functions, linear multi-port networks which are passive may be distinguished from those that are not. A possibly irrational matrix-valued function Z(s) is PR if and only if

• Each element of Z(s) is analytic in the open right half s-plane (Re[s] > 0)
• Each element of Z(s) is real when s is positive and real
• The Hermitian part (Z(s) + Z(s))/2 of Z(s) is positive semi-definite when Re[s] ≥ 0

## References

1. ^ a b c E. Cauer, W. Mathis, and R. Pauli, "Life and Work of Wilhelm Cauer (1900 – 1945)", Proceedings of the Fourteenth International Symposium of Mathematical Theory of Networks and Systems (MTNS2000), Perpignan, June, 2000. Retrieved online 19 September 2008.
2. ^ a b Brune, O, "Synthesis of a finite two-terminal network whose driving-point impedance is a prescribed function of frequency", Doctoral thesis, MIT, 1931. Retrieved online 3 June 2010.
3. ^ Bakshi, Uday; Bakshi, Ajay (2008). Network Theory. Pune: Technical Publications. ISBN 978-81-8431-402-1.
4. ^ a b Wing, Omar (2008). Classical Circuit Theory. Springer. ISBN 978-0-387-09739-8.
5. ^ Cauer, W, "Die Verwirklichung der Wechselstromwiderst ände vorgeschriebener Frequenzabh ängigkeit", Archiv für Elektrotechnik, vol 17, pp355–388, 1926.
6. ^ Brune, O, "Synthesis of a finite two-terminal network whose driving-point impedance is a prescribed function of frequency", J. Math. and Phys., vol 10, pp191–236, 1931.


text:
Episode #125 of the Stack Overflow podcast is here. We talk Tilde Club and mechanical keyboards. Listen now

# Tag Info

0

If the $n$-factor is what I assume it is (the basicity, or ‘how many moles of sodium hydroxide can one mole of acid neutralise?’) then … It is really hard to explain this without molecular structures. Sum formula are of little help to chemists once you reach more than four or five atoms or if there are two types of non-hydrogen atoms. Two compounds may have ...

0

I am afraid the answer if your teacher or textbook has told that the n-factor of phosphoric acid is 2, they are teaching it the wrong way. The concept of equivalent weight is totally dependent on how the reaction equation is written, therefore there is no fixed equivalent weight. Think of the permanganate ion- how many equivalent weights does it have! I ...

1

$n$-Factor is basically a method to find out a relationship between the compound and what it is equivalent to in terms of acidic nature or basic nature. Note that the two hydrogen atoms in $\ce{H2SO4}$ are both attached to oxygen. But in $\ce{H3PO3}$ two of the three hydrogens are attached to oxygen and the other hydrogen is directly attached to phosphorous. ...

2

It reacted with the baking soda and is no longer sulfuric acid, but much safer sodium sulfate. If you're unsure, add more baking soda until it no longer fizzes. The remaining white powder is a mix of baking soda or $\ce{NaHCO3}$, and sodium sulfate, $\ce{Na2SO4}$. It safe to wipe up with a damp rag and fairly harmless to touch. Just avoid getting the ...

0

The indicator may slowly disappear after the titration because it is reacting with the air. As it reacts with chemicals in the air (CO 2 for example) which causes the indicator to be slowly disappear from the solution.

2

$\ce{HSO4-}$ is acidic ($\mathrm pK_\mathrm a = 1.99$) yet also reacts as a nucleophile e.g. in the acid-catalysed hydration of olefins, which according to my mechanism lecture procedes via a sulphate diester intermediate. If I thought for long enough, I probably could come up with a basic compound that acts as an electrophile in a given situation. However,...

3

More resonance structures does not necessarily mean more stabilization. Phenolate has a negative charge at an oxygen atom which cannot be shared without "disturbing" the aromaticity of the ring. This is in part because in the resonance structures drawn, there are negative charged carbons within the aromatic ring, which either implies that: The ring is ...

1

Using the definition of $\mathrm{pH}$ (which is a log value) i.e. the difference of $\mathrm{pH}~1$ value is 10 folds difference in real terms. You have $\pu{100 ml}$ of $\ce{NaOH}$ @ $\mathrm{pH}~13.$ If you dilute by $10$ times, then the new $\mathrm{pH}$ will be $12,$ so you'll need to dilute it another $10$ times to reach $\mathrm{pH}~11.$ If you start ...

1

$$\ce{Mn^{2+} + Cr2O7^{2-} -> Mn^{3+} + Cr^{3+}}$$ If you were to balance this reaction, using the appropriate molecules of water and protons, you'd get: $$\ce{ Mn^{2+} + Cr2O7^{2-} + 14H+ + 5e- -> Mn^{3+} + 2Cr^{3+} + 7H2O }$$ The presence of protons the left side of equation suggests that this reaction has a higher rate when the concentration of ...

-2

In this solution, there are two sorts of H+ ions, those coming from added HCl, whose concentration is 10-9 M, and those coming from the water which are unknown, ans so described by x. But the concentration of OH- is also x. So Kw may be written : Kw = (x + 10-9)x = 10-14. If you solve this equation, you find : x = 0.999·10-7 The total concentration of H+ ...

1

Most of $\mathrm{pH}$ calculations make simplifying assumptions and one must determine if they are justified for the particular case. Simplification for strong acids $$\mathrm{pH}=-\log{[\mathrm{acid}]}$$ implies $\pu{[\mathrm{acid}] >> 10^{-7} mol/L}$. Because water dissociation constant $K_\mathrm{w}=\ce{[H+][OH-]}=\pu{10^{-14} mol^2/L^2}$, ...

1

The article text itself does go into details. The first thing to note: while aromaticity on the larger scale is retained in your structures 3 and 4, this is only true for the 10-π-electron system as a whole. In many of these types of 10 π systems made up of a six-membered and a five-membered ring, the two rings are not equal with the double bond ...

1

Chromic acid can well be stored in glass bottles with stopper/lid but should be kept away from any organic compounds. Chromic acid is a dangerous oxidising agent and incompatible with various organic compounds. Also, it should be well marked with proper label and hazard symbol to distinguish it from other mineral acids. It has a decent shelf life and can be ...

2

There are a so-called acid error and an alkaline error with glass electrodes. Which means that the pH electrode shows a systematic bias when your solution is too acidic or too basic and if it contains lot of sodium ions. The value which you quote for +13 are unreliable. So if you are bubbling carbon dioxide into NaOH solution, you are indeed consuming NaOH ...

1

It seems the most used way is reaction of sulphuric acid with salts of volatile mineral acids : $$\ce{H2SO4 + NaCl -> NaHSO4 + HCl ^}$$ Reaction with excess of the acid $$\ce{H2SO4 + NaOH ->[H2SO4] NaHSO4 +H2O}$$ has several drawbacks: It is much more exothermic than the former one It releases extra water It needs somewhat diluted solution not to ...

2

This is not the answer OP would expect because I couldn't find metal complexes of type $\ce{[M(H3O)_x]^{y+}}$ (the reason for non-existence(?) is mentioned in the comments) but I did find a complex of an organic compound where hydronium ion act as a ligand i.e a complex of 18-crown-6 and hydronium ion(emphasis mine): The observation that cyclic crown ...

3

In strongly basic solutions, phenolphthalein is converted to its $\ce{In(OH)^3-}$ form, and its pink color undergoes a rather slow fading reaction and becomes completely colorless above $\mathrm{pH}= 13$. wikipedia So the strong alkali is very probable reason, but bleachers or other compounds reacting with phph cannot be eliminated.

5

To answer this question, we have to assume that the given buffer solution is an ideal solution of which the total volume of it always equals to the summed volumes of sodium ethanoate ($\ce{NaOAc}$) and $\ce{HCl}$ solutions when they are added together. It should also be assumed that $\mathrm{p}K_\mathrm{a}$ of ethanoic acid (acetic acid; $\ce{HOAc}$) is ...

1

TLDR: Sulfuric acid act as catalyst in the isomerization reaction of parafin hydrocarbons and sometimes it produce undesired sideproducts like sulfonate salts, napthenic acid, bisulfite ions etc. from incomplete oxidation. It follows "hydrogen exchange mechanism. Long answer: @andselisk cited some papers that shows that sulfuric acid can act as a catalyst ...

0

Salts do not have any idea they are in a buffer solution. They interact with $\ce{H2O, H+, OH- }$ according to their respective equilibrium constants and $\mathrm{pH}$ of solution, that determine the concentration ratio of the respective conjugated acid/base pairs.

1

Water autohydrolyzes, or autodissociates, or autoionizes, to produce a very small amount of H+ and OH- ions. The product of their concentrations is 10e-14. Most salts that dissolve in water dissociate into ions. Often, one of these ions will have a strong attraction for H+ or OH- and tends to reduce the concentration of that ion in the solution. But since ...

1

At $\mathrm{pH}=\mathrm{p}K_\mathrm{a}=3.75$ , there will be 50% of non-dissociated formic acid and 50% of formiate. You need 50 % of stoichiometric amount of hydrochloric acid to convert 50% of the formiate to the acid. If we consider the percentage as (w/w) %, then $\pu{1 kg}$ of 75% formic acid ($M=\pu{46 g/mol}$) contains $\pu{16.3 mol}$ of formic ...

5

As the significance of the two contrasting factors cannot be easily evaluated, let us first look at some evidence. Wan, Modro & Yates (1980) reported the $\ce {pK_a}$ values of the conjugate acids of the various lactams, as shown in the table below: \begin{array}{|c|c|c|c|} \hline \text{Lactam} & pK_a\\ \hline \ce{$\beta$, 4} & -0.96\\ \hline \...

2

In simplistic terms, Arrhenius acids are molecules while Arrhenius bases are ionic compounds. Much like $\ce{BaCl2}$ – another ionic compound – it is a good first rule of thumb to assume it completely dissociates into its respective ions: one $\ce{Ba^2+}$ and two single-charge anions. You could say that barium were never really connected to either hydroxide; ...

-1

In3,5 dimethyl 4 nitro phenol , the nitro group is between two ortho methyl groups . Therefore due to this steric inhibition of resonance occurs and this nitro group is pushed out of plane by two methyl groups. Such that nitro group does not take part in the resonance. And there have no extra stable structure. That is why it is less acidic than2,6 dimethyl 4 ...

2

I think OP is confused with the colorimetric method of detection of piperine, which uses the concentrated sulfuric acid as one of reactant with chromotropic acid, 1,8-dihydroxynaphthalene-3,6-disulfonic acid (Chromotropic Acid Test; Ref.1 & 2). Accordingly, this colorimetric method, developed in 1965, relied on hydrolyzing the methylenedioxy group of ...

1

The explanation given to you is rather incorrect. Dilute sulfuric acid is added to iron(II) or even iron(III) salts is added to prevent hydrolysis of the salts. What happens when you don't add an acid to iron(II) solution in water. With time, you will see that a brown precipitate is forming and settling to the bottom. Iron forms hydroxides which are very ...

1

Resonance structures can only be drawn for single compounds. Rephrased: if a set of resonance structures are to be valid, then all atoms (nuclei) must be at exactly the same positions. Thus, when asked to draw resonance structures of e.g. phenol or nitric acid, you cannot deprotonate or transfer protons. All protons must be exactly at the same spot in all ...

Top 50 recent answers are included


text:
# Plz help....meeeee induction

• Nov 14th 2006, 07:40 PM
jenjen
Plz help....meeeee induction
I am studying for my 2nd midterm, but then I am stuck..plz help me..thank you so much.

1) prove by induction that for each nonnegative integer k, the integer k^2 + 5k is even.

2) The number n!, which is the number of permutations of the first n positive integers, is defined recursively by 0! = 1 and n! = n X (n-1)! for all n > o. Prove that n! < n ^n for all n > 0 and strict inequality holds if n > 1.
• Nov 14th 2006, 07:46 PM
ThePerfectHacker
Quote:

Originally Posted by jenjen
1) prove by induction that for each nonnegative integer k, the integer k^2 + 5k is even.

I let somebody else do this problem. It is just boring.
Quote:

2) The number n!, which is the number of permutations of the first n positive integers, is defined recursively by 0! = 1 and n! = n X (n-1)! for all n > o. Prove that n! < n ^n for all n > 0 and strict inequality holds if n > 1.
We see that,
$2!< 2^2$
Good.

Next there is a $k$ such as,
$k!

But,
$k!
Multiply by both sides by $k+1>0$
Thus,
$(k+1)!<(k+1)^k(k+1)=(k+1)^{k+1}$
Q.E.D.
• Nov 14th 2006, 07:49 PM
topsquark
Quote:

Originally Posted by jenjen
1) prove by induction that for each nonnegative integer k, the integer k^2 + 5k is even.

Let k = 0. Then $0^2 + 5 \cdot 0 = 0$ is even.

Let the theorem be true for some k = n.

Then we need to show that the theorem is true for k = n + 1. That is:
$(n+1)^2 + 5(n+1)$ is even.

So
$(n+1)^2 + 5(n+1) = n^2 + 2n + 1 + 5n + 5$ $= (n^2 + 5n) + (2n + 6) = (n^2 + 5n) + 2(n + 3)$

By hypothesis $n^2+5n$ is even. The second term 2(n + 3) is a number multiplied by 2, which is even by definition. Thus the problem is in the form of an even number plus an even number, which is always even.

-Dan
• Nov 14th 2006, 08:25 PM
jenjen
thank you so much ThePerfectHacker and Topsquark for the quick replies!!!!!


text:
# How many queries can you do (concretely, not asymptotically) for a balanced Feistel network with 7 rounds?

In this paper Patarin says that: "for every $$\epsilon > 0$$, when $$m \ll 2^{n(1 - \epsilon)}$$ ... for 7 rounds or more it is secure against all adaptive chosen plaintext attacks" where m is the number of queries that the adversary can evaluate.

What concretely is meant by $$m \ll 2^{n(1 - \epsilon)}$$?

For instance, to have statistical security $$2^{-\sigma}$$ (e.g. $$\sigma = 40$$) concretely how many queries can be evaluated?

• define statistical security – kodlu Apr 25 '20 at 4:15
• How would you even define << concretely? It's sort an asymptotic idea. – Adrian Self Apr 26 '20 at 1:00
• @kodlu Where statistical security is the advantage of an adversary has in the Chosen Plaintext Attack game, when they are able to make $m$ queries and have unlimited computational resources. – Daniel-耶稣活着 Apr 26 '20 at 2:24
• ok so you want the advantage to be upper bounded by $2^{-40}.$ – kodlu Apr 26 '20 at 3:22
• @kodlu Yes, exactly. – Daniel-耶稣活着 Apr 26 '20 at 3:32

The linked paper is missing a lot of proof details. In any case, it seems impossible to say anything concrete for finite $$n$$ due to the existence of terms in $$O(\cdot)$$ notation in the bounds. You simply do not know how large those implied constants are.
• Yeah--the missing proof details is a lot of the problem... In principle if the details of the proof were there I should be able to go through and figure out what the constants are in the $\mathcal{O}()$ notation. But since they are omitted, I was hoping someone out there might know of a place that someone had gone through and calculated the constants. Or if not, a place where I can find the full proof so I can go through and try to figure out the constants myself. – Daniel-耶稣活着 Apr 26 '20 at 2:28


text:
# Breathing Earth - visual display of world polluters

By Murray Bourne, 19 Aug 2007

Breathingearth.net has a nice simulation of the most polluting countries.

The legend is as follows:

It reminds me of my own World Population simulation, where you can get a sense of population growth in the past, and future.

See the 3 Comments below.

### 3 Comments on “Breathing Earth - visual display of world polluters”

1. Sierra says:

I think that that breathing map is really scary
because it shows just how stupid humanity is
to let it get this far and to do nothing about it. Making Hybrid cars? yea why don't the buisness and manufactures stop making everything else for about 100 years. Then maybe everyone won't die!!!

2. Murray says:

Hi Sierra and thanks for your comment.

We have to do something about reducing energy demand - and quick.

You are right:

"how stupid humanity is"

3. Bizz says:

We are so intelligent yet so naive to be so irresponsible and inconsiderate. May god forgive us for we do not know what we are doing.

### Comment Preview

HTML: You can use simple tags like <b>, <a href="...">, etc.

To enter math, you can can either:

1. Use simple calculator-like input in the following format (surround your math in backticks, or qq on tablet or phone):
a^2 = sqrt(b^2 + c^2)
(See more on ASCIIMath syntax); or
2. Use simple LaTeX in the following format. Surround your math with $$ and $$.
$$\int g dx = \sqrt{\frac{a}{b}}$$
(This is standard simple LaTeX.)

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text:
# Which Hadamard Products of Generating Functions Are Known?

The Hadamard product, Schur product, or entrywise product of two generating functions is computed as follows:

The Hadamard Product, H(x), given two generating functions f(x) and g(x) where

$$f(x) = c_0 + c_1x + c_2x^2 + c_3x^3 + \dots + c_nx^n + \dots$$ $$g(x) = d_0 + d_1x + d_2x^2 + d_3x^3 + \dots + d_nx^n + \dots$$

is defined as

$$H(x) = c_0d_0 + c_1d_1x + c_2d_2x^2 + c_3d_3x^3 + \dots + c_nd_nx^n + \dots$$

Simply put, it is the result of multiplying individual coefficients of two generating functions.

Results detailing certain types of functions are known. I would like to compile a list of results that are known.

For example, if we have two generating functions of the form $(1-x)^a$ and $(1-x)^b$, we obtain: ${}\_2F_1[-a,-b;1;x]$ where ${}\_2F_1$ represents a hypergeometric function of Gauss, according to

"Singularity Analysis, Hadamard Products, and Tree Recurrences", by Jim Fill, Philippe Flajolet, and Nevin Kapur In Journal of Computational and Applied Mathematics, volume 174 (February 2005), pages 271--313 (around page 289)

Again, which Hadamard products involving generating functions are known or solved? Additionally, solutions to the equations would be greatly appreciated.

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I'm not sure what you mean by "known." There are lots of closure results about classes of generating functions closed under Hadamard product - for example, rational and holonomic (D-finite) generating functions have this property. Is this the kind of result you're looking for? –  Qiaochu Yuan Feb 18 '10 at 7:07
A note to anyone who wants to write an answer: based on my correspondence with the OP, what he is really interested in is ways to compute the Hadamard product mod 2. –  Qiaochu Yuan Feb 19 '10 at 2:56
Not only is the holonomic class 'known', it is algorithmic with polynomial-time algorithms (essentially all linear algebra). They all have, for example, Maple implementations. So, if you can write down 2 generating functions, their Hadamard product can be computed, at least to the point of giving an explicit linear recurrence for the coefficients, which can be (very efficiently) unwound. That is as 'solved' as one could ask for, no? –  Jacques Carette Feb 19 '10 at 4:22
Again from my correspondence with the OP, he is actually only interested in the Hadamard products of (large) polynomials with 0-1 coefficients. –  Qiaochu Yuan Feb 19 '10 at 17:13
I would prefer to see the results of entire functions as a whole as opposed to equations involving coefficients. There are results on the web, but I thought it would be better to attempt to organize them in one place. Sorry if I was misleading. –  Generator Feb 22 '10 at 3:23