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I know that the derivative of $b^x$ is just $b^x \log{(b)}$, and I've seen it being derived using chain rule and such (not that I understand how it's done, I just learned about $e$ today so using the chain rule to derive $b^x$ is outside my scope as of now). In any case I was trying to find out how you could derive $b^x$ using the definition of the derivative, $[f(x+h) - f(x)]/h$. I couldn't find it on the internet so I was wondering if someone here could show me. Thanks in advance!
$$ \frac{d}{dx} b^x = \lim_{h \to 0} \frac{b^{x+h}-b^x}{h} = b^x \lim_{h \to 0} \frac{b^h-1}{h}. $$ At this point, you have to evaluate this limit. You have a couple of options here. Obviously L'Hôpital's not going to do. One way is, if your definition of $e^{y}$ is $$ \exp{y} = \sum_{k=0}^{\infty} \frac{y^k}{k!}, $$ then write $b^h = \exp{(h\log{b})} $, and you can evaluate the limit using the first few terms of the series.
On the other hand, if your definition is $$ \exp{y} = \lim_{n \to \infty} \left( 1 + \frac{y}{n} \right)^n, $$ you have a bit more of a problem. Essentially what this comes down to is the rearrangement $$ y = \left( 1+\frac{x}{n} \right)^n \to x = n(y^{1/n}-1) $$ Now, set $h=1/n$ and this looks like your limit, so you conclude that the limit is $\log{b}$. (I would like to see a cleaner version of this, should anyone have one to offer: the rearrangement to get the inverse doesn't look completely justified the way I normally do it.)
And if your definition of $\log{x}$ is an integral, and $\exp{x}$ as its inverse, then you're probably best proving it equivalent to one of the above first...
Here's how one might do it with the definition $\log b = \int_1^b \frac{\mathrm dx}{x}$. However I'm not sure if the exchange of derivative and limit in the first step is allowed.
On one hand we have: $$\frac{\mathrm d}{\mathrm db}\lim_{h\to 0}\frac{b^h-1}{h} = \lim_{h\to 0}\frac{\mathrm d}{\mathrm db}\frac{b^h-1}{h} = \lim_{h\to 0}\frac{hb^{h-1}}{h} = \frac{1}{b}$$ On the other hand we have: $$\frac{\mathrm d}{\mathrm db}\log b = \frac{\mathrm d}{\mathrm db}\int_1^b\frac{\mathrm dx}{x} = \frac1b$$ Therefore we have $$\lim_{h\to 0}\frac{b^h-1}{h} = \log b+C$$ To determine $C$ we note that with $b=1$, we have on one hand $$\lim_{h\to 0}\frac{1^h-1}{h} = 0$$ and on the other hand $$\int_1^1\frac{\mathrm dx}{x} = 0$$ and therefore $C=0$. |
If a function has a slant asymptote, it means that it is asymptotically equivalent to a certain line #y=mx+q#.
This means that
#lim_{x\to\infty} f(x)-(mx+q)=0#.
Divide by #x#:
#lim_{x\to\infty} f(x)/x-(m+q/x)=0#.
Since #q/x\to 0#, we have that
#lim_{x\to\infty} f(x)/x-m=0#.
And so #m=lim_{x\to\infty} f(x)/x#
Once #m# is known, we can find #q# by calculating
#lim_{x\to\infty} f(x)-mx=q#.
Now, let's do the calulations: in your case,
#lim_{x\to\pm\infty} f(x)/x = 3#, since we have
#f(x)/x=\frac{3x^3-28x^2+54x-24}{x^3-8x^2+7x}#
And the limits at #\pm\infty# are given by the ratio of the coefficients of the leading terms (i.e. #x^3#), which in this case are #3# and #1#.
So, #m=3#. Now let's compute #q#:
#\lim_{x\to\pm\infty} f(x)-mx = \frac{3x^3-28x^2+54x-24}{x^2-8x+7} - 3x#
And since
#\frac{3x^3-28x^2+54x-24}{x^2-8x+7} - 3x = \frac{3x^3-28x^2+54x-24-(3x^3-24x^2+21x)}{x^2-8x+7}#
#=\frac{-4x^2+33x-24}{x^2-8x+7}#
And this function tends to #-4# as #x\to\pm\infty# |
2018-09-11 04:29
Proprieties of FBK UFSDs after neutron and proton irradiation up to $6*10^{15}$ neq/cm$^2$ / Mazza, S.M. (UC, Santa Cruz, Inst. Part. Phys.) ; Estrada, E. (UC, Santa Cruz, Inst. Part. Phys.) ; Galloway, Z. (UC, Santa Cruz, Inst. Part. Phys.) ; Gee, C. (UC, Santa Cruz, Inst. Part. Phys.) ; Goto, A. (UC, Santa Cruz, Inst. Part. Phys.) ; Luce, Z. (UC, Santa Cruz, Inst. Part. Phys.) ; McKinney-Martinez, F. (UC, Santa Cruz, Inst. Part. Phys.) ; Rodriguez, R. (UC, Santa Cruz, Inst. Part. Phys.) ; Sadrozinski, H.F.-W. (UC, Santa Cruz, Inst. Part. Phys.) ; Seiden, A. (UC, Santa Cruz, Inst. Part. Phys.) et al. The properties of 60-{\mu}m thick Ultra-Fast Silicon Detectors (UFSD) detectors manufactured by Fondazione Bruno Kessler (FBK), Trento (Italy) were tested before and after irradiation with minimum ionizing particles (MIPs) from a 90Sr \b{eta}-source . [...] arXiv:1804.05449. - 13 p. Preprint - Full text 詳細記錄 - 相似記錄 2018-08-25 06:58
Charge-collection efficiency of heavily irradiated silicon diodes operated with an increased free-carrier concentration and under forward bias / Mandić, I (Ljubljana U. ; Stefan Inst., Ljubljana) ; Cindro, V (Ljubljana U. ; Stefan Inst., Ljubljana) ; Kramberger, G (Ljubljana U. ; Stefan Inst., Ljubljana) ; Mikuž, M (Ljubljana U. ; Stefan Inst., Ljubljana) ; Zavrtanik, M (Ljubljana U. ; Stefan Inst., Ljubljana) The charge-collection efficiency of Si pad diodes irradiated with neutrons up to $8 \times 10^{15} \ \rm{n} \ cm^{-2}$ was measured using a $^{90}$Sr source at temperatures from -180 to -30°C. The measurements were made with diodes under forward and reverse bias. [...] 2004 - 12 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 533 (2004) 442-453 詳細記錄 - 相似記錄 2018-08-23 11:31 詳細記錄 - 相似記錄 2018-08-23 11:31
Effect of electron injection on defect reactions in irradiated silicon containing boron, carbon, and oxygen / Makarenko, L F (Belarus State U.) ; Lastovskii, S B (Minsk, Inst. Phys.) ; Yakushevich, H S (Minsk, Inst. Phys.) ; Moll, M (CERN) ; Pintilie, I (Bucharest, Nat. Inst. Mat. Sci.) Comparative studies employing Deep Level Transient Spectroscopy and C-V measurements have been performed on recombination-enhanced reactions between defects of interstitial type in boron doped silicon diodes irradiated with alpha-particles. It has been shown that self-interstitial related defects which are immobile even at room temperatures can be activated by very low forward currents at liquid nitrogen temperatures. [...] 2018 - 7 p. - Published in : J. Appl. Phys. 123 (2018) 161576 詳細記錄 - 相似記錄 2018-08-23 11:31 詳細記錄 - 相似記錄 2018-08-23 11:31
Characterization of magnetic Czochralski silicon radiation detectors / Pellegrini, G (Barcelona, Inst. Microelectron.) ; Rafí, J M (Barcelona, Inst. Microelectron.) ; Ullán, M (Barcelona, Inst. Microelectron.) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Fleta, C (Barcelona, Inst. Microelectron.) ; Campabadal, F (Barcelona, Inst. Microelectron.) Silicon wafers grown by the Magnetic Czochralski (MCZ) method have been processed in form of pad diodes at Instituto de Microelectrònica de Barcelona (IMB-CNM) facilities. The n-type MCZ wafers were manufactured by Okmetic OYJ and they have a nominal resistivity of $1 \rm{k} \Omega cm$. [...] 2005 - 9 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 548 (2005) 355-363 詳細記錄 - 相似記錄 2018-08-23 11:31
Silicon detectors: From radiation hard devices operating beyond LHC conditions to characterization of primary fourfold coordinated vacancy defects / Lazanu, I (Bucharest U.) ; Lazanu, S (Bucharest, Nat. Inst. Mat. Sci.) The physics potential at future hadron colliders as LHC and its upgrades in energy and luminosity Super-LHC and Very-LHC respectively, as well as the requirements for detectors in the conditions of possible scenarios for radiation environments are discussed in this contribution.Silicon detectors will be used extensively in experiments at these new facilities where they will be exposed to high fluences of fast hadrons. The principal obstacle to long-time operation arises from bulk displacement damage in silicon, which acts as an irreversible process in the in the material and conduces to the increase of the leakage current of the detector, decreases the satisfactory Signal/Noise ratio, and increases the effective carrier concentration. [...] 2005 - 9 p. - Published in : Rom. Rep. Phys.: 57 (2005) , no. 3, pp. 342-348 External link: RORPE 詳細記錄 - 相似記錄 2018-08-22 06:27
Numerical simulation of radiation damage effects in p-type and n-type FZ silicon detectors / Petasecca, M (Perugia U. ; INFN, Perugia) ; Moscatelli, F (Perugia U. ; INFN, Perugia ; IMM, Bologna) ; Passeri, D (Perugia U. ; INFN, Perugia) ; Pignatel, G U (Perugia U. ; INFN, Perugia) In the framework of the CERN-RD50 Collaboration, the adoption of p-type substrates has been proposed as a suitable mean to improve the radiation hardness of silicon detectors up to fluencies of $1 \times 10^{16} \rm{n}/cm^2$. In this work two numerical simulation models will be presented for p-type and n-type silicon detectors, respectively. [...] 2006 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 53 (2006) 2971-2976 詳細記錄 - 相似記錄 2018-08-22 06:27
Technology development of p-type microstrip detectors with radiation hard p-spray isolation / Pellegrini, G (Barcelona, Inst. Microelectron.) ; Fleta, C (Barcelona, Inst. Microelectron.) ; Campabadal, F (Barcelona, Inst. Microelectron.) ; Díez, S (Barcelona, Inst. Microelectron.) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Rafí, J M (Barcelona, Inst. Microelectron.) ; Ullán, M (Barcelona, Inst. Microelectron.) A technology for the fabrication of p-type microstrip silicon radiation detectors using p-spray implant isolation has been developed at CNM-IMB. The p-spray isolation has been optimized in order to withstand a gamma irradiation dose up to 50 Mrad (Si), which represents the ionization radiation dose expected in the middle region of the SCT-Atlas detector of the future Super-LHC during 10 years of operation. [...] 2006 - 6 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 566 (2006) 360-365 詳細記錄 - 相似記錄 2018-08-22 06:27
Defect characterization in silicon particle detectors irradiated with Li ions / Scaringella, M (INFN, Florence ; U. Florence (main)) ; Menichelli, D (INFN, Florence ; U. Florence (main)) ; Candelori, A (INFN, Padua ; Padua U.) ; Rando, R (INFN, Padua ; Padua U.) ; Bruzzi, M (INFN, Florence ; U. Florence (main)) High Energy Physics experiments at future very high luminosity colliders will require ultra radiation-hard silicon detectors that can withstand fast hadron fluences up to $10^{16}$ cm$^{-2}$. In order to test the detectors radiation hardness in this fluence range, long irradiation times are required at the currently available proton irradiation facilities. [...] 2006 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 53 (2006) 589-594 詳細記錄 - 相似記錄 |
Edit: My original post assumed $p$-arity rather than $p$-regularity. If the $p$ child trees of the root, rather than being (infinite) trees similar to the $p$-regular parent, are of $(p-1)$-arity, then the recursion given needs to be adapted accordingly.
Note however this previous Question and Answer, which appears to give a closed form solution.
The recursion required here is a bit messy but seems to be fairly straightforward.
Let $T_p(m)$ denote the number of rooted (labelled) subtrees of the rooted infinite $p$-arity tree which have $m$ edges and share the same root $v_0$.
Note that the Question asks about an infinite $p$-regular tree, which has arity $p$ for root $v_0$ but all other nodes, having degree $p$, have arity $p-1$. We let $\widetilde{T}_p(m)$ denote this slightly different count and express it in terms of $T_p(m)$.
Essential idea of recursion: Since the root $v_0$ must appear in each subtree, we can choose the number $k$ of the $p$ edges from $v_0$ that will appear in the subtree, and then count possible subtrees extending from those edges.
This gives a recursion on $m$ involving the set $\mathscr{W}(m-k,k)$ of weak compositions of $m-k$ with $k$ summands.
For the basis case, define $T_p(0) = 1$. Then for $m \gt 0$:
$$ T_p(m) = \sum_{k = 1}^{\min(m,p)} \binom{p}{k} \sum_{\vec{w}\in \mathscr{W}(m-k,k)} T_p(w_1)\cdot T_p(w_2) \cdot \ldots \cdot T_p(w_k) $$
Here the inner summation is indexed by weak compositions $\vec{w} = (w_1,w_2,\ldots,w_k)$ of $m-k$ with $k$ summands:
$$ w_1 + w_2 + \ldots + w_k = m-k $$
where the summands are nonnegative integers.
Finally we express the desired $\widetilde{T}_p(m)$ in terms of $T_{p-1}(m)$:
$$ \widetilde{T}_p(m) = \sum_{k = 1}^{\min(m,p)} \binom{p}{k} \sum_{\vec{w}\in \mathscr{W}(m-k,k)} T_{p-1}(w_1)\cdot T_{p-1}(w_2) \cdot \ldots \cdot T_{p-1}(w_k) $$
Added:
Do the cases $m\lt p$ and $m\gt p$ make a difference?
They do in the immediate sense that when $m\gt p$ we are restricted at the root vertex $v_0$ from using up all the edges there (there simply aren't enough to exhaust the $m$ edges of our sought-after subtrees). This shows up in the recursion as the upper limit of the outer summation being given by $\min(m,p)$ rather than depending only on $m$. |
The definition of complex Fourier series of a function is always given as the limit of symmetric partial sums
This is not so, for two reasons.
Some textbook authors explicitly define the Fourier series as the sum of two infinite series, over $n\ge 0$ and $n<0$ (e.g., Strauss, PDE, page 116). The authors of more advanced textbooks separate the process of defining a series and studying its convergence. The reason for doing this is that convergence of a Fourier series is treated in multiple ways: the same series may converge in one sense and diverge in another sense. One should not have to immediately deal with all of this just to write down the series.
Following approach 2, one says that the Fourier series is
$$\sum_{n\in\mathbb{Z}} c_ne^{2\pi i nx}$$where $c_n$ is defined as $\int_0^1 f(x)e^{-2\pi i nx}\,dx$, and the symbol $\sum$ is just that — a symbol that I put at the beginning of that line because I felt like it. I don't have to explain in what order I'm adding the terms, because I'm not actually adding them: the notation expresses the vague idea that I might decide to add them some day. No claim is made about the convergence of the series in any sense.
When it comes to proving theorems about convergence, one has to specify exactly what kind of convergence is proved. It can be convergence of symmetric partial sums, or of Cesàro means, or unconditional convergence (independent of order) in the norm of some function space, like $L^2(\mathbb{R})$ or $C_b(\mathbb{R})$, or something else.
For some convergence theorems the order is immaterial: e.g., if one shows $(c_n)\in \ell^2$, then the series of partial sums converges in $L^2$ no matter how the terms are ordered. Similarly, if one proves $(c_n)\in \ell^1$, then the series converges uniformly, regardless of order.
When proving results about
pointwise convergence it is convenient to focus on symmetric sums, because they can be expected to converge to $f(x)$. We don't want just to prove that the series converges; we want it to converge to $f(x)$. And to do that, we need some partial sums that get closer to $f(x)$. The symmetric partial sums often do that; the one-sided partial sums do not, simply because they don't include half of the series.
Some results about symmetric partial sums (e.g., pointwise convergence for Hölder functions and Carleson's theorem about a.e. convergence for $L^2$ functions) imply corresponding statements for one-sided sums. This is because the Hilbert transform $H$ flips the sign of the Fourier coefficients with negative indices, while preserving the Hölder spaces (of fractional order) and the $L^2$ space. Adding the symmetric partial sums of $f$ and $Hf$ yields a one-sided partial sum for $2f$, and conclusion follows.
Summary:
the definition of Fourier series is agnostic of the methods of summation symmetric partial sums are convenient because they can be expected to converge to $f$, and also because they can be written as the convolution with a real-valued kernel (Dirichlet's kernel) |
Of course, the logarithm here is defined on the ring region $|z|>R\ge\max\{|p|,|q|\}$ as $$\log\frac{z-p}{z-q}=\int_{z_0}^z \left(\frac1{w-p}-\frac1{w-q}\right)\mathrm d w. $$ Here the integral is along an arbitrary curve connecting $z_0$, a fixed point, to $z$ in the ring region. It's noteworthy that this logarithm is actually well defined, although not appearing so at first glance.
All I know is $\log(1+z)=z-\frac12z^2+\frac13z^3+\cdots$ when $|z|<1$ and $\log$ is chosen to be the principal branch. With this idea I can work out a naive argument: for our fractional logarithm, first rewrite the expression as $\log\frac{1-p/z}{1-q/z}$ by dividing both the numerator and denominator simultaneously (I don't know how to justify this
from the integral definition, though); then it all reduces to $\log(1-p/z)-\log(1-q/z)$, with the two logarithm both chosen as suitable branches, to which the canonical power expansion applies.
I believe I'm almost on the right track, but can't get over that confusion. Could you help me? Thanks! |
Show that if a formula $\varphi$ is independent of a propositional symbol $p$, then there exists a formula $\varphi'$ such that $\varphi' \sim \varphi$ and $$VOC(\varphi') = VOC(\varphi)\setminus \{ p\}$$ where $VOC(\psi)$ is the set of all propositional symbols that occur in $\psi$.
We say that a formula $\varphi$ is independent of a symbol $p$ is for every valuation $v$ we have that: $$[[\varphi]]^{v[p/\bot]} = [[\varphi]]^{v[p/\top]} = [[\varphi]]^v$$ that is, $[[\varphi]]^v$ does not depend on whether $v(p)=0$ or $v(p)=1$.
I am having some trouble trying prove this but I came up with a "possible proof" by induction.
For the base case, if $\varphi = \top$, then $VOC(\varphi) = VOC(\top) = 0$, and we can take $\varphi' = \varphi$ since $\varphi$ does not depend on any $p$. If $\varphi = \bot$ we can do just the same thing. Again, if $\varphi = p$, $\varphi$ depends on $p$ the hypothesis $\varphi$ does not depend on $p$ doesn't apply. Finally, if $\varphi = q$ and $q \neq p$, we can take $\varphi' = \varphi$.
Now, assume that $\varphi = \neg \psi$ and that $\varphi$ is independent of $p$. Then we must have that $\psi$ is independent of $p$ as well, and by induction hypothesis there is some $\psi' \sim \psi$ such that $VOC(\psi')=VOC(\psi) \setminus \{ p\}$. Then we take $\varphi ' = \neg \psi '$. This works, since $$VOC(\varphi) = VOC(\psi)$$ and $$VOC(\varphi') = VOC(\psi') = VOC(\psi) \setminus \{ p \} = VOC(\varphi) \setminus \{ p \}$$ Now, if $\varphi = \psi_2 \square \psi_2$ where $\square \in \{ \land, \lor, \rightarrow, \leftrightarrow \}$ and $\varphi$ is independent of $p$, then $\psi_1$ and $\psi_2$ must be independent of $p$. By induction hypothesis, there are some $\psi_1' \sim \psi_1$ and $\psi_2' \sim \psi_2$ such that $$VOC(\psi_1') = VOC(\psi_1)\setminus \{p\}$$ and $$VOC(\psi_2') = VOC(\psi_2) \setminus \{ p \}$$ In that case, we take $\varphi' = \psi_1' \square \psi_2'$ and check that it works. Since $$VOC(\varphi) = VOC(\psi_1) \cup VOC(\psi_2)$$ we have that $$\begin{align*} VOC(\varphi') &= VOC(\psi_1') \cup VOC(\psi_2') = \left( VOC(\psi_1) \setminus \{p \} \right) \cup \left( VOC(\psi_2) \setminus \{p \}\right)\\ & = \left( VOC(\psi_1) \cup VOC(\psi_2) \right) \setminus \{ p \}\\ & = VOC(\varphi) \setminus \{ p \} \end{align*}$$ which ends the proof.
I would really thank someone who could point me out if I made any mistakes here and help me with a correct proof of the statement. Thank you in advance! |
I provide an approach which involves many identities from the classical theory of elliptic and theta functions. Also I present them using Ramanujan's notation. Although the theory of elliptic and theta functions needs complex analysis to understand all the details, the formulas which I give here can be established within limits of real analysis only (and the proofs are available in my blog posts, let me know if you need them).
In what follows $k$ is a real number with $0 < k < 1$ and $k' = \sqrt{1 - k^{2}}$. Further $$K = K(k) = \int_{0}^{\pi/2}\frac{dx}{\sqrt{1 - k^{2}\sin^{2}x}}, K' = K(k'), E = E(k) = \int_{0}^{\pi/2}\sqrt{1 - k^{2}\sin^{2}x}\,dx$$ and $$q = \exp\left(-\pi\cdot\frac{K'}{K}\right)$$
Let us use Ramanujan Function $$Q(q) = 1 + 240\sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 - q^{n}}\tag{1}$$ Then we know that $$Q(q^{2}) = \left(\frac{2K}{\pi}\right)^{4}(1 - k^{2} + k^{4})\tag{2}$$ where $q = e^{-\pi K'/K}$ and we can change $q$ to $q^{2}$ if $k$ is replaced by $(1 - k')/(1 + k')$ and $K$ is replaced by $K(1 + k')/2$ Thus we have $$Q(q^{4}) = \left(\frac{K(1 + k')}{\pi}\right)^{4}\left(1 - \frac{(1 - k')^{2}}{(1 + k')^{2}} + \frac{(1 - k')^{4}}{(1 + k')^{4}}\right)$$ or $$Q(q^{4}) = \left(\frac{2K}{\pi}\right)^{4}\left(1 - k^{2} + \frac{k^{4}}{16}\right)\tag{3}$$ Lets put $q = e^{-\pi}$ to get $K' = K$ so that $k = 1/\sqrt{2}$. Then $$\sum_{n = 1}^{\infty}\frac{n^{3}}{e^{2n\pi} - 1} = \frac{Q(q^{2}) - 1}{240}, \sum_{n = 1}^{\infty}\frac{n^{3}}{e^{4n\pi} - 1} = \frac{Q(q^{4}) - 1}{240}$$ so the desired sum is $$\frac{1}{240}\{5 + 11Q(q^{2}) - 16Q(q^{4})\} = \frac{1}{48} + \frac{(2K/\pi)^{4}}{240}\{11\cdot(3/4) - 16\cdot(33/64)\} = \frac{1}{48}$$
About the values of such sums being rational I don't see them as surprising. Clearly for $q = e^{-\pi}$ we have $$Q(q^{2}) = \frac{3}{4}\left(\frac{2K}{\pi}\right)^{4}$$ And we can connect it with many other formulas like $$P(q) = 1 - 24\sum_{n = 1}^{\infty}\frac{nq^{n}}{1 - q^{n}} = \left(\frac{2K}{\pi}\right)^{2}\left(\frac{6E}{K} + k^{2} - 5\right)$$ and $$P(q^{2}) = \left(\frac{2K}{\pi}\right)^{2}\left(\frac{3E}{K} + k^{2} - 2\right)$$ so that $$2P(q^{2}) - P(q) = \left(\frac{2K}{\pi}\right)^{2}\left(1 + k^{2}\right) = \frac{3}{2}\left(\frac{2K}{\pi}\right)^{2}$$ and hence finally $$3Q(q^{2}) - \{2P(q^{2}) - P(q)\}^{2} = 0$$ And this will give a polynomial equation with rational coefficients connecting $P(q), P(q^{2}), Q(q^{2})$. Many such relations can be derived via simple manipulation of the formulas for $P, Q, R$ in terms of $k, K, E$. Just to add one last example of such sums with rational value, it can be established using some of Ramanujan's results (or some other techniques which I am not aware of) that $$\sum_{n = 1}^{\infty}\frac{n^{s}}{e^{n\pi} - 1}$$ is rational for all positive integers $s$ of the form $s = 4k + 1$ (the sum has a closed form for odd values of $s$ but not for even values of $s$).
In his paper "Modular Equations and Approximations to $\pi$" Ramanujan does the ultimate form of cancellation to obtain various series for $1/\pi$ some of which consists purely of rational terms and the most spectacular one consists of just one irrational $\sqrt{2}$ namely $$\boxed{\displaystyle \frac{1}{2\pi\sqrt{2}} = \frac{1103}{99^{2}} + \frac{27493}{99^{6}}\frac{1}{2}\frac{1\cdot 3}{4^{2}} + \frac{53883}{99^{10}}\frac{1\cdot 3}{2\cdot 4}\frac{1\cdot 3\cdot 5\cdot 7}{4^{2}\cdot 8^{2}} + \cdots}$$ See the series on my blog starting with this post for more details. |
Find irreducible and connected components of $\operatorname{Spec}(\mathbb{C}[x] \times \mathbb{C}[y])$
That's the problem 11.1 from commutative algebra course
As answered here
we can see the bijection between $\operatorname{Spec}(A_1\times A_2)$ and $\operatorname{Spec}(A_1)\sqcup\operatorname{Spec}(A_2)$ as follows: the prime ideals of $A_1\times A_2$ are of the form $\mathfrak{p}\times A_2$ where $\mathfrak p$ is a prime ideal of $A_1$, or $A_1\times\mathfrak q$ where $\mathfrak q$ is a prime ideal of $A_2$.
That means that we should find irreducible and connected components of $\operatorname{Spec}(\mathbb{C}[x])$
As discussed here: Let $A$ be a commutative ring with unit, $X = \operatorname{Spec}(A) $ with the Zariski topology. The irreducible components are $\lbrace V(p) : p\subset A \ \text{minimal prime ideal} \rbrace$ where $V(P) =\lbrace q \ \text{prime ideal } \mid p\subset q\rbrace$.
An ideal is called
prime if the complement is a multiplicative set.
I know that prime ideals of $\mathbb{C}[x]$ are the principal ideals generated by $\lbrace (x-a) | a \in \mathbb{C}\rbrace$ and $0$. $0$ is contained in all the other ideals.
The prime ideals $\mathfrak{p}$ of $\mathbb{C}[x]$ except $0$ are all minimal, so the irreducible components of $\operatorname{Spec}\mathbb{C}[x]$ are all $\lbrace V(p) \rbrace$.
Is the final answer that the connected component are the same as irreducible components and are in the bijection with $I_{x-a} \times \mathbb{C}[y]$ and $\mathbb{C}[x] \times I_{y-a}$ ?
Could you explain how this topology can be seen "in a picture"? |
In the previous tutorial you got introduced to basic probability and the rules dealing with it. Now we are equipped with the ability to calculate probability of events when they are not dependent on any other events around them. But this definitely creates a practical limitation as many events are contingent on each other in reality. This tutorial dealing with conditional probability and bayes' theorem will answer these limitations.
Conditional Probability Conditional probability as the name suggests, comes into play when the probability of occurrence of a particular event changes when one or more conditions are satisfied (these conditions again are events). Speaking in technical terms, if X and Y are two events then the conditional probability of X w.r.t Y is denoted by $$P(X|Y)$$. So when we talk in terms of conditional probability, just for an example, we make a statement like "The probability of event X given that Y has already occurred". What if X and Y are independent events? From the definition of independent events, the occurrence of event X is not dependent on event Y. Therefore $$P(X|Y) = P(X)$$. What if X and Y are mutually exclusive? As X and Y are disjoint events, the probability that X will occurs when Y has already occurred is 0. Therefore, $$P(X|Y) = 0$$ Product rule Product rule states that,
\begin{equation} P(X \cap Y) = P(X|Y)*P(Y) \end{equation}
So the joint probability that both X and Y will occur is equal to the product of two terms:
Probability that event Y occurs.
Probability that X occurs given that Y has already occurred.
From the product rule, the following can be inferred,
$$X \subseteq Y$$ implies $$P(X|Y) = P(X)/P(Y)$$
$$Y \subseteq X$$ implies $$P(X|Y) = 1$$ Note: The distributive, associative and De Morgans laws are valid for probability calculation too. The following can be inferred using these laws, $$P(X \cup Y|Z) = P(X|Z) + P(Y|Z) - P(X \cap Y|Z)$$ $$P(X^{c}|Z) = 1-P(X|Z)$$ Chain rule Generalizing the product rule leads to the chain rule. Let $$E_{1}, E_{2},....E_{n}$$ be n events. The joint probability of all the n events is given by,
\begin{equation} P(\bigcap_{i=1,..,n}E_{i}) = P(E_{n}|\bigcap_{i=1,..,n-1}E_{i})*P(\bigcap_{i=1,..,n-1}E_{i}) \end{equation}
The chain rule can be used iteratively to calculate the joint probability of any no.of events.
Bayes' theorem From the product rule, $$P(X \cap Y) = P(X|Y)P(Y)$$ and $$P(Y \cap X) = P(Y|X)P(X)$$. As $$P(X \cap Y)$$ and $$P(Y \cap X)$$ are both same, \begin{equation} P(Y|X) = \frac{P(X|Y)*P(Y)}{P(X)} \end{equation}
where, $$P(X) = P(X \cap Y) + P(X \cap Y^{c})$$ from sum rule. |
The "iterated" limit$$ \lim_{x\to 0} \lim_{y\to 0} x\sin \Bigl(\frac{1}{y}\Bigr)$$does not exists, because it does not exist the inner limit.
EDIT: my first treatment of the limit for $(x,y)\to (0,0)$ was not correct. I think that now all works well.
We have to be careful about the following limit in $\mathbb{R}^2$$$\lim_{(x,y)\to (0,0)} x\sin \Bigl(\frac{1}{y}\Bigr)$$Call for simplicity $f(x,y) = x\sin \Bigl( \frac{1}{y}\Bigr)$.
The point $(0,0)$ is an accumulation point for pairs of the form $(x,0)$ where the inner function is not defined. It turns out that the limit does not exist, because the sequence $(\frac{1}{n},0)$ converges to $(0,0)$ as $n$ approaches $+\infty$, but $f(\frac{1}{n},0)$ does not converge to any value (since $f$ is not defined at every point of the sequence); while at the same time, there are infinitely many sequences $(x_n,y_n)$ converging to $(0,0)$ such that $f(x_n,y_n)$ converges to $0$.
However, it is meaningful to consider the limit$$\lim_{\substack{(x,y)\to (0,0) \\ y \neq 0}} x\sin \Bigl(\frac{1}{y}\Bigr)$$We are simply restricting ourselves to the domain of $f$. In this region, which is $D = \mathbb{R}^2 \setminus \{y=0\}$, the inner function is everywhere defined and we can argue in a simple way: the sine is bounded everywhere and the function $x$ is infinitesimal as $(x,y)\to (0,0)$. Since this estimate holds in the whole $D$, we conclude that the above limit exists and is equal to $0$. |
I will use the standard symbols $\lor$ for OR and $\lnot$ for NOT. Let’s look first at the example. You have $$\begin{align*}
&A\lor\lnot B\tag{1}\\
&A\lor\lnot C\tag{2}\\
&C\lor D\tag{3}\\
&A\lor\lnot D\tag{4}\;.
\end{align*}$$
You can resolve two of these if one of them contains $X$ and the other $\lnot X$ for some $X$. Here that gives you two possibilities, $(2)$ and $(3)$, where you have $C$ in one and $\lnot C$ in the other, and $(3)$ and $(4)$, where you have $D$ in one and $\lnot D$ in the other. Resolving $(2)$ and $(3)$ gives you $A\lor D$; it contains $D$, and $(4)$ contains $\lnot D$, so you can resolve these two to get $A\lor A$, or simply $A$. This shows that $A$ must be true.
What if we’d resolved $(3)$ and $(4)$ to get $C\lor A$? We could then have resolved that with $(2)$, getting rid of the $C$’s, to get $A\lor A$ and therefore $A$, so we’d have ended up in the same place.
Now let’s look at Problem (a). You have
$$\begin{align*}
&A\lor B\lor C\tag{5}\\
&\lnot A\lor\lnot B\tag{6}\\
&\lnot C\lor\lnot D\tag{7}\;.
\end{align*}$$
The proposition letter $D$ appears only once, so there’s no hope of resolving it away. On the other hand, $(5)$ and $(6)$ look like
very good candidates for resolution. Unfortunately, they’re too good in a sense: after we ‘resolve out’ $A$, we have $B\lor C\lor\lnot B$, which is a tautology: it’s always true, on account of the $B\lor\lnot B$. Resolving out $B$ also leads to a tautology. (This always happens when you have more than one complementary pair, like $A$ and $\lnot A$ or $B$ and $\lnot B$, in the two expressions.) A tautology doesn’t tell us anything about the truth of the individual propositions mentioned in it, so it’s useless to us here.
Since that didn’t get us very far, let’s resolve out $C$ in $(5)$ and $(7)$ instead, getting $A\lor B\lor\lnot D$. And that’s pretty much a dead end: we could resolve with $(6)$, but we’d just get another tautology, and no other resolutions are available. Thus, we conclude that none of $A,B,C$ and $D$ is
necessarily true or false.
If in doubt, you can verify this by brute force. Is it possible for $A$ to be true? Yes: if $A$ is true, $B$ is false, and $C$ is false, $(5)-(7)$ are true no matter what $D$ is. If $A$ is false, $B$ is true, and $D$ is false, $(5)-(7)$ are true no matter what $C$ is. Thus, any of the four atomic propositions can be either true or false when $(5)-(7)$ are true.
Problem (b) is actually a bit easier; since this is homework, I’ll give you a chance to try it on your own first. You should be able to show that one of the atomic propositions $A,B,C$, and $D$ is definitely true and that one is definitely false. |
Disclaimer. I already asked this question on math.stackexchange.com without any answers or comments as of yet.
In which weak sense does the series representation of the log-characteristic function of a probability distribution converge? More specific, if $\mu$ is a probability distribution on the real line and $\phi\colon \mathbb R\to\mathbb R$ is a test function, under which conditions on $\phi$ and $\mu$ do we have the convergence: $$\int_{\mathbb R}\log\left(\int_{\mathbb R} e^{i t x} \,d \mu(x)\right) \phi(t)\,d t = \sum_{n=0}^\infty \frac{\kappa_n}{n!} \int_{\mathbb R} (it)^n \phi(t)\,dt, $$
Here $\kappa_n$ are the Taylor coefficients of the log-characteristic function, i.e., the cumulants of $\mu$.
I guess for any such result there is a trade-off between decay and regularity of $\phi$ and the tail-behaviour of $\mu$. I think the effect I am asking for might be quite delicate: Even for bounded distribution $\mu$ we can have that $\kappa_n\sim n!$ and the integral on the rhs. can certainly also grow with $n$, so if such a convergence holds, it might rely on some subtle cancellation.
Out of the two possible regimes I am rather interested in the regime, where one assumes a lot of regularity on the distribution (say, sub-Gaussian decay or compact support) and less regularity on $\phi$ (maybe $C_0^\infty$ and not analytic). |
Infinitely many non-radial solutions for the prescribed curvature problem of fractional operator
1.
Department of Mathematics, Tsinghua University, Beijing, 100084
2.
Department of Mathematics, Tsinghua University, Beijing 100084, China
Keywords:infinitely many solutions, elliptic equation, reduction., Fractional Laplacian, critical exponent. Mathematics Subject Classification:35B09, 35J6. Citation:Yuxia Guo, Jianjun Nie. Infinitely many non-radial solutions for the prescribed curvature problem of fractional operator. Discrete & Continuous Dynamical Systems - A, 2016, 36 (12) : 6873-6898. doi: 10.3934/dcds.2016099
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B. Barrios, E. Colorado, A. de Pablo and U. Sánchez, On some critical problems for the fractional Laplacian operator,,
252 (2012), 6133.
doi: 10.1016/j.jde.2012.02.023.
Google Scholar
[3]
C. Brändle, E. Colorado, A. de Pablo and U. Sánchez, A concave-convex elliptic problem involving the fractional Laplacian,,
143 (2013), 39.
doi: 10.1017/S0308210511000175.
Google Scholar
[4] [5]
X. Cabré and Y. Sire, Nonlinear equations for fractional Laplacians I: regularity, maximum principles, and Hamiltonian estimates,,
31 (2014), 23.
doi: 10.1016/j.anihpc.2013.02.001.
Google Scholar
[6]
D. M. Cao, E. Noussair and S. S. Yan, On the scalar curvature equation $-\Delta u=(1+\epsilon K)u^{(N+2)/(N-2)}$ in $\mathbbR^N$,,
15 (2002), 403.
doi: 10.1007/s00526-002-0137-1.
Google Scholar
[7] [8]
C. C. Chen and C. S. Lin, Estimate of the conformal scalar curvature equation via the method of moving planes. II,,
49 (1998), 115.
Google Scholar
[9]
C. C. Chen and C. S. Lin, Prescribing scalar curvature on $S^N$, I. A priori estimates,,
57 (2001), 67.
Google Scholar
[10]
M. del Pino, P. Felmer and M. Musso, Two-bubble solutions in the super-critical Bahri-Coron's problem,,
16 (2003), 113.
doi: 10.1007/s005260100142.
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49 (1996), 541.
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Infinitely many radial
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involving critical exponents.
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Positive solutions of nonhomogeneous fractional Laplacian problem with critical exponent.
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№ 9
All Issues On the Bernstein - Walsh-type lemmas in regions of the complex plane Abstract
Let $G \subset C$ be a finite region bounded by a Jordan curve $L := \partial G,\quad \Omega := \text{ext} \; \overline{G}$ (respect to $\overline{C}$), $\Delta := \{z : |z| > 1\}; \quad w = \Phi(z)$ be the univalent conformal mapping of $\Omega$ ont $\Phi$ normalized by $\Phi(\infty) = \infty,\quad \Phi'(\infty) > 0$. Let $A_p(G),\; p > 0$, denote the class of functions $f$ which are analytic in $G$ and satisfy the condition $$||f||^p_{A_p(G)} := \int\int_G |f(z)|^p d \sigma_z < \infty,\quad (∗)$$ where $\sigma$ is a two-dimensional Lebesque measure. Let $P_n(z)$ be arbitrary algebraic polynomial of degree at most $n$. The well-known Bernstein – Walsh lemma says that $$P_n(z)k ≤ |\Phi(z)|^{n+1} ||P_n||_{C(\overline{G})}, \; z \in \Omega. \quad (∗∗)$$ Firstly, we study the estimation problem (∗∗) for the norm (∗). Secondly, we continue studying the estimation (∗∗) when we replace the norm $||P_n||_{C(\overline{G})}$ by $||P_n||_{A_2(G)}$ for some regions of complex plane.
English version (Springer): Ukrainian Mathematical Journal 63 (2011), no. 3, pp 337-350. Citation Example: Abdullayev F. G., Aral N. D. On the Bernstein - Walsh-type lemmas in regions of the complex plane // Ukr. Mat. Zh. - 2011. - 63, № 3. - pp. 291-302. Full text |
Title:
electromagnetic electronics vs propogation/board:29-100-
Post by:
High Hopes on July 11, 2010, 09:33:38 am
Hello all,
I am new to the forum. Some years ago, I wrote a Turbo C app that models a charge traveling on an antenna which develops a standing wave. I didn't know enough about electric magnetic radiation to understand the phase relationship between the E and B fields.
So here are my questions. I originally sent this as an email to Mr. Fu-Kwun, and he suggested I post this in the forum where all could share in the knowledge.
--------
I have an EE background which has taught me that currents and voltages are always 90 degrees out of phase when inductive of capacitive devices are involved. I understand that a peak change in current will introduce a peak change in voltage. For sinusoidal currents, the peak change is on the positive and negative going slopes. So the peak voltage is at the zero crossing of the current.
So do radio waves work the same way? And if so, why are the magnetic and electric fields shown in phase in the animated graphics?
Or is this a general relativity question? Does the transformation one energy from one field into another take time to occur? Is time for this to transfer just equal to the time it takes the wave (event) to travel forward, so even though the fields are transferring energy at 90 degrees apart, the time for the new field to form (90 degrees) is equal to the time if takes for the fields to exist (propagate) in space.
And if this is true, does this explain why the waves travel forward, each field pushing the other in the only direction it can go (orthogonal to the field fronts) – forward. Are they steered this way? Is this why they are waves in the first place? Is it because the magnetic and electric fields are in phase? And they can only exist in phase in space – and cannot exist in phase in a wire?
Thanks,
Steve
Title:
Re: electromagnetic electronics vs propogation/board:29-100-
Post by:
Fu-Kwun Hwang on July 11, 2010, 02:25:06 pm
The relation you remembered is for R-L-C circuit where
for resistor: $V_R=I R$
for inductance: $V_L= L\frac{dI}{dt}$
for capacitor: $V_C=\frac{1}{C}\int I dt$
That is the reason why for AC signal $I=I_0 sin\omega t$,the voltage for inductance or capacitor are 90 degree out of phase with current.
For electromagnetic wave, it is another sets of equations (it is also related to the above equations under some conditions)
$\nabla\times\vec{E}=-\frac{\partial \vec{B}}{\partial t}$
and
$\nabla\times{B}=\mu_0 I + \mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t}$
For EM wave in free space, I=0
Combine the above two equations will give us
$\nabla \vec{E}=\mu_0\epsilon_0\frac{\partial^2\vec{E}}{\partial t^2}$
and
$\nabla \vec{B}=\mu_0\epsilon_0\frac{\partial^2\vec{B}}{\partial t^2}$
which are standard form for wave equation.
You can chek out http://en.wikipedia.org/wiki/Electromagnetic_wave for derivation
The wave can be represented with $\vec{E}=\vec{E_0} sin (\vec{k}\cdot\vec{r}-\omega t)$
The change on electric field (flux) will product magnetic field distribution ($\nabla\times\vec{B}$)at near by space.
The time derivative will produce 90 degree phase differences.
However, the curl will also produce another 90 degree phase differences.
That is why the net effect is in phase. The electric field is in the same phase with magnetic field in free space.
It is possible to produce out of phase EM wave in wave guide.
You are welcomed to check out Propagation of Electromagnetic Wave (http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=35.0)
:=) |
I understand that in order to calculate the angle between two vectors one does the arccos of the results of the dot product divided by the product of the magnitude of the two vectors. However, this procedure always returns the smallest of the two angles. What I am interested in is the angle from vector
v to vector u. This angle may excede $\pi$. I am using this in a computer program with data that I cannot inspect myself to know whever I should add $\pi$ to the angle or not. Any help would be greatly appreciated.
I understand that in order to calculate the angle between two vectors one does the arccos of the results of the dot product divided by the product of the magnitude of the two vectors. However, this procedure always returns the smallest of the two angles. What I am interested in is the angle from vector
The question only makes sense if your vectors are from a two-dimensional vector space. In that case we also have the relation $$\sin\alpha =\frac{v_xu_y-v_yu_x}{|\mathbf u||\mathbf v|}. $$ From the combined signs of $\sin$ and $\cos$ you can determine the quadrant your angle is in (so in fact you only need the sign of $v_xu_y-v_yu_x$ as additional criterion). Thus we an say that the angle from $(1,0)$ to $(0,1)$ is $\frac\pi2$, but the angle from $(0,1)$ to $(1,0)$ or that from $(1,0)$ to $(0,-1)$ is $\frac{3\pi}2$ (or $-\frac\pi2$, if one prefers negative angles to angles $>\pi$).
To do the same in three or more dimensions, one would have to introduce the distinction between clockwise and anti-clockwise rotation in the plane spanned by two given vectors $\bf u$ amd $\bf v$. There is no poper way to introduce such an orientation (as we'd expect this way to be invariant under rotations). Therefore the range $[0,\pi]$ of angles as found per $\arccos$ is all we can hope for anyway. |
I had hard time to wrap my head around Reinforcement learning, still thinking of the possible implementations, implications and ramifications, thats why there will be alot of simplifications.
To spare you the mathematical baggage required to start with basic RL instead of using MDP, probability models and so on, I will try a different way of explaining the core of Reinforcement learning by using very simple math and some bold assumptions :). Also you will understand were the basic RL formula comes from, I hope.
For starters let me see if I can surmise what RL is in one line, ready ??
Exponentially weighted averaging of discounted future rewards.
got it ? if you do, no need to read further :), but please do ...
Basically the idea of RL is to find the best
ACTION to execute given the accumulated knowledge so far.The way to do it is to do that is by using rewards and punishments to guide the algorithm towards a general goal.RL works in uncertain environment, from us is expected to have some idea how to REWARD the agent when it goes in the right direction.That is the greatest strength and weakness of RL, you have to, need, must know your rewards, most of the other requirements are more or less optional.
Rewards are expressed as a number, indicating how good or bad the action the agent took, was!.Rewards are just immediate bonus, to solve the
RL problem we need to know the VALUE of an action i.e. the long term benefit of doing something.
Here is the obligatory pictorial representation :
So the
AGENT interacts with the ENVIRONMENT transitioning from STATE to STATE. By taking different ACTIONS the agent receives REWARDS and PUNISHMENTS (i.e. negative rewards). The agent make a decision between actions depending on the long term VALUE (of future accumulated rewards).(We would not talk about POLICIES in this article to simplify the discussion).
The question then is how do we calculate an
ACTION long term VALUE, so that we can make best decision how to act ?
The most complete way would be to sum all the future rewards (expected return : Gt), from our current state for all possible paths of actions and pick the best one.
The formula for Expected return for a sequence is:$$G_t = R_{t+1} + R_{t+2} + R_{t+3} +... $$
As you may suspect that is unfeasible from computational point of view.
Another hurdle is that if we have continuous process (T = infinity),
Gt will be infinity for all paths-of-actions i.e. we can not distinguish which one is the best. The solution is to instead find the best next action to take, this way we don't have to calculate all action-paths, but just care only about near term prospects.
We will using time series again. Like we did in Threensitions.
In Threensitions we approximated the y-values in the series by a number of states
S={s1,s2,s3, ... sn} i.e. we substitute real numbers with finite discrete values. (The lesson from our previous experiment is that in most of the cases 200-300 states are more than enough to represent the data range for the test signals we tried).
This time we would use a 2D table to represent transitions from one state to another, like in markov order-1 chain. There will be a twist on that but be patient.
The RL algorithm requires us to have
ACTIONS, we will use the state transition to be our action. (In normal RL, actions can encode much richer behavior, but we will stick with this simple notion for now)
So our goal is to award the transition/action with a better score if it happens often enough. In the treensitions we used
counting for that. But as we discussed there a model with a simple 2D counting table is not rich enough to represent complex time series, that's why we used 3D cube, so that the statistics can be more sparse.
The twist with RL is that instead of counting we will use
discounting. What does this mean ? It means we add smaller and smaller parts of every next reward, instead of simply summing all the rewards.
here
"gamma" is discount percentage (0,1] .
Why would we do that ? It provides mathematical tool to solve the infinity problem and as we will see in a moment allows us to do step by step calculation.The bigger
gamma is, the more influence we receive from actions further away.
So the first trick is to use
DISCOUNTING, rather than simple COUNTING.
In a normal RL the environment will award different rewards, but in our simple case we will award any successful transition/action with reward of
1.
$$ G_t = R_{t+1} + \gamma R_{t+2} + \gamma^2 R_{t+3} + \gamma^3 R_{t+4} + .... $$$$ G_t = R_{t+1} + \gamma ( R_{t+2} + \gamma R_{t+3} + \gamma^2 R_{t+4} + ....) $$$$ G_t = R_{t+1} + \gamma G_{t+1} $$
Now what is the goal of our model ? The goal is to predict correctly what is the next state. This is what a normal transition model does too, but by using discounting we don't just look at the transition itself, but the subsequent transitions also have a say on how valuable the action is.There is one small problem with our current formula of discounting, we have to wait several steps before we apply the discounts. So let's refactore it abit :
Now that is more like it, we have to wait just one step to apply the discounted rewards. There is one additional subtlety, Gt, Gt+1 are all predictions, we will mention this again later.
The second trick
RL uses is the so called temporal difference. What does it mean ?
Like we said every transition/action has accumulated
value depending of how often it is visited (counting) and how often states that follow are visited (discounting).There will be some tendency toward an average value on the long run, depending on which path-of-actions is winning.
One way to do this is apply the full reward at every step :$$Q_t = Q_t + R_t$$
( Qis state-action score i.e. this is the value we store for every transition between state-A =to=> state-B in the table).
But this is cumulative reward, which if you remember suffers from the infinity problem.
This formula in our reward=1 scenario is equivalent to counting. In Threensitions app it would overflow much earlier i.e. 65353 (uint16 type) to be more correct. In our simple scenario we didn't have this problem, but if we had continuous signal, eventually we would reach this limit and flip.
What if instead we adjust it slightly towards an average
Q value, like this : alpha is the learning rate and describe how much of the difference between the current Q-value (the one in the table) and current Reward is applied. If the rewards trend up/down the Q will slowly follow it.
What is left now is to plug in the place of the absolute
Reward the Discounted reward we mentioned earlier i.e..
So let substitute :$$Q_t = Q_t + \alpha * (G_t - Q_t)$$$$Q_t = Q_t + \alpha * (R_{t+1} + \gamma G_{t+1} - Q_t)$$
So now the state-action (Q-value) trends toward the discounted reward.
Did you see the something ikky ... yep, when we write the algorithm we have to wait one step (t+1) before we do the calculation and then apply it backward to the previous state in the table i.e. the update happens at time
t+1, rather than time t and we update Q(st, st+1) transition.
Also we use the
predicted discounted reward because we don't have the real discounted reward. In the table we store the temporary value that tends toward the real value.
So how to we get the prediction, then! ... easy :$$G \sim max(Q)$$
let's substitute :$$Q_t = Q_t + \alpha * (R_{t+1} + \gamma * max(Q_{t+1}) - Q_t)$$
shorter for programmers :$$Q_t \mathrel{+=} \alpha * (R_{t+1} + \gamma * max(Q_{t+1}) - Q_t)$$
The
max() in the formula is the prediction part i.e. we pick the biggest Q-value between all subsequent states. That is the most probable direction the signal will go according to information we currently have.
You can visualize it like this :
Every row in the table specifies the Q-value between state and the next state. Whichever column have the highest Q-value is the next predicted state (number).
Let's rewrite the formula with states included, so that you can see what I just explained :$$Q_t(s_{t}, s_{t+1}) = Q_t(s_{t}, s_{t+1}) + \alpha * ( R_{t+1}(s_{t}, s_{t+1}) + \gamma * max(Q_{t+1}(s_{t+1}) ) - Q_t(s_{t}, s_{t+1}))$$
or in our case where the reward is always ONE, so :$$Q_t(s_{t}, s_{t+1}) = Q_t(s_{t}, s_{t+1}) + \alpha * (1 + \gamma * max( Q_{t+1}(s_{t+1})) - Q_t(s_{t}, s_{t+1}) )$$
I want to point your attention to one more thing. There is second way to interpret the Temporal difference mechanism.
If you look at the formula :$$Q_t = Q_t + \alpha (R_t - Q_t)$$$$Q_t = Q_t + \alpha R_t - \alpha Q_t$$$$Q_t = (1 - \alpha) Q_t + \alpha * R_t$$
Is exactly the same as the formula of
Exponentially weighted moving average. Now the heading of this article make sense.
DEW = Discounted Exponentially weighted : Moving average
Now that we have the model let see the implementation.
It is split in two parts QLearn which is implementation of QLearn algorithm, the formula we saw above.
Then we have the wrapper around it which is the DEW average module which is mostly graphing, stats and handling of the data.
Here is how to use it :
import matplotlib.pyplot as plt%matplotlib inline plt.rcParams['figure.figsize'] = (20.0, 10.0)import syssys.path.append('../lib')from dew_avg import *#those DataSets are not available in the repo, it is external library and will take me some time to integrate it# ... for now what you need is to load your data-set in 1D numpy array by some other means. Sorry..from data_sources.data_sets import DataSetny_taxi = DataSet(fname='../data/nyc_taxi.csv', field='passenger_count')
ny_taxi.data[:1000].max()
29993
dew = DEWAvg(nstates=200, vmin=0, vmax=30000, learn_rate=0.5, gamma=0.9)
dew.batch_train(ny_taxi.data[:1000], log=False)
dew.plot(nope=True)
==== MODEL stats ============== mape: 11.594% mae : 1709.062 rmse: 2338.999 r2: 87.713% nll: 0.989 resolution: 150.0
We can also print the Q-table, darker colors means higher value.
dew.ql.qmap.max()
2.4380155439116087
dew.ql.show_map()
So the basic idea of RL is temporal defference learning of discounted rewards.
That's all folks ..... |
In a
Cartesian Closed Category ( CCC), there exist the so-called exponential objects, written $B^A$. When a CCC is considered as a model of the simply-typed $\lambda$-calculus, an exponential object like $B^A$ characterizes the function space from type $A$ to type $B$. An exponential object is introduced by an arrow called $curry : (A \times B \rightarrow C) \rightarrow (A \rightarrow C^B)$ and eliminated by an arrow called $apply : C^B \times B \rightarrow C$ (which unfortunately called $eval$ in most texts on category theory). My questions here is: is there any difference between the exponential object $C^B$, and the arrow $B \rightarrow C$?
In a
One is
internal and the other is external.
A category $\mathcal{C}$ consists of objects and morphisms. When we write $f : A \to B$ we mean that $f$ is a morphism from object $A$ to object $B$. We may collect all morphisms from $A$ to $B$ into a
set of morphisms $\mathrm{Hom}_{\mathcal{C}}(A,B)$, called the "hom-set". This set is not an object of $\mathcal{C}$, but rather an object of the category of sets.
In contrast, an exponential $B^A$ is an object in $\mathcal{C}$. It is how "$\mathcal{C}$ thinks of its hom-sets". Thus, $B^A$ must be equipped with whatever structure the objects of $\mathcal{C}$ have.
As an example, let us consider the category of topological spaces. Then $f : X \to Y$ is a continuous map from $X$ to $Y$, and $\mathrm{Hom}_{\mathsf{Top}}(X,Y)$ is the set of all such continuous maps. But $Y^X$, if it exists, is a topological space! You can prove that the points of $Y^X$ are (in bijective correspondence with) the continuous maps from $X$ to $Y$. In fact, this holds in general: the morphisms $1 \to B^A$ (which are "the global points of $B^A$") are in bijective correspondence with morphisms $A \to B$, because $$\mathrm{Hom}(1, B^A) \cong \mathrm{Hom}(1 \times A, B) \cong \mathrm{Hom}(A, B). $$
Sometimes we get sloppy about writing $B^A$ as opposed to $A \to B$. In fact, often these two are synonyms, with the understanding that $f : A \to B$ might mean "oh by the way here I meant the other notation, so this means $f$ is a morphism from $A$ to $B$." For example, when you wrote down the currying morphism $$\textit{curry}: (A \times B \to C) \to (A \to C^B)$$ you really should have written $$\textit{curry}: C^{A \times B} \to {(C^B)}^A.$$ So we cannot really blame anyone for getting confused here. The inner $\to$ is used in the internal sense, and the outer in the external.
If we work in simply typed $\lambda$-calculus then everything is internal, so to speak. We have just a basic typing judgment "$t$ has type $B$", written as $t : B$. Because here $B$ is a type, and types correspond to objects, then we clearly have to interpet any exponentials and arrows in $B$ in the internal sense. So then, if we understand$$\textit{curry}: (A \times B \to C) \to (A \to C^B)$$as a typing judgment in the $\lambda$-calculus,
all arrows are internal, so this is the same as$$\textit{curry}: ((C^B)^A)^{C^{A \times B}}.$$I hope by now it is clear why people use $B^A$ and $A \to B$ as synonyms. |
Q. A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is $f_1$. If the speed of the train is reduced to 17 m/s, the frequency registered is $f_2$. If speed of sound is 340 m/s, then the ratio $f_1/f_2$ is :
Solution:
$f_{\text{app}} = f_{0} \left[\frac{V_{2} \pm V_{0}}{V_{2} \mp V_{s}}\right] $ $ f_{1} =f_{0} \left[\frac{340}{340-34}\right] $ $f_{2} = f_{0} \left[\frac{340}{340-17}\right] $ $\frac{f_{1}}{f_{2}} = \frac{340 -17}{340-34} = \frac{323}{306} \Rightarrow \frac{f_{1}}{f_{2}} = \frac{19}{18} $ Questions from JEE Main 2019 5. One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the central of the loop $(B_L)$ to that at the centre of the coil $(B_C)$ , i.e., $R \frac{B_L}{B_C}$ will be 10. The magnetic field associated with a light wave is given, at the origin, by $B = B_0 [\sin(3.14 \times 10^7)ct + \sin(6.28 \times 10^7)ct]$.
If this light falls on a silver plate having a work
function of 4.7 eV, what will be the maximum
kinetic energy of the photo electrons ?
$(c = 3 \times 10^{8} ms^{-1}, h = 6.6 \times 10^{-34} J-s)$ Physics Most Viewed Questions 1. An AC ammeter is used to measure current in a circuit. When a given direct current passes through the circuit, the AC ammeter reads 3 A. When another alternating current passes through the circuit, the AC ammeter reads 4 A. Then the reading of this ammeter, if DC and AC flow through the circuit simultaneously, is 9. A cannon of mass I 000 kg, located at the base of an inclined plane fires a shelI of mass 100 kg in a horizontal direction with a velocity 180 $kmh^{-1}$. The angle of inclination of the inclined plane with the horizontal is 45$^{\circ}$. The coefficient of friction between the cannon and the inclined plane is 0.5. The height, in metre, to which the cannon ascends the inclined plane as a result of the recoiI is (g = 10 $ms^{-1})$ Latest Updates Top 10 Medical Entrance Exams in India Top 10 Engineering Entrance Exams in India JIPMER Results Announced NEET UG Counselling Started NEET Result Announced KCET Result Announced KCET College Predictor JEE Main Result Announced JEE Advanced Score Cards Available AP EAMCET Result Announced KEAM Result Announced UPSEE – Online Applications invited from NRI &Kashmiri Migrants MHT CET Result Announced NEET Rank Predictor Questions Tardigrade |
63 0 1. Homework Statement
Let u(x,y) be harmonic in a simply connected domain [itex]\Omega[/itex]. Use the Cauchy-Riemann equations to obtain the formula for the conjugate harmonic
[tex]v(x,y)=\int^{(x,y)}_{x_0,y_0} (u_xdy-u_ydx)[/tex]
where [itex](x_0,y_0)[/itex] is any fixed point of [itex]\Omega[/itex] and the integration is along any path in [itex]\Omega[/itex] joining [itex](x_0,y_0)[/itex] and (x,y).
2. Homework Equations
Cauchy Riemann eqns
[tex]u_x=v_y, u_y=v_x[/tex]
3. The Attempt at a Solution
At first this just looks like a simple bit of integration but for some reason I just cannot get the result. How do I get rid of the dependance on x and y of the constants of integration?
[tex]u_x=v_y [/tex]
[tex]\Rightarrow v(x,y) = \int^y_{y_0} u_xdy[/tex]
[tex] v(x,y) = \int^x_{x_0} u_ydx[/tex]
Differentiating each wrt the other variable in an attempt to link the two eqns doesnt seem to get me anywhere... |
[This question is an extension of my question Does a positive-measure subset of the unit interval almost surely intersect a random translation of some countable subgroup of $\mathbb{R}$?. I'm asking it to help me solve my question Do the Birkhoff averages of a measurable stationary homogeneous Markov process in continuous time "converge to the right limit"?. ]
Does there exist a measurable function $F\colon [0,1]^{\mathbb{Q}\cap [0,1]} \to [0,1]$ with the property that for every measurable $f\colon [0,\infty) \to [0,1]$, for Lebesgue-almost all $\tau \geq 0$ we have $$ \int_\tau^{\tau+1} f(t) \, dt \ = \ F\left( \, (f(\tau+q))_{q \in \mathbb{Q}\cap [0,1]} \, \right) \ ? $$
Given all the results to the effect that "measurable objects are approximately topological objects", it seems highly intuitive that the answer should be
yes. In fact, it even seems intuitive to me that the function$$ F\left( \, (r_q)_{q \in \mathbb{Q}\cap [0,1]} \, \right) \ := \ \limsup_{n \to \infty} \, \frac{1}{2^n} \sum_{k=0}^{2^n-1} r_{\!\frac{k}{2^n}} $$should work, but I have not managed to prove it. |
I know that by using a package like
xcolor I can use
$\color{<color>} <math symbols>$ to typeset math symbols in my preferred color. But how can I isolate the color to specific symbols only?
Say for instance the illustrations of commented equations in Howard Anton's Calculus book have colors for underbraces and the bounding text boxes but have none for the included text.
Consider the following MWE
\documentclass[]{article}\usepackage{amsmath}\usepackage{xcolor}\begin{document}\begin{equation}\dfrac{d}{dx}[\sin(3x^2+2)]=\underbrace{\cos(3x^2+2)}_{\text{\fbox{\parbox[b][]{2cm}{Derivative of the outise evaluated at the inside}}}}\cdot \underbrace{6x}_{\text{\color{blue}{\fbox{\parbox[b][]{1.25cm}{Derivative of the inside}}}}}\end{equation}\end{document}
which outputs
How can I isolate the coloring to the underbraces and the bounding box to blue without affecting the other symbols/text? |
Contents Homework 10, ECE438, Fall 2014, Prof. Boutin Question 1
Consider a model of the vocal tract consisting of three tubes of equal length l connected to a first tube that is infinitely thin. Assume that l=Tc where c is the speed of sound and T is the period at which you sample the airflow throughout the model.
a) Obtain the transfer function of this model of the vocal tract. (You may use the matrix equations for the tube junction/time delay obtained in class without justification.)
$ r_0=\frac{A_1-A_0}{A_1+A_0}=1 $
$ r_1=\frac{A_2-A_1}{A_2+A_1} $
$ r_2=\frac{A_3-A_2}{A_3+A_2} $
$ \left( \begin{array}{c} Ad\left(z \right) \\ Cd\left(z \right) \end{array} \right)= $$ \frac{1}{2}\left( \begin{array} {cc} 1&-1 \\ -1&1 \end{array} \right) z\left( \begin{array} {cc} 1&0 \\ 0&z^{-2} \end{array} \right) \frac{1}{1+r_1} \left( \begin{array}{cc} 1&-r_1 \\ -r_1&1 \end{array} \right) z\left( \begin{array} {cc} 1&0 \\ 0&z^{-2} \end{array} \right) \frac{1}{1+r_2} \left( \begin{array}{cc} 1&-r_2 \\ -r_2&1 \end{array} \right) z\left( \begin{array} {cc} 1&0 \\ 0&z^{-2} \end{array} \right) \frac{1}{2}\left( \begin{array} {cc} 1&-1 \\ -1&1 \end{array} \right) \left( \begin{array}{c} Bd\left(z \right) \\ 0 \end{array} \right) $
$ \left( \begin{array}{c} Ad\left(z \right) \\ Cd\left(z \right) \end{array} \right)= \frac{z^{3}}{4\left(1+r_1\right)\left(1+r_2\right)} \left(\begin{array} {cc} X&Y \\ Y&X \end{array} \right) \left( \begin{array}{c} Bd\left(z \right) \\ 0 \end{array} \right) $
where $ X=\frac{r_2 \left( \frac{r_1}{z^{2}}+1 \right) + \frac{r_1+\frac{1}{z^{2}}}{z^{2}}}{z^{2}} + \frac{r_1}{z^{2}} + \frac{r_2\left(r_1+\frac{1}{z^{2}} \right)}{z^{2}} + 1 $
$ Y=\frac{r_2 \left( \frac{r_1}{z^{2}}+1 \right) + \frac{r_1+\frac{1}{z^{2}}}{z^{2}}}{z^{2}} - \frac{r_1}{z^{2}} - \frac{r_2\left(r_1+\frac{1}{z^{2}} \right)}{z^{2}} - 1 $
$ H\left( z \right) = \frac{Bd\left( z \right)}{Ad \left( z \right)}=\frac{1}{X} $
Simplify X, we get
$ H\left( z \right) = \frac{4z^{3}\left( 1+r_1 \right) \left( 1+r_2 \right)}{z^{6}+r_1r_2z^{4}+\left(r_1+r_2 \right)z^{4} + r_1r_2z^{2} + \left(r_1+r2 \right)z^{2}+1} $
b) How many formants could one create with such a model? Explain.
The denominator is a polynomial with power of 6. In this case, one will get 6 roots from the equation that the polynomial equals 0. All 3 pairs of roots can be written in the format of a+(-)b*i, and b can be 0 to indicate a pair of real roots. Thus, we can find 3 formants.
c) Explain how one would control the location of the formants with such a model.
The location of the three formants are controlled by the value of the 6 roots. The values of them vary along with the coefficients of the polynomial mentioned in part (b), while the coefficients are simply functions of $ r_1 $ $ r_2 $, etc..This means that we can control the location of formants by changing the area ratio of adjacent areas.
Question 2
Why do the poles of the transfer function of the vocal tract always come in complex conjugate pairs? Explain.
Since real systems have transfer functions with real coefficients, the poles of the vocal tract should come in complex conjugate pairs. If we write the transfer function H(z) as H(z)=P(z)/Q(z), where P(z) and Q(z) are polynomials, then the poles of the transfer function are the zeros of the polynomial Q(z). But Q(z) has real coefficients (Since the system can be written as a difference equation with real coefficients). And the zeros of a polynomial with real coefficients always come in complex conjugate pairs.
Question 3
We have seen that the transfer function of the vocal tract for voiced phonemes has poles (which create the formants).
a) What does this imply regarding the difference equation representing the system (in discrete-time)?
This implies that the difference equation must has the form
$ y[n]=\sum_{i=0}^{N-1} b_i x[n-i] -\sum_{k=1}^{M} a_k x[n-k] $
where M is the number of poles and M>0
b) Could the vocal tract be modeled using an FIR filter? Explain.
No, it must be an IIR filter as it must have poles. As explained in (a), the difference equation describing the system involves values of the output y[n] at previous times.
Questions 4
Warning: do not confuse the period of the sampling with the period of the pulse train produced by the vocal tract (1/pitch). Use different variables!
A person is pronouncing a phoneme. The pitch of the person's voice is 250Hz. The phoneme has two formants: a large one at 500 Hz, a weak one at 1.25 kHz.
You are given a digital recording of that phoneme. The sampling rate for the recording is 5kHz.
a) From the information given, can you tell the gender of the person?
pitch period = 1/pitch = 1/250Hz = 4ms.
Males usually have pitch period around 8ms and females have around 4ms.
So, it is likely to be female voice.
b) How does the gender of the person influence the location of the local maxima of the magnitude of the frequency response of the vocal tract?
The gender will not influence the location of the local maxima. It only affects the pitch frequency.
c) Sketch the graph of the magnitude of the CT Fourier transform of the phoneme. (Put three dots "..." in the inaudible region of the spectrum.) How does it compare to the graph of the magnitude of the DT Fourier transform of the digital recording of the phoneme?
d) Sketch the approximate location of the poles of the transfer function H(z) corresponding to the vocal tract of that person when he/she is pronouncing the phoneme.
Discussion
You may discuss the homework below.
write comment/question here answer will go here |
Let me introduce to you the topic of modal model theory, injecting some ideas from modal logic into the traditional subject of model theory in mathematical logic.
For example, we may consider the class of all models of some first-order theory, such as the class of all graphs, or the class of all groups, or all fields or what have you. In general, we have $\newcommand\Mod{\text{Mod}}\Mod(T)$, where $T$ is a first-order theory in some language $L$.
We may consider $\Mod(T)$ as a potentialist system, a Kripke model of possible worlds, where each model accesses the larger models, of which it is a submodel. So $\newcommand\possible{\Diamond}\possible\varphi$ is true at a model $M$, if there is a larger model $N$ in which $\varphi$ holds, and $\newcommand\necessary{\Box}\necessary\varphi$ is true at $M$, if $\varphi$ holds in all larger models.
In this way, we enlarge the language $L$ to include these modal operators. Let $\possible(L)$ be the language obtained by closing $L$ under the modal operators and Boolean connectives; and let $L^\possible$ also close under quantification. The difference is whether a modal operator falls under the scope of a quantifier.
Recently, in a collaborative project with Wojciech Aleksander Wołoszyn, we made some progress, which I’d like to explain. (We also have many further results, concerning the potentialist validities of various natural instances of $\Mod(T)$, but those will wait for another post.)
Theorem. If models $M$ and $N$ are elementarily equivalent, that is, if they have the same theory in the language of $L$, then they also have the same theory in the modal language $\possible(L)$. Proof. We show that whenever $M\equiv N$ in the language of $L$, then $M\models\varphi\iff N\models\varphi$ for sentences $\varphi$ in the modal language $\possible(L)$, by induction on $\varphi$.
Of course, by assumption the statement is true for sentences $\varphi$ in the base language $L$. And the property is clearly preserved by Boolean combinations. What remains is the modal case. Suppose that $M\equiv N$ and $M\models\possible\varphi$. So there is some extension model $M\subset W\models\varphi$.
Since $M\equiv N$, it follows by the Keisler-Shelah theorem that $M$ and $N$ have isomorphic ultrapowers $\prod_\mu M\cong\prod_\mu N$, for some ultrafilter $\mu$. It is easy to see that isomorphic structures satisfy exactly the same modal assertions in the class of all models of a theory. Since $M\subset W$, it follows that the ultrapower of $M$ is extended to (a copy of) the ultrapower of $W$, and so $\prod_\mu M\models\possible\varphi$, and therefore also $\prod_\mu N\models\possible\varphi$. From this, since $N$ embeds into its ultrapower $\prod_\mu N$, it follows also that $N\models\possible\varphi$, as desired. $\Box$
Corollary. If one model elementarily embeds into another $M\prec N$, in the language $L$ of these structures, then this embedding is also elementary in the language $\possible(L)$. Proof. To say $M\prec N$ in language $L$ is the same as saying that $M\equiv N$ in the language $L_M$, where we have added constants for every element of $M$, and interpreted these constants in $N$ via the embedding. Thus, by the theorem, it follows that $M\equiv N$ in the language $\possible(L_M)$, as desired. $\Box$
For example, every model $M$ is elementarily embedding into its ultrapowers $\prod_\mu M$, in the language $\possible(L)$.
We’d like to point out next that these results do not extend to elementary equivalence in the full modal language $L^\possible$.
For a counterexample, let’s work in the class of all simple graphs, in the language with a binary predicate for the edge relation. (We’ll have no parallel edges, and no self-edges.) So the accessibility relation here is the induced subgraph relation.
Lemma. The 2-colorability of a graph is expressible in $\possible(L)$. Similarly for $k$-colorability for any finite $k$. Proof. A graph is 2-colorable if we can partition its vertices into two sets, such that a vertex is in one set if and only if all its neighbors are in the other set. This can be effectively coded by adding two new vertices, call them red and blue, such that every node (other than red and blue) is connected to exactly one of these two points, and a vertex is connected to red if and only if all its neighbors are connected to blue, and vice versa. If the graph is $2$-colorable, then there is an extension realizing this statement, and if there is an extension realizing the statement, then (even if more than two points were added) the original graph must be $2$-colorable. $\Box$
A slightly more refined observation is that for any vertex $x$ in a graph, we can express the assertion, “the component of $x$ is $2$-colorable” by a formula in the language $\possible(L)$. We simply make the same kind of assertion, but drop the requirement that every node gets a color, and insist only that $x$ gets a color and the coloring extends from a node to any neighbor of the node, thereby ensuring the full connected component will be colored.
Theorem. There are two graphs that are elementary equivalent in the language $L$ of graph theory, and hence also in the language $\possible(L)$, but they are not elementarily equivalent in the full modal language $L^\possible$. Proof. Let $M$ be a graph consisting of disjoint copies of a 3-cycle, a 5-cycle, a 7-cycle, and so on, with one copy of every odd-length cycle. Let $M^*$ be an ultrapower of $M$ by a nonprincipal ultrafilter.
Thus, $M^*$ will continue to have one 3-cycle, one 5-cycle, one 7-cycle and on on, for all the finite odd-length cycles, but then $M^*$ will have what it thinks are non-standard odd-length cycles, except that it cannot formulate the concept of “odd”. What it actually has are a bunch of $\mathbb{Z}$-chains.
In particular, $M^*$ thinks that there is an $x$ whose component is $2$-colorable, since a $\mathbb{Z}$-chain is $2$-colorable.
But $M$ does not think that there is an $x$ whose component is $2$-colorable, because an odd-length finite cycle is not $2$-colorable. $\Box$.
Since we used an ultrapower, the same example also shows that the corollary above does not generalize to the full modal language. That is, we have $M$ embedding elementarily into its ultrapower $M^*$, but it is not elementary in the language $L^\possible$.
Let us finally notice that the Łoś theorem for ultraproducts fails even in the weaker modal language $\possible(L)$.
Theorem. There are models $M_i$ for $i\in\mathbb{N}$ and a sentence $\varphi$ in the language of these models, such that every nonprincipal ultraproduct $\prod_\mu M_i$ satisfies $\possible\varphi$, but no $M_i$ satisfies $\possible\varphi$. . Proof. In the class of all graphs, using the language of graph theory, let the $M_i$ be all the odd-length cycles. The ultraproduct $\prod_\mu M_i$ consists entirely of $\mathbb{Z}$-chains. In particular, the ultraproduct graph is $2$-colorable, but none of the $M_i$ are $2$-colorable. $\Box$ |
In this answer I will not distinguish between elements/variables and finite tuples of elements/variables (so you may read every lowercase letter as if it denotes a finite tuple).
First I should say that your definition of "definable" it's not quite right. The way you defined it, every set would be definable. The correct definition is that $A$ is definable (in $M$) if there is a formula $\phi(x,y)$ and $b \in M$ such that$$A = \{a \in M : M \models \phi(a,b)\}.$$We call $b$ the
parameters.
Let me also recall the definition of a special structure here (to introduce some notation at the same time). A structure $M$ of infinite cardinality $\kappa$ is special if it is the union $\bigcup_{i < \kappa} M_i$ of a chain $(M_i)_{i < \kappa}$ such that for every $i < \kappa$ the structure $M_i$ is $|i|^+$-saturated.
Now that is out of the way, we can actually have a look at your question.Let $A \subseteq M$ be infinite and $|A| < |M|$. Suppose that $A$ is definable by some $\phi(x, y)$ with parameters $b$. We will aim for a contradiction. Since $b$ is just a finite tuple of parameters, there will be some $\lambda < \kappa$ such that $b \in M_\lambda$. We may assume that $|A| \leq \lambda$ (otherwise replace $\lambda$ by $|A|$). Now define$$\Sigma(x) = \{x \neq a : a \in A \cap M_\lambda\} \cup \{\phi(x, b)\},$$then clearly $\Sigma(x)$ is finitely satisfiable. So we can extend $\Sigma(x)$ to a complete type $p(x)$ over $(A \cap M_\lambda) \cup \{b\}$. Since $|(A \cap M_\lambda) \cup \{b\}| < \lambda^+$ and $M_\lambda$ is $\lambda^+$-saturated we then must be able to find a realisation $c \in M_\lambda$ of $p(x)$ and thus of $\Sigma(x)$. But that means $M_\lambda \models \phi(c, b)$, so $c \in A \cap M_\lambda$ while the first part of $\Sigma(x)$ says that $c$ must be different from every element in $A \cap M_\lambda$. We thus reach our desired contradiction, and we can conclude that of $A$ is infinite and $|A| < |M|$, then $A$ cannot be definable in $M$. |
A was reading a book with this question in it:
Q.Find a quadratic polynomial, the sum of whose zeroes is 7 and their product is 12. Hence find the zeores of the polynomial. Sol. Let the polynomial be $ax^2+bx+c$ and suppose its zeroes are $\alpha$ and $\beta$ Therefore, sum of zeroes $=\alpha+\beta=\frac{-b}{a}=7$ and product of zeores $=\alpha\beta=\frac{c}{a}=12$ Take $a=LCM(12,1)=12$ Therefore $b=-7a=-7\times12=-84$ and $c=12a=12\times12=144$ So the polynomial is $12x^2-84x+144$
$\cdots$
Why have we taken $a=LCM(12,1)$? If we had taken $a=1$ then we could have got the answer without any kind of calculation. Is there any real reason for taking $a=LCM(12,1)$? |
Let me add a bit of clarity to what's going on here and/or overcomplicate things needlessly. In the end, we will be able to construct a similar-problem-production-machine that produces these kinds of annoying problems effortlessly!
Okay. Most of the interesting stuff that is going on here can be understood using certain general principles that have nothing to do with polynomials or matrices and everything to do with basic order theory. We'll need a couple of basic order-theoretic concepts. Firstly, that of a
minimum.
Definition (minimum). Let $P$ denote a preordered set. Then $a \in P$ is a minimum of $P$ iff for all $b \in P$ we have $a \lesssim b$.
Observation. If $\lambda$ is a minimum of $P$, then for all $a \in P$, we have that $a$ is a minimum of $P$ iff $a \mid \lambda$.
Proposition A0. If $a$ is a minimum of $P$ and $a' \lesssim a$, then $a'$ is a minimum of $P$.
Secondly, that of an
inflationary endofunction.
Definition (inflationary). Let $P$ denote a preordered set and $f$ denote an endofunction of $P$. Then $f$ is inflationary iff for all
$x \in P$ we have $x \lesssim f(x).$
(This is sometimes called "expansive" rather than "inflationary").
Proposition A1. Inflationary maps reflect minima.
i.e. Suppose $f : P \leftarrow P$ is an inflationary endofunction. Then for all $x \in P$, if $f(x)$ is minimum, then so too is $x$.
Moving right along.
Let $M$ denote a monoid. Then we can make $M$ into a preordered set in the following way.
Definition (divisibility). Given $a,b \in M$, define that $a \mid b$ iff the following two conditions hold.
There exists $x$ such that $ax=b.$
There exists $x$ such that $xa=b.$
All our previous ideas apply, just under a different name:
Definition (invertible). Given $a \in M$, define that $a$ is invertible iff it is a minimum of $M$ with respect to divisibility.
Observation. Given $a \in M$, $a$ is invertible iff $a \mid 1$.
Proposition B0. Given $a,b \in M$, if $b$ is invertible and $a \mid b$, then $a$ is invertible.
In particular:
Proposition B1. Given an endofunction $f$ on $M$, if $f$ is inflationary with respect to divisibility, then $f$ reflects invertibility.
It turns out that inflationary endofunctions arise whenever we're dealing with univariate polynomials without a constant term. In particular:
Claim. Let $S$ denote a commutative semiring suppose $p \in S[x].$ Then if $p$ has no constant term, it follows that for all $S$-algebras
$A$, the function $[p]_A : A \leftarrow A$ induced by substitution is inflationary with respect to divisibility.
Let us now return to your question. We're given a matrix $C \in \mathbb{R}^{n \times n}$. Rearranging the information a little, we obtain:
$$C^3-3C^2+C=-I$$
So define a function $f : \mathbb{R}^{n \times n} \leftarrow \mathbb{R}^{n \times n}$ as follows: $$f(A) = A^3-3A^2+A$$
Then $f$ is induced by the polynomial $p=x^3-3x+x$. That is, $f = [p]_{\mathbb{R}^{n \times n}}$. But since $p$ has no constant term, hence by the previous claim, we deduce that is $f$ is inflationary with respect to divisibility.
Therefore by Proposition B1, $f$ reflects invertibility.
So the proof goes like so: since $f(C)=-I$, hence $f(C) \mid -I$ (since $\mid$ is a preorder and in particular reflexive), hence $f(C)$ is invertible by Proposition B0. But since $f$ reflects invertibility, hence $C$ is invertible.
Okay, what have we achieved? Well, we've now understood the
crap out of why this occurs, and it now becomes possible to produce many similar problem for unsuspecting students. In particular:
Similar-problem production-machine. Write $[p]_{\mathbb{R}^{n \times n}}(C)=U,$ where $p$ is your favorite non-constant univariate
polynomial and $U$ is your favorite $n \times n$ invertible matrix.
Now do some obnoxious rearranging and call the resulting equation $E$.
Problem for students: show that if $E$ holds, then $C$ is invertible. |
N-vortex equilibrium theory
1.
Department of Aerospace & Mechanical Engineering and Department of Mathematics, University of Southern California, Los Angeles, CA 90089-1191
configurationmatrix $A$ obtained by requiring that all interparticle distances remain fixed in time. If the determinant of the square, symmetric $N \times N$ covariance matrix$A^T A$ is zero, there is a non-trivial nullspace of $A$ and a basis set for this nullspace can be used to determine all vortex strengths $\vec{\Gamma} \in R^N$ for which the configuration remains rigid. Optimal basis sets are obtained by using the singular value decomposition of $A$ which allows one to categorize exact equilibria, approximate equilibria, and the distance between different equilibria in the appropriate vector space, as characterized by the Frobenius norm. Keywords:Point vortex equilibria; Relative equilibria; Hamiltonian systems; Singular value decomposition; Optimal approximate equilibria; Kelvin's variational principle.. Mathematics Subject Classification:76B47; 76F20; 76M2. Citation:P.K. Newton. N-vortex equilibrium theory. Discrete & Continuous Dynamical Systems - A, 2007, 19 (2) : 411-418. doi: 10.3934/dcds.2007.19.411
[1] [2]
James Montaldi.
Bifurcations of relative equilibria near zero momentum in Hamiltonian systems with spherical symmetry.
[3]
Frederic Laurent-Polz, James Montaldi, Mark Roberts.
Point vortices on the sphere: Stability of symmetric relative equilibria.
[4]
Lyudmila Grigoryeva, Juan-Pablo Ortega, Stanislav S. Zub.
Stability of Hamiltonian relative equilibria in symmetric magnetically confined rigid bodies.
[5]
Florian Rupp, Jürgen Scheurle.
Classification of a class of relative equilibria in three body coulomb systems.
[6]
Marshall Hampton, Anders Nedergaard Jensen.
Finiteness of relative equilibria in the planar generalized $N$-body problem with fixed subconfigurations.
[7] [8] [9] [10]
Florin Diacu, Shuqiang Zhu.
Almost all 3-body relative equilibria on $ \mathbb S^2 $ and $ \mathbb H^2 $ are inclined.
[11] [12] [13] [14] [15]
Alejandra Fonseca-Morales, Onésimo Hernández-Lerma.
A note on differential games with Pareto-optimal NASH equilibria: Deterministic and stochastic models
[16] [17]
Saulo R.M. Barros, Antônio L. Pereira, Cláudio Possani, Adilson Simonis.
Spatially periodic equilibria for a non local evolution equation.
[18] [19]
Paul Georgescu, Hong Zhang, Daniel Maxin.
The global stability of coexisting equilibria for three models of mutualism.
[20]
Fethallah Benmansour, Guillaume Carlier, Gabriel Peyré, Filippo Santambrogio.
Numerical approximation of continuous traffic congestion equilibria.
2018 Impact Factor: 1.143
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At the angle $\frac\pi3$ on the unit circle, the $x$ coordinateis not merely "about $\frac12$"; it is
exactly $\frac12$.So you have already found one solution, $\theta = \frac\pi6$.
If you're looking for points where $x$ is $\frac12$, a good place to lookis the set of all points that satisfy $x=\frac12$. That set of points isa vertical line that crosses the $x$ axis at the coordinate $\frac12$.If this is belaboring obvious facts, I apologize; but the next step isto ask, how many times does that line intersect the unit circle?You found one point of intersection already, and measured the angleto it and found it was $\frac\pi3$; are there any others?
(By the way, this works for solving problems like $\sin(\alpha) = \frac12$,too. If you want to find points with sine $\frac12$, you're looking forpoints on the horizontal line $y=\frac12$.)
Another fact that you probably know, but is worth repeating:whenever you add or subtract $2\pi$ to or from an angle, you go exactlyonce around the unit circle and end up exactly where you were.So if $\cos(2\cdot\frac\pi6) = \frac12$ then also $\cos(2\cdot\frac\pi6 + 2\pi) = \frac12$,because the angles $2\cdot\frac\pi6$ and $2\cdot\frac\pi6 + 2\pi$ point at exactly the same point. So do $2\cdot\frac\pi6 - 2\pi$, $2\cdot\frac\pi6 + 4\pi$, $2\cdot\frac\pi6 - 4\pi$, $2\cdot\frac\pi6 + 6\pi$, and so forth.
And if each of those numbers in the preceding list is a possible valueof $2\theta$, what possible values of $\theta$ do they represent? |
Suppose the universe is completely empty with one sole particle trapped in it. To simplify, I will only be looking at the one dimensional case. However, all arguments are applicable for three dimensions. The solution of the Schrödinger equation $\hat{H} \psi = E \psi $ with $\hat{H} = \frac{\hat{p}^2}{2m}$ and $\quad \hat{p} = -i\hbar \frac{\mathrm d}{\mathrm d x}$ for a free particle ($V=0$) is then given by $\psi(x,t)=A\exp{i(kx-wt)}$ with constants $w=\frac{E}{\hbar}$ and $E=\frac{\hbar^2k^2}{2m}$ which can easily be shown. In order to meet the criteria of QM, $\langle \psi|\psi\rangle$ needs to be normalized (=1). If that is not possible, there is no way such a quantity can be interpreted as a probability density. However, if you try to normalize the density you will find:
\begin{align*} \langle\psi|\psi\rangle &=\int_{-\infty}^{\infty} \psi \psi^* dx \\ &= |A|^2 \int_{-\infty}^{\infty}\exp{i(kx-wt)} \exp{-i(kx-wt)}\mathrm dx \\ &=|A|^2 \int_{-\infty}^{\infty}1\cdot\mathrm dx\\ &=\infty\,. \end{align*}
Thus, it is not possible to find said normalization. Asking around, I found that such cases are normally treated as an approximation of a very wide (yet finite) potential well, or in the 3-dimensional case, a box. This allows a fairly simple solution which can be normalized. However it only represents an
approximation. In a rigorous point of view there exist no such solutions which satisfy the postulates of Quantum Mechanics. Does that mean that the laws of QM fundamentally prohibit a infinite empty universe with only one sole particle trapped in it? |
I've just one question :
How can I write a model like y = x + w,( with w a white gaussian noise) with a fixed SNR and a noise variance equal to 1. What coefficient may I have before x ?
Thanks !
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If you define the SNR as the ratio of the signal power and the noise power in dB, you have
$$SNR_{dB}=10\log \left(\frac{P_s}{P_w}\right)\tag{1}$$
where $P_s$ is the power of the desired signal and $P_w$ is the noise poiwer. If the noise $w$ has a mean of zero, then $P_w=\sigma^2_w=1$. From (1) (with $P_w=1)$ you get the desired value of $P_s$ for a given value of $SNR_{dB}$:
$$P_s=10^{SNR_{dB}/10}$$
In order to normalize the signal $x$ such that it has the desired power $P_s$ you first need to know its power $P_x$. Dividing $x$ by $\sqrt{P_x}$ will give you a unity power signal, which can then be multiplied by $\sqrt{P_s}$ in order to obtain the desired SNR:
$$s=\sqrt{P_s}\frac{x}{\sqrt{P_x}}=10^{SNR_{dB}/20}\frac{x}{\sqrt{P_x}}$$
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead? |
I'm self-learning set theory, and as an exercise, I tried to prove one of DeMorgan's laws, i.e.,
$$(S \cap T)^c=S^c \cup T^c$$
So my proof goes as follows:
Let $x \in S^c$, and $x \in T^c$. Then, $x \notin S$ and $x \notin T$. Therefore, $x$ is neither a member of $S \cup T$, nor a member of $S \cap T$. As a result, it can be seen that $(S \cap T)^c \subset S^c \cup T^c$. Being that $x$ is a member of both $(S \cap T)^c$ and $S^c \cup T^c$, it follows that $S^c \cup T^c \subset (S \cap T)^c$. Therefore, being that $(A=B) \iff (A\subset B) \land (B\subset A)$, we can conclude that $(S \cap T)^c=S^c \cup T^c$. QED. My question is, does this prove DeMorgan's law? |
I really appreciate the physical explanations made in other answers, but I want to add that
Fourier transform of the Coulomb potential makes mathematical sense, too. This answer is meant to clarify on what sense the standard calculation is valid mathematically.
Firstly, and maybe more importantly, I want to emphasize that
The Fourier transform of f is not simply just $\int{f(x)e^{-ikx}dx}$.
For an $L^1$ function (a function which is norm integrable), this is always the case but Coulomb potential is definitely not in $L^1$. So the Fourier transform of it, if it ever exists, is not expected to be the integral above.
So here comes the second question:
can Fourier transformation be defined on functons other than $L^1$?
The answer is "yes", and there are many Fourier transformations. Here are two examples.
Fourier transformation on $L^2$ functions (i.e., square integrable functions).
It turns out that the Fourier transform behaves more nicely on $L^2$ than on $L^1$, thanks to the Plancherel's theorem. However, as we mentioned above, if an $L^2$ function is not in $L^1$, then the above integral may not exist and Fourier transform is not given by that integral, either. (However, it has a simple characterization theorem, saying that in this case the Fourier transform is given by the principle-value integration of the above integral.)
Fourier transform of distributions (generalized functions)
It is in this sense that the Forier transform of Coulomb potential holds. The Coulomb potential, although not an $L^1$ or $L^2$ function, is a
distribution. So we need to use the definition of the Fourier transform to distributions in this case. Indeed, one can check the definition and directly calculate the Fourier transform of it. However, the physicists' calculation illustrates another point.
Fourier transformation on distributions (however it is defined) is continuous (under a certain topology on the distribution space, but let's not be too specific about it).
Remeber that if f is continuous, then $x_\epsilon\rightarrow x$ implies that $f(x_\epsilon)\rightarrow f(x)$.
Now $\frac{1}{r}e^{-\mu r}\rightarrow\frac{1}{r}$ when $\mu\rightarrow 0$ (again, under the "certain topology" mentioned above), and therefore continuity implies
$$\operatorname{Fourier}\left\{\frac{1}{r}e^{-\mu r}\right\}\rightarrow \operatorname{Fourier}\left\{\frac{1}{r}\right\}.$$
However, $\frac{1}{r}e^{-\mu r}$ is in $L^1$ and therefore its Fourier transformation can be computed using the integral $\int{f(x)e^{-ikx}dx}$.
Therefore, those physicists' computations make perfect mathematical sense, but it's on Fourier transform of distributions, which is much more general than that on $L^1$ functions.
Wish this answer can build people's confidence that the Coulomb potential Fourier transformation problem is not only physically reasonable but also mathematically justifiable. |
From the very first example of a matrix presented in the "Matrix Library" section 57.1 of the TikZ & PGF manual, I expected the following to result in an arrow connecting the "Signatures" cell to the "Transition" cell in the picture.
\documentclass{standalone}\usepackage{tikz}\usetikzlibrary{matrix,fit,positioning,arrows}\begin{document}\begin{tikzpicture}[every node/.style={draw, anchor=west}] % added anchor=west\matrix [matrix of nodes,draw=red,column sep=1cm,name=m] { \node [xshift=1mm] {\textbf {Transition Record}};\\ % added xshift=1mm \node [align=left] {\textbf{Signatures} \\ signed by Ted};\\ \node [align=left] {\textbf{Transition} \\ \(\tau\)};\\ \node [align=left] {\textbf{End State} \\ \(\mathtt{Alice2}\mid\mathtt{Carol2}\mid\mathtt{Ted}\)};\\};\draw[-latex,bend right] (m-2-1.north west) to (m-4-1.south west);\end{tikzpicture}\end{document}
This results in the errors:
Package pgf Error: No shape named m-4-1 is known. ...] (m-2-1.north west) to (m-4-1.south west) Package pgf Error: No shape named m-2-1 is known. ...draw[-latex,bend right] (m-2-1.north west)
On the other hand, if I replace the \draw command with
\draw[-latex,bend right] (m.north west) to (m.south west);
I get exactly the arrow I would expect.
Given that the name "m" is known, it is not clear to me why derived names, "m-1-1", "m-2-1", etc. would not be. This is especially the case given the above cited example from the manual. Apart from getting some insight as to why I am unable to reference the "unknown" shape names, I would like to understand the rules regarding the introduction of names and the scopes within which they are available to reference.
Thank you in advance. |
Difference between revisions of "Fujimura.tex"
Line 6: Line 6:
(Kobon Fujimura is a prolific inventor of puzzles, and in [http://www.puzzles.com/PuzzlePlayground/CoinsAndTriangles/CoinsAndTriangles.htm this] puzzle asked the related question of eliminating all equilateral triangles.)
(Kobon Fujimura is a prolific inventor of puzzles, and in [http://www.puzzles.com/PuzzlePlayground/CoinsAndTriangles/CoinsAndTriangles.htm this] puzzle asked the related question of eliminating all equilateral triangles.)
−
The
+
The table was formed mostly by computer searches for optimal solutions. We also found human proofs for most of them (see {\tt http://michaelnielsen.org/polymath1/index.php?title=Fujimura's\_problem}).
\begin{figure}
\begin{figure}
Line 22: Line 22:
For any equilateral triangle $(a+r,b,c)$,$(a,b+r,c)$ and $(a,b,c+r)$, the value $y+2z$ forms an arithmetic progression of length 3. A Behrend set is a finite set of integers with no arithmetic progression of length 3 (see {\tt http://arxiv.org/PS\_cache/arxiv/pdf/0811/0811.3057v2.pdf}). By looking at those triples $(a,b,c)$ with $a+2b$ inside a Behrend set, one can obtain the lower bound of $\overline{c}^\mu_n \geq n^2 exp(-O(\sqrt{\log n}))$.
For any equilateral triangle $(a+r,b,c)$,$(a,b+r,c)$ and $(a,b,c+r)$, the value $y+2z$ forms an arithmetic progression of length 3. A Behrend set is a finite set of integers with no arithmetic progression of length 3 (see {\tt http://arxiv.org/PS\_cache/arxiv/pdf/0811/0811.3057v2.pdf}). By looking at those triples $(a,b,c)$ with $a+2b$ inside a Behrend set, one can obtain the lower bound of $\overline{c}^\mu_n \geq n^2 exp(-O(\sqrt{\log n}))$.
−
It can be shown by a
+
It can be shown by a corners theorem' of Ajtai and Szemeredi \cite{ajtai} that $\overline{c}^\mu_n = o(n^2)$ as $n \rightarrow \infty$.
An explicit lower bound is $3(n-1)$, made of all points in $\Delta_n$ with exactly one coordinate equal to zero.
An explicit lower bound is $3(n-1)$, made of all points in $\Delta_n$ with exactly one coordinate equal to zero.
An explicit upper bound comes from counting the triangles. There are $\binom{n+2}{3}$ triangles, and each point belongs to $n$ of them. So you must remove at least $(n+2)(n+1)/6$ points to remove all triangles, leaving $(n+2)(n+1)/3$ points as an upper bound for $\overline{c}^\mu_n$.
An explicit upper bound comes from counting the triangles. There are $\binom{n+2}{3}$ triangles, and each point belongs to $n$ of them. So you must remove at least $(n+2)(n+1)/6$ points to remove all triangles, leaving $(n+2)(n+1)/3$ points as an upper bound for $\overline{c}^\mu_n$.
Latest revision as of 06:46, 27 July 2009
\section{Fujimura's problem}\label{fujimura-sec}
Let $\overline{c}^\mu_n$ be the size of the largest subset of the trianglular grid $$\Delta_n := \{(a,b,c)\in {\mathbb Z}^3_+ : a+b+c = n\}$$ which contains no equilateral triangles $(a+r,b,c), (a,b+r,c), (a,b,c+r)$ with $r>0$. These are upward-pointing equilateral triangles. We shall refer to such sets as 'triangle-free'. (Kobon Fujimura is a prolific inventor of puzzles, and in this puzzle asked the related question of eliminating all equilateral triangles.)
The table in Figure \ref{lowFujimura} was formed mostly by computer searches for optimal solutions. We also found human proofs for most of them (see {\tt http://michaelnielsen.org/polymath1/index.php?title=Fujimura's\_problem}).
\begin{figure} \centerline{ \begin{tabular}{l|llllllllllllll} $n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13\\ \hline $\overline{c}^\mu_n$ & 1 & 2 & 4 & 6 & 9 & 12 & 15 & 18 & 22 & 26 & 31 & 35 & 40 & 46 \end{tabular} } \label{lowFujimura} \caption{Fujimura numbers} \end{figure}
For any equilateral triangle $(a+r,b,c)$,$(a,b+r,c)$ and $(a,b,c+r)$, the value $y+2z$ forms an arithmetic progression of length 3. A Behrend set is a finite set of integers with no arithmetic progression of length 3 (see {\tt http://arxiv.org/PS\_cache/arxiv/pdf/0811/0811.3057v2.pdf}). By looking at those triples $(a,b,c)$ with $a+2b$ inside a Behrend set, one can obtain the lower bound of $\overline{c}^\mu_n \geq n^2 exp(-O(\sqrt{\log n}))$.
It can be shown by a `corners theorem' of Ajtai and Szemeredi \cite{ajtai} that $\overline{c}^\mu_n = o(n^2)$ as $n \rightarrow \infty$.
An explicit lower bound is $3(n-1)$, made of all points in $\Delta_n$ with exactly one coordinate equal to zero.
An explicit upper bound comes from counting the triangles. There are $\binom{n+2}{3}$ triangles, and each point belongs to $n$ of them. So you must remove at least $(n+2)(n+1)/6$ points to remove all triangles, leaving $(n+2)(n+1)/3$ points as an upper bound for $\overline{c}^\mu_n$. |
I thought I was done writing about this topic, but it just keeps coming back. The internet just cannot seem to leave this sort of problem alone:
I don't know what it is about expressions of the form \(a\div b(c+d)\) that fascinates us as a species, but fascinate it does. I've written about this before (as well as why "PEMDAS" is terrible), but the more I've thought about it, the more sympathy I've found with those in the minority of the debate, and as a result my position has evolved somewhat.oomfies solve this pic.twitter.com/0RO5zTJjKk— em ♥︎ (@pjmdolI) July 28, 2019
So I'm going to go out on a limb, and claim that the answer
shouldbe \(1\).
Before you walk away shaking your head and saying "he's lost it, he doesn't know what he's talking about", let me assure you that I'm obviouly not denying the left-to-right convention for how to do explicit multiplication and division. Nobody's arguing that.* Rather, there's something much more subtle going on here.
What we may be seeing here is evidence of a mathematical "language shift".
It's easy to forget that mathematics did not always look as it does today, but has arrived at its current form through very human processes of invention and revision. There's an excellent page by Jeff Miller that catalogues the earliest recorded uses of symbols like the operations and the equals sign -- symbols that seem timeless, symbols we take for granted every day.
People also often don't realize that this process of invention and revision still happens to this day. The modern notation for the floor function is a great example that was only developed within the last century. Even today on the internet, you occasionally see discussions in which people debate on how mathematical notation can be improved. (I'm still holding out hope that my alternative notation for logarithms will one day catch on.)
Of particular note is the evolution of grouping symbols. We usually think only of parentheses (as well as their variations like square brackets and curly braces) as denoting grouping, but an even earlier symbol used to group expressions was the
vinculum-- a horizontal bar found over or under an expression. Consider the following expression: \[3-(1+2)\] If we wrote the same expression with a vinculum, it would look like this: \[3-\overline{1+2}\] Vincula can even be stacked: \[13-\overline{\overline{1+2}\cdot 3}=4\] This may seem like a quaint way of grouping, but it does in fact survive in our notation for fractions and radicals! You can even see both uses in the quadratic formula: \[x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\]
Getting back to the original problem, what I think we're seeing is evidence that
concatenation-- placing symbols next to each other with no sort of explicit symbol -- has become another way to represent grouping.
"But wait", you might say, "concatenation is used to represent
multiplication, not grouping!" That's certainly true in many cases, for example in how we write polynomials. However, there are a few places in mathematics that provide evidence that there's more to it than that.
First of all, as a beautifully-written Twitter thread by EnchantressOfNumbers (@EoN_tweets) points out, we use concatenation to show a special importance of grouping when we write out certain trigonometric expressions without putting their arguments in parentheses. Consider the following identity:
\[\sin 4u=2\sin 2u\cos 2u\] When we write such an equation, we're saying that not only do \(4u\) and \(2u\) represent multiplications, but that this grouping is so tight that they constitute the entire arguments of the sine and cosine functions. In fact, the space between \(\sin 2x\) and \(\cos 2x\) can also be seen as a somewhat looser form of concatention. Then again, so does the space between \(\sin\) and \(x\), which represents a different thing -- the connection of a function to its argument. Perhaps this is why the popular (and amazing) online graphing calculator Desmos is only so permissive when it comes to parsing concatenation:
An even more curious case is
mixed numbers. When writing mixed numbers, concatenation actually stands for addition, not multiplication. \[3\tfrac{1}{2}=3+\tfrac{1}{2}\] In fact, concatenation actually makes addition come beforemultiplication when we multiply mixed numbers! \[3\tfrac{1}{2}\cdot 5\tfrac{5}{6}=(3+\tfrac{1}{2})\cdot(5+\tfrac{5}{6})=20\tfrac{5}{12}\]
Now, you may feel that this example shows how mixed numbers are an inelegance in mathematical notation (and I would agree with you). Even so, I argue that this is evidence that we
fundamentallyview concatenation as a way to represent grouping. It just so happens that, since multiplication takes precedence over addition anyway in the absence of other grouping symbols, we use concatenation when we write it. This all stems from a sort of "laziness" in how we write things -- - laying out precedence rules allows us to avoid writing parentheses, and once we've established those precedence rules, we don't even need to write out the multiplication at all.
So how does the internet's favorite math problem fit into all this?
The most striking feature of the expression \(8\div 2(2+2)\) is that
it's written all in one line.
Mathematical typesetting is difficult. LaTeX is powerful, but has a steep learning curve, though various other editors have made it a bit easier, such as Microsoft Word's Equation Editor (which has much improved since when I first used it!). Calculators have also recognized this difficulty, which is why TI calculators now have MathPrint templates (though its entry is quite clunky compared to Desmos's "as-you-type" formatting via MathQuill).
Even so, all of these input methods exist in very specific applications. What about when you're writing an email? Or sending a text? Or a Facebook message? (If you're wondering "who the heck writes about math in a Facebook message", the answer at least includes "students who are trying to study for a test".) The evolution of these sorts of media has led to the importance of one-line representations of mathematics with easily-accessible symbols. When you don't have the ability (or the time) to neatly typeset a fraction, you're going to find a way to use the tools you've got. And that's even more important as we realize that
everybodycan (and should!) engage with mathematics, not just mathematicians or educators.
So that might explain why a physics student might type "hbar = h / 2pi", and others would know that this clearly means \(\hbar=\dfrac{h}{2\pi}\) rather than \(\hbar=\dfrac{h}{2}\pi\). Remember, mathematics is not about just answer-getting. It's about communication of those ideas. And when the medium of communication limits how those ideas can be represented, the method of communication often changes to accomodate it.
What the infamous problem points out is that while almost nobody has laid out any explicit rules for how to deal with concatenation, we seem to have developed some implicit ones, which we use without thinking about them. We just never had to deal with them until recently, as more "everyday" people communicate mathematics on more "everyday" media.
Perhaps it's time that we address this convention explicitly and admit that
concatenation really has become a way to represent grouping, just like parentheses or the vinculum. This is akin to taking a more descriptivist, rather than prescriptivist, approach to language: all we would be doing is recognizing that this is alreadyhow we do things everywhere else.
Of course, this would throw a wrench in PEMDAS, but that just means we'd need to actually talk about the mathematics behind it rather than memorizing a silly mnemonic. After all, as inane as these internet math problems can be, they've shown that (whether they admit it or not) people really
dowant to get to the bottom of mathematics, to truly understand it.
I'd say that's a good thing.
* If your argument for why the answer is \(16\) starts with "Well, \(2(2+2)\) means \(2\cdot(2+2)\), so...", then you have missed the point entirely. |
On positive solution to a second order elliptic equation with a singular nonlinearity
1.
Department of Applied Mathematics, Bauman Moscow State Technical University, 2-aya Baumanskaya, 5, 105005 Moscow, Russian Federation
$ (a_{ij}(x)u_{x_i})_{x_j}+p(x)|x|^su^{-\sigma}=0, x\in\Omega \setminus \{ O\}, $
where $\sigma >0$, $s$ is any real number, and
$\Omega\subset R^n$, $n\ge 3$ is a bounded domain, which
contains the origin $O$.
The aim of this paper is to establish existence, nonexistence and behavior of positive weak solutions near the isolated singularity $O$. Keywords:existence, singular point, positive solution, singular nonlinearity, Elliptic equation, nonexistence.. Mathematics Subject Classification:Primary: 35J15, 35J61, 35J75; Secondary: 35A01, 35Bx. Citation:Galina V. Grishina. On positive solution to a second order elliptic equation with a singular nonlinearity. Communications on Pure & Applied Analysis, 2010, 9 (5) : 1335-1343. doi: 10.3934/cpaa.2010.9.1335
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Positive solution to extremal Pucci's equations with singular and gradient nonlinearity.
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Baishun Lai, Qing Luo.
Regularity of the extremal solution for a fourth-order elliptic problem with singular nonlinearity.
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Zongming Guo, Xuefei Bai.
On the global branch of positive radial solutions of an elliptic problem with singular nonlinearity.
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Zongming Guo, Yunting Yu.
Boundary value problems for a semilinear elliptic equation with
singular nonlinearity.
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Existence, nonexistence and uniqueness of positive stationary solutions of a singular Gierer-Meinhardt system.
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Existence and uniqueness of singular solutions for elliptic equation on the hyperbolic space.
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Xiumei Deng, Jun Zhou.
Global existence and blow-up of solutions to a semilinear heat equation with singular potential and logarithmic nonlinearity.
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Yayun Li, Yutian Lei.
On existence and nonexistence of positive solutions of an elliptic system with coupled terms.
[13]
Zongming Guo, Zhongyuan Liu, Juncheng Wei, Feng Zhou.
Bifurcations of some elliptic problems with a singular
nonlinearity via Morse index.
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Emil Novruzov.
On existence and nonexistence of the
positive solutions of non-newtonian filtration equation.
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Kin Ming Hui, Sunghoon Kim.
Existence of Neumann and singular solutions of the fast diffusion equation.
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Existence results to a quasilinear and singular parabolic equation.
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Xiaomei Sun, Wenyi Chen.
Positive solutions for singular elliptic equations with critical Hardy-Sobolev exponent.
[19]
Tokushi Sato, Tatsuya Watanabe.
Singular positive solutions for a fourth order elliptic problem in $R$.
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Existence of nonnegative solutions to singular elliptic problems, a variational approach.
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Contents Practice Question on the Nyquist rate of a signal
Is the following signal band-limited? (Answer yes/no and justify your answer.)
$ x(t)= 7 \frac{\sin (3 \pi t)}{\pi t} \ $>
If you answered "yes", what is the Nyquist rate for this signal?
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
Yes, this signal is band limited. It is a sinc function, and its Fourier transform can be found using the table of formulas in the textbook on page 329.
$ \mathcal X (\omega) = \begin{cases} 1 & \Big|\omega\Big| < 3\pi \\ 0 & \mbox{otherwise} \end{cases} $
This is band limited.
In addition, the $ \omega_m $ is $ 3\pi $.
Therefore the Nyquist rate for this signal is $ 6\pi $.
--Cmcmican 23:11, 30 March 2011 (UTC)
Instructor's comments. You got the correct Nyquist rate, but there is a small mistake in the Fourier transform. Since the mistake is a non-zero constant factor, it does not change the bandwidth of the signal, and therefore you were able to obtain the correct max frequency of the signal. It would be ok to say that the Fourier transform is a non-zero constant multiple of a low-pass filter with gain 1 and cutoff $ 3 \pi $ and conclude from there; you would get full credit because finding the constant is not necessary to answer the question. -pm Answer 2
From observation we can see $ x(t)= 7 \frac{\sin (3 \pi t)}{\pi t} $ is a sinc function with $ W = 3 \pi $ and multiplied by a constant $ C = 7 $ Therefore the Fourier Transform can be taken from the table of transforms as
$ \mathcal X (\omega) =\left\{\begin{array}{ll}7, & \text{ if }|\omega| <3 \pi,\\ 0, & \text{else.}\end{array} \right. \ $
From here it is fairly evident this function is band limited, with $ \omega_m = 3\pi $.
Since $ \omega_s \ge 2\omega_m $ must hold true for Nyquist to be satisfied, the Nyquist rate must be $ 6\pi $
--Darichar 21:12, 12 April 2011 (UTC)
TA's comment: Your approach is correct. Your only mistake is that you transfered the constant unintentionally wrong. The constant should be 7 and not $ \frac{2}{7} $. Answer 3
Write it here. |
I'm calculating molar changes in thermodynamic properties due to reactions between gasses (assumed to be ideal gases). I can calculate $\Delta H$ easily enough, because it's just $\sum_i \nu_i\Delta_f H^\circ_i$, with $\nu_i$ the stoichiometric coefficients. $\Delta G$ (at standard pressure) can be calculated from $\sum_i \nu_i\mu_i$, with $\mu_i = \Delta_f G^\circ_i + RT\log[i]$. Since $\Delta G = \Delta H - T\Delta S$, I can calculate $\Delta S$ as $\frac{1}{T}(\Delta H - \Delta G)$.
So far so good. But I'm inexperienced in working with tabulated quantities (I'm a more of a theoretical physicist than a chemist) and keep making mistakes with signs and units. So I figured that as a sanity check I would calculate $\Delta S$ directly, using tabulated values of $S^\circ$. But then I realised I don't know how to do this, and it seems that none of the textbooks on my desk explain it either.
Based mostly on intuition, it seems like it should be $\Delta S = \sum_i\nu_i(S^\circ_i + R\log[i])$, with the $R\log[i]$ term having something to do with the entropy of mixing. For the example reaction I chose, this gave something with the approximately correct magnitude but the wrong sign (-15.6 instead of 14.2, which is what I get by calculating it from the Gibbs energy and the enthalpy).
So my question is, what is the correct way to calculate the entropy change due to an ideal gas reaction if, for some reason, you only have access to the concentrations and the standard entropies of the reactants? |
eISSN:
2163-2480 Evolution Equations & Control Theory
June 2015 , Volume 4 , Issue 2
Special issue dedicated to Professor Abdelhaq El Jai on the occasion of his retirement
Select all articles
Export/Reference:
Abstract:
This special issue of the Journal of Evolution Equations and Control Theory (EECT) is dedicated to Professor Abdelhaq El Jai on the occasion of his retirement and in celebration of his significant achievements in the field of control and distributed parameter systems theory, both in his own research and in his leadership in the development of a Moroccan research community in DPS.
For more information please click the “Full Text” above.
Abstract:
In this paper, we present a predator-prey model with modified Leslie-Gower and Beddington-DeAngelis functional response. The model is governed by a two dimensional reaction diffusion system defined on a disc domain. The conditions of boundedness, existence of a positively invariant and attracting set are proved. Sufficient conditions of local and global stability of the positive steady state are established. In the end, we carry out some numerical simulations in order to illustrate our theoretical results and to interpret how biological processes affect spatio-temporal pattern formation.
Abstract:
This paper investigates two important questions for a class of partially observed infinite-dimensional semilinear systems.The first one is the design of an exponential Luenberger-like observer for this class of systems. Then, the stabilization problem around a desired equilibrium profile of the system is solved, yielding a compensator based on the Luenberger-like observer. Finally, the main result is applied to a nonisothermal chemical plug flow reactor model. The approach is illustrated by numerical simulations. The paper also gives a short overview of selected works considering the same questions for linear distributed parameter systems.
Abstract:
This paper is devoted to the application of the input/state-invariant linear quadratic (LQ) problem in order to solve the problem of coexistence of species in competition in a chemostat. The methodology that is used has for objective to guarantee the local positive input/state-invariance of the nonlinear system which describes the chemostat model by ensuring the input/state-invariance of its linear approximation around an equilibrium. This is achieved by applying an appropriate LQ-optimal control to the system, following two different approaches, namely a receding horizon method and an inverse problem approach.
Abstract:
We consider linear lumped control systems of the form $y'(t)=Ay(t)+Bu(t)$ where $A \in \mathbb{R}^{m\times m}$, $B \in \mathbb{R}^{m\times p}$. Taking into account eventual control constraint (such as saturation), we study the problem of controllability by using a general variational approach. The results are applied to the following saturation constraints on the control $u(t)=(u_{1}(t), ..., u_{p}(t))$: (i) the quadratic one specified by $\underset{j=1}{\overset{p}\sum}\left|u_{j}(t)\right|^{2} \leq 1$ for all $0\leq t\leq T$ and (ii) the polyhedral one characterized by $\underset{1 \leq j \leq p}{\max}\left|u_{j}(t)\right| \leq 1$ for all $0\leq t\leq T$.
Abstract:
This paper aims to establish necessary conditions for sensors structure (number and location) in order to obtain regional boundary gradient observability for hyperbolic system. The obtained results are applied to a two-dimensional diffusion process considering various types of sensors. Also, we will explore an approach that can reconstruct the gradient on a part $\Gamma$ of the boundary $\partial\Omega$ of the evolution domain $\Omega$. The simulations illustrate the established results and lead to some conjectures.
Abstract:
The article studies the nutrient transfer mechanism and its control for mixed cropping systems. It presents a mathematical analysis and optimal control of the absorbed nutrient concentration, governed by a transport-diffusion equation in a bounded domain near the root system, satisfying to the Michaelis-Menten uptake law.
The existence, uniqueness and positivity of a solution (the absorbed concentration) is proved. We also show that for a given plant we can determine the optimal amount of required nutrients for its growth. The characterization of the optimal control leading to the desired concentration at the root surface is obtained. Finally, some numerical simulations to evaluate the theoretical results are proposed.
Abstract:
A first extension of the IDA-PBC control synthesis to infinite dimensional port Hamiltonian systems is investigated, using the same idea as for the finite dimensional case, that is transform the original model into a closed loop target Hamiltonian model using feedback control. To achieve this goal both finite rank distributed control and boundary control are used. The proposed class of considered port Hamiltonian distributed parameters systems is first defined. Then the matching equation is derived for this class before considering the particular case of damping assignment on the resistive diffusion example, for the radial diffusion of the poloidal magnetic flux in tokamak reactors.
Abstract:
The aim of this paper is to study the optimal control problem for finite dimensional bilinear systems with bounded controls. We characterize an optimal control that minimizes a quadratic cost functional using Pontryagin's minimum principle, we derive sufficient conditions of uniqueness from the fixed point theorem, and we develop an algorithm that allows to compute the optimal control and the associated states. Our approach is applied to a cancer treatment by chemotherapy in order to determine the optimal dose of a killing agent.
Abstract:
In this article we show that systems properties of the systems governed by the second order differential equation $\frac{d^{2}w}{dt^{2}}=-A_{0}w$ and the first order differential equation $\frac{dz}{dt}=iA_{0}z$ are related. This can be used to show that, for instance, exact observability of the $N$-dimensional wave equation implies the similar property for the $N$-dimensional Schrödinger equation.
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I only have a partial answer for 1. and a hopefully non-confusing answer to 2.
To start with, let us work with the fundamental groupoid, which is more, ahem, fundamental and better suited to generalisation. In particular, we can consider the set $\pi^J(X,a,b)$ of homotopy classes (rel endpoints) of maps $(J,0,1) \to (X,a,b)$, which is more natural in the setting you outline. This is, assuming it isn't empty, a torsor for the groups $\pi^J(X,a,a)$ and $\pi^J(X,b,b)$, so you're not really losing too much. But the more important structure is the whole groupoid.
The unit interval is at least weakly initial in the category of
path-connected bipointed spaces and homotopy classes of maps (and we always have a torsor as above). If you don't assume path-connected, then the two-point set (with any of its topologies) can be allowed, but is completely useless in measuring homotopy. This is an important fact using $[0,1]$, and this can't be derived from formal homotopy theory. One could define $J$-connectedness for other bipointed spaces $J$, but the utility of such a definition is debatable unless you put in extra conditions, like making it a cylinder object.
The 'reason' we get a fundamental groupoid is that $[0,1]$ is an $A_\infty$ topological cogroupoid, namely a groupoid object in $Top^{op}$, up to homotopy, and then coherence of that up to homotopy, and so on, all the way up. Woah, I hear you say, that's a bit extreme. But it is true, and we can just focus on the first few layers.
First, we have a cocomposition $[0,1] \to [0,1]\sqcup_{1,0}[0,1]$ and a coidentity $[0,1] \to \ast$. Then instead of coassociativity, which would be the equality of the to obvious maps$$[0,1] \to [0,1]\sqcup_{1,0}[0,1]\sqcup_{1,0}[0,1],$$we have a homotopy between these two maps. We also have a map$$[0,1] \sqcup_{1,0}[0] \to [0,1]$$expressing the identity on the right, and a similar one on the left. Again, these aren't equal to the identity maps of $[0,1]$, but are homotopic to them. And again, we have coinverses up to homotopy. The choices of all these homotopies aren't important (although you can look up representatives in any book on algebraic topology), because the spaces of such homotopies are contractible.
When we want to involve another space and actually get $\Pi_1(X)$, what we do is hom this topological $A_\infty$-cogroupoid into the space $X$, and get an $A_\infty$-groupoid, and then we truncate it to a groupoid, by quotienting out by these homotopies that we have chosen (but remember the choices are unimportant). It is important that $[0,1]$ is path-connected, because this makes the $A_\infty$-cogroupoid contractible in certain technical ways which are important for generalisations to higher categories (most of the ideas in this answer come from Todd Trimble's work). For instance, in my thesis I defined a certain sort of fundamental bigroupoid which could be applied to topological stacks, and I relied heavily on the $A_\infty$-cogroupoid structure, because it was the only way I could prove I even
had a fundamental bigroupoid (I confess I did have much more complicated interval objects than here). |
I was asked to prove that: $\lim_{x\to 2} (x^2-3x)=-2$ using the $\epsilon, \delta$ deifnition
I started to try and solve the epsilon inequality in this manner:
for every $\epsilon>0$ there exist $\delta >0$ such that if:
$0<|x-2|<\delta$ then $|(x^2-3x)-(-2)|<\epsilon$
Then I started to manipulate my epsilon inequality in order to get a delta in terms of epsilon by doing the following:
$$|(x^2-3x)-(-2)|=|x^2-3x+2|=|(x-2)(x-1)|=|x-2||x-1|$$
Now, we must see that
$$|x-1|=|x-2+1| ≤ |x-2|+1$$
And thus, that: $$|x-2||x-2+1|≤ (|x-2|+1)|x-2|$$
After this step I get stuck. I do not know how to proceed or how to use the equations I have in order to get delta in terms of epsilon.
Thanks in advance for the help. |
In this problem, we will find the unconstrained portfolio allocation where we introduce the weighting parameter $\lambda(0 \leq \lambda \leq$ 1)and minimize the $\lambda * risk - (1-\lambda)* return$. By varying the values of $\lambda$, we trace out the efficient frontier.
Suppose that we know the mean returns $R \in \mathbf{R}^n$ of each asset and the covariance $Q \in \mathbf{R}^{n \times n}$ between the assets. Our objective is to find a portfolio allocation that minimizes the
risk (which we measure as the variance $w^T Q w$) and maximizes the return ($w^T R$) of the portfolio of the simulataneously. We suppose further that our portfolio allocation must comply with some lower and upper bounds on the allocation, $w_\mbox{lower} \leq w \leq w_\mbox{upper}$ and also $w \in \mathbf{R}^n$ $\sum_i w_i = 1$.
This problem can be written as\begin{array}{ll} \mbox{minimize} & \lambda*w^T Q w - (1-\lambda)*w^T R \\ \mbox{subject to} & \sum_i w_i = 1 \\ & w_\mbox{lower} \leq w \leq w_\mbox{upper} \end{array}
where $w \in \mathbf{R}^n$ is the vector containing weights allocated to each asset in the efficient frontier.
We can solve this problem as follows.
using Convex, ECOS #We are using ECOS solver. Install using Pkg.add("ECOS")# generate problem datasrand(0); #Set the seedn = 5; # Assume that we have portfolio of 5 assets.R = 5 * randn(n);A = randn(n, 5);Q = A * A' + diagm(rand(n));w_lower = 0;w_upper = 1;risk = zeros(2000); # Initialized the risk and the return vectors. ret = zeros(2000); # lambda varies in the interval(0,1) in the steps of 1/2000.w = Variable(length(R));#Defining constraintsc1 = sum(w) == 1;c2 = w_lower <= w; c3 = w <= w_upper;for i in 1:2000 λ = i/2000; #Defining Objective function objective = λ * quadform(w,Q) - (1-λ)* w' *R; p = minimize(objective, c1,c2,c3); solve!(p, ECOSSolver(verbose = false)); risk[i] = (w.value' * Q * w.value)[1]; ret[i] = (w.value'R)[1]; #println("$i ","$(λ*risk[i] - (1-λ)*ret[i]) ","$p.optval"); endusing PyPlot #Install PyPlot if you don't have it installed. Pkg.add("PyPlot")plot(risk,ret)title("Markowitz Efficient Frontier");xlabel("Expected Risk-Variance");ylabel("Expected Return"); |
Let $T$ be a distribution and $\varphi$ a test function.
I have read the following statement :
statement : $\text{supp}\varphi \cap \text{supp}T = \emptyset$ does not imply that $T(\varphi)=0$
I think it is wrong and I think it should be : $\varphi ^{-1} (\mathbb{C}^*)\cap \text{supp}T = \emptyset$ does not imply that $T(\varphi)=0$
I think I can prove that the initial statement it wrong, that is, I can prove that $\text{supp}\varphi \cap \text{supp}T = \emptyset \implies T(\varphi)=0$.
Reminder : let $A= \lbrace U \subset \Omega \text { such that } U \text{ open and } (\text{supp} \varphi \subset U \implies T(\varphi)=0) \rbrace$.
Then $\text{supp} T$ is the complement of $\cup_{U\in A} U$.
Proof : Let $K=\text{supp}\varphi$. By assumption, $K$ can be covered by a family of elements of $A$, and by compactness, this family can be chosen finite. Then multiplying $\varphi$ by a partition of unity $\psi$ corresponding to those elements, we get that $\varphi$ is a combination of functions supported in elements of $A$, hence by definition of $\text{supp}T$ and linearity of $T$, we get that $T(\varphi)=0$.
Am I wrong about all this ? |
I want to prove that $(\alpha+\beta)+\gamma=\alpha+(\beta+\gamma)$. now the method I first attempted this was by transfinite induction which is what i'm trying to get a better understanding of, The addition of ordinal numbers definition that I am using is $\alpha+\beta = \begin{cases} \text{$\alpha$,} &\quad\text{if $\beta=0$}, \\ \text{$S(\alpha+\gamma)$,} &\quad\text{if $\beta=S(\gamma)$} \\ \text{$\sup_{\gamma<\beta}(\alpha+\gamma)$} &\quad\text{if $\beta$ is a limit ordinal}\\ \end{cases}$
Where $S(\gamma)$ is the successor of $\gamma$
So my attempt at the proof is
Using induction on $\gamma$ we get that when $\gamma=0$ the result is trivial so suppose that $\gamma=\delta+1$ then we get that $(\alpha+\beta)+\gamma=(\alpha+\beta)+(\delta+1)=((\alpha+\beta)+\delta)+1$ $=(\alpha+(\beta+\delta))+1$ $=(\alpha+(\beta+\delta)+1)$ $=\alpha+(\beta+(\delta+1)$ $=\alpha+(\beta+\gamma)$ so the case when $\gamma$ is a successor is satisfied.
Now when $\gamma$ is a limit, specifically when $\gamma>1$. Then $\beta+\gamma$ is a limit and therefore so is $\alpha+(\beta+\gamma)$ and $(\alpha+\beta)+\gamma$ so we have that; $(\alpha+\beta)+\gamma=\sup_{\epsilon<\gamma}((\alpha+\beta)+\epsilon)$ $=\sup_{\epsilon+\beta<\beta+\gamma}((\alpha+\beta)+\epsilon)$ $=\sup_{\beta+\epsilon<\beta+\gamma}(\alpha+(\beta+\epsilon))$ $=\sup_{\delta<\beta+\gamma}(\alpha+\delta)$ $=\alpha+(\beta+\gamma)$ So it is satisfied when $\gamma$ is a limit ordinal, Thus $(\alpha+\beta)+\gamma=\alpha+(\beta+\gamma)$ for all ordinals $\alpha,\beta,\gamma$.
My question is how/why in the limiting case do we change the inequality $\sup_{\epsilon<\gamma}((\alpha+\beta)+\epsilon))$ to $\sup_{\epsilon+\beta<\beta+\gamma}((\alpha+\beta)+\epsilon))$. Any clarification would be great (I kind of guessed for the limiting step, and it turned out to be correct, but I don't understand it properly).
In response to the comment, If $\gamma$ is a limit then $\forall \alpha$ $\alpha+\gamma $ is a limit
proof $\gamma \ne 0$ so $\alpha+\gamma \geq \gamma >0$, i,e $\alpha+\gamma \ne 0$. so let $x\in \alpha+\gamma$. then show that $x+1<\alpha+\gamma$ , $x\in \alpha+\gamma =\bigcup_{\beta<\gamma}(\alpha+\beta)$, i.e there is $\beta < \gamma$ such that $x \in \alpha+\beta$. by a previous lemma, $x+1\leq \alpha+\beta$. if $x+1 \in \alpha+\beta, x+1<\alpha+\gamma$. so suppose $\alpha+\beta=x+1$. then since $\gamma$ is a limit , $\beta+1<\gamma$ and by definition $\alpha+(\beta+1)=(\alpha+\beta)+1$ and $x+1 \in (\alpha+\beta)+1$, hence $x+1 \in \alpha+\gamma$. |
In my Differential Equations course, we defined the equilibrium point $x_0$ of a dynamical system $\frac{dx}{dt} = f(x(t))$ (for $f$ defined on an open subset of $\mathbb R^n$, say $\mathbb R^n$ itself) to be stable if it is:
Lyapunov Stable There is an $\epsilon$ ball around $x_0$ such that the solutions $\varphi$ of this differential equation with initial conditions in this ball satisfy $\lim_{t \to \infty} \varphi(t) = x_0$.
I am trying to find an example of the case where the property (2) holds while the point $x_0$ is not Lyapunov stable.
After some searching, I ran across Homoclinic Bifurcation, which is intuitively how I would expect Lyapunov Stability to fail, but have been unable to find examples of Homoclinic Bifurcation where property (2) holds as well.
Any help would be appreciated. |
Your hovercraft device has a mass of 2.25 kg and at max power can travel a speed of 0.2 m/s. You also learn that the target distance is 180 cm and you are trying to get there in 15 seconds.
How many rolls of pennies should you place on the hovercraft in order to get as close to 15 seconds as possible?
Assuming there is no change in lift (which affects friction) and the hovercraft gets to max speed instantly:One penny roll is around the mass of 50 (new) pennies, 50 * 3.11 g or 0.156 kg.Required speed is 1.8 m / 15 s = 0.12 m/s.(2.25 kg)(0.2 m/s) = (2.25 kg + n * 0.156 kg)(0.12 m/s)n = 9.6
Your hovercraft device has a mass of 2.25 kg and at max power can travel a speed of 0.2 m/s. You also learn that the target distance is 180 cm and you are trying to get there in 15 seconds.
How many rolls of pennies should you place on the hovercraft in order to get as close to 15 seconds as possible?
Assuming there is no change in lift (which affects friction) and the hovercraft gets to max speed instantly:One penny roll is around the mass of 50 (new) pennies, 50 * 3.11 g or 0.156 kg.Required speed is 1.8 m / 15 s = 0.12 m/s.(2.25 kg)(0.2 m/s) = (2.25 kg + n * 0.156 kg)(0.12 m/s)n = 9.6
10
Oh sorry, I forgot to mention that you should assume that each roll of pennies has a mass of 125 grams (it's mentioned in the rules manual). However, your answer is technically correct, so you can go ahead with the next one!
"The fault, dear Brutus, is not in our stars,But in ourselves, that we are underlings."
University of Texas at Austin '23Seven Lakes High School '19
Derive the formula for the distance a projectile travels given the angle of elevation of its launch, theta, and the initial speed, v. Assume the projectile encounters no air resistance and starts at ground level.
Derive the formula for the distance a projectile travels given the angle of elevation of its launch, theta, and the initial speed, v. Assume the projectile encounters no air resistance and starts at ground level.
First, find the individual components of the 2D velocity vector:[math]v_x = vcos\theta[/math][math]v_y = vsin\theta[/math]The amount of time the ball will be in the air can be found using[math]\Delta y = v_{y}t + \frac{at^2}{2}[/math]Since [math]\Delta y = 0[/math] and [math]a = -9.8[/math]:[math]0 = t(vsin\theta - 4.9t)[/math][math]t = 0, \frac{vsin\theta}{4.9}[/math]To then get the total distance traveled, we simply use [math]v_{x}t = \Delta x[/math].[math]vcos\theta*\frac{vsin\theta}{4.9} = \Delta x[/math][math]\frac{v^2sin(2\theta)}{9.8} = \Delta x[/math] is the final equation.
Conant 19' UIUC 23' Member of The Builder CultPhysics is the only real scienceChange my mind
Derive the formula for the distance a projectile travels given the angle of elevation of its launch, theta, and the initial speed, v. Assume the projectile encounters no air resistance and starts at ground level.
First, find the individual components of the 2D velocity vector:[math]v_x = vcos\theta[/math][math]v_y = vsin\theta[/math]The amount of time the ball will be in the air can be found using[math]\Delta y = v_{y}t + \frac{at^2}{2}[/math]Since [math]\Delta y = 0[/math] and [math]a = -9.8[/math]:[math]0 = t(vsin\theta - 4.9t)[/math][math]t = 0, \frac{vsin\theta}{4.9}[/math]To then get the total distance traveled, we simply use [math]v_{x}t = \Delta x[/math].[math]vcos\theta*\frac{vsin\theta}{4.9} = \Delta x[/math][math]\frac{v^2sin(2\theta)}{9.8} = \Delta x[/math] is the final equation.
There is a 20kg ball heading east at 26m/s. A 36kg ball is traveling southwest at 38m/s. If both balls undergo an inelastic collision, what is the speed and the direction of the balls after the collision?
Conant 19' UIUC 23' Member of The Builder CultPhysics is the only real scienceChange my mind
There is a 20kg ball heading east at 26m/s. A 36kg ball is traveling southwest at 38m/s. If both balls undergo an inelastic collision, what is the speed and the direction of the balls after the collision?
It doesn't seem like there's enough information given to solve the problem, but given that the balls "stick together", thenwhere east is positive.
There is a 20kg ball heading east at 26m/s. A 36kg ball is traveling southwest at 38m/s. If both balls undergo an inelastic collision, what is the speed and the direction of the balls after the collision?
It doesn't seem like there's enough information given to solve the problem, but given that the balls "stick together", thenwhere east is positive.
or
where south is positive.
or
That’s correct. You turn!
Conant 19' UIUC 23' Member of The Builder CultPhysics is the only real scienceChange my mind
1. A uniform disk is rolled down a hill of height 11m. What is the speed of the disk when it reaches the bottom of the hill?2. As soon as the disk reaches the bottom of the hill, it hits a horizontal surface of frictional coefficient 0.13. What is the acceleration of the disk?3. A ball of mass 2kg is attached to a string of length 0.75m and rotated vertically with an angular velocity of 7rad/s. What is the ratio of the string tension at the top of the loop to the string tension at the bottom?Ignore significant figures.
i wish i was goodEvents 2019: Expd, Water, HerpRip states 2019 |
Q. A resistance is shown in the figure. Its value and tolerance are given respectively by :
Solution:
Color code : Red violet orange silver $R = 27 \times 10^3 \; \Omega \pm 10 \%$ $= 27 \; K \Omega \pm \; 10 \%$ Questions from JEE Main 2019 4. One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the central of the loop $(B_L)$ to that at the centre of the coil $(B_C)$ , i.e., $R \frac{B_L}{B_C}$ will be 9. The magnetic field associated with a light wave is given, at the origin, by $B = B_0 [\sin(3.14 \times 10^7)ct + \sin(6.28 \times 10^7)ct]$.
If this light falls on a silver plate having a work
function of 4.7 eV, what will be the maximum
kinetic energy of the photo electrons ?
$(c = 3 \times 10^{8} ms^{-1}, h = 6.6 \times 10^{-34} J-s)$ Physics Most Viewed Questions 1. An AC ammeter is used to measure current in a circuit. When a given direct current passes through the circuit, the AC ammeter reads 3 A. When another alternating current passes through the circuit, the AC ammeter reads 4 A. Then the reading of this ammeter, if DC and AC flow through the circuit simultaneously, is 9. A cannon of mass I 000 kg, located at the base of an inclined plane fires a shelI of mass 100 kg in a horizontal direction with a velocity 180 $kmh^{-1}$. The angle of inclination of the inclined plane with the horizontal is 45$^{\circ}$. The coefficient of friction between the cannon and the inclined plane is 0.5. The height, in metre, to which the cannon ascends the inclined plane as a result of the recoiI is (g = 10 $ms^{-1})$ Latest Updates Top 10 Medical Entrance Exams in India Top 10 Engineering Entrance Exams in India JIPMER Results Announced NEET UG Counselling Started NEET Result Announced KCET Result Announced KCET College Predictor JEE Main Result Announced JEE Advanced Score Cards Available AP EAMCET Result Announced KEAM Result Announced UPSEE – Online Applications invited from NRI &Kashmiri Migrants MHT CET Result Announced NEET Rank Predictor Questions Tardigrade |
Here's how I first did it. First we make a few notes. We have $\displaystyle f(0) = f(0 +0) = f(0) +f(0)$, so $f(0)=0$. More interestingly, we have $f(2x) = f(x + x) = f(x) + f(x) = 2f(x)$, upon iterating, $f(nx) = nf(x)$. Looking at $-x$, we have $f(0) = f(x +(-x)) = f(x) + f(-x)$ so $f(-x) = - f(x)$. Finally, we have $f(x) = \displaystyle f\left(\frac x2 + \frac x2\right) = 2 f\left( \frac x2\right)$ and upon iterating, $\displaystyle f\left(\frac{x}{2^n}\right) = \frac{1}{2^n} f(x)$. Putting it all together, we have that $\displaystyle f\left(\frac{k}{2^n}x \right) = \frac{k}{2^n}f(x)$ for any $k \in \mathbb{Z}$, $n \in \mathbb{N}$. \\par Suppose $f$ is not identically $0$, for otherwise our function would be trivially continuous. Then there exists some $p$ such that $f(p) = a \neq 0$. Then by our observation above, every number of the form $\displaystyle \frac{k}{2^m} p $ is such that $\displaystyle f\left(\frac{k}{2^m} p\right) = \frac{k}{2^m}f(p) = \frac{k}{2^m} a \neq0$. Since there are countably such numbers of the form $\displaystyle\frac{k}{2^m} p$, enumerate them by $\{p_n\}_{n \in \mathbb{N}}$. Call $f(p_n) = a_n$. Since every $a_n$ is of the form $\displaystyle \frac{k}{2^m} a$ for some $k, m$, and $a\neq 0$, we have that $\{a_n\}_{n \in \mathbb{N}}$ is dense in $\mathbb{R}$, since dyadic rationals are dense in $\mathbb{R}$, and scaling a dense, infinite set in $\mathbb{R}$ will do nothing to change its density.
Now, fix $\epsilon >0$. Let $V= \displaystyle B\left(0, \frac\epsilon2\right)$. Since $\{a_n\}_{n \in \mathbb{N}}$ are dense, $\mathbb{R} = \displaystyle \bigcup_{n \in \mathbb{N}} V + a_n$. Now we have the following claim to show: $f^{-1}(V+a_n) = f^{-1}(V) + p_n$. First, suppose $x \in f^{-1}(V) +p_n$. Then $x = y+ p_n$ for some $y \in f^{-1}(V)$, so $f(x) = f(y) + f(p_n) = v + a_n$ for some $v \in V$, and thus $f^{-1}(V+a_n) \supset f^{-1}(V) + p_n$. Now we want to show $f^{-1}(V+a_n) \subset f^{-1}(V) + p_n$ where $f(p_n) =a_n$. So, suppose $f(x) = v+a_n$ for some $v \in V$. Then $f(x-p_n) = f(x) - f(p_n) = v+a_n - a_n = v$, so $x-p_n \in f^{-1}(V)$. Thus $x = (x-p_n) +p_n \in f^{-1}(V) + p_n$. Thus we have shown the equality we wanted.\\par From here, we see$$\mathbb{R} = f^{-1}(\mathbb{R}) = f^{-1}\left( \displaystyle \bigcup_{n \in \mathbb{N}} V+a_n\right) = \bigcup_{n \in \mathbb{N}} f^{-1}(V+a_n) = \bigcup_{n \in \mathbb{N}} f^{-1}(V) + p_n.$$
Now, since $V$ is open and $f$ measurable, $f^{-1}(V)$ is measurable, and by translation invariance of the Lebesgue measure (denoted by $m$), we have $m(f^{-1}(V) +p_n) = m(f^{-1}(V))$. Set $U = f^{-1}(V)$. The equality above implies $m(U) >0$, otherwise we'd have $\mathbb{R}$ as the countable union of null sets. Thus we can apply Steinhaus's theorem to $U-U$ to see that $U-U \supset B(0, \delta)$, that is, a ball of radius $\delta>0$ around the origin. Then for every $|x|< \delta$, we have $x= y-w$ for $y, w \in U$. But then $$|f(x)| = |f(y-w)| = |f(y) - f(w)| \le |f(y)| + |f(w)| < \frac\epsilon2 + \frac\epsilon2 = \epsilon$$since $f(y), f(w) \in V$.
This shows continuity of $f$ at zero, and by the pseudo-linearity property, we can get linearity at any point $z$. For take any sequence $z_n \to z$, then $z- z_n \to 0$, so we have $f(z) - f(z_n) = f(z-z_n) \to 0$ by continuity at zero. This concludes the proof. |
The answer to this question depends on what you mean by $\mathbf{R}^2$. You can write $\mathbf{R}\times\mathbf{R}$, but the "$\times$" can have several different meanings depending on which
category you are working in.
As sets:
If you view $\mathbf{R}\times\mathbf{R}$ and $\mathbf{C}$ as sets, this means you are ignoring any possible structure on these things besides their elements (so forget about multiplication, addition, anything.) From this perspective, $\mathbf{R}\times\mathbf{R}$ and $\mathbf{C}$ are the same object (in technical terms they are isomorphic in the category of sets) because there is a bijection$$(a,b)\ \leftrightarrow\ a+bi.$$A key thing to note about this is that "$\times$" refers to a direct product of sets, i.e. the Cartesian product. This will change as we add more structure.
As real vector spaces:
You can give each of $\mathbf{R}\times\mathbf{R}$ and $\mathbf{C}$ the structure of a real vector space, meaning you can add vectors and multiply by real numbers. Then from the theory of linear algebra, we know that $\mathbf{R}\times\mathbf{R}$ has a basis of $\{(1,0), (0,1)\}$ and $\mathbf{C}$ has a (real) basis of $\{1, i\}$. Since these real vector spaces both have dimension 2, they are isomorphic (in the linear algebra sense, i.e. in the category of $\mathbf{R}$-modules). So from this perspective they are again the same object.
Note here that $\mathbf{R}\times\mathbf{R}$ can be interpreted as $\mathbf{R}\oplus \mathbf{R}$ which may be more familiar to linear algebra students. The point is that now we are requiring more from the operation (it has to preserve addition of vectors now).
As rings:
Here is where the difference comes in. We can think of $\mathbf{R}$ and $\mathbf{C}$ as
rings, meaning we can add and multiply elements together according to some axioms. Then if you write $\mathbf{R}\times\mathbf{R}$, you mean a direct product in the category of rings, so now multiplication in $\mathbf{R}\times\mathbf{R}$ has to satisfy$$(a,b)\cdot (c,d)=(ac,bd).$$But in particular this means things like$$(1,0)\cdot (0,1)=(0,0),$$which means it is possible for two nonzero things to have a product zero. In contrast, if $z_1z_2=0$ in $\mathbf{C}$, then either $z_1=0$ or $z_2=0$. In this way, $\mathbf{R}\times\mathbf{R}$ and $\mathbf{C}$ have fundamentally different behavior as rings. Because of this, there is no isomorphism of rings between the two objects.
As fields:
A field is a commutative ring with more structure (we can invert multiplication for nonzero things). It turns out that $\mathbf{C}$ can be given the structure of a field because $z^{-1}$ exists for any nonzero $z\in\mathbf{C}$, but $\mathbf{R}\times\mathbf{R}$ cannot be a field because equations like $(1,0)\cdot (0,1)=(0,0)$ mess everything up (try to cancel something from the left side).
tl;dr -- You have to specify what you mean by "$\times$". $\mathbf{C}$ and $\mathbf{R}\times\mathbf{R}$ are exactly the same until you start saying you want to do things like multiply elements together. |
Mathematical Explanation
When examining the linear combination of atomic orbitals (LCAO) for the $\ce{H2+}$ molecular ion, we get two different energy levels, $E_+$ and $E_-$ depending on the coefficients of the atomic orbitals. The energies of the two different MO's are:$$\begin{align}E_+ &= E_\text{1s} + \frac{j_0}{R} - \frac{j' + k'}{1+S} \\E_- &= E_\text{1s} + \frac{j_0}{R} - \frac{j' - k'}{1-S}\end{align} $$
Note that $j_0 = \frac{e^2}{4\pi\varepsilon_0}$, $R$ is the internuclear distance, $S=\int \chi_\text{A}^* \chi_\text{B}\,\text{d}V$ the overlap integral, $j'$ is a coulombic contribution to the energy and $k'$ is a contribution to the
resonance integral, and it does not have a classical analogue. $j'$ and $k'$ are both positive and $j' > k'$. You'll note that $j'-k' > 0$.
This is why the energy levels of $E_+$ and $E_-$ are not symmetrical with respect to the energy level of $E_\text{1s}$.
Intuitive Explanation
The intuitive explanation goes along the following line: Imagine two hydrogen nuclei that slowly get closer to each other, and at some point start mixing their orbitals. Now, one very important interaction is the coulomb force between those two nuclei, which gets larger the closer the nuclei come together. As a consequence of this, the
energies of the molecular orbitals get shifted upwards, which is what creates the asymmetric image that we have for these energy levels.
Basically, you have two positively charged nuclei getting closer to each other. Now you have two options:
Stick some electrons between them. Don't stick some electrons between them.
If you follow through with option 1, you'll diminish the coulomb forces between the two nuclei somewhat in favor of electron-nucleus attraction. If you go with method 2 (remember that the $\sigma^*_\text{1s}$ MO has a node between the two nuclei), the nuclei feel each other's repulsive forces more strongly.
Further Information
I highly recommend the following book, from which most of the information above stems:
P. Atkins and R. Friedman: Molecular Quantum Mechanics, $5^\text{th}$ ed. Oxford University Press, 2011. |
This question already has an answer here:
Background: This is part b of problem 12.4.3 from Arfken, Weber, Harris Math Methods for Physicists to show that $\int_0^\infty \frac{\ln^2(z)}{1+z^2}$dz$=4(1-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\dots)$.
Part b of the question asks to show that this series evaluates to $\frac{\pi^3}{8}$ by contour integration. Where is my mistake: $\lim_{z \to 0}zf(z)=0$ and $\lim_{z \to \infty}zf(z)=0$ so the big and little circle equal 0.$
Assume $I=\int_0^\infty \frac{\ln^2(x)}{1+x^2}\text{dx}$
We can add the components of along the contour and set that equal to the value of $2\pi i \text{Res}[f(z),i]$ evaluated at the poles $\pm i$ $$\int_0^\infty \frac{\ln^2(z)}{1+z^2}\text{dz}+\int_{\infty}^0 \frac{\ln^2(z)}{1+z^2}\text{dz}=2\pi i \text{Res}[f(z),\pm i]\tag{1}$$
$$\int_0^\infty \frac{(\ln^2 \mid x\mid}{1+x^2}\text{dx}-\int_0^{\infty} \frac{(\ln\mid x\mid+2i\pi)^2}{1+x^2}\text{dx}=2\pi i \left (\lim_{z \to i}\frac{\ln^2(z)}{2z}+\lim_{z \to -i}\frac{\ln^2(z)}{2z}\right )\tag{2}$$
$$\int_0^\infty \frac{(\ln^2\mid x\mid}{1+x^2}\text{dx}-\int_0^{\infty} \frac{(\ln^2\mid x\mid+\color{red}{4\ln|x|i\pi}-4\pi^2)}{1+x^2}\text{dx}=2\pi i \left (\lim_{z \to i}\frac{\ln^2(z)}{2z}+\lim_{z \to -i}\frac{\ln^2(z)}{2z}\right )\tag{3}$$ $$0I+\color{red}{0}-\left[\tan^{-1}(x)\right]\mid^{\infty}_0(4\pi^2)\text{dx}=(2\pi i) \left (\frac{-\pi^2/4+9\pi^2/4}{2i}\right )\tag{4}$$ $$0I+2\pi^3=\frac{8\pi^3}{4}\tag{5}$$
I found my error. It was a negative sign, and the two sides cancel to zero so you can't evaluate it this way, but I found an answer which evaluates it from negative to positive infinity so I'm marking the question as a duplicate. See dustin's answer at the link for the contour integration. |
This is a proposition that I wanted to use but I can't seem to prove it. It also happens to be an exercise in Hatcher, ex. 0.13, but I couldn't find a correct solution online.
Here a 'deformation retraction' from $X$ onto a subspace $A$ means a continuous family $F$ of maps $f_t: X \rightarrow X$ for $t \in I$ with $f_0$ the identity on $X$, $f_1$ a retraction of $X$ onto $A$ and such that each $f_t$ restricts to the identity on $A$. This is sometimes called a strong deformation retraction.
The exercise says that given two such deformation retractions $F, G$ of $X$ onto $A$ there is a continuous family of deformation retractions $\mathcal{R} = \lbrace R^t$ such that the function $r(x)_s^t$ from $X \times I \times I$ to $X$ is continuous, with $R_0 = F$ and $R_1 = G$.
Appreciate any help! |
Let $ a,b,c$ positive integer such that $ a + b + c \mid a^2 + b^2 + c^2$.
Show that $ a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer $ n$.
(problem composed by
Laurentiu Panaitopol)
So far no idea.
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Let $ a,b,c$ positive integer such that $ a + b + c \mid a^2 + b^2 + c^2$.
Show that $ a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer $ n$.
(problem composed by
Laurentiu Panaitopol)
So far no idea.
Claim.$a+b+c\mid a^{2^n}+b^{2^n}+c^{2^n}$ for all $n\geq0$. Proof. By induction: True for $n=0,1$ $\checkmark$. Suppose it's true for $0,\ldots,n$. Note that $$a^{2^{n+1}}+b^{2^{n+1}}+c^{2^{n+1}}=(a^{2^n}+b^{2^n}+c^{2^n})^2-2(a^{2^{n-1}}b^{2^{n-1}}+b^{2^{n-1}}c^{2^{n-1}}+c^{2^{n-1}}a^{2^{n-1}})^2+4a^{2^{n-1}}b^{2^{n-1}}c^{2^{n-1}}(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})$$
and that
$$2(a^{2^{n-1}}b^{2^{n-1}}+b^{2^{n-1}}c^{2^{n-1}}+c^{2^{n-1}}a^{2^{n-1}})=(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})^2-(a^{2^n}+b^{2^n}+c^{2^n})$$
is divisible by $a+b+c$ by the induction hypothesis.
It seems that there's a partial solution.
Suppose that $\mathrm{gcd}(a,a+b+c)=\mathrm{gcd}(b,a+b+c)=\mathrm{gcd}(c,a+b+c)=1$. Then for $n=k\cdot \phi(a+b+c)+1 \, (k=1,2, \ldots )$, where $\phi$ is Euler's function, we have: $$ (a^n+b^n+c^n)-(a^2+b^2+c^2)=a^2 (a^{n-1}-1) + b^2 (b^{n-1}-1) + c^2 (c^{n-1}-1), $$ where all round brackets are divisible by $a+b+c$ according to Euler theorem. Therefore $(a+b+c) \mid (a^n+b^n+c^n)$ for all these $n$.
There's one more solution (it isn't mine). One can even prove that $(a + b + c) \mid (a^n + b^n + c^n)$ for all $n=3k+1$ and $n=3k+2$. It's enough to prove that $a + b + c \mid a^n + b^n + c^n$ => $a + b + c \mid a^{n+3} + b^{n+3} + c^{n+3}$. The proof is here: https://vk.com/doc104505692_416031961?hash=3acf5149ebfb5338b5&dl=47a3df498ea4bf930e (unfortunately, it's in Russian but it's enough to look at the formulae). One point which may need commenting: $(ab+bc+ca)(a^{n-2} + b^{n-2} + c^{n-2})$ is always divisible by $(a+b+c)$ (it's necessary to consider 2 cases: $(a+b+c)$ is odd and $(a+b+c)$ is even).
If $a,b,c,n\in\Bbb Z_{\ge 1}$, $a+b+c\mid a^2+b^2+c^2$, then $$a+b+c\mid a^n+b^n+c^n$$
is true when $n\nmid 3$, but not necessarily when $n\mid 3$.
$$x^2+y^2+z^2+2(xy+yz+zx)=(x+y+z)^2$$
$$\implies x+y+z\mid 2(xy+yz+zx)$$
$$\implies x+y+z\mid (x^k+y^k+z^k)(xy+yz+zx)$$
for all $k\ge 1$ (to see why, check cases when $x+y+z$ is even and when it's odd).
$$x^{n+3}+y^{n+3}+z^{n+3}=(x^{n+2}+y^{n+2}+z^{n+2})(x+y+z)$$
$$-(x^{n+1}+y^{n+1}+z^{n+1})(xy+yz+zx)+(x^n+y^n+z^n)xyz$$
for all $n\ge 1$. We know $$x+y+z\mid (x^{n+2}+y^{n+2}+z^{n+2})(x+y+z)$$
$$-(x^{n+1}+y^{n+1}+z^{n+1})(xy+yz+zx)$$
Now let $(x,y,z)=(x_1,y_1,z_1)=(1,3,9)$. $$x_1+y_1+z_1\nmid x_1^3+y_1^3+z_1^3$$ $$x_1+y_1+z_1\nmid \left(x_1^3+y_1^3+z_1^3\right)x_1y_1z_1$$ $$\implies x_1+y_1+z_1\nmid x_1^6+y_1^6+z_1^6$$
Since $x_1+y_1+z_1$ is coprime to $x_1,y_1,z_1$, we get $$x_1+y_1+z_1\nmid (x_1^6+y_1^6+z_1^6)x_1y_1z_1,$$
and so $x_1+y_1+z_1\nmid x_1^9+y_1^9+z_1^9$, etc.
Therefore $x+y+z$ cannot generally (for all $x,y,z\in\mathbb Z_{\ge 1}$) divide $x^{3m}+y^{3m}+z^{3m}$ for any given $m\ge 1$.
However, we easily get $x+y+z$ always divides $x^n+y^n+z^n$ for $n$ not divisible by $3$,
because $x+y+z\mid (x+y+z)xyz$ and $x+y+z\mid \left(x^2+y^2+z^2\right)xyz$,
because $x+y+z\mid x^2+y^2+z^2$ (given), so $x+y+z\mid x^4+y^4+z^4, x^5+y^5+z^5$,
so $x+y+z\mid \left(x^4+y^4+z^3\right)xyz, \left(x^5+y^5+z^5\right)xyz$,
so $x+y+z\mid x^7+y^7+z^7, x^8+y^8+z^8$, etc.
Here's a more intuitive way to get the idea of considering powers of $2$.
Added (below): in the same way we can prove that any $n=6k\pm1$ works.
Note that $a+b+c\mid(a+b+c)^2-(a^2+b^2+c^2)=2(ab+bc+ca)$. By The Fundamental Theorem of Symmetric Polynomials (FTSP), $a^n+b^n+c^n$ is an integer polynomial in $a+b+c$, $ab+bc+ca$ and $abc$. If $3\nmid n$, no term has degree divisible by $3$ so each term has at least one factor $a+b+c$ or $ab+bc+ca$. If we can find infinitely many $n$ such that the terms without a factor $a+b+c$ have a coefficient that is divisible by $2$, then we are done because $a+b+c\mid2(ab+bc+ca)$. This suggests taking a look to the polynomial $a^n+b^n+c^n$ over $\Bbb F_2$. Note that over $\Bbb F_2$, $a^{2^n}+b^{2^n}+c^{2^n}=(a+b)^{2^n}+c^{2^n}=(a+b+c)^{2^n}$ is divisible by $(a+b+c)$. Because the polynomial given by FTSP over $\Bbb F_2$ is the reduction modulo $2$ of that polynomial over $\mathbb Z$ (this is a consequence of the uniqueness given by the FTSP), this shows that the coefficients of those terms that have no factor $a+b+c$ is divisible by $2$, and we are done because $3\nmid2^n$. (In fact all coefficients except that of $(a+b+c)^{2^n}$ are divisible by $2$.)
Added later Ievgen's answer inspired me to generalise the above approach to $n=6k\pm1$. Consider again $a^n+b^n+c^n$ as an integer polynomial in $abc,ab+bc+ca,a+b+c$ (which we can do by FTSP). Because $3\nmid n$, no term has the form $(abc)^k$. It remains to handle the terms of the form $m\cdot(ab+bc+ca)^k(abc)^l$. If $a+b+c$ is odd, then $a+b+c\mid ab+bc+ca$ and we're done. If $a+b+c$ is even, at least one of $a,b,c$ is even so $2\mid abc$, and hence $a+b+c\mid m\cdot(ab+bc+ca)^k(abc)^l$. (Note that $l>0$ because $n$ is odd.)
For any positive integer x this is true: $x \leqslant x^2$ (From $1 \leqslant x$ for any positive ineger x ). So $a + b + c \leqslant a^2 + b^2 + c^2$. But for 2 positive integers $x$, $y$ $x$ is divisible by $y$ only if $x \geqslant y$. So $a + b + c \geqslant a^2 + b^2 + c^2$ if $a + b + c \mid a^2 + b^2 + c^2$. From these 2 inequalities: $a + b + c = a^2 + b^2 + c^2$
So $a + b - a^2 - b^2 = c^2 - c$
Evaluation for the right part $c^2 - c \geqslant 0$ (because $x \leqslant x^2$ for any positive integer x). Evaluation for the left part $a + b - a^2 - b^2 \leqslant 0$ (adding 2 inequalities $a - a^2 \leqslant 0$ and $b - b^2 \leqslant 0$)
So the left part is $\leqslant 0$ and the right part $\geqslant 0$. But they are equal so $a + b - a^2 - b^2 = c^2 - c = 0$ and $c^2 = c$. c is positive so we can divide both part of the last equation by c and get $c = 1$.
Similarly $b = 1$ and $a = 1$. So $a + b + c = 3$ and $a^n + b^n + c^n = 3$ for any positive $n$. 3 is divisible by 3 so $a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer n. |
Defining parameters
Level: \( N \) = \( 4000 = 2^{5} \cdot 5^{3} \) Weight: \( k \) = \( 1 \) Character orbit: \([\chi]\) = 4000.x (of order \(8\) and degree \(4\)) Character conductor: \(\operatorname{cond}(\chi)\) = \( 32 \) Character field: \(\Q(\zeta_{8})\) Newforms: \( 0 \) Sturm bound: \(600\) Trace bound: \(0\) Dimensions
The following table gives the dimensions of various subspaces of \(M_{1}(4000, [\chi])\).
Total New Old Modular forms 40 0 40 Cusp forms 0 0 0 Eisenstein series 40 0 40
The following table gives the dimensions of subspaces with specified projective image type.
\(D_n\) \(A_4\) \(S_4\) \(A_5\) Dimension 0 0 0 0 |
I have to find the limit $$\lim_{n\to\infty}\sum_{k=1}^n \frac{2k+1}{k^2(k+1)^2}.$$ I tried to make it into a telescopic series but it doesn't really work out...
$$\lim_{n\to\infty} \sum_{k=1}^n \frac{2k+1}{k^2(k+1)^2}=\sum_{k=1}^n \left(\frac{1-k}{k^2}+\frac1{k+1}-\frac1{(k+1)^2} \right)$$ so that is what I did using telescopic...
I said that:
$$\frac{2k+1}{k^2(k+1)^2}=\frac{Ak+B}{k^2}+\frac C{k+1}+\frac D{(k+1)^2}$$ but now as I look at it.. I guess I should "build up the power" with the ${k^2}$ too, right? |
I've been unable to find any references for the lagrangian density in presence of matter. It's a lot easier to find this expressions in vacuum, but, when it comes to matter, I couldn't find any clear information. I mean, when we're referring to this equations: $$\begin{cases} \nabla \cdot \mathbf{E}=\large{\frac{\rho}{\epsilon_0}} \\ \\ \nabla \cdot \mathbf{B}=0 \\ \\ \nabla \times \mathbf{E}=-\large{\frac{\partial \mathbf{B}}{\partial t}} \\ \\ \nabla \times \mathbf{B}=\mu_0 \mathbf{J}+\large{\frac{1}{c^2}\frac{\partial\mathbf{E}}{\partial t}} \end{cases}$$
I know the associated lagrangian density is:
$$\mathcal{L}=-\frac{1}{4\mu_0} F_{\mu \nu} F^{\mu \nu}-J_{\mu}A^{\mu}$$
Nevertheless, I was seeking how these other equations would be written in terms of the lagrangian formulation:
$$\begin{cases} \nabla \cdot \mathbf{D}=\rho \\ \\ \nabla \cdot \mathbf{B}=0 \\ \\ \nabla \times \mathbf{E}=-\large{\frac{\partial \mathbf{B}}{\partial t}} \\ \\ \nabla \times \mathbf{H}=\mathbf{J}+\large{\frac{\partial\mathbf{D}}{\partial t}} \end{cases}$$
And the only thing I got, is the link I leave below:
It says, " Separating the free currents from the bound currents, another way to write the Lagrangian density is as follows":
$$\mathcal{L}=-\frac{1}{4\mu_0} F_{\mu \nu} F^{\mu \nu}-J_{\mu}A^{\mu}+\frac{1}{2} F_{\mu \nu}\mathcal{M^{\mu \nu}}$$
Where $\mathcal{M^{\mu \nu}}$ is the magnetization-polarization tensor.
However, in this link, it's not explained why the term corresponding to the bound currents is the way it is, and I couldn't find anywhere information that clarifies my doubts.
So, my question is: Why the term corresponding to the bound currents is the way it is? |
Second Principle of Finite Induction/One-Based Theorem Suppose that: $(1): \quad 1 \in S$ $(2): \quad \forall n \in \N_{>0}: \paren {\forall k: 1 \le k \le n \implies k \in S} \implies n + 1 \in S$ Then: $S = \N_{>0}$ Proof
Define $T$ as:
$T = \set {n \in \N_{>0}: \forall k: 1 \le k \le n: k \in S}$
Since $n \le n$, it follows that $T \subseteq S$.
Therefore, it will suffice to show that:
$\forall n \ge 1: n \in T$ Firstly, we have that $1 \in T$ if and only if the following condition holds: $\forall k: 1 \le k \le 1 \implies k \in S$
Since $1 \in S$, it thus follows that $1 \in T$.
Now suppose that $n \in T$; that is: $\forall k: 1 \le k \le n \implies k \in S$
By $(2)$, this implies:
$n + 1 \in S$
Thus, we have:
$\forall k: 1 \le k \le n + 1 \implies k \in S$
Therefore, $n + 1 \in T$.
Hence, by the Principle of Finite Induction: $\forall n \ge 1: n \in T$
That is:
$T = \N_{>0}$
and as $S \subseteq \N_{>0}$ it follows that:
$S = N_{>0}$
$\blacksquare$
Sources 1971: Allan Clark: Elements of Abstract Algebra... (previous) ... (next): Chapter $1$: Properties of the Natural Numbers: $\S 20 \alpha$ 1980: David M. Burton: Elementary Number Theory(revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction 1982: P.M. Cohn: Algebra Volume 1(2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: $\mathbf{I}'$ |
Need help to understand LAMDA clustering algorithm, taken from some article.
Here is the algorithm from the article
LAMDA is a conceptual clustering and classification methodology that computes the degree of adequation of an object to a class with all the partial or marginal information available [17]. The difference between this algorithm and the classical clustering and classification approaches is that LAMDA models the total indistinguishability or homogeneity inside the context or universe from which the information is extracted. This is done by means of a special class, the so called non-informative class (NIC), which accepts all objects under the same status. Therefore, the adequation degree of the objects of NIC acts as a minimum threshold to assign an element to a significant class. Hence, the minimum threshold is not fixed arbitrarily but is auto- matically determined by the proper context. Algorithm 1 summarizes the main steps of LAMDA clustering. Given a set of objects X = { $\vec{x_1}$, $\vec{x_2}$,..., $\vec{x_n}$}, where an object is represented by an m-dimensional vector $\vec{x_k}$ = {$\vec{x_1^k}$, $\vec{x_2^k}$, ..., $\vec{x_m^k}$}, the algorithm starts by creating an initial class consisting of one of the objects selected at random.
For each remaining object $\vec{x_k}$ and for each existing class $C_j$, LAMDA computes for every descriptor the so-called marginal adequacy degree $MAD_{ij}$ ($x^k_i$ ) between the values that the ith descriptor takes over $\vec{x_k}$ and the class $C_j$ . Thus, a vector $MAD_j$($\vec{x_k}$) can be associated with object $\vec{x_k}$ for each class $C_j$ . $MAD_j$($\vec{x_k}$) is a membership function derived from a fuzzy generalization of a binomial probability law, as expressed in the algorithm. In the expression, ν ($x^k_i$ , $c_ij$ ) is a distance function between the descriptor $x^k_i$ and the attribute $c_{ij}$ of the center of the class $C_j$ ; $ρ_{ij}$ is the possibility of the descriptor $x^k_i$ to belong to class $C_j$ .
Algorithm: Data clustering with LAMDA
Input:A set of data objects X = { $\vec{x_1}$, $\vec{x_2}$,..., $\vec{x_n}$}, where $\vec{x_k}$ = {$\vec{x_1^k}$, $\vec{x_2^k}$, ..., $\vec{x_m^k}$}
Output: Γ = {$C_j$} set of classes where each class $C_j$ is represented by the parameters $c_{ij}$ and $ρ_{ij}$
Initialization: ν($x_i^k$) = 1 - $||$$x_i^k$ - $c_{ij}$$||$ and $ρ_{init}$ = 0.5, α ∈ ]0,1[, $T_{norm}$, $T_{conorm}$ and $C$ = 1
begin
for$k$ ← 1 to $n$ do
for$j$ ← 1 to $C$ do
for$i$ ← 1 to $m$ do
$MAD_{ij}$($x_i^k$) = $ρ_{ij}^{ν(x_i^k,c_{ij})}*(1 - ρ_{ij})^{1 - ν(x_i^k, c_{ij})}$
end
$MAD_j$($\vec{x_k}$) = {$MAD_{ij}$($\vec{x_i^k}$) | 1 ≤ i ≤ m}
$GAD_j$($\vec{x_k}$) = $L_α$($MAD_j$($\vec{x_k}$)) = α × $T_{norm}$($MAD_j$($\vec{x_k}$))+(1 − α) × $T_{conorm}$($MAD_j$($\vec{x_k}$))
end
j ← arg $max_{1≤l≤C}$($GAD_l$ ($\vec{x_k}$))
if($GAD_j$($\vec{x_k}$) > 0.5) then
//1 − Affect object $\vec{x_k}$ to class j
$\vec{x_k}$ → $C_j$
//2 − Update parameters ρ and c for class $C_j$
for$i$ ← 1 to $m$ do
$\sum_{i=0}^m$(δ/δ$c_{ij}$ )ν($\vec{x_k}$,$\hat{c}_{ij}$ )=0
$\hat{ρ}_{ij}$ =1/n $\sum_{i=0}^n$ν($x^k_i$,$\hat{c}_{ij}$)
end
else
//Create a new class
Γ ← Γ ∪ {$C_j$}
C ← C + 1
//Initialize the new class parameters
$ρ_{ij}$ = $ρ_{init}$
$c_{ij}$=$x^k_i$
end
end
returnΓ
end
When computing $GAD_j$ ($\vec{x_k}$ ), if the value is smaller or equal to 0.5, the object is considered as part of NIC and automatically assigned as the first element of the new class as a result. Otherwise, after comput- ing the GAD values corresponding to all the classes, the object will be assigned to the class with the greatest GAD value. Using sample data, the $ρ_{ij}$ and $c_{ij}$ values for each class are estimated by minimizing a maximum likelihood criterion as expressed in the algorithm.
Question 1:
What is ${c}_{ij}$ in this article? Is it center of the cluster?
Question 2:
What is $x_i^k$? Is it i-th component of k-th vector (scalar value)? If so, what does norm operation mean in the following statement:
ν($x_i^k$) = 1 - $||$$x_i^k$ - $c_{ij}$$||$
Question 3:
There is a statement in article:
Using sample data, the $ρ_{ij}$ and $c_{ij}$ values for each class are estimated by minimizing a maximum likelihood criterion as expressed in the algorithm.
It points out on the following statements:
for$i$ ← 1 to $m$ do
$\sum_{i=0}^m$(δ/δ$c_{ij}$ )ν($\vec{x_k}$,$\hat{c}_{ij}$ )=0
$\hat{ρ}_{ij}$ =1/n $\sum_{i=0}^n$ν($x^k_i$,$\hat{c}_{ij}$)
3.1 What does hat symbol in these statements mean?
3.2 There is cycle by i and sum by i (in article). How could it be? Is it a typo?
3.3 How would I calculate ${c}_{ij}$ and ${ρ}_{ij}$? Need some help with turning these statements into pseudocode. |
Patrick Suppes in his book Introduction to Logic on page 63 asks a reader to proof a statement$$\forall x\forall y\forall z(xPy\land yPz\to xPz)$$ from the theory which he calls "Theory of rational behavior". The statement is based on the notion of weak preference $xQy$, its two properties (lines 1 and 2) and a definition of strict preference $xPy$ (line 3):$$\begin{array}{p}\{1\}&(1)&\forall x\forall y\forall z(xQy\land yQz\to xQz)&\text{Transitive property} \\\{2\}&(2)&\forall x\forall y(xQy \lor yQx)&\text{Axiom of order}\\\{3\}&(3)&\forall x\forall y(xPy\leftrightarrow \neg yQx)&\text{Definition of strict preferece}\\\{4\}&(4) & xPy\land yPz & \text{Assumption} \\\{3,4\}&(5) & \neg yQx\land \neg zQy & \text{from (3)(4) using U.S.} \\\{2,3,4\}&(6) & xQy & \text{from (2)(5) using U.S.} \\\{2,3,4\}&(7) & yQz & \text{from (2)(5) using U.S.} \\\{1,2,3,4\}&(8) & xQz & \text{from (1)(6)(7)} \\\end{array}$$U.S. stands here for the
Rule of Universal Specification
$xPz$ is equal to $\neg zQx$ by the definition of strict preference on line $3$. So we want to show that $\neg zQx$ logically follows from the premises $\{1,2,3,4\}$ and then use conditioning on line $(4)$ and
The Rule of Universal Generalisation to prove the given statement. But from $xQz\land(xQz \lor zQx)$ we cannot conclude $\neg zQx$ because according to the Axiom of order both $xQz$ and $zQx$ can be true together.
I've tried the method of interpretations to check validity of the statement that has to be proven but haven't found any, such that its antecedent would be
true and conclusion would be false.
If my derivation is fine so far, I'm looking for tips which will help me to get to the finish line here. Will appreciate any feedback. |
Given a variable $X$ with states $\lbrace 1,2,\ldots,n \rbrace$ with probabilities $\lbrace p_1,p_2,\ldots,p_n \rbrace$ we can associate with the probability distribution an entropy defined as $$ H(\lbrace p_i \rbrace)= -\sum_{i=0}^n p_i \log p_i = \left\langle \log{\frac{1}{p}} \right\rangle $$ where $\langle\cdot\rangle$ denotes the average over the probabilites $\lbrace p_i \rbrace$.
In terms of information, then I can relate this to the number of questions I need to make in order to completely specify the configuration the variable is in (which is the same as the number of bits needed to store the configuration).
Why do the bits go with $\log$?
For a constant probability distribution the probability of each state is $p_i=1/n$ (with $1/p$ being the number of states) and the number of questions I need to make can be estimating visually by drawing the sample space:
Now I can first divide the sample space S into two regions and ask: is it contained in the first region or the second? If it is in the second, I ask: is it contained in the new first half or the second half? and so on until the regions I'm dividing into are roughly the same size as my probability. The number of questions $m$ are approximately (it is drawn as exact but $p_i$ needn't be exactly $1/16$ of the space as in the figure): $$ \frac{1}{2^m} \sim p_i \rightarrow m \sim \log \frac{1}{p_i} $$ The problem is that this is justified only for uniform distributions.
Now,
my question is: this is not the most efficient way to get information for non-uniform distributions. The best way, given an ordering of my probabilities from biggest to lowest $p_{i_1}^{1)}\ge p_{i_2}^{2)}\ge\ldots \ge p_{i_n}^{n)}$, is: "is it the most probable configuration, $p_{i_1}^{1)}$?" If it is not, "is it the second most probable configuration, $p_{i_2}^{2)}$? And so on. But I have no idea how to get that the entropy would go with $\log \frac{1}{p_i}$ given this more efficient way of getting information.
Thank you for you time. |
Baire's lemma says :
Let $X$ a complete metric space. Let $(X_n)_n$ a sequence of closed subset. Suppose $Int(X_n)=\emptyset$ for all $n$. Then $Int(\bigcup_{n\in\mathbb N}X_n)=\emptyset$.
And here an equivalent form :
Let $X$ a non-empty complete metric space. Let $(X_n)_n$ a sequence of closed set s.t. $\bigcup_{n=1}^\infty X_n=X$. Then there is a $n_0$ s.t. $Int(X_n)\neq \emptyset$.
For example, $\{1\}$ with the induced topology from $\mathbb R$ is a complete metric space, we can take $X_n=\{1\}$ for all $n$, we have that $\bigcup_{n=1}^\infty X_n=X$, but $Int(X_n)=\emptyset$ for all $n$. Where is my mistakes here ? |
If $\lambda_1,\dots,\lambda_n$ are distinct positive real numbers, then $$\sum_{i=1}^n \prod_{j\neq i} {\lambda_j\over \lambda_j-\lambda_i}=1.$$This identity follows from a probability calculation that you can find at the top of page 311 in the 10th edition of
Introduction to Probability Models by Sheldon Ross.
Is there a slick or obvious explanation for this identity?
This question is sort of similar to my previous problem; clearly algebra is not my strong suit! |
In Minkowski spacetime the one way light travel time to a galaxy at proper distance $\chi$ is just:
$$ t = \frac{\chi}{c} $$
so:
$$ \chi = ct $$
As you say. However in an FRW universe the travel time is given by a different equation so the proper distance is not simply $ct$. Let's assume all motion is in the $x$ direction, so the metric simplifies to:
$$ c^2ds^2 = -c^2dt^2 + a^2(t)dx^2 \tag{1} $$
We'll take our position to be $(0, 0)$ and the galaxy to be at $(0, \chi)$, and we'll adopt the usual convention that $a = 1$ at the current time. To get the proper distance we integrate $ds$, and since $dt = 0$ and $a = 1$ the proper distance is just:
$$ \Delta s = \int_0^\chi dx = \chi $$
Now let's calculate the time it takes the light beam to get from the galaxy back to us (i.e. one half of the journey). Light travels on a null geodesic so $ds = 0$ and putting this into the metric (1) and rearranging we get:
$$ \frac{dx}{dt} = \frac{c}{a(t)} $$
If the universe is static $a(t) = 1$ for all $t$, and we get $x = ct$ so you would be correct that the proper distance is equal to half the total travel time times $c$. But then with $a = 1$ we just have Minkowski spacetime so that's hardly surprising. To calculate the trajectory of the light we need to assume a form for $a(t)$ so let's make the approximation:
$$ a(t) = 1 + Ht $$
where $H$ is the current value of the Hubble constant. Then we get:
$$ \frac{dx}{dt} = \frac{c}{1 + Ht} $$
and this integrates to give us:
$$ \chi = \frac{c}{H} \log(H\tau + 1) $$
or rearranging this to get the travel time:
$$ \tau = \frac{exp(\frac{\chi H}{c}) -1}{H} \tag{2} $$
So the proper distance $\chi$ is not simply the travel time times $c$.
Just to reassure ourselves that we get the correct result in the limit of $H \rightarrow 0$, i.e. Minkowski spacetime, note that for small $H$:
$$ exp(\frac{\chi H}{c}) \approx 1 + \frac{\chi H}{c} $$
Put this back into equation (2) and we get:
$$ \tau \approx \frac{\chi}{c} $$
which is where we came in. |
I’d like to discuss my theorem that the collection of models $M[c]$ obtained by adding an $M$-generic Cohen real $c$ over a fixed countable transitive model of set theory $M$ is upwardly countably closed, in the sense that every increasing countable chain has an upper bound.
I proved this theorem back in 2011, while at the Young Set Theory Workshop in Bonn and continuing at the London summer school on set theory, in a series of conversations with Giorgio Venturi. The argument has recently come up again in various discussions, and so let me give an account of it.
We consider the collection of all forcing extensions of a fixed countable transitive model $M$ of ZFC by the forcing to add a Cohen real, models of the form $M[c]$, and consider the question of whether every countable increasing chain of these models has an upper bound. The answer is yes! (Actually, Giorgio wants to undertake forcing constructions by forcing over this collection of models to add a generic upward directed system of models; it follows from this theorem that this forcing is countably closed.) This theorem fits into the theme of my earlier post, Upward closure in the toy multiverse of all countable models of set theory, where similar theorems are proved, but not this one exactly.
Theorem. For any countable transitive model $M\models\text{ZFC}$, the collection of all forcing extensions $M[c]$ by adding an $M$-generic Cohen real is upward-countably closed. That is, for any countable tower of such forcing extensions $$M[c_0]\subset M[c_1]\subset\cdots\subset M[c_n]\subset\cdots,$$ we may find an $M$-generic Cohen real $d$ such that $M[c_n]\subset M[d]$ for every natural number $n$. Proof. $\newcommand\Add{\text{Add}}$Suppose that we have such a tower of forcing extensions $M[c_0]\subset M[c_1]\subset\cdots$, and so on. Note that if $M[b]\subset M[c]$ for $M$-generic Cohen reals $b$ and $c$, then $M[c]$ is a forcing extension of $M[b]$ by a quotient of the Cohen-real forcing. But since the Cohen forcing itself has a countable dense set, it follows that all such quotients also have a countable dense set, and so $M[c]$ is actually $M[b][b_1]$ for some $M[b]$-generic Cohen real $b_1$. Thus, we may view the tower as having the form: $$M[b_0]\subset M[b_0\times b_1]\subset\cdots\subset M[b_0\times b_1\times\cdots\times b_n]\subset\cdots,$$ where now it follows that any finite collection of the reals $b_i$ are mutually $M$-generic.
Of course, we cannot expect in general that the real $\langle b_n\mid n<\omega\rangle$ is $M$-generic for $\Add(\omega,\omega)$, since this real may be very badly behaved. For example, the sequence of first-bits of the $b_n$’s may code a very naughty real $z$, which cannot be added by forcing over $M$ at all. So in general, we cannot allow that this sequence is added to the limit model $M[d]$. (See further discussion in my post Upward closure in the toy multiverse of all countable models of set theory.)
We shall instead undertake a construction by making finitely many changes to each real $b_n$, resulting in a real $d_n$, in such a way that the resulting combined real $d=\oplus_n d_n$ is $M$-generic for the forcing to add $\omega$-many Cohen reals, which is of course isomorphic to adding just one. To do this, let’s get a little more clear with our notation. We regard each $b_n$ as an element of Cantor space $2^\omega$, that is, an infinite binary sequence, and the corresponding filter associated with this real is the collection of finite initial segments of $b_n$, which will be an $M$-generic filter through the partial order of finite binary sequences $2^{<\omega}$, which is one of the standard isomorphic copies of Cohen forcing. We will think of $d$ as a binary function on the plane $d:\omega\times\omega\to 2$, where the $n^{th}$ slice $d_n$ is the corresponding function $\omega\to 2$ obtained by fixing the first coordinate to be $n$.
Now, we enumerate the countably many open dense subsets for the forcing to add a Cohen real $\omega\times\omega\to 2$ as $D_0$, $D_1$, and so on. There are only countably many such dense sets, because $M$ is countable. Now, we construct $d$ in stages. Before stage $n$, we will have completely specified $d_k$ for $k<n$, and we also may be committed to a finite condition $p_{n-1}$ in the forcing to add $\omega$ many Cohen reals. We consider the dense set $D_n$. We may factor $\Add(\omega,\omega)$ as $\Add(\omega,n)\times\Add(\omega,[n,\omega))$. Since $d_0\times\cdots\times d_{n-1}$ is actually $M$-generic (since these are finite modifications of the corresponding $b_k$’s, which are mutually $M$-generic, it follows that there is some finite extension of our condition $p_{n-1}$ to a condition $p_n\in D_n$, which is compatible with $d_0\times\cdots\times d_{n-1}$. Let $d_n$ be the same as $b_n$, except finitely modified to be compatible with $p_n$. In this way, our final real $\oplus_n d_n$ will contain all the conditions $p_n$, and therefore be $M$-generic for $\Add(\omega,\omega)$, yet every $b_n$ will differ only finitely from $d_n$ and hence be an element of $M[d]$. So we have $M[b_0]\cdots[b_n]\subset M[d]$, and we have found our upper bound.
QED
Notice that the real $d$ we construct is not only $M$-generic, but also $M[c_n]$-generic for every $n$.
My related post, Upward closure in the toy multiverse of all countable models of set theory, which is based on material in my paper Set-theoretic geology, discusses some similar results. |
I think that the problem stems from the action of the operator $\hat p$. Please correct me if I am mistaken.
The action of the operator $\hat p$ in the quantum space is defined as $<x|\hat p|a>=-i \hbar \partial_x <x|a>$ if the state $|a>$ does not depend on x. In fact, if the state $|a>$ depended on $x$, like for instance $|a>=f(x)|b>$ for any scalar function $f(x)$, then clearly the equation $<x|\hat p|a>=<x|\hat p f(x)|b>=-i \hbar \partial_x <x|f(x)|b>= -i \hbar \partial_x (f(x) <x|b>) $ would be badly defined, as it could be evaluated in another different way:$<x|\hat p|a>=<x|\hat p f(x)|b>=f(x) <x|\hat p |b>=f(x)(-i \hbar) \partial_x <x|b>$ The second evaluation comes from the fact that, in Standard Quantum Mechanics, it is postulated that any operator acts on ket vectors and not on scalars (with the exception the Time reversal operator, which is not of any use here).
The commutator relation $\left[\hat x, \hat p\right]=i\hbar$ is obtained from the action of the operator $\hat p$ as defined above. Thus, it comes straightforwardly that such a commutation relation cannot be generally used in a scalar product ($<x|...|ket>$) if the ket state on the right depends on $x$.
Having said that, when you perform the trace of the commutator $\left[\hat x, \hat p\right]$, you are doing
$Tr\Big[\left[\hat x, \hat p\right]\Big]=\int dx <x|(\hat x\hat p-\hat p\hat x)|x>=\int dx <x|(x\hat p-\hat p x)|x>$, where in the last step above I have just extracted the eigenvalues from the eigenstates $|x>$.In the above equation you have a scalar product where the ket on the right depends on $x$. Thus, you'll have to be careful in the evaluation and you cannot use the $xp$-commutation relations straight away. With a little care, everyone can see from the above equation that, indeed, the trace gives zero $\int dx <x|(x\hat p-\hat p x)|x>=\int dx \,x<x|(\hat p-\hat p )|x>=0$, as it should. Whereas, if you had used the $xp$-commutation relations from the outset, you would have wrongly found $Tr\Big[\left[\hat x, \hat p\right]\Big]=Tr\Big[i\hbar\Big]=i\hbar$.
Edited after Joe's Comment In the last equation I forgot the dimensionality of the space. It must be modified as$Tr\Big[\left[\hat x, \hat p\right]\Big]=Tr\Big[i\hbar\Big]=i\hbar\,D$ where $D$ are the dimensions of the quantum space you are taking the trace in. Thanks Joe. |
I’d like to discuss my theorem that the collection of models $M[c]$ obtained by adding an $M$-generic Cohen real $c$ over a fixed countable transitive model of set theory $M$ is upwardly countably closed, in the sense that every increasing countable chain has an upper bound.
I proved this theorem back in 2011, while at the Young Set Theory Workshop in Bonn and continuing at the London summer school on set theory, in a series of conversations with Giorgio Venturi. The argument has recently come up again in various discussions, and so let me give an account of it.
We consider the collection of all forcing extensions of a fixed countable transitive model $M$ of ZFC by the forcing to add a Cohen real, models of the form $M[c]$, and consider the question of whether every countable increasing chain of these models has an upper bound. The answer is yes! (Actually, Giorgio wants to undertake forcing constructions by forcing over this collection of models to add a generic upward directed system of models; it follows from this theorem that this forcing is countably closed.) This theorem fits into the theme of my earlier post, Upward closure in the toy multiverse of all countable models of set theory, where similar theorems are proved, but not this one exactly.
Theorem. For any countable transitive model $M\models\text{ZFC}$, the collection of all forcing extensions $M[c]$ by adding an $M$-generic Cohen real is upward-countably closed. That is, for any countable tower of such forcing extensions $$M[c_0]\subset M[c_1]\subset\cdots\subset M[c_n]\subset\cdots,$$ we may find an $M$-generic Cohen real $d$ such that $M[c_n]\subset M[d]$ for every natural number $n$. Proof. $\newcommand\Add{\text{Add}}$Suppose that we have such a tower of forcing extensions $M[c_0]\subset M[c_1]\subset\cdots$, and so on. Note that if $M[b]\subset M[c]$ for $M$-generic Cohen reals $b$ and $c$, then $M[c]$ is a forcing extension of $M[b]$ by a quotient of the Cohen-real forcing. But since the Cohen forcing itself has a countable dense set, it follows that all such quotients also have a countable dense set, and so $M[c]$ is actually $M[b][b_1]$ for some $M[b]$-generic Cohen real $b_1$. Thus, we may view the tower as having the form: $$M[b_0]\subset M[b_0\times b_1]\subset\cdots\subset M[b_0\times b_1\times\cdots\times b_n]\subset\cdots,$$ where now it follows that any finite collection of the reals $b_i$ are mutually $M$-generic.
Of course, we cannot expect in general that the real $\langle b_n\mid n<\omega\rangle$ is $M$-generic for $\Add(\omega,\omega)$, since this real may be very badly behaved. For example, the sequence of first-bits of the $b_n$’s may code a very naughty real $z$, which cannot be added by forcing over $M$ at all. So in general, we cannot allow that this sequence is added to the limit model $M[d]$. (See further discussion in my post Upward closure in the toy multiverse of all countable models of set theory.)
We shall instead undertake a construction by making finitely many changes to each real $b_n$, resulting in a real $d_n$, in such a way that the resulting combined real $d=\oplus_n d_n$ is $M$-generic for the forcing to add $\omega$-many Cohen reals, which is of course isomorphic to adding just one. To do this, let’s get a little more clear with our notation. We regard each $b_n$ as an element of Cantor space $2^\omega$, that is, an infinite binary sequence, and the corresponding filter associated with this real is the collection of finite initial segments of $b_n$, which will be an $M$-generic filter through the partial order of finite binary sequences $2^{<\omega}$, which is one of the standard isomorphic copies of Cohen forcing. We will think of $d$ as a binary function on the plane $d:\omega\times\omega\to 2$, where the $n^{th}$ slice $d_n$ is the corresponding function $\omega\to 2$ obtained by fixing the first coordinate to be $n$.
Now, we enumerate the countably many open dense subsets for the forcing to add a Cohen real $\omega\times\omega\to 2$ as $D_0$, $D_1$, and so on. There are only countably many such dense sets, because $M$ is countable. Now, we construct $d$ in stages. Before stage $n$, we will have completely specified $d_k$ for $k<n$, and we also may be committed to a finite condition $p_{n-1}$ in the forcing to add $\omega$ many Cohen reals. We consider the dense set $D_n$. We may factor $\Add(\omega,\omega)$ as $\Add(\omega,n)\times\Add(\omega,[n,\omega))$. Since $d_0\times\cdots\times d_{n-1}$ is actually $M$-generic (since these are finite modifications of the corresponding $b_k$’s, which are mutually $M$-generic, it follows that there is some finite extension of our condition $p_{n-1}$ to a condition $p_n\in D_n$, which is compatible with $d_0\times\cdots\times d_{n-1}$. Let $d_n$ be the same as $b_n$, except finitely modified to be compatible with $p_n$. In this way, our final real $\oplus_n d_n$ will contain all the conditions $p_n$, and therefore be $M$-generic for $\Add(\omega,\omega)$, yet every $b_n$ will differ only finitely from $d_n$ and hence be an element of $M[d]$. So we have $M[b_0]\cdots[b_n]\subset M[d]$, and we have found our upper bound.
QED
Notice that the real $d$ we construct is not only $M$-generic, but also $M[c_n]$-generic for every $n$.
My related post, Upward closure in the toy multiverse of all countable models of set theory, which is based on material in my paper Set-theoretic geology, discusses some similar results. |
Speaker Description
The transverse momentum spectrum of $\eta$ meson in relativistic heavy-ion collisions is studied at the next-to-leading-order (NLO) within the perturbative QCD, where the jet quenching effect in the QGP is incorporated with the effectively medium-modified $\eta$ fragmentation functions using the higher-twist approach. We show that the theoretical simulations could give nice descriptions of PHENIX data on $\eta$ meson in both $\rm p+p$ and central $\rm Au+Au$ collisions at the RHIC, and also provide numerical predictions of $\eta$ spectra in central $\rm Pb+Pb$ collisions with $\sqrt{s_{NN}}=2.76$~TeV at the LHC. The ratios of $\eta/\pi^0$ in $\rm p+p$ and in central $\rm Au+Au$ collisions at $200$~GeV are found to overlap in a wide $p_T$ region, which matches well the measured ratio $\eta / \pi^0$ by PHENIX. We demonstrate that, at the asymptotic region when $p_{T} \rightarrow \infty$ the ratios of $\eta/\pi^{0}$ in both $\rm Au+Au$ and $\rm p+p$ are almost determined only by quark jets fragmentation and thus approach to the one in $e^{+} e^{-}$ scattering; in addition, the almost identical gluon (quark) contribution fractions to $\eta$ and to $\pi$ result in a rather moderate variation of $\eta/\pi^{0}$ distribution at intermediate and high $p_T$ region in $\rm A+A$ relative to that in $\rm p+p$; while a slightly higher $\eta/\pi^{0}$ at small $p_T$ in $\rm Au+Au$ can be observed due to larger suppression of gluon contribution fraction to $\pi^{0}$ as compared to the one to $\eta$. The theoretical prediction for $\eta / \pi^0$ at the LHC has also been presented.
Also, we present our further studies on vector mesons such as $\rho^0$ and $\phi$ within the same framework. The theoretical predictions based on pQCD are thus firstly given which math well with the experimental measurements. It paved the way to the uniformly understanding of the strong suppression of single hadron productions at large transverse momentum which is a convincing evidence of the jet quenching effect.
Summary
The transverse momentum spectrum of $\eta$ meson in relativistic heavy-ion collisions is studied at the next-to-leading-order (NLO) within the perturbative QCD, where the jet quenching effect in the QGP is incorporated with the effectively medium-modified $\eta$ fragmentation functions using the higher-twist approach. We show that the theoretical simulations could give nice descriptions of PHENIX data on $\eta$ meson in both $\rm p+p$ and central $\rm Au+Au$ collisions at the RHIC, and also provide numerical predictions of $\eta$ spectra in central $\rm Pb+Pb$ collisions with $\sqrt{s_{NN}}=2.76$~TeV at the LHC. The ratios of $\eta/\pi^0$ in $\rm p+p$ and in central $\rm Au+Au$ collisions at $200$~GeV are found to overlap in a wide $p_T$ region, which matches well the measured ratio $\eta / \pi^0$ by PHENIX. We demonstrate that, at the asymptotic region when $p_{T} \rightarrow \infty$ the ratios of $\eta/\pi^{0}$ in both $\rm Au+Au$ and $\rm p+p$ are almost determined only by quark jets fragmentation and thus approach to the one in $e^{+} e^{-}$ scattering; in addition, the almost identical gluon (quark) contribution fractions to $\eta$ and to $\pi$ result in a rather moderate variation of $\eta/\pi^{0}$ distribution at intermediate and high $p_T$ region in $\rm A+A$ relative to that in $\rm p+p$; while a slightly higher $\eta/\pi^{0}$ at small $p_T$ in $\rm Au+Au$ can be observed due to larger suppression of gluon contribution fraction to $\pi^{0}$ as compared to the one to $\eta$. The theoretical prediction for $\eta / \pi^0$ at the LHC has also been presented.
Also, we present our further studies on vector mesons such as $\rho^0$ and $\phi$ within the same framework. The theoretical predictions based on pQCD are thus firstly given which math well with the experimental measurements. It paved the way to the uniformly understanding of the strong suppression of single hadron productions at large transverse momentum which is a convincing evidence of the jet quenching effect.
Presentation type Oral |
I came across this question on another forum. The question is:
$$ \text{If $m,n\in \mathbb{Z}_+$ such that $3m^2+m=4n^2+n$, then $(m-n)$ is a perfect square.}$$
I have managed to partially prove this using this question as motivation as follows.
Let $m>n$ and $k^2 = m-n$. The problem then becomes to show $k$ is an integer. Making the substitution $m=n+k^2$ we get
$$3(n+k^2)^2+(n+k^2) = 4n^2+n$$
And solving for $n$ yields
$$n = 3k^2\pm |k|\sqrt{12k^2+1}$$
So $n$ will be an integer if and only if $12k^2+1$ is a perfect square. This is where the previous question comes in. We want all solutions $(k,N)$ to $12k^2+1=N^2$, i.e. $$N^2-12k^2=1$$ Using Pell's equation and Wikipedia (Pell Equation) as a guide we find the fundamental solution as $y_1=k=2, x_1=N=7$, and hence all other solutions are $x_i, y_i$ where $$x_i+y_i\sqrt{12} = \left(7+2\sqrt{12}\right)^i.$$
It is not hard to see $y_i$ is an integer for all $i$. My conclusion is then: If $(m,n)$ is a solution then $k^2=(m-n)\in S=\{y_i^2\}_{i=1}^{\infty} = \{2^2, 28^2, 390^2,...\}$.
My questions are:
$\ \ \ \bullet$ I made the assumption that $m>n$, is this easy to show?
$\ \ \ \bullet$ If $y\in S$, is there always a solution $(m,n)$ with $(m-n)=y$ ?
$\ \ \ \bullet$ More importantly: Is there an
easierway to prove this? |
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Now showing items 1-10 of 17
J/Ψ production and nuclear effects in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-02)
Inclusive J/ψ production has been studied with the ALICE detector in p-Pb collisions at the nucleon–nucleon center of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement is performed in the center of mass rapidity ...
Suppression of ψ(2S) production in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-12)
The ALICE Collaboration has studied the inclusive production of the charmonium state ψ(2S) in proton-lead (p-Pb) collisions at the nucleon-nucleon centre of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement was ...
Event-by-event mean pT fluctuations in pp and Pb–Pb collisions at the LHC
(Springer, 2014-10)
Event-by-event fluctuations of the mean transverse momentum of charged particles produced in pp collisions at s√ = 0.9, 2.76 and 7 TeV, and Pb–Pb collisions at √sNN = 2.76 TeV are studied as a function of the ...
Centrality, rapidity and transverse momentum dependence of the J/$\psi$ suppression in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2014-06)
The inclusive J/$\psi$ nuclear modification factor ($R_{AA}$) in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76TeV has been measured by ALICE as a function of centrality in the $e^+e^-$ decay channel at mid-rapidity |y| < 0.8 ...
Multiplicity Dependence of Pion, Kaon, Proton and Lambda Production in p-Pb Collisions at $\sqrt{s_{NN}}$ = 5.02 TeV
(Elsevier, 2014-01)
In this Letter, comprehensive results on $\pi^{\pm}, K^{\pm}, K^0_S$, $p(\bar{p})$ and $\Lambda (\bar{\Lambda})$ production at mid-rapidity (0 < $y_{CMS}$ < 0.5) in p-Pb collisions at $\sqrt{s_{NN}}$ = 5.02 TeV, measured ...
Multi-strange baryon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2014-01)
The production of ${\rm\Xi}^-$ and ${\rm\Omega}^-$ baryons and their anti-particles in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been measured using the ALICE detector. The transverse momentum spectra at ...
Measurement of charged jet suppression in Pb-Pb collisions at √sNN = 2.76 TeV
(Springer, 2014-03)
A measurement of the transverse momentum spectra of jets in Pb–Pb collisions at √sNN = 2.76TeV is reported. Jets are reconstructed from charged particles using the anti-kT jet algorithm with jet resolution parameters R ...
Two- and three-pion quantum statistics correlations in Pb-Pb collisions at √sNN = 2.76 TeV at the CERN Large Hadron Collider
(American Physical Society, 2014-02-26)
Correlations induced by quantum statistics are sensitive to the spatiotemporal extent as well as dynamics of particle-emitting sources in heavy-ion collisions. In addition, such correlations can be used to search for the ...
Exclusive J /ψ photoproduction off protons in ultraperipheral p-Pb collisions at √sNN = 5.02TeV
(American Physical Society, 2014-12-05)
We present the first measurement at the LHC of exclusive J/ψ photoproduction off protons, in ultraperipheral proton-lead collisions at √sNN=5.02 TeV. Events are selected with a dimuon pair produced either in the rapidity ...
Measurement of quarkonium production at forward rapidity in pp collisions at √s=7 TeV
(Springer, 2014-08)
The inclusive production cross sections at forward rapidity of J/ψ , ψ(2S) , Υ (1S) and Υ (2S) are measured in pp collisions at s√=7 TeV with the ALICE detector at the LHC. The analysis is based on a data sample corresponding ... |
Show that $\int_{-\infty}^{0^+}\frac{\mathrm{Log}(t)}{e^{-t}-1}dt=0$ where the integral is over a contour of the Hankel-type. What I mean is that the contour looks like this but reflected across the imaginary axis so that it comes from $-\infty$ below the real axis, loops around the origin in a counter clockwise way with radius $\varepsilon$, and then continues to $-\infty$ above the real axis.
I am meant to take the limit as $\varepsilon\rightarrow0$. What I have so far is to break the integral over the entire contour up into three integrals, one for each ray to $-\infty$ and one for the circular section near the origin:
$\int\limits_{-\infty}^{(0^+)}\frac{\mbox{Log}(t)}{e^{-t}-1}dt=\int\limits_{-\infty}^{-\varepsilon}\frac{\mbox{Log}(t)}{e^{-t}-1}dt+\int\limits_{-\varepsilon}^{-\infty}\frac{\mbox{Log}(t)}{e^{-t}-1}dt+\int_{\left|t\right|=\varepsilon}\frac{\mbox{Log}(t)}{e^{-t}-1}dt$
Now I know that for the first integral $t=\tau e^{i\pi}$ and for the second integral $t=\tau e^{-i\pi}$ with $0<\tau<\infty$. Substituting this gives $\int\limits_{-\infty}^{-\varepsilon}\frac{\mbox{Log}(t)}{e^{-t}-1}dt+\int\limits_{-\varepsilon}^{-\infty}\frac{\mbox{Log}(t)}{e^{-t}-1}dt=\int_{\infty}^{\varepsilon}\left(\ln|t|+i\pi-\ln|t|-i(-\pi)\right)(e^{\tau}-1)^{-1}(-d\tau)$
Not really sure where to go from here. |
Let $M$ and $N$ be manifolds with Riemannian metrics $g$ and $h$ respectively. A diffeomorphism $F: M\to N$ is an isometry if \begin{equation*} h_{F(x)}(T_x F(u), T_x F(v))=g_x(u,v) \end{equation*} for all $x\in M$ and $u,v\in T_x M$. A submanifold $\Sigma\subset M$ is called
totally geodesic if for every $x\in\Sigma$ and every $v\in T_x\Sigma\subset T_x M$, the geodesic in $M$ with initial point $x$ and initial velocity is contained in $\Sigma$.
I want to show
i) The intersection of $S^n$ with any vector subspace (at least 2D) is a connected complete totally geodesic sub manifold.
Then deduce:
ii) Every connected, complete, totally geodesic submanifold of $S^n$ is the intersection of $S^n$ with vector subspace of $\mathbb{R}^{n+1}$
Here is my thought for (1): The first of my claim is that the closed totally geodesic subspaces of the sphere $S^n\subset\mathbb{R}^{n+1}$ are precisely the intersection of $S^n$ with linear subspace of $\mathbb{R}^{n+1}$. This follows from the description of the geodesics on $S^n$, according to Jurgen Jost's 4th Edn. So one would naturally look at how to determine the geodesics of $S^n$, in this case, the idea is to show the intersection of $S^n$ with any linear subspaces of $\mathbb{R}^{n+1}$ determined a great circle The orthogonal group $O(n+1)$ operates isometrically on $\mathbb{R}^{n+1}$, and since it maps $S^n$ into $S^n$, it also operate isometrically on $S^n$. Now let take a point $x\in S^n$ and a vector $v\in T_x S^n$. Let $E$ be the 2D plane through the origin of $\mathbb{R}^{n+1}$ containing $v$. We claim that the geodesic $\gamma_v$ through $x$ with tangent vector $v$ is the great circle (intersection of 2-plane with $S^n$ through the origin) through $x$ with tangent vector $v$ (parametrized by arc length). For this, let $S\in O(n+1)$ be the reflection across that 2-plane $E$. Together with $\gamma_v$, $S\gamma_v$ is also a geodesic through $x$ with tangent vector $v$. Now the uniqueness of the solution to geodesic equations implies $\gamma_v=S\gamma_v$, and thus the image of $\gamma_v$ is the great circle we want.
Hence we've shown that every tangent vector on the plane uniquely determines a connected complete totally geodesic sub manifold. Then what do I need to do for (2), i.e. to conclude every connected complete totally geodesic complete totally geodesic submanifold of $S^n$ is the intersection of $S^n$. I am lost, any correction on what I've written is appreciated. |
How can I prove that the Cartier dual of $\alpha_p$ is again $\alpha_p$ (using the Yoneda lemma)? It should be something like $\alpha_p(R) \to (\alpha_p(R) \to \mu_p(R)), x \mapsto (y \mapsto \exp_{p-1}(x + y))$, where $\exp_{p-1}$ is the truncated exponential sequence. My problem is that this isn't a homomorphism.
I'm not sure which definition of Cartier dual you are using, but here are answers using two possibilities:
Definition 1 Let $A$ be a finite commutative group scheme over $k$. Let $\hat{A}$ be the dual vector space to $A$. Define $\Delta : \hat{A} \to \hat{A} \otimes \hat{A}$ to be dual to the multiplication map $A \otimes A \to A$, and define a multiplication on $\hat{A}$ by the dual of the comultiplication of $A$. Then $\hat{A}$ is the Cartier dual.
Application to $\alpha_p$: As a ring, $\alpha_p$ is $k[t]/t^p$. Set $e_i = t^i$, for $0 \leq i \leq p-1$. In this basis, multiplication is given by $e_i e_j = \sum \delta_{i+j}^k e_k$ and comultiplication is given by $\Delta(e_k) = \Delta(t)^k = (t \otimes 1 + 1 \otimes t)^k = \sum \frac{k!}{i! j!} \delta_{i+j}^k e_i \otimes e_j$. (Here $\delta_a^b$ is $1$ if $a=b$ and $0$ otherwise.) Taking $f_i$ to be the basis of $\hat{A}$ with $\langle e_i, f_j\rangle = \delta_i^j i!$, we see that $A \cong \hat{A}$.
Definition 2: For any $k$-scheme $S$, the $S$-points of $\hat{A}$ are $\mathrm{Hom}(A \times S, \mathbb{G}_m \times S)$, with group structure given by multiplication of characters.
Application to $\alpha_p$: By a happy coincidence, I computed $\mathrm{Hom}(\mathbb{G}_a \times S, \mathbb{G}_m \times S)$ in an earlier answer. Every homorphism is of the form $t \mapsto e^{tN}$ for $N$ a nilpotent of $\mathcal{O}(S)$. If we want that morphism to factor through $\alpha_p$, a simple modification of that argument gives a bijection $$\mathrm{Hom}(\mathbb{G}_a \times S, \mathbb{G}_m \times S) \to \{ N \in \mathcal{O}(S) : N^p=0 \}.$$ with the bijection between the two sides again being that $N$ corresponds to the map $N \mapsto e^{tN}$.
Obviously, there is also a bijection between $\{ N \in \mathcal{O}(S) : N^p=0 \}$ and $\alpha_p(S)$. I leave it to you to check the necessary compatibilities for the full Yoneda proof. |
Answering only the first question:
Yes, it is a realistic way.
Precision may be questionable however (so, she will figure out she is a few hundred million years ahead in time, but won't be sure whether it's 500M or 600M, if we presume present-day understanding of CMB, universe expansion and all the relevant stuff, and Planck-level precision of measurement. With future tech, she certainly can have precision under a million years.
So, you want to measure CMB. It would be useless to observe its patterns, as you won't know what those are in your new position, and those change on the scale of 100 thousand years anyway. So the only thing you can use is CMB temperature. Fortunately for you, we know how that behaves.$$T_{CMB}(t)=\frac{T_{CMB}(0)}{a(t)},$$where $T_{CMB}(0)$ is the present-day temperature, $T_{CMB}(t)$ is temperature you want to measure, and $a(t)$ is the scale factor (defined to be 1 at the present day).To properly describe how $a$ changes over time you would need to integrate Friedmann equation. Which is something your hero would do, but I'm too lazy for that. Fortunately, there is a good enough proportionality: in the current dark-energy-dominated era$$a(t) \propto exp(Ht),$$where $H$ is the Hubble constant. Plugging in $H=70km*s^{-1}*Mpc^{-1}$ and $t=570My$, we get$$a(t) \approx 1.042$$That means, as the first approximation, CMB temperature would drop by about 4%, or 0.11K. That's certainly a noticeable and measurable change even with present-day (if state of the art) detectors.
The problem with precision in our day arises from the uncertainty about Hubble constant. For example, we have two values ($67.66 m*s^{-1}*Mpc^{-1}$ and $74.03 m*s^{-1}*Mpc^{-1}$) which are supposed to have precision of less than 2%, but you can see that they differ way more. So far this discrepancy wasn't resolved. There is also the issue of Hubble constant not being really
constant. We know it changes over time, we know it changes not by much on smaller time scale, but we don't really know how it would change. All this factored in, you can have precision of about $\pm100My$ or so with current data. On the other hand, it is certainly not a stretch to say that even in 10 years, this would be improved to $\pm10My$, and with whatever future technology and science you have, precision can be plausibly at least a couple orders better. |
This question already has an answer here:
How to plot periodic function's graphic? 5 answers
Is it possible to set the following function
f[a], where
a can be any real number?
$$ f(\alpha) = \begin{cases} 1 & \text{if } 2\pi n < \alpha < \pi (2n+1) \\ 2 & \text{if } \pi (2n+1) \le \alpha \le 2\pi (n+1) \end{cases}, \ n \in \mathbb{Z}\ $$
The issue here is
n which I want to be substituted consecutively by an integer in order to produce propriate continuous conditions. I guess, it is not the case where
Assumptions should be used as my attempt underneath doesn't work.
f[a_] := \[Piecewise] { {1, 2 \[Pi]n < a && a < \[Pi] (2 n + 1)}, {2, \[Pi] (2 n + 1) <= a && a <= 2 \[Pi] (n + 1)} }, Assumptions -> n \[Element] Integers
Thanks. |
What is best practice for typesetting fractions with large objects in the numerator and denominator?
For example, fractions with limits, integrals, summations, etc... cause the expressions to look contracted and somewhat odd.
I understand that it is probably best to leave this as is when writing mathematics inline, but as a stand-alone fraction I feel it would be much less distracting if the symbols were typeset at normal size and the fraction was given more height in which to place itself.
Are there standards (for pure mathematics) regarding this problem? And what are the options - if it's generally permitted - for expanding the fraction size to incorporate large objects?
Example
\frac{\sum_{i=0}^{n} i^2}{\iint_{D}\frac{\partial (x,y)}{\partial (u,v)}\,\mathrm{d}u\,\mathrm{d}v}
Both the summation and the integral look somewhat cramped. |
The Infinite dihedral group $D_\infty\subset \operatorname{Sym}(\mathbb{Z})$ is defined as:
$D_\infty=\{x\mapsto rx+s\mid r=\pm 1,s\in\mathbb{Z}\}$
My thoughts:
We can define the translation around $s$ as $\tau_s:\mathbb{R}\rightarrow\mathbb{R}, x\mapsto x+s$ (with $r=1$)
and the reflexion at $\frac{s}{2}$ as $\sigma_s:\mathbb{R}\rightarrow\mathbb{R}, x\mapsto s-x$ (with $r=-1$).
I can see that $D_\infty$ is generated by $\langle\tau_1,\sigma_0 \mid \sigma_0^2=e, \sigma_0\tau_1\sigma_0=\tau_1^{-1}\rangle$.
Definition from Wikipedia:
Let $G$ be a group. Two elements $a$ and $b$ of $G$ are conjugate, if there exists an element $g$ in $G$ such that $gag^{−1} = b$. One says also that $b$ is a conjugate of $a$ and that $a$ is a conjugate of $b$ .
I already found out that $\sigma_s\circ\tau_s\circ\sigma_s^{-1}=\tau_s^{-1}$. Hence, $\tau_s$ and $\tau_s^{-1}$ are in the same conjugacy class. I also tried $\tau_s\circ\sigma_s\circ\tau_s^{-1}$ and got $\sigma_{3s}$, so $\{\sigma_s,\sigma_{3s}\}$ is a conjugacy class. Also $\sigma_s$ is a conjugate of itself because of $\tau_0\circ\sigma_s\circ\tau_0^{-1}=\sigma_s$. The same is true for $\tau_s$.
My conjugacy classes so far are $\{\sigma_s\}$,$\{\tau_s\}$,$\{\tau_s,\tau_s^{-1}\}$, $\{\sigma_s,\sigma_{3s}\}$. Is this right? Which conjugacy classes am I missing?
For the second part I have a theorem which says:
Let $G$ be a group and $H\subset G$. The following are equivalent:
a) $H$ is a normal subgroup.
b) $H$ is a union of conjugacy classes of elements of G.
For example $\{\sigma_s\}\cup\{\tau_s\}= \{\sigma_s,\tau_s\} $ would be a normal subgroup of $D_\infty$ right? Does this theorem means that all possible unions of conjugacy classes are a normal subgroup? |
ECE662: Statistical Pattern Recognition and Decision Making Processes
Spring 2008, Prof. Boutin
Collectively created by the students in the class
Contents Lecture 11 Lecture notes Derivation of Fischer's Linear Discriminant
Main article: Derivation of Fisher's Linear Discriminant
The derivation was completed in this lecture.
Recall from last lecture
Last time, we considered
$ J(\vec{w}) = \frac{\vec{w}^t S_B \vec{w}}{\vec{w}^t S_W \vec{w}} $
which is explicit function of $ \vec{w} $
One can do this because numerator of $ J(\vec{w}) $ can be written as
$ \mid \tilde m_1 - \tilde m_2 \mid^2 = \mid w \cdot (m_1 - m_2) \mid^2 = w^t (m_1 - m_2) (m_1^t - m_2^t) w $
$ \rightarrow S_B = (m_1 - m_2) (m_1^t - m_2^t) $
In a same way, denominator can be written as
$ \tilde s_1^2 + \tilde s_2^2 = \sum_{y_i \in class \ i} (w \cdot y_i - \tilde m_1)^2 = \sum w^t (y_i - m_i)(y_i^t - m_i^t) w $
$ = w^t \left[ \sum (y_i - m_i)(y_i^t - m_i^t) \right] w $
$ \rightarrow S_W = \sum_{y_i \in class \ i} (y_i - m_i)(y_i^t - m_i^t) $
Fisher Linear Discriminant
It is a known result that J is maximum at $ \omega_0 $ such that $ S_B\omega_0=\lambda S_W\omega_0 $. This is the "Generalized eigenvalue problem.
Note that if $ |S_W|\neq 0 $, then $ {S_W}^{-1}S_B\omega_0=\lambda\omega_0 $. It can be written as the "Standard eigenvalue problem". The only difficulty (which is a big difficulty when the feature space dimension is large) is that matrix inversion is very unstable.
Observe that $ S_B\omega_0=(\vec{m_1}-\vec{m_2})(\vec{m_1}-\vec{m_2})^T\omega_0=cst.(\vec{m_1}-\vec{m_2}) $. Therefore the standard eigenvalue problem as presented above becomes $ {S_W}^{-1}cst.(\vec{m_1}-\vec{m_2})=\lambda\omega_0 $. From this equation, value of $ \omega_0 $ can easily be obtained, as $ \omega_0={S_W}^{-1}(\vec{m_1}-\vec{m_2}) $ or any constant multiple of this. Note that magnitude of $ \omega_0 $ is not important, the direction it represents is important. Fischer's Linear Discriminant in Projected Coordinates Claim
$ \vec{c}=\omega_0={S_W}^{-1}(\vec{m_1}-\vec{m_2}) $
is the solution to $ \mathbf{Y}\vec{c}=\vec{b} $ with $ \vec{b}=(d/d_1, \cdots, <d_1 times>, d/(d-d_1), \cdots, <(d-d_1) times>)^T $
Here is an animation of the 1D example given in class on projections Explanation
starts with $ \vec{\omega} \cdot y_i + \omega_0 > 0 $ for class 1 and $ \vec{\omega} \cdot y_i + \omega_0 < 0 $ for class 2
the data points are then projected onto an axis at 1 which results in
$ \vec{\omega} \cdot y_i > 0 $ for class 1 and $ \vec{\omega} \cdot y_i < 0 $ for class 2
one class is then projected onto an axis at -1 which results in
$ \vec{\omega} \cdot y_i > 0 $ for all $ y_i $
Support Vector Machines (SVM)
A support vector for a hyperplane $ \vec{c} $ with margin $ b_i \geq b $ is a sample $ y_{io} $ such that $ c\cdot{y_{io}} = b $.
Support Vector Machines are a two step process:
1) Preprocessing - X1,...,Xd features in kth dimensional real space is mapped to features in n dimensional real space where n>>k.
2) Linear Classifier - separates classes in n dimensional real space via hyperplane. - Support Vectors - for finding the hyperplane with the biggest margins. - Kernel - to simplify computation (This is key for real world applications) |
I'm sorry this must be an elementary question. I spent a good deal of time searching through webs including this site for the problem but I got none.Here's the problem:Say we have a binomial tree ...
I wonder if anyone here has read the following paper by Paul Kupiec in which he approximates a loss rate distribution for a portfolio composed of (possibly) concentrated bond positions.https://www....
The JKY ABMC Model (taken from Jabbour, et al. 2001) parameterizes the binomial model (in a risk-neutral world) such that,$u = e^{r\Delta t} + e^{r\Delta t}\sqrt{e^{\sigma^2\Delta t} - 1}$$d = e^{...
I have a call option with expiry in two years. In my case the option is bermudan style with first 9 months w/o ability to exercise (i.e. European) and after exercise at any time (i.e. American), but I ...
I'm coursing a financial engineering course and must build a binomial tree for a security that is valued 1 in state and time 0. I dont get the formula given by the instructors, and trying to apply the ... |
Motivation. If we presume that there exists an odd perfect number $n$, then since the Euler's totient function satisfies $$\varphi(n)=n\cdot\prod_{p\mid n}\left(1-\frac{1}{p}\right),$$ then denoting with $\operatorname{rad}(m)=\prod_{p\mid m}p$ the radical of an integer $m\geq 1$ with $\operatorname{rad}(1)=1$, and $\sigma(m)=\sum_{d\mid m}d$ then sum of divisors function we get that our odd perfect number satisifies $$2\varphi(n)\operatorname{rad}(n)=\sigma(n)\varphi(\operatorname{rad}(n)).$$That is, our odd perfect number satisifies $$\varphi(n)\operatorname{rad}(n)=n\varphi(\operatorname{rad}(n))\tag{1}$$ and we conclude that our odd perfect number $n$ satisfies by application of the Fermat's little theorem $$2^{n\varphi(\operatorname{rad}(n))}\equiv 1\text{ mod }n,\tag{2}$$ and $$2^{\operatorname{rad}(n)\varphi(n)}\equiv 1\text{ mod }\operatorname{rad}(n).\tag{3}$$
Question 1.Is true or false the following conjecture:
If $n>1$ is an odd integer that satisfies $$m\varphi(n)=n\varphi(m)\tag{4}$$ where $m\mid n$ and $m<n$ then $n$ is an odd perfect number with $\operatorname{rad}(n)=m$.
Thus I am asking if you can provide me a proof of the statement or well a counterexample $(n,m)$ of odd integers such that satisfy $(4)$ and $m\mid n$, but or well $n$ is not an odd perfect nunber or well $m$ is such that $m\neq\operatorname{rad}(n)$.
Question 2.Do you know odd integers $n,m\geq 1$ such that $$2^{n\varphi(m)}\equiv 1\text{ mod }n,$$ and $$2^{m\varphi(n)}\equiv 1\text{ mod }m?$$ (If there are many examples and you want help me, I am especially interested in examples of odd integers $n$ of the form $n\equiv 1\text{ mod }12$ or well of the form $n\equiv 9\text{ mod }36$, and being $m\mid n$ with $m$ without repeated prime factors.) Thanks in advance. |
Asymptotics of the entire functions with $\upsilon$-density of zeros along the logarithmic spirals Abstract
Let $\upsilon$ be the growth function such that $r\upsilon'(r)/\upsilon (r) \to 0$ as $r \to +\infty$,$l_\varphi^c = \{z=te^{i(\varphi+c \ln t)}, 1 \leqslant t < +\infty\}$ be the logarithmic spiral,$f$ be the entire function of zero order.The asymptotics of $\ln f(re^{i(\theta +c \ln r)})$ along ordinary logarithmic spirals $l_\theta^c$ of the function $f$ with$\upsilon$-density of zeros along $l_\varphi^c$ outside of the $C_0$-set is found. The inverse statement is true just in casezeros of $f$ are placed on the finite logarithmic spirals system $\Gamma_m = \bigcup_{j=0}^m l_{\theta_j}^c$.
Keywords
entire function, density of zeros, logarithmic spiral
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Defining parameters
Level: \( N \) = \( 3600 = 2^{4} \cdot 3^{2} \cdot 5^{2} \) Weight: \( k \) = \( 1 \) Character orbit: \([\chi]\) = 3600.ea (of order \(20\) and degree \(8\)) Character conductor: \(\operatorname{cond}(\chi)\) = \( 25 \) Character field: \(\Q(\zeta_{20})\) Newforms: \( 0 \) Sturm bound: \(720\) Trace bound: \(0\) Dimensions
The following table gives the dimensions of various subspaces of \(M_{1}(3600, [\chi])\).
Total New Old Modular forms 192 8 184 Cusp forms 0 0 0 Eisenstein series 192 8 184
The following table gives the dimensions of subspaces with specified projective image type.
\(D_n\) \(A_4\) \(S_4\) \(A_5\) Dimension 0 0 0 0 |
Evaluate the integral:
$$\int \sin^4 2x \cos 2x\, dx$$
What I currently did is
$$\int \left(\frac{1-\cos 2x}{2}\right)^2 \cos2x\, dx$$
Honestly I'm not sure what to do, I was absent from class that day. If I try substitution I get $du$ with $\sin(2x)$ and you cant mix variables.. Any help would be appreciated. |
I wanted to better understand dfa. I wanted to build upon a previous question:Creating a DFA that only accepts number of a's that are multiples of 3But I wanted to go a bit further. Is there any way we can have a DFA that accepts number of a's that are multiples of 3 but does NOT have the sub...
Let $X$ be a measurable space and $Y$ a topological space. I am trying to show that if $f_n : X \to Y$ is measurable for each $n$, and the pointwise limit of $\{f_n\}$ exists, then $f(x) = \lim_{n \to \infty} f_n(x)$ is a measurable function. Let $V$ be some open set in $Y$. I was able to show th...
I was wondering If it is easier to factor in a non-ufd then it is to factor in a ufd.I can come up with arguments for that , but I also have arguments in the opposite direction.For instance : It should be easier to factor When there are more possibilities ( multiple factorizations in a non-ufd...
Consider a non-UFD that only has 2 units ( $-1,1$ ) and the min difference between 2 elements is $1$. Also there are only a finite amount of elements for any given fixed norm. ( Maybe that follows from the other 2 conditions ? )I wonder about counting the irreducible elements bounded by a lower...
How would you make a regex for this? L = {w $\in$ {0, 1}* : w is 0-alternating}, where 0-alternating is either all the symbols in odd positions within w are 0's, or all the symbols in even positions within w are 0's, or both.
I want to construct a nfa from this, but I'm struggling with the regex part |
It's strange, but there are mechanisms whereby this type of thing can happen. Basically you have a rate determining step that involves an intermediate species, and that rate determining step must have the usual positive temperature dependence. But the preceding equilibrium that produces the intermediate species could have a negative temperature dependence that dominates.
Let's translate this into concrete terms for the reaction given. A quite plausible mechanism would look like this:
$\ce{NO + O2 <=> ONOO}$ Fast, reversible
$\ce{ONOO + NO -> 2NO2}$ Rate determining step
The equilibrium in the first step is given by
$K_\mathrm{p1} =\dfrac{p^\mathrm{eq}(\ce{ONOO})}{p^\mathrm{eq}(\ce{NO})p^\mathrm{eq}(\ce{O2})}$
Assuming $\ce{ONOO}$ is a minor species, it's formation hardly affects the equilibrium partial pressures of the other species at any time, so we may just use $p(\ce{NO})$ as the nominal nitric oxide partial pressure and similarly for the molecular oxygen. Then we can solve for the partial pressure of the $\ce{ONOO}$ intermediate:
${p^\mathrm{eq}(\ce{ONOO})}=K_\mathrm{p1}(p(\ce{NO}))(p(\ce{O2})$
Now put this result into the rate determining step with rate
$r_2=k_2{p^\mathrm{eq}(\ce{ONOO})p^\mathrm{eq}(\ce{NO})}=k_2K_\mathrm{p1}(p(\ce{NO})^2)(p(\ce{O2})$
Now for the eye-opener: put in the temperature dependence of the equilibrium and rate constants
$K_\mathrm{p1}=K_\mathrm{p1}^0\exp(-\Delta H_1/RT)$
$k_2=k_2^0\exp(-E_{a2}/RT)$
So then
$r_2=k_2^0K_\mathrm{p1}^0\exp(-\dfrac{\color{blue}{E_{a2}+\Delta H_1}}{RT})p(\ce{NO})^2p(\ce{O2})$
The activation energy then contains two terms, one from the inherent rate in Step 2 and the other from the equilibrium in Step 1. The latter is negative if Step 1 is exothermic, as it might well be if a bond is being (temporarily) formed between the reactants. Then if this enthalpy term is the absolutely larger part of the sum ... we see a reversed temperature dependence! |
Given a fiber square of simplicial sets
$$\begin{array}{cc} & \hspace{-7mm} E \\ &\hspace{-7mm}\downarrow \\ \ast\longrightarrow &\hspace{-7mm} B \end{array}$$
and a homology theory $h(-)$, there is an associated Eilenberg-Moore spectral sequence converging to the homology of the homotopy fiber $F$. This is just the spectral sequence for a cosimplicial space, specifically a cobar construction $C^\bullet(E,B,\ast)$. It is claimed in this paper of Hopkins and Ravenel, in the middle of page 7, that if this Eilenberg-Moore spectral sequence converges strongly then $\Sigma_+^\infty Tot(C^\bullet(E,B,\ast))\simeq Tot(C^\bullet(\Sigma^\infty_+E,\Sigma^\infty_+ B,\mathbb{S}))$.
Why is this true? I know from an old paper of Bousfield that the EMSS for stable homotopy converges strongly in certain nice cases (e.g. if $B$ is simply connected), but does that give me an actual equivalence of objects? That only seems to tell me nice things about the algebra. For instance that there
exists a finite filtration of $\pi_n^S(F)$ whose filtration quotients are the $E_{i,j}^\infty$ for $i+j=n$. That seems several steps away from making a general statement about the suspension spectrum functor commuting with totalization.
More generally, are there other functors $Top\to Spectra$ which commute in this fashion? For instance, it seems like the Thom spectrum functor (a sort of twisted suspension spectrum functor) for suitable fibrations should also commute with totalization is suspension spectrum does.
EDIT----------------------------
I've also discovered this mysterious correspondence between Tom Goodwillie and Mike Hopkins (this is how people talked about this stuff before MO I guess!) that shows that suspension commutes with Tot in certain cases. |
Let $C_i$ be the execution time for task i, $T_i$ be the task period and utilization rate $U = \frac{C_i}{T_i}$
Then $U$ must be less or equal to $1$ for the task to be schedulable
Proof:
Let $\hat T$ be the lcm of task $T_1,...T_n$ i.e. $\hat T = \prod_i > T_i$. Define $\hat L_i = \hat T/T_i$ the number of times the task is run. Then the total number of execution is $\sum_i C_i/\hat L_i = \sum_i \frac{C_i \hat T} {T_i}$. Suppose utilization rate $U >1$, then $\sum_i \frac{C_i \hat T} {T_i} > \hat T$ which is impossible. End of proof.
I am confused about several things in the proof.
First, what is the physical quantity representing the product or lcm of all the task periods $T_i$? It would make more sense if it was the sum of all the $T_i$ which represents the total task period.
Second, $\sum_i C_i/\hat L_i$ represents the total time spent on execution, I do not get how if this time is larger than $\hat T$, then the tasks would not be schedulable. This goes back to my misunderstanding of what $\hat T$ is.
Can anyone help? |
I thought of this question whilst doing some physics.
There is a circle $C_1$ with radius $r$ and a circle $C_2$ with radius $\lambda$. The centre of $C_2$ is on the circumference of $C_1$. What value of $\lambda$ in terms of $r$ will give the area $A$ equal to the area $B$?
My attempt: Taking the centre of $C_2$ as the origin and thinking of them as circles on a graph gives an equation for each circle.
C1: $ (x+r)^2+y^2=r^2 $
C2: $x^2+y^2=\lambda^2$
If solve it as a simultaneous equation you can find that their point of intersection is $-\lambda^2/2r$
Then you can do $\int_{-r}^{-\lambda^2/2r} \sqrt{\lambda^2-x^2} dx$ to find area $D$ and $\int_{-\lambda^2/2r}^0 \sqrt{-x^2-2xr} dx$ to find area $E$. Adding these areas together and doubling them should give half the circle, which is $0.5\pi r^2$. Then you should be able to solve the equation for $\lambda$. However when I attempted this (after slogging through about two pages of algebra) I put the equation in Wolfram Alpha and got values for $\lambda$ which were either complex or smaller than $r$. I would like to know if my method is correct and I just made a mistake, or if there is another method to this problem. |
When you chase the definitions, the issues become trivial (although perhaps still unintuitive):
$\mathbb{E}(Y\mid X) = \mathbb{E}(Y\mid \mathcal{F}_X)$ by definition.
For any subalgebra $\mathcal{G}\subset \mathcal{F}$, $\mathbb{E}(Y\mid \mathcal{G})$ is defined to be any $\mathcal{G}$-measurable function for which
$$\int_G \mathbb{E}(Y\mid \mathcal{G})(\omega) \,\mathrm d \mathbb{P}(\omega) = \int_G Y(\omega) \,\mathrm d \mathbb{P}(\omega)$$
for every $G\in \mathcal{G}$.
Therefore, whenever $\mathcal{F}_X = \mathcal{F}$, it must be the case that
$\mathbb{E}(Y\mid X)$ is $\mathcal{F}$-measurable and
$\int_F \mathbb{E}(Y\mid X)(\omega) \,\mathrm d \mathbb{P}(\omega) = \int_F Y(\omega) \,\mathrm d \mathbb{P}(\omega)$ for every $F\in \mathcal{F}$.
The equality of the integrals for all measurable sets implies (as is well known and established early in any account of Lebesgue integration) that
$\mathbb{E}(Y\mid X)$ must equal $Y$ (a.s.): they can differ only on a set of measure zero. almost surely
The second part of the question requests an example. Let's construct a very simple but not entirely trivial one. It concerns a finite binomial process used to model (among other things) changes in prices of a financial asset over time. For simplicity, I have restricted it to a sequence of two times during which the price could go up ($+$) or down ($-$), whence
$\Omega$ can be identified with the set $\{++, +-, -+, --\}$.
$\mathcal{F}$ consists of all subsets of $\Omega$ (the discrete algebra).
$\mathbb{P}$ is determined by its values on the atoms, written $p_{++}=\mathbb{P}(\{++\})$,
etc.
Let $Y$ be the price of the asset after the first time and $X$ be its price after the second time. (These natural and meaningful descriptions show this is not some pathological construction we're about to review.)
The figure displays this model as a binary tree in which the individual (conditional!) probabilities label the branches, the elements of $\Omega$ are the four possible paths from left to right through the tree, and the values of $Y$ and $X$ are indicated at the points where they are determined.
Suppose all four prices assigned by $X$ are distinct. Then, since any individual price is measurable in $\mathcal{B}(\mathbb{R})$, $\mathcal{F}_X$ contains all the atoms, whence it consists of $\mathcal{F}$ itself. But $Y$ can assign at most two distinct prices, $Y_{+} = Y(++) = Y(+-)$ and $Y_{-} = Y(-+) = Y(--)$. The inverse images of these two prices then are the sets $+_1=\{++,+-\}$ and $-_1=\{-+,--\}$. They generate a strict subalgebra of $\mathcal{F}$: it has four measurable sets and does not include any of the atoms. It describes what is "known" after the first time but before the second one.
The definition of conditional expectation needs to be checked only on a basis for $\mathcal{F}_X$. The set of its atoms is most convenient. Here is an example of a calculation for the atom $\{-+\}$:
$$\int_{\{-+\}} \mathbb{E}(Y\mid X)(\omega) \,\mathrm d \mathbb{P}(\omega) = \int_{\{-+\}} Y(\omega) \,\mathrm d \mathbb{P}(\omega) = Y(-+)p_{-+}$$
The parallel calculations for the other atoms make it clear that for all $\omega\in\Omega$,
$$Y(\omega) p_\omega = \int_{\omega} \mathbb{E}(Y\mid X)(\omega) = \mathbb{E}(Y\mid X)(\omega) p_\omega,$$
where the second equality computes the integral directly. From this
we can construct two interesting examples:
Suppose every outcome has nonzero probability. Then we may always divide both sides by $p_\omega$, no matter what $\omega$ may be, and obtain
$$\mathbb{E}(Y\mid X)(\omega) = Y(\omega).$$
The conditional expectation of $Y$ is just $Y$ itself.
Suppose $p_{++}=p_{+-}=1/2$ and $p_{--}=p_{-+}=0$. (This models a situation where an initial decrease is impossible.) Then we may define $Y_{-}$ to be
any value, since it does not matter (due to the impossibility of this event): the defining equality for $\omega={--}$
$$0 = Y_{-} p_{--} = Y(--) p_{--} = \mathbb{E}(Y\mid X)(--) p_{--} = 0$$
and its counterpart for $\omega=-+$ automatically hold. Thus,
it is not necessarily the case that $\mathbb{E}(Y|\mathcal{F}) = Y$, but the set of $\omega$ where the two sides differ must have zero probability (and, of course, be measurable with respect to $Y$).
Looking back at the tree might supply some intuition: in conditioning $Y$ on $X$, whose values were determined later, we thereby have complete information about $Y$ along any sets of paths having nonzero probability of occurring. |
Tagged: normal subgroup If a Half of a Group are Elements of Order 2, then the Rest form an Abelian Normal Subgroup of Odd Order Problem 575
Let $G$ be a finite group of order $2n$.
Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup. Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities of $S$ and $H$ are both $n$.
Then prove that $H$ is an abelian normal subgroup of odd order.Add to solve later
Problem 470
Let $G$ be a finite group of order $p^n$, where $p$ is a prime number and $n$ is a positive integer.
Suppose that $H$ is a subgroup of $G$ with index $[G:P]=p$. Then prove that $H$ is a normal subgroup of $G$.
(
Michigan State University, Abstract Algebra Qualifying Exam) Problem 332
Let $G=\GL(n, \R)$ be the
general linear group of degree $n$, that is, the group of all $n\times n$ invertible matrices. Consider the subset of $G$ defined by \[\SL(n, \R)=\{X\in \GL(n,\R) \mid \det(X)=1\}.\] Prove that $\SL(n, \R)$ is a subgroup of $G$. Furthermore, prove that $\SL(n,\R)$ is a normal subgroup of $G$. The subgroup $\SL(n,\R)$ is called special linear group |
Let $w(a,b)$ be a word in two letter alphabet. Let $$A=\left(\begin{array}{lll}x_1 & x_2 & x_3\\\ x_4 &x_5 & x_6\\\ x_7 & x_8 & x_9\end{array}\right), B=\left(\begin{array}{lll}y_1 & y_2 & y_3\\\ y_4 &y_5 & y_6\\\ y_7 & y_8 & y_9\end{array}\right)$$ where $x_i,y_i$ are commuting variables. Let $f_w=\mathrm{trace}(w(A,B))$, a polynomial in 18 variables.
Question. Is it possible to reconstruct $w$ up to a cyclic shift from $f_w$?
Note that there exists a polynomial in one variable that encodes $w$: $x^{p_1}+...+x^{p_s}+x^{|w|}$ where $p_1,...,p_s$ are the places where $a$ occurs in $w$. Also note that for 2 by 2 matrices the answer is "no". For example if $w=abbaba$ and $w'=ababba$, then $f_w=f_{w'}$ for 2 by 2 matrices. The question is related to the study of the moduli space of representations (of degree 3) of the free group.
Update I think that as George suggested below, one can assume that $A=\mathrm{diag}(a,b,c)$ is a diagonal matrix (otherwise consider a conjugate of the pair $A,B$ over some algebraically complete field). After that the problem reduces to the following problem which seems longer but is in fact easier because we reduce the number of variables to from 18 to 3:
Pick a natural number $n\gg 1$. For every cyclic sequence $p$ (i.e. $p_{n+1}=p_1$) of $\{1,2,3\}$ of length $n$ consider a 9-vector $\phi(p)=($number of occurrences of 11, number of occurrences of 12, ..., number of occurrences of 33$)\in \mathbb{N}^9$. The sum of coordinates of $\phi(p)$ is $n$, so we get a partition of $n$, and the number of different $\phi(p)$ is at most the number of partitions of $n$ into 9 parts, so less than $n^9$. Thus the map $\phi$ has a non-trivial kernel $\mathrm{Ker}(\phi)$ (i.e. the equivalence relation $p\equiv q$ iff $\phi(p)=\phi(q)$ ). Let $S$ be a preimage of a point in $\mathbb{N}^9$ under $\phi$. Let $v=(v_1,...,v_n)$ be a cyclic vector of natural numbers (including 0). For every $p\in S$ consider the monomial $m_p=a^sb^tc^u$ in 3 variables where $s$ is the sum of $v_i$ such that $p_i=1$, $t$ is the sum of $v_i$ such that $p_i=2$, $u$ is the sum of $v_i$ such that $p_i=3$. The sum of all the monomials $m_p$, $p\in S$, is a polynomial $f_S(v)$ in $a,b,c$. That polynomial is the coefficient of the monomial $\prod_{(i,j)} B[i,j]^{\phi(p)[i,j]}$ in $f_w$.
Question Is the sequence $v$ determined by the sequence of polynomials $f_S(v)$ where $S$ runs over the equivalence classes of the partition $\mathrm{Ker}(\phi)$. |
Introduction
This post introduces multivariate adaptive regression splines (MARS). The focus of this post is to explain the algorithm in a regression context
1, and some background knowledge on stepwise linear regression is necessary. The Building Blocks
Like standard linear regression, MARS uses the ordinary least squares (OLS) method to estimate the coefficient of each term. However, instead of an original predictor, each term in a MARS model is a basis function derived from original predictors. A basis function takes one of the following forms:
a constant 1, which represents the intercept; an original predictor from the data set; a hinge function (see below) derived from a predictor; a product of two or more hinge functions derived from different predictors, which captures the interaction between/among the predictors.
MARS does not treat categorical predictors differently from standard linear regression. However, for each of all combinations of continuous predictors and their observed values, MARS creates a reflected pair of hinge functions in the form
\[ \{max(0, x - c), max(0, c - x)\} \quad (1) \]
where \(x\) denotes a continuous predictor; \(c\) denotes an observed value of that predictor which is referred to as a knot. Hinge functions are piecewise linear and each is used to model an isolated portion of the original data. For example, \(max(0, x - c)\), which is symmetrical to \(max(0, c - x)\), is a linear function of \(x\) when \(x > c\) but remains constantly at zero otherwise. Suppose there are \(p\) continuous predictors and they all have \(n\) distinct values, then there will be \(np\) such pairs of hinge functions. The model-training process will iteratively select and add some of them into the model (see the next section).
Let \(h_m(X)\) (\(m \in \{0, 1, ..., M\}\)) denote the basis functions
2, then a MARS model can be written as
\[ f(X) = \sum_{m = 0}^{M} \beta_m h_m(X) \]
where \(h_0(X) = 1\) is the only constant basis function.
The Training Process
The training process of MARS is similar to a forward stepwise linear regression: at each step, MARS selects new terms into the model that minimize the sum of squared error using OLS. However, the construction of new terms by MARS is more sophisticated and deserves further description.
Initially, the basis function \(h_0(X) = 1\) is added to the model and the result is a model with an intercept term \(\beta_0\). At each subsequent step, an original predictor or a reflected pair of hinge functions are selected and added to the model. The selected pair of hinge functions (or original predictor) can enter the model directly; alternatively, they can be multiplied by
an existing basis function that is already in the model (candidate functions are excluded) and become new basis functions. The second case allows the interaction between/among different predictors to be modeled. Note that a reflected pair of hinge functions always enter the model together (but may be removed separately in the pruning process; see the next section). The training process goes on until it meets one of many condition such as: 1. maximum number of model terms before pruning; 2. forward stepping threshold measured by \(R^2\). The Pruning Process
Although there are other methods, MARS typically applies a backward deletion procedure to prune the model. At each step, the algorithm removes a term in the model that results in the smallest increase in the sum of squared error, obtaining an optimal model \(\hat{f}\) at each size \(\lambda\). The final model can be determined using cross-validation (CV), but generalized cross-validation (GCV) may be preferred since it is much more computationally efficient.
Generalization error given by GCV is defined as
\[ GCV(\lambda) = \frac{\sum^N_{i = 1}(y_i - \hat{f}_\lambda(x_i))^2}{(1 - \frac{M'(\lambda)}{N})^2} \]
where \(M'(\lambda)\) is the effective number of parameters in the model. \(M'(\lambda)\) is in turn defined as
\[ M(\lambda) = M + cK \]
where \(M\) is the number of terms in the model, \(K\) is the number of knots in the model, \(c = 2\) if the model does not involve interaction terms and \(c = 3\) otherwise
3. The Parameters
The
earth package in R provides an implementation of MARS, on which
caret::train() with
method = earth is based. After conducting a simulation (see appendix), I found the following parameters to be particularly worth noting (please refer to the help page and associated documentation for more detailed information):
Parameters for the training process:
degree: Maximum degree of interaction.
nk: Maximum number of model terms before pruning.
thresh: Forward stepping threshold.
minspan: Minimum number of observations between knots.
endspan: Minimum number of observations before the first and after the final knot.
linpreds: Index vector specifying which predictors should enter linearly.
allowed: Function specifying which predictors can interact and how.
newvar.penalty: Penalty for adding a new variable in the forward pass.
Parameters for the pruning process:
nprune: Maximum number of terms (including intercept) in the pruned model.
nfold: Number of cross-validation folds (if CV is used to prune the model instead of GCV).
pmethod: Pruning method.
The
caret::train() function considers
degree and
nprune to be the only major hyperparameters that need to be tuned. However, my experience with the simulation indicate that, if the underlying pattern in training data is complicated,
nk should be set with a bigger value and
thresh a smaller value so that a more flexible model can be obtained from the training process; additionally, if the sample size is relatively small, it may also be necessary to use smaller values of
minspan and
endspan.
One thing I noticed about the training function provided by the
earth package (i.e.,
earth::earth()) is that it performs model selection by default and gives the best pruned model as the end result. This is indicated by the
nprune parameters, which specifies the maximum instead of exact number of terms in the final model. This feature can be turned off by specifying
penalty = -1 in the training function. This special value of
penalty causes
earth::earth() to set the GCV to \(RSS/nrow(x)\). Since the RSS on the training set always decreases with more terms, the pruning pass will choose the maximum allowable number of terms.
Appendix: a Simulation
A simulation can be conducted to show how different values of
nk,
thresh,
minspan and
endspan affects the model-training process. You can run the code below in R to see the results for yourself.
The results indicate that the final model is more flexible if:
nk is bigger;
thresh is smaller;
minspan and
endspan are smaller. In this simulation, the model that best captures the underlying pattern is the one with
nk = 20,
thresh = 0 and
minspan = endspan = 1 (i.e., the most flexible one which also contains the most number of terms).
A word of caution: this simulation does not imply a more flexible MARS model is always more accurate in terms of test error. It works well in this case perhaps because the random noise is relatively small and there are no correlated or redundant predictors. Additionally, a critical strength of MARS is that it can train models that are very interpretable
4; however, increasing the flexibility generally reduces the interpretability. If interpretability is not a key consideration, then perhaps a more flexible algorithm such as random forest should be used instead.
library(tidyverse)library(earth)# generate a training setset.seed(1)sample_size <- 100dat <- tibble( x = runif(sample_size, min = -3.5 * pi, max = 3.5 * pi), e = rnorm(sample_size, sd = 0.3), y = sin(x) + e)# the underlying pattern of the data setggplot(dat %>% mutate(y = sin(x))) + geom_line(aes(x = x, y = y))# train MARS models with different values of nk, thresh and span (including minspan and endspan)results <- crossing(nk = c(5, 10, 20), thresh = c(0, 0.01, 0.1), span = c(1, 5, 10)) %>% pmap(function(nk, thresh, span, dat) { fit <- earth(y ~ x, data = dat, degree = 1, nprune = NULL, nk = nk, thresh = thresh, minspan = span, endspan = span) mutate(dat, nk = nk, thresh = thresh, span = span, y_predicted = predict(fit)[,1], p = length(coef(fit))) }, dat = dat) %>% bind_rows()# predictions given by different modelsggplot(results) + geom_point(aes(x = x, y = y_predicted)) + facet_grid(nk ~ thresh + span, labeller = label_both)# the number of terms included in each model with different parameters (the output is attached below)results %>% select(nk, thresh, span, p) %>% distinct() %>% mutate(nk = as.factor(nk) %>% fct_relabel(function(x) {paste0("nk=", x)})) %>% split(.$nk) %>% map(function(x) { x %>% select(-nk) %>% spread(key = span, value = p, sep = "=") %>% as.data.frame() %>% `rownames<-`(paste0("thresh=", .$thresh)) %>% select(-thresh) })#> $`nk=5`#> span=1 span=5 span=10#> thresh=0 4 4 2#> thresh=0.01 4 4 2#> thresh=0.1 1 1 1#> #> $`nk=10`#> span=1 span=5 span=10#> thresh=0 5 5 4#> thresh=0.01 5 5 4#> thresh=0.1 1 1 1#> #> $`nk=20`#> span=1 span=5 span=10#> thresh=0 8 7 4#> thresh=0.01 5 7 4#> thresh=0.1 1 1 1
References
Hastie, Trevor, Robert Tibshirani, and Jerome Friedman. 2009.
The Elements of Statistical Learning: Data Mining, Inference, and Prediction. 2nd ed. New York: Springer.
Kuhn, Max, and Kjell Johnson. 2013.
Applied Predictive Modeling. 1st ed. New York: Springer.
MARS can also incorporate logistic regression to predict probabilities in a classification context.↩
Since there may be multiple predictors in a basis function, \(h_m(X)\) should be considered as a function over the entire predictor space.↩
According to Hastie, Tibshirani, and Friedman (2009), these values are suggested by some mathematical and simulation results.↩
This is the reason why the degree of interaction is usually limited to one or two but rarely above↩ |
OK, let's get back to the original question: Is the image $T(S)$ of a linearly dependent set $S$ under a linear transformation $T$ linearly dependent?
We must to notice that, there are
many possible definitions of span, linear combination and linearly dependence/independence, each of them again have several versions, one talking on $v_1,v_2,\cdots,v_k$(this vector can be required to be totally distinct, or not necessarily distinct, which means there are again two possible choices one may choose), one talking on a set, especially having the ability to talk on infinite set of vectors, such as $\{1,x,x^2,x^3,x^4,\cdots\}$.
First, suppose we adopt the version of
span, linear combination and linearly dependence/independence that talking about $v_1,\cdots,v_k$(not set) which are not necessarily distinct. Then in this case, for example, we may say $v,v$ is linearly dependent. Then it is true that for any $v_1,\cdots,v_k\in V$ not necessarily distinct, if they are linearly dependent, then $T(v_1),\cdots,T(v_k)$(possible duplicated here) are linearly dependent.
However, the important point is that, the original proposition in the title, as one might not expect, is
false. This is because the set language has an essence that repeating the same object in a set doesn't behave differently than the original set. I just thought and found an unbeatable counterexample.
Let $T:\Bbb R^3\to\Bbb R^3$ defined by $T(\mathbb{x})=\begin{bmatrix}3&0&0\\0&2&0\\0&0&0\end{bmatrix}\mathbb{x}$.
Let $S=\{(1,2,3),(1,2,1),(1,3,7),(1,3,5)\}$. Then since $S\subseteq \Bbb R^3$, then $S$ must be linearly dependent.
Next, by the definition of image of a set under a function, we see that $T(S)=\{(3,4,0),(3,6,0)\}$. Notice that this result followed by the definition of
image of a set under a function, which has nothing to do with linear combination or span or linearly (in)dependence. So it doesn't depend on what kind of defintion you wish to choose! Then, as you can see, $T(S)$ is literally and undoubtfully linearly independent. So the original proposition in the title is FALSE. |
I'm trying to get into Kalman filters. I've noticed an issue with Euler angles near -180°/180° (or -pi/pi) and wonder how to correctly resolve this. Its often said you need to normalize the angles into this range. However, this isn't as easy as it seems at first sight. Especially, when using a Kalman filter class from a library (e.g. OpenCV).
Let's take a very simple examle. We have only a compass sensor, which will give us a heading in the XY-plane. And our state keeps track of the current heading and its first derivative (speed of change).
Matrix Algebra
First, I'll give all information in the typical matrix form to show how this actually relates to the Kalman filter. Later I'll reduce this to simple formulas for better reasoning.
We define our state space vector as:
$x_{k} = \begin{bmatrix} \alpha \\ \dot{\alpha} \end{bmatrix}$
Where $\alpha$ is the current heading in degrees (for simplicity) and $\dot{\alpha}$ is the current change of the heading in degrees per second.
The new a priori state estimate (prediction) is calculated by:
$\hat{x}_{k|k+1} = F_k x_k$ where $F_k = \begin{bmatrix} 1 & \Delta t \\ 0 & 1\end{bmatrix}$
Our observations look like this:
$z_k = \begin{bmatrix} \alpha_{new} \end{bmatrix}$
And the correction / update step looks like this:
$\tilde{y}_k = z_k - H_k \hat{x}_{k|k+1}$ where $H_k = \begin{bmatrix} 1 & 0\end{bmatrix}$
Later the Kalman gain $K_k$ is calculated and influences the state estimate like this:
$\hat{x}_{k+1} = \hat{x}_{k|k+1} + K_k \tilde{y}_k$
Simplifications
$\hat{x}_{k|k+1} = \begin{bmatrix} \alpha + \Delta t \dot{\alpha} \\ \dot{\alpha}\end{bmatrix}$
$\tilde{y}_k = \alpha_{new} - (\alpha + \Delta t \dot{\alpha})$
$\hat{x}_{k+1} = \hat{x}_{k|k+1} + K_k \tilde{y}_k$
Now lets only focus only on the estimation of $\alpha$:
$\alpha_{k|k+1} = \alpha_k + \Delta t \dot{\alpha}_k$
$\alpha_{k+1} = \alpha_{k|k+1} + K_k^1 (a_{new} - \alpha_{k|k+1})$
The Issue and Semi-Solution
For simplicity I'll use angles in degrees. Let's assume, we've got the following situation:
$\alpha_k = 160°$, $\dot{\alpha}_k = 100°/s$, $\Delta t = 0.1 s$, $a_{new} = -170°$ ($\equiv$ 190°) and $K_k^1 = 0.1$.
Hence, $\alpha_{k|k+1} = 160° + 0.1 * 100° = 170°$ and therefore $a_{new} - \alpha_{k|k+1} = -170° - 170° = -340° \equiv 20°$.
The updated estimate would therefore (incorrect estimate) be:
$\alpha_{k+1} = 170° + 0.1 * (-340°) = 170° - 34° = 136°$
If we normalize $\alpha_{k|k+1}$ to [-180°, 180°] we get the equivalent 20° and the result is (correctly estimated):
$\alpha_{k+1} = 170° + 0.1 * 20° = 170° + 2° = 172°$
The normalization would fix the issue. However, if I use a normal Kalman filter class, I usually cannot influence how the innovation $\tilde{z}_k$ is calculated and the filter class cannot know that this is an Euler angle that needs special handling. Thus, I would need to write it my own... Or is there any way around this? A quick solution might be to use a unit vector $\begin{bmatrix} cos(\alpha) \\ sin(\alpha) \end{bmatrix}$ instead of the euler angle $\alpha$, which I would have to renormalize every now and then. I know that one could maybe use quaternions for this, but I currently can't grasp the math behind it to use it. Also I saw someone argue that it would be a bad idea to use quaternions in the state space of a Kalman filter (e.g. see comments to this question). This could also apply to my unit vector approach. Hence, I would like you to discuss the issue and give me some clues on how this is usually resolved. |
(a) If $AB=B$, then $B$ is the identity matrix. (b) If the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions. (c) If $A$ is invertible, then $ABA^{-1}=B$. (d) If $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix. (e) If $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.
Let $A$ and $B$ be $3\times 3$ matrices and let $C=A-2B$.If\[A\begin{bmatrix}1 \\3 \\5\end{bmatrix}=B\begin{bmatrix}2 \\6 \\10\end{bmatrix},\]then is the matrix $C$ nonsingular? If so, prove it. Otherwise, explain why not.
Let $A, B, C$ be $n\times n$ invertible matrices. When you simplify the expression\[C^{-1}(AB^{-1})^{-1}(CA^{-1})^{-1}C^2,\]which matrix do you get?(a) $A$(b) $C^{-1}A^{-1}BC^{-1}AC^2$(c) $B$(d) $C^2$(e) $C^{-1}BC$(f) $C$
Let $\calP_3$ be the vector space of all polynomials of degree $3$ or less.Let\[S=\{p_1(x), p_2(x), p_3(x), p_4(x)\},\]where\begin{align*}p_1(x)&=1+3x+2x^2-x^3 & p_2(x)&=x+x^3\\p_3(x)&=x+x^2-x^3 & p_4(x)&=3+8x+8x^3.\end{align*}
(a) Find a basis $Q$ of the span $\Span(S)$ consisting of polynomials in $S$.
(b) For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$.
(The Ohio State University, Linear Algebra Midterm)
Let $V$ be a vector space and $B$ be a basis for $V$.Let $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ be vectors in $V$.Suppose that $A$ is the matrix whose columns are the coordinate vectors of $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ with respect to the basis $B$.
After applying the elementary row operations to $A$, we obtain the following matrix in reduced row echelon form\[\begin{bmatrix}1 & 0 & 2 & 1 & 0 \\0 & 1 & 3 & 0 & 1 \\0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0\end{bmatrix}.\]
(a) What is the dimension of $V$?
(b) What is the dimension of $\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$?
(The Ohio State University, Linear Algebra Midterm)
Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers.Let\[W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix}a & b\\c& -a\end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.\]
(a) Show that $W$ is a subspace of $V$.
(b) Find a basis of $W$.
(c) Find the dimension of $W$.
(The Ohio State University, Linear Algebra Midterm)
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).The time limit was 55 minutes.
This post is Part 3 and contains Problem 7, 8, and 9.Check out Part 1 and Part 2 for the rest of the exam problems.
Problem 7. Let $A=\begin{bmatrix}-3 & -4\\8& 9\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}-1 \\2\end{bmatrix}$.
(a) Calculate $A\mathbf{v}$ and find the number $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$.
(b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$.
Problem 8. Prove that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular.
Problem 9.Determine whether each of the following sentences is true or false.
(a) There is a $3\times 3$ homogeneous system that has exactly three solutions.
(b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric.
(c) If $n$-dimensional vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent for any $n$-dimensional vector $\mathbf{v}_4$.
(d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).The time limit was 55 minutes.
This post is Part 2 and contains Problem 4, 5, and 6.Check out Part 1 and Part 3 for the rest of the exam problems.
Problem 4. Let\[\mathbf{a}_1=\begin{bmatrix}1 \\2 \\3\end{bmatrix}, \mathbf{a}_2=\begin{bmatrix}2 \\-1 \\4\end{bmatrix}, \mathbf{b}=\begin{bmatrix}0 \\a \\2\end{bmatrix}.\]
Find all the values for $a$ so that the vector $\mathbf{b}$ is a linear combination of vectors $\mathbf{a}_1$ and $\mathbf{a}_2$.
Problem 5.Find the inverse matrix of\[A=\begin{bmatrix}0 & 0 & 2 & 0 \\0 &1 & 0 & 0 \\1 & 0 & 0 & 0 \\1 & 0 & 0 & 1\end{bmatrix}\]if it exists. If you think there is no inverse matrix of $A$, then give a reason.
Problem 6.Consider the system of linear equations\begin{align*}3x_1+2x_2&=1\\5x_1+3x_2&=2.\end{align*}
(a) Find the coefficient matrix $A$ of the system.
(b) Find the inverse matrix of the coefficient matrix $A$.
(c) Using the inverse matrix of $A$, find the solution of the system.
(Linear Algebra Midterm Exam 1, the Ohio State University) |
Differential Equations of \(RLC\)-Circuits
Electric oscillations can be excited in a circuit containing resistance \(R\), inductance \(L\) and capacitance \(C\). In terms of topology, two types of circuits are often considered: series \(RLC\)-circuit (Figure \(1\)) and parallel \(RLC\)-circuit (Figure \(2\)).
We derive the differential equation describing the current change in a series \(RLC\) circuit.
The voltages \({V_R},{V_C},{V_L},\) respectively, on the resistor \(R,\) capacitor \(C\) and inductor \(L\) are given by
\[
{{V_R}\left( t \right) = RI\left( t \right),\;\;\;}\kern-0.3pt {{V_C}\left( t \right) = \frac{1}{C}\int\limits_0^t {I\left( \tau \right)d\tau } ,\;\;\;}\kern-0.3pt {{V_L}\left( t \right) = L\frac{{dI}}{{dt}}.} \]
It follows from the Kirchhoff’s voltage law \(\left(KVL\right)\) that
\[{{V_R}\left( t \right) + {V_C}\left( t \right) }+{ {V_L}\left( t \right) }={ E\left( t \right),}\]
where \(E\left( t \right)\) is the electromotive force (emf) of the power supply.
In the case of constant emf \(E,\) we obtain the following differential equation after substituting the expressions for \({V_R},\) \({V_C},\)\({V_L}\) and differentiation:
\[{\frac{{{d^2}I\left( t \right)}}{{d{t^2}}} + \frac{R}{L}\frac{{dI\left( t \right)}}{{dt}} }+{ \frac{1}{{LC}}I\left( t \right) }={ 0.}\]
If we denote \(2\beta = {\large\frac{R}{L}\normalsize},\) \(\omega _0^2 = {\large\frac{1}{{LC}}\normalsize},\) the equation can be written as
\[{\frac{{{d^2}I}}{{d{t^2}}} + 2\beta \frac{{dI}}{{dt}} + \omega _0^2I }={ 0.}\]
This differential equation coincides with the equation describing the damped oscillations of a mass on a spring. Hence, damped oscillations can also occur in series
RLC-circuits with certain values of the parameters.
Now we consider the parallel \(RLC\)-circuit and derive a similar differential equation for it.
By the Kirchhoff’s current law \(\left(KCL\right)\), the total current is equal to the sum of currents through a resistor \(R,\) inductor \(L\) and capacitor \(C\) (Figure \(2\)):
\[{{I_R}\left( t \right) + {I_L}\left( t \right) + {I_C}\left( t \right) }={ I\left( t \right).}\]
Given that
\[
{{I_R} = \frac{V}{R},\;\;\;}\kern-0.3pt {{I_L} = \frac{1}{L}\int\limits_0^t {Vd\tau } ,\;\;\;}\kern-0.3pt {{I_C} = C\frac{{dV}}{{dt}},} \]
for the case of constant total current \(I\left( t \right) = {I_0},\) we obtain the following differential equation of the second order with respect to the variable \(V:\)
\[
{{\frac{V}{R} + \frac{1}{L}\int\limits_0^t {Vd\tau } }+{ C\frac{{dV}}{{dt}} = {I_0},\;\;}}\Rightarrow {{C\frac{{{d^2}V}}{{d{t^2}}} + \frac{1}{R}\frac{{dV}}{{dt}} }+{ \frac{1}{L}V = 0.}} \]
As one can see, we again have the equation describing the damped oscillations. Thus, the oscillatory mode can also occur in parallel \(RLC\)-circuits.
Resonant Circuit. Thomson Formula
In the simplest case, when the ohmic resistance is zero \(\left(R = 0\right)\) and the source of emf is removed \(\left(E = 0\right)\), the resonant circuit consists only of a capacitor \(C\) and inductor \(L,\) and is described by the differential equation
\[{\frac{{{d^2}I}}{{d{t^2}}} + \omega _0^2I = 0,\;\; }\kern-0.3pt{\text{where}\;\;\omega _0^2 = \frac{1}{{LC}}.}\]
In this circuit there will be undamped electrical oscillations with a period
\[{T_0} = \frac{{2\pi }}{{{\omega _0}}} = 2\pi \sqrt {LC} .\]
This formula is called the Thomson formula in honor of British physicist William Thomson \(\left(1824-1907\right)\), who derived it theoretically in \(1853.\)
Damped Oscillations in Series \(RLC\)-Circuit
The second order differential equation describing the damped oscillations in a series \(RLC\)-circuit we got above can be written as
\[{\frac{{{d^2}I}}{{d{t^2}}} + \frac{R}{L}\frac{{dI}}{{dt}} }+{ \frac{1}{{LC}}I }={ 0.}\]
The corresponding characteristic equation has the form
\[{\lambda ^2} + \frac{R}{L}\lambda + \frac{1}{{LC}} = 0.\]
Its roots are calculated by the formulas:
\[
{{\lambda _{1,2}} }={ \frac{{ – \frac{R}{L} \pm \sqrt {\frac{{{R^2}}}{{{L^2}}} – \frac{4}{{LC}}} }}{2} } = { – \frac{R}{{2L}} \pm \sqrt {{{\left( {\frac{R}{{2L}}} \right)}^2} – \frac{1}{{LC}}} } = { – \beta \pm \sqrt {{\beta ^2} – \omega _0^2} ,} \]
where the value of \(\beta = {\large\frac{R}{{2L}}\normalsize}\) is called the damping coefficient, and \({\omega_0}\) is the resonant frequency of the circuit.
Depending on the values of \(R, L, C\) there may be three options.
Case \(1.\) Overdamping: \({R^2} \gt {\large\frac{{4L}}{C}\normalsize}\)
In this case, both roots of the characteristic equation \({\lambda_1}\) and \({\lambda_2}\) and real, distinct and negative. The general solution of the differential equation is given by
\[{I\left( t \right) }={ {C_1}{e^{{\lambda _1}t}} + {C_2}{e^{{\lambda _2}t}}.}\]
In this mode, the current decreases monotonically approaching zero (Figure \(3\)).
Case \(2.\) Critical Damping: \({R^2} = {\large\frac{{4L}}{C}\normalsize}\)
This mode can be called boundary or critical. Here, both roots of the characteristic equation are equal, real and negative. The general solution is expressed by the function
\[
{I\left( t \right) = \left( {{C_1}t + {C_2}} \right){e^{ – \beta t}} } = {\left( {{C_1}t + {C_2}} \right){e^{ – {\large\frac{R}{{2L}}\normalsize} t}}.} \]
At the beginning of the process, the current may even increase, but then it quickly decreases exponentially.
Case \(3.\) Underdamping: \({R^2} \lt {\large\frac{{4L}}{C}\normalsize}\)
In this case, the roots of the characteristic equation are complex conjugate, which leads to damped oscillations in the circuit. The change of current is given by
\[{I\left( t \right) }={ {e^{ – \beta t}}\left( {A\cos \omega t + B\sin \omega t} \right),}\]
where the value of \(\beta = {\large\frac{R}{{2L}}\normalsize}\) is as above the damping factor, \(\omega =\) \( \sqrt {{\large\frac{1}{{LC}}\normalsize} – {{\left( {\large\frac{R}{{2L}}\normalsize} \right)}^2}} \) is the frequency of oscillation, \(A, B\) are constants of integration, depending on initial conditions. Note that the frequency \(\omega\) of damped oscillations is less than the resonant frequency \({\omega_0}\) of the circuit. The typical shape of the curve \(I\left( t \right)\) in this mode is also shown in Figure \(3\) above.
Forced Oscillations and Resonance
If the resonant circuit includes a generator with periodically varying emf, the forced oscillations arise in the system. If the emf \(E\) of the source varies according to the law
\[E\left( t \right) = {E_0}\cos \omega t,\]
then the differential equation of forced oscillations in series \(RLC\)-circuit can be written as
\[
{{\frac{{{d^2}q\left( t \right)}}{{d{t^2}}} + \frac{R}{L}\frac{{dq\left( t \right)}}{{dt}} }+{ \frac{1}{{LC}}q\left( t \right) }={ \frac{1}{L}{E_0}\cos \omega t\;\;}}\kern-0.3pt {{\text{or}\;\;\frac{{{d^2}q}}{{d{t^2}}} + 2\beta \frac{{dq}}{{dt}} + \omega _0^2q }={ \frac{{{E_0}}}{L}\cos \omega t,}} \]
where \(q\) the charge of the capacitor, \(2\beta = \frac{R}{L},\) \(\omega _0^2 = \frac{1}{{LC}}.\)
This equation is analogous to the equation of forced oscillations of a spring pendulum, discussed on the page Mechanical Oscillations. Its general solution is the sum of two components: the general solution of the associated homogeneous equation and a particular solution of the nonhomogeneous equation. The first component describes the decaying transient process, after which the behavior of the system depends only on the external driving force. The forced oscillations will occur according to the law
\[
{q\left( t \right) \text{ = }}\kern0pt {{\frac{{{E_0}}}{{L\sqrt {{{\left( {\omega _0^2 – {\omega ^2}} \right)}^2} + 4{\beta ^2}{\omega ^2}} }} \cdot}}\kern0pt{{ \cos \left( {\omega t + \varphi } \right) }} = {{\frac{{{E_0}}}{{\omega \sqrt {{R^2} + {{\left( {\omega L – \frac{1}{{\omega C}}} \right)}^2}} }} \cdot}\kern0pt{ \cos \left( {\omega t + \varphi } \right),}} \]
where the phase \(\varphi\) is determined by the formula
\[
{\varphi = \arctan \left( { – \frac{{2\beta \omega }}{{\omega _0^2 – {\omega ^2}}}} \right) } = {\arctan \frac{R}{{\omega L – \frac{1}{{\omega C}}}}.} \]
Knowing the change of the charge \(q\left( t \right),\) it is easy to find the change of the current \(I\left( t \right):\)
\[
{I\left( t \right) = \frac{{dq\left( t \right)}}{{dt}} } = {{ – \frac{{{E_0}}}{{\sqrt {{R^2} + {{\left( {\omega L – \frac{1}{{\omega C}}} \right)}^2}} }} \cdot}\kern0pt{ \sin\left( {\omega t + \varphi } \right) }} = {{\frac{{{E_0}}}{{\sqrt {{R^2} + {{\left( {\omega L – \frac{1}{{\omega C}}} \right)}^2}} }} \cdot}\kern0pt{ \cos\left( {\omega t – \theta } \right),}} \]
where we have introduced the angle \(\theta\) such that \(\theta = – \left( {\varphi + \frac{\pi }{2}} \right).\) The angle \(\theta\) indicates the phase shift of the current oscillations \(I\left( t \right)\) with respect to oscillations in the supply voltage \(E\left( t \right) = {E_0}\cos \omega t.\)
The amplitude of the current \({I_0}\) and the phase shift \(\theta\) are given by
\[
{{{I_0} }={ \frac{{{E_0}}}{{\sqrt {{R^2} + {{\left( {\omega L – \frac{1}{{\omega C}}} \right)}^2}} }} }={ \frac{{{E_0}}}{Z},\;\;\;}}\kern-0.3pt {\theta = \arctan \frac{{\omega L – \frac{1}{{\omega C}}}}{R}.} \]
The quantity \(Z =\) \( \sqrt {{R^2} + {{\left( {\omega L – {\large\frac{1}{{\omega C}}\normalsize}} \right)}^2}} \) is called the impedance, or impedance of the circuit. It consists of an ohmic resistance \(R\) and a reactance \({\omega L – {\large\frac{1}{{\omega C}}}\normalsize}.\) Impedance of the resonant circuit in the complex form can be written as
\[Z = R + i\left( {\omega L – \frac{1}{{\omega C}}} \right).\]
We see from these formulas that the amplitude of steady-state oscillations of the current is maximum when
\[{\omega L = \frac{1}{{\omega C}}\;\;\text{or}\;\;}\kern-0.3pt{\omega = {\omega _0} = \frac{1}{{\sqrt {LC} }}.}\]
On this condition, resonance appears in the circuit. The resonant frequency \({\omega_0}\) is equal to the frequency of free oscillations in the circuit and does not depend on the resistance \(R.\)
We can transform the formula for the amplitude of the forced oscillations to get an explicit dependence on the frequency ratio \(\large\frac{\omega }{{{\omega _0}}}\normalsize,\) where \({\omega_0}\) is the resonant frequency. As a result, we obtain
\[\require{cancel}
{{I_0} = \frac{{{E_0}}}{{\sqrt {{R^2} + {{\left( {\omega L – \frac{1}{{\omega C}}} \right)}^2}} }} } = {\frac{{\frac{{{E_0}}}{{{\omega _0}}}}}{{\frac{{\sqrt {{R^2} + {{\left( {\omega L – \frac{1}{{\omega C}}} \right)}^2}} }}{{{\omega _0}}}}} } = {\frac{{\frac{{{E_0}}}{{{\omega _0}}}}}{{\sqrt {\frac{{{R^2}}}{{\omega _0^2}} + {{\left( {\frac{\omega }{{{\omega _0}}}L – \frac{1}{{\omega {\omega _0}C}}} \right)}^2}} }} } = {\frac{{{E_0}\sqrt {LC} }}{{\sqrt {{R^2}LC + {{\left( {\frac{\omega }{{{\omega _0}}}L – \frac{1}{{\frac{\omega }{{{\omega _0}}}\frac{\cancel{C}}{{L\cancel{C}}}}}} \right)}^2}} }} } = {\frac{{{E_0}\sqrt {LC} }}{{\sqrt {{R^2}LC + {{\left( {\frac{\omega }{{{\omega _0}}}L – \frac{L}{{\frac{\omega }{{{\omega _0}}}}}} \right)}^2}} }} } = {\frac{{{E_0}\sqrt C }}{{\sqrt {{R^2}C + {{\left( {\frac{\omega }{{{\omega _0}}} – \frac{1}{{\frac{\omega }{{{\omega _0}}}}}} \right)}^2}} }}.} \]
Dependencies of the current amplitude on the frequency ratio \(\large\frac{\omega }{{{\omega _0}}}\normalsize\) for different values of \(R\) and \(C\) are shown in Figures \(4\) and \(5.\) These graphs are built at \(E = 100\;\text{V},\) \(L = 1\;\text{mH},\) \(C = 10\;\mu\text{F}\) (Figure \(4\)), \(R = 10\;\text{ohms}\) (Figure \(5\)).
Resonance properties of a circuit are characterized by the quality factor \(Q,\) which is numerically equal to the ratio of the resonance frequency \({\omega_0}\) to the width \(\Delta\omega\) of the resonance curve at \(\large{\frac{1}{\sqrt 2}}\normalsize\) of the maximum value (Figure \(6\)).
The \(Q\) factor in a series \(RLC\) circuit is given by
\[Q = \frac{1}{R}\sqrt {\frac{L}{C}} .\]
For a parallel \(RLC\) circuit, the \(Q\) factor is determined by the inverse expression:
\[Q = R\sqrt {\frac{C}{L}} .\]
Solved Problems
Click a problem to see the solution.
Example 1An electrical circuit consists of a series-connected resistor \(R = 100\;\text{ohms}\) and a coil with inductance \(L = 50\;\text{H}.\) At time \(t = 0\) a \(DC\) source with the voltage \({V_0} = 200\;\text{V}\) is connected. Find: the current change \(I\left( t \right)\) in the circuit; the voltage change across the resistor \({V_R}\left( t \right)\) and the inductor \({V_L}\left( t \right)\). Example 2An electrical circuit consists of a series-connected resistor \(R = 100\;\text{ohms}\) and a capacitor \(C = 0.01\;\mu\text{F}.\) At the initial moment a \(DC\) source with the voltage \({V_0} = 200\;\text{V}\) is connected to the circuit. Find: the current change \(I\left( t \right)\) in the circuit; the voltage change across the resistor \({V_R}\left( t \right)\) and the capacitor \({V_C}\left( t \right)\). Example 3An electrical circuit consists of a series-connected resistor \(R = 1\;\text{ohms},\) a coil with inductance \(L = 0.25\;\text{H}\) and a capacitor \(C = 1\;\mu\text{F}.\) How many oscillations will it make before the amplitude of the current is reduced by a factor of \(e?\) Example 4An \(AC\) source with amplitude \({E_0} = 128\;\text{V}\) and frequency \(\omega = 250\;\text{Hz}\) is connected to a series circuit consisting of a resistance \(R = 100\;\text{ohms},\) a coil with inductance \(L = 0.4\;\text{H}\) and a capacitor \(C = 200\;\mu\text{F}.\) Find: the current amplitude in the circuit; the voltage amplitude on the capacitor. |
For Continuous time aperiodic signals, the duality property of Continuous Time Fourier Transform (CTFT) is following
$$\mathscr{F}\Big\{x(t)\Big\} = X(f), \qquad\text{then} \quad \mathscr{F}\Big\{X(t)\Big\} = x(-f)$$
Now we know while Dirichlet conditions are not satisfied for unit step function $u(t)$, so its CTFT analysis and synthesis cannot be done. However, we can still do it provided we are willing to accept occurrence of singularity functions like Dirac delta impulse in its Fourier transform equation.
i.e.
$$\begin{align} \mathscr{F}\Big\{u(t)\Big\} &= \mathscr{F}\Big\{\tfrac{1}{2} + \tfrac{1}{2}\operatorname{sgn}(t) \Big\} \\ &= \frac{\delta(f)}{2} + \frac{1}{j2\pi f} \\ \end{align}$$
where the signum function,
$$\operatorname{sgn}(t) \triangleq \begin{cases} -1 \qquad & t<0 \\ 0 \qquad & t=0 \\ +1 \qquad & t>0 \\ \end{cases}$$
However if I apply duality property to the above result, then I should get following:
$$\begin{align} \mathscr{F}\Big\{\frac{\delta(t)}{2} + \frac{1}{j2\pi t}\Big\} &= u(-f) \\ &= \tfrac{1}{2} + \tfrac{1}{2}\operatorname{sgn}(-f) \\ &= \tfrac{1}{2} - \tfrac{1}{2}\operatorname{sgn}(f) \\ \end{align}$$
However when I read at least some books on Fourier transform, I find that result is $u(f)$ and not $u(-f)$. Question is why ? |
Let\[\mathbf{v}_{1}=\begin{bmatrix}1 \\ 1\end{bmatrix},\;\mathbf{v}_{2}=\begin{bmatrix}1 \\ -1\end{bmatrix}.\]Let $V=\Span(\mathbf{v}_{1},\mathbf{v}_{2})$. Do $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ form an orthonormal basis for $V$?
For a set $S$ and a vector space $V$ over a scalar field $\K$, define the set of all functions from $S$ to $V$\[ \Fun ( S , V ) = \{ f : S \rightarrow V \} . \]
For $f, g \in \Fun(S, V)$, $z \in \K$, addition and scalar multiplication can be defined by\[ (f+g)(s) = f(s) + g(s) \, \mbox{ and } (cf)(s) = c (f(s)) \, \mbox{ for all } s \in S . \]
(a) Prove that $\Fun(S, V)$ is a vector space over $\K$. What is the zero element?
(b) Let $S_1 = \{ s \}$ be a set consisting of one element. Find an isomorphism between $\Fun(S_1 , V)$ and $V$ itself. Prove that the map you find is actually a linear isomorpism.
(c) Suppose that $B = \{ e_1 , e_2 , \cdots , e_n \}$ is a basis of $V$. Use $B$ to construct a basis of $\Fun(S_1 , V)$.
(d) Let $S = \{ s_1 , s_2 , \cdots , s_m \}$. Construct a linear isomorphism between $\Fun(S, V)$ and the vector space of $n$-tuples of $V$, defined as\[ V^m = \{ (v_1 , v_2 , \cdots , v_m ) \mid v_i \in V \mbox{ for all } 1 \leq i \leq m \} . \]
(e) Use the basis $B$ of $V$ to constract a basis of $\Fun(S, V)$ for an arbitrary finite set $S$. What is the dimension of $\Fun(S, V)$?
(f) Let $W \subseteq V$ be a subspace. Prove that $\Fun(S, W)$ is a subspace of $\Fun(S, V)$.
Let $\mathrm{P}_3$ denote the set of polynomials of degree $3$ or less with real coefficients. Consider the ordered basis\[B = \left\{ 1+x , 1+x^2 , x – x^2 + 2x^3 , 1 – x – x^2 \right\}.\]Write the coordinate vector for the polynomial $f(x) = -3 + 2x^3$ in terms of the basis $B$.
Let $V$ denote the vector space of $2 \times 2$ matrices, and $W$ the vector space of $3 \times 2$ matrices. Define the linear transformation $T : V \rightarrow W$ by\[T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = \begin{bmatrix} a+b & 2d \\ 2b – d & -3c \\ 2b – c & -3a \end{bmatrix}.\]
For an integer $n > 0$, let $\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \{ 1 , x , x^2 , \cdots , x^n \}$ is a basis of $\mathrm{P}_n$, called the standard basis.
Let $T : \mathrm{P}_n \rightarrow \mathrm{P}_{n+1}$ be the map defined by, for $f \in \mathrm{P}_n$,\[T (f) (x) = x f(x).\]
Prove that $T$ is a linear transformation, and find its range and nullspace.
Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for $\R^2$. Let $S:=[\mathbf{v}_1, \mathbf{v}_2]$.Note that as the column vectors of $S$ are linearly independent, the matrix $S$ is invertible.
Prove that for each vector $\mathbf{v} \in V$, the vector $S^{-1}\mathbf{v}$ is the coordinate vector of $\mathbf{v}$ with respect to the basis $B$.
Let $C[-2\pi, 2\pi]$ be the vector space of all real-valued continuous functions defined on the interval $[-2\pi, 2\pi]$.Consider the subspace $W=\Span\{\sin^2(x), \cos^2(x)\}$ spanned by functions $\sin^2(x)$ and $\cos^2(x)$.
(a) Prove that the set $B=\{\sin^2(x), \cos^2(x)\}$ is a basis for $W$.
(b) Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$.
Let $V$ be a vector space and $B$ be a basis for $V$.Let $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ be vectors in $V$.Suppose that $A$ is the matrix whose columns are the coordinate vectors of $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ with respect to the basis $B$.
After applying the elementary row operations to $A$, we obtain the following matrix in reduced row echelon form\[\begin{bmatrix}1 & 0 & 2 & 1 & 0 \\0 & 1 & 3 & 0 & 1 \\0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0\end{bmatrix}.\]
(a) What is the dimension of $V$?
(b) What is the dimension of $\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$?
(The Ohio State University, Linear Algebra Midterm)
Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers.Let\[W=\left\{\, A\in V \quad \middle | \quad A=\begin{bmatrix}a & b\\c& -a\end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.\]
(a) Show that $W$ is a subspace of $V$.
(b) Find a basis of $W$.
(c) Find the dimension of $W$.
(The Ohio State University, Linear Algebra Midterm)
Let $C[-1, 1]$ be the vector space over $\R$ of all continuous functions defined on the interval $[-1, 1]$. Let\[V:=\{f(x)\in C[-1,1] \mid f(x)=a e^x+b e^{2x}+c e^{3x}, a, b, c\in \R\}\]be a subset in $C[-1, 1]$.
(a) Prove that $V$ is a subspace of $C[-1, 1]$.
(b) Prove that the set $B=\{e^x, e^{2x}, e^{3x}\}$ is a basis of $V$.
(c) Prove that\[B’=\{e^x-2e^{3x}, e^x+e^{2x}+2e^{3x}, 3e^{2x}+e^{3x}\}\]is a basis for $V$. |
If symmetry conditions are met, FIR filters have a linear phase. This is not true for IIR filters.
However, for what applications is it bad to apply filters that do not have this property and what would be the negative effect?
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A linear phase filter will
preserve the waveshape of the signal or component of the input signal (to the extent that's possible, given that some frequencies will be changed in amplitude by the action of the filter).
This could be important in several domains:
coherent signal processing and demodulation, where the waveshape is important because a thresholding decision must be made on the waveshape (possibly in quadrature space, and with many thresholds, e.g. 128 QAM modulation), in order to decide whether a received signal represented a "1" or "0". Therefore, preserving or recovering the originally transmitted waveshape is of utmost importance, else wrong thresholding decisions will be made, which would represent a bit error in the communications system. radar signal processing, where the waveshape of a returned radar signal might contain important information about the target's properties audio processing, where some believe (although many dispute the importance) that "time aligning" the different components of a complex waveshape is important for reproducing or maintaining subtle qualities of the listening experience (like the "stereo image", and the like)
Let me add the following graphic to the great answers already given.
When a filter has
linear phase, then all the frequencies within that signal will be delayed the same amount in time (as described mathematically in Fat32's answer).
Any signal can be decomposed (via Fourier Series) into separate frequency components. When the signal gets delayed through any channel (such as a filter), as long as all of those frequency components get delayed the same amount, the same signal (signal of interest, within the passband of the channel) will be recreated after the delay.
Consider a square wave, which through the Fourier Series Expansion is shown to be made up of an infinite number of odd harmonic frequencies.
In the graphic above I show the summation of the first three components. If these components are all delayed the same amount, the waveform of interest is intact when these components are summed. However, significant group delay distortion will result if each frequency component gets delayed a different amount in time.
The following may help give additional intuitive insight for those with some RF or analog background.
Consider an ideal lossless broadband delay line (such as approximated by a length of coaxial cable), which can pass wideband signals without distortion.
The transfer function of such a cable is shown in the graphic below, having a magnitude of 1 for all frequencies and a phase negatively increasing in direct linear proportion to frequency. The longer the cable, the steeper the slope of the phase, but in all cases "linear phase".
This makes sense; the phase delay of 1 Hz signal passing through a cable with a 1 second delay will be 360°, while a 2 Hz signal with the same delay will be 720°, etc...
Bringing this back to the digital world, $z^{-1}$ is the z-transform of a 1 sample delay (therefore a delay line), with a similar frequency response to what is shown, just in terms of H(z); a constant magnitude = 1 and a phase that goes linearly from $0$ to $-2\pi$ from f = 0 Hz to f = fs (the sampling rate).
Just to add to what's already been said, you can see this intuitively by looking at the following sinusoid with monotonically increasing frequency.
Shifting this signal to the right or left will change its phase. But note also that the phase change will be larger for higher frequencies, and smaller for lower frequencies. Or in other words, the phase increases linearly with frequency. Thus a constant time shift corresponds to a linear phase change in the frequency domain.
The answer to this question is already been explained clearly in the previous replies. Yet I wish to give it a try to present a mathematical interpretation of the same
Consider a linear time invariant System whose frequency response is governed by $H(w)$.
i.e if the input to this system is $e^{jw_{0}t}$ the output will be $H(w_{0})e^{jw_{0}t}$
Here $H(w_{0})$ is a complex number which has a phase component denoted by $arg(H(w))$ and magnitude component denoted by $|H(w)|$
if the system has linear phase response then $$arg(H(w)) = Kw$$ where $K$ is a constant
If the phase is linear the output of the system for the input $e^{jw_{0}t}$ will be $$y(t) = |H(w)|*e^{jw_{0}t + jKw_{0}}$$ $$ = |H(w)|*e^{jw_{0}(t + K)}$$ which is nothing but a delayed version of input with some scaling applied.
So if the phase is linear then all the frequency components of the signal will undergo the same amount of delay in time-domain which results in shape preservation.
The essence and importance of linear phase property lies in the definition and the effect of
group delay $$\tau(\omega) = - \frac {d\phi(\omega)}{d\omega}$$ on the applied signal $x[n]$, where $\phi(\omega)$ is the phase response of the filter; (phase of its frequency response).
Assume that a filter, with a fixed group delay of $n_0$ samples, is applied a
narrowband input signal $x[n]$. Then the output signal will be (approximately) of the form $y[n] = K x[n-n_0]$ where $K$ is the filter gain evaluated at the center frequency of the narrowband input signal $x[n]$. This means that the input signal will be weighted and shifted intact as a whole by the group delay of the filter. And this can only happen when the group delay is independent of the frequency $\omega$. And this will be the case if the underlying filter has linear phase (or generalized linear phase). Note that if the input signal is of broadband type; i.e., its minimum and maximum frequencies are far from its center frequency, then the approximation is not valid and eventhough the group delay would still be the same for each sinusoidal component in the signal, their relative output amplitudes will differ by the frequency dependent filter gain $K(w)$.
Then what's the effect of a filter with nonlinear phase (or frequency dependent group delay) on the input signal? A simple example would be a complicated input signal considered as a sum of multiple wavepackets at different center frequencies. After the filtering, each packet with a particular center frequency will be
shifted (delayed) differently due to frequency dependent group delay. And this will be resulting in a change in the time-order (or space order) of those wave packets, sometimes drastically, depending on how nonlinear the phase is, which is called as dispersion in communications terminalology. Not only the composite waveshape, but also some event orders may be lost. This kind of dispersive channels have severe effects such as ISI (inter symbol interference) on transmitted data.
This property of linear phase filters, therefore, is also known as
waveform-preserving property, which is applicable to narrowband signals in particular. An example where waveform is important, other than ISI as mentioned above, is in processing of images, where the Fourier transform phase information is of paramount importance compared to magnitude of Fourier transform, for intelligibility of the image. The same, however, cannot be said for perception of sound signals due to a different kind of sensitivity of the ear to the stimulus.
I will just put a summary for these great answers mentioned above: |
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