text:
# Averaged dynamics and control for heat equations with random diffusion

Bárcena J.A., Zuazua E.. Averaged dynamics and control for heat equations with random diffusion (2021) Syst. Control. Lett. Vol. 158. https://doi.org/10.1016/j.sysconle.2021.105055

Abstract. This paper deals with the averaged dynamics for heat equations in the degenerate case where the diffusivity coefficient, assumed to be constant, is allowed to take the null value. First we prove that the averaged dynamics is analytic. This allows to show that, most often, the averaged dynamics enjoys the property of unique continuation and is approximately controllable. We then determine if the averaged dynamics is actually null controllable or not depending on how the density of averaging behaves when the diffusivity vanishes. In the critical density threshold the dynamics of the average is similar to the \frac{1}{2} -fractional Laplacian, which is wellknown to be critical in the context of the controllability of fractional diffusion processes. Null controllability then fails (resp. holds) when the density weights more (resp. less) in the null diffusivity regime than in this critical regime.

Read Full Paper


text:
# Find the value of k, if x – 1 is a factor of p(x) in the following case:- p(x) = 2x^2+kx+√2 - Mathematics

Find the value of k, if x – 1 is a factor of p(x) in the following case:-

p(x) = 2x^2+kx+sqrt2

#### Solution

If x − 1 is a factor of polynomial p(x), then p(1) must be 0.

p(x) = 2x^2+kx+sqrt2

p(1) = 0

⇒ 2(1)^2 + k(1) + sqrt2 = 0

⇒ 2 + k + sqrt2 = 0

⇒ k = -2 -sqrt2 = -(2+sqrt2)

Therefore, the value of k is -(2+sqrt2)" ."

Concept: Factorising the Quadratic Polynomial (Trinomial) of the type ax2 + bx + c, a ≠ 0.
Is there an error in this question or solution?

#### APPEARS IN

NCERT Class 9 Maths
Chapter 2 Polynomials
Exercise 2.4 | Q 3.2 | Page 44

format: plain


text:
# Make an Array Reactive (Observable)

I have written a simple JavaScript function to make an array reactive (i.e. observable). Is there anything I could improve? What the code does is take an array, a callback and a this value as arguments, then whenever the array gets updated or a value gets added to it, it runs the callback with the specified this value. It works similar to the obsolete Array.observe().

function arrayObserve(array, callback, thisArg){
var result = [];
result.push.apply(result, array);
function getterAndSetter() {
if (Array.isArray(array))
var currentArray = result;
else
var currentArray = Object.keys(result);
console.log(result);
currentArray.forEach( function(prop){
if (Array.isArray(result))
var currentProp = result.lastIndexOf(prop);
else
var currentProp = prop;

Object.defineProperty(array, currentProp, {
configurable: true,
enumerable: true,
set: function(val) {
result[currentProp] = val;
callback.call(thisArg, result);
},
get: function() {
return result[currentProp];
}
});
//}
if (typeof result[currentProp] == 'object') {
arrayObserve(result[currentProp], callback, result);
}
});
}
getterAndSetter();
['pop','push','reverse','shift','unshift','splice','sort','map','filter'].forEach( function(method){
Object.defineProperty(array, method, {
configurable: false,
enumerable: false,
writable: true,
value: function () {
var noReturnMethods = ['push', 'pop', 'reverse', 'shift', 'unshift'];
if (noReturnMethods.indexOf(method) > -1) {
Array.prototype[method].apply(result, arguments);
} else {
var results = Array.prototype[method].apply(result, arguments);
result = results;
}
callback.call(thisArg, result);
getterAndSetter();
return result;
}
});
});
return result;
}

With this block:

if (typeof result[currentProp] == 'object') {
arrayObserve(result[currentProp], callback, result);
}

It appears that the code is handling nested arrays. Is that supposed to work with nested objects (A.K.A. associative arrays)? If not, then perhaps it would be better to use Array.isArray() like the start of getterAndSetter().

The if-else logic within getterAndSetter() doesn't put the statements in a block (i.e. in curly brackets - { and }), whereas the if-else logic within the value function does put the statements in blocks. For the sake of readability (especially for others) it is best to be consistent with curly brackets style.

In the else block of the value function, when a return method is called:

var results = Array.prototype[method].apply(result, arguments);
result = results;

is results used to avoid side-effects of the method (e.g. splice)? If not, then there isn't really a need to use results - just re-assign result with the return value from calling apply on the method.

Perhaps you are already aware of this, but if I add a new element to an array using bracket notation after arrayObserve has already been called with the array as the first parameter, then the callback is not called - should that be the case?

var arr = [3, 5, [1]];
arrayObserve(arr, console.log.bind(null,'callback - arguments:'));
arr[3] = 4;

### Edit

One other thing to consider is using Array.from() when creating result, unless a shallow-copy won't work. That could simplify the following first two lines of arrayObserve():

var result = [];
result.push.apply(result, array);

To this:

var result = Array.from(array);

If a deep-copy is desired, then the trusty JSON methods can be utilized:

var result = JSON.parse(JSON.stringify(array));

• I have accepted this answer because it helped me the most, thank you for the answer. – Jacques Marais Jul 10 '18 at 16:55

Object.defineProperty(array, method, {
configurable: false,
enumerable: false,
writable: true,
value: function () {
// ...

Here, configurable and enumerable default to false anyway. You needn't set them explicitly.

if (noReturnMethods.indexOf(method) > -1) {
Array.prototype[method].apply(result, arguments);
}

With ES6, you can write the if condition more concisely as noReturnMethods.includes(method).

You could improve readability by storing ['pop','push','reverse','shift','unshift','splice','sort','map','filter'] in a variable. You could probably do

var noReturnMethods = ['push', 'pop', 'reverse', 'shift', 'unshift'];
var returnMethods = ['splice','sort','map','filter'];
var methods = [];
methods = methods.push.apply(noReturnMethods, returnMethods);

... to avoid duplication of the names of methods.

With ES6, you could do methods = [...noReturnMethods, ...returnMethods].

Assuming your environment supports ES6, you should probably make use of arrow functions and the let and const keywords.

• Thanks for the answer. My environment doesn't necessarily support ES6, but will keep those in mind. If it did, I would probably use Proxies to achieve this. – Jacques Marais Jul 8 '18 at 19:56


text:
# Cartesian Product

The cartesian product of n sets is the set of ordered n-tuples containing one element from each set.

For example, the cartesian product of the sets {a, b}, {p, q, r} and {w, x, y, z} is:

[a, p, w]
[a, p, x]
[a, p, y]
[a, p, z]
[a, q, w]
[a, q, x]
[a, q, y]
[a, q, z]
[a, r, w]
[a, r, x]
[a, r, y]
[a, r, z]
[b, p, w]
[b, p, x]
[b, p, y]
[b, p, z]
[b, q, w]
[b, q, x]
[b, q, y]
[b, q, z]
[b, r, w]
[b, r, x]
[b, r, y]
[b, r, z]

The algorithm uses an array of n integers. Each number in the array represents the index number within the corresponding set of that element.

The algorithm then simply needs to count upwards from n zeroes, with the cardinality of each set determining when a digit rolls over.

For example, these are the correspondences between the array and the n-tuple for the sets in the example:

[0, 0, 0] = [a, p, w]
[0, 0, 1] = [a, p, x]
[0, 0, 2] = [a, p, y]
[0, 0, 3] = [a, p, z]
[0, 1, 0] = [a, q, w]
[0, 1, 1] = [a, q, x]
[0, 1, 2] = [a, q, y]
[0, 1, 3] = [a, q, z]
[0, 2, 0] = [a, r, w]
[0, 2, 1] = [a, r, x]
[0, 2, 2] = [a, r, y]
[0, 2, 3] = [a, r, z]
[1, 0, 0] = [b, p, w]
[1, 0, 1] = [b, p, x]
[1, 0, 2] = [b, p, y]
[1, 0, 3] = [b, p, z]
[1, 1, 0] = [b, q, w]
[1, 1, 1] = [b, q, x]
[1, 1, 2] = [b, q, y]
[1, 1, 3] = [b, q, z]
[1, 2, 0] = [b, r, w]
[1, 2, 1] = [b, r, x]
[1, 2, 2] = [b, r, y]
[1, 2, 3] = [b, r, z]

unsigned int next_n_tuple(unsigned int *ar, size_t len, const size_t *sizes)
{
unsigned int changed = 0;
unsigned int finished = 0;
unsigned int i;

for (i = len - 1; !changed && !finished; i--) {
if (ar[i] < sizes[i] - 1) {
/* Increment */
ar[i]++;
changed = 1;
}
else {
/* Roll over */
ar[i] = 0;
}
finished = i == 0;
}

return changed;
}


text:
# Find all the elements in Dih(2n), n odd, which commute with all other elements

http://crazyproject.wordpress.com/2010/01/08/find-all-the-elements-in-dih2n-n-odd-which-commute-with-all-other-elements/

What is the “$q$” in the answer? And what "$0\leq a_0$" means? It's a typo. And I don’t understand the last 2 sentences: If $q≥2$ we have $2a<2n ≤ qn$, a contradiction. If $q=1$, we have $n=2n$ even, a contradiction.

-
 Under the hypothesis that $x$ is a power of $r$, then $a$ must be $0\leq a\leq n-1$. – Sigur Jan 23 at 0:49 That page is pretty bad. I wouldn't worry about it. – Gerry Myerson Jan 23 at 0:58 who can help me understand what the proof in that page means? – i_a_n Jan 23 at 4:20 Alexander has supplied a better proof. If you are not familiar with Alexander's notation $[x,y]$, it just means $x^{-1}y^{-1}xy$, and it yields the identity element if and only if $x$ commutes with $y$. – Gerry Myerson Jan 24 at 0:04

I don't think that proof is very good, so I will explain the general case (for arbitrary $n$). Then if you like you can go back and figure out what the link is doing.
First, the presentation of $D_{2n}$ provides $r^s=r^{-1}$. Thus we have $[r,s]=r^{-2}=r^{n-2}$, which, since $rs=sr[r,s]$, tells us that any given word in $r$ and $s$ can be written in the form $s^ar^b$, where $a=0$ or $1$ and $b=0,\ldots,n-1$. (Why?) Thus any $g\in D_{2n}$ has the form $g$ has the form $g=s^ar^b$.
You want to find which elements $z=s^{x}r^{y}$ so that $[z,g]=1$ for all $g\in D_{2n}$. Let's calculate: $$\begin{eqnarray*} [z,g]&=&z^{-1}g^{-1}zg\\ &=&(s^xr^y)^{-1}(s^ar^b)^{-1}(s^xr^y)(s^ar^b)\\ &=&r^{-y}s^{-x}r^{-b}s^{-a}s^xr^ys^ar^b\\ &=&r^{-y}s^xr^{-b}s^as^xr^ys^ar^b\\ \end{eqnarray*}$$ Note here that if $x=0$, $$\begin{eqnarray*} &=&r^{-y}r^{-b}s^ar^ys^ar^b\\ \end{eqnarray*}$$ Obviously this is $1$ when $a=0$. When $a\not=0$, $$\begin{eqnarray*} &=&r^{-y}r^{-b}sr^ysr^b\\ &=&r^{-y}r^{-b}r^{-y}r^b\\ &=&r^{-y-b-y+b}\\ &=&r^{-2y}\\ \end{eqnarray*}$$ which is $1$ only when $y=0$ or $n$ is even and $y=n/2$. A similar calculation for the $x=1$ case (which you should do) yields an expression which cannot be equal to $1$ for an arbitrary $x$ and $y$. Thus the central elements of $D_{2n}$ are $1$ and $r^{n/2}$ if $n$ is even, or only $1$ if $n$ is odd.


text:
# Math Scraps: 2 3 4 5 + =

I had an equation that made perfect mathematical sense, but my dog ripped it to shreds. Fortunately, I've managed to recover all of the pieces to the equation, but I can't seem to remember how they were arranged.

Everything in the braces below must be used in reconstructing the formula (other than the braces themselves). You can arrange them in any way you want, provided that you don't change their actual shape or orientation (for instance, a + is not the same as a x). What was the original equation?

{ 2 3 4 5 + = }

$3^2 = 4+5$
$\frac{4}{2} + 3 = 5$
$3 + 2 = 5$
(Otherwise $3 + 2_4 = 5$ works, where 2 is in base 4)


text:
# Is the sequence $f_n(x)=\sqrt{n}\chi_{[0,1/n]}(x)$ uniformly integrable? prove or disprove

If $X=[0,\infty)$, $m$ is the collection of Lebesgue measurable sets and $\mu$ is Lebesgue measure. Prove or disprove whether the sequence $$f_n(x)=\sqrt{n}\chi_{[0,1/n]}(x)$$ is uniformly integrable.

At a glance, my judgement is No, it is NOT uniformly integrable.

By definition,

Given a measure space $(X,m,\mu)$, A sequence $\{f_n\}_{n=1}$ is said to be uniformly integrable if $\forall \epsilon >0$ $\exists \delta > 0$ such that given $E \subset X$ ,measurable, if $\mu (E)< \delta$ then $\forall n$, $\int_E |f_n| d\mu < \epsilon$.

So I let $E \subset R$ such that $\mu (E)<\delta$. Hence I wish to show that $\int_E|f_n|<\epsilon$ $\forall n$

BUT $$\int_E|f_n|=\int_E \sqrt{n}\chi_{[0,1/n]} =\sqrt{n}\cdot \mu (E\cap[0,1/n])\leq\sqrt{n} \cdot min(\delta,1/n)$$

And Since $\sqrt{n} \delta$ is unbounded as $n \to \infty$, we conclude that the sequence of functions is not uniformly integrable

Does this make sense?

• You give away too much in passing from $\int_E\sqrt{n}\chi_{[0,1/n]}$ to $\sqrt{n}\mu(E)$. A sharper upper bound would be $\sqrt{n}\mu(E\cap[0,1/n])\le\sqrt{n}\min(\delta,1/n)$. – John Dawkins Jul 11 '17 at 21:14
• Any sequence $f_n$ in $L^1$ such that $\|f_n\|_1 \to 0$ is uniformly integrable. – zhw. Jul 11 '17 at 21:14
• What @zhw. said, or: bounded in $L^2$. – Did Jul 11 '17 at 21:21
• @zhw. very helpful. – J. Kyei Jul 11 '17 at 21:31
• @JohnDawkins. You're right, I jumped a couple relevant steps. But if I get you well, even though $\sqrt{n}min(\delta,1/n)$ depends on $n$, we can still conclude that it is less than some $\epsilon$? I thought because it's dependent on $n$ it increases past any $\epsilon$ and hence not uniformly integrable – J. Kyei Jul 11 '17 at 21:37

$$(i) \int|f_n|dm=\int_{[0,\frac{1}{n}]}\sqrt{n}=\frac{\sqrt{n}}{n} \le 1$$ $$(ii) \text{For each } A, m(A) \lt \delta, \int_A f_n=\int_{A \cap [0,\frac{1}{n}]}\sqrt{n}=\sqrt{n}m\left(A\cap\left[0,\frac{1}{n}\right]\right)\le \frac{1}{\sqrt{n}}$$
From above two, it can be seen that $\{f_n\}$ is uniformly integrable
• thank you. But Isn't our $\epsilon$ supposed to be a constant? – J. Kyei Jul 11 '17 at 21:39
• I don't understand what you are asking. If you want to choose $\delta$, you can take $\delta=\epsilon$ always in this case. – tattwamasi amrutam Jul 11 '17 at 21:40
• What I wanted to mean is that, Assume $\epsilon =1/2$. then we will wish to show that suppose $\exists \delta >0$ such that if $\mu (E)<\delta$ then $\int_E |f_n|<1/2$ $\forall n$. Now if we choose $n$ large enough so that $1/n=\mu ([,1/n]) <\delta$ then I should have my result. So let $E=[0,1/n]$ but $\int_E |f_n|\leq 1$ . To me me this is not the same as saying $\int_E |f_n| \leq 1/2$ – J. Kyei Jul 11 '17 at 22:10


text:
slice.index

0th

Percentile

Slice Indexes in an Array

Returns a matrix of integers indicating the number of their slice in a given array.

Keywords
array
Usage
slice.index(x, MARGIN)
Arguments
x

an array. If x has no dimension attribute, it is considered a one-dimensional array.

MARGIN

an integer giving the dimension number to slice by.

Value

An integer array y with dimensions corresponding to those of x such that all elements of slice number i with respect to dimension MARGIN have value i.

row and col for determining row and column indexes; in fact, these are special cases of slice.index corresponding to MARGIN equal to 1 and 2, respectively when x is a matrix.

• slice.index
Examples
library(base) # NOT RUN { x <- array(1 : 24, c(2, 3, 4)) slice.index(x, 2) # }
Documentation reproduced from package base, version 3.5.1, License: Part of R 3.5.1

Community examples

Looks like there are no examples yet.


text:
# Revision history [back]

The problem seems to be that packet-ethertype.c:dissect_ethertype() expects to be passed a pointer to an ethertype_data_t. Since there appears to be no way to pass this data to the dissector from a Lua dissector, dissect_ethertype() rejects it since the data is NULL.

I don't know any way around this problem except to open a bug report and see if someone can add support for this, or to rewrite the f2 shim as a built-in C dissector and ideally submit it for inclusion into Wireshark.

(Incidentally, the comment for dissect_ethertype() is wrong and should be fixed.)

So in theory, the f2 shim Lua dissector would look something more like so:

f2_shim = Proto ("f2_shim", "Cisco F2 shim header")
index = ProtoField.uint16("f2_shim.index", "Index", base.HEX)
data =  ProtoField.uint64("f2_shim.data", "Data", base.HEX)
f2_shim.fields = {index, data}

function f2_shim.dissector(buffer, pinfo, tree)
local etype_data = nil -- The "magic" piece we can't use from Lua

pinfo.cols.protocol = "f2_shim"
local subtree = tree:add(f2_shim, buffer(0, 10))
--[[
In theory, we'd initialize and pass a pointer to etype_data,
consisting of:
etype
fh_tree
trailer_fd
fcs_len
--]]
Dissector.get("ethertype"):call(buffer:range(10):tvb(), pinfo, tree, etype_data)
end

ether_table = DissectorTable.get("ethertype")

The problem seems to be that packet-ethertype.c:dissect_ethertype() expects to be passed a pointer to an ethertype_data_t. Since there appears to be no way to pass this data to the dissector from a Lua dissector, dissect_ethertype() rejects it since the data is NULL.

I don't know any way around this problem except to open a bug report and see if someone can add support for this, or to rewrite the f2 shim as a built-in C dissector and ideally submit it for inclusion into Wireshark.

(Incidentally, the comment for dissect_ethertype() is wrong and should be fixed.)

So in theory, the f2 shim Lua dissector would look something more like so:

f2_shim = Proto ("f2_shim", "Cisco F2 shim header")
index = ProtoField.uint16("f2_shim.index", "Index", base.HEX)
data =  ProtoField.uint64("f2_shim.data", "Data", base.HEX)
f2_shim.fields = {index, data}

function f2_shim.dissector(buffer, pinfo, tree)
local etype_data = nil -- The "magic" piece we can't use from Lua

pinfo.cols.protocol = "f2_shim"
local subtree = tree:add(f2_shim, buffer(0, 10))
--[[
In theory, we'd initialize and pass a pointer to etype_data,
consisting of:
etype
fh_tree
trailer_fd
fcs_len
--]]
Dissector.get("ethertype"):call(buffer:range(10):tvb(), pinfo, tree, etype_data)
end

ether_table = DissectorTable.get("ethertype")

Lastly, since there seems to be some confusion about this f2 shim, here's some packet data that can be converted to a pcap file using text2pcap for testing, e.g., text2pcap f2_shim.txt f2_shim.pcap. It adds an f2 shim with index=1 and data=0xdeadbeefdeadbeef, at least as I understand the format and placement of the f2 shim to be. The real payload should be dissected as IP (carrying UDP), but it isn't due to the limitations described above:
0000   00 0e b6 00 00 02 00 0e b6 00 00 01 f0 01 00 01


text:
# Evaluate the integral. int 8 sin(4t) sin(t/2)dt

Question
Applications of integrals
Evaluate the integral.
$$\displaystyle\int{8}{\sin{{\left({4}{t}\right)}}}{\sin{{\left(\frac{{t}}{{2}}\right)}}}{\left.{d}{t}\right.}$$

2021-03-05
Step 1
Let the given integral is,
$$\displaystyle\int{8}{\sin{{\left({4}{t}\right)}}}{\sin{{\left(\frac{{t}}{{2}}\right)}}}{\left.{d}{t}\right.}$$
By using the formula,
$$\displaystyle{\sin{{\left({a}\right)}}}{\sin{{\left({b}\right)}}}=\frac{{-{\cos{{\left({a}+{b}\right)}}}+{\cos{{\left({a}-{b}\right)}}}}}{{2}}$$
$$\displaystyle\int{8}{\left(\frac{{{\cos{{\left({4}{t}-\frac{{t}}{{2}}\right)}}}-{\cos{{\left({4}{t}+\frac{{t}}{{2}}\right)}}}}}{{2}}\right)}{\left.{d}{t}\right.}$$
$$\displaystyle\Rightarrow{8}\int{\left(\frac{{{\cos{{\left({7}\frac{{t}}{{2}}\right)}}}-{\cos{{\left({9}\frac{{t}}{{2}}\right)}}}}}{{2}}\right)}{\left.{d}{t}\right.}$$
Step 2
By separating the integrals,
$$\displaystyle\Rightarrow\frac{{8}}{{2}}\int{\left({\cos{{\left({7}\frac{{t}}{{2}}\right)}}}\right)}{\left.{d}{t}\right.}-\int{\left({9}\frac{{t}}{{2}}\right)}{)}{\left.{d}{t}\right.}$$
Simplifying this,
$$\displaystyle\Rightarrow\int{8}{\sin{{\left({4}{t}\right)}}}{\sin{{\left(\frac{{t}}{{2}}\right)}}}{\left.{d}{t}\right.}={4}{\left[\frac{{2}}{{7}}{\sin{{\left({7}\frac{{t}}{{2}}\right)}}}-\frac{{2}}{{9}}{\sin{{\left({9}\frac{{t}}{{2}}\right)}}}\right]}+{C}$$

### Relevant Questions

Evaluate the following integral.
$$\int_{0}^{1}t^{\frac{5}{2}}(\sqrt{t}-3t)dt$$
Evaluate the following integral.
$$\int \frac{3x^{2}+\sqrt{x}}{\sqrt{x}}dx$$
Evaluate the following integral.
$$\int 2x^{3}+3x-2dx$$
Use a change of variables to evaluate the following integral.
$$\int-(\cos^{7}x-5\cos^{5}x-\cos x)\sin x dx$$
Evaluate the following integral: $$\int \frac{(y-3)}{y^{2}-6y+1}$$
Evaluate the following integral: $$\int \frac{x+3}{x-1}dx$$
Evaluate the following integral: $$\int\frac{vdv}{6v^{2}-1}$$
Use the table of integrals at the back of the text to evaluate the integrals $$\displaystyle\int{8}{\sin{{\left({4}{t}\right)}}}{\sin{{\left({\frac{{{t}}}{{{2}}}}\right)}}}{\left.{d}{t}\right.}$$
Find the indefinite integral $$\int \ln(\frac{x}{3})dx$$ (a) using a table of integrals and (b) using the Integration by parts method.
Evaluate the integrals $$\int \cos ec^{4}0d0$$


text:
## Algebra Solutions: Hungerford I.4

1) Let G be a group and $\{H_i \mid i\in I\}$ a family of subgroups. Then for any $a\in G \text{, } (\cap_i H_i)a = \cap_i H_ia$.
Proof-
$(\cap_i H_i)a = \{x\in G \mid x \equiv_r a \pmod{\cap_i H_i}\} = \{ x\in G \mid xa^{-1} \in \cap_i H_i \}= \{ x\in G \mid xa^{-1} \in H_i \forall i \} = \{ x\in G \mid x \equiv_r a \pmod{H_i} \forall i \} = \{ x\in G \mid x \in H_ia \forall i \} = \cap_i H_ia$

3) The following conditions on a finite group G are equivalent.
(i) |G| is prime
(ii) $G \neq \langle e \rangle$ and G has no proper subgroups.
(iii) $G \cong \mathbb{Z}_p$ for some prime p.
Proof-
$(i) \Rightarrow (iii)$ Assume that the order of G is a prime, p. Then G has at least 2 elements. Pick an element a, which is not the identity and consider $\langle a \rangle$. By Lagrange, we know the order of a divides the order of the group. Thus the order of a is either 1 or p, but it is not 1 since a is not the identity. Thus a generates G, so G is cyclic of order p and hence isomorphic to $\mathbb{Z}_p$ since any finite cyclic group of a certain order is actually the same.
$(iii) \Rightarrow (ii)$ Assume that $G \cong \mathbb{Z}_p$ for some prime p. Then the order of G is p which is greater than or equal to 2, so G is nontrivial. Let $H \leq G$. By Lagrange again, the order of H divides the order of G, so that the order of H is either 1 or p. The first case gives H to be trivial, while the second gives H to be the entire group. Thus G has no proper subgroups.
$(ii) \Rightarrow (iii)$ Assume that G is not trivial and G has no proper subgroups. Since G is nontrivial, there is some a in G such that a is not the identity. Since G has no proper subgroups, $\langle a \rangle = G$ so that G is cyclic and of finite order, hence isomorphic to $\mathbb{Z}_m$ for some integer m. Suppose m is not prime then there is some natural number k such that k divides m and k is not 1 or m. Then because G is cyclic and k divides the order of G, there is exactly 1 subgroup of order k. However we know G has no proper subgroups, so this is a contradiction. Therefore m is prime.
$(iii) \Rightarrow (i)$ Trivial

4) Let $a \in \mathbb{Z}$ and p be a prime such that $p \nmid a$. Then $a^{p-1} \equiv 1 \pmod{p}$.
Proof-
By the division algorithm $a = pq + r$ for some $q,r \in \mathbb{Z}$ with $0 < r < p$. Thus taking the mod of both sides by p we get $a \equiv r \pmod{p}$. Recall that the nonzero elements of $\mathbb{Z}_p$ form a group of order p-1 under multiplication mod p. Since $0 r is in this group. Therefore the order of r, call it n, must divide p-1, the order of the group. Thus $p-1 = nm$ for some integer m. So we have:
$a^{p-1} \equiv r^{p-1} \pmod{p}$ but
$r^{p-1} = r^{nm} = (r^{n})^m = 1^m = 1$ so that
$a^{p-1} \equiv 1 \pmod{p}$

6) Let H, K be subgroups of a group G. Then $HK \leq H \text{ iff } HK = KH$
Proof-
$( \Rightarrow )$ Assume that $HK \leq G$. We prove that $HK = KH$ by subset inclusion. Let $g\in HK$ Then g can be written $g = hk$ for some $h \in H \text{, } k \in K$. Since $HK \leq G \text{ then } g^{-1} \in HK$ as well. Note, $g^{-1} = (hk)^{-1} = k^{-1}h^{-1}$ so that we know $k^{-1}h^{-1} \in HK$. This tells us that there is some $h' \in H \text{ and } k' \in K$ such that $k^{-1}h^{-1} = k'h'$. Since H and K are subgroups of G, $(h')^{-1} \in H \text{ and } (k')^{-1} \in K$. Putting everything together we have that:
$g = (g^{-1})^{-1} = (k^{-1}h^{-1})^{-1} = (k'h')^{-1} = (h')^{-1}(k')^{-1} \in KH$.
Now let $g \in KH$ so that $g=kh$ for some $k \in K \text{ and } h \in H$. Then we have that $g^{-1} = (kh)^{-1} = h^{-1} k^{-1}$. Since H and K are subgroups we have $h^{-1} \in H$ and $k^{-1} \in K$, so that $g^{-1} = h^{-1}k^{-1} \in HK$ and HK is a subgroup so that $g = (g^{-1})^{-1} \in HK$ also. Thus HK = KH
$( \Leftarrow )$ Assume now that HK = KH and show $HK \leq G$. Since H and K are subgroups of G, we have that the $e \in H \cap K$ so that $e = ee \in HK$ which makes HK nonempty. Now we use the subgroup test:
Let $hk,h'k' \in HK$ Then $(hk)(h'k')^{-1} = hk(k')^{-1}(h')^{-1}$ Note that $k(k')^{-1}(h')^{-1} \in KH = HK$ so that we can write $k(k')^{-1}(h')^{-1} = h''k''$ Then we have:
$(hk)(h'k')^{-1} = hh''k'' \in HK$. So HK is a subgroup of G.

7) Let G be a group s.t. $|G| = p^km$ where p is prime and gcd(p,m) = 1. Let H, K be supgroups of G s.t. $|H| = p^k \text{ and } |K| = p^d$ with $0 < d \leq k$ and K not a subgroup of H. Show that $HK \nleq G$
Proof-
(By contradiction) Assume that $HK \leq G$ then we get $|HK| = \frac{|H||K|}{|H \cap K|} = \frac{p^kp^d}{|H \cap K|}$. We know that the intersection of subgroups is a subgroup so $H \cap K \leq G$ so that in order for $|HK| \in \mathbb{N}$ it must be the case that $|H\cap k| = p^s$ for some $s \leq k+d$. However since K is not a subgroup of H, the intersection of H and K must contain strictly less elements than K. Thus:
$p^s = |H\cap K| < |K| = p^d$ so that s < d. Now we have that $|HK| = p^{k+d-s}$. By Lagrange the order of the subgroup must divide the order of G, so that $p^{k+d-s} \mid p^km$ which is the same as saying, for some $n \in \mathbb{N}$ that:
$p^km = p^{k+d-s}n \Rightarrow m = p^{d-s}n \Rightarrow m = p(p^{d-s-1}n) \Rightarrow p \mid m$ This is a contradiction because gcd(p,m) = 1. Thus HK is not a subgroup of G.

8) If H and K are subgroups of finite index of a group G such that $[G:H] \text{ and } [G:K]$ are relatively prime, then G = HK.
Proof-
First note two facts: $H\cap K \leq H \leq G$ and $H\cap K \leq K \leq G$ so by theorem 4.5 we can say:
$[G: H\cap K] = [G: K] [K: H\cap K]$ and
$[G: H\cap K] = [G: H] [H: H\cap K]$. This tells us that $[G: H\cap K]$ is a common multiple of both $[G:H] \text{ and } [G:K]$. However because $gcd([G:H],[G:K]) = 1$ we have that $lcm([G:H],[G:K]) = [G:H][G:K]$. Then clearly because $[G: H\cap K]$ is a common multiple it must be larger than or equal to the least common multiple so that $[G: H\cap K] \geq [G:H][G:K]$. Lastly, by proposition 4.9 we have $[G: H\cap K] \leq [G:H][G:K]$. Therefore $[G: H\cap K] = [G:H][G:K]$ so again by proposition 4.9 we can say that G = HK.

9) If H, K, and N are subgroups of a group G such that $H\leq N$, then $HK \cap N = H ( K\cap N)$
Proof-
By subset inclusion. Let $x\in HK \cap N$ Then we have $x\in HK \text{ and } x\in N$. From this we can say $x= hk$ for some $h\in H , k\in K$. So $k = h^{-1}x$ Since H is given as a subgroup of N this means $h^{-1} \in N$. so that $k = h^{-1}x \in N$ by closure. But $k\in K$ so that $k\in K \cap N$ Thus:
$x = hk\in H(K\cap N)$ since $h\in H, k \in K cap N$.
Now let $x \in H(K\cap N)$ This says that $x =hg$ where $h\in H \text{ and } g\in K\cap N$. Since this tells us g is in K we have $x\in HK$. Note H is a subgroup of N so that h is in N, and also g is in N. Thus their product is in N. So in conclusion we have $x\in HK \cap N$

10) Let H,K,N be subgroups of a group G such that $H \leq K, H\cap N = K\cap N, \text{ and } HN =KN$. Show that H = K.
Proof-
H is already a subgroup of K so we need only show that K is inside of H. Let $k\in K$. Then k can be written as $k = ke \in KN = HN$ which says that for some $h\in H \text{ and } n\in N$ we have $k = hn$. Then $h^{-1}k = n \in N$. Also we have $h^{-1} \in K$because H is a subgroup of K. Therefore we have $h^{-1}k \in N \text{ and } h^{-1}k \in K$ so that$h^{-1}k \in N\cap K = H\cap N$.Note the intersection of H and N is inside of H, so that $h^{-1}k \in H$. Thus $h^{-1}k = h' \Rightarrow k = hh' \in H$. So H = K.

11) Let G be a group of order 2n, then G contains an element of order 2. If n is odd and G is abelian, there is only one element of order 2.
Proof-
Let the order of G be 2n. Consider the set $\{e, a_1,a_1^{-1}, a_2,a_2^{-1}, ..., a_m,a_m^{-1}\}$. If there does not exist an element $a_i$ such that $a_i^{-1} = a_i$ then the order of this set is $2m+1$, which is a contradiction so there must exist at least one element of order 2.
If G is abelian and n is odd, suppose we have two elements $a,b \in G$ such that $a^2 = e = b^2$. G is abelian so that $(ab)^2 = e \Rightarrow \{e,a,b,ab\}$ is a subgroup of G. But because n is odd, we know that 4 does not divide 2n, so this subgroup cannot exist by Lagrange. Thus there can be only one element of order 2.


text:
## 4.1 Time series plots

Time series plots are an excellent way to begin the process of understanding what sort of process might have generated the data of interest. Traditionally, time series have been plotted with the observed data on the $$y$$-axis and time on the $$x$$-axis. Sequential time points are usually connected with some form of line, but sometimes other plot forms can be a useful way of conveying important information in the time series (e.g., bAR_p_coeflots of sea-surface temperature anomolies show nicely the contrasting El Niño and La Niña phenomena).

### 4.1.1ts objects and plot.ts()

The CO$$_2$$ data are stored in R as a data.frame object, but we would like to transform the class to a more user-friendly format for dealing with time series. Fortunately, the ts() function will do just that, and return an object of class ts as well. In addition to the data themselves, we need to provide ts() with 2 pieces of information about the time index for the data.

The first, frequency, is a bit of a misnomer because it does not really refer to the number of cycles per unit time, but rather the number of observations/samples per cycle. So, for example, if the data were collected each hour of a day then frequency = 24.

The second, start, specifies the first sample in terms of ($$day$$, $$hour$$), ($$year$$, $$month$$), etc. So, for example, if the data were collected monthly beginning in November of 1969, then frequency = 12 and start = c(1969, 11). If the data were collected annually, then you simply specify start as a scalar (e.g., start = 1991) and omit frequency (i.e., R will set frequency = 1 by default).

The Mauna Loa time series is collected monthly and begins in March of 1958, which we can get from the data themselves, and then pass to ts().

## create a time series (ts) object from the CO2 data
co2 <- ts(data = CO2$ppm, frequency = 12, start = c(CO2[1, "year"], CO2[1, "month"])) Now let’s plot the data using plot.ts(), which is designed specifically for ts objects like the one we just created above. It’s nice because we don’t need to specify any $$x$$-values as they are taken directly from the ts object. ## plot the ts plot.ts(co2, ylab = expression(paste("CO"[2], " (ppm)"))) Examination of the plotted time series (Figure 4.1) shows 2 obvious features that would violate any assumption of stationarity: 1) an increasing (and perhaps non-linear) trend over time, and 2) strong seasonal patterns. (Aside: Do you know the causes of these 2 phenomena?) ### 4.1.2 Combining and plotting multiple ts objects Before we examine the CO$$_2$$ data further, however, let’s see a quick example of how you can combine and plot multiple time series together. We’ll use the data on monthly mean temperature anomolies for the Northern Hemisphere (Temp). First convert Temp to a ts object. temp_ts <- ts(data = Temp$Value, frequency = 12, start = c(1880,
1))

Before we can plot the two time series together, however, we need to line up their time indices because the temperature data start in January of 1880, but the CO$$_2$$ data start in March of 1958. Fortunately, the ts.intersect() function makes this really easy once the data have been transformed to ts objects by trimming the data to a common time frame. Also, ts.union() works in a similar fashion, but it pads one or both series with the appropriate number of NA’s. Let’s try both.

## intersection (only overlapping times)
dat_int <- ts.intersect(co2, temp_ts)
## dimensions of common-time data
dim(dat_int)
[1] 682   2
## union (all times)
dat_unn <- ts.union(co2, temp_ts)
## dimensions of all-time data
dim(dat_unn)
[1] 1647    2

As you can see, the intersection of the two data sets is much smaller than the union. If you compare them, you will see that the first 938 rows of dat_unn contains NA in the co2 column.

It turns out that the regular plot() function in R is smart enough to recognize a ts object and use the information contained therein appropriately. Here’s how to plot the intersection of the two time series together with the y-axes on alternate sides (results are shown in Figure 4.2):

## plot the ts
plot(dat_int, main = "", yax.flip = TRUE)


text:
Suggested languages for you:

Americas

Europe

10

Expert-verified
Found in: Page 28

University Physics with Modern Physics

Book edition 14th edition
Author(s) Hugh D. Young, Roger A. Freedman
Pages 1596 pages
ISBN 9780321973610

The following conversions occur frequently in physics and are very useful. (a) Use 1 mi = 5280 ft and 1 h = 3600 s to convert 60 mph to units of ft/s. (b) The acceleration of a freely falling object is 32 ft/s2. Use 1 ft = 30.48 cm to express this acceleration in units of m/s2. (c) The density of water is 1.0 g/cm3. Convert this density to units of kg/m3.

(a) The velocity in ft/s is 88 ft/s.

(b) The acceleration in m/s2 is 18.3 m/s2.

(c) The density in kg/m3 is 1000 kg/m3.

See the step by step solution

Conversion in ft/s

Part (a)

The conversion in ft/s is shown below,

$60mph=60\frac{mi}{h}×\frac{5280ft}{1mile}×\frac{1h}{3600s}\phantom{\rule{0ex}{0ex}}60mph=88ft/s$

Thus, the value in ft/s is 88 ft/s.

Conversion in m/s2

Part (b)

The conversion in m/s2 is shown below,

$32ft/{s}^{2}=60\frac{ft}{h}×\frac{30.48cm}{1ft}×\frac{1m}{100cm}\phantom{\rule{0ex}{0ex}}32ft/{s}^{2}=18.3m/{s}^{2}$

Thus, the value in m/s2 is 18.3 m/s2

Conversion in m/s2

Part (c)

The conversion in kg/m3 is shown below,

$1g/c{m}^{3}=1\frac{g}{c{m}^{3}}×{\left(\frac{100cm}{1m}\right)}^{3}×\frac{1kg}{1000g}\phantom{\rule{0ex}{0ex}}1g/c{m}^{3}=1000kg/{m}^{3}$

Thus, the value of density in kg/m3 is 1000 kg/m3.


text:
MORE IN Signals & Systems
Total marks: --
Total time: --
INSTRUCTIONS
(1) Assume appropriate data and state your reasons
(2) Marks are given to the right of every question
(3) Draw neat diagrams wherever necessary

1 (a) Define signal and system with example. And briefly explain operations performed on independent variable of the signal.
6 M
1 (b) Determine whether the following signal is energy signal or power signal and calculate its energy of power $x(t)=rect \left ( \dfrac {t}{t_o} \right ) \cos \omega_o t.$
4 M
1 (c) Find the whether the following system is stable, memory less, linear and time invariant? y(t)=sin[x(t+2)]
4 M
1 (d) Two signals x(t) and g(t) as shown in Fig Q1(d). Express the signals x(t) in terms of g(t).

Fig.1 d (i)

Fig. 1 d (ii)

6 M

2 (a) Given the signal x(t) as shown in Fig Q2(a). Sketch the following: $i) \ x(-2t+3) \\ ii) \ x\left ( \dfrac {1}{2}-2 \right )$

4 M
2 (b) For a DT LTI system to be stable show that $S\overset {\Delta}= \sum^{k=+\infty}_{k=-\infty} |h(K)|<\infty$
5 M
2 (c) Two discrete time LTI systems are connected in cascade as shown in Fig Q2(c). Determine the unit sample response of this cascade connection.

6 M
2 (d) Find convolution of 2 finite duration sequences,
h(n)=anu(n) for all n and
x(n)=bnu(n) for all n
i) when a?b
ii) when a=b
5 M

3 (a) Determine the LTI system characterized by impulse response. $i) \ h(n)=n\left ( \dfrac {1}{2} \right )^n u(n) \\ ii) h(t)=e^{-t}u(t+100)$ Stable and causal.
6 M
3 (b) Find the forced response of the following system; $y(n)- \dfrac {1}{4}y(n-1)- \dfrac {1}{8} y(n-2)= x(n)+x(n-1) \ for \ x(n)= \left ( \dfrac {1}{8} \right )^n u(n)$
8 M
3 (c) Draw direct form II implementation for the system described by the following equation and indicate number of delay elements, adders, multipliers.
y(n)-0.25y(n-1)-0.125y(n-2)-x(n)-x(n-2)=0
6 M

4 (a) Prove the following properties of DTFS:
i) Convolution in time ii) Modulation theorem
6 M
4 (b) Determine the complex exponential Fourier series for periodic rectangular pulse train shown in Fig Q4(b). Plot its magnitude and phase spectrum.

8 M
4 (c) Determine the DTFS representation for the signal $x(n)=\cos \left ( \dfrac {n\pi}{3}$ Plot the spectrum of x(n).
6 M

5 (a) State and prove the following properties of DTFT:
(i) Parseval's theorem ii) Linearity
6 M
5 (b) Find the DTFT on the signals shown, $i) \ x(n) = \left ( \dfrac {1}{4} \right )^n u(n+4) \\ ii) x(n)= u(n)$
8 M
5 (c) Find the inverse Fourier transform of the rectangular shown below

6 M

6 (a) Consider the continuous time LTI system described by, $\dfrac {d}{dt} y(t)+2y(t)=x(t)$ Using FT, find the output y(t) to each of the following input signals.
i) x(t)=e-tu(t)
ii) x(t)=u(t)
8 M
6 (b) Find the Nyquist rate and Nyquist interval for each of the following signals:
i) x(t)=sin c2 (200t)
ii) x(t)=2 sin c(50t) sin(5000?t)
6 M
6 (c) An LTI system is described by $H(f)=\dfrac {4} {2+j2\pi f}$ find its response y(t) if the input is x(t)=u(t).
6 M

7 (a) Define ROC and list its properties.
4 M
7 (b) State and prove time reversal property of z-transform.
4 M
7 (c) Determine the inverse z-transform of $x(z)=\dfrac {1}{(1+z^{-1})(1-z^{-1})^2} ; ROC;|z|>1$
6 M
7 (d) Determine z-transform and ROC of $x(n)= \left ( \dfrac {1}{3} \right )^n \sin \left ( \dfrac {\pi}{4}n \right )u(n)$
6 M

8 (a) A causal, stable discrete time system is defined by $y(n)= \dfrac {5}{6} y(n-1)-\dfrac {1}{6}y(n-2)+x(n)-2x(n-1) Determine : i) System function H(z) and magnitude response at zero frequency ii) Impulse response of the system. iii) Output y(n) for x(n)=?(n)- 1/3 ?(n-1) 12 M 8 (b) Solve the following difference equation for the given initial conditions and input, \[ y(n)-\dfrac {1}{9} y(n-2)=x(n-1) \\ with \ y(-1)=0, y(-2)=1 \ and \ x(n)=3u(n)$
8 M

More question papers from Signals & Systems


text:
# Math Help - Calculus of Variations

1. ## Calculus of Variations

What is the Euler-Lagrange equation for:

$I[u] = \int _{- \infty} ^{ \infty} \frac{1}{2} u' ^{2} + \left( 1 - \cos{u} \right) \,dx$?

I got that it was u'' = sin(u) but I cannot solve this nicely and this is a question that I should be able to do.

2. Well, in general, if $L(u,u';t)$ is the integrand of your functional $I$, then the Euler-Lagrange equations are

$\displaystyle{\frac{\partial L}{\partial u}-\frac{d}{dt}\frac{\partial L}{\partial u'}=0.}$

I get

$\sin(u)-u''=0,$

the same as you. To integrate this, I would multiply through by $u'$ and then integrate once. You might be able to integrate the resulting first-order DE at that point. Make sense?

3. I understand this, and get:

$u' ^{2} = C - 2 \cos{u}$.

I have to find all solutions where the limit as x tends to infinity of u is 2pi and as x tends to minus infinity is 0. What can I do here? Clearly I can set the integral out for this equation however I don't think that this integral will have a nice expression.

4. Yeah, off-hand I'd say you're likely going to end up with an elliptic. (checking...) Yep, you've got yourself an elliptic integral there. Not very pretty.

In the original DE, can you make the small-angle approximation with u? That is, sin(u) is approximately u?

Do you have any initial conditions? Or are you dealing with asymptotic conditions?

5. The OP has given me a link to the original problem, which I will now type out in full.

Write down the Euler-Lagrange equation for the functional

$\displaystyle{I[u]=\int_{-\infty}^{\infty}\left[\frac{1}{2}\,u'^{2}+(1-\cos(u))\right]dx}$

and find all solutions which satisfy $\lim_{x\to-\infty}u(x)=0$ and $\lim_{x\to+\infty}u(x)=2\pi.$ Show that if $u\in C^{1}(\mathbb{R})$ satisfies $\lim_{x\to-\infty}u(x)=0$ and $\lim_{x\to+\infty}u(x)=2\pi$

$\displaystyle{I[u]=\frac{1}{2}\int_{-\infty}^{\infty}\left(u'-2\sin\left(\frac{u}{2}\right)\right)^{2}dx+8.}$

Deduce that a lower bound for $I[u]$ amongst such functions is $8$, and give a first order differential equation which $u$ must satisfy in order to realize this lower bound. Show that any solution of this first order equation solves the Euler-Lagrange equation you derived in the first part of the question. Give all the functions satisfying $I[u]=8.$
I guess one question I have is this: when you are finding all the solutions which satisfy the Euler-Lagrange equation, are you already allowed to assume they are $C^{1}(\mathbb{R})$? If so, I think you can simplify a bit. For a $C^{1}(\mathbb{R})$ function, I think it's true that

$\displaystyle{\lim_{x\to\pm\infty}u(x)=\text{const .}\to\lim_{x\to\pm\infty}u'(x)=0.}$

So, supposing it is allowed to assume that you're restricted to finding $C^{1}(\mathbb{R})$ functions as solutions, we might be able to get a handle on that first constant of integration.

The first integration yielded

$\displaystyle{\frac{(u')^{2}}{2}=C-\cos(u).}$

Suppose we take the limit of this equation as $x\to-\infty.$ On the LHS we'll get $0$, and we'll get $C-1$ on the RHS. Therefore, $C=1.$ Now, I think you can see that the integration becomes much more manageable. Can you continue from here?


text:
# Transformations involving singular points

1. Nov 6, 2008

### SW VandeCarr

Can a triangle be smoothly transformed to a circle?

2. Nov 16, 2008

### SW VandeCarr

Either this question is too easy, too dumb or too hard. I'd just like an answer. I've found two two possibly relevant theorems:

Hironaka: Every analytic space Y admits a resolution of singularities: there is a smooth manifold X and a proper map f:X->Y such that f is an isomorphism except over singular points.

Castelnuovo, Enriques: Every singular surface has a unique minimal resolution of singularities.

What exactly is a "minimal resolution"? It suggests to me that the singular points of a triangle cannot be "eliminated" by a smooth transformation, but only arbitrarily "minimized". Is this correct?

Last edited: Nov 16, 2008
3. Nov 20, 2008

### SW VandeCarr

People are reading, but not answering. I'll give a very simple special case example of how I think a singular point might be eliminated, not "minimally resolved" by means of a topological transformation:

Consider an equalateral triangle circumscribed by a circle such that the three singular points lie on the circle. If the triangle is transformed to coincide with the circle (each line segment is transformed into a 2pi/3 radian arc), the singular points are eliminated.

Can this idea be applied to the more general case of functions that are analytic almost everywhere, except for singular points?

Last edited: Nov 20, 2008


text:
How to prove that the equation $x^2-3y^2=17$ has no integer solutions?

How to prove that the equation $$x^2-3y^2=17$$ has no integer solutions? Can you help me?

-
–  lab bhattacharjee Jan 13 '13 at 13:05
@Ayman Hourieh Thanks a lot! –  Dao yi Peng Jan 13 '13 at 13:14

$$x^2-3y^2=17\implies x^2\equiv2\pmod 3$$

but $x$ can be $\equiv 0,\pm1\pmod 3\implies x^2\equiv0,1\pmod 3$

-
@bhattachaejee Thank you very much! –  Dao yi Peng Jan 13 '13 at 13:15
@DaoyiPeng, my pleasure. Don't miss the related question. –  lab bhattacharjee Jan 13 '13 at 13:16
bhattachaejee OK! I need to see some Pell's eqution. –  Dao yi Peng Jan 13 '13 at 13:18
@labbhattacharjee could you explain to me the step x can be $0, +- 1$? I'm not sure how that proves the answer.. –  user128914 Mar 7 at 1:08
@user128914, what are the possible remainders when divided by $3$? –  lab bhattacharjee Mar 7 at 3:20


text:
0
69views
Using Fourier Cosine integral prove that $e^{-x}\cos{x=\frac{2}{\pi{}}\\int_0^{\infty{}}\frac{\left({\omega{}}^2+2\right)}{\left({\omega{}}^4+4\right)}.\\cos{\omega{}\ x\ d\omega{}}}$
$Fourier \ Cosine\ Integral:\\[3ex] \displaystyle f(x)=\int_0^{\infty{}}A\left(w\right)\cos {wx\ dw} \\[2ex] \displaystyle where,\ A\left(w\right)=\frac{2}{\pi{}}\int_0^{\infty{}}f(x) \cos {wx\ }dx \\[2ex] \displaystyle A\left(w\right)=\frac{2}{\pi{}}\int_0^{\infty{}}e^{-x}\cos {x\ }\cos{wx\ }dx \\[2ex] \displaystyle \cos x\cos{wx=}\frac{1}{2}(\cos({x+wx)}+\cos{\left(x-wx\right)}) \\[2ex] \displaystyle A\left(w\right)=\frac{2}{2\pi{}}\int_0^{\infty{}}e^{-x}\cos{(x+wx)\ }dx+\ \frac{2}{2\pi{}}\int_0^{\infty{}}e^{-x}\cos{(x-wx)\ }dx \\[3ex] \displaystyle Using,\\[3ex] \displaystyle \int e^{ax}\cos{\left(bx\right)}dx=\frac{e^{ax}(a\cos{\left(bx\right)+b\sin{(bx)}})}{a^2+b^2}\\[2ex] \displaystyle A\left(w\right)=\frac{1}{\pi{}}\ {\left[\frac{e^{-x}\ (- \ cos{(1+w)x}+\left(1+w\right)\sin{(1+w)x)}}{{(-1)}^2+{\left(1+w\right)}^2}\right]}_0^{\infty{}}\\[2ex] \ \ \ \ \ \ \ \ \ \ \ \ \ \displaystyle +\ \frac{1}{\pi{}}\ {\left[\frac{e^{-x}\ (-\ cos{(1-w)x}+\left(1-w\right)\sin{(1-w)x)}}{{(-1)}^2+{\left(1-w\right)}^2}\right]}_0^{\infty{}} \\[4ex] \displaystyle Now \ since \ e^{-\infty{}}=0 \\[2ex] \displaystyle A\left(w\right)=-\frac{1}{\pi{}}\ \left[\frac{e^0\left(-1\right)}{1+w^2+2w+1\ }+\frac{e^0\left(-1\right)}{1+w^2-2w+1}\right]\\[2ex] \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{\pi{}}\left[\frac{1}{1+w^2+2w+1\ }+\frac{1}{1+w^2-2w+1}\right] \\[2ex] \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \frac{1}{\pi{}}[\frac{2+w^2+2w+2+w^2-2w}{(2+w^2+2w)(2+w^2-2w)}] \\[2ex] \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{2}{\pi{}}\left[\frac{w^2+2}{4+2w^2-4w+2w^2+w^4-2w^3+4w+2w^3-4w^2}\right] \\[2ex] \displaystyle \ \ A\left(w\right)\ \ \ =\frac{2}{\pi{}}\left[\frac{w^2+2\ }{w^4+4\ }\right] \\[2ex] \displaystyle f(x)=\int_0^{\infty{}}A\left(w\right)\cos{wx\ dw} \\[2ex] \displaystyle e^{-x}\cos{x\ }=\ \int_0^{\infty{}}\frac{2}{\pi{}}\left[\frac{w^2+2\ }{w^4+4\ }\right]\ \cos{wx\ dw}\ \ \ \ \\[2ex] \displaystyle e^{-x}\cos{x\ }=\frac{2}{\pi{}}\ \int_0^{\infty{}}\left[\frac{w^2+2\ }{w^4+4\ }\right]\ \cos{wx\ dw}\ \ \ \ \\[2ex]$


text:
$$\require{cancel}$$

# 7.1.3: The Decibel Scale

The human ear is an amazing instrument that can detect intensities as low as $$10^{-12}\text{ W/m}^{2}$$ and can hear intensities as high as $$10^{3}\text{ W/m}^{2}$$ (although this is loud enough to cause damage to the ear). To make this huge range easier to write down, a second scale of loudness was created called the sound intensity level, measured in decibels. The relationship between sound intensity, $$I$$ measured in watts per meter squared and sound intensity level ($$SIL$$) measured in decibels ($$\text{dB}$$), is given by $$SIL =10\log (I/I_{o})$$. Here $$\log$$ is the logarithm and $$I_{o}=10^{-12}\text{ W/m}^{2}$$ is a reference sound intensity at about the threshold of human hearing.

We know sound waves are longitudinal fluctuations in the pressure of the air (or other medium) in which the sound is traveling. This is a fluctuation above and below normal atmospheric pressure and is measured in pascals ($$\text{Pa}$$), the standard unit of pressure. If you tried to report an average value of the fluctuation you would get zero since the pressure varies from above to below normal atmospheric pressure. We could just give the (maximum) amplitude of the fluctuation but what is usually done is to report the rms value which is the square root of the average of the squares of the amplitude. By first squaring the amplitude and then taking an average and then a square root you have a kind of average value that is not zero. In the chart below the root mean square ($$\text{rms}$$) variation in pressure from normal atmospheric pressure (in pascals $$\text{Pa}$$), the sound intensity level ($$SIL$$ in $$\text{dB}$$) and the intensity ($$I$$ in $$\text{W/m}^{2}$$) are given for several sounds.

Source Pressure $$\text{rms (Pa)}$$ Sound Intensity Level $$\text{SIL (dB)}$$ Intensity $$(\text{W/m}^{2})$$
Jet engine at $$10\text{ m}$$ $$150$$ $$10^{3}$$
Jet engine $$200$$ $$140$$ $$100$$
Jack hammer $$60$$ $$130$$ $$10$$
Car horn $$20$$ $$120$$ (pain threshold) $$1$$
Rock band $$6$$ $$110$$ $$0.1$$
Machine shop $$2$$ $$100$$ $$0.01$$
Train $$0.6$$ $$90$$ $$10^{-3}$$
Vacuum cleaner $$0.2$$ $$80$$ $$10^{-4}$$
TV $$0.06$$ $$70$$ $$10^{-5}$$
Conversation $$0.02$$ $$60$$ $$10^{-6}$$
Office $$0.006$$ $$50$$ $$10^{-7}$$
Library $$0.002$$ $$40$$ $$10^{-8}$$
Hospital $$0.0006$$ $$30$$ $$10^{-9}$$
Broadcast studio $$0.0002$$ $$20$$ $$10^{-10}$$
Rustle of leaves $$0.00006$$ $$10$$ $$10^{-11}$$
Threshold of hearing $$0.00002$$ $$0$$ $$10^{-12}$$

Table $$\PageIndex{1}$$

Here are a few examples and rules of thumb for converting intensity ($$\text{W/m}^{2}$$) into intensity levels (in $$\text{dB}$$):

• A $$10$$ fold increase in intensity equals an addition of $$10\text{ dB}$$. So going from a car horn to a jackhammer multiplies the intensity by $$10$$ ($$1\text{ W/m}^{2}$$ to $$10\text{ W/m}^{2}$$) but adds $$10\text{ dB}$$ to the intensity level ($$120\text{ dB}$$ to $$130\text{ dB}$$).
• A two fold increase in intensity (twice as loud in $$\text{W/m}^{2}$$) equals and addition of $$3\text{ dB}$$ to the $$\text{SIL}$$. Suppose one trombone produces a sound level of $$40\text{ dB}$$. How loud are four trombones? Doubling the number of trombones to two adds $$3\text{ dB}$$, doubling again to four adds $$3\text{ dB}$$ more so the new sound level is $$46\text{ dB}$$.
• Suppose the sound intensity is $$100\text{ W/m}^{2}$$. What is the sound level? $$\text{SIL} = 10 \log (100/10^{-12}) = 10 \log (10^{14}) = 10\ast 14 = 140\text{ dB}$$.
• Suppose the sound level is $$110\text{ dB}$$. What is the sound intensity? $$110\text{ dB} = 10 \log (I/10^{-12})$$. Divide both sides by $$10$$ to get $$11 = \log (I/10^{-12})$$. Now take inverse $$\log\: 11$$ (same as $$10^{11}$$) to get $$10^{11} = I/10^{-12}$$. Multiply both sides by $$10^{-12}$$ to get $$0.1\text{ W/m}^{2} = I$$.
• A Sound Conversion web site that converts between sound level, sound pressure and sound intensity.

Both sound intensity ($$\text{W/m}^{2}$$) and sound intensity level ($$\text{SIL}$$) are numbers that can be measured precisely in the laboratory (objective). The human ear, however, is an imperfect measuring instrument. We hear better at a mid-range of frequencies than we do at very low or very high frequencies. The Loudness Level ($$L_{L}$$) scale is a subjective measurement of loudness. This scale is arrived at by asking real humans to compare the loudness of different notes and an average is taken for many people (subjective). The units of the Loudness Level are the phon.

The diagram below (modified from an MIT OpenCourseWare graph) relates sound intensity level ($$\text{SIL}$$, measured in $$\text{dB}$$ with laboratory instruments), intensity (measured in $$\text{W/m}^{2}$$ with laboratory instruments) and Loudness Level (measured in phons). A $$\text{SIL}$$ of $$110\text{ dB}$$ is considered painful while a $$\text{SIL}$$ of $$0$$ is at the threshold of hearing. If our ears were the same as laboratory instruments the lines would go straight across. The $$L_{L}$$ scale and the $$\text{SIL}$$ scale do give approximately the same number only for frequencies around $$1000\text{ Hz}$$. In other words our subjective perception of loudness (in phons) and the laboratory measurement (in $$\text{dB}$$) agree but only for sounds with a frequency of $$1000\text{ Hz}$$.

Notice there is a dip in all the curves between $$1000\text{ Hz}$$ and $$5000\text{ Hz}$$ indicating we are more sensitive to these frequencies and this is true for all loudness readings. For example suppose we perceive a sound at $$4000\text{ Hz}$$ to be $$45\text{ phons}$$ (labeled by a blue X in the diagram). The chart shows that at this loudness and frequency the $$\text{dB}$$ reading in the laboratory is actually around $$36\text{ dB}$$ (dotted line to the $$\text{SIL}$$ axis). So we perceive a sound of $$36\text{ dB}$$ (measured in the lab) as being much louder ($$45\text{ phons}$$) if it occurs at $$4000\text{ Hz}$$. Most other animals have a similar curve allowing them to hear better in a certain frequency range although the dip is usually much more narrow and does not go as low as for humans. Many animals can hear frequencies above and below what humans can hear, but we can hear much softer sounds in this range of frequencies than almost any other animal, owls being an exception. Your cat or dog can hear higher frequencies than you can but you can hear softer sounds in the $$1000\text{ Hz}$$ to $$5000\text{ Hz}$$ range.

This improved ability to hear softer sounds in the $$1000\text{ Hz}$$ to $$5000\text{ Hz}$$ range is not surprising once you realize these are important frequencies for human speech; our hearing mechanism is built to hear human voices better than sound with much higher or much lower frequencies. This greater sensitivity around $$3500\text{ Hz}$$ is due to the tube resonance of the auditory canal (see chapter 12 for tube resonance and chapter 10 for a picture of the auditory canal).

Figure $$\PageIndex{1}$$

It is also the case that intensity has an effect on perceived frequency; the same laboratory frequency will appear to be a slightly different frequency if the intensity is different. High frequencies are perceived to be a slightly higher pitch than normal if they are very loud. Low frequencies are perceived to be slightly lower than expected if they are very loud. Medium loudness doesn't change the perceived pitch very much.

The above curves are very much like the frequency response curves of microphones and speakers. No microphone has the same sensitivity to all frequencies and no speaker reproduces all frequencies equally well, as we will see in Chapter 18 on electronics. Likewise our hearing does not have the same sensitivity at all frequencies.


text:
Tag Info

43

This type of integral has appeared so many times and in so many places; for example, here, here and here. Basically, for each sample $\omega$, we can treat $\int_0^t W_s ds$ as a Riemann integral. Moreover, note that \begin{align*} d(tW_t) = W_t dt + tdW_t. \end{align*} Therefore, \begin{align*} \int_0^t W_s ds &= tW_t -\int_0^t sdW_s \tag{1}\\ &= \...

20

These are all examples on Ito Formula in its general form (with quadratic variations):

16

Brownian motion is simply the limit of a scaled (discrete-time) random walk and thus a natural candidate to use. It is very intuitive and arguably one of the simplest and best understood time-continuous stochastic processes. Also, don't forget that you obtain many more stochastic processes as functions of a (time-changed) Brownian motion. In many books on ...

14

Baxter and Rennie say it better than me, so I will summarize them. Suppose that $N_t$ is not stochastic and $f(.)$ is a smooth function then the Taylor expansion is $$df(N_t) = f'(N_t)dN_t + \frac{1}{2}f''(N_t)(dN_t)^2 + \frac{1}{3!} f'''(N_t)(dN_t)^3 + \ldots$$ and the term $(dN_T)^2$ and higher terms are zero. Ito showed that this is not the case in the ...

14

In the integral $$\int_0^t S_u dW^{*}_u \, ,$$ $dW^{*}_u \equiv W^{*}_{u+du} - W^{*}_u$ is independent from the integrand $S_u$. So, $\mathbb{E}\left[ \int_0^t S_u dW^{*}_u\middle\vert \mathcal{F}_0\right] = \int_0^t \mathbb{E}\left[S_u \middle\vert \mathcal{F}_0\right]\mathbb{E}\left[dW^{*}_u\middle\vert \mathcal{F}_0\right] = 0$, since \mathbb{E}\... 13 Let \begin{align*} Y_t = e^{(a+\frac{c^2}{2})t-cW_t}. \end{align*} Then \begin{align*} dY_t = Y_t\left[\big(a+c^2\big)dt -c dW_t \right]. \end{align*} Moreover, \begin{align*} d(X_tY_t) &= Y_t dX_t + X_t dY_t + d\langle X, Y\rangle_t\\ &=abY_tdt. \end{align*} That is, \begin{align*} X_t = Y_t^{-1}\left(X_0 + ab\int_0^t Y_sds\right). \end{align*} 12 The convexity of the exponential function of the stochastic variableW$makes its expectation greater than the exponentiation of the expectation of$W$. This is an example of Jensen's inequality,$E[e^{\sigma W}]> e^{\sigma E[W]}=1$.$\sigma$can be interpreted as the magnitude of the convexity of the exponential function. This can be seen by Taylor ... 11 My understanding is because the Ito's integration definition keeps the martingale property. With Brownian motion$W(t, \omega)$defined, to define stochastic integration in a Riemann–Stieltjes style: $$\int_0^t f(t, \omega) d W(t, \omega) = \lim_{\| \Delta_n\| \to 0 } \sum_{i=1}^{n} f(\tau_i,\omega) \left ( W(t_i, \omega) - W(t_{i-1}, \omega) \right )$$ , ... 11 From$(2), \begin{align*} \ln S_t &=\ln F_{t, t} \\ &= \ln F_{0, t}-\frac{1}{2}\int_0^t\sigma^2 e^{-2\lambda (t-s)}ds+\int_0^t \sigma e^{-\lambda(t-s)} dB_s\\ &=\ln F_{0, t}-\frac{\sigma^2}{4\lambda} \left(1-e^{-2\lambda t}\right)+e^{-\lambda t}\int_0^t \sigma e^{\lambda s} dB_s. \end{align*} Then, \begin{align*} \lambda e^{-\lambda t}\int_0^t \... 11 Quadratic variation and variance are two different concepts. LetX $be an Ito process and$t\geq 0$. Variance of$X_t$is a deterministic quantity where as quadratic variation at time$t $that you denoted by$[X,X]_t $is a random variable. What is confusing you is the fact that when$X $is a martingale then$X^2_t-[X,X]_t$is a martingale thus you ... 11 Of course making money is always the key issue. That (not completely facetious) comment aside: On the practical side, in many firms IT is struggling with being clear, transparent, and intuitive in their handling of multiple curves and their associated risks. Stumbling over your own systems is an annoying way to lose money. These risks can be surprisingly ... 10 I think this question has no easy answer but I'll give it a shot anyway (beware: oversimplification ahead!). The main idea of the Malliavin calculus is to be able to differentiate stochastic processes like Brownian motion (or more general martingales with bounded quadratic variation), which are not differentiable in the traditional sense (because of their ... 10 I will try to answer this a bit differently. The rigorous answer: because Ito calculus tells us that we need the second order term. Look at $$S_t = S_0\exp(\mu t + \sigma B_t).$$ Assume that$S_0$is known and fixed and look at by Ito's formula $$d(S_t/S_0) = \mu dt + \sigma B_t + \frac{\sigma^2}{2} dt.$$ Then with some abuse of notation: $$E[d(S_t/... 10 Shreve's theorem also called "Girsanov II" indeed represents a special case of the general "Girsanov I" from Wiki above, with$$Y_t:=W_t,X_t:=-\int_0^t\Theta_udW_u$$We can show:$$[Y,X]=-\int_0^t\Theta_udu$$by using general Stochastic Calculus rules (e.g. p.37, 6.6 here):$$[Y,X]=[W_t,-\int_0^t\Theta_udW_u]=-\int_0^t\Theta_ud[W_u,W_u]=-\int_0^t\... 10 For any$s \geq t, note that \begin{align*} r_s = r_t + \sigma\int_t^s dW_u + \int_t^s \theta_u du. \end{align*} Then, \begin{align*} \int_t^T r_s ds &= (T-t)r_t + \sigma\int_t^T\int_t^s dW_u ds + \int_t^T \int_t^s\theta_u du ds\\ &=(T-t)r_t + \sigma\int_t^T\int_u^T ds\, dW_u +\int_t^T\int_u^T\theta_u ds du\\ &=(T-t)r_t + \sigma\int_t^T (T-u)... 10 Because you can hedge. Once you have delta hedged, the pay-off is symmetric about up and down moves so drift doesn't matter. Also the delta-hedged call and the delta hedged put have to have the same value since they have the same pay-off. (Put-call parity) Yet any argument that the call should be worth more because of drift says that the put should be ... 10 Assume deterministic and constant interest rates. For an investor in the foreign economy i.e. a market participant that can only trade assets delivering a payout in the foreign currency, let us define $$\tilde{X}_t = \tilde{X}_0 \exp \left(\left(r_f-r_d-\frac{\sigma_\tilde{X}^2}{2}\right)+\sigma_\tilde{X} W_t^{\tilde{X},\mathbb{Q}^f} \right)$$ Y_t =Y_0\... 10 Just to add to the already nice answers, the result can also be obtained using the (stochastic) Fubini theorem. \begin{align} \int_0^t W_s ds &= \int_0^t \int_0^s dW_u\, ds \tag{W_s=\int_0^s dW_u}\\ &= \int_0^t \int_u^t ds\,dW_u \tag{Fubini} \\ &= \int_0^t (t-u) dW_u \tag{\int_u^t ds = t-u } \end{align} And we fall back on the same equation ... 10 Note that the Ito integral of a deterministic integrand f: \mathbb{R}_+ \rightarrow \mathbb{R} is normally distributed $$\int_0^t f(u) \mathrm{d}W_u \sim \mathcal{N} \left( 0, \int_0^t f^2(u) \mathrm{d}u \right).$$ In your case, we have f(t) = e^{-\lambda t} and thus \int_0^t f^2(u) \mathrm{d}u = \... 10 It is, of course, possible to price such a contract in a no-arbitrage market. Indeed, if f is a sufficiently smooth function, then you can price all contracts paying f(S_T). Note that your specific payoff has no optionality and that the payoff may be negative. Bakshi and Madan (2000) discuss the economic meaning of a derivative paying \cos(S_T) in the ... 10 I've seen that Gordon answer is more concise and to the point. Take this as a complementary answer. This is a general approach that will work for all this type of linear SDEs, not just this one. Assume we have the following linear SDEdX_t = (F_t X_t +f_t)dt + (G_t X_t +g_t)dB_t \tag*{(1)}$$where F, G, f and g are Borel measurable bounded ... 9 Okay so I'll take Jase answer and format it properly so that it answers your question and it will be useful for users in the future. For clarity, let me restate the dynamics of the Modified Ornstein-Uhlenbeck model using the more common notation:$$dS_t = \theta (\mu-S_t)dt + \sigma S_t dW_t$$This blog post provides a closed form solution:$$ S_t = S_0 \... 9 The part where you say that $$\frac{dS_t}{S_t} = d\ln(S_t)$$ is wrong, becauseS$is a stochastic variable. This is exactly what Itô tells you with his formula that you apply right do compute your$dZ$. The difference comes from the quadratic variation of the process$S$which you express as$(dS)^2$. If you don't add this term when the variable are ... 9 In fact Ito and Stratonovich calculus are both mathematically equivalent. In the following paper you can e.g. see that both derivations lead to the same result, i.e. the Black-Scholes equation: Black-Scholes option pricing within Ito and Stratonovich conventions by J. Perello, J. M. Porra, M. Montero and J. Masoliver From the abstract: Options financial ... 8 Let$P$be the risk-neutral measure. We define the measure$P_Ssuch that \begin{align*} \frac{dP_S}{dP}\big|_t &=\frac{S_t}{e^{rt}S_0}\\ &=e^{-\frac{1}{2}\sigma^2 t+\sigma W_t}. \end{align*} Then\{\widehat{W}_t \mid t \ge 0\}, where \begin{align*} \widehat{W}_t = W_t -\sigma t, \end{align*} is a standard Brownian motion under the measureP_S$. ... 8 We know that$(\tilde{W}_t) := (-W_t)$is also a Wiener process so $$E[W_pW_qW_r] = E[\tilde{W}_p\tilde{W}_q\tilde{W}_r] = (-1)^3E[W_pW_qW_r]$$ and that implies that$E[W_pW_qW_r] = 0\$.

Only top voted, non community-wiki answers of a minimum length are eligible


text:
#### need solution for RD Sharma maths class 12 chapter 5 Determinants exercise Fill in the blanks question 19

Hint: Here, we use basic concept of determinant of matrix

Given:

\begin{aligned} &b_{i 1}=2 a_{i 1} \\ &b_{i 2}=3 a_{i 2} \\ &b_{i 3}=4 a_{i 3} i, i=1,2,3 \end{aligned}

Let $\left | A \right |=\left [ a_{ij} \right ]$ and $B=b_{ij}$ be a square matrix of order 3 and $\left | A \right |=5$

Solution:

\begin{aligned} &|B|=\left(\begin{array}{lll} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array}\right) \\ &|B|=\left(\begin{array}{lll} 2 a_{11} & 3 a_{12} & 4 a_{13} \\ 2 a_{21} & 3 a_{22} & 4 a_{23} \\ 2 a_{31} & 3 a_{32} & 4 a_{33} \end{array}\right) \end{aligned}

Let’s transfer $\left | B \right |=\left | B^{T} \right |$ So,

\begin{aligned} &|B|=\left(\begin{array}{ccc} 2 a_{11} & 2 a_{12} & 2 a_{13} \\ 3 a_{21} & 3 a_{22} & 3 a_{23} \\ 4 a_{31} & 4 a_{32} & 4 a_{33} \end{array}\right) \\ &|B|=2 \times 3 \times 4\left(\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right) \end{aligned}

\begin{aligned} &=24 \times\left|A^{T}\right| \\ &=24 \times 5 \quad\left[|A|=\left|A^{T}\right|=5\right] \\ &=120 \end{aligned}


text:
# Using Biplots to Map Cluster Solutions

July 2, 2014
By

(This article was first published on Engaging Market Research, and kindly contributed to R-bloggers)

FactoMineR is a quick and easy R package for generating biplots, such as the following plot showing the columns as arrows with the rows to be added later as points. As you might recall from a previous post, a biplot maps a data matrix by plotting both the rows and columns in the same figure. Here the columns (variables) are arrows and the rows (individuals) will be points. By default, FactoMineR avoids cluttered maps by separating the variables and individuals factor maps into two plots. The variables factor map appears below, and the individuals factor map will be shown later in this post.
The dataset comes from David Wishart's book Whiskey Classified, Choosing Single Malts by Flavor. Some 86 whiskies from different regions of Scotland were rated on 12 aromas and flavors from "not present" (a rating of 0) to "pronounced" (a rating of 4). Luba Gloukhov ran a cluster analysis with this data and plotted the location where each whisky was distilled on a map of Scotland. The dataset can be retrieved as a csv file using the R function read.csv("clipboard'). All you need to do is go to the web site, select and copy the header and the data, and run the R function read.csv pointing to the clipboard. All the R code is presented at the end of this post.

Each arrow in the above plot represents one of the 12 ratings. FactoMineR takes the 86 x 12 matrix and performs a principal component analysis. The first principal component is labeled as Dim 1 and accounts for almost 27% of the total variation. Dim 2 is the second principal component with an additional 16% of the variation. One can read the component loadings for any rating by noting the perpendicular projection of the arrow head onto each dimension. Thus, Medicinal and Smoky have high loadings on the first principal component with Sweetness, Floral and Fruity anchoring the negative end. One could continue in the same manner with the second principal component, however, at some point we might notice the semi-circle that runs from Floral, Sweetness and Fruity through Nutty, Winey and Spicy to Smoky, Tobacco and Medicinal. That is, the features sweep out a one-dimensional arc, not unlike a multidimensional scaling of color perceptions (see Figure 1).
Now, we will add the 86 points representing the different whiskies. But first we will run a cluster analysis so that when we plot the whiskies, different colors will indicate cluster membership. I have included the R code to run both a finite mixture model using the R package mclust and a k-means. Both procedures yield four-cluster solutions that classify over 90% of the whiskies into the same clusters. Luba Gloukhov also extracted four clusters by looking for an "elbow" in the plot of the within-cluster sum-of-squares from two through nine clusters. By default, Mclust will test one through nine clusters and select the best model using the BIC as the selection criteria. The cluster profiles from mclust are presented below.

 Black Red Green Blue Total 27 36 6 17 86 31% 42% 7% 20% 100% Body 2.7 1.4 3.7 1.9 2.1 Sweetness 2.4 2.5 1.5 2.1 2.3 Smoky 1.5 1.0 3.7 1.9 1.5 Medicinal 0.0 0.2 3.3 1.0 0.5 Tobacco 0.0 0.0 0.7 0.3 0.1 Honey 1.9 1.1 0.2 1.0 1.3 Spicy 1.6 1.1 1.7 1.6 1.4 Winey 1.9 0.5 0.5 0.8 1.0 Nutty 1.9 1.3 1.2 1.4 1.5 Malty 2.1 1.7 1.3 1.7 1.8 Fruity 2.1 1.9 1.2 1.3 1.8 Floral 1.6 2.1 0.2 1.4 1.7

Finally, we are ready to look at the biplot with the rows represented as points and the color of each point indicating cluster membership, as shown below in what FactoMineR calls the individuals factor map. To begin, we can see clear separation by color suggesting that differences among the cluster reside in the first two dimensions of this biplot. It is important to remember that the cluster analysis does not use the principal component scores. There is no data reduction prior to the clustering.
The Green cluster contains only 6 whiskies and falls toward the right of the biplot. This is the same direction as the arrows for Medicinal, Tobacco and Smoky. Moreover, the Green cluster received the highest scores on these features. Although the arrow for Body does not point in that direction, you should be able to see that the perpendicular projection of the Green points will be higher than that for any other cluster. The arrow for Body is pointed upward because a second and larger cluster, the Black, also receives a relatively high rating. This is not the case for other three ratings. Green is the only cluster with high ratings on Smoky or Medicinal. Similarly, though none of the whiskies score high on Tobacco, the six Green whiskies do get the highest ratings.

You can test your ability to interpret biplots by asking on what features the Red cluster should score the highest. Look back up to the vector map, and identify the arrows pointing in the same direction as the Red cluster or pointing in a direction so that the Red points will project toward the high end of the arrow. Do you see at least Floral and Sweetness? The process continues in the same manner for the Black cluster, but the Blue cluster, like its points, fall in the middle without any distinguishing features.

Hopefully, you have not been troubled by my relaxed and anthropomorphic writing style. Vectors do not reposition themselves so that all the whiskies earning high scores will project themselves toward its high end, and points do not move around looking for that one location that best reproduces all their ratings. However, principal component analysis does use a singular value decomposition to factor data matrices into row and column components that reproduce the original data as closely as possible. Thus, there is some justification for such talk. Nevertheless, it helps with the interpretation to let these vectors and points come alive and have their own intentions.

What Did We Do and Why Did We Do It?

We began trying to understand a cluster analysis derived from a data matrix containing the ratings for 86 whiskies across 12 aroma and taste features. Although not a large data matrix, one still has some difficulty uncovering any underlying structure by looking one variable/column at a time. The biplot helps by creating a low-dimensional graphic display with ratings as vectors and whiskies as points. The ratings appeared to be arrayed along an arc from floral to medicinal, and the 86 whiskies were located as points in this same space.

Now, we are ready to project the cluster solution onto this biplot. By using separate ratings, the finite mixture model worked in the 12-dimensional rating space and not in the two-dimensional world of the biplot. Yet, we see relatively coherent clusters occupying different regions of the map. In fact, except for the Blue cluster falling in the middle, the clusters move along the arc from a Red floral to a Black malty/honey/nutty/winey to a Green medicinal. The relationships among the four clusters are revealed by their color coding on the biplot. They are no longer four qualitatively distinct entries, but a continuum of locally adjacent groupings arrayed along a nonlinear dimension from floral to medicinal.

R code needed to run all the analysis in this post.

# read data from external site# after copied into the clipboarddata <- read.csv("clipboard")ratings<-data[,3:14] # runs finite mixture modellibrary(mclust)fmm<-Mclust(ratings)fmmtable(fmm$classification)fmm$parameters$mean # compares with k-means solutionkcl<-kmeans(ratings, 4, nstart=25)table(fmm$classification, kcl$cluster) # creates biplotslibrary(FactoMineR)pca<-PCA(ratings)plot(pca, choix=c("ind"), label="none", col.ind=fmm$classification)

Created by Pretty R at inside-R.org


text:
# Figure 15

 Ratios of the production yields to pions for several particle species as a function of the midrapidity pion yields for pp, \pPb\ and \PbPb\ colliding systems compared to the THERMUS model prediction for the strangeness canonical suppression (black line), in which only the system size is varied. All values except for the \kstar\ are normalized to the high multiplicity limit (see text for details). The upper axis shows the radius $R$ of the correlation volume $V=R^{3}$ which corresponds to the predicted particle ratios. The width of the model prediction line corresponds to a variation of the chemical freeze-out temperature between 146 MeV and 166 MeV. Reference p-Pb and Pb-Pb data from


text:
# The Gradient and Hessian of the Smooth Max Over Functions

| Feedback

Suppose you have a set of functions over a vector space, and you are interested in taking the smooth-maximum over those functions. For example, maybe you are doing gradient descent, or convex optimization, etc, and you need a variant on “maximum” that has a defined gradient. The smooth maximum function has both a defined gradient and Hessian, and in this post I derive them.

I am using the logarithm-based definition of smooth-max, shown here:

I will use the second variation above, ignoring function arguments, with the hope of increasing clarity. Applying the chain rule gives the ith partial gradient of smooth-max:

Now that we have an ith partial gradient, we can take the jth partial gradient of that to obtain the (i,j)th element of a Hessian:

This last re-grouping of terms allows us to see that we can express the full gradient and Hessian in the following more compact way:

With a gradient and Hessian, we now have the tools we need to use smooth-max in algorithms such as gradient descent and convex optimization. Happy computing!


text:
# Math Insight

### An algebra trick for finding critical points

The algebra trick here goes back at least $350$ years. This is worth looking at if only as an additional review of algebra, but is actually of considerable value in a variety of hand computations as well.

The algebraic identity we use here starts with a product of factors each of which may occur with a fractional or negative exponent. For example, with $3$ such factors: $$f(x)=(x-a)^k\;(x-b)^\ell\;(x-c)^m$$ The derivative can be computed by using the product rule twice: \begin{align*} f'(x) &=k(x-a)^{k-1}(x-b)^\ell(x-c)^m+ (x-a)^k \ell(x-b)^{\ell-1}(x-c)^m\\ &\quad+ (x-a)^k(x-b)^\ell m(x-c)^{m-1} \end{align*} Now all three summands here have a common factor of $$(x-a)^{k-1}(x-b)^{\ell-1}(x-c)^{m-1}$$ which we can take out, using the distributive law in reverse: we have \begin{align*} f'(x)&=(x-a)^{k-1}(x-b)^{\ell-1}(x-c)^{m-1}\\ &\quad \times [k(x-b)(x-c)+\ell(x-a)(x-c)+m(x-a)(x-b)] \end{align*} The minor miracle is that the big expression inside the square brackets is a mere quadratic polynomial in $x$.

Then to determine critical points we have to figure out the roots of the equation $f'(x)=0$: If $k-1>0$ then $x=a$ is a critical point, if $k-1\le 0$ it isn't. If $\ell-1>0$ then $x=b$ is a critical point, if $\ell-1\le 0$ it isn't. If $m-1>0$ then $x=c$ is a critical point, if $m-1\le 0$ it isn't. And, last but not least, the two roots of the quadratic equation $$k(x-b)(x-c)+\ell(x-a)(x-c)+m(x-a)(x-b)=0$$ are critical points.

There is also another issue here, about not wanting to take square roots (and so on) of negative numbers. We would exclude from the domain of the function any values of $x$ which would make us try to take a square root of a negative number. But this might also force us to give up some critical points! Still, this is not the main point here, so we will do examples which avoid this additional worry.

#### Example

A very simple numerical example: suppose we are to find the critical points of the function $$f(x)=x^{5/2}(x-1)^{4/3}$$ Implicitly, we have to find the critical points first. We compute the derivative by using the product rule, the power function rule, and a tiny bit of chain rule: $$f'(x)={5\over 2}x^{3/2}(x-1)^{4/3}+x^{5/2}{4\over 3}(x-1)^{1/3}.$$ And now solve this for $x$? It's not at all a polynomial, and it is a little ugly.

But our algebra trick transforms this issue into something as simple as solving a linear equation: first figure out the largest power of $x$ that occurs in all the terms: it is $x^{3/2}$, since $x^{5/2}$ occurs in the first term and $x^{3/2}$ in the second. The largest power of $x-1$ that occurs in all the terms is $(x-1)^{1/3}$, since $(x-1)^{4/3}$ occurs in the first, and $(x-1)^{1/3}$ in the second. Taking these common factors out (using the distributive law ‘backward’), we rearrange to \begin{align*} f'(x)&={5\over 2}x^{3/2}(x-1)^{4/3}+x^{5/2}{4\over 3}(x-1)^{1/3}\\ &=x^{3/2}(x-1)^{1/3}\,\left( {5\over 2}(x-1)+{4\over 3}x\right)\\ &=x^{3/2}(x-1)^{1/3}\left({5\over 2}x-{5\over 2}+{4\over 3}x\right)\\ &=x^{3/2}(x-1)^{1/3}\left({23\over 6}x-{5\over 2}\right). \end{align*}

Now to see when this is $0$ is not so hard: first, since the power of $x$ appearing in front is positive, $x=0$ make this expression $0$. Second, since the power of $x+1$ appearing in front is positive, if $x-1=0$ then the whole expression is $0$. Third, and perhaps unexpectedly, from the simplified form of the complicated factor, if ${23\over 6}x-{5\over 2}=0$ then the whole expression is $0$, as well. So, altogether, the critical points would appear to be $$x=0,{15\over 23},1.$$ Many people would overlook the critical point ${15\over 23}$, which is visible only after the algebra we did.

#### Exercises

1. Find the critical points and intervals of increase and decrease of $f(x) = x^{10}(x-1)^{12}$.
2. Find the critical points and intervals of increase and decrease of $f(x) = x^{10}(x-2)^{11}(x+2)^{3}$.
3. Find the critical points and intervals of increase and decrease of $f(x)=x^{5/3}(x+1)^{6/5}$.
4. Find the critical points and intervals of increase and decrease of $f(x)=x^{1/2}(x+1)^{4/3}(x-1)^{-11/3}$.


text:
# Can't understand what's psi

1. Aug 26, 2014

### quawa99

I just started reading the quantum theory.I came across the wave particle duality experiments and just after that the book had the Schrödinger equation.I don't understand what psi is exactly in that equation and what a wave function is and how it differs from a wave equation mathematically.can anyone explain what it is in simple terms please?

2. Aug 26, 2014

### Staff: Mentor

Which book? You'll get better answers from us if we know which book you're working through.

The function $\psi(x,t)$ is the unknown in the Schrodinger equation. Once you know what it is, $\psi(x,t)\psi^*(x,t)$) is the probability of finding the particle at position $x$ at time $t$.

3. Aug 26, 2014

### JohnPrior3

Wavefunction describes the quantum state. When I took my first quantum mechanics course, I had a lot of trouble with stuff so I can probably understand it's ambiguity. So basically, the most important idea of quantum mechanics is measurements and how they change a system. The quantum state describes the information of the given particle i.e. spin. These mathematics behind the wave function deals with complex numbers and probability. The difference between the wave equation and the wave function is that the wave equation describes a wave in classic mechanics and the wave function is quantum mechanics. You get definitive answers through the wave equation, but the wave function has values that are probabilities. Have you studied the Stern Gerlach experiment? Tackling Schrödinger would be a little easier if you have those basics down.


text:
# polygon

This command draws a polygon in the geometry.

## Inline mode usage

polygon -h
Displays the help file that summarizes the parameters for this command.
polygon –n <name> -p <x1> <y1> <z1> <x2> <y2> <x3> <y3> ...
Draws a N-gon with the specified vertices.
polygon –n <name> -r -p <sides> <x y z> <radius>
Draws a regular polygon with <sides> number of sides.
Draws a regular star with <sides> number of sides.

## Interactive mode usage

Vertices method:

Invocation
polygon
Parameters
• The coordinates of each vertex of the polygon, in “x y z” format. When the desired points have been entered, the user needs to press Enter (without specifying any coordinates) to draw the polygon. A minimum of three vertices are required.

Regular polygon method:

Invocation
polygon -r
Parameters
• The number of sides of the polygon.
• The center point of the polygon.
• The radius of the circle that the polygon is circumscribed in.

Star method:

Invocation
polygon -s
Parameters:
• Number of spikes of the star.
• Center point of the star.
• Minor radius of the star.
• Major radius of the star.

## Example

> polygon
Point 0 [x y z] 1 1 0
Point 1 [x y] 1 3
Point 2 [x y] 3 3
Point 3 [x y] 4 1
Point 4 [x y]:
>


text:
# If a rocket with a mass of 700 kg vertically accelerates at a rate of  3/2 m/s^2, how much power will the rocket have to exert to maintain its acceleration at 4 seconds?

Jan 29, 2017

P=6.3 KW

#### Explanation:

see

MASS=700 kg
Acceleration=$\frac{3}{2} \frac{m}{s} ^ 2$

THEREFORE force exerted is $m \cdot a = 700 \cdot \frac{3}{2} = 1050 N$

Now

$a = \frac{v - u}{t}$ and we have U ,i.e initial velocity =0 as it started from rest and ofcourse every rockets before launching is at rest,lol.

so by this we can get $v$ and $t$ we are given 4 seconds

$\frac{3}{2} = \frac{v}{4}$
$v = 6 \frac{m}{s}$

therefore

$E = F \cdot D i s p l a c e m e n t$
dividing both sides of the equation by $t$

$\frac{E}{t} = F \cdot \frac{\mathrm{di} s p l a c e m e n t}{t}$

And by definition power is energy per unit time and velocity is displacement per unit time.so applying these

$p = F \cdot v$
$P = 1050 \cdot 6$ Watt

$P = 6300 W$
$P = 6.3 K W$


text:
NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### JEE Main 2019 (Online) 11th January Evening Slot

The standard reaction Gibbs energy for a chemical reaction at an absolute temperature T is given by $$\Delta$$rGo = A – BT

Where A and B are non-zero constants. Which of the following is TRUE about this reaction ?
A
Exothermic if B < 0
B
Endothermic if A > 0
C
Exothermic if A > 0 and B < 0
D
Endothermic if A < 0 and B > 0

## Explanation

$$\Delta$$Go = $$\Delta$$Ho - T$$\Delta$$So

Given that A and B are non-zero constants, i.e., A = $$\Delta$$Ho, B = $$\Delta$$So

If $$\Delta$$Ho is positive means reaction is endothermic.
2

### JEE Main 2019 (Online) 11th January Evening Slot

The reaction, MgO(s) + C(s) $$\to$$ Mg(s) + CO(g), for which $$\Delta$$rHo + 491.1 kJ mol–1 and $$\Delta$$rSo = 198.0 JK–1 mol–1, is not feasible at 298 K. Temperature above which reaciton will be feasible is :
A
2480.3 K
B
2040.5 K
C
2380.5 K
D
1890.0 K

## Explanation

We know,

$$\Delta$$Go = $$\Delta$$Ho - T$$\Delta$$So

For a reaction to be spontaneous $$\Delta$$Go must be negative i.e.,
T$$\Delta$$So > $$\Delta$$Ho

$$\Rightarrow$$ T > $${{\Delta H^\circ } \over {\Delta S^\circ }}$$

$$\Rightarrow$$ T > $${{491.1 \times 1000} \over {198}}$$

$$\Rightarrow$$ T > 2480.3 K
3

### JEE Main 2019 (Online) 11th January Morning Slot

Two blocks of the same metal having same mass and at temperature T1 and T2, respectively, are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, $$\Delta$$S, for this process is :
A
2Cp In $$\left[ {{{{{\left( {{T_1} + {T_2}} \right)}^{{1 \over 2}}}} \over {{T_1}{T_2}}}} \right]$$
B
2Cp In $$\left[ {{{\left( {{T_1} + {T_2}} \right)} \over {2{T_1}{T_2}}}} \right]$$
C
Cp In $$\left[ {{{{{\left( {{T_1} + {T_2}} \right)}^2}} \over {4{T_1}{T_2}}}} \right]$$
D
2Cp In $$\left[ {{{\left( {{T_1} + {T_2}} \right)} \over {4{T_1}{T_2}}}} \right]$$

## Explanation

When two blocks comes in contact with each other and attain thermal equilibrium then

final temperature of the blocks,

Tf = $${{{T_1} + {T_2}} \over 2}$$

$$\Delta$$S = $$\Delta$$S1 + $$\Delta$$S2

= Cp $$\ln \left( {{{{T_f}} \over {{T_1}}}} \right)$$ + Cp $$\ln \left( {{{{T_f}} \over {{T_2}}}} \right)$$

= Cp $$\ln \left( {{{{T_1} + {T_2}} \over {2{T_1}}}} \right)$$ + Cp $$\ln \left( {{{{T_1} + {T_2}} \over {2{T_2}}}} \right)$$

= Cp $$\ln \left( {{{{{\left( {{T_1} + {T_2}} \right)}^2}} \over {4{T_1}{T_2}}}} \right)$$
4

### JEE Main 2019 (Online) 11th January Morning Slot

For the chemical reaction X $$\rightleftharpoons$$ Y, the standard reaction Gibbs energy depends on temperature T (in K) as $$\Delta$$rGo (in kJ mol–1) = 120 $$- {3 \over 8}$$ T.

The major component of the reaction mixture at T is :
A
Y if T = 300 K
B
Y if T = 280 K
C
X if T = 350 K
D
X if T = 315 K

## Explanation

X $$\rightleftharpoons$$ Y

Keq = $${{\left[ Y \right]} \over {\left[ X \right]}}$$

If Keq > 1 $$\Rightarrow$$ [Y] > [X]

and Keq < 1 $$\Rightarrow$$ [Y] < [X]

We know, $$\Delta$$Go = -RT ln(Keq)

So when Keq > 1 then $$\Delta$$Go < 0 and Y is major.

And when Keq < 1 then $$\Delta$$Go > 0 and X is major.

Temperature at which $$Go = 0 is 120$$ - {3 \over 8}$$T = 0$$ \Rightarrow $$T = 320 K For T > 320 K$$\Delta $$Go < 0 and Y is major. For T < 320 K$$\Delta Go > 0 and X is major.

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

NEET

Class 12


text:
# 마른김의 유기산조성(有機酸組成)과 저장중의 변화

• Park, Young-Ho (Department of Food Science and Technology, Pusan Fisheries College)
• 박영호 (부산수산대학 식품공학과)
• Published : 1973.12.28
• 30 9

#### Abstract

Dried lavers, Porphyra yezoensis Ueda, were stored at $20^{\circ}C$ for 10, 25 and 45 days under a dry and humid conditions respectively, and changes in the contents of organic acids were studied by GLC and supplementary use of GC-MS. 1) Out of eight acids determined, i. e., pyruvic, lactic, oxalic, malonic, fumaric, succinic, malic, and citric acid, relatively abundant ones were oxalic, succinic, malic, and citric acid. 2) Oxalic, malic, and succinic acid did not show any notable changes in their contents under a dry condition, however, under a humid condition the contents of the former tow acids decreased rapidly showing contrast to that of succinic acid, of which content increased markedly. The content of citric acid rapidly· decreased independent of the storage conditions. 3) Among some unidentified components, two of them were presumed to be aromatic acids from their mass spectra. The contents of the two acids decreased slowly under a dry condition, however, rapidly decreased under a humid condition and after 45 days storage almost all of them disappeared. 4) The contents of total volatile acids were increased under the both storage conditions, however, the rate of increment was larger under a humid than under a dry condition.


text:
### Climbing Powers

$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$?

### Little and Large

A point moves around inside a rectangle. What are the least and the greatest values of the sum of the squares of the distances from the vertices?

### Approximating Pi

By inscribing a circle in a square and then a square in a circle find an approximation to pi. By using a hexagon, can you improve on the approximation?

# How Does Your Function Grow?

##### Stage: 5 Challenge Level:

Think about the meaning of each of the functions carefully. Try writing them out in full, explicit notation and comparing them. The log function is somewhat different in definition to the other three, but you can look at the effect of log on the other functions to develop a sense for its speed of growth.


text:
# Crystal plane

crystallography

## X-ray reflection

...incidence can be reflected. For X rays where the wavelengths are comparable to the lattice spacings in analyzing crystals, the radiation can be “Bragg reflected” from the crystal: each crystal plane acts as a weakly reflecting surface, but if the angle of incidence θ and crystal spacing d satisfy the Bragg condition, 2 d sin θ = nλ, where...
MEDIA FOR:
crystal plane
Previous
Next
Citation
• MLA
• APA
• Harvard
• Chicago
Email
You have successfully emailed this.
Error when sending the email. Try again later.


text:
The Leibniz integral rule in electrodynamics

[ physics ]

I have been slowly working through David Griffith’s widely-used textbook, Introduction to Electrodynamics. It is known for having a large number of reasonably difficult exercises, which are instrumental in conveying some concepts not directly addressed in the main text. I found an elegant shortcut to one of the questions, which was not noted in the solution manual. The shortcut involves (ab)using a very versatile, if slightly overlooked theorem in calculus – the Leibniz integral rule. I will briefly discuss this, before working through the problem.

The Leibniz integral rule

The Leibniz rule concerns derivatives of integrals; the most basic form is as follows:

$\frac{d}{dx} \left( \int_a^b f(x,t) dt \right) = \int_a^b \frac{\partial}{\partial x} f(x,t) dt$

This can be proven by letting $I(x) = \int_a^b f(x,t) dt$, then using the definition of the derivative to evaluate $I’(x)$, which will be left as an exercise for the reader. However I hope you can get an intuitive feel for this: the integral measures the total variation with respect to $t$, whereas the derivative measures the rate of change with respect to $x$, so it makes sense that the order of operation doesn’t really matter.

A slightly more general version concerns the case where the limits of the integral are variable:

$\frac{d}{dx} \left( \int_{a(x)}^{b(x)} f(x,t) dt \right) = f\left(x, b(x)\right)\frac{d b(x)}{dx} - f(x, a(x))\frac{d a(x)}{dx} + \int_{a(x)}^{b(x)}\frac{\partial}{\partial x} f(x,t) dt$

Effectively, it is the same as what we had before, plus some additions to take into account of the variable limits. This formula can be explained in terms of the fundamental theorem of calculus, but the derivation is besides the point. For now, I would urge you to just accept the above rule (it is surprisingly useful). Let’s go through an example to evaluate the following expression:

$\frac{d}{dx} \int_{\sin x}^{\tan x} \sqrt{1+ t^3}dt$

Hint: do not do it by evaluating the indefinite integral, substituting in the limits, then differentiating. Applying the Leibniz rule, we see that:

\begin{aligned} \frac{d}{dx} \int_{\sin x}^{\tan x} \sqrt{1+ t^3}dt &= \sqrt{1+ \tan^3 x}\frac{d}{dx}\tan x - \sqrt{1 + \sin^3 x} \frac{d}{dx} \sin x\\ &= \sec^2x\sqrt{1+ \tan^3 x} - \cos x \sqrt{1 + \sin^3 x} \end{aligned}

The electrodynamics problem

I say ‘electrodynamics’, but really it is an electrostatics problem. Below I have reproduced the question from Chapter 2 (Electrostatics) of Griffith’s Introduction to Electrodynamics. It has been edited slightly.

Problem 27 Find the electric field on the axis of a uniformly charged solid cylinder, at a distance z from the centre. The length of the cylinder is L, its radius is R, and the charge density is $\rho$.

Our general strategy will be to first find the potential (or rather, an expression for the potential), by decomposing the cylinder into simpler shapes. Then we will find the field using $\mathbf{E} = - \nabla V$.

A cylinder can be thought of as many discs; a disc can be thought of as many rings; a ring can be thought of as many point charges. We will thus build a cylinder.

1. A single point charge

Recall the formula for the potential at a point $P$:

$V = \frac{1}{4\pi \epsilon_0} \frac{q}{r}$

In this formula, $r$ is the displacement between the charge and the test point. Applying Pythagoras’ theorem to the diagram, we see that $r^2 = R^2 + z^2$.

Plugging values into the formula:

$V = \frac{1}{4\pi \epsilon_0} \frac{q}{\sqrt{R^2 + z^2}}$

2. A ring of charge

In order to find the potential for a ring of charge given the potential of a point charge, we consider a charge element $\lambda dl$. But the length of a segment on a circle is given by $R \theta$, so we can say that $dl = R d\theta$.

Instead of a positive point charge $q$, we now have a charge element $\lambda R d\theta$. Using our previous result, the potential at P from one of these charge elements is:

$dV = \frac{1}{4\pi \epsilon_0} \frac{\lambda R d\theta}{\sqrt{R^2 + z^2}}$

The total potential for the ring is the sum of the contributions from each of the charge elements.

$V = \int_0^{2\pi} \frac{1}{4\pi \epsilon_0} \frac{\lambda R d\theta}{\sqrt{R^2 + z^2}} = \frac{1}{4\pi \epsilon_0} \frac{\lambda R }{\sqrt{R^2 + z^2}} \int_0^{2\pi} d\theta$ $\therefore V = \frac{1}{4\pi \epsilon_0} \frac{2 \pi R \lambda}{\sqrt{R^2 + z^2}}$

3. A disc of charge

We can build up a disc of charge with radius R from an infinite number of rings of charge of radius r. Note that this r is different to the one briefly mentioned earlier, which was the separation between P and the charge.

Because we are going from 1D to 2D, we must let $\lambda = \sigma dr$, where $\sigma$ is the surface charge. Thus the potential from one ring is:

$dV = \frac{1}{4\pi \epsilon_0} \frac{2 \pi r \sigma dr}{\sqrt{r^2 + z^2}}$

We could arrive at this otherwise, by noting that the total charge in the ring is

$\sigma \cdot \text{area} = \sigma \cdot 2\pi r dr$

Anyway, let us continue.

$V = \frac{\sigma}{2 \epsilon_0} \int_0^R \frac{r}{\sqrt{r^2 + z^2}} dr$

As is rarely the case in electrodynamics, this integral is not at all difficult to evaluate. We use the ‘reverse chain rule’ to see that:

$\int_0^R \frac{r}{\sqrt{r^2 + z^2}} dr = \int_0^R r \left( r^2 + z^2\right)^{-1/2} dr =\left[ (r^2+z^2)^{1/2} \right]_0^R$

The total potential is then easily found.

$V = \frac{\sigma}{2\epsilon_0} \left(\sqrt{R^2+z^2} - z \right)$

4. A cylinder of charge

We are now ready to tackle the original problem. Here it is again:

Find the electric field on the axis of a uniformly charged solid cylinder, at a distance z from the centre. The length of the cylinder is L, its radius is R, and the charge density is $\rho$.

I’ll set up the solution in the same way Griffiths does – this image is taken from the solution manual.

A cylinder is made up of an infinite number of discs. We want an expression for the potential at the test point, a distance z from the centre. We will let x denote the distance between the test point and a given disc. The potential from one such disc is given by the formula we found earlier, with the substitution $z \rightarrow x$ and $\sigma \rightarrow \rho dx$:

$dV = \frac{\rho dx}{2 \epsilon_0} \left( \sqrt{R^2 + x^2} - x \right)$

Since L is the length of the cylinder, and z is the distance from the centre of the cylinder, it is clear that there are discs all the way from $x = z - \frac L 2$ to $x = z+ \frac L 2$. The total potential at our test point is just the integral over all of these discs:

$V = \frac{\rho}{2\epsilon_0} \int_{z-L/2}^{z+L/2}(\sqrt{R^2 + x^2} - x) dx$

Despite the simple-looking expression, this integral will soon become a nightmare to evaluate. And don’t forget – we are trying to find the electric field, so we will then have to take the gradient of the whole mess.

To be clear, because of the cylinder’s symmetry, ‘taking the gradient’ here just means differentiating with respect to z, but nevertheless this will not be a fun calculation.

$\mathbf{E} = - \nabla V = - \mathbf{\hat{z}} \frac{\partial V}{\partial z}$

Leibniz to the rescue

Instead of immediately evaluating the integral to calculate the potential, then differentiating, why don’t we write it in one expression first:

$\mathbf{E} = - \mathbf{\hat{z}} \frac{\rho}{2\epsilon_0} \frac{\partial}{\partial z} \int_{z-L/2}^{z+L/2}\left( \sqrt{R^2 + x^2} - x \right) dx$

This might look more complicated, but in fact it makes life much easier. Because we are taking the derivative of an integral, we can simply apply the Leibniz integral rule. The limits are not constant, but the integrand is independent of z, so the rule tells us that:

\begin{align*} \mathbf{E} = - \mathbf{\hat{z}} \frac{\rho}{2\epsilon_0} &\bigg[ \left( \sqrt{R^2 + (z+L/2)^2} - (z+L/2) \right) \frac{\partial}{\partial z}(z+L/2)\\ & - \left( \sqrt{R^2 + (z-L/2)^2} - (z-L/2) \right) \frac{\partial}{\partial z}(z-L/2) \bigg] \end{align*}

Intuitively, there is some kind of ‘cancellation’ between the differentiation and integration, so we simply substitute the limits into the integrand (and multiply by the derivative of the limit since it is not constant), then subtract. In this case it is even simpler because the derivatives of the limits are unity:

$\mathbf{E} = \frac{\rho}{2\epsilon_0} \left[L + \sqrt{R^2 + (z-L/2)^2} - \sqrt{R^2 + (z+L/2)^2} \right] \mathbf{\hat{z}}$

Conclusion

In many electrodynamics problems, one finds the electric field by first calculating the potential, then taking the gradient. I hope I’ve given you a reasonable motivation to not just do this mechanically – keep an eye out for shortcuts that will save you time, and reduce the chances of you making a minor algebraic error.


text:
# Homework Help: Showing a matrix is invertible

1. Feb 19, 2014

### noelo2014

1. The problem statement, all variables and given/known data

let A and B be nXn invertible matrices. Show that (A.B)T is invertible by finding it's inverse

2. Relevant equations

3. The attempt at a solution

(A.B)T=BT.AT

Last edited: Feb 19, 2014
2. Feb 19, 2014

### CAF123

You are looking for a matrix $C$ such that $(AB)^{T}C = C(AB)^T = I_{n}$. The question does not say that A and B are invertible, however given that the inverse of (AB)T exists you can show that they are indeed invertible.

3. Feb 19, 2014

### noelo2014

Sorry I left out the fact that A and B are invertible.

4. Feb 19, 2014

### HallsofIvy

You have already said that $(AB)^T= B^TA^T$. Okay, what will get rid of that $B^T$ and then what will get rid of that $A^T$

(Of course, you know that a matrix is invertible if and only if its determinant is not 0? And that the determinant of $X^T$ is the same as the determinant of $X$?)


text:
Definition Edit

Given a numeric set $\textstyle \mathcal{N}$, the diference distance is a distance function $\textstyle DiffDis: \mathcal{N}\times \mathcal{N}\longrightarrow \mathbb{R}^+$ such that $\textstyle \forall x,y\in \mathcal{N}$

$DiffDis(x,y)=|x-y|.$

Examples Edit

• $DiffDis(-2,4) = 6$
• $DiffDis(2.5, -1.2) = 3.7$

Normalization Edit

It is not possible when the range of $\textstyle \mathcal{N}$ is not bounded, otherwise

$DiffSim(x,y)=1-\frac{|x-y|}{\max(\mathcal{N})-\min(\mathcal{N})}.$

Examples Edit

• If $\textstyle \mathcal{N}=\{0,1,2,\ldots,100\} \quad DiffSim(4,14) = 1-\frac{|4-14|}{100-0} = 0.9$
• If $\textstyle \mathcal{N}=[-10,10] \quad DiffSim(2.5,-1.2) = \frac{|-1.2-2.5|}{10-(-10)} = 0.815$