Lagrangian, central forces and conservation of angular momentum When studying central forces it is possible to propose the Lagrangian:   $$ L = T-U=\frac{1}{2}m \dot{r}^2+\frac{1}{2}mr^2 \dot{\theta}^2 - U(r)$$   Then we can solve the equation of motions for $\theta$: $$ \frac{d}{dt}\frac{\partial}{\partial \dot{\theta}}L-\frac{\partial}{\partial \theta}L=0=\frac{d}{dt}\left( mr^2\dot{\theta}\right)  \rightarrow mr^2\dot{\theta} = constant = l$$  Then  $\dot{\theta} = l/mr^2$   Why can't I take this equality and put it into the Lagrangian $\mathcal{L}$ before solving the equations of motion for $r$? So:   $$ L = \frac{1}{2}m \dot{r}^2+\frac{l^2}{2 mr^2}  - U(r) $$   $$ \frac{d}{dt}\frac{\partial}{\partial \dot{r}}L-\frac{\partial}{\partial r}L=0\rightarrow\frac{d}{dt}(mr)+\frac{l^2}{mr^3}+\frac{\partial}{\partial r}U(r)=0$$   The result would be obviously different if I solve the equations for $r$ and then substitute $\dot{\theta} = l/mr^2$:  $$L = \frac{1}{2}m \dot{r}^2+\frac{1}{2}mr^2 \dot{\theta}^2 - U(r)$$  $$\frac{d}{dt}\frac{\partial}{\partial \dot{r}}L-\frac{\partial}{\partial r}L=0\rightarrow\frac{d}{dt}(mr)-\frac{l^2}{mr^3}+\frac{\partial}{\partial r}U(r)=0$$   In this case, the difference is the sign on the angular momentum term. But in general, it is not the same at all. Why is solving the Euler-Lagrange Equation first the right approach?
Lagrangian of an effective potential If there is a system, described by an Lagrangian $\mathcal{L}$ of the form   $$\mathcal{L} = T-V = \frac{m}{2}\left(\dot{r}^2+r^2\dot{\phi}^2\right) + \frac{k}{r},\tag{1}$$   where $T$ is the kinetic energy and $V$ the potential energy, it is also possible to define the total energy $E$ of the system   $$ E = T + V =\frac{m}{2}\left(\dot{r}^2+r^2\dot{\phi}^2\right) - \frac{k}{r}.\tag{2}$$  If the angular momentum $M$ is defined by $$ M = m r^{2} \dot{\phi},\tag{3}$$ then $E$ can be written as  $$ E = \frac{m}{2}\dot{r}^2 + \underbrace{\frac{M^2}{2mr^2}-\frac{k}{r}}_{V_\textrm{eff}\left(r\right)},\tag{4}$$  where the last two terms are written as new, "effective" potential $V_\textrm{eff}\left(r\right)$.    In addition, using the definition of $M$, the Lagrangian $\mathcal{L}$ can be written as  $$ \mathcal{L} = \frac{m}{2}\dot{r}^2 + \underbrace{\frac{M^2}{2mr^2}+\frac{k}{r}}_{-V_\textrm{eff}\left(r\right)},\tag{5}$$  where the sign of $V_\textrm{eff}\left(r\right)$ has been changed, since $\mathcal{L} = T-V$. But from this argument, it appears that there are two different possible ways to construct the same effective potential. This seems to me contradictory. Where is my mistake?
Supremum and Infimum Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A = -\sup(-A)$.   We know that $-A$ is bounded above. Hence $\sup(-A)$ exists. So to proceed, one would show that $\inf A \leq -\sup(-A)$ and $\inf A \geq -\sup(-A)$? Or $-\inf A &gt; \sup(-A)$ and $-\inf A &lt; \sup(-A)$.  I get the sense that one can use the following fact as well: (i) Let $\alpha = \sup S$. If $\beta &lt; \alpha$ then $\beta$ is not an upper bound for $S$. Singular verb after two noun phrases joined by "and" that can be thought of as a single thing I am having trouble with the following sentence:      "The complexity and diversity of the new platforms is worrying us."   If "complexity" and "diversity" constitute two subjects, then the verb should presumably be "are"; however, the sentence feels more natural with "is". Is this correct? Is there a linguistic term for this occurrence?    The following questions address the general topic of compound subjects, but not specific cases like this:             And the following question seems to have somewhat inconclusive answers:         So this question is not really a duplicate of these three. Probability that all three shirts were worn Mr. Jones has three shirts: Red, Green and White. Each day he picks randomly a red shirt with probability of $\frac{1}{2}$, green with probability of $\frac{1}{3}$ and white with probability of $\frac{1}{6}$.     What is the probability that he wears all three shirts after 6 days?   My try:   $$\binom{6}{1}\binom{5}{1}\binom{4}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}\cdot[\binom{3}{2}(\frac{1}{2})^2\cdot[\frac{1}{3}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{3})^2\cdot[\frac{1}{2}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{6})^2\cdot[\frac{1}{2}+\frac{1}{3}]+\binom{3}{1}\binom{2}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}]$$  First, we pick possessions for red, green and white. Then, we multiply it by the other combinations possible of the other three possessions left, and summing it up (I didn't count all possibilities). The problem is the answer is bigger than 1, so it must be wrong. Also, I guess there is more elegant way.  Thanks in advance. iPhone Apps section not showing up in iTunes Find the value of : $\cos x \cos 2x...\cos 999x$ given that $x=\frac {2\pi}{1999}$ Given $x=\frac {2\pi}{1999}$  Find the value of $$\cos x \cos 2x \cos 3x ...\cos 999x$$  So I tried expanding $\sin {2000x}=2\sin 1000x \cos 1000x$  Then rewriting  $\cos 1000x= \cos {(999x+x)}$   No luck so far. How to extend WIFI from ethernet For a continuous function $f$ show $\exists c\in (0,1)$ s.t $f(c)=3c^2$ Let $f: [0,1] \to \mathbb{R}$ be a continuous function such that $$\int_0^1 f (t) dt =1 .$$ Show that there exists a point $c \in (0,1)$ such that $ f(c)=3c^2$.  I tried using mean value theorem of integrals on $f (x)$, so $$\int_0^1 f (t) dt = f (c) =1,$$ but it should be $3 c^2$. What is going wrong? Is $\{0\}$ a field? Consider the set $F$ consisting of the single element $I$.  Define addition and multiplication such that $I+I=I$ and $I \times I=I$ .  This ring satisfies the field axioms:   Closure under addition.  If $x, y \in F$, then $x = y = I$, so $x + y = I + I = I \in F$. Closure under multiplication.  $x \times y = I \times I = I \in F$ Existence of additive identity.  $\forall x \in F$ (i.e., for $x=I$), $x + I = x$, so $I$ is the additive identity. Existence of mulitiplicative identity. $\forall x \in F, x \times I = x$, so $I$ is the multiplicative identity. Additive inverse. $\forall x \in F, \exists y = I \in F: x + y = I$ Multiplicative inverse. $\forall x \in F, \exists y = I \in F: x \times y = I$.  However, because the additive identity need not have a multiplicative inverse, this is a vacuous truth. Commutativity of addition.  $\forall x, y \in F, x + y = I = y + x$ Commutativity of multiplication.  $\forall x, y \in F, x \times y = I = y \times x$ Associativity of addition.  $\forall x, y, z \in F$, $(x + y) + z = I + I = I$ and $x + (y + z) = I + I = I$, so $(x + y) + z = x + (y + z)$ Associativity of multiplication.  $\forall x, y, z \in F$, $(x \times y) \times z = I \times I = I$ and $x \times (y \times z) = I \times I = I$, so $(x \times y) \times z = x \times (y \times z)$ Distributivity of multiplication over addition. $I \times (I + I) = I$ and $I \times I + I \times I = I$, so $\forall x,y,z \in F, x(y+z) = xy+xz$   Based on the above, $\{I\}$ seems to qualify as a field.  If $I$ is assumed to be a real number, then the unique solution of $I + I = I$ and $I \times I = I$ is, of course, $I = 0$.  So, is {0} a field, or is there generally considered to be an additional field axiom which would exclude it?  Specifically, is it required for the multiplicative identity to be distinct from the additive identity? Is there really any difference between the jackknife and leave one out cross validation? The procedure seems identical am I missing something? What is $n!$ when $n=0$? Let $a\ge1$ and define the sequence $(x_n)$ recursively by: $$x_1 = \sqrt{a}$$ $$x_{n+1}= \sqrt{a+x_n}$$  Here's what I did:  Plugging in some values makes it seem as if the sequence is increasing. I first solve the equation $t=\sqrt{a+t}$, equivalently $t^2 -t -a =0$, to get an idea what's going on. The roots are $$t_{1,2}= \frac{1 \pm \sqrt{1+4a} }{2}.$$ Therefore, between the roots, that is for $$t\in \left( \frac{1 - \sqrt{1+4a} }{2}, \frac{1 + \sqrt{1+4a} }{2}\right)$$  we have $t^2-t-a \le 0 \iff t \le \sqrt{a+t}$. Here's my attempt at showing inductively that $$\forall n\in \mathbb{N}, \ x_n \le \frac{1 + \sqrt{1+4a} }{2}$$  For $n=1$, the statement is true (simple arithmetic). Suppose it's true for $n$. Consider  $$x_{n+1}= \sqrt{a+x_n}\le \sqrt{a+\frac{1 + \sqrt{1+4a} }{2}}=\frac{1 + \sqrt{1+4a} }{2},$$  because $\frac{1 + \sqrt{1+4a} }{2}$ satisfies $x=\sqrt{x+a}$.  So this establishes the sequence $(x_n)$ is bounded. Now, by induction again, I try to show it's monotonically increasing. The base case $x_1\le x_2$ is obvious. Assume (strong induction) $x_{n-1}\le x_n$ for all $n&gt;1$. But since we have $$x_1\le x_n \le \frac{1 + \sqrt{1+4a} }{2}, $$   then $x_n$ lies in the appropriate interval, so $x_n \le \sqrt{a+x_n}=x_{n+1}.$  Is this proof correct? Can the condition on $a$ be relaxed to $a&gt;0$? Maybe even small negative $a$? And could someone provide me with another proof of convergence of $x_n$? I absolutely hate this one! biblatex: Cite commands with the same author prints authors name twice I'm using biblatex 1.4b, and Biber 0.9. However, I think the problem is with my settings, not with the versions of those programs.  I'm trying to cite a single author, and produce their name, followed by the dates of two papers, using the authoryear-ibid style, and natbib compatability. Here's my biblatex preamble:  \usepackage[style=authoryear-ibid,natbib=true,backend=biber]{biblatex}   What I want is for the following command: The works by \citet{Blau81, Blau88} ... to produce: The works by Blau (1981; 1988) ... however, what I'm getting is: The works by Blau (1981); Blau (1988)...  I know that biblatex knows they're the same author, since in the bibliography they fall under the same name (i.e. after the first instance of the name, there's just a dash instead of the name)  What am I doing wrong here? I've also tried it with the commands \cite and \cites, and neither produces what I want.   (crossposted with comp.text.tex) How old is the word "prolly"? Prolly is given this definition at :     Clipped pronunciation of probably.   I was reading an interesting article today that  from 1947 and that surprised me. Unfortunately, I wasn't able to find a listed source for the claim. So how old is prolly? What does it mean that GUIkit is being deprecated? I read about Drupal behaviors today, and I tried writing the following code.  (function ($) {    Drupal.behaviors.mymodule = {     attach: function (context, settings) {       $('#mymodule_id', context).change(function () {         alert('Handler for .change() called.');       });     }   };  }(jQuery));    Is Drupal.behaviors.mymodule the namespace? What are the context and settings parameters passed to the Drupal behavior? Is this the equivalent of document.ready()?   Can I attach any number of functions?  Can I define JavaScript functions which will be called somewhere? The name of the boy vs.The boy's name Which one is correct or are both correct?     The boy's name is nice.   The name of the boy is nice.     I'm really confused these two sentences?  Which one is more suitable? Can an abstract class have a constructor? Can an abstract class have a constructor?  If so, how can it be used and for what purposes? I can't update 12.04 LTS. The Update Manager shows this:  CD/DVD 'Ubuntu 12.04 LTS _Precise Pangolin_ - Release i386 (20120423)' is required Please insert the above CD/DVD into the drive '/media/cdrom/' to install software packages from it.   The image I downloaded and burned to CD is from  the PC (Intel x86) desktop CD version. I burned many CDs but the result is the same. Flash drive .docx disappeared files $n$-th digit in the sequence of natural numbers