text
stringlengths 256
16.4k
|
|---|
$\tan{(A+B)}$ $\,=\,$ $\dfrac{\tan{A}+\tan{B}}{1\,–\,\tan{A}\tan{B}}$
$\dfrac{\tan{A}+\tan{B}}{1\,–\,\tan{A}\tan{B}}$ $\,=\,$ $\tan{(A+B)}$
The tan of angle sum identity is called as tan of sum of two angles identity or tan of compound angle identity. It is mainly used in mathematics in two cases possibly.
The tan of angle sum identity is written in several ways in which $\tan{(A+B)}$, $\tan{(x+y)}$ and $\tan{(\alpha+\beta)}$ are popular in the world. You can write it in terms of any two angles.
$(1) \,\,\,\,\,\,$ $\tan{(A+B)}$ $\,=\,$ $\dfrac{\tan{A}+\tan{B}}{1\,–\,\tan{A}\tan{B}}$
$(2) \,\,\,\,\,\,$ $\tan{(x+y)}$ $\,=\,$ $\dfrac{\tan{x}+\tan{y}}{1\,–\,\tan{x}\tan{y}}$
$(3) \,\,\,\,\,\,$ $\tan{(\alpha+\beta)}$ $\,=\,$ $\dfrac{\tan{\alpha}+\tan{\beta}}{1\,–\,\tan{\alpha}\tan{\beta}}$
You have learned the tan of sum of two angles formula and let’s learn how the tan of angle sum identity is derived in mathematical form by the geometrical approach.
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
|
Combination Theorem for Cauchy Sequences/Quotient Rule Theorem
Let $\sequence {x_n}$ and $\sequence {y_n}$ be Cauchy sequences in $R$.
Suppose $\sequence {y_n}$ does not converge to $0$.
Then:
$\exists K \in \N : \forall n > K : y_n \ne 0$.
and the sequences
$\sequence { {x_{K+n}} \paren {y_{K+n}}^{-1} }_{n \in \N}$ and $\sequence { \paren {y_{K+n}}^{-1} {x_{K+n}} }_{n \in \N}$ are well-defined and Cauchy sequences. Proof $\exists K \in \N : \forall n > K : y_n \ne 0$.
and the sequence
$\sequence { \paren {x_{K+n}}^{-1} }_{n \in \N}$ is well-defined and a Cauchy sequence.
By Subsequence of a Cauchy Sequence is a Cauchy Sequence then $\sequence { x_{K+n} }_{n \in \N}$ is a Cauchy sequence.
the sequences $\sequence { {x_{K+n}} \paren {y_{K+n}}^{-1} }_{n \in \N}$ and $\sequence { \paren {y_{K+n}}^{-1} {x_{K+n}} }_{n \in \N}$ are Cauchy sequences.
$\blacksquare$
|
Existence of Logarithm Theorem
Let $b, y \in \R$ such that $b > 1$ and $y > 0$.
Then there exists a unique real $x \in \R$ such that $b^x = y$.
This $x$ is called the logarithm of $y$ to the base $b$. Also see the definition of a (general) logarithm. Proof
We start by establishing a lemma:
Lemma 1
Let $t \in \R$ be such that $t > 1$, and let $n \in \N$ be such that $n > \dfrac {b - 1} {t - 1}$.
Then $b^{\frac 1 n} < t$.
Proof of Lemma 1
Because $b > 1$, we have that $b^{\frac 1 n} > 1$.
From Sum of Geometric Progression and this observation, we find that:
$\displaystyle \frac {\left({b^{\frac 1 n} }\right)^n - 1} {b^{\frac 1 n} - 1} = \sum_{k \mathop = 0}^{n - 1} \left({ b^{\frac 1 n} }\right)^k > n$
Rewriting this inequality, we obtain:
$b - 1 > n \left({b^{\frac 1 n} - 1}\right)$
As $n > \dfrac {b - 1} {t - 1}$, this means that:
$t - 1 > b^{\frac 1 n} - 1$
We conclude that:
$b^{\frac 1 n} < t$
$\Box$
Now let $w \in \R$ be any number satisfying $b^w < y$.
Then $t = b^{-w} y > 1$, so we can apply the lemma 1.
It follows that for $n > \dfrac {b - 1} {t - 1}$:
$b^{\frac 1 n} < b^{-w} y$
and thus:
$b^{w + \frac 1 n} < y$
Similarly, when $b^w > y$ and $n > \dfrac {b - 1} {t - 1}$, we have:
$b^{w - \frac 1 n} > y$ Next, let $W = \left\{ {w \in \R: b^w < y }\right\}$.
Let $x = \sup W$. The following lemma helps us to verify existence of $x$.
Lemma 2
For any real number $z > 0$, we have $n \in \N$ such that:
$b^{-n} < z < b^n$ Proof of Lemma 2
Observe that, as $z > 0$, $b^{-n} < z$ whenever $b^n > \dfrac 1 z$.
Therefore, it suffices to show that there is an $n \in \N$ such that $b^n > z$ exists.
By Sum of Geometric Progression, we obtain for any $n \in \N$: $\displaystyle \left({b - 1}\right) \sum_{k \mathop = 0}^{n - 1} b^k = b^n - 1$
As $b > 1$, we obtain:
$n \left({b - 1}\right) < b^n - 1$
This implies that any $n$ with $n \left({b - 1}\right) > z - 1$ has the desired property.
$\Box$
We will now prove that $b^x = y$.
Suppose $b^x < y$.
We have seen that there exists an $n \in \N$ such that $b^{x + \frac 1 n} < y$.
That is:
$x + \dfrac 1 n \in W$
But this is impossible as $x = \sup W$.
Thus:
$b^x \ge y$ Now suppose $b^x > y$.
We have seen that there exists an $n \in \N$ such that $b^{x - \frac 1 n} > y$.
That is:
$x - \dfrac 1 n \not \in W$
As $x - \frac 1 n$ is not the supremum of $W$, we find a $w \in W$ such that:
$x - \dfrac 1 n < w < x$
But then we have, as $b > 1$:
$b^{x - \frac 1 n} - b^w = b^{x - \frac 1 n}\left({1 - b^{w - x + \frac 1 n}}\right) < 0$
This is contradicts our initial choice of $x - \dfrac 1 n$ outside $W$.
We conclude that $b^x = y$.
Lastly, suppose that $b^x = y = b^z$.
Then:
$b^{z-x} = \dfrac {b^z} {b^x} = 1 = \dfrac {b^x} {b^z} = b^{x-z}$
It is concluded that $x = z$.
$\blacksquare$
|
I’d like to give a simple account of what I call the hierarchy of logical expressivity for fragments of classical propositional logic. The idea is to investigate and classify the expressive power of fragments of the traditional language of propositional logic, with the five familiar logical connectives listed below, by considering subsets of these connectives and organizing the corresponding sublanguages of propositional logic into a hierarchy of logical expressivity.
conjunction (“and”), denoted $\wedge$ disjunction (“or”), denoted $\vee$ negation (“not”), denoted $\neg$ conditional (“if…, then”), denoted $\to$ biconditional (“if and only if”), denoted $\renewcommand\iff{\leftrightarrow}\iff$
With these five connectives, there are, of course, precisely thirty-two ($32=2^5$) subsets, each giving rise to a corresponding sublanguage, the language of propositional assertions using only those connectives. Which sets of connectives are equally as expressive or more expressive than which others? Which sets of connectives are incomparable in their expressivity? How many classes of expressivity are there?
Before continuing, let me mention that Ms. Zoë Auerbach (CUNY Graduate Center), one of the students in my logic-for-philosophers course this past semester, Survey of Logic for Philosophers, at the CUNY Graduate Center in the philosophy PhD program, had chosen to write her term paper on this topic. She has kindly agreed to make her paper, “The hierarchy of expressive power of the standard logical connectives,” available here, and I shall post it soon.
To focus the discussion, let us define what I call the (pre)order of logical expressivity on sets of connectives. Namely, for any two sets of connectives, I define that $A\leq B$ with respect to logical expressivity, just in case every logical expression in any number of propositional atoms using only connectives in $A$ is logically equivalent to an expression using only connectives in $B$. Thus, $A\leq B$ means that the connectives in $B$ are collectively at least as expressive as the connectives in $A$, in the sense that with the connectives in $B$ you can express any logical assertion that you were able to express with the connectives in $A$. The corresponding equivalence relation $A\equiv B$ holds when $A\leq B$ and $B\leq A$, and in this case we shall say that the sets are expressively equivalent, for this means that the two sets of connectives can express the same logical assertions.
The full set of connectives $\{\wedge,\vee,\neg,\to,\iff\}$ is well-known to be
complete for propositional logic in the sense that every conceivable truth function, with any number of propositional atoms, is logically equivalent to an expression using only the classical connectives. Indeed, already the sub-collection $\{\wedge,\vee,\neg\}$ is fully complete, and hence expressively equivalent to the full collection, because every truth function can be expressed in disjunctive normal form, as a disjunction of finitely many conjunction clauses, each consisting of a conjunction of propositional atoms or their negations (and hence altogether using only disjunction, conjunction and negation). One can see this easily, for example, by noting that for any particular row of a truth table, there is a conjunction expression that is true on that row and only on that row. For example, the expression $p\wedge\neg r\wedge s\wedge \neg t$ is true on the row where $p$ is true, $r$ is false, $s$ is true and $t$ is false, and one can make similar expressions for any row in any truth table. Simply by taking the disjunction of such expressions for suitable rows where a $T$ is desired, one can produce an expression in disjunctive normal form that is true in any desired pattern (use $p\wedge\neg p$ for the never-true truth function). Therefore, every truth function has a disjunctive normal form, and so $\{\wedge,\vee,\neg\}$ is complete.
Pressing this further, one can eliminate either $\wedge$ or $\vee$ by systematically applying the de Morgan laws
$$p\vee q\quad\equiv\quad\neg(\neg p\wedge\neg q)\qquad\qquad p\wedge q\quad\equiv\quad\neg(\neg p\vee\neg q),$$ which allow one to reduce disjunction to conjunction and negation or reduce conjunction to disjunction and negation. It follows that $\{\wedge,\neg\}$ and $\{\vee,\neg\}$ are each complete, as is any superset of these sets, since a set is always at least as expressive as any of it subsets. Similarly, because we can express disjunction with negation and the conditional via $$p\vee q\quad\equiv\quad \neg p\to q,$$ it follows that the set $\{\to,\neg\}$ can express $\vee$, and hence also is complete. From these simple observations, we may conclude that each of the following fourteen sets of connectives is complete. In particular, they are all expressively equivalent to each other. $$\{\wedge,\vee,\neg,\to,\iff\}$$ $$\{\wedge,\vee,\neg,\iff\}\qquad\{\wedge,\to,\neg,\iff\}\qquad\{\vee,\to,\neg,\iff\}\qquad \{\wedge,\vee,\neg,\to\}$$ $$\{\wedge,\neg,\iff\}\qquad\{\vee,\neg,\iff\}\qquad\{\to,\neg,\iff\}$$ $$\{\wedge,\vee,\neg\}\qquad\{\wedge,\to,\neg\}\qquad\{\vee,\to,\neg\}$$ $$\{\wedge,\neg\}\qquad \{\vee,\neg\}\qquad\{\to,\neg\}$$
Notice that each of those complete sets includes the negation connective $\neg$. If we drop it, then the set $\{\wedge,\vee,\to,\iff\}$ is not complete, since each of these four connectives is truth-preserving, and so any logical expression made from them will have a $T$ in the top row of the truth table, where all atoms are true. In particular, these four connectives collectively cannot express negation $\neg$, and so they are not complete.
Clearly, we can express the biconditional as two conditionals, via $$p\iff q\quad\equiv\quad (p\to q)\wedge(q\to p),$$ and so the $\{\wedge,\vee,\to,\iff\}$ is expressively equivalent to $\{\wedge,\vee,\to\}$. And since disjunction can be expressed from the conditional with $$p\vee q\quad\equiv\quad ((p\to q)\to q),$$ it follows that the set is expressively equivalent to $\{\wedge,\to\}$. In light of $$p\wedge q\quad\equiv\quad p\iff(p\to q),$$ it follows that $\{\to,\iff\}$ can express conjunction and hence is also expressively equivalent to $\{\wedge,\vee,\to,\iff\}$. Since
$$p\vee q\quad\equiv\quad(p\wedge q)\iff(p\iff q),$$ it follows that $\{\wedge,\iff\}$ can express $\vee$ and hence also $\to$, because $$p\to q\quad\equiv\quad q\iff(q\vee p).$$ Similarly, using $$p\wedge q\quad\equiv\quad (p\vee q)\iff(p\iff q),$$ we can see that $\{\vee,\iff\}$ can express $\wedge$ and hence also is expressively equivalent to $\{\wedge,\vee,\iff\}$, which we have argued is equivalent to $\{\wedge,\vee,\to,\iff\}$. For these reasons, the following sets of connectives are expressively equivalent to each other. $$\{\wedge,\vee,\to,\iff\}$$ $$\{\wedge,\vee,\to\}\qquad\{\wedge,\vee,\iff\}\qquad \{\vee,\to,\iff\}\qquad \{\wedge,\to,\iff\}$$ $$\{\wedge,\iff\}\qquad \{\vee,\iff\}\qquad \{\to,\iff\}\qquad \{\wedge,\to\}$$ And as I had mentioned, these sublanguages are strictly less expressive than the full language, because these four connectives are all truth-preserving and therefore unable to express negation.
The set $\{\wedge,\vee\}$, I claim, is unable to express any of the other fundamental connectives, because $\wedge$ and $\vee$ are each false-preserving, and so any logical expression built from $\wedge$ and $\vee$ will have $F$ on the bottom row of the truth table, where all atoms are false. Meanwhile, $\to,\iff$ and $\neg$ are not false-preserving, since they each have $T$ on the bottom row of their defining tables. Thus, $\{\wedge,\vee\}$ lies strictly below the languages mentioned in the previous paragraph in terms of logical expressivity.
Meanwhile, using only $\wedge$ we cannot express $\vee$, since any expression in $p$ and $q$ using only $\wedge$ will have the property that any false atom will make the whole expression false (this uses the associativity of $\wedge$), and $p\vee q$ does not have this feature. Similarly, $\vee$ cannot express $\wedge$, since any expression using only $\vee$ is true if any one of its atoms is true, but $p\wedge q$ is not like this. For these reasons, $\{\wedge\}$ and $\{\vee\}$ are both strictly weaker than $\{\wedge,\vee\}$ in logical expressivity.
Next, I claim that $\{\vee,\to\}$ cannot express $\wedge$, and the reason is that the logical operations of $\vee$ and $\to$ each have the property that any expression built from that has at least as many $T$’s as $F$’s in the truth table. This property is true of any propositional atom, and if $\varphi$ has the property, so does $\varphi\vee\psi$ and $\psi\to\varphi$, since these expressions will be true at least as often as $\varphi$ is. Since $\{\vee,\to\}$ cannot express $\wedge$, this language is strictly weaker than $\{\wedge,\vee,\to,\iff\}$ in logical expressivity. Actually, since as we noted above $$p\vee q\quad\equiv\quad ((p\to q)\to q),$$ it follows that $\{\vee,\to\}$ is expressively equivalent to $\{\to\}$.
Meanwhile, since $\vee$ is false-preserving, it cannot express $\to$, and so $\{\vee\}$ is strictly less expressive than $\{\vee,\to\}$, which is expressively equivalent to $\{\to\}$.
Consider next the language corresponding to $\{\iff,\neg\}$. I claim that this set is not complete. This argument is perhaps a little more complicated than the other arguments we have given so far. What I claim is that both the biconditional and negation are parity-preserving, in the sense that any logical expression using only $\neg$ and $\iff$ will have an even number of $T$’s in its truth table. This is certainly true of any propositional atom, and if true for $\varphi$, then it is true for $\neg\varphi$, since there are an even number of rows altogether; finally, if both $\varphi$ and $\psi$ have even parity, then I claim that $\varphi\iff\psi$ will also have even parity. To see this, note first that this biconditional is true just in case $\varphi$ and $\psi$ agree, either having the pattern T/T or F/F. If there are an even number of times where both are true jointly T/T, then the remaining occurrences of T/F and F/T will also be even, by considering the T’s for $\varphi$ and $\psi$ separately, and consequently, the number of occurrences of F/F will be even, making $\varphi\iff\psi$ have even parity. If the pattern T/T is odd, then also T/F and F/T will be odd, and so F/F will have to be odd to make up an even number of rows altogether, and so again $\varphi\iff\psi$ will have even parity. Since conjunction, disjunction and the conditional do not have even parity, it follows that $\{\iff,\neg\}$ cannot express any of the other fundamental connectives.
Meanwhile, $\{\iff\}$ is strictly less expressive than $\{\iff,\neg\}$, since the biconditional $\iff$ is truth-preserving but negation is not. And clearly $\{\neg\}$ can express only unary truth functions, since any expression using only negation has only one propositional atom, as in $\neg\neg\neg p$. So both $\{\iff\}$ and $\{\neg\}$ are strictly less expressive than $\{\iff,\neg\}$.
Lastly, I claim that $\iff$ is not expressible from $\to$. If it were, then since $\vee$ is also expressible from $\to$, we would have that $\{\vee,\iff\}$ is expressible from $\to$, contradicting our earlier observation that $\{\to\}$ is strictly less expressive than $\{\vee,\iff\}$, as this latter set can express $\wedge$, but $\to$ cannot, since every expression in $\to$ has at least as many $T$’s as $F$’s in its truth table.
These observations altogether establish the hierarchy of logical expressivity shown in the diagram displayed above.
It is natural, of course, to want to extend the hierarchy of logical expressivity beyond the five classical connectives. If one considers all sixteen binary logical operations, then Greg Restall has kindly produced the following image, which shows how the hierarchy we discussed above fits into the resulting hierarchy of expressivity. This diagram shows only the equivalence classes, rather than all $65536=2^{16}$ sets of connectives.
If one wants to go beyond merely the binary connectives, then one lands at Post’s lattice, pictured below (image due to Emil Jeřábek), which is the countably infinite (complete) lattice of logical expressivity for all sets of truth functions, using any given set of Boolean connectives. Every such set is finitely generated.
|
Forgot password? New user? Sign up
Existing user? Log in
Say anything you have in your mind! This is for an informal conversation!
Note by Ameya Salankar 5 years, 5 months ago
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
_italics_
**bold**
__bold__
- bulleted- list
1. numbered2. list
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"
# I indented these lines# 4 spaces, and now they show# up as a code block.print "hello world"
\(
\)
\[
\]
2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Sort by:
@milind prabhu, I don't know why, but my Gmail suddenly crashed! Sorry for the inconvenience!
Log in to reply
@Christopher Boo, did you try my set on RMO & INMO? I was inspired by you!
Nice "Trickster" set that you collected. Good collection of diverse problems.
Also, could you change the fire question to Which State Of Matter Is Fire? There is a better discussion on that problem.
Yes! I have done that. Thank You!
Good music.
Nirvana, Radiohead, Sigur Ros, Ratatat.
I don't usually listen to a certain bands, I listen to songs I think is great.
Can you tell me how to make a hash-tag @Sam Solomon? Also, in some problems posted by @Arron Kau, I saw a faint line after the problem below which, details & assumptions were written. Can you also tell me how to construct that line?
Edit: Calvin was here.
@Ameya Salankar – To construct a line to separate your paragraphs, place 5 "-" without spaces. I've edited your note so that you can see it.
@Calvin Lin – I am really thankful for your support @Calvin Lin. But can you tell me how to create a hash-tag?
@Ameya Salankar – You can add tags when you create a problem, note or set. We have not added hashtags to comments as yet, and may add more functionality later.
What?
Just thinking what to write!!
Well, don't just think, WRITE!
What came first chicken or egg!!??
@Satvik Golechha – No, better : Egg
@Satvik Golechha – I think chicken.
@Ameya Salankar – where did that chicken come from?
everything's falling back to place! as it always did!
What does that mean?
actually, i wasn't thinking when i wrote this.it just struck my mind.
Problem Loading...
Note Loading...
Set Loading...
|
I have system of two equations that describes position of robot end-effector ($X_C, Y_C, Z_C$), in accordance to prismatic joints position ($S_A, S_B$):
$S^2_A - \sqrt3(S_A + S_B)X_C = S^2_B + (S_A - S_B)Y_C$
$X^2_C + Y^2_C + Z^2_C = L^2 - S^2_A + S_A(\sqrt3X_C + Y_C)+M(S^2_A+S_BS_A + S^2_B)$
where M and L are constants.
In paper, author states that differentiating this system at given point ($X_C, Y_C, Z_C$) gives the "differential relationship" in form:
$a_{11}\Delta S_A + a_{12}\Delta S_B = b_{11}\Delta X_C + b_{12}\Delta Y_C + b_{13}\Delta Z_C$
$a_{21}\Delta S_A + a_{22}\Delta S_B = b_{21}\Delta X_C + b_{22}\Delta Y_C + b_{23}\Delta Z_C$
Later on, author uses those parameters ($a_{11}, a_{12}, b_{11}...$) to construct matrices, and by multiplying those he obtains Jacobian of the system.
Im aware of partial differentiation, but I have never done this for system of equations, neither I understand how to get those delta parameters.
Can anyone explain what are the proper steps to perform partial differentiation on this system, and how to calculate delta parameters?
EDIT
Following advice given by N. Staub, I differentiated equations w.r.t time.
First equation:
$S^2_A - \sqrt3(S_A + S_B)X_C = S^2_B + (S_A - S_B)Y_C$ $=>$
$2S_A \frac{\partial S_A}{\partial t} -\sqrt3S_A \frac{\partial X_C}{\partial t} -\sqrt3X_C \frac{\partial S_A}{\partial t} -\sqrt3S_B \frac{\partial X_C}{\partial t} -\sqrt3X_C \frac{\partial S_B}{\partial t} = 2S_B\frac{\partial S_B}{\partial t} + S_A\frac{\partial Y_C}{\partial t} + Y_C\frac{\partial S_A}{\partial t} - S_B\frac{\partial Y_C}{\partial t} - Y_C\frac{\partial S_B}{\partial t}$
Second equation:
$X^2_C + Y^2_C + Z^2_C = L^2 - S^2_A + S_A(\sqrt3X_C + Y_C)+M(S^2_A+S_BS_A + S^2_B)$ $=>$
$2X_C \frac{\partial X_C}{\partial t} + 2Y_C \frac{\partial Y_C}{\partial t} + 2Z_C \frac{\partial Z_C}{\partial t} = -2S_A \frac{\partial S_A}{\partial t} + \sqrt3S_A \frac{\partial X_C}{\partial t} +\sqrt3X_C \frac{\partial S_A}{\partial t} + S_A \frac{\partial Y_C}{\partial t} + Y_C \frac{\partial S_A}{\partial t} + 2MS_A \frac{\partial S_A}{\partial t} + MS_B \frac{\partial S_A}{\partial t} + MS_A \frac{\partial S_B}{\partial t} + 2MS_B \frac{\partial S_B}{\partial t}$
then, I multiplied by $\partial t$, and grouped variables:
First equation:
$(2S_A -\sqrt3X_C - Y_C)\partial S_A +(-2S_B -\sqrt3X_C + Y_C)\partial S_B = (\sqrt3S_A +\sqrt3S_B)\partial X_C + (S_A - S_B)\partial Y_C$
Second equation:
$(-2S_A+\sqrt3X_C+Y_C+2MS_A + MS_B)\partial S_A + (MS_A + 2MS_B)\partial S_B = (2X_C-\sqrt3S_A)\partial X_C + (2Y_C-S_A)\partial Y_C + (2Z_C)\partial Z_C$
therefore I assume that required parameters are:
$a_{11} = 2S_A -\sqrt3X_C - Y_C$
$a_{12} = -2S_B -\sqrt3X_C + Y_C$
$a_{21} = -2S_A + \sqrt3X_C + Y_C + 2MS_A + MS_B$
$a_{22} = MS_A + 2MS_B$
$b_{11} = \sqrt3S_A +\sqrt3S_B$
$b_{12} = S_A - S_B$
$b_{13} = 0$
$b_{21} = 2X_C - \sqrt3S_A$
$b_{22} = 2Y_C - S_A$
$b_{23} = 2Z_C$
Now. According to paper, jacobian of the system can be calculated as:
$J = A^{-1} B$, where
$A=(a_{ij})$
$B=(b_{ij})$
so I im thinking right, it means:
$$ A = \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{matrix} $$
$$ B = \begin{matrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ \end{matrix} $$
and Jacobian is multiplication of reverse A matrix and B matrix.
Next, author states that Jacobian at given point, where
$X_C = 0$
$S_A=S_B=S$
$Y_C = l_t-\Delta\gamma$
is equal to:
$$ J = \begin{matrix} \frac{\sqrt3S}{2S-l_t+\Delta\gamma} & -\frac{\sqrt3S}{2S-l_t+\Delta\gamma}\\ \frac{2(l_t-\Delta\gamma)-S}{6cS-2S+l_t-\Delta\gamma} & \frac{2(l_t-\Delta\gamma)-S}{6cS-2S+l_t-\Delta\gamma} \\ \frac{2Z_C}{6cS-2S+l_t-\Delta\gamma} & \frac{2Z_C}{6cS-2S+l_t-\Delta\gamma} \\ \end{matrix} ^T $$
Everything seems fine. BUT! After multiplicating my A and B matrices I get some monster matrix, that I am unable to paste here, becouse it is so frickin large! Substituting variables for values given by author does not give me proper jacobian (i tried substitution before multiplying matrices (on parameters), and after multiplication (on final matrix)). So clearly Im still missing something. Either Ive done error in differentiation, or Ive done error in matrix multiplication (I used maple) or I dont understand how to subsititue those values. Can anyone point me in to right direction?
EDITProblem solved! Parameters that I calculated were proper, I just messed up simplification of equations in final matrix. Using snippet from Petch Puttichai I was able to obtain full Jacobian of system. Thanks for help!
|
Let's take an aqueous solution of a salt $\ce{NaHA}$ with the initial concentration $C$ when added to water. It will completely dissociate according to the eaquation: $\ce{NaHA(s) \rightarrow Na^+ +HA^-}$.
$\ce{HA^-}$ will participate in three equilibria:
$\ce{2HA^- \leftrightarrows H2A +A^{2-}\quad \quad \quad }$ ${K_1^0=K_{A2}/K_{A1}}$
$\ce{HA^- +H2O\leftrightarrows H3O^+ +A^{2-}\quad }$ $K_2^0=K_{A2}$
$\ce{HA^- +H2O\leftrightarrows OH^- +H_2A^\quad }$ $K_3^0=K_{B1}={K_w}/{K_{A1}}$
In most cases, $K_1^0$ is far bigger than $K_2^0$ and $K_3^0$. So, the first equilibrium is
, and this reaction will impose the pH of the solution. the preponderant reaction
Let's now calculate the product $K_{A2}\times K_{A1}$:
$K_{A2}\times K_{A1}= \frac{[\ce{}A^{2-}].[\ce{H3O+}]}{\ce{[HA^-]}}.\frac{[\ce{}HA^{-}].[\ce{H3O+}]}{\ce{[H_2A]}}$
According to the stoichiometry of the preponderant reaction, we have $\ce{[A^{2-}]=[H_2A^]}$. So the product$K_{A2}\times K_{A1}= \ce{[H3O+}]^2}$ .i.e. $\ce{pH}=0.5(\ce{p}K_{A2} +\ce{p} K_{A1})$
|
I have encountered an expression for an inverse Laplace transform, and now I am wondering whether it is correct or maybe there is some error. More likely I don't understand it. so here goes:
$$ g(p)(e^{ap}+\beta)^{-c} \Leftrightarrow \sum_{0\leq n\leq t/a -c}\left(\begin{array}[c] - -c\\ n\end{array}\right)\beta^n f(t-ac-an) $$
Where $a,c>0$.
So what is the meaning of this? Anyone?
Thanks
|
Heat Transfer in Fully-Developed Internal Turbulent Flow From Thermal-FluidsPedia
Line 254: Line 254:
|{{EquationRef|(28)}}
|{{EquationRef|(28)}}
|}
|}
-
Since the buffer region is also very thin,
+
Since the buffer region is also very thin, <math>1-y/{{r}_{0}}</math> in eq. (5.321) is effectively equal to 1. Defining <math>{{T}^{+}}={(\bar{T}-{{T}_{w}})}/{\left( -\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}}\sqrt{\frac{\rho }{{{\tau }_{w}}}} \right)}\;</math> and Integrating eq. (5.321) yields
-
<math>1-y/{{r}_{0}}</math>
+ - + -
<math>{{T}^{+}}={(\bar{T}-{{T}_{w}})}/{\left( -\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}}\sqrt{\frac{\rho }{{{\tau }_{w}}}} \right)}\;</math>
+ - +
{| class="wikitable" border="0"
{| class="wikitable" border="0"
Revision as of 03:04, 8 July 2010
Heat transfer in fully-developed turbulent flow in a circular tube subject to constant heat flux (
q'' = const) will be considered in this subsection (Oosthuizen and Naylor, 1999). When the turbulent flow in the tube is fully developed, we have and the energy eq. (5.268) becomes w
After the turbulent flow is hydrodynamically and thermally fully developed, the time-averaged temperature profile is no longer a function of axial distance from the inlet, i.e.,
where is the time-averaged temperature at the centerline of the tube, and Tw is the wall temperature. Thus, is a function of r only, i.e.,
where f is independent from x. Differentiating (5.295) yields
At the wall, the contribution of eddy diffusivity on the heat transfer is negligible, and the heat flux at the wall becomes
Substituting eq. (5.296) into eq. (5.298), one obtains:
Since the heat flux is constant,
q'' = const, it follows that , i.e., w
Therefore, eq. (5.297) becomes:
For fully developed flow, the local heat transfer coefficient is:
where is the time-averaged mean temperature defined as:
Since
q'' = const, it follows from eq. (5.302) that , i.e., w
Combining eqs. (5.300), (5.301) and (5.304), the following relationships are obtained:
The time-averaged mean temperature, , changes with x as the result of heat transfer from the tube wall. By following the same procedure as that in Example 5.2, the rate of mean temperature change can be obtained as follows:
Substituting eq. (5.305) into eq. (5.294), the energy equation becomes:
where
y = r 0 − r is the distance measured from the tube wall. Equation (5.307) is subject to the following two boundary conditions:
(axisymmetric condition)
Integrating eq. (5.307) in the interval of (r0, r) and considering eq. (5.308), we have:
which can be rearranged to
where
Integrating eq. (5.311) in the interval of (0, y) and considering eq. (5.309), one obtains:
If the profiles of axial velocity and the thermal eddy diffusivity are known, eq. (5.313) can be used to obtain the correlation for internal forced convection heat transfer. With the exception of the very thin viscous sublayer, the velocity profile in the most part of the tube is fairly flat. Therefore, it is assumed that the time-averaged velocity, , in eq. (5.312) can be replaced by , and I(y) becomes:
Substituting eqs. (5.314) and (5.306) into eq. (5.313) yields:
which can be rewritten in terms of wall coordinate
where y+ is defined in eq. (5.277). To consider heat transfer in an internal turbulent flow, the entire turbulent boundary layer is divided into three regions: (1) inner region (
y + < 5), (2) buffer region (), and (3) outer region ( y + > 30). In the inner region and eq. (5.316) becomes
Since the inner region is very thin, and 1 −
y / r 0 is effectively equal to 1. Therefore, the temperature profile in the inner region becomes:
The temperature at the boundary between the inner and buffer regions (
y + = 5), , can be obtained from eq. (5.318) as
In the buffer region where , the eddy diffusivity in the buffer region is:
Substituting eq. (5.320) into eq. (5.316) and assuming the turbulent Prandtl number , the following expression is obtained:
Since the buffer region is also very thin, 1 −
y / r 0 in eq. (5.321) is effectively equal to 1. Defining and Integrating eq. (5.321) yields
i.e.,
The temperature at the top of the buffer region where
y + = 30, , becomes
For the outer region where , eq. (5.316) becomes
where the turbulent Prandtl number is assumed to be equal to 1. It is assumed that the Nikuradse equation (5.276) is valid in the outer region and the velocity gradient in this region becomes:
The expression of apparent shear stress in this region, eq. (5.280) , can be non-dimensionalized using eqs. (5.277) and (5.278) as:
Substituting eqs. (5.275) and (5.326) into eq. (5.327), the eddy diffusivity in the outer region is obtained as:
Substituting eq. (5.328) into eq. (5.325), the temperature distribution in this region becomes:
which is valid from
y + = 30 to the center of the tube where y = c r 0or
The temperature at the center of the tube, , can be obtained by letting
in eq. (5.329), i.e.
The overall temperature change from the wall to the center of the tube can be obtained by adding eqs. (5.319), (5.324) and (5.331):
It follows from the definition of friction factor, eq. (5.282), that
Substituting eq. (5.333) into eq. (5.332) and considering the definition of Reynolds number, , eq. (5.332) becomes:
(5.334) In order to obtain the heat transfer coefficient, , the temperature difference must be obtained. If the velocity profile can be approximated by eq. (5.287), and the temperature and velocity can also be approximated by the one-seventh law, i.e.,
it follows that
Substituting eq. (5.334) into eq. (5.336) results in:
which can be rearranged to the following empirical correlation
which can be used together with appropriate friction coefficient discussed in the previous subsection to obtain the Nusselt number.
|
How to determine the range of the following function $\frac{x}{1+ |x|}$?
when I calculated it, it was $\mathbb{R}$, but my professor said that the range is ]-1,1[, could anyone explain for me why?
thanks!
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
How to determine the range of the following function $\frac{x}{1+ |x|}$?
when I calculated it, it was $\mathbb{R}$, but my professor said that the range is ]-1,1[, could anyone explain for me why?
thanks!
Observe that $\large|\frac {x}{1+|x|}|=\frac {|x|}{1+|x|}=1-\frac {1}{1+|x|} \lt 1.$
$\therefore |\frac {x}{1+|x|}| \lt 1 \Rightarrow -1 \lt \frac {x}{1+|x|} \lt 1$.
For $x\geq 0$, $\dfrac{x}{1+|x|}=\dfrac{x}{1+x}=1-\dfrac{1}{1+x}$, it is increasing on $[0,\infty)$, so it maps onto $[0,1)$. Similarly you can deal with $(-\infty,0]$.
Let $f(x)=\frac{x}{1+ |x|}$. Then: $|f(x)|=\frac{|x|}{1+ |x|} \le 1$, hence
$f( \mathbb R) \subseteq [-1,1]$.
Furthermore: $\lim_{x \to \infty}f(x)=1$ and $\lim_{x \to -\infty}f(x)=-1$.
Show that $f(x) \ne 1$ and $f(x) \ne -1$ for all $x$.
Are you now in a position to derive $f( \mathbb R) =]-1,1[$ ?
check : $1+|x|\neq 0\Rightarrow |x|\neq-1$, which is true for all $x$.
Now,
when $0\leq x<\infty , y=\frac{x}{1+x}, x<x+1\Rightarrow 0\leq y<1...(I)$
when $-\infty <x<0 , y=\frac{x}{1-x}$ ,
|numerator|<|denominator|, so $|y|$ lies between $0$ and $1$, but since numerator is negative and denominator is positive,$-1<y<0...(II)$
combining $(I),(II)$ we get $-1<y<1$
|
I am pretty confident that the following limit is $0.5$:
$$\lim_{n\to\infty} \left[\frac{1}{n^{2}} + \frac{2}{n^{2}} + \frac{3}{n^{2}} + \cdots + \frac{n}{n^{2}}\right]=\lim_{n\to\infty} \left[\frac{1+2+3+ \cdots +n}{n^{2}}\right]=\lim_{n\to\infty} \left[\frac{n^2+n}{2n^{2}}\right]=\frac{1}{2}$$
However one of the students argued that if we write limit of sum as sum of individual limits, it will be zero. Why we cannot write limit of sum as sum of limits in this case?
I've been taught that if individual limits exist, the limit of sum is equal to the sum of limits. It would be helpful to get an explanation or a reference to similar rules for limits of series.
|
I'm going to write $X$ for $(X_1,\ldots,X_n)$ and $x$ for $(x_1,\ldots,x_n)$.
The likelihood function when $X$ is observed to have a certain value $x$ is$$
L_1(\theta) = \Pr(X=x \mid \theta) = E(\Pr(X=x\mid H(X),\theta)) = \sum_h \Pr(H(X)=h\mid\theta)\Pr(X=x\mid H(X)=h).
$$In the very last probability, we don't need to write $\Pr(X=x\mid H(X)=h,\theta)$, since lack of this dependence on $\theta$ is the very definition of sufficiency. The index $h$ of course runs through the discrete set of all possible values of the random variable $H(X)$.
Now $\Pr(X=x\mid H(X)=h) = 0$ for any value of $h$ except $h=H(x)$. So all but one of the terms in the last sum vanish and the sum is$$\Pr(H(X)=h\mid\theta)\Pr(X=x\mid H(X)=h) =\Pr(H(X)=H(x)\mid\theta)\Pr(X=x\mid H(X)=H(x)).$$The second factor in the last product does not depend on $\theta$. So as a function of $\theta$, it's constant. Therefore the likelihood function when $H(X)$ is observed to have a certain value $H(x)$ is$$
L_2(\theta) = \Pr(H(X) = h\mid \theta) = \mathrm{constant}\cdot \Pr(X=x\mid \theta) = \mathrm{constant} \cdot L_1(\theta).
$$
Bayes' theorem says if you multiply the likelihood function by the prior pmf and then normalize, you get the posterior pmf.
The "constant" factor referred to above goes away when you normalize; it appears as a factor in both the numerator and the denominator and cancels. Therefore you get the same posterior pmf if you observe $H(X)$ that you get if you observe $X$.
|
Exam-Style Questions on Graphs Problems on Graphs adapted from questions set in previous Mathematics exams.
1. GCSE Higher
The equation of the line L
1 is \(y = 2 - 5x\).
The equation of the line L
2 is \(3y + 15x + 17 = 0\).
Show that these two lines are parallel.
2. GCSE Higher
Show that line \(5y = 7x - 7\) is perpendicular to line \(7y = -5x + 55\).
3. GCSE Higher
A straight line goes through the points \((a, b)\) and \((c, d)\), where$$a + 3 = c$$ $$b + 6 = d$$
Find the gradient of the line.
4. GCSE Higher
The graph shows the height of water in a container over a period time during which the water enters the container at a constant rate.
Which of the following might be a diagram of the container?
a. b. c. d. e.
5. GCSE Higher
The diagram is of a container which is filled with water entering at a constant rate.
Which of the following might be the graph of height of the water in the container plotted against time?
a. b. c. d. e.
6. GCSE Higher
Match the equation with the letter of its graph
Equation Graph $$y=3-\frac{10}{x}$$ $$y=2^x$$ $$y=\sin x$$ $$y=x^2+7x$$ $$y=x^2-8$$ $$y=2-x$$
7. GCSE Higher
The graph of y = f(x) is drawn accurately on the grid.
(a) Write down the coordinates of the turning point of the graph.
(b) Write down estimates for the roots of f(x) = 0
(c) Use the graph to find an estimate for f(-5.5).
8. GCSE Higher
(a) By completing the square, solve \(x^2+8x+13=0\) giving your answer to three significant figures.
(b) From the completed square you found in part (a) find the minimum value of the curve \(y=x^2+8x+13\).
9. GCSE Higher
Suppose a rhombus ABCD is drawn on a coordinate plane with the point A situated at (4,7). The diagonal BD lies on the line \(y = 2x - 5 \)
Find the equation the line that passes through A and C.
10. GCSE Higher
The graph shows the temperature (\(T\)) of an unidentified flying object over a period of 10 seconds (\(t\)).
Use the graph to work out an estimate of the rate of decrease of temperature at 7 seconds. You must show your working.
11. GCSE Higher
The graph of the following equation is drawn and then reflected in the x-axis$$y = 2x^2 - 3x + 2$$
(a) What is the equation of the reflected curve?
The original curve is reflected in the y-axis.
(b) What is the equation of this second reflected curve?
12. GCSE Higher
(a) Find the interval for which \(x^2 - 9x + 18 \le 0\)
(b) The point (-4, -4) is the turning point of the graph of \(y = x^2 + ax + b\), where a and b are integers. Find the values of a and b.
13. GCSE Higher
(a) Write \(2x^2+8x+27\) in the form \(a(x+b)^2+c\) where \(a\), \(b\), and \(c\) are integers, by 'completing the square'
(b) Hence, or otherwise, find the line of symmetry of the graph of \(y = 2x^2+8x+27\)
(c) Hence, or otherwise, find the turning point of the graph of \(y = 2x^2+8x+27\)
14. IB Standard
A function is defined as \(f(x) = 2{(x - 3)^2} - 5\) .
(a) Show that \(f(x) = 2{x^2} - 12x + 13\).
(b) Write down the equation of the axis of symmetry of this graph.
(c) Find the coordinates of the vertex of the graph of \(f(x)\).
(d) Write down the y-intercept.
(e) Make a sketch the graph of \(f(x)\).
Let \(g(x) = {x^2}\). The graph of \(f(x)\) may be obtained from the graph of \(g(x)\) by the two transformations:
(f) Find the values of \(j\), \(k\) and \(s\).
15. IB Studies
Consider a straight line graph L1, which intersects the x-axis at A(8, 0) and the y-axis at B (0, 4).
(a) Write down the coordinates of C, the midpoint of line segment AB.
(b) Calculate the gradient of the line L1.
The line L2 is parallel to L1 and passes through the point (5 , 9).
(c) Find the equation of L2. Give your answer in the form \(ay + bx + c = 0\) where \(a, b \text{ and } c \in \mathbb{Z}\).
16. IB Standard
\(f\) and \(g\) are two functions such that \(g(x)=3f(x+2)+7\).
The graph of \(f\) is mapped to the graph of \(g\) under the following transformations:
A vertical stretch by a factor of \(a\) , followed by a translation \(\begin{pmatrix}b \\c \\ \end{pmatrix}\)
Find the values of
(a) \(a\);
(b) \(b\);
(c) \(c\).
(d) Consider two other functions \(h\) and \(j\). Let \(h(x)=-j(2x)\). The point A(8, 7) on the graph of \(j\) is mapped to the point B on the graph of \(h\). Find the coordinates of B.
17. IB Standard
Let \(f(x) = {x^2}\) and \(g(x) = 3{(x+2)^2}\) .
The graph of \(g\) can be obtained from the graph of \(f\) using two transformations.
(a) Give a full description of each of the two transformations.
(b) The graph of \(g\) is translated by the vector \( \begin{pmatrix}-4\\5\\ \end{pmatrix}\) to give the graph of \(h\).
The point \(( 2{\text{, }}-1)\) on the graph of \(f\) is translated to the point \(P\) on the graph of \(h\).
Find the coordinates of \(P\).
18. IB Standard
Let \(f\) and \(g\) be functions such that \(g(x) = 3f(x - 2) + 7\) .
The graph of \(f\) is mapped to the graph of \(g\) under the following transformations: vertical stretch by a factor of \(k\) , followed by a translation \(\left( \begin{array}{l} p\\ q \end{array} \right)\) .
Write down the value of:
(a) \(k\)
(b) \(p\)
(c) \(q\)
(d) Let \(h(x) = - g(2x)\) . The point A(\(8\), \(7\)) on the graph of \(g\) is mapped to the point \({\rm{A}}'\) on the graph of \(h\) . Find \({\rm{A}}'\)
19. IB Standard
Let \(f(x) = \frac{9x-3}{bx+9}\) for \(x \neq -\frac9b, b \neq 0\).
(a) The line \(x = 3\) is a vertical asymptote to the graph of \(f\). Find the value of b.
(b) Write down the equation of the horizontal asymptote to the graph of \(f\).
(c) The line \(y = c\) , where \(c\in \mathbb R\) intersects the graph of \( \begin{vmatrix}f(x) \end{vmatrix} \) at exactly one point. Find the possible values of \(c\).
20. IB Standard
Two functions are defined as follows: \(f(x) = 2\ln x\) and \(g(x) = \ln \frac{x^2}{3}\).
(a) Express \(g(x)\) in the form \(f(x) - \ln a\) , where \(a \in {{\mathbb{Z}}^ + }\) .
(b) The graph of \(g(x)\) is a transformation of the graph of \(f(x)\) . Give a full geometric description of this transformation.
|
I'm doing
Exercise I.11.8 from textbook Analysis I by Amann/Escher.
Show that the identity function and $z \mapsto \overline{z}$ are the only field automorphisms of $\mathbb{C}$ which leave the elements of $\mathbb{R}$ fixed.
Could you please verify if my attempt contains logical gaps/errors?
My attempt:
Let $\phi:\mathbb{C} \to \mathbb{C}$ be an automorphism of $\mathbb{C}$ which leaves the elements of $\mathbb{R}$ fixed.
Since group homomorphism maps identity elements to identity elements, $\phi(1) = 1$ and $\phi(0) = 0$. Then $(\phi(i))^2 = \phi(i^2) = \phi (-1) = - \phi(1) = -1$, so $\phi(i) = \pm i$.
For $z = a + ib \in \mathbb{C}$, we have $$\begin{aligned}\phi(z) &= \phi (a + ib) \\ &= \phi(a) + \phi(i) \phi(b) \\ &= a + \phi(i)b \\ &= a \pm ib \\ & = z \text{ or } \overline{z}\end{aligned}$$
This completes the proof.
|
Let $\mathcal H^p$, with $p \in [1,\infty)$, be the space of all (continuous-time) martingales $M$ such that $$ \|M\|_{\mathcal H^p} := \mathbb E\left[\sup_t \left| M_t\right|^p\right]^{1/p} < \infty. $$
I want to show that (identifying indistinguishable martingales) $\mathcal H^p$ is complete. Could someone help me complete the following proof?
Suppose that $\left\{ X^n \right\}$ is a Cauchy sequence in $\mathcal H^p$, so that, for every $\varepsilon > 0$, there is an $N$ such that $$ \| X^n -X^m \|_{\mathcal H^p} < \varepsilon $$ for all $n,m \ge N$. Since $$ \sup_t \mathbb E\left[\left| X^n_t - X^m_t \right|^p\right] \le \| X^n -X^m \|_{\mathcal H^p}^p, $$ we have that $\left\{ X^n_t \right\}$ is uniformly (in $t$) Cauchy in $L^p$, and thus must converge uniformly in $L^p$ to some $X_t \in L^p$.
It is straightforward to verify that $X$ must be a martingale. How do you show that $X^n \to X$ in the $\mathcal H^p$-norm?
By passing to a subsequence if necessary, we can assume that $X^n_t \to X_t$ uniformly almost surely. That is, $$ \sup_t | X_t^n -X_t |^p \to 0 \quad \text{a.s.} $$ One might then be able to appeal to an appropriate convergence theorem to show that $$ \mathbb E \left[ \sup_t | X_t^n -X_t |^p \right] \to 0. $$ We can then use the fact that $\{X^n\}$ is Cauchy to recover convergence for the original sequence. Equivalently, one can show that $\sup_t | X_t^n -X_t |^p$ is uniformly integrable. Unfortunately, I don't know how to show this.
Another thought is to use Doob's $L^p$ inequality, but that only works for $p > 1$.
Any suggestions? I suspect this should be rather simple, but my analysis is quite rusty.
|
I have the following problem emerged. Let's say we have $l$ finite sets $A_1, A_2, \ldots, A_l$ with cardinality of $n_1, n_2, \ldots\, n_l$, respectively. We know that $| A_i \cap A_j | \le a_{ij}$ (ibvously, $a_{ij}=a_{ji}$).
What could be a lower bound on $\left|\bigcup_{k=1}^l A_k \right|$?
The bound I would be satisfied with should be easy (polynomial) to compute and thus could be rough (but better than just maximum of $n_i$).
Some notes
From inclusion-exclusion principle we know exact expresion for that: $$ \left|\bigcup_{k=1}^l A_k\right| = \sum_{k=1}^l (-1)^{k+1} \left( \sum_{1 \le i_1 < i_2 < \ldots < i_k \le l} \left| A_{i_1} \cap A_{i_2} \cap \ldots \cap A_{i_k} \right| \right) $$ but it is an exponintioal computation and we don't know all the intersections anyway.
In the real problem I am facing,there are the following values $$ n_i = w_i \binom{n-w_i}{k-1}, \\ a_{ij} = \delta \binom{n-w_i-w_j+\delta}{k-1}, \\ w_i, n, k, \delta \mbox{ are some natural numbers.} $$ But it's getting more messy to work with these expressions.
UPD. The question could be re-phrased like the following.
Find an easy computable function $\beta(l; a_{ij})$ such that for every $l$, $A_j$ and $a_{ij} = a_{ji}$ it holds $$ \left|\bigcup_{k=1}^l A_k\right| \ge \beta(l; a_{ij}) $$
|
Communities (10) Top network posts 139 group by two columns in ggplot2 71 Calculate difference between values in consecutive rows by group 59 How can you read a CSV file in R with different number of columns 39 How to type logarithms in Wolfram|Alpha? 25 Every Function in a Finite Field is a Polynomial Function 21 How to show $\lim_{n \to \infty} a_n = \frac{ [x] + [2x] + [3x] + \dotsb + [nx] }{n^2} = x/2$? 21 Grep in R using OR and NOT View more network posts → Top tags (4)
76 How can non-experts best contribute? Apr 26 '11
7 Tag merging and synonyms Jun 26 '11
-3 Tag merging and synonyms Jun 23 '11
|
$\log_{b}{b}$ $\,=\,$ $1$
The logarithm of a nonzero positive number to the same quantity is called logarithm of base rule.
Logarithm of any number (nonzero positive number) is equal to one when the same number is taken as a base of the logarithm. There is a reason for this property. When a quantity is split as multiplying factors on the basis of same quantity, the total number of multiplying factors is one.
Therefore, it is called as logarithm of base rule. It is also simply called as log of base rule.
$m$ is a quantity and it is written as multiplying factors of another quantity $b$. Assume, the total number of multiplying factors of $b$ is equal to $x$.
$m$ $\,=\,$ $\underbrace{b \times b \times b \times \ldots \times b}_{\displaystyle x \, factors}$
Write product form of the quantity $m$ in exponential notation.
$\implies m \,=\ b^{\displaystyle x}$
The quantity ($m$) in exponential form can be expressed in logarithmic form as per the mathematical relation between exponents and logarithms.
$\implies \log_{b}{m} \,=\, x$
Write the mathematical relation between exponential form and logarithmic form of the quantity.
$m \,=\ b^{\displaystyle x}$ $\,\, \Leftrightarrow \,\,$ $\log_{b}{m} \,=\, x$
Assume, $x \,=\, 1$
$m \,=\ b^{\displaystyle 1}$ $\,\, \Leftrightarrow \,\,$ $\log_{b}{m} \,=\, 1$
The meaning of $b$ is raised to the power of $1$, is one time variable $b$.
$\implies m \,=\ b$ $\,\, \Leftrightarrow \,\,$ $\log_{b}{m} \,=\, 1$
When the value of $m$ is equal to $b$, then the logarithm of $m$ to the base $b$ is equal to one. In this case, $m \,=\, b$. Therefore, the variable $m$ can be replaced by $b$.
$\implies m \,=\ b$ $\,\, \Leftrightarrow \,\,$ $\log_{b}{b} \,=\, 1$
Therefore, it is proved that log of a quantity is equal to one when the same quantity is considered as a base of the logarithm.
$\,\,\, \therefore \,\,\,\,\,\, \log_{b}{b} \,=\, 1$
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
|
I am given the series:
$$\sum_{n=1}^{\infty} \frac{\sqrt{n}+\sin(n)}{n^2+5}$$
and I am asked to determine whether it is convergent or not. I know I need to use the comparison test to determine this. I can make a comparison with a harmonic p series ($a_n=\frac{1}{n^p}$ where p > 1, series converges). I argue that as the denominator grows more rapidly than the numerator, I need only look at the denominators:
$$\frac{1}{n^2+5}\le\frac{1}{n^2}$$
$\frac{1}{n^2}$ is a harmonic p series where $p>1$ which converges. As $\frac{\sqrt{n}+\sin(n)}{n^2+5}$ is less than that, by the comparison test, $\sum_{n=1}^{\infty} \frac{\sqrt{n}+\sin(n)}{n^2+5}$ is convergent.
Is this a valid argument for this question?
|
Using interact
I want to visualize these functions in the complex plane:
$$f_n(z)=\exp(z) - \sum_{k=0}^{n} \frac{z^k}{k!}$$
I tried this code, but it gives me an error.
z,k = var('z,k')@interact def _(n=(1..8)): complex_plot(exp(z)- sum(z^k/factorial(k), k, 0, n), (-5, 5), (-5, 5))
I am new to Sage (I previously used Mathematica). I wrote the code based on this example: http://wiki.sagemath.org/interact/ . How can I fix that?
EDIT: I would also like to know why doesn't this work either
def myPlot(n): complex_plot(exp(z)- sum(z^k/factorial(k), k, 0, n), (-5, 5), (-5, 5))
Then I type myPlot(2) in the notebook, but I get nothing. However, if I type:
complex_plot(exp(z)- sum(z^k/factorial(k), k, 0, 2), (-5, 5), (-5, 5))
I get the correct plot.
EDIT 2:
I tried
myPlot = lambda n: complex_plot(exp(z)- sum(z^k/factorial(k), k, 0, n), (-5, 5), (-5, 5))
And now I can evaluate it. However, I still can't use @Interact.
This doesn't make sense to me.
|
Searching for just a few words should be enough to get started. If you need to make more complex queries, use the tips below to guide you.
Purchase individual online access for 1 year to this journal.
Impact Factor 2019: 0.808
The journal
Asymptotic Analysis fulfills a twofold function. It aims at publishing original mathematical results in the asymptotic theory of problems affected by the presence of small or large parameters on the one hand, and at giving specific indications of their possible applications to different fields of natural sciences on the other hand. Asymptotic Analysis thus provides mathematicians with a concentrated source of newly acquired information which they may need in the analysis of asymptotic problems.
Authors: Zielinski, Lech
Article Type: Research Article
Abstract: We consider the Weyl formula describing the asymptotic behaviour of the number of eigenvalues N(\lambda) for elliptic boundary value problems. The remainder estimate of the form N(\lambda){\rm O}(\lambda^{-\mu}) is proved with \mu<r/(2m) in the case of operators of order 2m with Hölder coefficients of exponent r\in \,]0; 1] .
Citation: Asymptotic Analysis, vol. 16, no. 3,4, pp. 181-201, 1998
Authors: Kostin, I.N.
Article Type: Research Article
Abstract: The behaviour of trajectories of a nonlinear semigroup in the neighbourhood of a non‐hyperbolic stationary point is studied. The obtained results are used to derive an estimate for the rate of convergence of solutions of the Chafee–Infante problem to its attractor in the case of non‐hyperbolic zero stationary solution.
Citation: Asymptotic Analysis, vol. 16, no. 3,4, pp. 203-222, 1998
Article Type: Research Article
Abstract: Consider the domain \varOmega_\varepsilon =\varOmega -T_\varepsilon obtained by removing a closed set T_\varepsilon of \varepsilon ‐periodic holes of size \varepsilon from a bounded open set \varOmega . We study the homogenization of the nonlinear problem \cases{-{\rm div}(A({x/\varepsilon})Du_\varepsilon)+ \gamma u_\varepsilon = H( {x/\varepsilon}, u_\varepsilon , Du_\varepsilon )& $\hbox{in }\varOmega_\varepsilon ,$\cr (A({x/\varepsilon})Du_\varepsilon)\cdot\underline\nu =0 &$\hbox{on } \Ncurpartial T_\varepsilon ,$\cr u_\varepsilon =0&$\hbox{on } \Ncurpartial \varOmega,$\cr u_\varepsilon\in H^1(\varOmega_\varepsilon )\cap L^\infty(\varOmega_\varepsilon ),&\cr} where H( y,s,\xi) is ]0,1[^n ‐periodic in y and has quadratic growth with respect to \xi …. We prove that the linear part of the limit problem is the homogenized matrix of the linear problem and the nonlinear part is given by H^0( u, Du) , where H^0 is defined by H^0( s,\xi) =\int_{]0,1[^n-\overline T} H ( y,s,C(y)\xi)\, {\rm d}y \quad \forall( s,\xi)\in\NBbbR\times\NBbbR^n , C( {\cdot/\varepsilon}) being the corrector matrices of the linear problem and T the reference hole. Show more
Citation: Asymptotic Analysis, vol. 16, no. 3,4, pp. 223-243, 1998
Authors: Shubov, Marianna A.
Article Type: Research Article
Abstract: We consider an infinite sequence of radial wave equations obtained by the separation of variables in the spherical coordiantes from the 3‐dimensional damped wave equation with spacially nonhomogeneous spherically symmetric coefficients. The nonconservative boundary conditions are given on the sphere |x|=a . Our main objects of interest are the nonselfadjoint operators in the energy space of 2‐component initial data, which are the dynamics generators for the systems governed by the aforementioned equations and boundary conditions. Our main results are precise asymptotic formulas for the complex eigenvalues and eigenfunctions of these operators and the corresponding nonselfadjoint quadratic operator pencils. Based …on the asymptotic results of the present work, we will show in a forthcoming paper that the sets of root vectors of the above operators and of the dynamics generator corresponding to the full 3‐dimensional damped wave equation form Riesz bases in the appropriate energy spaces. Therefore, we will show that all these operators are spectral in the sense of Dunford. Show more
Citation: Asymptotic Analysis, vol. 16, no. 3,4, pp. 245-272, 1998
Article Type: Research Article
Abstract: The asymptotic behaviour, with respect to the small period, of the equilibrium displacements corresponding to the Koiter’s shell model of periodically perturbed plate is considered. In the limit as the period tends to zero a model describing the deformation of a wrinkled plate is obtained in the two‐scale form. The two‐scale problem can be decoupled; the longitudinal displacement satisfies the classical equations for longitudinal deformations of a plate, while the transverse displacement is a solution of a fourth‐order elliptic problem modeling the anisotropic plate. Corresponding convergence result with correctors is proved by use of two‐scale convergence method. In the case …of wavy plate the effective coefficients of anisotropic plate are explicitly computed. Show more
Citation: Asymptotic Analysis, vol. 16, no. 3,4, pp. 273-297, 1998
Authors: Ono, Kosuke
Article Type: Research Article
Abstract: We consider the global existence and asymptotic stability of solutions to the Cauchy problem for degenerate nonlinear wave equations of Kirchhoff type with a dissipative term in unbounded domain. We derive the sharp decay estimates of the solution and its derivatives. Moreover, we show that the solution has a lower decay estimate of some algebraic rate.
Citation: Asymptotic Analysis, vol. 16, no. 3,4, pp. 299-314, 1998
Article Type: Research Article
Abstract: Our aim in this article is to study the existence of finite dimensional attractors for a class of generalized Cahn–Hilliard equations based on a microforce balance introduced by M. Gurtin (Physica D 92 (1996), 178–192). We first consider some general results on the existence of finite dimensional attractors, and then apply these results to the generalized Cahn–Hilliard equations.
Citation: Asymptotic Analysis, vol. 16, no. 3,4, pp. 315-345, 1998
Article Type: Research Article
Abstract: Bellman–Isaacs equation of ergodic type is studied in paricular case. A solution of the equation related to the principal eigenfunction of the Schrödinger operator is obtained as the limit of the solution of a Bellman–Isaacs equation of discounted type. Furthermore, a singular limit of the equation is studied in relation to semi‐classical analysis.
Citation: Asymptotic Analysis, vol. 16, no. 3,4, pp. 347-362, 1998
Inspirees International (China Office)
Ciyunsi Beili 207(CapitaLand), Bld 1, 7-901 100025, Beijing China Free service line: 400 661 8717 Fax: +86 10 8446 7947 china@iospress.cn
For editorial issues, like the status of your submitted paper or proposals, write to editorial@iospress.nl
如果您在出版方面需要帮助或有任何建, 件至: editorial@iospress.nl
|
The angle sum cosine identity is used as a formula to expanded cosine of sum of two angles. For example, $\cos{(A+B)}$, $\cos{(x+y)}$, $\cos{(\alpha+\beta)}$, and so on. Here, you learn how cos of sum of two angles formula is derived in geometric method.
The $\Delta EDF$ is a right triangle and the angle of this triangle is divided as two angles. It’s useful in deriving the cosine of sum of two angles trigonometric identity geometrically.
$(1). \,\,\,$ Firstly, draw a straight line to side $\overline{EF}$ from point $D$ for dividing the $\angle EDF$ as two angles $x$ and $y$, and it intersects the side $\overline{EF}$ at point $G$.
$(2). \,\,\,$ Draw a straight line to side $\overline{DE}$ from point $G$ but it should be perpendicular to the side $\overline{DG}$. It means $\overline{DG} \perp \overline{HG}$
$(3). \,\,\,$ Similarly, draw a perpendicular line to side $\overline{DF}$ from point $H$, but it intersects the side $\overline{DG}$ at point $I$ and also perpendicularly intersects the side $\overline{DF}$ at point $J$.
$(4). \,\,\,$ Lastly, draw a perpendicular line to side $\overline{HJ}$ from point $G$ and it intersects the side $\overline{HJ}$ at point $K$.
Thus, the $\Delta GDF$ having $x$ as its angle, the $\angle EDG$ having $y$ as its angle and $\angle KHG$ with unknown angle are formed geometrically.
Now, let’s start the geometrical approach to derive the angle sum cosine identity in the form of $\cos{(x+y)}$ formula in trigonometry on the basis of the above constructed triangle.
The $\angle JDH$ is $x+y$ in the $\Delta JDH$ and write the cos of compound angle $x+y$ in its ratio from.
$\cos{(x+y)}$ $\,=\,$ $\dfrac{DJ}{DH}$
The side $\overline{HJ}$ divides the side $\overline{DF}$ as two parts. So, the sum of lengths of the sides $\overline{DJ}$ and $\overline{JF}$ is equal to the length of the side $\overline{DF}$.
$DF \,=\, DJ+JF$
$\implies DJ \,=\, DF-JF$
Substitute the length of the side $\overline{DJ}$ by its equivalent value in the expansion of $\cos{(x+y)}$.
$\implies$ $\cos{(x+y)}$ $\,=\,$ $\dfrac{DF-JF}{DH}$
$\implies$ $\cos{(x+y)}$ $\,=\,$ $\dfrac{DF}{DH}$ $\,-\,$ $\dfrac{JF}{DH}$
The lengths of sides $\overline{JF}$ and $\overline{KG}$ are equal because they both are parallel lines. Therefore, $JF = KG$
$\implies$ $\cos{(x+y)}$ $\,=\,$ $\dfrac{DF}{DH}$ $\,-\,$ $\dfrac{KG}{DH}$
The side $\overline{DF}$ is adjacent side of $\Delta FDG$ and angle of this triangle is $x$. It is possible to express the length of the side $\overline{DF}$ in a trigonometric function.
$\cos{x} \,=\, \dfrac{DF}{DG}$
$\implies DF \,=\, {DG}\cos{x}$
Replace the length of the side $\overline{DF}$ by its equivalent value in the $\cos{(x+y)}$ expansion.
$\implies$ $\cos{(x+y)}$ $\,=\,$ $\dfrac{{DG}\cos{x}}{DH}$ $\,-\,$ $\dfrac{KG}{DH}$
$\implies$ $\cos{(x+y)}$ $\,=\,$ $\cos{x} \times \dfrac{DG}{DH}$ $\,-\,$ $\dfrac{KG}{DH}$
$DG$ and $DH$ are lengths of the sides $\overline{DG}$ and $\overline{DH}$ respectively. They are sides of the $\Delta GDH$ and its angle is $y$. The ratio between them can be represented by cos of angle $y$.
$\cos{y} \,=\, \dfrac{DG}{DH}$
The ratio of $DG$ to $DH$ can be replaced by $\cos{y}$ in the expansion of $\cos{(x+y)}$.
$\implies$ $\cos{(x+y)}$ $\,=\,$ $\cos{x} \times \cos{y}$ $\,-\,$ $\dfrac{KG}{DH}$
$\implies$ $\cos{(x+y)}$ $\,=\,$ $\cos{x}\cos{y}$ $\,-\,$ $\dfrac{KG}{DH}$
Similarly, the ratio of $KG$ to $DH$ can also be expressed in terms of trigonometric functions but it is not possible at this time because it’s the side of the $\Delta KHG$ and its angle is unknown.
Now, it’s time to find the $\angle KHG$ and it is useful to us in expressing the length of the side $\overline{KG}$ in a trigonometric function. The sides $\overline{KG}$ and $\overline{DF}$ are parallel lines and their transversal is $\overline{DG}$. The $\angle FDG$ and $\angle KGD$ are alternate interior angles and they are equal geometrically.
$\angle KGD \,=\, \angle FDG$ but $\angle FDG = x$
$\therefore \,\,\,\,\,\, \angle KGD \,=\, x$
Actually $\overline{DG} \perp \overline{HG}$. So, $\angle DGH = 90^°$. The side $\overline{KG}$ splits the $\angle DGH$ as two angles $\angle KGD$ and $\angle KGH$.
$\angle DGH \,=\, \angle KGD + \angle KGH$
$\implies 90^° = x+\angle KGH$
$\,\,\, \therefore \,\,\,\,\,\, \angle KGH = 90^°-x$
Two angles are known in $\Delta KHG$ and its third angle $\angle KHG$ can be calculated from them by using sum of angles of a triangle rule.
$\angle KHG + \angle HKG + \angle KGH$ $\,=\,$ $180^°$
$\implies$ $\angle KHG + 90^° + 90^°-x$ $\,=\,$ $180^°$
$\implies$ $\angle KHG + 180^°-x$ $\,=\,$ $180^°$
$\implies$ $\angle KHG$ $\,=\,$ $180^°-180^°+x$
$\implies$ $\angle KHG$ $\,=\,$ $\require{cancel} \cancel{180^°}-\cancel{180^°}+x$
$\,\,\, \therefore \,\,\,\,\,\,$ $\angle KHG \,=\, x$
It is derived that $\angle KHG = x$ but the $\angle FDG = x$ and it clears that they both are congruent. Therefore, $\Delta FDG$ and $\Delta KHG$ are similar triangles.
Let’s continue deriving the expansion of the angle sum cosine identity.
$\cos{(x+y)}$ $\,=\,$ $\cos{x}\cos{y}$ $\,-\,$ $\dfrac{KG}{DH}$
$KG$ is length of the side $\overline{KG}$ in $\Delta KHG$ and it can be written in terms of trigonometric function.
$\sin{x} \,=\, \dfrac{KG}{HG}$
$\implies KG \,=\, {HG}\sin{x}$
Therefore, the length of the side $\overline{KG}$ by its equivalent value in the expansion of cos of compound angle.
$\implies$ $\cos{(x+y)}$ $\,=\,$ $\cos{x}\cos{y}$ $\,-\,$ $\dfrac{{HG}\sin{x}}{DH}$
$\implies$ $\cos{(x+y)}$ $\,=\,$ $\cos{x}\cos{y}$ $\,-\,$ $\sin{x} \times \dfrac{HG}{DH}$
$HG$ and $DH$ are lengths of the sides $\overline{HG}$ and $\overline{DH}$ in $\Delta GDH$. The angle of $\Delta GDH$ is $y$. So, the ratio between them can be represented by the trigonometric function $\sin{y}$.
$\sin{y}$ $\,=\,$ $\dfrac{HG}{DH}$
Therefore, the ratio of $HG$ to $DH$ can be replaced by $\sin{y}$ in the expansion of the $\cos{(x+y)}$ function.
$\implies$ $\cos{(x+y)}$ $\,=\,$ $\cos{x}\cos{y}$ $\,-\,$ $\sin{x} \times \sin{y}$
$\,\,\, \therefore \,\,\,\,\,\,$ $\cos{(x+y)}$ $\,=\,$ $\cos{x}\cos{y}$ $\,-\,$ $\sin{x}\sin{y}$
Therefore, it is proved that the cos of sum of two angles can be expanded as the subtraction of product of sines of angles from product of cosines of the angles.
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
|
I suggest to create a tag synonym analysis-and-odes $\to$ ca.classical-analysis-and-odes. The tag analysis-and-odes has only five questions and empty tag info - so it does not seem to have any distinction from ca.classical-analysis-and-odes. Four of the five questions in this tag were asked by t...
I'd like to propose creation of continuum-theory tag. To me (as an outsider, but still a bit interested in this topic) it seems that continuum theory is an area of general topology which enjoys some interest both among topologists and among mathematicians in general. (For example, one part of Op...
Let $P$ be a compact connected set in the plane and $x,y\in P$. Is it always possible to connect $x$ to $y$ by a path $\gamma$ such that the length of $\gamma\backslash P$ is arbitrary small? Comments: Be aware of pseudoarc --- it is a compact connected set which contains no nontrivial p...
Let $X$ be a connected compact metric space. Given a positive $\varepsilon$ and two points $x,y\in X$ we write $x\sim_\varepsilon y$ if there exists a sequence $C_1,\dots,C_n$ of connected subsets of diameter $<\varepsilon$ in $X$ such that $x\in C_1$, $y\in C_n$ and $C_i\cap C_{i+1}\ne\emptyset...
This problem was motivated by the MO problems: "Running most of the time in a connected set", "Is every metric continuum almost path connected?" and "Are $\varepsilon$-connected components dense?". Let $p$ be a positive real number. A metric space $(X,d)$ is called $\ell_p$-chain connected if fo...
Continuum $=$ compact connected metric space. Let $X$ be a continuum. $X$ is indecomposable means that every proper subcontinuum of $X$ is nowhere dense in $X$. It is easy to see that if $X$ is indecomposable then every connected open subset of $X$ is dense in $X$. Question. Are these two cond...
Definition. A closed subset $S$ of a topological space $X$ is called a separator between points $x,y\in X\setminus S$ if the points $x$ and $y$ belong to different connected components of $X\setminus S$. A separator $S$ is called an irreducible separator between $x$ and $y$ is $S$ coincides with ...
Let $\square=[0,1]\times[0,1]$ be the unit square and $f\colon\square\to \square$ is a continuous map that fixes the points on the boundary. Assume $f$ is a limit of homeomorphisms $\square\to \square$. (By Moore's theorem it is equivalent to the condition that for any point $p\in \square$ the i...
The question was motivated by this question of Anton Petrunin. By a metric continuum we understand a connected compact metric space. Let $p$ be a positive real number. A metric continuum $X$ is called $\ell_p$-almost path-connected if for any points $x,y\in X$ and any $\varepsilon>0$ here exist...
Let $\Sigma$ be a sphere topologically embedded into $\mathbb{R}^3$. Is it always possible to find a disc $\Delta\subset\Sigma$ which is bounded by a plane curve? It is easy to find an open disc which boundary lies in a plane, but the boundary might be crazy; for example it might be Polish ...
Let $X$ be a connected separable metrizable topological space. Call it a cut-point space if $X\setminus \{x\}$ is disconnected for every $x\in X$. Then does $X$ embed into the plane? My thoughts: (0) It is not difficult to see that $X$ must have dimension $1$, and therefore embeds into $\mathb...
This is a very general sort of question, and the use of the phrase 'shape operator' is a bit sloppy since there is already an established "shape theory." But what I have are topological spaces that have trivial shape in the classic Borsuk sense (and thus also in the sense of Cech homology). How...
Problem. Is each Peano continuum a topological fractal? A compact Hausdorff space $X$ is a topological fractal if $X=\bigcup_{i=1}^n f_i(X)$ for some continuous maps $f_1,\dots,f_n:X\to X$ such that for any sequence $(g_i)_{i\in\omega}\subset\{f_1,\dots,f_n\}^{\omega}$ the intersection $\big...
I found a strange attractor which looks a lot like a solenoid. The attractor continuum is the closure of a continuous line which limits onto itself, and it is locally homeomorphic to Cantor set times Reals. It sits in the Möbius strip. Does the Dyadic solenoid embed into the Möbius strip? Wha...
« first day (1939 days earlier) ← previous day next day → last day (280 days later) »
|
@egreg It does this "I just need to make use of the standard hyphenation function of LaTeX, except "behind the scenes", without actually typesetting anything." (if not typesetting includes typesetting in a hidden box) it doesn't address the use case that he said he wanted that for
@JosephWright ah yes, unlike the hyphenation near box question, I guess that makes sense, basically can't just rely on lccode anymore. I suppose you don't want the hyphenation code in my last answer by default?
@JosephWright anway if we rip out all the auto-testing (since mac/windows/linux come out the same anyway) but leave in the .cfg possibility, there is no actual loss of functionality if someone is still using a vms tex or whatever
I want to change the tracking (space between the characters) for a sans serif font. I found that I can use the microtype package to change the tracking of the smallcaps font (\textsc{foo}), but I can't figure out how to make \textsc{} a sans serif font.
@DavidCarlisle -- if you write it as "4 May 2016" you don't need a comma (or, in the u.s., want a comma).
@egreg (even if you're not here at the moment) -- tomorrow is international archaeology day: twitter.com/ArchaeologyDay , so there must be someplace near you that you could visit to demonstrate your firsthand knowledge.
@barbarabeeton I prefer May 4, 2016, for some reason (don't know why actually)
@barbarabeeton but I have another question maybe better suited for you please: If a member of a conference scientific committee writes a preface for the special issue, can the signature say John Doe \\ for the scientific committee or is there a better wording?
@barbarabeeton overrightarrow answer will have to wait, need time to debug \ialign :-) (it's not the \smash wat did it) on the other hand if we mention \ialign enough it may interest @egreg enough to debug it for us.
@DavidCarlisle -- okay. are you sure the \smash isn't involved? i thought it might also be the reason that the arrow is too close to the "M". (\smash[t] might have been more appropriate.) i haven't yet had a chance to try it out at "normal" size; after all, \Huge is magnified from a larger base for the alphabet, but always from 10pt for symbols, and that's bound to have an effect, not necessarily positive. (and yes, that is the sort of thing that seems to fascinate @egreg.)
@barbarabeeton yes I edited the arrow macros not to have relbar (ie just omit the extender entirely and just have a single arrowhead but it still overprinted when in the \ialign construct but I'd already spent too long on it at work so stopped, may try to look this weekend (but it's uktug tomorrow)
if the expression is put into an \fbox, it is clear all around. even with the \smash. so something else is going on. put it into a text block, with \newline after the preceding text, and directly following before another text line. i think the intention is to treat the "M" as a large operator (like \sum or \prod, but the submitter wasn't very specific about the intent.)
@egreg -- okay. i'll double check that with plain tex. but that doesn't explain why there's also an overlap of the arrow with the "M", at least in the output i got. personally, i think that that arrow is horrendously too large in that context, which is why i'd like to know what is intended.
@barbarabeeton the overlap below is much smaller, see the righthand box with the arrow in egreg's image, it just extends below and catches the serifs on the M, but th eoverlap above is pretty bad really
@DavidCarlisle -- i think other possible/probable contexts for the \over*arrows have to be looked at also. this example is way outside the contexts i would expect. and any change should work without adverse effect in the "normal" contexts.
@DavidCarlisle -- maybe better take a look at the latin modern math arrowheads ...
@DavidCarlisle I see no real way out. The CM arrows extend above the x-height, but the advertised height is 1ex (actually a bit less). If you add the strut, you end up with too big a space when using other fonts.
MagSafe is a series of proprietary magnetically attached power connectors, originally introduced by Apple Inc. on January 10, 2006, in conjunction with the MacBook Pro at the Macworld Expo in San Francisco, California. The connector is held in place magnetically so that if it is tugged — for example, by someone tripping over the cord — it will pull out of the socket without damaging the connector or the computer power socket, and without pulling the computer off the surface on which it is located.The concept of MagSafe is copied from the magnetic power connectors that are part of many deep fryers...
has anyone converted from LaTeX -> Word before? I have seen questions on the site but I'm wondering what the result is like... and whether the document is still completely editable etc after the conversion? I mean, if the doc is written in LaTeX, then converted to Word, is the word editable?
I'm not familiar with word, so I'm not sure if there are things there that would just get goofed up or something.
@baxx never use word (have a copy just because but I don't use it;-) but have helped enough people with things over the years, these days I'd probably convert to html latexml or tex4ht then import the html into word and see what come out
You should be able to cut and paste mathematics from your web browser to Word (or any of the Micorsoft Office suite). Unfortunately at present you have to make a small edit but any text editor will do for that.Givenx=\frac{-b\pm\sqrt{b^2-4ac}}{2a}Make a small html file that looks like<!...
@baxx all the convertors that I mention can deal with document \newcommand to a certain extent. if it is just \newcommand\z{\mathbb{Z}} that is no problem in any of them, if it's half a million lines of tex commands implementing tikz then it gets trickier.
@baxx yes but they are extremes but the thing is you just never know, you may see a simple article class document that uses no hard looking packages then get half way through and find \makeatletter several hundred lines of trick tex macros copied from this site that are over-writing latex format internals.
|
Searching for just a few words should be enough to get started. If you need to make more complex queries, use the tips below to guide you.
Purchase individual online access for 1 year to this journal.
Impact Factor 2019: 0.808
The journal
Asymptotic Analysis fulfills a twofold function. It aims at publishing original mathematical results in the asymptotic theory of problems affected by the presence of small or large parameters on the one hand, and at giving specific indications of their possible applications to different fields of natural sciences on the other hand. Asymptotic Analysis thus provides mathematicians with a concentrated source of newly acquired information which they may need in the analysis of asymptotic problems.
Article Type: Research Article
Abstract: We obtain the large time asymptotic expansion of small solutions to the Landau–Ginzburg type equations \[\cases{u_{t}-\alpha u_{xx}+\beta |u|^{2}u=0,\quad x\in {\mathbf{R}},\ t>0,\cr \noalign{\vskip3pt}u(0,x)=\phi (x),\quad x\in {\mathbf{R}},\cr}\] where α,β∈C,Rα>0, R(β/$\sqrt{2|\alpha |^{2}+\alpha ^{2}}$ )≥0, under the condition that the initial data ϕ are sufficiently small in a suitable weighted norm and the mean value ∫ϕ(x) dx≠0.
Keywords: dissipative nonlinear evolution equation, asymptotic expansions, Landau–Ginzburg equation
Citation: Asymptotic Analysis, vol. 32, no. 2, pp. 91-106, 2002
Authors: Hamouda, Makram
Article Type: Research Article
Abstract: In this article, we derive the first term in the asymptotic expansion of the regularised minimal surface solution in the radially symmetric case when the domain is a pair of concentric circles in R2 . General domains and time‐dependent problems will be considered elsewhere.
Keywords: minimal surfaces, singular perturbations, asymptotic analysis, boundary layer, small viscosity
Citation: Asymptotic Analysis, vol. 32, no. 2, pp. 107-130, 2002
Article Type: Research Article
Abstract: State‐of‐the‐art magnetic storage devices have head‐to‐disk distances of about 300 Angstrom, for which compressibility, slip‐flow and roughness effects are significant. Since the head and the disk are in relative motion, the air‐gap thickness when both surfaces are rough varies rapidly in both space and time. A rigorous homogenization of the transient compressible Reynolds equation appropriate for such situation is presented in this article. If ε is the roughness length and pε the pressure field for that roughness, the existence of p0 ∈L2 (0,T,H1 (Ω)) such that pε →p0 strongly in L2 (Ω×]0,T[) when ε→0 is proved. A …homogenized problem for p0 is introduced together with a uniqueness result under remarkably weak assumptions, i.e., p0 ∈L2 (0,T,H1 (Ω)) and ∂p0 /∂t∈L2 (0,T,H−1 (Ω)). Interestingly, no time derivatives appear in the auxiliary local problems, which are thus computed as in the steady state case. The role of the time variable is to parameterize the relative positions of the roughness shapes, and the homogenized coefficients result from averaging all such positions. To our knowledge, this is the first rigorous treatment of lubrication problems accounting for roughness on both surfaces. Show more
Citation: Asymptotic Analysis, vol. 32, no. 2, pp. 131-152, 2002
Article Type: Research Article
Abstract: We study the effective properties of an elastic composite medium under the assumption of small deformations. This composite is made of a periodic possibly disconnected subset filled up with a strong material surrounded by another material whose elastic coefficients are very small. The effective macroscopic behaviour obtained by homogenization turns out to be nonlocal and depends highly on the geometry of the strong component. In particular, second order derivative of the displacement appear in the limit energy when disconnected fibers are considered. These results were announced in [8].
Keywords: homogenization, linear elasticity, fiber reinforced structures, two‐scale convergence, Γ‐convergence
Citation: Asymptotic Analysis, vol. 32, no. 2, pp. 153-183, 2002
Inspirees International (China Office)
Ciyunsi Beili 207(CapitaLand), Bld 1, 7-901 100025, Beijing China Free service line: 400 661 8717 Fax: +86 10 8446 7947 china@iospress.cn
For editorial issues, like the status of your submitted paper or proposals, write to editorial@iospress.nl
如果您在出版方面需要帮助或有任何建, 件至: editorial@iospress.nl
|
$\large \log_{b}{(m)} = \dfrac{\log_{d}{(m)}}{\log_{d}{(b)}}$
The base of a logarithmic term can be changed mathematically by expressing it as a quotient of two logarithmic terms which contain another quantity as their base. It is usually called as change of base rule and used as a formula in logarithms.
The change of base log formula in quotient form is derived in algebraic form on the basis of rules of exponents and also mathematical relation between exponents and logarithms.
$\log_{b}{m}$ and $\log_{d}{b}$ are two logarithmic terms and it is taken that their values are $x$ and $y$ respectively.
$\log_{b}{m} = x$ and $\log_{d}{b} = y$
Express both logarithmic equations in exponential form by the mathematical relationship between them.
$(1) \,\,\,$ $\log_{b}{m} = x \,\Leftrightarrow\, m = b^{\displaystyle x}$
$(2) \,\,\,$ $\log_{d}{b} \,\,\, = y \,\Leftrightarrow\, b = d^{\displaystyle y}$
Eliminate the base $b$ in the equation $m = b^{\displaystyle x}$ by substituting the $b = d^{\displaystyle y}$.
$\implies$ $m = {(d^{\displaystyle y})}^{\displaystyle x}$
Apply power of power exponents rule to simplify this exponential equation.
$\implies$ $m = d^{\displaystyle xy}$
Express this exponential equation in logarithmic form.
$m = d^{\displaystyle xy} \Longleftrightarrow xy = \log_{d}{m}$
$\implies$ $xy = \log_{d}{m}$
In fact, $x = \log_{b}{m}$ and $y = \log_{d}{b}$. So, replace them.
$\implies$ $\log_{b}{m} \times \log_{d}{b} = \log_{d}{m}$
$\,\,\, \therefore \,\,\,\,\,\,$ $\log_{b}{m} = \dfrac{\log_{d}{m}}{\log_{d}{b}}$
Thus, the change of base formula is derived in mathematics for changing base of any log term.
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
|
I'm trying to choose a group that is hard under the Chosen-Target Computational Diffie-Hellman assumption, according to the definition in this paper, in order to implement the oblivious transfer scheme defined in the top box on page 10(=406).
The (intimidating, to me) CT-CDH assumption is defined as follows (page 7=403):
Let $\mathbb{G}_q$ be a group of prime order $q$, $g$ be a generator of $\mathbb{G}_q$, $x\in \mathbb{Z}^*_q$. Let $H_1 : \{0, 1\}^∗ \rightarrow \mathbb{G}_q$ be a cryptographic hash function. The adversary $A$ is given input $(q, g, g^x, H_1)$ and two oracles: target oracle $TG(\cdot)$ that returns a random element $w_i \in \mathbb{G}_q$ at the $i$-th query and helper oracle $HG(\cdot)$ that returns $(\cdot)^x$. Let $q_T$ and $q_H$ be the number of queries $A$ made to the target oracle and helper oracle respectively.
Assumption: The probability that $A$ outputs $k$ pairs $((v_1, j_1), (v_2, j_2), \dots, (v_k, j_k))$, where $v_i = (w_{j_i})^x$ for $i \in \{1, 2, \dots , k\}$, $q_H \lt k \leq q_T$, is negligible.
It should be noted that this assumption is equivalent to the standard Computational Diffie-Hellman assumption when $q_T=1$, according to this paper.
Can anyone give an example of a group that fits the bill? I tried $\mathbb{Z}^*_q$ for a prime $q$ under multiplication, but that's of order $q-1$, which is clearly not prime. However, the complexity analysis on page 12 of the paper is in terms of modular exponentiations.
Additionally (I can make a new question for this, if scolded), how would one implement the $(D_j)^{a_j^{-1}}$ operation in step 5 of the protocol? I can't figure out if it's equivalent to the discrete log problem.
|
Equivalence of Definitions of Separated Sets/Definition 2 implies Definition 1 Theorem
Let $T = \struct{S, \tau}$ be a topological space.
Let $A, B \subseteq S$.
Let $U,V \in \tau$ satisfy:
$A \subset U$ and $U \cap B = \empty$ $B \subset V$ and $V \cap A = \empty$
where $\empty$ denotes the empty set.
Then $A^- \cap B = A \cap B^- = \empty$
where $A^-$ denotes the closure of $A$ in $T$.
Proof
From Empty Intersection iff Subset of Relative Complement, $B \subseteq S \setminus U$.
From Set Closure is Smallest Closed Set, $B^- \subseteq S \setminus U$.
From Empty Intersection iff Subset of Relative Complement, $B^- \cap U = \empty$.
Then
\(\displaystyle B^- \cap A\) \(=\) \(\displaystyle B^- \cap \paren {U \cap A}\) Intersection with Subset is Subset \(\displaystyle \) \(=\) \(\displaystyle \paren {B^- \cap U} \cap A\) Intersection is Associative \(\displaystyle \) \(=\) \(\displaystyle \empty \cap A\) As $B^- \cap U = \empty$ \(\displaystyle \) \(=\) \(\displaystyle \empty\) Intersection with Empty Set Similarly, $A^- \cap B = \empty$.
$\blacksquare$
|
As the OP cross-posted this question from Math stackexchange, I have also cross-posted the answer that I wrote there.
The simplest traditional solution to the $d$-dimensional which provides quite good results in 3-dimensions is to use the Halton sequence based on the first three primes numbers (2,3,5). The Halton sequence is a generalization of the 1-dimensional Van der Corput sequence and merely requires that the three parameters are pairwise-coprime. Further details can be found at the Wikipedia article: "Halton Sequence".
An alternative sequence you could use is the generalization of the Weyl / Kronecker sequence. This sequence also typically uses the first three prime numbers, however, in this case they are chosen merely because the square root of these numbers is irrational.
However, I have recently written a detailed blog post, "The Unreasonable Effectiveness of Quasirandom Sequences", on how to easily create an open-ended low discrepancy sequences in arbitrary dimensions, that is:
algebraically simpler faster to compute produces more consistent outputs suffers less technical issues
than existing existing low discrepancy sequences, such as the Halton and Kronecker sequences.
The solution is an additive recurrence method (modulo 1) which generalizes the 1-Dimensional problem whose solution depends on the Golden Ratio. The solution to the $d$-dimensional problem, depends on a special constant $\phi_d$, where $\phi_d$ is the value of smallest, positive real-value of $x$ such that $$ x^{d+1}\;=x+1$$
For $d=1$, $ \phi_1 = 1.618033989... $, which is the canonical golden ratio.
For $d=2$, $ \phi_2 = 1.3247179572... $, which is often called the plastic constant, and has some beautiful properties. This value was conjectured to most likely be the optimal value for a related two-dimensional problem [Hensley, 2002]. Jacob Rus has posted a beautiful visualization of this 2-dimensional low discrepancy sequence, which can be found here.
And finally specifically relating to your question, for $d=3$, $ \phi_3 = 1.2207440846... $
With this special constant in hand, the calculation of the $n$-th term is now extremely simple and fast to calculate:
$$ R: \mathbf{t}_n = \pmb{\alpha}_0 + n \pmb{\alpha} \; (\textrm{mod} \; 1), \quad n=1,2,3,... $$$$ \textrm{where} \quad \pmb{\alpha} =(\frac{1}{\phi_d}, \frac{1}{\phi_d^2},\frac{1}{\phi_d^3},...\frac{1}{\phi_d^d}), $$
Of course, the reason this is called a recurrence sequence is because the above definition is equivalent to $$ R: \mathbf{t}_{n+1} = \mathbf{t}_{n} + \pmb{\alpha} \; (\textrm{mod} \; 1) $$
In nearly all instances, the choice of $\pmb{\alpha}_0 $ does not change the key characteristics, and so for reasons of obvious simplicity, $\pmb{\alpha}_0 =\pmb{0}$ is the usual choice. However, there are some arguments, relating to symmetry, that suggest that $\pmb{\alpha}_0=\pmb{1/2}$ is a better choice.
Specifically for $d=3$, $\phi_3 = 1.2207440846... $ and so for $\pmb{\alpha}_0= (1/2,1/2,1/2) $, $$\pmb{\alpha} = (0.819173,0.671044,0.549700) $$ and so the first 5 terms of the canonical 3-dimensional sequence are:
(0.319173, 0.171044, 0.0497005) (0.138345, 0.842087, 0.599401) (0.957518, 0.513131, 0.149101) (0.77669, 0.184174, 0.698802) (0.595863, 0.855218, 0.248502)...
Of course, this sequence ranges between [0,1], and so to convert to a range of [-1,1], simply apply the linear transformation $ x:= 2x+1 $. The result is
(-0.361655, -0.657913, -0.900599) (-0.72331, 0.684174, 0.198802) (0.915035, 0.0262616, -0.701797) (0.55338, -0.631651, 0.397604) (0.191725, 0.710436, -0.502995),...
The Mathematica Code for creating this sequence is as follows:
f[n_] := N[Root[-1 - # + #^(n + 1) &, 2 - Boole[EvenQ[n]]]];
d = 3;
n = 5
gamma = 1/f[d];
alpha = Table[gamma^k , {k, Range[d]}]
ptsPhi = Map[FractionalPart, Table[0.5 + i alpha, {i, Range[n]}], {2}]
Similar Python code is
# Use Newton-Rhapson-Method
def gamma(d):
x=1.0000
for i in range(20):
x = x-(pow(x,d+1)-x-1)/((d+1)*pow(x,d)-1)
return x
d=3
n=5
g = gamma(d)
alpha = np.zeros(d)
for j in range(d):
alpha[j] = pow(1/g,j+1) %1
z = np.zeros((n, d))
for i in range(n):
z = (0.5 + alpha*(i+1)) %1
print(z)
Hope that helps!
|
I have a much younger sister who is currently taking calculus as a high school junior. Given that I have a little more mathematical training than anyone else in my family, she sometimes asks me for help with her homework. This is almost invariably a humbling experience.
Last week, she was having trouble with the following indefinite integral:
\(\displaystyle
\int \frac{{\rm csch}(\ln(t))\coth(\ln(t))}{t}\,dt \)
If you are unfamiliar with calculus, don’t worry about it—I don’t plan to go that deep into the topic today. The point of today’s exercise is to demonstrate how helpful it can be to keep in mind the old axiom, “Keep it simple, stupid!”
Here was my first thought: “Hyperbolic trig functions? Hrm… let’s unpack those.” The functions in the numerator are hyperbolic trigonometric functions. These functions are actually defined in terms of exponential functions (i.e. \(e^x\)), but they behave in a manner similar to trigonometric functions with respect the derivatives, so there is some logic in calling them “trigonometric” functions. My first impulse was to unpack the definitions, and rewrite the numerator in terms of exponential functions.
This wasn’t actually a terrible idea. The argument of the two hyperbolic trig functions are natural logarithms, and (for instance) the hyperbolic cosecant of a natural logarithm simplifies down to a rational function (i.e. a fraction with polynomials in the numerator and denominator). In some ways, rational functions are easier to work with, so my instincts weren’t horrible.
After replacing the hyperbolic trig functions with rational functions, I was able to do some simplifications, and reduced the integrand to a rational function. This was not a nice function, and there really wasn’t an easy way of working with it, but I had already committed myself to one approach, and I was damned if I wasn’t going to keep going.
Rational functions aren’t too bad to integrate if the denominator is “nice” in some fashion. The usual technique for working with such functions is to decompose the fraction by way of partial fractions. It has been a while since I took calculus, so I had to look up the technique online, but after a few minutes, I had it down.
Of course, it turned out that there were several possible decompositions of the function that I was working with, so I made many false starts before gaining ground. However, after banging my head against the wall for a good 45 minutes, I finally had a decomposition that was useful. Success!
Finishing the problem required a couple of quick substitutions and some simplification. I ended up with a neat little rational function, and all it took was three hours of false starts and complex computations. Now, being a clever student, I looked up the answer in the solutions manual, and discovered that my answer looked nothing like the answer given (which was \(-{\rm csch}(\ln(t))\)).
Frustration! Back to the drawing board. But first, a quick check: after unpacking the “correct” solution, I was able to confirm that my answer matched. Whew! Crisis averted.
It was at this moment, after completely filling my whiteboard and struggling with the problem for hours, that one of my colleagues walked by my office, and made the following comment: “Why didn’t you just do a substitution in the first step?” After five minutes of work and four lines of computation, I had the correct answer. All that
sturm und drang, and the solution was simple, elegant, and obvious. Why hadn’t I seen that?
At the very beginning of the process, I managed to convince myself that the problem was far more complicated than it actually was, and let that conviction lead me down a garden path. As always, the correct route was to keep it simple. D’oh!
|
Consider the problem $$ (*)\begin{cases} x''(t)= F(x(t)) \\x(0)=P, x(1)=Q \end{cases}$$ where $P,Q\in\mathbb{R}^3$, $x=(x_1,x_2,x_3)$ and $F=-\nabla U$ for some potential $U:\mathbb{R}^3\rightarrow\mathbb{R}$.
With the assumption of $U$ is upper bounded, I want to show that $(*)$ admits weak solution. This could be done by minimizing the functional $$\phi (x)= \int_0^1 \frac{1}{2} \Vert x' \Vert ^2 - U(x)\ dt $$on $A=\lbrace x\in H^1(0,1)^3: x(0)=P, x(1)=Q\rbrace$, where $\Vert x'\Vert^2=x_1'^2+ x_2'^2+ x_3'^2 $. One classical result in minimizing functional is essentially: $\phi$ coercive and weakly lower semicontinuous $\Rightarrow$ $\phi$ attains its minimum. But I don't know how to prove this two conditions for $\phi$. Any help/hint would be appreciated. Thanks in advance!
Consider the problem $$ (*)\begin{cases} x''(t)= F(x(t)) \\x(0)=P, x(1)=Q \end{cases}$$ where $P,Q\in\mathbb{R}^3$, $x=(x_1,x_2,x_3)$ and $F=-\nabla U$ for some potential $U:\mathbb{R}^3\rightarrow\mathbb{R}$.
I'm going to assume that $U$ is continuous (which is not that obvious because the condition $-\nabla U = f$ could be in the sense of distributions). Let's start by making the boundary conditions homogeneous: Take $x_0$ a fixed function that satisfies $$x_0(0) = P, x_0(1) = Q$$ for instance $x_0(t) = P + (P - Q)t$ will do. Now we consider the problem in the new variable $y = x - x_0$ which is in $H^1_0(0,1)^3$ $$\phi(y) = \int_0^1 \frac12 \|y' - (P - Q)\|^2 - U(y - x_0)dt$$
For the lower semi-continuity: the first part $$\int_0^1 \frac12 \|y' - (P - Q)\|^2dt$$ is clearly continuous in the strong topology and convex, so it is lower semi continuous in the weak topology. For the second part $$\int_0^1 -U(y - x_0)dx$$ we can simply use Fatou's lemma. Let say that $U$ is bounded by $U_0$, hence $U_0 - U$ is positive. Lets take $(y_n)_{n\in \mathbb N}\subseteq H^1_0(0,1)^3$ that converges weakly to $y\in H^1_0(0,1)^3$. Then by compact embedding, $y_n$ converges strongly in $C^0([0,1])^3$ so
$$\liminf \int_0^1 U_0 - U(y_n - x_0) dt \geq \int_0^1 \liminf U_0 - U(y_n - x_0) dt = \int_0^1 U_0 - U(y - x_0) dt $$ which proves sequential lower semi-continuity(which is enough for existence of minimisers). (Here you actually can prove that $U(y_n - x_0$ also converge uniformly so you can use the usual riemman convergece theorem instead of Fatou, but either way, it is good to now this way of proving it because sometimes the only thing that you can proof is point-wise convergence (a.e.))
The coercivity is trivial from the inequality $$ \phi(y) \geq C\|y\|_{H_0^1}^2 + \|P - Q\|^2 - 2\|y\|_{H_0^1} \|P - Q\|- U_0$$ where we used the Poincare inequality.
|
So I have a given lets say $(x+1)^{2x}$ in addition to $\frac{\mathrm dy}{\mathrm dx}a^u=a^u\log(a)u'$. I still have to multiply this by the derivative of the inside function $x+1$ correct?
This is what logarithmic differentiation is for. You start with writing the function as an equation $$y = (x + 1)^{2x},$$ then take the natural log of both sides: $$\ln y = \ln\left[(x + 1)^{2x}\right] = 2x \ln(x+1).$$ We then implicitly differentiate both sides with respect to $x$. By chain rule (remember, $y$ is a function of $x$), the left side comes to $$\frac{1}{y} \cdot y'.$$ The right side can be differentiated as normal: $$\frac{2x}{x + 1} + 2\ln(x + 1).$$ So, \begin{align*} &\frac{1}{y} \cdot y' = \frac{2x}{x + 1} + 2\ln(x + 1) \\ \implies \, &y' = y\left(\frac{2x}{x + 1} + 2\ln(x + 1)\right) \\ \implies \, &y' = (x + 1)^{2x}\left(\frac{2x}{x + 1} + 2\ln(x + 1)\right). \end{align*}
Since $(x+1)^{2x}=e^{2x\ln(x+1)}=e^{u(x)}$, its derivate is $u'(x)e^{u(x)}$. Notice that if $a:I\longrightarrow\mathbb{R}^{+*}$ and $b:I\longrightarrow\mathbb{R}$, then we define $a(x)^{b(x)}$ as $e^{b(x)\ln a(x)}$ for all $x\in I$
That whole $a^u$ thing works when $a$ is a constant, not another expression in terms of $x$.
To take the derivative of this, you would have to convert it to $\displaystyle e^{2x\ln(x+1)}$ and THEN use the Chain Rule.
This would be $\displaystyle e^{2x\ln(x+1)}\left(2\ln(x+1)+\frac{2x}{x+1}\right)=(x+1)^{2x}\left(2\ln(x+1)+\frac{2x}{x+1}\right)$
A nice way to do this is to use a somewhat stronger fact than the other answers: as a function of
two variables, the expression $a^b$ is differentiable. Simply put, what this means is that, if $f(x)$ and $g(x)$ are functions, and $h(x)=f(x)^{g(x)}$, then $h'(x)$ is the sum of how fast this expression changes when we treat $f(x)$ as a constant plus how fast it changes when we treat $g(x)$ as a constant. You already know how to differentiate polynomials and exponentials, so this suffices - just apply both rules and add them!
Since we know that the derivative of $f^{g(x)}$ is $\log(f)\cdot f^{g(x)}\cdot g'(x)$ and the derivative of $f(x)^g$ is $g\cdot f(x)^{g-1}\cdot f'(x)$, we get $$h'(x)=\underbrace{\log(f(x))\cdot f(x)^{g(x)}\cdot g'(x)}_{\text{Derivative treating $f$ as constant}}+\underbrace{g(x)\cdot f(x)^{g(x)-1}\cdot f'(x)}_{\text{Derivative treating $g$ as constant}}.$$ You would get the same result by using logarithmic differentiation as other answers suggest*, but I generally find this is a bit easier to remember and more generalizable - for instance, note that differentiating a product $f(x)\cdot g(x)$ can be done by this same method.
(*Logarithmic differentiation is a good way to prove this result, since you write $$\log(h(x)) = \log(f(x))\cdot g(x)$$then differentiate both sides to get $$\frac{h'(x)}{h(x)}=\frac{f'(x)\cdot g(x)}{f(x)}+\log(f(x))\cdot g'(x)$$ and moving $h(x)$ to the other side and substituting it for its formula gives the formula I claim)
Start with $$f(x) = a(x)^{b(x)}$$.
Now, we can approach this a few ways. The one way is to go into multidimensional derivatives. Define
$$h(x,y) = a(x)^{b(y)}$$
$$h'(x,y) = \left[ {\begin{array}{cc} \frac{ \partial h_x }{ \partial x } & \frac{ \partial h_x }{ \partial y } \\ \frac{ \partial h_y }{ \partial x } & \frac{ \partial h_y }{ \partial y } \\ \end{array} } \right] = \left[ {\begin{array}{cc} a'(x) b(y) a(x)^{b(y)-1} & 0 \\ 0 & ln(a(x)) b'(y) a(x)^{b(y)}\\ \end{array} } \right]$$
Let $g(x) = \left[ \begin{array}{c} 1 \\ 1 \end{array} \right]$. Then $f(x) = h(g(x))$. By the chain rule, $$f'(x) = h'(g(x)) g'(x) = h'(x,x) \left[ \begin{array}{c} 1 \\ 1 \end{array} \right]$$
or $$f'(x) = a'(x) b(x) a(x)^{b(x)-1} + ln(a(x)) b'(x) a(x)^{b(x)}$$
$$ f'(x) = f(x) * ( b(x) \frac{a'(x)}{a(x)} + b'(x) ln(a(x)))$$
We can then do a sanity check using logarithmic derivatives. Also we can look at what happens when $a(x) = k$:
$$ f'(x) = f(x) * ( b(x) \frac{0}{a(x)} + b'(x) ln(k))$$ $$ f'(x) = ( b'(x) ln(k) ) k^{b(x)} $$
or $b(x)=k$:
$$ f'(x) = f(x) * ( k \frac{a'(x)}{a(x)} + 0 ln(a(x)))$$ $$ f'(x) = k a'(x) a(x)^{k-1}$$
which are both right.
Conceptually, this has two terms. You add them to find out what the instant "percent change" in $f(x)$ will be.
$b \frac{a'}{a}$ is $b$ times the instant percent change in $a$: If $a$ is growing by $1\%$, and you raise this to the power $b$, you grow by $b\%$.
$b' ln a$ is the change in $b$ wrt $x$ scaled by the log-scale of $a$. With exponentials, a given linear-scale unit of exponent increase causes a scale by a factor in the base. $ln(a)$ is the ratio between the two (the linear effect in the exponent to the exponential effect in the result), and we multiply that by $f$ because we are scaling the entire value.
|
Equivalence of Definitions of Separated Sets/Definition 1 implies Definition 2 Theorem
Let $T = \struct{S, \tau}$ be a topological space.
Let $A, B \subseteq S$ satisfy:
$A^- \cap B = A \cap B^- = \empty$ Then there exist $U,V \in \tau$ with: $A \subset U$ and $U \cap B = \empty$ $B \subset V$ and $V \cap A = \empty$ Proof
Let $U = S \setminus B^-$ be the relative complement of $B^-$.
From Empty Intersection iff Subset of Relative Complement, $A \subseteq S \setminus B^- = U$
From Relative Complement of Relative Complement, $S \setminus U = B^-$.
From Empty Intersection iff Subset of Relative Complement, $U \cap B = \empty$.
Similarly, let $V = S \setminus A^-$ then $V \in \tau$ with: $B \subset V$ and $V \cap A = \empty$
$\blacksquare$
|
I know the following type of questions has been asked many times on Math SE. Please let me apologies if my questions seems to be of a repeating nature. For the purpose of my question, we won't consider metric spaces at all.
The notion of open and closed sets between general topological spaces are just called open sets in the three criteria within the definition of a topological spaces.
Often time, when we are asked to prove a set is open, we can try to prove the set's complement is closed, and hence take the set's complement. In vice versa manner, similar strategy can be used for the case of a closed set. Also, we are told, an open set is not the opposite of a closed set. So a set that is not open is not the same as saying that a set is closed. Here are my points of confusions or questions. There are a collection of results stating necessary and sufficient conditions for when one can consider sets to be either open or closed, and they are stated involving notions of interior, closure, limit point and boundary of the set. This also applies to both cases of a function being an open or a closed map.
Below, I have listed an example of a theorem respectively for the cases of open sets, closed sets, and open and closed map. I have stated a standard theorem for each case, and in the case of say a set is not open, from the theorem stated, does how I change the phrasing reflect the consequences of how the theorem would not hold. Basically, what I am interested in is, to say a set B is not open, how does it affect the collection of results stating when a set can be consider to be an open set mathematically speaking, and also with respect to the relations to each other between set B's boundary, closure, interior and limit points. So in the case of set B being not open, does it mean it is not a subset of its interior and as a result, it contain some of its boundary. Similarly for a closed set, what else can we say about it. I ask this because if in the case of trying to prove that a set is open, but you are having difficulty proving its complement is closed. Well, what about assuming that is is not open, then what can be stated about a set that is known only to be not open and nothing else.
A set $B$ is not open means $B\subsetneqq \mathring{B}$
A set $C$ is closed iff $C=\overline{C}$
A set $B$ is not closed means $\overline{B}\subsetneqq C$
Also for the cases of both an open and a closed maps; we define a function to be an open map as:
Let $(X,\mathcal{T})$ and $(Y,\mathcal{U})$ be topological spaces, A function $f:X\rightarrow Y$ is open provided if $U \in \mathcal{T}$ then $f(U)\in\mathcal{U}$
hence when we talk about $f$ is not an open map, does it mean either one of the following:
A function $f:X\rightarrow Y$ is not open provided if $U \in \mathcal{T}$ is an open subset in $X$, then $f(U)\notin\mathcal{U}$, meaning $f(U) \subsetneq \mathring{f(U)}$ a)$f(U)$ is not open means: if $U \in \mathcal{T}$ is an open subset in $X$, then $f(U) \subsetneqq \mathring{f(U)}$ or b)if $U \notin \mathcal{T}$, which means $U \subsetneq \mathring{U},$ and $f(U) \subsetneqq \mathring{f(U)}$
Similarly, we define a function to be a closed map to mean:
$C$ is a closed subset of $(X,\mathcal{T})$, then $f(C)$ is a closed subset of $(Y,\mathcal{U})$, meaning $f(C)=\overline{f(C)}$ So to speak of a function being not a closed map, does it mean either: a)A function $f:X\rightarrow Y$ is not closed provided if $C$ is a closed subset of $(X,\mathcal{T})$, and $f(C)$ is not a closed subset of $(Y,\mathcal{U})$, meaning $\overline{f(C)} \subsetneqq f(C)$ or b)A function $f:X\rightarrow Y$ is not closed provided if $C$ is not a closed subset of $(X,\mathcal{T})$, $\overline{C} \subsetneqq C$, and $f(C)$ is not a closed subset of $(Y,\mathcal{U})$, meaning $\overline{f(C)} \subsetneqq f(C)$
Thank you in advance
|
Distance-Preserving Mapping is Injection of Metric Spaces Theorem
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.
Let $\phi: M_1 \to M_2$ be a distance-preserving mapping.
Then $\phi$ is an injection.
Proof
Let $a, b \in A_1$ and suppose that $\phi \paren {a} = \phi \paren{b}$.
Then by the definition of a metric space:
$d_2 \tuple{ \phi \paren a, \phi \paren b } = 0$ By the definition of a distance-preserving mapping then: $d_1 \tuple{ a, b } = 0$
Thus by the definition of a metric space:
$a = b$ Hence $\phi$ is injective.
$\blacksquare$
|
MU First Year Engineering (Semester 1)
Engineering Mechanics May 2016
Engineering Mechanics
May 2016
Total marks: --
Total time: --
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) 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) Determine the resultant of the forces acting as given in figure below. Find the angle which the resultant makes with the positive x-axis.
4 M
1(b) Two spheres A and B are kept in a horizontal channel. Determine the reactions coming from all the contact surfaces. Consider the radius of A and B are 40mm and 30mm respectively. Take W
A= 500 N and W B= 200N
4 M
1(c) Define Angle of Friction and Angle of Repose
4 M
1(d) Car A starts from rest & accelerates uniformly on a straight road. Another car B starts from the same place 5 seconds later with initial velocity zero & it accelerates uniformly at 5 m/sec
2. If both the cars overtake at 500 m from the starting place, find the acceleration of car A.
4 M
1(e) Find the angle the force P makes with horizontal such that the block of mass 4 kg has an acceleration of 10m/sec
2, when it is subjected to a force of 35 N. μs=0.7, μk=0.6
4 M
2(a) Replace the force system by a single force w.r.to point C
6 M
2(b) A uniform beam AB hinged at A is kept horizontal by supporting & setting a 100 n weight by using a string tied at B & passing over a smooth pulley at C. The beam also loaded as shown in figure below. Find the reactions at A & C.
8 M
2(c) Prove that for a perfectly elastic body two equal masses participating in collision exchange their velocities.
6 M
3(a) Find Centroid of shaded area with reference to X and Y axes.
8 M
3(b) Find the resultant of the spatial concurrent force system concurrent at A (1,0,0) and passing through points B(-1,3,5), C(3,5,7), D(0,4,0). Magnitude of forces F
AB= 100N, F AC= 150N, F AD= 200N.
6 M
3(c) A collar of mass 1 kg is attached to a spring ad slides without friction along a circular rod which lies in a horizontal plane. The spring is underformed when the collar is at B. Knowing that the collar is passing through the point D with a speed of 1.8 m/s, determine the speed of the collar when it passes through point C and B. Take Stiffness of the spring, k=250 N/m, Radius of the circular path = 300 mm and distance OA=125 mm.
6 M
4(a) Find the reactions at supports A and E for the beam loaded as shown in figure below.
8 M
4(b) A fighter Plane Moving horizontally with a constant velocity of 200 m/seconds releases a bomb from an altitude of 400 m. Find the velocity and direction of the bomb just before it strikes the ground. Also determine the distance travelled by the plane before the bomb just strikes the ground.
6 M
4(c) Find velocity of C and point D at the instant shown ω
AB= 3 rad/sec clockwise. AB = 400mm.
6 M
5(a) Find the forces in CF and CD by method of section and the remaining by Method of Joints.
8 M
5(b) For a vehicle moving along a straight line, v-t diagram is as shown in figure below. Plt a-t & s-t diagrams for the given time period.
6 M
5(c) A wheel is rolling along a straight path without slipping. Determine velocity of points A, B and p. OP = 600mm, ω=4 rad/sec, Vo=4m/s
6 M
6(a) A force of magnitude 500N is acting from A(2,3,6) and passes through a point B(6,2,6). Compute its moment about point C(4,6,3).
4 M
6(b) A stone is thrown with a velocity (u) m/sec at an angle of 20° horizontal from a point 2 m above the ground. The stone strikes the ground 5 m away from the original position. The motion of stone is subjected to gravitational acceleration & wind resistance of 0.82 m/sec
2, opposing the horizontal motion. Determine the time of flight of the stone.
4 M
6(c) A heavy metal bar AB rests with its lower end A on a rough horizontal floor having coefficient of friction μ
F& the other end B on a rough vertical wall having coefficient of friction μ W. If the centre of gravity of the bar is at distances a & b from the ends A & B respectively, show that at impending motion, the inclination of the bar with the horizontal will be : \(\theta=\tan^{-1}\left ( \dfrac{1}{\mu_F}\dfrac{a-b\mu_F \mu_w}{a+b} \right )\)
8 M
6(d) Two masses are interconnected with the pulley system Neglecting inertial & frictional effect of pulleys & cord, determine the acceleration of the mass m
take m 2.
take m
1= 50kg & m 2= 40kg
4 M
More question papers from Engineering Mechanics
|
$x$ is an angle of the right angled triangle, $\tan{x}$ is a trigonometric function. A special trigonometric function is formed to represent a value in mathematics as $x$ approaches $\tan^{-1}{3}$.
$\displaystyle \large \lim_{x \,\to\, \tan^{-1}{3}} \normalsize {\dfrac{\tan^2{x}-2\tan{x}-3}{\tan^2{x}-4\tan{x}+3}}$
The function is in terms of $\tan{x}$. So, try to change the limit value in the same form.
If $x \to \tan^{-1}{3}$, then $\tan{x} \to 3$.
$\implies \displaystyle \large \lim_{\tan{x} \,\to\, 3} \normalsize {\dfrac{\tan^2{x}-2\tan{x}-3}{\tan^2{x}-4\tan{x}+3}}$
Substitute $\tan{x} = 3$ to find the value of the function as $\tan{x}$ approaches $3$.
$= \dfrac{{(3)}^2-2 \times 3-3}{{(3)}^2-4 \times 3+3}$
$= \dfrac{9-6-3}{9-12+3}$
$= \dfrac{9-9}{12-12}$
$= \dfrac{0}{0}$
It is undefined. So, the limit of trigonometric function should be solved in another method to obtain value of the function as $x$ approaches $\tan^{-1}{3}$
It is a trigonometric function which contains $\tan{x}$ terms in different forms. The trigonometric function cannot be simplified by only using trigonometric identities. So, alternative method should be searched to evaluate it.
Observe the trigonometric expressions in both numerator and denominator. They both are representing a quadratic equation. So, try to simplify the expression in both numerator and denominator by quadratic equation system.
Express each trigonometric expression as factors by using factoring method of quadratics.
$= \displaystyle \large \lim_{\tan{x} \,\to\, 3} \normalsize {\dfrac{\tan^2{x}-3\tan{x}+\tan{x}-3}{\tan^2{x}-3\tan{x}-\tan{x}+3}}$
$= \displaystyle \large \lim_{\tan{x} \,\to\, 3} \normalsize {\dfrac{\tan{x}{(\tan{x}-3)}+\tan{x}-3}{\tan{x}{(\tan{x}-3)}-\tan{x}+3}}$
$= \displaystyle \large \lim_{\tan{x} \,\to\, 3} \normalsize {\dfrac{\tan{x}{(\tan{x}-3)}+1{(\tan{x}-3)}}{\tan{x}{(\tan{x}-3)}-1{(\tan{x}-3)}}}$
$= \displaystyle \large \lim_{\tan{x} \,\to\, 3} \normalsize {\dfrac{(\tan{x}-3)(\tan{x}+1)}{(\tan{x}-3)(\tan{x}-1)}}$
Now, simplify the function to the possible level.
$= \displaystyle \large \lim_{\tan{x} \,\to\, 3} \normalsize \require{cancel} {\dfrac{\cancel{(\tan{x}-3)}(\tan{x}+1)}{\cancel{(\tan{x}-3)}(\tan{x}-1)}}$
$= \displaystyle \large \lim_{\tan{x} \,\to\, 3} \normalsize {\dfrac{\tan{x}+1}{\tan{x}-1}}$
Substitute $\tan{x} = 3$ to find the value of the function as $\tan{x}$ approaches $3$.
$= \dfrac{3+1}{3-1}$
$= \dfrac{4}{2}$
$= \require{cancel} \dfrac{\cancel{4}}{\cancel{2}}$
$= 2$
It is the solution for this limits problem of trigonometric function.
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
|
So I have two functions. $f(x) = e^{-x^2+1}$ and $g(x)=\sqrt{x^2-4x+3}$. I am then asked to determine the domain and range of
$a)f∘g,$
$b)g∘f$
I already did part $a)$ and the domain for part $b)$.
For part $a)$, the domain was $(-\infty,1)\cup(3,\infty)$ and the range was $(0,e^2)$.
For part $b$, I figured out that the domain was $(-\infty,-1]\cup[1,\infty)$. I am not sure how to find the range though. Normally, I would take the inverse of g∘f and find the domain of that, and although I can do it, I don't think I did it correctly.
Currently, I did figure out that $g∘f$ is $\sqrt{e^{-2x^2+2}-4e^{-x^2+1}+3}$. How do I find the range of this mess though? I attempted to take the inverse which I believe is:
$y=\pm\sqrt{1-\ln(2\pm\sqrt{1+y^2})}$
Although I know that Wolfram Alpha is not the arbitrator of what correct is, it's generally been right and my answer disagrees with what Wolfram alpha has obtained (As seen here). In addition, the range is something that I am not sure how Wolfram obtained (As seen here). This also looks REALLY messy.
Can anyone guide me as to how this was obtained? That would be much appreciated!
|
On a data set I inherited, I calculated the standardized difference between the means of 2 groups,
d, for each of 5 variables. All the data was collected from same set of about 175 people, with 35 in the smaller group. What test might I use to compare the d's? One of the 5 d's is noticeably different from the others, but it could be due to chance. Thanks.
On a data set I inherited, I calculated the standardized difference between the means of 2 groups,
This can be easily done with methods described in chapter 19 by Gleser and Olkin in the
Handbook of Research Synthesis and Meta-Analysis (2009).
First, put the 5 d-values into a vector, $\vec{d}$.
Next, we need to construct the $5 \times 5$ variance-covariance matrix for $\vec{d}$. The diagonal elements (the variances) are given by $$Var[d_i] = \frac{1}{n_1} + \frac{1}{n_2} + \frac{d_i^2}{2(n_1 + n_2)},$$ where $n_1$ and $n_2$ are the group sizes (see equation 19.26 in the chapter). The off-diagonal elements (the covariances) are given by $$Cov[d_i, d_j] = \left(\frac{1}{n_1} + \frac{1}{n_2}\right) r_{ij} + \left(\frac{d_i d_j}{2(n_1 + n_2)}\right) r_{ij}^2,$$ where $r_{ij}$ is the correlation between variable $i$ and variable $j$ in your data (see equation 19.27 in the chapter). Let's call the resulting matrix $V$.
Now you can do several different things, including:
Test the null hypothesis that the true standardized mean differences underlying these 5 observed effects are the same (i.e., $H_0: \delta_1 = \delta_2 = \ldots = \delta_5$). This is commonly called a 'test for heterogeneity' in the meta-analytic literature and can be done with the so-called $Q$-test (see equation 19.31). Test whether the true standardized mean difference for one particular effect is different from the rest (which are assumed to share the same common true effect). For this, you can fit a model that includes a dummy variable for that one effect believed to be significantly different and then test that dummy (see section 19.4.2 for regression models with such data).
I'll illustrate all of this with some made-up data and R code.
### group sizesn1 <- 35n2 <- 175-35### vector with the observed d valuesd <- c(.24, .10, .38, .86, .29)### construct the var-cov matrix (R is the correlation matrix of the 5 variables)R <- matrix(c( 1, .52, .35, .68, .44, NA, 1, .48, .27, .33, NA, NA, 1, .56, .25, NA, NA, NA, 1, .49, NA, NA, NA, NA, 1), nrow=5)R[upper.tri(R)] <- t(R)[upper.tri(R)]V <- (1/n1 + 1/n2) * R + (outer(d, d, '*') / (2*(n1 + n2))) * R^2### load metafor packagelibrary(metafor)### fit model assuming homogeneous effectsres <- rma.mv(d, V)### examine results (esp. Q-test for heterogeneity)res### test if the 4th effect is significantly different from the rest### note: I(1:5 == 4) gives me a dummy variable that is equal to FALSE (0) ### for effects 1, 2, 3, and 5, and equal to TRUE (1) for effect 4res <- rma.mv(d ~ I(1:5 == 4), V)### examine results (esp. the p-value for the dummy variable)res
So, in these data, we would reject the null hypothesis that the true effects are homogeneous ($Q(4) = 20.97$, $p = .0003$) and we would conclude that the 4th effect is significantly larger than the rest ($p < .0001$). In fact, the test for residual heterogeneity is not significant ($Q(3) = 2.23$, $p = .53$), which implies in this case that there is no significant amount of heterogeneity among effects 1, 2, 3, and 5.
One word of caution: You are picking out that one effect to test that appears to be different from the rest. But you did not choose it a priori -- you picked it after examining the effects. So, a better approach would be to consider all 5 possible tests you could have run and apply a correction for multiple testing. In the example above, the test easily survives a Bonferroni correction (just multiply the p-value for the dummy by 5), but that may not be the case in other data. Addition: Here is code that allows you to fit all 5 models, extract the p-values, and then apply some correction for multiple testing, such as Holm's method.
pvals <- rep(NA, 5)for (i in 1:5) { res <- rma.mv(d ~ I(1:5 == i), V) pvals[i] <- res$pval[2]}round(p.adjust(pvals, method="holm"), 4)
The results are:
[1] 0.0305 0.6274 0.3228 0.0001 0.3835
So, for these data, there is some evidence that the first and fourth values could be considered to be significantly different from the rest.
I don't know any statistical test to check the difference of effect sizes. But nevertheless there is a possible solution to the problem: confidence intervals.
You can calculate the confidence interval for each of your
d. If the confidence intervals have some large intersection it is OK to claim that they are similar. If they have no intersection, they are definitely not similar.
I'm a little confused at your terminology (in particular, I
think you mean d is an index for the variables, but I'm not 100% sure. You also don't provide a lot of context for how the d's are related. I'm going to assume the variables are continuous because you standardized them and compare means. Without making any other assumptions, you could try Multivariate Analysis of Variance, followed by ANOVA or t-tests for each variable assuming the MANOVA is significant.
If your data are repeated observations, MANOVA will work but you could do a longitudinal model, where you nest observations within the people. This can be done in a multilevel regression model.
If your data are reflective of an underlying latent variable, use structural equation modeling or confirmatory factor analysis, and include each variable a manifestation or indicator of your underlying factor.
You bring up the smaller sample size of one of the groups. That imbalance will affect the power of your test (i.e. the ability to detect a difference when it is really there), but shouldn't affect the validity of the tests.
|
The question Compute $ \lim\limits_{n \to \infty }\sin \sin \dots\sin n$ was recently closed (then reopened and historical-locked). I agree that, by modern standards, the question is not a good one. It is simply a problem statement question, and would rapidly be closed if it were posted today.
However, the question has proved to be quite useful to the site. It is highly upvoted, with a highly upvoted answer (for whatever upvotes are worth) and at the list of linked questions, there includes 20 other questions, most of which are duplicates. This seems to imply that the question of iterating the sine function comes up with some frequency, and that the older question has proved useful in the sense that it gives a place to direct new askers of the same question. Thus I believe that the question should be preserved for its "historical significance" to the site, as well as its continued utility.
Additionally, Martin Sleziak suggested in chat that some older low quality questions have become the target of links from offsite, and have therefore proved useful to the outside world. These questions, too, might benefit from a historical lock–they are poor questions, but useful and deserving of preservation.
I would like to suggest that we discuss a policy regarding the use of historical locks to preserve such questions. So that there is a point of view to discuss:
There are (typically quite old) questions on MSE that are of very low quality but have proved to be useful as duplicate targets or as reference material for other websites, such as Wikipedia. These questions should be preserved for their utility, but should be locked by moderators, as they are not examples of the kinds of questions that we want to see asked today.
|
$$\int_1^2 f(x)\,\mathrm{d}x$$ where $$f(x)=\begin{cases} 1 & \text{if $x$ is rational,} \\ 0 & \text{if $x$ is irrational} \end{cases}$$
How does one interpret this to find the upper and lower sums on a regular partition
$$\Delta x={1\over n}$$
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
$$\int_1^2 f(x)\,\mathrm{d}x$$ where $$f(x)=\begin{cases} 1 & \text{if $x$ is rational,} \\ 0 & \text{if $x$ is irrational} \end{cases}$$
How does one interpret this to find the upper and lower sums on a regular partition
$$\Delta x={1\over n}$$
The values of x are 0, 1/n, 2/n, 3/n, ..., (n-1)/n, 1. Each pair of successive x values, 0 to 1/n, 1/n to 2/n, etc. gives an interval. To find the "upper sum" take the largest value of f in each interval. To find the "lower sum", take the lowest value of f in each interval. For this particular interval that is very easy- in ANY interval there exist both rational and irrational numbers so the largest value of f in any interval is 1 and the smallest is 0, 1 times 1/n is 1/n and 0 times 1/n is 0. Adding 1/n n times gives 1 and adding 0 n times gives 0.
|
Current browse context:
math.DS
Change to browse by: References & Citations Bookmark(what is this?) Mathematics > Dynamical Systems Title: Polynomial dynamical systems and Korteweg--de Vries equation
(Submitted on 13 May 2016)
Abstract: In this work we explicitly construct polynomial vector fields $\mathcal{L}_k,\;k=0,1,2,3,4,6$ on the complex linear space $\mathbb{C}^6$ with coordinates $X=(x_2,x_3,x_4)$ and $Z=(z_4,z_5,z_6)$. The fields $\mathcal{L}_k$ are linearly independent outside their discriminant variety $\Delta \subset \mathbb{C}^6$ and tangent to this variety. We describe a polynomial Lie algebra of the fields $\mathcal{L}_k$ and the structure of the polynomial ring $\mathbb{C}[X, Z]$ as a graded module with two generators $x_2$ and $z_4$ over this algebra. The fields $\mathcal{L}_1$ and $\mathcal{L}_3$ commute. Any polynomial $P(X,Z) \in \mathbb{C}[X, Z]$ determines a hyperelliptic function $P(X,Z)(u_1, u_3)$ of genus $2$, where $u_1$ and $u_3$ are coordinates of trajectories of the fields $\mathcal{L}_1$ and $\mathcal{L}_3$. The function $2 x_2(u_1, u_3)$ is a 2-zone solution of the KdV hierarchy and $\frac{\partial}{\partial u_1}z_4(u_1, u_3)=\frac{\partial}{\partial u_3}x_2(u_1, u_3)$. Submission historyFrom: Elena Bunkova [view email] [v1]Fri, 13 May 2016 06:34:40 GMT (21kb)
|
The exact value of tan of 30 degrees in fraction form is $1/\sqrt{3}$. It can be derived mathematically in three approaches in which two of them are geometric approaches and third one is trigonometric approach. In trigonometry, the tan $30$ degrees value is derived on the basis of values of sin and cos of $30$ degrees.
The exact value of $\tan{(30^°)}$ can be derived in trigonometry on the basis of geometrical relations between sides of the right triangle when the angle of the triangle is $\dfrac{\pi}{6}$. According to properties of right triangle, if the angle of right triangle is $30^°$, then the length of adjacent side is $\dfrac{\sqrt{3}}{2}$ times of length of hypotenuse.
Therefore, $OQ = \dfrac{\sqrt{3}}{2} \times {OP}$ in the case of $\Delta POQ$
Now, express relation between all three sides of the triangle in mathematical form by Pythagorean Theorem.
${OP}^2 = {PQ}^2+{OQ}^2$
If $OQ = \dfrac{\sqrt{3}}{2} \times {OP}$ then $OP = \dfrac{2}{\sqrt{3}} \times {OQ}$. Now, substitute the length of hypotenuse by its equivalent value in the above equation.
$\implies$ ${\Bigg(\dfrac{2}{\sqrt{3}} \times {OQ}\Bigg)}^2 = {PQ}^2+{OQ}^2$
$\implies$ $\dfrac{4}{3}{OQ}^2 = {PQ}^2+{OQ}^2$
$\implies$ $\dfrac{4}{3}{OQ}^2-{OQ}^2 = {PQ}^2$
$\implies$ ${\Bigg(\dfrac{4}{3}-1\Bigg)}{OQ}^2 = {PQ}^2$
$\implies$ ${\Bigg(\dfrac{4-3 \times 1}{3}\Bigg)}{OQ}^2 = {PQ}^2$
$\implies$ ${\Bigg(\dfrac{4-3}{3}\Bigg)}{OQ}^2 = {PQ}^2$
$\implies$ ${\Bigg(\dfrac{1}{3}\Bigg)}{OQ}^2 = {PQ}^2$
$\implies$ $\dfrac{1}{3} = \dfrac{{PQ}^2}{{OQ}^2}$
$\implies$ $\dfrac{1}{3} = {\Bigg(\dfrac{PQ}{OQ}\Bigg)}^2$
$\implies$ ${\Bigg(\dfrac{PQ}{OQ}\Bigg)}^2 = \dfrac{1}{3}$
$\implies$ $\dfrac{PQ}{OQ} = \sqrt{\dfrac{1}{3}}$
$\implies$ $\dfrac{PQ}{OQ} = \dfrac{1}{\sqrt{3}}$
$PQ$ and $OQ$ are lengths of opposite and adjacent sides of the right triangle.
$\implies$ $\dfrac{Length \, of \, Opposite \, side}{Length \, of \, Adjacent \, side}$ $=$ $\dfrac{1}{\sqrt{3}}$
The angle of $\Delta POQ$ is $30$ degrees. So, the ratio represents tan of angle $30$ degrees according to definition of tan function.
$\,\,\, \therefore \,\,\,\,\,\, \tan{(30^°)} \,=\, \dfrac{1}{\sqrt{3}}$
$\implies \tan{(30^°)}$ $\,=\,$ $0.5773502691\ldots$
The exact value of tan of $30$ degrees is equal to $\dfrac{1}{\sqrt{3}}$ in fraction form and its value in decimal is $0.5773502691\ldots$
You can even derive the value of tan of angle $30$ degrees by constructing a right triangle with $\dfrac{\pi}{6}$, using geometric tools but it is not possible to obtain the exact value of $\tan{{\Big(33\dfrac{1}{3}}^g\Big)}$ due to measuring lengths of sides approximately. However, its value approximately equals to the actual value.
In this way, the right triangle ($\Delta JHI$) with $\dfrac{\pi}{6}$ radians is constructed geometrically. Now, let’s find the value of $\tan{\Big(\dfrac{\pi}{6}\Big)}$ experimentally.
$\tan{(30^°)} = \dfrac{Length \, of \, Opposite \, side}{Length \, of \, Adjacent \, side}$
$\implies \tan{(30^°)} \,=\, \dfrac{IJ}{HJ}$
Actually, the lengths of opposite side ($IJ$) and adjacent side ($HJ$) are unknown but they can be measured by ruler.
You will be observed that the length of opposite side is $3.5 \, cm$ exactly and the length of adjacent side lies between $6$ to $6.1 \, cm$. So, its length is considered as $6.05 \, cm$ approximately.
$\implies \tan{(30^°)} \,=\, \dfrac{IJ}{HJ} = \dfrac{3.5}{6.05}$
$\implies \tan{(30^°)} \,=\, \require{cancel} \dfrac{\cancel{3.5}}{\cancel{6.05}}$
$\,\,\, \therefore \,\,\,\,\,\, \tan{(30^°)} \,=\, 0.5785123967\ldots$
The exact value of $\tan{(30^°)}$ can also be evaluated in trigonometry by the quotient or ratio identity of sin and cos functions. In this case, the $\tan{\Big(\dfrac{\pi}{6}\Big)}$ value is actually calculated by the values of sin 30 degrees and cos 30 degrees.
$\tan{(30^°)} \,=\, \dfrac{\sin{(30^°)}}{\cos{(30^°)}}$
$\implies \tan{(30^°)} \,=\, \dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}$
$\implies \tan{(30^°)} \,=\, \dfrac{1}{2} \times \dfrac{2}{\sqrt{3}}$
$\implies \tan{(30^°)} \,=\, \dfrac{1 \times 2}{2 \times \sqrt{3}}$
$\implies \tan{(30^°)} \,=\, \dfrac{2}{2\sqrt{3}}$
$\implies \tan{(30^°)} \,=\, \require{cancel} \dfrac{\cancel{2}}{\cancel{2}\sqrt{3}}$
$\,\,\, \therefore \,\,\,\,\,\, \tan{(30^°)} \,=\, \dfrac{1}{\sqrt{3}}$
The value of tan of $30$ degrees is $\dfrac{1}{\sqrt{3}}$ or $0.5773502691\ldots$ and is same according to both theoretical geometric and trigonometric methods but its value is $0.5785123967\ldots$ as per practical geometrical approach. They both are approximately equal but the value of $\tan{\Big(\dfrac{\pi}{6}\Big)}$ obtained from practical geometrical approach slightly differs with the other methods. It is due to measuring lengths of the sides with parallax error. It can’t be considered as actual value because it changes if you construct a right triangle practically with different lengths of the sides.
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
|
Being a geek and having gone to high school in Los Alamos, I knew about Stanislaw Ulam. At least, I knew a bit about the work he had done during the Manhattan project at Los Alamos. And that he was among the greats of mathematics. But it seemed totally implausible that the person behind that door could really be Stanislaw Ulam. Ulam was a creature of myth to me. Not just from another generation, but practically from another world. There was no way that it could be him. No way.
So I had to find out.
I knocked.
And was invited in.
So I asked, "Are you Stanislaw Ulam?". He seemed amused. "Yes". "Uh..." I said.
Just as a note to anyone meeting a god who steps down onto the earth to walk among us, I suggest you have a question figured out ahead of time because you aren't going to come up with anything very good in the moment. At least I couldn't. In a similar vein, figure out your three wishes ahead of time as well in case you meet a genie.
After a moment of hurried thought, I came up with "Could we talk?". Seriously. That is the best I could do.
He said that would be fine, but that he had about 20 minutes of work to do first. To keep me busy, he gave me a problem to work on during that time. Not surprisingly, that problem was a really good one for assessing my mathematical sophistication.
That sophistication was not very high (and really still isn't decades later) partly because of youth and partly because of attitude and partly because schools can't deal with students who learn quickly. My history with math was that during an explosive few weeks in the 8th grade I had gotten the bug and worked my way through all of the high school math curriculum including calculus. It was a wonderful time and didn't feel like learning so much as just remembering how things worked. But there wasn't much anything to be done beyond that because there wasn't much anybody to talk to who knew what lay past that because we lived on a military base in Germany. I could wander through books, but I didn't really have enough understanding to go much further than a bit of differential geometry and multivariate calculus. Later, when we moved to Los Alamos, there were all kinds of people I could have learned from but I was engrossed in electronics and computing and German and Russian.
The problem that he gave me was to prove a simple fixed point theorem. Given a set $S$ and a continuous function $f: S \rightarrow S$ such that $f$ is a contraction, show that $f$ has a fixed point in $S$. A contraction is a function that maps distinct points closer together, $x_1 \ne x_2 \implies \rho(x_1, x_2) > \rho(f(x_1), f(x_2))$ where $\rho$ is some metric on $S$.
As with all good math questions, the first response should always be clarifications to make sure that the problem is actually well understood. Even though it has been a very long time, I remember asking if I could assume that $S$ is compact and that $f$ is continuous. As I remember, I didn't actually finish a proof during the short time I had then, but I did sketch out an approach that started by noting that $\rho(x, f(x)) > \rho(f(x), f(f(x)))$. Unfortunately, I took a wrong turn at that point and thought about limits. That apparently didn't matter and I had some great talks with Professor Ulam and his friend Jan Micielski.
One of the cool things about mathematics is that questions like this wind up in the back of your head and can pop out at any time. That just happened as I was reading a review of the latest edition of Proofs From the Book. This looks like a great book with all kinds of interesting insights to be had, but I just looked at their wonderfully short proof of the fundamental theorem of algebra. That proof goes something like this
Every real-valued function $f$ over a compact set $S$ has a minimum in $S$ Every point $x$ such that a complex polynomial is non-zero, $|p(x)|>0$, has a nearby point $x'$ such that $|p(x')| < |p(x)|$ Thus, the minimum of $|p(x)|$ exists, but all points $x$ such that $|p(x)|>0$ are not the minimum. Therefore, for some point $x_0$ we have $|p(x_0)|=0$
Point 2 takes a tiny bit of algebra, but overall this proof is stunningly simple and beautiful. But in the way that these things happen, after reading this proof I remembered the contraction theorem that I hadn't quite come up with so long ago and it occurred to me that this same trick would apply there.
The outline is
For any fixed point $x^*$, we have $\rho(x^*,f(x^*)) = 0$ For any other point $x$ we have $\rho(x, f(x)) > 0$ The function $d(x) = \rho(x, f(x))$ is continuous and has a minimum in $S$ For any point $x$ such that $d(x) \ne 0$, we have $d(x) > d(f(x))$ and thus $d(x)$ is not minimized at $x$ Thus, there exists some point $x^*$ where $d(x^*)=0$, that is to say $x^*$ is a fixed point.
Isn't it just the way of the world that you think of the perfect answer just a little bit too late?
|
$\dfrac{d}{dx}{\, \Big(k.f(x)\Big)} \,=\, k \times \dfrac{d}{dx}{\, f(x)}$
The derivative of product of a constant and a function is equal to the product of constant and the derivative of the function. This property of differentiation is called the constant multiple rule of derivatives.
Let’s take $x$ is a variable, $k$ is a constant and $f(x)$ is a function in terms of $x$. If the constant $k$ is multiplied by the function $f(x)$, then the product of them is $k.f(x)$, which is called as the constant multiple function.
The derivative of the constant multiple function with respect to $x$ is written in mathematical form as follows.
$\dfrac{d}{dx}{\, \Big(k.f(x)\Big)}$
The differentiation of the constant multiple function with respect to $x$ is equal to the product of the constant $k$ and the derivative of the function $f(x)$.
$\implies$ $\dfrac{d}{dx}{\, \Big(k.f(x)\Big)} \,=\, k \times \dfrac{d}{dx}{\, f(x)}$
This property is called the constant multiple rule of differentiation and it is used as a formula in differential calculus.
Look at the following examples to understand the use of the constant multiple rule in differential calculus.
$(1) \,\,\,$ $\dfrac{d}{dx}{\, \Big(6x^2\Big)} \,=\, 6 \times \dfrac{d}{dx}{\, x^2}$
$(2) \,\,\,$ $\dfrac{d}{dy}{\, \Bigg(\dfrac{\log_{e}{y}}{4}\Bigg)} \,=\, \dfrac{1}{4} \times \dfrac{d}{dy}{\, \log_{e}{y}}$
$(3) \,\,\,$ $\dfrac{d}{dz}{\, \Big(-0.7\sin{3z}\Big)} \,=\, -0.7 \times \dfrac{d}{dx}{\, \sin{3z}}$
Learn how to derive the constant multiple rule in differential calculus.
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
|
ä is in the extended latin block and n is in the basic latin block so there is a transition there, but you would have hoped \setTransitionsForLatin would have not inserted any code at that point as both those blocks are listed as part of the latin block, but apparently not.... — David Carlisle12 secs ago
@egreg you are credited in the file, so you inherit the blame:-)
@UlrikeFischer I was leaving it for @egreg to trace but I suspect the package makes some assumptions about what is safe, it offers the user "enter" and "exit" code for each block but xetex only has a single insert, the interchartoken at a boundary the package isn't clear what happens at a boundary if the exit of the left class and the entry of the right are both specified, nor if anything is inserted at boundaries between blocks that are contained within one of the meta blocks like latin.
Why do we downvote to a total vote of -3 or even lower? Weren't we a welcoming and forgiving community with the convention to only downvote to -1 (except for some extreme cases, like e.g., worsening the site design in every possible aspect)?
@Skillmon people will downvote if they wish and given that the rest of the network regularly downvotes lots of new users will not know or not agree with a "-1" policy, I don't think it was ever really that regularly enforced just that a few regulars regularly voted for bad questions to top them up if they got a very negative score. I still do that occasionally if I notice one.
@DavidCarlisle well, when I was new there was like never a question downvoted to more than (or less?) -1. And I liked it that way. My first question on SO got downvoted to -infty before I deleted it and fixed my issues on my own.
@DavidCarlisle I meant the total. Still the general principle applies, when you're new and your question gets donwvoted too much this might cause the wrong impressions.
@DavidCarlisle oh, subjectively I'd downvote that answer 10 times, but objectively it is not a good answer and might get a downvote from me, as you don't provide any reasoning for that, and I think that there should be a bit of reasoning with the opinion based answers, some objective arguments why this is good. See for example the other Emacs answer (still subjectively a bad answer), that one is objectively good.
@DavidCarlisle and that other one got no downvotes.
@Skillmon yes but many people just join for a while and come form other sites where downvoting is more common so I think it is impossible to expect there is no multiple downvoting, the only way to have a -1 policy is to get people to upvote bad answers more.
@UlrikeFischer even harder to get than a gold tikz-pgf badge.
@cis I'm not in the US but.... "Describe" while it does have a technical meaning close to what you want is almost always used more casually to mean "talk about", I think I would say" Let k be a circle with centre M and radius r"
@AlanMunn definitions.net/definition/describe gives a websters definition of to represent by drawing; to draw a plan of; to delineate; to trace or mark out; as, to describe a circle by the compasses; a torch waved about the head in such a way as to describe a circle
If you are really looking for alternatives to "draw" in "draw a circle" I strongly suggest you hop over to english.stackexchange.com and confirm to create an account there and ask ... at least the number of native speakers of English will be bigger there and the gamification aspect of the site will ensure someone will rush to help you out.
Of course there is also a chance that they will repeat the advice you got here; to use "draw".
@0xC0000022L @cis You've got identical responses here from a mathematician and a linguist. And you seem to have an idea that because a word is informal in German, its translation in English is also informal. This is simply wrong. And formality shouldn't be an aim in and of itself in any kind of writing.
@0xC0000022L @DavidCarlisle Do you know the book "The Bronstein" in English? I think that's a good example of archaic mathematician language. But it is still possible harder. Probably depends heavily on the translation.
@AlanMunn I am very well aware of the differences of word use between languages (and my limitations in regard to my knowledge and use of English as non-native speaker). In fact words in different (related) languages sharing the same origin is kind of a hobby. Needless to say that more than once the contemporary meaning didn't match a 100%.
However, your point about formality is well made. A book - in my opinion - is first and foremost a vehicle to transfer knowledge. No need to complicate matters by trying to sound ... well, overly sophisticated (?) ...
The following MWE with showidx and imakeidx:\documentclass{book}\usepackage{showidx}\usepackage{imakeidx}\makeindex\begin{document}Test\index{xxxx}\printindex\end{document}generates the error:! Undefined control sequence.<argument> \ifdefequal{\imki@jobname }{\@idxfile }{}{...
@EmilioPisanty ok I see. That could be worth writing to the arXiv webmasters as this is indeed strange. However, it's also possible that the publishing of the paper got delayed; AFAIK the timestamp is only added later to the final PDF.
@EmilioPisanty I would imagine they have frozen the epoch settings to get reproducible pdfs, not necessarily that helpful here but..., anyway it is better not to use \today in a submission as you want the authoring date not the date it was last run through tex
and yeah, it's better not to use \today in a submission, but that's beside the point - a whole lot of arXiv eprints use the syntax and they're starting to get wrong dates
@yo' it's not that the publishing got delayed. arXiv caches the pdfs for several years but at some point they get deleted, and when that happens they only get recompiled when somebody asks for them again
and, when that happens, they get imprinted with the date at which the pdf was requested, which then gets cached
Does any of you on linux have issues running for foo in *.pdf ; do pdfinfo $foo ; done in a folder with suitable pdf files? BMy box says pdfinfo does not exist, but clearly do when I run it on a single pdf file.
@EmilioPisanty that's a relatively new feature, but I think they have a new enough tex, but not everyone will be happy if they submit a paper with \today and it comes out with some arbitrary date like 1st Jan 1970
@DavidCarlisle add \def\today{24th May 2019} in INITEX phase and recompile the format daily? I agree, too much overhead. They should simply add "do not use \today" in these guidelines: arxiv.org/help/submit_tex
@yo' I think you're vastly over-estimating the effectiveness of that solution
(and it would not solve the problem with 20+ years of accumulated files that do use it)
@DavidCarlisle sure. I don't know what the environment looks like on their side so I won't speculate. I just want to know whether the solution needs to be on the side of the environment variables, or whether there is a tex-specific solution
@yo' that's unlikely to help with prints where the class itself calls from the system time.
@EmilioPisanty well th eenvironment vars do more than tex (they affect the internal id in teh generated pdf or dvi and so produce reproducible output, but you could as @yo' showed redefine \today oor teh \year, \month\day primitives on teh command line
@EmilioPisanty you can redefine \year \month and \day which catches a few more things, but same basic idea
@DavidCarlisle could be difficult with inputted TeX files. It really depends on at which phase they recognize which TeX file is the main one to proceed. And as their workflow is pretty unique, it's hard to tell which way is even compatible with it.
"beschreiben", engl. "describe" comes from the math. technical-language of the 16th Century, that means from Middle High German, and means "construct" as much. And that from the original meaning: describe "making a curved movement". In the literary style of the 19th to the 20th century and in the GDR, this language is used.
You can have that in englisch too: scribe(verb) score a line on with a pointed instrument, as in metalworking https://www.definitions.net/definition/scribe
@cis Yes, as @DavidCarlisle pointed out, there is a very technical mathematical use of 'describe' which is what the German version means too, but we both agreed that people would not know this use, so using 'draw' would be the most appropriate term. This is not about trendiness, just about making language understandable to your audience.
Plan figure. The barrel circle over the median $s_b = |M_b B|$, which holds the angle $\alpha$, also contains an isosceles triangle $M_b P B$ with the base $|M_b B|$ and the angle $\alpha$ at the point $P$. The altitude of the base of the isosceles triangle bisects both $|M_b B|$ at $ M_ {s_b}$ and the angle $\alpha$ at the top. \par The centroid $S$ divides the medians in the ratio $2:1$, with the longer part lying on the side of the corner. The point $A$ lies on the barrel circle and on a circle $\bigodot(S,\frac23 s_a)$ described by $S$ of radius…
|
This is due to the mass-energy equivalence and a phenomenon called binding energy.
Forming a nucleus releases energy because the nucleons are falling into a potential energy well. Due to Einstein's mass energy equivalence this results in the mass of the new nucleus being less than that of the particles that formed it.
The binding energy of carbon-12 is quoted on Wikipedia as $\pu{92.162 MeV}$. Therefore we can estimate the mass defect of a carbon-12 atom, $\Delta m$, using $ E = (\Delta m)c^2$:
$$\Delta m = \frac{(\pu{92.162 \times 10^6 eV})(\pu{1.6022 \times 10^-19 J eV-1})}{(\pu{2.9979 \times 10^8 m s-1})^2} = \pu{1.6430 \times 10^-28 kg} = \pu{0.098943 u}$$
The difference in the mass of carbon-12 to the mass of its constituent particles is $\pu{0.08940 u}$, so we can see that our calculation is a reasonable estimate of the mass defect. The slight difference is due to other more complicated factors that I have not taken into account, but it still illustrates the main reason for the mass defect.
|
$\sin{2\theta} \,=\, 2\sin{\theta}\cos{\theta}$
$\cos{2\theta} \,=\, \cos^2{\theta}-\sin^2{\theta}$
$\tan{2\theta} \,=\, \dfrac{2\tan{\theta}}{1-\tan^2{\theta}}$
$\cot{2\theta} \,=\, \dfrac{\cot^2{\theta}-1}{2\cot{\theta}}$
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
|
The integration of $\sin^3{x}$ function with respect to $x$ can be calculated in integral calculus in two methods.
The sin and cos functions both have direct relation in integral calculus. If the $sin^3{x}$ function is expressed in terms of both sine and cosine functions. The integral of $\sin^3{x}$ with respect to $x$ can be done easily in mathematics.
The sin cubed function can be written in terms of cos and sin functions by splitting them as two factors.
$= \,\,\,$ $\displaystyle \int{\sin^2{x} \times \sin{x}}dx$
Actually, the square of sin function can be written in terms of cos function by the sin squared formula.
$= \,\,\,$ $\displaystyle \int{(1-\cos^2{x}) \times \sin{x}}dx$
$= \,\,\,$ $\displaystyle \int{(1-\cos^2{x})\sin{x}}dx$
The differentiation of cos function is equal to minus of sin function. Hence, take $t = \cos{x}$, then $dt = -\sin{x}dx$. Therefore, $\sin{x}dx = -dt$.
Now, eliminate the $x$ terms by $t$ terms.
$= \,\,\,$ $\displaystyle \int{(1-t^2})(-dt)$
$= \,\,\,$ $-\displaystyle \int{(1-t^2})dt$
$= \,\,\,$ $\displaystyle \int{(-1+t^2})dt$
There is no straight integral rule to find the integration of difference of the two terms. So, use difference rule of integration.
$= \,\,\,$ $\displaystyle \int{(-1)}dt+\displaystyle \int{t^2}dt$
$= \,\,\,$ $-\displaystyle \int dt+\displaystyle \int{t^2}dt$
$= \,\,\,$ $-t+\dfrac{t^3}{3}+C$
The actual function is not in terms of $t$. So, replace back the $t$ by its assumed value.
$= \,\,\,$ $-\cos{x}+\dfrac{\cos^3{x}}{3}+C$
$= \,\,\,$ $\dfrac{\cos^3{x}}{3}-\cos{x}+C$
The integration of $\sin^3{x}$ with respect to $x$ can be done by transforming the sin cubed function as per sin triple angle formula.
Actually, there is no formula for finding the integral of sin cubed function in integral calculus. So, an alternative approach must be used to start finding the sine cubed function. In this method, sin triple angle formula is used to express sin cubed function in terms of sine and sin triple angle functions.
$\sin{3x} \,=\, 3\sin{x}-4\sin^3{x}$
$\implies 4\sin^3{x} \,=\, 3\sin{x}-\sin{3x}$
$\implies \sin^3{x} \,=\, \dfrac{3\sin{x}-\sin{3x}}{4}$
$= \,\,\,$ $\displaystyle \int{\Bigg[\dfrac{3\sin{x}-\sin{3x}}{4}\Bigg]}dx$
$= \,\,\,$ $\displaystyle \int{\Bigg[\dfrac{3}{4}\sin{x}-\dfrac{1}{4}\sin{3x}\Bigg]}dx$
$= \,\,\,$ $\displaystyle \int{\dfrac{3}{4}\sin{x}}dx-\displaystyle \int{\dfrac{1}{4}\sin{3x}}dx$
$= \,\,\,$ $\displaystyle \dfrac{3}{4}\int{\sin{x}}dx-\displaystyle \dfrac{1}{4}\int{\sin{3x}}dx$
The integration of first term can be calculated as per integral formula of sin function but it is not possible to find the integration of $\sin{3x}$ function directly by this formula. However, it can be done by transforming the angle $3x$ in terms of a variable.
If $t = 3x$, then $dt = 3dx$. Therefore, $dx = \dfrac{dt}{3}$
$= \,\,\,$ $\displaystyle \dfrac{3}{4}\int{\sin{x}}dx-\displaystyle \dfrac{1}{4}\int{\sin{t}}\dfrac{dt}{3}$
$= \,\,\,$ $\displaystyle \dfrac{3}{4}\int{\sin{x}}dx-\displaystyle \dfrac{1}{4 \times 3}\int{\sin{t}}dt$
$= \,\,\,$ $\displaystyle \dfrac{3}{4}\int{\sin{x}}dx-\displaystyle \dfrac{1}{12}\int{\sin{t}}dt$
There are two integral terms in which one term is in terms of $x$ and other term is in terms of $t$ but both terms represents the indefinite integral of sin function.
$= \,\,\,$ $\dfrac{3}{4}(-\cos{x})-\dfrac{1}{12}(-\cos{t})+C$
$= \,\,\,$ $-\dfrac{3}{4}\cos{x}+\dfrac{1}{12}\cos{t}+C$
The second term in this expression is in terms of $t$ but our actual function is in terms of $x$. So, replace the value of $t$ in terms $x$.
$= \,\,\,$ $-\dfrac{3}{4}\cos{x}+\dfrac{1}{12}\cos{3x}+C$
The first term is cos function but second term is cos triple angle function in the expression. It can be simplified further by expanding $\cos{3x}$ function in terms of $\cos{x}$ by cos triple angle formula.
$= \,\,\,$ $-\dfrac{3}{4}\cos{x}+\dfrac{1}{12}\Bigg[4\cos^3{x}-3\cos{x}\Bigg]+C$
$= \,\,\,$ $-\dfrac{3}{4}\cos{x}+\dfrac{4}{12}\cos^3{x}-\dfrac{3}{12}\cos{x}+C$
$= \,\,\,$ $-\dfrac{3}{4}\cos{x}+\require{cancel} \dfrac{\cancel{4}}{\cancel{12}}\cos^3{x}-\require{cancel} \dfrac{\cancel{3}}{\cancel{12}}\cos{x}+C$
$= \,\,\,$ $-\dfrac{3}{4}\cos{x}+\dfrac{1}{3}\cos^3{x}-\dfrac{1}{4}\cos{x}+C$
$= \,\,\,$ $-\dfrac{3}{4}\cos{x}-\dfrac{1}{4}\cos{x}+\dfrac{1}{3}\cos^3{x}+C$
$= \,\,\,$ $\Bigg[-\dfrac{3}{4}-\dfrac{1}{4}\Bigg]\cos{x}+\dfrac{1}{3}\cos^3{x}+C$
$= \,\,\,$ $-\dfrac{4}{4}\cos{x}+\dfrac{1}{3}\cos^3{x}+C$
$= \,\,\,$ $\require{cancel} -\dfrac{\cancel{4}}{\cancel{4}}\cos{x}+\dfrac{1}{3}\cos^3{x}+C$
$= \,\,\,$ $-\cos{x}+\dfrac{\cos^3{x}}{3}+C$
$= \,\,\,$ $\dfrac{\cos^3{x}}{3}-\cos{x}+C$
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
|
Basically 2 strings, $a>b$, which go into the first box and do division to output $b,r$ such that $a = bq + r$ and $r<b$, then you have to check for $r=0$ which returns $b$ if we are done, otherwise inputs $r,q$ into the division box..
There was a guy at my university who was convinced he had proven the Collatz Conjecture even tho several lecturers had told him otherwise, and he sent his paper (written on Microsoft Word) to some journal citing the names of various lecturers at the university
Here is one part of the Peter-Weyl theorem: Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$.
What exactly does it mean for $\rho$ to split into finite dimensional unitary representations? Does it mean that $\rho = \oplus_{i \in I} \rho_i$, where $\rho_i$ is a finite dimensional unitary representation?
Sometimes my hint to my students used to be: "Hint: You're making this way too hard." Sometimes you overthink. Other times it's a truly challenging result and it takes a while to discover the right approach.
Once the $x$ is in there, you must put the $dx$ ... or else, nine chances out of ten, you'll mess up integrals by substitution. Indeed, if you read my blue book, you discover that it really only makes sense to integrate forms in the first place :P
Using the recursive definition of the determinant (cofactors), and letting $\operatorname{det}(A) = \sum_{j=1}^n \operatorname{cof}_{1j} A$, how do I prove that the determinant is independent of the choice of the line?
Let $M$ and $N$ be $\mathbb{Z}$-module and $H$ be a subset of $N$. Is it possible that $M \otimes_\mathbb{Z} H$ to be a submodule of $M\otimes_\mathbb{Z} N$ even if $H$ is not a subgroup of $N$ but $M\otimes_\mathbb{Z} H$ is additive subgroup of $M\otimes_\mathbb{Z} N$ and $rt \in M\otimes_\mathbb{Z} H$ for all $r\in\mathbb{Z}$ and $t \in M\otimes_\mathbb{Z} H$?
Well, assuming that the paper is all correct (or at least to a reasonable point). I guess what I'm asking would really be "how much does 'motivated by real world application' affect whether people would be interested in the contents of the paper?"
@Rithaniel $2 + 2 = 4$ is a true statement. Would you publish that in a paper? Maybe... On the surface it seems dumb, but if you can convince me the proof is actually hard... then maybe I would reconsider.
Although not the only route, can you tell me something contrary to what I expect?
It's a formula. There's no question of well-definedness.
I'm making the claim that there's a unique function with the 4 multilinear properties. If you prove that your formula satisfies those (with any row), then it follows that they all give the same answer.
It's old-fashioned, but I've used Ahlfors. I tried Stein/Stakarchi and disliked it a lot. I was going to use Gamelin's book, but I ended up with cancer and didn't teach the grad complex course that time.
Lang's book actually has some good things. I like things in Narasimhan's book, but it's pretty sophisticated.
You define the residue to be $1/(2\pi i)$ times the integral around any suitably small smooth curve around the singularity. Of course, then you can calculate $\text{res}_0\big(\sum a_nz^n\,dz\big) = a_{-1}$ and check this is independent of coordinate system.
@A.Hendry: It looks pretty sophisticated, so I don't know the answer(s) off-hand. The things on $u$ at endpoints look like the dual boundary conditions. I vaguely remember this from teaching the material 30+ years ago.
@Eric: If you go eastward, we'll never cook! :(
I'm also making a spinach soufflé tomorrow — I don't think I've done that in 30+ years. Crazy ridiculous.
@TedShifrin Thanks for the help! Dual boundary conditions, eh? I'll look that up. I'm mostly concerned about $u(a)=0$ in the term $u'/u$ appearing in $h'-\frac{u'}{u}h$ (and also for $w=-\frac{u'}{u}P$)
@TedShifrin It seems to me like $u$ can't be zero, or else $w$ would be infinite.
@TedShifrin I know the Jacobi accessory equation is a type of Sturm-Liouville problem, from which Fox demonstrates in his book that $u$ and $u'$ cannot simultaneously be zero, but that doesn't stop $w$ from blowing up when $u(a)=0$ in the denominator
|
In evolutionary psychology, it's a given that genes are selfish, and our genes determine what we are and set the limits on who we can be. (The human brain is very malleable, so those limits aren't always clear in humans but they're nevertheless there.) So we look at human universals like altruism, a sense of justice, honor, trust, guilt, friendship, sympathy, suspicion and hypocrisy, and many of these things defy explanation if you can only think that "everyone is out for themselves."
The long-standing explanation for these traits is based in Kin Selection, which in a nutshell (I recommend reading the linked paper or Robert Wright's
The Moral Animal) is the idea that kin have more or less the same genes, and when a gene arises that increases the likelihood of altruism (through yelling a warning cry when a predator is spotted for squirrels and possibly sacrificing itself, or through sharing a banana), the individuals most likely to benefit from that altruism are closely related and may have that gene as well. Hence the genes still propagate and are selected for; the benefits to an individual's kin who share the gene outweigh any individual loss from the altruism that might make that individual lose out in the gene pool normally. This doesn't mean that conscious altruism goes away, or that conscious minds are therefore automatically entirely selfish, just that at the gene level, which is all that matters to evolution, high-level altruism got there from a low-level selfish process. The mechanisms of an altruistic brain are altruistic: altruistic people will feel like they're being altruistic, others will call them altruistic, and for all intents and purposes they are altruistic. At the gene level this altruism is a selfish benefit for that person's (and likely their kin's) genes.
See Full Post and Comments
Horrible joke I know, and I suck at drawing, especially "anatomy". Computer Engineers have a trick to convert a bunch of resistors in a delta (triangle) formation to a Y (three prongs) formation and vice versa, and with that trick one can reduce the "hair circuit" shown (that some people apparently call a Christmas tree since it contains both a delta and a Y) to the nice single-resistor trim.
The "Christmas tree" appears in Giancoli, because usually Computer Engineers don't have to deal with it, and Giancoli's "solution" is to write down several equations of Kirchhoff current and voltage laws to reduce the circuit. Yuck! Here is the diagram, they want to know the equivalent resistance between points A and C.
See Full Post and Comments
----
I just tried to read the same thing six times and still didn't parse it, so instead of going to sleep I'm going to do this instead! Tagged by :iconTechnologic-Skies: here are Teh Rules:
See Full Post and Comments
curl -d @t.json -u un:pw -H "Content-Type: application/json" -X PUT http://dynamobi.cloudant.com/sw/_design/rar
This will take the contents of the file t.json (which is below) and send them off to CouchDB. It will create a new document called "rar" under the "sw" database. Here is t.json:
See Full Post and Comments
There are two obvious, killer problems with that, though. One, it severely hurts poor people. 50% of $1 is a lot more than 50% of $1000 if those two amounts are all two people have, but the cost of a loaf of bread is still 99 cents and so one person has to go hungry. The second problem, is that the government then doesn't have enough income to support itself unless the percentage is so high. (And as you raise the percentage, the poor suffer more and more.)
So the crafters of our tax laws realized these two problems, and they went a step further by introducing tax brackets. You pay 20% on all income under $X, but every dollar you make above $X, you might get taxed 30% until you hit $Y, and so on. Obviously someone with $1000 does not need $1 as much as someone who only has $2 and wants a gallon of milk, so the idea of the tax bracket again makes sense if you want to ease the suffering of the poor.
See Full Post and Comments
"I quarrel not with far-off foes, but with those who, near at home, co-operate with, and do the bidding of, those far away, and without whom the latter would be harmless." ~Thoreau
In the post-war Nazi trials, soldiers would justify their horrible actions as just "following orders." Employees of big corporations do less horrible but still offensive actions and justify them by saying "I'm just doing my job." (Recent case in point: TSA Agent frisking a 6 year old girl.) Through the Milgram experiment, psychology tells us that these people may even have a point. For whatever reasons, we evolved to respond differently to authority than we might under our own direction. Can we really blame these perpetrators for their actions when they're just victims of the same human malfunction we all have? See Full Post and Comments
In the post-war Nazi trials, soldiers would justify their horrible actions as just "following orders." Employees of big corporations do less horrible but still offensive actions and justify them by saying "I'm just doing my job." (Recent case in point: TSA Agent frisking a 6 year old girl.)
Through the Milgram experiment, psychology tells us that these people may even have a point. For whatever reasons, we evolved to respond differently to authority than we might under our own direction. Can we really blame these perpetrators for their actions when they're just victims of the same human malfunction we all have?
See Full Post and Comments
[math]f(x) = e^{-x^2}[/math]
and how it looks really neat when plotted, but in a normal calculus course you'll probably be told that you can't integrate it. We then noted that:
[math]\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}[/math]
See Full Post and Comments
|
Search
Now showing items 1-10 of 19
J/Ψ production and nuclear effects in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-02)
Inclusive J/ψ production has been studied with the ALICE detector in p-Pb collisions at the nucleon–nucleon center of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement is performed in the center of mass rapidity ...
Suppression of ψ(2S) production in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-12)
The ALICE Collaboration has studied the inclusive production of the charmonium state ψ(2S) in proton-lead (p-Pb) collisions at the nucleon-nucleon centre of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement was ...
Event-by-event mean pT fluctuations in pp and Pb–Pb collisions at the LHC
(Springer, 2014-10)
Event-by-event fluctuations of the mean transverse momentum of charged particles produced in pp collisions at s√ = 0.9, 2.76 and 7 TeV, and Pb–Pb collisions at √sNN = 2.76 TeV are studied as a function of the ...
Centrality, rapidity and transverse momentum dependence of the J/$\psi$ suppression in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2014-06)
The inclusive J/$\psi$ nuclear modification factor ($R_{AA}$) in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76TeV has been measured by ALICE as a function of centrality in the $e^+e^-$ decay channel at mid-rapidity |y| < 0.8 ...
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
Multiplicity Dependence of Pion, Kaon, Proton and Lambda Production in p-Pb Collisions at $\sqrt{s_{NN}}$ = 5.02 TeV
(Elsevier, 2014-01)
In this Letter, comprehensive results on $\pi^{\pm}, K^{\pm}, K^0_S$, $p(\bar{p})$ and $\Lambda (\bar{\Lambda})$ production at mid-rapidity (0 < $y_{CMS}$ < 0.5) in p-Pb collisions at $\sqrt{s_{NN}}$ = 5.02 TeV, measured ...
Multi-strange baryon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2014-01)
The production of ${\rm\Xi}^-$ and ${\rm\Omega}^-$ baryons and their anti-particles in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been measured using the ALICE detector. The transverse momentum spectra at ...
Measurement of charged jet suppression in Pb-Pb collisions at √sNN = 2.76 TeV
(Springer, 2014-03)
A measurement of the transverse momentum spectra of jets in Pb–Pb collisions at √sNN = 2.76TeV is reported. Jets are reconstructed from charged particles using the anti-kT jet algorithm with jet resolution parameters R ...
Two- and three-pion quantum statistics correlations in Pb-Pb collisions at √sNN = 2.76 TeV at the CERN Large Hadron Collider
(American Physical Society, 2014-02-26)
Correlations induced by quantum statistics are sensitive to the spatiotemporal extent as well as dynamics of particle-emitting sources in heavy-ion collisions. In addition, such correlations can be used to search for the ...
Exclusive J /ψ photoproduction off protons in ultraperipheral p-Pb collisions at √sNN = 5.02TeV
(American Physical Society, 2014-12-05)
We present the first measurement at the LHC of exclusive J/ψ photoproduction off protons, in ultraperipheral proton-lead collisions at √sNN=5.02 TeV. Events are selected with a dimuon pair produced either in the rapidity ...
|
Being a geek and having gone to high school in Los Alamos, I knew about Stanislaw Ulam. At least, I knew a bit about the work he had done during the Manhattan project at Los Alamos. And that he was among the greats of mathematics. But it seemed totally implausible that the person behind that door could really be Stanislaw Ulam. Ulam was a creature of myth to me. Not just from another generation, but practically from another world. There was no way that it could be him. No way.
So I had to find out.
I knocked.
And was invited in.
So I asked, "Are you Stanislaw Ulam?". He seemed amused. "Yes". "Uh..." I said.
Just as a note to anyone meeting a god who steps down onto the earth to walk among us, I suggest you have a question figured out ahead of time because you aren't going to come up with anything very good in the moment. At least I couldn't. In a similar vein, figure out your three wishes ahead of time as well in case you meet a genie.
After a moment of hurried thought, I came up with "Could we talk?". Seriously. That is the best I could do.
He said that would be fine, but that he had about 20 minutes of work to do first. To keep me busy, he gave me a problem to work on during that time. Not surprisingly, that problem was a really good one for assessing my mathematical sophistication.
That sophistication was not very high (and really still isn't decades later) partly because of youth and partly because of attitude and partly because schools can't deal with students who learn quickly. My history with math was that during an explosive few weeks in the 8th grade I had gotten the bug and worked my way through all of the high school math curriculum including calculus. It was a wonderful time and didn't feel like learning so much as just remembering how things worked. But there wasn't much anything to be done beyond that because there wasn't much anybody to talk to who knew what lay past that because we lived on a military base in Germany. I could wander through books, but I didn't really have enough understanding to go much further than a bit of differential geometry and multivariate calculus. Later, when we moved to Los Alamos, there were all kinds of people I could have learned from but I was engrossed in electronics and computing and German and Russian.
The problem that he gave me was to prove a simple fixed point theorem. Given a set $S$ and a continuous function $f: S \rightarrow S$ such that $f$ is a contraction, show that $f$ has a fixed point in $S$. A contraction is a function that maps distinct points closer together, $x_1 \ne x_2 \implies \rho(x_1, x_2) > \rho(f(x_1), f(x_2))$ where $\rho$ is some metric on $S$.
As with all good math questions, the first response should always be clarifications to make sure that the problem is actually well understood. Even though it has been a very long time, I remember asking if I could assume that $S$ is compact and that $f$ is continuous. As I remember, I didn't actually finish a proof during the short time I had then, but I did sketch out an approach that started by noting that $\rho(x, f(x)) > \rho(f(x), f(f(x)))$. Unfortunately, I took a wrong turn at that point and thought about limits. That apparently didn't matter and I had some great talks with Professor Ulam and his friend Jan Micielski.
One of the cool things about mathematics is that questions like this wind up in the back of your head and can pop out at any time. That just happened as I was reading a review of the latest edition of Proofs From the Book. This looks like a great book with all kinds of interesting insights to be had, but I just looked at their wonderfully short proof of the fundamental theorem of algebra. That proof goes something like this
Every real-valued function $f$ over a compact set $S$ has a minimum in $S$ Every point $x$ such that a complex polynomial is non-zero, $|p(x)|>0$, has a nearby point $x'$ such that $|p(x')| < |p(x)|$ Thus, the minimum of $|p(x)|$ exists, but all points $x$ such that $|p(x)|>0$ are not the minimum. Therefore, for some point $x_0$ we have $|p(x_0)|=0$
Point 2 takes a tiny bit of algebra, but overall this proof is stunningly simple and beautiful. But in the way that these things happen, after reading this proof I remembered the contraction theorem that I hadn't quite come up with so long ago and it occurred to me that this same trick would apply there.
The outline is
For any fixed point $x^*$, we have $\rho(x^*,f(x^*)) = 0$ For any other point $x$ we have $\rho(x, f(x)) > 0$ The function $d(x) = \rho(x, f(x))$ is continuous and has a minimum in $S$ For any point $x$ such that $d(x) \ne 0$, we have $d(x) > d(f(x))$ and thus $d(x)$ is not minimized at $x$ Thus, there exists some point $x^*$ where $d(x^*)=0$, that is to say $x^*$ is a fixed point.
Isn't it just the way of the world that you think of the perfect answer just a little bit too late?
|
Let's assume we restrict consideration to symmetric distributions where the mean and variance are finite (so the Cauchy, for example, is excluded from consideration).
Further, I'm going to limit myself initially to continuous unimodal cases, and indeed mostly to 'nice' situations (though I might come back later and discuss some other cases).
The relative variance depends on sample size. It's common to discuss the ratio of ($n$ times the) the asymptotic variances, but we should keep in mind that at smaller sample sizes the situation will be somewhat different. (The median sometimes does noticeably better or worse than its asymptotic behaviour would suggest. For example, at the normal with $n=3$ it has an efficiency of about 74% rather than 63%. The asymptotic behavior is generally a good guide at quite moderate sample sizes, though.)
The asymptotics are fairly easy to deal with:
Mean: $n\times$ variance = $\sigma^2$.
Median: $n\times$ variance = $\frac{1}{[4f(m)^2]}$ where $f(m)$ is the height of the density at the median.
So if $f(m)>\frac{1}{2\sigma}$, the median will be asymptotically more efficient.
[In the normal case, $f(m)=\frac{1}{\sqrt{2\pi}\sigma}$, so $\frac{1}{[4f(m)^2]}=\frac{\pi\sigma^2}{2}$, whence the asymptotic relative efficiency of $2/\pi$)]
We can see that the variance of the median will depend on the behaviour of the density very near the center, while the variance of the mean depends on the variance of the original distribution (which in some sense is affected by the density everywhere, and in particular, more by the way it behaves further away from the center)
Which is to say, while the median is less affected by outliers than the mean, and we often see that it has lower variance than the mean when the distribution is heavy tailed (which does produce more outliers), what really drives the performance of the median is
inliers. It often happens that (for a fixed variance) there's a tendency for the two to go together.
That is, broadly speaking, as the tail gets heavier, there's a tendency for (at a fixed value of $\sigma^2$) the distribution to get "peakier" at the same time (more kurtotic, in a loose sense). This is not, however, a certain thing - it tends to be the case across a broad range of commonly considered densities, but it doesn't always hold. When it does hold, the variance of the median will reduce (because the distribution has more probability in the immediate neighborhood of the median), while the variance of the mean is held constant (because we fixed $\sigma^2$).
So across a variety of common cases the median will often tend to do "better" than the mean when the tail is heavy, (but we must keep in mind that it's relatively easy to construct counterexamples). So we can consider a few cases, which can show us what we often see, but we shouldn't read too much into them, because heavier tail doesn't universally go with higher peak.
We know the median is about 63.7% as efficient (for $n$ large) as the mean at the normal.
What about, say a logistic distribution, which like the normal is approximately parabolic about the center, but has heavier tails (as $x$ becomes large, they become exponential).
If we take the scale parameter to be 1, the logistic has variance $\pi^2/3$ and height at the median of 1/4, so $\frac{1}{4f(m)^2}=4$. The ratio of variances is then $\pi^2/12\approx 0.82$ so in large samples, the median is roughly 82% as efficient as the mean.
Let's consider two other densities with exponential-like tails, but different peakedness.
First, the hyperbolic secant ($\text{sech}$) distribution, for which the standard form has variance 1 and height at the center of $\frac{1}{2}$, so the ratio of asymptotic variances is 1 (the two are equally efficient in large samples). However, in small samples the mean is more efficient (its variance is about 95% of that for the median when $n=5$, for example).
Here we can see how, as we progress through those three densities (holding variance constant), that the height at the median increases:
Can we make it go still higher? Indeed we can. Consider, for example, the double exponential. The standard form has variance 2, and the height at the median is $\frac{1}{2}$ (so if we scale to unit variance as in the diagram, the peak is at $\frac{1}{\sqrt{2}}$, just above 0.7). The asymptotic variance of the median is half that of the mean.
If we make the distribution peakier still for a given variance, (perhaps by making the tail heavier than exponential), the median can be far more efficient (relatively speaking) still. There's really no limit to how high that peak can go.
If we had instead used examples from say the t-distributions, broadly similar effects would be seen, but the progression would be different; the crossover point is a little below $\nu=5$ df (actually around 4.68) -- for smaller df the median is more efficient, for large df the mean is.
...
At finite sample sizes, it's sometimes possible to compute the variance of the distribution of the median explicitly. Where that's not feasible - or even just inconvenient - we can use simulation to compute the variance of the median (or the ratio of the variance*) across random samples drawn from the distribution (which is what I did to get the small sample figures above).
* Even though we often don't actually need the variance of the mean, since we can compute it if we know the variance of the distribution, it may be more computationally efficient to do so, since it acts like a control variate (the mean and median are often quite correlated).
|
my new favorite equation a couple months ago, but I neglected to mention an insight that has been revealed to me!
It's Tau Day today! Long live Tau! What does this have to do with my equation? Well, my equation was: [math]\int_{-\infty}^{\infty} e^{-x^2} = \sqrt{\pi}[/math] See Full Post and Comments
It's Tau Day today! Long live Tau! What does this have to do with my equation? Well, my equation was:
[math]\int_{-\infty}^{\infty} e^{-x^2} = \sqrt{\pi}[/math]
See Full Post and Comments
crab canon appeared on Hacker News recently, I thought it was really pretty. I started writing my own (I'll post it whenever I finish if I think it's good enough), but noticed that it was fairly hard to do in my head. The amount of context you have to keep track of... Anyway, I took 5-10 minutes or so to write a Canon Maker in Flex that just copies what you wrote and duplicates it to mirror itself.
Source for anyone interested (also below) See Full Post and Comments
Source for anyone interested (also below)
See Full Post and Comments
|
Equivalence of Definitions of Arborescence Contents Theorem
Let $G = \struct {V, A}$ be a directed graph.
Let $r \in V$.
$G$ is an
arborescence of root $r$ if and only if: For each $v \in V$ there is exactly one directed walk from $r$ to $v$.
$G$ is an
arborescence of root $r$ if and only if: $(1): \quad$ $G$ is an orientation of a tree $(2): \quad$ For each $v \in V$, $v$ is reachable from $r$.
$G$ is an
arborescence of root $r$ if and only if: $(1): \quad$ Each vertex $v \ne r$ is the final vertex of exactly one arc. $(2): \quad$ $r$ is not the final vertex of any arc. $(3): \quad$ For each $v \in V$ such that $v \ne r$ there is a directed walk from $r$ to $v$. Proof Definition 1 implies Definition 3
Let $G$ be an $r$-arborescence by definition 1.
Let $v \in V$ such that $v \ne r$.
Then there is exactly one directed walk $w$ from $r$ to $v$.
Since $v \ne r$, either:
$w = \tuple {r, v}$
or:
$\exists m \in V: w = \tuple {r, \ldots, m, v}$
Then there exist directed walks $w_1$ and $w_2$ from $r$ to $x$ and $r$ to $y$ respectively.
$\Box$
But then $w$ appended to $w$ is a directed walk from $r$ to $x$ which is not equal to $w$.
Thus $G$ is an $r$-arborescence by Definition 3.
$\Box$
Definition 3 implies Definition 1
Suppose that $G$ is an $r$-arborescence by Definition 3.
Let $v \in V$.
If $v = r$, then $\tuple r$ is a directed walk from $r$ to $v$.
If $v \ne r$, then there exists some directed walk $w$ from $r$ to $v$.
Suppose that $z$ is a directed walk from $r$ to $v$.
If $x = r$, then $z$ must end with $\tuple {r, v}$.
So in fact:
$z = \tuple {r, v}$
If $x \ne r$, then:
$z = \tuple {r, \ldots, x, v}$
Continuing inductively from $x$ proves that $z = w$.
So $G$ is an $r$-arborescence by Definition 1.
$\Box$
Definitions 1 and 3 imply Definition 2
Let $G$ be an $r$-arborescence by Definition 1.
From the above, $G$ is then also an $r$-arborescence by Definition 3.
Let $T = \struct {V, E}$ be the simple graph corresponding to $G$.
That is, for $x, y \in V$, let $\set {x, y} \in E$ if and only if $\tuple {x, y} \in A$ or $\tuple {y, x} \in A$.
Let $x, y \in V$ and suppose for the sake of contradiction that $\tuple {x, y} \in A$ and $\tuple {y, x} \in A$.
Let $w_x$ be the directed walk from $r$ to $x$.
Then appending $y$ to $w_x$ yields a directed walk $w_x + y$ from $r$ to $y$.
But then appending $x$ to $w_x + y$ yields a directed walk $w_x + y + x$ from $r$ to $x$, contradicting the fact that $w_x$ is unique.
Thus $A$ is asymmetric.
So $G$ is an orientation of $T$.
We must now show that $T$ is a tree.
Let $x, y \in V$.
Reversing $w_x$ and then appending $w_y$ to it (eliding the duplicate $r$) yields a walk from $x$ to $y$.
Thus $T$ is connected.
Thus $T$ has no cycles.
$\Box$
Definition 2 implies Definition 1
Let $G$ be an $r$-arborescence by Definition 2.
Let $v \in V$.
Then either $z$ extends $w$, $w$ extends $z$, or there is some $k$ such that $w_k \ne z_k$.
First suppose that $z$ extends $w$.
So $w = \tuple {x_0, \ldots, x_n}$ and $z = \tuple {x_0, \ldots, x_n, \ldots, x_m}$.
Thus there exists $k$ such that $w_k \ne z_k$.
Since $w$ and $z$ both end at $v$, it follows from the Well-Ordering Principle that:
there must be some smallest $n > k$ such that there exists $m > k$ such that $w_n = z_m$ and that there exists a smallest such $m$. $\blacksquare$
|
Newform invariants
Coefficients of the \(q\)-expansion are expressed in terms of a basis \(1,\beta_1,\beta_2\) for the coefficient ring described below. We also show the integral \(q\)-expansion of the trace form.
Basis of coefficient ring in terms of \(\nu = \zeta_{14} + \zeta_{14}^{-1}\):
\(\beta_{0}\) \(=\) \( 1 \) \(\beta_{1}\) \(=\) \( \nu \) \(\beta_{2}\) \(=\) \( \nu^{2} - 2 \)
\(1\) \(=\) \(\beta_0\) \(\nu\) \(=\) \(\beta_{1}\) \(\nu^{2}\) \(=\) \(\beta_{2}\mathstrut +\mathstrut \) \(2\) Character Values
We give the values of \(\chi\) on generators for \(\left(\mathbb{Z}/2008\mathbb{Z}\right)^\times\).
\(n\) \(257\) \(503\) \(1005\) \(\chi(n)\) \(-1\) \(1\) \(-1\)
For each embedding \(\iota_m\) of the coefficient field, the values \(\iota_m(a_n)\) are shown below.
For more information on an embedded modular form you can click on its label.
This newform can be constructed as the kernel of the linear operator \(T_{11}^{3} \) \(\mathstrut -\mathstrut T_{11}^{2} \) \(\mathstrut -\mathstrut 2 T_{11} \) \(\mathstrut +\mathstrut 1 \) acting on \(S_{1}^{\mathrm{new}}(2008, [\chi])\).
|
A fair die is thrown 12,000 times. Use the central limit theorem to find values of $a$ and $b$ such that $$ \mathbb P(1900<S<2200)\approx\int_a^b\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\,\mathrm dx, $$ where $S$ is the total number of sixes thrown.
Right, so the central limit theorem goes as follows:
Let $X_1,X_2,\dots$ be independent and identically distributed random variables, each with mean $\mu$ and non-zero variance $\sigma^2$. The standardised version $$ Z_n=\frac{S_n-n\mu}{\sigma\sqrt{n}} $$ of the sum $S_n=X_1+\dots+X_n$ satisfies, as $n\to\infty$ $$ \mathbb P(Z_n\leq x)\to\int_{-\infty}^x\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}u^2}\,\mathrm du\quad\text{for }x\in\mathbb R. $$
Okay, so it's obvious that I need to express $Z_n$ in terms of known quantities. For $S<2200$, we basically want $$ Z_n<\frac{2200-12,000\cdot12,000\cdot\frac{1}{6}}{\sqrt{12,000\cdot\frac{1}{6}\cdot\frac{5}{6}\cdot12,000}}, $$ where I used $\mu=np$ and $\sigma^2=np(1-p)$, where $p=\frac{1}{6}$.
This approach gives me a numerical value of -5366.07.
I'm doing something wrong here. Could someone point out my mistake?
Edit;
OK, I got the solution.
|
After looking around for a while, having checked the Not-So-Short Introduction to LaTeX2e (2.5.5) and even after having found some sources on the internet, I still could not get this to work. I have been fighting with the problem for a while and it almost seems futile to continue. It must work though, as I have serious need of it. I would hope that a great relevant (and easy) answer would come that will end the issue once and for all.
English and Greek:
In a document wherein English is the main language, I wish to occasionally use Greek
without resorting to commands such as
$\gamma$ and
$\acute{omikron}$ and the like. Even if occasional, spelling a whole word in Greek in this manner is rather time-consuming.
I normally use
\usepackage[utf8]{inputenc}. I thought this would allow me to easily input Greek as easily as one would type or paste English. Then I learned one must also have a fitting font for Greek. How must one go about achieving the desired result?
English, Greek and Hebrew:
Once that is done, how does one go about adding Hebrew to it? I would want to type or paste it as easily as one would type or paste any language into a file.
Since I have lost too much time on it already, what would be the right template to have the whole matter taken care of?
Thank you.
EDIT (adding a small test file):
Below is the beginning of the file:
\documentclass[12pt, a4paper, titlepage]{article}\usepackage[utf8]{inputenc}\usepackage[hidelinks]{hyperref}\usepackage{amssymb}\usepackage{amsmath}\usepackage{verbatim}\usepackage[english]{babel}\usepackage{afterpage}\usepackage[top=1.00in, bottom=1.00in, left=1.00in, right=1.00in]{geometry}\usepackage{setspace}% for Greek and such\usepackage{fontspec}\usepackage[T1]{fontenc}\newfontfamily\greekfont[Script=Greek, Scale=MatchUppercase, Ligatures=TeX]{linux libertine o}\newcommand{\textgreek}[1]{\bgroup\greekfont\emph{#1}\egroup}% fancyhdr\usepackage{fancyhdr}\pagestyle{fancy}% \newcommand\blankpage{% \null \thispagestyle{empty}% \addtocounter{page}{-1}% \newpage}%;\begin{document}\tableofcontents\thispagestyle{empty}\newpage\section{Intro}This is a word in Greek: νους.\end{document}
I installed the font Linux Libertine O.
EDIT 2 (15-11-'16):
@Ulrike Fischer Thank you. Ever since I registered, I cannot add a comment directly under your response, so I must respond here.
I report that your first code block for use with pdflatex works, and that the second for use with xelatex or lualatex does not. When trying to compile, the earlier problem returns: it does not compile and seems to show no message or progress at all.
Regarding the first code block for use with pdflatex: if you were to enhance it to also include Hebrew, what would it look like?
|
[Note: if you are using smartphone or portable device to browser this post, some math formula might not appear properly. To see the math in correct form, scroll down to the bottom and click " View web version"]
After finishing the mini CNC laser engraver, the next thing in my mind is 3D printer. One of the most important part of a 3D printer is the plastic filament extruder, composed by a cold end constantly supplying plastic filament, and a hot end melting the plastic and feeding the liquid plastic through a small nozzle.
For PLA or ABS plastic filaments (most common materials used in home-made 3D printers), the hot end needs to be at a temperature of about 200C (or 180C for PLA, 220C for ABS). It is important to keep this temperature (roughly) constant, so is the plastic melting speed and then a controllable plastic feeding rate.
Therefore two components are essential: a heater, and a temperature sensor. The heater heats up the hot end nozzle, and the sensor monitors the temperature. The heater is on when the nozzle temperature is too low, and is switched off when the nozzle is too hot.
I plan to build a home-made 3D printer controlled by my Raspberry Pi, which, unfortunately, does not have any analog data acquisition pin. Therefore I decide to build a stand-alone temperature control system.
The idea is very simple. I use a power resistor as the heater and a thermistor as the temperature sensor. The system contains an LM324 quad op-amps chip. One op-amp is used as a comparator to compare the thermistor resistance with a nominal resistance and output LOW or HIGH as the comparison result. The other three op-amps inside the LM324 are used to perform some linear transformation and output a voltage that is proportional to the thermistor temperature. This voltage is applied to a 0-30V voltmeter so one can read the temperature. A N-Channel MOSFET transistor is used to control on/off of the heater.
This is the controller
The finished controller box. About 8cm by 5cm by 4cm
Here is the circuit diagram and some key parts used in the system.
Schematics of the temperature control system
LM324 is 14-pin integrated circuit (IC) containing four identical low-profile low-power operational amplifiers (op-amp). The op-amps have a bandwidth of 1MHz, and can be powered by a single power supply with a wide range 3-32V. It is one of the most widely used and lowest cost op-amp ICs. The datasheet can be found here [Note: there is nothing too special about this chip except its compact size. It can be certainly replaced by essentially any other type of op-amps. For the task carried by pin 1, 2 and 3 of LM324 shown in the schematics above, a comparator, instead of an op-amp, is more suitable than LM324.]
12V 40W ceramic cartridge heater resistor I used. It is the R_hot in the schematics. They are sold on ebay an other place at a price of about $2/each. For example, here
100k ohm NTC 3950 1% thermistor with glass enclosure. It is the R_T in the schematics. They have a wide temperature range from -50C to 260C. Perfect for 3D print (200C). The They cost about $1/each. E.g., here
0-30V digital voltmeter. The red lead is for power, black lead for ground, and the blue lead goes to the voltage to be measured. They are about $1.5 each (like this one). Make sure the meter has three wires. There are meters with only two wires (no blue wire) which are not suitable here.
IRFP150 N Channel MOSFET transistor. About $1.5/each (like this one). From left to right are the pins of G, D and S. Data sheet can be downloaded here. It can be replaced by similar high power N MOSFET transistors like IRFP140 or IRFZ24 Choice of the resistorsThe choice for the resistors R0, R1, R2, R3 and R4 are crucial. They need to be carefully calculated so that the meter can show correct temperature.
I choose:
R0=600 ohm
R1=10k ohm
R2=3.9k ohm
R3=6.19k ohm
R4=2.4k ohm
R5=10k ohm
I will show how these numbers come out.
About the thermistorThe NTC thermistor I use has a beta value of 3950 and resistance of 100k ohm at room temperature (~25C, or rounghly 300Kelvin).
NTC stands for Negative Temperature Coefficient, meaning that the resistance of the thermistor decreases as the temperature increases.
The Steinhart Hart equation for NTC thermistor is
$$\frac{1}{T}=\frac{1}{T_0}+\frac{1}{B}ln\left(\frac{R}{R_0}\right)$$
where \(T_0\) is room temperature in Kelvin (~300), B=3950 is the coefficient, \(R_0\) is the resistance of the thermistor in ohm at room temperature, i.e., 100000. Then given temperature \(T\) in Kelvin we can calculate the resistance of the thermistor, or given a resistance we can calculate the temperature.
The S-H equation is a highly non-linear equation. It is, however, possible to use a simple linear function to approximate the S-H equation within some narrow temperature range.
How the temperature meter worksThe system has two modes: normal working mode, and temperature setting mode. The switch SW controls which mode is selected. When SW is at 1 (default), the system is at normal working mode. When SW is at 2 (temporal), the system allows you to set the desired temperature.
The LM324 has four op-amps.Pin 4 and 11 are the Vcc and Vee. I use a single 12V power supply to power the whole system so pin 4 is connected to +12V and 11 is connected to ground. Pin [1,2,3], [5,6,7],[8,9,10] and [12,13,14] are the four op-amps.
Op-amp [1,2,3] is used as a voltage comparator. I will discuss this later.
(It is usually not recommended to use an op-amp as a comparator. But it does work here because LM324 accepts true differential input signals into its inverting and non-inverting inputs).
Op-amp [5,6,7] behaves simply as a voltage follower. Pin 6 and 7 repeats whatever voltage Pin 5 has without draining current from pin 5. In normal working mode (SW at 1), the voltage of Pin 5 is the voltage between the thermistor and R0. We define this voltage, the voltage of the junction of thermistor R_T and R0, as \(Vin\).
With R0, there is
$$Vin=\frac{R0}{R0+R_T}12 V=Pin5=Pin6=Pin7$$
Then let's move to resistor R1 and R2. Clearly, the voltage at pin 10 is
$$ \frac{R2}{R1+R2}\cdot12+\frac{R1}{R1+R2}\cdot Vin=\frac{R1}{R1+R2}\left(\frac{R2}{R1}\cdot12+Vin\right) $$
Op-amp [8, 9, 10] is another voltage follower. Hence Voltage of Pin 8 = Pin 9 = Pin 10.
Resistor R3 and R4 form a voltage divider. The voltage between R3 and R4 is
$$\frac{R4}{R3+R4}\cdot Pin9=\frac{R4}{R3+R4}\frac{R1}{R1+R2}\left(\frac{R2}{R1}\cdot 12+Vin\right)$$
Op-amp [12,13,14] is another follower. There is Pin 14=Pin 13 = Pin 12= voltage between R3 and R4.
We define the voltage at Pin 14 is \(Vout\). Hence
$$Vout=\frac{R4}{R3+R4}\frac{R1}{R1+R2}\left(\frac{R2}{R1}\cdot 12 + Vin \right) $$
Now plug into the numbers: R1=10K ohm, R2=3.9k ohm, R3=6.19k ohm, R4=2.4k ohm, we get
$$Vout=0.279\times 0.719\times (0.39\times 12+Vin)=0.2(4.7+Vin) $$
All the three op-amps do is to perform this linear scaling.
Here is the
magic. I claim that, with R0=600ohm and \(Vin=600/(600+R_T)\times12\)V, there is $$Vout=T/100$$ where T is the temperature of the thermistor in Celsius. This equation is true for a quite wide temperature range 120C-260C.
The figure below is the proof. It is shown that with R0=600ohm, \(Vin\) is quite linear in the range of 120C to 260C. Therefore we can perform some linear transformation to map \(Vin\) linearly to T/100.
To map Vin to T/100, we perform linear regression between Vin and T/100, and obtain
$$\frac{T}{100}=0.2Vin+0.94=0.2(Vin+4.7)$$
Recall that
$$Vout=\frac{R4}{R3+R4}\frac{R1}{R1+R2}\left(\frac{R2}{R1}\cdot 12 + Vin \right) $$
If we want to have \(T/100=Vout\), there must be
$$\frac{R2}{R1}=4.7/12=0.39$$
and
$$\frac{R4}{R3+R4}\frac{R1}{R1+R2}=0.2$$
Since \(R2/R1=0.39\) gives \(R1/(R1+R2)=0.719\), the second requirement is reduced to
$$\frac{R4}{R3+R4}=0.2/0.719=0.278$$
Therefore [R1=10K ohm, R2=3.9k ohm, R3=6.19k ohm, R4=2.4k ohm] is an obvious solution (not the only one though).
Here is a list for key parameters at different temperatures
T(C) R_T(k ohm) Vin (V) Vout (V)
120 4.0594 1.5453 1.2491
140 2.4950 2.3263 1.4053
160 1.6041 3.2667 1.5933
180 1.0723 4.3055 1.8011
200 0.7416 5.3666 2.0133
220 0.5285 6.3801 2.2160
240 0.3867 7.2969 2.3994
260 0.2897 8.0928 2.5586
Note that the \(Vout=T/100\) relation is preserved very good within the range of 140C to 240C. Detailed calculation shows that the averaged errors from 140C to 240C is less than 0.5%. Very good!
We feed this Vout to a voltmeter so the voltmeter reading is T/100.
How the temperature control worksThe temperature control is done by the op-amp [1,2,3]. Toggle the SW to position 2. The meter is reading a temperature assuming the potentiometer R is R_T and R0. We set the potentiometer R so that the meter reads the desired value, say 2V (corresponding to 200C). Then toggle the SW back to position 1.
The voltage of Pin 3 is then a constant. The voltage of Pin 2 equals to Pin 3 when and only when the temperature is 200C.
Below 200C, R_T is too large. So the Pin 2<Pin 3. The op-amp is trying to amplify the voltage difference between its non-inverting (pin 3) and inverting pins (pin 2) by hundreds of thousands times. But it can't output any voltage above the power supply voltage-1.5V Therefore it outputs 12-1.5=10.5V through pin 1. The Gate-Source voltage of the MOSFET transistor is then large enough and the D-S of the MOSFET become conductive. The heater is then switched on.
When the temperature hits above 200C, R_T is too small. So the Pin 2>Pin 3. The op-amp outputs zero at Pin 1. The MOSFET is switched off and so is the heater.
After the temperature drops below 200C again, the cycle repeats.
So it is a negative feedback system. The temperature is maintained around 200C.
Note that when Pin 1 outputs 10.5V, the LED is also switched on to indicate that the heater is working.
Some remarkThe system accomplishes two major tasks: displaying the temperature correctly and maintaining a constant temperature at the location of thermistor.
The first task is accomplished by using three op-amps. The limitation is that the system can only display temperature correctly within a certain range. Here, in our case, is 120C-260C. Below 120C or above 260C the system behaves very poor.
Mathematically it is guaranteed that the op-amps idea will definitely work essentially for any system, but within some limited range. This is because in most times the Taylor-expansion of any smooth function \(f(x)=f(x_0)+(x-x_0)f'(x_0)+...\) always has a linear term \((x-x_0)f'(x_0)\). One can always use a linear function to approximate the real function. On the other hand, this approximation only works within a narrow range.
I feel very lucky that the linear range in my case is so wide (120C-260C).
To properly display the temperature in wider range, say, 0C to 300C, it is necessary to use some micro controller to perform the complicated T-R conversion calculation. But micro controller is is far more complicated and expensive than LM324.
My idea can be easily extended to other system. For example, a constant temperature water-filling container (what does it for? Say, 45C Baby milk warmer). One need to linearize the system around 45C and find proper R0-R4.
A note is that LM324 cannot output voltage below 0 or above power supply voltage-1.5V. So every step of the linearization shouldn't involve any voltage beyond this range.
|
According to the IUPAC Goldbook a
solution is:
A liquid or solid phase containing more than one substance, when for convenience one (or more) substance, which is called the solvent, is treated differently from the other substances, which are called solutes. When, as is often but not necessarily the case, the sum of the mole fractions of solutes is small compared with unity, the solution is called a dilute solution. A superscript attached to the ∞ symbol for a property of a solution denotes the property in the limit of infinite dilution.
There is no mentioning of that the solute may not react with the solvent. If you investigate a few common solutions, you will also find that it is indeed quite common, that the solvent reacts with the solution. Take for example the dissolution of hydrochloride in water, which is often expressed with the following equation:$$\ce{HCl (g) + H2O (l) <=> Cl- (aq) + H3O+ (aq)}$$You can see here, that the proton is transferred onto a water molecule and from one point of view this is a displacement reaction; the chloride is displaced by water.
Another example is the dissolution of acetic acid. It is often described with the following equation:$$\ce{H3C-COOH (s) + H2O (l) <=> H3C-COO- (aq) + H3O+ (aq)}$$The same happens here. We hide the fact that the dissolution process includes getting the solvent into a highly flexible network of hydrogen bonds in the addendum of $\ce{(aq)}$.
When you look at common salts that get dissolved, it is often that the metal ion reacts as a weak acid, co-coordinating the solvent molecules. In many cases this is so flexible, like for the alkaline metals, that we cannot describe even the first solvation sphere approximately.
For other salts this is quite possible. Take for example the dissolution of copper sulfate. We often write for short$$\ce{CuSO4 (s) <=> Cu^2+ (aq) + SO4^2- (aq)}$$when we actually mean the formation of a complex$$\ce{CuSO4.5H2O (s) + H2O (l) <=> [Cu(H2O)6]^2+ (aq) + SO4^2- (aq)}.$$A chemical reaction actually takes place. And since everything is in equilibrium, there are many chemical reactions happening in solutions all the time.
Now the dissolution of bromine in water is no different. In fact the disproportionation $$\ce{Br2 (l) + 2 H2O <=> BrOH (aq) + Br- (aq) + H3O+ (aq)}$$is quite necessary to dissolve bromine in the first place. While I do not even slightly agree to the causality of having a dipole means being a polar molecule, I can still see, that the bromine molecule is less polar than water. It is however quite well polarizable; it has a quadrupole moment.
[1] This is what also leads to the above equation. The bromine hydrolysis has been investigated a couple of times.$$\begin{align}\ce{Br2 + H2O &<=>[K_1] HBrO + H+ + Br- }&K_1 &= \ce{\frac{[HBrO][H+][Br- ]}{[Br2]}}\end{align}$$The temperature dependency of $K_1$ has been studied by Liebhafsky as early as 1934. [2] In their studies they also considered the effects of the tribromide equilibrium $$\begin{align}\ce{Br3^- &<=>[K_3] Br- + Br2}K_3 &= \ce{\frac{[Br- ][Br2]}{[Br3^- ]}}.\end{align}$$And later continue to state that the existence of the pentabromide ion is highly probable.$$\begin{align}\ce{Br5^- &<=>[K_5] Br- + 2 Br2}K_5 &= \ce{\frac{[Br- ][Br2]^2}{[Br5^- ]}}\end{align}$$From the equilibrium constants you can see, that the main dissolved species is still $\ce{Br2}$.
$$\begin{array}{lrrrrr}T/~\mathrm{^\circ C} & 0 & 10 & 25 & 30 & 35 \\\hlineK_1\times10^{-9} & 0.70 & 1.78 & 5.8 & 8.3 & 11.3\\\end{array}$$
Conclusion Solutions do not exist without chemical reactions. There are often multiple different species in solution, that have very low concentration, so for various purposes a valid approximation is to neglect those species' concentrations. Jacek Bieroń, Pekka Pyykkö, Dage Sundholm, Vladimir Kellö, and Andrzej J. Sadlej; Phys. Rev. A 2001, 64, 052507. Herman A. Liebhafsky; J. Am. Chem. Soc. 1934, 56 (7), 1500–1505.
|
When answering such problems, I recommend writing everything know down, then use the appropriate formula to solve for the unknown. For the first part you already did that correctly.
In the laboratory you dissolve $\pu{24.7 g}$ of iron(III) chloride in a volumetric flask in water to a total volume of $\pu{375 ml}$.
\begin{align} m(\ce{FeCl3 (s)}) &= \pu{24.7 g}\\ M(\ce{FeCl3}) &= \pu{162.3 g//mol}\\ V(\ce{FeCl3 (aq)}) &= \pu{375 mL}\\\end{align}
What is the molarity of the solution?
Molarity is another word for amount concentration, so you are looking for $c(\ce{FeCl3 (aq)})$.\begin{align} c &= \frac{n}{V}\\ n &= \frac{m}{M}\\ c(\ce{FeCl3 (aq)}) &= \frac{m(\ce{FeCl3 (s)})}{M(\ce{FeCl3})\cdot V(\ce{FeCl3 (aq)})}\\ c(\ce{FeCl3 (aq)}) &= \frac{\pu{24.7 g}}{\pu{162.3 g//mol}\cdot\pu{375E-3L}}\\ c(\ce{FeCl3 (aq)}) &= \pu{0.406 mol//L} \end{align}
What is the concentration of iron(III) cation in M?
And what is the concentration of chloride anion in M?
You are looking for the ion (amount) concentrations in your solution, i.e. $c(\ce{Fe^3+ (aq)})$ and $c(\ce{Cl- (aq)})$.For these two questions you need to know how the substance dissolves. The wording here is already very explicit, as it says iron(III) cation, i.e. $\ce{Fe^3+ (aq)}$, and chloride anion, i.e. $\ce{Cl-}$. Otherwise, with some more experience you will know that this compound is a salt and it dissociates into ions when dissolved in polar solvents like water. Therefore the reaction equation is$$\ce{FeCl3 (s) ->[H2O] Fe^3+ (aq) + 3 Cl- (aq)}.$$From this you can derive the actual relationships between the ion concentrations.You can clearly see that there is one iron(III) ion and three chloride ions per formula of $\ce{FeCl3}$; therefore\begin{align} c(\ce{Fe^3+ (aq)})&= c(\ce{FeCl3 (aq)}) &&=\pu{0.406 mol//L},\\ c(\ce{Cl- (aq)}) &= 3\times c(\ce{FeCl3 (aq)}) &&=\pu{1.22 mol//L}.\end{align}
The other equation would probably be appropriate when you (thermally) decompose the substance (in absence of everything) as it is the reverse equation which can be used to prepare the molecule:$$\ce{2 Fe (s) + 3 Cl2 (g) -> 2 FeCl3 (s)}.$$
|
"Super-exponential" just means more than exponential, so a function is super-exponential if it grows faster than any exponential function. More formally, this means that it is $\omega(c^n)$ for every constant $c$, i.e., if $\lim_{n\to\infty} f(n)/c(n)=\infty$ for all constants $c$.
Conversely, a function is "sub-exponential" if it is $o(c^n)$ for every constant $c>1$, i.e., $\lim_{n\to\infty} f(n)/c(n)=\infty$ for all constants $c>1$.
Asymptotically, the $n$th Catalan number is $\Theta(4^n\, n^{-3/2})$. This is $o(4^n)$, so the Catalan numbers are not super-exponential; it is $\omega(2^n)$, so they're not subexponential either. The Catalan numbers are just exponential.
An exception to the above definitions is that, in some contexts, functions of the form $b^{n^k}$ for constants $b,k>1$ are considered to be exponential, even though$$\lim_{n\to\infty}\frac{b^{n^k}}{c^n}=\lim_{n\to\infty}b^{n^k-n\log_b c}=\infty\,.$$For example, the complexity class
EXP is defined as the class of languages decided by Turing machines running in time $O(2^{n^k})$ for any $k$. Thanks to Yuval Filmus for pointing this out.
|
The distance between the center of the circle to the touching point should be equal to 4 and perpendicular to it.
The equation of the perpendicular line will be:
\begin{equation}g(x)=-1x+b\end{equation}\begin{equation}g(1)=-1*1+b\end{equation}\begin{equation}2=-1*1+b\end{equation}\begin{equation}b=2+1\end{equation}\begin{equation}b=3\end{equation} \begin{equation}g(x)=-x+3\end{equation}
The distance between the center of the circle and the point (1,2) is exactly 4 (the radius).Using the formula of distance:
$$ (1-x_c)^2+(2-y_c)^2=4^2 $$
Where $ x_c $ and $ y_c$ are the coordinates of the center of the circle. Since the slope of the perpendicular line is -1 it means that the ratio between $\bigtriangleup x$ and $\bigtriangleup y$ is 1. Thus $1-x_c=2-y_c$. Let $a$ be $1-x_c and 2-y_c$, then$$ a^2+a^2=4^2 $$$$ 2a^2=4^2 $$$$ a^2=8 $$$$ a=\pm2\surd2$$
Then $1-x_c=\pm2\surd2$ and $2-y_c=\pm2\surd2$.$$ 1-x_c=\pm2\surd2 $$ $$ x_c=1\pm2\surd2 $$ Same with $2-y_c=\pm2\surd2$. $$ y_c=2\pm2\surd2 $$
The final equations are:$$Circle1: (x-(1+2\surd2))^2+(y-(2-2\surd2))^2=4^2$$$$Circle2: (x-(1-2\surd2))^2+(y-(2+2\surd2))^2=4^2$$
|
Possible Duplicate: Evaluating $\\int P(\\sin x, \\cos x) \\text{d}x$
Hi,
My question is: How can I solve the following integral question? $$\int(\sin ^4 x ) dx$$
Thanks in advance.
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
Possible Duplicate: Evaluating $\\int P(\\sin x, \\cos x) \\text{d}x$
Hi,
My question is: How can I solve the following integral question? $$\int(\sin ^4 x ) dx$$
Thanks in advance.
There are standard techniques for solving integrals that consist of powers of sines and cosines.
One is to use reduction formulas when you simply have a power of sines or a power of cosines. These can be obtained by performing integration by parts, followed by using a trigonometric identity. This is done in pretty much every single calculus textbook I have ever encountered.
Another is to use trigonometric power reduction formulas to change the fourth power of the sine into a sum of multiples of simple cosines, which can then be solved with an easy substitution.
Sometimes handling trigonometric functions is easier in complex exponential form:
$$\sin ^4(x)$$ $$\frac{1}{16} \left(e^{-i x}-e^{i x}\right)^4$$ $$-\frac{1}{4} e^{-2 i x}-\frac{1}{4} e^{2 i x}+\frac{1}{16} e^{-4 i x}+\frac{1}{16} e^{4 i x}+\frac{3}{8}$$ $$-\frac{1}{2} \cos (2 x)+\frac{1}{8} \cos (4 x)+\frac{3}{8}$$
Note that $$\int \sin^{n} \ dx = -\frac{\cos(x) \sin^{n-1}(x)}{n}+ \frac{n-1}{n} \int \sin^{n-2}(x) \ dx$$ and $$\sin^{2}(x) = \frac{1}{2}-\frac{1}{2} \cos(2x)$$
This can be viewed as almost the same thing as exponential form, only in this case I assume that you remember multiple angles formula. (I.e. you do not need to use complex numbers.) $\cos 4x=\cos^4 x- 6\cos^2x\sin^2x + \sin^4 x = (1-\sin^2x)^2-6(1-\sin^2x)\sin^2x+\sin^4x=\\=1-2\sin^2x+\sin^4x-6\sin^2x+6\sin^4x+\sin^4x-8\sin^4x-8\sin^2x+1$
$\cos 2x = \cos^2x - \sin^2x=1-2\sin^2x$
$\sin^4x=\sin^2x+\frac{\cos4x-1}8=\frac{1-\cos2x}2+\frac{\cos4x-1}8$
|
D K Choudhury
Articles written in Pramana – Journal of Physics
Volume 60 Issue 3 March 2003 pp 563-567
The exponent λ of the structure function F
2 ∼ −λ is calculated using the solution of the DGLAP equation for gluon at low 2 and as a function of 2 at fixed
Volume 61 Issue 5 November 2003 pp 979-985
The transversity distribution of quarks in a nucleon is one of the three fundamental distributions, that characterize nucleon's properties in hard scattering processes at leading twist (twist 2). It measures the distribution of quark transverse spin in a nucleon polarized transverse to its (infinite) momentum. It is a chiral-odd twist-two distribution function — gluons do not couple to it. Quarks in a nucleon/hadron are relativistically bound and transversity is a measure of the relativistic nature of bound quarks in a nucleon. In this work, we review some important aspects of this less familiar distribution function which has not been measured experimentally so far.
Volume 65 Issue 2 August 2005 pp 193-213
At low
2 ( 2)/ 2 2( 2)/
Volume 75 Issue 3 September 2010 pp 423-438 Research Articles
We used variationally improved perturbation theory (VIPT) in calculating the slope and curvature of Isgur–Wise (I–W) function with the Cornell potential $− \dfrac{4\alpha_{s}}{3r} br + c$ instead of the usual stationary state perturbation theory as done earlier. We used $−(4\alpha_{s} /3r)$, i.e. the Coulombic potential, as the parent and the linear one, i.e. $br +c$ as the perturbed potential in the theory and calculated the slope and curvature of Isgur–Wise function including three states in the summation involved in the first-order correction to wave function in the method.
Volume 78 Issue 4 April 2012 pp 555-564 Research Articles
We have recently reported the calculation of slope and curvature of Isgur–Wise function based on variationally improved perturbation theory (VIPT) in a quantum chromodynamics (QCD)-inspired potential model. In that work, Coulombic potential was taken as the parent while the linear one as the perturbation. In this work, we choose the linear one as the parent with Coulombic one as the perturbation and see the consequences.
Volume 79 Issue 4 October 2012 pp 833-837 Poster Presentations
An analytical solution of the non-singlet polarized parton distribution $\Delta q^{NS} (x, Q^{2}) = (\Delta u(x, Q^{2}) − \Delta d(x, Q^{2}))$ is obtained by solving the DGLAP (Gribov and Lipatov,
Volume 79 Issue 6 December 2012 pp 1385-1393
We modify the mesonic wave function by using a short distance scale $r_{0}$ in analogy with hydrogen atom and estimate the values of masses and decay constants of the open flavour charm mesons 𝐷, $D_{s}$ and $B_{c}$ within the framework of a
Volume 84 Issue 1 January 2015 pp 69-85
We use variationally improved perturbation theory (VIPT) for calculating the elastic form factors and charge radii of $D$, $D_{s}$, $B$, $B_{s}$ and $B_{c}$ mesons in a quantum chromodynamics (QCD)-inspired potential model. For that, we use linear-cum-Coulombic potential and opt the Coulombic part first as parent and then the linear part as parent. The results show that charge radii and form factors are quite small for the Coulombic parent compared to the linear parent. Also, the analysis leads to a lower as well as upper bounds on the four-momentum transfer $Q^{2}$, hinting at a workable range of $Q^{2}$ within this approach, which may be useful in future experimental analyses. Comparison of both the options shows that the linear parent is the better option.
Current Issue
Volume 93 | Issue 5 November 2019
Click here for Editorial Note on CAP Mode
|
I'm interested in numerically finding the maximum likelihood estimator of a parameter $\theta$, as well as the confidence interval of this estimator. First I'll describe the method I've been trying, then I'll ask my specific questions about this method.
I perform $N$ binary trials at each possible value of $\theta$ (in my case, $\theta$ is integer-valued and known to fall in a certain range). Given these results, I can choose the maximum likelihood estimator $\hat \theta$ to be the value of $\theta$ that results in the most positive trials.
To find a confidence interval for $\hat \theta$, I can take subsets of my trials and use these to find estimators $\hat\theta_i$. Let the fraction of the trials that I take for each subset be $d$. Let $\hat\theta_\mu$ be the mean of all these estimators $\hat\theta_i$, and $\hat\theta_\sigma$ be the standard deviation.
My questions are: For my estimator, is it better to use the original estimator $\hat\theta$ based on all the data, or $\hat\theta_\mu$ based on my subset estimators? Is the confidence interval just equal to $\hat\theta \pm \hat\theta_\sigma$, or can I take into account the fact that the subset estimators will have more variance than the overall estimator, and use $\hat\theta \pm \sqrt{d} \hspace{3pt}\hat\theta_\sigma$? Is this overall a sound method, or are there other ways to improve it, or another method I should use entirely? I can always run more trials, for example I could run many sets of $N$ trials independently and then get a confidence interval from that, but of course that gets costly.
|
Given $2.14\ \mathrm g$ of $\ce{K(s)}$, determine $\Delta H^\circ_\mathrm{f,m}$ and $\Delta U^\circ_\mathrm{f,m}$ for $\ce{K2O}$.
We know:
The calorimeter's constant: $1849\ \mathrm{J\cdot K^{-1}}$ The mass of water inside it: $1450\ \mathrm g$ The change in temperature: $2.62\ \mathrm K$ The end product is $\ce{K2O}$
The process should be: determining mol of $\ce{K(s)}$, which is $0.054\ \mathrm{mol}$. Then we obtain the amount of energy absorbed by the calorimeter/water. Here are the issues. I don't know which one changed its temperature by $2.62\ \mathrm K$ or if the calorimeter's constant already considers the water. Either way, the amount of energy released by each mol of potassium is
extravagant; according to the result it is $322\ \mathrm{kJ}$ or $96.8\ \mathrm{kJ}$.
Continuing with this reasoning, the $\Delta U^\circ_\mathrm{f,m}$ should be $755\ \mathrm{kJ}$ or $193\ \mathrm{kJ}$, right? What do we need to obtain $\Delta H^\circ_\mathrm{f,m}$?
|
Searching for just a few words should be enough to get started. If you need to make more complex queries, use the tips below to guide you.
Purchase individual online access for 1 year to this journal.
Impact Factor 2019: 0.808
The journal
Asymptotic Analysis fulfills a twofold function. It aims at publishing original mathematical results in the asymptotic theory of problems affected by the presence of small or large parameters on the one hand, and at giving specific indications of their possible applications to different fields of natural sciences on the other hand. Asymptotic Analysis thus provides mathematicians with a concentrated source of newly acquired information which they may need in the analysis of asymptotic problems.
Authors: Frénod, E.
Article Type: Research Article
Abstract: We show that the solution to an oscillatory–singularly perturbed ordinary differential equation may be asymptotically expanded into a sum of oscillating terms. Each of those terms writes as an oscillating operator acting on the solution to a nonoscillating ordinary differential equation with an oscillating correction added to it. The expression of the nonoscillating ordinary differential equations are defined by a recurrence relation. We then apply this result to problems where charged particles are submitted to large magnetic field.
Citation: Asymptotic Analysis, vol. 46, no. 1, pp. 1-28, 2006
Article Type: Research Article
Abstract: We obtain large time decay of the Lp norms, 2<p<+∞, of solutions to the wave equation with a real-valued potential which decays slowly at infinity. This extends previous results by Beals and Strauss [Comm. Partial Differential Equations 18 (1993), 1365–1397] and Georgiev and Visciglia [Comm. Partial Differential Equations 28 (2003), 1325–1369].
Citation: Asymptotic Analysis, vol. 46, no. 1, pp. 29-42, 2006
Authors: François, Gilles
Article Type: Research Article
Abstract: This paper deals with a spectral problem for a second-order elliptic operator A stemming from a parabolic problem under a dynamical boundary condition. The discrete character and the convergence to infinity of the eigenvalue sequence of the problem \[\cases{Au=\lambda u& \mbox{in }\varOmega,\cr \noalign{\vskip3pt}\curpartial_{\nu_{A}}u=\lambda\sigma u& \mbox{on }\curpartial\varOmega,}\] are shown. By means of min–max formulae, a comparison of the eigenvalue sequence with the spectra of the Dirichlet and Neumann problem is obtained and yields an upper bound for λk . On the other hand, comparing with the Steklov problem leads to a lower bound. In the two-dimensional case, this …yields the exact growth order of the eigenvalue sequence and leads to inequalities about the constant of the leading term in its asymptotic behavior. For higher dimensions, the same arguments hold for a sufficiently smooth domain, while for a Lipschitz boundary, lower and upper bounds for the growth exponent are obtained. Finally, the variational characterization of the eigenvalues yields an upper bound for the number of nodal domains of the k-th eigenfunction. The one-dimensional case is also discussed. Show more
Keywords: elliptic operator, eigenvalue problems, asymptotic behavior of eigenvalues, dynamical boundary conditions for parabolic problems
Citation: Asymptotic Analysis, vol. 46, no. 1, pp. 43-52, 2006
Authors: Amadori, Debora
Article Type: Research Article
Abstract: We consider the scalar conservation law with oscillatory, periodic source term \[u^{\varepsilon}_{t}+f(u^{\varepsilon})_{x}=\dfrac{1}{\varepsilon}V'\big(\dfrac{x}{\varepsilon}\big),\quad x\in \mathbb{R},\ t>0,\ \varepsilon>0,\] and with initial data \[u^{\varepsilon}(x,0)=u_{o}\big(x,\dfrac{x}{\varepsilon}\big).\] For possibly resonant initial data, we prove a corrector-type result for this problem, extending a previous one by E and Serre [Asymptotic Anal. 5 (1992), 311–316]: an asymptotic representation \[$U(x,t,\tfrac{x}{\varepsilon})$ is identified for the sequence uε (x,t), and the strong convergence of the asymptotic expansion is shown.
Keywords: conservation laws, periodic source term, oscillations, homogenization
Citation: Asymptotic Analysis, vol. 46, no. 1, pp. 53-79, 2006
Article Type: Research Article
Abstract: In this paper we consider a nonlocal integro-differential model as it was discussed by Bates and Chen [Electron. J. Differential Equations 1999(26) (1999), 1–19]. It is known that unique, stable traveling waves exist for the classical reaction–diffusion model as well as for the nonlocal model and for combinations of both for certain bistable nonlinearities. Here we are concerned with the traveling wave speed and how small perturbations with a nonlocal term affect the speed of the original reaction–diffusion problem. We show that an expansion for the wave speed of the perturbed problem exists and calculate the sign of the first-order …coefficient. Show more
Keywords: nonlocal integro-differential equation, reaction–diffusion equation, traveling wave speed
Citation: Asymptotic Analysis, vol. 46, no. 1, pp. 81-91, 2006
Inspirees International (China Office)
Ciyunsi Beili 207(CapitaLand), Bld 1, 7-901 100025, Beijing China Free service line: 400 661 8717 Fax: +86 10 8446 7947 china@iospress.cn
For editorial issues, like the status of your submitted paper or proposals, write to editorial@iospress.nl
如果您在出版方面需要帮助或有任何建, 件至: editorial@iospress.nl
|
In my textbook (
Introduction to Electrodynamics, D. Griffiths), we derive the equation for some strange potential function. Eventually, we get to this (for $n \in \mathbb{Z}^+$):
$$ V_0(y) = \sum_{n=0}^{\infty} C_n\sin{\frac{n\pi}{a}y} \tag{3.31}$$
Here's where things go awry for me.
... how do we actually
determinethe coefficients $C_n$, buried as they are in that infinite sum? The device for accomplishing this is so lovely it deserves a name—I call it Fourier's trick, though it seems Euler had used essentially the same idea somewhat earlier. Here's how it goes: Multiply Eq. 3.31 by $\sin{n'\pi y/a}$ (where $n'$ is a positive integer), and integrate from 0 to a:
$$ \displaystyle \sum_{n=0}^{\infty} C_n \int_0^a\sin{\frac{n\pi}{a}y} \sin{\frac{n'\pi}{a}y} dy ~~~=~~~ \int_0^a V_0(y)\sin{\frac{n'\pi}{a}y} dy$$
The answer understandably comes out to something very nice and convenient. But... why is this something you can do? There's no obvious reason for why that doesn't intrinsically change the problem (in the same way that I can say "Multiply both sides by $0$. You've successfully reduced the problem to zero. Well done!)
(While typing out the above, I suspect that it has something to do with the inner product of a function and an orthonormal basis? The infinite $\sin$ functions create an orthonormal basis, and taking that integral over all possible values effectively
extracts the coefficients for each basis function. When it is suggested that we multiply by $\sin{\frac{n'\pi}{a}y}$ and integrate, this isn't changing the basis at all, it's just (sneakily) extracting the coefficients, which only exist when $n = n'$ (because the $\sin$ functions are all orthogonal). It's like taking the coefficients of a basis with itself... right?
I think this may be one of those cases where, in the process of asking the question, I figure out the answer—but this is all fairly new to me, and I'd like to ask it anyway for confirmation and, possibly, a clearer explanation).
|
I have a problem:
Let $a \in \mathbb{Z}$, $a \geq 3$, and set $\xi= \sum_{n=0}^\infty 10^{-a^{2n}}>0$. Then the inequality
$$\Big|\,\xi - \dfrac{x}{y}\,\Big| \leq \dfrac{1}{y^a}$$ has infinitely many solutions with $x,y \in \mathbb{Z}$, $y>0$ and $\gcd(x,y)=1$.
I would imitate the proof of Dirichlet's theorem as follow, I claim my lemma
Lemma: Let $\xi=\sum_{n=0}^\infty 10^{-a^{2n}}$, for every integer $Q \geq 2$, there are integers $x,y$ which are not both equal to $0$, such that $$|x - \xi y| \leq \dfrac{1}{Q^{a-1}}$$ with $0<y \leq Q$ and $\gcd(x,y) =1$ Try to prove this lemma:
Partition the interval $[0,1]$ into $Q^{a-1}$ subintervals of length $\dfrac{1}{Q^{a-1}}$. Consider $Q^{a-1}+1$ numbers $\xi-[\xi]$,..,$Q^{a-1}\xi-[Q^{a-1}\xi]$ and $1$. By the Dirichlet principle, two among these numbers must lie in the same subinterval of length $\dfrac{1}{Q^{a-1}}$. Hence we can find $x,y \in \mathbb{z}$ such that $|x-y \xi| \leq \frac{1}{Q^{a-1}}$. But now my trouble is $y$ is not smaller than $Q$.
Does anyone have other ideas?
|
In signal processing, cross-correlation is a measure of similarity of two waveforms as a function of a time-lag applied to one of them. This is also known as a sliding dot product or sliding inner-product. It is commonly used for searching a long signal for a shorter, known feature. It has applications in pattern recognition, single particle analysis, electron tomographic averaging, cryptanalysis, and neurophysiology.For continuous functions f and g, the cross-correlation is defined as:: (f \star g)(t)\ \stackrel{\mathrm{def}}{=} \int_{-\infty}^{\infty} f^*(\tau)\ g(\tau+t)\,d\tau,whe...
That seems like what I need to do, but I don't know how to actually implement it... how wide of a time window is needed for the Y_{t+\tau}? And how on earth do I load all that data at once without it taking forever?
And is there a better or other way to see if shear strain does cause temperature increase, potentially delayed in time
Link to the question: Learning roadmap for picking up enough mathematical know-how in order to model "shape", "form" and "material properties"?Alternatively, where could I go in order to have such a question answered?
@tpg2114 For reducing data point for calculating time correlation, you can run two exactly the simulation in parallel separated by the time lag dt. Then there is no need to store all snapshot and spatial points.
@DavidZ I wasn't trying to justify it's existence here, just merely pointing out that because there were some numerics questions posted here, some people might think it okay to post more. I still think marking it as a duplicate is a good idea, then probably an historical lock on the others (maybe with a warning that questions like these belong on Comp Sci?)
The x axis is the index in the array -- so I have 200 time series
Each one is equally spaced, 1e-9 seconds apart
The black line is \frac{d T}{d t} and doesn't have an axis -- I don't care what the values are
The solid blue line is the abs(shear strain) and is valued on the right axis
The dashed blue line is the result from scipy.signal.correlate
And is valued on the left axis
So what I don't understand: 1) Why is the correlation value negative when they look pretty positively correlated to me? 2) Why is the result from the correlation function 400 time steps long? 3) How do I find the lead/lag between the signals? Wikipedia says the argmin or argmax of the result will tell me that, but I don't know how
In signal processing, cross-correlation is a measure of similarity of two waveforms as a function of a time-lag applied to one of them. This is also known as a sliding dot product or sliding inner-product. It is commonly used for searching a long signal for a shorter, known feature. It has applications in pattern recognition, single particle analysis, electron tomographic averaging, cryptanalysis, and neurophysiology.For continuous functions f and g, the cross-correlation is defined as:: (f \star g)(t)\ \stackrel{\mathrm{def}}{=} \int_{-\infty}^{\infty} f^*(\tau)\ g(\tau+t)\,d\tau,whe...
Because I don't know how the result is indexed in time
Related:Why don't we just ban homework altogether?Banning homework: vote and documentationWe're having some more recent discussions on the homework tag. A month ago, there was a flurry of activity involving a tightening up of the policy. Unfortunately, I was really busy after th...
So, things we need to decide (but not necessarily today): (1) do we implement John Rennie's suggestion of having the mods not close homework questions for a month (2) do we reword the homework policy, and how (3) do we get rid of the tag
I think (1) would be a decent option if we had >5 3k+ voters online at any one time to do the small-time moderating. Between the HW being posted and (finally) being closed, there's usually some <1k poster who answers the question
It'd be better if we could do it quick enough that no answers get posted until the question is clarified to satisfy the current HW policy
For the SHO, our teacher told us to scale$$p\rightarrow \sqrt{m\omega\hbar} ~p$$$$x\rightarrow \sqrt{\frac{\hbar}{m\omega}}~x$$And then define the following$$K_1=\frac 14 (p^2-q^2)$$$$K_2=\frac 14 (pq+qp)$$$$J_3=\frac{H}{2\hbar\omega}=\frac 14(p^2+q^2)$$The first part is to show that$$Q \...
Okay. I guess we'll have to see what people say but my guess is the unclear part is what constitutes homework itself. We've had discussions where some people equate it to the level of the question and not the content, or where "where is my mistake in the math" is okay if it's advanced topics but not for mechanics
Part of my motivation for wanting to write a revised homework policy is to make explicit that any question asking "Where did I go wrong?" or "Is this the right equation to use?" (without further clarification) or "Any feedback would be appreciated" is not okay
@jinawee oh, that I don't think will happen.
In any case that would be an indication that homework is a meta tag, i.e. a tag that we shouldn't have.
So anyway, I think suggestions for things that need to be clarified -- what is homework and what is "conceptual." Ie. is it conceptual to be stuck when deriving the distribution of microstates cause somebody doesn't know what Stirling's Approximation is
Some have argued that is on topic even though there's nothing really physical about it just because it's 'graduate level'
Others would argue it's not on topic because it's not conceptual
How can one prove that$$ \operatorname{Tr} \log \cal{A} =\int_{\epsilon}^\infty \frac{\mathrm{d}s}{s} \operatorname{Tr}e^{-s \mathcal{A}},$$for a sufficiently well-behaved operator $\cal{A}?$How (mathematically) rigorous is the expression?I'm looking at the $d=2$ Euclidean case, as discuss...
I've noticed that there is a remarkable difference between me in a selfie and me in the mirror. Left-right reversal might be part of it, but I wonder what is the r-e-a-l reason. Too bad the question got closed.
And what about selfies in the mirror? (I didn't try yet.)
@KyleKanos @jinawee @DavidZ @tpg2114 So my take is that we should probably do the "mods only 5th vote"-- I've already been doing that for a while, except for that occasional time when I just wipe the queue clean.
Additionally, what we can do instead is go through the closed questions and delete the homework ones as quickly as possible, as mods.
Or maybe that can be a second step.
If we can reduce visibility of HW, then the tag becomes less of a bone of contention
@jinawee I think if someone asks, "How do I do Jackson 11.26," it certainly should be marked as homework. But if someone asks, say, "How is source theory different from qft?" it certainly shouldn't be marked as Homework
@Dilaton because that's talking about the tag. And like I said, everyone has a different meaning for the tag, so we'll have to phase it out. There's no need for it if we are able to swiftly handle the main page closeable homework clutter.
@Dilaton also, have a look at the topvoted answers on both.
Afternoon folks. I tend to ask questions about perturbation methods and asymptotic expansions that arise in my work over on Math.SE, but most of those folks aren't too interested in these kinds of approximate questions. Would posts like this be on topic at Physics.SE? (my initial feeling is no because its really a math question, but I figured I'd ask anyway)
@DavidZ Ya I figured as much. Thanks for the typo catch. Do you know of any other place for questions like this? I spend a lot of time at math.SE and they're really mostly interested in either high-level pure math or recreational math (limits, series, integrals, etc). There doesn't seem to be a good place for the approximate and applied techniques I tend to rely on.
hm... I guess you could check at Computational Science. I wouldn't necessarily expect it to be on topic there either, since that's mostly numerical methods and stuff about scientific software, but it's worth looking into at least.
Or... to be honest, if you were to rephrase your question in a way that makes clear how it's about physics, it might actually be okay on this site. There's a fine line between math and theoretical physics sometimes.
MO is for research-level mathematics, not "how do I compute X"
user54412
@KevinDriscoll You could maybe reword to push that question in the direction of another site, but imo as worded it falls squarely in the domain of math.SE - it's just a shame they don't give that kind of question as much attention as, say, explaining why 7 is the only prime followed by a cube
@ChrisWhite As I understand it, KITP wants big names in the field who will promote crazy ideas with the intent of getting someone else to develop their idea into a reasonable solution (c.f., Hawking's recent paper)
|
I am giving a popular talk on LIGO in 90 minutes and Tristan du Pree has offered me a distraction via Twitter. How do you get distracted if you think about LIGO too much? Yes, by hearing about the LHC: TV:John Oliver gave a totally sensible 20-minute tirade explaining why "scientific study says" stories in the media are mostly bullšit. Did you find already a good model for a possible \(375/750/1500\) tower of \(Z\gamma/\gamma \gamma\)?Well, I didn't, I wrote him: it seemed increasingly clear to me that the invariant masses in the \(Z\gamma\) and \(\gamma\gamma\) decays should better be the same. So the numerological explanation of the coincidence doesn't work.
But then I decided that I haven't carefully enough investigated a loophole that could explain why the \(\gamma\gamma\) signal isn't observed near \(375\GeV\): the Landau-Yang theorem. A massive spin-one boson cannot decay to two identical massless spin-one bosons – or, if you wish, a \(Z\)-boson or \(Z'\)-boson cannot decay to two photons.
The reason for or the proof of the theorem? Well, there's no trilinear function of the three polarization vectors \(\vec \epsilon_{1,2,3}\) that may also depend on the massless particle's momentum \(\vec k\) but that is also symmetric under the exchange of the two final photons.
That seems to be the only possible explanation of the absence of the \(\gamma\gamma\) signal at \(375\GeV\) given the assumption that the excesses of \(Z\gamma\) at \(375\GeV\) are real new physics. So if that's the case, there has to be a new \(Z'\)-boson at \(375\GeV\). Or maybe even a composite particle, such as a toponium, could be OK?
And if the coincidence \(750/2=375\) is more than just a coincidence, then the \(750\GeV\) cernette should better be a bound state of the two \(375\GeV\) \(Z'\)-bosons. Probably a tightly bound state, indeed, but the large width observed especially by ATLAS could potentially be explained by this composite character of the object.
I realize that the interactions felt by the new \(Z'\)-boson would have to be immensely strong and it probably doesn't work but I am running out of time so I expect some commenters to tell me whether it could work.
|
Pfleiderer, C.; Fai\ss{}t, A.; {von L̈ohneysen}, H.; Hayden, S. M.; Lonzarich, G. G.
Title:
Field Dependence of the Specific Heat of Single-Crystalline {{ZrZn2}}
Abstract:
We present measurements of the specific heat C of a single crystal of ZrZn2 in the range 2\textendash{}30K, at magnetic field B up to 14T. For B=0 and low temperature the specific heat varies as C$\approx\gamma$T+$\beta$T3, where $\gamma\approx$45mJ/molK2 and $\beta$ corresponds to a Debye temperature $\Theta$D$\approx$340K. Magnetic field reduces $\gamma$ by up to 30% at 14T. The variation of $\gamma$ with B is compared with predictions of a self-consistent model of the magnetic equation of state, where phenomenological parameters are taken from the DC magnetization and neutron scattering. «
We present measurements of the specific heat C of a single crystal of ZrZn2 in the range 2\textendash{}30K, at magnetic field B up to 14T. For B=0 and low temperature the specific heat varies as C$\approx\gamma$T+$\beta$T3, where $\gamma\approx$45mJ/molK2 and $\beta$ corresponds to a Debye temperature $\Theta$D$\approx$340K. Magnetic field reduces $\gamma$ by up to 30% at 14T. The variation of $\gamma$ with B is compared with predictions of a self-consistent model of the magnetic equation of sta... »
Keywords:
Weak ferromagnetism,Specific heat — low temperature
|
I'm trying to maximize a firm's profit given the production function $F(L,K)=L^\alpha K^\beta$ (where $L$ is labor and $K$ is capital) and that $\alpha + \beta \neq 1$.
So, I know that this maximization problem can be written as $\text{max }pF(L,K)-w_1 L-w_2 K$.
Since $pMP_L (L^*,K^*)=w_1$, $p\alpha(L^*)^{\alpha-1}(K^*)^\beta=w_1$. And since $pMP_K (L^*,K^*)=w_2$, $p\beta(L^*)^{\alpha}(K^*)^{\beta-1}=w_2$.
By dividing these functions and simplifying, we get $\displaystyle\frac{\alpha K^*}{\beta L^*}=\displaystyle\frac{w_1}{w_2}$.
I'm unsure how to proceed from here, though. Should I solve for $L^*$ by separating $K^*$ from the equation and plugging into $pMP_L$? Wouldn't this yield a very complicated solution?
|
For an abelian gauge field, the field strength $G_{\mu \nu}$ is gauge-invariant. This means it is a physically observable quantity, e.g. we can build an apparatus to measure electromagnetic field strength.
For a non-abelian gauge field, $G_{\mu \nu}$ transforms non-trivially under infinitesimal gauge transformations $\xi^a$:
$G_{\mu \nu} \to G'_{\mu \nu} = G_{\mu \nu} + [\xi^a t_a, G_{\mu \nu}]$
What are the implications for the observability of $G_{\mu \nu}$? Could we in principle build a physical apparatus to measure the field strength of a non-abelian field? Let's ignore practical difficulties such as short range of non-abelian fields actually found in nature.
|
$\def\A{{\bf A}}\def\B{{\bf B}}\def\C{{\bf C}}\def\R{{\bf R}}\def\D{{\bf D}}\def\f{\phi}$As others have mentioned, if the task is to find the coordinates of $B$ and $C$, this problem is underdetermined.
Let's first consider the triangle with point $A$ located at the origin and point $C$ lying along the positive $x$-axis.Then the coordinates of the points can be found with simple trigonometry, $$\begin{eqnarray*}\A_0 &=& (0,0) \\\B_0 &=& (c \cos A, c\sin A) \\\C_0 &=& (b,0).\end{eqnarray*}$$The angle $A$, and the sides $b$ and $c$ have been given. This is an SAS triangle. The triangle can be solved by finding $a$ with the law of cosines and one of the other angles with the law of sines.
( Added: For completeness, $A = 72^\circ$, $b = 2.61r$, and $c=r$. The law of cosines gives $a = \sqrt{b^2+c^2-2bc\cos A} = 2.49r$.Then $\frac{\sin A}{a} = \frac{\sin B}{b}$ implies $B = 86^\circ$.Lastly, $A+B+C = 180^\circ$ implies $C = 22^\circ$.)
The collection of triangles you are interested in have vertices of the form $$\begin{equation*}\D = \A + \R(\f)\D_0 \tag{1} \end{equation*}$$where $\A = (A_x,A_y)$ is the given location of $A$, and where $\R(\f)$ is a rotation matrix. The transformation (1) is a counterclockwise rotation by the angle $\f$, followed by a shift so the point $A$ has the given coordinates. In components, $$\begin{eqnarray*}A_x &=& A_x \\A_y &=& A_y \\ B_x &=& A_x + c\cos A\cos\f - c\sin A\sin\f \\ &=& A_x + c\cos(A+\f) \\B_y &=& A_y + c\cos A\sin\f + c\sin A\cos\f \\ &=& A_y + c\sin(A+\f) \\ C_x &=& A_x + b\cos \f \\C_y &=& A_y + b\sin \f. \end{eqnarray*}$$
Below we plot the triangle before rotation and translation in black.(We set $r = 1$ in the figure.) The dotted triangle has been rotated counterclockwise by $\f = 30^\circ$, and then translated so the new location of point $A$ is $(2,1)$.For reference,
$$\begin{eqnarray*}A_x &=& 2 \\A_y &=& 1 \\c &=& r = 1 \\b &=& 2.61 \\A &=& 72^\circ \\\f &=& 30^\circ.\end{eqnarray*}$$
|
The mathematical formula for the integration by parts can be derived in integral calculus by the concepts of differential calculus.
$f{(x)}$ and $g{(x)}$ are two functions in terms of $x$. As per the product rule of differentiation, the derivative of the product of the functions $f{(x)}$ and $g{(x)}$ can be written into its equivalent form.
$\dfrac{d}{dx}{\, \Big({f{(x)}}{g{(x)}}\Big)}$ $\,=\,$ $f{(x)}\dfrac{d}{dx}{\, {g{(x)}}}$ $+$ $g{(x)}\dfrac{d}{dx}{\, {f{(x)}}}$
Take $u = f{(x)}$ and $v = g{(x)}$. Now, express the derivative product rule in differential form.
$\implies$ $\dfrac{d}{dx}{\, (uv)}$ $\,=\,$ $u\dfrac{d}{dx}{\, v}$ $+$ $v\dfrac{d}{dx}{\, u}$
$\implies$ $\dfrac{d(uv)}{dx}$ $\,=\,$ $u\dfrac{dv}{dx}$ $+$ $v\dfrac{du}{dx}$
Multiply both sides of the equation by the differential element $dx$.
$\implies$ $\dfrac{d(uv)}{dx} \times dx$ $\,=\,$ $\Big(u\dfrac{dv}{dx}$ $+$ $v\dfrac{du}{dx}\Big) \times dx$
$\implies$ $\dfrac{d(uv)}{dx} \times dx$ $\,=\,$ $u\dfrac{dv}{dx} \times dx$ $+$ $v\dfrac{du}{dx} \times dx$
$\implies$ $d(uv) \times \dfrac{dx}{dx}$ $\,=\,$ $udv \times \dfrac{dx}{dx}$ $+$ $vdu \times \dfrac{dx}{dx}$
$\implies$ $d(uv) \times 1$ $\,=\,$ $udv \times 1$ $+$ $vdu \times 1$
$\implies$ $d(uv)$ $\,=\,$ $udv$ $+$ $vdu$
Now, integrate both sides of the differential equation.
$\implies$ $\displaystyle \int d(uv)$ $\,=\,$ $\displaystyle \int udv$ $+$ $\displaystyle \int vdu$
$\implies$ $uv$ $\,=\,$ $\displaystyle \int udv$ $+$ $\displaystyle \int vdu$
$\implies$ $uv$ $-$ $\displaystyle \int vdu$ $\,=\,$ $\displaystyle \int udv$
$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \int udv$ $\,=\,$ $uv$ $-$ $\displaystyle \int vdu$
Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.
|
Partly based on the 438 Fall 2015 lecture material of Professor Boutin.
Contents 1. Introduction
One of the fundamental tools of digital signal processing is the Discrete Time Fourier Transform (DTFT). Unlike the Continuous Time Fourier Transform (CTFT), the DTFT of a signal is periodic with period $ 2\pi $. Although it is easy to show this property mathematically, it's not often explained in an intuitive manner
why this repetition occurs, and how it is fundamentally linked to the phenomenon of aliasing that occurs when converting a continuous signal to a discrete signal via sampling.
Since I personally learn far better visually and through intuitive reasoning than via memorization or analytical proofs, I wanted to attempt an in-depth intuitive and visual explanation of repetition and aliasing with the DTFT in an effort to provide another perspective and help other students gain a deeper understanding of one of the most fundamental tools in DSP.
2. Background
Let's start with a brief review. The DTFT of a signal is a transformation applied to an aperiodic, discrete signal in order to represent the signal in terms of it's various spectral (frequency) components. It is defined as:
$ X(\omega) \overset{\underset{\mathrm{def}}{}}{=} \sum_{n=-\infty}^\infty x[n]e^{-j\omega n} $
For completeness, we'll include the inverse transform (IDTFT) as well:
$ x[n] = \frac{1}{2\pi}\int\limits_{-\pi}^\pi X(\omega)e^{j\omega n} d\omega $
Although there is much to be said about intuitively reasoning about Fourier transforms in general, we'll assume the reader has a good intuitive understanding of the CTFT and Fourier series, and is generally familiar with the DTFT.
What we want to try to understand here is the question:
why does repetition occur when moving from continuous space to discrete space?
If we look at the plot of an arbitrary DTFT and interpret it directly, trying to forget the periodic repetition as a "rule," we can see that the plot quite clearly shows something interesting about discrete signals: that they are
always composed of infinitely many frequencies.
So, asked another way,
why are all discrete signals composed of an infinite number of frequencies? 3. Discrete signals in the frequency domain
In order to explain this, let's start with a continuous signal $ x_1(t) $, a simple 1Hz cosine:
$ x_1(t) = \cos(2\pi t) $
It is easy to see that this signal is composed of a single frequency (well, technically, two: -1Hz and 1Hz, since $ \cos(-x) = \cos(x) $). If we were to plot the CTFT of the above signal, we would get impulses at -1Hz and 1Hz. There is no other possible set of frequencies that can produce the above signal.
Now, what about this discrete signal $ x_2[n] $?
$ x_2[n] $
If asked to draw a function to fit the discrete set of points, most people will draw the 1Hz cosine above. This is not incorrect; the discrete points are indeed a sampling of the continuous cosine:
$ x_2[n] = \cos\left(2\pi\frac{n}{N}\right) = x_1\left(\frac{n}{N}\right), N = 6 $
However, with a little bit of thinking, we can actually see that there are
many cosines that can fit this set of discrete points:
We can see that 1Hz, 5Hz, 7Hz, 11Hz, and 13Hz (and thus -13Hz, -11Hz, -7Hz, -5Hz, and -1Hz) cosines all fit this set of points! Of course, we don't need to stop at 13Hz: there are infinitely many more higher frequencies that also fit.
Although increasing the number of discrete points (sampling with a smaller period) would seem to restrict the number of frequencies that "match," no matter how many points we use, the fact that the points are ultimately discrete and that there is "space" between them means that there will
always be an infinite number of frequency components. Only when the signal becomes truly continuous (you can imagine this as the limit $ N \to \infty $, where $ N $ is the number of samples, although this is hardly rigorous) is there a single frequency that produces the resulting signal (the frequency of the continuous cosine itself, obviously).
$ x_3[n] $
This presents us with somewhat of a problem if we want to try to transform the discrete signal into the frequency domain! Clearly a 1Hz cosine produces the discrete signal, but so does a 5Hz, 7Hz, etc. cosine. We don't have any reason to claim any single frequency should be weighted more than the others, they are all valid! Really, the discrete signal consists of
all of these frequencies!
We now understand why the DTFT repeats: there are an infinite number of higher frequency components that can match the same discrete data, so the frequency domain of a discrete signal contains
all of these frequencies.
To better visualize this, let's plot the frequencies we've found from Figure 3.4 above in the frequency domain:
Look familiar? We can now see quite clearly the structure of the frequency domain of our discrete signal. It consists of impulses at -1Hz and 1Hz, which are repeated every 6Hz to produce components at 5Hz, 7Hz, 11Hz, 13Hz, etc. The plot is periodic with a period of 6Hz! Of course, this plot is in the $ f $ domain, not $ \omega $, and we haven't worried about the scaling of the impulses. If we were to compute the DTFT of the signal using the definition, we would obtain this graph:
$ X_3(\omega) $
Getting from Figure 3.6 to Figure 3.7 is a simple matter of scaling, and from this it's easy to see the relationship between the sampling period and the scaling of the DTFT: the impulses in the $ \omega $ domain are located at $ 2\pi \cdot f \cdot T $.
This is why the DTFT repeats with period $ 2\pi $: because there are infinitely many frequencies that can produce the discrete points, no matter how closely they are spaced!
4. Sampling and a matter of perception
This understanding gives us an insight into how and why aliasing occurs when sampling a continuous signal. Although the discrete signal $ x_2[n] $ above can be produced by sampling a 1Hz cosine, it can also be produced by sampling a 5Hz cosine, a 7Hz cosine, etc.!
The discrete points are a sampling not only of the original 1Hz cosine, but also of infinitely many other cosines. When sampled with a small enough period, the 1Hz, 5Hz, 7Hz, etc. cosines will be indistinguishable from one another, the phenomenon known as aliasing.
We can gain a little bit more insight into exactly how this occurs by looking at a 5Hz cosine sampled at various periods to produce a discrete signal $ x_4[n] = \cos(2\pi\cdot 5 \cdot T) $.
The above graphic shows $ T $ ranging from 1/30 to 1/6. The gray continuous function is the function being sampled, the blue points are the sampled points, and the blue continuous function is a continuous representation of the lowest frequency component: the signal that you would hear if you were to listen to $ x_4[n] $. We can quite clearly see that for $ T \geq \frac{1}{10} $, aliasing occurs because the repetitions centered at $ 2\pi $ and $ -2\pi $ "spillover" into the $ [-\pi, \pi] $ range. At $ T = 1/6 $, it's clear to see that the DTFT looks identical to Figure 3.7, and the sampled points are identical to those in Figure 3.2!
This is the fundamental link between aliasing and repetition with the DTFT: with a large-enough sampling period, repeated "copies" of the frequency domain of higher-frequency signals "spillover" into the $ [-\pi, \pi] $ range, resulting in
an identical DTFT and discrete signal, no matter the original continuous signal being sampled!
So, why do we perceive discrete signals as consisting of the lowest possible frequency, instead of one of the higher-frequency components? Why, when looking at the set of discrete points in Figure 3.2 above, does the 1Hz cosine jump out at you instead of the 3Hz, 5Hz, or 7Hz waves? Or why, when listening to a sampled audio signal, do you perceive the lowest possible frequency instead of a higher frequency sound?
This is somewhat interesting to think about, because mathematically the lowest possible frequency is no more "correct" than any other frequency component. It seems that our visual and auditory systems are just optimized for simplicity. After all, there are few things in the natural world that oscillate fast enough for aliasing to occur with the human visual system.
However, it is easy to find examples of aliasing in today's world, even in situations that we typically think of as inherently "analog." If you've ever watched the wheels on a passing car, or the propeller on an airplane, you've probably experienced seeing them standing still, moving very slowly, or even appearing to move backwards, as in Figure 4.2 and 4.3. This is due to the fact that even the human visual system performs some form of "sampling" of the world [2], and temporal aliasing occurs as a result!
5. Conclusion
Although it's rather easy to show certain properties mathematically, or memorize a formula, in my opinion it's much harder to understand what is truly going on with many fundamental tools of digital signal processing, as with many areas of mathematics.
We've spent a lot of space here discussing the DTFT, sampling, and aliasing, but hopefully it is now easier to understand how aliasing and the periodic nature of the DTFT are fundamentally linked:
The DTFT of a discrete signal is periodic with period $ 2\pi $ because there are infinitely many higher frequency multiples (i.e. copies shifted in the frequency domain) that can also produce the discrete signal; the discrete signal is said to contain
all of these frequency components. Likewise, all of these frequency components, if represented as a continuous cosine that is then sampled with the appropriate sampling period, can produce the discrete signal; hence, aliasing.
In summary, graphically:
$ \begin{array}{rcl} \fbox{Discrete signal $x[n]$} & \xrightarrow{\text{is composed of}} & \fbox{Infinitely many frequencies} \\ \fbox{Infinitely many frequencies} & \xrightarrow{\text{with sampling, alias to}} & \fbox{Discrete signal $x[n]$} \end{array} $
5. References
[1] Mireille Boutin, "ECE 438: Digital Signal Processing," Purdue University. Dec. 2015.
If you have any questions, comments, etc. please post them here.
|
[1003.0299] The local B-polarization of the CMB: a very sensitive probe of cosmic defects
Authors: Juan Garcia-Bellido, Ruth Durrer, Elisa Fenu, Daniel G. Figueroa, Martin Kunz Abstract: We present a new and especially powerful signature of cosmic strings and other topological or non-topological defects in the polarization of the cosmic microwave background (CMB). We show that even if defects contribute 1% or less in the CMB temperature anisotropy spectrum, their signature in the local $\tilde{B}$-polarization correlation function at angular scales of tens of arc minutes is much larger than that due to gravitational waves from inflation, even if the latter contribute with a ratio as big as $r\simeq 0.1$ to the temperature anisotropies. Proposed B-polarization experiments, with a good sensitivity on arcminute scales, may either detect a contribution from topological defects produced after inflation or place stringent limits on them. Even Planck should be able to improve present constraints on defect models by at least an order of magnitude, to the level of $\ep <10^{-7}$. A future full-sky experiment like CMBpol, with polarization sensitivities of the order of $1\mu$K-arcmin, will be able to constrain the defect parameter $\ep=Gv^2$ to a few $\times10^{-9}$, depending on the defect model. [PDF] [PS] [BibTex] [Bookmark]
Discussion related to specific recent arXiv papers
Post Reply
3 posts • Page
1of 1
Topological defects can source scalar, vector and tensor modes in the early universe. The vector modes have power on small scales and can generate E and B polarization; the B signal can be quite distinctive, and used to constrain defect models with future data.
This paper appears to take some previous results for the B-mode power spectrum and multiply them by l^4, so e.g. in Fig 1 the power is very blue. Of course to be consistent you also have to multiply the noise and the any other spectrum of interest by l^4 as well, so you seem to gain nothing by doing this. Is there some point I have missed? The paper also defines a 'local' scalar [tex]\tilde{B}[/tex] by taking two derivatives of the polarization tensor. However you gain nothing by doing this; with noisy or non-band-limited data you cannot calculate derivatives on a scale L without having data available over a scale L - the non-locality just hits you in a different form (see astro-ph/0305545 and refs).
This paper appears to take some previous results for the B-mode power spectrum and multiply them by l^4, so e.g. in Fig 1 the power is very blue. Of course to be consistent you also have to multiply the noise and the any other spectrum of interest by l^4 as well, so you seem to gain nothing by doing this. Is there some point I have missed?
The paper also defines a 'local' scalar [tex]\tilde{B}[/tex] by taking two derivatives of the polarization tensor. However you gain nothing by doing this; with noisy or non-band-limited data you cannot calculate derivatives on a scale L without having data available over a scale L - the non-locality just hits you in a different form (see astro-ph/0305545 and refs).
The main point is that vector components of defects' contribution to CMB polarization anisotropies peak at scales smaller than those from inflation.
On the other hand, the ordinary E- and B-modes depend non-locally on the Stokes parameters, so they cannot be used to put constraints on causal sources like defects using the angular correlation function of E- and B-modes on small scales. That is the reason why Baumann and Zaldarriaga [0901.0958] suggested using instead the local modes. Those are the true causal modes, written in terms of derivatives of the Stokes parameters. These local B-modes then have power spectra that are much bluer than the non-local ones, and hence enhance the small scale (high-l) end of the spectrum. It is by looking at the angular correlation functions at small separations (tens of arcmin) that one has a chance to measure the defect's contribution to the local B-modes, and distinguish it from the one of inflation. Of course, the usual white noise power spectrum for polarization will also be modified by this [tex]\ell^4[/tex] factor, but by a suitable gaussian smoothing of the data (following Baumann&Zaldarriaga), we can indeed obtain large signal to noise ratios for binned data at small angular scales. Baumann&Zaldarriaga looked at the model-independent signature of inflation at angles [tex]\theta>2[/tex] degrees. What we have realiazed is that, although model-dependent, the signal at angles [tex]\theta < 1[/tex] degrees can be much more significant. In fact, the feature at small angles is rather universal. The differences between defect models (and we considered four different ones) is just in the height and width of the first and second oscillations in the angular correlation functions (related to the heigth and position of the angular power spectrum). Therefore, with sufficient angular resolution one could not only detect defects (if they are there) but also differentiate between different models.
On the other hand, the ordinary E- and B-modes depend non-locally on the Stokes parameters, so they cannot be used to put constraints on causal sources like defects using the angular correlation function of E- and B-modes on small scales. That is the reason why Baumann and Zaldarriaga [0901.0958] suggested using instead the local modes. Those are the true causal modes, written in terms of derivatives of the Stokes parameters.
These local B-modes then have power spectra that are much bluer than the non-local ones, and hence enhance the small scale (high-l) end of the spectrum. It is by looking at the angular correlation functions at small separations (tens of arcmin) that one has a chance to measure the defect's contribution to the local B-modes, and distinguish it from the one of inflation.
Of course, the usual white noise power spectrum for polarization will also be modified by this [tex]\ell^4[/tex] factor, but by a suitable gaussian smoothing of the data (following Baumann&Zaldarriaga), we can indeed obtain large signal to noise ratios for binned data at small angular scales.
Baumann&Zaldarriaga looked at the model-independent signature of inflation at angles [tex]\theta>2[/tex] degrees. What we have realiazed is that, although model-dependent, the signal at angles [tex]\theta < 1[/tex] degrees can be much more significant. In fact, the feature at small angles is rather universal. The differences between defect models (and we considered four different ones) is just in the height and width of the first and second oscillations in the angular correlation functions (related to the heigth and position of the angular power spectrum). Therefore, with sufficient angular resolution one could not only detect defects (if they are there) but also differentiate between different models.
I think it is clear from the normal power spectra that the sourced vector mode B-polarization peaks at much smaller scales than the gravitational wave spectrum: mostly scales sub-horizon at recombination as opposed to tensor modes which decay on sub-horizon scales. I agree that with low enough noise this is an interesting signal (and has been calculated many times before), though it needs to be distinguished from other possible vector mode sources like magnetic fields.
I thought the point of the Baumann paper was to make a nice picture showing visually the structure of the correlations. The E and B modes contain exactly the same information as the tilde versions; in the same way the WMAP7 papers make some nice plots of the polarization-temperature correlation to visually show a physical effect, but these constrain the same information as the usual power spectra. In the Gaussian limit the usual E/B spectra contain all the information on the defect power spectrum. Only Q and U can actually be measured locally on the sky (in one pixel you cannot calculate any spatial derivatives). The two-point Q/U correlations can be calculated from the usual E and B spectra.
I thought the point of the Baumann paper was to make a nice picture showing visually the structure of the correlations. The E and B modes contain exactly the same information as the tilde versions; in the same way the WMAP7 papers make some nice plots of the polarization-temperature correlation to visually show a physical effect, but these constrain the same information as the usual power spectra. In the Gaussian limit the usual E/B spectra contain all the information on the defect power spectrum.
Only Q and U can actually be measured locally on the sky (in one pixel you cannot calculate any spatial derivatives). The two-point Q/U correlations can be calculated from the usual E and B spectra.
|
Help:Wikitext Examples Contents Basic text formatting
You can format the page using Wikitext special characters.
What it looks like What you type
You can
3 apostrophes will
5 apostrophes will
(Using 4 apostrophes doesn't do anythingspecial --
You can ''italicize text'' by putting 2 apostrophes on '''each''' side. 3 apostrophes will '''embolden the text'''. 5 apostrophes will '''embolden''' and ''italicize'' '''''the text'''''. (Using 4 apostrophes doesn't do anything special -- <br /> the last pair are just ''''left over ones'''' that are included as part of the text.)
A single newlinegenerally has no effect on the layout.These can be used to separatesentences within a paragraph.Some editors find that this aids editingand improves the
But an empty line starts a new paragraph.
When used in a list, a newline
A single newline generally has no effect on the layout. These can be used to separate sentences within a paragraph. Some editors find that this aids editing and improves the ''diff'' function (used internally to compare different versions of a page). But an empty line starts a new paragraph. When used in a list, a newline ''does'' affect the layout ([[#lists|see below]]).
You can break lines
Please do not start a link or
You can break lines<br/> without a new paragraph.<br/> Please use this sparingly. Please do not start a link or ''italics'' or '''bold''' text on one line and end on the next. You should "sign" your comments on talk pages: You should "sign" your comments on talk pages: - Three tildes gives your signature: ~~~ - Four tildes give your signature plus date/time: ~~~~ - Five tildes gives the date/time alone: ~~~~~
You can use some HTML tags, too. However, you should avoid HTML in favor of Wiki markup whenever possible.
What it looks like What you type
Put text in a
Put text in a <tt>typewriter font</tt>. The same font is generally used for <code> computer code</code>. <strike>Strike out</strike> or <u>underline</u> text, or write it <span style= "font-variant:small-caps"> in small caps</span>.
Superscripts and subscripts:X
Superscripts and subscripts: X<sup>2</sup>, H<sub>2</sub>O <center>Centered text</center> * Please note the American spelling of "center". <blockquote> The '''blockquote''' command ''formats'' block quotations, typically by surrounding them with whitespace and a slightly different font. </blockquote>
Invisible comments to editors (<!-- -->) appear only while editing the page.
Invisible comments to editors (<!-- -->) appear only while editing the page. <!-- Note to editors: blah blah blah. --> Organizing your writing
What it looks like What you type Section headings Subsection
Using more "equals" (=) signs creates a subsection.
A smaller subsection
Don't skip levels, like from two to four equals signs.
Start with 2 equals signs not 1 because 1 creates H1 tags which should be reserved for page title.
== Section headings == ''Headings'' organize your writing into sections. The ''Wiki'' software can automatically generate a [[Help:table of contents|table of contents]] from them. === Subsection === Using more "equals" (=) signs creates a subsection. ==== A smaller subsection ==== Don't skip levels, like from two to four equals signs. Start with 2 equals signs not 1 because 1 creates H1 tags which should be reserved for page title.
marks the end of the list.
* ''Unordered lists'' are easy to do: ** Start every line with a star. *** More stars indicate a deeper level. *: Previous item continues. ** A newline * in a list marks the end of the list. *Of course you can start again.
A newline marks the end of the list.
# ''Numbered lists'' are: ## Very organized ## Easy to follow A newline marks the end of the list. # New numbering starts with 1.
Here's a
Begin with a semicolon. One item per line; a newline can appear before the colon, but using a space before the colon improves parsing.
Here's a ''definition list'': ; Word : Definition of the word ; A longer phrase needing definition : Phrase defined ; A word : Which has a definition : Also a second one : And even a third Begin with a semicolon. One item per line; a newline can appear before the colon, but using a space before the colon improves parsing. * You can even do mixed lists *# and nest them *# inside each other *#* or break lines<br>in lists. *#; definition lists *#: can be *#:; nested : too
A newline starts a new paragraph.
: A colon (:) indents a line or paragraph. A newline starts a new paragraph. Should only be used on talk pages. For articles, you probably want the blockquote tag. : We use 1 colon to indent once. :: We use 2 colons to indent twice. ::: 3 colons to indent 3 times, and so on.
You can make horizontal dividing lines (----) to separate text.
But you should usually use sections instead, so that they go in the table of contents.
You can make [[w:horizontal dividing line|horizontal dividing line]]s (----) to separate text. ---- But you should usually use sections instead, so that they go in the table of contents.
You can add footnotes to sentences using the
You can add footnotes to sentences using the ''ref'' tag -- this is especially good for citing a source. :There are over six billion people in the world.<ref>CIA World Factbook, 2006.</ref> References: <references/> For details, see [[Wikipedia:Footnotes]] and [[Help:Footnotes]]. Links
You will often want to make clickable
links to other pages.
What it looks like What you type Here's a link to a page named [[Official position]]. You can even say [[official position]]s and the link will show up correctly.
You can put formatting around a link.Example:
You can put formatting around a link. Example: ''[[Wikipedia]]''. The ''first letter'' of articles is automatically capitalized, so [[wikipedia]] goes to the same place as [[Wikipedia]]. Capitalization matters after the first letter.
Intentionally permanent red link is a page that doesn't exist yet. You could create it by clicking on the link.
[[Intentionally permanent red link]] is a page that doesn't exist yet. You could create it by clicking on the link.
You can link to a page section by its title:
If multiple sections have the same title, add a number. #Example section 3 goes to the third section named "Example section".
You can link to a page section by its title: * [[Doxygen#Doxygen Examples]]. If multiple sections have the same title, add a number. [[#Example section 3]] goes to the third section named "Example section".
You can make a link point to a different place with a piped link. Put the link target first, then the pipe character "|", then the link text.
Or you can use the "pipe trick" so that a title that contains disambiguation text will appear with more concise link text.
You can make a link point to a different place with a [[Help:Piped link|piped link]]. Put the link target first, then the pipe character "|", then the link text. * [[Help:Link|About Links]] * [[List of cities by country#Morocco|Cities in Morocco]] Or you can use the "pipe trick" so that a title that contains disambiguation text will appear with more concise link text. * [[Spinning (textiles)|]] * [[Boston, Massachusetts|]]
You can make an external link just by typing a URL: http://www.nupedia.com
You can give it a title: Nupedia
Or leave the title blank: [1]
External link can be used to link to a wiki page that cannot be linked to with [[page]]: http://meta.wikimedia.org/w/index.php?title=Fotonotes&oldid=482030#Installation
You can make an external link just by typing a URL: http://www.nupedia.com You can give it a title: [http://www.nupedia.com Nupedia] Or leave the title blank: [http://www.nupedia.com] External link can be used to link to a wiki page that cannot be linked to with <nowiki>[[page]]</nowiki>: http://meta.wikimedia.org/w/index.php?title=Fotonotes &oldid=482030#Installation Linking to an e-mail address works the same way: mailto:someone@example.com or [mailto:someone@example.com someone]
You can redirect the user to another page.
#REDIRECT [[Official position]]
Category links do not show up in linebut instead at page bottom
Add an extra colon to
[[Help:Category|Category links]] do not show up in line but instead at page bottom ''and cause the page to be listed in the category.'' [[Category:English documentation]] Add an extra colon to ''link'' to a category in line without causing the page to be listed in the category: [[:Category:English documentation]]
The Wiki reformats linked dates to match the reader's date preferences. These three dates will show up the same if you choose a format in your Preferences:
The Wiki reformats linked dates to match the reader's date preferences. These three dates will show up the same if you choose a format in your [[Special:Preferences|]]: * [[1969-07-20]] * [[July 20]], [[1969]] * [[20 July]] [[1969]] Just show what I typed
A few different kinds of formatting will tell the Wiki to display things as you typed them.
What it looks like What you type
The nowiki tag ignores [[Wiki]] ''markup''. It reformats text by removing newlines and multiple spaces. It still interprets special characters: →
<nowiki> The nowiki tag ignores [[Wiki]] ''markup''. It reformats text by removing newlines and multiple spaces. It still interprets special characters: → </nowiki> The pre tag ignores [[Wiki]] ''markup''. It also doesn't reformat text. It still interprets special characters: → <pre> The pre tag ignores [[Wiki]] ''markup''. It also doesn't reformat text. It still interprets special characters: → </pre>
Leading spaces are another way to preserve formatting.
Putting a space at the beginning of each line stops the text from being reformatted. It still interprets Wiki Leading spaces are another way to preserve formatting. Putting a space at the beginning of each line stops the text from being reformatted. It still interprets [[Wiki]] ''markup'' and special characters: → Source code
If the syntax highlighting extension is installed, you can display programming language source code in a manner very similar to the HTML
<pre> tag, except with the type of syntax highlighting commonly found in advanced text editing software.
Here's an example of how to display some C# source code:
<source lang="csharp"> // Hello World in Microsoft C# ("C-Sharp"). using System; class HelloWorld { public static int Main(String[] args) { Console.WriteLine("Hello, World!"); return 0; } } </source>
Results in:
// Hello World in Microsoft C# ("C-Sharp").using System;class HelloWorld{ public static int Main(String[] args) { Console.WriteLine("Hello, World!"); return 0; }}
Images, tables, video, and sounds This is a very quick introduction. For more information, see: Help:Images and other uploaded files, for how to upload files; w:en:Wikipedia:Extended image syntax, for how to arrange images on the page; Help:Table, for how to create a table.
After uploading, just enter the filename, highlight it and press the "embedded image"-button of the edit_toolbar.
This will produce the syntax for uploading a file
[[Image:filename.png]]
What it looks like What you type
A picture, including alternate text:
You can put the image in a frame with a caption:
A picture, including alternate text: [[Image:Wiki.png|This is Wiki's logo]] You can put the image in a frame with a caption: [[Image:Wiki.png|frame|This is Wiki's logo]]
A link to Wikipedia's page for the image: Image:Wiki.png
Or a link directly to the image itself: Media:Wiki.png
A link to Wikipedia's page for the image: [[:Image:Wiki.png]] Or a link directly to the image itself: [[Media:Wiki.png]] Use media: links to link
directly to sounds or videos: A sound file
Use '''media:''' links to link directly to sounds or videos: [[media:Classical guitar scale.ogg|A sound file]] Provide a spoken rendition of some text in a template: Provide a spoken rendition of some text in a template: {{listen |title = Flow my tears |filename = Flow my tears.ogg |filesize = 583KB }}
{| border="1" cellspacing="0" cellpadding="5" align="center" ! This ! is |- | a | table |} Galleries Main article: w:Gallery tag
Images can also be grouped into galleries using the
<gallery> tag, such as the following:
Wiki.png Wiki.png
Captioned
Wiki.png Wiki.png
Links can be put in captions.
Mathematical formulae
You can format mathematical formulae with TeX markup.
What it looks like What you type
<math>\sum_{n=0}^\infty \frac{x^n}{n!}</math>
<math>\sum_{n=0}^\infty \frac{x^n}{n!}</math>
|
Rational Equations
A rational expression is a quotient whose numerator and denominator are polynomials, where the denominator cannot equal zero. For example:
$$\frac{2x^2 +4x -7}{x^2 -3x+8}$$
A rational equation is one that involves only a rational expression. For example:
$$\frac{2x^2 +4x -7}{x^2 -3x+8}=0$$
Solving By Factoring
Factoring is often an important step in solving rational equations. To solve the following rational equation, the numerator must be factored. The solutions of the equation are the solutions that result when the numerator has been set to zero. The denominator is not set equal to zero, because if you remember, one of the conditions of a rational expression is that the denominator cannot equal zero. Once the numerator has been factored, set each of the factors to zero and solve for x.
Example: Solve for x
$$\frac{x^2-5x+6}{x^2+3x+2}=0$$
Solution
$$\begin{align}
\frac{x^2-5x+6}{x^2+3x+2} & =0 \\ x^2-5x+6 & =0 \\ (x-3)(x-2) & =0 \\ \end{align}$$ $$\smash{x-3=0 \qquad \text{or} \qquad x -2 = 0}$$ $$\smash{x=3 \qquad \qquad \qquad x =2}$$
Therefore, the above rational equation has two solutions, one when x = 3 and the other when x = 2.
A Review of the Lowest Common Denominator
Finding a common denominator is a math technique that is often required to solve rational equations. To find the lowest common denominator of a rational expression, you basically just multiply all of the unique factors in the denominator together. To change each term in the expression to have that denominator, figure out what you multiplied it's denominator by to make it become the lowest common denominator, and then multiply the numerator by that same factor (otherwise, you're changing the expression). For example, if you need to multiply the denominator by (x - 1), then the numerator must also be multiplied by (x -1). Be sure to collect the like terms of the numerator for your final answer.
Example: Combine \(\frac{2}{x+4}+\frac{3}{x-1}\) into a single rational expression. Solution
$$\begin{align}
\frac{2}{x+4}+\frac{3}{x-1}& = \frac{2}{x+4} \cdot \frac{x-1}{x-1} + \frac{3}{x-1}\cdot\frac{x+4}{x+4} \\ & = \frac{2(x-1)+3(x+4)}{(x+4)(x-1)} \\ & = \frac{2x-2+3x+12}{(x+4)(x-1)} \\ & = \frac{5x+10}{(x+4)(x-1)} \end{align} $$ Solving By Multiplying By a Common Denominator
When solving a rational equation, a common denominator for all terms may have to be found. The terms must then be rearranged through addition and subtraction so that only zero remains on one side of the equation. Once this has been done, the numerator can be set equal to zero and it becomes easy to solve.
Example: Solve \(\frac{5x}{x+2}=7\) for x. Solution
$$\begin{align}
\frac{5x}{x+2}&=7 \\ \frac{5x}{x+2} &=\frac{7(x+2)}{x+2} \\ \frac{5x}{x+2} - \frac{7(x+2)}{x+2} &=0 \\ \frac{5x-7x-14}{x+2} &= 0 \\ 5x-7x-14 &=0 \\ -2x &= 14 \\ x &= -7 \end{align}$$
Another way to solve the equation above is to simply multiply both sides of the equation by the common denominator.
$$\begin{align}
\frac{5x}{x+2}&=7 \\ 5x &=7(x+2)\\ 5x &= 7x+14 \\ 5x-7x &=14 \\ -2x &= 14 \\ x &= -7 \end{align}$$ Example: Example (Harder):
|
Rutherford stated that deflection of alpha particles is due to repulsive positive charge of nucleus. Why can't alpha particles be deflected by attraction of so many electrons revolving around nucleus?
See, what the Geiger-Marsden-Rutherford experiment achieved was the following: by bombarding (with alpha particles) a one-atom thick gold sheet and counting how many alpha particles passed through, they were able to relate the already known atomic radius with the actual area that could get collided by alpha particles. Figure 1 shows a sketch of the apparatus.
Figure 1: depiction of the apparatus (courtesy of Wikipedia). Now follows a simplified rationalisation.
Assume you have an atom-thick sheet of some material. By knowing the element's molar mass and the mass of the sheet you can calculate how many atoms $n$ are there, and knowing the sheet's surface area $A$ allows you to calculate the atomic surface density $\sigma = n/A$.
Now if the atoms have radius $\rho$, an area $A_\rho = n \pi \rho^2$ of the sheet should be covered by atoms. This means a ratio
$$r = \frac{A_\rho}{A} = \pi \rho^2 \sigma$$
is actually covered by atoms and $f = 1 - r$ is the ratio of "free space".
Rutherford actually tested if this made sense by throwing particles at the sheet: those scattered simply hit something, the rest passed through. Imagine $M$ particles were thrown and $m$ of those passed through. If $M$ is big enough we can set $f_R = m/M$ as Rutherford's "free space" estimate. Going backwards we may set the coverage ratio
$$r_R = 1 - f_R = 1 - \frac{m}{M} = \frac{A_R}{A} = \pi \rho_R^2 \sigma$$
were $\rho_R$ above is
the measured radius of the collision cross section of an atom.The experiment has shown that $r_R \ll r$ and thus $\rho_R \ll \rho$, which implies that most of the atomic mass is buried inside the atom.In fact (Wikipedia),
The diameter of the nucleus is in the range of $1.75$ fm ($1.75 \times 10^{-15}$ m) for hydrogen (the diameter of a single proton) to about $15$ fm for the heaviest atoms, such as uranium. These dimensions are much smaller than the diameter of the atom itself (nucleus + electron cloud), by a factor of about $23,000$ (uranium) to about $145,000$ (hydrogen).
It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you. On consideration, I realized that this scattering backward must be the result of a single collision, and when I made calculations I saw that it was impossible to get anything of that order of magnitude unless you took a system in which the greater part of the mass of the atom was concentrated in a minute nucleus. It was then that I had the idea of an atom with a minute massive centre, carrying a charge.
--- Ernest Rutherford
The simple reason for this being the ratio of masses of alpha particles and electrons.....Both experience equal electrostatic forces which leaves no option for the electron but to get displaced from its current position ( it gains potential energy to do so) whereas in the case of alpha-nucleus interaction,the size of a nucleus is comparable (in the case of the gold foil he used...way more than that of an alpha particle) to that of the projected alpha particle hence causes a measureable deflection on the alpha particle....Hopefully you got your answer
I was also wondering upon the fact that it could be the electrons which caused the sway, but you see if a certain force were acting on two particles, one lighter than the other, the lighter one will be the one feeling the "force". For more clarity refer to Newtons second law in acceleration format. So the only left out thing was a nucleus, which is way more massive than the alpha particle, again apply our first logic here(stated above) and you see, this time its the He(2+) that has to do the displacement. Hope it resolves your issue.
Well gentlemen we can't deny that fact that there might be some forces of attraction between the alpha particles as well as the electrons,because the alpha particles passed through the atom by small distance margin from nucleus and stroked the zinc sulphide coating but it wasn't clearly stated by Ernest Rutherford that whether the deflection was also caused due to alpha particle attraction and election attraction. Also, attraction (interaction)is between comparable things like a heavy nucleus and heavy alpha particle but it's silly to think about alpha particle and an electron , the alpha particle is approximately 4000 times heavier than electron and it is travelling with a high speed (energy) so we can say that the interaction time might be very much less in order to show its effect and we can underestimate it . though most of the physics practicals are based on assumptions and we have to underestimate those assumptions that's way because the reason for performing the experiments was to propose a model of an atom (accidentally) , to find where is the mass of the atom is concentrated. that's why we got the so we don't have anything to do with the alpha particle and electron attraction.
|
Probability Theory: The Logic of Science, by the great E.T. Jaynes, has been in my reading queue for quite some time now. Unfortunately for me it's a dense book after the first few chapters, so I've kind of plateaued around chapter 3 while reading from a bunch of other sources.
I've found that my brain is like boiling soup, in a sense, with different things coming up to my attention almost randomly but the important ones usually coming up just-in-time. So now a Jaynes bubble has reappeared and I'm going to review what I've read! If you're interested I highly recommend the actual book, since I'm here going to be sometimes more verbose, sometimes less, sometimes tangential, and always less organized than Jaynes; leave your email in the comment form and I'll send you a PDF copy if you want.
Chapter one begins with this thought-provoking quote:
The actual science of logic is conversant at present only with things either certain, impossible, or entirely doubtful, none of which (fortunately) we have to reason on. Therefore the true logic for this world is the calculus of Probabilities, which takes account of the magnitude of the probability which is, or ought to be, in a reasonable man’s mind.
James Clerk Maxwell (1850)
1.1 - Deductive and Plausible (Inductive) Reasoning (Inference)
Not long afterwards, quantum mechanics was born and if you go back and listen/read Feynman he'll tell you that science has been "reduced" to probabilities.
Jaynes asks us to consider a scenario: it's a dark night, a cop is walking down a street which he thought seemed deserted. He suddenly hears a burglar alarm, looks across the street, and sees a jewelry store with a broken window. A moment later, a gentleman wearing a mask comes crawling out through the broken window, carrying a bag which is full of expensive jewelry.
Any cop wouldn't hesitate at all deciding that this gentleman is dishonest. How does he come to this conclusion? Without going too far into the depths of the brain, what, in general, is his mind probably doing?
Was this a deductive decision by the cop, assuming just the evidence alone? (You can of course assume the conclusion but that's not helpful.) Clearly not, because with a little thought one might come up with an entirely innocent explanation for the observed evidence. Jaynes offers: perhaps the gentleman is the store owner who was walking home from a masquerade party, and he didn't have his store key with him. When he was walking by his store, a passing truck may have thrown a brick through the window, and the owner was just protecting his own property. The alarm may have had a delay since the cop never heard the window break.
So clearly,
because a reasonable alternative exists, the cop cannot logicallydeduce the dishonesty of the masked person. Yet we still think the cop has validreasoning to suspect dishonesty. We think the evidence available, while not presenting a logically certainconclusion, nevertheless presents an extremely plausibleconclusion. Why do we think this? Our brains are doing this all the time: we're constantly faced with situations where we simply don't have enough information available to allow ourselves enough assumptions to perform deductive reasoning. Premature deduction can lead you off a cliff.
It turns out that Probability Theory is a concrete, axiomatic, deductive theory for describing what roughly goes on in the cop's mind. It's a theory to explain plausible reasoning. What sort of power does this give us? Through human inductive (plausible) reasoning that we do with intuition and general, qualitative ideas, we can come up with a couple rules to assume as axioms that should govern
induction in general.
All of us can think of cases where humans reasoned poorly, can we actually quantify different kinds of errors and produce a better alternative that the person in error could have done instead? Yup, we can. It's also known that humans can reason very poorly even while under the delusion they're doing just fine. I know I've convinced myself of many falsehoods that had to be corrected. If we can define a minimal set of rules, which should be simple enough not to let us make any errors, then we can deductively produce more theorems from those rules, which give us fences to help us combat our failures when our intuitive induction leads us astray.
I've gotten ahead, though. (This is why you should read Jaynes, especially if you're finding it difficult to follow me.) What do I even mean by deduction? Why do we even want deduction?
Deductive reasoning, or classical two-value logic, typically credited to Aristotle, basically boils down to two strong
syllogismsrepeated over and over. The first, symbolically, follows the form: $$A \to B; A\ is\ true;
\lower0.1ex\hbox{$\bullet$}
\kern-0.2em\raise0.7ex\hbox{$\bullet$}
\kern-0.2em\lower0.2ex\hbox{$\bullet$}
\thinspace B\ is\ true$$. In English:
1:We are given (or assume) the fact that if some proposition A is true, then automatically we know proposition B to be true as well. Some time through our day, we stumble upon an instance of A that we discover is true. We can automatically deduce that the other proposition B must also be true. 2:The second syllogism is just the inverse. Throughout our day, we find an instance of B and learn that it is false. Therefore we can automatically deduce that proposition A must also be false.
It should be obvious why we should prefer these two rules of logic: they're simple, and it's incredibly hard to make a mistake. Indeed, we can even program computers to check mistakes for us automatically! These rules also give us the power of Laziness. Suppose we know that if we ever find a golden apple today, we know that it will be raining above the Empire State Building for the day. So if we stumble upon a golden apple, we
knowit's raining above the Empire State Building. You can go look if you want, but it's not necessary. Inversely, if we're on top of the Empire State Building and it's sunny, we know that we won't find a golden apple today no matter how hard we look and we don't even need to try. Notice, however, that if we're at the Empire State Building and it's raining, that doesn't guarantee we'll find a golden apple. The if-then works one way.
Of course, rarely can you get away with saying you know such an absurd thing as golden apples implying rain. This isn't a knock against
logic, though: logic only requires that you suppose, or assume. If your assumption turns out to be true, because you bothered to check it enough times until you were convinced, then hey, you get a nice bonus of having something useful. Nothing is stopping you from proving whatever you want, apart from people questioning your assumptions. Your deductive conclusion is only as valid as the place you started, and any wrong move in the dance destroys it all.
But it's not even those absurd things like my example that we generally can't get away with. Even the cop example doesn't let us get away with it, unless you want to assume that if someone is wearing a mask then they're dishonest! What'll you do at Halloween?
For many $$A \to B$$ relationships, it's often the case that we simply don't have enough information about A which makes the first syllogism useless to us. So in general, for everyday life, we have to fall back on a weaker syllogism:
3:Again we assume that if A is true, then B is true. We learn that B is true, therefore A becomes more plausible.
Jaynes gives another example here: suppose proposition A is "it will start to rain by 10am at the latest", and B is "the sky will become cloudy before 10am".
If we notice clouds before 10am, in other words we observe that B is true, that doesn't mean it will start to rain. But our common sense, our intuition, follow this weak syllogism in that, even if we can't
logically prove certainthat it will rain, we nevertheless find the idea that it's going to rain more plausible. This process of obeying the weak syllogism is a form of inductive reasoning.
Note that $$A \to B$$ does not need to be causal. Clearly, as far as most people's conceptions of causality go, clouds in the past cause rain in the future. Rain in the future doesn't cause clouds in the past! But rain in the future
doesmean that there wereclouds in the past, while clouds in the past does not necessarilymean rain in the future. And so it is that if we're trying to be careful in what we assume, only the implication of $$Future\ Rain \to Past\ Clouds$$ seems certain enough to let by.
We are not proving
causality, we are only proving valid implication according to some rules we intuitively like and agree on. Proving causality is tricky business, for years most people assumed it couldn't be done. Nowadays we have Judea Pearl. 4:There is one more related weak syllogism, using the same premise of "if A, then B." If we discover that A is false, then ( precisely because we're not talking about causality) we can't say decisively whether B is true or false. The syllogism then is that if A is false, B becomes less plausible.
We say it's less plausible because, while there may be another reason for B to be true, we've at least eliminated one possible reason out of an unknown number.
These last two weak syllogisms account for almost all scientists. Is that surprising? The two strong syllogisms are so rare and hard to justify premises for that they're reserved largely for the realm of mathematics. (And even then, what mathematics
actuallyuse initially is often these weaker forms and the next just like everyone else. It's only when they want a formal proof do they try hard to use the strongest syllogisms.) So what does our cop use? He uses a still weaker syllogism, that has a different premise entirely: 5:We assume that if A is true, then B becomes more plausible. We observe B as true, therefore A becomes more plausible.
Even though this seems weak when stated formally, our brains nevertheless
feellike a policeman's reaction of taking the masked man to be a dishonest criminal has almost all the power of deductive logic! What this means is that this word "plausible" comes in degrees. This will be important later. In the specific cop case, we're assuming that "If a man is a dishonest thief robbing a jewelry store, then it is more plausible for him to be wearing a mask and coming out of a broken window of an alarm-sounding jewelry store." We observe the mask, broken window, and alarm, therefore we become almost entirely certain that this man is also a dishonest thief.
In the case of the clouds before 10am, how much we believe that it will rain depends very much on the darkness of the clouds. Why is this so? Because as humans with memory we have past experiences. Imagine that there's a city where the above cop scenario happens all the time,
exceptall the time it actually isjust the partying store owner trying to protect his property finding his window was broken. After a while the cops would just start ignoring such circumstances since they aren't needed: they're making use of prior informationto determine what degree of plausibilitysomething should have.
This reasoning about determining how plausible something sounds from past information we usually just call
common sense, and much of it is unconscious and very rapid. What's something that goes against your common sense? Maybe that you'll be fine after jumping off a roof, that doesn't sound very plausible. Imagine your shock and subsequent arguments at finding someone who says "Pfft, of courseyou'll be fine. It's common sense." What explains the differences in plausibilities you each have? Maybe the other person has jumped off many roofs him or herself without harm. In other words, you both have different prior information that you're making use of.
If you want to resolve a debate, you need to first get to common ground with respect to prior information. I lamented in my previous post about how difficult it is to have a debate with someone not at my level in terms of understanding information and probability theory: my arguments rely on prior information they don't have, and it is
a lotof prior information. Imagine you're trying to prove to a member of the Piraha tribe that for any equation of the form $$Ax^2 + Bx + C = 0$$, there are roots at $$x = \frac{-B \pm \sqrt{B^2 - 4(A)(C)}}{2A}$$. To even state the problem, in the US kids take at least 6 school years to learn the symbols and operations! To actually proveit, should it be disputed in debate, requires prior knowledge of things like "Completing the Square".
Jaynes mentions George Pólya as an author who from 1945 to 1954 wrote three books about plausible reasoning and how, we can learn through experiment, humans tend to follow certain
qualitativerules. (And as later experiments have shown, humans also tend to violatecertain rules.) Subsequent work by Jaynes and others have been attempts at quantitativelydefining such rules, with great success. 1.2
Join me next time.
Posted on 2011-10-01 by Jach Permalink: https://www.thejach.com/view/id/209
|
Introduction
In the
Subsymbolic Methods in AI course at NTNU, one of the assignments is to evolve a fully connected feed-forward artificial neural network (ANN) which represents an agent strategy for a Flatland environment. This Flatland environment is a \(10 \times 10\) toroidal grid with cells containing either food, poison, or nothing. By default a third of the cells are filled with food and a third of the remaining cells are filled with poison. For fun; a fairly pixel-perfect version of the original Pac-Man was built through recording screenshots in a MAME-emulator (enemies represent poison -- delicious cherries, melons, apples and oranges represent food). The extra dots in the visualization are purely cosmetic.
An agent network is fed six binary inputs with indications of whether there is food or poison for any of the cells in its left, forward, and right direction. It can respond with one of the actions
move left, which turns it \(90^\circ\) to the left and moves it forward one step, move forward, which moves the agent one step forward, or move right, which turns it \(90^\circ\) to the right and moves it forward one step. An agent can also choose to do nothing, but as it has no internal state, choosing to do nothing for a set of inputs will cause it to keep doing nothing until time runs out. By default there are 60 time steps in the game, with each action taking 1 time step.
Functionally, there are three possible outcomes for each of the three cells which the agent receives information about; they can either be empty, contain food, or contain poison. Discounting non-action, there are three possible actions an agent can respond with. Even in this simple environment where there are \(3^3=27\) possible input cases, this means that the number of functionally distinct agent strategies is \(3^{3^3} \approx 7.6 \cdot 10^{12}\), since one out of three actions must be chosen for each of the input cases.
Note the special case where the agent is surrounded by poison on the left, in front, and on the right. In this case, the agent must either forfeit any future possible reward by remaining immobile, or eat poison to progress. Given that the agent does not known the remaining time and that eating poison is non-fatal, the rational choice is to eat poison.
Fitness
When formulating a fitness function, it is important that eating poison is not punished too harshly, as exploration will be deterred, giving less room for constructive evolution. If it is not punished enough, evolution will only value food eaten, no matter how much poison is also eaten. Given a \(10 \times 10\) grid, \(\frac{1}{3}\) food coverage, 60 time steps, and awarding one point for each food eaten; the normalized upper fitness boundary is 0.55. In practise, a normalized fitness score of around 0.45 or greater results in good performance.
Agent fitness is either evaluated
statically or dynamically. When evaluated statically, a given number of random environments are generated, and the same environments are used to evaluate all individuals in all generations. This gives the evolution process little opportunity to generalize well as individuals may find limited strategies which work well in this small number of environments, but which may lack adaptation to important edge cases. When evaluated dynamically, each individual's fitness is evaluated using a unique set of randomly generated environments. In addition, the fitness of all individuals is retested using new random environments for every generation. This provides great opportunity to find general behavior, as individuals continually needs to perform well in new environments to maintain their fitness within the population. Implementation
The implementation uses a bit-vector genotype, single uniform random crossover, and full generational replacement with elitism. Switch bits can be used which adds one bit per weight which can enable or disable the weight, which is equivalent to a weight value of zero. This allows for neutral evolution of the weight value when disabled, and easier changes to topology. Note that this implementation is not efficient as it is appropriately abstracted for educational purposes. It is also my first time using Javascript.
To start testing, click
Evolve, then when the evolution is complete the buttons directly below the Flatland widget can be used to run the agent in the available environments.
|
Huge cardinal
Huge cardinals (and their variants) were introduced by Kenneth Kunen in 1972 as a very large cardinal axiom. Kenneth Kunen first used them to prove that the consistency of the existence of a huge cardinal implies the consistency of $\text{ZFC}$+"there is a $\aleph_2$-saturated ideal over $\omega_1$". [1]
Contents Definitions
Their formulation is similar to that of the formulation of superstrong cardinals. A huge cardinal is to a supercompact cardinal as a superstrong cardinal is to a strong cardinal, more precisely. The definition is part of a generalized phenomenon known as the "double helix", in which for some large cardinal properties $n$-$P_0$ and $n$-$P_1$, $n$-$P_0$ has less consistency strength than $n$-$P_1$, which has less consistency strength than $(n+1)$-$P_0$, and so on. This phenomenon is seen only around the $n$-fold variants as of modern set theoretic concerns. [2]
Although they are very large, there is a first-order definition which is equivalent to $n$-hugeness, so the $\theta$-th $n$-huge cardinal is first-order definable whenever $\theta$ is first-order definable. This definition can be seen as a (very strong) strengthening of the first-order definition of measurability.
Elementary embedding definitions
The elementary embedding definitions are somewhat standard. Let $j:V\rightarrow M$ be an elementary embedding with critical point $\kappa$ such that $M$ is a standard inner model of $\text{ZFC}$. Then:
$\kappa$ is almost $n$-huge with target $\lambda$iff $\lambda=j^n(\kappa)$ and $M$ is closed under all of its sequences of length less than $\lambda$ (that is, $M^{<\lambda}\subset M$). $\kappa$ is $n$-huge with target $\lambda$iff $\lambda=j^n(\kappa)$ and $M$ is closed under all of its sequences of length $\lambda$ ($M^\lambda\subset M$). $\kappa$ is almost $n$-hugeiff it is almost $n$-huge with target $\lambda$ for some $\lambda$. $\kappa$ is $n$-hugeiff it is $n$-huge with target $\lambda$ for some $\lambda$. $\kappa$ is super almost $n$-hugeiff for every $\gamma$, there is some $\lambda>\gamma$ for which $\kappa$ is almost $n$-huge with target $\lambda$ (that is, the target can be made arbitrarily large). $\kappa$ is super $n$-hugeiff for every $\gamma$, there is some $\lambda>\gamma$ for which $\kappa$ is $n$-huge with target $\lambda$. $\kappa$ is almost huge, huge, super almost huge, and superhugeiff it is almost $1$-huge, $1$-huge, etc. respectively. Ultrafilter definition
The first-order definition of $n$-huge is somewhat similar to measurability. Specifically, $\kappa$ is measurable iff there is a nonprincipal $\kappa$-complete ultrafilter, $U$, over $\kappa$. A cardinal $\kappa$ is $n$-huge iff there is some cardinal $\lambda$, a nonprincipal $\kappa$-complete ultrafilter, $U$, over $\mathcal{P}(\lambda)$, and cardinals $\kappa=\lambda_0<\lambda_1<\lambda_2...<\lambda_{n-1}<\lambda_n=\lambda$ such that:
$$\forall i<n\forall x\subseteq\lambda(ot(x\cap\lambda_{i+1})=\lambda_i\rightarrow x\in U)$$
Where $ot(X)$ is the order-type of the poset $(X,\in)$. [1] This definition is, more intuitively, making $U$ very large, like most ultrafilter characterizations of large cardinals (supercompact, strongly compact, etc.).
Consistency strength and size
Hugeness exhibits a phenomenon associated with similarly defined large cardinals (the $n$-fold variants) known as the
double helix. This phenomenon is when for one $n$-fold variant, letting a cardinal be called $n$-$P_0$ iff it has the property, and another variant, $n$-$P_1$, $n$-$P_0$ is weaker than $n$-$P_1$, which is weaker than $(n+1)$-$P_0$. [2] In the consistency strength hierarchy, here is where these lay (top being weakest): measurable = $0$-superstrong = almost $0$-huge = super almost $0$-huge = $0$-huge = super $0$-huge $n$-superstrong $n$-fold supercompact $(n+1)$-fold strong, $n$-fold extendible $(n+1)$-fold Woodin, $n$-fold Vopěnka $(n+1)$-fold Shelah almost $n$-huge super almost $n$-huge $n$-huge super $n$-huge $(n+1)$-superstrong
All huge variants lay at the top of the double helix restricted to some natural number $n$, although each are bested by I3 cardinals (the critical points of the I3 elementary embeddings). In fact, every I3 is preceeded by a stationary set of $n$-huge cardinals, for all $n$. [1]
Similarly, every huge cardinal $\kappa$ is almost huge, and there is a normal measure over $\kappa$ which contains every almost huge cardinal $\lambda<\kappa$. Every superhuge cardinal $\kappa$ is extendible and there is a normal measure over $\kappa$ which contains every extendible cardinal $\lambda<\kappa$. Every $(n+1)$-huge cardinal $\kappa$ has a normal measure which contains every cardinal $\lambda$ such that $V_\kappa\models$"$\lambda$ is super $n$-huge". [1]
In terms of size, however, the least $n$-huge cardinal is smaller than the least supercompact cardinal. Assuming both exist, for any $\kappa$ which is supercompact and has an $n$-huge cardinal above it, there are $\kappa$ many $n$-huge cardinals less than $\kappa$. [1]
Every $n$-huge cardinal is $m$-huge for every $m\leq n$. Similarly with almost $n$-hugeness, super $n$-hugeness, and super almost $n$-hugeness. Every almost huge cardinal is Vopěnka (therefore the consistency of the existence of an almost-huge cardinal implies the consistency of Vopěnka's principle). [1]
References Kanamori, Akihiro. Second, Springer-Verlag, Berlin, 2009. (Large cardinals in set theory from their beginnings, Paperback reprint of the 2003 edition) www bibtex The higher infinite. Kentaro, Sato. Double helix in large large cardinals and iteration ofelementary embeddings., 2007. www bibtex
|
Search
Now showing items 1-1 of 1
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
|
Yes, this should be solvable and should be doable in a reasonable amount of computation time, using a pretty cool homomorphic cryptosystem.
Here is one approach: the participants jointly pick a random number $y$, publicly commit to $y$, and then they all prove/check in zero knowledge that $y$ is different from their numbers. If it isn't, they go back to the beginning and try again, until they have found a choice that differs from all their numbers.
The zero-knowledge proof part can be made efficient using the following paper:
Boneh, Goh, and Nissim build a public-key encryption algorithm $E(\cdot)$ that allows you to evaluate any 2-DNF formula on encrypted values. In other words, if you have $E(x_1),\dots,E(x_m)$, where $x_1,\dots,x_m$ are boolean values, and if you have a 2-DNF formula $\Psi$, then you can compute $E(\Psi(x_1,\dots,x_m))$ from $E(x_1),\dots,E(x_m)$. The computation can be done quite efficiently. This is a very cool kind of homomorphic property, and it is very useful here.
In particular, if we have two numbers $y,z$, notice that the condition $y \ne z$ is a 2-DNF formula on the bits of $y,z$: $y\ne z$ is equivalent to
$$(y_1 \land \neg z_1) \lor (\neg y_1 \land z_1) \lor \dots \lor (y_\ell \land \neg z_\ell) \lor (\neg y_\ell \land z_\ell),$$
where $y_i$ is the $i$th bit of $y$ and $z_i$ the $i$th bit of $z$.
Moreover, if we have $n+1$ numbers $y,z_1,\dots,z_n$, and the ciphertexts of each bit of each number were encrypted separately using the Boneh-Goh-Nissim, we can check the condition $(y \ne z_1) \land \dots \land (y \ne z_n)$. Here is how. Let the bit $b_i$ be $1$ if $y \ne z_i$, and $0$ otherwise. Notice that we can compute $E(b_i)$ and the encryptions of the bits of $y,z_i$, using the ideas above and the homomorphicity of $E(\cdot)$. Define $k=b_1+b_2+\dots+b_n$. Notice that we can also compute the ciphertext $E(k)$ from $E(b_1),\dots,E(b_n)$, by the homomorphism properties of $E(\cdot)$. Now the condition $(y \ne z_1) \land \dots \land (y \ne z_n)$ holds if and only if $k=n$. Thus, to check the condition, we can decrypt $E(k)$ and check whether we get $n$ or not.
In addition, in the case where we know $z_1,\dots,z_n$ are distinct, the procedure does not reveal anything about the underlying $z_1,\dots,z_n$, beyond whether $y$ is equal to one of them. That's because $k$ will be either $n$ or $n-1$; it will be $n$ if $y$ is different from all of them, or $n-1$ if $y$ is equal to one of them. So the decryption of $E(k)$ does not disclose anything beyond whether $y$ is equal to one of the $z_i$'s (and if it is, it does not disclose which $z_i$ it is equal to).
Therefore, we immediately obtain a solution, using a threshold version of the Boneh-Goh-Nissim scheme:
The parties jointly generate a public/private keypair for the Boneh-Goh-Nissim cryptosystem, such that each party ends up with a share of the private key, no party (or coalition of parties) knows the private key, and everyone knows the public key. Their paper describes how to do that. This only needs to be done once.
Each party writes his/her private number in binary notation, separately encrypts each bit with the Boneh-Goh-Nissim cryptosystem, and then publishes all of those ciphertexts. In more detail, call party $i$'s secret number $z_i$. Write it in binary as the bits $z_{i,1},\dots,z_{i,\ell}$, then publish $E(z_{i,1}),\dots,E(z_{i,\ell})$, where $E(\cdot)$ represents Boneh-Goh-Nissim encryption under the public key established in step 1. Each party separately does this. This is also a one-time step that never needs to be repeated.
The parties now jointly generate a random number $y$, using standard methods. They encrypt each bit of $y$ separately, using the Boneh-Goh-Nissim scheme. (For instance, you can have one party do the encryption and then prove that he/she did it correctly.)
Now, check whether $y$ is equal to any of $z_1,\dots,z_n$. This can be done using the techniques described above. They can all compute $E(k)$, then they use the threshold decryption procedure to jointly decrypt $E(k)$ and learn $k$. The value of $k$ (either $n$ or $n-1$) reveals whether $y$ is different. If $y$ is different, they are done, and they output $y$. Otherwise, they go back to step 3 and try again.
How long will this procedure take? If there are $n$ parties, and the range of values covers $x$ possible values, then each iteration has a $1 - n/x$ probability of success and a $n/x$ probability of leading to a retry. Therefore, on average we need to retry $x/n$ times. For instance, if you're trying to pick a random number from $[0..200]$, then as long as there are no more than 100 parties, on average only 2 iterations are needed.
How much side information is disclosed? None of the parties ever discloses their secret number. However, any time we have to retry the procedure, an eavesdropper does learn that the number $y$ we were using must be the secret number of one of the parties (but the eavesdropper doesn't learn which party it was). This is a downside. This downside can be removed using more sophisticated methods (e.g., a way for the parties to jointly choose the random number $y$ without any of them knowing the value of $y$, but they all know the encryption of the bits of $y$), if you care.
As DrLecter helpfully points out:
BGN requires pairings over composite order bilinear groups (which are terribly slow to work with), [so] it may be worth to mention Freemans' product group approach, that among others allows to transfer BGN into a prime order setting and nicely increases performance by several orders of magnitude.
|
This question already has an answer here:
I'm a beginning student of Probability and Statistics and I've been reading the book
Elementary Probability for Applications by Rick Durret.
In this book, he outlines the 4 Axioms of Probability.
For any event $A$, $0 \leq P (A) \leq 1$. If $\Omega $ is the sample space then $P (\Omega) =1$. If $A$ and $B$ are disjoint, that is, the intersection $A \cap B = \emptyset$, then $$P(A\cup B) = P(A) + P(B)$$ If $A_1, A_2,\ldots$, is an infinite sequence of pairwise disjoint events(that is, $A_i\cap A_j = \emptyset$ when $i \neq j $) then $$P\left(\bigcup_{i=1}^\infty A_i\right)=\sum_{i=1}^\infty P(A_i).$$
The book fails to explain why we need Axiom 4. I have tried searching on Wikipedia but I haven't had any luck. I don't understand how we can have a probability of disjoint infinite events. The book states that when the you have infinitely many events, the last argument breaks down and this is now a new assumption. But then the book states that we need this or else the theory of Probability becomes useless.
I was wondering if there were any intuitive examples of situations where this fourth axiom applies.
Why is it so important for probabilty theory? And why does the author state that not everyone believes we should use this axiom.
|
Let $1\leq p < \infty$ and consider a sequence $(x_{n})_{n}\subseteq L^{p}[0,1]$. Show the equivalence of:
$1.$ $x_{n} \xrightarrow{ w} 0$
$2.$ $\sup\limits_{n \in \mathbb N} \vert \vert x_{n}\vert\vert_{p}<\infty $ and $\int_{A}x_{n}(t)dt\xrightarrow{n \to \infty} 0$ for any borel sets $A$ on $[0,1]$.
for $1. \Rightarrow 2.$
note that for any $\ell \in (L^{p})^{*}$ we have $\ell(x_{n})\xrightarrow{n \to \infty} 0$ and hence:
$\sup\limits_{n \in \mathbb N}\vert \ell(x_{n})\vert<\infty$ for any $\ell \in (L^{p})^{*}$ but now for $1<p < \infty$ we know that $L^{p}$ is reflexive, so we can write: $\sup\limits_{n \in \mathbb N} \vert \vert x_{n}\vert\vert_{p}=\sup\limits_{n \in \mathbb N} \vert \vert Jx_{n}\vert\vert_{*}<\infty$ by the uniform boundedness principle and the fact that $\sup\limits_{n \in \mathbb N} \vert \ell (x_{n})\vert= \sup\limits_{n \in \mathbb N}\vert Jx_{n}(\ell)\vert$ by reflexiveness. But we have now only shown this for $1< p < \infty$, since $L^{1}$ is not reflexive. How do we show this when $p=1$?
For "$\int_{A}x_{n}(t)dt\xrightarrow{n \to \infty} 0$ for any borel sets $A$ on $[0,1]$", consider any $1 \in L^{q}[0,1]$, note that because of Riesz there has to exist a $\ell \in (L^{p})^{*}$ so that $\ell(x)=\int 1 x(t)dt $ for any $x\in L^{q}$. Then it is clear that $\ell(x_{n})=\int_{0}^{1}x_{n}(t)dt\xrightarrow{n \to \infty}0$ by assumption. But how can I show that $\int_{A}x_{n}(t)dt\xrightarrow{n \to \infty}0$ for any Borel set $A$? Particularly since I have no assumption on whether the $(x_{n})_{n}$ are positive functions?
for $2. \Rightarrow 1.$
note that for any $\ell \in (L^{p})^{*}$ we can find a unique $y \in L^{q}$ where $\frac{1}{p}+\frac{1}{q}=1$ so that:
Note that $\vert \ell(x_{n})\vert =\vert\int_{0}^{1}y(t)x_{n}(t)dt\vert\leq \sup\limits_{n \in \mathbb N} \vert \vert x_{n}\vert\vert_{p}\vert\vert y\vert\vert_{q}<\infty$ and can I state from $\int_{A}x_{n}(t)dt\xrightarrow{n \to \infty} 0$ for any borel sets $A$ that $x_{n} \to 0$ almost everywhere, but I am not sure as the assumption of positivity of $x_{n}$ is once again missing. If I could use this and dominated convergence, then
$\lim\limits_{n\to \infty}\ell(x_{n})=\lim\limits_{n\to \infty}\int_{0}^{1}y(t)x_{n}(t)dt=0$ and we are done.
|
Today it was told me that wave properties of a particle increase if the temperature decreases. I'm surprised because I have never listened a similar thing, but I think that it's very interesting.
Could you explain me why it happens?
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up.Sign up to join this community
Today it was told me that wave properties of a particle increase if the temperature decreases. I'm surprised because I have never listened a similar thing, but I think that it's very interesting.
Could you explain me why it happens?
I think that your teacher (?) asked you about thermal de Broglie wavelength, where $$\lambda_T \propto\frac{1}{\sqrt{T}}.$$ You get this expression when you express the momentum in $\lambda=h/p$ in in terms of kinetic energy and the kinetic energy itself in terms of the energy due to temperature. (The derivation is also in the wikipedia article...) Indeed, when you drop the temperature the thermal de Broglie wavelength increases.
As far as I understand it, the thermal de Broglie wavelength indicitates
roughly when you have to treat a gas clasically or quantum mechanically. In the sense that when $\lambda_T$ is much smaller than the particle separation then the particles "have to do with each other" and if the wavelength is approximately the same as the particle separation then the particles "have to do with each other". Depending on how they "have to do with each other" you then have to treat the gas as a Bose- or a Fermi gas.
Some time ago I read about the interference of Bose-Einstein-condensates. I thought that in this context this concept could be applicable and in fact this is true: when you, for example, quickly read through this paper, which describes work which was awarded the Nobel prize in 2001, then you will find the sentence (page three)
The critical number of atoms ... is determined by the condition that the number of atoms per cubic thermal de broglie wavelength exceeds... .
So even if it's just for estimation purposes, this concept is surely used in research and definitely not "nonsense".
|
I am analyzing the asymptotic runtime of a randomized algorithm in expectation. The algorithm has the following properties:
Given input size $n$, with probability $3/4$ it moves on to solve an instance of size $n-1$ With probability $1/8$ it moves on to solve an instance of size $n-2$ With probability $1/16$ it moves on to solve an instance of size $n-3$ With probability $1/2^i$ it moves on to solve an instance of size $n-i$ Each instance pays a cost of $O(n)$, where $n$ is the input size of that instance
Over expectation, the runtime can be defined recursively as follows:
$T(n) = O(n) + \sum\limits_{i=0}^{n-1} (\dfrac{1}{2^{n-i+1}} T(i)) + \frac{1}{2}T(i-1)$
$T(0) = O(1)$
I have calculated that the expected number of "jumps" at each stage is $\leq 1$. I did this by showing $\sum\limits_{i=0}^\infty \dfrac{i}{2^{i+1}}= 1$ by using telescoping and geometric series. However, since the complexity at each stage diminishes as $n$ gets smaller, although this hints the runtime is $O(n^2)$, it does not prove it. Anyone have any ideas to prove a runtime for the less relaxed version?
EDIT: Slight gap in my formulation. The "$3/4$" probability for moving onto an instance of size $n-1$ should actually be larger than $3/4$ since the probabilities $1/2, 1/4, 1/8, ...$ only go on till $1/2^{n+1}$. If no jumps were made, the algorithm deterministically moves on to an instance of size $n-1$.
|
Equivalence of Definitions of Local Basis Contents Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $x$ be an element of $S$.
$\forall U \in \tau: x \in U \implies \exists H \in \mathcal B: H \subseteq U$ Proof $\forall U \in \tau: x \in U \implies \exists H \in \mathcal B: H \subseteq U$
Let $N$ be a neighborhood of $x$.
Then there exists $U \in \tau$ such that $x \in U$ and $U \subseteq N$ by definition.
By assumption, there exists $H \in \mathcal B$ such that $H \subseteq U$.
From Subset Relation is Transitive, $H \subseteq N$.
The result follows.
$\Box$
every neighborhood of $x$ contains a set in $\mathcal B$.
Let $U \in \tau$ such that $x \in U$.
By assumption, there exists $H \in \mathcal B$ such that $H \subseteq U$.
The result follows.
$\blacksquare$
|
As with most people that have studied mathematics beyond high school, I first encountered implicit differentiation in my initial semester of calculus. It seemed like magic then. In the intervening years, I have seen the idea and the underlying theory pop up again a few times, and it still seems like some kind of magic. I have worked through the appropriate proofs, and am continually astonished that it works. I probably shouldn’t admit this in public, but even today I feel like I have no intuition regarding the proof. The whole argument is a bit of a black box.
On the bright side, I didn’t have to prove the theorem to my calc I students (and, frankly, the proof probably shouldn’t be done in public—it should only be performed behind closed doors by small groups of hooded monks as a kind of hazing ritual for young math majors). For them, it is sufficient to see the power of the result.
What I Taught
To get some idea of the incredibly fast pacing of the class, here is a very brief synopsis of what went on: I gave a quiz on trig functions (20 minutes), discussed homework problems which addressed the chain rule (25 minutes), introduced implicit differentiation (30 minutes), and used the theorem to derive the derivatives of a couple of inverse trig functions (20 minutes). In the normal course of a semester-long class, this probably would have been split over two or three days, with both the quiz and the homework questions taking place in a recitation section run by a grad student. Whee!
The main thrust of the day was introducing implicit differentiation. The basic idea is that an equation like \(x^2+y^2=1\) defines a curve in the plane (in this case, a circle of radius 1 centered at the origin). It is reasonable to ask what the slope of a line tangent to that curve at a particular point will be. Using the techniques developed prior to this moment, this involves solving the equation for \(y\) (which gives two results in this case—we have to choose the correct result before continuing), finding the derivative function, and evaluating that function for the value of \(x\) that is of interest.
This can be quite tedious, and can also be essentially impossible (consider a fifth degree polynomial in \(x\) and \(y\), for instance, which cannot be solved by radicals). Thus, instead of solving for \(y\) as an explicit function of \(x\), we note that the original equation implicitly defines a function in terms of \(x\). That is, the equation tells us that \(y=f(x)\) for some function \(f\), even if we don’t know exactly what the function is. (The existence and differentiability of \(f\) is the part that is pure magic, by the way.) To find the slope, we note that if \(x^2 + y^2 = 1\), then the derivatives of the left- and right-hand sides must also be equal, i.e.
\[ \frac{d}{dx}(x^2+y^2) = \frac{d}{dx}1. \] Since \(y\) is a function of \(x\), we apply chain rule to obtain \[ 2x + 2yy’ = 0, \] then solve for the derivative, getting \[ y’ = \frac{-x}{y}. \] Now finding the slope is a “simple” matter of substitution.
In addition to this example, I also worked through the folium of Descarte and an example involving an astroid (not an ast
eroid, but an astroid; namely \(x^{2/3}+y^{2/3} = 4\)). I finished up by showing that \[ \frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1-x^2}} \qquad\text{and}\qquad \frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}. \] I stated the derivatives for the remaining inverse trig functions without further derivation. What Worked
When discussing the folium of Descarte, I worked through much of the problem explicitly using Maple. The goal was to impress upon the students how difficult a problem it is without implicit differentiation. When I issued the command for Maple to solve the equation for \(y\), the old computer that I was working on chugged for a bit, then displayed 10 lines of output, all gnarly notation with lots of radicals. It looked quite impressive, and made my point quite well, I think.
I am also pleased with the time management. While there was a lot to do, I never felt rushed, had time to address questions and concerns, and still managed to wrap up everything with two or three minutes to spare at the end of class to go over administrative stuff.
What Didn’t Work
I am concerned that the lecture was a little scattered. I generally try to model clear, linear derivations and arguments, as those are the kinds of things that I want to see on quizzes and exams. However, there are several points in the lecture when I seem to have done too much algebra all at once, or skipped some minor simplification detail to the confusion of the students. This often involved backtracking to fill in details in the margins of the board, which was neither clear nor linear.
The biggest problem was taking the derivative of functions of \(y\), where \(y\) is understood to be a function of \(x\). In the first problem, I began by very explicitly writing \(y = f(x)\), where \(f\) is some unknown function. Then \(y^2 = [f(x)]^2\), and, using the chain rule, we have
\[ \frac{d}{dx} y^2 = \frac{d}{dx} [f(x)]^2 = 2f(x)\cdot\frac{d}{dx}f(x) = 2f(x)f'(x). \] Recalling that \(f(x) = y\), this implies that \(\frac{d}{dx} y^2 = 2yy’\). I had planned to go through all of these details only on one problem, then elide them a bit (with more elision as the class went on) on future problems. Turns out, this was a bad plan. Every time that we took the derivative of some function of \(y\), we had to work through all of these details. Every. Single. Time.
Had I planned for it, I would have left space on the board. I didn’t, and it became marginalia, probably causing as much confusion as it cured. On the bright side, there will be homework questions to answer during the next class, so hopefully some of that confusion can be cleared up.
|
State functions such as $G$ only depend on the state of the system and are not dependent on the "path" that took the system to that state (which would be the case for work, for example, which is not a state function.
We know that:
$$G=V\mathrm{d}p-S\mathrm{d}T$$ So... $$G=\left(\frac{\partial G}{\partial p}\right)_T\mathrm{d}p+\left(\frac{\partial G}{\partial T}\right)_p\mathrm{d}T$$ Consequently, by comparing coefficients:
$$V=\left(\frac{\partial G}{\partial p}\right)_T$$ and $$-S=\left(\frac{\partial G}{\partial T}\right)$$
Just taking the equation involving $V$ now to save time and space: $$\int_{p_1}^{p_2}V\mathrm{d}p=\int_{p_1}^{p_2}\left(\frac{\partial G}{\partial p}\right)_T\mathrm{d}p$$
Using the perfect gas equation and integrating leaves the result:
$$G(p_2)-G(p_1)=n\mathcal{R}T\ln\left(\frac{p_2}{p_1}\right)$$
But if $G$ is independent of the path taken to get to the final state why shouldn't I use the equation: $$\Delta G=V(p_2-p_1)$$
|
I am wondering to what degree we can define an RSA variant, with a security argument that it is as safe as regular RSA with a given modulus size $m$ (e.g. $m=2048$), in which the
public key has a compact representation of $k \ll m$ bits.
We can fix the public exponent to our favorite customary value, e.g. $e=2^{16}+1$ or $e=3$, thus need to store only the public modulus $n$. We need not store the leftmost bit of $n$, which is always set by definition; nor the rightmost bit, which is always set since $n$ is odd. With a little effort, we could save (very) few more bits noticing $n$ has no small divisors, but that will still be $k \sim m$ bits.
We can do better by forcing the $\lfloor m/2-log_2(m)-2\rfloor$ high bits of $n$ to some arbitrary constant such as $\lfloor\pi \cdot 2^{\lfloor m/2-log_2(m)-4\rfloor}\rfloor$. Observe that we can chose the smallest prime factor $p$ of $n$ just as we would do in regular RSA, then find the maximum integer interval $[q0,q1]$ such that any $q$ in that interval cause $n=p\cdot q$ to have the right high bits, then pick a random prime $q$ in that interval (most often there will be at least one, if not we try another $p$). Some of the security argument is that
generating an RSA key $(p,q')$ using a regular method, with random huge primes in some appropriate range and no other criteria beside the number of bits in $n'=p\cdot q'$, and $p<q$; then deciding the high bits of $n$ from those in $n'$; then finding $[q0,q1]$, generating $q$ as a random prime in that interval, and setting $n=p\cdot q$;
demonstrably gives the same distribution of $(p,q)$ as said regular generation method, hence is as secure; then we remark that the high bits of $n$ are random (with some distribution not too far from uniform), and public, thus fixing it can't much help an attack (I think this can be made rigorous).
This is now $k \sim m/2+log_2(m)$ bits to express the public key. We can save a few more bits, each one at worse doubles the amount of work to generate the private key (we can repeat the generation process outlined above until we find a key with these bits equal to some public arbitrary constant; or equal to bits from a hash of the other bits if we want a tighter assurance that the scheme is not weakened).
Can we do better, and what's the practical limit?
Update: In RSA Moduli with a Predetermined Portion: Techniques and Applications, Marc Joye discuses this, and reaches $k \sim m/3$ (without claim of optimality). I'm worried that I do not see an argument that the manner in which $(p,q)$ are selected in
Algorithm 3 does not weaken $n$ against a dedicated factoring algorithm.
|
I want to store a subset of $\{1,2,\dots,n\}$ in a data structure such that the total space used is $O(n)$ bits and accessing a particular element can be done in $O(1)$ time. I have tried binary search tree and array data structures but they are too slow.
In short I need a data structure with $O\big(\frac {n } { \log \log n}\big)$ many cells and size of each cell should be $O(\log \log n)$ bits.
Model of computation: Input memory and a write only output memory. The computation proper takes place in a working memory of limited size. When stating a problem can be solved with in a certain bits, what we mean is that the working memory comprises that many bits. The input and working memories are divided into words of $w$ bits for a fixed parameter $w$, arithmetic and logical operations on $w$-bit words take constant time, and random access to the input and working memories is provided. In the context of inputs of $n$ words,$w= \theta(\log n)$. Operations: Insert($i$), access($i$), where $i \in [n]$. The access operation must run in $O(1)$ time. Motivation: I'd like to store the stack in DFS using this data structure.
I am looking for an deterministic approach, but to me it does not seems working, so probabilistic approach is also fine.
|
Let $n,m \in \mathbb{N}$ $$n=\prod_{i=1}^{r}p_{i}^{a_i}$$ where $p_i$ are prime factors and $f$ , $g$ and $h$ are the functions $$f(n,m)=\sum_{j=1}^{n}j^m$$ And $$g(n)=\sum_{i=1}^{r}a_i.p_i$$ If we put $m=1,n=21$ then $$g(f(21,1))=g(231)=21.$$
21 is only number satisfy $g(f(n,1))=n$.
Now let
$$h(m) = \sum_{g(f(n,m))=n}1$$
So $h(1)=1$.
Question
If $m$ have finite $n$ satisfied $g(f(n,m))=n$then what is formula for $h(m)$?
Can we prove there are infinitely many $m$ satisfying above statement?
|
Combination Theorem for Sequences/Normed Division Ring/Product Rule/Proof 2 Theorem
Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring.
Let $\sequence {x_n}$, $\sequence {y_n} $ be sequences in $R$.
$\displaystyle \lim_{n \mathop \to \infty} x_n = l$ $\displaystyle \lim_{n \mathop \to \infty} y_n = m$
Then:
$\sequence {x_n y_n}$ is convergent to the limit $\displaystyle \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$ Proof
Because $\sequence {x_n}$ is convergent, it is a bounded by Convergent Sequence in Normed Division Ring is Bounded.
Suppose $\norm {x_n} \le K$ for $n = 1, 2, 3, \ldots$.
Then for $n = 1, 2, 3, \ldots$:
\(\displaystyle \norm {x_n y_n - l m}\) \(=\) \(\displaystyle \norm {x_n y_n - x_n m + x_n m - l m}\) \(\displaystyle \) \(\le\) \(\displaystyle \norm {x_n y_n - x_n m} + \norm {x_n m - l m}\) Axiom (N3) of norm (Triangle Inequality) \(\displaystyle \) \(=\) \(\displaystyle \norm {x_n} \norm {y_n - m} + \norm {x_n - l} \norm m\) Axiom (N2) of norm (Multiplicativity) \(\displaystyle \) \(\le\) \(\displaystyle K \norm {y_n - m} + \norm m \norm {x_n - l}\) since $\sequence {x_n}$ is bounded by $K$ \(\displaystyle \) \(=:\) \(\displaystyle z_n\)
We note that $\sequence {z_n}$ is a real sequence.
But $x_n \to l$ as $n \to \infty$.
So $\norm {x_n - l} \to 0$ as $n \to \infty$ from Definition:Convergent Sequence in Normed Division Ring.
Similarly $\norm {y_n - m} \to 0$ as $n \to \infty$.
From the Combined Sum Rule for Real Sequences:
$\displaystyle \lim_{n \mathop \to \infty} \paren {\lambda x'_n + \mu y'_n} = \lambda l' + \mu m'$, $z_n \to 0$ as $n \to \infty$
By applying the Squeeze Theorem for Sequences of Complex Numbers (which applies as well to real as to complex sequences):
$\sequence {\norm {x_n y_n - l m}}$ converges to $0$ in $\R$.
By definition of a convergent sequence in a normed division ring:
$\sequence{x_n y_n}$ is convergent in $R$
It follows that:
$\displaystyle \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$
$\blacksquare$
|
So here in the picture \$e^{jwt}\$ is the input to the system and \$h(t)\$ is the impulse response. So, by convolution integral shouldn't the response be \$h(t-T)e^{jwT} dT\$? But here it is \$h(T)e^{jw(t-T)}dT\$. What am I missing here?
Nothing. Just do a variable change \$\tau'=t-\tau\$ and you'll get your integral.
In the convolution integral we're used to seeing \$h(t - \tau)\$ when talking about impulse response \$h(t)\$, however:
$$\int_{-\infty}^\infty{f(\tau)}g(t - \tau)d\tau = \int_{-\infty}^\infty{f(t - \tau)}g(\tau)d\tau $$
In this case the left side is used with \$ f(\tau) = h(\tau)\$ and \$ g(t - \tau) = e^{j\omega_k(t - \tau)}\$ so you have:
$$\int_{-\infty}^\infty{f(\tau)}g(t - \tau)d\tau = \int_{-\infty}^\infty{h(\tau)}e^{j\omega_k(t - \tau)}d\tau = e^{j\omega_kt}\int_{-\infty}^\infty{h(\tau)}e^{-j\omega_k\tau}d\tau$$
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.