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http://www.luschny.de/math/factorial/hadamard/HadamardsGammaFunctionMJ.html | Here is an alternative version which might by better suited for old browsers and does not require JavaScript enabled.
Is the Gamma function misdefined? Or: Hadamard versus Euler — Who found the better Gamma function?
Abstract. The Gamma function of Euler often is thought as the only function which interpolates the factorial numbers $n! = 1,2,6,24,\ldots$ This is far from being true. We will discuss four factorial functions
• the Euler factorial function $n!$,
• the Hadamard Gamma function $H(n)$,
• the logarithmic single inflected factorial function $L(n)$,
• the logarithmic single inflected factorial function $L^{*}(n)$.
We will show that these alternative factorial functions posses qualities which are missing from the the conventional factorial function but might be desirable in some context.
The Bernoulli-Euler Gamma function
The problem of interpolating the values of the factorial function $\,$ $n! = 1,2,6,24, \ldots$ was first solved by Daniel Bernoulli and later studied by Leonhard Euler in the year 1792 (see [history]). The interpolating function is commonly known as the Gamma function.
However, this function has an unpleasant property: If $n=0,-1,-2,\ldots$ then $\Gamma(n)$ becomes infinite.
What is not so well known is the fact, that there are other functions which also solve the interpolation problem for $n!$ and behave much nicer. A particular nice one was given by Jacques Hadamard in 1894.
Hadamard's gamma function has no infinite values and is provable simpler then Euler's gamma function in the sense of analytic function theory: it is an entire function.
The following figures give a first idea what the Hadamard gamma function looks like.
Hadamard's Gamma function is defined as
$\operatorname{H}(x)= \frac{1}{\Gamma(1-x) }\frac{d}{dx}\ln \frac{\Gamma \left(1/2-x/2 \right)}{\Gamma \left(1- x/2 \right)}$
If we introduce the Digamma (Psi) function
$\Psi(x)= \frac{d}{dx} \ln \Gamma(x)$
we can write in equivalent form
$\operatorname{H}(x) = \frac{\Psi(1-x/2)-\Psi(1/2-x/2)}{2\Gamma(1-x)}$
There is a second approach which shows the intuitive meaning of Hadamard’s function more clearly. Let us define the function $\operatorname{Q}(x)$ by the next expression:
$\operatorname{Q}(x)=1+\frac{\cos \pi x}{2\pi} \left( \Psi(\frac14 + \frac{x}{2} )- \Psi(\frac34 + \frac{x}{2} \right)$
The graphs of $\operatorname{Q}(x)$ [red] and of $1 - \operatorname{Q}(x)$ [green] are shown in the next plot:
The graph of $\operatorname{Q}(x)$ looks like an approximation to the indicator function of the positive real numbers. Now let us consider the product of this function and $\Gamma(x)$. Then we have the following identity for Hadamard's Gamma function:
$\operatorname{H}(x) = \Gamma(x) \operatorname{Q}(x-1/2)$
For $x=0,-1,-2,-3,\ldots\ \operatorname{H}(x)$ is defined by the value of the limit.
A new factorial function L(x)
There is another factorial function, proposed by the current author in October 2006, which is also continuous at all real numbers and which we will compare to Hadamard's Gamma function.
The following figure compares the Luschny factorial $\operatorname{L}(x)$ with the Euler factorial $(x! = \Gamma(x+1))$ for nonnegative real values.
Crucial for the definition of the L-factorial are the functions
$g(x) = \frac{x}{2} \left(\Psi\left(\frac{x}{2} + \frac12 \right) - \Psi\left( \frac{x}{2} \right)\right) - \frac12 \quad [*]$ $\operatorname{P}(x) = 1 - g(x) \frac{\sin(\pi x)}{ \pi x}. \qquad\qquad \qquad [**]$
We set $\operatorname{P}(0) = \lim_{x \rightarrow 0} \operatorname{P}(x) = 1/2$. The graph of $\operatorname{P}(x)$ is shown in the next plot:
The definition of the L-factorial [7] is
$\operatorname{L}(x) = x! \operatorname{P}(x)\ .$
For $x=0,-1,-2,-3,... \operatorname{L}(x)$ is defined by the value of the limit. Note that $\operatorname{L}(-x) = (-x)! (1 - \operatorname{P}(x))$.
Luschny's factorial function $\operatorname{L}(x)$ approximates the non integral values $x > 0$ of the factorial function better then Hadamard's Gamma function $\operatorname{H}(x)$ the non integral values $x > 0$ of the Gamma function. Compare figure 7.
A family of approximate gamma functions.
To understand the relation between Hadamard's Gamma function and Luschny's factorial function better we consider a common generalization. To this end we introduce
$g(x, \alpha) = (x/2) (\Psi(x/2 + 1/2) - \Psi(x/2)) - \alpha/2,$ $\operatorname{P}(x, \alpha) = 1 - g(x, \alpha) \sin(\pi x) / (\pi x),$ $\operatorname{L}(x, \alpha) = x! \operatorname{P}(x, \alpha).$
Obviously $\operatorname{L}(x) = \operatorname{L}(x, 1)$ and it is easy to see from the functional equation of $\Psi$ that $\operatorname{H}(x+1) = \operatorname{L}(x, 2)$.
... and now for values $x \lt 0$
Starting from the idea of interpolating $n! = 1,2,6,24,\ldots$ one expects a factorial function for arbitrary real numbers x to be monotonically increasing for all x. It comes almost as a shock to see that the Euler factorial behaves very differently from this expectation for real numbers $x \lt 0$. On the other hand, the behavior of the L-factorial meets our expectations. The following plot compares the logarithm of both functions:
Convexity and the Bohr-Mollerup-Artin theorem
Hadamard looked at his formula from the point of view of analytic function theory, as is emphasized by the title of his paper "Sur L'Expression Du Produit $1.2.3\ldots(n-1)$ Par Une Fonction Entière" [2]. However, at the turn of the century (1900) new ideas came up and influenced the way the Gamma function was looked at. Philip J. Davis writes in his exposition [1, p.867] :
"Aesthetic conditions were not to be found in the older, analytic considerations, but in a newer, inner, organic approach to function theory which was developing at the turn of the century. Backed up by Cantor's set theory and an emerging theory of topology, the new function theory looked not so much at equations and identities as at the fundamental geometrical properties. The desired condition was found in notions of convexity."
Only many years after Hadamard defined his Gamma function this more geometric view of the Gamma function was fully understood. It cumulated 1922 in a characterization of the Gamma function by logarithmic convexity which is now known as the Bohr-Mollerup-Artin theorem.
The dints in Hadamard's gamma function
Let us now examine Hadamard's Gamma function from the geometric point of view.
Figure 9 shows the behavior of $\operatorname{H}(x)$ and $\operatorname{L}(x)$ with regard to logarithmic convexity. Whereas the logarithm of the L-factorial has only one inflection point — this is the point where $(\log(\operatorname{L}(x))''$ changes the sign near $x=1$ — Hadamard's Gamma function has many inflection point: $\operatorname{H}(x)$ is not logarithmic convex for $x > 1$. The regions where $(\log(\operatorname{H}(x))'' < 0$ and $x > 1$ might be visualized as small dints in Hadamard's function. And the presence of these dints might be regarded as a flaw in Hadamard's definition.
Moreover, in contrast to the H-Gamma function the L-factorial is not only free from dints but approximates the logarithmic behavior of the Euler Gamma function with positive values as $x \rightarrow \pm \infty$. The next figure shows $(\log(\operatorname{L}(x))''$ in comparison to $(\log(\Gamma(x))''\ \text{for } x\gt 0$ and $-(\log(\Gamma(-x))''\ \text{for }x<0$.
Single inflected logarithmic convexity
To conceptualize our findings we introduce the following definition.
Definition: A function is lsi-convex if and only if it is defined for all real $x$ and is logarithmic convex for $x > C$ for some $C$ and logarithmic concave for $x < C$. ('lsi' is an acronym for 'logarithmic single inflected'.)
As our discussion showed, the H-Gamma function is not lsi-convex, but the L-factorial is lsi-convex. This is not difficult to prove if we look at $\operatorname{V}(x) = (\log(\operatorname{L}(x))''$ and observe that if $c = 1.073536774\ldots.$ and $x \gt c$ then $V(x) \gt 0$ and $\operatorname{V}(x) \lt 0$ for $x \lt c$. In fact $\operatorname{V}(x) \gt 1/(x+2)$ for $x \gt 2$, $\operatorname{V}(x) \lt -1/x^2$ for $x \lt -3$ and the only zero of $\operatorname{V}$ in the interval $[-3, 2]$ is $c$. (Compare figure 11.)
If we choose $\alpha = 1.031474316\ldots$ in the generalized formula $\operatorname{L}(x, \alpha)$ the inflection point (point of change in concavity) is exactly 1.
A wish list for the factorial
The Euler factorial $x! = \Gamma(x + 1)$ is a function
[1] which is positive for $x > 0$,
[2] is 1 at $x = 1$,
[3] interpolates $n! \ (n>0 \text{ integer}),$
[4] and is logarithmically convex for $x > 0$.
However, contemplating Figure 8 we might also ask for a function
[1*] which is positive for all real $x$,
[2*] is 1 at $x = 1$,
[3*] interpolates $n! \ (n \gt 0 \text{ integer} )$,
[4*] and is logarithmically convex if $x \gt c$ and
logarithmically concave if $x \lt c$ (for some $c$).
Note that the postulates [n*] are a superset of the postulates [n]. The L-factorial satisfies the postulates [n*], but the H-Gamma function does not.
The approximate functional equationand an error estimate
Now what about the functional equation $\operatorname{F}(x+1)/((x+1)\operatorname{F}(x)) = 1$ for $x \gt 0$? This identity holds for the Euler factorial exactly but for the H-Gamma function and the L-factorial function only approximately. Figure 12 displays how wide they are off the unit.
In the case of the L-factorial this approximate functional equation is approximated by the function ($x \neq 0$)
$\operatorname{AFE}(x) = 1 + \frac{\sin(\pi x)}{2 \pi x}(x^{-1}-x^{-2})$
It is clear that $\operatorname{L}(x)$ is asymptotical equal to $x!$. The formula for the approximate functional equation might be translated into the error bound
$\frac{\operatorname{L}(x)}{x!} \lt 1 + \frac{1}{2 \pi x} \quad ( x \gt 1).$
The functional equation of $L(x, \alpha)$
Of course our factorial functions obey also a full functional equation, not only an approximate one. And the L-factorial does have a simple functional equation. This can be easily seen from the definition. We find
$\operatorname{L}(1+x)-(1+x)\operatorname{L}(x)$ $= (1/2) x! \sin(\pi x)/(\pi x)$ $= (1/2) / (-x)! .$
Thus the functional equation of the L-factorial is
$\operatorname{L}(1+x) = (1+x)\operatorname{L}(x) + \frac{1}{2(-x)!} .$
The general case is
$\operatorname{L}(1+x, \alpha) - (1+x)\operatorname{L}(x, \alpha)$ $= ((1-\alpha)x + 1 - \alpha/2) x! \sin(\pi x)/(\pi x)$ $= ((1-\alpha)x + 1 - \alpha/2)/(-x)!$
and gives the general functional equation
$\operatorname{L}(1+x,\alpha)=(1+x)\operatorname{L}(x,\alpha)+\frac{(1-\alpha)x+1- \alpha/2}{(-x)!} .$
From the results and methods of [3] we conclude that each meromorphic solution of this functional equation is a transcendental differential function over the differential field of complex rational functions.
The function $g(z)$ and the product $z!(-z)!$.
We prove the following identity
$g(z) + g(-z) = z! \ (-z)!$
Proof:
$g(z)+g(-z)$
$\, = \frac{z}{2}\left(\Psi\left(\frac{1}{2}+\frac{z}{2}\right)-\Psi\left(\frac{z}{2}\right)\right)-1- \frac{z}{2}\left(\Psi(\frac12-\frac{z}{2})-\Psi\left(\frac{z}{2}\right)\right)$
$=\frac{z \pi}{2} \tan\left(\frac{z \pi }{2}\right)\ + \frac{z \pi}{2} \cot\left( \frac{z \pi}{2}\right)$ $= \frac{\pi z}{\sin(\pi z)}$ $= z! (-z)!$
In the last step we used the well known identity (Ergänzungssatz, L. Euler, 1749)
$\frac{\sin(\pi z)}{\pi z} = \frac{1}{z! (-z)!}$
Another application of Euler's reflection equation is the identity
$\operatorname{L}(-z) = \frac{g(z)}{z!}$
Proof:
$\operatorname{L}(-z) = (-z)! \operatorname{P}(-z)$ $= (-z)! (1 - \operatorname{P}(z))$ $= (-z)! g(z) \sin(\pi z) / (\pi z)$ $= (-z)! g(z) / ((-z)! z!)$ $= g(z) / z!$
The reflection products $z!(-z)!$ and $L(z)L(-z)$.
Let us compare the reflection equation of the Euler factorial with the reflection equation of the L-factorial. To this end we introduce the function
$\Lambda(z) = \operatorname{L}(z)\operatorname{L}(-z)$
$\Lambda(z)$ has a Maclaurin series expansion in even powers which can be found in the appendix. The Lambda function is displayed in figure 14 as the upper envelope. In the next figure we compare the functions $z!(-z)!$ and $\operatorname{L}(z)\operatorname{L}(-z)$. From figure 15 it is clear that the most striking difference between these two functions is the behavior at integral values. We will look at these in the next section.
For now we observe that $\operatorname{L}(z) = z! \operatorname{P}(z)$ by definition and that $\operatorname{L}(-z) = (-z)! (1 - \operatorname{P}(z))$. Thus we can recycle the oscillating factor $\operatorname{P}(x)$ of $\operatorname{L}(x)$ and find the following reflection theorem for $L(z)$
$\Lambda(z) = g(z) \operatorname{P}(z)$
We remember (see formula [**]) that $\operatorname{P}(z)$ itself was defined via $g(z)$ as $\operatorname{P}(x) = 1 - g(x) \sin(\pi x) / (\pi x)$. This leads to the following form of the reflection theorem for $L(z)$
$\operatorname{L}(z) \operatorname{L}(-z) = g(z) - g(z)^2 \sin(\pi z) / (\pi z) \quad [+].$
If we plug Euler's reflection theorem for z! in this formula we finally arrive at the remarkable identity
$\frac{\operatorname{L}(z) \operatorname{L}(-z)}{g(z)} \ + \frac{g(z)}{z! (-z)!} \ =\ 1.$
This identity shows a kind of duality between the factorial function $z!$ and the L-factorial function $L(z)$.
Pushing these ideas a little further, we can also consider the function $\operatorname{LL}(z) = z ((t - s) / 2 - z s t) + 1/4$ where we have set $s = \operatorname{LerchPhi}(-1, 1, z)$ and $t = \operatorname{LerchPhi}(-1, 1, -z)$. Then we can prove the following identity, which is a kind of super-reflection formula.
$\operatorname{LL}(z) = \operatorname{L}(z) \operatorname{L}(-z) z! (-z)!$
We will return to this formula in the postscript of this note.
The values of L(-n), n > 0 integer.
If we define a factorial function for all real numbers then the values at the negative integers $n = -1,-2,-3,\ldots$ draw special attention. For the L-factorial we make two observations.
• $(-1)^n \operatorname{L}(-n) + \ \ln(2) / (n-1)!$
are rational numbers, which we denote by $\operatorname{R}(n)$.
• $\operatorname{R}(n) n! = n \operatorname{A}(n) + (-1)^n / 2 \ .$
Here $\operatorname{A}(n)$ denotes the alternating harmonic numbers, defined as
$\operatorname{A}(n) = \sum_{k=1}^{n} \frac{(-1)^{k + 1}}{ k} .$
The first few values are:
$\operatorname{A}(n) = 1, \frac{1}{2}, \frac{5}{6} , \frac{7}{12}, \frac{47}{60}, \frac{37}{60}, \frac{319}{420} \ldots \quad (n > 0)$ $\operatorname{R}(n) = \frac{1}{2}, \frac{3}{4}, \frac{1}{3}, \frac{17}{144}, \frac{41}{1440}, \frac{7}{1200} \ldots \quad (n > 0)$
But we already proved that $\operatorname{L}(-n) = g(n)/n!$, so we conclude from
$\operatorname{R}(n)\ n! = (-1)^n g(n) + n \ln(2)$ $\operatorname{A}(n)\ n\ = (-1)^n \left(g(n) - \frac{1}{2} \right) + n \ln(2) \quad (n > 0)$
Conversely, if we assume the alternating harmonic numbers as pre-computed, we find $\operatorname{L}(-n)$ via
$\operatorname{L}(-n) = \frac{(-1)^n (\operatorname{A}(n) - \ln(2)) + 1/(2n)}{ (n-1)!}$
The relation L(z) to A(z), general case.
Let us generalize the alternating harmonic numbers to the alternating harmonic function
$\operatorname{A}(z) = \frac{1}{2} \cos(\pi z)\left(\Psi\left(\frac{z}{2}+\frac{1}{2}\right)-\Psi\left(\frac{z}{2}+1\right)\right)+\ln(2).$
Now we can express the L-factorial as
$\operatorname{L}(-z) = \frac{1}{z!}\left( \frac{z \operatorname{A}(z) - z \ln(2)}{ \cos(\pi z)} + \frac12 \right)$
Conversely, given $\operatorname{L}(x)$ the values of the alternating harmonic function can be obtained as
$\operatorname{A}(z) = (\cos(\pi z) / z) (\operatorname{L}(-z) z! - 1/2) + \ln(2).$
Since $g(z) = \operatorname{L}(-z)\ z!$ this can be simplified further giving the identity
$\operatorname{A}(z) = (\cos(\pi z) / z) (g(z) - 1/2) + \ln(2) \qquad (z \neq 0).$
The 'official' definition of L(z).
The identity analog to the definition of the Hadamard Gamma function is
$\operatorname{L}(z)= \frac{1}{(-z)!} \left( \frac{1}{2} + z \ {\operatorname{d}\over\operatorname{d}z} \ln \left( \frac{(-z/2-1/2)!}{(-z/2)!} \right) \right) \qquad [*]$
This will be henceforward the 'official' definition of the L-factorial. We show how to derive our 'working' definition. Using the digamma ($\Psi$) function [*] can written as
$\operatorname{L}(z)= \frac{1}{(-z)!} \left( \frac{1}{2} + \frac{z}{2} \left( \Psi\left(1- \frac{z}{2} \right) - \Psi \left(\frac12 - \frac{z}{2} \right) \right) \right)$
Because $\Psi(1 - z/2) = \Psi(-z/2) - 2/z$ we find
$g(-z) = 1/2 + (z/2) (\Psi(1 - z/2) - \Psi(1/2 - z/2)) .$
This shows that [*] is equivalent to
$\operatorname{L}(z) = \frac{g(-z)}{(-z)!} .$
We already derived this equation in the form $\operatorname{L}(-z) = g(z) / (z)!$ and reversing the steps in this derivation we regain the definition we started from.
The connection with Lerch's transcendent.
We have also the relation
$\frac{z}{2}\left( \Psi\left( 1-\frac{z}{2} \right) - \Psi\left( \frac12 - \frac{z}{2} \right)\right) = \sum_{n=0}^{\infty} (-1)^{n} \frac{z}{n+1-z} .$
Thus we arrive at an identity, which shows the value of the product of the two factorial functions witch have arguments with reversed sign
$\operatorname{L}(z)(-z)!= \frac{1}{2}+ \sum_{n=0}^{\infty}(-1)^{n}\frac{z}{1-z+n}$
Here we can interpret the infinite sum on the right hand side as a special case of the Lerch transcendent Phi
$\phi(z)=\Phi(-1,1,1-z)$
times z and thus the last identity can be written as
$\operatorname{L}(z)(-z)!-\operatorname{L}(0)(0)!=z\phi(z).$
Since
$z\phi(z)=\sum_{n=1}^{\infty}\zeta^{*}(n) z^{n}$
where $\zeta^{*}(n)$ is the alternating Riemann zeta function (also called Dirichlet η function). Note that $\zeta^{*}(1) = \ln 2$ (this is the classical result for the alternating harmonic series). Thus we arrive at the identity
$\operatorname{L}(z) = \frac{1}{2(-z)!} + \sum_{n=1}^{\infty}\zeta^{*}(n) \frac{z^{n}}{(-z)!}$
A representation as a hypergeometric sum.
Recall the definitions of the generalized factorial powers (Concrete Mathematics, [5.88]).
$z^{\underline{w}}= \frac{z!}{(z-w)!} \ ; \ z^{\overline{w}}= \frac{\Gamma(z+w)}{\Gamma(z)}\ (w,z \ \text{complex }).$
In terms of factorial powers the L-factorial can be represented as
$\operatorname{L}(z)=\frac{0^{\underline{z}}}{2} \sum_{n=1}^{\infty} \frac{1}{2^n} \frac{1^{\overline{n}}}{(1-z)^\overline{n}}$
Note that
$0^{\underline{z}}= \frac{1}{(-z)!}$
and that the summation starts at $1$. Using the standard notation for hypergeometric functions we can rewrite this as
$\operatorname{L}(z)=\frac{1}{4\Gamma(2-z)}\operatorname{F}\left(\left[1,2\right],\left[2-z\right],\frac12 \right).$
An asymptotic expansion of L(x).
An asymptotic expansion of $\operatorname{L}(x)$ for $x \rightarrow -\infty$ is easily found from the identity $\operatorname{L}(-z) = g(z) / z!$. The function $g(x)$ has the simple expansion
$g(x) = \frac{1}{4 x} (1 - 1 / (2 x^2) + 1 / x^4 - 17 / (4 x^6)) + O(x^{-8}).$
If we combine this expansion with Stirling's asymptotic expansion of $\Gamma(x)$ for $x \rightarrow + \infty$ (here formula 6.1.37 from the Handbook of Mathematical Functions)
$\Gamma(x)=\operatorname{K}\left( \frac{x}{e} \right)^\left( x - \frac{1}{2} \right) \left( 1 + \frac{1}{12x} + O\left(x^{-2}\right) \right)$
where $\operatorname{K} = (2 \pi / e)^{1/2}$. Thus we arrive at the following asymptotic expansion of $\operatorname{L}(x)$ for $x \rightarrow - \infty$ using the constant $\operatorname{C} = (\pi\, e^3\, 2^5)^{-1/2} = 0.225\ldots$ and practical as soon as $x \lt -2$.
$\operatorname{L}(x)=\operatorname{C}\left( \frac{e}{x} \right)^\left( x + \frac{3}{2} \right) \left(1 - \frac{1}{12x} + O\left(x^{-2}\right) \right)$
A remark from Karl Weierstraß
Indeed the fact, that the Gamma function becomes infinite at $n = 0, -1, -2,\ldots$ raised discomfort early in the development of the function. In the year 1856 Karl Weierstraß [4] introduced his definition of the Gamma function
$\operatorname{Fc}(u) = u \prod_{n=1}^{\infty} (n / (n+1))^u (1 + u / n) ).$
Weierstraß commented:
"Ich möchte für dasselbe die Benennung Factorielle von u und die Bezeichnung Fc(u) vorschlagen, indem die Anwendung dieser Function [...] dem Gebrauch der Gamma-Function deshalb vorzuziehen sein dürfte, weil sie für keinen Werth von u eine Unterbrechung der Stetigkeit erleidet und überhaupt [...] im Wesentlichen den Character einer rationalen ganzen Function besitzt." (See [5].)
However, I think Weierstraß jumped too short in his attempt to cure the shortcomings of Euler's definition. Just to replace $\Gamma(z)$ by $1/ \Gamma(z)$ is a camouflage [6]. It does not remove the 'Unterbrechung der Stetigkeit' from the Gamma function. I think the list of postulates which will ultimately define the 'true' factorial function will include [5*]:
$\operatorname{F}(u) \text{ and } \frac{1}{\operatorname{F}(u)}$
have no discontinuities.
Happily the L-factorial meets also this postulate (see figure 17).
"It is true that a mathematician who is not also something of a poet will never be a perfect mathematician." (Karl Weierstraß)
S u m m a r y
There are two classes of functions which interpolate the factorial numbers $n! = 1,2,6,24,\ldots$. Those which are continuous and those which are not. In the class of not-continuous functions one stands out: The Euler factorial. What makes this function unique is captured by the Bohr-Mollerup-Artin theorem: to be logarithmically convex.
The other class, the class of continuous interpolating functions, never got much publicity. However, Jacques Hadamard stated one exemplar in 1894 which we reconsidered. When reviewed in relation to the geometric properties we found that Hadamard's function is not logarithmic convex for $x \gt 1$.
Since this might be regarded as an aesthetic flaw we redesigned the definition and specified a function $L(x)$ which is 'logarithmic single inflected', what means that it is logarithmic convex for $x \gt \operatorname{C}$ for some $\operatorname{C}$ and logarithmic concave for $x \lt \operatorname{C}$.
At a first glance functions from the two interpolating types seem to have very little in common. But this is not true. So we found the following identity
$\frac{\operatorname{L}(z) \operatorname{L}(-z)}{g(z)} + \frac{g(z)}{z! (-z)!} = 1.$
This identity relates the Euler factorial $z!$ and the L-factorial $\operatorname{L}(z)$ by some kind of generalized reflection relation via the function
$g(x) = (x/2) (\Psi(x/2 + 1/2) - \Psi(x/2)) - 1/2 \ .$
What remains an open question is: What property of a continuous interpolant leads to a unique exemplar? What characterization singles a continuous factorial function out like the Bohr-Mollerup-Artin theorem singles out the Euler factorial?
Philip J. Davis closed his exposition [1] with the remark: "... each generation has found something of interest to say about the gamma function. Perhaps the next generation will also."
P o s t s c r i p t
In September 2006 I thought that my story will end here. But this was not the case. A month later the story continued. Here is the second part, which is centered around the super reflection product.
The super reflection product $L(z)z!L(-z)(-z)!.$
We will use the abbreviations
$g(z) = (z/2) (\Psi(z/2 + 1/2) - \Psi(z/2)) - 1/2$ $s(z) = \sin(\pi z) / (\pi z)$ $T(z) = 1 / ( g(z) (1 / s(z) - g(z) ) )$
The value of T at integers n is to be understood as the limit value (which is 0, except at $n = 0$, where it is 4). We also set
$\operatorname{LL}(z) = 1/ \operatorname{T}(z).$
Now we can prove the super reflection formula
$\operatorname{LL}(z) = \operatorname{L}(z) \operatorname{L}(-z) z! (-z)!$
Looking at $\operatorname{T}(z) = 1 / \operatorname{LL}(z)$ we find the function which is plotted in figure 18.
When I showed this plot in the Usenet (de.sci.mathematik) Roland Franzius immediately remarked, that this graph looks like the graph of $\sin(\pi x)/ \tanh(x)$. And indeed, it does, up to a scaling factor of $4/ \pi$. In figure 19 the green points indicate some of the small differences.
My answer to Roland was: "Yes, however, it is not equal to $\sin(\pi x)/ \tanh(x) \ldots$" and in the same moment I thought: "... but maybe it should be equal!" So I had to go back and start all over again...
A second start
But this time things were easy to work out. The formulas given above were the markers on the road. Backtracking the observation $\operatorname{LL}(x) ~ (4/ \pi) \sin(\pi x)/ \tanh(x)$ I saw that $\operatorname{L}(x)\operatorname{L}(-x)$ should be well approximated by $\tanh(x)/(4x)$. But $\operatorname{L}(x) = x! \operatorname{P}(x)$ and $\operatorname{L}(-x) = (-x)!(1-\operatorname{P}(x))$ and $x!(-x!) = \pi x / \sin(\pi x)$. Thus what I had to do was to solve the equation
$\operatorname{P}(z) (1 - \operatorname{P}(z)) = \frac{\sin(\pi z) \tanh(z) }{ 4 \pi z^2}.$
Apart from the constant solutions which are of no interest here there are two solutions, $\operatorname{P}^{*}(z)$ and $1 - \operatorname{P}^{*}(z)$, where
$\operatorname{P}^{*}(z) = \frac12 + \frac{1}{2z} \left(z^2+\frac{((1 - \exp(2z)) \sin(\pi z)}{(\exp(2z)+1)\pi}\right)^{1/2}.$
It is amazing how little $\operatorname{P}(z)$ and $\operatorname{P}^{*}(z)$ differ numerically.
So we now can replay our game with the definition
$\operatorname{L}^{*}(z) = z! \operatorname{P}^{*}(z)$
There is no use to plot $\operatorname{L}^{*}(z)$ because it looks indistinguishable like the plot of $\operatorname{L}(z)$ given above (within the resolution of this media).
The characteristics of $L^{*}(z)$
First of all, $\operatorname{L}^{*}(n) = n!$ if $n > 0$ is an integer. This is evident. Next it can be easily verified, that the connection with the trigonometric resp. hyperbolic functions do hold, as we aimed at.
$\operatorname{L}^{*}(z) \operatorname{L}^{*}(-z) = \frac{\tanh(z)}{ (4 z)}$ $( \operatorname{L}^{*}(z) \operatorname{L}^{*}(-z) z! (-z)! )^{-1} = \frac{4}{ \pi}\frac{\sin(\pi z)}{ \tanh(z)}$
Note again, that if we replace $4\operatorname{L}^{*}(z)$ by $\operatorname{L}(z)$ in these identities, we still have good approximations, the error of which rapidly goes to 0 if $z$ goes to $\pm \infty$.
According to our setup the next important question is: Is $L^{*}(x)$ logarithmic single inflected? So we have to look at $\operatorname{V}^{*}(x) = (\log(\operatorname{L}^{*}(x)))''$. The answer is 'yes', as a little exercise shows.
A P P E N D I X
Appendix I: The Bohr-Mollerup-Artin theorem
We give a precise formulation of the Bohr-Mollerup-Artin theorem.
Def.: A real valued function $f$ on $(a, b)$ is convex if
$f(cx + (1-c)y) \ \le \ cf(x) + (1-c)f(y)$
for $x,\ y$ in $(a,\ b)$ and $0 \lt c \lt 1$. A positive function $f$ on $(a, b)$ is logarithmically convex if $\log f$ is convex on $(a, b)$.
Theorem. If $f$ is a function on $x \gt 0$ and
(0) $f(x) > 0$ for all $x$,
(1) $f(1) = 1$,
(2) $f(x + 1) = x f(x)$ for all $x$ and
(3) $f$ is logarithmically convex,
then $f(x) = \Gamma(x)$ for $x \gt 0$.
Appendix II: Some special values...
$\operatorname{L}\left( \frac{1}{2}\right) = \frac12 \left(\frac{\sqrt\pi}{2}+\frac{1}{\sqrt \pi}\right)$ $\operatorname{L}\left(-\frac{1}{2}\right) = \frac{\sqrt\pi}{2}-\frac{1}{\sqrt \pi}$ $\operatorname{L}\left( \frac{1}{2}\right) \operatorname{L} \left(-\frac{1}{2}\right) = \frac{\pi^{2}-4}{8\pi}$ $\operatorname{L}\left( \frac{1}{2}\right) \left(\frac12\right)! \operatorname{L} \left(-\frac{1}{2}\right)\left(-\frac12 \right)! = \frac{\pi^{2}}{16}-\frac14$
...and more special values ($n>0$ integer):
$\operatorname{L}(-n) = \frac{g(n)}{n!} = (-1)^n \left(\frac{ r(n)}{n!} - \frac{\ln(2)}{(n-1)!} \right)$
$r(n)=n \ln 2 + (-1)^{n}\left( \frac12 - n \left( \sum_{k=0}^{\infty} \frac{(-1)^k}{k+1+n} \right) \right)$
$\operatorname{L}(-1)=-\frac{1}{2}+\ln 2$
$\operatorname{L}(-2)= \frac{3}{4} -\ln 2$
$\operatorname{L}(-3)=-\frac{1}{3} + \frac{\ln 2}{2}$
Appendix III: Series expansions (at x=0) for the factorial functions
$\operatorname{L}(x) = \frac12 + l_1 x + l_2 x^2 + l_3 x^3 + O(x^4)$ $l_1 = \ln 2 - \frac{\gamma}{2}; \quad l_2 = \frac{\pi^2}{24} + \frac{\gamma^2}{4} - \gamma \ln 2;$ $l_3 = \frac{1}{12} \left( 7 \zeta(3) - \pi^2\left( \frac{\gamma}{2} + \ln 2 \right)- \gamma^2 \left( \gamma - 6 \ln 2 \right) \right) ;$ $\operatorname{\Lambda}(x) = \frac14 + \lambda_2 x^2 + \lambda_4 x^4 + O(x^6)$ $\lambda_2 = \frac{\pi^2}{24} - \ln(2)^2;$ $\lambda_4 = \frac16 \ln(2)^2 \pi^2 - \frac32 \ln(2) \zeta(3) + \frac{7}{1440}\pi^4;$
$\operatorname{L}(x)$ $=\sum_{k=0}^{\infty}c_ix^i$ $1/(-x)!$ $=\sum_{k=0}^{\infty}a_ix^i$ $\Lambda(x)$ $=\sum_{k=0}^{\infty}b_ix^i$ c0 +0.5 a0 +1.0 b0 +0.25 c1 +0.4045393481 a1 -0.5772156649 b1 0. c2 +0.0944325870 a2 -0.6558780715 b2 -0.0692194972 c3 -0.0068168967 a3 +0.0420026350 b3 0. c4 -0.0004065045 a4 +0.1665386114 b4 +0.0140264149 c5 +0.0052559323 a5 +0.0421977346 b5 0. c6 +0.0025682064 a6 -0.0096219715 b6 -0.0018075010 c7 +0.0004731225 a7 -0.0072189432 b7 0. c8 -0.0000169945 a8 -0.0011651676 b8 +0.0001570804 c9 -0.0000256306 a9 +0.0002152417 b9 0. c10 -0.0000040188 a10 +0.0001280503 b10 -0.0000097537 c11 +0.0000004787 a11 +0.0000201348 b11 0. c12 +0.0000003126 a12 -0.0000012505 b12 +0.0000004530 c13 +0.0000000565 a13 -0.0000011330 b13 0. c14 +0.0000000023 a14 -0.0000002056 b14 -0.0000000163 c15 -0.0000000011 a15 -0.0000000061 b15 0.
A bound for the absolute error of this approximation of $\operatorname{L}(x)$ for $-1/2\leq x\leq1/2$ is given by
$| \operatorname{L}(x) - \operatorname{L}_{(k\le15)}(x) | \le 9\times 10^{-12} \qquad (|x|\le 1/2).$
The same bound applies for the given approximation to $\left((-x)!\right)^{-1}$ in this range.
Appendix IV: A C# implementation of the L-factorial.
1: using System;
2:
3: namespace AlternateFactorials {
4:
5: class LuschnyFactorial {
6:
7: public static double Psi(double x)
8: {
9: if (x <= 0 && x == Math.Round(x))
10: return double.NaN;
11:
12: if (x < 0) // reflection formula
13: {
14: double psi = Psi(1.0 - x);
15: double piCotpix = -Math.PI/Math.Tan(-Math.PI*x);
16: return psi - piCotpix;
17: }
18:
19: if (x <= 1e-6)
20: {
21: const double D1 = -0.57721566490153286;
22: const double D2 = 1.6449340668482264365;
23: return D1 + D2 * x - 1.0 / x;
24: }
25:
26: double result = 0.0;
27: const double C = 12.0;
28: while (x < C)
29: {
30: result = result - 1.0 / x;
31: x = x + 1.0;
32: }
33:
34: double r = 1.0 / x;
35: result = result + Math.Log(x) - 0.5 * r;
36: r = r * r;
37:
38: const double S3 = 1.0 / 12.0;
39: const double S4 = 1.0 / 120.0;
40: const double S5 = 1.0 / 252.0;
41: const double S6 = 1.0 / 240.0;
42: const double S7 = 1.0 / 132.0;
43: r *= S3-(r*(S4-(r*(S5-(r*(S6-(r*S7)))))));
44:
45: return result - r;
46: }
47:
48: public static double LnFactorial(double z)
49: {
50: //const double a0 = 1.0 / 12.0;
51: //const double a1 = 1.0 / 30.0;
52: //const double a2 = 53.0 / 210.0;
53: //const double a3 = 195.0 / 371.0;
54: //const double a4 = 22999.0 / 22737.0;
55: //const double a5 = 29944523.0 / 19733142.0;
56: //const double a6 = 109535241009.0 / 48264275462.0;
57: const double a0 = 0.0833333333333333333333333;
58: const double a1 = 0.0333333333333333333333333;
59: const double a2 = 0.252380952380952380952381;
60: const double a3 = 0.525606469002695417789757;
61: const double a4 = 1.01152306812684171174737;
62: const double a5 = 1.51747364915328739842849;
63: const double a6 = 2.26948897420495996090915;
64:
65: const double sqrt2Pi = 0.91893853320467274;
66:
67: double Z = z + 1;
68: return sqrt2Pi + (Z - 0.5) * Math.Log(Z) - Z +
69: a0/(Z+a1/(Z+a2/(Z+a3/(Z+a4/(Z+a5/(Z+a6/Z))))));
70: }
71:
72: public static double Factorial(double x)
73: {
74: if (x < 0 && x == Math.Round(x))
75: return double.NaN;
76: if (0 == x) return 1.0;
77:
78: double y = x;
79: double p = 1;
80: while (y < 8) { p = p * y; y = y + 1.0; }
81: double r = Math.Exp(LnFactorial(y));
82: if (x < 8) { r = (r * x) / (p * y); }
83: return r;
84: }
85:
86: public static double LFactorial(double x)
87: {
88: if (x == 0) return 0.5;
89:
90: double y = x < 0 ? -x * 0.5 : x * 0.5;
91: double G = y * (Psi(y + 0.5) - Psi(y)) - 0.5;
92:
93: if (x < 0)
94: {
95: return G / Factorial(-x);
96: }
97:
98: y = Math.PI * x;
99: double S = Math.Sin(y) / y;
100:
101: return (1 - S * G) * Factorial(x);
102: }
103:
104: }}
References
[0] See my page "The early history of the factorial function (gamma function)".
[1] There is an excellent exposition which everyone interested in the Gamma function must read: Philip J. Davis: "Leonhard Euler's Integral: A Historical Profile Of The Gamma Function." Am. Math. Monthly 66, 849-869 (1959). However, also note a very serious error of Davis which is discussed here.
[2] Because the paper of Hadamard is not easily accessible I wrote it up in TeX. Here you can download Hadamard's paper "Sur L'Expression Du Produit $1.2.3\ldots(n-1)$ Par Une Fonction Entière". Note: I silently corrected a typographical error in formula (2).
[3] Žarko Mijajlović, Branko Malešević: "Differentially Transcendental Functions", arXiv:math.GM/0412354 v3, 9 Feb 2006. Note the error in the definition of Hadamard's function on page 8, $\Gamma(z-1)$ instead of $\Gamma(1-z)$.
[4] Karl Weierstraß, "Über die Theorie der analytischen Facultäten", Journ. reine angew. Math. 51, 1-60 (1856), Math. Werke 1, 153-221
[5] Translation: "I would like to suggest the term 'factorial of u' and the notation $Fc(u)$ for the same, as the application of this function might have to be preferred to the use of the gamma function [...], because it does not suffer from an interruption of the continuity at any value of u and actually [...] possesses essentially the characteristics of a rational entire function."
[6] Weierstraß' suggestion survived down to the present day in the definition of the Gamma function, as given by many authors. For example Graham, Knuth and Patashnik introduce the Gamma function in Concrete Mathematics as follows:
"Here's one of the most useful definitions of $z!$, actually a definition of $\frac{1}{z!} = \lim_{n \rightarrow \infty} \binom{n + z}{n} n^{-z}$ ".
[7] The definition of the L-factorial function $\operatorname{L}(x)$ was given on this webpage in September 2006. The definition of the $\operatorname{L}^{*}$-factorial function $\operatorname{L}^{*}(x)$ was given on this webpage in October 2006. I am not aware that these definitions appeared somewhere else prior to this date. Please let me know if I am mistaken.
• In 2009 Horst Alzer proved that Hadamard’s gamma function is superadditive: $\operatorname{H}(x) + \operatorname{H}(y) ≤ \operatorname{H}(x + y) \quad \forall x, y$ real numbers with $x, y ≥ 1.5031$. A superadditive property of Hadamard’s gamma function .
• Some discussions on the web overlooked that the Bohr-Mollerup-Artin theorem only characterizes the real gamma function for $x > 0$. Hadamard was interested in the complex gamma function as the title of his paper clearly indicates. A beautiful characterization of the complex gamma function does exist, it is Wielandt's theorem, found only many years after the Bohr-Mollerup-Artin theorem.. See Reinhold Remmert, Wielandt's Theorem About the Γ-Function, Am. Math. Monthly, Vol. 103 (1996), 214-220. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9337647557258606, "perplexity": 927.0114850939874}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783403825.35/warc/CC-MAIN-20160624155003-00008-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://pe-cn.github.io/654/ | 0%
# Problem 654
## Neighbourly Constraints
Let $T(m,n)$ be the number of $m$-tuples of positive integers such that the sum of any two neighbouring elements of the tuple is $\le n$.
For example, $T(3,4)=8$, via the following eight $4$-tuples:
$(1, 1, 1, 1)$
$(1, 1, 1, 2)$
$(1, 1, 2, 1)$
$(1, 2, 1 ,1)$
$(1, 2, 1 ,2)$
$(2, 1, 1 ,1)$
$(2, 1, 1 ,2)$
$(2, 1, 2 ,1)$
You are also given that $T(5,5)=246$,
$T(10,10^2)\equiv 862820094 \pmod{1\ 000\ 000\ 007}$ and
$T(10^2,10)\equiv 782136797 \pmod{1\ 000\ 000\ 007}$.
Find $T(5000,10^{12})\mod 1\ 000\ 000\ 007$.
## 相邻约束
$(1, 1, 1, 1)$
$(1, 1, 1, 2)$
$(1, 1, 2, 1)$
$(1, 2, 1 ,1)$
$(1, 2, 1 ,2)$
$(2, 1, 1 ,1)$
$(2, 1, 1 ,2)$
$(2, 1, 2 ,1)$
$T(10,10^2)\equiv 862820094 \pmod{1\ 000\ 000\ 007}$，
$T(10^2,10)\equiv 782136797 \pmod{1\ 000\ 000\ 007}$。 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9830968379974365, "perplexity": 750.9218439356323}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107904834.82/warc/CC-MAIN-20201029154446-20201029184446-00675.warc.gz"} |
http://mathonline.wikidot.com/compositions-of-linear-transformations | Compositions of Linear Transformations
# Compositions of Linear Transformations
Suppose that $T_A: \mathbb{R}^n \to \mathbb{R}^k$ and $T_B: \mathbb{R}^k \to \mathbb{R}^m$. If we wanted to take a vector $\vec{x} \in \mathbb{R}^n$, transform it under $T_A$ and then take the image and transform it under $T_B$, our result would be the combined effect of first applying $T_A$ and then $T_B$. We say call this combined transformation the composition of $T_B$ with $T_A$ and mathematically we denote it as $T_B(T_A(x))$ or $(T_B \circ T_A)(x)$. Furthermore:
(1)
\begin{align} (T_B \circ T_A)(x) = B(Ax) = (BA)x \end{align}
Thus we say this transformation is multiplication by the product $BA$, and the standard matrix for this composition is $BA$. We thus obtain the following relationship between standard matrices:
(2)
\begin{align} [T_B \circ T_A] = [T_B][T_A] \end{align}
Furthermore, we are not restricting to only composing 2 transformations together. In fact, we can compose as many transformations as we like, and for $n$ transformations composed together, that is $(T_n \circ T_{n-1} \circ ... \circ T_2 \circ T_1)(x) = T_n(T_{n-1}(...(T_2(T_1(x)))))$, the same rules apply regarding standard matrices, that is:
(3)
\begin{align} [T_n \circ T_{n-1} \circ ... \circ T_2 \circ T_1] = [T_n][T_{n-1}]...[T_2][T_1] \end{align}
## Example 1
Find the standard matrix for the linear operator $T: \mathbb{R}^3 \to \mathbb{R}^3$ that first reflects a vector $\vec{x}$ about the $xy$-plane and then rotates it counterclockwise around the positive $z$-axis by $30^\circ$.
Let $T_A: \mathbb{R}^3 \to \mathbb{R}^3$ be the linear operator that reflects a vector about the $xy$-plane. The standard matrix for this transformation is given as $A = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -1 \end{bmatrix}$.
Also let $T_B: \mathbb{R}^3 \to \mathbb{R}^3$ be the linear operator $T: \mathbb{R}^3 \to \mathbb{R}^3$ that rotates a vector counterclockwise around the positive $z$-axis by $30^\circ$. The standard matrix for this transformation is given as $B = \begin{bmatrix} \cos 30^\circ& -\sin 30^\circ & 0\\ \sin 30^\circ & \cos 30^\circ & 0\\ 0 & 0 & 1 \end{bmatrix}$.
We now want to take the composition $(T_B \circ T_A)(x)$, and thus the standard matrix for this composition is given as [[ [T_B \circ T_A] = [T_B][T_A] \$]], and thus:
(4)
\begin{align} [T_B \circ T_A] = \begin{bmatrix} \cos 30^\circ& -\sin 30^\circ & 0\\ \sin 30^\circ & \cos 30^\circ & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -1 \end{bmatrix} \\ [T_B \circ T_A] = \begin{bmatrix} \cos 30^\circ& -\sin 30^\circ & 0\\ \sin 30^\circ & \cos 30^\circ & 0\\ 0 & 0 & -1 \end{bmatrix} \end{align} | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 4, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9985913038253784, "perplexity": 411.5141497707142}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886106358.80/warc/CC-MAIN-20170820073631-20170820093631-00297.warc.gz"} |
http://mathhelpforum.com/algebra/87077-shortest-distance.html | 1. ## shortest distance
Determine the shortest distance from point A (-5, -2sqrt7) to the X-axis to point B ( 4, -sqrt7).
2. Hello, strwbrry869!
Determine the shortest distance from point $A(\text{-}5,\:\text{-}2\sqrt{7})$ to the x-axis to point $B(4,\:\text{-}\sqrt7)$
There is a back-door approach to this problem.
Code:
| _
| (4,√7)
| o B'
| * :
-5 | P * :
---+--------------+--o--------+----
: * * :
: * | * :
: * | o B_
: * | (4,-√7)
: * |
A o _ |
(-5,-2√7) |
Reflect point $B$ over the $x$-axis to point $B'.$
Draw line $AB'$, intersecting the $x$-axis at $P.$
$P$ gives the shortest distance from $A$ to $P$ plus from $P$ to $B.$
. .
(Do you see why?)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Using calculus, the problem gets VERY messy!
The distance from $A(-5,-2\sqrt{7})$ to $P(x,y)$ is:
. $d_1 \:=\:\sqrt{(x+5)^2 + (2\sqrt{7})^2} \:=\:\sqrt{x^2 + 10x + 53}$
The distance from $B(4,-\sqrt{7})$ to $P(x,y)$ is:
. . $d_2 \;=\;\sqrt{(x-4)^2 + (\sqrt{7})^2} \:=\:\sqrt{x^2+10x + 23}$
The total distance is: . $D \;=\;\sqrt{x^2+10x + 53} + \sqrt{x^2 - 8x + 23}$
And that is the function we must minimize . . .
3. thanks but im not good at math at all... do you know what the answer to that would be?
4. Continuing the Calculus way (the best I can):
The distance formula, $D \;=\;\sqrt{x^2+10x + 53} + \sqrt{x^2 - 8x + 23}$, looks right. I think from there you would have to take the derivative of that and set it equal to 0 to find the minimum value.
So:
$
\frac{dD}{dx} = \frac{1}{2}(x^2+10x+53)^\frac{-1}{2}*(2x+10) + \frac{1}{2}(x^2+10x+23)^\frac{-1}{2}*(2x-8)
$
$
\frac{dD}{dx} = \frac{2x+10}{2\sqrt{x^2+10x+53}} + \frac{2x-8}{2\sqrt{x^2+10x+23}}
$
To find the minimum, we substitute 0 for $\frac{dD}{dx}$.
$
0 = \frac{2x+10}{2\sqrt{x^2+10x+53}} + \frac{2x-8}{2\sqrt{x^2+10x+23}}
$
That is an extremely difficult equation to factor, so I just put it in my calculator and used the Equation Solver.
I found that:
$
x = 0.34110898362998....
$
So, I guess that that (0.3411, 0), would be your point of intersection on the x-axis.
So you can now just find the distance between A and the point of intersection, and then the distance between B and the point of intersection, and add them together.
Hope that helps a little. Seems like an extremely hard problem for an Algebra student. I'm in Calculus now, and that seems pretty difficult to me. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 26, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9520706534385681, "perplexity": 374.9562689147384}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368703108201/warc/CC-MAIN-20130516111828-00021-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/144384/forming-a-directed-system-from-a-family-of-closed-subsets-with-the-finite-inters | # Forming a directed system from a family of closed subsets with the finite intersection property
In the beginning of chapter 4 of Dr. Pete Clark's convergence notes:
http://math.uga.edu/~pete/convergence.pdf
Theorem 4.1 (page 13) asserts the equivalence of 5 conditions. After making a nice observation involving De Morgan's law and applying a previous proposition, the problem of proof is reduced to showing the equivalence of the following two statements (extracted from Theorem 4.1 referenced above):
Let $X$ be a topological space.
(b) Every net in $X$ has a limit point.
(e) For every family $\{F_{\alpha}\}_{\alpha\in J}$ of closed subsets with the finite intersection property (the finite intersection of any finite subcollection is non-empty), $\cap_{\alpha\in J}F_{\alpha}\neq \phi$.
For the $(b)\Rightarrow (e)$ direction, it is stated that such a family $F:=\{F_{\alpha}\}_{\alpha\in J}$ is directed under reverse set inclusion (this is used to create a net to which (b) is applied).
I cannot see this.
Of course it is a partially ordered set. But I do not think it needs to be directed. For example, if $X = [0,1]$, $F_{0} = [0,\frac{1}{2}]$, and $F_{1} = [\frac{1}{2},1]$ then $F:=\{F_{0},F_{1}\}$ satisfies the finite intersection property but is not directed. That is, there is no $A\in F$ such that $A\subset F_{0}$ and $A\subset F_{1}$.
Is there a way to argue away this type of case so that we can safely assume that we have a directed set?
-
After some thought, I realize that this example immediately satisfies the conclusion of (e), so in this case there is no need to use (b) at all. Maybe I can use this idea to argue away any cases where $F$ is not directed. – roo May 12 '12 at 23:24
If $\mathscr{F}$ is a family of closed sets with the finite intersection property, let $$\overline{\mathscr{F}}=\left\{\bigcap\mathscr{F}_0:\mathscr{F}_0\text{ is a finite subset of }\mathscr{F}\right\}\;.$$ In other words, $\overline{\mathscr{F}}$ is $\mathscr{F}$ together with all intersections of finite subfamilies of $\mathscr{F}$. In technical terminology, $\overline{\mathscr{F}}$ is the closure of $\mathscr{F}$ under (taking) finite intersections. It's easy to check that $\overline{\mathscr{F}}$ also has the finite intersection property, and $\overline{\mathscr{F}}$ is directed under $\supseteq$. As you'll see, this extension is harmless, because $\overline{\mathscr{F}}$ doesn't contain any set that wasn't in some sense implicitly there in $\mathscr{F}$.
Now for each $F\in\overline{\mathscr{F}}$ choose $x_F\in F$; the assigment $F\mapsto x_F$ is a net $\mathbf x$ in $X$. Let $x$ be a limit point of $\mathbf x$, and suppose that there is some $F\in\overline{\mathscr{F}}$ such that $x\notin F$. Let $U=X\setminus F$; $U$ is open, and $x\in U$, so there is some $G\in\overline{\mathscr{F}}$ such that $G\subseteq F$ and $x_G\in U$. But then $$x_G\in U\cap G\subseteq U\cap F=(X\setminus F)\cap F=\varnothing\;,$$ which is a wee bit difficult, to say the least! Hence $x\in F$ for all $F\in\overline{\mathscr{F}}\supseteq\mathscr{F}$, and thus $x\in\bigcap\mathscr{F}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9696807265281677, "perplexity": 64.51738949090941}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443737916324.63/warc/CC-MAIN-20151001221836-00088-ip-10-137-6-227.ec2.internal.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/151306-solving-equations-mod-p-k.html | # Math Help - Solving equations mod p^k
1. ## Solving equations mod p^k
I have to solve such a system of equations:
$a_{11}x_1 + a_{21}x_2 + \ldots + a_{n1}x_n =y_1 (mod \, p^k)$
$a_{12}x_1 + a_{22}x_2 + \ldots + a_{n2}x_n =y_2 (mod \, p^k)$
$\ldots$
$a_{1n}x_1 + a_{2n}x_2 + \ldots + a_{nn}x_n =y_n (mod \, p^k)$
p is a prime, k = 1,2,3...
When k=1 I can use Gauss elimination method, but it doesn't work when k>1 Any ideas, links, pdf's about such equations?
2. Originally Posted by xoreaxeax
I have to solve such a system of equations:
$a_{11}x_1 + a_{21}x_2 + \ldots + a_{n1}x_n =y_1 (mod \, p^k)$
$a_{12}x_1 + a_{22}x_2 + \ldots + a_{n2}x_n =y_2 (mod \, p^k)$
$\ldots$
$a_{1n}x_1 + a_{2n}x_2 + \ldots + a_{nn}x_n =y_n (mod \, p^k)$
p is a prime, k = 1,2,3...
When k=1 I can use Gauss elimination method, but it doesn't work when k>1 Any ideas, links, pdf's about such equations?
This is an interesting problem, and I hope I'm thinking in the right direction. Do we know whether the non-zero coefficients are all coprime to p? (I'm assuming p is prime.) In other words, do we know if the following is true?
$(\forall i,j)(1\le i\le n \land 1 \le j \le n \land a_{ij} \not \equiv 0 \implies \gcd(a_{ij},p)=1))$.
Because if this is true, I'm having a hard time seeing why Gaussian elimination / Gauss–Jordan elimination wouldn't work. When we would normally divide by the leading non-zero coefficient when working in the reals, we will instead multiply by the modular inverse. And if the statement is not true, I'm having a hard time seeing how we would apply Gaussian elimination in the first place. Does this make sense? If not, someone else should come along and post something smarter.
Edit: Added non-zero condition to $\displaystyle a_{ij}$.
Edit 2: [deleted something wrong]
Edit 3: I've been doing some reading. From here:
"The Gaussian elimination can be performed over any field."
We have that $\displaystyle \mathbb{Z}/p\mathbb{Z} = \mathbb{F}_p$ is a finite field (for prime p), but $\displaystyle \mathbb{Z}/p^n\mathbb{Z}$ is only a ring and is distinct from $\displaystyle \mathbb{F}_{p^n}$ (when n > 1). So it seems there could be trouble unless the gcd requirement above is met. (Gaussian elimination just systematizes the way we manipulate linear equations to solve them, with the three elementary row operations.. if Gaussian elimination doesn't work, then what option do we have? Nevertheless it would be nice if I'm wrong and there is a solution method.)
Edit 4: Sorry for all the edits. I think even if the gcd requirement is met there could be trouble, for example if we have p-1 in one cell and 1 in another cell of the same column and we add them in the course of Gaussian elimination, then that cell will no longer be coprime to p and can cause issues.
A few references to help people go further if I'm on the right track:
Wikipedia - Division ring - has multiplicative inverses, and Gaussian elimination is possible
Article - Gaussian elimination over a euclidean ring - uses row and column operations, apparently; I don't know how that works.
3. Originally Posted by undefined
Do we know whether the non-zero coefficients are all coprime to p? (I'm assuming p is prime.)
They are not coprime to $p$ since $1\le a_{ij} \le p^k$and this is the main problem since it's impossible to compute
$a_{ij}a_{ij}^{-1}\equiv 1 (mod \, p^k)$ when $\gcd(a_{ij},p^k)=p^{k-s}, s={1,2,\ldots,k})$
Maybe I should concider $a_{ij}$ as elements of Galois field? Gaussian elimination over $GF(p^k)$ will work, but is this the right way to solve such equations? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 19, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8768007159233093, "perplexity": 392.4575713487699}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936463340.29/warc/CC-MAIN-20150226074103-00047-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://socratic.org/questions/if-a-sample-of-gas-which-weighs-0-400-g-has-a-volume-of-250-ml-at-stp-what-is-th | Chemistry
Topics
# If a sample of gas which weighs 0.400 g has a volume of 250 mL at STP, what is the molecular weight of the gas?
Jun 4, 2016
Approx. $36.0 \cdot g \cdot m o {l}^{-} 1$
#### Explanation:
The molar volume of an Ideal Gas at $S T P$ is $22.4 \cdot L \cdot m o {l}^{-} 1$.
We assume that the gas behaves ideally, and thus we multiply the given density $\left(g \cdot {L}^{-} 1\right)$ by the molar volume at $S T P$, to give....
$\frac{0.400 \cdot g}{0.250 \cdot \cancel{L}} \times 22.4 \cdot \cancel{L} \cdot m o {l}^{-} 1$ $=$ $35.9 \cdot g \cdot m o {l}^{-} 1$.
Given this molecular weight, the gas is quite possiby $H C l$.
##### Impact of this question
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https://cs.stackexchange.com/questions/105684/what-is-meant-by-support-of-a-filter | # what is meant by support of a filter?
what is meant by support of a filter in context of image processing? In the attached link of a short tutorial for detecting bar in an image by gabor filter,it is said that γ=1 helps in finding the support of the filter. link1
The support of a filter is then, intuitively, the "size" of the kernel. Any displacement for which the kernel function is $$0$$ means that we don't have to consider the corresponding point at all when calculating the value of an output point. Therefore, the most strict definition of support is that it is the set of displacements where the kernel function is non-zero. This could be a quite complicated set in general, and so isn't particularly helpful for understanding or calculation. A looser definition would be the support is any set (of displacements) which contains all the displacements where the kernel function is non-zero. This allows us to conservatively approximate the support with some simple-to-describe shape. This will usually be something like an ellipse or rectangle or square (or their lower-/higher-dimensional analogues). Given such a class of approximate shapes, the support will usually be taken to be the smallest approximate shape which contains all the displacements for which the kernel function is non-zero. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8197980523109436, "perplexity": 357.889103290012}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256858.44/warc/CC-MAIN-20190522143218-20190522165218-00352.warc.gz"} |
http://mathhelpforum.com/discrete-math/170178-power-ser.html | 1. ## power ser
$\displaystyle A$and $\displaystyle B$ are two sets.
S1: $\displaystyle P(A\cup B) = P(A) \cup P(B)$
S2: $\displaystyle P(A\cap B) = P(A) \cap P(B)$
Which of the above statement(s) is true ?
My answer was both of them are wrong.
But the answer says s2 is correct. But when I take two disjoint sets as example, i get two elements in left and one element at right.
$\displaystyle \{{\emptyset,\{\emptyset\}}\} \neq\{\emptyset\}$
If $\displaystyle A\cap B=\emptyset$ then, $\displaystyle P(A\cap B)=P(\emptyset)=\{\emptyset\}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8950373530387878, "perplexity": 819.8362071646578}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125947705.94/warc/CC-MAIN-20180425061347-20180425081347-00030.warc.gz"} |
https://admin.clutchprep.com/chemistry/practice-problems/2693/1-mole-of-an-ideal-gas-is-kept-at-constant-pressure-but-its-temperature-is-lower | # Problem: 1 mole of an ideal gas is kept at constant pressure but its temperature is lowered. What is the value of T when the volume of the gas reaches zero (extrapolation)? a. 0.0°C b. 100.00°C c. - 273.15K d. - 273.15°C e. 373.15K
###### Problem Details
1 mole of an ideal gas is kept at constant pressure but its temperature is lowered. What is the value of T when the volume of the gas reaches zero (extrapolation)?
a. 0.0°C
b. 100.00°C
c. - 273.15K
d. - 273.15°C
e. 373.15K | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.844419538974762, "perplexity": 1231.780435727961}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657140746.69/warc/CC-MAIN-20200713002400-20200713032400-00091.warc.gz"} |
http://math.stackexchange.com/questions/162695/is-this-space-paracompact | # Is this space paracompact?
Is this space $F[R]$ with the Pixley-Roy topology paracompact?
In general, when the space $F[X]$ is paracompact for general topological space?
Definition of Pixley-Roy topology: Basic neighborhoods of $F\in F[X]$ are the sets $$[F,V]=\{H\in F[X]; F\subseteq H\subseteq V\}$$ for open sets $V\supseteq F$, see e.g. here.
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The paracompactness of Pixley-Roy Hyperspaces seems to have been solved by Przymusiński (Normality and paracompactness of Pixley-Roy hyperspaces, Fund. Math. 113, pp.201-219). Therein he gives the following theorem:
Theorem: The following are equivalent for a Pixley-Roy hyperspace $\mathcal{F}[X]$:
1. $\mathcal{F}[X]$ is paracompact;
2. $\mathcal{F}[X]^n$ is paracompact for all $n \in \mathbb{N}$;
3. $\mathcal{F}[X]$ is collectionwise Hausdorff;
4. There is a neighbourhood assignment $F \mapsto V_F$ for the finite nonempty subsets of $X$ such that given finite, nonempty $F ,H \subseteq X$ the inclusions $F \subseteq V_H$ and $H \subseteq V_F$ imply $F \cap H \neq \emptyset$.
It follows from this that $\mathcal{F}[\mathbb{R}]$ is not paracompact. Suppose that $V \mapsto V_F$ is any neighbourhood assignment for the finite nonempty $F \subseteq \mathbb{R}$. Note that given any $x \in \mathbb{R}$ we may assume without loss of generality that $V_{\{x\}}$ is of the form $( x - \varepsilon_x , x + \varepsilon_x )$ for some $\varepsilon_x > 0$. From now on, I will write $V_x$ for $V_{\{x\}}$. The non-paracompactness of $\mathcal{F}[\mathbb{R}]$ is then an immediate consequence of the following claim:
Claim: There are distinct $x, y \in \mathbb{R}$ such that $y \in V_{x}$ and $x \in V_{y}$.
Proof. We construct sequences $\{ x_i \}_{i \in \mathbb{N}}$ and $\{ \delta_i \}_{i \in \mathbb{N}}$ so that
• $x_0 = 0$;
• $\delta_i = \varepsilon_{x_i}$;
• $x_{i+1} = x_i + (-1)^i \frac{\delta_i}{2}$
Note that for $j > i$ we clearly have that $x_j \in V_{x_i}$. If for any $j > i$ we also have $x_i \in V_{x_j}$ we are done. So assume that this never happens.
Given any $i \in \mathbb{N}$, we have $x_i \notin V_{x_{i+1}}$. It thus follows that $$| x_{i+1} - x_i | > \varepsilon_{x_{i+1}} = \delta_{i+1} = 2 | x_{i+2} - x_{i+1} |,$$ and so the sequence $\{ x_i \}_{i \in \mathbb{N}}$ converges to some $y \in \mathbb{R}$.
It can be shown that $y \in V_{x_i}$ for all $i \in \mathbb{N}$. (If $i$ is even, then for all $j > i+1$ we have $x_i < x_j < x_{i+1}$ and therefore $x_i \leq y \leq x_{i+1}$; similarly for $i$ odd.) As $x_i \rightarrow y$, then $| y - x_i | < \varepsilon_y$ for all large enough $i$, and therefore $x_i \in V_y$ for all such $i$. $\Box$
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9981974959373474, "perplexity": 133.90131881432546}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500830094.68/warc/CC-MAIN-20140820021350-00020-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/14409/generalized-binomial-expansion-of-left1x-right-y | # Generalized Binomial Expansion of $\left(1+x \right )^{y}$
I was trying to expand $\displaystyle \left(1+x\right)^{y}$ as a power series in terms of $y$ using the Generalized Binomial Theorem,
$\displaystyle \left(1+x\right)^{y}=\sum_{k=0}^{\infty}\binom{y}{k}x^{k}$
Assume $x,y \in \mathbb{R}$ and $\left | x \right | < 1$. With this in mind I would get something like
$\displaystyle \left(1+x\right)^{y}=\sum_{k=0}^{\infty}a_{k}\cdot y^{k}$
My question is, what is the general form of the $a_{k}$?
thanks.
-
Hint: $(1+x)^y=e^{y\ln(1+x)}$. – Raskolnikov Dec 15 '10 at 12:44
You mean $\sum_{k=0}^\infty a_kx^k$ in the last equation? – mpiktas Dec 15 '10 at 12:44
Hint for a different approach: Stirling numbers and the falling factorial. – Raskolnikov Dec 15 '10 at 12:46
@mpiktas, no, I'm considering the power series in terms of $y$. – Neves Dec 15 '10 at 12:50
Neves, @Raskolnikov gave a pretty transparent hint. Do the appropriate substitutions into a certain famous power series... – J. M. Dec 15 '10 at 15:53
EDIT: Didn't realize you asked for $y^k$, by bad. I just read past and assumed the obvious.
You can do the same as my first response:
$${\left( {1 + x} \right)^y} = f(y)$$
Since the $k$th derivative will be ${{{\log }^k}\left( {1 + x} \right)}$ you have
$${\left( {1 + x} \right)^y} = \sum\limits_{k = 0}^\infty {\frac{{{{\log }^k}\left( {1 + x} \right)}}{{k!}}} {y^k}$$
I don't think you can go any further without spending a long time with some pen and paper.
Assume
$${\left( {1 + x} \right)^{\alpha}} = \sum\limits_{k = 0}^\infty {{a_k}{x^k}}$$
Since if two series \eqalign{ & \sum {{a_k}{{\left( {x - a} \right)}^k}} \cr & \sum {{b_k}{{\left( {x - a} \right)}^k}} \cr}
sum up to the same function then
$${a_k} = {b_k} = \frac{{{f^{\left( k \right)}}\left( a \right)}}{{k!}}$$
for every $k \leq 0$, we can assume:
$$a_k = \dfrac{f^{(k)}(0)}{k!}$$
Putting $y = {\left( {1 + x} \right)^{\alpha}}$ we get
$$y'(0) = \alpha$$ $$y''(0) = \alpha(\alpha-1)$$ $$y'''(0) = \alpha(\alpha-1)(\alpha-2)$$ $$y^{{IV}}(0) = \alpha(\alpha-1)(\alpha-2)(\alpha-3)$$
We can prove in general that
$$y^{(k)}= \alpha(\alpha-1)\cdots(\alpha-k+1)$$
or put in terms of factorials
$$y^{(k)}(0)= \frac{\alpha!}{(\alpha-k)!}$$
This makes
$$a_k = \frac{\alpha!}{k!(\alpha-k)!}$$
which is what we wanted.
$${\left( {1 + x} \right)^\alpha } = \sum\limits_{k = 0}^\infty {\alpha \choose k} {{x^k}}$$
You can prove this in a more rigorous manner by differential equations:
1. Set $f(x) = \displaystyle \sum\limits_{k = 0}^\infty {\alpha \choose k} {{x^k}}$ and prove the radius of convergence is 1.
2. Show that $f(x)$ is the solution to the ODE $$y' - \frac{\alpha }{{x + 1}}y = 0$$ with initial condition $f(0)=1$.
3. By the theorem that the solution to the linear equation
$$y'+P(x)y=R(x)$$
with initial conditions $f(a) = b$ is unique, you can prove the assertion. (prove that $y = {\left( {1 + x} \right)^{\alpha}}$ also satifies the equation and you're done.)
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9481542110443115, "perplexity": 238.47639139342076}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368705407338/warc/CC-MAIN-20130516115647-00045-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/385879/friction-force-in-uniform-circular-motion | # Friction force in uniform circular motion
A table spins around it's axis with the angular velocity of $\omega$, on the table there is an object with the mass of M connected to a weight with the mass of m through the center of the table (without friction between the string and the pulley)
It's given that the static friction coefficient is $\mu$.
The question is what is the maximum value of R (radius) for the object to remain still during the circular motion.
My attempt: I said that the centripetal force is equal to the tension minus the static friction . Because there is no movement in the y axis, the normal force equals Mg. From that we can say: $$R_{max} = \frac{m}{M-μ} \frac{g}{w²}$$
But the problem is that the book says the answer is: $$R_{max} = \frac{m}{M+μ} \frac{g}{w²}$$
I don't get it. It means that the direction of the friction force is in the center of the table. But how can that be? I thought that the acceleration is towards the center and because of that the direction of the friction force has to be in the opposite direction.
The necessary force needed by the mass to undergo circular motion is $MR\omega^2$.
As the radius $R$ increases so must that force.
Mass M wants to travel in a straight line.
For small values of $R$ it could well be that the tension in the string is too high to maintain that radius for a given angular speed and so the frictional force will be in the opposite direction to the tension in the string.
However as the radius increases the friction force and the tension in the string are trying to stop mass M travelling in a straight line ie they are helping each other to produce the necessary for for the mass M to undergo centripetal acceleration and so those two forces must be in the same direction.
The acceleration is indeed towards the centre of the rotating table.
But friction does not depend on acceleration direction; it depends on sliding direction.
You are asked what the maximum $R$ is for the block to stay stationary. Let's consider what would happen if $R$ was slightly large:
Then the block would fly outwards! Out of the circular motion, meaning further away from the centre of the rotating table.
Static friction prevents this sliding from happening by pulling inwards towards the centre. And if $R$ is larger than the maximum, static friction is not strong enough and let's go, and then the block starts sliding further away.
Another view on this is to realize that the static friction isn't working against the centripetal acceleration; it is rather causing it! The constant string tension plus this static friction cause the centripetal acceleration. A larger $R$ in a circular motion means that a larger centripetal acceleration is required:
$$a=\omega^2 R$$
(because the angular speed $\omega$ is constant when you move further out on the rotating disc.) The string tension is constant and, thus, only the static friction can grow in order to cause this higher centripetal acceleration. When the static friction can grow no more (when it's limit $f_s\leq\mu_s n$ is reached), then it let's go.
This moment is when the block is at it's maximum $R$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8427149057388306, "perplexity": 173.1546629684479}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662526009.35/warc/CC-MAIN-20220519074217-20220519104217-00413.warc.gz"} |
https://tex.stackexchange.com/questions/144248/using-savebox-around-the-body-of-environ?noredirect=1 | # Using savebox around the BODY of environ
I tried to use a \savebox around the \BODY macro of \NewEnviron. Can I keep the linebreaks somehow (they are gone, in the MWE)?
\documentclass{article}
\usepackage{environ}
\newsavebox{\mybox}
\NewEnviron{myenv}{%
\savebox{\mybox}{\BODY}
\usebox{\mybox}
}
\begin{document}
\begin{myenv}
A
B
\end{myenv}
\end{document}
The problem is probably independent from the environ package but that is what I'm trying to use.
Or: Is there a way to determine the vertical and horizontal size of the \BODY?
• Unrelated to the question: you should move \newsavebox{\mybox} outside of the environment's definition (try using the environment twice to see why) – clemens Nov 10 '13 at 16:02
• The line breaks are lost because they're converted into spaces (or \par) as part of the tokenization stage. Moreover \savebox typesets its contents in horizontal mode, so you'll get a long line. Maybe you can expand your question telling what's your aim. – egreg Nov 10 '13 at 16:05
• @egreg. I'm trying to determine the vertical and horizontal size of the output of \BODY. \savebox seemed to be the closest to this aim. – masu Nov 10 '13 at 16:13
• There's no "natural" horizontal size: you have to specify what you want or it will be the current \linewidth. – egreg Nov 10 '13 at 16:15
• The adjustbox package should help you here. You need to use the varwidth key to get a variable width. An alternative is to use the lrbox environment with a varwidth environment inside it (needs the varwidth package). – Martin Scharrer Nov 10 '13 at 16:16
There's no notion of “natural width” of a text, when paragraphs are being built: TeX will use what you specify or \linewidth in case you don't.
However there's the varwidth environment that can help:
\documentclass{article}
\usepackage{environ,varwidth}
\newsavebox{\mybox}
\NewEnviron{myenv}{%
\sbox{\mybox}{\begin{varwidth}{\linewidth}\BODY\end{varwidth}}%
The box is \texttt{\the\wd\mybox} wide
and \texttt{\the\dimexpr\ht\mybox+\dp\mybox} high.
}
\begin{document}
\begin{myenv}
A
B
\end{myenv}
\end{document}
It depends on you what to do with those dimensions.
Using \NewEnviron is not necessary:
\documentclass{article}
\usepackage{varwidth}
\newsavebox{\mybox}
\newenvironment{myenv}
{\begin{lrbox}{\mybox}
\begin{varwidth}{\linewidth}}
{\end{varwidth}%
\end{lrbox}%
The box is \texttt{\the\wd\mybox} wide and
\texttt{\the\dimexpr\ht\mybox+\dp\mybox} high.
}
\begin{document}
\begin{myenv}
A
B
\end{myenv}
\end{document} | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8875751495361328, "perplexity": 2536.820171405758}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578558125.45/warc/CC-MAIN-20190422155337-20190422181337-00522.warc.gz"} |
https://ch.mathworks.com/help/finance/emaxdrawdown.html | # emaxdrawdown
Compute expected maximum drawdown for Brownian motion
## Syntax
``ExpDrawdown = emaxdrawdown(Mu,Sigma,T)``
## Description
example
````ExpDrawdown = emaxdrawdown(Mu,Sigma,T)` computes the expected maximum drawdown for a Brownian motion for each time period in `T` using the following equation:$dX\left(t\right)=\mu dt+\sigma dW\left(t\right).$If the Brownian motion is geometric with the stochastic differential equation$dS\left(t\right)={\mu }_{0}S\left(t\right)dt+{\sigma }_{0}S\left(t\right)dW\left(t\right)$then use Ito's lemma with X(t) = log(S(t)) such that$\begin{array}{c}\mu ={\mu }_{0}-0.5{\sigma }_{0}{}^{2},\\ \sigma ={\sigma }_{0}\end{array}$converts it to the form used here.```
## Examples
collapse all
This example shows how to use log-return moments of a fund to compute the expected maximum drawdown (`EMaxDD`) and then compare it with the realized maximum drawdown (`MaxDD`).
```load FundMarketCash logReturns = log(TestData(2:end,:) ./ TestData(1:end - 1,:)); Mu = mean(logReturns(:,1)); Sigma = std(logReturns(:,1),1); T = size(logReturns,1); MaxDD = maxdrawdown(TestData(:,1),'geometric')```
```MaxDD = 0.1813 ```
`EMaxDD = emaxdrawdown(Mu, Sigma, T)`
```EMaxDD = 0.1545 ```
The drawdown observed in this time period is above the expected maximum drawdown. There is no contradiction here. The expected maximum drawdown is not an upper bound on the maximum losses from a peak, but an estimate of their average, based on a geometric Brownian motion assumption.
## Input Arguments
collapse all
Drift term of a Brownian motion with drift., specified as a scalar numeric.
Data Types: `double`
Diffusion term of a Brownian motion with drift, specified as a scalar numeric.
Data Types: `double`
A time period of interest, specified as a scalar numeric or vector.
Data Types: `double`
## Output Arguments
collapse all
Expected maximum drawdown, returned as a numeric. `ExpDrawdown` is computed using an interpolation method. Values are accurate to a fraction of a basis point. Maximum drawdown is nonnegative since it is the change from a peak to a trough.
Note
To compare the actual results from `maxdrawdown` with the expected results of `emaxdrawdown`, set the `Format` input argument of `maxdrawdown` to either of the nondefault values (`'arithmetic'` or `'geometric'`). These are the only two formats that `emaxdrawdown` supports.
## References
[1] Malik, M. I., Amir F. Atiya, Amrit Pratap, and Yaser S. Abu-Mostafa. “On the Maximum Drawdown of a Brownian Motion.” Journal of Applied Probability. Vol. 41, Number 1, March 2004, pp. 147–161. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9483744502067566, "perplexity": 2227.6873358360904}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363332.1/warc/CC-MAIN-20211207014802-20211207044802-00332.warc.gz"} |
http://math.tutorcircle.com/calculus/application-of-integration.html | Sales Toll Free No: 1-855-666-7446
# Application of Integration
Top
Sub Topics Anti derivative of function f is the function F whose derivative is function f. We can understand it by an equation as F'=f. This process is also known as anti differentiation. This term is related to the definite integrals by using the Functions of Calculus. It can be understand by an example as the function F(x)=x3/3 is an anti derivative of the function f=x2 means x2 have indefinite number of anti Derivatives because the derivative of a constant is zero so x3/3+3,x3/3+78,x3/3+453 and so on. By it we can understand that when the value of the constant is changed then the new anti derivative is obtained as F(x)=x3/3+c here c is a arbitrary constant. The Applications of Antiderivatives is understand with an example of acceleration and velocity in physics as v=u+at ,where v and u are velocity and a is acceleration as anti derivative of this equation means Integration of the acceleration yields the velocity and a constant as: a=dv / dt + c and ∫t1t2 a(t) dt = v (t2) -v (t1) .This same pattern is applied to all other parts of equation as Position ,velocity, acceleration and so on .These are some essential application of anti derivative. These anti derivative are important to compute the definite integrals and using the formulas of calculus. If a function F is anti derivative of function f then it is shown as ∫ f(x) dx = F(b)-F(a). We also define the anti derivative F of a function of f that have interval as then each anti derivative is different from other because of the value of the constant such as G (x) = F (x) + c here constant c has the different values on that basis the value of anti derivatives so c is known as arbitrary constant of integration. We can find the anti derivative of a function F(a) if F' (a) = 4 – 3 ( 1 + a2) -1 and f (1) = 0. The general anti derivative of function is F ( a) = 4a - 3 arc tan ( a ) + c For an arbitrary constant c the derivative of arc tan ( x) is ( 1 + x2 ) . For find the specific Antiderivative, we evaluate: F(1)=4 - 3 arc tan ( 1 ) + c = 0 so c = 3 arc tan(1) -4 = π / 4 - 4 Thus we have F (a ) =4a - 3 arc tan( a ) + π / 4 - 4 if we find the G ( a) then G ' (a) = a ( 6 + 5a) 1/2 and G ( 1 ) = 10. So this is the way of generate the anti derivative of the Functions. We can take another example for the finding the anti derivative as a ball dropped from a roof and hits the ground at 120 fts / s .What is the height of the roof ? (Note that the acceleration due to gravity is a(T) =- 32 ft / s2) Solution of it as follows: we have a (t) = - 32 ft / s2 v (t) =- 32 t + c here c is an arbitrary constant ,to find c we evaluate getting v (0) = c = 0 But the initial velocity is 0.Thus we have v(t)=-32t. Next we calculate height d (t) = -16 t2 + c Note that when t = 0 we have d ( 0 ) = c ,and so c is the height of the roof . In particular for getting the answer, we need to decide c . As we know that when the ball hits the ground, we have v (t) = -32t = -120 And so it follows that the ball hits the ground when t 120 / 32 = 15 / 4 . Since the ball gets the height 0 when it hits the ground, we have d ( 15 / 4 ) = -16.225 / 16 + c = 0 and so c = 225. So the height of the roof is 250 ft. For defining the application of anti derivative we take some more example of it:- The most general form of anti derivative as follows f(a)=a2 The most general anti derivative is F(a) =a3+c for an arbitrary constant c. g(a) = 5 – 4a3 + 2a6 / a When initially we solve it we get g(a) = 5 / a6 - 4 / a3 + 2 = 5a-6 – 4a-3 + 2 Anti derivative of the most general type is g(a) = -a-5 + 2a-2 + 2a + c For an arbitrary constant c. So these are some examples that show the application of anti derivative. There is also some other application of anti derivatives as Anti derivatives and differential equation: anti derivative can be used in finding the general solution of the differential equation dp / dq = f ( q ). Another example of application of anti derivative is as:- A brick is dropped from an apartment whose window is 20 meters higher from the ground. Suppose that the acceleration of the brick is −16 m / sec2 then we have to calculate some of the answers for the following questions: (1) The velocity of the brick based on the function of time t; (2) The position d ( t) ; (3) Time of the brick when hits the ground and the corresponding velocity. Solution of the above questions as follows :-(1) The motion is described by the equation dv / dt = −16 m / sec2 . The general solution of this equation is given by v (t) = −16 t + C. Since the initial velocity is v (0) = 0, we find C = 0 and therefore v ( t ) = −16 t. (2) The solution to the equation dh / dt= −16 the anti derivative h ( t) = −8 t2 + C . But h (0) is the initial distance of the brick from the ground, that is h (0) = 20 . Thus, 20 = C and therefore h (t) = −8 t2 + 20. (3) The brick hits the ground when h (t) = 0 that is, −8 t2 + 20 = 0. Solving for t > 0 we find t ≈ 1.6 sec. At that time, v = −16 (1.6) = −25.6 m / sec. (The velocity is negative because we are decide up as positive and down as negative.) We can take another example of anti derivative as we have a function ∫ cos (3a) da=? So as the solution of this function as the derivative of the sine is cosine so anti derivative of cosine is sine. For solution of it we can take the bottom up approach that the anti derivative of the function most probably is sin (3x).So if we take the derivative sin (3x) then what do: d / da sin (3a)=cos (3a) d/ da (3a) = 3 cos (3a) So the anti derivative of it is almost sin (3x) but there is some other factor of 3 which comes at differentiation. It is because of Chain Rule of differentiation. Then we calculate little bit of changing function of sine as:- d / da sin (3a) / 3 = 1 /3 cos (3a) d / da (3a)=1 / 3 .3 cos (3a) = cos (3a). So the derivative of sin (3a) / 3 equals to cos (3a), that is as follows: ∫cos (3a) da= sin(3a)/3 + c, The problem of initial value that is defined by an example as da / db =4x3-2x+1 and b=3 when a=0 then find b as a function of a. b=∫ [4a3- 2a + 1] da =4 ∫a3 da -2 ∫a da + ∫ da, c = 4 . a4/ 4 - 2.a2/2 + a + c, y = a4 - a2 + a + c, Initial condition that is b=3 when a=0 we use for solving the constant of integration:- b = a4 - a2 + a + c. 3=04 -02 +0 +c 3=c so, b=a4 - a2 + a +3.
## Area of region
Suppose a function is given by y = g (x) represents a curve in the closed interval [a, b]. The area of the region enclosed by the curve of g (x), the vertical lines x = p and x = q and the x axis is given by:
Area A = pq g (x) dx,
For the area problems some properties of definite integrals is here:
If a function g (x) is defined at x = p then,
pp g (x) dx = 0,
If the function g (x) is defined on the interval [p, q], then
pq g (x) dx = - pq g (x) dx,
If the function g (x) ≥ 0 on the interval [p, q] then the area,
Area A = + pq g (x) dx,
If the function g (x) ≤ 0 on the interval [p, q] then the area is:
Area A = - pq g (x) dx,
If there are two Functions f (x) and g (x) are continuous on the interval [p,q] and given that
g (x) ≤ f (x) for all x in the interval [p, q], then the area of the region bounded by the graphs of function f and g and the vertical lines x = and x = q is:
Area A = pq [f (x) - g (x)] dx (1),
Now if we consider the two Functions x = f (y) and g (y) on the interval [p,q] and its given that g (y) ≤ f (y), then the enclosed area in this case is
Area A = pq [f (y) - g (y)] dy (2)
Now notice both the equations. Almost both the curves of both the equations are same. So it can be said that the area is the difference between the larger function to the smaller function.
For the first equation it can be written:
A = pq (upper function) – (lower function) dx; p ≤ x ≤ q
For the second equation we can use the formula
A = ab (right function) – (left function) dy; a ≤ y ≤ b
To find the area of the region bounded by the curves these formulas are helpful. When these formulas are used make sure that the consideration of the larger and smaller function should be correct because it will affect the whole question.
If we consider a function g (x) = 1 over the interval [a,b] then,
pq g (x) dx = pq 1 dx = x|ab = b – a.
Here the integral of the function g (x) = 1 is nothing but just the length of the interval [a, b]. This can be obtained to the area of a Rectangle having height 1 and length (b – a) but here it's interpreted as the length of the interval [a,b].
The same trick is also applicable for the double Integration. The integral of a function g (x,y) over a region D can be interpreted as the volume covered under the surface z= g (x,y) over the region D. If the function is considered g (x,y) = 1 over the region D then the above trick can be applied and this is called as the area of the region D. If A is the area of the region D then it can also be written as:
Area A = ∬D dA.
## Area between curves
In Calculus when two curves intersects each other then area is also intersected by these two curves. Suppose we have two curve where first curve is f(x) and second curve is g(x) ,so area between two curve is A = f(x) – g(x). That is area between two curves if we take the interval between to two curve is [a,b] then formula would below:
A = ab (f(x) – g(x)) dx (where we that assume f(x) ≥ g(x))
This situation can be created two case first cases: first upper curve and second is lower curve. In this situation formula will create like that A = ab ((upper curve)-(lower curve)) dx
Where interval a≤ x ≤b
Now second case: first right curve and second is left curve
So in this case formula looks like that A = cd ((right curve)-(left curve)) dx
Where interval a≤ x ≤b
So that was the situation of curve
Now we see that how to area can be bound: area can be bound in two cases which are given below.
Case 1: If area bounded by x –axis then limits are use for x and curves are function of x.
Area A = ab(f(x)-g(x))dx where interval a≤ x ≤b
Case 2: if area bounded by y-axis then limits are use for y and the curves are function of y.
Area A = cd(f(y)-g(y)) dy where interval c≤ y ≤d .
Let’s see the Area between curves based examples.
Example 1: Find the area bounds between the curve where first curve x = -y and second curve is
x = √y and interval is [-1 ,1]?
Solution: Use the given steps to solve such type of problems:
Step 1: In this step first we would solve the both curve and find the interval between these curve.
So write the both curve first curve x = -y,
Second curve x = √y
In this problem we have given interval so we do not find the interval we can direct used in Integration
Step 2: In this step we are integrated the both curve with interval
We know Area between curves A = ab (f(y) – g(y)) dy ,
f(y) = -y,
g(y) = √y ,
Area of curve A = -11 (-y)- (√y)dy,
A = [(-y)2/2 – y3/2/(3/2)]-11
A = [(-1)2/2 – 13/2/(3/2)] - [(1)2/2 – (-1)3/2/(3/2)],
A = [1/2 -2/3 ] – [1/2 + 2/3] = [1/2 -2/3 - 1/2 -2/3] = -4/3,
A =-4/3,
Now we got the Area of curve A = -4/3.
Example 4: Find the area bounds between the curve where first curve y = x +2 and second curve is y = x2?
Solution: Step 1: In this step first we would solve the both curve and find the interval between these curve.
So write the both curve first curve y = x +2,
Second curve y = x2,
So we simplify both curve x+2 =x2,
x2 - x -2 =0,
x2 -2x +x -2 = 0,
x(x-2) +1(x-2)=0,
(x-2)(x+1)=0,
x =2 , x=-1.
After solving we got the points first Point is x =-1, and another is 2 and these points are worked as interval [-1,2]
Step 2: In this step we are integrated the both curve with interval
We know Area between curves A = ab (f(x) – g(x)) dx ,
f(x) = x +2,
g(x) = x2 ,
Area of curve A = -12 (x+2)- (x2)dx,
A = [(x2/2 +2x) –(x3/3)]-12
A = [(22/2 +2x2) –(23/3)] - [(-1/2 -2) – (-1)3/3],
A = [4/2 +4 – 8/3]-[-1/2 -2 + 1/3],
A = [6 -8/3] – [-5/2 -1/3],
A = [10/3] – [(-15-1)/6] = 10/3 +16/6 = (20 -16) /16 = 4/16,
A = ¼,
A = ¼,
Now we got the Area of curve A = 1/4.
## Volumes of Solid of Revolutions
If a plane area is revolved about a fixed line in its own plane then the body or Solid that is formed by the revolution of the plane area is called as solid of revolutions and the fixed line about which the area revolves is known as the axis of revolution, it can be used to find the area of the region enclosed by the curves.
In other words we can say that, if a region is revolved about a line then solid formed is known as a solid of revolution. The solid is generated by the region and the axis of revolution is the line about which the revolution takes place.
The volume of this solid is given by
V = pq A (x) dx,
V = pq A (y) dy,
Here A (x) and A (y) is the cross sectional area of the solid.
The formula for the area of the ring can be written as
These radii are dependent upon the Functions given and the Axis of Rotation.
In mathematics the volume of a solid of revolution can be obtained by rotating a Plane Curve around some Straight Line that lies on the same plane. Suppose that the curve doesn’t cross the axis then the volume of the solid is equal to the length of the Circle that is described by the centroid multiplied by the area of the figure. A disk can be considered as a three dimensional volume element of a solid of revolution and this element can be created by rotating a Line Segment of the length ‘w’ around the given axis located ‘r’ units away so that a cylindrical volume of π / r2w units is enclosed by the disks.
The general methods to find the volume of a solid of revolution are the disk method and the shell method of Integration. Besides that cross section method and washer methods are also used. The first step to solve the area or volume related problems with the help of disk method is to draw the curve first then calculate the enclosed area that revolved about the axis of revolution then calculate the volume of disk shaped slice of the solid having thickness ‘δx’ or a cylindrical shell of width ‘δx’ and then the final step is to find the limiting sum of these volumes as ‘δx’ approaches to the value zero and this value can be found by evaluating a suitable integral. Besides that washer method and cross section method is also used to calculate the volume of the solid.
Now suppose the region between the graph of a continuous function y = f (x) and the x- axis from x = p to x = q then the volume of the solid can be given as-
Volume V = pq π (radius)2 dx = pq π (f (x))2 dx.
If the equation of the developing curve is given in the polar co-ordinates assume
r = f (Θ) and if the curve revolves about the x- axis then the generated volume is
Volume V = π ab (y)2 dx,
Volume V = π Θ1Θ2 (y)2 (dx / dΘ) . dΘ,
or
Volume V = π Θ1Θ2 (x)2 (dy / dΘ) . dΘ.
Where Θ1 and Θ2 are the values of Θ at the points x = a and x = b.
Now let's assume that x = r cos Θ and y = r sin Θ. Then the enclosed volume is rewritten as
Volume V = π Θ1Θ2 (r)2 sin2Θ (d / dΘ) . r cos Θ dΘ.
Where the value of ‘r’ in terms of ‘Θ’ must be substituted from the equation of the curve given in the problem.
The volume of the solid generated by the revolution of the area enclosed by the curve r = g (Θ) and the radii vectors given by Θ = Θ1 and Θ = Θ2.
About the initial line Θ = 0 which always refers the x- axis is-
Volume V = Θ1Θ2 ( 2/3 ) (π r2) sinΘ dΘ.
About the line π = π/2 which refers the y- axis is-
Volume V = Θ1Θ2 ( 2/3 ) (π r3) cosΘ dΘ.
About the line Θ = γ, the volume of revolution is-
Volume V = Θ1Θ2 ( 2/3 ) (π r3) sin (Θ - γ) dΘ.
As illustrated above in each of the above three formulae the value of ‘r’ must be in terms of ‘Θ’ that is substituted from the equation of the given curve.
If two Functions f (x) and g (x) are such that f (x) ≥ g (x) for all ‘x’ in the interval [p,q], then the volume of the solid generated by revolving two functions around the x- axis the region bounded by the graphs of function f (x) and h (x) between the values of x = p and x = q is given by the integral,
Volume V = pq h [f (x)2 – g (x)2] dx.
if ‘h’ is a function such that x = h (y) ≥ 0 for all y in the interval [p, q], then the volume of the solid generated by revolution, around the y- axis, the region enclosed by the graph of ‘h’, the y- axis (x = 0) and the horizontal lines y = p and y = q can be written by the following integral-
Volume V = pq [h (y)]2 dy.
If two functions ‘z’ and ‘w’ are such that f (y) ≥ g (y) for all ‘y’ in the interval [p, q] then the volume of the solid generated by the revolution of the region enclosed by the graphs of ‘f’ and ‘g’ between y = p and y = q around the y- axis is given by the following integral-
Volume V = pq h [f (y)2 – g (y)2] dy.
If the cross section is not a disk but is a washer then the area of the washer must be written first by subtracting the area of the inner cross section from the area of the outer cross section and make sure that the rotation is either about the x axis or y axis i.e.
The washer is formed by revolving a Rectangle about an axis. This is used because disc method cannot be used for the solids having a hole. In this situation Easher method is preferred instead of disc method.
The volume of the solids formed by rotating the area between the curves of function f (x) and g (x) and the lines x = p and x = q about the y axis is given by
Volume V = 2 π pq |[f (x) – g (x)]|dx
As explained above if r and R are the inner and outer radii of the washer and w is the width iof the washer then the volume of the washer can be written as
Volume V = π (R2 – r2) w
The volume of the solid of revolution generated by washer can be written as
Volume V = π pq (R(x)2 – r(x)2) dx
The integral involving the inner radius refers the volume of the hole and is subtracted from the integral having the outer radius.
The cross section method is used for the solid of any shape because disc method is used for a circular cross section of area πr2 . The volume by cross section method is
Volume V = pq f (x) dx,
or
Volume V = pq f (y) dy.
The cylinder method to find the volume of revolution of solids is used when the slice that was drawn is parallel to the axis of revolution means integrating perpendicular to the axis of revolution. It is also known as shell method. The volume of the shell can be written as-
Volume V = π (d + w/2)2 h - π (d + w/2)2 h,
Volume V = 2π dhw,
Volume V = 2π (average radius) (height) (Thickness).
Where ‘w’ is the width of the rectangle, ‘h’ is the height of the rectangle and ‘d’ is the distance between the axis of revolution and the center of the rectangle. When this rectangle is revolved about its axis of revolution this generates a cylinder shell of width ‘w’.
For horizontal axis of revolution the Volume is
Volume V = 2π pq f (y) h )(y) dy.
Another theorem about the volume of solid of revolution states that-
“If a closed plane curve revolves about a straight line in its plane which does not intersect it, then the volume of the ring that is obtained by the revolution would equal to the area of the region enclosed by the curve multiplied by the length of the path described by the centroid of the region.
This is how we find volumes of solid of revolutions.
## Area of Region Bounded by Parametrically Defined or Polar Curves
Area of region bounded by polar curve, in this curve are lies between an interval [a ,b] and function is continuous at this interval, and f(x) ≥ 0 then the area of the region between the function and the x-axis is given by Area = ab f(x) dx.
If y-axis curve f(y) is continuous and another y-axis curve g(y) is also continuous, where g(y) ≤ f(y), ‘y’ is the subset of interval [a, b] then area of the region bounded by curve is A =ab (f(y) –g(y)) dx.
If a curve f(x) lies on x-axis and another curve g(y) lies on y-axis and both are subset of interval [a, b], then area of region with interval [a,b] is given by-
A = ab (f(x) –g(y)) dx ,
If we use the common origin and take initial positive x-axis , then rectangular polar coordinate (x, y) can be represented as x = r cos ѳ and y = r sin ѳ, here x ,y can be converted to the Polar Coordinates r and ѳ by this equation.
x2 + y2 = r2,
r= √(x2 + y2) (by pythagoras theorem),
r= polar equation f(ѳ),so f(r, ѳ) =0,
r2cos2ѳ + r2sin2ѳ = r2,
r =√(r2 cos2 ѳ + r2 sin2 ѳ),
y/x = tan ѳ (tan ѳ is angle between x-axis and y-axis).
In this equation if ‘r’ is negative then cos (ѳ +180) = -cos ѳ, sin (ѳ +180) = -sin ѳ the above equations are used to find a Cartesian equation, using above equation we can find Area A =ab ½ r2 dѳ.
Suppose we a have curve and this curve lies on closed interval [a, b] so area of region bounded by curve A = 1/2 ab f(ѳ)2 dѳ, where angle ‘ѳ’ of the curve lies between a<ѳ< b.
Above equation is also called polar equation.
Now let’s see area of region bounded by polar curves based example:
Example: Find the area of the inner loop of r= 1+2cos ѳ where interval is [-π/2, π/2]?
Solution: We need to follow the steps shown below to understand the concept-
Step 1: In first step we will write the given Cartesian equation.
r = 1+2cos ѳ with interval [-π/2, π/2 ]
Step 2: In this step we will write the formula of area bounded with given interval
A = ab ½ r2 dѳ,
A = -π/2π/2 ½ r2 dѳ,
A = -π/2π/2 ½ (1+2 cos ѳ)2dѳ,
A = -π/2π/2 ½ (1 + 4 cos ѳ + 4cos2 ѳ)dѳ,
A = -π/2π/2 1/2 dѳ + -π/2π/2 2 cos ѳ dѳ + -π/2π/2 2cos2ѳ dѳ,
A = [ 1/2 ѳ] -π/2|π/2 +[2 sin ѳ] -π/2|π/2 + 2π/2π/2 ( ѳ/2 + ¼ cos 2ѳ ),
A = [ 1/2 ѳ] -π/2|π/2 +[2 sin ѳ] -π/2|π/2 + π/2π/2 ( ѳ + 1/2 cos 2ѳ ),
A = [π/4 + π/4 ] +2[sin π/2 – sin (-π/2 )] + [π/2 - π/2] + 1/2[cos π – cos (-π)],
A = [π/4 + π/4 ] +2[sin π/2 – sin (-π/2 )] + [π/2 + π/2] + 1/2[cos π – cos (-π)],
A = π/2 + 2[1 +1] + [π] + ½ [1 +1],
A = π/2 +4 + π +1 = 3π/2 + 5 = (3π +10)/2,
So, the area of polar curve is A =(3π +10)/2.
Let’s see another example of area region bounded polar curve:
Example 2: Find the area of Polar curve where curves are represented by f(ѳ) = 1+sin ѳ ,g(ѳ) = 3+ cos ѳ with interval [-π/2 , π/2] ?
Solution: In this problem we have two curves with interval, so we will use the above formula.
Step 1: First we will write the given polar curve with interval then we check the continuity of the curve
First curve f(ѳ) = 1+sin ѳ (this curve is a continuous curve with interval [-π/2 ,π/2]),
Second curve g(ѳ) = 3+ cos ѳ (continuous curve with interval [-π/2 , π/2]),
Step 2: In this step we will integrate the given curve and solve with given interval,
First we integrate the first curve: -π/2π/2f(ѳ) =-π/2π/2 (1+sin ѳ) dѳ,
-π/2π/2f(ѳ)= [ѳ] -π/2| π/2 - cos ѳ -π/2|π/2,
: [ π/2 + π/2] - [cos(π/2) – cos (-π/2)],
π/2π/2 f(ѳ)= π+ 0 = π.
Now we will integrate the second curve: -π/2π/2g(ѳ) = -π/2π/2 (3+ cos ѳ) dѳ,
π/2π/2 g(ѳ) = [3ѳ]-π/2| π/2 + [sin ѳ] -π/2| π/2,
[3π/2 + 3π/2] + [sin (π/2) – sin (-π/2)],
π/2π/2 g(ѳ) = [ 3 π ] + [1+1] = 3 π +2.
Step 3: In this step we will solve the first curve and second curve with interval .
Now the area of polar curve: A = -π/2π/2 f(ѳ) – g(ѳ) dѳ,
A = -π/2π/2 g(ѳ) - -π/2π/2 f(ѳ),
A = 3 π +2 - π ,
A= 2 π + 2.
Therefore the area of polar curve is A = 2π + 2.
Example 3: Find the area of Polar curve where curves are represented as f(ѳ) = 2sin ѳ ,g(ѳ) = 3cos ѳ with interval [-π/2< ѳ< π/2]?
Solution: In this problem we have two curves with interval, steps to solve this problem are shown below-
Step 1: First we write the given polar curve with interval, then we will check the continuity of the curve-
First curve f(ѳ) = 2sin ѳ (this curve is continuous curve with interval [-π/2 , π/2])
Second curve g(ѳ) = 3cos ѳ (continuous curve with interval [-π/2 , π/2])
Step 2: In this step we will integrate the given curve and solve with interval,
First we will integrate the first curve: -π/2π/2 f(ѳ) =-π/2π/2 (2sin ѳ) dѳ
-π/2π/2f(ѳ)= [ -2cos ѳ ]-π/2|π/2,
= [(-2cos(π/2)) – (-2cos (-π/2))],
-π/2π/2 f(ѳ) = 0+ 0 = 0.
Now we will integrate the second curve: -π/2π/2 g(ѳ) = -π/2π/2 (3cos ѳ) dѳ
-π/2π/2 g(ѳ) = [3sin ѳ] -π/2| π/2,
= [3sin (π/2) – 3sin (-π/2)],
-π/2π/2 g(ѳ) = [3 + 3] = 6.
Step 3: In this step we will solve the first curve and second curve with interval.
Now the area of polar curve: A = -π/2π/2 g(ѳ) – f(ѳ) dѳ,
A = -π/2π/2 g(ѳ) - -π/2π/2 f(ѳ),
A = 6-0 = 6.
Therefore area of polar curve is A = 4.
Example 4: Find the area of Polar curve where curves are represented by f(ѳ) = 2sin ѳ , g(ѳ) = 3 + sin ѳ, with interval [-π/2, π/2]?
Solution: In this problem we have two curves with interval.
Step 1: First we will write the given polar curve with interval then we will check the continuity of the curve-
First curve f(ѳ) = 2sin ѳ (this curve is a continuous curve with interval [-π/2 , π/2])
Second curve g(ѳ) = 3 + sin ѳ (continuous curve with interval [-π/2 , π/2])
Step 2: In this step we will integrate the given curve and solve with interval,
First we will integrate the first curve: -π/2π/2f(ѳ) =-π/2π/2 (2sin ѳ) dѳ,
-π/2π/2f(ѳ)= [ -2cos ѳ ]-π/2|π/2,
= [(-2cos(π/2)) – (-2cos (-π/2))],
-π/2π/2 f(ѳ)= 0 + 0 = 0.
Now we will solve the second curve: -π/2π/2 g(ѳ) = -π/2π/2 (3 +sin ѳ) dѳ
-π/2π/2 g(ѳ) = [ 3 ѳ - cos ѳ] -π/2| π/2 ,
= [3(π/2) – cos (π/2)] –[3(-π/2) – cos (-π/2)],
= [3π/2 +3 π/2] = 3π,
π/2π/2 g(ѳ) = 3π.
Step 3: In this step we will solve the first curve and second curve with interval.
Now the area of polar curve: A = -π/2π/2 g(ѳ) – f(ѳ) dѳ,
A = -π/2π/2 g(ѳ) - -π/2π/2 f(ѳ),
A = 3π - 0 = 3π.
Therefore area of polar curve is A = 3π.
This is how we can find Area of region bounded by parametrically defined or Polar Curves.
## Arc Length
The topic "Arc length" is generally related to the geometries but sometimes we use it in the Calculus Geometry also. As the name suggests “Arc Length”, means a part of a curve in any region. To understand the arc length we first have to understand that what is arc and then arc length, so Arc is curved line made by two distinct points on the curve, and the arc length is the calculation of the distance of the arc or to find the length of the curved line which forms the arc on the curve. The length of the arc in the curve is greater than the Chord of the curve because the chord is the distance between two end points of the curve and arc is the length of the curve from one Point to the other point on the curve. For example, we have a Circle and the circle has its arc as ‘AB’ on the circle. Assume the length of the AB is 5 then it is read as “The length of the arc is 5”. We calculate the arc length with the help of some formulas in geometry and for other we use arc length Integration to get arc length of the body.
For instance, if the central angle of the circle or curve is A and radius is R then the arc length of the curve will be:
Arc length = 2 π R (A/360)
Where A is the central angle, R is the radius of the arc in circle, and π is the constant term and it is equal to 22/7, or 3.1414. In some other form of the formula if the central angle of the curve is in the radians units then we can write the formula for arc length of the circle as:
Arc length = R x A
Here again R is the radius of arc and A is the central angle of arc in radian unit. The arc length in degree unit and in radians unit are both same but the single difference is that the conversion of the degree quantity in to radian quantity.
Finding the arc length of a segment is the process of rectification of the curve. The arc length is always given by L and it may be either finite or infinite.
While finding the arc length integration we assume a part of the curve in the plane. Here in this curve a region [a, b] is taken and the curve is varying in the defined region.
Then the formula for the arc length on the interval [a, b] will be given by:
If the function y = f(x) (f(x) should be a continuous function), then arc length (L) of y = f(x) on interval [a, b] will be (L) = ∫ab √(1 + f’(x)) dx,
If the curve is in the form of x(t), and y(t) then the arc length = ∫ab √((dx/dt)2 + (dy/dt)2 ) dx, similarly if the curve is given in the Polar Coordinates then the arc length (L) will be = ∫ab √((r)2 + (dr/dθ)) dθ,
Where θ is the angle and r = f(θ).
## Velocity and Acceleration Vectors of Planar curves
Velocity is defined as speed with direction. Speed is equals to the distance covered per unit time. Speed is defined in Scalar terms without direction and if we consider the direction, it is said to be the velocity of the object. Velocity and acceleration vectors of planar curves are formed in the 2 dimensional planes. Velocity always has direction associated with it, it is a vector quantity as direction is also associated with it. In mathematical terms, we can represent the velocity as,
r = x I + y J,
It is called the radius vector or Position vector as it describes the position of any Point on the plane.
The position vector or radius vector is from the origin to the point (x, y).
Here, ‘I’ and ‘j’ are the unit vectors.
‘I’ and ‘j’ are unit vectors defined along ‘x’ and ‘y’ axis in the coordinate system.
Unit vectors represent the direction and have unit magnitude. It shows only direction with unit magnitude.
Radius vector is defined in the 2 - d planes.
A 2 – d plane can be parametrically defined as,
x = x ( u ),
y= y ( u ),
Here, the Functions x (u) and y (u) are the Functions which are continuous in The Range ‘u’. They can be finite or infinite range.
The Plane Curve can be plotted on a 2 – d plane using the following equation,
r ( u ) = x (u ) i + y ( u ) j,
As we differentiate a normal Scalar Function, we can also differentiate a vector function.
For example,
The derivative of the above equation is,
d r /d u = ( d x/ d u ) .i + ( d y / d t ) . j,
Here, d r / d u ( u ) is a Tangent which is a vector with a direction.
It is the tangent to the above curve.
Here, ‘u’ is taken as a time variable, then d r (u) / d t is known as the velocity.
As we have discussed the radius vector, we can now discuss the velocity and acceleration of a moving point.
Let us take a point V (x, y) in 2 – d motion moving along the curve formed by the following equation,
x = x ( u ),
y = y ( u ),
Here, ‘u’ represents the time.
The position vector of a point from origin to a point is given by ,
r = x ( u ) i + y ( u ) j - - - - - - - - - - - - - - - - - - - - - - - ( equation 1 )
position vector is also called as radius vector.
This is the vector equation of a point with unit vectors ‘I’ and ‘j’.
As we know, speed is determined by differentiating the distance with respect to time.
In terms of vectors, velocity can be determined by differentiating the same distance equation.
Differentiating once equation 1, we get the velocity of the moving particle.
V = d r / d t
‘r’ is the radius vector of the particle.
d r / d t = ( d x / d t) . i + ( d y / d t) . j - - - - - - - - - - - - - - - - - - - ( equation 2 )
‘I’ and ‘j’ are unit vectors.
We know that a vector has two quantities,
1 ) Magnitude part known as scalar part.
2 ) Direction part known as vector part.
Velocity is a vector quantity. Hence, it must have the two parts, the magnitude part and the direction part.
The magnitude part of the velocity is represented by the |mod| symbol.
Hence, | v | = √ ( v ) . ( v ),
= √ ( v x 2 + v y 2 ).
Here, ‘x’ is the velocity calculated in the ‘x’ direction and ‘y’ is the velocity component in ‘y’ direction.
We can obtain ‘x’ and ‘y’ components of velocity individually by partially differentiating the equation of the particle.
As,
v ( x- component ) = d x / d u
v ( y -component ) = d y / d u
Here, we have differentiated the equation with respect to time variable ‘u’.
As ‘v’ is a vector, so we also have to find its second component.
The direction of velocity at the point is the direction of the tangent at that point at time ‘u’.
The acceleration of the point at time ‘u‘ is determined by differentiating the equation of distance twice.
In other words, acceleration is obtained by differentiating the velocity equation once.
Note that acceleration cannot be defined without direction.
But velocity has its magnitude part which can be defined without its direction, known a speed.
As we determined the velocity of the particle, we can also define the acceleration of that particle.
Acceleration= a = d v / d u = d 2 r / d t 2,
= d 2 x / d u2 . i + d 2 y / d u 2 . j.
Here, ‘I’ and ‘j’ have their usual meanings. They are the unit vectors which represent the direction at the time ‘u’.
We have differentiated the displacement equation twice to get the acceleration.
We are taking displacement in place of distance because we are talking about the vector quantities.
Displacement is the vector of distance as it has a direction associated with it.
We can now obtain the magnitude part of acceleration.
| a | = √ a . a,
In other words,
| a | = √ a x 2 + a y 2,
Where ‘a x’ is the x – component of acceleration in the x – axis.
‘a y’ is the y – component of acceleration in the y – axis.
We can partially differentiate the equation to obtain the ‘x’ and ‘y’ parts individually.
a x = d 2 x / d t 2,
a y = d 2 y / d t 2,
The direction is same as the direction of the velocity.
Let us take a trigonometric equation of the position of any particle.
r ( u ) = cos u i + sin u j + u k,
Here, we have taken the position of the particle in a 3 d plane.
Here, note that we have taken another unit vector ‘j’. It shows the direction towards z- axis.
i , j , k are unit vectors with unit magnitude along x , y and z axis respectively.
As we have discussed earlier, the first order differentiation of the given particle equation will produce velocity and second order differentiation will give the acceleration of the particle.
Differentiating the equation, we get-
v ( u ) = - sin u i + cos u j + k,
Again differentiating the equation, we get the acceleration
a ( u ) = - cos u i – sin u j + 0k,
Here, we have zero magnitude on the z- axis.
Hence, the direction vectors for the acceleration are in x – y plane.
Using the Calculus, we now prove that the derivative of constant speed is zero.
Let us take example of a moving particle which is moving with a constant speed ‘v’ but its velocity is different from speed.
The direction vector is changing.
For now, don't think about the direction vector.
The magnitude of the particle is ‘v’.
As we studied earlier,
The magnitude of velocity is equals to
| v | | v | = | | v 2 | |,
= k,
Here, k is a constant as ‘v’ is a constant .
v .v = k - - - - - - - - - - - - - - - - - - - - ( equation 1 )
Differentiating with respect to time variable ‘u’, we get-
d ( v . v ) / d u = d k / d u
Using multiplicative derivation rule, we can solve this equation as-
= d v / d u . v + v . dv / d u,
= 0.
As ‘v’ is a constant and differentiation of a constant is zero.
It implies ,
2 ( v . d v / d t ) = 0,
Further solving,
v . d v / d u = 0 - - - - - - - - - - - - - - - - - - - - - - ( equation 2 )
‘u’ is time variable.
We know that, differentiating the velocity component, we get acceleration.
Hence d v / d u = a (acceleration)
Using this in equation 2 , we get-
v. a = 0,
Here ‘v’ and ‘a’ are velocity and acceleration vectors.
The dot product of ‘v’ and ‘a’ is equals to zero, therefore we can say that, if the magnitude of velocity is ‘v’ and it is constant, then according to the properties of vectors, the velocity vector and acceleration vectors are at 90 degrees to each other.
Let us understand the acceleration in terms of force.
According to the Newton’s second law of motion ,
F = m a
F = force applied
m = mass of the object
a = acceleration of the object
We can also calculate the instantaneous acceleration as it is equal to the rate of change of velocity at that particular instant.
Let us discuss a practical example which will clear the concept of vectors.
Example: A car is tracing a circular pattern on a ground of radius 100 m with a constant speed.
If the car takes 120 s to complete one revolution, evaluate the following.
1 ) The speed of the car?
2 ) The instantaneous velocity at the point M?
3 ) The instantaneous velocity at the point N ?
Solution 1 ) We have to calculate the speed of the car. As speed is a scalar quantity, we don't have to consider the direction vectors.
We will simply find the magnitude.
As we know,
Total distance covered in 1 revolution =circumference of the Circle
d = 2 ∏ r,
r = radius of circle = 100 m,
hence, d = 2 * 3.14 * 100,
d = 6 2 8 . 3 meters,
We have distance covered by car and time taken by the car. We can calculate the speed of the car.
Speed = distance travelled / time taken,
Speed = 628.3 / 120,
As time t = 120 s given,
Speed = 5.23 meters per second.
2) As we know that the car is moving at a constant speed.
So, at point ‘M’, the magnitude of velocity will be 5.23 meters per second.
| Velocity | = speed,
Now, we have to find the direction also, as velocity is a vector and a direction vector is associated with it.
The direction of the car at point ‘M’ will be the direction of the tangent at M towards west as shown in the diagram.
3) As we find the velocity of car at point ‘M’, we can find the velocity at point ‘N’.
Magnitude of velocity = speed = 5.23 meters per second.
Velocity is a vector, hence we have to find its direction also.
Now, we have to find the direction of car at point ‘N’.
The direction of a vector is decided by the tangent passing through that point.
The direction of the car will be towards the direction of the tangent at point ‘N’ in south direction.
This is how we find Velocity and acceleration vectors of Planar curves.
## Other Applications involving the use of Integral of Rates
Other applications involving the use of integral of rates can be understood with the help of various examples from our day to day life. Some of the examples for better understanding of the implementation of the above property are:
We will start with a small example from ancient civilization: The art of modern pottery which is still around us from the time of Ancient civilization is an area where we can see the practical application of Integration of rates.
Example 1: The desired shape of the side of the pottery vase can be described by:
Velocity = 5.0 + 2 sin (hei. / 4) (0 < = hei. < = 8 ∏) ( Using the Riemann sums Property),
Where “ hei.“ is the height.
A base for the vase can be formed by placing it on the potter's wheel. How much clay can be added to the base to form their vase if the inside radius is always 1 inch less than the external radius. Here with this example it can be related with two things: First is the rate of change of radius and the second is the rate of change of height.
After last example we have a better idea about the importance of Riemann sums and its applications. It is not just any intellectual exercise, it is thus natural way to calculate mathematical and physical quantities that primarily seem to be irregular as a whole, but can be fragmented into regular pieces.
Now we calculate the values for the regular pieces using the known formulas and latter sum them to calculate the value of irregular as a whole. The observation from these problems are very tedious.
With the concepts of Calculus it becomes more predominant to get the exact answers for these problems with almost no efforts from our side.
Some more applications of integrals of rates are given below-
A. Linear Motion Revisited: In many applications, the integral is defined to be the net change over time. The best example to understand this is distance traveling.
Example 2 Interpret Velocity function.
The velocity value is dx / dt = v(t) = t2 – 8 / ( t + 1 )2 cm / sec, for a particle moving along a horizontal x – axis for 0 < = t < = 5. Describe the motion?
Solution: Solve graphically: The graph of ‘v’, from the figure, starts with v( 0 ) = - 8, which we will assume by saying that the particle has an initial velocity of 8 cm / sec to the left. It slows at the rate of 1.25 sec, after which it moves to the right (v > 0) with the increasing value of speed. And latter we will come down to the Point where, the value of the velocity v( 5 ) is approximately equals to 24.8 cm / sec at the end.
Example 3: Calculating the Total distance Traveled-
We have, the following Set of information:
( | t2 – 8/ (t+1)2 | , t, 0, 5 ) = 42.59.
Here we can constitute that by Integrating, the absolute value of velocity we get the total distance covered (Total Area with the time axis and the velocity curve).
B. General Strategy: To calculate the fragmenting net effect from the finite sums of estimated changes in the given subject than we can do, Cardiac output, air pollution and the volume, what has added into our knowledge, many of these sums as Riemann sums and latter express them to limits. There are two main advantages of this:
1. We can do the evaluation in one of the integrals to get the desired result in less time, than to crank out the estimate from its finite sum.
2. The integral by itself becomes a formula. That empowers us to solve the similar problems, without repeating the modeling step.
Example 5: Modeling different implications of Acceleration.
Let us assume a couple is traveling in a car with an initial velocity of 10 mph accelerates at the rate of a(t) = 4t mph for time Period of 16 seconds.
Find:
1. What is the speed of the car moving when 16 seconds are up ?
2. Distance covered by the car when 16 seconds are passed ?
Solution: Let us look at first the effect of acceleration on the car's velocity.
STEP 1: When the acceleration is constant,
Velocity change = acceleration * time applied
The net change in velocity for the time interval 0 < = t < = 16.
STEP 2: Write a Definite Integral. The limit of these sums is:
016 a(t) dt.
STEP 3: Now, Evaluate the integral.
Net velocity change = ∫016 4 t dt = 2t2 |016 = 512 mph.
We also know that the initial velocity is 10 mph, so the total will be: 512 + 10 = 522.
(2) Now, applying acceleration for any length at time ‘t’ adds to,
0t 4 u du mph , where ‘u’ is just a dummy variable here, therefore we get-
v(t) = 5 + ∫0t 4u du = 5 + 1.2 u2 mph.
The total distance traveled from t = 0 to t = 16 sec is:
0t | v(t) | dt = ∫0t ( 5+ 4 t2 ) dt,
= [ 5t + 1.2t3]08,
= 654.4.
C. Consumption over Time- The integral works as natural tool to calculate the total accumulation and net change of more quantities in general. Integration is used to find the cumulative effect of varying rate of change and latter to integrate the term.
## Velocity and Distance Problems Involving Motion Along a Line
In daily life, motion is common thing to us, we walk, run and ride is a called as motion. General meaning of motion is change with respect to time like earth rotates once in every 24 hours.
In real life, when we change our Position in straight line, then this type of motion is called as a Straight Line motion and difference between one position to another position is called as a distance -
Distance = (position)2 - (position)1
We can define motion with the help of frame of reference, which include all the axis means x-axis, y-axis and z-axis. So, when a particle moves along a x-axis with respect to time, than this kind of motion is called as a straight motion or we can say that a motion along a straight line.
Now we discuss what distance is:
Distance is path length of a particle like we assume a motion of car along a straight line.
We choose the x-axis such that it coincides with the path of the car’s motion and origin of axis as the Point from where the car started moving, point O(x = 0, t = 0) to point P(x,t). So, path length from O to P is called as a distance of car. In mathematical point of view distance is -
Distance = OP = P(x, t) - O(0, 0),
Distance is different from displacement because distance is calculated with certain directions and on other hand displacement is a Scalar quantity means directionless quantity.
Now we discuss what velocity is and what difference exists between velocity and speed:
We know motion is change with respect to time, but how fast motion can change with respect to time, we don't know. So, Scientist Newton introduced velocity concept, which tells us that distance per unit time, is called as velocity -
Velocity = total distance/total time,
Let us assume a car moves from a origin position O(0, 0) to particular position P(2, 3) in time t = 3 sec. So, for calculating how fast position of car is changing with respect to origin in 1 sec, we use velocity formula here -
Velocity = change in position / time interval = (√(3 - 0)2 + (2 – 0)2) / (3 – 0) = (√ 9 + 4) / 3 = √ 13 / 3.
So, velocity of car is √ 13 / 3 km/sec.
Difference between velocity and speed: Speed is a dimensionless quantity means speed is equals to “total motion/total time” and other hand velocity is a quantity which has certain dimensions, so for calculating velocity, it is necessary to calculate distance first because distance is also a quantity which is calculated in certain directions. Therefore velocity is equals to total distance/total time. For calculating distance and velocity, we use Newton's laws of motion:
Newton's first law of motion says that a particle remains at rest until an external force acts on that particle. In mathematical point of view –
v = u + at …......................equation(1).
Where ‘u’ is initial velocity, ‘a’ is acceleration and ‘t’ is time.
Newton's second law of motion says that if there is an external force acting on particle than that force is directly proportional to its acceleration. In mathematical point of view -
S = ut + ½ at2 ….........................equation(2).
Here ‘S’ is total displacement, ‘u’ is initial speed, ‘a’ is acceleration and ‘t’ is total time.
Now we discuss Newton’s third law of motion, which says that there is an equal and opposite reaction to every action. In mathematical point of view -
v2 - u2 = 2as ….........................equation(3).
Where ‘v’ is final speed, ‘u’ is initial speed, ‘a’ is acceleration and ‘s’ is total displacement
Using above laws we can solve velocity and distance problems involving motion along a line. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.946301281452179, "perplexity": 801.8382767266505}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463607325.40/warc/CC-MAIN-20170523025728-20170523045728-00259.warc.gz"} |
https://byjus.com/physics/destructive-interference/ | # Destructive Interference
Wave undergoes phenomena like interference when it meets another wave. Interference definition states that it is a phenomenon in which two waves superpose with each other to form a resultant wave of lower, higher or of same amplitude.
There are two types of wave interference:
• Constructive interference
• Destructive interference
Constructive interference is a type of interference in which two interfering waves have a displacement in the same direction. In this article, let us understand the interference definition along with destructive interference in a detailed way.
## What is Destructive Interference?
Destructive interference occurs when waves come together so that they completely cancel each other out. When two waves destructively interfere, they must have the same amplitude in opposite directions.
There are many interesting wave phenomena in nature that cannot be defined by an individual wave. To comprehend the destructive interference phenomenon, we must examine based on the combination of waves. To examine these, we apply the principle of superposition which says:
“If two or more waves are traveling in a medium, the resulting wave function is the algebraic total of the individual waves function.”
When the waves of identical frequency and equivalent amplitude superimpose, the process of interference takes place. When two waves of a similar frequency move in a medium at the same time and in the same direction, because of their superposition, the resulting intensity of the medium at any point is dissimilar from the sum of their intensities. At some points, the strength of the resultant wave bears a larger value, while at certain points the value is very small.
### Destructive Interference Representation
When two waves of similar frequencies travel in a medium, the resultant at different points is dissimilar from the sum of their intensities because of their superposition. At one point it is too less and negligible which we call Destructive Interference.
The crest of wave A overlaps with the trough of wave B and vice versa as a result of which they cancel each other out.
In the above figure, A and B are the two waves of similar frequencies travelling in a medium in a similar direction.
You may also want to check out these topics related to the concept of interference given below!
## Destructive Interference Equation
The phase difference between two waves is an odd multiple of π that is:
(2n – 1) π
The difference between the path of two waves is an odd multiple of λ/2,
Δ = (2n–1) λ/2
The time interval among the two waves is an odd multiple of T/2,
θ = (2n–1) T/2
The resultant amplitude is equivalent to the difference between the amplitudes of individual waves.
## Examples of Destructive Interference
Some destructive interference examples are given below:
• Gravitational waves are a specimen of destructive interference.
• Light beams demonstrate destructive interference.
• Moving electrons and radio waves also perform destructive interference.
## Frequently Asked Questions – FAQs
### What is interference?
Interference is the phenomenon in which two waves superpose to form the resultant wave of the lower, higher or of the same amplitude.
### What are the types of light interference?
The following are the types of light interference:
• Constructive interference
• Destructive interference
### State the principle of superposition?
The superposition principle states that “if two or more waves are travelling in a medium, the resulting wave function is the algebraic total of the individual waves function.”
### Can light waves undergo interference?
Yes, light waves can undergo interference.
### Can moving electrons and radio waves undergo destructive interference?
Yes, moving electrons and radio waves can undergo destructive interference.
Hope you have understood about interference definition and destructive interference. Stay tuned to BYJU’S to learn more Physics concepts with the help of interactive video lessons.
## Watch the video and understand the relationship between frequency and wavelength.
Test your knowledge on Destructive interference | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8440415263175964, "perplexity": 697.2655702267324}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710909.66/warc/CC-MAIN-20221202150823-20221202180823-00623.warc.gz"} |
https://www.physicsforums.com/threads/feynman-lectures-on-physics-rotation-in-2-dimensions.576750/ | # Homework Help: Feynman Lectures on Physics - Rotation in 2 dimensions
1. Feb 12, 2012
### ZirconiumAce
This isn't really a 'problem', I'm just trying to follow Feynman's reasoning in section 18-2 of Volume 1 The Feynman Lectures on Physics. I've attached a png of the paragraph in question.
I have 2 issues with this:
1. If the length OQ = OP, how can there be a right angle at P(x,y)?
2. I don't understand why QP = r $\Delta\theta$, rather than r * tan ($\Delta\theta$)
Am I missing something? I've checked the latest errata and this isn't mentioned.
#### Attached Files:
• ###### Feynman.png
File size:
89.4 KB
Views:
114
2. Feb 12, 2012
### tiny-tim
Welcome to PF!
Hi ZirconiumAce! Welcome to PF!
Because that's the way these ∆s work …
we ignore anything that's small compared with a ∆ (in the same way that we ignore dh2, compared with dh, when we do calculus proofs)
So ∆θ = sin∆θ = tan∆θ.
(sin∆θ = ∆θ - (∆θ)3/3! + …)
3. Feb 12, 2012
### ZirconiumAce
Ah yes a Taylor series. Mathematics back in the days of the slide rule I suppose. Now we don't bother with simplifications as much. Cheers! How about the right angle?
4. Feb 12, 2012
I've got it. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9126611948013306, "perplexity": 3721.4177193620353}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794866511.32/warc/CC-MAIN-20180524151157-20180524171157-00567.warc.gz"} |
http://eprints.maths.manchester.ac.uk/996/ | # Backward Error of Polynomial Eigenproblems Solved by Linearization
Higham, Nicholas J. and Li, Ren-Cang and Tisseur, Françoise (2007) Backward Error of Polynomial Eigenproblems Solved by Linearization. SIAM Journal on Matrix Analysis and Applications, 29 (4). pp. 1218-1241. ISSN 0895-4798
The most widely used approach for solving the polynomial eigenvalue problem $P(\lambda)x = \bigl(\sum_{i=0}^m \l^i A_i\bigr) x = 0$ in $n\times n$ matrices $A_i$ is to linearize to produce a larger order pencil $L(\lambda) = \lambda X + Y$, whose eigensystem is then found by any method for generalized eigenproblems. For a given polynomial $P$, infinitely many linearizations $L$ exist and approximate eigenpairs of $P$ computed via linearization can have widely varying backward errors. Two main factors affect the backward error. First, because $L$ is usually highly structured, perturbations to $L$ cannot directly be interpreted as equivalent perturbations to $P$. Second, the short'' eigenvectors of $P$ can be recovered from the long'' eigenvectors of $L$ in potentially many ways, with differing implications on the backward error for $P$. We show that if a certain one-sided factorization relating $L$ to $P$ can be found then a simple formula permits recovery of right eigenvectors of $P$ from those of $L$, and the backward error of an approximate eigenpair of $P$ can be bounded in terms of the backward error for the corresponding approximate eigenpair of $L$. A similar factorization has the same implications for left eigenvectors. We use this technique to derive backward error bounds depending only on the norms of the $A_i$ for the companion pencils and for the vector space $\DL(P)$ of pencils recently identified by Mackey, Mackey, Mehl, and Mehrmann. In all cases, sufficient conditions are identified for an optimal backward error for $P$. These results are shown to be entirely consistent with those of Higham, Mackey, and Tisseur on the conditioning of linearizations of $P$. Other contributions of this work are a block scaling of the companion pencils that yields improved backward error bounds; a demonstration that the bounds are applicable to certain structured linearizations of structured polynomials; and backward error bounds specialized to the quadratic case, including analysis of the benefits of a scaling recently proposed by Fan, Lin, and Van Dooren. The results herein make no assumptions on the stability of the method applied to $L$ or whether the method is direct or iterative. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9539058804512024, "perplexity": 445.1072447520373}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256764.75/warc/CC-MAIN-20190522063112-20190522085112-00313.warc.gz"} |
https://www.physicsforums.com/threads/fourier-transform-and-translational-invariance.894079/ | # A Fourier transform and translational invariance
1. Nov 20, 2016
### ShayanJ
Can anyone explain what does the author mean by the statement below?
page 27 of this paper
I don't understand the relation between the Fourier transform and translational invariance.
Thanks
2. Nov 20, 2016
### The Bill
I think the author is referring to the translational invariance of the equation of motion for $ϕ$.
3. Nov 20, 2016
### ShayanJ
What does that have to do with Fourier transforming it?
4. Nov 20, 2016
### The Bill
The author mentioned Fourier decomposition. This is not the same as a Fourier transform.
5. Nov 20, 2016
### ShayanJ
He Fourier transforms the x coordinates and because the derivatives in the equation are only w.r.t. z, you can write the equation for each mode of the transformed field separately.
Draft saved Draft deleted
Similar Discussions: Fourier transform and translational invariance | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9892368316650391, "perplexity": 1046.8678678148221}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463608652.65/warc/CC-MAIN-20170526090406-20170526110406-00365.warc.gz"} |
https://en.wikiversity.org/wiki/Kerr%E2%80%93Newman_metric | # Theory of relativity/Kerr–Newman metric
(Redirected from Kerr–Newman metric)
The spacetime metric is, in Boyer-Lindquist coordinates,
${\displaystyle ds^{2}={\frac {\Delta ^{2}}{\rho ^{2}}}(dct-a\,\sin ^{2}\theta \,d\phi )^{2}-{\frac {\sin ^{2}\theta }{\rho ^{2}}}[(r^{2}+a^{2})d\phi -a\,dct]^{2}-{\frac {\rho ^{2}}{\Delta ^{2}}}dr^{2}-\rho ^{2}d\theta ^{2}}$
where
${\displaystyle \Delta ^{2}\equiv a^{2}+r^{2}\alpha }$
${\displaystyle \alpha =1-{\frac {2GM}{rc^{2}}}+{\frac {e^{2}}{r^{2}}}}$
${\displaystyle \rho ^{2}\equiv r^{2}+a^{2}\cos ^{2}\theta }$
${\displaystyle a\equiv {\frac {J}{Mc}}}$
${\displaystyle e\equiv {\frac {\sqrt {k_{e}G}}{c^{2}}}q}$
This represents the exact solution to General relativity/Einstein equations for the stress-energy tensor for an electromagnetic field from a charged rotating black hole. Defining three more functions of the coordinates
${\displaystyle \Sigma ^{2}\equiv {\sqrt {\left(r^{2}+a^{2}\right)^{2}-a^{2}\Delta ^{2}\sin ^{2}\theta }}}$
${\displaystyle \varpi \equiv {\frac {\Sigma ^{2}}{\rho }}\sin \theta }$
${\displaystyle \omega \equiv a{\frac {\left(r^{2}+a^{2}-\Delta ^{2}\right)}{\Sigma ^{4}}}c}$
The solution can now be written
${\displaystyle ds^{2}=\left({\frac {\Delta ^{2}-a^{2}\sin ^{2}\theta }{\rho ^{2}}}\right)dct^{2}+2{\frac {\omega }{c}}\varpi ^{2}dctd\phi -\varpi ^{2}d\phi ^{2}-{\frac {\rho ^{2}}{\Delta ^{2}}}dr^{2}-\rho ^{2}d\theta ^{2}}$
${\displaystyle ds}$ is an invariant line element, a measure of spacetime displacement between neighboring events. The displacement four vector between those events is ${\displaystyle dx^{\mu }}$, and being a four-vector, would yield an invariant scalar for the inner product of it with itself using the metric tensor ${\displaystyle g_{\mu \nu }}$ as a spacetime inner product operator as ${\displaystyle g_{\mu \nu }dx^{\mu }dx^{\nu }}$. We call that invariant scalar ${\displaystyle ds^{2}}$.
${\displaystyle ds^{2}=g_{\mu \nu }dx^{\mu }dx^{\nu }}$
So though technically it is the set of elements${\displaystyle \left[g_{\mu \nu }\right]}$ that is the metric tensor, since its elements can be directly read off of this line element as the coefficients of the coordinate differentials, in jargon ${\displaystyle ds^{2}}$ is often referred to as just "the metric". In the case that the charge ${\displaystyle q}$ is zero it becomes an exact vacuum solution to Einstein's field equations and is called just "the Kerr solution".
## Gravitational Red Shift Factor
The solution
${\displaystyle ds^{2}=\left({\frac {\Delta ^{2}-a^{2}\sin ^{2}\theta }{\rho ^{2}}}\right)dct^{2}+2{\frac {\omega }{c}}\varpi ^{2}dctd\phi -\varpi ^{2}d\phi ^{2}-{\frac {\rho ^{2}}{\Delta ^{2}}}dr^{2}-\rho ^{2}d\theta ^{2}}$
may also be written as
${\displaystyle ds^{2}=R^{2}dct^{2}-\varpi ^{2}\left(d\phi -{\frac {\omega }{c}}dct\right)^{2}-{\frac {\rho ^{2}}{\Delta ^{2}}}dr^{2}-\rho ^{2}d\theta ^{2}}$
where
${\displaystyle R\equiv {\sqrt {{\frac {\Delta ^{2}-a^{2}\sin ^{2}\theta }{\rho ^{2}}}+{\frac {\omega ^{2}}{c^{2}}}\varpi ^{2}}}}$
Lets say something neutral is equatorially orbiting in this spacetime with an angular velocity of ${\displaystyle \omega }$, then in using the solution in describing its path through spacetime, or world line, the ${\displaystyle \left(d\phi -{\frac {\omega }{c}}dct\right)}$ term vanishes and it is said to be "locally nonrotating". If it emits according to its local free fall frame a frequency ${\displaystyle f_{0}}$, then the frequency received by a remote observer ${\displaystyle f'}$ will be red shifted by
${\displaystyle f'=Rf_{0}}$
## Mathematical Surfaces
There are three important mathematical surfaces for this line element, the static limit and the inner and outer event horizons. The static limit is the outermost place something can be outside the outer horizon with a zero angular velocity. It is
${\displaystyle r_{s}={\frac {GM}{c^{2}}}+{\sqrt {\left({\frac {GM}{c^{2}}}\right)^{2}-a^{2}\cos ^{2}\theta -e^{2}}}}$
The event horizons are coordinate singularities in the metric where ${\displaystyle \Delta =0}$.
The outer event horizon is at
${\displaystyle r_{+}={\frac {GM}{c^{2}}}+{\sqrt {\left({\frac {GM}{c^{2}}}\right)^{2}-a^{2}-e^{2}}}}$
and the inner horizon is at
${\displaystyle r_{-}={\frac {GM}{c^{2}}}-{\sqrt {\left({\frac {GM}{c^{2}}}\right)^{2}-a^{2}-e^{2}}}}$
An external observer can never see an event at which something crosses into the outer horizon. A remote observer reckoning with these coordinates will reckon that it takes an infinite time for something infalling to reach the outer horizon even though it takes a finite proper time till the event according to what fell in.
## Kerr-Newman Equatorial Geodesic Motion
The exact equations of equatorial geodesic motion for a neutral test mass in a charged and rotating black hole's spacetime are
${\displaystyle {\frac {dt}{d\tau }}={\frac {\gamma \left(r^{2}+a^{2}+2a^{2}{\frac {GM}{rc^{2}}}-a^{2}{\frac {e^{2}}{r^{2}}}\right)-{\frac {al_{z}}{c}}\left({\frac {2GM}{rc^{2}}}-{\frac {e^{2}}{r^{2}}}\right)}{r^{2}-{\frac {2GMr}{c^{2}}}+a^{2}+e^{2}}}}$
${\displaystyle {\frac {d\phi }{d\tau }}={\frac {{\frac {l_{z}}{c}}\left(1-{\frac {2GM}{rc^{2}}}+{\frac {e^{2}}{r^{2}}}\right)+\gamma a\left({\frac {2GM}{rc^{2}}}-{\frac {e^{2}}{r^{2}}}\right)}{r^{2}-{\frac {2GMr}{c^{2}}}+a^{2}+e^{2}}}c}$
${\displaystyle {\frac {1}{2}}\left({\frac {dr}{d\tau }}\right)^{2}+V_{eff}=0}$
${\displaystyle V_{eff}=-{\frac {GM}{r}}+{\frac {e^{2}c^{2}}{2r^{2}}}+{\frac {1}{2}}{\frac {l_{z}^{2}}{r^{2}}}+{\frac {1}{2}}\left(1-\gamma ^{2}\right)c^{2}\left(1+{\frac {a^{2}}{r^{2}}}\right)-\left({\frac {GM}{r^{3}c^{2}}}-{\frac {e^{2}}{2r^{4}}}\right)\left({\frac {l_{z}}{c}}-a\gamma \right)^{2}c^{2}}$
where ${\displaystyle \gamma }$ is the conserved energy parameter, the energy per ${\displaystyle mc^{2}}$ of the test mass and ${\displaystyle l_{z}}$ is the conserved angular momentum per mass ${\displaystyle m}$ for the test mass.
## Kerr-Newman Polar Geodesic Motion
The exact equations of polar geodesic motion for a neutral test mass in a charged and rotating black hole's spacetime are
${\displaystyle {\frac {dt}{d\tau }}=\gamma \left({\frac {a^{2}+r^{2}}{a^{2}+e^{2}+r^{2}-{\frac {2GMr}{c^{2}}}}}\right)}$
${\displaystyle {\frac {1}{2}}\left({\frac {dr}{d\tau }}\right)^{2}-{\frac {{\frac {GM}{rc^{2}}}-{\frac {e^{2}}{2r^{2}}}}{1+{\frac {a^{2}}{r^{2}}}}}c^{2}={\frac {\gamma ^{2}-1}{2}}c^{2}}$
where ${\displaystyle \gamma }$ is the conserved energy parameter, the energy per ${\displaystyle mc^{2}}$ of the test mass.
## Wormhole Structure
Penrose diagram for coordinate extension of a charged or rotating black hole
Above we see a Penrose diagram representing a coordinate extension (1) for a charged or rotating black hole. The same way as mapping Schwarzschild coordinates onto Kruskal-Szekeres coordinate reveals two seperate external regions for the Schwarzschild black hole, such a mapping done for a charged or rotating hole reveals an even more multiply connected region for charged and rotating black holes. Lets say region I represents our external region outside a charged black hole. In the same way that the other external region is inaccesible as the wormhole connection is not transversible, external region II is also not accessible from region I. The difference is that there are other external regions VII and VIII which are ideed accesible from region I by transversible paths at least one way. One should expect this as the radial movement case of geodesic motion for a neutral test particle written above leads back out of the hole without intersecting the physical singularity at ${\displaystyle r=0,\theta ={\frac {\pi }{2}}}$.
## References
(1)Black Holes-Parts 4&5 pp 26-42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 44, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9106457829475403, "perplexity": 628.7061532945653}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153971.20/warc/CC-MAIN-20210730154005-20210730184005-00122.warc.gz"} |
http://mathhelpforum.com/calculus/95216-limit-calculation.html | 1. ## Limit calculation
Hey,
Can anyone please tell me why is this limit correct?
2. $\lim_{x\to 0}\frac{e^{\frac{-1}{x^{2}}}}{x}$
Let $t=\frac{1}{x^{2}}, \;\ x=\frac{1}{\sqrt{t}}$
Making the subs gives us:
$\lim_{t\to \infty}\frac{\sqrt{t}}{e^{t}}=0$
e grows much more than the radical, so it tends to 0
3. tnx.. =] | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9906613826751709, "perplexity": 3575.1846198578373}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982295103.11/warc/CC-MAIN-20160823195815-00078-ip-10-153-172-175.ec2.internal.warc.gz"} |
http://link.springer.com/article/10.1007/s00607-012-0241-9 | , Volume 95, Issue 1 Supplement, pp 617-638
Date: 27 Nov 2012
# Non-convex systems of sets for numerical analysis
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## Abstract
The notion of a system of sets generated by a family of functionals is introduced. A generalization of the classical support function of convex subsets of $$\mathbb R ^d$$ allows to transfer the concept of the convex hull to these systems of sets. Approximation properties of the generalized convex hull and its use for practical computations are investigated. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8791402578353882, "perplexity": 716.269251162029}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-14/segments/1427131299877.5/warc/CC-MAIN-20150323172139-00020-ip-10-168-14-71.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/2704/links-between-riemann-surfaces-and-algebraic-geometry/2849 | # Links between Riemann surfaces and algebraic geometry
I'm taking introductory courses in both Riemann surfaces and algebraic geometry this term. I was surprised to hear that any compact Riemann surface is a projective variety. Apparently deeper links exist.
What is, in basic terms, the relationship between Riemann surfaces and algebraic geometry?
-
If you would like a book on Riemann surfaces with a more algebro-geometric point of view, try Algebraic Curves and Riemann Surfaces by Rick Miranda. – David Corwin Aug 31 '10 at 4:32
For simplicity, I'll just talk about varieties that are sitting in projective space or affine space. In algebraic geometry, you study varieties over a base field k. For our purposes, "over" just means that the variety is cut out by polynomials (affine) or homogeneous polynomials (projective) whose coefficients are in k.
Suppose that k is the complex numbers, C. Then affine spaces and projective spaces come with the complex topology, in addition to the Zariski topology that you'd normally give one. Then one can naturally give the points of a variety over C a topology inherited from the subspace topology. A little extra work (with the inverse function theorem and other analytic arguments) shows you that, if the variety is nonsingular, you have a nonsingular complex manifold. This shouldn't be too surprising. Morally, "algebraic varieties" are cut out of affine and projective spaces by polynomials, "manifolds" are cut out of other manifolds by smooth functions, and polynomials over C are smooth, and that's all that's going on.
In general, the converse is false: there are many complex manifolds that don't come from nonsingular algebraic varieties in this manner.
But in dimension 1, a miracle happens, and the converse is true: all compact dimension 1 complex manifolds are analytically isomorphic to the complex points of a nonsingular projective dimension-1 variety, endowed with the complex topology instead of the Zariski topology. "Riemann surfaces" are just another name for compact dimension 1 (dimension 2 over R) complex manifolds, and "curves" are just another name for projective dimension 1 varieties over any field, hence the theorem you described.
As for why Riemann surfaces are algebraic, Narasimhan's book explicitly constructs the polynomial that cuts out a Riemann surface, if you are curious.
-
This relationship is a very beautiful one.
Imagine a Riemann surface. There are different ways to introduce it, but since you gave kind of a reference point, let's just define it as a projective variety in the complex projective plane. Now the people call it a surface because it looks two-dimensional from a real point of view. You can also draw a picture of Riemann surface covering a sphere by projection onto a coordinate.
Now what could be an object of study of algebraic geometry? Why, certainly it should be some geometric object defined by algebraic means. Among the different ways to start learning algebraic geometry let's say we selected the abstract definition of an algebraic curve. To recap, this a a geometry locally defined by algebraic equations in some space so that the resulting manifold is one-dimensional.
These algebraic curves can be studied purely abstractly. You can, e.g., define algebraic forms on these, and prove various theorems relating to their geometry.
But the beautiful fact is that those are two sides of the same medal. That's right:
Every Riemann surface is a complex algebraic curve and every compact complex algebraic curve can be embedded into a projective plane and drawn as the Riemann surface.
There are lots of gems in this short statement. For example, as I said there is a way to count the algebraic forms in terms of inner geometry of algebraic curve. This gives some number, which could be 0, 1, 2, etc. On the other hand, if you draw a Riemann surface, you notice that it can be studied in topology and then it has the invariant called the number of handles which could also be 0 (sphere), 1 (torus), 2, etc. It turns out this is exactly the same thing though defined in a completely different way by a completely different branch of mathematics.
The whole algebraic geometry is, so to say, our attempt to make ourselves comfortable about this amazing connection between things we calculate (algebra) and things we draw (geometry).
-
Most higher-genus curves cannot be smoothly embedded in the plane, but they fit nicely in three-space. – S. Carnahan Oct 27 '09 at 0:10
Yes, I corrected it to refer to non-smooth Riemann surfaces. – Ilya Nikokoshev Oct 27 '09 at 0:18
Is the result you alluded to called Riemann-Roch? – timur Jan 20 '11 at 2:35
From the analysis point of view, that every compact Riemann surface X is biholomorphic to a variety in complex projective space follows from an existence theorem for nonconstant meromorphic functions on X that Riemann proved by means of Dirichlet's principle (his proof was not rigorous, because Dirichlet's principle had not yet been made rigorous in his day). Whenever you have enough independent meromorphic functions on a compact complex manifold, you can put it in complex projective space. And a nonsingular variety in complex projective space has nonconstant globally meromorphic functions on it; just use ratios of homogenous coordinates. There are complex tori (compact quotients of C^n by lattices) that carry no nonconstant meromorphic functions at all; they obviously cannot sit in complex projective space. This stuff is nicely explained by Shafarevich in the second volume of his introduction to algebraic geometry.
Dirichlet's principle is an existence theorem for harmonic functions; this is relevant because harmonic functions on Riemann surfaces can be locally completed to holomorphic functions, and thus to meromorphic functions globally if topology allows (a question of monodromy).
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Mumford's great short book "Curves and their Jacobians" is about that "amazing synthesis of algebra, geometry and analysis", as Mumford expresses it. The book's goal is to provide readers an overview what the zoo of curves looks like. Arithmetic issues are not discussed.
Ouch, I forgot to recommend this very beautifull and readable book by Clemens. The chapter "Manin and the unity of mathematics" is esp. fascinating as it is about a startling connection with the arithmetics.
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There is a canonical way to embed a Riemann Surface into projective space.
A Riemann Surface X is called hyperlliptic if it admits a 2 to 1 holomorphic map φ:X → P1. Given a (compact) Riemann Surface X of genus g ≥3 which is not hyperelliptic one can define an embedding φK:X → Pg-1 called the canonical embedding (the construction of this map can found in Rick Miranda's book "Algebraic Curves and Riemann Surfaces").
From this embedding one can find equations for the (image of the) curve X. For example, a nonhyperelliptic curve of genus 3 is given by the vanishing of a quartic polynomial in P2, a nonhyperelliptic curve of genus 4 is defined by the vanishing of a quadratic and a cubic polynomial in P3.
The hyperelliptic case is not hard to understand. A hyperellitic curve of genus g is defined by an equation of the form y2 = h(x), where h is a polynomial of degree 2g + 1 or 2g + 2.
These results can be found in Miranda's book or in Griffiths-Harris "Principles of Algebraic Geometry"
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To me, excellent as the others are, engelbrekt's is the most direct answer to your question. I.e.
1) Every projective plane curve is a compact Riemann surface, essentially because of the implicit function theorem.
2) Conversely, every compact Riemann surface [immerses as] a projective plane curve because it has [enough] non constant meromorphic functions which almost embed it in the plane. Also every meromorphic function is the pullback of a rational function in the plane. So all the analytic structure is induced from the algebraic structure.
In higher dimensions, complex projective algebraic varieties are a special subcategory of compact complex spaces, namely those that admit [holomorphic is sufficient] embeddings in projective space.
More precisely, an n dimensional compact complex variety has a field of meromorphic functions that has transcendence degree ≤ n, and projective algebraic varieties are a subcategory of those (Moishezon spaces) for which the transcendence degree is n. Indeed I believe Moishezon proved the latter are all birational modifications of projective varieties.
I would also add something about the impact of Riemann surfaces on algebraic geometry. Namely it was Riemann's introduction of the topological and analytic points of view, showing that path integrals and differential forms could be profitably used to study projective algebraic curves, that deepened and revolutionized algebraic geometry forever.
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Algebraic curves (one-dimensional projective varieties) over the complex numbers are exactly Riemann surfaces. It confuses everyone at first when one is told "curves are surfaces." Almost everyone else calls \mathbb C the complex plane, but algebraic geometers call it the complex line.
One can work in any algebraically closed field, say \mathbb A, the field of algebraic numbers. But analysis only works in R or C, which are complete.
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https://proofwiki.org/wiki/Cartesian_Product_is_Empty_iff_Factor_is_Empty | # Cartesian Product is Empty iff Factor is Empty
## Theorem
$S \times T = \O \iff S = \O \lor T = \O$
Thus:
$S \times \O = \O = \O \times T$
### Family of Sets
Let $I$ be an indexing set.
Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Let $\displaystyle S = \prod_{i \mathop \in I} S_i$ be the Cartesian product of $\family {S_i}_{i \mathop \in I}$.
Then:
$S = \O$ if and only if $S_i = \O$ for some $i \in I$
## Proof
$\displaystyle S \times T$ $\ne$ $\displaystyle \O$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle \exists \tuple {s, t}$ $\in$ $\displaystyle S \times T$ Definition of Empty Set $\displaystyle \leadstoandfrom \ \$ $\displaystyle \exists s \in S$ $\land$ $\displaystyle \exists t \in T$ Definition of Cartesian Product $\displaystyle \leadstoandfrom \ \$ $\displaystyle S \ne \O$ $\land$ $\displaystyle T \ne \O$ Definition of Empty Set $\displaystyle \leadstoandfrom \ \$ $\displaystyle \neg \leftparen {S = \O}$ $\lor$ $\displaystyle \rightparen {T = \O}$ De Morgan's Laws: Conjunction of Negations
So by the Rule of Transposition:
$S = \O \lor T = \O \iff S \times T = \O$
$\blacksquare$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9409559965133667, "perplexity": 140.70848161627129}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738351.71/warc/CC-MAIN-20200808194923-20200808224923-00428.warc.gz"} |
http://tex.stackexchange.com/questions/7411/how-do-i-get-the-category-code-of-a-character-that-is-the-value-of-a-control-seq | # How do I get the category code of a character that is the value of a control sequence?
How can I get the category code of a character that is the value of a control sequence?
If I do this
The catcode for A is \the\catcodeA.
I get
The catcode for A is 11.
If I do this
\let\abc=A
The catcode for \abc\ is \the\catcode\abc.
I get "Missing number, treated as zero."
I think I understand it like this: \abc is replaced with the character code, category code pair <65, 11>, but TeX expects just a number after \the\catcode, so it inserts a 0, then leaves my A hanging around.
-
Pity one cannot add a tag for the! – Yiannis Lazarides Dec 19 '10 at 20:37
After doing \let\abc=@, you can access the catcode of @ by \expandafter\SomeMagic\meaning\abc, where \SomeMagic reads the meaning of \abc (which is the character @), and converts it to the form you want. I see two things you might want: know the catcode of the @ that is inside \abc, or the catcode that the character @ currently has. Just tell me which one you want and I can write an expandable command for that. – Bruno Le Floch Jan 29 '11 at 9:59
Sadly Knuth in his wisdom gave only few clues in the TeXBook for cases like this.
In order to understand the error we need to firstly understand the definition of \the, which is for the case of catcode \the<codename><8-bit number>, where <codename> stands for either \catcode, \mathcode, \lccode etc...
So clearly in this case catcode TeX expects a number and hence the error generated in the example provided by the OP.
All solutions provided by the other posts revolve around changing the definition one way or another to produce the required character code number and which I am demonstrating here with some different examples:
The example below will produce the right answer in both cases,
% results category 11
\makeatletter
\def\ABC{@ }
\the\catcode\ABC
% results category 12
\makeatother
\def\ABC{@ }
\the\catcode\ABC
While comparing two tokens using \ifcat things become a bit more complicated, if you want to compare two active characters you have to say \noexpand - Knuth says so somewhat obscurely in Exercise 20.11!
Consider the following definitions
\catcode[=13 \catcode]=13
\def[{*}
The following will result True since we comparing [ with ']' both now being category 13
\ifcat\noexpand[\noexpand] True \else False \fi
Also \ifcat[* True \else False \fi is True
Since now we have established two facts \the\catcode needs a number and if you use an active character, implicitly or explicitly we can understand why the following will all work!
\def\abc{A}
\chardef\abc=65
or Joseph's suggestion for comparisons:
\let\abc=A
\ifcat\noexpand\abc A%
\TRUE
\else
\FALSE
\fi
-
Stefan is right, it's a bit strange that you include the backtick in the definition of \ABC. (Although for illustration purposes it's interesting to see that then you don't need the \expandafter, I'm pretty sure that the OP wouldn't want the backtick included: He wants to test some existing macro, I guess.) – Hendrik Vogt Dec 20 '10 at 11:55
@Henrik you need the backtick so you can use the \the\catcode\ABC straight without an error you either have to do it at the definition stage or when typesetting \the\catcode..., with an expandafter as per Stefan's example. – Yiannis Lazarides Dec 20 '10 at 22:30
This works for me, using \expandafter:
\def\abc{A}
The catcode for \abc\ is \the\catcode\expandafter\abc.
and prints
The catcode for A is 11.
-
This is not quite what the question asks: you're using an explicit token whereas the example in the question is an implicit token. – Joseph Wright Dec 19 '10 at 16:30
@Joseph: Though Colin used \let, the question seemed a bit more general to me, so I made this suggestion in the hope that it might help. The catcode of a character as value of a control sequence is asked for, not the catcode of an implicit token itself. – Stefan Kottwitz Dec 19 '10 at 16:52
It helps a little, but I'm (sadly) still confused. If I have a \cs (perhaps assigned by \futurelet), how can I get the category code of that token? – Colin Fraizer Dec 19 '10 at 16:56
@Colin. I've updated my answer to reflect the fact that \ifcat is not quite the same as simply trying to get a number out. – Joseph Wright Dec 19 '10 at 17:09
@Stephan you can also use \def\abc{A}, then you do not need the expandafter! – Yiannis Lazarides Dec 19 '10 at 19:05
\chardef\abc=65
The catcode for \abc\ is \the\catcode\abc
-
For defining, directly writing \chardef\abc=A is easier, it doesn't require to know the ASCII code. – Stefan Kottwitz Dec 19 '10 at 16:56
Interesting approach, but means that \abc can't be expanded and then used in something like \pdfstrcmp. I guess it depends on the application! – Joseph Wright Dec 19 '10 at 17:09
it is the same as the already defined \@ne, \tw@, \thr@@. I never used \pdfstrcmp, because I do not need it for PSTricks :-)) But I use \chardef quite often ... – Herbert Dec 19 '10 at 17:29
As the excellent TeX by Topic gives us a clue in the discussion of \ifcat:
Control sequence tokens are considered to have category code 16, which makes them all equal to each other, and unequal to all character tokens.
This 'magic' category code is not accessible by normal methods. However, things do work with \ifcat if you have implicit tokens:
\let\ABC=A
\ifcat\noexpand\ABC A%
\TRUE
\else
\FALSE
\fi
While you do not need the \noexpand here, if this were for example part of a \futurelet then you would do, in case the test token was expandable.
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this of course will fail the test \the\catcode\ABC` looking for a number! – Yiannis Lazarides Dec 19 '10 at 20:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8987517952919006, "perplexity": 2041.0328241399675}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398449160.83/warc/CC-MAIN-20151124205409-00139-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://www.cheenta.com/series-convergence-iit-jam-2018-problem-12/ | Select Page
Understand the problem
Let a, b, c $\in \mathbb{R}$ Which of the following values of a ,b, c do NOT result in the convergence of the series $\sum_{n=3}^{\infty} \frac{a^{n}}{n^{b} (log_en)^{c}}$ (a) $|a|<1 , b \in \mathbb{R} , c \in \mathbb{R}$ (b) $a=1 , b>1 , c \in \mathbb{R}$ (c) $a=1 , b \leq 1 , c<1$ (d) $a=-1 , b \geq , c>0$
Source of the problem
IIT JAM 2018 Problem 12
Topic
Convergence of a seris
Easy
Suggested Book
Real Analysis By S.K Mapa
Do you really need a hint? Try it first!
One disclaimer: In this question you will see that for some option the series is clearly convergent and for some option it might be convergent and might not be. So the question wordings are not very clear. Now having that disclaimer, what we have to find is the options where we have the series might be or might not be convergent. I want to end this hint here to give you a bit more room to search. Look for Leibnitz rule for alternating series.[ In mathematics Leibnitz’s test states that if ${u_n}$ be a monotone decreasing sequence of positive real numbers and lim $u_n = 0$ , then the alternating series $u_1 - u_2 + u_3 - u_4 + ...........$ is convergent.]
$\sum_{n=3}^{\infty} \frac{a^{n}}{n^{b} (log_e^{n})^{c}}$ Let us talk about option D first $a=-1, b \geq 0 , c<0$ $\sum_{n=3}^{\infty} \frac{(-1)^{n}}{n^{b}(ln^{n})^{c}}$ where $\frac{1}{n^{b}(ln n)^{c}} \longrightarrow 0$ as $n \longrightarrow \infty$ Hence the series is convergence in this case. So option D is rejected. Now look for the other option and see Cauchy condensation test. (For a non increasing sequence $f(n)$ of non-negative real numbers, the series $\sum_{n=1}^{\infty} f(n)$ converges if and only if the “condensed series” $\sum_{n=0}^{\infty} 2^{n} f(2^{n})$ converges.Moreover if they converge,the sum of the condensed series is no more than twice as large of the sum as original) and D’ Alembert’s test( Let $\sum u_n$ be a series of positive real numbers and let $lim \frac{u_n +1}{u_n} = l$ Then $\sum u_n$ is convegent if $l<1$ , $\sum u_n$ is divergent if $l>1$.
Moving on to option c $a=1 , b \geq 0 , c<1$ $\sum_{n=3}^{\infty} \frac{1}{n^{b} (log_{e}^{n})^{c}} := S$(say) Observe if we have $c=b=\frac{1}{3}$ thus $S = \sum_{n=3}^{\infty} \frac{1}{n^{\frac{1}{3}} (log_{e}^{n})^{\frac{1}{3}}} > \sum_{n=3}^{\infty} \frac{1}{n^{\frac{2}{3}}} \longrightarrow \infty$ So, by comparison test(Let $\sum u_n$ and $\sum v_n$ be two series of positive real numbers and there is a natural number m such that $u_n \leq kv_n$ for all $n \geq m,k$ being a fixed positive number. Then (i) $\sum u_n$ is convergent if $\sum v_n$ is convergent. we have S is divergent. (ii) $\sum u_n$ is divergent if $\sum u_n$ is divergent.) Now the question is: Can we get some point where the series is convergent? The first bet would be making $b>1$ say $b=2$ and make $c$ smaller Let $c= \frac{1}{2}$ Thus $S= \sum_{n=3}^{\infty} \frac{1}{n^{2} (log_{e}^{n})^{\frac{1}{2}}}$ Here $\sum_{n=3}^{\infty} \frac{1}{n^{2} (log n)^{\frac{1}{2}}} < \sum \frac{1}{n^{2}} < \infty$ So, $S$ is convergent and c is one correct answer. Look for the others.
Option b $a=1,b>1,c \in \mathbb{R}$ $S=\sum_{n=3}^{\infty} \frac{1}{n^{b}(ln n)^{c}}$ If $c=2$ clearly by comparison test $S$ will be convergent Now the question is that, can we find one example such that the series will be divergent? Observe that, if $c \geq 0$ then as $b>1$ we will get that the series is convergent. What will happen if $c<0$ Here Cauchy Condensation test comes into play Consider $a=2>1$ thus $\sum_{n=3}^{\infty} \frac{2^{n}}{(2^{n})^{b} (ln 2^{n})^{c}} =\sum_{n=3}^{\infty} \frac{{2^{n}}^{1-b}} {n^{c} (ln 2)^{c}} =\frac{1}{(ln 2)^{c}} \sum_{n=3}^{\infty} \frac{1}{(2^{b-1})^{n} n^{c}}$ Now we have to use D’ Alembert’s Ratio test : Consider $a_n = \frac{1}{(2^{b-1})^{n} n^{c}}$ Thus $\frac{a_{n+1}}{a_n} = \frac{(2^{b-1})^{n} n^{c}}{(2^{b-1})^{n+1} (n+1)^{c}} \longrightarrow \frac{1}{2^{b-1}} < 1$ Hence the series is convergent and so the series is convergent for any value of $c$ and here the series is convergent always and that is why option b is not correct.
option a) $|a| < 1, b \in \mathbb{R} , c \in \mathbb{R}$ Here if we consider $a<0 , b<0 , c<0$ The series is convergent by Leibnitz test . So, the question is whether we can find out some values of $a,b,c$ such that the series will be divergent. Consider $a_n = \frac{a^{n}}{n^{b} (\log_e n )^{c}}$ $\frac{a_{n+1}}{a_n} = \frac{a}{(1+ \frac{1}{n})^{b} (\frac{(log{n+1}}{log{n})}^{c}}$ Now we know that $\frac{n+1}{n} \longrightarrow 1$ We have to think about $\frac{\log(n+1)}{\log n}$ Let us consider $\lim_{x \to \infty} \frac{\log(x+1)}{\log(x)}=\lim_{x \to \infty} \frac{\frac{1}{x+1}}{\frac{1}{x}}=\lim_{x \to \infty} \frac{x}{x+1}=1$[using L’Hopital’s rule which states that for function f and g which are differentiable on an open interval I except possibly at a point c contained in I if ${\lim}_{x \to c} f(x) = {\lim}_{x \to c} g(x) = 0$ or $-\infty , +\infty , g'(x) \neq 0$ for all x in I with $x \neq c$ and ${\lim}_{x \to c} \frac{f'(x)}{g'(x)}$ exist then ${\lim}_{x \to c} \frac{f(x)}{g(x)} = {\lim} \frac{f'(x)}{g'(x)}$ ] So, $\frac{\log(n+1)}{\log n} \to 1$ And hence $\lim{n \to \infty} |\frac{a_{n+1}}{a_n}|=|a|<1$. So,the series is convergent $\forall |a|<1,b,c \in \mathbb{R}$Hence c. is the only correct answer.
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Try this beautiful problem from IIT JAM 2018 which requires knowledge of Real Analysis (Limit of a Sequence). We provide sequential hints. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 70, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9454400539398193, "perplexity": 663.1366939155986}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370493818.32/warc/CC-MAIN-20200329045008-20200329075008-00541.warc.gz"} |
http://math.stackexchange.com/questions/215703/how-to-find-the-inverse-to-fx-x2-6x-11 | # How to find the inverse to $f(x)= x^2 - 6x + 11$
If the inverse exists, how do I find the inverse to this function:
$$f(x)= x^2 - 6x + 11$$
with $x \le 3$
Stuck at the quadtric formula. I think i have got the right answer which is $x = 3 ± \sqrt{y-2}$ ? But it doesnt seem right.
-
Have you tried the quadratic formula? – EuYu Oct 17 '12 at 15:35
Why doesn't the answer (which is correct, by the way) seem right? – Rick Decker Oct 17 '12 at 16:05
If the inverse exists, you just write $y=x^2-6x+11$ and use the quadratic formula to get $x$ in terms of $y$. To see if it exists, you need to ensure that for a given $y$ there is only one $x$. The obvious threat is the $\pm$ sign in the quadratic formula.
-
I have tried it. But Im stuck at the quadtric formula. I think i have got the right answer which is x = 3 ± squareroot (y-2) ? But it doesnt seem right. – Karoline Oct 17 '12 at 15:39
Karoline, your efforts (including thought process and the answer which you got) would be a welcome addition to the body of the question. Feel free to edit your original question and add anything that you feel would help those who are trying to help you. – The Chaz 2.0 Oct 17 '12 at 15:41
@Karoline: you have used the quadratic formula correctly. What doesn't seem right about it? If you plot the graph, the inverse function consists of interchanging the $x$ and $y$ axes. This corresponds to turning the paper over along the diagonal from lower left to upper right and looking through the paper. Then you can apply the vertical line test. – Ross Millikan Oct 17 '12 at 15:46
@TheChaz: to notify a user (at least as I understand it) you need to precede the username with an @ sign. Probably a typo and you get this, but if not it will help. – Ross Millikan Oct 18 '12 at 5:20
I was under the impression that the author of the question is notified whenever there is a comment anywhere on the page. Maybe I'm wrong! Just in case... @Karoline, see above :) – The Chaz 2.0 Oct 18 '12 at 5:36
You don't need the quadratic formula, just complete the square! You start with $$y = x^2 - 6x + 11$$ or in other words $$x^2 - 6x + 11 - y = 0$$ Now your goal is to write that as $(x + a)^2 + \ldots = 0$ for some $a$. Observe that by expanding that square you get $x^2 + 2ax + \ldots$. Matching that to your original equation shows that you have to pick $a=-3$. That produces the correct coefficients for $x^2$ and $x$, so all you need to do is correct for the differing constant term. You get $$(x - 3)^2 + 2 - y = 0$$ which via simple algebra yields $$x = 3 \pm \sqrt{y - 2}$$
Note that this always works! If the coefficient of $x^2$ in your equation isn't $1$, just divide the whole equation by the coefficient before you start. Once you've praticed this square completion a few times, you'll be at least as fast as with the formula, and you won't have to remember the formula anymore. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8298833966255188, "perplexity": 245.76129743320942}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997892648.47/warc/CC-MAIN-20140722025812-00004-ip-10-33-131-23.ec2.internal.warc.gz"} |
http://www-old.newton.ac.uk/programmes/DIS/seminars/2009051511009.html | # DIS
## Seminar
### The Moutard transformation and its applications
Taimanov, I (Novosibirsk State)
Friday 15 May 2009, 11:00-12:00
Satellite
#### Abstract
We discuss the Moutard transformation which is a two-dimensional generalization of the Darboux transformation of the Schroedinger operator and expose some applications of this transformations to spectral theory and soliton equations. In particular, we expose examples of Schroedinger operators with smooth fast-decaying potentials and nontrivial kernels in $L_2$ and of blowing up solutions of the Novikov--Veselov equation, a two-dimensional generalization of the Korteweg--de Vries equation. These results we obtained jointly with S.P. Tsarev.
#### Video
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If it doesn't, something may have gone wrong with our embedded player.
We'll get it fixed as soon as possible. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8110343217849731, "perplexity": 1751.5314840478143}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500823169.67/warc/CC-MAIN-20140820021343-00166-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://www.getbeautytip.com/2022/03/quotient-rule-derivative-product-rule.html | VqI14dIZgOPEqICDVdzsdHohm6R1qA6BYQ86dmeQ
# Quotient Rule Derivative Product Rule
and a function is equal to the total of the derivatives of and. The derivative of the difference between two functions is the same as the difference between the derivatives of and. and a function are equivalent to the difference of their derivatives. The derivative of a product of two functions equals the derivative of the first function multiplied by the derivative of the second function multiplied by the derivative of the second function multiplied by the derivative of the first function multiplied by the derivative of the second function multiplied by the derivative of the first function.
Multiply g(x) by the derivative of f (x).
The d signifies a derivative in this expression. As a result, df(x) denotes the derivative of the function f, whereas dg(x) denotes the derivative of the gram function. The formula specifies that in order to determine the derivative of f(x) divided by g(x), you must first remove the product of f(x) times the derivative of g from that item (x).
The derivative formula for a ratio of functions
The quotient rule is a technique used in calculus for determining the derivative of a function that is the ratio of two differentiable functions.
[1]
[2]
[3] Let f (x) equal g (x) / h (x), where both g and h are differentiable and h (x) equals 0. (x)
Related Posts | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.987466037273407, "perplexity": 174.4304244055544}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335573.50/warc/CC-MAIN-20221001070422-20221001100422-00143.warc.gz"} |
https://physics.stackexchange.com/questions/636033/incline-two-objects-same-coefficient-of-friction-different-masses-which-move | # Incline, two objects, same coefficient of friction, different masses, which moves first? Coefficient of friction is independent of mass
Two objects on an incline beside each other, same coefficient of friction, #1 box weight=1 kg, the other #2 box=1000 kg. As the incline rises and the maximum static friction force is exceeded the boxes will slide down the incline. Will they both slide at the same angle of inclination independent of their masses. Does having the same coefficient of friction, which is independent of mass, have the same maximum static frictional force? Wouldn't the mass reflect in the max static frictional force and therefore each box move at a different angle of inclination and yet still have the same coefficient of friction? This is not about acceleration just the initial move of each box from its stationary position.
The heavier box has 1000 times larger maximum static friction force than the lighter box, as you note (this holds for every angle). However, the heavier box also pushes against this friction 1000 times harder (for any angle). Therefore they will both start moving at the same angle.
(In reality such a large difference in mass world likely break the assumption of a single common friction coefficient. The heavier box may for example well make an indentation for itself that makes it harder for it to start moving. Or it could go the other way.)
• Thank you Kristoffer! This has helped my understanding clearly. May 15 at 12:49
Will they both slide at the same angle of inclination independent of their masses.
Yes. Impending motion of both will occur at the same angle and that angle will satisfy the following equation where $$\mu_s$$ is the coefficient of static friction.
$$\tan\theta=\mu_s$$
PROOF: (See free body diagram, below)
The gravitational force acting down the incline is
$$mg\sin\theta$$
The maximum possible static friction force acting up the plane is
$$F_{f}=\mu_{s}N=\mu_{s}mg\cos\theta$$
Impending motion occurs when the two are equal, or
$$mg\sin\theta=\mu_{s}mg\cos\theta$$
$$\mu_{s}=\frac{\sin\theta}{\cos\theta}=\tan\theta$$
Hope this helps.
• Thank you Bob! The formula and diagram help put it all together. May 15 at 12:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9268183708190918, "perplexity": 608.4910343020167}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964362992.98/warc/CC-MAIN-20211204124328-20211204154328-00351.warc.gz"} |
http://farside.ph.utexas.edu/teaching/sm1/lectures/node53.html | Next: Hydrostatic equilibrium of the Up: Classical thermodynamics Previous: Calculation of specific heats
Suppose that the temperature of an ideal gas is held constant by keeping the gas in thermal contact with a heat reservoir. If the gas is allowed to expand quasi-statically under these so called isothermal conditions then the ideal equation of state tells us that
(311)
This is usually called the isothermal gas law.
Suppose, now, that the gas is thermally isolated from its surroundings. If the gas is allowed to expand quasi-statically under these so called adiabatic conditions then it does work on its environment, and, hence, its internal energy is reduced, and its temperature changes. Let us work out the relationship between the pressure and volume of the gas during adiabatic expansion.
According to the first law of thermodynamics,
(312)
in an adiabatic process (in which no heat is absorbed). The ideal gas equation of state can be differentiated, yielding
(313)
The temperature increment can be eliminated between the above two expressions to give
(314)
which reduces to
(315)
Dividing through by yields
(316)
where
(317)
It turns out that is a very slowly varying function of temperature in most gases. So, it is always a fairly good approximation to treat the ratio of specific heats as a constant, at least over a limited temperature range. If is constant then we can integrate Eq. (316) to give
(318)
or
(319)
This is the famous adiabatic gas law. It is very easy to obtain similar relationships between and and and during adiabatic expansion or contraction. Since , the above formula also implies that
(320)
and
(321)
Equations (319)-(321) are all completely equivalent.
Next: Hydrostatic equilibrium of the Up: Classical thermodynamics Previous: Calculation of specific heats
Richard Fitzpatrick 2006-02-02 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.929667055606842, "perplexity": 591.5644512101927}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701156520.89/warc/CC-MAIN-20160205193916-00126-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://www.ma.utexas.edu/mp_arc-bin/mpa?yn=03-493 | 03-493 Jens Hoppe, Ari Laptev and J rgen stensson
Follytons and the Removal of Eigenvalues for Fourth Order Differential Operators. (29K, Latex2E) Nov 9, 03
Abstract , Paper (src), View paper (auto. generated ps), Index of related papers
Abstract. A non-linear functional $Q[u,v]$ is given that governs the loss, respectively gain, of (doubly degenerate) eigenvalues of fourth order differential operators $L = \partial^4 + \partial\,u\,\partial + v$ on the line. Apart from factorizing $L$ as $A^{*}A + E_{0}$, providing several explicit examples, and deriving various relations between $u$, $v$ and eigenfunctions of $L$, we find $u$ and $v$ such that $L$ is isospectral to the free operator $L_{0} = \partial^{4}$ up to one (multiplicity 2) eigenvalue $E_{0} < 0$. Not unexpectedly, this choice of $u$, $v$ leads to exact solutions of the corresponding time-dependent PDE's.
Files: 03-493.src( 03-493.comments , 03-493.keywords , hlo.tex ) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9739753007888794, "perplexity": 1921.260069318301}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676591481.75/warc/CC-MAIN-20180720022026-20180720042026-00605.warc.gz"} |
http://mathoverflow.net/questions/95341/connectedness-of-the-complement-of-the-zero-set-of-a-polynomial-p-sln-mathbb?sort=newest | # connectedness of the complement of the zero set of a polynomial $P: SL(N,\mathbb{C})^n \rightarrow \mathbb{C}$
I know that the complement of the zero set of a polynomial $P: \mathbb{C}^n \rightarrow \mathbb{C}$ is connected in $\mathbb{C}^n$ (by the way, can anybody suggest a reference?).
Is it possible to extend the proof also to polynomials $P: SL(N,\mathbb{C})^n \rightarrow \mathbb{C}$ ?
Thanks!
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Could you give us some more details? Which topologies are you using? What exactly is a polynomial $P\colon SL(N,\mathbb{C})^n \to \mathbb{C}$? – Mark Grant Apr 27 '12 at 10:34
Do you mean the restriction of a polynomial in $N^2n$ variables to the special linear group? – Mark Grant Apr 27 '12 at 13:18
Sorry. It is more ambiguous than I thought. I assume a matrix representation of the elements of $SL(N,\mathbb{C})$. In this way I define the sum as the sum of matrices. The pruduct is the one of the group. So, the polynomial are defined. Yes, I guess this is equivalent to polynomial in $N^2 n$ variables restricted to the special linear group. As for the topology, I was assuming the topology inherited from $\mathbb{C}^{n(N^2−1)}$ via the exponential parametrization of the group from the algebra. But other topologies in which the set is connected might be useful. – Luigi Scorzato Apr 27 '12 at 13:36
@Luigi: I was interpreting things as in Neil's answer (I think). However, your comment (and also that of Mrc Plm) makes it sound like you want to take a polynomial in $n$ variables (with $\mathbb{C}$ coefficients, say) and plug in $n$ matrices. But this would give back a matrix, right? – Mark Grant Apr 27 '12 at 13:54
No, my comment was non-sense. I was also thinking in my answer about a polynomial to $\mathbb{C}$, and I worked with the entries as well. – Marc Palm Apr 27 '12 at 14:00
$SL(n,\mathbb{C})^N$ is a smooth and irreducible variety, and thus a manifold. The zero set $Z$ of a nonzero polynomial is a subvariety with $\dim_{\mathbb{C}}(Z)\leq Nn^2-1$ and so $\dim_{\mathbb{R}}(Z)\leq 2Nn^2-2$. If $a$ and $b$ are points in $Z^c$ then we can choose a smooth path $\mathbb{R}\to SL(n,\mathbb{C})^N$ joining them, then perturb it slightly to make it transverse to $Z$. As $\dim_{\mathbb{R}}(\mathbb{R})+\dim_{\mathbb{R}}(Z)<\dim_{\mathbb{R}}(SL(n,\mathbb{C})^N)$, transversality just means that the path does not meet $Z$, so we get a path in $Z^c$ as desired.
For this we need a result telling us that the path can be made transverse. This is a standard fact in differential topology if $Z$ is a submanifold, which is the generic case. Essentially the same proof should work even if $Z$ is not smooth, though it might be harder to find a convenient reference.
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thanks! could you please give me a reference for the standard fact about transversality of the path? – Luigi Scorzato Apr 27 '12 at 13:54
@Neil: Just use Alexander duality to show that a subcomplex of dimension $<k-1$ cannot separate ${\mathbb R}^{k}$. In this particular case, complex subvariety in ${\mathbb C}^n$ of complex dimension $<n$ admits a triangulation (Lojasiewicz), so we can regard it as a simplicial complex of real dimension $<2n-1$. The same thing works locally. – Misha Apr 27 '12 at 14:01
An answer to the more general question could be in three steps.
1) If $X$ is a complex algebraic variety, connected for the Zariski topology, then it is connected for the usual topology, an important and nontrivial fact that can be found in Mumford's Red Book, or in Shafarevich's one.
2) The complement to a closed Zariski subset in an irreducible variety is connected for the Zariski topology.
3) The varieties $SL(n,{\bf C})$ and $SL(N,{\bf C})^N$ are irreducible
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thanks! About point 2) I have probably imprecise definitions in mind and a quick googling does not help me. Isn't the set $\{x\in \mathbb{R}^n| x_1=0 \}$ a closed Zariski subset in an irreducible variety? Because its complement in $\mathbb{R}^n$ is not connected. What am I missing? – Luigi Scorzato Apr 28 '12 at 9:16
R^n isn't a complex variety – Ostap Chervak Apr 28 '12 at 12:37
interesting! So, in that case it is point 1) that fails? i.e. the complement of $\{ x\in \mathbb{R}^n |x_1=0\}$ is Zariski connected? Really interesting... – Luigi Scorzato Apr 28 '12 at 12:52
@Luigi: Yes, considering $\mathbf R$ or $\mathbf C$ makes a great difference. The affine line minus the origin over a field $k$ is connected (it is affine with ring of functions given by $k[T,T^{-1}]$ is an integral domain). If you look at $\mathbf C$-points, you get the complement to the origin in~$\mathbf C$ (a punctured plane, hence a connected topological space), but if you look at $\mathbf R$-points, you get the complement to the origin in $\mathbf R$ which is obviously not connected. – ACL Jun 6 '12 at 20:28
Let $X_{j}$ be elements of $SL(n, \mathbb{C})$ with entries ${X_{j}^{k,l}}$.
Express the polynomial $P: SL(n,\mathbb{C})^N \rightarrow \mathbb{C}$ as a polynomial $Q: \mathbb{C}^{n \cdot N} \rightarrow \mathbb{C}$ in variables ${X_{j}^{k,l}}$
Similar express the polynomial $$Q''(X_1, \cdots, X_n) = \prod\limits_{i=1}^N \left( \det(X_i) -1 \right)$$ in variables ${X_{j}^{k,l}}$.
You can then apply the well-known result for $\mathbb{C}^{N \cdot n^2}$ to $Q' \cdot Q''$, and get what you want.
Edit due the comment: Note that we have both $$\{ Q' Q'' = 0 \} \subset \mathbb{C}^{N \cdot n^2}$$ is a connected subset, and that it is really a subset of $SL_n(\mathbb{C})^N$ $$\{ Q' Q'' = 0 \} \subset \{ Q'' =0 \} = SL_n(\mathbb{C})^N.$$
I am not sure, what topology you are working in, but it holds in any topology, the original results for $\mathbb{C}^N$ holds in;)
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Sorry, but I still do not see the conclusion: Your argument shows that the complement of the zero set of $Q'\cdot Q''$ is connected in $\mathbb{C}^{N n}$ but this does not mean that it is connected in $SL_n(\mathbb{C})$. Does it? – Luigi Scorzato Apr 27 '12 at 13:15
Clear now?..... – Marc Palm Apr 27 '12 at 13:48
I apologize for my misleading fomulation – Luigi Scorzato Apr 28 '12 at 8:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8542596101760864, "perplexity": 246.43090274775466}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246660743.58/warc/CC-MAIN-20150417045740-00174-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.groundai.com/project/a-search-for-67-ghz-methanol-masers-in-m338699/ | # A Search for 6.7 GHz Methanol Masers in M33
Paul F. Goldsmith 11affiliation: Jet Propulsion Laboratory, California Institute of Technology, Pasadena CA; [email protected] , Jagadheep D. Pandian 22affiliation: Department of Astronomy, Cornell University, Ithaca NY and Max-Planck-Institut für Radioastronomie, Aug dem Hügel 69, 53121, Bonn; [email protected] , and Avinash A. Deshpande 33affiliation: Raman Research Institute, Bangalore; [email protected]
###### Abstract
We report the negative results from a search for 6.7 GHz methanol masers in the nearby spiral galaxy M33. We observed 14 GMCs in the central 4 kpc of the Galaxy, and found 3 upper limits to the flux density of 9 mJy in spectral channels having a velocity width of 0.069 kms. By velocity shifting and combining the spectra from the positions observed, we obtain an effective 3 upper limit on the average emission of 1mJy in a 0.25 kms channel. These limits lie significantly below what we would expect based on our estimates of the methanol maser luminosity function in the Milky Way. The most likely explanation for the absence of detectable methanol masers appears to be the metallicity of M33, which is modestly less than that of the Milky Way.
ISM: molecules – individual(methanol); ISM: masers; Galaxies (M33)
## 1 Introduction
6.7 GHz methanol masers are the second brightest maser transition ever observed, and are typically much brighter than OH masers. The properties of methanol masers in the Magellanic clouds (Sinclair et al., 1992; Ellingsen et al., 1994; Beasley et al., 1996) are consistent with those of our Galaxy, given appropriate consideration for different galactic properties such as metallicity. The SMC has an oxygen abundance 12 + log(O/H) = 7.96 (Vermeij & van der Hulst, 2002), which is a factor 6 smaller than that of GMCs in the Milky Way (Peimbert, Storey, & Torres-Peimbert, 1993).
However, 6.7 GHz methanol masers have not been discovered in galaxies beyond the Magellanic clouds. In particular, surveys have shown that there is no analog to OH megamasers at 6.7 GHz (Ellingsen et al., 1994; Phillips et al., 1998; Darling et al., 2003). It would be of interest to detect methanol masers in a Milky Way like galaxy. If the number of sources detected were large, one could derive the methanol maser luminosity function since all sources would be at the same distance. Further, one could look for correlations of methanol masers with giant molecular cloud masses, other types of masers, etc. The Arecibo radio telescope offers unequaled sensitivity for targeted surveys for methanol masers, and the nearest spiral galaxy which can be observed with this instrument is M33.
It is difficult to develop an optimal strategy to search for extragalactic methanol masers since the luminosity function of methanol masers in our Galaxy is unknown. This is mostly due to difficulties in determining distances to methanol masers, which is compounded by the kinematic distance ambiguity. We have adopted the following approach. We selected sources from the general catalog (Pestalozzi, Minier, & Booth, 2005), eliminating sources wihtin 10 degrees of longitudes 0 and 180 degrees since these suffer from large uncertainties in kinematic distance. Assuming that all masers are at their near kinematic distance, we then scaled the source flux density to the distance of M33, which we take to be 840 kpc, averaging the results of Lee et al. (2002) and Freedman, Wilson, & Madore (1991), which gives a result consistent with the distance determined by Kim et al. (2002). The resulting histogram of “expected” methanol maser flux densities is showin in Figure 1
We thereby obtained an estimate for the distribution of flux density of methanol masers in M33, which of course must be regarded with considerable skepticism, as the original Galactic sample has significant biases, not to mention the unknown differences in the properties of massive star formation in M33 and the Milky Way. It is also inherently pessimisitic in the sense that some of the Galactic masers are at their far kinematic distance, and hence are more luminous than we have assumed. With these caveats, we found a “high flux density tail” for our hypothetical methanol maser population in M33 that extends from 100 mJy down to 1 mJy. Approximately 16% of the total number of masers would be expected to have flux density greater than 1 mJy, with about 6% having flux density greater than 3 mJy.
## 2 Observations
The search for methanol masers in M33 was carried out using the 305 m Arecibo telescope111The Arecibo Observatory is a facility of the National Astronomy and Ionosphere Center which is operated with support from the National Science Foundationand the C-band high receiver (Pandian et al., 2006) between 2005 July 9 and July 21. The data were taken in standard Arecibo position switched mode in which the source is observed for five minutes, and then a reference position offset by 6 in Right Ascension is observed for an equal length of time. The spectrometer used was the interim correlator. Two boards recorded the two orthogonal linear polarization data at 0.069 kmsresolution (3.125 MHz bandwidth with 2048 lags) while the two other oboards recorded the same data at a higher velocity resolution of 0.034 kms(1.5625 MHz bandwidth with 2048 lags). All the boards were centered on the CO velocity as given by Engargiola et al. (2003). The FWHM beam width of the telescope at 6.7 GHz is 40, which corresponds to a linear size of 160 pc at a distance of 840 kpc. System temperatures varied between 23 K and 34 K, with approximately 20 percent difference between the two polarizations. The data were converted to flux density using the elevation dependent gain curve for this frequency, with a typical conversion factor of 5 K/Jy.
Based on the correlation between methanol masers, massive star formation, and Giant Molecular Clouds GMC) in the Milky Way, the positions in M33 observed were the most massive clouds given in Table 1 of Engargiola et al. (2003). Each of the first 14 clouds, including all of the most massive GMCs in M33 (M 410 ), was observed for a total of 30 minutes on plus 30 minutes for the reference position. The resulting rms for the low resolution data after averaging the two linear polarizations is 3 mJy, and for the high resolution data is a factor of higher. No spectral features having greater than 3 significance were observed.
## 3 Data Analysis
The observation of 14 massive GMCs in M33 has yielded no methanol maser detections at a 3 level of 9 mJy. Based on comparison with the expected distribution of flux densities based on 6.7 GHz methanol masers in the Milky Way, this is surprising. There are estimated to be approximately 1200 methanol masers in the Milky Way (van der Walt, 2005), while the number of GMCs out to 8 kpc from the Galactic Center is 4000 to 5000 (Solomon & Sanders 1980; Scoville & Sanders 1987). Having observed 14 GMCs in M33 with mass between 2 and 7 10 solar masses, if the relationship between methanol masers and GMCs were the same as in the Milky Way, we would have expected to have made several detections.
Assuming that all the GMCs do exhibit methanol maser emission at a weak level, one can combine the data for all GMCs to get a better upper limit on the emission. This assumes that the emission occurs at the same velocity offset with respect to the velocity of the central channel in any given spectrum, which is not necessarily true. To allow for different velocity offsets for the location of emission in different GMCs, we follow the following procedure. We first select a section of the spectrum around the velocity of CO emission for each GMC ( km s), which includes the largest offset one would expect between the methanol maser and the molecular line emission (expected to be less than 15 km s). Each sub-spectrum is then cross-correlated with a Gaussian whose width represents the expected linewidth of the maser emission ( kms). Each sub-spectrum is then shifted by the offset between the central channel and the channel of peak cross-correlation. Following this procedure, the shifted data for all GMCs are added together. We denote the resulting spectrum as “A”, and is shown in Figure 2. The above procedure will create a line feature resembling the correlation template (a Gaussian signal in this case) from the constructive alignment of purely random noise. Hence, the procedure is repeated for a different portion of the spectrum, where one does not expect to see any methanol maser emission, and the resulting spectrum is denote “B” (Figure 3). We then compare the A and B spectra. A weak emission feature would be manifest by the peak in the spectrum A being higher than the peak in spectrum B by a statistically significant amount. Since this is not the case as can be seen by comparing Figures 2 and 3, we conclude that there is no detectable methanol maser emission in the data. The resultant 3 upper limit on the average emission is mJy. This is a full order of magnitude below the flux density that we would expect to be common in M33 from the extrapolation of our admittedly uncertain results in the Milky Way. This difference, if real, would certainly suggest some large-scale difference in the molecular clouds or massive young stars in M33 compared to the Milky Way. A much more meaningful result could be obtained if we had better knowledge of the methanol maser luminosity function in our Galaxy.
## 4 Discussion and Conclusions
It is difficult to be highly quantiative about the lack of detection of methanol maser emission in M33 beyond the statements given above. Given that the 40 Arecibo beam subtends a region 160 pc in size at the distance of M33, we should be sensitive to a methanol maser located anywhere within the individual giant molecular clouds being observed. Based on our estimate of the Milky Way methanol maser luminosity function, we would expect a significant fraction of masers to have a flux density 10 mJy, significantly higher than the individual limits we have obtained, and an order of magnitude greater than the averaged limit derived by combining the results from the 14 GMCs observed.
Explanations of the rarity of methanol masers in external galaxies have focused on (1) insufficient methanol density over path where amplification could take place and (2) insufficient pumping to invert the methanol transition in question (Phillips et al., 1998). Both of these can result from low metallicity. A reduced abundance of oxygen, for example, will likely reduce both the abuance of methanol (CHOH) and of dust, which is required to convert the short wavelength radiation from massive young stars to the infrared wavelengths required for maser pumping. The rarity of masers in the Magellanic clouds has been discussed in similar terms by Beasley et al. (1996). The O/H ratio in the Magellanic clouds is dramatically lower than that of the Milky Way (see discussion in Beasley et al. 1996 and the more recent measurement by Vermeij & van der Hulst 2002). For M33, the O/H ratio is only slightly less than that of the Milky Way. Vilchez et al. (1988) determined that 12 + log(O/H) = 9.0 at the center of M33, falling to 8.5 at distances between 2 and 5 kpc from the center of the galaxy. The best fit line of Crockett et al. (2006) to their new data plus previous data has a very small slope of -0.12 dex kpc and 12 + log(O/H) = 8.3 at 5 kpc from the center of M33. These values may be compared to Galactic values of 12 + log(O/H) = 8.51 for Orion and 8.78 for M17 (Peimbert, Storey, & Torres-Peimbert, 1993). The GMCs we have observed are located between 0.5 and 4 kpc from the center of M33, with a mean distance of 1.9 kpc. We conclude that the relevant O/H ratio in M33 is between 0.1 and 0.4 dex below that of the Milky Way. This difference is much smaller than that between the SMC and the Milky Way, for which log(O/H) differs by 0.8 dex. If the relatively small difference between the metallicity in the Milky wand M33 is responsible for the lack of methanol masers in the latter, it suggests that the maser luminosity must be a very senstive function of the galactic metallicity.
Henkel et al. (1987) have detected thermal methanol emission from two galaxies, IC342 and NGC253, finding that the fractional abundance of methanol is 10. This is similar to that found in GMCs in the Milky Way, but is a factor of at least 100 lower than that required for high brightness methanol maser luminosity as discussed by Sobolev, Cragg, & Godfrey (1997). However, this difference is also found in comparing hot cores in the Milky Way, presumed to be the sites of methanol masers, with more extended molecular cloud regions. The methanol abundance may be greatly increased in regions near hot stars by thermal desorption of molecules frozen onto grain surfaces. We have no direct evidence regarding the methanol abundance in M33, so it is possible that the lower O/H ratio does yield an insufficient methanol abundance to produce highly luminous masers.
There are individual regions within M33, presumably powered by massive young stars (Hinz et al., 2004), which are powerful far–infrared sources. The infrared flux is a critical requirement for maser pumping as elaborated in models of Cragg et al. (1992), Sobolev & Deguchi (1994), and Sobolev, Cragg, & Godfrey (1997). The transition to the second torsional excited state occurs at a wavelength of 30 m, so that dust temperatures of at least 150 K are required to achieve high maser brightness (Sobolev, Cragg, & Godfrey, 1997). Among our targets was number 8 of Engargiola et al. (2003), which lies within 15of the optical nebula NGC 604. It has an IRAS luminosity of 6.810 (Rice et al., 1990) and the associated GMC has a mass derived from CO of 4.410 (Engargiola et al., 2003). By Galactic standards this region would seem likely to harbour a high luminosity methanol maser, but no such emisson was detected.
Fix & Mutel (1985) were unsuccessful in a search using the VLA for highly luminous OH masers in M33. Their limit of 25 mJy (5) was sufficient to elminate the presence of any type I maser as luminous as the brightest type I masers in the Milky Way The present work thus reemphasizes the mystery of the lack of luminous masers in M33.
We are grateful to Phil Perillat for many discussions regarding the performance of the Arecibo telescope, and routines for supplying routines for data reduction. We thank German Cortes–Medellin and Lynn Baker for their contributions to the receiver system development, Kurt Kabelac and David Overbaugh for machining most of the dewar components, and Rajagopalan Ganesan and Lisa Locke for calibrating the C-band high receiver used in this work. We are also grateful to numerous other members of staff at the Arecibo Observatory who helped us in the receiver installation and calibration, and in setting up our observations. This work was supported in part by the Jet Propulsion Laboratory, California Institute of Technology. This research has made use of NASA’s Astrophysics Data System.
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The feedback must be of minimum 40 characters and the title a minimum of 5 characters | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9180637001991272, "perplexity": 3586.696945254583}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250597230.18/warc/CC-MAIN-20200120023523-20200120051523-00302.warc.gz"} |
https://portonmath.wordpress.com/page/2/ | I’ve published in my blog a new theorem.
The proof was with an error (see the previous edited post)!
The below is wrong! The proof requires $\langle g^{-1}\rangle J$ to be a principal filter what does not necessarily hold.
I knew that composition of two complete funcoids is complete. But now I’ve found that for $g\circ f$ to be complete it’s enough $f$ to be complete.
The proof which I missed for years is rather trivial:
$\bigsqcup S \mathrel{[g \circ f]} J \Leftrightarrow J \mathrel{[f^{- 1} \circ g^{- 1}]} \bigsqcup S \Leftrightarrow \langle g^{- 1} \rangle J \mathrel{[f^{- 1}]} \bigsqcup S \Leftrightarrow \bigsqcup S \mathrel{[f]} \langle g^{- 1} \rangle J \Leftrightarrow \exists \mathcal{I} \in S : \mathcal{I} \mathrel{[f]} \langle g^{- 1} \rangle J \Leftrightarrow \exists \mathcal{I} \in S : \mathcal{I} \mathrel{[g \circ f]} J$
Thus $g\circ f$ is complete.
I will amend my book when (sic!) it will be complete.
I have proved (see new version of my book) the following proposition. (It is basically a special case of my erroneous theorem which I proposed earlier.)
Proposition For $f \in \mathsf{FCD} (A, B)$, a finite set $X \in \mathscr{P} A$ and a function $t \in \mathscr{F} (B)^X$ there exists (obviously unique) $g \in \mathsf{FCD} (A, B)$ such that $\langle g\rangle p = \langle f \rangle p$ for $p \in \mathrm{atoms}^{\mathscr{F} (A)} \setminus \mathrm{atoms}\, X$ and $\langle g\rangle @\{ x \} = t (x)$ for $x \in X$.
This funcoid $g$ is determined by the formula
$g = (f \setminus (@X \times^{\mathsf{FCD}} \top)) \sqcup \bigsqcup_{x \in X} (@\{ x \} \times^{\mathsf{FCD}} t (x)) .$
and its corollary:
Corollary If $f \in \mathsf{FCD} (A, B)$, $x \in A$, and $\mathcal{Y} \in \mathscr{F} (B)$, then there exists an (obviously unique) $g \in \mathsf{FCD} (A, B)$ such that $\langle g\rangle p = \langle f \rangle p$ for all ultrafilters $p$ except of $p = @\{ x \}$ and $\langle g \rangle @\{ x \} = \mathcal{Y}$.
This funcoid $g$ is determined by the formula
$g = (f \setminus (@\{ x \} \times^{\mathsf{FCD}} \top)) \sqcup (\{ x \} \times^{\mathsf{FCD}} \mathcal{Y})$.
I found the exact error noticed in Error in my theorem post.
The error was that I claimed that infimum of a greater set is greater (while in reality it’s lesser).
I will delete the erroneous theorem from my book soon.
It seems that there is an error in proof of this theorem.
Alleged counter-example:
$f=\bot$ and $z(p)=\top$ for infinite sets $A$ and $B$.
I am now attempting to locate the error in the proof.
I have proved (and added to my online book) the following theorem:
Theorem Let $f \in \mathsf{FCD} (A ; B)$ and $z \in \mathscr{F} (B)^A$. Then there is an (obviously unique) funcoid $g \in \mathsf{FCD} (A ; B)$ such that $\langle g\rangle x = \langle f\rangle x$ for nontrivial ultrafilters $x$ and $\langle g\rangle @\{ p \} = z (p)$ for $p \in A$
After I started to prove it, it took about a hour or like this to finish the proof.
I proved the following (in)equalities, solving my open problem which stood for a few months:
$\lvert \mathbb{R} \rvert_{>} \sqsubset \lvert \mathbb{R} \rvert_{\geq} \sqcap \mathord{>}$
$\lvert \mathbb{R} \rvert_{>} = \lvert \mathbb{R} \rvert_{>} \sqcap \mathord{>}$
The proof is currently available in the section “Some inequalities” of this PDF file.
Note that earlier I put online some erroneous proof related to this. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 38, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9584916234016418, "perplexity": 427.2682740793217}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948609934.85/warc/CC-MAIN-20171218063927-20171218085927-00580.warc.gz"} |
http://eprints.maths.manchester.ac.uk/1218/ | # Computing the Fréchet Derivative of the Matrix Exponential, with an application to Condition Number Estimation
Al-Mohy, Awad H. and Higham, Nicholas J. (2009) Computing the Fréchet Derivative of the Matrix Exponential, with an application to Condition Number Estimation. SIAM Journal On Matrix Analysis and Applications., 30 (4). pp. 1639-1657. ISSN 1095-7162
The matrix exponential is a much-studied matrix function having many applications. The Fr\'echet derivative of the matrix exponential describes the first order sensitivity of $e^A$ to perturbations in $A$ and its norm determines a condition number for $e^A$. Among the numerous methods for computing $e^A$ the scaling and squaring method is the most widely used. We show that the implementation of the method in [N.~J. Higham. The scaling and squaring method for the matrix exponential revisited. {\em SIAM J. Matrix Anal. Appl.}, 26(4):1179--1193, 2005] can be extended to compute both $e^A$ and the Fr\'echet derivative at $A$ in the direction $E$, denoted by $L(A,E)$, at a cost about three times that for computing $e^A$ alone. The algorithm is derived from the scaling and squaring method by differentiating the Pad\'e approximants and the squaring recurrence, re-using quantities computed during the evaluation of the Pad\'e approximant, and intertwining the recurrences in the squaring phase. To guide the choice of algorithmic parameters an extension of the existing backward error analysis for the scaling and squaring method is developed which shows that, modulo rounding errors, the approximations obtained are $e^{A+\Delta A}$ and $L(A+\Delta A, E + \Delta E)$, with the same $\Delta A$ in both cases, and with computable bounds on $\|\Delta A\|$ and $\|\Delta E\|$. The algorithm for $L(A,E)$ is used to develop an algorithm that computes $e^A$ together with an estimate of its condition number. In addition to results specific to the exponential, we develop some results and techniques for arbitrary functions. We show how a matrix iteration for $f(A)$ yields an iteration for the Fr\'echet derivative and show how to efficiently compute the Fr\'echet derivative of a power series. We also show that a matrix polynomial and its Fr\'echet derivative can be evaluated at a cost at most three times that of computing the polynomial itself and give a general framework for evaluating a matrix function and its Fr\'echet derivative via Pad\'e approximation. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9231498837471008, "perplexity": 194.1852536687182}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202640.37/warc/CC-MAIN-20190322074800-20190322100800-00519.warc.gz"} |
https://www.gradesaver.com/textbooks/science/physics/essential-university-physics-volume-1-3rd-edition/chapter-15-exercises-and-problems-page-281/61 | ## Essential University Physics: Volume 1 (3rd Edition)
a) We know the following equation: $\Delta P = .5(v_1)^2(\rho)((\frac{A_1}{A_2})^2-1)$ Thus, we see that pressure varies by: $\Delta P = |((\frac{.04}{.0015})^2-1)|$ $\Delta P = .98 = \fbox{98%}$ b) The lowest possible height would be as the flow speed approaches 0, so we plug in 0 for $v_1$ and solve: $\Delta h_{possible}=\frac{p_{boy}}{\rho g}$ Plugging in the known values, we find: $h_{possible}\approx .17 \ m$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9761127233505249, "perplexity": 352.9637113624292}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986684854.67/warc/CC-MAIN-20191018204336-20191018231836-00455.warc.gz"} |
http://mathhelpforum.com/algebra/146785-solved-completing-square-print.html | [SOLVED] Completing the Square
• May 28th 2010, 02:54 PM
Zero17
[SOLVED] Completing the Square
Hi all, this is my first post here.
Ok, I'm using a math program in college (I forgot everything I learned in highschool math and am retaking courses to get up tp speed) for some algebra level math. I'm currently working on a section called "completing the square". Everything was going fine until this problem:
http://img294.imageshack.us/img294/2425/whatm.png
I wasn't sure how to go about doing the problem, and this is how the program responded. Normally I will pick up on how to do something after it shows me, but I'm not sure how it went about doing this one. First off, the problem started as:
X^2+(x+2)^2=10^2. Where did the 4X come from? The other part I'm a bit confused about is, a few lines down, the program goes from:
x^2+2x+2=50
to:
X^2+2x+?=48+?
It then changes it to a 1. What's up with that? I don't recall ever doing that in highschool. The program isn't explaining how it did any of this either, other than what you see here. Can someone enlighten me?
• May 28th 2010, 03:21 PM
11rdc11
Quote:
Originally Posted by Zero17
Hi all, this is my first post here.
Ok, I'm using a math program in college (I forgot everything I learned in highschool math and am retaking courses to get up tp speed) for some algebra level math. I'm currently working on a section called "completing the square". Everything was going fine until this problem:
http://img294.imageshack.us/img294/2425/whatm.png
I wasn't sure how to go about doing the problem, and this is how the program responded. Normally I will pick up on how to do something after it shows me, but I'm not sure how it went about doing this one. First off, the problem started as:
X^2+(x+2)^2=10^2. Where did the 4X come from? The other part I'm a bit confused about is, a few lines down, the program goes from:
x^2+2x+2=50
to:
X^2+2x+?=48+?
It then changes it to a 1. What's up with that? I don't recall ever doing that in highschool. The program isn't explaining how it did any of this either, other than what you see here. Can someone enlighten me?
$(x+2)^2 = (x+2)(x+2)$
Now just use foil and you get
$x^2 +4x +4$
• May 28th 2010, 03:21 PM
bigwave
Quote:
Originally Posted by Zero17
x^2+(x+2)^2=10^2. Where did the 4X come from? The other part I'm a bit confused about is, a few lines down, the program goes from:
x^2+2x+2=50
to:
X^2+2x+?=48+?
$
x^2+(x+2)^2=10^2
$
the $4x$ comes from expanding $(x+2)^2$
which become $x^2 + 4x + 4$
after simplifing you get $x^2 + 2x + 2 = 50$
however this is not in the format of $( )^2 = ()^2$so you subtract 1 from both sides.
$x^2 + 2x + 2 -1 = 50 -1$
now you have
$
x^2 + 2x +1 = 49 \rightarrow (x+1)^2 = 49 = 7^2
$
• May 28th 2010, 03:49 PM
Zero17
I see, now it makes sense. Thanks guys! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8103978633880615, "perplexity": 546.7666018626059}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609526252.40/warc/CC-MAIN-20140416005206-00637-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://www.globalsecurity.org/military/library/policy/army/fm/6-2/Ch6.htm | # Military
## TRIANGULATION
Triangulation is a method of conventional survey used when the traverse method is impractical or impossible. This method is ideally suited to rough or mountainous terrain. The triangulation method employs oblique triangular figures and enables the surveyor to cross obstacles and long distances. This method is time-consuming and requires careful planning and extensive reconnaissance.
### METHODS AND OPERATIONS
The triangulation method of survey uses triangular figures to determine survey data. If the values of certain elements of a triangle are known, the values of other elements of the triangle can be computed.
#### 6-1. SURVEY METHODS USING TRIANGLES
a. In FA surveys, the term triangulation is restricted to operations that involve the measurement of all angles within a triangle. (See Figure 6-1.) Other methods of survey, however, use the triangular figure, but the procedures used in the fieldwork and in the computations differ somewhat from the methods used in triangulation. Two of these methods used in FA survey are intersection and resection.
(1) In the intersection method, two angles are measured. If the length and azimuth of one side and two angles are known, the intersection can be computed. (See Figure 6-2.)
(2) In resection, the coordinates of an unknown point are obtained by determining the horizontal angles at the unknown point between three unoccupied points of known coordinates. (See Figure 6-3.)
b. Survey methods using the triangular figure may be used at all levels of FA survey to establish the locations of survey control points. Generally, it is better to use some method of survey that employs these procedures if the distance or terrain involved makes traverse difficult or impossible.
c. Triangulation involves single triangles (Figure 6-1) and chains or schemes of triangles (Figure 6-4). Whether a particular triangle is a single triangle or part of a scheme of triangles, the angles in the triangle are determined in the same manner, and the unknown elements of the triangle are computed in the same manner. In a single triangle, the base (known side) is measured with electronic distance-measuring devices or a 30-meter steel tape, or it is computed from known coordinates of the required accuracy. In a chain of triangles, the base of the first triangle is determined in the same manner as a single triangle. The base for the second triangle is the side of the first triangle that is common to both triangles. This side is computed, which establishes a base distance for computation of the second triangle. The same procedure is used to determine the base for each triangle in a triangulation scheme.
#### 6-2. TRIANGULATION PARTIES
a. Triangulation operations are not confined to one area at any one time. Several operations are involved in triangulation. The personnel involved in each phase are usually separated from the personnel performing other phases of the survey. Since different lengths of time are required for the various operations, no standard division of duties can be made. There are general operations that must be performed in triangulation, and each of the survey personnel is assigned to one or more of these functions. The general operations are as follows:
• Reconnaissance and planning.
• Determination of angles.
• Base measurement.
• Computation.
b. The recon party normally consists of the survey officer and/or the chief surveyor and such other personnel as deemed necessary. The recon party selects and marks each station to be used in the triangulation scheme. It also may erect a target over each station. If considerable clearing of trees and underbrush is required near the station, clearing may be performed by either the recon party or the angle-measuring personnel. Additional personnel may be required in the party that does the clearing. (Usually, the time and personnel available to perform artillery surveys preclude extensive clearing of trees and underbrush.)
c. The angle-measuring party consists of the instrument operator, the recorder, and (when necessary) personnel for clearing.
d. When the base is measured with the SEDME-MR, a team of two men is required. This team consists of one instrument operator-recorder and one man to set up the reflector prism at the end of the base. If taping teams are used, the base is double taped to the specified comparative accuracy. If only one taping team is available, it must make at least two independent measurements of the base to the specified comparative accuracy. A tension handle with a 25-pound pull should be used, and horizontal alignment must be maintained to ensure the required accuracy of a taped base measurement. If SEDME-MR devices are used to measure the base, the base will be measured three times and the mean reading used to compute the base. After completing all required measurements, these personnel may be used as directed by the chief of party or the survey and recon officer.
e. The computing party consists of two computers (one has a dual role of computer-recorder). The computers make independent computations and compare their results. They make their computations at any convenient location specified by the party chief or the survey officer. Data from the various points in the survey are reported to the computer by some prearranged method (for example, by radiotelephone).
#### 6-3. TRIANGULATION FIELD NOTES
a. As the bases are measured, each tapeman of the two taping teams records all the base measurements in the taping notebook. As soon as practical, these recorded distances are transferred to the field notes kept by the recorder. When the base length is measured with the SEDME-MR, the recorder at the end of the base at which the SEDME is located enters the distance in the field notebook.
b. The field notes kept by the recorder in an angle-measuring party are discussed in Chapter 4. Figures 4-7 through 4-9 are examples of notes kept by a recorder for 1:1,000 (T16 theodolite) and 1:3,000 (T2 theodolite) triangulation.
#### 6-4. STANDARDS AND SPECIFICATIONS
In triangulation, fieldwork and computations adhere to certain standards and specifications to produce surveys of the desired accuracy. These standards and specifications are described in Appendix B.
### TRIANGULATION COMPUTATION
A triangle is defined as a closed three-sided geometric figure containing three interior angles the sum of which is 3,200 mils. Each triangle is solved separately whether it is a single triangle or a triangle in a scheme. The only type of triangulation problem (excluding intersection and resection) involved in artillery surveys is the solution of a triangle in which the values of all three angles and the length and azimuth of one side are known. This type of problem is solved by using DA Form 5592-R (Computation of Plane Triangle Coordinates and Height From One Side, Three Angles, and Vertical Angles (BUCS)). (A reproducible copy of this form is included in the Blank Forms section of this book.)
Note. If BUCS or form is not available, the triangle can be solved by using the law of sines. For an explanation on the law of sines, see paragraph 6-16.
#### 6-5. SURVEY APPLICATION OF BASIC TRIANGLE
To determine the coordinates and height of a point from another point of known coordinates and height requires three items of information--vertical angle, azimuth, and distance. To associate the basic triangle in Figure 6-5 with these objectives, assume that Point A of this triangle is a point at which the coordinates and height are to be determined by triangulation. To do this, take the steps discussed below.
a. Select two other points, B and C, intervisible to each other and Point A. The coordinates and height of at least one of these two points must be known.
b. Measure the horizontal interior angles and vertical angles at Points A, B, and C with an instrument. With one more item, a known side (the base), the triangle can be solved. The specifications for angle measurement for fourth- and fifth-order accuracies are shown in Appendix B.
c. The length of the base can be obtained in either of two ways--by direct measurement or by computation (see (1) below). As mentioned earlier, the coordinates and height of Point B or C must be known. The specifications for baseline measurement for fourth- and fifth-order accuracies are shown in Appendix B.
(1) If the coordinates and height of both B and C are known, the azimuth and length of the base (the line joining B and C) can be computed on DA Form 5590-R. These known coordinates and heights must be of an equal or higher accuracy than that of the survey being performed. (If a fourth-order survey is being conducted, the coordinates used to compute the azimuth and length of the base should be of third-order accuracy or higher but must be at least fourth-order accuracy.)
(2) If the coordinates of only one of the points are known, the base length must be taped or measured electronically. The azimuth must be obtained by astronomic observation, by gyroscopic means, or by sighting on a point of known azimuth (azimuth mark).
Note. If the distance is determined electronically, it must be converted to horizontal distance before computation. To convert a slope to horizontal distance, see paragraph 2-23d.
d. When the three horizontal interior angles, vertical angles, azimuth of the base (thus, the azimuths of all three sides), and length of the base have been determined, the triangle can be solved to determine coordinates and height of A. The decision as to which side to compute will be based on distance angles.
#### 6-6. DISTANCE ANGLES OF A SINGLE TRIANGLE
a. Only strong figures should be used in triangulation to minimize the effects of small measurement errors. The ideal figure is an equilateral triangle. However, field conditions generally make the use of the ideal figures impractical, and often figures must be selected that only approximate the ideal.
b. Computing the length of a side in a triangle involves two of the three angles in the triangle. The angles involved in the computation of the length of a side are called distance angles. Distance angles are defined as those angles in a triangle opposite the known side (base) and the required side (side common to an adjacent triangle). Since in a single triangle there is no specific required side, the distance angles in a single triangle are the angles opposite the known side and the stronger (closest to 1,600 mils) of the two remaining angles. (See Figure 6-5.)
c. The difference between the sines of angles near 0 mils or 3,200 mils is quite large for very small differences between angles. Because of this, a small error in the value of an angle near 0 mils or 3,200 mils will cause a relatively large error in the sine of the angle and a corresponding error in the computed length of the side opposite the angle. Therefore, for best results, distance angles must be at least 400 mils and preferably 533 mils. For this reason, in the case of a single triangle, the side opposite the stronger angle is the side computed.
d. Vertical angles at each end of the side of a triangle should be measured reciprocally to the height of instrument. Often, because of distances involved, the instrument operator must measure the vertical angle to a target erected and plumbed over the forward station. When the triangulation stations are greater than 1,000 meters, the vertical angle is measured reciprocally. In any triangulation scheme, the coordinates and height of at least one station must be known. This station is used as the starting point to obtain the height of the next station. The height control is extended along the forward line of each triangle in the scheme. In Figure 6-6, use the height of Point Dave as a starting point, since the height of Point John must be computed along the forward line (Dave-John). Using the height of Point John, compute the height of Point Bill and the height of Point Mike. If side Bill-Mike had been the forward line, then the computation would have been from Point Bill to Point Mike.
### TRIANGULATION SCHEMES
A chain (scheme) of triangles is a series of single triangles connected by common sides. (See Figure 6-4.) In a chain of triangles, only the length of the first, or original, base and the length of each check base are measured. The lengths of all other sides are computed.
#### 6-7. DESCRIPTION, SOLUTION, AND CHECKS OF A CHAIN OF TRIANGLES
a. In Figure 6-7, side John-Bill is common to triangles John-Bill-Joe and Mike-Bill-John. Each of these triangles may be solved individually if one side and the interior angles are known. In Figure 6-7, assume there is a requirement to locate an additional point, Mike, outside the single triangle John-Bill-Joe. While the interior and vertical angles at John and Bill are being measured, the angles for the new triangle would also be measured. Then the horizontal and vertical angles at Point Mike would be determined. Since side Joe-Bill is a known side, all the information needed to compute both triangles would then be available. Side John-Bill would be the required side (side John-Bill is common to both triangles), and the angle at Point Joe would be its distance angle, regardless of its value. A distance angle is an angle in a triangle opposite the known side and the angle opposite the required side (side common to an adjacent triangle). The required side is also known as the forward line or forward base, because it will become the base for the next triangle in the scheme. In a chain of triangles, the last triangle in the scheme is computed as a single triangle, and its distance angles are determined as discussed in paragraph 6-6b. For example, in Figure 6-7, the distance angles in triangle John-Bill-Joe are at points John (angle opposite known side Joe-Bill) and Joe (angle opposite the required side). For triangle Mike-Bill-John, the distance angles are at Point Mike (angle opposite known side John-Bill, which has been solved in triangle John-Bill-Joe) and Point Bill (the stronger of angles at John and Bill).
b. The size of the distance angles in a triangle is used as the measure of relative strength of the figure. The strength factor of a triangle is determined by use of a strength of figure factor table. (See Table 6-1.) The distance angles of the triangle serve as arguments for entering the table, the smaller distance angle dictates the column; the larger distance angle, the row. The smaller the factor, the greater the relative strength of the triangle.
c. When the sum of the strength factors of a chain of triangles exceeds 200 or at every fifth triangle, a check base (the required side) and azimuth must be determined. If the difference between the computed length and azimuth and the measured length and azimuth is within prescribed tolerances, the scheme may be continued with the measured data. For fifth-order surveys, the computed length of the check base must agree with the measured length within 1:1,000 (comparative accuracy), and the computed azimuth must agree with the astronomic (astro) or gyroscopic (gyro) azimuth within 0.1 times the number of main scheme angles used to carry the azimuth to the check base. For fourth-order surveys, the check comparisons are 1:3,000 (comparative accuracy) for the base and 0.04 x N, where N is the number of main scheme angles used to carry the azimuth to the check base.
Note. The accuracy of the check azimuth must be considered.
d. A chain of triangles does not provide enough internal checks for an estimation of the accuracy of the work performed. As a check, the length of the last computed side of the final triangle is measured and the computed and measured lengths are compared. The results of this comparison must produce a comparative accuracy as shown in c above. The azimuth of the last computed side must be determined by astro or gyro observation as soon as possible. Error in azimuth is determined by comparing the astro or gyro azimuth with the azimuth computed through the scheme as described in c above. If the scheme closes on a known point, an accuracy ratio must be determined. The total length used for computing the accuracy ratio is the sum of the lengths of the sides of the triangles used to compute coordinates in the scheme from starting station to closing station. The azimuth is verified by turning a closing angle to an azimuth mark. The sum of the closing angle and the azimuth of the base must agree with the known azimuth within the accuracies stated in c above. (See Figures 6-8 and 6-9.)
#### 6-8. TRIANGLE CLOSURE
When interior angles of a triangle are measured in the field, the sum of the angles may vary from 3,200 mils by a small amount. The term used to describe this variance is triangle closure. If the variance is within tolerance (Appendix B), the angles are adjusted to equal 3,200. The BUCS will automatically do this by distributing the closure variance equally among the three angles. The BUCS will display the closure. (See Figures 6-10, 6-11, and 6-12 for triangulation computations.)
#### 6-9. DA FORM 5592-R
a. DA Form 5592-R is used for computing triangulation schemes. The front of the form is designed for the solution of one triangle and for entering the field data for a second triangle.
b. DA Form 5592-R has five major sections used for computations. These sections are described in (1) through (5) below in the sequence in which they should be computed.
(1) The SKETCH: block is provided for the user to draw a sketch of the triangulation scheme or the single triangle. The sketch should be properly labeled with the station angles, station names and numbers, the starting base, and all the required sides for determining data.
(a) Label triangle 1 as follows: Station A is always located at the first unknown point (station opposite the starting base), and then stations B and C are labeled in a clockwise manner. In Figure 6-10, station Sue is labeled "A"; station Rob, "B;" and station Ben, "C." For succeeding triangles, the station that is opposite the side common to both triangles (see paragraph 6-7a) is labeled "A," and so forth.
(b) Before starting computations, the stations must be numbered. Station 1 is located at one end of the starting base (BC). This is the station where the known side and required side meet. Station 2 is the first unknown point (Station A), the succeeding unknown stations will be labeled in order of computation. In Figure 6-10, Station Ben is number one because it is the point where the known and required side meet. Station Sue is labeled "# 2," since it is the first unknown point. The next unknown station (Bill) is labeled "# 3."
(2) Steps 1 through 8 of the DATA RECORD section are used to record the data for the starting known point (B or C), to include the azimuth and distance of the starting base. If there is more than one triangle to be computed, data from subsequent triangles are also recorded in this section. (See Figure 6-11.)
Note. When you are entering the base distance, BUCS will prompt for grid or horizontal distance. If the base is measured, the base is a horizontal base. If the base is computed by using known coordinates, it is a grid base. If the base distance was determined by SEDME-MR, it must be converted to a horizontal distance before entering the data into the BUCS.
(3) Steps 9 through 12 of the DATA RECORD section are provided for the field data as determined by the angle-measuring party (horizontal and vertical angles).
Note. Specify whether the vertical angle is reciprocal or nonreciprocal. Refer to the note in paragraph 6-5d.
(4) Steps 1 through 13 of the DATA RECORD section on the reverse of the form are used when the triangulation scheme is closed on a known point. The known data of the point must be provided so the BUCS can compute the azimuth error, height error, radial error and the accuracy ratio for the triangulation scheme. Refer to paragraph 6-7d.
(5) Steps 1 through 6 of the DATA RECORD section on the bottom reverse of the form are used when the triangulation scheme requires a check base. For guidance on when to compute a check base, refer to paragraphs 6-7c and d.
### INTERSECTION
Intersection is a method of survey used to locate an unknown point by determining azimuths from two or more known points. This method of survey is used as a means of establishing control to desired positions and of checking the locations of points established by other survey methods. A point established by the intersection method should be observed from at least two known stations of equal or higher order of survey than the survey being conducted. One of the points is designated as 01. The height of 01 must also be known. The location and height of the unknown are computed from 01.
#### 6-10. LIMITATIONS
Limitations for the apex angle are the same as the limitations for distance angles discussed in paragraph 6-6c. (The exception to this is when intersection is used in target area survey. In this case, the apex angle must be at least 150 mils and preferably 300 mils.)
#### 6-11. POINT VISIBILITY
The known points can be either intervisible or nonintervisible.
a. If the points are intervisible (Figure 6-13), measure a horizontal angle from each point to the unknown point, using the other point as the rear station. Measure a vertical angle from 01 to the unknown point. Compute the azimuth between the two points by using DA Form 5590-R. Then separately add each angle to the azimuth or back-azimuth to determine the azimuths from 01 and 02 to the unknown point.
b. Determine if the apex angle meets the requirements (see paragraph 6-10) by doing the procedure below.
(1) Determine the back-azimuths from the unknown point back to the known points by adding or subtracting 3,200 mils.
(2) Imagine yourself standing at the unknown point looking back at the known points. (See Figure 6-13.) Point 02 is located to the left side and Point 01 to the right. Subtract the azimuths from left to right.
c. If the points are nonintervisible (Figure 6-14), do the procedure below. (See the example below.)
(1) Measure a horizontal angle from both points to the unknown point by using a point with a known azimuth as a rear station. Each angle is then added separately to the known azimuth to determine the azimuths from 01 and 02 to the unknown point.
(2) Determine if the apex angle meets the requirements as described in paragraph 6-11b.
#### 6-12. INTERSECTION COMPUTATIONS
Intersection computations are done on DA Form 5604-R (Computation of Coordinates and Height by Intersection (BUCS)). (A reproducible copy of this form is in the Blank Forms section of this book.) (See Figure 6-15.) (Table 6-5 gives the instructions for computing DA Form 5604-R.)
#### 6-13. INTERSECTION ACCURACY
To determine the accuracy of an intersection, do the procedure below.
a. Compute another intersection to the unknown (unk) point by using a different Point 01, Point 02, or Points 01 and 02.
b. Compute a DA Form 5590-R from the points designated as 01 to the unknown point.
c. Then compute a DA Form 5590-R from the computed coordinates of the first intersection to the computed coordinates of the second intersection. (This will be the radial error.)
d. Divide the radial error into the shorter of the two distances (dis), 01A to the unknown point or 01B to the unknown point. This will be the accuracy ratio of the intersection. The accuracy ratio must agree within the prescribed accuracy limits for the type of survey being performed (1:1,000 for fifth order, 1:3,000 for fourth order).
AR = 1/ Distance 01A to unk or 01B to unk (whichever is shortest) radial error (dis between the unk)
e. Once the accuracy ratio has been computed and meets specifications described in d above, determine the mean coordinates and elevation for the unknown point. (See Figure 6-15, REMARKS: block.)
### THREE-POINT RESECTION
Three-point resection is a method of survey used to obtain control for an unknown point on the basis of three inaccessible known points. However, before the fieldwork is begun, several factors must be considered. In Figure 6-16, Stations A, B, and C are the known points and Station P is the occupied station for which coordinate are to be determined. All points must be selected so that Angles P1, P2, C, and B (Figure 6-16) are at least 400 mils. The preferred value for the angles is 533 mils. Also, if the sum of the Angles P1, P2, and A1 is between 2,845 mils and 3,555 mils, no valid solution is possible. This simply means that Station P lies on or near a circle passing through Stations A, B, and C. To eliminate the possibility of this occurring, a map reconnaissance must be made. The fieldwork consists of measuring horizontal Angles P1 and P2 and the vertical angle to the center station. Resection field notes are recorded in the field notebook in the same basic format as triangulation field notes except that the height of target (known or estimated) and the height of instrument (measured to the nearest 0.1 meter) are recorded in the remarks section. A three-point resection may be considered a closed survey if a fourth known point is used to compute a second resection, the solutions are compared, and the accuracy ratio meets the specified position closure requirement. The points used for three-point resection should be fourth-order or better. For an example of the field notes of a three-point resection, see Figure 4-11 in Chapter 4.
#### 6-14. DA FORM 5593-R
a. DA Form 5593-R (Computation of Coordinates and Height by Three-Point Resection (BUCS)) is shown in Figure 6-17. (A reproducible copy of this form is included in the Blank Forms section of this book.) The top of the form is for recording administrative data, These data include the following:
• Computer's name.
• Checker's name.
• Area in which survey was performed.
• Notebook reference.
• Date the computations were performed.
• Identification of the sheet number.
b. The next part of the form provides notes for specific operations of the program and other notes necessary for completion of this form.
c. The form is divided into two major sections. The section on the left provides instructions for the computer. The right section (DATA RECORD) is for recording data--both field and computed. Two resections can be computed on each form. The left section is divided into three columns--STEP, PROMPT, and PROCEDURE. The STEP column is the numerical sequence the operator uses as he proceeds down the form. The PROMPT column tells the operator what will appear on the BUCS display at each particular step. The PROCEDURE column tells the computer the action he will take at each particular step or prompt.
d. Entries required are the coordinates of Points A, B, and C; the horizontal and vertical angles measured at Point P; and the height of Point A. The points used for three-point resection should be of fourth-order or higher accuracy.
#### 6-15. CHECKS AND CLOSURE
When possible, the checks below are made to verify data obtained by three-point resection.
a. A horizontal angle to a fourth known station is measured. This will give another three-point resection from which the coordinates of Point P1 (a second location of P) may be determined. (See Figure 6-18.) An accuracy ratio is computed by the following formula:
AR = 1/ Distance P to CENTER or P1 to CENTER 1 (whichever is shortest) radial error (P - P1)
(1) Compute DA Form 5590-R by using the coordinates of the established point to the center station of the first resection (P-CENTER). (See Figure 6-17, REMARKS: block.)
(2) Then compute DA Form 5590-R from the computed coordinates of the second resection to the center station (P1-CENTER 1).
(3) Compute DA Form 5590-R from the computed coordinates of the first resection to the computed coordinates of the second resection (P-P1). The distance will be the radial error.
(4) Divide the radial error into the shorter of the two distances--P-CENTER or P1-CENTER 1. The result is the accuracy ratio of the intersection. The accuracy ratio must agree within the prescribed accuracy limits for the type of survey being performed (1:1,000 for fifth order, 1:3,000 for fourth order).
b. An astronomic azimuth or gyroscopic azimuth is determined to check the azimuth.
c. If a fourth point (a above) is not known, then Point P must be verified by some other method of survey before it can be used to extend control.
#### 6-16. LAW OF SINES
As shown in Figure 6-19, the law of sines is a straight proportion-type formula which states that the sine of the angle at A is to its opposite side as the sine of the angle at B is to its opposite side or the sine of the angle at C is to its opposite side. If the value of each of the interior angles of the triangle were known and the length of side a were known, the law of sines would be transposed as shown in b and c below.
a. Ensure the BUCS is operating in degrees. (Type DEGREES, and press the END LINE key.)
b. To determine the length of side b, use the following formula:
side b = side a x sin Angle B
sin Angle A
The formula when using the BUCS is as follows:
side b = side a * SIN(Angle B * .05625)/SIN(Angle A * .05625) END LINE
c. To determine the length of side c, use the following formula:
side c = side a x sin Angle C
sin Angle A
The formula when using the BUCS is as follows:
side c = side a * SIN(Angle C * .05625)/SIN(Angle A * .05625) END LINE. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8577578067779541, "perplexity": 1036.50306566933}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413507443451.12/warc/CC-MAIN-20141017005723-00231-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/partial-fraction-integration-set-up-problems.258507/ | # Partial Fraction Integration - Set up problems?
1. Sep 22, 2008
### demersal
1. The problem statement, all variables and given/known data
$$\int$$$$\frac{2x^{2}}{(x^{2}-1)}$$ dx
2. Relevant equations
Partial Fractions
3. The attempt at a solution
2$$\int$$$$\frac{x^{2}}{(x^{2}-1)}$$ dx
2$$\int$$$$\frac{x^{2}}{(x-1)(x+1)}$$ dx
$$\frac{A}{x-1}$$ + $$\frac{B}{x+1}$$ = $$\frac{x^{2}}{(x+1)(x-1)}$$
A(x+1) + B(x-1) = x$$^{2}$$
Solving for coefficients does not work because there is no x$$^{2}$$ term of A or B and also when solving for the x and the constant contradictory answers occur. Did I set this up wrong??
2. Sep 22, 2008
### danago
Perhaps try polynomial long divison first
3. Sep 22, 2008
### demersal
I can't believe I didn't see that; thank you!!
4. Sep 22, 2008
No problem | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8850542902946472, "perplexity": 4699.9775700644905}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982949773.64/warc/CC-MAIN-20160823200909-00023-ip-10-153-172-175.ec2.internal.warc.gz"} |
http://www.chegg.com/homework-help/questions-and-answers/glass-ball-radius-rb-held-height-h-directlyabove-aluminum-foil-disk-radius-rf-thicknesst---q145499 | A glass ball of radius RB is held at a height h directlyabove an aluminum foil (disk, radius = RF and thicknesst). The ball is charged with QB. To make thecalculations easier, assume that RF <<RB, RF << h, and t <<RF. These assumptions imply that the electric field ofthe ball is nearly constant over the volume of the disk.
a) What is the ball's electric field at the position of the foil?Express in terms of QB and h.
b) Because of the field, the foil becomes polarized with + and -surfaces. Determine QF (induced charges).
c) Explain how the charged ball can lift the neutral foil.
d) Suppose RB = 5mm, QB = 100nC,RF = 1mm, and t = 0.1mm. The density of Al is 2700kg/m3. How close must the ball be to lift the foil?
You may find this useful: (1+u)n = 1 + nu + ... (neglect ... when u<< 1)
My professor told me that this question isasked on high school students so I assume most of you people herecan answer this question. Hehehe! | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9152147173881531, "perplexity": 2313.0124478302805}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783399425.79/warc/CC-MAIN-20160624154959-00071-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://infoscience.epfl.ch/record/234502?ln=en | ## Locating multi energy systems for a neighborhood in Geneva using K-means clustering
To determine the optimum location for non dispatchable renewable energy systems, this study comes up with an integrated tool to place energy systems considering distributed energy demand and renewable energy potential. The Citysim urban energy planning software is used to compute the hourly heating demand of 371 buildings in Jonction, a neighborhood in Geneva. Hourly solar irradiation on the roof tops of each building is computed using the Citysim model. The electricity demand profile for each building is generated using the hourly profile for Geneva using the databases of Swissgrid and Swiss building database. Subsequently, k-means algorithm is used to cluster the buildings based on spatial distribution. An energy economic model is used to evaluate the losses in the thermal and electrical distribution networks and initial investment. Sensitivity of cluster size is evaluated using the energy economic model to obtain an optimum number of clusters and the locations for the energy systems.
Published in:
Energy Procedia Special Issue CISBAT 2017 International Conference Future Buildings & Districts - Energy Efficiency from Nano to Urban Scale, 169-174
Presented at:
CISBAT 2017 - Future Buildings & Districts - Energy Efficiency from Nano to Urban Scale, Lausanne, Switzerland, 6-8 September 2017
Year:
2017
Publisher:
Elsevier
ISSN:
1876-6102
Keywords:
Laboratories:
Note: The status of this file is: Anyone | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8539477586746216, "perplexity": 2942.280479856539}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439739134.49/warc/CC-MAIN-20200814011517-20200814041517-00482.warc.gz"} |
https://nrich.maths.org/2422 | ### Golden Powers
You add 1 to the golden ratio to get its square. How do you find higher powers?
### 2^n -n Numbers
Yatir from Israel wrote this article on numbers that can be written as $2^n-n$ where n is a positive integer.
### Poly Fibs
A sequence of polynomials starts 0, 1 and each poly is given by combining the two polys in the sequence just before it. Investigate and prove results about the roots of the polys.
# And So on - and on -and On
##### Age 16 to 18Challenge Level
Let $$f_{0}(x)= \frac{1}{1-x}$$ and $$f_{n}(x)=f_{0}(f_{n-1}(x))$$ where $n = 1,2, 3, 4, ...$
Evaluate $f_{2000}(2000)$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9217464327812195, "perplexity": 1104.6397888465249}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337531.3/warc/CC-MAIN-20221005011205-20221005041205-00036.warc.gz"} |
http://s141453.gridserver.com/ooylwe7j/thermodynamics-study-adda-physics-1bb892 | Categorías
# thermodynamics study adda physics
question_answer63) In the following P-V adiabatic cut two isothermals at temperatures ${{T}_{1}}$and ${{T}_{2}}$(fig). Thermodynamics, science of the relationship between heat, work, temperature, and energy. If the heating is done at constant pressure, the external work done by the gas is (${{C}_{p}}=0.219cal/g{}^\circ C$ and ), Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move while that of B is held fixed. The amount of heat supplied to the gas in the process 1, 2, 3 and 4 are ${{Q}_{1}}$, ${{Q}_{2}}$, ${{Q}_{3}}$ and ${{Q}_{4}}$ respectively, then correct order of heat supplied to the gas is [AB is process-1, AC is process-2, AD is adiabatic process-3 and AE is process-4]. In particular, it describes how thermal energy is converted to and from other forms of energy and how it affects matter. During another process it does 25J of work and its temperature decreases by 5K. The gas has a higher ternperature $\,{{T}_{2}}$ at the final states a and b, which it can reach the paths shown. If the pressure in the state B is ${{P}_{0}}/2$ then the pressure of the gas in state C is, Certain perfect gas is found to obey $P{{V}^{3/2}}=$ constant during an adiabatic process. Amazing Facts question_answer45) Starting with the same initial conditions, an ideal gas expands from volume ${{V}_{1}}$ to ${{V}_{2}}$ in three different ways. If Q is positive, the system has gained energy (by heating).. Test Series The relation between U, P and V for an ideal gas in an adiabatic process is given by relation $U=\text{ }a+bPV.$ Find the value of adiabatic exponent $\left( \gamma \right)$of this gas. CBSE Class 11 Chemistry , CBSE Class 11 Physics. FAQ question_answer31) The relation between U, P and V for an ideal gas in an adiabatic process is given by relation $U=\text{ }a+bPV.$ Find the value of adiabatic exponent $\left( \gamma \right)$of this gas. Lalit Sardana Sir The heat absorbed by the gas is-, question_answer39) During an isothermal expansion, a confined ideal gas does -150 J of work against its surroundings. This unit is part of the Physics library. Then it is heated at constant volume by supplying 150 J of energy. | | question_answer59) The temperature of source and sink of a heat engine are $127{}^\circ C$ and $27{}^\circ C$ respectively. Videos The gas has a higher ternperature $\,{{T}_{2}}$ at the final states a and b, which it can reach the paths shown. ... Thermodynamics part 3: Kelvin scale and Ideal gas law example (Opens a modal) Thermodynamics part 4: Moles and the ideal gas law (Opens a modal) The work done by gas in going from state B to C is double the work done by gas in going from state A to B. Thermodynamics definition, the science concerned with the relations between heat and mechanical energy or work, and the conversion of one into the other: modern thermodynamics deals with the properties of systems for the description of which temperature is a necessary coordinate. | Its volume is 0.50 ml and it contains hydrogen at a pressure of 4.5 atm. When ${{T}_{2}}$ is lowered by 62 K its efficiency increases to $\frac{1}{3}$. On P-V coordinates, the slope of an isothermal curve of a gas at a pressure $P=1MPa$ and volume $F=0.0025{{m}^{3}}$ is equal to$-400\text{ }MPa/{{m}^{3}}$ . | Heat required to change the temperature of a given gas at a constant pressure is different from that required to change the temperature of same gas through same amount at constant volume. Media If the heating is done at constant pressure, the external work done by the gas is (${{C}_{p}}=0.219cal/g{}^\circ C$ and ), question_answer17) Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move while that of B is held fixed. The change in its internal energy is $({{C}_{v}}=10kJ/kg.K$and atm pressure $=1\times 105N/{{m}^{2}})$, When a system is taken a from state i to state f along the path iaf, it is found that $Q=50$cal and $W=20cal.$, Along the path ibf $Q=36$cal. If ${{C}_{p}}/{{C}_{v}}=1.4,$ the slope of the adiabatic curve passing through this point is: A refrigerator with coefficient of performance releases 200 J of heat to a hot reservoir. Small scale gas interactions are described by the kinetic theory of gases. The pressure inside a tyre is 4 times that of atmosphere. The Carnot cycle, another important topic of physical thermodynamics, is a popular choice for exam setters of competitive exams. | If the system is returned to state 1 and ${{Q}_{1\to 2}}+10kJ$ the work done ${{W}_{1\to 2}}$is, question_answer30) During an adiabatic process of an ideal gas, if P is proportional to $\frac{1}{{{V}^{1.5}}}$, then the ratio of specific heat capacities at constant pressure to that at constant volume for the gas is. Figure shows the variation of internal energy (U) with the pressure (P) of 2.0 mole gas in cyclic process abcda. Which two of them should be part of a closed cycle if the net work done by the gas is to be its maximum positive value? Study Packages Media Sample Papers question_answer25) Choose the incorrect statement related to an isobaric process. Then, choose the correct option from the following. If the rise in temperature of the gas in A is 30 K, then the rise in temperature of the gas in B is, The specific heat of a gas at constant pressure is greater than the specific heat of the same gas at constant volume because, The specific heat of a gas in an isothermal process is, The work done in an adiabatic change in a particular gas depends only upon. A tiny glass tube lies at the bottom of the cylinder. | This unit is part of the Physics library. An ideal gas has temperature${{T}_{1}}$, at the initial state shown in the P- V diagram. Purchase Courses Free Videos, Contact Us A system changes from the state $({{P}_{1}},\text{ }{{V}_{1}})$ to $({{P}_{2}},\text{ }{{V}_{2}})$ as shown in the figure. W along the path ibf is. https://www.studyadda.com If the pressure in the state B is ${{P}_{0}}/2$ then the pressure of the gas in state C is, question_answer52) Certain perfect gas is found to obey $P{{V}^{3/2}}=$ constant during an adiabatic process. | The amount of heat supplied to the gas in the process 1, 2, 3 and 4 are ${{Q}_{1}}$, ${{Q}_{2}}$, ${{Q}_{3}}$ and ${{Q}_{4}}$ respectively, then correct order of heat supplied to the gas is [AB is process-1, AC is process-2, AD is adiabatic process-3 and AE is process-4]. Solved Papers The glass tube is broken so that hydrogen also fills the cylinder. question_answer66) On P-V coordinates, the slope of an isothermal curve of a gas at a pressure $P=1MPa$ and volume $F=0.0025{{m}^{3}}$ is equal to$-400\text{ }MPa/{{m}^{3}}$ . If during the adiabatic expansion part of the cycle, volume of the gas increases from V to 32 V, the efficiency of the engine is, question_answer70) A Carnot engine operating between temperatures ${{T}_{1}}$and ${{T}_{2}}$ has efficiency$\frac{1}{6}$. Choose from 500 different sets of thermodynamics physics flashcards on Quizlet. Refund Policy, You need to login to perform this action.You will be redirected in For an ideal gas graph is shown for three processes, Process 1, 2 and 3 are respectively. Thermodynamics in physics follows the same basic principles, studying how thermal energy is converted into other different forms of energy. question_answer26) A thermodynamic system undergoes cyclic process ABCDA as shown in fig. | | | JEE NEET Study Material : Notes , Assignment. The glass tube is broken so that hydrogen also fills the cylinder. Heat & Thermodynamics is a major topic of Physics which includes various important concepts like entropy, enthalpy, pressure etc. | Process 1 is isobaric process, process 2 is isothermal and process is adiabatic. During another process it does 25J of work and its temperature decreases by 5K. The initial pressure and volume of the gas are ${{10}^{5}}N/{{m}^{2}}$ and 6 liter respectively. The key concept is that heat is a form of energy corresponding to a definite amount of mechanical work. We keep the library up-to-date, so you may find new or improved material here over time. There are two processes ABC and DBF. Thermodynamics Resources Videos. Choose the incorrect statement related to an isobaric process. (1) Heat $\mathbf{(\Delta Q)}$: It is the energy that is transferred between a system and its environment because of the temperature difference between them. In broad terms, thermodynamics deals with the transfer of energy from one place to another and from one form to another. Articles The gas is allowed to expand adiabatically to a temperature ${{T}_{2}}\,$ by releasing the piston suddenly. In the following P-V adiabatic cut two isothermals at temperatures ${{T}_{1}}$and ${{T}_{2}}$(fig). | See more. CBSE Class 12 Chemistry , CBSE Class 12 Physics. A Carnot engine works between a source and a sink maintained at constant temperatures ${{T}_{1}}$ and ${{T}_{2}}$, For efficiency to be the greatest. In the P-V diagram, I is the initial state and F is the final state. Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. Current Affairs The temperature of the gas is (use ln 9 = 2 and$R=\frac{25}{3}J/mol-K$), A gas is expanded from volume ${{V}_{0}}$ to $2{{V}_{0}}$ under three different processes. The final volume of the gas is 2 liters. https://www.studyadda.com In a thermodynamic process, pressure of a fixed mass of gas is changed in such a manner that the gas molecules gives out 30 joules of heat and 10 J of work is done on the gas. W along the path ibf is. The cycle consists of three steps: (i) an isothermal expansion $(a\to b),$ (ii) an isobaric compression $(b\to c),$ and (iii) a constant volume increases in pressure $(c\to a).$ If ${{T}_{a}}=400K,$ ${{P}_{a}}=4atm,$and ${{P}_{b}}={{P}_{c}},=1atm,$ the work done by the gas per cycle is, A 500 ml sealed cylinder contains nitrogen at a pressure of 1 atm. question_answer53) Figure shows the variation of internal energy (U) with the pressure (P) of 2.0 mole gas in cyclic process abcda. Michio Kaku and Morgan Freeman Explain Entropy The Second Law of Thermodynamics is a death sentence for our planet. Physics is the basic physical science.Until rather recent times physics and natural philosophy were used interchangeably for the science whose aim is the discovery and formulation of the fundamental laws of nature. If the rise in temperature of the gas in A is 30 K, then the rise in temperature of the gas in B is, question_answer18) The specific heat of a gas at constant pressure is greater than the specific heat of the same gas at constant volume because, question_answer19) The specific heat of a gas in an isothermal process is, question_answer20) The work done in an adiabatic change in a particular gas depends only upon. Learn thermodynamics physics with free interactive flashcards. | The temperature of gas at c and d are 300 K and 500 K. calculate the heat absorbed by the gas during the process. | [Latent heat of vaporization of steam = 540 cal, If an air conditioner is put in the middle of a room and started working. At what volume will the entropy of gas be maximum? The change in its internal energy is $({{C}_{v}}=10kJ/kg.K$and atm pressure $=1\times 105N/{{m}^{2}})$, question_answer24) When a system is taken a from state i to state f along the path iaf, it is found that $Q=50$cal and $W=20cal.$, Along the path ibf $Q=36$cal. However, if you hone in on the most important thermodynamic formulas and equations, get comfortable converting from one unit of physical measurement to another, and become familiar with the physical constants related to thermodynamics, youâll be at the head of the class. Welcome to the Physics library! The volume changes from $2\times {{10}^{-3}}{{m}^{3}}$ to $3.34{{m}^{3}}.$The work done by the system is about. The branch of science called thermodynamics deals with systems that are able to transfer thermal energy into at least one other form of energy (mechanical, electrical, etc.) A thermodynamic system undergoes cyclic process ABCDA as shown in fig. question_answer80) A system changes from the state $({{P}_{1}},\text{ }{{V}_{1}})$ to $({{P}_{2}},\text{ }{{V}_{2}})$ as shown in the figure. View Study Package on Thermodynamics.pdf from PHYSICS 32003 at National University. question_answer90) An ideal gas is subjected to cyclic process involving four thermodynamic states, the amounts of heat (Q) and work (W) involved in each of these states ${{Q}_{1}}=6000J;{{Q}_{2}}=-5500J;{{Q}_{3}}=-3000J$ ${{Q}_{4}}=+\,3500J$ ${{W}_{1}}=2500J;{{W}_{2}}=-1000J;{{W}_{3}}=-1200J$ ${{W}_{4}}=xJ$ The ratio of the net work done by the gas to the total heat absorbed by the gas is $\eta$. question_answer34) During an adiabatic process an object does 100 J of work and its temperature decreases by 5K. | | The pressure-volume diagram shows six curved paths that can be followed by the gas (connected by vertical paths). Thermodynamics sounds intimidating, and it can be. A gas is compressed from a volume of $2{{m}^{3}}$ to a volume of $1{{m}^{3}}$ at a constant pressure of$100\text{ }N/{{m}^{2}}$. | question_answer87) A monoatomic ideal gas goes through a process $p={{p}_{0}}-\alpha V$ where ${{p}_{0}}$ and $\alpha$ are positive constants and V is its volume. The First Law can be expressed as the change in internal energy of a system equals the amount of energy added to a system (Q), such as heat, minus the work expended by the system on its surroundings (W).. Its efficiency is, question_answer73) A Carnot engine whose efficiency is 50% has an exhaust temperature of 500 K. If the efficiency is to be 60% with the same intake temperature, the exhaust temperature must be (in K), question_answer74) For an ideal gas four processes are marked as 1, 2, 3 and 4 on P-V diagram as shown in figure. question_answer22) The slopes of isothermal and adiabatic curves are related as, question_answer23) 2 k mol of hydrogen at NTP expands isobarically to twice its initial volume. If such a gas at initial temperature T is adiabatically compressed to half the initial volume, its final temperature will be. It is found that work done on the gas is 70% higher than what would be on an ideal gas. Let$\Delta {{U}_{1}}$, $\Delta {{U}_{2}}$and $\Delta {{U}_{3}}$be the change in internal energy of the gas is these three processes. | The new pressure in the cylinder is $(1\text{ }atm=1\times 105N/{{m}^{2}})$, question_answer56) In a reversible cyclic process of a gaseous system. Free Question Bank for NEET Chemistry Thermodynamics. | question_answer36) The pressure inside a tyre is 4 times that of atmosphere. Questions Bank One of the chambers has volume ${{V}_{1}}$ and contains ideal gas at pressure ${{P}_{1}}$, and temperature ${{T}_{1}}$ The other chamber has volume ${{V}_{2}}$ and contains ideal gas at pressure ${{P}_{2}}$ and temperature ${{T}_{2}}$ If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be, question_answer15) A cube of side 5 cm made of iron and having a mass of 1500 g is heated from $25{}^\circ C$ to $400{}^\circ C.$The specific heat for iron is $0.12\text{ }cal/g{}^\circ C$and the coefficient of volume expansion is $3.5\times {{10}^{-5}}/{}^\circ C,$the change in the internal energy of the cube is (atm pressure $1\times {{10}^{5}}N/{{m}^{2}}$), question_answer16) 4 kg of oxygen gas is heated so as to raise its temperature from 20 to$120{}^\circ C$. question_answer33) For an ideal gas graph is shown for three processes, Process 1, 2 and 3 are respectively. | Molar specific heat of the gas at constant volume is 3R/2. question_answer32) An ideal gas is initially at ${{P}_{1}}$, ${{V}_{1}}$ is expanded to${{P}_{2}}$, ${{V}_{2}}$ and then compressed adiabatically to the same volume f[ and pressure ${{P}_{3}}$ If W is the net work done by the gas in complete process which of the following is true? Thermodynamics deals only with the large scale response of a system which we can observe and measure in experiments. One of the chambers has volume ${{V}_{1}}$ and contains ideal gas at pressure ${{P}_{1}}$, and temperature ${{T}_{1}}$ The other chamber has volume ${{V}_{2}}$ and contains ideal gas at pressure ${{P}_{2}}$ and temperature ${{T}_{2}}$ If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be, A cube of side 5 cm made of iron and having a mass of 1500 g is heated from $25{}^\circ C$ to $400{}^\circ C.$The specific heat for iron is $0.12\text{ }cal/g{}^\circ C$and the coefficient of volume expansion is $3.5\times {{10}^{-5}}/{}^\circ C,$the change in the internal energy of the cube is (atm pressure $1\times {{10}^{5}}N/{{m}^{2}}$), 4 kg of oxygen gas is heated so as to raise its temperature from 20 to$120{}^\circ C$. | If the gas is taken from state A to B via process in which the net heat absorbed by the system is 12 cal, then the net work done by the system is $\left( 1\text{ }cal=4.19J \right)$, A Carnot engine is working between $127{}^\circ C$ and $27{}^\circ C.~$The increase in efficiency will be maximum when the temperature of, When 1 kg of ice at $0{}^\circ C$ melts to water at $0{}^\circ C,$the resulting change in its entropy, taking latent heat of ice to be 80 cal/${}^\circ C$, is, If the energy input to a Carnot engine is thrice the work it performs then, the fraction of energy rejected to the sink is, The efficiency of an ideal gas with adiabatic exponent $'\gamma '$ for the shown cyclic process would be, In a Carnot engine, the temperature of reservoir is $927{}^\circ C$ and that of sink is$27{}^\circ C$. Thermodynamics is a highly specialized area of physics that deals with the relationship between heat and energy. Here, you can browse videos, articles, and exercises by topic. question_answer50) If a given mass of an ideal gas followed a relation VT = constant during a process, then which of the following graphs correctly represents the process? Study Packages | The work done during isobaric compression is, question_answer79) 1 gm of water at a pressure of $1.01\times {{10}^{5}}$ Pa is converted into steam without any change of temperature. The change in the internal energy, Î U , of a system relates to energy, Q , in the form of heat, added to the system, and in the form of mechanical work, W , that leaves, or is done by the system. It is found that work done on the gas is 70% higher than what would be on an ideal gas. If the gas is taken from state A to B via process in which the net heat absorbed by the system is 12 cal, then the net work done by the system is $\left( 1\text{ }cal=4.19J \right)$, question_answer82) A Carnot engine is working between $127{}^\circ C$ and $27{}^\circ C.~$The increase in efficiency will be maximum when the temperature of, question_answer83) When 1 kg of ice at $0{}^\circ C$ melts to water at $0{}^\circ C,$the resulting change in its entropy, taking latent heat of ice to be 80 cal/${}^\circ C$, is, question_answer84) If the energy input to a Carnot engine is thrice the work it performs then, the fraction of energy rejected to the sink is, question_answer85) The efficiency of an ideal gas with adiabatic exponent $'\gamma '$ for the shown cyclic process would be, question_answer86) In a Carnot engine, the temperature of reservoir is $927{}^\circ C$ and that of sink is$27{}^\circ C$. | | question_answer51) The state of an ideal gas is changed through an isothermal process at temperature ${{T}_{0}}$ as shown in figure. Thermodynamics is the field of physics that successfully explains the properties of matter visible in our everyday, ... CliffsNotes study guides are written by real teachers and professors, so no matter what you're studying, CliffsNotes can ease your homework headaches and help you score high on exams. Then the work done on the working substance is, A Carnot engine takes $3\times {{10}^{6}}cal.$ of heat from a reservoir at $\,627{}^\circ C,$and gives it to a sink at $27{}^\circ C.$ The work done by the engine is, A diatomic ideal gas is used in a car engine as the working substance. Then. The state of an ideal gas is changed through an isothermal process at temperature ${{T}_{0}}$ as shown in figure. It is now transferred between the same two states in another process in which it absorbs ${{10}^{5}}J$ of heat. Sample Papers All the topics of Heat and Thermodynamics have been prepared in accordance with the elite exams like IIT JEE (Main and Advanced). The change in internal energy due to evaporation of 1 gm of water is. Solved Papers Which of the following processes is irreversible? An ideal gas can be expanded from an initial state to a certain volume through two different processes, $P{{V}^{2}}=K$ and $P=K{{V}^{2}},$ where K is a positive constant. | question_answer37) A monatomic ideal gas, initially at temperature ${{T}_{1}}$ is enclosed in a cylinder fitted with a frictionless piston. The laws of thermodynamics were developed over the years as some of the most fundamental rules which are followed when a thermodynamic system goes through some sort of energy change. question_answer62) The pressure-volume diagram shows six curved paths that can be followed by the gas (connected by vertical paths). The heat absorbed by the gas is-, During an isothermal expansion, a confined ideal gas does -150 J of work against its surroundings. question_answer47) There are two processes ABC and DBF. question_answer27) What will be the final pressure if an ideal gas in a cylinder is compressed adiabatically to $\frac{1}{3}rd$ of its volume? branch of physics which is concerned with the relationship between other forms of energy and heat | Amazing Facts question_answer88) An ideal gas has temperature${{T}_{1}}$, at the initial state shown in the P- V diagram. The temperature of source and sink of a heat engine are $127{}^\circ C$ and $27{}^\circ C$ respectively. The final volume of the gas is 2 liters. It was born in the 19th century as scientists were first discovering how to build and operate steam engines. The same amount of heat is given to the gas in each cylinder. In which of the process is the amount of work done by the gas greater? About Chemistry Notes , Chemistry Assignment , Chemistry Quiz , NCERT Solution What will be the final pressure if an ideal gas in a cylinder is compressed adiabatically to $\frac{1}{3}rd$ of its volume? The work done in the process ABC is. The volume of 1 g of steam is 1671 cc and the latent heat of evaporation is 540 cal. Notification 3 sec, OTP has been sent to your mobile number and is valid for one hour. question_answer75) An ideal gas can be expanded from an initial state to a certain volume through two different processes, $P{{V}^{2}}=K$ and $P=K{{V}^{2}},$ where K is a positive constant. Its heat capacity for 2nd process is, question_answer35) Heat energy absorbed by a system in going through a cyclic process shown in the given figure is. At what volume will the entropy of gas be maximum? Then-, question_answer78) The P-V diagram of a gas system undergoing cyclic process is shown here. 6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by ⦠(i) Heat is a path dependent quantity e.g. The temperature of gas at c and d are 300 K and 500 K. calculate the heat absorbed by the gas during the process. The work done by the gas is ${{W}_{1}}$ if the process is purely isothermal, ${{W}_{2}}$ if purely isobaric and ${{W}_{3}}$ if purely adiabatic. Molar specific heat of the gas at constant volume is 3R/2. Privacy Policy Then the work done on the working substance is, question_answer68) A Carnot engine takes $3\times {{10}^{6}}cal.$ of heat from a reservoir at $\,627{}^\circ C,$and gives it to a sink at $27{}^\circ C.$ The work done by the engine is, question_answer69) A diatomic ideal gas is used in a car engine as the working substance. Suppose 0.5 mole of an ideal gas undergoes an isothermal expansion as energy is added to its heat Q. Graph shows the final volume ${{V}_{f}}$ versus Q. question_answer28) In the P-V diagram, I is the initial state and F is the final state. Videos Then-, The P-V diagram of a gas system undergoing cyclic process is shown here. The change in entropy: question_answer89) In Carnot engine efficiency is 40% at hot reservoir temperature T. For efficiency 50% what will be temperature of hot reservoir? hi the second process, question_answer13) Some of the thermodynamic parameters are state variables while some are process variables. In which of the process is the amount of work done by the gas greater? An ideal gas is initially at ${{P}_{1}}$, ${{V}_{1}}$ is expanded to${{P}_{2}}$, ${{V}_{2}}$ and then compressed adiabatically to the same volume f[ and pressure ${{P}_{3}}$ If W is the net work done by the gas in complete process which of the following is true? Which of the following graphs correctly represents the variation of $\beta =\frac{dV/dP}{V}$ with P for an ideal gas at constant temperature? A monatomic ideal gas, initially at temperature ${{T}_{1}}$ is enclosed in a cylinder fitted with a frictionless piston. If the initial internal energy of the gas was 40 J, then final internal energy will be, Calculate the work done when 1 mole of a perfect gas is compressed adiabatically. Thermodynamics touches on virtually every field of physics, from astrophysics to biophysics, because they all deal in some fashion with the change of energy in a system. [Latent heat of vaporization of steam = 540 cal, question_answer61) If an air conditioner is put in the middle of a room and started working. question_answer48) In a thermodynamic process, pressure of a fixed mass of gas is changed in such a manner that the gas molecules gives out 30 joules of heat and 10 J of work is done on the gas. Process 1 is isobaric process, process 2 is isothermal and process is adiabatic. Purchase Courses Which two of them should be part of a closed cycle if the net work done by the gas is to be its maximum positive value? Privacy Policy If the tyre bursts suddenly at temperature 300 K, what will be the new temperature? The volume of 1 g of steam is 1671 cc and the latent heat of evaporation is 540 cal. When ${{T}_{2}}$ is lowered by 62 K its efficiency increases to $\frac{1}{3}$. The new pressure in the cylinder is $(1\text{ }atm=1\times 105N/{{m}^{2}})$, In a reversible cyclic process of a gaseous system. | [Given ${{(3)}^{5/3}}=\text{ }6.19$]. The slopes of isothermal and adiabatic curves are related as, 2 k mol of hydrogen at NTP expands isobarically to twice its initial volume. If ${{C}_{p}}/{{C}_{v}}=1.4,$ the slope of the adiabatic curve passing through this point is: question_answer67) A refrigerator with coefficient of performance releases 200 J of heat to a hot reservoir. BBC: How a sandcastle reveals the end of all things â Wonders of the Universe Entropy via sandcastles: children of all ages channel their inner preschooler to probe the Second Law of Thermodynamics. Free Question Bank for JEE Main & Advanced Physics Thermodynamical Processes OTP has been sent to your mobile number and is valid for one hour question_answer54) A one mole sample of an ideal gas is carried around the thermodynamic cycle shown in the figure. A certain diatomic gas has the same specific heats as an ideal gas but a slightly different equation of state:$PV=R(T+\alpha {{T}^{2}}),$$\alpha =0.001{{K}^{-1}}.$The temperature of the gas is raised from ${{T}_{1}}=300K$to ${{T}_{2}}$ at constant pressure. Then. The gas is allowed to expand adiabatically to a temperature ${{T}_{2}}\,$ by releasing the piston suddenly. The work done by the system in the cycle is. Why Studyadda? Measure in experiments Notes, physics Quiz, HC Verma Solution, NCERT Solution thermodynamics study adda physics ) heat a. The entropy of gas at initial temperature T is adiabatically compressed to half initial... Is 2 liters, question_answer13 ) Some of the gas is 70 % higher than what would be an. Thermal energy is converted into other different forms of energy its temperature decreases by 5K how it matter! Dependent quantity e.g ) for an isochoric process volume by supplying 150 J of work done by kinetic! For our planet so that hydrogen also fills the cylinder compressed to half the initial,... Branch of physics which includes various important concepts like entropy, enthalpy, pressure.... By an insulating partition tiny glass tube is broken so that hydrogen also fills the thermodynamics study adda physics. Another important topic studied under the thermodynamics of physics which deals with the energy and work of a system... Initial volume, its final temperature will be 1 g of steam is 1671 cc and the latent are. Of physics formulas thermodynamics conceptual flashcards on Quizlet, I is the amount of heat is given the. Tiny glass tube is broken so that hydrogen also fills the cylinder, what will.... Graph is shown here 6.19\ ] ] ) the pressure ( P ) of 2.0 mole gas final! Heat are concepts which are studied temperature 300 K and 500 K. calculate the heat absorbed by gas! Thermodynamics conceptual with free interactive flashcards are 300 K, what will.... Heat and thermodynamics are branches of physics from other forms of energy time... Hydrogen also fills the cylinder the 19th century as scientists were first discovering to! Exams like IIT JEE ( Main and Advanced ) discovering how to build and operate steam engines following holds for. Process variables setters of competitive exams how thermal energy is conserved in a closed system, Quiz... And Morgan Freeman Explain entropy the second process, process 1, 2 and 3 are respectively material! Were first discovering how to build and operate steam engines U ) with the energy and of. Pressure etc figure shows the variation of internal energy ( by heating ) isobaric! Up-To-Date, so you may find new or improved material here over time measure experiments! An insulated container of gas has two chambers separated by an insulating partition is heated at constant volume is ml... And thermodynamics are branches of physics which deals with the energy and of! Same basic principles, studying how thermal energy is converted into other different forms of energy been... Entropy, enthalpy, pressure etc century as scientists were first discovering to. 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Cbse Class 12 Chemistry, cbse Class 12 physics other forms of energy and work of a system by ). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8305044770240784, "perplexity": 567.1164217365572}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057584.91/warc/CC-MAIN-20210924231621-20210925021621-00681.warc.gz"} |
https://www.usgs.gov/publications/map-projections-used-us-geological-survey | # Map projections used by the U.S. Geological Survey
January 1, 1994
After decades of using only one map projection, the Polyconic, for its mapping program, the U.S. Geological Survey (USGS) now uses sixteen of the more comnon map projections for its published maps. For larger scale maps, including topographic quadrangles and the State Base Map Series, conformal projections such as the Transverse Mercator and the Lambert Conformal Conic are used. On these, the shapes of small areas are shown correctly, but scale is correct only along one or two lines. Equal-area projections, especially the Albers Equal-Area Conic, and equidistant projections which have correct scale along many lines appear in the National Atlas. Other projections, such as the Miller Cylindrical and the Van der Grinten, are chosen occasionally for convenience, sometimes making use of existing base maps prepared by others. Some projections treat the Earth only as a sphere, others as either ellipsoid or sphere.
The USGS has also conceived and designed several new projections, icluding the Space Oblique Mercator, the first map projection designed to permit mapping of the Earth continuously from a satellite with low distortion. The mapping of extraterrestrial bodies has resulted in the use of standard projections in completely new settings.
With increased computerization, it is important to realize that rectangular coordinates for all these projections may be mathematically calculated with formulas which would have seemed too complicated in the past, but which now may be programed routinely, if clearly delineated with numerical examples. A discussion of appearance, usage, and history is given together with both forward and inverse equations for each projection involved.
## Citation Information
Publication Year 1982 Map projections used by the U.S. Geological Survey 10.3133/b1532 John Parr Snyder Report USGS Numbered Series Bulletin 1532 b1532 USGS Publications Warehouse | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8420477509498596, "perplexity": 2995.3239369104485}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296950373.88/warc/CC-MAIN-20230402012805-20230402042805-00044.warc.gz"} |
https://www.physicsforums.com/threads/force-and-motion.63444/ | # Force and Motion
1. Feb 10, 2005
### Kp0684
A car traveling at 53km/h hits a bridge abutment. A passenger in the car moves forward a distance of 65cm (with respect to the road) while being brought to rest by an inflated air bag. What magnitude of force (assumed constant) acts on the passenger's upper torso, which has a mass of 41kg? ............. i get 14.7m/s2 for the cars speed which is my (a), and my mass is 41kg.....F=ma gives me 602Newtons...........i am lost on which equation to use....need help.....
2. Feb 11, 2005
### Davorak
I would use one of the kinematic equations to find the acceleration and then multiple that by the mass of the torso myself.
Edit:Should probably post questions like these in the homework section.
Last edited: Feb 11, 2005
3. Feb 11, 2005
### Staff: Mentor
14.7m/s is the car's initial speed, not acceleration. To find the acceleration, follow Davorak's advice. (Hint: Find a kinematics equation that relates acceleration, distance, and speed.)
(I am moving this to the Homework forum, where it belongs.)
4. Feb 11, 2005
### xanthym
However, it's also true the original kinetic energy E=(1/2)(41 kg)(14.7 m/s)^2 was dissipated to 0 over 0.65 meters by the assumed constant force F=(E/0.65).
~~
Last edited: Feb 11, 2005
5. Feb 12, 2005
### Kp0684
i have the magnitude force equal to (6.82 x 10^3 N).........
6. Feb 12, 2005
### xanthym
Similar Discussions: Force and Motion | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9362332820892334, "perplexity": 4714.478027046802}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128322320.8/warc/CC-MAIN-20170628032529-20170628052529-00408.warc.gz"} |
https://annals.math.princeton.edu/2004/160-1/p07 | # Removability of point singularities of Willmore surfaces
### Abstract
We investigate point singularities of Willmore surfaces, which for example appear as blowups of the Willmore flow near singularities, and prove that closed Willmore surfaces with one unit density point singularity are smooth in codimension one. As applications we get in codimension one that the Willmore flow of spheres with energy less than $8 \pi$ exists for all time and converges to a round sphere and further that the set of Willmore tori with energy less than $8 \pi – \delta$ is compact up to Möbius transformations.
## Authors
Ernst Kuwert
Mathematisches Institut, Albert-Ludwigs-Universität, D-79104 Freiburg, Germany
Reiner Schätzle
Mathematisch-Naturwissenschaftliche Fakultät, Rheinische Friedrich-Wilhelms-Universität Bonn, 53115 Bonn, Germany | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8818312287330627, "perplexity": 1229.399911215698}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400204410.37/warc/CC-MAIN-20200922063158-20200922093158-00374.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-1-introduction-to-algebraic-expressions-mid-chapter-review-mixed-review-page-37/12 | ## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$6m+15n+30$
$\bf{\text{Solution Outline:}}$ Use the Distributive Property to find the product of the given expression, $3(2m+5n+10) .$ $\bf{\text{Solution Details:}}$ Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 3(2m+5n+10) \\\\= 3(2m)+3(5n)+3(10) \\\\= 6m+15n+30 .\end{array} | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9590123295783997, "perplexity": 1358.5310766192558}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655911896.73/warc/CC-MAIN-20200710175432-20200710205432-00438.warc.gz"} |
https://www.arxiv-vanity.com/papers/1504.07263/ | # Cern-Ph-Th-2015-093 Gravitational Waves From a Dark (Twin) Phase Transition
Pedro Schwaller CERN, Theory Division, CH-1211 Geneva 23, Switzerland
February 20, 2021
###### Abstract
In this work, we show that a large class of models with a composite dark sector undergo a strong first order phase transition in the early universe, which could lead to a detectable gravitational wave signal. We summarise the basic conditions for a strong first order phase transition for SU(N) dark sectors with flavours, calculate the gravitational wave spectrum and show that, depending on the dark confinement scale, it can be detected at eLISA or in pulsar timing array experiments. The gravitational wave signal provides a unique test of the gravitational interactions of a dark sector, and we discuss the complementarity with conventional searches for new dark sectors. The discussion includes Twin Higgs and SIMP models as well as symmetric and asymmetric composite dark matter scenarios.
## 1 Introduction
Violent phenomena in the early universe can lead to large anisotropic fluctuations in the energy momentum tensor, which can act as sources for gravitational waves (GW). Strong first order phase transitions (PT) are an example of such a phenomenon, and it is well known that they can produce GWs [1, 2, 3, 4]. Once produced, GWs propagate through space almost undisturbed, and can therefore serve as a unique probe of phenomena in the early universe.
Phase transitions in particle physics are usually associated with symmetry breaking, i.e. with the transition of the universe from a symmetric phase to a phase of broken symmetry, as the temperature drops below a critical temperature . Within the standard model (SM) of particle physics, at least two phase transitions should take place, associated with the breaking of electroweak symmetry around GeV and with the breaking of chiral symmetry at the time of the QCD phase transition, GeV.
Today we know that both the QCD and the electroweak phase transition are not first order, but proceed through a smooth cross-over [5, 6, 7, 8], and can therefore not produce a strong GW signal through the usual mechanism (see however [9]). This can however be changed in models beyond the SM. Extensions of the SM which lead to a strong first order electroweak phase transition are particularly attractive since they can provide one missing ingredient for generating the observed baryon asymmetry of the universe. GW signals from such models have for example been studied in [10, 11, 12, 13, 14, 15, 16]. It is more difficult to modify the QCD phase transition, although a large neutrino chemical potential could be sufficient to provide a strong first order PT [17]. The resulting signal was studied in [18].
The aim of this work is to point out that gravitational waves could also be produced by a strong PT in a dark or hidden sector. The particular scenario we have in mind is a dark sector with a new SU() gauge interaction which confines at some scale . Such models have recently received renewed interest either as models of dark matter [28, 29, 30, 19, 20, 31, 32, 21, 22, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 23, 24, 25, 43, 44, 26, 27] or as part of the low energy sector of so called Twin Higgs models [45, 46, 47, 48, 49, 50, 51]. Different from generic hidden sectors [52], these models provide a preferred mass range and some restrictions on the particle content, such that the frequency range of the potential GW signal can be predicted.
Given that the SM QCD transition is not first order, we will review the known results on the order of the PT in strongly coupled gauge theories in the next section, followed by a discussion of models that fall into this category. In Sec. 3 we calculate the GW spectra that can be produced in these models, and compare them to the sensitivity of current and planned GW detection experiments in Sec. 4. We discuss the complementarity of GW experiments with other searches for dark sectors in Sec. 5, before presenting our conclusions.
## 2 Models with First Order Phase Transition
Near the QCD confinement scale , the dynamics of QCD is governed by three flavours, two of which are almost massless, while the strange quark mass is of order . Lattice studies [5, 53, 6] have shown that for these values of the quark masses, the QCD PT is a weak cross-over.
However this is not a generic result for QCD and similar theories, but more a consequence of the precise values of and in the SM. The QCD phase diagram for arbitrary and can be summarised in the so called Columbia plot, which is reproduced in Fig. 1, based on [54]. The pure Yang-Mills limit is known to have a strong first order PT [55] from the restoration of a global center symmetry at low temperatures. The opposite limit, i.e. the theory with three exactly massless quarks, also features a strong first order transition, related to the breakdown of the chiral symmetry [56].
Here we are interested in theories with and massless111More precisely, is the number of fermions with . Dirac fermions in the fundamental representation, and with a confinement scale . To guarantee the existence of a confining phase we further impose , in order to stay outside of the conformal window.
For pure Yang-Mills theories, , the confinement phase transition is related to the restoration of a global center symmetry, which is broken in the high temperature phase. Lattice simulations have shown that this PT is strong first order for [57]. For the case there is also the classic argument of [55]: In the case of a second order PT, the critical behaviour of the theory should be described by a symmetric effective theory at an infrared fixed point. Since no fixed points are known for theories with symmetry, the PT must be first order.
The case of non-zero for and is discussed in [56]. For there is no symmetry breaking, and therefore no phase transition. For the PT is first order222The behaviour can be understood by studying the effective theory near the confinement scale, which consists of massless Goldstone bosons from the breaking of the chiral symmetry. For , the anomaly is negligible, and the PT is first order due to the absence of infrared fixed points in the effective theory, provided that . For finite and the anomaly contribution to the effective lagrangian (i.e. the mass) is cubic in the Goldstone fields, and therefore alone sufficient to render the PT first order. For the anomaly does not affect the PT behaviour, so the large result applies and the PT is first order. What remains unresolved is the case for finite . For more details the reader is referred to [56]. for any value of . In the large limit, the PT is also first order for . The only case that is currently not understood is for finite , where no analytic arguments can be applied and lattice results are difficult to obtain.
To summarise, theories with massless flavours have a first order PT if either or .
The above discussion hopefully convinced the reader that strong PTs are generic in extensions of the SM which feature new confining gauge symmetries. In the following we will discuss a few examples which are well motivated either from a dark matter perspective or by naturalness arguments. These physically motivated scenarios furthermore provide a preferred range for the confinement scale , which allows us to estimate the temperature of the PT, , and therefore make a prediction for the GW spectrum.
### Composite Dark Matter I (CDM1)
In this class of models, the dark matter candidate is the lightest baryonic bound state of a dark sector with dark quarks which are neutral with respect to all SM interactions. This allows the DM to be light, since it will only communicate with the SM through heavy mediators. The most natural realisation of these models is in the context of asymmetric dark matter, where the DM number density is related to the baryon asymmetry (see e.g. [58, 59, 60] for reviews). The measured DM density then implies a mass range of GeV for the DM particle, and therefore motivates GeV.
As a concrete benchmark we will consider the dark QCD model [33, 43], which, as the name suggests, consists of a theory with dark quarks, such that a strong PT is guaranteed.
### Composite Dark Matter II (CDM2)
Similar to the previous case, the DM is a baryonic bound state of a new theory. The important difference now is that the dark quarks carry electroweak quantum numbers. Models of these type were considered for example in [24, 25], where mass spectra and form factors are calculated from first principles on the lattice. Constraints on the invisible width of the -boson immediately suggest a confinement scale at or above the GeV scale, while unitarity arguments on DM freeze-out place a limit of TeV. For fermionic DM (i.e. for odd ), direct detection furthermore imposes a bound of TeV [24]. For scalar DM (even ) lower DM masses are allowed [25]. The benchmark model considered in [25] has with flavours and should therefore feature a strong first order PT.333In [25] the quark masses were taken close to the confinement scale to allow a faster simulation. In this case the general arguments for the order of the PT do not apply, and a lattice simulation might be necessary to determine the order of the PT. However the model is also viable phenomenologically with lighter quark masses.
### Twin Higgs (TH)
These models attempt to solve the hierarchy problem without introducing coloured partners for the top quark [45, 46], and have recently received renewed attention [48, 49, 50]. This is achieved by adding a twin sector to the SM with an approximate parity symmetry, with the minimal requirement that there should be fermionic partners for the top and bottom quarks which are charged under a new interaction. The approximate symmetry constrains to lie in the GeV range. Furthermore the twin top and bottom partner masses are necessarily much larger than , such that the theory at the PT scale is with , which again predicts a strong first order PT.
### Strongly Interacting Massive Particle (SIMP)
The key idea behind SIMP DM is that the relic density is determined through freeze-out of instead of annihilations [61]. The correct relic abundance is then obtained for DM masses in the 100 MeV range. One simple realisation of this mechanism is a dark sector with dark pions as DM candidates. The processes can then be induced by the Wess-Zumino-Witten term [62, 63]. The condition for the existence of the WZW term is , which again allows for a strong first order PT, this time in the range.
All these models should give rise to a GW signal from the strong PT, and the resulting spectra and detection possibilities will be discussed in the following sections. Before moving on, we should mention a few other scenarios which also contain new strongly coupled sectors. The first group consists of Hidden Valley [52] and vector-like confinement models [64], which propose the existence of new confining sectors, communicating with the SM either through heavy mediators or directly via SM gauge interactions. If the particle content is such that the models fall into one of the strong PT regions, then also these models will give rise to a GW signal. Hidden sectors also play a role in conformally invariant extensions of the SM [65, 66], and the presence of a strong PT in these models was demonstrated using NJL methods in [67].
Furthermore there are so called mirror-world models [68], where the dark sector consists of an exact copy of the SM. Here one does not expect a GW signal for two reasons. First, if the quark content is an exact copy of QCD, there will not be a first order PT. If for some reason the quark masses are modified, such that a strong PT happens, cosmological constraints on the mirror photon require the mirror sector to have a lower temperature. Since the energy density is proportional to the fourth power of the temperature, and the intensity of a GW signal is proportional to the energy density, this will strongly suppress any GW signal from a mirror world.
## 3 Gravitational Wave Spectra
Gravitational waves produced at a time (or equivalently at a temperature ) will propagate undisturbed in the expanding universe, therefore their frequency and their fraction of the critical energy density will decrease as and , respectively, where is the scale factor [69, 10]. Denoting by and the scale factors at time of production and today, entropy conservation () implies
a∗a0=(g0,sg∗,s)13T0T∗. (1)
Here GeV is the temperature of the CMB, () is the effective number of relativistic degrees of freedom contributing to the total entropy at the time of production (today), and the entropy density at temperature is given by , where
gs(T) =∑i=bosonsgi(TiT)3+78∑i=fermionsgi(TiT)3, (2)
and counts the internal degrees of freedom of the i-th particle. It follows that the frequency today can be expressed as
f =a∗a0H∗f∗H∗=1.59×10−7 Hz×(g∗80)16×(T∗1 GeV)×f∗H∗, (3)
where we have used the Hubble rate at time of production, , and assumed that all species are in thermal equilibrium at , i.e. . For the fraction of energy density in gravitational waves today we similarly obtain
ΩGW =ρρcrit=(a∗a0)4H2∗H20Ω∗GW=1.77×10−5h−2(80g∗)13Ω∗GW, (4)
where we used that and GeV. It is noteworthy that the intensity of the GW signal is independent of the production temperature (except for the implicit dependence of on ).
The most sensitive frequency regions of pulsar timing arrays and satellite based experiments are in the nano-Hz and milli-Hz range, respectively. To get an idea about the detectability of GWs from a strong dark PT we will therefore need to understand the spectrum of the produced GWs. For this, we will closely follow the discussion of [18].
Gravitational Waves are sourced by tensor fluctuations of the energy momentum tensor of the primordial plasma. During first order phase transitions both bubble collisions [70, 71, 72] and magnetohydrodynamical (MHD) turbulence [73, 74, 75, 76, 77, 78, 79] provide sources of gravitational waves. As functions of the conformal wave number , the GW spectra produced by either source can be approximated by [18]
dΩ(B)GWh2dlogk ≃23πh2Ωr0(H∗β)2Ω2S∗v3(k/β)31+(k/β)4, (5) dΩ(MHD)GWh2dlogk ≃8π6h2Ωr0(H∗β)Ω3/2S∗v4(k/β)3(1+4k/H∗)(1+(v/π2)(k/β))11/3. (6)
Eqn. (5) is based on [80, 81] while Eqn. (6) is a fit to the numerical results of [82]. Here is the conformal Hubble parameter at , and is the radiation energy density today. The quantities that determine the GW spectrum are the bubble nucleation rate (the duration of the PT is ), the bubble velocity and the relative energy density in the source, . The temperature of the PT enters through the dependence of on .
The duration of the PT is usually taken as % of a Hubble time, and therefore [2]. To understand the relation with the physical frequency, remember that the conformal frequency is related to the conformal wave number via . Furthermore using we see that , which together with Eqn. (3) allows us to translate the GW spectra into physical frequencies.
In a given theory, the dynamics of the phase transition, and therefore the parameters , and , are in principle calculable. For the strongly coupled models considered here they are however not known, and can only be estimated using lattice simulations. We will therefore take , and as additional input parameters, with values motivated by results of analyses in weakly coupled models.
Following [18], we will use and , but instead of 0.7. We are now in a position to study the location of the peaks of the GW signals from bubble collisions and MHD turbulence. The bubble collision signal is triangular shaped with a maximum at , while the MHD turbulence peaks at somewhat larger wave numbers . To obtain physical frequencies, we use Eqn. (3) and . Then the peak locations are
f(B)peak =3.33×10−8 Hz×(g∗80)16(T∗1 GeV)(βH∗),f(MHD)peak≈10f(B)peak. (7)
In Fig. 2 we show the location of the frequency peaks as function of the PT temperature and . As expected from Eqn. (3), the peak frequencies increase linearly with the transition temperature and with .
The source term can be different for bubble collision and turbulence. Here we will assume that equal amounts of energy act as source for and . In this case the turbulence signal dominates over the one from bubble collisions over most of the relevant frequency range, see Fig. 2. The intensity of both signals decreases as , therefore smaller values of are preferable. From Eqn. (6) it might appear that the turbulence signal only decreases as , however the term in the denominator gives another power of for .
Recent simulations of first order PTs suggest that sound waves generated by the expansion of bubbles could be the dominant source of GWs from these transitions [83, 84, 85]. Sound waves continue propagating through the early universe after the PT is finished, and decay on a timescale . Compared to the above discussed spectra, they will therefore not be suppressed as much by the velocity of the transition , and the signal could be increased by a factor compared to the bubble collision signal, but with a spectrum decaying as . This could potentially boost the signal, in particular for cases where the PT is fast, i.e. .
## 4 Detectability
In the previous section, we have seen that the peak frequencies of GW signals from GeV-TeV scale PTs are of order Hz. Furthermore it is important to note that a broad spectral region around the peak is populated by GWs, from Hz.
GWs with frequencies down to Hz can be probed by satellite based experiments like eLISA [86], however the sensitivity quickly degrades below Hz. On the other end of the spectrum, pulsar timing arrays (PTA) can probe frequencies in the Hz range. In Fig. 3 we overlay the expected GW signal for different model parameters with the expected sensitivities of current and planned GW detection experiments (based on [87]).
Clearly the most promising signals are those from models with a PT temperature in the TeV range, where the peak region of the GW spectrum falls right into the most sensitive frequency range of satellite experiments. Here the signal should even be detectable for choices of the parameters that are less optimistic than those used for Fig. 3. Models of the CDM2 type naturally fall into this region, but also the CDM1 models can be viable with a confinement scale in that region.
The fact that TeV dark sectors predict an observable GW signal is not surprising, since here we are in the energy range of the electroweak PT or beyond, and the observability in particular of TeV scale strong PTs has been noted before [89, 90, 91]. The novelty here is that the dynamics leading to this strong PT does not have to be connected to the electroweak sector of the SM, and is therefore not in tension with the non-observation of new physics at the LHC.
For models with GeV the situation is a bit more difficult, since the signal peak ends up in a frequency region where neither PTA nor (e)LISA are sensitive. Looking at the GeV curve in Fig. 3 more closely, we see that in the best case scenario, for , both PTAs and LISA would be able to detect parts of the GW spectrum. For larger the signal quickly drops out of the PTA sensitivity region, however LISA remains sensitive. This is due to the increase of the observed frequency with , which partially compensates the overall drop of the signal in the LISA sensitivity region. Therefore there is a chance to detect a GW signal from the CDM1 and TH models, even if the predicted frequency range is not optimal.
Recently the EPTA experiment has reported the first limit on a stochastic GW background [88], which probes values of of order in the () Hz range. Their limit can not directly be displayed in Fig. 3 since it depends on the assumed spectrum, but it is nevertheless interesting to note that GWs from a 100 MeV scale dark sector could already be detectable.
Another way of interpreting these results is that GW searches have the potential to observe signals of completely unknown physics in the early universe, while the absence of a signal in a certain frequency range would place (very weak) bounds on the dynamics of new physics at the corresponding energy scale.
## 5 Complementarity
Signals of the new physics models introduced in Sec. 2 are being searched for at collider experiments and in direct and indirect dark matter searches. The possibility to observe such a signal in those experiments always relies on sufficiently strong non-gravitational interactions of the dark sector with the SM. Instead the GW signal is unique in the sense that it probes the gravitational effects of a model at very early times, and, while the detectability of a signal eventually still depends on model parameters, those are relatively independent of whether the model is detectable at colliders or in dark matter searches.
In CDM1 type models, or more general, hidden valley models with a new strong interaction, the dark sector is neutral under all SM interactions, and communicates only through a heavy mediator. If that mediator is in the TeV range, both direct detection experiments and the LHC could discover these models, but for larger masses it becomes increasingly difficult. From the DM perspective on the other hand, mediator masses of 10s of TeV are acceptable, such that part of the parameter space will remain unexplored in the near future. The GW signal instead is completely independent of the mediator mass, such that it could be detected even if the model was not discovered before, and thanks to the straightforward connection of mass scales and GW frequencies, could even motivate future experimental efforts.
The situation is similar for TH models, which mainly communicate with the SM through the Higgs portal. Here the parameter that controls detectability is not a mass scale but the smallness of the mixing angle of the Higgs with its twin partner. First hints for such a model could come from deviations in Higgs couplings and from exotic Higgs decays, but even then it would be difficult to uncover the whole structure. A GW signal in the right energy range could provide much information about the dynamics of these models near their confinement scale.
Finally the CDM2 type models have many detectable features at hadron colliders. However the upper limit on their mass scale is of order 100 TeV, which comes from the requirement that dark matter is not overproduced. Such high scales are not in reach of the current or next generation of collider experiments, but seem very accessible by GW searches, since the signal moves into the most sensitive frequency region of LISA and other satellite based experiments.
Overall we see that GW experiments provide a unique window to explore the dynamics of these models in the early universe, even if they are not discovered in the near future. Of course it would be even more interesting if such a dark sector was discovered at the LHC. The GW signal would then provide an important and unique probe of the physics of those models in the early universe.
Moving beyond the concrete models discussed above, perturbative unitarity constrains the mass of thermal DM to be below 110 TeV [92, 93]. For composite non-perturbative DM this limit does not apply directly444We thank T. Cohen for discussion of this point., instead a lower bound on the radius of the extended object can be obtained, . It is reasonable to expect the radius to be of order of the inverse mass, which again implies an upper bound on the DM mass of order 100 TeV. GW signals could therefore be a unique probe of the thermal DM paradigm.
## 6 Conclusions
Models beyond the standard model with a confining dark sector can lead to unexpected phenomenological signatures. Here we have explored the possibility to detect gravitational waves due to a first order phase transition at the confinement scale . The main messages from this paper are as follows:
• Different from QCD, dark sectors with QCD-like interactions can undergo strong first order phase transitions, with only mild constraints on the particle content of the theories.
• Several classes of new physics models that are currently being explored fulfil the criteria for first order PTs. The physics problems these models are trying to address, either dark matter or naturalness, constrain the confinement scales and therefore the temperature range of the phase transition.
• The GW signals originating from these dark phase transitions are in the detectable frequency range of future GW experiments, either at (E)LISA and BBO for high scale models, or in PTA experiments for the lower end of the spectrum.
Depending on other aspects of the model, GW signals will either provide complementary information about the models in question, or might even be the the best option to find evidence for these models of new physics. Different from the electroweak PT, a null result at LHC will not strongly disfavour a strong PT in a dark sector, although of course a confirmation of their existence would be more exciting.
A shortcoming of the present work is a lack of precise quantitative predictions. The bubble velocity as well as the time scale of the phase transition and the energy fraction are currently unknown, and are set to optimistic (but not unrealistic) values. Two approaches seem possible to improve upon this situation: On one side, lattice simulations could be used to measure quantities like the latent heat and the surface tension, which are related to the above parameters and can be used to obtain a more quantitative prediction for the GW spectra. Alternatively, one could attempt to construct a holographic dual for some of these theories, and analyse the PT in that setup.
### Acknowledgements
I would like to thank M. Laine for very valuable discussions regarding the order of the PT and for encouragement and comments on the manuscript. Furthermore I would like to thank A. Kurkela, T. Konstandin, M. Mccullough, D. Stolarski and A. Weiler for useful discussions and comments on the manuscript, and the organisers and participants of the eLISA and Neutral Naturalness workshops at CERN for providing a stimulating environment to finalise this work. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9376546740531921, "perplexity": 561.2951928335294}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178374217.78/warc/CC-MAIN-20210306004859-20210306034859-00022.warc.gz"} |
https://xianblog.wordpress.com/tag/viva/ | ## scalable Metropolis-Hastings, nested Monte Carlo, and normalising flows
Posted in Books, pictures, Statistics, University life with tags , , , , , , , , , , , , , , , , , , , , , , , , , on June 16, 2020 by xi'an
Over a sunny if quarantined Sunday, I started reading the PhD dissertation of Rob Cornish, Oxford University, as I am the external member of his viva committee. Ending up in a highly pleasant afternoon discussing this thesis over a (remote) viva yesterday. (If bemoaning a lost opportunity to visit Oxford!) The introduction to the viva was most helpful and set the results within the different time and geographical zones of the Ph.D since Rob had to switch from one group of advisors in Engineering to another group in Statistics. Plus an encompassing prospective discussion, expressing pessimism at exact MCMC for complex models and looking forward further advances in probabilistic programming.
Made of three papers, the thesis includes this ICML 2019 [remember the era when there were conferences?!] paper on scalable Metropolis-Hastings, by Rob Cornish, Paul Vanetti, Alexandre Bouchard-Côté, Georges Deligiannidis, and Arnaud Doucet, which I commented last year. Which achieves a remarkable and paradoxical O(1/√n) cost per iteration, provided (global) lower bounds are found on the (local) Metropolis-Hastings acceptance probabilities since they allow for Poisson thinning à la Devroye (1986) and second order Taylor expansions constructed for all components of the target, with the third order derivatives providing bounds. However, the variability of the acceptance probability gets higher, which induces a longer but still manageable if the concentration of the posterior is in tune with the Bernstein von Mises asymptotics. I had not paid enough attention in my first read at the strong theoretical justification for the method, relying on the convergence of MAP estimates in well- and (some) mis-specified settings. Now, I would have liked to see the paper dealing with a more complex problem that logistic regression.
The second paper in the thesis is an ICML 2018 proceeding by Tom Rainforth, Robert Cornish, Hongseok Yang, Andrew Warrington, and Frank Wood, which considers Monte Carlo problems involving several nested expectations in a non-linear manner, meaning that (a) several levels of Monte Carlo approximations are required, with associated asymptotics, and (b) the resulting overall estimator is biased. This includes common doubly intractable posteriors, obviously, as well as (Bayesian) design and control problems. [And it has nothing to do with nested sampling.] The resolution chosen by the authors is strictly plug-in, in that they replace each level in the nesting with a Monte Carlo substitute and do not attempt to reduce the bias. Which means a wide range of solutions (other than the plug-in one) could have been investigated, including bootstrap maybe. For instance, Bayesian design is presented as an application of the approach, but since it relies on the log-evidence, there exist several versions for estimating (unbiasedly) this log-evidence. Similarly, the Forsythe-von Neumann technique applies to arbitrary transforms of a primary integral. The central discussion dwells on the optimal choice of the volume of simulations at each level, optimal in terms of asymptotic MSE. Or rather asymptotic bound on the MSE. The interesting result being that the outer expectation requires the square of the number of simulations for the other expectations. Which all need converge to infinity. A trick in finding an estimator for a polynomial transform reminded me of the SAME algorithm in that it duplicated the simulations as many times as the highest power of the polynomial. (The ‘Og briefly reported on this paper… four years ago.)
The third and last part of the thesis is a proposal [to appear in ICML 20] on relaxing bijectivity constraints in normalising flows with continuously index flows. (Or CIF. As Rob made a joke about this cleaning brand, let me add (?) to that joke by mentioning that looking at CIF and bijections is less dangerous in a Trump cum COVID era at CIF and injections!) With Anthony Caterini, George Deligiannidis and Arnaud Doucet as co-authors. I am much less familiar with this area and hence a wee bit puzzled at the purpose of removing what I understand to be an appealing side of normalising flows, namely to produce a manageable representation of density functions as a combination of bijective and differentiable functions of a baseline random vector, like a standard Normal vector. The argument made in the paper is that imposing this representation of the density imposes a constraint on the topology of its support since said support is homeomorphic to the support of the baseline random vector. While the supporting theoretical argument is a mathematical theorem that shows the Lipschitz bound on the transform should be infinity in the case the supports are topologically different, these arguments may be overly theoretical when faced with the practical implications of the replacement strategy. I somewhat miss its overall strength given that the whole point seems to be in approximating a density function, based on a finite sample.
## Viva in Toronto (not really!)
Posted in Statistics, Travel, University life with tags , , , on July 22, 2011 by xi'an
This was the second viva of the week, for the thesis of Madeleine Thompson, but as it was in Toronto, I took part in it by a phone connection. This was rather ineffective as the connection was rather poor and I could not follow most of the questions… I had previously read (and commented) two papers, Slice Sampling with Adaptive Multivariate Steps: The Shrinking-Rank Method, and Graphical Comparison of MCMC Performance, co-written by Madeleine so I was well-aware of a part of the contents of the thesis, which I read in toto a few weeks ago. It was an interesting thesis with diversified threads in the various chapter, but I found frustrating to be unable to fully take part in the thesis debate… In retrospect, I should have flown to Toronto from Manchester yesterday or abstained from taking part in the viva!
## Viva and talk in Lancaster [back]
Posted in Books, Statistics, Travel, University life with tags , , , , , , , , , , , on July 20, 2011 by xi'an
Both viva and talk went on well (even though I was a bit too tired to give a good talk, I fear!), with interesting outcomes in both cases. The viva lasted over two hours with an exciting exchange over the increase in overall error linked with the increase in dimension and over handling HMMs with four parameters to calibrate in parallel. At some point I got confused with Dennis’ result that
$\mathbb{E}[\theta|x] = \mathbb{E}\{\theta|\mathbb{E}[\theta|x]\}$
which I though was contradicting my favourite example of the non-central chi-square domination of the regular normal mean, namely that
$\mathbb{E}[||\theta||^2|x]\text{ is doing worse than }\mathbb{E}[||\theta||^2|||x||^2]$
under squared error loss. (This is Example 3.35 in The Bayesian Choice.) I had completely forgotten that the Jeffreys’ priors associated with both posterior expectations were different! The above equality is thus not invalidated by this example. It is further quite interesting in that it shows the posterior expectation is a sort of weak sufficient statistics for the estimation of the parameter, even though I remain in favour of using more summary statistics in ABC than a posterior expectation or a pseudo-MLE. In any case, the discussion of the corresponding Read Paper at the Royal Statistical Society next December 14 promises to be interesting and well-attended… Overall, the trip was quite pleasant (nice hotel, nice run in the countryside, where I took the attached pictures) and profitable, with discussions with Paul Fearnhead gearing me towards taking advantage of my colleagues’ expertise on indirect inference at CREST. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 2, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8426172137260437, "perplexity": 1096.8750329782545}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780053657.29/warc/CC-MAIN-20210916145123-20210916175123-00251.warc.gz"} |
https://tex.stackexchange.com/questions/423821/how-can-i-minimize-the-distance-between-a-symbol-and-a-word-above-that | # How can I minimize the distance between a symbol and a word above that?
I use the command \underset to write a symbol or word below of a symbol. For instance, $\underset{\mbox{iso}}{\cong}$, but the distance between them looks too big. How can I minimize that?
• Welcome to TeX.SX! Please help us help you and add a minimal working example (MWE) that illustrates your problem. Reproducing the problem and finding out what the issue is will be much easier when we see compilable code, starting with \documentclass{...} and ending with \end{document}. – Skillmon likes topanswers.xyz Mar 29 '18 at 10:21
• @Skillmon : thank you very much! And I will write the next question with the appropiate beginning and the end (I am sorry for this case). – joseabp91 Mar 29 '18 at 10:28
You can reduce the height of the \hbox{iso} to 0pt via \smash:
\documentclass[border=2mm]{standalone}
\usepackage{mathtools}
\begin{document}
$\underset{\smash{\hbox{iso}}}{\cong}$
\end{document}
• @joseabp91 perhaps you shouldn't accept my answer right now. There is a good chance that in the next few hours there will be a better answer. If that's not the case you can accept mine. – Skillmon likes topanswers.xyz Mar 29 '18 at 10:47
• I have accepted it as I have found the thing which I was searching in your answer – joseabp91 Mar 29 '18 at 10:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8432545065879822, "perplexity": 727.8594583078253}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875143079.30/warc/CC-MAIN-20200217175826-20200217205826-00114.warc.gz"} |
http://www.physicsforums.com/showthread.php?s=401c0fd83ebfae76a9198b81b11224d9&p=4652619 | # Q about electric field between two parallel plates
by asdff529
Tags: electric, field, parallel, plates
P: 32 There are two expressions of electric field between two parallel plates,say one carries Q and another carries -Q Then the electric field between them=σ/ε0 But there is another expression that E=V/d where d is their distance of separation What are the differences between them?And what are the conditions when using either of one? Thank you!
P: 866 Consider Gauss's law $\oint \vec{E}\cdot\hat{n}da=\frac{q}{\varepsilon_0}$. Now take, as a Gaussian surface, a rectangular cube which includes only part of one of the planes and is far enough from the edges. The charge that it includes is $q=lw\sigma$ where l and w are dimensions of the part of the cube which is parallel to the charged planes.From the symmetry, we know that the electric field is perpendicular to the plane of the cube parallel to the charged planes and is constant all over it and so the surface integral is just $E lw$ and so we have $E=\frac{\sigma}{\varepsilon_0}$. Now consider $V=-\int_a^b \vec{E}\cdot\vec{dr}$ which is used for finding the potential difference between points a and b from the electric field.If electric field is constant along the way and isn't changing direction,the integral will be just the product of electric field and path length and we will have $E=\frac V d$. As you saw in the last paragraph,the electric field between the planes was constant along their separation and so the formula $E=\frac V d$ can be used in that case.For example you can have $V=\frac{d\sigma}{\varepsilon_0}$ for the potential difference between two points between the charged planes with separation d. Its not that they are two different formulas.They're just in terms of different things.
P: 32
Quote by Shyan Consider Gauss's law $\oint \vec{E}\cdot\hat{n}da=\frac{q}{\varepsilon_0}$. Now take, as a Gaussian surface, a rectangular cube which includes only part of one of the planes and is far enough from the edges. The charge that it includes is $q=lw\sigma$ where l and w are dimensions of the part of the cube which is parallel to the charged planes.From the symmetry, we know that the electric field is perpendicular to the plane of the cube parallel to the charged planes and is constant all over it and so the surface integral is just $E lw$ and so we have $E=\frac{\sigma}{\varepsilon_0}$. Now consider $V=-\int_a^b \vec{E}\cdot\vec{dr}$ which is used for finding the potential difference between points a and b from the electric field.If electric field is constant along the way and isn't changing direction,the integral will be just the product of electric field and path length and we will have $E=Vd$. As you saw in the last paragraph,the electric field between the planes was constant along their separation and so the formula $E=Vd$ can be used in that case.For example you can have $V=\frac{\sigma}{d\varepsilon_0}$ for the potential difference between two points between the charged planes with separation d. Its not that they are two different formulas.They're just in terms of different things.
but my teacher said if we use E=σ/ε0 and Q is kept constant,the electric field is independent of d.
it seems that there is a contradiction,because E=V/d as well,where am i wrong?
P: 866
Q about electric field between two parallel plates
Quote by asdff529 but my teacher said if we use E=σ/ε0 and Q is kept constant,the electric field is independent of d. it seems that there is a contradiction,because E=V/d as well,where am i wrong?
There is no contradiction. E=V/d doesn't mean E depends on d! Because V can be a function of d as well.
Thanks
PF Gold
P: 12,185
Quote by asdff529 but my teacher said if we use E=σ/ε0 and Q is kept constant,the electric field is independent of d. it seems that there is a contradiction,because E=V/d as well,where am i wrong?
There is no contradiction. The one expression can be re-arranged into the other.
The quantity in this relationship is Capacitance (C) and Q = CV
You can replace this by σ=c0V
where c0 is the capacitance per unit area.
Keeping Q constant and increasing d will require work, so V will have increased. The Volts per Meter will remain the same. Alternatively, separating the plates will decrease the Capacitance, which implies an increase in V.
Related Discussions Introductory Physics Homework 2 Classical Physics 2 Introductory Physics Homework 7 Introductory Physics Homework 5 Introductory Physics Homework 1 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.96595299243927, "perplexity": 337.3995266798471}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1409535920694.0/warc/CC-MAIN-20140909043153-00076-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/stimulated-and-spotaneous-emission.716880/ | # Stimulated and spotaneous emission
1. Oct 16, 2013
### ajayguhan
Why the light emitted in spontaneous emission is poly chromatic whereas the light in stimulated emissions is monochromanti?
If E1 and E2 be two energy level such that E2 >E1, in both emission the energy difference is fixed, so the frequency and so the wavelength thus the light emitted in both case should be monochromatic but why the light emitted in stimulidated emission is alone monochromatic ?
2. Oct 16, 2013
### Simon Bridge
It's not. They may both be polychromatic if you want it to be.
A transition between a particular pair of energy states will always give the same wavelength ... but you don't normally get just one pair of energy states or just one photon.
You get a variety of colors when there are many different pairs of states involved producing lots of photons.
We usually build the devices relying on stimulated emission to be as monochromatic as possible.
With stimulated emission you can select which pair of states to use - the effect is particularly strong when your feed some of the stimulated photons back through the medium - as in a laser.
He-Ne gas (for instance) can be made to lase at a variety of different frequencies and steps have to be taken to suppress lasing at some of them.
3. Oct 16, 2013
### MikeGomez
Yes, that is true. Assuming a jump between two specific energy levels, the frequency of the light be identical in both cases (monochromatic).
In spontanious emission the phase and the direction will be random. In simulated emission they are the same.
Maybe you can get various other frequencies from spontanious emission because other energy levels are involved. This does not happen with stimulated emission because the parameters are set (controlled) for only a single frequency.
Didn't see Simon's post when I replied. What he said... | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9345474243164062, "perplexity": 1002.7306395192298}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464051299749.12/warc/CC-MAIN-20160524005459-00201-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://community.cisco.com/t5/other-network-architecture/difference-between-gain-of-different-frequencies-with-integrated/m-p/4078039 | cancel
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## Difference Between Gain of Different Frequencies with Integrated Omnidirectional Antennas
Here is a quote from some ENCOR material:
Integrated omnidirectional antennas typically have a gain of 2 dBi in the 2.4 GHz
band and 5 dBi in the 5 GHz band.
Why is it that the lower frequency has less gain than the higher frequency?
Any input is appreciated.
Everyone's tags (7)
4 REPLIES 4
Highlighted
VIP Expert
## Re: Difference Between Gain of Different Frequencies with Integrated Omnidirectional Antennas
Might have to due with signal absorption and/or signal reflection from other items in the environment.
Highlighted
Beginner
## Re: Difference Between Gain of Different Frequencies with Integrated Omnidirectional Antennas
I do not think so. This is just the baseline gain from the antenna radiation.
Highlighted
VIP Expert
## Re: Difference Between Gain of Different Frequencies with Integrated Omnidirectional Antennas
Well that's possible too assuming the antenna is more suitable for 5 GHz vs. for 2.4 GHz (recall any antenna is only really optimal for one wavelength/frequency [https://en.wikipedia.org/wiki/Standing_wave_ratio]).
Highlighted
Hall of Fame Community Legend
## Re: Difference Between Gain of Different Frequencies with Integrated Omnidirectional Antennas
When the power gain are set equally, 2.4 Ghz can push further than 5.0 Ghz.
NOTE: I am talking about when the setup is in a wide open field and the nearest WiFi interference is 100 metres away.
The probably reason why the antennas power gain is/was designed like that is to ensure that at a certain distance, the RSSI of the 2.4 Ghz and 5.0 Ghz are equal. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8618665337562561, "perplexity": 3264.430681084983}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655902377.71/warc/CC-MAIN-20200709224746-20200710014746-00147.warc.gz"} |
https://www.math.kyoto-u.ac.jp/ja/event/seminar/3152 | # Mackey’s formula for cyclotomic Hecke algebras and rational Cherednik algebras of type $G(r, 1, n)$
2017/12/22 Fri 13:00 - 14:30
RIMS402号室
The restriction/induction functors play an important role for the representation theory
of cyclotomic Hecke algebras and rational Cherednik algebras of type $G(r, 1, n)$.
In this talk, we discuss an analog of Mackey’s formula for two parabolic subalgebras
of the cyclotomic Hecke algebras and the rational Cherednik algebras. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.899734377861023, "perplexity": 556.0894467981036}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027321696.96/warc/CC-MAIN-20190824194521-20190824220521-00112.warc.gz"} |
https://ypei.org/posts/2014-04-01-q-robinson-schensted-symmetry-paper.html | # Symmetry property of $$q$$-weighted Robinson-Schensted algorithms and branching algorithms Published on 2014-04-01 | Comments
In this paper a symmetry property analogous to the well known symmetry property of the normal Robinson-Schensted algorithm has been shown for the $$q$$-weighted Robinson-Schensted algorithm. The proof uses a generalisation of the growth diagram approach introduced by Fomin. This approach, which uses "growth graphs", can also be applied to a wider class of insertion algorithms which have a branching structure.
Above is the growth graph of the $$q$$-weighted Robinson-Schensted algorithm for the permutation $${1 2 3 4\choose1 4 2 3}$$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9660911560058594, "perplexity": 682.9731727811294}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662561747.42/warc/CC-MAIN-20220523194013-20220523224013-00174.warc.gz"} |
https://marcoriello.wordpress.com/research/supernova-rates/ | ## Supernova Rates
Back to the Reseach page
The evolution of the rate of Supernovae with redshift contains unique information on the galaxy star formation history and initial mass function. In particular, the rate of core collapse (CC) SNe (Type II + Ib/c), originating from the collapse of the core of massive stars $(M > 8M_{\odot})$, is a direct tracer of the ongoing star formation rate and of the chemical enrichment of elements like O and Ca. By measuring the evolution of the CC SN rate with redshift it is possible to reconstruct the history of star formation and to put constraints on different models of galaxy evolution. On the other hand, the thermonuclear explosions of accreting white dwarf stars in close binary systems, which originate Type Ia SNe, give a major contribution to the iron-peak elements and echo the long term star formation history. The actual configuration of the precursor binary system is still an open issue that can be investigated by studying the evolution of the Type Ia SN rate with redshift.
Despite the obvious interest, so far an accurate estimate of the rate of SNe has been obtained only for the local Universe. For more distant galaxies the current estimates are mainly by-products of surveys optimised to tackle other scientific purposes. The vast majority of SN rate estimates at intermediate and high redshifts have been obtained in the framework of programs aimed at exploiting Type Ia SNe as distance indicators to probe the geometry of the Unverse. To achieve their goals, these programs need a large sample of well observed, un-reddened Type Ia SNe spanning a large redshift range. The limiting factor is mainly due to the requirements imposed by obtaining a well sampled SN light curve, which is the key ingredient in the procedure used to make Type Ia SNe, actually, standard candles.
For this reason the sample of objects collected by these programs are affected by a severe, intrinsic bias. This is even more the case for CC SNe which are normally considered as “contamination” by such programs. The candidate screening process is indeed tuned to avoid false positives getting through to the expensive spectroscopic follow-up stage.
Addressing these issues was among the main goal of the southern intermediate redshifts ESO supernova search (STRESS) program. The project run between 2000 and 2005 discovering several SNe between redshift 0.1 and 0.7. The novelties of the study was represented by the completely unbiased selection of both CC and Type Ia SNe, by the adoption of the photometric redshift technique to characterise the galaxy sample and by the first quantification of the actual AGN/QSO contamination in a SN search at intermediate redshifts. Preliminary results based on a subset of the data where presented in Cappellaro, Riello, Altavilla et al. 2005 providing the first direct measurement of the rate of CC SNe at z=0.26 and evidence for an increase of a factor of three (+/- 50%) with respect to local values.
Currently I am contributing to the preparation of the final paper (Botticella, Riello, Cappellaro, et al. 2007 – hopefully). This new work is not a mere copy of the previous work with just a larger sample but it actually includes some novelties. First, we used dust extinction models developed by Riello & Patat 2006 to correct the observed rates for the effect of host galaxy extinction. Typically a SN search of this kind is mainly limited by the limited telescope time that can be allocated for the spectroscopic follow-up that provides the most reliable evidence to actually prove a candidate to be a SN rather than an AGN or a QSO. In the attempt to overcome such a limitation we performed a short, post-mortem program aimed at obtaining spectra of the host galaxies of those SN candidates that could not make it to the spectroscopic follow up because of the lack of telescope time. In this way we have been able to put a reliable constraint on the actual AGN/QSO contamination and to use the unconfirmed candidates (in a statistical fashion).
Watch out for more results soon !
Back to the Reseach page | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 1, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8121040463447571, "perplexity": 1057.3270572914262}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128329372.0/warc/CC-MAIN-20170629154125-20170629174125-00480.warc.gz"} |
http://mathhelpforum.com/geometry/158819-area-semi-circle.html | # Math Help - area of semi circle?
1. ## area of semi circle?
hi i need to find the area of this:
im not sure how to do it because there isn't a radius there. if anyone can help and explain i will be very thankful!
2. is 26 the diameter of the semicircle so i half it to make the radius so i would do, pie x 13 x13 to make the area of a circle then divide by 2 to make area of a semicircle? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.929905116558075, "perplexity": 473.8057472064344}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1430459917448.46/warc/CC-MAIN-20150501055837-00037-ip-10-235-10-82.ec2.internal.warc.gz"} |
http://mathhelpforum.com/differential-geometry/130832-supremums-sup-ac-supa-supc.html | # Thread: Supremums: sup AC = supA supC
1. ## Supremums: sup AC = supA supC
If A and C are subsets of R, let AC={ac: a E A, c E C}. If A and C are bounded and the sets consist of strictly positive elements, prove that sup AC = supA supC.
Attempt:
a(sup A) and c(sup C)
=> ac supA supC for all (ac) E (AC)
Hence supA supC is an upper bound for the set AC.
But how to prove that sup AC = supA supC?
Any help is appreciated!
2. Originally Posted by kingwinner
If A and C are subsets of R, let AC={ac: a E A, c E C}. If A and C are bounded and the sets consist of strictly positive elements, prove that sup AC = supA supC.
Attempt:
a(sup A) and c(sup C)
=> ac supA supC for all (ac) E (AC)
Hence supA supC is an upper bound for the set AC.
But how to prove that sup AC = supA supC?
Any help is appreciated!
Let $a\in A$ and $c\in C$ then $a\leqslant \sup\text{ }A$ and $c\leqslant \sup\text{ }C$ and so $ac\leqslant\sup\text{ }A\cdot\sup\text{ }C$ it follows that $\sup\text{ }AC\leqslant\sup\text{ }A\cdot\sup\text{ }C$.
That finishes up the rest of your initial thought, care to try the second part? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9817613959312439, "perplexity": 2702.5990887932617}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501172831.37/warc/CC-MAIN-20170219104612-00476-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://listserv.tau.ac.il/cgi-bin/wa?A2=ind0311&L=ivritex&T=0&P=6346 | Date: Sun, 30 Nov 2003 12:40:54 +0200 Reply-To: Hebrew TeX list <[log in to unmask]> Sender: Hebrew TeX list <[log in to unmask]> From: Ron Artstein <[log in to unmask]> Subject: Re: Fonts needed for mathematical text in a Hebrew paper. In-Reply-To: <[log in to unmask]> Content-Type: TEXT/PLAIN; charset=US-ASCII > This was my concern as well. And you have not answered it. How > should \textbf{\emph boldface-emphathized text}} look? It should be a bold version of whatever \emph gives. Anything else would be confusing and counterintuitive. In general, I believe NFSS attributes should have the same meanings in Hebrew as in other scripts, that is n = upright, it = italics, sl = slanted/oblique, sc = small caps; m = medium, b = bold, bx = bold extended. Not all of these attributes are used in Hebrew, but if we use these attributes with different meanings it would just cause confusion. If we want to use a different font family for emphasis, this should be defined at a higher level. When a font doesn't exist for a combination of attributes, LaTeX can provide a substitution. > BTW: while we discuss the subject of various emphasis methods, > is there any equivalent to small-caps? Is it needed? IMHO, the answer to the above two questions is no. For technical reasons it's better to have the sc attribute defined even in Hebrew, but with font substitutions. > Note: with Babel each letter has a predefined language and thus > encoding. If I change settings in the configuration of > 'hebrew', they will still not affect the English parts. > > Thus if you have some English paragraphs, emphasis will still > use normal italic. Tzafrir, I'm not quite sure I follow. The association between NFSS attributes and actual fonts depends on the font encoding; this is because font encoding is one of the NFSS attributes :-) However, the macro \em does not work with the attributes directly, but is rather defined through the high-level commands \itshape and \upshape. \DeclareRobustCommand\em {\@nomath\em \ifdim \fontdimen\@ne\font >\z@ \upshape \else \itshape \fi} Explanation: \em checks the slant of the font (\fontdimen1); in an oblique environment (positive slant) \em expands to \upshape, and in an upright environment (non-positive slant) \em expands to \itshape. It is possible to redefine \emph so that the redefinition will only take place in a particular language environment (e.g. Hebrew), and revert to the default when you exit this environment. It is also possible to redefine it globally. > The base classes don't seem to use \it , \textit and \itshape Actually, \itshape is explicitly used for at least two purposes (code below taken from latex.ltx): for typesetting theorems \def\@begintheorem#1#2{\trivlist \item[\hskip \labelsep{\bfseries #1\ #2}]\itshape} \def\@opargbegintheorem#1#2#3{\trivlist \item[\hskip \labelsep{\bfseries #1\ #2\ (#3)}]\itshape} and for the footnote designator inside minipages \def\thempfootnote{\itshape\@alph\c@mpfootnote} -Ron. Back to: Top of message | Previous page | Main IVRITEX page
LISTSERV.TAU.AC.IL | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9736128449440002, "perplexity": 3674.9219039667496}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719416.57/warc/CC-MAIN-20161020183839-00245-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://www.math-only-math.com/division-sharing-and-grouping.html | # Division Sharing and Grouping
We will learn division sharing and grouping.
Share eight strawberries between four children.
Let us distribute strawberries equally to all the four children one by one.
After distributing equally among four children, four strawberries were left. They can be further distributed among four children.
Now, there are no more strawberries left in the box. Each child got 2 strawberries. So, division is an operation of repeated subtracted of the same number.
8 – 4 = 4
4 – 4 = 0
Here, we have subtracted 4 repeatedly 2 times until we reached 0.
Hence, 8 strawberries ÷ 4 children = 2 strawberries for each child.
We can also say that 8 strawberries have been divided into 4 equal groups of 2.
So, 8 ÷ 4 = 2. This is called a division fact.
Questions and Answers on Division Sharing and Grouping:
I. Divide the following:
(i) Put flowers in vases equally and write the division sentence.
(ii) Put fishes equally in the tanks and write the division sentence.
I. (i) 4 ÷ 4 = 1
(ii) 3 ÷ 3 = 1
II. Make groups of 2 and complete the division fact.
………… bananas have been divided into ………… groups.
………… ÷ ………… = …………
………… flowers have been divided into ………… groups.
………… ÷ ………… = …………
………… balloons have been divided into ………… groups.
………… ÷ ………… = …………
Snswer
III. Divide and complete the division fact.
(i) Distribute 12 candies equally among 4 kids.
Each kid has ……….. candies.
So, 12 ÷ 4 = …………
(ii) Distribute 18 oranges equally among 3 boxes.
Each box has ……….. oranges.
So, 18÷3 = …………
(iii) Distribute 20 pencils in 4 pencil stands.
Each pencil stand has ……….. pencils.
So, 20÷ 4 = …………
III. (i) 3
(ii) 6
(iii) 5
IV. Divide by repeated subtraction:
(i) 40 ÷ 5
(ii) 64 ÷ 8
(iii) 42 ÷ 7
(iv) 54 ÷ 9
IV. (i) 8
(ii) 8
(iii) 6
(iv) 6 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8278580904006958, "perplexity": 4987.590757729543}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153521.1/warc/CC-MAIN-20210728025548-20210728055548-00560.warc.gz"} |
http://gabarro.org/ccn/cauchy_basis.html | The Cauchy basis
In this note we describe the construction of the Cauchy basis, whose functions satisfy at the same time Dirichlet and Neumann boundary conditions. Besides being beautiful, it is useful to solve fourth-order boundary problems.
1. Introduction: Fourier, cosine and sine basis
Everybody knows the Fourier basis of $L^2([0,2\pi])$, defined by sine and cosine functions of integer frequencies. $$1,\cos(x),\cos(2x),\ldots,\sin(x),\sin(2x),\ldots$$
This was the original basis used by Joseph Fourier to study the heat equation. Notice that cosines are symmetric around the center of the interval, and sines are anti-symmetric. Thus, all the functions of this basis are necessary to represent arbitrary functions on the whole interval.
Less well-known are the cosine basis, defined by cosines of half-integer frequencies $$1,\cos\left(\frac{1}{2}x\right), \cos(x), \cos\left(\frac{3}{2}x\right), \cos(2x),\ldots$$
and the sine basis, defined by sines of half-integer frequencies $$\sin\left(\frac{1}{2}x\right), \sin(x), \sin\left(\frac{3}{2}x\right), \sin(2x), \ldots$$
Each one of these three sequences of functions is a Hilbert basis of the space $L^2([0,2\pi])$. In particular, you can express a sine as a linear combination of cosines, and vice-versa. This is a favourite exam question of mine. Of course, the convergence is not uniform on the boundaries of the interval.
The cosine and sine bases are very useful when you want to express solutions of a differential equation (or a variational problem) satisfying particular boundary conditions. Notice that if $f(x)$ is a finite linear combination of the sine basis, then $f(0)=f(2\pi)=0$ and if it is a finite linear combination of the cosine basis, then $f'(0)=f'(2\pi)=0$. The same relationships hold for series as long as the coefficients decrease fast enough. Thus, the sine basis is useful for Dirichlet boundary conditions, and the cosine basis is useful for Neumann boundary conditions.
2. The Cauchy basis
The functions of the Cauchy basis, defined below, satisfy simultaneously the four conditions $f(0)=f(1)=f'(0)=f'(1)=0$.
A simple construction of such a set of functions is obtained by considering eigenfunctions of the fourth derivative that satisfy the four boundary conditions.
Eigenfunctions of the fourth derivative are of the form $\exp(\rho x)$ where $\rho$ is a fourth-root of $1$, thus $\rho\in\{1,-1,i,-i\}$. Equivalently, they are linear combinations of $\sin$, $\cos$, $\sinh$ and $\cosh$ $$\varphi(x)=\alpha\cos(\lambda x)+\beta\sin(\lambda x)+\gamma\cosh(\lambda x)+\delta\sinh(\lambda x)$$
This is an eigenfunction of the fourth derivative with eigenvalue $\lambda^4$, that we can assume to be real positive.
By imposing the four boundary conditions, we obtain a linear system on the coefficients $(\alpha,\beta,\gamma,\delta)$, with a parameter $\lambda$. The matrix of this linear system is singular, and its determinant vanishes when $\lambda$ satisfies the equation $$\cos\lambda\cosh\lambda=1$$ The solutions of this equation (which are slight perturbations of even multiples of $pi/2$) are the eigenvalues of our basis.
A good enough approximation is given by $$\lambda_n = \frac{2n+1}{2}\pi - (-1)^n2\exp\left(-\frac{(2n+1)}{2}\pi\right)$$ and the eigenfunctions are thus $$\varphi_n(x)=\sin(\lambda_nx)-\sinh(\lambda_nx)+\beta_n(\cos(\lambda_nx)-\cosh(\lambda_nx))$$ where $$\beta_n=\frac{\sinh\lambda_n-\sin\lambda_n}{\cos\lambda_n-\cosh\lambda_n}.$$
Notice that, although the functions $\varphi_n$ are beautiful and symmetric inside the interval $[0,1]$, they explode in wild ways outside this interval:
3. Properties and open problems
We have just defined a set of functions $\varphi_n(x)$. By construction, they are $C^\infty$ functions on the interval $[0,1]$ that satisfy $0=\varphi_n(0)=\varphi_n(1)=\varphi_n'(0)=\varphi_n'(1)$. Since these functions are eigenvectors of a linear operator with different eigenvalues, they must be orthogonal. The following remains to be done:
1. Prove that $\{\varphi_n(x)\}$ is a Hilbert basis of $L^2([0,1])$
2. Prove that $|\varphi_n(x)|\le 2$ for $x\in[0,1]$
3. Prove that $\varphi_{2n}(x)=-\varphi_{2n}(1-x)$ for $x\in[0,1]$ (so that $\varphi_{2n}(1/2)=0$).
4. Prove that $\varphi_{2n+1}(x)=\varphi_{2n+1}(1-x)$ for $x\in[0,1]$ (so that $\varphi'_{2n+1}(1/2)=0$).
5. Prove that $\varphi_n$ has $n-1$ zeros on $(0,1)$, and identify them.
6. Compute the exact normalization coefficients required so that the basis is orthonormal.
7. Obtain an effective algorithm/formula to evaluate $\varphi_n(x)$ for large values of $n$, avoiding the numerical cancellations that appear with the current expression. Notice that, thanks to the symmetry properties above, the formula only needs to be stable for $0\le x\le\frac{1}{2}$ (the difficult case being near $\frac{1}{2}$).
8. Rewrite the definition so that the interval is centered around 0, and the symmetries are more visible.
9. Study how the regularity of a function can be measured from the rate of decrease of its "Cauchy coefficients"
10. Develop a theory of sampling using "Cauchy polynomials" as an interpolation model
11. Develop a "Fast Cauchy Transform", to obtain the coefficients of these polynomials from their samples.
12. Extend this basis to the case of a square.
13. Extend this basis to the case of a disk.
Some of these propositions are easy, others may not actually be possible (especially the last one).
4. Detailed calculation
In this section we detail the construction leading to the Cauchy basis. The computation has two parts. First, we explain how the equation $\cos\lambda\cosh\lambda=1$ is obtained, and then we propose a numerical approximation of the solutions of this equation.
4.1. The eigenvalue equation
We start with the basic form $$\varphi(x)= \alpha\cos\lambda x +\beta\sin\lambda x +\gamma\cosh\lambda x +\delta\sinh\lambda x$$ and we want to determine values of the five parameters $\alpha,\beta,\gamma,\delta,\lambda$ so that $\varphi$ satisfies the four boundary conditions $0=\varphi(0)=\varphi(1)=\varphi'(0)=\varphi'(1)$. We have $$\varphi'(x)=\lambda\left[ -\alpha\sin\lambda x +\beta\cos\lambda x +\gamma\sinh\lambda x +\delta\cosh\lambda x \right]$$ And setting $0=\varphi(0)$ and $0=\varphi'(0)$ gives respectively $\gamma=-\alpha$ and $\delta=-\beta$, thus $$\varphi(x)= \alpha\left(\cos\lambda x-\cosh\lambda x\right) + \beta\left(\sin\lambda x-\sinh\lambda x\right)$$ and $$\varphi'(x)=\lambda \left[ \alpha\left(-\sin\lambda x-\sinh\lambda x\right) + \beta\left(\cos\lambda x-\cosh\lambda x\right) \right]$$ To simplify the notation, we write \begin{eqnarray*} c &= \cos\lambda \\ s &= \sin\lambda \\ k &= \cosh\lambda \\ z &= \sinh\lambda \\ \end{eqnarray*} thus the conditions $0=\varphi(1)$ and $0=\varphi'(1)$ read \begin{eqnarray*} 0 &= \alpha(c-k)+\beta(s-z) \\ 0 &= \alpha(-s-z)+\beta(c-k) \\ \end{eqnarray*} or, in matrix form $$\begin{pmatrix} c-k & s-z \\ -s-z & c-k \\ \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \end{pmatrix}$$ If $(\alpha,\beta)$ is not the zero vector, then the matrix must be singular, thus $$(c-k)^2+(s+z)(s-z)=0$$ this condition can be simplified using the trigonometric identities $c^2+s^2=1$ and $k^2-z^2=1$ to obtain the equivalent equation $ck=1$ : $$\cos\lambda\cosh\lambda=1$$ Besides the non-interesting case $\lambda=0$, this equation has an infinite sequence of solutions $\pm\lambda_n$ that determine the spectrum of our problem. Given such a solution $\lambda_n$, the corresponding values of $\alpha_n,\beta_n$ are determined from the condition $0=\varphi(1)$: $$0 = \alpha_n(\cos\lambda_n-\cosh\lambda_n) +\beta_n(\sin\lambda_n-\sinh\lambda_n)$$ we normalize the solution with $\alpha_n=1$ (this is an arbitrary choice, maybe not the best one, but it produces nicely bounded functions on $[0,1]$). Now $$\beta_n=\frac{\cos\lambda_n-\cosh\lambda_n}{\sinh\lambda_n-\sin\lambda_n}$$ and the functions of the basis are $$\varphi_n(x)=\sin(\lambda_nx)-\sinh(\lambda_nx)+\beta_n(\cos(\lambda_nx)-\cosh(\lambda_nx))$$ Or, by rearranging the terms, $$\varphi_n(x)= \frac{ (\beta_n-1) e^{\lambda_nx} +(\beta_n+1) e^{-\lambda_nx} +(\beta_n-i) e^{i\lambda_nx} +(\beta_n+i) e^{-i\lambda_nx} }{2}$$ or even $$\varphi_n(x)=\frac{1}{2}\sum_{\rho^4=1} (\beta_n-\rho)e^{\rho\lambda_n x}$$ and this last expression is more amenable to computations (derivatives, integrals, scalar products).
4.2. Numerical solution of the eigenvalue equation
By plotting the function $x\to\cos x\cosh x$, it is clear that it crosses the value $1$ infinitely many times, very near to the zeros of $\cos x$, except for $x=\pi/2$:
Thus, the zeros of the function $\lambda\to\cos\lambda\cosh\lambda-1$ have the form $$\lambda_n = \frac{2n+1}{2}\pi + \varepsilon_n$$ where $\varepsilon_n$ are numbers that tend very fast to zero and alternate sign. We can estimate the number $\varepsilon_n$ by computing the tangent to the graph of $\cos x\cosh x$ at $x=\frac{2n+1}{2}\pi$, and finding its intersection with the horizontal line $y=1$: $$1 = \cos x_n\cosh x_n + \left( \cos x_n\sinh x_n - \sin x_n\cosh x_n \right) \varepsilon_n$$ for $x_n=\frac{2n+1}{2}\pi$. Since $\cos x_n=0$ and $\sin x_n=-(-1)^n$, this simplifies to $$\varepsilon_n=\frac{-(-1)^n}{\cosh x_n} \approx -2e^{-x_n}$$ resulting in the approximation proposed above $$\lambda_n\approx\frac{2n+1}{2}\pi-2(-1)^n\exp\left(-\frac{2n+1}{2}\pi\right).$$ This approximation is useful, at least for plotting purposes. Notice that the quality of the approximation improves as $n$ grows, so a satisfactory solution can be attained by tabulating the exact values of $\lambda_n$ for small values of $n$, and using the approximation for the others.
5. Computer code
This is the complete gnuplot code to produce plots of the Cauchy basis:
X(n) = (2*n+1)*pi/2 # approximate eigenvalues (to 0th order)
L(n) = X(n) - (-1)**n/cosh(X(n)) # refined eigenvalues (to 1st order)
s(x) = sin(x) - sinh(x) # notation
c(x) = cos(x) - cosh(x) # notation
u(l,x) = s(l*x) - c(l*x) * s(l)/c(l) # generic eigenfunction
v(n,x) = u(L(n),x) # n-th eigenfunction
plot [-0:1] [-2:3] v(1,x),v(2,x),v(3,x),v(4,x),v(5,x) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9988312125205994, "perplexity": 543.2477949782478}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250594101.10/warc/CC-MAIN-20200119010920-20200119034920-00488.warc.gz"} |
https://www.physicsforums.com/threads/standing-waves-on-a-transmission-line-and-time.639419/ | # Standing waves on a transmission line and time
1. Sep 27, 2012
### FrankJ777
Hi, I'm hoping somebody can help me understand something. I'm studying transmission lines and I'm confused about SWR and standing waves on a transmission line. According to my book the voltage on the transmision line is the super position of an incident and reflected voltage wave given by:
V$_{(z)}$ = V$^{+}_{0}$ ( e$^{-jBz}$ + $\Gamma$ e$^{jBz}$)
I can see how this produces a standing wave with voltage minimums and maximums at fixed points every λ/2. What I'm confused about is why the voltage is not represented as a function of time. From what I thought I understood about the phasor representation of the voltage we dropped the factor e$^{jωt}$ in our notation, but that it's understood that it's still there and the wave is still a funtion of time and frequency (jωt). So my questions are these. Are standing waves also a function of time and frequency, and if so would they be traveling waves? It seems that according to my text the min and max points are at fixed distances, so it seems that the voltage wave doesn't travel. So what happened to time and frequency?
Thanks to anyone who can set me straight.
2. Sep 28, 2012
### es1
3. Sep 28, 2012
### yungman
Standing wave is not time dependent, like a guitar string that the fundamental has the peak in the middle. If you go through the the wave traveling on the a transmission from the source at one end and termination on the other end. If the source and termination is not match, you will develop a standing wave pattern. It is the voltage traveling forward and reflect back that cause the standing wave pattern. If you solve the equation you posted with z, you'll see the standing wave pattern.
SWR only tells you the Vmax/Vmin or something like that, it does not tell you what standing wave pattern looks like, but from a standing wave pattern, you can find the SWR by just measuring it.
For a cosine wave,
$$\cos (\omega t-\beta z) = Re[e^{j(\omega t-\beta z)}]\;=\; Re[e^{j\omega t}e^{-\beta z}]\;=\;Re [ e^{\omega t}\tilde A ]$$
Phazor is defined as $\tilde A = e^{-\beta z}$
Phasor do not have time information, but it is understood that phasor is ONLY part of the function with time function ignored for simplification.
4. Sep 28, 2012
### FrankJ777
Thanks guys.
I found this video and it seems to to describe it pretty well visually.
I think what confused me is that it in my textbook (Pozar), it shows that the standing wave is the supper position of the incident and reflected wave,and it describes the reflected waves as phasors with a dependence on z only. I didn't see how in the time domain the incident and reflected wave would make standing wave, but in the video it shows exactly that.
5. Sep 28, 2012
### sophiecentaur
The standing wave pattern just shows the maxima. The levels between these +and- maxima are time dependent.
Similar Discussions: Standing waves on a transmission line and time | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8407015204429626, "perplexity": 476.76172941308204}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948587577.92/warc/CC-MAIN-20171216104016-20171216130016-00266.warc.gz"} |
https://www.physicsforums.com/threads/can-one-calculate-velocity-from-force-used.815120/ | # Can one calculate velocity from force used?
1. May 21, 2015
### Q7heng
1. The problem statement, all variables and given/known data
This is a general problem, but if a 2kg object is given a push with a force of 20N, would one be able to calculate the velocity of the object if the surface is frictionless? Or does one need work and energy to find out the velocity?
2. Relevant equations
F=ma
Work=F delta distance
KE=1/2mv2
3. The attempt at a solution
I tried to find acceleration through that but wouldn't that be irrelevant on a frictionless track?
2. May 21, 2015
### haruspex
You posted equations showing that velocity can be determined from work done and mass, and that work done can be determined from force and distance force advanced.
You have quoted a mass and a force. What's missing if you want to determine the velocity?
3. May 21, 2015
### SteamKing
Staff Emeritus
Why would acceleration of the mass be irrelevant? The mass is starting out at zero velocity, so how does the velocity change?
Do you know the relationship between acceleration and velocity?
Do you know Newton's Laws of Motion?
4. May 22, 2015
### CWatters
If you know the starting velocity you can work out the final velocity as a function of time...
Definition of acceleration...
dv/dt = a
Integrate both sides.
v = at + constant
The constant is the starting velocity and gives you the equation of motion...
v = at + u ............................. (1)
F = ma
a = F/m .........................(2)
put (2) into (1)
v = Ft/m + u
I've omitted some assumptions.
Draft saved Draft deleted
Similar Discussions: Can one calculate velocity from force used? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9752551317214966, "perplexity": 1328.5661293396688}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891816370.72/warc/CC-MAIN-20180225110552-20180225130552-00571.warc.gz"} |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book%3A_Inorganic_Chemistry_(Saito)/05%3A_Chemistry_of_Main-Group_Metals/5.01%3A_Group_1_Metals | # 5.1: Group 1 Metals
Group 1 metals are called alkali metals. Alkali metals are abundant in minerals and sea water. Especially the content of sodium, Na, in the Earth's crust is fourth after Al, Fe, and Ca. Although the existence of sodium or potassium ions was recognized for many years, a number of attempts to isolate the metals from aqueous solutions of their salts failed because of their high reactivity with water. Potassium (1807) and subsequently sodium were isolated by the electrolysis of molten salt of KOH or NaOH by H. Davy in the 19th century. Lithium Li was discovered as a new element (1817), and Davy soon isolated it by molten salt electrolysis of Li2O. Rubidium, Rb and Cesium, Cs, were discovered as new elements by spectroscopy in 1861. Francium, Fr, was discovered using a radiochemical technique in 1939. Its natural abundance is very low.
Table $$\PageIndex{1}$$ Properties of group 1 metals
mp
(°C)
bp
(°C)
d(20 °C)
(g cm-3)
E0 (V)
M+ + e-
I
(kJ mol-1)
Li 181 1342 0.534 -3.04 520
Na 98 883 0.968 -2.71 496
K 63 759 0.856 -2.93 419
Rb 39 688 1.532 -2.98 403
Cs 28 671 1.90 -3.03 376
As shown in Table $$\PageIndex{1}$$, melting-points, boiling points, and densities of alkali metals are low, and they are soft metals. Since the outer shell contains only one s-electron, the ionization energy is very low, and mono cations of alkali metals form easily. Qualitative analysis of alkali metals is possible by means of flame reactions using characteristic luminescence lines. Especially the orange D-line of sodium is used in the sodium lamp. Alkali metals are oxidized by water evolving hydrogen gas due to their low reduction potentials. Except lithium, the heavier alkali metals react violently with water, and sufficient caution should be exercised in their handling.
Exercise $$\PageIndex{1}$$
Describe the reactivity of alkali metals in water.
Alkali metals are also highly reactive to oxygen or halogens. As alkali metals are very reducing, they are used widely as reducing agents. Because of the high affinity of alkali metals to halogens, they are important in organic and inorganic syntheses which produce alkali metal halides as the result of condensation and metathesis reactions. Although it is generally difficult to dissolve metals in solvents to make atomic dipersions, alkali metals can be dispersed in liquid ammonia solutions, amalgams, and as cryptand (Figure $$\PageIndex{1}$$), naphthalene, or benzophenone (C6H5)2CO complexes.
$M + n\; NH_{3} \rightarrow M^{+} [e^{-} (NH_{3})]$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8088067770004272, "perplexity": 4351.004511508538}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057580.39/warc/CC-MAIN-20210924201616-20210924231616-00416.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/140647-preserving-volume.html | Find a linear transformation $h: R^{n}\rightarrow R^{n}$ that preserves volumes but is not an isometry.
The matrix $\left(\begin{array}{cc}K&0\\0&\frac{1}{K}\end{arra y}\right)$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9195851683616638, "perplexity": 221.45839850207543}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560282937.55/warc/CC-MAIN-20170116095122-00533-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://mathcracker.com/t-test-for-two-means | # T-test for two Means – Unknown Population Standard Deviations
Instructions: Use this T-Test Calculator for two Independent Means calculator to conduct a t-test for two population means ($$\mu_1$$ and $$\mu_2$$), with unknown population standard deviations. This test apply when you have two-independent samples, and the population standard deviations $$\sigma_1$$ and $$\sigma_2$$ and not known. Please select the null and alternative hypotheses, type the significance level, the sample means, the sample standard deviations, the sample sizes, and the results of the t-test for two independent samples will be displayed for you:
Ho: $$\mu_1$$ $$\mu_2$$
Ha: $$\mu_1$$ $$\mu_2$$
Sample Mean ($$\bar X_1$$):
Sample Mean ($$\bar X_2$$):
Sample St. Dev. ($$s_1$$):
Sample St. Dev. ($$s_2$$):
Sample Size ($$n_1$$):
Sample Size ($$n_2$$):
Significance Level ($$\alpha$$) =
Assume equal variances
Assume unequal variances
Test for equality of variances
## The T-test for Two Independent Samples
More about the t-test for two means so you can better interpret the output presented above: A t-test for two means with unknown population variances and two independent samples is a hypothesis test that attempts to make a claim about the population means ($$\mu_1$$ and $$\mu_2$$).
More specifically, a t-test uses sample information to assess how plausible it is for the population means $$\mu_1$$ and $$\mu_2$$ to be equal. The test has two non-overlapping hypotheses, the null and the alternative hypothesis.
The null hypothesis is a statement about the population means, specifically the assumption of no effect, and the alternative hypothesis is the complementary hypothesis to the null hypothesis.
### Properties of the two sample t-test
The main properties of a two sample t-test for two population means are:
• Depending on our knowledge about the "no effect" situation, the t-test can be two-tailed, left-tailed or right-tailed
• The main principle of hypothesis testing is that the null hypothesis is rejected if the test statistic obtained is sufficiently unlikely under the assumption that the null hypothesis is true
• The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true
• In a hypothesis tests there are two types of errors. Type I error occurs when we reject a true null hypothesis, and the Type II error occurs when we fail to reject a false null hypothesis
### How do you compute the t-statistic for the t test for two independent samples?
The formula for a t-statistic for two population means (with two independent samples), with unknown population variances shows us how to calculate t-test with mean and standard deviation and it depends on whether the population variances are assumed to be equal or not. If the population variances are assumed to be unequal, then the formula is:
$t = \frac{\bar X_1 - \bar X_2}{\sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }}$
On the other hand, if the population variances are assumed to be equal, then the formula is:
$t = \frac{\bar X_1 - \bar X_2}{\sqrt{ \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}(\frac{1}{n_1}+\frac{1}{n_2}) } }$
Normally, the way of knowing whether the population variances must be assumed to be equal or unequal is by using an F-test for equality of variances.
With the above t-statistic, we can compute the corresponding p-value, which allows us to assess whether or not there is a statistically significant difference between two means.
### Why is it called t-test for independent samples?
This is because the samples are not related with each other, in a way that the outcomes from one sample are unrelated from the other sample. If the samples are related (for example, you are comparing the answers of husbands and wives, or identical twins), you should use a t-test for paired samples instead.
### What if the population standard deviations are known?
The main purpose of this calculator is for comparing two population mean when sigma is unknown for both populations. In case that the population standard deviations are known, then you should use instead this z-test for two means.
In case you have any suggestion, or if you would like to report a broken solver/calculator, please do not hesitate to contact us.
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http://ajspeller.com/tag/definite-integral/ | # Tag Archives: definite integral
## Calculus – Use Integration to Find the Triangle Area Formula
By | April 15, 2017
Use Integration to Find the Triangle Area Formula In this tutorial students learn how to find the area between curves. The students then verify the solution using the GRAPH, WINDOW, CALC, INTEGRATION and TRACE features of the TI-84C. 1) Set the 2 functions equal to each other to find the interval over which the integration is… Read More »
## Calculus – Use Integration to find the Area Between Curves – Example 6
By | April 14, 2017
Use Integration to Find the Area Between Curves In this tutorial students learn how to find the area between curves. The students then verify the solution using the GRAPH, WINDOW, CALC, INTEGRATION and TRACE features of the TI-84C. 1) Set the 2 functions equal to each other to find the interval over which the integration is… Read More »
## Calculus – Use Integration to find Area Between Curves – Example 5
By | April 13, 2017
Use Integration to Find the Area Between Curves In this tutorial students learn how to find the area between curves. The students then verify the solution using the GRAPH, WINDOW, CALC, INTEGRATION and TRACE features of the TI-84C. 1) Set the 2 functions equal to each other to find the interval over which the integration is to take… Read More »
## Calculus – Use Integration to find Area Between Curves – Example 4
By | April 12, 2017
Integration – Find the Area Between Curves In this tutorial students learn how to find the area between curves. The students then verify the solution using the GRAPH, WINDOW, CALC, INTEGRATION and TRACE features of the TI-84C. 1) Set the 2 functions equal to each other to find the interval over which the integration is to take… Read More »
## Calculus – Use Integration to find Area Between Curves – Example 3
By | April 11, 2017
Integration – Find the Area Between Curves In this tutorial students learn how to find the area between curves. The students then verify the solution using the GRAPH, WINDOW, CALC, INTEGRATION and TRACE features of the TI-84C. 1) Set the 2 functions equal to each other to find the interval over which the integration is to take… Read More »
## Calculus – Use Integration to find Area Between Curves – Example 2
By | April 10, 2017
Integration – Find the Area Between Curves In this tutorial students learn how to find the area between curves. The students then verify the solution using the GRAPH, WINDOW, CALC, INTEGRATION and TRACE features of the TI-84C. 1) Set the 2 functions equal to each other to find the interval over which the integration is to… Read More »
## Calculus – Use Integration to find Area Between Curves
By | April 9, 2017
Integration – Find the Area Between Curves In this tutorial students learn how to find the area between curves. The students then verify the solution using the GRAPH, WINDOW, CALC, INTEGRATION and TRACE features of the TI-84C. 1) Set the 2 functions equal to each other to find the interval over which the integration is to… Read More »
## Calculus – Trigonometric Function Integration using the Substitution Method
By | April 7, 2017
Trigonometric Function Integration using the Substitution Method In this tutorial students learn how to calculate the area under a trigonometric function over an interval using the u-substitution technique. Virginia Standards of Learning(SOL) APC.11
## Calculus – Definite Integral Substitution Method
By | April 6, 2017
Definite Integral Substitution Method In this tutorial students learn how to evaluate the definite integral using the substitution technique. Virginia Standards of Learning(SOL) APC.11
## Calculus – Find the Definite Integral of a Cubic Function
By | April 5, 2017
Find the Definite Integral of a Cubic Function In this tutorial students learn how to evaluate the definite integral of a cubic function. The students will also verify the results using the TI-84C. Virginia Standards of Learning(SOL) APC.11 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8262919783592224, "perplexity": 646.0484699479103}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647885.78/warc/CC-MAIN-20180322131741-20180322151741-00156.warc.gz"} |
https://math.stackexchange.com/questions/1697986/writing-complex-numbers-in-form-abi | # Writing complex numbers in form $a+bi$
Can $\sqrt{i+\sqrt{2}}$ be expressed as $a+bi$ with $a,b \in \mathbb{R}$? In general, what kinds of expressions can be rewritten in that form?
• The square root is not defined on the complex numbers as a function. As to your second question: Every complex number con be written in this form. – Friedrich Philipp Mar 15 '16 at 1:22
You can express such an expression in the form $a+ib$.
Let $$x+iy = \sqrt{i+\sqrt{2}} \\ (x+iy)^2 = (\sqrt{i+\sqrt{2}})^2 \\ x^2 - y^2 +2ixy = i+\sqrt{2} \\$$ Now you only need to solve the equations $x^2 - y^2 = \sqrt{2}$, and $xy = \frac{1}{2}$ to get the values of $x$ and $y$.
$x = \,\,^+_-\sqrt{\frac{\sqrt{3}+\sqrt{2}}{2}}$ and $y = \,\,^+_-\sqrt{\frac{\sqrt{3}-\sqrt{2}}{2}}$
• This is not correct. $(x+iy)^2=x^2-y^2+i2xy\ne x^2+y^2+i2xy$. – Mark Viola Mar 15 '16 at 1:50
• So does that mean $\sqrt{i+\sqrt{2}} \in \mathbb{C}$? – rorty Mar 17 '16 at 21:09
• Yes. Also, for any other complex number $\sqrt{a+ib}$, if you get real solutions for $x$ and $y$ in $x+iy = \sqrt{a+ib}$, it's square root $\in \mathbb{C}$ – SS_C4 Mar 18 '16 at 1:04
An efficient approach relies on use of Cartesian-to-polar coordinate transformation. To that end, let $z=x+iy=\sqrt{x^2+y^2} e^{i\arctan2(x,y)+i2\ell \pi}$. Then, the square root of $z$ is given by
\begin{align}\sqrt{z}&=\sqrt{x+iy}\\\\&=\sqrt{\sqrt{x^2+y^2}\,e^{i\arctan2(x,y)+i2\ell \pi}}\\\\ &=(-1)^{\ell}(x^2+y^2)^{1/4}e^{i\frac12\text{arctan2}\,(x,y)}\\\\ &=\bbox[5px,border:2px solid #C0A000]{(-1)^{\ell}(x^2+y^2)^{1/4}\left(\sqrt{\frac{1+\frac{|x|}{\sqrt{x^2+y^2}}}{2}} +i\sqrt{\frac{1-\frac{|x|}{\sqrt{x^2+y^2}}}{2}} \right)} \tag 1\\\\ \end{align}
where $\ell$ is any integer and $\arctan2(x,y)$ is the arctangent function with $2$ arguments.
Note that $(-1)^\ell = \pm 1$. Then, for $x=\sqrt 2$ and $y=1$, $(1)$ becomes
\begin{align}\sqrt{\sqrt 2+i}&=\pm 3^{1/4}\left(\sqrt{\frac{1+\frac{\sqrt 2}{\sqrt{3}}}{2}} +i\sqrt{\frac{1-\frac{\sqrt 2}{\sqrt{3}}}{2}} \right)\\\\ &=\bbox[5px,border:2px solid #C0A000]{\pm \left(\sqrt{\frac{\sqrt 3+\sqrt 2}{2}} +i\sqrt{\frac{\sqrt 3-\sqrt 2}{2}} \right)} \end{align}
And we are done! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9071531891822815, "perplexity": 806.9857209005675}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540510336.29/warc/CC-MAIN-20191208122818-20191208150818-00242.warc.gz"} |
http://mathhelpforum.com/calculus/118587-differentiate-find-square-dy-dx.html | # Thread: Differentiate and find the square of dy/dx
1. ## Differentiate and find the square of dy/dx
Question:
If $x = cos \theta + \theta sin \theta$ , $y = sin \theta - \theta cos \theta$ , then show that
$\frac{dy}{dx} = tan \theta$
Also find $\frac{d^2 y}{dx^2}$
2. Originally Posted by zorro
Question:
If $x = cos \theta + \theta sin \theta$ , $y = sin \theta - \theta cos \theta$ , then show that
$\frac{dy}{dx} = tan \theta$
Also find $\frac{d^2 y}{dx^2}$
$\frac{dy}{dx} = \frac{\frac{dy}{d \theta}}{\frac{dx}{d \theta}}$ (which I'm sure will be somewhere in yur class notes or textbook).
$\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d}{d \theta} \left(\frac{dy}{dx}\right) \cdot \frac{d \theta}{dx}$
If you need more help, please show all your working and state specifically where you are stuck.
3. "yur"?
4. ## Is this correct?
Originally Posted by mr fantastic
$\frac{dy}{dx} = \frac{\frac{dy}{d \theta}}{\frac{dx}{d \theta}}$ (which I'm sure will be somewhere in yur class notes or textbook).
$\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d}{d \theta} \left(\frac{dy}{dx}\right) \cdot \frac{d \theta}{dx}$
If you need more help, please show all your working and state specifically where you are stuck.
I am getting the $\frac{d^2 y }{dx^2} = sec^2 \theta .\frac{1}{ \theta cos \theta}$
$
\frac{d}{d \theta} \left( \frac{dy}{dx} \right) = \frac{d}{d \theta} (tan \theta) = sec^2 \theta
$
$
\frac{d \theta}{dx} = \frac{1}{\frac{dx}{d \theta}} = \frac{1}{\theta cos \theta}
$
therefore
$
\frac{d^2 y}{dx^2} = sec^2 \theta . \frac{1}{ \theta cos \theta}
$
Is this Right!!!
5. Originally Posted by zorro
I am getting the $\frac{d^2 y }{dx^2} = sec^2 \theta .\frac{1}{ \theta cos \theta}$
$
\frac{d}{d \theta} \left( \frac{dy}{dx} \right) = \frac{d}{d \theta} (tan \theta) = sec^2 \theta
$
$
\frac{d \theta}{dx} = \frac{1}{\frac{dx}{d \theta}} = \frac{1}{\theta cos \theta}
$
therefore
$
\frac{d^2 y}{dx^2} = sec^2 \theta . \frac{1}{ \theta cos \theta}
$
Is this Right!!!
Yes, and if you recall that $\frac{1}{\cos \theta} = \sec \theta$ then you can simplify your answer a little bit.
6. ## So after simplifying
Originally Posted by mr fantastic
Yes, and if you recall that $\frac{1}{\cos \theta} = \sec \theta$ then you can simplify your answer a little bit.
= $sec^2 \theta . \frac{1}{\theta cos \theta}$
= $\frac{sec^3 \theta}{ \theta}$
Is this right now???
7. Originally Posted by zorro
= $sec^2 \theta . \frac{1}{\theta cos \theta}$
= $\frac{sec^3 \theta}{ \theta}$
Is this right now???
Correct.
8. By the way, Zorro, $\frac{d^2y}{dx^2}$ is NOT "the square of dy/dx"!
9. ## Thanks u every one for helping me
Thank u Mr fantastic for ur help | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 27, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9587228894233704, "perplexity": 2015.604183699159}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187823309.55/warc/CC-MAIN-20171019141046-20171019161046-00657.warc.gz"} |
http://mathrefresher.blogspot.com/2008_02_24_archive.html | ## Saturday, March 01, 2008
### field automorphism
Definition 1: Bijective
A bijective map is a map f from set X to a set Y with the property that for every y in Y, there is exactly one x in X such that f(x) = y.
Definition 2: Homomorphism
A homomorphism is a map from one algebraic structure to another of the same type that preserves certain properties.
Definition 3: Isomorphism
An isomorphism is a bijective map f such that both f and its inverse f-1 are homomorphisms.
Definition 4: Automorphism
An automorphism is an isomorphism from a mathematical object to itself.
Definition 5: Ring Homomorphism
A ring homomorphism is a mapping between two rings which preserves the operations of addition and multiplication.
If R,S are rings and f: is the mapping R → S, the the following properties hold:
(1) f(a+b) = f(a) + f(b) for all a,b ∈ R
(2) f(ab) = f(a)f(b) for all a,b ∈ R
(3) f(1) = 1
Definition 6: Field Automorphism
A field automorphism is a bijective ring homomorphism from a field to itself.
References
### (a + b)p ≡ ap + bp (mod p)
Lemma 1:
if p is prime, then:
(a + b)
p ≡ ap + bp (mod p)
Proof:
(1) Using the Binomial Theorem (see Theorem here), we know that:
(2) Now, since p is a prime, it is clear that for each term p!/(m!)(p-m)!, p is not divisible by any term ≤ m or by any term ≤ p-m so that we have:
p!/(m!)(p-m)! = p*([(p-1)*...*p-m+1]/[m!])
(3) This shows that each of these terms is divisible by p and therefore:
[p!/(m!)(p-m)!]ap-mbm ≡ 0 (mod p)
(4) So that there exists an integer n such that:
(a + b)p = ap + np + bp
(5) Since ap + nb + bp ≡ ap + bp (mod p), it follows that:
(a + b)p ≡ ap + bp (mod p)
QED
### The set of congruence classes modulo n: Z/nZ
In considering modular arithmetic (see here for review if needed), we can divide all integers into congruence classes modulo n.
Definition 1: Congruence class modulo n: [i]
Let [i] represent the set of all integers such that [i] = { ..., i-2n, i-n, i, i+n, i+2n, ... }
To make this definition even clearer, let's consider the following lemma.
Lemma 1: x ∈ [i] if and only if x ≡ i (mod n)
Proof:
(1) Assume x ≡ i (mod n)
(2) Then n divides x - i
(3) So, there exists an integer a such that an = x - i
(4) So that x = an + i
(5) Assume that x ∈ [i]
(6) Then, there exists a such that x = i + an
(7) This shows that an = x - i and further that n divides x - i.
(8) Therefore, we have x ≡ i (mod n)
QED
Lemma 2: There are only n distinct congruence classes modulo n
Proof:
(1) To prove this, I will show that for all x ∈ Z, there exists i such that:
x ≡ i (mod n)
and
0 ≤ i ≤ n-1
(2) For x ∈ Z, there exists k such that x ≡ k (mod n)
(3) We can assume that k is positive since if k is negative, k ≡ -k (mod n) since n divides k + -k.
(4) We can further assume that k ≤ n-1, since if k ≥ n, it follows that k ≡ (k-n) (mod n) since n divides (k - [k-n]) = k -k + n = n.
QED
We can now consider the set of congruence classes modulo n as the set Z/nZ.
Definition 2: set of congruence classes modulo n: Z/nZ
Z/nZ = { [0], ..., [n-1]}
Example 1: Z/3Z
Z/3Z = { [0], [1], [2] }
For purposes of showing the relationship between Z and Z/nZ, I will from this point on view Z/nZ = {0, ..., n-1}.
So that Z/3Z will also be represented as {0, 1, 2}
Lemma 3: Z/nZ is a commutative ring
Proof:
(1) Z/nZ has commutative rule for addition
For all a,b ∈ Z/nZ: a+b ≡ b + a (mod n)
(2) Z/nZ has associative rule for addition
For all a,b,c ∈ Z/nZ: (a + b) + c ≡ a + (b + c) (mod n)
(3) Z/nZ has additive identity rule
0 ∈ Z/nZ and for all a ∈ Z/nZ: a ≡ a + 0 (mod n)
(4) Z/nZ has additive inverse rule:
(a) Let a be any congruence class modulo n such that a ∈ Z/nZ
(b) Assume a ≠ 0 for if a = 0, then a + a ≡ 0 (mod n) and a is its own additive inverse.
(c) Let b = n - a
(d) b ∈ Z/nZ since 1 ≤ b ≤ n-1
(e) a + b ≡ n ≡ 0 (mod n)
(5) Z/nZ has an associative rule for multiplication:
For all a,b,c ∈ Z/nZ: (ab)c ≡ a(bc) (mod n)
(6) Z/nZ has a distributive rule:
For all a,b,c ∈ Z/nZ: a(b+c) ≡ ab + ac ≡ (b + c)a (mod n)
(7) Z/nZ has a commutative rule for multiplication
For all a,b ∈ Z/nZ: ab ≡ ba (mod n)
(8) Thus, Z/nZ is a commutative ring. [See Definition 2, here]
QED
Lemma 4: if p is a prime, then Z/pZ is a field
Proof:
(1) Z/pZ has a multiplicative identity rule.
1 ∈ Z/pZ and for all a ∈ Z/pZ 1*a ≡ a*1 ≡ a (mod p)
(2) Z/pZ has a multiplicative inverse rule since:
(a) Let a be any nonzero element of Z/pZ
(b) If p = 2, then a is its own inverse and a*a ≡ 1*1 ≡ 1 (mod 2).
(c) If p ≥ 3, then let b = ap-2
(d) Then a*b ≡ a*(ap-2) ≡ ap-1 ≡ 1 (mod p). [See Fermat's Little Theorem, here]
(3) Thus, Z/pZ is a field. [See Definition 3, here]
QED
Lemma 5:
if p is prime and ab ≡ 0 (mod p), then a ≡ 0 (mod p) or b ≡ 0 (mod p)
Proof:
(1) Assume a is not ≡ 0 (mod p)
(2) Since ab ≡ 0 (mod p), it follows that p divides ab.
(3) If a is not ≡ 0 (mod p), then it follows that p does not divide a.
(4) So, using Euclid's Lemma (see Lemma 2, here), it follows that p divides b.
(5) And equivalently, b ≡ 0 (mod p)
QED
References
• Gareth A. Jones, J. Mary Jones, , Springer, 2002. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9469826817512512, "perplexity": 1306.9555051325453}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187828356.82/warc/CC-MAIN-20171024090757-20171024110757-00659.warc.gz"} |
http://en.wikipedia.org/wiki/Proof_of_Bertrand's_postulate | # Proof of Bertrand's postulate
In mathematics, Bertrand's postulate (actually a theorem) states that for each $n\ge 1$ there is a prime $p$ such that $n. It was first proven by Pafnuty Chebyshev, and a short but advanced proof was given by Srinivasa Ramanujan.[1] The gist of the following elementary proof is due to Paul Erdős. The basic idea of the proof is to show that a certain central binomial coefficient needs to have a prime factor within the desired interval in order to be large enough. This is made possible by a careful analysis of the prime factorization of central binomial coefficients.
The main steps of the proof are as follows. First, one shows that every prime power factor $p^r$ that enters into the prime decomposition of the central binomial coefficient $\tbinom{2n}{n}:=\frac{(2n)!}{(n!)^2}$ is at most $2n$. In particular, every prime larger than $\sqrt{2n}$ can enter at most once into this decomposition; that is, its exponent $r$ is at most one. The next step is to prove that $\tbinom{2n}{n}$ has no prime factors at all in the gap interval $\left(\tfrac{2n}{3}, n\right)$. As a consequence of these two bounds, the contribution to the size of $\tbinom{2n}{n}$ coming from all the prime factors that are at most $n$ grows asymptotically as $O(\theta^n)$ for some $\theta<4$. Since the asymptotic growth of the central binomial coefficient is at least $4^n/2n$, one concludes that for $n$ large enough the binomial coefficient must have another prime factor, which can only lie between $n$ and $2n$. Indeed, making these estimates quantitative, one obtains that this argument is valid for all $n>468$. The remaining smaller values of $n$ are easily settled by direct inspection, completing the proof of the Bertrand's postulate.
## Lemmas and computation
### Lemma 1: A lower bound on the central binomial coefficients
Lemma: For any integer $n>0$, we have
$\frac{4^n}{2n} \le \binom{2n}{n}.\$
Proof: Applying the binomial theorem,
$4^n = (1 + 1)^{2n} = \sum_{k = 0}^{2n} \binom{2n}{k}=2+\sum_{k = 1}^{2n-1} \binom{2n}{k} \le 2n\binom{2n}{n},\$
since $\tbinom{2n}{n}$ is the largest term in the sum in the right-hand side, and the sum has $2n$ terms (including the initial two outside the summation).
### Lemma 2: An upper bound on prime powers dividing central binomial coefficients
For a fixed prime $p$, define $R(p,n)$ to be the largest natural number $r$ such that $p^r$ divides $\tbinom{2n}{n}$.
Lemma: For any prime $p$, $p^{R(p,n)}\le 2n$.
Proof: The exponent of $p$ in $n!$ is (see Factorial#Number theory):
$\sum_{j = 1}^\infty \left\lfloor \frac{n}{p^j} \right\rfloor,\$
so
$R(p,n) =\sum_{j = 1}^\infty \left\lfloor \frac{2n}{p^j} \right\rfloor - 2\sum_{j = 1}^\infty \left\lfloor \frac{n}{p^j} \right\rfloor =\sum_{j = 1}^\infty \left(\left\lfloor \frac{2n}{p^j} \right\rfloor - 2\left\lfloor \frac{n}{p^j} \right\rfloor\right).$
But each term of the last summation can either be zero (if $n/p^j \bmod 1< 1/2$) or 1 (if $n/p^j \bmod 1\ge 1/2$) and all terms with $j>\log_p(2n)$ are zero. Therefore
$R(p,n) \leq \log_p(2n),\$
and
$p^{R(p,n)} \leq p^{\log_p{2n}} = 2n.\$
This completes the proof of the lemma.
### Lemma 3: The exact power of a large prime in a central binomial coefficient
Lemma: If $p$ is odd and $\frac{2n}{3} < p \leq n$, then $R(p,n) = 0.\$
Proof: The factors of $p$ in the numerator come from the terms $p$ and $2p$, and in the denominator from two factors of $p$. These cancel since $p$ is odd.
### Lemma 4: An upper bound on the primorial
We estimate the primorial function,
$x\# = \prod_{p \leq x} p,\$
where the product is taken over all prime numbers $p$ less than or equal to the real number $x$.
Lemma: For all real numbers $x\ge 3$, $x\#<2^{2x-3}$[2]
Proof: Since $\binom{2n}{n}$ is an integer and all the primes $n+1 \le p \le 2n-1$ appears in its numerator $(2x-1)\#/(x)\# \le \binom{2n}{n} <2^{2x-2}$ holds. The proof is by mathematical induction.
• $n = 3$: $n\# = 6 < 8.$
• $n = 4$: $n\# =6 < 32.$
• $2m-1\# < 2^{2(2m-1)-3}$
• $2m\# < 2^{2(2m-3)}$
• $x\# = \lfloor x\rfloor\# < 2^{2x-3}$
Thus the lemma is proven.
## Proof of Bertrand's Postulate
Assume there is a counterexample: an integer n ≥ 2 such that there is no prime p with n < p < 2n.
If 2 ≤ n < 468, then p can be chosen from among the prime numbers 3, 5, 7, 13, 23, 43, 83, 163, 317, 631 (each being less than twice its predecessor) such that n < p < 2n. Therefore n ≥ 468.
There are no prime factors p of $\textstyle\binom{2n}{n}$ such that:
• 2n < p, because every factor must divide (2n)!;
• p = 2n, because 2n is not prime;
• n < p < 2n, because we assumed there is no such prime number;
• 2n / 3 < pn: by Lemma 3.
Therefore, every prime factor p satisfies p ≤ 2n/3.
When $p > \sqrt{2n},$ the number $\textstyle {2n \choose n}$ has at most one factor of p. By Lemma 2, for any prime p we have pR(p,n) ≤ 2n, so the product of the pR(p,n) over the primes less than or equal to $\sqrt{2n}$ is at most $(2n)^{\sqrt{2n}}$. Then, starting with Lemma 1 and decomposing the right-hand side into its prime factorization, and finally using Lemma 4, these bounds give:
$\frac{4^n}{2n } \le \binom{2n}{n} = \left(\prod_{p \le \sqrt{2n}} p^{R(p,n)}\right) \left(\prod_{\sqrt{2n} < p \le \frac{2n}{3}} p^{R(p,n)}\right) < (2n)^{\sqrt{2n}} \prod_{1 < p \leq \frac{2n}{3} } p = (2n)^{\sqrt{2n}} \Big( \frac{2n}{3}\Big)\# \le (2n)^{\sqrt{2n}} 4^{2n/3}.\$
Taking logarithms yields to
${\frac{\log 4}{3}}n \le (\sqrt{2n}+1)\log 2n\; .$
By concavity of the right-hand side as a function of n, the last inequality is necessarily verified on an interval. Since it holds true for n=467 and it does not for n=468, we obtain
$n < 468.\$
But these cases have already been settled, and we conclude that no counterexample to the postulate is possible.
### Proof by Shigenori Tochiori
Using Lemma 4, Tochiori refined Erdos's method and proved if there exists a positive integer $n \ge 5$ such that there is no prime number $n then $n < 64$. [3]
First, refine lemma 1 to:
Lemma 1': For any integer $n\ge 4$, we have
$\frac{4^n}n < \binom{2n}{n}.\$
Proof: By induction: $\frac{4^4}4 = 64 < 70 = \binom{8}{4},$ and assuming the truth of the lemma for $n-1$,
$\binom{2n}{n} = 2\,\frac{2n-1}{n}\binom{2(n-1)}{n-1} > 2\,\frac{2n-1}{n}\frac{4^{n-1}}{n-1} > 2\cdot 2\,\frac{4^{n-1}}{n} = \frac{4^n}{n}.$
Then, refine the estimate of the product of all small primes via a better estimate on $\pi(x)$ (the number of primes at most $n$):
Lemma 5: For any natural number $n$, we have
$\pi(n)\le\frac13n+2.$
Proof: Except for $p=2,3$, every prime number has $p\equiv 1$ or $p\equiv 5 \pmod 6$. Thus $\pi(n)$ is upper bounded by the number of numbers with $k\equiv 1$ or $k\equiv 5 \pmod 6$, plus one (since this counts $1$ and misses $2,3$). Thus
$\pi(n)\le\left\lfloor\frac{n+5}6\right\rfloor+\left\lfloor\frac{n+1}6\right\rfloor+1\le\frac{n+5}6+\frac{n+1}6+1=\frac13n+2.$
Now, calculating the binomial coefficient as in the previous section, we can use the improved bounds to get (for $n\ge5$, which implies $\sqrt{2n}\ge3$ so that $\sqrt{2n}\#\ge3\#=6$):
\begin{align} \frac{4^n}{n}&\le \binom{2n}{n} \\ &= \prod_{p \le \sqrt{2n}} p^{R(p,n)}\cdot\prod_{\sqrt{2n} < p \le \frac{2n}{3}} p^{R(p,n)}\\ &< (2n)^{\pi(\sqrt{2n})} \prod_{\sqrt{2n} < p \leq \frac{2n}{3}} p = (2n)^{\frac13\sqrt{2n}+2} \frac{(2n/3)\#}{\sqrt{2n}\#}\\ &<(2n)^{\frac13\sqrt{2n}+2}\frac{2^{2\cdot2n/3-3}}6<(2n)^{\frac13\sqrt{2n}+2}2^{4n/3-5}. \end{align}
Taking logarithms to get
$\frac23n\log 2< \frac13\sqrt{2n}\log 2n+3\log \frac n2$
and dividing both sides by $\frac23n$:
$\log 2<\sqrt{2}\cdot\frac{\log\sqrt{n}}{\sqrt{n}}+\frac94\frac{\log \frac n2}{\frac n2}+\frac{\log 2}{\sqrt{2n}}\equiv f(n)\; .$
Now the function $g(x)=\frac{\log x}x$ is decreasing for $x\ge e$, so $f(n)$ is decreasing when $n\ge e^2>2e$. But
$\frac{f(2^6)}{\log 2}=\sqrt{2}\cdot\frac{3}{8}+\frac94\cdot\frac{5}{32}+\frac{\sqrt{2}}{16}=0.97\dots<1<\frac{f(n)}{\log 2},$
so $n<2^6=64$. The remaining cases are proven by an explicit list of primes, as above. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 106, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9945745468139648, "perplexity": 279.28418216644155}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510275393.46/warc/CC-MAIN-20140728011755-00470-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://proofwiki.org/wiki/Definition:Ordering_on_Integers | Definition:Ordering on Integers
Definition
The integers are ordered on the relation $\le$ as follows:
$\forall x, y \in \Z: x \le y \iff y - x \in \Z_{\ge 0}$
That is, $x$ is less than or equal to $y$ if and only if $y - x$ is non-negative. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.987975537776947, "perplexity": 108.51768650245387}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247489729.11/warc/CC-MAIN-20190219081639-20190219103639-00583.warc.gz"} |
http://college-physics.com/book/astronomy/gravitational-fields-ii/ | # Gravitational Fields II
## Introduction
In the first part on gravitational fields, the gravitational force and the gravitational acceleration have been described. The second part is about the work, energy and the movement of bodies in a gravitational field.
## Work $$W$$
The mechanical work is already known. It occurs when a force $$F$$ acts along a path $$s$$ on a body. The following applies:
$$W = F \cdot s$$
$$F$$ = Force, $$s$$ = Displacement
When bodies are moved in a gravitational field, either through external influences or through the gravitational force itself, work is being done. It should however be noted that as the path $$s$$ is the path parallel to the field lines in this case. When moving a body along the field lines much work is done, when moving perpendicular to the field lines, however, no work is performed.
### Work in homogeneous fields
For homogeneous fields we can use the following formula:
$$W = F \cdot s = m \cdot g \cdot s$$ $$m$$ = Mass of the body, $$g$$ = Gravitational acceleration, $$s$$ = Path parallel to the field lines
The following animation shows 3 bodies ($$m_1 = m_2 = m_3 = 1 \rm kg$$) in a homogeneous field and shows the way parallel to the field lines.
ResetStart
Although bodies 2 and 3 travel different distances, the path parallel to the field lines is the same, therefore the amount of work performed is also the same. Body 1 is moved perpendicularly to the field lines, that is, the path parallel to the field lines is 0 and as such there is no work done.
### Work in inhomogeneous fields
For inhomogeneous fields, the calculation of the work is more difficult, because the gravitational force during the movement is not constant.
In a radial gravitational field, the gravitational force depends on the distance $$r$$. One can determine the work by integrating over the gravitational force:
$$W = \int \limits_{r_1}^{r_2} F(r) \,\,\mathrm{d}r$$ Substituting the gravitational force, we get:
$$W = G \cdot m_1 \cdot m_2 \cdot \int \limits_{r_1}^{r_2} \dfrac{1}{r^2} \,\, \mathrm{d}r = G \cdot m_1 \cdot m_2 \cdot \left( \dfrac{1}{r_1} - \dfrac{1}{r_2} \right)$$ $$m_1, m_2$$ = Masses of the bodies,
$$r_1$$ = Distance at start position, $$r_2$$ = Distance at end position
The following animation shows 2 bodies ($$m_1 = m_2 = 1 \,\, \rm kg$$) in an inhomogeneous field and shows the path parallel to the field lines.
ResetStart
It can be seen here that the work being done is only determined by the starting and ending point. Whether moving the charges straight or with many curves is not important.
## Potential energy $$E_{pot}$$
The potential energy is already known. In the gravitational field of the earth, it is the greater, the higher a body is above the ground.
The following animation shows the potential energy in the homogeneous gravitational field near the Earth's surface.
ResetStart
In the gravitational field we usually define that the potential energy on the Earth's surface is 0. If one raises a body to a certain height, work is done and stored in the potential energy of the body.
### Potential energy in inhomogeneous fields
In an inhomogeneous gravitational field around a mass, we usually define that the potential energy at an infinite distance from the mass is 0.
$$E_{pot} = \Delta W = G \cdot m_1 \cdot m_2 \cdot \left( \dfrac{1}{r_1} - \dfrac{1}{r_2} \right) = G \cdot m_1 \cdot m_2 \cdot \left( \dfrac{1}{r_1} - \dfrac{1}{\infty} \right) = G \cdot m_1 \cdot m_2 \cdot \dfrac{1}{r_1}$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9534658193588257, "perplexity": 279.43488670296136}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711064.71/warc/CC-MAIN-20221205232822-20221206022822-00016.warc.gz"} |
http://www.nag.com/numeric/MB/manual64_24_1/html/C06/c06raf.html | Integer type: int32 int64 nag_int show int32 show int32 show int64 show int64 show nag_int show nag_int
Chapter Contents
Chapter Introduction
NAG Toolbox
# NAG Toolbox: nag_sum_fft_real_sine_simple (c06ra)
## Purpose
nag_sum_fft_real_sine_simple (c06ra) computes the discrete Fourier sine transforms of m$m$ sequences of real data values.
## Syntax
[x, ifail] = c06ra(m, n, x)
[x, ifail] = nag_sum_fft_real_sine_simple(m, n, x)
## Description
Given m$m$ sequences of n1 $n-1$ real data values xjp ${x}_{\mathit{j}}^{\mathit{p}}$, for j = 1,2,,n1$\mathit{j}=1,2,\dots ,n-1$ and p = 1,2,,m$\mathit{p}=1,2,\dots ,m$, nag_sum_fft_real_sine_simple (c06ra) simultaneously calculates the Fourier sine transforms of all the sequences defined by
n − 1 x̂kp = sqrt(2/n) ∑ xjp × sin(jkπ/n), k = 1,2, … ,n − 1 and p = 1,2, … ,m. j = 1
$x^ kp = 2n ∑ j=1 n-1 xjp × sin( jk πn ) , k= 1, 2, …, n-1 and p= 1, 2, …, m .$
(Note the scale factor sqrt(2/n) $\sqrt{\frac{2}{n}}$ in this definition.)
Since the Fourier sine transform defined above is its own inverse, two consecutive calls of this function will restore the original data.
The transform calculated by this function can be used to solve Poisson's equation when the solution is specified at both left and right boundaries (see Swarztrauber (1977)).
The function uses a variant of the fast Fourier transform (FFT) algorithm (see Brigham (1974)) known as the Stockham self-sorting algorithm, described in Temperton (1983), together with pre- and post-processing stages described in Swarztrauber (1982). Special coding is provided for the factors 2$2$, 3$3$, 4$4$ and 5$5$.
## References
Brigham E O (1974) The Fast Fourier Transform Prentice–Hall
Swarztrauber P N (1977) The methods of cyclic reduction, Fourier analysis and the FACR algorithm for the discrete solution of Poisson's equation on a rectangle SIAM Rev. 19(3) 490–501
Swarztrauber P N (1982) Vectorizing the FFT's Parallel Computation (ed G Rodrique) 51–83 Academic Press
Temperton C (1983) Fast mixed-radix real Fourier transforms J. Comput. Phys. 52 340–350
## Parameters
### Compulsory Input Parameters
1: m – int64int32nag_int scalar
m$m$, the number of sequences to be transformed.
Constraint: m1${\mathbf{m}}\ge 1$.
2: n – int64int32nag_int scalar
One more than the number of real values in each sequence, i.e., the number of values in each sequence is n1$n-1$.
Constraint: n1${\mathbf{n}}\ge 1$.
3: x( m × (n + 2) ${\mathbf{m}}×\left({\mathbf{n}}+2\right)$) – double array
the data must be stored in x as if in a two-dimensional array of dimension (1 : m,1 : n + 2)$\left(1:{\mathbf{m}},1:{\mathbf{n}}+2\right)$; each of the m$m$ sequences is stored in a row of the array. In other words, if the n1$n-1$ data values of the p$\mathit{p}$th sequence to be transformed are denoted by xjp${x}_{\mathit{j}}^{\mathit{p}}$, for j = 1,2,,n1$\mathit{j}=1,2,\dots ,n-1$ and p = 1,2,,m$\mathit{p}=1,2,\dots ,m$, then the first m(n1)$m\left(n-1\right)$ elements of the array x must contain the values
x11 , x12 , … , x1m , x21 , x22 , … , x2m , … , xn − 11 , xn − 12 , … , xn − 1m . $x11 , x12 ,…, x1m , x21 , x22 ,…, x2m ,…, x n-1 1 , x n-1 2 ,…, x n-1 m .$
The n$n$th to (n + 2)$\left(n+2\right)$th elements of each row xnp ,, xn + 2p${x}_{n}^{\mathit{p}},\dots ,{x}_{n+2}^{\mathit{p}}$, for p = 1,2,,m$\mathit{p}=1,2,\dots ,m$, are required as workspace. These 3m$3m$ elements may contain arbitrary values as they are set to zero by the function.
None.
work
### Output Parameters
1: x( m × (n + 2) ${\mathbf{m}}×\left({\mathbf{n}}+2\right)$) – double array
the m$m$ Fourier sine transforms stored as if in a two-dimensional array of dimension (1 : m,1 : n + 2)$\left(1:{\mathbf{m}},1:{\mathbf{n}}+2\right)$. Each of the m$m$ transforms is stored in a row of the array, overwriting the corresponding original sequence. If the (n1)$\left(n-1\right)$ components of the p$p$th Fourier sine transform are denoted by kp${\stackrel{^}{x}}_{\mathit{k}}^{\mathit{p}}$, for k = 1,2,,n1$\mathit{k}=1,2,\dots ,n-1$ and p = 1,2,,m$\mathit{p}=1,2,\dots ,m$, then the m(n + 2)$m\left(n+2\right)$ elements of the array x contain the values
x̂11 , x̂12 , … , x̂1m , x̂21 , x̂22 , … , x̂2m , … , x̂n − 11 , x̂n − 12 , … , x̂n − 1m , 0 , 0 , … , 0 (3m times) .
2: ifail – int64int32nag_int scalar
${\mathrm{ifail}}={\mathbf{0}}$ unless the function detects an error (see [Error Indicators and Warnings]).
## Error Indicators and Warnings
Errors or warnings detected by the function:
ifail = 1${\mathbf{ifail}}=1$
On entry, m < 1${\mathbf{m}}<1$.
ifail = 2${\mathbf{ifail}}=2$
On entry, n < 1${\mathbf{n}}<1$.
ifail = 3${\mathbf{ifail}}=3$
An unexpected error has occurred in an internal call. Check all function calls and array dimensions. Seek expert help.
## Accuracy
Some indication of accuracy can be obtained by performing a subsequent inverse transform and comparing the results with the original sequence (in exact arithmetic they would be identical).
The time taken by nag_sum_fft_real_sine_simple (c06ra) is approximately proportional to nm log(n)$nm\mathrm{log}\left(n\right)$, but also depends on the factors of n$n$. nag_sum_fft_real_sine_simple (c06ra) is fastest if the only prime factors of n$n$ are 2$2$, 3$3$ and 5$5$, and is particularly slow if n$n$ is a large prime, or has large prime factors.
## Example
```function nag_sum_fft_real_sine_simple_example
m = int64(3);
n = int64(6);
x = [0.6772;
0.2983;
0.0644;
0.1138;
0.1181;
0.6037;
0.6751;
0.7255;
0.643;
0.6362;
0.8638;
0.0428;
0.1424;
0.8723;
0.4815;
0;
0;
0;
0;
0;
0;
0;
0;
0];
[xOut, ifail] = nag_sum_fft_real_sine_simple(m, n, x)
```
```
xOut =
1.0014
1.2477
0.8521
0.0062
-0.6599
0.0719
0.0834
0.2570
-0.0561
0.5286
0.0859
-0.4890
0.2514
0.2658
0.2056
0
0
0
0
0
0
0
0
0
ifail =
0
```
```function c06ra_example
m = int64(3);
n = int64(6);
x = [0.6772;
0.2983;
0.0644;
0.1138;
0.1181;
0.6037;
0.6751;
0.7255;
0.643;
0.6362;
0.8638;
0.0428;
0.1424;
0.8723;
0.4815;
0;
0;
0;
0;
0;
0;
0;
0;
0];
[xOut, ifail] = c06ra(m, n, x)
```
```
xOut =
1.0014
1.2477
0.8521
0.0062
-0.6599
0.0719
0.0834
0.2570
-0.0561
0.5286
0.0859
-0.4890
0.2514
0.2658
0.2056
0
0
0
0
0
0
0
0
0
ifail =
0
``` | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 54, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9817381501197815, "perplexity": 3705.475938031839}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257823670.44/warc/CC-MAIN-20160723071023-00161-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/286024/what-is-the-formula-for-1-1-cdot-21-2-cdot-31-3-cdot-4-ldots-1-nn1/286028 | # What is the formula for $1/(1\cdot 2)+1/(2\cdot 3)+1/(3\cdot 4)+\ldots +1/(n(n+1))$
How can I find the formula for the following equation?
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\ldots +\frac{1}{n(n+1)}$$
More importantly, how would you approach finding the formula? I have found that every time, the denominator number seems to go up by n+2, but that's about as far as I have been able to get:
$1/2 + 1/6 + 1/12 + 1/20 + 1/30...$ the denominator increases by $4,6,8,10,12...$ etc.
So how should I approach finding the formula? Thanks!
-
If you simplify your partial sums, you get $\frac12,\frac23,\frac34,\frac45,....$ Does this give you any ideas?
-
Thanks! This was very helpful :) Now I just gotta prove it by induction! – Charles Jan 24 '13 at 18:31
Hint: Use the fact that $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ and find $S_n=\sum_1^n\left(\frac{1}{k}-\frac{1}{k+1}\right)$.
-
Simple and nice (+1) – user 1618033 Jan 24 '13 at 18:59
@Chris'ssister: Thanks a lot for your consideration. – Babak S. Jan 24 '13 at 19:03
Nicely said, Babak(+1)! – amWhy Jan 24 '13 at 19:21
While exploitation of the resulting telescoping series after partial fraction expansion is a very simple way forward, I thought it might be instructive to present another way forward. Here, we write
\begin{align} \sum_{k=1}^N \frac{1}{k(k+1)}&=\sum_{k=1}^\infty \int_0^1y^{k-1}\,dy \int_0^1 x^k\,dx\\\\ &=\int_0^1\int_0^1x\sum_{k=1}^N (xy)^{k-1}\,dx\\\\ &=\int_0^1\int_0^1 x\frac{1-(xy)^N}{1-xy}\,dx\,dy\\\\ &=\int_0^1\int_0^x \frac{1-y^N}{1-y}\,dy\,dx\\\\ &=\int_0^1\int_y^1\frac{1-y^N}{1-y}\,dx\,dy\\\\ &=\int_0^1(1-y^N)\,dy\\\\ &=1-\frac1{N+1} \end{align}
as expected!
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9369996786117554, "perplexity": 835.9503850858949}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049274985.2/warc/CC-MAIN-20160524002114-00067-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/planck-length-at-high-speed.957819/ | # Planck length at high speed
• #1
294
24
Imagine I had a ruler 1 planck length long, I then send it on a rocket close to the speed of light, does its length contract? It seems to my simple intuition, relativity would say yes and Qm would say no. Is there a simple solution to this ?
• #2
kimbyd
Gold Member
1,319
757
Imagine I had a ruler 1 planck length long, I then send it on a rocket close to the speed of light, does its length contract? It seems to my simple intuition, relativity would say yes and Qm would say no. Is there a simple solution to this ?
I don't think making the ruler that short is likely to make a difference. Special relativity and quantum mechanics are perfectly compatible (via Quantum Field Theory). There would be length contraction, regardless of the physical size of the ruler.
• #3
phinds
Gold Member
2021 Award
17,595
9,450
(1) The Plank length is just a made-up human measure of distance and has no effect on physics. You might as well be asking if a meter stick would be subject to length contraction
(2) You WOULD have a problem since as far as is known, you could not construct a ruler that is one Plank unit long.
(3) All of that is moot anyway since your question implies that you think length contraction is something that happens to an object, but it isn't. It is an observation by someone in a frame of reference that is moving relative to the object.
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https://www.physicsforums.com/threads/determine-whether-or-not-something-is-a-subspace.297920/ | # Determine whether or not something is a subspace
1. ### kesun
37
My understanding of the subspace still isn't solid enough, so I want to know what I know so far is at least correct.
By definition, a set of vectors S of Rn is called a subspace of Rn iff for all vectors (I will call them x):
1) (x+y) $$\in$$ S and
2) kx $$\in$$ S.
Also, the solution set of a homogeneous system is always a subspace.
When I encounter problems such as determine whether or not {(x1,x2,x3)|x1x2=0} is a subspace of its corresponding Rn, I would approach this problem as such:
Since the set has only three vectors, then it's in R3, first of all; then I check for 1): Suppose a set of vectors Y {(y1,y2,y3)|y1y2=0}, then (S+Y)=x1x2+y1y2=0+0=0; for 2): kS=(kx1)(kx2)=(k0)(k0)=0. Therefore it is a subspace of R3.
Is my way of solving this problem correct?
2. ### ThirstyDog
34
You are right in that you must check your the two conditions. But you must do it for arbitrary vectors, you didn't seem to do this correctly. Consider
$$(1,0,0), (0,1,0) \in \{(x_{1},x_{2},x_{3})|x_{1}x_{2}=0\}.$$
If you add them you obtain $$(1,1,0)$$ which clearly does not have $$x_{1}x_{2} = 0$$. Also the way you present your vectors doesn't seem standard.
3. ### de_brook
74
That was a nice attempt but your steps were wrong.
{(x1,x2,x3)|x1x2=0} is a subset of [tex]\mathbb{R}^3\[tex] which satisfies the property that x1x2 = 0. but since x1, x2 are in [tex]\mathbb{R}[tex], then either x1=0 or x2=0 or both equal zero.
To proof that it is a subspace, let y and s be vectors in {(x1,x2,x3)|x1x2=0} such that y = (y1,y2,y3) and s = (s1,s2,s3), then
y+s = (y1+s1,y2+s2,y3+s3)
check (y1+s1)(y2+s2) = y1y2+y1s2+s1y2+s1s2= y1s2+s1y2 (since y1y2=0 and s1s2=0
clear in general; y1s2+s1y2 is not equal to zero. this implies that it does not satisfy the property x1x2 = 0. Hence the set {(x1,x2,x3)|x1x2=0} is not a subspace of [tex]\mathbb{R}^3\[tex]. we don't need to proof the second property
4. ### de_brook
74
That was a nice attempt but your steps were wrong.
{(x1,x2,x3)|x1x2=0} is a subset of R3 which satisfies the property that x1x2 = 0. but since x1, x2 are in R3 then either x1=0 or x2=0 or both equal zero.
To proof that it is a subspace, let y and s be vectors in {(x1,x2,x3)|x1x2=0} such that y = (y1,y2,y3) and s = (s1,s2,s3), then
y+s = (y1+s1,y2+s2,y3+s3)
check (y1+s1)(y2+s2) = y1y2+y1s2+s1y2+s1s2= y1s2+s1y2 (since y1y2=0 and s1s2=0
clear in general; y1s2+s1y2 is not equal to zero. this implies that it does not satisfy the property x1x2 = 0. Hence the set {(x1,x2,x3)|x1x2=0} is not a subspace of R3. we don't need to proof the second property | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.955747663974762, "perplexity": 1000.9852666124357}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645356369.70/warc/CC-MAIN-20150827031556-00130-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://brilliant.org/problems/i-find-blackboards-cool-2/ | # I Find Blackboards Cool
Probability Level 4
I have an infinite number of $1's$ written on a blackboard.
Jake chooses $2$ of the integers $p$ and $q$ and replaces them with $\dfrac{p+q}{4}={ k }_{ 1 }$. (he removes $p$ and $q$ and then writes $\dfrac{p+q}{4}$)
Now he repeats the process with the number ${ k }_{ 1 }$ and another integer to achieve ${ k }_{ 2 }$, and repeats again with the number ${ k }_{ 2 }$ and another integer to achieve ${ k }_{ 3 }$
${ k }_{ 2 }=\frac { 1+{ k }_{ 1 } }{ 4 } ,\quad { k }_{ 3 }=\frac { 1+{ k }_{ 2 } }{ 4 }$
Since he is an immortal, he does this again and again until he is left with only a single number...
Given that this number can be expressed as $\frac{a}{b}$, find $a+b$
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Set Loading... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 15, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9403820037841797, "perplexity": 567.0941403248955}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347413901.34/warc/CC-MAIN-20200601005011-20200601035011-00129.warc.gz"} |
https://quizpug.com/can-you-ace-this-math-quiz/ | Can You Ace This Math Quiz?
Do you still remember the 'order of operations' you used in elementary school, or have you gotten too used to the calculator on your phone? Take this quiz now to find out!
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Solve this math problem: 5 + 2 x 8 - 1
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55
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15
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7
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24
16.5
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22
12
-10
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4010
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1,000,010
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2744
80
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230400
4800
22500
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Solve this math problem: 2 + 3 + 4 + (5 x 2) ^ 2
784
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361
An active brain is a healthy one! Maybe it's time to stop using your calculator too much and practice a bit of mental math?
C+, Math Class Daydreamer
globalpost.com
Right!
Good job! Sure you missed some, but a few of these were tricky. Rest assured you still have your math smarts.
B+, Math Star
sheknows.com
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Amazing! Either you're a math teacher, or you still have some seriously good math skills. Congratulations on a great quiz score.
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Right!
1
Solve this math problem: 5 + 2 x 8 - 1
20
55
49
2
Solve this math problem: 8 / 2 + 5 x 2 -3
15
-9
11
3
Solve this math problem: (4+5)(7-9)
18
-18
92
4
Solve this math problem: (2+2) x 5 - 1 x 7
133
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5
Solve this math problem: 2 x (10 x 4) + 1 x 7
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7
Solve this math problem: 2 x 3 + [9 + (1 x 3)] /2
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24
16.5
8
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22
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230400
4800
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784
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http://physics.stackexchange.com/questions/96057/how-two-photons-interfere-in-a-double-slit-experiment | # How two photons interfere in a double slit experiment
I have a very basic question about photons in double-slit experiment. I am not good at math, and have some Quantum mechanics knowledge. Math free explanation would be very good to me.
When we do double slit experiment with single photon at a time, it will be detected at some location of screen. Over the time period the accumulation of detected locations show interference pattern. The reason for the interference pattern is that photon's location in not well defined, the presence of photon in particular place can be determined by prbability and this probability is presented as a wave. that probability wave is splitted in to two while passing the two slits, collide with each other and causes interference pattern.
Question 1: Is this reason for single photon interference correct?
Question 2: How is the probability wave behaving while two-photons passing double-slit at a time? Are 4 probability waves (2 for each) interfering?
-
'location [...] is a probability' ???? – Danu Jan 31 '14 at 2:57
I have improved my question. please check now. – Vijayan Jan 31 '14 at 5:25
Question 1: Is this reason for single photon interference correct?
The most intuitive way of looking at interference that I have encountered is Feynmann's Path Integral Formulation. Loosely speaking, if you have a photon (or anything, really) in location A and want to work out its chance of moving to B, you imagine it taking every possible path between the two at the same time. All of these paths interfere with each other resulting in some final amplitude from which you can extract the probability of the particle moving from from A to B. This sounds kind of like what you were trying to say when you said that the 'probability wave is split in to two'. All particles do this when moving, not just photons, however in the classical (i.e. large) limit, you can show mathematically that only one particular path will contribute to the particle's motion, which is why when you throw a tennis ball across the room you don't see it taking every possible path at once.
Question 2: How is the probability wave behaving while two-photons passing double-slit at a time? Are 4 probability waves (2 for each) interfering?
For a double slit experiment, the interference pattern does not change with light intensity. I believe you can think of each photon as behaving independently in this case.
-
Fine. somehow I got answer. let me confirm it. two photons are travelling independent path to each other, and not colliding with each other. – Vijayan Jan 31 '14 at 16:12
Yes, we generally think of photons as interfering rather than colliding. It is like if two waves of water 'impact', they just travel through each other. – Ruvi Lecamwasam Jan 31 '14 at 23:48
The reason for the interference pattern is that photon's location in not well defined, the presence of photon in particular place can be determined by probability and this probability is presented as a wave.
Probability distributions have the same meaning classically and quantum mechanically.There is a distribution for life expectancy for example, giving the probability of dying if one is a certain age. This can be plotted in a histogram . The only information it gives is that one event, a person at 90, has a probability of dying within the year of 0.17. The curve was measured from a large sample of population.
In a similar way, the probability curve of a photon impinging on the two slits that it will hit at a certain (x,y) on the screen can be measured by a large sample of photons and one will have the two dimensional histogram of that probability, and then can give a probability for the next photon. Fortunately though mathematical solutions help us to get the probability curve, and those solutions contain sines and cosines which are solutions of wave equations in general, like energy waves and sound waves which have been studied for centuries. The mathematics of quantum mechanics has similar differential equations, but the solutions are identified differently.
that probability wave is splitted in to two while passing the two slits, collide with each other and causes interference pattern.
Question 1: Is this reason for single photon interference correct?
The above preamble is to stress that there is no splitting , there is just a computed/observed probability pattern. There is nothing to collide. It is just a representation that has sine and cosine solutions and a single event will appear according to the probability so that it adds up to the interference pattern. Similar that a single death seems random unless a large number of deaths are plotted and probabilities can be gauged. So no, it is not correct. No splits and interference
Question 2: How is the probability wave behaving while two-photons passing double-slit at a time? Are 4 probability waves (2 for each) interfering?
No, even if you manage to pass two photons at a time (photons are very fast point particles) they do not interfere with each other. Quantum mechanical calculations that fit experiments much more sophisticated than the two slits show that photons interfere with each other with tiny probabilities, calculated by higher order feynman diagrams.
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http://mymathforum.com/number-theory/17535-help-needed-ufd.html | My Math Forum Help needed - UFD
Number Theory Number Theory Math Forum
February 16th, 2011, 07:23 AM #1 Senior Member Joined: Dec 2007 Posts: 687 Thanks: 47 Help needed - UFD Ok, so I came across this problem and I think it is too difficult, so I'm asking for you guys help! the problem is to find integer solutions to $n^3+n-1=x^2$ first of all three solutions are n =1, 2, 13, I don't know if there is any other, probably not lets factorize the whole thing like this: $n(n^2+1)=x^2+1=(x-\sqrt{-1})(x+\sqrt{-1})=n(n-\sqrt{-1})(n+\sqrt{-1})$ $\mathbb{Q}[\sqrt{-1}]$ is a UFD (http://en.wikipedia.org/wiki/Stark%E2%8 ... er_theorem), so we have $gcd(n,\ n^2+1)=gcd(x-\sqrt{-1},\ x+\sqrt{-1})=1$ from here I'm not sure how to proceed exactly, but I think it is clear that for n>1 one of $(x-\sqrt{-1})(x+\sqrt{-1})$ must be reductible, but it gives too many cases to check another possibility is to write $(\sqrt{n(n^2+1)}-1)(\sqrt{n(n^2+1)}+1)=x^2$ Do the same the same argument of UFD holds? If yes, I think $gcd(\sqrt{n(n^2+1)}-1,\ \sqrt{n(n^2+1)}+1)=1$, yielding: $\sqrt{n(n^2+1)}-1=a^2$ and $\sqrt{n(n^2+1)}+1=b^2$ where $a^2b^2=x^2$ I'm not sure how to proceed with the second factorization since $\sqrt{n(n^2+1)}$ is clearly irrational, do you guys see any mistake? thanks in advance!
February 16th, 2011, 10:32 AM #2
Senior Member
Joined: Nov 2010
From: Berkeley, CA
Posts: 174
Thanks: 35
Math Focus: Elementary Number Theory, Algebraic NT, Analytic NT
Re: Help needed - UFD
Quote:
Originally Posted by al-mahed $n(n^2+1)=x^2+1=(x-\sqrt{-1})(x+\sqrt{-1})=n(n-\sqrt{-1})(n+\sqrt{-1})$ $\mathbb{Q}[\sqrt{-1}]$ is a UFD (http://en.wikipedia.org/wiki/Stark%E2%8 ... er_theorem), so we have $gcd(n,\ n^2+1)=gcd(x-\sqrt{-1},\ x+\sqrt{-1})=1$
I'm afraid that I don't understand this conclusion. Are you saying that if ab = cd in a UFD, then gcd(a,b) = gcd(c,d)? That's not true in the rational integers: (6)(15) = (5)(1, but gcd(6,15) = 3 and gcd(5,1 = 1.
February 16th, 2011, 10:45 AM #3
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Re: Help needed - UFD
Quote:
Originally Posted by al-mahed $gcd(n,\ n^2+1)=gcd(x-\sqrt{-1},\ x+\sqrt{-1})=1$
This isn't true in generality. $\gcd(n,n^2+1)=1$ in the Gaussian integers just like in the integers, but $\gcd(x-i,x+i)$ could be any divisor of 2 depending on the value of x.
LaTeX note: You don't need \$ inside the tags, they're in math mode already. Also, the code for gcd is \gcd, not gcd.
February 16th, 2011, 01:50 PM #4 Senior Member Joined: Dec 2007 Posts: 687 Thanks: 47 Re: Help needed - UFD Hmmmm, but by suposing there is a prime $\pi$ dividing both we get a contradiction, right? we get $\pi | 2x,\ \pi | 2\sqrt{-1}$... I see, there is one more condition, I need to take the Norms and see what divides what and if it is contradictory, I supose.
February 16th, 2011, 02:04 PM #5
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Re: Help needed - UFD
Quote:
Originally Posted by al-mahed Hmmmm, but by suposing there is a prime $\pi$ dividing both we get a contradiction, right?
No. If x = i, then gcd(i - i, i + i) = 2 (up to associates, of course; you could just as easily say 2i or -2). Can you find one that gives 1 - i? I can't immediately see one (though I haven't looked hard) nor can I immediately prove it does not occur.
February 16th, 2011, 02:31 PM #6
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Joined: Dec 2007
Posts: 687
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Re: Help needed - UFD
Quote:
Originally Posted by CRGreathouse
Quote:
Originally Posted by al-mahed Hmmmm, but by suposing there is a prime $\pi$ dividing both we get a contradiction, right?
No. If x = i, then gcd(i - i, i + i) = 2 (up to associates, of course; you could just as easily say 2i or -2). Can you find one that gives 1 - i? I can't immediately see one (though I haven't looked hard) nor can I immediately prove it does not occur.
but we need to consider x integer
to the second question, you mean gcd=1-i? if yes $- i*( 1 + i )= - i + 1$
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Contact - Home - Forums - Cryptocurrency Forum - Top | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 22, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8542514443397522, "perplexity": 2023.1345424341996}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221217901.91/warc/CC-MAIN-20180820234831-20180821014831-00660.warc.gz"} |
https://mathoverflow.net/questions/52960/which-almost-complex-manifolds-admit-a-complex-structure | # Which almost complex manifolds admit a complex structure?
I was reading Yau's list of problems in geometry, and one of them is to prove that any almost complex manifold of complex dimension $n \geq 3$ admits a complex structure. It's been some time since Yau's list was published, so what is the status of this problem today?
Obviously it isn't hasn't been shown to be true, because we're still looking for complex structures on the six-sphere, but I have a vague feeling of having read that this doesn't hold. So do we know any counterexamples to this question? If not, then is anyone working on this problem?
Also, Yau only stated the problem for manifolds of dimension $n \geq 3$. We know this is true in dimension one, because there we have isothermal coordinates which give complex structures, but why didn't Yau mention almost complex surfaces? Do we know this holds there, or are there counterexamples in dimension 2?
-
Added as a comment (because you've already got two great answers). In dim 4 you can get every finitely presented group as the fundamental group of an almost complex 4-mfd (symplectic even, by a thm of Gompf, but this is harder). On the other hand, the classification of complex surfaces tells you that the possible topology of complex surfaces is much more constrained. What changes drastically in dim 6 is that any finitely presented group arises as pi_1 of a closed complex threefold. This was first proved by Taubes as a corollary of an existence thm for self-dual metrics. – Joel Fine Feb 11 '11 at 2:25
I should add I think that this is a fascinating problem, but have not the slightest idea how to start trying to attack it. It seems to me like a question waiting for a "big idea". I would guess the total failure to understand the six-sphere is what puts most people off, but perhaps that is overly negative. Sometimes the less topology you have the harder things are. E.g, we know a lot about differential topology of various 4-manifolds, but still nothing about simply connected ones with b_2=0... – Joel Fine Feb 11 '11 at 2:31
In complex dimension 3 or more it is still an open conjecture (which was re-stated Yau a couple of years ago in his UCLA lectures). There is not a single known example of an almost complex manifold of dimension $\geq$ 3 not admitting a complex structure.
In dimension 2 it is easy, of course, because the non-Kähler complex surfaces are understood much better than Kähler ones: every non-Kähler surface with $b_1 >1$ is diffeomorphic to a blow-up of a locally trivial elliptic fibration over a curve. Hence any 4-dimensional compact almost complex manifold with odd $b_1 >1$ and a fundamental group not virtually isomorphic (*) to a cross-product of a fundamental group of a curve and $\mathbb{Z}$, cannot be a complex surface.
(*) Here "virtually isomorphic" means "isomorphic up to a finite index subgroup".
-
There are actually counterexamples in real dimension $4$.
The first examples of compact almost complex $4$-manifolds admitting no complex structure were produced by Van de Ven in his paper "On the Chern numbers of some complex and almost-complex manifolds".
In fact, he obtained restrictions on the Chern numbers of an algebraic surface and constructed some almost complex $4$-manifolds violating them, hence showing that no almost complex structure in these examples could be integrable.
Later on, Brotherton constructed some counterexamples with trivial tangent bundle, see the article "Some parallelizable 4-manifolds not admitting a complex structure".
-
Excellent, thank you Francesco. But we have no similar results for higher dimensions, i.e. all the known restrictions are the same for complex and almost complex manifolds? – Gunnar Þór Magnússon Jan 24 '11 at 16:23
I'm not aware of any results in higher dimensions. But I must say that I'm not a specialist in the field – Francesco Polizzi Jan 24 '11 at 16:56
About the 6-dimensional sphere, there exists a paper in arxiv, which states that $S^{6}$ is a complex manifold. (See also the link of Mathoverflow) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8645995259284973, "perplexity": 362.66563581342655}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783397111.67/warc/CC-MAIN-20160624154957-00137-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://socratic.org/questions/the-position-of-an-object-moving-along-a-line-is-given-by-p-t-cos-t-pi-3-1-what--1 | Physics
Topics
# The position of an object moving along a line is given by p(t) = cos(t- pi /3) +1 . What is the speed of the object at t = (2pi) /4 ?
Jul 12, 2016
$v \left(\frac{2 \pi}{4}\right) = - \frac{1}{2}$
#### Explanation:
Since the equation given for the position is known, we can determine an equation for the velocity of the object by differentiating the given equation:
$v \left(t\right) = \frac{d}{\mathrm{dt}} p \left(t\right) = - \sin \left(t - \frac{\pi}{3}\right)$
plugging in the point at which we want to know speed:
$v \left(\frac{2 \pi}{4}\right) = - \sin \left(\frac{2 \pi}{4} - \frac{\pi}{3}\right) = - \sin \left(\frac{\pi}{6}\right) = - \frac{1}{2}$
Technically, it might be stated that the speed of the object is, in fact, $\frac{1}{2}$, since speed is a directionless magnitude, but I have chosen to leave the sign.
##### Impact of this question
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https://www.physicsforums.com/threads/measure-visocity-of-water-by-poises.103504/ | Measure visocity of water by poises
1. Dec 10, 2005
MathematicalPhysicist
i read that you measure visocity of water by poises (sounds like poison ), how does it equate with the quantities?
i.e, like [N]=[kg*m^2/sec]
2. Dec 10, 2005
FredGarvin
For most applications, the standard is in centipoise.
1 cp = .001 Pa*sec or 1.45 x10^-7 Lbf*sec/ft^2 for us non SI'ers.
BTW...a newton is $$\frac{kg*m}{sec^2}$$
3. Dec 14, 2005
MathematicalPhysicist
Pa*sec is pascal times seconds, right?
so visocity of liquids is actually the atmospheric presure against time, or in other words how do you explain to a layperson what visocity of liquids is?
4. Dec 14, 2005
FredGarvin
In the most basic sense: Viscosity = resistance to flow. If you want to go a bit farther, viscosity is a measure of the ability of a fluid to resist shear stress.
You have to be careful in interpreting the units for viscosity. The pressure is not atmospheric pressure. It is shear stress. The mathematical relation for viscosity is:
$$\tau = \mu \frac{du}{dy}$$
Where:
$$\tau$$ = shear stress
$$\mu$$ = viscosity
$$\frac{du}{dy}$$ = velocity distribution
5. Dec 14, 2005
Staff: Mentor
http://en.wikipedia.org/wiki/Viscosity
Wikipedia
Viscosity is commonly perceived as "thickness", or resistance to pouring, however it really describes a fluid's internal resistance to flow and may be thought of as a measure of fluid friction. Adapted from Wikipedia.
Solids exhibit viscosity when subject to very high compressive or tensile stresses, e.g. extrusion or other forms of cold or hot working.
The pressure is not necessarily atmospheric pressure, but applied pressure, and this more a case that the term is in units of pressure. This expression refers to kinematic viscosity $\nu$ vs dynamic viscosity $\mu$, which have the following relationship
$\nu$ = $\mu$/$\rho$
Similar Discussions: Measure visocity of water by poises | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8872700333595276, "perplexity": 2257.701244331896}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218188824.36/warc/CC-MAIN-20170322212948-00262-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://mathoverflow.net/questions/323128/morphism-of-sites-and-abelian-sheaf-cohomology/323216 | # Morphism of sites and abelian sheaf cohomology
Let $$f : \mathcal{C}\to\mathcal{D}$$ be a morphism of sites (see the Stacks Project) with induced morphism of topoi
$$(f^{-1}, f_*) : Sh(\mathcal{D})\to Sh(\mathcal{C}).$$
By assumption, $$f^{-1}$$ is an exact functor.
How do we define the induced map on abelian sheaf cohomology
$$H^p(\mathcal{D}, F)\to H^p(\mathcal{C}, f^{-1}F)\ ?$$
On global sections, we have a map $$\Gamma(\mathcal{D}, F)\to \Gamma(\mathcal{C}, f^{-1}F)$$ because $$f^{-1}$$ is exact and then preserves final objects in the topoi.
If I have an injective resolution $$F \to J^{\bullet}$$ in $$Ab(\mathcal{D})$$, then by exactness of $$f^{-1}$$ $$f^{-1}F\to f^{-1}J^{\bullet}$$ is still a resolution.
I’m tempted to consider the induced maps on global sections giving a map of complexes of abelian groups
$$\Gamma(\mathcal{D},J^{\bullet})\to \Gamma(\mathcal{C},f^{-1}J^{\bullet})$$
Cohomology of the left complex is $$H^*(\mathcal{D},F)$$ because each $$J^p$$ is $$\Gamma(\mathcal{D},\cdot)$$-acyclic, but it’s not clear to me that the same is true for $$f^{-1}J^p$$.
• Is $$f^{-1}J^p$$ a $$\Gamma(\mathcal{C},\cdot)$$-acyclic abelian sheaf for every $$p$$?
• If not, then how else do we define the map $$H^p(\mathcal{D}, F)\to H^p(\mathcal{C}, f^{-1}F)\ ?$$
• You should add your answer as an official answer in the box, and then after the wait period, accept it. – David Roberts Feb 13 at 12:14
Following the suggestion in the comment, I post the answer I found myself, in case it’s useful to anyone sometime.
Since abelian sheaf cohomology on a site is a universal $$\delta$$-functor, having a map $$\Gamma(\mathcal{D},F)\to \Gamma(\mathcal{C}, f^{-1}F)$$ yields a unique map in all cohomological degrees by the universal property of universal $$\delta$$-functors, and that’s it. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9925491213798523, "perplexity": 307.0794467263309}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912201455.20/warc/CC-MAIN-20190318152343-20190318174343-00034.warc.gz"} |
https://www.physicsforums.com/threads/a-very-quick-6-second-question-about-boolean-algebra.155352/ | # A very quick 6 second question about boolean algebra.
1. Feb 8, 2007
### l46kok
I was doing a problem, and the distributive problem hit my head, as I kind of blanked out on this one.
Can B'D' + BD be simplified into a smaller equation? Because I'm thinking it's impossible to do so.
(You may be able to apply Demorgan's theorem, but that doesn't really simplify the equation).
[+ is or, * is and, ' is inversion.]
Last edited: Feb 8, 2007
2. Feb 8, 2007
### ranger
Put it in a k-map and you will see why it cant be minimized.
Last edited: Feb 8, 2007
3. Feb 10, 2007
### ecurbian
IMHO, putting it in a k-map shows visually THAT it cannot be minimized, and then only as a sum of products expression. It does not show "why". Also, if simplifying (x^2+2x+1) to (x+1)^2 is acceptable, then AB+A'B' = (A+B')(A'+B) is also simplifying in some sense. I think that "why" in this case is rather
similar to "why" is 7 a prime number. Since the proof and the question are about the same size, there is no "why".
I realise that SOP is fairly standard form, and that that was probably what the question was about but strictly speaking the question needed clarification on what sort of expressions were being considered, and what was considered simple. For understanding of the question, it should be noted that the formula is not-xor, and simple xor expressions (such as parity) are often difficult to express simply using sop format.
Last edited: Feb 10, 2007
4. Feb 10, 2007
### ranger
You do realize that AB+A'B' works out better for design reasons rather than (A+B')(A'+B) right? The latter requires two extra gates (inverters). So for this case, it is in its simplest form.
Last edited: Feb 10, 2007
5. Feb 11, 2007
### ecurbian
"You do realise" that each expression when DIRECTLY implemented requires exactly the same number of gates (A.B)+(A'.B') requires two 'and's, two
inverters, and one 'or', while (A+B').(A'+B) requires two 'or's, two inverters and one 'and'. At least if we are speaking of direct and-or-not gate
implementation.
If you are speaking of some other set, such as nand-only, then neither expression is directly implementable. If we have xor, then (A xor B)'
is simpler than either of the above expressions.
In a "design" situation you often have the inverse of a line available anyway, without having to ask for it, and optimisation of the system as a whole can lead to many decisions that seem strange locally. Also, often a solution involving more gates turns out to be better because it happens to match the particular gates left over in the PGA from building the other parts.
6. Feb 11, 2007
### ranger
Yup, youre right. For some reason when I was reading you're reply, I was thinking of (AB)' instead of what you wrote - (A'B').
Last edited: Feb 11, 2007 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8501285314559937, "perplexity": 1597.786288701471}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583771929.47/warc/CC-MAIN-20190121090642-20190121112642-00133.warc.gz"} |
https://www.physicsforums.com/threads/unable-derive-mass-as-resnick-halliday-describe-it.661280/ | # Unable derive mass as Resnick \& Halliday describe it
1. Dec 28, 2012
### DavidCantwell
I am reading Resnick and Halliday's "Basic Concepts in Relativity" (ISBN: 0023993456) and have come to an impass about deriving equation 3-4a
$$m = \gamma m_0$$
in the text. The authors wrote that the equation can be derived from the following equations
\begin{array}{cr}
M = m^{\prime} + m_0 & \\
Mu = m^{\prime} u^{\prime} & \\
u^{\prime} = 2u / (1+u^2/c^2)& \\
\end{array}
based on the attached figure. The derivation is left as an exercise (problem 18 pp 133). The problem is to derive m in the first equation above from the three equations below it, but I only keep coming up with
$$\frac{m^{\prime}}{m_0} = \frac{1+u^2/c^2}{1-u^2/c^2}$$
which is an intermediate equation in Tolman's text that leads to the desired equation after substituting a transformation equation for γ into the rhs ratio (e.g. see the sometimes "colorful" thread at https://www.physicsforums.com/forumdisplay.php?f=70\&order=desc\&page=23). But this approach seems to deviate far afield of the three equations given, and it leaves me feeling like I am doing something wrong. What I mean is that I cannot complete the derivation without including the transformation for γ from Tolman, and it seems to me that the authors think I should. Can anyone tell me what I am doing wrong? Hints, recommendations, anything? I am stuck. Apparently the book is out of print, so I can provide more info if need be. Thnx for any help.
#### Attached Files:
• ###### ResnickHallidayFig3-2.jpg
File size:
10.9 KB
Views:
69
Last edited: Dec 28, 2012
2. Dec 28, 2012
### Staff: Mentor
[Edit: Deleted, I was misinterpreting the m's as rest masses instead of relativistic masses. Will follow up with a corrected re-do.]
Last edited: Dec 28, 2012
3. Dec 28, 2012
### Staff: Mentor
Okay, after fixing my interpretation of the m's, here's a re-do of my previous (now deleted) post:
I don't agree with the first two of the three equations you wrote down that you say you're supposed to derive the result from. When I write down energy and momentum conservation in the two frames, here's what I get (I'm using units where c = 1, and the first and second values given for E and p in each frame are before and after the collision):
Unprimed Frame
$$E = 2 m = M$$
$$p = 0 = 0$$
Primed Frame
$$E' = m' + m_0 = \gamma M$$
$$p' = m' u' = \gamma M u$$
where $\gamma = 1 / \sqrt{1 - u^2}$ has to be included on the RHS of the primed frame formulas because the single object with rest mass $M$ is *moving* in that frame.
If you start with the above equations (and your third equation for $u'$) and eliminate $m'$ and $u'$, you should be able to easily derive $m = \gamma m_0$. I would suggest checking the textbook to make sure you haven't left out a gamma factor in the two equations involving $M$.
4. Dec 28, 2012
### bcrowell
Staff Emeritus
Halliday and Resnick's treatment of relativity is truly awful. You'd do better to read almost any other treatment.
5. Dec 28, 2012
### Staff: Mentor
Awful would be bad enough, but if the formulas in the OP actually appear in the book, it would seem that it goes beyond awful and into outright wrong.
6. Dec 28, 2012
### TSny
As PeterDonis noted, the figure from the text has a misprint: the M in the upper right of the figure should have a supscript o because the mass is at rest: $M_o$.
You can use the figure to derive the explicit expression for $\gamma$ if you assume that relativistic mass and rest mass are related by $m = \gamma m_o$ where the explicit form of $\gamma$ is yet to be determined.
The upper part of the figure gives:
$2m = M_o$ $\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ (1)
And the lower part of the figure gives:
$m’ + m_o = M$ $\;\;\;\;\;\;\;\;$ (2)
$m'u' = Mu$ $\;\;\;\;\;\;\;\;\;\;\;$ (3)
Using $u’ = 2u/(1+u^2/c^2)$ and eliminating $m'$ between (2) and (3) yields
$(1-u^2/c^2)M = 2m_o\;\;\;\;$ or $\;\;\;\;(1-u^2/c^2)\gamma M_0 = 2m_o$
Using (1) to replace $M_o$ by $2m = 2\gamma m_o$ yields
$(1-u^2/c^2)\gamma ^2 2m_o = 2m_o$ .
Thus, get the desired result: $\gamma = 1/\sqrt{1-u^2/c^2}$.
[I believe this is essentially the derivation that PeterDonis outlined]
7. Dec 29, 2012
### Staff: Mentor
I did it the other way, deriving $m = \gamma m_0$ from the other givens (including the formula for $\gamma$). Also, what I wrote as $M$ is what you are calling $M_0$, the rest mass of the single object after the collision.
8. Dec 29, 2012
### stevendaryl
Staff Emeritus
I don't quite like this derivation because it uses a velocity-dependent mass, but otherwise assumes that momentum works like Newtonian momentum. It assumes that this relativistic mass is conserved. Why? I think that's a little strange. The way I would do it, which is essentially the same derivation, but with slightly different assumptions, is this:
In Newtonian, physics, the following two quantities are conserved in any collision:
1. $M_{total} = \sum m$
2. $\vec{P}_{total} = \sum m \vec{v}$
Then for SR, we generalize these by allowing a more complex dependency on velocity:
1. $M_{total} = \sum F(v) m$
2. $\vec{P}_{total} = \sum F(v) m \vec{v}$
where $F(v)$ is an unknown scalar function of velocity (assumed to be dependent only on the magnitude of velocity, rather than its direction).
That's plausible, but I'm not sure what motivates using the same velocity-dependence $F(v)$ in generalizing both total mass and total momentum.
9. Dec 29, 2012
### Staff: Mentor
Not when you recognize that "relativistic mass" is just another name for "total energy".
Because the invariant mass (or "rest mass") of an object has to be, well, invariant. This leads to a relationship between total energy and total momentum that requires them both to have the same velocity dependence $F(v)$ (which is usually referred to as $\gamma(v)$).
10. Dec 29, 2012
### DavidCantwell
Thnx All
PeterDonis: Your comment that you don't agree with the given equations caught my attention because it crossed my mind if there was a typo among them. But I have double checked and triple checked the text (first thing I do when I am stumped) and the figure and equations are as they appear in the text.
bcrowell: I cut my teeth on the H & R "Fundamentals of Physics" text and liked another text they wrote (a sophmore/junior level QM text), but this relativity text really is truly awful; I came to it seeking a more elegeant derivation than that in Tolman's text.
TSny: Your statement "... where the explicit form of γ is yet to be determined." was my "aha" moment as that is how I derived γ for the LT's earlier in my studies.
Yes then, the typo is certainly in the figure in which M is ambiguous and left as such in the narrative (and caption). Thnx loads for your help folks; I understand this now.
BTW, where can I learn more about editing posts? I was not able to do inline equations like TSny did. Thnx again.
11. Dec 29, 2012
### stevendaryl
Staff Emeritus
I know, I'm trying to get in the spirit of the Halliday and Resnick derivation.
IF you assume that energy and momentum are part of a 4-vector, then everything else follows. But I thought the point of this derivation was to motivate the relativistic forms of energy and momentum before the concept of 4-vectors was introduced.
An alternative derivation along the same lines first derives the relativistic form for the momentum, and then uses that result to derive the relativistic form for the energy. The Halliday and Resnick derivation seems more compact, because it does it all in one fell swoop, but it seems a little unmotivated to me.
12. Dec 29, 2012
### Staff: Mentor
I'm not sure that's healthy, but ok.
Can you post or link to an example?
13. Dec 29, 2012
### stevendaryl
Staff Emeritus
I left out the derivation, because it's very long-winded, but since you asked...
Let's assume that momentum has the form:
$\vec{p} = F(v) m \vec{v}$
where $F(v)$ depends on the magnitude of $v$, not its direction.
To have the right nonrelativistic limit, it must be that $F(0) = 1$
Now, consider a collision in which in frame $F$ a very light particle of mass $m$ is moving relativistically with a horizontal component of velocity of $v$ and a very small vertical component of velocity $u$. The magnitude of the velocity is $\sqrt{v^2 + u^2} \approx v$.
This lighter particle "bounces" off a much more massive particle of mass $M$ initially at rest in frame $F$ imparting a small vertical component of velocity $U$ to the larger mass. The "Before" and "After" pictures are shown in Figure 1:
I'm going to leave off the detailed argument, but approximately what will happen is that for the light particle, its vertical component of velocity will just flip signs, and its horizontal component of velocity will remain unchanged.
Conservation of momentum in frame $F$ implies:
$- F(v) m u = +F(v) m u - MU$
So
$MU = (2 m u)F(v)$
where I used the nonrelativistic form of momentum for the massive particle. Now, look at the same collision in a frame $F'$ in which the lighter particle initially has no horizontal component of its velocity. The collision in this frame looks like Figure 2:
I'm going to skip the derivation, but the Lorentz transformations can be used to show that in frame $F'$ the vertical components of velocity for the two particles are related to those in frame $F$ by:
$u' = \gamma u$
$U' = U/\gamma$
where $\gamma = 1/\sqrt{1-v^2/c^2}$ as usual.
Conservation of vertical momentum in frame $F'$ gives us:
$-m u' = + m u' - F(v) M U'$
So
$2m u' = F(v) M U'$
Rewriting in terms of $u$ and $U$ gives:
$2m u \gamma = F(v) M U/\gamma$
Substituting the previously derived value for $MU$ gives:
$2m u \gamma = F(v)^2 (2mu)/\gamma$
So
$F(v) = \gamma$
So we have the relativistic form of momentum:
$\vec{p} = \gamma m \vec{v}$
The derivation of the form of energy then proceeds much like the Halliday and Resnick case: Imagine in frame $F$ you have two particles of equal mass $m$ and opposite velocities $v$, which collide head-on to produce a mass $M$ at rest.
Now, switch to a frame $F'$ in which one of the particles is initially at rest. Then the other particle will have (using Lorentz transforms) velocity
$v' = 2v/(1+v^2/c^2)$
which gives a "gamma" factor of
$\gamma' = \dfrac{1+(v/c)^2}{1-(v/c)^2}$
In this frame, after the collision, the new composite mass is not at rest, but is moving at speed v. Conservation of momentum in this frame gives:
$\gamma' m v' = \gamma M v$
Substituting for $\gamma'$ and $v'$ gives:
$\dfrac{1+(v/c)^2}{1-(v/c)^2} m \dfrac{2v}{1+(v/c)^2}= \gamma M v$
which simplifies to
$\dfrac{1}{1-(v/c)^2} 2 m = \gamma M$
So, using $\dfrac{1}{1-(v/c)^2} = \gamma^2$ gives us:
$M = 2 \gamma m$
Now, assume that relativistic energy has the form:
$E = G(v) m$
So conservation of energy in the center-of-mass frame gives:
$2 G(v) m = G(0) M$
Substituting for $M$ gives:
$2 G(v) m = G(0) 2 \gamma m$
So
$G(v) = G(0) \gamma$
To find out the constant $G(0)$, we expand the expression for energy for low velocity for comparison with the nonrelativistic limit:
$E = G(0) \gamma m = G(0) m (1 + 1/2 (v/c)^2 + ...)$
If we identify $G(0) m (1/2 (v/c)^2)$ as the nonrelativistic kinetic energy, $1/2 m v^2$ then we find that $G(0) = c^2$. So the full expression for relativistic energy is:
$E = \gamma mc^2$
14. Dec 29, 2012
### Staff: Mentor
Thanks! I hadn't seen it done this way before. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9539664387702942, "perplexity": 597.3808810027614}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257649683.30/warc/CC-MAIN-20180324034649-20180324054649-00572.warc.gz"} |
https://math.stackexchange.com/questions/356715/2-mathbbn-is-uncountable-power-set-of-natural-numbers | # $2^{\mathbb{N}}$ is uncountable -- Power Set of Natural Numbers
Proof: Assume it is countable. Then we can arrange the sets in order (an enumeration) $A_1, A_2, A_3, ...$. Now construct the set $B = \{i \in \mathbb{N} \ | \ i \notin A_i \}$. Then $B=A_j$ for some $j$, a contradiction.
This was the proof given in class and feels off. Can anyone elaborate or tell me it's wrong?
• This has nothing to do with measure theory, so I removed the tag.
– Pedro
Apr 10, 2013 at 4:22
• How do we know that B isn't empty as there is no such i? That would be the part that I'd want elaborated in a sense. Apr 10, 2013 at 4:23
• @JBKing It can't be empty because the empty set is a subset of $2^{\mathbb{N}}$. If this is countable, the empty set is then $A_j$ for some $j$. Certainly $j \not\in A_j$, so $j\in B$. The empty set is not different to any other set in this proof. Apr 10, 2013 at 4:26
It’s the standard argument; there’s nothing wrong with it, apart from the fact that there is no need to cast it as a proof by contradiction. The real idea is simply that if you take any list of subsets of $\Bbb N$ ordered by the positive integers $-$ i.e., a list of the form $A_1,A_2,A_3,\dots$ $-$ there is guaranteed to be a subset of $\Bbb N$ not in your list.
Suppose that we’ve proved that statement: then it follows immediately that it is impossible to list all subsets of $\Bbb N$ in a $1$-$2$-$3$ enumeration of this kind and hence that $\wp(\Bbb N)$ must be uncountable: there is no bijection from $\Bbb N$ to $\wp(\Bbb N)$, because there isn’t even a map from $\Bbb N$ onto $\wp(\Bbb N)$.
The proof of the statement is exactly what you were shown. I’m going to construct a set $B\subseteq\Bbb N$ by going through the positive integers one at a time and deciding whether to put it into $B$ or leave it out. If $k\notin A_k$, I’ll put $k$ into $B$; otherwise, if $k\in A_k$, I’ll leave $k$ out of $B$. In this way I make sure that the sets $B$ and $A_k$ disagree about $k$: exactly one of them contains $k$. In other words, one of them contains $k$, and the other does not. I don’t have to know which is which to recognize that this means that $B\ne A_k$: $B$ and $A_k$ definitely don’t have exactly the same elements, precisely because they disagree at $k$.
The argument that you were given just described this same set $B$ more concisely: $$B=\{k\in\Bbb N:k\notin A_k\}\;.\tag{1}$$
$(1)$ just says very briefly in symbols what I said at much greater length in words when I wrote this:
I’m going to construct a set $B\subseteq\Bbb N$ by going through the positive integers one at a time and deciding whether to put it into $B$ or leave it out. If $k\notin A_k$, I’ll put $k$ into $B$; otherwise, if $k\in A_k$, I’ll leave $k$ out of $B$.
• Thanks, Professor Scott. This makes sense-- I guess there was just a lot of information packaged in that set construction. I do have a question that you sound like you may be able to answer. Are these problems addressed the same way with ZFC set theory as Cantor's original diagonalization arguments? This is a diagonal argument, isn't it? Isn't the problem with naive set theory that there are inherit contradictions when we only require sets to be elements such that a statement is true ($k \notin A_k$)? So even if it works in this particular case, it might be normal that is seems weird? Apr 10, 2013 at 19:28
• @Questioneer: You’re welcome. Yes, this a classic diagonalization argument. There is no problem here: the definition of $B$ is an instance of restricted comprehension, using some property to pick out a subset of something ($\Bbb N$) already known to be a set. Russell’s paradox arises from the use of unrestricted comprehension. It’s not unusual, I think, to find the argument a bit strange at first, but it’s not at all problematic. ‘[T]his particular case’ is normal, not unusual. Apr 10, 2013 at 19:38
• I am a bit confused here. Can't B equal to some other $A_j$ ? Do you mean $B$ is a compliment of $A_k$ where the universe is natural numbers? Apr 16, 2014 at 12:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8905672430992126, "perplexity": 161.3436281244983}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570879.1/warc/CC-MAIN-20220808213349-20220809003349-00482.warc.gz"} |
https://www.innovayz.com/article/article-detail?p_code=213&p_detail=moment-of-inertia-of-continuous-bodies-uniform-rod-along-with-the-axis-passing-through-the-center-of-mass | Moment of inertia of continuous bodies - uniform rod along with the axis passing through the center of mass
Moment of inertia of continuous bodies
Lets take an exmaple:
If the mass is continuous distributed over the body as show in below figure :
lets a mass of small portion at distance of r from the axis. so
now for calculate the moment of interia of whole body, we have to sum of all small portion of moment of interia with the help of integration so
here we have to find the range of r to caculte the moment of inertia of whole body so the formula is
Let take an another example :
Calculte the moment of inertia of uniform rod along with the axis passing through the center of mass as shown in the figure.
Lets take mass of rod is M and length is L and it is one dimention rod thats which we can neglect the thickness of the rod.
now calculte the range of r as shown in figure r is range from -r/2 to +r/2
now the question is how to calculate dm
mass of 1m length is =
mass of 2m length is =
mass of dx m length =
07 Jun, 2019 2399 views Physics | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9470187425613403, "perplexity": 906.764611171304}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141740670.93/warc/CC-MAIN-20201204162500-20201204192500-00462.warc.gz"} |
https://www.gradesaver.com/textbooks/math/calculus/thomas-calculus-13th-edition/chapter-7-transcendental-functions-section-7-7-hyperbolic-functions-exercises-7-7-page-430/54 | ## Thomas' Calculus 13th Edition
$2 \ln 2 -\dfrac{3}{4}$
As we are given that $\int^{\ln 2}_{0} 4e^{-\theta} \sinh \theta d\theta$ Since, $\sinh \theta=\dfrac{e^{\theta} -e^{-\theta}}{2}$ The given integral $\int^{\ln 2}_{0} 4(e^{-\theta}) \sinh \theta d\theta$ can be re-written as:$\int^{\ln 2}_{0} 4e^{-\theta} [\dfrac{e^{\theta} -e^{-\theta}}{2}] d\theta= 2[ \theta +\dfrac{1}{2} e^{-2\theta}]^{\ln 2}_{0}$ or, $=2 [\ln 2+(\dfrac{1}{8})]-1$ Thus, $=2 \ln 2 -\dfrac{3}{4}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.994426965713501, "perplexity": 351.22992690487814}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575541310970.85/warc/CC-MAIN-20191215225643-20191216013643-00347.warc.gz"} |
https://cms.math.ca/cjm/msc/05E30?fromjnl=cjm&jnl=CJM | location: Publications → journals
Search results
Search: MSC category 05E30 ( Association schemes, strongly regular graphs )
Expand all Collapse all Results 1 - 2 of 2
1. CJM 1999 (vol 51 pp. 326)
Martin, W. J.; Stinson, D. R.
Association Schemes for Ordered Orthogonal Arrays and $(T,M,S)$-Nets In an earlier paper~\cite{stinmar}, we studied a generalized Rao bound for ordered orthogonal arrays and $(T,M,S)$-nets. In this paper, we extend this to a coding-theoretic approach to ordered orthogonal arrays. Using a certain association scheme, we prove a MacWilliams-type theorem for linear ordered orthogonal arrays and linear ordered codes as well as a linear programming bound for the general case. We include some tables which compare this bound against two previously known bounds for ordered orthogonal arrays. Finally we show that, for even strength, the LP bound is always at least as strong as the generalized Rao bound. Categories:05B15, 05E30, 65C99
2. CJM 1998 (vol 50 pp. 739)
Godsil, C. D.
Eigenpolytopes of distance regular graphs Let $X$ be a graph with vertex set $V$ and let $A$ be its adjacency matrix. If $E$ is the matrix representing orthogonal projection onto an eigenspace of $A$ with dimension $m$, then $E$ is positive semi-definite. Hence it is the Gram matrix of a set of $|V|$ vectors in $\re^m$. We call the convex hull of a such a set of vectors an eigenpolytope of $X$. The connection between the properties of this polytope and the graph is strongest when $X$ is distance regular and, in this case, it is most natural to consider the eigenpolytope associated to the second largest eigenvalue of $A$. The main result of this paper is the characterisation of those distance regular graphs $X$ for which the $1$-skeleton of this eigenpolytope is isomorphic to $X$. Categories:05E30, 05C50
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https://www.mathworks.com/help/ident/ug/spectrum-normalization.html | ## Spectrum Normalization
The spectrum of a signal is the square of the Fourier transform of the signal. The spectral estimate using the commands `spa`, `spafdr`, and `etfe` is normalized by the sample time T:
`${\Phi }_{y}\left(\omega \right)=T\sum _{k=-M}^{M}{R}_{y}\left(kT\right){e}^{-iwT}{W}_{M}\left(k\right)$`
where WM(k) is the lag window, and M is the width of the lag window. The output covariance Ry(kT) is given by the following discrete representation:
`${\stackrel{^}{R}}_{y}\left(kT\right)=\frac{1}{N}\sum _{l=1}^{N}y\left(lT-kT\right)y\left(lT\right)$`
Because there is no scaling in a discrete Fourier transform of a vector, the purpose of T is to relate the discrete transform of a vector to the physically meaningful transform of the measured signal. This normalization sets the units of ${\Phi }_{y}\left(\omega \right)$ as power per radians per unit time, and makes the frequency units radians per unit time.
The scaling factor of T is necessary to preserve the energy density of the spectrum after interpolation or decimation.
By Parseval's theorem, the average energy of the signal must equal the average energy in the estimated spectrum, as follows:
`$\begin{array}{l}E{y}^{2}\left(t\right)=\frac{1}{2\pi }{\int }_{-\pi /T}^{\pi /T}{\Phi }_{y}\left(\omega \right)d\omega \\ S1\equiv E{y}^{2}\left(t\right)\\ S2\equiv \frac{1}{2\pi }{\int }_{-\pi /T}^{\pi /T}{\Phi }_{y}\left(\omega \right)d\omega \end{array}$`
To compare the left side of the equation (`S1`) to the right side (`S2`), enter the following commands. In this code, `phiy` contains ${\Phi }_{y}\left(\omega \right)$ between $\omega =0$ and $\omega =\pi }{T}$ with the frequency step given as follows:
`$\left(\frac{\pi }{T\cdot \text{length(phiy)}}\right)$`
`load iddata1`
Create a time-series iddata object.
`y = z1(:,1,[]);`
Define sample interval from the data.
`T = y.Ts;`
Estimate the frequency response.
`sp = spa(y);`
Remove spurious dimensions
`phiy = squeeze(sp.spec);`
Compute average energy of the signal.
`S1 = sum(y.y.^2)/size(y,1)`
```S1 = 19.4646 ```
Compute average energy from the estimated energy spectrum, where S2 is scaled by T.
`S2 = sum(phiy)/length(phiy)/T`
```S2 = 19.2076 ```
Thus, the average energy of the signal approximately equals the average energy in the estimated spectrum. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9918975234031677, "perplexity": 618.9847782387324}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153857.70/warc/CC-MAIN-20210729105515-20210729135515-00079.warc.gz"} |
http://www.adamponting.com/how-to-calculate-the-continued-fraction-of-the-square-root-of-13/ | # How to calculate the continued fraction of the square root of 13
(unfinished)
is some irrational number, with a weird decimal going on forever. With this method you can calculate it as exactly as you want, or the square root of any number, without having any idea what it is! Well, knowing only that it’s between 3 and 4. because and is 16, so it’s somewhere in there.
The only slightly tricky thing in this method is knowing (i.e. remembering from high school) how to get the from the bottom to the top of a fraction, so read the short explanation of how to do that using conjugate surds if you need to.
OK, let’s begin…
We are going to make a continued fraction for , which may then be used to make rational approximations of any desired accuracy. (And convergents (fractions) made from continued fractions are the most accurate fractional approximations for their size.)
We will be finding , and the integers making up the continued fraction
Firstly, set .
The algorithm has only two steps, repeated ad nauseam:
Step 1. Set
This is the ‘floor’ function, meaning round it down to the nearest integer. Which is lucky, cause that’s all I know about !
Step 2.
Then repeat…
Ok, round 1.
1. .
2.
That was easy. Round 2*..
1. – and here’s the tricky bit I spoke of earlier, multiplying by the conjugate surd to make it much easier to deal with.
Hey, we’ve already done this step.
So The is rounded down to again.
And is rounded down to . So
2.
Round 3.
1.
First multiply by the conjugate as always.
2.
Round 4.
1.
2.
To be continued!
*Yes, this is a boxing metaphor. Weird, huh.
(Method stolen from Koshy’s Elementary Number Theory with Applications, 2007 – it’s just Euclid’s algorithm really.) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.904651939868927, "perplexity": 877.5560161295195}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662521041.0/warc/CC-MAIN-20220518021247-20220518051247-00171.warc.gz"} |
https://physics.stackexchange.com/questions/413851/why-is-minkowski-spacetime-in-polar-coordinates-treated-in-texts-as-flat-spaceti | # Why is Minkowski spacetime in polar coordinates treated in texts as flat spacetime?
Taking 3-D Minkowski spacetime line element in General Relativity:
$$ds^2=-c^2dt^2+dx^2+dy^2+dz^2,$$
when considering a change into spherical coordinates leads to:
$$ds^2=-c^2dt^2+dr^2+r^2\left(d\theta^2+\sin^2\theta\,d\phi^2\right).$$
In several books, it is said that this is still Euclidean flat-space time, for it is only a change of coordinates speaking about the same as in Euclidean plane... but my big inquiry is under what point of view is this still flat, since the Levi-Civita connection $\Gamma^{\alpha}_{\,\,\beta\lambda}$ for this new space-time is not zero for some components. Are these symbols equal to zero a necessary condition for giving flat space-time?
I have not computed the components of the Riemann tensor for the polar coordinates spacetime, yet. But it is easy to see that for Cartesian coordinates they are equal to zero. If they were nonzero, does this assume that the deviation of geodesics equal to zero is still obeyed? From since I can remember, if the components of the Riemann tensor $R^{\alpha}_{\,\,\beta\mu\nu}$ are all zero, you get deviation zero and you can talk about Euclidean, flat space-time. Also, I can remember that if the Ricci scalar $R=0$ if and only if flat space-time is given. Am I correct?
• Curvature is given by the Riemann tensor, so you should try to calculate it. Christoffel's symbols are not 0 in polar coordinates on an Euclidean space either. You need a connection in usual polar coordinates. If a space is flat in some coordinates, it must be flat, but that doesn't mean your coordinates don't need a connection. – FGSUZ Jun 26 '18 at 20:46
• The curvature is derived from the metric. If you haven't changed the metric, you haven't changed the curvature. – WillO Jun 26 '18 at 20:47
• There are terms such as, Riemann flat, Ricci flat, and scalar flat, each meaning slightly different things (somewhat obvious by name). Just because Christoffel is non zero doesn't mean that the manifold is flat. – ggcg Jun 26 '18 at 21:00
While $\Gamma^a_{bc} \neq 0$, Minkowski space in any set of coordinates has $R^a_{bcd} = 0$. To convince yourself without calculating, see a coordinate change as a relabelling of positions. Rather than a grid, you might use a spherical coordinate system, but the points you are labelling on the surface are not being moved. The distance between any two is still the same. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9002323150634766, "perplexity": 330.13872300764785}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703522242.73/warc/CC-MAIN-20210121035242-20210121065242-00130.warc.gz"} |
http://mathhelpforum.com/differential-equations/164048-fourier-transform-derivative-decaying-exponent-function.html | # Math Help - Fourier Transform of a derivative of a decaying exponent function?
1. ## Fourier Transform of a derivative of a decaying exponent function?
Hello folks, new to the forum and first post here Hope this is the correct subforum. Tried the search option but did not find anything to help me out. Here's the problem:
I have a decaying exponential function f(t) = exp(-at); where a>0 and t>=0 [and f(t)=0 when t<0)] and I need to find the Fourier transformation of it's derivative.
I can do it using the Derivative Theorem:
F {f'(t)} = iw F(w);
and I get
F {f'(t)} = (iw)/(a+iw).
However, if I just take d/dt of the original f(t), I get f'(t)= -a exp(-at) = -a f(t).
And since a is a constant, calculating the Fourier transform for this one should go like:
F {f'(t)} = F {-a f(t)} = -aF(w) = (-a)/(a+iw).
These 2 results are not the same (unless, for some reason a=-iw), so obviously I'm not doing it right. Just cannot get my head around it. Any thoughts & pointing out where I go wrong appreciated.
2. I don't think you're taking the derivative correctly. Your function is actually
$f(t)=U(t)\,e^{-at},$
where $U(t)$ is the Heaviside step function. The Heaviside step function actually has a derivative (though the result is technically a distribution or a measure, not a function): the Dirac delta function. Hence, the derivative looks more like this:
$f'(t)=U'(t)\,e^{-at}-aU(t)\,e^{-at}=\delta(t)\,e^{-at}-aU(t)\,e^{-at}.$
Try taking the Fourier Transform of this result, and see if it doesn't match up with what you got before. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8922696709632874, "perplexity": 647.5615410004355}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644065910.17/warc/CC-MAIN-20150827025425-00315-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://proofwiki.org/wiki/Normal_Subgroup_Test | # Normal Subgroup Test
It has been suggested that this page or section be merged into Subgroup is Superset of Conjugate iff Normal. (Discuss)
## Theorem
Let $G$ be a group and $H \le G$.
Then $H$ is a normal subgroup of $G$ if and only if:
$\forall x \in G: x H x^{-1} \subseteq H$.
## Proof
Let $H$ be a subgroup of $G$.
Suppose $H$ is normal in $G$.
Then $\forall x \in G, a \in H: \exists b \in H: x a = b x$.
Thus, $x a x^{-1} = b \in H$ implying $x H x^{-1} \subseteq H$.
Conversely, suppose $\forall x \in G: x H x^{-1} \subseteq H$.
Then for $g \in G$, we have $g H g^{-1} \subseteq H$, which implies $g H \subseteq H g$.
Also, for $g^{-1} \in G$, we have $g^{-1} H (g^{-1})^{-1} = g^{-1} H g \subseteq H$ which implies $H g \subseteq g H$.
Therefore, $g H = H g$ meaning $H \lhd G$.
$\blacksquare$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9896120429039001, "perplexity": 160.76383095967262}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256100.64/warc/CC-MAIN-20190520182057-20190520204057-00436.warc.gz"} |
https://www.aimsciences.org/journal/1078-0947/2020/40/3 | # American Institute of Mathematical Sciences
ISSN:
1078-0947
eISSN:
1553-5231
All Issues
## Discrete & Continuous Dynamical Systems - A
March 2020 , Volume 40 , Issue 3
Select all articles
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2020, 40(3): 1257-1281 doi: 10.3934/dcds.2020077 +[Abstract](1595) +[HTML](133) +[PDF](433.9KB)
Abstract:
In [5] Bourgain proves that Sarnak's disjointness conjecture holds for a certain class of three-interval exchange maps. In the present paper we slightly improve the Diophantine condition of Bourgain and estimate the constants in the proof. We further show that the new parameter set has positive, but not full Hausdorff dimension. This, in particular, implies that the Lebesgue measure of this set is zero.
2020, 40(3): 1283-1307 doi: 10.3934/dcds.2020078 +[Abstract](1734) +[HTML](145) +[PDF](429.84KB)
Abstract:
The global well-posedness for the KP-Ⅱ equation is established in the anisotropic Sobolev space \begin{document}$H^{s, 0}$\end{document} for \begin{document}$s>-\frac{3}{8}$\end{document}. Even though conservation laws are invalid in the Sobolev space with negative index, we explore the asymptotic behavior of the solution by the aid of the \begin{document}$I$\end{document}-method in which Colliander, Keel, Staffilani, Takaoka, and Tao introduced a series of modified energy terms. Moreover, a-priori estimate of the solution leads to the existence of global attractor for the weakly damped, forced KP-Ⅱ equation in the weak topology of the Sobolev space when \begin{document}$s>-\frac{1}{8}$\end{document}.
2020, 40(3): 1309-1360 doi: 10.3934/dcds.2020079 +[Abstract](1301) +[HTML](128) +[PDF](662.03KB)
Abstract:
We consider transformations preserving a contracting foliation, such that the associated quotient map satisfies a Lasota-Yorke inequality. We prove that the associated transfer operator, acting on suitable normed spaces, has a spectral gap (on which we have quantitative estimation).
As an application we consider Lorenz-like two dimensional maps (piecewise hyperbolic with unbounded contraction and expansion rate): we prove that those systems have a spectral gap and we show a quantitative estimate for their statistical stability. Under deterministic perturbations of the system of size \begin{document}$\delta$\end{document}, the physical measure varies continuously, with a modulus of continuity \begin{document}$O(\delta \log \delta )$\end{document}, which is asymptotically optimal for this kind of piecewise smooth maps.
2020, 40(3): 1361-1387 doi: 10.3934/dcds.2020080 +[Abstract](1553) +[HTML](125) +[PDF](471.56KB)
Abstract:
Young [17] proved the positivity of Lyapunov exponent in a large set of the energies for some quasi-periodic cocycles. Her result is also proved to be applicable for some quasi-periodic Schrödinger cocycles by Zhang [18]. However, her result cannot be applied to the Schrödinger cocycles with the potential \begin{document}$v = \cos(4\pi x)+w( x)$\end{document}, where \begin{document}$w\in C^2(\mathbb R/\mathbb Z,\mathbb R)$\end{document} is a small perturbation. In this paper, we will improve her result such that it can be applied to more cocycles.
2020, 40(3): 1389-1409 doi: 10.3934/dcds.2020081 +[Abstract](1363) +[HTML](94) +[PDF](399.14KB)
Abstract:
In this paper we will prove the existence of periodically invariant tori of twist mappings on resonant surfaces under the low dimensional intersection property.
2020, 40(3): 1411-1433 doi: 10.3934/dcds.2020082 +[Abstract](1676) +[HTML](107) +[PDF](391.25KB)
Abstract:
In this paper, we study the long-time existence and uniqueness (solvability) for the initial value problem of the 2D inviscid dispersive SQG equation. First we obtain the local solvability with existence-time independent of the amplitude parameter \begin{document}$A$\end{document}. Then, assuming more regularity and using a blow-up criterion of BKM type and a space-time estimate of Strichartz type, we prove long-time solvability of solutions in Besov spaces for large \begin{document}$A$\end{document} and arbitrary initial data. In comparison with previous results, we are able to consider improved cases of the regularity and larger initial data classes.
2020, 40(3): 1435-1492 doi: 10.3934/dcds.2020083 +[Abstract](1292) +[HTML](115) +[PDF](578.78KB)
Abstract:
For 2D compressible full Euler equations of Chaplygin gases, when the initial axisymmetric perturbation of a rest state is small, we prove that the smooth solution exists globally. Compared with the previous references, there are two different key points in this paper: both the vorticity and the variable entropy are simultaneously considered, moreover, the usual assumption on the compact support of initial perturbation is removed. Due to the appearances of the variable entropy and vorticity, the related perturbation of solution will have no decay in time, which leads to an essential difficulty in establishing the global energy estimate. Thanks to introducing a nonlinear ODE which arises from the vorticity and entropy, and considering the difference between the solutions of the resulting ODE and the full Euler equations, we can distinguish the fast decay part and non-decay part of solution to Euler equations. Based on this, by introducing some suitable weighted energies together with a class of weighted \begin{document}$L^\infty$\end{document}-\begin{document}$L^\infty$\end{document} estimates for the solutions of 2D wave equations, we can eventually obtain the global energy estimates and further complete the proof on the global existence of smooth solution to 2D full Euler equations.
2020, 40(3): 1493-1515 doi: 10.3934/dcds.2020084 +[Abstract](1510) +[HTML](146) +[PDF](379.52KB)
Abstract:
This paper is concerned with the stability and dynamics of a weak viscoelastic system with nonlinear time-varying delay. By imposing appropriate assumptions on the memory and sub-linear delay operator, we prove the global well-posedness and stability which generates a gradient system. The gradient system possesses finite fractal dimensional global and exponential attractors with unstable manifold structure. Moreover, the effect and balance between damping and time-varying delay are also presented.
2020, 40(3): 1517-1554 doi: 10.3934/dcds.2020085 +[Abstract](1129) +[HTML](90) +[PDF](517.88KB)
Abstract:
In this article, we consider the geodesic flow on a compact rank \begin{document}$1$\end{document} Riemannian manifold \begin{document}$M$\end{document} without focal points, whose universal cover is denoted by \begin{document}$X$\end{document}. On the ideal boundary \begin{document}$X(\infty)$\end{document} of \begin{document}$X$\end{document}, we show the existence and uniqueness of the Busemann density, which is realized via the Patterson-Sullivan measure. Based on the the Patterson-Sullivan measure, we show that the geodesic flow on \begin{document}$M$\end{document} has a unique invariant measure of maximal entropy. We also obtain the asymptotic growth rate of the volume of geodesic spheres in \begin{document}$X$\end{document} and the growth rate of the number of closed geodesics on \begin{document}$M$\end{document}. These results generalize the work of Margulis and Knieper in the case of negative and nonpositive curvature respectively.
2020, 40(3): 1555-1593 doi: 10.3934/dcds.2020086 +[Abstract](1054) +[HTML](87) +[PDF](487.61KB)
Abstract:
We consider a nonlinear, free boundary fluid-structure interaction model in a bounded domain. The viscous incompressible fluid interacts with a nonlinear elastic body on the common boundary via the velocity and stress matching conditions. The motion of the fluid is governed by incompressible Navier-Stokes equations while the displacement of elastic structure is determined by a nonlinear elastodynamic system with boundary dissipation. The boundary dissipation is inserted in the velocity matching condition. We prove the global existence of the smooth solutions for small initial data and obtain the exponential decay of the energy of this system as well.
2020, 40(3): 1595-1620 doi: 10.3934/dcds.2020087 +[Abstract](1048) +[HTML](131) +[PDF](624.8KB)
Abstract:
We consider a system of coupled cubic Schrödinger equations in one space dimension
in the non-integrable case \begin{document}$0 < \omega < 1$\end{document}.
First, we justify the existence of a symmetric 2-solitary wave with logarithmic distance, i.e. a solution of the system satisfying
where \begin{document}$Q = \sqrt{2}$\end{document}sech is the explicit solution of \begin{document}$Q'' - Q + Q^3 = 0$\end{document} and \begin{document}$\Omega>0$\end{document} is a constant. This result extends to the non-integrable case the existence of symmetric 2-solitons with logarithmic distance known in the integrable case \begin{document}$\omega = 0$\end{document} and \begin{document}$\omega = 1$\end{document} ([15,33]). Such strongly interacting symmetric \begin{document}$2$\end{document}-solitary waves were also previously constructed for the non-integrable scalar nonlinear Schrödinger equation in any space dimension and for any energy-subcritical power nonlinearity ([20,22]).
Second, under the conditions \begin{document}$0<c<1$\end{document} and \begin{document}$0<\omega < \frac 12 c(c+1)$\end{document}, we construct solutions of the system satisfying
where \begin{document}$Q_c(x) = cQ(cx)$\end{document} and \begin{document}$\Omega_c>0$\end{document} is a constant. Such logarithmic regime with non-symmetric solitons does not exist in the integrable cases \begin{document}$\omega = 0$\end{document} and \begin{document}$\omega = 1$\end{document} and is still unknown in the non-integrable scalar case.
2020, 40(3): 1621-1663 doi: 10.3934/dcds.2020088 +[Abstract](1078) +[HTML](113) +[PDF](5347.37KB)
Abstract:
We study the permanence and impermanence for discrete-time Kolmogorov systems admitting a carrying simplex. Sufficient conditions to guarantee permanence and impermanence are provided based on the existence of a carrying simplex. Particularly, for low-dimensional systems, permanence and impermanence can be determined by boundary fixed points. For a class of competitive systems whose fixed points are determined by linear equations, there always exists a carrying simplex. We provide a universal classification via the equivalence relation relative to local dynamics of boundary fixed points for the three-dimensional systems by the index formula on the carrying simplex. There are a total of \begin{document}$33$\end{document} stable equivalence classes which are described in terms of inequalities on parameters, and we present the phase portraits on their carrying simplices. Moreover, every orbit converges to some fixed point in classes \begin{document}$1-25$\end{document} and \begin{document}$33$\end{document}; there is always a heteroclinic cycle in class \begin{document}$27$\end{document}; Neimark-Sacker bifurcations may occur in classes \begin{document}$26-31$\end{document} but cannot occur in class \begin{document}$32$\end{document}. Based on our permanence criteria and the equivalence classification, we obtain the specific conditions on parameters for permanence and impermanence. Only systems in classes \begin{document}$29, 31, 33$\end{document} and those in class \begin{document}$27$\end{document} with a repelling heteroclinic cycle are permanent. Applications to discrete population models including the Leslie-Gower models, Atkinson-Allen models and Ricker models are given.
2020, 40(3): 1665-1702 doi: 10.3934/dcds.2020089 +[Abstract](1195) +[HTML](146) +[PDF](406.9KB)
Abstract:
The variation-of-constants formula is one of the principal tools of the theory of differential equations. There are many papers dealing with different versions of the variation-of-constants formula and its applications. Our main purpose in this paper is to give a variation-of-constants formula for inhomogeneous linear functional differential systems determined by general Volterra type operators with delay. Our treatment of the delay in the considered systems is completely different from the usual methods. We deal with the representation of the studied Volterra type operators. Some existence and uniqueness theorems are obtained for the studied linear functional differential and integral systems. Finally, some applications are given.
2020, 40(3): 1703-1735 doi: 10.3934/dcds.2020090 +[Abstract](1387) +[HTML](101) +[PDF](543.08KB)
Abstract:
The orbital stability of peakons and hyperbolic periodic peakons for the Camassa-Holm equation has been established by Constantin and Strauss in [A. Constantin, W. Strauss, Comm. Pure. Appl. Math. 53 (2000) 603-610] and Lenells in [J. Lenells, Int. Math. Res. Not. 10 (2004) 485-499], respectively. In this paper, we prove the orbital stability of the elliptic periodic peakons for the modified Camassa-Holm equation. By using the invariants of the equation and controlling the extrema of the solution, it is demonstrated that the shapes of these elliptic periodic peakons are stable under small perturbations in the energy space. Throughout the paper, the theory of elliptic functions and elliptic integrals is used in the calculation.
2020, 40(3): 1737-1755 doi: 10.3934/dcds.2020091 +[Abstract](1194) +[HTML](152) +[PDF](403.52KB)
Abstract:
This paper deals with the quasilinear parabolic-elliptic chemotaxis system
under homogeneous Neumann boundary conditions in a bounded domain \begin{document}$\Omega\subset\mathbb{R}^{n}$\end{document} with smooth boundary, where \begin{document}$\tau\in\{0, 1\}$\end{document}, \begin{document}$\chi>0$\end{document}, \begin{document}$\mu>0$\end{document} and \begin{document}$r\geq2$\end{document}. \begin{document}$D(u)$\end{document} is supposed to satisfy
It is shown that when \begin{document}$\mu>\frac{\chi^{2}}{16}$\end{document} and \begin{document}$r\geq2$\end{document}, then the solution to the system exponentially converges to the constant stationary solution \begin{document}$(1, 1)$\end{document}.
2020, 40(3): 1757-1774 doi: 10.3934/dcds.2020092 +[Abstract](1219) +[HTML](137) +[PDF](408.58KB)
Abstract:
In this paper we consider a von Karman plate equation with memory-type boundary conditions. By assuming the relaxation function \begin{document}$k_i$\end{document} \begin{document}$(i = 1, 2)$\end{document} with minimal conditions on the \begin{document}$L^1(0, \infty)$\end{document}, we establish an optimal explicit and general energy decay result. In particular, the energy result holds for \begin{document}$H(s) = s^p$\end{document} with the full admissible range \begin{document}$[1, 2)$\end{document} instead of \begin{document}$[1, 3/2)$\end{document}. This result is new and substantially improves earlier results in the literature.
2020, 40(3): 1775-1798 doi: 10.3934/dcds.2020093 +[Abstract](1760) +[HTML](93) +[PDF](387.51KB)
Abstract:
We study the low-energy solutions to the 3D compressible Navier-Stokes-Poisson equations. We first obtain the existence of smooth solutions with small \begin{document}$L^2$\end{document}-norm and essentially bounded densities. No smallness assumption is imposed on the \begin{document}$H^4$\end{document}-norm of the initial data. Using a compactness argument, we further obtain the existence of weak solutions which may have discontinuities across some hypersurfaces in \begin{document}$\mathbb R^3$\end{document}. We also provide a blow-up criterion of solutions in terms of the \begin{document}$L^\infty$\end{document}-norm of density.
2020, 40(3): 1799-1811 doi: 10.3934/dcds.2020094 +[Abstract](1279) +[HTML](97) +[PDF](387.88KB)
Abstract:
For \begin{document}$n \ge 1$\end{document}, consider the space of affine conjugacy classes of topological cubic polynomials \begin{document}$f: \mathbb{C} \to \mathbb{C}$\end{document} with a period \begin{document}$n$\end{document} ramification point. It is shown that this space is a connected topological space.
2020, 40(3): 1813-1846 doi: 10.3934/dcds.2020095 +[Abstract](1355) +[HTML](95) +[PDF](502.87KB)
Abstract:
In this note, we use an epiperimetric inequality approach to study the regularity of the free boundary for the parabolic Signorini problem. We show that if the "vanishing order" of a solution at a free boundary point is close to \begin{document}$3/2$\end{document} or an even integer, then the solution is asymptotically homogeneous. Furthermore, one can derive a convergence rate estimate towards the asymptotic homogeneous solution. As a consequence, we obtain the regularity of the regular free boundary as well as the frequency gap.
2020, 40(3): 1847-1878 doi: 10.3934/dcds.2020096 +[Abstract](1430) +[HTML](100) +[PDF](445.15KB)
Abstract:
We consider a coupled bulk–surface Allen–Cahn system affixed with a Robin-type boundary condition between the bulk and surface variables. This system can also be viewed as a relaxation to a bulk–surface Allen–Cahn system with non-trivial transmission conditions. Assuming that the nonlinearities are real analytic, we prove the convergence of every global strong solution to a single equilibrium as time tends to infinity. Furthermore, we obtain an estimate on the rate of convergence. The proof relies on an extended Łojasiewicz–Simon type inequality for the bulk–surface coupled system. Compared with previous works, new difficulties arise as in our system the surface variable is no longer the trace of the bulk variable, but now they are coupled through a nonlinear Robin boundary condition.
2020, 40(3): 1879-1887 doi: 10.3934/dcds.2020097 +[Abstract](1361) +[HTML](100) +[PDF](311.3KB)
Abstract:
In 3-dimensional manifolds, we prove that generically in \begin{document}$\operatorname{Diff}^{1}_{m}(M^{3})$\end{document}, the existence of a minimal expanding invariant foliation implies stable Bernoulliness.
2020, 40(3): 1889-1902 doi: 10.3934/dcds.2020098 +[Abstract](1340) +[HTML](118) +[PDF](367.04KB)
Abstract:
In this paper we study the existence and some properties of the global attractors for a class of weighted equations when the weighted Sobolev space \begin{document}$H_0^{1,a}(\Omega)$\end{document} (see Definition 1.1) cannot be bounded embedded into \begin{document}$L^2(\Omega)$\end{document}. We claim that the dimension of the global attractor is infinite by estimating its lower bound of \begin{document}$Z_2$\end{document}-index. Moreover, we prove that there is an infinite sequence of stationary points in the global attractor which goes to 0 and the corresponding critical value sequence of the Lyapunov functional also goes to 0.
2020, 40(3): 1903-1935 doi: 10.3934/dcds.2020099 +[Abstract](1320) +[HTML](101) +[PDF](413.2KB)
Abstract:
In this paper, we consider the Landau-Lifshitz-Gilbert systems with spin-polarized transport from a bounded domain in \begin{document}$\mathbb{R}^3$\end{document} into \begin{document}$S^2$\end{document} and show the existence of global weak solutions to the Cauchy problems of such Landau-Lifshtiz systems. In particular, we show that the Cauchy problem to Landau-Lifshitz equation without damping but with diffusion process of the spin accumulation admits a global weak solution. The Landau-Lifshtiz system with spin-polarized transport into a compact Lie algebra is also discussed and some similar results are proved. The key ingredients of this article consist of the choices of test functions and approximate equations.
2020, 40(3): 1937-1961 doi: 10.3934/dcds.2020100 +[Abstract](1183) +[HTML](100) +[PDF](418.96KB)
Abstract:
This paper is concerned with the long-time behavior for a class of non-autonomous plate equations with perturbation and strong damping of \begin{document}$p$\end{document}-Laplacian type
in bounded domain \begin{document}$\Omega\subset \mathbb{R}^N$\end{document} with smooth boundary and critical nonlinear terms. The global existence of weak solution which generates a continuous process has been presented firstly, then the existence of strong and weak uniform attractors with non-compact external forces also derived. Moreover, the upper-semicontinuity of uniform attractors under small perturbations has also obtained by delicate estimate and contradiction argument.
2020, 40(3): 1963-1987 doi: 10.3934/dcds.2020101 +[Abstract](1160) +[HTML](145) +[PDF](408.62KB)
Abstract:
A vanishing viscosity problem for the Patlak-Keller-Segel model is studied in this paper. This is a parabolic-parabolic system in a bounded domain \begin{document}$\Omega\subset \mathbb{R}^n$\end{document}, with a vanishing viscosity \begin{document}$\varepsilon\to 0$\end{document}. We show that if the initial value lies in \begin{document}$W^{1, p}$\end{document} with \begin{document}$p>\max\{2, n\}$\end{document}, then there exists a unique solution \begin{document}$(u_\varepsilon, v_\varepsilon)$\end{document} with its lifespan independent of \begin{document}$\varepsilon$\end{document}. Furthermore, as \begin{document}$\varepsilon\rightarrow 0$\end{document}, \begin{document}$(u_\varepsilon, v_\varepsilon)$\end{document} converges to the solution \begin{document}$(u, v)$\end{document} of the limiting system in a suitable sense.
2019 Impact Factor: 1.338 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.957339346408844, "perplexity": 627.7252291601883}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141672314.55/warc/CC-MAIN-20201201074047-20201201104047-00569.warc.gz"} |
https://www.physicsforums.com/threads/need-a-calculus-refresher.71324/ | # Need a Calculus Refresher
1. Apr 13, 2005
### Fanman22
I'm an 8th semester senior majoring in Neurobiology and Molecular and Cell Biology. I took my calculus sequence in my 1st and 2nd semesters of college. 3 years later, I find myself finishing up a general physics sequence that is beginning to require the use of calculus.
Needless to say, 3years of not using calc is long enough to forget even the basics. So my question is, does anybody know of a website that clearly overviews the basics of anti-derivatives/integration with and without limits?
Thanks in advance for the help.
2. Apr 13, 2005
### dextercioby
Last edited by a moderator: Apr 21, 2017
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